lecture 20

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@ -15,6 +15,7 @@ sketch of \yaref{thm:l16:3} in the case of $|Z| = 1$.
Let $(X,T)$ be a distal flow.
Then $G \coloneqq E(X,T)$ is a group.
\begin{definition}
\label{def:F}
For $x, x' \in X$ define
\[
F(x,x') \coloneqq \inf \{d(gx, gx') : g \in G\}.

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@ -274,6 +274,7 @@ More generally we can show:
$(X'_{\xi+1}, T)$ is a factor of $(X_{\xi+1}, T)$.
\end{proof}
\begin{example}[{\cite[p. 513]{Furstenberg}}]
\label{ex:19:inftorus}
Let $X$ be the infinite torus
\[
X \coloneqq \{(\xi_1, \xi_2, \ldots) : \xi_i \in \C, |\xi_i| = 1\}.

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\lecture{20}{2024-01-09}{The infinite Torus}
\begin{example}
\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 $\alpha$.
\end{example}
\begin{remark}+
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.
\end{remark}
We will be studying projections to the first $d$ coordinates,
i.e.
\[
\tau_d \colon (x_1,\ldots,x_d) \mapsto (e^{\i \alpha} x_1, x_1x_2, \ldots, x_{d-1}x_d).
\]
$\tau_d$ is called the \vocab{$d$-skew shift}.
For $d = 1$ we get the circle rotation $x \mapsto e^{\i \alpha} x$.
\begin{fact}
\label{fact:tau1minimal}
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 $m \in \Z$.
\end{fact}
\todo{Homework!}
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 $n$
coordinates.
\begin{lemma}
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 $k$
and
\[
F(x_k, x) \xrightarrow{k \to \infty} 0,
F(x_k, x') \xrightarrow{k \to \infty} 0,
\]
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
\end{lemma}
\begin{proof}
Let
\begin{IEEEeqnarray*}{rCl}
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).\\
\end{IEEEeqnarray*}
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 of such $\beta_k$.
Then
\[d(x_k,x) = 2^{-n} |e^{\i \beta_k} - 1| < 2^{-n-k} \xrightarrow{k\to \infty} 0.\]
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 $1$.
Consider
\[
w_k \coloneqq \frac{x_k}{x'} = (\underbrace{1,\ldots,1}_{n-1}, e^{\i \beta_k}, \overbrace{\frac{\alpha_{n+1}}{\alpha'_{n+1}}, \frac{\alpha_{n+2}}{\alpha'_{n+2}}, \ldots}^{\mathclap{\text{not interesting}}}).
\]
\begin{claim}
$F(x_k, x') = \inf_m d(\sigma^m(w_k), 1)$,
where $\sigma(\xi_1, \xi_2, \ldots) = (\xi_1, \xi_1\xi_2, \xi_2\xi_3, \ldots)$.
\end{claim}
\begin{subproof}
We have
\begin{IEEEeqnarray*}{rCl}
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).
\end{IEEEeqnarray*}
\end{subproof}
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
$d(\sigma^m w_k, w^\ast) < \epsilon$.
% TODO Think about this
Then
\begin{IEEEeqnarray*}{rCl}
\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}.
\end{IEEEeqnarray*}
\end{proof}
\begin{definition}
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
$x \mapsto x^{n}$.
\end{definition}
\begin{remark}
Note that for
\begin{IEEEeqnarray*}{rCl}
\sigma\colon (S^1)^d &\longrightarrow & S^1 \\
(x_1,\ldots,x_d) &\longmapsto & x_d
\end{IEEEeqnarray*}
we have that $T = \tau_{d+1}$,
where
\begin{IEEEeqnarray*}{rCl}
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}).
\end{IEEEeqnarray*}
\end{remark}
\begin{theorem}
\label{thm:taudminimal:help}
For every $d$ if $\tau_d$\footnote{more formally $((S^1)^d, \langle \tau_d \rangle)$}
is minimal, then $\tau_{d+1}$ is minimal.
\end{theorem}
\begin{corollary}
$\tau_d$ is minimal for all $d$.
\end{corollary}
\begin{proof}
$\tau_1$ is minimal (\yaref{fact:tau1minimal}).
Apply \yaref{thm:taudminimal:help}.
\end{proof}
\begin{corollary}
Since all the $\tau_d$ are minimal,
$\tau$ is minimal.
\end{corollary}
\begin{proof}
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.
\end{proof}
\begin{refproof}{thm:taudminimal:help}
Let $s \coloneqq \tau_d$ and $Y \coloneqq (S^1)^d$.
Consider
\begin{IEEEeqnarray*}{rCl}
\gamma\colon S^1 &\longrightarrow & Y \\
x &\longmapsto & (x,x,\ldots,x)
\end{IEEEeqnarray*}
\begin{enumerate}[(a)]
\item $\gamma$ and $s \circ \gamma$ are homotopic
via
\begin{IEEEeqnarray*}{rCl}
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})
\end{IEEEeqnarray*}
\item For all $m \in \Z \setminus \{0\}$, we have
$[x \mapsto \left(\sigma(\gamma(x))\right)^m] = m \neq 0$,
since $\sigma(\gamma(x)) = \sigma((x,\ldots,x)) = x$.
\end{enumerate}
[to be continued]
\phantom\qedhere
\end{refproof}

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\tutorial{11}{2024-01-09}{}
An equivalent definition of subflows can be given as follows:
\begin{definition}
Let $(X,T)$ be a flow with action $\alpha_x$.
Let $Y \subseteq X$ be a compact subspace of $X$.
If $Y$ is invariant under $\alpha_x$, we say that
$(Y,T)$ (with action $\alpha_x\defon{T \times Y}$
is a subflow of $(X,T)$.
\end{definition}
\begin{example}[Flows with a non-closed orbit]
\begin{enumerate}[1.]
\item Consider $(S^1, \Z)$
with action given by $1 \cdot x = x + c$ for
a fixed $c \in \R\setminus\Q$.\footnote{We identify $S^1$ and $\faktor{\R}{\Z}$.}
Then the orbit of $0$, $\{nc : n \in \Z\}$ is dense but consists only of irrationals
(except $0$),
so it is not closed.
\item Consider $(S^1, \Q)$ with action $qx \coloneqq x + q$.
The orbit of $0$, $\faktor{\Q}{\Z} \subseteq S^1$,
is dense but not closed.
$(S^1,\Q)$ is minimal.
\end{enumerate}
\end{example}
\begin{example}[\vocab{Left Bernoulli shift}]
Consider $(\{0,1\}^{\Z}, T)$,
where $T = \Z$ and the action is given by
\begin{IEEEeqnarray*}{rCl}
\Z \times \{0,1\}^{\Z}&\longrightarrow & \Z \\
(m, (x_n)_{n \in \Z})&\longmapsto & (x_{n+m})_{n \in \Z}.
\end{IEEEeqnarray*}
The orbit of $z \coloneqq (0)_{n \in \Z}$ consist of only on point.
In particular it is closed.
Let $x \coloneqq ( [n = 0])_{n \in \Z}$.
Then $Tx = \{([n = m])_{n \in \Z} | m \in \Z\}$.
Clearly $z \not\in Tx$.
\begin{claim}
$z \in \overline{Tx}$
\end{claim}
\begin{proof}
Consider a basic open $z \in U_I = \{y : y_i = 0, i \in I\}$
where $I \subseteq \Z$ is finite.
Then $U_I \cap Tx \neq \emptyset$
as we can shift the $1$ out of $I$,
i.e.~$(\max I + 1) x \in U_I$.
\end{proof}
\end{example}
Flows are always on non-empty spaces $X$.
\begin{fact}
Consider a flow $(X,T)$.
The following are equivalent:
\begin{enumerate}[(i)]
\item Every $T$-orbit is dense.
\item There is no proper subflow,
\end{enumerate}
If these conditions hold, the flow is called \vocab{minimal}.
\end{fact}
\begin{proof}
(i) $\implies$ (ii):
Let $(Y,T)$ be a subflow of $(X,T)$.
take $y \in Y$. Then $Ty$ is dense in mKX.
But $Ty \subseteq Y$, so $Y$ is dense in $X$.
Since $Y$ is closed, we get $Y = X$.
(ii) $\implies$ (i):
Take $x \in X$. Consider $Tx$.
It suffices to show that $\overline{Tx}$ is a subflow.
Clearly $\overline{Tx}$ is closed,
so it suffices to show that it is $T$-invariant.
Let $y \in \overline{Tx}$ and $t \in T$.
Take $ty \in U \overset{\text{open}}{\subseteq} X$.
Since $t^{-1}$ acts as a homeomorphism
we have $y \in t^{-1} U \overset{\text{open}}{\subseteq} X$.
We find some $t'x \in t^{-1}U$ since $y \in \overline{Tx}$.
So $tt'x \in Tx \cap U$.
\end{proof}
\begin{fact}
Every flow $(X,T)$ contains a minimal subflow.
\end{fact}
\begin{proof}
We use Zorn's lemma:
Let $S$ be the set of all subflows of $(X,T)$
ordered by $Y \le Y' :\iff Y \supseteq Y'$.
We need to show that for a chain $\langle Y_i : i \in I \rangle$,
there exists a lower bound.
Consider $\bigcap_{i \in I} Y_i$. This a subflow:
\begin{itemize}
\item It is closed as it is an intersection of closed sets.
\item It is $T$-invariant, since each of the $Y_i$ is.
\item It is non-empty by \yaref{tut10fact}.
\end{itemize}
\end{proof}
\begin{fact}
\label{tut10fact}
Let $X$ be a topological space.
Then $X$ is compact iff every family of closed sets with
FIP\footnote{finite intersection property, i.e.~the intersection of every finite sub-family is non-empty}
has non-empty intersection.
\end{fact}
\begin{proof}
Note that families of
closed sets correspond to families of open sets by taking complements.
A family of open sets is a cover iff the corresponding family
has empty intersection,
and is admits a finite subcover iff the corresponding family
has the FIP.
\end{proof}

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@ -44,10 +44,7 @@
\input{inputs/lecture_17}
\input{inputs/lecture_18}
\input{inputs/lecture_19}
\input{inputs/lecture_20}
\cleardoublepage
@ -64,6 +61,7 @@
\input{inputs/tutorial_08}
\input{inputs/tutorial_09}
\input{inputs/tutorial_10}
\input{inputs/tutorial_11}
\section{Facts}
\input{inputs/facts}