w23-logic-3/inputs/lecture_21.tex

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\lecture{21}{2024-01-12}{Iterated Skew Shift}
\begin{refproof}{thm:taudminimal:help}
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 $(\theta \cdot Z)\cap Z = \emptyset$.
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 $Z = Y \times S^1$),
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\todo{Why is this 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)}.
% \end{IEEEeqnarray*}
Note that $f(Z)$ is homeomorphic to $Y$.\todo{Why?}
\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) $\lightning$
\end{refproof}
Let $X_n \coloneqq (S^1)^n$ and $X \coloneqq (S^1)^{\N}$.
\begin{theorem}
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\label{thm:21:xnmaxiso}
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$(X_n, \tau_n)$ is the maximal isometric extension of $(X_{n-1}, \tau_{n-1})$
in $(X,\tau)$.
\end{theorem}
\begin{corollary}
The order of $(X,\tau)$ is $\omega$.
\end{corollary}
\todo{I could not attend lecture 21 as I was sick. The official notes on the lecture are very short.
Is something missing in the official notes?}