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

\begin{refproof}{thm:taudminimal:help}
\gist{%
    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 $(\theta \cdot  Z)\cap Z = \emptyset$.%
    \footnote{actually $(1,\ldots,1, \theta) \cdot Z$, we identify $S^1$ and $\{0\}^d \times S^1 \subseteq Y \times S^1$.}

    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 $Z = Y \times S^1$, so this cannot be the case),
    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
    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)})^m
    \end{IEEEeqnarray*}
    Z is isomorphic to $m$ copies of the graph of that function, hence
    the graph is closed, so the function is continuous.

    Note that $f(Z)$ is homeomorphic to $Y$
    (for every $y \in Y$, $\phi(y)$ is the unique element such that $(y,\phi(y)) \in f(Z)$).

    \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) $\lightning$
}{
    \begin{itemize}
        \item $S \coloneqq  \tau_d$, $T \coloneqq  \tau_{d+1}$
            ($T(y, \xi) = (S(y), \sigma(y) + \xi)$),
            $Y \coloneqq  (S^1)^d$,
            $\gamma: S^1 \to Y, x \mapsto (x,x,\ldots,x)$.
        \item (a)  $\gamma \simeq S \circ \gamma$.
        \item (b) $\forall m \in \Z.~ [m \cdot \sigma \circ \gamma] = m$
            ($\sigma(\gamma(x)) = x$).
        \item Suppose $Y \times S^1$ is not minimal.
            Let $Z$ be a proper minimal subflow,
            $\pi\colon Y \times S^1 \to Y$.
        \item $\pi(Z) = Y$ ($Y$ minimal)
        \item $Z + (0,\ldots,0,\theta)$ is minimal, so $\theta + Z = Z$ 
            or $\theta + Z \cap Z = \emptyset$.
        \item $H \coloneqq  \{\theta \in S^1 : \theta + Z = Z\}$ (closed subgroup of $S^1$).
        \begin{itemize}
            \item $H \neq S^1$ as $Z$ is a proper subflow.
            \item $H = \langle \frac{1}{m} \rangle$.
        \end{itemize}
        \item $(y, \beta) \in Z$, then $(y, \beta + t) \in Z \iff m \cdot t = 0$.
            Pick $\beta^{(y)}$ such that $(y, \xi) \in Z \iff \xi \in \{\beta^{(y)} + \frac{n}{m} : n \in \N\}$.
        \item Consider $(y, \xi) \mapsto (y, m \cdot \xi)$.
            Get
            \begin{IEEEeqnarray*}{rCl}
                \phi\colon Y &\longrightarrow & S^1 \\
                 y &\longmapsto & m \beta^{(y)}
            \end{IEEEeqnarray*}
            continuous.
        \item $\phi \circ S = \phi + m \sigma$:
            $Z$ is $T$-invariant.
        \item $[\phi \circ S \circ \gamma] = [\phi \circ \gamma] + [m \sigma \circ \gamma]$
            $\implies [m \sigma \circ \gamma] = 0 \lightning$.
    \end{itemize}
}
\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}