\subsection{The Order of a Flow} \lecture{19}{2023-12-19}{Orders of Flows} See also \cite[\href{https://terrytao.wordpress.com/2008/01/24/254a-lecture-6-isometric-systems-and-isometric-extensions/}{Lecture 6}]{tao}. \begin{definition}+ Let $X,Y$ be metric spaces. A family $F$ of functions $X \to Y$ is called \vocab{equicontinuous} at $x_0 \in X$ iff \[ \forall \epsilon > 0.~\exists \delta > 0.~ \forall f \in F.~d_X(x_0, x) < \delta \implies d_Y(f(x_0),f(x)) < \epsilon. \] It is called equicontinuous iff it is equicontinuous at every point. It is called \vocab{uniformly equicontinuous} iff \[ \forall \epsilon > 0.~\exists \delta > 0.~ \forall x_0 \in X.~\forall f \in F.~d_X(x_0, x) < \delta \implies d_Y(f(x_0),f(x)) < \epsilon. \] A flow $(X,T)$ is called equicontinuous iff $T$ is equicontinuous. \end{definition} Note that since $X$ compact the notions of equicontinuity and uniform equicontinuity coincide. \begin{fact}+[{\cite[Lecture 6, Exercise 1]{tao}}] \label{fact:isometriciffequicontinuous} A flow $(X,T)$ is isometric iff it is equicontinuous. \end{fact} \begin{proof} Clearly an isometric flow is equicontinuous. On the other hand suppose that $T$ is uniformly equicontinuous. Define a metric $\tilde{d}$ on $X$ by setting $\tilde{d}(x,y) \coloneqq \sup_{t \in T} d(tx,ty) \le 1$ (wlog.~$d \le 1$). By equicontinuity of $T$ we get that $\tilde{d}$ and $d$ induce the same topology on $X$. \end{proof} \gist{ Recall that we defined the order of a quasi-isometric flow to be the minimal number of steps required when building the tower to reach the flow with a quasi-isometric system (cf.~\yaref{thm:l16:3}, \yaref{def:floworder}). }{} \begin{theorem}[Maximal isometric factor] \label{thm:maxisomfactor} For every flow $(X,T)$ there is a maximal factor $(Y,T)$, $\pi\colon X\to Y$, i.e.~if $(Y',T), \pi'\colon X \to Y'$ is any isometric factor of $(X,T)$, then $(Y',T)$ is a factor of $(Y,T)$. % https://q.uiver.app/#q=WzAsMyxbMCwwLCIoWCxUKSJdLFsxLDEsIiBcXHN1YnN0YWNreyhZLFQpXFxcXFxcdGV4dHttYXhpbWFsIGlzb21ldHJpY319Il0sWzAsMiwiXFxzdWJzdGFja3soWScsVClcXFxcXFx0ZXh0e2lzb21ldHJpY319Il0sWzAsMl0sWzAsMV0sWzEsMiwiXFxleGlzdHMiLDAseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XV0= \[\begin{tikzcd} {(X,T)} \\ & { \substack{(Y,T)\\\text{maximal isometric}}} \\ {\substack{(Y',T)\\\text{isometric}}} \arrow[from=1-1, to=3-1] \arrow[from=1-1, to=2-2] \arrow["\exists", dashed, from=2-2, to=3-1] \end{tikzcd}\] \end{theorem} \begin{proof} We want to apply Zorn's lemma. If suffices to show that isometric flows are closed under inverse limits,% \footnote{This seems to be an inverse limit in the category theory sense.} i.e.~if $(Y_\alpha, f_{\alpha,\beta})$, $\beta < \alpha \le \Theta$ are isometric, then the inverse limit $Y$ is isometric.% % https://q.uiver.app/#q=WzAsNCxbMSwxLCJZX1xcYWxwaGEiXSxbMSwyLCJZX1xcYmV0YSJdLFswLDEsIlkiXSxbMSwwLCJYIl0sWzAsMSwiZl97XFxhbHBoYSwgXFxiZXRhfSIsMV0sWzIsMCwiZl9cXGFscGhhIl0sWzIsMSwiZl9cXGJldGEiLDJdLFszLDAsIlxccGlfXFxhbHBoYSIsMV0sWzMsMSwiXFxwaV9cXGJldGEiLDEseyJjdXJ2ZSI6LTN9XV0= \[\begin{tikzcd} & X \\ Y & {Y_\alpha} \\ & {Y_\beta} \arrow["{f_{\alpha, \beta}}"{description}, from=2-2, to=3-2] \arrow["{f_\alpha}", from=2-1, to=2-2] \arrow["{f_\beta}"', from=2-1, to=3-2] \arrow["{\pi_\alpha}"{description}, from=1-2, to=2-2] \arrow["{\pi_\beta}"{description}, curve={height=-18pt}, from=1-2, to=3-2] \end{tikzcd}\] \gist{% Consider \[ \Delta_\alpha \coloneqq \{(y,y') \in Y^2 : f_{\alpha}(y) = f_\alpha(y')\}. \] Let $d$ be a metric on $Y$ and $d_{\alpha}$ a metric on $Y_{\alpha}$, wlog.~$d, d_\alpha \le 1$. Note that $\beta < \alpha \implies \Delta_\beta \supseteq \Delta_\alpha$ and \[ \bigcap_{\alpha \le \theta}\Delta_\alpha = \{(y,y) : y \in Y\}. \] Fix $\epsilon > 0$ and consider \[\{(y,y') \in \Delta_\alpha : d(y,y') \ge \epsilon\}.\] By the finite intersection property we get \[ \exists \alpha.~f_\alpha(y) = f_\alpha(y') \implies d(y,y') < \epsilon, \] i.e.~$\forall z \in Y_\alpha.~\diam(f^{-1}_\alpha(z)) \le \epsilon$. Towards a contradiction assume that $Y$ is not isometric, i.e.~not equicontinuous. Then there are $(y_j), (y'_j) \in Y$ such that $d(y_j,y'_j) \to 0$ and $\epsilon > 0, t_j \in T$ such that $d(t_jy_j, t_jy'_j) > \epsilon$. By compactness wlog.~$(y_j)$ and $(y'_j)$ converge (to the same point). Find $\alpha$ such that $f_\alpha(y) = f_{\alpha}(y') \implies d(y,y') < \frac{\epsilon}{4}$. Let $z_j \coloneqq f_{\alpha}(y_j)$ and $z'_j \coloneqq f_\alpha(y'_j)$. Then $(z_j)$ and $(z'_j)$ converge to the same point $z \in Y_\alpha$. By equicontinuity of $(Y_\alpha, T)$, $d_{Y_{\alpha}}(t_jz_j, t_jz'_j) \to 0$. Wlog.~$(t_jz_j)$ and $(t_jz'_j)$ converge. Let $z^\ast$ be their limit. On the one hand, by the triangle inequality we get \[ d(f^{-1}_\alpha(t_jz_j), f^{-1}_\alpha(t_jz_j')) > \underbrace{\epsilon}_{\mathclap{< d(t_jy_j, t_jy_j')}} - \overbrace{\frac{\epsilon}{4}}^{\mathclap{\text{Diameter of fiber}}}- \frac{\epsilon}{4} = \frac{\epsilon}{2}. \] On the other hand, from \begin{IEEEeqnarray*}{rCl} d(f^{-1}_\alpha(t_jz_j), f^{-1}_{\alpha}(z^\ast)) &\to & 0,\\ d(f^{-1}_\alpha(t_jz'_j), f^{-1}_{\alpha}(z^{\ast})) &\to & 0,\\ \diam f^{-1}_\alpha(\{z^\ast\}) & <& \frac{\epsilon}{4} \end{IEEEeqnarray*} we obtain \[ d(f^{-1}_\alpha(t_jz_j), f^{-1}_\alpha(t_jz'_j)) < \frac{\epsilon}{2} \lightning. \] }{ \begin{itemize} \item Consider $\Delta_\alpha = \{(y,y') \in Y \times Y : f_\alpha(y) = f_\alpha(y')\} = Y \times_{Y_\alpha} Y$. \begin{itemize} \item $\beta < \alpha \implies \Delta_\beta \supseteq \Delta_\alpha$, \item $\bigcap_{\alpha < \theta} \Delta_\alpha = \{(y,y) : y \in Y\}$. \end{itemize} \item Fix $\epsilon > 0$. Consider $M^{\epsilon}_{\alpha}\{(y,y') \in \Delta_\alpha : d(y,y') \ge \epsilon\}$. FIP $\implies$ $\exists \alpha.~M^{\epsilon}_{\alpha} = \emptyset$, i.e.~$\forall z \in Y_\alpha.~\diam(f^{-1}_\alpha(z)) \le \epsilon$. \item Suppose $Y$ is not isometric (i.e.~not equicontinuous). Then $\exists (y_j), (y'_j)$ in $Y$ with $d(y_j,y_j') \to 0$ and $\epsilon > 0, t_j \in T$ s.t.~$d(t_jy_j, t_jy_j') > \epsilon$. \item Wlog.~$y_j \to y$, $y'_j \to y$. Fix $\alpha$ s.t.~$M^{\frac{\epsilon}{4}}_\alpha = \emptyset$. \item $f_\alpha(y_j), f_{\alpha}(y'_j)$ converge $ z \in Y_\alpha$, equicontinuity of $Y_{\alpha} \implies d_{Y_\alpha}(t_jf_\alpha(y_j), t_jf_\alpha(y'_j)) \to 0$. Wlog.~$t_jf_\alpha(y_j^{(')})$ converge to same point. \item Consider $d(f^{-1}_\alpha(t_jf_\alpha(y_j)), f^{-1}_\alpha(t_jf_{\alpha}(y_{j}'))) \lessgtr \frac{\epsilon}{2}$ $\lightning$ \end{itemize} } \end{proof} More generally we can show: \begin{theorem}[{\cite[Prop.~13.1]{Furstenberg}}] Let $(X,T)$ be a distal flow and $(Z,T) = \pi(X,T)$ a factor. Then there exists an isometric extension $(Y,T)$ of $(Z,T)$ which is a factor of $(X,T)$, such that $(Y,T)$ is maximal among such extensions, i.e.~if $(Y',T)$ is any flow with these two properties, then $(Y',T)$ is a factor of $(Y,T)$. % https://q.uiver.app/#q=WzAsNCxbMCwwLCJcXHN1YnN0YWNreyhYLFQpXFxcXFxcdGV4dHtkaXN0YWx9fSJdLFswLDMsIihaLFQpIl0sWzEsMSwiKFksVCkiXSxbMiwyLCIoWScsVCkiXSxbMCwyXSxbMCwxLCJcXHBpIl0sWzIsMSwiXFx0ZXh0e21heC5+aXNvLn0iLDFdLFswLDMsIiIsMix7ImN1cnZlIjotM31dLFszLDEsIlxcdGV4dHtpc28ufSIsMV0sWzIsMywiIiwxLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV1d \[\begin{tikzcd} {\substack{(X,T)\\\text{distal}}} \\ & {(Y,T)} \\ && {(Y',T)} \\ {(Z,T)} \arrow[from=1-1, to=2-2] \arrow["\pi", from=1-1, to=4-1] \arrow["{\text{max.~iso.}}"{description}, from=2-2, to=4-1] \arrow[curve={height=-18pt}, from=1-1, to=3-3] \arrow["{\text{iso.}}"{description}, from=3-3, to=4-1] \arrow[dashed, from=2-2, to=3-3] \end{tikzcd} \] Such a factor $(Y,T)$ is called a \vocab{maximal isometric extension} of $(Z,T)$. \end{theorem} \begin{lemma} \label{lec19:lem1} Let four flows be given as in % https://q.uiver.app/#q=WzAsNCxbMSwwLCIoWSxUKSJdLFsyLDEsIihaXzIsIFQpIl0sWzAsMSwiKFpfMSwgVCkiXSxbMSwyLCIoVyxUKSJdLFsyLDMsIndfMSJdLFsxLDMsIndfMiIsMl0sWzAsMiwiXFxwaV8xIl0sWzAsMSwiXFxwaV8yIiwyXV0= \[\begin{tikzcd} & {(Y,T)} \\ {(Z_1, T)} && {(Z_2, T)} \\ & {(W,T)} \arrow["{w_1}", from=2-1, to=3-2] \arrow["{w_2}"', from=2-3, to=3-2] \arrow["{\pi_1}", from=1-2, to=2-1] \arrow["{\pi_2}"', from=1-2, to=2-3] \end{tikzcd}\] Suppose that whenever $y \neq y' \in Y$, then % either % TODO REALLY? $\pi_1(y) \neq \pi(y')$ or $\pi_2(y) \neq \pi_2(y')$. If $(Z_1,T)$ is an isometric extension of $(W,T)$, then $(Y,T)$ is an isometric extension of $(Z_2, T)$. \end{lemma} \begin{proof} \gist{% For $z_1,z_1' \in Z_1$ with $w_1(z_1) = w_1(z_1')$ let $\rho(z_1,z_1')$ be the metric on the fiber of $Z_1$ over $W$. Set $\sigma(y,y') \coloneqq \rho(\pi_1(y), \pi_1(y'))$ whenever $\pi_2(y) = \pi_2(y')$. In this case $w_2 \circ \pi_2(y) = w_2 \circ \pi_2(y')$ and $w_1 \circ \pi_1(y) = w_1 \circ \pi_1(y')$, so $\sigma$ is well defined. $\sigma$ is a semi-metric\footnote{Like a metric, but the distinct points can have distance $0$.} on the fibers of $Y$ over $Z_2$ and invariant under $T$. $\sigma$ is a metric on fibers, since if $\pi_2(y) = \pi_2(y')$ and $\sigma(y,y') = 0$, then $\pi_1(y) = \pi_1(y')$ or $y = y'$. }{% \begin{itemize} \item Let $\rho\colon Z_1 \times_W Z_1 \to \R$. \item Consider $\sigma\colon Y \times_{Z_2} Y \to \R$ given by $\sigma(y,y') \coloneqq \rho(\pi_1(y), \pi_1(y'))$. \end{itemize} } \end{proof} \begin{definition} A quasi-isometric system $\{(X_\xi, T) : \xi \le \eta\}$ is called \vocab{normal} if $(X_{\xi+1}, T)$ is the maximal isometric extension of $(X_\xi,T)$ in $(X_\eta, T)$ for all $\xi < \eta$. \end{definition} \begin{theorem}[{\cite[{}13.2]{Furstenberg}}] If $\{(X_\xi, T), \xi \le \eta\}$ is a normal quasi-isometric system, then $(X_\eta, T)$ has order $\eta$. \end{theorem} \begin{proof} We only sketch the proof here. Details can be found in \cite{Furstenberg}, section 13. Let $\{(X_\xi', T), \xi \le \eta'\} $ be another quasi-isometric system terminating with $(X_\eta, T) = (X'_{\eta'}, T)$. We want to show that $\eta' \ge \eta$. For this, we show that for all $\xi < \eta$, $(X_\xi', T)$ is a factor of $(X_\xi ,T)$ using transfinite induction. % https://q.uiver.app/#q=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 \[\begin{tikzcd} {X'_{\eta'}} & \dots & {X'_3} & {X'_2} & {X_1'} \\ X \\ {X_\eta} & \dots & {X_3} & {X_2} & {X_1} \arrow[Rightarrow, no head, from=1-1, to=2-1] \arrow[Rightarrow, no head, from=2-1, to=3-1] \arrow[from=3-3, to=3-4] \arrow[from=3-4, to=3-5] \arrow[from=1-4, to=1-5] \arrow[from=1-3, to=1-4] \arrow[dotted, from=3-3, to=1-3] \arrow[dotted, from=3-4, to=1-4] \arrow[dotted, from=3-5, to=1-5] \arrow["{\pi_3}"', curve={height=6pt}, from=3-1, to=3-3] \arrow["{\pi_2}"', curve={height=18pt}, from=3-1, to=3-4] \arrow["{\pi_1}"', curve={height=30pt}, from=3-1, to=3-5] \arrow["{\pi'_3}", curve={height=-6pt}, from=1-1, to=1-3] \arrow["{\pi'_2}", curve={height=-18pt}, from=1-1, to=1-4] \arrow["{\pi'_1}", curve={height=-30pt}, from=1-1, to=1-5] \end{tikzcd}\] We'll only show the successor step: Suppose we have $(X'_\xi, T) = \theta((X_\xi, T)$. Let $\pi_\xi$ and $\pi'_\xi$ denote the maps from $X$ to $X_\xi$ resp.~$X'_\xi$. Set \[Y \coloneqq \{(\pi_\xi(x), \pi'_{\xi+1}(x)) \in X_\xi \times X'_{\xi+ 1}: x \in X\}\] Then % https://q.uiver.app/#q=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 \begin{tikzcd} {(X_{\xi+1},T)} && {(Y,T)} \\ & {(X_\xi,T)} && {(X'_{\xi+1},T)} \\ && {(X'_\xi,T)} \arrow["{\text{max.~iso}}"{description}, from=1-1, to=2-2] \arrow["\theta"{description}, from=2-2, to=3-3] \arrow[""{name=0, anchor=center, inner sep=0}, "{{\color{orange}\text{iso}}}"{description}, from=2-4, to=3-3] \arrow[""{name=1, anchor=center, inner sep=0}, "{{\color{orange}\text{iso}}}"{description}, from=1-3, to=2-2] \arrow[from=1-3, to=2-4] \arrow[curve={height=-6pt}, dashed, from=1-1, to=1-3] \arrow["{\pi'}", draw=none, from=1-3, to=2-4] \arrow["{\theta'}", draw=none, from=0, to=3-3] \arrow["\pi"', draw=none, from=1-3, to=1] \end{tikzcd} The diagram commutes, since all maps are the induced maps. By definition of $Y$ is clear that $\pi$ and $\pi'$ separate points in $Y$. Thus \yaref{lec19:lem1} can be applied. Since $\theta'$ is an isometric extension, so is $\pi$. Then $(Y,T)$ is a factor of $(X_{\xi+1}, T)$ by the maximality of the isometric extension $(X_{\xi+1 }, T) \to (X_\xi, T)$. In particular, $(X'_{\xi+1}, T)$ is a factor of $(X_{\xi+1}, T)$. \end{proof} \begin{example}[{\cite[p. 513]{Furstenberg}}] \label{ex:19:inftorus} Let $X$ be the infinite torus \[ X \coloneqq \{(\xi_1, \xi_2, \ldots) : \xi_i \in \C, |\xi_i| = 1\}. \] Let $\pi_n$ be the projection to the first $n$ coordinates and $X_n \coloneqq \pi_n(X)$. Let $\tau_1(\xi_1,\xi_2, \ldots, \xi_n, \ldots) = (e^{\i \alpha} \xi_1, \xi_1\xi_2, \ldots, \xi_{n-1}\xi_n, \ldots)$ where $\frac{\alpha}{\pi}$ is irrational. Let $T = \langle \tau_1 \rangle \cong \Z$. We will show that $(X_n,T)$ is minimal for all $n$, and so $(X,T)$ is minimal. Furthermore $(X_{n+1},T)$ is the maximal isometric extension of $(X_n,T)$ so $(X,T)$ has order $\omega$. \end{example}