Josia Pietsch 458dd9ab1f
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\lecture{20}{2024-01-09}{The Infinite Torus}
\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 $\alpha$.
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.
We will be studying projections to the first $d$ coordinates,
\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 $x \mapsto e^{\i \alpha} x$.
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 $m \in \Z$.%
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 $n$
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 $k$
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
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).\\
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 $\beta_k$.
\[d(x_k,x) = 2^{-n} |e^{\i \beta_k} - 1| < 2^{-n-k} \xrightarrow{k\to \infty} 0.\]
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 $1$.
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}}}).
It is
\[F(x_k, x') = \inf_m d(\sigma^m(w_k), 1),\]
where $\sigma(\xi_1, \xi_2, \ldots) = (\xi_1, \xi_1\xi_2, \xi_2\xi_3, \ldots)$.
We have
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).
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
$d(\sigma^m w_k, w^\ast) < \epsilon$.
\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}.
}{[some technical details omitted]}
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
$x \mapsto x^{n}$.
Note that for
\sigma\colon (S^1)^d &\longrightarrow & S^1 \\
(x_1,\ldots,x_d) &\longmapsto & x_d
we have that $T = \tau_{d+1}$,
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}).
For every $d$ if $\tau_d$\footnote{more formally $((S^1)^d, \langle \tau_d \rangle)$}
is minimal, then $\tau_{d+1}$ is minimal.
$\tau_d$ is minimal for all $d$.
$\tau_1$ is minimal (\yaref{fact:tau1minimal}).
Apply \yaref{thm:taudminimal:help}.
Since all the $\tau_d$ are minimal,
$\tau$ is minimal.
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.}
Let $S \coloneqq \tau_d$, $T \coloneqq \tau_{d+1}$ and $Y \coloneqq (S^1)^d$.
\gamma\colon S^1 &\longrightarrow & Y \\
x &\longmapsto & (x,x,\ldots,x).
Note that
\item $\gamma$ and $S \circ \gamma$ are homotopic
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})
\item For all $m \in \Z \setminus \{0\}$, we have
$[x \mapsto \left(\sigma(\gamma(x))\right)^m] = m \neq 0$,
since $\sigma(\gamma(x)) = \sigma((x,\ldots,x)) = x$.