improved lecture 17
Some checks failed
Build latex and deploy / checkout (push) Failing after 18m19s
Some checks failed
Build latex and deploy / checkout (push) Failing after 18m19s
This commit is contained in:
parent
f4d64527c4
commit
6c2a76d838

@ 154,7 +154,7 @@ We will see later that $\Sigma^1_1(X) \cap \Pi^1_1(X) = \cB(X)$.


\begin{proof}


Take $\cU \subseteq Y \times X \times \cN$


which is $Y$universal for $\Pi^0_1(X \times \cN)$.


Let $\cV \coloneqq \proj_{Y \times Y}(\cU)$.


Let $\cV \coloneqq \proj_{Y \times X}(\cU)$.


Then $\cV$ is $Y$universal for $\Sigma^1_1(X)$:


\begin{itemize}


\item $\cV \in \Sigma^1_1(Y \times X)$





@ 140,7 +140,7 @@ By Zorn's lemma, this will follow from


\begin{theorem}[Furstenberg]


\label{thm:l16:3}


Let $(X, T)$ be a minimal distal flow


and let $(Y, T)$ be a proper factor.


and let $(Y, T)$ be a proper factor.%


\footnote{i.e.~$(X,T)$ and $(Y,T)$ are not isomorphic}


Then there is another factor $(Z,T)$ of $(X,T)$


which is a proper isometric extension of $Y$.





@ 15,7 +15,7 @@ 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.


$X^{X}$ is a compact Hausdorff space.\footnote{cf.~\href{https://en.wikipedia.org/wiki/Tychonoff's_theorem}{Tychonoff's theorem}}


\begin{remark}%


\footnote{cf.~\yaref{s11e1}}


Let $f_0 \in X^X$ be fixed.



@ 34,16 +34,20 @@ $X^{X}$ is a compact Hausdorff space.


\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 $t \in T$ with $x \mapsto tx$


and take the closure in $X^X$.


\gist{%


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 $E(X,T) \coloneqq \overline{T} \subseteq X^X$.


}


\end{definition}


$E(X,T)$ is compact and Hausdorff,


since $X^X$ has these properties.


% TODO THINK ABOUT THIS




\gist{


Properties of $(X,T)$ translate to properties of $E(X,T)$:



@ 58,15 +62,14 @@ since $X^X$ has these properties.


i.e.~closed under composition.


\end{proposition}


\begin{proof}


\gist{


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 $th \in G$.




\gist{


We have that $t^{1}G$ is compact,


since $t^{1}$ is continuous


and $G$ is compact.


}{$t^{1}G$ is compact.}


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 $T \ni s = t^{1}\underbrace{(ts)}_{\in G}$.





@ 93,6 +96,21 @@ since $X^X$ has these properties.


Since $G$ compact,


and $Tg \subseteq G$,


we have $ \overline{Tg} \subseteq G$.


}{


$G \coloneqq E(X,T)$.


\begin{itemize}


\item $\forall t \in T. ~ tG \subseteq G$:


\begin{itemize}


\item $t^{1}G$ is compact.


\item $T \subseteq t^{1}G$,


\item $\leadsto G = \overline{T} \subseteq t^{1}G$,


i.e.~$tG \subseteq G$.


\end{itemize}


\item $\forall g \in G.~\overline{T}g = \overline{Tg}$ (cts.~map from compact to Hausdorff)


\item $\forall g \in G.~Gg \subseteq G$ :


$Gg = \overline{T}g = \overline{Tg} \overset{G \text{ compact}, Tg \subseteq G}{\subseteq} G$.


\end{itemize}


}


\end{proof}




\begin{definition}



@ 101,9 +119,11 @@ since $X^X$ has these properties.


with a compact Hausdorff topology,


such that $S \ni x \mapsto xs$ is continuous for all $s$.


\end{definition}


\gist{


\begin{example}


The Ellis semigroup is a compact semigroup.


\end{example}


}{}




\begin{lemma}[Ellis–Numakura]


\yalabel{EllisNumakura Lemma}{EllisNumakura}{lem:ellisnumakura}



@ 112,6 +132,7 @@ since $X^X$ has these properties.


i.e.~$f$ such that $f^2 = f$.


\end{lemma}


\begin{proof}


\gist{


Using Zorn's lemma, take a $\subseteq$minimal


compact subsemigroup $R$ of $S$


and let $s \in R$.



@ 128,25 +149,39 @@ since $X^X$ has these properties.


Thus $P = R$ by minimality,


so $s \in P$,


i.e.~$s^2 = s$.


}{


\begin{itemize}


\item Take $R \subseteq S$ minimal compact subsemigroup (Zorn),


$s \in R$.


\item $Rs \subseteq R \implies Rs = R$.


\item $P \coloneqq \{x \in R : xs = s\}$:


\begin{itemize}


\item $P \neq \emptyset$, since $s \in Rs$


\item $P$ compact, since $P = \alpha^{1}(s) \cap R$,


$\alpha: x \mapsto xs$ cts.


\item $P = R \implies s^2 = s$.


\end{itemize}


\end{itemize}%


}


\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 might have chosen $R = \{1\}$ in the proof.


But it is interesting for other semigroups.


\gist{


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.


}{}






\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 $X$.




Let $G \coloneqq E(X,T)$ and let $d$ be a metric on $X$.


\gist{


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,


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 $\Gamma \coloneqq Gg$.



@ 164,19 +199,39 @@ But it is interesting for other semigroups.


Hence $g'$ is bijective


and $x = gg'(x)$,


i.e.~$g g' = \id$.


}{


\begin{itemize}


\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 $g \in G$.


\begin{itemize}


\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 $g'g = f$. $g'gg' = g' \implies gg' = \id$.


\end{itemize}


\end{itemize}


}




\todo{The other direction is left as an easy exercise.}


\gist{


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 $g$.}


\end{proof}




% TODO ANKIMARKER




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.


% 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:



@ 189,6 +244,7 @@ However if we pick $y \in Y$, $Ty$ might not be dense.


\end{theorem}


\begin{proof}


Let $G = E(X,T)$.


\gist{


Note that for all $x \in X$,


we have that $Gx \subseteq X$ is compact


and invariant under the action of $G$.



@ 196,17 +252,18 @@ However if we pick $y \in Y$, $Ty$ might not be dense.


Since $G$ is a group, 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 $Gx \subseteq Gy$


% TODO: WHY?


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 $Gx$.


}{


\begin{itemize}


\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.


\end{itemize}


}


\end{proof}


\begin{corollary}


If $(X,T)$ is distal and minimal,





@ 1,16 +1,11 @@


\subsection{Sketch of proof of \yaref{thm:l16:3}}


\lecture{18}{20231215}{Sketch of proof of \yaref{thm:l16:3}}




% TODO ANKIMARKER






The goal for this lecture is to give a very rough


sketch of \yaref{thm:l16:3} in the case of $Z = 1$.


% \begin{theorem}[Furstenberg]


% Let $(X, T)$ be a minimal distal flow


% and let $(Z,T)$ be a proper factor of $X$%


% \footnote{i.e.~$(X,T)$ and $(Z,T)$ are not isomorphic.}


% Then three is another factor $(Y,T)$ of $(X,T)$


% which is a proper isometric extension of $Z$.


% \end{theorem}






Let $(X,T)$ be a distal flow.


Then $G \coloneqq E(X,T)$ is a group.





@ 78,7 +78,7 @@


is Borel for all $\alpha < \omega_1$.


\end{enumerate}


\end{theorem}


\todo{This was already stated as \yaref{thm:beleznayforeman} in lecture 16


\todo{This was already stated as \yaref{thm:beleznayforeman} in lecture 16


and should not have two numbers.}




A few words on the proof:





@ 108,13 +108,13 @@ This requires the use of the axiom of choice.


\item Let $B = \{x \in \R  (F_i)_x \text{ is meager}\}$.


Then $B$ is comeager in $\R$ and $B = \fc$.




We have $B = \fc$, since $B$ contains a comeager


$G_\delta$ set, $B'$:


We have $B = \fc$:


$B$ contains a comeager $G_\delta$ set, say $B'$.


$B'$ is Polish,


hence $B' = P \cup C$


for $P$ perfect and $C$ countable,


and $P \in \{\fc, 0\}$.


But $B'$ can't contain isolated point.


But $B'$ can't contain an isolated point.


\item We use $B$ to find a suitable point $a_i$:




To ensure that (i) holds, it suffices to chose





@ 102,7 +102,7 @@ Let $d$ be a compatible metric on $X$.


Consider $\ev_x \colon X^X \to X$.


$X^X$ is compact and $X$ is Hausdorff.


Hence we can apply


\label{fact:t12:2}.


\yaref{fact:t12:2}.




\item Let $x_0 \neq x_1 \in X$.


Then $(x_0,x_1)$ is a proximal pair iff $d(gx_0,gx_1) = 0$




Loading…
Reference in New Issue
Block a user