\documentclass{beamer}
% \documentclass[handout]{beamer}
\usetheme{Boadilla}
\usepackage{graphicx}
\usepackage{pgfpages}
% \pgfpagesuselayout{4 on 1}[letterpaper,border shrink=5mm, landscape]
\theoremstyle{definition}
\newtheorem{thm}{Theorem}[section]
\newtheorem{lem}[thm]{Lemma}
\newtheorem{prp}[thm]{Proposition}
\newtheorem{dfn}[thm]{Definition}
\newtheorem{cor}[thm]{Corollary}
\newtheorem{cnj}[thm]{Conjecture}
\newtheorem{cnv}[thm]{Convention}
\newtheorem{rmk}[thm]{Remark}
\newtheorem{ntn}[thm]{Notation}
\newtheorem{exa}[thm]{Example}
\newtheorem{pbm}[thm]{Problem}
\newtheorem{wrn}[thm]{Warning}
\newtheorem{qst}[thm]{Question}
\newtheorem{exr}[thm]{Exercise}
\newcommand{\af}{\alpha}
\newcommand{\bt}{\beta}
\newcommand{\gm}{\gamma}
\newcommand{\dt}{\delta}
\newcommand{\ep}{\varepsilon}
\newcommand{\zt}{\zeta}
\newcommand{\et}{\eta}
\newcommand{\ch}{\chi}
\newcommand{\io}{\iota}
\newcommand{\te}{\theta}
\newcommand{\ld}{\lambda}
\newcommand{\sm}{\sigma}
\newcommand{\kp}{\kappa}
\newcommand{\ph}{\varphi}
\newcommand{\ps}{\psi}
\newcommand{\rh}{\rho}
\newcommand{\om}{\omega}
\newcommand{\ta}{\tau}
\newcommand{\Gm}{\Gamma}
\newcommand{\Dt}{\Delta}
\newcommand{\Et}{\Eta}
\newcommand{\Th}{\Theta}
\newcommand{\Ld}{\Lambda}
\newcommand{\Sm}{\Sigma}
\newcommand{\Ph}{\Phi}
\newcommand{\Ps}{\Psi}
\newcommand{\Om}{\Omega}
\newcommand{\Q}{{\mathbb{Q}}}
\newcommand{\Z}{{\mathbb{Z}}}
\newcommand{\R}{{\mathbb{R}}}
\newcommand{\C}{{\mathbb{C}}}
% \newcommand{\N}{{\mathbb{N}}}
\newcommand{\N}{{\mathbb{Z}}_{> 0}}
\newcommand{\Nz}{{\mathbb{Z}}_{\geq 0}}
\pagenumbering{arabic}
\newcommand{\inv}{{\mathrm{inv}}}
\newcommand{\tsr}{{\mathrm{tsr}}}
\newcommand{\RR}{{\mathrm{RR}}}
\newcommand{\Tr}{{\mathrm{Tr}}}
\newcommand{\id}{{\mathrm{id}}}
\newcommand{\ev}{{\mathrm{ev}}}
\newcommand{\sint}{{\mathrm{int}}}
\newcommand{\diam}{{\mathrm{diam}}}
\newcommand{\dist}{{\mathrm{dist}}}
\newcommand{\sa}{{\mathrm{sa}}}
\newcommand{\spec}{{\mathrm{sp}}}
\newcommand{\Prim}{{\mathrm{Prim}}}
\newcommand{\diag}{{\mathrm{diag}}}
\newcommand{\supp}{{\mathrm{supp}}}
\newcommand{\rank}{{\mathrm{rank}}}
\newcommand{\sgn}{{\mathrm{sgn}}}
\newcommand{\spn}{{\mathrm{span}}}
\newcommand{\card}{{\mathrm{card}}}
\newcommand{\Aut}{{\mathrm{Aut}}}
\newcommand{\Ad}{{\mathrm{Ad}}}
\newcommand{\Aff}{{\mathrm{Aff}}}
\newcommand{\dirlim}{\varinjlim}
% \newcommand{\dirlim}{\displaystyle \lim_{\longrightarrow}}
\newcommand{\invlim}{\varprojlim}
\newcommand{\Mi}{M_{\infty}}
\newcommand{\limi}[1]{\lim_{{#1} \to \infty}}
\newcommand{\SL}{{\mathrm{SL}}}
\newcommand{\Dirlim}{\varinjlim}
\newcommand{\andeqn}{\,\,\,\,\,\, {\mbox{and}} \,\,\,\,\,\,}
\newcommand{\QED}{\rule{0.4em}{2ex}}
\newcommand{\ts}[1]{{\textstyle{#1}}}
\newcommand{\ds}[1]{{\displaystyle{#1}}}
\newcommand{\ssum}[1]{{\ts{ {\ds{\sum}}_{#1} }}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\newcommand{\wolog}{without loss of generality}
\newcommand{\Wolog}{Without loss of generality}
\newcommand{\Tfae}{The following are equivalent}
\newcommand{\tfae}{the following are equivalent}
\newcommand{\ifo}{if and only if}
\newcommand{\wrt}{with respect to}
\newcommand{\ca}{C*-algebra}
\newcommand{\uca}{unital C*-algebra}
\newcommand{\usa}{unital subalgebra}
\newcommand{\hm}{homomorphism}
\newcommand{\uhm}{unital homomorphism}
\newcommand{\am}{automorphism}
\newcommand{\fd}{finite dimensional}
\newcommand{\tst}{tracial state}
\newcommand{\hsa}{hereditary subalgebra}
\newcommand{\pj}{projection}
\newcommand{\mops}{mutually orthogonal \pj s}
\newcommand{\nzp}{nonzero projection}
\newcommand{\mvnt}{Murray-von Neumann equivalent}
\newcommand{\mvnc}{Murray-von Neumann equivalence}
\newcommand{\ct}{continuous}
\newcommand{\cfn}{continuous function}
\newcommand{\nbhd}{neighborhood}
\newcommand{\cms}{compact metric space}
\newcommand{\cpt}{compact Hausdorff}
\newcommand{\chs}{compact Hausdorff space}
\newcommand{\lchs}{locally compact Hausdorff space}
\newcommand{\cg}{compact group}
\newcommand{\lcg}{locally compact group}
\newcommand{\hme}{homeomorphism}
\newcommand{\mh}{minimal homeomorphism}
\newcommand{\mf}{manifold}
\newcommand{\cmf}{compact smooth \mf}
\newcommand{\cp}{crossed product}
\newcommand{\tgca}{transformation group \ca}
\newcommand{\tggp}{transformation group groupoid}
\renewcommand{\S}{\subset}
\newcommand{\ov}{\overline}
\newcommand{\SM}{\setminus}
\newcommand{\I}{\infty}
\newcommand{\E}{\varnothing}
\title[SNU crossed products course: Lecture 2]{Seoul National University
short course:
An introduction to the structure of crossed product C*-algebras.}
\subtitle{Lecture~3: Crossed products by minimal homeomorphisms}
\author{N.~Christopher Phillips}
\institute[U.~of Oregon]{University of Oregon}
\date{15~December 2009}
\begin{document}
\maketitle
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Minimal actions}
We will prove that reduced crossed products by free minimal
actions of countable discrete groups on compact metric spaces
are simple.
\pause
\vspace{2ex}
\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}}
\pause
if
whenever $T \S X$ is a closed subset such that $g T \S T$
for all $g \in G,$
\pause
then $T = \varnothing$ or $T = X.$
\end{dfn}
\pause
\vspace{2ex}
In short, there are no nontrivial invariant closed subsets.
\pause
For those knowing some ergodic theory,
this is the topological analog of an ergodic action
on a measure space.
\pause
\vspace{2ex}
If the action of $G$ on $X$ is not minimal,
then there is a nontrivial invariant closed subset $T \S X,$
\pause
and $C^*_{\mathrm{r}} (G, \, X \setminus T)$
turns out to be a nontrivial
ideal in $C^*_{\mathrm{r}} (G, X).$
\pause
Thus $C^*_{\mathrm{r}} (G, X)$ is not simple.
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Minimal homeomorphisms}
For the case $G = \Z,$
the conventional terminology is a bit different.
\pause
\vspace{2ex}
\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}}
\pause
if
whenever $T \S X$ is a closed subset such that $h (T) = T,$
\pause
then $T = \varnothing$ or $T = X.$
\end{dfn}
\pause
\vspace{2ex}
Almost all work on \mh s has been on compact spaces.
For these, we have the following equivalent conditions.
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Minimal homeomorphisms (continued)}
\begin{lem}\label{L:MHC}
Let $X$ be a \chs,
and let $h \colon X \to X$ be a \hme.
Then \tfae:
\pause
\begin{enumerate}
\item\label{L:MHC:Min}
$h$ is minimal.
\pause
\item\label{L:MHC:FMin}
Whenever $T \S X$ is a closed subset such that $h (T) \subset T,$
then $T = \varnothing$ or $T = X.$
\pause
\item\label{L:MHC:UMin}
Whenever $U \S X$ is an open subset such that $h (U) = U,$
then $U = \varnothing$ or $U = X.$
\pause
\item\label{L:MHC:UFMin}
Whenever $U \S X$ is an open subset such that $h (U) \subset U,$
then $U = \varnothing$ or $U = X.$
\pause
\item\label{L:MHC:DOrb}
For every $x \in X,$ the orbit $\{ h^n (x) \colon n \in \Z \}$
is dense in $X.$
\pause
\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}
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Minimal homeomorphisms (continued)}
Conditions (\ref{L:MHC:Min}), (\ref{L:MHC:UMin}), and~(\ref{L:MHC:DOrb})
are equivalent even when $X$ is only locally compact,
\pause
and in fact there is an analog for actions of arbitrary groups.
\pause
Minimality does not imply the other three conditions
without compactness,
\pause
as can be seen by considering the \hme\ $n \mapsto n + 1$ of $\Z.$
\pause
Also, even for compact $X,$
it isn't good enough to merely have the existence of some
dense orbit,
\pause
as can be seen by considering the \hme\ $n \mapsto n + 1$
on the two point compactification $\Z \cup \{ \pm \I \}$ of $\Z.$
\pause
\vspace{2ex}
\begin{exr}\label{P:MHC}
Prove the lemma.
\end{exr}
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Examples of minimal actions}
\begin{exa}\label{E:MinTr}
Let $G$ be a locally compact group,
let $H \S G$ be a closed subgroup,
and let $G$ act on $G / H$ by translation.
\pause
This action is minimal:
there are no nontrivial invariant subsets, closed or not.
\end{exa}
\pause
\vspace{2ex}
This is a ``trivial'' example
of a minimal action.
\pause
Here are several more interesting ones.
\pause
\vspace{2ex}
\begin{exa}\label{E:MinRot}
Irrational rotations of the circle are minimal.
\end{exa}
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Minimality of irrational rotations}
Minimality of irrational rotations follows from the following lemma.
\pause
\vspace{2ex}
\begin{lem}\label{L:IrratRotDense}
Let $\te \in \R \setminus \Q.$
Then $\big\{ e^{2 \pi i n \te} \colon n \in \Z \big\}$
is dense in~$S^1.$
\end{lem}
\pause
\vspace{2ex}
\begin{proof}
It suffices to prove that $\Z + \te \Z$ is dense in $\R.$
\pause
Suppose not.
Because we are dealing with groups,
there is an open set $U \S \R$
such that $U \cap (\Z + \te \Z) = \{ 0 \}.$
\pause
Let
$t = \inf \big( \big\{ x \in {\overline{\Z + \te \Z}}
\colon x > 0 \big\} \big).$
\pause
Then $t > 0.$
\pause
Since ${\overline{\Z + \te \Z}}$ is a group,
one checks that ${\overline{\Z + \te \Z}} = \Z t.$
\pause
Then we must have $\Z + \te \Z = \Z t.$
\pause
It follows that both $1$ and $\te$ are integer multiples of $t,$
so that $\te \in \Q.$
\end{proof}
\pause
\vspace{2ex}
There are other proofs of minimality.
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{More examples of minimal homeomorphisms}
\begin{exa}\label{E:MinpAdic}
The \hme\ $x \mapsto x + 1$ on the $p$-adic integers
is minimal.
\pause
The orbit of $0$ is $\Z,$
which is dense, essentially by definition.
\pause
Every other orbit is a translate of this one, so also dense.
\end{exa}
\pause
\vspace{2ex}
\begin{exa}\label{E:NonMin}
The shift \hme\ on $\{ 0, 1\}^{\Z}$
and the action of ${\mathrm{SL}}_2 (\Z)$ on $S^1 \times S^1$
are not minimal.
\pause
In fact, they have fixed points.
\end{exa}
\pause
\vspace{2ex}
Other examples of \mh s include Furstenberg transformations,
\pause
restrictions of Denjoy \hme s of the circle to their minimal sets,
\pause
and certain irrational time maps of suspension flows.
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Minimal actions are plentiful}
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.
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Free actions}
The \tgca\ of a minimal action need not be simple.
\pause
Consider, for example, the trivial action of a group $G$
(particularly an abelian group) on a one point space,
for which the \tgca\ is $C^* (G).$
\pause
\vspace{2ex}
\begin{dfn}\label{D:Free}
Let a locally compact group $G$ act \ct ly on a locally
compact space $X.$
The action is called {\emph{free}}
\pause
if
whenever $g \in G \setminus \{ 1 \}$ and $x \in X,$
then $g x \neq x.$
\pause
The action is called {\emph{essentially free}}
\pause
if
whenever $g \in G \setminus \{ 1 \},$
the set $\{ x \in X \colon g x = x \}$ has empty interior.
\end{dfn}
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Remarks on freeness}
Let $X$ be an infinite \chs, and let $h \colon X \to X$ be a \mh.
Then one easily checks that
the corresponding action of $\Z$ on $X$ is free.
\pause
\vspace{2ex}
An essentially free minimal action of an abelian group is free.
\pause
This is because the fixed point set of any group element
is invariant under the action.
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Simplicity of the transformation group C*-algebra}
\begin{thm}[Archbold-Spielberg]\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}
\pause
\vspace{2ex}
Archbold and Spielberg actually proved something stronger,
involving actions on not necessarily commutative \ca s.
\pause
See the notes,
and also see the notes for discussions of other proofs and
related results.
\pause
\vspace{2ex}
% \begin{rmk}\label{R:OnASProof}
The usual proof for $G = \Z$
% (see, for example, Theorem VIII.3.9 of Davidson's book)
depends on Rokhlin type arguments.
\pause
See the proof of Lemma VIII.3.7 of Davidson's book.
\pause
We have avoided such arguments here, but will use them later.
\pause
To obtain more information about simple \tgca s,
such arguments are necessary,
at least with the current state of knowledge.
\pause
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.
% \end{rmk}
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{First lemma}
The proof of the theorem on simplicity needs several lemmas,
which are special cases of the corresponding lemmas
in the paper of Archbold and Spielberg.
\pause
\vspace{2ex}
\begin{lem}\label{L:MState}
Let $A$ be a \ca,
let $B \S A$ be a subalgebra,
and let $\om$ be a state on $A$ such that $\om |_B$ is multiplicative.
\pause
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}
\pause
\vspace{2ex}
This is also a special case of a result of Choi.
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Proof of the first lemma}
We prove $\om (a b) = \om (a) \om (b).$
The other equation will follow by using adjoints and
the relation $\om (c^*) = {\overline{\om (c)}}.$
\pause
\vspace{2ex}
If $A$ is not unital,
then $\om$ extends to a state on the unitization $A^+.$
\pause
Thus, we may assume that $A$ is unital.
\pause
Also, if $\om$ is multiplicative on $B,$
one easily checks that $\om$ is multiplicative on $B + \C \cdot 1.$
\pause
Thus, we may assume that $1 \in B.$
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Proof of the first lemma (continued)}
We recall from the Cauchy-Schwarz inequality
that $| \om (x^* y) |^2 \leq \om (y^* y) \om (x^* x).$
\pause
Replacing $x$ by $x^*,$ we get
$| \om (x y) |^2 \leq \om (y^* y) \om (x x^*).$
\pause
Now let $a \in A$ and $b \in B.$
\pause
Then
\begin{align*}
| \om (a b) - \om (a) \om (b) |^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^*).
\end{align*}
\pause
Since $\om$ is multiplicative on $B,$
we have
\pause
\[
\om \big( (b - \om (b) \cdot 1)^* (b - \om (b) \cdot 1) \big)
= \om \big( (b - \om (b) \cdot 1)^* \big)
\om \big( b - \om (b) \cdot 1) \big)
= 0.
\]
\pause
So $| \om (a b) - \om (a) \om (b) |^2 = 0.$
This completes the proof.
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Second lemma}
\begin{lem}\label{L:Distinct}
Let a discrete group $G$ act on a locally compact space $X.$
\pause
Let $x \in X,$ let $g \in G,$ and assume that $g x \neq x.$
\pause
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 $\ev_x.$
\pause
Then $\om (f u_g) = 0$ for all $f \in C_0 (X).$
\end{lem}
\pause
\vspace{2ex}
For the 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 $x \in X.$
\pause
Choose $f_0 \in C_0 (X)$ such that $f_0 (x) = 1$ and $f_0 (g x) = 0.$
\pause
Applying the first lemma to $\om$ at the second and fourth steps,
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 ( \af_g^{-1} (f_0) )
= \om (f u_g) f_0 (g x)
= 0.
\end{align*}
\pause
This completes the proof.
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Reminder: The conditional expectation}
Recall the conditional expectation:
\pause
Let $\af \colon G \to \Aut (A)$
be an action of a discrete group $G$ on a \ca~$A.$
\pause
Then the conditional expectation
$E \colon C^*_{\mathrm{r}} (G, A, \af) \to A$
does the following:
if $a = \sum_{g \in G} a_g u_g \in C^*_{\mathrm{r}} (G, A, \af)$
with $a_g = 0$ for all but finitely many~$g,$
then $E (a) = a_1.$
\pause
Moreover, $E$ has the following properties:
\pause
\begin{enumerate}
\item\label{P:CondExpt:Idem-R}
$E (E (b)) = E (b)$ for all $b \in C^*_{\mathrm{r}} (G, A, \af).$
\pause
\item\label{P:CondExpt:Pos-R}
If $b \geq 0$ then $E (b) \geq 0.$
\pause
\item\label{P:CondExpt:Norm-R}
$\| E (b) \| \leq \| b \|$ for all $b \in C^*_{\mathrm{r}} (G, A, \af).$
\pause
\item\label{P:CondExpt:Mod-R}
If $a \in A$ and $b \in C^*_{\mathrm{r}} (G, A, \af),$
then $E (a b) = a E (b)$ and $E (b a) = E (b) a.$
\pause
\item\label{P:CondExpt:Fth-R}
If $a \in C^*_{\mathrm{r}} (G, A, \af)$ and $E (a^* a) = 0,$
then $a = 0.$
\end{enumerate}
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Proof of the simplicity theorem}
Let $I \S C^*_{\mathrm{r}} (G, X)$ be a nonzero closed ideal.
\pause
\vspace{2ex}
First suppose $I \cap C_0 (X) = 0.$
Choose $a \in I$ with $a \neq 0.$
\pause
Let
\[
E \colon C^*_{\mathrm{r}} (G, X) \to C_0 (X)
\]
be the standard conditional expectation.
\pause
Then $E (a^* a) \neq 0$
because $E$ is faithful.
\pause
Choose $b \in C_{\mathrm{c}} (G, \, C_0 (X), \, \af)$ such that
$\| b - a^* a \| < \tfrac{1}{4} \| E (a^* a) \|.$
\pause
We can write $b = \sum_{g \in S} b_g u_g$ for some finite set $S \S G$
and with $b_g \in C_0 (X)$ for $g \in S.$
\Wolog\ $1 \in S.$
\pause
Since $E (a^* a)$ is a positive element of $C_0 (X),$
there is $x_0 \in X$ such that $E (a^* a) (x_0) = \| E (a^* a) \|.$
\pause
Essential freeness implies that
\[
\{ x \in X \colon {\mbox{$g x \neq x$
for all $g \in S \setminus \{ 1 \}$}} \}
\]
is dense in $X.$
\pause
In particular,
there is $x \in X$ so close to $x_0$ that
\pause
$E (a^* a) (x) > \tfrac{3}{4} \| E (a^* a) \|,$
and also satisfying $g x \neq x$ for all $g \in S.$
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Proof of the simplicity theorem (continued)}
The set $C_0 (X) + I$ is a C*-subalgebra of $C^*_{\mathrm{r}} (G, X).$
\pause
Let $\om_0 \colon C_0 (X) + I \to \C$ be the following composition:
\[
C_0 (X) + I
\longrightarrow (C_0 (X) + I) / I
\stackrel{\cong}{\longrightarrow} C_0 (X) / (C_0 (X) \cap I)
= C_0 (X)
\stackrel{\ev_x}{\longrightarrow} \C.
\]
\pause
Then $\om_0$ is a \hm.
\pause
Use the Hahn-Banach Theorem in the usual way to get a state
$\om \colon C^*_{\mathrm{r}} (G, X) \to \C$ which extends $\om_0.$
\pause
Note that $\om (a^* a) = 0.$
\pause
\vspace{2ex}
We now have, using the second lemma 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|
\\
& \hspace*{-2em} \mbox{} = | \om (b_1) |
= | \om_0 (b_1) |
= | b_1 (x) |
\geq E (a^* a) (x) - \| E (a^* a) - b_1 \|
\\
& \hspace*{-2em} \mbox{} \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*}
\pause
This contradiction shows that $I \cap C_0 (X) \neq 0.$
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Proof of the simplicity theorem (continued)}
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 $U \S X.$
\pause
We claim that $U$ is $G$-invariant.
\pause
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 $C^*_{\mathrm{r}} (G, X),$
\pause
and we have
$(e_{\ld} u_g) f (e_{\ld} u_g)^* = e_{\ld} \af_g (f) e_{\ld},$
\pause
which converges to $\af_g (f).$
\pause
We also have $(e_{\ld} u_g) f (e_{\ld} u_g)^* \in I \cap C_0 (X),$
since $I$ is an ideal.
\pause
So $\af_g (C_0 (U)) \S C_0 (U)$ for all $g \in G,$
and the claim follows.
\pause
\vspace{2ex}
Since $U$ is open, invariant, and nonempty, we have $U = X.$
\pause
One easily checks that an approximate identity for $C_0 (X)$
\pause
is also an approximate identity for $C^*_{\mathrm{r}} (G, X),$
so $I = C^*_{\mathrm{r}} (G, X),$ as desired.
\pause
This completes the proof.
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Kishimoto's result}
The following theorem, originally due to Kishimoto,
also follows from the full Archbold-Spielberg theorem.
\pause
\vspace{2ex}
\begin{thm}\label{T:Ks}
Let $\af \colon G \to \Aut (A)$ be an action of a
discrete group $G$ on a simple \ca~$A.$
\pause
Suppose that $\af_g$ is outer for every $g \in G \setminus \{ 1 \}.$
\pause
Then $C^*_{\mathrm{r}} (G, A, \af)$ is simple.
\end{thm}
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{The main theorem on minimal homeomorphisms}
Definitions (with explanation) and the
main corollary are below.
\vspace{2ex}
\begin{thm}[with H.~Lin; to appear]\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.
\pause
Suppose that $\rh (K_0 (C^* (\Z, X, h)))$
is dense in $\Aff (T (C^* (\Z, X, h))).$
\pause
Then $C^* (\Z, X, h)$ is a simple unital \ca\ %
with tracial rank zero
which satisfies the Universal Coefficient Theorem.
\end{thm}
\pause
\vspace{2ex}
We do not give a full proof here.
Instead, we assume $X$ is the Cantor set.
\pause
The full proof
would require three lectures about recursive subhomogeneous algebras
and their direct limits,
\pause
as well as several lectures
on technical computations involving KK-theory.
\pause
\vspace{2ex}
Note: Some of the discussion here is not in the notes,
and some of the results are in a slightly different order.
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{The main corollary}
\begin{cor}[with H.~Lin; to appear]\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 $\rh (K_0 (C^* (\Z, X, h)))$
is dense in $\Aff (T (C^* (\Z, X, h))).$
\pause
Then $C^* (\Z, X, h)$ is a simple AH~algebra with no dimension
growth and with real rank zero.
\end{cor}
\pause
\vspace{2ex}
An AH~algebra
% with no dimension growth
is a direct limit
$\Dirlim A_n,$
in which each $A_n$ is a finite direct sum of \ca s
of the form $C (Y, M_r),$
\pause
for varying compact metric spaces~$Y$ and positive integers~$r.$
\pause
``No dimension growth'' means that there is a finite upper bound
on the (covering) dimensions of the spaces~$X$
which occur throughout in the direct system.
\pause
\vspace{2ex}
The theorem also implies that $C^* (\Z, X, h)$ is classifiable
in the sense of the Elliott program.
\pause
(This is used in the proof of the corollary.)
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Covering dimension}
The covering dimension $\dim (X)$
is defined for compact metric spaces~$X$
(and more generally).
\pause
Rather than giving the (somewhat complicated) definition,
here are examples:
\pause
\begin{itemize}
\item
The Cantor set has covering dimension zero.
\pause
\item
If $X$ is a compact manifold with dimension~$d$ as a manifold,
then $\dim (X) = d.$
\pause
\item
If $X$ is a finite CW~complex,
then $\dim (X)$ is the dimension of the highest dimensional cell.
\pause
\item
$\dim \big( [0, 1]^{\Z} \big) = \infty.$
\end{itemize}
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{The map from $K_0$ to the affine functions on
the tracial state space}
For any unital \ca~$A,$
we let $T (A)$ be the tracial state space of~$A.$
\pause
For any compact convex set~$\Dt,$
we let $\Aff (\Dt)$ be the space of real valued \ct\ affine
functions on $\Dt,$
\pause
with the supremum norm.
\pause
Further let $\rh = \rh_A \colon K_0 (A) \to \Aff (T (A))$
be the \hm\ determined by
\pause
$\rh ([p]) (\ta) = \ta (p)$
for $\ta \in T (A)$ and $p$ a \pj\ in some matrix algebra over~$A.$
\pause
\vspace{2ex}
In the situation at hand,
$T (C^* (\Z, X, h))$ is affinely homeomorphic to
the simplex of $h$-invariant Borel probability measures on~$X.$
(See the end of the lecture.)
\pause
\vspace{2ex}
There is machinery available to compute the range of $\rh$
in the above theorem without computing $C^* (\Z, X, h).$
\pause
See, for example, Ruy Exel's Ph.D.\ thesis.
(Reference to the published version in the notes.)
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{The definition of tracial rank zero}
We use the notation $[a, b]$ for the commutator $a b - b a.$
\pause
\vspace{2ex}
\begin{dfn}\label{D:TR0}
Let $A$ be a simple unital \ca.
Then $A$ has tracial rank zero if
\pause
for every finite subset $F \S A,$
\pause
every $\ep > 0,$
\pause
and every nonzero positive element $c \in A,$
\pause
there exists a \pj\ $p \in A$
and a unital finite dimensional
subalgebra $D \S p A p$ such that:
\pause
\begin{enumerate}
\item\label{D:TR0:Comm}
$\| [a, p] \| < \ep$ for all $a \in F.$
\pause
\item\label{D:TR0:Close}
$\dist (p a p, \, D) < \ep$ for all $a \in F.$
\pause
\item\label{D:TR0:Small}
$1 - p$ is Murray-von Neumann equivalent
to a \pj\ in ${\overline{c A c}}.$
\end{enumerate}
\end{dfn}
\pause
\vspace{2ex}
(This is equivalent to the original definition for simple \ca s.
See the notes.)
\pause
\vspace{2ex}
The condition was originally called ``tracially~AF''.
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Restricting to the Cantor set}
The notes contain a complete proof of the main theorem
when $X$ is the Cantor set,
using the same methods as in the proof of the full theorem.
\pause
\vspace{2ex}
The restriction to the Cantor set simplifies the argument
by avoiding recursive subhomogeneous \ca s
and some K-theory computations.
\pause
\vspace{2ex}
However, for the Cantor set, there is an older and shorter
proof,
of which the main part is due to Putnam.
\pause
This proof gives directly the result that $C^* (\Z, X, h)$
is a direct limit of finite direct sums of \ca s
of the form $C (S^1, M_r)$ (for varying~$r$).
\pause
\vspace{2ex}
One can get this result for the Cantor set by combining
the main theorem with known classification results,
so Putnam's argument doesn't give any more in the end.
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Other known results: Minimal diffeomorphisms}
Let $X$ be a compact manifold,
and let $h \colon X \to X$ be a minimal diffeomorphism.
\pause
Then $C^* (\Z, X, h)$
is a direct limit, with no dimension growth,
of recursive subhomogeneous \ca s.
\pause
\vspace{2ex}
There is no condition on
$\rh (K_0 (C^* (\Z, X, h))) \subset \Aff (T (C^* (\Z, X, h))).$
\pause
In particular, $C^* (\Z, X, h)$ sometimes has no nontrivial \pj s.
\pause
\vspace{2ex}
This is joint with Qing Lin (unpublished).
It uses Putnam's methods, but is very long.
\pause
As a corollary,
$C^* (\Z, X, h)$ has stable rank one.
\pause
Applicable classification results are just now starting to appear.
\pause
\vspace{2ex}
It should be possible to generalize to
minimal homeomorphisms of finite dimensional compact metric spaces.
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Other known results: Infinite dimensional spaces}
The condition $\dim (X) < \infty$ is needed.
\pause
Giol and Kerr have an example of a
minimal homeomorphism $h$
of an infinite dimensional compact metric space~$X$
\pause
such that $C^* (\Z, X, h)$ does not have stable rank one.
\pause
\vspace{2ex}
However, probably $h$ having ``mean dimension zero''
is enough.
\pause
Uniquely ergodic implies mean dimension zero,
\pause
and $\dim (X) < \infty$ implies mean dimension zero.
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Other known results: Actions of $\Z^d,$
groups which are not discrete, and actions on simple C*-algebras}
Let $X$ be the Cantor set,
and suppose $\Z^d$ acts freely and minimally on~$X.$
\pause
Then $C^* (\Z^d, X)$
has stable rank one,
\pause
real rank zero,
\pause
and the order on \pj s over $C^* (\Z^d, X)$ is determined
by traces.
\pause
\vspace{1ex}
All these follow from tracial rank zero,
but do not imply tracial rank zero.
\pause
It should be true that $C^* (\Z^d, X)$
has tracial rank zero,
\pause
but this is still open.
\pause
\vspace{1ex}
Let $X$ be a finite dimensional compact metric space,
and suppose $\Z^d$ acts freely and minimally on~$X.$
\pause
Then $C^* (\Z^d, X)$ has strict comparison of positive
elements
\pause
(the Cuntz semigroup analog of
the order on \pj s over $C^* (\Z^d, X)$ being determined
by traces).
\pause
\vspace{1ex}
There is work in progress which will probably
give positive results about $C^* (\R, X)$
when $\dim (X) < \I$ and the action is free and minimal.
\pause
\vspace{1ex}
There is also a collection of related results
on crossed products of simple \ca s by actions
of $\Z$ and of finite groups which have the tracial Rokhlin
property,
and generalizations.
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Lemma~1}
The proof of the main theorem,
even for just the Cantor set, requires a number of lemmas.
\pause
The first essentially
says one can use AF~algebras in place of \fd\ algebras
in the definition of tracial rank zero.
\pause
The proof is omitted; see the notes.
\pause
\vspace{2ex}
\begin{lem}\label{L:TRAF}
Let $A$ be a simple unital \ca.
\pause
Suppose that
for every finite subset $F \S A,$ every $\ep > 0,$
and every nonzero positive element $c \in A,$
\pause
there exists a \pj\ $p \in A$ and a unital AF~subalgebra $B \S A$
with $p \in B$
such that:
\pause
\begin{enumerate}
\item\label{L:TRAF:Comm}
$\| [a, p] \| < \ep$ for all $a \in F.$
\pause
\item\label{L:TRAF:Close}
$\dist (p a p, \, p B p) < \ep$ for all $a \in F.$
\pause
\item\label{L:TRAF:Small}
$1 - p$ is Murray-von Neumann equivalent
to a \pj\ in ${\overline{c A c}}.$
\end{enumerate}
\pause
Then $A$ has tracial rank zero.
\end{lem}
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Lemma~2}
We need only consider finite subsets of a generating set:
\pause
\vspace{2ex}
\begin{lem}\label{Ex:TRGen}
Let $A$ be a unital \ca, and let $S \S A$ be a subset
which generates $A$ as a \ca.
\pause
Assume that the condition of the previous lemma
holds for all finite subsets $F \S S.$
\pause
Then $A$ has tracial rank zero.
\end{lem}
\pause
\vspace{2ex}
The proof is an exercise.
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{$C^* (\Z, X, h)_Y$}
A key point,
in this proof and others,
is the construction of a ``large''
and more tractable subalgebra of the \cp.
\pause
\vspace{2ex}
\begin{ntn}\label{N:A6}
Let $X$ be a compact metric space,
and let $h \colon X \to X$ be a homeomorphism.
\pause
In the \tgca\ $C^* (\Z, X, h),$ we write
$u$ for the standard unitary representing the generator of $\Z.$
\pause
For a closed subset $Y \subset X,$ we define
the C*-subalgebra $C^* (\Z, X, h)_Y$ to be
\[
C^* (\Z, X, h)_Y
= C^* ( C (X), \, u C_0 (X \setminus Y)) \subset C^* (\Z, X, h).
\]
% We will sometimes let $A$ denote the
% \tgca\ $C^* (\Z, X, h),$ in which case we refer to $A_Y.$
\end{ntn}
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{$C^* (\Z, X, h)_Y$ is the C*-algebra of a groupoid}
Although we will not use formally groupoids in these notes,
it should be pointed out that
$C^* (\Z, X, h)_Y$ is the \ca\ of a subgroupoid of the
transformation group groupoid $\Z \ltimes X$
made from the action of $\Z$ on $X$ generated by~$h.$
\pause
Informally,
we ``break'' every orbit each time it goes through~$Y.$
\pause
More formally,
for $n < 0$
the pair $(n, x)$ is in the subgroupoid
only if all of $h^n (x), \, h^{n + 1} (x), \, \ldots, \, h^{-1} (x)$
are in $X \setminus Y,$
\pause
and for $n > 0$
the pair $(n, x)$ is in the subgroupoid
only if all of $x, \, h (x), \, \ldots, \, h^{n - 1} (x)$
are in $X \setminus Y.$
\pause
\vspace{2ex}
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.
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Lemma~3}
The following lemma is due to Putnam.
\pause
\vspace{2ex}
\begin{lem}\label{L:AF}
Let $X$ be the Cantor set,
and let $h \colon X \to X$ be a \mh.
Let $Y \S X$ be a nonempty compact open subset.
\pause
Then $C^* (\Z, X, h)_Y$ is an AF~algebra.
\end{lem}
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Proof of Lemma~3}
The proof depends on the construction of Rokhlin towers,
which is a crucial element of many structure results for \cp s.
\pause
\vspace{2ex}
We first claim that there is $N \in \N$ such that
$\bigcup_{n = 1}^{N} h^{-n} (Y) = X.$
Set $U = \bigcup_{n = 1}^{\infty} h^{-n} (Y),$
which is a nonempty open subset of~$X$ such that $U \subset h (U).$
\pause
Then $Z = X \setminus \bigcup_{n = 1}^{\infty} h^{-n} (Y)$
\pause
is a closed subset of~$X$ such that $h (Z) \S Z,$
and $Z \neq X.$
\pause
Therefore $Z = \varnothing.$
\pause
So $U = X,$ and the claim now follows from compactness of~$X.$
\pause
\vspace{2ex}
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.
\pause
Define the {\emph{first return time}} $r (y)$ to be
\pause
\[
r (y) = \min \{ n \geq 1 \colon h^n (y) \in Y \} \leq N.
\]
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Proof of Lemma~3 (continued)}
Let $n (0) < n (1) < \cdots < n (l) \leq N$ be the values of~$r.$
\pause
Set
\[
Y_k = \{ y \in Y \colon r (y) = n (k) \}.
\]
\pause
Then the sets $Y_k$ are compact, open, and partition $Y,$
\pause
and the sets $h^j (Y_k),$
for $1 \leq j \leq n (k),$ partition $X$:
\pause
\[
Y = \coprod_{k = 0}^l Y_k
\andeqn
X = \coprod_{k = 0}^l \coprod_{j = 1}^{n (k)} h^j (Y_k).
\]
\pause
Each finite sequence
$h (Y_k), \, h^2 (Y_k), \ldots, h^{n (k)} (Y_k)$
is a {\emph{Rokhlin tower}}
\pause
with base $h (Y_k)$ and height~$n (k).$
\pause
(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 $C^* (\Z, X, h)_Y.$)
\pause
Further set $X_k = \bigcup_{j = 1}^{n (k)} h^j (Y_k).$
The sets $X_k$ then also partition~$X.$
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Rokhlin towers, with part of an orbit}
We take $x \in Y,$ in fact, $x \in Y_0.$
\vspace{2ex}
The bases of the towers are
$h (Y_0), h (Y_1), \ldots, h (Y_l),$
and the heights are $n (0), n (1), \ldots, n (l).$
The tower over $h (Y_k)$ corresponds to a summand of
$C^* (\Z, X, h)_Y$ isomorphic to $M_{n (k)} \otimes C (Y_k).$
% \vspace{1ex}
\centerline{ \includegraphics[width=10cm]{TowerWOrbit_NL} }
\vspace*{-1.7ex}
\hspace*{1.6cm}$h (x)$\hspace*{0.6cm}$h^{m_2} (x)$\hspace*{0.5cm}$h^{m_3} (x)$\hspace*{4.3cm}$h^{m_1} (x)$
\vspace{-1ex}
\[
m_1 = n (0) + 1, \,\,\,\,\,\,
m_2 = n (0) + n (l) + 1, \,\,\,\,\,\,
m_3 = n (0) + n (l) + n (1) + 1
\]
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Proof of Lemma~3 (continued)}
% From here to end of proof: Put in notes file.
% Watch references! ???
Define $p_k \in C (X) \S C^* (\Z, X, h)_Y$
by $p_k = \ch_{X_k}.$
\pause
Then $p_k$ trivially commutes with every element of $C (X).$
\pause
Moreover,
suppose $f \in C (X)$ vanishes on $Y.$
\pause
Since $Y_k, \, h^{n (k)} (Y_k) \S Y,$
we have
\pause
\[
\ch_{Y_k} f = 0
\andeqn
\ch_{ h^{n (k)} (Y_k) } f = 0
\]
\pause
Use the action of $h$ on the levels of the
tower at the first step, and the equations above at the second step,
to get
\pause
\[
p_k u f = u \big( p_k - \ch_{ h^{n (k)} (Y_k) } + \ch_{Y_k} \big) f
= u p_k f
= u f p_k.
\]
\pause
It follows that $p_k$ commutes with all elements of $C^* (\Z, X, h)_Y.$
\pause
So it suffices to prove that $p_k C^* (\Z, X, h)_Y p_k$
is~AF for each $k.$
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Proof of Lemma~3 (continued)}
Now $p_k C^* (\Z, X, h)_Y p_k$ is the \ca\ generated by
$C (X_k)$ and
\begin{align*}
u (\ch_{X \setminus Y}) p_k
= u (\ch_{X_k \setminus h^{n (k)} (Y_k) })
& = \sum_{j = 1}^{n (k) - 1}
u ( \ch_{ h^{j} (Y_k) } )
\\
& = \sum_{j = 1}^{n (k) - 1}
( \ch_{ h^{j + 1} (Y_k) } ) u ( \ch_{ h^{j} (Y_k) } ).
\end{align*}
\pause
One can now check,
although it is a bit tedious to write out the details,
% Expand? ??? Check Putnam paper.
that there is an isomorphism
$\ps_k \colon p_k C^* (\Z, X, h)_Y p_k \to M_{n (k)} \otimes C (Y_k)$
\pause
such that for $f \in C (X_k)$ we have
\[
\ps_k (f)
= \diag \big( f \circ h |_{Y_k}, \, f \circ h^2 |_{Y_k}, \,
\ldots, \, f \circ h^{n (k)} |_{Y_k} \big)
\]
\pause
and for $1 \leq j \leq n (k) - 1$ we have
\[
\ps_k \big( \ch_{ h^{j + 1} (Y_k) } u \ch_{ h^{j} (Y_k) } \big)
= e_{j + 1, \, j} \otimes 1.
\]
\pause
The algebra $M_{n (k)} \otimes C (Y_k)$ is~AF because $Y_k$
is totally disconnected.
\pause
This completes the proof.
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Lemma~4}
\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 $A,$
\pause
that is, there exists $v \in A$ such that $v^* v = p$ and $v v^* = q.$
\pause
We write $p \precsim q$ if $p$ is \mvnt\ to a subprojection of~$q.$
\end{ntn}
\pause
\begin{lem}\label{L:PjComp}
Let $X$ be the Cantor set,
and let $h \colon X \to X$ be a \mh.
\pause
Let $c \in C^* (\Z, X, h)$ be a nonzero positive element.
\pause
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}
\pause
% \vspace{2ex}
This shows that we can control the size of the leftover
in the definition of tracial rank zero using \pj s in $C (X)$
instead of in the crossed product.
\pause
This is a big advantage.
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{What we have so far}
Combining our results so far,
we have reduced our problem to proving the following:
\pause
\vspace{2ex}
Let $X$ be the Cantor set,
and let $h \colon X \to X$ be a \mh.
\pause
Let $S = C (X) \cup \{ u \}$
(clearly a generating set for $C^* (\Z, X, h)$).
\pause
Then
for every finite subset $F \S C (X),$
\pause
every $\ep > 0,$
\pause
and every nonempty open set $U \subset X,$
\pause
there exists a compact open set $Y \S X$
\pause
and a projection $p \in C^* (\Z, X, h)_{Y}$ such that:
\begin{itemize}
\item[(1)]
$\| p a - a p \| < \ep$ for all $a \in F \cup \{u\}.$
\pause
\item[(2)]
$p a p \in p C^* (\Z, X, h)_Y p$ for all $a \in F \cup \{u\}.$
\pause
\item[(3)]
There is a compact open set $Z \S U$ such that
$1 - p \precsim \ch_Z$ in $C^* (\Z, X, h).$
\end{itemize}
\pause
\vspace{2ex}
The point is that $C^* (\Z, X, h)_Y$ is an AF~algebra,
and (3)~says, in view of Lemma~4, that $1 - p$ is ``small''.
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Proof of Lemma~4}
Let $E \colon C^* (\Z, X, h) \to C (X)$ be the
standard conditional expectation.
\pause
Then $E (c)$ is a nonzero positive element of $C (X).$
\pause
Choose a nonempty compact open subset $K_0 \S X$ and $\dt > 0$
such that the function $E (c)$ satisfies $E (c) (x) > 4 \dt$
for all $x \in K_0.$
\pause
Choose a finite sum $b = \sum_{n = -N}^N b_n u^n \in C^* (\Z, X, h)$
such that $\| b - c \| < \dt.$
\pause
Since the action of $\Z$ induced by $h$ is free,
there is a nonempty compact open subset $K \S K_0$ such that
the sets
\[
h^{-N} (K), \, h^{- N + 1} (K), \, \ldots, \, h^N (K)
\]
% Change in notes. ???
are disjoint.
\pause
Set $p = \ch_K \in C (X).$
\pause
For $n \in \{ -N, \, - N + 1, \, \ldots, \, N \} \setminus \{ 0 \},$
\pause
the disjointness condition implies that $p u^n p = 0.$
Therefore
\[
p b p = p b_0 p = p E (b) p.
\]
% Change in notes. ???
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Proof of Lemma~4 (continued)}
Using this equation at the first 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}
\pause
Since $K \S K_0,$
the function $p E (c) p$ is invertible in $p C (X) p.$
\pause
In the following, we take inverses in $p C^* (\Z, X, h) p.$
\pause
Then, in fact,
$\| [ p E (c) p ]^{-1} \| < \frac{1}{4} \dt^{-1}.$
\pause
The estimate~(\ref{Est13}) now implies that $p c p$ is invertible in
$p C^* (\Z, X, h) p.$
\pause
Let $a = (p c p)^{-1/2},$ calculated in $p C^* (\Z, X, h) p.$
\pause
Set $v = a p c^{1/2}.$
\pause
Then
\[
v v^* = a p c p a = (p c p)^{-1/2} (p c p) (p c p)^{-1/2} = p
\]
\pause
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.
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Lemma~5}
\begin{lem}\label{L:Canc}
Let $X$ be the Cantor set,
and let $h \colon X \to X$ be a \mh.
\pause
Let $Y \S X$ be a nonempty compact open subset.
Let $N \in \N,$
and suppose that $Y, \, h (Y), \, \ldots, h^N (Y)$ are disjoint.
\pause
Then the \pj s $\ch_Y$ and $\ch_{h^N (Y)}$
are \mvnt\ in $C^* (\Z, X, h)_Y.$
\end{lem}
\pause
\begin{proof}
First, observe that if $Z \S X$ is a compact open subset
such that $Y \cap Z = \varnothing,$
\pause
then $v = u \ch_Z \in C^* (\Z, X, h)_Y$
\pause
and satisfies $v^* v = \ch_Z$ and $v v^* = \ch_{h (Z)}.$
\pause
Thus $\ch_Z \sim \ch_{h (Z)}$ in $C^* (\Z, X, h)_Y.$
% Change in notes ???
\pause
\vspace{2ex}
An induction argument now shows that
$\ch_{h (Y)} \sim \ch_{h^N (Y)}$ in $C^* (\Z, X, h)_Y.$
% Change in notes ???
\pause
Also, $\ch_{X \setminus Y} \sim \ch_{X \setminus h (Y)}$
in $C^* (\Z, X, h)_Y.$
% Change in notes ???
\pause
Since $C^* (\Z, X, h)_Y$ is an AF~algebra,
it follows that $\ch_Y \sim \ch_{h (Y)}.$
\pause
The result follows by transitivity.
\end{proof}
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Lemma~6}
\begin{lem}\label{L:L3}
Let $X$ be the Cantor set,
and let $h \colon X \to X$ be a \mh.
\pause
Let $y \in X.$
Then for any
$\ep > 0,$
\pause
any nonempty open set $U \S X,$
\pause
and any finite subset $F \subset C (X),$
\pause
there
is a compact open set $Y \S X$ containing $y$
\pause
and a projection $p \in C^* (\Z, X, h)_{Y}$
\pause
such that:
\begin{itemize}
\item[(1)]
$\| p a - a p \| < \ep$ for all $a \in F \cup \{u\}.$
\pause
\item[(2)]
$p a p \in p C^* (\Z, X, h)_Y p$ for all $a \in F \cup \{u\}.$
\pause
\item[(3)]
There is a compact open set $Z \S U$ such that
$1 - p \precsim \ch_Z$ in $C^* (\Z, X, h).$
\end{itemize}
\end{lem}
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Lemma~6 finishes the proof}
Recall that we reduced our problem to proving the following:
\pause
\vspace{2ex}
Let $X$ be the Cantor set,
and let $h \colon X \to X$ be a \mh.
Let $S = C (X) \cup \{ u \}$
Then
for every finite subset $F \S S,$ every $\ep > 0,$
and every nonempty open set $U \subset X,$
there exists compact open set $Y \S X$
and a projection $p \in C^* (\Z, X, h)_{Y}$ such that:
\begin{itemize}
\item[(1)]
$\| p a - a p \| < \ep$ for all $a \in F \cup \{u\}.$
\item[(2)]
$p a p \in p C^* (\Z, X, h)_Y p$ for all $a \in F \cup \{u\}.$
\item[(3)]
There is a compact open set $Z \S U$ such that
$1 - p \precsim \ch_Z$ in $C^* (\Z, X, h).$
\end{itemize}
\pause
\vspace{2ex}
So Lemma~6 finishes the proof.
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Comment on the proof of Lemma~6: Berg's technique}
The basic idea behind the proof of Lemma~6 is Berg's technique.
\pause
It was originally used to prove that the direct sum
of the bilateral shift
and a finite cyclic shift
is approximately unitarily equivalent,
\pause
to within an error depending on the length of the finite
cyclic shift,
\pause
to the bilateral shift.
\pause
The error goes to zero as the length of the finite
cyclic shift goes to infinity.
\pause
\vspace{1ex}
We effectively do this in reverse:
the bilateral shift
\pause
(in this case, the standard generating unitary of the crossed product
by~$\Z$)
\pause
can be approximated by something which has a (long) finite cyclic shift
as a direct summand.
% Notes should include more discussion of the original version.
\pause
\vspace{1ex}
In this proof,
we apply Berg's technique ``over'' a small subset of the space~$X.$
\pause
For some other related results on crossed products,
the methods of proof known so far require
an application ``over'' subsets which fill up most of~$X.$
\pause
\vspace{1ex}
For actions of more general groups on compact metric spaces,
to get the strongest expected results
one needs a generalization or replacement,
\pause
so far not found.
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Proof of Lemma~6}
% Proof not properly proofread. ???
Let $d$ be the metric on $X.$
\pause
\vspace{2ex}
Choose $N_0 \in \N$ so large that $4 \pi / N_0 < \ep.$
\pause
\vspace{2ex}
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 $f \in F.$
\pause
\vspace{2ex}
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
$d (h^{-k} (x_1), \, h^{-k} (x_2)) < \dt_0.$
\pause
\vspace{2ex}
Since $h$ is minimal, there is $N > N_0 + 1$ such that
$d (h^N (y), \, y) < \dt.$
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Proof of Lemma~6 (continued)}
Choose $N + N_0 + 1$ disjoint nonempty open subsets
$U_{- N_0}, U_{- N_0 + 1}, \ldots, U_{N} \S U.$
\pause
Using minimality again,
choose $r_{- N_0}, r_{- N_0 + 1}, \ldots, r_{N} \in \Z$
\pause
such that $h^{r_l} (y) \in U_l$
for $- N_0 \leq l \leq N.$
\pause
Since $h$ is free, there is a compact open set $Y \S X$
containing $y$ such that
\[
h^{- N_0} (Y), \, h^{- N_0 + 1} (Y), \, \ldots,
\, Y, \, h (Y), \, \ldots, \, h^N (Y)
\]
are disjoint and all have diameter less than~$\dt.$
\pause
We may also require that $h^{r_l} (Y) \S U_l$
for ${- N_0} \leq l \leq N.$
\pause
\vspace{2ex}
Set $q_0 = \ch_Y.$
For $- N_0 \leq n \leq N$ set
\[
T_n = h^n (Y)
\andeqn q_n = u^n q_0 u^{- n} = \ch_{h^n (Y)}.
\]
\pause
Then the $q_n$ are \mops\ in $C (X).$
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Proof of Lemma~6 (continued)}
We now have a sequence of \pj s, in principle going to infinity
in both directions:
\[
\ldots, \, q_{- N_0}, \, \ldots, \, q_{-1}, \, q_0, \, q_1, \,
\ldots, \, q_{N - N_0}, \, \ldots, \, q_{N - 1}, \, q_N, \, \ldots.
\]
\pause
The ones shown are orthogonal, and conjugation by $u$ is the shift.
\pause
The \pj s $q_0$ and $q_N$ are the characteristic
functions of compact open sets which are disjoint
but close to each other,
\pause
and similarly for the pairs
$q_{-1}$ and $q_{N - 1}$ down to $q_{- N_0}$ and $q_{N - N_0}.$
\pause
We are now going to use Berg's technique
to splice this sequence along the pairs of indices
$(- N_0, \, N - N_0)$ through $(0, N),$
\pause
obtaining a loop of length $N$
on which conjugation by $u$ is approximately the cyclic shift.
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Proof of Lemma~6 (continued)}
Lemma~5 provides a partial isometry
$w \in C^* (\Z, X, h)_Y$
such that $w^* w = q_0$ and $w w^* = q_N.$
\pause
For $t \in [0, 1]$ define
\[
v (t) = \cos (\pi t / 2) (q_0 + q_N) + \sin (\pi t / 2) (w - w^*).
\]
% Change in notes ???
\pause
Then $v (t)$ is a unitary in the corner
\[
(q_0 + q_N) C^* (\Z, X, h)_{Y} (q_0 + q_N)
\]
% Change in notes ???
\pause
whose matrix with respect to
the obvious block decomposition is
\pause
\[
v (t)
= \left( \begin{array}{cc} \cos (\pi t / 2) & - \sin (\pi t / 2) \\
\sin (\pi t / 2) & \cos (\pi t / 2) \end{array} \right).
\]
\pause
For $0 \leq k \leq N_0$ define $z_k = u^{- k} v (k / N_0) u^k.$
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Proof of Lemma~6 (continued)}
We claim that $z_k \in C^* (\Z, X, h)_Y$
for $0 \leq k \leq N_0.$
\pause
For $k = 0$ this is true by construction.
\pause
For $1 \leq k \leq N_0,$
with
\[
a_k = q_0 u^k
= (u q_{- 1}) (u q_{- 2}) \cdots (u q_{- k})
\in C^* (\Z, X, h)_{Y}
\]
\pause
and
\[
b_k = q_N u^k
= (u q_{N - 1}) (u q_{N - 2}) \cdots (u q_{N - k})
\in C^* (\Z, X, h)_{Y}
\]
(because $N_0 < N$),
\pause
and using $T_{- k} \cap T_{N - k} = T_0 \cap T_{N} = \varnothing,$
\pause
we can write
\[
z_k = (a_k + b_k)^* v (k / N_0) (a_k + b_k) \in C^* (\Z, X, h)_{Y}.
\]
\pause
Therefore $z_k$ is a unitary in the corner
\[
(q_{- k} + q_{N - k}) C^* (\Z, X, h)_{Y} (q_{- k} + q_{N - k}).
\]
% Change in notes ???
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Proof of Lemma~6 (continued)}
Moreover,
adding estimates on the differences of the
matrix entries at the second step,
\pause
\[
\| u z_{k + 1} u^* - z_k \|
= \| v (k / N_0) - v ((k - 1) / N_0) \|
\leq 2 \pi / N_0
< \tfrac{1}{2} \ep.
\]
\pause
\vspace{2ex}
Now define $e_n = q_n$ for $0 \leq n \leq N - N_0,$
\pause
and for $N - N_0 \leq n \leq N$ write $k = N - n$ and set
$e_n = z_k q_{- k} z_k^*.$
\pause
The two definitions for $n = N - N_0$ agree
because $z_{N_0} q_{- N_0} z_{N_0}^* = q_{N - N_0},$
\pause
and moreover $e_N = e_0.$
\pause
Therefore $u e_{n - 1} u^* = e_n$ for $1 \leq n \leq N - N_0,$
and also $u e_N u^* = e_1,$
\pause
while for $N - N_0 < n \leq N$ we have
\[
\| u e_{n - 1} u^* - e_n \|
\leq 2 \| u z_{N - n + 1} u^* - z_{N - n} \|
< \ep.
\]
\pause
Also, clearly $e_n \in C^* (\Z, X, h)_{Y}$ for all $n.$
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Proof of Lemma~6 (continued)}
Set $e = \sum_{n = 1}^{N} e_n$ and $p = 1 - e.$
\pause
We verify that $p$ satisfies~(1) through~(3):
\begin{enumerate}
\item
$\| p a - a p \| < \ep$ for all $a \in F \cup \{u\}.$
\item
$p a p \in p C^* (\Z, X, h)_Y p$ for all $a \in F \cup \{u\}.$
\item
There is a compact open set $Z \S U$ such that
$1 - p \precsim \ch_Z$ in $C^* (\Z, X, h).$
\end{enumerate}
\pause
\vspace{2ex}
First,
\[
p - u p u^*
= u e u^* - e
= \sum_{n = N_0 + 1}^{N} (u e_{n - 1} u^* - e_n ).
\]
\pause
The terms in the sum are orthogonal and have norm less than $\ep,$
\pause
so $\| u p u^* - p \| < \ep.$
\pause
Furthermore,
since $p \leq 1 - q_0 = 1 - \ch_Y,$
\pause
we get $p u p \in C^* (\Z, X, h)_{Y}.$
\pause
This is~(1) and~(2) for the element $u \in F \cup \{u\}.$
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Proof of Lemma~6 (continued)}
Next, let $f \in F.$
\pause
The sets $T_0, T_1, \ldots, T_N$ all have diameter less than~$\dt.$
\pause
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 $N - N_0 \leq n \leq N.$
\pause
Also, $T_{n - N} = h^{n - N} (T_0)$
has diameter less than~$\dt.$
\pause
Therefore $T_{n - N} \cup T_n$ has diameter less than
$2 \dt + \dt_0 \leq 3 \dt_0.$
\pause
Since $f$ varies by at most $\frac{1}{4} \ep$ on any set with
diameter less than $4 \dt_0,$
\pause
and since the sets
\[
S_1 = T_1, \,\,\, S_2 = T_2, \,\,\, \ldots, \,\,\,
S_{N - N_0 - 1} = T_{N - N_0 - 1},
\]
\[
S_{N - N_0} = T_{N - N_0} \cup T_{- N_0}, \,\,\,
S_{N - N_0 + 1} = T_{N - N_0 + 1} \cup T_{- N_0 + 1},
\,\,\, \ldots, \,\,\,
\]
\[
\,\,\, \ldots, \,\,\,
S_N = T_N \cup T_0
\]
are disjoint,
\pause
there is $g \in C (X)$
which is constant on each of these sets
\pause
and satisfies
$\| f - g \| < \frac{1}{2} \ep.$
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Proof of Lemma~6 (continued)}
Let the values of $g$ on these sets be
$\ld_1$ on $S_1$ through $\ld_N$ on $S_N.$
\pause
Then $g e_n = e_n g = \ld_n e_n$
for $0 \leq n \leq N - N_0.$
\pause
For $N - N_0 < n \leq N$ we use
$e_n \in (q_{n - N} + q_{n}) C^* (\Z, X, h)_{Y} (q_{n - N} + q_{n})$
to get,
% \pause
% using the same calculations as above
% at the third and fourth steps,
\[
g e_n
= g (q_{n - N} + q_{n}) e_n
= \ld_n (q_{n - N} + q_{n}) e_n
= e_n (q_{n - N} + q_{n}) g
= e_n g.
\]
\pause
Since $\| f - g \| < \frac{1}{2} \ep$ and $g e = e g,$
it follows that
\[
\| p f - f p \| = \| f e - e f \| < \ep.
\]
\pause
This is~(1) for $f.$
\pause
That $p f p \in C^* (\Z, X, h)_{Y}$ follows from the fact that
$f$ and $p$ are in this subalgebra.
\pause
So we also have~(2) for~$f.$
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Proof of Lemma~6 (continued)}
It remains only to verify~(3).
\pause
Using $h^{r_l} (Y) \S U_l$ for ${- N_0} \leq l \leq N$
at the third step,
\pause
and with $Z = \bigcup_{l = - N_0}^N h^{r_l} (Y) \S U,$
\pause
we get (with Murray-von Neumann equivalence in $C^* (\Z, X, h)$)
\pause
\[
1 - p
= e \leq \sum_{l = - N_0}^N q_l
\sim \sum_{l = - N_0}^N \ch_{h^{r_l} (Y)}
= \ch_Z.
\]
\pause
This completes the proof.
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Some comments on the general case}
In the general case,
one major complication
is that one can't choose the Rokhlin towers to
consist of compact open sets.
\pause
It turns out that one must take $Y$ to be closed
with nonempty interior,
\pause
and replace the sets $Y_k$ by their closures.
\pause
Then they are no longer disjoint.
\pause
The algebra $C^* (\Z, X, h)_Y$
is now a very complicated subalgebra
of $\bigoplus_{k = 0}^l M_{n (k)} \otimes C (Y_k).$
\pause
It is what is known as a ``recursive subhomogeneous algebra''.
\pause
Such algebras are generally not~AF,
and may have few or no nontrivial \pj s.
\pause
The hypothesis on the range of $\rh$
\pause
(which was not used above, although it is automatic when $X$
is the Cantor set)
\pause
must be used to produce sufficiently many \nzp s
and approximating \fd\ subalgebras.
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Some comments on the general case (continued)}
Here are a few more details.
\pause
The algebra $C^* (\Z, X, h)_Y$ is not~AF.
So we choose closed sets $Y_n$ with
\[
Y_0 \supset Y_1 \supset Y_2 \supset \cdots
\andeqn
\bigcap_{n = 0}^{\I} Y_n = \{ y_0 \}
\]
for a suitable $y_0 \in X,$
and such that $\sint (Y_n) \neq \varnothing$ for all~$n.$
\pause
Then
\[
C^* (\Z, X, h)_{Y_0} \subset C^* (\Z, X, h)_{Y_1}
\subset C^* (\Z, X, h)_{Y_2} \subset \cdots
\]
and
\[
{\overline{\bigcup_{n = 0}^{\I} C^* (\Z, X, h)_{Y_n} }}
= C^* (\Z, X, h)_{ \{ y_0 \} }.
\]
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Some comments on the general case (continued)}
$C^* (\Z, X, h)_{ \{ y_0 \} }$ is simple and has the same
tracial states and $K_0$-group as $C^* (\Z, X, h).$
\pause
(These facts require proof.
The one about $K_0$ is hard,
but had already been done by Putnam.)
\pause
Using this,
and some results on dymanics
(essentially ``mean dimension zero''),
\pause
the theory of direct limits of recursive subhomogeneous
algebras can be used to prove that
\pause
$C^* (\Z, X, h)_{ \{ y_0 \} }$ has tracial rank zero.
\pause
This turns out to be a sufficient substitute
for $C^* (\Z, X, h)_Y$ being an AF~algebra.
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{The tracial state space of $C^* (\Z, X, h)_{ \{y\} }$}
We describe the proof of one of the pieces needed.
\pause
\vspace{2ex}
\begin{lem}\label{L:TABij}
Let $X$ be an infinite compact metric space,
and let $h \colon X \to X$ be a \mh.
Let $y \in X.$
\pause
Then the restriction map
$T (C^* (\Z, X, h)) \to T \big( C^* (\Z, X, h)_{ \{y\} } \big)$
is a bijection.
\end{lem}
\pause
\vspace{2ex}
The proof follows from the following two results.
\pause
The first is well known,
and holds in much greater generality.
\pause
The proofs are roughly the same,
but the proof of the second is more complicated,
since the algebra is smaller.
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Tracial states and invariant measures}
\begin{prp}\label{P:TA}
Let $X$ be an infinite compact metric space,
and let $h \colon X \to X$ be a \mh.
\pause
Then the restriction map
$T (C^* (\Z, X, h)) \to T ( C (X))$
\pause
is a bijection from $T (C^* (\Z, X, h))$ to the set
of $h$-invariant Borel probability measures on~$X.$
\pause
Moreover, if $E \colon C^* (\Z, X, h) \to C (X)$ is the
standard conditional expectation,
and $\mu$ is an $h$-invariant Borel probability measure on~$X,$
\pause
then the \tst\ $\ta_{\mu}$ on $C^* (\Z, X, h)$
which restricts to $\mu$
\pause
is given by the formula
$\ta_{\mu} (a) = \int_X E (a) \, d \mu$ for $a \in C^* (\Z, X, h).$
\end{prp}
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Proof of the proposition}
It is immediate that $\ta_{\mu}$ is positive,
and that $\ta_{\mu} (1) = 1.$
So $\ta_{\mu}$ is a state.
\pause
\vspace{2ex}
We prove $\ta_{\mu} (a b) = \ta_{\mu} (b a)$
for all $a, b \in C^* (\Z, X, h).$
\pause
Since $\ta_{\mu}$ is \ct,
and since $C^* (\Z, X, h)$ is the closed linear span
of all elements $f u^m$ with $f \in C (X)$ and $m \in \Z,$
\pause
it suffices to prove
$\ta_{\mu} \big( (f u^m) (g u^n) \big)
= \ta_{\mu} \big( (g u^n) (f u^m) \big)$
for all $f, g \in C (X)$ and $m, n \in \Z.$
\pause
For $n \neq - m,$ we have
\[
\ta_{\mu} \big( (f u^m) (g u^n) \big)
= \ta_{\mu} \big( (g u^n) (f u^m) \big) = 0,
\]
\pause
while for $n = - m,$ we have,
using $h$-invariance of $\mu$ at the second step,
\pause
\begin{align*}
\ta_{\mu} \big( (f u^m) (g u^n) \big)
& = \ta_{\mu} \big( f (u^m g u^{- m}) \big)
= \int_X f \big( g \circ h^{-m} \big) \, d \mu \\
& = \int_X \big( f \circ h^{m} \big) g \, d \mu
= \ta_{\mu} \big( g (u^{- m} f u^{m}) \big)
= \ta_{\mu} \big( (g u^n) (f u^m) \big).
\end{align*}
\pause
This completes the proof that $\ta_{\mu}$ is a tracial state on
$C^* (\Z, X, h).$
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Proof of the proposition (continued)}
Now let $\ta$ be any \tst\ on $C^* (\Z, X, h).$
\pause
Let $\mu$ be the Borel probability measure on $X$
determined by $\ta (f) = \int_X f \, d \mu$ for $f \in C (X).$
\pause
We complete the proof by showing that $\mu$ is $h$-invariant,
and that $\ta = \ta_{\mu}.$
\pause
\vspace{2ex}
For the first, for every $f \in C (X),$
we use the trace property at the second step to get
\pause
\[
\int_X (f \circ h^{-1}) \, d \mu
= \ta (u f u^*)
= \ta (u^* (u f))
= \ta (f)
= \int_X f \, d \mu.
\]
\pause
Since $f \in C (X)$ is arbitrary,
it follows that $\mu$ is $h$-invariant.
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Proof of the proposition (continued)}
For the second,
since $C^* (\Z, X, h)$ is the closed linear span
of all elements $f u^n$ with $f \in C (X)$ and $m \in \Z,$
\pause
it suffices to prove
$\ta (f u^n) = 0$ for $f \in C (X)$ and $n \in \Z \SM \{ 0 \}.$
\pause
Since $h^n$ has no fixed points,
there is an open cover of $X$ consisting of sets $U$
such that $h^n (U) \cap U = \varnothing.$
\pause
Choose $g_1, g_2, \ldots, g_m \in C (X)$ which form a
partition of unity subordinate to this cover.
\pause
In particular, the supports of $g_j$ and $g_j \circ h^{-n}$
are disjoint for all~$j.$
\pause
For $1 \leq j \leq m$ we have,
using the trace property at the second step,
and the
relation $u^n g u^{-n} = g \circ h^{-n}$ at the third step,
\[
\ta (g_j f u^n)
= \ta \big( g_j^{1/2} f u^n g_j^{1/2} \big)
= \ta \big( g_j^{1/2} f \big( g_j^{1/2} \circ h^{-n} \big) u^n \big)
= \ta (0)
= 0.
\]
\pause
Summing over $j$ gives $\ta (f u^n) = 0.$
This completes the proof.
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Tracial states on the subalgebra}
\begin{lem}\label{L:TAy}
Let $X$ be an infinite compact metric space,
and let $h \colon X \to X$ be a \mh.
Let $y \in X.$
\pause
Then the restriction map
$T \big( C^* (\Z, X, h)_{ \{y\} } \big) \to T ( C (X))$
\pause
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}
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Proof of the lemma}
Applying the previous proposition
and restricting from $C^* (\Z, X, h)$ to $C^* (\Z, X, h)_{ \{y\} },$
\pause
we see that every $h$-invariant Borel probability measure on~$X$
gives a \tst\ on $C^* (\Z, X, h)_{ \{y\} }.$
\pause
\vspace{2ex}
Now let $\ta$ be any \tst\ on $C^* (\Z, X, h)_{ \{y\} }.$
\pause
Let $\mu$ be the Borel probability measure on $X$
determined by $\ta (f) = \int_X f \, d \mu$ for $f \in C (X).$
\pause
As in the proof of the previous proposition,
we complete the proof by showing that $\mu$ is $h$-invariant,
and that $\ta = \ta_{\mu} |_{C^* (\Z, X, h)_{ \{y\} }}.$
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Proof of the lemma (continued)}
For the first, we again show that
$\int_X (f \circ h^{-1}) \, d \mu = \int_X f \, d \mu$
for every $f \in C (X).$
\pause
This is clearly true for constant functions~$f.$
\pause
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 $f_1 (y) = f_2 (y) = 0.$
\pause
(For example, take $f_1 = f | f |^{- 1/2}$ and $f_2 = | f |^{1/2}.$)
\pause
Then $u f_1, \, u f_2 \in C^* (\Z, X, h)_{ \{y\} }.$
So
\[
f \circ h^{-1}
= u f u^*
= (u f_1) (u f_2)^*
\in C^* (\Z, X, h)_{ \{y\} }.
\]
\pause
We now use the trace property at the second step to get
\[
\int_X (f \circ h^{-1}) \, d \mu
= \ta \big( (u f_1) (u f_2)^* \big)
= \ta \big( (u f_2)^* (u f_1) \big)
= \ta (f)
= \int_X f \, d \mu.
\]
Thus $\mu$ is $h$-invariant.
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Proof of the lemma (continued)}
For the second,
we first claim that $C^* (\Z, X, h)_{ \{y\} }$
is the closed linear span of all elements of the form $f u^m,$
with $f \in C (X)$ and $m \in \Z,$
which actually happen to be in $C^* (\Z, X, h)_{ \{y\} }.$
\pause
The claim follows from three facts:
all the generators
listed in the definition of $C^* (\Z, X, h)_{ \{y\} }$
have this form
(using $u f = (f \circ h^{-1}) u$),
\pause
the product of two things of this form again has this form
(use $(f u^m) (g u^n) = [f (g \circ h^{-m})] u^{m + n},$
\pause
which is again in $C^* (\Z, X, h)_{ \{y\} }$ because
$C^* (\Z, X, h)_{ \{y\} }$ is closed under multiplication),
\pause
and the adjoint of something of this form again has this form
(by a similar argument).
\pause
\vspace{2ex}
It now suffices to prove that if $f u^n \in C^* (\Z, X, h)_{ \{y\} }$
and $n \neq 0,$
then $\ta (f u^m) = 0.$
\pause
The argument is the same as in the proof of
the previous Proposition;
\pause
one merely needs to note that the elements $g_j$ used there are
in $C (X) \S C^* (\Z, X, h)_{ \{y\} }.$
\pause
This completes the proof.
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\end{document}
\begin{frame}
\frametitle{???}
The following is a generalization of Lemma~\ref{L:PjComp}.
It is both more elementary and more general than the
corresponding argument in~\cite{LP}
(the main part of the proof of Theorem~4.5 there).
% Improve ???
% {\tt{Proof of the following needs to be rechecked.}} % ???
\begin{lem}\label{L:PjComp2}
Let $X$ be a \cms,
and let $h \colon X \to X$ be a \mh.
Let $B \S C^* (\Z, X, h)$ 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}
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{Proof ???}
Let $E \colon C^* (\Z, X, h) \to C (X)$ be the conditional
expectation $E_1$ of Proposition~\ref{P:CondExpt}.
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.
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{???}
Choose a finite sum $a = \sum_{n = -N}^N a_n u^n \in C^* (\Z, X, h)$
such that $\| a - c \| < \dt.$
Let $d$ be the metric on $X.$
Choose $\ep > 0$ so small that whenever $x_1, x_2 \in X$
satisfy $d (x_1, x_2) < \ep,$
then
\[
| a_n (x_1) - a_n (x_2) | < \frac{\dt}{2 N + 1}
\]
for $- N \leq n \leq N.$
Using freeness of the action of $\Z$ induced by $h,$
choose a nonempty open subset $U \S U_0$ such that
the sets $h^{-N} (U), \, h^{- N + 1} (U), \, \ldots, \, h^N (U)$
are disjoint,
and such that ${\overline{U}}$ 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 $- N \leq n \leq N.$
Set $b = \sum_{n = -N}^N b_n u^n,$
and note that $\| b - a \| < \dt.$
Thus, $\| b - c \| < 2 \dt.$
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{???}
Choose \cfn s $g_0, g_1 \colon X \to [0, 1]$
such that $\supp (g_0) \S U_0,$ $g_0 g_1 = g_1,$ and $g_1 \neq 0.$
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 $f p = f g_0 p = 6 \dt g_0 p = 6 \dt p,$
and similarly $p f = 6 \dt p.$
The same reasoning shows that $p b_n = b_n p$ for $- N \leq n \leq N.$
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
\[
p b p = \sum_{n = -N}^N p b_n u^n p
= \sum_{n = -N}^N b_n p u^n p
= b_0 p = p E (b) p.
\]
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\frametitle{???}
Using this last equation at the second step
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 $a = (p c p)^{-1/2},$ 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.
% \pause
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%