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\begin{document}
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\title{\Huge The Smooth/Non-Smooth Distinction} % Enter your title between curly braces
\author{\LARGE Greg Hjorth\\ \LARGE Department of Mathematics\\ \LARGE UCLA CA90095-1555\\\LARGE
USA\\\LARGE greg@math.ucla.edu}
\date{\LARGE February 2001\\Rutgers} % Enter your date or \today between curly braces
\maketitle
\huge
\noindent{\bf $\S$ 0 Definitions}
\bigskip
%\footnote{\large This view has been attributed to
%G.W. Mackey's {\it Infinite dimensional group representations,}
%{\bf Bulletin of the AMS,} vol. 69(1963), pp. 628-686.}
\noindent {\bf 0.1 Definition} A topological space $X$ is said to be {\it Polish} if
it is separable and allows a complete metric. The {\it Borel sets} are those subsets
of $X$ appearing in the $\sigma$-algebra generated by the open sets.
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Frequently we are interested in question relating to the Borel structure alone. Thus a
{\it standard Borel space} is a Polish space stripped down to its Borel structure.
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In particular, $(X, {\cal B})$ is a standard Borel space if there is some complete
separable metric on $X$ which gives rise to ${\cal B}$ as its $\sigma$-algebra of Borel sets.
\newpage
\noindent {\bf 0.2 Examples:} (i) $\R$; $\C$; $C([0,1])$ in the sup norm topology
are Polish spaces; $C(\R)$ is not.
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(ii) $\N$ in the discrete topology is Polish; $\aleph_1$ is not.
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(iii) $L^2([0,1],{\rm Lebesgue})$, and indeed any separable Banach space is Polish.
\newpage
\noindent {\bf 0.3 Examples:} (i) $U_\infty$ the ``unitary group"
[i.e. collection of all transformations which respect the inner product]
of
Hilbert space is a Polish group, in the strong operator topology, with
subbasic open sets of the form $\{T: |\!|T (\xi)-\zeta|\! |<\epsilon\}$.
In fact, this is a {\it Polish group}: A topological group that is Polish as
a space.
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(ii) $M_\infty$, the group of measure preserving transformations of the unit
interval. This can be thought of as a closed subgroup of
$U_\infty(L^2([0,1],{\rm Lebesgue})$, the group of unitary transformations of
$L^2([0,1],{\rm Lebesgue})$; thus it is Polish by the last example.
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(iii) Group structures on $\N$ in the topology of pointwise convergence for
the group operations. (Subbasic open sets look like
$\{G: n\cdot^G m= k\}$.) Can be shown to be Polish.
\newpage
\noindent {\bf 0.4 Examples:} (i) Open subsets of the complex plane,
${\cal O}(\C)$; with say
the $\sigma$-algebra
generated by sets of the form $\{U: U\cap V\neq \emptyset\}$ for some
$V\subset \C$ open. This turns out to be a standard Borel space: we can give
a Polish topology with this above $\sigma$-algebra
as its Borel algebra.
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Equally, ${\cal F}(\C)$, the closed subsets of $\C$, or any Polish space in
fact, is a standard Borel space in the natural Borel structure.
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(ii) Any Borel subset of a Polish space is again a standard Borel space, in
the algebra of all Borel subsets.
\newpage
\noindent{\bf 0.5 Theorem} (Classical) Any standard Borel space is either finite,
of size $\aleph_0$, or of size $2^{\aleph_0}$. Any two standard Borel spaces of
the same cardinality are isomorphic -- that is to say there is a Borel bijection
between them which allows a Borel inverse.
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\noindent{\bf 0.6 Definition} A function between standard Borel spaces is said
to be {\it Borel} if it is Borel measurable -- i.e. it pulls Borel sets back to
Borel sets.
\newpage
\noindent{\bf $\S$1 Smooth/Non-smooth}
\bigskip
\noindent{\bf 1.1 Definition} Let $X$ be a Polish space, or even just a standard
Borel space. An equivalence relation $E$ on $X$ is said to be {\it smooth}
if
\[X/E=\{[x]_E: x\in X\},\]
the set of equivalence classes, in the quotient Borel structure generated by all
$E$-invariant Borel sets, is a subset of some standard Borel space.
\bigskip
\bigskip
\noindent{\bf 1.2 Lemma} The following are equivalent for $E$ on $X$:
(i) $E$ is smooth;
(ii) there is some standard Borel space $Y$ and Borel function $\rho: X\rightarrow Y$
such that
\[x_1 E x_2 \Leftrightarrow \rho(x_1) =\rho(x_2);\]
(iii) there is a countable sequence of Borel $E$-invariant subsets of $X$,
$(B_n)_n$, such that
\[x_1 E x_2\Leftrightarrow \forall n(x_1\in B_n\Leftrightarrow x_2\in B_n).\]
\bigskip
\bigskip
\noindent{\bf 1.3 Remark} Using that any two uncountable Borel spaces are isomorphic,
in (ii) we can actually assume $Y=\R$; thus smoothness
is equivalent to the existence of some Borel $\rho$ which ``assigns reals as complete
invariants."
\newpage
\noindent {\bf $\S$2 Examples}
\bigskip
\noindent{\bf 2.1 Example} $X=R^2$ and $(x_1, y_1) E (x_2, y_2)$ if $y_1=y_2$.
Then $X/E$ can naturally be identified with $\R$. So it is certainly smooth.
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\noindent {\bf 2.2 Example} $U(\C^n)$, unitary operators on $\C^n$. $E$=similarity.
(I.e. they are the same operator ``up to a change in basis").
Here we can completely classify by eigenvalues, listed with multiplicities in
increasing order. It can all be calculated in a Borel way, and thus we obtain a Borel
function
\[{\cal I}: U(\C^n)\rightarrow \C^n\]
such that $T_1$ similar $T_2$ if and only if
\[{\cal I}(T_1)={\cal I}(T_2).\]
\newpage
\noindent {\bf 2.3 Example:} $\Z$ acting on $\R/\Z$ by
\[n\cdot \bar x = n\surd 2 + \bar x \:\: {\rm mod} 2.\]
This is in fact not smooth.
\bigskip
\bigskip
\noindent {\bf Proof} Let $\theta: \R/\Z\rightarrow \R$ be a Borel
function with
\[\bar x_1 E_\Z \bar x_2\Rightarrow \theta(\bar x_1) =\theta(\bar x_2).\]
Let $C\subset \R/\Z$ be a comeager set on which $\theta|_C$ is continuous.
(Any Borel function is continuous on a comeager set; this follows from
all Borel sets having property of Baire.)
Let
\[C_\infty =\bigcup_{n\in\Z} n\cdot C.\]
Again a comeager set, but now invariant under the $\Z$-action.
Choose some $\bar x\in C_\infty$. Then $[\bar x]_\Z=\Z\cdot \bar x$ is
dense in $C_\infty$ and $\theta|_{\Z\cdot \bar x}$ is constant.
Hence by continuity of $\theta|_{C_\infty}$ and density $\bar x$'s orbit,
$\theta$ is constant on the comeager set $C_\infty$.
\hfill $\Box$
\newpage
\noindent{\bf 2.4 Fact} Let $G$ be a group acting by homeomorphisms on a Polish space
$X$ with orbit equivalence relation $E_G$.
If every orbit dense and meager, then $E_G$ is not smooth.
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It turns out there is a funny kind of converse. ({\small More on this later, if time
allows.})
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\noindent{\bf 2.5 Fact} Let $G$ be a Polish group acting continuously on a Polish space
$X$. Assume the orbit equivalence relation $E_G$ is Borel as a subset of $X\times X$.
Then exactly one of the following two things happens:
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(I) $E_G$ is smooth;
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(II) there is a Polish $G$-space $Y$ (action of $G$ on $Y$ continuous), with every orbit
dense and meager, and a continuous $G$-embedding from $Y$ to $X$.
\newpage
\noindent{\bf 2.6 Definition} For $G$ a countable group and $H$ a separable
Hilbert space, let
Rep$(G, H)$ be the space of representations of $G$ on $H$ -- that is to say
the space of all homomorphisms from $G$ to $U(H)$, the unitary group of $H$, with the
topology of pointwise convergence.
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Irr($G, H$) the subspace of irreducible representations. (Can be shown to be Polish).
\bigskip
$\sim$ the equivalence relation of unitary equivalence --
that is to say, two representations are
equivalent if they are the same up to some unitary ``change of coordinates".
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\noindent{\bf 2.7 Example} Here one has $\sim$ on Irr$({\mathbb F}_2, H)$, the
irreducible representations of the free group,
non-smooth in general.
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But for $G$ abelian-by-finite, the ``dual" is always smooth. In fact, one can show that
the irreducible representations are all finite dimensional, and hence the unitary group
responsible for any possible change in coordinates is compact, and such compact groups always
give rise to smooth equivalence relations.
\newpage
\noindent{\bf 2.9 Fact} Let $G$ be a compact Polish group acting continuously on a Polish
space $X$. Then $E_G$ is smooth.
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\noindent{\bf Idea} Every orbit is compact, and hence closed. Then can show
\[\rho: X\rightarrow {\cal F}(X) \:\: (={\rm standard \: Borel \: space \: of \: closed \: subsets})\]
\[x\mapsto [x]_G\]
is Borel.
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\noindent{\bf 2.10 Aside} Slawek Solecki has recently shown a converse: The above fact characterizes
compactness for Polish groups: A Polish group is compact if and only if all its orbit equivalence
relations are smooth.
\newpage
\noindent {\bf $\S$3 Smoothness as a measure of classifiability}
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\noindent {\bf 3.1 Example} George W. Mackey is sometimes associated with the
view that we can only hope to classify the irreducible representations (up to
equivalence, where as before we consider conjugation by the unitary group) when
they are smooth.
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His published writings seem to be somewhat more equivocal than this. For instance,
in his survey paper {\it Infinite dimensional group representations,}
{\bf Bulletin of the AMS,} vol. 69(1963), he writes only
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\noindent {\it ...there is no known case in which one has nontype I representations
{\rm [this corresponds to the non-smooth case]} and can describe (in any reasonable
sense) the set of equivalence classes of the irreducibles...}[p. 633.]
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\newpage
\noindent{\bf 3.2 Example} A parallel problem
is that of classifying measure preserving transformations,
considered up to conjugacy --
i.e. a measure respecting ``relabellings" of the underlying space. ({\small See Matt
Foreman's talk.})
In {\it Borel structure and invariants for measurable
transformations}\footnote{\large Proceedings of the AMS, vol. 46(1974)} Jacob Feldman
writes:
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\noindent {\it One view is that a reasonable invariant for a set $X$ of objects
must be realizable as a real-valued Borel function for some naturally arising
Borel structure on $X$...By a {\rm complete} set of conjugacy invariants will be
meant a countable family of conjugate invariants which, between them, can distinguish
between any pair of nonconjugate transformations. (Obviously, one wants countability;
else, why not use indicator functions for the conjugacy classes?)} [p. 390]
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This notion amounts to smoothness. Feldman goes on to show conjugacy of m.p.t.s is
non-smooth, and then therefore concludes they are unclassifiable.
%In fact if we assign ${\cal D}$, the space of
%open subsets of ${\C}$, with the {\it Effros standard Borel structure} -- under
\newpage
{\bf $\S$4. Harrington-Kechris-Louveau}
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\noindent{\bf 4.1 Definition} We consider the Polish space of infinite
binary sequences:
\[\{0,1\}^\N\]
in the product topology. This is a compact metric space, and hence Polish.
We define $E_0$ to be the equivalence of eventual agreement:
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\noindent{\bf 4.2 Fact} $E_0$ is non-smooth.
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One way to see this is to apply our earlier criterion for non-smoothness. If we
take the group $G=\Z_2^{<\N}$, elements of the infinite product group
\[\prod_\N Z_2\]
with finite support, and we let this group act by addition mod 2 on each coordinate,
then $E_0$ is the resulting orbit equivalence relation. Every equivalence class is
dense and countable $\therefore$ meager.
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\newpage
\noindent{\bf 4.3 Theorem} (Harrington-Kechris-Louveau) Let $E$ be a Borel
equivalence relation on a standard Borel space $X$. Then exactly one of:
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\noindent (I) $E$ is smooth; or
\noindent (II) ``$E_0$ embeds into $E$".
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By (II) I mean there is a Borel $\rho: \{0,1\}^\N\rightarrow X$ such that
for all $\vec x, \vec y\in \{0,1\}^\N$
\[\vec x E_0 \vec y \Leftrightarrow \rho(\vec x) E \rho(\vec y).\]
\newpage
\noindent {\bf $\S$5 Life without smoothness}
\bigskip
Some examples of classification theorems without smoothness.
These suggest that it might make sense to consider other
{\it degrees of classifiability} other than the smooth/non-smooth distinction.
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\noindent {\bf 5.1 Example} Consider Hom$^+([0,1])$, the orientation
preserving homeomorphisms of the unit interval. Here there is a kind of
folklore theorem describing these homeomorphisms up to conjugacy.
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For $\pi: [0,1]\rightarrow [0,1]$ a homeo with $\pi(0)=(0), \pi(1)=1$, we
take the maximal open intervals on which $\pi$ is increasing $(\pi(x)>x)$,
decreasing ($\pi(x)< x$), or the identity ($\pi(x)=x$).
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A complete invariant
is given symbolically by describing the ordering between these intervals along
with the behavior (increasing, decreasing, identity) for $\pi$.
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Here the conjugacy equivalence relation is non-smooth. Yet we have in effect classified
the homeomorphisms by countable linear orderings equipped with some unary predicates
($P_{\rm increasing}, P_{\rm decreasing}, P_{\rm identity}$).
\newpage
\noindent {\bf 5.2 Example} $U_\infty(L^2([0,1], {\rm Lebesgue})$.
\bigskip
Infinite dimensional unitary operators considered up
to conjugacy.
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Unlike the finite dimensional case, we can not in general diagonalize.
However it turns out that any such operator will be unitary equivalent
to the rotation operator on the disjoint sum of infinitely sum of the
unit circle in $\C$ equipped with some measure; and two operators will be
conjugate if and only if the measures agree up to absolute continuity.
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This ``classification by measure up to absolute continuity" is generally considered
to be some kind of solution to the classification problem for $U_\infty$, despite the
non-smoothness of the conjugacy equivalence relation.
\newpage
\noindent {\bf 5.3 Example} A more recent example
(from $C^*$ algebras and topological dynamics).
Consider minimal (no non-trivial closed invariant sets) homeomorphisms of $\{0,1\}^\N$
considered up to conjugacy of orbit equivalence relations: Thus $f_1 E f_2$ if there is
some homeomorphism $g$ conjugating their orbits:
\[\forall \vec x(g[\{f_1^\ell(\vec x): \ell \in\Z\}]
=\{f_2^\ell (g(\vec x)): \ell \in \Z\}).\]
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T. Giordano, I.F. Putnam, and C.F. Skau
\footnote{\large{{\it Topological orbit equivalence
and $C^*$-crossed products} ({\bf Journal Reine und Angewandt Mathematische,}
vol. 469(1995), pp. 51-111)}}
produce countable ordered abelian groups which, considered up to isomorphism,
act as complete invariants.
\newpage
\noindent{\bf $\S$6 Other notions of classifiability}
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\noindent {\bf 6.1 Definition} For ${\cal L}$ a countable language
we consider the standard Borel space of all ${\cal L}$-structures on
$\N$, Mod$({\cal L})$.
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Thus in the case that ${\cal L}$ consists of a single binary relation $R$ we
have Mod$({\cal L})$ the collection of all directed
graphs on $\N$.
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It turns out that all the plausible suggestions for a standard Borel structure
on Mod$({\cal L})$ are the same.
\newpage
\noindent{\bf 6.2 Definition} $E$ on $X$ is said to admit {\it classification by countable
structures} if there is a Borel function
\[\theta: X\rightarrow {\rm Mod}({\cal L})\]
(some countable language ${\cal L}$) such that
\[x_1 E x_2 \Leftrightarrow \theta(x_1)\cong \theta (x_2).\]
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Equally we can ask when $E$ can be classified by measures up to absolute continuity (Gelfand,
the spectral theorem
for infinite dimensional operators);
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or countable sets of reals (Halmos and von Neumann, for
``discrete spectrum" measure preserving transformations);
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\bigskip or reduction to an equivalence
relation in which every equivalence class is countable (compare Kechris, and Feldman and Moore,
for locally compact group actions);
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or reduction to the continuous action of a Polish group
(and for this last one we know just about nothing).
\newpage
\noindent
\noindent {\bf 6.3 Theorem} (Farah) There is nothing at all like Harrington-Kechris-Louveau
for Borel equivalence relations.
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(a) No canonical bad guy $F$ such that $E$ classifiable by countable structures if and only
if no way to ``embed" $F$ into $E$;
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(b) no small collection $\{F_1,...,F_k\}$ of disjunctively canonical baddies such that
$E$ classifiable by countable structures if and only
if no $F_i$ embeds into $E$;
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(c) in fact no countable or even less than continuum size set of canonical villains, even
for $E$ induced by the continuous action of a Polish group.
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(d) And worse. (There are just not even enough ``minimally" non-classifiable Borel equivalence
relations.)
\newpage
\noindent {\bf $\S$7. A dynamical analysis of classification by countable structures}
\bigskip
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\noindent {\bf 7.1 Theorem}
Let $G$ be a Polish group and $X$ a Polish space. Suppose that
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(i) every orbit is dense;
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(ii) every orbit is meager;
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(iii) for $x\in X$, {\it the local orbits of $x$} are all somewhere dense; that is to
say, if $V$ is an open neighborhood of $1_G$, $U$ is an open set containing $x$, and
if $O(x,U,V)$ is the set of all $\hat{x}\in [x]_G$ such that there is a finite sequence
$(x_i)_{i\leq k}\subset U$ such that $x_0=x$, $x_k=\hat{x}$, and each $x_{i+1}\in V\cdot x_i$,
then the closure of $O(x,U,V)$ contains an open set.
\bigskip
Then there is no
Borel (or even Baire measurable)
\[\theta : X\rightarrow {{\mbox{ Mod}}({\cal L})}\]
such that for all $x, y\in X$
\[xE_Gy\Leftrightarrow \theta(x)\cong \theta(y).\]
\newpage
This seems to be the right way to analyze classification by countable structures (at least
for equivalence relations induced by Polish group actions). ({\small Note parallel to
folklore theorem for non-smoothness.})
\bigskip
For instance:
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\noindent {\bf 7.2 Theorem}
Let $G$ be a Polish group and $X$ a Polish space with $E_G$ Borel.
Then exactly one of
(I) $E_G$ is classifiable by countable structures (and in fact, by a reduction
$\theta$ where the saturation of the image of $\theta$ is Borel, and [other desirable
properties] BLAH BLAH BLAH etc etc etc);
(II) there is a Polish $G$-space $Y$ satisfying (i), (ii), and (iii) from 7.1
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Remark: Here, unlike most theorems of this kind, it is open whether we
can prove the same result even for $E_G$ non-Borel. If this {\it were} true then
it would for instance reduce the ``topological Vaught conjecture" to ``Vaught's
conjecture" for ${\cal L}_{\omega_1, \omega}$ [countably infinitary logic].
\newpage
\noindent{\bf 7.3 Aside} Since then it has proved possible to provide a
similar kind of analysis for when an orbit equivalence relation allows
reduction to an equivalence relation where all the class are countable.
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On the other hand, for various other natural dividing lines virtually nothing
is known. For instance, reducibility of an arbitrary Borel equivalence relation
to a Polish group action. Or reduction to measures up to absolute continuity.
\end{document}
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