lecture 17

inputs/lecture_17.tex

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+\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}
+
+
+
+
+
+
+
+

logic3.tex

@@ -41,6 +41,7 @@
 \input{inputs/lecture_14}
 \input{inputs/lecture_15}
 \input{inputs/lecture_16}
+\input{inputs/lecture_17}