Josia Pietsch
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172 lines
6 KiB
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\lecture{23}{2024-01-19}{More sketches of ideas of Beleznay and Foreman}
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\begin{notation}
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Let $X$ be a Polish space and $\cP$ a property of elements of $X$,
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then we say that $x_0 \in X$ is \vocab{generic}
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if
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\[
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A_\cP \coloneqq \{x \in X \colon \cP(x)\}
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\]
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is comeager
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and $x_0 \in A_\cP$.
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\end{notation}
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For example let $X = \mathbb{K}_I$
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and $\cP$ the property of being a distal minimal flow.
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\begin{abuse}
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We will usually omit $\cP$.
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\end{abuse}
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Let $I$ be a linear order
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\begin{theorem}[Beleznay and Foreman]
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The set of distal minimal flows is $\Pi_1^1$-complete.
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\end{theorem}
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\begin{proof}[sketch]
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Consider $\WO(\N) \subset \LO(\N)$.
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We know that this is $\Pi_1^1$-complete. % TODO ref
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Let
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\begin{IEEEeqnarray*}{rCll}
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S & \coloneqq & \{ x \in \LO(\N) :& x \text{ has a least element},\\
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&&& \text{for any $t$, there is $t \oplus 1$, the successor of $t$.}\}
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\end{IEEEeqnarray*}
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\todo{Exercise sheet 12}
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$S$ is Borel.
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We will % TODO ?
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construct a reduction
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\begin{IEEEeqnarray*}{rCl}
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M \colon S &\longrightarrow & C(\mathbb{K}^\N,\mathbb{K})^\N. %\\
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% \alpha &\longmapsto & M(\alpha)
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\end{IEEEeqnarray*}
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We want that $\alpha \in \WO(\N) \iff M(\alpha)$
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codes a distal minimal flow of rank $\alpha$.
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\begin{enumerate}[1.]
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\item For any $\alpha \in S$, $M(\alpha)$ is a code for
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a flow which is coded by a generic $(f_i)_{i \in I}$.
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Specifically we will take a flow
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corresponding to some $(f_i)_{i \in I}$
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which is in the intersection of all
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$U_n$, $V_{j,m,n,\frac{p}{q}}$
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(cf.~proof of \yaref{thm:distalminimalofallranks}).
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\item If $\alpha \in \WO(\N)$,
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then additionally $(f_i)_{i \in I}$ will code
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a distal minimal flow of ordertype $\alpha$.
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\end{enumerate}
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One can get a Borel map $S \ni \alpha \mapsto \{T_n^{\alpha} : n \in \N\}$,
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such that $T^{\alpha}_n$ is closed,
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$T^{\alpha}_n \neq \emptyset$, $\diam(T^\alpha_n) \xrightarrow{n \to \infty} 0$,
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$T^\alpha_{n+1} \subseteq T^\alpha_n$,
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$T^{\alpha}_n \subseteq W^{\alpha}_n$,
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where $W^{\alpha}_n$ is an enumeration of $U_m^\alpha$,$V^\alpha_{j,m,n,\frac{p}{q}}$.
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Then $(f_i)_{i \in I} \in \bigcap_{n} T_{n}^\alpha$.
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\end{proof}
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\begin{lemma}
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Let $\{(X_\xi, T) : \xi \le \eta\}$ be a normal
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quasi-isometric system
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and $\{(Y_i, T) : i \in I\}$
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such that
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\begin{enumerate}[(i)]
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\item $I \in S$ and additionally $I$ has a largest element.
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\item $Y_0$ is the trivial flow and $Y_\infty = X_\eta$,
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where $0$ and $\infty$ denote the minimal
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resp.~maximal element of $I$.
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\item $\forall i < j$
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% https://q.uiver.app/#q=WzAsMyxbMCwwLCIoWF9cXGV0YSwgVCkiXSxbMSwwLCJZX2oiXSxbMSwxLCJZX2kiXSxbMCwxLCJcXHBpX2oiXSxbMCwyLCJcXHBpX2kiLDJdLFsxLDIsIlxccGleal9pIl1d
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\[\begin{tikzcd}
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{(X_\eta, T)} & {Y_j} \\
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& {Y_i}
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\arrow["{\pi_j}", from=1-1, to=1-2]
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\arrow["{\pi_i}"', from=1-1, to=2-2]
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\arrow["{\pi^j_i}", from=1-2, to=2-2]
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\end{tikzcd}\]
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\item If $i \in I$ is a limit (i.e.~there does not exist
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an immediate predecessor),
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then $(Y_i,T)$ is the inverse limit
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of $\{(Y_j,T) : j < i\}$
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with respect to the factor maps.
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\item $(Y_{i\oplus 1}, T)$ is a maximal isometric
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extension of $(Y_i, T)$
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in $(X_\eta, T)$.
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\end{enumerate}
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Then $I$ is well-ordered with $\otp(Y) = \eta + 1$.
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\end{lemma}
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\begin{theorem}[Beleznay Foreman]
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The order %TODO (Furstenberg rank)
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is a $\Pi^1_1$-rank.
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\end{theorem}
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For the proof one shows that $\le^\ast$ and $<^\ast$
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are $\Pi^1_1$, where
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\begin{enumerate}[(1)]
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\item $p_1 \le^\ast p_2$ iff $p_1$ codes
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a distal minimal flow and if
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$p_2$ also codes a distal minimal flow,
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then $\mathop{order}(p_1) \le \mathop{order}(p_2)$.
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\item $p_1 <^\ast p_2$ iff $p_1$ codes
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a distal minimal flow and if
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$p_2$ also codes a distal minimal flow,
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then $\mathop{order}(p_1) < \mathop{order}(p_2)$.
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\end{enumerate}
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One uses that $(Y_{i+1}, T)$ is a maximal
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isometric extension of $(Y_i,T)$
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ind $(X,T)$
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iff for all $x_1,x_2$ from a fixed countable dense set
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in $X$,
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for all $i$ with $\pi_{i\oplus 1}(x_1) = \pi_{i \oplus 1}(x_2)$,
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there is a sequence $(z_k)$ such that $\pi_i(z_k) = \pi_i(x_1)$,
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$F(z_k, x_1) \to 0$, $F(z_k, x_2) \to 0$.
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\begin{proposition}
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The order of a minimal distal flow on a separable,
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metric space is countable.
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\end{proposition}
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\begin{proof}
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Let $(X,\Z)$ be such a flow,
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i.e.~ $X$ is separable, metric and compact.
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Produce a normal quasi-isometric system
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\[
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\{(X_\alpha, \Z) : \alpha \le \beta\}
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\]
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with $(X_\beta, \Z) = (X,\Z)$.
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We need to show that $\beta < \omega_1$.
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Let $\pi_\alpha\colon (X,\Z) \to (X_\alpha, \Z)$.
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Fix $x_0 \in X$.
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For every $\alpha$
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consider $\pi_\alpha^{-1}\left( \pi_\alpha(x_0) \right)
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= F_\alpha \overset{\text{closed}}{\subseteq} X$.
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\begin{itemize}
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\item For $\alpha_1 < \alpha_2 \le \beta$
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we have that $F_{\alpha_1} \supseteq F_{\alpha_2}$.
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\item For limits $\gamma \le \beta$,
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we have that $F_\gamma = \bigcap_{\alpha < \gamma} F_\alpha$,
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since $(X_\gamma,\Z)$ is the inverse limit of
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$\{(X_{\alpha}, \Z):\alpha < \gamma\}$.
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\item For all $\alpha < \beta$, $F_{\alpha+1} \subsetneq F_\alpha$,
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because $\pi^{\alpha+1}_\alpha \colon (X_{\alpha+1},\Z) \to (X_\alpha,\Z)$
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is not a bijection
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and all the fibers are isomorphic.
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\end{itemize}
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So $(F_\alpha)_{\alpha \le \beta}$ is a strictly
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increasing chain of closed subsets.
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But $X$ is second countable,
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so $\beta$ is countable:
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Let $\{U_n\} = \cB$ be a countable basis
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and for $\alpha$ let $U_\alpha \in \cB$
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be such that $U_\alpha \cap F_\alpha = \emptyset$
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and $U_\alpha \cap F_{\alpha+1} \neq \emptyset$.
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Then $\alpha \mapsto U_\alpha$ is an injection.
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\end{proof}
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