Josia Pietsch
bc8b5a8b6c
Some checks failed
Build latex and deploy / checkout (push) Failing after 15m14s
240 lines
8.9 KiB
TeX
240 lines
8.9 KiB
TeX
\subsection{Sketch of proof of \yaref{thm:l16:3}}


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




The goal for this lecture is to give a very rough


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




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


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


\begin{definition}


\label{def:F}


For $x, x' \in X$ define


\[


F(x,x') \coloneqq \inf \{d(gx, gx') : g \in G\}.


\]


\end{definition}


\begin{fact}


\begin{enumerate}[(a)]


\item $F(x,x') = F(x', x)$,


\item $F(x,x') \ge 0$ and $F(x,x') = 0$ iff $x = x'$.


\item $F(gx, gx') = F(x,x')$ since $G$ is a group.


\item $F$ is an \vocab{upper semicontinuous}\footnote{%


Wikimedia has a nice \href{https://upload.wikimedia.org/wikipedia/commons/c/c0/Upper_semi.svg}{picture}.}


function on $X^2$,


i.e.~$\forall a \in R.~\{(x,x') \in X^2 : F(x,x') < a\} \overset{\text{open}}{\subseteq} X^2$.




This holds because $F$ is the infimum of continuous functions


\begin{IEEEeqnarray*}{rCl}


f_g\colon X^2 &\longrightarrow & \R \\


(x,x') &\longmapsto & d(gx, gx')


\end{IEEEeqnarray*}


for $g \in G$.


\end{enumerate}


\end{fact}


\begin{theoremdef}


\label{def:ftop}


The sets


\[


U_a(x) \coloneqq \{x' : F(x,x') < a\}


\]


form the basis of a topology in $X$.


This topology is called the \vocab{Ftopology} on $X$.


In this setting, the original topology


is also called the \vocab{Etopology}.


\end{theoremdef}


This will follow from the following lemma:


\begin{lemma}


\label{lem:ftophelper}


Let $F(x,x') < a$.


\gist{%


Then there exists $\epsilon > 0$ such that


whenever $F(x',x'') < \epsilon$, then $F(x,x'') < a$.


}{Then $\exists \epsilon > 0.~\forall x''.~F(x',x'') < \epsilon \implies F(x,x'') < a$.}


\end{lemma}


\begin{refproof}{def:ftop}


\gist{%


We have to show that if $U_a(x_1) \cap U_b(x_2) \neq \emptyset$,


then this intersection is the union


of sets of this kind.


}{}


Let $x' \in U_a(x_1) \cap U_b(x_2)$.


Then by \yaref{lem:ftophelper},


there exists $\epsilon_1 > 0$ with $U_{\epsilon_1}(x') \subseteq U_a(x_1)$.


Similarly there exists $\epsilon_2 > 0$\gist{


such that $U_{\epsilon_2}(x') \subseteq U_b(x_2)$.}{.}


So for $\epsilon \le \epsilon_1, \epsilon_2$,


we get $U_{\epsilon}(x') \subseteq U_a(x_1) \cap U_b(x_2)$.


\end{refproof}


\begin{refproof}{lem:ftophelper}%


\notexaminable{\footnote{This was not covered in class.}


% TODO: maybe learn?




Let $T = \bigcup_n T_n$,% TODO Why does this exist?


$T_n$ compact, wlog.~$T_n \subseteq T_{n+1}$, and


let $G(x,x') \coloneqq \{(gx,gx') : g \in G\} \subseteq X \times X$.


Take $b$ such that $F(x,x') < b < a$.


Then $U = \{(u,u') \in G(x,x') : d(u,u') < b\}$


is open in $G(x,x')$


and since $F(x,x') < b$ we have $U \neq \emptyset$.


\begin{claim}


There exists $n$ such that


\[


\forall (u,u') \in G(x,x').~T_n(u,u')\cap U \neq \emptyset.


\]


\end{claim}


\begin{subproof}


Suppose not.


Then for all $n$, there is $(u_n, u_n') \in G(x,x')$


with


\[T_n(u_n, u_n') \subseteq G(x,x') \setminus U.\]


Note that the RHS is closed.


For $m > n$ we have


$T_n(u_m, u'_m) \subseteq G(x,x') \setminus U$


since $T_n \subseteq T_m$.


By compactness of $X$,


there exists $v,v'$ and some subsequence


such that $(u_{n_k}, u'_{n_k}) \to (v,v')$.




So for all $n$ we have $T_n(v,v') \subseteq G(x,x') \setminus U$,


hence $T(v,v') \cap U = \emptyset$,


so $G(v,v') \cap U = \emptyset$.


But this is a contradiction as $\emptyset\neq U \subseteq G(v,v')$.


\end{subproof}


The map


\begin{IEEEeqnarray*}{rCl}


T\times X&\longrightarrow & X \\


(t,x) &\longmapsto & tx


\end{IEEEeqnarray*}


is continuous.


Since $T_n$ is compact,


we have that $\{(x,t) \mapsto tx : t \in T_n\}$


is equicontinuous.\todo{Sheet 11}


So there is $\epsilon > 0$ such that


$d(x_1,x_2) < \epsilon \implies d(tx_1, tx_2) < a b$


for all $t \in T_n$.




Suppose now that $F(x', x'') < \epsilon$.


Then there is $t_0 \in T$ such that $d(t_0x', t_0x'') < \epsilon$,


hence $d(t t_0x', t t_0 x'') < ab$ for all $t \in T_n$.


Since $(t_0x, t_0x') \in G(x,x')$,


there is $t_1 \in T_n$


with $(t_1t_0x, t_1t_0x') \in U$,


i.e.~$d(t_1t_0x, t_1t_0x') < b$


and therefore


$F(x,x'') = d(t_1t_0x, t_1t_0x'') < a$.


}


\end{refproof}




Now assume $Z = \{\star\}$.


We want to sketch a proof of \yaref{thm:l16:3} in this case,


i.e.~show that if $(Z,T)$ is a proper factor of a minimal distal flow


$(X,T)$ then there is another factor $(Y,T)$ of $(X,T)$


which is a proper isometric extension of $Z$.




\begin{proof}[sketch] % TODO: Think about this


\leavevmode


\begin{enumerate}[1.]


\item For $x \in X$ define


\begin{IEEEeqnarray*}{rCl}


F_x\colon X &\longrightarrow & \R \\


x' &\longmapsto & F(x,x').


\end{IEEEeqnarray*}


\item Define an equivalence relation on $X$,


by $x_1 \sim x_2 :\iff \{x \in X : F_{x_1}(x) = F_{x_2}(x)\}$


is comeager in $X$\footnote{with respect to the Etopology}.


Then for all $g \in G$ we have


$x_1 \sim x_2 \implies gx_1 \sim ~ gx_2$.




Let $M \coloneqq \{[x]_{\sim } : x \in X\} = \faktor{X}{\sim}$


bet the quotient space.


It is compact, second countable and Hausdorff.


Let $\pi\colon X\to M$ denote the quotient map.


\gist{%


\item $(Y,T) \mathbin{\text{\reflectbox{$\coloneqq$}}} (M,T)$


is an isometric flow:


\begin{enumerate}


\item For $a > 0$, $x,x' \in X$ let


\[


W(x,x') \coloneqq \{g \in G : F(x, gx') < a\}.


\]


This turns out to be a subbasis of a topology


which is coarser than the original topology on $G$.


The new topology makes $G$ compact.


\item Let $\theta(g)$ be the transformation of $M$


defined by $\theta(g) \pi(x) = \pi(gx)$.


This is well defined.


Let $H = \theta(G)$.


This is just a quotient of $G$, $g \mapsto \theta(g)$


may not be injective.


\item One can show that $H$ is a topological group and $(M,H)$


is a flow.\footnote{This is nontrivial.}


\item Since $H$ is compact,


$(M,H)$ is equicontinuous, i.e.~it is isometric.


In particular, $(M,T)$ is isometric.


\end{enumerate}


\item $M \neq \{\star\}$, i.e.~$(M,T)$ is nontrivial:




Suppose towards a contradiction that $M = \{\star\}$,


i.e.~$x_1 \sim x_2$ for all $x_1,x_2 \in X$.


Fix $x_2$. For every $x_1 \in X$


we have that


\[


\{x : F(x_1,x) = F(x_2,x)\}


\]


is comeager.


Let $x_1$ be a point of continuity of $F_{x_2}$.


Let $\langle a_n : n < \omega \rangle$ be a sequence


of elements that set, i.e.~$F(x_1, a_n) = F(x_2, a_n)$,


such that $a_n \to x_1$.


So by the continuity of $F_{x_2}$ at $x_1$


\begin{IEEEeqnarray*}{rCl}


\lim_{n \to \infty} F(x_2, a_n) &=& F(x_2, x_1)


\end{IEEEeqnarray*}


and by the definition of $F$


\begin{IEEEeqnarray*}{rCl}


\lim_{n \to \infty} F(x_1,a_n) &=& F(x_1,x_1) = 0.


\end{IEEEeqnarray*}


So


\[


F(x_2,x_1) = \lim_{n \to \infty} F(x_2, a_n) = \lim_{n \to \infty}


F(x_1,a_n) = 0


\]


and by distality we get $x_1 = x_2$.


Since almost all points of $X$


are points of continuity of $F_{x_2}$


(\yaref{thm:usccomeagercont})


this implies that $X \setminus \{x_2\}$ is meager.


But then $X = \{\star\} \lightning$.


}{}


\end{enumerate}


\end{proof}




\begin{theorem}\footnote{Not covered in class}


\label{thm:usccomeagercont}


Let $X$ be a metric space


and $\Gamma\colon X \to \R$ be upper semicontinuous.


Then the set of continuity points of $\Gamma$ is comeager.


\end{theorem}


\begin{proof}


\notexaminable{


Take $x$ such that $\Gamma$ is not continuous at $x$.


Then there is an $\epsilon > 0$


and $x_n \to x$ such that


$\Gamma(x_n) + \epsilon \le \Gamma(x)$.


Take $q \in \Q$ such that $\Gamma(x)  \epsilon < q < \Gamma(x)$.


Then let


\[


B_q \coloneqq \{a \in X : \Gamma(a) \ge q\}.


\]


$X \setminus B_q = \{a \in X : \Gamma(a) < q\}$


is open, i.e.~$B_q$ is closed.


Note that $x \in F_q \coloneqq B_q \setminus \inter(B_q)$


and $B_q \setminus \inter(B_q)$ is nwd


as it is closed and has empty interior,


so $\bigcup_{q \in \Q} F_q$ is meager.


}


\end{proof}






