Josia Pietsch 82f0dfd4de
Some checks failed
Build latex and deploy / checkout (push) Failing after 15m5s
fixed important typo
2024-02-07 16:04:28 +01:00

271 lines
8.4 KiB
Raw Permalink Blame History

This file contains ambiguous Unicode characters

This file contains Unicode characters that might be confused with other characters. If you think that this is intentional, you can safely ignore this warning. Use the Escape button to reveal them.

\subsection{The Ellis semigroup}
\lecture{17}{2023-12-12}{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.\footnote{cf.~\href{https://en.wikipedia.org/wiki/Tychonoff's_theorem}{Tychonoff's theorem}}
Let $f_0 \in X^X$ be fixed.
\item $X^X \ni f \mapsto f \circ f_0$
is continuous:
Consider $\{f : f \circ f_0 \in U_{\epsilon}(x,y)\}$.
We have $f \circ f_0 \in U_{\epsilon}(x,y)$
iff $f \in U_\epsilon(x,f_0(y))$.
\item Fix $x_0 \in X$.
Then $f \mapsto f(x_0)$ 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.
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 $t \in T$ with $x \mapsto tx$
and take the closure in $X^X$.
The \vocab{Ellis semigroup} of a flow $(X,T)$
is $E(X,T) \coloneqq \overline{T} \subseteq X^X$.
$E(X,T)$ is compact and Hausdorff,
since $X^X$ has these properties.
Properties of $(X,T)$ translate to properties of $E(X,T)$:
We want to show that if $(X,T)$ is distal,
then $E(X,T)$ is a group.
$E(X,T)$ is a semigroup,
i.e.~closed under composition.
Let $G \coloneqq E(X,T)$.
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.
It is $T \subseteq t^{-1}G$ since $T \ni s = t^{-1}\underbrace{(ts)}_{\in G}$.
So $G = \overline{T} \subseteq t^{-1}G$.
Hence $tG \subseteq G$.
If $g \in G$, then
\overline{T} g = \overline{Tg}.
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$.
$G \coloneqq E(X,T)$.
\item $\forall t \in T. ~ tG \subseteq G$:
\item $t^{-1}G$ is compact.
\item $T \subseteq t^{-1}G$,
\item $\leadsto G = \overline{T} \subseteq t^{-1}G$,
i.e.~$tG \subseteq G$.
\item $\forall g \in G.~\overline{T}g = \overline{Tg}$ (cts.~map from compact to Hausdorff)
\item $\forall g \in G.~Gg \subseteq G$ :
$Gg = \overline{T}g = \overline{Tg} \overset{G \text{ compact}, Tg \subseteq G}{\subseteq} G$.
A \vocab{compact semigroup} $S$
is a nonempty semigroup\footnote{may not contain inverses or the identity}
with a compact Hausdorff topology,
such that $S \ni x \mapsto xs$ is continuous for all $s$.
The Ellis semigroup is a compact semigroup.
\yalabel{Ellis-Numakura Lemma}{Ellis-Numakura}{lem:ellisnumakura}
Every non-empty compact semigroup
contains an \vocab{idempotent} element,
i.e.~$f$ such that $f^2 = f$.
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.~$s^2 = s$.
\item Take $R \subseteq S$ minimal compact subsemigroup (Zorn),
$s \in R$.
\item $Rs \subseteq R \implies Rs = R$.
\item $P \coloneqq \{x \in R : xs = s\}$:
\item $P \neq \emptyset$, since $s \in Rs$
\item $P$ compact, since $P = \alpha^{-1}(s) \cap R$,
$\alpha: x \mapsto xs$ cts.
\item $P = R \implies s^2 = s$.
The \yaref{lem:ellisnumakura} is not very interesting for $E(X,T)$,
since we already know that it has an identity.
%in fact we might have chosen $R = \{1\}$ in the proof.
But it is interesting for other semigroups.
$(X,T)$ is distal iff $E(X,T)$ is a group.
Let $G \coloneqq E(X,T)$ and let $d$ be a metric on $X$.
For all $g \in G$ we need to show that $x \mapsto gx$ is injective.
If we had $gx = gy$, then $d(gx,gy) = 0$.
Then $\inf_{t \in T} d(tx,ty) = 0$, 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.~$f = \id$.
Since $f \in Gg$, there exists $g' \in G$ such that $f = g' \circ g$.
It is $g' = g'gg'$,
so $\forall x .~g'(x) = g'(g g'(x))$.
Hence $g'$ is injective
and $x = gg'(x)$,
i.e.~$g g' = \id$.
\item $x \mapsto gx$ injective for all $g \in G$:
\[gx = gy
\implies d(gx,gy) = 0
\implies \inf_{t \in T} d(tx, ty) = 0
\overset{\text{distal}}{\implies} x = y.
\item Fix $g \in G$.
\item $\Gamma \coloneqq Gg$ is a compact semigroup.
\item $\exists f\in \Gamma.~f^2 = f$ (\yaref{lem:ellisnumakura})
\item $f$ is injective, hence $f = \id$.
\item Take $g'$ such that $g'g = f$. $g'gg' = g' \implies gg' = \id$.
On the other hand if $(x_0,x_1)$ is proximal,
then there exists $g \in G$ such that $gx_0 = gx_1$.%
\footnote{cf.~\yaref{s11e1} (e)}
It follows that an inverse to $g$ can not exist.
}{If $(x_0,x_1)$ is proximal, there is $g \in G$ with $gx_0 = gx_1$, i.e.~no inverse to $g$.}
% 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: question!
% TODO: think about this!
% We want to a minimal subflow in a nice way:
If $(X,T)$ is distal, then $X$ is the disjoint union of minimal subflows.
In fact those disjoint sets
will be orbits of $E(X,T)$.
Let $G = E(X,T)$.
Note that for all $x \in X$,
we have that $Gx \subseteq X$ is compact
and invariant under the action of $G$.
Since $G$ is a group, the orbits partition $X$.%
\footnote{Note that in general this does not hold for semigroups.}
We need to show that $(Gx, T)$ is minimal.
Suppose that $y \in Gx$, i.e.~$Gx = Gy$.
Since $g \mapsto gy$ is continuous,
we have $Gx = Gy = \overline{T}y = \overline{Ty}$,
so $Ty$ is dense in $Gx$.
\item $G$ is a group, so the $G$-orbits partition $X$.
\item Let $y \in Gx$, then $Gx \overset{y \in Gx}{=} Gy \overset{\text{def}}{=} \overline{T}y \overset{g \mapsto gy \text{ cts.}}{=} \overline{Ty}$,
i.e.~$(Gx,T)$ is minimal.
If $(X,T)$ is distal and minimal,
then $E(X,T) \acts X$ is transitive.