w23-logic-3/inputs/lecture_19.tex
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tutorial 12
2024-01-16 23:14:56 +01:00

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\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=WzAsMTEsWzAsMSwiWCJdLFswLDIsIlhfXFxldGEiXSxbMCwwLCJYJ197XFxldGEnfSJdLFsxLDIsIlxcZG90cyJdLFsxLDAsIlxcZG90cyJdLFs0LDAsIlhfMSciXSxbNCwyLCJYXzEiXSxbMywwLCJYJ18yIl0sWzMsMiwiWF8yIl0sWzIsMCwiWCdfMyJdLFsyLDIsIlhfMyJdLFsyLDAsIiIsMix7ImxldmVsIjoyLCJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV0sWzAsMSwiIiwyLHsibGV2ZWwiOjIsInN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XSxbMTAsOF0sWzgsNl0sWzcsNV0sWzksN10sWzEwLDksIiIsMSx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRvdHRlZCJ9fX1dLFs4LDcsIiIsMSx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRvdHRlZCJ9fX1dLFs2LDUsIiIsMSx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRvdHRlZCJ9fX1dLFsxLDEwLCJcXHBpXzMiLDIseyJjdXJ2ZSI6MX1dLFsxLDgsIlxccGlfMiIsMix7ImN1cnZlIjozfV0sWzEsNiwiXFxwaV8xIiwyLHsiY3VydmUiOjV9XSxbMiw5LCJcXHBpJ18zIiwwLHsiY3VydmUiOi0xfV0sWzIsNywiXFxwaSdfMiIsMCx7ImN1cnZlIjotM31dLFsyLDUsIlxccGknXzEiLDAseyJjdXJ2ZSI6LTV9XV0=
\[\begin{tikzcd}
{X'_{\eta'}} & \dots & {X'_3} & {X'_2} & {X_1'} \\
X \\
{X_\eta} & \dots & {X_3} & {X_2} & {X_1}
\arrow[Rightarrow, no head, from=1-1, to=2-1]
\arrow[Rightarrow, no head, from=2-1, to=3-1]
\arrow[from=3-3, to=3-4]
\arrow[from=3-4, to=3-5]
\arrow[from=1-4, to=1-5]
\arrow[from=1-3, to=1-4]
\arrow[dotted, from=3-3, to=1-3]
\arrow[dotted, from=3-4, to=1-4]
\arrow[dotted, from=3-5, to=1-5]
\arrow["{\pi_3}"', curve={height=6pt}, from=3-1, to=3-3]
\arrow["{\pi_2}"', curve={height=18pt}, from=3-1, to=3-4]
\arrow["{\pi_1}"', curve={height=30pt}, from=3-1, to=3-5]
\arrow["{\pi'_3}", curve={height=-6pt}, from=1-1, to=1-3]
\arrow["{\pi'_2}", curve={height=-18pt}, from=1-1, to=1-4]
\arrow["{\pi'_1}", curve={height=-30pt}, from=1-1, to=1-5]
\end{tikzcd}\]
% 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
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\[\begin{tikzcd}
{(X_{\xi+1},T)} && {(Y,T)} \\
& {(X_\xi,T)} && {(X'_{\xi+1},T)} \\
&& {(X'_\xi,T)}
\arrow["{\text{max.~iso}}"{description}, from=1-1, to=2-2]
\arrow["\theta"{description}, from=2-2, to=3-3]
\arrow["{{\color{orange}\text{iso}}}"{description}, from=2-4, to=3-3]
\arrow["{{\color{orange}\text{iso}}}"{description}, from=1-3, to=2-2]
\arrow[from=1-3, to=2-4]
\arrow[curve={height=-6pt}, dashed, from=1-1, to=1-3]
\end{tikzcd}\]
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}