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.}
|
||||
|
||||
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{Ellis-Numakura Lemma}{Ellis-Numakura}{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 ANKI-MARKER
|
||||
|
||||
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}{2023-12-15}{Sketch of proof of \yaref{thm:l16:3}}
|
||||
|
||||
% TODO ANKI-MARKER
|
||||
|
||||
|
||||
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:beleznay-foreman} 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