lecture 20
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@ -15,6 +15,7 @@ sketch of \yaref{thm:l16:3} in the case of $|Z| = 1$.
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Let $(X,T)$ be a distal flow.
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Then $G \coloneqq E(X,T)$ is a group.
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\begin{definition}
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\label{def:F}
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For $x, x' \in X$ define
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\[
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F(x,x') \coloneqq \inf \{d(gx, gx') : g \in G\}.
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@ -274,6 +274,7 @@ More generally we can show:
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$(X'_{\xi+1}, T)$ is a factor of $(X_{\xi+1}, T)$.
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\end{proof}
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\begin{example}[{\cite[p. 513]{Furstenberg}}]
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\label{ex:19:inftorus}
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Let $X$ be the infinite torus
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\[
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X \coloneqq \{(\xi_1, \xi_2, \ldots) : \xi_i \in \C, |\xi_i| = 1\}.
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196
inputs/lecture_20.tex
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196
inputs/lecture_20.tex
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\lecture{20}{2024-01-09}{The infinite Torus}
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\begin{example}
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\footnote{This is the same as \yaref{ex:19:inftorus},
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but with new notation.}
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Let $X = (S^1)^{\N}$\footnote{We identify $S^1$ and $\faktor{\R}{\Z}$.}
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and consider $\left(X, \Z \right)$
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where the action is generated by
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\[
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\tau\colon (x_1,x_2,x_3,\ldots) \mapsto(x_1 + \alpha, x_1 + x_2, x_2 + x_3, \ldots)
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\]
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for some irrational $\alpha$.
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\end{example}
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\begin{remark}+
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Note that we can identify $S^1$ with a subset of $\C$ (and use multiplication)
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or with $\faktor{\R}{\Z}$ (and use addition).
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In the lecture both notations were used.% to make things extra confusing.
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Here I'll try to only use multiplicative notation.
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\end{remark}
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We will be studying projections to the first $d$ coordinates,
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i.e.
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\[
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\tau_d \colon (x_1,\ldots,x_d) \mapsto (e^{\i \alpha} x_1, x_1x_2, \ldots, x_{d-1}x_d).
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\]
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$\tau_d$ is called the \vocab{$d$-skew shift}.
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For $d = 1$ we get the circle rotation $x \mapsto e^{\i \alpha} x$.
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\begin{fact}
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\label{fact:tau1minimal}
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The circle rotation $x \mapsto e^{\i \alpha} x$ is minimal.
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In fact, every subgroup of $S^1$ is either dense in $S^1$
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or it is of the form
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\[
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H_m \coloneqq \{x \in S^1 : x^m = 0\}
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\]
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for some $m \in \Z$.
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\end{fact}
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\todo{Homework!}
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We will show that $\tau_d$ is minimal for all $d$,
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i.e.~every orbit is dense.
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From this it will follow that $\tau$ is minimal.
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Let $\pi_n\colon X \to (S^1)^n$ be the projection to the first $n$
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coordinates.
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\begin{lemma}
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Let $x,x' \in X$ with $\pi_n(x) = \pi_n(x')$
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for some $n$.
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Then there is a sequence of points $x_k$ with
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\[\pi_{n-1}(x_k) = \pi_{n-1}(x) = \pi_{n-1}(x')\]
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for all $k$
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and
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\[
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F(x_k, x) \xrightarrow{k \to \infty} 0,
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F(x_k, x') \xrightarrow{k \to \infty} 0,
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\]
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where $F$ is as in \yaref{def:F},
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i.e.~$F(a,b) = \inf_{n \in \Z} d(\tau^n a, \tau^n b)$,
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where $d$ is the metric on $X$,
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$d((x_i), (y_i)) = \max_n \frac{1}{2^n} | x_n - y_n|$.% TODO use multiplicative notation
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\end{lemma}
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\begin{proof}
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Let
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\begin{IEEEeqnarray*}{rCl}
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x &=& (\alpha^0_1, \alpha^0_2, \ldots, \alpha^0_{n-1}, \alpha_n, \alpha_{n+1}, \alpha_{n+2},\ldots)\\
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x' &=& (\alpha^0_1, \alpha^0_2, \ldots, \alpha^0_{n-1}, \alpha_n, \alpha'_{n+1}, \alpha'_{n+2},\ldots).\\
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\end{IEEEeqnarray*}
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We will choose $x_k$ of the form
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\[
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(\alpha^0_1, \alpha^0_2, \ldots, \alpha^0_{n-1} \alpha_n e^{\i \beta_k}, \alpha_{n+1}, \alpha_{n+2}, \ldots),
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\]
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where $\beta_k$ is such that $\frac{\beta_k}{\pi}$ is irrational
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and $|\beta_k| < 2^{-k}$.
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Fix a sequence of such $\beta_k$.
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Then
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\[d(x_k,x) = 2^{-n} |e^{\i \beta_k} - 1| < 2^{-n-k} \xrightarrow{k\to \infty} 0.\]
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In particular $F(x_k, x) \to 0$.
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We want to show that $F(x_k, x') < 2^{-n-k}$.
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For $u, u' \in X$,
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$u = (\xi_n)_{n \in \N}$,
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$u' = (\xi'_n)_{n \in \N}$,
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let $\frac{u}{u'} = (\frac{\xi_n}{\xi'_n})_{n \in \N}$
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($X$ is a group).
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We are interested in $F(x_k, x') = \inf_m d(\tau^m x_k, \tau^m x')$,
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but it is easier to consider the distance between
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their quotient and $1$.
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Consider
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\[
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w_k \coloneqq \frac{x_k}{x'} = (\underbrace{1,\ldots,1}_{n-1}, e^{\i \beta_k}, \overbrace{\frac{\alpha_{n+1}}{\alpha'_{n+1}}, \frac{\alpha_{n+2}}{\alpha'_{n+2}}, \ldots}^{\mathclap{\text{not interesting}}}).
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\]
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\begin{claim}
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$F(x_k, x') = \inf_m d(\sigma^m(w_k), 1)$,
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where $\sigma(\xi_1, \xi_2, \ldots) = (\xi_1, \xi_1\xi_2, \xi_2\xi_3, \ldots)$.
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\end{claim}
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\begin{subproof}
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We have
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\begin{IEEEeqnarray*}{rCl}
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F(u,u') &=& \inf_m d(\tau^m u, \tau^m u')\\
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&=& \inf_m d(\frac{\tau^m u}{\tau^m u'}, 1)\\
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&=& \inf_m d(\sigma^m\left( \frac{u}{u'} \right), 1).
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\end{IEEEeqnarray*}
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\end{subproof}
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Fix $k$. Let $w^\ast = (1,\ldots,1, e^{\i \beta_k}, 1, \ldots)$.
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By minimality of $(X,T)$ for any $\epsilon >0$,
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there exists $m \in \Z$ such that
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$d(\sigma^m w_k, w^\ast) < \epsilon$.
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% TODO Think about this
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Then
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\begin{IEEEeqnarray*}{rCl}
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\inf_m d(\sigma^m w_k, 1) &\le & \inf_m d(\sigma^m w_k, w^\ast) + d(w^\ast, 1)\\
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&\le & 2^{-n} | e^{\i \beta_k}- 1|\\
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&<& 2^{-n-k}.
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\end{IEEEeqnarray*}
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\end{proof}
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\begin{definition}
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For every continuous $f\colon S^1 \to S^1$, the
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\vocab{winding number} $[f] \in \Z$
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is the unique integer such that $f$ is homotopic%
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\footnote{$f\colon Y \to Z$ and $g\colon Y \to Z$ are homotopic
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iff there is $H\colon Y \times [0,1] \to \Z$
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continuous such that $H(\cdot ,0) = f$ and $H(\cdot ,1) = g$.}
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to the map
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$x \mapsto x^{n}$.
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\end{definition}
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\begin{remark}
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Note that for
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\begin{IEEEeqnarray*}{rCl}
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\sigma\colon (S^1)^d &\longrightarrow & S^1 \\
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(x_1,\ldots,x_d) &\longmapsto & x_d
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\end{IEEEeqnarray*}
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we have that $T = \tau_{d+1}$,
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where
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\begin{IEEEeqnarray*}{rCl}
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T\colon (S^1)^d \times S^1 &\longrightarrow & (S^1)^d \times S^1 \\
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(y, x_{d+1}) &\longmapsto & (\tau_d(y), \sigma(y) x_{d+1}).
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\end{IEEEeqnarray*}
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\end{remark}
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\begin{theorem}
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\label{thm:taudminimal:help}
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For every $d$ if $\tau_d$\footnote{more formally $((S^1)^d, \langle \tau_d \rangle)$}
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is minimal, then $\tau_{d+1}$ is minimal.
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\end{theorem}
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\begin{corollary}
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$\tau_d$ is minimal for all $d$.
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\end{corollary}
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\begin{proof}
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$\tau_1$ is minimal (\yaref{fact:tau1minimal}).
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Apply \yaref{thm:taudminimal:help}.
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\end{proof}
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\begin{corollary}
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Since all the $\tau_d$ are minimal,
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$\tau$ is minimal.
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\end{corollary}
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\begin{proof}
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This follows from the definition of the product topology,
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since for a basic open set $U = U_1 \times \ldots \times U_d \times (S^1)^{\infty}$
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it suffices to analyze the first $d$ coordinates.
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\end{proof}
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\begin{refproof}{thm:taudminimal:help}
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Let $s \coloneqq \tau_d$ and $Y \coloneqq (S^1)^d$.
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Consider
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\begin{IEEEeqnarray*}{rCl}
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\gamma\colon S^1 &\longrightarrow & Y \\
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x &\longmapsto & (x,x,\ldots,x)
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\end{IEEEeqnarray*}
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\begin{enumerate}[(a)]
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\item $\gamma$ and $s \circ \gamma$ are homotopic
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via
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\begin{IEEEeqnarray*}{rCl}
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H\colon S^1 \times [0,1] &\longrightarrow & (S^1)^d \\
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(x, t)&\longmapsto & (x e^{\i t \alpha}, x^{t+1}, x^{t+1}, x^{t+1},\ldots, x^{t+1})
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\end{IEEEeqnarray*}
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\item For all $m \in \Z \setminus \{0\}$, we have
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$[x \mapsto \left(\sigma(\gamma(x))\right)^m] = m \neq 0$,
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since $\sigma(\gamma(x)) = \sigma((x,\ldots,x)) = x$.
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\end{enumerate}
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[to be continued]
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\phantom\qedhere
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\end{refproof}
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113
inputs/tutorial_11.tex
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113
inputs/tutorial_11.tex
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@ -0,0 +1,113 @@
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\tutorial{11}{2024-01-09}{}
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An equivalent definition of subflows can be given as follows:
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\begin{definition}
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Let $(X,T)$ be a flow with action $\alpha_x$.
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Let $Y \subseteq X$ be a compact subspace of $X$.
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If $Y$ is invariant under $\alpha_x$, we say that
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$(Y,T)$ (with action $\alpha_x\defon{T \times Y}$
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is a subflow of $(X,T)$.
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\end{definition}
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\begin{example}[Flows with a non-closed orbit]
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\begin{enumerate}[1.]
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\item Consider $(S^1, \Z)$
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with action given by $1 \cdot x = x + c$ for
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a fixed $c \in \R\setminus\Q$.\footnote{We identify $S^1$ and $\faktor{\R}{\Z}$.}
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Then the orbit of $0$, $\{nc : n \in \Z\}$ is dense but consists only of irrationals
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(except $0$),
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so it is not closed.
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\item Consider $(S^1, \Q)$ with action $qx \coloneqq x + q$.
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The orbit of $0$, $\faktor{\Q}{\Z} \subseteq S^1$,
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is dense but not closed.
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$(S^1,\Q)$ is minimal.
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\end{enumerate}
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\end{example}
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\begin{example}[\vocab{Left Bernoulli shift}]
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Consider $(\{0,1\}^{\Z}, T)$,
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where $T = \Z$ and the action is given by
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\begin{IEEEeqnarray*}{rCl}
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\Z \times \{0,1\}^{\Z}&\longrightarrow & \Z \\
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(m, (x_n)_{n \in \Z})&\longmapsto & (x_{n+m})_{n \in \Z}.
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\end{IEEEeqnarray*}
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The orbit of $z \coloneqq (0)_{n \in \Z}$ consist of only on point.
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In particular it is closed.
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Let $x \coloneqq ( [n = 0])_{n \in \Z}$.
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Then $Tx = \{([n = m])_{n \in \Z} | m \in \Z\}$.
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Clearly $z \not\in Tx$.
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\begin{claim}
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$z \in \overline{Tx}$
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\end{claim}
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\begin{proof}
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Consider a basic open $z \in U_I = \{y : y_i = 0, i \in I\}$
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where $I \subseteq \Z$ is finite.
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Then $U_I \cap Tx \neq \emptyset$
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as we can shift the $1$ out of $I$,
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i.e.~$(\max I + 1) x \in U_I$.
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\end{proof}
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\end{example}
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Flows are always on non-empty spaces $X$.
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\begin{fact}
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Consider a flow $(X,T)$.
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The following are equivalent:
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\begin{enumerate}[(i)]
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\item Every $T$-orbit is dense.
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\item There is no proper subflow,
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\end{enumerate}
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If these conditions hold, the flow is called \vocab{minimal}.
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\end{fact}
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\begin{proof}
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(i) $\implies$ (ii):
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Let $(Y,T)$ be a subflow of $(X,T)$.
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take $y \in Y$. Then $Ty$ is dense in mKX.
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But $Ty \subseteq Y$, so $Y$ is dense in $X$.
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Since $Y$ is closed, we get $Y = X$.
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(ii) $\implies$ (i):
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Take $x \in X$. Consider $Tx$.
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It suffices to show that $\overline{Tx}$ is a subflow.
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Clearly $\overline{Tx}$ is closed,
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so it suffices to show that it is $T$-invariant.
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Let $y \in \overline{Tx}$ and $t \in T$.
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Take $ty \in U \overset{\text{open}}{\subseteq} X$.
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Since $t^{-1}$ acts as a homeomorphism
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we have $y \in t^{-1} U \overset{\text{open}}{\subseteq} X$.
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We find some $t'x \in t^{-1}U$ since $y \in \overline{Tx}$.
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So $tt'x \in Tx \cap U$.
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\end{proof}
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\begin{fact}
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Every flow $(X,T)$ contains a minimal subflow.
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\end{fact}
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\begin{proof}
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We use Zorn's lemma:
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Let $S$ be the set of all subflows of $(X,T)$
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ordered by $Y \le Y' :\iff Y \supseteq Y'$.
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We need to show that for a chain $\langle Y_i : i \in I \rangle$,
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there exists a lower bound.
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Consider $\bigcap_{i \in I} Y_i$. This a subflow:
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\begin{itemize}
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\item It is closed as it is an intersection of closed sets.
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\item It is $T$-invariant, since each of the $Y_i$ is.
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\item It is non-empty by \yaref{tut10fact}.
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\end{itemize}
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\end{proof}
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\begin{fact}
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\label{tut10fact}
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Let $X$ be a topological space.
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Then $X$ is compact iff every family of closed sets with
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FIP\footnote{finite intersection property, i.e.~the intersection of every finite sub-family is non-empty}
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has non-empty intersection.
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\end{fact}
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\begin{proof}
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Note that families of
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closed sets correspond to families of open sets by taking complements.
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A family of open sets is a cover iff the corresponding family
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has empty intersection,
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and is admits a finite subcover iff the corresponding family
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has the FIP.
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\end{proof}
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@ -44,10 +44,7 @@
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\input{inputs/lecture_17}
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\input{inputs/lecture_18}
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\input{inputs/lecture_19}
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\input{inputs/lecture_20}
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\cleardoublepage
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@ -64,6 +61,7 @@
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\input{inputs/tutorial_08}
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\input{inputs/tutorial_09}
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\input{inputs/tutorial_10}
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\input{inputs/tutorial_11}
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\section{Facts}
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\input{inputs/facts}
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