w23-logic-3/inputs/lecture_19.tex

\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}


\begin{question}
    What is the minimal number of steps required
    when building the tower to reach the flow
    as in \yaref{thm:l16:3}?
\end{question}
\begin{definition}[{\cite[{}13.1]{Furstenberg}}]
    Let $(X,T)$ be a quasi isometric flow,
    and let $\eta$ be the smallest ordinal
    such that there exists a quasi-isometric system $\{(X_\xi, T), \xi \le \eta\}$
    with $(X,T) = (X_\xi, T)$.
    Then $\eta$ is called the \vocab{order}  of the flow.
\end{definition}
\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=WzAsMyxbMCwwLCIoWCxUKSJdLFsxLDEsIihZLFQpXFxcXFxcdGV4dHttYXhpbWFsIGlzb21ldHJpY30iXSxbMCwyLCIoWScsVCkiXSxbMCwyLCJcXHRleHR7aXNvbWV0cmljfSIsMl0sWzAsMV0sWzEsMiwiXFxleGlzdHMiLDAseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XV0=
\[\begin{tikzcd}
	{(X,T)} \\
	& {\substack{(Y,T)\\\text{maximal isometric}}} \\
	{(Y',T)}
	\arrow["{\text{isometric}}"', from=1-1, to=3-1]
	\arrow[from=1-1, to=2-2]
	\arrow["\exists", dashed, from=2-2, to=3-1]
\end{tikzcd}\]
\end{theorem}
\begin{proof}
    % TODO Think about this
    We want to apply Zorn's lemma.
    If suffices to show that isometric flows are closed under inverse limits,
    i.e.~if $(Y_\alpha, f_{\alpha,\beta})$,
    $\beta < \alpha \le  \Theta$
    are isometric, then the inverse limit $Y$ is isometric.%
    \todo{Why does an inverse limit exist?}
    % https://q.uiver.app/#q=WzAsNCxbMSwwLCJZX1xcYWxwaGEiXSxbMSwxLCJZX1xcYmV0YSJdLFswLDAsIlkiXSxbMiwwLCJYIl0sWzAsMSwiZl97XFxhbHBoYSwgXFxiZXRhfSJdLFsyLDAsImZfXFxhbHBoYSJdLFsyLDEsImZfXFxiZXRhIiwyXSxbMywwLCJcXHBpX1xcYWxwaGEiLDJdLFszLDEsIlxccGlfXFxiZXRhIl1d
\[\begin{tikzcd}
	Y & {Y_\alpha} & X \\
	& {Y_\beta}
	\arrow["{f_{\alpha, \beta}}", from=1-2, to=2-2]
	\arrow["{f_\alpha}", from=1-1, to=1-2]
	\arrow["{f_\beta}"', from=1-1, to=2-2]
	\arrow["{\pi_\alpha}"', from=1-3, to=1-2]
	\arrow["{\pi_\beta}", from=1-3, to=2-2]
\end{tikzcd}\]
    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
    \[
    \bigcap_{\alpha \le \theta}\Delta_\alpha = \{(y,y) : y \in Y\}.
    \]
    Consider
    \[\{(y,y') \in \Delta_\alpha : d(y,y') \ge  \epsilon\}\]
    for any $\epsilon > 0$.
    By the finite intersection property % TODO WHY? TODO what is this TODO for compact?
    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
    \[
    d(f^{-1}_\alpha(t_jz_j), f^{-1}_\alpha(t_jz'_j)) < \frac{\epsilon}{2} \lightning.
    \]
\end{proof}

More generally we can show:
\begin{theorem}[{\cite[13.1]{Furstenberg}}]
    Let $(X,T)$ be a distal flow
    and $(Y,T) = \pi(X,T)$ 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=WzAsMyxbMCwwLCIoWCxUKSJdLFswLDIsIihaLFQpIl0sWzEsMSwiKFksVCkiXSxbMCwyXSxbMCwxLCJcXHBpIl0sWzIsMV1d
\[\begin{tikzcd}
	{(X,T)} \\
	& {(Y,T)} \\
	{(Z,T)}
	\arrow[from=1-1, to=2-2]
	\arrow["\pi", from=1-1, to=3-1]
	\arrow[from=2-2, to=3-1]
\end{tikzcd}\]
\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}
    % TODO TODO TODO Think about this
    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, since if
    if $\pi_2(y) = \pi_2(y')$ and $\sigma(y,y') = 0$,
    then $\pi_1(y) = \pi_1(y')$ or $y = y'$.
\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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
\[\begin{tikzcd}
	{X'_{\eta'}} & \dots & {X'_3} & {X'_2} & {X_1'} \\
	X \\
	{X_\eta} & \dots & {X_3} & {X_2} & {X_1}
	\arrow[Rightarrow, no head, from=1-1, to=2-1]
	\arrow[Rightarrow, no head, from=2-1, to=3-1]
	\arrow[from=3-3, to=3-4]
	\arrow[from=3-4, to=3-5]
	\arrow[from=1-4, to=1-5]
	\arrow[from=1-3, to=1-4]
	\arrow[dotted, from=3-3, to=1-3]
	\arrow[dotted, from=3-4, to=1-4]
	\arrow[dotted, from=3-5, to=1-5]
	\arrow["{\pi_3}"', curve={height=6pt}, from=3-1, to=3-3]
	\arrow["{\pi_2}"', curve={height=18pt}, from=3-1, to=3-4]
	\arrow["{\pi_1}"', curve={height=30pt}, from=3-1, to=3-5]
	\arrow["{\pi'_3}", curve={height=-6pt}, from=1-1, to=1-3]
	\arrow["{\pi'_2}", curve={height=-18pt}, from=1-1, to=1-4]
	\arrow["{\pi'_1}", curve={height=-30pt}, from=1-1, to=1-5]
\end{tikzcd}\]
    % TODO: induction start?

    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=WzAsNSxbMCwwLCIoWF97XFx4aSsxfSxUKSJdLFsyLDAsIihZLFQpIl0sWzMsMSwiKFgnX3tcXHhpKzF9LFQpIl0sWzIsMiwiKFgnX1xceGksVCkiXSxbMSwxLCIoWF9cXHhpLFQpIl0sWzAsNCwiXFx0ZXh0e21heC5+aXNvfSIsMV0sWzQsMywiXFx0aGV0YSIsMV0sWzIsMywie1xcY29sb3J7b3JhbmdlfVxcdGV4dHtpc299fSIsMV0sWzEsNCwie1xcY29sb3J7b3JhbmdlfVxcdGV4dHtpc299fSIsMV0sWzEsMl0sWzAsMSwiIiwwLHsiY3VydmUiOi0xLCJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XV0=
\[\begin{tikzcd}
	{(X_{\xi+1},T)} && {(Y,T)} \\
	& {(X_\xi,T)} && {(X'_{\xi+1},T)} \\
	&& {(X'_\xi,T)}
	\arrow["{\text{max.~iso}}"{description}, from=1-1, to=2-2]
	\arrow["\theta"{description}, from=2-2, to=3-3]
	\arrow["{{\color{orange}\text{iso}}}"{description}, from=2-4, to=3-3]
	\arrow["{{\color{orange}\text{iso}}}"{description}, from=1-3, to=2-2]
	\arrow[from=1-3, to=2-4]
	\arrow[curve={height=-6pt}, dashed, from=1-1, to=1-3]
\end{tikzcd}\]

    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}