Josia Pietsch
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240 lines
8.9 KiB
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240 lines
8.9 KiB
TeX
\subsection{Sketch of proof of \yaref{thm:l16:3}}
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\lecture{18}{2023-12-15}{Sketch of proof of \yaref{thm:l16:3}}
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The goal for this lecture is to give a very rough
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sketch of \yaref{thm:l16:3} in the case of $|Z| = 1$.
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Let $(X,T)$ be a distal flow.
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Then $G \coloneqq E(X,T)$ is a group.
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\begin{definition}
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\label{def:F}
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For $x, x' \in X$ define
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\[
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F(x,x') \coloneqq \inf \{d(gx, gx') : g \in G\}.
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\]
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\end{definition}
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\begin{fact}
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\begin{enumerate}[(a)]
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\item $F(x,x') = F(x', x)$,
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\item $F(x,x') \ge 0$ and $F(x,x') = 0$ iff $x = x'$.
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\item $F(gx, gx') = F(x,x')$ since $G$ is a group.
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\item $F$ is an \vocab{upper semi-continuous}\footnote{%
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Wikimedia has a nice \href{https://upload.wikimedia.org/wikipedia/commons/c/c0/Upper_semi.svg}{picture}.}
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function on $X^2$,
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i.e.~$\forall a \in R.~\{(x,x') \in X^2 : F(x,x') < a\} \overset{\text{open}}{\subseteq} X^2$.
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This holds because $F$ is the infimum of continuous functions
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\begin{IEEEeqnarray*}{rCl}
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f_g\colon X^2 &\longrightarrow & \R \\
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(x,x') &\longmapsto & d(gx, gx')
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\end{IEEEeqnarray*}
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for $g \in G$.
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\end{enumerate}
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\end{fact}
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\begin{theoremdef}
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\label{def:ftop}
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The sets
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\[
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U_a(x) \coloneqq \{x' : F(x,x') < a\}
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\]
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form the basis of a topology in $X$.
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This topology is called the \vocab{F-topology} on $X$.
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In this setting, the original topology
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is also called the \vocab{E-topology}.
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\end{theoremdef}
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This will follow from the following lemma:
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\begin{lemma}
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\label{lem:ftophelper}
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Let $F(x,x') < a$.
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\gist{%
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Then there exists $\epsilon > 0$ such that
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whenever $F(x',x'') < \epsilon$, then $F(x,x'') < a$.
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}{Then $\exists \epsilon > 0.~\forall x''.~F(x',x'') < \epsilon \implies F(x,x'') < a$.}
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\end{lemma}
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\begin{refproof}{def:ftop}
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\gist{%
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We have to show that if $U_a(x_1) \cap U_b(x_2) \neq \emptyset$,
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then this intersection is the union
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of sets of this kind.
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}{}
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Let $x' \in U_a(x_1) \cap U_b(x_2)$.
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Then by \yaref{lem:ftophelper},
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there exists $\epsilon_1 > 0$ with $U_{\epsilon_1}(x') \subseteq U_a(x_1)$.
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Similarly there exists $\epsilon_2 > 0$\gist{
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such that $U_{\epsilon_2}(x') \subseteq U_b(x_2)$.}{.}
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So for $\epsilon \le \epsilon_1, \epsilon_2$,
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we get $U_{\epsilon}(x') \subseteq U_a(x_1) \cap U_b(x_2)$.
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\end{refproof}
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\begin{refproof}{lem:ftophelper}%
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\notexaminable{\footnote{This was not covered in class.}
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% TODO: maybe learn?
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Let $T = \bigcup_n T_n$,% TODO Why does this exist?
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$T_n$ compact, wlog.~$T_n \subseteq T_{n+1}$, and
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let $G(x,x') \coloneqq \{(gx,gx') : g \in G\} \subseteq X \times X$.
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Take $b$ such that $F(x,x') < b < a$.
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Then $U = \{(u,u') \in G(x,x') : d(u,u') < b\}$
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is open in $G(x,x')$
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and since $F(x,x') < b$ we have $U \neq \emptyset$.
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\begin{claim}
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There exists $n$ such that
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\[
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\forall (u,u') \in G(x,x').~T_n(u,u')\cap U \neq \emptyset.
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\]
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\end{claim}
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\begin{subproof}
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Suppose not.
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Then for all $n$, there is $(u_n, u_n') \in G(x,x')$
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with
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\[T_n(u_n, u_n') \subseteq G(x,x') \setminus U.\]
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Note that the RHS is closed.
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For $m > n$ we have
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$T_n(u_m, u'_m) \subseteq G(x,x') \setminus U$
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since $T_n \subseteq T_m$.
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By compactness of $X$,
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there exists $v,v'$ and some subsequence
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such that $(u_{n_k}, u'_{n_k}) \to (v,v')$.
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So for all $n$ we have $T_n(v,v') \subseteq G(x,x') \setminus U$,
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hence $T(v,v') \cap U = \emptyset$,
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so $G(v,v') \cap U = \emptyset$.
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But this is a contradiction as $\emptyset\neq U \subseteq G(v,v')$.
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\end{subproof}
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The map
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\begin{IEEEeqnarray*}{rCl}
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T\times X&\longrightarrow & X \\
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(t,x) &\longmapsto & tx
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\end{IEEEeqnarray*}
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is continuous.
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Since $T_n$ is compact,
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we have that $\{(x,t) \mapsto tx : t \in T_n\}$
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is equicontinuous.\todo{Sheet 11}
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So there is $\epsilon > 0$ such that
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$d(x_1,x_2) < \epsilon \implies d(tx_1, tx_2) < a -b$
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for all $t \in T_n$.
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Suppose now that $F(x', x'') < \epsilon$.
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Then there is $t_0 \in T$ such that $d(t_0x', t_0x'') < \epsilon$,
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hence $d(t t_0x', t t_0 x'') < a-b$ for all $t \in T_n$.
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Since $(t_0x, t_0x') \in G(x,x')$,
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there is $t_1 \in T_n$
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with $(t_1t_0x, t_1t_0x') \in U$,
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i.e.~$d(t_1t_0x, t_1t_0x') < b$
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and therefore
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$F(x,x'') = d(t_1t_0x, t_1t_0x'') < a$.
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}
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\end{refproof}
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Now assume $Z = \{\star\}$.
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We want to sketch a proof of \yaref{thm:l16:3} in this case,
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i.e.~show that if $(Z,T)$ is a proper factor of a minimal distal flow
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$(X,T)$ then there is another factor $(Y,T)$ of $(X,T)$
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which is a proper isometric extension of $Z$.
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\begin{proof}[sketch] % TODO: Think about this
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\leavevmode
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\begin{enumerate}[1.]
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\item For $x \in X$ define
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\begin{IEEEeqnarray*}{rCl}
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F_x\colon X &\longrightarrow & \R \\
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x' &\longmapsto & F(x,x').
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\end{IEEEeqnarray*}
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\item Define an equivalence relation on $X$,
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by $x_1 \sim x_2 :\iff \{x \in X : F_{x_1}(x) = F_{x_2}(x)\}$
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is comeager in $X$\footnote{with respect to the E-topology}.
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Then for all $g \in G$ we have
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$x_1 \sim x_2 \implies gx_1 \sim ~ gx_2$.
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Let $M \coloneqq \{[x]_{\sim } : x \in X\} = \faktor{X}{\sim}$
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bet the quotient space.
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It is compact, second countable and Hausdorff.
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Let $\pi\colon X\to M$ denote the quotient map.
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\gist{%
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\item $(Y,T) \mathbin{\text{\reflectbox{$\coloneqq$}}} (M,T)$
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is an isometric flow:
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\begin{enumerate}
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\item For $a > 0$, $x,x' \in X$ let
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\[
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W(x,x') \coloneqq \{g \in G : F(x, gx') < a\}.
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\]
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This turns out to be a subbasis of a topology
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which is coarser than the original topology on $G$.
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The new topology makes $G$ compact.
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\item Let $\theta(g)$ be the transformation of $M$
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defined by $\theta(g) \pi(x) = \pi(gx)$.
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This is well defined.
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Let $H = \theta(G)$.
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This is just a quotient of $G$, $g \mapsto \theta(g)$
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may not be injective.
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\item One can show that $H$ is a topological group and $(M,H)$
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is a flow.\footnote{This is non-trivial.}
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\item Since $H$ is compact,
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$(M,H)$ is equicontinuous, i.e.~it is isometric.
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In particular, $(M,T)$ is isometric.
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\end{enumerate}
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\item $M \neq \{\star\}$, i.e.~$(M,T)$ is non-trivial:
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Suppose towards a contradiction that $M = \{\star\}$,
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i.e.~$x_1 \sim x_2$ for all $x_1,x_2 \in X$.
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Fix $x_2$. For every $x_1 \in X$
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we have that
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\[
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\{x : F(x_1,x) = F(x_2,x)\}
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\]
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is comeager.
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Let $x_1$ be a point of continuity of $F_{x_2}$.
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Let $\langle a_n : n < \omega \rangle$ be a sequence
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of elements that set, i.e.~$F(x_1, a_n) = F(x_2, a_n)$,
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such that $a_n \to x_1$.
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So by the continuity of $F_{x_2}$ at $x_1$
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\begin{IEEEeqnarray*}{rCl}
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\lim_{n \to \infty} F(x_2, a_n) &=& F(x_2, x_1)
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\end{IEEEeqnarray*}
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and by the definition of $F$
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\begin{IEEEeqnarray*}{rCl}
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\lim_{n \to \infty} F(x_1,a_n) &=& F(x_1,x_1) = 0.
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\end{IEEEeqnarray*}
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So
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\[
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F(x_2,x_1) = \lim_{n \to \infty} F(x_2, a_n) = \lim_{n \to \infty}
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F(x_1,a_n) = 0
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\]
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and by distality we get $x_1 = x_2$.
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Since almost all points of $X$
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are points of continuity of $F_{x_2}$
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(\yaref{thm:usccomeagercont})
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this implies that $X \setminus \{x_2\}$ is meager.
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But then $X = \{\star\} \lightning$.
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}{}
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\end{enumerate}
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\end{proof}
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\begin{theorem}\footnote{Not covered in class}
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\label{thm:usccomeagercont}
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Let $X$ be a metric space
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and $\Gamma\colon X \to \R$ be upper semicontinuous.
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Then the set of continuity points of $\Gamma$ is comeager.
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\end{theorem}
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\begin{proof}
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\notexaminable{
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Take $x$ such that $\Gamma$ is not continuous at $x$.
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Then there is an $\epsilon > 0$
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and $x_n \to x$ such that
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$\Gamma(x_n) + \epsilon \le \Gamma(x)$.
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Take $q \in \Q$ such that $\Gamma(x) - \epsilon < q < \Gamma(x)$.
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Then let
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\[
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B_q \coloneqq \{a \in X : \Gamma(a) \ge q\}.
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\]
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$X \setminus B_q = \{a \in X : \Gamma(a) < q\}$
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is open, i.e.~$B_q$ is closed.
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Note that $x \in F_q \coloneqq B_q \setminus \inter(B_q)$
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and $B_q \setminus \inter(B_q)$ is nwd
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as it is closed and has empty interior,
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so $\bigcup_{q \in \Q} F_q$ is meager.
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}
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\end{proof}
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