Josia Pietsch
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86 lines
3.2 KiB
TeX
86 lines
3.2 KiB
TeX
\lecture{21}{2024-01-12}{Iterated Skew Shift}
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\begin{refproof}{thm:taudminimal:help}
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Suppose towards a contradiction that
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$Y \times S^1$ contains a proper minimal subflow $Z$.
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Consider the projection $\pi\colon Y \times S^1 \to Y$.
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By minimality of $Y$, we have $\pi(Z) = Y$.
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Note that for every $\theta \in S^1$, $\theta \cdot Z$ is minimal,
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so either $\theta \cdot Z = Z$ or $(\theta \cdot Z)\cap Z = \emptyset$.
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Let $H = \{\theta \in S^1 : \theta \cdot Z = Z\}$.
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$H$ is a closed subgroup of $S^1$.
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% H is a rotation of Z containing 1 (?)
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Therefore either $H = S^1$ (but in that case $Z = Y \times S^1$),
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or there exists $m \in \Z$ such that $H = \{ \xi \in S^1 : \xi^m = 1 \}$
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by \yaref{fact:tau1minimal}.
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Note that if $(y, \beta) \in Z$ then for $t \in S^1$,
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we have
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\[
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(y, \beta \cdot t) \in Z \iff t^m = 1.
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\]
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Therefore for every $y \in Y$, there are exactly $m$ many
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$\xi \in S^1$
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such that $(y, \xi) \in Z$.
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Specifically for all $y$ there exists $\beta^{(y)} \in S^1$
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such that $(y,\xi) \in Z$ iff
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\[
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\xi \in \{\beta^{(y)} \cdot t_1, \beta^{(y)} \cdot t_2, \ldots,\beta^{(y)} \cdot t_m\},
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\]
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where the $t_i \in S^1$
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are such that
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$t_i^m = 1$ for all $i$ and $i \neq j \implies t_i \neq t_j$,
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i.e.~the $t_i$ are the $m$\textsuperscript{th} roots of unity.
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Consider $f \colon (y,\xi) \mapsto (y, \xi^m)$.
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Since $(\beta^{(y)} \cdot t_i)^m = (\beta^{(y)})^m$
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we get a continuous\todo{Why is this continuous?}
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function $\phi\colon Y \to S^1$
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such that
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\[
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Z = \{(y,\xi) \in Y \times S^1 : \xi^m = \phi(y)\}.
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\]
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% namely
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% \begin{IEEEeqnarray*}{rCl}
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% \phi\colon Y &\longrightarrow & S^1 \\
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% y &\longmapsto & \beta^{(y)}.
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% \end{IEEEeqnarray*}
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Note that $f(Z)$ is homeomorphic to $Y$.\todo{Why?}
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\begin{claim}
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$\phi(S(y)) = \phi(y) \cdot (\sigma(y))^m$.
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\end{claim}
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\begin{subproof}
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We have $T(y, \xi) = (S(y), \sigma(y) \cdot \xi)$
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(cf.~\yaref{rem:l20:sigma}).
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$Z$ is invariant under $T$.
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So for $(y, \xi) \in Z$ we get $T(y, \xi) = ({\color{red}S(y)}, {\color{blue}\sigma(y) \cdot \xi}) \in Z$.
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Thus
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\begin{IEEEeqnarray*}{rCl}
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\phi({\color{red}S(y)}) &=& ({\color{blue}\sigma(y) \cdot \xi})^m\\
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&=& (\sigma(y))^m \cdot \xi^m\\
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&=& (\sigma(y))^m \cdot \phi(y).
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\end{IEEEeqnarray*}
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\end{subproof}
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Applying $\gamma$ we obtain
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\[
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[\phi \circ S \circ \gamma] = [\phi \circ \gamma] + [x \mapsto (\sigma(\gamma(x))^n].
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\]
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$S\circ \gamma$ is homotopic to $\gamma$,
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so $[\phi \circ S \circ \gamma] = [\phi \circ \gamma]$.
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Thus $[x \mapsto (\sigma(\gamma(x))^n] = 0$,
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but that is a contradiction to (b) $\lightning$
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\end{refproof}
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Let $X_n \coloneqq (S^1)^n$ and $X \coloneqq (S^1)^{\N}$.
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\begin{theorem}
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\label{thm:21:xnmaxiso}
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$(X_n, \tau_n)$ is the maximal isometric extension of $(X_{n-1}, \tau_{n-1})$
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in $(X,\tau)$.
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\end{theorem}
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\begin{corollary}
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The order of $(X,\tau)$ is $\omega$.
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\end{corollary}
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