Josia Pietsch bc8b5a8b6c
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\subsection{Sketch of proof of \yaref{thm:l16:3}}
\lecture{18}{2023-12-15}{Sketch of proof of \yaref{thm:l16:3}}
The goal for this lecture is to give a very rough
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.
For $x, x' \in X$ define
F(x,x') \coloneqq \inf \{d(gx, gx') : g \in G\}.
\item $F(x,x') = F(x', x)$,
\item $F(x,x') \ge 0$ and $F(x,x') = 0$ iff $x = x'$.
\item $F(gx, gx') = F(x,x')$ since $G$ is a group.
\item $F$ is an \vocab{upper semi-continuous}\footnote{%
Wikimedia has a nice \href{https://upload.wikimedia.org/wikipedia/commons/c/c0/Upper_semi.svg}{picture}.}
function on $X^2$,
i.e.~$\forall a \in R.~\{(x,x') \in X^2 : F(x,x') < a\} \overset{\text{open}}{\subseteq} X^2$.
This holds because $F$ is the infimum of continuous functions
f_g\colon X^2 &\longrightarrow & \R \\
(x,x') &\longmapsto & d(gx, gx')
for $g \in G$.
The sets
U_a(x) \coloneqq \{x' : F(x,x') < a\}
form the basis of a topology in $X$.
This topology is called the \vocab{F-topology} on $X$.
In this setting, the original topology
is also called the \vocab{E-topology}.
This will follow from the following lemma:
Let $F(x,x') < a$.
Then there exists $\epsilon > 0$ such that
whenever $F(x',x'') < \epsilon$, then $F(x,x'') < a$.
}{Then $\exists \epsilon > 0.~\forall x''.~F(x',x'') < \epsilon \implies F(x,x'') < a$.}
We have to show that if $U_a(x_1) \cap U_b(x_2) \neq \emptyset$,
then this intersection is the union
of sets of this kind.
Let $x' \in U_a(x_1) \cap U_b(x_2)$.
Then by \yaref{lem:ftophelper},
there exists $\epsilon_1 > 0$ with $U_{\epsilon_1}(x') \subseteq U_a(x_1)$.
Similarly there exists $\epsilon_2 > 0$\gist{
such that $U_{\epsilon_2}(x') \subseteq U_b(x_2)$.}{.}
So for $\epsilon \le \epsilon_1, \epsilon_2$,
we get $U_{\epsilon}(x') \subseteq U_a(x_1) \cap U_b(x_2)$.
\notexaminable{\footnote{This was not covered in class.}
% TODO: maybe learn?
Let $T = \bigcup_n T_n$,% TODO Why does this exist?
$T_n$ compact, wlog.~$T_n \subseteq T_{n+1}$, and
let $G(x,x') \coloneqq \{(gx,gx') : g \in G\} \subseteq X \times X$.
Take $b$ such that $F(x,x') < b < a$.
Then $U = \{(u,u') \in G(x,x') : d(u,u') < b\}$
is open in $G(x,x')$
and since $F(x,x') < b$ we have $U \neq \emptyset$.
There exists $n$ such that
\forall (u,u') \in G(x,x').~T_n(u,u')\cap U \neq \emptyset.
Suppose not.
Then for all $n$, there is $(u_n, u_n') \in G(x,x')$
\[T_n(u_n, u_n') \subseteq G(x,x') \setminus U.\]
Note that the RHS is closed.
For $m > n$ we have
$T_n(u_m, u'_m) \subseteq G(x,x') \setminus U$
since $T_n \subseteq T_m$.
By compactness of $X$,
there exists $v,v'$ and some subsequence
such that $(u_{n_k}, u'_{n_k}) \to (v,v')$.
So for all $n$ we have $T_n(v,v') \subseteq G(x,x') \setminus U$,
hence $T(v,v') \cap U = \emptyset$,
so $G(v,v') \cap U = \emptyset$.
But this is a contradiction as $\emptyset\neq U \subseteq G(v,v')$.
The map
T\times X&\longrightarrow & X \\
(t,x) &\longmapsto & tx
is continuous.
Since $T_n$ is compact,
we have that $\{(x,t) \mapsto tx : t \in T_n\}$
is equicontinuous.\todo{Sheet 11}
So there is $\epsilon > 0$ such that
$d(x_1,x_2) < \epsilon \implies d(tx_1, tx_2) < a -b$
for all $t \in T_n$.
Suppose now that $F(x', x'') < \epsilon$.
Then there is $t_0 \in T$ such that $d(t_0x', t_0x'') < \epsilon$,
hence $d(t t_0x', t t_0 x'') < a-b$ for all $t \in T_n$.
Since $(t_0x, t_0x') \in G(x,x')$,
there is $t_1 \in T_n$
with $(t_1t_0x, t_1t_0x') \in U$,
i.e.~$d(t_1t_0x, t_1t_0x') < b$
and therefore
$F(x,x'') = d(t_1t_0x, t_1t_0x'') < a$.
Now assume $Z = \{\star\}$.
We want to sketch a proof of \yaref{thm:l16:3} in this case,
i.e.~show that if $(Z,T)$ is a proper factor of a minimal distal flow
$(X,T)$ then there is another factor $(Y,T)$ of $(X,T)$
which is a proper isometric extension of $Z$.
\begin{proof}[sketch] % TODO: Think about this
\item For $x \in X$ define
F_x\colon X &\longrightarrow & \R \\
x' &\longmapsto & F(x,x').
\item Define an equivalence relation on $X$,
by $x_1 \sim x_2 :\iff \{x \in X : F_{x_1}(x) = F_{x_2}(x)\}$
is comeager in $X$\footnote{with respect to the E-topology}.
Then for all $g \in G$ we have
$x_1 \sim x_2 \implies gx_1 \sim ~ gx_2$.
Let $M \coloneqq \{[x]_{\sim } : x \in X\} = \faktor{X}{\sim}$
bet the quotient space.
It is compact, second countable and Hausdorff.
Let $\pi\colon X\to M$ denote the quotient map.
\item $(Y,T) \mathbin{\text{\reflectbox{$\coloneqq$}}} (M,T)$
is an isometric flow:
\item For $a > 0$, $x,x' \in X$ let
W(x,x') \coloneqq \{g \in G : F(x, gx') < a\}.
This turns out to be a subbasis of a topology
which is coarser than the original topology on $G$.
The new topology makes $G$ compact.
\item Let $\theta(g)$ be the transformation of $M$
defined by $\theta(g) \pi(x) = \pi(gx)$.
This is well defined.
Let $H = \theta(G)$.
This is just a quotient of $G$, $g \mapsto \theta(g)$
may not be injective.
\item One can show that $H$ is a topological group and $(M,H)$
is a flow.\footnote{This is non-trivial.}
\item Since $H$ is compact,
$(M,H)$ is equicontinuous, i.e.~it is isometric.
In particular, $(M,T)$ is isometric.
\item $M \neq \{\star\}$, i.e.~$(M,T)$ is non-trivial:
Suppose towards a contradiction that $M = \{\star\}$,
i.e.~$x_1 \sim x_2$ for all $x_1,x_2 \in X$.
Fix $x_2$. For every $x_1 \in X$
we have that
\{x : F(x_1,x) = F(x_2,x)\}
is comeager.
Let $x_1$ be a point of continuity of $F_{x_2}$.
Let $\langle a_n : n < \omega \rangle$ be a sequence
of elements that set, i.e.~$F(x_1, a_n) = F(x_2, a_n)$,
such that $a_n \to x_1$.
So by the continuity of $F_{x_2}$ at $x_1$
\lim_{n \to \infty} F(x_2, a_n) &=& F(x_2, x_1)
and by the definition of $F$
\lim_{n \to \infty} F(x_1,a_n) &=& F(x_1,x_1) = 0.
F(x_2,x_1) = \lim_{n \to \infty} F(x_2, a_n) = \lim_{n \to \infty}
F(x_1,a_n) = 0
and by distality we get $x_1 = x_2$.
Since almost all points of $X$
are points of continuity of $F_{x_2}$
this implies that $X \setminus \{x_2\}$ is meager.
But then $X = \{\star\} \lightning$.
\begin{theorem}\footnote{Not covered in class}
Let $X$ be a metric space
and $\Gamma\colon X \to \R$ be upper semicontinuous.
Then the set of continuity points of $\Gamma$ is comeager.
Take $x$ such that $\Gamma$ is not continuous at $x$.
Then there is an $\epsilon > 0$
and $x_n \to x$ such that
$\Gamma(x_n) + \epsilon \le \Gamma(x)$.
Take $q \in \Q$ such that $\Gamma(x) - \epsilon < q < \Gamma(x)$.
Then let
B_q \coloneqq \{a \in X : \Gamma(a) \ge q\}.
$X \setminus B_q = \{a \in X : \Gamma(a) < q\}$
is open, i.e.~$B_q$ is closed.
Note that $x \in F_q \coloneqq B_q \setminus \inter(B_q)$
and $B_q \setminus \inter(B_q)$ is nwd
as it is closed and has empty interior,
so $\bigcup_{q \in \Q} F_q$ is meager.