Josia Pietsch 195ac2c378
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\lecture{21}{2024-01-12}{Iterated Skew Shift}
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)\},
\phi\colon Y &\longrightarrow & S^1 \\
y &\longmapsto & (\beta^{(y)})^m
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 $(y,\phi(y)) \in f(Z)$).
$\phi(S(y)) = \phi(y) \cdot (\sigma(y))^m$.
We have $T(y, \xi) = (S(y), \sigma(y) \cdot \xi)$
$Z$ is invariant under $T$.
So for $(y, \xi) \in Z$ we get $T(y, \xi) = ({\color{red}S(y)}, {\color{blue}\sigma(y) \cdot \xi}) \in Z$.
\phi({\color{red}S(y)}) &=& ({\color{blue}\sigma(y) \cdot \xi})^m\\
&=& (\sigma(y))^m \cdot \xi^m\\
&=& (\sigma(y))^m \cdot \phi(y).
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) $\lightning$
\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 $S^1$).
\item $H \neq S^1$ as $Z$ is a proper subflow.
\item $H = \langle \frac{1}{m} \rangle$.
\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 $(y, \xi) \mapsto (y, m \cdot \xi)$.
\phi\colon Y &\longrightarrow & S^1 \\
y &\longmapsto & m \beta^{(y)}
\item $\phi \circ S = \phi + m \sigma$:
$Z$ is $T$-invariant.
\item $[\phi \circ S \circ \gamma] = [\phi \circ \gamma] + [m \sigma \circ \gamma]$
$\implies [m \sigma \circ \gamma] = 0 \lightning$.
Let $X_n \coloneqq (S^1)^n$ and $X \coloneqq (S^1)^{\N}$.
$(X_n, \tau_n)$ is the maximal isometric extension of $(X_{n-1}, \tau_{n-1})$
in $(X,\tau)$.
The order of $(X,\tau)$ is $\omega$.