w23-logic-3/inputs/lecture_17.tex

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\subsection{The Ellis semigroup}
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\lecture{17}{2023-12-12}{The Ellis semigroup}
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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:
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Consider $\{f : f \circ f_0 \in U_{\epsilon}(x,y)\}$.
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We have $ff_0 \in U_{\epsilon}(x,y)$
iff $f \in U_\epsilon(x,f_0(y))$.
\item Fix $x_0 \in X$.
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Then $f \mapsto f(x_0)$ is continuous.
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\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
$E(X,T) \coloneqq \overline{T} \subseteq X^X$,
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i.e.~identify $t \in T$ with $x \mapsto tx$
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and take the closure in $X^X$.
\end{definition}
$E(X,T)$ is compact and Hausdorff,
since $X^X$ has these properties.
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Properties of $(X,T)$ translate to properties of $E(X,T)$:
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\begin{goal}
We want to show that if $(X,T)$ is distal,
then $E(X,T)$ is a group.
\end{goal}
\begin{proposition}
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$E(X,T)$ is a semigroup,
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i.e.~closed under composition.
\end{proposition}
\begin{proof}
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Let $G \coloneqq E(X,T)$.
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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}
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The Ellis semigroup is a compact semigroup.
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\end{example}
\begin{lemma}[EllisNumakura]
\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}
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\label{thm:distalflowpartition}
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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 $x \in X$,
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we have that $Gx \subseteq X$ is compact
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and invariant under the action of $G$.
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Since $G$ is a group, the orbits partition $X$.%
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\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 $Gx \subseteq Gy$
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% TODO: WHY?
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\end{proof}
\begin{corollary}
If $(X,T)$ is distal and minimal,
then $E(X,T) \acts X$ is transitive.
\end{corollary}