\lecture{16}{2023-12-08}{}


% \begin{definition}
%     % TODO
%     Isomorphism from $T \acts X$ to $T \acts Y$ :
%     Bijection $X \xrightarrow{b}  Y$
%     such that $b(tx) = t b(x)$.
% \end{definition}
$X$ is always compact metrizable.

\begin{theorem}
    Every minimal isometric flow $(X,\Z)$
    for $X$ a compact metrizable space%
    \footnote{Such a flow is uniquely determined by $h\colon X \to  X, x \mapsto 1\cdot x$.}
    is isomorphic to an abelian group rotation
    $(K, \Z)$, with
    $K$ an abelian compact group
    and $h(x) = x + \alpha$ for all $x \in K$
\end{theorem}
\begin{example}
    Let $\alpha \in S^1$
    and $h\colon x \mapsto x + \alpha$.\footnote{Again $x + \alpha$ denotes $x \cdot \alpha$ in $\C$.}
\end{example}
\begin{proof}
    The action of $1$ determines $h$
    and $n \in  \Z \leadsto h^n$.
    Consider
    \[
        \{h^n : n \in \Z\}  \subseteq  \cC(X,X) = \{f\colon  X \to X : f \text{continuous}\},
    \]
    where the topology is the uniform convergence topology.
    Let $G = \overline{ \{h^n : n \in \Z\} } \subseteq \cC(X,X)$.
    Since
    \[
    \forall \epsilon > 0.~\exists \delta > 0.~d(x,y) < \delta \implies \forall n.~d(h^n(x), h^n(y)) < \epsilon
    \]
    we have by the Arzela-Ascoli-Theorem % TODO REF
    that $G$ is compact.

    $G$ is a closure of a of a topological group,
    hence it is a topological group,
    i.e.~$g \mapsto g^{-1}$ and $(g,h) \mapsto gh$ are continuous.
    % TODO THINK ABOUT THIS
    Moreover since $\Z$ is abelian, 
    $\forall n,m \in \Z.~h^n \cdot h^m = h^m \cdot h^n$,
    so $G$ is abelian.
    % TODO THINK ABOUT THIS
    
    Take any $x \in X$ and consider
    the orbit % TODO DEFINITION
    $G \cdot x = \{f(x) : f \in G\}$.
    Since $\Z \acts X$ is minimal,
    i.e.~every orbit is dense,
    we have that $G \cdot x$ is dense in $X$.

    \begin{claim}
        $G \cdot x$ is compact.
    \end{claim}
    \begin{subproof}
        Since $\Z \acts X$ is continuous,
        $g \mapsto g x$ is continuous:

        Let $g_n$ be a sequence in $G$
        such that $g_n \to g$.
        Then $g_n x \to  gx$,
        since the topology on $\cC(X,X)$
        is the uniform convergence topology.


        Therefore the compactness of $G$ implies
        that the orbit  $Gx$ is compact.
    \end{subproof}

    Since $G\cdot x$ is compact and dense,
    we get $G \cdot x = X$.
    % TODO THINK ABOUT THIS

    Let $\Gamma = \{f \in G : f(x) = x\} <  G$ % TODO \triangleright?
    be the stabilizer group. % TODO DEFINITION
    Then $\Gamma \subseteq G$ is closed.
    Take $K \coloneqq \faktor{G}{\Gamma}$ with the quotient topology.

    $K$ is an abelian compact group
    and $G \to Gx$ gives
    a homeomorphism $K = \faktor{G}{\Gamma} \to Gx = X$.

    Conclusion:
    $\Z \acts K \equiv \Z \acts X$
    % and $h$ is a claimed.
    \todo{Copy from official notes}
    % TODO Definition transitive group action.
\end{proof}
\begin{definition}
    Let $(X,T)$ be a flow
    and  $(Y,T)$ a factor of  $(X,T)$.%
    \footnote{i.e. there exists a continuous surjection $\pi\colon X \twoheadrightarrow Y$
        commuting with the action, i.e.~$\forall t \in T. x \in X.~\pi(tx) = t \pi(x)$.
        Warning: Fürstenberg called factors subflows.
        % TODO: Definition
    }
    Suppose there is $\eta \in \Ord$
    such that for any $\xi < \eta$
    there is a factor $(X_\xi, T)$ of  $(X,T)$
    \begin{enumerate}[(a)]
        \item $(X_0, T) = (Y,T)$ and $(X_\eta, T) = (X,T)$.
        \item If  $\xi < \xi'$, then $(X_\xi, T)$ is a factor of  $(X_{\xi'}, T)$
            ``inside $(X,T)$'', i.e.~$\pi_\xi = \pi_{\xi, \xi'} \circ \pi_{\xi'}$.
        \item $\forall \xi <  \eta.~ (X_{\xi + 1}, T)$ is an isometric extension of $(X_\xi, T)$.
        \item $\xi \le \eta$ is a limit, then $(X_\xi, T)$
            is a limit of  $\{(X_\alpha,T), \alpha < \xi\}$.
    \end{enumerate}
    % https://q.uiver.app/#q=WzAsNCxbMCwwLCJYIl0sWzAsNSwiWSJdLFsxLDIsIlhfe1xceGknfSJdLFsxLDMsIlhfXFx4aSJdLFswLDEsIlxccGkiLDAseyJjdXJ2ZSI6NH1dLFswLDIsIlxccGlfe1xceGknfSIsMix7ImN1cnZlIjotMX1dLFswLDMsIlxccGlfe1xceGl9IiwyLHsiY3VydmUiOjJ9XSxbMiwzLCJcXHBpX3tcXHhpLCBcXHhpJ30iXV0=
\[\begin{tikzcd}
	X \\
	\\
	& {X_{\xi'}} \\
	& {X_\xi} \\
	\\
	Y
	\arrow["\pi", curve={height=24pt}, from=1-1, to=6-1]
	\arrow["{\pi_{\xi'}}"', curve={height=-6pt}, from=1-1, to=3-2]
	\arrow["{\pi_{\xi}}"', curve={height=12pt}, from=1-1, to=4-2]
	\arrow["{\pi_{\xi, \xi'}}", from=3-2, to=4-2]
\end{tikzcd}\]

    Then we say that $(X,T)$ is a \vocab{quasi-isometric extension}
    of $(Y,T)$.
\end{definition}
\begin{definition}
    If $(Y,T)$ is trivial, i.e.~$|Y| = 1$,
    then a quasi-isometric extension $(X,T)$ of $(Y,T)$
    is called a \vocab{quasi-isometric flow}.
\end{definition}

\begin{corollary}
    Every quasi-isometric flow is distal.
\end{corollary}
\begin{proof}
    \todo{TODO}
    % The trivial flow is distal.
\end{proof}

\begin{theorem}[Fürstenberg]
    Every minimal distal flow is quasi-isometric.
\end{theorem}
Therefore one can talk about ranks of distal minimal flows.
\begin{definition}
    Let $(X, \Z)$ be distal minimal.
    Then $\rank((X,\Z)) \coloneqq  \min \{\eta : (X, \Z) \cong (X_\eta, \Z)\}$
    where $(X_{\eta}, \Z)$ is as from the definition of quasi-isometric flows,
    i.e.~$\rank((X,\Z))$ is the minimal height such 
    that a tower as in the definition exists.
\end{definition}

\begin{theorem}[Beleznay-Foreman]
    Let $T = \Z$.
    \begin{itemize}
        \item For any $\alpha < \omega_1$,
            there is a distal minimal flow of rank $\alpha$.
        \item Distal flows form a $\Pi^1_1$-complete set:
            \todo{Move the explanations to a remark}
            For a fixed compact metric space $X$,
            view the flows $(X,\Z)$
            as a subset of $\cC(X,X)$.
            Note that $\cC(X,X)$ is Polish.
            Then the minimal flows on $X$ are a Borel subset of $\cC(X,X)$.
            But we want to look all flows at the same time.
            The Hilbert cube $[0,1]^{\N}$
            embeds all compact metric spaces.
            Thus we consider $K([0,1]^{\N})$,
            the space of compact subsets of $[0,1]^{\N}$.\todo{move definition}
            $K([0,1]^{\N})$ is a Polish space.

            Consider $K(([0,1]^\N)^2)$.
            A flow $\Z \acts X$ corresponds to the graph of
            \begin{IEEEeqnarray*}{rCl}
                X &\longrightarrow & X \\
                 1&\longmapsto & 1 \cdot x
            \end{IEEEeqnarray*}
            and this graph is an element of $K(([0,1]^{\N})^2)$.
        \item Moreover, this rank is a $\Pi^1_1$-rank.
    \end{itemize}
\end{theorem}
\fi
