\lecture{20}{2024-01-09}{The Infinite Torus}

\gist{
\begin{example}
    \footnote{This is the same as \yaref{ex:19:inftorus},
        but with new notation.}
    Let $X = (S^1)^{\N}$\footnote{We identify $S^1$ and $\faktor{\R}{\Z}$.}
    and consider $\left(X, \Z \right)$
    where the action is generated by
    \[
        \tau\colon (x_1,x_2,x_3,\ldots) \mapsto(x_1 + \alpha, x_1 + x_2, x_2 + x_3, \ldots)
    \]
    for some irrational $\alpha$.
\end{example}
\begin{remark}+
    Note that we can identify $S^1$ with a subset of  $\C$ (and use multiplication)
    or with  $\faktor{\R}{\Z}$ (and use addition).
    In the lecture both notations were used. % to make things extra confusing.
    Here I'll try to only use multiplicative notation.
\end{remark}
}{}

We will be studying projections to the first $d$ coordinates,
i.e.
\[
    \tau_d \colon (x_1,\ldots,x_d) \mapsto (e^{\i \alpha} x_1, x_1x_2, \ldots, x_{d-1}x_d).
\]
$\tau_d$ is called the \vocab{$d$-skew shift}.
For $d = 1$ we get the circle rotation $x \mapsto e^{\i \alpha} x$.
\begin{fact}
    \label{fact:tau1minimal}
    The circle rotation $x \mapsto e^{\i \alpha} x$ is minimal.
    In fact, every subgroup of $S^1$ is either dense in $S^1$
    or it is of the form
    \[
    H_m \coloneqq  \{x \in S^1 : x^m = 0\}
    \]
    for some $m \in \Z$.%
    \footnote{cf.~\yaref{s12e2}}
\end{fact}
We will show that $\tau_d$ is minimal for all $d$,
i.e.~every orbit is dense.
From this it will follow that $\tau$ is minimal.

Let $\pi_n\colon X \to  (S^1)^n$ be the projection to the first $n$
coordinates.


\begin{lemma}
    \label{lem:lec20:1}
    Let $x,x' \in X$ with $\pi_n(x) = \pi_n(x')$
    for some $n$.
    Then there is a sequence of points $x_k$ with
    \[\pi_{n-1}(x_k) = \pi_{n-1}(x) = \pi_{n-1}(x')\]
    for all $k$
    and
    \[
    F(x_k, x) \xrightarrow{k \to \infty} 0,
    F(x_k, x') \xrightarrow{k \to  \infty} 0,
    \]
    where $F$ is as in \yaref{def:F},
    i.e.~$F(a,b) = \inf_{n \in \Z} d(\tau^n a, \tau^n b)$,
    where $d$ is the metric on $X$,
    $d((x_i), (y_i)) = \max_n \frac{1}{2^n} | x_n - y_n|$.% TODO use multiplicative notation
\end{lemma}
\begin{refproof}{lem:lec20:1}
    Let
    \begin{IEEEeqnarray*}{rCl}
        x &=& (\alpha^0_1, \alpha^0_2, \ldots, \alpha^0_{n-1}, \alpha_n, \alpha_{n+1}, \alpha_{n+2},\ldots)\\
        x' &=& (\alpha^0_1, \alpha^0_2, \ldots, \alpha^0_{n-1}, \alpha_n, \alpha'_{n+1}, \alpha'_{n+2},\ldots).\\
    \end{IEEEeqnarray*}
    We will choose $x_k$ of the form
    \[
        (\alpha^0_1, \alpha^0_2, \ldots, \alpha^0_{n-1}, \alpha_n e^{\i \beta_k}, \alpha_{n+1}, \alpha_{n+2}, \ldots),
    \]
    where $\beta_k$ is such that $\frac{\beta_k}{\pi}$ is irrational
    and $|\beta_k| < 2^{-k}$.
    Fix a sequence'(b)). of such $\beta_k$.
    Then
    \[d(x_k,x) = 2^{-n} |e^{\i \beta_k} - 1| < 2^{-n-k} \xrightarrow{k\to \infty} 0.\]
    In particular $F(x_k, x) \to 0$.

    We want to show that $F(x_k, x') < 2^{-n-k}$.
    For $u, u' \in X$,
    $u = (\xi_n)_{n \in \N}$,
    $u' = (\xi'_n)_{n \in \N}$,
    let $\frac{u}{u'} = (\frac{\xi_n}{\xi'_n})_{n \in \N}$
    ($X$ is a group).
    We are interested in $F(x_k, x') = \inf_m d(\tau^m x_k, \tau^m x')$,
    but it is easier to consider the distance between
    their quotient and $1$.
    Consider
    \[
        w_k \coloneqq \frac{x_k}{x'} = (\underbrace{1,\ldots,1}_{n-1}, e^{\i \beta_k}, \overbrace{\frac{\alpha_{n+1}}{\alpha'_{n+1}}, \frac{\alpha_{n+2}}{\alpha'_{n+2}}, \ldots}^{\mathclap{\text{not interesting}}}).
    \]
\gist{%
    \begin{claim}
        $F(x_k, x') = \inf_m d(\sigma^m(w_k), 1)$,
        where  $\sigma(\xi_1, \xi_2, \ldots) = (\xi_1, \xi_1\xi_2, \xi_2\xi_3, \ldots)$.
    \end{claim}
    \begin{subproof}
        We have
        \begin{IEEEeqnarray*}{rCl}
            F(u,u') &=& \inf_m d(\tau^m u, \tau^m u')\\
                    &=& \inf_m d(\frac{\tau^m u}{\tau^m u'}, 1)\\
                    &=& \inf_m d(\sigma^m\left( \frac{u}{u'} \right), 1).
        \end{IEEEeqnarray*}
    \end{subproof}

    Fix $k$. Let $w^\ast = (1,\ldots,1, e^{\i \beta_k}, 1, \ldots)$.
    By minimality of $(X,T)$ for any $\epsilon >0$,
    there exists $m \in \Z$ such that 
    $d(\sigma^m w_k, w^\ast) < \epsilon$.
    % TODO Think about this
    Then
    \begin{IEEEeqnarray*}{rCl}
        \inf_m d(\sigma^m w_k, 1) &\le & \inf_m d(\sigma^m w_k, w^\ast) + d(w^\ast, 1)\\
                                  &\le & 2^{-n} | e^{\i \beta_k}- 1|\\
                                  &<& 2^{-n-k}.
    \end{IEEEeqnarray*}
}{[some technical details omitted]}
\end{refproof}


\begin{definition}
    For every continuous  $f\colon S^1 \to  S^1$, the
    \vocab{winding number} $[f] \in \Z$
    is the unique integer such that $f$ is homotopic%
    \footnote{$f\colon Y \to  Z$ and $g\colon  Y \to  Z$ are homotopic
        iff there is $H\colon Y \times [0,1] \to  \Z$
        continuous such that $H(\cdot ,0) = f$ and $H(\cdot ,1) = g$.}
    to the map
    $x \mapsto x^{n}$.
\end{definition}

\gist{%
\begin{remark}
    \label{rem:l20:sigma}
    Note that for
    \begin{IEEEeqnarray*}{rCl}
        \sigma\colon (S^1)^d &\longrightarrow & S^1 \\
        (x_1,\ldots,x_d) &\longmapsto & x_d
    \end{IEEEeqnarray*}
    we have that $T = \tau_{d+1}$,
    where
    \begin{IEEEeqnarray*}{rCl}
        T\colon (S^1)^d \times S^1 &\longrightarrow & (S^1)^d \times S^1 \\
        (y, x_{d+1}) &\longmapsto & (\tau_d(y), \sigma(y) x_{d+1}).
    \end{IEEEeqnarray*}
\end{remark}
}{}
\begin{theorem}
    \label{thm:taudminimal:help}
    For every $d$ if $\tau_d$\footnote{more formally $((S^1)^d, \langle \tau_d \rangle)$}
    is minimal, then $\tau_{d+1}$ is minimal.
\end{theorem}
\begin{corollary}
    $\tau_d$ is minimal for all $d$.
\end{corollary}
\begin{proof}
    $\tau_1$ is minimal (\yaref{fact:tau1minimal}).
    Apply \yaref{thm:taudminimal:help}.
\end{proof}
\begin{corollary}
    Since all the $\tau_d$ are minimal,
    $\tau$ is minimal.
\end{corollary}
\begin{proof}
    \gist{%
    We need to show that every orbit is dense.
    This follows from the definition of the product topology,
    since for a basic open set $U = U_1 \times \ldots \times U_d \times (S^1)^{\infty}$
    it suffices to analyze the first $d$ coordinates.
    }{We need to show that every orbit is dense. For this it suffices to analyze finitely many coordinates.}
\end{proof}
\gist{%
\begin{refproof}{thm:taudminimal:help}
    Let $S \coloneqq  \tau_d$, $T \coloneqq \tau_{d+1}$ and $Y \coloneqq (S^1)^d$.
    Consider
    \begin{IEEEeqnarray*}{rCl}
        \gamma\colon S^1 &\longrightarrow & Y \\
        x &\longmapsto & (x,x,\ldots,x).
    \end{IEEEeqnarray*}
    Note that
    \begin{enumerate}[(a)]
        \item $\gamma$ and $S \circ \gamma$ are homotopic
            via
            \begin{IEEEeqnarray*}{rCl}
                H\colon S^1 \times [0,1] &\longrightarrow & (S^1)^d \\
            (x, t)&\longmapsto & (x e^{\i t  \alpha}, x^{t+1}, x^{t+1}, x^{t+1},\ldots, x^{t+1})
            \end{IEEEeqnarray*}
        \item For all $m \in  \Z \setminus \{0\}$, we have
            $[x \mapsto \left(\sigma(\gamma(x))\right)^m] = m \neq 0$,
            since $\sigma(\gamma(x)) = \sigma((x,\ldots,x)) = x$.
    \end{enumerate}

    \phantom\qedhere
\end{refproof}
}{}