22

inputs/lecture_16.tex

@@ -207,6 +207,7 @@ A flow $\Z \acts X$ corresponds to the graph of
 and this graph is an element of $K(\bH^2)$.
 
 \begin{theorem}[Beleznay-Foreman]
+    \label{thm:beleznay-foreman}
     Consider $\Z$-flows.
     \begin{itemize}
         \item For any $\alpha < \omega_1$,

inputs/lecture_17.tex

@@ -173,6 +173,7 @@ However if we pick $y \in Y$, $Ty$ might not be dense.
 % We want to a minimal subflow in a nice way:
 
 \begin{theorem}
+    \label{thm:distalflowpartition}
     If $(X,T)$ is distal, then $X$ is the disjoint union of minimal subflows.
     In fact those disjoint sets
     will be orbits of $E(X,T)$.

inputs/lecture_20.tex

@@ -44,9 +44,8 @@ coordinates.
 
 
 
-
-
 \begin{lemma}
+    \label{lem:lec20:1}
     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
@@ -74,7 +73,7 @@ coordinates.
     \]
     where $\beta_k$ is such that $\frac{\beta_k}{\pi}$ is irrational
     and $|\beta_k| < 2^{-k}$.
-    Fix a sequence of such $\beta_k$.
+    Fix a sequence'(b)). 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 $F(x_k, x) \to 0$.

inputs/lecture_21.tex

@@ -77,6 +77,7 @@
 
 Let $X_n \coloneqq  (S^1)^n$ and $X \coloneqq (S^1)^{\N}$.
 \begin{theorem}
+    \label{thm:21:xnmaxiso}
     $(X_n, \tau_n)$ is the maximal isometric extension of  $(X_{n-1}, \tau_{n-1})$
     in $(X,\tau)$.
 \end{theorem}

inputs/lecture_22.tex

@@ -0,0 +1,194 @@
+\lecture{22}{2024-01-16}{}
+
+\begin{refproof}{thm:21:xnmaxiso}
+    We have the following situation:
+    % https://q.uiver.app/#q=WzAsNCxbMCwwLCJYIl0sWzEsMSwiWF97bn0iXSxbMSwyLCJYX3tuLTF9Il0sWzIsMSwiWSJdLFswLDEsIlxccGlfe259Il0sWzEsMiwiXFx0ZXh0e2lzb21ldHJpY30iLDFdLFswLDIsIlxccGlfe24tMX0iLDIseyJjdXJ2ZSI6Mn1dLFswLDMsIlxccGknIiwwLHsiY3VydmUiOi0zfV0sWzMsMSwiaCIsMix7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFszLDIsIlxcb3ZlcmxpbmV7Z30sIFxcdGV4dHsgbWF4LiBpc29tLn0iLDAseyJjdXJ2ZSI6LTJ9XV0=
+    \[\begin{tikzcd}
+    	X \\
+    	& {X_{n}} & Y \\
+    	& {X_{n-1}}
+    	\arrow["{\pi_{n}}", from=1-1, to=2-2]
+    	\arrow["{\text{isometric}}"{description}, from=2-2, to=3-2]
+    	\arrow["{\pi_{n-1}}"', curve={height=12pt}, from=1-1, to=3-2]
+    	\arrow["{\pi'}", curve={height=-18pt}, from=1-1, to=2-3]
+    	\arrow["h"', dashed, from=2-3, to=2-2]
+    	\arrow["{\overline{g}, \text{ max. isom.}}", curve={height=-12pt}, from=2-3, to=3-2]
+    \end{tikzcd}\]
+    
+    We want to show that this tower is normal,
+    i.e.~the isometric extensions are maximal isometric extension.
+    
+    Let $Y$ be a maximal isometric extension of $X_{n-1}$ in $X$
+    and let $\overline{g} = \pi^n_{n-1} \circ h$. % factor map?
+    We need to show that $h$ is an isomorphism.
+    Towards a contradiction assume that $h$ is not an isomorphism.
+    Then there are $x,x' \in X$ with
+    $\pi'(x) \neq  \pi'(x')$ but $\pi_n(x) = \pi_n(x') =t \in X_n$.
+    Then $h^{-1}(t) \ni \pi'(x), \pi'(x')$.
+    
+    By a \yaref{lem:lec20:1}
+    there is a sequence $(x_k)$ in $X$
+    with $\pi_{n-1}(x_k) = \pi_{n-1}(x) = \pi_{n-1}(x')$ for all $k$,
+    such that $F(x_k, x) \to 0$ and $F(x_k, x') \to 0$.
+    
+    Let $\rho$ be a metric witnessing that $\overline{g}$
+    is an isometric extension,
+    i.e.~
+    $\rho$ is defined on $\bigcup_{x \in X_{n-1}} (\overline{g}^{-1}(x))^2 \overset{\text{closed}}{\subseteq} Y \times Y$,
+    continuous and $\rho(Ta, Tb) = \rho(a,b)$ for $\overline{g}(a) = \overline{g}(b)$.
+    
+    For $a,b \in X$ such that
+    \[
+    \overline{g}(\pi'(a)) = \overline{g}(\pi'(b))
+    \]
+    define
+    \[
+    R(a,b) \coloneqq  \rho(\pi'(a), \pi'(b)).
+    \]
+    
+    \begin{itemize}
+        \item For any two out of $x,x',(x_k)$, $R$ is defined.
+        \item $R(x,x_k) = R(\tau^m x, \tau^m x_k)$ for all  $m$.
+        \item $F(x,x_k) \xrightarrow{k\to \infty} 0$,
+            so there is a sequence $(m_k)$
+            such that
+            $d(\tau^{m_k}x, \tau^{m_k} x_k) \xrightarrow{k \to \infty} 0$.
+    \end{itemize}
+    By continuity of $\rho$,
+    we have that  $R(x,x_k) = R(\tau^{m_k} x, \tau^{m_k} x_k) \xrightarrow{k \to  \infty} 0$,
+    and similarly $R(x',x_k) \to 0$.
+    Hence $R(x,x') \xrightarrow{k \to \infty} 0$
+    by the triangle inequality.
+    But $x$ and $x'$ don't depend on $k$,
+    hence $R(x,x') = 0$.
+    It follows that $\pi'(x) = \pi'(x')$ $\lightning$.
+\end{refproof}
+
+
+\begin{theorem}[Beleznay-Foreman]
+    \begin{enumerate}[(1)]
+        \item For every $\eta < \omega_1$,
+            there is a distal minimal flow
+            of order $\eta$.%\footnote{For second countable spaces this is the best we can get.}
+        \item The set of distal minimal flows
+            is $\Pi^1_1$-complete.
+        \item The order is a $\Pi^1_1$-rank.
+            In particular
+            $\{\text{distal minimal flows of rank } < \alpha\}$
+            is Borel for all $\alpha < \omega_1$.
+    \end{enumerate}
+\end{theorem}
+\todo{This was already stated as \yaref{thm:beleznayforeman} in lecture 16
+    and should not have two numbers.}
+
+A few words on the proof:
+Let $\mathbb{K} = S^1$
+and $I$ a countable linear order.
+Let $\mathbb{K}^I$ be the product of $|I|$ many $\mathbb{K}$,
+$\mathbb{K}^{<i} \coloneqq \mathbb{K}^{\{j : j < i\}}$
+and $\pi_{i}\colon \mathbb{K}^{I} \to  \mathbb{K}^{<i}$\todo{maybe call it $\pi_{<i}$?}
+the projection.
+
+Let $\mathbb{K}_I \coloneqq  \prod_{i \in I} C(\mathbb{K}^{<i}, \mathbb{K})$.
+
+Fix some $(f_i)_{i \in I} \in \mathbb{K_I}$.
+We build a flow acting on $\mathbb{K}$ from $(f_i)_{i \in I}$.
+
+For this we define
+\begin{IEEEeqnarray*}{rCl}
+    E_I \colon  \mathbb{K}_I&\longrightarrow &  C(\mathbb{K}^I, \mathbb{K}^I)\\
+    (f_i)_{i \in I}&\longmapsto & \begin{pmatrix*}[l]
+        \mathbb{K}^I&\longrightarrow & \mathbb{K}^I \\
+        x &\longmapsto & (f_i(\pi_i(x)) \cdot  x_i)_{i \in I}
+    \end{pmatrix*}
+\end{IEEEeqnarray*}
+
+
+\begin{example}
+    Consider the following flow:
+    \begin{IEEEeqnarray*}{rCl}
+        \tau\colon \mathbb{K}^3 &\longrightarrow & \mathbb{K}^3 \\
+        (x,y,z)&\longmapsto & (x \cdot \alpha, x^2y, xy^3z).
+    \end{IEEEeqnarray*}
+    Using
+    \begin{IEEEeqnarray*}{rCl}
+        f_1\colon \mathbb{K}^0 &\longrightarrow & \mathbb{K} \\
+        x &\longmapsto & \alpha,\\\\
+        f_2\colon \mathbb{K}^1 &\longrightarrow & \mathbb{K}\\
+        x &\longmapsto & x^2,\\ \\
+        f_3\colon \mathbb{K}^2 &\longrightarrow & \mathbb{K}\\
+        (x,y) &\longmapsto & xy^3.
+    \end{IEEEeqnarray*}
+    we can write this as $\tau(x,y,z) = (x \cdot  f_1 \circ \pi_1(x,y,z), y \cdot f_2\pi_2(x,y,z), z \cdot f_3\pi_3(x,y,z))$
+\end{example}
+% \begin{example}
+%     The skew shift can be written in this form as well:
+%     TODO
+% \end{example}
+
+\begin{theorem}[Beleznay Foreman]
+    Whenever $I = \eta$ for some $\eta  < \omega_1$,
+    then
+    \[
+    \{\overline{f} \in \mathbb{K}_I : E_I(\overline{f}) \text{ is distal, minimal and of rank$\eta$}\} % TODO rank = order
+    \]
+    is comeager in $\mathbb{K}_I$.
+    In particular such flows exist.
+\end{theorem}
+\begin{proof}[sketch]
+    \begin{itemize}
+        \item Distality:
+            For all $\overline{f} \in \mathbb{K}_I$,
+            the flow $E_I \overline{f}$is distal.
+            This is the same as for iterated skew shifts.
+            %  TODO since for $\overline{x}, \overline{y} \in \mathbb{K}^I$,
+            % $d(x_\alpha, y_\alpha) = d((f(\overline{x})_\alpha, (f(\overline{y})_\alpha))$.
+        \item Minimality:
+
+            Let $\langle E_n : n < \omega \rangle$
+            be an enumeration of a countable basis for $\mathbb{K}^I$.
+
+            For all $n$ let
+            \[
+            U_n \coloneqq  \{\overline{f} \in \mathbb{K}_I : \exists k \in \Z.~ f^k(\overline{1}) \in E_n\}
+            \]
+            where $f = E_I \overline{f}$ and $\overline{1} = (1,1,1,\ldots)$.
+
+            Beleznay and Foreman showed that $U_n$ is open
+            and dense for all $n$.
+
+            So if $\overline{f} \in  \bigcap_{n} U_n$, then $\overline{1}$
+            is dense in $\overline{x} \mapsto f(\overline{x})$.
+            Since the flow is distal, it suffices to show
+            that one orbit is dense (cf.~\yaref{thm:distalflowpartition}).
+            % TODO REF Distal flow can be decomposed into disjoint minimal flows
+
+        \item Let $\overline{f} = (f_i)_{i \in I} \in \mathbb{K}_I$.
+            Consider the flows we get from $(f_i)_{i < j}$
+            resp.~$(f_i)_{i \le j}$
+            denoted by $X_{<j}$ resp.~$X_{\le j}$.
+            We aim to show that $X_{\le j} \to X_{<j}$
+            is a maximal isometric extension for comeagerly many $\overline{f}$.
+
+            The following open dense sets are used to make sure that all isometric extensions
+            are maximal and hence the order of the flow is $\eta$:
+
+            Fix a countable dense set $(\overline{x_n})$ in $\mathbb{K}^I$.
+            For $\epsilon \in \Q$ let
+            \begin{IEEEeqnarray*}{rCl}
+                V_{j,m,n,\epsilon} &\coloneqq & \{\overline{f} \in \mathbb{K}_I : \\
+                &&\text{if } \Pi_{j+1}(\overline{x}_n) = \Pi_{j+1}(\overline{x_m}),\\
+                &&\text{then there are $k_m$, $k_n$, $\overline{z}$ such that}\\
+                &&\pi_j(\overline{x_n}) = \pi_j(\overline{z}), \forall k> j+1.~z_k = 1,\\
+                &&d(f^{k_m}(\overline{x_m}), f^{k_m}(\overline{z})) < \epsilon \text{ and }\\
+                &&d(f^{k_n}(\overline{x_n}), f^{k_n}(\overline{z})) < \epsilon
+                &&\} 
+            \end{IEEEeqnarray*}
+            Beleznay and Foreman show that this is open and dense.%
+            \footnote{This is not relevant for the exam.}
+            % TODO similarities to the lemma used today
+    \end{itemize}
+\end{proof}
+
+
+

logic3.tex

@@ -46,6 +46,7 @@
 \input{inputs/lecture_19}
 \input{inputs/lecture_20}
 \input{inputs/lecture_21}
+\input{inputs/lecture_22}
 
 \cleardoublepage