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\begin{document}
\providecommand{\M}{\mathcal{M}}
\providecommand{\N}{\mathcal{N}}
\providecommand{\A}{\mathcal{A}}
\providecommand{\R}{\mathbb{R}}
\providecommand{\Aut}{\operatorname{Aut}}
\providecommand{\Iso}{\operatorname{Iso}}
\providecommand{\Bij}{\operatorname{Bij}}
\providecommand{\supp}{\operatorname{supp}}
\providecommand{\id}{\operatorname{id}}
\providecommand{\MALG}{\operatorname{MALG}}
\renewcommand{\d}{\partial}
\providecommand{\symdiff}{\Delta}

\author{Martin Bays}
\title{Notes on \em{Polish topometric groups}}
\maketitle

Notes from a seminar series on automatic continuity, M\"unster, Nov 2017,
on Ben Yaacov, Berenstein, Melleray \em{Polish topometric groups}.

\section{$\Aut([0,1],\lambda)$}

Let $\MALG$ be the Lebesgue measure algebra
\begin{align*} \MALG &:= \{ X \subseteq  [0,1] : X \textrm{Lebesgue measurable} \} /
  (\lambda(X\symdiff Y)=0) \\
    &= \{ X \subseteq  [0,1] : X \textrm{Borel} \} /
  (\lambda(X\symdiff Y)=0) .
\end{align*}

Let $\Aut([0,1],\lambda) := \Aut((\MALG;\lambda, \vee , \neg))$.
This has two natural topologies:
\begin{itemize}\item $\Aut([0,1],\lambda)_p$: topology of pointwise convergence,
  i.e.\ the topology inherited from the product topology on $\MALG^{\MALG}$;
\item $\Aut([0,1],\lambda)_u$: topology of uniform convergence,
  defined by the metric
  \[ \d(\alpha,\beta) := \sup_{x \in M} \lambda(\alpha(x)\symdiff\beta(x)) .\]
\end{itemize}

\begin{fact}
    $f \mapsto  (X \mapsto  f(X))$ 
    induces a group isomorphism
    \[ \Bij_\lambda([0,1]) / (\lambda(\supp(fg^{-1}))=0) \xrightarrow{\sim} \Aut([0,1],\lambda) ,\]
    where $\Bij_\lambda([0,1])$ is the group of measure-preserving bijections.
\end{fact}

\begin{theorem}
  Any homomorphism from $\Aut([0,1],\lambda)_p$ to a separable topological
  group is continuous.
\end{theorem}
\begin{proof}[Outline of proof]
  \begin{enumerate}[(I)]\item (Ryzikov,Kittrell-Tsankov,BBM): Using a result of Ryzikov that
    every element of $\Aut([0,1],\lambda)$ is a product of 3 involutions,
    $\Aut([0,1],\lambda)_u$ has the (38)-Steinhaus property and so has this
    automatic continuity property;
  \item (Kechris-Rosendal): The automorphism group of the countable dense
    measure subalgebra $\A \subseteq  \MALG$ generated by dyadic rational intervals
    has ample generics;
  \item (BBM):
    Hence $\Aut([0,1],\lambda)$ has ample generics as a ``topometric
    structure'', and so the automatic continuity of $\Aut([0,1],\lambda)_u$
    lifts to $\Aut([0,1],\lambda)_p$.
  \end{enumerate}
\end{proof}

\section{Metric structures}
\begin{definition}
  $\M = (M;d,(P_i)_{i \in I},(f_i)_{i \in I'})$ is a {\bf metric structure}
  if
  \begin{itemize}\item $(M;d)$ is a complete bounded metric space,
  \item each $P_i : M^{k_i} \rightarrow  \R$ is a uniformly continuous bounded map,
  \item each $f_i : M^{k'_i} \rightarrow  M$ is a uniformly continous map.
  \end{itemize}

  $\M$ is {\bf Polish} if $(M;d)$ is Polish (i.e.\ if $(M;d)$ is separable).

  An isomorphism of metric structures is an isometry preserving the
  predicates and functions.
\end{definition}

\begin{example}
  $\MALG := (\MALG;d, \vee , \neg)$ 
  with metric $d(A,B) := \lambda(A\symdiff B)$
  is a Polish metric structure.
\end{example}


\section{Topometric groups}
\begin{definition}
  A {\bf topometric space} $(X;\tau,\d)$ is a topological space $(X;\tau)$
  equipped with a metric $\d$ which refines the topology $\tau$,
  such that $\d$ is topologically lower semi-continuous,
  i.e.\ $\{(x,y) : \d(x,y) \leq  r\}$ is $\tau$-closed for $r \in \R$.
  
  A {\bf topometric group} $(G;\tau,\d)$ is a topometric space with a group
  structure such that $(G;\tau)$ is a topological group and $\d$ is
  bi-invariant.
\end{definition}

\begin{fact}
  If $(M;d)$ is a Polish metric space, then $\Iso((M;d))$, with the topology
  of pointwise convergence is a Polish group.
\end{fact}
\begin{proof}[Proof sketch]
  Say $A \subseteq  M$ is countable dense.
  Then $\Iso((M;d))$ is homeomorphic to the space of isometric embeddings of
  $A$ into $M$,
  which embeds as a closed subspace of $M^A$,
  which is Polish since it is the countable product of Polish spaces
  (explicitly: the product measure
  $\d(\alpha,\beta) := \Sigma_i 2^{-1} \gamma(d(\alpha(a_i),\beta(a_i))$,
  where $\gamma : \R_{\geq 0} \rightarrow  [0,1); x \mapsto  \frac{x}{1+x}$,
  is a complete metrisation of $M^A$, and $M^A$ is second countable by
  definition of the product topology).
\end{proof}
\begin{fact}
  Let $\M = (M;d,\ldots )$ be a Polish metric structure.
  Consider $\Aut(\M)$ with the topology $\tau$ of pointwise convergence.

  Then $\Aut(\M)$ is a closed subgroup of $\Iso((M;d))$, so is Polish.

  Let $\d^\M(\alpha,\beta) := \sup_{x \in M} d(\alpha(x),\beta(x))$
  be the metric of uniform convergence on $\Aut(\M)$.
  
  Then $\d^\M$ is bi-invariant, lower semi-continuous, and refines $\tau$.
\end{fact}
\begin{definition}
  For $\M$ as in the previous fact,
  we consider $\Aut(\M)$ as a Polish topometric group
  with the topometric structure $(\Aut(\M);\tau,\d^\M)$.
\end{definition}

\begin{example}
  $\Aut([0,1];\lambda) := \Aut(\MALG)$ as a topometric group.
  (So $\Aut([0,1];\lambda)_p$ is the $\tau$ topology,  
  $\Aut([0,1];\lambda)_u$ is the $\d$ topology.
\end{example}

\begin{remark}
  Any Polish group is isomorphic as a topological group to $\Aut(\M)$ for
  some Polish metric structure $\M$ in a countable language (and $\M$ can be
  taken approximately ultrahomogeneous).

  (No analagous statement given for Polish topometric groups).
\end{remark}

\section{Topometric groups with ample generics}

\begin{definition}
  A Polish topometric group $(G,\tau,\d)$ has {\bf ample generics} if for each
  $n$, the diagonal conjugacy action on $G^n$ has an orbit with comeagre
  $\d$-closure.

  %A Polish group $(G,\tau)$ has {\bf ample metric generics} if for some $\d$,
  %$(G,\tau,\d)$ is a topometric group with ample generics.
\end{definition}

\begin{theorem} \label{liftCont}
  If $(G;\tau,\d)$ is a topometric group with ample generics
  and $\phi : (G;\d) \rightarrow  H$ is a morphism of topological groups,
  where $H$ is separable
  (or, more generally, $H$ has uniform Souslin number\footnote{
    the {\bf uniform Souslin number} of a group $G$ is the least
    $\kappa$ such that if $1 \in V \subseteq _{op} G$ then $\kappa$ left-translates
    of $V$ cover $G$.}
  $\leq  2^{\aleph_0}$),
  then $\phi : (G;\tau) \rightarrow  H$ is continuous.
\end{theorem}

The proof is analagous to Kechris-Rosendal.

\begin{remark}
  Metric version of the small index property: if $(G;\tau,\d)$ is a Polish
  topometric group with ample generics, then any $\d$-closed subgroup of
  index $<2^{\aleph_0}$ is open.
\end{remark}

\section{Lifting ample generics}

\begin{definition}
    If $(X;\tau,\d)$ is a topometric space, $A \subseteq  X$, and $\epsilon>0$, define
    \[ (A)_\epsilon := \{ x : \d(x,A) < \epsilon \} .\]

    $A \subseteq  X$ has {\bf open enlargements} in $X$
    if $(A)_\epsilon \subseteq  X$ is open for any $\epsilon>0$.
\end{definition}

\begin{definition}
  Let $\M=(M;d,\ldots )$ be a Polish metric structure. 
  A {\bf good countable approximating substructure} is a classical countable
  structure $\N$ such that
  \begin{enumerate}[(i)]\item the universe $N$ of $\N$ is a dense subset of $(M;d)$;
  \item any automorphism of $\N$ extends (necessarily uniquely) to an
    automorphism of $\M$;
  \item with this embedding, $\Aut(\N) \subseteq  \Aut(\M)$ is dense;
  \item Consider $\Aut(\N)$ with the topology of pointwise convergence where
    $N$ has the discrete topology.

    If $U \subseteq _{op} \Aut(\N)$ is open as a subset of $\Aut(\N)$,
    then $U$ has open enlargements in $\Aut(\M)$.
  \end{enumerate}
\end{definition}

\begin{lemma} \label{AApprox}
  Let $\A = (A; \vee , \neg) \subseteq  \MALG$ be the measure subalgebra generated
  as a boolean algebra by the dyadic intervals, being the subintervals of
  $[0,1]$ whose endpoints are dyadic rationals (i.e.\ of form $m2^{-n}$).
  
  Then $\A$ is a good countable approximating substructure.
\end{lemma}

\begin{theorem} \label{liftAmple}
  Suppose $\N$ is a good countable approximating substructure of a Polish
  metric structure $\M$, and the Polish group $\Aut(\N)$ has ample generics.
  Then the Polish topometric group $\Aut(\M)$ has ample generics.
\end{theorem}

\subsection{Banach-Mazur}

Let $X$ be a topological space, $A \subseteq  X$.
The {\bf Banach-Mazur game} for $A \subseteq  X$ is as follows:
\begin{itemize}\item Players P1 and P2 alternate making moves.
\item At any time, the {\bf game state} is a non-empty open subset of $X$.
\item a move consists of playing a non-empty open subset of the current game state;
  the game state is then replaced by this set.
\item The initial game state is $X$, and P1 plays first.
\item P2 {\bf wins} a game whose successive game states are $X = V_0 \supseteq V_1 \supseteq \ldots $
  if $\bigcap_{i \in \omega} V_i \subseteq  A$.
\end{itemize}


A {\bf winning strategy} for P2 is a deterministic choice of what to play on a
given P2 move given the history of the game up to that point,
such that if P2 plays according to the strategy, then P2 wins however
P1 plays.

\begin{theorem}
  P2 has a winning strategy iff $A$ is comeagre.
\end{theorem}
\begin{proof}[Proof in the case that $X$ is second countable]
  \newcommand{\B}{\mathcal{B}}
  (Adapted from a math.stackexchange.org post by Andreas Blass.)

  Suppose $A$ is comeagre, so say $A \supseteq \bigcap_{i \in \omega} U_i$ with
  $U_i$ open dense.
  Then the following is a winning strategy for P2:
  if on P2's $i$th move the state is $V$, play $V \cap U_i$.

  Conversely, suppose P2 has a winning strategy.
  Let $\B$ be a countable base for $X$;
  we may assume $\emptyset  \notin \B$.

  Suppose it is P1 to move,
  and the game state is $V$,
  so far P1 has always played sets from $\B$,
  and P2 has played according to the strategy.
  Let $K \subseteq  V$ be the set of points $k \in V$ such that
  for any set from $\B$ which P1 could play,
  P2's response (playing according to P2's strategy) doesn't contain $k$.

  Then $K$ is nowhere dense:
  if it were dense in an open,
  then it would be dense in some $V \supseteq U \in \B$,
  but then $K$ would intersect P2's response to $U$.

  Now there are countably many such $K$, and they cover $X \setminus A$:
  if $x \in X \setminus A$ is not covered by the $K$ then P1 can play such that $x$
  survives throughout the game,
  contradicting P2's strategy being winning.
\end{proof}

\subsection{Lifting comeagreness}

\begin{theorem}[BBM] \label{liftComeagre}
  Let $(X,\tau,\d)$ be a Polish topometric space.
  Let $Y \subseteq  X$ be a dense subset equipped with a Polish topology refining
  the subspace topology, with complete metric $d_Y$.
  Suppose that any open (i.e.\ $d_Y$-open) $U \subseteq  Y$
  has open enlargements in $X$,
  and so does any open (i.e.\ $\tau$-open) $U \subseteq  X$.

  Let $A \subseteq  Y$ be comeagre in $Y$.
  Then $\overline{A}^{\d}$ is comeagre in $X$.
\end{theorem}
\begin{proof}
  Say $A \supseteq \bigcap_{1 \leq  n < \omega} O_n$ with $O_n \subseteq  Y$ open dense in $Y$.
  Since $Y$ is dense in $X$, also $O_n$ is dense in $X$.

  $\overline{A}^{\d} = \bigcap_n (A)_{\frac 1n}$,
  so it suffices to show that for any $\epsilon>0$,
  $(A)_{2\epsilon}$ is comeagre in $X$.
  We give a winning strategy for P2 in the Banach-Mazur game for
  $(A)_{2\epsilon} \subseteq  X$.
  
  Set $W_0 := Y$.
  On P2's $i$th move ($i\geq 1$), if the game state is $U_i$,
  the strategy tells P2 to
  choose $W_i \subseteq  Y$ an open non-empty $d_Y$-ball of radius $< 2^{-i}$
  such that $W_i \subseteq  W_{i-1} \cap (U_i)_\epsilon \cap O_i$,
  then play $V_i := U_i \cap (W_i)_\epsilon$.

  \begin{claim}
    Such a $W_i$ exists.
  \end{claim}
  \begin{proof}
    $A_i := W_{i-1} \cap (U_i)_\epsilon$ is non-empty;
    this holds for $i=1$ because $W_0=Y$ is dense,
    and for $i>1$ because $\emptyset  \neq  U_i \subseteq  (W_{i-1})_\epsilon$ by the rules of 
    the game.

    Now $A_i$ is open in $Y$ since $U_i$ has open enlargements and the topology
    on $Y$ refines the subspace topology,
    so $A_i$ intersects the dense open $O_i$ in a non-empty open set.
  \end{proof}

  Now $V_i$ is open since $W_i$ has open enlargements,
  and is non-empty since $\emptyset  \neq  W_i \subseteq  (U_i)_\epsilon$.

  Now say $x \in \bigcap U_i$ for a game where P2 plays according to this
  strategy.
  Then for each $i$,
  $x \in U_{i+1} \subseteq  (W_i)_\epsilon$,
  so say $y_i \in W_i$ with $\d(x,y_i) < \epsilon$.
  Let $y := \lim_i^{d_Y} y_i$.
  Then $A \supseteq \bigcap O_n \supseteq \bigcap W_i = \{y\}$.

  Now by topological lower semi-continuity of $\d$,
  $\d(x,y) \leq  \epsilon$. So $x \in (A)_{2\epsilon}$.
  So the strategy is winning.
\end{proof}

\subsection{Proof of Theorem~\ref{liftAmple}}
\begin{proof}[Proof of Theorem~\ref{liftAmple}]
  Let $A \subseteq  \Aut(\N)^n$ be a comeagre orbit of the diagonal conjugation
  action of $\Aut(\N)$.
  Let $A' \subseteq  \Aut(\M)^n$ be the orbit of the diagonal conjugation action of
  $\Aut(\M)$ containing $A$.

  It suffices to show that the conditions of Theorem~\ref{liftComeagre} hold
  for the pair $\Aut(\N)^n \subseteq  \Aut(\M)^n$, since then $\overline{A'}^\d \supseteq
  \overline{A}^\d$ is comeagre as required.

  \begin{claim}
    The topology on $\Aut(\N)^n \subseteq  \Aut(\M)^n$ refines the subspace topology.
  \end{claim}
  \begin{proof}
    Assume $n=1$; the same proof goes through in general.

    Let $U \subseteq  \Aut(\M)$ be a subbasic open set of the form
    \[ U = \{ \beta \in \Aut(\M) : d(\beta(a),b) < \epsilon \} .\]

    Let $\beta \in U \cap \Aut(\N)$.
    Let $a' \in N$ such that $r := \epsilon - d(\beta(a'),b) > 0$
    and $d(a,a') < \frac r2$.

    Let $V := \{ \beta' \in \Aut(\N) : d(\beta'(a'),\beta(a')) <
    \frac r2 \} \subseteq _{op} \Aut(\N)$.
    Then $\beta \in V$, and $V \subseteq  U \cap \Aut(\N)$
    since if $\beta' \in V$ then
    \begin{align*} d(\beta'(a),b) &\leq 
    d(\beta'(a),\beta'(a')) + d(\beta'(a'),\beta(a')) + d(\beta(a'),b) \\
        &= d(a,a') + d(\beta'(a'),\beta(a')) + d(\beta(a'),b) \\
        &< \frac r2 + \frac r2 + (\epsilon - r) \\
        &= \epsilon .\end{align*}
    So $U\cap\Aut(\N) \subseteq _{op} \Aut(\N)$.
  \end{proof}

  \begin{claim}
    Any open subset $U \subseteq  \Aut(\M)^n$ has open enlargements.
  \end{claim}
  \begin{proof}
    $(U)_\epsilon = (\{1\})_\epsilon \cdot U$.
  \end{proof}

  The other conditions follow immediately from the definition of a good
  approximating substructure.
\end{proof}

\section{Automatic contiuity for $\Aut([0,1],\lambda)_p$}
\begin{fact}[Kechris-Rosendal]
  $\A$ is the Fraïss\'e limit of the class of finite measured boolean algebras
  where the measure takes dyadic rational values.
\end{fact}
\begin{proof}[Proof of Lemma~\ref{AApprox}]
  \begin{enumerate}[(i)]\item
    By regularity of Lebesgue measure, every $X \in \MALG$ is of the form $X = 
    \bigwedge_{i<\omega} U_i$ where $U_i$ is open,
    so $X = \bigwedge_{i<\omega} \bigvee_{j<\omega} I_{ij}$ where $I_{ij}$ is a dyadic 
    interval.

  \item Extend $\sigma \in \Aut(\A)$ to $\MALG$ by continuity,
    $\sigma \lim_i X_i := \lim_i \sigma X_i$ if $X_i \in \A$.

    Then $\sigma$ preserves $\lambda$, and $\sigma$ preserves boolean 
    operations since they are continuous,
    e.g.\ $\sigma \neg \lim_i X_i = \sigma \lim_i \neg X_i = \lim_i \sigma \neg 
    X_i = \lim_i \neg \sigma X_i = \neg \lim_i \sigma X_i = \neg \sigma \lim_i 
    X_i$.
  
  \item
    It suffices to show that if $\beta \in \Aut(\MALG)$
    and $B_1,\ldots ,B_n \in \MALG$
    and $\epsilon>0$,
    then there exists $\beta' \in \Aut(\A)$ with
    $d(\beta(B_i),\beta'(B_i)) < \epsilon$.

    We may assume $B_1,\ldots ,B_n$ are the atoms of a finite subalgebra of $\MALG$.

    Take approximations from below $B_i \supseteq B_i' \in \A$ for $i>1$
    with $\lambda(B_i \setminus B_i') < \epsilon/n$.
    Set $B_1' := \neg (\bigvee_{i>1} B_i') \in \A$, so $\lambda(B_1' \setminus B_1) <
    \epsilon$.

    Similarly obtain disjoint approximations $C_i' \in \A$ to $\beta(B_i)$
    with $\lambda(B_i') = \lambda(C_i')$.

    Then $B_i' \mapsto  C_i'$ extends to an isomorphism of the generated finite dyadic
    rational subalgebras, so by homogeneity extends to $\beta' \in \Aut(\A)$.
  \item
    In fact we show this not for $\d$ but for the equivalent metric $\d'$
    \[ \d'(\alpha,\beta) := \lambda(\supp(\alpha^{-1}\beta)) .\]

    Take $U \subseteq  \Aut(\A)$ a basic open neighbourhood of $\id$,
      \[ U = \{ \alpha : \bigwedge_{i=1}^n \alpha(B_i) = B_i \} .\]
    WLOG $B_1,\ldots ,B_n$ are the atoms of a finite subalgebra of $\A$.
    \begin{claim}
      \[ (U)_\epsilon =
      \{ \beta :
        \Sigma_{i=1}^n \lambda(B_i \setminus \beta^{-1}(B_i)) < \epsilon \}
        =: U'_\epsilon .\]
    \end{claim}
    \begin{proof}
      $\subseteq $ is clear.

      For the converse, note
      $\lambda(B_i \setminus \beta^{-1}(B_i)) = \lambda(B_i \setminus \beta(B_i))$;
      indeed, 
      \begin{align*} \lambda(B_i \setminus \beta^{-1}(B_i))
      &= \lambda(B_i) - \lambda(B_i \cap \beta^{-1}(B_i)) \\
      &= \lambda(\beta^{-1}(B_i)) - \lambda(B_i \cap \beta^{-1}(B_i)) \\
      &= \lambda(\beta^{-1}(B_i) \setminus B_i) \\
      &= \lambda(B_i \setminus \beta(B_i)) .\end{align*}

      Let $\beta \in U'_\epsilon$. Define $\beta'$ setting it on $B_i \setminus
      \beta^{-1}(B_i)$ to be an arbitrary isomorphism with $B_i \setminus
      \beta(B_i)$, and otherwise to agree with $\beta$. Then $\beta' \in U$.
      So $\beta \in (U)_\epsilon$ by definition of $U'_\epsilon$ and of $\d'$.
    \end{proof}
    So $(U)_\epsilon$ is open, as required.
  \end{enumerate}
\end{proof}
\begin{fact}[Kechris-Rosendal]
  $\Aut(\A)$ has ample generics.
\end{fact}

\begin{fact}
  Any homomorphism from $\Aut([0,1],\lambda)_p$ to a separable group is
  continuous.
\end{fact}

\begin{corollary}
  The topometric group $(\Aut(\MALG);\tau,\d')$ has ample generics,
  hence by Theorem~\ref{liftCont},
  any homomorphism from $\Aut([0,1],\lambda)_u$ to a separable group is
  continuous.
\end{corollary}
\end{document}
