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\title[Group actions]
{Group actions and\\
and countable models}
\author{Greg Hjorth}
\revauthor{Hjorth, G.}
\address{Department of Mathematics\\
UCLA\\
Los Angeles CA90095-1555, USA}
\email{greg@math.ucla.edu}
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\begin{document}
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\maketitle
\tableofcontents
\section{Introduction}
The field of Polish group actions has emerged in the last couple
of years as a kind of
sub-sub-discipline in its own right.
To some extent one can think of the study of countable models as
being a special case of a Polish group action -- since the infinite
symmetric group, consisting of all permutations of the natural
numbers, is a Polish group whose action on countable models gives
rise to the isomorphism relation as its orbit equivalence relation.
In these talks I will try to emphasize the connections between
recent work on Polish groups and some basic concepts one might see in a first
course in logic or model theory.
I will begin the discussion at a very general level, which
hopefully might make some sense to almost any kind of logician.
Of course this means that the initial pace will seem tedious to
an expert. On the other hand as the paper progresses I want to increasingly
mention connections to other areas and recent results
and include stick figure sketches of their
proofs. Perhaps the later parts of the paper will only be completely
clear for someone who already knows much of this material or is willing
to spend hours in the library digging up references.
Thus the criticism may be raised that the level of exposition is inconsistent;
and by the end perhaps no one will be completely happy.
But short of turning this paper into a book that problem strikes me as
an inevitable consequence of its goals.
My own outlook on the history of this sub-area would really be to
place {\bf The descriptive set theory of Polish group actions},
Becker-Kechris[1996], as being central. This book represented the intersection
of two seemingly unrelated schools of thought.
On the one hand was a sequence of papers treating topics in what one
might call ``invariant descriptive set theory." Perhaps the most well known
papers here are those by Vaught, in which the development of the notion
of Vaught transform led to unified proofs for
results by Ryll-Nardzewski (every orbit is Borel) and Lopez-Escobar (invariant
Borel sets in the space of countable structures are described
by ${\cal L}_{\omega_1, \omega}$ sentences).
For this Vaught was awarded the first Karp prize.
A decade later, in work that was completed in the 1980's but not published
until 1994, Sami used Vaught's work to obtain results on changing
topologies for the ``logic action" of $S_\infty$ and showed that the topological
Vaught conjecture holds for abelian Polish groups. (More on this below.)
The second sequence of papers comes from researchers in operator algebras and the study of
infinite dimensional group representations. Although the topic of Polish
group actions for its own sake was somewhat
incidental to the main direction of their
research, the group responsible for
conjugacy on infinite dimensional representations is the infinite dimensional
unitary group and thus Polish but not locally
compact. By the time of Effros[1982] one has a completely general
result for Polish groups acting on Polish spaces, largely divorced from
the original context of Effros[1965], Glimm[1961], and
Mackey[1957].
%\newpage
%\begin{figure}[ht]
%{\psfig{figure=people.ps}}
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\begin{figure}
$$\xymatrix@=10pt{
& & *+[F]{\txt{ Mackey}} \ar[d] &
\\
& & *+[F]{\txt{ (Glimm) }} \ar[dd] &
\\
*+[F]{\txt{(Ryll-Nardzewski)[1964]}} \ar[dd] & & &
\\
& & *+[F]{\txt{Effros[1965, 1982]}} \ar[ddddl]
&
\\
*+[F]{\txt{Vaught[1974, 1975]}}\ar[d] & &
&
\\
*+[F]{\txt{Miller[1977, 1978]}}\ar[d] & &
&
\\
*+[F]{\txt{Sami[1994]}}\ar[dr] & &
&
\\
& *+[F]{\txt{Becker-Kechris[1996]}} \ar[d] & &
\\
& *+[F]{\txt{Lots of people (Becker,\\ Camerlo, Dougherty,\\
Gao, Farah, Kechris,\\ Louveau, Sami,\\ Solecki, and others) }} &
}$$
\end{figure}
Both these traditions were very much in the mind of Becker and Kechris
by the time of
{\bf The descriptive set theory of Polish group actions}. I have drawn up
a small flow chart to try to illustrate how these papers might
be viewed as interrelating, but it goes without saying that this
interpretation is personal and subjective. Certainly there are papers
off to one side of this chart (such as Friedman-Stanley[1989] and
Burgess[1978]) which have proved extremely influential, but do not
easily fall into the cartoon like summary of the history I have tried
to give above.
\section{Dynamic changes in topologies}
The main issue here is to discuss some of the tricks that
enable us to change topologies and remain in the category
of Polish spaces.
\subsection{Polish spaces and Polish groups [definitions; the
statement of the topological Vaught conjecture]}
\begin{Definition} A topological space is said to be {\it Polish}
if it is separable and there is some complete metric which generates
its topology.
\end{Definition}
\no{\bf Examples} (i) $\R$, the reals equipped with the
topology generated by the open intervals.
(The usual
euclidean distance gives a complete metric.)
(ii) $C([0, 1])$, the continuous functions from the unit
interval to $\R$ with the topology generated by the sup
norm, $d(f_1, f_2)=$ sup$_{0\leq x\leq 1}|f_1(x)-f_2(x)|.$
(iii) $\N$ in the discrete topology. As a metric,
\leftskip 0.6in
\noindent $d(x_1, x_2)=1$ if $x_1\neq x_2$,
\noindent $\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=0$ if $x_1=x_2$.
\leftskip 0in
(iv) $\N^{\N}$ in the product topology.
(v) More generally if $(X_i)_{i\in\N}$ is a
sequence of Polish spaces, then their infinite
product
\[\prod_\N X_i\]
is a Polish space in the product topology.
\begin{proof} After possibly replacing each
metric $d_i$ with
\[\hat{d}_i(x, y)=\frac{d_i(x, y)}{d_i(x, y)+1}\]
we may assume that each $d_i$ is bounded by 1.
Then we can obtain a compatible complete metric
on the product given by
\[d(\vec x, \vec y)=\sum_{i\in \N}2^{-i}d_i(x_i, y_i)\]
where
\[\vec x=(x_0, x_1, x_2, \ldots, x_i, \ldots)\]
\[\vec y= (y_0, y_1, y_2, \ldots, y_i, \ldots)\]
are both in $\prod_\N X_i$.
\end{proof}
\begin{lemma} \label{G_d}
Let $X$ be a Polish space and $A\subset X$ a $G_{\delta}$
subset -- that is to say, defined by a countable intersection
of open sets. Then $A$ in the subspace topology is again a
Polish space.
\end{lemma}
\begin{proof} First for open ${\cal O}\subset X$.
Let $d$ be complete metric for $X$, and find continuous
\[f: {\cal O}\rightarrow \R\]
with $f(x)\rightarrow \infty$ as $x\rightarrow X\setminus {\cal O}$.
(For instance, $f(x)=$ inf$\{\frac{1}{d(x, c)}: c\notin {\cal O} \}$.)
Then define $d_{\cal O}$ on ${\cal O}$
by
\[d_{\cal O}(x, y)=d(x, y)+|f(x)-f(y)|.\]
Suppose $(x_i)$ is a sequence in
${\cal O}$ that is $d_{\cal O}$-Cauchy. From the definition
of $d_{\cal O}$ it must be $d$-Cauchy, and hence by assumption that
$d$ witnesses $X$ Polish, there is a limit $x_\infty$. But we must
also have that $(f(x_i))_{i\in\N}$ will be Cauchy in $\R$, and thus
$x_\infty\in {\cal O}$.
For the general case
\[A=\bigcap_{i\in\N}{\cal O}_i\]
we can take
\[f_n: {\cal O}_n\rightarrow\R\]
as above and let
\[d_A(x, y)=d(x, y)+\sum_{n\in\N} 2^{-n}|f_n(x)-f_n(y)|.\]
\end{proof}
\begin{Definition} A topological group is said to be {\it Polish} if
it is Polish as a space.
\end{Definition}
\begin{examples} (i) $(\R, +)$, the reals equipped with its
usual additive structure.
(ii) Any discrete group.
Countable products of discrete groups, and more generally the class
of Polish groups is closed under countable products.
(iii) $S_\infty$ the group of all permutations of $\N$, equipped with the
topology of pointwise convergence. So as a basis we may take all
sets of the form
\[\{\pi\in S_\infty: \pi(\ell_1)=k_1, ..., \pi(\ell_n)=k_n\}.\]
(iv) $U_\infty$ the unitary group of $\ell^2$, infinite dimensional
(separable) Hilbert space. Thus $U_\infty$ is the group of all
bijective linear
operators
\[T:\ell^2\rightarrow \ell^2\]
such that for all $\xi, \zeta\in \ell^2$
\[\langle \xi, \zeta\rangle=\langle T(\xi), T(\zeta)\rangle.\]
We give this the topology generated by the subbasis consisting of
sets of the form
\[\{T\in U_\infty: |\langle T(\xi), \zeta\rangle -
\langle T_0(\xi), \zeta\rangle|<\epsilon\}.\]
\end{examples}
\begin{definition} Let $G$ be a Polish group. We say that a Polish space
$X$ is a {\it Polish $G$-space} if it is a Polish space which is equipped
with a continuous action by $G$.
We then let $E_G$ denote the orbit equivalence relation: $x_1 E_G x_2$ if
\[\exists g\in G(g\cdot x_1=x_2).\]
For $x\in X$ let $[x]_G$ indicate the orbit of $x$:
\[\{y\in X: \exists g\in G(g\cdot x=y)\}.\]
$X/G$ indicates the set of orbits.
\end{definition}
\begin{examples} (i) Let $\Z$ act on the complex unit
circle $=\{e^{ix}: x\in \R\}$ in the
following manner: For $e^{ix}$ and $k\in\Z$ we let
\[k\cdot e^{ix}=e^{i(x+2\pi k\surd 2)}.\]
In other words the action is induced by a
rotation of $\surd 2$ radians. In this way, among many
others, we could view the complex unit circle as a Polish $\Z$-space.
(ii) Given a countable set $X$ we can consider the space
$(U_\infty)^X$ of function from $X$ to
the unitary group of Hilbert space; this
space is Polish since it is naturally isomorphic to a countable
product of $U_\infty$ with itself. We can let $U_\infty$ act on
this space by point-wise conjugation:
\[(T\cdot f)(x)=T\circ f(x)\circ T^{-1}\]
for any $x\in X$, $f\in (U_\infty)^X$.
$U_\infty$ is a Polish group and under this action $(U_\infty)^X$
is a Polish $U_\infty$-space.
(iii) Perhaps the most natural example for a logician of a
Polish group acting on a Polish space is the action of
$S_\infty$ on the space of countable models. There are a few
fiddley points here related to precisely which topology we would
want to place on this space, so I will postpone the details to
section \ref{2.2}.
\end{examples}
\begin{conjecture}{\bf Topological Vaught Conjecture}\footnote{
This conjecture is very much still open.}:
Whenever $G$ is a
Polish group and $X$ a Polish $G$-space, then either
\begin{enumerate}
\item $X/G\leq \aleph_0$ (countably many orbits\footnote{In general
I use
``countably many" to mean
either finitely many or exactly $\aleph_0$ many.}), or
\item $X/G=2^{\aleph_0}$ (continuum many orbits).
\end{enumerate}
\end{conjecture}
This conjecture should be compared with its counterparts for
first order logic and countably infinitary logic.
\begin{conjecture} {\bf Original Vaught Conjecture} Let $T$ be a first
order theory. Then either $T$ has countably many countable models up to isomorphism
or it has continuum many countable models up to isomorphism.
\end{conjecture}
\begin{definition} For ${\cal L}$ a language we
let ${\cal L}_{\omega_1, \omega}$ be the collection of
formulas obtained by closing under countable disjunctions and
conjunctions, as well as the usual first operations of existential
and universal quantification, substitutions, and negation.
\end{definition}
\begin{conjecture} {\bf Vaught Conjecture for ${\cal L}_{\omega_1, \omega}$} Let
${\cal L}$ be a countable language and $\sigma$ a sentence in
${\cal L}_{\omega_1, \omega}$. Then either $\sigma$ has countably many countable
models up to isomorphism
or it has continuum many countable models up to isomorphism.
\end{conjecture}
As stated here, the topological Vaught conjecture is trivially true
under CH (the continuum hypothesis). Frequently people use a reformulation
of this conjecture that is equivalent under the negation of CH by Burgess' theorem
\ref{burgess}
below. And in fact this equivalence is local, action by action, not just
global: For each Polish group $G$ and Polish $G$-space $X$, if CH fails then
there are perfectly many orbits if and only if there are continuum many.
In fact I will find it convenient to refer to the following
as {\it the} topological Vaught conjecture.
\begin{conjecture}{\bf Topological Vaught Conjecture}
(Modern form):
Whenever $G$ is a
Polish group and $X$ a Polish $G$-space, then either
\begin{enumerate}
\item $X/G\leq \aleph_0$ (countably many orbits), or
\item there is a perfect set $P\subset X$ (that is to say, a
set which is closed, non-empty, and has no isolated points) such that
any two points in $P$ are orbit inequivalent under the action of
$G$.
\end{enumerate}
\end{conjecture}
Since any perfect set necessarily has size $2^{\aleph_0}$ it
is easily seen that this alternative form of the topological
Vaught conjecture implies the earlier. In fact it turns out
that these two conjectures are equiprovable over ZFC -- one can
prove the first using only the axioms of ZFC if and only if
one can prove the second using only the axioms of ZFC. This
equiprovability follows by a short absoluteness argument. If the
stronger form of the topological Vaught conjecture failed, then
we can add lots of reals to blow up the size of the continuum;
in the resulting model there must still be uncountably many orbits,
by
Shoenfield absoluteness; but equally there cannot be a perfect set
of orbit inequivalent points, again by Shoenfield absoluteness; and
thus we arrive at a contradiction with the local equivalence of the
two forms of the topological Vaught conjecture under $\neg$CH.
We sketch this argument due to Sami
at \ref{borel} below.
The proof
of this equivalence for the model theoretic versions of the Vaught
conjecture has also been written up precisely in
Steel[1978].
One way to think of this alternate form of the topological Vaught conjecture
is this: It is the conjecture that there cannot be ``exactly $\aleph_1$
many orbits", where we understand ``exactly" to mean that there are
$\aleph_1$ many orbits but not perfectly many orbits.
It might be helpful before continuing with the discussion of
the painfully open topological Vaught conjecture to look at
some parallel problems.
One can pose similar kinds of questions for other classes of equivalence
relations, and in almost every case the answer is known.
I will need a couple more definitions from descriptive set theory.
If you have not seen these before they may appear technical, but
the motivation is a natural one: We wish to describe certain classes of
equivalence relations which are not intractably complicated and compare
them with the kind that arise for Polish group actions.
\begin{definition} A subset $B$ of a Polish space $X$ is said to
be {\it Borel} if it appears in the smallest collection ${\cal B}$ of
subsets of $X$ such that
\begin{enumerate}
\item ${\cal B}$ includes the open sets in $X$;
\item ${\cal B}$ is an algebra -- that is to say it is closed under
complementation in $X$ and finite unions and intersections;
\item moreover ${\cal B}$ is a $\sigma$-algebra -- that is to say, an
algebra which is also closed under countable intersections.
\end{enumerate}
(Thus the collection of
{\it the Borel sets} is the smallest $\sigma$-algebra
containing the open sets.)
An equivalence relation $E\subset X\times X$ is {\it Borel} if it is Borel
in the product topology.
A subset $A$ of a Polish space $X$ is said to be {\it analytic}
or $\Ubf {\Sigma}^1_1$ if there is a Polish space $Y$ and
a Borel set $B\subset X\times Y$ (that is to say, Borel in the product topology)
with
\[A=\{x\in X| \exists y\in Y((x, y)\in B)\}.\]
$P\subset X$ is $\Ubf {\Pi}^1_1$ or {\it co-analytic} if its complement is
analytic.
\end{definition}
Clearly every Borel set is analytic, and then since the Borel sets are
closed under complementation we equally have that every Borel set is
co-analytic. It is not hard to see that if a Polish group $G$ acts continuously
on a Polish space $X$ then the set
\[\{(x_1, g, x_2)| g\cdot x_1=x_2\}\]
is closed in $X\times G\times X$, and thus the resulting orbit equivalence
relation $E_G$ is analytic.
In the context of model theory Borel sets have an intuitive meaning. They
correspond roughly to sets which can be defined by sentences in
${\cal L}_{\omega_1, \omega}$. (See corollary \ref{lopez} below.) Analytic sets
can be thought of as like a generalization of the set of models in a
language which admit an expansion to a model of some given theory in a larger
language.
\begin{theorem} (Silver[1980]) Let $E$ be a Borel (or even $\Ubf {\Pi}^1_1$)
equivalence relation on a Polish space $X$. Then either
\begin{enumerate}
\item $X/E\leq \aleph_0$, or
\item $X/E=2^{\aleph_0}$ (continuum many equivalence classes),
and in fact in this
case there is a perfect set of $E$-inequivalent points in $X$.
\end{enumerate}
\end{theorem}
This would be very nice for those of us who slave away trying to prove the
topological Vaught conjecture if it were the case that the equivalence
relations arising from Polish group actions were always Borel. Unfortunately
while they must be analytic they can in general be non-Borel.
\begin{theorem} (Burgess[1978])
\label{burgess} Let $E$ be a $\Ubf {\Sigma}^1_1$ equivalence
relation on a Polish space $X$.
\begin{enumerate}
\item $X/E\leq \aleph_1$, or
\item $X/E=2^{\aleph_0}$, and in fact in this
case there is a perfect set of $E$-inequivalent points in $X$.
\end{enumerate}
\end{theorem}
Burgess' theorem is optimal. If one considers the space $2^\N$ and sets
$x_1Ex_2$ if $\omega_1^{{\rm ck}(x_1)}=\omega_1^{{\rm ck}(x_2)}$
(the supremum of the lengths of the $x_1$-recursive well orders of $\N$
equals the supremum of the lengths of the $x_2$-recursive well orders of $\N$)
then we obtain an equivalence relation which has exactly $\aleph_1$
many classes and is in the
complexity class $\Ubf {\Sigma}^1_1$.
By Cohen we know that $2^{\aleph_0}$ may be bigger than $\aleph_1$, and
thus we cannot hope to extend Burgess' result in ZFC.
Thus any proof of the topological Vaught conjecture must use more than
the simple fact that orbit equivalence relations being considered are
analytic. It is known from Ryll-Nardzewski[1964] that the
orbit equivalence
relations arising from continuous actions of Polish groups on Polish spaces
are a special kind of $\Ubf {\Sigma}^1_1$ equivalence relation in
which every orbit is equivalence class is Borel,
but unfortunately the counterexample
above with
$\aleph_1$ many equivalence classes is also one in which every
class is Borel.
It does not seem as if any easily stated fact about the kinds
of equivalence relations arising from Polish group actions would allow
some general descriptive set theoretic proof that there cannot
be exactly $\aleph_1$ many orbits. The proof, if the conjecture is
indeed true, would need to use very specific facts about the context of a
Polish group action. It is unlikely to be soft.
\subsection{The space of countable models [isomorphism of countable models
viewed as a kind of Polish group action; different
Polish topologies we may place on
spaces of countable models]}
\label{2.2}
\begin{definition} Let ${\cal L}$ be a countable language, and for notational
simplicity assume that ${\cal L}$ is generated by a single binary relation
$R$. We then let Mod$({\cal L})$ be the space of all ${\cal L}$-structures
whose underlying set is $\N$, equipped with the topology generated by sets of the
form
\[\{{\cal M}\in {\rm Mod}({\cal L}): {\cal M}\models \varphi(k_1, \ldots, k_n)\}\]
where $\varphi$ is quantifier free and $k_1, \dots, k_n$ are in $\N$.
There is then a natural homeomorphism
\[{\rm Mod}({\cal L})\rightarrow \{0, 1\}^{\N\times \N}\]
\[{\cal M}\mapsto \chi_{R^{\cal M}}\]
associating to each ${\cal M}$ the characteristic
function of its interpretation of $R$.
Thus Mod$({\cal L})$ is a Polish space.
We let $S_\infty$ act on Mod$({\cal L})$ by
\[\pi\cdot {\cal M}\models R(k_1, k_2)\] if and only if
\[{\cal M}\models R(\pi^{-1}(k_1), \pi^{-1}(k_2)).\]
Then Mod$({\cal L})$ is a Polish $S_\infty$ space and the orbit
equivalence relation is just the usual isomorphism relation.
\end{definition}
Our next goal is to see that the topological Vaught conjecture
really does generalize the model theoretic versions of the Vaught
conjecture. To do this we need to consider alternative Polish
topologies we may place on the space of countable models.
\begin{definition}
(i) Let $\tau_{\rm FO}$ be the topology on ${\rm Mod}({\cal L})$ whose
basic open sets have the form
\[\{{\cal M}\in {\rm Mod}({\cal L}): {\cal M}\models \varphi(k_1, \ldots, k_n)\}\]
where $\varphi$ is first order and $k_1, \dots, k_n$ are in $\N$.
(ii) For $F\subset {\cal L}_{\omega_1, \omega}$ a countable fragment
(closed under subformulas, instantiations, and first order operations),
let $\tau_F$ be the topology on ${\rm Mod}({\cal L})$ whose
basic open sets have the form
\[\{{\cal M}\in {\rm Mod}({\cal L}): {\cal M}\models \varphi(k_1, \ldots, k_n)\}\]
where $\varphi\in F$.
\end{definition}
All these
provide Polish topologies on ${\rm Mod}({\cal L})$. (Proof for (i) below).
They are all finer that the topology of quantifier free formulas.
They all give rise to the same Borel structure as the topology of
quantifier free formulas, which I will continue to use as my default
choice of topology on the space.
Thus for any theory $T$ the set
\[\{{\cal M}\in {\rm Mod}({\cal L}): {\cal M}\models T\}\]
is a closed $S_\infty$-invariant subspace of $({\rm Mod}({\cal L}),
\tau_{\rm FO})$, and hence a Polish $S_\infty$ space in its own right.
Sets of the form
$\{{\cal M}\in {\rm Mod}({\cal L}): {\cal M}\models T\}$ correspond exactly to
the closed invariant subspaces; the subspace is minimal iff $T$ complete.
A key step in the proof of Becker's
theorem (which is coming up in
section \ref{3.3}) is to pass from a not so good topology,
like the quantifier free topology, to a better topology, like $\tau_{\rm FO}$.
\begin{lemma} (Gregorczyk, Mostowski, Ryll-Nardzewski[1960])
$({\rm Mod}({\cal L}),
\tau_{\rm FO})$ is a Polish space.
\end{lemma}
\begin{proof} Let $B$ be the set of substitutions of the form
$\varphi(k_1, \ldots, k_n)$ where $\varphi$ is a first order ${\cal L}$ formula.
Let $X=\{0, 1\}^B$, the space of all functions from $B$ to $\{0, 1\}$.
We may naturally identify $({\rm Mod}({\cal L}),
\tau_{\rm FO})$ with the functions in $f\in X$ which preserve the ``inductive
unraveling of truth" -- for instance, $f(\varphi_1(k)\wedge \varphi_2(k))=1$ if and
only if $f(\varphi_1(k))=f(\varphi_2(k))=1$. Most of these requirements correspond
to closed conditions in $X$, except for the rule governing existential quantifiers,
\[f(\exists x\varphi(k, x))=1\Leftrightarrow\exists l\in \N f(\varphi(k, l))=1{\rm ,}\]
which is still no worse than $G_\delta$.
Thus $({\rm Mod}({\cal L}),
\tau_{\rm FO})$ is homeomorphic to a $G_\delta$ subspace of a Polish space and
therefore Polish.
\end{proof}
In many respects $\tau_{\rm FO}$ would seem to be a nicer topology than
the topology generated by quantifier free formulas.
This becomes more apparent after we
introduce the notion of {\it Vaught transform} below and
note that $\tau_{\rm FO}$ has a basis that is closed under the Vaught
transforms.
The next subsection discusses an important theorem due to Becker and Kechris which states that
among other things that for any Polish $G$-space there are ``cofinally" many
Polish topologies in which that action is still continuous and that are
closed under the Vaught transforms. This was arguably the main result of
Becker-Kechris[1996]. Sami[1994] had previously shown the same thing for
a general class of continuous actions of $S_\infty$, including its natural action
on $({\rm Mod}({\cal L}),
\tau_{\rm FO})$.
\subsection{Becker-Kechris theorem [Vaught transforms;
adding to the topology of a Polish
$G$-space and keeping the action continuous]}
\begin{definition} Let $G$ be a Polish group and $X$ a Polish $G$-space.
For $V\subset G$ open and $B\subset X$ (preferably Borel), we let
\[B^{\Delta V}\] be the set of
$x\in X$ for which there is a non-meager\footnote{A set is
meager if it is included in a countable union of closed sets each
of which contain no non-empty open subset.} set of $g\in V$ for which
$g\cdot x\in B$. We symmetrically define $B^{* V}$
to be the set of $x\in X$ for which there is a relatively
comeager set of $g\in V$ with $g\cdot x\in B$.
\end{definition}
\begin{lemma} (Vaught[1973]) For $\varphi\in{\cal L}_{\omega_1, \omega}$, $B$ the set
of ${\cal M}\in {\rm Mod}({\cal L})$ with $ {\cal M}\models \varphi(k_1, \ldots, k_n,
a_1, \ldots, a_m)\}{\rm ,}$
and
$V=\{\pi\in S_\infty: \pi(a_1)=b_1,\ldots, \pi(a_m)=b_m\}$,
\[B^{\Delta V}=\{{\cal M}\in {\rm Mod}({\cal L}): {\cal M}\models \exists x_1\ldots\exists x_n
\varphi(\vec x,\vec b)\}{\rm ,}\]
\[B^{* V}=\{{\cal M}\in {\rm Mod}({\cal L}): {\cal M}\models \forall x_1\ldots \forall x_n
\varphi(\vec x, \vec b)\}{\rm .}\]
\end{lemma}
Using this one can show by an induction on
the Borel complexity of $B$ that:
\begin{corollary} (Lopez-Escobar[1965]) \label{lopez}
If $B\subset {\rm Mod}({\cal L})$
is Borel and invariant then there is a sentence $\varphi\in
{\cal L}_{\omega_1, \omega}$ with
\[B=\{{\cal M}\in {\rm Mod}({\cal L}): {\cal M}\models \varphi\}.\]
\end{corollary}
\begin{theorem} (Becker-Kechris[1996]) \label{change}
Let $G$ be a Polish group and $X$ a Polish $G$-space. Let $B\subset X$ be
Borel. Let ${\cal B}$ be a countable basis for $G$.
Then there is a Polish topology $\tau$ on $X$ such that:
\begin{enumerate}
\item $\tau$ is richer than the original topology on $X$;
\item for any $V\in {\cal B}$
\[B^{\Delta V}\in \tau{\rm ;}\]
\item $\tau$ has a countable basis ${\cal C}$ such that for all $U_1, U_2\in {\cal C}$
and $V_1, V_2\in {\cal B}$,
\[(U_1\cap (X\setminus U_2)^{*V_1})^{\Delta V_2}\in {\cal C}{\rm ;}\]
{\rm (This rather ominous
looking condition plays a central role in the proof of Becker's theorem
\ref{beckertheorem} below; it represents an approximation to the
basis having the kind of closure conditions presented by the
natural basis for $\tau_{\rm FO}$ on Mod$({\cal L})$.)}
\item the action of $G$ on $X$ is still continuous with respect to $\tau$
(and thus $(X, \tau)$ is still a Polish $G$-space).
\end{enumerate}
\end{theorem}
In the case that $B$ is invariant under the action of $G$ we obtain that
$B^{\Delta G}=B^{*G}=B$, and thus $B$ is an invariant clopen subset of $(X,\tau)$.
In particular, since for any sentence $\sigma\in {\cal L}_{\omega_1, \omega}$
we have that
\[\{{\cal M}: {\cal M}\models \sigma \}\]
is Borel and invariant, Vaught's conjecture for ${\cal L}_{\omega_1, \omega}$ is implied by
the topological Vaught conjecture for $S_\infty$.
It would take us too far afield to go into the details of their proof. Instead
I will present a lemma, which by a transfinite induction can be used to prove
the Becker Kechris theorem.
\begin{lemma} $X$ a Polish $G$-space, $V\subset G$ open and $C\subset X$ closed.
Then there is a richer Polish topology on $X$ for which the action is still continuous
and $C^{\Delta V}$ is now open.
\end{lemma}
\begin{proof} By a theorem of Birkhoff and Kakutani (see 1.1.1 Becker-Kechris[1996])
we may find a compatible right invariant metric
$d$ on $G$ that is bounded by 1. Let ${\cal L}(G)$ be the space of $f: G\rightarrow
[0, 1]$ such that for all $g_1, g_2 \in G$
\[|f(g_1)-f(g_2)|\leq d(g_1, g_2){\rm .}\]
Then ${\cal L}(G)$ is a Polish $G$-space under the natural action $(g\cdot f)(h)=
f(hg).$
For $x\in X$ let $f_x\in {\cal L}(G)$ be given by
\[f_x(g)={\rm \: inf}\{d(g, h): h\cdot x\notin C\}{\rm .}\]
$X_C=\{(x, f_x): x\in X\}$
is an invariant subset of $X\times {\cal L}(G)$ and $\pi_C:X\rightarrow X_C$
\[x\mapsto (x, f_x)\]
is a bijection which respects the $G$-action.
One then verifies, as in Hjorth[1999],
that $X_C$ is a $G_\delta$ subset of $X\times {\cal L}(G)$ and
thus a Polish $G$-space. The pullback of the topology on $X_C$ to $X$ is as
required.
\end{proof}
\section{Smoothness and the Glimm-Effros dichotomy}
\subsection{Smoothness [Effros lemma; $G_\delta$ orbits as the analogue
of atomic models]}
\begin{definition} An orbit equivalence relation $E_G$ on Polish
space $X$ is said to be {\it smooth} if there is a Borel
function\footnote{That is
to say, for any open $O\subset \R$ the set
$f^{-1}(O)$ appears in the $\sigma$-algebra on $X$ generated by the open
sets --i.e., $f^{-1}[O]$ is Borel.} $f: X\rightarrow \R$
such that for all $x_1, x_2\in X$
\[x_1 E_G x_2 \Leftrightarrow f(x_1)=f(x_2).\]
\end{definition}
Burgess[1979] showed that for
$E_G$ given by the
continuous action of a Polish group action we have
$E_G$ smooth if and only if there is a Borel set meeting every
orbit in exactly one point.
\begin{example} For ${\cal M}, {\cal N}\in {\rm Mod}({\cal L})$ set
\[{\cal M}\equiv {\cal N}\]
if they have the same (first order) theory.
Then $\equiv$ is smooth. To see this, let $(\varphi_n)_{n\in\N}$ enumerate
the first order sentences of ${\cal L}$. For ${\cal M}\in {\rm Mod}({\cal L})$
associate the real $x_{\cal M}\in (0, 1)$ in whose decimal expansion we have
a 5 at the $n^{\rm th}$ position if ${\cal M}\models \varphi_n$ and 6 if
it does not. ${\cal M}\mapsto x_{\cal M}$ is a Borel function witnessing
smoothness.
\end{example}
\begin{example} Let $\Z_2^{<\N}$ be the infinite binary sequences
which are eventually constantly 0. This is a countable set, and we
give it the discrete topological structure. We further
introduce the group structure
obtained by pointwise addition mod 2; thus for $\vec g, \vec h\in
\Z_2^{<\N}$ and $n\in\N$ we have
\leftskip 0.5in
\no $(\vec g +\vec h)(n)=0$ if $\vec h(n)=
\vec g(n)$ and
\no $\:\:\:\:\:\:\:\: \:\: \:\: \:\:\:\:\:=1$ otherwise.
\leftskip 0in
We then let it act on $2^\N$ in the natural manner: For $\vec x\in
2^\N=_{df}{\{0, 1\}}^\N$,
$\vec g\in \Z_2^{<\N}$, $n\in\N$, we again have $(\vec g\cdot \vec x)(n)=0$
if and only if $\vec x(n)=\vec g(n)$.
\end{example}
\begin{lemma} \label{non-smooth} The above orbit equivalence relation
$E_{\Z_2^{<\N}}$ is non-smooth.
\end{lemma}
\begin{proof} Suppose instead (by Burgess[1979]) $B$ is a
Borel selector. Let $\mu$ be the
product (``coin flipping") measure on $2^\N$. Note that $(2^\N, \mu)$ is a
probability space and ${\Z_2^{<\N}}$ acts by measure preserving
transformations on $(2^\N, \mu)$.
Since countably many translates of $B$ cover $2^\N$ we must have $\mu(B)>0$.
But then there are $\aleph_0$ many {\it disjoint} translates
of $B$, each having the same measure, and hence a contradiction to
$\mu(2^\N)=1$.
\end{proof}
There are many theorems giving necessary conditions for
non-smoothness in terms of being able to {\it embed} this orbit equivalence
relation $E_{\Z_2^{<\N}}$. For a wide variety of groups
(but not $S_\infty$!)
one can show
that $E_G$ is smooth if and only if we cannot so embed this canonical
example of a non-smooth equivalence relation.
For instance:
\begin{theorem} (Becker-Kechris) \label{E_0} Let $G$ be a Polish
group and $X$ a Polish $G$-space, $x\in X$.
Suppose that no orbit $[y]_G$ with the same closure as $[x]_G$
is
comeager in
\[\overline{[x]_G}{\rm ,}\]
the closure of the orbit.
Then
\begin{enumerate}
\item $E_G$ is not smooth and
\item there is a continuous injection
\[\rho: 2^\N\rightarrow X\]
such that for all $\vec x, \vec y\in 2^\N$
\[\vec x E_{\Z_2^{<\N}} \vec y\Leftrightarrow (\rho(\vec x) E_G
\rho(\vec y)).\]
\empty
\end{enumerate}
\end{theorem}
In fact 2. implies 1. in the statement of the theorem.
After this Becker-Kechris theorem one might conclude that it
would be nice to understand when an orbit is comeager in its own
closure. Here a theorem was proved some 30 years earlier by Effros:
\begin{theorem} (Effros[1965]) Let $G$ be a Polish group and $X$ a Polish
$G$-space. Then for $x\in X$ the following are equivalent:
\begin{enumerate}
\item $[x]_G$ is comeager in $\overline{[x]_G}$;
\item $[x]_G$ is $G_\delta$;
\item the map
\[G\rightarrow [x]_G\]
\[g\mapsto g\cdot x\]
is open.
\end{enumerate}
\end{theorem}
In the case of the logic action of $S_{\infty}$ on
$({\rm Mod}({\cal L}), \tau_{\rm FO})$ -- the ${\cal L}$-structures
on $\N$ with the topology generated by first order logic --
one has that the $G_\delta$ orbits correspond to atomic
models, see Miller[1978]. (The omitting types theorem shows that the
realization of any non-principal type is meager.)
\subsection{Examples [algebraic
characterization of discrete groups with
smooth dual (Thoma); Bernoulli shifts up to isomorphism
are smooth (Ornstein); but not general mpt's (Feldman)]}
\begin{example} As before let $U_\infty$ be the infinite
dimensional unitary group, consisting of all isomorphisms
of Hilbert space. For $H$ a countable (discrete) group we define
Rep$(H)$ to be the space of all homomorphism
\[\rho: H\rightarrow U_\infty.\] This may naturally be
identified with a closed subspace of
the $H$-fold product of $U_\infty$,
\[\prod_H U_\infty{\rm ,}\]
and thus is a Polish space.
We let $U_\infty$ act on Rep$(H)$ by ``pointwise
conjugation",
\[(T\cdot \rho)(h)=T\circ\rho (h) \circ T^{-1}.\]
We then let Irr$(H)$ be the {\it irreducible} representations
in Rep$(H)$ -- that is to say those $\rho$ for which
there are no non-trivial closed subspaces of Hilbert space
closed under the operators $\{\rho(h): h\in H\}$.
It turns out that Irr$(H)$ is a $G_\delta$ subset of Rep$(H)$,
and thus a Polish $U_\infty$-space in its own right.
\end{example}
\begin{theorem} (Thoma[1964]) \label{thoma}
The orbit equivalence $E_{U_\infty}$ on {\rm Irr}$(H)$ is smooth
if and only if $H$ is abelian-by-finite.
\end{theorem}
One can extend the definition of Rep$(H)$ and Irr$(H)$ to
a general locally compact group. Here one demands that the
representations all be continuous, and we give these spaces
the ``compact-open" topology, generated by sets of the
form
\[\{\rho: \rho[C]\subset O\}{\rm ,}\]
where $C\subset H$ is compact and $O\subset U_\infty$ is open.
Again we obtain that Rep$(H)$ and Irr$(H)$ are $U_\infty$
Polish spaces.
\begin{theorem} (Harish-Chandra[1953], more or less\footnote{I
fib terribly. In fact he proved that the representations of
semi-simple connected Lie groups are all ``type I". While at the time it
was apparently felt that this should turn into a proof of the smoothness
of $E_{U_\infty}|_{{\rm Irr}(H)}$, this does not seem to have been
confirmed until the proof of the ``Mackey conjecture" a few years
later by Glimm}) If $H$ is a semi-simple
connected Lie group, then the
orbit equivalence $E_{U_\infty}$ on {\rm Irr}$(H)$ is smooth.
\end{theorem}
\begin{example} Let $M_\infty$
be the group of measure preserving transformations
of the unit interval, subject to identification of transformations
agreeing almost everywhere. We give this the topology generated by sets
of the form
\[\{\pi\in M_\infty: \mu(\pi(A)\Delta \pi_0(A))<\epsilon\}{\rm ,}\]
for $\pi_0\in M_\infty$, $A\subset [0,1]$ measurable, $\epsilon>0$.
In this topology and the operation of composition it becomes a Polish
group.
It seems natural to think of measure preserving transformations as
really being ``isomorphic" if they are the same up to some isomorphism
of the underlying standard Borel probability space.
Thus we are led to the orbit equivalence
relation of $M_\infty$ acting on itself by conjugation:
\[\pi\cdot \rho=\pi\circ \rho\circ \pi^{-1}.\]
\end{example}
\begin{theorem} (Feldman\footnote{Actually something stronger and
rather more
specific is
shown in Feldman[1974]. He obtains non-smoothness even for the
property K transformations. See a Foreman[to appear] for a
thorough discussion of this and related results.}) The orbit equivalence relation
$E_{M_\infty}$ obtained above is not smooth.
\end{theorem}
For some special classes of measure preserving transformations
one obtains smoothness of the restriction of $E_{M_\infty}$. For instance
Ornstein showed that the isomorphism relation on {\it Bernoulli
shifts} is smooth and that {\it entropy} provides a complete
invariant.
\subsection{Becker's theorem [Glimm-Effros dichotomy, and hence
Vaught's conjecture, for solvable groups]}
\label{3.3}
\begin{definition} (Becker) A Polish group $G$ satisfies the
{\it Glimm-Effros dichotomy} if whenever $G$ acts continuously on a
Polish space $X$ either:
\begin{enumerate}
\item $E_G$ is smooth; or
\item there is a continuous injection
\[\rho: 2^\N\rightarrow X\]
such that for all $\vec x, \vec y\in 2^\N$
\[\vec x E_{\Z_2^{<\N}} \vec y\Leftrightarrow (\rho(\vec x) E_G
\rho(\vec y)).\]
\end{enumerate}
\end{definition}
It is not hard to see that the Glimm-Effros property for a group
$G$ implies that it does not provide a counterexample for the topological
Vaught conjecture.
\begin{lemma} Let $G$ be a Polish group and $X$ a Polish $G$-space.
If $G$ satisfies the Glimm-Effros dichotomy then $X$ does
not have exactly $\aleph_1$ many orbits.
\end{lemma}
\begin{proof} Let $X$ be a Polish $G$-space. We wish to show that
either there are perfectly many orbits or (at most)
countably many orbits.
Apply the definition of the Glimm-Effros
dichotomy. Suppose
first we are in case 2, and so we have
a continuous embedding
$\rho$ of $E_{\Z_2^{<\N}}$ into $E_G$. We can consider the compact set
\[C=\{\vec x| \forall n(n \: {\rm not\: a\: prime\: power \:}\Rightarrow x(n)=0) \]
\[\wedge \forall p \: {\rm prime}\: \forall k>0 (x(p)=x(p^k))\}.\]
Any two elements of $C$ are $ E_{\Z_2^{<\N}}$-inequivalent. $\rho[C]$ is again
compact and without isolated points. And thus $\rho[C]$ provides a perfect set
of
orbit inequivalent points.
Alternatively suppose we are in case 1. Then there is Borel
$G$-invariant
\[f: X\rightarrow \R{\rm ,}\] such that for all $x_1, x_2 \in X$
\[x_1E_G x_2 \Leftrightarrow f(x_1)=f(x_2).\] A routine calculation
shows that $E_G$ must consequently be Borel, and we can finish
by applying Silver's theorem.
\end{proof}
Ever expanding classes of groups have been shown
to satisfy the Glimm-Effros
dichotomy:
\begin{enumerate}
\item Locally compact Polish groups (Effros, after Glimm).
\item Abelian Polish groups (Solecki).
\item Nilpotent (Hjorth).
\item Solvable (Becker).
\end{enumerate}
Notably $S_\infty$ fails this dichotomy.
\begin{theorem} (Becker) Solvable Polish
groups satisfy the Glimm-Effros
dichotomy.
\label{beckertheorem}
\end{theorem}
\begin{proof} Let $G$ be a solvable Polish group and $X$ a Polish
$G$-space. Assume that 2. fails, that is to say, we are unable
to continuously inject $E_{\Z_2^{<\N}}$ into $E_G$.
Since $G$ is solvable it has a {\it complete} compatible
right invariant metric (Gao[1998], Hjorth-Solecki[1999]), call it $d_G$.
Following theorem \ref{change} we may assume that
we have a basis ${\cal C}$ for the topology
$\tau$ on $X$ and
${\cal B}$ on $G$ such that:
\begin{enumerate}
\item for all $U_1, U_2\in {\cal C}$
and $V_1, V_2\in {\cal B}$,
\[(U_1\cap (X\setminus U_2)^{*V_1})^{\Delta V_2}\in {\cal C}{\rm ;}\]
\item the action of $G$ on $X$ is still continuous with respect to $\tau$
(and thus $(X, \tau)$ is still a Polish $G$-space).
\end{enumerate}
It is not hard to show that there is a
Borel function
\[\theta: X\rightarrow \R\]
such that for all $x_1, x_2 \in X$
\[\theta(x_1)=\theta(x_2)\Leftrightarrow \overline{[x_1]_G}=
\overline{[x_2]_G}.\]
(We may code the closure of an orbit by a real number, and the coding
can be completed in a Borel manner.)
Thus it suffices to show that $\overline{[x]_G}$ is a complete
invariant of $[x]_G$; that is to say, $\overline{[x_1]_G}=
\overline{[x_2]_G}$ if and only if $[x_1]_G =[x_2]_G$.
Fix some $x\in X$.
Now by theorem \ref{E_0}, the Becker-Kechris theorem on embedding
$E_{\Z_2^{<\N}}$, we know that there must an
$[y]_G$ with the same closure as $[x]_G$ which is comeager in
$\overline{[x]_G}=\overline{[y]_G}$.
Using the Effros theorem, we can obtain for each open
\[V_n=\{g\in G: d_G(1_G, g)<2^{-n}\}\]
some open $U_n\in{\cal C}$ containing $y$ with diameter less than $2^{-n}$
such that
\[V_n\cdot y\supset U_n\cap [y]_G.\]
Then using the closure assumptions on the the basis ${\cal C}$
for the topology and a
calculation which we do not present here,
it can be argued that whenever $U\in {\cal C}$ and
\[U\cap U_n\cap [y]_G\neq\emptyset\]
\[\hat{x}\in U_n\cap [x]_G\]
then there are $g$ arbitrarily close to the identity
with
\[g\cdot \hat{x}\in U^{\Delta V_n^2},\]
and thus in particular we may find
$g,h\in V_n^2$ with $hg\cdot \hat{x}\in U$.
Repeating this observation infinitely often we may find
a sequence $(h_n)_{n\in\N}$ in $G$ and $(x_n)_{n\in\N}$
in $[x]_G$ with:
\begin{enumerate}
\item $x_n\in U_n$;
\item $h_n\in V_n^4$, and thus $d_G(h_n, 1_G)<2^{-n+2}$;
\item $x_{n+1}=h_n\cdot x_n$.
\end{enumerate}
We find $h_{n+1}$ by applying the preceding observation with
$U_{n+1}$ taking the place of the set $U$.
Thus if we let $g_n=h_nh_{n-1}\ldots h_1h_0$ then we have
$x_n=g_n\cdot x_0\in [x]_G$ and by 1. we have
\[x_n\rightarrow y.\]
On the other hand 2. and right invariance of the metric
gives
\[d_G(g_n, g_{n-1})< 2^{-n+2}\]
and thus $(g_n)$ is Cauchy. But for $g_\infty$ the limit
we obtain from continuity of the action,
\[g_\infty\cdot x_0=y.\]
Thus we have established that $[x]_G=[y]_G$ is comeager in
$\overline{[x]_G}$.
Generalizing we therefore have that whenever $\overline{[x_1]_G}=
\overline{[x_2]_G}$ we must have both $[x_1]_G$ and $[x_2]_G$ are
comeager in this common closure. Hence the orbits overlap. Hence
they are equal.
\end{proof}
\begin{example} $S_\infty$ does not satisfy
the Glimm-Effros dichotomy.
Let ${\cal L}$ be generated by binary function $+$, unary
function $-(\:)$, and constant symbol $0$. Let AG be the
subset of Mod$({\cal L})$ consisting of ${\cal G}$ in
which $(+^{\cal G}, -(\:)^{\cal G}, 0^{\cal G})$ defines an
abelian group structure on $\N$. Certainly this corresponds to
a first order definable property on ${\cal G}$, and so
AG is a closed invariant subspace of Mod$({\cal L})$.
Then for $p$ a prime
let
TAG$_p$ be the set of all $p$-groups -- that is to say, ${\cal G}$
such that for all $g\in {\cal G}$ there exists $n>0$ with
\[{\cal G}\models p^n\cdot g=0.\]
TAG$_p$ is an invariant $G_\delta$ set, and hence a Polish
$S_\infty$-space in its own right.
The Ulm analysis of abelian $p$-groups shows that we may find
a {\it reasonably nice} (e.g. absolutely $\Delta^1_2$ in the codes, in the
sense of Hjorth[2000])
function $U:{\rm TAG}_p\rightarrow 2^{<\omega_1}$
assigning bounded subsets of $\aleph_1$ as complete invariants.
{\bf Claim:} ${\rm TAG}_p/S_\infty$ is not smooth.
Roughly this follows because in general there can be no {\it reasonably
simply definable} (e.g. absolutely $\Delta^1_2$!) $\omega_1$ sequence
of reals.
{\bf Claim:} We cannot embed $E_{\Z_2^{<\N}}$ into
${\rm TAG}_p/S_\infty$.
Otherwise by postcomposing with $U$ we would obtain an
assignment $\hat{U}$ of bounded subsets of $\aleph_1$ as complete
invariants for $2^\N/{\Z_2^{<\N}}$. Then we would obtain some
fixed $\alpha<\omega_1$ and
$\mu$-measure\footnote{$\mu$ the ``coin flipping" measure
from before} one set on which the
reduction $\hat{U}$ takes values inside $2^\alpha$. (To see this,
we can take $(M; \in)$ some countable transitive model containing the real
parameter used in the embedding of $E_{\Z_2^{<\N}}$;
then there will be a $\mu$-measure 1 set of $x\in 2^\N$ that are
``random" generic for $M$ -- and for any such $x$ there will be some
$\alpha< \omega_1^M=\omega_1^{M[x]}<\omega_1^V$ with $\hat{U}(x)\in 2^\alpha$).
But then since $2^\alpha$ is naturally isomorphic to $2^\N$, and
hence Borel isomorphic to $\R$, we obtain a contradiction to our
proof before that $2^\N/{\Z_2^{<\N}}$ is non-smooth on any
$\mu$-measure one set.
(Here we have skipped over technicalities relating to the fact that
$\hat{U}$ is not necessarily Borel. The issue here however is that it
can be chosen to be universally measurable in the codes; hence
on a measure one set it will equal a Borel function, or alternatively
the proof of \ref{non-smooth} can be seen to work in the more
general context of $\mu$-measurable functions.)
\end{example}
\section{Vaught's conjecture on analytic sets}
\subsection{Variations on the conjecture}
\begin{notation} For $H$ a Polish group let TVC$(H)$
be the statement that whenever $X$ is a Polish $H$-space then
either there are only countably many orbits or there is a perfect set $P\subset X$
of orbit inequivalent points:
\[\forall x_1, x_2 \in P(x_1\neq x_2 \Rightarrow [x_1]_H\neq [x_2]_H).\]
Let WVC$(H)$ be the statement that when $X$ is a Polish $H$-space then
either there are only countably many orbits or there are $2^{\aleph_0}$ many
orbits.
\end{notation}
\begin{lemma} (Sami) \label{borel}
If $H$ is a Polish group and $X$ is a Polish $H$-space
witnessing the failure of TVC$(H)$, then $X$ continues to witness the
failure of TVC$(H)$ through all generic extensions.
\end{lemma}
\begin{proof}
{\bf Claim:} In no generic extension can $X$ contain a perfect set of
orbits.
Proof of Claim: The statement that there are perfectly many orbits may
be expressed as:
\[\exists P\subset X(P {\rm \: perfect\:} \wedge\]
\[\forall x_1, x_2 \in
P\forall g\in G(x_1\neq x_2\Rightarrow g\cdot x_1\neq x_2)).\]
This is $\Ubf{\Sigma}^1_2$, and hence absolute by Shoenfield.
\hfill (Claim$\square$)\footnote{This proof also works
to show that if a $\Ubf{\Sigma}^1_1$ $A$ does not contain a perfect
set of orbits then in no generic extension it will contain a
perfect set of orbits. The point here is that if $P$ is a
perfect subset of $A$ then by say an appeal to the Baire
category theorem and the Jankov-von Neumann uniformization we
may find a perfect $P_0\subset P$ and a continuous function
$f$ on $P_0$ such that for all $x\in P_0$ we that $f(x)$ witnesses
$x\in A$.}
{\bf Claim:} In every generic extension $X/H$ is uncountable.
Proof of Claim: For conceptual simplicity let us do the case
that
\[X=\{{\cal M}\in {\rm Mod}({\cal L}):{\cal M}\models T\}\]
and $H=S_\infty$. Then for any
${\cal M}\in$ Mod$({\cal L})$ we have that $[{\cal M}]_{S_\infty}$
is uniformly Borel in (any code for) the Scott sentence of ${\cal M}$.
Moreover, in some natural parameter space $Y$ we have that
\[{\cal S}=\{({\cal M}, y): y {\rm \: codes \: the \: Scott \: sentence
\: of \: }{\cal M}\}\]
is $\Ubf{\Pi}^1_1.$ Then the statement that the models of $T$ has
uncountably many isomorphism types among its countable models
becomes the statement that for all
sequences $({\cal M}_i), (y_i)$, either:
\begin{enumerate}
\item there exists $i$ with $({\cal M}_i, y_i)\notin {\cal S}$; or
\item there exists ${\cal N}\models T$ with ${\cal N}$ not satisfying any
of the sentences coded by the $\{y_i: i\in\N\}$.
\end{enumerate}
Thus it is $\Ubf{\Pi}^1_2$ and subject to Shoenfield absoluteness.
The general case for arbitrary Polish group actions
follows by working not with the Scott sentence, but
the {\it stabilizer} of the points in question; again $[x]_H$ is
uniformly Borel in (any code for) the stabilizer of $x$
(see the proof of 1.2.4, 7.1.2 Becker-Kechris[1996]).
\hfill (Claim$\square$)
\end{proof}
\begin{corollary} It is provable in ZFC that every Polish group $H$ satisfies
WVC$(H)$ (i.e. the original
topological Vaught conjecture) if and only
it is provable that every Polish group satisfies TVC$(H)$.
\end{corollary}
\begin{proof} $\Leftarrow$ since any perfect set has size $2^{\aleph_0}$.
$\Rightarrow$ since if we have a group $H$ and a Polish $H$-space $X$ showing
the failure of TVC$(H)$ then we may go to a forcing extension in
which $2^{\aleph_0}=\aleph_2$. In the forcing extension
Sami's lemma and
Burgess' theorem from \ref{burgess}
on analytic
equivalence relations imply that
we must have $|X/H|=\aleph_1$.
\end{proof}
\begin{notation} For $H$ a Polish group let
TVC$(H, \Ubf{\Sigma}^1_1)$ be the
statement that whenever $X$ is a Polish $H$-space and $A\subset X$ is $\Ubf{\Sigma}^1_1$
then either
\[A/H\leq \aleph_0\]
or there is perfect $P\subset A$ such that for all $x_1, x_2\in P$
\[x_1\neq x_2\Rightarrow [x_1]_H\neq [x_2]_H.\]
\end{notation}
Sami's argument also shows absoluteness of TVC$(H, \Ubf{\Sigma}^1_1)$.
\begin{lemma} (H. Friedman[1973]) Up to isomorphism there are
exactly $\aleph_1$ many possible order types for the ordinals
in countable $\omega$-models of Kripke-Platek set theory.
\end{lemma}
\begin{corollary} TVC$(S_\infty, \Ubf{\Sigma}^1_1)$ fails.
\end{corollary}
\subsection{Some sketches
around the proof for characterizing
Vaught's conjecture on analytic sets}
%\begin{definition} Let us say that $S_\infty$ {\it divides} a
%Polish group $H$ if there is a closed subgroup $K