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