From de4897f59ac50e630d3f3b6738f1da066f73a82e Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Tue, 12 Dec 2023 12:10:37 +0100 Subject: [PATCH] lecture 17 --- inputs/lecture_17.tex | 212 ++++++++++++++++++++++++++++++++++++++++++ logic3.tex | 1 + 2 files changed, 213 insertions(+) create mode 100644 inputs/lecture_17.tex diff --git a/inputs/lecture_17.tex b/inputs/lecture_17.tex new file mode 100644 index 0000000..7c7f253 --- /dev/null +++ b/inputs/lecture_17.tex @@ -0,0 +1,212 @@ +\lecture{17}{2023-12-12}{The Ellis semigroup} + + +\subsection{The Ellis semigroup} + +Let $(X, d)$ be a compact metric space +and $(X, T)$ a flow. + +Let $X^{X} \coloneqq \{f\colon X \to X\}$ +be the set of all functions.\footnote{We take all the functions, + they need not be continuous.} +We equip this with the product topology, +i.e.~a subbasis +is given by sets +\[ +U_{\epsilon}(x,y) \coloneqq \{f \in X^X : d(x,f(y)) < \epsilon\}. +\] +for all $x,y \in X$, $\epsilon > 0$. + +$X^{X}$ is a compact Hausdorff space. +\begin{remark} + \todo{Copy from exercise sheet 10} + Let $f_0 \in X^X$ be fixed. + \begin{itemize} + \item $X^X \ni f \mapsto f \circ f_0$ + is continuous: + + Consider $\{f : f f_0 \in U_{\epsilon}(x,y)\}$. + We have $ff_0 \in U_{\epsilon}(x,y)$ + iff $f \in U_\epsilon(x,f_0(y))$. + \item Fix $x_0 \in X$. + Then $f \mapsto f(x)$ is continuous. + \item In general $f \mapsto f_0 \circ f$ is not continuous, + but if $f_0$ is continuous, then the map is continuous. + \end{itemize} +\end{remark} + +\begin{definition} +Let $(X,T)$ be a flow. +Then the \vocab{Ellis semigroup} +is defined by +$E(X,T) \coloneqq \overline{T} \subseteq X^X$, +i.e.~identify $T$ with $x \mapsto tx$ +and take the closure in $X^X$. +\end{definition} +$E(X,T)$ is compact and Hausdorff, +since $X^X$ has these properties. + +Properties of $(X,T)$ translate to +\begin{goal} + We want to show that if $(X,T)$ is distal, + then $E(X,T)$ is a group. +\end{goal} + +\begin{proposition} + $G$ is a semigroup, + i.e.~closed under composition. +\end{proposition} +\begin{proof} + Take $t \in T$. We want to show that $tG \subseteq G$, + i.e.~for all $h \in G$ we have $th \in G$. + + We have that $t^{-1}G$ is compact, + since $t^{-1}$ is continuous + and $G$ is compact. + + Then $T \subseteq t^{-1}G$ since $T \ni s = t^{-1}\underbrace{(ts)}_{\in G}$. + So $G = \overline{T} \subseteq t^{-1}G$. + Hence $tG \subseteq G$. + + \begin{claim} + If $g \in G$, then + \[ + \overline{T} g = \overline{Tg}. + \] + \end{claim} + \begin{subproof} + \todo{Homework} + \end{subproof} + + Let $g \in G$. + We need to show that $Gg \subseteq G$. + + It is + \[ + Gg = \overline{T}g = \overline{Tg}. + \] + Since $G$ compact, + and $Tg \subseteq G$, + we have $ \overline{Tg} \subseteq G$. +\end{proof} + +\begin{definition} + A \vocab{compact semigroup} $S$ + is a nonempty semigroup with a compact + Hausdorff topology, + such that $S \ni x \mapsto xs$ is continuous for all $s$. +\end{definition} +\begin{example} + Ellis semigroup is a compact semigroup. +\end{example} + +\begin{lemma}[Ellis–Numakura] + \yalabel{Ellis-Numakura Lemma}{Ellis-Numakura}{lem:ellisnumakura} + Every compact semigroup + contains an \vocab{idempotent} element, + i.e.~$f$ such that $f^2 = f$. +\end{lemma} +\begin{proof} + Using Zorn's lemma, take a $\subseteq$-minimal + compact subsemigroup $R$ of $S$ + and let $s \in R$. + + Then $Rs$ is also a compact subsemigroup + and $Rs \subseteq R$. + By minimality of $R$, $R = Rs$. + Let $P \coloneqq \{ x \in R : xs = s\}$. + Then $P \neq \emptyset$, + since $s \in Rs$ + and $P$ is a compact semigroup, + since $x \overset{\alpha}\mapsto xs$ + is continuous and $P = \alpha^-1(s) \cap R$. + Thus $P = R$ by minimality, + so $s \in P$, + i.e.~$s^2 = s$. +\end{proof} + +The \yaref{lem:ellisnumakura} is not very interesting for $E(X,T)$, +since we already know that it has an identity, +in fact we have chosen $R = \{1\}$ in the proof. +But it is interesting for other semigroups. + + +\begin{theorem}[Ellis] + $(X,T)$ is distal iff $E(X,T)$ is a group. +\end{theorem} +\begin{proof} + + Let $G \coloneqq E(X,T)$. + Let $d$ be a metric on $X$. + + For all $g \in G$ we need to show that $x \mapsto gx$ is bijective. + If we had $gx = gy$, then $d(gx,gy) = 0$. + Then $\inf d(tx,ty) = 0$, but the flow is distal, + hence $x = y$. + + Let $g \in G$. Consider the compact semigroup $\Gamma \coloneqq Gg$. + By the \yaref{lem:ellisnumakura}, + there is $f \in \Gamma$ such that $f^2 = f$, + i.e.~for all $x \in X$ we have $f^2(x) = f(x)$. + Since $f$ is injective, we get that $x = f(x)$, + i.e.~$f = \id$. + + Take $g' \in G$ such that $f = g' \circ g$.% + %\footnote{This exists since $f \in Gg$.} + + It is $g' = g'gg'$, + so $\forall x .~g'(x) = g'(g g'(x))$. + Hence $g'$ is bijective + and $x = gg'(x)$, + i.e.~$g g' = \id$. + + \todo{The other direction is left as an easy exercise.} +\end{proof} + +Let $(X,T)$ be a flow. +Then by Zorn's lemma, there exists $X_0 \subseteq X$ +such that $(X_0, T)$ is minimal. +In particular, +for $x \in X$ and $\overline{Tx} = Y$ +we have that $(Y,T)$ is a flow. +However if we pick $y \in Y$, $Ty$ might not be dense. +% TODO: think about this! +% We want to a minimal subflow in a nice way: + +\begin{theorem} + If $(X,T)$ is distal, then $X$ is the disjoint union of minimal subflows. + In fact those disjoint sets + will be orbits of $E(X,T)$. +\end{theorem} +\begin{proof} + Let $G = E(X,T)$. + Note that for all $x \in X$, + we have $Gx \subseteq X$ is compact + and invariant under the action of $G$. + + Since $G$ is a group, we have that the orbits partition $X$.% + \footnote{Note that in general this does not hold for semigroups.} + + % Clearly the sets $Gx$ cover $X$. We want to show that they + % partition $X$. + % It suffices to show that $y \in Gx \implies Gy = Gx$. + +% Take some $y \in Gx$. +% Recall that $\overline{Ty} = \overline{T} y = Gy$. +% We have $\overline{Ty} \subseteq Gx$, +% so $Gy \subseteq Gx$. +% Since $y = g_0 x \implies x = g_0^{-1}y$, we also have $x \in Gy$, +% hence $Gx \subseteq Gy$ +\end{proof} +\begin{corollary} + If $(X,T)$ is distal and minimal, + then $E(X,T) \acts X$ is transitive. +\end{corollary} + + + + + + + + diff --git a/logic3.tex b/logic3.tex index 5a3f644..ba544f9 100644 --- a/logic3.tex +++ b/logic3.tex @@ -41,6 +41,7 @@ \input{inputs/lecture_14} \input{inputs/lecture_15} \input{inputs/lecture_16} +\input{inputs/lecture_17}