\lecture{22}{2024-01-16}{}

\begin{refproof}{thm:21:xnmaxiso}
    We have the following situation:
    % https://q.uiver.app/#q=WzAsNCxbMCwwLCJYIl0sWzEsMSwiWF97bn0iXSxbMSwyLCJYX3tuLTF9Il0sWzIsMSwiWSJdLFswLDEsIlxccGlfe259Il0sWzEsMiwiXFx0ZXh0e2lzb21ldHJpY30iLDFdLFswLDIsIlxccGlfe24tMX0iLDIseyJjdXJ2ZSI6Mn1dLFswLDMsIlxccGknIiwwLHsiY3VydmUiOi0zfV0sWzMsMSwiaCIsMix7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFszLDIsIlxcb3ZlcmxpbmV7Z30sIFxcdGV4dHsgbWF4LiBpc29tLn0iLDAseyJjdXJ2ZSI6LTJ9XV0=
    \[\begin{tikzcd}
    	X \\
    	& {X_{n}} & Y \\
    	& {X_{n-1}}
    	\arrow["{\pi_{n}}", from=1-1, to=2-2]
    	\arrow["{\text{isometric}}"{description}, from=2-2, to=3-2]
    	\arrow["{\pi_{n-1}}"', curve={height=12pt}, from=1-1, to=3-2]
    	\arrow["{\pi'}", curve={height=-18pt}, from=1-1, to=2-3]
    	\arrow["h"', dashed, from=2-3, to=2-2]
    	\arrow["{\overline{g}, \text{ max. isom.}}", curve={height=-12pt}, from=2-3, to=3-2]
    \end{tikzcd}\]

    We want to show that this tower is normal,
    i.e.~the isometric extensions are maximal isometric extension.
\gist{%
    Let $Y$ be a maximal isometric extension of $X_{n-1}$ in $X$
    and let $\overline{g} = \pi^n_{n-1} \circ h$. % factor map?
    We need to show that $h$ is an isomorphism.
    Towards a contradiction assume that $h$ is not an isomorphism.
    Then there are $x,x' \in X$ with
    $\pi'(x) \neq  \pi'(x')$ but $\pi_n(x) = \pi_n(x') =t \in X_n$.
    Then $h^{-1}(t) \ni \pi'(x), \pi'(x')$.

    By a \yaref{lem:lec20:1}
    there is a sequence $(x_k)$ in $X$
    with $\pi_{n-1}(x_k) = \pi_{n-1}(x) = \pi_{n-1}(x')$ for all $k$,
    such that $F(x_k, x) \to 0$ and $F(x_k, x') \to 0$.

    Let $\rho$ be a metric witnessing that $\overline{g}$
    is an isometric extension,
    i.e.~
    $\rho$ is defined on $\bigcup_{x \in X_{n-1}} (\overline{g}^{-1}(x))^2 \overset{\text{closed}}{\subseteq} Y \times Y$,
    continuous and $\rho(Ta, Tb) = \rho(a,b)$ for $\overline{g}(a) = \overline{g}(b)$.

    For $a,b \in X$ such that
    \[
    \overline{g}(\pi'(a)) = \overline{g}(\pi'(b))
    \]
    define
    \[
    R(a,b) \coloneqq  \rho(\pi'(a), \pi'(b)).
    \]

    \begin{itemize}
        \item For any two out of $x,x',(x_k)$, $R$ is defined.
        \item $R(x,x_k) = R(\tau^m x, \tau^m x_k)$ for all  $m$.
        \item $F(x,x_k) \xrightarrow{k\to \infty} 0$,
            so there is a sequence $(m_k)$
            such that
            \[d(\tau^{m_k}x, \tau^{m_k} x_k) \xrightarrow{k \to \infty} 0.\]
    \end{itemize}
    By continuity of $\rho$,
    we have that  $R(x,x_k) = R(\tau^{m_k} x, \tau^{m_k} x_k) \xrightarrow{k \to  \infty} 0$,
    and similarly $R(x',x_k) \to 0$.
    Hence $R(x,x') \xrightarrow{k \to \infty} 0$
    by the triangle inequality.
    But $x$ and $x'$ don't depend on $k$,
    hence $R(x,x') = 0$.
    It follows that $\pi'(x) = \pi'(x')$ $\lightning$.
}{
    \begin{itemize}
        \item $Y$ max.~isometric extension of $X_{n-1}$ in $X$
            and $\overline{g} = \pi^n_{n-1} \circ h$.
        \item $h$ isomorphism.
            Suppose not, then $\exists  y_0,y_1 \in X.~\pi'(y_0) \neq  \pi'(y_1),
            \pi_n(y_0) = \pi_n(y_1) = t$.

        \item Apply \yaref{lem:lec20:1} $\leadsto$ sequence $(x_k)$ in  $X$,
            such that $\pi_{n-1}(x_k) = \pi_{n-1}(y_i)$,
            $F(x_k,y_i) \to 0$.
        \item $\rho\colon  Y \times_{X_{n-1}} Y \to \R$ witnessing isometric.
        \item $R(a,b) \coloneqq  \rho(\pi'(a), \pi'(b))$ for $a,b \in X$ with
            $\overline{g}(\pi'(a)) = \overline{g}(\pi'(b))$.
            (defined for any two of $x_k$, $y_0$, $y_1$, $\tau$-equivariant)
        \item $F(y_0,x_{k}) \to 0$, so $d(\tau^{m_k} y_0, \tau^{m_k} x_k) \to 0$.
        \item $R(y_0,x_k) \to 0$, hence $\underbrace{R(y_0,y_1)}_{\text{no } k} \to 0$ $\lightning$.
    \end{itemize}
}
\end{refproof}


\begin{theorem}[Beleznay-Foreman]
    \begin{enumerate}[(1)]
        \item For every $\eta < \omega_1$,
            there is a distal minimal flow
            of order $\eta$.%\footnote{For second countable spaces this is the best we can get.}
        \item The set of distal minimal flows
            is $\Pi^1_1$-complete.
        \item The order is a $\Pi^1_1$-rank.
            In particular
            $\{\text{distal minimal flows of rank } < \alpha\}$
            is Borel for all $\alpha < \omega_1$.
    \end{enumerate}
\end{theorem}
\todo{This was already stated as \yaref{thm:beleznay-foreman} in lecture 16
    and should not have two numbers.}

A few words on the proof:
Let $\mathbb{K} = S^1$
and $I$ a countable linear order.
Let $\mathbb{K}^I$ be the product of $|I|$ many $\mathbb{K}$,
$\mathbb{K}^{<i} \coloneqq \mathbb{K}^{\{j : j < i\}}$
and $\pi_{i}\colon \mathbb{K}^{I} \to  \mathbb{K}^{<i}$% \todo{maybe call it $\pi_{<i}$?}
the projection.

Let $\mathbb{K}_I \coloneqq  \prod_{i \in I} C(\mathbb{K}^{<i}, \mathbb{K})$.

Fix some $(f_i)_{i \in I} \in \mathbb{K_I}$.
We build a flow acting on $\mathbb{K}$ from $(f_i)_{i \in I}$.

For this we define
\begin{IEEEeqnarray*}{rCl}
    E_I \colon  \mathbb{K}_I&\longrightarrow &  C(\mathbb{K}^I, \mathbb{K}^I)\\
    (f_i)_{i \in I}&\longmapsto & \begin{pmatrix*}[l]
        \mathbb{K}^I&\longrightarrow & \mathbb{K}^I \\
        x &\longmapsto & (f_i(\pi_i(x)) \cdot  x_i)_{i \in I}
    \end{pmatrix*}
\end{IEEEeqnarray*}

\gist{%
\begin{example}
    Consider the following flow:
    \begin{IEEEeqnarray*}{rCl}
        \tau\colon \mathbb{K}^3 &\longrightarrow & \mathbb{K}^3 \\
        (x,y,z)&\longmapsto & (x \cdot \alpha, x^2y, xy^3z).
    \end{IEEEeqnarray*}
    Using
    \begin{IEEEeqnarray*}{rCl}
        f_1\colon \mathbb{K}^0 &\longrightarrow & \mathbb{K} \\
        x &\longmapsto & \alpha,\\\\
        f_2\colon \mathbb{K}^1 &\longrightarrow & \mathbb{K}\\
        x &\longmapsto & x^2,\\ \\
        f_3\colon \mathbb{K}^2 &\longrightarrow & \mathbb{K}\\
        (x,y) &\longmapsto & xy^3.
    \end{IEEEeqnarray*}
    we can write this as $\tau(x,y,z) = (x \cdot  f_1 \circ \pi_1(x,y,z), y \cdot f_2\pi_2(x,y,z), z \cdot f_3\pi_3(x,y,z))$
\end{example}
\begin{example}
    The skew shift can be written in this form as well.
    Consider $f_1\colon x \mapsto \alpha$
    and $f_n\colon (x_0,\ldots, x_{n-2}) \mapsto x_{n-2}$.
\end{example}
}{}

\begin{theorem}[Beleznay Foreman]
    \label{thm:distalminimalofallranks}
    Whenever $I = \eta$ for some $\eta  < \omega_1$,
    then
    \[
    \{\overline{f} \in \mathbb{K}_I : E_I(\overline{f}) \text{ is distal, minimal and of rank $\eta$}\} % TODO rank = order
    \]
    is comeager in $\mathbb{K}_I$.
    In particular such flows exist.
\end{theorem}
\begin{proof}[sketch]
    \leavevmode
    \begin{itemize}
        \item Distality:
            For all $\overline{f} \in \mathbb{K}_I$,
            the flow $E_I \overline{f}$ is distal.
            This is the same as for iterated skew shifts.

        \item Minimality:
            \notexaminable{%
                Let $\langle E_n : n < \omega \rangle$
                be an enumeration of a countable basis for $\mathbb{K}^I$.

                For all $n$ let
                \[
                U_n \coloneqq  \{\overline{f} \in \mathbb{K}_I : \exists k \in \Z.~ f^k(\overline{1}) \in E_n\}
                \]
                where $f = E_I \overline{f}$ and $\overline{1} = (1,1,1,\ldots)$.

                \textsc{Beleznay} and \textsc{Foreman} showed that $U_n$ is open
                and dense for all $n$.

                So if $\overline{f} \in  \bigcap_{n} U_n$, then $\overline{1}$
                is dense in $\overline{x} \mapsto f(\overline{x})$.
                Since the flow is distal, it suffices to show
                that one orbit is dense (cf.~\yaref{thm:distalflowpartition}).
            }

        \item The order of the flow is $\eta$:
            \notexaminable{%
                Let $\overline{f} = (f_i)_{i \in I} \in \mathbb{K}_I$.
                Consider the flows we get from $(f_i)_{i < j}$
                resp.~$(f_i)_{i \le j}$
                denoted by $X_{<j}$ resp.~$X_{\le j}$.
                We aim to show that $X_{\le j} \to X_{<j}$
                is a maximal isometric extension for comeagerly many $\overline{f}$.

                The following open dense sets are used to make sure that all isometric extensions
                are maximal and hence the order of the flow is $\eta$:

                Fix a countable dense set $(\overline{x_n})$ in $\mathbb{K}^I$.
                For $\epsilon \in \Q$ let
                \begin{IEEEeqnarray*}{rCl}
                    V_{j,m,n,\epsilon} &\coloneqq  \{\overline{f} \in \mathbb{K}_I :& \\
                    &&\text{if } \Pi_{j+1}(\overline{x}_n) = \Pi_{j+1}(\overline{x_m}),\\
                    &&\text{then there are $k_m$, $k_n$, $\overline{z}$ such that}\\
                    &&\pi_j(\overline{x_n}) = \pi_j(\overline{z}), \forall k> j+1.~z_k = 1,\\
                    &&d(f^{k_m}(\overline{x_m}), f^{k_m}(\overline{z})) < \epsilon \text{ and }\\
                    &&d(f^{k_n}(\overline{x_n}), f^{k_n}(\overline{z})) < \epsilon\\
                    &&\}
                \end{IEEEeqnarray*}
                Beleznay and Foreman show that this is open and dense.%
                % TODO similarities to the lemma used today
            }
    \end{itemize}
\end{proof}