w23-logic-3/inputs/lecture_21.tex

\lecture{21}{2024-01-12}{Iterated Skew Shift}

\begin{refproof}{thm:taudminimal:help}
    Suppose towards a contradiction that
    $Y \times S^1$ contains a proper minimal subflow $Z$.
    Consider the projection $\pi\colon Y \times S^1 \to Y$.
    By minimality of $Y$, we have $\pi(Z) = Y$.
    Note that for every $\theta \in S^1$, $\theta \cdot  Z$ is minimal,
    so either $\theta \cdot  Z = Z$ or $(\theta \cdot  Z)\cap Z = \emptyset$.

    Let $H = \{\theta \in S^1 : \theta \cdot  Z = Z\}$.
    $H$ is a closed subgroup of $S^1$.
    % H is a rotation of Z containing 1 (?)
    Therefore either $H = S^1$ (but in that case $Z = Y \times S^1$),
    or there exists $m \in \Z$ such that $H = \{ \xi \in S^1 : \xi^m = 1 \}$
    by \yaref{fact:tau1minimal}.

    Note that if $(y, \beta) \in Z$ then for $t \in S^1$,
    we have
    \[
        (y, \beta \cdot  t) \in  Z \iff t^m = 1.
    \]
    Therefore for every $y \in Y$, there are exactly $m$ many
    $\xi \in S^1$
    such that $(y, \xi) \in Z$.

    Specifically for all $y$ there exists $\beta^{(y)} \in S^1$
    such that $(y,\xi) \in Z$ iff
    \[
        \xi \in  \{\beta^{(y)} \cdot  t_1, \beta^{(y)} \cdot  t_2, \ldots,\beta^{(y)} \cdot  t_m\},
    \]
    where the $t_i \in S^1$
    are such that
    $t_i^m = 1$ for all $i$ and $i \neq j \implies t_i \neq t_j$,
    i.e.~the $t_i$ are the $m$\textsuperscript{th} roots of unity.

    Consider $f \colon  (y,\xi) \mapsto (y, \xi^m)$.
    Since $(\beta^{(y)} \cdot  t_i)^m = (\beta^{(y)})^m$
    we get a continuous\todo{Why is this continuous?}
    function $\phi\colon Y \to S^1$
    such that
    \[
    Z = \{(y,\xi) \in Y \times S^1 : \xi^m = \phi(y)\}.
    \]
    % namely
    % \begin{IEEEeqnarray*}{rCl}
    %     \phi\colon Y &\longrightarrow & S^1 \\
    %     y &\longmapsto & \beta^{(y)}.
    % \end{IEEEeqnarray*}

    Note that $f(Z)$ is homeomorphic to $Y$.\todo{Why?}

    \begin{claim}
        $\phi(S(y)) = \phi(y) \cdot  (\sigma(y))^m$.
    \end{claim}
    \begin{subproof}
        We have $T(y, \xi) = (S(y), \sigma(y) \cdot \xi)$
        (cf.~\yaref{rem:l20:sigma}).
        $Z$ is invariant under $T$.
        So for $(y, \xi) \in Z$ we get $T(y, \xi) = ({\color{red}S(y)}, {\color{blue}\sigma(y) \cdot  \xi}) \in Z$.
        Thus
        \begin{IEEEeqnarray*}{rCl}
            \phi({\color{red}S(y)}) &=& ({\color{blue}\sigma(y) \cdot \xi})^m\\
                                    &=& (\sigma(y))^m \cdot \xi^m\\
                                    &=& (\sigma(y))^m \cdot  \phi(y).
        \end{IEEEeqnarray*}
    \end{subproof}
    Applying $\gamma$ we obtain
    \[
        [\phi \circ S \circ \gamma] = [\phi \circ \gamma] + [x \mapsto (\sigma(\gamma(x))^n].
    \]
    $S\circ \gamma$ is homotopic to $\gamma$,
    so $[\phi \circ S \circ \gamma] = [\phi \circ \gamma]$.
    Thus $[x \mapsto (\sigma(\gamma(x))^n] = 0$,
    but that is a contradiction to (b) $\lightning$
\end{refproof}

Let $X_n \coloneqq  (S^1)^n$ and $X \coloneqq (S^1)^{\N}$.
\begin{theorem}
    \label{thm:21:xnmaxiso}
    $(X_n, \tau_n)$ is the maximal isometric extension of  $(X_{n-1}, \tau_{n-1})$
    in $(X,\tau)$.
\end{theorem}
\begin{corollary}
    The order of $(X,\tau)$ is $\omega$.
\end{corollary}