Josia Pietsch
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299 lines
9.6 KiB
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299 lines
9.6 KiB
TeX
\lecture{15}{2023-12-05}{}
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\begin{theorem}[Boundedness Theorem]
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\yalabel{Boundedness Theorem}{Boundedness}{thm:boundedness}
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\gist{%
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Let $X$ be Polish, $C \subseteq X$ coanalytic,
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$\phi\colon C \to \omega_1$ a coanalytic rank on $C$,
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$A \subseteq C$ analytic, i.e.~$A \in \Sigma^1_1(X)$.
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}{%
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Let $X$ Polish, $C \in \Pi^1_1(X)$, $A \in \Sigma^1_1(X)$,
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$A \subseteq C$, $\phi\colon C \to \omega_1$ a coanalytic rank.
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}
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Then $\sup \{\phi(x) : x \in A\} < \omega_1$.
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Moreover for all $\xi < \omega_1$,
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\[
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D_\xi \coloneqq \{x \in C : \phi(x) < \xi\}
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\]
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and
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\[
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E_\xi \coloneqq \{x \in C : \phi(x) \le \xi\}
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\]
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are Borel subsets of $X$.
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\end{theorem}
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\begin{proof}
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Let
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\begin{IEEEeqnarray*}{rCl}
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x \prec y&:\iff& x,y \in A \land \phi(x) < \phi(y)\\
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&\iff& x,y \in A \land y \not\le_\phi^\ast x.
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\end{IEEEeqnarray*}
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Since $A$ is analytic,
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this relation is analytic and wellfounded on $X$.
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By \yaref{thm:kunenmartin}
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we get $\rho(\prec) < \omega_1$.
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Thus $\sup \{\phi(x) : x \in A\} < \omega_1$.
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Since $D_\xi = \bigcup_{\eta < \xi} E_\xi$,
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it suffices to check $E_\xi \in \Sigma_1^1(X)$.
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Let $\alpha \coloneqq \sup \{\phi(x) : x \in C\}$.
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Then $E_\xi = E_\alpha$ for all $\alpha \le \xi < \omega_1$.
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Consider $\xi \le \alpha$.
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\begin{itemize}
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\item If there exists $x_0 \in C$ with $\phi(x_0) \ge \xi$,
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pick such $x_0$ of minimal rank.
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Then for all $y \in X$ we have
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\begin{IEEEeqnarray*}{rClr}
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y \in E_\xi &\iff& y \in C \land \phi(y) \le \xi\\
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&\iff& y \le^\ast_\phi x_0 & ~ \text{ coanalytic}\\
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&\iff& x_0 \not<^\ast_\phi y & ~ \text{ analytic}\\
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\end{IEEEeqnarray*}
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So $E_\xi$ is Borel.
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% TODO If $\alpha < \omega_1$, this also shows that $E_\alpha$ is Borel?
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\item If there exists no such $x_0$
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then $\xi = \alpha$
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and
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\[
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E_\xi = E_\alpha = \bigcup_{\eta < \alpha} E_\eta
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\]
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is a countable union of Borel sets by the previous case.
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\end{itemize}
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\end{proof}
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\pagebreak
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\section{Abstract Topological Dynamics}
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% \subsection*{Basic Definitions}
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% TODO: move to appendix?
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\gist{%
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Recall:
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\begin{definition}+
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Let $X$ be a set.
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A \vocab{group action} of a group $G$ on $X$
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is a function
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$\alpha\colon G \times X \to X$
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such that
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\begin{itemize}
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\item $\forall x \in X.~\alpha(1_G,x) = x$,
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\item $\forall g,h \in G, x \in X.~\alpha(gh,x) = \alpha(g,\alpha(h,x))$.
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\end{itemize}
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Often we will abbreviate $\alpha(g,x)$ as $g\cdot x$.
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For $x \in X$,
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the \vocab{orbit} of $x$ is defined as
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\[
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G\cdot x \coloneqq \{g\cdot x : g \in G\}.
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\]
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A group action is called \vocab{transitive}
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iff $g \mapsto g \cdot x$ is surjective for all $x \in X$,
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i.e.~iff the action has exactly one orbit.
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For $x \in X$,
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the \vocab{stabilizer subgroup}
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of $G$ with respect to $x$ is
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\[
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G_x \coloneqq \{g \in G : g\cdot x = x\}.
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\]
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\end{definition}
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\begin{remark}+
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Group actions of a group $G$ on a set $X$
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correspond to group homomorphisms
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$G \to \Sym(X)$.
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Indeed for a group action $\alpha\colon G \times X \to X$
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consider
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\begin{IEEEeqnarray*}{rCl}
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G&\longrightarrow & \Sym(X) \\
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g&\longmapsto & (x \mapsto g \cdot x).
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\end{IEEEeqnarray*}
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\end{remark}
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}{}
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\begin{definition}+
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A group $G$ with a topology
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is a \vocab{topological group}
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iff
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\begin{IEEEeqnarray*}{rCl}
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G \times G&\longrightarrow & G \\
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(x,y) &\longmapsto & x \cdot y
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\end{IEEEeqnarray*}
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and
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\begin{IEEEeqnarray*}{rCl}
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G&\longrightarrow & G \\
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x&\longmapsto & x^{-1}
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\end{IEEEeqnarray*}
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are continuous.
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\end{definition}
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\begin{definition}
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\label{def:flow}
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Let $T$ be a topological group\footnote{usually $T = \Z$ with the discrete topology}
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and let $X$ be a compact metrizable space.
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A \vocab{flow} $(X, T)$, sometimes denoted $T \acts X$
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is a continuous action
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\begin{IEEEeqnarray*}{rCl}
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T \times X&\longrightarrow & X \\
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(t,x) &\longmapsto & tx.
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\end{IEEEeqnarray*}
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A flow is \vocab{minimal} iff every orbit is dense.
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$(Y,T)$ is a \vocab{subflow} of $(X,T)$ if $Y \subseteq X$
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and $Y$ is invariant under $T$,
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i.e.~$\forall t \in T,y \in Y.~ty \in Y$.
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A flow $(X,T)$ is \vocab{isometric}
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iff there is a metric $d$ on $X$ such that
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for all $t \in T$ the map
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\begin{IEEEeqnarray*}{rCl}
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a_t\colon X &\longrightarrow & X \\
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x &\longmapsto & tx
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\end{IEEEeqnarray*}
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is an \vocab{isometry},
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i.e.~$\forall t \in T.~\forall x,y \in X.~d(a_t(x),a_t(y)) = d(x,y)$.
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If $(X,T)$ is a flow, then a pair $(x,y)$, $x \neq y$
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is \vocab{proximal} iff
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\[
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\exists z \in X.~\exists (t_n)_{n < } \in T^{\omega}.~t_n x \xrightarrow{n \to \infty} z \land t_n y \xrightarrow{n \to \infty} z.
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\]
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A flow is \vocab{distal} iff
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it has no proximal pair.
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\end{definition}
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\begin{remark}
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Note that a flow is minimal iff it has no proper subflows.
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\end{remark}
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\begin{definition}+
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Let $(T,X)$ and $(T,Y)$ be flows.
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A \vocab{factor map} $\pi\colon (T,X) \to (T,Y)$
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is a continuous surjection $X \twoheadrightarrow Y$
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that is $T$-equivariant,
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i.e.~$\forall t \in T, x \in X.~\pi(t\cdot x) = t\cdot \pi(x)$.
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If such a factor map exists,
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we also say that $(T,Y)$ is a \vocab{factor}
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of $(T,X)$.
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An \vocab{isomorphism} from $(T,X)$ to $(T,Y)$ is
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a homeomorphism $X \leftrightarrow Y$
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commuting with the group action.
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\end{definition}
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\gist{%
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\begin{warning}+
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What is called ``factor'' here is called ``subflow''
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by Furstenberg.
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\end{warning}
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\begin{example}
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Recall that $S_1 = \{z \in \C : |z| = 1\}$.
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Let $X = S_1$, $T = S_1$
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$(\alpha,\beta) \mapsto \alpha + \beta$ is isometric.%
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\footnote{Note that here we consider the abelian group structure of $S^1$
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and $\alpha + \beta$ denotes the addition of \emph{angles},
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i.e.~$\alpha \cdot \beta$ in complex numbers.}
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\end{example}
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}{}
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\begin{definition}
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\label{def:isometricextension}
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Let $X,Y$ be compact metric spaces
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and $\pi\colon (X,T) \to (Y,T)$ a factor map.
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Then $(X,T)$ is an \vocab{isometric extension}
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of $(Y,T)$ if there is
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$\rho\colon X\times_Y X \to \R$%
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\footnote{Recall that in the category of topological spaces
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the \vocab{fiber product} of
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$A \xrightarrow{f} C$, $B \xrightarrow{g} C$
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is $A \times_C B = \{(a,b) \in A \times B: f(a) = g(b)\}$,
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i.e. $X \times_Y X = \{(x_1,x_2) \in X^2 : \pi(x_1) = \pi(x_2)\}$.}
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such that
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\begin{enumerate}[(a)]
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\item $\rho$ is continuous.
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\item For each $y \in Y$, $\rho$ is a metric on the fiber
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$X_y \coloneqq \{x \in X: \pi(x) = y\}$.
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\item $\forall t \in T.~\rho(tx_1,tx_2) = \rho(x_1,x_2)$.
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\item $\forall y,y' \in Y.~$
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the metric spaces $(X_y, \rho)$ and $(X_{y'}, \rho)$
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are isometric.
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\end{enumerate}
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\end{definition}
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\begin{remark}
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A flow is isometric iff it is an isometric extension
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of the trivial flow,
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i.e.~the flow acting on a singleton.
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Indeed maps $\rho\colon X\times_\star X = X^2 \to \R$
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as in \yaref{def:isometricextension}
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correspond to metrics witnessing that the flow is isometric.
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\end{remark}
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\begin{proposition}
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\label{prop:isomextdistal}
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An isometric extension of a distal flow is distal.
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\end{proposition}
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\begin{proof}
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\gist{%
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Let $\pi\colon X\to Y$ be an isometric extension.
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Towards a contradiction,
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suppose that $x_1,x_2 \in X$ are proximal.
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Take $z \in X$ and a sequence $(g_n)_{n < \omega}$ in $T$
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such that $g_n x_1 \to z$ and $g_n x_2 \to z$.
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Then $g_n \pi(x_1) \to \pi(z)$
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and $g_n \pi(x_2) \to \pi(z)$,
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so by distality of $Y$
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we have $\pi(x_1) = \pi(x_{2})$.
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Then $\rho(g_n x_1, g_n x_2)$
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is defined and equal to $\rho(x_1,x_2)$.
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By the continuity of $\rho$,
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we get $\rho(g_n x_1, g_n x_2) \to \rho(z,z) = 0$.
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Therefore $\rho(x_1,x_2) = 0$.
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Hence $x_1 = x_2$ $\lightning$.
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}{Let $\pi\colon X \to Y$ isometric extension.
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Suppose $x_1,x_2 \in X$ is proximal.
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Then $\pi(x_1) = \pi(x_2)$.
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But there exists a $T$-equivariant metric on the fibers.
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}
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\end{proof}
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\begin{definition}
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Let $\Sigma = \{(X_i, T) : i \in I\} $
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be a collection of factors of $(X,T)$.
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Let $\pi_i\colon (X,T) \to (X_i, T)$ denote the factor map.
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Then $(X, T)$ is a \vocab{limit}%
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\footnote{This is not a limit in the category theory sense and not uniquely determined.}
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% TODO THE inverse limit is A limit
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of $\Sigma$ iff
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\[
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\forall x_1 \neq x_2 \in X.~\exists i \in I.~\pi_i(x_1) \neq \pi_i(x_2).
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\]
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\end{definition}
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\begin{proposition}
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\label{prop:limitdistal}
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A limit of distal flows is distal.
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\end{proposition}
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\begin{proof}
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\gist{%
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Let $(X,T)$ be a limit of $\Sigma = \{(X_i, T) : i \in I\}$.
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Suppose that each $(X_i, T)$ is distal.
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If $(X,T)$ was not distal,
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then there were $x_1, x_2, z \in X$
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and a sequence $(g_n)$ in $T$
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with $g_n x_1 \to z$ and $g_n x_2 \to z$.
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Take $i \in I$ such that $\pi_i(x_1) \neq \pi_i(x_2)$.
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But then $g_n \pi_i(x_1) \to \pi_i(z)$
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and $g_n \pi_i(x_2) \to \pi_i(z)$,
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which is a contradiction since $(X_i, T)$ is distal.
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}{Suppose there is a proximal pair $x_1,x_2$.
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Take $i$ such that $\pi_i(x_1) \neq \pi_i(x_2) \lightning$.
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}
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\end{proof}
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