Josia Pietsch 458dd9ab1f
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We have the following situation:
% https://q.uiver.app/#q=WzAsNCxbMCwwLCJYIl0sWzEsMSwiWF97bn0iXSxbMSwyLCJYX3tuLTF9Il0sWzIsMSwiWSJdLFswLDEsIlxccGlfe259Il0sWzEsMiwiXFx0ZXh0e2lzb21ldHJpY30iLDFdLFswLDIsIlxccGlfe24tMX0iLDIseyJjdXJ2ZSI6Mn1dLFswLDMsIlxccGknIiwwLHsiY3VydmUiOi0zfV0sWzMsMSwiaCIsMix7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFszLDIsIlxcb3ZlcmxpbmV7Z30sIFxcdGV4dHsgbWF4LiBpc29tLn0iLDAseyJjdXJ2ZSI6LTJ9XV0=
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]
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,
$\rho$ 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))
R(a,b) \coloneqq \rho(\pi'(a), \pi'(b)).
\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
\[d(\tau^{m_k}x, \tau^{m_k} x_k) \xrightarrow{k \to \infty} 0.\]
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 $\pi'(x) = \pi'(x')$ $\lightning$.
\item $Y$ max.~isometric extension of $X_{n-1}$ in $X$
and $\overline{g} = \pi^n_{n-1} \circ h$.
\item $h$ isomorphism.
Suppose not, then $\exists y_0,y_1 \in X.~\pi'(y_0) \neq \pi'(y_1),
\pi_n(y_0) = \pi_n(y_1) = t$.
\item Apply \yaref{lem:lec20:1} $\leadsto$ sequence $(x_k)$ in $X$,
such that $\pi_{n-1}(x_k) = \pi_{n-1}(y_i)$,
$F(x_k,y_i) \to 0$.
\item $\rho\colon Y \times_{X_{n-1}} Y \to \R$ witnessing isometric.
\item $R(a,b) \coloneqq \rho(\pi'(a), \pi'(b))$ for $a,b \in X$ with
$\overline{g}(\pi'(a)) = \overline{g}(\pi'(b))$.
(defined for any two of $x_k$, $y_0$, $y_1$, $\tau$-equivariant)
\item $F(y_0,x_{k}) \to 0$, so $d(\tau^{m_k} y_0, \tau^{m_k} x_k) \to 0$.
\item $R(y_0,x_k) \to 0$, hence $\underbrace{R(y_0,y_1)}_{\text{no } k} \to 0$ $\lightning$.
\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 $\alpha < \omega_1$.
\todo{This was already stated as \yaref{thm:beleznay-foreman} 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
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}
Consider the following flow:
\tau\colon \mathbb{K}^3 &\longrightarrow & \mathbb{K}^3 \\
(x,y,z)&\longmapsto & (x \cdot \alpha, x^2y, xy^3z).
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.
we can write this as $\tau(x,y,z) = (x \cdot f_1 \circ \pi_1(x,y,z), y \cdot f_2\pi_2(x,y,z), z \cdot f_3\pi_3(x,y,z))$
The skew shift can be written in this form as well.
Consider $f_1\colon x \mapsto \alpha$
and $f_n\colon (x_0,\ldots, x_{n-2}) \mapsto x_{n-2}$.
\begin{theorem}[Beleznay Foreman]
Whenever $I = \eta$ for some $\eta < \omega_1$,
\{\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.
\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.
\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 $\overline{1} = (1,1,1,\ldots)$.
\textsc{Beleznay} and \textsc{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}).
\item The order of the flow is $\eta$:
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
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\\
Beleznay and Foreman show that this is open and dense.%
% TODO similarities to the lemma used today