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\title[Crossed Product C*Algebras]{An Introduction to Crossed Product
C*Algebras and Minimal Dynamics}
\author{N.~Christopher Phillips}
\date{5~February 2017} % ???
\address{Department of Mathematics, University of Oregon,
Eugene OR 974031222, USA.}
\email[]{ncp@darkwing.uoregon.edu}
\subjclass[2010]{Primary 46L40, 46L55;
Secondary 46L05, 54H20.}
\thanks{This material is based upon work supported by the
US National Science Foundation under
Grants DMS0302401, DMS0701076, DMS1101742, and DMS1501144.}
\begin{document}
% \begin{abstract}
% We present an introduction to the theory
% of crossed product C*algebras,
% with emphasis on some of the background needed for recent work on
% the classification of crossed products.
% \end{abstract}
\maketitle
% \setcounter{section}{1}
\tableofcontents
\section{Introduction and Motivation}\label{Sec:Dfn}
\indent
These notes are an introduction to group actions on \ca{s}
and their crossed products,
primarily by discrete groups
and with emphasis on situations in which
the crossed products are simple
and at least close to the class of \ca{s}
expected to be classifiable in the sense of the Elliott program.
They are aimed at graduate students who have had
a one semester or one year course on the general theory of \ca{s}.
(We give more details on the prerequisites later in this section.)
These notes are not intended as a reference work.
Our emphasis is on explaining ideas and methods,
rather than on giving complete proofs.
For some results, different proofs are given at different locations
in these notes,
or special cases are proved of results
which are proved later in greater generality
by quite different methods.
For others, some of the main ideas
are explained and simpler versions of some of the relevant lemmas
are proved,
but we refer to the research papers for the full proofs.
Other results and calculations are left as exercises;
the reader is strongly encouraged to do many of these,
to develop facility with the material.
Yet other results,
needed for the proofs of the theorems described here
but not directly related to dynamics,
are quoted with only some general description,
or with no background at all.
Before giving a general outline, we describe some of the
highlights of our treatment.
We give a very large collection of examples of actions of groups
on \ca{s} (Part~\ref{Part_Basic}),
and we give a number of explicit computations
of crossed products (Section~\ref{Sec_Comp}).
We give most or all of the proofs of the following results,
including background:
\begin{itemize}
\item
The reduced \ca{} of a finitely generated nonabelian free group
is simple (\Thm{T_3301_CStRFnSimp})
and has a unique tracial state (\Thm{T_3301_CStRFnUniqT}).
\item
If $G$ is an amenable locally compact group,
then the map $C^* (G, A, \af) \to C^*_{\mathrm{r}} (G, A, \af)$
is an isomorphism
(\Thm{T_4131_CPFToRed}; proved using the F{\o}lner condition).
\item
If $G$ is a discrete group,
then the standard conditinal expectation
from $C^*_{\mathrm{r}} (G, A, \af)$ to $A$ is faithful
(Proposition \ref{P:Faithful}(\ref{P:Faithful:1});
this is hard to find in the literature).
\item
The crossed product of an AF~algebra by a Rokhlin action of a finite
group is AF (\Thm{T_RokhAF}).
\item
The crossed product of a simple tracially AF C*algebra
by a tracial Rokhlin action of a finite
group is tracially AF (\Thm{RokhTAF}).
\item
The reduced crossed product of a locally compact Hausdorff space by
a minimal and essentially free action of a discrete group is simple
(\Thm{T:AS}).
\end{itemize}
We give substantial parts of the proofs of the following results,
including the relevant dynamics background:
\begin{itemize}
\item
Let $X$ be a finite dimensional infinite compact metric space,
and let $h \colon X \to X$ be a minimal homeomorphism.
Suppose that the image of $K_0 (C^* (\Z, X, h))$
is dense in $\Aff \big( \T (C^* (\Z, X, h)) \big)$.
Then $C^* (\Z, X, h)$ has tracial rank zero.
(See \Thm{T:TM}.)
This includes the proof of \Thm{T_7121_AY_rshd},
giving the recursive subhomogeneous structure
of the orbit breaking subalgebra
$C^* (\Z, X, h)_Y$ of \Def{D_5421_VSubalg}
when $Y \subset X$ is closed and $\sint (Y) \neq \E$,
for which as far as we know a detailed proof has not been published.
\item
Let $h \colon X \to X$ be a minimal homeomorphism
of a compact metric space.
Assume that
there is a continuous surjective map from $X$ to the Cantor set.
Then the radius of comparison of $C^* (\Z, X, h)$
is at most half the mean dimension of~$h$.
(See \Thm{T_5526_rcmdim_Rpt}.)
\end{itemize}
We give a brief description of the contents.
Parts \ref{Part_Basic} and~\ref{Part_BasicCP}
(Sections \ref{Sec_CommEx}\ref{Sec_Comp})
are quite elementary in nature.
Part~\ref{Part_Basic} gives many examples
of group actions on \ca{s}.
Part~\ref{Part_BasicCP} develops the theory
of full and reduced group \ca{s}
and full and reduced crossed products,
with full details given for discrete groups
and some indications of the theory for general
locally compact groups.
This part ends with a number of explicit computations
of crossed products by discrete groups.
Part~\ref{Part_Finite}
(Sections \ref{Sec_FGStr}\ref{Sec_TRPFG})
is about structure theory for crossed products
of simple \ca{s} by finite groups.
Section~\ref{Sec_FGStr} discusses
(giving some proofs, but not a complete presentation)
some of the relevant structural properties of \ca{s}.
In the rest of this part,
we consider crossed products,
primarily under the assumption that the
action has the \rp{}
or the \trp.
The presentation of the crossed product related machinery
is fairly detailed but not complete,
and a few results from other parts of the theory of \ca{s}
are used with little indication of proof.
Part~\ref{Part_MH}
(Sections \ref{Sec_MH}\ref{Sec_More})
is a first look at \mh{s} of \cms{s}
and their crossed products.
We give a complete proof of simplicity
of reduced crossed products by essentially free minimal actions
of discrete groups.
When we turn to stronger structure theorems,
for the case of actions of~$\Z$,
much more outside material is needed,
and our presentation accordingly becomes much more sketchy.
In Part~\ref{Part_Large}
(Sections \ref{Sec_Cuntz}\ref{Sec_OP}),
we discuss the machinery of large subalgebras,
which is used to prove further results about the
structure of crossed products by \mh{s}
(and by free minimal actions of some other groups,
as well as automorphisms of some noncommutative \ca{s}).
Large subalgebras are motivated by the proofs
in Section~\ref{Sec:Class}
and those sketched in Section~\ref{Sec_More}.
The theory here is considerably more technical,
and uses considerably more material
from outside the theory of crossed products.
In particular, the Cuntz semigroup plays a key role
in the statements of some results,
and in the proofs of some results whose statement
does not mention the Cuntz semigroup.
Our presentation here is accordingly much less complete.
In a number of cases,
we give direct proofs of results which in the original papers
are derived from stronger results with more complicated proofs,
or we prove only special cases or simplified statements.
These proofs are simpler, but are still not simple.
The hope is that the presentation here can serve as an
introduction to the machinery of large subalgebras,
and enable beginners in the area
to better understand the research papers
using this method.
These notes are a greatly expanded version of lectures on crossed
product \ca{s} given at the
Ottawa Summer School in Operator Algebras,
2024 August 2007.
It contains additional material from lectures given at the
Fields Institute in Fall 2007,
from graduate courses given at the University of Oregon
in Spring 2008 and Spring 2013
and at the University of Toronto in Winter 2014,
and from lecture series given in Lisbon, Seoul, Shanghai, Barcelona,
Kyoto, and Laramie.
These notes are still rough.
There are surely many remaining misprints and some more serious
errors.
Some references are incomplete or missing entirely.
There is no index.
Even given the omissions discussed below,
there should have been,
as just one example,
enough discussion of groupoids and their \ca{s}
to identify the orbit breaking subalgebras
(\Def{D_5421_VSubalg}) of crossed products,
used in Parts \ref{Part_MH} and~\ref{Part_Large},
as \ca{s} of open subgroupoids of the transformation group groupoid.
The author plans to keep a list of misprints,
and a corrected and possibly expanded version of these notes,
on his website.
Developments in the theory of
Part~\ref{Part_Large},
and even to some extent in the theory of
Part~\ref{Part_Finite},
are quite rapid,
and are faster than it is possible to keep up with in
writing these notes.
In particular,
four extremely important developments are barely mentioned here.
One is the use of versions of the \trp{}
(for both finite and countable amenable groups)
which do not require the presence of projections.
Several more sections could be written in Part~\ref{Part_Finite}
based on these developments.
The second is the importance of stability under tensoring
with the JiangSu algebra~$Z$
as a regularity condition.
This condition is barely mentioned in
Part~\ref{Part_Finite},
and deserves a much more substantial treatment there.
Third,
essentially nothing is said about higher dimensional
Rokhlin properties,
despite their importance even for finite groups
and also as a competing method for obtaining results of
some of the same kinds as in Part~\ref{Part_Large}.
Finally,
essentially nothing is said about classifiability and
related weaker conditions for crossed products
of simple \ca{s} by infinite discrete groups,
not even by~$\Z$.
Our discussion of crossed products
of simple \ca{s}
stops after considering finite groups,
and the actions of infinite groups
we consider almost all come from actions on compact metric spaces.
% Some discussion about prerequisites:
These notes assume the basic theory of \ca{s},
including:
\begin{itemize}
\item
The basics of representation theory
(including states and
the GelfandNaimarkSegal construction).
\item
Type~I \ca{s}.
\item
Some familiarity with nuclear \ca{s}.
\item
Direct limits and the usual examples
constructed with them,
such as UHF algebras, AF~algebras,
AT~algebras, and AH~algebras.
\item
Tensor products of Hilbert spaces.
\item
Some familiarity with
minimal and maximal tensor products of \ca{s}.
\item
The basics of \ca{s} given by generators and relations
and the usual elementary examples
(such as $M_n$, $C (S^1)$,
$C (S^1, M_n)$, the Toeplitz algebra,
and the Cuntz algebras).
\item
Multiplier algebras.
\item
The Double Commutant Theorem.
% \item
% \item
% \item
\end{itemize}
We will give some exposition of the following topics,
but not enough to substitute
% on serious reading
for a thorough presentation:
\begin{itemize}
\item
Stable rank one.
% Topological stable rank.
\item
Real rank zero.
% Real rank.
\item
Tracial rank zero.
% Tracial rank.
\item
The Cuntz semigroup.
\item
Recursive subhomogeneous \ca{s}.
\item
Dimension theory for compact metric spaces.
\item
The mean dimension of a \hme.
\item
Graph \ca{s}.
\end{itemize}
There will be occasional comments assuming other material,
but which are not essential to the development:
\begin{itemize}
\item
Larger values of topological stable rank,
real rank, and tracial rank.
\item
Ktheory.
(\ca{s} satisfying the Universal Coefficient Theorem will be
mentioned moderately often.)
\item
Morita equivalence.
\item
Groupoids and their \ca{s}.
\item
Partial actions and their crossed products.
\item
Free products and reduced free products.
\item
Quasitraces.
\end{itemize}
In a number of places,
we make comments which refer to later material.
We encourage the reader to jump back and forth.
Some statements are given without proof:
the proofs are either left as exercises
or are beyond the scope of these notes.
In Part~\ref{Part_BasicCP},
although the definitions related to group \ca{s}
and crossed products are presented for actions of
general locally compact groups,
most of the proofs and examples are restricted
to the discrete case,
which is often considerably easier.
% If the group is not discrete,
% one needs a theory of integration of Banach space valued functions,
% which we do not discuss.
% Some things go through in general,
% changing sums to integrals and using approximate identities,
% but some is unknown or fails.
% In the later parts, we will use tracial rank zero;
% a definition will be given,
% but little will be said about its properties.
% 999 If add a new section, more will need to be said.
Items labelled ``Exercise'' are intended to be done by the reader.
Items labelled ``Problem'' or ``Question'' are open questions.
By convention,
all topological groups will be assumed to be Hausdorff.
Homomorphisms of \ca{s} will be *homomorphisms.
We also use the following terminology.
\begin{dfn}\label{D_6X13_Kbg}
A {\emph{Kirchberg algebra}}
is a separable nuclear purely infinite simple \ca.
\end{dfn}
We don't assume that a Kirchberg algebra
satisfies the Universal Coefficient Theorem.
We now give enough of the basic definitions related to
group actions on \ca{s}
and locally compact spaces
that the discussion in the rest of this section
will make sense.
\begin{dfn}\label{D:Action}
Let $G$ be a topological group, and let $A$ be a \ca.
An {\emph{action of $G$ on $A$}}
is a group \hm{} $\af \colon G \to \Aut (A)$,
usually written $g \mapsto \af_g$,
such that, for every $a \in A$, the
function $g \mapsto \af_g (a)$,
from $G$ to $A$, is norm \ct.
\end{dfn}
The continuity condition
is the analog of requiring that a unitary representation
of $G$ on a Hilbert space be \ct{} in the strong operator topology.
It is usually much too strong a condition to
require that $g \mapsto \af_g$ be a norm \ct{} map from
$G$ to the bounded operators on~$A$.
For example, let $G$ be a locally compact group,
and let $\af \colon G \to \Aut ( C_0 (G))$
be the action given by $\af_g (f) (k) = f (g^{1} k)$
for $f \in C_0 (G)$ and $g, k \in G$.
We certainly want this action to be \ct.
Suppose $g, h \in G$ with $g \neq h$.
Then $\ \af_g  \af_h \ \geq 2$,
as can be seen by choosing $f \in C_0 (G)$
such that $f (g^{1}) = 1$, $f (h^{1}) =  1$,
and $\ f \ = 1$.
Indeed,
one gets
\[
\ \af_g  \af_h \
\geq \ \af_g (f)  \af_h (f) \
\geq  \af_g (f) (1)  \af_h (f) (1) 
=  f (g^{1})  f (h^{1}) 
= 2.
\]
(The inequality $\ \af_g  \af_h \ \leq 2$ is easy.)
Thus, if $G$ is not discrete,
then $g \mapsto \af_g$ is
never norm \ct.
Of course, if $G$ is discrete,
there is no difference
between the continuity conditions.
Isomorphism of actions is called conjugacy.
\begin{dfn}\label{D_6X01_Conjugate}
Let $G$ be a group,
let $A$ and $B$ be \ca{s},
and let $\af \colon G \to \Aut (A)$
and $\bt \colon G \to \Aut (B)$
be actions of $G$ on $A$ and~$B$.
A \hm{} $\ph \colon A \to B$ is called {\emph{equivariant}}
if $\ph \circ \af_g = \bt_g \circ \ph$
for all $g \in G$.
The actions $\af$ and $\bt$ are called {\emph{conjugate}}
if there is an equivariant isomorphism
$\ph \colon A \to B$.
\end{dfn}
Equivariance means that the following diagram commutes
for all $g \in G$:
\[
\begin{CD}
A @>{\af_g}>> A \\
@VV{\ph}V @VV{\ph}V \\
B @>{\bt_g}>> B.
\end{CD}
\]
Given $\af \colon G \to \Aut (A)$,
we will construct
in Section~\ref{Sec:CP} below a crossed product \ca{} $C^* (G, A, \af)$
and a reduced crossed product \ca{} $C^*_{\mathrm{r}} (G, A, \af)$.
(There are many other commonly used notations.
See Remark~\ref{R:CPNtn}.
We may omit $\af$ if it is understood.)
If $A$ is unital and $G$ is discrete,
the crossed products are a suitable
completion of the algebraic skew group ring $A [G]$,
with multiplication determined by $g a g^{1} = \af_g (a)$
for $g \in G$ and $a \in A$.
The main subject of these notes
is some aspects of the structure of crossed products.
Earlier sections give a large collection of examples of
group actions on \ca{s},
and discuss the full and reduced group \ca{s},
which are the crossed products gotten from
the trivial action of the group on~$\C$.
Just as locally compact spaces give commutative \ca{s},
group actions on locally compact spaces
give group actions on commutative \ca{s}.
\begin{dfn}\label{DActionOnSp}
Let $G$ be a topological group,
and let $X$ be a topological space.
An {\emph{action}} of~$G$ on~$X$
is a \ct{} function $G \times X \to X$,
usually written $(g, x) \mapsto g \cdot x$ or $(g, x) \mapsto g x$,
such that $(g h) x = g (h x)$ for all $g, h \in G$ and $x \in X$
and $1 \cdot x = x$ for all $x \in X$.
\end{dfn}
Discontinuous actions on spaces are of course also possible,
but we will encounter very few of them.
\begin{dfn}\label{DFromSpToCStar}
Let $G$ be a topological group,
let $X$ be a \lchs,
and let $(g, x) \mapsto g x$ be an action of~$G$ on~$X$.
We define the induced action of $G$ on $C_0 (X)$,
say~$\af$,
by $\af_g (f) (x) = f (g^{1} x)$ for $g \in G$, $f \in C_0 (X)$,
and $x \in X$.
(Exercise~\ref{Ex:OnSp} asks for a proof that we really get an action.)
\end{dfn}
% For a \ct{} action of a locally compact group $G$ on a \lchs~$X$,
% there is a corresponding action $\af \colon G \to \Aut (C_0 (X))$,
% given by $\af_g (f) (x) = f (g^{1} x)$.
The inverse appears for the same reason it does in the
formula for the left regular representation of a group.
If $G$ is not abelian, the inverse is necessary
to get $\af_g \circ \af_h$ to be $\af_{g h}$ rather than $\af_{h g}$.
If $K \subset X$ is a compact open set,
so that its characteristic function $\ch_K$ is in $C_0 (X)$,
then $\af_g (\ch_K) = \ch_{g K}$,
not $\ch_{g^{1} K}$.
We write $C^* (G, X)$ for the \cp{} \ca{}
and $C^*_{\mathrm{r}} (G, X)$ for the reduced \cp{} \ca.
We call them the \tgca{} and the reduced \tgca.
\begin{exr}\label{Ex:OnSp}
Let $G$ be a topological group,
and let $X$ be a \lchs.
Prove that the formulas given above
determine a one to one correspondence between \ct{} actions of $G$
on $X$ and \ct{} actions of $G$ on $C_0 (X)$.
(The main point is to show that
an action on $X$ is \ct{}
\ifo{} the corresponding action on $C_0 (X)$ is \ct.)
\end{exr}
For the special case $G = \Z$,
the same notation is often used for the action and for the
automorphism which generates it.
Thus, if $A$ is a \ca{} and $\af \in \Aut (A)$,
one often writes $C^* (\Z, A, \af)$.
For a \hme{} $h$ of a locally compact Hausdorff space~$X$,
one gets an automorphism $\af \in \Aut (C_0 (X))$,
and thus an action of $\Z$ on $C_0 (X)$.
We abbreviate this crossed product to $C^* (\Z, X, h)$.
We give some motivation for
studying group actions on \ca{s} and their crossed products.
\begin{enumerate}
\item\label{SemiDirPrd}
Let $G$ be a locally compact group obtained as a
semidirect product $G = N \rtimes H$.
The action of $H$ on $N$ gives actions of $H$ on the full and reduced
group \ca{s}
$C^* (N)$ and $C^*_{\mathrm{r}} (N)$,
and one has
$C^* (G) \cong C^* (H, \, C^* (N))$ and
$C^*_{\mathrm{r}} (G) \cong C^*_{\mathrm{r}} (H, \, C^* (N))$.
\item\label{Time}
Probably the most important group action is time evolution:
if a \ca~$A$ is supposed to represent the possible states
of a physical system in some manner,
then there should be an action $\af \colon \R \to \Aut (A)$
which describes the time evolution of the system.
Actions of $\Z$,
which are easier to study,
can be though of as ``discrete time evolution''.
\item\label{GetSimp}
Crossed products are a common way of constructing simple \ca{s}.
Here are some of the more famous examples.
\begin{itemize}
\item
The irrational rotation algebras.
See Example~\ref{E:C:IrrRot} below.
They were not originally defined as crossed products.
\item
The BunceDeddens algebras.
See~\cite{BD} or Section V.3 of~\cite{Dvd};
one crossed product realization is Theorem VII.4.1 of~\cite{Dvd},
and another, for a specific choice of BunceDeddens algebra,
and using an action of the dyadic rationals on the circle,
can be found at the beginning of Section VIII.9 of~\cite{Dvd}.
% 999 eventually add this example and forward reference
\item
The reduced \ca{} of the free group on two generators.
See~Section VII.7 of~\cite{Dvd};
simplicity is proved in \Thm{T_3301_CStRFnSimp}.
\end{itemize}
We will see other examples later.
\item\label{FromSpace}
If one has a \hme~$h$ of a locally compact Hausdorff space~$X$,
the crossed product $C^* (\Z, X, h)$
sometimes carries considerable information
about the dynamics of $h$.
The best known example is the result of~\cite{GPS}
on \mh{s} of the Cantor set:
isomorphism of the \tgca{s} is equivalent to strong orbit equivalence
of the \hme{s}.
\item\label{EqKThy}
% 999 Rewrite, add references
For compact groups,
equivariant indices take values in the equivariant Ktheory
of a suitable \ca\ with an action of the group.
When the group is not compact,
one usually needs instead the Ktheory of the crossed product \ca,
or of the reduced crossed product \ca.
(When the group is compact, this is the same thing.)
In other situations as well,
the Ktheory of the full or reduced \cp{} is the appropriate
substitute for equivariant Ktheory.
\end{enumerate}
There are many directions in the theory of crossed products.
These notes are biased towards the general problem of
understanding the structure of
crossed products
by finite groups, by~$\Z$, and by more complicated groups,
in cases in which these crossed products are expected to be simple,
and, in good cases, classifiable in the sense of the Elliott program.
% (However, we will not get very far
% in the direction of classification.)
I should at least mention some of the other directions.
Some of these are large and very active areas of research,
some are small but active areas of research,
in some it seems that most of the theory has been worked out,
and some are just beginning.
The list is not complete,
and there is also interesting work which doesn't fit
under any of these directions.
Directions of work on group actions
which don't involve crossed products
(such as work on classification of actions)
are mostly not mentioned.
The references provided are not necessarily recent
or representative of work in the subject;
they are often just ones I have managed to find,
sometimes with the help of people in the area.
Moreover, some very active areas have very few references listed,
perhaps only one or two books or survey articles.
\begin{itemize}
\item
% Done
The relation between the structure of a nonminimal homeomorphism
and the structure of its crossed product.
See \cite{Tmy1}, \cite{Tmy2}, and~\cite{Tmy3}.
The article~\cite{ST} is one example
of more recent work in this direction.
\item
% Maybe eventually more % 999
The structure of crossed products of continuous trace C*algebras
by actions for which the induced action on the primitive ideal space
is proper.
See the textbook~\cite{RbnWms}.
\item
% Done
Extensions of the notion of crossed product to coactions
and actions of C*~Hopf algebras (``quantum groups''),
and the associated duality theory.
The textbook~\cite{Tmmn} on quantum groups
has a chapter on this subject.
One of the classic papers is~\cite{BjSk},
which uses the formalism of multiplicative unitaries and,
among other things,
give a version of Takai duality for crossed products by quantum groups.
For a recent survey of this area,
see~\cite{DCmm}.
For one application
(imprimitivity theorems,
in connection with induction and restriction of
representations of quantum groups),
see~\cite{Vaes5},
and the earlier paper~\cite{EKQR}.
\item
% Done
Crossed products twisted by cocycles.
Cocycles can be untwisted by stabilization,
so such crossed products are stably isomorphic to
ordinary crossed products.
See Corollary~3.7 of~\cite{PkrRbn},
with further applications in~\cite{PkrRbn2}.
But for some purposes,
one doesn't want to stabilize.
\item
% Done
Von Neumann algebra crossed products.
There are several chapters on group actions and crossed products
in Volume~2 \cite{Tk_Bk2}
and Volume~3 \cite{Tk_Bk3}
of Takesaki's three volume work on operator algebras.
One direction with major recent activity
is the classification of von Neumann algebra
\cp{s} by ergodic measure preserving actions
of countable nonamenable groups on probability spaces,
including cases in which the group and the action can be
recovered from the von Neumann algebra.
See~\cite{Ion7} for a recent survey.
The papers \cite{Ppa6}, \cite{Vaes7}, and~\cite{Vaes11}
are older surveys.
Two of the important early papers
in this direction are \cite{Ppa3} and~\cite{Ppa4}.
Two of many more recent important papers
are \cite{PpVs} and~\cite{Ion1}.
\item
% Done
Smooth crossed products.
See \cite{Swz1} and~\cite{Swz2}
for some of the foundations.
See \cite{Nst} and~\cite{ElNtNs1}
for cyclic cohomology of crossed products
by ${\mathbb{Z}}$ and ${\mathbb{R}}$,
and see~\cite{PhSch} for their Ktheory.
\item
% Maybe eventually more % 999
C*algebras of groupoids,
and crossed products by actions of groupoids on C*algebras.
The original book is~\cite{Rn};
a more recent book is~\cite{Ptn}.
There is much more work in this direction.
\item
% Done
Computation of the Ktheory of crossed products,
from the PimsnerVoi\cu\les\cu exact sequences
\cite{PV}, \cite{PV}
their generalization~\cite{Pmnr}
and the
Connes isomorphism \cite{Cnns} through the BaumConnes conjecture.
See~\cite{LcRc} for a survey of the BaumConnes conjecture
and related conjectures.
\item
% Done.
The Connes spectrum and its generalizations.
See \cite{OPd1}, \cite{OPd2}, \cite{OPd3}, and \cite{Kshm1}
for some of the early work for abelian groups.
The Connes spectrum for compact nonabelian groups
was introduced in~\cite{GtLzPg},
and for actions of compact quantum groups in~\cite{DmtPgd}.
These ideas have even been extended into ring theory,
in which there is no topology~\cite{OstPgd}.
\item
% Done.
The ideal structure of crossed products,
without assuming analogs of freeness or properness.
Much of Williams' book~\cite{Wlms}
is related to this subject.
A generalization to groupoids can be found in~\cite{Rnlt}.
See~\cite{Srk}, \cite{EctEmr}, and~\cite{EctWms}
for examples of more recent work.
The Connes spectrum is also relevant here.
\item
% Done.
Structural properties of crossed products which are inspired
by those related to the Elliott program,
but in cases in which neither the original algebra
nor the crossed product is expected to be simple.
(See \cite{PsnPh} and~\cite{PsnPh2}
for some recent work,
and \cite{RrSr}, \cite{GrSr}, \cite{KbgSrk1}, and~\cite{KbgSrk2}
for a related direction.)
\item
% More ???
Crossed products by endomorphisms, semigroups, and partial actions.
The book~\cite{ExlBk} will appear soon,
and is already available on the arXiv.
A recent paper with some relation to problems considered here
is~\cite{GrSr}.
\item
% Done.
Semicrossed products:
nonselfadjoint crossed products gotten from semigroup actions
on \ca{s}.
This area has a long history,
starting with Arveson in the weak operator closed case~\cite{Avsn}
and with Arveson and Josephson in the norm closed case~\cite{AvsJpn}.
See \cite{DvdKts} and~\cite{DvFlKd2} for two much more recent survey
articles in the area,
and~\cite{DvFlKd1} for a recent substantial paper.
\item
% Done.
Crossed products by actions of locally compact groups
on nonselfadjoint Hilbert space operator algebras.
This is a very new field,
in effect started in~\cite{KtsRmy}.
It already has applications to \cp{s}
of \ca{s}; see~\cite{Ktsl}.
\item
% Done.
$L^1$~crossed products, so far mostly of $C (X)$ by~$\Z$.
See \cite{dJT}, \cite{dJST}, \cite{KshTmy}, and references
in these papers.
\item
% Done.
Algebraic crossed products of \ca{s} by discrete groups,
so far mostly of $C (X)$ by~$\Z$.
See \cite{SSdJ1}, \cite{SSdJ2}, and~\cite{SSdJ3}.
\item
% Done.
General Banach algebra crossed products.
The beginnings of a general theory appear in~\cite{DDW}.
\item
% Done.
Crossed products of algebras of operators on $L^p$~spaces.
This is very recent. See~\cite{PhLp3}.
\end{itemize}
We will not touch at all on many of these directions.
However, work on the structure and classification of simple
crossed products does not occur in isolation,
and we will need some information from some of the other directions,
including Ktheory, groupoids, and partial actions.
% 999
The textbook references on crossed products that I know are
Chapters~7 and~8 of~\cite{Pd1}
(very condensed; the primary emphasis is on properties
of group actions rather than of crossed products),
\cite{Wlms} (quite detailed; the primary emphasis is on ideal structure
of general crossed products),
and
Chapter~8 of~\cite{Dvd} (the primary emphasis is on crossed products,
especially by $\Z$, as
a means of constructing interesting examples of \ca{s}).
There are no textbooks with primary emphasis on
classification of crossed products
or on crossed products by \mh{s}.
The ``further reading'' section in
the introduction of~\cite{Wlms} gives a number of references
for various directions in the theory of crossed products
which are treated neither in~\cite{Wlms} nor here.
I am grateful to Ken Davidson,
Ruy Exel,
Eusebio Gardella,
Adrian Ioana,
Elias Katsoulis,
Jae Hyup Lee,
Hutian Liang,
Adam Skalski,
Stuart White,
Dana Williams,
Guoliang Yu,
% References to Hatcher, and to Wolfgang Lueck and Holger Reich
and many others for comments, suggestions, answering questions,
finding misprints
(many of which remain),
and providing solutions to problems left open in earlier versions.
% Timothy Rainone (one misprint).
% ???
\part{Group Actions}\label{Part_Basic}
\section{Examples of Group Actions on Locally Compact
Spaces}\label{Sec_CommEx}
\indent
This is the first of three sections devoted to examples
of group actions.
In this section,
we give examples of actions on commutative \ca{s}.
In Section~\ref{Sec_NCEx} we give a variety
of examples of actions on noncommutative \ca{s},
and in Section~\ref{Sec_AddGauge}
we give an additional collection of examples
of actions that are similar to gauge actions.
Some general comments are in order.
The main focus of the later part of these notes is
group actions $\af \colon G \to \Aut (A)$
for a locally compact group $G$ on a \ca~$A$
such that the \cp{} $C^* (G, A, \af)$
or reduced \cp{} $C^*_{\mathrm{r}} (G, A, \af)$
(as defined in Sections \ref{Sec:CP} and~\ref{Sec:RedCP})
is at least as complicated as $A$ itself.
In particular, we usually want the (reduced) \cp{}
to be simple,
and to be purely infinite if~$A$ is.
There are many interesting and sometimes very important actions
whose nature is quite different,
and in our examples
we do not discriminate:
we give a very broad collection.
We make some comments (without proof)
about the kinds of crossed products one gets.
These don't make sense without knowing at least a little
about \cp{s} (Sections \ref{Sec:CP} and~\ref{Sec:RedCP}),
so it is useful to come back to the examples
after reading much farther into these notes.
Some of the comments made will be proved in the later
part of these notes,
but for many no proof will be given at all.
For actions of compact groups,
the crossed product is often closely related to
the fixed point algebra~$A^G$
(or $A^{\af}$ when necessary to avoid confusion),
given by
\[
A^G = \big\{ a \in A \colon
{\mbox{$\af_g (a) = a$ for all $g \in G$}} \big\}.
\]
Instead of commenting on the \cp,
we therefore sometimes comment on the fixed point algebra.
There is one way in which we do discriminate.
Crossed products only exist for actions of locally compact groups,
because the group must have a Haar measure.
With very few exceptions,
we therefore only give examples of actions of locally compact groups.
We will also sometimes mention the Rokhlin property or
related conditions on actions.
Some of these are defined later.
(The Rokhlin property for actions of finite groups
is in \Def{D_RPDfn},
and the tracial Rokhlin property for actions of finite groups
is in \Def{D_TRP}.)
For some, however, no definition will be given in these notes.
Turning specifically to the commutative case,
recall from \Def{DFromSpToCStar}
and Exercise~\ref{Ex:OnSp}
that giving an action of a topological group~$G$
on a commutative \ca{} $C_0 (X)$
is the same as giving an action of~$G$ on the underlying space~$X$.
When $G$ is locally compact,
the crossed product \ca{}
$C^* (G, \, C_0 (X))$ is usually abbreviated to $C^* (G, X)$.
(See \Def{DTGCA}.)
As noted above,
we are mostly interested in the case in which $C^* (G, X)$ is simple.
So as to be able to make meaningful comments,
we discuss several easy to state conditions on an action on
a locally compact space
which are related to simplicity of the \cp.
We will say more about these conditions in
Section~\ref{Sec_MH}.
\begin{dfn}\label{D_3331_1Min}
Let a topological group $G$ act \ct{ly}
on a topological space~$X$.
The action is called {\emph{minimal}} if
whenever $T \subset X$ is a closed subset such that $g T \subset T$
for all $g \in G$,
then $T$ is trivial,
that is, $T = \varnothing$ or $T = X$.
\end{dfn}
\begin{lem}\label{L_3415_MinOrb}
Let a topological group $G$ act \ct{ly}
on a topological space~$X$.
The action is minimal \ifo{} for every
$x \in X$, the orbit $G x = \{ g x \colon g \in G \}$ is dense in~$X$.
\end{lem}
\begin{proof}
If there is $x \in X$ such that
$G x$ is not dense,
then ${\ov{G x}}$ is a nontrivial $G$invariant closed subset
of~$X$.
For the converse,
let $T \subset X$ be a nontrivial $G$invariant closed subset
of~$X$.
Choose any $x \in T$.
Then $G x \subset T$ and is therefore not dense.
\end{proof}
\Lem{L:MHC} gives a number of weaker equivalent conditions
for minimality for the special case $G = \Z$ and $X$ is compact.
As shown by the action of $\Z$ on its one point compactification
(in Example~\ref{E_3415_OnePtCpt} below),
it is not enough to require that one orbit be dense.
There are special circumstances under which density of
one orbit is sufficient,
such as for an action of a subgroup
by translation on the whole group.
See Proposition~\ref{P_3408_DnsOrb} below.
It follows from Theorem~\ref{T_3331_RedCP}(\ref{T_3331_RedCP_NZI})
that minimality is a necessary condition
for simplicity of $C^*_{\mathrm{r}} (G, X)$,
and from Theorem~\ref{T_CPExact}
that minimality is a necessary condition
for simplicity of $C^* (G, X)$.
(As we will see in Section~\ref{Sec:RedCP},
$C^*_{\mathrm{r}} (G, X)$ is a quotient of $C^* (G, X)$,
so we really only need to cite
Theorem~\ref{T_3331_RedCP}(\ref{T_3331_RedCP_NZI}).)
\begin{dfn}\label{D_3331_Free}
Let a locally compact group $G$ act \ct{ly} on a locally
compact space $X$.
The action is called {\emph{free}} if
whenever $g \in G \setminus \{ 1 \}$ and $x \in X$,
then $g x \neq x$.
The action is called {\emph{essentially free}} if
whenever $g \in G \setminus \{ 1 \}$,
the set $\{ x \in X \colon g x = x \}$ has empty interior.
\end{dfn}
Essential freeness makes sense in general,
but for nonminimal actions it is not the most useful condition.
One should at least insist that the restriction of the action
to any closed invariant subset be essentially free
in the sense of \Def{D_3331_Free}.
The action of $\Z$ on its one point compactification
by translation is essentially free in the sense of \Def{D_3331_Free},
but but does not satisfy the stronger condition,
and its transformation group \ca{}
does not behave the way that a good version of
essential freeness for nonminimal actions should imply.
\begin{prp}\label{P_3421_FreeEssF}
Let $G$ be an abelian group.
Then every minimal and essentially free action of
$G$ on a topological space~$X$
is free.
\end{prp}
\begin{proof}
Let $(g, x) \mapsto g x$ be a minimal action of~$G$
which is not free.
Then there is $h \in G \setminus \{ 1 \}$
such that the closed set $T = \{ x \in X \colon h x = x \}$
is not empty.
We claim that $T$ is invariant.
To see this,
let $g \in G$ and let $x \in T$.
Then $h (g x) = g (h x) = g x$,
so $g x \in T$.
This proves the claim.
By minimality, $T = X$.
Therefore the action is not essentially free.
\end{proof}
The actions in Example~\ref{E_3302_MHOn2Tor}
and Example~\ref{EFnBdy} below
are minimal and essentially free but not free.
The following theorem
(to be proved in Section~\ref{Sec_MH})
provides a very useful sufficient condition
for simplicity of $C^*_{\mathrm{r}} (G, X)$.
\begin{thm}[Theorem~\ref{T:AS}]\label{T_3331_1AS}
Let a discrete group $G$ act minimally and essentially freely
on a locally compact space~$X$.
Then $C^*_{\mathrm{r}} (G, X)$ is simple.
\end{thm}
This condition is not necessary;
it follows from Theorem~\ref{T_3301_CStRFnSimp}
that the trivial action of the free group on two generators
on a one point space has a simple reduced crossed product.
The analog of minimality for actions on measure spaces
is ergodicity.
\begin{dfn}\label{D_3408_Ergod}
Let $(X, {\mathcal{B}}, \mu)$
be a measure space,
let $G$ be a group,
and let $(g, x) \mapsto g x$
be an action of $G$ on~$X$.
For each $g \in G$,
assume that the map $h^g \colon X \to X$,
given by $h^g (x) = g x$,
is measurable and preserves the measure~$\mu$.
We say that the action is
{\emph{ergodic}}
if whenever a measurable set $E \subset X$ satisfies
$g E = E$ for all $g \in G$,
then $\mu (E) = 0$ or $\mu (X \setminus E) = 0$.
\end{dfn}
The conditions on the action are just that
the $\sm$algebra ${\mathcal{B}}$ and the measure $\mu$
are both $G$invariant.
That is,
for all $g \in G$ and all $E \in {\mathcal{B}}$,
we have $g E \in {\mathcal{B}}$
and $\mu (g E) = \mu (E)$.
(Actually, all that one needs is that the measure class of $\mu$
is $G$invariant,
that is, that $\mu (g E) = 0$ \ifo{} $\mu (E) = 0$
for $E \in {\mathcal{B}}$ and $g \in G$.)
\begin{dfn}\label{D_3408_UniqErg}
Let $X$ be a compact metric space,
let $G$ be a topological group,
and let $(g, x) \mapsto g x$
be an action of $G$ on~$X$.
We say that the action is
{\emph{uniquely ergodic}}
if there is a unique $G$invariant Borel probability measure
on~$X$.
\end{dfn}
In \Def{D_3408_UniqErg},
it turns out that the measure $\mu$ is necessarily ergodic.
More generally,
the $G$invariant Borel probability measures on~$X$
form a (possibly empty) weak* compact convex subset~$K$
of the dual space $C (X)^*$.
We prove the standard result that such a measure $\mu$ is ergodic \ifo{}
it is an extreme point of~$K$,
under the assumption that the group is discrete and countable.
This hypothesis is stronger than necessary,
but avoids some technicalities.
The proof that extreme points are ergodic measures
works in complete generality,
in particular,
no matter what the group is.
% 999 Find ref for more general result.
\begin{thm}\label{T_6X11_ErgodExtr}
Let $X$ be a compact metric space,
let $G$ be a countable discrete group,
and let $(g, x) \mapsto g x$
be an action of $G$ on~$X$.
Then a $G$invariant Borel probability measure $\mu$ on $X$
is ergodic \ifo{} it is an extreme point
in the set of all $G$invariant Borel probability measures on~$X$.
\end{thm}
\begin{proof}
First assume that $\mu$ is not ergodic.
Choose a $G$invariant Borel set $F \subset X$ such that
$0 < \mu (F) < 1$.
Define $G$invariant Borel probability measures
$\mu_1$ and $\mu_2$ on $X$ by
\[
\mu_1 (E) = \frac{\mu (E \cap F)}{\mu (F)}
\andeqn
\mu_2 (E) = \frac{\mu (E \cap (X \setminus F))}{\mu (X \setminus F)}
\]
for every Borel set $E \subset X$.
Taking $\af = \mu (F)$,
we have
$\af \mu_1 + (1  \af) \mu_2 = \mu$,
$\mu_1 \neq \mu_2$, and $\af \in (0, 1)$.
So $\mu$ is not an extreme point.
Now assume that $\mu$ is ergodic.
Suppose that $\mu_1$ and $\mu_2$
are $G$invariant Borel probability measures,
that $\af \in (0, 1)$,
and that $\af \mu_1 + (1  \af) \mu_2 = \mu$.
We prove that $\mu_1 = \mu$.
We have $\mu_1 \leq \af^{1} \mu$,
so $\mu_1 \ll \mu$.
Let $f_0 \colon X \to [0, \I]$
be a RadonNikodym derivative of $\mu_1$ with respect to $\mu$.
Since $\mu_1$ and $\mu$ are $G$invariant,
for every $g \in G$
the function $x \mapsto f_0 (g^{1} x)$
is also a RadonNikodym derivative of $\mu_1$ with respect to $\mu$,
and is therefore equal to $f_0 (x)$
almost everywhere with respect to~$\mu$.
Now define
\[
f (x) = \sup_{g \in G} f_0 (g^{1} x).
\]
Since $G$ is countable,
this function is equal to $f_0 (x)$
almost everywhere with respect to~$\mu$,
so
\[
\mu_1 (E) = \int_E f \, d \mu
\]
for every Borel set $E \subset X$.
Also,
$f$ is exactly $G$invariant.
For $\bt \in [0, \I)$
set
\[
E_{\bt} = \big\{ x \in X \colon f (x) \geq \bt \big\}.
\]
Then $E_{\bt}$ is a $G$invariant Borel set,
so $\mu (E_{\bt}) \in \{ 0, 1 \}$.
Whenever $\bt, \gm \in [0, \I)$ satisfy $\gm \geq \bt$,
we have $E_{\gm} \subset E_{\bt}$, so $\mu (E_{\bt}) \geq \mu (E_{\gm})$.
Also, $\mu (E_0) = 1$ and $\mu (E_{\af^{1} + 1}) = 0$.
Define
\[
r = \sup \big( \big\{ \bt \in [0, \I) \colon \mu (E_{\bt}) = 1 \big\} \big).
\]
If $r = 0$ then $f = 0$ so $\mu_1 = 0$,
which is clearly impossible.
So there is a strictly increasing sequence $(\bt_n)_{n \in \N}$
in $[0, \I)$ such that $\limi{n} \bt_n = r$.
We have
\[
X \setminus E_r = \bigcup_{n = 1}^{\I} (X \setminus E_{\bt_n}),
\]
so $\mu (E_r) = 1$.
It follows that $f$ is equal to the constant function~$r$
almost everywhere with respect to~$\mu$.
Since $\mu_1 (X) = 1$,
we get $r = 1$.
So $\mu_1 = \mu$,
as desired.
\end{proof}
\begin{thm}\label{T_6X11_ExistInv}
Let $X$ be a compact metric space,
let $G$ be an amenable locally compact group,
and let $(g, x) \mapsto g x$
be an action of $G$ on~$X$.
Then there exists a $G$invariant Borel probability measure
on~$X$.
\end{thm}
See the discussion before \Thm{T_4131_GpFToRed}
for more on amenable groups.
\begin{proof}
In~\cite{Grf},
combine Theorem 3.3.1 and Theorem 2.2.1.
\end{proof}
\begin{cor}\label{C_6X11_ExistErgod}
Let $X$ be a compact metric space,
let $G$ be an amenable locally compact group,
and let $(g, x) \mapsto g x$
be an action of $G$ on~$X$.
Then there exists an ergodic
$G$invariant Borel probability measure $\mu$ on $X$.
\end{cor}
\begin{proof}
\Thm{T_6X11_ExistInv} shows that
the set of $G$invariant Borel probability measures on $X$
is not empty.
It is easily seen to be a weak* compact convex subset of the
dual space of $C (X)$.
Therefore it has an extreme point, by Alaoglu's Theorem.
Any extreme point is an ergodic measure by
\Thm{T_6X11_ErgodExtr}.
(This direction of the proof of \Thm{T_6X11_ErgodExtr}
did not need any hypotheses on the group.)
\end{proof}
Part of the significance of $G$invariant Borel probability measures
is that,
when $G$ is discrete,
they give tracial states (\Def{D_Trace})
on the crossed product \ca.
See Example~\ref{EInvMeas}.
Moreover, if the action is free,
then sometimes all tracial states on the crossed product
arise this way.
See Theorem~\ref{T11202Traces}.
% We will compute the \cp{s} for some of these examples
% in Section~\ref{Sec_Comp},
% but there are many more examples here than there are
% computations of examples in Section~\ref{Sec_Comp}.
Now we give examples.
\begin{exa}\label{E:Triv}
The group $G$ is arbitrary locally compact,
the space $X$ consists of just one point,
and the action is trivial.
This action is minimal, but is as far from being free as possible.
It gives the trivial action of $G$ on the \ca~$\C$.
The full and reduced crossed products are
the usual full and reduced group \ca{s}
$C^* (G)$ and $C^*_{\mathrm{r}} (G)$,
discussed in Section~\ref{Sec_GpCSt}
(when $G$ is discrete)
and Section~\ref{Sec_LCGpCSt}.
As we will see,
this is essentially immediate by comparing definitions.
See Example~\ref{E:C:Triv} below.
More generally,
any group has a trivial action on any space.
\end{exa}
\begin{exa}\label{E:Tr}
The group $G$ is arbitrary locally compact,
$X = G$,
and the action is given by the group operation:
$g \cdot x = g x$.
The full and reduced crossed products are both
isomorphic to $K (L^2 (G))$.
We will prove this for the discrete case
in Example~\ref{E_C_Trans} below.
This action is called (left) translation.
It is clearly free.
It is also minimal, but in a rather trivial way:
there are no nontrivial invariant subsets,
closed or not.
As we will see,
in the interesting examples,
with more interesting crossed products,
the orbits are dense but not equal to the whole space.
See the irrational rotations in Example~\ref{E_Rot}.
Also see \Prp{P_3408_DnsOrb},
Example~\ref{E_3302_Furst},
Example~\ref{E:pAdic},
\Def{D_6827_Odometer},
and
Example~\ref{E_3408_SumProd}.
Many further examples will appear.
More generally, if $H \subset G$ is a closed subgroup,
then $G$ acts \ct{ly} on $G / H$ by translation.
Example~\ref{E:Triv} is the case $H = G$.
See Example~\ref{E:C:NonEff} below for the computation
of the \cp{} when
$G = \Z$ and $H = n \Z$,
and for the description of the \cp{} in the general case.
This action is still minimal (in the same trivial way as before),
but for $H \neq \{ 1 \}$ it is no longer free.
\end{exa}
\begin{exa}\label{E_3408_SubgpActs}
We can generalize left translation in Example~\ref{E:Tr}
in a different way.
Again let $G$ be an arbitrary locally compact group,
set $X = G$,
and let $H \subset G$ be a closed subgroup.
Then $H$ acts on $X = G$ by left translation.
The action is still free,
but is now no longer minimal (unless $H = G$).
% For $n \in \N$,
% the case
% \[
% G = S^1
% \andeqn
% H = \{ \exp (2 \pi i k / n) \colon k = 0, 1, \ldots, n \}
% \]
% appears in Example~\ref{E_Rot}.
The crossed product
$C^* (H, G)$ turns out to be
stably isomorphic to $K (L^2 (H)) \otimes C_0 (G / H)$.
Stably, there is no ``twisting'',
even though $G$ may be a nontrivial bundle over $G / H$.
See Theorem 14 and Corollary~15 in Section~3 of~\cite{Gr}.
\end{exa}
\begin{exa}\label{E_3415_Conj}
Let $G$ be any locally compact group.
Then $G$ acts on itself by conjugation:
$g \cdot k = g k g^{1}$ for $g, k \in G$.
Unless $G = \{ 1 \}$,
this action is neither free nor minimal,
since $1$ is a fixed point.
There is also a conjugation action of $G$
on any normal subgroup of~$G$.
\end{exa}
\begin{exa}\label{E_3415_OnePtCpt}
Let $G = \Z$ and let $X = \Z^{+}$,
the one point compactification $\Z \cup \{ \I \}$ of~$\Z$.
Then $\Z$ acts on $\Z^{+}$ by translation,
fixing~$\I$.
This action has a dense orbit (namely~$\Z$),
but is not minimal (since $\{ \I \}$ is invariant)
and not free.
Here are some related examples.
The group $\Z$ acts on $\Z \cup \{  \I, \I \}$
by translation,
fixing $ \I$ and~$\I$.
Both $\R$ and $\Z \subset \R$
act on both $\R^{+} \cong S^1$ and
$[ \I, \I]$ by translation,
fixing the point or points at infinity.
None of these actions is either free or minimal.
\end{exa}
\begin{exa}\label{E_Rot}
Take $X = S^1 = \{ \zt \in \C \colon  \zt  = 1 \}$.
Taking $G = S^1$, acting by translation, gives a special case of
Example~\ref{E:Tr}.
But we can also take $G$ to be the finite subgroup
of $S^1$ of order~$n$ generated by $\exp (2 \pi i / n)$,
still acting by translation (in this case, usually called rotation).
This is a special case of Example~\ref{E_3408_SubgpActs}.
The computation of the \cp{}
for this case is in Example~\ref{E:C:FiniteRot}.
% This action is still free, but not minimal.
Or we can fix $\te \in \R$, and take $G = \Z$,
with $n \in \Z$ acting by $\zt \mapsto \exp (2 \pi i n \te) \zt$.
(The use of $\exp (2 \pi i \te)$ rather than
$\exp (i \te)$ is standard here.)
These are {\emph{rational rotations}} (for $\te \in \Q$) or
{\emph{irrational rotations}} (for $\te \not\in \Q$).
% See Example~\ref{E:C:IrrRot}.
The rational rotations are neither free nor minimal.
(Their \cp{s}
are discussed in Example~\ref{E_4302_RatRot}.)
The irrational rotations are free (easy)
and minimal (intuitively clear but slightly tricky;
see \Lem{L:IrratRotDense}
and Proposition~\ref{P_3408_DnsOrb} below).
Irrational rotations are also uniquely ergodic.
(Theorem~1.1 of~\cite{Frst} gives unique ergodicity for a
class of \hme{s} of the circle which contains
the irrational rotations.)
\end{exa}
\begin{lem}\label{L:IrratRotDense}
Let $\te \in \R \setminus \Q$.
Then $\big\{ e^{2 \pi i n \te} \zt \colon n \in \Z \big\}$
is dense in~$S^1$.
\end{lem}
There are a number of ways to prove this lemma.
In~\cite{Wlms} (see the proof of Lemma~3.29),
the basic idea is that all proper closed subgroups of $S^1$ are finite.
One can also get the result from number theory:
there is a constant $c$ such that there are pairs $(p, q)$ of integers,
with $q$ arbitrarily large,
such that $ \te  p / q  < c q^{2}$.
(See Corollary~1B of~\cite{Smdt}.
The best general constant is $1/\sqrt{5}$; see Theorem~2F of~\cite{Smdt}.
We thank Shabnam Akhtari for pointing out this reference.)
We give here a proof close to that of~\cite{Wlms}.
\begin{proof}[Proof of Lemma~\ref{L:IrratRotDense}]
It suffices to prove that $\Z + \te \Z$ is dense in~$\R$.
Suppose not.
Let
$t = \inf \big( \big\{ x \in {\overline{\Z + \te \Z}}
\colon x > 0 \big\} \big)$.
We will show that $t = 0$.
So suppose $t > 0$.
We claim that ${\overline{\Z + \te \Z}} = \Z t$.
First, ${\overline{\Z + \te \Z}}$ is clearly a subgroup of~$\R$.
So $\Z t \subset {\overline{\Z + \te \Z}}$.
Suppose the reverse inclusion is false.
Then there are $m \in \Z$ and $r \in {\overline{\Z + \te \Z}}$
such that $m t < r < (m + 1) t$.
But then $r  m t \in {\overline{\Z + \te \Z}} \cap (0, t)$.
This contradiction proves the claim.
It is clear that the only subset of $\R$ whose closure is $\Z t$
is $\Z t$ itself.
So $\Z + \te \Z = \Z t$.
Therefore there are $m, n \in \Z$
with $\te = m t$ and $1 = n t$.
So $n \neq 0$ and $\te = \frac{m}{n} \in \Q$.
This contradiction shows that $t = 0$.
Now let $r \in \R$.
We claim that $r \in {\overline{\Z + \te \Z}}$.
Let $\ep > 0$.
Choose $s \in {\overline{\Z + \te \Z}}$
such that $0 < s < \ep$.
Choose $n \in \Z$ such that $n s \leq r < (n + 1) s$.
Then $n s \in {\overline{\Z + \te \Z}}$
and $ r  n s  < s < \ep$.
So the closure of ${\overline{\Z + \te \Z}}$ contains~$r$.
The claim follows.
\end{proof}
Here is a second proof,
based on part of a lecture by David Kerr.
Again, it suffices to prove that $\Z + \te \Z$ is dense in~$\R$.
Suppose this fails.
Choose $\ld_1, \ld_2 \in \R$ such that
$(\ld_1, \ld_2)$ is a connected component
of $\R \setminus {\overline{\Z + \te \Z}}$.
Let $F$ be the image of $\Z + \te \Z$ in $\R / \Z$, which we
identify with~$S^1$.
Since $\te$ is irrational,
$F$ is infinite,
so that for every $\ep > 0$ there are distinct points in $F$
whose arc length distance is less than~$\ep$.
Equivalently,
for every $\ep > 0$ there are $r, s \in \Z + \te \Z$
such that $0 < s  r < \ep$.
Choose such numbers $r$ and $s$ for $\ep = \ld_2  \ld_1$.
We have $\ld_1 \in {\overline{\Z + \te \Z}}$,
so there is $t \in \Z + \te \Z$ such that
$ t  \ld_1  < s  r$.
Since $t \not\in (\ld_1, \ld_2)$ and $s  r < \ld_2  \ld_1$,
we have $\ld_1  (s  r) < t \leq \ld_1$.
Therefore
\[
\ld_1 < t + (s  r) < t + \ld_2  \ld_1 \leq \ld_2.
\]
It follows that $t + (s  r) \in (\Z + \te \Z) \cap (\ld_1, \ld_2)$,
which is a contradiction.
\begin{prp}\label{P_3408_DnsOrb}
Let $G$ be a topological group,
and let $H \subset G$ be a dense subgroup.
Then the action of $H$ on $G$ be left translation
(in which $h \cdot g$ is just the group product $h g$
for $h \in H$ and $g \in G$)
is a free minimal action of $H$ on~$G$.
\end{prp}
\begin{proof}
That this formula defines an action is obvious,
as is freeness.
For minimality,
let $T \subset G$ be a nonempty closed $H$invariant subset.
Choose $g_0 \in T$.
Then $H g_0 \subset T$.
Moreover, $H$ is dense in~$G$
and right multiplication by $g_0$ is a \hme,
so $H g_0$ is dense in~$G$.
Therefore $T = G$.
\end{proof}
\begin{exa}\label{E_3302_Furst}
Let $\gm \in \R$, let $d \in \Z$,
and let $f \colon S^1 \to \R$ be \ct.
The associated Furstenberg transformation
$h_{\gm, d, f} \colon S^1 \times S^1 \to S^1 \times S^1$
(introduced in Section~2 of~\cite{Frst})
is defined by
\[
h_{\gm, d, f} (\zt_1, \zt_2)
= \big( e^{2 \pi i \gm} \zt_1, \,
\exp (2 \pi i f (\zt_1)) \zt_1^d \zt_2 \big)
\]
for $\zt_1, \zt_2 \in S^1$.
The inverse is given by
\[
h_{\gm, d, f} (\zt_1, \zt_2)
= \big( e^{ 2 \pi i \gm} \zt_1, \,
\exp \big( 2 \pi i [d \gm  f ( e^{ 2 \pi i \gm} \zt_1)] \big)
\zt_1^{d} \zt_2 \big)
\]
for $\zt_1, \zt_2 \in S^1$.
If $\gm \not\in \Q$ and $d \neq 0$,
Furstenberg proved that $h_{\gm, d, f}$ is minimal.
(See the discussion after Theorem~2.1 of~\cite{Frst}.)
By Theorem~2.1 of~\cite{Frst},
if $f$ is in addition smooth
(weaker conditions suffice),
then $h_{\gm, d, f}$ is uniquely ergodic.
For arbitrary \ct~$f$, Theorem~2 in Section~4 of~\cite{ILR}
shows that $h_{\gm, d, f}$
need not be uniquely ergodic.
\end{exa}
These homeomorphisms,
and higher dimensional analogs
(which also appear in~\cite{Frst}),
have attracted significant interest in operator algebras.
See, for example, \cite{Pc86}, \cite{Ji}, \cite{Kd}, and~\cite{Reh}.
The higher dimensional version has the general form
\[
(\zt_1, \zt_2, \ldots, \zt_n)
\mapsto \big( e^{2 \pi i \gm} \zt_1, \, g_2 (\zt_1) \zt_2, \,
g_3 (\zt_1, \zt_2) \zt_3, \, \ldots, \,
g_n (\zt_1, \zt_2, \ldots, \zt_{n  1}) \zt_n \big)
\]
for fixed $\gm \in \R$ and \cfn{s}
\[
g_2 \colon S^1 \to S^1,
\qquad
g_3 \colon S^1 \times S^1 \to S^1,
\qquad
\ldots,
\qquad
g_n \colon (S^1)^{n  1} \to S^1.
\]
There are further generalizations, called skew products.
Furstenberg transformations and their generalizations
have also attracted interest
in parts of dynamics not related to \ca{s};
as just two examples, we mention \cite{ILR} and~\cite{Rohn}.
Examples \ref{E_3302_MHOn2Tor} and~\ref{E_3302_Milnes3Tor}
are related but more complicated.
``Noncommutative'' Furstenberg transformations
(Furstenberg transformations on noncommutative analogs
of $S^1 \times S^1$)
are given in Example~\ref{E_3303_NCFurst}.
\begin{exa}\label{E:Shift}
Take $X = \{ 0, 1 \}^{\Z}$,
with elements being described as $x = (x_n)_{n \in \Z}$
with $x_n \in \{ 0, 1 \}$ for all $n \in \Z$.
(This space is homeomorphic to the Cantor set.)
Take $G = \Z$,
with action generated
by the {\emph{shift}} \hme{} $h (x)_n = x_{n + 1}$
for $x \in X$ and $n \in \Z$.
This action is neither free nor minimal;
in fact, it has fixed points.
One can replace $\{ 0, 1 \}$ by some other \cms~$K$.
(See \Def{D_5611_Shift}.)
Further examples (``subshifts'')
can be gotten by restricting to closed invariant subsets of~$X$.
Some of these are minimal.
For example, substitution minimal systems
and Toeplitz flows (mentioned after Example~\ref{E:pAdic})
can be obtained this way,
using a general finite set in place of $\{ 0, 1 \}$.
\end{exa}
\begin{exa}\label{E:pAdic}
Fix a prime $p$,
and let $X = \Z_p$, the group of $p$adic integers.
This group can be defined as the completion of $\Z$
in the metric $d (m, n) = p^{d}$ when $p^d$ is the largest
power of $p$ which divides $n  m$.
Alternatively, it is $\invlim \Z / p^d \Z$.
It is a compact topological group,
and as a metric space it is homeomorphic to the Cantor set.
Let $h \colon X \to X$ be the \hme{} defined
using the group operation in the completion
by $h (x) = x + 1$ for $x \in X$.
The resulting action is free and minimal
by Proposition~\ref{P_3408_DnsOrb}.
%
% 999
% Many generalizations are possible in the inverse limit
% version of the construction.
% One need not use a prime, nor even the same number at each stage of
% the inverse limit;
% thus, one could consider $X = \invlim \Z / k_d \Z$,
% % with $k_1  k_2  \cdots$
% under the assumption that $k_1$ divides $k_2$,
% $k_2$ divides $k_3$, etc.
% and $\limi{d} k_d = \infty$.
% These generalizations all give compact totally disconnected
% groups
% with a dense subgroup isomorphic to~$\Z$,
% so Proposition~\ref{P_3408_DnsOrb} applies.
\end{exa}
Next, we consider odometers.
They are a generalization of Example~\ref{E:pAdic}.
See Example~(i) on page 210 of~\cite{Skau},
Section VIII.4 of~\cite{Dvd},
and the first example in Section~2 of~\cite{Pt1}.
We refer to these sources for more information,
including minimality.
\begin{dfn}\label{D_6827_Odometer}
Let $d = \Sq{d}{n}$ be a sequence in $\N$
with $d_n \geq 2$ for all~$n \in \N$.
The {\emph{$d$odometer}}
is the minimal system $(X_d, h_d)$
defined as follows.
Set
\[
X_d = \prod_{n = 1}^{\infty} \{ 0, 1, 2, \ldots, d_n  1 \},
\]
which is homeomorphic to the Cantor set.
For $x = \Sq{x}{n} \in X_d$,
let
\[
n_0 = \inf \big( \big\{ n \in \N \colon x_n \neq d_n  1 \big\} \big).
\]
If $n_0 = \infty$
set $h_d (x) = (0, 0, \ldots)$.
Otherwise, $h_d (x) = \Sq{h_d (x)}{n}$ is
\[
h_d (x)_n = \begin{cases}
0 & n < n_0
\\
x_n + 1 & n = n_0
\\
x_n & n > n_0.
\end{cases}
\]
\end{dfn}
The \hme{} is
``addition of $(1, 0, 0, \ldots)$ with carry to the right''.
When $n_0 \neq \infty$, we have
\[
h (x) = \big( 0, 0, \ldots, 0, x_{n_0} + 1, x_{n_0 + 1}, x_{n_0 + 2},
\ldots \big).
\]
\begin{exr}\label{Ex_7129_OdMin}
Prove that the odometer \hme{}
of \Def{D_6827_Odometer} is minimal.
\end{exr}
See Theorem VIII.4.1 of~\cite{Dvd} for the computation
of the crossed product by an odometer action.
% 999 Add this example.
There are many other classes
of interesting minimal \hme{s} of the Cantor set,
such as substitution minimal systems
(Section~5 of~\cite{Skau}),
Toeplitz flows (Section~6 of~\cite{Skau}),
topological versions of interval exchange transformations
(the second example in Section~2 of~\cite{Pt1}),
and restrictions to their minimal sets of
Denjoy \hme{s},
which are nonminimal \hme{s} of the circle
whose rotation numbers are irrational
(\cite{PSS}).
The relation of strong orbit equivalence
of minimal \hme{s} of the Cantor set is
defined in~\cite{GPS}, where
it is shown to be equivalent to
isomorphism of the \tgca{s}.
Sugisaki has shown (\cite{Sg1}, \cite{Sg2}, and~\cite{Sg3})
that all possible values of
entropy in $[0, \I]$
occur in all strong orbit equivalence classes
of minimal \hme{s} of the Cantor set.
One can make various other kinds of examples
of free minimal actions
using Proposition~\ref{P_3408_DnsOrb}.
Here is one such example.
\begin{exa}\label{E_3408_SumProd}
Let $k_1, k_2, \ldots \in \{ 2, 3, \ldots \}$.
Set $X = \prod_{n = 1}^{\I} \Z / k_n \Z$,
which is a compact group.
Take $G = \bigoplus_{n = 1}^{\I} \Z / k_n \Z$,
which is a dense subgroup of~$X$.
Give $G$ the discrete topology,
so that $G$ becomes a locally compact group.
Then the action of $G$ on~$X$
by left translation is free and minimal,
by Proposition~\ref{P_3408_DnsOrb}.
The crossed product turns out to be the UHF algebra
$\bigotimes_{n = 1}^{\I} M_{k_n}$.
See \Exr{Ex_6X11_InfTProd}.
\end{exa}
\begin{exa}\label{ENoncptCantor}
The locally compact (but noncompact) Cantor set~$X$ is a
metrizable totally disconnected locally compact space
with no isolated points and which is not compact.
This description determines it uniquely up to homeomorphism,
by Proposition~2.1 of~\cite{Dnk}.
Minimal \hme{s} of~$X$
have been studied in \cite{Dnk} and~\cite{Mti}.
Section~3 of~\cite{Dnk} contains a good sized collection
of easy to construct examples,
although the construction is slightly more complicated than
we want to present here.
The comment after Theorem~2.11 of~\cite{Mti}
proves the existence of a much larger class of examples.
\end{exa}
% \begin{exa}\label{ENonminCantor}
% At least briefly mention nonminimal \hme s
% of the Cantor set.
% (Details to be filled in.)
% \end{exa}
% 999
\begin{exa}\label{E_6913_GJ}
For each \mh{} $h_0 \colon X_0 \to X_0$ of the Cantor set~$X_0$,
Gjerde and Johansen construct in~\cite{GjJh}
a \mh{} $h \colon X \to X$ of a \cms~$X$
which has $h_0 \colon X_0 \to X_0$ as a factor
(see Definition~\ref{D_6827_Factor} below),
and in which some of the connected components of~$X$ are
points (as for the Cantor set)
but some are compact intervals.
Among other things,
these examples show that if $h \colon X \to X$
is a \mh,
then the space~$X$ need not be ``homogeneous'':
different points can give different local properties of the space,
and, in particular,
for $x, y \in X$ there need not be a \hme{}
from any \nbhd{} of~$x$ to any \nbhd{} of~$y$
which sends $x$ to~$y$.
\end{exa}
\begin{dfn}\label{D_6827_Factor}
Let $G$ be a group,
let $X$ and $Y$ be \chs{s},
and assume $G$ acts \ct{ly} on $X$ and $Y$.
We say that the dynamical system $(G, Y)$ is a {\emph{factor}}
of the dynamical system $(G, X)$ if there is a
a surjective \ct{} map $f \colon X \to Y$
(the {\emph{factor map}})
such that $f (g x) = g f (x)$
for all $g \in G$ and $x \in X$.
\end{dfn}
If we take $G = \Z$,
then the actions are given by
\hme{s} $h \colon X \to X$
and $k \colon Y \to Y$.
Then we are supposed to
have a surjective \ct{} map $f \colon X \to Y$
such that $g \circ h = k \circ g$.
That is, the following diagram commutes:
\[
\begin{CD}
X @>{h}>> X \\
@V{f}VV @VV{f}V \\
Y @>{k}>> Y.
\end{CD}
\]
In general, there should be such a diagram
for the action of every group element $g \in G$
(always using the same choice of~$f$).
Essentially,
$(G, Y)$ is supposed to be a topological quotient of $(G, X)$.
Without compactness,
presumably one should ask that $f$ be a quotient map
of topological spaces.
\begin{exa}\label{E:Sign}
Take $X = S^n = \{ x \in \R^{n + 1} \colon \ x \_2 = 1 \}$.
Then the \hme{} $x \mapsto  x$ has order~$2$,
and so gives an action of $\Z / 2 \Z$ on $S^n$.
This action is free but is far from minimal.
See Example~\ref{E:C:Sign} below for a description of the \cp{}
(without proof).
\end{exa}
\begin{exa}\label{E:Conj}
Take $X = S^1 = \{ \zt \in \C \colon  \zt  = 1 \}$,
and consider the order~$2$ \hme{}
$\zt \mapsto {\overline{\zt}}$.
We get an action of $\Z / 2 \Z$ on $S^1$.
This action is neither free nor minimal.
See Example~\ref{E:C:Conj} below for the computation of the \cp.
\end{exa}
\begin{exa}\label{E:CommSL2}
The group
$\SL_2 (\Z)$ acts on $S^1 \times S^1$
as follows.
For
\[
n = \left( \begin{matrix} n_{1, 1} & n_{1, 2} \\
n_{2, 1} & n_{2, 2} \end{matrix} \right) \in \SL_2 (\Z),
\]
let $n$ act on $\R^2$ via the usual matrix multiplication.
Since $n$ has integer entries,
one gets $n \Z^2 \subset \Z^2$,
and thus the action is well defined on
$\R^2 / \Z^2 \cong S^1 \times S^1$.
Similarly, $\SL_d (\Z)$ acts on $(S^1)^d$.
In fact, the larger group $\GL_2 (\Z)$ acts on $S^1 \times S^1$
in the same way,
and the larger group $\GL_d (\Z)$ acts on $(S^1)^d$
in the same way.
We have emphasized the action of $\SL_2 (\Z)$
because it extends much more easily
to noncommutative deformations.
See Example~\ref{E:SL2}.
These actions are neither free nor minimal,
because the image of $0 \in \R^2$ (or $\R^d$)
is a fixed point.
% 999 Close to free?
\end{exa}
\begin{exa}\label{E:CommPerm}
Let $G$ be the symmetric group~$S_n$,
consisting of all permutations of $\{ 1, 2, \ldots, n \}$.
Let $X$ be any compact metric space.
Let $S_n$ act on $X^n$ by permuting the coordinates:
\[
\sm \cdot (x_1, x_2, \cdot, x_n)
= \big( x_{\sm^{1} (1)}, \, x_{\sm^{1} (2)}, \, \ldots, \,
x_{\sm^{1} (n)} \big).
\]
(One must use $\sm^{1}$ in the formula in order to get
$\sm \cdot ( \ta \cdot x) = (\sm \ta) \cdot x$
rather than $(\ta \sm) \cdot x$.)
These actions are not free.
Unless $X$ has only one point,
they are also far from minimal.
\end{exa}
\begin{exa}\label{E:UnOnS2n1}
The unitary group $U (M_n)$ of the $n \times n$ matrices
acts on the unit sphere $S^{2 n  1} \subset \C^n$,
since $S^{2 n  1}$ is invariant under the action
of $U (M_n)$ on~$\C^n$.
This is actually a special case of Example~\ref{E:Tr},
gotten by taking $G = U (M_n)$
and $H = U (M_{n  1})$,
embedded as a closed subgroup of~$G$
via the map
$h \mapsto \left( \begin{smallmatrix}
h & 0 \\
0 & 1
\end{smallmatrix} \right)$.
The action is thus minimal in a trivial way,
but not free.
Restricting to the scalar multiples of the identity,
we get an action of $S^1$ on $S^{2 n  1}$.
This action is free but not minimal.
Similarly, $U (M_n)$ and $S^1$ act on the closed unit ball in~$\C^n$.
\end{exa}
\begin{exa}\label{E1205Pi1}
Let $Z$ be a compact manifold,
or a connected finite complex.
(Much weaker conditions on~$Z$ suffice,
but $Z$ must be path connected.)
Let $X = {\widetilde{Z}}$ be the universal cover of~$Z$,
and let $G = \pi_1 (Z)$ be the fundamental group of~$Z$.
Then there is a standard action of $G$ on~$X$.
The space $X$ is locally compact
when $Z$ is locally compact,
and compact when $Z$ is compact and $\pi_1 (Z)$ is finite.
Spaces with finite fundamental groups include real
projective spaces
(in which case this example is really just Example~\ref{E:Sign})
and lens spaces (Example 2.43 of~\cite{Htch}).
In Example 1.43 of~\cite{Htch},
there is some discussion of spaces
with nonabelian finite fundamental groups whose universal covers
are spheres,
equivalently,
free actions of nonabelian finite groups on spheres.
There are also many spaces with interesting infinite fundamental group.
Any (discrete) group~$G$ is the fundamental group
of a two dimensional CW~complex~$X$
(Corollary 1.28 of~\cite{Htch}),
and (as is clear from the proof),
if $G$ is finitely presented
then $X$ can be taken to be a finite complex.
These actions are all free but are far from minimal.
\end{exa}
One can get free minimal actions of $\Z^2$
on compact metric spaces
by letting $h_1 \colon X_1 \to X_1$ and $h_2 \colon X_2 \to X_2$
be \mh{s} of infinite compact metric spaces,
setting $X = X_1 \times X_2$,
letting one generator of $\Z^2$ act on $X$
via $h_1 \times \id_{X_2}$,
and letting the other generator of $\Z^2$ act on $X$
via $\id_{X_1} \times h_2$.
A few other examples are known,
but examples seem to be hard to find.
Here is one, taken from~\cite{Mln1}.
\begin{exa}[Item~2 on page~311 of~\cite{Mln1}]\label{E_3421_Z2T3}
Fix $\te \in \R \setminus \Q$.
Then the \hme{s}
$h_1, h_2 \colon (S^1)^3 \to (S^1)^3$
(called $\af_1$ and $\af_2$ in~\cite{Mln1})
determined by
\[
h_1 (\zt_1, \zt_2, \zt_3)
= (\zt_1, \, e^{2 \pi i \te} \zt_2, \, \zt_1 \zt_3)
\andeqn
h_2 (\zt_1, \zt_2, \zt_3)
= (e^{2 \pi i \te} \zt_1, \, \zt_2, \, \zt_2 \zt_3)
\]
for $\zt_1, \zt_2, \zt_3 \in S^1$,
commute and generate a free minimal action of $\Z^2$ on $(S^1)^3$.
\end{exa}
The next two examples
have some similarity with Example~\ref{E_3302_Furst},
but are more complicated.
\begin{exa}\label{E_3302_MHOn2Tor}
Let $H$ be the discrete Heisenberg group,
that is,
\[
H = \left\{ \left( \begin{matrix}
1 & n & k \\
0 & 1 & m \\
0 & 0 & 1
\end{matrix} \right) \colon
k, m, n \in \Z \right\}.
\]
Equivalently,
$H = \Z^3$ as a set,
and the group operation is
\[
(k_1, m_1, n_1) (k_2, m_2 ,n_2)
= (k_1 + k_2 + n_1 m_2, m_1 + m_2, n_1 + n_2)
\]
for $k_1, m_1, n_1, k_2, m_2 ,n_2 \in \Z$.
This formula comes from the assignment
\[
(k, m, n) \mapsto
\left( \begin{matrix}
1 & n & k \\
0 & 1 & m \\
0 & 0 & 1
\end{matrix} \right)
\]
for $k, m, n \in \Z$.
The proof of Theorem~1 of~\cite{Mln1}
uses a minimal action of $H$ on $(S^1)^2$
which depends on a parameter $\te \in \R \setminus \Q$.
It is given by
\[
(k, m, n) \cdot (\zt_1, \zt_2)
= \big( e^{ 2 \pi i n \te} \zt_1, \,
e^{2 \pi i (m n  k) \te} \zt_1^{ m} \zt_2 \big)
\]
for $k, m, n \in \Z$ and $\zt_1, \zt_2 \in S^1$.
This action is not free.
For example,
% \[
% (0, 0, 1) \cdot ( e^{2 \pi i \te}, \, 1)
% = (0, 1, 1) \cdot ( e^{2 \pi i \te}, \, 1)
% = (1, 1).
% \]
\[
(1, 1, 0) \cdot ( e^{2 \pi i \te}, \, 1) = ( e^{2 \pi i \te}, \, 1).
\]
However, it is essentially free.
\end{exa}
\begin{exa}\label{E_3302_Milnes3Tor}
Let $H$ be the discrete Heisenberg group,
as in Example~\ref{E_3302_MHOn2Tor}.
The proofs of Theorem~2 and Theorem~4 of~\cite{Mln2}
use free minimal actions of $H$ on $(S^1)^3$
which depend on a parameter $\te \in \R \setminus \Q$
and (for Theorem~2) on relatively prime integers $p$ and~$q$.
The action used in Theorem~2 of~\cite{Mln2}
is given by
\[
(k, m, n) \cdot (\zt_1, \zt_2, \zt_3)
= \big( e^{2 \pi i q m \te} \zt_1, \,
e^{2 \pi i p n \te} \zt_2, \,
e^{ 2 \pi i [ (p + q) k  q m n ] \te }
\zt_1^{ n} \zt_2^m \zt_3
\big)
\]
% \begin{align*}
% & (k, m, n) \cdot (\zt_1, \zt_2, \zt_3)
% \\
% & = \big( \exp (2 \pi i q m \te) \zt_1, \,
% \exp (2 \pi i p n \te) \zt_2, \,
% \exp ( 2 \pi i [ (p + q) k  q m n ] \te )
% \zt_1^{ n} \zt_2^m \zt_3
% \big)
% \end{align*}
for $k, m, n \in \Z$ and $\zt_1, \zt_2, \zt_3 \in S^1$.
The action used in Theorem~4 of~\cite{Mln2}
is given by
\[
(k, m, n) \cdot (\zt_1, \zt_2, \zt_3)
= \big( e^{2 \pi i m \te} \zt_1, \,
e^{2 \pi i (m + n) \te} \zt_2, \,
e^{2 \pi i [2 k  m n + m (m  1) / 2 ] \te }
\zt_1^{ n} \zt_2^m \zt_3
\big)
\]
% \begin{align*}
% & (k, m, n) \cdot (\zt_1, \zt_2, \zt_3)
% \\
% & = \big( \exp (2 \pi i m \te) \zt_1, \,
% \exp (2 \pi i (m + n) \te) \zt_2, \,
% \exp ( 2 \pi i [2 k  m n + m (m  1) / 2 ] \te )
% \zt_1^{ n} \zt_2^m \zt_3
% \big)
% \end{align*}
for $k, m, n \in \Z$ and $\zt_1, \zt_2, \zt_3 \in S^1$.
\end{exa}
\begin{exa}\label{E_3305_StoneCech}
Let $G$ be a discrete group.
Then the action of $G$ on itself by translation
(\Ex{E:Tr})
extends to an action of $G$
on the Stone\v{C}ech compactification $\bt G$ of~$G$,
and thus to an action of $G$
on the remainder $\bt G \setminus G$.
% Comments on action: Ask Guoliang Yu. Is it free?
% 999
\end{exa}
\begin{exa}\label{EFnBdy}
Let $n \in \{ 2, 3, \ldots \}$.
The Gromov boundary $\partial F_n$ of $F_n$
% 999 Ref needed.
consists of all right infinite
reduced words in the generators and their inverses,
with the topology (given in detail below)
in which two words are close if they have the same
long finite initial segment.
The group $F_n$ acts on it by left translation.
We claim that this action is minimal and
essentially free,
but not free.
Call the standard generators $g_1, g_2, \ldots, g_n$.
Set
\[
S = \big\{ g_1, g_1^{1}, g_2, g_2^{1}, \ldots, g_n, g_n^{1} \big\},
\]
with the discrete topology.
Then $\partial F_n$ is the subset of the
compact set $S^{\N}$ consisting of those sequences
$x = (x_1, x_2, \ldots) \in S^{\N}$
such that $x_{n + 1} \neq x_n^{1}$
for all $n \in \Nz$.
This set is an intersection of closed sets,
hence compact.
The element $x = g_1 \cdot g_1 \cdot \cdots$ is a right infinite word
such that $g_1 x = x$.
Thus the action is not free.
(More generally, if $h \in F_n \setminus \{ 1 \}$ is arbitrary,
then the reduced form of $h \cdot h \cdot \cdots$
is in $\partial F_n$ and is a fixed point for~$h$.)
We show that the action is minimal.
We use \Lem{L_3415_MinOrb}.
Let $x, y \in \partial F_n$.
Use sequence notation as above.
It suffices to show that for every $n \in \Nz$
there is $g \in F_n$
such that $(g x)_k = y_k$ for $k = 1, 2, \ldots, n$.
Let $g_0 = y_1 y_2 \cdots y_n \in F_n$.
Choose $h \in S$ such that $h \not\in \{ y_n^{1}, x_1^{1} \}$.
Then
$z = (y_1, y_2, \ldots, y_n, h, x_1, x_2, \ldots)$
is a right infinite reduced word
which agrees with $y$ in positions $1, 2, \ldots, n$.
Moreover, with $g = g_0 h$,
we get $g x = z$.
This completes the proof of minimality.
It remains to show that the action is essentially free.
By \Def{D_3331_Free},
it suffices to show that if $h \in F_n \setminus \{ 1 \}$
and $x \in \partial F_n$,
there is $y \in \partial F_n$
such that $y_j = x_j$ for $j = 1, 2, \ldots, n$
and such that $h y \neq y$.
If $h x \neq x$,
there is nothing to prove.
So suppose $h x = x$.
Write $h$ as a reduced word $h = h_1 h_2 \cdots h_l$
with $h_1, h_2, \ldots, h_l \in S$.
There is $k \in \{ 0, 1, \ldots, l \}$
such that, in reduced form,
we have
%
\begin{equation}\label{Eq_4305_hx}
h x = \big( h_1, \, h_2, \, \ldots, \, h_k,
\, x_{l  k + 1}, \, x_{l  k + 2}, \, \ldots \big).
\end{equation}
%
That is,
$h x$ is one of
\[
\big( h_1, \, h_2, \, \ldots, \, h_l, \, x_1, \, x_2, \ldots \big),
\quad
\big( h_1, \, h_2, \, \ldots, \, h_{l  1}, \, x_2, \, x_3, \ldots \big),
\]
\[
\big( h_1, \, h_2, \, \ldots, \, h_{l  2}, \, x_3, \, x_4, \ldots \big),
\quad
\ldots,
\quad
\big( x_{l + 1}, \, x_{l + 2}, \, x_{l + 3}, \ldots \big).
\]
We claim that $l \neq 2 k$.
(This means that passing from $x$ to $h x$ actually
shifts the sequence~$x$,
so that $x$ is eventually periodic.)
Suppose that $l = 2 k$.
The cancellations which occur to make the formula for $h x$
correct imply that
\[
h_l = x_1^{1},
\qquad
h_{l  1} = x_2^{1},
\qquad
\ldots,
\qquad
h_{l  k + 1} = x_k^{1}.
\]
Looking at the first $k$ positions of the equation $h x = x$,
we get
\[
h_1 = x_1,
\qquad
h_2 = x_2,
\qquad
\ldots,
\qquad
h_k = x_k.
\]
Combine these (in the opposite order) and use $l  k = k$ to get
\[
h_{k + 1} = h_k^{1},
\qquad
h_{k + 2} = h_{k  1}^{1},
\qquad
\ldots,
\qquad
h_{1} = h_l^{1}.
\]
Therefore $h = 1$.
This is a contradiction,
and the claim follows.
By the definition of~$k$,
we have $(h x)_j = x_{j  2 k + l}$
for $j = k + 1, \, k + 2, \, \ldots$.
Set $m = n + l + 1$.
Choose
$y_m \in
S \setminus \big\{ x_{m  1}^{1}, x_{m + 1}^{1}, x_m \big\}$.
Then setting $y_j = x_j$ for $j \in \N \setminus \{ m \}$
gives a reduced right infinite word $y \in \partial F_n$.
Clearly $y_j = x_j$ for $j = 1, 2, \ldots, n$.
Since $m > l  k  1$,
we have
%
\begin{equation}\label{Eq_4305_hyForx}
h y = \big( h_1, \, h_2, \, \ldots, \, h_k,
\, y_{l  k + 1}, \, y_{l  k + 2}, \, \ldots \big).
\end{equation}
%
Therefore, using $2 k  l \neq 0$ at the first step,
(\ref{Eq_4305_hx}) at the third step,
and (\ref{Eq_4305_hyForx}) at the fifth step,
we get
\[
y_{m + 2 k  l}
= x_{m + 2 k  l}
= (h x)_{m + 2 k  l}
= x_m
\neq y_m
= (h y)_{m + 2 k  l}.
\]
Thus $h y \neq y$.
\end{exa}
% 999
% (What is the crossed product?)
A related example,
in which $G$ is a finite free product of
at least two nontrivial cyclic groups
(excluding $\Z / 2 \Z \star \Z / 2 \Z$),
acting on the Cantor set,
is given in Definition~2.1 of~\cite{Spl1}.
Essential freeness is a consequence of Lemma 3.12 of~\cite{Spl1}.
Minimality isn't explicitly stated,
but it is shown in some cases that the crossed products
are simple,
by explicitly computing them.
(See Example~2.8 and Remark 2.9 of~\cite{Spl1}.)
Actions of a subclass of these groups on the Cantor set
are given in Definition~2.1 of~\cite{Spl2}
and the comment afterwards,
and are shown to be minimal and essentially free
in Theorem 3.3 of~\cite{Spl2}.
The crossed products are proved to be CuntzKrieger algebras
in Theorem 2.2 of~\cite{Spl2}.
Some further examples of this general nature
are given in Section~3 of~\cite{LcSp},
and some more are in Section~3 of~\cite{AnDl}.
We mention a few other examples very briefly.
% 999 more details on first, maybe second.
A general construction known as the flow under a ceiling function
starts with a \hme{} $h$ of, say, a \cms~$X$,
and yields an action of $\R$ on a space that looks
like the mapping cylinder of~$X$.
One can consider this action or the action of~$\Z$
generated by the time $t$ map of this action for
a fixed $t \in \R$.
The crossed products by some interesting examples
of this construction are considered in~\cite{It1},
with $X$ taken to be the Cantor set.
We refer to~\cite{It1} for further details.
Let $X$ be the Cantor set.
There are interesting classes of \mh{s}
of $S^1 \times X$
and of $S^1 \times S^1 \times X$.
See \cite{LnMti1}, \cite{LnMti2}, and~\cite{LnMti3} for $S^1 \times X$
and~\cite{Sn} for $S^1 \times S^1 \times X$.
(The spaces $S^1 \times X$ are locally
homeomorphic to those of~\cite{It1}.)
The geodesic flow on a compact Riemannian manifold $M$ is an action
of~$\R$ on the unit sphere bundle $X$ over~$M$.
At $v \in T_x M$ it follows the geodesic starting at $x$
in the direction~$v$ at unit speed,
carrying $v$ with it.
Various dynamical properties of this flow are
considered in Chapter~12 of~\cite{BrrPsn}.
For example, under suitable conditions on~$M$,
it is topologically transitive (Theorem 12.2.10 of~\cite{BrrPsn}).
If the Riemannian metric on~$M$ is $C^3$
and the sectional curvatures are all strictly negative,
then the geodesic flow is ergodic
with respect to the standard measure.
See Theorem~5.5 in the appendix to~\cite{Blmn}.
We would also like to mention several existence theorems
for actions.
\begin{thm}[Theorem~1.1 of~\cite{HjMl}]\label{T_3305_HjMl}
Let $G$ be an infinite countable discrete group.
Then there exists a free action of $G$ on the Cantor set
which has an invariant Borel probability measure.
\end{thm}
By passing to a minimal set for such an action,
one obtains:
\begin{cor}[Corollary~1.5 of~\cite{HjMl}]\label{C_3305_MinHjMl}
Let $G$ be an infinite countable discrete group.
Then there exists a free minimal action of $G$ on the Cantor set.
\end{cor}
The action in Corollary~\ref{C_3305_MinHjMl} need not
have an invariant Borel probability measure.
However, if $G$ is amenable,
then, by \Thm{T_6X11_ExistInv},
every action on a compact metric space
has an invariant Borel probability measure.
\begin{thm}[Theorem~6.11 of~\cite{RrSr}]\label{T_3305_Kerr}
Let $G$ be an infinite countable discrete group
which is exact but not amenable.
Then there exists a free minimal action of $G$ on the Cantor set~$X$
such that the \tgca{}
$C^* (G, X)$ is a Kirchberg algebra
satisfying the Universal Coefficient Theorem.
\end{thm}
The following result is a special case of the
combination of Theorem~1 and Theorem~3 of~\cite{FtHr}.
In~\cite{FtHr},
freeness of the action of $S^1$ is weakened to
the requirement that the stabilizers of all points be finite
and that the action be effective.
\begin{thm}\label{T_3305_ExistMinZ}
Let $M$ be a connected compact $C^{\infty}$~manifold
which admits a free $C^{\infty}$~action of~$S^1$.
Then there exists a uniquely ergodic minimal diffeomorphism of~$M$.
\end{thm}
The most obvious examples are spheres $S^{2 n  1}$
for $n \in \N$.
The free action of $S^1$ is the one in \Ex{E:UnOnS2n1}.
By contrast, there are no minimal homeomorphisms of even spheres.
This can be easily proved using the Lefschetz fixed point theorem.
The general result is as follows
(a special case of Theorem~3 of~\cite{Fll}).
\begin{thm}\label{T_3305_NoMinHme}
Let $X$ be a finite complex with nonzero Euler characteristic.
Then every homeomorphism of $X$ has a periodic point.
\end{thm}
The Euler characteristic of an even sphere is~$2$.
The proof of Theorem~\ref{T_3305_ExistMinZ}
uses a Baire category argument.
For $n > 1$,
there is no known explicit formula for even a \mh{}
of $S^{2 n  1}$.
\begin{pbm}\label{Pb_3305_MinS3}
Find an explicit formula for a \mh{} of~$S^3$.
\end{pbm}
\begin{thm}[\cite{Wnd}]\label{T_3305_ExistkErgZ}
Let $M$ be a connected compact $C^{\infty}$~manifold
which admits a free $C^{\infty}$~action of~$S^1$,
and let $k \in \N$.
Then there exists a minimal diffeomorphism of~$M$
which admits exactly $k$ ergodic invariant Borel probability measures.
\end{thm}
The following result is a special case of the
combination of Theorem~2 and Theorem~4 of~\cite{FtHr}.
In~\cite{FtHr},
freeness of the action of $S^1$ is weakened in the same
way as for \Thm{T_3305_ExistMinZ}.
\begin{thm}\label{T_3305_MinFlow}
Let $n \in \N$,
and let $M$ be a connected compact $C^{\infty}$~manifold
which admits a free $C^{\infty}$~action of~$(S^1)^{n + 1}$.
Then there exists a uniquely ergodic free minimal action
of $\R^n$ on~$M$.
\end{thm}
By embedding $\Z^d$ in~$\R$ as a dense subgroup,
one gets:
\begin{cor}\label{C_3305_ExistMinZd}
Let $d \in \N$ with $d \geq 2$,
and let $M$ be a connected compact $C^{\infty}$~manifold
which admits a free $C^{\infty}$~action of~$S^1 \times S^1$.
Then there exists a uniquely ergodic minimal action
of $\Z^d$ on~$M$.
\end{cor}
% 999
% Missing examples:
% Horocycle flow.
\section{Examples of Group Actions on Noncommutative
C*Algebras}\label{Sec_NCEx}
\indent
In this section, we turn to
examples of group actions on noncommutative \ca{s}.
Along with a number of miscellaneous examples,
we give an assortment of examples from each of several fairly
general classes of actions:
``gauge type'' actions,
shifts and other permutations of
the factors in various kinds of tensor products and free products,
and highly nontrivial actions obtained as direct limits
of various much simpler (even inner) actions
on smaller \ca{s}.
Section~\ref{Sec_AddGauge}
contains many more examples of ``gauge type'' actions.
The most elementary action is the trivial action.
\begin{exa}\label{E_3408_TrivOnA}
Let $G$ be a locally compact group,
let $A$ be a \ca,
and define an action $\af \colon G \to \Aut (A)$
by $\af_g (a) = a$ for all $g \in G$ and all $a \in A$.
This is the trivial action of $G$ on~$A$.
The crossed products turn out to be
\[
C^* (G, A, \af) = C^* (G) \otimes_{\mathrm{max}} A
\andeqn
C^*_{\mathrm{r}} (G, A, \af)
= C^*_{\mathrm{r}} (G) \otimes_{\mathrm{min}} A.
\]
See \Ex{E:C:Triv}.
\end{exa}
Before we go farther,
the following notation is convenient.
\begin{ntn}\label{N_3302_Adu}
Let $A$ be a unital \ca,
and let $u \in A$ be unitary.
We denote by $\Ad (u)$ the automorphism of~$A$
given by $a \mapsto u a u^*$.
We use the same notation when $A$ is not unital
and $u$ is a unitary in its multiplier algebra $M (A)$.
\end{ntn}
\begin{dfn}\label{D_6320_InnerAut}
Let $A$ be a \ca{} and let $\af \in \Aut (A)$.
Then $\af$ is {\emph{inner}}
if there is $u \in M (A)$ such that $\af = \Ad (u)$.
Otherwise, $\af$ is {\emph{outer}}.
\end{dfn}
\begin{exa}\label{E:Inner}
Let $G$ be a locally compact group,
let $A$ be a unital \ca,
and let $g \mapsto z_g$ be a norm \ct{} group \hm{} from $G$ to the
unitary group $U (A)$ of~$A$.
Then the formula $\af_g = \Ad (z_g)$,
% \[
% \af_g (a) = z_g a z_g^*,
% \]
for $g \in G$ and $a \in A$,
defines an action of $G$ on $A$.
Actions obtained this way are called {\emph{inner actions}}.
If $A$ is not unital,
let $M (A)$ be its multiplier algebra,
and use $U (M (A))$ with the strict topology in place of $U (A)$
with the norm topology.
As a special case,
let $g \mapsto u_g$ be a unitary representation of~$G$
on a Hilbert space~$H$,
which is continuous in the strong operator topology
(the conventional topology in this situation;
the formal definition is in \Def{D_2236_UnitRpn} below).
Then $g \mapsto \Ad (u_g)$
defines a \ct{} action of~$G$ on the compact operators $K (H)$.
(The map $g \mapsto \Ad (u_g)$ is generally not a
continuous action, in the \ca{} sense,
of $G$ on the bounded operators $L (H)$.)
The crossed product by an inner action
is isomorphic to the \cp{} by the trivial action.
See Example~\ref{E:C:Inner} below for the computation
of the \cp{} when $G$ is discrete.
\end{exa}
An action via inner automorphisms is not necessarily an
inner action in the sense of Example~\ref{E:Inner}.
There are no counterexamples with $G = \Z$ (trivial)
or when $G$ finite cyclic and $A$ is simple
(easy; see Exercise~\ref{P:Inn} below).
Here is the smallest counterexample.
\begin{exa}\label{E_PtInn}
Let $A = M_{2}$, let $G = (\Z / 2 \Z)^{2}$ with
generators $g_{1}$ and $g_{2}$,
and set
\[
\af_1 = \id_A,
\quad
\af_{g_{1}}
= \Ad \left(
\begin{smallmatrix} 1 & 0 \\ 0 & 1 \end{smallmatrix} \right),
\quad
\af_{g_{2}}
= \Ad \left(
\begin{smallmatrix} 0 & 1 \\ 1 & 0 \end{smallmatrix} \right),
\quad {\mbox{and}} \quad
\af_{g_1 g_{2}}
= \Ad \left(
\begin{smallmatrix} 0 & 1 \\ 1 & 0 \end{smallmatrix} \right).
\]
These define an action $\af \colon G \to \Aut (A)$
such that $\af_{g}$ is inner for all $g \in G$,
but for which
there is no homomorphism $g \mapsto z_{g} \in U (A)$ such that
$\af_{g} = \Ad (z_{g})$ for all $g \in G$.
The point is that the implementing unitaries
for $\af_{g_1}$ and $\af_{g_2}$ commute up to a scalar,
but can't be appropriately modified to commute exactly.
See Exercise~\ref{E:C:Conj} below for the computation
of the \cp.
\end{exa}
\begin{exr}\label{P:PtInn}
Prove the statements made in Example~\ref{E_PtInn}.
\end{exr}
\begin{exr}\label{P:Inn}
Let $A$ be a simple unital \ca, and
let $\af \colon \Z / n \Z \to \Aut (A)$ be an action such that
each automorphism $\af_g$, for $g \in \Z / n \Z$,
is an inner automorphism.
Prove that $\af$ is an inner action
in the sense of Example~\ref{E:Inner}.
\end{exr}
The result of Exercise~\ref{P:Inn}
fails when $A$ is not assumed simple.
The following example is due to Jae Hyup Lee.
% Lee is the family name.
\begin{exa}\label{E:Cyclic}
Let $A = C (S^1, M_2)$,
and define $u \in A$ by
\[
u (\zt) = \frac{1}{2}
\left( \begin{matrix} \zt + 1 & i (\zt  1) \\
i (\zt  1) &  (\zt + 1) \end{matrix} \right)
\]
for $\zt \in S^1$.
Then one can check that $u$ is unitary,
and that $u^2$ is the function
%
\begin{equation}\label{Eq:ECyclic1}
u (\zt)^2
= \left( \begin{matrix} \zt & 0 \\ 0 & \zt \end{matrix} \right),
\end{equation}
%
which is in the center of~$A$.
Therefore $\Ad (u) \in \Aut (A)$
is an automorphism of order~$2$,
and so gives an action $\af$ of $\Z / 2 \Z$ on~$A$.
We claim that this action is not inner
in the sense of Example~\ref{E:Inner}.
That is, there is no unitary $z \in A$
such that $z^2 = 1$ and $\Ad (z) = \Ad (u)$.
Suppose $z$ is such a unitary.
Then $\Ad (z^* u) = \id_A$,
so $z^* u$ is in the center of~$A$.
Thus, there is a \cfn{} $\ld \colon S^1 \to S^1$
such that
\[
z (\zt)^* u (\zt) = \ld (\zt) \cdot 1_{M_2}
\]
for all $\zt \in S^1$.
We can rearrange this equation to get
%
\begin{equation}\label{Eq:ECyclic2}
\ld (\zt) z (\zt) = u (\zt)
\end{equation}
%
for all $\zt \in S^1$.
Squaring both sides of~(\ref{Eq:ECyclic2}),
and using~(\ref{Eq:ECyclic1}) and $z^2 = 1$, we get
\[
\ld (\zt)^2 \cdot 1_{M_2} = u (\zt)^2 = \zt \cdot 1_{M_2}
\]
for all $\zt \in S^1$.
Thus, $\ld (\zt)$ is a \ct{} square root of $\zt$ on~$S^1$,
which is well known not to exist.
This contradiction shows that $\af$ is not an inner action.
\end{exa}
\begin{rmk}\label{R:ExtEq}
There is a generalization of inner actions that should be mentioned.
Actions $\af$ and $\bt$ of a locally compact group $G$ on a unital
C*algebra $A$ are called {\emph{exterior equivalent}}
if there is a \ct{} map $g \mapsto z_g$ from $G$ to the unitary
group of $A$ such that $z_{g h} = z_g \af_g ( z_h)$ and
% $\bt_g ( a ) = z_g \af_g ( a ) {z_g^*}$
$\bt_g = \Ad (z_g) \circ \af_g$
for $g, h \in G$.
If $A$ is not unital, use a strictly \ct{} map to the unitary
group of the multiplier algebra.
(See 8.11.3 of~\cite{Pd1}.)
An action is inner \ifo{} it is exterior equivalent to the trivial
action,
and it turns out that exterior equivalent actions
give isomorphic crossed products.
See Exercise~\ref{P:ExtEqIsom} below.
\end{rmk}
Since they play such a prominent role in our examples,
we explicitly recall the rotation algebras.
\begin{exa}\label{E_3303_RotAlg}
Let $\te \in \R$.
The rotation algebra $A_{\te}$ is
the universal \ca{} generated by two unitaries $u$ and $v$
satisfying the commutation relation $v u = \exp (2 \pi i \te) u v$.
(The convention $e^{2 \pi i \te}$ instead of $e^{i \te}$
has become so standard that it can't be changed.)
The algebra $A_{\te}$ is often considered to be a noncommutative
analog of the torus $S^1 \times S^1$
(more accurately, of $A_0 \cong C (S^1 \times S^1)$).
It turns out to be the crossed product
by the corresponding rotation $\zt \mapsto e^{2 \pi i \te} \zt$
of the circle,
the integer action version of Example~\ref{E_Rot}.
If $\te \not\in \Q$, then $A_{\te}$ is known to be simple.
This follows from
Example~\ref{E:C:IrrRot} and Theorem~\ref{T:AS} below.
Thus, one may take {\emph{any}} \ca{} generated by two
unitaries satisfying the appropriate commutation relation.
If $\te \in \Q$,
then
$A_{\te}$ is the section algebra of a locally trivial bundle over
$S^1 \times S^1$ whose fiber is a single matrix algebra.
Its structure is determined in~\cite{HS}.
(See Example~8.46 of~\cite{Wlms}.
Some further discussion is given in Example~\ref{E_4302_RatRot}.)
In the special case $\te \in \Z$,
one just gets $C (S^1 \times S^1)$.
\end{exa}
There are also versions with more generators.
\begin{exa}\label{E_3303_NCTorus}
Let $d \in \N$ with $d \geq 2$.
Let $\te$ be a skew symmetric real $d \times d$ matrix.
Recall (\cite{Rf2})
that the (higher dimensional) noncommutative torus $A_{\te}$ is
the universal \ca{} generated by unitaries $u_1, u_2, \dots, u_d$
subject to the relations
\[
u_k u_j = \exp (2 \pi i \te_{j, k} ) u_j u_k
\]
for $j, k = 1, 2, \ldots, d$.
Of course, if $\te_{j, k} \in \Z$ for $j, k = 1, 2, \ldots, d$,
it is not really
noncommutative.
Some authors use $\te_{k, j}$ in the commutation relation instead.
See for example Section~6 of~\cite{Ks2}.
The algebra $A_{\te}$ is simple \ifo{}
$\te$ is nondegenerate,
which means that whenever $x \in \Z^d$ satisfies
$\exp (2 \pi i \langle x, \te y \rangle ) = 1$ for all $y \in \Z^d$,
then $x = 0$.
That nondegeneracy implies simplicity
is Theorem~3.7 of~\cite{Slw}.
(Note the standing assumption of nondegeneracy throughout
Section~3 of~\cite{Slw}.)
The converse is essentially 1.8 of~\cite{EllI};
see Theorem~1.9 of~\cite{PhT2}
for the explicit statement.
It seems worth pointing out that there is a coordinate free
way to obtain a higher dimensional noncommutative torus.
The algebra $A_{\te}$ is the universal
\ca{} generated by unitaries $u_x$, for $x \in \Z^d$,
subject to the relations
\[
u_y u_x = \exp (\pi i \langle x, \te (y) \rangle ) u_{x + y}
\]
for $x, \, y \in \Z^d$.
(See the beginning of Section~4 of~\cite{Rf1} and
the introduction to~\cite{RS}.)
It follows that if $b \in {\mathrm{GL}}_d (\Z)$, and if
$b^{\mathrm{t}}$ denotes the transpose of $b$, then
$A_{b^{\mathrm{t}} \te b} \cong A_{\te}$.
That is, $A_{\te}$ is unchanged if $\te$ is rewritten in terms of
some other basis of $\Z^d$.
\end{exa}
\begin{exa}\label{E:SL2}
Let $\te \in \R$,
and let $A_{\te}$ be the rotation algebra,
as in Example~\ref{E_3303_RotAlg}.
The group $\SL_2 (\Z)$ acts on $A_{\te}$ by sending the
matrix
\[
n = \left( \begin{matrix} n_{1, 1} & n_{1, 2} \\
n_{2, 1} & n_{2, 2} \end{matrix} \right)
\]
to the automorphism determined by
\[
\af_n (u)
= \exp (\pi i n_{1, 1} n_{2, 1} \te) u^{n_{1, 1}} v^{n_{2, 1}}
\qquad
{\mbox{and}}
\qquad
\af_n (v)
= \exp (\pi i n_{1, 2} n_{2, 2} \te) u^{n_{1, 2}} v^{n_{2, 2}}.
\]
To see that there is such an automorphism,
one checks that the intended values of $\af_n (u)$ and $\af_n (v)$
are unitaries which satisfy the relation
\[
\af_n (v) \af_n (u) = e^{2 \pi i \te} \af_n (u) \af_n (v).
\]
The extra scalar factors in the definition are
present in order to get $\af_{m n} = \af_m \circ \af_n$
for $m, n \in \SL_2 (\Z)$.
If we view $A_{\te}$ as a noncommutative
analog of the torus $S^1 \times S^1$
as in Example~\ref{E_3303_RotAlg},
this action is the analog of the action
of $\SL_2 (\Z)$ on $S^1 \times S^1$ in Example~\ref{E:CommSL2}.
The group $\SL_2 (\Z)$ has finite subgroups of orders
$2$, $3$, $4$, and~$6$.
They can be taken to be generated by
\[
\begin{split}
&\left(\begin{matrix} 1 & 0 \\ 0 &  1 \end{matrix} \right)
\;\; {\text{(for $\Z / 2 \Z$)}}, \quad \quad\quad \quad\quad
\left(\begin{matrix} 1 & 1 \\1 & 0 \end{matrix} \right)
\;\; {\text{(for $\Z / 3 \Z$)}},
\\
&\left(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right)
\;\; \;\;{\text{(for $\Z / 4 \Z$)}}, \quad \quad {\text{and}}
\quad \quad
\left(\begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix} \right)
\;\; {\text{(for $\Z / 6 \Z$)}}.
\end{split}
\]
Restriction of the action gives actions of these groups
on rotation algebras.
The crossed products by these actions have been intensively studied.
Recently, it has been proved~\cite{ELPW} that for $\te \not\in \Q$
they are all AF~algebras.
In at least some of these cases,
the
extra scalar factors are equal to~$1$.
Thus, the action of $\Z / 2 \Z$ on $A_{\te}$ is generated
by the automorphism determined by
\[
u \mapsto u^*
\andeqn
v \mapsto v^*,
\]
and the action of $\Z / 4 \Z$ on $A_{\te}$ is generated
by the automorphism determined by
\[
u \mapsto v
\andeqn
v \mapsto u^*.
\]
\end{exa}
% Known earlier for Z/2Z. 999
Although we will not prove it here
(see~\cite{ELPW}),
for $\te \not\in \Q$ these actions have the \trp{}
of \Def{D_TRP}.
It seems to be unknown whether the action of $\GL_2 (\Z)$
on $S^1 \times S^1$ in \Ex{E:CommSL2}
can be deformed to an action on a rotation algebra.
This can be done for the subgroup consisting of the diagonal
matrices in $\GL_2 (\Z)$,
an order~$4$ subgroup isomorphic to $\Z / 2 \Z \times \Z / 2 \Z$.
\begin{exa}\label{E_3303_IrrRotZ2xZ2}
Let $\te \in \R \setminus \Q$,
and let $A_{\te}$ be the rotation algebra,
as in Example~\ref{E_3303_RotAlg}.
Let $G = \{ \diag (\pm 1, \, \pm 1) \} \subset \GL_2 (\Z)$.
Then there is (Theorem~1.1 of~\cite{Stc1})
an action
$\bt^{(\te)} \colon G \to \Aut (A_{\te})$
such that, for $g \in G$,
we have $(\bt^{(\te)}_g)_* = g$ on $K_1 (A_{\te}) \cong \Z^2$.
\end{exa}
The Ktheory condition matches the action of this
subgroup on $K^1 (S^1 \times S^1)$.
The construction is an existence proof using a direct
limit decomposition,
and it is not clear how close the action
of the diagonal subgroup is to the action of $\Z / 2 \Z$
in \Ex{E:SL2}
(although, by Theorem~1.1 of~\cite{Stc1},
it does have the right fixed point algebra).
There is no claim that the actions on the different algebras $A_{\te}$
can be chosen to vary continuously with~$\te$
in a reasonable sense.
This is probably not possible.
The results of~\cite{Stc2} probably imply
(although this has not been checked in detail)
that for
$\te
\in \big[ (0, 1) \cap \Q \big] \setminus \big\{ \tfrac{1}{2} \big\}$,
there is no $\af \in \Aut (A_{\te})$ whose induced map on
$K_1 (A_{\te})$ is in $\GL_2 (\Z) \setminus \SL_2 (\Z)$.
% There is a ``higher dimensional'' analog of $A_{\te}$,
% given by the universal \ca{} generated by $d$ unitaries
% $u_1, u_2, \ldots, u_d$ which
% commute up to fixed scalars.
% The resulting \ca{} is called a
% ``higher dimensional noncommutative torus''.
Unfortunately, there is in general no action of
$\SL_d (\Z)$ on the higher dimensional noncommutative torus
of \Ex{E_3303_NCTorus}
analogous to
the action of $\SL_2 (\Z)$ on~$A_{\te}$.
That is, there is no general noncommutative deformation
of the action of $\SL_n (\Z)$ on $(S^1)^d$
of \Ex{E:CommSL2}.
In Example~\ref{E:SL2},
we had a \ca~$A$
given in terms of generators and relations,
and we defined an action of a discrete group on~$A$
by specifying what the group elements are supposed to do to the generators.
We want to define actions of not necessarily discrete
groups in the same way.
We will obviously only do this when the action on the generators
is \ct.
We need the following lemma to ensure that this method gives
an action which is \ct{}
on the entire algebra.
\begin{lem}\label{LCtGen}
Let $X$ be a topological space,
let $A$ be a \ca,
and let $x \mapsto \af_x$ be a function from $X$ to the
endomorphisms of~$A$.
Suppose there is a subset $S \subset A$ which generates $A$ as a \ca{}
and such that $x \mapsto \af_x (a)$ is \ct{} for all $a \in S$.
Then $x \mapsto \af_x (a)$ is \ct{} for all $a \in A$.
\end{lem}
The proof is an $\frac{\ep}{3}$~argument.
The key point is that $\sup_{x \in X} \ \af_x \$
is finite.
As far as we know,
without explicitly including this condition in the
hypotheses,
there are no analogous results for Banach algebras,
even in the situation of group actions.
\begin{proof}[Proof of \Lem{LCtGen}]
Let $A_0 \subset A$
be the complex *subalgebra of $A$ generated by~$S$.
Then $x \mapsto \af_x (a)$ is \ct{} for all $a \in A_0$.
Now let $a \in A$ be arbitrary,
let $x_0 \in X$, and let $\ep > 0$.
We have to find an open set $U \subset X$ with $x_0 \in U$
such that for all $x \in U$,
we have $\ \af_x (a)  \af_{x_0} (a) \ < \ep$.
Choose $a_0 \in A_0$ such that $\ a  a_0 \ < \frac{\ep}{3}$.
Since $x \mapsto \af_x (a_0)$ is \ct,
there is an open set $U \subset X$ with $x_0 \in U$
such that for all $x \in U$,
we have $\ \af_x (a_0)  \af_{x_0} (a_0) \ < \frac{\ep}{3}$.
For $x \in X$, since $\af_x$ is a \hm{} of \ca{s},
we have $\ \af_x (b) \ \leq \ b \$ for all $b \in A$.
For $x \in U$, we thus get
\begin{align*}
\ \af_x (a)  \af_{x_0} (a) \
& \leq \ \af_x (a)  \af_{x} (a_0) \
+ \ \af_x (a_0)  \af_{x_0} (a_0) \
+ \ \af_{x_0} (a_0)  \af_{x_0} (a) \
\\
& < \frac{\ep}{3} + \frac{\ep}{3} + \frac{\ep}{3}
< \ep.
\end{align*}
This completes the proof.
\end{proof}
\begin{exa}\label{E_RotGauge}
Let $\te \in \R$,
and let $A_{\te}$ be the rotation algebra,
as in Example~\ref{E_3303_RotAlg}.
For $\zt_1, \zt_2 \in S^1$,
the unitaries $\zt_1 u$ and $\zt_2 v$ satisfy the same
commutation relation.
Therefore there is an action
$\af \colon S^1 \times S^1 \to \Aut (A_{\te})$
determined by
$\af_{(\zt_1, \zt_2)} (u) = \zt_1 u$ and
$\af_{(\zt_1, \zt_2)} (v) = \zt_2 v$.
Continuity of the action follows from Lemma~\ref{LCtGen}.
If we fix $\zt_1, \zt_2 \in S^1$,
then $\af_{(\zt_1, \zt_2)}$ generates an action of~$\Z$.
The crossed product by this action turns out to be a three
dimensional noncommutative torus
as in Example~\ref{E_3303_NCTorus},
namely
the universal \ca{} generated by unitaries $u, v, w$
such that
\[
v u = \exp (2 \pi i \te) u v,
\qquad
w u = \zt_1 u w,
\andeqn
w v = \zt_2 w w.
\]
Repeating the construction,
one realizes an arbitrary higher dimensional noncommutative torus
as an iterated crossed product.
See Example~\ref{E_3302_HDNC} below.
If both $\zt_1$ and $\zt_2$ have finite order,
we get an action of a finite cyclic group.
For example, there is an action of $\Z / n \Z$
generated by the automorphism which sends $u$ to $\exp (2 \pi i / n) u$
and $v$ to~$v$.
\end{exa}
\begin{pbm}\label{P:AutHDNCT}
Find examples of actions of finite groups on
higher dimensional noncommutative tori
with interesting \cp{s}.
For this purpose,
the actions one gets from the higher dimensional version
of Example~\ref{E_RotGauge} are not very interesting,
because the \cp{} is closely related to another
higher dimensional noncommutative torus.
The only known general example that is interesting in this sense
is the ``flip'' action of $\Z / 2 \Z$,
generated by $u_k \mapsto u_k^*$ for $1 \leq k \leq d$.
Whenever the higher dimensional noncommutative torus is simple,
the \cp{} by this action is known to be AF~\cite{ELPW}.
\end{pbm}
There is recent work in this direction in~\cite{JngLee},
and some further work has been done.
Also see~\cite{Glpy} for some related work.
The following example gives the one related general family
of finite group actions that we know of.
It isn't on quite the same algebras as in Problem~\ref{P:AutHDNCT},
but more examples like this one would also be interesting.
\begin{exa}\label{E_3302_MilnesNCFT}
Fix $\te \in \R \setminus \Q$,
and let $A_{\te}^{6, 4}$ be the universal
unital \ca{} generated by unitaries
$u, v, w, x, y$ satisfying the following
commutation relations
(the relations~(CR) at the beginning of Section~2
of~\cite{Mln1}):
\[
u v = x v u,
\qquad
u w = w u,
\qquad
u x = x u,
\qquad
u y = e^{2 \pi i \te} y u,
\qquad
v w = y w v,
\]
\[
v x = x v,
\qquad
v y = y v,
\qquad
w x = e^{ 2 \pi i \te} x w,
\qquad
w y = y w,
\qquad
x y = y x.
\]
This algebra is simple,
and in fact it is the crossed product of the
action of the discrete Heisenberg group on $S^1 \times S^1$
in Example~\ref{E_3302_MHOn2Tor}.
(See Theorem~1 of~\cite{Mln1}.
The motivation is that $A_{\te}^{6, 4}$ is a simple quotient
of a discrete cocompact subgroup of a particular
nilpotent Lie group.)
Then there is an automorphism $\af \in \Aut \big( A_{\te}^{6, 4} \big)$
of order~$4$,
given by
\[
\af (y) = x,
\qquad
\af (w) = u,
\qquad
\af (u) = w^*,
\qquad
\af (x) = y^*,
\andeqn
\af (v) = v^*.
\]
See
Remark~2 on page 312 of~\cite{Mln1}.
\end{exa}
As far as we know,
nothing is known about the crossed products by the actions
of $\Z / 4 \Z$ in Example~\ref{E_3302_MilnesNCFT}.
We hope,
for example,
that this action has the \trp{}
(\Def{D_TRP}),
and that this can be used to help identify the crossed product,
perhaps by methods similar to those of~\cite{ELPW}.
The following example is a noncommutative version of
Example~\ref{E_3302_Furst}.
The automorphisms in this example were introduced in Definition~1.1
of~\cite{OsPh2}.
Several special cases were considered earlier,
in \cite{Pc87} and \cite{Pc88}.
% (Kodaka and Osaka; in preparation at the time
% of~\cite{OsPh2}: This seems not to exist.) 999
\begin{exa}\label{E_3303_NCFurst}
Let $\te \in \R$,
and let $A_{\te}$ be the rotation algebra,
as in Example~\ref{E_3303_RotAlg}.
Let $\gm \in \R$, let $d \in \Z$,
and let $f \colon S^1 \to \R$ be a \cfn.
The Furstenberg transformation on $A_{\te}$
determined by $(\te, \gm, d, f)$ is the automorphism
$\af_{\te, \gm, d, f}$ of $A_{\te}$ such that
\[
\af_{\te, \gm, d, f} (u) = e^{2 \pi i \gm} u
\andeqn \af_{\te, \gm, d, f} (v) = \exp (2 \pi i f (u)) u^d v.
\]
The parameter $\te$ does not appear in the formulas;
its only role is to specify the algebra on which the automorphism
acts.
When $\te = 0$,
we get the action determined by the
\hme{} of Example~\ref{E_3302_Furst}.
When $\te \not\in \Q$, the automorphism
$\af_{\te, \gm, d, f}$ is the most general automorphism
$\af$ of $A_{\te}$
for which $\af (u)$ is a scalar multiple of~$u$.
(See Proposition~1.6 of~\cite{OsPh2}.)
\end{exa}
\begin{exr}[Lemma~1.2 of~\cite{OsPh2}]\label{Ex_3303_NCFTPf}
Prove that the formula
for $\af_{\te, \gm, d, f}$
in Example~\ref{E_3303_NCFurst}
does in fact define an automorphism of~$A_{\te}$.
\end{exr}
\begin{exa}\label{E:CuntzGauge}
Let $n \in \N$ satisfy $n \geq 2$.
Recall that the Cuntz algebra
${\mathcal{O}}_n$ is the universal unital \ca{} on generators
$s_1, s_2, \ldots, s_n$,
subject to the relations $s_j^* s_j = 1$ for $1 \leq j \leq n$
and $\sum_{j = 1}^n s_j s_j^* = 1$.
(It is in fact simple,
so any \ca{} generated by elements satisfying
these relations is isomorphic to~${\mathcal{O}}_n$.)
There is an action of $(S^1)^n$ on~${\mathcal{O}}_n$
such that $\af_{(\zt_1, \zt_2, \ldots, \zt_n)} (s_j) = \zt_j s_j$
for $1 \leq j \leq n$.
(Check that the elements $\zt_j s_j$ satisfy the required
relations.)
The restriction to
the diagonal elements of $(S^1)^n$
gives an action of $S^1$ on~${\mathcal{O}}_n$,
sometimes called the gauge action.
In fact, regarding $(S^1)^n$ as the diagonal unitary matrices,
this action extends to an action of the unitary group
$U (M_n)$ on ${\mathcal{O}}_n$,
defined as follows.
If $u = ( u_{j, k} )_{j, k = 1}^n \in M_n$ is unitary, then
define an automorphism $\af_u$ of ${\mathcal{O}}_n$
by the following action
on the generating isometries $s_1, s_2, \dots, s_n$:
\[
\af_u (s_j) = \sum_{k = 1}^n u_{k, j} s_k.
\]
The assignment $u \mapsto \af_u$
determines a \ct{} action of the compact group
$U (M_n)$ on ${\mathcal{O}}_n$.
(This action is described, in a different form,
in Section~2 of~\cite{Ev}.)
Any individual automorphism from this action gives an
action of $\Z$ on ${\mathcal{O}}_n$.
More generally,
if $G$ is a topological group,
and $\rh \colon G \to U (M_n)$
is a \ct{} \hm{}
(equivalently,
a unitary representation of $G$ on~$\C^n$),
then the composition $\af \circ \rh$
is an action of $G$ on~${\mathcal{O}}_n$.
Such actions are called {\emph{quasifree actions}}.
\end{exa}
Several specific quasifree actions are used for
counterexamples in the discussion after \Thm{T_Ks}.
The action of $U (M_n)$ on ${\mathcal{O}}_n$
in \Ex{E:CuntzGauge}
is actually a special case of a much more general
(and natural looking) construction.
See Example~\ref{EGaugeCP} and Exercise~\ref{Ex_3311_GCP_On}.
\begin{exr}\label{P:CuntzGauge}
Verify that the formula given in Example~\ref{E:CuntzGauge}
does in fact define a \ct{} action of $U (M_n)$ on~${\mathcal{O}}_n$.
\end{exr}
The actions of $(S^1)^2$ in
Example~\ref{E_RotGauge}
and of $U (M_n)$ in Example~\ref{E:CuntzGauge}
are examples of what we think of as ``gauge type'' actions.
(The actions usually called gauge actions are the restrictions
of these to~$S^1$,
embedded diagonally.
Thus, in Example~\ref{E:CuntzGauge},
this is the action $\bt_{\zt} (s_j) = \zt s_j$
for $\zt \in S^1$ and $j = 1, 2, \ldots, n$.)
There are many more actions of this same general type,
and we give a collection of such actions in Section~\ref{Sec_AddGauge}.
Here we mention only the dual action on a crossed product
by an abelian group.
\begin{exa}\label{EZDual}
Let $A$ be a \ca, and let $\af \in \Aut (A)$ be an automorphism.
Then the dual action of ${\widehat{\Z}} = S^1$ is a \ct{} action
of $S^1$ on the crossed product $C^* (\Z, A, \af)$.
We will describe this action in Remark~\ref{R_6814_DualAction} below,
after we have given the construction of crossed products.
\end{exa}
\begin{exa}\label{EGenDual}
More generally, let $G$ be any locally compact group,
let $A$ be a \ca,
and
let $\af \colon G \to \Aut (A)$ be a \ct{} action of $G$ on~$A$.
Then there is a dual action
${\widehat{\af}} \colon
{\widehat{G}} \to \Aut \big( C^* (G, A, \af) \big)$.
Again,
we will describe this action in Remark~\ref{R_6814_DualAction} below,
after we have given the construction of crossed products.
\end{exa}
Although we will not give any details here,
there are several kinds of more general dual actions.
Crossed products by partial automorphisms,
and more generally by partial actions of groups,
are defined in~\cite{ExlBk}.
When the group $G$ which acts partially is abelian,
such a crossed product has an action of~${\widehat{G}}$.
In a somewhat different direction,
there are coactions of (not necessarily abelian)
locally compact groups on \ca{s},
and (full and reduced) crossed products by coactions are defined.
The full and reduced crossed products by a coaction of
a locally compact group~$G$ have a dual action,
which is an action of the (not necessarily abelian) group~$G$.
% {\tt{(Reference to be filled in.)}} % ???
The following result,
giving actions on direct limits of equivariant direct systems,
is useful for the next several examples.
We state it in general,
but in most of its applications,
the directed set~$I$
is $\N$ or $\Nz$ with its usual order,
and the maps of the direct system are all injective.
Then we can think of $\Dirlim A_n$
as being made by
arranging to have $A_1 \subset A_2 \subset \cdots$
and taking ${\overline{\bigcup_{n = 1}^{\I} A_n}}$.
Equivariance is then the condition
that the restriction to $A_n$
of the action on $A_{n + 1}$ is the action on $A_n$.
The action on $\bigcup_{n = 1}^{\I} A_n$
is then defined in the obvious way,
and is extended to ${\overline{\bigcup_{n = 1}^{\I} A_n}}$
by continuity.
\begin{prp}\label{P_3305_DirLimAction}
Let $G$ be a locally compact group.
Let
\[
\big( ( G, A_i, \af^{(i)} )_{i \in I},
\, (\ph_{j, i})_{i \leq j} \big)
\]
be a direct system of $G$algebras.
Let $A = \Dirlim A_i$.
Then there exists a unique action
$\af \colon G \to \Aut (A)$
such that $\af_g = \Dirlim \af^{(i)}_g$ for all $g \in G$.
\end{prp}
\begin{proof}
Existence of the automorphisms $\af_g$ for $g \in G$,
and their algebraic properties,
is easily obtained from the universal property of the
direct limit.
Continuity of the action follows from Lemma~\ref{LCtGen}.
\end{proof}
\begin{exa}\label{E_PrdType}
Let $k_1, k_2, \ldots$
be integers with $k_n \geq 2$ for all $n \in \N$.
Consider the UHF~algebra $A$ of type $\prod_{n = 1}^{\I} k_n$.
We construct it as $\bigotimes_{n = 1}^{\I} M_{k_n}$,
or, in more detail, as $\Dirlim A_n$
with $A_n = M_{k_1} \otimes M_{k_2} \otimes \cdots \otimes M_{k_n}$.
Thus $A_n = A_{n  1} \otimes M_{k_n}$,
and the map $\ph_n \colon A_{n  1} \to A_n$
is given by $a \mapsto a \otimes 1_{M_{k_n}}$.
Let $G$ be a locally compact group,
and for $n \in \N$
let $\bt^{(n)} \colon G \to \Aut (M_{k_n})$ be an
action of $G$ on $M_{k_n}$.
(The easiest way to get such an action is to use
an inner action as in Example~\ref{E:Inner}.
That is,
choose
a unitary representation $g \mapsto u_n (g)$ on $\C^{k_n}$,
and set $\bt^{(n)}_g (a) = u_n (g) a u_n (g)^*$
for $g \in G$ and $a \in M_{k_n}$.)
Then there is a unique
action $\af^{(n)} \colon G \to \Aut (A_n)$
such that
\[
\af^{(n)}_g (a_1 \otimes a_2 \otimes \cdots \otimes a_n)
= \bt^{(1)}_g (a_1) \otimes \bt^{(2)}_g (a_2)
\otimes \cdots \otimes \bt^{(n)}_g (a_n)
\]
for
\[
a_1 \in M_{k_1},
\qquad
a_2 \in M_{k_2},
\qquad
\ldots,
\qquad
a_n \in M_{k_n},
\andeqn
g \in G.
\]
One checks immediately that
$\ph_n \circ \af^{(n  1)}_g = \af^{(n)}_g \circ \ph_n$
for all $n \in \N$ and $g \in G$,
so, by Proposition~\ref{P_3305_DirLimAction},
there is a direct limit action
$g \mapsto \af_g$ of $G$ on $A = \Dirlim A_n$.
It is written $\af_g = \bigotimes_{n = 1}^{\I} \bt^{(n)}_g$.
We call such actions {\emph{infinite tensor product actions}}.
If each $\bt^{(n)}$ is inner,
the resulting action was
originally called a {\emph{product type action}}.
The general case of such actions was first seriously investigated
in \cite{HR1} and~\cite{HR2}.
As a specific example, take $G = \Z / 2 \Z$,
and for every $n$ take $k_n = 2$ and take
$\bt^{(n)}$ to be generated by
$\Ad \left(
\begin{smallmatrix} 1 & 0 \\ 0 & 1 \end{smallmatrix} \right)$.
For another specific example, take $G = S^1$,
and for every $n$ take $k_n = 2$ and for $\zt \in S^1$ take
\[
\bt^{(n)}_{\zt}
= \Ad \left(
\begin{matrix} 1 & 0 \\ 0 & \zt \end{matrix} \right)
\qquad {\mbox{or}} \qquad
\bt^{(n)}_{\zt}
= \Ad \left(
\begin{matrix} 1 & 0 \\ 0 & \zt^{2^{n  1}} \end{matrix} \right).
\]
The second choice gives
\[
\bt^{(n)}_{\zt}
= \Ad \big( \diag \big( 1, \zt, \zt^2, \ldots,
\zt^{2^n  1} \big) \big)
\]
for $n \in \N$ and $\zt \in S^1$.
\end{exa}
In Example~\ref{E_PrdType},
even if all the actions $\bt^{(n)}$
(and hence also the actions $\af^{(n)}$)
in the construction are inner,
one does not expect the action $\af$ to be inner.
It is often easy to compute the crossed product
(see Example~\ref{E:C:PTypeAgain} for an illustration of the method),
and the result is often not the same as the crossed product by
an inner action.
Here, though, we prove that the action is not inner
in one case for which a direct proof is easy.
\begin{lem}\label{L_3416_NotInn}
In Example~\ref{E_PrdType},
assume that $k_n \geq 2$ for all $n \in \N$,
take $G = \Z / 2 \Z$,
for $n \in \N$
choose $r_n, s_n \in \N$
such that $r_n + s_n = k_n$,
set $z_n = \diag (1_{r_n}, \,  1_{s_n} ) \in M_{k_n}$,
and let $\bt^{(n)} \colon G \to \Aut (M_{k_n})$ be the
action generated by $\Ad (z_n)$.
Let $A$ and $\af \colon G \to \Aut (A)$
be as in the construction of Example~\ref{E_PrdType}.
Then $\af$ is not an inner action.
\end{lem}
\begin{proof}
Let $\gm \in \Aut (A)$ be the automorphism given by
the nontrivial element of $\Z / 2 \Z$.
Assume that there is $v \in U (A)$ such that
$\gm = \Ad (v)$.
Choose $n \in \N$ and $c \in A_n \subset A$
such that $\ c  v \ < \frac{1}{2}$.
Define \pj{s} $e_0, e_1 \in M_{k_{n + 1}}$ by
\[
e_0 = \frac{1}{2}
\left( \begin{matrix}
1 & 0 & \cdots & 0 & 1 \\
0 & 0 & \cdots & 0 & 0 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \cdots & 0 & 0 \\
1 & 0 & \cdots & 0 & 1
\end{matrix} \right)
\andeqn
e_1 = \frac{1}{2}
\left( \begin{matrix}
1 & 0 & \cdots & 0 &  1 \\
0 & 0 & \cdots & 0 & 0 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \cdots & 0 & 0 \\
 1 & 0 & \cdots & 0 & 1
\end{matrix} \right).
\]
Then $e_0$ and $e_1$ are orthogonal \pj{s}
which are exchanged by the nontrivial
element of $\Z / 2 \Z$ under the action $\bt^{(n + 1)}$.
Therefore
\[
p_0 = 1_{A_n} \otimes e_0
\andeqn
p_1 = 1_{A_n} \otimes e_1,
\]
regarded as elements of~$A$,
are orthogonal \pj{s} such that $\gm (p_0) = p_1$ and $\gm (p_1) = p_0$.
Also $c$ commutes with~$p_0$.
Therefore, using $v p_0 v^* = \gm (p_0) = p_1$ at the third step,
\[
1 = \ p_0  p_1 \
= \ ( p_0  p_1 ) v \
= \ p_0 v  v p_0 \
\leq \ p_0 c  c p_0 \ + 2 \ c  v \
= 2 \ c  v \
< 1.
\]
This is a contradiction,
and we have proved that $\gm$ is not inner.
\end{proof}
The following example is taken from the beginning
of Section~4 of~\cite{PhVl},
and is a special case of the adaptation to \ca{s}
of the construction of Proposition~1.6 of~\cite{Connes3},
where an analogous example is constructed on the
hyperfinite factor of type~${\mathrm{II}}_1$.
We give the formulas for the action and the beginning of the proof
that it is an action,
but we refer to the proof of Proposition~1.6 of~\cite{Connes3}
for details.
\begin{exa}\label{E_3412_Connes}
Let $D = \bigotimes_{m = 1}^{\I} M_d$
be the $d^{\I}$~UHF algebra.
We describe an action $\af \colon \Z / d^2 \Z \to \Aut (D)$
such that,
writing $\Z / d^2 \Z = \{ 0, 1, 2, \ldots, d^2  1 \}$,
the automorphism $\af_d$ is inner,
but every unitary $v$ such that $\af_d (v) = v a v^*$
for all $a \in D$ satisfies $\af (v) = \exp (2 \pi i / d) v$.
Thus the image $\gm$ of $\af_1$ in the outer automorphism group
${\operatorname{Out}} (A) = \Aut (A) / {\operatorname{Inn}} (A)$
(the quotient of $\Aut (A)$ by the inner automorphisms)
has order~$d$,
but $\gm$ can't be lifted to an order~$d$ element of $\Aut (A)$.
We identify $D$ as the closed linear span of all
elements of the form
%
\begin{equation}\label{Eq_3413_ConnesSt}
a = a_1 \otimes a_2 \otimes \cdots \otimes a_n
\otimes 1 \otimes 1 \otimes \cdots
\end{equation}
%
with $n \in \Nz$ and $a_1, a_2, \ldots, a_n \in M_d$.
For $n \in \Nz$,
set $D_n = \bigotimes_{k = 1}^{n} M_d$,
and let $\ps_n \colon D_n \to D_{n + 1}$
be the unique \hm{}
such that $\ps_n (a) = a \otimes 1_{M_d}$
for all $a \in D_n$.
Thus $D = \Dirlim_n D_n$.
For $n \in \Nz$ let $\gm_n \colon D_n \to D$
be the map obtained from the direct limit.
For $n \in \N$ let $\pi_n \colon M_d \to D$
be the embedding of $M_d$ as the tensor factor in position~$n$.
Thus,
$\pi_n (a) = \gm_n (1_{D_{n  1}} \otimes a)$.
Equivalently,
\[
\pi_n (x) = 1 \otimes 1 \otimes \cdots \otimes 1 \otimes a
\otimes 1 \otimes 1 \otimes \cdots,
\]
with $a$ in position~$n$.
Let $\ld \colon D \to D$
be the shift endomorphism of~$D$,
that is, using the notation~(\ref{Eq_3413_ConnesSt}),
the endomorphism
given by $\ld (a) = 1 \otimes a$ for $a \in D$.
Then $\ld \circ \pi_n = \pi_{n + 1}$ for all $n \in \N$.
Let $(e_{j, k})_{j, k = 1, 2, \ldots, d}$
be the standard system of matrix units for~$M_d$.
Define unitaries $v, u \in D$ by
\[
v = \pi_1 \Bigg( \sum_{j = 1}^{d} e^{2 \pi i j / d} e_{j, j} \Bigg)
\andeqn
u = \pi_1 (e_{d, 1}) \ld (v^*)
+ \sum_{j = 1}^{d  1} \pi_1 (e_{j, j + 1} ).
\]
Then define
$\af_n \in \Aut (D)$
by
$\af_n = \Ad \big( u \ld (u) \ld^2 (u) \cdots \ld^{n  1} (u) \big)$.
We claim that there is $\af \in \Aut (D)$
such that $\af (a) = \limi{n} \af_n (a)$ for all $a \in D$.
Moreover, we claim that
$\af^d = \Ad (v)$, that $\af (v) = e^{2 \pi i/d} v$,
and that $\af^l$ is an outer automorphism of~$D$
for $l = 1, 2, \ldots, d  1$.
Finally, we claim that for every unitary $w \in D$,
there is a unitary $z \in D$
such that $( \Ad (w) \circ \af)^d = \Ad (z)$,
and that for every such $z$ we have
$( \Ad (w) \circ \af)^d (z) = e^{2 \pi i/d} z$.
We prove only the first part of this,
and refer to the calculations
in the proof of Proposition~1.6 of~\cite{Connes3}
for the rest.
% CCC
We start by proving the existence of a \hm~$\af \colon D \to D$
such that $\af (a) = \limi{n} \af_n (a)$ for all $a \in D$.
By a standard $\frac{\ep}{3}$~argument,
it suffices to prove that
$\limi{n} \af_n (a)$ exists for every $a$ in a dense
subset $S \subset D$.
Our choice for $S$ is $S = \bigcup_{n = 0}^{\I} \gm_n (D_n)$.
For every $m \in \N$,
every element of $\gm_m (D_m)$
commutes with every element in the range of~$\ld_m$,
and in particular with $\ld^n (u)$ for every $n \geq m$.
If $a \in \Gm_m (D_m)$,
it therefore follows that
$\af_n (a) = \af_m (a)$ for all $n \geq m$,
so that $\limi{n} \af_n (a)$ certainly exists.
Since $D$ is simple, $\af$ is injective.
The next step is to prove that $\af^d (a) = \Ad (v) (a)$
for every $a \in D$.
It then follows that $\af$ is surjective.
Thus $\af \in \Aut (D)$.
We omit the rest of the proof.
% CCC To be filled in.
\end{exa}
\begin{exa}\label{E_3412_UHFShift}
Let $d \in \{ 2, 3, \ldots \}$.
Let $D = \bigotimes_{m = 1}^{\I} M_d$
be the $d^{\I}$~UHF algebra.
We describe an action of $\Z$ on $K \otimes D$
which scales the trace on~$D$,
by describing its generating automorphism~$\af$.
We will identify $D$ as the closed linear span of all
elements of the form
\[
a = a_1 \otimes a_2 \otimes \cdots \otimes a_m
\otimes 1 \otimes 1 \otimes \cdots
\]
with $m \in \Nz$ and $a_1, a_2, \ldots, a_m \in M_d$.
To help keep the notation straight,
we use the isomorphism $\mu \colon M_d \otimes D \to D$
given by, for $a$ as above and $x \in M_d$,
\[
\mu (x \otimes a)
= x \otimes a_1 \otimes a_2 \otimes \cdots \otimes a_m
\otimes 1 \otimes 1 \otimes \cdots.
\]
To be explicit,
on the right hand side $x$ is in the first tensor factor
of $M_d$ in $D = \bigotimes_{m = 1}^{\I} M_d$,
the factor~$a_1$,
which previously was in the first tensor factor,
is now in the second, etc.
Set $C_n = \big( \bigotimes_{k = n}^{0} M_d \big) \otimes D$
for $n \in \Nz$.
(The indexing is chosen so that we can think of $C_n$
as $\bigotimes_{m =  n}^{\I} M_d$.)
Let $(e_{j, k})_{j, k = 1, 2, \ldots, d}$
be the standard system of matrix units for~$M_d$.
For $n \in \Nz$,
there are \hm{s}
\[
\ps_n \colon C_n \to C_{n + 1}
\andeqn
\af_n \colon C_n \to C_n
\]
such that,
for $x_{n}, x_{n + 1}, \ldots, x_0 \in M_d$ and $a \in D$,
we have
\[
\ps_n \big( x_{n} \otimes x_{n + 1} \otimes \cdots
\otimes x_0 \otimes a \big)
= e_{1, 1} \otimes x_{n} \otimes x_{n + 1} \otimes \cdots
\otimes x_0 \otimes a
\]
and
\[
\af_n \big( x_{n} \otimes x_{n + 1} \otimes \cdots
\otimes x_0 \otimes a \big)
= e_{1, 1} \otimes x_{n} \otimes x_{n + 1} \otimes \cdots
\otimes \mu (x_0 \otimes a).
\]
For all $n \in \Nz$,
one checks that the diagram
\[
\begin{CD}
\bigotimes_{k = n}^{0} M_d \otimes D @>{\af_n}>>
\bigotimes_{k = n}^{0} M_d \otimes D \\
@V{\ps_n}VV @VV{\ps_n}V \\
\bigotimes_{k = n  1}^{0} M_d \otimes D @>>{\af_{n + 1}}>
\bigotimes_{k = n  1}^{0} M_d \otimes D
\end{CD}
\]
commutes.
Indeed,
both possible maps from the top left to the bottom right
are given by
\[
x_{n} \otimes x_{n + 1} \otimes \cdots
\otimes x_0 \otimes a
\mapsto e_{1, 1} \otimes e_{1, 1} \otimes x_{n}
\otimes x_{n + 1} \otimes \cdots
\otimes \mu (x_0 \otimes a)
\]
for $x_{n}, x_{n + 1}, \ldots, x_0 \in M_d$ and $a \in D$.
Set $C = \Dirlim_n C_n$,
using the maps $\ps_n \colon C_n \to C_{n + 1}$,
and for $n \in \Nz$ let $\gm_n \colon C_n \to C$
be the associated map.
Then there is a \hm{} $\af \colon C \to C$
% given by $\af_n$ on the image of
% $C_n = \bigotimes_{k = n}^{0} M_d \otimes D$ in~$C$.
such that $\af \circ \gm_n = \gm_n \circ \af_n$
for all $n \in \Nz$.
The map $\af$ is injective because $C$ is simple.
Also, for $n \in \Nz$,
$\af (C)$ contains $\gm_n (C_n)$,
since $\ps_n (C_n) = \af_{n + 1} ( C_{n + 1})$.
Thus $\af$ is an automorphism.
It is easy to see that $C \cong K \otimes D$.
The crossed product $C^* (\Z, \, K \otimes D, \, \af)$
turns out to be
the stabilized Cuntz algebra $K \otimes {\mathcal{O}}_d$.
See Section~2.1 of~\cite{Cu77}.
\end{exa}
We can't quite use Proposition~\ref{P_3305_DirLimAction} here,
because $\af_n \colon C_n \to C_n$ is not surjective.
Many stable Kirchberg algebras~$A$
satisfying the Universal Coefficient Theorem
can be realized as crossed products by
actions of $\Z$ on stable AF~algebras of the same general type
as in Example~\ref{E_3412_UHFShift}.
The group $K_1 (A)$ must be torsion free;
then see Corollary~4.6 of~\cite{Rrd1}.
(The statement there is for unital algebras obtained as
crossed products by ``corner endomorphisms''.
See Proposition~2.1 of~\cite{Rrd1}
for the relation to our construction.)
For general $K_1 (A)$,
suitable actions on AT~algebras are given
in Theorem~3.6 of~\cite{Rrd1}.
It isn't proved there that the crossed products are Kirchberg algebras.
However, they are certainly nuclear
and satisfy the Universal Coefficient Theorem.
It is presumably easy to show that they are purely infinite and simple,
and it would follow from the classification theorem
that they are Kirchberg algebras
satisfying the Universal Coefficient Theorem.
We now give several examples of direct limit actions on~AH algebras
in which homeomorphisms of the spaces in the construction are used
to define the actions.
\begin{exa}\label{E_3502_Blackadar}
In~\cite{Bl0},
Blackadar gives an action~$\af$
of $\Z / 2 \Z$ on the $2^{\infty}$~UHF algebra~$D$
such that $C^* (\Z/ 2 \Z, \, D, \, \af)$ is not an AF~algebra.
We refer to that paper for the details,
which require a fair amount of description.
The action is obtained by realizing $D$ as a direct limit
$D = \Dirlim_{n} C (S^1, M_{4^n})$,
with the maps $\ph_n \colon C (S^1, M_{4^{n  1}}) \to C (S^1, M_{4^n})$
of the system
being described as follows.
Choose a unitary path $t \mapsto s_t \in M_2$, for $t \in [0, 1]$,
such that
\[
s_0 = 1
\andeqn
s_1 = \left( \begin{matrix}
0 & 1 \\
1 & 0
\end{matrix} \right).
\]
Define (justification afterwards)
$\ps \colon C (S^1) \to C (S^1, M_2)$
by
\[
\ps (f) ( e^{2 \pi i \te})
= s_t \left( \begin{matrix}
f (e^{\pi i t}) & 0 \\
0 & f ( e^{\pi i (t + 1)} )
\end{matrix} \right) s_t^*
\]
for $t \in [0, 1]$
and $f \in C (S^1)$.
The only point requiring justification
is that the values at $t = 0$ and at $t = 1$
(both corresponding to the point $1 \in S^1$) are equal,
and this is easily checked.
(This kind of map will implicitly reappear
in the computations in Example~\ref{E:C:FiniteRot}.)
The map $\ph_0 \colon C (S^1) \to C (S^1, M_4)$
is then given by
\[
\ph_0 (f) (\zt) = \diag \big( \ps (f) (\zt), \, \ps (f) ( \zt^{1} ) \big)
\]
for $f \in C (S^1)$ and $\zt \in S^1$,
and $\ph_n$ is obtained by tensoring $\ph_0$ with $\id_{M_{4^{n}}}$.
Of course,
one must prove that the resulting direct limit
is in fact the $2^{\infty}$~UHF algebra.
These days, the isomorphism is an immediate consequence
of standard classification theorems.
(At the time this example was constructed,
no applicable classification theorems were known.)
\end{exa}
The action of \Ex{E_3502_Blackadar}
is also an ingredient in the construction of
the action in \Ex{E_6X02_Blackadar}.
The following example is adapted from~\cite{Grd}.
\begin{exa}\label{E_3305_Gardella}
Let $G$ be a compact metrizable group.
Let $(k_n)_{n \in \Nz}$ be a sequence in $G$ such that
$\{ k_n \colon n \geq N \}$ is dense in~$G$ for all $N \in \N$.
(The only use of density in the construction is to ensure
that the algebra we get at the end is simple.
Everything else works for an arbitrary sequence $(k_n)_{n \in \Nz}$.)
For $n \in \Nz$,
define $\ph_n \colon C (G, M_{2^{n  1}} ) \to C (G, M_{2^{n}} )$
by $\ph_n (a) (g) = \diag (a (g), \, a (g k_n))$
for $a \in C (G, M_{2^n} )$ and $g \in G$.
Define an action $\af^{(n)} \colon G \to \Aut (C (G, M_{2^n} ))$
by $\af^{(n)}_g (a) (h) = a (g^{1} h)$
for $a \in C (G, M_{2^n} )$ and $g, h \in G$.
It is easy to check that
$\ph_n \circ \af^{(n  1)}_g = \af^{(n)}_g \circ \ph_n$
for all $n \in \N$ and $g \in G$,
so, by Proposition~\ref{P_3305_DirLimAction},
there is a direct limit action
$g \mapsto \af_g$ of $G$ on $A = \Dirlim C (G, M_{2^n} )$.
The direct limit $A$ is a simple AH~algebra.
(Use Proposition~2.1 of~\cite{DNNP}.)
The resulting direct limit action of~$G$
(Proposition~\ref{P_3305_DirLimAction})
turns out to have the Rokhlin property
for actions of compact groups,
as in Definition~3.2 of~\cite{HrWn}.
When $G = S^1$,
one gets an action of $S^1$ on a simple AT~algebra
with the Rokhlin property.
\end{exa}
Actions of compact groups with the Rokhlin property
seem to be hard to find.
\begin{exr}\label{Ex_3305_Gdl}
Prove the statements made in \Ex{E_3305_Gardella}.
\end{exr}
Several further examples of this general type are
found in Exercise~\ref{Ex:RCFG4_K1}
and Exercise~\ref{Ex:RCFG4_Torsion}.
\begin{exa}\label{E:TensFlip}
Let $A$ be a \ca.
The {\emph{tensor flip}}
is the automorphism
$\ph \in \Aut ( A \otimes_{\mathrm{max}} A )$ of order~$2$
determined by the formula
$\ph (a \otimes b) = b \otimes a$ for $a, b \in A$.
To prove the existence of such an automorphism
in the unital case,
use the universal property of $A \otimes_{\mathrm{max}} A$.
Reduce the nonunital case to the unital case.
This gives an action of $\Z / 2 \Z$ on $A \otimes_{\mathrm{max}} A$.
The same formula also defines a tensor flip action
of $\Z / 2 \Z$ on $A \otimes_{\mathrm{min}} A$.
To prove the existence of such an automorphism,
choose an injective representation $\pi \colon A \to L (H)$,
and consider $\pi \otimes \pi$ as a
representation of $A \otimes_{\mathrm{min}} A$ on $H \otimes H$.
Let $u \in L (H \otimes H)$
be the unitary which exchanges the two tensor factors.
Then the required automorphism is given by conjugation by~$u$.
In a similar manner, the symmetric group $S_n$ acts on
the $n$fold maximal and minimal tensor products of $A$ with itself.
This is a noncommutative generalization of Example~\ref{E:CommPerm}.
\end{exa}
\begin{exa}\label{E:JiangSuFlip}
The JiangSu algebra~$Z$,
introduced in~\cite{JS},
is an in\fd{} simple separable nuclear
\ca{} with no nontrivial \pj{s} whose Ktheory is the same as
that of~$\C$,
and such that $Z \otimes Z \cong Z$.
Thus, the tensor flip action of $\Z / 2 \Z$ on $Z \otimes Z$,
as in Example~\ref{E:TensFlip},
gives an action of $\Z / 2 \Z$ on~$Z$.
Similarly, tensor permutation as in Example~\ref{E:TensFlip}
gives an action of the symmetric group $S_n$ on~$Z$.
\end{exa}
The JiangSu algebra plays a key role in classification theory,
but will appear in only a few places in these notes.
\begin{exa}\label{E:TensShift}
Let $A$ be a unital \ca.
Let $B = \bigotimes_{n \in \Z} A$ be the infinite
minimal tensor product of copies of~$A$.
We define the {\emph{minimal shift}} on $B$
as follows.
Set $B_n = A^{\otimes (2 n)}$, the (minimal) tensor product
of $2 n$ copies of $A$.
(Take $B_0 = \C$.)
For $n \in \Nz$,
define $\ph_n \colon B_n \to B_{n + 1}$
by $\ph_n (a) = 1_A \otimes a \otimes 1_A$ for $a \in B_n$.
Identify $B$ with $\Dirlim B_n$, using the maps $\ph_n$
in the direct system.
Then take $\sm \colon B \to B$ to be the direct limit of
the maps $\sm_n \colon B_n \to B_{n + 1}$ defined
by $\sm_n (a) = 1_A \otimes 1_A \otimes a$ for $a \in B_n$.
We define the {\emph{maximal shift}} on the infinite
maximal tensor product in the same manner.
These are called {\emph{tensor shifts}}
or {\emph{Bernoulli shifts}} over~$\Z$.
There are Bernoulli shifts over any discrete group~$G$.
\end{exa}
Example~\ref{E:TensShift} is the noncommutative analog
of Example~\ref{E:Shift}.
Indeed, using the notation there,
if $A = \C^2$, then $\bigotimes_{n \in \Z} A \cong C (X)$,
and the tensor shift is the automorphism induced by the shift on~$X$.
\begin{exa}\label{E_3303_FreeFlip}
Let $A$ be a \ca.
The {\emph{free flip}}
on the (full) free product $A \star A$
is the automorphism
$\ph \in \Aut ( A \star A )$ of order~$2$
given as follows.
Let $\io_1, \io_2 \colon A \to A \star A$
be the inclusions of the two free factors.
Then $\ph$ is
determined by the formula
$\ph (\io_1 (a)) = \io_2 (a)$ and $\ph (\io_2 (a)) = \io_1 (a)$
for $a \in A$.
(To see that it exists,
use the universal property of $A \star A$.)
This gives an action of $\Z / 2 \Z$ on $A \star A$.
The same formula also defines a free flip action
of $\Z / 2 \Z$ on the reduced free product $A \star_{\mathrm{r}} A$,
taken with respect to the same state on both copies of~$A$.
One also gets a flip action of $\Z / 2 \Z$
on the amalgamated free product $A \star_B A$ over
a subalgebra $B \subset A$,
taking the same inclusion of $B$ into both copies of~$A$.
If $A$ is unital,
one important choice is $B = \C \cdot 1_A$,
giving a unital amalgamated free product.
One can also used reduced amalgamated free products.
In a similar manner, the symmetric group $S_n$ acts on
the $n$fold full and reduced (amalgamated) free products
of $A$ with itself.
This is a different
noncommutative generalization of Example~\ref{E:CommPerm}.
There are (reduced or amalgamated) free Bernoulli shifts on free
products of copies of~$A$ indexed by~$\Z$
(the free analog of \Ex{E:TensShift}),
free Bernoulli shifts over other discrete groups,
and more general versions of the same kind of construction.
\end{exa}
Free Bernoulli shifts are used in Section~2 of~\cite{PhSrTh}
to give (initially surprising)
examples of actions of noncompact groups which are equivariantly
semiprojective.
Our next example involves graph algebras.
We take~\cite{Rbn} as our main reference.
However, we warn that there are two conflicting conventions,
both in common use,
for the relation
between the direction
of the arrows in the graph and the definition of its C*algebra.
(For example, the papers \cite{KmPs} and~\cite{Spl}, cited below,
use the opposite convention from~\cite{Rbn}.)
In the definition below, the other convention
exchanges $s_e$ and $s_e^*$.
When reading papers about graph algebras,
one must therefore always check which convention is being used.
We also warn that the graph terminology
commonly used in this subject conflicts with graph terminology
used in some other parts of graph theory.
The following definition is
from the beginning of Chapter~5 of~\cite{Rbn}.
See Proposition~1.21 of~\cite{Rbn} for the case of a rowfinite graph.
We emphasize that graphs are allowed to have parallel edges
and edges which begin and end at the same vertex,
and that the edges are oriented.
\begin{dfn}\label{D_3306_GraphAlg}
Let $E = \big( E^{(0)}, E^{(1)}, r, s \big)$ be a directed graph,
with vertex set $E^{(0)}$, edge set $E^{(1)}$,
and range and source maps
$r, s \colon E^{(1)} \to E^{(0)}$.
That is, if $e \in E^{(1)}$ is an edge,
then $e$ begins at $s (e)$ and ends at $r (e)$.
The graph \ca{} $C^* (E)$
is the universal \ca{} on generators
$p_v$ for $v \in E^{(0)}$ and $s_e$ for $e \in E^{(1)}$,
subject to the following relations:
\begin{enumerate}
\item\label{D_3306_GraphAlg_Pj}
The elements $p_v$ for $v \in E^{(0)}$ are
\mops.
\item\label{D_3306_GraphAlg_Isom}
The elements $s_e$ for $e \in E^{(1)}$ are partial isometries.
\item\label{D_3306_GraphAlg_Sc}
$s_e^* s_e = p_{s (e)}$ for all $e \in E^{(1)}$.
\item\label{D_3306_GraphAlg_Rg}
$p_{r (e)} s_e s_e^* = s_e s_e^*$ for all $e \in E^{(1)}$.
\item\label{D_3306_GraphAlg_Sum}
For every $v \in E^{(0)}$ for which
$r^{1} (v) = \big\{ e \in E^{(1)} \colon r (e) = v \big\}$
is finite but not empty,
we have
$\sum_{e \in r^{1} (v)} s_e s_e^* = p_v$.
\end{enumerate}
\end{dfn}
We give brief descriptions of some examples.
For $n \in \Nz$,
let $E_n$ be the graph with one vertex~$v$ and
$n$~edges $e_1, e_2, \ldots, e_n$.
Here is the picture:
\centerline{
\includegraphics[width=3cm]{OnGraph2}
}
\noindent
The relations guarantee that $C^* (E_n)$ is unital,
with identity~$p_v$.
The graph $E_0$ has no edges,
so $C^* (E_0)$ has no other generators,
and is isomorphic to~$\C$.
The graph $E_1$ gives one additional generator,
namely
an element $s_e$ such that $s_e^* s_e = s_e s_e^* = p_v = 1$.
Thus $C^* (E_1) \cong C (S^1)$.
For the graph $E_n$,
with $n \geq 2$,
the additional generators are $s_{e_1}, e_{e_2}, \ldots, s_{e_n}$,
and the relations are
$s_{e_j}^* s_{e_j} = 1$ for $j = 1, 2, \ldots, n$
and $\sum_{j = 1}^n s_{e_j} s_{e_j}^* = 1$.
Under the identification
\[
s_1 = s_{e_1},
\,\,\,\,\,
s_2 = s_{e_2},
\,\,\,\,\,
\ldots,
\,\,\,\,\,
s_n = s_{e_n},
\]
these obviously generate the Cuntz algebra~${\mathcal{O}}_n$
which was used in Example~\ref{E:CuntzGauge}.
The following well known \ca{s} are also isomorphic
to \ca{s} of suitable graphs:
the Toeplitz \ca{} (Example~1.23 of~\cite{Rbn}),
CuntzKrieger algebras (Remark~2.8 of~\cite{Rbn}),
$M_n$ (this is essentially contained
in Proposition~1.18 of~\cite{Rbn}),
and many AF~algebras
(Proposition~2.12 and Remark~2.13 of~\cite{Rbn}).
Automorphisms of graphs give automorphisms of the
corresponding graph algebras.
This is essentially immediate
from \Def{D_3306_GraphAlg}.
See the discussion before Lemma~3.1 and before
Example~3.2 in~\cite{KmPs}.
Here are some specific examples.
\begin{exa}\label{E_3512_OnGraphAuto}
For $n \in \N$ with $n \geq 2$,
let $E_n$ be the graph above
(with one vertex and $n$ edges).
Then the permutation group $S_n$ acts on $E_n$
by permuting the edges.
The corresponding action $\af \colon S_n \to \Aut ({\mathcal{O}}_n)$
is given on the generators $s_1, s_2, \ldots, s_n$
by $\af_{\sm} (s_j) = s_{\sm (j)}$ for $j = 1, 2, \ldots, n$.
This action is a special case of the quasifree
actions in Example~\ref{E:CuntzGauge},
obtained by restricting from the unitary group $U (M_n)$
to the permutation matrices.
\end{exa}
\begin{exa}\label{E_3512_QIGraphAuto}
Consider the following graph~$Q$:
\centerline{
\includegraphics[width=12cm]{QInftyGraph}
}
\noindent
It is taken from the proof of Theorem~2.2 of~\cite{Spl}.
We have reversed the arrows, because the convention
used in~\cite{Spl}
is the opposite to that of \Def{D_3306_GraphAlg}.
We have also used different names for the vertices.
It is shown in~\cite{Spl}
that $C^* (Q)$
(called ${\mathcal{O}} (Q)$ in the notation of~\cite{Spl})
is the nonunital Kirchberg algebra
satisfying the Universal Coefficient Theorem,
$K_0 ( C^* (Q) ) = 0$,
and $K_1 ( C^* (Q) ) \cong \Z$.
(The algebra is nonunital since the graph has infinitely many vertices.)
We derive the computation of
$K_1 ( C^* (Q) )$ from Theorem~6.1 of~\cite{BHRS}.
(The corresponding formula in~\cite{Spl},
in Equation~(2.2) there, has a misprint:
in the formula for $K_1 ( C^* (E) )$,
the first condition on $f (x)$ there should be required to hold for
all $x \in E^{(0)}$,
not just the vertices~$x$ which emit a nonzero finite number of edges.)
Accordingly,
$K_1 ( C^* (Q) )$ can be identified with the set of functions
\[
f \colon \{ x_0, x_1, x_2, \ldots \} \cup \{ y_0, y_1, y_2, \ldots \}
\to \Z
\]
which have finite support,
such that
%
\begin{equation}\label{Eq_3512_OIFirst}
f (x_0) + f (y_0) = 0
\end{equation}
%
and
\[
f (x_j)  [f (x_j) + f (x_{j + 1})] = 0
\andeqn
f (y_j)  [f (y_j) + f (y_{j + 1})] = 0
\]
for $j = 0, 1, 2, \ldots$.
These simplify to
%
\begin{equation}\label{Eq_3512_OISecond}
f (x_j) = f (y_j) = 0
\qquad {\text{for $j = 1, 2, \ldots$.}}
\end{equation}
%
One checks immediately that
there is an injective \hm{} $\ld \colon \Z \to K_1 ( C^* (Q) )$
defined by $\ld (n) (x_0) = n$,
$\ld (n) (y_0) =  n$,
and $\ld (n) (x_j) = \ld (n) (y_j) = 0$ for $j = 1, 2, \ldots$.
It follows easily from
(\ref{Eq_3512_OIFirst}) and~(\ref{Eq_3512_OISecond})
that $\ld$ is surjective.
There is a unique automorphism $\af$ of $Q$ of order~$2$
such that
$\af (x_j) = y_j$ and $\af (y_j) = x_j$ for $j = 0, 1, 2, \ldots$,
and $\af (v) = v$.
It gives rise to an automorphism of $C^* (Q)$
of order~$2$,
which we also call~$\af$,
such that $\af_* \colon K_1 ( C^* (Q) ) \to K_1 ( C^* (Q) )$
is multiplication by~$1$.
This is an example of
the conclusion of Corollary~\ref{C_3305_KtsCyc} below.
(The paper~\cite{Spl} is a predecessor of~\cite{Kts}.
Its Corollary~2.3 is Corollary~\ref{C_3305_KtsCyc} below
when $n$ is prime.
The method is to construct a suitable automorphism
of a suitable graph.)
\end{exa}
\begin{exa}\label{E_3306_GraphAuto}
Consider the following graph~$F$:
\centerline{
\includegraphics[width=12cm]{ComplexGraph}
}
\noindent
(This graph appears as an example in~\cite{PsnPh},
in the discussion after Example~4.12 of~\cite{PsnPh}.
Its \ca{} is a nonsimple purely infinite \ca{}
with a composition series
whose subquotients have finite primitive ideal spaces.)
There is an automorphism $h \colon F \to F$ of order~$2$
which acts on the vertices by
\[
h (v_n) = w_n,
\qquad
h (w_n) = v_n,
\qquad
h (x_n) = y_n,
\andeqn
h (y_n) = x_n
\]
for $n \in \Z$, and which sends the inner loop at
each vertex $z$ to the inner loop at $h (z)$
and the outer loop at
$z$ to the outer loop at $h (z)$.
This automorphism induces an automorphism
$\ph$ of $C^* (F)$ of order~$2$,
which was used as an example for a theorem in~\cite{PsnPh}.
The corresponding action of $\Z_2$ on~$F$ is free.
Free actions on graphs are the subject of a very nice result,
Theorem~1.1
of~\cite{KmPs},
according to which the reduced crossed product
is stably isomorphic to the \ca{} of the quotient graph.
One can easily write down many other examples of actions of
finite or infinite groups on this graph,
or on others.
\end{exa}
We also give some theorems on the existence of actions.
\begin{thm}[Theorem~3.5 of~\cite{Kts}]\label{T_3305_Katsu}
Let $G$ be a finite group such that every Sylow subgroup
of $G$ is cyclic.
Let $A$ be a Kirchberg algebra
(separable nuclear purely infinite simple \ca;
\Def{D_6X13_Kbg})
which satisfies the Universal Coefficient Theorem.
Let $\sm \colon G \to \Aut (K_* (A))$
be an action of $G$ on the Ktheory of~$A$.
If $A$ is unital,
also assume that $\sm_g ([1_A]) = [1_A]$ for all $g \in G$.
Then there exists an action $\af \colon G \to \Aut (A)$
such that $(\af_g)_* = \sm_g$ for all $g \in G$.
\end{thm}
\begin{cor}[Corollary~3.6 of~\cite{Kts}]\label{C_3305_KtsCyc}
Let $A$ be a Kirchberg algebra
which satisfies the Universal Coefficient Theorem.
Let $n \in \N$,
and let $\sm \in \Aut (K_* (A))$ be an automorphism
such that $\sm^n = \id_{K_* (A)}$.
If $A$ is unital,
also assume that $\sm ([1_A]) = [1_A]$.
Then there exists an automorphism $\af \in \Aut (A)$
such that $\af_* = \sm$ and $\af^n = \id_A$.
\end{cor}
\begin{thm}[Theorem 4.8(3) of~\cite{Izm}]\label{T_3305_Izumi}
Let $\Gm_0$ and $\Gm_1$ be countable abelian groups which are
uniquely $2$divisible.
Then there exists an action
$\af \colon \Z / 2 \Z \to \Aut ( {\mathcal{O}}_2 )$
such that $B = C^* ( \Z / 2 \Z, \, {\mathcal{O}}_2, \, \af)$
satisfies the Universal Coefficient Theorem,
$K_0 (B) \cong \Gm_0$, and $K_1 (B) \cong \Gm_1$.
\end{thm}
\section{Additional Examples of Generalized Gauge
Actions}\label{Sec_AddGauge}
\indent
In this section,
we give further examples of what we think of as
``gauge type'' actions.
Example~\ref{E_RotGauge} (on the rotation algebras),
Example~\ref{E:CuntzGauge} (on Cuntz algebras),
Example~\ref{EZDual},
Example~\ref{EGenDual},
and the actions in the discussion after
Example~\ref{EGenDual} (dual actions),
are all of this type.
In many of the examples,
there is an action on the \ca{}
which is conventionally referred to as a gauge action.
Usually this is an action of~$S^1$.
(For the \ca{s} of rank~$k$ graphs,
discussed in Example~\ref{EGaugeHighRkGraph},
it is an action of $(S^1)^k$.)
In most cases,
we give actions of a larger group~$G$,
but which is still usually compact.
(For ${\mathcal{O}}_{\infty}$ (Example~\ref{P_3306_OIGauge})
and CuntzPimsner algebras (Example~\ref{EGaugeCP}),
our larger group~$G$ is not even locally compact.)
There is usually an obvious embedding of $S^1$ in~$G$
as a diagonal in some sense,
and the action usually called the gauge action is
the restriction to this subgroup.
If $\af \colon G \to \Aut (A)$
is an action of a compact group~$G$ on a \ca~$A$,
then the fixed point algebra $A^G$
and the \cp{} $C^* (G, A, \af)$
are, in suitable senses, not more complicated than~$A$.
Often they are in fact less complicated;
indeed, for some of the applications of gauge actions,
this is an important feature.
Since the main thrust of the later part of these notes
is situations in which the crossed products
are more complicated than the original algebra,
these examples are thus less relevant than some of the others.
However,
one can often get more relevant examples by considering
actions of other groups which factor through a gauge action
or an action of one of the larger groups in the examples
of this section.
As a very elementary example,
let $\af \colon S^1 \to \Aut (C (S^1))$
be the rotation action
(Example~\ref{E:Tr} with $G = S^1$).
This action is the dual action from
the identification of $C (S^1)$ as $C^* (\Z, \C)$
using the trivial action of $\Z$ on~$\C$.
It is also the gauge action of $S^1$
obtained from Example~\ref{EGaugeGraph}
using the realization of $C (S^1)$ as the \ca{}
of the graph with one vertex and one edge,
as in the discussion after \Def{D_3306_GraphAlg}.
The fixed point algebra is clearly~$\C$.
The crossed product is $K (L^2 (S^1))$.
(See the discussion at the beginning of Example~\ref{E_C_Trans}.)
However,
for $\te \in \R \setminus \Q$,
the irrational rotation action of~$\Z$
(see Example~\ref{E_Rot})
is the composition of $\af$ with the \hm{} $\Z \to S^1$
given by $n \mapsto \exp (2 \pi i n)$ for $n \in \Z$.
By Example~\ref{E:C:IrrRot},
the crossed product is the well known irrational rotation algebra
of Example~\ref{E_3303_RotAlg}.
This algebra is a simple infinite dimensional \ca{}
not of type~I.
The action of Example~\ref{E_RotGauge}
generalizes to an arbitrary higher dimensional
noncommutative torus.
\begin{exa}\label{E_3302_HDNC}
Let $d \in \N$ with $d \geq 2$.
Let $\te$ be a skew symmetric real $d \times d$ matrix.
Let $A_{\te}$ be the (higher dimensional) noncommutative torus
of \Ex{E_3303_NCTorus}.
By similar reasoning as in Example~\ref{E_RotGauge},
there is an action $\af \colon (S^1)^d \to \Aut (A_{\te})$
determined by
$\af_{(\zt_1, \zt_2, \ldots, \zt_d)} (u_j) = \zt_j u_j$
for $j = 1, 2, \ldots, d$.
Also, as in Example~\ref{E_RotGauge},
each individual element $(\zt_1, \zt_2, \ldots, \zt_d) \in (S^1)^d$
gives an automorphism of~$A_{\te}$,
and hence an action of $\Z$ on~$A_{\te}$.
As mentioned in Example~\ref{E_RotGauge},
using these automorphisms,
it is possible to realize an arbitrary
higher dimensional noncommutative torus
as an iterated crossed product by~$\Z$,
starting with a rotation algebra.
Finite subgroups of $(S^1)^d$
give actions of finite abelian groups on~$A_{\te}$.
Although we will not prove it in these notes,
their crossed products turn out to be strongly Morita equivalent
to other higher dimensional noncommutative tori.
\end{exa}
\begin{exa}\label{EToeplitz}
Recall that the {\emph{unilateral shift}} is the operator~$s$
on $l^2 (\Nz)$ which sends a sequence
$\xi = (\xi_0, \xi_1, \xi_2, \ldots)$ to the sequence
$s \xi = (0, \xi_0, \xi_1, \xi_2, \ldots)$.
One checks that
\[
s^* (\xi_0, \xi_1, \xi_2, \ldots)
= (\xi_1, \xi_2, \xi_3, \ldots).
\]
(This operator is called the backward shift.)
The C*subalgebra $T \subset L (H)$ generated by $s$ is called
the {\emph{Toeplitz algebra}}.
We recall that there is an exact sequence
\[
0 \longrightarrow K (l^2 (\Nz))
\longrightarrow T
\longrightarrow C (S^1)
\longrightarrow 0,
\]
in which the map $T \to C (S^1)$ sends $s$ to the function
$f (\zt) = \zt$ for $\zt \in S^1$.
The algebra $T$ can also be obtained as the universal \ca{}
generated by an isometry (which is~$s$).
(See Example 1.3(e)(6) of~\cite{Blk2}.)
There is a unique action $\gm \colon S^1 \to \Aut (T)$
such that $\af_{\zt} (s) = \zt s$ for all $\zt \in S^1$.
Uniqueness follows from the fact that $s$ generates~$T$.
Existence is immediate
from the description of~$T$ as a universal \ca,
but we can also give a direct proof using the description
as a subalgebra of $L (l^2 (\Nz))$.
For $\zt \in S^1$, define a unitary $u_{\zt} \in L (l^2 (\Nz))$
by
\[
u_{\zt} (\xi_0, \xi_1, \xi_2, \ldots)
= (\xi_0, \zt \xi_1, \zt^2 \xi_2, \ldots).
\]
Then one checks that $u_{\zt} s u_{\zt}^* = \zt s$.
It follows that $u_{\zt} T u_{\zt}^* \subset T$.
Moreover, since $\zt s$ generates $T$ just as well as $s$ does,
we get $u_{\zt} T u_{\zt}^* = T$.
Since
$u_{\zt_1} u_{\zt_2} = u_{\zt_1 \zt_2}$ for $\zt_1, \zt_2 \in S^1$,
it follows that the formula $\af_{\zt} (a) = u_{\zt} a u_{\zt}^*$
defines a \hm{} from $S^1$ to $\Aut (T)$.
Continuity of this action follows from Lemma~\ref{LCtGen}.
\end{exa}
\begin{exa}\label{EMoreCuntzGauge}
Let $n \in \N$.
Recall that the extended Cuntz algebra
$E_n$ is the universal unital \ca{} on generators
$s_1, s_2, \ldots, s_n$,
subject to the relations
stating that $s_j^* s_j = 1$ for $1 \leq j \leq n$
and $s_1 s_1^*, \, s_2 s_2^*, \, \ldots, \, s_n s_n^*$
are orthogonal \pj{s}.
(The difference from the relations in Example~\ref{E:CuntzGauge}
is that we no longer require that
$\sum_{j = 1}^n s_j s_j^* = 1$.
It follows that ${\mathcal{O}}_n$ is a quotient of~$E_n$.
The kernel is~$K$.)
The same formula as in Example~\ref{E:CuntzGauge}
defines an action of $U (M_n)$ on $E_n$.
That is,
if $u = ( u_{j, k} )_{j, k = 1}^n \in M_n$ is unitary, then
there is an automorphism $\bt_u$ of $E_n$
such that
\[
\bt_u (s_j) = \sum_{k = 1}^n u_{k, j} s_k
\]
for $j = 1, 2, \ldots, n$.
The restriction to~$S^1$,
realized as the scalar multiples of the identity
in $U (M_n)$,
is the gauge action on~$E_n$.
The case $n = 1$ makes sense.
The algebra is then the Toeplitz algebra of \Ex{EToeplitz},
and the action is the same action as in \Ex{EToeplitz}.
\end{exa}
\begin{exa}\label{E_3306_OIGauge}
Recall that the Cuntz algebra
${\mathcal{O}}_{\infty}$ is the universal unital \ca{} on generators
$s_1, s_2, \ldots$,
subject to the relations
stating that $s_j^* s_j = 1$ for $j \in \N$
and $s_1 s_1^*, \, s_2 s_2^*, \, \ldots$
are orthogonal \pj{s}.
(Like ${\mathcal{O}}_n$, it is in fact simple,
so any \ca{} generated by elements satisfying
these relations is isomorphic to~${\mathcal{O}}_{\infty}$.)
Now let $u \in L (l^2 (\N))$ be unitary.
Write $u$ in infinite matrix form,
as $u = ( u_{j, k} )_{j, k = 1}^n$.
Then
there is an automorphism $\af_u$ of ${\mathcal{O}}_{\infty}$
such that
\[
\af_u (s_j) = \sum_{k = 1}^{\I} u_{k, j} s_k
\]
for $j \in \N$.
By \Exr{P_3306_OIGauge} below,
$u \mapsto \af_u$ a \ct{} action of the unitary group $U (l^2 (\N))$
on~${\mathcal{O}}_{\infty}$.
Its restriction to~$S^1$,
realized as the scalar multiples of the identity
in $U (l^2 (\N))$,
is the gauge action on~${\mathcal{O}}_{\infty}$.
\end{exa}
The restriction of this action to~$S^1$
is used for a
counterexample in the discussion after \Thm{T_Ks}.
\begin{exr}\label{P_3306_OIGauge}
Verify that the formula given in Example~\ref{E_3306_OIGauge}
does in fact define a \ct{} action
of $U (l^2 (\N))$ on~${\mathcal{O}}_{\infty}$.
(Among other things, one must show that the series in the definition
of $\af_u (s_j)$ actually converges.)
\end{exr}
\begin{exa}\label{EGaugeGraph}
Recall from \Def{D_3306_GraphAlg}
that the \ca{} $C^* (E)$
of a directed graph $E = \big( E^{(0)}, E^{(1)} \big)$
is generated by \pj{s} $p_v$ for $v \in E^{(0)}$
and partial isometries $s_e$ for $e \in E^{(1)}$.
There is a gauge action $\af$ of $S^1$ on $C^* (E)$,
defined by $\af_{\zt} (p_v) = p_v$ for $v \in E^{(0)}$
and $\af_{\zt} (s_e) = \zt s_e$ for $e \in E^{(1)}$.
See Proposition~2.1 of~\cite{Rbn} for the case of a rowfinite graph.
The gauge action plays a fundamental role
in the theory of graph \ca{s},
as can be seen from~\cite{Rbn}.
This action generalizes
the gauge actions in
Example~\ref{E:CuntzGauge},
Example~\ref{EToeplitz},
and Example~\ref{EMoreCuntzGauge}.
The action extends to an action $\bt$
of $G = \prod_{e \in E^{(1)}} S^1$.
For $\zt = ( \zt_e)_{e \in E^{(1)}}$,
we take $\bt_{\zt} (p_v) = p_v$ for $v \in E^{(0)}$
and $\bt_{\zt} (s_e) = \zt_e s_e$ for $e \in E^{(1)}$.
\end{exa}
\begin{exa}\label{EGaugeHighRkGraph}
Higher rank graphs and their \ca{s}
are a generalization of graph \ca{s}.
They are described in Chapter~10 of~\cite{Rbn},
the \ca{} being defined under the assumption that
the graph is row finite and has no sources.
(Weaker conditions are also considered.)
We do not repeat the definitions of higher rank graphs
and their \ca{s} here, but we give some of the ideas.
A graph of rank~$k$ has edges of $k$ colors,
and there are specific conditions relating edges of different colors.
The C*algebra $C^* (E)$ of a row finite higher rank graph~$E$
with no sources
is generated by a family of \pj{s},
one for each vertex,
and a family of partial isometries,
one for each finite path in the graph.
A finite path in a rank~$k$ graph~$E$
has a degree $n = (n_1, n_2, \ldots, n_k) \in (\Nz)^k$,
in which $n_j$ is the number of edges in the path of color~$j$.
There is a gauge action $\af \colon (S^1)^k \to \Aut (C^* (E))$,
described after Corollary 10.13~\cite{Rbn}.
For $\zt = (\zt_1, \zt_2, \ldots, \zt_k) \in (S^1)^k$,
the automorphism $\af_{\zt}$ fixes the \pj{s}
corresponding to the vertices,
and multiplies the partial isometry corresponding
to a finite path of degree~$n$
by $\zt_1^{n_1} \zt_2^{n_2} \cdots \zt_k^{n_k}$.
This action generalizes the gauge action
of $S^1$ on a graph \ca{} in Example~\ref{EGaugeGraph}.
It plays a role in the theory of \ca{s}
of higher rank graphs similar to the role of a gauge action
of $S^1$ in the theory of ordinary directed graphs.
\end{exa}
\begin{exa}\label{EGaugeCP}
The algebras now known as CuntzPimsner algebras were
introduced in~\cite{Pms}.
Also see Chapter~8 of~\cite{Rbn}.
We don't give details here,
but we give a brief outline.
One starts with a \ca~$A$
and a Hilbert bimodule $E$ over~$A$,
that is,
a right Hilbert module $E$ over~$A$ with a \hm{}
from $A$ to the algebra $L (E)$
of adjointable right $A$module \hm{s} of~$E$.
(The perhaps more descriptive term
``correspondence'' is used instead
of ``Hilbert bimodule'' in~\cite{Rbn}.
See the discussion after Example~8.4 of~\cite{Rbn}.)
One constructs a Toeplitz algebra~${\mathcal{T}}_E$,
which is described in Definition~1.1 of~\cite{Pms}
and after Proposition~8.8 of~\cite{Rbn}.
It is generated by creation and annihilation operators
on the Fock space made from~$E$.
There is further a CuntzPimsner algebra~${\mathcal{O}}_E$,
given in Definition~1.1 of~\cite{Pms}
and after Proposition~8.11 of~\cite{Rbn}.
It is a suitable quotient of~${\mathcal{T}}_E$.
The gauge action $\ld \colon S^1 \to \Aut ( {\mathcal{T}}_E )$
is described on page 198 of~\cite{Pms}.
(The algebra~${\mathcal{P}}_E$ which appears there
is described at the beginning of Section~3~\cite{Pms}.)
The associated $\Z$grading is given in Proposition~8.9 of~\cite{Rbn}.
The automorphism $\ld_{\zt}$
multiplies the creation operator coming from
an element of $E^{\otimes n}$ by~$\zt^n$.
This action descends to a gauge action of $S^1$ on~${\mathcal{O}}_E$.
As described in the Examples starting on page 192 of~\cite{Pms},
CuntzPimsner algebras generalize Cuntz algebras,
CuntzKrieger algebras,
crossed products by actions of~$\Z$,
and crossed products by partial actions of~$\Z$.
The corresponding gauge actions of~$S^1$ turn out to be the
usual gauge actions
on the Cuntz algebras
(Example~\ref{E:CuntzGauge})
and CuntzKrieger algebras
and the dual actions on the crossed products
(Example~\ref{EZDual} for an action of~$\Z$).
Graph \ca{s} (\Def{D_3306_GraphAlg}) are special cases
of CuntzPimsner algebras (Example 8.13 of~\cite{Rbn}),
and this example generalizes Example~\ref{EGaugeGraph}.
\end{exa}
\begin{exa}\label{EGaugePlusCP}
In the situation of \Ex{EGaugeCP},
as with various other examples of gauge actions,
there is in fact an action of a much bigger group.
Again let $A$ be a \ca,
let $E$ be a Hilbert bimodule (or correspondence)
over~$A$,
and let ${\mathcal{T}}_E$ and~${\mathcal{O}}_E$
be the associated Toeplitz and CuntzPimsner algebras.
As described in Remark 4.10(2) of~\cite{Pms},
the whole automorphism group $\Aut (E)$ of~$E$
acts on ${\mathcal{T}}_E$ and~${\mathcal{O}}_E$.
In fact, consider the group $\Aut (A, E)$
of automorphisms of the pair $(A, E)$,
that is, pairs $(\af, \sm)$ consisting of an automorphism
$\af \in \Aut (A)$ and an automorphism of $E$ as a Banach space
which is compatible with $\af$ in a suitable sense.
Then $\Aut (A, E)$ acts on ${\mathcal{T}}_E$ and~${\mathcal{O}}_E$.
\end{exa}
\begin{exr}\label{Ex_3311_GCP_On}
In \Ex{EGaugePlusCP},
take $A = \C$ and $E = \C^n$.
Then ${\mathcal{T}}_E$ is the extended Cuntz algebra~$E_n$
of \Ex{EMoreCuntzGauge}
and ${\mathcal{O}}_E$ is the Cuntz algebra~${\mathcal{O}}_n$
as in Example~\ref{E:CuntzGauge}.
Prove that the action of $\Aut (E)$
on ${\mathcal{O}}_n$ can be identified
with the action of $U (M_n)$ on ${\mathcal{O}}_n$
given in \Ex{E:CuntzGauge},
and that action of $\Aut (E)$
on $E_n$ can be identified
with the action of $U (M_n)$ on $E_n$
given in \Ex{EMoreCuntzGauge}.
\end{exr}
Examples~\ref{EGaugeFullFn},
\ref{EGaugeRedFn},
and~\ref{EFundGpSurf}
use full and reduced group \ca{s}
of discrete groups,
which are formally introduced in Section~\ref{Sec_GpCSt},
and Exercise~\ref{ExFnCrPrd}
and part of Example~\ref{EFundGpSurf}
use full crossed products (Section~\ref{Sec:CP})
and reduced crossed products (Section~\ref{Sec:RedCP}).
\begin{exa}\label{EGaugeFullFn}
Let $F_n$ be the free group on $n$~generators.
Then $C^* (F_n)$ is the universal \ca{} generated by $n$ unitaries
$u_1, u_2, \ldots, u_n$,
with no other relations.
(Use the description of the group \ca{}
in Exercise~\ref{Ex_3401_CstGRel}.)
It follows that for $\zt = (\zt_1, \zt_2, \ldots, \zt_n) \in (S^1)^n$,
there is a \hm{} $\af_{\zt} \colon C^* (F_n) \to C^* (F_n)$
such that $\af_{\zt} (u_k) = \zt_k u_k$ for $k = 1, 2, \ldots, n$.
Lemma~\ref{LCtGen} implies that these \hm{s} define a \ct{} action
$\af \colon (S^1)^n \to \Aut (C^* (F_n))$.
An analogous procedure works for the full \ca{} of the free group
on countably many generators,
or even on an arbitrary set of generators.
\end{exa}
\begin{exa}\label{EGaugeRedFn}
The action $\af$ of $(S^1)^n$ on $C^* (F_n)$
in Example~\ref{EGaugeFullFn}
descends to an action of $(S^1)^n$ on the reduced
group \ca{} $C^*_{\mathrm{r}} (F_n)$.
(See Definition~\ref{D_3317_RedGpCStar}.)
That is, letting $\pi \colon C^* (F_n) \to C^*_{\mathrm{r}} (F_n)$
be the quotient map,
there is an action
$\bt \colon (S^1)^n \to \Aut (C^*_{\mathrm{r}} (F_n))$
such that for every $\zt \in (S^1)^n$,
we have $\pi \circ \af_{\zt} = \bt_{\zt} \circ \pi$.
We prove this by exhibiting unitaries in $L (l^2 (F_n))$
which implement the action~$\bt$.
For $g \in F_n$, let $\dt_g$ denote the corresponding
element of the standard Hilbert basis for $l^2 (F_n)$.
Let $g_1, g_2, \ldots, g_n$ denote the
standard generators of~$F_n$.
Then the unitaries $u_1, u_2, \ldots, u_n$
of Example~\ref{EGaugeFullFn} are the
standard unitaries $u_{g_1}, u_{g_2}, \ldots, u_{g_n}$
of the group \ca.
% Denote the standard generators of the free abelian group $\Z^n$
% on $n$~generators by $h_1, h_2, \ldots, h_n$.
% Let $\gm \colon F_n \to \Z^n$
% be the \hm{} with sends $g_k$ to~$h_k$ for $k = 1, 2, \ldots, n$.
Let $\gm_k \colon F_n \to \Z$ be the \hm{} determined by
$\gm_k (g_k) = 1$ and $\gm_k (g_j) = 0$ for $j \neq k$.
For $\zt = (\zt_1, \zt_2, \ldots, \zt_n) \in (S^1)^n$,
define a unitary
$v_{\zt} \in L (l^2 (F_n))$ by
\[
v_{\zt} \dt_g
= \zt_1^{\gm_1 (g)} \zt_2^{\gm_2 (g)} \cdots \zt_n^{\gm_n (g)} \dt_g
\]
for $g \in F_n$.
Then one can check that
\[
v_{\zt} \pi (u_k) v_{\zt}^* = \pi ( \af_{\zt} (u_k))
\]
for $k = 1, 2, \ldots, n$ and all $\zt \in (S^1)^n$.
This proves the existence of~$\bt$.
Continuity follows easily from continuity of~$\af$.
\end{exa}
\begin{exr}\label{ExFnCrPrd}
Show that
the constructions in Examples \ref{EGaugeFullFn} and~\ref{EGaugeRedFn}
work not just for the full and reduced \ca{s} of~$F_n$,
but for full and reduced crossed products by~$F_n$.
For the full crossed product,
use the description of the crossed product
in Theorem~\ref{L:UnivCP}.
For the reduced crossed product,
see Definition~\ref{D:RedCP}.
\end{exr}
There are other groups for which
there is a construction similar to
that of
Example~\ref{EGaugeFullFn}, Example~\ref{EGaugeRedFn},
and Exercise~\ref{ExFnCrPrd}.
Here is one such example.
\begin{exa}\label{EFundGpSurf}
Recall (see the discussion before Corollary 1.27 of~\cite{Htch})
that the fundamental group $\Gm_n$ of a compact
orientable surface of genus~$n$ is generated by $2 n$
elements
\[
g_1, g_2, \ldots, g_n, h_1, h_2, \ldots, h_n
\]
subject to the single relation (with $[g, h]$ denoting the group
commutator $[g, h] = g h g^{1} h^{1}$)
\[
[g_1, h_1] [g_2, h_2] \cdots [g_n, h_n] = 1.
\]
It follows that $C^* (\Gm_n)$ is the universal \ca{} generated
by unitaries
\[
u_1, u_2, \ldots, u_n, v_1, v_2, \ldots, v_n
\]
(with, following Notation~\ref{N_3317_GinAlg} below,
$u_j = u_{g_j}$ and $v_j = u_{h_j}$
for $j = 1, 2, \ldots, n$),
subject to the single additional relation
\[
(u_1 v_1 u_1^* v_1^*) (u_2 v_2 u_2^* v_2^*)
\cdots (u_n v_n u_n^* v_n^*) = 1.
\]
If
\[
\ld = (\ld_1, \ld_2, \ldots, \ld_n) \in (S^1)^n
\andeqn
\zt = ( \zt_1, \zt_2, \ldots, \zt_n) \in (S^1)^n,
\]
then the elements
\[
\ld_1 u_1, \, \ld_2 u_2, \, \ldots, \, \ld_n u_n, \,
\zt_1 v_1, \, \zt_2 v_2, \, \ldots, \, \zt_n v_n \in C^* (\Gm_n)
\]
are also unitaries satisfying the same additional relation.
Therefore there is a unique
endomorphism $\af_{\ld, \zt} \colon C^* (\Gm_n) \to C^* (\Gm_n)$
such that
\[
\af_{\ld, \zt} (u_1) = \ld_1 u_1, \qquad
\af_{\ld, \zt} (u_2) = \ld_2 u_2, \qquad \ldots, \qquad
\af_{\ld, \zt} (u_n) = \ld_n u_n,
\]
and
\[
\af_{\ld, \zt} (v_1) = \zt_1 v_1, \qquad
\af_{\ld, \zt} (v_2) = \zt_2 v_2, \qquad \ldots, \qquad
\af_{\ld, \zt} (v_n) = \zt_n v_n.
\]
Lemma~\ref{LCtGen} implies that these endomorphisms
actually form a \ct{} action
$\af \colon (S^1)^{2 n} \to \Aut (C^* (\Gm_n))$.
An argument similar to that in Example~\ref{EGaugeRedFn}
shows that the action~$\af$ descends to an action
$\bt \colon (S^1)^{2 n} \to \Aut (C^*_{\mathrm{r}} (\Gm_n))$.
For $g \in \Gm_n$, let $\dt_g$ denote the corresponding
element of the standard Hilbert basis for $l^2 (\Gm_n)$.
For $k = 1, 2, \ldots, n$,
there is a unique group \hm{}
$\gm_k \colon \Gm_n \to \Z$ such that
$\gm_k (g_k) = 1$, $\gm_k (g_j) = 0$ for $j \neq k$,
and $\gm_k (h_j) = 0$ for $j = 1, 2, \ldots, n$,
and there is a unique group \hm{}
$\rh_k \colon \Gm_n \to \Z$ such that
$\rh_k (g_j) = 0$ for $j = 1, 2, \ldots, n$,
$\rh_k (h_k) = 1$,
and $\rh_k (h_j) = 0$ for $j \neq k$.
For
\[
\ld = (\ld_1, \ld_2, \ldots, \ld_n) \in (S^1)^n
\andeqn
\zt = ( \zt_1, \zt_2, \ldots, \zt_n) \in (S^1)^n,
\]
define a unitary
$v_{\ld, \zt} \in L (l^2 (F_n))$ by
\[
v_{\ld, \zt} \dt_g
= \ld_1^{\gm_1 (g)} \ld_2^{\gm_2 (g)} \cdots \ld_n^{\gm_n (g)}
\zt_1^{\rh_1 (g)} \zt_2^{\rh_2 (g)} \cdots \zt_n^{\rh_n (g)} \dt_g
\]
for $g \in \Gm_n$.
Then one can check that
\[
v_{\ld, \zt} \pi (u_k) v_{\ld, \zt}^* = \pi ( \af_{\ld, \zt} (u_k))
\andeqn
v_{\ld, \zt} \pi (v_k) v_{\ld, \zt}^* = \pi ( \af_{\ld, \zt} (v_k))
\]
for $k = 1, 2, \ldots, n$ and all $\ld, \zt \in (S^1)^n$.
This proves the existence of~$\bt$.
One also checks,
in the same way as for Exercise~\ref{ExFnCrPrd},
that the same thing works for full and reduced
crossed products by~$\Gm_n$.
We omit the details.
\end{exa}
\begin{exa}\label{EUnc}
The \ca{} $U^{\mathrm{nc}}_n$ is defined to be the
universal \uca{} generated by elements $u_{j, k}$,
for $1 \leq j, k \leq n$,
subject to the relation that the matrix
\[
u = \left( \begin{matrix}
u_{1, 1} & u_{1, 2} & \cdots & u_{1, n} \\
u_{2, 1} & u_{2, 2} & \cdots & u_{2, n} \\
\vdots & \vdots & \ddots & \vdots \\
u_{n, 1} & u_{n, 2} & \cdots & u_{n, n}
\end{matrix} \right)
\in M_n ( U^{\mathrm{nc}}_n )
\]
is unitary.
This amounts to $2 n^2$ relations on the generators $u_{j, k}$,
namely
\[
\sum_{k = 1}^n u_{j, k} u_{l, k}^* = \dt_{j, l}
\andeqn
\sum_{k = 1}^n u_{k, j}^* u_{k, l} = \dt_{j, l}
\]
for $1 \leq j, k \leq n$.
(This \ca{} was introduced in (2b) in Section~3 of~\cite{Bwn81}.)
There is an action~$\af$
of the unitary group $U (M_n)$ on $U^{\mathrm{nc}}_n$,
defined as follows.
Let
\[
g = \left( \begin{matrix}
g_{1, 1} & g_{1, 2} & \cdots & g_{1, n} \\
g_{2, 1} & g_{2, 2} & \cdots & g_{2, n} \\
\vdots & \vdots & \ddots & \vdots \\
g_{n, 1} & g_{n, 2} & \cdots & g_{n, n}
\end{matrix} \right)
\in U (M_n).
\]
Regard $g$ as an element of $M_n ( U^{\mathrm{nc}}_n )$
via the unital inclusion of $\C$ in~$U^{\mathrm{nc}}_n$.
Then the product $g u$ is defined and is unitary.
Thus its entries
\[
(g u)_{j, l} = \sum_{k = 1}^n g_{j, k} u_{k, l}
\]
form an $n \times n$ unitary matrix.
So there exists a unique unital \hm{}
$\af_g \colon U^{\mathrm{nc}}_n \to U^{\mathrm{nc}}_n$
such that $\af_g (u_{j, l}) = (g u)_{j, l}$
for $j, l = 1, 2, \ldots, n$.
The proof that this gives a \ct{}
action is requested in Exercise~\ref{Ex_6X13_Uncn}.
Any unitary representation of a group~$G$ in $M_n$
therefore also gives an action of $G$ on $U^{\mathrm{nc}}_n$.
Here is a special case,
coming from the representation
\[
\zt \mapsto \left( \begin{matrix}
1 & 0 \\
0 & \zt
\end{matrix} \right)
\]
of $S^1$ on~$\C^2$.
For $\zt \in S^1$, we take $\sm_{\zt} \in \Aut (U^{\mathrm{nc}}_n)$
to be the automorphism determined by
\[
\sm_{\zt} (u_{1, 1}) = u_{1, 1},
\quad
\sm_{\zt} (u_{1, 2}) = u_{1, 2},
\quad
\sm_{\zt} (u_{2, 1}) = \zt u_{2, 1},
\quad {\mbox{and}} \quad
\sm_{\zt} (u_{2, 2}) = \zt u_{2, 2}.
\]
There is a second action~$\bt$ of $U (M_n)$ on $U^{\mathrm{nc}}_n$,
determined by $\bt_g (u_{j, l}) = (u g^*)_{j, l}$
for $g \in U (M_n)$ and $j, l = 1, 2, \ldots, n$.
These actions are different,
as can be checked with $n = 2$ and
% \[
% g = \frac{1}{\sqrt{2}} \left( \begin{matrix}
% 1 & 1 \\
%  1 & 1
% \end{matrix} \right).
% \]
\[
u = \left( \begin{matrix}
1 & 0 \\
0 & \zt
\end{matrix} \right)
\]
(as above):
one now gets
\[
u_{1, 1} \mapsto u_{1, 1},
\quad
u_{1, 2} \mapsto u_{1, 2},
\quad
u_{2, 1} \mapsto \zt u_{2, 1},
\quad {\mbox{and}} \quad
u_{2, 2} \mapsto \zt u_{2, 2}.
\]
A third action comes from letting $U (M_n)$ act on $M_n$
by conjugation.
The same matrix $u$ as above now gives the automorphism
determined by
\[
u_{1, 1} \mapsto u_{1, 1},
\quad
u_{1, 2} \mapsto \zt^{1} u_{1, 2},
\quad
u_{2, 1} \mapsto \zt u_{2, 1},
\quad {\mbox{and}} \quad
u_{2, 2} \mapsto u_{2, 2}.
\]
\end{exa}
% Conjugation by $\diag (1, \zt)$.
% \[
% \sm_{\zt} (u_{1, 1}) = u_{1, 1},
% \qquad
% \sm_{\zt} (u_{1, 2}) = \zt^{1} u_{1, 2},
% \qquad
% \sm_{\zt} (u_{2, 1}) = \zt u_{2, 1},
% \andeqn
% \sm_{\zt} (u_{2, 2}) = u_{2, 2}.
% \]
% This gives the conjugation action, not the one claimed.
% In Proposition~2.2 of~\cite{McK1},
% it is observed that $U^{\mathrm{nc}}_n$ can be identified
% with the relative commutant of $M_n$ in the
% amalgamated free product $M_n \star_{\C} C (S^1)$,
% the amalgamation identifying the subalgebras $\C \cdot 1$
% in both factors.
% The isomorphism
% \[
% \ph \colon U^{\mathrm{nc}}_n \to M_n' \cap (M_n \star_{\C} C (S^1))
% \]
% is defined as follows.
% We let $e_{j, k} \in M_n$ be the standard matrix units,
% and we let $z \in C (S^1)$
% be the function $z (\zt) = \zt$ for $\zt \in S^1$.
% Then
% \[
% \ph (u_{j, k}) = \sum_{l = 1}^n e_{l, j} z e_{k, l}.
% \]
% for $j, k = 1, 2, \ldots, n$.
% The action $\af$ is then presumably % 999
% the free product of the obvious inner action of $U (M_n)$
% on $M_n$ and the trivial action on $C (S^1)$.
% This action makes sense on
% $M_n' \cap (M_n \star_{\C} C (S^1))$ because it sends
% $M_n$ to itself.
% (This has not been checked.) % 999
\begin{exr}\label{Ex_6X13_Uncn}
Prove that the definition of
the action of $U (M_n)$ on $U^{\mathrm{nc}}_n$
given in \Ex{EUnc}
actually gives a \ct{} action.
\end{exr}
\begin{exa}\label{EFcnlCalcUnivUnitary}
Let $U^{\mathrm{nc}}_n$, its generators
$u_{j, k}$ for $1 \leq j, k \leq n$,
and the unitary matrix
\[
u = \left( \begin{matrix}
u_{1, 1} & u_{1, 2} & \cdots & u_{1, n} \\
u_{2, 1} & u_{2, 2} & \cdots & u_{2, n} \\
\vdots & \vdots & \ddots & \vdots \\
u_{n, 1} & u_{n, 2} & \cdots & u_{n, n}
\end{matrix} \right)
\in M_n ( U^{\mathrm{nc}}_n ),
\]
% and the isomorphism
% \[
% \ph \colon U^{\mathrm{nc}}_n \to M_n' \cap (M_n \star_{\C} C (S^1))
% \]
be as in Example~\ref{EUnc}.
Let $h \colon S^1 \to S^1$ be a \ct{} map.
Then functional calculus
gives an element $h (u) \in M_n ( U^{\mathrm{nc}}_n )$,
so that there is an endomorphism
$\af_h \colon U^{\mathrm{nc}}_n \to U^{\mathrm{nc}}_n$
such that $\af_h (u_{j, k}) = h (u)_{j, k}$
for $1 \leq j, k \leq n$.
% Presumably,
% on $M_n' \cap (M_n \star_{\C} C (S^1))$ it is the free product
% of $\id_{M_n}$ and the endomorphism $f \mapsto f \circ h^{1}$
% of $C (S^1)$.
This endomorphism is uniquely determined by the relation
$(\id_{M_n} \otimes \af_h) (u) = h (u)$.
Suppose $h_1, h_2 \colon S^1 \to S^1$
are \ct.
We prove that $\af_{h_1 \circ h_2} = \af_{h_2} \circ \af_{h_1}$,
that is,
that $h \mapsto \af_h$
is an anti\hm{}
from the semigroup of \ct{} maps $S^1 \to S^1$
to the semigroup of endomorphisms of $U^{\mathrm{nc}}_n$.
To prove the claim,
first observe that
for any \ca{s} $A$ and $B$,
any unital \hm{} $\ph \colon A \to B$,
and any unitary $v \in A$,
we have $\ph (h_2 (v)) = h_2 (\ph (v))$.
Apply this fact with $\ph = \id_{M_n} \otimes \af_{h_1}$
and $v = u$ at the third step
in the following calculation:
% we get
\begin{align*}
(\id_{M_n} \otimes \af_{h_1 \circ h_2}) (u)
& = h_1 (h_2 (u))
= h_1 \big( (\id_{M_n} \otimes \af_{h_2}) (u) \big)
\\
& = (\id_{M_n} \otimes \af_{h_2}) (h_1 (u))
= \big[ (\id_{M_n} \otimes \af_{h_2})
\circ (\id_{M_n} \otimes \af_{h_1}) \big] (u).
\end{align*}
The claim follows.
The claim implies that,
in particular,
$h \mapsto \af_{h^{1}}$ is a well defined action of
the group of \hme{s} of $S^1$ on $U^{\mathrm{nc}}_n$.
Some special cases:
take the rotations by all $\zt \in S^1$ to get an action of~$S^1$;
take a rotation by a fixed $\zt \in S^1$ to get
an action of $\Z$ which is
a noncommutative analog of a rational or irrational rotation;
take a rotation by $e^{2 \pi i l / m}$ to get an action of $\Z / m \Z$.
In general,
if $h \colon S^1 \to S^1$ is any fixed \hme,
then $n \mapsto \af_h^n$ is an action of $\Z$ on $U^{\mathrm{nc}}_n$.
\end{exa}
There is a reduced version $U^{\mathrm{nc}}_{n, {\mathrm{red}} }$
of the algebra $U^{\mathrm{nc}}_{n}$
used in Examples \ref{EUnc} and~\ref{EFcnlCalcUnivUnitary},
and presumably there are reduced versions of some
of the actions above.
The algebra is defined
in the discussion after Proposition~3.1 of~\cite{McK1}.
To describe it,
start with the fact (Proposition~2.2 of~\cite{McK1})
that $U^{\mathrm{nc}}_n$
can be identified
with the relative commutant of $M_n$ in the
amalgamated free product $M_n \star_{\C} C (S^1)$,
the amalgamation identifying the subalgebras $\C \cdot 1$
in both factors.
The isomorphism
\[
\ph \colon U^{\mathrm{nc}}_n \to M_n' \cap (M_n \star_{\C} C (S^1))
\]
is defined as follows.
We let $e_{j, k} \in M_n$ be the standard matrix units,
and we let $z \in C (S^1)$
be the function $z (\zt) = \zt$ for $\zt \in S^1$.
Then
\[
\ph (u_{j, k}) = \sum_{l = 1}^n e_{l, j} z e_{k, l}.
\]
for $j, k = 1, 2, \ldots, n$.
Now take $U^{\mathrm{nc}}_{n, {\mathrm{red}} }$
to be the relative commutant of $M_n$ in the
reduced amalgamated free product
$M_n \star_{\C, {\mathrm{r}}} C (S^1)$
with respect to the unique tracial state on $M_n$ and
Lebesgue measure on~$S^1$,
the amalgamation identifying the subalgebras $\C \cdot 1$
in both factors as above.
It is thus a quotient of the algebra
$U^{\mathrm{nc}}_n$.
The actions of Example~\ref{EUnc}
presumably
descend to actions on $U^{\mathrm{nc}}_{n, {\mathrm{red}} }$.
Similarly,
the automorphism $\af_h$ of Example~\ref{EFcnlCalcUnivUnitary}
presumably
descends to an automorphism of $U^{\mathrm{nc}}_{n, {\mathrm{red}} }$
provided $h$ preserves Lebesgue measure on~$S^1$.
In particular, the rotation action of~$S^1$,
the rational and irrational rotations,
and the rotation actions of $\Z / m \Z$ presumably all descend to
actions on $U^{\mathrm{nc}}_{n, {\mathrm{red}} }$.
As far as we know,
nobody has checked that any of these presumed actions
really exists.
% \end{exa}
\begin{exa}\label{E_3303_Uncmn}
Example~\ref{EUnc} can be generalized as follows.
Let $m, n \in \N$.
The \ca{} $U^{\mathrm{nc}}_{m, n}$,
introduced in Section~2 of~\cite{McK2},
is defined to be the
universal \uca{} generated by elements $u_{j, k}$,
for $1 \leq j \leq m$ and $1 \leq k \leq n$,
subject to the relation that the $m \times n$ matrix
\[
u = \left( \begin{matrix}
u_{1, 1} & u_{1, 2} & \cdots & u_{1, n} \\
u_{2, 1} & u_{2, 2} & \cdots & u_{2, n} \\
\vdots & \vdots & \ddots & \vdots \\
u_{m, 1} & u_{m, 2} & \cdots & u_{m, n}
\end{matrix} \right)
\]
is unitary,
that is, $u^* u$ is the identity matrix in $M_n (U^{\mathrm{nc}}_{m, n})$
and $u u^*$ is the identity matrix in $M_m (U^{\mathrm{nc}}_{m, n})$.
The constructions of Example~\ref{EUnc} now
give actions of $U (M_m)$ and $U (M_n)$
on $U^{\mathrm{nc}}_{m, n}$.
If $m = 1$,
then the relations are exactly those for the Cuntz algebra
${\mathcal{O}}_n$,
and the action of $U (M_n)$
generalizes the action of $U (M_n)$ on ${\mathcal{O}}_n$
of Example~\ref{E:CuntzGauge}.
\end{exa}
A free product description of $U^{\mathrm{nc}}_{m, n}$ is given
in Section~2 of~\cite{McK2},
but it is of a different form from the free product
description after Example~\ref{EFcnlCalcUnivUnitary}
for the case $m = n$.
As far as we know,
no reduced version of $U^{\mathrm{nc}}_{m, n}$
(analogous to the algebra $U^{\mathrm{nc}}_{n, {\mathrm{red}} }$
discussed after Example~\ref{EFcnlCalcUnivUnitary})
% in Example~\ref{EUncRed})
has been proposed.
One does not expect automorphisms or actions like those
of Example~\ref{EFcnlCalcUnivUnitary},
because the matrix $u$ here,
not being square, isn't an element of a \ca.
\part{Group C*algebras and Crossed Products}\label{Part_BasicCP}
\section{C*Algebras of Discrete Groups}\label{Sec_GpCSt}
% Joey Iverson has notes on typos in this section and the next. 999
% May need to add reminders of multiple meanings of $u_g$
% several places. % 999
\indent
The main focus of these notes is the structure of
certain kinds of crossed products.
The \ca{} of a group is a special case of a crossed productit
comes from the trivial action of the group on $\C$but not
one of the ones we are mainly concerned with.
We devote this section and Section~\ref{Sec_LCGpCSt}
to group \ca{s} anyway,
in order to provide an introduction to \cp{s}
in a simpler case,
and because understanding the group \ca{} is helpful,
at least at a heuristic level, for
understanding more general crossed products.
Section~\ref{Sec:CP} treats crossed product \ca{s}
and Section~\ref{Sec:RedCP} treats
reduced crossed product \ca{s}.
In Section~\ref{Sec_Comp} we give a number of explicit computations
of crossed product \ca{s}.
The brief Section~\ref{Sec_CStrFn} contains a
proof that the reduced \ca{} of a
finitely generated nonabelian free group is simple.
We recall that, by convention,
all topological groups will be assumed to be Hausdorff.
We start with discrete groups
(groups with the discrete topology),
because this case avoids many technicalities.
Moreover,
in the later part of these notes,
almost all groups will be discrete.
(The term ``discrete'' could be considered redundant.
We routinely include it anyway for clarity.)
\ca{s} of locally compact groups will
be discussed in Section~\ref{Sec_LCGpCSt},
but in less detail and without full proofs.
However,
some of the elementary definitions in this section,
and some theorems
(in particular, the summary of duality and the Fourier transform
for locally compact abelian groups),
are given for general locally compact groups,
to avoid later repetition.
The notation we use is chosen to avoid conflicts
with later notation for crossed products
and other \ca{s}.
The letters most commonly used for unitary representations
of locally compact groups are $\pi$ and $\sm$
(which we use for representations of \ca{s})
and $u$
(we use $u_g$ for the image of the group element $g$
in the group ring and various \ca{s}
made from it).
Our notation for group rings is designed
to be compatible with commonly used notation
for crossed product \ca{s},
and is not the same as the notation usually used in algebra.
The common notation $\ld$ for the left regular representation
of a locally compact group also conflicts
with notation we use elsewhere.
The construction of $C^* (G)$ is designed so that the
representations of $C^* (G)$ are the ``same''
as the unitary \rpn{s} of~$G$.
\begin{ntn}\label{N_3326_UH}
Let $H$ be a Hilbert space
We denote by $U (H)$ the unitary group of~$H$.
\end{ntn}
We repeat for reference the standard definition
of a unitary representation.
Our main reason is to emphasize the topology in
which continuity is required.
\begin{dfn}\label{D_2236_UnitRpn}
Let $G$ be a topological group
and let $H$ be a nonzero Hilbert space.
A {\emph{unitary representation}}
of~$G$ on~$H$
is a group \hm{} $w \colon G \to U (H)$
which is \ct{} in the strong operator topology on $L (H)$,
that is,
such that for every $\xi \in H$,
the function $g \mapsto w (g) \xi$
is a \cfn{} from $G$ to $H$ with the norm topology on~$H$.
\end{dfn}
Norm continuity of representations is much too strong
a condition to be useful.
For example,
it follows from Exercise~\ref{Ex_3318_Norm2}
that the left regular representation
(\Def{D_3407_DRegRep} below)
of a locally compact group
which is not discrete is never norm \ct.
Of course, if $G$ is discrete,
the main subject of this section,
there is no difference.
Since representations of groups are not the main subject of
these notes,
we won't give a list of examples.
But we want to mention at least two:
the one dimensional trivial representation,
which sends every group element to the identity operator
on a one dimensional Hilbert space,
and the left regular representation.
\begin{dfn}\label{D_3407_DRegRep}
Let $G$ be a discrete group.
The {\emph{left regular representation}}
of $G$ is the \rpn{} $v \colon G \to U (l^2 (G))$
given by
$(v (g) \xi) (h) = \xi (g^{1} h)$
for $g, h \in G$ and $\xi \in l^2 (G)$.
\end{dfn}
\begin{exr}\label{Ex_3407_DRgRpn}
Prove that the formula of \Def{D_3407_DRegRep}
gives a unitary \rpn{}
$v \colon G \to U (l^2 (G))$.
\end{exr}
The main point of this exercise is to see
why $g^{1}$ appears in the formula.
Here is an alternative description
of the left regular representation.
For $h \in G$,
let $\dt_h \in l^2 (G)$ be the standard basis vector
corresponding to~$h$.
Then $v$ is determined by $v (g) \dt_h = \dt_{g h}$
for $g, h \in G$.
There is also a right regular \rpn{}
$w \colon G \to U (l^2 (G))$,
given by $(w (g) \xi) (h) = \xi (h g)$
for $g, h \in G$ and $\xi \in l^2 (G)$.
It is determined by $w (g) \dt_h = \dt_{h g^{1}}$
for $g, h \in G$.
\begin{rmk}\label{R_3401_RpnFacts}
The elementary theory of unitary representations of topological groups
is very much like the elementary theory of representations of \ca{s}.
Unitary equivalence,
invariant subspaces,
irreducible representations,
subrepresentations,
direct sums (not necessarily finite) of representations,
and cyclic vectors and cyclic representations,
are all defined just as for representations of \ca{s}.
The same proofs as for \ca{s}
show that the orthogonal complement of an invariant subspace
is again invariant,
so that every subrepresentation is a direct summand,
and that every representation is a direct sum of cyclic representations.
All of this can be found in Section~3.1 of~\cite{Fld}.
\end{rmk}
\begin{exr}\label{Ex_3407_GpRpPf}
Supply the definitions
and prove the statements in Remark~\ref{R_3401_RpnFacts}.
\end{exr}
\begin{rmk}\label{R_4201_TensorRpn}
There is one significant construction
for unitary representations of topological groups
which does not make sense for representations of general \ca{s},
namely the tensor product of two representations.
Let $G$ be a topological group,
let $H_1$ and $H_2$ be Hilbert spaces,
and let $w_1 \colon G \to U (H_1)$ and $w_2 \colon G \to U (H_2)$
be unitary representations.
Then there is a unitary representation
\[
w_1 \otimes w_2 \colon G \to U (H_1 \otimes H_2)
\]
(using the Hilbert space tensor product)
such that $(w_1 \otimes w_2) (g) = w_1 (g) \otimes w_2 (g)$
for all $g \in G$.
The construction can be found in Section~7.3 of~\cite{Fld},
which starts with the construction of the Hilbert space tensor product
of Hilbert spaces.
Our $w_1 \otimes w_2$ is what is called the inner tensor product
before Theorem 7.20 of~\cite{Fld}.
(Section~7.3 of~\cite{Fld} is mainly about the tensor product
of representations of two groups as a representation of the product
of the groups,
a construction for which there is an analog
for representations of general \ca{s}.)
\end{rmk}
We make only a little use of tensor products of representations,
because there is no analog in the context of crossed products.
However, some parts of the representation theory of compact groups
are primarily concerned with how a tensor product of two
irreducible representations decomposes as a direct sum
of other irreducible representations.
One consequence of the properties of the \ca{}
of a locally compact group is that the elementary representation theory
of locally compact groups
is a special case
of the elementary representation theory of \ca{s}.
We start with a purely algebraic construction,
the group ring.
\begin{dfn}\label{D_3402_StarAlg}
A {\emph{*algebra}}
over the complex numbers is a complex algebra~$A$
with an adjoint operation $a \mapsto a^*$
satisfying the following properties:
\begin{enumerate}
\item\label{D_3402_StarAlg_Add}
$(a + b)^* = a^* + b^*$ for all $a, b \in A$.
\item\label{D_3402_StarAlg_Scal}
$(\ld a)^* = {\ov{\ld}} a^*$ for all $a \in A$ and $\ld \in \C$.
\item\label{D_3402_StarAlg_Mult}
$(a b)^* = b^* a^*$ for all $a, b \in A$.
\item\label{D_3402_StarAlg_Idem}
$a^{**} = a$ for all $a \in A$.
\end{enumerate}
If $A$ and $B$ are complex *algebras,
then a {\emph{*homomorphism}} from $A$ to~$B$
is an algebra \hm{} $\ph \colon A \to B$
such that $\ph (a^*) = \ph (a)^*$ for all $a \in A$.
\end{dfn}
That is, a *algebra has all the structure of a Banach *algebra
or a \ca{} except for the norm.
\begin{dfn}\label{D_3317_GpRing}
Let $G$ be a discrete group.
We define its (complex)
{\emph{group ring}} $\C [G]$
to be the set of formal linear combinations
of elements of~$G$ with coefficients in~$\C$.
We write $u_g$ for the element of $\C [G]$
corresponding to~$g \in G$.
Thus, for every $b \in \C [G]$
there is a unique family $(b_g)_{g \in G}$
of complex numbers such that $b_g = 0$ for all but finitely
many $g \in G$
and such that $b = \sum_{g \in G} b_g u_g$.
Multiplication is determined by specifying that
$u_g u_h = u_{g h}$ for all $g, h \in G$,
and extending linearly.
Justified by Exercise~\ref{Ex_3317_PfGpRing} below,
we make $\C [G]$ into a *algebra
by
%
\begin{equation}\label{Eq_3317_AdjCG}
\left( \ssum{g \in G} b_g u_g \right)^*
= \ssum{g \in G} {\ov{b_g}} \cdot u_{g^{1}}.
\end{equation}
%
\end{dfn}
\begin{rmk}\label{R_3317_MultGpRing}
The product in $\C [G]$ as defined above can be written in
the following equivalent ways:
\[
\left( \ssum{g \in G} a_g u_g \right)
\left( \ssum{g \in G} b_g u_g \right)
= \ssum{g, h \in G} a_g b_h u_{g h}
\]
or
%
\begin{equation}\label{Eq_3317_Prod}
\left( \ssum{g \in G} a_g u_g \right)
\left( \ssum{g \in G} b_g u_g \right)
= \ssum{g \in G} \left( \ssum{h \in G} a_h b_{h^{1} g} \right) u_g.
\end{equation}
%
\end{rmk}
\begin{rmk}\label{R_3317_AlgGpRing}
The convention in algebra seems to be that
our $\sum_{g \in G} b_g u_g$ is just written
$\sum_{g \in G} b_g \cdot g$.
Also, algebraists have no reason to always choose $\C$
as the coefficients.
Indeed, for any field $K$ they routinely construct $K [G]$
in the same way,
except that usually there is nothing quite like the adjoint
operation we defined above.
More generally, for any ring~$R$,
one can form $R [G]$ in the same way,
and we will do this when we consider \cp{s}
by discrete groups.
(See Remark~\ref{R:Discrete}.)
One needs an adjoint on $R$ in order to get an adjoint on $R [G]$.
\end{rmk}
\begin{exr}\label{Ex_3317_PfGpRing}
Let $G$ be a discrete group.
Prove that the product given in \Def{D_3317_GpRing}
makes $\C [G]$ into a unital algebra over~$\C$.
Further prove that the operation~(\ref{Eq_3317_AdjCG})
makes $\C [ G ]$ a *algebra
as in \Def{D_3402_StarAlg}.
\end{exr}
\begin{dfn}\label{D_3317_CGRepDfn}
Let $G$ be a discrete group,
let $H$ be a Hilbert space,
and let $v \colon G \to U (H)$ be a unitary representation of~$G$.
We define $\rh_v \colon \C [G] \to L (H)$
as follows.
For a family $(b_g)_{g \in G}$
of complex numbers such that $b_g = 0$ for all but finitely
many $g \in G$,
we set
%
\begin{equation}\label{Eq_3317_CGRpFormula}
\rh_v \left( \ssum{g \in G} b_g u_g \right)
= \ssum{g \in G} b_g v (g).
\end{equation}
%
\end{dfn}
\begin{prp}\label{P_3317_CGRep}
Let $G$ be a discrete group,
and let $H$ be a Hilbert space.
For any unital *representation $\pi$ of $\C [G]$ on~$H$,
we define a unitary representation $w_{\pi} \colon G \to U (H)$
by $w_{\pi} (g) = \pi (u_g)$.
Then $\pi \mapsto w_{\pi}$ is a bijection from
unital representations of $\C [G]$ on~$H$
to unitary representations of $G$ on~$H$.
The inverse is given by $v \mapsto \rh_v$
as in \Def{D_3317_CGRepDfn}.
\end{prp}
\begin{exr}\label{Ex_3317_CGRepPf}
Prove \Prp{P_3317_CGRep}.
\end{exr}
To demonstrate that this really is easy,
we prove that if $\pi \colon \C [G] \to L (H)$
is a unital *representation,
then $w_{\pi}$ is a group \hm.
Let $g, h \in G$.
Then,
using $u_{g h} = u_g u_h$ at the second step
and the fact that $\pi$ is a \hm{}
at the third step,
we have
\[
w_{\pi} (g h)
= \pi (u_{g h})
= \pi (u_g u_h)
= \pi (u_g) \pi (u_h)
= w_{\pi} (g) w_{\pi} (h).
\]
One must also prove that $\rh_v$ is in fact a unital *\hm;
this is just algebra.
We recall the universal representation
of a discrete group~$G$.
The construction is essentially the same as that of
the universal representation of a \ca.
We would like it to be a representation~$z$
such that every unitary representation is unitarily
equivalent to a subrepresentation of~$z$,
and the obvious way to do this is to take $z$ to be
the direct sum of all possible unitary representations of~$G$.
Unfortunately,
there are set theoretic problems with this definition.
First, there are representations on arbitrarily large Hilbert spaces,
and there is no set whose elements include sets
with arbitrarily large cardinality.
Second, even the collection of all one dimensional Hilbert spaces
is not a set.
We therefore proceed as follows.
\begin{dfn}\label{D_3317_UnivRep}
Let $G$ be a discrete group.
Choose a fixed Hilbert space $M$ with dimension
(cardinality of an orthonormal basis)
equal to $\card (G)$.
Let $z$ be the unitary representation
obtained as the direct sum of all possible
unitary representations of $G$ on closed subspaces of~$M$.
We call it the {\emph{universal representation}} of~$G$.
\end{dfn}
We really need only make sure that $\dim (M) \geq \card (G)$.
But then the unitary equivalence class
of our choice of universal representation
would depend $\dim (M)$.
This dependence would not matter in any essential way,
but would be annoying.
\begin{rmk}\label{R_3317_URepRmk}
With the construction of \Def{D_3317_UnivRep},
the universal representation $z$ is unique up to unitary equivalence.
It has the property that every unitary representation
with a cyclic vector is unitarily
equivalent to a subrepresentation of~$z$.
Since every representation is a direct sum of cyclic subrepresentations,
it follows that every unitary representation
of~$G$
is a direct sum of subrepresentations
which are unitarily equivalent to subrepresentations of~$z$.
\end{rmk}
We can now define two standard \ca{s} associated to a discrete
group.
\begin{dfn}\label{D_3317_GpCStar}
Let $G$ be a discrete group.
Let $z \colon G \to U (M)$
be the universal unitary representation of~$G$,
as in \Def{D_3317_UnivRep}.
Using the notation of \Def{D_3317_CGRepDfn},
we define the {\emph{group C*algebra}} $C^* (G)$
to be the closure ${\overline{\rh_z ( \C [G] )}}$
of $\rh_z ( \C [G] ) \subset L (M)$ in the norm topology on $L (M)$.
When $z$ is the universal \rpn{} of~$G$,
we write $\rh_z$ for both the map $\C [G] \to L (M)$
and for the same map with restricted codomain $C^* (G)$.
\end{dfn}
Equivalently,
$C^* (G) = {\ov{\spn}} \big( \{ z (g) \colon g \in G \} \big)$.
The \ca{} $C^* (G)$ has the following description in terms
of generators and relations.
\begin{exr}\label{Ex_3401_CstGRel}
Let $G$ be a discrete group.
Prove that $C^* (G)$ is the universal unital \ca{}
with generators $u_g$ for $g \in G$
and relations $u_g u_g^* = u_g^* u_g = 1$
for $g \in G$
and $u_g u_h = u_{g h}$ for $g, h \in G$.
\end{exr}
\begin{dfn}\label{D_3317_RedGpCStar}
Let $G$ be a discrete group.
Let $v \colon G \to U (l^2 (G))$
be the left regular representation
(\Def{D_3407_DRegRep}).
Using the notation of \Def{D_3317_CGRepDfn},
we define the {\emph{reduced group C*algebra}} $C^*_{\mathrm{r}} (G)$
to be the closure ${\overline{\rh_v ( \C [G] )}}$
of $\rh_v ( \C [G] ) \subset L (l^2 (G))$
in the norm topology on $L (l^2 (G))$.
When $v$ is the left regular \rpn{} of~$G$,
we write $\rh_v$ for both the map $\C [G] \to L (l^2 (G))$
and for the same map with restricted codomain $C^*_{\mathrm{r}} (G)$.
\end{dfn}
\begin{ntn}\label{N_3317_GinAlg}
Let $G$ be a discrete group.
For $g \in G$,
we also write $u_g$ for the images of $u_g \in \C [G]$
in both $C^* (G)$ and $C^*_{\mathrm{r}} (G)$.
(No confusion should arise.
In effect, in Exercise~\ref{Ex_3401_CstGRel},
we already used this notation in $C^* (G)$.)
\end{ntn}
The algebra $C^* (G)$ is sometimes called the full \ca{} of~$G$.
Sometimes the notation $C^*_{\mathrm{max}} (G)$ is used,
and correspondingly
$C^*_{\mathrm{min}} (G)$ for $C^*_{\mathrm{r}} (G)$.
The reduced \ca{} is also sometimes written $C^*_{\ld} (G)$,
based on the traditional notation $\ld$
for the left regular representation.
Besides the full and reduced group \ca{s},
there are ``exotic'' group \ca{s},
completions of $\C [G]$ with convolution multiplication
in norms which lie between
those giving the full and reduced \ca{s}.
The first systematic study of such algebras
seems to be the recent paper~\cite{BrnGnt}.
Further work on such algebras
appears in~\cite{Okys}
(where uncountably many such algebras are given for
nonabelian free groups),
\cite{RnWrs}, and~\cite{Wrsm}.
We don't discuss these algebras in these notes.
The next theorem shows that the full \ca{} of a group
plays a role for unitary representations
analogous to the role of the group ring for representations
in the purely algebraic situation.
\begin{thm}\label{T_3317_UProp}
Let $G$ be a discrete group,
and let $H$ be a Hilbert space.
For any unital representation $\pi$ of $C^* (G)$ on~$H$,
we define a unitary representation $w_{\pi} \colon G \to U (H)$
by $w_{\pi} (g) = \pi (u_g)$.
Then $\pi \mapsto w_{\pi}$ is a bijection from
unital representations of $C^* (G)$ on~$H$
to unitary representations of $G$ on~$H$.
In addition,
if $v \colon G \to U (H)$ is a unitary \rpn{} of $G$
on a Hilbert space~$H$,
if $\rh_v$ is as in \Def{D_3317_CGRepDfn},
and if $\pi \colon C^* (G) \to L (H)$
is the corresponding \rpn{} of $C^* (G)$,
then:
\begin{enumerate}
\item\label{T_3317_UProp_Back}
Let $z$ be the universal unitary representation of~$G$.
Then $\pi$ is uniquely determined by the relation
$\rh_v (a) = \pi ( \rh_z (a))$ for all $a \in C^* (G)$.
\item\label{T_3317_UProp_Surj}
$\pi (C^* (G))
= C^* (v (G))
= {\ov{\spn}} (v (G))
= {\ov{\rh_v (\C [G])}}$.
\end{enumerate}
\end{thm}
We have used the same notation $\pi \mapsto w_{\pi}$
as in Proposition~\ref{P_3317_CGRep}.
We don't quite get the formula (\ref{Eq_3317_CGRpFormula})
of \Def{D_3317_CGRepDfn}
for the inverse correspondence.
The sums in~(\ref{Eq_3317_CGRpFormula})
are finite,
and one might hope that one could simply replace them
with convergent series,
and proceed in the obvious way.
However,
not all elements of $C^* (G)$
can be represented by convergent series
which directly generalize the finite sums
in~(\ref{Eq_3317_CGRpFormula}).
See Remark~\ref{R:WhatIsIn}(\ref{R:WhatIsIn3}),
Remark~\ref{R:WhatIsIn}(\ref{R:WhatIsIn4}),
and Remark~\ref{R_3408_NotAmen}
for further discussion.
Given a unitary \rpn~$v$,
the best we can do is to extend $\rh_v$ by continuity,
which is what part~(\ref{T_3317_UProp_Back}) of the theorem amounts to.
\begin{proof}[Proof of \Thm{T_3317_UProp}]
If $\pi$ is a unital representation of $C^* (G)$ on~$H$,
it is easy to check that $w_{\pi}$
is a unitary representation of $G$ on~$H$.
(The proof is the same as the proof of the corresponding
part of Proposition~\ref{P_3317_CGRep}.)
Suppose $\pi$ and $\sm$ are unital representations of $C^* (G)$ on~$H$,
and that $w_{\pi} = w_{\sm}$.
The definition
immediately implies that $\pi (u_g) = \sm (u_g)$ for all $g \in G$.
Since $\{ u_g \colon g \in G \}$
spans a dense subset of $C^* (G)$,
it follows that $\pi = \sm$.
Now let $v$ be any unitary representation of $G$ on~$H$.
Then there are an index set $I$ and
orthogonal invariant subspaces $H_i \subset H$ for $i \in I$
such that $H = \bigoplus_{i \in I} H_i$
and such that for $i \in I$ the restriction $v_i$
of $v$ to $H_i$ is a cyclic representation.
Let $z \colon G \to U (M)$
be the universal representation of~$G$,
on the Hilbert space~$M$,
as described in \Def{D_3317_UnivRep}.
By construction,
for every $i \in I$ there is a direct summand $M_i \subset M$
such that the restriction $z_i$ of $z$ to $M_i$
is unitarily equivalent to~$v_i$.
That is,
there is a unitary $c_i \in L (M_i, H_i)$
such that $c_i z_i (g) c_i^* = v_i (g)$
for all $g \in G$.
Now define a unital representation $\pi_i \colon C^* (G) \to L (H_i)$
by $\pi (a) = c_i (a _{M_i} ) c_i^*$.
Define a unital representation $\pi \colon C^* (G) \to L (H)$
by $\pi = \bigoplus_{i \in I} \pi_i$.
It is immediate that $w_{\pi} = v$.
We prove~(\ref{T_3317_UProp_Back}).
Let $v$ be given.
Since $w_{\pi} = v$,
we have $\pi (u_g) = v (g)$ for all $g \in G$.
It follows from linearity that
$\rh_v (a) = \pi ( \rh_z (a))$ for all $a \in \C [G]$.
By definition,
$\rh_z ( \C [G] )$ is dense in $C^* (G)$,
so this equation determines $\pi$ uniquely.
For~(\ref{T_3317_UProp_Surj}),
the equality $C^* (v (G)) = {\ov{\rh_v (\C [G])}}$
follows from the fact that $\rh_v (\C [G])$ is a *subalgebra
of $L (H)$.
The equality ${\ov{\spn}} (v (G)) = {\ov{\rh_v (\C [G])}}$
follows from the fact that $\rh_v (\C [G]) = \spn (v (G))$.
The relation
$C^* (v (G)) \subset \pi (C^* (G))$ holds because
$\pi (C^* (G))$ is closed and
$v (g) = \pi (u_g) \in \pi (C^* (G))$ for all $g \in G$.
The relation $\pi (C^* (G)) \subset {\ov{\rh_v (\C [G])}}$
follows from
$\rh_v (a) = \pi ( \rh_z (a))$ for $a \in \C [G]$
and density of $\rh_z ( \C [G] )$ in $C^* (G)$.
\end{proof}
\begin{cor}\label{C_3317_FToRed}
Let $G$ be a discrete group.
Then there is a unique surjective \hm{}
$\kp \colon C^* (G) \to C^*_{\mathrm{r}} (G)$
determined (following Notation~\ref{N_3317_GinAlg})
by $u_g \mapsto u_g$ for $g \in G$.
\end{cor}
\begin{proof}
It follows from \Thm{T_3317_UProp}
that $u_g \mapsto u_g$
determines a unique \hm{} $\kp_0 \colon C^* (G) \to L (l^2 (G))$,
and from \Thm{T_3317_UProp}(\ref{T_3317_UProp_Surj})
and \Def{D_3407_DRegRep}
that $\kp_0 ( C^* (G)) = C^*_{\mathrm{r}} (G)$.
\end{proof}
The map $\kp \colon C^* (G) \to C^*_{\mathrm{r}} (G)$
of Corollary~\ref{C_3317_FToRed} need not be injective.
In fact, $\kp$ is injective \ifo{} $G$ is amenable
(\Thm{T_3320_FRIso}, for which we do not give a proof,
and \Thm{T_4131_GpFToRed}).
Amenability is an important property,
which we mostly do not treat in these notes;
we refer to the discussion before \Thm{T_4131_GpFToRed}
for more information.
% We do prove (\Thm{T_3322_DiscAbRed})
% that $\kp$ is injective when $G$ is discrete abelian.
% And maybe locally compact abelian. 999
We do include enough in these notes to see that $\kp$ is not
injective when $G$ is a countable nonabelian free group.
Indeed,
we show in \Thm{T_3301_CStRFnSimp}
that $C^*_{\mathrm{r}} (G)$ is simple.
However, $C^* (G)$ is never simple unless $G$ has only one element.
To see this,
let $H$ be a one dimensional Hilbert space,
and let $v \colon G \to U (H)$
be the trivial \rpn,
that is, $v (g) = 1$ for all $g \in G$.
Applying \Thm{T_3317_UProp}(\ref{T_3317_UProp_Surj})
to this representation,
we obtain a nonzero \hm{} $\pi \colon C^* (G) \to L (H) = \C$.
It follows from Corollary~\ref{C_3317_InjFull} below
that $\pi$ is not injective,
so ${\operatorname{Ker}} (\pi)$ is a nontrivial ideal in $C^* (G)$.
\begin{prp}\label{P_3317_InjRed}
Let $G$ be a discrete group.
Then the map $\rh_v \colon \C [G] \to C^*_{\mathrm{r}} (G)$
of \Def{D_3317_RedGpCStar}
is injective.
\end{prp}
\begin{proof}
As in \Def{D_3317_RedGpCStar},
let $v \colon G \to U (l^2 (G))$
be the left regular representation.
% (\Def{D_3407_DRegRep}).
% Let $\rh_v \colon \C [G] \to L (l^2 (G))$
% be as in \Def{D_3317_CGRepDfn}.
Let $b \in \C [G]$.
Then there is a family $(b_g)_{g \in G}$
of complex numbers such that $b_g = 0$ for all but finitely
many $g \in G$
and such that $b = \sum_{g \in G} b_g u_g$.
For $g \in G$,
let $\dt_g \in l^2 (G)$ be the standard basis vector
corresponding to~$g$.
Then $\rh_v (b) \dt_1 = \sum_{g \in G} b_g \dt_g$.
If $b \neq 0$,
then there is $g \in G$ such that $b_g \neq 0$,
so $\langle \rh_v (b) \dt_1, \, \dt_g \rangle = b_g \neq 0$.
Thus $\rh_v (b) \neq 0$.
\end{proof}
\begin{cor}\label{C_3317_InjFull}
Let $G$ be a discrete group.
Then the map $\rh_z \colon \C [G] \to C^* (G)$
of \Def{D_3317_GpCStar}
is injective.
\end{cor}
\begin{proof}
This follows from \Prp{P_3317_InjRed}
and Corollary~\ref{C_3317_FToRed}.
\end{proof}
We are primarily interested in crossed products,
and the sort of functoriality we are most interested in
is what happens for a suitable \hm{}
between algebras on which a fixed group $G$ acts.
But functoriality of group \ca{s}
is a sufficiently obvious question
that we should at least describe what happens.
\begin{exr}\label{Ex_4220_FunctGpCSt}
Let $G_1$ and $G_2$ be discrete groups,
and let $\ph \colon G_1 \to G_2$ be a \hm.
Prove that there is a unique \hm{}
$C^* (\ph) \colon C^* (G_1) \to C^* (G_2)$
such that $C^* (\ph) (u_g) = u_{\ph (g)}$ for all $g \in G_1$.
Prove that, with this definition of the action on morphisms,
$G \mapsto C^* (G)$
is a functor
from the category of discrete group and group \hm{s}
to the category of unital \ca{s} and unital \hm{s}.
\end{exr}
The main point is that if $w$ is a unitary \rpn{} of~$G_2$,
then $w \circ \ph$ is a unitary \rpn{} of~$G_1$.
\begin{exr}\label{Ex_4220_FunctRedCSt}
Let $G_1$ and $G_2$ be discrete groups,
and let $\ph \colon G_1 \to G_2$ be an injective \hm.
Prove that there is a unique \hm{}
$C^*_{\mathrm{r}} (\ph) \colon
C^*_{\mathrm{r}} (G_1) \to C^*_{\mathrm{r}} (G_2)$
such that
$C^*_{\mathrm{r}} (\ph) (u_g) = u_{\ph (g)}$ for all $g \in G_1$.
Prove that, with this definition of the action on morphisms,
$G \mapsto C^*_{\mathrm{r}} (G)$
is a functor
from the category of discrete group and injective group \hm{s}
to the category of unital \ca{s} and unital \hm{s}.
\end{exr}
The main point here is that
if $v$ is the regular representation of $G_2$,
then $v \circ \ph$ is a direct sum of copies
of the regular representation of $G_1$.
(The number of copies is the cardinality of the coset
space $G_2 / \ph (G_1)$.)
Without injectivity of~$\ph$,
there might be no nonzero \hm{} from
$C^*_{\mathrm{r}} (G_1)$ to $C^*_{\mathrm{r}} (G_2)$.
As an example,
let $n \in \{ 2, 3, \ldots, \I \}$,
and let $F_n$ be the free group on $n$ generators.
Theorem~\ref{T_3301_CStRFnSimp} implies that $C^*_{\mathrm{r}} (F_n)$
is simple.
Therefore there is no nonzero \hm{} from
$C^*_{\mathrm{r}} (G_1)$ to $\C$
to go with the \hm{} from $F_n$ to the group with one element.
We warn that when the groups are not discrete,
there is much less functoriality.
See the discussion after Proposition~\ref{P_6X15_FToRed}.
We so far haven't given any justification for the use
of $C^*_{\mathrm{r}} (G)$.
% In these notes, we will not do much with reduced group \ca{s}
% (or their generalization,
% reduced crossed products),
% because for most of the situations we study in detail,
% the full and reduced crossed products are the same.
Here is one reason for its importance.
Recall that a state $\om$ on a \ca~$A$
is said to be {\emph{faithful}} if whenever
$a \in A$ satisfies $\om (a^* a) = 0$,
then $a = 0$.
A state $\om$ on a \ca~$A$ is {\emph{tracial}}
if $\om (a b) = \om (b a)$
for all $a, b \in A$.
(We state this formally as \Def{D_Trace} below.)
\begin{thm}\label{T_3318_RedTr}
Let $G$ be a discrete group.
Then there is a unique continuous linear functional
$\ta \colon C^*_{\mathrm{r}} (G) \to \C$
such that $\ta (u_1) = 1$
and $\ta (u_g) = 0$ for $g \in G \setminus \{ 1 \}$.
Moreover, $\ta$ is a faithful tracial state.
\end{thm}
The condition on $\ta$ means that if $(b_g)_{g \in G}$
is a family
of complex numbers such that $b_g = 0$ for all but finitely
many $g \in G$,
then
%
\begin{equation}\label{Eq_3319_TrFormula}
\ta \left( \ssum{g \in G} b_g u_g \right) = b_1.
\end{equation}
%
Our main application of \Thm{T_3318_RedTr}
will be to the existence of ``coefficients''
for elements of $C^*_{\mathrm{r}} (G)$.
See Proposition~\ref{P_3408_Coeffs}
and Proposition~\ref{P_3408_MeaningCffs},
and see Remark~\ref{R:WhatIsIn} for warnings about the use
of these coefficients.
Remark~\ref{R_3408_NotAmen}
explains one thing which goes wrong in $C^* (G)$
when $C^* (G) \neq C^*_{\mathrm{r}} (G)$.
\begin{proof}[Proof of \Thm{T_3318_RedTr}]
Since $\C [G]$ is dense in $C^*_{\mathrm{r}} (G)$,
there can be at most one such
\ct{} linear functional.
We now prove existence.
As before,
for $g \in G$,
let $\dt_g \in l^2 (G)$ be the standard basis vector
corresponding to~$g$.
Define $\ta \colon C^*_{\mathrm{r}} (G) \to \C$
by $\ta (a) = \langle a \dt_1, \dt_1 \rangle$.
We immediately check
that $\ta (u_1) = \langle \dt_1, \dt_1 \rangle = 1$
and that if $g \in G \setminus \{ 1 \}$
then $\ta (u_g) = \langle \dt_g, \dt_1 \rangle = 0$.
It is obvious that $\ta$ is a state on $C^*_{\mathrm{r}} (G)$.
To prove that $\ta$ is tracial,
by linearity and continuity it suffices to prove that
$\ta (u_g u_h) = \ta (u_h u_g)$
for all $g, h \in G$.
This reduces immediately to the fact that
$g h \neq 1$ \ifo{} $h g \neq 1$.
It remains to show that $\ta$ is faithful.
Identify $\C [G]$ with its image in $C^*_{\mathrm{r}} (G)$.
We first claim that $\C [G] \dt_1$
is dense in $l^2 (G)$.
It suffices to show that if $(b_g)_{g \in G}$
is a family
of complex numbers such that $b_g = 0$ for all but finitely
many $g \in G$,
then
$\sum_{g \in G} b_g \dt_g \in \C [G] \dt_1$.
Set $b = \sum_{g \in G} b_g u_g$,
which is in $\C [G]$,
and observe that $\sum_{g \in G} b_g \dt_g = b \dt_1 \in \C [G] \dt_1$.
This proves the claim.
Now let $a \in C^*_{\mathrm{r}} (G)$ satisfy $\ta (a^* a) = 0$.
Let $b, c \in \C [G] \dt_1$.
Using the CauchySchwarz inequality at the fourth step,
we have
\[
 \langle a b \dt_1, c \dt_1 \rangle 
=  \langle c^* a b \dt_1, \dt_1 \rangle 
=  \ta (c^* a b) 
=  \ta (b c^* a) 
\leq \ta (a^* a)^{1/2} \ta (b c^* c b^*)^{1/2}
= 0.
\]
So $\langle a b \dt_1, c \dt_1 \rangle = 0$.
Since $b \dt_1$ and $c \dt_1$ are arbitrary elements
of a dense subset of $l^2 (G)$,
it follows that $a = 0$.
\end{proof}
We now look at two easy classes of examples:
finite groups and discrete abelian groups.
\begin{exa}\label{E_3319_CStFinGp}
Let $G$ be a finite group.
Then $\C [G]$ is \fd,
hence already complete in any norm.
Therefore $C^* (G) = C^*_{\mathrm{r}} (G) = \C [G]$
is a \fd{} \ca,
with dimension equal to $\card (G)$.
So there are $m \in \N$
and $r (1) \leq r(2) \leq \cdots \leq r (m)$
such that $C^* (G) \cong \bigoplus_{j = 1}^m M_{r (j)}$
and $\sum_{j = 1}^m r (j)^2 = \card (G)$.
The numbers $r (1), r(2), \ldots, r (m)$
are the dimensions of the distinct equivalence classes of
irreducible representations of $C^* (G)$,
equivalently, of~$G$.
Since the one dimensional trivial representation of~$G$
is irreducible,
we must have $r (1) = 1$.
A standard theorem from algebra
(Theorem~7 in Section~2.5 of~\cite{Srr}) asserts that the number
of distinct equivalence classes of
irreducible representations of~$G$
is equal to the number of conjugacy classes in~$G$.
\end{exa}
The book~\cite{Srr} contains much more information about
the representation theory of finite groups.
We turn to discrete abelian groups.
We will need Pontryagin duality
and various related results,
which we state without proof.
To avoid later repetition,
we give the statements for the case of locally compact abelian groups.
A discussion of the very beginnings of
this subject (including the identification
of the dual ${\widehat{G}}$
with the maximal ideal space of the Banach algebra $L^1 (G)$,
but not including the Pontryagin duality theorem)
appears in Section~1.4 of~\cite{Wlms}.
There is a more extensive discussion in Chapter~4 of~\cite{Fld},
and there are thorough presentations
in Chapter~6 of~\cite{HwRs} and Chapter~8 of~\cite{HwRs2}.
\begin{dfn}[Definition 1.74 of~\cite{Wlms};
beginning of Section~4.1 of~\cite{Fld};
Definition 23.3 of~\cite{HwRs}]\label{D_3319_DualDfn}
Let $G$ be a locally compact abelian group.
Its {\emph{Pontryagin dual}}
(or just dual)
${\widehat{G}}$
is the set of \ct{} \hm{s} $\ch \colon G \to S^1$,
with the topology of uniform convergence on compact sets.
\end{dfn}
There are two motivations for this definition.
One is the duality theorem (\Thm{T_3319_PDuality}),
in condensed form
\[
{\widehat{\widehat{G{}}}} = G.
\]
The other is that ${\widehat{G}}$ is essentially
the set of one dimensional \rpn{s} of~$G$
(see Proposition~\ref{P_3320_Dual1dRep} below),
and that the irreducible \rpn{s}
are exactly the one dimensional \rpn{s}.
For this,
we recall Schur's Lemma for unitary \rpn{s}
of topological groups.
The proofs of Schur's Lemma and the corollary
are essentially the same as that of the analogous statements
for \ca{s}.
If $G$ is a topological group and $v_1 \colon G \to U (H_1)$
and $v_2 \colon G \to U (H_2)$ are unitary \rpn{s} of~$G$
on Hilbert spaces $H_1$ and~$H_2$,
then we let $C (v_1, v_2)$ be the set of intertwining operators,
that is,
\[
C (v_1, v_2)
= \big\{ a \in L (H_1, H_2) \colon
{\mbox{$a v_1 (g) = v_2 (g) a$ for all $g \in G$}} \big\}.
\]
\begin{thm}[3.5(b) in~\cite{Fld}]\label{T_3408_Schur}
Let $G$ be a topological group and
let $v_1 \colon G \to U (H_1)$
and $v_2 \colon G \to U (H_2)$ be irreducible unitary \rpn{s} of~$G$
on Hilbert spaces $H_1$ and~$H_2$.
If $v_1$ and $v_2$ are unitarily equivalent,
then there is a unitary $u \in L (H_1, H_2)$
such that $C (v_1, v_2) = \C u$.
Otherwise, $C (v_1, v_2) = 0$.
\end{thm}
\begin{cor}[Corollary~3.6 of~\cite{Fld}]\label{C_3408_1dim}
Let $G$ be an abelian topological group.
Then every irreducible unitary \rpn{} of~$G$
is one dimensional.
\end{cor}
\begin{proof}
Let $v \colon G \to U (H)$ be an irreducible unitary \rpn{}
of $G$ on a Hilbert space~$H$.
It follows from \Thm{T_3408_Schur}
that $C (v, v)$ is one dimensional.
Since $1 \in C (v, v)$,
we get $C (v, v) = \C \cdot 1$.
Since $G$ is abelian,
for every $h \in G$ we have $v (h) \in C (v, v)$.
Therefore $v (h) \in \C \cdot 1$.
It follows that every closed subspace of $H$ is invariant.
Since $v$ is irreducible,
this is only possible if $\dim (H) = 1$.
\end{proof}
The set ${\widehat{G}}$ exactly parametrizes the
one dimensional representations of~$G$:
\begin{prp}\label{P_3320_Dual1dRep}
Let $G$ be a locally compact abelian group.
Identify $S^1$ with the unitary group of the one dimensional
Hilbert space~$\C$ in the obvious way.
Then:
\begin{enumerate}
\item\label{P_3320_Dual1dRep_Exist}
Every one dimensional representation of~$G$
is unitarily equivalent to some element of~${\widehat{G}}$.
\item\label{P_3320_Dual1dRep_Uniq}
If $\ch_1, \ch_2 \in {\widehat{G}}$ are unitarily equivalent,
then $\ch_1 = \ch_2$.
\end{enumerate}
\end{prp}
\begin{proof}
Both parts are immediate.
\end{proof}
In the following theorem,
local compactness is Corollary 1.79 of~\cite{Wlms},
the discussion after Theorem~4.2 in~\cite{Fld},
or Theorem 23.13 of~\cite{HwRs}.
Duality (the statement that $\ep_G$ is an isomorphism)
is Theorem 4.31 of~\cite{Fld},
or Theorem 24.8 of~\cite{HwRs}.
The fact that $G \mapsto {\widehat{G}}$ is a contravariant
functor is clear
(and is in Theorem 24.38 of~\cite{HwRs}),
naturality of $\ep_G$ is obvious,
and that $G \mapsto {\widehat{G}}$ is a category equivalence
follows from
duality and naturality of~$\ep_G$.
\begin{thm}\label{T_3319_PDuality}
Let $G$ be a locally compact abelian group.
Then ${\widehat{G}}$
is a locally compact abelian group.
The assignment $G \mapsto {\widehat{G}}$ is the map
on objects of a contravariant category equivalence
from the category of locally compact abelian groups
and \ct{} group \hm{s} to itself,
for which the map on morphisms assigns to a
\ct{} group \hm{} $\ph \colon G \to H$
the \hm{} $\ch \mapsto \ch \circ \ph$
from ${\widehat{H}}$ to ${\widehat{G}}$.
There is a natural isomorphism
of locally compact abelian groups
\[
\ep_G \colon G \to {\widehat{\widehat{G{}}}}
\]
(Pontryagin duality),
given by $\ep_G (g) (\ch) = \ch (g)$
for $g \in G$ and $\ch \in {\widehat{G}}$.
\end{thm}
In the following collection of examples,
the one we care most about is ${\widehat{\Z}} = S^1$.
\begin{exa}\label{E_3327_Duals}
We give the examples of dual groups which are most important for
our purposes.
\begin{enumerate}
\item\label{E_3327_Duals_Fin}
Let $G$ be a finite abelian group.
Then there is a (noncanonical) isomorphism ${\widehat{G}} \cong G$.
See Corollary~4.7 of~\cite{Fld}, or 23.27(d) of~\cite{HwRs}.
% or apply Theorem~4.5(d) of~\cite{Fld},
% (\ref{E_3327_Duals_Prod}) below,
% and the structure theorem for finitely generated abelian groups.
\item\label{E_3327_Duals_Z}
For $\zt \in S^1$,
define $\ch_{\zt} \in {\widehat{\Z}}$ by $\ch_{\zt} (n) = \zt^n$
for $n \in \Z$.
Then $\zt \mapsto \ch_{\zt}$ defines an isomorphism
$S^1 \to {\widehat{\Z}}$.
See Theorem~4.5(c) of~\cite{Fld}, or 23.27(b) of~\cite{HwRs}.
\item\label{E_3327_Duals_S1}
For $n \in \Z$,
define $\ch_{n} \in {\widehat{\Z}}$ by $\ch_{n} (\zt) = \zt^n$
for $\zt \in S^1$.
Then $n \mapsto \ch_{n}$ defines an isomorphism
$\Z \to {\widehat{S^1}}$.
See Theorem~4.5(b) of~\cite{Fld}, or 23.27(a) of~\cite{HwRs}.
\item\label{E_3327_Duals_R}
For $t \in \R$,
define $\ch_{t} \in {\widehat{\R}}$ by $\ch_{t} (x) = \exp (i x t)$
for $x \in \R$.
Then $t \mapsto \ch_{t}$ defines an isomorphism
$\R \to {\widehat{\R}}$.
See Theorem~4.5(a) of~\cite{Fld}, or 23.27(e) of~\cite{HwRs}.
(In Theorem~4.5(a) of~\cite{Fld},
the slightly different formula $\ch_t (x) = \exp (2 \pi i x t)$
is used, but clearly one formula gives an isomorphism
\ifo{} the other does.
The difference shows up in formulas for Fourier transforms
and related objects.)
\item\label{E_3327_Duals_Prod}
Let $G_1, G_2, \ldots, G_n$
be locally compact abelian groups,
and let $G = \prod_{k = 1}^n G_k$.
Then ${\widehat{G}} \cong \prod_{k = 1}^n {\widehat{G_k}}$.
The isomorphism sends
$(\ch_1, \ch_2, \ldots, \ch_n) \in \prod_{k = 1}^n {\widehat{G_k}}$
to the function
\[
(g_1, g_2, \ldots, g_n) \mapsto
\ch_1 (g_1) \ch_2 (g_2) \cdots \ch_n (g_n)
\]
for $(g_1, g_2, \ldots, g_n) \in \prod_{k = 1}^n G_k$.
See Proposition~4.6 of~\cite{Fld}, or Theorem 23.18 of~\cite{HwRs}.
\item\label{E_3327_Duals_PrdCpt}
Let $I$ be an index set,
and for $i \in I$ let $G_i$ be a compact abelian group.
Let $G = \prod_{i \in I} G_i$.
Then ${\widehat{G}} \cong \bigoplus {\widehat{G_i}}$.
(The direct sum is the algebraic direct sum of the discrete
abelian groups~${\widehat{G_i}}$.)
The map is the obvious generalization of
that of~(\ref{E_3327_Duals_Prod});
the product is well defined because the factors commute
and all but finitely many of them are equal to~$1$.
See Proposition~4.8 of~\cite{Fld}, or Theorem 23.21 of~\cite{HwRs}.
\end{enumerate}
\end{exa}
For many further results about the relations
between $G$ and~${\widehat{G}}$,
we refer to Chapter~4 of~\cite{Fld}
and particularly to Chapter~6 of~\cite{HwRs}.
Here we point out just a few facts.
\begin{thm}\label{T_3319_PropOfDual}
Let $G$ be a locally compact abelian group.
Then:
\begin{enumerate}
\item\label{T_3319_PropOfDual_DsCpt}
$G$ is discrete if and only if ${\widehat{G}}$ is compact.
\item\label{T_3319_PropOfDual_CptDs}
$G$ is compact if and only if ${\widehat{G}}$ is discrete.
% \item\label{T_3319_PropOfDual_999}
% $G$ is 999 if and only if ${\widehat{G}}$ is 999.
% \item\label{T_3319_PropOfDual_999}
% $G$ is 999 if and only if ${\widehat{G}}$ is 999.
% \item\label{T_3319_PropOfDual_999}
% $G$ is 999 if and only if ${\widehat{G}}$ is 999.
% \item\label{T_3319_PropOfDual_999}
% $G$ is 999 if and only if ${\widehat{G}}$ is 999.
\end{enumerate}
\end{thm}
\begin{proof}
The forward implication in each of the two parts
is in Proposition~4.4 of~\cite{Fld}
or Theorem 23.17 of~\cite{HwRs}.
The reverse direction in each part
follows from the forward implication in the other part
by duality (\Thm{T_3319_PDuality}).
\end{proof}
Further statements of this general nature can be found
in Theorems 24.23, 24.25, 24.26, and 24.28 of~\cite{HwRs}.
In the statements of all these results,
${\textsf{X}} = {\widehat{G}}$,
and additional related theorems can be obtained by
using duality (\Thm{T_3319_PDuality})
to exchange $G$ and~${\widehat{G}}$.
The first part of the following result is known as
Plancherel's Theorem.
The element $y \xi$ is a generalized Fourier transform,
and is often written ${\widehat{\xi}}$.
\begin{thm}\label{T_3322_Fourier}
Let $G$ be a locally compact abelian group.
For any choice of Haar measure on~$G$,
there is a choice of Haar measure on~${\widehat{G}}$
such that there is a unitary
$y \in L \big( L^2 (G), \, L^2 \big( {\widehat{G}} \big) \big)$
such that
\[
(y \xi) (\ch) = \int_G {\ov{ \ch (g) }} \xi (g) \, d \mu (g)
\]
for all $\ch \in {\widehat{G}}$ and $\xi \in L^1 (G) \cap L^2 (G)$.
Further, let $v$ be the left regular representation of~$G$
on $L^2 (G)$,
and let $w$ be the unitary representation of~$G$
on $L^2 \big( {\widehat{G}} \big)$ defined by
$(w (g) \et) (\ch) = {\ov{ \ch (g) }} \et (\ch)$
for $g \in G$, $\ch \in {\widehat{G}}$,
and $\et \in L^2 \big( {\widehat{G}} \big)$.
Then $y$ intertwines $v$ and~$w$,
that is,
$y v (g) y^* = w (g)$ for all $g \in G$.
\end{thm}
The first part is Theorem 4.25 of~\cite{Fld},
or Theorem 31.18 of~\cite{HwRs2}.
Given the first part of the theorem,
the second part is easily justified.
For $g \in G$, $\ch \in {\widehat{G}}$,
and $\xi \in L^1 (G) \cap L^2 (G)$,
we have
\begin{align*}
( y v (g) \xi) (g)
& = \int_G {\ov{ \ch (h) }} (v (g) \xi) (h) \, d \mu (h)
= \int_G {\ov{ \ch (h) }} \xi (g^{1} h) \, d \mu (h)
\\
& = \int_G {\ov{ \ch (g h) }} \xi (h) \, d \mu (h)
= {\ov{ \ch (g) }} {\widehat{\xi}} (\ch)
= (w (g) y \xi) (\ch).
\end{align*}
Since $L^1 (G) \cap L^2 (G)$ is dense in $L^2 (G)$,
the second part follows.
We are now ready to calculate the \ca{} of a discrete abelian group.
The answer is essentially the same without discreteness:
$C^* (G) \cong C_0 \big( {\widehat{G}} \big)$
for every locally compact abelian group~$G$.
We also point out that,
according to some presentations of the theory,
what we are doing here is backwards:
${\widehat{G}}$ is (almost) defined as the
maximal ideal space of $C^* (G)$.
(The common version of this approach
is to define ${\widehat{G}}$
to be the maximal ideal space of $L^1 (G)$.)
\begin{thm}\label{T_3319_CStDiscAb}
Let $G$ be a discrete abelian group.
Then there is an isomorphism
$\gm \colon C^* (G) \to C \big( {\widehat{G}} \big)$
determined by the following formula.
If $(b_g)_{g \in G}$ is a family
of complex numbers such that $b_g = 0$ for all but finitely
many $g \in G$,
then
\[
\gm \left( \ssum{g \in G} b_g u_g \right) (\ch)
= \ssum{g \in G} \ch (g) b_g
\]
for all $\ch \in {\widehat{G}}$.
\end{thm}
\begin{proof}
Since $C^* (G)$ is a commutative unital \ca,
we can let $X$ be its
maximal ideal space ${\operatorname{Max}} (C^* (G))$,
which we think of as the set of
unital \hm{s} from $C^* (G)$ to~$\C$.
Then there is a canonical
isomorphism $\ph \colon C^* (G) \to C (X)$.
Proposition~\ref{P_3320_Dual1dRep}
identifies ${\widehat{G}}$
with the set of \rpn{s} of $G$ on the one dimensional
Hilbert space~$\C$,
and
\Thm{T_3317_UProp} provides a bijection from
such \rpn{s}
to the unital \hm{s} from $C^* (G)$ to~$\C$.
Combining them, we obtain a bijection $h \colon {\widehat{G}} \to X$
such that $h (\ch) (u_g) = \ch (g)$
for all $\ch \in {\widehat{G}}$ and $g \in G$.
We claim that $h$ is \ct.
Let $(\ch_i)_{i \in I}$ be a net in ${\widehat{G}}$
which converges uniformly on compact sets
to $\ch \in {\widehat{G}}$.
Then for all $g \in G$ we have
\[
\lim_{i \in I} h (\ch_i) (u_g)
= \lim_{i \in I} \ch_i (g)
= \ch (g)
= h (\ch) (u_g).
\]
It follows that
$\lim_{i \in I} h (\ch_i) (a) = h (\ch) (a)$
for all
$a \in \spn \big ( \{ u_g \colon g \in G \} \big) \subset C^* (G)$.
It now follows from an $\frac{\ep}{3}$~argument
that $\lim_{i \in I} h (\ch_i) (a) = h (\ch) (a)$
for all
$a \in {\ov{\spn}} \big ( \{ u_g \colon g \in G \} \big)
= C^* (G)$.
By the definition of the topology on
${\operatorname{Max}} (C^* (G))$,
this means that $\lim_{i \in I} h (\ch_i) = h (\ch)$.
Continuity of~$h$ follows.
We now know that $h$ is a \ct{} bijection of \chs{s}.
Therefore $h$ is a \hme.
So $h$ determines an
isomorphism ${\operatorname{Max}} (C^* (G)) \to {\widehat{G}}$.
The theorem follows.
\end{proof}
The following theorem holds in much greater generality
(for arbitrary amenable locally compact groupssee
Theorem~\ref{T_4131_GpFToRed} and Theorem~\ref{T_4131_CPFToRed} below),
but this special case has an easy proof, which we give here.
\begin{thm}\label{T_3322_DiscAbRed}
Let $G$ be a discrete abelian group.
Then the canonical \hm{} $\kp \colon C^* (G) \to C^*_{\mathrm{r}} (G)$
is an isomorphism.
\end{thm}
\begin{proof}
For any unitary representation $w$ of $G$ on a Hilbert space~$H$,
let $\rh_w \colon C^* (G) \to L (H)$
be the corresponding representation of $C^* (G)$
as in \Thm{T_3317_UProp}.
We have to prove that $\ \kp (b) \ = \ b \$
for all $b \in C^* (G)$.
% It suffices to do this for $b \in \C [G]$.
As in Theorem~\ref{T_3322_Fourier},
let $v$ be the left regular representation of~$G$
on $L^2 (G)$,
and let $w \colon G \to U \big(L^2 \big( {\widehat{G}} \big) \big)$
be
$(w (g) \et) (\ch) = {\ov{ \ch (g) }} \et (\ch)$
for $g \in G$, $\ch \in {\widehat{G}}$,
and $\et \in L^2 \big( {\widehat{G}} \big)$.
Also let
$y \in L \big( L^2 (G), \, L^2 \big( {\widehat{G}} \big) \big)$
be as in Theorem~\ref{T_3322_Fourier}.
By definition,
$\kp = \rh_v$.
Since $y$ intertwines $v$ and~$w$,
it is immediate that $y$ intertwines $\rh_v$ and~$\rh_w$.
Therefore $\ \kp (b) \ = \ \rh_w (b) \$.
Let $\gm \colon C^* (G) \to C \big( {\widehat{G}} \big)$
be as in \Thm{T_3319_CStDiscAb}.
For $g \in G$,
the operator $\rh_w (u_g)$ is multiplication by the function
$\ch \mapsto {\ov{\ch (g)}} = \gm (u_g) (\ch^{1})$.
Therefore $\rh_w (b)$ is multiplication by the function
$\ch \mapsto \gm (b) (\ch^{1})$.
Since Haar measure on ${\widehat{G}}$ has full support,
we get $\ \rh_w (b) \ = \ \gm (b) \$.
Combining this with the result of the previous paragraph,
and with $\ \gm (b) \ = \ b \$
(from \Thm{T_3319_CStDiscAb}),
we get $\ \kp (b) \ = \ b \$.
\end{proof}
The following remark and problem are not directly related
to the main topic of these notes,
but they seem interesting enough to include.
\begin{rmk}\label{R_3402_GpSameCst}
Neither $C^* (G)$ nor $C^*_{\mathrm{r}} (G)$ determines~$G$,
not even for $G$ discrete abelian.
One example that is easy to get from what has already been
done is that the full and reduced \ca{s}
of all second countable infinite compact groups are the same,
namely $C_0 (S)$ for a countable infinite set~$S$.
Any two finite abelian groups with the
same cardinality have isomorphic \ca{s},
since if $\card (G) = n$ then $\card \big( {\widehat{G}} \big) = n$
and $C^* (G) = C^*_{\mathrm{r}} (G) \cong \C^n$.
Among nonabelian groups,
the simplest example is that both the nonabelian groups
of order~$8$
have both full and reduced \ca{s}
isomorphic to $\C^4 \oplus M_2$.
\end{rmk}
However,
the following problem,
from the introduction to~\cite{IPV},
seems to be open.
(We are grateful to Narutaka Ozawa for this reference.)
\begin{pbm}\label{Pb_3402_TFrSameC}
Let $G$ and $H$ be countable torsion free groups
such that $C^*_{\mathrm{r}} (G) \cong C^*_{\mathrm{r}} (H)$.
Does it follow that $G \cong H$?
\end{pbm}
As discussed in the introduction to~\cite{IPV},
the answer is yes if $G$ and $H$ are abelian.
In much of what we have done,
one can use the algebra $l^1 (G)$ in place of $\C [G]$.
\begin{dfn}\label{D_3326_l1GpAlg}
Let $G$ be a discrete group.
We write elements of $l^1 (G)$ as functions $a \colon G \to \C$
(such that $\sum_{g \in G}  a (g)  < \I$).
We make $l^1 (G)$ into a Banach *algebra as follows.
The Banach space structure is as usual.
Multiplication is given by convolution:
for $a, b \in l^1 (G)$,
\[
(a b) (g) = \sum_{h \in G} a (h) b ( h^{1} g).
\]
The adjoint is
\[
a^* (g) = {\ov{a (g^{1})}}
\]
for $a \in l^1 (G)$.
For $g \in G$, we define $u_g \in l^1 (G)$
by $u_g (g) = 1$ and $u_g (h) = 0$ for $h \neq g$.
\end{dfn}
We give the properties of $l^1 (G)$ as a series of easy exercises.
\begin{exr}\label{Ex_3326_l1GIsAlg}
Let $G$ be a discrete group.
Prove that the operations in \Def{D_3326_l1GpAlg}
make $l^1 (G)$
into a unital Banach *algebra
whose identity is~$u_1$.
\end{exr}
The following result justifies the use of the notation $u_g$
for elements of both $\C [G]$ and $l^1 (G)$.
Using it, we normally regard $\C [G]$
as a dense subalgebra of $l^1 (G)$.
\begin{exr}\label{Ex_3326_CGTol1}
Let $G$ be a discrete group.
Prove that there is a unique algebra \hm{}
$\io \colon \C [G] \to l^1 (G)$
such that the image of the element $u_g \in \C [G]$
of \Def{D_3317_GpRing}
is the element $u_g \in l^1 (G)$
of \Def{D_3326_l1GpAlg}.
Prove that $\io$ is injective,
preserves the adjoint operation,
and has dense range.
\end{exr}
\begin{dfn}\label{D_3326_l1GRepDfn}
Let $G$ be a discrete group,
let $H$ be a Hilbert space,
and let $w \colon G \to U (H)$ be a unitary representation of~$G$.
We define ${\ov{\rh}}_w \colon l^1 (G) \to L (H)$ by
%
\begin{equation}\label{Eq_3326_l1GRpFormula}
{\ov{\rh}}_w (b)
= \sum_{g \in G} b (g) w (g)
\end{equation}
%
for $b \in l^1 (G)$.
\end{dfn}
\begin{exr}\label{Ex_3326_l1RegFromURep}
Let $G$ be a discrete group,
let $H$ be a Hilbert space,
and let $w \colon G \to U (H)$ be a unitary representation of~$G$.
Prove that the map ${\ov{\rh}}_w$
of \Def{D_3326_l1GRepDfn}
is a well defined unital *\hm{}
from $l^1 (G) \to L (H)$.
Prove that the \rpn{} $\rh_w$ of \Def{D_3317_CGRepDfn}
and the map $\io$ of Exercise~\ref{Ex_3326_CGTol1}
satisfy ${\ov{\rh}}_w \circ \io = \rh_w$.
\end{exr}
\begin{exr}\label{Ex_3326_AutoCt}
Let $G$ be a discrete group,
let $H$ be a Hilbert space,
and let $\pi \colon l^1 (G) \to L (H)$
be a unital *\hm{} (no continuity is assumed).
Prove that $\ \pi (b) \ \leq \ b \$
for all $b \in l^1 (G)$.
\end{exr}
\begin{exr}\label{Ex_3326_l1GReps}
Let $G$ be a discrete group,
and let $H$ be a Hilbert space.
Prove that the
assignment $w \mapsto {\ov{\rh}}_w$
of \Def{D_3326_l1GRepDfn}
defines a bijection from unitary representations $w \colon G \to U (H)$
to unital *\hm{s} $l^1 (G) \to L (H)$.
\end{exr}
\begin{exr}\label{Ex_3415_l1toCst}
Let $G$ be a discrete group.
Prove that the map which for $g \in G$ sends $u_g \in l^1 (G)$
to $u_g \in C^* (G)$
extends to a contractive *\hm{}
$\ld \colon l^1 (G) \to C^* (G)$
with dense range.
Further prove that
if $w \colon G \to U (H)$ is a unitary \rpn{} of $G$
on a Hilbert space~$H$,
${\ov{\rh}}_w$ is as in \Def{D_3326_l1GRepDfn},
and $\pi \colon C^* (G) \to L (H)$
is the \rpn{} of $C^* (G)$ corresponding to~$w$
(as in \Thm{T_3317_UProp}),
then $\pi \circ \ld = {\ov{\rh}}_w$.
\end{exr}
We state three important theorems about $C^* (G)$ and
$C^*_{\mathrm{r}} (G)$.
In the first and second, we consider arbitrary locally compact groups;
their full and reduced \ca{s}
are discussed in Section~\ref{Sec_LCGpCSt}.
We give a proof only for the first.
We restrict here to the case of a discrete group,
in which the ideas are exposed with less distraction,
but the proof of the crossed product generalization
(Theorem~\ref{T_4131_CPFToRed} below)
includes the case of a general locally compact group
in \Thm{T_4131_GpFToRed}.
All three involve amenability of a group.
For information on amenable groups,
including many equivalent conditions for amenability,
we refer to~\cite{Grf} or to Section~A.2 of~\cite{Wlms}.
We will use the F{\o}lner set criterion.
A discrete group $G$ is amenable \ifo{}
for every finite set $F \subset G$ and every $\ep > 0$
there is a nonempty
finite set $S \subset G$ such that
for all $g \in F$ the symmetric difference
$g S \bigtriangleup S$ satisfies
$\card (g S \bigtriangleup S) < \ep \card (S)$.
(See Theorem 3.6.1 of~\cite{Grf}.)
When $G$ is locally compact,
one uses Haar measure instead of cardinality:
if $\mu$ is a left Haar measure on~$G$,
then $G$ is amenable \ifo{}
for every compact set $F \subset G$ and every $\ep > 0$
there is a compact set $S \subset G$ such that
$\mu (S) > 0$
and $\mu (g S \bigtriangleup S) < \ep \mu (S)$
for all $g \in F$.
(See Theorem 3.6.2 of~\cite{Grf}.)
It is easy to show that the condition for a discrete group
is equivalent if the conclusion is rewritten to require
that $\card (F S \bigtriangleup S) < \ep \card (S)$.
(This will be implicit in the proof below of the
discrete case of Theorem~\ref{T_4131_GpFToRed} below.)
It is true,
but nontrivial to prove,
that the condition for a locally compact group
is equivalent if the conclusion is rewritten to require
that $\mu (F S \bigtriangleup S) < \ep \mu (S)$.
The equivalence is in neither \cite{Grf}
nor Section~A.2 of~\cite{Wlms},
but it is the main result of~\cite{EmGr}.
(Also see Theorem 3.1.1 there.)
Locally compact abelian groups are amenable.
(Combine Theorems 1.2.1 and 2.2.1 of~\cite{Grf}.)
From the conditions involving invariant means,
it is obvious that compact groups
(in particular, finite groups) are amenable.
The class of amenable locally compact groups
is closed under passage to closed subgroups
(Theorem 2.3.2 of~\cite{Grf}),
quotients by closed normal subgroups
(Theorem 2.3.1 of~\cite{Grf}),
extensions
(Theorem 2.3.3 of~\cite{Grf}),
and increasing unions
(Theorem 2.3.4 of~\cite{Grf}).
In particular,
all solvable locally compact groups are amenable,
and direct limits of discrete amenable groups are amenable.
\begin{thm}[One direction of Theorem~A.18 of~\cite{Wlms} and
Theorem 7.3.9 of~\cite{Pd1}]\label{T_4131_GpFToRed}
Let $G$ be a an amenable locally compact group.
Then the map
$\kp \colon C^* (G) \to C^*_{\mathrm{r}} (G)$
(in Corollary~\ref{C_3317_FToRed} for discrete groups;
in \Prp{P_6X15_FToRed} for general locally compact groups)
is an isomorphism.
\end{thm}
We will give a direct proof for discrete groups from the
F{\o}lner set criterion described above.
The proof for the locally compact case is very similar.
In fact,
essentially the same proof shows that for an amenable group,
the map from a full crossed product to the corresponding
reduced crossed product is an isomorphism.
See Theorem~\ref{T_4131_CPFToRed} below,
for which we do give a full proof.
Our proof does not use the machinery of positive definite functions.
This machinery is very important,
but doing without it has the advantage
that one sees the role of amenability very clearly in the proof.
It is instructive to specialize our proof
to the case of a finite group,
in which $\ep$ is not needed and the finite subsets $F$ and $S$
in the proof can both be taken to be~$G$.
\begin{proof}[Proof of Theorem~\ref{T_4131_GpFToRed}
for discrete groups]
For any unitary representation $w$ of $G$ on a Hilbert space~$H$,
let $\rh_w \colon C^* (G) \to L (H)$
be the corresponding representation of $C^* (G)$
as in \Thm{T_3317_UProp}.
We have to prove that,
for any unitary representation $w$ of $G$ on a Hilbert space~$H$,
and any $b \in C^* (G)$,
we have $\ \rh_w (b) \ \leq \ \kp (b) \$.
Let $v$ be the left regular representation of~$G$
on $l^2 (G)$.
We can rewrite the relation to be proved as
$\ \rh_w (b) \ \leq \ \rh_v (b) \$.
The main tool is the tensor product representation $v \otimes w$
as in Remark~\ref{R_4201_TensorRpn}.
It acts on the Hilbert space $l^2 (G) \otimes H$.
Throughout the proof, we identify $l^2 (G) \otimes H$
with the space $l^2 (G, H)$ of $l^2$~functions from $G$ to~$H$.
We first claim that $v \otimes w$ is unitarily equivalent
to the tensor product of $v$ and the trivial representation
of $G$ on~$H$.
Let $z \in U (l^2 (G, H) )$ be the unitary determined by
$(z \xi) (g) = w_g^* (\xi (g))$
for $\xi \in l^2 (G, H)$ and $g \in G$.
Now let $\xi \in l^2 (G, H)$
and let $g, h \in G$.
Then
\begin{align*}
\big( z (v_h \otimes w_h) \xi \big) (g)
& = w_g^* \big( [ (v_h \otimes w_h) \xi ] (g) \big)
= w_g^* \big( w_h ( \xi (h^{1} g) ) \big)
\\
& = w_{g^{1} h} (\xi (h^{1} g))
= (z \xi) (h^{1} g)
= \big( (v_h \otimes 1) z \xi \big) (g),
\end{align*}
which is the statement of the claim.
It follows that
$z \rh_{v \otimes w} (b) z^* = \rh_v (b) \otimes 1$,
so
\[
\ \rh_{v \otimes w} (b) \
= \ \rh_v (b) \otimes 1 \
= \ \rh_v (b) \.
\]
It remains to prove that
$\ \rh_w (b) \ \leq \ \rh_{v \otimes w} (b) \$.
It suffices to prove this for $b$ in the subalgebra $\C [G]$.
Thus,
there is a finite set $S \subset G$ and a family $(b_g)_{g \in S}$
of complex numbers such that $b = \sum_{g \in S} b_g u_g$.
Let $\ep > 0$.
We prove that $\ \rh_w (b) \  \ep < \ \rh_{v \otimes w} (b) \$.
\Wolog{} $\rh_w (b) \neq 0$ and $\ep < \ \rh_w (b) \$.
Choose $\xi_0 \in H$
such that
\[
\ \xi_0 \ = 1
\andeqn
\ \rh_w (b) \xi_0 \
> \ \rh_w (b) \  \frac{\ep}{2}.
\]
Set
\[
\dt = \frac{1}{\card (S)}
\left( 1  \left( \frac{\ \rh_w (b) \  \ep}{
\ \rh_w (b) \  \frac{\ep}{2}} \right)^{2} \right).
\]
Then $\dt > 0$.
The F{\o}lner set condition for amenability
(Theorem 3.6.1 of~\cite{Grf})
provides a nonempty finite subset $K \subset G$
such that
\[
\card \big( g K \bigtriangleup K \big) < \dt \card (K)
\]
for all $g \in S$.
Define $\xi \in l^2 (G, H)$
by
\[
\xi (g) = \begin{cases}
\xi_0 & g \in K
\\
0 & g \not\in K.
\end{cases}
\]
Then
$\ \xi \ = \card (K)^{1/2}$.
We estimate
$\big\ \rh_{v \otimes w} (b) \xi \big\$.
Set
\[
E = \big\{ g \in K \colon
{\mbox{$h^{1} g \in K$ for all $h \in S$}} \big\}.
\]
Then
\[
\card (K \setminus E)
\leq \sum_{h \in S} \card (K \setminus h K)
< \card (S) \dt \card (K).
\]
So $\card (E) > \big(1  \card (S) \dt \big) \card (K)$.
Moreover,
for $g \in E$ we have,
using the definition of $E$ at the third step,
\begin{align*}
\big( \rh_{v \otimes w} (b) \xi \big) (g)
& = \sum_{h \in S} b_h \big( (v_h \otimes w_h) \xi \big) (g)
\\
& = \sum_{h \in S} b_h w_h ( \xi (h^{1} g))
\\
& = \sum_{h \in S} b_h w_h \xi_0
= \rh_w (b) \xi_0.
\end{align*}
Therefore
\[
\big\ \rh_{v \otimes w} (b) \xi \big\
\geq \card (E)^{1/2} \ \rh_w (b) \xi_0 \
> \card (E)^{1/2}
\left( \ \rh_w (b) \  \frac{\ep}{2} \right),
\]
from which it follows that
\begin{align*}
\big\ \rh_{v \otimes w} (b) \big\
& \geq \frac{\card (E)^{1/2}
\left( \ \rh_w (b) \  \frac{\ep}{2} \right)}{\card (K)^{1/2}}
\\
& > \big( 1  \card (S) \dt \big)^{1/2}
\left( \ \rh_w (b) \  \frac{\ep}{2} \right)
= \ \rh_w (b) \  \ep,
\end{align*}
as desired.
\end{proof}
\begin{thm}[The other direction of Theorem~A.18 of~\cite{Wlms} and
Theorem 7.3.9 of~\cite{Pd1}]\label{T_3320_FRIso}
Let $G$ be a locally compact group.
If the standard \hm{} $C^* (G) \to C^*_{\mathrm{r}} (G)$
is an isomorphism,
then $G$ is amenable.
\end{thm}
\begin{thm}\label{T_3320_RedNuc}
Let $G$ be a discrete group.
Then \tfae:
\begin{enumerate}
\item\label{T_3320_RedNuc_Amen}
$G$ is amenable.
\item\label{T_3320_RedNuc_Red}
$C^*_{\mathrm{r}} (G)$ is nuclear.
\item\label{T_3320_RedNuc_Full}
$C^* (G)$ is nuclear.
\end{enumerate}
\end{thm}
The equivalence of the first two conditions
(and many others) is contained in Theorem 2.6.8 of~\cite{BrOz}.
(The definition of amenability used there is existence
of an invariant mean.
See Definition 2.6.1 of~\cite{BrOz}.)
If $G$ is amenable,
then $C^* (G)$ is nuclear
because $C^* (G) \cong C^*_{\mathrm{r}} (G)$.
If $C^* (G)$ is nuclear,
then $C^*_{\mathrm{r}} (G)$ is nuclear
because it is a quotient of $C^* (G)$.
Theorem~\ref{T_3320_RedNuc} does not hold without discreteness.
Even the full group \ca{s} of connected semisimple Lie groups
are not only type~I but even CCR:
the image of every irreducible representation
is exactly the compact operators.
This fact follows from Theorem~5 on page 248 of~\cite{Hrc}.
Not only are most such groups not amenable;
many even have Kazhdan's Property~(T).
Example: ${\operatorname{SL}}_3 (\R)$.
Theorem~2 on page~47 of~\cite{Pkn}
describes exactly when the full group \ca{}
of a connected simply connected Lie group is CCR,
and Theorem~1 on page~39 of~\cite{Pkn}
gives some conditions under which the full group \ca{}
of a connected simply connected Lie group has type~I.
Although we say very little about von Neumann algebras
in these notes,
we want to at least mention the group von Neumann algebra.
\begin{dfn}\label{D_3401_vNG}
Let $G$ be a discrete group.
Regard $C^*_{\mathrm{r}} (G)$ as a subalgebra of $L (l^2 (G))$,
as in \Def{D_3317_RedGpCStar}.
We define the {\emph{group von Neumann algebra}}
$W^*_{\mathrm{r}} (G)$ to be the closure of $C^*_{\mathrm{r}} (G)$
in the weak operator topology on $L (l^2 (G))$.
\end{dfn}
Equivalently,
using the notation of \Def{D_3317_RedGpCStar}
and taking $v$ to be the left regular \rpn{} of~$G$,
the algebra $W^*_{\mathrm{r}} (G)$
is the closure of $\rh_v (\C [G])$
in the weak operator topology on $L (l^2 (G))$.
(This is the definition given in the introduction
to Section VII.3 of~\cite{Tk_Bk2}.
Also see Definition V.7.4 of~\cite{Tks}.)
We can also write $W^*_{\mathrm{r}} (G) = \rh_v (\C [G])''$.
The notation follows a suggestion of Simon Wassermann.
It was previously common to write $W^* (G)$,
which unfortunately suggests a relation with $C^* (G)$
instead of with $C^*_{\mathrm{r}} (G)$.
These days, the notation $L (G)$
(or ${\mathcal{L}} (G)$)
is much more common.
The group von Neumann algebra carries much less information about the
group than its full or reduced \ca.
For example,
although we will not prove this here,
it is not difficult to show that if $G$ is discrete abelian,
then $W^*_{\mathrm{r}} (G) \cong L^{\I} \big( {\widehat{G}} \big)$.
In particular,
these algebras are the same
for every countable infinite discrete abelian group.
This is much worse than the situation
for group \ca{s},
as described in Remark~\ref{R_3402_GpSameCst}.
We can give some description of the elements of the
reduced \ca{} and von Neumann algebra of a discrete group.
The term in the following definition is motivated by the case
$G = \Z$,
and the ideas are based on a lecture of Nate Brown.
We think of elements of $L (l^2 (\Z))$ as being given by
infinite matrices $a = (a_{j, k})_{j, k \in \Z}$.
The main diagonal consists of the elements
$a_{j, j}$ for $j \in \Z$,
and the other diagonals are gotten by fixing $m \in \Z$
and taking the elements
$a_{j, \, j + m}$ for $j \in \Z$.
That is, they are the elements $a_{j, k}$ with $j  k$
constant.
We begin with notation for matrix elements of an operator
$a \in L (l^2 (S))$.
\begin{dfn}\label{D_3416_MatElt}
Let $S$ be a set.
For $s \in S$ let $\dt_s \in l^2 (S)$
be the standard basis vector associated with~$s$.
For $a \in L (l^2 (S))$
and $s, t \in S$,
we define the {\emph{matrix coefficient}}
$a_{s, t}$ of $a$ by
$a_{s, t} = \langle a \dt_t, \dt_s \rangle$.
\end{dfn}
\begin{rmk}\label{R_3416_OnMatC}
The indexing in \Def{D_3416_MatElt}
is consistent with the usual conventions
for entries of finite matrices.
For example,
let $s_0, t_0 \in S$,
and let $v \in L (l^2 (S))$
be the partial isometry determined by $v \dt_{t_0} = \dt_{s_0}$
and $v \dt_t = 0$ for $t \in S \setminus \{ t_0 \}$.
Then $v_{s_0, t_0} = 1$ and $v_{s, t} = 0$ for all other
pairs $(s, t) \in S \times S$.
Moreover,
for general $a \in L (l^2 (S))$ and $t \in S$,
the element $\xi = a \dt_t \in l^2 (S)$
is determined by the relations
$\xi (s) = \langle \xi, \dt_s \rangle = a_{s, t}$
for all $s \in S$.
In particular,
$a \dt_t = \sum_{s \in S} a_{s, t} \dt_s$
with convergence in norm in $l^2 (S)$.
Finally,
we note that $(a^*)_{s, t} = {\ov{a_{t, s}}}$
for all $s, t \in S$.
\end{rmk}
\begin{dfn}\label{D_3317_ConstOnDiag}
Let $G$ be a discrete group.
Let $a \in L (l^2 (G))$,
and write $a = (a_{g, h})_{g, h \in G}$.
We say that $a$ is
{\emph{constant on diagonals}}
if $a_{g, h} = a_{s, t}$
whenever $g, h, s, t \in G$ satisfy $g h^{1} = s t^{1}$.
\end{dfn}
\begin{thm}\label{T_3317_WStCOnD}
Let $G$ be a discrete group.
Then
\[
W^*_{\mathrm{r}} (G)
= \big\{ a \in L (l^2 (G)) \colon
{\mbox{$a$ is constant on diagonals}} \big\}.
\]
\end{thm}
\begin{proof}
Let $M \subset L (l^2 (G))$ be the set of all $a \in L (l^2 (G))$
which are constant on diagonals.
% in the sense of \Def{D_3317_ConstOnDiag}.
Let $N \subset L (l^2 (G))$ be the set of all $b \in L (l^2 (G))$
such that
$b_{g, h} = b_{s, t}$
whenever $g, h, s, t \in G$ satisfy $g^{1} h = s^{1} t$.
(Note the different placement of the inverses.)
Let $v \colon G \to U (l^2 (G))$ be the left regular representation
(\Def{D_3407_DRegRep}),
and let $w \colon G \to U (l^2 (G))$
be the right regular \rpn,
given by $(w (g) \xi) (h) = \xi (h g)$
for $g, h \in G$ and $\xi \in l^2 (G)$.
We first claim that if $a \in M$ and $b \in N$,
then $a b = b a$.
Fix $s, t \in G$;
we prove that
$\langle a b \dt_s, \dt_t \rangle = \langle b a \dt_s, \dt_t \rangle$.
Using several parts of Remark~\ref{R_3416_OnMatC}
at the second step,
we get
\[
\langle a b \dt_s, \dt_t \rangle
= \langle b \dt_s, a^* \dt_t \rangle
= \left\langle \ssum{g \in G} b_{g, s} \dt_g, \,
\ssum{g \in G} {\ov{a_{t, g}}} \dt_g \right\rangle
= \ssum{g \in G} a_{t, g} b_{g, s}.
\]
Similarly,
we get the first step of the following calculation.
The second step follows from the definitions of $a \in M$
and $b \in N$, and the third step is a change of variables:
\[
\langle b a \dt_s, \dt_t \rangle
= \sum_{g \in G} a_{g, s} b_{t, g}
= \sum_{g \in G} a_{t, s g^{1} t} b_{s g^{1} t, s}
= \sum_{g \in G} a_{t, g} b_{g, s}
= \langle a b \dt_s, \dt_t \rangle.
\]
This proves the claim.
We next claim that $w (G)' = M$.
Let $a \in L (l^2 (G))$.
We have to show that $a w (g) = w (g) a$ for all $g \in G$
\ifo{} $a$ is constant on diagonals.
For $g, h, k \in G$, we compute
\[
(a w (g))_{h, k}
= \langle a w (g) \dt_k, \, \dt_h \rangle
= \langle a \dt_{k g^{1}}, \, \dt_h \rangle
= a_{h, k g^{1}}
\]
and similarly
\[
(w (g) a)_{h, k}
= \langle w (g) a \dt_k, \, \dt_h \rangle
= \langle a \dt_k, \, w (g)^* \dt_h \rangle
= a_{h g, k}.
\]
It is easy to check that $a_{h, k g^{1}} = a_{h g, k}$
for all $g, h, k \in G$
\ifo{} $a$ is constant on diagonals.
The claim follows.
Similarly, one proves that $v (G)' = N$.
Since $w (G)$ and $v (G)$ are both closed under adjoints,
it follows that $M$ and $N$ are von Neumann algebras.
It is immediate from the definitions that $v (g) w (h) = w (h) v (g)$
for all $g, h \in G$.
Therefore,
using the first claim at the second step,
\[
v (G)' = N \subset M'.
\]
Take commutants throughout to get
\[
v (G)'' = N' \supset M'' = M.
\]
Since also $v (G) \subset w (G)' = M$,
we use the definition at the first step to get
$W^*_{\mathrm{r}} (G) = v (G)'' = M$,
as was to be proved.
\end{proof}
The following proposition gives ``coefficients'' of elements
of $C^*_{\mathrm{r}} (G)$.
It actually works not just for $C^*_{\mathrm{r}} (G)$
but for $W^*_{\mathrm{r}} (G)$,
once one has extended the tracial state on $C^*_{\mathrm{r}} (G)$
to $W^*_{\mathrm{r}} (G)$.
\begin{prp}\label{P_3408_Coeffs}
Let $G$ be a discrete group,
let $b \in C^*_{\mathrm{r}} (G) \subset L (l^2 (G))$,
and let $g \in G$.
For $g \in G$,
let $\dt_g \in l^2 (G)$ be the standard basis vector
corresponding to~$g$.
Let $\ta \colon C^*_{\mathrm{r}} (G) \to \C$
be the tracial state of \Thm{T_3318_RedTr}.
Then the following three numbers are equal:
\begin{enumerate}
\item\label{P_3408_Coeffs_Tau}
$\ta (b u_g^*)$.
\item\label{P_3408_Coeffs_ScPrd}
$\langle b \dt_1, \dt_g \rangle$.
\item\label{P_3408_Coeffs_MCoeff}
The constant value $\ld_g$ that the matrix of $b \in L (l^2 (G))$
has on the diagonal consisting of those elements $b_{s, t}$
for $s, t \in G$ such that $s t^{1} = g$.
\end{enumerate}
\end{prp}
\begin{proof}
The equation $\ld_g = \langle b \dt_1, \dt_g \rangle$
comes from the formula for the coefficients
$b_{s, t}$,
namely $b_{s, t} = \langle b \dt_t, \dt_s \rangle$
for $s, t \in G$.
We prove that $\ta (b u_g^*) = \langle b \dt_1, \dt_g \rangle$.
By linearity and continuity,
we may assume that $b \in \C [G]$.
Thus, we may assume that
$b = \sum_{h \in G} b_h u_h$ with $b_g = 0$ for all but finitely
many $h \in G$.
Then $\ta (b u_g^*) = b_g$.
Also,
letting $v \colon G \to U (l^2 (G))$ be the left regular representation
(\Def{D_3407_DRegRep}),
the operator
$\rh_v (b) \in L (l^2 (G))$ acts as
$\sum_{h \in G} b_h v (h)$,
so
\[
\langle b \dt_1, \dt_g \rangle
= \left\langle \ssum{h \in G} b_h v (h) \dt_1, \, \dt_g \right\rangle
= \sum_{h \in G} b_h \langle \dt_h, \dt_g \rangle
= b_g.
\]
This completes the proof.
\end{proof}
The last part of the proof above is simpler if we remember
the proof of \Thm{T_3318_RedTr}.
We defined $\ta$ by the formula
$\ta (b) = \langle b \dt_1, \dt_1 \rangle$.
So, using the trace property at the first step,
we have
\[
\ta (b u_g^*)
= \ta (u_g^* b)
= \langle v (g)^* b \dt_1, \dt_1 \rangle
= \langle b \dt_1, v (g) \dt_1 \rangle
= \langle b \dt_1, \dt_g \rangle.
\]
We can now think of an element $b \in C^*_{\mathrm{r}} (G)$
as a formal sum ``$b = \sum_{g \in G} b_g u_g$''.
We emphasize that,
in general, this sum is {\textbf{only formal}}.
It does have one good feature.
\begin{prp}\label{P_3408_MeaningCffs}
Let $G$ be a discrete group,
let $\ta \colon C^*_{\mathrm{r}} (G) \to \C$
be the tracial state of \Thm{T_3318_RedTr},
and let $b \in C^*_{\mathrm{r}} (G)$.
Suppose $\ta (b u_g) = 0$ for all $g \in G$.
Then $b = 0$.
\end{prp}
\begin{proof}
Recall from \Thm{T_3318_RedTr}
that if $a \in C^*_{\mathrm{r}} (G)$
and $\ta (a^* a) = 0$, then $a = 0$.
It therefore suffices to show that for all $a \in C^*_{\mathrm{r}} (G)$
and all $g \in G$,
we have $\ta (a^* a) \geq  \ta (a u_g) ^2$.
By continuity of~$\ta$ and density of $\C [G]$ in $C^*_{\mathrm{r}} (G)$,
it suffices to prove this inequality for $a \in \C [G]$.
So assume that $a = \sum_{h \in G} a_h u_h$
with $a_h \in \C$ for all $h \in G$ and $a_h = 0$ for all but finitely
many $h \in G$.
Then,
using $\ta (u_h^* u_k) \neq 0$ only if $h = k$ at the second step,
\[
\ta ( a^* a)
= \ta \left( \ssum{h, k \in G}
{\ov{ a_{h} }} a_k u_h^* u_k \right)
= \sum_{k \in G}  a_k ^2
\geq  a_{g^{1}} ^2
=  \ta (a u_g) ^2.
\]
This completes the proof.
\end{proof}
Proposition~\ref{P_3408_MeaningCffs} is useful,
but it is quite weak.
There are, in fact, many difficulties
in understanding group \ca{s}.
\begin{rmk}\label{R:WhatIsIn}
Consider the special case $G = \Z$.
Then
$C^* (G)$
is isomorphic to $C (S^1)$,
and the map $\ld \colon l^1 (\Z) \to C (S^1)$
of Exercise~\ref{Ex_3415_l1toCst}
is the Fourier series map:
for $a = (a_n)_{n \in \N} \in l^1 (\Z)$,
its image $\ld (a)$ is the function
\[
\ld (a) (\zt) = \sum_{n \in \Z} a_n \zt^n
\]
for $\zt \in S^1$.
This looks more familiar
when we identify $C (S^1)$ with the
set of $2 \pi$periodic \ct{} functions on $\R$:
it is
\[
\ld (a) (t) = \sum_{n \in \Z} a_n e^{i n t}
\]
for $t \in \R$.
Every $f \in C (S^1)$ has a Fourier series.
Letting $\mu$ be normalized arc length measure on~$S^1$,
its coefficients are given by
\[
a_n = \int_{S^1} f (\zt) \zt^{n} \, d \mu (\zt).
\]
It is well known that $\limi{n} a_n = \lim_{n \to  \I} a_n = 0$,
whence $a = (a_n)_{n \in \N} \in C_0 (\Z)$.
However:
\begin{enumerate}
\item\label{R:WhatIsIn1}
We know of no good description of which sequences $a \in C_0 (\Z)$
are the Fourier coefficients of some $f \in C (S^1)$.
Since the Fourier series map is a bijection from $l^2 (\Z)$ to
$L^2 (S^1)$,
we do know that any such $a$ must be in $l^2 (\Z)$.
But in fact
the Fourier coefficients of every element of $L^{\I} (S^1)$,
which is the group von Neumann algebra of~$\Z$,
are also in $l^2 (\Z)$,
for the same reason.
We get essentially no useful information out of a criterion
for membership in a group \ca{} which is satisfied by all
elements in the group von Neumann algebra.
\item\label{R:WhatIsIn2}
For $a \in l^1 (\Z)$,
or even in $\C [ \Z ]$,
we know of no general way
to compute the norm $\ \ld (a) \$ in terms of~$a$,
except by directly carrying out the computation of
\[
\sup_{\zt \in S^1} \left \ssum{n \in \Z} a_n \zt^n \right.
\]
There are of course a few specific cases
in which computations can be done.
For example, let $\dt_n \in \C [ \Z ] \subset l^1 (\Z)$
be the element which takes the value $1$ at~$n$
and is zero elsewhere.
Then $\dt_n$ is unitary in $l^1 (\Z)$ and therefore also in $C^* (\Z)$.
So $\ \ld (\dt_n) \ = 1$.
Computations of norms of some special elements of
reduced group \ca{s} can be found in~\cite{AO}.
\item\label{R:WhatIsIn3}
Let $\dt_n$ be as in~(\ref{R:WhatIsIn2}),
and set $z_n = \ld (\dt_n)$,
which is the function $z_n (\zt) = \zt^n$
for $\zt \in S^1$.
For $f \in C (S^1)$, its sequence $a$ of Fourier coefficients
gives a formal series $\sum_{n \in \Z} a_n z_n$ for~$f$.
However, this series need not converge to~$f$
(or, indeed, to anything) in $C (S^1)$.
In more familiar terms,
this is the statement that the Fourier series of a \cfn{} need not
converge uniformly.
\item\label{R:WhatIsIn4}
In fact, the series in~(\ref{R:WhatIsIn3})
need not even converge in the weak operator topology
on the von Neumann algebra,
which here is isomorphic to $L^{\infty} (S^1)$.
See Proposition~1 and the following remark in~\cite{Mr}.
We warn the reader that erroneous claims for the convergence
of this series have been made in some well known textbooks,
such as in 7.11.2 of~\cite{Pd1}
and before Proposition V.7.6 of~\cite{Tks},
as well as in some papers.
(See~\cite{Mr} for details.)
There is a topology, described in~\cite{Mr},
in which one does have convergence.
(We are grateful to Stuart White for pointing out this issue
and providing the reference to~\cite{Mr}.)
Note, though,
that the Ces\`{a}ro means of the Fourier series
of a \cfn{} $f$ do converge uniformly to~$f$.
See~2.5 and Theorem 2.11 in Chapter~1 of~\cite{Ktzn}.
This idea can be generalized substantially,
to countable amenable groups and somewhat beyond,
and to reduced crossed products rather than just
reduced group \ca{s}.
In~\cite{BdsCti},
see Sections~5,
and for example Theorem 5.6,
which considers reduced crossed products
by general countable amenable groups.
\end{enumerate}
\end{rmk}
When $G$ is abelian,
the description of $C^* (G)$ as $C_0 \big( {\widehat{G}} \big)$
is a concrete description of a different sort which is
extremely useful.
There are other groups,
particularly various semisimple Lie groups,
for which there are descriptions
of $C^* (G)$ or $C^*_{\mathrm{r}} (G)$
which might be considered similar in spirit
(although they are much more complicated).
However, for many groups,
including many countable amenable groups,
no concrete description of $C^* (G)$ or $C^*_{\mathrm{r}} (G)$
is known.
\begin{rmk}\label{R_3408_NotAmen}
The situation for $C^* (G)$ when $G$ is not amenable is even worse
than is suggested by Remark~\ref{R:WhatIsIn}.
For $a \in C^* (G)$,
we can still use
the \hm{} $\kp \colon C^* (G) \to C^*_{\mathrm{r}} (G)$
to define ``coefficients'' $a_g$ for $g \in G$,
by $a_g = \ta (\kp (a) u_g^*)$.
However, since there are nonzero elements $a \in C^* (G)$
such that $\kp (a) = 0$,
these coefficients no longer even determine $a$ uniquely.
\end{rmk}
When we get to them,
we will see that the situation can be worse
for crossed products.
See Remark~\ref{R_6X11_ComputeCrPrd}.
It seems appropriate to point out that,
despite the issues presented in Remark~\ref{R:WhatIsIn}
and Remark~\ref{R_3408_NotAmen},
in some ways $C^* (G)$
(in which we don't know the elements as functions on~$G$,
and where the natural convergence can fail)
is better behaved that $l^1 (G)$.
For example,
again take $G = \Z$.
We can certainly write down an explicit description
of all the elements of $l^1 (\Z)$.
However, the (closed) ideal structure of $l^1 (\Z)$
is very complicated,
and not completely known,
while the ideal structure of $C^* (\Z)$
is very simple:
the closed ideals are in bijective
order reversing correspondence with the closed subsets of~$S^1$.
According to Theorem 42.21 of~\cite{HwRs2}
(see Definition 39.9 of~\cite{HwRs2} for the terminology),
and the additional statements in 42.26 of~\cite{HwRs2},
the phenomenon of intractable ideal structure
occurs in $L^1 (G)$ for every locally compact but noncompact
abelian group~$G$.
Another example is the computation
of the Ktheory for crossed products.
% 999
% which we say a little about later.
% Need Forward reference if do this
It turns out that the computation of the Ktheory of crossed products by~$\Z$,
and even by nonabelian free groups,
is easier than the computation of the Ktheory
of crossed products by $\Z / 2 \Z$.
\section{Simplicity of the Reduced C*Algebra of a Free
Group}\label{Sec_CStrFn}
\indent
In this short section,
we prove that $C^*_{\mathrm{r}} (F_n)$ is simple
and has a unique tracial state for $n \in \{ 2, 3, \ldots, \I \}$.
We follow the original proof of Powers~\cite{Pw},
with a slight simplification.
A differently organized proof
can be found in Section VII.7 of~\cite{Dvd}.
This result is not in the main direction of these notes,
which are mainly concerned with the structure
of crossed products by much smaller
(in particular, amenable)
groups in situations in which the action is free
in some sense.
It is included to provide a contrast to \Thm{T:AS},
a simplicity theorem
which requires that the action be essentially free,
and the observation that if $G$ has more than one element,
then $C^* (G)$ is never simple
(since the one dimensional trivial representation gives a
nontrivial \hm{} $C^* (G) \to \C$).
This result is the original one of its type.
Simplicity of $C^*_{\mathrm{r}} (G)$ is now known
for many (nonamenable) countable groups~$G$.
For recent definitive results,
see~\cite{BKKO}.
\begin{ntn}\label{L_3301_FreeGpNtn}
Let $n \in \{ 2, 3, \ldots, \I \}$.
We let $F_n$ denote the free group on $n$ generators,
and we call the generators $\gm_1, \gm_2, \ldots, \gm_n$
(or $\gm_1, \gm_2, \ldots$ when $n = \I$).
We let $\ta \colon C^*_{\mathrm{r}} (F_n) \to \C$
be the tracial state
of \Thm{T_3318_RedTr}.
For $g \in F_n$,
we let $\dt_g \in l^2 (F_n)$ be the corresponding standard
basis vector.
We take a {\emph{reduced word}} in the generators
to be an expression of the form
%
\begin{equation}\label{Eq_3302_Word}
\gm_{j (1)}^{l (1)} \cdot \gm_{j (2)}^{l (2)}
\cdots \gm_{j (m)}^{l (m)}
\end{equation}
%
with
%
\begin{equation}\label{Eq_3302_mCond}
m \in \Nz,
\qquad
j (1) \neq j (2),
\qquad
j (2) \neq j (3),
\qquad
\ldots,
\qquad
j (m  1) \neq j (m),
\end{equation}
%
and
%
\begin{equation}\label{Eq_3302_lj}
l (1), \, l (2), \, \ldots, \, l (m) \in \Z \setminus \{ 0 \}.
\end{equation}
%
When $m = 0$,
we get the empty word,
representing $1 \in F_n$.
We recall that every element of $F_n$ is represented by a
unique reduced word.
For $m \neq 0$,
we say that the reduced word~(\ref{Eq_3302_Word})
begins with $\gm_{j (1)}^{l (1)}$
and ends with $\gm_{j (m)}^{l (m)}$.
\end{ntn}
\begin{lem}[Lemma~4 of~\cite{Pw}]\label{L_3301_Words}
Let $n \in \{ 2, 3, \ldots, \I \}$.
Let $s \in \N$
and let $g_1, g_2, \ldots, g_s \in F_n \setminus \{ 1 \}$.
Then there exists $k \in \Z$ such that,
for $r = 1, 2, \ldots, s$,
the reduced word representing $\gm_1^k g_r \gm_1^{k}$
begins and ends with nonzero powers of $\gm_1$.
\end{lem}
\begin{proof}
We renumber the elements $g_1, g_2, \ldots, g_s$
so that there is $s_0 \leq s$
such that $g_1, g_2, \ldots, g_{s_0}$
are not powers of $\gm_1$
and $g_{s_0 + 1}, g_{s_0 + 2}, \ldots, g_s$
are powers of $\gm_1$.
For $r = 1, 2, \ldots, s_0$,
the element $g_r$ is then given by a reduced word of the form
\[
g_r
= \gm_1^{\mu_r} \gm_{j_r (1)}^{l_r (1)} \cdot \gm_{j_r (2)}^{l_r (2)}
\cdots \gm_{j_r (m_r)}^{l (m_r)} \gm_1^{\nu_r}
\]
with
$\gm_{j_r (1)}^{l_r (1)} \cdot \gm_{j_r (2)}^{l_r (2)}
\cdots \gm_{j_r (m_r)}^{l (m_r)}$
as in (\ref{Eq_3302_Word}), (\ref{Eq_3302_mCond}),
and~(\ref{Eq_3302_lj}),
with $m_r \geq 1$,
with $j_r (1) \neq 1$ and $j_{m_r} (1) \neq 1$,
and with $\mu_r, \nu_r \in \Z$.
If $\mu_r = 0$ or $\nu_r = 0$,
the corresponding term in~$g_r$ is absent.
For $r = s_0 + 1, \, s_0 + 2, \, \ldots, \, s$,
there is $\nu_r \in \Z$ such that $g_r = \gm_1^{\nu_r}$,
and $\nu_r \neq 0$ since $g_r \neq 1$.
It is immediate that any $k \in \Z$ such that
\[
k \not\in \big\{  \mu_1, \,  \mu_2, \, \ldots, \,  \mu_{s_0}, \,
\nu_1, \, \nu_2, \, \ldots, \, \nu_{s_0} \big\}
\]
will satisfy the conclusion of the lemma.
\end{proof}
The following lemma generalizes Lemma~3 of~\cite{Pw}.
\begin{lem}\label{L_3301_MvSubsp}
Let $H$ be a Hilbert space,
let $E \subset H$ be a closed subspace,
let $p \in L (H)$ be the orthogonal projection onto~$E$,
and let $a \in L (H)$ satisfy $a (E^{\perp}) \subset E$.
Then for all $\xi, \et \in H$ we have
\[
 \langle \xi, a \et \rangle 
\leq \ a \ \big( \ p \xi \ \cdot \ p \et \
+ \ p \xi \ \cdot \ (1  p) \et \
+ \ (1  p) \xi \ \cdot \ p \et \ \big).
\]
\end{lem}
\begin{proof}
We expand
\[
 \langle \xi, a \et \rangle 
\leq  \langle p \xi, a p \et \rangle 
+  \langle p \xi, a (1  p) \et \rangle 
+  \langle (1  p) \xi, a p \et \rangle 
+  \langle (1  p) \xi, a (1  p) \et \rangle .
\]
By hypothesis,
the last term is zero.
Estimate
\[
 \langle p \xi, a p \et \rangle 
\leq \ p \xi \ \cdot \ p \et \ \cdot \ a \,
\qquad
 \langle p \xi, a (1  p) \et \rangle 
\leq \ p \xi \ \cdot \ (1  p) \et \ \cdot \ a \,
\]
and
\[
 \langle (1  p) \xi, a p \et \rangle 
\leq \ (1  p) \xi \ \cdot \ p \et \ \cdot \ a \
\]
to complete the proof.
\end{proof}
\begin{lem}\label{L_3301_SqEst}
Let $M \in \N$
and let
\[
\ld_1, \ld_2, \ldots, \ld_M, \mu_1, \mu_2, \ldots, \mu_M \in \R
\]
be positive numbers such that $\sum_{m = 1}^M \ld_m^2 \leq 1$
and $\sum_{m = 1}^M \mu_m^2 \leq 1$.
Then
\[
\sum_{m = 1}^M \ld_m \leq \sqrt{M},
\qquad
\sum_{m = 1}^M \mu_m \leq \sqrt{M},
\andeqn
\sum_{m = 1}^M \ld_m \mu_m \leq 1.
\]
\end{lem}
\begin{proof}
Define $\ld, \mu, \xi \in \C^M$ by
\[
\ld = (\ld_1, \ld_2, \ldots, \ld_M),
\,\,\,\,\,
\mu = (\mu_1, \mu_2, \ldots, \mu_M),
\andeqn
\xi = (1, 1, \ldots, 1).
\]
Using the CauchySchwarz inequality at the second step,
we have
\[
\sum_{m = 1}^M \ld_m
= \langle \ld, \xi \rangle
\leq \ \ld \_2 \cdot \ \xi \_2
= \left( \sum_{m = 1}^M \ld_m^2 \right)^{1/2} \cdot \sqrt{M}
\leq \sqrt{M}.
\]
This proves the first inequality.
The proof of the second is the same,
and the third follows by applying the CauchySchwarz inequality
to $\langle \ld, \mu \rangle$.
\end{proof}
The following result is our substitute
for Lemma~5 of~\cite{Pw},
and the proof is essentially the same.
However, we need not restrict to selfadjoint elements.
Our statement includes that of Theorem~1 of~\cite{Pw},
without using the iteration
step in Lemma~6 of~\cite{Pw}.
The proof obviously implies that the result holds
simultaneously for all elements of any finite set
in $C^*_{\mathrm{r}} (F_n)$.
\begin{lem}\label{L_3301_RedNorm}
Let $n \in \{ 2, 3, \ldots, \I \}$.
Let $a \in C^*_{\mathrm{r}} (F_n)$
and let $\ep > 0$.
Then there exist $M \in \N$
and $h_1, h_2, \ldots, h_m \in F_n$
such that the linear map
$T \colon C^*_{\mathrm{r}} (F_n) \to C^*_{\mathrm{r}} (F_n)$,
defined by
\[
T (b) = \frac{1}{M} \sum_{m = 1}^M u_{h_m} b u_{h_m}^*,
\]
satisfies $\ T (a)  \ta (a) \cdot 1 \ < \ep$.
\end{lem}
\begin{proof}
We first suppose that
$a \in
\spn \big( \big\{ u_g \colon g \in F_n \setminus \{ 1 \} \big\} \big)$.
That is, there are $s \in \N$,
$g_1, g_2, \ldots, g_s \in F_n \setminus \{ 1 \}$,
and $\ld_1, \ld_2, \ldots, \ld_s \in \C$
such that $a = \sum_{r = 1}^s \ld_j u_{g_r}$.
Then $\ta (a) = 0$,
and we must find $T$ of the form described in
the conclusion such that $\ T (a) \ < \ep$.
We may clearly assume $a \neq 0$.
Choose $M \in \N$
such that
\[
M > \frac{\ep^2}{9 \ a \^2}.
\]
Choose $k \in \Z$ as in Lemma~\ref{L_3301_Words},
with $g_1, g_2, \ldots, g_s$ as given.
For $m = 1, 2, \ldots, M$,
define $h_m = \gm_2^m \gm_1^k$.
Then,
for $r = 1, 2, \ldots, s$,
the reduced word representing $h_m g_r h_m^{1}$
begins with $\gm_2^m$ and ends with $\gm_2^{m}$.
Let $S_m \subset F_n$ be the set of
all $g \in F_n$
for which the reduced word representing $g$ begins with $\gm_2^m$.
Let $E_m \subset l^2 (F_n)$ be
$E_m = {\ov{\spn}} ( \{ \dt_g \colon g \in S_m \} )$.
For any $g \in F_n \setminus S_m$,
in the product $h_m g_r h_m^{1} g$ the factor
$\gm_2^{m}$ at the end of $h_m g_r h_m^{1}$ does not completely
cancel,
so the immediately preceding nonzero power of $\gm_1$
is still present in the reduced word representing $h_m g_r h_m^{1} g$.
One can check that this word must then still begin with $\gm_2^m$.
We have shown that $h_m g_r h_m^{1} (F_n \setminus S_m) \subset S_m$.
It follows that
$u_{h_m} u_{g_r} u_{h_m}^* (E_m^{\perp}) \subset E_m$.
Since this is true for $r = 1, 2, \ldots, s$,
it follows that $u_{h_m} a u_{h_m}^* (E_m^{\perp}) \subset E_m$.
Let
$T \colon C^*_{\mathrm{r}} (F_n) \to C^*_{\mathrm{r}} (F_n)$
be defined as in the statement of the lemma,
with this choice of $M$ and $h_1, h_2, \ldots, h_m$.
Let $\xi, \et \in l^2 (F_n)$ satisfy $\ \xi \, \ \et \ \leq 1$.
Let $p_m \in L (H)$ be the orthogonal projection onto~$E_m$.
The spaces $E_1, E_2, \ldots, E_M$ are orthogonal,
so
%
\begin{equation}\label{Eq_3302_N1}
\sum_{m = 1}^M \ p_m \xi \^2 \leq \ \xi \^2 = 1
\andeqn
\sum_{m = 1}^M \ p_m \et \^2 \leq \ \et \^2 = 1.
\end{equation}
%
Using \Lem{L_3301_MvSubsp} at the second step,
and (\ref{Eq_3302_N1}) and \Lem{L_3301_SqEst} at the fifth step,
we then have
\begin{align*}
 \langle \xi, T (a) \et \rangle 
& = \frac{1}{M} \left \sum_{m = 1}^M
\langle \xi, u_m a u_m^* \et \rangle \right
\\
& \leq \frac{1}{M} \sum_{m = 1}^M
\ a \ \big( \ p_m \xi \ \cdot \ p_m \et \
+ \ p_m \xi \ \cdot \ (1  p_m) \et \
+ \ (1  p_m) \xi \ \cdot \ p_m \et \ \big)
\\
& \leq \frac{1}{M} \sum_{m = 1}^M
\ a \ \big( \ p_m \xi \ \cdot \ p_m \et \
+ \ p_m \xi \ + \ p_m \et \ \big)
\\
& = \frac{\ a \}{M} \left(
\sum_{m = 1}^M \ p_m \xi \ \cdot \ p_m \et \
+ \sum_{m = 1}^M \ p_m \xi \
+ \sum_{m = 1}^M \ p_m \et \ \right)
\\
& \leq \frac{\ a \}{M} \big( 1 + \sqrt{M} + \sqrt{M} \big)
% = \ a \ \left( \frac{1}{M} + \frac{2}{\sqrt{M}} \right).
\leq \frac{3 \ a \}{\sqrt{M}}.
\end{align*}
Since $\xi, \et \in l^2 (F_n)$ are arbitrary elements of norm~$1$,
it follows that
\[
\ T (a) \
\leq \frac{3 \ a \}{\sqrt{M}}
< \ep.
\]
The special case
$a \in
\spn \big( \big\{ u_g \colon g \in F_n \setminus \{ 1 \} \big\} \big)$
has been proved.
Next, suppose that
$a \in \spn \big( \{ u_g \colon g \in F_n \} \big)$.
Then $b = a  \ta (a) \cdot 1$
is in $\spn \big( \{ u_g \colon g \in F_n \setminus \{ 1 \} \big)$,
so there is $T$ of the form in the conclusion
such that $\ T (b) \ < \ep$.
One easily checks that $T (1) = 1$.
Therefore
\[
\ T (a)  \ta (a) \cdot 1 \
= \ T (a  \ta (a) \cdot 1) \
< \ep.
\]
Finally,
we consider an arbitrary element $a \in C^*_{\mathrm{r}} (F_n)$.
Choose $b \in \spn \big( \{ u_g \colon g \in F_n \big)$
such that $\ b  a \ < \frac{\ep}{3}$.
The previous paragraph provides $T$ of the form in the conclusion
such that $\ T (b)  \ta (b) \cdot 1 \ < \frac{\ep}{3}$.
It is easy to check that $\ T \ \leq 1$.
Therefore
\[
\ T (a  \ta (a) \cdot 1) \
\leq \ T (a  b) \
+ \big\ T ( (\ta (a)  \ta (b)) \cdot 1 ) \big\
+ \ T (b)  \ta (b) \cdot 1 \
< \frac{\ep}{3} + \frac{\ep}{3} + \frac{\ep}{3}
= \ep.
\]
This completes the proof of the lemma.
\end{proof}
\begin{thm}[Theorem~2 of~\cite{Pw}]\label{T_3301_CStRFnSimp}
Let $n \in \{ 2, 3, \ldots, \I \}$.
Then $C^*_{\mathrm{r}} (F_n)$ is simple.
\end{thm}
\begin{proof}
Let $I \subset C^*_{\mathrm{r}} (F_n)$ be a nonzero ideal.
Choose $a \in I$ such that $a \neq 0$.
Then $\ta (a^* a) \neq 0$
by \Thm{T_3318_RedTr}.
\Lem{L_3301_RedNorm}
provides $M \in \N$
and $h_1, h_2, \ldots, h_m \in F_n$
such that the element
\[
c = \frac{1}{M} \sum_{m = 1}^M u_{h_m} a^* a u_{h_m}^*
\]
satisfies $\ c  \ta (a^* a) \cdot 1 \ < \frac{1}{2} \ta (a^* a)$.
Clearly $c \in I$.
Then $b = \ta (a^* a)^{1} c$ is also in~$I$,
and $\ b  1 \ < \frac{1}{2}$,
so $b$ is invertible.
Therefore $I = C^*_{\mathrm{r}} (F_n)$.
\end{proof}
The following result (for $n = 2$) is proved at the end of~\cite{Pw}.
\begin{thm}[\cite{Pw}]\label{T_3301_CStRFnUniqT}
Let $n \in \{ 2, 3, \ldots, \I \}$.
Then $C^*_{\mathrm{r}} (F_n)$ has a unique tracial state.
\end{thm}
\begin{proof}
Let $\sm$ be any tracial state on $C^*_{\mathrm{r}} (F_n)$.
We prove that $\sm = \ta$.
Let $a \in C^*_{\mathrm{r}} (F_n)$ and let $\ep > 0$.
Use \Lem{L_3301_RedNorm} to find
$M \in \N$
and $h_1, h_2, \ldots, h_m \in F_n$
such that the element
\[
c = \frac{1}{M} \sum_{m = 1}^M u_{h_m} a u_{h_m}^*
\]
satisfies $\ c  \ta (a) \cdot 1 \ < \ep$.
We clearly have $\sm (c) = \sm (a)$
and $\sm (\ta (a) \cdot 1) = \ta (a)$.
So
\[
 \sm (a)  \ta (a) 
=  \sm ( c  \ta (a) \cdot 1 ) 
< \ep.
\]
Since $\ep > 0$ is arbitrary, we conclude that $\sm (a) = \ta (a)$.
\end{proof}
\section{C*Algebras of Locally Compact Groups}\label{Sec_LCGpCSt}
\indent
In this section,
we consider the \ca{s} of general locally compact groups.
Since our later focus will be mostly on discrete groups,
we omit a number of proofs.
In the discrete case,
in Section~\ref{Sec_GpCSt},
we constructed
the group \ca{} as the closed linear span
of the group elements in a suitable norm,
and we constructed the group von Neumann algebra as
the closed linear span
of the group elements in a suitable (much weaker) topology.
For the von Neumann algebra,
this definition turns out to still work,
but it does not give a reasonable outcome for the \ca.
For example,
if the group~$G$ is second countable,
one wants the group \ca{} to be separable.
However,
if $g, h \in G$ with $g \neq h$,
then $\ u_g  u_h \ = 2$.
See \Exr{Ex_3318_Norm2}.
Throughout this section,
we let $\mu$ be a fixed left Haar measure on~$G$.
The following definition is the generalization of
\Def{D_3407_DRegRep}.
\begin{dfn}\label{D_3402_LCLReg}
% Let $\mu$ be left Haar measure on~$G$.
The {\emph{left regular representation of~$G$}}
is the \rpn{} $v \colon G \to U (L^2 (G, \mu))$
given by $(v (g) \xi) (h) = \xi (g^{1} h)$
for $g, h \in G$ and $\xi \in L^2 (G, \mu)$.
\end{dfn}
\begin{exr}\label{Ex_3318_LCRgRp}
Let $G$ be a locally compact group.
Prove that $v$ as in
\Def{D_3402_LCLReg}
is a unitary \rpn{}
of $G$ on~$L^2 (G)$.
\end{exr}
The main point beyond Exercise~\ref{Ex_3407_DRgRpn}
(the case of a discrete group)
is to prove continuity.
Left invariance of the measure will be needed to show that $v (g)$
is unitary.
As for discrete groups, there is also a right regular \rpn.
There is one new feature:
one must use right Haar measure,
or else correct the formula by including suitable
RadonNikodym derivatives
(here, a suitable power of the modular function
of \Thm{T_3401_ModFcn} below).
\begin{exr}\label{Ex_3318_Norm2}
Let $G$ be a locally compact group,
let $v \colon G \to U (L^2 (G))$
be the left regular representation,
and let $g, h \in G$ with $g \neq h$.
Prove that $\ v (g)  v (h) \ = 2$.
Use this fact to prove that if $G$ is not discrete,
then ${\ov{\spn}} \big ( \{ v (g) \colon g \in G \} \big)$
is not separable.
\end{exr}
When we have constructed $C^* (G)$
and $C^*_{\mathrm{r}} (G)$,
it will turn out that the group elements $u_g$
are in the multiplier algebras
$M ( C^* (G) )$ and $M ( C^*_{\mathrm{r}} (G) )$.
(We will not prove this.)
In particular,
the naive analog of Exercise~\ref{Ex_3401_CstGRel}
certainly does not hold,
and we know of no general method of describing
either $C^* (G)$ or $C^*_{\mathrm{r}} (G)$
in terms of generators and relations.
Instead of $\C [G]$,
we will use the space $C_{\mathrm{c}} (G)$
of compactly supported \cfn{s} on~$G$,
with the convolution defined by the analog of~(\ref{Eq_3317_Prod})
in Remark~\ref{D_3317_GpRing}.
\begin{ntn}\label{N_3328_CcX}
Let $X$ be a locally compact Hausdorff space.
We denote by $C_{\mathrm{c}} (X)$ the complex vector space
of all \cfn{s} from $X$ to~$\C$
which have compact support,
with pointwise addition and scalar multiplication.
Unless otherwise specified,
we make this space a complex *algebra using pointwise
complex conjugation and pointwise multiplication,
but we will frequently use other operations;
in particular, if $G$ is a group,
the operations will usually be as in
\Def{D_3328_CcConv} below.
If $E$ is any Banach space,
we further denote by $C_{\mathrm{c}} (X, E)$
the vector space
of all \cfn{s} from $X$ to~$E$
which have compact support.
\end{ntn}
Another complication which appears for general locally compact groups
is the possible failure of unimodularity.
We recall for reference the basic properties of the modular function.
\begin{thm}\label{T_3401_ModFcn}
Let $G$ be a locally compact group.
Make $(0, \I)$ into a locally compact abelian group
by taking the group operation to be multiplication.
Then there is a unique \ct{} \hm{}
$\Dt \colon G \to (0, \I)$,
called the {\emph{modular function}} of~$G$,
such that,
for every choice of Haar measure $\mu$ on~$G$,
for every $g \in G$,
and every measurable set $E \subset G$,
we have $\mu (E g) = \Dt (g) \mu (E)$.
Moreover, for $g \in G$ and every $a \in C_{\mathrm{c}} (G)$,
we have
%
\begin{equation}\label{Eq_3401_RtTr}
\int_G a (g h) \, d \mu (g) = \Dt (h)^{1} \int_G a (g) \, d \mu (g),
\end{equation}
%
and
for every $a \in C_{\mathrm{c}} (G)$
we have
%
\begin{equation}\label{Eq_3401_Inv}
\int_G \Dt (g)^{1} a (g^{1}) \, d \mu (g) = \int_G a (g) \, d \mu (g).
\end{equation}
%
\end{thm}
For the proof,
see Lemma 1.61 and Lemma 1.67 of~\cite{Wlms}.
\begin{dfn}\label{D_3328_CcConv}
Let $G$ be a locally compact group.
Using left Haar measure $\mu$ on~$G$,
for $a, b \in C_{\mathrm{c}} (G)$ we define
%
\begin{equation}\label{Eq_3318_ConvCc}
(a b) (g) = \int_G a (h) b ( h^{1} g) \ d \mu (h)
\andeqn
a^* (g) = \Dt (g)^{1} {\ov{a (g^{1})}}.
\end{equation}
%
\end{dfn}
We will need Fubini's Theorem several times,
and the following lemma will be used to verify its hypotheses.
\begin{lem}\label{L_3401_JustFub}
Let $G$ be a locally compact group,
and let $a, b \in C_{\mathrm{c}} (G)$.
Define $f_{a, b} \colon G \times G \to \C$
by $f_{a, b} (g, h) = a (h) b ( h^{1} g)$
for $g, h \in G$.
Then $f_{a, b} \in C_{\mathrm{c}} (G \times G)$,
and $\supp (f_{a, b})$ is contained in the compact set
$(\supp (a) \cdot \supp (b)) \times \supp (a)$.
\end{lem}
\begin{proof}
It is immediate that $f_{a, b}$ is \ct.
To see that $f_{a, b}$
has compact support,
define $K \subset G$
by
\[
K = \supp (a) \cdot \supp (b)
= \big\{ g h \colon {\mbox{$g \in \supp (a)$ and
$h \in \supp (b)$}} \big\}.
\]
Then $K$ is compact because $K$ is the image of the
compact set $\supp (a) \times \supp (b) \subset G \times G$
under the multiplication map.
We show that $\supp (f_{a, b}) \subset K \times \supp (a)$.
So suppose $f_{a, b} (g, h) \neq 0$.
Obviously $h \in \supp (a)$ and $h^{1} g \in \supp (b)$.
Therefore $g = h \cdot h^{1} g \in K$.
\end{proof}
\begin{prp}\label{P_3328_Alg}
Let $G$ be a locally compact group.
Equipped with the operations in \Def{D_3328_CcConv},
the space $C_{\mathrm{c}} (G)$
is a complex *algebra.
\end{prp}
\begin{proof}
Let $\mu$ be left Haar measure on~$G$.
For $a, b \in C_{\mathrm{c}} (G)$,
let $f_{a, b} \in C_{\mathrm{c}} (G \times G)$
be as in \Lem{L_3401_JustFub}.
We then have
\[
(a b) (g) = \int_G f_{a, b} (g, h) \, d \mu (h).
\]
Therefore $(a b) (g)$ can be nonzero
only for $g \in \supp (a) \cdot \supp (b)$.
We next prove that $a b$ is \ct.
This is a standard argument,
which we give for completeness.
We need only consider the case $a \neq 0$.
Set $M = \mu ( \supp (a)) > 0$.
Let $\ep > 0$ and let $g_0 \in G$.
For $h \in G$ choose open sets $U (h), V (h) \subset G$
such that $g_0 \in U (h)$,
$h \in V (h)$,
and for all $g \in U (h)$ and $k \in V (h)$,
we have
\[
\big f_{a, b} (g, k)  f_{a, b} (g_0, h) \big < \frac{\ep}{4 M}.
\]
Choose $n \in \N$ and $h_1, h_2, \ldots, h_n \in G$
such that the sets $V (h_1), \, V (h_2), \, \ldots, \, V (h_n)$
cover $\supp (a)$.
Set $U = \bigcap_{j = 1}^n U (h_j)$,
which is an open set containing~$g_0$.
Let $g \in U$.
For $h \in \supp (a)$,
we claim that
\[
\big f_{a, b} (g, h)  f_{a, b} (g_0, h) \big < \frac{\ep}{2 M}.
\]
Choose $j \in \{ 1, 2, \ldots, n\}$
such that $h \in V (h_j)$.
Then $g \in U (h_j)$, so
\begin{align*}
\big f_{a, b} (g, h)  f_{a, b} (g_0, h) \big
& \leq \big f_{a, b} (g, h)  f_{a, b} (g_0, h_j) \big
+ \big f_{a, b} (g_0, h_j)  f_{a, b} (g_0, h) \big
\\
& < \frac{\ep}{4 M} + \frac{\ep}{4 M}
= \frac{\ep}{2 M},
\end{align*}
as desired.
It now follows that
\begin{align*}
\big (a b) (g)  (a b) (g_0) \big
& \leq \int_{\supp (a)}
\big f_{a, b} (g, h)  f_{a, b} (g_0, h) \big \, d \mu (h)
\\
& \leq \left( \frac{\ep}{2 M} \right) \mu ( \supp (a) )
= \frac{\ep}{2}
< \ep.
\end{align*}
This completes the proof that $a b$ is \ct.
We have now shown that $(a, b) \mapsto a b$
is a well defined map
$C_{\mathrm{c}} (G) \times C_{\mathrm{c}} (G) \to C_{\mathrm{c}} (G)$.
It is obviously bilinear.
It remains only to prove associativity and the properties of the adjoint.
Let $a, b, c \in C_{\mathrm{c}} (G)$.
We compute as follows,
with the second step being an application of Fubini's Theorem
which is justified afterwards.
The third step is a change of variables in the inner integral,
replacing $h$ with $k h$.
For $g \in G$,
we have
\begin{align*}
[ (a b) c] (g)
& = \int_G \left( \int_G a (k) b (k^{1} h) \, d \mu (k) \right)
c (h^{1} g) \, d \mu (h)
\\
& = \int_G a (k) \left( \int_G b (k^{1} h) c (h^{1} g) \, d \mu (h)
\right)
\, d \mu (k)
\\
& = \int_G a (k) \left( \int_G b (h) c (h^{1} k^{1} g) \, d \mu (h)
\right)
\, d \mu (k)
\\
& = \int_G a (k) (b c) (k^{1} g) \, d \mu (k)
= [a (b c)] (g).
\end{align*}
To justify the application of Fubini's Theorem
at the second step,
we observe that the integrand
as a function of both variables is
$(h, k) \mapsto f_{a, b} (h, k) c (h^{1} g)$,
which is a \cfn{} on $G \times G$ with support in the
compact set $(\supp (a) \cdot \supp (b)) \times \supp (a)$.
Therefore it is integrable with respect to $\mu \times \mu$.
It is obvious that $a \mapsto a^*$ is conjugate linear,
and easy to check that $a^{* *} = a$
for all $a \in C_{\mathrm{c}} (G)$.
It remains only to check that
$(a b)^* = b^* a^*$ for $a, b \in C_{\mathrm{c}} (G)$.
For $g \in G$,
using the change of variables from $h$ to $g h$ at the third step,
we have
\begin{align*}
(b^* a^*) (g)
& = \int_G \Dt (h)^{1} {\ov{b ( h^{1} )}}
\Dt (h^{1} g)^{1} {\ov{a ( (h^{1} g)^{1} )}}
\, d \mu (h)
\\
& = \int_G \Dt (g)^{1} {\ov{a ( g^{1} h)}} \cdot {\ov{b ( h^{1} )}}
\, d \mu (h)
\\
& = \Dt (g)^{1} \int_G {\ov{a (h)}} \cdot {\ov{b ( h^{1} g^{1})}}
\, d \mu (h)
= (a b)^* (g).
\end{align*}
This completes the proof.
\end{proof}
\begin{exr}\label{Ex_3328_CGToCc}
Let $G$ be a discrete group.
Prove that there is a complex *algebra isomorphism
of $\C [G]$ as in \Def{D_3317_GpRing}
with $C_{\mathrm{c}} (G)$
as in \Def{D_3328_CcConv}
and Proposition~\ref{P_3328_Alg}.
\end{exr}
The main point is to make sure that the definitions of the
product and adjoint match.
We will need a topology on $C_{\mathrm{c}} (G)$.
To follow what we did for discrete~$G$
as closely as possible,
we would use the direct limit topology.
Continuity of linear functionals in this topology
is determined by testing on nets
$(b_i)_{i \in I}$ in $C_{\mathrm{c}} (G)$
and elements $b \in C_{\mathrm{c}} (G)$
such that $b_i \to b$ uniformly and there is some
common compact set $K \subset G$ with $\supp (b_i) \subset K$
for all $i \in I$.
See Remark~1.86 of~\cite{Wlms} for more on this topology.
(It can have other convergent nets.)
Here,
it seems simpler to just use the $L^1$ norm,
and to complete $C_{\mathrm{c}} (G)$ in this norm,
getting the convolution algebra $L^1 (G)$.
When $G$ is discrete,
this definition specializes to the algebra $l^1 (G)$
of Definition~\ref{D_3326_l1GpAlg}.
\begin{dfn}\label{D_3328_L1Norm}
Let $G$ be a locally compact group.
Using Haar measure in the integral,
we define a norm on $C_{\mathrm{c}} (G)$
by $\ b \_1 = \int_G  b (g)  \, d \mu (g)$.
We define $L^1 (G)$ to be the completion of $C_{\mathrm{c}} (G)$
in this norm.
Justified by Proposition~\ref{P_3328_L1IsBAlg} below,
we make $L^1 (G)$ into a Banach *algebra
by extending the operations of \Def{D_3328_CcConv}
by continuity.
\end{dfn}
There is never any problem with the integral,
because we need only integrate \cfn{s} on compact sets.
When $G$ is second countable,
so that Haar measure is $\sm$finite
and all Borel sets are Baire sets,
the resulting space is just the usual space $L^1 (G)$
of integrable Borel functions on~$G$.
In our presentation,
we avoid technicalities of measure theory
(including but not limited to dealing with
measures which are not $\sm$finite)
by defining $L^1 (G)$ to be the completion of $C_{\mathrm{c}} (G)$.
\begin{prp}\label{P_3328_L1IsBAlg}
Let $G$ be a locally compact group.
Then for $a, b \in C_{\mathrm{c}} (G)$,
we have $\ a b \_1 \leq \ a \_1 \ b \_1$
and $\ a^* \_1 = \ a \_1$.
\end{prp}
\begin{proof}
For the first part,
let $a, b \in C_{\mathrm{c}} (G)$.
Let $f_{a, b} \in C_{\mathrm{c}} (G \times G)$
be as in \Lem{L_3401_JustFub},
that is, $f_{a, b} (g, h) = a (h) b (h^{1} g)$.
Since $f_{a, b}$ is integrable with respect to $\mu \times \mu$,
we can apply Fubini's Theorem at the third step
in the following calculation:
\begin{align*}
\ a b \_1
& = \int_G \left \int_G f_{a, b} (g, h) \, d \mu (h) \right
\, d \mu (g)
\leq \int_G \left( \int_G  f_{a, b} (g, h)  \,
d \mu (h) \right)
\, d \mu (g)
\\
& = \int_G \left( \int_G  f_{a, b} (g, h)  \,
d \mu (g) \right)
\, d \mu (h)
\\
& = \int_G  a (h)  \left( \int_G  b (h^{1} g)  \,
d \mu (g) \right)
\, d \mu (h)
= \int_G  a (h)  \cdot \ b \_1 \, d \mu (h)
= \ a \_1 \ b \_1.
\end{align*}
% \begin{align*}
% \ a b \_1
% & = \int_G \left \int_G a (h) b (h^{1} g) \, d \mu (h) \right
% \, d \mu (g)
% \leq \int_G \left( \int_G  a (h) b (h^{1} g)  \,
% d \mu (h) \right)
% \, d \mu (g)
% \\
% & = \int_G  a (h)  \left( \int_G  b (h^{1} g)  \,
% d \mu (g) \right)
% \, d \mu (h)
% = \int_G  a (h)  \cdot \ b \_1 \, d \mu (h)
% = \ a \_1 \ b \_1.
% \end{align*}
For the second part,
we apply~(\ref{Eq_3401_Inv}) in Theorem~\ref{T_3401_ModFcn}
at the second step to get
\[
\ a^* \_1
= \int_G \Dt (g)^{1} \big {\ov{a (g^{1})}} \big \, d \mu (g)
= \int_G \big {\ov{a (g)}} \big \, d \mu (g)
= \ a \_1.
\]
This completes the proof.
\end{proof}
\begin{exr}\label{Ex_3328_L1l1}
Let $G$ be a discrete group,
and take Haar measure on~$G$ to be counting measure.
Prove that there is a Banach *algebra isomorphism
of $L^1 (G)$
as in \Def{D_3328_L1Norm}
and $l^1 (G)$ as in \Def{D_3326_l1GpAlg}.
\end{exr}
Given Exercise~\ref{Ex_3328_CGToCc},
this exercise is essentially trivial.
We now give the analog of the construction of \Def{D_3317_CGRepDfn}.
At this point,
we want to integrate \cfn{s} with compact support
which have values in a Banach space.
In principle,
the ``right'' approach to Banach space valued integration
is to define measurable
Banach space valued functions and their integrals.
This has been done;
one reference is Appendix B of~\cite{Wlms}.
(Note the systematic misprint there:
``separatelyvalued'' should be ``separablyvalued''.)
Things simplify considerably if $G$ is second countable
and $E$ is separable,
but neither of these conditions is necessary for the constructions
we carry out,
either here or in Section~\ref{Sec:CP}.
For \cfn{s} with compact support,
it is easy to avoid this theory,
and this is the route we take.
An integration theory sufficient for this purpose
is developed in Section~1.5 of~\cite{Wlms}.
We summarize the properties.
We could avoid such integrals here by always working in terms of scalar
products below.
This seems pointless since
we won't be able to do something similar when
defining multiplication in crossed products
in \Def{D:L1}.
In the following theorem,
the relation in~(\ref{T_3402_BSpInt_Scalar})
is (1.23) in~\cite{Wlms},
existence is in Lemma 1.91 of~\cite{Wlms},
and uniqueness is in the discussion
at the beginning of Section 1.5 of~\cite{Wlms}.
\begin{thm}\label{T_3402_BSpInt}
Let $G$ be a locally compact group
with left Haar measure~$\mu$,
and let $E$ be a Banach space.
Then there is a unique linear map
$I_E \colon C_{\mathrm{c}} (G, E) \to E$
with the following properties:
\begin{enumerate}
\item\label{T_3402_BSpInt_Norm}
$\ I_E (\xi) \ \leq \int_G \ \xi (g) \ \, d \mu (g)$
for all $\xi \in C_{\mathrm{c}} (G, E)$.
\item\label{T_3402_BSpInt_Scalar}
For $\et \in E$ and $f \in C_{\mathrm{c}} (G)$,
the function $\xi (g) = f (g) \et$ for $g \in G$
satisfies
$I_E (\xi) = \left( \int_G f (g) \, d \mu (g) \right) \et$.
\end{enumerate}
\end{thm}
\begin{dfn}\label{D_3402_BanSpI}
Let $G$ be a locally compact group
with left Haar measure~$\mu$,
and let $E$ be a Banach space.
With $I_E$ as in \Thm{T_3402_BSpInt},
we define
$\int_G \xi (g) \, d \mu (g) = I_E ( \xi)$
for $\xi \in C_{\mathrm{c}} (G, E)$.
\end{dfn}
The next lemma is part of Lemma 1.91 of~\cite{Wlms},
but we give a direct proof
directly from the properties of the integral
given in \Thm{T_3402_BSpInt}.
\begin{lem}\label{L_3402_LinOfInt}
Let $G$ be a locally compact group
with left Haar measure~$\mu$,
let $E$ and $F$ be Banach spaces,
and let $a \in L (E, F)$.
Then for all $\xi \in C_{\mathrm{c}} (G, E)$,
we have
\[
a \left( \int_G \xi (g) \, d \mu (g) \right)
= \int_G a (\xi (g)) \, d \mu (g).
\]
\end{lem}
\begin{proof}
Let $I_E \colon C_{\mathrm{c}} (G, E) \to E$
and $I_F \colon C_{\mathrm{c}} (G, F) \to F$
be as in \Thm{T_3402_BSpInt}.
Define $T \colon C_{\mathrm{c}} (G, E) \to C_{\mathrm{c}} (G, F)$
by $T (\xi) (g) = a (\xi (g))$
for $\xi \in C_{\mathrm{c}} (G, E)$ and $g \in G$.
We must prove that $a \circ I_E = I_F \circ T$.
Using \Thm{T_3402_BSpInt}(\ref{T_3402_BSpInt_Scalar}) twice,
it is easy to check that
if $\xi_0 \in E$, $f \in C_{\mathrm{c}} (G)$,
and we define $\xi \in C_{\mathrm{c}} (G, E)$
by $\xi (g) = f (g) \xi_0$ for $g \in G$,
then
\[
(a \circ I_E) (\xi)
= a \left( \left[ \int_G f \mu \right] \xi_0 \right)
= \left( \int_G f \, d \mu \right) a \xi_0
= (I_F \circ T) (\xi).
\]
Now let $\xi \in C_{\mathrm{c}} (G, E)$ be arbitrary.
Let $\ep > 0$.
We use a partition of unity argument
to prove that
$\big\ (a \circ I_E) (\xi)  (I_F \circ T) (\xi) \big\ < \ep$.
Choose an open set $U \subset G$ such that
$\supp (\xi) \subset U$ and the set $L = {\ov{U}}$ is compact.
Set
\[
\dt = \frac{\ep}{3 (\ a \ + 1) ( \mu (L) + 1)}.
\]
Use compactness of $L$ and continuity of $\xi$
to find $n \in \N$, open sets $V_1, V_2, \ldots, V_n \subset G$
which cover~$L$,
and $g_j \in V_j$ for $j = 1, 2, \ldots, n$
such that $\ \xi (g)  \xi (g_j) \ < \dt$
for $j = 1, 2, \ldots, n$ and $g \in V_j$.
Choose \cfn{s} $f_j \colon L \to [0, 1]$
which form a partition of unity on~$L$
and such that $\supp (f_j) \subset V_j \cap L$
for $j = 1, 2, \ldots, n$.
We may extend the functions $f_1, f_2, \ldots, f_n$
so that they are \cfn{s} defined on all of~$G$,
take values on $[0, 1]$,
satisfy $\supp (f_j) \subset V_j$ for $j = 1, 2, \ldots, n$,
and satisfy $\sum_{j = 1}^n f_j (g) \leq 1$ for all $g \in G$.
Further choose a \cfn{} $f \colon G \to [0, 1]$
such that $f (g) = 1$ for all $g \in \supp (\xi)$
and $\supp (f) \subset U$.
Define $\et \in C_{\mathrm{c}} (G, E)$ by
\[
\et (g) = f (g) \sum_{j = 1}^n f_j (g) \xi (g_j)
\]
for $g \in G$.
Then
\[
\ \et (g)  \xi (g) \
\leq f (g) \sum_{j = 1}^n f_j (g) \ \xi (g)  \xi (g_j) \.
\]
We have $\ \xi (g)  \xi (g_j) \ < \dt$
whenever $f_j (g) \neq 0$,
and $0 \leq f (g) \sum_{j = 1}^n f_j (g) \leq 1$,
so
$\ \et (g)  \xi (g) \ < \dt$.
Moreover $\et (g) = \xi (g)$ for all $g \in G \setminus L$.
\Thm{T_3402_BSpInt}(\ref{T_3402_BSpInt_Norm}) therefore
implies that $\ I_E (\xi)  I_E (\et) \ \leq \mu (L) \dt$.
So
\[
\big\ (a \circ I_E) (\xi)  (a \circ I_E) (\et) \big\
\leq \ a \ \mu (L) \dt.
\]
Also
\[
\ T (\xi) (g)  T (\et) (g) \
= \ a (\et (g))  a (\xi (g)) \
< \ a \ \dt
\]
for all $g \in G$,
so \Thm{T_3402_BSpInt}(\ref{T_3402_BSpInt_Norm})
implies
\[
\big\ (I_F \circ T) (\xi)  (I_F \circ T) (\et) \big\
\leq \ a \ \mu (L) \dt.
\]
The first paragraph of the proof implies that
$(a \circ I_E) (\et) = (I_F \circ T) (\et)$,
so
\[
\big\ (a \circ I_E) (\xi)  (I_F \circ T) (\xi) \big\
\leq \ a \ \mu (L) \dt + \ a \ \mu (L) \dt
< \ep,
\]
as desired.
\end{proof}
The formula in the following definition
should be compared with~(\ref{Eq_3326_l1GRpFormula})
in \Def{D_3326_l1GRepDfn}.
\begin{dfn}\label{D_3328_IntForm}
Let $G$ be a locally compact group
with left Haar measure~$\mu$,
let $H$ be a Hilbert space,
and let $v \colon G \to U (H)$ be a unitary representation.
Then the {\emph{integrated form}} of $v$ is the representation
$\rh_v \colon C_{\mathrm{c}} (G) \to L (H)$
given by
\[
\rh_v (b) \xi = \int_G b (g) v (g) \xi \, d \mu (g)
\]
for $b \in C_{\mathrm{c}} (G)$.
Justified by Proposition~\ref{P_3328_IFIsRep} below,
we extend this representation by continuity to
a representation $L^1 (G) \to L (H)$,
which we also denote by $\rh_v$ and call
the {\emph{integrated form}} of~$v$.
\end{dfn}
We want to think of $\rh_v (b)$
as $\int_G b (g) v (g) \, d \mu (g)$.
Defining $\rh_v$ directly by this formula causes
technical problems,
because $g \mapsto b (g) v (g)$ is only a strong operator
\cfn{} to $L (H)$,
not a norm \cfn.
The definition given is the easiest solution to these difficulties.
\begin{prp}[Part of Theorem 3.9 of~\cite{Fld};
part of Proposition 13.3.4 of~\cite{Dx}]\label{P_3328_IFIsRep}
Let $G$ be a locally compact group,
let $H$ be a Hilbert space,
and let $w \colon G \to U (H)$ be a unitary representation.
Then $\rh_w \colon C_{\mathrm{c}} (G) \to L (H)$
is a *homomorphism
and $\ \rh_w (b) \ \leq \ b \_1$
for all $b \in C_{\mathrm{c}} (G)$.
\end{prp}
We want to make one point explicitly.
Even though the function $g \mapsto w (g)$
is not required to be norm \ct,
the representation $\rh_w$ {\emph{is}} norm \ct.
In particular,
if $a \in C_{\mathrm{c}} (G)$
and we define $a_g \in C_{\mathrm{c}} (G)$
by $a_g (h) = a (g^{1} h)$
for $g, h \in G$,
then $g \mapsto a_g$
is a \cfn{} from $G$ to $L^1 (G)$,
and $g \mapsto \rh_w (a_g)$
is a norm \cfn{} from $G$ to $L (H)$.
\begin{proof}[Proof of Proposition~\ref{P_3328_IFIsRep}]
The expression for $\rh_w (b) \xi$ is defined,
by
Definition~\ref{D_3402_BanSpI}
and Theorem~\ref{T_3402_BSpInt}.
Moreover, $\rh_w (b) \xi$ is obviously linear
in both $b \in C_{\mathrm{c}} (G)$ and $\xi \in H$.
Now let $b \in C_{\mathrm{c}} (G)$ and $\xi \in H$.
Using Theorem~\ref{T_3402_BSpInt}(\ref{T_3402_BSpInt_Norm})
at the first step,
and $\ w (g) \ = 1$ at the second step,
we get
\[
\ \rh_w (b) \xi \
\leq \int_G  b (g)  \cdot \ w (g) \xi \ \, d \mu (g)
\leq \ b \_1 \ \xi \.
\]
Thus $\rh_w (b) \in L (H)$ for all $b \in C_{\mathrm{c}} (G)$.
It remains to prove that $\rh_w$ preserves products and adjoints.
Let $a, b \in C_{\mathrm{c}} (G)$,
and let $\xi, \et \in H$.
We prove that
\[
\langle \rh_w (a b) \xi, \, \et \rangle
= \langle \rh_w (a) \rh_w (b) \xi, \, \et \rangle
\andeqn
\langle \rh_w (b^*) \xi, \, \et \rangle
= \langle \xi, \, \rh_w (b) \et \rangle.
\]
For the first,
we use \Lem{L_3402_LinOfInt}
at the first, third, and fifth steps,
Fubini's Theorem
(justified by \Lem{L_3401_JustFub}
and continuity of
$(g, h) \mapsto f_{a, b} (g, h) \langle w (g) \xi, \et \rangle$)
at the second step,
\Lem{L_3402_LinOfInt}
and left translation invariance of~$\mu$ at the fourth step,
getting
\begin{align*}
\langle \rh_w (a b) \xi, \, \et \rangle
& = \int_G \left( \int_G a (h) b (h^{1} g) \, d \mu (h) \right)
\langle w (g) \xi, \et \rangle \, d \mu (g)
\\
& = \int_G \left( \int_G a (h) b (h^{1} g)
\langle w (g) \xi, \et \rangle \, d \mu (g) \right) \, d \mu (h)
\\
% & = \int_G \left( \int_G \big\langle [a (h) w (h)]
% [ b (h^{1} g) w (h^{1} g) ] \xi, \, \et \big\rangle \, d \mu (g)
% \right) \, d \mu (h)
% \\
& = \int_G \left\langle a (h) w (h) \left( \int_G
b (h^{1} g) w (h^{1} g) \xi \, d \mu (g) \right), \,
\et \right\rangle
\, d \mu (h)
\\
& = \int_G \big\langle a (h) w (h) \rh_w (b) \xi, \, \et \big\rangle
\, d \mu (h)
= \langle \rh_w (a) \rh_w (b) \xi, \, \et \rangle.
\end{align*}
For the second,
we use \Lem{L_3402_LinOfInt}
at the first step,
$w (g)^* = w (g^{1})$
at the second step,
(\ref{Eq_3401_Inv}) (in Theorem~\ref{T_3401_ModFcn})
at the third step,
and \Lem{L_3402_LinOfInt}
at the fourth step,
getting
\begin{align*}
\langle \rh_w (b^*) \xi, \, \et \rangle
& = \int_G \Dt (g)^{1}
\big\langle {\ov{b (g^{1})}} w (g) \xi, \, \et \big\rangle \,
d \mu (g)
\\
& = \int_G \Dt (g)^{1}
\big\langle \xi, \, b (g^{1}) w (g^{1}) \et \big\rangle \,
d \mu (g)
\\
& = \int_G \langle \xi, \, b (g) w (g) \et \rangle \, d \mu (g)
= \langle \xi, \, \rh_w (b) \et \rangle.
\end{align*}
This completes the proof.
\end{proof}
The following theorem is the analog for locally compact groups
of Theorem~\ref{T_3317_UProp} (for discrete groups).
\begin{thm}[Theorems 3.9 and 3.11 of~\cite{Fld};
Proposition 7.1.4 of~\cite{Pd1};
Proposition 13.3.4 of~\cite{Dx}]\label{P_3328_IntFBij}
Let $G$ be a locally compact group,
and let $H$ be a Hilbert space.
Then the integrated form construction defines a bijection
from the set of unitary representations of $G$ on~$H$
to the set of nondegenerate \ct{} *representations
of $L^1 (G)$ on~$H$.
\end{thm}
Since our main subject is discrete groups,
we will not give a proof here.
We do mention one key technical point.
The proof can't be done the same way as the
proof of Proposition~\ref{P_3317_CGRep},
because there is no analog in $C_{\mathrm{c}} (G)$,
or even in $L^1 (G)$,
of the images $u_g$ of the group elements in $\C [G]$.
The analogs of the elements $u_g$
can only be found in the multiplier algebra of $L^1 (G)$.
Since integrated form representations of $L^1 (G)$
are necessarily contractive,
{\emph{all}} \ct{} representations of $L^1 (G)$
are necessarily contractive.
We now give the analog of
\Def{D_3317_UnivRep}.
\begin{dfn}\label{D_3328_LCUnivRp}
Let $G$ be a locally compact group.
Choose a fixed Hilbert space $H_0$ with dimension
$\card (G)$,
and define a unitary representation $w$
of $G$ to be the direct sum of all possible
unitary representations of $G$ on subspaces of~$H_0$.
We call $w$ the {\emph{universal representation}} of~$G$.
\end{dfn}
\begin{dfn}\label{D_3328_CstGDfn}
Let $G$ be a locally compact group,
and let $w \colon G \to U (H)$ be its universal unitary representation,
as in \Def{D_3328_LCUnivRp}.
Using the notation of \Def{D_3328_IntForm},
we define $C^* (G)$ to be the norm closure
in $L (H)$ of $\rh_w ( C_{\mathrm{c}} (G) )$.
\end{dfn}
Equivalently,
one can take $C^* (G) = {\ov{\rh_w ( L^1 (G) )}}$.
\begin{thm}[13.9.3 of~\cite{Dx}]\label{T_3328_IntFCSt}
Let $G$ be a locally compact group,
and let $H$ be a Hilbert space.
Then the integrated form construction defines a bijection
from the set of unitary representations of $G$ on~$H$
to the set of nondegenerate representations
of $C^* (G)$ on~$H$.
\end{thm}
Given \Thm{P_3328_IntFBij},
the proof is similar to the first part of the proof of
Theorem~\ref{T_3317_UProp}.
If we were able to take the universal representation of~$G$ to be
the direct sum of all possible representations of~$G$,
the proof would be clear.
Given any representation of~$G$,
it would be the restriction of the universal representation of~$G$
to some invariant subspace,
and we would simply restrict the corresponding representation
of $C^* (G)$ to the same subspace.
\begin{dfn}\label{D_3328_RedCstG}
Let $G$ be a locally compact group,
and let $v \colon G \to U (L^2 (G))$ be its left regular representation
(\Def{D_3402_LCLReg}).
Using the notation of \Def{D_3328_IntForm},
we define the {\emph{reduced group C*algebra}} $C^*_{\mathrm{r}} (G)$
to be the closure ${\overline{\rh_v ( C_{\mathrm{c}} (G) )}}$
in the norm topology on $L ( L^2 (G))$.
\end{dfn}
\begin{prp}\label{P_6X15_FToRed}
Let $G$ be a locally compact group.
Then there is a surjective \hm{}
$\kp \colon C^* (G) \to C^*_{\mathrm{r}} (G)$
obtained from \Thm{T_3328_IntFCSt}
by taking the nondegenerate representation used
there to be the left regular representation of~$G$.
It is uniquely determined by the property that
if $f \in C_{\mathrm{c}} (G)$
and $a \in C^* (G)$ and $b \in C^*_{\mathrm{r}} (G)$
are the images of $f$ in those two algebras,
then $\kp (a) = b$.
\end{prp}
\begin{proof}
The result is immediate from \Thm{T_3328_IntFCSt}
as soon as one knows that the left regular representation of~$G$
is \ct.
This fact is \Exr{Ex_3318_LCRgRp}.
\end{proof}
Recall (\Thm{T_4131_GpFToRed} and \Thm{T_3320_FRIso};
both stated for the general case)
that $\kp \colon C^* (G) \to C^*_{\mathrm{r}} (G)$
is an isomorphism \ifo{} $G$ is amenable.
The proof given after \Thm{T_4131_GpFToRed}
covers only the discrete case.
However,
the proof of the general case is contained in
the proof of the corresponding result for crossed products,
\Thm{T_4131_CPFToRed} below.
We give that proof in full below.
Evaluation at $1 \in G$ gives a tracial linear
functional from $C_{\mathrm{c}} (G)$ to~$\C$.
However, this functional is not \ct{}
with respect to $\ \cdot \_1$.
Thus,
unlike in \Thm{T_3318_RedTr},
we do not get a tracial state on $C^*_{\mathrm{r}} (G)$.
Functoriality as in
Exercise~\ref{Ex_4220_FunctGpCSt}
and Exercise~\ref{Ex_4220_FunctRedCSt}
does not generalize very well.
In particular,
the the full group \ca{} is not a functor from locally
compact group and group \hm{s}
to \ca{s} and \hm{s}.
If $G_2$ is discrete
and $\ph \colon G_1 \to G_2$ is the inclusion of a subgroup,
then the map in Exercise~\ref{Ex_4220_FunctGpCSt}
is given at the level of
$C_{\mathrm{c}} (G_1) \to C_{\mathrm{c}} (G_2)$
by extending a function on $G_1$ to all of $G_2$
by having it take the value zero on $G_2 \setminus G_1$.
However,
suppose $\ph \colon G_1 \to G_2$
is the inclusion of the subgroup $G_1 = \{ 1 \}$
in~$G_2$,
and assume that $G_2$ is not discrete.
There is no related \hm{}
$C_{\mathrm{c}} (G_1) \to C_{\mathrm{c}} (G_2)$.
If we try an analogous definition of a map
$L^1 (G_1) \to L^1 (G_2)$,
since $\{ 1 \}$ has measure zero in~$G_2$,
we get the zero map.
The same kind of thing goes wrong for the inclusion of,
for example, the subgroup $\R \times \{ 0 \}$ in $\R^2$.
Things still work if the range of $\ph$ is open in $G_2$.
We omit the proof.
There are other things that can be done instead,
but we do not discuss them here.
% 999 Rieffel induction
% Additional material: More discussion of the
% C*algebra of a not necessarily discrete group.
% 999
% Additional material:
% C*algebra of a locally compact abelian group.
% 999
There is an approach to the theory of locally compact abelian groups
which starts out by defining
${\widehat{G}}$ to be
the maximal ideal space ${\operatorname{Max}} (C^* (G))$.
Chapter~4 of~\cite{Fld}
comes close to following this approach.
Remark~\ref{R:WhatIsIn}
describes some of the difficulties with
understanding and working with $C^* (\Z)$.
When the group is not discrete,
everything that can go wrong before can still go wrong,
although,
since the canonical unitaries associated to the group elements
are no longer in the group \ca{}
(only in its multiplier algebra),
the situation is harder to describe.
The obvious analogous case to consider is $G = \R$.
We make explicit just one issue.
The analog of Remark \ref{R:WhatIsIn}(\ref{R:WhatIsIn1})
is to ask exactly which functions on~$\R$
have Fourier transforms (in the distributional sense)
which are in $C_0 (\R) = C^* (\R)$.
This is certainly at least as hard as,
and probably harder than,
asking which functions on~$\Z$
are the sequence
of Fourier coefficients of functions in $C (S^1) = C^* (\Z)$.
If $G$ is not amenable,
the situation for $C^* (G)$
is of course also at least as bad as described
in Remark~\ref{R_3408_NotAmen},
and it is harder to even formulate the problem.
There is also a group von Neumann algebra.
The following definition is the analog for
locally compact groups
of \Def{D_3401_vNG} for discrete groups.
\begin{dfn}\label{D_3402_vNG2}
Let $G$ be a locally compact group.
Regard $C^*_{\mathrm{r}} (G)$ as a subalgebra of $L (L^2 (G))$,
as in \Def{D_3328_RedCstG}.
We define the {\emph{group von Neumann algebra}}
$W^*_{\mathrm{r}} (G)$ to be the closure of $C^*_{\mathrm{r}} (G)$
in the weak operator topology on $L (L^2 (G))$.
\end{dfn}
See Section VII.3 of~\cite{Tk_Bk2},
and Definition V.7.4 of~\cite{Tks} for the discrete case.
The notation used in \cite{Tks} and~\cite{Tk_Bk2}
(${\mathcal{R}} (G)$ and ${\mathcal{R}}_{\mathrm{r}} (G)$)
is not common.
The most frequently used notation seems to be
$L (G)$, ${\mathcal{L}} (G)$, and $W^* (G)$.
Although we will not prove this here,
the unitaries $u_g$ corresponding to the group elements
$g \in G$ are in $W^*_{\mathrm{r}} (G)$.
In fact,
taking $v \colon G \to U (L^2 (G))$
to be the left regular representation
(as in \Def{D_3328_RedCstG}),
one has
\[
W^*_{\mathrm{r}} (G)
= \big\{ v (g) \colon g \in G \big\}''.
\]
\section{Crossed Products}\label{Sec:CP}
\indent
In this section, we define (full) \cp{s},
and prove a few results closely related to the construction.
We omit some of the details, especially in the case that the group
is not discrete.
See Sections 7.4 and~7.6 of~\cite{Pd1},
and, for considerably more detail,
Sections 2.4 and~2.5 of~\cite{Wlms}.
\begin{dfn}\label{D:CvRep}
Let $\af \colon G \to \Aut (A)$
be an action of a locally compact group $G$ on a \ca~$A$.
A {\emph{covariant representation}}
of $(G, A, \af)$ on a Hilbert space $H$
is a pair $(v, \pi)$
consisting of a unitary representation $v \colon G \to U (H)$
(the unitary group of $H$)
and a representation $\pi \colon A \to L (H)$
(the algebra of all bounded operators on $H$),
satisfying the {\emph{covariance condition}}
\[
v (g) \pi (a) v (g)^* = \pi (\af_g (a))
\]
for all $g \in G$ and $a \in A$.
It is called {\emph{nondegenerate}} if $\pi$ is nondegenerate.
\end{dfn}
Recall that,
by convention, unitary representations are strong operator \ct.
% (This is the usual continuity condition.)
By convention,
representations of \ca{s},
and of other *algebras
(such as the algebras $L^1 (G, A, \af)$ and
$C_{\mathrm{c}} (G, A, \af)$ introduced below)
will be *representations
(and, similarly, \hm{s} are *\hm s).
The crossed product \ca{} $C^* (G, A, \af)$
is the universal \ca{} for
covariant representations of $(G, A, \af)$,
in essentially the same way that the (full) group \ca{} $C^* (G)$ is
the universal \ca{} for unitary representations of~$G$,
as in \Thm{P_3328_IntFBij}
(Theorem~\ref{T_3317_UProp} when $G$ is discrete).
We construct it in a similar way to the group \ca.
We start with the analogs
of $C_{\mathrm{c}} (G)$ (\Def{D_3328_CcConv})
and of $L^1 (G)$ (\Def{D_3328_L1Norm}).
To define the crossed product by a general \lcg,
one needs an integration theory for Banach space valued functions.
This theory was not needed to define the convolution multiplication
in $C_{\mathrm{c}} (G)$,
but it was needed for later work involving $C^* (G)$,
such as the integrated form of a representation
(\Def{D_3328_IntForm}).
Here, we already need it for the definition of the product
in Definition~\ref{D:L1}.
A sufficient theory for our purposes
is discussed before Theorem~\ref{T_3402_BSpInt},
and the main facts we need are in
Theorem~\ref{T_3402_BSpInt}, Definition~\ref{D_3402_BanSpI},
and \Lem{L_3402_LinOfInt}.
As in Section~\ref{Sec_LCGpCSt},
we let $\mu$ be a fixed left Haar measure on~$G$.
\begin{dfn}\label{D:L1}
Let $\af \colon G \to \Aut (A)$
be an action of a locally compact group $G$ on a \ca~$A$.
We let $C_{\mathrm{c}} (G, A, \af)$ be the *algebra of
compactly supported \cfn{s} $a \colon G \to A$,
with pointwise addition and scalar multiplication.
Using Haar measure in the integral,
we define multiplication by the
following ``twisted convolution'':
\[
(a b) (g) = \int_G a (h) \af_h (b ( h^{1} g)) \, d \mu (h).
\]
Let $\Dt$ be the modular function of~$G$.
We define the adjoint by
\[
a^* (g) = \Dt (g)^{1} \af_g ( a (g^{1})^*).
\]
This does in fact make $C_{\mathrm{c}} (G, A, \af)$ a *algebra;
see \Exr{Ex:CkConv} below.
We define a norm $\ \cdot \_1$ on $C_{\mathrm{c}} (G, A, \af)$
by $\ a \_1 = \int_G \ a (g) \ \, d \mu (g)$.
One checks (\Exr{Ex:CkConv})
that $\ a b \_1 \leq \ a \_1 \ b \_1$
and $\ a^* \_1 = \ a \_1$.
Then $L^1 (G, A, \af)$ is the Banach *algebra obtained by
completing $C_{\mathrm{c}} (G, A, \af)$ in $\ \cdot \_1$.
\end{dfn}
The next exercise is the analog of \Prp{P_3328_Alg}.
It needs Fubini's Theorem
for Banach space valued integrals
of \cfn{s} with compact support.
See Proposition 1.105 of~\cite{Wlms}.
Since such functions are automatically integrable,
the required result can be gotten from the usual scalar valued
Fubini's Theorem
by applying \ct{} linear functionals
and using the HahnBanach Theorem.
\begin{exr}\label{Ex:CkConv}
In the situation of Definition~\ref{D:L1},
and assuming a suitable version of Fubini's Theorem
for Banach space valued integrals,
prove that that
multiplication in $C_{\mathrm{c}} (G, A, \af)$ is associative.
Further prove
for $a, b \in C_{\mathrm{c}} (G, A, \af)$
that $\ a b \_1 \leq \ a \_1 \ b \_1$,
that $(a b)^* = b^* a^*$,
and that $\ a^* \_1 = \ a \_1$.
Finally, prove that $L^1 (G, A, \af)$ is a Banach *algebra.
\end{exr}
\begin{rmk}\label{R:TGCA}
Suppose $A = C_0 (X)$, and $\af$ comes from an action of $G$ on~$X$.
Since we complete in a suitable norm later on,
it suffices to use only the dense subalgebra $C_{\mathrm{c}} (X)$
in place of $C_0 (X)$.
There is an obvious identification of
$C_{\mathrm{c}} (G, \, C_{\mathrm{c}} (X))$
with $C_{\mathrm{c}} (G \times X)$.
On $C_{\mathrm{c}} (G \times X)$, the formulas for multiplication
and adjoint become
\[
(f_1 f_2) (g, x)
= \int_G f_1 (h, x) f_2 (h^{1} g, \, h^{1} x) \, d \mu (h)
\]
and
\[
f^* (g, x) = \Dt (g)^{1} {\overline{f (g^{1}, \, g^{1} x)}}.
\]
\end{rmk}
\begin{exr}\label{Ex:CkTGCA}
Prove the formulas in Remark~\ref{R:TGCA}.
\end{exr}
\begin{rmk}\label{R:Discrete}
If $G$ is discrete, we choose Haar measure to be counting measure.
In this case,
$C_{\mathrm{c}} (G, A, \af)$ is, as a vector space,
the group ring $A [G]$,
consisting of all finite formal linear combinations of elements in $G$
with coefficients in $A$.
The multiplication and adjoint are given by
\[
(a \cdot g) (b \cdot h) = (a [ g b g^{1}]) \cdot (g h)
= (a \af_g (b)) \cdot (g h)
\andeqn
(a \cdot g)^* = \af_g^{1} (a^*) \cdot g^{1}
\]
for $a, b \in A$ and $g, h \in G$,
extended linearly.
This definition makes sense in the purely algebraic situation,
where it is called the {\emph{skew group ring}}.
When $G$ is discrete, we also often write $l^1 (G, A, \af)$ instead of
$L^1 (G, A, \af)$.
\end{rmk}
\begin{ntn}\label{N:ug}
Let $\af \colon G \to \Aut (A)$
be an action of a discrete group $G$ on a \ca~$A$.
In these notes, we will adopt the following fairly
commonly used notation.
First, suppose $A$ is unital.
For $g \in G$,
we let $u_g$ be the element of $C_{\mathrm{c}} (G, A, \af)$
which takes the value $1_A$ at $g$
and $0$ at the other elements of~$G$.
We use the same notation for its image in $l^1 (G, A, \af)$
(Definition~\ref{D:L1} above)
and in $C^* (G, A, \af)$
and $C^*_{\mathrm{r}} (G, A, \af)$ (Definitions~\ref{D:CP}
and~\ref{D:RedCP} below).
It is unitary,
and we call it the canonical unitary associated with~$g$.
If $A$ is not unital, extend the action to an action
$\af^+ \colon G \to \Aut (A^+)$ on the unitization $A^+$ of $A$
by
$\af_g^+ (a + \ld \cdot 1) = \af_g (a) + \ld \cdot 1$.
Then write $u_g$ as above.
Products $a u_g$,
with $a \in A$, are still in $C_{\mathrm{c}} (G, A, \af)$,
$l^1 (G, A, \af)$,
$C^* (G, A, \af)$,
or $C^*_{\mathrm{r}} (G, A, \af)$,
as appropriate.
\end{ntn}
\begin{rmk}\label{R_6320_Sumagug}
In particular,
$l^1 (G, A, \af)$ is the set of all sums
$\sum_{g \in G} a_g u_g$ with $a_g \in A$
and $\sum_{g \in G} \ a_g \ < \infty$.
These sums converge in $l^1 (G, A, \af)$,
and hence also in
$C^* (G, A, \af)$ and $C^*_{\mathrm{r}} (G, A, \af)$.
A general element of $C^*_{\mathrm{r}} (G, A, \af)$
has such an expansion, but unfortunately the series
one writes down generally does not converge.
See Remark~\ref{R_6X11_ComputeCrPrd};
as in Remark~\ref{R:WhatIsIn},
there is usually no convergence even when $A = \C$
and $G$ is amenable.
\end{rmk}
\begin{dfn}\label{D:IntForm}
Let $\af \colon G \to \Aut (A)$
be an action of a locally compact group $G$ on a \ca~$A$,
and let $(v, \pi)$ be a covariant representation
of $(G, A, \af)$ on a Hilbert space~$H$.
(We do not assume that $\pi$ is nondegenerate.)
Then the {\emph{integrated form}} of $(v, \pi)$ is the representation
$\sm \colon C_{\mathrm{c}} (G, A, \af) \to L (H)$
given by
\[
\sm (a) \xi = \int_G \pi (a (g)) v (g) \xi \, d \mu (g).
\]
(This representation is sometimes called
$v \times \pi$ or $\pi \times v$.
We will sometimes use the notation $v \ltimes \pi$.)
\end{dfn}
One needs to be more careful with the integral here,
just as in \Def{D_3328_IntForm}
and the remark afterwards,
because $v$ is generally only strong operator \ct,
not norm \ct.
Nevertheless,
one gets $\ \sm (a) \ \leq \ a \_1$,
so $\sm$ extends to a representation of $L^1 (G, A, \af)$.
We use the same notation $\sm$ for this extension.
Of course, one also needs to check that $\sm$ is a representation.
When $G$ is discrete,
and using Notation~\ref{N:ug},
the formula for $\sm$ comes down to $\sm (a u_g) = \pi (a) v (g)$
for $a \in A$ and $g \in G$.
Then
\begin{align*}
\sm (a u_g) \sm (b u_h)
& = \pi (a) v (g) \pi (b) v (g)^* v (g) v (h)
= \pi (a) \pi (\af_g (b)) v (g) v (h) \\
& = \pi (a \af_g (b)) v (g h)
= \sm \big( [a \af_g (b)] u_{g h} \big)
= \sm \big( (a u_g) (b u_h) \big).
\end{align*}
\begin{exr}\label{P:IntFIsRep}
Let $\af \colon G \to \Aut (A)$
be an action of a discrete group $G$ on a \ca~$A$,
and let $(v, \pi)$ be a nondegenerate covariant representation
of $(G, A, \af)$ on a Hilbert space $H$.
Starting from the computation above,
fill in the details of
the proof that the integrated form representation $\sm$ of
Definition~\ref{D:IntForm}
really is a nondegenerate representation
of $C_{\mathrm{c}} (G, A, \af)$.
\end{exr}
\begin{thm}[Proposition~7.6.4 of~\cite{Pd1}]\label{T:IntFBij}
Let $\af \colon G \to \Aut (A)$
be an action of a locally compact group $G$ on a \ca~$A$.
Then the integrated form construction defines a bijection
from the set of nondegenerate covariant representations
of $(G, A, \af)$ on a Hilbert space $H$
to the set of nondegenerate \ct{} representations
of $L^1 (G, A, \af)$ on the same Hilbert space.
\end{thm}
(There is a misprint
in the statement of Proposition~7.6.4 of~\cite{Pd1}:
it omits the nondegeneracy condition on the covariant representation,
but includes nondegeneracy for the integrated form.)
Also see Propositions 2.39 and~2.40 of~\cite{Wlms}.
These are stated in terms of $C^* (G, A, \af)$,
but, by Definition~\ref{D:CP} below,
that is the same thing.
(The \ca{} result is stated as
Theorem~\ref{P_6X20_CPIntFBij} below.)
\begin{rmk}\label{R:NRed}
Since integrated form representations of $L^1 (G, A, \af)$
are necessarily contractive,
{\emph{all}} \ct{} representations of $L^1 (G, A, \af)$
are necessarily contractive.
\end{rmk}
If $G$ is discrete and $A$ is unital, then
there are homomorphic images of both $G$ and $A$
inside $C_{\mathrm{c}} (G, A, \af)$,
given (following Notation~\ref{N:ug})
by $g \mapsto u_g$ and $a \mapsto a u_1$,
so it is clear how to get a covariant representation
of $(G, A, \af)$ from a nondegenerate representation
of $C_{\mathrm{c}} (G, A, \af)$.
In general, one must use the multiplier algebra of
$L^1 (G, A, \af)$,
which contains copies of $M (A)$ and $M (L^1 (G))$.
The point is that $M (L^1 (G))$
is the measure algebra of $G$,
and therefore contains the group elements as point masses.
\begin{exr}\label{P:IntFBij1}
Prove Theorem~\ref{T:IntFBij} when $G$ is discrete and $A$ is unital.
\end{exr}
For a small taste of the general case,
use approximate identities in $A$ to do the following exercise.
\begin{exr}\label{P:IntFBij2}
Prove Theorem~\ref{T:IntFBij} when $G$ is discrete
but $A$ is not necessarily unital.
\end{exr}
In the following definition,
we ignore the set theoretic problem,
that the collection of all nondegenerate representations
of $L^1 (G, A, \af)$ is not a set.
\Exr{P:FixSet} afterwards
asks for a set theoretically correct definition,
and a proof from this definition that one still has
the correct universal property.
The case of $C^* (G)$ for a discrete group~$G$
was done carefully in \Def{D_3317_GpCStar}
and the first part of the proof of \Thm{T_3317_UProp}.
For locally compact~$G$,
see \Def{D_3328_CstGDfn}
and \Thm{T_3328_IntFCSt}
(for which we did not give a proof).
It suffices to use a fixed Hilbert space
whose dimension is at least $\card (G) \card (A)$.
\begin{dfn}\label{D:CP}
Let $\af \colon G \to \Aut (A)$
be an action of a locally compact group $G$ on a \ca~$A$.
We define the {\emph{universal representation}}
$\sm$ of $L^1 (G, A, \af)$
to be the direct sum of all nondegenerate representations
of $L^1 (G, A, \af)$ on Hilbert spaces.
Then we define the {\emph{\cp}} $C^* (G, A, \af)$
to be the norm closure of $\sm (L^1 (G, A, \af))$.
\end{dfn}
One could of course equally well use
the norm closure of $\sm (C_{\mathrm{c}} (G, A, \af))$.
\begin{exr}\label{P:FixSet}
Give a set theoretically correct definition of the \cp.
The important point is to preserve the universal
property in \Thm{P_6X20_CPIntFBij};
prove that your definition does this.
\end{exr}
It follows that
every nondegenerate covariant representation of $(G, A, \af)$
gives a representation of $C^* (G, A, \af)$.
(Take the integrated form,
and restrict elements of $C^* (G, A, \af)$ to the
appropriate summand in the direct sum in Definition~\ref{D:CP}.)
The \cp{} is,
essentially by construction,
the universal \ca{} for covariant representations of $(G, A, \af)$,
in the same sense that if $G$ is a locally compact group,
then $C^* (G)$ is
the universal \ca{} for unitary representations of~$G$.
\Thm{T:IntFBij} then becomes the following result,
which is the analog for crossed products
of Theorem~\ref{T_3328_IntFCSt} (for group \ca{s}).
\begin{thm}[Propositions 2.39 and 2.40 of~\cite{Wlms};
Theorem 7.6.6 of~\cite{Pd1}]\label{P_6X20_CPIntFBij}
Let $\af \colon G \to \Aut (A)$
be an action of a locally compact group $G$ on a \ca~$A$,
and let $H$ be a Hilbert space.
Then the integrated form construction defines a bijection
from the set of nondegenerate covariant representations
of $(G, A, \af)$ on~$H$
to the set of nondegenerate representations
of $C^* (G, A, \af)$ on~$H$.
\end{thm}
\begin{exr}\label{Ex_6X20_IntFBijCst}
Prove Theorem~\ref{P_6X20_CPIntFBij}
when $G$ is discrete and $A$ is unital.
\end{exr}
\begin{rmk}\label{R:CPNtn}
There are many notations in use for crossed products,
and for related objects called reduced \cp s
(to be constructed in
Section~\ref{Sec:RedCP} below).
Here are most of the most common ones,
listed in pairs (notation for the full \cp{} first):
\begin{itemize}
\item
$C^* (G, A, \af)$ and $C^*_{\mathrm{r}} (G, A, \af)$.
\item
$C^* (A, G, \af)$ and $C^*_{\mathrm{r}} (A, G, \af)$.
\item
$A \rtimes_{\af} G$ and $A \rtimes_{\af, {\mathrm{r}}} G$
(used in the book~\cite{Wlms}).
\item
$A \times_{\af} G$ and $A \times_{\af, {\mathrm{r}}} G$
(used in the book~\cite{Dvd}).
\item
$G \times_{\af} A$ and $G \times_{\af, {\mathrm{r}}} A$
(used in the book~\cite{Pd1}).
\end{itemize}
In all of them, we may omit $\af$ if it is understood.
In the notation for the reduced crossed products (especially
the first two versions), the letter ``r'' (``reduced'')
is sometimes replaced by ``$\ld$''
(the conventional name for the left regular representation of a group).
The symbol in the third comes from the relation
$C^* (N \rtimes H) \cong C^* (H, \, C^* (N))$,
and is meant to suggest a generalized semidirect product.
The first two make it easy to distinguish \ca{} crossed products
from other sorts,
such as von Neumann algebra \cp{s},
smooth \cp{s},
$L^1$ \cp{s},
$L^p$~operator \cp{s},
and purely algebraic \cp s
(all of which will receive short shrift in these notes,
but are important in their own right,
sometimes in the same paper).
I use the order $C^* (G, A, \af)$ because it matches the
natural order in $C_{\mathrm{c}} (G, A, \af)$ and $L^1 (G, A, \af)$.
\end{rmk}
\begin{dfn}\label{DTGCA}
Let $G$ be a locally compact group,
let $X$ be a \lchs,
and let $(g, x) \mapsto g x$ be an action of~$G$ on~$X$.
The {\emph{transformation group C*algebra}} of $(G, X)$,
written $C^* (G, X)$, is the \cp{} \ca{} $C^* (G, \, C_0 (X))$.
\end{dfn}
\begin{thm}\label{L:UnivCP}
Let $\af \colon G \to \Aut (A)$
be an action of a discrete group $G$ on a unital \ca~$A$.
Then $C^* (G, A, \af)$ is the universal \ca{} generated by
a unital copy of $A$
(that is, the identity of $A$ is supposed to be the identity of the
generated \ca)
and unitaries $u_g$, for $g \in G$,
subject to the relations $u_g u_h = u_{g h}$ for $g, h \in G$
and $u_g a u_g^* = \af_g (a)$ for $a \in A$ and $g \in G$.
\end{thm}
\begin{exr}\label{P:PfUniv}
Based on the discussion above,
write down a careful proof of Theorem~\ref{L:UnivCP}.
\end{exr}
\begin{cor}\label{C:UnivCPZ}
Let $A$ be a \uca, and let $\af \in \Aut (A)$.
Then the \cp{} $C^* (\Z, A, \af)$ is the universal \ca{} generated by
a copy of $A$ and a unitary $u$,
subject to the relations
$u a u^* = \af (a)$ for $a \in A$.
\end{cor}
We now discuss functoriality of crossed products.
% For our discussion of functoriality,
The locally compact group~$G$
will be treated as fixed.
Since we have not included full proofs earlier
in this section when $G$ is not discrete,
we are not giving self contained proofs of
the functoriality results.
However,
given the results stated earlier,
the functoriality proofs are the same even when $G$ is not discrete.
\begin{dfn}\label{D:GAlg}
Let $G$ be a \lcg.
A \ca~$A$ equipped with an action $G \to \Aut (A)$
will be called a {\emph{$G$algebra}},
or a {\emph{$G$\ca}}.
We sometimes refer to $(G, A, \af)$ as a $G$algebra or $G$\ca.
\end{dfn}
Recall from \Def{D_6X01_Conjugate}
that if $(G, A, \af)$ and $(G, B, \bt)$ are $G$algebras,
then a \hm{} $\ph \colon A \to B$ is said to be
% {\emph{equivariant}}
equivariant
if for every $g \in G$, we have $\ph \circ \af_g = \bt_g \circ \ph$.
We say that
$\ph$ is {\emph{$G$equivariant}} if the group must be specified.
\begin{prp}\label{L_GAlgCat}
For a fixed \lcg~$G$,
the $G$algebras and equivariant \hm{s} form a category.
\end{prp}
\begin{proof}
This is obvious.
\end{proof}
We will need to use degenerate covariant representations
when considering functoriality for \hm{s}
whose ranges are ``too small''
(such as being contained in proper ideals).
We recall the following standard lemma on
degenerate representations of \ca{s}.
We omit the easy proof.
\begin{lem}\label{L_4217_DegRepFacts}
Let $A$ be a \ca,
let $H_0$ be a Hilbert space,
and let $\pi_0 \colon A \to L (H_0)$ be a representation.
Let $H$ be the closed linear span of $\pi_0 (A) H_0$.
Then:
\begin{enumerate}
\item\label{L_4217_DegRepFacts_Inv}
The subspace $H$ is invariant for~$\pi$.
\item\label{L_4217_DegRepFacts_Ndg}
The representation $\pi = \pi_0 () _H$ is nondegenerate.
\item\label{L_4217_DegRepFacts_Perp}
We have
\[
H^{\perp} = \big\{ \xi \in H_0 \colon
{\mbox{$\pi (a) \xi = 0$ for all $a \in A$}} \big\}.
\]
\item\label{L_4217_DegRepFacts_DSum}
The representation $\pi_0$ is the direct sum
of $\pi$ and the zero representation on $H^{\perp}$.
\item\label{L_4217_DegRepFacts_Norm}
For all $a \in A$,
we have $\ \pi (a) \ = \ \pi_0 (a) \$.
\item\label{L_4217_DegRepFacts_Ker}
We have ${\operatorname{Ker}} (\pi) = {\operatorname{Ker}} (\pi_0)$.
\item\label{L_4217_DegRepFacts_GivenD}
If $\pi_0$ is given as the direct sum of a nondegenerate
representation on a Hilbert space~$H_1$
and the zero representation on a Hilbert space~$H_2$,
then $H = H_1$ and $H^{\perp} = H_2$.
\end{enumerate}
\end{lem}
\begin{lem}\label{L_4217_IntFDSum}
Let $\af \colon G \to \Aut (A)$
be an action of a locally compact group $G$ on a \ca~$A$.
Let $I$ be a set,
and for $i \in I$ let $(v_i, \pi_i)$ be a covariant representation
of $(G, A, \af)$ on a Hilbert space~$H_i$,
with integrated form $\sm_i$.
Then
$\left( \bigoplus_{i \in I} v_i, \, \bigoplus_{i \in I} \pi_i \right)$ is a covariant representation
of $(G, A, \af)$ on $\bigoplus_{i \in I} H_i$,
and its integrated form is $\bigoplus_{i \in I} \sm_i$.
\end{lem}
\begin{proof}
The proof is routine.
\end{proof}
\begin{lem}\label{L_4217_IntFZero}
Let $\af \colon G \to \Aut (A)$
be an action of a locally compact group $G$ on a \ca~$A$.
Let $(v, \pi)$ be a covariant representation
of $(G, A, \af)$ on a Hilbert space~$H$.
If $\pi$ is the zero representation,
then the integrated form $\sm$ of $(v, \pi)$
is the zero representation.
\end{lem}
\begin{proof}
It is immediate that $\sm (a) \xi = 0$
for all $a \in C_{\mathrm{c}} (G, A, \af)$
and $\xi \in H$.
\end{proof}
\begin{lem}\label{L_4215_DegRpn}
Let $\af \colon G \to \Aut (A)$
be an action of a locally compact group $G$ on a \ca~$A$.
Let $(v_0, \pi_0)$ be a covariant representation
of $(G, A, \af)$ on a Hilbert space~$H_0$.
(We do not assume that $\pi_0$ is nondegenerate.)
Let $\sm_0 \colon C_{\mathrm{c}} (G, A, \af) \to L (H_0)$
be the integrated form of $(v_0, \pi_0)$,
as in \Def{D:IntForm}.
Then $\pi_0 (A) H_0$ and $\sm_0 (C^* (G, A, \af)) H_0$
have the same closed linear spans.
\end{lem}
\begin{proof}
% \begin{proof}[Proof of \Lem{L_4215_DegRpn}]
Let $H$ be the closed linear span of $\pi_0 (A) H_0$.
Then $H$ is invariant under $\pi_0$
by \Lem{L_4217_DegRepFacts}(\ref{L_4217_DegRepFacts_Inv}).
We claim that $H$ is invariant under~$v_0$.
It is enough to prove invariance of $\pi_0 (A) H_0$.
Let $g \in G$, let $a \in A$, and let $\xi \in H_0$.
Then
\[
v_0 (g) \pi_0 (a) \xi
= \pi_0 (\af_g (a)) v_0 (g) \xi
\in \pi_0 (A) H_0.
\]
The claim is proved.
Set $\pi = \pi_0 () _H$.
Then $\pi_0 = \pi \oplus 0$
by \Lem{L_4217_DegRepFacts}(\ref{L_4217_DegRepFacts_DSum}),
and the claim implies the existence of
a representation $w$ of $G$ on $H^{\perp}$
such that $v_0 = v \oplus w$.
Let $\sm$ be the integrated form of $(v, \pi)$.
Use \Lem{L_4217_IntFDSum}
and then \Lem{L_4217_IntFZero}
to get $\sm_0 = \sm \oplus 0$,
the zero representation being on $H^{\perp}$.
It follows from
Theorem~\ref{T:IntFBij}
and \Def{D:CP}
that $\sm$ is nondegenerate.
Therefore the conclusion follows from
\Lem{L_4217_DegRepFacts}(\ref{L_4217_DegRepFacts_GivenD}).
\end{proof}
\begin{cor}\label{C_4217_DegNorm}
Let $\af \colon G \to \Aut (A)$
be an action of a locally compact group $G$ on a \ca~$A$.
Let $a \in C^* (G, A, \af)$.
Then
\begin{align*}
\ a \
& = \sup \big( \big\{ \ \sm (a) \ \colon
{\mbox{$\sm$ is the integrated form of a
possibly}}
\\
& \hspace*{5em} {\mbox{degenerate covariant representation
of $(G, A, \af)$}} \big\} \big).
\end{align*}
\end{cor}
\begin{proof}
It follows from Lemma~\ref{L_4215_DegRpn},
\Lem{L_4217_DegRepFacts}(\ref{L_4217_DegRepFacts_Ndg}),
and \Lem{L_4217_DegRepFacts}(\ref{L_4217_DegRepFacts_Norm})
that the supremum on the right
is unchanged if we restrict to nondegenerate
covariant representations of $(G, A, \af)$.
\end{proof}
\begin{thm}\label{T:CPFunct}
Let $G$ be a \lcg.
If $(G, A, \af)$ and $(G, B, \bt)$ are $G$algebras
and $\ph \colon A \to B$ is an equivariant \hm,
then there is a \hm{}
$\ps \colon C_{\mathrm{c}} (G, A, \af) \to C_{\mathrm{c}} (G, B, \bt)$
given by the formula
$\ps (b) (g) = \ph (b (g))$ for $b \in C_{\mathrm{c}} (G, A, \af)$
and $g \in G$,
and this \hm{} extends by continuity to a \hm{}
$L^1 (G, A, \af) \to L^1 (G, B, \bt)$,
and then to a \hm{}
$C^* (G, A, \af) \to C^* (G, B, \bt)$.
This construction makes the \cp{} construction
a functor from the category of $G$algebras to the category of \ca{s}.
\end{thm}
\begin{proof}
One checks directly that
$\ps$ preserves multiplication and adjoint,
and that
$\ \ps (a) \_1 \leq \ a \_1$
for all $a \in C_{\mathrm{c}} (G, A, \af)$.
The extension to the $L^1$algebras
is now immediate.
To prove that $\ps$ extends by continuity to a \hm{}
$C^* (G, A, \af) \to C^* (G, B, \bt)$,
we let $\ \cdot \$ denote restrictions
to $C_{\mathrm{c}} (G, A, \af)$ and $C_{\mathrm{c}} (G, B, \bt)$
of the norms on $C^* (G, A, \af)$ and $C^* (G, B, \bt)$.
We have to prove that
$\ \ps (b) \ \leq \ b \$
for all $b \in C_{\mathrm{c}} (G, A, \af)$.
So let $(w, \rh)$ be a nondegenerate covariant representation
of $(G, B, \bt)$ on a Hilbert space~$H$,
and let $\nu \colon C_{\mathrm{c}} (G, B, \bt) \to L (H)$
be the integrated form of $(w, \rh)$,
as in \Def{D:IntForm}.
We have to prove that
$\ \nu ( \ps (b)) \ \leq \ b \$.
Clearly $(w, \, \rh \circ \ph)$
is a covariant representation
of $(G, A, \af)$ on~$H$,
with integrated form $\sm = \nu \circ \ps$.
There is no reason to suppose that $(w, \, \rh \circ \ph)$
is nondegenerate,
but,
even without nondegeneracy,
Corollary~\ref{C_4217_DegNorm}
gives $\ \sm (b) \ \leq \ b \$.
Thus $\ \nu ( \ps (b)) \ = \ \sm (b) \ \leq \ b \$,
as desired.
\end{proof}
\begin{thm}[Lemma~2.8.2 of~\cite{Ph1}; Theorem~2.6 of~\cite{Sh};
Proposition~3.9 of~\cite{Wlms}]\label{T_CPExact}
Let $G$ be a locally compact group.
Let
\[
0 \longrightarrow J
\stackrel{\io_0}{\longrightarrow} A
\stackrel{\kp_0}{\longrightarrow} B
\longrightarrow 0
\]
be an exact sequence of $G$algebras,
with actions $\gm$ on $J$, $\af$ on $A$, and $\bt$ on $B$.
Then the sequence
\[
0 \longrightarrow C^* (G, J, \gm)
\stackrel{\io}{\longrightarrow} C^* (G, A, \af)
\stackrel{\kp}{\longrightarrow} C^* (G, B, \bt)
\longrightarrow 0
\]
of crossed products and induced maps is exact.
\end{thm}
Theorem~\ref{T_CPExact} implies in particular that
if $(G, J, \gm)$ and $(G, A, \af)$ are $G$algebras,
and $\ph \colon J \to A$ is an injective equivariant \hm{}
whose image is an ideal,
then the corresponding \hm{}
$C^* (G, J, \gm) \to C^* (G, A, \af)$ is injective.
% If the image is merely a subalgebra,
% presumably this map need {\emph{not}} be injective,
% but we do not know a counterexample.
% (Reference for counterexample to be found 999)
% The difficulty in the proof occurs when trying to
If the image is merely a subalgebra,
the proof fails.
The difficulty occurs when we
extend a covariant representation of $(G, J, \gm)$
to a covariant representation of~$(G, A, \af)$.
If $J$ is not an ideal,
to extend a representation of $J$ to one of~$A$
one usually needs a bigger Hilbert space,
and one has trouble with how to extend the representation of~$G$
to a representation on the larger space.
The (full) crossed product should be thought of as
somehow analogous to the maximal tensor product
of \ca{s}.
Similarly, the reduced crossed product
(discussed in Section~\ref{Sec:RedCP} below)
should be thought of as
somehow analogous to the minimal tensor product
of \ca{s}.
Compare with Example~\ref{E:C:Triv},
where it is observed that if the action of $G$ on~$A$
is trivial,
then
\[
C^* (G, A) \cong C^* (G) \otimes_{\mathrm{max}} A
\andeqn
C^*_{\mathrm{r}} (G, A)
\cong C^*_{\mathrm{r}} (G) \otimes_{\mathrm{min}} A.
\]
Theorem~\ref{T_CPExact}
should then be compared with
Proposition 3.7.1 of~\cite{BrOz},
according to which $A \otimes_{\mathrm{max}} {}$
is an exact functor.
The proof of Theorem~\ref{T_CPExact}
requires at least the first part of the following exercise.
For this part,
% part~(\ref{E_6X20_CXASurj_App}),
one can use a partition of unity argument
similar to that in the proof of \Lem{L_3402_LinOfInt}.
For part~(\ref{E_6X20_CXASurj_Exact}),
one can then apply part~(\ref{E_6X20_CXASurj_App})
to the error ${\ov{\kp}} (a)  b$,
with a smaller error,
repeat, and sum the results.
One can also reduce to the case in which $X$ is compact,
where one can apply C*algebraic tensor products
and the isomorphism $C (X, A) \cong C (X) \otimes A$.
\begin{exr}\label{E_6X20_CXASurj}
Let $X$ be a locally \chs,
let $A$ and $B$ be \ca{s},
and let $\kp \colon A \to B$ be a surjective \hm.
Let
${\ov{\kp}}
\colon C_{\mathrm{c}} (X, A) \to C_{\mathrm{c}} (X, B)$
be the linear map given by
${\ov{\kp}} (a) (x) = \kp (a (x))$
for $a \in C_{\mathrm{c}} (X, A)$ and $x \in X$.
Let $b \in C_{\mathrm{c}} (X, B)$.
\begin{enumerate}
\item\label{E_6X20_CXASurj_App}
Prove that there is a compact set $K \subset X$
such that for every $\ep > 0$
there is $a \in C_{\mathrm{c}} (X, A)$
satisfying
\[
\ a \_{\I} \leq \ b \_{\I} + \ep,
\qquad
\supp (a) \subset K,
\andeqn
\ {\ov{\kp}} (a)  b \_{\I} < \ep.
\]
\item\label{E_6X20_CXASurj_Exact}
Prove that there is $a \in C_{\mathrm{c}} (X, A)$
such that ${\ov{\kp}} (a) = b$
and $\supp (a) = \supp (b)$.
\end{enumerate}
\end{exr}
\begin{proof}[Proof of Theorem~\ref{T_CPExact}]
We prove that $\kp$ is surjective.
Since $\kp$ is a \hm,
it suffices to prove that $\kp$ has dense range.
It follows from Exercise \ref{E_6X20_CXASurj}(\ref{E_6X20_CXASurj_Exact})
that the range of $\kp$ contains the image
of $C_{\mathrm{c}} (G, B, \bt)$,
and we know that the image
of $C_{\mathrm{c}} (G, B, \bt)$ is dense.
(Actually,
Exercise \ref{E_6X20_CXASurj}(\ref{E_6X20_CXASurj_App})
is good enough here,
since it implies that
the closure of the range of $\kp$
contains the image
of $C_{\mathrm{c}} (G, B, \bt)$.)
This proves surjectivity of~$\kp$.
It is immediate that $\kp \circ \io = 0$.
We prove that $\io$ is injective.
For this,
it is convenient to identify $J$ with the ideal $\io_0 (J) \subset A$.
Let $y \in C^* (G, J, \gm)$ be nonzero.
Choose a nondegenerate covariant representation $(v, \pi_0)$
of $(G, J, \gm)$ on a Hilbert space~$H$
such that the integrated form $\pi \colon C^* (G, J, \gm) \to L (H)$
satisfies $\pi (y) \neq 0$.
Since $\pi_0$ is nondegenerate and $J \subset A$ is an ideal,
a standard result in the representation theory of \ca{s}
shows that there is a unique representation $\rh_0 \colon A \to L (H)$
such that $\rh_0 _J = \pi_0$.
We claim that $(v, \rh_0)$ is covariant.
Let $g \in G$.
Since $(v, \pi_0)$ is covariant,
$a \mapsto v (g) \rh (\af_g^{1} (a)) v (g)^*$
is a representation whose restriction to $J$ is~$\pi_0$.
By uniqueness of $\rh_0$,
we have $v (g) \rh_0 (\af_g^{1} (a)) v (g)^* = \rh_0 (a)$
for all $a \in A$,
which is covariance.
Let $\rh \colon C^* (G, A, \af) \to L (H)$
be the integrated form of $(v, \rh_0)$.
Then $\rh \circ \io = \pi$,
so $\rh ( \io (y)) = \pi (y) \neq 0$.
Therefore $\io (y) \neq 0$.
It remains to prove that if $y \in C^* (G, A, \af)$
and $\kp (y) = 0$,
then $y$ is in the range of~$\io$.
We again identify $J$ with the ideal $\io_0 (J) \subset A$.
Since $\io$ is injective,
we may use $\io$ to identify $C^* (G, J, \gm)$
with a subalgebra of $C^* (G, A, \af)$.
Since $C_{\mathrm{c}} (G, J, \gm)$
is an ideal in $C_{\mathrm{c}} (G, A, \af)$
and since $C_{\mathrm{c}} (G, J, \gm)$ and $C_{\mathrm{c}} (G, A, \af)$
are dense in $C^* (G, J, \gm)$ and $C^* (G, A, \af)$,
it follows that $C^* (G, J, \gm)$ is an ideal in $C^* (G, A, \af)$.
Let $y \in C^* (G, A, \af)$ and suppose that
$y \not\in C^* (G, J, \gm)$.
We show that $\kp (y) \neq 0$.
Use a nondegenerate representation
of $C^* (G, A, \af) / C^* (G, J, \gm)$ which does not vanish on~$y$
to find a Hilbert space $H$
and a nondegenerate representation
$\sm \colon C^* (G, A, \af) \to L (H)$ such that $\sm (y) \neq 0$
but $\sm _{C^* (G, J, \gm)} = 0$.
Then $\sm$ is the integrated form
of a nondegenerate covariant representation $(w, \sm_0)$
of $(G, A, \af)$.
Since $\sm _{C^* (G, J, \gm)} = 0$,
\Lem{L_4215_DegRpn}
implies that $\sm_0 _J = 0$.
So $\sm_0$ induces a representation
$\pi_0 \colon B \to L (H)$.
Clearly $(w, \pi_0)$ is a nondegenerate covariant representation
of $(G, B, \bt)$
whose integrated form $\pi$ satisfies $\pi \circ \kp = \sm$.
So $\pi (\kp (y)) \neq 0$.
Thus $\kp (y) \neq 0$.
\end{proof}
\begin{thm}\label{T:DLim}
Let $G$ be a locally compact group.
Let
$\big( \big( G, A_i, \af^{(i)} \big)_{i \in I},
\, (\ph_{j, i})_{i \leq j} \big)$
be a direct system of $G$algebras.
Let $A = \Dirlim A_i$,
with action $\af \colon G \to \Aut (A)$ given by
$\af_g = \Dirlim \af^{(i)}_g$ for all $g \in G$.
(See Proposition~\ref{P_3305_DirLimAction}.)
Let
\[
\ps_{j, i}
\colon C^* \big( G, A_i, \af^{(i)} \big)
\to C^* \big( G, A_j, \af^{(j)} \big)
\]
be the map obtained from $\ph_{j, i}$.
Using these maps in the direct system of \cp{s},
there is a natural isomorphism
$C^* (G, A, \af) \cong \Dirlim C^* \big( G, A_i, \af^{(i)} \big)$.
\end{thm}
\begin{proof}
We show that $C^* (G, A, \af)$ satisfies the universal property
which defines $\Dirlim C^* \big( G, A_i, \af^{(i)} \big)$.
First, for $i \in I$ let $\ph_i \colon A_i \to A$ be the canonical map
for the direct limit of the system
$\big( (A_i )_{i \in I}, \, (\ph_{j, i})_{i \leq j} \big)$.
We have maps
\[
\ps_i \colon C^* \big( G, A_i, \af^{(i)} \big) \to C^* (G, A, \af)
\]
obtained from the maps $\ph_i$ by forming \cp{s}.
Clearly $\ps_j \circ \ps_{j, i} = \ps_i$
whenever $i, j \in I$ satisfy $i \leq j$.
Now suppose we have a \ca{} $B$ and \hm s
$\nu_i \colon C^* \big( G, A_i, \af^{(i)} \big) \to B$
such that $\nu_j \circ \ps_{j, i} = \nu_i$
whenever $i, j \in I$ satisfy $i \leq j$.
We need to prove that there is a unique \hm{}
$\nu \colon C^* (G, A, \af) \to B$
such that $\nu \circ \ps_i = \nu_i$ for all $i \in I$.
\Wolog, $B$ is a nondegenerate subalgebra of $L (H)$
for some Hilbert space~$H$.
For each $i \in I$,
set
\[
H_i = {\ov{ \nu_i \big( C^* \big( G, A_i, \af^{(i)} \big) \big) H}}.
\]
Keeping \Lem{L_4217_DegRepFacts} in mind for the next several
paragraphs,
observe that there is a nondegenerate covariant representation
$(v_i, \pi_i)$ of $\big( G, A_i, \af^{(i)} \big)$
on $H_i$
whose integrated form is $\nu_i () _{H_i}$.
Extend $\pi_i$ to a representation on~$H$
by forming the direct sum with the zero representation on $H_i^{\perp}$.
Let $i, j \in I$ satisfy $i \leq j$.
Then $H_i \subset H_j$.
Moreover,
$H_i$ is an invariant subspace for $v_j$
and, by uniqueness of the nondegenerate covariant representation
determined by a nondegenerate representation
of the crossed product
(Theorem~\ref{P_6X20_CPIntFBij}),
we have
\[
v_j () _{H_i} = v_i ()
\andeqn
(\pi_j \circ \ph_{j, i}) () _{H_i} = \pi_i ().
\]
Moreover,
both $(\pi_j \circ \ph_{j, i}) ()$ and $\pi_i ()$
are zero on $H_j \cap H_i^{\perp}$
and on $H_j^{\perp}$,
so $\pi_j \circ \ph_{j, i} = \pi_i$.
Since $B$ is nondegenerate,
we have ${\ov{\bigcup_{i \in I} H_i}} = H$.
It is then easy to see that there is a unique
unitary representation $v$ of $G$ on~$H$
such that $v () _{H_i} = v_i$ for all $i \in I$.
By the universal property of $\Dirlim A_i$,
there is a unique representation $\pi \colon A \to L (H)$
such that $\pi \circ \ph_i = \pi_i$ for all $i \in I$,
and moreover (using uniqueness)
$(v, \pi)$ is a covariant representation.
Let $\nu \colon C^* (G, A, \af) \to L (H)$
be the integrated form of $(v, \pi)$.
Then one gets $\nu \circ \ps_i = \nu_i$ for all $i \in I$.
Since $A$ is generated by the images of the algebras~$A_i$,
it follows that $\nu ( C^* (G, A, \af) ) \subset B$.
Uniqueness of $\nu$ follows from uniqueness of
the integrated form of a covariant representation.
\end{proof}
\section{Reduced Crossed Products}\label{Sec:RedCP}
\indent
So far, it is not clear that a \ga{} $\GAa$
has any covariant
representations at all.
In this section, we exhibit a large
easily constructed class of them,
called regular covariant representations.
We then study the reduced crossed product,
which is defined by using the universal regular representation
in place of the universal representation.
We will concentrate on the case of discrete groups.
As in Sections \ref{Sec_LCGpCSt} and~\ref{Sec:CP},
we let $\mu$ be a fixed left Haar measure on~$G$.
We will need Hilbert spaces of the form $L^2 (G, H_0)$.
The easy way to construct $L^2 (G, H_0)$ is to take it to be the
completion of $C_{\mathrm{c}} (G, H_0)$ in the norm
coming from the scalar product
\[
\langle \xi, \et \rangle
= \int_G \langle \xi (g), \et (g) \rangle \, d \mu (g).
\]
\begin{dfn}[7.7.1 of~\cite{Pd1}]\label{D:RegRpn}
Let $\af \colon G \to \Aut (A)$
be an action of a locally compact group $G$ on a \ca~$A$.
Let $\pi_0 \colon A \to L (H_0)$ be a representation.
We define the {\emph{regular covariant representation}}
$(v, \pi)$ of $(G, A, \af)$ on the Hilbert space $H = L^2 (G, H_0)$
of $L^2$ functions from $G$ to $H_0$
as follows.
For $g, h \in G$, set
\[
(v (g) \xi) (h) = \xi (g^{1} h).
\]
For $a \in A$ and $g \in G$, set
\[
(\pi (a) \xi) (h) = \pi_0 (\af_{h^{1}} (a)) ( \xi (h)).
\]
(\Exr{P:RegIsRep} asks you to prove that $(v, \pi)$
really is covariant.)
The integrated form of~$\sm$,
as in Definition~\ref{D:IntForm},
will be called a regular representation of
any of $C_{\mathrm{c}} (G, A, \af)$,
$L^1 (G, A, \af)$,
$C^* (G, A, \af)$,
and (when we have defined it;
see Definition~\ref{D:RedCP})
$C^*_{\mathrm{r}} (G, A, \af)$.
Justified by \Lem{L_4215_RegIsND} below,
we will refer to $(v, \pi)$
as a nondegenerate covariant representation
when $\pi_0$ is nondegenerate.
\end{dfn}
\begin{exr}\label{P:RegIsRep}
In Definition~\ref{D:RegRpn},
prove that $(v, \pi)$ really is a covariant representation.
\end{exr}
If $A = \C$, $H_0 = \C$,
and $\pi_0$ is the obvious representation of $A$ on $H_0$,
then the representation of Definition~\ref{D:RegRpn} is
the usual left regular representation of~$G$
(\Def{D_3402_LCLReg}; \Def{D_3407_DRegRep} in the discrete case).
\begin{lem}\label{L_4215_RegIsND}
In Definition~\ref{D:RegRpn}, the representation $\pi$ is nondegenerate
\ifo{} $\pi_0$ is nondegenerate.
\end{lem}
\begin{proof}
Suppose $\pi_0$ is degenerate.
Choose a nonzero element $\xi_0 \in (\pi_0 (A) H_0)^{\perp}$.
\Lem{L_4217_DegRepFacts}(\ref{L_4217_DegRepFacts_Perp})
implies that $\pi_0 (a) \xi = 0$ for all $a \in A$.
Choose a nonzero function $f \in C_c (G)$.
Define $\xi (g) = f (g) \xi_0$ for $g \in G$.
Then $\xi$ is a nonzero element of $L^2 (G, H_0)$,
and $\pi (a) \xi = 0$ for all $a \in A$.
So $\pi$ is degenerate.
Now assume that $\pi_0$ is nondegenerate.
It suffices to show that ${\overline{\pi (A) L^2 (G, H_0)}}$
contains all elements $\xi \in C_{\mathrm{c}} (G, H_0)$
which are elementary tensors,
that is,
for which there exist $f \in C_{\mathrm{c}} (G)$
and $\xi_0 \in H_0$
such that $\xi (h) = f (h) \xi_0$
for all $h \in G$.
Let $\xi$, $f$, and $\xi_0$ be as above,
and let $\ep > 0$.
Recall that $\mu$ is a left Haar measure on~$G$.
Set
\[
M = \big( \mu ( \supp (f)) + 1 \big)^{1/2} (\ f \ + 1).
\]
Since $\pi_0$ is nondegenerate,
there are $a \in A$ and $\et_0 \in H_0$
such that
\[
\ \pi_0 (a) \et_0  \xi_0 \ < \frac{\ep}{2 M}.
\]
Since
$\big\{ \af_h (a) \colon h \in \supp (f) \big\}$
is compact,
there is $b \in A$
such that
\[
\ b \af_h (a)  \af_h (a) \ < \frac{\ep}{2 M (\ \et_0 \ + 1)}
\]
for all $h \in \supp (f)$.
Then
\[
\ \af_{h}^{1} (b) a  a \ < \frac{\ep}{2 M (\ \et_0 \ + 1)}
\]
for all $h \in \supp (f)$.
Define $\et \in C_{\mathrm{c}} (G)$
by $\et (h) = f (h) \pi_0 (a) \et_0$ for $h \in G$.
For all $h \in \supp (f)$,
we then have
\begin{align*}
\ (\pi (b) \et) (h)  \xi (h) \
& =  f (h)  \cdot
\big\ \pi_0 (\af_{h}^{1} (b)) (\pi_0 (a) \et_0)  f (h) \xi_0 \big\
\\
& \leq \ f \ \cdot \ \af_{h}^{1} (b) a  a \ \cdot \ \et_0 \
+ \ f \ \cdot \ \pi_0 (a) \et_0  \xi_0 \
\\
& < \frac{\ep \ f \ \cdot \ \et_0 \}{2 M (\ \et_0 \ + 1)}
+ \frac{\ep \ f \}{2 M}
\\
& < \frac{\ep}{2 \mu ( \supp (f)) + 1)^{1/2}}
+ \frac{\ep}{2 \mu ( \supp (f)) + 1)^{1/2}}
\\
& = \frac{\ep}{\mu ( \supp (f)) + 1)^{1/2}}.
\end{align*}
Therefore
\[
\ \pi (b) \et  \xi \^2
\leq \mu ( \supp (f))
\left( \frac{\ep}{\mu ( \supp (f)) + 1)^{1/2}} \right)^2
< \ep^2,
\]
so $\ \pi (b) \et  \xi \ < \ep$.
\end{proof}
In the following definition,
we ignore a set theoretic problem analogous to
those encountered previously,
for example in \Def{D:CP}.
\begin{dfn}\label{D:RedCP}
Let $\af \colon G \to \Aut (A)$
be an action of a locally compact group $G$ on a \ca~$A$.
Let $\ld \colon L^1 (G, A, \af) \to L (H)$
be the direct sum of all regular
representations of $L^1 (G, A, \af)$
coming from nondegenerate representations of~$A$.
We define the {\emph{reduced crossed product}}
$C^*_{\mathrm{r}} (G, A, \af)$
to be the norm closure of $\ld (L^1 (G, A, \af))$.
\end{dfn}
\begin{exr}\label{Ex_6X20_SetRedCP}
Give a set theoretically correct definition of the reduced \cp.
\end{exr}
We use notation analogous to that of Definition~\ref{DTGCA}
in the case of an action on a locally compact space.
\begin{dfn}\label{DredTGCA}
Let $G$ be a locally compact group,
let $X$ be a \lchs,
and let $(g, x) \mapsto g x$ be an action of~$G$ on~$X$.
The {\emph{reduced transformation group C*algebra}} of $(G, X)$,
written $C^*_{\mathrm{r}} (G, X)$,
is the reduced \cp{} \ca{} $C^*_{\mathrm{r}} (G, \, C_0 (X))$.
\end{dfn}
Implicit in the definition of $C^*_{\mathrm{r}} (G, A, \af)$
is a representation of $L^1 (G, A, \af)$,
hence of $C^* (G, A, \af)$.
Thus, there is a \hm{}
$C^* (G, A, \af) \to C^*_{\mathrm{r}} (G, A, \af)$.
By construction, it has dense range,
and is therefore surjective.
Moreover, by construction,
any regular representation of $L^1 (G, A, \af)$
extends to a representation of $C_{\mathrm{r}}^* (G, A, \af)$.
In the context of the next theorem,
see the comments before \Thm{T_4131_GpFToRed}
for a discussion of amenability.
\begin{thm}[Theorem~7.13 of~\cite{Wlms};
Theorem 7.7.7 of~\cite{Pd1}]\label{T_4131_CPFToRed}
Let $\af \colon G \to \Aut (A)$
be an action of a locally compact group $G$ on a \ca~$A$.
If $G$ is amenable,
then $C^* (G, A, \af) \to C^*_{\mathrm{r}} (G, A, \af)$
is an isomorphism.
\end{thm}
The converse is true for $A = \C$:
if $C^* (G) \to C^*_{\mathrm{r}} (G)$
is an isomorphism,
then $G$ is amenable.
See Theorem 7.3.9 of~\cite{Pd1}.
But it is not true in general.
For example, if $G$ acts on itself by translation,
then $C^* (G, \, C_0 (G)) \to C^*_{\mathrm{r}} (G, \, C_0 (G))$
is an isomorphism for every~$G$.
See Example~\ref{E_C_Trans} for the case of a discrete group.
The proof of \Thm{T_4131_CPFToRed}
is similar to that of \Thm{T_4131_GpFToRed},
with the algebra~$A$
just carried along.
\begin{proof}[Proof of Theorem~\ref{T_4131_CPFToRed}]
Let
$\io \colon C_{\mathrm{c}} (G, A, \af) \to C^* (G, A, \af)$
and $\kp \colon C^* (G, A, \af) \to C^*_{\mathrm{r}} (G, A, \af)$
be the standard maps.
We have to prove that
$\ \kp ( \io (b)) \ \geq \ \io (b) \$
for all $b \in C_{\mathrm{c}} (G, A, \af)$.
It suffices to prove the following.
Let $b \in C_{\mathrm{c}} (G, A, \af)$,
let $H$ be a Hilbert space,
let $(w, \sm)$ be a nondegenerate
covariant \rpn{} of $(G, A, \af)$ on $H$,
and let $\ep > 0$.
Then there is a Hilbert space~$E$
and a nondegenerate \rpn{}
$\pi_0 \colon A \to L (E)$
such that, if we let $(y, \pi)$ be the associated
regular covariant \rpn{} of Definition~\ref{D:RegRpn},
then
\[
\ (w \ltimes \sm) (b) \  \ep
< \ (y \ltimes \pi) (b) \.
\]
We will in fact take $\pi_0 = \sm$.
As usual,
let $\mu$ be a left Haar measure on~$G$.
Let $v$ be the left regular representation of $G$ on $L^2 (G)$.
Let $(y, \pi)$ be the regular covariant \rpn{}
associated to $\sm$,
which acts on $L^2 (G, H) = L^2 (G, \mu) \otimes H$.
Thus $y_g = v_g \otimes 1$ for all $g \in G$.
It is easy to check that there is a unique unitary
$z \in L ( L^2 (G, H) )$
such that
$(z \xi) (g) = w_g^{1} (\xi (g))$
for $\xi \in L^2 (G, H)$ and $g \in G$.
We claim that $z (v_h \otimes w_h) z^{1} = v_h \otimes 1$
for all $h \in G$
and that $z (1 \otimes \sm (a)) z^{1} = \pi (a)$
for all $a \in A$.
To check these,
let $\xi \in L^2 (G, H)$
and let $g \in G$.
Then
\begin{align*}
\big( z (v_h \otimes w_h) \xi \big) (g)
& = w_g^{1} \big( [ (v_h \otimes w_h) \xi ] (g) \big)
= w_g^{1} \big( w_h ( \xi (h^{1} g) ) \big)
\\
& = w_{g^{1} h} (\xi (h^{1} g))
= (z \xi) (h^{1} g)
= \big( (v_h \otimes 1) z \xi \big) (g)
\end{align*}
and,
using covariance of $(w, \sm)$ at the third step
and the definition of $\pi$ at the fifth step,
\begin{align*}
\big( z [1 \otimes \sm (a)] \xi \big) (g)
& = w_g^{1} \big( [(1 \otimes \sm (a)) \xi] (g) \big)
= w_g^{1} \sm (a) ( \xi ( g) )
\\
& = \sm \big( \af_g^{1} (a) \big) w_g^{1} ( \xi ( g) )
= \sm \big( \af_g^{1} (a) \big) \big( (z \xi) (g) \big)
= \big( \pi (a) z \xi \big) (g).
\end{align*}
This proves the claim.
Writing $1 \otimes \sm$ for the representation
$a \mapsto 1 \otimes \sm (a)$ on
$L^2 (G) \otimes H = L^2 (G, H)$,
and recalling the notation in \Def{D:IntForm}
for integrated forms of covariant representations,
the claim implies that
$(v \otimes w, \, 1 \otimes \sm )$
is a covariant representation and
\[
\big\ \big( (v \otimes w) \ltimes (1 \otimes \sm) \big) (b) \big\
= \ (y \ltimes \pi) (b) \.
\]
We finish the proof by showing that
\[
\big\ \big( (v \otimes w) \ltimes (1 \otimes \sm) \big) (b) \big\
> \ (w \ltimes \sm) (b) \  \ep.
\]
We may assume that $(w \ltimes \sm) (b) \neq 0$
and $\ep < \ (w \ltimes \sm) (b) \$.
Choose $\xi_0 \in H$
such that
\[
\ \xi_0 \ = 1
\andeqn
\ (w \ltimes \sm) (b) \xi_0 \
> \ (w \ltimes \sm) (b) \  \frac{\ep}{2}.
\]
Set
\[
\dt = \left( \frac{\ (w \ltimes \sm) (b) \  \frac{\ep}{2}}{
\ (w \ltimes \sm) (b) \  \ep} \right)^{2}
 1.
\]
Then $\dt > 0$.
Set $S = \supp (b) \cup \{ 1 \}$.
Then $S$ and $S^{1}$ are compact subsets of~$G$.
Since $G$ is amenable,
the main result of~\cite{EmGr}
(also see Theorem 3.1.1 there)
provides a compact subset $K \subset G$
such that
\[
0 < \mu (K) < \infty
\andeqn
\mu \big( S^{1} K \bigtriangleup K \big) < \dt \mu (K).
\]
Since $1 \in S^{1}$,
the second condition implies that
$\mu \big( S^{1} K \setminus K \big) < \dt \mu (K)$.
In particular, $\mu (S^{1} K) < (1 + \dt) \mu (K)$.
Define $\xi \in L^2 (G, H)$
by
\[
\xi (g) = \begin{cases}
\xi_0 & g \in S^{1} K
\\
0 & g \not\in S^{1} K.
\end{cases}
\]
Then
%
\begin{equation}\label{Eq_6X20_NStar}
\ \xi \
= \mu (S^{1} K)^{1/2} \ \xi_0 \
< (1 + \dt)^{1/2} \mu (K)^{1/2}.
\end{equation}
%
We estimate
$\big\ \big( (v \otimes w) \ltimes (1 \otimes \sm) \big) (b)
\xi \big\$.
For $g \in K$ we have,
at the fourth step using
$\xi (h^{1} g) = \xi_0$ whenever $b (h) \neq 0$,
\begin{align*}
\big(
\big[ (v \otimes w) \ltimes (1 \otimes \sm) \big] (b) \xi \big) (g)
& = \int_G \big( \big[ (1 \otimes \sm) (b (h) ) \big]
(v_h \otimes w_h) \xi \big) (g) \, d \mu (h)
\\
& = \int_G \sm (b (h)) w_h ( \xi (h^{1} g)) \, d \mu (h)
\\
& = \int_G \sm (b (h)) w_h \xi_0 \, d \mu (h)
= (w \ltimes \sm) (b) \xi_0.
\end{align*}
Therefore
\[
\big\ \big( (v \otimes w) \ltimes (1 \otimes \sm) \big) (b) \xi \big\
\geq \mu (K)^{1/2} \ (w \ltimes \sm) (b) \xi_0 \
> \mu (K)^{1/2}
\left( \ (w \ltimes \sm) (b) \  \frac{\ep}{2} \right),
\]
from which it follows using~(\ref{Eq_6X20_NStar}) that
\begin{align*}
\big\ \big( (v \otimes w) \ltimes (1 \otimes \sm) \big) (b) \big\
& > \frac{\mu (K)^{1/2}
\left( \ (w \ltimes \sm) (b) \  \frac{\ep}{2} \right)}{
(1 + \dt)^{1/2} \mu (K)^{1/2}}
\\
& = (1 + \dt)^{ 1/2}
\left( \ (w \ltimes \sm) (b) \  \frac{\ep}{2} \right)
= \ (w \ltimes \sm) (b) \  \ep,
\end{align*}
as desired.
\end{proof}
\begin{thm}\label{T:InjOnL1}
Let $\af \colon G \to \Aut (A)$
be an action of a locally compact group $G$ on a \ca~$A$.
Then $C_{\mathrm{c}} (G, A, \af) \to C^*_{\mathrm{r}} (G, A, \af)$
is injective.
\end{thm}
We will prove this below in the case of a discrete group.
The proof of the general case
can be found in Lemma~2.26 of~\cite{Wlms}.
It is, I believe, true that
$L^1 (G, A, \af) \to C^*_{\mathrm{r}} (G, A, \af)$ is injective,
and this can probably be proved by working a little harder
in the proof of Lemma~2.26 of~\cite{Wlms},
but I have not carried out the details and I do not know a reference.
% Improve if possible 999
\begin{thm}[Theorem~7.7.5 of~\cite{Pd1}]\label{T_4217_AnyRep}
Let $\af \colon G \to \Aut (A)$
be an action of a locally compact group $G$ on a \ca~$A$.
Let $\pi_0 \colon A \to L (H_0)$ be any nondegenerate injective
representation.
Then the integrated form of the regular representation
associated to~$\pi_0$ is injective on $C^*_{\mathrm{r}} (G, A, \af)$.
\end{thm}
We will not prove this in general,
but we will obtain the result when $G$ is discrete,
as a special case
of Proposition~\ref{P:Faithful}(\ref{P:Faithful:Inj})
below.
We now further analyze the reduced \cp{} $C^*_{\mathrm{r}} (G, A, \af)$
when $G$ is discrete.
One of the consequences will
be the discrete group case of Theorem~\ref{T_4217_AnyRep},
but some of what we do does not have a good analog
for groups which are not discrete.
The main tool is the structure of regular representations
of $C^*_{\mathrm{r}} (G, A, \af)$.
When $G$ is discrete, we can write $L^2 (G, H_0)$
as a Hilbert space direct sum $\bigoplus_{g \in G} H_0$,
and elements of it can be thought of as families
$( \xi_g)_{g \in G}$.
The main result is Proposition~\ref{P:Faithful},
which in particular contains the faithfulness
of the conditional expectation
from the reduced crossed product to the original algebra.
Faithfulness is proved in Theorem~4.12 of~\cite{ZM};
also see some of the preceding results there.
The development there differs somewhat from ours.
We have not found a reference for the following development,
although we presume that there is one.
The closest we have come is Section~1.2 of~\cite{OD},
especially Lemma 1.2.3 and Lemma 1.2.5 there,
where it is specifically assumed that $G = \Z$.
The proofs in~\cite{OD} are more complicated than
what we give here.
Since~\cite{OD} treats the full rather than the reduced
crossed product,
the proofs there must also implicitly prove that the map
$C^* (\Z, A, \af) \to C^*_{\mathrm{r}} (\Z, A, \af)$
is an isomorphism.
(We are grateful to Sriwulan Adji for calling our attention
to this reference.)
\begin{lem}\label{L:StructRR}
Let $\af \colon G \to \Aut (A)$
be an action of a discrete group $G$ on a \ca~$A$.
Let $\pi_0 \colon A \to L (H_0)$ be a representation,
and let $\sm \colon C^*_{\mathrm{r}} (G, A, \af) \to H = L^2 (G, H_0)$
be the integrated form
of the associated regular representation.
Let
\[
a = \sum_{g \in G} a_g u_g \in C^*_{\mathrm{r}} (G, A, \af),
\]
with $a_g = 0$ for all but finitely many~$g$.
For $\xi \in H$, we then have
\[
( \sm (a) \xi) (h)
= \sum_{g \in G} \pi_0 (\af_h^{1} (a_g)) \big( \xi (g^{1} h ) \big)
\]
for all $h \in G$.
\end{lem}
\begin{proof}
This is a calculation.
\end{proof}
In particular, picking off coordinates in $L^2 (G, H_0)$
gives the following result.
\begin{cor}\label{C:StructRR}
Let the hypotheses be as in Lemma~\ref{L:StructRR},
and let
\[
a = \sum_{g \in G} a_g u_g \in C^*_{\mathrm{r}} (G, A, \af)
\]
as there.
For $g \in G$,
let $s_g \in L (H_0, H)$ be the isometry which sends
$\et \in H_0$ to the function $\xi \in L^2 (G, H_0)$
% which takes the value $\et$ at $g$ and is zero elsewhere.
given by
\[
\xi (h) = \begin{cases}
\et & h = g
\\
0 & h \neq g.
\end{cases}
\]
Then
\[
s_h^* \sm (a) s_k = \pi_0 \big( \af_h^{1} ( a_{h k^{1} }) \big)
\]
for all $h, k \in G$.
\end{cor}
\begin{proof}
This is an easy calculation from Lemma~\ref{L:StructRR}.
\end{proof}
\begin{lem}\label{L:FGpCP}
Let $\af \colon G \to \Aut (A)$
be an action of a discrete group $G$ on a \ca~$A$.
Let $\ \cdot \$
be the \ca{} norm on $C^* (G, A, \af)$
restricted to $C_{\mathrm{c}} (G, A, \af)$,
let $\ \cdot \_{\mathrm{r}}$
be the \ca{} norm on $C^*_{\mathrm{r}} (G, A, \af)$
restricted to $C_{\mathrm{c}} (G, A, \af)$,
and let $\ \cdot \_{\I}$
be the supremum norm.
Then for every $a \in C_{\mathrm{c}} (G, A, \af)$,
we have
$\ a \_{\I} \leq \ a \_{\mathrm{r}} \leq \ a \ \leq \ a \_1$.
\end{lem}
\begin{proof}
The middle of this inequality follows from the definitions.
The last part follows from the
observation in Remark~\ref{R:NRed}
that all \ct{} representations of $L^1 (G, A, \af)$
are norm reducing.
Here is a direct proof:
for $a = \sum_{g \in G} a_g u_g \in C_{\mathrm{c}} (G, A, \af)$,
with all but finitely many of the $a_g$ equal to zero, we have
\[
\left\ \ssum{g \in G} a_g u_g \right\
\leq \ssum{g \in G} \ a_g \ \cdot \ u_g \
= \ssum{g \in G} \ a_g \
= \left\ \ssum{g \in G} a_g u_g \right\_1.
\]
We prove the first part of the inequality.
Let $a = \sum_{g \in G} a_g u_g$,
with all but finitely many of the $a_g$ equal to zero,
and let $g \in G$.
Let $\pi_0 \colon A \to L (H_0)$
be an injective nondegenerate representation.
With the notation of Corollary~\ref{C:StructRR},
we have
\[
\ a_g \
= \ \pi_0 (a_g) \
= \ s_1^* \sm (a) s_{g^{1}} \
\leq \ \sm (a) \
\leq \ a \_{\mathrm{r}}.
\]
This completes the proof.
\end{proof}
\begin{rmk}\label{R:Ident}
Lemma~\ref{L:FGpCP} implies that
the map $a \mapsto a u_1$,
from $A$ to $C^*_{\mathrm{r}} (G, A, \af)$,
is injective.
We routinely identify $A$
with its image in $C^*_{\mathrm{r}} (G, A, \af)$ under this map,
thus treating it as a subalgebra of $C^*_{\mathrm{r}} (G, A, \af)$.
Of course, we can do the same with the full \cp{} $C^* (G, A, \af)$.
\end{rmk}
\begin{cor}\label{C_FGpCP}
Let $\af \colon G \to \Aut (A)$
be an action of a finite group $G$ on a \ca~$A$.
Then the maps
$C_{\mathrm{c}} (G, A, \af)
\to C^* (G, A, \af) \to C^*_{\mathrm{r}} (G, A, \af)$
are bijective.
\end{cor}
\begin{proof}
When $G$ is finite,
$\ \cdot \_1$ is equivalent to $\ \cdot \_{\I}$ as
defined in Lemma~\ref{L:FGpCP},
and $C_{\mathrm{c}} (G, A, \af)$ is complete in both.
Lemma~\ref{L:FGpCP} implies that both C*~norms are
equivalent to these norms,
so $C_{\mathrm{c}} (G, A, \af)$ is complete in both C*~norms.
\end{proof}
When $G$ is discrete but not finite,
things are much more complicated.
We can get started.
\begin{prp}\label{P:CondExpt}
Let $\af \colon G \to \Aut (A)$
be an action of a discrete group $G$ on a \ca~$A$.
Then for each $g \in G$,
there is a linear map $E_g \colon C^*_{\mathrm{r}} (G, A, \af) \to A$
with $\ E_g \ \leq 1$
such that if
\[
a = \sum_{g \in G} a_g u_g \in C_{\mathrm{c}} (G, A, \af),
\]
then $E_g (a) = a_g$.
Moreover,
for every representation $\pi_0$ of~$A$,
and with $s_g$ as in Corollary~\ref{C:StructRR},
we have
\[
s_h^* \sm (a) s_k = \pi_0 \big( \af_h^{1} (E_{h k^{1}} (a)) \big)
\]
for all $h, k \in G$.
% Need to say: What is $\pi_0$? 999 [Doneneed to check, though.]
\end{prp}
\begin{proof}
The first part is immediate from the first inequality
in Lemma~\ref{L:FGpCP}.
The last statement follows from
Corollary~\ref{C:StructRR} by continuity.
\end{proof}
Thus, for any $a \in C^*_{\mathrm{r}} (G, A, \af)$,
and therefore also for $a \in C^* (G, A, \af)$,
it makes sense to talk about its coefficients~$a_g$.
As we have already seen in Remark~\ref{R:WhatIsIn},
even when $A = \C$ the obvious series
made with these coefficients need not converge to~$a$
(or to anything).
See Remark~\ref{R_6X11_ComputeCrPrd} for more information.
If
$C^* (G, A, \af) \neq C^*_{\mathrm{r}} (G, A, \af)$
(which can happen if $G$ is not amenable,
but not if $G$ is amenable; see Theorem~\ref{T_4131_CPFToRed}),
the coefficients $(a_g)_{g \in G}$ do not even uniquely determine
the element~$a$.
(See further discussion of the case $A = \C$
in Remark~\ref{R_3408_NotAmen}.)
This is why we only consider reduced \cp{s} here.
\begin{prp}\label{P:Faithful}
Let $\af \colon G \to \Aut (A)$
be an action of a discrete group $G$ on a \ca~$A$.
Let the maps $E_g \colon C^*_{\mathrm{r}} (G, A, \af) \to A$
be as in Proposition~\ref{P:CondExpt}.
Then:
\begin{enumerate}
\item\label{P:Faithful:G}
If $a \in C^*_{\mathrm{r}} (G, A, \af)$
and $E_g (a) = 0$ for all $g \in G$,
then $a = 0$.
\item\label{P:Faithful:Inj}
If $\pi_0 \colon A \to L (H_0)$ is a nondegenerate representation
such that $\bigoplus_{g \in G} \pi_0 \circ \af_g$ is injective,
then the regular representation $\sm$ of $C^*_{\mathrm{r}} (G, A, \af)$
associated to $\pi_0$ is injective.
\item\label{P:Faithful:Step1}
If $a \in C^*_{\mathrm{r}} (G, A, \af)$ and $g \in G$,
then $\ E_g (a) \^2 \leq \ E_1 (a^* a) \$.
\item\label{P:Faithful:1}
If $a \in C^*_{\mathrm{r}} (G, A, \af)$
and $E_1 (a^* a) = 0$, then $a = 0$.
\end{enumerate}
\end{prp}
Proposition~\ref{P:Faithful}(\ref{P:Faithful:Inj})
implies the discrete group case of Theorem~\ref{T_4217_AnyRep}.
\begin{proof}[Proof of Proposition~\ref{P:Faithful}]
We prove~(\ref{P:Faithful:G}).
Let $\pi_0 \colon A \to L (H_0)$ be a representation,
and let the notation be as in Corollary~\ref{C:StructRR}.
If $a \in C^*_{\mathrm{r}} (G, A, \af)$ satisfies $E_g (a) = 0$
for all $g \in G$,
then $s_h^* \sm (a) s_k = 0$ for all $h, k \in G$,
whence $\sm (a) = 0$.
Since $\pi_0$ is arbitrary, it follows that $a = 0$.
This proves~(\ref{P:Faithful:G}).
For~(\ref{P:Faithful:Inj}),
suppose $a \in C^*_{\mathrm{r}} (G, A, \af)$ and $\sm (a) = 0$.
Fix $l \in G$.
Taking $h = g^{1}$ and $k = l^{1} g^{1}$ in
Proposition~\ref{P:CondExpt},
we get $(\pi_0 \circ \af_g) (E_l (a)) = 0$
for all $g \in G$.
So $E_l (a) = 0$.
This is true for all $l \in G$,
so $a = 0$.
We now prove~(\ref{P:Faithful:Step1}).
As before,
let
\[
a = \sum_{g \in G} a_g u_g \in C_{\mathrm{c}} (G, A, \af).
\]
Then
\[
a^* a = \sum_{g, h \in G} u_g^* a_g^* a_h u_h
= \sum_{g, h \in G} \af_g^{1} (a_g a_h^*) u_{g^{1} h},
\]
so
\[
E_1 (a^* a)
% = \sum_{g \in G} u_g^* a_g^* a_g u_g
= \sum_{g \in G} \af_g^{1} \big( E_g (a)^* E_g (a) \big).
\]
In particular, for each fixed~$g$, we have
$E_1 (a^* a) \geq \af_g^{1} \big( E_g (a)^* E_g (a) \big)$.
By continuity, this inequality holds
for all $a \in C^*_{\mathrm{r}} (G, A, \af)$.
So
\[
\ E_1 (a^* a) \
\geq \big\ \af_g^{1} \big( E_g (a)^* E_g (a) \big) \big\
= \big\ E_g (a)^* E_g (a) \big\
= \ E_g (a) \^2,
\]
as desired.
Part~(\ref{P:Faithful:1})
now follows easily.
If $E_1 (a^* a) = 0$,
then by~(\ref{P:Faithful:Step1})
we have $E_g (a)^* E_g (a) = 0$ for all~$g$.
Therefore $a = 0$ by Part~(\ref{P:Faithful:G}).
\end{proof}
The map $E_1$
used in Proposition~\ref{P:Faithful}(\ref{P:Faithful:1})
is an example of what is called a
{\emph{conditional expectation}}
(from $C^*_{\mathrm{r}} (G, A, \af)$ to $A$)
that is, it has the properties given in the following exercise.
(Some of them are redundant.)
Proposition~\ref{P:Faithful}(\ref{P:Faithful:1})
asserts that this conditional expectation is faithful.
\begin{exr}\label{Pb_CondExpt}
Let $\af \colon G \to \Aut (A)$
be an action of a discrete group $G$ on a \ca~$A$.
Let $E = E_1 \colon C^*_{\mathrm{r}} (G, A, \af) \to A$
be as in Proposition~\ref{P:CondExpt}.
Prove that $E$ has the following properties:
\begin{enumerate}
\item\label{P:CondExpt:Ident}
$E (a) = a$ for all $a \in A$.
\item\label{P:CondExpt:Idem}
$E (E (b)) = E (b)$ for all $b \in C^*_{\mathrm{r}} (G, A, \af)$.
\item\label{P:CondExpt:Pos}
If $b \geq 0$ then $E (b) \geq 0$.
\item\label{P:CondExpt:Norm}
$\ E (b) \ \leq \ b \$
for all $b \in C^*_{\mathrm{r}} (G, A, \af)$.
\item\label{P:CondExpt:Mod}
If $a \in A$ and $b \in C^*_{\mathrm{r}} (G, A, \af)$,
then $E (a b) = a E (b)$ and $E (b a) = E (b) a$.
\end{enumerate}
\end{exr}
\begin{dfn}\label{D_StdCond}
Let $\af \colon G \to \Aut (A)$
be an action of a discrete group $G$ on a \ca~$A$.
The map $E = E_1 \colon C^*_{\mathrm{r}} (G, A, \af) \to A$
of Proposition~\ref{P:CondExpt},
determined by
\[
E \left( \ssum{g \in G} a_g u_g \right) = a_1
\]
when $\sum_{g \in G} a_g u_g \in C_{\mathrm{c}} (G, A, \af)$,
is called the
{\emph{standard conditional expectation
from $C^*_{\mathrm{r}} (G, A, \af)$ to~$A$.}}
It is usually written~$E$.
When $A = \C$
(and the action is trivial),
we obtain a tracial state on $C^*_{\mathrm{r}} (G)$,
which we call the {\emph{standard tracial state}}.
\end{dfn}
The standard tracial state already appeared in \Thm{T_3318_RedTr}.
\begin{rmk}\label{R_6X11_ComputeCrPrd}
Unfortunately,
in general the series $\sum_{g \in G} a_g u_g$
does not converge in $C^*_{\mathrm{r}} (G, A, \af)$.
Indeed, we saw in
Remark~\ref{R:WhatIsIn}(\ref{R:WhatIsIn3})
that this already fails for the trivial action of $\Z$ on~$\C$.
As suggested by Remark~\ref{R:WhatIsIn}(\ref{R:WhatIsIn1}),
it can be very difficult to determine exactly which families
$(a_g)_{g \in G}$
correspond to elements of $C^*_{\mathrm{r}} (G, A, \af)$.
If $G$ is discrete abelian,
then there is a good alternate description of~$C^* (G)$.
Since $C^* (G)$ is commutative and unital,
it must be isomorphic to $C (X)$ for some \chs~$X$,
and the right choice is the Pontryagin dual~${\widehat{G}}$.
This was already proved in \Thm{T_3319_CStDiscAb}.
In general, the computation of~$C^* (G)$ and $C^*_{\mathrm{r}} (G)$
is a difficult problem,
as is suggested by Remark~\ref{R:WhatIsIn}.
Answers are known for some groups,
particularly semisimple Lie groups (which of course are not discrete).
Even if one understands completely what all the elements
of $C^*_{\mathrm{r}} (G)$ are,
and even if the action $\af \colon G \to \Aut (A)$ is trivial,
understanding the elements of the reduced crossed product requires that
one understand all the elements of the completed
tensor product $C^*_{\mathrm{r}} (G) \otimes_{\mathrm{min}} A$.
If $G$ is abelian,
one gets $C \big( {\widehat{G}}, A \big)$.
However,
as far as I know, this problem is also in general intractable.
When the group is not amenable,
for full \cp{s}
instead of reduced \cp{s},
one of course has the generalization
of the difficulty described in Remark~\ref{R_3408_NotAmen}
with the full group \ca.
There is just one bright spot,
although we will not prove it here.
The Ces\`{a}ro means of the Fourier series of a \cfn{} always
converge uniformly to the function,
and,
as already mentioned in
Remark \ref{R:WhatIsIn}(\ref{R:WhatIsIn4}),
this fact has generalizations to crossed products
by discrete amenable groups and even some cases beyond that.
See Section~5 of~\cite{BdsCti}.
The case $G = \Z$ is Theorem VIII.2.2 of~\cite{Dvd}.
\end{rmk}
Remark~\ref{R:WhatIsIn} is meant to point out
the difficulties in dealing with crossed products by infinite groups.
Despite all this, for some problems, finite groups are harder.
As suggested after Remark~\ref{R_3408_NotAmen},
we have excellent information about the Ktheory
of crossed products by $\Z$~\cite{PV} and by~$\R$~\cite{Cnns},
and even for both reduced crossed products by
free groups $F_n$~\cite{PV2}
and the corresponding full crossed products
(in~\cite{Ctz7} see Theorem 2.1(c), Definition 2.2,
and Theorem 2.4(c)).
See~\cite{Pmnr} for a generalization
of the result on reduced crossed products by~$F_n$.
The result for full crossed products by $F_n$ holds
despite the fact that the the conditional expectation
of \Def{D_StdCond}
is usually not faithful on full crossed products,
so that an element is not even uniquely determined by its
``coefficients''.
All these formulas imply, in particular,
that if the Ktheory of the original algebra is zero,
then so is the Ktheory of the crossed product.
There is no such formula for the Ktheory of crossed
products by the two element group $\Z / 2 \Z$,
in which not even any completion is needed.
There even exists a \ca~$A$ which is contractible
(a much stronger condition than $K_* (A) = 0$)
and an action $\af \colon \Z / 2 \Z \to \Aut (A)$
such that $K_* \big( C^* ( \Z / 2 \Z, \, A, \, \af ) \big) \neq 0$.
(Examples can be constructed using some of the examples
in Section~3 of~\cite{Ph89a}.
We omit the details.)
% computing the Ktheory of a crossed product by $\Z / 2 \Z$
% is harder than computing the Ktheory of a crossed product
% or reduced \cp{} by any
% of $\Z$, $\R$, or even a (nonabelian) free group!
We now discuss functoriality of reduced \cp{s}.
Again,
given the results above,
the the proofs are no harder for general locally compact groups~$G$
than in the discrete case,
but we can't claim that the presentation
of the general case is self contained.
\begin{lem}\label{L_4217_RegDSum}
Let $\af \colon G \to \Aut (A)$
be an action of a locally compact group $G$ on a \ca~$A$.
Let $I$ be a set,
and for $i \in I$ let $\rh_i$ be a representation
of $A$ on a Hilbert space~$H_i$.
Let $(v_i, \pi_i)$ be the associated regular representation,
and let $\sm_i$
be its integrated form.
Then
the regular representation associated with
$\bigoplus_{i \in I} \rh_i$ is
$\left( \bigoplus_{i \in I} v_i, \, \bigoplus_{i \in I} \pi_i \right)$.
Its integrated form is $\bigoplus_{i \in I} \sm_i$.
\end{lem}
\begin{proof}
The proof of the first statement is routine.
The second statement follows from \Lem{L_4217_IntFDSum}.
\end{proof}
\Lem{L_4215_DegRpn} has the following analog for regular
covariant representations.
\begin{lem}\label{L_4217_RegDegRpn}
Let $\af \colon G \to \Aut (A)$
be an action of a locally compact group $G$ on a \ca~$A$.
Let $\rh_0 \colon A \to L (H_0)$ be a representation
of $A$ on a Hilbert space~$H_0$.
(We do not assume that $\rh_0$ is nondegenerate.)
Let $\sm_0 \colon C^* (G, A, \af) \to L (L^2 (G, H_0))$
be the integrated form (as in \Def{D:IntForm})
of the regular covariant representation
associated to $\rh_0$ (as in \Def{D:RegRpn}).
Let $H$ be the closed linear span of $\pi_0 (A) H_0$.
Then the closed linear span
of $\sm_0 (C^* (G, A, \af)) L^2 (G, H_0)$
is $L^2 (G, H)$.
\end{lem}
\begin{proof}
% Let $H$ be the closed linear span of $\rh_0 (A) H_0$.
The subspace $H$ is invariant under $\rh_0$
by \Lem{L_4217_DegRepFacts}(\ref{L_4217_DegRepFacts_Inv}).
Set $\rh = \rh_0 () _H$.
Then $\rh_0 = \rh \oplus 0$
by \Lem{L_4217_DegRepFacts}(\ref{L_4217_DegRepFacts_DSum}).
Let $(v_0, \pi_0)$, $(v, \pi)$, and $(w, \zt)$
be the regular covariant representations of $(G, A, \af)$
on $L (L^2 (G, H_0))$, $L (L^2 (G, H))$, and $L (L^2 (G, H)^{\perp})$
associated to $\rh_0$, $\rh$, and the zero representation
on~$H^{\perp}$.
Let $\sm$ and $\nu$ be the integrated forms of $(v, \pi)$
and $(w, \zt)$.
\Lem{L_4217_RegDSum}
gives $v_0 = v \oplus w$, $\pi_0 = \pi \oplus \zt$,
and $\sm_0 = \sm \oplus \nu$.
Clearly $\zt$ is the zero representation of~$A$,
so $\nu$ is the zero representation of $C^* (G, A, \af)$,
while $\pi$ is nondegenerate
by \Lem{L_4215_RegIsND}.
Apply
\Lem{L_4217_DegRepFacts}(\ref{L_4217_DegRepFacts_GivenD})
to the direct sum decomposition $\sm_0 = \sm \oplus \nu$,
and then apply
\Lem{L_4217_DegRepFacts}(\ref{L_4217_DegRepFacts_DSum}).
\end{proof}
\begin{cor}\label{C_4217_RegDegNorm}
Let $\af \colon G \to \Aut (A)$
be an action of a locally compact group $G$ on a \ca~$A$.
Let $\kp \colon C^* (G, A, \af) \to C^*_{\mathrm{r}} (G, A, \af)$
be the quotient map.
Let $a \in C^* (G, A, \af)$.
Then
\begin{align*}
\ \kp (a) \
& = \sup \big( \big\{ \ \sm (a) \ \colon
{\mbox{$\sm$ is the integrated form of a
possibly}}
\\
& \hspace*{5em} {\mbox{degenerate regular covariant representation
of $(G, A, \af)$}} \big\} \big).
\end{align*}
\end{cor}
\begin{proof}
It follows from Lemma~\ref{L_4217_RegDegRpn},
\Lem{L_4217_DegRepFacts}(\ref{L_4217_DegRepFacts_Ndg}),
and \Lem{L_4217_DegRepFacts}(\ref{L_4217_DegRepFacts_Norm})
that the supremum on the right
is unchanged if we restrict to
covariant representations of $(G, A, \af)$
coming from nondegenerate representations of~$A$.
\end{proof}
\begin{thm}\label{T_4215_RedCPFunct}
Let $G$ be a \lcg.
If $(G, A, \af)$ and $(G, B, \bt)$ are $G$algebras
and $\ph \colon A \to B$ is an equivariant \hm,
then the \hm{}
$C^* (G, A, \af) \to C^* (G, B, \bt)$
of Theorem~\ref{T:CPFunct}
induces a \hm{}
$C^*_{\mathrm{r}} (G, A, \af) \to C^*_{\mathrm{r}} (G, B, \bt)$.
This construction makes the reduced \cp{} construction
a functor from the category of $G$algebras to the category of \ca{s}.
\end{thm}
\begin{proof}
One observes that if $\pi_0 \colon B \to L (H_0)$
is a representation,
and if
\[
\sm \colon C_{\mathrm{c}} (G, B, \bt) \to L (L^2 (G, H_0))
\]
is the associated regular representation,
then $\sm \circ \ps$
is the regular representation
associated with the representation
$\pi_0 \circ \ph \colon A \to L (H_0)$.
In view of Corollary~\ref{C_4217_RegDegNorm},
it follows that
$C^* (G, A, \af) \to C^* (G, B, \bt)$
induces a well defined \hm{}
$C^*_{\mathrm{r}} (G, A, \af) \to C^*_{\mathrm{r}} (G, B, \bt)$.
The properties of a functor are easy to check.
\end{proof}
The analog of Theorem~\ref{T_CPExact} for reduced \cp{s}
is in general false.
Counterexamples are hard to find,
and the history is confusing;
we refer to the (brief) discussion
in the introduction to~\cite{BGW}.
Since the reduced \cp{} is functorial,
the maps in the sequence are defined.
In fact, exactness can only fail in the middle.
Indeed,
we have the following result.
\begin{thm}\label{T_3331_RedCP}
Let $G$ be a locally compact group,
let $(G, A, \af)$ and $(G, B, \bt)$ be $G$algebras,
and let $\ph \colon A \to B$
be an equivariant \hm.
Let
\[
\ps \colon
C^*_{\mathrm{r}} (G, A, \af) \to C^*_{\mathrm{r}} (G, B, \bt)
\]
be the corresponding \hm{} of the reduced \cp{s}.
\begin{enumerate}
\item\label{T_3331_RedCP_Subalg}
If $\ph$ is injective then so is~$\ps$.
\item\label{T_3331_RedCP_Ideal}
If $\ph (A)$ is an ideal in~$B$,
then $\ps \big( C^*_{\mathrm{r}} (G, A, \af) \big)$
is an ideal in $C^*_{\mathrm{r}} (G, B, \bt)$.
\item\label{T_3331_RedCP_Quot}
If $\ph$ is surjective then so is~$\ps$.
\item\label{T_3331_RedCP_NZI}
If $\ph (A)$ is a nonzero proper ideal in~$B$,
then $\ps \big( C^*_{\mathrm{r}} (G, A, \af) \big)$
is a nonzero proper ideal in $C^*_{\mathrm{r}} (G, B, \bt)$.
\end{enumerate}
\end{thm}
\begin{proof}
For~(\ref{T_3331_RedCP_Subalg}),
choose a nondegenerate injective representation
$\pi_0$ of $B$ on a Hilbert space~$H$,
let $(v, \pi)$ be the associated regular covariant representation
of $(G, B, \bt)$
(\Def{D:RegRpn}),
and let $\sm \colon C^*_{\mathrm{r}} (G, B, \bt) \to L (L^2 (G, H))$
be its integrated form
(Definition~\ref{D:IntForm}).
Then $\pi_0 \circ \ph$
is an injective representation of $A$ on~$H$
(not necessarily nondegenerate).
Use \Lem{L_4217_DegRepFacts}(\ref{L_4217_DegRepFacts_DSum})
to find a closed subspace $M \subset H$
and a nondegenerate representation $\rh \colon A \to L (M)$
such that $\pi_0 \circ \ph$ is the direct sum
of $\rh$ and the zero representation on~$M^{\perp}$.
Then $\rh$ is injective by
\Lem{L_4217_DegRepFacts}(\ref{L_4217_DegRepFacts_Ker}).
Let $\mu \colon C^*_{\mathrm{r}} (G, A, \af) \to L (L^2 (G, M))$
be the integrated form of the
regular covariant representation associated to~$\rh$.
Theorem~\ref{T_4217_AnyRep} implies that
$\mu$ is injective,
and \Lem{L_4217_RegDegRpn}
implies that $\mu$ is a direct summand in the
representation $\sm \circ \ps$.
Therefore $\ps$ must be injective.
In the next two parts,
we let
\[
\io_A \colon
C_{\mathrm{c}} (G, A, \af) \to C^*_{\mathrm{r}} (G, A, \af)
\andeqn
\io_B \colon
C_{\mathrm{c}} (G, B, \bt) \to C^*_{\mathrm{r}} (G, B, \bt)
\]
be the standard maps.
We prove~(\ref{T_3331_RedCP_Ideal}).
Since $\io_A$ and $\io_B$ have dense ranges,
to show that $\ps \big( C^*_{\mathrm{r}} (G, A, \af) \big)$
is an ideal,
it suffices to prove that that for $a \in C_{\mathrm{c}} (G, A, \af)$
and $b \in C_{\mathrm{c}} (G, B, \bt)$,
we have
$\ps (\io_A (a)) \io_B (b) \in C^*_{\mathrm{r}} (G, B, \bt)$
and $\io_B (b) \ps (\io_A (a)) \in C^*_{\mathrm{r}} (G, B, \bt)$.
In fact, both are obviously in
$\io_B (C_{\mathrm{c}} (G, B, \bt))$.
We prove~(\ref{T_3331_RedCP_Quot}).
One checks that $\ph$ induces a surjective map
$C_{\mathrm{c}} (G, A, \af) \to C_{\mathrm{c}} (G, B, \bt)$.
Therefore the range of $\ps$ contains
$\io_B (C_{\mathrm{c}} (G, B, \bt))$.
So $\ps$ has dense range, and is therefore surjective.
Finally, we prove~(\ref{T_3331_RedCP_NZI}).
The subalgebra $\ps \big( C^*_{\mathrm{r}} (G, A, \af) \big)$
is an ideal by~(\ref{T_3331_RedCP_Ideal}),
and is nonzero by~(\ref{T_3331_RedCP_Subalg}).
Let $\pi \colon B \to B / \ph (A)$
be the quotient map,
and let
\[
\sm \colon
C^*_{\mathrm{r}} (G, B, \af)
\to C^*_{\mathrm{r}} (G, \, B / \ph (A), \, \bt)
\]
be the corresponding \hm{} of the reduced \cp{s}.
Clearly $\sm \circ \ps = 0$,
so
$\ps \big( C^*_{\mathrm{r}} (G, A, \af) \big)
\subset {\operatorname{Ker}} (\sm)$.
Since $\sm$ is surjective by~(\ref{T_3331_RedCP_Quot}),
it follows that
$\ps \big( C^*_{\mathrm{r}} (G, A, \af) \big)
\neq C^*_{\mathrm{r}} (G, B, \bt)$.
\end{proof}
\begin{rmk}\label{R_6814_DualAction}
We describe dual actions without proof;
see~\cite{Tki} for details.
Let $A$ be any \ca, let $G$ be a locally compact abelian group,
and let $\af \colon G \to \Aut (A)$ be an action.
Let ${\widehat{G}}$ be the Pontryagin dual of~$G$
(\Def{D_3319_DualDfn}).
For $\sm \in {\widehat{G}}$,
there is an automorphism ${\widehat{\af}}_{\sm}$ of $C^* (G, A, \af)$
given on $C_{\mathrm{c}} (G, A, \af)$
by ${\widehat{\af}}_{\sm} (a) (g) = {\overline{\sm (g)}} a (g)$
for $a \in C_{\mathrm{c}} (G, A, \af)$, $\sm \in {\widehat{G}}$,
and $g \in G$.
Moreover,
${\widehat{\af}}
\colon {\widehat{G}} \to \Aut \big( C^* (G, A, \af) \big)$
is a \ct{} action of ${\widehat{G}}$ on $C^* (G, A, \af)$,
called the dual action.
One can also use $\sm (g)$ in place of ${\ov{\sm (g)}}$.
The choice ${\ov{\sm (g)}}$
seems to be more common.
It agrees with the conventions
in~\cite{Tki} (see the beginning of Section~3 there)
and~\cite{Wlms} (see the beginning of Section~7.1 there),
but disagrees with the choice in~\cite{Pd1}
(see Proposition 7.8.3 there).
To see the reason for the choice ${\ov{\sm (g)}}$,
consider the case $G = S^1$ and $A = \C$.
For $f \in C_{\mathrm{c}} (G)$,
we have ${\widehat{\af}}_{n} (f) (\zt) = \zt^{n}$
for $n \in \Z$ and $\zt \in S^1$.
The $n$th Fourier coefficient of $f$ is then
${\widehat{f}} (n) = \int_G {\widehat{\af}}_{n} (f) (\zt) \, d \zt$,
giving the corresponding Fourier series
$f (\zt) = \sum_{n \in \Z} {\widehat{f}} (n) \zt^n$.
If one uses $\sm (g)$ in the definition of the
dual action,
then some extra signs are required in these formulas.
If $G$ is discrete then ${\widehat{G}}$ is compact,
and the conditional expectation $E \colon C^* (G, A, \af) \to A$
of \Def{D_StdCond}
is given by
$E (a) = \int_{{\widehat{G}}} {\widehat{\af}}_{\sm} (a) \, d \sm$,
using normalized Haar measure in the integral.
Whether or not $G$ is discrete,
the \cp{} by the dual action is $K (L^2 (G)) \otimes A$.
This result is Takai duality
(\cite{Tki}; Theorem~7.1 of~\cite{Wlms}; Theorem 7.9.3 of~\cite{Pd1}).
It is a generalization of
the abelian case of Example~\ref{E_C_Trans} below:
if $A = \C$,
then $C^* (G, A, \af) = C^* (G) \cong C_0 \big( {\widehat{G}} \big)$,
and the dual action is just translation on ${\widehat{G}}$.
\end{rmk}
\begin{exr}\label{Er_7130_ExistDualAction}
Adopt the notation of Remark~\ref{R_6814_DualAction}.
Prove that the formula
${\widehat{\af}}_{\sm} (a) (g) = {\overline{\sm (g)}} a (g)$,
for $a \in C_{\mathrm{c}} (G, A, \af)$, $\sm \in {\widehat{G}}$,
and $g \in G$,
extends to a \ct{} action of ${\widehat{G}}$
on $C^* (G, A, \af)$.
\end{exr}
Exercise~\ref{Er_7130_ExistDualAction}
is easiest when $G$ is discrete and $A$ is unital,
in which case one can use the description in \Thm{L:UnivCP}
of $C^* (G, A, \af)$ in terms of generators and relations.
\begin{exr}\label{Er_7130_GetCondExpt}
Adopt the notation of Remark~\ref{R_6814_DualAction},
and assume that $G$ is discrete.
Let $\nu$ be normalized Haar measure on~${\widehat{G}}$.
Prove that for all $a \in C^* (G, A, \af)$,
the the conditional expectation $E \colon C^* (G, A, \af) \to A$
of \Def{D_StdCond}
satisfies
\[
E (a) = \int_{{\widehat{G}}} {\widehat{\af}}_{\sm} (a) \, d \nu (\sm)
\]
for all $a \in C^* (G, A, \af)$,
as claimed in Remark~\ref{R_6814_DualAction}.
Hint: Prove this for $a \in C_{\mathrm{c}} (G, A, \af)$ first.
\end{exr}
\section{Computation of Some Examples of Crossed
Products}\label{Sec_Comp}
\indent
We give some explicit elementary computations of \cp{s},
mostly involving finite groups.
These examples serve several purposes.
First, they give, in a comparatively elementary context,
an explicit sense of what crossed products look like.
In particular, our calculations motivate
the statements of various general theorems,
some of which we give without proof.
Second, a number of interesting examples of actions and
their crossed products have been constructed by taking
direct limits of some of the kinds of examples we consider.
The computation of some of these \cp{s} depends on
knowing enough detail in examples of some of the types
discussed here that one can calculate direct limits of them.
We include in this section several examples
of computations of crossed products by actions
constructed using direct limits.
Some of our examples can be found in Section~2.5 of~\cite{Wlms}.
For the most part, however,
we have not found calculations in the literature in the
explicit form which we give here.
Throughout this section,
we will use the *algebra $C_{\mathrm{c}} (G, A, \af)$
of compactly supported \cfn{s} $a \colon G \to A$,
with pointwise addition and scalar multiplication,
with multiplication given by convolution
as in \Def{D:L1},
and with the adjoint defined there.
By construction
(\Def{D:CP}, together with density of $C_{\mathrm{c}} (G, A, \af)$
in $L^1 (G, A, \af)$, as in \Def{D:L1}),
the image of this algebra in $C^* (G, A, \af)$ is dense,
and by \Thm{T:InjOnL1}
the map $C_{\mathrm{c}} (G, A, \af) \to C^*_{\mathrm{r}} (G, A, \af)$
is injective.
It follows that the map
$C_{\mathrm{c}} (G, A, \af) \to C^* (G, A, \af)$
is injective.
We therefore routinely identify $C_{\mathrm{c}} (G, A, \af)$
with a dense subalgebra of $C^* (G, A, \af)$
and also,
depending on context,
with a dense subalgebra of $C^*_{\mathrm{r}} (G, A, \af)$.
Our group $G$ will almost always be discrete.
In this case and if $A$ is unital,
for $g \in G$ we let $u_g \in C_{\mathrm{c}} (G, A, \af)$
be the canonical unitary corresponding to~$g$,
as in Notation~\ref{N:ug}.
Also following Notation~\ref{N:ug},
we use the same notation for the corresponding unitaries
in $C^* (G, A, \af)$ and $C^*_{\mathrm{r}} (G, A, \af)$.
When $A$ is not unital,
we follow the conventions for the
nonunital case in Notation~\ref{N:ug}.
(In particular,
$u_g$ denotes corresponding elements in
the multiplier algebras of
$C^* (G, A, \af)$ and $C^*_{\mathrm{r}} (G, A, \af)$.)
When $G$ is discrete,
$C_{\mathrm{c}} (G, A, \af)$ is the set of functions
from $G$ to~$A$
which have finite support.
Following the notation of
Lemma~\ref{L:StructRR} and later results in Section~\ref{Sec:RedCP},
and as suggested by Remark~\ref{R_6320_Sumagug},
in both the unital and nonunital cases
we regularly identify $C_{\mathrm{c}} (G, A, \af)$
with the set of sums
$a = \sum_{g \in G} a_g u_g$ in which $a_g \in A$ for all $g \in G$
and $a_g = 0$ for all but finitely many $g \in G$.
When $G$ is finite,
as is the case in many of our examples,
$C_{\mathrm{c}} (G, A, \af)$ is then just the
set of all sums
$a = \sum_{g \in G} a_g u_g$ in which $a_g \in A$ for all $g \in G$,
and the map from
$C_{\mathrm{c}} (G, A, \af)$ to $C^* (G, A, \af)$
is bijective (Corollary~\ref{C_FGpCP}).
\begin{exa}\label{E:C:Triv}
If $G$ acts trivially on the \ca~$A$,
then
\[
C^* (G, A) \cong C^* (G) \otimes_{\mathrm{max}} A
\andeqn
C^*_{\mathrm{r}} (G, A)
\cong C^*_{\mathrm{r}} (G) \otimes_{\mathrm{min}} A.
\]
The case of the full crossed product,
in fact, the generalization to the case of an inner action,
is Example~2.53 of~\cite{Wlms}.
For the full crossed product,
first assume $G$ is discrete and $A$ is unital.
Then Theorem~\ref{L:UnivCP} implies that
$C^* (G, A)$ is the universal unital \ca{} generated by a unital copy
of $A$ and a commuting unitary representation of $G$ in the algebra.
Since $C^* (G)$ is the universal unital \ca{} generated by
a unitary representation of $G$ in the algebra,
this is exactly the universal property of the maximal tensor product.
The proof for the general case is essentially the same.
The basic point (omitting the technicalities) is that
a covariant representation
consists of commuting representations of $A$ and $G$,
and hence of $A$ and $C^* (G)$.
For the reduced crossed product,
the point is that
a regular nondegenerate covariant representation of $(G, A)$
has the form $(\ld \otimes 1_{H_0}, \, 1_{L^2 (G)} \otimes \pi_0)$
for an arbitrary nondegenerate
representation $\pi_0 \colon A \to L (H_0)$
and with $\ld \colon G \to U (L^2 (G))$
being the left regular representation.
By Proposition~\ref{P:Faithful}(\ref{P:Faithful:Inj}),
it suffices to take $\pi_0$ to be a single injective representation.
Now we are looking at $C^*_{\mathrm{r}} (G)$ on one Hilbert space
and $A$ on another,
and taking the tensor product of the Hilbert spaces.
This is exactly how one gets the minimal tensor product of two \ca{s}.
\end{exa}
Note how full and reduced crossed products parallel
maximal and minimal tensor products.
\begin{rmk}\label{R:CPTens}
More generally,
let $A$ and $B$ be \ca{s},
let $\af \colon G \to \Aut (A)$ be any action,
and let $\bt \colon G \to \Aut (B)$ be the trivial action.
Even if $\bt$ is not trivial,
one gets actions $\af \otimes_{\mathrm{max}} \bt$
of $G$ on $A \otimes_{\mathrm{max}} B$
and $\af \otimes_{\mathrm{min}} \bt$
of $G$ on $A \otimes_{\mathrm{min}} B$
which,
interpreting tensor products of elements of $A$ and $B$
as being in $A \otimes_{\mathrm{max}} B$
or $A \otimes_{\mathrm{min}} B$ as appropriate,
are uniquely determined by
\[
(\af \otimes_{\mathrm{max}} \bt)_g (a \otimes b)
= \af_g (a) \otimes \bt_g (b)
\andeqn
(\af \otimes_{\mathrm{min}} \bt)_g (a \otimes b)
= \af_g (a) \otimes \bt_g (b)
\]
for $a \in A$, $b \in B$, and $g \in G$.
If $\bt$ is trivial,
these formulas become
\[
(\af \otimes_{\mathrm{max}} \bt)_g (a \otimes b)
= \af_g (a) \otimes b
\andeqn
(\af \otimes_{\mathrm{min}} \bt)_g (a \otimes b)
= \af_g (a) \otimes b,
\]
and one has
\[
C^* (G, \, A \otimes_{\mathrm{max}} B,
\, \af \otimes_{\mathrm{max}} \bt)
\cong C^* (G, A, \af) \otimes_{\mathrm{max}} B
\]
and
\[
C^*_{\mathrm{r}} (G, \, A \otimes_{\mathrm{min}} B,
\, \af \otimes_{\mathrm{min}} \bt)
\cong C^*_{\mathrm{r}} (G, A, \af) \otimes_{\mathrm{min}} B.
\]
\end{rmk}
\begin{exr}\label{P:Triv}
Prove Remark~\ref{R:CPTens} when $G$ is discrete
and $A$ and $B$ are both unital.
\end{exr}
Exercise~\ref{Ex_6X11_TProdAction},
Exercise~\ref{Ex_6X11_TProdCrPrd},
and Exercise~\ref{Ex_6X11_TProdRdCrPrd}
contain a generalization.
\begin{exa}\label{E:C:Inner}
Let $\af \colon G \to \Aut (A)$
be an inner action of a discrete group $G$ on a \uca~$A$.
Thus, there is a \hm{} $g \mapsto z_g$ from $G$ to $U (A)$
such that $\af_g (a) = z_g a z_g^*$ for all $g \in G$ and $a \in A$.
(See Example~\ref{E:Inner}.)
We claim that $C^* (G, A, \af) \cong C^* (G) \otimes_{\mathrm{max}} A$.
(This is true even if $G$ is not discrete.
See Exercise~\ref{P:ExtEqIsom}, or Example 2.53 of~\cite{Wlms}.)
It is also true that
$C^*_{\mathrm{r}} (G, A, \af)
\cong C^*_{\mathrm{r}} (G) \otimes_{\mathrm{min}} A$.
% Give more details. 999
We prove the claim.
Let $\io \colon G \to \Aut (A)$
be the trivial action of $G$ on $A$.
As in Notation~\ref{N:ug}, for $g \in G$
let $u_g \in C_{\mathrm{c}} (G, A, \af)$ be the standard unitary,
but let $v_g \in C_{\mathrm{c}} (G, A, \io)$ be the standard
unitary in the crossed product by the trivial action.
Define
$\ph_0 \colon
C_{\mathrm{c}} (G, A, \af) \to C_{\mathrm{c}} (G, A, \io)$
by $\ph_0 (a u_g) = a z_g v_g$ for $a \in A$ and $g \in G$,
and extend linearly.
This map is obviously bijective
(the inverse sends $a v_g$ to $a z_g^* u_g$)
and isometric for $\ \cdot \_1$.
For multiplicativity, it suffices to check the following,
for $a, b \in A$ and $g, h \in H$,
using the fact that $v_g$ commutes with all elements of $A$:
\begin{align*}
\ph_0 (a u_g) \ph_0 (b u_h)
& = a z_g v_g b z_h v_h
= a z_g b z_g^* z_{g h} v_g v_h
\\
& = a \af_g (b) z_{g h} v_{g h}
= \ph_0 \big(a \af_g (b) u_{g h} \big)
= \ph_0 \big( (a u_g) (b u_h) \big).
\end{align*}
Also,
\begin{align*}
\ph_0 (a u_g)^*
& = (a z_g v_g)^*
= v_g^* z_g^* a^*
= (z_g^* a^* z_g) z_g^* v_g^*
\\
& = \af_{g^{1}} (a^*) z_{g^{1}} v_{g^{1}}
= \ph_0 \big( \af_{g^{1}} (a^*) u_{g^{1}} \big)
= \ph_0 \big( (a u_g)^* \big).
\end{align*}
So $\ph_0$ is an isometric isomorphism of *algebras,
and therefore extends to an isomorphism of the universal
\ca{s} as in Theorem~\ref{L:UnivCP}.
Now use Example~\ref{E:C:Triv}.
For use in Example~\ref{E:C:PType},
we write out explicitly what happens
when $G = \Z / 2 \Z$.
Let $v_0 \in C^* (\Z / 2 \Z)$ be the image of the nontrivial element
of the group.
Then $\ld + \mu v_0 \mapsto (\ld + \mu, \, \ld  \mu)$
is an isomorphism from $C^* (\Z / 2 \Z)$ to $\C \oplus \C$.
(The algebra $C^* (\Z / 2 \Z)$ is the universal \ca{} generated
by a unitary with square~$1$,
and the corresponding unitary in $\C \oplus \C$ is $(1, \,  1)$.
But one can check directly that the map above is an isomorphism.)
For the crossed product of a unital \ca~$A$
by the trivial action $\io$ of $\Z / 2 \Z$,
let $v \in C^* (\Z / 2 \Z, \, A, \, \io)$
be the standard unitary
associated to the nontrivial element of the group.
Then $a + b v \mapsto (a + b, \, a  b)$
is an isomorphism from $C^* (\Z / 2 \Z, \, A, \, \io)$ to $A \oplus A$.
This map is a \hm{}
because the copy
$\{ (a, a) \colon a \in A \} \subset A \oplus A$ of $A$
and the unitary $(1, \,  1) \in A \oplus A$
satisfy the appropriate commutation relations.
One proves that this map is an isomorphism
{}from $C_{\mathrm{c}} (\Z / 2 \Z, \, A, \, \io)$ to $A \oplus A$
by explicitly writing down an inverse.
Corollary~\ref{C_FGpCP} now shows it is an isomorphism
{}from $C^* (\Z / 2 \Z, \, A, \, \io)$ to $A \oplus A$.
(For a faster proof, just tensor the isomorphism of the previous
paragraph with $\id_A$.)
Now suppose that $z \in A$ is a unitary of order~$2$.
Let $g_0 \in \Z / 2 \Z$ be the nontrivial group element,
and let $\af \colon \Z / 2 \Z \to \Aut (A)$
be the action such that $\af_{g_0} = \Ad (z)$.
Let $u = u_{g_0} \in C^* (\Z / 2 \Z, \, A, \, \af)$.
Then $a + b u \mapsto (a + b z, \, a  b z)$
is an isomorphism
{}from $l^1 (\Z / 2 \Z, \, A, \, \af)$ to $A \oplus A$.
(Of course, once one has the formula, one can prove this directly.)
\end{exa}
\begin{exr}\label{P:ExtEqIsom}
Prove the following generalization of Example~\ref{E:C:Inner}.
Let $\af, \bt \colon G \to \Aut (A)$
be two actions of a \lcg{} $G$ on a \ca~$A$
which are exterior equivalent in the sense of Remark~\ref{R:ExtEq}.
Prove that
\[
C^* (G, A, \af) \cong C^* (G, A, \bt)
\andeqn
C_{\mathrm{r}}^* (G, A, \af) \cong C_{\mathrm{r}}^* (G, A, \bt).
\]
\end{exr}
The case of the full crossed product
is done in the proof of Theorem 2.8.3(5) of~\cite{Ph1}.
(The compactness hypothesis in the theorem is
not needed for the relevant part of the proof.)
\begin{exr}\label{E:C:PtInn}
Let $\af \colon (\Z / 2 \Z)^{2} \to \Aut ( M_2)$
be as in Example~\ref{E_PtInn}.
Prove that the \cp{} $C^* ((\Z / 2 \Z)^{2}, \, M_2, \, \af)$
is isomorphic to~$M_4$.
\end{exr}
Since the group is finite and the algebra is finite dimensional,
this exercise can be done with linear algebra.
% 999 Do it.
It shows that the hypothesis in \Ex{E:C:Inner}
can't be weakened from ``inner'' to ``pointwise inner''.
For the next example,
we need notation for standard matrix units.
(We have already used the usual version of this notation,
when $S = \{ 1, 2, \ldots, n \}$, a number of times.)
\begin{ntn}\label{N_MatUnit}
For any index set~$S$,
let $\dt_s \in l^2 (S)$ be the standard basis vector,
determined by
% $\dt_s (s) = 1$ and $\dt_s (t) = 0$ for $t \neq s$.
\[
\dt_s (t) = \begin{cases}
1 & \hspace{3em} t = s
\\
0 & \hspace{3em} t \neq s.
\end{cases}
\]
For $j, k \in S$,
we let the ``matrix unit'' $e_{j, k}$ be the rank one operator
on $l^2 (S)$ given by
$e_{j, k} \xi = \langle \xi, \dt_k \rangle \dt_j$.
This gives the product formula
$e_{j, k} e_{l, m} = \dt_{k, l} e_{j, m}$.
Conventional matrix units for $M_n$ are obtained by
taking $S = \{ 1, 2, \ldots, n \}$,
but we will
sometimes want to take $S$ to be a discrete (even finite) group.
For $S = \{ 1, 2 \}$,
with the obvious choice of matrix representation, we get
\[
e_{1, 1}
= \left( \begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix} \right),
\quad
e_{1, 2}
= \left( \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right),
\quad
e_{2, 1}
= \left( \begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix} \right),
\smandeqn
e_{2, 2}
= \left( \begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix} \right).
\]
\end{ntn}
\begin{exa}\label{E_C_Trans}
We prove that if $G$ is discrete and acts on itself by translation,
then the \cp{} is $K (l^2 (G))$.
When $G$ is finite, this result is proved in Lemma~2.50 of~\cite{Wlms}.
(The conclusion is true for general locally compact groups.
See Theorem~4.24 of~\cite{Wlms}.)
More generally
(compare with Remark~\ref{R:CPTens},
but we will not give a proof),
if $G$ acts on $G \times X$ by translation on the first factor
and trivially on the second factor,
then
\[
C^* (G, \, G \times X)
\cong K (l^2 (G)) \otimes C_0 (X)
\cong C_0 (X, \, K (l^2 (G)) ).
\]
In fact,
the action on $X$ need not be trivial.
The map $(h, x) \mapsto (h, \, h^{1} x)$ is an isomorphism
{}from $G \times X$ with a general action of $G$ on $X$
to $G \times X$ with the trivial action of $G$ on~$X$.
(For those familiar with the appropriate part of
the representation theory of locally compact groups,
this fact is related to the fact that the tensor product
of the regular representation of a group and any other representation
is a direct sum of copies of the regular representation.)
Let $\af \colon G \to \Aut (C_0 (G))$ denote the action.
For $g \in G$, we let $u_g$ be
the standard unitary as in Notation~\ref{N:ug},
and we let $\dt_g \in C_0 (G)$ be the function
$\ch_{ \{ g \} }$.
Then $\af_g (\dt_h) = \dt_{g h}$ for $g, h \in G$.
Also, $\spn \big( \{ \dt_g \colon g \in G \} \big)$ is dense
in $C_0 (G)$.
For $g, h \in G$, the element
%
\begin{equation}\label{Eq_6X08_DefineMatUnit}
v_{g, h} = \dt_g u_{g h^{1}}
\end{equation}
%
is in $C^* (G, \, C_0 (G), \, \af)$.
Moreover,
for $g_1, h_1, g_2, h_2 \in G$, we have
\begin{align*}
v_{g_1, h_1} v_{g_2, h_2}
& = \dt_{g_1} u_{g_1 h_1^{1}} \dt_{g_2} u_{g_2 h_2^{1}} \\
& = \dt_{g_1} \af_{g_1 h_1^{1}} (\dt_{g_2})
u_{g_1 h_1^{1}} u_{g_2 h_2^{1}}
= \dt_{g_1} \dt_{g_1 h_1^{1} g_2}
u_{g_1 h_1^{1} g_2 h_2^{1}}.
\end{align*}
Thus, if $g_2 \neq h_1$, the answer is zero,
while if $g_2 = h_1$, the answer is $v_{g_1, h_2}$.
Similarly, $v_{g, h}^* = v_{h, g}$.
That is, the elements $v_{g, h}$ satisfy the relations for a system
of matrix units indexed by~$G$.
Also, $\spn \big( \{ v_{g, h} \colon g, h \in G \} \big)$ is dense
in $l^1 (G, \, C_0 (G), \, \af)$,
and hence in $C^* (G, \, C_0 (G), \, \af)$.
For any finite set $F \subset G$,
we thus get a \hm{}
\[
\ps_F \colon
L (l^2 (F)) \to C_{\mathrm{c}} (G, \, C_0 (G), \, \af)
\]
sending the matrix unit $e_{g, h} \in L (l^2 (F))$
(Notation~\ref{N_MatUnit}) to $v_{g, h}$.
Let
\[
\ph_F \colon L (l^2 (F)) \to C^* (G, \, C_0 (G), \, \af)
\]
be the result of composing with the map
from $C_{\mathrm{c}} (G, \, C_0 (G), \, \af)$
to $C^* (G, \, C_0 (G), \, \af)$.
Set
\[
K_0 = \bigcup_{\mbox{$F \subset G$ finite}} L (l^2 (F)).
\]
Putting our \hm{s} together gives a \hm{}
\[
\ph^{(0)} \colon K_0 \to C^* (G, \, C_0 (G), \, \af).
\]
Since for each $F$ the restriction to $L (l^2 (F))$ is a \hm{} of \ca{s},
it follows that $\big\ \ph^{(0)} (x) \big\ \leq \ x \$
for all $x \in K_0$.
Therefore $\ph^{(0)}$ extends by continuity to a \hm{}
\[
\ph \colon K (l^2 (G)) \to C^* (G, \, C_0 (G), \, \af).
\]
The \hm{} $\ph$ is surjective because it has dense range,
and it is injective because $K (l^2 (G))$ is simple.
It follows that $C^* (G, \, C_0 (G), \, \af)$ is simple.
The natural map
$C^*_{\mathrm{r}} (G, \, C_0 (G), \, \af)
\to C^* (G, \, C_0 (G), \, \af)$
is then necessarily an isomorphism.
\end{exa}
We point out that in Example~\ref{E_C_Trans},
the full and reduced \cp{s} are the same
even if the group is not amenable.
\begin{exa}\label{E:C:FiniteRot}
Fix $n \in \N$, and consider the action of
$G = \Z / n \Z$ on $S^1$ generated by rotation by $2 \pi / n$,
that is, the \hme{} $h (\zt) = e^{2 \pi i / n} \zt$ for $\zt \in S^1$.
(This action is from Example~\ref{E_Rot}.)
We describe what to expect.
Every point in $S^1$ has a closed invariant \nbhd{} which
is equivariantly homeomorphic to $G \times I$
for some closed interval $I \subset \R$,
with the translation action on $G$
and the trivial action on~$I$.
This leads to quotients of $C^* (G, S^1, h)$
isomorphic to $M_n \otimes C (I)$.
(See Theorem~\ref{T_CPExact} and the general version of
Example~\ref{E_C_Trans}.)
Since $S^1$ itself is not such a product,
one does not immediately get an isomorphism
$C^* (G, S^1, h) \cong M_n \otimes C (Y)$ for any~$Y$.
Instead, one gets the section algebra of a locally trivial bundle
over $Y$ with fiber~$M_n$.
However, the appropriate
space $Y$ is the orbit space $S^1 / G \cong S^1$,
and all locally trivial bundles
over $S^1$ with fiber $M_n$ are in fact trivial.
Thus, one gets $C^* (G, S^1, h) \cong C (S^1, M_n)$ after all.
We carry out the details.
Let $\af \in \Aut (C (S^1))$ be the order~$n$ automorphism
given by
$\af (f) = f \circ h^{1}$ for $f \in C (S^1)$.
Thus, $\af (f) (\zt) = f (e^{2 \pi i / n} \zt)$ for $\zt \in S^1$.
Let $s \in M_n$ be the shift unitary
\[
s = \left( \begin{matrix}
0 & 0 & \cdots & \cdots & 0 & 0 & 1 \\
1 & 0 & \cdots & \cdots & 0 & 0 & 0 \\
0 & 1 & \cdots & \cdots & 0 & 0 & 0 \\
\vdots & \vdots & \ddots & & \vdots & \vdots & \vdots \\
\vdots & \vdots & & \ddots & \vdots & \vdots & \vdots \\
0 & 0 & \cdots & \cdots & 1 & 0 & 0 \\
0 & 0 & \cdots & \cdots & 0 & 1 & 0
\end{matrix} \right).
\]
The key computation,
which we leave to the reader, is
\begin{equation}\label{FiniteRot:Key}
s \, \diag (\ld_1, \ld_2, \ld_3, \ldots, \ld_n) \, s^*
= \diag (\ld_n, \ld_1, \ld_2, \ldots, \ld_{n  1})
\end{equation}
for $\ld_1, \ld_2, \ldots, \ld_n \in \C$.
Set
\[
B = \big\{ f \in C ([0, 1], \, M_n) \colon f (0) = s f (1) s^* \big\}.
\]
Define $\ph_0 \colon C (S^1) \to B$ by sending
$f \in C (S^1)$ to the continuously varying diagonal matrix
\[
\ph_0 (f) (t)
= \diag \big( f \big( e^{2 \pi i t / n} \big), \,
f \big( e^{2 \pi i (t + 1) / n} \big), \,
\ldots, \,
f \big( e^{2 \pi i (t + n  1) / n} \big) \big).
\]
(For fixed~$t$, the diagonal entries are obtained by evaluating
$f$ at the points in the orbit of $e^{2 \pi i t / n}$.)
The diagonal entries of $f (0)$ are gotten from those of $f (1)$
by a forwards cyclic shift,
so $\ph_0 (f)$ really is in~$B$.
For the same reason,
we get
\begin{align*}
\ph_0 (\af (f)) (t)
& = \diag \big( f \big( e^{2 \pi i (t  1) / n} \big), \,
f \big( e^{2 \pi i t / n} \big), \,
\ldots, \,
f \big( e^{2 \pi i (t + n  2) / n} \big) \big)
\\
& = s \ph_0 (f) (t) s^*.
\end{align*}
Now let $v \in C ([0, 1], \, M_n)$
be the constant function with value~$s$.
Then $v \in B$.
The calculation just done implies that
\[
\ph_0 \big( \af^k (f) \big) = v^k \ph_0 (f) v^{k}
\]
for $0 \leq k \leq n  1$.
Also clearly $v^n = 1$.
We write the group elements as $0, \, 1, \, \ldots, \, n  1$,
by abuse of notation treating them as integers when convenient.
The universal property of the crossed product therefore
implies that there is a \hm{} $\ph \colon C^* (G, S^1, h) \to B$
such that $\ph _{C (S^1)} = \ph_0$ and
(with $u_k$ as in Notation~\ref{N:ug}) $\ph (u_k) = v^k$
for $0 \leq k \leq n  1$.
% (We identify $k$ with its image in $G = \Z / n \Z$.)
We prove directly that $\ph$ is bijective.
By Corollary~\ref{C_FGpCP},
we can rewrite $\ph$ as the map
$C (\Z / n \Z \times S^1) \to B$
given by
\[
\ph (f)
= \sum_{k = 0}^{n  1} \ph_0 (f (k, )) v^k.
\]
Injectivity now reduces to the fact
that if $a_0, a_1, \ldots, a_{n  1} \in M_n$ are diagonal matrices,
and $\sum_{k = 0}^{n  1} a_k s^k = 0$,
then $a_0 = a_1 = \cdots = a_{n  1} = 0$.
To see this explicitly,
suppose that for $k = 0, 1, \ldots, n  1$,
we have
\[
a_k = \diag \big( \ld_1^{(k)}, \ld_2^{(k)}, \ldots, \ld_n^{(k)} \big)
\]
with $\ld_1^{(k)}, \ld_2^{(k)}, \ldots, \ld_n^{(k)} \in \C$.
Then
\[
\sum_{k = 0}^{n  1} a_k s^k = \left( \begin{matrix}
\ld_1^{(0)} & \ld_1^{(n1)} & \ld_1^{(n2)} & \cdots & \ld_1^{(1)} \\
\ld_2^{(1)} & \ld_2^{(0)} & \ld_2^{(n1)} & \cdots & \ld_2^{(2)} \\
\ld_3^{(2)} & \ld_3^{(1)} & \ld_3^{(0)} & \cdots & \ld_3^{(3)} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
\ld_n^{(n1)}& \ld_n^{(n2)} & \ld_n^{(n3)} & \cdots & \ld_n^{(0)}
\end{matrix} \right).
\]
For surjectivity,
let $a \in B$,
and write
\[
a (t) = \left( \begin{matrix}
a_{1, 1} (t) & a_{1, 2} (t) & \cdots & a_{1, n} (t) \\
a_{2, 1} (t) & a_{2, 2} (t) & \cdots & a_{2, n} (t) \\
\vdots & \vdots & \ddots & \vdots \\
a_{n, 1} (t) & a_{n, 2} (t) & \cdots & a_{n, n} (t)
\end{matrix} \right)
\]
with $a_{j, k} \in C ([0, 1])$ for $1 \leq j, k \leq n$.
The condition $a \in B$ implies that, taking the indices mod~$n$
in $\{ 1, 2, \ldots, n \}$,
we have $a_{j, k} (1) = a_{j + 1, \, k + 1} (0)$ for all $j$ and $k$.
Therefore the formula
\[
f \big( l, \, e^{2 \pi i (t + j) / n} \big) = a_{j + 1, \, j + 1  l} (t)
\]
for $t \in [0, 1]$, $j = 1, 2, \ldots, n$, and $l = 0, 1, \ldots, n  1$,
with $j + 1  l$ taken mod~$n$ in $\{ 1, 2, \ldots, n \}$,
gives a well defined element of $C (\Z / n \Z \times S^1)$.
One checks that $\ph (f) = a$.
It remains to prove that $B \cong C (S^1, M_n)$.
Since $U (M_n)$ is connected,
there is a unitary path $t \mapsto s_t$,
defined for $t \in [0, 1]$,
such that $s_0 = 1$ and $s_1 = s$.
Define $\ps \colon C (S^1, M_n) \to B$
by $\ps (f) (t) = s_t^* f (e^{2 \pi i t}) s_t$.
For $f \in C (S^1, M_n)$,
we have
\[
\ps (f) (1) = s^* f (1) s = s^* \ps (f) (0) s,
\]
so $\ps (f)$ really is in~$B$.
It is easily checked that $\ps$ is bijective.
\end{exa}
\begin{exa}\label{E:C:Sign}
Let $X = S^n = \{ x \in \R^{n + 1} \colon \ x \_2 = 1 \}$,
and let $\Z / 2 \Z$ act by sending the nontrivial group element to
the order~$2$ \hme{}
$x \mapsto  x$.
(This is Example~\ref{E:Sign}.)
The ``local structure'' of the \cp{} $C^* (\Z / 2 \Z, \, X)$
is the same as in Example~\ref{E:C:FiniteRot}.
However, for $n \geq 2$ the resulting bundle is no longer trivial.
The \cp{} is isomorphic to the section algebra of a
locally trivial but nontrivial bundle over the
real projective space $\R P^n = S^n / ( \Z / 2 \Z)$
with fiber~$M_2$.
See Proposition~4.15 of~\cite{Wlms}.
\end{exa}
The bundles one gets from free proper actions are,
however, often stably trivial.
Theorem~14 of~\cite{Gr} implies that the bundle
always comes from a bundle of Hilbert spaces,
and, if the algebra is separable,
Theorem~10.7.15 of~\cite{Dx} implies that the DixmierDouady
invariant is zero.
If the fibers are infinite dimensional,
and if the quotient space has finite covering dimension
or if the map $X \to X / G$ is locally trivial,
then the crossed product is $K \otimes C_0 (X / G)$.
See Theorems 10.8.4 and 10.8.8 of~\cite{Dx},
and Corollary~15 of~\cite{Gr}.
Proposition~2.52 of~\cite{Wlms}
gives a fairly explicit description of the crossed product
by a free action of $\Z / 2 \Z$ on a compact space~$X$,
although the question of triviality of the resulting bundle
is not addressed.
\begin{exa}\label{E:C:NonEff}
Let $X = \Z / n \Z$,
and let $\Z$ act on $X$ by translation.
We will give a direct proof that
that $C^* (\Z, X) \cong M_n \otimes C (S^1)$.
This is a special case of Example~\ref{E:Tr}.
In the general case (see Theorem~\ref{T:Homog} below),
it turns out that
\[
C^* (G, \, G / H) \cong K (L^2 (G / H)) \otimes C^* (H).
\]
There is no twisting.
Identify $\Z / n \Z$ with $\{ 1, 2, \ldots, n \}$.
(We start at $1$ instead of $0$
to be consistent with common matrix unit notation.)
Let $\af \in \Aut ( C (\Z / n \Z) )$ be
$\af (f) (k) = f (k  1)$,
with the argument taken mod~$n$ in $\{ 1, 2, \ldots, n \}$.
(Equivalently, $\af (\ch_{ \{ k \} } ) = \ch_{ \{ k + 1 \} }$,
with $k + 1$ taken to be $1$ when $k = n$.)
In $C (S^1)$ let $z$ be the function $z (\zt) = \zt$ for all~$\zt$.
In $M_n (C (S^1)) \cong M_n \otimes C (S^1)$,
abbreviate $e_{j, k} \otimes 1$ to $e_{j, k}$,
and let $v$ be the unitary
\[
v = \left( \begin{matrix}
0 & 0 & \cdots & \cdots & 0 & 0 & z \\
1 & 0 & \cdots & \cdots & 0 & 0 & 0 \\
0 & 1 & \cdots & \cdots & 0 & 0 & 0 \\
\vdots & \vdots & \ddots & & \vdots & \vdots & \vdots \\
\vdots & \vdots & & \ddots & \vdots & \vdots & \vdots \\
0 & 0 & \cdots & \cdots & 1 & 0 & 0 \\
0 & 0 & \cdots & \cdots & 0 & 1 & 0
\end{matrix} \right).
\]
(This unitary differs from the unitary $s$
in Example~\ref{E:C:FiniteRot}
only in that here the upper right corner entry is $z$ instead of~$1$.)
Define $\ph_0 \colon C (\Z / n \Z) \to M_n \otimes C (S^1)$
by $\ph_0 (\ch_{ \{ k \} } ) = e_{k, k}$
for $k = 1, 2, \ldots, n$,
and extending linearly.
Then one checks that $v \ph_0 (f) v^* = \ph_0 ( \af (f))$
for all $f \in C (\Z / n \Z)$.
Letting $u$ be the standard unitary in $C^* (\Z, \Z / n \Z)$
from the generator $1 \in \Z$
(called $u_1$ in Notation~\ref{N:ug}),
there is therefore a \hm{}
$\ph \colon C^* (\Z, \Z / n \Z) \to M_n \otimes C (S^1)$
such that $\ph _{C (\Z / n \Z)} = \ph_0$ and $\ph (u) = v$.
We claim that $\ph$ is an isomorphism.
The following description of $M_n \otimes C (S^1)$ will be useful:
it is the universal unital \ca{} generated by a system
$( e_{j, k} )_{1 \leq j, k \leq n}$ of matrix units
such that $\sum_{j = 1}^n e_{j, j} = 1$ and a central unitary~$y$.
The generators $e_{j, k}$ are the matrix units we have already used,
and the central unitary is $1 \otimes z$.
(Proof: Exercise~\ref{E:MnCSGenRel} below.)
To prove that $\ph$ is surjective,
it suffices to prove that its image contains
$1 \otimes z$ and contains $e_{j, k}$ for $j, k = 1, 2, \ldots, n$.
The image contains $1 \otimes z$ because $v^n = 1 \otimes z$.
For $j = 1, 2, \ldots, n$,
the image contains $e_{j, j} = \ph_0 (\ch_{ \{ j \} } )$.
The image therefore also contains
$e_{j + 1, \, j} = e_{j + 1, \, j + 1} v e_{j, j}$
for $j = 1, 2, \ldots, n  1$.
It now easily follows that the image contains
$e_{j, k}$ for all $j$ and $k$.
To prove injectivity,
we claim that
it suffices to prove that whenever $A$ is a unital \ca,
$\ps_0 \colon C (\Z / n \Z) \to A$ is a unital \hm,
and $w \in A$ is a unitary
such that $w \ps_0 (f) w^* = \ps_0 ( \af (f))$
for all $f \in C (\Z / n \Z)$,
then there is a \hm{} $\gm \colon M_n \otimes C (S^1) \to A$
such that $\gm \circ \ph_0 = \ps_0$
and $\gm (v) = w$.
To prove the claim, we use the
% existence part of the
universal property
of the \cp{}
(\Thm{L:UnivCP}).
% Suppose $\ph$ is not injective.
Take $A = C^* (\Z, \Z / n \Z)$,
let $\ps_0$ be the inclusion
of $C (\Z / n \Z)$ in $C^* (\Z, \Z / n \Z)$,
and let $w = u$ (the standard unitary in $C^* (\Z, \Z / n \Z)$).
Let $\gm \colon M_n \otimes C (S^1) \to C^* (\Z, \Z / n \Z)$
be the corresponding \hm.
Then
\[
\gm \circ \ph \colon C^* (\Z, \Z / n \Z) \to C^* (\Z, \Z / n \Z)
\]
satisfies $(\gm \circ \ph) (a) = a$
for $a \in C (\Z / n \Z)$
and $(\gm \circ \ph) (u) = u$.
So $\gm \circ \ph = \id_{C^* (\Z, \Z / n \Z)}$.
Therefore $\ph$ is injective.
It remains to construct~$\gm$,
and it suffices to define $\gm$ on the generators.
For $j = 1, 2, \ldots, n$,
we define $f_{j, j} = \ps_0 (\ch_{ \{ j \} } )$.
For $1 \leq k < j \leq n$,
we define $f_{j, k} = f_{j, j} w^{j  k} f_{k, k}$
and $f_{k, j} = f_{j, k}^*$.
One easily checks that
$( f_{j, k} )_{1 \leq j, k \leq n}$ is a system of matrix units
such that $\sum_{j = 1}^n f_{j, j} = 1$,
and that $w^n$ is a unitary which commutes with
$f_{j, k}$ for $j, k = 1, 2, \ldots, n$.
Accordingly, we may define $\gm$ by $\gm (1 \otimes z) = w^n$
and $\gm (e_{j, k}) = f_{j, k}$ for $j, k = 1, 2, \ldots, n$.
It is obvious that $\gm \circ \ph_0 = \ps_0$.
To compute $( \gm \circ \ph )(u)$,
we observe that
\[
\ph (u)
= v
= (1 \otimes z) e_{1, n} + \sum_{j = 1}^{n  1} e_{j + 1, \, j}.
\]
Therefore, using the definitions of the $f_{j, k}$ for $j \neq k$
at the third step
and the relations
$f_{j + 1, \, j + 1} = w f_{j, j} w^*$ for $j = 1, 2, \ldots, n  1$
and $f_{1, 1} = w^{(n  1)} f_{n, n} w^{n  1}$
at the fourth step,
we get
\begin{align*}
( \gm \circ \ph )(u)
& = \gm (v)
= w^n f_{1, n} + \sum_{j = 1}^{n  1} f_{j + 1, \, j}
\\
& = w^n f_{1, 1} w^{(n  1)} f_{n, n}
+ \sum_{j = 1}^{n  1} f_{j + 1, \, j + 1} w f_{j, j}
= w f_{n, n} + \sum_{j = 1}^{n  1} w f_{j, j}
= w.
\end{align*}
This completes the proof.
\end{exa}
\begin{exr}\label{E:MnCSGenRel}
Prove the description of $M_n \otimes C (S^1)$ in terms of
generators and relations used in Example~\ref{E:C:NonEff}.
\end{exr}
The relations are essentially the ones which define
$M_n \otimes_{\max} C (S_1)$.
The outcome of Example~\ref{E:C:NonEff}
holds much more generally.
\begin{thm}[Corollary~2.10 of~\cite{Gr2};
Theorem~4.30 of~\cite{Wlms}]\label{T:Homog}
Let $G$ be a locally compact group,
let $H \subset G$ be a closed subgroup,
and let $G$ act on $G / H$ by translation.
(This is Example~\ref{E:Tr}.)
Assume that there is a measurable cross section from $G / H$ to~$G$.
Then
\[
C^* (G, \, G / H) \cong K (L^2 (G / H)) \otimes C^* (H).
\]
\end{thm}
There is no twisting.
Several generalizations are worth mentioning.
\begin{thm}[Corollary~2.8 of~\cite{Gr2}]\label{T:Homog2}
Let $G$ be a locally compact group,
let $H \subset G$ be a closed subgroup,
and let $G$ act on $G / H$ by translation.
Assume that there is a measurable cross section from $G / H$ to~$G$.
Let $G$ also act on a \ca~$A$.
Then, using the diagonal action,
\[
C^* (G, \, C_0 (G / H) \otimes A)
\cong K (L^2 (G / H)) \otimes C^* (H, A).
\]
\end{thm}
\begin{thm}[Theorem~4.1 of~\cite{Gr2}]\label{T:Homog3}
Let $G$ be a locally compact group,
let $H \subset G$ be a closed subgroup,
and let $G$ act on $G / H$ by translation.
Let $X$ be a locally compact $G$space
such that there is a surjective continuous equivariant
map $p \colon X \to G / H$.
Assume that there is a measurable cross section from $G / H$ to~$G$.
Let $Y$ be the inverse image under~$p$ of the point $H \in G / H$.
Then
\[
C^* (G, X) \cong K (L^2 (G / H)) \otimes C^* (H, Y).
\]
\end{thm}
\begin{exa}\label{E_4302_RatRot}
The following example (not done in detail here)
combines the features
of Examples \ref{E:C:FiniteRot} and~\ref{E:C:NonEff}.
Regard the action of Example~\ref{E:C:FiniteRot}
as an action of $\Z$ rather than of $\Z / n \Z$.
(This action of $\Z$ also appears in Example~\ref{E_Rot},
where it is called a rational rotation.)
That is, fix $n \in \N$, and consider the action of
$G = \Z$ on $S^1$ generated by rotation by $2 \pi / n$,
equivalently,
generated by the \hme{}
$h (\zt) = e^{2 \pi i / n} \zt$ for $\zt \in S^1$.
The crossed product is a special case of what is known
as a rational rotation algebra.
(The general case uses generating rotations by $2 \pi k / n$,
not just $2 \pi / n$.)
The heuristic argument of Example~\ref{E:C:FiniteRot}
and the outcome of Example~\ref{E:C:NonEff}
suggest that the crossed product should be
the section algebra of a locally trivial bundle
over $S^1$ with fiber $C (S^1, M_n)$.
It is not hard to show that this is in fact what happens.
(Exercise: Do it.)
The resulting bundle is not trivial.
In fact, it can be easily seen that it is also
the section algebra of a locally trivial bundle
over $S^1 \times S^1$ with fiber~$M_n$.
This bundle is also nontrivial.
The bundles for general rational rotation algebras
are computed in~\cite{HS}.
(See Example~8.46 of~\cite{Wlms}.)
\end{exa}
\begin{rmk}\label{R:GR}
In Examples \ref{E:C:FiniteRot} and~\ref{E:C:NonEff},
we have seen two sources of ideals in a
reduced \cp{} $C^*_{\mathrm{r}} (G, A, \af)$:
invariant ideals in $A$,
and group elements which act trivially on~$A$.
There is a theorem
due to Gootman and Rosenberg
which gives a description of the primitive ideals
of any \cp{} $C^* (G, A)$ with $G$ amenable,
and which, very roughly, says that they all
come from some combination of these two sources.
(One does not even need to restrict to discrete groups.)
To be a little more precise,
every primitive ideal in $C^* (G, A)$
is ``induced'' from an ideal $J$ in a \cp{} by the
stabilizer subgroup of some primitive ideal $P$ of $A$,
with $J$ closely related to~$P$.
The theorem is Theorem~8.21 of~\cite{Wlms};
see Definition~8.18 of~\cite{Wlms} for the terminology.
The proof of the GootmanRosenberg Theorem is quite long.
(Starting from about the same assumed background as these notes,
it occupies a large part of the book~\cite{Wlms}.)
\end{rmk}
\begin{exa}\label{E:C:Conj}
Take $X = S^1 = \{ \zt \in \C \colon  \zt  = 1 \}$,
and let $\Z / 2 \Z$ act by sending the nontrivial group element to
the order two \hme{}
$\zt \mapsto {\overline{\zt}}$.
(This is Example~\ref{E:Conj}.)
Let $\af \in \Aut (C (S^1))$ be the corresponding automorphism.
We compute the crossed product, but we first describe what to expect.
By considering Theorem~\ref{T_CPExact}
and Examples \ref{E:C:Triv} and~\ref{E:C:FiniteRot},
we should expect that the points $1$ and $1$ contribute
quotients isomorphic to $\C \oplus \C$,
and that for $\zt \neq \pm 1$,
the pair of points $\big( \zt, {\overline{\zt}} \big)$ contributes
a quotient isomorphic to~$M_2$.
We will in fact show that $C^* (\Z / 2 \Z, \, X)$ is isomorphic to the
\ca{}
% \[
% B = \{ f \colon [1, \, 1] \to M_2 \colon
% {\mbox{$f$ is \ct{} and $f (1)$ and $f (1)$ are diagonal matrices}}
% \}.
% \]
\[
B = \big\{ f \in C ( [1, \, 1], \, M_2) \colon
{\mbox{$f (1)$ and $f (1)$ are diagonal matrices}} \big\}.
\]
First, let $C_0 \subset M_2$ be the subalgebra consisting of all
matrices of the form
$\left( \begin{smallmatrix} \ld & \mu \\
\mu & \ld \end{smallmatrix} \right)$
with $\ld, \mu \in \C$.
(The reader should check that $C_0$ is actually a subalgebra.)
Then define
\[
C = \big\{ f \colon [1, \, 1] \to M_2 \colon
{\mbox{$f$ is \ct{} and $f (1), \, f (1) \in C_0$}} \big\}.
\]
Let $v \in C$ be the constant function
$v (t) = \left(\begin{smallmatrix} 0 & 1 \\
1 & 0 \end{smallmatrix} \right)$
for all $t \in [1, \, 1]$.
Define $\ph_0 \colon C (S^1) \to C$ by
\[
\ph_0 (f) (t)
= \left( \begin{matrix} f \big( t + i \sqrt{1  t^2} \big) & 0 \\
0 & f \big( t  i \sqrt{1  t^2} \big) \end{matrix} \right)
\]
for $f \in C (S^1)$ and $t \in [1, \, 1]$.
One checks that the conditions at $\pm 1$ for membership in $C$
are satisfied.
Moreover,
$v^2 = 1$ and $v \ph_0 (f) v^* = \ph_0 (\af (f))$ for $f \in C (S^1)$.
Therefore there is a \hm{} $\ph \colon C^* (\Z / 2 \Z, \, X) \to C$
such that $\ph _{C (S^1)} = \ph_0$ and $\ph$ sends the standard
unitary $u$ in $C^* (\Z / 2 \Z, \, X)$ to~$v$.
It is given by the formula
\[
\ph (f_0 + f_1 u) (t)
= \left( \begin{matrix} f_0 \big( t + i \sqrt{1  t^2} \big)
& f_1 \big( t + i \sqrt{1  t^2} \big) \\
f_1 \big( t  i \sqrt{1  t^2} \big)
\rule{0em}{2.7ex}
& f_0 \big( t  i \sqrt{1  t^2} \big) \end{matrix} \right)
\]
for $f_1, f_2 \in C (S^1)$ and $t \in [1, \, 1]$.
We claim that $\ph$ is an isomorphism.
Since
\[
C^* (\Z / 2 \Z, \, X)
= \big\{ f_0 + f_1 u \colon f_1, f_2 \in C (S^1) \big\}
\]
by Corollary~\ref{C_FGpCP},
it is easy to check injectivity.
For surjectivity,
let
\[
a (t)
= \left( \begin{matrix} a_{1, 1} (t) & a_{1, 2} (t) \\
a_{2, 1} (t) & a_{2, 2} (t) \end{matrix} \right)
\]
define an element $a \in C$.
Then
\begin{equation}\label{Match1}
a_{1, 1} (1) = a_{2, 2} (1) \andeqn
a_{2, 1} (1) = a_{1, 2} (1),
\end{equation}
and
\begin{equation}\label{Match2}
a_{1, 1} (1) = a_{2, 2} (1) \andeqn
a_{2, 1} (1) = a_{1, 2} (1).
\end{equation}
Now set
\[
f_0 (\zt) = \begin{cases}
a_{1, 1} ({\mathrm{Re}} (\zt))
& \hspace{3em} {\mathrm{Im}} (\zt) \geq 0 \\
a_{2, 2} ({\mathrm{Re}} (\zt))
& \hspace{3em} {\mathrm{Im}} (\zt) \leq 0
\end{cases}
\]
and
\[
f_1 (\zt) = \begin{cases}
a_{1, 2} ({\mathrm{Re}} (\zt))
& \hspace{3em} {\mathrm{Im}} (\zt) \geq 0 \\
a_{2, 1} ({\mathrm{Re}} (\zt))
& \hspace{3em} {\mathrm{Im}} (\zt) \leq 0
\end{cases}
\]
for $\zt \in S^1$.
The relations (\ref{Match1}) and~(\ref{Match2})
ensure that $f_0$ and $f_1$
are well defined at $\pm 1$, and are \ct.
One easily checks that $\ph (f_0 + f_1 u) = a$.
This proves surjectivity.
The algebra $C$ is not quite what was promised.
Set
\[
w =
\left( \begin{matrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
 \frac{1}{\sqrt{2}} \rule{0em}{2.7ex}
& \frac{1}{\sqrt{2}} \end{matrix} \right),
\]
which is a unitary in~$M_2$.
Then the required isomorphism $\ps \colon C^* (\Z / 2 \Z, \, X) \to B$
is given by $\ps (a) (t) = w \ph (a)(t) w^*$.
(Check this!)
\end{exa}
In this example, one choice of matrix units in $M_2$
was convenient for the free orbits,
while another choice was convenient for the fixed points.
It seemed better to compute everything in terms of the choice
convenient for the free orbits,
and convert afterwards.
\begin{exr}\label{E:C:FlipInt}
Let $\Z / 2 \Z$ act on $[1, \, 1]$ via $x \mapsto  x$.
Compute the \cp.
\end{exr}
\begin{exr}\label{E:C:FlipSph}
Let $\Z / 2 \Z$ act on
\[
S^n = \{ (x_1, x_2, \ldots, x_{n + 1} ) \colon
x_1^2 + x_2^2 + \cdots + x_{n + 1}^2 = 1 \}
\]
via
$(x_1, x_2, \ldots, x_n, x_{n + 1} )
\mapsto (x_1, \, x_2, \, \ldots, \, x_n, \,  x_{n + 1} )$.
Compute the \cp.
\end{exr}
% \begin{exa}\label{E:C:ConjZ}
% (To go after Example~\ref{E:C:Conj}.)
% Take $X = S^1 = \{ \zt \in \C \colon  \zt  = 1 \}$,
% and let $\Z$ act by sending the standard generator to
% the order two \hme{}
% $\zt \mapsto {\overline{\zt}}$.
% (The \hme{} is the same as the that in Example~\ref{E:C:Conj},
% but now we are considering it as generating an action of~$\Z$.)
%
% There is supposed to be some nontriviality in the structure
% of the \cp.
% % 999
% \end{exa}
In~\cite{Evn},
there is a detailed analysis of the structure of
crossed products of compact spaces by compact groups,
in terms of sections of suitable bundles
of \ca{s}, usually not locally trivial
but locally trivial over suitable subspaces of the base space.
The crossed products and fixed point algebras of
the actions of finite subgroups of $\SL_2 (\Z)$
(discussed in Example~\ref{E:SL2})
on the rational rotation algebras
(take $\te \in \Q$ in Example~\ref{E:SL2};
the case $\te = 0$
is the action on $S^1 \times S^1$ in Example~\ref{E:CommSL2})
have been computed in
Theorems 6.1, 1.2, and~1.3 of~\cite{BEEK} (for $\Z / 2 \Z$),
in the theorem at the end of Section~1 of~\cite{FW3} (for $\Z / 3 \Z$),
in Theorem 6.2.1 of~\cite{FW4} (for $\Z / 4 \Z$),
and in the theorem at the end of Section~1 of~\cite{FW6} (for $\Z / 6 \Z$).
(For $\Z / 3 \Z$ and $\Z / 6 \Z$,
the proofs are only given
for the corresponding computation of the fixed point algebras.)
The rational rotation algebras are not commutative,
but they are close to commutative,
being section algebras of locally trivial bundles over
$S^1 \times S^1$ whose fiber is a single matrix algebra.
\begin{exa}\label{E:C:PType}
We compute the crossed product by one
of the specific examples at the end of Example~\ref{E_PrdType},
namely the action of $\Z / 2 \Z$ on the $2^{\infty}$~UHF algebra~$A$
generated by
$\bigotimes_{n = 1}^{\I}
\Ad \left(
\begin{smallmatrix} 1 & 0 \\ 0 & 1 \end{smallmatrix} \right)$.
We simply write $\af$ for the automorphism given by the nontrivial
group element.
(In Example~\ref{E:Z2RP},
this action is shown to have the Rokhlin property.)
Write $A = \Dirlim M_{2^n}$,
with maps $\ph_n \colon M_{2^n} \to M_{2^{n + 1}}$
given by
$a \mapsto
\left( \begin{smallmatrix} a & 0 \\ 0 & a \end{smallmatrix} \right)$
for $a \in M_{2^n}$ and $n \in \Nz$.
Define unitaries $z_n \in M_{2^n}$ inductively by $z_0 = 1$
and
$z_{n + 1} = \left(
\begin{smallmatrix} z_n & 0 \\ 0 & z_n \end{smallmatrix} \right)$.
(In tensor product notation, and with an appropriate choice of
isomorphism $M_{2^n} \otimes M_2 \to M_{2^{n + 1}}$,
these are $\ph_n (a) = a \otimes 1_{M_2}$ and
$z_{n + 1} = z_n \otimes \left(
\begin{smallmatrix} 1 & 0 \\ 0 & 1 \end{smallmatrix} \right)$.)
Let
\[
{\overline{\ph}}_{n}
\colon C^* \big( \Z / 2 \Z, \, M_{2^n}, \, \Ad (z_n) \big)
\to C^* \big( \Z / 2 \Z, \, M_{2^{n + 1}}, \, \Ad (z_{n + 1}) \big)
\]
be the corresponding map on the crossed products.
By \Thm{T:DLim},
the crossed product $C^* (\Z / 2 \Z, \, A, \, \af)$
is the direct limit of the resulting direct system.
In the crossed product
$C^* \big( \Z / 2 \Z, \, M_{2^n}, \, \Ad (z_n) \big)$,
let $u_n$ be the standard unitary corresponding to the nontrivial
group element.
(This notation is not entirely consistent with Notation~\ref{N:ug}.)
{}From the discussion at the end of Example~\ref{E:C:Inner},
we get the isomorphisms
\[
\sm_n \colon C^* \big( \Z / 2 \Z, \, M_{2^n}, \, \Ad (z_n) \big)
\to M_{2^n} \oplus M_{2^n}
\]
given by
$a + b u_n \mapsto (a + b z_n, \, a  b z_n)$.
We now need a map
\[
\ps_n \colon M_{2^n} \oplus M_{2^n}
\to M_{2^{n + 1}} \oplus M_{2^{n + 1}}
\]
which makes the following diagram commute:
\[
\begin{CD}
C^* \big( \Z / 2 \Z, \, M_{2^n}, \, \Ad (z_n) \big)
@>{\sm_n}>> M_{2^n} \oplus M_{2^n}
\\
@V{{\overline{\ph}}_{n}}VV @VV{\ps_n}V
\\
C^* \big( \Z / 2 \Z, \, M_{2^{n + 1}}, \, \Ad (z_{n + 1}) \big)
@>{\sm_{n + 1}}>> M_{2^{n + 1}} \oplus M_{2^{n + 1}}.
\end{CD}
\]
That is, $\ps_n$~sends
\[
\sm_n (a + b u_n) = (a + b z_n, \, a  b z_n)
\]
to
\begin{align*}
& \sm_{n + 1} (\ph_n (a) + \ph_n (b) u_{n + 1})
\\
& \hspace*{3em} {\mbox{}}
= \left(
\left( \begin{matrix} a & 0 \\ 0 & a \end{matrix} \right)
+
\left( \begin{matrix} b & 0 \\ 0 & b \end{matrix} \right)
\left(
\begin{matrix} z_n & 0 \\ 0 & z_n \end{matrix} \right),
\,\,
\left( \begin{matrix} a & 0 \\ 0 & a \end{matrix} \right)

\left( \begin{matrix} b & 0 \\ 0 & b \end{matrix} \right)
\left(
\begin{matrix} z_n & 0 \\ 0 & z_n \end{matrix} \right)
\right)
\\
& \hspace*{3em} {\mbox{}}
= \left(
\left(
\begin{matrix} a + b z_n & 0 \\ 0 & a  b z_n \end{matrix} \right),
\,\,
\left(
\begin{matrix} a  b z_n & 0 \\ 0 & a + b z_n \end{matrix} \right)
\right).
\end{align*}
So we take
\[
\ps_n (b, c)
= \left( \left( \begin{matrix} b & 0 \\ 0 & c \end{matrix} \right),
\,\,
\left( \begin{matrix} c & 0 \\ 0 & b \end{matrix} \right)
\right)
\]
for $b, c \in M_{2^n}$.
Those familiar with Bratteli diagrams will now be able to write
down the Bratteli diagram for the crossed product.
Here, we give a direct identification
of the direct limit.
Inductively define unitaries $x_n, y_n \in M_{2^n}$
by $x_0 = y_0 = 1$ and
\[
x_{n + 1}
= \left( \begin{matrix} x_n & 0 \\ 0 & y_n \end{matrix} \right)
\andeqn
y_{n + 1}
= \left( \begin{matrix} 0 & y_n \\ x_n & 0 \end{matrix} \right)
\]
for $n \in \Nz$.
Then define $\ld_n \colon M_{2^n} \to M_{2^n} \oplus M_{2^n}$
by $\ld_n (a) = (x_n a x_n^*, \, y_n a y_n^*)$ for $a \in M_{2^n}$,
and define $\mu_n \colon M_{2^n} \oplus M_{2^n} \to M_{2^{n + 1}}$
by
\[
\mu_n (b, c)
= \left(
\begin{matrix} x_n^* b x_n & 0 \\ 0 & y_n^* c y_n \end{matrix} \right)
\]
for $b, c \in M_{2^n}$.
Then one checks that $\mu_n \circ \ld_n = \ph_n$
and $\ld_{n + 1} \circ \mu_n = \ps_n$ for all $n$.
It follows that the direct limit of the system
\[
\C \oplus \C \stackrel{\ps_0}{\longrightarrow}
M_2 \oplus M_2 \stackrel{\ps_1}{\longrightarrow}
M_4 \oplus M_4 \stackrel{\ps_2}{\longrightarrow}
M_8 \oplus M_8 \stackrel{\ps_3}{\longrightarrow} \cdots,
\]
which is the crossed product
$C^* (\Z / 2 \Z, \, A, \, \af)$,
is isomorphic to the direct limit of the system
\[
\C \stackrel{\ph_0}{\longrightarrow}
M_2 \stackrel{\ph_1}{\longrightarrow}
M_4 \stackrel{\ph_2}{\longrightarrow}
M_8 \stackrel{\ph_3}{\longrightarrow} \cdots,
\]
which is the original algebra~$A$.
\end{exa}
It follows from \Lem{L_3416_NotInn}
that the action in Example~\ref{E:C:PType}
is not inner.
The result of the computation of the crossed product
implies this as well.
Indeed, the \cp{} is simple,
so comparison with Example~\ref{E:C:Inner}
shows that the action is not inner.
The theorem
of Gootman and Rosenberg described in Remark~\ref{R:GR}
gives no information here.
The fact that we got the same algebra back in Example~\ref{E:C:PType}
is somewhat special,
but the general principle of the computation is much more
generally applicable.
We sketch a slightly different example
in which we do not get the same algebra back.
\begin{exa}\label{E:C:PTypeAgain}
Let $\af$ the action of $\Z / 2 \Z$ on the $3^{\infty}$~UHF
algebra~$A$
generated by
\[
\bigotimes_{n = 1}^{\I}
\Ad \left(
\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}
\right)
\]
for $n \in \N$ and $a \in M_{3^n}$.
Again, we also write $\af$ for the automorphism
given by the nontrivial
group element.
(This action has the \trp{}
but not the Rokhlin property.
See Remark~\ref{R_6X17_IsConj} and Example~\ref{E_1211_NoRPUHF_2On3}.)
Write $A = \Dirlim M_{3^n}$,
with maps $\ph_n \colon M_{3^n} \to M_{3^{n + 1}}$
given by
$a \mapsto \diag (a, a, a)$
for $n \in \Nz$
and $a \in M_{3^n}$.
Define unitaries $z_n \in M_{3^n}$ inductively by $z_0 = 1$
and
\[
z_{n + 1} = \left(
\begin{matrix} z_n & 0 & 0 \\ 0 & z_n & 0 \\ 0 & 0 & z_n \end{matrix}
\right).
\]
Let
\[
{\overline{\ph}}_{n}
\colon C^* \big( \Z / 2 \Z, \, M_{3^n}, \, \Ad (z_n) \big)
\to C^* \big( \Z / 2 \Z, \, M_{3^{n + 1}}, \, \Ad (z_{n + 1}) \big)
\]
be the corresponding map on the crossed products,
so that $C^* (\Z / 2 \Z, \, A, \, \af)$
is the direct limit of the resulting direct system.
Let $u_n \in C^* \big( \Z / 2 \Z, \, M_{3^n}, \, \Ad (z_n) \big)$
be the standard unitary,
as in \Ex{E:C:PType}.
The isomorphism
\[
\sm_n \colon C^* \big( \Z / 2 \Z, \, M_{3^n}, \, \Ad (z_n) \big)
\to M_{3^n} \oplus M_{3^n}
\]
is still
$a + b u_n \mapsto (a + b z_n, \, a  b z_n)$.
Using calculations similar to those of Example~\ref{E:C:PType},
one sees that the map
\[
\ps_n \colon M_{3^n} \oplus M_{3^n}
\to M_{3^{n + 1}} \oplus M_{3^{n + 1}}
\]
should now be given by
\[
\ps_n (b, c)
= \big( \diag (b, b, c),
\,\,
\diag (c, c, b)
\big).
\]
Again, one can immediately write
down the Bratteli diagram for the crossed product.
Instead, we directly calculate the (unordered) $K_0$group
of the crossed product.
It is the direct limit $\Dirlim K_0 (M_{3^n} \oplus M_{3^n})$,
with the maps being
\[
(\ps_n)_* \colon K_0 (M_{3^n} \oplus M_{3^n})
\to K_0 (M_{3^{n + 1}} \oplus M_{3^{n + 1}}).
\]
The calculation is based on the observation that
the map $(\ps_n)_* \colon \Z^2 \to \Z^2$ is given by the
matrix
$(\ps_n)_*
= \left( \begin{smallmatrix} 2 & 1 \\ 1 & 2 \end{smallmatrix} \right)$,
which has eigenvector $(1, 1)$ with eigenvalue~$1$
and eigenvector $(1, 1)$ with eigenvalue~$3$.
(Usually one will not be so lucky:
the calculations will be messier.)
We claim that
we can identify $K_0 \big( C^* (\Z / 2 \Z, \, A, \, \af) \big)$
with
\[
H = \left\{ (k, l) \in \Z \oplus \Z \left[ \tfrac{1}{3} \right]
\colon k + l \in 2 \cdot \Z \left[ \tfrac{1}{3} \right] \right\},
\]
and with the class $[1]$ being sent to $(0, 1)$.
For $n \in \Nz$,
define $f_n \colon \Z^2 \to \Z \oplus \Z \left[ \tfrac{1}{3} \right]$
by
\[
f_n (r, s) = \left( r  s, \, \frac{r + s}{3^n} \right)
\]
for $r, s \in \Z$.
One checks immediately that $f_n = f_{n + 1} \circ (\ps_n)_*$.
Therefore the group \hm{s} $f_n$
combine to yield a \hm{}
$f \colon \Dirlim \Z^2 \to \Z \oplus \Z \left[ \tfrac{1}{3} \right]$,
whose range is easily seen to be in~$H$.
This \hm{} is injective because $f_n$ is injective
for all $n \in \Nz$.
It remains only to show that if $(r, s) \in H$
then there exist $n \in \Nz$ and $k, l \in \Z$
such that $f_n (k, l) = (r, s)$.
Choose $n \in \Nz$ such that $3^n s \in \Z$.
Set
\[
k = \frac{3^n s + r}{2}
\andeqn
l = \frac{3^n s  r}{2}.
\]
It is easy to see that either
$r \in 2 \Z$ and $s \in 2 \cdot \Z \left[ \tfrac{1}{3} \right]$
or $r \not\in 2 \Z$
and $s \not\in 2 \cdot \Z \left[ \tfrac{1}{3} \right]$,
and that in either case $k, l \in \Z$.
Thus $(k, l) \in \Z^2$,
and clearly $f_n (k, l) = (r, s)$.
This completes the calculation.
\end{exa}
The following two exercises are much harder than most of
the exercises in these notes.
The first combines the methods
of Example~\ref{E:C:Conj} (see Exercise~\ref{E:C:FlipSph})
and the methods of Example~\ref{E:C:PTypeAgain},
and the second uses Example~\ref{E:C:Sign}
in place of Exercise~\ref{E:C:FlipSph}.
The computations asked for in the exercises are
an important part of Propositions~4.6 and~4.2 of~\cite{PhT4},
which describe the properties of two significant examples
of \cp{s}.
Both actions are shown in~\cite{PhT4} to have the \trp,
but do not have the \rp.
\begin{exr}\label{Ex:RCFG4_K1}
Let $m \in \N$.
Define $h \colon S^{2 m} \to S^{2 m}$ by
\[
h (x_0, x_1, \ldots, x_{2 m}) = (x_0, x_1, \ldots, x_{2 m})
\]
for $x = (x_0, x_1, \ldots, x_{2 m}) \in S^{2 m}$,
and let
$\bt \in \Aut (C (S^{2 m} ))$ be the corresponding automorphism of
order~$2$.
For $r \in \N$ and $b \in S^{2 m}$, define
$\ps_{r, b} \colon C (S^{2 m}) \to M_{2 r + 1} \otimes C (S^{2 m})$
by
\[
\ps_{r, b} (f) (x)
= \diag \big( f (x), \, f (b), \, f (h (b)), \, f (b), \, f (h (b)),
\, \ldots, \, f (b), \, f (h (b)) \big)
\]
for $x \in S^{2 m}$, where $f (b)$ and $f (h (b))$ each occur $r$ times.
Choose a dense sequence $(x (n))_{n \in \N}$ in $S^{2 m}$,
such that no point $x_n$ is a fixed point of $h$,
and choose a
sequence $(r (n))_{n \in \N}$ of strictly positive integers.
Set
\[
s (n) = [2 r (1) + 1] [2 r (2) + 1] \cdots [2 r (n) + 1],
\]
and set
$A_n = M_{s (n)} \otimes C (S^{2 m})$,
which, when appropriate, we think of as
\[
M_{2 r (1) + 1} \otimes M_{2 r (2) + 1}
\otimes \cdots \otimes M_{2 r (n) + 1}
\otimes C (S^{2 m}).
\]
Define $\ph_n \colon A_{n  1} \to A_n$ by
$\ph_n = \id_{M_{s (n  1)}} \otimes \ps_{r (n), \, x (n)}$.
Then set $A = \Dirlim A_n$.
For $r \in \N$ define a unitary $w_r \in M_{2 r + 1}$ by
\[
w_r = \diag \left( 1, \,\,
\left( \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right), \,\,
\left( \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right), \,\,
\ldots, \,\,
\left( \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right)
\right).
\]
Then define an automorphism $\af_n \in \Aut (A_n)$ of order~$2$ by
\[
\af_n
= \Ad (w_{r (1)} \otimes w_{r (2)} \otimes \cdots \otimes w_{r (n)} )
\otimes \bt.
\]
One checks that
$\ph_n \circ \af_{n  1} = \af_n \circ \ph_n$,
so that the automorphisms $\af_n$
define an automorphism $\af \in \Aut (A)$ of order~$2$.
Compute the \cp{} $C^* (\Z / 2 \Z, \, A, \af)$,
at least sufficiently well to determine its Ktheory.
\end{exr}
\begin{exr}\label{Ex:RCFG4_Torsion}
Repeat Exercise~\ref{Ex:RCFG4_K1},
with just one change:
the formula for $h$ is now $h (x) = x$ for all $x \in S^{2 m}$.
The algebras in the direct system
for the \cp{} are harder to describe,
since they are section algebras of nontrivial bundles
(see \Ex{E:C:Sign}),
but the full description is not needed in order to compute the
Ktheory of the resulting direct limit.
\end{exr}
\begin{exa}\label{E:C:IrrRot}
Let $\te \in \R$.
Recall from Example~\ref{E_3303_RotAlg}
that the rotation algebra $A_{\te}$
is the universal \ca{} generated by unitaries
$u$ and $v$ satisfying $v u = e^{2 \pi i \te} u v$.
Let $h_{\te} \colon S^1 \to S^1$ be the \hme{}
$h_{\te} (\zt) = e^{2 \pi i \te} \zt$.
(Recall Example~\ref{E_Rot}.)
We claim that there is an isomorphism
$\ph \colon A_{\te} \to C^* (\Z, S^1, h_{\te})$
which sends $u$ to the standard unitary $u_1$ in the \cp{}
(see Notation~\ref{N:ug}),
and sends $v$ to the function $z \in C (S^1)$
defined by $z (\zt) = \zt$ for all $\zt \in S^1$.
(In Corollary~\ref{C:UnivCPZ},
the unitary $u_1$ was called~$u$,
so we are essentially sending $u$ to $u$.)
The proof of the claim is by comparison of universal properties.
First, one checks that $z u_1 = e^{2 \pi i \te} u_1 z$,
so at least there is a \hm~$\ph$
with the properties claimed.
Next,
define a \hm{} $\ps_0 \colon C (S^1) \to A_{\te}$
by $\ps_0 (f) = f (v)$ (\ct{} functional calculus)
for $f \in C (S^1)$.
For $n \in \Z$, we have,
using $v u = e^{2 \pi i \te} u v$ at the second step,
\[
u \ps_0 (z^n) u^*
= (u v u^*)^n
= e^{ 2 \pi i n \te} v^n
= \ps_0 \big( e^{ 2 \pi i n \te} z^n \big)
= \ps_0 \big( z^n \circ h_{\te}^{1} \big).
\]
Since the functions $z^n$ span a dense subspace of $C (S^1)$,
it follows that
$u \ps_0 (f) u^* = \ps_0 \big( f \circ h_{\te}^{1} \big)$
for all $f \in C (S^1)$.
By Corollary~\ref{C:UnivCPZ},
there is a \hm{} $\ps \colon C^* (\Z, S^1, h_{\te}) \to A_{\te}$
such that $\ps _{C (S^1)} = \ps_0$ and $\ps (u_1) = u$.
We have $(\ps \circ \ph) (u) = u$ and $(\ps \circ \ph) (v) = v$.
Since $u$ and $v$ generate~$A_{\te}$,
we conclude that $\ps \circ \ph = \id_{A_{\te}}$.
Similarly,
$(\ph \circ \ps) (z) = z$ and $(\ph \circ \ps) (u_1) = u_1$,
the elements $z$ and $u_1$ generate $C^* (\Z, S^1, h_{\te})$
(since $z$ generates $C (S^1)$),
and therefore $\ph \circ \ps = \id_{ C^* (\Z, S^1, h_{\te}) }$.
We will see below that for $\te \in \R \setminus \Q$,
the algebra $C^* (\Z, S^1, h_{\te})$ is simple.
(See Theorems~\ref{T:AS} and~\ref{T:IrrRot} below,
and also Proposition~2.56 of~\cite{Wlms}.)
On the other hand,
if $\te = p / q$ in lowest terms, with $q > 0$,
then $A_{\te}$ turns out to be the section algebra
of a locally trivial bundle over $S^1 \times S^1$ with fiber $M_q$.
(See Example~8.46 of~\cite{Wlms}.)
The bundles have trivial DixmierDouady class,
so are stably trivial,
but they are not trivial.
They are analyzed in~\cite{HS}.
\end{exa}
We finish this section with several further results
on crossed products by tensor products of actions,
given as exercises.
Remark~\ref{R:CPTens}
and Exercise~\ref{P:Triv}
can be obtained from Exercise~\ref{Ex_6X11_TProdAction}
and Exercise~\ref{Ex_6X11_TProdCrPrd}
by taking $H = G$
and restricting to the diagonal subgroup
$\{ (g, g) \colon g \in G \} \subset G \times G$,
or (for Exercise~\ref{P:Triv})
taking $H$ to be trivial.
\begin{exr}\label{Ex_6X11_TProdAction}
Let $G$ and $H$ be topological groups,
let $A$ and $B$ be \ca{s},
and let $\aGA$ and $\bt \colon H \to \Aut (B)$
be actions of $G$ and $H$ on $A$ and~$B$.
\begin{enumerate}
\item\label{Ex_6X11_TProdAction_Max}
Prove that there is a unique action
$\gm \colon G \times H \to \Aut (A \otimes_{\mathrm{max}} B)$
such that for all $g \in G$, $h \in H$,
$a \in A$, and $b \in B$,
we have $\gm_{(g, h)} (a \otimes b) = \af_g (a) \otimes \bt_h (b)$.
\item\label{Ex_6X11_TProdAction_Min}
Prove that there is a unique action
$\rh \colon G \times H \to \Aut (A \otimes_{\mathrm{min}} B)$
such that for all $g \in G$, $h \in H$,
$a \in A$, and $b \in B$,
we have $\rh_{(g, h)} (a \otimes b) = \af_g (a) \otimes \bt_h (b)$.
\end{enumerate}
\end{exr}
When $G$ and $H$ are locally compact,
the full crossed product
in Exercise \ref{Ex_6X11_TProdAction}(\ref{Ex_6X11_TProdAction_Max})
and the reduced crossed product
in Exercise \ref{Ex_6X11_TProdAction}(\ref{Ex_6X11_TProdAction_Min})
are
\[
C^* (G, A, \af) \otimes_{\mathrm{max}} C^* (H, B, \bt)
\andeqn
C^*_{\mathrm{r}} (G, A, \af)
\otimes_{\mathrm{min}} C^*_{\mathrm{r}} (H, B, \bt).
\]
See Exercise~\ref{Ex_6X11_TProdCrPrd}
and Exercise~\ref{Ex_6X11_TProdRdCrPrd}
for the case in which $G$ and $H$ are discrete.
\begin{exr}\label{Ex_6X11_TProdCrPrd}
Let $G$ and $H$ be discrete groups,
let $A$ and $B$ be \ca{s},
and let $\aGA$ and $\bt \colon H \to \Aut (B)$
be actions of $G$ and $H$ on $A$ and~$B$.
Let $\gm \colon G \times H \to \Aut (A \otimes_{\mathrm{max}} B)$
be the action of
Exercise \ref{Ex_6X11_TProdAction}(\ref{Ex_6X11_TProdAction_Max}),
satisfying $\gm_{(g, h)} (a \otimes b) = \af_g (a) \otimes \bt_h (b)$
for $g \in G$, $h \in H$,
$a \in A$, and $b \in B$.
Prove that
\[
C^* (\gm, \, G \times H, \, A \otimes_{\mathrm{max}} B)
\cong C^* (G, A, \af) \otimes_{\mathrm{max}} C^* (H, B, \bt).
\]
\end{exr}
\begin{exr}\label{Ex_6X11_TProdRdCrPrd}
Let $G$ and $H$ be discrete groups,
let $\aGA$ and $\bt \colon H \to \Aut (B)$
be as in Exercise~\ref{Ex_6X11_TProdCrPrd},
and let
$\rh \colon G \times H \to \Aut (A \otimes_{\mathrm{min}} B)$
be the action of
Exercise \ref{Ex_6X11_TProdAction}(\ref{Ex_6X11_TProdAction_Min}),
satisfying $\gm_{(g, h)} (a \otimes b) = \af_g (a) \otimes \bt_h (b)$
for $g \in G$, $h \in H$,
$a \in A$, and $b \in B$.
Prove that
\[
C^*_{\mathrm{r}} (\rh, \, G \times H, \, A \otimes_{\mathrm{min}} B)
\cong C^*_{\mathrm{r}} (G, A, \af)
\otimes_{\mathrm{min}} C^*_{\mathrm{r}} (H, B, \bt).
\]
\end{exr}
\begin{exr}\label{Ex_6X11_InfTProd}
Let $G$, $X$, and the action of $G$ on~$X$
be as in Example~\ref{E_3408_SumProd}.
(That is, $X$ is the group $\prod_{n = 1}^{\I} \Z / k_n \Z$,
and
$G$ is the subgroup $\bigoplus_{n = 1}^{\I} \Z / k_n \Z$,
taken as discrete and acting by translation.)
Prove that $C^* (G, X) \cong \bigotimes_{n = 1}^{\I} M_{k_n}$.
\end{exr}
If there were only finitely many factors in the product,
this computation would follow from \Exr{Ex_6X11_TProdCrPrd}.
With infinitely many factors,
one must take a direct limit.
We mention
some explicit computations of crossed products that are found
elsewhere:
VIII.4.1 of~\cite{Dvd}
(crossed products of the Cantor set by odometer actions);
Section VIII.9 of~\cite{Dvd}
(the crossed product of $S^1 = \R / \Z$ by the group
$\Z [\frac{1}{2}] \subset \R$ regarded as a discrete group
and acting by translation,
and also the crossed product of a particular BunceDeddens algebra
by a particular action of $\Z / 2 \Z$).
\part{Some Structure Theory for
Crossed Products by Finite Groups}\label{Part_Finite}
\section{Introductory Remarks on the Structure of
C*Algebras}\label{Sec_FGStr}
\indent
Our main interest is in structural results for crossed products.
We want simplicity, but we really want much more than that.
We particularly want theorems which show that certain crossed
products are in classes of \ca{s} known to be covered by the
Elliott classification program,
so that the \cp{} can be identified up to isomorphism
by computing its Ktheory and other invariants.
In many cases, one settles for related weaker structural results,
such as stable rank one,
real rank zero,
order on traces determined by \pj{s},
strict comparison of positive elements,
or $Z$stability.
Some results with conclusions of this sort are stated in these
notes, but mostly without proof.
We provide definitions of some of these conditions here:
stable rank one, real rank zero, order on traces determined by \pj{s},
and property~(SP).
% Some others will appear later.
(Strict comparison of positive elements will be discussed later.
See \Def{7121_StrCmp}.)
%
We also define tracial rank zero.
We state various results relating these conditions,
and prove some of them.
For use in these proofs,
and some later proofs,
we prove an assortment of standard lemmas
on Murrayvon Neumann equivalence of \pj{s}.
The proofs mostly consist of repeated application
of \ct{} functional calculus.
Many of these results are in Section~2.5 of~\cite{LnBook}.
\begin{dfn}\label{D:SR1}
Let $A$ be a unital \ca.
We say that $A$ has {\emph{stable rank one}}
if the invertible elements in $A$ are dense in~$A$.
If $A$ is not unital,
we say that $A$ has stable rank one if its unitization $A^{+}$ does.
\end{dfn}
The (topological) stable rank $\tsr (A)$
of a general \ca~$A$
(not necessarily unital) was introduced in~\cite{Rf}.
(See Definition~1.4 there.)
It can take arbitrary values in $\N \cup \{ \I \}$.
Definition~\ref{D:SR1}
gives the value most relevant for classification,
since,
apart from the purely infinite case,
almost all known classification results apply only to \ca{s}
with stable rank one.
For further information,
see Section V.3.1 of~\cite{BlEC} (without proofs),
and for the case of stable rank one,
including some consequences,
see Sections 3.1 and~3.2 of~\cite{LnBook}.
% Check refs.
The topological stable rank of $C (X)$
is related to the covering dimension of~$X$,
which is discussed after Corollary~\ref{C:C1}.
It is clear that $M_n$ has stable rank one.
\begin{thm}\label{T_6X01_ConsTsr1}
Let $A$ be a \ca.
Then \tfae:
\begin{enumerate}
\item\label{T_6X01_ConsTsr1_A}
$A$ has stable rank one.
\item\label{T_6X01_ConsTsr1_SomeMnA}
There is $n \in \N$ such that $M_n (A)$ has stable rank one.
\item\label{T_6X01_ConsTsr1_AllMnA}
For all $n \in \N$, the algebra $M_n (A)$ has stable rank one.
\item\label{T_6X01_ConsTsr1_KTensA}
$K \otimes A$ has stable rank one.
\end{enumerate}
\end{thm}
\begin{proof}
See Theorem 3.3 and Theorem 3.6 of~\cite{Rf}.
(Theorem 3.3 actually only does the unital case.
To get the nonunital case,
one needs Theorem 4.4 and Theorem 4.11 of~\cite{Rf}.)
% 999 Proof of equivalence of conditions for tsr = 1.
\end{proof}
Usually the stable rank of~$M_n (A)$ is smaller
than that of~$A$.
(There is an exact formula.
See Theorem~6.1 of~\cite{Rf}.)
It is easily checked that $C (X)$ has stable rank one
if $X$ is the Cantor set, $[0, 1]$, or~$S^1$.
In fact,
$C (X)$ has stable rank one \ifo{} the covering dimension
of~$X$ is at most one.
More generally, by Proposition~1.7 of~\cite{Rf},
the algebra $C (X)$ has stable rank~$n$
\ifo{} the covering dimension
of~$X$ is $2 n  1$ or~$2 n$.
(We will say more about covering dimension
near the beginning of Section~\ref{Sec:Class}.
The formal definition is \Def{D_6827_dimX}.)
\begin{dfn}\label{D:RR0}
Let $A$ be a \ca.
We say that $A$ has {\emph{real rank zero}}
if the selfadjoint elements with finite spectrum are
dense in the selfadjoint part of~$A$.
\end{dfn}
Again, this is the bottom case of a rank which takes
arbitrary values in $\Nz \cup \{ \I \}$.
The general version is a kind of generalization
of having the invertible selfadjoint elements be dense
in the selfadjoint part of~$A$.
See the beginning of Section~1 of~\cite{BP}.
The case real rank zero is discussed in Section~V.7 of~\cite{Dvd},
with various examples,
although one of the basic results
($A$ has real rank zero \ifo{} $M_n (A)$ has real rank zero)
is not explicitly stated.
% improve ref 999
% Nonunital 999
% Mention inv selfadj elts. 999
For further information,
see Section V.3.2 of~\cite{BlEC} (without proofs),
and for the case of real rank zero,
including some consequences,
see Sections 3.1 and~3.2 of~\cite{LnBook}.
The following \ca{s} all have real rank zero:
$M_n$, $C (X)$ when $X$ is the Cantor set,
$K (H)$, all AF~algebras,
all von Neumann algebras,
and all purely infinite simple \ca{s}.
(See Theorem V.7.4 of~\cite{Dvd} for the purely infinite simple case.)
The real rank of $C (X)$
is the covering dimension of~$X$.
(See Proposition~1.1 of~\cite{BP}.)
The real rank of $M_n (C (X))$
is usually smaller
than that of $C (X)$
(again, there is an exact formula; see Corollary~3.2 of~\cite{BE}),
but the behavior is unknown when $C (X)$
is replaced by a general \ca~$A$.
Property~(SP) is a condition which is considerably weaker
than real rank zero,
but which will play an important role later.
\begin{dfn}\label{D:SPD}
Let $A$ be a \ca.
Then $A$ is said to have {\emph{property~(SP)}} if every nonzero
\hsa{} in $A$ contains a nonzero \pj.
\end{dfn}
It is fairly easy to show that real rank zero implies
property~(SP).
See Proposition~\ref{P_5Y21_RRzSP} below.
The converse is known to be false, even in the simple case.
The examples $A_2$ and $A_3$ in~\cite{BlkKmj}
are counterexamples.
We state here some results about simple \ca{s} with property~(SP)
which will be needed later.
They involve \mvnc{} of \pj{s},
so we start by giving our notation for \mvnc{}
and proving some standard results.
Murrayvon Neumann equivalence will also often be needed
later.
\begin{ntn}\label{N:MvN}
Let $A$ be a \ca, and let $p, q \in A$ be \pj{s}.
We write $p \sim q$ to mean that $p$ and $q$ are \mvnt{} in $A$,
that is, there exists $v \in A$ such that $v^* v = p$ and $v v^* = q$.
We write $p \precsim q$ if $p$ is \mvnt{} to a subprojection of~$q$.
\end{ntn}
There are two other commonly used equivalence relations on \pj{s},
namely homotopy and unitary equivalence,
so one needs to be careful with the meaning of $p \sim q$
when reading papers.
There is also a relation on positive elements,
used in connection with the Cuntz semigroup,
which is commonly written with the same symbol.
(See \Def{D_5511_CzSGp}(\ref{D_5511_CzSGp:2}) below.)
This relation does not always agree
with \mvnc{}
on \pj{s}.
However, the most common meaning of $\sim$ is \mvnc.
We give several standard functional calculus lemmas for working
with projections.
Proofs are included for the convenience of the reader.
For convenience, we recall polar decomposition in \uca{s}.
\begin{lem}\label{L_6320_PolD}
Let $A$ be a \uca,
and let $a \in A$ be invertible.
Then $a (a^* a)^{ 1/2}$ and $(a a^*)^{ 1/2} a$ are unitary.
\end{lem}
\begin{proof}
We only prove the first;
the second is similar.
Set $u = a (a^* a)^{ 1/2}$.
Then
\[
u^* u = (a^* a)^{ 1/2} a^* a (a^* a)^{ 1/2} = 1
\]
and
\[
u u^* = a (a^* a)^{ 1/2} (a^* a)^{ 1/2} a^*
= a (a^* a)^{ 1} a^*
= 1.
\]
Thus $u$ is unitary.
\end{proof}
The following lemma is contained in Proposition 4.6.6 of~\cite{Blk_2}.
See Chapter~4 of~\cite{Blk_2} for much other related material.
\begin{lem}\label{L_5Y21_ClosePj}
Let $A$ be a \ca,
and let $p, q \in A$ be \pj{s}
such that $\ p  q \ < 1$.
Then $p \sim q$.
\end{lem}
In fact,
$p$ is unitarily equivalent to~$q$:
the unitary $u$ in the proof satisfies $u^* p u = q$.
\begin{proof}[Proof of \Lem{L_5Y21_ClosePj}]
Define
\[
a = (2 p  1) (2 q  1) + 1 \in A^{+}.
\]
Using $\ 1  2 p \ \leq 1$ at the third step,
we get
\[
\ a  2 \
= \ 4 p q  2 p  2 q \
\leq 2 \ 1  2 p \ \ p  q \
< 2.
\]
Therefore $a$ is invertible.
Then $u = a (a^* a)^{ 1/2}$
is unitary by Lemma~\ref{L_6320_PolD}.
We have
\[
p a
= p (2 q  1) + p
= 2 p q
= (2 p  1) q + q
= a q.
\]
Taking adjoints gives
\[
a^* p = q a^*.
\]
Combining these equations gives
\[
(a a^*) q = q (a a^*).
\]
Therefore
\[
(a a^*)^{ 1/2} q = q (a a^*)^{ 1/2}.
\]
So
\[
u q = a (a a^*)^{ 1/2} q
= a q (a a^*)^{ 1/2}
= p a (a a^*)^{ 1/2}
= p u.
\]
Now $v = u q$
satisfies $v^* v = q$ and $v v^* = p$.
\end{proof}
\begin{lem}\label{L_6X07_ClSub}
Let $A$ be a \ca,
and let $p, q \in A$ be \pj{s}.
Suppose that $\ p q  q \ < 1$.
Then $q \precsim p$.
\end{lem}
\begin{proof}
We have
\[
\ q p q  q \ \leq \ q \ \ p q  q \ < 1.
\]
Therefore $q p q$ is an invertible element of $q A q$.
Let $x$ be the inverse of $q p q$ in $q A q$.
Set $s = x^{1/2} q p$.
Then $s s^* = q$.
Therefore $s^* s$ is a \pj.
Clearly $s^* s \in p A p$, so $s^* s \leq p$.
\end{proof}
At one point,
we will need a quantitative version of the argument in
\Lem{L_5Y21_ClosePj}.
The estimate is not the best possible,
but is chosen for convenience.
(All we really need is that for all $\ep > 0$ there is $\dt > 0$
such that if $\ p  q \ < \dt$ then there is a unitary $u$
such that $u q u^* = p$ and $\ u  1 \ < \ep$.)
\begin{lem}\label{L_6X02_QuantUE}
Let $A$ be a \uca,
and let $p, q \in A$ be \pj{s}
such that $\ p  q \ \leq \frac{1}{6}$.
Then there is a unitary $u \in A$ such that
\[
\ u  1 \ \leq 10 \ p  q \
\andeqn
u q u^* = p.
\]
\end{lem}
\begin{proof}
We follow the proof of \Lem{L_5Y21_ClosePj}
with a slight change.
Define
\[
b = \frac{1}{2} \big[ (2 p  1) (2 q  1) + 1 \big].
\]
Then the calculation in the proof of \Lem{L_5Y21_ClosePj}
shows that $\ b  1 \ \leq \ p  q \ \leq \frac{1}{6}$.
So $b$ is invertible,
and we define $u = b (b^* b)^{1/2}$.
This element is the same unitary
as in the proof of \Lem{L_5Y21_ClosePj},
so $u q = p u$ as there,
whence $u q u^* = p$.
Since $\ b  1 \ \leq \frac{1}{6}$,
we certainly have $\ b \ \leq 2$.
Therefore
\[
\ b^* b  1 \
\leq \ b^*  1 \ \ b \ + \ b  1 \
\leq 3 \ b  1 \
\leq \frac{1}{2}.
\]
One can check that if
$\ld \in \R$ satisfies $ \ld  1  \leq \frac{1}{2}$,
then $\big \ld^{ 1/2}  1 \big \leq \sqrt{2}  \ld  1 $,
so that
\begin{align*}
\ u  1 \
& \leq \ b \ \big\ (b^* b)^{1/2}  1 \big\ + \ b  1 \
\leq 2 \sqrt{2} \ b^* b  1 \ + \ b  1 \
\\
& \leq 2 \sqrt{2} \cdot 3 \ b  1 \ + \ b  1 \
= \big( 6 \sqrt{2} + 1 \big) \ b  1 \
\leq \big( 6 \sqrt{2} + 1 \big) \ p  q \.
\end{align*}
Since $6 \sqrt{2} + 1 < 10$,
this completes the proof.
\end{proof}
\begin{lem}\label{L_5Y21_CloseToPj}
Let $A$ be a \ca,
and let $a \in A_{\sa}$ satisfy $\ a^2  a \ < \frac{1}{4}$.
Then there is a \pj{} $p \in A$
such that
\[
\ p  a \
\leq \frac{2 \ a^2  a \}{1 + \sqrt{1  4 \ a^2  a \}}.
\]
\end{lem}
\begin{proof}
Set $r = \ a^2  a \$.
Since $r < \frac{1}{4}$,
the sets
\[
S_0 = \big\{ \ld \in \big( \I, \tfrac{1}{2} \big) \colon
 \ld^2  \ld  \leq r \big\}
\]
and
\[
S_1 = \big\{ \ld \in \big( \tfrac{1}{2}, \I \big) \colon
 \ld^2  \ld  \leq r \big\}
\]
are disjoint.
Moreover $\spec (a) \subset S_0 \cup S_1$.
Therefore we can define a \pj{} $p \in A$ by $p = \ch_{S_1} (a)$.
We need to estimate $\ p  a \$.
Clearly
%
\begin{equation}\label{Eq_5Y21_Sup}
\ p  a \
\leq \max \left( \sup_{\ld \in S_0}  \ld , \, \,
\sup_{\ld \in S_1}  \ld  1  \right).
\end{equation}
%
By inspection of the shape of the graph of the function
$\ld \mapsto \ld^2  \ld$ on~$\R$,
it is easy to see that both the supremums in~(\ref{Eq_5Y21_Sup})
are equal to $\sup (S_0)$,
and that moreover the number $s = \sup (S_0)$
is completely determined by the relations
$s \in \big[0, \frac{1}{2} \big)$
and $s  s^2 = r$.
It is easily checked directly that the number
\[
s = \frac{2 r}{1 + \sqrt{1  4 r}}
\]
satisfies both these conditions.
\end{proof}
\begin{cor}\label{C_4Y21_EpClose}
For every $\ep > 0$ there is $\dt > 0$
such that whenever
$A$ is a \ca{} and $a \in A_{\sa}$ satisfies $\ a^2  a \ < \dt$,
then there is a \pj{} $p \in A$
such that $\ p  a \ < \ep$.
\end{cor}
\begin{proof}
Using
\[
\lim_{r \to 0^{+}} \frac{2 r}{1 + \sqrt{1  4 r}} = 0,
\]
this is immediate from Lemma~\ref{L_5Y21_CloseToPj}.
\end{proof}
\begin{lem}\label{L_5Y21_PjInSbalg}
For every $\ep > 0$ there is $\dt > 0$
such that whenever
$A$ is a \ca,
$B \subset A$ is a subalgebra,
and $p \in A$ is a \pj{}
such that $\dist (p, B) < \dt$,
then there is a \pj{} $q \in B$
such that $\ p  q \ < \ep$.
\end{lem}
\begin{proof}
Choose $\dt_0 > 0$ following Corollary~\ref{C_4Y21_EpClose}
with $\frac{\ep}{2}$ in place of~$\ep$.
Set $\dt = \min \big( 1, \frac{\ep}{2}, \frac{\dt_0}{4} \big)$.
Let $A$ be a \ca,
let $B \subset A$ be a subalgebra,
and $p \in A$ be a \pj{}
such that $\dist (p, B) < \dt$.
Choose $c \in B$ such that $\ p  c \ < \dt$.
Set $b = \frac{1}{2} (c + c^*)$.
Then $b \in B_{\sa}$ and $\ p  b \ < \dt$.
Using $p^2 = p$ at the first step and $\dt \leq 1$ at the third step,
we have
\[
\ b^2  b \
\leq \ b \ \ b  p \ + \ b  p \ \ p \ + \ b  p \
= (\ b \ + 2) \ b  p \
\leq 4 \ b  p \
< \dt_0.
\]
The choice of $\dt_0$ provides a \pj{} $q \in B$
such that $\ p  b \ < \frac{\ep}{2}$.
Now
\[
\ q  p \
\leq \ q  b \ + \ b  p \
< \dt + \frac{\ep}{2}
\leq \frac{\ep}{2} + \frac{\ep}{2}
= \ep.
\]
This completes the proof.
\end{proof}
\begin{prp}\label{P_5Y21_RRzSP}
Let $A$ be a \ca{} with real rank zero.
Then $A$ has property~(SP).
\end{prp}
Much more is true:
every \hsa{} in~$A$
has an approximate identity consisting of \pj{s}
(in the nonseparable case,
not necessarily increasing).
See Theorem~2.6 of~\cite{BP}.
\begin{proof}[Proof of Proposition~\ref{P_5Y21_RRzSP}]
Let $B \subset A$ be a nonzero hereditary subalgebra.
Choose $b \in B_{+}$ such that $\ b \ = 1$.
Choose $\dt_0 > 0$ as in Lemma~\ref{L_5Y21_PjInSbalg}
for $\ep = 1$.
Set $\dt = \min \big( 1, \frac{\dt_0}{4} \big)$.
Choose $c \in A_{\sa}$ with finite spectrum such that
$\ c  b \ < \dt$.
Write $c = \sum_{j = 1}^n \ld_j p_j$
for nonzero orthogonal \pj{s} $p_1, p_2, \ldots, p_n \in A$
and numbers $\ld_1, \ld_2, \ldots, \ld_n \in \R$
such that $\ld_1 < \ld_2 < \cdots < \ld_n$.
Then
\[
\ c \ < 1 + \dt \leq 2,
\qquad
 \ld_n  1  < \dt,
\andeqn
c p_n c = \ld_n p_n.
\]
Therefore
\[
\ b p_n b  p_n \
\leq \ b  c \ \ p_n \ \ b \
+ \ c \ \ p_n \ \ b  c \
+  \ld_n  1  \ p_n \
< 4 \dt
\leq \dt_0.
\]
Therefore the choice of~$\dt_0$
provides a \pj{} $p \in B$
such that $\ p  p_n \ < 1$.
Since $p_n \neq 0$,
we deduce from Lemma~\ref{L_5Y21_ClosePj}
that $p \neq 0$.
\end{proof}
\begin{lem}\label{L_6125_PjSbq}
Let $A$ be a \ca,
let $a \in A_{+}$,
and let $p \in A$ be a \pj.
Suppose that there is $v \in A$
such that $\ v^* a v  p \ < 1$.
Then there is a \pj{} $q \in {\ov{a A a}}$
such that $q$ is \mvnt{} to~$p$.
\end{lem}
Although we have not yet defined Cuntz subequivalence
(see Definition \ref{D_5511_CzSGp}(\ref{D_5511_CzSGp:1}) below),
we state a consequence in these terms.
If $a \in A_{+}$, $p \in A$ is a \pj,
and $p \precsim a$,
then ${\ov{a A a}}$ contains a \pj{}
which is \mvnt~$p$.
\begin{proof}[Proof of Lemma~\ref{L_6125_PjSbq}]
Define $b \in A$ by $b = a^{1/2} v p$.
Then $b^* b \in p A p$
and
\[
\ b^* b  p \
= \ p (v^* a v  p) p \
\leq \ p \ \cdot \ v^* a v  p \ \cdot \ p \
< 1.
\]
Therefore $b^* b$ is an invertible element of $p A p$,
and,
taking functional calculus in $p A p$,
we can form $(b^* b)^{ 1/2}$.
Define $s \in A$ by $s = b (b^* b)^{ 1/2}$.
Then
\[
s^* s
= (b^* b)^{ 1/2} b^* b (b^* b)^{ 1/2}
= p.
\]
Therefore $s s^*$ is a \pj.
Since (with $(b^* b)^{1}$ evaluated in $p A p$)
we have
\[
s s^*
= a^{1/2} v p (b^* b)^{1} p v^* a
\in {\ov{a A a}},
\]
the result follows.
\end{proof}
\begin{lem}\label{L_6125_FSwitch}
Let $r \in (0, \I)$,
and let $f \colon [0, r] \to \C$
be a \cfn.
Then for any \ca~$C$ and any $c \in C$
with $\ c \ \leq r^{1/2}$,
we have $c f (c^* c) = f (c c^*) c$.
\end{lem}
\begin{proof}
We first observe that for
any \ca~$C$, any $c \in C$,
and any $n \in \Nz$,
we have
$c (c^* c)^n = (c c^*)^{n} c$.
Therefore $c h (c^* c) = h (c c^*) c$
whenever $h$ is a polynomial.
Now let $f$ be arbitrary.
If $C$ is not unital,
we work in~$C^{+}$.
Let $\ep > 0$;
we prove that $\ c f (c^* c)  f (c c^*) c \ < \ep$.
Choose a polynomial $h$
such that
$ h (\ld)  f (\ld)  < \ep / (3 r^{1/2})$
for all $\ld \in [0, r]$.
Then
\[
\ h (c^* c)  f (c^* c) \
\leq \sup_{\ld \in [0, r]}  h (\ld)  f (\ld) 
\leq \frac{\ep}{3 r^{1/2}},
\]
so
\[
\ c h (c^* c)  c f (c^* c) \
\leq \ c \ \cdot \left( \frac{\ep}{3 r^{1/2}} \right)
\leq \frac{\ep}{3}.
\]
Similarly,
\[
\ h (c c^*) c  f (c c^*) c \
\leq \frac{\ep}{3}.
\]
Therefore
\[
\ c f (c^* c)  f (c c^*) c \
\leq \ c f (c^* c)  c h (c^* c) \
+ \ h (c c^*) c  f (c c^*) c \
\leq \frac{\ep}{3} + \frac{\ep}{3}
< \ep.
\]
This completes the proof.
\end{proof}
The following lemma is essentially in Section~1 of~\cite{Cn1}.
(Also see the proof of Lemma~4.1 of~\cite{OP1}.)
\begin{lem}\label{LSwitch}
Let $A$ be a \ca, and let $c \in A$.
Then for any projection $p \in {\overline{c A c^*}}$,
there exists a \pj{} $q \in {\overline{c^* A c}}$
such that $p \sim q$.
\end{lem}
Much more is true.
There is an isomorphism
$\ph \colon {\overline{c^* A c}} \to {\overline{c A c^*}}$
(this is in 1.4 of~\cite{Cn1})
such that $\ph (p) \sim p$ for all \pj{s} $p \in {\overline{c^* A c}}$
(this is easily deduced from~\cite{Cn1}).
In fact,
using Cuntz equivalence (which we have not defined),
for every $a \in \big( {\overline{c^* A c}} )_{+}$,
$\ph (a)$ is Cuntz equivalent in~$A$ to~$a$.
This fact is made explicit
in Lemma~3.8 of~\cite{PsnPh2}.
\begin{proof}[Proof of Lemma~\ref{LSwitch}]
For each $\ep > 0$
define \cfn{s} $f_{\ep}, g_{\ep} \colon [0, \infty) \to [0, 1]$ by
\[
f_{\ep} (\ld) = \begin{cases}
0 & \hspace{1em} \ld \leq \frac{\ep}{2} \\
\frac{2}{\ep \ld} \left( \ld  \frac{\ep}{2} \right)
& \hspace{1em} \frac{\ep}{2} \leq \ld \leq \ep \\
\frac{1}{\ld} & \hspace{1em} \ep \leq \ld
\end{cases}
\andeqn
g_{\ep} (\ld) = \begin{cases}
0 & \hspace{1em} \ld \leq \frac{\ep}{2} \\
\frac{2}{\ep} \left( \ld  \frac{\ep}{2} \right)
& \hspace{1em} \frac{\ep}{2} \leq \ld \leq \ep \\
1 & \hspace{1em} \ep \leq \ld.
\end{cases}
\]
Then $g_{\ep} (\ld) = \ld f_{\ep} (\ld)$
for all $\ld \in [0, \I)$.
The net $(g_{\ep} (c c^*) )_{\ep > 0}$
is an approximate identity for ${\overline{c A c^*}}$.
In particular,
there is $\ep > 0$ such that
\[
\big\ g_{\ep} (c c^*) p g_{\ep} (c c^*)  p \big\ < 1.
\]
Define $a = f_{\ep} (c^* c) c^* p c f_{\ep} (c^* c)$,
which is a positive element in ${\overline{c A c^*}}$.
Then,
using Lemma~\ref{L_6125_FSwitch} twice at the second step,
\begin{align*}
\ c a c^*  p \
& = \big\ c f_{\ep} (c^* c) c^* p c f_{\ep} (c^* c) c^*  p \big\
\\
& = \big\ f_{\ep} (c c^*) c c^* p f_{\ep} (c c^*) c c^*  p \big\
= \big\ g_{\ep} (c c^*) p g_{\ep} (c c^*)  p \big\
< 1.
\end{align*}
Now Lemma~\ref{L_6125_PjSbq}
provides a \pj{} $q$ in the hereditary subalgebra generated by~$a$,
and hence in the hereditary subalgebra
generated by $c c^*$, such that $q \sim p$.
The hereditary subalgebra
generated by $c c^*$ is ${\overline{c A c^*}}$.
\end{proof}
The following lemma is essentially Lemma~3.1 of~\cite{Ln1},
but no proof is given there.
\begin{lem}[Lemma~1.9 of~\cite{PhT1}]\label{L:CompSP}
Let $A$ be a simple \ca{} with property~(SP).
Let $B \subset A$ be a nonzero \hsa,
and let $p \in A$ be a nonzero \pj.
Then there is a nonzero \pj{} $q \in B$
such that $q \precsim p$.
\end{lem}
\begin{proof}
Choose a nonzero positive element $a \in B$.
Since $A$ is simple,
there exists $x \in A$ such that $c = a x p$ is nonzero.
Since $A$ has property~(SP),
there is a nonzero \pj{} $q \in {\overline{c A c^*}}$.
Then $q \in B$,
and by
Lemma~\ref{LSwitch} there is a \pj{} $e \in {\overline{c^* A c}}$
such that $e \sim q$.
We have ${\overline{c^* A c}} \subset p A p$, so $e \leq p$.
\end{proof}
We need to know that an \idsuca{}
contains an arbitrarily large finite number of nonzero
orthogonal positive elements.
In the literature,
this is usually derived from
a result on page~61 of~\cite{AkSh},
according to which a \ca{} which is not ``scattered''
contains a selfadjoint element whose spectrum is $[0, 1]$.
In the next three lemmas,
we give instead an elementary proof
of the statement we need,
which applies to any in\fd{} \ca.
\begin{lem}\label{L_3228_Dim1}
Let $A$ be a \ca,
let $e, f \in A$ be \pj{s},
and suppose that
\[
e A e = \big\{ \ld e \colon \ld \in \C \big\}
\andeqn
f A f = \big\{ \ld f \colon \ld \in \C \big\}.
\]
Then $\dim (e A f) \leq 1$.
\end{lem}
\begin{proof}
We may assume that $e, f \neq 0$ and $e A f \neq \{ 0 \}$.
Choose a nonzero element $c \in e A f$.
Then $c^* c$ is a nonzero element of $f A f$,
so there is $\gm \in (0, \I)$ such that $c^* c = \gm f$.
Define $s = \gm^{1/2} c$.
Then $s^* s = f$.
We show that $e A f = \spn (s)$.
Let $a \in e A f$.
Then $a s^* \in e A e$,
so there is $\ld \in \C$ such that $a s^* = \ld e$.
Now
\[
a = a f
= a s^* s
= \ld s,
\]
as desired.
\end{proof}
\begin{lem}\label{L_3228_fdcorner}
Let $A$ be a \uca,
and let $p \in A$ be a \pj{}
such that $p A p$ and $(1  p) A (1  p)$ are \fd.
Then $A$ is \fd.
\end{lem}
\begin{proof}
Since $p A p$ and $(1  p) A (1  p)$ are finite direct
sums of matrix algebras,
we can find mutually orthogonal rank one \pj{}
$e_1, e_2, \ldots, e_m \in p A p$
and $f_1, f_2, \ldots, f_n \in (1  p) A (1  p)$
such that
\[
\sum_{j = 1}^m e_j = p
\andeqn
\sum_{k = 1}^n f_k = 1  p.
\]
In particular,
\[
e_j A e_j = \big\{ \ld e_j \colon \ld \in \C \big\}
\andeqn
f_k A f_k = \big\{ \ld f_k \colon \ld \in \C \big\}
\]
for $j = 1, 2, \ldots, m$
and $k = 1, 2, \ldots, n$.
Now
\[
p A (1  p)
= \sum_{j = 1}^m \sum_{k = 1}^n e_j A f_k,
\]
so $\dim (p A (1  p)) \leq m n$
by \Lem{L_3228_Dim1}.
Similarly $\dim ((1  p) A p) \leq m n$.
This completes the proof.
\end{proof}
\begin{lem}\label{L_3228_InfOrth}
Let $A$ be an in\fd{} \ca.
Then there exists a sequence $a_1, a_2, \ldots$ in $A$
consisting of nonzero positive orthogonal elements.
\end{lem}
In the proof,
the case dealt with at the end,
in which $\spec (a)$ is finite for all $a \in A_{\sa}$,
can't actually occur.
\begin{proof}[Proof of Lemma~\ref{L_3228_InfOrth}]
We first observe that it suffices to prove the result
when $A$ is unital.
Indeed,
if $A$ is not unital,
$a_1, a_2, \ldots$ is such a sequence in~$A^{+}$,
and $\pi \colon A^{+} \to \C$
is the map associated with the unitization,
then there can be at most one $n \in \N$
such that $\pi (a_n) \neq 0$.
We therefore assume that $A$ is unital.
Suppose that there is $a \in A_{\sa}$
such that $\spec (a)$ is infinite.
Choose a sequence in $\spec (a)$ whose terms are all
distinct,
choose a convergent subsequence,
and (deleting at most one term) choose a subsequence
$(\ld_n)_{n \in \N}$
such that the limit is not one of the terms.
Then there are disjoint open sets $U_1, U_2, \ldots \subset \R$
such that $\ld_n \in U_n$ for all $n \in \N$.
For $n \in \N$,
choose a nonzero \cfn{} $f_n \colon \R \to [0, 1]$
with compact support contained in~$U_n$,
and set $a_n = f_n (a)$.
Then the sequence $a_1, a_2, \ldots$
satisfies the conclusion of the lemma.
Now suppose that $\spec (a)$ is finite for all $a \in A_{\sa}$.
We claim that if $B$ is a unital \ca{} with $B \not\cong \C$
and such that every element of~$B_{\sa}$
has finite spectrum,
then $B$ has a nontrivial \pj.
Indeed, there must be an element $b \in B_{\sa}$
which is not a scalar,
so $\spec (b)$ is a finite set with more than one element.
Therefore functional calculus produces a nontrivial \pj.
In particular,
there is a nontrivial \pj{} $p_1 \in A$.
By \Lem{L_3228_fdcorner},
and replacing $p_1$ with $1  p_1$ if necessary,
we can assume that $p_1 A p_1$ is in\fd.
Clearly $\spec (a)$ is finite
for all $a \in ( p_1 A p_1 )_{\sa}$.
Therefore there is a nontrivial \pj{} $p_2 \in p_1 A p_1$,
and we may assume that
$(p_1  p_2) A (p_1  p_2 )$ is in\fd.
Proceed by induction.
Then taking $p_0 = 1$ and $a_n = p_{n  1}  p_n$
for $n \in \N$ gives a sequence $a_1, a_2, \ldots$
as in the conclusion of the lemma.
\end{proof}
\begin{lem}[Lemma~1.10 of~\cite{PhT1};
Lemma~3.2 of~\cite{Ln1}]\label{OrthInSP}
Let $A$ be an \idsuca{} with property~(SP).
Let $B \subset A$ be a nonzero \hsa,
and let $n \in \N$.
Then there exist nonzero \mvnt{} \mops{} $p_1, p_2, \ldots, p_n \in B$.
\end{lem}
\begin{proof}
Use Lemma~\ref{L_3228_InfOrth}
to choose nonzero positive orthogonal elements
\[
a_1, a_2, \ldots, a_n \in A.
\]
Choose a nonzero \pj{} $e_1 \in {\ov{a_1 A a_1}}$.
Inductively use \Lem{L:CompSP}
to find \nzp{s}
\[
e_2 \in {\ov{a_2 A a_2}}, \,
e_3 \in {\ov{a_3 A a_3}}, \,
\ldots, \,
e_n \in {\ov{a_n A a_n}}
\]
such that $e_j \precsim e_{j  1}$ for $j = 2, 3, \ldots, n$.
Set $p_n = e_n$.
Since $p_n \precsim e_{n  1}$,
there is $p_{n  1} \leq e_{n  1}$ such that $p_{n  1} \sim p_n$.
Then $p_{n  1} \precsim e_{n  2}$,
so the same reasoning gives $p_{n  2} \leq e_{n  2}$
such that $p_{n  2} \sim p_{n  1}$.
Construct $p_{n  3}, p_{n  4}, \ldots, p_1$ similarly.
\end{proof}
\begin{lem}[Lemma~1.11 of~\cite{PhT1}]\label{L:MnSP}
Let $A$ be an \idsuca, and let $n \in \N$.
Then $A$ has property~(SP) \ifo{} $M_n \otimes A$ has property~(SP).
Moreover, in this case,
for every nonzero \hsa{} $B \subset M_n \otimes A$,
there exists a nonzero \pj{} $p \in A$ such that
$1 \otimes p$ is \mvnt{} to a \pj{} in $B$.
\end{lem}
\begin{proof}
Let $( e_{j, k} )_{1 \leq j, k \leq n}$ be a system of matrix units
for $M_n$.
If $M_n \otimes A$ has property~(SP),
then so do all its hereditary subalgebras,
including $\C e_{1, 1} \otimes A \cong A$.
Now assume that $A$ has property~(SP),
and let $B \subset M_n \otimes A$ be a nonzero \hsa.
Choose $x \in B \setminus \{ 0 \}$.
There is $j \in \{ 1, 2, \ldots, n \}$
such that $(e_{j, j} \otimes 1) x \neq 0$.
Then
$C = (e_{j, j} \otimes 1) x (M_n \otimes A) x^* (e_{j, j} \otimes 1)$
is a nonzero \hsa{} in
$(e_{j, j} \otimes 1) (M_n \otimes A) (e_{j, j} \otimes 1) \cong A$.
Because $A$ has property~(SP),
there is a \pj{} $f \in A \setminus \{ 0 \}$
such that $e_{j, j} \otimes f \in C$.
Use Lemma~\ref{OrthInSP}
to choose mutually orthogonal nonzero \mvnt{} \pj{s}
$f_1, f_2, \ldots, f_n \in A$
such that $f_j \leq f$ for all $k$.
Then
\[
1 \otimes f_1
= \sum_{k = 1}^n e_{k, k} \otimes f_1
\sim \sum_{k = 1}^n e_{j, j} \otimes f_k
\leq e_{j, j} \otimes f.
\]
Furthermore,
Lemma~\ref{LSwitch}
tells us that $e_{j, j} \otimes f$ is \mvnt{} to a \pj{}
in
\[
{\ov{x^* (e_{j, j} \otimes 1) (M_n \otimes A)
(e_{j, j} \otimes 1) x}}
\subset B.
\]
This completes the proof.
\end{proof}
Now we consider tracial states.
See the beginning of Section~6.2 of~\cite{Mph}.
% Blackadar encycl. 999
\begin{dfn}\label{D_Trace}
Let $A$ be a \ca.
A {\emph{tracial state}} on $A$ is a state $\ta \colon A \to \C$
with the additional property that $\ta (b a) = \ta (a b)$
for all $a, b \in A$.
We define the {\emph{tracial state space}} $\T (A)$ of~$A$
to be the set of all \tst{s} on~$A$,
equipped with the relative weak* topology inherited from
the Banach space dual of~$A$.
\end{dfn}
That is, a \tst{} is a normalized trace.
Tracial states have actually already occurred,
in \Thm{T_3318_RedTr}
and in \Thm{T_3301_CStRFnUniqT}.
Recall (Corollary 3.3.4 of~\cite{Mph})
that if $A$ is a \uca{} and $\om \colon A \to \C$
is a linear functional such that $\om (1) = 1$ and $\ \om \ = 1$,
then $\om$ is automatically positive, and hence a state.
In particular,
if $\ta \colon A \to \C$
is a linear functional such that $\ta (1) = 1$, $\ \ta \ = 1$,
and $\ta (b a) = \ta (a b)$ for all $a, b \in A$,
then $\ta$ is a \tst.
\begin{exa}\label{ETMn}
Let $n \in \N$.
Define $\ta \colon M_n \to \C$ by
\[
\ta \left( \left( \begin{matrix}
a_{1, 1} & a_{1, 2} & \cdots & a_{1, n} \\
a_{2, 1} & a_{2, 2} & \cdots & a_{2, n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{n, 1} & a_{n, 2} & \cdots & a_{n, n}
\end{matrix} \right) \right)
= \frac{1}{n} \sum_{k = 1}^{n} a_{k, k}.
\]
Then $\ta$ is a tracial state.
\end{exa}
The tracial state on $M_n$ in Example~\ref{ETMn}
is just a normalization of the usual trace
on the $n \times n$ matrices.
The following example generalizes Example~\ref{ETMn}.
\begin{exa}\label{ETMnOfA}
Let $A$ be a \ca,
let $\ta_0$ be a \tst{} on~$A$,
and let $n \in \N$.
Define $\ta \colon M_n (A) \to \C$ by
\[
\ta \left( \left( \begin{matrix}
a_{1, 1} & a_{1, 2} & \cdots & a_{1, n} \\
a_{2, 1} & a_{2, 2} & \cdots & a_{2, n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{n, 1} & a_{n, 2} & \cdots & a_{n, n}
\end{matrix} \right) \right)
= \frac{1}{n} \sum_{k = 1}^{n} \ta_0 (a_{k, k}).
\]
Then $\ta$ is a tracial state.
\end{exa}
For consistency with Ktheory,
we usually want to use the unnormalized version,
namely $\sum_{k = 1}^{n} \ta_0 (a_{k, k})$.
For example,
see \Def{DRhoA}.
\begin{exa}\label{ETrMeas}
Let $X$ be a compact metric space,
and let $\mu$ be a Borel probability measure on~$X$.
Then the formula
\[
\ta (f) = \int_X f \, d \mu
\]
defines a \tst{} on $C (X)$.
\end{exa}
Of course, all that is really happening in Example~\ref{ETrMeas}
is that every state on a commutative \ca{} is automatically tracial.
Given Example~\ref{ETrMeas},
the following is a special case of Example~\ref{ETMn}.
\begin{exa}\label{ETrMnCX}
Let $n \in \N$,
let $X$ be a compact metric space,
and let $\mu$ be a Borel probability measure on~$X$.
Let $\ta_0 \colon M_n \to \C$ be the \tst{} of Example~\ref{ETMn}.
Then the formula
\[
\ta (a) = \int_X \ta_0 (a (x)) \, d \mu (x)
\]
defines a \tst{} on $C (X, M_n)$.
Explicitly, if
\[
a (x) = \left( \begin{matrix}
a_{1, 1} (x) & a_{1, 2} (x) & \cdots & a_{1, n} (x) \\
a_{2, 1} (x) & a_{2, 2} (x) & \cdots & a_{2, n} (x) \\
\vdots & \vdots & \ddots & \vdots \\
a_{n, 1} (x) & a_{n, 2} (x) & \cdots & a_{n, n} (x)
\end{matrix}
\right)
\]
for $x \in X$,
then
\[
\ta (a)
= \int_X \frac{1}{n}
\Bigg( \sum_{k = 1}^{n} a_{k, k} (x) \Bigg) \, d \mu (x).
\]
\end{exa}
\begin{exr}\label{ExProveTrMnCX}
Let $n \in \N$,
let $X$ be a compact metric space,
and let $\ta$ be a \tst{} on $C (X, M_n)$.
Prove that there exists a Borel probability measure $\mu$ on~$X$
such that $\ta$ is obtained from $\mu$ as in Example~\ref{ETrMnCX}.
\end{exr}
\begin{exa}\label{ETrCStG}
Let $G$ be a discrete group.
Then the continuous linear functional
$\ta \colon C^*_{\mathrm{r}} (G) \to \C$
such that $\ta (u_1) = 1$
and $\ta (u_g) = 0$ for $g \in G \setminus \{ 1 \}$
(see \Thm{T_3318_RedTr}) is proved there to be a \tst.
\end{exa}
\begin{exa}\label{EInvTrCndExpt}
This example is a generalization of Example~\ref{ETrCStG}.
Let $\af \colon G \to \Aut (A)$
be an action of a discrete group $G$ on a \ca~$A$.
Let $\ta_0$ be a \tst{} on~$A$,
which is $G$invariant in the sense that
$\ta_0 (\af_g (a)) = \ta_0 (a)$ for all $a \in A$ and $g \in G$.
Let $E \colon C^*_{\mathrm{r}} (G, A, \af) \to A$
be the standard conditional expectation (Definition~\ref{D_StdCond}).
Define $\ta \colon C^*_{\mathrm{r}} (G, A, \af) \to \C$
by $\ta = \ta_0 \circ E$.
Then $\ta$ is a \tst{} on $C^*_{\mathrm{r}} (G, A, \af)$.
Using Exercise~\ref{Pb_CondExpt},
it is easy to check that $\ta$ is a state.
It remains to prove that
$E (a b) = E (b a)$ for $a, b \in C^*_{\mathrm{r}} (G, A, \af)$.
By continuity, it suffices to prove this when
\[
a = \sum_{g \in G} a_g u_g \in C_{\mathrm{c}} (G, A, \af)
\andeqn
b = \sum_{g \in G} b_g u_g \in C_{\mathrm{c}} (G, A, \af)
\]
with all but finitely many of the $a_g$ and $b_g$ equal to zero.
% The computation looks similar to that of Example~\ref{ETrCStG}.
We have, changing variables at the third step,
\[
a b = \sum_{g, h \in G} a_g u_g b_h u_h
= \sum_{g, h \in G} a_g \af_g (b_h) u_{g h}
= \sum_{g \in G} \left( \sum_{k \in G}
a_{k} \af_k ( b_{ k^{1} g } ) u_g \right)
\]
and similarly
\[
b a = \sum_{g \in G} \left( \sum_{k \in G}
b_{k} \af_k ( a_{ k^{1} g } ) u_g \right).
\]
Therefore
\[
\ta (a b)
= \sum_{k \in G} \ta_0 \big( a_{k} \af_k ( b_{ k^{1} }) \big)
\andeqn
\ta (b a)
= \sum_{k \in G} \ta_0 \big( b_{k} \af_k (a_{ k^{1} }) \big).
\]
Starting with the second expression,
we change $k$ to $k^{1}$ at the first step,
use the trace property of~$\ta_0$ at the second step,
and then
$G$invariance of~$\ta_0$ at the third step, to get
\begin{align*}
\ta (b a)
% & = \sum_{k \in G} \ta_0 \big( b_{k} \af_k ( a_{ k^{1} } \big)
& = \sum_{k \in G} \ta_0 \big( b_{k^{1}} \af_{k^{1}} ( a_{k}) \big)
\\
& = \sum_{k \in G} \ta_0 \big( \af_{k^{1}} ( a_{k}) b_{k^{1}} \big)
= \sum_{k \in G} \ta_0 \big( a_{k} \af_k ( b_{k^{1}} ) \big)
= \ta (a b).
\end{align*}
This completes the proof.
\end{exa}
\begin{exa}\label{EInvMeas}
As a special case of Example~\ref{EInvTrCndExpt},
let $G$ be a discrete group,
and let $X$ be a compact metric space with an action of~$G$.
Then every $G$invariant Borel probability measure on~$X$
induces a \tst{} $\ta$ on $C^*_{\mathrm{r}} (G, X)$.
On elements of $C_{\mathrm{c}} (G, C (X))$,
written as finite sums $\sum_{g \in G} f_g u_g$ with
$f_g$ in $C (X)$ for $g \in G$
and $f_g = 0$ for all but finitely $g \in G$,
it is given by the formula
\[
\ta \Bigg( \sum_{g \in G} f_g u_g \Bigg)
= \int_X f_1 \, d \mu.
\]
\end{exa}
\begin{lem}[Remark 6.2.3 of~\cite{Mph}]\label{LTrIdeal}
Let $A$ be a unital \ca, and let $\ta$ be a \tst{} on~$A$.
Then the set
\[
\big\{ a \in A \colon \ta (a^* a) = 0 \big\}
\]
is a closed ideal in~$A$.
\end{lem}
\begin{proof}
Set
\[
I = \big\{ a \in A \colon \ta (a^* a) = 0 \big\}.
\]
Since $\ta$ is a state,
it follows from the GelfandNaimarkSegal construction that
$I$ is a closed left ideal in~$A$.
To show that $I$ is in fact a two sided ideal,
it suffices to show that $I^* = I$.
But $\ta (a a^*) = 0$ \ifo{} $\ta (a^* a) = 0$
by the trace property.
\end{proof}
Traces are related to \mvnc{} in the following way.
\begin{lem}\label{LTraceAndMvN}
Let $A$ be a unital \ca, let $\ta$ be a \tst{} on~$A$,
and let $p, q \in A$ be \pj{s}.
\begin{enumerate}
\item\label{LTraceAndMvN1}
If $p \sim q$, then $\ta (p) = \ta (q)$.
\item\label{LTraceAndMvN2}
If $p \precsim q$, then $\ta (p) \leq \ta (q)$.
\item\label{LTraceAndMvN3}
If $A$ is simple and there is a \pj{} $e \in A$
such that
\[
p \sim e,
\qquad
e \leq q,
\andeqn
e \neq q,
\]
then $\ta (p) < \ta (q)$.
\end{enumerate}
\end{lem}
\begin{proof}
For~(\ref{LTraceAndMvN1}),
the hypotheses imply that there is
$v \in A$ such that $v^* v = p$ and $v v^* = q$.
Therefore
\[
\ta (p) = \ta (v^* v) = \ta (v v^*) = \ta (q).
\]
Part~(\ref{LTraceAndMvN2}) follows from~(\ref{LTraceAndMvN1})
because positivity of $\ta$ implies that
if $e$ is a subprojection of~$q$,
then $\ta (e) \leq \ta (q)$.
For~(\ref{LTraceAndMvN3}),
we have $\ta (p) = \ta (e) \leq \ta (q)$ by~(\ref{LTraceAndMvN2}).
It remains to show that $\ta (e) \neq \ta (q)$.
Now $q  e$ is a nonzero positive element.
By Lemma~\ref{LTrIdeal},
if $\ta (q  e)$ were zero,
then $A$ would contain the nontrivial ideal
\[
I = \big\{ a \in A \colon \ta (a^* a) = 0 \big\}.
\]
This completes the proof.
\end{proof}
Under good conditions (some of which we will see later),
there is a kind of converse to
Lemma~\ref{LTraceAndMvN}.
For simple unital exact \ca{s},
the right notion is given in the following definition.
It is a version of Blackadar's
Second Fundamental Comparability Question (FCQ2).
See 1.3.1 of~\cite{Blc4}.
\begin{dfn}\label{D:OrdDetD}
Let $A$ be a unital \ca.
We say that the
{\emph{order on \pj{s} over $A$ is determined by traces}}
if whenever $p, q \in \Mi (A)$ are \pj{s} such that
$\ta (p) < \ta (q)$ for every tracial state $\ta$ on~$A$,
then $p \precsim q$.
\end{dfn}
In general,
one should use quasitraces in place of tracial states.
See Definition II.1.1 of~\cite{BH} or Definition 2.31 of~\cite{APT}
for the definition of a quasitrace.
When $A$ is exact,
every quasitrace is a trace; see Theorem 5.11 of~\cite{Hgp}.
For general \ca{s},
it is an open question whether every quasitrace is a trace.
Algebras with this property include $M_n$,
finite factors,
and simple unital AF~algebras.
(The case of simple unital AF~algebras is a special case of
Theorem 5.2.1 of~\cite{Blc4}.)
Another useful condition on the relation
between traces and Ktheory
is presented in
Remark~\ref{RTACptCnv},
\Def{DAffDelta},
\Def{DRhoA},
and Remark~\ref{RDenseRange}.
We use the following definition of tracial rank zero.
Tracial rank was first defined in Definition~3.1 of~\cite{Ln2},
and tracial rank zero is equivalent (by Theorem~7.1(a) of~\cite{Ln2})
to being tracially AF in the sense of Definition~2.1 of~\cite{Ln1}
(at least for simple \ca{s}).
We use the version in Definition 3.6.2 of~\cite{LnBook},
with $k$ there taken to be zero.
(See Definition 2.4.1 of~\cite{LnBook},
where it is stated that
equivalence means Murrayvon Neumann equivalence.)
The original version
(Definition~2.1 of~\cite{Ln1}) omitted the requirement that $p \neq 0$,
but required unitary equivalence in~(\ref{D:TR0:Small}).
One warning: the condition $p \neq 0$ was omitted
in Proposition~2.3 of~\cite{PhT1}.
Without this condition,
purely infinite simple \uca{s}
would have tracial rank zero,
by taking $p = 0$.
We use the notation $[a, b]$ for the commutator $a b  b a$.
\begin{dfn}[Definition 3.6.2 of~\cite{LnBook}]\label{D_TR0}
Let $A$ be a simple unital \ca.
Then $A$ has tracial rank zero if
for every finite subset $F \subset A$, every $\ep > 0$,
and every nonzero positive element $c \in A$,
there exist a nonzero \pj{} $p \in A$
and a unital finite dimensional
subalgebra $D \subset p A p$ such that:
\begin{enumerate}
\item\label{D:TR0:Comm}
$\ [a, p] \ < \ep$ for all $a \in F$.
\item\label{D:TR0:Close}
$\dist (p a p, \, D) < \ep$ for all $a \in F$.
\item\label{D:TR0:Small}
$1  p$ is Murrayvon Neumann equivalent
to a \pj{} in ${\overline{c A c}}$.
\end{enumerate}
\end{dfn}
The word ``nonzero'' is missing in
Proposition~2.3 of~\cite{PhT1}.
Without this requirement,
all purely infinite simple unital \ca{s}
would have tracial rank zero.
When checking whether a \ca{} has tracial rank zero,
it is only necessary to use finite subsets
of a fixed generating set.
\begin{lem}\label{L_6X08_TAFGenerators}
Let $A$ be a simple unital \ca,
and let $T \subset A$ be a subset
which generates $A$ as a \ca.
Assume that
for every finite subset $F \subset T$, every $\ep > 0$,
and every nonzero positive element $c \in A$,
there exists a nonzero \pj{} $p \in A$
and a unital finite dimensional
subalgebra $D \subset p A p$ such that:
\begin{enumerate}
\item\label{L_6X08_TAFGenerators_Comm}
$\ [a, p] \ < \ep$ for all $a \in F$.
\item\label{L_6X08_TAFGenerators_Close}
$\dist (p a p, \, D) < \ep$ for all $a \in F$.
\item\label{L_6X08_TAFGenerators_Small}
$1  p$ is Murrayvon Neumann equivalent
to a \pj{} in ${\overline{c A c}}$.
\end{enumerate}
Then $A$ has tracial has rank zero.
\end{lem}
The only change from \Def{D_TR0}
is that we only use finite subsets of~$T$.
\begin{exr}\label{Ex_6X08_TAFGenerators}
Prove \Lem{L_6X08_TAFGenerators}.
\end{exr}
The proof is related to the proof of \Lem{L_5601_CommEst},
which is given in full,
and also to the proofs of similar statements earlier.
For example, see the proof of \Lem{LCtGen},
although that proof is easier.
\Lem{L:TRAF} gives another slightly weaker condition
which implies tracial rank zero.
Higher values of the tracial rank also exist
(Definition 3.6.2 of~\cite{LnBook}),
and there is a definition for algebras which are not simple.
See Definition~3.1 of~\cite{Ln2} for both generalizations.
AF~algebras have tracial rank zero; indeed,
one can always take $p = 1$.
Other examples are less obvious.
The condition looks hard to check.
One of the important points in the theory is that in fact
there are a number of cases in which the condition can be checked.
(See \Thm{RokhTAF} for the case most relevant here.)
For our purposes,
the most important consequence of tracial rank zero
is that,
together with simplicity, separability, nuclearity,
and the Universal Coefficient Theorem,
it implies classification.
See Theorem~5.2 of~\cite{Ln3}.
\begin{thm}\label{T_6X02_TRZToRR}
Let $A$ be an \idsuca{} with tracial rank zero.
Then $A$ has real rank zero and stable rank one,
and the order on \pj{s} over $A$ is determined by traces.
\end{thm}
\begin{proof}
Real rank zero and stable rank one
are part of Theorem~3.4 of~\cite{Ln1}.
Order on \pj{s} over $A$ determined by traces is
Theorem~6.8 of~\cite{Ln2},
which applies by Theorem 6.13 of~\cite{Ln2}.
\end{proof}
These results are also found in~\cite{LnBook}:
Theorem 3.6.11 (for stable and real rank),
and Theorem 3.7.2 (for order on \pj{s} determined by traces;
to get from $A$ to $\Mi (A)$, see Lemma~\ref{L_6X08_TAFCM_n}
below).
At least the first two parts can fail in the nonsimple case.
% More details needed. 999
\begin{cor}[Lemma 3.6.6 of~\cite{LnBook}]\label{C_6X08_TAFSP}
Let $A$ be an \idsuca{} with tracial rank zero.
Then $A$ has property~(SP).
\end{cor}
\begin{proof}
Combine Theorem~\ref{T_6X02_TRZToRR}
and Proposition~\ref{P_5Y21_RRzSP}.
\end{proof}
\begin{lem}[Lemma 3.6.5 of~\cite{LnBook}]\label{L_6X08_TAFCorner}
Let $A$ be an \idsuca{} with tracial rank zero,
and let $e \in A$ be a nonzero \pj.
Then $e A e$ has tracial has rank zero.
\end{lem}
We omit the proof,
although it is not hard with what we now have.
% \begin{proof}
% \end{proof}
\begin{lem}[Special case of Theorem 3.7.3
of~\cite{LnBook}]\label{L_6X08_TAFCM_n}
Let $A$ be an \idsuca{} with tracial rank zero,
and let $n \in \N$.
Then $M_n (A)$ tracial has rank zero.
\end{lem}
\begin{proof}
We use standard matrix unit notation.
Let $F \subset M_n \otimes A$ be finite, let $\ep > 0$,
and let $c \in (M_n \otimes A)_{+} \setminus \{ 0 \}$.
Let $S \subset A$ be the set of all matrix entries of all elements of~$F$.
By Lemma~\ref{L:MnSP} and Corollary~\ref{C_6X08_TAFSP},
$M_n \otimes A$ has property~(SP).
So there is a \nzp{} $q \in {\overline{c (M_n \otimes A) c}}$.
Use Lemma~\ref{L:CompSP}
to find a \nzp{} $q_0 \in A$
such that $e_{1, 1} \otimes q_0 \precsim q$.
Use Lemma~\ref{OrthInSP}
to find nonzero \mvnt{} \mops{} $e_1, e_2, \ldots, e_n \in q_0 A q_0$.
Apply Definition~\ref{D_TR0}
with $\ep / n^2$ in place of $\ep$,
with $e_1$ in place of $c$,
and with $S$ in place of~$F$,
getting a nonzero \pj{} $p_0 \in A$
and a unital finite dimensional
subalgebra $D_0 \subset p_0 A p_0$.
Set $p = 1 \otimes p_0$ and $D = M_n \otimes D_0$.
Then
\[
1  p \precsim \sum_{j = 1}^n e_{j, j} \otimes e_1
\sim \sum_{j = 1}^n e_{1, 1} \otimes e_j
\leq q_0
\precsim q.
\]
Also,
for $a \in F$ we can find $a_{j, k} \in S$ for $j, k = 1, 2, \ldots, n$
such that $a = \sum_{j, k = 1}^n e_{j, k} \otimes a_{j, k}$,
and $b_{j, k} \in D_0$ for $j, k = 1, 2, \ldots, n$ such that
$\ p_0 a_{j, k} p_0  b_{j, k} \ < \ep / n^2$.
Then $b = \sum_{j, k = 1}^n e_{j, k} \otimes b_{j, k} \in D$
and
\[
\ p a p  b \
= \Bigg\ \sum_{j, k = 1}^n e_{j, k} \otimes p_0 a_{j, k} p_0
 \sum_{j, k = 1}^n e_{j, k} \otimes b_{j, k} \Bigg\
\leq \sum_{j, k = 1}^n \ p_0 a_{j, k} p_0  b_{j, k} \
< \ep.
\]
Finally,
\[
\ p a  a p \
= \Bigg\ \sum_{j, k = 1}^n
e_{j, k} \otimes (p_0 a_{j, k}  a_{j, k} p_0 ) \Bigg\
\leq \sum_{j, k = 1}^n \ p_0 a_{j, k}  p_0 a_{j, k} \
< \ep.
\]
This completes the proof.
\end{proof}
\section{Crossed Products by Finite Groups}\label{Sec:CrPrdFinGp}
\indent
In this section,
we look briefly at some of the general theory of crossed products
by finite groups,
mostly in the simple case.
In Section~\ref{Sec_RPFG}
we will consider the structure of crossed products
when the action has the Rokhlin property,
and in Section~\ref{Sec_TRPFG}
we will consider the structure of crossed products
when the action has the tracial Rokhlin property.
A version of the tracial Rokhlin property using positive elements
instead of \pj{s}
seems to be the weakest hypothesis for good structure
theorems for crossed products,
but in these notes we will only consider the version using
projections.
In this section,
we give a fairly short proof that if $G$ is finite,
$A$ is simple,
and $\af \colon G \to \Aut (A)$ is pointwise outer,
then $\CGAa$ is simple.
From the point of view of these notes,
one can't say more without a stronger hypothesis on the action,
presumably some version of the tracial Rokhlin property.
The various examples and problems we discuss in this section
indicate how things can go wrong
if one assumes less.
See~\cite{PhSvy} for a much more extensive discussion,
with the defect that higher dimensional Rokhlin properties
are not mentioned;
they were not known at the time that \cite{PhSvy} was written.
Recall (Corollary~\ref{C_FGpCP})
that if $\af \colon G \to \Aut (A)$
is an action of a finite group $G$ on a \ca~$A$,
then the maps
\begin{equation}\label{Eq_6Y26_Star}
C_{\mathrm{c}} (G, A, \af)
\to C^* (G, A, \af) \to C^*_{\mathrm{r}} (G, A, \af)
\end{equation}
are bijective.
This means that,
unlike all other cases,
one can explicitly write down all elements of
$C^* (G, A, \af)$:
as in Remark~\ref{R_6320_Sumagug},
they are the sums $\sum_{g \in G} a_g u_g$
with $a_g \in A$ for $g \in G$.
The next strongest condition
after versions of the tracial Rokhlin property
without \pj{s} is pointwise outerness.
\begin{dfn}\label{D_6320_PtwOut}
Let $A$ be a \ca,
let $G$ be a group,
and let $\af \colon G \to \Aut (A)$
be an action of $G$ on~$A$.
The action $\af$ is called {\emph{pointwise outer}}
if $\af_g$ is not inner (Definition~\ref{D_6320_InnerAut})
for all $g \in G \setminus \{ 1 \}$.
\end{dfn}
Such actions are often simply called outer.
This designation can lead to confusion
because of the temptation to say that
an action is outer if it is not inner
(as in Example~\ref{E:Inner}).
There are many actions~$\af$
for which $\af_g$ is inner for some choices
of $g \in G \setminus \{ 1 \}$ but outer for other choices.
There are even actions $\af$
for which $\af_g$ is inner for all $g \in G$
but $\af$ itself is not inner.
(See Example~\ref{E_PtInn}.)
\begin{thm}\label{T_6320_FgCP}
Let $G$ be a finite group,
let $A$ be a simple unital \ca,
and let $\aGA$ be a pointwise outer action
of $G$ on~$A$.
Then $\CGAa$ is simple.
\end{thm}
The proof we give for Theorem~\ref{T_6320_FgCP}
is based on that of Theorem~1.1 of~\cite{RfFG},
but is much simpler,
since we prove much less.
In fact, what we actually prove was known long before.
Apart from a small piece of operator algebra theory
(isolated in Lemma~\ref{L_6320_AlgInn}),
it is entirely algebraic,
and proves that the skew group ring
for a pointwise outer action of a finite group
on a simple unital ring is again simple.
Rieffel was in fact motivated by arguments from algebra,
but the algebraic version of the result we prove
was already proved in Theorem~4 of~\cite{Azm}.
The result generalizes in at least two directions.
By Theorem~\ref{T_Ks} below,
the reduced \cp{} of a simple \ca{} by a pointwise
outer action of a discrete group is simple.
Thus, provided we use the reduced \cp,
we can replace ``finite'' by ``discrete'' in Theorem~\ref{T_6320_FgCP}.
The proof of Theorem~\ref{T_Ks} is quite different,
requiring much more machinery.
We do not give it in these notes,
although we give a proof of a special case of a theorem
which implies this result
(not, however, the case needed for this result).
The other direction is that taken in~\cite{RfFG}.
For example,
one part of Theorem~4.1 of~\cite{RfFG}
states that if $G$ is finite
and $A^G$ is type~I,
then $A$ is type~I
(without assuming that $\af$ is pointwise outer,
but results on outerness are used in the proof).
This is false for both compact and infinite discrete groups.
Also see Section~2 of~\cite{RfFG},
about primeness of \cp{s} by finite groups.
We isolate the C*algebraic part as a general lemma.
\begin{lem}\label{L_6320_AlgInn}
Let $A$ be a \ca{}
and let $\af \in \Aut (A)$.
Suppose there is $x$ in the multiplier algebra $M (A)$
such that for all $a \in A$
we have $\af (a) = x a x^{1}$.
Then $\af$ is inner
(Definition~\ref{D_6320_InnerAut}),
that is, there is a unitary $u \in M (A)$ such that
$\af (a) = u a u^*$ for all $a \in A$.
\end{lem}
The proof is essentially the same as part of the
proof of Lemma~\ref{L_5Y21_ClosePj}.
It depends (as it must)
on the relation $\af (a^*) = \af (a)^*$ for all $a \in A$.
\begin{proof}[Proof of Lemma~\ref{L_6320_AlgInn}]
We immediately get $x a = \af (a) x$ for all $a \in A$.
In this equation,
take adjoints and replace $a$ by~$a^*$,
getting $x^* \af (a) = a x^*$ for all $a \in A$.
Combine these two equations,
getting $x^* x a = a x^* x$ for all $a \in A$.
Therefore
%
\begin{equation}\label{Eq_6320_Comm}
(x^* x)^{ 1/2} a = a (x^* x)^{ 1/2}
\end{equation}
%
for all $a \in A$.
Now $u = x (x^* x)^{ 1/2}$
is unitary by Lemma~\ref{L_6320_PolD}.
Combining $\af (a) = x a x^{1}$ for all $a \in A$
with~(\ref{Eq_6320_Comm}),
we get
$\af (a) = u a u^*$ for all $a \in A$.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{T_6320_FgCP}]
As discussed at the beginning of this section,
we have
\[
\CGAa
= \left\{ \ssum{g \in G} a_g u_g \colon
{\mbox{$a_g \in A$ for $g \in G$}} \right\}.
\]
Thinking of
$\CGAa$ as $C_{\mathrm{c}} (G, A, \af)$
(bijectivity of the maps in~(\ref{Eq_6Y26_Star});
see Corollary~\ref{C_FGpCP}),
we define the support of an element
$a = \sum_{g \in G} a_g u_g \in \CGAa$
by
\[
\supp (a)
= \big\{ g \in G \colon a_g \neq 0 \big\}.
\]
Now let $I \subset \CGAa$ be a nonzero ideal.
We will eventually show that $I = \CGAa$.
Choose $b \in I \setminus \{ 0 \}$
such that $\card (\supp (b))$ is minimal
among all nonzero elements of~$I$.
There is $h \in G$ such that $b_h \neq 0$.
Setting $a = b u_h^*$,
we get $a = \sum_{g \in G} a_g u_g \in I \setminus \{ 0 \}$
such that $\card (\supp (a))$ is minimal
among all nonzero elements of~$I$
and such that $a_1 \neq 0$.
We want to show that $a \in A$.
Suppose not.
Then there is $h \in \GMI$
such that $a_h \neq 0$.
We claim that there is a well defined bijective linear map
$T \colon A \to A$
such that, whenever $n \in \N$
and $x_j, y_j \in A$ for $j = 1, 2, \ldots, n$,
we have
%
\begin{equation}\label{Eq_6320_DfnT}
T \Bigg( \sum_{j = 1}^n x_j a_1 y_j \Bigg)
= \sum_{j = 1}^n x_j a_h \af_h (y_j).
\end{equation}
%
To prove this claim, we first observe that
\[
\Bigg\{ \sum_{j = 1}^n x_j a_1 y_j \colon
{\mbox{$n \in \N$
and $x_j, y_j \in A$ for $j = 1, 2, \ldots, n$}} \Bigg\}
\]
is equal to $A$ because $A$ is simple and unital.
So $T$ is defined on all of~$A$.
Next,
we show that if
$\sum_{j = 1}^n x_j a_1 y_j = 0$
then $\sum_{j = 1}^n x_j a_h \af_h (y_j) = 0$.
So let $n \in \N$
and for $j = 1, 2, \ldots, n$ let $x_j, y_j \in A$.
Suppose $\sum_{j = 1}^n x_j a_1 y_j = 0$.
Define
\[
s = \sum_{j = 1}^n x_j a y_j \in \CGAa.
\]
Then $s \in I$.
Moreover,
we can calculate
\[
s = \sum_{j = 1}^n \sum_{g \in G} x_j a_g u_g y_j
= \sum_{j = 1}^n \sum_{g \in G} x_j a_g \af_g (y_j) u_g
= \sum_{g \in G}
\Bigg( \sum_{j = 1}^n x_j a_g \af_g (y_j) \Bigg) u_g.
\]
It is clear from this formula that $\supp (s) \subset \supp (a)$.
Moreover,
$s_1 = 0$.
The fact that $\card (\supp (a))$ is minimal
among all nonzero elements of~$I$
therefore implies $s = 0$.
In particular,
$\sum_{j = 1}^n x_j a_h \af_h (y_j) = s_h = 0$.
This proves the desired implication.
By considering differences of two expressions of the form
$\sum_{j = 1}^n x_j a_1 y_j$, it follows that $T$ is well defined.
With this in hand, $T$ is obviously linear,
and it now also follows that $T$ is injective.
We finish the proof of the claim by showing that $T$ is surjective.
Let $d \in A$.
Since $a_h \neq 0$ and $A$ is simple and unital,
there are $n \in \N$
and $x_j, y_j \in A$ for $j = 1, 2, \ldots, n$
such that $\sum_{j = 1}^n x_j a_h y_j = d$.
Set $c = \sum_{j = 1}^n x_j a_1 \af_h^{1} (y_j)$.
Then $T (c) = d$.
This completes the proof of the claim.
It is immediate from~(\ref{Eq_6320_DfnT})
that for all $a, c \in A$ we have
%
\begin{equation}\label{Eq_6X08_ModMap}
T (c a) = c T (a)
\andeqn
T (a c) = T (a) \af_h (c).
\end{equation}
%
Thus $T (c) = c T (1)$ for all $c \in A$,
and using surjectivity to choose $c \in A$ such that $T (c) = 1$,
we see that $T (1)$ is left invertible.
Similarly,
$T (c) = T (1) \af_h (c)$ for all $c \in A$,
and using surjectivity to choose $c \in A$ such that $T (c) = 1$,
we see that $T (1)$ is right invertible.
So $T (1)$ is invertible.
Combining the two parts of~(\ref{Eq_6X08_ModMap}),
we get $T (1) \af_h (c) = T (c) = c T (1)$ for all $c \in A$.
Applying Lemma~\ref{L_6320_AlgInn} with $x = T (1)^{1}$
shows that $\af_h$ is inner.
This contradiction shows that $a \in A$.
We have shown that $I \cap A \neq \{ 0 \}$.
Since $I \cap A$ is an ideal in the simple \ca~$A$,
it follows that $1 \in I \cap A$.
So $1 \in I$,
and $I = \CGAa$.
\end{proof}
% Add more? 999 [What was intended?]
Pointwise outerness is not good enough for
the kind of structural results we have in mind
for \cp{s} by finite groups.
\begin{exa}\label{E_Elliott}
Example~9 of~\cite{Ell3}
contains a pointwise outer action~$\af$
of $\Z / 2 \Z$
% (in the sense of Definition~\ref{D:Out} below) of $\Zqt$
on a simple unital AF~algebra~$A$
such that $C^* (\Z / 2 \Z, \, A, \, \af)$ does not have real rank zero.
However, AF~algebras have real rank zero for fairly trivial reasons.
\end{exa}
The first example of an action of a finite group on an AF~algebra
such that the crossed product is not AF was given in~\cite{Bl0}.
The action is in \Ex{E_3502_Blackadar}.
The actions in Exercise~\ref{Ex:RCFG4_K1}
and Exercise~\ref{Ex:RCFG4_Torsion}
are also examples of this phenomenon.
Among the known examples,
the one that is easiest to construct
is in Section VIII.9 of~\cite{Dvd}.
It is the dual action to
an action of $\Z / 2 \Z$ on a BunceDeddens algebra
whose crossed product is~AF.
% Put this example in somewhere? 999
\begin{exa}\label{E_6X02_Blackadar}
Example~8.2.1 of~\cite{Bl0}
gives an example of a pointwise outer action~$\af$
of $\Z / 2 \Z$
% (in the sense of Definition~\ref{D:Out} below) of $\Zqt$
on a separable unital \ca~$A$
such that $A$ has stable rank one
but $C^* (\Z / 2 \Z, \, A, \, \af)$ has stable rank two.
\end{exa}
\Ex{E_Elliott} and \Ex{E_6X02_Blackadar}
are both accessible via the methods of Section~\ref{Sec_Comp}
(although we need to appeal to classification theorems).
% but we have not yet had time to put them in the file.
% Add? 999
The action in \Ex{E_6X02_Blackadar}
is the tensor product of the action in \Ex{E_3502_Blackadar}
with the trivial action on $C ([0, 1])$.
The following is a long standing open problem.
\begin{pbm}\label{Pb_6X02_SimpTsr1}
Let $A$ be a simple \uca{} with stable rank one.
Let $G$ be a finite group,
and let $\aGA$ be an action of $G$ on~$A$.
Does it follow that $\CGAa$ has stable rank one?
\end{pbm}
A positive answer is not known even if $G = \Z / 2 \Z$ and $A$ is~AF.
We do have
(using methods not considered here) the following theorem,
which improves earlier known estimates.
\begin{thm}[Theorem~2.4 of~\cite{JOPT}]\label{T:JOPT}
Let $A$ be a \ca, and let $\af \colon G \to \Aut (A)$
be an action of a finite group $G$ on $A$.
Then
\[
\tsr ( C^* (G, A, \af)) \leq \tsr (A) + \card (G)  1.
\]
\end{thm}
Crossed products by finite groups do preserve type~I \ca{s}
and nuclear \ca{s}.
Both statements are true more generally:
preservation of type~I holds for compact groups,
at least when the algebra is separable
and the group is second countable
(this can be gotten from Theorem 6.1 of~\cite{Tks67})
% 999 ref needs improvement; smoothness could be gotten from my thesis
and preservation of nuclearity holds for amenable groups
(Theorem 4.2.6 of~\cite{BrOz}).
% 999 The nuclearity reference isn't good,
% since the proof there isn't cleanly organized.
Crossed products by finite groups presumably do not preserve the
Universal Coefficient Theorem,
although,
as far as we know,
no example has been published.
The idea is as follows.
Let $A$ be the \ca{} in the example in Section~4 of~\cite{Skds}.
It is not KKequivalent to a nuclear \ca,
and therefore does not satisfy the Universal Coefficient Theorem.
% Call it~$A$.
Choose (see below) a contractible nuclear \ca~$B$
in the bootstrap class,
with an action $\bt$ of a finite group~$G$ on $B$
such that $C^* (G, B, \bt)$
is also in the bootstrap class and $K_* (C^* (G, B, \bt)) \neq 0$.
(Preferably $K_* (C^* (G, B, \bt))$
should have a summand isomorphic to~$\Z$.)
Set $C = A \otimes B$ and define
$\gm \colon G \to \Aut (C)$
by $\gm_g = \id_A \otimes \bt_g$ for $g \in G$.
Then $C$ is contractible,
so satisfies the Universal Coefficient Theorem
for trivial reasons.
However, $C^* (G, C, \gm) \cong A \otimes C^* (G, B, \bt)$.
(See Remark~\ref{R:CPTens}
and Exercise~\ref{P:Triv}.)
So the K\"{u}nneth formula~\cite{Scct2}
% 999 Theorem number missing.
relates $K_* (C^* (G, C, \gm))$
to $K_* (A)$ and $K_* (C^* (G, B, \bt))$.
For example,
if
\[
K_0 (C^* (G, B, \bt)) \cong \Z
\andeqn
K_1 (C^* (G, B, \bt)) = 0,
\]
then
\[
K_* (C^* (G, C, \gm)) \cong K_* (C^* (G, B, \bt)).
\]
This should transfer failure of the
Universal Coefficient Theorem for~$A$
to failure of the
Universal Coefficient Theorem for $C^* (G, C, \gm)$.
In Section~3 of~\cite{Ph89a},
there are examples of homotopies
$t \mapsto \af^{(t)}$ of actions of a finite group~$G$
on a nuclear \ca~$D$
(even a commutative \ca)
such that
$K_* \big( G, D, \af^{(0)} \big)
\not\cong K_* \big( G, D, \af^{(1)} \big)$.
Such a homotopy defines an action on $C ([0, 1], \, D)$,
for which the cone $C_0 ((0, 1], \, D)$
is invariant,
and for which the Ktheory of the crossed product
of the cone is sometimes nonzero.
This can actually happen in at least some of the examples
in~\cite{Ph89a},
but it isn't clear whether one can arrange
to have the Ktheory of the crossed product
isomorphic to~$\Z$.
(It can be made isomorphic to~$\Z [\frac{1}{2}]$.)
Despite all that seems to go wrong with crossed products
by pointwise outer actions of finite groups
without stronger assumptions,
there are no examples in which the crossed product
of a classifiable \ca{}
by a pointwise outer action of a finite group
is known not to be classifiable.
\section{The Rokhlin Property for Actions of Finite
Groups}\label{Sec_RPFG}
\indent
What is needed for good results on the structure of \cp{s} is
some notion of freeness of the action.
Free actions on spaces are well known;
see Definition~\ref{D_3331_Free}.
There are many versions of freeness for actions on \ca{s}
even when the group is finite.
See~\cite{PhSvy} for an extensive discussion
(which, however, makes no mention of higher dimensional
Rokhlin properties; these were introduced after \cite{PhSvy}
was written).
Versions of freeness
range from free action on the primitive ideal space
(impossible when the group is nontrivial and the algebra is simple)
to conditions even weaker than pointwise outerness.
The conditions which seem to be most useful for theorems on the
structure of \cp{s} are the Rokhlin property,
the tracial Rokhlin property,
and various higher dimensional Rokhlin properties.
In this section, we consider the Rokhlin property,
and in the next section we consider the tracial Rokhlin property.
Higher dimensional Rokhlin properties,
which we don't discuss,
were introduced in~\cite{HWZ},
and generalized (along with the ordinary Rokhlin property)
to the nonunital case in~\cite{HrsPh1}.
The paper~\cite{HrsPh1}
also defines a related property called the ``$X$Rokhlin property''.
Although we will not discuss them in these notes,
there are versions of the Rokhlin property and
the tracial Rokhlin property (including versions
using positive elements instead of \pj{s})
for actions of suitable not necessarily finite groups.
In Definition~\ref{D_RPDfn}
(and in Definition~\ref{D_TRP} below),
one must use finite subsets of~$G$ instead of the whole group;
these finite subsets should be approximately invariant
under translation by a given finite set of group elements.
(They should be F{\o}lner sets in the sense used in the
F{\o}lner condition for amenability.
See Theorem 3.6.1 of~\cite{Grf};
F{\o}lner sets were used in the proof of the discrete case
of Theorem~\ref{T_4131_GpFToRed}
and in the proof of \Thm{T_4131_CPFToRed}.)
We give only a few references:
\cite{Izmi} for Rokhlin actions of~$\Z$,
\cite{OP1} for actions of~$\Z$ with the tracial Rokhlin property,
\cite{HirOr2013} for a tracial Rokhlin property
for finite groups and~$\Z$ in terms of positive elements,
and
\cite{OvPhWn} for a tracial Rokhlin property
for countable amenable groups in terms of positive elements.
The Rokhlin property was first introduced by Rokhlin,
in measurable dynamics for an action of~$\Z$
of a measure space.
See the discussion at the top of page~611 of~\cite{Yzv}.
The original Rokhlin Lemma is
given in Lemma VIII.3.4 of~\cite{Dvd}.
The Rokhlin property for actions of finite groups
was defined for von Neumann algebras before \ca{s},
in~\cite{Jns},
but not under that name and in a slightly different formulation.
The Rokhlin property and higher dimensional Rokhlin properties
are also useful in the nonsimple case.
We don't know how to define the tracial Rokhlin property
in the nonsimple case.
At first sight, the Rokhlin property looks strange.
We explain how it can be used in Remark~\ref{R_6310_UseRP}
and in Lemma~\ref{L_1216PartR}
and the discussion before its proof.
The interested reader can skip the discussion of
examples of actions with (and without)
the Rokhlin property
and look first at this remark and lemma.
\begin{dfn}\label{D_RPDfn}
Let $A$ be a \uca,
and let $\af \colon G \to \Aut (A)$
be an action of a finite group $G$ on~$A$.
We say that $\af$ has the
{\emph{Rokhlin property}} if for every finite set
$S \subset A$ and every $\ep > 0$,
there are \mops{} $e_g \in A$ for $g \in G$ such that:
\begin{enumerate}
\item\label{D_RPDfn:1}
$\ \af_g (e_h)  e_{g h} \ < \ep$ for all $g, h \in G$.
\item\label{D_RPDfn:2}
$\ e_g a  a e_g \ < \ep$ for all $g \in G$ and all $a \in S$.
\item\label{D_RPDfn:3}
$\sum_{g \in G} e_g = 1$.
\end{enumerate}
We call $(e_g)_{g \in G}$ a
{\emph{family of Rokhlin projections}} for $\af$, $S$, and~$\ep$.
\end{dfn}
One can strengthen the statement.
\begin{thm}[Proposition 5.26 of~\cite{PhEqSj}]\label{T_ExactRP}
Let $A$ be a separable unital \ca,
and let $\af \colon G \to \Aut (A)$
be an action of a finite group $G$ on~$A$.
Then $\af$ has the
Rokhlin property \ifo{} for every finite set
$F \subset A$ and every $\ep > 0$,
there are \mops{} $e_g \in A$ for $g \in G$ such that:
\begin{enumerate}
\item\label{ERPDfnE:1}
$\af_g (e_h) = e_{g h}$ for all $g, h \in G$.
\item\label{ERPDfnE:2}
$\ e_g a  a e_g \ < \ep$ for all $g \in G$ and all $a \in F$.
\item\label{ERPDfnE:3}
$\sum_{g \in G} e_g = 1$.
\end{enumerate}
\end{thm}
The difference is that in~(\ref{ERPDfnE:1})
we ask for exact rather than approximate equality.
Theorem~\ref{T_ExactRP} simplifies some proofs by replacing some
approximate equalities by equalities.
In particular,
\Lem{L_SemiProjMn} becomes unnecessary.
However, the proof uses methods which are not standard
and are not related to those here,
and moreover is more complicated than the work it would save here.
In the interest of completeness,
we therefore give proofs without using this result.
\begin{exa}\label{E_6X01_Translate}
Let $G$ be a finite group,
let $B$ be any unital \ca,
set $A = C (G, B)$,
and define
$\aGA$
by $\af_g (a) (h) = a (h^{1} g)$
for $a \in A$ and $g, h \in G$.
The algebra $A$ is the direct sum of copies of $B$,
indexed by~$G$,
and the action permutes the summands.
The projections required for the Rokhlin property
can be taken to be given by
\[
e_g (h) = \begin{cases}
1 & h = g
\\
0 & h \neq g
\end{cases}
\]
for $g, h \in G$.
\end{exa}
The algebra in Example~\ref{E_6X01_Translate} is not simple.
The Rokhlin property is very rare for actions on simple \ca{s}.
We give in the next section some examples of actions with the \trp{}
but not the \rp.
Here we mention just a few examples
of nonexistence of actions with the Rokhlin property,
based on elementary Ktheoretic obstructions
(not all of which require explicit use of Ktheory).
There is no action of any nontrivial finite group
on~${\mathcal{O}}_{\infty}$
or any irrational rotation algebra which has the \rp{}
(Proposition~\ref{E_1211_NoRPOI}; Example~\ref{E_1211_NoRPAT}),
there is no action of any finite group whose order
is divisible by~$2$ on~${\mathcal{O}}_{3}$ which has the \rp{}
(Proposition~\ref{E_1211_NoRPO3}),
and there is no action of any finite group whose order
is divisible by any prime other than~$2$
on the $2^{\infty}$~UHF algebra which has the \rp{}
(Example~\ref{E_1211_NoRPUHF}).
Even more obviously,
there is no action of a nontrivial
finite group
on the JiangSu algebra
(briefly described in \Ex{E:JiangSuFlip})
which has the \rp,
since the algebra has no nontrivial \pj{s}.
See Example 3.12 in~\cite{PhSvy}
and the surrounding discussion for more examples.
On the other hand,
actions with the \rp{} do exist on suitable simple \ca{s},
and can be obtained using suitable choices in Example~\ref{E_PrdType}.
We describe the details for $\Z / 2 \Z$ in Example~\ref{E:Z2RP}.
This example is, in slightly different notation,
the special case at the end of Example~\ref{E_PrdType},
whose crossed product is treated in Example~\ref{E:C:PTypeAgain}.
See Exercise~\ref{E_RPRegRep}
for a more general case.
We look at the commutative case first.
\begin{prp}\label{P_6310_RPForSpace}
Let $G$ be a finite group,
and let $X$ be a compact Hausdorff $G$space.
Then the corresponding action of $G$ on $C (X)$
has the Rokhlin property
\ifo{} there are a compact Hausdorff space $Y$
% with the trivial action of $G$
and an equivariant homeomorphism from
$X$ to $G \times Y$,
with $G$ acting on $G$ by translation,
trivially on~$Y$,
and via the product action on $G \times Y$.
\end{prp}
\begin{proof}
Assume first that
there is an equivariant homeomorphism $X \to G \times Y$.
We may then assume that $X = G \times Y$.
For any finite set
$F \subset A$ and any $\ep > 0$,
we can take $e_g = \ch_{ \{ g \} \times Y}$ for $g \in G$.
Now assume that the action on $C (X)$ has the Rokhlin property.
Apply Definition~\ref{D_RPDfn}
with $F = \E$ and $\ep = \frac{1}{2}$,
obtaining a family $(e_g)_{g \in G}$ of Rokhlin projections.
In $C (X)$,
if $p$ and $q$ are \pj{s}
with $\ p  q \ < 1$,
then $p = q$.
Therefore we get $\af_g (e_h) = e_{g h}$ for all $g, h \in G$.
There is a closed and open subset $Y \subset X$
such that $e_1 = \ch_Y$.
Define a \cfn{} $m \colon G \times Y \to X$
by $m (g, y) = g y$.
Since the projections $\af_g (\ch_Y)$ for $g \in G$ are orthogonal,
the sets $g Y$ are disjoint,
so $m$ is injective.
Since $\sum_{g \in G} \af_g (\ch_Y) = 1$,
we have $\bigcup_{g \in G} g Y = X$,
so $m$ is surjective.
Therefore $m$ is a \hme.
Giving $Y$ the trivial action of $G$ and $G$ the translation action,
it is immediate to check that $m$ is equivariant.
\end{proof}
The following result is a restatement of Theorem~1.2 of~\cite{PhSvy}.
\begin{thm}\label{CT:Cantor}
Let $G$ be a finite group,
and let $X$ be a totally disconnected $G$space.
Then the corresponding action of $G$ on $C (X)$
has the Rokhlin property
\ifo{} the action of $G$ on $X$ is free.
\end{thm}
\begin{proof}
If the action of $G$ on $C (X)$
has the Rokhlin property,
then freeness of the action of $G$ on $X$
is is immediate from Proposition~\ref{P_6310_RPForSpace}.
So assume the action of $G$ on $X$ is free.
We first claim that for every $x \in X$,
there is a compact open set $L \subset X$ such that
$x \in L$ and the sets
$g L$, for $g \in G$, are disjoint.
To prove the claim,
for $g \in G$ choose disjoint compact open sets
$L_g$ and $M_g$ such that $x \in L_g$ and $g x \in M_g$.
Then take
\[
L = \bigcap_{g \in G \setminus \{ 1 \}} (L_g \cap g^{1} M_g).
\]
This proves the claim.
Since $X$ is compact,
we can now find compact open sets $L_1, L_2, \ldots, L_n \subset X$
which cover $X$ and such that, for each~$m$,
the sets $g L_m$, for $g \in G$, are disjoint.
Set $K_1 = L_1$
and for $m = 2, 3, \ldots, n$ set
\[
K_m = L_m \cap \left( X \setminus \bigcup_{g \in G}
g (L_1 \cup L_2 \cup \cdots \cup L_{m  1}) \right).
\]
(This set may be empty.)
One verifies by induction on~$m$
that the sets $g K_j$, for $g \in G$ and $j = 1, 2, \ldots, m$,
are disjoint
and cover $\bigcup_{g \in G} g (L_1 \cup L_2 \cup \cdots \cup L_m)$.
For $m = n$,
these sets form a partition of~$X$.
Set $Y = K_1 \cup K_2 \cup \cdots \cup K_n$.
Then the sets $g Y$, for $g \in G$, form a partition of~$X$.
The conclusion follows.
\end{proof}
The Rokhlin property is a strong form of freeness.
Not all free actions of finite groups on compact spaces
have the Rokhlin property.
The actions of finite subgroups of $S^1$ on $S^1$ be translation
(given in Example~\ref{E_Rot})
are free but don't have the Rokhlin property.
(See~\cite{PhSvy} for an extensive discussion of notions of
freeness of actions of finite groups on \ca{s},
but note that higher dimensional Rokhlin properties
had not yet been introduced when this article was written.)
\begin{exa}\label{E:Z2RP}
Let $\af$ be the action of $\Z / 2 \Z$
on the $2^{\infty}$~UHF algebra~$A$
generated by
$\bigotimes_{n = 1}^{\I}
\Ad \left(
\begin{smallmatrix} 0 & 1 \\ 1 & 0 \end{smallmatrix} \right)$.
(The $2 \times 2$ matrix in the above formula is unitarily
equivalent to the $2 \times 2$ matrix
$\left( \begin{smallmatrix} 1 & 0 \\ 0 & 1 \end{smallmatrix} \right)$
used in Example~\ref{E:C:PTypeAgain}.
One checks that this implies that the actions are conjugate.
See Exercise~\ref{E_6X01_PrTypCnj} below.)
We simply write $\af$ for the automorphism given by the nontrivial
group element.
In Example~\ref{E:C:PTypeAgain}, we wrote $A = \Dirlim M_{2^n}$,
with maps $\ph_n \colon M_{2^n} \to M_{2^{n + 1}}$
given by
$a \mapsto
\left( \begin{smallmatrix} a & 0 \\ 0 & a \end{smallmatrix} \right)$.
Here, we identify $M_{2^n}$
as the tensor product of $n$ copies of $M_2$,
which we write for short as $(M_2)^{\otimes n}$.
We identify the maps of the direct system
\[
(M_2)^{\otimes n}
\stackrel{\ph_n}{\longrightarrow}
(M_2)^{\otimes (n + 1)} = (M_2)^{\otimes n} \otimes M_2
\]
as
$a \mapsto a \otimes 1$.
We also identify $(M_2)^{\otimes n}$ with its image in~$A$.
We claim that $\af$ has the \rp.
Let $S \subset A$ be finite and let $\ep > 0$.
We have to find orthogonal \pj{s} $e_0, e_1 \in A$
such that:
\begin{enumerate}
\item\label{Z2:RPDfn:1}
$\ \af (e_0)  e_1 \ < \ep$ and $\ \af (e_1)  e_0 \ < \ep$.
\item\label{Z2:RPDfn:2}
$\ e_0 a  a e_0 \ < \ep$ and
$\ e_1 a  a e_1 \ < \ep$ for all $a \in S$.
\item\label{Z2:RPDfn:3}
$e_0 + e_1 = 1$.
\end{enumerate}
Write $S = \{ a_1, a_2, \ldots, a_N \}$.
% Set $M = \sup_{a \in S} \ a \$.
Since $\bigcup_{n \in \Nz} (M_2)^{\otimes n}$ is dense in~$A$,
there are $n$ and
$b_1, b_2, \ldots, b_N \in (M_2)^{\otimes n} \subset A$
such that
\[
\ b_1  a_1 \ < \tfrac{1}{2} \ep,
\qquad
\ b_2  a_2 \ < \tfrac{1}{2} \ep,
\qquad
\ldots,
\qquad
\ b_N  a_N \ < \tfrac{1}{2} \ep.
\]
Define
\[
e_0, e_1
\in (M_2)^{\otimes n} \otimes M_2 = (M_2)^{\otimes (n + 1)} \subset A
\]
by
\[
e_0 = 1_{(M_2)^{\otimes n}} \otimes
\left( \begin{smallmatrix} 1 & 0 \\ 0 & 0 \end{smallmatrix} \right)
\andeqn
e_1 = 1_{(M_2)^{\otimes n}} \otimes
\left( \begin{smallmatrix} 0 & 0 \\ 0 & 1 \end{smallmatrix} \right).
\]
It is obvious that $e_0 + e_1 = 1$,
which is~(\ref{Z2:RPDfn:3}),
and we easily check that
$\af (e_0) = e_1$ and $\af (e_1) = e_0$,
which implies~(\ref{Z2:RPDfn:1}).
It remains to check~(\ref{Z2:RPDfn:2}).
For $k \in \{ 1, 2, \ldots, N \}$,
the element $b_k$ actually commutes with $e_0$ and $e_1$,
so
\[
\ e_0 a_k  a_k e_0 \
\leq \ e_0 \ \cdot \ a_k  b_k \
+ \ a_k  b_k \ \cdot \ e_0 \
< \tfrac{1}{2} \ep + \tfrac{1}{2} \ep
= \ep.
\]
This completes the proof that $\af$ has the \rp.
\end{exa}
\begin{exr}\label{E_6X01_PrTypCnj}
Let $G$ be a locally compact group,
let $A$ be a UHF algebra,
and let $\aGA$ and $\rh \colon G \to \Aut (A)$
be two infinite tensor product actions as in \Ex{E_PrdType},
using the same infinite tensor product decomposition.
That is,
let $k_1, k_2, \ldots$
be integers with $k_n \geq 2$ for all $n \in \N$,
assume that $A = \bigotimes_{n = 1}^{\I} M_{k_n}$,
and that there are actions
$\bt^{(n)}, \, \sm^{(n)} \colon G \to \Aut (M_{k_n})$ for $n \in \N$
such that for all $g \in G$,
we have
$\af_g = \bigotimes_{n = 1}^{\I} \bt^{(n)}_g$
and $\rh_g = \bigotimes_{n = 1}^{\I} \sm^{(n)}_g$.
Now suppose that for every $n \in \N$,
the actions $\bt^{(n)}$ and $\sm^{(n)}$
are conjugate.
Prove that the actions $\af$ and $\rh$
are conjugate.
\end{exr}
\begin{exr}\label{E_RPRegRep}
In Example~\ref{E_PrdType},
let $G$ be a finite group,
for each $n \in \N$
let $g \mapsto u_n (g)$ be a unitary representation
of $G$ on $\C^{k_n}$ which is unitarily equivalent to a finite
direct sum of copies of the regular representation,
and set $\bt^{(n)}_g (a) = u_n (g) a u_n (g)^*$
for $g \in G$ and $a \in M_{k_n}$.
Prove that the corresponding action
$g \mapsto \bigotimes_{n = 1}^{\I} \bt^{(n)}_g$ of $G$
on $\bigotimes_{n = 1}^{\I} M_{k_n}$ has the \rp.
\end{exr}
% 999 Exercise above should be done.
One use of
the Rokhlin property is to ``average'' over the group
in ways not normally possible.
This construction is more related
to its use in classification of actions
than its use for structural properties of crossed products.
Some cases have an interpretation as
``cohomology vanishing lemmas'',
about which we say nothing more here.
The next remark gives an example of the method.
\begin{rmk}\label{R_6310_UseRP}
Let
$A$ be a unital \ca{}
and let $\af \colon G \to \Aut (A)$
be an action of a finite group $G$ on~$A$
which has the Rokhlin property.
Let $u \in A$ be a unitary.
The usual average
\[
\frac{1}{\card (G)} \sum_{g \in G} \af_g (u)
\]
will almost never be a unitary.
Suppose,
however, we choose Rokhlin projections $e_g \in A$ for $g \in G$
as in Theorem~\ref{T_ExactRP}.
In particular,
$\af_g (e_h) = e_{g h}$ for $g, h \in G$.
Assume, first,
that they exactly commute with~$u$.
Then the element
\[
v = \sum_{g \in G} \af_g (e_1 u e_1)
= \sum_{g \in G} e_g \af_g (u) e_g
\]
is a $G$invariant unitary in~$A$.
Having $e_g u = u e_g$ for all $g \in G$ is far too much to
hope for.
If, however, $\ e_g u  u e_g \$ is small enough,
then
\[
b = \sum_{g \in G} \af_g (e_1 u e_1)
= \sum_{g \in G} e_g \af_g (u) e_g
\]
will be $G$invariant
and approximately unitary.
So
$b (b^* b)^{ 1/2}$ will be a $G$invariant unitary
which is close to~$b$,
and thus close to $\sum_{g \in G} e_g \af_g (u) e_g$.
\end{rmk}
One doesn't really need the stronger condition in Theorem~\ref{T_ExactRP};
the approximation argument works nearly as well
using \Def{D_RPDfn} as it stands.
The next exercise gives an example
of what one can do with the ideas in Remark~\ref{R_6310_UseRP}.
It does not have much connection with the main ideas
in these notes,
but is very important elsewhere.
Let $A$ and $B$ be unital \ca{s}.
Two \hm{s} $\ph, \ps \colon A \to B$
are said to be {\emph{approximately unitarily equivalent}}
if for every $\ep > 0$ and every finite set $F \subset A$,
there is a unitary $u \in B$
such that $\ u \ph (a) u^*  \ps (a) \ < \ep$
for all $a \in F$.
(This concept is very important in the Elliott classification program.
As just one example,
if $\ph$ and $\ps$ are approximately unitarily equivalent,
then $\ph_*, \ps_* \colon K_* (A) \to K_* (B)$
are equal.)
Suppose now that $G$ is a finite group,
and $\af \colon G \to \Aut (A)$
and $\bt \colon G \to \Aut (B)$
are actions of~$G$ on $A$ and~$B$.
One can easily imagine that one would want
equivariant \hm{s} $\ph, \ps \colon A \to B$
to be not just approximately unitarily equivalent
but in fact equivariantly approximately unitarily equivalent,
that is, the unitaries $u$ above can be chosen to be
$G$invariant.
(If this is true,
then, for example,
$\ph_*, \ps_* \colon K_*^G (A) \to K_*^G (B)$
are equal.)
\begin{exr}\label{Ex_6310_AppUEq}
Let $G$ be a finite group,
let $A$ and $B$ be unital \ca{s},
and let $\af \colon G \to \Aut (A)$
and $\bt \colon G \to \Aut (B)$
be actions of~$G$ on $A$ and~$B$.
Let $\ph, \ps \colon A \to B$ be equivariant unital \hm{s},
and assume that $\ph$ and $\ps$ are approximately unitarily equivalent
(ignoring the group actions).
\begin{enumerate}
\item\label{Ex_6310_AppUEq_Cod}
Suppose that $\bt$ has the Rokhlin property.
Prove that $\ph$ and $\ps$
are equivariantly approximately unitarily equivalent.
\item\label{Ex_6310_AppUEq_Dom}
Suppose that $\af$ has the Rokhlin property.
Prove that $\ph$ and $\ps$
are equivariantly approximately unitarily equivalent.
\end{enumerate}
\end{exr}
We will now return to ideas more directly related to the structure
of crossed products.
We first give several results whose proofs are more direct than
that of Theorem~\ref{T_RokhAF}
(the main result of this section).
\begin{thm}[Proposition 4.14 of~\cite{OP3}]\label{T1211RpTst}
Let $A$ be a unital \ca,
let $G$ be a finite group,
and let $\af \colon G \to \Aut (A)$
be an action of $G$ on $A$ which has the \rp.
Then the restriction map defines a bijection
from $\T (C^* (G, A, \af))$ (see Definition~\ref{D_Trace})
to the set $\T (A)^G$ of $G$invariant tracial states on~$A$.
\end{thm}
\begin{proof}
For $g \in G$, following Notation~\ref{N:ug},
let $u_g \in C^* (G, A, \af)$ be the standard unitary
in the crossed product.
Let $E \colon C^* (G, A, \af) \to A$ be the standard conditional
expectation (\Def{D_StdCond}),
which is given by $E \left( \sum_{g \in G} a_g u_g \right) = a_1$
when $a_g \in A$ for $g \in G$.
We will show that the map $\ta \mapsto \ta \circ E$
is an inverse of the restriction map.
First, let $\ta \in \T (A)^G$.
Then $\ta \circ E$ is a tracial state
on $C^* (G, A, \af)$ by \Ex{EInvTrCndExpt}.
It is immediate that $(\ta \circ E) _A = \ta$.
Now let $\ta \in \T (C^* (G, A, \af))$.
We claim that for all $g \in G \setminus \{ 1 \}$
and $a \in A$,
we have $\ta (a u_g) = 0$.
Let $\ep > 0$.
Choose Rokhlin \pj{s} $e_h \in A$ for $h \in G$
according to Definition~\ref{D_RPDfn},
using
\[
\dt = \frac{\ep}{(1 + \ a \) \card (G)}
\]
in place of $\ep$ and with $F = \{ a \}$.
For $h \in G$,
using at the second step $g \neq 1$
(so that $e_h e_{g h} = 0$),
we get
\[
e_h u_g e_h
= e_h (u_g e_h u_g^*  e_{g h}) u_g + e_h e_{g h} u_g
= e_h (u_g e_h u_g^*  e_{g h}) u_g.
\]
Therefore
\[
\ e_h u_g e_h \
\leq \ e_h \ \ \af_h (e_h)  e_{g h} \ \ u_g \
< \dt.
\]
So,
using $\sum_{h \in G} e_h = 1$ at the first step
and the trace property at the second step,
\begin{align*}
 \ta (a u_g) 
& \leq \sum_{h \in G}  \ta ( a u_g e_h^2 ) 
= \sum_{h \in G}  \ta (e_h a u_g e_h) 
\leq \sum_{h \in G}
\big( \ e_h a  a e_h \ +  \ta ( a e_h u_g e_h )  \big)
\\
& \leq \sum_{h \in G}
\big( \ e_h a  a e_h \ + \ a \ \cdot \ e_h u_g e_h \ \big)
< \card (G) (1 + \ a \) \dt
= \ep.
\end{align*}
Since $\ep > 0$ is arbitrary, the claim follows.
Now let $a \in \CGAa$,
and choose $a_g \in A$ for $g \in G$
such that $a = \sum_{g \in G} a_g u_g$.
Then $E (a) = a_1$,
so, remembering that $u_1 = 1$ and $\ta (a u_g) = 0$ for $g \neq 1$,
we get
\[
(\ta _A) \circ E
= \ta (a_1)
= \sum_{g \in G} \ta (a u_g)
= \ta (a).
\]
This completes the proof.
\end{proof}
\begin{prp}\label{P_6316_RPOut}
Let $A$ be a unital \ca,
let $G$ be a finite group,
and let $\af \colon G \to \Aut (A)$
be an action of $G$ on $A$ which has the \rp.
Then $\af_g$ is outer for every $g \in G \setminus \{ 1 \}$.
\end{prp}
Proposition~4.16 of~\cite{PhSvy} has a stronger statement,
with a closely related but more complicated proof:
$\af$ is strongly pointwise outer
in the sense of Definition~4.11 of~\cite{PhSvy}.
\begin{proof}[Proof of Proposition~\ref{P_6316_RPOut}]
Let $h \in G \setminus \{ 1 \}$,
and suppose that $\af_h$ is inner.
Thus,
there is a unitary $u \in A$ such that
such that $\af_h (a) = u a u^*$ for all $a \in A$.
Choose \pj{s}
$e_g \in A$ for $g \in G$ as in
Definition~\ref{D_RPDfn},
with $S = \{ u \}$
and $\ep = \frac{1}{3}$.
Then calculate as follows,
using orthogonality of $e_1$ and $e_h$ at the first step
and $\af_h (e_1) = u e_1 u^*$ at the second step:
\begin{align*}
1 = \ e_1  e_h \
& \leq \ e_1  u e_1 u^* \ + \ \af_h (e_1)  e_h \
\\
& = \ e_1 u  u e_1 \ + \ \af_h (e_1)  e_h \
< \frac{1}{3} + \frac{1}{3}
= \frac{2}{3}.
\end{align*}
This is a contradiction.
\end{proof}
Thus, if $A$ is simple,
then Theorem~\ref{T_Ks} below
implies that $C^* (G, A, \af)$ is simple.
Actually, more can be proved directly,
although with a bit of work.
\begin{prp}[Corollary~2.5 of~\cite{PsnPh}]\label{P_RPCPSimp}
Let $A$ be a unital \ca,
let $G$ be a finite group,
and let $\af \colon G \to \Aut (A)$
be an action of $G$ on $A$ which has the \rp.
Let $J$ be an ideal in $C^* (G, A, \af)$.
Then there is a $G$invariant ideal $I \subset A$
such that $J = C^* (G, I, \af)$.
\end{prp}
The proof in~\cite{PsnPh} uses other results not
directly related to the Rokhlin property.
We give a direct proof.
A direct proof for the same result for integer actions
with the \rp{}
is given for Theorem~2.2 of~\cite{PsnPh2}.
We state a lemma separately,
which is the finite group version of
Lemma~2.1 of~\cite{PsnPh2}.
It is in the proof of the lemma that the \rp{}
is actually used.
\begin{lem}\label{L_6311_RPcsq323}
Let $A$ be a unital \ca,
let $G$ be a finite group,
and let $\af \colon G \to \Aut (A)$
be an action of $G$ on $A$ which has the \rp.
Let $E \colon C^* (G, A, \af) \to A$
be the standard conditional expectation
(Definition~\ref{D_StdCond}).
Then for every finite set $F \subset C^* (G, A, \af)$
and every $\ep > 0$,
there exist \mops{}
$e_g \in A$ for $g \in G$
such that $\sum_{g \in G} e_g = 1$ and
\[
\left\ E (a)  \ssum{g \in G} e_g a e_g \right\ < \ep
\]
for all $a \in F$.
\end{lem}
\begin{proof}
% Let $\ep > 0$.
Write $F = \{ b_1, b_2, \ldots, b_n \}$.
For $j = 1, 2, \ldots, n$
write $b_j = \sum_{h \in G} a_{j, h} u_h$
with $a_{j, h} \in A$ for $h \in G$.
Set
\[
F_0 = \big\{ a_{j, h} \colon
{\mbox{$h \in G$ and $j \in \{ 1, 2, \ldots, n \}$}} \big\}
\andeqn
M = \sup_{a \in F_0} \ a \.
\]
Choose \pj{s} $e_g \in A$ for $g \in G$
according to Definition~\ref{D_RPDfn},
with
\[
\dt = \frac{\ep}{(1 + M) \card (G)^2}
\]
in place of $\ep$ and with $F_0$ in place of~$F$.
Let $j \in \{ 1, 2, \ldots, n \}$.
The key estimate is as follows:
for $g, h \in G$,
we have
%
% \begin{equation}\label{Eq_6311_KeyRE}
\begin{align*}
&
\big\ e_g a_{j, h} u_h e_g  e_g e_{h g} a_{j, h} u_h \big\
\\
& \hspace*{3em} {\mbox{}}
\leq \ e_g \ \ a_{j, h} \ \big\ u_h e_g  e_{h g} u_h \big\
+ \ e_g \
\big\ a_{j, h} e_{h g}  e_{h g} a_{j, h} \big\ \ u_h \
\\& \hspace*{3em} {\mbox{}}
< M \dt + \dt.
\end{align*}
% \end{equation}
%
For $h \neq 1$ we have $e_g e_{h g} = 0$,
so $\ e_g a_{j, h} u_h e_g \ < (M + 1) \dt$.
For $h = 1$ we have $e_g e_{h g} = e_g$,
so
$\big\ e_g a_{j, h} u_h e_g  e_g a_{j, h} u_h \big\ < (M + 1) \dt$.
(Since $u_h = 1$ here,
one actually gets the estimate~$\dt$.)
Summing over $g, h \in G$,
we get
\[
\left\ \ssum{g, h \in G} e_g a_{j, h} u_h e_g
 \sum_{g \in G} e_g a_{j, 1} u_1 \right\
< \card (G)^2 (M + 1) \dt
= \ep.
\]
Using
\[
b_j = \sum_{h \in G} a_{j, h} u_h,
\qquad
\sum_{g \in G} e_g = 1,
\qquad
u_1 = 1,
\andeqn
E (b_j) = a_{j, 1},
\]
we can rewrite this inequality as
\[
\left\ \ssum{g \in G} e_g b_j e_g  E (b_j) \right\
< \ep,
\]
which is the desired estimate.
\end{proof}
\begin{proof}[Proof of Proposition~\ref{P_RPCPSimp}]
For $g \in G$, following Notation~\ref{N:ug},
let $u_g \in C^* (G, A, \af)$ be the standard unitary
in the crossed product.
Let $J \subset C^* (G, A, \af)$ be an ideal.
Set $I = J \cap A$.
We first claim that $I$ is a $G$invariant ideal in~$A$.
It is obvious that $I$ is an ideal in~$A$.
Let $g \in G$ and $a \in I$.
Certainly $\af_g (a) \in A$,
and $\af_g (a) = u_g a u_g^*$,
which is in $J$ since $a \in J$.
The claim is proved.
By Theorem~\ref{T_CPExact},
we can identify $C^* ( G, I, \af )$ with an ideal in $C^* (G, A, \af)$.
We next claim that $C^* ( G, I, \af ) \subset J$.
So let $a \in C^* ( G, I, \af )$.
Since $G$ is finite,
there are $a_g \in I$ for $g \in G$
such that $a = \sum_{g \in G} a_g u_g$.
Since the elements $a_g$ are in~$J$ and $J$ is an ideal
in $C^* (G, A, \af)$,
it follows that $a = \sum_{g \in G} a_g u_g \in J$.
The claim is proved.
Let $E \colon C^* (G, A, \af) \to A$
be the standard conditional expectation.
We claim that $E (J) \subset I$.
So let $a \in J$, and let $\ep > 0$.
Lemma~\ref{L_6311_RPcsq323} provides \mops{}
$e_g \in A$ for $g \in G$
such that $\sum_{g \in G} e_g = 1$ and
\[
\left\ E (a)  \ssum{g \in G} e_g a e_g \right\ < \ep.
\]
Since
$\sum_{g \in G} e_g a e_g \in J$ and $\ep > 0$ is arbitrary,
it follows that $E (a) \in J$.
The claim follows.
We finish the proof by showing that $J \subset C^* ( G, I, \af )$.
Let $a \in J$.
Choose $a_g \in A$ for $g \in G$
such that $a = \sum_{g \in G} a_g u_g$.
For $g \in G$,
we have $a u_g^* \in I$,
so $a_g = E (a u_g^*) \in I$
by the previous claim.
Therefore $a = \sum_{g \in G} a_g u_g \in C^* ( G, I, \af )$.
\end{proof}
The \rp{} for finite groups
was used in noncommutative von Neumann algebras
before it was used in noncommutative \ca{s}.
It was first used there for the
purpose of classification
of actions on the hyperfinite factor
of type ${\mathrm{II}}_1$~\cite{Jns}.
In that situation,
pointwise outerness implies the \rp,
which is far from the case for \ca{s}.
The \rp{} has been used for classification of actions on \ca{s};
for a brief survey,
in~\cite{PhSvy} see Theorems 2.102.13 and the preceding
discussion.
That it has strong consequences for classification
of crossed products was only
realized very late.
We show (Theorem 2.2 of~\cite{PhT1})
that if $G$ is finite, $A$ is a unital AF~algebra,
and $\af \colon G \to \Aut (A)$ has the \rp,
then $C^* (G, A, \af)$ is~AF.
Thus, crossed products by actions with the \rp{} preserve
classifiability in the sense of Elliott's original
AF~algebra classification
theorem~\cite{Ell3}.
\begin{thm}[Theorem~2.2 of~\cite{PhT1}]\label{T_RokhAF}
Let $A$ be a unital AF~algebra.
Let $\af \colon G \to \Aut (A)$
be an action of a finite group $G$ on $A$ which has the \rp.
Then $C^* (G, A, \af)$ is an AF~algebra.
\end{thm}
The basic idea is as follows.
Let $e_g \in A$, for $g \in G$,
be Rokhlin \pj{s}.
Let $u_g \in C^* (G, A, \af)$
be the canonical unitary implementing the automorphism $\af_g$
(Notation~\ref{N:ug}).
Then $w_{g, h} = u_{g h^{1}} e_h$ defines an
approximate system of matrix units in $C^* (G, A, \af)$.
(This formula is derived
from the formula~(\ref{Eq_6X08_DefineMatUnit})
for $v_{g, h}$ in Example~\ref{E_C_Trans}.)
Let $(v_{g, h})_{g, h \in G}$ be a nearby true system
of matrix units.
Using the \hm{} $M_n \otimes e_1 A e_1 \to C^* (G, A, \af)$
given by $v_{g, h} \otimes d \mapsto v_{g, 1} d v_{1, h}$,
one can approximate $C^* (G, A, \af)$ by matrix algebras
over corners of~$A$.
A more detailed discussion is given after the statement
of Lemma~\ref{L_1216PartR}.
We begin with a semiprojectivity lemma
(Lemma~2.1 of~\cite{PhT1}),
whose proof we omit.
The proof uses the kinds of methods
(functional calculus) that go into the proof
of \Lem{L_5Y21_ClosePj},
but is more work.
\begin{lem}\label{L_SemiProjMn}
Let $n \in \N$.
For every $\ep > 0$ there is $\dt > 0$ such that,
whenever
$(e_{j, k})_{1 \leq j, k \leq n}$ is a
system of matrix units for $M_n$,
whenever $B$ is a unital \ca,
and whenever $w_{j, k}$, for $1 \leq j, k \leq n$,
are elements of $B$ such that
$\ w_{j, k}^*  w_{k, j} \ < \dt$ for $1 \leq j, k \leq n$,
such that
$\ w_{j_1, k_1} w_{j_2, k_2}  \dt_{j_2, k_1} w_{j_1, k_2} \ < \dt$
for $1 \leq j_1, j_2, k_1, k_2 \leq n$,
and such that the $w_{j, j}$ are orthogonal \pj{s}
with $\sum_{j = 1}^n w_{j, j} = 1$,
then there exists a unital \hm{} $\ph \colon M_n \to B$
such that $\ph (e_{j, j}) = w_{j, j}$ for $1 \leq j \leq n$
and $\ \ph (e_{j, k})  w_{j, k} \ < \ep$ for $1 \leq j, k \leq n$.
\end{lem}
\begin{exr}\label{E_6X01_ProveSj}
Prove \Lem{L_SemiProjMn}.
\end{exr}
Whenever we have a unital \hm{} $\ps \colon M_n \to A$,
then $A$ has a tensor factorization as $M_n \otimes B$,
in which $B$ is the corner of~$A$ corresponding to
the image under~$\ps$ of a rank one \pj{} in~$M_n$.
\begin{lem}\label{L0108MnSplit}
Let $A$ be a unital \ca,
let $S$ be a finite set,
and let $\ph_0 \colon L (l^2 (S)) \to A$ be a unital \hm.
Let $(v_{s, t})_{s, t \in S}$
be the standard system of matrix units in $L (l^2 (S))$
(as in Notation~\ref{N_MatUnit},
except that they were called $e_{j, k}$ there).
Let $s_0 \in S$, and set $e = \ph (v_{s_0, s_0})$.
Then there is an isomorphism
$\ph \colon L (l^2 (S)) \otimes e A e \to A$
such that for all $s, t \in S$ and $a \in e A e$,
we have
\[
\ph (v_{s, t} \otimes a) = \ph_0 (v_{s, s_0}) a \ph_0 (v_{s_0, t}).
\]
\end{lem}
\begin{proof}
To check that there is such a \hm,
it suffices to show that
\[
\big[ \ph_0 (v_{s, s_0}) a \ph_0 (v_{s_0, t}) \big]^*
= \ph_0 (v_{t, s_0}) a^* \ph_0 (v_{s_0, s})
\]
for $s, t \in S$ and $a \in e A e$,
and that
\begin{align*}
&
\big[ \ph_0 (v_{s_1, s_0}) a_1 \ph_0 (v_{s_0, t_1}) \big]
\big[ \ph_0 (v_{s_2, s_0}) a_2 \ph_0 (v_{s_0, t_2}) \big]
\\
& \hspace*{8em} {\mbox{}}
= \begin{cases}
\big[ \ph_0 (v_{s_1, s_0}) a_1 a_2 \ph_0 (v_{s_0, t_2}) \big]
& s_2 = t_1
\\
0 & s_2 \neq t_1
\end{cases}
\end{align*}
for $s_1, s_2, t_1, t_2 \in S$ and $a_1, a_2 \in e A e$.
Both these are immediate.
For surjectivity,
let $a \in A$.
Then one easily checks that
\[
\ph \left( \ssum{s, t \in S}
v_{s, t} \otimes \ph_0 (v_{s_0, s}) a \ph_0 (v_{t, s_0}) \right)
= a.
\]
For injectivity,
suppose that $a_{s, t} \in e A e$ for $s, t \in S$,
and that
\[
\ph \left( \ssum{s, t \in S} v_{s, t} \otimes a_{s, t} \right) = 0.
\]
For $s, t \in S$,
multiply this equation on the left by $\ph_0 (v_{s_0, s})$
and on the right by $\ph_0 (v_{t, s_0})$
to get
\[
0 = \ph (v_{s_0, s_0}) a_{s, t} \ph (v_{s_0, s_0})
= a_{s, t}.
\]
Since this is true for all $s, t \in S$,
injectivity of~$\ph$ follows.
\end{proof}
\begin{lem}\label{L_1216PartR}
Let $G$ be a finite group, and set $n = \card (G)$.
Then for every $\ep > 0$ there is $\dt > 0$
such that the following holds.
Let $\GAa$ be a unital \ga,
let $(e_g)_{g \in G}$ be a family of orthogonal \pj{s} in~$A$,
and let $F \subset A$ be a finite set
such that $\ a \ \leq 1$ for all $a \in F$.
Suppose that:
\begin{enumerate}
\item\label{L_1216PartR:1}
$\ \af_g (e_h)  e_{g h} \ < \dt$ for all $g, h \in G$.
\item\label{L_1216PartR:2}
$\ e_g a  a e_g \ < \dt$ for all $g \in G$ and all $a \in F$.
\item\label{L_1216PartR:3}
$\sum_{g \in G} e_g = 1$.
\end{enumerate}
For $g \in G$
let $u_g$ be the standard unitary of Notation~\ref{N:ug}.
Then there exists a unital \hm{}
$\ph \colon M_n \otimes e_1 A e_1 \to \CGAa$
such that for every $a \in F \cup \{ u_g \colon g \in G \}$,
we have
$\dist \big( a, \, \ph \big( L (l^2 (G)) \otimes e_1 A e_1 \big) \big)
< \ep$,
and such that (using standard matrix unit notation)
for every $a \in e_1 A e_1$ we have $\ph (e_{1, 1} \otimes a) = a$.
\end{lem}
To make clear what is happening,
suppose that in the hypotheses of \Lem{L_1216PartR}
we actually had:
\begin{enumerate}
\item\label{L_1216PartRSpecial:1}
$\af_g (e_h) = e_{g h}$ for all $g, h \in G$.
\item\label{L_1216PartRSpecial:2}
$e_g a = a e_g$ for all $g \in G$ and all $a \in F$.
\item\label{L_1216PartRSpecial:3}
$\sum_{g \in G} e_g = 1$.
\end{enumerate}
We use $L (l^2 (G))$ instead of~$M_n$.
The same computation as in Example~\ref{E_C_Trans}
(where we showed that if $G$ is discrete then
$C^* (G, \, C_0 (G)) \cong K (l^2 (G))$)
shows that if we define $w_{g, h} \in C^* (G, A, \af)$
by $w_{g, h} = e_g u_{g h^{1}}$
for $g, h \in G$
(compare with equation~(\ref{Eq_6X08_DefineMatUnit})),
then the $w_{g, h}$ form a system of matrix units in $C^* (G, A, \af)$.
That is, letting $(v_{g, h})_{g, h \in G}$
be the standard system of matrix units in $L (l^2 (G))$
(as in Notation~\ref{N_MatUnit},
except that they were called $e_{j, k}$ there),
there is a unital \hm{} $\ph_0 \colon L (l^2 (G)) \to C^* (G, A, \af)$
such that $\ph_0 (v_{g, h}) = w_{g, h}$ for all $g, h \in G$.
The elements $u_g$ are already in the range of~$\ph_0$.
Indeed,
we have $w_{h, \, g^{1} h} = e_h u_g$,
so
\[
u_g
= \sum_{h \in G} e_h u_g
= \ph_0 \left( \ssum{h \in G} v_{h, \, g^{1} h} \right).
\]
Since $\ph_0 (v_{1, 1}) = e_1 \in A$,
we have $e_1 A e_1 \subset e_1 C^* (G, A, \af) e_1$,
and we can apply \Lem{L0108MnSplit}
to get a unital \hm{}
$\ph \colon L (l^2 (G)) \otimes e_1 A e_1 \to C^* (G, A, \af)$.
Suppose now that
$a \in A$ commutes with $e_g$ for all $g \in G$.
Then
\[
a = \sum_{g \in G} e_g a e_g
= \sum_{g \in G} \af_g \big( e_1 \af_g^{1} (a) e_1 \big).
\]
Applying the formula for $\ph$ in \Lem{L0108MnSplit},
we get
\begin{align*}
\ph \left( \ssum{g \in G}
v_{g, g} \otimes e_1 \af_g^{1} (a) e_1 \right)
& = \sum_{g \in G} e_g u_g e_1 \af_g^{1} (a) e_1 u_g^* e_g
\\
& = \sum_{g \in G} e_g \af_g \big( e_1 \af_g^{1} (a) e_1 \big) e_g
\\
& = \sum_{g \in G} \af_g \big( e_1 \af_g^{1} (a) e_1 \big)
= a.
\end{align*}
In the actual proof,
many of the equations in the computations
above become statements that the norm of the
difference between the two sides is small.
\begin{proof}[Proof of \Lem{L_1216PartR}]
We will use $L (l^2 (G))$ instead of~$M_n$;
the lemma as stated will follow by choosing a bijection
from $G$ to $\{ 1, 2, \ldots, n \}$.
As will be seen later,
we require that this bijection
send the identity of~$G$ to~$1$.
Set $\ep_0 = \ep / (4 n)$.
Choose $\dt > 0$ according to Lemma~\ref{L_SemiProjMn}
for $n$ as given and for $\ep_0$ in place of $\ep$.
Also require $\dt \leq \ep / [2 n (n + 1)]$.
Assume that $(e_g)_{g \in G}$ is a family of orthogonal \pj{s} in~$A$
and that $F \subset A$ is a finite set
such that the hypotheses
(\ref{L_1216PartR:1}), (\ref{L_1216PartR:2}),
and~(\ref{L_1216PartR:3})
of the lemma
hold for this value of~$\dt$.
Define $w_{g, h} = e_g u_{g h^{1}}$ for $g, h \in G$.
We claim that the $w_{g, h}$ form a $\dt$approximate system of
$n \times n$ matrix units in $C^* (G, A, \af)$.
We estimate:
\[
\ w_{g, h}^*  w_{h, g} \
= \ u_{g h^{1}}^* e_g  e_h u_{h g^{1}} \
= \ e_g  u_{g h^{1}} e_h u_{g h^{1}}^* \
= \ e_g  \af_{g h^{1}} (e_h) \
< \dt.
\]
Also, using $e_g e_h = \dt_{g, h} e_h$ for $g, h \in G$
at the second step,
\begin{align*}
\big\ w_{g_1, h_1} w_{g_2, h_2}  \dt_{g_2, h_1} w_{g_1, h_2} \big\
& = \big\ e_{g_1} u_{g_1 h_1^{1}} e_{g_2} u_{g_2 h_2^{1}}
 \dt_{g_2, h_1} e_{g_1} u_{g_1 h_2^{1}} \big\ \\
& = \big\ e_{g_1} u_{g_1 h_1^{1}} e_{g_2} u_{g_2 h_2^{1}}
 e_{g_1} e_{g_1 h_1^{1} g_2} u_{g_1 h_1^{1} g_2 h_2^{1}} \big\
\\
& = \big\ e_{g_1}
\big( u_{g_1 h_1^{1}} e_{g_2} u_{g_1 h_1^{1}}^*
 e_{g_1 h_1^{1} g_2} \big)
u_{g_1 h_1^{1} g_2 h_2^{1}} \big\
\\
& = \big\ e_{g_1}
\big( \af_{g_1 h_1^{1}} (e_{g_2})
 e_{g_1 h_1^{1} g_2} \big)
u_{g_1 h_1^{1} g_2 h_2^{1}} \big\
< \dt.
\end{align*}
Finally, $\sum_{g \in G} w_{g, g} = \sum_{g \in G} e_{g} = 1$.
This proves the claim.
Let $(v_{g, h})_{g, h \in G}$
be the standard system of matrix units in $L (l^2 (G))$
(as in Notation~\ref{N_MatUnit},
except that they were called $e_{j, k}$ there).
By the choice of $\dt$,
there exists a unital \hm{}
$\ph_0 \colon L (l^2 (G)) \to C^* (G, A, \af)$
such that
$\ \ph_0 (v_{g, h})  w_{g, h} \ < \ep_0$ for all $g, h \in G$,
and $\ph_0 (v_{g, g}) = e_g$ for all $g \in G$.
Since $\ph_0 (v_{1, 1}) = e_1 \in A \subset C^* (G, A, \af)$,
we can restrict the \hm{} of \Lem{L0108MnSplit}
from $L (l^2 (G)) \otimes e_1 C^* (G, A, \af) e_1$
to the subalgebra $L (l^2 (G)) \otimes e_1 A e_1$.
We get a unital \hm{}
$\ph \colon L (l^2 (G)) \otimes e_1 A e_1 \to C^* (G, A, \af)$
such that
$\ph (v_{g, h} \otimes a) = \ph_0 (v_{g, 1}) a \ph_0 (v_{1, h})$
for $g, h \in G$ and $a \in e_1 A e_1$.
Since $\ph_0 (v_{1, 1}) = e_1$
and we identify $1 \in G$ with $1 \in \{ 1, 2, \ldots, n \}$,
the relation $\ph \big( e_{1, 1} \otimes a \big) = a$ for $a \in e_1 A e_1$
is immediate.
We complete the proof by
showing that every element of~$S$
is within~$\ep$ of an element of
the algebra $D = \ph \big( L (l^2 (G)) \otimes e_1 A e_1 \big)$.
For $g \in G$ we have $\sum_{h \in G} \ph_0 (v_{g h, h}) \in D$
and, using $\sum_{h \in G} e_h = 1$,
\begin{align*}
\left\ u_g  \ssum{h \in G} \ph_0 (v_{h, g^{1} h}) \right\
& \leq \sum_{h \in G} \ e_h u_g  \ph_0 (v_{h, g^{1} h}) \ \\
& = \sum_{h \in G} \ w_{h, g^{1} h}  \ph_0 (v_{h, g^{1} h}) \
< n \ep_0
\leq \ep.
\end{align*}
Now let $a \in F$.
Set
\[
b = \sum_{g \in G} v_{g, g} \otimes e_1 \af_g^{1} (a) e_1
\in M_n \otimes e_1 A e_1.
\]
Using
$\ e_g a e_h \
\leq \ e_g a  a e_g \ + \ a e_g e_h \
= \ e_g a  a e_g \$
at the third step,
we get
%
\begin{align}\label{Eq_6X08_Star}
\left\ a  \ssum{g \in G} e_g a e_g \right\
& = \left\ \ssum{g, h \in G} e_g a e_h
 \ssum{g \in G} e_g a e_g \right\
\\
& \leq \sum_{g \neq h} \ e_g a e_h \
< n (n  1) \dt.
\notag
\end{align}
%
Combining $\ e_1  \af_g^{1} (e_g)  < \dt$ for all $g \in G$
and $\ a \ \leq 1$ for all $a \in F$,
we get
%
\begin{equation}\label{Eq_6X08_StSt}
\big\ e_1 \af_g^{1} (a) e_1  \af_g^{1} (e_g a e_g) \big\ < 2 \dt.
\end{equation}
%
Also,
for $g \in G$ we have, taking adjoints at the first step,
%
\begin{equation}\label{Eq_6X08_StStSt}
\ \ph_0 (v_{g, 1}) e_1  u_g e_1 \
= \ e_1 \ph_0 (v_{1, g})  e_1^2 u_g^* \
\leq \ e_1 \ \ \ph_0 (v_{1, g})  w_{1, g} \
< \ep_0.
\end{equation}
%
For $a \in F$,
we use (\ref{Eq_6X08_StStSt}) and $\ a \ \leq 1$
at the second step,
(\ref{Eq_6X08_StSt}) at the third step,
and (\ref{Eq_6X08_Star}) at the fifth step,
to get
\begin{align*}
\ a  \ph (b) \
& = \left\ a  \ssum{g \in G}
\ph_0 (v_{g, 1}) e_1 \af_g^{1} (a) e_1 \ph_0 (v_{1, g}) \right\
\\
& < 2 n \ep_0
+ \left\ a  \ssum{g \in G}
u_g e_1 \af_g^{1} (a) e_1 u_g^* \right\
\\
& < 2 n \ep_0 + 2 n \dt
+ \left\ a  \ssum{g \in G}
u_g \af_g^{1} (e_g a e_g) u_g^* \right\
\\
& = 2 n \ep_0 + 2 n \dt
+ \left\ a  \ssum{g \in G} e_g a e_g \right\
\\
& < 2 n \ep_0 + 2 n \dt + n (n  1) \dt
\leq \ep.
\end{align*}
This completes the proof.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{T_RokhAF}]
We prove that for every finite set $S \subset C^* (G, A, \af)$
and every $\ep > 0$,
there is an AF~subalgebra $D \subset C^* (G, A, \af)$
such that every element of~$S$ is within~$\ep$ of an element of~$D$.
It is then easy to use Theorem~2.2 of~\cite{Brt} to show that
$C^* (G, A, \af)$ is~AF.
It suffices to fix a set $T$ which generates $\CGAa$
as a \ca,
and to consider only finite subsets $S \subset T$.
Thus, we need only consider
$S$ of the form
$S = F \cup \{ u_g \colon g \in G \}$,
where $F$ is a finite subset of the unit ball of $A$ and
$u_g \in C^* (G, A, \af)$
is the canonical unitary implementing the automorphism $\af_g$
(Notation~\ref{N:ug}).
So let $F \subset A$ be a finite subset
with $\ a \ \leq 1$ for all $a \in F$ and let $\ep > 0$.
Choose $\dt > 0$ as in \Lem{L_1216PartR}
for $\ep$ as given.
Apply the \rp{} to $\af$ with $F$ as given
and with $\dt$ in place of $\ep$,
obtaining \pj{s} $e_g \in A$ for $g \in G$.
Define $w_{g, h} = e_g u_{g h^{1}}$ for $g, h \in G$.
Set $n = \card (G)$,
and let $\ph \colon M_n \otimes e_1 A e_1 \to A$
be the \hm{} of \Lem{L_1216PartR}.
It is well known that a corner of an AF~algebra is~AF,
and that a quotient of an AF~algebra is~AF,
so $D = \ph ( M_n \otimes e_1 A e_1 )$
is an AF~subalgebra of $C^* (G, A, \af)$
such that for every $a \in F \cup \{ u_g \colon g \in G \}$,
we have
$\dist \big( a, \, \ph ( M_n \otimes e_1 A e_1 ) \big) < \ep$.
This completes the proof.
\end{proof}
We summarize a number of other theorems
related to Theorem~\ref{T_RokhAF}
which have been proved in~\cite{OP3}
and elsewhere.
Crossed products by actions of finite groups with the Rokhlin property
preserve the following classes of \ca{s}:
\begin{enumerate}
\item\label{CRP_6310_DLim}
Various other classes of unital but not necessarily simple countable
direct limit \ca{s} using semiprojective building blocks,
and in which the maps of the direct system need not be injective:
\begin{enumerate}
\item\label{CRP_6310_DL_AI}
AI~algebras (Corollary~3.6(1) of~\cite{OP3}).
\item\label{CRP_6310_DL_AT}
AT~algebras (Corollary~3.6(2) of~\cite{OP3}).
\item\label{CRP_6310_DL_NCCW}
Unital direct limits of one dimensional noncommutative CW~complexes
(Corollary~3.6(4) of~\cite{OP3}).
\item\label{CRP_6310_DL_Tp}
Unital direct limits of Toeplitz algebras,
a special case
of the sort studied in~\cite{LnSu}
except not necessarily of real rank zero
(Example~2.10 and Theorem~3.5 of~\cite{OP3}).
\item\label{CRP_6310_DL_Other}
Various other classes;
see Section~2 and Theorem~3.5 of~\cite{OP3} for details.
\end{enumerate}
\item\label{CRP_6310_SAH}
Simple unital AH~algebras with slow dimension growth and real rank zero
(Theorem~3.10 of~\cite{OP3}).
\item\label{CRP_6310_Abs}
$D$absorbing separable unital \ca{s} for a strongly selfabsorbing
\ca~$D$ (Theorem~1.1(1) and Corollary~3.4(i) of~\cite{HrWn}).
(See~\cite{HrWn} for the definition of
a strongly selfabsorbing \ca.)
\item\label{CRP_6310_RR0}
Unital \ca{s} with real rank zero (Proposition~4.1(1) of~\cite{OP3}).
\item\label{CRP_6310_tsr1}
Unital \ca{s} with stable rank one (Proposition~4.1(2) of~\cite{OP3}).
\item\label{CRP_6310_UCT}
Separable nuclear unital \ca{s} whose quotients all satisfy the
Universal Coefficient Theorem (Proposition~3.7 of~\cite{OP3}).
\item\label{CRP_6310_Kbg}
Unital Kirchberg algebras satisfying the Universal Coefficient Theorem
(Corollary~3.11 of~\cite{OP3}).
\item\label{CRP_6310_ADiv}
Separable unital approximately divisible \ca{s}
(Corollary~3.4(2) of \cite{HrWn},
which also covers actions of compact groups;
also see Proposition~4.5 of~\cite{OP3}).
\item\label{CRP_6310_PjProp}
Unital \ca{s} with the ideal property and unital \ca{s} with
the projection property
(\cite{PsnPh}; also see~\cite{PsnPh}
for the definitions of these properties).
\item\label{CRP_6310_KThy}
Simple unital \ca{s} whose Ktheory:
\begin{enumerate}
\item\label{CRP_6310_KThy_1}
Is torsion free.
\item\label{CRP_6310_KThy_2}
Is a torsion group.
\item\label{CRP_6310_KThy_3}
Is zero.
\end{enumerate}
(Theorem 2.6(11) of~\cite{PhSvy};
the proof of this part
is in the discussion after Theorem~2.7 of~\cite{PhSvy}.
The main part of the proof comes from Theorem~3.13 of~\cite{Izm}.)
\end{enumerate}
% (More from~\cite{PsnPh}?) 999
We now give some examples to show that the \rp{} is rare.
The proofs depend implicitly or explicitly on Ktheory.
For the first three,
we use the restriction in the next lemma.
\begin{lem}\label{C_1211_RpU}
Let $A$ be a unital \ca{} with a unique \tst~$\ta$,
let $G$ be a finite group,
and let $\af \colon G \to \Aut (A)$
be an action of $G$ on $A$ which has the \rp.
Then there exists a \pj{} $p \in A$
such that $\ta (p) = \card (G)^{1}$.
\end{lem}
\begin{proof}
In Definition~\ref{D_RPDfn},
take $\ep = 1$ and $F = \varnothing$.
We get \pj{s} $e_g$ for $g \in G$ such that, in particular:
\begin{enumerate}
\item\label{C1211RpU1}
$\ \af_g (e_1)  e_g \ < 1$ for all $g \in G$.
\item\label{C1211RpU2}
$\sum_{g \in G} e_g = 1$.
\end{enumerate}
It follows from~(\ref{C1211RpU1}) and \Lem{L_5Y21_ClosePj}
that $\af_g (e_1)$ is \mvnt{} to~$e_g$ for all $g \in G$.
Therefore $\ta (e_g) = \ta ( \af_g (e_1) )$
by Lemma \ref{LTraceAndMvN}(\ref{LTraceAndMvN1}).
Since $\ta$ is unique,
we have $\ta \circ \af_g = \ta$ for all $g \in G$.
So $\ta (e_g) = \ta (e_1)$ for all $g \in G$.
It follows that
\[
\ta (e_1) = \frac{1}{\card (G)}.
\]
This completes the proof.
\end{proof}
Existence of an action of $G$ with the \rp{} implies much
stronger restrictions on the Ktheory
than are suggested by this result,
or by the methods used in \Prp{E_1211_NoRPOI}
and \Prp{E_1211_NoRPO3} below.
See Theorem~3.2 of~\cite{Iz2}.
\begin{exa}\label{E_1211_NoRPAT}
Let $\te \in \R \setminus \Q$,
and let $A_{\te}$ be the rotation algebra,
as in Example~\ref{E_3303_RotAlg}.
It is known (Proposition VI.1.3 of~\cite{Dvd})
that $A_{\te}$ has a unique \tst~$\ta$.
Moreover, for every \pj{} $p \in A_{\te}$,
one has $\ta (p) \in \Z + \te \Z \subset \R$.
(See Theorem VI.5.2 of~\cite{Dvd};
for less technical proofs
relying on the PimsnerVoiculescu exact sequence in Ktheory,
see the Appendix in~\cite{PV}
and the general theory developed in~\cite{Ex2},
specifically Example IX.12 there.)
In particular,
there is no $n \in \N$ with $n \geq 2$ such that there is
a \pj{} $p \in A_{\te}$ with $\ta (p) = \tfrac{1}{n}$.
It follows from Lemma~\ref{C_1211_RpU}
that there is no action of any nontrivial finite group
on~$A_{\te}$ which has the \rp.
\end{exa}
\begin{exa}\label{E_1211_NoRPUHF}
Let $A$ be the $2^{\infty}$~UHF algebra.
Then $A$ has a unique \tst~$\ta$.
Moreover, for every \pj{} $p \in A$,
one has $\ta (p) \in \Z \big[ \tfrac{1}{2} \big] \subset \R$.
(This is really a statement in Ktheory,
but there is enough in Section~\ref{Sec_FGStr}
to give a direct proof.
See below.)
It follows from Lemma~\ref{C_1211_RpU}
that the only finite groups~$G$
which can possibly have actions on~$A$ with the Rokhlin property
are groups whose order is a power of~$2$.
In particular, there is no action of $\Z / 3 \Z$
on~$A$ which has the Rokhlin property.
We prove the statement about traces of projections.
Let $p \in A$ be a \pj.
Choose $\dt > 0$ as in \Lem{L_5Y21_PjInSbalg} for $\ep = 1$.
By the direct limit description of~$A$,
there are $n \in \Nz$
and a unital subalgebra $B \subset A$
with $B \cong M_{2^n}$
such that $\dist (p, B) < \dt$.
By the choice of $\dt$ using \Lem{L_5Y21_PjInSbalg},
there is a \pj{} $q \in B$
such that $\ p  q \ < 1$.
We have $p \sim q$ by \Lem{L_5Y21_ClosePj},
so $\ta (p) = \ta (q)$
by Lemma \ref{LTraceAndMvN}(\ref{LTraceAndMvN1}).
The restriction of $\ta$ to $B$ must be the
normalized trace on $M_{2^n}$,
so $\ta (q)$ is an integer multiple of $\frac{1}{2^n}$.
Therefore so is $\ta (p)$, as desired.
\end{exa}
\begin{exa}\label{E_1211_NoRPUHF_2On3}
The same reasoning as in \Ex{E_1211_NoRPUHF}
shows that there is no action of $\Z / 2 \Z$
on the $3^{\infty}$~UHF algebra which has the Rokhlin property.
\end{exa}
The next two examples depend much more heavily on Ktheory,
and we therefore assume basic knowledge of Ktheory.
\begin{prp}\label{E_1211_NoRPOI}
There is no action of any nontrivial finite group
on~${\mathcal{O}}_{\infty}$ which has the \rp.
\end{prp}
\begin{proof}
Let $G$ be a nontrivial finite group,
and let $\af \colon G \to \Aut ( {\mathcal{O}}_{\infty} )$
be an action with the \rp.
The computation $K_0 ( {\mathcal{O}}_{\infty} ) \cong \Z$
is Corollary 3.11 of~\cite{Ctz}.
The fact that $[1]$ is a generator can be read from the
proof there and the proof of Proposition 3.9 of~\cite{Ctz}.
It follows that every automorphism of ${\mathcal{O}}_{\infty}$
is the identity on $K_0 ( {\mathcal{O}}_{\infty} )$.
Now apply Definition~\ref{D_RPDfn}
with $F = \E$ and $\ep = \frac{1}{2}$.
We get \pj{s} $e_g \in {\mathcal{O}}_{\infty}$
for $g \in G$ such that
$\ e_g  \af_g (e_1) \ < \frac{1}{2}$ for $g \in G$
and $\sum_{g \in G} e_g = 1$.
By Lemma~\ref{L_5Y21_ClosePj},
the inequality implies $[e_g] = [\af_g (e_1)]$
in $K_0 ( {\mathcal{O}}_{\infty} )$;
since $\af_g$ is the identity on $K_0 ( {\mathcal{O}}_{\infty} )$,
it follows that $[e_g] = [e_1]$.
{}From $\sum_{g \in G} e_g = 1$ we therefore get
$\card (G) [e_1] = [1]$ in $K_0 ( {\mathcal{O}}_{\infty} )$.
Since $K_0 ( {\mathcal{O}}_{\infty} ) \cong \Z$
via $n \mapsto n [1]$,
this is a contradiction.
\end{proof}
\begin{prp}\label{E_1211_NoRPO3}
The only finite groups~$G$
which can possibly have actions on~${\mathcal{O}}_3$
with the Rokhlin property
are groups of odd order.
\end{prp}
\begin{proof}
The proof is similar to that of Proposition~\ref{E_1211_NoRPOI}.
Let $G$ be a finite group with even order,
and let $\af \colon G \to \Aut ( {\mathcal{O}}_{3} )$
be an action with the \rp.
The computation $K_0 ( {\mathcal{O}}_{n} ) \cong \Z / (n  1) \Z$
is Theorem 3.7 of~\cite{Ctz};
in the proof it is shown that $[1]$ is a generator.
It follows that every automorphism of ${\mathcal{O}}_{n}$
is the identity on $K_0 ( {\mathcal{O}}_{n} )$.
In particular,
$K_0 ( {\mathcal{O}}_{3} ) \cong \Z / 2 \Z$
and $(\af_g)_{*}$ is the identity on $K_0 ( {\mathcal{O}}_{3} )$
for all $g \in G$.
Apply Definition~\ref{D_RPDfn}
with $F = \E$ and $\ep = \frac{1}{2}$.
As in the proof of Proposition~\ref{E_1211_NoRPOI},
one gets $\card (G) [e_1] = [1]$ in $K_0 ( {\mathcal{O}}_{3} )$.
Since $\card (G)$ is even,
this implies $[1] = 0$ in $K_0 ( {\mathcal{O}}_{3} )$,
a contradiction.
\end{proof}
\section{The Tracial Rokhlin Property for Actions of Finite
Groups}\label{Sec_TRPFG}
\indent
As was discussed in Section~\ref{Sec_RPFG},
there are very few actions of finite groups
which have the Rokhlin property.
The \trp{} (Definition~\ref{D_TRP} below)
is much more common.
The differences are discussed in several places
in Section~3 of~\cite{PhSvy},
and an illuminating example is given in Exercise~\ref{Ex:Z2Struct}.
The \trp{} is still very useful in classification.
Indeed, if $G$ is finite, $A$ is a simple unital \ca{} with
tracial rank zero in the sense of Lin
(originally called ``tracially AF''; see Definition~\ref{D_TR0} above),
and $\af \colon G \to \Aut (A)$ has the \trp,
then $C^* (G, A, \af)$ has tracial rank zero.
This is Theorem~2.6 of~\cite{PhT1}
(\Thm{RokhTAF} below).
Lin has proved (Theorem~5.2 of~\cite{Ln3})
that simple separable unital nuclear \ca{s} with tracial rank zero
and which satisfy the Universal Coefficient Theorem
are classifiable.
Thus, this result can be used for classification purposes.
For example, it played a key role in the proof~\cite{ELPW} that the
crossed products by the actions of Example~\ref{E:SL2}
are AF
(except that this was known earlier for the action of $\Z / 2 \Z$).
It is not true the crossed products of simple unital AF~algebras
by actions of finite groups with the \trp{}
are~AF.
See Sections 3 and~4 of~\cite{PhT4}.
The actions used are those in \Ex{E_3502_Blackadar},
Exercise~\ref{Ex:RCFG4_K1},
and Exercise~\ref{Ex:RCFG4_Torsion}.
Ironically,
Theorem~2.6 of~\cite{PhT1}
was proved {\emph{before}}
Theorem~2.2 of~\cite{PhT1}
(the AF algebra and Rokhlin property version).
It is presumably not sufficient for classification
purposes
to just consider pointwise outer actions.
Example~\ref{E_Elliott}
shows that the \cp{} by a pointwise outer action of a
finite group on a simple unital AF~algebra
need not have real rank zero;
in particular, by Theorem~\ref{T_6X02_TRZToRR},
it need not have tracial rank zero.
Example~\ref{E_6X02_Blackadar}
shows that the \cp{} by a pointwise outer action of a
finite group on a nonsimple unital \ca{} with stable rank one
need not have stable rank one.
These results suggest that \cp{s}
by pointwise outer actions of finite
groups might well not respect classifiability,
although no examples are known.
Although we will not pursue this direction in these notes,
there are useful weakenings of the \trp{}
which are stronger than pointwise outerness.
For example, see Definition~5.2 of~\cite{HirOr2013}
(and also Definition~6.1 of~\cite{HirOr2013}
for actions of~$\Z$ and~\cite{OvPhWn}
for actions of countable amenable groups).
\begin{dfn}[Definition~1.2 of~\cite{PhT1}]\label{D_TRP}
Let $G$ be a finite group,
let $A$ be an infinite dimensional simple \uca,
and let $\af \colon G \to \Aut (A)$ be an
action of $G$ on~$A$.
We say that $\af$ has the
{\emph{tracial Rokhlin property}} if for every finite set
$F \subset A$, every $\ep > 0$, and every positive element $x \in A$
with $\ x \ = 1$,
there are nonzero \mops{} $e_g \in A$ for $g \in G$
such that:
\begin{enumerate}
\item\label{D_TRP_1}
$\ \af_g (e_h)  e_{g h} \ < \ep$ for all $g, h \in G$.
\item\label{D_TRP_2}
$\ e_g a  a e_g \ < \ep$ for all $g \in G$ and all $a \in F$.
\item\label{D_TRP_3}
With $e = \sum_{g \in G} e_g$, the \pj{} $1  e$ is \mvnt{} to a
\pj{} in the \hsa{} of $A$ generated by~$x$.
\item\label{D_TRP_4}
With $e$ as in~(\ref{D_TRP_3}), we have $\ e x e \ > 1  \ep$.
\end{enumerate}
\end{dfn}
When $A$ is finite, the last condition is redundant.
(See Lemma~1.16 of~\cite{PhT1},
which is \Lem{TRPForFinite} below.)
However, without it, the trivial action on ${\mathcal{O}}_2$
would have the \trp.
(It is, however,
not clear that this condition is really the right extra
condition to impose.)
Without the requirement that the algebra be in\fd,
the trivial action on $\C$ would have the \trp{}
(except for the condition~(\ref{D_TRP_4})),
for the rather silly reason that
the \hsa{} in Condition~(\ref{D_TRP_3})
can't be ``small''.
% For unital Kirchberg algebras,
% the \trp{} is equivalent to pointwise outerness.
% See~\cite{?}.
% 999
\begin{rmk}[Remark~1.4 of~\cite{PhT1}]\label{RSRPImpTRP}
Let $G$ be a finite group,
let $A$ be an \idssuca,
and let $\af \colon G \to \Aut (A)$ be an action of $G$ on~$A$.
If $\af$ has the
Rokhlin property, then $\af$ has the \trp.
\end{rmk}
\begin{lem}[Lemma~1.13 of~\cite{PhT1}]\label{RImpSP}
Let $G$ be a finite group,
let $A$ be an \idssuca,
and let $\af \colon G \to \Aut (A)$ be an action
which has the \trp.
If $A$ does not have property~(SP),
then $\af$ has the Rokhlin property.
\end{lem}
\begin{proof}
Suppose $A$ does not have property~(SP).
Then there is $x \in A_{+} \setminus \{ 0 \}$
which generates a \hsa{} which contains no nonzero \pj{s}.
So the \pj{} $e$ in condition~(\ref{D_TRP_3}) must be equal to~$1$.
\end{proof}
As with the Rokhlin property
(see Theorem~\ref{T_ExactRP}),
one can strengthen the statement.
\begin{thm}[Proposition 5.27 of~\cite{PhEqSj}]\label{T_ExactTRP}
Let $G$ be a finite group,
let $A$ be an \idssuca,
and let $\af \colon G \to \Aut (A)$ be an action of $G$ on~$A$.
Then $\af$ has the
tracial Rokhlin property \ifo{} for every finite set
$F \subset A$ and every $\ep > 0$,
there are \mops{} $e_g \in A$ for $g \in G$ such that:
\begin{enumerate}
\item\label{D:TRP:E1}
$\af_g (e_h) = e_{g h}$ for all $g, h \in G$.
\item\label{D:TRP:E2}
$\ e_g a  a e_g \ < \ep$ for all $g \in G$ and all $a \in F$.
\item\label{D:TRP:E3}
With $e = \sum_{g \in G} e_g$, the \pj{} $1  e$ is \mvnt{} to a
\pj{} in the \hsa{} of $A$ generated by~$x$.
\item\label{D:TRP:E4}
With $e$ as in~(\ref{D:TRP:E3}), we have $\ e x e \ > 1  \ep$.
\end{enumerate}
\end{thm}
The proof uses the same methods as that of Theorem~\ref{T_ExactRP},
and,
for the same reasons as given in the discussion there,
we do not use the stronger condition in these notes.
We give two other conditions for the tracial Rokhlin property.
We omit both proofs.
The first uses an an assumption on comparison
of \pj{s} using traces
to substitute an estimate on the trace of the error \pj{}
for condition~(\ref{D_TRP_3})
in \Def{D_TRP},
and finiteness and \Lem{TRPForFinite} below
to omit condition~(\ref{D_TRP_4}).
It is the motivation for the term ``tracial Rokhlin property''.
\begin{prp}[Lemma~5.2 of~\cite{ELPW}]\label{P_6X01_DfnUsingTrace}
Let $G$ be a finite group,
let $A$ be an \idssuca,
and let $\af \colon G \to \Aut (A)$ be an action of $G$ on~$A$.
Assume that $A$ is finite, that $A$ has property~(SP),
and that the order on \pj{s} over $A$ is determined by traces.
Then $\af$ has the
tracial Rokhlin property \ifo{}
for every finite set $S \subset A$ and every $\ep > 0$, there exist
orthogonal \pj{s} $e_g \in A$ for $g \in G$ such that:
\begin{enumerate}
\item\label{P_6X01_DfnUsingTrace_1}
$\ \af_g (e_h)  e_{g h} \ < \ep$ for all $g, h \in G$.
\item\label{P_6X01_DfnUsingTrace_2}
$\ e_g a  a e_g \ < \ep$ for all $g \in G$ and all $a \in S$.
\item\label{P_6X01_DfnUsingTrace_3}
With $e = \sum_{g \in G} e_g$, we have $\ta (1  e) < \ep$
for all $\ta \in \T (A)$.
\end{enumerate}
\end{prp}
The second applies to \ca{s}
with tracial rank zero (\Def{D_TR0})
and a unique \tst.
(There should be an analog without requiring uniqueness
of the \tst,
but we don't know what it is.)
It relates the \trp{} for~$\af$
to the corresponding action~$\af''$
on the factor of type ${\mathrm{II}}_1$
gotten by applying the GelfandNaimarkSegal construction
to the \tst.
Since this factor is hyperfinite,
and since pointwise outer actions of finite groups on
the hyperfinite type ${\mathrm{II}}_1$ factor
necessarily have the von Neumann algebra version of the
Rokhlin property,
it says that $\af$ has the tracial Rokhlin property
\ifo{} $\af''$ has the Rokhlin property.
Its proof proceeds via Theorem~5.3 of~\cite{ELPW},
a criterion for the \trp{}
for an action of a finite group on
a simple \ca{} with tracial rank zero
which looks very similar to \Def{D_TRP}
except that it uses trace norms in place of the usual norm.
This criterion does not depend on uniqueness of the \tst.
\begin{thm}[Theorem~5.5 of~\cite{ELPW}]\label{T_6X01_OuterImpTRP}
Let $G$ be a finite group,
let $A$ be an \idssuca,
and let $\af \colon G \to \Aut (A)$ be an action of $G$ on~$A$.
Assume that $A$ has tracial rank zero
and has a unique \tst{} $\ta$.
Let $\pi_{\ta} \colon A \to B (H_{\ta})$
be the GelfandNaimarkSegal representation associated with $\ta$,
and for $\bt \in \Aut (A)$ let $\bt''$
denote the
automorphism of $\pi_{\ta} (A)''$ determined by $\bt$.
Then $\af$ has the \trp{} \ifo{} %
$\af_g''$ is an outer automorphism of
$\pi_{\ta} (A)''$ for every $g \in G \setminus \{ 1 \}$.
\end{thm}
\Prp{P_6316_RPOut}
is also valid for actions with the \trp,
with essentially the same proof.
\begin{lem}[Lemma~1.5 of~\cite{PhT1}]\label{TRPImpOuter}
Let $G$ be a finite group,
let $A$ be an \idssuca,
and let $\af \colon G \to \Aut (A)$ be an action
which has the \trp.
Then $\af$ is pointwise outer (\Def{D_6320_PtwOut}).
\end{lem}
\begin{proof}
Let $g \in G \setminus \{ 1 \}$;
we prove that $\af_g$ is outer.
So let $u \in A$ be unitary.
Apply Definition~\ref{D_TRP} with $F = \{ u \}$,
with $\ep = \frac{1}{2}$, and with $x = 1$.
Then $e_1$ and $e_g$ are orthogonal nonzero \pj{s}, so
\[
\ \af_g (e_1)  u e_1 u^* \
\geq \ e_g  e_1 \
 \ \af_g (e_1)  e_{g} \  \ u e_1 u^*  e_1 \
> 1  \frac{1}{2}  \frac{1}{2}
= 0.
\]
Therefore $\af_g \neq \Ad (u)$.
Since $u$ is arbitrary, this shows that $\af_g$ is outer.
\end{proof}
We will prove results about the \trp{} below.
First, we give an example for which it is easy to see
(using several of the results below) that the
\trp{} holds,
but where the Rokhlin property fails.
\begin{exa}\label{EEasyTRP}
For $k \in \N$,
define $v_k \in M_{3^k}$ to be the unitary
\[
v_k = \diag
\big(1, \, 1, \, \ldots, \, 1, \, 1, \, 1, \, \ldots, \, 1 \big)
\in M_{3^k},
\]
in which the diagonal entry $1$ occurs $\tfrac{1}{2} (3^k + 1)$ times
and the diagonal entry $ 1$ occurs $\tfrac{1}{2} (3^k  1)$ times.
Set $A = \bigotimes_{k = 1}^{\I} M_{3^k}$,
which is just a somewhat different expression for
the $3^{\infty}$~UHF algebra.
Define
\[
\mu = \bigotimes_{n = 1}^{\I} \Ad (v_k) \in \Aut (A).
\]
Then $\mu$ is an automorphism of order~$2$.
Let $\af \colon \Z / 2 \Z \to \Aut (A)$
be the action generated by~$\mu$.
Then $\af$ is a product type action,
as in Example~\ref{E_PrdType}.
It follows from \Ex{E_1211_NoRPUHF_2On3}
that $\af$ does not have the \rp.
However, we will show that $\af$ does have the \trp.
Set $r (k) = \tfrac{1}{2} (3^k  1)$.
It is easy to check that
$v_k$ is unitarily equivalent to the block unitary
\[
w_k = \left( \begin{matrix}
0 & 1_{M_{r (k)}} & 0 \\
1_{M_{r (k)}} & 0 & 0 \\
0 & 0 & 1_{\C}
\end{matrix} \right)
\in M_{3^k}.
\]
It follows (Exercise~\ref{E_6X01_PrTypCnj})
that
$\mu$ is conjugate to the automorphism
\[
\nu = \bigotimes_{n = 1}^{\I} \Ad (w_k),
\]
and therefore that $\af$ is conjugate to
the action $\bt \colon \Z / 2 \Z \to \Aut (A)$ generated by~$\nu$.
We claim that $\bt$ has the \trp.
It will follow that $\af$ does too.
Let $S \subset A$ be finite and let $\ep > 0$.
Let $\ta$ be the unique tracial state on~$A$.
Appealing to Proposition~\ref{P_6X01_DfnUsingTrace},
we have to find orthogonal \pj{s} $e_0, e_1 \in A$
such that:
\begin{enumerate}
\item\label{Z2Ex:TRPDfn:1}
$\ \nu (e_0)  e_1 \ < \ep$ and $\ \nu (e_1)  e_0 \ < \ep$.
\item\label{Z2Ex:TRPDfn:2}
$\ e_0 a  a e_0 \ < \ep$ and
$\ e_1 a  a e_1 \ < \ep$ for all $a \in S$.
\item\label{Z2Ex:TRPDfn:3}
$\ta (1  e_0  e_1) < \ep$.
\end{enumerate}
Write $S = \{ a_1, a_2, \ldots, a_N \}$.
For $n \in \N$ set
$A_n = \bigotimes_{k = 1}^{n} M_{3^k}$
and identify $A_n$
with its image in~$A$.
Since $\bigcup_{n \in \Nz} A_n$
is dense in~$A$,
there are $n$ and
$b_1, b_2, \ldots, b_N \in A_n$
such that
\[
\ b_1  a_1 \ < \frac{\ep}{2},
\,\,\,\,\,\,
\ b_2  a_2 \ < \frac{\ep}{2},
\,\,\,\,\,\,
\ldots,
\,\,\,\,\,\,
\ b_N  a_N \ < \frac{\ep}{2}.
\]
We can increase~$n$,
so we may also assume that $3^{ n  1} < \ep$.
Using subscripts to indicate block sizes on the diagonals,
set
\[
p_0 = \left( \begin{matrix}
1_{r (n + 1)} & 0 & 0 \\
0 & 0_{r (n + 1)} & 0 \\
0 & 0 & 0_1
\end{matrix} \right)
\in M_{3^{n + 1}}
\]
and
\[
p_1 = \left( \begin{matrix}
0_{r (n + 1)} & 0 & 0 \\
0 & 1_{r (n + 1)} & 0 \\
0 & 0 & 0_1
\end{matrix} \right)
\in M_{3^{n + 1}}.
\]
Then
\[
w_{n + 1} p_0 w_{n + 1}^* = p_1
\andeqn
w_{n + 1} p_1 w_{n + 1}^* = p_0,
\]
and the normalized trace of $1  p_0  p_1$ is
\[
\frac{1}{2 r (n + 1) + 1} = \frac{1}{3^{n + 1}} < \ep.
\]
Set
\[
e_0 = 1_{A_n} \otimes p_0
\andeqn
e_1 = 1_{A_n} \otimes p_1,
\]
so
\[
e_0, e_1
\in A_n \otimes M_{3^{n + 1}} = A_{n + 1} \subset A.
\]
On $A_n \otimes M_{3^{n + 1}}$,
the automorphism
$\nu$ has the form
$\Ad \big( \left( \bigotimes_{k = 1}^{n} w_k \right)
\otimes w_{n + 1} \big)$,
so
\[
\nu (e_0) = e_1
\andeqn
\nu (e_1) = e_0.
\]
Condition~(\ref{Z2Ex:TRPDfn:1}) follows trivially.
The \tst~$\ta$ restricts to the unique \tst{} on $M_{3^{n + 1}}$,
so $\ta (1  e_0  e_1)$
is the normalized trace of $1  p_0  p_1$,
and is thus equal to $3^{ n  1} < \ep$.
This is condition~(\ref{Z2Ex:TRPDfn:3}).
It remains to check~(\ref{Z2Ex:TRPDfn:2}).
For $k \in \{ 1, 2, \ldots, N \}$,
the element $b_k$ actually commutes with $e_0$ and $e_1$,
so
\[
\ e_0 a_k  a_k e_0 \
\leq \ e_0 \ \ a_k  b_k \ + \ a_k  b_k \ \ e_0 \
< \frac{\ep}{2} + \frac{\ep}{2}
= \ep.
\]
This completes the proof that $\af$ has the \trp.
\end{exa}
\begin{rmk}\label{R_6X17_IsConj}
The matrix sizes in \Ex{EEasyTRP} grow rapidly,
and the number of diagonal entries equal to~$1$
and the number of diagonal entries equal to $1$ are very close.
This looks special.
It actually isn't.
It turns out that the action $\af$ of \Ex{EEasyTRP} is conjugate to
the action generated by
\[
\rh = \bigotimes_{n = 1}^{\I}
\Ad \left( \begin{matrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{matrix} \right)
\qquad
{\mbox{on}}
\qquad
\bigotimes_{n = 1}^{\I} M_3.
\]
So the action generated by~$\rh$ has the \trp.
\end{rmk}
\begin{exr}\label{E_6X01_3Conj}
Prove the conjugacy statement
in Remark~\ref{R_6X17_IsConj}
% after \Ex{EEasyTRP},
by combining suitable finite collections of tensor factors
in the definition of~$\rh$.
\end{exr}
The following exercise (which requires work,
and also requires some of the results below)
gives some idea of the differences between
the \rp, the \trp, and pointwise outerness.
\begin{exr}[Section~2 of~\cite{PhT4}]\label{Ex:Z2Struct}
Let $D$ be a UHF algebra
and let $\af \in \Aut (D)$
be an automorphism of order two, of the form
\[
D = \bigotimes_{n = 1}^{\I} M_{k (n)} \andeqn
\af = \bigotimes_{n = 1}^{\I}
\Ad ( p_n  q_n ),
\]
with $k (n) \in \N$ and where $p_n, \, q_n \in M_{k (n)}$
are \pj{s} with $p_n + q_n = 1$
and $\rank (p_n) \geq \rank (q_n)$.
Define
\[
\ld_n
= \frac{\rank (p_n)  \rank (q_n)}{\rank (p_n) + \rank (q_n)}
\geq 0
\]
for $n \in \N$,
and, for $m \leq n$,
define
\[
\Ld (m, n) = \ld_{m + 1} \ld_{m + 2} \cdots \ld_{n}
\andeqn
\Ld (m, \infty) = \limi{n} \Ld (m, n).
\]
Prove the following:
\begin{enumerate}
\item\label{Ex:Z2Struct1}
The action $\af$ has the \rp{} \ifo{}
there are infinitely many $n \in \N$
such that $\rank (p_n) = \rank (q_n)$
(that is, $\ld_n = 0$).
\item\label{Ex:Z2Struct2}
The action $\af$ has the \trp{} \ifo{}
$\Ld (m, \infty) = 0$ for all $m$.
\item\label{Ex:Z2Struct3}
The action $\af$ is pointwise outer \ifo{}
there are infinitely many $n \in \N$ such that $\ld_n < 1$
(that is, $q_n \neq 0$).
\end{enumerate}
Further prove that if $(k (n))_{n \in \N}$ is bounded
and $q_n \neq 0$ for all $n \in \N$,
then $\af$ has the \trp.
\end{exr}
In Section~2 of~\cite{PhT4},
in each case various other equivalent conditions are proved,
involving tracial states, Ktheory, or the dual action.
Example~3.12 of~\cite{PhSvy}
contains a list of other interesting actions
which have the \trp{} but not the \rp.
They include the actions in \Ex{E_3502_Blackadar}, Exercise~\ref{Ex:RCFG4_K1},
and Exercise~\ref{Ex:RCFG4_Torsion}.
We now show that when $A$ is finite,
the last condition in \Def{D_TRP}
(part~(\ref{D_TRP_4}), the requirement that $\ e x e \ > 1  \ep$)
can be omitted.
The next lemma comes from an argument
that goes back to Cuntz, in the proof of Lemma~1.7 of~\cite{Ctz}.
\begin{lem}[Lemma~1.14 of~\cite{PhT1}]\label{PjAndNorm}
Let $A$ be a \ca{} with property~(SP),
let $x \in A_{+} \setminus \{ 0 \}$ satisfy $\ x \ = 1$,
and let $\ep > 0$.
Then there is a nonzero \pj{} $p \in {\overline{x A x}}$
such that,
for every \nzp~$q$ satisfying
$q \leq p$,
we have
\[
\ q x  x q \ < \ep,
\qquad
\ q x q  q \ < \ep,
\andeqn
\ q x q \ > 1  \ep.
\]
\end{lem}
\begin{proof}
Choose \cfn{s} $g_1, g_2 \colon [0, 1] \to [0, 1]$ satisfying
$g_1 (0) = 0$, $g_1 (t) = 1$ for $t \geq 1  \tfrac{1}{4} \ep$,
and $ g_1 (t)  t  \leq \tfrac{\ep}{4}$ for all $t$,
and such that $g_2 (1) = 1$ and $g_1 g_2 = g_2$.
Define $y = g_1 (x)$ and $z = g_2 (x)$.
These elements satisfy $\ x  y \ \leq \tfrac{\ep}{4}$ and $y z = z$.
Since $1 \in \spec (x)$, we have $z \neq 0$.
Property~(SP)
provides a nonzero \pj{} $p \in {\overline{z A z}}$.
Now suppose that $q$ is a nonzero \pj{}
such that $q \leq p$.
Since $q \in {\overline{z A z}}$,
we have $y q = q y = q$.
So
\[
\ q x  x q \
\leq 2 \ x  y \
\leq \frac{\ep}{2}
< \ep.
\]
Moreover,
\[
\ q x q  q \
= \ q x q  q y q \
\leq \ x  y \
\leq \frac{\ep}{4}
< \ep,
\]
whence also $\ q x q \ > 1  \ep$.
This completes the proof.
\end{proof}
\begin{lem}[Lemma~1.15 of~\cite{PhT1}]\label{FPjNorm}
Let $A$ be an infinite dimensional finite unital \ca{}
with property~(SP).
Let $x \in A_{+}$ satisfy $\ x \ = 1$,
and let $\ep > 0$.
Then there is a nonzero \pj{} $q \in {\overline{x A x}}$
such that,
for every \pj{} $e \in A$
satisfying $1  e \precsim q$,
we have $\ e x e \ > 1  \ep$.
\end{lem}
\begin{proof}
We apply Lemma~\ref{PjAndNorm} with $x^{1/2}$ in place of $x$
and with $\tfrac{\ep}{5}$ in place of $\ep$.
Since ${\overline{x^{1/2} A x^{1/2}}} = {\overline{x A x}}$,
this gives a nonzero \pj{} $p \in {\overline{x A x}}$
such that for every nonzero \pj{} $q \leq p$
we have, in particular,
\[
\big\ q x^{1/2} q  q \big\ < \frac{\ep}{5}
\andeqn
\big\ q x^{1/2}  x^{1/2} q \big\ < \frac{\ep}{5}.
\]
Combining these estimates gives
%
\begin{equation}\label{Eq_6X08_qxhq}
\big\ q x^{1/2}  q \big\ < \frac{2 \ep}{5}.
\end{equation}
%
Using Lemma~\ref{OrthInSP},
choose a nonzero \pj{} $q \leq p$ such that $p  q \neq 0$.
Now let $e \in A$ be a \pj{} satisfying $1  e \precsim q$
and $\ e x e \ \leq 1  \ep$.
Using $\ a^* a \ = \ a a^* \$
at the first and fourth steps
and~(\ref{Eq_6X08_qxhq}) at the second step, we get
\[
\ e p e \
= \ p e p \
< \big\ p x^{1/2} e x^{1/2} p \big\ + \frac{4 \ep}{5}
\leq \big\ x^{1/2} e x^{1/2} \big\ + \frac{4 \ep}{5}
= \ e x e \ + \frac{4 \ep}{5}
< 1  \frac{\ep}{5}.
\]
So
\[
\ e  e (1  p) \
= \ e p \
= \ e p e \^{1/2}
< \left( 1  \tfrac{\ep}{5} \right)^{1/2}
< 1.
\]
By Lemma~\ref{L_6X07_ClSub},
we now get $e \precsim 1  p$.
We have $1  e \precsim q$ by assumption,
so it follows that $1 \precsim 1  (p  q)$.
We have contradicted finiteness of~$A$.
\end{proof}
\begin{lem}[Lemma~1.16 of~\cite{PhT1}]\label{TRPForFinite}
Let $A$ be an \idfssuca,
and let $\af \colon G \to \Aut (A)$
be an action of a finite group $G$ on $A$.
Then $\af$ has the
tracial Rokhlin property \ifo{} for every finite set
$F \subset A$, every $\ep > 0$,
and every $x \in A_{+} \setminus \{ 0 \}$,
there are \mops{} $e_g \in A$ for $g \in G$ such that:
\begin{enumerate}
\item\label{TRPForFinite:1} %
$\ \af_g (e_h)  e_{g h} \ < \ep$ for all $g, h \in G$.
\item\label{TRPForFinite:2} %
$\ e_g a  a e_g \ < \ep$ for all $g \in G$ and all $a \in F$.
\item\label{TRPForFinite:3} %
With $e = \sum_{g \in G} e_g$,
the \pj{} $1  e$ is \mvnt{} to a
\pj{} in the \hsa{} of $A$ generated by~$x$.
\end{enumerate}
\end{lem}
\begin{proof}
Since (\ref{TRPForFinite:1}), (\ref{TRPForFinite:2}),
and~(\ref{TRPForFinite:3})
are all part of \Def{D_TRP},
the \trp{} certainly implies the condition in the lemma.
So assume the condition in the lemma holds.
If $A$ does not have property~(SP),
we can choose $x \in A_{+} \setminus \{ 0 \}$
so that the hereditary subalgebra it generates
contains no \nzp{s}.
Then the \pj{} $e$ in condition~(\ref{D_TRP_3}) must be equal to~$1$.
This shows that $\af$ has the Rokhlin property.
(This is the same proof as for Lemma~\ref{RImpSP}.)
Accordingly, we assume that $A$ has property~(SP).
Let $F \subset A$ be finite,
let $\ep > 0$,
and let $x \in A_{+}$ satisfy $\ x \ = 1$.
Lemma~\ref{FPjNorm}
gives us a nonzero \pj{} $q \in {\overline{x A x}}$
such that for all \pj{s} $e \in A$ with $1  e \precsim q$,
we have $\ e x e \ > 1  \ep$.
Apply the hypothesis of the lemma
with $F$ and $\ep$ as given
and with $q$ in place of $x$.
We get \pj{s} $e_g \in A$ for $g \in G$.
As in~(\ref{TRPForFinite:3}),
define $e = \sum_{g \in G} e_g$.
Then $\ e x e \ > 1  \ep$
by the relation $1  e \precsim q$
and the choice of $q$ using Lemma~\ref{FPjNorm}.
This completes the proof.
\end{proof}
It is convenient to have a formally stronger version of the \trp,
in which the defect \pj{} is $\af$invariant.
This is a weaker statement than \Thm{T_ExactTRP},
but is much easier to prove.
\begin{lem}[Lemma~1.17 of~\cite{PhT1}]\label{StTRPDfn}
Let $G$ be a finite group,
let $A$ be an \idssuca,
and let $\af \colon G \to \Aut (A)$ be an action of $G$ on~$A$
which has the \trp.
Let $F \subset A$ be finite,
let $\ep > 0$,
and let $x \in A$ be a positive element with $\ x \ = 1$.
Then there are \mops{} $e_g \in A$ for $g \in G$ such that:
\begin{enumerate}
\item\label{StTRPDfn:1} %
$\ \af_g (e_h)  e_{g h} \ < \ep$ for all $g, h \in G$.
\item\label{StTRPDfn:2} %
$\ e_g a  a e_g \ < \ep$ for all $g \in G$ and all $a \in F$.
\item\label{StTRPDfn:4} %
With $e = \sum_{g \in G} e_g$,
the \pj{} $1  e$ is \mvnt{} to a
\pj{} in the \hsa{} of $A$ generated by $x$.
\item\label{StTRPDfn:5} %
With $e$ as in~(\ref{StTRPDfn:4}), we have $\ e x e \ > 1  \ep$.
\item\label{StTRPDfn:3} %
The \pj{} $e$ of~(\ref{StTRPDfn:4}) is $\af$invariant.
\end{enumerate}
\end{lem}
\begin{proof}
\Wolog{} $\ a \ \leq 1$ for all $a \in F$.
Set
\[
\ep_0 = \min \left( \frac{\ep}{41}, \frac{1}{20} \right).
\]
Choose $\dt$ as in \Lem{L_5Y21_PjInSbalg}
with $\ep_0$ in place of $\ep$,
and also require $\dt \leq \frac{\ep}{2}$.
Set
\[
\dt_0 = \frac{\dt}{\card (G)}.
\]
Apply Definition~\ref{D_TRP} to $\af$,
with $F$ and $x$ as given,
and with $\dt_0$ in place of $\ep$.
Let $(p_g)_{g \in G}$ be the resulting family of \pj{s}.
Define $p = \sum_{h \in G} p_h$.
For $g \in G$ we have
\[
\ \af_g (p)  p \
\leq \sum_{h \in G} \ \af_g (p_h)  p_{g h} \
< \card (G) \dt_0
\leq \dt.
\]
Set
\[
b = \frac{1}{\card (G)} \sum_{g \in G} \af_g (p).
\]
Then $b$ is in the fixed point algebra $A^{G}$ and
\[
\ b  p \
\leq \frac{1}{\card (G)} \sum_{g \in G} \ \af_g (p)  p \
< \dt.
\]
The choice of $\dt$ using \Lem{L_5Y21_PjInSbalg}
means that there is a \pj{} $e \in A^G$
such that $\ e  p \ < \ep_0$.
Since $\ep_0 \leq \frac{1}{20}$,
\Lem{L_6X02_QuantUE}
provides a unitary $v \in A$
such that $\ v  1 \ \leq 10 \ e  p \ < 10 \ep_0$
and $v p v^* = e$.
Now define $e_g = v p_g v^*$ for $g \in G$.
Clearly $\ e_g  p_g \ < 20 \ep_0$.
So, for $g, h \in G$,
\[
\ \af_g (e_h)  e_{g h} \
\leq \ e_h  p_h \ + \ e_{g h}  p_{g h} \
+ \ \af_g (p_h)  p_{g h} \
< 20 \ep_0 + 20 \ep_0 + \dt_0
% \leq 41 \ep_0
\leq \ep.
\]
For $g \in G$ and $a \in F$,
and using $\ a \ \leq 1$, we similarly
get
\[
\ e_g a  a e_g \ < 20 \ep_0 + 20 \ep_0 + \dt_0 \leq \ep.
\]
We have
\[
\ (1  e)  (1  p) \ < 20 \ep_0 \leq 1,
\]
so $1  e \sim 1  p$, and is hence \mvnt{} to a
\pj{} in the \hsa{} of $A$ generated by $x$.
Finally,
\[
\ e x e \ \geq \ p x p \  2 \ e  p \
> 1  \dt_0  2 \ep_0
\geq 1  \frac{\ep}{41}  \frac{2 \ep}{41}
> 1  \ep.
\]
This completes the proof.
\end{proof}
We adapt \Lem{L_1216PartR},
the key step in the proof that
crossed products of AF~algebras by Rokhlin actions are AF
(\Thm{T_RokhAF}),
to the \trp.
\begin{lem}\label{L_1216PartTR}
Let $G$ be a finite group,
and set $n = \card (G)$.
Identify $M_n$ with $L (l^2 (G))$,
and let for $g, h \in G$ let $e_{g, h}$ be the rank one operator
on $l^2 (G)$ given by
$e_{g, h} \xi = \langle \xi, \dt_h \rangle \dt_g$,
as in Notation~\ref{N_MatUnit}.
Also, for $g \in G$
let $u_g$ be the standard unitary of Notation~\ref{N:ug}.
Then for every $\ep > 0$ there is $\dt > 0$
such that the following holds.
Let $\GAa$ be a \ga,
let $(e_g)_{g \in G}$ be a family of orthogonal \pj{s},
and let $F \subset A$ be a finite set
such that $\ a \ \leq 1$ for all $a \in F$.
Suppose that:
\begin{enumerate}
\item\label{L_1216PartTR:1}
$\ \af_g (e_h)  e_{g h} \ < \dt$ for all $g, h \in G$.
\item\label{L_1216PartTR:2}
$\ e_g a  a e_g \ < \dt$ for all $g \in G$ and all $a \in F$.
\item\label{L_1216PartTR:3}
The \pj{} $e = \sum_{g \in G} e_g$ is $\af$invariant.
\end{enumerate}
Then there exists a unital \hm{}
$\ph \colon M_n \otimes e_1 A e_1 \to e A e$
such that for every $a \in F \cup \{ u_g \colon g \in G \}$
there are
\[
x \in M_n \otimes e_1 A e_1
\andeqn
y \in (1  e) A (1  e)
\]
with $\ [ \ph (x) + y ]  a \ < \ep$,
and such that (using standard matrix unit notation)
for every $a \in e_1 A e_1$ we have $\ph (e_{1, 1} \otimes a) = a$.
\end{lem}
\begin{proof}
Apply \Lem{L_1216PartR}
with $\frac{\ep}{2}$ in place of~$\ep$,
getting a number $\dt > 0$,
and further require that $\dt \leq \ep / (4 n)$.
% \[
% \dt \leq \frac{\ep}{4 \card (G)}.
% \]
Now let $\GAa$ be a \ga,
let $(e_g)_{g \in G}$ be a family of orthogonal \pj{s},
let $F \subset A$ is a finite set
such that $\ a \ \leq 1$ for all $a \in F$,
and suppose that the conditions
(\ref{L_1216PartTR:1}), (\ref{L_1216PartTR:2}),
and~(\ref{L_1216PartTR:3})
hold.
Define $e = \sum_{g \in G} e_g$.
Using $e_g e  e e_g = e_g$ for $g \in G$,
it is easy to check that $\ e_g e a e  e a e e_g \ < \dt$
for all $g \in G$ and all $a \in F$.
Since $e$ is $\af$invariant,
$G$ acts on the algebra $e A e$.
Call this action $\bt$,
and for $g \in G$ let $v_g \in C^* (G, e A e, \bt)$
be the standard unitary of Notation~\ref{N:ug}.
As usual, we let $u_g \in \CGAa$
be the standard unitary in this \cp.
One immediately checks that
$C^* (G, e A e, \bt)$ is a subalgebra of $\CGAa$,
in fact,
that $C^* (G, e A e, \bt) = e \CGAa e$,
that $u_g$ commutes with $e$ for all $g \in G$,
and that $v_g = e u_g e$.
With this in mind,
apply the choice of $\dt$ using \Lem{L_1216PartR}
to the algebra $e A e$ and the finite set
$\{ e a e \colon a \in F \}$.
The result is a unital \hm{}
\[
\ph \colon M_n \otimes e_1 A e_1 \to C^* (G, e A e, \bt)
\]
such that for every $a \in F \cup \{ u_g \colon g \in G \}$,
we have
%
\begin{equation}\label{Eq_6X02_DistEst}
\dist
\big( e a e, \, \ph ( M_n \otimes e_1 A e_1 ) \big)
< \frac{\ep}{2},
\end{equation}
%
and $\ph (e_{1, 1} \otimes a) = a$
for all $a \in e_1 A e_1$.
This last condition
is the last part of the conclusion of the lemma.
We next claim that for all $a \in F \cup \{ u_g \colon g \in G \}$,
we have
%
\begin{equation}\label{Eq_6X02_CutEst}
\big\ a  [ e a e + (1  e) a (1  e) ] \big\ < \frac{\ep}{2}.
\end{equation}
%
For $a = u_g$ with $g \in G$,
this is immediate since $e$ commutes with~$u_g$.
To prove the claim for $a \in F$, first
estimate
\[
\ e a  a e \
\leq \sum_{g \in G} \ e_g a  a e_g \
< \card (G) \dt
\leq \frac{\ep}{4}.
\]
Therefore
\[
\ e a (1  e) \ < \frac{\ep}{4}
\andeqn
\ (1  e) a e \ < \frac{\ep}{4},
\]
so
\[
\big\ a  [ e a e + (1  e) a (1  e) ] \big\
\leq \ e a (1  e) \ + \ (1  e) a e \
< \frac{\ep}{2}.
\]
Now let $a \in F \cup \{ u_g \colon g \in G \}$.
Use~(\ref{Eq_6X02_DistEst}) to choose
$x \in M_n \otimes e_1 A e_1$
such that $\ \ph (x)  e a e \ < \frac{\ep}{2}$.
Set $y = (1  e) a (1  e)$.
Then,
using~(\ref{Eq_6X02_CutEst})
at the second step,
we get
\[
\ [ \ph (x) + y ]  a \
\leq \ \ph (x)  e a e \
+ \big\ [ e a e + (1  e) a (1  e) ]  a \big\
< \frac{\ep}{2} + \frac{\ep}{2}
= \ep,
\]
as desired.
This completes the proof.
\end{proof}
\begin{thm}[Theorem~2.6 of~\cite{PhT1}]\label{RokhTAF}
Let $G$ be a finite group,
and let $A$ be an infinite dimensional simple
separable \uca{} with tracial rank zero.
Let $\af \colon G \to \Aut (A)$
be an action of $G$ on $A$ which has the \trp.
Then $C^* (G, A, \af)$ has tracial rank zero.
\end{thm}
The proof will be given at the end of this section.
As mentioned above,
in~\cite{PhT1} the condition $p \neq 0$ was omitted
in one of the ingredients, Proposition~2.3 of~\cite{PhT1}.
% Without this condition,
% purely infinite simple \uca{s}
% would have tracial rank zero,
% by taking $p = 0$.
The basic idea is the same as that of
the proof of Theorem~\ref{T_RokhAF}.
The main difference is that there is a small ``error projection''
in both the definition of the \trp{} and the definition of
tracial rank zero.
The main technical complication is that when one carries out
the obvious modification of the proof of Theorem~\ref{T_RokhAF},
what one gets is that the ``error projection''
in the definition
of tracial rank zero for the \cp,
which is supposed to be \mvnt{}
to a \pj{} in a previously
specified hereditary subalgebra of $C^* (G, A, \af)$,
actually comes out to be \mvnt{}
to a \pj{} in a previously
specified hereditary subalgebra of~$A$.
A priori,
this is not good enough.
The day is saved by the following theorem,
which is a special case of Theorem~4.2 of~\cite{JO}.
\begin{thm}[see Theorem~4.2 of~\cite{JO}]\label{T:JO}
Let $G$ be a finite group.
Let $A$ be a simple unital \ca{} with property~(SP).
Let $\af \colon G \to \Aut (A)$ be a pointwise outer action.
Let $B \subset C^* (G, A, \af)$ be a nonzero \hsa.
Then there exists a \nzp{} $p \in B$
which is \mvnt{} to a \pj{} in~$A$.
\end{thm}
We omit the proof of Theorem~\ref{T:JO}.
% 999 (It might be filled in later.)
Instead,
we give a proof of a special case which is good enough for
the purposes of this section,
with some of the lemmas given in greater generality.
Our proof requires less work
and uses methods closer to those of these notes.
\begin{dfn}\label{DGSimple}
Let $\af \colon G \to \Aut (A)$
be an action of a locally compact group $G$ on a \ca~$A$.
We say that $\af$ is {\emph{minimal}},
or that $A$ is {\emph{$G$simple}},
if the only $G$invariant (closed) ideals in~$A$ are $\{ 0 \}$
and~$A$.
\end{dfn}
This definition generalizes the usual definition of
minimality of a group action on a locally compact Hausdorff space,
which is given in \Def{D_3331_1Min}.
We make three brief comments.
First, if $A$ is simple
(the case of most interest to us now),
then clearly $A$ is $G$simple.
However, in Theorem~\ref{TAKCImpSimp},
where we assume $G$simplicity,
there is no simplification in the proof by assuming simplicity instead.
Second, $G$simplicity is an elementary necessary condition
for simplicity
of the reduced crossed product $C^*_{\mathrm{r}} (G, A, \af)$,
since if $I$ is a proper $G$invariant ideal in~$A$,
then $C^*_{\mathrm{r}} (G, I, \af)$
is a proper ideal in $C^*_{\mathrm{r}} (G, A, \af)$.
(See Theorem~\ref{T_3331_RedCP}(\ref{T_3331_RedCP_NZI}).)
Third, $G$simplicity is not a sufficient condition
for simplicity
of the reduced crossed product.
Indeed, the trivial action of a locally compact group~$G$
on~$\C$ is obviously minimal,
but the reduced crossed product is $C^*_{\mathrm{r}} (G)$,
which is usually not simple
(in particular, never simple if $G$ is amenable and nontrivial).
We introduce a property of actions,
not previously named,
which we call Kishimoto's condition after the paper~\cite{Ks1}
in which it appeared in close to this form.
We proceed via Kishimoto's condition in this section for two reasons.
First, it is the first step in the proofs of two different results
which we need here.
Second, we will want it again later,
for proofs of these same results under weaker hypotheses.
\begin{dfn}\label{DKshCond}
Let $\af \colon G \to \Aut (A)$
be an action of a discrete group $G$ on a \ca~$A$.
We say that $\af$ satisfies {\emph{Kishimoto's condition}}
if for every positive element $x \in A$ with $\ x \ = 1$,
every finite set $F \subset G \setminus \{ 1 \}$,
every finite set $S \subset A$,
and every $\ep > 0$,
there is a positive element $c \in A$ with $\ c \ = 1$
such that:
\begin{enumerate}
\item\label{DKshCond1}
$\ c x c \ > 1  \ep$.
\item\label{DKshCond2}
$\ c b \af_g (c) \ < \ep$ for all $g \in F$ and $b \in S$.
\end{enumerate}
\end{dfn}
This condition is essentially the conclusion
of Lemma~3.2 of~\cite{Ks1}.
It is a kind of freeness condition.
For example,
if $A = C (X)$ and $f \in C (X)$ is a function such that
$\supp (f) \cap g \cdot \supp (f) = \E$,
then $f b \af_g (f) = 0$ for every $y \in C (X)$.
It is shown in~\cite{Ks1}
that if $A$ is simple and $\af$ is pointwise outer,
then $\af$ satisfies Kishimoto's condition.
In fact, as discussed there, weaker hypotheses suffice.
(In~\cite{Ks1},
see Lemma~3.2 and the second part of Remark~2.2.)
We give here a much easier proof of a special case
of this fact,
strengthening the hypotheses to the tracial Rokhlin property.
The fact that our hypotheses are unnecessarily strong
is suggested by the fact that we never use
the condition in the tracial Rokhlin property
which requires that $1  e$ be ``small''.
We point out that a condition related to
Kishimoto's condition has been generalized
in~\cite{Os2} to conditional expectations on unital \ca{s},
with the reduced crossed product situation
corresponding to the standard conditional expectation
(Definition~\ref{D_StdCond}).
The definition is near the beginning of Section~2 of~\cite{Os2}.
In general,
outerness of the conditional expectation is stronger
than pointwise outerness of the action.
% 999 Say more here.
\begin{lem}\label{LTRPImpAKC}
Let $G$ be a finite group,
let $A$ be an infinite dimensional simple \uca,
and let $\af \colon G \to \Aut (A)$ be an
action of $G$ on~$A$.
Assume that $\af$ has the tracial Rokhlin property
(\Def{D_TRP}).
Then $\af$ satisfies Kishimoto's condition
(\Def{DKshCond}).
In fact, the element $a$ in the conclusion can be taken to be a \pj.
\end{lem}
\begin{proof}
Let $x \in A$ be a positive element with $\ x \ = 1$,
let $S \subset A$ be finite,
and let $\ep > 0$.
We may as well take the finite set $F \subset G \setminus \{ 1 \}$
in Kishimoto's condition to be $G \setminus \{ 1 \}$ itself.
\Wolog{} $\ep < 1$.
Set
\[
n = \card (G),
\qquad
M = \max \left( 1, \, \sup_{b \in S} \ b \ \right),
\andeqn
\ep_0 = \min \left( \frac{\ep}{M + 1}, \, \frac{\ep}{n^2} \right).
\]
Apply the tracial Rokhlin property
(\Def{D_TRP})
with $S \cup \{ x \}$ in place of~$F$,
with $\ep_0$ in place of~$\ep$,
and with $x$ as given.
Call the resulting family of \pj{s} $(e_g)_{g \in G}$,
and set $e = \sum_{g \in G} e_g$.
We have
\[
\left\ e x e  \ssum{g \in G} e_g x e_g \right\
%\leq \sum_{g \in G} \sum_{h \in G \setminus \{ h \}} }
% \ e_g x e_h \.
\leq \ssum{g \neq h} \ e_g x e_h \.
\]
Since $e_g e_h = 0$ for $g \neq h$,
the term $\ e_g x e_h \$ on the right
is dominated by $\ x e_h  e_h x \ < \ep_0$.
Therefore
\[
\left\ e x e  \ssum{g \in G} e_g x e_g \right\
< n (n  1) \ep_0,
\]
and
\[
\left\ \ssum{g \in G} e_g x e_g \right\
\geq \ e x e \  \left\ e x e  \ssum{g \in G} e_g x e_g \right\
> 1  \ep_0  n (n  1) \ep_0
% > 1  n^2 \ep_0
\geq 1  \ep.
\]
Since the elements
$e_g x e_g$,
for $g \in G$,
are orthogonal,
it follows that there is $g_0 \in G$ such that
$\ e_{g_0} x e_{g_0} \ > 1  \ep$.
Set $a = e_{g_0}$.
Since $\ep < 1$, we have $e_{g_0} \neq 0$,
so $\ a \ = 1$.
Now let $b \in S$ and let $h \in G \setminus \{ 1 \}$.
Then $e_{g_0} e_{h g_0} = 0$,
so
\begin{align*}
\ a b \af_h (a) \
& = \ e_{g_0} b \af_h (e_{g_0}) \
= \ e_{g_0} b \af_h (e_{g_0})  e_{g_0} e_{h g_0} b \
\\
& \leq \ e_{g_0} \ \cdot \ b \
\cdot \ \af_h (e_{g_0})  e_{h g_0} \
+ \ e_{g_0} \ \cdot \ b e_{h g_0}  e_{h g_0} b \
\\
& < M \ep_0 + \ep_0
\leq \ep.
\end{align*}
This completes the proof of Kishimoto's condition.
\end{proof}
The following two results are stated for discrete groups
rather than merely for finite groups.
The finite group case is all that is needed here.
The proofs when $G$ is finite are a bit simpler,
because one can omit the step
in which an element of $C^*_{\mathrm{r}} (G, A, \af)$
is approximated by an element of $C_{\mathrm{c}} (G, A)$,
eliminating some of the estimates,
but otherwise the proofs are the same.
The next theorem is contained in
Theorem~3.1 of~\cite{Ks1},
and follows the proof of Theorem~3.2 of~\cite{El0}.
\begin{thm}\label{TAKCImpSimp}
Let $A$ be a \ca,
and let $\af \colon G \to \Aut (A)$ be a minimal
action (\Def{DGSimple})
of a discrete group $G$ on~$A$.
Assume that $\af$ satisfies Kishimoto's condition
(\Def{DKshCond}).
Then $C^*_{\mathrm{r}} (G, A, \af)$ is simple.
\end{thm}
\begin{proof}
Let $J \subset C_{\mathrm{c}} (G, A)$ be a proper ideal.
For $g \in G$ let $u_g$
be the standard unitary of Notation~\ref{N:ug}.
We first claim that $J \cap A$ is a $G$invariant ideal in~$A$.
That it is an ideal is clear.
Let $g \in G$.
If $A$ is unital,
then $u_g \in C^*_{\mathrm{r}} (G, A, \af)$,
and for $a \in J \cap A$ we have
$\af_g (a) = u_g a u_g^* \in J \cap A$.
In the general case,
let $(e_{\ld})_{\ld \in \Ld}$ be an approximate identity for~$A$.
Then for $\ld \in \Ld$,
the elements $e_{\ld} u_g$ and $u_g e_{\ld} = \af_g (e_{\ld}) u_g$
are in $C^*_{\mathrm{r}} (G, A, \af)$,
so
\[
\af_g (a)
= \lim_{\ld} \af_g ( e_{\ld} a e_{\ld} )
= \lim_{\ld} u_g e_{\ld} a e_{\ld} u_g
\in J \cap A.
\]
The claim is proved.
We next claim that $J \cap A = \{ 0 \}$.
Since $\af$ is minimal,
we need only rule out $A \subset J$.
Suppose $A \subset J$.
Let $(e_{\ld})_{\ld \in \Ld}$ be an approximate identity for~$A$.
For $a \in A$ and $g \in G$,
we have
$a u_g = \lim_{\ld} a e_{\ld} u_g \in J$.
Therefore $C_{\mathrm{c}} (G, A) \subset J$,
whence $J = C^*_{\mathrm{r}} (G, A, \af)$.
This contradiction proves the claim.
Let $E \colon C^*_{\mathrm{r}} (G, A, \af) \to A$
be the standard conditional expectation,
as in Definition~\ref{D_StdCond}.
We now claim that if $a \in J$ then $E (a^* a) = 0$.
Given the claim, since $E$ is faithful
(Proposition~\ref{P:Faithful}(\ref{P:Faithful:1})),
this implies that $a^* a = 0$,
whence $a = 0$.
So $J = \{ 0 \}$,
proving the theorem.
We prove the claim.
Let $a \in J$ and let $\ep > 0$.
We show that $\ E (a^* a) \ < \ep$.
Choose $y \in C_{\mathrm{c}} (G, A)$ with $\ y  a \$
so small that $\ y^* y  a^* a \ < \tfrac{\ep}{5}$.
Then there are a finite set $F \subset G$ and elements $b_g \in A$
for $g \in F$ such that
$y^* y = \sum_{g \in F} b_g u_g$.
\Wolog{} $1 \in F$.
We must have $b_1 = E (y^* y) \geq 0$.
Also
%
\begin{equation}\label{Eq:1112071}
\ b_1  E (a^* a) \
\leq \ y^* y  a^* a \
< \tfrac{\ep}{5}.
\end{equation}
%
Suppose $b_1 = 0$.
Then~(\ref{Eq:1112071}) implies $\ E (a^* a) \ < \ep$,
as desired.
So we may assume $b_1 \neq 0$.
Set
\[
x = \ b_1 \^{ 1} b_1
\andeqn
\ep_0
= \min \left( \frac{1}{2}, \, \frac{\ep}{5 \cdot \card (F)} \right).
\]
Apply Kishimoto's condition
with $F \setminus \{ 1 \}$ in place of~$F$,
with
\[
S = \{ b_g \colon g \in F \setminus \{ 1 \} \},
\]
with $x$ as given,
and with $\ep_0$ in place of~$\ep$.
Let $c$ be the resulting element.
We can now estimate
\begin{align*}
\ c y^* y c  c b_1 c \
& = \left\ \ssum{g \in F \setminus \{ 1 \} } c b_g u_g c \right\
= \left\
\ssum{g \in F \setminus \{ 1 \} } c b_g \af_g (c) u_g \right\
\\
& \leq \sum_{g \in F \setminus \{ 1 \} } \ c b_g \af_g (c) \
< \card (F) \ep_0
\leq \frac{\ep}{5}.
\end{align*}
Therefore, using $\ c \ \leq 1$,
\[
\ c a^* a c  c b_1 c \
\leq \ a^* a  y^* y \ + \ c y^* y c  c b_1 c \
< \frac{\ep}{5} + \frac{\ep}{5}
= \frac{2 \ep}{5}.
\]
Let
$\pi \colon C^*_{\mathrm{r}} (G, A, \af)
\to C^*_{\mathrm{r}} (G, A, \af) / J$
be the quotient map.
Since $J \cap A = \{ 0 \}$,
the restriction $\pi _A$ is injective,
so $\ \pi (c b_1 c) \ = \ c b_1 c \$.
On the other hand,
$c a^* a c \in J$,
so $\pi (c a^* a c) = 0$.
Thus
\[
\ c b_1 c \
= \ \pi (c b_1 c) \
= \ \pi (c b_1 c  c a^* a c) \
\leq \ c b_1 c  c a^* a c \
< \frac{2 \ep}{5}.
\]
The choice of~$c$ and
the relation $\ep_0 \leq \tfrac{1}{2}$ imply that
\[
\ c b_1 c \ > \ b_1 \ \big( 1  \ep_0 \big)
\geq \tfrac{1}{2} \ b_1 \.
\]
Thus
$\ b_1 \ \leq 2 \ c b_1 c \ < \tfrac{4 \ep}{5}$.
% Since $\ b_1  E (a^* a) \ < \tfrac{\ep}{5}$,
Combining this with~(\ref{Eq:1112071}),
we get $\ E (a^* a) \ < \ep$.
This completes the proof.
\end{proof}
The next theorem is contained in
Theorem~4.2 of~\cite{JO}.
\begin{thm}\label{TAKCImpSP}
Let $A$ be a \ca{} which has property~(SP),
and let $\af \colon G \to \Aut (A)$ be an
action of a discrete group $G$ on~$A$.
Assume that $\af$ satisfies Kishimoto's condition
(\Def{DKshCond}).
Then for every nonzero hereditary subalgebra
$D \subset C^*_{\mathrm{r}} (G, A, \af)$,
there is a \nzp{} $p \in D$
which is \mvnt{}
to a \pj{} in~$A$.
\end{thm}
The proof of \Lem{L:PjComp} is very similar but done in an
easier context,
so one may want to read the proof of that lemma first.
\begin{proof}[Proof of \Thm{TAKCImpSP}]
Let $E \colon C^*_{\mathrm{r}} (G, A, \af) \to A$
be the standard conditional expectation
(Definition~\ref{D_StdCond}).
Choose $a \in D_{+} \setminus \{ 0 \}$.
Since $E$ is faithful
(Proposition~\ref{P:Faithful}(\ref{P:Faithful:1})),
we have
$E (a) \neq 0$.
By scaling, we may assume $\ E (a) \ = 1$.
Choose $y \in C_{\mathrm{c}} (G, A)$ with $\big\ y  a^{1 / 2} \big\$
so small that $\ y^* y  a \ < \tfrac{1}{4}$.
Then there are a finite set $F \subset G$ and elements $b_g \in A$
for $g \in F$ such that
$y^* y = \sum_{g \in F} b_g u_g$.
\Wolog{} $1 \in F$.
Set
\[
\dt = \frac{1}{2 ( \card (F) + 2 )}.
\]
Apply Kishimoto's condition
with $F \setminus \{ 1 \}$ in place of~$F$,
with $S = \{ b_g \colon g \in F \setminus \{ 1 \} \}$,
with $x = E (a)$,
and with $\dt$ in place of~$\ep$.
Let $c$ be the resulting element.
For $g \in G$ let $u_g$
be the standard unitary of Notation~\ref{N:ug}.
We can now estimate
\begin{align*}
\ c y^* y c  c E (y^* y) c \
& = \left\ \ssum{g \in F \setminus \{ 1 \} } c b_g u_g c \right\
= \left\
\ssum{g \in F \setminus \{ 1 \} } c b_g \af_g (c) u_g \right\
\\
& \leq \sum_{g \in F \setminus \{ 1 \} } \ c b_g \af_g (c) \
< \card (F) \dt.
\end{align*}
Therefore, using $\ c \ \leq 1$,
\begin{equation}\label{Eq:TAKCImpSPEst}
\ c a c  c E (a) c \
\leq 2 \ a  y y^* \ + \ c y^* y c  c E (y^* y) c \
< \tfrac{1}{2} + \card (F) \dt.
\end{equation}
Let $f, f_0 \colon [0, 1] \to [0, 1]$
be the \ct{} functions which are
linear on $[0, \, 1  2 \dt]$ and $[ 1  2 \dt, \, 1]$,
and satisfy
\[
f (0) = f_0 (0) = 0,
\quad
f_0 (1  2 \dt) = 0,
\quad
f (1  2 \dt) = 1,
\quad {\mbox{and}} \quad
f (1) = f_0 (1) = 1.
\]
Then $f f_0 = f_0$.
Also $\ c E (a) c \ > 1  \dt$,
so $f_0 (c E (a) c) \neq 0$.
Use property~(SP)
to choose a \nzp{} $e$ in the hereditary subalgebra
of $A$ generated by $f_0 (c E (a) c)$.
Since $f f_0 = f_0$,
we have
$f (c E (a) c) e = e$.
Since $ f (t)  t  \leq 2 \dt$ for $t \in [0, 1]$,
we get $\ c E (a) c e  e \ < 2 \dt$.
Combining this estimate with~(\ref{Eq:TAKCImpSPEst}),
we get
\[
\ e c a c e  e \
\leq \ c a c  c E (a) c \ + \ e \ \cdot \ c E (a) c e  e \
< \tfrac{1}{2} + \card (F) \dt + 2 \dt
= 1.
\]
Set $z_0 = a^{1 / 2} c e$.
Then $z_0^* z_0 = e c a c e \in e C^*_{\mathrm{r}} (G, A, \af) e$,
and moreover satisfies $\ z_0^* z_0  e \ < 1$.
Evaluating functional calculus in $e C^*_{\mathrm{r}} (G, A, \af) e$,
we may therefore set
$r = (z_0^* z_0)^{ 1 / 2}$.
Then $z = z_0 r$ satisfies $z^* z = e$.
Also $p = z z^*$ is a projection such that
\[
p = a^{1 / 2} c e r^2 e c a^{1 / 2}
\in a^{1 / 2} C^*_{\mathrm{r}} (G, A, \af) a^{1 / 2}
\subset D.
\]
Since $p$ is \mvnt{} to~$e$,
the proof is complete.
\end{proof}
We are now ready for the proof of \Thm{RokhTAF}.
\begin{proof}
We will use \Lem{L_6X08_TAFGenerators},
taking $T$ to be (following Notation~\ref{N:ug} for the
standard unitaries in the \cp)
\[
T = \big\{ u_g \colon g \in G \big\}
\cup \big\{ a \in A \colon \ a \ \leq 1 \big\}.
\]
Accordingly,
let $S \subset T$ be finite,
let $\ep > 0$,
and let $c \in \CGAa_{+} \setminus \{ 0 \}$.
We may take
\[
S = \big\{ u_g \colon g \in G \big\} \cup F
\]
with $F \subset A$ finite and $\ a \ \leq 1$ for all $a \in F$.
We further write
\[
S = \{ a_1, a_2, \ldots, a_N \}.
\]
In \Lem{L_1216PartTR},
choose $\dt > 0$ for the number $\frac{\ep}{4}$ in place of~$\ep$.
% Set
% \[
% \dt = \min \left( \dt_0, \frac{\ep}{2 \card (G)^2} \right).
% \]
Since $A$ has property~(SP)
by Corollary~\ref{C_6X08_TAFSP}
and $\af$ satisfies Kishimoto's condition
by \Lem{LTRPImpAKC},
we can apply \Thm{TAKCImpSP}
to find a \nzp{}
$q \in A$ which is \mvnt{} to a \pj{}
in ${\ov{c \CGAa c}}$.
Again using the fact that $A$ has property~(SP),
use \Lem{OrthInSP}
to choose nonzero orthogonal \pj{s} $q_1, q_2 \in q A q$.
Apply the strengthening of the \trp{}
in \Lem{StTRPDfn},
with $F$ as given,
with $\dt$ in place of~$\ep$,
and with $q_1$ in place of~$x$,
getting \pj{s} $e_g \in A$ for $g \in G$
as there.
In particular,
the \pj{} $e = \sum_{g \in G} e_g$
is $G$invariant
and satisfies $1  e \precsim q_1$.
The choice of $\dt$ using \Lem{L_1216PartTR}
implies that there is a unital \hm{}
$\ph \colon M_n \otimes e_1 A e_1 \to e A e$
such that for $j = 1, 2, \ldots, N$
there are
\[
x_j \in M_n \otimes e_1 A e_1
\andeqn
y_j \in (1  e) A (1  e)
\]
with $\ [ \ph (x_j) + y_j ]  a_j \ < \frac{\ep}{4}$.
Moreover, we have
$\ph (e_{1, 1} \otimes a) = a$
for all $a \in e_1 A e_1$.
Use \Lem{L:CompSP} to choose a nonzero \pj{} $f \in e_1 A e_1$
such that $f \precsim q_2$.
Since $A$ has tracial rank zero,
so does $e_1 A e_1$ (by \Lem{L_6X08_TAFCorner})
and therefore also so does $M_n \otimes e_1 A e_1$
(by \Lem{L_6X08_TAFCM_n}).
Therefore there exist a \pj{} $p_0 \in M_n \otimes e_1 A e_1$,
a unital finite dimensional
subalgebra $D_0 \subset p_0 (M_n \otimes e_1 A e_1) p_0$,
and $d_1, d_2, \ldots, d_N \in D_0$ such that:
\begin{enumerate}
\item\label{6X08_Pf:TR0:Comm}
$\ [x_j, p_0] \ < \frac{\ep}{2}$ for $j = 1, 2, \ldots, N$.
\item\label{6X08_Pf:TR0:Close}
$\ p_0 x_j p_0  d_j \ < \frac{\ep}{2}$ for $j = 1, 2, \ldots, N$.
\item\label{6X08_Pf:TR0:Small}
$1  p_0 \precsim e_{1, 1} \otimes f$.
\end{enumerate}
Set $p = \ph (p_0)$ and $D = \ph (D_0)$.
Then
\begin{align*}
1  p
& = (1  e) + (e  p)
\\
& = (1  e) + \ph (1  p_0)
\precsim (1  e) + \ph (e_{1, 1} \otimes f)
= (1  e) + f
\precsim q_1 + q_2
\leq q,
\end{align*}
so $1  p$ is \mvnt{} to a \pj{} in ${\ov{c \CGAa c}}$.
Next,
for $j = 1, 2, \ldots, N$,
we have $p y_j = y_j p = 0$.
So
\[
\ [ p, \, \ph (x_j) + y_j ] \
= \ [p, \ph (x_j) ] \
\leq \ [x_j, p_0] \ < \frac{\ep}{2},
\]
whence
\[
\ [ p, a_j ] \
\leq 2 \ a_j  [ \ph (x_j) + y_j ] \
+ \ [ p, \, \ph (x_j) + y_j) ] \
< 2 \left( \frac{\ep}{4} \right) + \frac{\ep}{2}
= \ep.
\]
Moreover, $\ph (d_j) \in D$
and
\begin{align*}
\ p a_j p  \ph (d_j) \
& \leq \ a_j  [ \ph (x_j) + y_j ] \
+ \big\ p [ \ph (x_j) + y_j ] p  \ph (d_j) \big\
\\
& = \ a_j  [ \ph (x_j) + y_j ] \
+ \ p \ph (x_j) p  \ph (d_j) \
\\
& \leq \ a_j  [ \ph (x_j) + y_j ] \
+ \ p_0 x_j p_0  d_j \
< \frac{\ep}{4} + \frac{\ep}{2}
< \ep.
\end{align*}
This completes the proof.
\end{proof}
\part{An Introduction to Crossed Products by Minimal
Homeomorphisms}\label{Part_MH}
\section{Minimal Actions and their Crossed
Products}\label{Sec_MH}
\indent
In this section,
we discuss free and essentially free minimal actions
of countable discrete groups on compact metric spaces,
with emphasis on \mh{s}
(actions of~$\Z$).
We give two simplicity proofs,
using very different methods.
One works for free minimal actions,
and the method gives further information,
as well as some information when the action is not minimal.
See \Thm{TFreeMinSimp} and \Thm{T11202Traces}.
The second proof is a special case of a more general
simplicity theorem;
the case we prove allows some simplification of the argument.
Our theorem is \Thm{T:AS},
and its proof is given before \Thm{T:FullAS}.
The full theorem is stated as \Thm{T:FullAS}.
Both proofs end with an argument related to
the proof that Kishimoto's condition (\Def{DKshCond})
implies simplicity of the \cp{}
(Theorem~\ref{TAKCImpSimp}),
but the two proofs use quite different routes to get there.
We recall Definition~\ref{D_3331_1Min},
specialized to the case of locally compact groups and spaces.
It is also the specialization of \Def{DGSimple}
to the commutative case.
\begin{dfn}\label{D:Min}
Let a locally compact group $G$ act \ct{ly} on a locally
compact space $X$.
The action is called {\emph{minimal}} if
whenever $T \subset X$ is a closed subset such that $g T \subset T$
for all $g \in G$,
then $T = \varnothing$ or $T = X$.
\end{dfn}
In short, there are no nontrivial invariant closed subsets.
This is the topological analog of an ergodic action
on a measure space (\Def{D_3408_Ergod}).
It is equivalent (\Lem{L_3415_MinOrb})
that every orbit be dense.
If the action of $G$ on $X$ is not minimal,
then there is a nontrivial invariant closed subset $T \subset X$,
and $C^* (G, \, X \setminus T)$ is a nontrivial
ideal in $C^* (G, X)$.
See Theorem~\ref{T_CPExact}.
Thus $C^* (G, X)$ is not simple.
In fact,
$C^*_{\mathrm{r}} (G, X)$ is not simple,
by Theorem~\ref{T_3331_RedCP}(\ref{T_3331_RedCP_NZI}).
For the case $G = \Z$,
the conventional terminology is a bit different.
\begin{dfn}\label{D:MH}
Let $X$ be a locally \chs, and let $h \colon X \to X$
be a \hme.
Then $h$ is called {\emph{minimal}} if
whenever $T \subset X$ is a closed subset such that $h (T) = T$,
then $T = \varnothing$ or $T = X$.
\end{dfn}
Almost all work on \mh{s} has been on compact spaces.
For these, we have the following equivalent conditions.
\begin{lem}\label{L:MHC}
Let $X$ be a \chs,
and let $h \colon X \to X$ be a \hme.
Then \tfae:
\begin{enumerate}
\item\label{L:MHC:Min}
$h$ is minimal.
\item\label{L:MHC:FMin}
Whenever $T \subset X$ is a closed subset such that $h (T) \subset T$,
then $T = \varnothing$ or $T = X$.
\item\label{L:MHC:UMin}
Whenever $U \subset X$ is an open subset such that $h (U) = U$,
then $U = \varnothing$ or $U = X$.
\item\label{L:MHC:UFMin}
Whenever $U \subset X$ is an open subset such that $h (U) \subset U$,
then $U = \varnothing$ or $U = X$.
\item\label{L:MHC:DOrb}
For every $x \in X$, the orbit $\{ h^n (x) \colon n \in \Z \}$
is dense in $X$.
\item\label{L:MHC:DFOrb}
For every $x \in X$,
the forward orbit $\{ h^n (x) \colon n \geq 0 \}$
is dense in $X$.
\end{enumerate}
\end{lem}
Conditions (\ref{L:MHC:Min}), (\ref{L:MHC:UMin}), and~(\ref{L:MHC:DOrb})
are equivalent even when $X$ is only locally compact,
and in fact there is an analog for actions of arbitrary groups.
Minimality does not imply the other three conditions
without compactness,
as can be seen by considering the \hme{} $n \mapsto n + 1$ of~$\Z$.
(This is the case $G = \Z$ of \Ex{E:Tr}.)
Also, even for compact $X$,
it isn't good enough to merely have the existence of some
dense orbit,
as can be seen by considering the \hme{} $n \mapsto n + 1$
on the two point compactification $\Z \cup \{ \pm \I \}$ of~$\Z$.
(This action is one of those described in
\Ex{E_3415_OnePtCpt}.)
\begin{exr}\label{P:MHC}
Prove Lemma~\ref{L:MHC}.
\end{exr}
We recall a few examples.
\begin{exa}\label{E:MinTr}
Let $G$ be a locally compact group,
let $H \subset G$ be a closed subgroup,
and let $G$ act on $G / H$ be translation,
as in Example~\ref{E:Tr}.
This action is minimal:
there are no nontrivial invariant subsets, closed or not.
\end{exa}
Example~\ref{E:MinTr} is a ``trivial'' example
of a minimal action.
Here are several more interesting ones.
\begin{exa}\label{E:MinRot}
The irrational rotations in Example~\ref{E_Rot}
are \mh{s}.
\end{exa}
\begin{exa}\label{E:MinpAdic}
The \hme{} $x \mapsto x + 1$ on the $p$adic integers
(Example~\ref{E:pAdic})
is minimal.
The orbit of $0$ is~$\Z$,
which is dense, essentially by definition.
Every other orbit is a translate of this one, so is also dense.
(This is a special case of \Prp{P_3408_DnsOrb}.)
\end{exa}
\begin{exa}\label{E:NonMin}
The shift \hme{} of $\{ 0, 1 \}^{\Z}$ (Example~\ref{E:Shift})
and the action of $\SL_2 (\Z)$ on $S^1 \times S^1$
(Example~\ref{E:CommSL2})
are not minimal.
In fact, they have fixed points.
\end{exa}
Other examples of \mh{s} include Furstenberg transformations
(\Ex{E_3302_Furst})
and generalizations
(some of which are discussed after \Ex{E_3302_Furst}),
odometers (\Def{D_6827_Odometer};
see Exercise~\ref{Ex_7129_OdMin}),
restrictions of Denjoy \hme{s} of the circle to their minimal sets
(\cite{PSS}),
% 9999 This example should be added. So should others.
and certain irrational time maps of suspension flows,
studied in~\cite{It1}.
There are many others,
such as those discussed after \Ex{EFnBdy}
and those of \Thm{T_3305_ExistMinZ}
and \Thm{T_3305_ExistkErgZ}.
The \ca{s} associated with
many \mh{s} have been studied:
Furstenberg transformations and generalizations
in \cite{Pc86}, \cite{Ji}, \cite{Kd}, Example~4.9 of~\cite{Ph7},
Sections~2 and~3 of~\cite{PhX}, and \cite{Reh},
restricted Denjoy \hme{s} in~\cite{PSS},
irrational time maps of suspension flows in~\cite{It1},
certain classes of \mh{s}
of $S^1 \times X$
in \cite{LnMti1}, \cite{LnMti2}, and~\cite{LnMti3},
and certain classes of \mh{s}
of $S^1 \times S^1 \times X$ in~\cite{Sn}.
Again, there are others not mentioned here.
Minimal actions are plentiful:
a Zorn's Lemma argument shows that every nonempty compact $G$space~$X$
contains a nonempty invariant closed subset on which the restricted
action is minimal.
The \tgca{} of a minimal action need not be simple.
Consider, for example, the trivial action of a group $G$
(particularly an abelian group) on a one point space,
for which the \tgca{} is $C^* (G)$.
Let a locally compact group $G$ act \ct{ly} on a locally
compact space~$X$.
Recall from \Def{D_3331_Free}
that the action is free if
whenever $g \in G \setminus \{ 1 \}$ and $x \in X$,
then $g x \neq x$,
and is essentially free if
whenever $g \in G \setminus \{ 1 \}$,
the set $\{ x \in X \colon g x = x \}$ has empty interior.
\begin{rmk}\label{R:MHIsFree}
Let $X$ be an infinite \chs, and let $h \colon X \to X$ be a \mh.
Then the corresponding action of $\Z$ on $X$ is free.
Indeed, if for some $n \neq 0$ and $x \in X$,
we have $h^n (x) = x$,
then the orbit of $x$ is finite, hence closed,
and is clearly invariant.
Now minimality contradicts infiniteness of~$X$.
\end{rmk}
Of course, nothing like Remark~\ref{R:MHIsFree} is true for
general groups.
For example, let $G$ act freely and minimally on $X$,
let $H$ be some other group,
and let $G \times H$ act on $X$ via $(g, h) x = g x$.
Recall from Proposition~\ref{P_3421_FreeEssF}
that an essentially free minimal action of an abelian group is free,
and from the discussion after \Def{D_3331_Free}
that essential freeness is not the right concept
for nonminimal actions.
\Ex{E_3302_MHOn2Tor}
gives an action of a countable discrete group
which is minimal and essentially free, but not free.
Let a locally compact group $G$ act \ct{ly} on a locally
compact space~$X$.
Recall from \Def{DFromSpToCStar}
that the corresponding action $\af \colon G \to \Aut (C_0 (X))$
is given by
$\af_g (f) (x) = f (g^{1} x)$ for $g \in G$, $f \in C_0 (X)$,
and $x \in X$.
Also recall (\Def{DTGCA} and \Def{DredTGCA})
that we abbreviate $C^* (G, \, C_0 (X), \, \af)$ to $C^* (G, X)$
and $C^*_{\mathrm{r}} (G, \, C_0 (X), \, \af)$
to $C^*_{\mathrm{r}} (G, X)$.
The following result is essentially a special case
of the corollary at the end of~\cite{ArSp};
see the discussion before the corollary and the Remark
before Lemma~1 of~\cite{ArSp}.
We state a much more general result from~\cite{ArSp} below
(Theorem~\ref{T:FullAS}).
\begin{thm}\label{T:AS}
Let a discrete group $G$ act minimally and essentially freely
on a locally compact space~$X$.
Then $C^*_{\mathrm{r}} (G, X)$ is simple.
\end{thm}
Essential freeness of the action is not necessary.
The reduced \tgca{} for the trivial action of the free
group on two generators on a one point space is simple,
by \Thm{T_3301_CStRFnSimp}.
However, minimality is certainly necessary.
This follows from \Thm{T_3331_RedCP}(\ref{T_3331_RedCP_NZI}).
\begin{cor}\label{C1128A}
Let $X$ be an infinite \chs,
and let $h \colon X \to X$ be a \mh.
Then $C^* (\Z, X, h)$ is simple.
\end{cor}
\begin{proof}
This follows from Theorem~\ref{T:AS}
and the fact that $\Z$ is amenable,
so that the full and reduced crossed products are equal
by Theorem~\ref{T_4131_CPFToRed}.
\end{proof}
Our proof of Theorem~\ref{T:AS} will follow~\cite{ArSp},
and will be given at the end of this section.
We first discuss some special cases and different proofs.
First, we point out that,
when $G$ is amenable and the action is free,
and probably even when the action is only essentially free,
Theorem~\ref{T:AS} can be derived from the theorem
of Gootman and Rosenberg described in Remark~\ref{R:GR}.
See Corollary~8.22 of~\cite{Wlms} for the free case.
Next, we give a simple proof for the special case
of an irrational rotation on the circle.
It introduces some important ideas which we,
regretfully, will not develop further.
(Also see Proposition~2.56 of~\cite{Wlms}.)
\begin{thm}\label{T:IrrRot}
Let $\te \in \R \setminus \Q$.
Let $h_{\te} \colon S^1 \to S^1$ be the \hme{}
$h_{\te} (\zt) = e^{2 \pi i \te} \zt$.
Then $C^* (\Z, S^1, h_{\te})$ is simple.
\end{thm}
\begin{proof}
Following Example~\ref{E:C:IrrRot},
we identify $C^* (\Z, S^1, h_{\te})$
with the universal \ca{} $A_{\te}$
in Example~\ref{E_3303_RotAlg} generated by unitaries
$u$ and $v$ satisfying $v u = e^{2 \pi i \te} u v$,
by identifying $v$ with the function $\zt \mapsto \zt$ on $S^1$
and identifying $u$ with the standard unitary of the \cp.
Following Example~\ref{E_RotGauge},
let $\bt \colon S^1 \to \Aut (A_{\te})$
be the action
such that $\bt_{\zt} (u) = \zt u$ and $\bt_{\zt} (v) = v$
for $\zt \in S^1$.
Using normalized Haar measure in the integral,
we define a linear map $E \colon A_{\te} \to A_{\te}$
by $E (a) = \int_{S^1} \bt_{\zt} (a) \, d \zt$.
(The special case
of Banach space valued integration theory needed here,
essentially for continuous functions on a compact interval
with respect to Lebesgue measure,
is easily treated by elementary methods.)
One checks that $E (v^n u^m) = v^n$ for $m, n \in \Z$.
Since the elements $v^n u^m$ span a dense subset of $A_{\te}$,
it follows that $E$ is equal to
the the standard conditional expectation
coming from the crossed product structure
(Definition~\ref{D_StdCond}).
Now let $I \subset A_{\te}$ be a nonzero closed ideal.
We claim that $E (I) \subset I$.
First, check that, for $\zt = e^{2 \pi i k \te}$ with $k \in \Z$,
and for $m, n \in \Z$,
we have
\[
\bt_{\zt} (v^n u^m)
= e^{2 \pi i k m \te} v^n u^m = v^k (v^n u^m) v^{k}.
\]
Therefore $\bt_{\zt} (a) = v^k a v^{k}$ for all $a \in A_{\te}$.
In particular, $\bt_{\zt} (I) \subset I$.
Since $\{ e^{2 \pi i k \te} \colon k \in \Z \}$
is dense in $S^1$ (by Lemma~\ref{L:IrratRotDense}),
it follows from continuity of the action that $\bt_{\zt} (I) \subset I$
for all $\zt \in S^1$.
The claim now follows by integration.
We finish the proof by showing that $I = A_{\te}$.
Choose a nonzero positive element $a \in I$.
Let $f = E (a)$,
which is a nonzero nonnegative function in $I \cap C (S^1)$.
Then
$u^k f u^{k}$,
which is the function in $C (S^1)$ given by
$\zt \mapsto f \big( e^{ 2 \pi i k \te} \zt)$,
is also in $I \cap C (S^1)$.
Let $U = \{ \zt \in S^1 \colon f (\zt) \neq 0 \}$.
Then $u^k f u^{k}$ is strictly positive on $e^{2 \pi i k \te} U$.
The set $\bigcup_{k \in \Z} e^{2 \pi i k \te} U$
is a nonempty invariant open subset of $S^1$,
and it therefore equal to $S^1$.
By compactness, there is a finite set $S \subset \Z$
such that $\bigcup_{k \in S} e^{2 \pi i k \te} U = S^1$.
Then $\sum_{k \in S} u^k f u^{k}$ is a strictly positive
function on $S^1$,
and is hence invertible.
Since it is in $I$, we conclude that $I = A_{\te}$.
\end{proof}
\begin{rmk}\label{R_6814_IsDualAction}
The action $\bt \colon S^1 \to \Aut (A_{\te})$ used in the proof
of Theorem~\ref{T:IrrRot} is a special case of the dual action
on a \cp{} by an abelian group,
as described in Remark~\ref{R_6814_DualAction}.
\end{rmk}
The proof of Theorem~\ref{T:AS} for $G = \Z$
given in~\cite{Dvd}
(see Theorem VIII.3.9 of~\cite{Dvd})
is similar to the proof given for Theorem~\ref{T:IrrRot} above.
However, it is harder to prove that $E (I) \subset I$,
since there is no analog of the automorphism $\Ad (v)$.
The proof in~\cite{Dvd} uses the Rokhlin Lemma.
(See the proof of Lemma VIII.3.7 of~\cite{Dvd}.)
We have avoided Rokhlin type arguments in this section.
To obtain more information about simple \tgca{s},
such arguments are necessary,
at least with the current state of knowledge.
Examples show that,
in the absence of some form of the Rokhlin property,
stronger structural properties of \cp{s} of noncommutative \ca{s}
need not hold,
even when they are simple.
However, the Rokhlin Lemma is not actually needed
for the proof in~\cite{Dvd},
and, in fact, the proof works for reduced crossed products
by arbitrary (not necessarily amenable) discrete groups.
We give a version of this proof here.
The method has the added advantage of providing
information about the tracial states on the \cp,
and of being easily adaptable to at least some
Banach algebra versions of \cp{s}.
However, it requires that the space $X$ be compact.
% The method we use will prove simplicity
% of reduced transformation group \ca{s} for essentially free
% minimal actions,
% but to get other things out of it,
% we must apparently restrict to free actions.
% (Forward reference needed.) % 999
The following definition is intended only for use
in the proof of Proposition~\ref{PRmvEst}
and the lemmas leading up to it.
For the definition to make sense,
and for some of the lemmas,
we do not need to require that the subset $F$ be finite.
\begin{dfn}\label{DRemov}
Let $G$ be a discrete group,
let $X$ be a compact $G$space,
let $U \subset X$ be open,
and let $F \subset G \setminus \{ 1 \}$ be finite.
We say that $(F, U)$ is {\emph{inessential}}
if there exist $n \in \N$ and $s_1, s_2, \ldots, s_n \in C (X)$
such that $ s_k (x)  = 1$
for $k = 1, 2, \ldots, n$ and all $x \in X$,
and such that for all $x \in U$ and $g \in F$,
we have
\[
\frac{1}{n} \sum_{k = 1}^n s_k (x) {\overline{s_k (g^{1} x) }} = 0.
\]
\end{dfn}
\begin{lem}\label{LExistRmv}
Let $G$ be a discrete group,
let $X$ be a compact $G$space,
let $g \in G \setminus \{ 1 \}$,
and let $x \in X$ be a point such that $g x \neq x$.
Then there exists an open set $U \subset X$ with $x \in U$
such that $( \{ g \}, U )$ is inessential in the sense
of Definition~\ref{DRemov}.
\end{lem}
\begin{proof}
Choose an open set $U \subset X$ with $x \in U$
such that ${\overline{U}} \cap g^{1} {\overline{U}} = \varnothing$.
Take $n = 2$,
and take $s_1$ to be the constant function~$1$.
Choose a \cfn{} $r \colon X \to \R$
such that $r (x) = 0$ for $x \in {\overline{U}}$
and $r (x) = \pi$ for $x \in g^{1} {\overline{U}}$.
Set $s_2 (x) = \exp (i r (x))$ for $x \in X$.
For $x \in U$,
we have
\[
\frac{1}{n} \sum_{k = 1}^n s_k (x) {\overline{s_k (g^{1} x) }}
= \frac{1}{2} \big[ 1 \cdot 1 + 1 \cdot (1) \big]
= 0.
\]
Thus $( \{ g \}, U )$ is inessential.
\end{proof}
The next two lemmas are based on the same calculation,
namely~(\ref{Eq:RmvProd})
in the proof of Lemma~\ref{LRmvUnion}.
\begin{lem}\label{LRmvUnion}
Let $G$ be a discrete group,
let $X$ be a compact $G$space,
let $U, V \subset X$ be open,
and let $F \subset G \setminus \{ 1 \}$ be finite.
If $(F, U)$ and $(F, V)$ are both inessential,
then so is $(F, \, U \cup V)$.
\end{lem}
\begin{proof}
By definition,
there exist $m, n \in \N$
and \cfn{s}
\[
r_1, r_2, \ldots, r_m, s_1, s_2, \ldots, s_n \colon X \to S^1
\]
such that for every $g \in F$,
we have
\begin{equation}\label{Eq:LRmvUnionEq}
{\mbox{${\displaystyle{
\frac{1}{m} \sum_{j = 1}^m r_j (x) {\overline{r_j (g^{1} x) }} = 0}}$
for $x \in U$}}
\andeqn
{\mbox{${\displaystyle{
\frac{1}{n} \sum_{k = 1}^n s_k (x) {\overline{s_k (g^{1} x) }} = 0}}$
for $x \in V$}}.
\end{equation}
The functions $r_j s_k$ are \cfn{s} from $X$ to~$S^1$,
and we have
\begin{align}\label{Eq:RmvProd}
\lefteqn{
\frac{1}{m n} \sum_{j = 1}^m \sum_{k = 1}^n
(r_j s_k) (x) {\overline{(r_j s_k) (g^{1} x) }} }
\\
& \hspace*{3em}
% = \left(
% \frac{1}{m} \sssum{j = 1}{m} r_j (x) {\overline{r_j (g^{1} x) }}
% \right)
% \left(
% \frac{1}{n} \sssum{k = 1}{n} s_k (x) {\overline{s_k (g^{1} x) }}
% \right).
% \notag
= \Bigg(
\frac{1}{m} \sum_{j = 1}^m r_j (x) {\overline{r_j (g^{1} x) }}
\Bigg)
\left(
\frac{1}{n} \sum_{k = 1}^n s_k (x) {\overline{s_k (g^{1} x) }}
\right).
\notag
\end{align}
By~(\ref{Eq:LRmvUnionEq}),
this product vanishes for $x \in U$ and also for $x \in V$.
\end{proof}
\begin{lem}\label{LRmvGpU}
Let $G$ be a discrete group,
let $X$ be a compact $G$space,
let $U \subset X$ be open,
and let $E, F \subset G \setminus \{ 1 \}$ be finite.
If $(E, U)$ and $(F, U)$ are both inessential,
then so is $(E \cup F, \, U)$.
\end{lem}
\begin{proof}
By definition,
there exist $m, n \in \N$
and \cfn{s}
\[
r_1, r_2, \ldots, r_m, s_1, s_2, \ldots, s_n \colon X \to S^1
\]
such that for every $x \in U$,
we have
\[
% \begin{equation}\label{Eq:LRmvGpUEq}
{\mbox{${\displaystyle{
\frac{1}{m} \sum_{j = 1}^m r_j (x) {\overline{r_j (g^{1} x) }} = 0}}$
for $g \in E$}}
\andeqn
{\mbox{${\displaystyle{
\frac{1}{n} \sum_{k = 1}^n s_k (x) {\overline{s_k (g^{1} x) }} = 0}}$
for $g \in F$}}.
% \end{equation}
\]
The calculation in~(\ref{Eq:RmvProd}) in the proof
of Lemma~\ref{LRmvUnion}
shows that for all $x \in U$ and $g \in E \cup F$,
we have
\[
\frac{1}{m n} \sum_{j = 1}^m \sum_{k = 1}^n
(r_j s_k) (x) {\overline{(r_j s_k) (g^{1} x) }} = 0.
\]
This completes the proof.
\end{proof}
\begin{lem}\label{LRmvOneElt}
Let $G$ be a discrete group,
let $X$ be a free compact $G$space,
and let $F \in G \setminus \{ 1 \}$ be finite.
Then $(F, X)$ is inessential.
\end{lem}
\begin{proof}
Let $g \in G \setminus \{ 1 \}$.
Use compactness of~$X$ and Lemma~\ref{LExistRmv}
to find $n$ and open sets $U_1, U_2, \ldots, U_n \subset X$
such that $( \{ g \}, U_k)$ is inessential
for $k = 1, 2, \ldots, n$
and such that $\bigcup_{k = 1}^n U_k = X$.
Then $n  1$ applications of Lemma~\ref{LRmvUnion}
show that $( \{ g \}, X)$ is inessential.
Since $F$ is finite,
repeated application of Lemma~\ref{LRmvGpU}
implies that $(F, X)$ is inessential.
\end{proof}
% 999 Show we now get Kishimoto's condition?
% (Is the above even useful?)
\begin{prp}\label{PRmvEst}
Let $G$ be a discrete group,
let $X$ be a free compact $G$space,
and let $E \colon C^*_{\mathrm{r}} (G, X) \to C (X)$
be the standard conditional expectation (Definition~\ref{D_StdCond}),
viewed as a map $C^*_{\mathrm{r}} (G, X) \to C^*_{\mathrm{r}} (G, X)$.
Then for every $a \in C^*_{\mathrm{r}} (G, X)$
and $\ep > 0$,
there exist $n \in \N$ and $s_1, s_2, \ldots, s_n \in C (X)$
such that $ s_k (x)  = 1$
for $k = 1, 2, \ldots, n$ and all $x \in X$,
and such that
\[
\Bigg\ E (a)  \frac{1}{n} \sum_{k = 1}^n s_k a s_k^* \Bigg\
% \left\ E (a)  \frac{1}{n} \sssum{k = 1}{n} s_k a s_k^* \right\
< \ep.
\]
\end{prp}
\begin{proof}
Let $\af \colon G \to \Aut (C (X))$
be the induced action (Definition~\ref{DFromSpToCStar}),
that is, $\af_g (f) (x) = f (g^{1} x)$
for $g \in G$, $f \in C (X)$, and $x \in X$.
Also, for $g \in G$ let $u_g \in C^*_{\mathrm{r}} (G, X)$
be the standard unitary (Notation~\ref{N:ug}).
Choose a finite set $F \subset G$
and elements $b_g \in C (X)$ for $g \in G$
such that,
with $b = \sum_{g \in F} b_g u_g$,
we have
\[
\left\ a  b \right\ < \frac{\ep}{2}.
\]
\Wolog{} $1 \in F$.
By Lemma~\ref{LRmvOneElt}
and Definition~\ref{DRemov},
there exist $n \in \N$ and $s_1, s_2, \ldots, s_n \in C (X)$
such that $ s_k (x)  = 1$
for $k = 1, 2, \ldots, n$ and all $x \in X$,
and such that for all $x \in U$ and $g \in F \setminus \{ 1 \}$,
we have
\[
\frac{1}{n} \sum_{k = 1}^n s_k (x) {\overline{s_k (g^{1} x) }} = 0.
\]
Define $P \colon C^*_{\mathrm{r}} (G, X) \to C^*_{\mathrm{r}} (G, X)$
by
\[
P (c) = \frac{1}{n} \sum_{k = 1}^n s_k c s_k^*
\]
for $c \in C^*_{\mathrm{r}} (G, X)$.
We have to show that $\ E (a)  P (a) \ < \ep$.
Since $\ s_k \ = 1$ for all~$k$,
we have $\ P \ \leq 1$.
Therefore
\begin{align*}
\ E (a)  P (a) \
& \leq \ E (a)  E (b) \ + \ E (b)  P (b) \ + \ P (b)  P (a) \
\\
& < \frac{\ep}{2} + \ E (b)  P (b) \ + \frac{\ep}{2}
= \ E (b)  P (b) \ + \ep.
\end{align*}
So it suffices to prove that $P (b) = E (b)$.
Let $g \in F \setminus \{ 1 \}$.
Then
\[
P (b_g u_g)
= \frac{1}{n} \sum_{k = 1}^n s_k b_g u_g s_k^*
= b_g \left( \frac{1}{n} \sum_{k = 1}^n
s_k \af_g (s_k^*) \right) u_g.
\]
Moreover, for $x \in X$,
we have
\[
\frac{1}{n} \sum_{k = 1}^n [s_k \af_g (s_k^*)] (x)
= \frac{1}{n} \sum_{k = 1}^n s_k (x) {\overline{s_k (g^{1} x) }}
= 0.
\]
Thus $P (b_g u_g) = 0$.
Also,
\[
P (b_1 u_1)
% = b_1 \left( \frac{1}{n} \right) \sum_{k = 1}^n s_k s_k^*
= b_1 \cdot \frac{1}{n} \sum_{k = 1}^n s_k s_k^*
= b_1
= E (b).
\]
Thus, $P (b) = E (b)$,
as desired.
\end{proof}
\begin{thm}\label{TFreeMinSimp}
Let $G$ be a discrete group,
and let $X$ be a free minimal compact $G$space.
Then $C^*_{\mathrm{r}} (G, X)$ is simple.
\end{thm}
\begin{proof}
Let $I \subset C^*_{\mathrm{r}} (G, X)$ be a proper closed ideal.
We first claim that $I \cap C (X) = \{ 0 \}$.
If not,
let $f \in I \cap C (X)$ be nonzero.
Choose a nonempty open set $U \subset X$ on which $f$ does not vanish.
By minimality,
we have $\bigcup_{g \in G} g U = X$.
Since $X$ is compact, there is a finite set $S \subset G$
such that $\bigcup_{g \in S} g U = X$.
Define $b \in C (X)$
by
\[
b (x) = \sum_{g \in S} f (g^{1} x) {\overline{ f (g^{1} x) }}
\]
for $x \in X$.
Then $b (x) > 0$ for all $x \in X$, so $b$ is invertible.
For $g \in G$ let $u_g \in C^*_{\mathrm{r}} (G, X)$
be the standard unitary (Notation~\ref{N:ug}).
Then $b = \sum_{g \in S} u_g f f^* u_g^* \in I$.
So $I$ contains an invertible element,
contradicting the assumption that $I$ is proper.
This proves the claim.
Let $E \colon C^*_{\mathrm{r}} (G, X) \to C (X)$
be the standard conditional expectation (Definition~\ref{D_StdCond}),
viewed as a map $C^*_{\mathrm{r}} (G, X) \to C^*_{\mathrm{r}} (G, X)$.
We claim that $E (a) = 0$ for all $a \in I$.
% It suffices to prove that $E (a) \in I \cap C (X)$.
% We already know that $E (a) \in C (X)$.
It suffices to show that $E (a) \in I$.
To prove this,
let $\ep > 0$.
Use Proposition~\ref{PRmvEst} to choose
$n \in \N$ and $s_1, s_2, \ldots, s_n \in C (X)$
such that $ s_k (x)  = 1$
for $k = 1, 2, \ldots, n$ and all $x \in X$,
and such that
\[
\Bigg\ E (a)  \frac{1}{n} \sum_{k = 1}^n s_k a s_k^* \Bigg\
< \ep.
\]
We have $\frac{1}{n} \sum_{k = 1}^n s_k a s_k^* \in I$.
Since $\ep > 0$ is arbitrary,
this implies that $E (a) \in {\overline{I}} = I$.
The claim is proved.
Now let $a \in I$.
For all $g \in G$,
we have $a u_g \in I$, so $E (a u_g) = 0$.
Proposition~\ref{P:Faithful}(\ref{P:Faithful:G})
now implies that $a = 0$.
\end{proof}
We can use the same methods to identify all the tracial states
on $C^*_{\mathrm{r}} (G, X)$.
This result requires that the action be free,
but not necessarily minimal.
The main point is contained in the following proposition.
The proof is taken from the proof of Corollary VIII.3.8 of~\cite{Dvd}.
\begin{prp}\label{PTracesOnSubalg}
Let $G$ be a discrete group,
let $X$ be a free compact $G$space,
and let $A \subset C^*_{\mathrm{r}} (G, X)$
be a subalgebra such that $C (X) \subset A$.
% and such that $C_{\mathrm{c}} (G \times X) \cap A$ is dense in~$A$.
Let $E \colon C^*_{\mathrm{r}} (G, X) \to C (X)$
be the standard conditional expectation (Definition~\ref{D_StdCond}).
Then for every \tst{} $\ta \colon A \to \C$,
there exists a Borel probability measure $\mu$ on~$X$
such that for all $a \in A$ we have
\[
\ta (a) = \int_X E (a) \, d \mu.
\]
% Moreover, $\ta$~is $G$invariant.
\end{prp}
\begin{proof}
We prove that $\ta = (\ta _{C (X)} ) \circ E$.
The statement then follows by applying the Riesz Representation Theorem
to $\ta _{C (X)}$.
Let $a \in A$ and let $\ep > 0$.
We prove that
$ \ta (a)  \ta (E (a))  < \ep$.
Use Proposition~\ref{PRmvEst} to choose
$n \in \N$ and $s_1, s_2, \ldots, s_n \in C (X)$
such that $ s_k (x)  = 1$
for $k = 1, 2, \ldots, n$ and all $x \in X$,
and such that
\[
\Bigg\ E (a)  \frac{1}{n} \sum_{k = 1}^n s_k a s_k^* \Bigg\
< \ep.
\]
Since $s_1, s_2, \ldots, s_n \in A$,
we have $\ta (s_k a s_k^*) = \ta (a)$ for $k = 1, 2, \ldots, n$.
Therefore
\[
\Bigg \ta (a)  \ta (E (a)) \Bigg
= \Bigg \ta \Bigg( \frac{1}{n} \sum_{k = 1}^n s_k a s_k^* \Bigg)
 \ta (E (a)) \Bigg
\leq \Bigg\ E (a)  \frac{1}{n} \sum_{k = 1}^n s_k a s_k^* \Bigg\
< \ep.
\]
This completes the proof.
\end{proof}
\begin{thm}\label{T11202Traces}
Let $G$ be a discrete group,
and let $X$ be a free compact metrizable $G$space.
Let $E \colon C^*_{\mathrm{r}} (G, X) \to C (X)$
be the standard conditional expectation (Definition~\ref{D_StdCond}).
For a $G$invariant Borel probability measure $\mu$ on~$X$,
define a linear functional $\ta_{\mu}$ on $C^*_{\mathrm{r}} (G, X)$
by
\[
\ta_{\mu} (a) = \int_X E (a) \, d \mu.
\]
Then $\mu \mapsto \ta_{\mu}$ is an affine bijection
from the $G$invariant Borel probability measures on~$X$
to the \tst{s} on $C^*_{\mathrm{r}} (G, X)$.
Its inverse sends $\ta$ to the measure obtained from the
functional $\ta _{C (X)}$ via the Riesz Representation Theorem.
\end{thm}
The only reason for restricting to metrizable spaces~$X$
is to avoid the technicalities surrounding
regularity and the uniqueness part of the Riesz Representation Theorem
on spaces which are not second countable.
\begin{proof}[Proof of Theorem~\ref{T11202Traces}]
By Example~\ref{EInvMeas},
if $\mu$ is a $G$invariant Borel probability measure on~$X$,
then $\ta_{\mu}$ is a \tst{} on $C^*_{\mathrm{r}} (G, X)$.
Clearly $\ta_{\mu} (f) = \int_X f \, d \mu$ for $f \in C (X)$.
This implies that $\mu \mapsto \ta_{\mu}$ is injective
and that the description of its inverse is correct on the range
of this map.
It remains only to prove that $\mu \mapsto \ta_{\mu}$ is surjective.
Let $\ta$ be a \tst{} on $C^*_{\mathrm{r}} (G, X)$.
Proposition~\ref{PTracesOnSubalg} provides
a Borel probability measure $\mu$ on~$X$
such that $\ta (a) = \int_X E (a) \, d \mu$
for all $a \in C^*_{\mathrm{r}} (G, X)$.
For $g \in G$ and $f \in C (X)$,
using the fact that $\ta$ is a trace at the second step,
we have
\[
\int_X f (g^{1} x) \, d \mu (x)
= \ta (u_g f u_g^*)
= \ta (f)
= \int_X f \, d \mu.
\]
Uniqueness in the Riesz Representation Theorem
now implies that $\mu$ is $G$invariant.
This completes the proof.
\end{proof}
We now turn to the direct proof of Theorem~\ref{T:AS}.
We need several lemmas,
which are special cases of the corresponding lemmas in~\cite{ArSp}.
\begin{lem}\label{L:MState}
Let $A$ be a \ca,
let $B \subset A$ be a subalgebra,
and let $\om$ be a state on $A$ such that $\om _B$ is multiplicative.
Then for all $a \in A$ and $b \in B$,
we have $\om (a b) = \om (a) \om (b)$
and $\om (b a) = \om (b) \om (a)$.
\end{lem}
This is a special case of Theorem~3.1 of~\cite{Ch}.
(The corresponding lemma in~\cite{ArSp}
also follows from Theorem~3.1 of~\cite{Ch}.)
\begin{proof}[Proof of Lemma~\ref{L:MState}]
We prove $\om (a b) = \om (a) \om (b)$.
The other equation will follow by using adjoints and
the relation $\om (c^*) = {\overline{\om (c)}}$.
If $A$ is not unital,
then $\om$ extends to a state on the unitization $A^+$.
Thus, we may assume that $A$ is unital.
Also, if $\om$ is multiplicative on $B$,
one easily checks that $\om$ is multiplicative on $B + \C \cdot 1$.
Thus, we may assume that $1 \in B$.
We recall from the CauchySchwarz inequality
that $ \om (x^* y) ^2 \leq \om (y^* y) \om (x^* x)$.
Replacing $x$ by $x^*$, we get
$ \om (x y) ^2 \leq \om (y^* y) \om (x x^*)$.
Now let $a \in A$ and $b \in B$.
Then
\[
\big \om (a b)  \om (a) \om (b) \big^2
= \big \om \big( a [b  \om (b) \cdot 1] \big) \big^2
\leq \om \big( [b  \om (b) \cdot 1]^* [b  \om (b) \cdot 1] \big)
\om (a a^*).
\]
Since $\om$ is multiplicative on $B$,
we have
\[
\om \big( [b  \om (b) \cdot 1]^* [b  \om (b) \cdot 1] \big)
= \om \big( [b  \om (b) \cdot 1]^* \big)
\om \big( b  \om (b) \cdot 1) \big)
= 0.
\]
So $\big \om (a b)  \om (a) \om (b) \big^2 = 0$.
\end{proof}
\begin{lem}\label{L:Distinct}
Let $G$ be a discrete group,
and let $X$ be a locally compact $G$space.
Let $x \in X$, let $g \in G$, and assume that $g x \neq x$.
Let $\ev_x \colon C_0 (X) \to \C$ be the evaluation map
$\ev_x (f) = f (x)$ for all $f \in C_0 (X)$,
and let $\om$ be a state on $C^*_{\mathrm{r}} (G, X)$
which extends $\ev_x$.
Then $\om (f u_g) = 0$ for all $f \in C_0 (X)$.
\end{lem}
\begin{proof}
Let $\af \colon G \to \Aut (C_0 (X))$ be $\af_g (f) (x) = f (g^{1} x)$
for $f \in C_0 (X)$, $g \in G$, and $x \in X$
(as in \Def{DFromSpToCStar};
recalled before \Thm{T:AS}).
Choose $f_0 \in C_0 (X)$ such that $f_0 (x) = 1$ and $f_0 (g x) = 0$.
Applying Lemma~\ref{L:MState} to $\om$ at the second and fourth steps,
with $A = C^*_{\mathrm{r}} (G, X)$ and $B = C_0 (X)$,
and using $\om (f_0) = 1$ at the first step,
we have
\begin{align*}
\om (f u_g)
& = \om (f_0) \om (f u_g)
= \om (f_0 f u_g)
= \om \big( f u_g \af_g^{1} (f_0) \big) \\
& = \om (f u_g) \om \big( \af_g^{1} (f_0) \big)
= \om (f u_g) f_0 (g x)
= 0.
\end{align*}
This completes the proof.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{T:AS}]
Let $I \subset C^*_{\mathrm{r}} (G, X)$ be a nonzero closed ideal.
First suppose $I \cap C_0 (X) = 0$.
Choose $a \in I$ with $a \neq 0$.
Let
\[
E \colon C^*_{\mathrm{r}} (G, X) \to C_0 (X)
\]
be the standard conditional expectation
(Definition~\ref{D_StdCond}).
Then $E (a^* a) \neq 0$
by Proposition~\ref{P:Faithful}(\ref{P:Faithful:1}).
Choose $b \in C_{\mathrm{c}} (G, \, C_0 (X), \, \af)$ such that
$\ b  a^* a \ < \tfrac{1}{4} \ E (a^* a) \$.
We can write $b = \sum_{g \in S} b_g u_g$
for some finite set $S \subset G$
and with $b_g \in C_0 (X)$ for $g \in S$.
\Wolog{} $1 \in S$.
Since $E (a^* a)$ is a positive element of $C_0 (X)$,
there is $x_0 \in X$ such that $E (a^* a) (x_0) = \ E (a^* a) \$.
Essential freeness implies that
\[
\big\{ x \in X \colon {\mbox{$g x \neq x$
for all $g \in S \setminus \{ 1 \}$}} \big\}
\]
is the intersection of finitely many dense open subsets of~$X$,
and is therefore a dense open subset of~$X$.
In particular,
there is $x \in X$ so close to $x_0$ that
$E (a^* a) (x) > \tfrac{3}{4} \ E (a^* a) \$,
and also satisfying $g x \neq x$ for all $g \in S \setminus \{ 1 \}$.
The set $C_0 (X) + I$ is a C*subalgebra of $C^*_{\mathrm{r}} (G, X)$.
Let $\om_0 \colon C_0 (X) + I \to \C$ be the following composition:
\[
C_0 (X) + I
\longrightarrow (C_0 (X) + I) / I
\stackrel{\cong}{\longrightarrow} C_0 (X) / (C_0 (X) \cap I)
= C_0 (X)
\stackrel{\ev_x}{\longrightarrow} \C.
\]
Then $\om_0$ is a \hm.
Use the HahnBanach Theorem in the usual way to get a state
$\om \colon C^*_{\mathrm{r}} (G, X) \to \C$ which extends~$\om_0$.
Since $a^* a \in I$, we have $\om (a^* a) = 0$.
We now have, using Lemma~\ref{L:Distinct} at the fifth step,
\begin{align*}
\tfrac{1}{4} \ E (a^* a) \
& > \ b  a^* a \
\geq  \om (b  a^* a) 
=  \om (b) 
= \left \ssum{g \in S} \om ( b_g u_g ) \right
\\
& =  \om (b_1) 
=  \om_0 (b_1) 
=  b_1 (x) 
\geq E (a^* a) (x)  \ E (a^* a)  b_1 \
\\
& \geq E (a^* a) (x)  \ a^* a  b \
> \tfrac{3}{4} \ E (a^* a) \  \tfrac{1}{4} \ E (a^* a) \
= \tfrac{1}{2} \ E (a^* a) \.
\end{align*}
This contradiction shows that $I \cap C_0 (X) \neq 0$.
Since $I \cap C_0 (X)$ is an ideal in $C_0 (X)$,
it has the form $C_0 (U)$ for some nonempty open set $U \subset X$.
We claim that $U$ is $G$invariant.
Let $g \in G$ and let $f \in C_0 (U)$.
Let $(e_{\ld})_{\ld \in \Ld}$ be an approximate identity for $C_0 (X)$.
Then the elements $e_{\ld} u_g$ are in $C^*_{\mathrm{r}} (G, X)$,
and we have
$(e_{\ld} u_g) f (e_{\ld} u_g)^* = e_{\ld} \af_g (f) e_{\ld}$,
which converges to $\af_g (f)$.
We also have $(e_{\ld} u_g) f (e_{\ld} u_g)^* \in I \cap C_0 (X)$,
since $I$ is an ideal.
So $\af_g (C_0 (U)) \subset C_0 (U)$ for all $g \in G$,
and the claim follows.
Since $U$ is open, invariant, and nonempty, we have $U = X$.
One easily checks that an approximate identity for $C_0 (X)$
is also an approximate identity for $C^*_{\mathrm{r}} (G, X)$,
so $I = C^*_{\mathrm{r}} (G, X)$, as desired.
\end{proof}
Theorem~\ref{T:AS} generalizes,
with essentially the same proof,
to crossed products of actions of discrete groups on
noncommutative \ca{s} $A$ satisfying a kind of essential
freeness condition for the action on the
space of unitary equivalence classes of irreducible representations
of $A$.
Here is the general statement;
it is the corollary after Theorem~1 in~\cite{ArSp}.
\begin{thm}\label{T:FullAS}
Let $\af \colon G \to \Aut (A)$ be an action of a
discrete group $G$ on a \ca~$A$.
Suppose that $\af$ is minimal
(\Def{DGSimple}),
that is, $A$ has no nontrivial $\af$invariant ideals.
Suppose further that $\af$ is topologically free,
that is, with ${\widehat{A}}$ being the space of unitary
equivalence classes of irreducible representations of $A$
with the hullkernel topology,
the following property holds:
for every finite set $F \subset G \setminus \{ 1 \}$,
the set
\[
\big\{ x \in {\widehat{A}}
\colon {\mbox{$g x \neq x$ for all $g \in F$}} \big\}
\]
is dense in ${\widehat{A}}$.
Then $C^*_{\mathrm{r}} (G, A, \af)$ is simple.
\end{thm}
The changes to the proof include using irreducible representations
in place of the maps $\ev_x$,
and completely positive maps to $L (H)$
in place of states.
As discussed in~\cite{ArSp}
(see the remark after the corollary after Theorem~1),
this result implies Theorem~3.1 of~\cite{Ks1}.
We state the following important special case:
\begin{thm}\label{T_Ks}
Let $\af \colon G \to \Aut (A)$ be an action of a
discrete group $G$ on a simple \ca~$A$.
Suppose that $\af_g$ is outer for every $g \in G \setminus \{ 1 \}$.
Then $C^*_{\mathrm{r}} (G, A, \af)$ is simple.
\end{thm}
The original proof of Theorem~3.1 of~\cite{Ks1}
proceeded via Kishimoto's condition
(\Def{DKshCond})
and a generalization of Theorem~\ref{TAKCImpSimp}.
Theorem~\ref{T_Ks} fails for actions of~$\R$ and~$S^1$.
We give examples based on calculations in~\cite{Kt1}
(originally Theorem~4.4 of~\cite{Kshm1},
but the statement in~\cite{Kt1} is more explicit).
See the beginning of Section~2
and Definition 2.1 of~\cite{Kt1} for the notation.
Theorem~4 of~\cite{ETW} shows that the automorphisms
which appear there are all outer unless they are trivial.
We will use Theorem~4.8 of~\cite{Kt1},
verifying condition (iii) there.
(See Definition 4.7 of~\cite{Kt1} for the notation.)
For $S^1$ take $n = 2$, take $G = S^1$ (so that $\Gm = \Z$),
and take $\om_1 = 1$ and $\om_2 = 0$.
Then $\Om_2 = \Nz \neq \Z$,
so the crossed product is not simple.
The action one gets this way is the action
$\af \colon S^1 \to \Aut ({\mathcal{O}}_2)$
determined by $\af_{\zt} (s_1) = \zt s_1$
and $\af_{\zt} (s_2) = s_2$ for $\zt \in S^1$.
It is the restriction of an action from
Example~\ref{E:CuntzGauge} to a subgroup.
For~$\R$,
take $n = 3$, take $G = \R$ (so that $\Gm = \R$),
and take $\om_1 = 1$, $\om_2 = \sqrt{2}$, and $\om_3 = 0$.
Then $\Om_3 \subset \{ 0 \} \cup [1, \I) \neq \R$,
so again the crossed product is not simple.
The action one gets this way is the action
$\bt \colon \R \to \Aut ({\mathcal{O}}_3)$
determined by
\[
\bt_{t} (s_1) = \exp (i t) s_1,
\qquad
\bt_{t} (s_2) = \exp (i \sqrt{2} t) s_2,
\andeqn
\bt_{t} (s_3) = s_3
\]
for $t \in \R$.
It is the restriction of an action from
Example~\ref{E:CuntzGauge} to a nonclosed subgroup.
For~$S^1$,
alternatively,
consider of the usual gauge action of $S^1$ on~${\mathcal{O}}_{\I}$
(the restriction of the action of Example~\ref{E_3306_OIGauge}
to the scalar multiples of the identity).
Its strong Connes spectrum
is $\N$
(Remark~5.2 of~\cite{Kshm1}),
so its crossed product is not simple (Theorem~3.5 of~\cite{Kshm1}).
This action is pointwise outer by Theorem~4 of~\cite{ETW}.
\section{Classifiability: Introduction and a Special
Case}\label{Sec:Class}
% There is discussion of other results on the Lisbon lecture
% slides, which should be added here. 999
\indent
We discuss the structure and classification
of \tgca{s} of \mh{s}.
We will later say a little about free minimal actions of
more complicated groups, but less is known.
Our first main goal is the main result of~\cite{LhP}
(Theorem~4.6 there),
which gives conditions under which $C^* (\Z, X, h)$ has
tracial rank zero.
(See Definition~\ref{D_TR0}.)
Such \tgca{s} are automatically nuclear and satisfy
the Universal Coefficient Theorem,
so the conditions we give imply that $C^* (\Z, X, h)$
is in a class covered by a classification theorem.
Here is the statement;
the map $\rh$ will be explained afterwards (Definition~\ref{DRhoA}),
along with a reformulation
of the condition involving it which does not
mention Ktheory (Remark~\ref{RDenseRange}).
For any compact metric space~$X$,
we let $\dim (X)$ be its covering dimension.
(We sometimes just refer to dimension.)
See the discussion starting after Corollary~\ref{C:C1}.
% For example, see Definition~3.1.1 of~\cite{Prs}.
This theorem is not the best known result;
by now, classifiability and related results
are known under much more general
conditions.
For example,
see Theorem~0.1 and Theorem~0.2 of~\cite{TW},
and also Theorem~\ref{T_5526_ENZStab} (from~\cite{EN2}).
Classifiability is known to fail for some minimal homeomorphisms,
such as those of~\cite{GlKr}.
% 999 More ref
\begin{thm}[Theorem~4.6 of~\cite{LhP}]\label{T:TM}
Let $X$ be an infinite compact metric space
with finite covering dimension,
and let $h \colon X \to X$ be a minimal homeomorphism.
Suppose that $\rh \big( K_0 (C^* (\Z, X, h)) \big)$
is dense in $\Aff \big( \T (C^* (\Z, X, h)) \big)$.
Then $C^* (\Z, X, h)$ is a simple unital \ca{}
with tracial rank zero
which satisfies the Universal Coefficient Theorem.
\end{thm}
% 999 Never said anything specifically about simplicity or
% the UCT here.
There is machinery available to compute the range of $\rh$
in the above theorem without computing $C^* (\Z, X, h)$.
See, for example,~\cite{Ex2}.
\begin{cor}[Corollary~4.7 of~\cite{LhP}]\label{C:C1}
Let $X$ be an infinite compact metric space
with finite covering dimension,
and let $h \colon X \to X$ be a minimal homeomorphism.
Suppose that $\rh \big( K_0 (C^* (\Z, X, h)) \big)$
is dense in $\Aff \big( \T (C^* (\Z, X, h)) \big)$.
Then $C^* (\Z, X, h)$ is a simple AH~algebra with no dimension
growth and with real rank zero.
\end{cor}
We give a brief explanation of dimension for compact spaces,
with definitions but without proofs,
to put the finite dimensionality hypothesis
of \Thm{T:TM}
in context.
This material is also background for the discussion of
the mean dimension of a \hme{}
(\Def{D_6827_MDim}).
Dimension theory attempts to assign a dimension to
each topological space (usually in some restricted class)
in such a way as to generalize the dimension of a manifold,
in particular,
the relation $\dim (\R^n) = n$,
and to preserve expected properties of the dimension.
There are a number of books on dimension theory;
the one I have so far found most useful is~\cite{Prs}.
(A warning on terminology there:
``bicompact'' is used for ``compact Hausdorff''.
See Definition 1.5.4 of~\cite{Prs}.)
The mean dimension of a \hme~$h$ of a space~$X$
should perhaps be thought of as saying
how much more of the space $X$ one sees
with every iteration of~$h$,
with ``how much one sees'' being measured in some sense
by dimension.
There are at least two quite different general approaches
to the problem of assigning dimensions to spaces.
One assumes the existence of a metric,
and attempts to quantify how the ``size'' of a ball in the
space shrinks with its radius.
This approach leads to the Hausdorff dimension and its relatives.
The result depends on the metric, need not be an integer,
and can be strictly positive for the Cantor set
(depending on the metric one uses).
Such dimensions have so far played no role in the structure
theory of \ca{s},
which is not surprising since $C (X)$ does not depend on the metric
on~$X$.
The approach more useful here relies entirely
on topological properties of~$X$,
takes integer values,
and is zero on the Cantor set,
regardless of the metric.
The three most well known dimension theories of this
kind are the covering dimension $\dim (X)$
(Section~3.1 of~\cite{Prs}),
the small inductive dimension ${\operatorname{ind}} (X)$
(Section~4.1 of~\cite{Prs}),
and the large inductive dimension ${\operatorname{Ind}} (X)$
(Section~4.2 of~\cite{Prs}).
There are three others that should be mentioned:
for compact~$X$,
the topological stable rank $\tsr (C (X, \R))$
of the algebra $C (X, \R)$
of \ct{} {\emph{real}} valued functions on~$X$
(topological stable rank is discussed briefly after \Def{D:SR1}
but for complex \ca{s});
% mainly just the case $\tsr (A) = 1$),
for metrizable~$X$ the infimum, over
all metrics $\rh$ defining the topology, of the Hausdorff
dimension of $(X, \rh)$;
and for compact metrizable~$X$ the cohomological dimension
as described in~\cite{Drnv} (with integer coefficients).
For nonempty compact metrizable~$X$,
these all agree
(except that one must use $\tsr (C (X, \R))  1$,
and in the case of cohomological dimension with integer coefficients
require that $\dim (X) < \I$),
and for specific pairs of dimension theories,
it is often known that they agree under much weaker conditions.
For $\dim (X)$, ${\operatorname{ind}} (X)$,
and ${\operatorname{Ind}} (X)$
see Corollary 4.5.10 of~\cite{Prs}.
Agreement with $\tsr (C (X, \R))  1$
is essentially Proposition 3.3.2 of~\cite{Prs}
(not stated in that language).
Agreement with the infimum of the
Hausdorff
dimensions of $(X, \rh)$
is in Section~7.4 of~\cite{HrwWlm}.
When $\dim (X) < \I$,
agreement with cohomological dimension with integer coefficients
is Theorem~1.4 of~\cite{Drnv};
without the condition $\dim (X) < \I$,
Theorem 7.1 of~\cite{Drnv} shows that agreement can fail.
We give a two warnings about dimension theories.
First,
they find the dimension of the highest dimensional part of the space.
A space like $\R^n$ is homogeneous
(in a very strong sense: the diffeomorphism group acts transitively),
as is a connected compact manifold without boundary.
Even for a connected compact manifold with boundary,
it seems intuitively clear that the dimension as seen
at any point should be the same.
However,
a finite complex
or a disconnected compact manifold
may well have parts which should be considered to have
different dimensions.
All dimension theories I know of
assign to a finite simplicial complex the dimension
given by the largest standard
(combinatorial) dimension of any of its simplices,
even if there are other simplices of much lower dimension
which are not contained in any higher dimensional simplex.
There are at least some notions of ``local dimension'' at a point,
which attempt to account for this kind of behavior,
but the theory seems to be much less well developed.
We will primarily be interested in spaces~$X$
which admit minimal homeomorphisms,
or minimal actions of other countable groups.
Such spaces clearly have at least a weak form of homogeneity,
since each orbit is dense
and the action is transitive on orbits.
We know little about what one can really get from this,
but Example~\ref{E_6913_GJ} shows
that it does not imply that the local dimension
is the same at every point.
Second,
they do not necessarily have the properties one expects,
or the properties they have are weaker than what one expects.
Some such examples are presented or at least mentioned~\cite{Prs}.
We list just a few.
The notes to Chapter~8 of~\cite{Prs}
mention an example due to Filippov:
if $1 \leq m \leq n \leq 2 m  1$,
there is a \chs~$X$ such that
\[
\dim (X) = 1,
\qquad
{\operatorname{ind}} (X) = m,
\andeqn
{\operatorname{Ind}} (X) = n.
\]
The conventions usually take $\dim (\E) =  1$,
so that the standard inequality
%
\begin{equation}\label{Eq_7131_DimProd}
\dim (X \times Y) \leq \dim (X) + \dim (Y)
\end{equation}
%
fails
if $X = \E$ or $Y = \E$.
However,
this inequality can fail even if $X \neq \E$ and $Y \neq \E$.
See Example 9.3.7 of~\cite{Prs}.
If $X$ and $Y$ are nonempty \chs{s},
then (\ref{Eq_7131_DimProd}) does hold
% then one does get $\dim (X \times Y) \leq \dim (X) + \dim (Y)$
(Proposition 3.2.6 of~\cite{Prs};
see Section 9.3 of~\cite{Prs}
for an assortment of weaker conditions
under which (\ref{Eq_7131_DimProd}) holds).
There are nonempty \cms{s}
$X$ and $Y$ such that
$\dim (X \times Y) < \dim (X) + \dim (Y)$;
in~\cite{Drnv}
combine Example 1.3(1), Example 1.9,
and the example after Corollary~3.8.
There is even a nonempty \cms~$X$
such that $\dim (X \times X) < 2 \dim (X)$;
for example, combine \cite{McRb} and~\cite{Krsz}.
Two expected properties that are true are given
in Proposition~\ref{P_7122_DimSubsp}
and Proposition~\ref{P_7122_ClCover}.
The dimension theory most useful so far for \mh{s}
is the covering dimension,
which is defined using open covers.
We thus start by stating the basic concepts
used to define the covering dimension.
We make all our definitions for finite open covers of
\chs{s},
although the earlier ones make sense in much greater generality
(for more general spaces,
not requiring that the covers be open,
and sometimes not even requiring that
the covers be finite).
By a finite open cover $\cU$ of a \chs~$X$,
we mean a finite collection $\cU$ of open subsets of~$X$
such that $X = \bigcup_{U \in \cU} U$.
(This convention follows~\cite{LndWs}.)
Possibly
(following Section~3.1 of~\cite{Prs})
one should instead use indexed families
$(U_i)_{i \in I}$ of open subsets,
for a finite index set~$I$;
this formulation allows repetitions among the sets.
We will not need this refinement.
(It is easy to check that it makes no difference
in the definition of covering dimension,
since one can simply delete repeated sets.)
\begin{ntn}\label{N_6827_Cov}
Let $X$ be a \chs.
We write ${\operatorname{Cov}} (X)$
for the set of all finite open covers of~$X$.
\end{ntn}
\begin{dfn}\label{D_6827_Ord}
Let $X$ be a \chs,
and let $\cU$ be a finite open cover of~$X$.
The {\emph{order}} $\ord (\cU)$ of $\cU$
is the least number $n \in \N$ such that the intersection
of any $n + 2$ distinct elements of~$\cU$ is empty.
\end{dfn}
That is,
$\ord (\cU)$ is the largest $n \in \N$
such that there are $n + 1$ distinct sets in~$\cU$
whose intersection is not empty.
An alternative formulation is
\[
\ord (\cU) =  1 + \sup_{x \in X} \sum_{U \in \cU} \chi_U (x).
\]
The normalization is chosen so that
if $\cU$ is cover of $X$ by disjoint open sets,
and $X \neq \E$,
then $\ord (\cU) = 0$:
the intersection of any two distinct sets
in $\cU$ is empty,
but the sets themselves need not be empty.
\begin{dfn}\label{D_6827_Rfn}
Let $X$ be a \chs,
and let $\cU$ and $\cV$ be finite open covers of~$X$.
Then $\cV$ {\emph{refines}} $\cU$ (written $\cV \prec \cU$)
if for every $V \in \cV$ there is $U \in \cU$
such that $V \subset U$.
\end{dfn}
That is,
every set in $\cV$ is contained in some set in~$\cU$.
\begin{dfn}\label{D_6827_DimU}
Let $X$ be a \chs,
and let $\cU$ be a finite open cover of~$X$.
We define the {\emph{dimension}} $\cD (\cU)$ of $\cU$
by
\[
\cD (\cU)
= \inf \big( \big\{ \ord (\cV) \colon
{\mbox{$\cV \in {\operatorname{Cov}} (X)$ and $\cV \prec \cU$}}
\big\} \big).
\]
\end{dfn}
That is,
$\cD (\cU)$ is the least possible order
of a finite open cover which refines~$\cU$.
\begin{dfn}[Definition 3.1.1 of~\cite{Prs}]\label{D_6827_dimX}
Let $X$ be a nonempty \chs.
The {\emph{covering dimension}} $\dim (X)$
is
\[
\dim (X)
= \sup \big( \big\{ \cD (\cU) \colon
\cU \in {\operatorname{Cov}} (X) \big\} \big).
\]
By convention,
$\dim (\E) =  1$.
\end{dfn}
That is,
$\dim (X)$ is the supremum of $\cD (\cU)$
over all finite open covers $\cU$ of~$X$.
We will say that a \chs~$X$
is {\emph{totally disconnected}}
if there is a base for the topology of~$X$ consisting
of compact open sets.
(This seems to be the standard definition for this class of spaces.
In~\cite{Prs},
a different definition is used,
but for \chs{s} it is equivalent.
See Proposition 3.1.3 of~\cite{Prs}.)
\begin{exr}[Proposition 3.1.3 of~\cite{Prs}]\label{Ex_6913_ZeroDim}
Let $X$ be a nonempty \chs.
Prove that $\dim (X) = 0$ \ifo{} $X$ is totally disconnected.
\end{exr}
\begin{exr}\label{Ex_6913_OneDim}
Prove that $\dim ([0, 1]) = 1$.
\end{exr}
We have $\dim ([0, 1]) \neq 0$
by Exercise~\ref{Ex_6913_ZeroDim}.
To show $\dim ([0, 1]) \leq 1$,
consider open covers of $[0, 1]$
consisting of intervals
\[
[0, \bt_0), \,
(\af_1, \bt_1), \,
(\af_2, \bt_2), \,
\ldots, \,
(\af_{n  1}, \bt_{n  1}), \,
(\af_n, 1]
\]
such that
$\af_j \leq \bt_{j  1}$
but $\bt_{j  1} < \af_{j + 1}$,
and $\bt_j  \af_j$ is small,
for all~$j$.
The intervals this cover $[0, 1]$,
but $[0, \bt_0)$ is disjoint from $(\af_2, \bt_2)$, etc.
One sees that $\dim ([0, 1]^2) \leq 2$
by using open covers consisting of small
neighborhoods of the tiles in a fine hexagonal tiling.
In general,
one has $\dim (\R^n) = n$
(Theorem 3.2.7 of~\cite{Prs}),
but proving this is nontrivial.
Most proofs rely on some version of the
Brouwer Fixed Point Theorem,
and thus, in effect, on algebraic topology.
\begin{prp}[Proposition 3.1.5 of~\cite{Prs}]\label{P_7122_DimSubsp}
Let $X$ be a topological space
and let $Y \subset X$ be closed.
Them $\dim (Y) \leq \dim (X)$.
\end{prp}
\begin{prp}[Special case of
Theorem 3.2.5 of~\cite{Prs}]\label{P_7122_ClCover}
Let $X$ be a \chs{}
and let $Y_1, Y_2, \ldots, Y_n \subset X$ be closed
subsets such that $X = \bigcup_{k = 1}^n Y_k$.
Them $\dim (X) \leq \max_{1 \leq k \leq n} \dim (Y_k)$.
\end{prp}
% 999 Move this material on traces to Section 11,
% where traces are defined?
We now give the definitions related to the map~$\rh$
which appears in the statement of Theorem~\ref{T:TM}.
Recall from \Def{D_Trace}
that a \tst\ on $A$ is a state $\ta$ on~$A$
such that $\ta (b a) = \ta (a b)$
for all $a, b \in A$,
and
that $\T (A)$
is the set of all \tst{s} on~$A$,
equipped with the relative weak* topology inherited from
the Banach space dual of~$A$.
\begin{rmk}\label{RTACptCnv}
Let $A$ be a unital \ca.
Then $\T (A)$ is a compact convex subset of
the Banach space dual of~$A$ (with the weak* topology).
Convexity is immediate,
and compactness follows from the fact that $\T (A)$ is closed
in the set of all states on~$A$.
\end{rmk}
If $A$ is not unital,
then compactness can fail.
\begin{dfn}\label{DAffDelta}
Let $E$ be a topological vector space,
and let $\Dt \subset E$ be a compact convex set.
We let $\Aff (\Dt)$ be
the real Banach space of real valued \ct{} affine
functions on~$\Dt$,
with the supremum norm.
\end{dfn}
We will need a condition which is normally expressed using
the following map involving the $K_0$group of a \ca.
However, the condition can be stated without using Ktheory,
and we give the explanation afterwards.
\begin{dfn}\label{DRhoA}
Let $A$ be a unital \ca.
We let $\rh_A \colon K_0 (A) \to \Aff (\T (A))$
be the \hm{} determined by $\rh ([p]) (\ta) = \ta (p)$
for $\ta \in \T (A)$ and $p$ a \pj\ in some matrix algebra over~$A$.
The trace $\ta$ is taken to be defined on
$M_n (A)$ via the unnormalized version of the \tst{}
in \Ex{ETMnOfA}.
That is,
we define $\ta$ on $M_n (A)$ by
\[
\ta \left( \left( \begin{matrix}
a_{1, 1} & a_{1, 2} & \cdots & a_{1, n} \\
a_{2, 1} & a_{2, 2} & \cdots & a_{2, n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{n, 1} & a_{n, 2} & \cdots & a_{n, n}
\end{matrix} \right) \right)
= \sum_{k = 1}^{n} \ta (a_{k, k}).
\]
The map $\rh$ is well defined by
Lemma \ref{LTraceAndMvN}(\ref{LTraceAndMvN1})
(extended with the same proof to traces which don't necessarily
have norm~$1$).
\end{dfn}
When $A$ is clear from the context,
we often abbreviate $\rh_A$ to~$\rh$.
\begin{rmk}\label{RDenseRange}
We will often use the hypothesis
that the map $\rh_A \colon K_0 (A) \to \Aff (\T (A))$
of Definition~\ref{DRhoA} have dense range.
This hypothesis can be stated without using Ktheory as follows.
Let $R \subset \Aff (\T (A))$ be the set
all functions $\ta \mapsto \ta (p)$,
as $p$ runs through all the \pj{s} in $M_n (A)$
for all~$n$
(using the notation of \Def{DRhoA}).
Then the condition is that the additive subgroup of $\Aff (\T (A))$
generated by $R$ be dense in $\Aff (\T (A))$.
\end{rmk}
In the rest of this section,
we give the proof of Theorem~\ref{T:TM}
in the special case in which $\dim (X) = 0$,
following the method of~\cite{LhP}.
Since $X$ is assumed to be infinite and to admit a \mh,
it can have no isolated points,
and therefore must be the Cantor set.
This restriction simplifies the argument greatly.
In particular,
one need not deal with recursive subhomogeneous \ca{s},
KKtheory,
or subsets of~$X$ with ``small boundary''.
We will give some parts of the proof of the general case
in the next section,
but we will have to cite several theorems without giving proofs.
It is implicit in Section~8 of~\cite{HPS}, with the main step
having been done in~\cite{Pt2},
that these \tgca{s} are AT~algebras
(direct limits of circle algebras) with real rank zero.
The result we prove is weaker and the proof is longer.
General theory
(Lin's classification theorem for \ca{s} with tracial rank zero,
Theorem~5.2 of~\cite{Ln3},
a Ktheory calculation
using the PimsnerVoiculescu exact sequence~\cite{PV},
and results on the range of the Elliott invariant)
shows that the AT~algebra result follows from the theorem
we prove here.
Our reason for giving this proof is to illustrate a technical method.
\begin{lem}\label{L:TRAF}
Let $A$ be a simple unital \ca.
Suppose that
for every finite subset $F \subset A$, every $\ep > 0$,
and every nonzero positive element $c \in A$,
there exists a nonzero \pj{} $p \in A$
and a unital AF~subalgebra $B \subset A$
with $p \in B$
such that:
\begin{enumerate}
\item\label{L:TRAF:Comm}
$\ [a, p] \ < \ep$ for all $a \in F$.
\item\label{L:TRAF:Close}
$\dist (p a p, \, p B p) < \ep$ for all $a \in F$.
\item\label{L:TRAF:Small}
$1  p$ is Murrayvon Neumann equivalent
to a \pj{} in ${\overline{c A c}}$.
\end{enumerate}
Then $A$ has tracial rank zero (\Def{D_TR0}).
\end{lem}
\begin{proof}
Let $F \subset A$ be a finite subset, let $\ep > 0$,
and let $c \in A$ be a nonzero positive element.
Choose $p$ and $B$ as in the hypotheses,
with $F$ and $c$ as given and with $\frac{1}{2} \ep$ in place of $\ep$.
Let $F_0 \subset p B p$ be a finite set such that
$\dist (p a p, \, F_0) < \frac{1}{2} \ep$ for all $a \in F$.
Since $p \in B$, the algebra $p B p$ is also~AF.
Choose a unital finite dimensional
subalgebra $D \subset p B p$ such that
$\dist (b, D) < \frac{1}{2} \ep$ for all $b \in F_0$.
Then $\dist (p a p, \, D) < \ep$ for all $a \in F$.
\end{proof}
\begin{exr}\label{Ex:TRGen}
Let $A$ be a unital \ca, and let $S \subset A$ be a subset
which generates $A$ as a \ca.
Assume that the condition of Lemma~\ref{L:TRAF}
holds for all finite subsets $F \subset S$.
Prove that $A$ has tracial rank zero.
\end{exr}
\begin{dfn}\label{D_5421_VSubalg}
Let $X$ be a compact metric space,
and let $h \colon X \to X$ be a homeomorphism.
In the \tgca{} $C^* (\Z, X, h)$, we normally write
$u$ for the standard unitary representing the generator of~$\Z$.
(This unitary is called $u_1$ in Notation~\ref{N:ug}.)
For a closed subset $Y \subset X$, we define
the C*subalgebra $C^* (\Z, X, h)_Y$ to be
\[
C^* (\Z, X, h)_Y
= C^* ( C (X), \, C_0 (X \setminus Y) u) \subset C^* (\Z, X, h).
\]
% We will sometimes let $A$ denote the
% \tgca{} $C^* (\Z, X, h)$, in which case we refer to $A_Y$.
We call it the {\emph{$Y$orbit breaking subalgebra}}
of $C^* (\Z, X, h)$.
\end{dfn}
This subalgebra was introduced by Putnam in Section~3 of~\cite{Pt1},
specifically in the case that $X$ is the Cantor set.
There, and in all subsequent papers,
the definition
\[
C^* (\Z, X, h)_Y
= C^* ( C (X), \, u C_0 (X \setminus Y)) \subset C^* (\Z, X, h).
\]
was used.
As will be seen in the course of the proof of \Lem{L:AF},
and later,
the analysis of the structure of $C^* (\Z, X, h)_Y$
for $\sint (Y) \neq \E$
depends on Rokhlin towers constructed from~$Y$.
When $Y$ is compact and open,
the Rokhlin towers take a standard form,
given in~(\ref{Eq_4315_Rokhlin}) below:
there are positive integers $n (0) < n (1) < \cdots < n (l)$
and subsets $Y_0, Y_1, \ldots, Y_l \subset Y$
such that
\[
Y = \coprod_{k = 0}^l Y_k
\andeqn
X = \coprod_{k = 0}^l \coprod_{j = 0}^{n (k)  1} h^j (Y_k).
\]
The sequences $Y_k, \, h (Y_k), \ldots, h^{n (k)  1} (Y_k)$
are called {\emph{Rokhlin towers}}.
The sets $Y_k$ are the {\emph{bases}} of the towers,
and the numbers $n (k)$ are their {\emph{heights}}.
The reason for changing the convention in the definition
of $C^* (\Z, X, h)_Y$ is that the old convention
leads to Rokhlin towers with bases
$h (Y_0), h (Y_1), \ldots, h (Y_l)$ instead of
$Y_0, Y_1, \ldots, Y_l$,
so that the useful partition of~$X$ becomes
\[
X = \coprod_{k = 0}^l \coprod_{j = 1}^{n (k)} h^j (Y_k).
\]
We do not need groupoids at this stage,
but they do seem to be needed for
useful analogs of $C^* (\Z, X, h)_Y$
in more general situations.
They are also used in the usual computation of the
Ktheory of $C^* (\Z, X, h)_{ \{ y \} }$
for $y \in X$;
see the discussion of the proof of
Theorem~\ref{T_7122_KThyAy}
(after the proof of Lemma~\ref{L:TABij}).
We therefore describe briefly how to realize
$C^* (\Z, X, h)_Y$ in terms of groupoids.
Readers not familiar with groupoids should skip this description.
% referring ahead to the sections on groupoids
% and their \ca{s},
% starting with Section~\ref{Sec:Groupoids}.
\begin{rmk}\label{R:Gpoid}
The algebra
$C^* (\Z, X, h)_Y$ is the \ca{} of a subgroupoid of the
transformation group groupoid $\Z \ltimes X$
% (see Example~\ref{EGpdTGG} below)
made from the action of $\Z$ on $X$ generated by~$h$.
Informally,
we ``break'' every orbit each time it goes through~$Y$.
Here is a more formal description.
% We follow the notation of Example~\ref{EGpdTGG}.
% (See Remark~\ref{RTggNotation_Old} for the more common notation.)
The notation here differs slightly from the most common notation.
We take $\Z \ltimes X$
to be the set
\[
\big\{ (h^n (x), \, n, \, x) \colon
{\mbox{$x \in X$ and $n \in \Z$}} \big\}
\subset X \times \Z \times X,
\]
with the groupoid operation determined by
\[
(h^{m + n} (x), \, m, \, h^n (x)) \cdot (h^n (x), \, n, \, x)
= (h^{m + n}, \, m + n, \, x)
\]
and with other products undefined.
This is the transformation group groupoid
made from the action of $\Z$ on~$X$ generated by~$h$.
See Example 1.2a of~\cite{Rn},
which, as well as having different notation,
uses a right action instead of a left action.
With this notation,
$C^* (\Z, X, h)_Y$ is the \ca{} of the open subgroupoid
$G \subset \Z \ltimes X$
which contains $(x, 0, x)$ for all $x \in X$,
and such that for $n \in \N$ and $x \in X$ we have
$(h^n (x), \, n, \, x) \in G$
\ifo{} $h (x), h^2 (x), \ldots, h^n (x) \in X \setminus Y$
and
$(h^{n} (x), \, n, \, x) \in G$
\ifo{} $x, h^{1} (x), \ldots, h^{ n + 1} (x) \in X \setminus Y$.
% Forward reference to definition of orbit of a groupoid needed. % 999
If $Y$ has nonempty interior,
then all the orbits are finite,
and the orbit of $x \in X$ is as follows.
Let $j_0 \leq 0$ be the greatest nonpositive integer such that
$h^{j_0} (x) \in Y$,
and let $j_1 > 0$ be the least strictly positive integer such that
$h^{j_1} (x) \in Y$.
Then the orbit of $x$ is
\[
h^{j_0} (x), \, h^{j_0 + 1} (x), \, \ldots, \,
h^{j_1  2} (x), \, h^{j_1  1} (x).
\]
%
% For actions of $\Z^d$,
% it appears to be necessary to use subalgebras of the crossed product
% for which the only nice description is in terms of
% subgroupoids of the transformation group groupoid.
% See~\cite{Ph10}.
\end{rmk}
% Expand: Put in something from the groupoid notes? 999
The following lemma is a special case of Theorem~3.3 of~\cite{Pt1}.
Note the standing assumption of minimality throughout~\cite{Pt1},
stated in Section~1 there.
(Theorem~3.3 of~\cite{Pt1} does not assume that $Y$ is open.
The requirement that $Y$ be open is easily removed
by choosing compact open sets $Y_1 \supset Y_2 \supset \cdots$
in $X$ such that $\bigcap_{n = 1}^{\infty} Y_n = Y$,
and observing that
$C^* (\Z, X, h)_Y$ is the closure of the increasing union of the
subalgebras $C^* (\Z, X, h)_{Y_n}$.
Compare with Remark~\ref{R_7121_LimOfSubalgs} below.)
\begin{lem}\label{L:AF}
Let $X$ be the Cantor set,
and let $h \colon X \to X$ be a \mh.
Let $Y \subset X$ be a nonempty compact open subset.
Then $C^* (\Z, X, h)_Y$ is an AF~algebra.
\end{lem}
\begin{proof}
The proof depends on the construction of Rokhlin towers,
which is a crucial element of many structure results for \cp{s}.
(For the construction in more general spaces~$X$,
see \Def{D_4319_FirstRet},
\Lem{L_4319_RokhFin},
\Def{D_4319_ModRokh},
and \Lem{RokhCov}.)
% We first claim that there is $N \in \N$ such that
% $\bigcup_{n = 1}^{N} h^{n} (Y) = X$.
% Set $Z = X \setminus \bigcup_{n = 1}^{\infty} h^{n} (Y)$.
% Then $Z$ is closed, and $h (Z) \subset Z$.
% By (\ref{L:MHC:Min}) implies~(\ref{L:MHC:FMin}) of Lemma~\ref{L:MHC},
% we have $Z = \varnothing$.
% The claim now follows from compactness of~$X$.
We first claim that there is $N \in \N$ such that
$\bigcup_{n = 1}^{N} h^{n} (Y) = X$.
(This is \Lem{L_4319_RokhFin} in the general case.)
Set $U = \bigcup_{n = 1}^{\infty} h^{n} (Y)$,
which is a nonempty open subset of~$X$ such that $U \subset h (U)$.
Then $Z = X \setminus \bigcup_{n = 1}^{\infty} h^{n} (Y)$
is a closed subset of~$X$ such that $h (Z) \subset Z$,
and $Z \neq X$.
Therefore $Z = \varnothing$ by \Lem{L:MHC}.
So $U = X$, and the claim now follows from compactness of~$X$.
It follows that for each fixed $y \in Y$, the sequence of iterates
$h (y), \, h^2 (y), \dots$ of $y$ under $h$
must return to $Y$ in at most $N$ steps.
Define the {\emph{first return time}} $r (y)$ to be
\[
r (y) =
\min \big( \big\{ n \geq 1 \colon h^n (y) \in Y \big\} \big)
\leq N.
\]
(This is \Def{D_4319_FirstRet} in the general case.)
Let $n (0) < n (1) < \cdots < n (l) \leq N$
be the distinct values of~$r$.
Set
\[
Y_k = \big\{ y \in Y \colon r (y) = n (k) \big\}.
\]
Then the sets $Y_k$ are compact, open, and partition $Y$,
and the sets $h^j (Y_k)$,
for $1 \leq j \leq n (k)$, partition $X$:
%
\begin{equation}\label{Eq_4315_Rokhlin}
Y = \coprod_{k = 0}^l Y_k
\andeqn
X = \coprod_{k = 0}^l \coprod_{j = 0}^{n (k)  1} h^j (Y_k).
\end{equation}
%
% Each finite sequence
% $Y_k, \, h (Y_k), \ldots, h^{n (k)  1} (Y_k)$
% is a {\emph{Rokhlin tower}} with base $Y_k$ and height~$n (k)$.
% (It is more common to let the power of $h$
% run from $0$ to $n (k)  1$.
% The choice made here,
% effectively taking the base of the collection of Rokhlin
% towers to be $h (Y)$ rather than~$Y$,
% is more convenient for use
% with our definition of $C^* (\Z, X, h)_Y$.)
Further set $X_k = \bigcup_{j = 0}^{n (k)  1} h^j (Y_k)$.
The sets $X_k$ then also partition~$X$.
(This part is much messier in the general case;
see \Def{D_4319_ModRokh}
and \Lem{RokhCov}.)
Define $p_k \in C (X) \subset C^* (\Z, X, h)_Y$
by $p_k = \ch_{X_k}$.
We claim that $p_k$ commutes with all elements of $C^* (\Z, X, h)_Y$.
It suffices to prove that $p_k$
commutes with all elements of $C (X)$
and with all elements $f u$
with $f \in C_0 (X \setminus Y)$.
Clearly $p_k$ commutes with every element of $C (X)$.
Next, for any compact open subset $Z \subset X$,
we have $u \ch_Z u^* = \ch_{h (Z)}$.
In particular,
\[
u p_k u^*
= \sum_{j = 0}^{n (k)  1} u \ch_{h^j (Y_k)} u^*
= \sum_{j = 1}^{n (k)} \ch_{h^j (Y_k)}
= p_k  \ch_{Y_k} + \ch_{ h^{n (k)} (Y_k)}.
\]
So
%
\begin{equation}\label{Eq_4315_Comm}
u p_k = \big( p_k + \ch_{Y_k}  \ch_{ h^{n (k)} (Y_k) } \big) u.
\end{equation}
%
Now let $f \in C (X)$ vanish on~$Y$.
Multiply~(\ref{Eq_4315_Comm}) on the left by~$f$.
Since $Y_k \subset Y$ and $h^{n (k)} (Y_k) \subset Y$,
we get $f \ch_{Y_k} = f \ch_{ h^{n (k)} (Y_k) } = 0$,
whence $f u p_k = f p_k u = p_k f u$.
The claim is proved.
We now have
\[
C^* (\Z, X, h)_Y = \bigoplus_{k = 0}^l p_k C^* (\Z, X, h)_Y p_k.
\]
It therefore suffices to prove that $p_k C^* (\Z, X, h)_Y p_k$
is~AF for each $k$.
It is easy to see that $C^* (\Z, X, h)_Y$
is the \ca{} generated by $C (X)$ and $(\ch_{X \setminus Y}) u$.
Therefore $p_k C^* (\Z, X, h)_Y p_k$ is the \ca{} generated by
$C (X_k)$ and
(using~(\ref{Eq_4315_Comm}) at the first step of the calculation)
\begin{align*}
p_k (\ch_{X \setminus Y}) u p_k
= (\ch_{X_k \setminus Y_k}) (\ch_{X_k \setminus h^{n (k)} (Y_k) }) u
& = \sum_{j = 1}^{n (k)  1}
( \ch_{ h^{j} (Y_k) } ) u
\\
& = \sum_{j = 1}^{n (k)  1}
( \ch_{ h^{j + 1} (Y_k) } ) u ( \ch_{ h^{j} (Y_k) } ).
\end{align*}
One can now check,
although it is a bit tedious to write out the details
(see Exercise~\ref{Er_7171_CheckDecomp}),
that there is an isomorphism
%
\begin{equation}\label{Eq_7126_IsoForAF}
\ps_k \colon p_k C^* (\Z, X, h)_Y p_k \to M_{n (k)} \otimes C (Y_k)
\end{equation}
%
such that for $f \in C (X_k)$ we have
\[
\ps_k (f)
= \diag \big( f _{Y_k}, \, f \circ h _{Y_k}, \,
\ldots, \, f \circ h^{n (k)  1} _{Y_k} \big)
\]
and
(using matrix units in $M_{n (k)}$
labelled as $(e_{i, j})_{i, j = 0}^{n (k)  1}$)
for $1 \leq j \leq n (k)  1$ we have
\[
\ps_k \big( \ch_{ h^{j} (Y_k) } u \ch_{ h^{j  1} (Y_k) } \big)
= e_{j, \, j  1} \otimes 1.
\]
(Theorem~\ref{T_7121_AY_rshd} below gives the much messier
statement needed when the space $X$ is not totally disconnected,
and its proof is given in full.)
The algebra $M_{n (k)} \otimes C (Y_k)$ is~AF because $Y_k$
is totally disconnected.
\end{proof}
\begin{exr}\label{Er_7171_CheckDecomp}
Prove that $\ps_k$ as in~(\ref{Eq_7126_IsoForAF})
in the proof above is in fact an isomorphism.
\end{exr}
This exercise is preparation for reading the proof of
Theorem~\ref{T_7121_AY_rshd}.
The proof of \Lem{L:AF} shows,
in the notation used in it,
that
%
\begin{equation}\label{Eq_4319_FormOfIso}
C^* (\Z, X, h)_Y
\cong \bigoplus_{k = 0}^l C (Y_k, M_{n (k)}),
\end{equation}
%
via an isomorphism constructed from the system
of Rokhlin towers associated with~$Y$.
See Theorem~\ref{T_7121_AY_rshd} for what happens
for more general spaces~$X$.
% 999 If prove free implies Kishimoto's condition, refer to it here.
\begin{lem}\label{L:Canc}
Let $X$ be the Cantor set,
and let $h \colon X \to X$ be a \mh.
Let $Y \subset X$ be a nonempty compact open subset.
Let $N \in \N$,
and suppose that $Y, \, h (Y), \, \ldots, h^{N  1} (Y)$ are disjoint.
Then the \pj{s} $\ch_{h^{1} (Y)}$ and $\ch_{h^{N  1} (Y)}$
are \mvnt{} in $C^* (\Z, X, h)_Y$.
\end{lem}
The proof is short,
but we explain in terms of the Rokhlin towers
and the decomposition
\[
C^* (\Z, X, h)_Y \cong \bigoplus_{k = 0}^l M_{n (k)} \otimes C (Y_k)
\]
why one should expect it to be true.
First,
all the towers have height at least~$N$.
So passing from $\ch_Y$ to $\ch_{h^{N  1} (Y)}$
amounts to replacing,
in each summand $M_{n (k)} \otimes C (Y_k)$
and using the indexing in the proof of \Lem{L:AF},
the projection $e_{0, 0} \otimes 1$
with $e_{N  1, \, N  1} \otimes 1$.
These are certainly \mvnt.
Passing from $\ch_Y$ to $\ch_{h^{1} (Y)}$
corresponds to going off the bottoms of the Rokhlin towers.
This need not send $Y_k$ to $h^{n (k)  1} (Y_k)$,
so need not send $e_{0, 0} \otimes 1$
to $e_{n (k)  1, \, n (k)  1} \otimes 1$.
But $Y$ is also equal to $\coprod_{k = 0}^l h^{n (k)} (Y_k)$,
so it {\emph{does}} send
$Y = \coprod_{k = 0}^l Y_k$
to $\coprod_{k = 0}^l h^{n (k)  1} (Y_k) = h^{1} (Y)$.
The \pj{} corresponding to $\ch_Y$ is
\[
\big( e_{0, 0} \otimes 1, \, e_{0, 0} \otimes 1, \,
\ldots, \, e_{0, 0} \otimes 1 \big)
\in
\bigoplus_{k = 0}^l M_{n (k)} \otimes C (Y_k)
\]
and the identification of $h^{1} (Y)$
shows that \pj{} corresponding to $\ch_{h^{1} (Y)}$
is
\[
\big( e_{n (0)  1, \, n (0)  1} \otimes 1, \,
e_{n (1)  1, \, n (1)  1} \otimes 1, \,
\ldots, \, e_{n (l)  1, \, n (l)  1} \otimes 1 \big).
% \in \bigoplus_{k = 0}^l M_{n (k)} \otimes C (Y_k).
\]
These clearly {\emph{are}} \mvnt.
\begin{proof}[Proof of \Lem{L:Canc}]
We use the notation for \mvnc{}
in Notation~\ref{N:MvN}.
First, observe that if $Z \subset X$ is a compact open subset
such that $Y \cap Z = \varnothing$,
then $v = \ch_Z u \in C^* (\Z, X, h)_Y$
and satisfies $v v^* = \ch_Z$ and $v^* v = \ch_{h^{1} (Z)}$.
Thus $\ch_Z \sim \ch_{h^{1} (Z)}$.
An induction argument,
taking successively
\[
Z = h (Y), \,\,\,\,\, Z = h^2 (Y),
\,\,\,\,\, Z = h^N (Y),
\]
now shows that $\ch_{Y} \sim \ch_{h^{N  1} (Y)}$.
Also, taking $Z = X \setminus Y$ gives
$\ch_{X \setminus Y} \sim \ch_{X \setminus h^{1} (Y)}$.
Since $C^* (\Z, X, h)_Y$ is an AF~algebra,
it follows that $\ch_Y \sim \ch_{h^{1} (Y)}$.
The result follows by transitivity.
\end{proof}
The proof of the next lemma is closely related to
the first part of the proof of Theorem~\ref{T:AS}
(which is at the end of Section~\ref{Sec_MH})
and especially to the proof of Theorem~\ref{TAKCImpSP}.
Indeed,
we could get the result from Theorem~\ref{TAKCImpSP}
by proving that $\af$ satisfies
Kishimoto's condition
(\Def{DKshCond}).
\begin{lem}\label{L:PjComp}
Let $X$ be the Cantor set,
and let $h \colon X \to X$ be a \mh.
Let $c \in C^* (\Z, X, h)$ be a nonzero positive element.
Then there exists a \nzp{} $p \in C (X)$ such that
$p$ is \mvnt{} in $C^* (\Z, X, h)$
to a \pj{} in ${\overline{c C^* (\Z, X, h) c}}$.
\end{lem}
\begin{proof}
Let $E \colon C^* (\Z, X, h) \to C (X)$
be the standard conditional expectation
(Definition~\ref{D_StdCond}).
It follows from Proposition~\ref{P:Faithful}(\ref{P:Faithful:1})
and Exercise~\ref{Pb_CondExpt}(\ref{P:CondExpt:Pos})
that $E (c)$ is a nonzero positive element of $C (X)$.
Choose a nonempty compact open subset $K_0 \subset X$ and $\dt > 0$
such that the function $E (c)$ satisfies $E (c) (x) > 4 \dt$
for all $x \in K_0$.
Choose a finite sum $b = \sum_{n = N}^N b_n u^n \in C^* (\Z, X, h)$
such that $\ b  c \ < \dt$.
Since the action of $\Z$ induced by $h$ is free,
there is a nonempty compact open subset $K \subset K_0$ such that
the sets $h^{N} (K), \, h^{ N + 1} (K), \, \ldots, \, h^N (K)$
are disjoint.
Set $p = \ch_K \in C (X)$.
For $n \in \{ N, \,  N + 1, \, \ldots, \, N \} \setminus \{ 0 \}$,
the disjointness condition implies that $p u^n p = 0$.
Therefore $p b p = p b_0 p = p E (b) p$.
Using this equation at the first step
and Exercise~\ref{Pb_CondExpt}(\ref{P:CondExpt:Norm})
at the second step, we get
\begin{equation}\label{Est13}
\ p c p  p E (c) p \
\leq \ p c p  p b p \ + \ p E (b) p  p E (c) p \
\leq 2 \ c  b \
< 2 \dt.
\end{equation}
Since $K \subset K_0$,
the function $p E (c) p$ is invertible in $p C (X) p$.
In the following calculation,
inverses are taken in $p C^* (\Z, X, h) p$.
With this convention,
$[ p E (c) p ]^{1}$ exists and satisfies
$\ [ p E (c) p ]^{1} \ < \frac{1}{4} \dt^{1}$.
The estimate~(\ref{Est13}) now implies that $p c p$ is invertible in
$p C^* (\Z, X, h) p$.
Let $a = (p c p)^{1/2}$, calculated in $p C^* (\Z, X, h) p$.
Set $v = a p c^{1/2}$.
Then
\[
v v^* = a p c p a = (p c p)^{1/2} (p c p) (p c p)^{1/2} = p
\]
and
\[
v^* v = c^{1/2} p a^2 p c^{1/2} \in {\overline{c C^* (\Z, X, h) c}}.
\]
This completes the proof.
\end{proof}
Most of the proof of the following lemma is taken from~\cite{LhP}.
The definition of $C^* (\Z, X, h)_Y$ is different,
as explained after Definition~\ref{D_5421_VSubalg},
and the notation in the proof has been changed accordingly.
\begin{lem}\label{L:L3}
Let $X$ be the Cantor set,
and let $h \colon X \to X$ be a \mh.
Let $y \in X$.
Then for any
$\ep > 0$,
any nonempty open set $U \subset X$,
and any finite subset $F \subset C (X)$, there
is a compact open set $Y \subset X$ containing $y$
and a projection $p \in C^* (\Z, X, h)_{Y}$ such that:
\begin{enumerate}
\item\label{L:L31}
$\ p a  a p \ < \ep$ for all $a \in F \cup \{u\}$.
\item\label{L:L32}
$p a p \in p C^* (\Z, X, h)_Y p$ for all $a \in F \cup \{u\}$.
\item\label{L:L33}
There is a compact open set $Z \subset U$ such that
$1  p \precsim \ch_Z$ in $C^* (\Z, X, h)$.
\end{enumerate}
\end{lem}
The key point in the proof of \Lem{L:L3}
is the estimate~(\ref{Eq_4317_ueuStar}) below.
We outline the method.
For a suitable small compact open set $Y \subset X$ which contains~$y$,
set $q_n = \ch_{h^n (Y)}$.
There will be a large number $N \in \N$
such that $Y, \, h (Y), \, \ldots, \, h^{N  1} (Y)$
are disjoint,
and such that $h^N (y)$ is close to~$y$.
The \pj{} $1  p$ will be called $e$ in the proof.
The naive choice for~$e$
will turn out to be
$f = \sum_{n = 0}^{N  1} q_n$.
This projection commutes exactly with all elements of $C (X)$.
Also, it is easy to check that
$f C^* (\Z, X, h) f \subset C^* (\Z, X, h)_Y$.
If we had $h^N (Y) = Y$,
it would also commute with~$u$.
It is of course not possible to have $h^N (Y) = Y$,
since then $\bigcup_{n = 0}^{N  1} h^n (Y)$
would be a nontrivial $h$invariant closed set.
The main idea of the proof is to
modify this naive choice to get
a \pj{} $e$ which approximately commutes with~$u$
but which still has approximate versions
of the other good properties of~$f$.
There is a big difference between what we need to do
and what would happen if we were working in von Neumann algebras.
In the von Neumann algebra setting,
we would be given an ergodic probability measure~$\mu$ on~$X$,
and
it would be enough to ask that $q_N  q_0$ be small
in trace derived from~$\mu$.
Thus it would be sufficient to have $\mu (Y)$ small,
which is extremely easy to arrange.
The analog of \Lem{L:L3} would be essentially trivial:
just take $Y$ small enough,
pay no attention to how close $h^N (y)$ is to~$y$,
and take $e = \sum_{n = 0}^{N  1} q_n$.
In the C*~setting,
unless $h^N (Y)$ is exactly equal to~$Y$,
we get $\ q_N  q_0 \ = 1$.
Therefore we must work much harder.
\begin{proof}[Proof of \Lem{L:L3}]
We abbreviate $A = C^* (\Z, X, h)$
and $A_Y = C^* (\Z, X, h)_Y$.
Let $d$ be the metric on~$X$.
Choose $N_0 \in \N$ so large that $4 \pi / N_0 < \ep$.
Choose $\dt_0 > 0$ with $\dt_0 < \frac{1}{2} \ep$ and so small
that $d (x_1, x_2) < 4 \dt_0$ implies
$ f (x_1)  f (x_2) < \frac{1}{4} \ep$ for all $f \in F$.
Choose $\dt > 0$ with $\dt \leq \dt_0$ and such that
whenever $d (x_1, x_2) < \dt$ and $0 \leq k \leq N_0$, then
$d (h^{k} (x_1), \, h^{k} (x_2)) < \dt_0$.
Since $h$ is minimal, there is $N > N_0 + 1$ such that
$d (h^N (y), \, y) < \dt$.
Choose $N + N_0 + 1$ disjoint nonempty open subsets
$U_{ N_0}, U_{ N_0 + 1}, \ldots, U_{N} \subset U$.
Using minimality again,
choose $r_{ N_0}, r_{ N_0 + 1}, \ldots, r_{N} \in \Z$
such that $h^{r_l} (y) \in U_l$
for $l =  N_0, \,  N_0 + 1, \, \ldots, \, N$.
Since $h$ is has no periodic points,
there is a compact open set $Y \subset X$
such that:
\begin{enumerate}
\item\label{Y_6Y26_y}
$y \in Y$.
\item\label{Y_6Y26_Disj}
The sets
\[
h^{ N_0} (Y), \, h^{ N_0 + 1} (Y), \, \ldots,
\, Y, \, h (Y), \, \ldots, \, h^N (Y)
\]
are disjoint.
\item\label{Y_6Y26_lDt}
The sets
\[
h^{ N_0} (Y), \, h^{ N_0 + 1} (Y), \, \ldots,
\, Y, \, h (Y), \, \ldots, \, h^N (Y)
\]
all have diameter less than~$\dt$.
\item\label{Y_6Y26_InU}
$h^{r_l} (Y) \subset U_l$
for $l =  N_0, \,  N_0 + 1, \, \ldots, \, N$.
\end{enumerate}
Set $q_0 = \ch_Y$.
For $n =  N_0, \,  N_0 + 1, \, \ldots, \, N$ set
\[
T_n = h^n (Y)
\andeqn q_n = u^n q_0 u^{ n} = \ch_{h^n (Y)} = \ch_{T_n}.
\]
% Then the $q_n$ are \mops{} in $C (X)$.
We now have a sequence of \pj{s}, in principle going to infinity
in both directions:
\[
\ldots, \, q_{ N_0}, \, \ldots, \, q_{1}, \, q_0, \, q_1, \,
\ldots, \, q_{N  N_0}, \, \ldots, \, q_{N  1}, \, q_N, \, \ldots.
\]
The ones shown are orthogonal, and conjugation by $u$ is the shift
on this sequence.
The \pj{s} $q_0$ and $q_N$ are the characteristic
functions of compact open sets which are disjoint
but close to each other, and similarly
for the \pj{s} $q_{1}$ and $q_{N  1}$,
for the \pj{s} $q_{2}$ and $q_{N  2}$,
down to the \pj{s} $q_{ N_0}$ and $q_{N  N_0}$.
We are now going to use Berg's technique~\cite{Bg}
to splice this sequence along the pairs of indices
$( N_0, \, N  N_0)$ through $(0, N)$,
obtaining a loop of length $N$
on which conjugation by $u$ is approximately the cyclic shift.
Lemma~\ref{L:Canc} provides a partial isometry
$w \in A_Y$
such that $w^* w = q_{1}$ and $w w^* = q_{N  1}$.
% @@@
For $t \in [0, 1]$ define
%
\begin{equation}\label{Eq_4317_DfnV}
v (t)
= \cos (\pi t / 2) (q_{1} + q_{N  1})
+ \sin (\pi t / 2) (w  w^*)
\in A_Y.
\end{equation}
%
Then $v (t)$ is a unitary in the corner
\[
(q_{1} + q_{N  1}) A_{Y} (q_{1} + q_{N  1}).
\]
To see what is happening,
we write elements of this corner in $2 \times 2$ matrix form,
with the $(1, 1)$ entry corresponding to
$q_{1} A_{Y} q_{1}$.
That is,
there is a \hm{}
\[
\ph \colon M_2 \to (q_{1} + q_{N  1}) A_{Y} (q_{1} + q_{N  1})
\]
such that
\[
\ph (e_{1, 1}) = q_{1},
\qquad
\ph (e_{1, 2}) = w^*,
\qquad
\ph (e_{2, 1}) = w,
\andeqn
\ph (e_{2, 2}) = q_{N  1}.
\]
If we identify $M_2$ with its image under~$\ph$,
we get
\[
q_{1} = \left( \begin{matrix} 1 & 0 \\
0 & 0 \end{matrix} \right),
\qquad
w = \left( \begin{matrix} 0 & 0 \\
1 & 0 \end{matrix} \right),
\andeqn
q_{N  1} = \left( \begin{matrix} 0 & 0 \\
0 & 1 \end{matrix} \right)
\]
(these are just the definitions),
and
\[
w  w^*
= \left( \begin{matrix} 0 &  1 \\
1 & 0 \end{matrix} \right)
\andeqn
v (t)
= \left( \begin{matrix} \cos (\pi t / 2) &  \sin (\pi t / 2) \\
\sin (\pi t / 2) & \cos (\pi t / 2) \end{matrix} \right).
\]
For $k = 0, 1, \ldots, N_0$ define
%
\begin{equation}\label{Eq_6Y26_zk}
z_k = u^{ k + 1} v (k / N_0) u^{k  1},
\end{equation}
%
which is in
\[
(q_{k} + q_{N  k}) A (q_{k} + q_{N  k})
\]
and is a unitary in this corner.
We claim that $z_k \in A_Y$
for $k = 0, 1, \ldots, N_0$.
We have $z_0 = q_0 + q_N \in C (X) \subset A_Y$.
Also,
$z_1 \in A_Y$ by construction.
For $k = 2, 3 \ldots, N_0$,
set $a_k = q_{1} u^{k  1}$ and $b_k = q_{N  1} u^{k  1}$.
Since $u q_n u^* = q_{n + 1}$ for all~$n$,
we can write these as
%
\begin{align}\label{Eq_4317_Factor}
a_k
& = q_{ 1} (u q_{ 2} u^{1}) (u^2 q_{ 3} u^{2})
\cdots (u^{k  2} q_{ k + 1} u^{k + 2}) u^{k  1}
\\
& = (q_{ 1} u) (q_{ 2} u) \cdots (q_{ k + 1} u)
\notag
\end{align}
%
and
%
\begin{align}\label{Eq_4317_FactorPrime}
b_k
& = q_{N  1} (u q_{N  2} u^{1}) (u^2 q_{N  3} u^{2})
\cdots (u^{k  2} q_{N  k + 1} u^{k + 2}) u^{k  1}
\\
& = (q_{N  1} u) (q_{N  2} u) \cdots (q_{N  k + 1} u).
\notag
\end{align}
%
Since $N_0 < N$,
the \pj{s}
\[
q_{ 1}, q_{ 2}, \ldots, q_{ N_0 + 1},
q_{N  1}, q_{N  2}, \ldots, q_{N  N_0 + 1}
\]
are all characteristic functions of sets disjoint from~$Y$.
The factorizations in~(\ref{Eq_4317_Factor})
and~(\ref{Eq_4317_FactorPrime})
therefore show that $a_k, b_k \in A_{Y}$.
Now one checks that
\[
z_k = (a_k + b_k)^* v (k / N_0) (a_k + b_k),
\]
which is in $A_{Y}$.
This is the claim.
Thus
\[
z_k \in (q_{ k} + q_{N  k}) A_{Y} (q_{ k} + q_{N  k})
\]
and is a unitary in this corner.
{}From~(\ref{Eq_4317_DfnV}),
it is easy to get $\ v (t_1)  v (t_2) \ \leq 2 \pi  t_1  t_2 $
for $t_1, t_2 \in [0, 1]$.
Using~(\ref{Eq_6Y26_zk}),
for $k = 0, 1, \ldots, N_0$
we therefore get
%
\begin{equation}\label{Eq_4317_LessThan}
\ u z_{k + 1} u^*  z_k \
= \big\ v \big( (k + 1) / N_0 \big)  v (k / N_0) \big\
\leq \frac{2 \pi}{N_0}
< \frac{\ep}{2}.
\end{equation}
%
Now define $e_n = q_n$ for $n = 0, 1, \ldots, N  N_0$.
For $n = N  N_0, \, N  N_0 + 1, \, \ldots, N$,
define $k$ by $n = N  k$,
and set
$e_n = z_k q_{ k} z_k^*$.
These are clearly all elements of $A_{Y}$.
The two definitions for $n = N  N_0$ agree
because,
in the obvious block decomposition
(similar to that used above) of
\[
(q_{ N_0} + q_{N  N_0}) A_{Y} (q_{ N_0} + q_{N  N_0}),
\]
we get
\[
z_{N_0} = \left( \begin{matrix}
0 & 1 \\
1 & 0
\end{matrix} \right),
\]
so that
$z_{N_0} q_{ N_0} z_{N_0}^* = q_{N  N_0}$.
(One can check this formula by a direct calculation.)
Moreover, $z_0 = q_0 + q_N$,
so $e_N = e_0$.
Putting things together,
we have
%
\begin{equation}\label{Eq_6Y26_StSt}
u e_{n  1} u^* = e_n
\end{equation}
%
for $n = 1, 2, \ldots, N  N_0$,
and also $u e_N u^* = e_1$.
For $N  N_0 < n \leq N$ we define $k$ by $n = N  k$
and use~(\ref{Eq_4317_LessThan}) and $u q_{ k  1} u^* = q_{ k}$
to get
\begin{align}\label{Eq_6Y26_3St}
\ u e_{n  1} u^*  e_n \
& = \big\ u z_{k + 1} q_{ k  1} z_{k + 1} u^*
 z_k q_{ k} z_k \big\
\\
& = \big\ (u z_{k + 1} u^*) q_{ k} (u z_{k + 1} u^*)  q_{ k} \big\
\notag
\\
& \leq 2 \big\ u z_{k + 1} u^*  z_{k} \big\
< \ep.
\notag
\end{align}
Set
\[
e = \sum_{n = 1}^{N} e_n
\andeqn
p = 1  e,
\]
both of which are in $A_{Y}$.
We verify that $p$ satisfies (\ref{L:L31}), (\ref{L:L32}),
and~(\ref{L:L33}).
We verify (\ref{L:L31}) and~(\ref{L:L32}).
Consider $u$ first.
Since $e_N = e_0$,
we have
\[
u e u^*  e
= \sum_{n = 1}^{N} (u e_{n  1} u^*  e_n ).
\]
For $n = 1, 2, \ldots, N  N_0$,
equation~(\ref{Eq_6Y26_StSt}) applies,
so that in fact
\[
u e u^*  e
= \sum_{n = N  N_0 + 1}^{N} (u e_{n  1} u^*  e_n ).
\]
For the indices used in this sum,
the inequality~(\ref{Eq_6Y26_3St}) applies,
so the terms in the sum have norm less than~$\ep$.
They are orthogonal since, with $k$ determined by
$n = N  k$,
\[
u e_{n  1} u^*  e_n
\in (q_{ k} + q_{N  k}) A_{Y} (q_{ k} + q_{N  k}).
\]
Therefore
%
\begin{equation}\label{Eq_4317_ueuStar}
\ u e u^*  e \ < \ep.
\end{equation}
%
So $\ u p u^*  p \ = \ { u e u^* + e} \ < \ep$.
Furthermore,
since $p \in A_{Y}$
and $p \leq 1  q_0 = 1  \ch_Y$,
we get
\[
p u p
= p (1  \ch_Y) u p \in A_{Y}.
\]
This is (\ref{L:L31}) and~(\ref{L:L32})
for the element $u \in F \cup \{ u \}$.
Next, let $f \in F$.
Define sets $S_n$ for $n = 1, 2, \ldots, N$ by
\[
S_1 = T_1, \,\,\, S_2 = T_2, \,\,\, \ldots, \,\,\,
S_{N  N_0  1} = T_{N  N_0  1}
\]
and
\[
S_{N  N_0} = T_{N  N_0} \cup T_{ N_0}, \,\,\,
S_{N  N_0 + 1} = T_{N  N_0 + 1} \cup T_{ N_0 + 1},
\,\,\, \ldots, \,\,\,
S_N = T_N \cup T_0.
\]
These sets are disjoint.
The sets $T_0, T_1, \ldots, T_N$ all have diameter less than~$\dt$.
We have $d (h^N (y), \, y) < \dt$,
so the choice of $\dt$ implies that
$d (h^n (y), \, h^{n  N} (y)) < \dt_0$
for $n =  N_0, \,  N_0 + 1, \, \ldots, \, N$.
Also, $T_{n  N} = h^{n  N} (T_0)$
has diameter less than~$\dt$.
Therefore $T_{n  N} \cup T_n$ has diameter less than
$2 \dt + \dt_0 \leq 3 \dt_0$.
It follows that
$S_n$ has diameter less than $3 \dt_0$ for $n = 1, 2, \ldots, N$.
Since $f$ varies by at most $\frac{1}{4} \ep$ on any set with
diameter less than $4 \dt_0$,
and since the sets
$S_{1}, S_{2}, \ldots, S_N$
are disjoint,
there is $g \in C (X)$
which is constant on each of these sets and satisfies
$\ f  g \ < \frac{1}{2} \ep$.
Let the values of $g$ on these sets be
$\ld_1$ on $S_1$ through $\ld_N$ on $S_N$.
Then $g e_n = e_n g = \ld_n e_n$
for $0 \leq n \leq N  N_0$.
For $N  N_0 < n \leq N$ we use
\[
e_n \in (q_{n  N} + q_{n}) A_{Y} (q_{n  N} + q_{n})
\]
to get, using the same calculations as above
at the third and fourth steps,
\[
g e_n
= g (q_{n  N} + q_{n}) e_n
= \ld_n (q_{n  N} + q_{n}) e_n
= e_n (q_{n  N} + q_{n}) g
= e_n g.
\]
Since $\ f  g \ < \frac{1}{2} \ep$ and $g e = e g$,
it follows that
\[
\ p f  f p \ = \ f e  e f \ < \ep.
\]
This is~(\ref{L:L31}) for~$f$.
That $p f p \in A_{Y}$ follows from the fact that
$f$ and $p$ are in this subalgebra.
So we also have~(\ref{L:L32}) for~$f$.
It remains only to verify~(\ref{L:L33}).
Using $h^{r_l} (Y) \subset U_l$
for $l =  N_0, \,  N_0 + 1, \, \ldots, \, N$
and disjointness of the sets
$U_{ N_0}, \, U_{ N_0 + 1}, \, \ldots, \, U_N$
at the third step,
and defining $Z = \bigcup_{l =  N_0}^N h^{r_l} (Y) \subset U$,
we get (with Murrayvon Neumann equivalence in $A$)
\[
1  p
= e \leq \sum_{l =  N_0}^N q_l
\sim \sum_{l =  N_0}^N \ch_{h^{r_l} (Y)}
= \ch_Z.
\]
This completes the proof.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{T:TM} when $X$ is the Cantor set]
We use Exercise~\ref{Ex:TRGen}.
Let $u \in C^* (\Z, X, h)$
be the standard unitary
(called $u_1$ in Notation~\ref{N:ug}).
Take the set $S$ in Exercise~\ref{Ex:TRGen}
to be $S = C (X) \cup \{ u \}$.
We verify the conditions
(\ref{L:TRAF:Comm}), (\ref{L:TRAF:Close}), and (\ref{L:TRAF:Small})
in Lemma~\ref{L:TRAF}
for finite sets $F \subset S$.
We may clearly assume that $u$ is in our finite subset,
so let $F_0 \subset C (X)$
be finite, let $c \in A_{+} \setminus \{ 0 \}$,
let $\ep > 0$, and take $F = F_0 \cup \{ u \}$.
Use \Lem{L:PjComp}
to find a nonempty compact open set $U \subset X$
such that $\ch_U$ is \mvnt{}
to a \pj{} $q \in {\ov{c C^* (\Z, X, h) c}}$.
Choose any $y \in X$.
Apply Lemma~\ref{L:L3}
with $U$, $\ep$, and~$y$ as given,
and with $F_0$ in place of~$F$.
Let $p$ and $Y$ be the resulting \pj{} and compact open set.
Let $Z$ be as in part~(\ref{L:L33}) of Lemma~\ref{L:L3}.
Then $\ a p  p a \ < \ep$ for all $a \in F$,
which is (\ref{L:TRAF:Comm}) of Lemma~\ref{L:TRAF}.
Also, $C^* (\Z, X, h)_Y$ is an AF~algebra by Lemma~\ref{L:AF},
so $p C^* (\Z, X, h)_Y p$ is a corner of an AF~algebra,
hence~AF.
For $a \in F$,
part~(\ref{L:L32}) of Lemma~\ref{L:L3}
gives $\dist (a, \, p C^* (\Z, X, h)_Y p) = 0 < \ep$,
which is (\ref{L:TRAF:Comm}) of Lemma~\ref{L:TRAF}.
Finally,
using part~(\ref{L:L33}) of Lemma~\ref{L:L3}
at the first step,
we get
\[
1  p
\precsim \ch_Z
\leq \ch_U
\sim q
\in {\ov{c C^* (\Z, X, h) c}}.
\]
This is (\ref{L:TRAF:Small}) of Lemma~\ref{L:TRAF}.
\end{proof}
\section{Minimal Homeomorphisms of Finite Dimensional
Spaces}\label{Sec_More}
\indent
In this section,
we describe what needs to be done to prove the general case
of Theorem~\ref{T:TM}.
We do not give full details,
since to do so would require substantial excursions
into parts of \ca{s}
which have little to do with dynamics,
namely the structure of direct limits of subhomogeneous \ca{s}
and Ktheory.
We do give proofs of some the parts which are related to dynamics.
In this section,
we need to assume some familiarity with Ktheory.
We don't discuss Ktheory here.
Instead, we refer to \cite{WO} for a gentle introduction
and \cite{Blk_2} for a more extensive treatment.
For applications to classification,
the following generalization of Theorem~\ref{T:TM}
is useful.
See~\cite{TW};
the statement is essentially Proposition~4.6 of~\cite{TW}.
Since some parts of the proof involve essentially no extra work,
we describe parts of the proof of the generalization.
\begin{thm}[\cite{TW}]\label{T_4317_MinUHF}
Let $X$ be an infinite compact metric space
with finite covering dimension,
and let $h \colon X \to X$ be a minimal homeomorphism.
Let $D$ be $\C$ or a UHF~algebra
of the form $\bigotimes_{n = 1}^{\I} M_l$
for some prime~$l$.
Suppose that,
following the notation of \Def{DRhoA},
$\rh \big( K_0 (D \otimes C^* (\Z, X, h)) \big)$
is dense in $\Aff \big( \T (D \otimes C^* (\Z, X, h)) \big)$.
Then $D \otimes C^* (\Z, X, h)$ is a simple unital \ca{}
with tracial rank zero
which satisfies the Universal Coefficient Theorem.
\end{thm}
In the description we give of the proof,
presumably $D$ can be any UHF algebra.
If $X$ is an odd sphere of dimension at least~$3$
and $h$ is uniquely ergodic
(see \Thm{T_3305_ExistMinZ}),
then $h$ satisfies the hypotheses of
\Thm{T_4317_MinUHF}
when $D$ is any UHF~algebra but not when $D = \C$.
More generally,
it follows from Proposition 3.12(b) of~\cite{BKR}
that the hypotheses of
\Thm{T_4317_MinUHF}
are satisfied whenever $D$ is a UHF~algebra
and the \pj{s} in $\bigcup_{n = 1}^{\infty} M_n ( C^* (\Z, X, h))$
distinguish the traces on $C^* (\Z, X, h)$.
The first complication involves the construction
of Rokhlin towers,
as in the proof of \Lem{L:AF}
and the discussion after Definition~\ref{D_5421_VSubalg}.
The sets $Y$ and $Y_k$ used there
can't be chosen to be compact and open
(indeed, if $X$ is connected, there will be no nontrivial
compact open sets),
so that the projections $\ch_{Y_k}$ are not in $C (X)$
(and not in $C^* (\Z, X, h)$ either).
It turns out that one must take $Y$ to be closed
with nonempty interior,
and replace the sets $Y_k$ by their closures.
Then they are no longer disjoint.
The algebra $C^* (\Z, X, h)_Y$
is now a very complicated subalgebra
of $\bigoplus_{k = 0}^l M_{n (k)} \otimes C (Y_k)$.
It is what is known as a recursive subhomogeneous algebra.
See \Def{D_4317_RSHA} and \Thm{T_4317_AYIsRsha} below.
We give a complete proof of \Thm{T_4317_AYIsRsha} since,
as far as we know,
no complete proof has yet been published.
It is taken with little change from
the unpublished paper~\cite{LqPh_dr}.
This modification will lead to further difficulties.
Such algebras are generally not~AF,
and may have few or no nontrivial \pj{s}.
The hypothesis on the range of $\rh$
(which was not used in Section~\ref{Sec:Class},
although it is automatic when $X$
is the Cantor set)
must be used to produce sufficiently many \nzp{s}
and approximating \fd{} subalgebras.
The following definition and lemma
formalize the first return time used in the proof
of \Lem{L:AF}.
\begin{dfn}\label{D_4319_FirstRet}
Let $X$ be an infinite compact metric space
and let $h \colon X \to X$ be a minimal homeomorphism.
Let $Y \subset X$, and let $x \in Y$.
The {\emph{first return time}} $r_Y (x)$
of $x$ to $Y$
is the smallest integer $n \geq 1$ such that $h^n (x) \in Y$.
We set $r_Y (x) = \infty$ if no such $n$ exists.
If $Y$ is understood, we may simply write $r (x)$.
\end{dfn}
\begin{lem}\label{L_4319_RokhFin}
Let $X$ be an infinite compact metric space,
let $h \colon X \to X$ be a minimal homeomorphism,
and let $Y \subset X$.
If $\sint (Y) \neq \E$,
then $\sup_{x \in Y} r_Y (x) < \infty$.
\end{lem}
\begin{proof}
Set $U = \bigcup_{n = 1}^{\I} h^{n} (\sint (Y))$.
Clearly $h^{1} (U) \subset U$.
Applying \Lem{L:MHC}(\ref{L:MHC:UFMin})
to $h^{1}$,
and using $U \neq \E$,
we get $U = X$.
Therefore the sets $h^{n} (\sint (Y))$, for $n \geq 1$,
form an open cover of the compact set $\overline{Y}$.
Choose a finite subcover.
The largest value of $n$ used is an upper bound for
$\{ r_Y (x) \colon x \in Y \}$.
\end{proof}
\begin{dfn}\label{D_4319_ModRokh}
Let $Y \subset X$ be closed with $\sint (Y) \neq \emptyset$.
The {\emph{modified Rokhlin tower}} associated with $Y$
consists of the subsets and numbers
\[
Y_0, Y_1, \dots, Y_l \subset Y,
\quad
Y_0^{\bullet}, Y_1^{\bullet}, \dots, Y_l^{\bullet} \subset Y,
\smandeqn
1 \leq n (0) < n (1) < \cdots < n (l),
\]
defined as follows.
We let $n (0) < n (1) < \cdots < n (l)$ be the distinct values of
the first return time to $Y$ (there are only finitely many, by
Lemma~\ref{L_4319_RokhFin}), and we define
\[
Y_k = {\overline{ \{ x \in Y \colon r (x) = n (k) \} }}
\andeqn
Y_k^{\bullet} = \sint \big( \{ x \in Y \colon r (x) = n (k) \} \big)
\]
for $k = 0, 1, \ldots, l$.
\end{dfn}
We warn that there is no reason to expect
$Y_k^{\bullet}$ to be dense in~$Y_k$,
or even that $Y_k^{\bullet} = \sint (Y_k)$.
\begin{lem}\label{RokhCov}
Let $Y \subset X$ be closed with $\sint (Y) \neq \varnothing$.
Then (following the notation
of Definition~\ref{D_4319_ModRokh}):
\begin{enumerate}
\item\label{RokhCov_1}
Suppose $0 \leq k, \, k' \leq l$ and $0 \leq j, \, j' \leq n (k)  1$,
with $(k, j) \neq (k', j')$.
Then $h^{j} ( Y_k^{\bullet} ) \cap h^{j'} ( Y_{k'} ) = \varnothing$.
\item\label{RokhCov_2}
$X = \bigcup_{k = 0}^l \bigcup_{j = 0}^{n (k)  1} h^{j} (Y_k)$.
\item\label{RokhCov_3}
$X = \bigcup_{j = 1}^{n (l)} h^{j} (Y)$.
% \item\label{RokhCov_4}
% $Y = \bigcup_{k = 0}^l h^{n (k)} (Y_k)$.
\item\label{RokhCov_4b}
$Y = \bigcup_{k = 0}^l Y_k$.
\item\label{RokhCov_5}
For $k = 0, 1, \ldots, l$ and $y \in Y_k$,
if $r (y) < n (k)$
then $y \in Y_k \setminus Y_k^{\bullet}$.
\end{enumerate}
\end{lem}
\begin{proof}
% As in the proof of \Lem{L_4319_RokhFin},
% the assumption $\sint (Y) \neq \varnothing$,
% % Lemma~\ref{ForOrbDense} (applied to $h^{1}$)
% implies that
% $\bigcup_{j = 1}^{\infty} h^{ j} (Y) = X$.
% Applying the argument with $h^{1}$ in place of~$h$,
% we also get
% $\bigcup_{j = 0}^{\infty} h^{j} (Y) = X$.
We begin with an argument from the proof of \Lem{L_4319_RokhFin}.
Set $U = \bigcup_{j = 1}^{\I} h^{j} (\sint (Y))$.
Clearly $h (U) \subset U$.
Since $U \neq \E$,
\Lem{L:MHC}(\ref{L:MHC:UFMin})
implies $U = X$.
In particular, $\bigcup_{j = 1}^{\infty} h^{j} (Y) = X$.
It is now essentially immediate from the construction that
\[
X = \coprod_{k = 0}^l \coprod_{j = 0}^{n (k)  1}
h^{j} \left( \{ x \in Y \colon r (x) = n (k) \} \right).
\]
Part~(\ref{RokhCov_2}) follows since
\[
\{ x \in Y \colon r (x) = n (k) \} \subset Y_k.
\]
Part (\ref{RokhCov_3}) follows
by applying $h^{ n (l)}$ to part (\ref{RokhCov_2}),
since $n (l)$ is the largest of the $n (k)$.
For part~(\ref{RokhCov_1}),
apply the disjointness part of the above together
with the observation that $R \cap S = \varnothing$ implies
$\sint (R) \cap \overline{S} = \varnothing$.
In part~(\ref{RokhCov_4b}),
the inclusion $Y \subset \bigcup_{k = 0}^l Y_k$
is immediate from
\[
\{ x \in Y \colon r (x) = n (k) \} \subset Y_k
\]
for $k = 0, 1, \ldots, l$.
The reverse inclusion follows from
continuity of $h^{n (k)}$,
the fact that $Y$ is closed,
and the relation
\[
\{ x \in Y \colon r (x) = n (k) \} \subset Y.
\]
% For part~(\ref{RokhCov_4}),
% we first show that $Y \subset \bigcup_{k = 0}^l h^{n (k)} (Y_k)$.
% Let $y \in Y$.
% Since $\bigcup_{j = 0}^{\infty} h^{j} (Y) = X$,
% we can choose the least $j \in \N$
% such that $h^{ j} (y) \in Y$.
% Then $r ( h^{ j} (y) ) = j$.
% So there is $k \in \{ 0, 1, \ldots, l \}$
% such that $n (k) = j$.
% Now $h^{ n (k)} (y) \in Y_k$,
% so $y \in h^{n (k)} (Y)$.
%
% The reverse inclusion follows from the fact that
% $h^{n (k)} ( \{ x \in Y \colon r (x) = n (k) \} ) \subset Y$.
% continuity, and
% Definition~\ref{D_4319_FirstRet}.
Part (\ref{RokhCov_5}) follows from
the relations
\[
y \not\in \{ x \in Y \colon r (x) = n (k) \}
\andeqn
Y_k^{\bullet} \subset \{ x \in Y \colon r (x) = n (k) \}.
\]
This completes the proof.
\end{proof}
There is a \hm{} from $C^* (\Z, X, h)_Y$ to
$\bigoplus_{k = 0}^l C (Y_k, M_{n (k)})$,
like the isomorphism of~(\ref{Eq_4319_FormOfIso}),
but in general it is not surjective.
To describe it,
we need a description of $C^* (\Z, X, h)_Y$,
which we provide in the following proposition.
It is valid whether or not $\sint (Y) = \E$.
\begin{prp}[Proposition~7.5 of~\cite{Ph40}]\label{P_4319_CharOB}
Let $X$ be a \chs{} and let $h \colon X \to X$ be a \hme.
Let $u \in C^* (\Z, X, h)$ be the standard unitary generator
($u_1$ in Notation~\ref{N:ug}),
and let $E \colon C^* (\Z, X, h) \to C (X)$
be the standard conditional expectation
($E_1$ in Definition~\ref{D_StdCond}).
Let $Y \subset X$ be a nonempty closed subset.
For $n \in \Z$, set
\[
Z_n = \begin{cases}
\bigcup_{j = 0}^{n  1} h^j (Y) & \hspace{3em} n > 0
\\
\varnothing & \hspace{3em} n = 0
\\
\bigcup_{j = 1}^{ n} h^{j} (Y) & \hspace{3em} n < 0.
\end{cases}
\]
Then
%
\begin{equation}\label{Eq:2816CharOB}
C^* (\Z, X, h)_Y
= \big\{ a \in C^* (\Z, X, h) \colon
{\mbox{$E (a u^{n}) \in C_0 (X \setminus Z_n)$
for all $n \in \Z$}} \big\}
\end{equation}
%
and
%
\begin{equation}\label{Eq:2816Dense}
{\overline{C^* (\Z, X, h)_Y \cap C (X) [\Z]}} = C^* (\Z, X, h)_Y.
\end{equation}
\end{prp}
\begin{proof}
Define
\[
B = \big\{ a \in C^* (\Z, X, h) \colon
{\mbox{$E (a u^{n}) \in C_0 (X \setminus Z_n)$
for all $n \in \Z$}} \big\}
\]
and
\[
B_0 = B \cap C (X) [\Z].
\]
We claim that $B_0$ is dense in~$B$.
We would like to write an element of~$B$
as $\sum_{k =  \I}^{\I} b_k u^k$
with $b_k \in C_0 (X \setminus Z_k)$ for $k \in \Z$.
Unfortunately,
in general,
such series need not converge.
(See Remark~\ref{R_6X11_ComputeCrPrd}.
If $\sint (T) \neq \E$,
then the series is necessarily finite and therefore does converge.)
Instead, we use the Ces\`{a}ro means.
So let $b \in B$ and for $k \in \Z$ define
$b_k = E (b u^{k}) \in C_0 (X \setminus Z_k)$.
Then for $n \in \N$, the element
\[
a_n = \sum_{k =  n + 1}^{n  1}
\left( 1  \frac{k}{n} \right) b_k u^k.
\]
is clearly in~$B_0$,
and Theorem VIII.2.2 of~\cite{Dvd}
implies that $\limi{n} a_n = b$.
The claim follows.
In particular,
(\ref{Eq:2816Dense})~will now follow from~(\ref{Eq:2816CharOB}),
so we need only prove~(\ref{Eq:2816CharOB}).
For $0 \leq m \leq n$ and $0 \geq m \geq n$,
we clearly have $Z_m \subset Z_n$.
We claim that for all $n \in \Z$,
we have
%
\begin{equation}\label{Eq:2818hnyn}
h^{n} (Z_n) = Z_{n}.
\end{equation}
%
The case $n = 0$ is trivial,
the case $n > 0$ is easy,
and the case $n < 0$ follows from the case $n > 0$.
We next claim that for all $m, n \in \Z$,
we have
\[
Z_{m + n} \subset Z_m \cup h^m (Z_n).
\]
The case $m = 0$ or $n = 0$ is trivial.
For $m , n > 0$ and also for $m, n < 0$, it is easy to check that
$Z_{m + n} \subset Z_m = h^m (Z_n)$.
Now suppose $m > 0$ and $ m \leq n < 0$.
Then $0 \leq m + n \leq m$,
so
\[
Z_{m + n} \subset Z_m \subset Z_m \cup h^m (Z_n).
\]
If $m > 0$ and $n <  m$,
then
\[
Z_{m + n}
= \bigcup_{j = m + n}^{ 1} h^{j} (Y)
\subset \bigcup_{j = m + n}^{m  1} h^{j} (Y)
= \bigcup_{j = 0}^{m  1} h^j (Y)
\cup \bigcup_{j = m + n}^{m  1} h^{j} (Y)
= Z_m \cup h^m (Z_n).
\]
Finally, suppose $m < 0$ and $n > 0$.
Then, using~(\ref{Eq:2818hnyn}) at the first and third steps,
and the already done case $m > 0$ and $n < 0$ at the second step,
we get
\[
Z_{m + n}
= h^{m + n} (Z_{ m  n})
\subset h^{m + n} \big( Z_{ m} \cup h^{ m} (Z_{ n}) \big)
= h^n (Z_m) \cup Z_n.
\]
This completes the proof of the claim.
We now claim that $B_0$ is a *algebra.
It is enough to prove that if $f \in C_0 (X \setminus Z_m)$
and $g \in C_0 (X \setminus Z_n)$,
then $(f u^m) (g u^n) \in B_0$
and $(f u^m)^* \in B_0$.
For the first,
we have $(f u^m) (g u^n) = f \cdot (g \circ h^{ m}) \cdot u^{m + n}$.
Now $f \cdot (g \circ h^{ m})$ vanishes on $Z_m \cup h^m (Z_n)$,
so the previous claim implies that
$f \cdot (g \circ h^{ m}) \in C_0 (X \setminus Z_{m + n})$.
Also,
$(f u^m)^* = u^{ m} {\overline{f}}
= \big( {\overline{f \circ h^m}} \big) u$,
and, using~(\ref{Eq:2818hnyn}),
the function $f \circ h^m$ vanishes on $h^{ m} (Z_m) = Z_{ m}$,
so $(f u^m)^* \in B_0$.
This proves the claim.
Since $C (X) \subset B_0$ and $C_0 (X \setminus Y) u \subset B_0$,
it follows that $C^* (\Z, X, h)_Y \subset {\overline{B_0}} = B$.
We next claim that for all $n \in \Z$,
we have
$C_0 (X \setminus Z_n) \subset C^* (\Z, X, h)_Y$.
For $n = 0$ this is trivial.
Let $n > 0$,
and let $f \in C_0 (X \setminus Z_n)$.
Define $f_0 = (\sgn \circ f)  f ^{1/n}$
and for $j = 1, 2, \ldots, n  1$
define
$f_j =  f \circ h^{j} ^{1/n}$.
Then $f_0, f_1, \ldots, f_{n  1} \in C_0 (X \setminus Y)$.
Therefore the element
\[
a = (f_0 u) (f_1 u) \cdots (f_{n  1} u)
\]
is in $C^* (\Z, X, h)_Y$.
Moreover,
we can write
\begin{align*}
a
& = f_0 (u f_1 u^{1}) (u^2 f_2 u^{2})
\cdots (u^{n  1} f_{n  1} u^{ (n  1)}) u^n
\\
& = f_0 (f_1 \circ h^{1}) (f_2 \circ h^{2})
\cdots (f_{n  1} \circ h^{ (n  1)}) u^n
= (\sgn \circ f) \big(  f ^{1/n} \big)^n u^n
= f u^n.
\end{align*}
Finally,
suppose $n < 0$,
and let $f \in C_0 (X \setminus Z_n)$.
It follows from~(\ref{Eq:2818hnyn})
that $f \circ h^n \in C_0 (X \setminus Z_{ n})$,
whence also ${\overline{f \circ h^n}} \in C_0 (X \setminus Z_{ n})$.
Since $ n > 0$,
we therefore get
\[
f u^n = \big( u^{n} {\overline{f}} \big)^*
= \big( \big( {\overline{f \circ h^n}} \big) u^{ n} \big)^*
\in C^* (\Z, X, h)_Y.
\]
The claim is proved.
It now follows that $B_0 \subset C^* (\Z, X, h)_Y$.
Combining this result with ${\overline{B_0}} = B$
and $C^* (\Z, X, h)_Y \subset B$,
we get $C^* (\Z, X, h)_Y = B$.
\end{proof}
\begin{cor}\label{C_7120_DirSum}
Let $X$ be an infinite \chs{} and let $h \colon X \to X$ be a \mh.
As in Proposition~\ref{P_4319_CharOB},
let $u \in C^* (\Z, X, h)$ be the standard unitary generator.
Let $Y \subset X$ be a closed subset
such that $\sint (Y) \neq \E$.
Let $Z_n$ be as in Proposition~\ref{P_4319_CharOB}.
Then there exists $N \in \Nz$ such that $C^* (\Z, X, h)_Y$ has
the Banach space direct sum decomposition
\[
C^* (\Z, X, h)_Y
= \bigoplus_{n =  N}^N C_0 (X \setminus Z_n) u^n.
\]
\end{cor}
\begin{proof}
Define $N = \sup_{x \in Y} r_Y (x)$.
Then $N$ is finite by \Lem{L_4319_RokhFin}.
Proposition~\ref{P_4319_CharOB} implies that
\[
C^* (\Z, X, h)_Y
= \sum_{n =  N}^N C_0 (X \setminus Z_n) u^n.
\]
The sum on the right is algebraically a direct sum,
the subspaces are closed,
and there are finitely many of them,
so it is a Banach space direct sum
by the Open Mapping Theorem.
\end{proof}
\begin{ntn}\label{AYShiftNtn}
Assume $n (0), \, n (1), \, \dots, \, n (l)$ are positive integers.
(They will be the first return times associated with a
\mh{} $h \colon X \to X$ and a closed
subset $Y \subset X$ with nonempty interior.)
Define $s_k^{(0)}$ and $s_k$ in $M_{n (k)}$, or in
$C ( Z, M_{n (k)} )$
for any $Z$, by
\[
s_k^{(0)} = \left( \begin{matrix}
0 & 0 & \cdots & \cdots & 0 & 0 & 0 \\
1 & 0 & \cdots & \cdots & 0 & 0 & 0 \\
0 & 1 & \cdots & \cdots & 0 & 0 & 0 \\
\vdots & \vdots & \ddots & & \vdots & \vdots & \vdots \\
\vdots & \vdots & & \ddots & \vdots & \vdots & \vdots \\
0 & 0 & \cdots & \cdots & 1 & 0 & 0 \\
0 & 0 & \cdots & \cdots & 0 & 1 & 0
\end{matrix} \right)
% \]
\smandeqn
% \[
s_k = \left( \begin{matrix}
0 & 0 & \cdots & \cdots & 0 & 0 & 1 \\
1 & 0 & \cdots & \cdots & 0 & 0 & 0 \\
0 & 1 & \cdots & \cdots & 0 & 0 & 0 \\
\vdots & \vdots & \ddots & & \vdots & \vdots & \vdots \\
\vdots & \vdots & & \ddots & \vdots & \vdots & \vdots \\
0 & 0 & \cdots & \cdots & 1 & 0 & 0 \\
0 & 0 & \cdots & \cdots & 0 & 1 & 0
\end{matrix} \right).
\]
The only difference is in the upper right corner,
where $s_k$ has the entry~$1$.
\end{ntn}
The formula for $\gm$ in the following proposition is based on
a formula for the Cantor set case in~\cite{Pt1}.
Recall that our definition of $C^* (\Z, X, h)_Y$
differs from that in~\cite{Pt1}.
If we took
\[
C^* (\Z, X, h)_Y
= C^* \big( C (X), \, u C_0 (X \setminus Y) \big)
\subset C^* (\Z, X, h),
\]
as in~\cite{Pt1},
the correct formulas would use
$Z_m = \bigcup_{j = 0}^{m  1} h^{j} (Y)$
for $m \in \Nz$,
and would be
\[
\gm_k (u^m f) = s_k^m
\diag \big( f \circ h _{Y_k}, \, f \circ h^2 _{Y_k},
\, \dots, \, f \circ h^{n (k)} _{Y_k} \big)
\]
and
\[
\gm_k (f u^{m}) =
\diag \big( f \circ h _{Y_k}, \, f \circ h^2 _{Y_k},
\, \dots, \, f \circ h^{n (k)} _{Y_k} \big)
\cdot s_k^{m}
\]
for $f \in C_0 (X \setminus Z_m)$.
This proposition and its proof
are based on~\cite{LqPh_dr}.
\begin{prp}\label{P_4319_AYhom}
Let $X$ be an infinite \chs{} and let $h \colon X \to X$ be a \mh.
Let $Y \subset X$ be closed with $\sint (Y) \neq \varnothing$.
Adopt the notation of \Def{D_4319_ModRokh}
and Notation~\ref{AYShiftNtn},
and let $Z_m$ be as in Proposition~\ref{P_4319_CharOB}.
For $k = 0, 1, \ldots, l$
there a unique linear map
$\gm_k \colon C^* (\Z, X, h)_Y \to C (Y_k, M_{n (k)} )$
such that
\[
\gm_k ( f u^m ) =
\diag \big( f _{Y_k}, \, f \circ h _{Y_k},
\, \dots, \, f \circ h^{n (k)  1} _{Y_k} \big) \cdot s_k^m
\]
and
\[
\gm_k ( u^{m} f ) =
s_k^{m} \cdot \diag \big( f _{Y_k}, \, f \circ h _{Y_k},
\, \dots, \, f \circ h^{n (k)  1} _{Y_k} \big)
\]
for $f \in C_0 (X \setminus Z_m)$.
Moreover, the map
\[
\gm \colon
C^* (\Z, X, h)_Y
\to \bigoplus_{k = 0}^l C ( Y_k, M_{n (k)} )
\]
given by
\[
\gm (a) = ( \gm_0 (a), \, \gm_1 (a), \, \dots, \, \gm_l (a) ).
\]
is a \hm{} of \ca{s}.
\end{prp}
\begin{proof}
We first claim that if $f \in C (X)$ and $m \in \Nz$,
then $u^{m} f \in C^* (\Z, X, h)_Y$
\ifo{} $f \in C_0 (X \setminus Z_m)$.
Since $u^{m} f = (f \circ h^m) u^m$,
the claim follows from Proposition~\ref{P_4319_CharOB}
and the fact that $Z_{m} = h^{m} (Z_m)$
(so that $f$ vanishes on $Z_m$
\ifo{} $f \circ h^m$ vanishes on $Z_{m}$).
Existence and uniqueness of the linear map $\gm_k$
now follows from the Banach space direct sum decomposition of
Corollary~\ref{C_7120_DirSum}.
It remains to check that $\gm$ is a \hm.
We check that $\gm_k$ is a \hm{} for $k = 0, 1, \ldots, l$.
It is obvious that $\gm_k (a^*) = \gm_k (a)^*$
for $a \in C^* (\Z, X, h)_Y$.
So we only need to prove multiplicativity.
Define $\sm_k \colon C (X) \to C ( Y_k, M_{n (k)} )$ by
\[
\sm_k (f) =
\diag \big( f _{Y_k}, \, f \circ h _{Y_k}, \, \dots, \,
f \circ h^{n (k)  1} _{Y_k} \big).
\]
Let $Z_m$ be as in Proposition~\ref{P_4319_CharOB}.
We claim that if $f \in C (X \setminus Z_m)$ and $g \in C (X)$,
then
%
\begin{equation}\label{Eq_7121_sComm}
s_k^m \sm_k (f \circ h^m) \sm_k (g)
= \sm (f) s_k^m \sm_k (g) = \sm (f) \sm_k (g \circ h^{m}) s_k^m.
\end{equation}
%
Define
\begin{align*}
& b_1 = \diag \big( ( f \circ h^{n (k)} _{Y_k} )
( g \circ h^{n (k)  m} _{Y_k} ), \,
( f \circ h^{n (k) + 1} _{Y_k} )
( g \circ h^{n (k)  m + 1} _{Y_k} ),
\, \dots, \,
\\
& \hspace{9em}
( f \circ h^{n (k) + m  1} _{Y_k} )
( g \circ h^{n (k)  1} _{Y_k} ) \big),
\end{align*}
\begin{align*}
& b_2 = \diag \big( ( f _{Y_k} )
( g \circ h^{n (k)  m} _{Y_k} ), \,
( f \circ h _{Y_k} )
( g \circ h^{n (k)  m + 1} _{Y_k} ), \, \dots, \,
\\
& \hspace{9em}
( f \circ h^{m  1} _{Y_k} )
( g \circ h^{n (k)  1} _{Y_k} ) \big),
\end{align*}
\begin{align*}
& b_3 = \diag \big( ( f _{Y_k} )
( g \circ h^{  m} _{Y_k} ), \,
( f \circ h _{Y_k} )
( g \circ h^{  m + 1} _{Y_k} ), \, \dots, \,
\\
& \hspace{9em}
( f \circ h^{m  1} _{Y_k} ) g \circ h^{1} _{Y_k} \big),
\end{align*}
and
\begin{align*}
& c = \diag \big( ( f \circ h^{m} _{Y_k} )
( g _{Y_k} ), \,
( f \circ h^{m + 1} _{Y_k} )
( g \circ h _{Y_k} ), \, \dots, \,
\\
& \hspace{9em}
( f \circ h^{n (k)  1} _{Y_k} )
( g \circ h^{n (k)  m  1} _{Y_k} ) \big).
\end{align*}
Carrying out the matrix multiplications gives, for any
$f, g \in C (X)$, the block matrix forms
(in which the off diagonal
blocks are square, $m \times m$ in the upper right and
$(n (k)  m) \times (n (k)  m)$ in the lower left):
\[
s_k^m \sm_k (f \circ h^m) \sm_k (g)
= \left( \begin{matrix}
0 & b_1 \\ c & 0
\end{matrix} \right), \qquad
\sm (f) s_k^m \sm_k (g)
= \left( \begin{matrix}
0 & b_2 \\ c & 0
\end{matrix} \right),
\]
and
\[
\sm (f) \sm_k (g \circ h^{m}) s_k^m
= \left( \begin{matrix}
0 & b_3 \\ c & 0
\end{matrix} \right).
\]
Now we recall that $f$ is required to vanish on
\[
Z_m = Y \cup h (Y) \cup \cdots \cup h^{m  1} (Y).
\]
Since $h^{n (k)} (Y_k) \subset Y$,
it follows that $b_1 = b_2 = b_3 = 0$.
Thus, all three products agree.
This proves the claim.
Now let $p, \, q \in \Nz$,
let $f \in C (X \setminus Z_p)$, and let $g \in C (X \setminus Z_q)$.
We claim that
%
\begin{equation}\label{Eq_7121_fugu}
\gm_k \big( (f u^p) (g u^q) \big)
= \gm_k (f u^p) \gm_k (g u^q)
\end{equation}
%
for any such $p$ and~$q$,
that
%
\begin{equation}\label{Eq_7121_ufgu}
\gm_k \big( (u^{ p} f) (g u^q) \big)
= \gm_k (u^{ p} f) \gm_k (g u^q)
\end{equation}
%
whenever $p \leq q$,
and that
%
\begin{equation}\label{Eq_7121_fuug}
\gm_k \big( (f u^p) (u^{ q} g) \big)
= \gm_k (f u^p) \gm_k (u^{ q} g)
\end{equation}
%
whenever $q \leq p$.
Given the claim,
to prove multiplicativity
it suffices to
prove (\ref{Eq_7121_ufgu}) when $p \geq q$,
(\ref{Eq_7121_fuug}) when $q \geq p$,
and
\[
\gm_k \big( (u^{ p} f) (u^{ q} g) \big)
= \gm_k (u^{ p} f) \gm_k (u^{ q} g)
\]
for arbitrary $p, \, q \in \Nz$.
The first can be deduced
by taking adjoints in~(\ref{Eq_7121_ufgu}),
the second can be deduced
by taking adjoints in~(\ref{Eq_7121_fuug}),
and the third can be deduced
by taking adjoints in~(\ref{Eq_7121_fugu}).
Using the first part of~(\ref{Eq_7121_sComm}) with $m = p$
at the second step,
we get
\begin{align*}
\gm_k (f u^p) \gm_k (g u^q)
& = \sm_k (f) s_k^p \sm_k (g) s_k^q
= \sm_k (f) \sm_k (g \circ h^{p}) s_k^{p + q}
\\
& = \gm_k \big( f (g \circ h^{p}) u^{p + q} \big)
= \gm_k \big( (f u^p) (g u^q) \big),
\end{align*}
which is~(\ref{Eq_7121_fugu}).
Assume that $p \leq q$.
Then ${\overline{g}}$ vanishes on $Z_p$ since $Z_p \subset Z_q$.
Applying~(\ref{Eq_7121_sComm}) with $m = p$
at the third step,
we get
\begin{align*}
\gm_k (u^{ p} f) \gm_k (g u^q)
& = s_k^{ p} \sm_k (f) \sm_k (g) s_k^q
= \big[\sm_k ( {\overline{g}} )
\sm_k ({\overline{f}}) s_k^p \big]^*
s_k^q
\\
& = \big[ s_k^p \sm_k ( {\overline{g}} \circ h^p )
\sm_k ({\overline{f}} \circ h^p ) \big]^*
s_k^q
= \sm_k (f \circ h^{p}) \sm_k (g \circ h^{p}) s_k^{q  p}
\\
& = \gm_k \big( (f \circ h^{p}) (g \circ h^{p}) u^{q  p} \big)
= \gm_k \big( (u^{ p} f) (g u^q) \big),
\end{align*}
which is~(\ref{Eq_7121_ufgu}).
Now assume that $q \leq p$.
Since $p  q \leq p$,
we have $Z_{p  q} \subset Z_p$.
So $f$ vanishes on $Z_{p  q}$,
and we can apply~(\ref{Eq_7121_sComm}) with $m = p  q$
at the second step
to get
\begin{align*}
\gm_k (f u^p) \gm_k (u^{ q} g)
& = \sm_k (f) s_k^{p  q} \sm_k (g)
= \sm_k (f) \sm_k (g \circ h^{q  p}) s_k^{p  q}
\\
& = \gm_k \big( f (g \circ h^{q  p}) u^{p  q} \big)
= \gm_k \big( (f u^p) (u^{ q} g) \big).
\end{align*}
This is~(\ref{Eq_7121_fuug}).
The proof of the claim, and therefore of the proposition,
is complete.
\end{proof}
What does the range of the \hm~$\gm$
of Proposition~\ref{P_4319_AYhom}
look like?
To give a good answer,
we start with the definition of a recursive subhomogeneous algebra.
Essentially, it is a generalization
of an algebra of the form
$\bigoplus_{k = 0}^l C (X_k, \, M_{n (k)} )$,
in which one is allowed to glue the summands
together along the ``boundaries'' of the spaces~$X_k$.
As a very simple example,
let $M_3 \oplus M_4 \subset M_7$ be the subalgebra consisting of
all block diagonal matrices $a \oplus b$
with $a \in M_3$ and $b \in M_4$,
and
consider the \ca{}
\[
\big\{ f \in C ([1, \, 1], \, M_7) \colon
{\mbox{$f (t) \in M_3 \oplus M_4$ for $t \in [1, \, 0]$}} \big\},
\]
which is made by gluing together the algebras
\[
C ([1, \, 0], \, M_3),
\qquad
C ([1, \, 0], \, M_4),
\andeqn
C ([0, \, 1], \, M_7)
\]
at the point~$0$.
Here is an example with no nontrivial \pj{s},
using the same direct sum notation:
\[
\big\{ f \in C ([0, 1], \, M_7) \colon
{\mbox{$f (0) \in M_3 \oplus M_4$
and $f (1) \in M_2 \oplus M_5$}} \big\}.
\]
\begin{dfn}[Definition~1.1 of~\cite{Ph_RSHA1}]\label{D_4317_RSHA}
The class of {\emph{recursive subhomogeneous algebras}}
is the smallest class ${\mathcal{R}}$
of \ca{s} such that:
\begin{enumerate}
\item\label{D_4317_RSHA_Base}
If $X$ is a \cpt\ space and $n \in \N$, then
$C (X, M_n) \in {\mathcal{R}}$.
\item\label{D_4317_RSHA_Recursion}
${\mathcal{R}}$ is closed under the following pullback construction.
Let $A \in {\mathcal{R}}$,
let $X$ be a \cpt\ space,
let $X^{(0)} \subset X$ be closed,
let $\ph \colon A \to C \big( X^{(0)}, \, M_n \big)$ be any
unital \hm,
and let
$\rh \colon C (X, M_n) \to C \big( X^{(0)}, \, M_n \big)$
be the restriction \hm.
Then the pullback
\[
A \oplus_{C ( X^{(0)}, \, M_n )} C (X, M_n) =
\{ (a, f) \in A \oplus C (X, M_n) \colon \ph (a) = \rh (f) \}
\]
(compare with Definition~2.1 of~\cite{Pd3}) is in~${\mathcal{R}}$.
\end{enumerate}
\end{dfn}
In~(\ref{D_4317_RSHA_Recursion})
the choice $X^{(0)} = \varnothing$ is allowed
(in which case $\ph = 0$ is allowed).
Thus the pullback could be an ordinary direct sum.
\begin{rmk}\label{R_7121_rsha_rem}
{}From the definition,
it is clear that any recursive subhomogeneous algebra
can be written in
the form
\[
R = \Big[ \cdots \big[ \big[
C_0 \oplus_{C_1^{(0)}} C_1 \big]
\oplus_{C_2^{(0)}} C_2 \big] \cdots \Big]
\oplus_{C_l^{(0)}} C_l,
\]
% \[
% R = \left[ \cdots \rule{0em}{3ex} \left[ \left[
% C_0 \oplus_{C_1^{(0)}} C_1 \right]
% \oplus_{C_2^{(0)}} C_2 \right] \cdots \right]
% \oplus_{C_l^{(0)}} C_l,
% \]
with $C_k = C (X_k, \, M_{n (k)})$ for \cpt\ spaces $X_k$ and positive
integers $n (k)$, with
$C_k^{(0)} = C {\ts{ \left( \rsz{ X_k^{(0)} }, \, M_{n (k)} \right) }}$
for compact
subsets $X_k^{(0)} \subset X_k$ (possibly empty), and where the maps
$C_k \to C_k^{(0)}$ are always the restriction maps.
An expression of this type will be referred to as a
{\emph{recursive subhomogeneous decomposition}} of~$R$.
(The decomposition is very far from unique.)
\end{rmk}
We give parts of Definition~1.2 of~\cite{Ph_RSHA1}.
\begin{dfn}\label{D_4317_Struct}
Let $R$ be a recursive subhomogeneous algebra,
with a decomposition as in Remark~\ref{R_7121_rsha_rem}.
We associate with this decomposition:
\begin{enumerate}
% \item\label{D_4317_Struct_Len}
% Its {\emph{length}}~$l$.
\item\label{D_4317_Struct_Stage}
For $k = 0, 1, \ldots, l$,
the {\emph{$k$th stage algebra}}
\[
R^{(k)} = \Big[ \cdots \big[ \big[
C_0 \oplus_{C_1^{(0)}} C_1 \big]
\oplus_{C_2^{(0)}} C_2 \big] \cdots \Big]
\oplus_{C_k^{(0)}} C_k,
\]
% \[
% R^{(k)} = \left[ \cdots \rule{0em}{3ex} \left[ \left[
% C_0 \oplus_{C_1^{(0)}} C_1 \right]
% \oplus_{C_2^{(0)}} C_2 \right] \cdots \right]
% \oplus_{C_k^{(0)}} C_k,
% \]
obtained by using only the first $k + 1$ algebras
$C_0, C_1, \dots, C_k$.
\item\label{D_4317_Struct_BSp}
Its {\emph{base spaces}} $X_0, X_1, \dots, X_l$ and
{\emph{total space}} $X = \coprod_{k = 0}^l X_k$.
\item\label{D_4317_Struct_TDim}
Its {\emph{topological dimension}} $\dim (X)$
(following \Def{D_6827_dimX};
here equal to $\max_k \dim (X_k)$).
\item\label{D_4317_Struct_MS}
Its {\emph{matrix sizes}} $n (0), \dots, n (l)$.
\item\label{D_4317_Struct_MinS}
Its {\emph{minimum matrix size}} $\min_k n (k)$.
\item\label{D_4317_Struct_StdRep}
Its {\emph{standard representation}}
$\sm = \sm_R \colon R \to \bigoplus_{k = 0}^l C (X_k, \, M_{n (k)} )$,
defined by forgetting the restriction to a subalgebra in each of
the pullbacks in the decomposition.
\end{enumerate}
\end{dfn}
By abuse of language,
we will often refer to the base spaces,
topological dimension,
etc.\ of a recursive subhomogeneous algebra~$A$,
when they in fact apply to a
particular recursive subhomogeneous decomposition.
The minimum matrix size actually does not
depend on the decomposition,
since it is the smallest dimension of an irreducible
representation of~$A$.
The base spaces certainly do,
and even their dimensions do,
as can be seen by considering the following example.
\begin{exa}\label{E_4317_ChDim}
Let $X$ be an arbitrary \chs,
set $X^{(0)} = X$,
and define $\ph \colon \C \to C (X)$
by $\ph (\ld) = \ld \cdot 1$
for $\ld \in \C$.
Let $\rh \colon C (X) \to C (X)$ be $\rh = \id_{C (X)}$.
Then
$\C \oplus_{C (X)} C (X) \cong \C$,
so we have a recursive subhomogeneous decomposition
for $\C$ whose topological dimension is $\dim (X)$.
\end{exa}
\begin{thm}\label{T_4317_AYIsRsha}
Let $X$ be an infinite compact metric space,
let $h \colon X \to X$ be a minimal homeomorphism,
and let $Y \subset X$ be a closed subset with $\sint (Y) \neq \E$.
Then the algebra $C^* (\Z, X, h)_Y$ of Definition~\ref{D_5421_VSubalg}
is a recursive subhomogeneous algebra
with topological dimension equal to $\dim (X)$,
and whose base spaces are closed subsets of~$X$.
\end{thm}
The proof of this theorem
is based on~\cite{LqPh_dr};
also see Section~2 of the survey~\cite{LqPh}.
It proceeds via several further lemmas.
\begin{lem}\label{L_7121_AYsubst}
Let $X$ be an infinite \chs{} and let $h \colon X \to X$ be a \mh.
Let $Y \subset M$ be closed with $\sint (Y) \neq \varnothing$.
Then the \hm{} $\gm$ of Proposition~\ref{P_4319_AYhom}
is unchanged
if, for $k = 0, 1, \ldots, l$,
we replace $s_k$ by $s_k^{(0)}$ and
$s_k^{1}$ by $\big( s_k^{(0)} \big)^*$
in the definition.
That is, in the notation there,
for $k = 0, 1, \ldots, l$, $m \in \Nz$,
and $f \in C_0 (M \setminus Z_m)$,
we have
\[
\gm_k (f u^m) =
\diag \big( f _{Y_k}, \, f \circ h _{Y_k}, \, \dots, \,
f \circ h^{n (k)  1} _{Y_k} \big)
\big( s_k^{(0)} \big)^m
\]
and
\[
\gm_k (u^{m} f) = \big( \big( s_k^{(0)} \big)^* \big)^m
\diag \big( f _{Y_k}, \, f \circ h _{Y_k}, \, \dots, \,
f \circ h^{n (k)  1} _{Y_k} \big).
\]
\end{lem}
\begin{proof}
This follows by matrix multiplication from the fact that
$f$ vanishes on the sets
$Y_k, \, h (Y_k), \, \dots, \, h^{m  1} (Y_k)$.
\end{proof}
\begin{cor}\label{ImDirSum}
Let $X$ be an infinite \chs{} and let $h \colon X \to X$ be a \mh.
Let $Y \subset X$ be closed with $\sint (Y) \neq \varnothing$.
Let $E_k^{(m)} \colon C (Y_k, M_{n (k)} )
\to C (Y_k, M_{n (k)} )$
be the projection on the $m$th subdiagonal, that is,
identifying $C (Y_k, M_{n (k)} )$ with $M_{n (k)} (C (Y_k))$,
we have $E_k^{(m)} (b)_{m + j, j} = b_{m + j, j}$ for
$j = 1, 2, \ldots, n (k)  m$ (if $m \geq 0$) and
for $j =  m + 1, \,  m + 2, \, \ldots, \, n (k)$
(if $m \leq 0$), while
$E_k^{(m)} (b)_{i, j} = 0$ for all other pairs $(i, j)$.
(In particular, if $m > n (k)$, then $E_k^{(m)} = 0$.)
Set
\[
D_m
= \bigoplus_{k = 0}^l E_k^{(m)} \big( C ( Y_k, M_{n (k)} )
\big).
\]
Let
\[
\gm_k \colon C^* (\Z, X, h)_Y \to C (Y_k, M_{n (k)} )
\smandeqn
\gm \colon
C^* (\Z, X, h)_Y \to \bigoplus_{k = 0}^l C ( Y_k, M_{n (k)} )
\]
be as in Proposition~\ref{P_4319_AYhom}.
Then:
\begin{enumerate}
\item\label{ImDirSum_1}
There is a Banach space direct sum decomposition
\[
\bigoplus_{k = 0}^l C (Y_k, M_{n (k)})
= \bigoplus_{m = n (l)}^{n (l)} D_m.
\]
\item\label{ImDirSum_2}
For $k = 0, 1, \ldots, l$, $m \in \Nz$,
$f \in C_0 (M \setminus Z_m)$,
and $x \in Y_k$,
the expression
$\gm_k ( f u^m) (x)$ is given by the following
matrix,
in which the first nonzero entry is in row $m + 1$.
\[
\gm_k ( f u^m) (x) = \left( \begin{matrix}
0 & 0 & \cdots & \cdots & \cdots & \cdots & 0 \\
\vdots & \vdots & & & & & \vdots \\
0 & 0 & & & & & \vdots \\
f \circ h^m (x) & 0 & & & & & \vdots \\
0 & f \circ h^{m + 1} (x) & & & & & \vdots \\
\vdots & & \ddots & & & & \vdots\\
0 & \cdots & \cdots &
f \circ h^{n (k)  1} (x) & 0 & \cdots & 0
\end{matrix} \right).
\]
\item\label{ImDirSum_3}
For $m \geq 0$ and $f \in C_0 (X \setminus Z_m)$,
\[
\gm_k (f u^m)
\in E_k^{(m)} \big( C ( Y_k, M_{n (k)} ) \big)
\andeqn \gm ( f u^m) \in D_m
\]
and
\[
\gm_k ( u^{m} f )
\in E_k^{(m)} \big( C ( Y_k, M_{n (k)} ) \big)
\andeqn \gm ( u^{m} f ) \in D_{m}.
\]
\item\label{ImDirSum_4}
The \hm{} $\gm$ is compatible with the
vector space direct sum decomposition
of Proposition~\ref{P_4319_AYhom}
on its domain and the vector space direct sum decomposition
of part~(\ref{ImDirSum_1}) on its codomain.
\end{enumerate}
\end{cor}
\begin{proof}
The direct sum decomposition of part~(\ref{ImDirSum_1}) is easy.
The rest is all essentially
immediate from Proposition~\ref{P_4319_AYhom}
and Lemma~\ref{L_7121_AYsubst}.
\end{proof}
\begin{lem}\label{AYinj}
Let $X$ be an infinite \chs{} and let $h \colon X \to X$ be a \mh.
Let $Y \subset X$ be closed with $\sint (Y) \neq \varnothing$.
Then the \hm{} $\gm$ of Proposition~\ref{P_4319_AYhom} is injective.
\end{lem}
\begin{proof}
By Corollary~\ref{ImDirSum} and Corollary~\ref{C_7120_DirSum},
it suffices to show that if $\gm (f u^m) = 0$, with $m \geq 0$ and
$f \in C_0 (X \setminus Z_m)$, then $f = 0$.
By the definition of $\gm$, if $\gm (f u^m) = 0$ then
$f$ vanishes on all sets $h^j (Y_k)$ for
$k = 0, 1, \ldots, l$ and $j = 0, 1, \ldots, n (k)  1$.
These sets cover $X$ by Lemma \ref{RokhCov}(\ref{RokhCov_2}),
so $f = 0$.
\end{proof}
\begin{lem}\label{AYMembCond}
Let $X$ be an infinite \chs{} and let $h \colon X \to X$ be a \mh.
Let $Y \subset X$ be closed with $\sint (Y) \neq \varnothing$.
Adopt the notation
of Definition~\ref{D_4319_ModRokh},
and let $\gm$ be as in
Proposition~\ref{P_4319_AYhom}.
An element
\[
b = (b_0, b_1, \dots, b_l ) \in \bigoplus_{k = 0}^l C (Y_k, M_{n (k)})
\]
is in $\gm (C^* (\Z, X, h)_Y)$ if and only if, whenever
\[
r \in \N,
\qquad
k, t_1, t_2, \ldots, t_r \in \{ 0, 1, \ldots, l \},
\qquad
n ({t_1}) + n ({t_2}) + \cdots + n ({t_r}) = n (k),
\]
% \[
% t_1, t_2, \ldots, t_r \in \{ 0, 1, \ldots, l \},
% \]
% \[
% n ({t_1}) + n ({t_2}) + \cdots + n ({t_r}) = n (k),
% \]
and
\[
x \in (Y_k \setminus Y_k^{\bullet} ) \cap Y_{t_1}
\cap h^{n ({t_1})} (Y_{t_2} ) \cap \cdots \cap
h^{[n ({t_1}) + n ({t_2}) + \cdots + n ({t_{r1}}) ]}(Y_{t_r} ),
\]
then $b_k (x)$ is given by the block diagonal matrix
% \[
% b_k (x) = \left( \begin{matrix}
% b_{t_1} (x) & & & & \\
% & b_{t_2} (h^{n ({t_1})} (x)) & & & \\
% & & b_{t_3} (h^{n ({t_1}) + n ({t_2})} (x)) & & \\
% & & & \ddots & \\
% & & & & b_{t_r} (h^{n ({t_1}) + \cdots + n ({t_{r1}})} (x))
% \end{matrix} \right).
% \]
\begin{align}\label{Eq_7202_Rel}
b_k (x)
& = \diag \big( b_{t_1} (x), \,\,
b_{t_2} (h^{n ({t_1})} (x)), \,\,
\\
& \hspace*{3em} {\mbox{}}
b_{t_3} (h^{n ({t_1}) + n ({t_2})} (x)), \,\, \ldots, \,\,
b_{t_r} (h^{n ({t_1}) + \cdots + n ({t_{r1}})} (x)) \big).
\notag
\end{align}
\end{lem}
\begin{proof}
For $b$, $r$, $k$, $t_1, t_2, \ldots, t_r$, and $x$
as in the statement,
let
\[
\af_{x, r, k, t_1, t_2, \ldots, t_r} (b_0, b_1, \ldots, b_{k  1})
\]
denote the block diagonal matrix
on the right hand side of~(\ref{Eq_7202_Rel}).
% We omit the subscripts $t_1, t_2, \ldots, t_m$ when they
% are understood, and
We write
\[
b^{(k  1)} = (b_0, b_1, \ldots, b_{k  1}).
\]
Let $E_k^{(m)}$ be as in Corollary~\ref{ImDirSum}.
By Corollary~\ref{ImDirSum},
it suffices to verify, for each
fixed $m$,
the statement of the lemma for elements $(b_0, b_1, \ldots, b_l)$
such that $b_k$ is in the range of $E_k^{(m)}$
for $k = 0, 1, \ldots, l$.
Using the adjoint, we may in fact restrict to
the case $m \geq 0$.
We verify that elements of the range of $\gm$ satisfy the
required relations.
Let
\[
r \in \N,
\qquad
t_1, t_2, \ldots, t_r \in \{ 0, 1, \ldots, l \},
\qquad
n ({t_1}) + n ({t_2}) + \cdots + n ({t_r}) = n (k),
\]
% \[
% t_1, t_2, \ldots, t_r \in \{ 0, 1, \ldots, l \},
% \]
% \[
% n ({t_1}) + n ({t_2}) + \cdots + n ({t_r}) = n (k),
% \]
and
\[
x \in (Y_k \setminus Y_k^{\bullet} ) \cap Y_{t_1}
\cap h^{n ({t_1})} (Y_{t_2} ) \cap \cdots \cap
h^{[n ({t_1}) + n ({t_2}) + \cdots + n ({t_{r1}}) ]}(Y_{t_r} ).
\]
Let $f \in C_0 (X \setminus Z_m)$
and define
\[
(b_0, b_1, \ldots, b_l) \in \bigoplus_{k = 0}^l C (Y_k, M_{n (k)})
\]
by $(b_0, b_1, \ldots, b_l) = \gm (f u^m)$.
By Corollary \ref{ImDirSum}(\ref{ImDirSum_2}),
the $m$th subdiagonal of $b_k (x)$ is
%
\begin{equation}\label{Eq_7202_b}
\big( f \circ h^m (x), \, f \circ h^{m + 1} (x), \, \ldots, \,
f \circ h^{n (k)  1} (x) \big).
\end{equation}
%
Similarly, the $m$th subdiagonal of
$\af_{x, r, k, t_1, t_2, \ldots, t_r} (b^{(k  1)} )$ is
given by the following formula (explanations afterwards):
\begin{align}\label{Eq_7202_mdiag}
& \big(
f \circ h^m (x), \, f \circ h^{m + 1} (x), \, \ldots, \,
f \circ h^{n (t_1)  1} (x), \, 0, \, 0, \, \ldots, \, 0,
\\
& \hspace*{2em} \mbox{}
f \circ h^{n (t_1) + m} (x), \, f \circ h^{n (t_1) + m + 1} (x),
\, \ldots, \, f \circ h^{n (t_1) + n (t_2)  1} (x), \,
0, \, 0, \, \ldots, \, 0,
\notag
\\
& \hspace*{2em} \mbox{}
\ldots \ldots \ldots \ldots \ldots \ldots,
\notag
\\
& \hspace*{2em} \mbox{}
f \circ h^{n ({t_1}) + \cdots + n ({t_{r1}}) + m} (x), \,
f \circ h^{n ({t_1}) + \cdots + n ({t_{r1}}) + m + 1} (x),
\, \ldots,
\notag
\\
& \hspace*{10em} \mbox{}
f \circ h^{n ({t_1}) + \cdots + n ({t_{r1}}) + n (t_r)  1} (x)
\big).
\notag
\end{align}
The sequences of zeros all have length~$m$,
except that if $n (t_i) \leq m$
then the subsequence
\begin{align}\label{Eq_7202_SbSq}
& f \circ h^{n ({t_1}) + \cdots + n ({t_{i1}}) + m} (x), \,
f \circ h^{n ({t_1}) + \cdots + n ({t_{i1}}) + m + 1} (x),
\\
& \hspace*{2em} \mbox{}
\, \ldots, \, f \circ h^{n (t_1) + \cdots + n (t_i)  1} (x), \,
0, \, 0, \, \ldots, \, 0
\notag
\end{align}
should be read as a sequence of $n (t_i)$ zeros.
Thus, for $i = 1, 2, \ldots, r  1$,
the subsequence of the form~(\ref{Eq_7202_SbSq})
has total length $n (t_i)$,
while for $i = r$ the corresponding subsequence
(without zeros at the end) has total length $n (t_r)  m$.
The terms in (\ref{Eq_7202_b})
which have been replaced by zero in (\ref{Eq_7202_mdiag})
are exactly those containing values of $f$ at points of
the form $h^{n ({t_1}) + \cdots + n (t_i) + j} (x)$
for $i = 1, 2, \ldots r  1$ and $j = 0, 1, \ldots m  1$.
Since $h^{n ({t_1}) + \cdots + n (t_i) } (x) \in Y$,
all these points are in $Z_m$, so that $f$ is zero on them anyway.
Therefore the sequences (\ref{Eq_7202_b})
and (\ref{Eq_7202_mdiag}) are equal.
We have show that
elements of the range of $\gm$ satisfy the required relations.
For the converse, let $m \geq 0$, let
\[
(b_0, b_1, \ldots, b_l) \in \bigoplus_{k = 0}^l C (Y_k, M_{n (k)})
\]
satisfy the relations in the statement of the lemma,
and assume that
$b_k$ is in the range of $E_k^{(m)}$ for $k = 0, 1, \ldots, l$.
Define \cfn{s} $f_k^{(j)} \colon h^j (Y_k) \to \C$,
for $k = 0, 1, \ldots, l$ and $j = 0, 1, \ldots, n (k)  1$,
as follows.
When $n (k) > m$,
we specify that the $m$th subdiagonal of $b_k$
(starting at $(b_k)_{m + 1, 1}$)
be given by
\[
\big( f_k^{(m)} \circ h^m, \, f_k^{(m + 1)} \circ h^{m + 1},
\, \ldots, \,
f_k^{(n (k)  1)} \circ h^{n (k)  1} \big).
\]
That is,
\[
f_k^{(j)} = ( b_{j + 1, \, j  m + 1} \circ h^{ j} ) _{Y_k}
\]
for $j = m, m + 1, \ldots, n (k)  1$
We further set $f_k^{(j)} = 0$
for $k = 0, 1, \ldots, l$
and $j = 0, 1, \ldots, \min ( n (k)  1, \, m  1 )$.
We claim that there is a \cfn{} $f \colon X \to \C$ such that
$f _{h^j (Y_k)} = f_k^{(j)}$ for all $j$ and $k$, and that
$f$ vanishes on $Z_m$.
Given this, it is clear that $f u^m \in C^* (\Z, X, h)_Y$ and
$\gm (f u^m) = (b_0, b_1, \ldots, b_l)$.
Assume that $f$ exists.
We claim that $f$ vanishes on $Z_m$.
Recall that (Proposition~\ref{P_4319_CharOB} for the first equality
and Lemma \ref{RokhCov}(\ref{RokhCov_4b}) for the second)
\[
Z_m = \bigcup_{j = 0}^{m  1} h^{j} (Y)
= \bigcup_{j = 0}^{m  1} \bigcup_{k = 0}^l h^{j} (Y_k).
\]
Let $j \in \{ 0, 1, \ldots, m \}$,
let $k \in \{ 0, 1, \ldots, l \}$,
and
let $x \in h^{j} (Y_k)$.
We need to show that $f (x) = 0$.
If $j \leq n (k)  1$,
then $f_k^{(j)} = 0$ is immediate from the definition,
so $f (x) = 0$.
So assume $j > n (k)  1$.
Let $s \in \Nz$ be the least nonnegative integer
such that $h^{ s} (x) \in Y$.
Then $s \leq j$.
Set $x_0 = h^{ s} (x)$.
Choose $i \in \{ 0, 1, \ldots, l \}$ such that $r (x_0) = n (i)$.
Then $h (x_0), \, h^2 (x_0), \, \ldots, h^{s} (x_0) \not\in Y$,
so
\[
s \leq \min (n (i)  1, \, j)
\leq \min (n (i)  1, \, m  1)
\andeqn
x = h^s (x_0) \in h^s (Y_{n (i)}).
\]
So $f (x) = 0$ by the case considered first.
This proves the claim.
% JJJ
It remains only to prove that $f$ exists and is \ct.
Since the sets $h^j (Y_k)$ are closed, it suffices to prove that if
\[
0 \leq k_1, k_2 \leq l,
\qquad
0 \leq j_1 \leq n (k_1)  1,
\qquad
0 \leq j_2 \leq n (k_2)  1,
\]
and
\[
x \in h^{j_1} (Y_{k_1}) \cap h^{j_2} (Y_{k_2}),
\]
then $f_{k_1}^{(j_1)} (x) = f_{k_2}^{(j_2)} (x)$.
\Wolog{} $j_1 \leq j_2$.
First assume $j_1 = j_2$.
Call this number~$j$.
Then \wolog\ $k_1 \leq k_2$.
If $k_1 = k_2$, there is nothing to prove,
so we may assume that $k_1 < k_2$.
So $n (k_1) < n (k_2)$.
Let $x_0 = h^{j} (x)$.
Then $x_0 \in Y_{k_1} \cap Y_{k_2}$.
Since $n (k_1) < n (k_2)$ is a return time for $x_0$, we have
$x_0 \in Y_{k_2} \setminus Y_{k_2}^{\bullet}$
by Lemma \ref{RokhCov}(\ref{RokhCov_5}).
Choose $t_1 \in \{ 0, 1, \ldots, l \}$
such that $n (t_1)$ is the first return time
of $h^{n (k_1)} (x_0)$ to~$Y$.
So $h^{n (k_1)} (x_0) \in Y_{t_1}$.
If $n (k_1) + n (t_1) < n (k_2)$,
we can choose $t_2 \in \{ 0, 1, \ldots, l \}$
such that $n (t_2)$ is the first return
time of $h^{n (k_1) + n (t_1)} (x_0)$ to~$Y$.
So $h^{n (k_1) + n (t_1)} (x_0) \in Y_{t_2}$.
Proceed inductively.
Since the numbers $n (t_1), \, n (t_1) + n (t_2), \, \ldots$
are successive
return times of $h^{n (k_1)} (x_0)$ to $Y$, and since
$h^{n (k_2)} (x_0) \in Y$, there is $r$ such that
\[
n (k_1) + n (t_1) + n (t_2) + \cdots + n (t_r) = n (k_2).
\]
Then
\begin{align*}
x_0 \in & (Y_{k_2} \setminus Y_{k_2}^{\bullet} ) \cap Y_{k_1}
\cap h^{n ({k_1})} (Y_{t_1} )
\\
& \hspace*{1em} \mbox{} \cap
h^{[n (k_1) + n ({t_1})]} (Y_{t_2})
\cap \cdots \cap
h^{[n (k_1) + n ({t_1}) + n ({t_2}) + \cdots
+ n ({t_{r1}}) ]}(Y_{t_r} ),
\end{align*}
so
$\af_{x_0, r + 1, k_2, k_1, t_1, t_2, \ldots, t_r} (b^{(k_2  1)})
= b_{k_2} (x_0)$.
If $0 \leq j \leq m  1$,
then $f_{k_1}^{(j)} (x)$ and $f_{k_2}^{(j)} (x)$
are both zero.
Otherwise,
$f_{k_1}^{(j)} (x) = (f_{k_1}^{(j)} \circ h^j) (x_0)$
is the $(j + 1, \, j  m + 1)$ entry
of $b_{k_1} (x_0)$
and
$f_{k_2}^{(j)} (x) = (f_{k_2}^{(j)} \circ h^j) (x_0)$
is the $(j + 1, \, j  m + 1)$ entry
of $b_{k_2} (x_0)$.
The relations in the statement of the lemma,
with $k_1, t_1, t_2, \ldots, t_r$ in place of $t_1, t_2, \ldots, t_r$,
therefore imply
that $f_{k_1}^{(j)} (x) = f_{k_2}^{(j)} (x)$,
as desired.
Now suppose $j_1 < j_2$.
We split this case in two subcases, the first of which is
%
\begin{equation}\label{Eq_7121_nkj}
n (k_1)  j_1 \leq n (k_2)  j_2.
\end{equation}
%
Suppose $j_2 < m$.
Then also $j_1 < m$,
so
$f_{k_1}^{(j_1)} (x) = f_{k_2}^{(j_2)} (x) = 0$,
as desired.
So we can assume $m \leq j_2 \leq n (k_2)  1$.
Define $x_0 = h^{j_2} (x)$,
giving
\[
x_0 \in Y_{k_2} \cap h^{ (j_2  j_1)} (Y_{k_1}).
\]
Now $j_2  j_1 < n (k_2)$
and is a return time of $x_0$ to $Y$,
so $x_0 \in Y_{k_2} \setminus Y_{k_2}^{\bullet}$
by Lemma \ref{RokhCov}(\ref{RokhCov_5}).
Using the same argument as in the previous case, choose
\[
t_1, t_2, \ldots, t_{\mu} \in \{ 0, 1, \ldots, l \}
\]
such that
$n (t_1), \, n (t_1) + n (t_2), \, \ldots$
are successive return times of $x_0$ to $Y$, and such that
\[
n (t_1) + n (t_2) + \cdots + n (t_{\mu}) = j_2  j_1.
\]
Similarly, using~(\ref{Eq_7121_nkj})
to get $n (k_1) + j_2  j_1 \leq n (k_2)$,
choose
\[
t'_1, t'_2, \ldots, t'_{\nu} \in \{ 0, 1, \ldots, l \}
\]
such that
$n (t'_1), \, n (t'_1) + n (t'_2), \, \ldots$
are successive return times of $h^{n (k_1) + j_2  j_1} (x_0)$ to $Y$,
and such that
\[
n (t'_1) + n (t'_2) + \cdots + n (t'_{\nu}) = n (k_1)  (j_2  j_1).
\]
Then
\begin{align*}
x_0 \in & (Y_{k_2} \setminus Y_{k_2}^{\bullet} ) \cap Y_{t_1}
\cap h^{n ({t_1})} (Y_{t_2} )
\cap \cdots \cap
h^{[n ({t_1}) + \cdots + n ({t_{\mu  1}}) ]} (Y_{t_{\mu}})
\\
& \hspace*{1em} \mbox{} \cap
h^{[n ({t_1}) + \cdots + n ({t_{\mu}}) ]} (Y_{k_1})
\cap
h^{[ n ({t_1}) + \cdots + n ({t_{\mu}}) + n (k_1) ]}(Y_{t'_1} )
\cap \cdots
\\
& \hspace*{1em} \mbox{} \cap
h^{[ n ({t_1}) + \cdots + n ({t_{\mu}}) + n (k_1)
+ n ({t'_1}) + \cdots + n ({t'_{\nu  1}}) ]}(Y_{t'_{\nu}} )
\end{align*}
and
\[
n ({t_1}) + \cdots + n ({t_{\mu}}) + n (k_1)
+ n ({t'_1}) + \cdots + n ({t'_{\nu}}) = n (k_2).
\]
so the hypothesized relations give
%
\begin{equation}\label{Eq_7204_HypRel}
\af_{x_0, \mu + 1 + \nu, k_2,
t_1, t_2, \ldots, t_{\mu}, k_1, t_1', t_2', \ldots, t_{\nu}'}
(b^{(k_2  1)}) = b_{k_2} (x_0).
\end{equation}
%
We compare the $(j_2 + 1, \, j_2  m + 1)$ entries
of the two sides of~(\ref{Eq_7204_HypRel}).
The $(j_2 + 1, \, j_2  m + 1)$ entry of $b_{k_2} (x_0)$
is
$\big( f_{k_2}^{(j_2)} \circ h^{j_2} \big) (x_0)
= f_{k_2}^{(j_2)} (x)$.
We examine the $(j_2 + 1, \, j_2  m + 1)$ entry of the left hand side.
By considering the row number $j_2 + 1$,
and using the relations
\[
j_2 = n (t_1) + n (t_2) + \cdots + n (t_{\mu}) + j_1
\andeqn
0 \leq j_1 \leq n (k_1)  1,
\]
we see that the $(j_2 + 1, \, j_2  m + 1)$ entry of the left hand side
of~(\ref{Eq_7204_HypRel})
must be either in the diagonal block
\[
(b_{k_1} \circ h^{j_2  j_1}) (x_0)
= (b_{k_1} \circ h^{n (t_1) + n (t_2) + \cdots + n (t_{\mu})}) (x_0)
\]
in the formula for
$\af_{x_0, \mu + 1 + \nu, k_2,
t_1, t_2, \ldots, t_{\mu}, k_1, t_1', t_2', \ldots, t_{\nu}'}
(b^{(k_2  1)})$,
or in none of the diagonal blocks.
If $j_1 < m$,
then
\[
j_2  m + 1
< j_2  j_1 + 1
= n (t_1) + n (t_2) + \cdots + n (t_{\mu}) + 1,
\]
so the $(j_2 + 1, \, j_2  m + 1)$ entry of the left hand side
of~(\ref{Eq_7204_HypRel})
is in none of the diagonal blocks.
Thus (\ref{Eq_7204_HypRel}) implies that $f_{k_2}^{(j_2)} (x) = 0$,
while $f_{k_1}^{(j_1)} (x) = 0$ by the definition of
$f_{k_1}^{(j_1)}$.
Thus $f_{k_1}^{(j_1)} (x) = f_{k_2}^{(j_2)} (x)$,
as desired.
% JJJ
If instead $m \leq j_1 \leq n (k_1)  1$,
then
\[
n (t_1) + n (t_2) + \cdots + n (t_{\mu}) + 1
\leq j_2  m + 1
\leq j_2
< n (t_1) + n (t_2) + \cdots + n (t_{\mu}) + n (t_k),
\]
so the
$(j_2 + 1, \, j_2  m + 1)$ entry of the of the left hand side
of~(\ref{Eq_7204_HypRel})
is in the diagonal block $(b_{k_1} \circ h^{j_2  j_1}) (x_0)$.
In fact, it is the $(j_1 + 1, \, j_1  m + 1)$ entry of
$(b_{k_1} \circ h^{j_2  j_1}) (x_0)$.
So~(\ref{Eq_7204_HypRel}) implies that
\[
f_{k_2}^{(j_2)} (x)
= \big( f_{k_1}^{(j_1)} \circ h^{j_1} \circ h^{j_2  j_1} \big) (x_0)
= f_{k_1}^{(j_1)} (x),
\]
as desired.
Now suppose that $n (k_1)  j_1 > n (k_2)  j_2$,
the opposite of~(\ref{Eq_7121_nkj}).
We reduce this case to a strictly smaller value of $n (k_1) + n (k_2)$
together with instances of the cases already done, so that the
desired equality $f_{k_1}^{(j_1)} (x) = f_{k_2}^{(j_2)} (x)$ follows
by a finite descent argument.
Set $x_0 = h^{ j_1} (x) \in Y_{k_1}$.
Using the same argument as before, choose
\[
t_1, t_2, \ldots, t_{r} \in \{ 0, 1, \ldots, l \}
\]
such that
$n (t_1), \, n (t_1) + n (t_2), \, \ldots$
are successive return times of $x_0$ to~$Y$, and such that
\[
n (t_1) + n (t_2) + \cdots + n (t_{r}) = n (k_1).
\]
Then
\[
h^{ n (k_2)  (j_2  j_1)} (x_0) = h^{ n (k_2)  j_2 } (x)
\in h^{n (k_2)} (Y_{k_2}) \subset Y
\]
and $n (k_2)  (j_2  j_1) < n (k_1)$, so $r \geq 2$.
Choose $i \in \{ 0, 1, \ldots, l \}$ such that
\[
n (t_1) + n (t_2) + \cdots + n (t_{i  1}) \leq j_1
< n (t_1) + n (t_2) + \cdots + n (t_{i}),
\]
and let
\[
k_3 = t_i
\andeqn
j_3 = j_1  [n (t_1) + n (t_2) + \cdots + n (t_{i  1}) ].
\]
Then $0 \leq j_3 \leq n (k_3)  1$.
Define
\[
y = h^{n (t_1) + n (t_2) + \cdots + n (t_{i  1})} (x_0) \in Y_{k_3}.
\]
Then $h^{j_3} (y) = x$, so
\[
x \in h^{j_3} (Y_{k_3}) \cap h^{j_1} (Y_{k_1}) \andeqn
x \in h^{j_3} (Y_{k_3}) \cap h^{j_2} (Y_{k_2}).
\]
We have
\[
j_3 \leq j_1 \andeqn n (k_3)  j_3 \leq n (k_1)  j_1,
\]
so the cases we have already done give
$f_{k_1}^{(j_1)} (x) = f_{k_3}^{(j_3)} (x)$.
Therefore it suffices to replace $k_1$ and $j_1$ by $k_3$ and $j_3$
in the statement to be proved.
We have $n (k_3) < n (k_1)$ because
$n (t_1) + n (t_2) + \cdots + n (t_{r}) = n (k_1)$ and $r \geq 2$.
Meanwhile, $n (k_2)$ is the same as before.
This is the required reduction.
The proof that $f$ is well defined and \ct{} is now complete.
\end{proof}
We now give a more precise statement of Theorem~\ref{T_4317_AYIsRsha}.
It is the generalization
of the isomorphism~(\ref{Eq_4319_FormOfIso})
gotten from the proof of \Lem{L:AF}.
\begin{thm}\label{T_7121_AY_rshd}
Let $Y \subset X$ be closed with $\sint (Y) \neq \varnothing$.
Let
\[
\gm \colon C^* (\Z, X, h)_Y \to \bigoplus_{k = 0}^l C (Y_k, M_{n (k)} )
\]
be the \hm\ of Proposition~\ref{P_4319_AYhom}.
Then $\gm$ induces an isomorphism of $C^* (\Z, X, h)_Y$ with the
recursive subhomogeneous algebra defined,
in the notation of Remark~\ref{R_7121_rsha_rem}
and \Def{D_4317_Struct},
as follows.
\begin{enumerate}
\item\label{T_7121_AY_rshd_1}
$l$ and $n (0), \, n (1), \, \ldots, \, n (l)$
are as in \Def{D_4319_ModRokh}.
\item\label{T_7121_AY_rshd_2}
$X_k = Y_k$ for $k = 1, 2, \ldots, l$.
\item\label{T_7121_AY_rshd_3}
$X_k^{(0)} = Y_k \cap \bigcup_{j = 0}^{k  1} Y_j$
for $k = 1, 2, \ldots, l$.
\item\label{T_7121_AY_rshd_4}
For $k = 0, 1, \ldots, l$,
$x \in Y_k \cap \bigcup_{j = 0}^{k  1} Y_j$
and $(b_0, b_1, \dots , b_{k  1} )$ in the image in
$\bigoplus_{j = 0}^{k  1} C (Y_j, M_{n (j)})$ of
the $k  1$ stage algebra $R^{(k  1)}$, whenever
\[
x \in (Y_k \setminus Y_k^{\bullet} ) \cap Y_{t_1}
\cap h^{n ({t_1})} (Y_{t_2} ) \cap \cdots \cap
h^{[n ({t_1}) + n ({t_2}) + \cdots + n ({t_{r1}}) ]}(Y_{t_r} ),
\]
with
\[
n ({t_1}) + n ({t_2}) + \cdots + n ({t_r}) = n (k),
\]
then
\begin{align*}
& \ph_k (b_0, b_1, \dots , b_{k  1} ) (x) =
\\
& \hspace*{1em} {\mbox{}}
\left( \begin{matrix}
b_{t_1} (x) & & & & \\
& b_{t_2} (h^{n ({t_1})} (x)) & & & \\
& & b_{t_3} (h^{n ({t_1}) + n ({t_2})} (x)) & & \\
& & & \ddots & \\
& & & & b_{t_r} (h^{n ({t_1}) + \cdots + n ({t_{r1}})} (x))
\end{matrix} \right).
\end{align*}
\item\label{T_7121_AY_rshd_5}
For $k = 0, 1, \ldots, l$,
$\ps_k$ is the restriction map.
\end{enumerate}
Moreover, the standard representation of
$\gm (C^* (\Z, X, h)_Y)$ is the inclusion map in
$\bigoplus_{k = 0}^l C (Y_k, M_{n (k)} )$.
\end{thm}
\begin{proof}
The main point is to show that
the formula in~(\ref{T_7121_AY_rshd_4})
actually gives a well defined \hm{}
\[
\ph_k \colon
R^{(k  1)} \to C \big( Y_k^{(0)}, M_{n (k)} \big).
\]
We do this by induction on~$k$.
Once it is known that $\ph_1, \ph_2, \ldots, \ph_{k  1}$ are
well defined, it follows that $R^{(k  1)}$ as described is
a recursive subhomogeneous algebra,
and that its elements are exactly those sequences
$(b_0, b_1, \dots , b_{k  1} )$ which satisfy the conditions of
Lemma~\ref{AYMembCond} up to $l = k  1$
(that is, the number $k$ there
is at most the number $k  1$ here).
Having $R^{(k  1)}$, it makes sense to consider a \hm{}
$\ph_k \colon R^{(k  1)} \to C \big( Y_k^{(0)}, M_{n (k)} \big)$.
If the one described in (\ref{T_7121_AY_rshd_4}) is well defined,
it will follow that
$R^{(k)}$ is a recursive subhomogeneous algebra,
and that its elements are exactly those sequences
$(b_0, b_1, \dots , b_{k  1} )$ which satisfy the conditions of
Lemma~\ref{AYMembCond} up to $l = k$.
We start by showing that $\ph_1$ is well defined.
We have $Y_1^{(0)} = Y_1 \cap Y_0$.
For $x \in Y_1^{(0)}$, let $0 = \nu_0, \nu_1, \dots, \nu_r = n (1)$
be the successive return times of $x$ to~$Y$.
We have $\nu_1 = n (0) < n (1)$, so $r \geq 2$.
For $i = 0, 1, \ldots, r  1$,
the first return time of $h^{\nu_i} (x)$ is strictly
less than $n (1)$, so can only be $n (0)$.
Therefore $n (1) = r n (0)$.
Thus, if $Y_1^{(0)} \neq \varnothing$, then $n (1) = r n (0)$ and
\[
Y_1^{(0)} = Y_1 \cap Y_0 \cap h^{n (0)} (Y_0) \cap
h^{2 n (0)} (Y_0) \cap \cdots \cap h^{[n (1)  n (0)]} (Y_0).
\]
If $Y_1^{(0)} = \varnothing$ then $\ph_1$ is trivially well defined,
and if $Y_1^{(0)} \neq \varnothing$,
then $\ph_1$ is well defined by the formula
\[
\ph_1 (b)
= \diag \big( b _{ Y_1^{(0)} }, \, b \circ h^{n (0)} _{ Y_1^{(0)} },
\, \dots, \, b \circ h^{n (1)  n (0)} _{ Y_1^{(0)} } \big).
\]
Now assume we have $R^{(k  1)}$.
Let $S$ be the set of all sequences $(t_1, t_2, \dots, t_r)$
such that $n (t_1) + n (t_2) + \cdots + n (t_r) = n (k)$,
with $r \geq 2$.
In such a sequence,
we have $t_i < k$ for $i = 1, 2, \ldots, r$.
For $\sm = (t_1, t_2, \dots, t_r) \in S$, define
\[
Y_k^{(\sm)} = (Y_k \setminus Y_k^{\bullet} ) \cap Y_{t_1}
\cap h^{n ({t_1})} (Y_{t_2} ) \cap \cdots \cap
h^{[n ({t_1}) + n ({t_2}) + \cdots + n ({t_{r1}}) ]}(Y_{t_r} ).
\]
(Note that the intersection is the same if one uses $Y_k$ in place of
$Y_k \setminus Y_k^{\bullet}$.)
By considering successive return times as in the initial step of the
induction, one checks that
\[
Y_k^{(0)} = \bigcup_{\sm \in S} Y_k^{(\sm)}.
\]
Showing that $\ph_k$ is well defined is therefore equivalent to showing
that if $\sm \ta \in S$ and $x \in Y_k^{(\sm)} \cap Y_k^{(\ta)}$,
then the corresponding
two formulas in (\ref{T_7121_AY_rshd_4}) agree at $x$.
For $b \in R^{(k  1)}$, call these expressions
$\ph_k^{(\sm)} (b)(x)$ and $\ph_k^{(\ta)} (b)(x)$.
Given $\ta = (t_1, t_2, \dots, t_{\nu}) \in S$, define
\[
R (\ta) = \{ 0, \, n (t_1), \, n (t_1) + n (t_2), \, \dots,
n (t_1) + \cdots + n (t_{\nu  1}), \, n (k) \},
\]
the set of return times associated with $\ta$.
Let $\sm, \, \ta \in S$, and let $x \in Y_k^{(\sm)} \cap Y_k^{(\ta)}$.
Let $\rh = (r_1, r_2, \dots, r_{\nu}) \in S$ be the sequence using all
return times of $x$.
That is, $n (r_1)$ is the first return time of $x$,
$n (r_2)$ is the first return time of $h^{n (r_1)} (x)$, etc.
Then $x \in Y_0^{(\rh)}$ and $R (\rh)$ contains both $R (\sm)$
and $R (\ta)$.
It therefore suffices to prove agreement of the two formulas when
$x \in Y_k^{(\sm)} \cap Y_k^{(\ta)}$ and $R (\sm) \subset R (\ta)$.
Assuming this, write $\ta = (t_1, t_2, \dots, t_{\nu})$ and
\[
R (\sm) = \{ 0, \, n (t_1) + \cdots + n (t_{j (1)}), \,
n (t_1) + \cdots + n (t_{j (2)}), \, \dots, \,
n (t_1) + \cdots + n (t_{j (\mu) }) \},
\]
with
\[
j (1) < j (2) < \cdots < j (\mu) \andeqn
n (t_1) + n (t_2) + \cdots + n (t_{j (\mu) }) = n (k).
\]
Then $\sm = (s_1, s_2, \dots, s_{\mu})$, with
\[
n (s_i) = n (t_{j (i  1) + 1} ) + n (t_{j (i  1) + 2} )
+ \cdots + n (t_{j (i)})
\]
for $i = 1, 2, \ldots, \mu$.
Now $\ph_k^{(\sm)} (b) (x)$ is a block diagonal matrix, with blocks
\[
b_{s_1} (x), \, b_{s_2} (h^{n ({s_1})} (x)), \,
b_{s_3} (h^{n ({s_1}) + n ({s_2})} (x)), \, \dots, \,
b_{s_{\mu}} (h^{n ({s_1}) + n ({s_2})  \cdots + n (s_{\mu  1})}
(x)).
\]
The induction hypothesis implies that
\[
b_{s_1} (y) = \left( \begin{matrix}
b_{t_1} (y) & & & & \\
& b_{t_2} (h^{n ({t_1})} (y)) & & & \\
& & & \ddots & \\
& & & & b_{t_{j (1)}} (h^{n ({t_1}) +
\cdots + n (t_{j (1)  1})} (y))
\end{matrix} \right)
\]
for
\[
y \in (Y_{s_1} \setminus Y_{s_1}^{\bullet} ) \cap Y_{t_1}
\cap h^{n ({t_1})} (Y_{t_2} ) \cap \cdots \cap
h^{[n ({t_1}) + n ({t_2})
+ \cdots + n ({t_{j (1)  1}}) ]}(Y_{t_{j (1)}} ),
\]
that
\[
b_{s_2} (y) = \left( \begin{matrix}
b_{t_{j (1) + 1}} (y) & & & & \\
& b_{t_{j (1) + 2}} (h^{n ({t_{j (1) + 1}})} (y)) & & & \\
& & & \ddots & \\
& & & & b_{t_{j (2)}} (h^{n ({t_{j (1) + 1}})
+ \cdots + n (t_{j (2)  1})} (y))
\end{matrix} \right)
\]
for
\begin{align*}
& y \in (Y_{s_2} \setminus Y_{s_2}^{\bullet} ) \cap Y_{t_{j (1) + 1}}
\cap h^{n ({t_{j (1) + 1} } ) } (Y_{t_{j (1) + 2} } )
\\
& \hspace{7em} \mbox{} \cap \cdots
\cap h^{[n ({t_{j (1) + 1}}) + n ({t_{j (1) + 2}})
+ \cdots + n ({t_{j (2)  1}}) ]}(Y_{t_{j (2)}} ),
\end{align*}
etc.
Taking $y = x$ in the first,
\[
y = h^{n (s_1)} (x) = h^{n ({t_1}) + \cdots + n (t_{j (1)} ) } (x)
\]
in the second of these,
\[
y = h^{n ({s_1}) + n ({s_2})} (x)
= h^{n ({t_1}) + \cdots + n (t_{j (2)} ) } (x)
\]
in the third, etc., we get
$\ph_k^{(\sm)} (b) (x) = \ph_k^{(\ta)} (b) (x)$, as desired.
This completes the induction, and the proof.
\end{proof}
\begin{proof}[Proof of \Thm{T_4317_AYIsRsha}]
The only part of the statement of \Thm{T_4317_AYIsRsha}
which is not in \Thm{T_7121_AY_rshd}
is the statement that $C^* (\Z, X, h)_{Y}$ has
topological dimension equal to $\dim (X)$.
That the topological dimension is at most $\dim (X)$
follows from \Thm{T_7121_AY_rshd}
and Proposition~\ref{P_7122_DimSubsp}.
That it is at least $\dim (X)$
follows from \Thm{T_7121_AY_rshd},
Proposition~\ref{P_7122_ClCover},
and Lemma \ref{RokhCov}(\ref{RokhCov_2}).
\end{proof}
The subalgebras we really want are of the form
$C^* (\Z, X, h)_{ \{y\} }$ for suitable $y \in X$,
not $C^* (\Z, X, h)_Y$ with $\sint (Y) \neq \E$.
\begin{rmk}\label{R_7121_LimOfSubalgs}
Let $X$ be an infinite \chs{} and let $h \colon X \to X$ be a \mh.
Let $Y \subset X$ be closed.
The case of immediate interest is $Y = \{ y \}$
for some $y \in X$,
but other choices are important,
as for example in the discussion after
Proposition~\ref{P_6827_CantorFactor}.
We can choose a decreasing sequence
$Y_1 \supset Y_2 \supset \cdots$
of closed subsets of $X$ with nonempty interiors such that
%
\begin{equation}\label{Eq_7122_Inter}
\bigcap_{n = 1}^{\infty} Y_n = Y.
\end{equation}
%
Then
%
% \begin{equation}\label{Eq_7122_Union}
\[
C^* (\Z, X, h)_{Y_1} \subset C^* (\Z, X, h)_{Y_2}
\subset \cdots
\]
and
\[
{\ov{\bigcup_{n = 1}^{\I} C^* (\Z, X, h)_{Y_n} }}
= C^* (\Z, X, h)_{Y}.
\]
% \end{equation}
%
That is,
%
\begin{equation}\label{Eq_7122_DLimit}
C^* (\Z, X, h)_{Y} = \Dirlim_{n} C^* (\Z, X, h)_{Y_n}.
\end{equation}
%
When $\dim (X) < \I$,
by \Thm{T_4317_AYIsRsha}
we have expressed $C^* (\Z, X, h)_{Y}$
as the direct limit of a direct system
of recursive subhomogeneous algebras
which has no dimension growth,
in the sense of Corollary~1.9 of~\cite{Ph7}.
\end{rmk}
Moreover, we have the following result.
\begin{prp}[Proposition~2.5 of~\cite{LhP}]\label{T_7122_Simple}
Let $X$ be an infinite \chs{} and let $h \colon X \to X$ be a \mh.
Let $y \in X$.
Then $C^* (\Z, X, h)_{ \{y\} }$ is
infinite dimensional and simple.
\end{prp}
Infinite dimensionality is obvious.
We refer to \cite{LhP} for the proof of simplicity.
% 999 Add simplicity proof?
We point out that a generalization of this result follows from
Proposition~\ref{P_2627_NoSmp},
Proposition~\ref{P_6928_InfDim},
and
Theorem~\ref{T_5421_AYStabLg},
whose proofs we give sketches of below.
Direct limits of direct systems
of recursive subhomogeneous algebras
with no dimension growth have a number of good properties,
originally developed in \cite{Ph7}
and~\cite{Ph_rshaRR}.
By now, it is known that all such algebras are classifiable
in the sense of the Elliott classification program.
% 999 Ref missing.
Here, we want to use the density of the range
of the map
%
\begin{equation}\label{Eq_7122_DR_Cpy}
K_0 \big( C^* (\Z, X, h)_{ \{y\} } \big)
\to \Aff \big( \T \big( C^* (\Z, X, h)_{ \{y\} } \big) \big)
\end{equation}
%
to conclude that $C^* (\Z, X, h)_{ \{y\} }$ has tracial rank zero
(\Def{D_TR0}).
More generally,
if $D$ is a UHF algebra,
we want the map
%
\begin{equation}\label{Eq_7122_DR_DCpy}
K_0 \big( D \otimes C^* (\Z, X, h)_{ \{y\} } \big)
\to \Aff \big( \T \big( D \otimes C^* (\Z, X, h)_{ \{y\} } \big) \big)
\end{equation}
%
to have dense range.
Our hypotheses state that
%
\begin{equation}\label{Eq_7122_DR_Cp}
K_0 ( C^* (\Z, X, h) )
\to \Aff \big( \T ( C^* (\Z, X, h) ) \big)
\end{equation}
%
has dense range,
or that
%
\begin{equation}\label{Eq_7122_DR_DCp}
K_0 \big( D \otimes C^* (\Z, X, h) \big)
\to \Aff \big( \T \big( D \otimes C^* (\Z, X, h) \big) \big)
\end{equation}
%
has dense range.
The following results take care of the differences.
We need to deal with both \tst{s} and Ktheory.
We state the results,
and discuss the proofs of the main ingredients afterwards.
\begin{lem}[Proposition~2.5 of~\cite{LhP}]\label{L:TABij}
Let $X$ be an infinite compact metric space,
and let $h \colon X \to X$ be a \mh.
Let $y \in X$.
Then the restriction map
$\T (C^* (\Z, X, h)) \to \T \big( C^* (\Z, X, h)_{ \{y\} } \big)$
is a bijection and an affine \hme.
\end{lem}
\begin{cor}\label{C_7122_DTrRestr}
Let $X$ be an infinite compact metric space,
and let $h \colon X \to X$ be a \mh.
Let $y \in X$.
Let $D$ is a UHF algebra.
Then the restriction map
\[
\T (D \otimes C^* (\Z, X, h))
\to \T \big( D \otimes C^* (\Z, X, h)_{ \{y\} } \big)
\]
is a bijection and an affine \hme.
\end{cor}
The proof of Corollary~\ref{C_7122_DTrRestr}
needs the following well known result.
\begin{exr}\label{Ex_7122_UniqTrTens}
Let $A$ be a unital \ca,
and let $D$ be a unital \ca{} with a unique tracial state.
Then there is an affine homeomorphism
\[
R \colon \T (D \otimes_{\min} A) \to \T (A)
\]
such that for $\ta \in \T (D \otimes_{\min} A)$,
the \tst{} $R (\ta)$
is determined by $R (\ta) (a) = \ta (1 \otimes a)$ for $a \in A$.
\end{exr}
% 999 Not really fair as exercise: Define trace on tensor product?
\begin{proof}[Proof of Corollary~\ref{C_7122_DTrRestr}]
The result is immediate from
\Lem{L:TABij} and Exercise~\ref{Ex_7122_UniqTrTens}.
\end{proof}
\begin{thm}[Theorem 4.1(3) of~\cite{Ph7}]\label{T_7122_KThyAy}
Let $X$ be an infinite compact metric space,
and let $h \colon X \to X$ be a \mh.
Let $y \in X$.
Then the inclusion map
$\io \colon C^* (\Z, X, h)_{ \{y\} } \to C^* (\Z, X, h)$
induces an isomorphism
\[
\io_* \colon K_0 \big( C^* (\Z, X, h)_{ \{y\} } \big)
\to K_0 \big( C^* (\Z, X, h) \big).
\]
\end{thm}
We won't use this fact,
but it is also true that
\[
\io_* \colon K_1 \big( C^* (\Z, X, h)_{ \{y\} } \big)
\to K_1 \big( C^* (\Z, X, h) \big)
\]
is injective,
with cokernel isomorphic to~$\Z$,
generated by the image in the cokernel
of the $K_1$class of the standard unitary~$u$
in $C^* (\Z, X, h)$.
See Theorem 4.1(4) of~\cite{Ph7}.
\begin{cor}\label{C_7122_KThyDAy}
Let $X$ be an infinite compact metric space,
and let $h \colon X \to X$ be a \mh.
Let $y \in X$.
Let $D$ is a UHF algebra.
Then the inclusion map
\[
\id_D \otimes \io
\colon D \otimes C^* (\Z, X, h)_{ \{y\} } \to D \otimes C^* (\Z, X, h)
\]
induces an isomorphism
\[
(\id_D \otimes \io)_* \colon
K_0 \big( D \otimes C^* (\Z, X, h)_{ \{y\} } \big)
\to K_0 \big( D \otimes C^* (\Z, X, h) \big).
\]
\end{cor}
\begin{proof}
This result follows from the K\"{u}nneth formula~\cite{Scct2}
% 999 Theorem number missing.
and Theorem~\ref{T_7122_KThyAy}.
\end{proof}
One doesn't actually need the K\"{u}nneth formula.
If $D = \Dirlim_{n} M_{d (n)}$
with $d (1)  d (2)  \cdots$,
then
$\id_D \otimes \io$ is the direct limit
of the maps
\[
\id_{M_{d (n)}} \otimes \io
\colon M_{d (n)} \otimes C^* (\Z, X, h)_{ \{y\} }
\to M_{d (n)} \otimes C^* (\Z, X, h),
\]
which are all isomorphisms on~$K_0$.
\begin{cor}\label{C_7123_PassToSub}
Let $X$ be an infinite compact metric space,
and let $h \colon X \to X$ be a \mh.
Let $y \in X$.
Let $D$ be $\C$ or a UHF~algebra.
Suppose that $\rh \big( K_0 (D \otimes C^* (\Z, X, h)) \big)$
is dense in $\Aff \big( \T (D \otimes C^* (\Z, X, h)) \big)$.
Then $\rh \big( K_0 (D \otimes C^* (\Z, X, h)_{ \{y\} }) \big)$
is dense in $\Aff \big( \T (D \otimes C^* (\Z, X, h)_{ \{y\} }) \big)$.
\end{cor}
\begin{proof}
If $D = \C$,
combine Lemma~\ref{L:TABij}
and Theorem~\ref{T_7122_KThyAy}.
If $D$ is a UHF algebra,
combine
Corollary~\ref{C_7122_DTrRestr}
and Corollary~\ref{C_7122_KThyDAy}.
\end{proof}
\Lem{L:TABij}
is originally due to Qing Lin
(via the closely related Proposition~16 of~\cite{Lq2}),
and its proof is sketched in the proof of Theorem 1.2 of~\cite{LqP}.
We give the full proof here.
We also point out that a more general result follows
from Theorem~\ref{T_5522_SameT}
and
Theorem~\ref{T_5421_AYStabLg},
whose proofs we give sketches of below.
That route directly uses the properties of
$C^* (\Z, X, h)_{ \{y\} }$ as a subalgebra of $C^* (\Z, X, h)$,
but the proof we give here instead
compares traces on both algebras to
the set of invariant Borel probability measures on~$X$.
The following lemma is a special case of Theorem~\ref{T11202Traces}.
\begin{lem}\label{P:TA}
Let $X$ be an infinite compact metric space,
and let $h \colon X \to X$ be a \mh.
Then the restriction map
$\T (C^* (\Z, X, h)) \to \T ( C (X))$
is a bijection from $\T (C^* (\Z, X, h))$ to the set
of $h$invariant Borel probability measures on~$X$.
% Moreover, if $E \colon C^* (\Z, X, h) \to C (X)$
% is the standard conditional expectation
% (Definition~\ref{D_StdCond}),
% and $\mu$ is an $h$invariant Borel probability measure on~$X$,
% then the \tst{} $\ta_{\mu}$ on $C^* (\Z, X, h)$
% which restricts to $\mu$ is given by the formula
% $\ta_{\mu} (a) = \int_X E (a) \, d \mu$ for $a \in C^* (\Z, X, h)$.
\end{lem}
We need the analogous result for $C^* (\Z, X, h)_{ \{y\} }$.
\begin{lem}[Proposition 16 of~\cite{Lq2}]\label{L:TAy}
Let $X$ be an infinite compact metric space,
and let $h \colon X \to X$ be a \mh.
Let $y \in X$.
Then the restriction map
$\T \big( C^* (\Z, X, h)_{ \{y\} } \big) \to \T ( C (X))$
is a bijection from $\T \big( C^* (\Z, X, h)_{ \{y\} } \big)$
to the set of $h$invariant Borel probability measures on~$X$.
\end{lem}
The proof is simpler than the original because
we use Proposition~\ref{P_4319_CharOB}.
\begin{proof}[Proof of Lemma~\ref{L:TAy}]
Applying Proposition~\ref{P:TA}
and restricting from $C^* (\Z, X, h)$ to $C^* (\Z, X, h)_{ \{y\} }$,
we see that every $h$invariant Borel probability measure on~$X$
gives a \tst{} on $C^* (\Z, X, h)_{ \{y\} }$.
Now let $\ta$ be any \tst{} on $C^* (\Z, X, h)_{ \{y\} }$.
Let $\mu$ be the Borel probability measure on $X$
determined by $\ta (f) = \int_X f \, d \mu$ for $f \in C (X)$.
The rest of the proof has two steps.
The first step is to show that
$\mu$ is $h$invariant.
Then \Lem{L:TAy}
provides a \tst{} $\ta_{\mu}$ on $C^* (\Z, X, h)$.
The formula,
from Example~\ref{EInvMeas},
is
\[
\ta_{\mu} \Bigg( \sum_{n =  N}^N f_n u^n \Bigg)
= \int_X f_0 \, d \mu
\]
for $N \in \N$
and $f_{_N}, \, f_{ N + 1}, \, \ldots, \, f_N \in C (X)$.
The second step of the proof is to show that
$\ta_{\mu} _{C^* (\Z, X, h)_{ \{y\} }} = \ta$.
For the first step, we show that
$\int_X (f \circ h) \, d \mu = \int_X f \, d \mu$
for every $f \in C (X)$.
This is clearly true for constant functions~$f$.
Therefore, it suffices to consider functions $f$ such that $f (y) = 0$.
For such a function~$f$,
write $f = f_1^* f_2$
with $f_1, f_2 \in C (X)$ such that $f_1 (y) = f_2 (y) = 0$.
(For example, take $f_1 =  f ^{1/2}$
and $f_2 = (\sgn \circ f)  f ^{1/2}$.)
Then $f_1 u, \, f_2 u \in C^* (\Z, X, h)_{ \{y\} }$.
So
\[
f \circ h
= u^* f u
= (f_1 u)^* (f_2 u)
\in C^* (\Z, X, h)_{ \{y\} }.
\]
We now use the trace property at the second step to get
\[
\int_X (f \circ h) \, d \mu
= \ta \big( (f_1 u)^* (f_2 u) \big)
= \ta \big( (f_2 u) (f_1 u)^* \big)
= \ta (f)
= \int_X f \, d \mu.
\]
Thus $\mu$ is $h$invariant.
For the second step,
it follows from Proposition~\ref{P_4319_CharOB}
that $C^* (\Z, X, h)_{ \{y\} }$
is the closed linear span of all elements of the form $f u^n$,
with $f \in C (X)$ and $n \in \Z$,
which actually happen to be in $C^* (\Z, X, h)_{ \{y\} }$.
So it suffices to prove that if $f u^n \in C^* (\Z, X, h)_{ \{y\} }$
and $n \neq 0$,
then $\ta (f u^n) = 0$.
Since $h^n$ has no fixed points,
there is an open cover of $X$ consisting of sets $U$
such that $h^n (U) \cap U = \varnothing$.
Choose
\[
g_1, g_2, \ldots, g_m \in C (X) \subset C^* (\Z, X, h)_{ \{y\} }
\]
which form a
partition of unity subordinate to this cover.
In particular, the supports of $g_j$ and $g_j \circ h^{n}$
are disjoint for all~$j$.
For $j = 1, 2, \ldots, m$ we have,
using the trace property at the first step
and the
relation $u^n g u^{n} = g \circ h^{n}$ for any $g \in C (X)$
at the second step,
\[
\ta (g_j f u^n)
= \ta \big( g_j^{1/2} f u^n g_j^{1/2} \big)
= \ta \big( g_j^{1/2} f \big( g_j^{1/2} \circ h^{n} \big) u^n \big)
= \ta (0)
= 0.
\]
Summing over $j$ gives $\ta (f u^n) = 0$.
\end{proof}
\begin{proof}[Proof of Lemma~\ref{L:TABij}]
Let $M$ be the set of
$h$invariant Borel probability measures on~$X$.
Lemma~\ref{P:TA} shows that the restriction map
$\T (C^* (\Z, X, h)) \to M$
is a bijection.
Lemma~\ref{L:TAy} shows that the restriction map
$\T \big( C^* (\Z, X, h)_{ \{y\} } \big) \to M$
is a bijection.
So the restriction map
$\T (C^* (\Z, X, h)) \to \T \big( C^* (\Z, X, h)_{ \{y\} } \big)$
is a bijection.
The restriction map is clearly affine and \ct.
Since its domain and codomain are compact Hausdorff,
it is a \hme.
\end{proof}
We now say something about the proof of Theorem~\ref{T_7122_KThyAy}.
The usual proof uses
Theorem~2.4 of~\cite{Pt4},
which relates the Ktheory of the \ca{} of a groupoid
to the Ktheory of the \ca{} of a particular
kind of subgroupoid,
and is based on KKtheory computations in~\cite{Pt3}.
Example~2.6 of~\cite{Pt4} contains the application to
the Ktheory of $C^* (\Z, X, h)_{ \{y\} }$.
Groupoids enter because of the interpretation of $C^* (\Z, X, h)$
as the \ca{} of a transformation group groupoid
(called $\Z \ltimes X$)
and of $C^* (\Z, X, h)_{Y}$
(for an arbitrary closed subset $Y \subset X$)
as the \ca{} of an open subgroupoid of $\Z \ltimes X$.
This interpretation is briefly outlined in Remark~\ref{R:Gpoid}.
The philosophy is that $\Z \ltimes X$
has many more (open) subgroupoids
than $(\Z, X, h)$ has subobjects in the category of dynamical
systems.
The groupoid picture is not needed for the rest of
what we do here,
because of the concrete description of $C^* (\Z, X, h)_{Y}$,
but it is needed for the generalization of the construction
to actions of $\Z^d$.
Unfortunately, we will not be able to discuss
the relevant construction in these notes.
See \cite{Ph10} for the special case in which $X$ is the Cantor set.
We outline (without proofs) an alternate approach
to the proof of Theorem~\ref{T_7122_KThyAy},
using partial actions.
It is based on discussions with Ruy Exel.
The partial action approach to this problem
seems closely related to the subgroupoid approach.
It is known, but, as far as we know,
has not appeared in the literature.
We presume it also generalizes to
actions by groups other than~$\Z$.
For~$\Z$, this approach avoids~\cite{Pt3}
and puts the Ktheory computations
in a somewhat more familiar context,
namely a generalization
(\Thm{TPVForParAct} below)
of the PimsnerVoiculescu exact sequence
for crossed products by~$\Z$~\cite{PV}
to crossed products by partial actions.
We will follow~\cite{ExlBk}
until we get to the point where Ktheory appears,
but we do not reproduce the definitions and statements
of most of the theorems.
Let $X$ be a compact Hausdorff space,
let $h \colon X \to X$ be a \hme,
and let $Y \subset X$ be closed.
We start with the topological partial action
of $\Z$ on~$X$ obtained from the restriction and corestriction
of $h$ to a \hme{} from $X \setminus Y$ to $X \setminus h (Y)$.
Topological partial actions are defined in
Definition~5.1 of~\cite{ExlBk},
referring back to Definition~2.1 of~\cite{ExlBk}
for partial actions on sets.
Following the notation of~\cite{ExlBk},
we define open subsets $D_n \subset X$ by,
for $n \in \Nz$,
\[
D_n
= X \setminus
\big[ Y \cup h (Y) \cup \cdots \cup h^{n  1} (Y) \big]
\]
and
\[
D_{n} = X \setminus \big[ h^{ n} (Y) \cup h^{ n + 1} (Y) \cup
\cdots \cup h^{ 1} (Y) \big].
\]
As a sign of what is to come,
we point out that,
with $Z_n$ as in Proposition~\ref{P_4319_CharOB},
we have $D_n = X \setminus Z_n$
for all $n \in \Z$.
We further take $\te_n \colon D_{n} \to D_n$
to be the restriction and corestriction
of $h^n$ to $D_{n}$ and~$D_n$ for $n \in \Z$.
One easily checks that
$\big( (D_n)_{n \in \Z}, \, (\te_n)_{n \in \Z} \big)$
is in fact a topological partial action
of $\Z$ on~$X$.
This partial action gives a C*~partial action of $\Z$ on $C (X)$.
Definition~6.4 of~\cite{ExlBk}
gives the conditions for a partial action on an algebra,
Definition 11.4 of~\cite{ExlBk}
gives the additional conditions for a C*~partial action,
and the fact that a topological partial action
on a locally compact Hausdorff space
gives a C*~partial action
is Corollary 11.6 of~\cite{ExlBk}.
Now form the algebraic crossed product by this partial action,
as in Definition~8.3 of~\cite{ExlBk},
and complete it to a \ca{}
as in Definition 11.11 of~\cite{ExlBk}.
Call this \ca{} $C^* (\Z, X, \te)$.
(In~\cite{ExlBk}, the notation
$C (X) \rtimes_{\te} \Z$ is used.)
\begin{lem}\label{L_7125_PartCrPrd}
Let $X$ be a compact Hausdorff space,
let $h \colon X \to X$ be a \hme,
and let $Y \subset X$ be closed.
Let $\big( (D_n)_{n \in \Z}, \, (\te_n)_{n \in \Z} \big)$
and $C^* (\Z, X, \te)$
be as in the discussion above.
Let $\pi \colon C (X) \to C^* (\Z, X, h)$
be the standard inclusion of $C (X)$ in the ordinary
C*~crossed product
(see the discussion after Remark~\ref{R:NRed}),
and let $n \mapsto u_n$
be the map from $\Z$ to the unitary group of $C^* (\Z, X, h)$
(Notation~\ref{N:ug}).
Then $(\pi, u)$ is a covariant representation of
$\big( (D_n)_{n \in \Z}, \, (\te_n)_{n \in \Z} \big)$
in $C^* (\Z, X, h)$
in the sense of Definition 9.10 of~\cite{ExlBk},
and the associated \hm{}
$\gm \colon C^* (\Z, X, \te) \to C^* (\Z, X, h)$
of Proposition 13.1 of~\cite{ExlBk}
is injective and has range $C^* (\Z, X, h)_{Y}$.
\end{lem}
\begin{proof}
The proof that $(\pi, u)$ is a covariant representation
is immediate.
Let $B$ be the algebraic partial crossed product
of $C (X)$ by the partial action
$\big( (D_n)_{n \in \Z}, \, (\te_n)_{n \in \Z} \big)$
(see Definition 8.3 of~\cite{ExlBk}).
Following the notation there, write its elements as
formal sums $\sum_{n \in \Z} a_n \dt_n$,
with $a_n \in D_n$ for all $n \in \Z$
and $a_n = 0$ for all but finitely many $n \in \Z$.
By construction
(see Definition 11.11 of~\cite{ExlBk}),
$B$ is dense in $C^* (\Z, X, \te)$.
One checks,
again directly from its definition
and the definition of the \hm{}
determined by a covariant representation,
that the image in $C^* (\Z, X, h)$ under~$\gm$
of $B$ is,
in the notation in the statement of Proposition~\ref{P_4319_CharOB},
exactly
\[
\big\{ a \in C (X) [\Z] \colon
{\mbox{$E (a u^{n}) \in C_0 (X \setminus Z_n)$
for all $n \in \Z$}} \big\}.
\]
So Proposition~\ref{P_4319_CharOB}
implies that the range of $\gm$ is $C^* (\Z, X, h)_{Y}$.
It remain to prove that $\gm$ is injective.
We construct a dual action of $S^1$ on $C^* (\Z, X, \te)$.
This is known, and works for any partial crossed product by~$\Z$.
For $\zt \in S^1$,
one checks that there is a covariant representation
$(\sm, v)$
of $\big( (D_n)_{n \in \Z}, \, (\te_n)_{n \in \Z} \big)$ in
$C^* (\Z, X, \te)$
such that $\sm (a) = a \dt_0$ for $a \in C (X)$
and $v_n = \zt^{n} \dt_n$.
(This is essentially the same formula
as that of Remark~\ref{R_6814_DualAction}
for the dual action of $S^1$ on a crossed product by~$\Z$.)
Let $\bt_{\zt} \colon C^* (\Z, X, \te) \to C^* (\Z, X, \te)$
be the corresponding \hm{}
(Proposition 13.1 of~\cite{ExlBk}).
Then $\bt_1 = \id_{C^* (\Z, X, \te)}$
(this is clear),
and $\bt_{\zt_1} \circ \bt_{\zt_2} = \bt_{\zt_1 \zt_2}$
for $\zt, \zt_2 \in S^1$
(this is easily checked by looking at what they do to
elements of~$B$).
Therefore $\bt_{\zt}$ is an automorphism for $\zt \in S^1$
and $\zt \mapsto \bt_{\zt}$
is an action of $S^1$ on $C^* (\Z, X, \te)$.
Using \Lem{LCtGen},
in the same way as in Example~\ref{E_RotGauge},
one checks that this action is \ct.
It is clear that $\gm \colon C^* (\Z, X, \te) \to C^* (\Z, X, h)$
is equivariant
when $C^* (\Z, X, \te)$ is equipped with the action~$\bt$
and $C^* (\Z, X, h)$ is equipped with the dual action as in
Remark~\ref{R_6814_DualAction}.
The fixed point algebras of both actions are easily
checked to be the standard copies of $C (X)$,
and thus the restriction of $\gm$ to
the fixed point algebra $C^* (\Z, X, \te)^{\bt}$ is injective.
So $\gm$ is injective by Proposition 2.9 of~\cite{Exl94}.
\end{proof}
The dual action argument in the proof of \Lem{L_7125_PartCrPrd}
can be replaced, with appropriate preparation,
by Theorem 19.1(c) of~\cite{ExlBk},
which applies to much more general situations.
The following result is a generalization of the PimsnerVoiculescu
exact sequence for the Ktheory of crossed products by~$\Z$~\cite{PV}.
\begin{thm}[Theorem~7.1 of~\cite{Exl94}]\label{TPVForParAct}
Let $\big( (D_n)_{n \in \Z}, \, (\te_n)_{n \in \Z} \big)$
be a partial action of $\Z$ on a \ca~$A$.
Then there is a natural six term exact sequence
\[
\begin{CD}
K_0 (D_{1}) @>{}>> K_0 (A) @>{\io_*}>> K_0 (C^* (\Z, A, \af)) \\
@A{}AA & & @VV{}V \\
K_1 (C^* (\Z, A, \af)) @<<{\io_*}< K_1 (A) @<<{}< K_1 (D_{1}).
\end{CD}
\]
\end{thm}
By naturality one gets a commutative diagram with exact rows,
in which the bottom row is the usual
PimsnerVoiculescu
exact sequence~\cite{PV}:
\[
\begin{CD}
@>>> K^0 ( X \setminus Y ) @>>> K^0 (X)
@>>> K_0 \big( C^* (\Z, X, h)_Y \big)
@>>> K^1 ( X \setminus Y ) @>>> \\
& & @VVV @VVV @VVV @VVV \\
@>>> K^0 (X) @>>> K^0 (X)
@>>> K_0 \big( C^* (\Z, X, h) \big) @>>> K^1 (X) @>>>.
\end{CD}
\]
Taking $Y$ to be a one point set,
it is now not hard to derive \Thm{T_7122_KThyAy}
and the corresponding result for
$K_1 \big( C^* (\Z, X, h)_{\{ y \}} \big)$.
We omit the details,
but point out that the Five Lemma
is not quite enough.
One needs to show that the horizontal maps
$K^0 ( X \setminus Y ) \to K^0 (X)$
and $K^0 (X) \to K^0 (X)$ shown have the same range,
and that the vertical map $K^1 ( X \setminus Y ) \to K^1 (X)$
is an isomorphism.
% 999 Should have proof.
% range of $K^0 ( X \setminus Y ) \to K^0 (X)$
% is the same as range of $K^0 (X) \to K^0 (X)$
We return to the description of steps
in the proof of Theorem~\ref{T:TM}.
The projections used in the proof
when $X$ is the Cantor set
(the main part of the proof being \Lem{L:L3})
are in $C (X)$,
and were gotten from \Lem{L:PjComp}.
When $X$ is connected,
there are no nontrivial \pj{s} in $C (X)$,
and a different approach is required.
The following is a generalization of Lemma~\ref{L:PjComp}.
It is both more elementary and more general than the
corresponding argument in~\cite{LhP}
(the main part of the proof of Theorem~4.5 there).
% 999 Improve?
% 999 Crossrefs to similar arguments wanted.
\begin{lem}\label{L:PjComp2}
Let $X$ be an infinite \cms,
and let $h \colon X \to X$ be a \mh.
Let $B \subset C^* (\Z, X, h)$ be a unital subalgebra
which contains $C (X)$ and has property~(SP).
Let $c \in C^* (\Z, X, h)$ be a nonzero positive element.
Then there exists a \nzp{} $p \in B$ such that
$p$ is \mvnt{} in $C^* (\Z, X, h)$
to a \pj{} in ${\overline{c C^* (\Z, X, h) c}}$.
\end{lem}
\begin{proof}
Let $E \colon C^* (\Z, X, h) \to C (X)$
be the standard conditional expectation
(Definition~\ref{D_StdCond}).
It follows from Proposition~\ref{P:Faithful}(\ref{P:Faithful:1})
and Exercise~\ref{Pb_CondExpt}(\ref{P:CondExpt:Pos})
that $E (c)$ is a nonzero positive element of $C (X)$.
Set $\dt = \tfrac{1}{7} \ E (c) \$.
Let $f \in C (X)$ be the pointwise minimum
$f (x) = \min \big( 6 \dt, \, E (c) (x) \big)$.
Then
\begin{equation}\label{Est13a}
\ f  E (c) \ = \dt.
\end{equation}
Set
\[
U_0 = \{ x \in X \colon E (c) (x) > 6 \dt \},
\]
which is a nonempty open set
such that $f (x) = 6 \dt$ for all $x \in U_0$.
Choose a finite sum $a = \sum_{n = N}^N a_n u^n \in C^* (\Z, X, h)$
with $a_n \in C (X)$ for $n =  N, \,  N + 1, \, \ldots, \, N$
such that $\ a  c \ < \dt$.
Let $\rh$ be the metric on $X$.
Choose $\ep > 0$ so small that whenever $x_1, x_2 \in X$
satisfy $\rh (x_1, x_2) < \ep$,
then
\[
 a_n (x_1)  a_n (x_2)  < \frac{\dt}{2 N + 1}
\]
for $n =  N, \,  N + 1, \, \ldots, \, N$.
Using freeness of the action of $\Z$ induced by $h$,
choose open subsets $U, V \S X$
such that $U \neq \E$,
such that
$U \subset {\overline{U}}
\subset V \subset {\overline{V}} \subset U_0$,
such that
the sets $h^{N} (V), \, h^{ N + 1} (V), \, \ldots, \, h^N (V)$
are disjoint,
and such that ${\overline{V}}$ has diameter less than~$\ep$.
Then there are $b_{N}, \, b_{ N + 1}, \, \ldots, \, b_N \in C (X)$
such that $b_n$ is constant on ${\overline{U}}$ and
\[
\ b_n  a_n \ < \frac{\dt}{2 N + 1}
\]
for $n =  N, \,  N + 1, \, \ldots, \, N$.
Set $b = \sum_{n = N}^N b_n u^n$.
Then $\ b  a \ < \dt$.
So
%
\begin{equation}\label{Eq_7202_17StSt}
\ b  c \ < 2 \dt.
\end{equation}
%
Choose \cfn{s} $g_0, g_1 \colon X \to [0, 1]$
such that
\[
\supp (g_0) \subset U,
\qquad
g_0 g_1 = g_1,
\andeqn
g_1 \neq 0.
\]
Use the hypotheses on $B$
to choose a \nzp{} $p \in {\overline{g_1 B g_1}}$.
We clearly have $g_0 p = p g_0 = p$.
It follows that
%
\begin{equation}\label{Eq_7202_17St}
f p = f g_0 p = 6 \dt g_0 p = 6 \dt p,
\end{equation}
%
and similarly
%
\begin{equation}\label{Eq_7203_17NSt}
p f = 6 \dt p.
\end{equation}
%
The same reasoning shows that $p b_n = b_n p$
for $n =  N, \,  N + 1, \, \ldots, \, N$.
Also,
for $n \in \{ N, \,  N + 1, \, \ldots, \, N \} \setminus \{ 0 \}$,
the disjointness condition implies that $g_0 u^n g_0 = 0$,
whence $p u^n p = p g_0 u^n g_0 p = 0$.
It follows that
\[
p b p = \sum_{n = N}^N p b_n u^n p
= \sum_{n = N}^N b_n p u^n p
= b_0 p = p E (b) p.
\]
Using (\ref{Eq_7202_17St}) and~(\ref{Eq_7203_17NSt})
at the first step,
this last equation at the second step,
and (\ref{Est13a}), (\ref{Eq_7202_17StSt}),
and Exercise~\ref{Pb_CondExpt}(\ref{P:CondExpt:Norm})
at the third step, we get
\begin{align*}
\ p c p  6 \dt p \
& = \ p c p  p f p \ \\
& \leq \ p c p  p b p \ + \ p E (b) p  p E (c) p \
+ \ p E (c) p  p f p \ \\
& \leq 2 \ c  b \ + \ E (c)  f \
< 5 \dt.
\end{align*}
It follows that $p c p$ is invertible in $p C^* (\Z, X, h) p$.
Let $d = (p c p)^{1/2}$, calculated in $p C^* (\Z, X, h) p$.
Set $v = d p c^{1/2}$.
Then
\[
v v^* = d p c p d = (p c p)^{1/2} (p c p) (p c p)^{1/2} = p
\]
and
\[
v^* v = c^{1/2} p d^2 p c^{1/2} \in {\overline{c C^* (\Z, X, h) c}}.
\]
This completes the proof.
\end{proof}
For the full statement of Theorem~\ref{T_4317_MinUHF}.
(involving the tensor product with a UHF algebra~$D$),
one needs a generalization of \Lem{L:PjComp2},
which we omit.
We can now describe the proof of Theorem~\ref{T_4317_MinUHF}.
For simplicity, we omit~$D$,
thus really dealing only with Theorem~\ref{T:TM}.
The main part is the substitute for \Lem{L:PjComp},
but we also refer to
the proof of the Cantor set case of Theorem~\ref{T:TM},
given at the end of Section~\ref{Sec:Class}.
Combining the hypothesis and
Corollary~\ref{C_7123_PassToSub},
for any $y \in X$,
$\rh \big( K_0 (D \otimes C^* (\Z, X, h)_{ \{y\} }) \big)$
is dense in $\Aff \big( \T (D \otimes C^* (\Z, X, h)_{ \{y\} }) \big)$.
Since $C^* (\Z, X, h)_{ \{y\} }$ is simple and infinite dimensional
(Proposition~\ref{T_7122_Simple})
and a direct limit of a direct systems
of recursive subhomogeneous algebras
with no dimension growth,
classification results imply it has tracial rank zero
(\Def{D_TR0}).
(Actually,
in~\cite{LhP},
$y \in X$ is chosen with a little care,
to allow a direct proof
that $C^* (\Z, X, h)_{ \{y\} }$ has tracial rank zero.)
This algebra usually isn't AF,
so Lemma~\ref{L:TRAF}
must be replaced as follows,
allowing a simple subalgebra with tracial rank zero
in place of an AF~algebra.
The algebra $B$ in the statement takes the place
of the AF~algebra $p B p$ in Lemma~\ref{L:TRAF}.
\begin{lem}[Lemma~4.4 of~\cite{LhP}]\label{L_7202_TRTAF}
Let $A$ be a simple unital \ca.
Suppose that
for every finite subset $F \subset A$, every $\ep > 0$,
and every nonzero positive element $c \in A$,
there exists a nonzero \pj{} $p \in A$
and a simple unital subalgebra $B \subset p A p$
with tracial rank zero
such that:
\begin{enumerate}
\item\label{L_7202_TRTAF_Comm}
$\ [a, p] \ < \ep$ for all $a \in F$.
\item\label{L_7202_TRTAF_Close}
$\dist (p a p, \, B) < \ep$ for all $a \in F$.
\item\label{L_7202_TRTAF_Small}
$1  p$ is Murrayvon Neumann equivalent
to a \pj{} in ${\overline{c A c}}$.
\end{enumerate}
Then $A$ has tracial rank zero (\Def{D_TR0}).
\end{lem}
The word ``nonzero'' is missing in
Lemma~4.4 of~\cite{LhP}.
This leads to the same issue
as discussed after \Def{D_TR0},
although this condition is not needed if $A$ is already
known to be finite.
We must therefore verify the hypotheses of
\Lem{L_7202_TRTAF}.
This is Lemma~4.2 of~\cite{LhP}.
Following the proof of the Cantor set case of Theorem~\ref{T:TM},
and using the analog of Exercise~\ref{Ex:TRGen},
we take the finite set in \Lem{L_7202_TRTAF}
to be $F_0 \cup \{ u \}$
for a finite subset $F_0 \subset C (X)$.
The role of the compact open set $U$ used in
the proof of
the proof of the Cantor set case of Theorem~\ref{T:TM}
will be played by a nonzero \pj{} $r \in C^* (\Z, X, h)_{ \{ y \} }$
gotten from \Lem{L:PjComp2},
but getting $1  p \precsim r$
will require a different argument.
In the replacement for the proof of \Lem{L:PjComp},
we will not use $C^* (\Z, X, h)_Y$
(only $C^* (\Z, X, h)_{ \{ y \} }$),
so we choose $Y$ to be a small open set
containing~$y$.
(This set is called $U$ in the proof of Lemma~4.2 of~\cite{LhP}.)
To specify how small,
we choose $N_0$ and $N$ as in the proof of \Lem{L:PjComp},
and require
that conditions (\ref{Y_6Y26_y}), (\ref{Y_6Y26_Disj}),
and~(\ref{Y_6Y26_lDt})
in that proof hold,
plus a substitute
(see below) for condition~(\ref{Y_6Y26_InU}).
(This substitute will depend only on $\ep$ and the \pj~$r$.)
We don't have anything like $\ch_Y$,
so we proceed as follows.
Choose \cfn{s} $g_0, g_1, g_2, f_0 \colon X \to [0, 1]$ such that
\[
g_0 (y) = 1, \qquad
g_1 g_0 = g_0, \qquad
g_2 g_1 = g_1, \qquad
f_0 g_2 = g_2, \andeqn
\supp (f_0) \subset Y.
\]
Since $C^* (\Z, X, h)_{ \{ y \} }$
has real rank zero
(a consequence of tracial rank zero, by Theorem~\ref{T_6X02_TRZToRR}),
one can find a \pj{} $q_0 \in C^* (\Z, X, h)_{ \{ y \} }$ such that
\[
g_1 q_0 = q_0 g_1 = g_1 \andeqn f_0 q_0 = q_0 f_0 = q_0.
\]
(See Lemma~4.1 of~\cite{LhP};
the main part of the proof actually comes from
Theorem~1 of~\cite{Br_int}.)
We still get orthogonality
for the same list of projections as in the proof of \Lem{L:PjComp}.
The projection $q_0$ must commute with any function $f \in C (X)$
which is constant on~$Y$,
which is good enough for the parts of the proof of \Lem{L:PjComp}
involving commutators with and approximation of functions in~$F_0$.
The part about commutators with and approximation of~$u$
needs little change.
Moreover,
the relation $q_0 g_1 = g_1$ says,
heuristically,
that $q_0$ dominates the characteristic
function of a neighborhood of~$Y$,
and this relation is in fact good enough to prove
that the required cutdowns
by the new version of the \pj~$e$
in the proof of \Lem{L:PjComp}
are in fact in $C^* (\Z, X, h)_{ \{ y \} }$.
Also, $p C^* (\Z, X, h)_{ \{y\} } p$
has tracial rank zero by \Lem{L_6X08_TAFCorner}.
It remains only to show how to arrange to get $1  p \precsim r$.
Define
\[
\bt = \inf \big( \big\{ \ta (r)
\colon \ta \in \T (C^* (\Z, X, h)_{ \{y\} }) \big\} \big).
\]
Then $\bt > 0$
because $\T (C^* (\Z, X, h)_{ \{y\} })$
is weak* compact (Remark~\ref{RTACptCnv}),
$\ta \mapsto \ta (r)$
is weak* \ct,
and $\ta (r)$ can never by zero (\Lem{LTrIdeal}).
Choose $R \in \N$ such that $N / R < \bt$.
In place of condition~(\ref{Y_6Y26_InU})
in the proof of \Lem{L:PjComp},
we require that the sets
\[
Y, \, h (Y), \, h^2 (Y), \, \ldots, \, h^R (Y)
\]
be disjoint.
It follows that
for every $h$invariant Borel probability measure $\mu$ on~$X$,
we have $\mu (Y) < 1 / R$,
so $\int_X f_0 \, d \mu < 1 / R$.
Using \Lem{L:TAy},
we deduce that $\ta (f_0) < 1 / R$
for all $\ta \in \T (C^* (\Z, X, h)_{ \{y\} })$,
so $\ta (q_0) < 1 / R$
for all $\ta \in \T (C^* (\Z, X, h)_{ \{y\} })$.
The \pj{} $1  p$
is the sum of $N$ \pj{s}
which are \mvnt{} to~$q_0$ in $C^* (\Z, X, h)$.
Every \tst{} on $C^* (\Z, X, h)$ therefore takes
the same value on all of them.
By \Lem{L:TABij},
every \tst{} on $C^* (\Z, X, h)_{ \{y\} }$ takes
the same value on all of them.
It follows that for all $\ta \in \T (C^* (\Z, X, h)_{ \{y\} })$
we have $\ta (1  p) < N / R \leq \bt$.
By Theorem~\ref{T_6X02_TRZToRR},
tracial rank zero implies that the
order on \pj{s} is determined by traces as in
\Def{D:OrdDetD}.
So $1  p \precsim r$ in $C^* (\Z, X, h)_{ \{y\} }$,
and thus also in $C^* (\Z, X, h)$.
\part{An Introduction to Large Subalgebras and Applications to
Crossed Products}\label{Part_Large}
\section{The Cuntz Semigroup}\label{Sec_Cuntz}
In this part,
we give an introduction to large subalgebras of \ca{s}
and some applications.
Much of the text of this part is taken directly from~\cite{Ph_lgsvy},
which is a survey of applications of large subalgebras
based on lectures given at the University of Wyoming
in the summer of 2015.
That survey assumes much more background than these notes
(it starts with the material here),
there are some differences in the organization,
and it contains some open problems and other discussion
omitted here because they are too far off the topic of these notes.
Large subalgebras are an abstraction of the Putnam subalgebras
$C^* (\Z, X, h)_{ \{y\} }$
(see \Def{D_5421_VSubalg})
used in the proof of Theorem~\ref{T:TM}
(and in other places).
This abstraction was first introduced in~\cite{Ph40}.
We give some very brief motivation here,
but postpone a more systematic discussion to
the beginning of Section~\ref{Sec_Intro}.
The applications discussed in these notes mostly involve
$C^* (\Z, X, h)_{ \{y\} }$ and $C^* (\Z, X, h)_Y$
for other subsets $Y \subset X$
such that $h^n (Y) \cap Y = \E$ for every $n \in \Z \setminus \{ 0 \}$.
However, the real motivation for the abstraction
(given very short shrift in these notes)
is the construction of analogous subalgebras in
\ca{s} such as $C^* (\Z^d, X)$
for a free minimal action of $\Z^d$ on~$X$.
Such subalgebras are used in a crucial way in~\cite{Ph10}
when $X$ is the Cantor set,
although with an axiomatization
useful only for actions on the Cantor set.
In general,
there seems to be no useful concrete formula for such subalgebras.
Instead,
one specifies a list of properties
and proves the existence of a subalgebra which has these properties
and is otherwise accessible
(perhaps being a direct limit of the recursive subhomogeneous \ca{s}
of \Def{D_4317_RSHA}).
The two lists of properties which have been most useful
so far make up the definitions of a large subalgebra
(\Def{D_5421_Large})
and of a centrally large subalgebra
(\Def{D_5421_CentLarge}).
The definitions and proofs of the theorems make essential use
of Cuntz comparison, and to a lesser extent of the Cuntz semigroup.
We therefore begin with a summary of what we need to know
about Cuntz comparison and the Cuntz semigroup.
The reader is encouraged to just read the introductory
discussion and basic definitions,
and then skip to Section~\ref{Sec_Intro},
referring back to this section later as needed.
(In particular,
there is nothing about dynamics in this section.)
We refer to~\cite{APT}
for an extensive introduction
(which does not include all the results that we need).
The material we need is either summarized or proved
in the first two sections of~\cite{Ph40}.
The Cuntz semigroup can be thought of as being a version of
the $K_0$group
based on positive elements instead of \pj{s}.
For that reason,
we will occasionally make comparisons
with Ktheory.
The reader not familiar with Ktheory can ignore these remarks.
For a \ca~$A$,
we let $M_{\infty} (A)$ denote the algebraic direct limit of the
system $(M_n (A))_{n = 1}^{\infty}$
using the usual embeddings $M_n (A) \to M_{n+1} (A)$,
given by
\[
a \mapsto \left( \begin{array}{cc} a & 0 \\ 0 & 0 \end{array} \right).
\]
If $a \in M_m (A)$ and $b \in M_n (A)$,
we write $a \oplus b$ for the diagonal direct sum
\[
a \oplus b
= \left( \begin{array}{cc} a & 0 \\ 0 & b \end{array} \right).
\]
By abuse of notation,
we will also write $a \oplus b$ when $a, b \in M_{\infty} (A)$
and we do not care about the precise choice of $m$ and $n$
with $a \in M_m (A)$ and $b \in M_n (A)$.
Parts (\ref{D_5511_CzSGp:1}) and~(\ref{D_5511_CzSGp:2})
of the following definition are originally from~\cite{Cz1}.
Since we will frequently need to relate Cuntz subequivalence
in a \ca~$B$ to Cuntz subequivalence in a \ca~$A$
containing~$B$,
we include (contrary to the usual convention)
the algebra $A$ in the notation.
The notation $a \sim_A b$
in Definition \ref{D_5511_CzSGp}(\ref{D_5511_CzSGp:2})
conflicts with the notation $p \sim q$
in Notation~\ref{N:MvN},
except that in the context of Cuntz subequivalence
we include the algebra~$A$ as a subscript.
\begin{dfn}\label{D_5511_CzSGp}
Let $A$ be a \ca.
\begin{enumerate}
\item\label{D_5511_CzSGp:1}
For $a, b \in (K \otimes A)_{+}$,
we say that $a$ is {\emph{Cuntz subequivalent to~$b$
over~$A$}},
written $a \precsim_A b$,
if there is a sequence $(v_n)_{n = 1}^{\infty}$ in $K \otimes A$
such that
$\limi{n} v_n b v_n^* = a$.
This relation is transitive:
$a \precsim_A b$ and $b \precsim_A c$ imply $a \precsim_A c$.
\item\label{D_5511_CzSGp:2}
We say that $a$ and $b$ are {\emph{Cuntz equivalent over~$A$}},
written $a \sim_A b$,
if $a \precsim_A b$ and $b \precsim_A a$.
This relation is an equivalence relation,
and we write $\langle a \rangle_A$ for the equivalence class of~$a$.
\item\label{D_5511_CzSGp:3}
The {\emph{Cuntz semigroup}} of~$A$ is
\[
{\operatorname{Cu}} (A) = (K \otimes A)_{+} / \sim_A,
\]
together with the commutative semigroup operation,
gotten from an isomorphism $M_2 (K) \to K$,
\[
\langle a \rangle_A + \langle b \rangle_A
= \langle a \oplus b \rangle_A
\]
(the class does not depend on the choice of the isomorphism)
and the partial order
\[
\langle a \rangle_A \leq \langle b \rangle_A
\Longleftrightarrow a \precsim_A b.
\]
It is taken to be an object of the category~${\mathbf{Cu}}$
given in Definition~4.1 of~\cite{APT}.
We write $0$ for~$\langle 0 \rangle_A$.
\item\label{D_5511_CzSGp:4}
We also define the subsemigroup
\[
W (A) = M_{\infty} (A)_{+} / \sim_A,
\]
with the same operations and order.
(It will follow from Remark~\ref{R2727MnI} that the obvious
map $W (A) \to {\operatorname{Cu}} (A)$ is injective.)
\item\label{D_5511_CzSGp:5}
Let $A$ and $B$ be C*algebras,
and let $\ph \colon A \to B$ be a \hm.
We use the same letter for the induced maps
$M_n (A) \to M_n (B)$
for $n \in \N$,
$\Mi (A) \to \Mi (B)$,
and $K \otimes A \to K \otimes B$.
We define
${\operatorname{Cu}} (\ph)
\colon {\operatorname{Cu}} (A) \to {\operatorname{Cu}} (B)$
and $W (\ph) \colon W (A) \to W (B)$
by $\langle a \rangle_A \mapsto \langle \ph (a) \rangle_B$
for $a \in (K \otimes A)_{+}$
or $M_{\infty} (A)_{+}$ as appropriate.
\end{enumerate}
\end{dfn}
It is easy to check that
% in Definition~\ref{D_5511_CzSGp},
the maps ${\operatorname{Cu}} (\ph)$ and $W (\ph)$ are well defined
homomorphisms of ordered semigroups which send
$0$ to~$0$.
Also,
it follows from
Lemma \ref{L:CzBasic}(\ref{L:CzBasic:CmpDSum}) below
that if $\et_1, \et_2, \mu_1, \mu_2 \in {\operatorname{Cu}} (A)$
satisfy $\et_1 \leq \mu_1$ and $\et_2 \leq \mu_2$,
then $\et_1 + \et_2 \leq \mu_1 + \mu_2$.
The semigroup ${\operatorname{Cu}} (A)$
generally has better properties than $W (A)$.
For example, certain supremums exist
(Theorem~4.19 of~\cite{APT}),
and, when understood as an object of the category~${\mathbf{Cu}}$,
it behaves properly
with respect to direct limits
(Theorem~4.35 of~\cite{APT}).
In this exposition,
we mainly use $W (A)$ because,
when $A$ is unital,
the dimension function $d_{\ta}$
associated to a normalized quasitrace~$\ta$
(Definition~\ref{D:dtau} below)
is finite on $W (A)$
but usually not on ${\operatorname{Cu}} (A)$.
In particular,
the radius of comparison
(Definition~\ref{D_5519_ROfComp} below)
is easier to deal with in terms of $W (A)$.
We will not need the definition of the category~${\mathbf{Cu}}$.
\begin{rmk}\label{R2727MnI}
We make the usual identifications
%
\begin{equation}\label{Eq_5510_Chain}
A \subset M_n (A) \subset M_{\infty} (A) \subset K \otimes A.
\end{equation}
%
It is easy to check,
by cutting down to corners,
that if $a, b \in (K \otimes A)_{+}$
satisfy $a \precsim_A b$,
then the sequence $(v_n)_{n = 1}^{\infty}$
such that $\limi{n} v_n b v_n^* = a$
(as in Definition \ref{D_5511_CzSGp}(\ref{D_5511_CzSGp:1}))
can be taken to be
in the smallest of the algebras in~(\ref{Eq_5510_Chain})
which contains both $a$ and~$b$.
See Remark~1.2 of~\cite{Ph40} for details.
% \label{R2727MnI}
\end{rmk}
The Cuntz semigroup
of a separable \ca{} can be very roughly thought of as
Ktheory using open projections in matrices over $A''$,
that is, open supports of positive elements in matrices over~$A$,
instead of projections in matrices over~$A$.
% (We are implicitly assuming $A$ is separable.)
As justification for this heuristic,
we note that if $X$ is a \chs{}
and $f, g \in C (X)_{+}$,
then $f \precsim_{C (X)} g$ \ifo
\[
\big\{ x \in X \colon f (x) > 0 \big\}
\subset \big\{ x \in X \colon g (x) > 0 \big\}.
\]
A version of this can be made rigorous,
at least in the separable case.
See~\cite{OrRrThl}.
There is
a description of ${\operatorname{Cu}} (A)$
using Hilbert modules over~$A$
in place of finitely generated projective modules
as for Ktheory.
See~\cite{CEI}.
Unlike Ktheory,
the Cuntz semigroup is not discrete.
If $p, q \in A$ are \pj{s}
such that $\ p  q \ < 1$,
then $p$ and $q$ are \mvnt{}
(\Lem{L_5Y21_ClosePj}).
However, for $a, b \in A_{+}$,
the relation $\ a  b \ < \ep$
says nothing about the classes of $a$ and $b$
in ${\operatorname{Cu}} (A)$ or $W (A)$,
however small $\ep > 0$ is.
We can see this in ${\operatorname{Cu}} ( C (X) )$.
Even if $\big\{ x \in X \colon g (x) > 0 \big\}$ is a very
small subset of~$X$,
for every $\ep > 0$
the function $f = g + \frac{\ep}{2}$
has $\langle f \rangle_{C (X)} = \langle 1 \rangle_{C (X)}$.
What is true when $\ f  g \ < \ep$ is that
\[
\big\{ x \in X \colon f (x) > \ep \big\}
\subset \big\{ x \in X \colon g (x) > 0 \big\},
\]
so that
the function $\max (f  \ep, \, 0)$ satisfies
$\max (f  \ep, \, 0) \precsim_{C (X)} g$.
This motivates the systematic use of the elements
$(a  \ep)_{+}$, defined as follows.
\begin{dfn}\label{D:MinusEp}
Let $A$ be a \ca, let $a \in A_{+}$,
and let $\ep \geq 0$.
Let $f \colon [0, \infty) \to [0, \infty)$ be the function
\[
f (\lambda)
= (\lambda  \ep)_{+}
= \begin{cases}
0 & \hspace{3em} 0 \leq \lambda \leq \ep \\
\lambda  \ep & \hspace{3em} \ep < \lambda.
\end{cases}
\]
Then define $(a  \ep)_{+} = f (a)$
(using \ct{} functional calculus).
\end{dfn}
One must still be much more careful than with Ktheory.
First,
$a \leq b$
does not imply $(a  \ep)_{+} \leq (b  \ep)_{+}$
(although one does get
$(a  \ep)_{+} \precsim_A (b  \ep)_{+}$;
see Lemma~\ref{L:CzBasic}(\ref{L:CzBasic:L:CzCompIneq}) below).
Second,
$a \precsim_A b$
does not imply any relation between
$(a  \ep)_{+}$ and $(b  \ep)_{+}$.
For example,
if $A = C ([0, 1])$
and $a \in C ([0, 1])$ is $a (t) = t$
for $t \in [0, 1]$,
then for any $\ep \in (0, 1)$
the element $b = \ep a$ satisfies
$a \precsim_A b$.
But $(a  \ep)_{+} \not\precsim_A (b  \ep)_{+}$,
since $(a  \ep)_{+}$ has open support $(\ep, 1]$
while $(b  \ep)_{+} = 0$.
The best one can do is in
Lemma~\ref{L:CzBasic}(\ref{L:CzBasic:LMinusEp}) below.
We now list a collection of basic results about Cuntz comparison
and the Cuntz semigroup.
There are very few such results about \pj{s} and the $K_0$group,
the main ones being that if $\ p  q \ < 1$,
then $p$ and $q$ are \mvnt;
that $p \leq q$ \ifo{} $p q = p$;
the relations between
homotopy, unitary equivalence, and
\mvnc;
and the fact that addition of equivalence classes
respects orthogonal sums.
There are many more for Cuntz comparison.
We will not use all the facts listed below
in these notes (although they are all used in~\cite{Ph40});
we include them all so as to give a fuller picture
of Cuntz comparison.
Parts (\ref{L:CzBasic:Her}) through~(\ref{L:CzBasic:CmpDSum})
of Lemma~\ref{L:CzBasic}
are mostly taken from \cite{KR},
with some from \cite{Cu0}, \cite{ERS}, \cite{PsnPh2},
and~\cite{Rd0},
and are summarized in Lemma~1.4 of~\cite{Ph40};
% \label{L:CzBasic}
we refer to~\cite{Ph40} for more on the attributions
(although not all the attributions there are to
the original sources).
Part~(\ref{L:CzBasic:L:CzSumEp}) is Lemma~1.5 of~\cite{Ph40};
% \label{L:CzSumEp}
part~(\ref{L:CzBasic:C:MMvsM}) is Corollary~1.6 of~\cite{Ph40};
% \label{C:MMvsM}
part~(\ref{L:CzBasic:L:CzCompIneq}) is Lemma~1.7 of~\cite{Ph40};
% \label{L:CzCompIneq}
and part~(\ref{L:CzBasic:L2720KToMn}) is Lemma~1.9 of~\cite{Ph40}.
% \label{L2720KToMn}
As we have done earlier,
we denote by $A^{+}$ the unitization of a \ca~$A$.
(We add a new unit even if $A$ is already unital.)
\begin{lem}\label{L:CzBasic}
Let $A$ be a \ca.
\begin{enumerate}
\item\label{L:CzBasic:Her}
Let $a, b \in A_{+}$.
Suppose $a \in {\overline{b A b}}$.
Then $a \precsim_A b$.
\item\label{L:CzBasic:LCzOneWay}
Let $a \in A_{+}$ and let $f \colon [0, \, \ a \ ] \to [0, \infty)$
be a \cfn{} such that $f (0) = 0$.
Then $f (a) \precsim_A a$.
\item\label{L:CzBasic:LCzFCalc}
Let $a \in A_{+}$ and let $f \colon [0, \, \ a \ ] \to [0, \infty)$
be a \cfn{} such that $f (0) = 0$ and $f (\ld) > 0$ for $\ld > 0$.
Then $f (a) \sim_A a$.
\item\label{L:CzBasic:LCzComm} %Comment after Def 2.3 of \cite{KR}
Let $c \in A$.
Then $c^* c \sim_A c c^*$.
\item\label{L:CzBasic:3904_UE}
Let $a \in A_{+}$,
and let $u \in A^{+}$ be unitary.
Then $u a u^* \sim_A a$.
\item\label{L:CzBasic:LCzCommEp}
Let $c \in A$ and let $\af > 0$.
Then $(c^* c  \af)_{+} \sim_A (c c^*  \af)_{+}$.
\item\label{L:CzBasic:N5}
Let $v \in A$.
Then there is an isomorphism
$\ph \colon {\overline{v^* v A v^* v}}
\to {\overline{v v^* A v v^*}}$
such that, for every positive element
$z \in {\overline{v^* v A v^* v}}$,
we have
$z \sim_A \ph (z)$.
\item\label{L:CzBasic:MinIter}
Let $a \in A_{+}$ and let $\ep_1, \ep_2 > 0$.
Then
\[
\big( ( a  \ep_1)_{+}  \ep_2 \big)_{+}
= \big( a  ( \ep_1 + \ep_2 ) \big)_{+}.
\]
\item\label{L:CzBasic:N6}
Let $a, b \in A_{+}$ satisfy $a \precsim_A b$
and let $\dt > 0$.
Then there is $v \in A$
such that $v^* v = (a  \dt)_{+}$
and $v v^* \in {\overline{b A b}}$.
\item\label{L:CzBasic:LCzWithinEp} %Prop 2.2 of \cite{Rd0}
Let $a, b \in A_{+}$.
Then $\ a  b \ < \ep$ implies $(a  \ep)_{+} \precsim_A b$.
\item\label{L:CzBasic:LMinusEp} %Prop 2.4 of \cite{Rd0}
Let $a, b \in A_{+}$.
Then \tfae:
\begin{enumerate}
\item\label{L:CzBasic:LMinusEp:1}
$a \precsim_A b$.
\item\label{L:CzBasic:LMinusEp:2}
$(a  \ep)_{+} \precsim_A b$ for all $\ep > 0$.
\item\label{L:CzBasic:LMinusEp:3}
For every $\ep > 0$ there is $\dt > 0$ such that
$(a  \ep)_{+} \precsim_A (b  \dt)_{+}$.
\item\label{L:CzBasic:LMinusEp:4}
For every $\ep > 0$ there are $\dt > 0$ and $v \in A$ such that
\[
(a  \ep)_{+} = v [ (b  \dt)_{+}] v^*.
\]
\end{enumerate}
\item\label{L:CzBasic:LCzCmpSum} %Lemma 2.8 of \cite{KR}
Let $a, b \in A_{+}$.
Then $a + b \precsim_A a \oplus b$.
\item\label{L:CzBasic:Orth}
Let $a, b \in A_{+}$ be orthogonal (that is, $a b = 0$).
Then $a + b \sim_A a \oplus b$.
\item\label{L:CzBasic:CmpDSum}
Let $a_1, a_2, b_1, b_2 \in A_{+}$,
and suppose that $a_1 \precsim_A a_2$ and $b_1 \precsim_A b_2$.
Then $a_1 \oplus b_1 \precsim_A a_2 \oplus b_2$.
\item\label{L:CzBasic:L:CzSumEp}
% \label{L:CzSumEp}
Let $a, b \in A$ be positive,
and let $\af, \bt \geq 0$.
Then
\[
\big( (a + b  (\af + \bt) \big)_{+}
\precsim_A (a  \af)_{+} + (b  \bt)_{+}
\precsim_A (a  \af)_{+} \oplus (b  \bt)_{+}.
\]
\item\label{L:CzBasic:C:MMvsM}
% \label{C:MMvsM}
Let $\ep > 0$ and $\ld \geq 0$.
Let $a, b \in A$ satisfy $\ a  b \ < \ep$.
Then $(a  \ld  \ep)_{+} \precsim_A (b  \ld)_{+}$.
\item\label{L:CzBasic:L:CzCompIneq}
% \label{L:CzCompIneq}
Let $a, b \in A$ satisfy $0 \leq a \leq b$.
Let $\ep > 0$.
Then $(a  \ep)_{+} \precsim_A (b  \ep)_{+}$.
\item\label{L:CzBasic:L2720KToMn}
% \label{L2720KToMn}
Let $a \in (K \otimes A)_{+}$.
Then for every $\ep > 0$
there are $n \in \N$ and $b \in (M_n \otimes A)_{+}$
such that $(a  \ep)_{+} \sim_A b$.
% \item\label{L:CzBasic:XXX}
\end{enumerate}
\end{lem}
The following result is sufficiently closely tied to
the ideas behind large subalgebras that we include the proof.
\begin{lem}[Lemma~1.8 of~\cite{Ph40}]\label{L:C2}
Let $A$ be a \ca,
let $a \in A_{+}$, let $g \in A_{+}$ satisfy $0 \leq g \leq 1$,
and let $\ep \geq 0$.
Then
\[
(a  \ep)_{+}
\precsim_A \big[ (1  g) a (1  g)  \ep \big]_{+} \oplus g.
\]
\end{lem}
\begin{proof}
Set $h = 2 g  g^2$,
so that $(1  g)^2 = 1  h$.
We claim that $h \sim_A g$.
Since $0 \leq g \leq 1$,
this follows from Lemma~\ref{L:CzBasic}(\ref{L:CzBasic:LCzFCalc}),
using the \cfn{} $\ld \mapsto 2 \ld  \ld^2$ on $[0, 1]$.
Set $b = \big[ (1  g) a (1  g)  \ep \big]_{+}$.
Using Lemma~\ref{L:CzBasic}(\ref{L:CzBasic:L:CzSumEp})
at the second step,
Lemma~\ref{L:CzBasic}(\ref{L:CzBasic:LCzCommEp})
and Lemma~\ref{L:CzBasic}(\ref{L:CzBasic:LCzComm}) at the third step,
and Lemma~\ref{L:CzBasic}(\ref{L:CzBasic:CmpDSum}) at the last step,
we get
\begin{align*}
( a  \ep )_{+}
& = \big[ a^{1/2} (1  h) a^{1/2} + a^{1/2} h a^{1/2}  \ep \big]_{+}
\\
& \precsim_A \big[ a^{1/2} (1  h) a^{1/2}  \ep \big]_{+}
\oplus a^{1/2} h a^{1/2}
\\
& \sim_A \big[ (1  g) a (1  g)  \ep \big]_{+}
\oplus h^{1/2} a h^{1/2}
\\
& = b \oplus h^{1/2} a h^{1/2}
\leq b \oplus \ a \ h
\precsim_A b \oplus g.
\end{align*}
This completes the proof.
\end{proof}
The definition of the radius of comparison
(Definition~\ref{D_5519_ROfComp} below)
is stated in terms of quasitraces.
We don't discuss quasitraces here.
Instead,
we refer to the fact
(Theorem 5.11 of~\cite{Hgp})
that all normalized $2$quasitraces on exact \ca{s}
are tracial states.
For the purpose of all the applications discussed in these notes,
the reader can therefore substitute tracial states for quasitraces,
and the tracial state space $\T (A)$
for the space ${\operatorname{QT}} (A)$ defined below.
It is still open whether every quasitrace on any \ca{} is
a trace.
\begin{ntn}\label{N_4620_QT}
For a \uca~$A$,
we denote by ${\operatorname{QT}} (A)$
the set of normalized $2$quasitraces
on~$A$ (Definition II.1.1 of~\cite{BH};
Definition 2.31 of~\cite{APT}).
\end{ntn}
\begin{dfn}\label{D:dtau}
Let $A$ be a stably finite unital \ca,
and let $\ta \in {\operatorname{QT}} (A)$.
Define $d_{\ta} \colon \Mi (A)_{+} \to [0, \infty)$
by $d_{\ta} (a) = \lim_{n \to \infty} \ta (a^{1/n})$
for $a \in \Mi (A)_{+}$.
Further (the use of the same notation should cause no confusion)
define $d_{\ta} \colon (K \otimes A)_{+} \to [0, \infty]$
by the same formula, but now
for $a \in (K \otimes A)_{+}$.
We also use the same notation for the corresponding functions
on ${\operatorname{Cu}} (A)$ and $W (A)$,
as in Proposition~\ref{P_4819_dtau} below.
\end{dfn}
\begin{prp}\label{P_4819_dtau}
Let $A$ be a stably finite unital \ca,
and let $\ta \in {\operatorname{QT}} (A)$.
Then $d_{\ta}$ as in Definition~\ref{D:dtau} is well defined
on ${\operatorname{Cu}} (A)$ and $W (A)$.
That is, if $a, b \in (K \otimes A)_{+}$
satisfy $a \sim_A b$,
then $d_{\ta} (a) = d_{\ta} (b)$.
\end{prp}
\begin{proof}
This is part of Proposition~4.2 of~\cite{ERS}.
\end{proof}
Also see the beginning of Section~2.6 of~\cite{APT},
especially the proof of Theorem 2.32 there.
It follows that $d_{\ta}$ defines a state on $W (A)$.
Thus (see Theorem II.2.2 of~\cite{BH},
which gives the corresponding bijection
between $2$quasitraces and dimension functions
which are not necessarily normalized but are finite everywhere),
the map $\ta \mapsto d_{\ta}$
is a bijection from ${\operatorname{QT}} (A)$
to the set of lower semi\ct{} dimension functions on~$A$.
We now present some results related to Cuntz comparison
specifically for simple \ca{s}.
\begin{lem}[Proposition 4.10 of~\cite{KR}]\label{L_6X04_ConeEmbed}
Let $A$ be a \ca{} which is not of type~I
and let $n \in \N$.
Then there exists an injective \hm{}
from the cone $C M_n$ over $M_n$ to~$A$.
\end{lem}
The proof uses heavy machinery,
namely Glimm's result that there is a subalgebra $B \subset A$
and an ideal $I \subset B$ such that the $2^{\infty}$~UHF algebra embeds
in $B / I$.
Some of what we use this result for can be proved by
more elementary methods,
but for Lemma~\ref{L:Smaller} we don't know such a proof.
\begin{lem}[Lemma~2.1 of~\cite{Ph40}]\label{L:DivInSmp}
Let $A$ be a simple \ca{} which is not of type~I.
Let $a \in A_{+} \setminus \{ 0 \}$,
and let $l \in \N$.
Then there exist $b_1, b_2, \ldots, b_l \in A_{+} \setminus \{ 0 \}$
such that $b_1 \sim_A b_2 \sim_A \cdots \sim_A b_l$,
such that $b_j b_k = 0$ for $j \neq k$,
and such that $b_1 + b_2 + \cdots + b_l \in {\overline{a A a}}$.
\end{lem}
\begin{proof}
Replacing $A$ by ${\overline{a A a}}$,
we can ignore the requirement
$b_1 + b_2 + \cdots + b_l \in {\overline{a A a}}$ of the conclusion.
Now fix~$n \in \N$.
For $j, k = 1, 2, \ldots, n$,
we let $e_{j, k} \in M_n$ be the standard matrix unit.
In
\[
C M_n = \big\{ f \in C ([0, 1], \, M_n) \colon f (0) = 0 \big\},
\]
take $b_j (\ld) = \ld e_{j, j}$
for $\ld \in [0, 1]$
and $j = 1, 2, \ldots, n$.
Use Lemma~\ref{L_6X04_ConeEmbed}
to embed $C M_n$ in~$A$.
\end{proof}
This lemma has the following corollary.
\begin{cor}[Corollary~2.2 of~\cite{Ph40}]\label{C2718CuDiv}
Let $A$ be a simple unital infinite dimensional \ca.
Then for every $\ep > 0$ there is $a \in A_{+} \setminus \{ 0 \}$
such that for all $\ta \in {\operatorname{QT}} (A)$
we have $d_{\ta} (a) < \ep$.
\end{cor}
\begin{lem}[Lemma~2.4 of~\cite{Ph40}]\label{L2718CuSub}
Let $A$ be a simple \ca,
and let $B \subset A$ be a nonzero \hsa.
Let $n \in \N$,
and let $a_1, a_2, \ldots, a_n \in A_{+} \setminus \{ 0 \}$.
Then there exists $b \in B_{+} \setminus \{ 0 \}$
such that $b \precsim_A a_j$ for $j = 1, 2, \ldots, n$.
\end{lem}
\begin{proof}[Sketch of proof]
The proof is by induction.
The case $n = 0$ is trivial.
The induction step requires that
for $a, b_0 \in A_{+} \setminus \{ 0 \}$
one find $b \in A_{+} \setminus \{ 0 \}$
such that $b \in {\ov{b_0 A b_0}}$
(so that $b \precsim_A b_0$ by
Lemma~\ref{L:CzBasic}(\ref{L:CzBasic:Her}))
and $b \precsim_{A} a$.
Use simplicity to find $x \in A$
such that the element $y = b_0 x a$ is nonzero,
and take $b = y y^* \in {\ov{b_0 A b_0}}$.
Using
Lemma~\ref{L:CzBasic}(\ref{L:CzBasic:3904_UE})
and Lemma~\ref{L:CzBasic}(\ref{L:CzBasic:Her}),
we get
$b \sim_A y^* y \precsim_A a$.
\end{proof}
The following lemma says, roughly,
that a nonzero element of $W (A)$ can be approximated arbitrarily well
by elements of $W (A)$ which are strictly smaller.
\begin{lem}[Lemma~2.3 of~\cite{Ph40}]\label{L:Smaller}
Let $A$ be a simple infinite dimensional \ca{} which is not of type~I.
Let $b \in A_{+} \setminus \{ 0 \}$, let $\ep > 0$, and let $n \in \N$.
Then there are $c \in A_{+}$ and $y \in A_{+} \setminus \{ 0 \}$
such that, in $W (A)$,
we have
\[
n \langle (b  \ep)_{+} \rangle_A \leq (n + 1) \langle c \rangle_A
\andeqn
\langle c \rangle_A + \langle y \rangle_A \leq \langle b \rangle_A.
\]
\end{lem}
\begin{proof}[Sketch of proof]
We divide the proof into two cases.
First assume that $\spec (b) \cap (0, \ep) \neq \varnothing$.
Then there is a \cfn{} $f \colon [0, \infty) \to [0, \infty)$
which is zero on $\{ 0 \} \cup [\ep, \infty)$
and such that $f (b) \neq 0$.
We take $c = (b  \ep)_{+}$ and $y = f (b)$.
Now suppose that $\spec (b) \cap (0, \ep) = \varnothing$.
In this case,
we might as well assume that $b$ is a projection,
and that $\langle (b  \ep)_{+} \rangle_A$,
which is always dominated by $\langle b \rangle_A$,
is equal to~$\langle b \rangle_A$.
Cutting down by~$b$,
we can assume that $b = 1$
(in particular, $A$ is unital),
and it is enough to find
$c \in A_{+}$ and $y \in A_{+} \setminus \{ 0 \}$
such that
$n \langle 1 \rangle_A \leq (n + 1) \langle c \rangle_A$
and
$\langle c \rangle_A + \langle y \rangle_A \leq \langle 1 \rangle_A$.
Take the unitized cone over $M_{n + 1}$
to be
$C = (C M_{n + 1})^{+} = [C_0 ( (0, 1]) \otimes M_{n + 1}]^{+}$,
and use the usual notation for matrix units.
By Lemma~\ref{L_6X04_ConeEmbed},
we can assume that $C \subset A$.
Let $t \in C_0 ( (0, 1])$
be the function $t (\ld) = \ld$ for $\ld \in (0, 1]$.
Choose \cfn{s} $g_1, g_2, g_3 \in C ( [0, 1] )$
such that
\[
0 \leq g_3 \leq g_2 \leq g_1 \leq 1,
\quad
g_1 (0) = 0,
\quad
g_3 (1) = 1,
\quad
g_1 g_2 = g_2,
\quad {\mbox{and}} \quad
g_2 g_3 = g_3.
\]
Define
\[
x = g_2 \otimes e_{1, 1},
\,\,\,\,\,\,
c = 1  x,
\andeqn
y = g_3 \otimes e_{1, 1}.
\]
Then $x y = y$ so $c y = 0$.
It follows
from Lemma \ref{L:CzBasic}(\ref{L:CzBasic:Orth}) that
$\langle c \rangle_A + \langle y \rangle_A \leq \langle 1 \rangle_A$.
It remains to prove that
$n \langle 1 \rangle_A \leq (n + 1) \langle c \rangle_A$,
and it is enough to prove that
in $W (C)$ we have
$n \langle 1 \rangle_C
\leq (n + 1) \langle 1  g_2 \otimes e_{1, 1} \rangle_C$,
that is,
in $M_{n + 1} (C)$,
%
\begin{equation}\label{Eq_5519_InMC}
\diag (1, 1, \ldots, 1, 0)
\precsim_C \diag \big( 1  g_2 \otimes e_{1, 1}, \,
1  g_2 \otimes e_{1, 1}, \,
\ldots, \,
1  g_2 \otimes e_{1, 1} \big).
\end{equation}
%
To see why this should be true,
view $M_{n + 1} (C)$
as a set of functions from $[0, 1]$ to $M_{(n + 1)^2}$
with restrictions on the value at zero.
Since $g_1 g_2 = g_2$,
the function $1  g_2 \otimes e_{1, 1}$
is constant equal to~$1$ on a neighborhood $U$ of~$0$,
and at $\ld \in U$
the right hand side of~(\ref{Eq_5519_InMC})
therefore dominates the left hand side.
Elsewhere,
both sides of~(\ref{Eq_5519_InMC}) are diagonal,
with the right hand side being a constant \pj{} of rank~$n (n + 1)$
and the left hand side dominating
\[
\diag \big( 1  e_{1, 1}, \,
1  e_{1, 1}, \,
\ldots, \,
1  e_{1, 1} \big),
\]
which is a (different) constant \pj{} of rank~$n (n + 1)$.
It is not hard to construct
an explicit formula for a unitary
$v \in M_{n + 1} (C)$ such that
\[
\diag (1, 1, \ldots, 1, 0)
\leq v \cdot \diag \big( 1  g_2 \otimes e_{1, 1}, \,
1  g_2 \otimes e_{1, 1}, \,
\ldots, \,
1  g_2 \otimes e_{1, 1} \big) \cdot v^*.
\]
See~\cite{Ph40}
for the details (arranged a little differently).
\end{proof}
\section{Large Subalgebras}\label{Sec_Intro}
\indent
Large and centrally large subalgebras
are a technical tool which has played a key role
in work on the structure of
the C*algebras of minimal dynamical systems and some related algebras.
In this section,
we outline some old and new applications as motivation.
We then give the definitions
and several useful reformulations of them.
We next state some general theorems
(for some of which we give partial proofs in
Section~\ref{Sec_Prelim}
and Section~\ref{Sec_rc}).
Finally, we give further information on some recent applications.
Large subalgebras are a generalization and abstraction
of a construction introduced by Putnam in~\cite{Pt1}
(see \Def{D_5421_VSubalg}),
where it was used to prove that if $h$ is a \mh{}
of the Cantor set~$X$,
then $K_0 (C^* (\Z, X, h))$ is order isomorphic
to the $K_0$group of a simple AF algebra
(Theorem~4.1 and Corollary~5.6 of~\cite{Pt1}).
Putnam's construction and some generalizations
(almost all of which are centrally large subalgebras in our sense)
also played key roles in proofs of other many other results.
We list some of the them,
starting with older ones
(which in many cases have been superseded,
and which were proved before there was a formal definition
of a large subalgebra).
We then give some recent results for which no
proofs not using large subalgebras are known.
Here is a selection of the older results.
\begin{itemize}
\item
Let $h \colon X \to X$ be a \mh{} of the Cantor set.
Then $C^* (\Z, X, h)$ is an AT~algebra.
(Local approximation by circle algebras
was proved in Section~2 of~\cite{Pt2}.
Direct limit decomposition follows from semiprojectivity of
circle algebras.)
\item
Let $h \colon X \to X$ be a \mh{} of a \fd{} \cms.
Then $C^* (\Z, X, h)$ satisfies the following Ktheoretic version of
\Def{D:OrdDetD}
(Blackadar's Second Fundamental Comparability Question):
if $\et \in K_0 (A)$ satisfies $\ta_* (\et) > 0$
for all tracial states $\ta$ on~$A$, then there is a
\pj\ $p \in \Mi (A)$ such that $\et = [p]$.
(See \cite{LqP} and Theorem 4.5(1) of~\cite{Ph7}).
\item
Let $X$ be a finite dimensional infinite compact metric space,
and let $h \colon X \to X$ be a \mh{}
such that the map
\[
\rh \colon K_0 \big( C^* (\Z, X, h) \big)
\to \Aff \big( \T \big( C^* (\Z, X, h) \big) \big)
\]
of \Def{DRhoA}
has dense range.
Then $C^* (\Z, X, h)$ has tracial rank zero
(\Def{D_TR0}).
This was proved in~\cite{LhP};
much of the method is described in Section~\ref{Sec:Class}
and Section~\ref{Sec_More}.
\item
Let $X$ be the Cantor set
and let $h \colon X \times S^1 \to X \times S^1$
be a minimal \hme.
For any $x \in X$, the set $Y = \{ x \} \times S^1$
intersects each orbit at most once.
The algebra $C^* (\Z, \, X \times S^1, \, h)_Y$
(see Definition~\ref{D_5421_VSubalg} for the notation)
is introduced before Proposition~3.3 of~\cite{LnMti1},
where it is called~$A_x$.
It is a centrally large subalgebra
which plays a key role in the proofs of some of
the results there.
For example,
the proofs that the crossed products considered there
have stable rank one (as in \Def{D:SR1};
see Theorem 3.12 of~\cite{LnMti1})
and order on \pj{s} determined by traces
(as in \Def{D:OrdDetD};
see Theorem 3.13 of~\cite{LnMti1})
rely directly on the use of this subalgebra.
\item
A similar construction,
with $X \times S^1 \times S^1$ in place of $X \times S^1$
and with $Y = \{ x \} \times S^1 \times S^1$,
appears in Section~1 of~\cite{Sn}.
It plays a role in that paper similar to the role
of the algebra $C^* (\Z, \, X \times S^1, \, h)_Y$
in the previous item.
\item
Let $h \colon X \to X$ be a \mh{}
of an infinite \cms.
The large subalgebras $C^* (\Z, X, h)_Y$
of $C^* (\Z, X, h)$
(as in Definition~\ref{D_5421_VSubalg}),
with several choices of $Y$
(several one point sets as well as $\{ x_1, x_2 \}$
with $x_1$ and $x_2$ on different orbits),
have been used by Toms and Winter~\cite{TW}
to prove that $C^* (\Z, X, h)$ has finite decomposition rank.
\end{itemize}
Here are some newer results.
For most of them,
we give more information later in this section.
\begin{itemize}
\item
The extended irrational rotation algebras are~AF
(Elliott and Niu~\cite{EN1}; Theorem~\ref{T_5526_ENExtIrr}).
\item
Let $X$ be an infinite compact metric space,
and let $h \colon X \to X$ be a \mh{}
with mean dimension zero.
Then $C^* (\Z, X, h)$ is $Z$stable
(Elliott and Niu~\cite{EN2}; Theorem~\ref{T_5526_ENZStab}).
\item
Let $X$ be a compact metric space
such that there is a continuous surjection from $X$
to the Cantor set.
Then $\rc \big( C^* (\Z, X, h) \big) \leq \frac{1}{2} \mdim (h)$
(\cite{HPT}; Theorem~\ref{T_5526_rcmdim}).
\item
Let $X$ be a compact metric space
such that there is a continuous surjection from $X$
to the Cantor set.
Then $C^* (\Z, X, h)$ has stable rank one
(Theorem~7.1 of~\cite{ArPh}; Theorem~\ref{T_5413_TsrCP}).
(There are examples in which this holds but
$C^* (\Z, X, h)$ does not have strict comparison of positive elements
and is not $Z$stable.)
\item
Let $X$ be a compact metric space and let $h \colon X \to X$ be a \mh.
Then $\rc \big( C^* (\Z, X, h) \big) \leq 1 + 2 \cdot \mdim (h)$
(Theorem~\ref{T_6928_rcmdim}; this strengthens
Corollary~4.8 in the current version of of~\cite{Ph47}).
\item
We give an example involving a crossed product
$C^* (\Z, \, C (X, D), \, \af)$
in which $D$ is simple and $\af$ ``lies over''
a \mh{} of~$X$.
Let $F_{\infty}$ be the free group on
generators indexed by~$\Z$,
and for $n \in \Z$ let $u_n \in C^*_{\mathrm{r}} (F_{\infty})$
be the unitary which is the image of the corresponding
generator of~$F_{\infty}$.
Let $h \colon X \to X$ be the restriction
of a Denjoy \hme{}
(a nonminimal \hme{} of the circle
whose rotation number is irrational)
to its unique minimal set.
(See~\cite{PSS}.)
Thus $X$ is homeomorphic to the Cantor set,
and there are $\te \in \R \setminus \Q$
and a surjective map $\zt \colon X \to S^1$
such that $\zt (h (x)) = e^{2 \pi i \te} \zt (x)$
for all $x \in X$.
For $x \in X$ let $\af_x \in \Aut ( C^*_{\mathrm{r}} (F_{\infty}) )$
be determined by $\af_x (u_n) = \zt (x) u_n$
for $n \in \Z$.
Define a kind of noncommutative
Furstenberg transformation
$\af \in \Aut \big( C (X, \, C^*_{\mathrm{r}} (F_{\infty}) ) \big)$
by $\af (a) (x) = \af_x (a (x))$
for $a \in C (X, \, C^*_{\mathrm{r}} (F_{\infty}) )$
and $x \in X$.
Then
$C^* \big( \Z, \, C (X, \, C^*_{\mathrm{r}} (F_{\infty}) ),
\, \af \big)$
has stable rank one.
\end{itemize}
Large subalgebras were also used to give the first proof
that if $X$ is a finite dimensional compact metric space
with a free minimal action of $\Z^d$,
then $C^* (\Z^d, X)$
has strict comparison of positive elements.
Almost all the examples above involve actions of~$\Z$
(although not necessarily on an algebra of the form $C (X)$).
In the known applications of this type,
there are explicit formulas for the large subalgebras involved.
See \Def{D_5421_VSubalg} and \Def{D2623VSubalg}.
The real importance of the abstraction of the idea is
in applications to actions of groups such as $\Z^d$,
in which there are no known formulas for useful large subalgebras.
Instead, subalgebras with useful properties
must be shown to exist by more abstract methods.
These applications are barely touched on in these notes.
There is a competing approach,
the method of Rokhlin dimension of group actions~\cite{HWZ},
which can be used for some of the same problems
large subalgebras are good for.
When it applies,
it often gives stronger results.
For example,
Szab\'{o} has used this method successfully for
free minimal actions of $\Z^d$ on finite dimensional
compact metric spaces~\cite{Szb}.
For many problems involving crossed products for which
large subalgebras are a plausible approach,
Rokhlin dimension methods should also be considered.
Rokhlin dimension has also been successfully applied to
problems involving actions on simple \ca{s},
a context in which no useful large subalgebras are known.
(But see~\cite{OP1}
and~\cite{OvPhWn},
where what might be called large systems of subalgebras
are used effectively.)
On the other hand,
finite Rokhlin dimension requires freeness of the action
(in a suitable heuristic sense when the algebra is simple),
while some form of essential freeness
seems likely to be good enough for large subalgebra methods.
(This is suggested by the examples in~\cite{OvPhWn}.)
Finite Rokhlin dimension also requires some form of
topological finite dimensionality.
It seems plausible that there might be a generalization
of finite Rokhlin dimension which captures actions on infinite
dimensional spaces which have mean dimension zero.
Such a generalization
might be similar to the progression
from the study of simple AH~algebras with no dimension growth
to those with slow dimension growth.
It looks much less likely that
Rokhlin dimension methods can be usefully applied
to \mh{s} which do not have mean dimension zero.
Large subalgebras have been used to estimate
the radius of comparison of $C^* (\Z, X, h)$
when $h$ does not have mean dimension zero
(and the radius of comparison is nonzero);
see Theorem~\ref{T_5526_rcmdim}
and Theorem~\ref{T_6928_rcmdim},
both discussed in Section~\ref{Sec_rcCrPrd}.
These results do not seem to be accessible
via Rokhlin dimension methods.
Rokhlin dimension methods can also potentially be used to prove
regularity properties of crossed products
$C^* \big( \Z, \, C (X, D), \, \af \big)$
when $D$ is simple,
the automorphism $\af \in \Aut (C (X, D))$
``lies over'' a \mh{} of~$X$
with large mean dimension,
and the regularity properties of the crossed product
come from~$D$ rather than from the action of $\Z$ on~$X$.
See~\cite{Bk}.
Unfortunately,
we are not able to discuss Rokhlin dimension here.
In these notes,
we mostly limit ourselves to applications to
crossed products by \mh{s}.
By convention,
if we say that $B$ is a unital subalgebra of a C*algebra~$A$,
we mean that $B$ contains the identity of~$A$.
\begin{dfn}[Definition~4.1 of~\cite{Ph40}]\label{D_5421_Large}
% \label{D_Large}
Let $A$ be an infinite dimensional simple unital C*algebra.
A unital subalgebra $B \subset A$ is said to be
{\emph{large}} in~$A$ if
for every $m \in \N$,
$a_1, a_2, \ldots, a_m \in A$,
$\ep > 0$, $x \in A_{+}$ with $\ x \ = 1$,
and $y \in B_{+} \setminus \{ 0 \}$,
there are $c_1, c_2, \ldots, c_m \in A$ and $g \in B$
such that:
\begin{enumerate}
\item\label{D_5421_Large_Cut1}
$0 \leq g \leq 1$.
\item\label{D_5421_Large_Cut2}
For $j = 1, 2, \ldots, m$ we have
$\ c_j  a_j \ < \ep$.
\item\label{D_5421_Large_Cut3}
For $j = 1, 2, \ldots, m$ we have
$(1  g) c_j \in B$.
\item\label{D_5421_Large_Cut4}
$g \precsim_B y$ and $g \precsim_A x$.
\item\label{D_5421_Large_Cut5}
$\ (1  g) x (1  g) \ > 1  \ep$.
\end{enumerate}
\end{dfn}
We emphasize that the Cuntz subequivalence
involving $y$ in~(\ref{D_5421_Large_Cut4})
is relative to~$B$,
not~$A$.
Condition~(\ref{D_5421_Large_Cut5})
is needed to avoid triviality when $A$ is purely infinite and simple.
With $B = \C \cdot 1$,
we could then satisfy all the other conditions
by taking $g = 1$.
In the stably finite case,
we can dispense with~(\ref{D_5421_Large_Cut5})
(see Proposition~\ref{P:FinLarge} below),
but we still need $g \precsim_A x$
in~(\ref{D_5421_Large_Cut4}).
Otherwise, even if we require that $B$ be simple
and that the restriction maps
${\operatorname{T}} (A) \to {\operatorname{T}} (B)$
and ${\operatorname{QT}} (A) \to {\operatorname{QT}} (B)$
on traces and quasitraces be bijective,
we can take $A$ to be any UHF algebra
and take $B = \C \cdot 1$.
The choice $g = 1$ would always work.
It is crucial to the usefulness of large subalgebras
that $g$ in Definition~\ref{D_5421_Large} need not be a projection.
Also, one can do a lot without any kind of approximate
commutation condition.
Such a condition does seem to be needed for
some results.
Here is the relevant definition,
although we will not make full use of it in these notes.
% CCC
\begin{dfn}[Definition~3.2 of~\cite{ArPh}]\label{D_5421_CentLarge}
% \label{RokLargeSub}
Let $A$ be an in\fd{} simple unital C*algebra.
A unital subalgebra $B \subset A$ is said to be
{\emph{centrally large}} in~$A$ if
for every $m \in \N$,
$a_1, a_2, \ldots, a_m \in A$,
$\ep > 0$, $x \in A_{+}$ with $\ x \ = 1$,
and $y \in B_{+} \setminus \{ 0 \}$,
there are $c_1, c_2, \ldots, c_m \in A$ and $g \in B$
such that:
\begin{enumerate}
\item\label{D_5421_CentLarge_Cut1}
$0 \leq g \leq 1$.
\item\label{D_5421_CentLarge_Cut2}
For $j = 1, 2, \ldots, m$ we have
$\ c_j  a_j \ < \ep$.
\item\label{D_5421_CentLarge_Cut3}
For $j = 1, 2, \ldots, m$ we have
$(1  g) c_j \in B$.
\item\label{D_5421_CentLarge_Cut4}
$g \precsim_B y$ and $g \precsim_A x$.
\item\label{D_5421_CentLarge_Cut5a}
$\ (1  g) x (1  g) \ > 1  \ep$.
\item\label{D_5421_CentLarge_Cut5}
For $j = 1, 2, \ldots, m$ we have
$\ g a_j  a_j g \ < \ep$.
\end{enumerate}
\end{dfn}
The difference between Definition~\ref{D_5421_CentLarge}
and Definition~\ref{D_5421_Large}
is the approximate commutation condition
in Definition \ref{D_5421_CentLarge}(\ref{D_5421_CentLarge_Cut5}).
The following strengthening of Definition~\ref{D_5421_CentLarge}
will be more important in these notes.
% CCC
\begin{dfn}[Definition~5.1 of~\cite{Ph40}]\label{D_4619_StLg_N}
% \label{D_4619_StLg}
Let $A$ be an infinite dimensional simple unital \ca.
A unital subalgebra $B \subset A$ is said to be
{\emph{stably large}} in~$A$ if
$M_n (B)$ is large in $M_n (A)$ for all $n \in \N$.
\end{dfn}
\begin{prp}[Proposition~5.6 of~\cite{Ph40}]\label{P_4624_TLarge}
Let $A_1$ and $A_2$ be infinite dimensional simple unital \ca{s},
and let $B_1 \subset A_1$ and $B_2 \subset A_2$ be
large subalgebras.
Assume that $A_1 \otimes_{\min} A_2$
is finite.
Then $B_1 \otimes_{\min} B_2$
is a large subalgebra of $A_1 \otimes_{\min} A_2$.
\end{prp}
In particular,
if $A$ is stably finite and $B \subset A$ is large,
then $B$ is stably large.
We will give a direct proof (Proposition~\ref{P_5519_StFinStLg} below).
We don't know whether stable finiteness of~$A$
is needed
(Question~\ref{Q_5421_LgImpStb} below).
The main example used in these notes is the
$Y$orbit breaking subalgebra
(generalized Putnam subalgebra)
\[
C^* (\Z, X, h)_Y = C^* \big( C (X), \, C_0 (X \setminus Y) u \big)
\subset C^* (\Z, X, h).
\]
of \Def{D_5421_VSubalg},
for a \cms~$X$,
a \mh{} $h \colon X \to X$,
and a ``sufficiently small''
nonempty closed subset $Y \subset X$.
\begin{thm}\label{T_5421_AYStabLg}
Let $X$ be an infinite \chs{}
and let $h \colon X \to X$ be a minimal homeomorphism.
Let $Y \subset X$ be a compact subset
such that $h^n (Y) \cap Y = \varnothing$
for all $n \in \Z \setminus \{ 0 \}$.
Then $C^* (\Z, X, h)_Y$ (as above) is a
centrally large subalgebra of $C^* (\Z, X, h)$
in the sense of Definition~\ref{D_5421_CentLarge}.
\end{thm}
We give a proof in Section~\ref{Sec_AY},
along with proofs or sketches of proofs
of the lemmas which go into the proof.
The key fact about $C^* (\Z, X, h)_Y$ which makes this theorem
useful is that it is a direct limit of recursive
subhomogeneous \ca{s}
(as in Definition~1.1 of~\cite{Ph_RSHA1})
whose base spaces are closed subsets of~$X$.
This follows from Theorem~\ref{T_4317_AYIsRsha}
(or Theorem~\ref{T_7121_AY_rshd})
and Remark~\ref{R_7121_LimOfSubalgs}.
The structure of $C^* (\Z, X, h)_Y$
is therefore much more accessible than
the structure of crossed products.
We now state the main known results about large subalgebras
and some recent applications.
\begin{prp}[Proposition~5.2 and Proposition~5.5
of~\cite{Ph40}]\label{P_2627_NoSmp_0}
% \label{P_2627_NoSmp}, \label{P2729LgInfD}
Let $A$ be an infinite dimensional simple unital \ca,
and let $B \subset A$ be a large subalgebra.
Then $B$ is simple and infinite dimensional.
\end{prp}
The special case $C^* (\Z, X, h)_{ \{y\} }$ is stated
without proof as
Proposition~\ref{T_7122_Simple}.
In the next section,
we prove the simplicity statement
(see Proposition~\ref{P_2627_NoSmp} below)
and the stably finite case of the infinite dimensionality
statement (see Proposition~\ref{P_6928_InfDim} below).
\begin{thm}[Theorem~6.2 and Proposition~6.9
of~\cite{Ph40}]\label{L_4622_SameT_0}
% \label{L_4622_SameT} (?) and \label{P2724QTBij}
Let $A$ be an infinite dimensional simple unital \ca,
and let $B \subset A$ be a large subalgebra.
Then the restriction maps
${\operatorname{T}} (A) \to {\operatorname{T}} (B)$
and
${\operatorname{QT}} (A) \to {\operatorname{QT}} (B)$,
on traces and quasitraces
(see \Def{D_Trace} and Notation~\ref{N_4620_QT}),
are bijective.
\end{thm}
The special case involving
$\T \big(C^* (\Z, X, h)_{ \{y\} } \big)$ is
in \Lem{L:TABij},
but the proof given for \Lem{L:TABij}
is quite different.
The proofs for ${\operatorname{T}} (A)$
and for ${\operatorname{QT}} (A)$ are very different.
We prove that
${\operatorname{T}} (A) \to {\operatorname{T}} (B)$ is bijective
below (Theorem~\ref{T_5522_SameT}).
Let $A$ be a \ca.
Recall the Cuntz semigroup ${\operatorname{Cu}} (A)$ from
Definition~\ref{D_5511_CzSGp}(\ref{D_5511_CzSGp:3}).
Let ${\operatorname{Cu}}_{+} (A)$ denote
the set of elements $\et \in {\operatorname{Cu}} (A)$
which are not the classes of projections.
(Such elements
are sometimes called {\emph{purely positive}}.)
\begin{thm}[Theorem~6.8 of~\cite{Ph40}]\label{T_5522_IsoOnPure}
% \label{C_3X08_IsoOnPure}
Let $A$ be a stably finite infinite dimensional simple unital \ca,
and let $B \subset A$ be a large subalgebra.
Let $\io \colon B \to A$ be the inclusion map.
Then ${\operatorname{Cu}} (\io)$
defines an order and semigroup isomorphism
from ${\operatorname{Cu}}_{+} (B) \cup \{ 0 \}$
to ${\operatorname{Cu}}_{+} (A) \cup \{ 0 \}$.
\end{thm}
It is not true that ${\operatorname{Cu}} (\io)$ defines an isomorphism
from ${\operatorname{Cu}} (B)$ to ${\operatorname{Cu}} (A)$.
% Example~\ref{E_4819_NoKIso}
Example 7.13 of~\cite{Ph40}
shows that
${\operatorname{Cu}} (\io) \colon
{\operatorname{Cu}} (B) \to {\operatorname{Cu}} (A)$
need not be injective.
We suppose this map can also fail to be surjective,
but we don't know an example.
\begin{thm}[Theorem~6.14 of~\cite{Ph40}]\label{T_5421_RCEq_0}
% \label{T_4814_RCEq}
Let $A$ be an infinite dimensional stably finite
simple separable unital C*algebra.
Let $B \subset A$ be a large subalgebra.
Let ${\operatorname{rc}} ()$ be the radius of comparison
(Definition~\ref{D_5519_ROfComp} below).
Then ${\operatorname{rc}} (A) = {\operatorname{rc}} (B)$.
\end{thm}
We will prove this result in Section~\ref{Sec_rc}
when $A$ is exact.
See Theorem~\ref{T_5421_RCEq} below.
\begin{prp}[Proposition~6.15, Corollary~6.16,
and Proposition~6.17 of~\cite{Ph40}]\label{P_4626_Fin}
% \label{P_4626_Fin}, \label{C_4626_StFin}, \label{P2724PI}
Let $A$ be an infinite dimensional simple unital \ca,
and let $B \subset A$ be a large subalgebra.
Then:
\begin{enumerate}
\item\label{P_4626_Fin_N}
$A$ is finite \ifo{} $B$ is finite.
\item\label{C_4626_StFin_N}
If $B$ is stably large in~$A$,
then $A$ is stably finite \ifo{} $B$ is stably finite.
\item\label{P2724PI_N}
$A$ is purely infinite \ifo{} $B$ is purely infinite.
\end{enumerate}
\end{prp}
\begin{prp}[Theorem~6.18 of~\cite{Ph40}]\label{P_6X05_SP}
Let $A$ be an infinite dimensional simple unital \ca,
and let $B \subset A$ be a large subalgebra.
Suppose that $B$ has property~(SP) (\Def{D:SPD}).
Then $A$ has property~(SP).
\end{prp}
\begin{thm}[Theorem~6.3 and
Theorem~6.4 of~\cite{ArPh}]\label{T_5519_tsr}
% \label{tsr1GoesUpExtraLarge}, \label{T_4Y11_RRZ}
Let $A$ be an infinite dimensional simple unital \ca,
and let $B \subset A$ be a centrally large subalgebra.
Then:
\begin{enumerate}
\item\label{T_5519_tsr_tsr}
If $B$ has stable rank one
(\Def{D:SR1}), then so does~$A$.
\item\label{T_5519_tsr_RR}
If $B$ has real rank zero (\Def{D:RR0}) and stable rank one,
then so does~$A$.
\end{enumerate}
\end{thm}
In the next theorem, $Z$ is the JiangSu algebra
(briefly described in \Ex{E:JiangSuFlip}).
The condition that a given \ca~$A$
tensorially absorb the JiangSu algebra,
that is, $Z \otimes A \cong A$
($A$ is said to be ``$Z$stable'' or ``$Z$absorbing''),
is one of the regularity conditions
in the TomsWinter conjecture.
For simple separable nuclear \ca{s}
it is hoped, and known in many cases,
that $Z$stability implies classifiability
in the sense of the Elliott program.
\begin{thm}[Theorem~2.3 of~\cite{ArBcPh0}]\label{T_5519_ZStab}
Let $A$ be an infinite dimensional simple nuclear unital \ca,
and let $B \subset A$ be a centrally large subalgebra.
If $B$ tensorially absorbs the JiangSu algebra~$Z$,
then so does~$A$.
\end{thm}
If $A$ isn't nuclear,
the best we can say so far is that $A$ is
tracially $Z$absorbing in the sense
of Definition~2.1 of~\cite{HirOr2013}.
The following two key technical results
are behind many of the theorems stated above.
In particular, they are the basis for proving
\Thm{T_5522_IsoOnPure},
which is used to prove many of the other results.
\begin{lem}[Lemmas~6.3 and~6.5 of~\cite{Ph40}]\label{L2720SDomToB}
% \label{L2720SDomToB}, \label{L2721AToB}
Let $A$ be an infinite dimensional simple unital \ca,
and let $B \subset A$ be a stably large subalgebra.
\begin{enumerate}
\item\label{L2720SDomToB_Down}
Let $a, b, x \in (K \otimes A)_{+}$
satisfy $x \neq 0$ and $a \oplus x \precsim_A b$.
Then for every $\ep > 0$ there are
$n \in \N$,
$c \in (M_n \otimes B)_{+}$,
and $\dt > 0$
such that $(a  \ep)_{+} \precsim_A c \precsim_A (b  \dt)_{+}$.
\item\label{L2720SDomToB_Up}
Let $a, b \in (K \otimes B)_{+}$ and $c, x \in (K \otimes A)_{+}$
satisfy $x \neq 0$,
$a \precsim_A c$,
and $c \oplus x \precsim_A b$.
Then $a \precsim_B b$.
\end{enumerate}
\end{lem}
We state some of the applications.
In the following theorem,
$\rc (A)$ is the radius of comparison of~$A$
(see \Def{D_5519_ROfComp} below),
and $\mdim (h)$ is the mean dimension of~$h$
(see Definition~\ref{D_6827_MDim} below).
\begin{thm}[\cite{HPT}]\label{T_5526_rcmdim}
Let $X$ be a compact metric space.
Assume that
there is a continuous surjective map from $X$ to the Cantor set.
Let $h \colon X \to X$ be a minimal homeomorphism.
Then $\rc (C^* (\Z, X, h)) \leq \frac{1}{2} \mdim (h)$.
\end{thm}
It is conjectured that
$\rc (C^* (\Z, X, h)) = \frac{1}{2} \mdim (h)$
for all \mh{s}.
In~\cite{HPT},
we also prove that
$\rc (C^* (\Z, X, h)) \geq \frac{1}{2} \mdim (h)$
for a reasonably large class of \hm{s}
constructed using the methods of Giol and Kerr~\cite{GlKr},
including the ones in that paper.
For all \mh{s} of this type,
there is a continuous surjective map from the space to the Cantor set.
The proof of \Thm{T_5526_rcmdim} uses
\Thm{T_5421_AYStabLg},
\Thm{T_5421_RCEq_0},
the fact that we can arrange that
$C^* (\Z, X, h)_Y$ be the direct limit of an AH~system
with diagonal maps,
and methods of~\cite{Niu}
(see especially Theorem~6.2 there)
to estimate radius of comparison of
simple direct limits of AH~systems
with diagonal maps.
We would like to use Theorem~6.2 of~\cite{Niu} directly.
Unfortunately,
the definition of mean dimension of an AH direct system
in~\cite{Niu}
requires that the base spaces be connected.
See Definition~3.6 of~\cite{Niu},
which refers to the setup described
after Lemma~3.4 of~\cite{Niu}.
\begin{thm}\label{T_6928_rcmdim}
Let $X$ be a compact metric space.
Let $h \colon X \to X$ be a minimal homeomorphism.
Then $\rc (C^* (\Z, X, h)) \leq 1 + 2 \cdot \mdim (h)$.
\end{thm}
Corollary~4.8 of~\cite{Ph47}
states that $\rc (C^* (\Z, X, h)) \leq 1 + 36 \cdot \mdim (h)$.
A key ingredient is
Theorem~5.1 of~\cite{Lnd},
an embedding result for \mh{s}
in shifts on cubes,
the dimension of the cube depending on the
mean dimension of the \hme.
The improvement,
to appear in a revised version of~\cite{Ph47},
is based on the use of a stronger embedding result
for minimal dynamical systems,
Theorem~1.4 of~\cite{GtTs}.
% In Theorem~\ref{T_6928_rcmdim},
We really want $\rc (C^* (\Z, X, h)) \leq \frac{1}{2} \mdim (h)$,
as in Theorem~\ref{T_5526_rcmdim}.
\begin{thm}[Theorem~7.1 of~\cite{ArPh}]\label{T_5413_TsrCP}
Let $X$ be a compact metric space.
Assume that
there is a continuous surjective map from $X$ to the Cantor set.
Let $h \colon X \to X$ be a minimal homeomorphism.
Then $C^* (\Z, X, h)$ has stable rank one
(\Def{D:SR1}).
\end{thm}
There is no finite dimensionality
assumption on~$X$.
We don't even assume that $h$ has mean dimension zero.
In particular,
this theorem holds for the examples
of Giol and Kerr~\cite{GlKr},
for which the crossed products are known not to be $Z$stable
and not to have strict comparison of positive elements.
(For such systems, it is shown in~\cite{HPT}
that $\rc (C^* (\Z, X, h)) = \frac{1}{2} \mdim (h)$,
and in~\cite{GlKr}
that $\mdim (h) \neq 0$.
See the discussion in Section~7 of~\cite{ArPh} for details.)
The proof uses
\Thm{T_5421_AYStabLg},
\Thm{T_5519_tsr}(\ref{T_5519_tsr_tsr}),
the fact that we can arrange that
$C^* (\Z, X, h)_Y$ be the direct limit of an AH~system
with diagonal maps,
and Theorem~4.1 of~\cite{EHT},
according to which
simple direct limits of AH~systems
with diagonal maps
always have stable rank one,
without any dimension growth hypotheses.
\begin{thm}[Elliott and Niu~\cite{EN1}]\label{T_5526_ENExtIrr}
The ``extended'' irrational rotation algebras,
obtained by ``cutting'' each of the standard unitary generators
at one or more points in its spectrum,
are AF algebras.
\end{thm}
We omit the precise descriptions of these algebras.
If one cuts just one of the generators,
the resulting algebra is a crossed product by a \mh{}
of the Cantor set,
with the other unitary playing the role
of the image of a generator of the group~$\Z$.
If both are cut,
the algebra is no longer an obvious crossed product.
\begin{thm}[Elliott and Niu~\cite{EN2}]\label{T_5526_ENZStab}
Let $X$ be an infinite \cms,
and let $h \colon X \to X$ be a \mh{}
with mean dimension zero.
Then $C^* (\Z, X, h)$ is $Z$stable.
\end{thm}
\section{Basic Properties of Large Subalgebras}\label{Sec_Prelim}
In this section,
we give some equivalent versions
of the definition of a large subalgebra.
Then we state some of the basic properties of large subalgebras.
Recall that,
by convention,
if we say that $B$ is a unital subalgebra of a \ca~$A$,
we mean that $B$ contains the identity of~$A$.
The change from the definition in the following lemma
is that we only require the usual conclusions
of \Def{D_5421_Large}
to hold for $a_1, a_2, \ldots, a_m$ in a subset
of~$A$ whose linear span is dense.
\begin{lem}\label{L_4627_Dense}
Let $A$ be an infinite dimensional simple unital \ca,
let $B \subset A$ be a unital subalgebra,
and let $S \subset A$ be a subset whose linear span
is dense in~$A$.
Suppose that
for every $m \in \N$,
$a_1, a_2, \ldots, a_m \in S$,
$\ep > 0$, $x \in A_{+}$ with $\ x \ = 1$,
and $y \in B_{+} \setminus \{ 0 \}$,
there are $c_1, c_2, \ldots, c_m \in A$ and $g \in B$
such that:
\begin{enumerate}
\item\label{D_Large:Cut1}
$0 \leq g \leq 1$.
\item\label{D_Large:Cut2}
For $j = 1, 2, \ldots, m$ we have
$\ c_j  a_j \ < \ep$.
\item\label{D_Large:Cut3}
For $j = 1, 2, \ldots, m$ we have
$(1  g) c_j \in B$.
\item\label{D_Large:Cut4}
$g \precsim_B y$ and $g \precsim_A x$.
\item\label{D_Large:Cut5}
$\ (1  g) x (1  g) \ > 1  \ep$.
\end{enumerate}
Then $B$ is a large subalgebra of~$A$
in the sense of \Def{D_5421_Large}.
\end{lem}
As before, the Cuntz subequivalence
involving $y$ in~(\ref{D_Large:Cut4})
is relative to~$B$,
not~$A$.
\begin{exr}\label{Er_7131_ProveSpam}
Prove Lemma~\ref{L_4627_Dense}.
\end{exr}
Unlike other approximation properties
(such as tracial rank),
it seems not to be possible to take $S$
in Lemma~\ref{L_4627_Dense}
to be a generating subset,
or even a selfadjoint generating subset.
(We can do this for the definition of a centrally large subalgebra,
\Def{D_5421_CentLarge}.
See Proposition 3.10 of~\cite{ArPh}.)
% \label{P_4917_ZGen}
By Proposition~4.4 of~\cite{Ph40},
% \label{P_4624_Approx}
in \Def{D_5421_Large}
we can omit mention of $c_1, c_2, \ldots, c_m$,
and replace (\ref{D_Large:Cut2}) and~(\ref{D_Large:Cut3})
by the requirement that
$\dist \big( (1  g) a_j, \, B \big) < \ep$
for $j = 1, 2, \ldots, m$.
So far,
however,
most verifications of \Def{D_5421_Large}
proceed by
constructing elements $c_1, c_2, \ldots, c_m$
as in \Def{D_5421_Large}.
When $A$ is finite,
we do not need condition~(\ref{D_Large:Cut5})
of Definition~\ref{D_5421_Large}.
\begin{prp}[Proposition~4.5 of~\cite{Ph40}]\label{P:FinLarge}
% {P:FinLarge}
Let $A$ be a finite infinite dimensional simple unital \ca,
and let $B \subset A$ be a unital subalgebra.
Suppose that for $m \in \N$,
$a_1, a_2, \ldots, a_m \in A$,
$\ep > 0$, $x \in A_{+} \setminus \{ 0 \}$,
and $y \in B_{+} \setminus \{ 0 \}$,
there are $c_1, c_2, \ldots, c_m \in A$ and $g \in B$
such that:
\begin{enumerate}
\item\label{D:FinLarge:Cut1}
$0 \leq g \leq 1$.
\item\label{D:FinLarge:Cut2}
For $j = 1, 2, \ldots, m$ we have
$\ c_j  a_j \ < \ep$.
\item\label{D:FinLarge:Cut3}
For $j = 1, 2, \ldots, m$ we have
$(1  g) c_j \in B$.
\item\label{D:FinLarge:Cut4}
$g \precsim_B y$ and $g \precsim_A x$.
\end{enumerate}
Then $B$ is large in~$A$.
\end{prp}
The proof of Proposition~\ref{P:FinLarge}
needs Lemma~\ref{L2726Big} below,
which is a version for Cuntz comparison of
Lemma~1.15 of~\cite{PhT1}.
We describe the idea of the proof
of Proposition~\ref{P:FinLarge}.
(Most of the details are given below.)
Given $x \in A_{+}$ with $\ x \ = 1$,
we want $x_0 \in A_{+} \setminus \{ 0 \}$
such that $g \precsim_A x_0$
and otherwise as above
implies $\ (1  g) x (1  g) \ > 1  \ep$.
(We then use $x_0$ in place of~$x$
in the definition of a large subalgebra.)
Choose a sufficiently small number $\ep_0 > 0$.
(It will be much smaller than~$\ep$.)
Choose $f \colon [0, 1] \to [0, 1]$
such that $f = 0$ on $[0, \, 1  \ep_0 ]$ and $f (1) = 1$.
Construct $a, b_1, b_2, c_1, c_2, d_1, d_2 \in {\ov{f (x) A f (x)}}$
such that for $j = 1, 2$ we have
\[
0 \leq d_j \leq c_j \leq b_j \leq a \leq 1,
\,\,\,\,\,\,
a b_j = b_j,
\,\,\,\,\,\,
b_j c_j = c_j,
\,\,\,\,\,\,
c_j d_j = d_j,
\andeqn
d_j \neq 0,
\]
and $b_1 b_2 = 0$.
Take $x_0 = d_1$.
If $\ep_0$ is small enough,
$g \precsim_A d_1$,
and $\ (1  g) x (1  g) \ \leq 1  \ep$,
this gives
\[
\ (1  g) (b_1 + b_2) (1  g) \ < 1  \frac{\ep}{3}.
\]
One then gets $c_1 + c_2 \precsim_A d_1$.
(This is the calculation~(\ref{Eq_4819_Star})
in the proof below.)
Now $r = (1  c_1  c_2) + d_1$
satisfies $r \precsim_A 1$,
so there is $v \in A$ such that $\ v r v^*  1 \ < \frac{1}{2}$.
Then $v r^{1/2}$ is right invertible,
but $v r^{1/2} d_2 = 0$,
so $v r^{1/2}$ is not left invertible.
This contradicts finiteness of~$A$.
We now give a more detailed argument.
\begin{lem}[Lemma~2.5 of~\cite{Ph40}]\label{L2726Norm1}
Let $A$ be a \ca,
let $x \in A_{+}$ satisfy $\ x \ = 1$,
and let $\ep > 0$.
Then there are positive elements $a, b \in {\overline{x A x}}$
with $\ a \ = \ b \ = 1$,
such that $a b = b$,
and such that whenever $c \in {\overline{b A b}}$
satisfies $\ c \ \leq 1$,
then
$\ x c  c \ < \ep$.
\end{lem}
\begin{proof}[Sketch of proof]
Choose \cfn{s} $f_0, f_1 \colon [0, 1] \to [0, 1]$
such that $f_1 (1) = 1$,
$f_1$ is supported near~$1$,
$ f_0 (\ld)  \ld  < \ep$ for all $\ld \in [0, 1]$,
and $f_0 = 1$ near~$1$
(so that $f_0 f_1 = f_1$).
Take $a = f_0 (x)$ and $b = f_1 (x)$.
Then $\ x  a \ < \ep$ and $a b = b$.
\end{proof}
\begin{lem}[Lemma~2.6 of~\cite{Ph40}]\label{L2726Big}
Let $A$ be a finite simple infinite dimensional unital \ca.
Let $x \in A_{+}$ satisfy $\ x \ = 1$.
Then for every $\ep > 0$
there is $x_0 \in \big( {\overline{x A x}} \big)_{+} \setminus \{ 0 \}$
such that whenever $g \in A_{+}$ satisfies $0 \leq g \leq 1$
and $g \precsim_A x_0$,
then $\ (1  g) x (1  g) \ > 1  \ep$.
\end{lem}
\begin{proof}
Choose positive elements $a, b \in {\overline{x^{1/2} A x^{1/2}}}$
as in Lemma~\ref{L2726Norm1},
with $x^{1/2}$ in place of~$x$
and $\frac{\ep}{3}$ in place of~$\ep$.
Then $a, b \in {\overline{x A x}}$
since ${\overline{x^{1/2} A x^{1/2}}} = {\overline{x A x}}$.
Since $b \neq 0$,
Lemma~\ref{L:DivInSmp}
provides nonzero positive orthogonal elements
$z_1, z_2 \in {\overline{b A b}}$
(with $z_1 \sim_A z_2$).
We may require $\ z_1 \ = \ z_2 \ = 1$.
Choose \cfn{s} $f_0, f_1, f_2 \colon [0, \infty) \to [0, 1]$
such that
\[
f_0 (0) = 0,
\,\,\,\,\,\,
f_0 f_1 = f_1,
\,\,\,\,\,\,
f_1 f_2 = f_2,
\andeqn
f_2 (1) = 1.
\]
For $j = 1, 2$ define
\[
b_j = f_0 (z_j),
\,\,\,\,\,\,
c_j = f_1 (z_j),
\andeqn
d_j = f_2 (z_j).
\]
Then
\[
0 \leq d_j \leq c_j \leq b_j \leq 1,
\,\,\,\,\,\,
a b_j = b_j,
\,\,\,\,\,\,
b_j c_j = c_j,
\,\,\,\,\,\,
c_j d_j = d_j,
\andeqn
d_j \neq 0.
\]
Also $b_1 b_2 = 0$.
Define $x_0 = d_1$.
Then $x_0 \in \big( {\overline{x A x}} \big)_{+}$.
Let $g \in A_{+}$ satisfy $0 \leq g \leq 1$ and $g \precsim_A x_0$.
We want to show that
\[
\ (1  g) x (1  g) \ > 1  \ep,
\]
so suppose that $\ (1  g) x (1  g) \ \leq 1  \ep$.
The choice of $a$ and~$b$,
and the relations $(b_1 + b_2)^{1/2} \in {\overline{b A b}}$
and $\big\ (b_1 + b_2)^{1/2} \big\ = 1$,
imply that
\[
\big\ x^{1/2} (b_1 + b_2)^{1/2}  (b_1 + b_2)^{1/2} \big\
< \frac{\ep}{3}.
\]
Using this relation and its adjoint at the second step, we get
\begin{align*}
\big\ (1  g) (b_1 + b_2) (1  g) \big\
& = \big\ (b_1 + b_2)^{1/2} (1  g)^2 (b_1 + b_2)^{1/2} \big\
\\
& < \big\ (b_1 + b_2)^{1/2} x^{1/2} (1  g)^2
x^{1/2} (b_1 + b_2)^{1/2} \big\
+ \frac{2 \ep}{3}
\\
& \leq \big\ x^{1/2} (1  g)^2 x^{1/2} \big\ + \frac{2 \ep}{3}
\\
& = \ (1  g) x (1  g) \ + \frac{2 \ep}{3}
\leq 1  \frac{\ep}{3}.
\end{align*}
Using the equation $(b_1 + b_2) (c_1 + c_2) = c_1 + c_2$
and taking $C$ to be the commutative \ca{}
generated by $b_1 + b_2$ and $c_1 + c_2$,
one easily sees that for every $\bt \in [0, 1)$
we have
$c_1 + c_2 \precsim_C [(b_1 + b_2)  \bt]_{+}$.
Take $\bt = 1  \frac{\ep}{3}$,
use this fact and Lemma~\ref{L:C2} at the first step,
use the estimate above at the second step,
and use $g \precsim_A x_0 = d_1$ at the third step,
to get
%
\begin{equation}\label{Eq_4819_Star}
c_1 + c_2
% \precsim_A [(b_1 + b_2)  \bt]_{+}
\precsim_A \big[ (1  g) (b_1 + b_2) (1  g)  \bt \big]_{+}
\oplus g
= 0 \oplus g
\precsim_A d_1.
\end{equation}
%
Set $r = (1  c_1  c_2) + d_1$.
Use Lemma \ref{L:CzBasic}(\ref{L:CzBasic:LCzCmpSum})
at the first step,
(\ref{Eq_4819_Star})~at the second step,
and Lemma \ref{L:CzBasic}(\ref{L:CzBasic:Orth})
and $d_1 (1  c_1  c_2) = 0$ at the third step,
to get
\[
1 \precsim_A (1  c_1  c_2) \oplus (c_1 + c_2)
\precsim_A (1  c_1  c_2) \oplus d_1
\sim_A (1  c_1  c_2) + d_1
= r.
\]
Thus
there is $v \in A$ such that $\ v r v^*  1 \ < \tfrac{1}{2}$.
It follows that $v r^{1/2}$ has a right inverse.
But $v r^{1/2} d_2 = 0$, so $v r^{1/2}$ is not invertible.
We have contradicted finiteness of~$A$,
and thus proved the lemma.
\end{proof}
\begin{proof}[Proof of Proposition~\ref{P:FinLarge}]
Let $a_1, a_2, \ldots, a_m \in A$,
let $\ep > 0$,
let $x \in A_{+} \setminus \{ 0 \}$,
and let $y \in B_{+} \setminus \{ 0 \}$.
\Wolog{} $\ x \ = 1$.
Apply Lemma~\ref{L2726Big},
obtaining
$x_0 \in \big( {\overline{x A x}} \big)_{+} \setminus \{ 0 \}$
such that whenever $g \in A_{+}$ satisfies $0 \leq g \leq 1$
and $g \precsim_A x_0$,
then $\ (1  g) x (1  g) \ > 1  \ep$.
Apply the hypothesis
with $x_0$ in place of $x$
and everything else as given,
getting $c_1, c_2, \ldots, c_m \in A$ and $g \in B$.
We need only prove that $\ (1  g) x (1  g) \ > 1  \ep$.
But this is immediate from the choice of~$x_0$.
\end{proof}
The following strengthening of the definition is
often convenient.
First, we can always require $\ c_j \ \leq \ a_j \$.
Second, if we cut down on both sides instead of
on one side,
and the elements $a_j$ are positive,
then we may take the elements $c_j$ to be positive.
\begin{lem}[Lemma~4.8 of~\cite{Ph40}]\label{L:LargeStaysPositive}
% {L:LargeStaysPositive}
Let $A$ be an infinite dimensional simple unital \ca,
and let $B \subset A$ be a large subalgebra.
Let $m, n \in \Nz$,
let $a_1, a_2, \ldots, a_m \in A$,
let $b_1, b_2, \ldots, b_n \in A_{+}$,
let $\ep > 0$,
let $x \in A_{+}$ satisfy $\ x \ = 1$,
and let $y \in B_{+} \setminus \{ 0 \}$.
Then there are
$c_1, c_2, \ldots, c_m \in A$,
$d_1, d_2, \ldots, d_n \in A_{+}$,
and $g \in B$
such that:
\begin{enumerate}
\item\label{L:LargeStaysPositive:Cut1}
$0 \leq g \leq 1$.
\item\label{L:LargeStaysPositive:Cut2a}
For $j = 1, 2, \ldots, m$ we have
$\ c_j  a_j \ < \ep$,
and for $j = 1, 2, \ldots, n$ we have
$\ d_j  b_j \ < \ep$.
\item\label{L:LargeStaysPositive:Cut2b}
For $j = 1, 2, \ldots, m$ we have
$\ c_j \ \leq \ a_j \$,
and for $j = 1, 2, \ldots, n$ we have
$\ d_j \ \leq \ b_j \$.
\item\label{L:LargeStaysPositive:Cut3}
For $j = 1, 2, \ldots, m$ we have
$(1  g) c_j \in B$,
and for $j = 1, 2, \ldots, n$ we have
$(1  g) d_j (1  g) \in B$.
\item\label{L:LargeStaysPositive:Cut4}
$g \precsim_B y$ and $g \precsim_A x$.
\item\label{L:LargeStaysPositive:Cut5}
$\ (1  g) x (1  g) \ > 1  \ep$.
\end{enumerate}
\end{lem}
\begin{proof}[Sketch of proof]
To get $\ c_j \ \leq \ a_j \$ for $j = 1, 2, \ldots, m$,
one takes $\ep > 0$ to be a bit smaller in the definition,
and scales down $c_j$ for any $j$ for which $\ c_j \$ is too big.
Given that one can do this,
following the definition,
approximate
\[
a_1, \, a_2, \, \ldots, \, a_m,
\, b_1^{1/2}, \, b_2^{1/2}, \, \ldots, \, b_n^{1/2}
\]
sufficiently well by
\[
c_1, c_2, \ldots, c_m, r_1, r_2, \ldots, r_n,
\]
and take $d_j = r_j r_j^*$ for $j = 1, 2, \ldots, n$.
\end{proof}
In Definition~4.9 of~\cite{Ph40}
% \label{D2717CPType}
we defined a ``large subalgebra of crossed product type'',
a strengthening of the definition of a large subalgebra,
and in Proposition~4.11 of~\cite{Ph40}
% \label{P2729AltCPT}
we gave a convenient way to verify
that a subalgebra is a large subalgebra of crossed product type.
The large subalgebras we have constructed in crossed products
are of crossed product type.
Theorem~4.6 of~\cite{ArPh}
% \label{T2717CptImpRq}
shows that a large subalgebra of crossed product type
is in fact centrally large.
We will show directly (proof of Theorem~\ref{T_5421_AYStabLg},
in Section~\ref{Sec_AY} below)
that if $X$ is an infinite \chs,
$h \colon X \to X$ is a minimal homeomorphism,
and $Y \subset X$ is a compact subset
such that $h^n (Y) \cap Y = \varnothing$
for all $n \in \Z \setminus \{ 0 \}$,
then the orbit breaking subalgebra $C^* (\Z, X, h)_Y$
of \Def{D_5421_VSubalg} is
centrally large in $C^* (\Z, X, h)$.
This procedure is easier than using
large subalgebras of crossed product type.
The abstract version is more useful for subalgebras
of crossed products by more complicated groups,
but we don't consider these in these notes.
We now give proofs of two of the basic properties of
large subalgebras above:
if $B$ is large in~$A$,
then $B$ is simple
(part of Proposition~\ref{P_2627_NoSmp_0})
and has the ``same'' tracial states as~$A$
(part of Theorem~\ref{L_4622_SameT_0}).
We start with
the simplicity statement in
Proposition~\ref{P_2627_NoSmp_0}.
% The infinite dimensionality statement is easier to prove
% (provided it is done afterwards),
% and we refer to the proof of
% Proposition~5.5 in~\cite{Ph40}.
\begin{prp}[Proposition~5.2 of~\cite{Ph40}]\label{P_2627_NoSmp}
Let $A$ be an infinite dimensional simple unital \ca,
and let $B \subset A$ be a large subalgebra.
Then $B$ is simple.
\end{prp}
We need some preliminary work.
\begin{lem}[Lemma~1.12 of~\cite{Ph40}]\label{L_2627_SumDom}
% \label{L_2627_SumDom}
Let $A$ be a \ca,
let $n \in \N$,
and let $a_1, a_2, \ldots, a_n \in A$.
Set $a = \sum_{k = 1}^n a_k$.
Then $a^* a \leq 2^{n  1} \sum_{k = 1}^n a_k^* a_k$.
\end{lem}
\begin{proof}
We prove this by induction on~$n$.
For $n = 1$,
the statement is immediate.
Suppose it is known for~$n$;
we prove it for $n + 1$.
Set $x = \sum_{k = 1}^n a_k$.
Then,
expanding and cancelling at the third step,
using the induction hypothesis at the fourth step,
and using $n \geq 1$ at the fifth step,
we get
\begin{align*}
a^* a
& = (x + a_{n + 1})^* (x + a_{n + 1})
\leq (x + a_{n + 1})^* (x + a_{n + 1})
+ (x  a_{n + 1})^* (x  a_{n + 1})
\\
& = 2 x^* x + 2 a_{n + 1}^* a_{n + 1}
\leq 2^n \sum_{k = 1}^n a_k^* a_k + 2 a_{n + 1}^* a_{n + 1}
\leq 2^n \sum_{k = 1}^{n + 1} a_k^* a_k.
\end{align*}
This completes the induction step and the proof.
\end{proof}
\begin{lem}[Lemma~1.13 of~\cite{Ph40}]\label{L_4619_PosIdeal}
% \label{L_4619_PosIdeal}
Let $A$ be a \ca{} and let $a \in A_{+}$.
Let $b \in {\overline{A a A}}$ be positive.
Then for every $\ep > 0$ there exist $n \in \N$
and $x_1, x_2, \ldots, x_n \in A$
such that
$\left\{\rule{0em}{2ex}} b  \sum_{k = 1}^n x_k^* a x_k \right\
< \ep$.
\end{lem}
This result is used without proof in the proof
of Proposition 2.7(v) of~\cite{KR}.
We prove it when $A$ is unital and $b = 1$,
which is the case needed here.
In this case,
we can get $\sum_{k = 1}^n x_k^* a x_k = 1$.
In particular,
we get Corollary~1.14 of~\cite{Ph40} this way.
% \label{C_4619_PosSmp}
(This result can also be obtained from Proposition~1.10 of~\cite{Cu0},
as pointed out after the proof of that proposition.)
\begin{proof}[Proof of Lemma~\ref{L_4619_PosIdeal} when $b = 1$]
Choose
\[
n \in \N
\andeqn
y_1, y_2, \ldots, y_n, z_1, z_2, \ldots, z_n \in A
\]
such that
the element $c = \sum_{k = 1}^n y_k a z_k$ satisfies
$\ c  1 \ < 1$.
Set
\[
r = \sum_{k = 1}^n z_k^* a y_k^* y_k a z_k,
\,\,\,\,
M = \max \big( \ y_1 \, \, \ y_2 \, \, \ldots, \, \ y_n \ \big),
\,\,\,\,
{\mbox{and}}
\,\,\,\,
s = M^2 \sum_{k = 1}^n z_k^* a^2 z_k.
\]
Lemma~\ref{L_2627_SumDom}
implies that $c^* c \in {\ov{r A r}}$.
% $c^* c$ is in the hereditary subalgebra generated by~$r$.
The relation $\ c  1 \ < 1$
implies that $c$ is invertible,
so $r$ is invertible.
Since $r \leq s$,
it follows that $s$ is invertible.
Set $x_k = M a^{1/2} z_k s^{1/2}$
for $k = 1, 2, \ldots, n$.
Then
$\sum_{k = 1}^n x_k^* a x_k = s^{1/2} s s^{1/2} = 1$.
\end{proof}
\begin{proof}[Sketch of proof of Proposition~\ref{P_2627_NoSmp}]
Let $b \in B_{+} \setminus \{ 0 \}$.
We show that there are $n \in \N$
and $r_1, r_2, \ldots, r_n \in B$
such that $\sum_{k = 1}^n r_k b r_k^*$ is invertible.
Since $A$ is simple,
Lemma~\ref{L_4619_PosIdeal}
provides $m \in \N$
and $x_1, x_2, \ldots, x_m \in A$
such that $\sum_{k = 1}^m x_k b x_k^* = 1$.
Set
\[
M = \max \big( 1, \, \ x_1 \, \, \ x_2 \, \, \ldots, \, \ x_m \,
\, \ b \ \big)
\andeqn
\dt = \min \left( 1, \, \frac{1}{3 m M (2 M + 1)} \right).
\]
By definition,
there are $y_1, y_2, \ldots, y_m \in A$
and $g \in B_{+}$
such that
$0 \leq g \leq 1$,
such that $\ y_j  x_j \ < \dt$
and $(1  g) y_j \in B$
for $j = 1, 2, \ldots, m$,
and such that $g \precsim_B b$.
Set $z = \sum_{k = 1}^m y_j b y_j^*$.
The number $\dt$ has been chosen
to ensure that $\ z  1 \ < \frac{1}{3}$;
the estimate is carried out in~\cite{Ph40}.
It follows that
$\big\ (1  g) z (1  g)  (1  g)^2 \big\ < \frac{1}{3}$.
Set $h = 2 g  g^2$.
Lemma \ref{L:CzBasic}(\ref{L:CzBasic:LCzFCalc}),
applied to the function $\ld \mapsto 2 \ld  \ld^2$,
implies that $h \sim_B g$.
Therefore $h \precsim_B b$.
So there is $v \in B$ such that $\ v b v^*  h \ < \frac{1}{3}$.
Now
take $n = m + 1$,
take $r_j = (1  g) y_j$ for $j = 1, 2, \ldots, m$,
and take $r_{m + 1} = v$.
Then $r_1, r_2, \ldots, r_n \in B$.
One can now check,
using $(1  g)^2 + h = 1$,
that $\left\ 1  \sum_{k = 1}^n r_k b r_k^* \right\ < \frac{2}{3}$.
Therefore $\sum_{k = 1}^n r_k b r_k^*$ is invertible,
as desired.
\end{proof}
The following is a special case of the
infinite dimensionality statement in
Proposition~\ref{P_2627_NoSmp_0}
(Proposition~5.5 of~\cite{Ph40}),
which is easier to prove.
\begin{prp}[Stably finite case of Proposition~5.5
of~\cite{Ph40}]\label{P_6928_InfDim}
Let $A$ be a stably finite infinite dimensional simple unital \ca{}
and let $B \subset A$ be a large subalgebra.
Then $B$ is infinite dimensional.
\end{prp}
\begin{proof}
Suppose $B$ is finite dimensional.
Proposition~\ref{P_2627_NoSmp}
tells us that $B$ is simple,
so there is $n \in \N$ such that $B \cong M_n$.
It follows from the discussion after Theorem~3.3 of~\cite{BlkRr}
that there is a quasitrace $\ta$ on~$A$.
Apply Corollary~\ref{C2718CuDiv}
to get $x \in A_{+} \setminus \{ 0 \}$
such that $d_{\ta} (x) < (n + 1)^{1}$.
We may assume that $\ x \ = 1$.
Clearly $B \neq A$,
so there is $a \in A$ such that $\dist (a, B) > 1$.
Apply Definition~\ref{D_5421_Large},
getting $g \in B$ and $c \in A$
such that
\[
0 \leq g \leq 1,
\qquad
\ a  c \ < \frac{1}{2},
\qquad
(1  g) c \in B,
\andeqn
g \precsim_A x.
\]
Then $c \not\in B$, so $g \neq 0$.
Also,
$d_{\ta} (g) \leq d_{\ta} (x) < (n + 1)^{1}$.
Now $\sm = \ta _B$ is a quasitrace on~$B$,
so must be the normalized trace on~$B$,
and $0 < d_{\sm} (g) = d_{\ta} (g) < (n + 1)^{1}$.
There are no elements $g \in (M_n)_{+}$
with $0 < d_{\sm} (g) < (n + 1)^{1}$,
so we have a contradiction.
\end{proof}
\begin{prp}[Corollary~5.8 of~\cite{Ph40}]\label{P_5519_StFinStLg}
% \label{C_4619_StFinStLg}
Let $A$ be a stably finite infinite dimensional simple unital \ca,
and let $B \subset A$ be a large subalgebra.
Let $n \in \N$.
Then $M_n (B)$ is large in~$M_n (A)$.
\end{prp}
In~\cite{Ph40},
this result is obtained as a corollary of a more general result
(Proposition~\ref{P_4624_TLarge} here).
% the minimal tensor product of large subalgebras is large
% whenever the tensor product of the containing algebras is finite.
A direct proof is easier,
and we give it here.
\begin{proof}[Proof of Proposition~\ref{P_5519_StFinStLg}]
Let $m \in \N$,
let $a_1, a_2, \ldots, a_m \in M_n (A)$,
let $\ep > 0$,
let $x \in M_n (A)_{+} \setminus \{ 0 \}$,
and let $y \in M_n (B)_{+} \setminus \{ 0 \}$.
There are $b_{k, l} \in A$
for $k, l = 1, 2, \ldots, n$
such that
\[
x^{1/2}
= \sum_{k, l = 1}^n e_{k, l} \otimes b_{k, l} \in M_n \otimes A.
\]
Choose $k, l \in \{ 1, 2, \ldots, n \}$
such that $b_{k, l} \neq 0$.
Set $x_0 = b_{k, l}^* b_{k, l} \in A_{+} \setminus \{ 0 \}$.
Using selfadjointness of~$x^{1/2}$,
we find that
\[
e_{1, 1} \otimes x_0
= (e_{l, 1} \otimes 1)^* x^{1/2} (e_{k, k} \otimes 1) x^{1/2}
(e_{l, 1} \otimes 1)
\leq (e_{l, 1} \otimes 1)^* x (e_{l, 1} \otimes 1)
\precsim_A x.
\]
Similarly,
there is $y_0 \in B_{+} \setminus \{ 0 \}$
such that $e_{1, 1} \otimes y_0 \precsim_B y$.
Use \Lem{L:DivInSmp}
and simplicity (Proposition~\ref{P_2627_NoSmp})
and infinite dimensionality (Proposition~\ref{P_6928_InfDim}) of~$B$
to find systems of
nonzero mutually orthogonal and mutually Cuntz equivalent
positive elements
\[
x_1, x_2, \ldots, x_n \in {\ov{x_0 A x_0}}
\andeqn
y_1, y_2, \ldots, y_n \in {\ov{y_0 B y_0}}.
\]
For $j = 1, 2, \ldots, m$,
choose elements
$a_{j, k, l} \in A$
for $k, l = 1, 2, \ldots, n$
such that
\[
a_j
= \sum_{k, l = 1}^n e_{k, l} \otimes a_{j, k, l} \in M_n \otimes A.
\]
Apply Proposition~\ref{P:FinLarge}
with $m n^2$ in place of~$m$,
with the elements $a_{j, k, l}$ in place of $a_1, a_2, \ldots, a_m$,
with $\ep / n^2$ in place of~$\ep$,
with $x_1$ in place of~$x$,
and with $y_1$ in place of~$y$,
getting $g_0 \in A_{+}$
and $c_{j, k, l} \in A$
for $j = 1, 2, \ldots, m$ and $k, l = 1, 2, \ldots, n$.
Define
$c_j = \sum_{k, l = 1}^n e_{k, l} \otimes c_{j, k, l}$
for $j = 1, 2, \ldots, m$ and define $g = 1 \otimes g_0$.
It is clear that
$0 \leq g \leq 1$,
that $\ c_j  a_j \ < \ep$ and $(1  g) c_j \in M_n (B)$
for $j = 1, 2, \ldots, m$.
We have
$g \precsim_A 1 \otimes x_1$
and
$g \precsim_B 1 \otimes y_1$,
so
Lemma~\ref{L:CzBasic}(\ref{L:CzBasic:Her})
and Lemma~\ref{L:CzBasic}(\ref{L:CzBasic:Orth})
imply that
$g \precsim_A x_0$ and $g \precsim_B y_0$.
Therefore $g \precsim_A x$ and $g \precsim_B y$.
\end{proof}
We prove the statement about traces in
Theorem~\ref{L_4622_SameT_0},
assuming that the algebras are stably finite
(the interesting case).
\begin{thm}[Stably finite case of
Theorem~6.2 of~\cite{Ph40}]\label{T_5522_SameT}
% \label{L_4622_SameT}
Let $A$ be an infinite dimensional stably finite simple unital \ca,
and let $B \subset A$ be a large subalgebra.
Then the restriction map
${\operatorname{T}} (A) \to {\operatorname{T}} (B)$
is bijective.
\end{thm}
Again,
we need a lemma.
\begin{lem}\label{L_4622_ExtTr}
% \label{L_4622_ExtTr}
Let $A$ be an infinite dimensional simple unital \ca,
and let $B \subset A$ be a large subalgebra.
Let $\ta \in {\operatorname{T}} (B)$.
Then there exists a unique state $\om$ on~$A$
such that $\om _B = \ta$.
\end{lem}
\begin{proof}
Existence of $\om$ follows from the HahnBanach Theorem.
For uniqueness,
let $\om_1$ and $\om_2$ be states on~$A$ such that
$\om_1 _B = \om_2 _B = \ta$,
let $a \in A_{+}$,
and let $\ep > 0$.
We prove that $ \om_1 (a)  \om_2 (a)  < \ep$.
\Wolog{} $\ a \ \leq 1$.
It follows from Proposition~\ref{P_2627_NoSmp}
and Proposition~\ref{P_6928_InfDim}
that $B$ is simple and infinite dimensional.
So Corollary~\ref{C2718CuDiv}
provides $y \in B_{+} \setminus \{ 0 \}$
such that $d_{\ta} (y) < \frac{\ep^2}{64}$
(for the particular choice of $\ta$ we are using).
Use Lemma~\ref{L:LargeStaysPositive}
to find $c \in A_{+}$ and $g \in B_{+}$
such that
\[
\ c \ \leq 1,
\,\,\,\,\,\,
\ g \ \leq 1,
\,\,\,\,\,\,
\ c  a \ < \frac{\ep}{4},
\,\,\,\,\,\,
(1  g) c (1  g) \in B,
\andeqn
g \precsim_B y.
\]
For $j = 1, 2$,
the CauchySchwarz inequality
gives
%
\begin{equation}\label{Eq_4624_CauSch}
 \om_j (r s) 
\leq \om_j (r r^*)^{1/2} \om_j (s^* s)^{1/2}
\end{equation}
%
for all $r, s \in A$.
Also,
by Lemma~\ref{L:CzBasic}(\ref{L:CzBasic:LCzFCalc})
we have $g^2 \sim_B g \precsim_B y$.
Since $\ g^2 \ \leq 1$ and $\om_j _{B} = \ta$
is a \tst,
it follows that
$\om_j (g^2) \leq d_{\ta} (y) < \frac{\ep^2}{64}$.
Using $\ c \ \leq 1$
and the CauchySchwarz inequality,
we then get
\[
 \om_j (g c) 
\leq \om_j (g^2)^{1/2} \om_j (c^2)^{1/2}
< \frac{\ep}{8}
\]
and
\[
 \om_j ((1  g) c g) 
\leq \om_j \big( (1  g) c^2 (1  g) \big)^{1/2} \om_j (g^2)^{1/2}
< \frac{\ep}{8}.
\]
So
\begin{align*}
\big \om_j (c)  \ta ( (1  g) c (1  g) ) \big
& = \big \om_j (c)  \om_j ( (1  g) c (1  g) ) \big
\\
& \leq  \om_j (g c)  +  \om_j ((1  g) c g) 
< \frac{\ep}{4}.
\end{align*}
Also $ \om_j (c)  \om_j (a)  < \frac{\ep}{4}$.
So
\[
\big \om_j (a)  \ta ( (1  g) c (1  g) ) \big < \frac{\ep}{2}.
\]
Thus $ \om_1 (a)  \om_2 (a)  < \ep$.
\end{proof}
The uniqueness statement in \Lem{L_4622_ExtTr}
is used to prove that the restriction map
$\T (A) \to \T (B)$ is injective.
One might hope that \Lem{L_4622_ExtTr} would enable the following
idea for the proof that $\T (A) \to \T (B)$ is surjective.
% of Theorem~\ref{T_5522_SameT}.
We first observe that a state $\om$ is tracial whenever
$\om (u a u^*) = \om (u)$
for all $a \in A$ and all unitaries $u \in A$.
Indeed,
putting $a u$ for $a$
gives $\om (u a) = \om (a u)$
for all $a \in A$ and all unitaries $u \in A$.
Since $A$ is the linear span of its unitaries,
it follows that $\om (b a) = \om (a b)$ for all $a, b \in A$.
Now let $A$ and $B$ be as in Theorem~\ref{T_5522_SameT},
let $\ta \in \T (B)$, and $u \in A$.
Let $\om$ be the unique state on $A$ which extends~$\ta$
(\Lem{L_4622_ExtTr}).
We would like to argue that the state
$\rh (a) = \om (u a u^*)$ for $a \in A$
is equal to~$\om$
because it
also extends~$\ta$.
The first thing which goes wrong is that if $b \in B$
and $u \in A$ is unitary,
then $u b u^*$ need not even be in~$B$.
So the is no immediate reason to
think that $\rh$ extends~$\ta$.
If the unitary~$u$ is actually in~$B$,
then $\rh$ does indeed extend~$\om$.
Thus, the uniqueness statement in \Lem{L_4622_ExtTr}
implies that $\om (u a u^*) = \om (a)$
for all $a \in A$ and all unitaries $u \in B$.
We can still replace $a$ by $a u$ as above,
and deduce that $\om (b a) = \om (a b)$
for all $a \in A$ and $b \in B$.
In particular,
$\om (v b) = \om (b v)$
for all $b \in B$ and unitaries $v \in A$.
But to get $\om (v b v^*) = \om (b)$
from this requires putting $b v^*$ in place of~$b$,
and $b v^*$ isn't in~$B$.
\begin{proof}[Proof of Theorem~\ref{T_5522_SameT}]
Let $\ta \in {\operatorname{T}} (B)$.
We show that there is a unique $\om \in {\operatorname{T}} (A)$
such that $\om _B = \ta$.
Lemma~\ref{L_4622_ExtTr}
shows that there is a unique state $\om$ on~$A$
such that $\om _B = \ta$,
and it suffices to show that $\om$ is a trace.
Thus let $a_1, a_2 \in A$ satisfy $\ a_1 \ \leq 1$
and $\ a_2 \ \leq 1$,
and let $\ep > 0$.
We show that $ \om (a_1 a_2)  \om (a_2 a_1)  < \ep$.
It follows from Proposition~\ref{P_2627_NoSmp}
and Proposition~\ref{P_6928_InfDim}
(without stable finiteness, we must appeal to
Proposition~5.5 of~\cite{Ph40})
that $B$ is simple and infinite dimensional.
So Corollary~\ref{C2718CuDiv}
provides $y \in B_{+} \setminus \{ 0 \}$
such that $d_{\ta} (y) < \frac{\ep^2}{64}$.
Use Lemma~\ref{L:LargeStaysPositive}
to find $c_1, c_2 \in A$ and $g \in B_{+}$
such that
\[
\ c_j \ \leq 1,
\,\,\,\,\,\,
\ c_j  a_j \ < \frac{\ep}{8},
\andeqn
(1  g) c_j \in B
\]
for $j = 1, 2$,
and such that $\ g \ \leq 1$ and $g \precsim_B y$.
By Lemma~\ref{L:CzBasic}(\ref{L:CzBasic:LCzFCalc}),
we have $g^2 \sim g \precsim_B y$.
Since $\ g^2 \ \leq 1$ and $\om _{B} = \ta$
is a \tst,
it follows that
$\om (g^2) \leq d_{\ta} (y) < \frac{\ep^2}{64}$.
We claim that
\[
\big \om ( (1  g) c_1 (1  g) c_2 )  \om ( c_1 c_2 ) \big
< \frac{\ep}{4}.
\]
Using the CauchySchwarz inequality
((\ref{Eq_4624_CauSch}) in the previous proof),
we get
\[
 \om (g c_1 c_2) 
\leq \om (g^2)^{1/2} \om (c_2^* c_1^* c_1 c_2)^{1/2}
\leq \om (g^2)^{1/2}
< \frac{\ep}{8}.
\]
Similarly,
and also at the second step using $\ c_2 \ \leq 1$,
$(1  g) c_1 g \in B$,
and the fact that $\om _{B}$ is a \tst,
\begin{align*}
\big \om ( (1  g) c_1 g c_2) \big
& \leq \om \big( (1  g) c_1 g^2 c_1^* (1  g) \big)^{1/2}
\om (c_2^* c_2)^{1/2}
\\
& \leq \om \big( g c_1^* (1  g)^2 c_1 g \big)^{1/2}
\leq \om (g^2)^{1/2}
< \frac{\ep}{8}.
\end{align*}
The claim now follows from the estimate
\[
\big \om ( (1  g) c_1 (1  g) c_2 )  \om ( c_1 c_2 ) \big
\leq \big \om ( (1  g) c_1 g c_2) \big +  \om (g c_1 c_2) .
\]
Similarly
\[
\big \om ( (1  g) c_2 (1  g) c_1 )  \om ( c_2 c_1 ) \big
< \frac{\ep}{4}.
\]
Since $(1  g) c_1, \, (1  g) c_2 \in B$
and $\om _{B}$ is a \tst,
we get
\[
\om ( (1  g) c_1 (1  g) c_2 ) = \om ( (1  g) c_2 (1  g) c_1 ).
\]
Therefore $ \om ( c_1 c_2 )  \om ( c_2 c_1 )  < \frac{\ep}{2}$.
One checks that
$\ c_1 c_2  a_1 a_2 \ < \frac{\ep}{4}$
and
$\ c_2 c_1  a_2 a_1 \ < \frac{\ep}{4}$.
It now follows that $ \om (a_1 a_2)  \om (a_2 a_1)  < \ep$.
\end{proof}
\section{Large Subalgebras and the Radius of Comparison}\label{Sec_rc}
\indent
Let $A$ be a simple unital \ca.
Recall (\Def{D:OrdDetD})
that the order on projections over $A$ is
determined by traces if,
as happens for type ${\mathrm{II}}_1$ factors,
whenever $p, q \in M_{\infty} (A)$ are \pj{s}
such that for all $\ta \in \T (A)$
we have $\ta (p) < \ta (q)$,
then $p$ is \mvnt{} to a subprojection of~$q$.
Without knowing whether every quasitrace is a trace
(see the discussion before Notation~\ref{N_4620_QT}),
it is more appropriate to use a condition involving quasitraces.
For exact \ca{s},
every quasitrace is a trace (Theorem 5.11 of~\cite{Hgp}),
so it makes no difference.
Simple \ca{s} need not have very many \pj{s},
so a more definitive version of this condition is to ask
for the condition in the following definition.
\begin{dfn}\label{7121_StrCmp}
Let $A$ be a simple unital \ca.
Then $A$ has {\emph{strict comparison of positive elements}}
if
whenever $a, b \in M_{\infty} (A)$
satisfy $d_{\ta} (a) < d_{\ta} (b)$ for all $\ta \in \QT (A)$,
then $a \precsim_A b$.
\end{dfn}
By Proposition 6.12 of~\cite{Ph40},
% \label{P_4817_rCompCu}
one can use $K \otimes A$ in place of $M_{\infty} (A)$,
but this is not as easy to see as with \pj{s}.
Simple AH~algebras with slow dimension growth
have strict comparison,
but other simple AH~algebras need not.
(For example, see~\cite{Tms6}.)
Strict comparison seems to be necessary for any reasonable
hope of classification in the sense of the Elliott program.
According to the TomsWinter Conjecture,
when $A$ is simple, separable, nuclear, unital,
and stably finite,
strict comparison should imply $Z$stability,
and this is known to hold in a number of cases.
The radius of comparison $\rc (A)$ of~$A$
(for a \ca{} which is unital and stably finite
but not necessarily simple)
measures the failure of strict comparison.
(See \cite{BRTTW} for what to do in more general \ca{s}.)
For additional context,
we point out the following special case of Theorem~5.1 of~\cite{Tm2}
(which will be needed in Section~\ref{Sec_rcCrPrd},
where it is restated as Theorem~\ref{T_6827_RCCX}):
if $X$ is a \cms{} and $n \in \N$,
then
\[
\rc (M_n \otimes C (X)) \leq \frac{\dim (X)  1}{2 n}.
\]
Under some conditions on~$X$
(being a finite complex is enough),
this inequality is at least approximately an equality.
See~\cite{EllNiu2}.
The following definition
of the radius of comparison
is adapted from Definition~6.1 of~\cite{Tm1}.
\begin{dfn}\label{D_5519_ROfComp}
Let $A$ be a stably finite unital C*algebra.
\begin{enumerate}
\item\label{D_5519_ROfComp:1}
Let $r \in [0, \I)$.
We say that $A$ has {\emph{$r$comparison}} if whenever
$a, b \in M_{\infty} (A)_{+}$ satisfy
$d_{\ta} (a) + r < d_{\ta} (b)$
for all $\ta \in \QT (A)$,
then $a \precsim_A b$.
\item\label{D_5519_ROfComp:2}
The {\emph{radius of comparison}} of~$A$,
denoted ${\operatorname{rc}} (A)$, is
\[
{\operatorname{rc}} (A)
= \inf \big( \big\{ r \in [0, \I) \colon
{\mbox{$A$ has $r$comparison}} \big\} \big).
\]
(We take ${\operatorname{rc}} (A) = \infty$ if there is no $r$
such that $A$ has $r$comparison.)
\end{enumerate}
\end{dfn}
Definition~6.1 of~\cite{Tm1} actually uses
lower semi\ct{} dimension functions on~$A$
instead of $d_{\ta}$ for $\ta \in \QT (A)$,
but these are the same functions by Theorem II.2.2 of~\cite{BH}.
It is also stated in terms of the order on the Cuntz
semigroup $W (A)$
rather than in terms of Cuntz subequivalence;
this is clearly equivalent.
We also note (Proposition~6.3 of~\cite{Tm1})
that if every element of $\QT (A)$ is faithful,
then the infimum in \Def{D_5519_ROfComp}(\ref{D_5519_ROfComp:2})
is attained,
that is, $A$ has ${\operatorname{rc}} (A)$comparison.
In particular,
this is true when $A$ is simple.
(See Lemma 1.23 of~\cite{Ph40}.)
% \label{L_4822_infdTauPos}
We warn that $r$comparison and ${\operatorname{rc}} (A)$
are sometimes defined using tracial states
rather than quasitraces.
It is equivalent to use $K \otimes A$
in place of $M_{\infty} (A)$.
See Proposition 6.12 of~\cite{Ph40}.
% \label{P_4817_rCompCu}
We prove here the following special case
of Theorem~\ref{T_5421_RCEq_0}.
\begin{thm}\label{T_5421_RCEq}
Let $A$ be an infinite dimensional stably finite
simple separable unital exact C*algebra.
Let $B \subset A$ be a large subalgebra.
Then ${\operatorname{rc}} (A) = {\operatorname{rc}} (B)$.
\end{thm}
The extra assumption is that $A$ is exact,
so that every quasitrace is a trace by Theorem 5.11 of~\cite{Hgp}.
We will give a proof directly from the definition
of a large subalgebra.
We describe the heuristic argument,
using the following simplifications:
\begin{enumerate}
\item\label{Smp_5522_UniqT}
The algebra~$A$,
and therefore also~$B$,
has a unique \tst~$\ta$.
\item\label{Smp_5522_JustA}
We consider elements of $A_{+}$ and $B_{+}$
instead of elements of $M_{\infty} (A)_{+}$ and $M_{\infty} (B)_{+}$.
\item\label{Smp_5522_Exact}
For $a \in A_{+}$,
when applying the definition of a large subalgebra
(Definition~\ref{D_5421_Large}),
instead of getting $(1  g) c (1  g) \in B$
for some $c \in A_{+}$ which is close to~$a$,
we can actually get $(1  g) a (1  g) \in B$.
Similarly,
for $a \in A$
we can get $(1  g) a \in B$.
\item\label{Smp_5522_CzE}
For $a, b \in A_{+}$ with $a \precsim_A b$,
we can find $v \in A$ such that $v^* b v = a$
(not just such that $\ v^* b v  a \$ is small).
\item\label{Smp_5522_NoPj}
None of the elements we encounter are
Cuntz equivalent to \pj{s},
that is, we never encounter anything for which $0$
is an isolated point of, or not in, the spectrum.
\end{enumerate}
The most drastic simplification is~(\ref{Smp_5522_Exact}).
In the actual proof,
to compensate for the fact that we only get approximation,
we will need to make systematic use of elements
$(a  \ep)_{+}$ for carefully chosen,
and varying,
values of $\ep > 0$.
Avoiding this complication
gives a much better view of the idea behind the argument,
and the usefulness of large subalgebras in general.
We first consider the inequality
${\operatorname{rc}} (A) \leq {\operatorname{rc}} (B)$.
So let $a, b \in A_{+}$ satisfy
$d_{\ta} (a) + {\operatorname{rc}} (B) < d_{\ta} (b)$.
The essential idea is to replace $b$ by something slightly
smaller which is in~$B_{+}$, say~$y$,
and replace $a$ by something slightly
larger which is in~$B_{+}$, say~$x$,
in such a way that we still have
$d_{\ta} (x) + {\operatorname{rc}} (B) < d_{\ta} (y)$.
Then use the definition of ${\operatorname{rc}} (B)$.
With $g$ sufficiently small in the sense of Cuntz comparison,
we will take $y = (1  g) b (1  g)$
and (following \Lem{L:C2}) $x = (1  g) a (1  g) \oplus g$.
Choose $\dt > 0$
such that
%
\begin{equation}\label{Eq_5601_Chdt}
d_{\ta} (a) + {\operatorname{rc}} (B) + \dt \leq d_{\ta} (b).
\end{equation}
%
Applying~(\ref{Smp_5522_Exact}) of our simplification,
we can find $g \in B$ with $0 \leq g \leq 1$,
such that
\[
(1  g) a (1  g) \in B
\andeqn
(1  g) b (1  g) \in B,
\]
and so small in $W (A)$ that
%
\begin{equation}\label{Eq_5522_gdt3}
d_{\ta} (g) < \frac{\dt}{3}.
\end{equation}
%
Using Lemma~\ref{L:CzBasic}(\ref{L:CzBasic:LCzComm}) at the first step,
we get
\[
(1  g) b (1  g)
\sim_A b^{1/2} (1  g)^2 b^{1/2}
\leq b,
\]
so
%
\begin{equation}\label{Eq_5522_gb1b1}
(1  g) b (1  g)
\precsim_A b.
\end{equation}
%
%
Similarly,
$(1  g) a (1  g) \precsim_A a$,
and this relation implies
%
\begin{equation}\label{Eq_5522_ga}
d_{\ta} \big( (1  g) a (1  g) \big) \leq d_{\ta} (a).
\end{equation}
%
Also,
$b \precsim_A (1  g) b (1  g) \oplus g$
by \Lem{L:C2},
so
%
\begin{equation}\label{Eq_5522_bgm}
d_{\ta} \big( (1  g) b (1  g) \big) + d_{\ta} (g)
\geq d_{\ta} ( b ).
\end{equation}
%
Using (\ref{Eq_5522_ga}) at the first step,
using (\ref{Eq_5601_Chdt}) at the second step,
using (\ref{Eq_5522_bgm}) at the third step,
and using (\ref{Eq_5522_gdt3}) at the fourth step,
we get
\begin{align*}
d_{\ta} \big( (1  g) a (1  g) \oplus g \big) + \rc (B)
+ \frac{\dt}{3}
& \leq d_{\ta} (a) + d_{\ta} (g) + \rc (B) + \frac{\dt}{3}
\\
& \leq d_{\ta} (b) + d_{\ta} (g)  \frac{2 \dt}{3}
\\
& \leq d_{\ta} \big( (1  g) b (1  g) \big) + 2 d_{\ta} (g)
 \frac{2 \dt}{3}
\\
& \leq d_{\ta} \big( (1  g) b (1  g) \big).
\end{align*}
So, by the definition of ${\operatorname{rc}} (B)$,
\[
(1  g) a (1  g) \oplus g \precsim_B (1  g) b (1  g).
\]
Therefore,
using \Lem{L:C2} at the first step
and (\ref{Eq_5522_gb1b1}) at the third step,
we get
\[
a \precsim_A (1  g) a (1  g) \oplus g
\precsim_B (1  g) b (1  g)
\precsim_A b,
\]
that is, $a \precsim_A b$, as desired.
Now we consider the inequality
${\operatorname{rc}} (A) \geq {\operatorname{rc}} (B)$.
Let $a, b \in B_{+}$ satisfy
$d_{\ta} (a) + {\operatorname{rc}} (A) < d_{\ta} (b)$.
Choose $\dt > 0$
such that
$d_{\ta} (a) + {\operatorname{rc}} (A) + \dt \leq d_{\ta} (b)$.
By lower semicontinuity of~$d_{\ta}$,
we always have
\[
d_{\ta} (b) = \sup_{\ep > 0} d_{\ta} \big( (b  \ep)_{+} \big).
\]
So there is $\ep > 0$ such that
% with $r = (b  \ep)_{+}$,
%
\begin{equation}\label{Eq_5522_dtaub1}
d_{\ta} \big( (b  \ep)_{+} \big)
> d_{\ta} (a) + {\operatorname{rc}} (A).
\end{equation}
%
Define a \cfn{}
$f \colon [0, \I) \to [0, \I)$
by $f (\ld) = \max ( 0, \, \ep^{1} \ld (\ep  \ld))$
for $\ld \in [0, \I)$.
Then $f (b)$ and $(b  \ep)_{+}$ are orthogonal positive elements
such that $f (b) \neq 0$
(by~(\ref{Smp_5522_NoPj}))
and $f (b) + (b  \ep)_{+} \leq b$.
We have
$a \precsim_A (b  \ep)_{+}$
by~(\ref{Eq_5522_dtaub1})
and the definition of ${\operatorname{rc}} (A)$.
Applying~(\ref{Smp_5522_CzE}) of our simplification,
we can find $v \in A$ such that $v^* (b  \ep)_{+} v = a$.
Applying~(\ref{Smp_5522_Exact}) of our simplification,
we can find $g \in B$ with $0 \leq g \leq 1$
such that $(1  g) v^* \in B$ and $g \precsim_B f (b)$.
Since
\[
v (1  g) \in B
\andeqn
[v (1  g)]^* (b  \ep)_{+} [v (1  g)] = (1  g) a (1  g),
\]
we get
$(1  g) a (1  g) \precsim_B (b  \ep)_{+}$.
Therefore,
using \Lem{L:C2} at the first step,
\[
a \precsim_B (1  g) a (1  g) \oplus g
\precsim_B (b  \ep)_{+} \oplus g
\precsim_B (b  \ep)_{+} \oplus f (b)
\precsim_B b,
\]
as desired.
The original proof of
Theorem~\ref{T_5421_RCEq}
followed the heuristic arguments above,
and this is the proof we give below.
The proof in~\cite{Ph40}
uses the same basic ideas,
but gives much more.
The heuristic arguments above are the basis for
the technical results in \Lem{L2720SDomToB}.
In~\cite{Ph40}, these are used to prove \Thm{T_5522_IsoOnPure},
which states that,
after deleting the classes of the nonzero projections
from the Cuntz semigroups ${\operatorname{Cu}} (B)$
and ${\operatorname{Cu}} (A)$,
the inclusion of $B$ in~$A$
is an order isomorphism on what remains.
(The inclusion need not be an isomorphism
if the classes of the nonzero projections are included.
See Example 7.13 of~\cite{Ph40}.)
In Section~3 of~\cite{Ph40}, it is shown that,
in our situation,
the part of the Cuntz semigroup
without the classes of the nonzero projections
is enough to determine the quasitraces,
so that the restriction map $\QT (A) \to \QT (B)$
is bijective.
It follows that the radius of comparison in this part of the
Cuntz semigroup is the same for both $A$ and~$B$,
and it turns out that the radius of comparison in this part of the
Cuntz semigroup is the same as in the entire Cuntz semigroup.
We will use the characterizations
of ${\operatorname{rc}} (A)$ in the following theorem,
which is a special case of results in~\cite{BRTTW}.
The difference between (\ref{T_5522_RCCrit:1S})
and~(\ref{T_5522_RCCrit:2}) is that
(\ref{T_5522_RCCrit:2})~has $n + 1$ in one of the places where
(\ref{T_5522_RCCrit:1S}) has~$n$.
This result substitutes
for the observation
that if $a, b \in A_{+}$
satisfy $\ta (a) < \ta (b)$ for all $\ta \in \QT (A)$,
then,
by compactness of $\QT (A)$ and continuity,
we have
$\inf_{\ta \in \QT (A)} [\ta (b)  \ta (a)] > 0$.
The difficulty is that we need an analog
using $d_{\ta}$ instead of~$\ta$,
and $\ta \mapsto d_{\ta} (a)$ is in general only lower semicontinuous,
so that $\ta \mapsto d_{\ta} (b)  d_{\ta} (a)$
may be neither upper nor lower semicontinuous.
Unfortunately,
the results in~\cite{BRTTW}
are stated in terms of ${\operatorname{Cu}} (A)$
rather than $W (A)$.
\begin{thm}\label{T_5522_RCCrit}
Let $A$ be a stably finite simple unital C*algebra.
Then:
\begin{enumerate}
\item\label{T_5522_RCCrit:1S}
The radius of comparison ${\operatorname{rc}} (A)$
is the least number $s \in [0, \infty]$
such that whenever $m, n \in \N$ satisfy
$m / n > s$,
and
$a, b \in M_{\infty} (A)_{+}$ satisfy
\[
n \langle a \rangle_A + m \langle 1 \rangle_A
\leq n \langle b \rangle_A
\]
in $W (A)$,
then $a \precsim_A b$.
\item\label{T_5522_RCCrit:2}
The radius of comparison ${\operatorname{rc}} (A)$
is the least number $t \in [0, \infty]$
such that whenever $m, n \in \N$ satisfy
$m / n > t$,
and
$a, b \in M_{\infty} (A)_{+}$ satisfy
\[
(n + 1) \langle a \rangle_A + m \langle 1 \rangle_A
\leq n \langle b \rangle_A
\]
in $W (A)$,
then $a \precsim_A b$.
\end{enumerate}
\end{thm}
\begin{proof}
It is easy to check that there is in fact a least
$s \in [0, \infty]$
satisfying the condition in~(\ref{T_5522_RCCrit:1S}),
and similarly that there is a least
$t \in [0, \infty]$
as in~(\ref{T_5522_RCCrit:2}).
We will first prove this
for $K \otimes A$ and ${\operatorname{Cu}} (A)$
in place of $M_{\infty} (A)$ and $W (A)$.
So let $s_0$ and $t_0$ be the numbers
defined as in (\ref{T_5522_RCCrit:1S}) and~(\ref{T_5522_RCCrit:2}),
except
with $K \otimes A$ and ${\operatorname{Cu}} (A)$
in place of $M_{\infty} (A)$ and $W (A)$.
Again, it is clear that there are least such numbers $s_0$ and~$t_0$.
Clearly $s_0 \geq t_0$.
Since $A$ is simple and stably finite and $\langle 1 \rangle_A$
is a full element of ${\operatorname{Cu}} (A)$,
Proposition 3.2.3 of~\cite{BRTTW},
the preceding discussion in~\cite{BRTTW},
and Definition 3.2.2 of~\cite{BRTTW}
give $t_0 = {\operatorname{rc}} (A)$.
So we need to show that $s_0 \leq t_0$.
We thus assume $m, n \in \N$ and $m / n > t_0$,
and that $a, b \in (K \otimes A)_{+}$ satisfy
$n \langle a \rangle_A + m \langle 1 \rangle_A
\leq n \langle b \rangle_A$
in ${\operatorname{Cu}} (A)$.
We must prove that $a \precsim_A b$.
For any functional $\om$ on ${\operatorname{Cu}} (A)$
(as at the beginning of Section~2.4 of~\cite{BRTTW}),
we have
$n \om (\langle a \rangle_A) + m \om (\langle 1 \rangle_A)
\leq n \om (\langle b \rangle_A)$,
so
$\om (\langle a \rangle_A) + (m/n) \om (\langle 1 \rangle_A)
\leq \om (\langle b \rangle_A)$.
Since $m / n > t_0$,
Proposition 3.2.1 of~\cite{BRTTW}
implies that $a \precsim_A b$.
It remains to prove that
$s_0 = s$ and $t_0 = t$.
We prove that $s_0 = s$; the proof that $t_0 = t$ is the same.
Let $m, n \in \N$.
We have to prove the following.
Suppose that $m$ and $n$ have the property that
whenever $a, b \in M_{\infty} (A)_{+}$ satisfy
\[
n \langle a \rangle_A + m \langle 1 \rangle_A
\leq n \langle b \rangle_A
\]
in $W (A)$,
then $a \precsim_A b$.
Then whenever $a, b \in (K \otimes A)_{+}$ satisfy
\[
n \langle a \rangle_{K \otimes A} + m \langle 1 \rangle_{K \otimes A}
\leq n \langle b \rangle_{K \otimes A}
\]
in ${\operatorname{Cu}} (A)$,
we have $a \precsim_{K \otimes A} b$.
We also need to prove the reverse implication.
The reverse implication is easy,
so we prove the forwards implication.
Let $a, b \in (K \otimes A)_{+}$ satisfy
\[
n \langle a \rangle_A + m \langle 1 \rangle_A
\leq n \langle b \rangle_A
\]
in ${\operatorname{Cu}} (A)$.
Let $\ep > 0$;
by \Lem{L:CzBasic}(\ref{L:CzBasic:LMinusEp}),
it suffices to prove that $(a  \ep)_{+} \precsim_{K \otimes A} b$.
We may clearly assume that $\ep < 1$.
Using an isomorphism $K \otimes K \to K$,
let $x \in (K \otimes A)_{+}$ be the direct sum of $n$ copies of $a$,
let $y \in (K \otimes A)_{+}$ be the direct sum of $n$ copies of $b$,
and let $q \in (K \otimes A)_{+}$
be the direct sum of $m$ copies of the identity of~$A$.
The relation
$n \langle a \rangle_{K \otimes A} + m \langle 1 \rangle_{K \otimes A}
\leq n \langle b \rangle_{K \otimes A}$
means that
$x \oplus q \precsim_{K \otimes A} y$.
By \Lem{L:CzBasic}(\ref{L:CzBasic:LMinusEp:3}),
there exists $\dt > 0$ such that
\[
\big( (x \oplus q)  \ep \big)_{+}
\precsim_{K \otimes A} (y  \dt)_{+}.
\]
Since $\ep < 1$ and $q$ is a \pj,
this relation is equivalent to
\[
( x  \ep )_{+} \oplus q \precsim_{K \otimes A} (y  \dt)_{+}.
\]
Since $( x  \ep )_{+}$
is the direct sum of $n$ copies of $( a  \ep )_{+}$
and $(y  \dt)_{+}$
is the direct sum of $n$ copies of $(b  \dt)_{+}$,
we therefore have
\[
n \langle ( a  \ep )_{+} \rangle_{K \otimes A}
+ m \langle 1 \rangle_{K \otimes A}
\leq n \langle (b  \dt)_{+} \rangle_{K \otimes A}.
\]
It follows from
Lemma~1.9 of~\cite{Ph40}
% \label{L2720KToMn}
that $\langle ( a  \ep )_{+} \rangle_{K \otimes A}$
and $\langle (b  \dt)_{+} \rangle_{K \otimes A}$
are actually classes of elements
$c, d \in M_{\infty} (A)_{+}$,
and it is easy to check that inequalities among classes in $W (A)$
which hold in
${\operatorname{Cu}} (A)$
must also hold in $W (A)$.
The assumption therefore implies that
$c \precsim_A d$.
Thus
\[
( a  \ep )_{+}
\sim_{K \otimes A} c
\precsim_A d
\sim_{K \otimes A} (b  \dt)_{+}
\leq b,
\]
whence $(a  \ep)_{+} \precsim_{K \otimes A} b$, as desired.
\end{proof}
\begin{lem}\label{L:FCF}
Let $M \in (0, \I)$,
let $f \colon [0, \infty) \to \C$ be a \cfn{}
such that $f (0) = 0$,
and let $\ep > 0$.
Then there is $\dt > 0$ such that whenever
$A$ is a C*algebra and
$a, b \in A_{\mathrm{sa}}$ satisfy $\ a \ \leq M$
and $\ a  b \ < \dt$,
then $\ f (a)  f (b) \ < \ep$.
\end{lem}
This is a standard polynomial approximation argument.
We have not found it written in the literature.
There are similar arguments in~\cite{Ph40}
and many other places.
It is also stated (in a slightly different form)
as Lemma 2.5.11(2) of~\cite{LnBook};
the proof there is left to the reader
(although a related proof is given).
We therefore give it for completeness.
\begin{proof}[Proof of \Lem{L:FCF}]
Choose $n \in \N$
and $\af_1, \af_2, \ldots, \af_n \in \C$
such that the polynomial function
$g (\ld) = \sum_{k = 1}^n \af_k \ld^k$
satisfies
$ g (\ld)  f (\ld)  < \frac{\ep}{3}$
for $\ld \in [ M  1, \, M + 1]$.
Define
\[
\dt = \min \left( 1, \,
\frac{\ep}{1 + 3 \sum_{k = 1}^n  \af_k  k (M + 1)^{k  1}} \right).
\]
Now let $A$ be a C*algebra and
let $a, b \in A_{\mathrm{sa}}$ satisfy $\ a \ \leq M$
and $\ a  b \ < \dt$.
Then $\ b \ \leq M + 1$.
So for $m \in \N$ we have
\[
\ a^m  b^m \
\leq \sum_{k = 1}^m
\ a^{k  1} \ \cdot \ a  b \ \cdot \ b^{m  k} \
% \leq m (M + 1)^{m  1} \ a  b \
< m (M + 1)^{m  1} \dt.
\]
Therefore
\[
\ g (a)  g (b) \
\leq \sum_{k = 1}^n  \af_k  k (M + 1)^{k  1} \dt
< \frac{\ep}{3}.
\]
So
\[
\ f (a)  f (b) \
\leq \ f (a)  g (a) \ + \ g (a)  g (b) \ + \ g (b)  f (b) \
< \frac{\ep}{3} + \frac{\ep}{3} + \frac{\ep}{3}
= \ep.
\]
This completes the proof.
\end{proof}
\begin{prp}\label{T:RCAndTL}
Let $A$ be an in\fd{} stably finite simple separable unital
exact C*algebra.
Let $B \subset A$ be a large subalgebra.
Then ${\operatorname{rc}} (A) \leq {\operatorname{rc}} (B)$.
\end{prp}
\begin{proof}
We use the criterion
of Theorem \ref{T_5522_RCCrit}(\ref{T_5522_RCCrit:1S}).
Thus, let $m, n \in \N$ satisfy $m / n > {\operatorname{rc}} (B)$,
and let $a, b \in M_{\infty} (A)_{+}$ satisfy
$n \langle a \rangle_A + m \langle 1 \rangle_A
\leq n \langle b \rangle_A$
in $W (A)$.
We want to prove that $a \precsim_A b$.
\Wolog{} $\ a \, \ b \ \leq 1$.
It suffices to prove that $(a  \ep )_{+} \precsim_A b$
for every $\ep > 0$.
So let $\ep > 0$.
We may assume $\ep < 1$.
Let $x \in M_{\infty} (A)_{+}$ be the direct sum of $n$ copies of $a$,
let $y \in M_{\infty} (A)_{+}$ be the direct sum of $n$ copies of $b$,
and let $q \in M_{\infty} (A)_{+}$
be the direct sum of $m$ copies of the identity of~$A$.
The relation
$n \langle a \rangle_A + m \langle 1 \rangle_A
\leq n \langle b \rangle_A$
means that
$x \oplus q \precsim_A y$.
By \Lem{L:CzBasic}(\ref{L:CzBasic:LMinusEp:2}),
there exists $\dt > 0$ such that
\[
\big( (x \oplus q)  \tfrac{1}{3} \ep \big)_{+}
\precsim_A (y  \dt)_{+}.
\]
Since $\ep < 3$, this is equivalent to
%
\begin{equation}\label{Eq_5512_Star}
\big( x  \tfrac{1}{3} \ep \big)_{+} \oplus q \precsim_A (y  \dt)_{+}.
\end{equation}
%
Choose $l \in \N$ so large that $a, b \in M_l \otimes A$.
Since $m / n > {\operatorname{rc}} (B)$,
there is $k \in \N$ such that
\[
{\operatorname{rc}} (B) < \frac{m}{n}  \frac{2}{k}.
\]
Set
\[
\ep_0 = \min \big( \tfrac{1}{3} \ep, \tfrac{1}{2} \dt \big).
\]
Using \Lem{L:FCF},
choose $\ep_1 > 0$ with $\ep_1 \leq \ep_0$ and so small
that whenever $D$ is a \ca{}
and $z \in D_{+}$ satisfies $\ z \ \leq 1$,
then $\ z_0  z \ < \ep_1$
implies
\[
\ (z_0  \ep_0 )_{+}  (z  \ep_0 )_{+} \ < \ep_0,
\,\,\,\,\,\,
\big\ \big( z_0  \tfrac{1}{3} \ep \big)_{+}
 \big( z  \tfrac{1}{3} \ep \big)_{+}\big \
< \ep_0,
\]
and
\[
\big\ \big( z_0  \big( \ep_0 + \tfrac{1}{3} \ep \big) \big)_{+}
 \big( z  \big( \ep_0 + \tfrac{1}{3} \ep \big) \big)_{+} \big \
< \ep_0.
\]
Since $A$ is infinite dimensional and simple,
\Lem{L:DivInSmp}
provides $z \in A_{+} \setminus \{ 0 \}$
such that $(k + 1) \langle z \rangle_A \leq \langle 1 \rangle_A$.
Using Proposition~\ref{P_5519_StFinStLg}
and Lemma~\ref{L:LargeStaysPositive},
choose $g \in M_l (B)_{+}$
and $a_0, b_0 \in M_l (A)_{+}$
satisfying
\[
0 \leq g, a_0, b_0 \leq 1,
\,\,\,\,\,\,
\ a_0  a \ < \ep_1,
\,\,\,\,\,\,
\ b_0  b \ < \ep_1,
\,\,\,\,\,\,
g \precsim_A z,
\]
and such that
\[
(1  g) a_0 (1  g), \, (1  g) b_0 (1  g) \in M_l \otimes B.
\]
{}From $g \precsim_A z$ and
$(k + 1) \langle z \rangle_A \leq \langle 1 \rangle_A$
we get
%
\begin{equation}\label{Eq_5512_StSt}
\sup_{\ta \in \T (A)} d_{\ta} (g) < \frac{1}{k}.
\end{equation}
%
Set
\[
a_1 = \big[ (1  g) a_0 (1  g)
 \big( \ep_0 + \tfrac{1}{3} \ep \big) \big]_{+}
\andeqn
b_1 = \big[ (1  g) b_0 (1  g)  \ep_0 \big]_{+},
\]
which are in~$M_l \otimes B$.
We claim that $a_0$, $a_1$, $b_0$, and $b_1$
have the following properties:
\begin{enumerate}
\item\label{T:FPPImpSCPE:1a}
$(a  \ep )_{+}
\precsim_A
\big[ a_0  \big( \ep_0 + \tfrac{1}{3} \ep \big) \big]_{+}$.
\item\label{T:FPPImpSCPE:2a}
$\big[ a_0  \big( \ep_0 + \tfrac{1}{3} \ep \big) \big]_{+}
\precsim_B a_1 \oplus g$.
\item\label{T:FPPImpSCPE:3a}
$a_1 \precsim_A \big( a  \tfrac{1}{3} \ep \big)_{+}$.
\item\label{T:FPPImpSCPE:1b}
$(b  \dt )_{+} \precsim_A (b_0  \ep_0 )_{+}$.
\item\label{T:FPPImpSCPE:2b}
$(b_0  \ep_0 )_{+} \precsim_B b_1 \oplus g$.
\item\label{T:FPPImpSCPE:3b}
$b_1 \precsim_A b$.
\end{enumerate}
We give full details of the proofs for
(\ref{T:FPPImpSCPE:1a}), (\ref{T:FPPImpSCPE:2a}),
and~(\ref{T:FPPImpSCPE:3a})
(involving $a_0$ and $a_1$).
The proofs for
(\ref{T:FPPImpSCPE:1b}), (\ref{T:FPPImpSCPE:2b}),
and~(\ref{T:FPPImpSCPE:3b}) (involving $b_0$ and $b_1$)
are a bit more sketchy.
We prove~(\ref{T:FPPImpSCPE:1a}).
Since $\ a_0  a \ < \ep_1$,
the choice of $\ep_1$ implies
\[
\big\ \big[ a_0  \big( \tfrac{1}{3} \ep + \ep_0 \big) \big]_{+}
 \big[ a  \big( \tfrac{1}{3} \ep + \ep_0 \big) \big]_{+} \big\
< \ep_0
\leq \tfrac{1}{3} \ep.
\]
At the last step in the following computation use this inequality
and \Lem{L:CzBasic}(\ref{L:CzBasic:LCzWithinEp}),
at the first step use $\ep_0 \leq \tfrac{1}{3} \ep$,
and at the second step use
\Lem{L:CzBasic}(\ref{L:CzBasic:MinIter}),
to get
\begin{align*}
(a  \ep )_{+}
& \leq \big[ a  \big( \tfrac{2}{3} \ep + \ep_0 \big) \big]_{+}
\\
& = \big[ \big( a  \big( \tfrac{1}{3} \ep + \ep_0 \big) \big)_{+}
 \tfrac{1}{3} \ep \big]_{+}
\precsim_A
\big[ a_0  \big( \tfrac{1}{3} \ep + \ep_0 \big) \big]_{+}.
\end{align*}
For~(\ref{T:FPPImpSCPE:1b})
(the corresponding argument for $b_0$),
we use $\ep_0 \leq \tfrac{1}{2} \dt$ at the first step;
since
\[
\ (b  \ep_0 )_{+}  (b_0  \ep_0 )_{+} \ < \ep_0,
\]
we get
\[
(b  \dt)_{+}
\leq (b  2 \ep_0)_{+}
= \big[ (b  \ep_0)_{+}  \ep_0 \big]_{+}
\precsim_A (b_0  \ep_0 )_{+}.
\]
For~(\ref{T:FPPImpSCPE:2a}),
we use \Lem{L:C2} with $a_0$ in place of $a$
and with $\tfrac{1}{3} \ep + \ep_0$ in place of~$\ep$.
For~(\ref{T:FPPImpSCPE:2b}),
we use \Lem{L:C2} with $b_0$ in place of $a$
and with $\ep_0$ in place of~$\ep$.
For~(\ref{T:FPPImpSCPE:3a}),
begin by recalling that $\ a_0  a \ < \ep_1$, whence
\[
\ (1  g) a_0 (1  g)  (1  g) a (1  g) \ < \ep_1.
\]
Therefore
\[
\big\ \big[ (1  g) a_0 (1  g)  \tfrac{1}{3} \ep \big]_{+}
 \big[ (1  g) a (1  g)  \tfrac{1}{3} \ep \big]_{+} \big\
< \ep_0.
\]
Using Lemma~\ref{L:CzBasic}(\ref{L:CzBasic:MinIter})
at the first step,
this fact
and \Lem{L:CzBasic}(\ref{L:CzBasic:LCzWithinEp})
at the second step,
\Lem{L:CzBasic}(\ref{L:CzBasic:LCzCommEp})
at the third step,
and Lemma~\ref{L:CzBasic}(\ref{L:CzBasic:L:CzCompIneq})
and $a^{1/2} (1  g)^2 a^{1/2} \leq a$
at the last step,
we get
\begin{align*}
a_1
& = \big[ \big[ (1  g) a_0 (1  g)
 \tfrac{1}{3} \ep \big]_{+}  \ep_0 \big]_{+}
\\
& \precsim_A \big[ (1  g) a (1  g)  \tfrac{1}{3} \ep \big]_{+}
\sim_A \big[ a^{1/2} (1  g)^2 a^{1/2}  \tfrac{1}{3} \ep \big]_{+}
\precsim_A \big( a  \tfrac{1}{3} \ep \big)_{+},
\end{align*}
as desired.
For~(\ref{T:FPPImpSCPE:3b})
(the corresponding part involving $b_1$),
just use
\[
\ (1  g) b_0 (1  g)  (1  g) b (1  g) \ < \ep_1 \leq \ep_0
\]
to get,
using \Lem{L:CzBasic}(\ref{L:CzBasic:LCzComm})
at the second step,
\[
b_1
\precsim_A (1  g) b (1  g)
\sim_A b^{1/2} (1  g)^2 b^{1/2}
\leq b.
\]
The claims (\ref{T:FPPImpSCPE:1a})(\ref{T:FPPImpSCPE:3b})
are now proved.
Now let $\ta \in \T (A)$.
Recall that $x$ and $y$
are the direct sums of $n$ copies of $a$ and~$b$.
Therefore $\big( x  \tfrac{1}{3} \ep \big)_{+}$
is the direct sum of $n$ copies of
$\big( a  \tfrac{1}{3} \ep \big)_{+}$
and $(y  \dt)_{+}$
is the direct sum of $n$ copies of $(b  \dt)_{+}$.
So the relation~(\ref{Eq_5512_Star})
implies
%
\begin{equation}\label{Eq_5513_nm}
n \cdot d_{\ta} \big( \big( a  \tfrac{1}{3} \ep \big)_{+} \big) + m
\leq n \cdot d_{\ta} \big( (b  \dt)_{+} \big).
\end{equation}
%
Using (\ref{T:FPPImpSCPE:1b}) and~(\ref{T:FPPImpSCPE:2b})
at the first step and~(\ref{Eq_5512_StSt})
at the third step, we get the estimate
%
\begin{equation}\label{Eq_5513_bk}
d_{\ta} \big( (b  \dt)_{+} \big)
\leq d_{\ta} (b_1) + d_{\ta} (g)
< d_{\ta} (b_1) + k^{1}.
\end{equation}
%
The relation~(\ref{T:FPPImpSCPE:3a}) implies
%
\begin{equation}\label{Eq_5513_a1a}
d_{\ta} (a_1)
\leq d_{\ta} \big( \big( a  \tfrac{1}{3} \ep \big)_{+} \big).
\end{equation}
%
Using~(\ref{Eq_5513_a1a}) and
(\ref{Eq_5512_StSt}) at the second step,
(\ref{Eq_5513_nm}) at the third step,
and (\ref{Eq_5513_bk}) at the fourth step,
we get
\begin{align*}
n \cdot d_{\ta} (a_1 \oplus g) + m
& = n \cdot d_{\ta} (a_1) + m + n \cdot d_{\ta} (g)
% \\
% & \leq n \cdot d_{\ta} (a_1) + m + n k^{1}
\\
& \leq n \cdot d_{\ta} \big( \big( a  \tfrac{1}{3} \ep \big)_{+} \big)
+ m + n k^{1}
\\
& \leq n \cdot d_{\ta} \big( (b  \dt)_{+} \big) + n k^{1}
\\
& \leq n \cdot d_{\ta} (b_1) + 2 n k^{1}.
\end{align*}
It follows that
\[
d_{\ta} (a_1 \oplus g) + \frac{m}{n}  \frac{2}{k}
\leq d_{\ta} (b_1).
\]
This holds for all $\ta \in \T (A)$,
and therefore, by \Thm{T_5522_SameT},
for all $\ta \in \T (B)$.
Subalgebras of exact \ca{s} are exact
(by Proposition 7.1(1) of~\cite{Krh2}),
so Theorem 5.11 of~\cite{Hgp}
implies that $\QT (B) = \T (B)$.
Since
\[
\frac{m}{n}  \frac{2}{k} > {\operatorname{rc}} (B),
\]
and since $a_1, b_1, g \in M_l \otimes B$,
it follows that $a_1 \oplus g \precsim_B b_1$.
Using this relation at the third step,
(\ref{T:FPPImpSCPE:1a}) at the first step,
(\ref{T:FPPImpSCPE:2a}) at the second step,
and (\ref{T:FPPImpSCPE:3b}) at the last step,
we then get
\[
(a  \ep )_{+}
\precsim_A \big[ a_0  \big( \ep_0 + \tfrac{1}{3} \ep \big) \big]_{+}
\precsim_A a_1 \oplus g
\precsim_B b_1
\precsim_A b.
\]
This completes the proof that
${\operatorname{rc}} (A) \leq {\operatorname{rc}} (B)$.
\end{proof}
\begin{prp}\label{P_5523_RCAndTL2}
Let $A$ be an in\fd{} stably finite simple separable unital
exact C*algebra.
Let $B \subset A$ be a large subalgebra.
Then ${\operatorname{rc}} (A) \geq {\operatorname{rc}} (B)$.
\end{prp}
\begin{proof}
We use Theorem~\ref{T_5522_RCCrit}(\ref{T_5522_RCCrit:2}).
Thus, let $m, n \in \N$ satisfy $m / n > {\operatorname{rc}} (A)$.
Let $l \in \N$, and let $a, b \in (M_l \otimes B)_{+}$ satisfy
\[
(n + 1) \langle a \rangle_B + m \langle 1 \rangle_B
\leq n \langle b \rangle_B
\]
in $W (B)$.
We must prove that $a \precsim_B b$.
\Wolog{} $\ a \ \leq 1$.
Moreover, by \Lem{L:CzBasic}(\ref{L:CzBasic:LMinusEp}),
it is enough to show that for every $\ep > 0$
we have $(a  \ep)_{+} \precsim_B b$.
So let $\ep > 0$.
\Wolog{} $\ep < 1$.
Choose $k \in \N$ such that
\[
\frac{k m}{k n + 1} > {\operatorname{rc}} (A).
\]
Then in $W (B)$ we have
\[
(k n + 1) \langle a \rangle_B + k m \langle 1 \rangle_B
\leq k (n + 1) \langle a \rangle_B + k m \langle 1 \rangle_B
\leq k n \langle b \rangle_B.
\]
Let $x \in M_{\infty} (B)_{+}$
be the direct sum of $k n + 1$ copies of~$a$,
let $z \in M_{\infty} (B)_{+}$
be the direct sum of $k n$ copies of~$b$,
and let $q \in M_{\infty} (B)_{+}$
be the direct sum of $k m$ copies of~$1$.
Then, by definition,
$x \oplus q \precsim_B z$.
Therefore \Lem{L:CzBasic}(\ref{L:CzBasic:LMinusEp})
provides $\dt > 0$ such that
$\big( x \oplus q  \tfrac{1}{4} \ep \big)_{+}
\precsim_B (z  \dt)_{+}$.
Since $\ep < 4$, we have
\[
\big( x \oplus q  \tfrac{1}{4} \ep \big)_{+}
= \big( x  \tfrac{1}{4} \ep \big)_{+}
\oplus \big( q  \tfrac{1}{4} \ep \big)_{+}
\sim_B \big( x  \tfrac{1}{4} \ep \big)_{+} \oplus q,
\]
so
\[
(k n + 1)
\big\langle \big( a  \tfrac{1}{4} \ep \big)_{+} \big\rangle_B
+ k m \langle 1 \rangle_B
\leq k n \langle (b  \dt)_{+} \rangle_B.
\]
\Lem{L:Smaller} provides $c \in (M_l \otimes B)_{+}$
and $y \in (M_l \otimes B)_{+} \setminus \{ 0 \}$
such that
%
\begin{equation}\label{Eq_5513_cyb}
k n \langle ( b  \dt )_{+} \rangle_B
\leq (k n + 1) \langle c \rangle_B
\andeqn
\langle c \rangle_B + \langle y \rangle_B \leq \langle b \rangle_B
\end{equation}
%
in $W (B)$.
Then
\[
(k n + 1)
\big\langle \big( a  \tfrac{1}{4} \ep \big)_{+} \big\rangle_B
+ k m \langle 1 \rangle_B
\leq (k n + 1) \langle c \rangle_B.
\]
Applying the map $W (A) \to W (B)$, we get
\[
(k n + 1)
\big\langle \big( a  \tfrac{1}{4} \ep \big)_{+} \big\rangle_A
+ k m \langle 1 \rangle_A
\leq (k n + 1) \langle c \rangle_A.
\]
For $\ta \in \T (A)$,
we apply $d_{\ta}$ and divide by $k n + 1$
to get
\[
d_{\ta} \big( \big\langle
\big( a  \tfrac{1}{4} \ep \big)_{+} \big\rangle \big)
+ \frac{k m}{k n + 1}
\leq d_{\ta} (c).
\]
Since $\QT (A) = \T (A)$
(by Theorem 5.11 of~\cite{Hgp})
and
\[
\frac{k m}{k n + 1} > {\operatorname{rc}} (A),
\]
it follows that $\big( a  \tfrac{1}{4} \ep \big)_{+} \precsim_A c$.
In particular,
there is $v \in M_l \otimes A$
such that
\[
\big\ v c v^*  \big( a  \tfrac{1}{4} \ep \big)_{+} \big\
< \tfrac{1}{4} \ep.
\]
Since $B$ is large in~$A$,
we can apply Proposition~\ref{P_5519_StFinStLg}
and \Lem{L:LargeStaysPositive} to find
$v_0 \in M_l \otimes A$ and $g \in M_l \otimes B$
with $0 \leq g \leq 1$ and
such that
\[
g \precsim_B y,
\,\,\,\,\,\,
\ v_0 \ \leq \ v \,
\,\,\,\,\,\,
\ v_0  v \ < \frac{\ep}{4 \ v \ \ c \ + 1},
\andeqn
(1  g) v_0 \in M_l \otimes B.
\]
It follows that
$\ v_0^* c v_0  v^* c v \ < \frac{\ep}{2}$,
so
%
\[
% begin{equation}\label{Eq_5513_gby}
\big\ (1  g) v_0 c [(1  g) v_0]^*
 (1  g) \big( a  \tfrac{1}{4} \ep \big)_{+} (1  g) \big\
< \tfrac{3}{4} \ep.
\]
% end{equation}
%
Therefore,
using Lemma~\ref{L:CzBasic}(\ref{L:CzBasic:LCzWithinEp})
at the first step,
%
\begin{equation}\label{Eq_5513_gac}
\big[ (1  g) \big( a  \tfrac{1}{4} \ep \big)_{+} (1  g)
 \tfrac{3}{4} \ep \big]_{+}
\precsim_B (1  g) v_0 c [(1  g) v_0]^*
\precsim_B c.
\end{equation}
%
Using \Lem{L:C2} at the first step,
with $\big( a  \tfrac{1}{4} \ep \big)_{+}$ in place of $a$ and
$\tfrac{3}{4} \ep$ in place of~$\ep$,
as well as Lemma~\ref{L:CzBasic}(\ref{L:CzBasic:MinIter}),
using~(\ref{Eq_5513_gac}) at the second step,
using the choice of~$g$ at the third step,
and using the second part of~(\ref{Eq_5513_cyb}) at the fourth step,
we get
\[
(a  \ep)_{+}
\precsim_B
\big[ (1  g) \big( a  \tfrac{1}{4} \ep \big)_{+} (1  g)
 \tfrac{3}{4} \ep \big]_{+}
\oplus g
\precsim_B c \oplus g
\precsim_B c \oplus y
\precsim_B b.
\]
This is the relation we need,
and the proof is complete.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{T_5421_RCEq}]
Combine Proposition~\ref{T:RCAndTL}
and Proposition~\ref{P_5523_RCAndTL2}.
\end{proof}
\section{Large Subalgebras in Crossed Products
by~${\mathbb{Z}}$}\label{Sec_AY}
\indent
In this section, we prove that if $X$ is an infinite
\cms,
$h \colon X \to X$ is a \mh,
and $Y \S X$ is closed and intersects each orbit of~$h$ at most once,
then the $Y$orbit breaking subalgebra $C^* (\Z, X, h)_Y$
of \Def{D_5421_VSubalg}
is a centrally large subalgebra of $C^* (\Z, X, h)$.
For easy reference,
we summarize the relevant crossed product notation.
This summary combines parts of
\Def{DFromSpToCStar},
Notation~\ref{N:ug},
and \Def{D_StdCond}.
\begin{ntn}\label{NOld32}
Let $X$ be a \cms,
and let $h \colon X \to X$ be a \hme.
We take the corresponding automorphism
$\af \in \Aut (C (X))$ to be given by
$\af (f) (x) = f (h^{1} (x))$ for $f \in C (X)$
and $x \in X$.
The crossed product is $C^* (\Z, X, h)$.
(Since $\Z$ is amenable,
the full and reduced crossed products are the same,
by Theorem~\ref{T_4131_CPFToRed}.)
We let $u \in C^* (\Z, X, h)$
be the standard unitary
corresponding to the generator $1 \in \Z$.
Thus, $u f u^* = f \circ h^{1}$ for all $f \in C (X)$,
and for $n \in \Z$
the standard unitary $u_n$ (following Notation~\ref{N:ug})
is $u_n = u^n$.
The dense subalgebra $C (X) [\Z]$
is the set of all finite sums
%
\begin{equation}\label{Eq_6930_CXZ}
a = \sum_{k =  n}^n f_k u^k
\end{equation}
%
with $n \in \Nz$ and $f_{n}, f_{ n + 1}, \ldots, f_n \in C (X)$.
We identify $C (X)$ with a subalgebra of $C^* (\Z, X, h)$
in the standard way
(as discussed after Remark~\ref{R:NRed}):
it is all $a$ as in~(\ref{Eq_6930_CXZ})
such that $f_k = 0$ for $k \neq 0$.
The standard conditional expectation
$E \colon C^* (\Z, X, h) \to C (X)$
is given on $C (X) [\Z]$ by
$E (a) = f_0$ when $a$ is as in~(\ref{Eq_6930_CXZ}).
\end{ntn}
In order to state more general results,
we generalize the construction of
\Def{D_5421_VSubalg}.
Notation~\ref{N2818C0U}
and
\Def{D2623VSubalg}
below
differ from
Notation~\ref{NOld32}
and
\Def{D_5421_VSubalg}
in that they consider $C_0 (X, D)$
for a \ca~$D$ instead of just $C_0 (X)$.
\begin{ntn}\label{N2818C0U}
For a locally \chs~$X$,
a \ca~$D$,
and an open subset $U \subset X$,
we use the abbreviation
\[
C_0 (U, D)
= \big\{ f \in C_0 (X, D) \colon
{\mbox{$f (x) = 0$ for all $x \in X \setminus U$}} \big\}
\subset C_0 (X, D).
\]
This subalgebra is of course canonically isomorphic to
the usual algebra $C_0 (U, D)$ when $U$ is considered
as a locally \chs{} in its own right.
If $D = \C$ we omit it from the notation.
\end{ntn}
In particular,
if $Y \subset X$ is closed, then
%
\begin{equation}\label{Eq:2818C0XY}
C_0 (X \setminus Y, \, D)
= \big\{ f \in C_0 (X, D) \colon
{\mbox{$f (x) = 0$ for all $x \in Y$}} \big\}.
\end{equation}
%
\begin{dfn}\label{D2623VSubalg}
Let $X$ be a locally \chs,
let $D$ be a unital \ca, and let $h \colon X \to X$ be a \hme.
Let $\af \in \Aut (C (X, D) )$
be an automorphism which
``lies over~$h$'',
in the sense that there exists a function
$x \mapsto \af_{x}$ from $X$ to $\Aut (D)$
such that $\af (a) (x) = \af_{x} \big( a (h^{1} ( x ) ) \big)$
for all $x \in X$
and $a \in C_0 (X, D)$.
Let $Y \subset X$ be a nonempty closed subset,
and, following~(\ref{Eq:2818C0XY}), define
\[
C^* \big( \Z, \, C_0 (X, D), \, \af \big)_Y
= C^* \big( C_0 (X, D), \, C_0 (X \setminus Y, \, D) u \big)
\subset C^* \big( \Z, \, C_0 (X, D), \, \af \big).
\]
We call it the {\emph{$Y$orbit breaking subalgebra}}
of $C^* \big( \Z, \, C_0 (X, D), \, \af \big)$.
\end{dfn}
We describe the proof of Theorem~\ref{T_5421_AYStabLg},
namely that if $h \colon X \to X$ is a minimal homeomorphism
and $Y \subset X$ is a compact subset
such that $h^n (Y) \cap Y = \varnothing$
for all $n \in \Z \setminus \{ 0 \}$,
then $C^* (\Z, X, h)_Y$ is a
centrally large subalgebra of $C^* (\Z, X, h)$
in the sense of Definition~\ref{D_5421_CentLarge}.
Our presentation differs from that of \cite{Ph40} and~\cite{ArPh}
in that we prove the result directly rather than via
large subalgebras of crossed product type.
Under some technical conditions on $\af$ and~$D$,
similar methods can be used to prove the analogous result for
$C^* \big( \Z, \, C (X, D), \, \af \big)_Y$.
The following theorem is a consequence
of results in~\cite{ArBcPh}.
\begin{thm}[\cite{ArBcPh}]\label{T_5418_AYLg}
Let $X$ be an infinite \cms,
let $h \colon X \to X$ be a minimal homeomorphism,
let $D$ be a simple \uca{} which has a tracial state,
and let $\af \in \Aut (C (X, D))$ lie over~$h$.
Assume that $D$ has strict comparison of positive elements,
or that the automorphisms $\af_{x}$
in \Def{D2623VSubalg}
are all approximately inner.
Let $Y \subset X$ be a compact subset
such that $h^n (Y) \cap Y = \varnothing$
for all $n \in \Z \setminus \{ 0 \}$.
Then $C^* \big( \Z, \, C (X, D), \, \af \big)_Y$
is a centrally large subalgebra
of $C^* \big( \Z, \, C (X, D), \, \af \big)$
in the sense of Definition~\ref{D_5421_CentLarge}.
\end{thm}
The ideas of the proof of Theorem~\ref{T_5421_AYStabLg}
are all used in the proof of the general theorem
behind Theorem~\ref{T_5418_AYLg},
but additional work is needed to deal with the presence
of~$D$.
We now describe the proof of Theorem~\ref{T_5421_AYStabLg},
omitting a few details.
We will let $X$ be an infinite \chs{}
with a \mh{} $h \colon X \to X$.
We follow Notation~\ref{NOld32}.
We will fix a nonempty closed subset $Y \subset X$.
For $n \in \Z$, set
\[
Z_n = \begin{cases}
\bigcup_{j = 0}^{n  1} h^j (Y) & \hspace{3em} n > 0
\\
\varnothing & \hspace{3em} n = 0
\\
\bigcup_{j = 1}^{ n} h^{j} (Y) & \hspace{3em} n < 0.
\end{cases}
\]
Recall from Proposition~\ref{P_4319_CharOB}
that
%
\[
C^* (\Z, X, h)_Y
= \big\{ a \in C^* (\Z, X, h) \colon
{\mbox{$E (a u^{n}) \in C_0 (X \setminus Z_n)$
for all $n \in \Z$}} \big\}
\]
%
and
%
\[
{\overline{C^* (\Z, X, h)_Y \cap C (X) [\Z]}} = C^* (\Z, X, h)_Y.
\]
\begin{lem}[Corollary 7.6 of~\cite{Ph40}]\label{C2816AYOpp}
% \label{C2816AYOpp}
Let $X$ be a \chs{} and let $h \colon X \to X$ be a \hme.
Let $Y \subset X$ be a nonempty closed subset.
Let $u \in C^* (\Z, X, h)$ be the standard unitary,
as in Notation~\ref{NOld32},
and let $v \in C^* (\Z, X, h^{1})$ be the analogous standard unitary
in $C^* (\Z, X, h^{1})$.
Then there exists
a unique \hm{} $\ph \colon C^* (\Z, X, h^{1}) \to C^* (\Z, X, h)$
such that $\ph (f) = f$ for $f \in C (X)$ and $\ph (v) = u^*$,
the map $\ph$ is an isomorphism,
and
\[
\ph \big( C^* (\Z, X, h^{1})_{h^{1} (Y)} \big) = C^* (\Z, X, h)_Y.
\]
\end{lem}
See~\cite{Ph40} for the straightforward proof,
based on Proposition~\ref{P_4319_CharOB}.
\begin{lem}[Lemma~7.4 of~\cite{Ph40}]\label{L2623ZPart}
% \label{L2623ZPart}
Let $X$ be an infinite \chs{}
and let $h \colon X \to X$ be a minimal homeomorphism.
Let $K \subset X$ be a compact set
such that $h^n (K) \cap K = \varnothing$
for all $n \in \Z \setminus \{ 0 \}$.
Let $U \subset X$ be a nonempty open subset.
Then there exist $l \in \Nz$, compact sets
$K_1, K_2, \ldots, K_l \subset X$,
and $n_1, n_2, \ldots, n_l \in \N$,
such that
$K \subset \bigcup_{j = 1}^l K_j$
and such that
$h^{n_1} (K_1), \, h^{n_2} (K_2), \, \ldots, \, h^{n_l} (K_l)$
are disjoint subsets of~$U$.
\end{lem}
\begin{proof}[Sketch of proof]
Choose a nonempty open subset $V \subset X$
such that ${\overline{V}}$ is compact and contained in~$U$.
Use minimality of $h$ to cover $K$ with the
images of $V$ under finitely many negative powers of~$h$,
say $h^{ n_1} (V), \, h^{ n_2} (V), \, \ldots, \, h^{ n_l} (V)$.
Set $K_j = h^{ n_j} \big( {\overline{V}} \big) \cap K$
for $j = 1, 2, \ldots, l$.
\end{proof}
The next lemma
is straightforward if one only asks that
$f \precsim_{C^* (\Z, X, h)} g$
(Cuntz subequivalence in the crossed product),
and then doing it for one value of~$n$ is equivalent
to doing it for any other.
Getting
$f \precsim_{C^* (\Z, X, h)_Y} g$
for both positive $n$ and negative $n$
is a key step in showing $C^* (\Z, X, h)_Y$
a large subalgebra of $C^* (\Z, X, h)$.
This result is related to the statement about equivalence of
projections in \Lem{L:Canc}.
\begin{lem}[Lemma~7.7 of~\cite{Ph40}]\label{L2623CuSub}
% \label{L2623CuSub}
Let $X$ be an infinite \chs{}
and let $h \colon X \to X$ be a minimal homeomorphism.
Let $Y \subset X$ be a compact subset
such that $h^n (Y) \cap Y = \varnothing$
for all $n \in \Z \setminus \{ 0 \}$.
Let $U \subset X$ be a nonempty open subset
and let $n \in \Z$.
Then there exist $f, g \in C (X)_{+}$
such that
\[
f _{h^n (Y)} = 1,
\,\,\,\,\,\,
0 \leq f \leq 1,
\,\,\,\,\,\,
\supp (g) \subset U,
\andeqn
f \precsim_{C^* (\Z, X, h)_Y} g.
\]
\end{lem}
\begin{proof}
We first prove this when $n = 0$.
Apply Lemma~\ref{L2623ZPart} with $Y$ in place of~$K$,
obtaining $l \in \Nz$, compact sets
$Y_1, Y_2, \ldots, Y_l \subset X$,
and $n_1, n_2, \ldots, n_l \in \N$.
Set $N = \max (n_1, n_2, \ldots, n_l)$.
Choose disjoint open sets $V_1, V_2, \ldots, V_l \subset U$
such that $h^{n_j} (Y_j) \subset V_j$ for $j = 1, 2, \ldots, l$.
Then $Y_j \subset h^{ n_j} (V_j)$,
so the sets
$h^{ n_1} (V_1), \, h^{ n_2} (V_2), \, \ldots, \, h^{ n_l} (V_l)$
cover~$Y$.
For $j = 1, 2, \ldots, l$,
define
\[
W_j
= h^{ n_j} (V_j)
\cap \left( X \setminus \bigcup_{n = 1}^N h^{n} (Y) \right).
\]
Then $W_1, W_2, \ldots, W_l$ form an open cover of~$Y$.
Therefore there are $f_1, f_2, \ldots, f_l \in C (X)_{+}$
such that for $j = 1, 2, \ldots, l$
we have $\supp (f_j) \subset W_j$ and $0 \leq f_j \leq 1$,
and such that the function $f = \sum_{j = 1}^l f_j$
satisfies $f (x) = 1$ for all $x \in Y$ and $0 \leq f \leq 1$.
Further define $g = \sum_{j = 1}^l f_j \circ h^{ n_j}$.
Then $\supp (g) \subset U$.
Let $u \in C^* (\Z, X, h)$
be as in Notation~\ref{NOld32}.
For $j = 1, 2, \ldots, l$,
set $a_j = f_j^{1/2} u^{n_j}$.
Since $f_j$ vanishes on $\bigcup_{n = 1}^{n_j} h^{n} (Y)$,
Proposition~\ref{P_4319_CharOB}
implies that $a_j \in C^* (\Z, X, h)_Y$.
Therefore, in $C^* (\Z, X, h)_Y$ we have
\[
f_j \circ h^{ n_j}
= a_j^* a_j \sim_{C^* (\Z, X, h)_Y} a_j a_j^* = f_j.
\]
Consequently,
using Lemma~\ref{L:CzBasic}(\ref{L:CzBasic:LCzCmpSum})
at the second step
and Lemma~\ref{L:CzBasic}(\ref{L:CzBasic:Orth})
and disjointness of the supports of the
functions $f_j \circ h^{ n_j}$ at the last step,
we have
\[
f = \sum_{j = 1}^l f_j
\precsim_{C^* (\Z, X, h)_Y} \bigoplus_{j = 1}^l f_j
\sim_{C^* (\Z, X, h)_Y} \bigoplus_{j = 1}^l f_j \circ h^{ n_j}
\sim_{C^* (\Z, X, h)_Y} g.
\]
This completes the proof for $n = 0$.
Now suppose that $n > 0$.
Choose functions $f$ and $g$ for the case $n = 0$,
and call them $f_0$ and~$g$.
Since $f_0 (x) = 1$ for all $x \in Y$,
and since $Y \cap \bigcup_{l = 1}^n h^{ l} (Y) = \varnothing$,
there is $f_1 \in C (X)$ with $0 \leq f_1 \leq f_0$,
$f_1 (x) = 1$ for all $x \in Y$,
and $f_1 (x) = 0$
for $x \in \bigcup_{l = 1}^n h^{ l} (Y)$.
Set $v = f_1^{1/2} u^{ n}$
and $f = f_1 \circ h^{n}$.
Then $f (x) = 1$ for all $x \in h^n (Y)$ and $0 \leq f \leq 1$.
Proposition~\ref{P_4319_CharOB} implies that $v \in C^* (\Z, X, h)_Y$.
We have
\[
v^* v = u^n f_1 u^{ n} = f_1 \circ h^{ n} = f
\andeqn
v v^* = f_1.
\]
Using Lemma~\ref{L:CzBasic}(\ref{L:CzBasic:LCzComm}),
we thus get
\[
f \sim_{C^* (\Z, X, h)_Y} f_1
\leq f_0
\precsim_{C^* (\Z, X, h)_Y} g.
\]
This completes the proof for the case $n > 0$.
Finally,
we consider the case $n < 0$.
In this case, we have $ n  1 \geq 0$.
Apply the cases already done
with $h^{1}$ in place of~$h$.
We get $f, g \in C^* (\Z, X, h^{1})_{h^{1} (Y)}$
such that $f (x) = 1$
for all $x \in (h^{1})^{ n  1} (h^{1} (Y)) = h^n (Y)$,
such that $0 \leq f \leq 1$,
such that $\supp (g) \subset U$,
and such that $f \precsim_{C^* (\Z, X, h^{1})_{h^{1} (Y)}} g$.
Let $\ph \colon C^* (\Z, X, h^{1}) \to C^* (\Z, X, h)$
be the isomorphism of Lemma~\ref{C2816AYOpp}.
Then
\[
\ph (f) = f,
\,\,\,\,\,\,
\ph (g) = g,
\andeqn
\ph \big( C^* (\Z, X, h^{1})_{h^{1} (Y)} \big) = C^* (\Z, X, h)_Y.
\]
Therefore $f \precsim_{C^* (\Z, X, h)_Y} g$.
\end{proof}
The following result is a special case of Lemma~7.9
of~\cite{Ph40}.
The basic idea has been used frequently;
related arguments can be found, for example, in the proofs
of Theorem~3.2 of~\cite{El0},
Lemma~2 and Theorem~1 in~\cite{ArSp},
Lemma~10 of~\cite{KsKu},
and
Lemma~3.2 of~\cite{PsnPh2}.
(The papers listed are not claimed to be
representative or to be the original sources;
they are ones I happen to know of.)
\begin{lem}\label{L_5524_CXSubEq}
% \label{L:CXSubEq}
Let $X$ be an infinite compact space,
and let $h \colon X \to X$ be a \mh.
Let $B \subset C^* (\Z, X, h)$ be a unital
subalgebra such that
$C (X) \subset B$ and $B \cap C (X)[\Z]$ is dense in~$B$.
Let $a \in B_{+} \setminus \{ 0 \}$.
Then there exists $b \in C (X)_{+} \setminus \{ 0 \}$
such that $b \precsim_{B} a$.
\end{lem}
\begin{proof}[Sketch of proof]
\Wolog{} $\ a \ \leq 1$.
The conditional expectation
$E_{\af} \colon C^*_{\mathrm{r}} (G, X) \to C (X)$
is faithful.
Therefore $E_{\af} (a) \in C (X)$
is a nonzero positive element.
Set $\ep = \tfrac{1}{6} \ E_{\af} (a) \$.
Choose $c \in B \cap C (X) [\Z]$ such that
$\ c  a^{1/2} \ < \ep$ and $\ c \ \leq 1$.
One can check that $\ E_{\af} (c^* c) \ > 4 \ep$.
There are $n \in \Nz$ and
$g_{n}, g_{ n + 1}, \ldots, g_n \in C (X)$
such that $c^* c = \sum_{k =  n}^n g_k u^k$.
We have $g_0 = E_{\af} (c^* c) \in C (X)_{+}$
and $\ g_0 \ > 4 \ep$.
Therefore there is $x \in X$ such that
$g_0 (x) > 4 \ep$.
Choose $f \in C (X)$
such that $0 \leq f \leq 1$,
$f (x) = 1$,
and the sets $h^k ( \supp (f) )$
are disjoint for $k = n, \,  n + 1, \, \ldots, \, n$.
One can then check that $f c^* c f = f g_0 f$,
so that $\ f c^* c f \ > 4 \ep$.
Therefore
$(f c^* c f  2 \ep)_{+}$ is a nonzero element of $C (X)$.
Using Lemma~\ref{L:CzBasic}(\ref{L:CzBasic:LCzCommEp})
at the first step,
Lemma~\ref{L:CzBasic}(\ref{L:CzBasic:L:CzCompIneq})
and $c f^2 c^* \leq c c^*$ at the second step,
and Lemma~\ref{L:CzBasic}(\ref{L:CzBasic:LCzWithinEp})
and $\ c c^*  a \ < 2 \ep$
at the last step,
we then have
\[
(f c^* c f  2 \ep)_{+}
\sim_{B} (c f^2 c^*  2 \ep)_{+}
\precsim_{B} (c c^*  2 \ep)_{+}
\precsim_{B} a.
\]
This completes the proof.
\end{proof}
\begin{cor}\label{C_5531_DblCmp}
% \label{L:CXSubEq}
Let $X$ be an infinite \chs,
and let $h \colon X \to X$ be a \mh.
Let $B \subset C^* (\Z, X, h)$ be a unital
subalgebra such that
$C (X) \subset B$ and $B \cap C (X)[\Z]$ is dense in~$B$.
Let $a \in A_{+} \setminus \{ 0 \}$
and let $b \in B_{+} \setminus \{ 0 \}$.
Then there exists $f \in C (X)_{+} \setminus \{ 0 \}$
such that $f \precsim_{C^* (\Z, X, h)} a$ and $f \precsim_{B} b$.
\end{cor}
\begin{proof}
Applying Lemma~\ref{L_5524_CXSubEq}
to both $a$ (with $C^* (\Z, X, h)$ in place of~$B$)
and~$b$ (with $B$ as given),
we see that it is enough to prove the corollary
for $a, b \in C (X)_{+} \setminus \{ 0 \}$.
Also,
\wolog{} $\ a \ \leq 1$.
Choose $x_0 \in X$ such that $b (x_0) \neq 0$.
Since the orbit of~$x_0$ is dense,
there is $n \in \Z$ such that $a (h^n (x_0)) \neq 0$.
Define $f \in C (X)$ by
$f (x) = b (x) a (h^{n} (x))$ for $x \in X$.
Then $f \neq 0$ since $f (x_0) \neq 0$.
We have $f \precsim_{B} b$
because $\ a \ \leq 1$ implies $f \leq b$.
Also,
$f = (b^{1/2} u^{n}) a (b^{1/2} u^{n})^*$
so $f \precsim_{C^* (\Z, X, h)} a$.
\end{proof}
The next result is a standard type of approximation lemma.
\begin{lem}\label{L_5601_CommEst}
Let $A$ be a \ca,
and let $S \subset A$ be a subset which generates $A$ as a \ca{}
and such that $a \in S$ implies $a^* \in S$.
Then for every finite subset $F \subset A$ and every $\ep > 0$
there are a finite subset $T \subset S$
and $\dt > 0$
such that whenever $c \in A$ satisfies $\ c \ \leq 1$
and $\ c b  b c \ < \dt$
for all $b \in T$,
then $\ c a  a c \ < \ep$ for all $a \in F$.
\end{lem}
\begin{proof}
Let $B \subset A$ be the set of all $a \in A$
such that
for every $\ep > 0$
there are $T (a, \ep) \subset S$ and $\dt (a, \ep) > 0$
as in the statement of the lemma,
that is, $T (a, \ep)$ is finite and
whenever $c \in A$ satisfies $\ c \ \leq 1$
and $\ c b  b c \ < \dt (a, \ep)$
for all $b \in T (a, \ep)$,
then $\ c a  a c \ < \ep$.
We have $S \subset B$,
as is seen by taking $T (a, \ep) = \{ a \}$
and $\dt (a, \ep) = \ep$.
If $a \in B$,
then also $a^* \in B$,
as is seen by taking
\[
T (a^*, \ep) = \big\{ b^* \colon b \in T (a, \ep) \big\}
\andeqn
\dt (a^*, \ep) = \dt (a, \ep).
\]
We show that $B$ is closed under addition.
So let $a_1, a_2 \in B$ and let $\ep > 0$.
Define
\[
T = T \big( a_1, \tfrac{\ep}{2} \big)
\cup T \big( a_2, \tfrac{\ep}{2} \big)
\andeqn
\dt = \min \big( \dt \big( a_1, \tfrac{\ep}{2} \big), \,
\dt \big( a_2, \tfrac{\ep}{2} \big) \big).
\]
Suppose $c \in A$ satisfies $\ c \ \leq 1$
and $\ c b  b c \ < \dt$
for all $b \in T$.
Then
\[
\ c a_1  a_1 c \ < \tfrac{\ep}{2}
\andeqn
\ c a_2  a_2 c \ < \tfrac{\ep}{2},
\]
so
\[
\ c (a_1 + a_2)  (a_1 + a_2) c \
\leq \ c a_1  a_1 c \ + \ c a_2  a_2 c \
< \ep.
\]
This shows that $a_1 + a_2 \in B$.
To show that if $a_1, a_2 \in B$ then $a_1 a_2 \in B$,
we use a similar argument,
taking
\[
\ep_0 = \frac{\ep}{1 + \ a_1 \ + \ a_2 \}
\]
and
using the choices
\[
T = T ( a_1, \ep_0 )
\cup T ( a_2, \ep_0 )
\andeqn
\dt = \min ( \dt ( a_1, \ep_0 ), \,
\dt ( a_2, \ep_0 ) \big),
\]
and the estimate
\[
\ c a_1 a_2  a_1 a_2 c \
\leq \ c a_1  a_1 c \ \ a_2 \ + \ a_1 \ \ c a_2  a_2 c \.
\]
Finally,
we claim that $B$ is closed.
So let $a \in {\overline{B}}$
and let $\ep > 0$.
Choose $a_0 \in B$ such that $\ a  a_0 \ < \tfrac{\ep}{3}$.
Define
\[
T = T \big( a_0, \tfrac{\ep}{3} \big)
\andeqn
\dt = \dt \big( a_0, \tfrac{\ep}{3} \big).
\]
Suppose $c \in A$ satisfies $\ c \ \leq 1$
and $\ c b  b c \ < \dt$
for all $b \in T$.
Then $\ c a_0  a_0 c \ < \tfrac{\ep}{3}$,
so
\[
\ c a  a c \
\leq 2 \ c \ \ a  a_0 \ + \ c a_0  a_0 c \
< \frac{2 \ep}{3} + \frac{\ep}{3}
= \ep.
\]
The claim is proved.
Since $S$ generates $A$ as a \ca,
we have $B = A$.
Now let $F \subset A$ be finite
and let $\ep > 0$.
The conclusion of the lemma follows by taking
\[
T = \bigcup_{a \in F} T (a, \ep)
\andeqn
\dt = \min_{a \in F} \dt (a, \ep).
\]
This completes the proof.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{T_5421_AYStabLg}]
Set $A = C^* (\Z, X, h)$ and $B = C^* (\Z, X, h)_Y$.
Since $h$ is minimal,
it is well known that $A$ is simple and finite.
Also clearly $A$ is infinite dimensional.
We claim that the following holds.
Let $m \in \N$,
let $a_1, a_2, \ldots, a_m \in A$,
let $\ep > 0$,
and let $f \in C (X)_{+} \setminus \{ 0 \}$.
Then there are $c_1, c_2, \ldots, c_m \in A$ and $g \in C (X)$
such that:
\begin{enumerate}
\item\label{D:FinLarge:Cut1_Pf31}
$0 \leq g \leq 1$.
\item\label{D:FinLarge:Cut2_Pf31}
For $j = 1, 2, \ldots, m$ we have
$\ c_j  a_j \ < \ep$.
\item\label{D:FinLarge:Cut3_Pf31}
For $j = 1, 2, \ldots, m$ we have
$(1  g) c_j \in B$.
\item\label{D:FinLarge:Cut4_Pf31}
$g \precsim_B f$.
\item\label{D:FinLarge:Cut5_Pf31}
$\ g u  u g \ < \ep$.
\end{enumerate}
Suppose the claim has been proved;
we show that the theorem follows.
Let $m \in \N$,
let $a_1, a_2, \ldots, a_m \in A$,
let $\ep > 0$,
let $r \in A_{+} \setminus \{ 0 \}$,
and let $s \in B_{+} \setminus \{ 0 \}$.
(The elements $r$ and $s$ play the roles of $x$ and~$y$
in Definition~\ref{D_5421_CentLarge}.
Here, we use $x$ and $y$ for elements of~$X$.)
Apply Lemma~\ref{L2726Big} with $r$ in place of~$x$,
getting $r_0 \in A_{+} \setminus \{ 0 \}$.
In Lemma~\ref{L_5601_CommEst},
take $S = C (X) \cup \{ u, u^* \}$,
take $F = \{ a_1, a_2, \ldots, a_m \}$,
and let $\ep > 0$ be as given.
Let the finite set $T \subset S$ and $\dt > 0$
be as in the conclusion.
We may assume that $u, u^* \in T$.
Apply Corollary~\ref{C_5531_DblCmp}
with $r_0$ in place of~$a$
and $s$ in place of~$b$,
getting $f \in C (X)_{+} \setminus \{ 0 \}$
such that $f \precsim_{A} r_0$ and $f \precsim_{B} s$.
Apply the claim with $a_1, a_2, \ldots, a_m$ as given,
and with $\min (\ep, \dt)$ in place of~$\ep$.
We can now verify the conditions of
Definition~\ref{D_5421_CentLarge}.
Conditions (\ref{D_5421_CentLarge_Cut1}),
(\ref{D_5421_CentLarge_Cut2}),
and (\ref{D_5421_CentLarge_Cut3}) there
are conditions (\ref{D:FinLarge:Cut1_Pf31}),
(\ref{D:FinLarge:Cut2_Pf31}),
and (\ref{D:FinLarge:Cut3_Pf31}) here.
Condition~(\ref{D_5421_CentLarge_Cut4}) there
follows from condition (\ref{D:FinLarge:Cut4_Pf31})
here and the relations
$f \precsim_{A} r_0 \precsim_{A} r$ and $f \precsim_{B} s$.
Condition~(\ref{D_5421_CentLarge_Cut5a}) there
follows from $g \precsim_A r_0$ and the choice of~$r_0$.
It remains only to verify
condition~(\ref{D_5421_CentLarge_Cut5}) there,
namely $\ g a_j  a_j g \ < \ep$ for $j = 1, 2, \ldots, m$.
It suffices to check that
$\ g b  b g \ < \dt$ for all $b \in T$.
We have $\ g u  u g \ < \dt$ by construction.
Also,
$g u^*  u^* g =  u^* (g u  u g) u^*$,
so $\ g u^*  u^* g \ < \dt$.
Finally, if $b \in T$ is any element other than $u$ or~$u^*$,
then $b \in C (X)$,
so $g b = b g$.
This completes the proof that the claim implies the conclusion
of the theorem.
We now prove the claim.
Choose $c_1, c_2, \ldots, c_m \in C (X) [\Z]$
such that $\ c_j  a_j \ < \ep$
for $j = 1, 2, \ldots, m$.
(This estimate is condition~(\ref{D:FinLarge:Cut2_Pf31}).)
Choose $N \in \N$
such that
there are $c_{j, l} \in C (X)$
for $j = 1, 2, \ldots, m$
and $l = N, \,  N + 1, \, \ldots, \, N  1, \, N$
with
\[
c_j = \sum_{l = N}^N c_{j, l} u^l.
\]
Choose $N_0 \in \N$ such that $\frac{1}{N_0} < \ep$.
Define
\[
I = \big\{  N  N_0, \,  N  N_0 + 1, \, \ldots,
\, N + N_0  1, \, N + N_0 \big\}.
\]
Set $U = \{ x \in X \colon f (x) \neq 0 \}$,
and choose nonempty disjoint open sets
$U_l \subset U$ for
$l \in I$.
For each such~$l$,
use Lemma~\ref{L2623CuSub}
to choose $f_l, r_l \in C (X)_{+}$
such that $r_l (x) = 1$ for all $x \in h^l (Y)$,
such that $0 \leq r_l \leq 1$,
such that $\supp (f_l) \subset U_l$,
and such that $r_l \precsim_{B} f_l$.
Choose an open set $W$ containing~$Y$
such that the sets $h^l (W)$ are disjoint
for $l \in I$,
and choose $r \in C (X)$ such that
\[
0 \leq r \leq 1,
\,\,\,\,\,\,
r _Y = 1,
\andeqn
\supp (r) \subset W.
\]
Set
\[
% g_0 = r \cdot \prod_{l =  N  N_0}^{N + N_0} r_l \circ h^{l}.
g_0 = r \cdot \prod_{l \in I} r_l \circ h^{l}.
\]
Set $g_l = g_0 \circ h^{ l}$
for
% $l = N  N_0, \,  N  N_0 + 1, \, \ldots,
% \, N + N_0  1, \, N + N_0$.
$l \in I$.
Then $0 \leq g_l \leq r_l \leq 1$.
Define $\ld_l$ for $l \in I$ by
\[
\ld_{ N  N_0} = 0,
\,\,\,\,\,\,
\ld_{ N  N_0 + 1} = \frac{1}{N_0},
\,\,\,\,\,\,
\ld_{ N  N_0 + 2} = \frac{2}{N_0},
\,\,\,\,\,\,
\ldots,
\,\,\,\,\,\,
\ld_{ N  1} = 1  \frac{1}{N_0},
\]
\[
\ld_{ N} = \ld_{ N + 1} = \cdots = \ld_{N  1} = \ld_N = 1,
\]
\[
\ld_{N + 1} = 1  \frac{1}{N_0},
\,\,\,\,\,\,
\ld_{N + 2} = 1  \frac{2}{N_0},
\,\,\,\,\,\,
\ldots,
\,\,\,\,\,\,
\ld_{N + N_0  1} = \frac{1}{N_0},
\,\,\,\,\,\,
\ld_{N + N_0 } = 0.
\]
Set $g = \sum_{l \in I} \ld_l g_l$.
The supports of the functions $g_l$ are disjoint,
so $0 \leq g \leq 1$.
This is condition~(\ref{D:FinLarge:Cut1_Pf31}).
Using Lemma~\ref{L:CzBasic}(\ref{L:CzBasic:Orth})
at the first and fourth steps
and Lemma~\ref{L:CzBasic}(\ref{L:CzBasic:CmpDSum})
at the third step,
we get
\[
g \precsim_B \bigoplus_{l \in I} g_l
\leq \bigoplus_{l \in I} r_l
\precsim_B \bigoplus_{l \in I} f_l
\sim_{C (X)} \sum_{l \in I} f_l
\precsim_{C (X)} f.
\]
This is condition~(\ref{D:FinLarge:Cut4_Pf31}).
We check condition~(\ref{D:FinLarge:Cut5_Pf31}).
We have
\[
\ g u  u g \
= \ g  u g u^* \
= \ g  g \circ h^{1} \
= \left\ \sum_{l \in I} \ld_l g_0 \circ h^{l}
 \sum_{l \in I} \ld_l g_0 \circ h^{ l  1} \right\.
\]
In the second sum in the last term,
we change variables
to get $\sum_{l + 1 \in I} \ld_{l  1} g_0 \circ h^{ l}$.
Use $\ld_{ N  N_0} = \ld_{N + N_0} = 0$
and combine terms to get
\[
\ g u  u g \
= \left\ \sum_{l =  N  N_0 + 1}^{N + N_0} (\ld_l  \ld_{l  1})
g_0 \circ h^{ l} \right\.
\]
The expressions $g_0 \circ h^{ l}$ are orthogonal and have
norm~$1$,
so
\[
\ g u  u g \
= \max_{ N  N_0 + 1 \leq l \leq N + N_0}  \ld_l  \ld_{l  1} 
= \frac{1}{N_0}
< \ep.
\]
It remains to verify condition~(\ref{D:FinLarge:Cut3_Pf31}).
Since $1  g$ vanishes on the sets
\[
h^{N} (Y), \, h^{ N + 1} (Y),
\, \ldots, \, h^{N  2} (Y), \, h^{N  1} (Y),
\]
Proposition~\ref{P_4319_CharOB}
implies that $(1  g) u^l \in B$
for
$l = N, \,  N + 1, \, \ldots, \, N  1, \, N$.
For $j = 1, 2, \ldots, m$,
since $c_{j, l} \in C (X) \subset B$
for
$l = N, \,  N + 1, \, \ldots, \, N  1, \, N$,
we get
\[
(1  g) c_j = \sum_{l = N}^N c_{j, l} \cdot (1  g) u^l \in B.
\]
This completes
the verification of condition~(\ref{D:FinLarge:Cut3_Pf31}),
and the proof of the theorem.
\end{proof}
\section{Application to the Radius of Comparison of Crossed Products
by Minimal Homeomorphisms}\label{Sec_rcCrPrd}
\indent
The purpose of this section
is to describe some of the ideas involved in
Theorem~\ref{T_5526_rcmdim}
and its proof.
We describe the mean dimension of a \hme,
and we give proofs of simple special cases or related statements
for some of the steps in its proof.
We will need simplicial complexes.
See Section 2.6 of~\cite{Prs} for a presentation of
the basics.
Following a common abuse of terminology,
we say here that a topological space is a simplicial complex
when,
formally,
we mean that it is homeomorphic to the geometric realization
of a simplicial complex.
An explanation of mean dimension starts with dimension theory;
see the discussion after Corollary~\ref{C:C1}.
The mean dimension of a \hme{}
$h \colon X \to X$
was introduced in~\cite{LndWs}
For best behavior,
$h$ should not have ``too many'' periodic points.
It is designed so that if $K$ is a sufficiently nice
\cms{}
(in particular,
$\dim (K^n)$ should equal $n \cdot \dim (K)$ for all~$n \in \N$),
then the shift on $X = K^{\Z}$ should have mean dimension
equal to~$\dim (K)$.
Given this heuristic,
it should not be surprising that
if $\dim (X) < \I$ then $\mdim (h) = 0$.
We recall
that ${\operatorname{Cov}} (X)$
is the set of finite open covers of~$X$ (Notation~\ref{N_6827_Cov}),
that $\cV \prec \cU$ means that $\cV$ refines $\cU$
(\Def{D_6827_Rfn}),
the order $\ord (\cU)$ of a finite open cover~$\cU$
(\Def{D_6827_Ord}),
and that $\cD (\cU)$ is the least order
of a refinement of~$\cU$.
\begin{dfn}\label{D_6827_Join}
Let $X$ be a \chs,
and let $\cU$ and $\cV$ be two finite open covers of~$X$.
Then the {\emph{join}} $\cU \vee \cV$ of $\cU$ and $\cV$ is
\[
\cU \vee \cV
= \big\{ U \cap V \colon {\mbox{$U \in \cU$ and $V \in \cV$}} \big\}.
\]
\end{dfn}
\begin{dfn}\label{D_6827_CovImage}
Let $X$ be a \chs,
let $\cU$ be a finite open cover of~$X$,
and let $h \colon X \to X$ be a \hme.
We define
\[
h ( \cU) = \big\{ h (U) \colon U \in \cU \big\}.
\]
\end{dfn}
\begin{dfn}[Definition~2.6 of~\cite{LndWs}]\label{D_6827_MDim}
Let $X$ be a \cms{}
and let $h \colon X \to X$ be a \hme.
Then the
{\emph{mean dimension}} of~$h$
is
(see Corollary~\ref{C_6913_LimExist}
below for existence of the limit)
\[
\mdim (h) = \sup_{\cU \in {\operatorname{Cov}} (X)} \lim_{n \to \infty}
\frac{\cD
\big( \cU \vee h^{1} (\cU) \vee \cdots \vee h^{n + 1} (\cU)
\big)}{n}.
\]
\end{dfn}
The expression in the definition
uses the join of $n$ covers.
Existence of the limit depends on the following result.
\begin{prp}[Corollary~2.5 of~\cite{LndWs}]\label{L_6913_Subadd}
Let $X$ be a \cms,
and let $\cU$ and $\cV$ be two finite open covers of~$X$.
Then $\cD (\cU \vee \cV) \leq \cD (\cU) + \cD (\cV)$.
\end{prp}
We omit the proof,
but the idea is similar to that of the proof
of Proposition 3.2.6 of~\cite{Prs}
($\dim (X \times Y) \leq \dim (X) + \dim (Y)$
for nonempty \chs{s} $X$ and~$Y$).
The point is that an open cover $\cU$
has $\ord (\cU) \leq m$
\ifo{} there is a finite simplicial complex $K$
of dimension at most~$m$
which approximates $X$ ``as seen by $\cU$'',
and if $K$ and $L$ are
finite simplicial complexes
which approximate $X$ as seen by $\cU$ and by~$\cV$,
then $K \times L$
is a finite simplicial complex
with dimension $\dim (K) + \dim (L)$
which approximates $X$ as seen by $\cU \vee \cV$.
\begin{lem}\label{L_6913_SubaddSeq}
Let $(\af_n)_{n \in \N}$ be a sequence in $[0, \I)$
which is subadditive, that is,
$\af_{m + n} \leq \af_m + \af_n$
for all $m, n \in \N$.
Then $\limi{n} n^{1} \af_n$ exists
and is equal to $\inf_{n \in \N} n^{1} \af_n$.
\end{lem}
\begin{proof}
We follow part of the proof of Theorem~6.1 of~\cite{LndWs}.
Define $\bt = \inf_{n \in \N} n^{1} \af_n$.
Let $\ep > 0$.
Choose $N_0 \in \N$
such that $N_0^{1} \af_{N_0} < \bt + \frac{\ep}{2}$.
Choose $N \in \N$ so large that
\[
N \geq N_0
\andeqn
\frac{N_0 \af_1}{N} < \frac{\ep}{2}.
\]
Let $n \geq N$.
Since $N \geq N_0$,
there are $r \in \N$ and $s \in \{ 0, 1, \ldots, N_0  1 \}$
such that $n = r N_0 + s$.
Then,
using subadditivity at the first step,
\[
\frac{\af_n}{n}
\leq \frac{r \af_{N_0} + s \af_1}{n}
= \frac{r \af_{N_0}}{r N_0 + s} + \frac{s \af_1}{n}
< \frac{\af_{N_0}}{N_0} + \frac{N_0 \af_1}{n}
< \bt + \frac{\ep}{2} + \frac{\ep}{2}
= \bt + \ep.
\]
This completes the proof.
\end{proof}
\begin{cor}\label{C_6913_LimExist}
Let $X$ be a \cms{}
let $\cU$ be a finite open cover of~$X$,
and let $h \colon X \to X$ be a \hme.
Then the limit
\[
\lim_{n \to \infty}
\frac{\cD
\big( \cU \vee h^{1} (\cU) \vee \cdots \vee h^{n + 1} (\cU)
\big)}{n}
\]
in Definition~\ref{D_6827_MDim}
exists.
\end{cor}
\begin{proof}
Combine Lemma~\ref{L_6913_SubaddSeq}
and Proposition~\ref{L_6913_Subadd}.
\end{proof}
The following result is immediate.
\begin{prp}\label{P_6913_Xfd}
Let $X$ be a \cms{} with finite covering dimension,
and let $h \colon X \to X$ be a \hme.
Then $\mdim (h) = 0$.
\end{prp}
\begin{proof}
Let $\cU$ be a finite open cover of~$X$.
Then,
by definition,
\[
\cD \big( \cU \vee h^{1} (\cU) \vee \cdots \vee h^{n + 1} (\cU)
\big) \leq \dim (X),
\]
so
\[
\mdim (h) \leq \lim_{n \to \infty} \frac{\dim (X)}{n} = 0.
\]
This completes the proof.
\end{proof}
The following result is less obvious,
but not difficult
(although we refer to~\cite{LndWs}
for the proof).
In particular,
it shows that every uniquely ergodic \mh{}
has mean dimension zero.
\begin{prp}\label{P_6913_UniqErg}
Let $X$ be a \cms,
let $h \colon X \to X$ be a \hme,
and assume that $h$ has at most countably many ergodic
invariant Borel probability measures.
Then $\mdim (h) = 0$.
\end{prp}
\begin{proof}
In~\cite{LndWs},
see Theorem~5.4
and the discussion after Definition~5.2.
\end{proof}
Proposition~\ref{P_6913_Xfd}
covers most of the common examples of \mh{s}.
However, not all \mh{s} have mean dimension zero.
We start with the standard nonminimal example.
the shift,
as in Example~\ref{E:Shift}.
\begin{dfn}\label{D_5611_Shift}
Let $K$ be a set.
The {\emph{shift}} $h_K \colon K^{\Z} \to K^{\Z}$
is the bijection
given by
$h_K (x)_k = x_{k + 1}$
for $x = (x_k)_{k \in \Z} \in K^{\Z}$
and $k \in \Z$.
% We write $h$ when $K$ is understood.
\end{dfn}
\begin{thm}[Proposition~3.1 of~\cite{LndWs}]\label{T_6913_ShiftEst}
Let $K$ be a \cms,
and let $h_K$ be as in Definition~\ref{D_5611_Shift}.
Then $\mdim (h_K) \leq \dim (K)$.
\end{thm}
\begin{thm}[Proposition~3.3 of~\cite{LndWs}]\label{T_6913_ShiftEq}
Let $d \in \N$,
set $K = [0, 1]^d$,
and let $h_K$ be as in Definition~\ref{D_5611_Shift}.
Then $\mdim (h_K) = d$.
\end{thm}
We omit the proofs.
To understand the result heuristically,
in Definition~\ref{D_6827_MDim}
consider a finite open cover $\cU_0$ of~$K$,
for $n \in \Z$ let $p_n \colon K^{\Z} \to K$
be the projection on the $n$th coordinate,
and consider the finite open cover
\[
\cU = \big\{ p_0^{1} (U) \colon U \in \cU_0 \big\}.
\]
Then the cover
$\cU \vee h_K^{1} (\cU) \vee \cdots \vee h_K^{n + 1} (\cU)$
sees only $n$ of the coordinates in $K^{\Z}$,
so that
\[
\cD \big( \cU \vee h^{1} (\cU) \vee \cdots \vee h^{n + 1} (\cU)
\big)
\leq \dim (K^n)
\leq n \dim (K).
\]
The proof of Theorem~\ref{T_6913_ShiftEst}
requires only one modification of this idea,
namely that the original cover~$\cU$
must be allowed to depend on an arbitrary finite number of
coordinates
rather than just one.
The proof of Theorem~\ref{T_6913_ShiftEq}
requires more work.
One does not expect $\mdim (h_K) = \dim (K)$ in general,
because of the possibility of having
$\dim (K^n) < n \dim (K)$.
(This is the possibility of
having strict inequality in (\ref{Eq_7131_DimProd});
see the discussion after (\ref{Eq_7131_DimProd}).)
When $\dim (K) < \I$,
by combining Theorem~1.4 of~\cite{Drnv}
and Theorem 3.16(b) of~\cite{Drnv} and the discussion afterwards,
one sees that $\dim (K^n)$ is either always
$n \dim (K)$ or always $n \dim (K)  n + 1$.
In the first case,
\[
\limi{n} \frac{\dim (K^n)}{n} = \dim (K),
\]
while in the second case,
\[
\limi{n} \frac{\dim (K^n)}{n} = \dim (K)  1.
\]
Moreover, the second case actually occurs.
(For example, combine \cite{McRb} and~\cite{Krsz}.)
A modification of the proof of Theorem~\ref{T_6913_ShiftEst}
should easily give the upper bound
$\mdim (h_K) \leq \dim (K)  1$ in the second case.
This suggests the following question,
which, as far as we know,
has not been addressed.
\begin{qst}\label{Q_6913_DimShift}
Let $K$ be a \cms,
and let $h_K$ be as in Definition~\ref{D_5611_Shift}.
Does it follow that $\mdim (h_K) = \dim (K)$
or $\mdim (h_K) = \dim (K)  1$?
\end{qst}
% We expect that the answer should be yes,
% but, as far as we know, this has not been proved.
Shifts are not minimal
(unless $K$ has at most one point),
but one can construct minimal subshifts with large
mean dimension.
A basic construction of this type is given in~\cite{LndWs}.
\begin{thm}[Proposition~3.5 of~\cite{LndWs}]\label{T_6913_MinLg}
There exists a minimal invariant subset $X \subset ([0, 1]^2)^{\Z}$
such that $\mdim (h_{[0, 1]^2} _X) > 1$.
\end{thm}
A related construction is used in~\cite{GlKr}
to produce many more examples,
including ones with arbitrarily large mean dimension.
We now recall the statement of Theorem~\ref{T_5526_rcmdim}.
\begin{thm}[\cite{HPT}]\label{T_5526_rcmdim_Rpt}
Let $X$ be a compact metric space.
Assume that
there is a continuous surjective map from $X$ to the Cantor set.
Let $h \colon X \to X$ be a minimal homeomorphism.
Then $\rc (C^* (\Z, X, h)) \leq \frac{1}{2} \mdim (h)$.
\end{thm}
It is hoped that
$\rc (C^* (\Z, X, h)) = \frac{1}{2} \mdim (h)$
for any \mh{}
of an infinite compact metric space~$X$.
This has been proved in~\cite{HPT}
for some special systems
covered by Theorem~\ref{T_5526_rcmdim_Rpt},
slightly generalizing
the construction of~\cite{GlKr}.
The hypothesis on existence of a surjective map to
the Cantor set has other equivalent formulations,
one of which is the existence
of an equivariant surjective map to
the Cantor set.
\begin{prp}\label{P_6827_CantorFactor}
Let $X$ be a compact metric space,
and let $h \colon X \to X$ be a minimal homeomorphism.
Then \tfae:
\begin{enumerate}
\item\label{P_5406_CantorFactor_Decr}
There exists a decreasing sequence
$Y_0 \supset Y_1 \supset Y_2 \supset \cdots$
of nonempty compact open subsets of~$X$ such that the subset
$Y = \bigcap_{n = 0}^{\infty} Y_n$
satisfies
$h^r (Y) \cap Y = \varnothing$ for all $r \in \Z \setminus \{ 0 \}$.
\item\label{P_5406_CantorFactor_Factor}
There is a minimal homeomorphism of the Cantor set
which is a factor of $(X, h)$
(\Def{D_6827_Factor}).
\item\label{P_5406_CantorFactor_Surj}
There is a continuous surjective map from $X$ to the Cantor set.
\item\label{P_5406_CantorFactor_Partition}
For every $n \in \N$ there is a partition $\cP$
of $X$ into at least $n$ nonempty compact open subsets.
\end{enumerate}
\end{prp}
We omit the proof.
Assume $h$ is minimal and $h^n (Y) \cap Y = \E$
for $n \in \Z \setminus \{ 0 \}$.
As in Remark~\ref{R_7121_LimOfSubalgs},
write $Y = \bigcap_{n = 0}^{\I} Y_n$
with $Y_0 \supset Y_1 \supset \cdots$
and $\sint (Y_n) \neq \E$
for all $n \in \Nz$,
getting
\[
C^* (\Z, X, h)_Y = \varinjlim_{n} C^* (\Z, X, h)_{Y_n}.
\]
The algebras $C^* (\Z, X, h)_{Y_n}$
are recursive subhomogeneous \ca{s}
whose base spaces are closed subsets of~$X$.
(See Theorem~\ref{T_4317_AYIsRsha}.)
The effect of requiring a Cantor system factor
is that one can choose $Y$ and $(Y_n)_{n \in \Nz}$
so that $Y_n$ is both closed and open for all $n \in \Nz$.
Doing so ensures that $C^* (\Z, X, h)_{Y_n}$ is a
homogeneous \ca{}
whose base spaces are closed subsets of~$X$.
Thus $C^* (\Z, X, h)_Y$ is
a simple AH~algebra.
We get such a set $Y$ by taking the inverse image of a point
in the Cantor set.
To keep things simple,
in these notes
we will assume that $h$ has a particular
\mh{} of the Cantor set as a factor,
namely an odometer system
(\Def{D_6827_Odometer}).
The further simplification of
assuming an odometer factor
is that
one can arrange $C^* (\Z, X, h)_{Y_n} \cong M_{p_n} (C (Y_n))$,
that is, there is only one summand.
This simplifies the notation but
otherwise makes little difference.
We omit the proof of the following lemma.
Some work is required,
most of which consists of keeping notation straight.
A more general version
(assuming an arbitrary \mh{}
of the Cantor set as a factor)
is in~\cite{HPT}.
\begin{lem}\label{L_6913_OBifodometerfactor}
Let $X$ be a compact metric space,
and let $h \colon X \to X$ be a minimal homeomorphism.
We assume that $(X, h)$
has as a factor system the odometer on
$X_d = \prod_{n = 1}^{\infty} \{ 0, 1, 2, \ldots, d_n  1 \}$
(\Def{D_6827_Odometer})
for a sequence $d = \Sq{d}{n}$ of integers
with $d_n \geq 2$ for all~$n \in \N$.
Let
$Y$ be the inverse image of
$(0, 0, \ldots)$ under the factor map,
and let $Y_n$
be the inverse image of
\[
\{ 0 \}^n \times
\prod_{k = n + 1}^{\infty} \{ 0, 1, 2, \ldots, d_k  1 \}.
\]
For $n \in \N$ set $p_n = \prod_{k = 1}^n d_k$.
For $m, n \in \Nz$ with $n \geq m$,
define
\[
\ps_{n, m} \colon C (Y_m, M_{p_m}) \to C (Y_n, M_{p_n})
\]
by
\[
\ps_{n, m} (f)
= \diag \big( f _{Y_n}, \, f \circ h^{p_m} _{Y_n},
\, f \circ h^{2 p_m} _{Y_n}, \, \ldots,
\, f \circ h^{(p_n / p_m  1) p_m} _{Y_n}
\big)
\]
for $f \in C (Y_m, M_{p_m})$.
Then
\[
C^* (\Z, X, h)_{Y}
\cong \varinjlim_{n} C (Y_n, M_{p_n}).
\]
\end{lem}
The map $\ps_{n, 0}$ in the statement of the lemma
has the particularly suggestive formula
\[
\ps_{n, 0} (f)
= \diag \big( f _{Y_n}, \, f \circ h _{Y_n},
\, f \circ h^{2} _{Y_n}, \, \ldots,
\, f \circ h^{p_n  1} _{Y_n}
\big).
\]
The problem is now reduced to showing that
if
$A = \varinjlim_{n} C (Y_n, M_{p_n})$,
with maps
\[
\ps_{n, m} (f)
= \diag \big( f _{Y_n}, \, f \circ h^{p_m} _{Y_n},
\, f \circ h^{2 p_m} _{Y_n}, \, \ldots,
\, f \circ h^{(p_n / p_m  1) p_m} _{Y_n}
\big),
\]
then
$\rc (A) \leq \frac{1}{2} \mdim (h)$.
We will make a further simplification,
and prove instead the following theorem,
also from~\cite{HPT}.
\begin{thm}[\cite{HPT}]\label{T_6914_1SummAHModel}
Let $X$ be an infinite \cms.
Let $d = \Sq{d}{n}$ be a sequence of integers
with $d_n \geq 2$ for all~$n \in \N$.
For $n \in \N$ set $p_n = \prod_{k = 1}^n d_k$.
Let $h \colon X \to X$ be a homeomorphism,
and suppose that $h^{p_n}$ is minimal for all~$n \in N$.
For $m, n \in \N$ with $m \leq n$,
define
$\ps_{n, m} \colon C (X, M_{p_m}) \to C (X, M_{p_n})$
by
\[
\ps_{n, m} (f)
= \diag \big( f, \, f \circ h^{p_m},
\, f \circ h^{2 p_m}, \, \ldots,
\, f \circ h^{(p_n / p_m  1) p_m}
\big)
\]
for $f \in C (X, M_{p_m})$.
Using these maps,
define
\[
B = \varinjlim_{n} C (X, M_{p_n}).
\]
Then $\rc (B) \leq \frac{1}{2} \mdim (h)$.
\end{thm}
The following lemma (whose easy proof is left as an exercise)
ensures that the direct system in Theorem~\ref{T_6914_1SummAHModel}
actually makes sense.
\begin{lem}[\cite{HPT}]\label{L_6914_Consist}
Let $X$, $h$, $d$, and $\ps_{n, m}$ for $m, n \in \N$
with $m \leq n$, be as in Theorem~\ref{T_6914_1SummAHModel},
but without any minimality assumptions on~$h$.
Then for $k, m, n \in \N$ with $k \leq m \leq n$,
we have $\ps_{n, m} \circ \ps_{m, k} = \ps_{n, k}$.
\end{lem}
\begin{exr}\label{Ex_6914_Consist}
Prove Lemma~\ref{L_6914_Consist}.
\end{exr}
The algebra $B$ in Theorem~\ref{T_6914_1SummAHModel} is a kind of
AH~model for the crossed product $C^* (\Z, X, h)$.
In particular,
it is always an AH~algebra,
while we needed the assumption of a Cantor set factor system
to find a large subalgebra of $C^* (\Z, X, h)$
which is an AH~algebra.
This model has the defect that
we must now assume that $h^{p_n}$ is minimal for all~$n \in \N$.
Otherwise,
it turns out that the direct limit isn't simple.
(This minimality condition on the powers
actually excludes systems with odometer factors.)
The proof of the following lemma is a fairly direct consequence
of the simplicity criterion in Proposition 2.1(iii) of~\cite{DNNP}.
\begin{lem}[\cite{HPT}]\label{L_6914_Simple}
Let $X$, $h$, $d$, and $\ps_{n, m}$ for $m, n \in \N$
with $m \leq n$, be as in Theorem~\ref{T_6914_1SummAHModel},
but without any minimality assumptions on~$h$.
Set $B = \varinjlim_{n} C (X, M_{p_n})$.
Then $B$ is simple \ifo{}
$h^{p_n}$ is minimal for all~$n \in N$.
\end{lem}
\begin{exr}\label{Ex_6914_Simple}
Prove Lemma~\ref{L_6914_Simple}.
\end{exr}
The main effect
of passing to the situation of Theorem~\ref{T_6914_1SummAHModel}
is to further simplify the notation.
For \mh{s} without Cantor set factor systems,
the replacement of a direct limit of
recursive subhomogeneous algebras with an AH~algebra
of the sort appearing in Theorem~\ref{T_6914_1SummAHModel}
is a
much more substantial simplification.
There are difficulties (presumably technical)
in the more general context which we don't (yet) know how to solve.
We would like to use Theorem~6.2 of~\cite{Niu}
to prove Theorem~\ref{T_6914_1SummAHModel}
(and also
Theorem~\ref{T_5526_rcmdim_Rpt}).
Unfortunately,
the definition there of the mean dimension of an AH~direct system
requires that the base spaces be connected,
or at least have only finitely many connected components.
If $(X, h)$ has a Cantor set factor system,
the base spaces in the AH~model
(and also in the direct system
in Lemma~\ref{L_6913_OBifodometerfactor})
have surjective maps to the Cantor set.
So we proceed more directly,
although the arguments are closely related.
\begin{lem}\label{L_6827_DLimEst}
Let $X$ be a \cms{}
and let $h \colon X \to X$ be a homeomorphism with no periodic points.
Then for every $\ep > 0$
and every finite subset $F \subset C (X)$
there exists $N \in \N$ such that for all $n \geq N$
there is a \cms{} $K$ and a surjective map
$i \colon X \to K$
satisfying:
\begin{enumerate}
\item\label{L_5328_UseMdim_Dim}
$\dim (K) < n [ \mdim (h) + \ep]$.
\item\label{L_5328_UseMdim_Est}
For $m = 0, 1, \ldots, n  1$ and $f \in F$
there is $g \in C (K)$
such that $\ f \circ h^m  g \circ i \ < \ep$.
\end{enumerate}
\end{lem}
The argument depends on nerves of covers
and their geometric realizations.
See Section~2.6 of~\cite{Prs},
especially Definition 2.6.1,
Definition 2.6.2,
Definition 2.6.7,
and the proof of Proposition 2.6.8,
for more details of the theory than are presented here.
\begin{dfn}\label{D_6914_Nerve}
Let $X$ be a topological space,
and let $\cV$ be a finite open cover of~$X$,
with $\E \not\in \cV$.
The {\emph{nerve}} $K (\cV)$ is the finite simplicial complex
with vertices $[V]$ for $V \in \cV$,
and in which there is a simplex in $K (\cV)$ with vertices
$[V_0], [V_1], \ldots, [V_n]$
\ifo{} $V_0 \cap V_1 \cap \cdots \cap V_n \neq \varnothing$.
\end{dfn}
The points $z \in K (\cV)$
(really, points $z$ in its geometric realization)
are thus exactly the formal convex combinations
%
\begin{equation}\label{Eq_5328_Coords}
z = \sum_{V \in \cV} \af_V [V]
\end{equation}
%
in which $\af_V \geq 0$ for all $V \in \cV$,
$\sum_{V \in \cV} \af_V = 1$,
and
$\big\{ [V] \colon \af_V \neq 0 \big\}$
is a simplex in $K (\cV)$,
that is,
\[
\bigcap \big\{ V \in \cV \colon \af_V \neq 0 \big\} \neq \varnothing.
\]
\begin{lem}\label{L_6914_Dim}
Let $X$ be a topological space,
and let $\cV$ be a finite open cover of~$X$,
with $\E \not\in \cV$.
Then $\dim (K (\cV)) = \ord (\cV)$.
\end{lem}
\begin{proof}
It is immediate that $\ord (\cV)$ is the largest
(combinatorial) dimension
of a simplex occurring in $K (\cV)$.
It follows from standard results in dimension theory
(in~\cite{Prs}, see Proposition 3.1.5, Theorem 3.2.5,
and Theorem 3.2.7)
that this dimension is equal to $\dim (K (\cV))$.
\end{proof}
\begin{lem}\label{L_6914_Map}
Let $X$ be a topological space,
and let $\cV$ be a finite open cover of~$X$,
with $\E \not\in \cV$.
Let $(g_V)_{V \in \cV}$
be a partition of unity on~$X$ such that
$\supp (g_V) \subset V$ for all $V \in \cV$.
Then there is a \ct{} map
$i \colon X \to K (\cV)$
determined, using~(\ref{Eq_5328_Coords}), by
\[
i (x) = \sum_{V \in \cV} g_V (x) [V]
\]
for $x \in X$.
\end{lem}
\begin{exr}\label{Ex_6914_Map}
Prove Lemma~\ref{L_6914_Map}.
\end{exr}
This exercise is straightforward.
At this point,
we leave traditional topology.
\begin{lem}\label{L_6914_CndExpt}
Let $X$ be a \chs,
and let $\cV$, $(g_V)_{V \in \cV}$,
and $i \colon X \to K (\cV)$
be as in Lemma~\ref{L_6914_Map}.
Let $(x_V)_{V \in \cV}$
be a collection of points in~$X$
such that $x_V \in V$ for $V \in \cV$.
Then there is a linear map $P \colon C (X) \to C (K (\cV))$
(not a \hm)
defined, following~(\ref{Eq_5328_Coords}),
by
\[
P (f) \left( \sum_{V \in \cV} \af_V [V] \right)
= \sum_{V \in \cV} \af_V f (x_V)
\]
for $f \in C (X)$.
Moreover:
\begin{enumerate}
\item\label{L_6914_CndExpt_Ct}
$\ P \ \leq 1$.
\item\label{L_6914_CndExpt_App}
For all $f \in C (X)$,
we have
\[
\ P (f) \circ i  f \
\leq \sup_{V \in \cV} \sup_{x, y \in V}  f (x)  f (y) .
\]
\end{enumerate}
\end{lem}
The key point is part~(\ref{L_6914_CndExpt_App}):
if $f \in C (X)$ varies by at most $\dt > 0$
over each set $V \in \cV$,
then $P (f)$ is a function on $K (\cV)$
whose pullback to~$X$ is close to~$f$.
That is, if $\cV$ is sufficiently fine,
then we can approximate a finite set of functions on~$X$
by functions on the finite (in particular, \fd{})
simplicial complex $K (\cV)$.
Moreover,
the dimension of $K (\cV)$ is controlled by the order of~$\cV$.
\begin{proof}[Proof of Lemma~\ref{L_6914_CndExpt}]
It is easy to check that $P (f)$ is \ct,
that $P$ is linear,
and that $\ P \ \leq 1$.
For~(\ref{L_6914_CndExpt_App}),
let $r > 0$ and
suppose that for all $V \in \cV$ and $x, y \in V$ we have
$ f (x)  f (y)  \leq r$.
Let $x \in X$ and estimate:
\begin{align*}
\big P (f) (i (x))  f (x) \big
% & = \left P (f) \left( \sum_{V \in \cV} g_V (x) [V] \right)
%  f (x) \right
% = \left \sum_{V \in \cV} g_V (x) [f (x_V)  f (x)] \right
& = \left P (f) \left( \sum_{V \in \cV} g_V (x) [V] \right)
 \sum_{V \in \cV} g_V (x) f (x) \right
\\
& \leq \sum_{V \in \cV} g_V (x)  f (x_V)  f (x) 
\leq \sum_{V \in \cV} g_V (x) r
= r.
\end{align*}
This completes the proof.
\end{proof}
\begin{proof}[Proof of Lemma~\ref{L_6827_DLimEst}]
Choose a finite open cover $\cU$ of~$X$
such that for all $U \in \cU$, $x, y \in U$,
and $f \in F$,
we have $ f (x)  f (y)  < \frac{\ep}{2}$.
By definition,
we have
\[
\lim_{n \to \infty}
\frac{\cD \big(
\cU \vee h^{1} (\cU) \vee \cdots \vee h^{n + 1} (\cU)
\big)}{n}
\leq \mdim (h).
\]
Therefore there exists $N \in \N$ such that for all $n \geq N$ we have
\[
\frac{\cD \big( \cU \vee h^{1} (\cU) \vee \cdots \vee h^{n + 1} (\cU)
\big)}{n}
< \mdim (h) + \ep.
\]
Let $n \geq N$.
Then there is a finite open cover $\cV$ of $X$
which refines
\[
\cU \vee h^{1} (\cU) \vee \cdots \vee h^{n + 1} (\cU)
\]
and such that
%
\begin{equation}\label{Eq_5405_NLabel}
\ord (\cV) < n [\mdim (h) + \ep].
\end{equation}
%
Since $X$ is a \cms,
we can choose a partition of unity $(g_V)_{V \in \cV}$ on~$X$
such that $\supp (g_V) \subset V$ for all $V \in \cV$.
Apply Lemma~\ref{L_6914_Map},
getting $i \colon X \to K (\cV)$,
and let $P \colon C (X) \to C (K (\cV))$
be as in Lemma~\ref{L_6914_CndExpt}.
Let $f \in F$
and let $m \in \{ 0, 1, \ldots, n  1 \}$.
Since $\cV$ refines $h^{m} (\cU)$,
it follows that
for all $V \in \cV$ and $x, y \in V$ we have
$\big (f \circ h^m) (x)  (f \circ h^m) (y) \big < \frac{\ep}{2}$.
So
\[
\big\ P (f \circ h^m) \circ i  f \circ h^m \big\
\leq \frac{\ep}{2} < \ep.
\]
We are done with the proof except for the fact that
$i$ might not be surjective.
So define $K = i (X) \subset K (\cV)$.
Since the dimension of a subspace can't be larger
than the dimension of the whole space
(see Proposition 3.1 5 of~\cite{Prs}),
\[
\dim (K)
\leq \dim (K (\cV))
= \ord (\cV)
< n [\mdim (h) + \ep].
\]
In place of $P (f \circ h^m)$ we use
$P (f \circ h^m) _K$.
This completes the proof.
\end{proof}
The proof of Theorem~\ref{T_6914_1SummAHModel}
requires two further results.
For both proofs,
we refer to the original sources.
The first is a special case of Theorem~5.1 of~\cite{Tm2}.
\begin{thm}[see Theorem~5.1 of~\cite{Tm2}]\label{T_6827_RCCX}
Let $X$ be a \cms{} and let $n \in \N$.
Then
\[
\rc (M_n \otimes C (X)) \leq \frac{\dim (X)  1}{2 n}.
\]
\end{thm}
\begin{lem}[Lemma~6.1 of~\cite{Niu}]\label{L_6827_RCEst}
Let $B$ be a simple unital exact C*algebra
and let $r \in [0, \I)$.
Suppose:
\begin{enumerate}
\item\label{AppMD}
For every finite subset $S \subset B$ and every $\ep > 0$,
there is a \uca~$D$
such that $\rc (D) < r + \ep$
and an injective unital \hm{} $\rh \colon D \to B$
such that $\dist (a, \rh (D)) < \ep$
for all $a \in S$.
\item\label{UnifRank}
For every $s \in [0, 1]$ and every $\ep > 0$,
there exists a projection $p \in B$ with $ \ta (p)  s  < \ep$
for all $\ta \in \T (B)$.
\end{enumerate}
Then $\rc (B) \leq r$.
\end{lem}
\begin{proof}[Proof of Theorem~\ref{T_6914_1SummAHModel}]
We use Lemma~\ref{L_6827_RCEst}.
Certainly $B$ is simple, unital, and exact.
Since $C (X, M_{p_n}) \hookrightarrow B$
and $C (X, M_{p_n})$ has projections of constant rank~$k$
for any $k \in \{ 0, 1, \ldots, p_n \}$,
condition~(\ref{UnifRank}) in Lemma~\ref{L_6827_RCEst}
is satisfied.
We need to show that
for every finite subset $S \subset B$ and every $\ep > 0$,
there is a \uca~$D$
such that $\rc (D) < \frac{1}{2} \mdim (h) + \ep$
and an injective unital \hm{}
$\rh \colon D \to B$ such that $\dist (a, \rh (D)) < \ep$
for all $a \in S$.
For $n \in \Nz$
let $\ps_n \colon C (X, M_{p_n}) \to B$
be the map obtained from the direct limit description
of $B$.
Let $S \subset B$ be finite
and let $\ep > 0$.
Choose $m \in \N$ and a finite set
\[
S_0 \subset C (X, M_{p_m}) = M_{p_m} (C (X))
\]
such that for every $a \in S$ there is $b \in S_0$ with
$\ \ps_m (b)  a \ < \frac{1}{2} \ep$.
Let $F \subset C (X)$ be the set of all matrix entries
of elements of~$S_0$.
Use Lemma~\ref{L_6827_DLimEst}
% \ref{L_5328_UseMdim}
to find $N \in \Nz$ such that for all $l \geq N$
there are a \cms{} $K$ and a surjective map
$i \colon X \to K$
such that $\dim (K) < l [ \mdim (h) + \ep]$
and for $r = 0, 1, \ldots, l  1$ and $f \in F$
there is $g \in C (K)$
with
\[
\ f \circ h^r  g \circ i \ < \frac{\ep}{2 p_m^2}.
\]
Choose $n \geq m$ such that $p_n \geq N$.
Choose $K$ and $i$ for $l = p_n$,
so that
\[
\dim (K) < p_n [ \mdim (h) + \ep ]
\]
and for $r = 0, 1, \ldots, p_n  1$ and $f \in F$
there is $g \in C (K)$
with
\[
\ f \circ h^r  g \circ i \ < \frac{\ep}{2 p_m^2}.
\]
Define an injective homomorphism
$\rh_0 \colon C (K) \to C (X)$
by $\rh_0 (f) = f \circ i$ for $f \in C (K)$.
Set
$D = M_{p_n} ( C (K) )$
and define
\[
\rh = \ps_n \circ (\id_{M_{p_n}} \otimes \rh_0)
\colon D \to B.
\]
Then $\rh$ is also injective.
By Theorem~\ref{T_6827_RCCX},
\[
\rc (D)
\leq \frac{\dim (K)  1}{2 p_n}
< \frac{\mdim (h) + \ep}{2}
< \frac{\mdim (h)}{2} + \ep.
\]
It remains to prove that $\dist (a, \rh (D)) < \ep$
for all $a \in S$.
Let $a \in S$.
Choose $b \in S_0$ such that
$\ \ps_m (b)  a \ < \frac{\ep}{2}$.
For $j, k \in \{ 0, 1, \ldots, p_m  1 \}$,
we let $e_{j, k} \in M_{p_m}$ be the standard matrix unit
(except that we start the indexing at $0$ rather than~$1$).
Then there are $b_{j, k} \in F$
for $j, k \in \{ 0, 1, \ldots, p_m  1 \}$
such that $b = \sum_{j, k = 0}^{p_m  1} e_{j, k} \otimes b_{j, k}$.
By construction,
for $r = 0, 1, \ldots, p_n  1$
there is $g_{j, k, r} \in C (K)$
such that
\[
\big\ g_{j, k, r} \circ i  b_{j, k} \circ h^r \big\
< \frac{\ep}{2 p_m^2}.
\]
For $t = 0, 1, \ldots, p_n / p_m  1$,
define
\[
c_t = \sum_{j, k = 0}^{p_m  1}
e_{j, k} \otimes g_{j, k, t p_m}
\in M_{p_m} (C (K)).
\]
Then define
\[
c = \diag \big( c_0, c_1, \ldots, c_{p_n / p_m  1} \big)
\in M_{p_n} (C (K)).
\]
We claim that $\ \rh (c)  a \ < \ep$,
which will finish the proof.
We have,
using the definition of $\ps_{n, m}$ at the third step,
\begin{align*}
\ \rh (c)  a \
& \leq \ a  \ps_m (b) \ + \ \ps_m (b)  \rh (c) \
\\
& < \frac{\ep}{2} + \ \ps_{n, m} (b)  c \
\\
& < \frac{\ep}{2}
+ \big\ \diag \big( f, \, f \circ h^{p_m},
\, f \circ h^{2 p_m}, \, \ldots,
\, f \circ h^{(p_n / p_m  1) p_m} \big)
\\
& \hspace*{9em} {\mbox{}}
 \diag \big( c_0, c_1, c_2, \ldots, c_{p_n / p_m  1} \big) \big\
\\
& \leq \frac{\ep}{2}
+ \max_{0 \leq t \leq p_n / p_m  1} \ f \circ h^{p_m r}  c_t \
\\
& \leq \frac{\ep}{2}
+ \max_{0 \leq t \leq p_n / p_m  1}
\sum_{j, k = 1}^{p_m}
\big\ g_{j, k, r} \circ i  b_{j, k} \circ h^r \big\
\\
& \leq \frac{\ep}{2}
+ p_m^2 \left( \frac{\ep}{2 p_m^2} \right)
= \ep.
\end{align*}
This completes the proof.
\end{proof}
\section{Open Problems on Large Subalgebras
and their Applications to Crossed Products}\label{Sec_OP}
We discuss some open problems
related to large subalgebras,
some (but not all)
of which have some connection with dynamical systems.
We start with some which are motivated by particular
applications,
and then give some which are suggested by results
already proved but for which
we don't have immediate applications.
Not all the problems in~\cite{Ph_lgsvy} appear here.
In particular,
the ones about $L^p$~operator crossed products
have been omitted.
The first question is motivated by the hope that
large subalgebras can be used to get more information
about crossed products than we now know how to get.
In most parts,
we expect that positive answers would require
special hypotheses,
if they can be gotten at all.
We omit definitions of most of the terms.
\begin{qst}\label{Q_5519_TRZ}
Let $A$ be an infinite dimensional
simple separable unital C*algebra,
and let $B \subset A$ be a large
(or centrally large) subalgebra.
\begin{enumerate}
\item\label{Q_5519_TRZ_TRZ}
Suppose that $B$ has tracial rank zero
(\Def{D_TR0}).
Does it follow that $A$ has tracial rank zero?
\item\label{Q_5519_TRZ_QD}
Suppose that $B$ is quasidiagonal.
Does it follow that $A$ is quasidiagonal?
\item\label{Q_5519_TRZ_dr}
Suppose that $B$ has finite decomposition rank.
Does it follow that $A$ has finite decomposition rank?
\item\label{Q_5519_TRZ_NucDim}
Suppose that $B$ has finite nuclear dimension.
Does it follow that $A$ has finite nuclear dimension?
\end{enumerate}
\end{qst}
It seems likely that ``tracial'' versions of these
properties
pass from a large subalgebra to the containing algebra,
at least if the tracial versions are defined
using cutdowns by positive elements rather than by \pj{s}.
But we don't know how useful such properties are.
As far as we know,
they have not been studied.
Next, we ask whether being stably large is automatic.
\begin{qst}\label{Q_5421_LgImpStb}
Let $A$ be an infinite dimensional
simple separable unital C*algebra,
and let $B \subset A$ be a large
(or centrally large) subalgebra.
Does it follow that $M_n (B)$ is large
(or centrally large) in $M_n (A)$
for $n \in \N$?
\end{qst}
We know that this is true if $A$ is stably finite,
by Proposition~\ref{P_5519_StFinStLg}.
Not having the general statement is a technical annoyance.
This result would be helpful
when dealing with large subalgebras of
$C^* (\Z, \, C (X, D), \, \af)$
when $D$ is simple unital,
$X$ is compact metric,
and the \hme{} of $\Prim (C (X, D)) \cong X$
induced by $\af$ is minimal.
Some results on large subalgebras of such crossed products
can be found in~\cite{ArBcPh};
also see \Thm{T_5418_AYLg}.
More generally, does
Proposition~\ref{P_4624_TLarge}
still hold without the finiteness assumption?
\begin{qst}\label{Q_5421_ZTrp}
Let $A$ be an infinite dimensional
simple separable unital C*algebra,
and let $\af \colon \Z \to \Aut (A)$
have the tracial Rokhlin property.
Is there a useful large or centrally large subalgebra
of $C^* (\Z, A, \af)$?
\end{qst}
We want a centrally large subalgebra of $C^* (\Z, A, \af)$
which ``locally looks like matrices over corners of~$A$''.
The paper~\cite{OP1}
proves that crossed products by automorphisms
with the tracial Rokhlin property
preserve the combination of
real rank zero,
stable rank one,
and order on projections determined by traces.
The methods were inspired by those of~\cite{Ph10},
which used large subalgebras (without the name).
The proof in~\cite{OP1} does not, however,
construct a single large subalgebra.
Instead,
it constructs a suitable subalgebra
(analogous to $C^* (\Z, X, h)_Y$
for a small closed subset $Y \subset X$
with $\sint (Y) \neq \E$)
for every choice of finite set $F \subset C^* (\Z, A, \af)$
and every choice of $\ep > 0$.
It is far from clear how to choose these subalgebras
to form an increasing sequence so that a direct limit
can be built.
Similar ideas, under weaker hypotheses (without \pj{s}),
are used in~\cite{OvPhWn},
and there it is also far from clear how to choose the subalgebras
to form an increasing sequence.
The first intended application is simplification of~\cite{OP1}.
\begin{pbm}\label{Pb_5519_Zd}
Let $X$ be a compact metric space,
and let $G$ be a countable amenable group
which acts minimally and essentially freely on~$X$.
Construct a (centrally) large subalgebra $B \subset C^* (G, X)$
which is a direct limit of recursive subhomogeneous C*algebras
as in~\cite{Ph_RSHA1}
whose base spaces are closed subsets of~$X$,
and which is the (reduced) \ca{}
of an open subgroupoid of the transformation group groupoid
obtained from the action of $G$ on~$X$.
\end{pbm}
In a precursor to the theory of large subalgebras,
this is in effect done in~\cite{Ph10}
when $G = \Z^d$ and $X$ is the Cantor set,
following ideas of~\cite{Fr}.
The resulting centrally large subalgebra is used in~\cite{Ph10}
to prove that
$C^* (\Z^d, X)$ has stable rank one,
real rank zero,
and order on projections determined by traces.
(More is now known.)
We also know how to construct a centrally large subalgebra
of this kind
when $G = \Z^d$ and $X$ is finite dimensional (unpublished).
This gave the first proof that,
in this case,
$C^* (\Z^d, X)$
has stable rank one and strict comparison
of positive elements.
(Again, more is now known.)
Unlike for actions of~$\Z$,
there are no known explicit formulas like that in
\Thm{T_5421_AYStabLg};
instead, centrally large subalgebras
must be proved to exist via constructions involving many choices.
They are direct limits of
\ca{s} of open subgroupoids of the transformation group groupoid
as in Problem~\ref{Pb_5519_Zd}.
In each open subgroupoid,
there is a finite upper bound on the size of the orbits;
this is why they are recursive subhomogeneous C*algebras
(homogeneous when $X$ is the Cantor set, as in~\cite{Ph7}).
In fact,
the original motivation for the definition
of a large subalgebras
was to describe the essential properties of these subalgebras,
as a substitute for an explicit description.
We presume,
as suggested in Problem~\ref{Pb_5519_Zd},
that the construction can be done in much greater generality.
\begin{pbm}\label{Pb_5519_NUnital}
Develop the theory of large subalgebras of not necessarily
simple \ca{s}.
\end{pbm}
One can't just copy Definition~\ref{D_5421_Large}.
Suppose $B$ is a nontrivial large subalgebra of~$A$.
We surely want $B \oplus B$ to be a large subalgebra of $A \oplus A$.
Take $x_0 \in A_{+} \setminus \{ 0 \}$,
and take the element $x \in A \oplus A$
in Definition~\ref{D_5421_Large}
to be $x = (x_0, 0)$.
Writing $g = (g_1, g_2)$,
we have forced $g_2 = 0$.
Thus,
not only would $B \oplus B$ not be large in $A \oplus A$,
but even $A \oplus B$ would not be large in $A \oplus A$.
In this particular case,
the solution is to require that $x$ and~$y$
be full elements in $A$ and~$B$.
What to do is much less clear if,
for example,
$A$ is a unital extension of the form
\[
0 \longrightarrow K \otimes D
\longrightarrow A
\longrightarrow E
\longrightarrow 0,
\]
even if $D$ and $E$ are simple,
to say nothing of the general case.
The following problem goes just a small step away
from the simple case,
and just asking that $x$ and~$y$ be full might possibly work for it,
although stronger hypotheses may be necessary.
\begin{qst}\label{Q_5421_NonSimp}
Let $X$ be an infinite \cms{}
and let $h \colon X \to X$ be a homeomorphism
which has a factor system which is a
minimal homeomorphism
of an infinite \cms{}
(or, stronger,
a minimal homeomorphism of the Cantor set).
Can one use large subalgebra methods to relate
the mean dimension of $h$
to the radius of comparison
of $C^* (\Z, X, h)$?
\end{qst}
We point out that
Lindenstrauss's embedding result
for systems of finite mean dimension
in shifts built from finite dimensional spaces
(Theorem~5.1 of~\cite{Lnd})
is proved for \hme{s}
having a factor system which is a
minimal homeomorphism of an infinite \cms.
\begin{pbm}\label{Pb_5519_NonU}
Develop the theory of large subalgebras of simple but
not necessarily unital \ca{s}.
\end{pbm}
One intended application is to
crossed products
$C^* \big( \Z, \, C (X, D), \, \af \big)$
when $X$ is an infinite compact metric space,
$D$ is simple but not unital,
and the induced action on $X$ is given by a \mh.
(Compare with Theorem~\ref{T_5418_AYLg}.)
Another possible application
is to the structure of crossed products
$C^* (\Z, X, h)$
when $h$ is a \mh{}
of a noncompact version of the Cantor set.
Minimal \hme{s} of noncompact Cantor sets have
been studied in \cite{Mti} and~\cite{Mt2},
but, as far as we know,
almost nothing is known about their
transformation group \ca{s}.
For a large subalgebra $B \subset A$,
the proofs of most of the relations between $A$ and $B$
do not need $B$ to be centrally large.
The exceptions so far are for stable rank one
and $Z$stability.
Do we really need centrally large for these results?
\begin{qst}\label{Q_5519_NeedZForTsr}
Let $A$ be an infinite dimensional
simple separable unital C*algebra,
and let $B \subset A$ be a large subalgebra
(not necessarily centrally large).
If $B$ has stable rank one,
does it follow that $A$ has stable rank one?
\end{qst}
That is,
can \Thm{T_5519_tsr}
be generalized from centrally large subalgebras
to large subalgebras?
\begin{qst}\label{Q_5719_NeedZForZStab}
Let $A$ be an infinite dimensional
simple separable nuclear unital C*algebra,
and let $B \subset A$ be a large subalgebra
(not necessarily centrally large).
If $B$ is $Z$stable,
does it follow that $A$ is $Z$stable?
\end{qst}
That is,
can \Thm{T_5519_ZStab}
be generalized from centrally large subalgebras
to large subalgebras?
It is not clear how important these questions are.
In all applications so far,
with the single exception of~\cite{EN1}
(on the extended irrational rotation algebras),
the large subalgebras used are known to be centrally large.
In particular, all known useful large subalgebras
of crossed products are already known to be centrally large.
\begin{qst}\label{Q_5519_LargeNotZ}
Does there exist a large subalgebra
which is not centrally large?
Are there natural examples?
\end{qst}
The results of~\cite{EN1} depend on large subalgebras
which are not proved there to be centrally large,
but it isn't known that they are not centrally large.
\begin{qst}\label{Q_5421_RRZ}
Let $A$ be an infinite dimensional
simple separable unital C*algebra,
and let $B \subset A$ be a large
subalgebra.
If $\RR (B) = 0$,
does it follow that $\RR (A) = 0$?
What about the converse?
Does it help to assume that $B$ is centrally large
in the sense of Definition~\ref{D_5421_CentLarge}?
\end{qst}
If $B$ has both stable rank one and real rank zero,
and is centrally large in~$A$,
then $A$ has real rank zero
(as well as stable rank one)
by Theorem~\ref{T_5519_tsr}.
The main point of Question~\ref{Q_5421_RRZ} is to ask what happens
if $B$ is not assumed to have stable rank one.
The proof in~\cite{Ph10}
of real rank zero for the crossed product $C^* (\Z^d, X)$
of a free minimal action of $\Z^d$ on the Cantor set~$X$
(see Theorem 6.11(2) of~\cite{Ph10};
the main part is Theorem 4.6 of~\cite{Ph10})
gives reason to hope that
if $B$ is large in~$A$ and $\RR (B) = 0$,
then one does indeed get $\RR (A) = 0$.
Proposition~\ref{P_6X05_SP} could also be considered evidence
in favor.
Nothing at all is known about conditions under which
$\RR (A) = 0$ implies $\RR (B) = 0$.
Applications to crossed products may be unlikely.
It seems possible that
$C^* (G, X)$ has stable rank one for every
minimal essentially free action of a countable
amenable group~$G$ on a compact metric space~$X$.
\begin{qst}\label{Q_5421_K0}
Let $A$ be an infinite dimensional
simple separable unital C*algebra.
Let $B \subset A$ be centrally large
in the sense of Definition~\ref{D_5421_CentLarge}.
Does it follow that $K_0 (B) \to K_0 (A)$
is an isomorphism mod infinitesimals?
\end{qst}
In other places where this issue occurs
(in connection with tracial approximate innerness;
see Proposition~6.2 and Theorem~6.4 of~\cite{PhT1}),
it seems that everything in $K_1$ should be considered to be
infinitesimal.
A six term exact sequence
for the Ktheory of some orbit breaking subalgebras
is given in Example~2.6 of~\cite{Pt4}.
Related computations for some special more complicated
orbit breaking subalgebras
can be found in~\cite{Pt5}.
See \Thm{TPVForParAct} and the discussion afterwards.
\Thm{T_7122_KThyAy},
according to which the inclusion
of $C^* (\Z, X, h)_{\{ y \}}$ in $C^* (\Z, X, h)$
is an isomorphism on~$K_0$,
is an important consequence.
A positive answer to Question~\ref{Q_5421_K0}
would shed some light on both directions
in Question~\ref{Q_5421_RRZ}.
\begin{qst}\label{Q_5421_Tsr1Dn}
Let $A$ be an infinite dimensional stably finite
simple separable unital C*algebra.
Let $B \subset A$ be centrally large
in the sense of Definition~\ref{D_5421_CentLarge}.
If $A$ has stable rank one,
does it follow that $B$ has stable rank one?
\end{qst}
That is, does \Thm{T_5519_tsr} have a converse?
In many other results in Section~\ref{Sec_Intro},
$B$ has an interesting property \ifo{} $A$ does.
\begin{qst}\label{Q_5421_tsrnUp}
Let $A$ be an infinite dimensional
simple separable unital C*algebra,
and let $B \subset A$ be a
centrally large subalgebra.
Let $n \in \N$.
If $\tsr (B) \leq n$,
does it follow that $\tsr (A) \leq n$?
If $\tsr (B)$ is finite,
does it follow that $\tsr (A)$ is finite?
\end{qst}
That is, can \Thm{T_5519_tsr}
be generalized to other values
of the stable rank?
The proof of \Thm{T_5519_tsr}
uses $\tsr (B) = 1$ in two different places,
one of which is not directly related to $\tsr (A)$,
so an obvious approach seems unlikely to succeed.
As with Question~\ref{Q_5421_RRZ},
applications to crossed products seem unlikely.
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\end{document}