\subsection{Sketch of proof of \yaref{thm:l16:3}} \lecture{18}{2023-12-15}{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 semi-continuous}\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{F-topology} on $X$. In this setting, the original topology is also called the \vocab{E-topology}. \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 $d(t t_0x', t t_0 x'') < a-b$ 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 E-topology}. 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 non-trivial.} \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 non-trivial: 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}