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\begin{document}

\title{The complexity of the classification of Riemann surfaces
and complex manifolds
\footnote{AMS Subject classification:  04A15,
03A15, 32C10; Key words and phrases: {\it Borel equivalence relations,
classification problems, complex manifolds, Riemann surfaces,
conformal equivalence}}}

\author{G. Hjorth\thanks
{Research partially supported by NSF Grants DMS 96-22977 (G.H.) and 
DMS 96-19880 (A.S.K.)} 
$\ $ and A.S. Kechris$^\dag$}
\date{}
\maketitle

\begin{abstract} 




In answer to a question by Becker, Rubel, and Henson,
we show that countable subsets of ${\mathbb C}$ can be
used as complete invariants for Riemann surfaces considered
up to conformal equivalence, and that this equivalence
relation is itself Borel in a natural Borel structure
on the space of all such surfaces. We further proceed to
precisely calculate the classification difficulty
of this equivalence relation in terms of the modern
theory of Borel equivalence relations.
 
On the other hand we show that the analog of Becker, Rubel,
and Henson's question has a negative solution in (complex)
dimension $n\geq 2$.
 


\end{abstract} 

\section{Introduction}

In this paper we consider the problem of classifying various classes
of complex manifolds.  The investigation is completely abstract,
since we are not so much concerned with specific schemes of
classification and their success or failure, but rather in considering
what {\it kinds} of complete invariants could in principle be 
produced.  Here we have in mind that there is a hierarchy of levels
of difficulty of classifying various mathematical objects and
this paper joins Dougherty-Jackson-Kechris [94], 
Harrington-Kechris-Louveau [90], Hjorth [9?a], Hjorth-Kechris [95],
Kechris [92], Kechris [98], and others, as one more piece
in a general project to compare the classification problems
across a variety of mathematical disciplines and obtain a
language that can contrast their various forms.

At perhaps the simplest level are schemes of classification which
provide a single point in some highly concrete space as a complete
invariant.  In ergodic theory the real number corresponding to the
entropy of a Bernoulli shift is a complete invariant for this
class of measure preserving transformations.  One similarly finds in
the theory of Riemann surfaces that {\it compact} complex surfaces
considered up to conformal equivalence may be cataloged as points
in highly concrete spaces.

The work below came about after seeing the following theorem:

\medskip
\noindent{\bf Theorem 1.1.}
(Becker-Rubel-Henson [80])  {\it There is no ``reasonably concrete space"
$X$ and ``reasonably intrinsic" or ``reasonably definable" assignment

$$f:\d\rightarrow X$$
from $\d$, the collection of complex domains, such that for any
two $R_1, R_2\in \d$

$$R_1\cong R_2\Leftrightarrow f(R_1)=f(R_2).$$

In other words, there is no reasonable way to assign points in $X$
as complete invariants for complex domains considered up to conformal 
equivalence.}

\medskip
Of course the phrases ``reasonably concrete space" and ``reasonably
intrinsic assignment" are deliberately vague.  For the purposes
of this introduction we simply ask that the reader accept that this
can and will be made precise, that the various ways in which one
might do so are not subject to serious controversy, and that all the
competing explications of these phrases give rise to similar outcomes.
Indeed while one may have doubts about how a definition of such
concepts should be crafted, it should certainly be clear that for
instance the space of {\it all} subsets of $\bbC$ is not sufficiently
concrete and that a function obtained by invoking the axiom of choice
to well order the complex domains and then thus armed producing an
injection from $\d$ into the ordinals or into $\bbC$ or into
$\p _{\aleph_0}(\bbC )$ (the set of all countable subsets of $\bbC$) can
not be considered ``reasonably intrinsic" or ``reasonably definable".

Perhaps in passing we can mention that for us ``reasonably concrete
space" means something like a Polish space or a standard Borel space,
and that for us a "reasonably definable function" means something
like a function that is Borel measurable from some standard Borel
space of parametrizations.  More generously one may consider, as in
the Ulm invariants from abelian group theory (see Kaplansky [69]),
spaces such as the set of all countable subsets of the first
uncountable ordinal, or equivalently, countable transfinite sequences
from $\bbC$, and functions that are universally Baire measurable,
or projective, or even ordinal definable from reals.  As we discuss in section 5.C,
Theorem 1.1 survives in some form even in these contexts.

Our first concern is the extent to which classification may be 
obtained if we consider invariants more general than a {\it single}
point in some space.  In direct response to a question by Becker,
Rubel, and Henson, we show that {\it countable unordered subsets}
of $\bbC$ {\it can} provide complete invariants for $\r$, the class
of all Riemann surfaces:

\medskip
\noindent{\bf Theorem 1.2.}  {\it There is a "reasonably definable"
function
$$ f:\r\rightarrow\p_{\aleph_0}(\bbC ),$$
from the Riemann surfaces to countable subsets of $\bbC$, such that
for any two $R_1, R_2$
$$R_1\cong R_2\Leftrightarrow f(R_1)=f(R_2).$$
In other words, there is a ``reasonably definable" way to assign countable subsets
of a concrete space as complete invariants for Riemann surfaces
considered up to conformal equivalence.}

\medskip
One might draw an analogy between this result and the Halmos-von 
Neumann [42] invariants for {\it discrete spectrum} ergodic measure
preserving transformations.  There, as here, there is no reasonable
method to assign points in say $\bbC$ as complete invariants, but the
countable subset of $\bbC$ corresponding to the eigenvalues completely
classifies an ergodic discrete spectrum measure preserving transformation
up to isomorphism. The proof of 1.2 in 4.A below is based on an abstract method and does not seem to provide ``geometrically meaningful" invariants $f(R)$. One can wonder if it is possible to sharpen this theorem by providing such invariants.

The method of proof in 4.A below actually gives a much better upper
bound on the complexity of conformal equivalence for Riemann surfaces
and in 4.B we show that this is precise.

\medskip
\noindent{\bf Theorem 1.3.}  {\it  The classification problem for
Riemann surfaces considered up to conformal equivalence is ``equal
in difficulty" to that of the universal countable Borel equivalence
relation, $E_\infty$.}

\medskip
Here ``equal in difficulty" indicates that each equivalence relation
can be embedded in the other using a function that is Borel
measurable in some suitable space of parameters. $E_\infty$
is known to have a number of instantiations.  For instance, it can
be realized as the orbit equivalence relation produced by the shift
action of $F_2$ (the free group on 2 generators) on $2^{F_2}=
\{ f\vert f: F_2\rightarrow \{ 0,1\}\}$, or as isomorphism on 
finitely branching trees (Jackson-Kechris-Louveau [9?]),
or (very recently, Thomas-Velickovic [98]) isomorphism on finitely
generated groups.

Put another way, this means that the ``moduli space" of all Riemann surfaces is ``Borel equivalent" to the very complicated quotient space $2^{F_2} / E_\infty$. Actually, as it follows from the proof of 1.3 given in section 4 below, this holds as well for Riemann surfaces homeomorphic to the infinitely punctured plane, i.e., $\bbC \setminus S$, where $S\subseteq\bbC$ is infinite discrete. Thus this gives a precise measure of the set theoretic complexity of the moduli space of these Riemann surfaces. It is much more complex than the moduli spaces of finitely punctured compact Riemann surfaces, which are fairly ``concrete" and admit a rich geometrical structure.

Finally, the higher dimensional case is discussed in section 6.  Here
we provide a new lower bound on the complexity of biholomorphism,
and indicate a sense -- which when translated into the theory of Borel
equivalence relations can be made totally precise -- in which the passage from
complex dimension 1 to complex dimension 2 brings an increase in 
classification difficulty. The following is a simple corollary of 6.1 below.

\medskip
\noindent{\bf Theorem 1.4.}  {\it Let $\m^2$ be the class of two
dimensional complex manifolds.  Then there is no ``reasonably definable" assignment
$$f:\m^2\rightarrow\p_{\aleph_0}(\bbC )$$
such that for any $M_1, M_2\in \m^2$,
$$M_1\cong M_2\Leftrightarrow f(M_1)=f(M_2).$$}

\section{Borel equivalence relations}

\noindent{\bf Definition 2.1.}  A topological space is said to
{\it Polish} if it is separable and admits a complete metric.  We
then define the {\it Borel} sets to be those appearing in the 
smallest $\sigma$-algebra containing the open sets.  A function
between two Polish spaces is said to be {\it Borel} if the inverse
image of any open set under $f$ is Borel.

If $X$ is a Polish space and $E$ is an equivalence relation, then $E$
is said to be {\it Borel} if it is Borel as a subset of $X\times X$
(in the product topological structure).  For $x\in X$, we let
$[x]_E=\{ y\in X:xEy\}$.  An equivalence relation is
{\it countable} if every $[x]_E$ is countable.

\medskip
\noindent{\bf Examples of Polish spaces}
\begin{enumerate}
\item[(i)] $\bbR ,\bbC ,$ in their usual topologies; $2^\bbN =_{df}
\{f\vert f:\bbN\rightarrow \{0,1\}\}$ equipped with the metric
$$d(x,y) = 2^{-\text{ least }n\text{ such that }x(n)\neq y(n)}$$
is a compact metric space, and so certainly the underlying topological
space is Polish.
\item[(ii)] Polish spaces are closed under countable products, and
thus we obtain that $\bbR^n,\bbC^n,\bbR^\bbN , \bbC^\bbN , \bbN^\bbN$
are all Polish.
\item[(iii)] Any $G_\delta$ subset of a Polish space is Polish 
(see Kechris [95, 3C]).
\item[(iv)] The space of all countable models of a given countable
language is a Polish space; this is an especially important example from the point of view of logic.  Let $\l$ be a countable (relational) language and let
Mod($\l$) be the set of all $\l$-structures with underlying set $\bbN$
and equipped with the topology induced by taking as subbasic open
sets those of the form
$$\{\m\in\text{Mod}(\l ):\m\models R(n_1,\cdots , n_k)\},$$
$$\{\m\in\text{Mod}(\l ): M\models\neg R(n_1,\cdots ,n_k)\},$$
for $n_1,\cdots ,n_k$ in $\bbN$ and $R\in\l$ a relation of arity $k$. This space is Polish since it is homeomorphic to
$$\prod_{R\in\l}2^{\bbN^{a(\bbR)}},$$
where $a(R)$ is the arity of the relation symbol $R$ (see 
Becker-Kechris [96]).  (A similar definition works even for 
languages with function symbols, since we can show that the space
Mod($\l$) is homeomorphic to a $G_\delta$ subset of Mod($\l'$) for
a suitably chosen relational $\l'$.)
\end{enumerate}

Many mathematical objects can be thought of as points in some
appropriately chosen Polish space considered up to some notion of
isomorphism or equivalence.  For instance, we may think of countable
groups as elements of some appropriate Mod($\l$) considered up to
isomorphism.  A fundamental notion used in comparing equivalence
relations is that of Borel reducibility.
\medskip

\noindent{\bf Definition 2.2.}  For $E$ and $F$ equivalence relations
on Polish spaces $X$ and $Y$, we write $E\leq_BF$, and say that $E$ is {\it Borel
reducible to} $F$, to indicate that there is a Borel function $f:
X\rightarrow Y$ such that for all $x_1, x_2\in X$
$$x_1Ex_2\Leftrightarrow f(x_1)Ff(x_2).$$
We use $E\sim_BF$ for $E\leq_BF\leq_B; E<_BF$ indicates that
$E\leq_BF$ but it is not the case that $F\leq_B E$.
\medskip

\noindent{\bf Examples of equivalence relations}
\begin{enumerate}
\item[(i)] id($\bbC$), the equality relation on $\bbC$; more 
generally for any Polish space $X$ we can consider the identity
relation id($X$) on $X$.  Since any two uncountable Polish spaces
are Borel isomorphic, one obtains for any two uncountable
Polish spaces $X$ and $Y$ that id$(X)\sim_B$id($Y$).
\item[(ii)] We let $E_0$ be the equivalence relation of eventual
agreement on $2^\bbN$. This means that $x_1E_0x_2$ if and only if there
is some $N\in\bbN$ such that for all $m>\bbN,\; x_1(m)=x_2(m)$.
Here it is well known that id$(\bbC )<_BE_0$ (see Hjorth [9?a,
Chapter 3], or Hjorth-Kechris [95], or Kechris [98]).
\item[(iii)] Given a countable group $G$, we can let it act on the
Polish space $2^G (=_{df}\{ f\vert f:G\rightarrow \{0, 1\}\}$
in the product topology) by {\it shift}: for $g\in G$ and $f\in
2^G$ we define $g\cdot f$ by
$$(g\cdot f)(h)=f(g^{-1}h)$$
for $h\in G$.  We denote the orbit equivalence by $E(G,2)$.
\item[(iv)] In specific cases the complexity under $\leq_B$ of
$E(G,2)$ is exactly understood.  For instance it is known that
$$E_0\sim_BE(\bbZ , 2)$$
(Weiss, Slaman-Steel; see Dougherty-Jackson-Kechris [94]) and that 
for any Borel equivalence relation $E$ one has either
$$E\leq_B \text{ id}(\bbC )$$
or
$$E_0\leq_BE,$$
but not both (this was proved in Harrington-Kechris-Louveau [90]).
\item[(v)] Let $F_2$ be the free group on 2 generators.  Here it
is known that for any countable Borel equivalence relation $E$
$$E\leq_BE(F_2, 2).$$
In this sense, we say that $E(F_2, 2)$ is a {\it universal countable
Borel equivalence relation}, and we will in the future denote it by
$E_\infty$.
\item[(vi)]  More generally, we consider orbit equivalence relations
induced by the continuous actions of {\it Polish groups} (that is,
topological groups whose underlying space is Polish). If $G$ is a Polish
group acting continuously on a Polish space $X$ we will use $E^X_G$ for the
orbit equivalence relation given by
$$x_1E^X_Gx_2\Leftrightarrow \exists g\in G(g\cdot x_1=x_2).$$
\item[(vii)] Finally, for $\l$ a countable language, we let $\cong_{Mod(\l )}$
be the equivalence relation of isomorphism on elements of Mod($\l )$.
Thus $\m\cong_{Mod(\l )}\n$ iff there is some permutation $\pi$ of the
natural numbers such that for all relations $R$ in the language
$$\m\models R(n_1,\cdots , n_k)\Leftrightarrow\n\models R(\pi (n_1),\cdots 
, \pi (n_k)).$$
\end{enumerate}

\medskip
\noindent {\bf Definition 2.3.}  We say that an equivalence relation $E$
is {\it smooth} if $E\leq_B \text{id}(X)$ for some Polish space $X$;
note that here we may assume without loss of generality that $X=\bbC$.
We say that $E$ {\it admits classification by countable structures}
if there is some countable language $\l$ with
$$E\leq_B\cong_{Mod(\l )}.$$
Let us further say that $E$ {\it admits classification by countable
subsets of} $\bbC$ if there is a countable sequence of Borel functions
$(f_n)_{n\in\bbN}$ into $\bbC$ such that for all $x_1,x_2$
$$x_1Ex_2\Leftrightarrow \{ f_n(x_1): n\in\bbN\}=\{ f_n(x_2):n\in\bbN\}.$$

Note that in the definition of classification by countable subsets we ask
that $\{f_n(x_1):n\in\bbN\}$ and $\{f_n(x_2):n\in\bbN\}$ are equal as
unordered sets; if we were to alternatively require something along the
lines of the equality of the {\it sequences} $(f_n(x_1))_{n\in\bbN}$ and
$(f_n(x_2))_{n\in\bbN}$ then this would be nothing other than the notion
of smoothness, since $\bbC^\bbN$ is itself a Polish space.

The next couple of lemmas clarify these definitions.
\medskip

\noindent {\bf Lemma 2.4.}  If $E$ is a countable Borel equivalence
relation, then $E$ admits classification by countable subsets of $\bbC$.
\medskip

\noindent {\bf Proof}.  Let $X$ by the Polish space on which $E$ is defined.
Following the uniformization theorem for Borel sets in the plane with
countable sections (see Kechris [95, 18.C]) we may find a countable
sequence of Borel functions
$$h_n:X\rightarrow X$$
such that for all $x\in X$

$$[x]_E=\{h_n(x):n\in\bbN\}.$$
Then if $h:X\hookrightarrow \bbC$ is a Borel injection we may let $f_n=h
\circ h_n$ witness the definition at 2.3\hfill$\dashv$

\medskip
This implication does {\it not} reverse (see for instance, Hjorth [9?a,
\S 2.3]).
\medskip

\noindent {\bf Lemma 2.5.}  If $E$ admits classification by countable
subsets of $\bbC$, then $E$ admits classification by countable structures.
\medskip

\noindent {\bf Proof}.  Let $(P_n)_{n\in\bbN}$ be a sequence of unary 
predicates, $\{U_n:n\in\bbN\}$ a countable basis for the topology of
$\bbC$, and let
$$\langle \cdot ,\cdot\rangle :\bbN^2\rightarrow\bbN$$
be a bijection.  Let $(f_n)_{n\in\bbN}$ be the sequence of Borel functions
witnessing the classifiability by countable subsets of $\bbC$.  Let
$\l$ be the countable language determined by $(P_n)_{n\in\bbN}$.

Then for any point $x$ in the space on which $E$ is defined we let $\m_x$ be
the $\l$-model with underlying set $\bbN$ defined by
$$\m_x\models P_n(\langle k, l\rangle )$$
if and only if $f_k(x)\in U_n$.  (Here each $f_k(x)$ is quite deliberately
repeated infinitely often; allowance needs to be made for the fact that
the {\it sets} $\{ f_n(x_1):n\in\bbN\}$ and $\{f_n(x_2):n\in\bbN\}$ may
be equal, but the {\it sequences} $(f_n(x_1))_{n\in\bbN}$ and 
$(f_n(x_2))_{n\in\bbN}$ may list a given complex number with different
frequency.)\hfill$\dashv$

\medskip
The implication here also fails to reverse (see, for example,
Hjorth-Kechris-Louveau [98]).

Although we confine ourselves to Borel functions, a similar reducibility
theory arises for any general class of functions including the Borel,
closed under composition, and satisfying reasonable regularity properties
(such as Lebesgue measurability and the pullbacks of open sets having
the Baire property).  This is an important point.  Key theorems are not
held hostage to circumstantial and subjective choices used in our
explication of {\it reasonably definable}.

For instance we could equally work with the $C$-measurable functions, i.e., those that pull back open sets to sets lying in the smallest $\sigma$-algebra
containing the open sets and closed under Souslin's $\a$-operation
(compare Hjorth-Kechris [95]).  Alternatively we might use the {\it absolutely}
$\bDe^1_2$ functions.

For functions of higher complexity, it is necessary to work in systems stronger than ZFC to
avoid the whole pursuit dissolving into a quick sand of indistinguishable
independence results.  For instance, we can develop a similar theory for
functions in $L(\bbR )$ (the smallest class of sets containing all the
reals and closed, in some sense, under transfinite iterations of
rudimentarily definable operations) assuming sufficient {\it determinacy}
or the existence of {\it large cardinals}.  The authors of 
Becker-Henson-Rubel [80] choose to use the collection of all ``ordinal
definable functions'' and in this context one again obtains a similar
theory by working in the very specific model of set theory known as the
{\it Solovay model}.

\section{Parametrizing complex manifolds}

In order to precisely formulate the problem of classifying species of
complex manifolds we need a method of parametrizing complex manifolds.
Here the method must be completely general, and in some sense given in
advance of any specific classification result.  We will use points in
a standard Borel space -- indeed without loss of generality we can take
the space $\bbR$ -- to describe or {\it parametrize} complex manifolds.
The obvious properties will become Borel in the space of parametrizations
and the equivalence relation of biholomorphism will be 
$\bSi^1_1$ (analytic) -- that is to say, defined by
the Borel image of a Borel set.  It should not be felt that there is any great
mystery or presumption in the method of parameterization.  It is an
empirical fact that most classes of concrete mathematical objects can be
represented by points in a standard Borel space and all such methods of
representation tend to be ``Borel equivalent".  

We recount some basic definitions, all detailed in Kodaira [86].

\medskip
\noindent {\bf Definition 3.1.}~~A {\it complex manifold of dimension
n} is a connected Hausdorff second countable topological space $M$
along with a chart $\{ U_{\alpha}, \varphi_\alpha \}_{\alpha\in\Lambda}$,
where:
\renewcommand\theenumi{\roman{enumi}}
\begin{enumerate}
\item[(i)] $\bigcup_{\alpha\in\Lambda}U_\alpha=M$;
\item[(ii)] each $\varphi_\alpha:U_\alpha \rightarrow V_\alpha$ is a
homeomorphism between $U_\alpha \subseteq M$, open in $M$, and
$V_\alpha \subseteq \bbC^n$, open in $\bbC^n$;
\item[(iii)] the overlap maps
$$\varphi_\beta \circ \varphi^{-1}_\alpha:\varphi_\alpha(U_\alpha
\cap U_\beta) \rightarrow \varphi_\beta
(U_\alpha \cap U_\beta),\,\,\,\text{when}
\,\,\,U_\alpha \cap U_\beta \ne \emptyset , $$
are holomorphic.
\end{enumerate}

A complex manifold of dimension 1 is called a {\it Riemann surface}.
For $M$ and $N$ complex manifolds, with charts $\{U_\alpha,
\varphi_\alpha \},\ \{W_\beta, \psi_\beta \}$, we say that
$\tau:M\rightarrow N$ is {\it holomorphic} if each $\psi_\beta
\tau\varphi_\alpha^{-1}$ is holomorphic.  A 
{\it biholomorphic} map is a holomorphic map with holomorphic inverse.
We shall write
$$ M \cong N $$
if they are {\it biholomorphically equivalent}, that is to say, there is
some biholomorphic
$$\pi : M \rightarrow N.$$

For $n=1$, i.e., for Riemann surfaces, one usually says that $M,N$ are
{\it conformally equivalent} if $M\cong N$.

A manifold is a ``separable" object, and thus we might hope to represent
or parametrize a manifold by describing the arrangement of a countable
dense subset.  There is some judgment regarding the kind of objects to
take as satisfactory parameterizations; plainly the class of all manifolds
would be absurdly complicated and the set of all subsets of say
$\bbC\times\bbC^2\times\bbC^3\ldots$ only slightly less so.  Our idea of
a {\it reasonable} space is something like $\bbR$, or the set of all
subsets of $\bbN$, or $C([0,1])$, or infinite strings from a finite 
alphabet, or any of the classical Banach spaces.  In each case we are
less interested in the topological structure of these spaces than 
their {\it Borel structure}.

\medskip
\noindent {\bf Definition 3.2.}~~A set $S$ equipped with a 
$\sigma$-algebra $\b$ of subsets is said to be a {\it standard Borel
space} if there is a completely metrizable separable topology $\tau$ on
$S$ which gives rise to $\b$ as the $\sigma$-algebra generated by its
open sets.
\medskip

As a remark on terminology, we say that a completely metrizable separable
topological space is {\it Polish}, and that the {\it Borel sets} for a
Polish topology are those appearing in the smallest $\sigma$-algebra
containing its open sets.  Thus we may say that $(S,\b)$ is a standard
Borel space if there is a Polish topology $\tau$ on $S$ that has $\b$
as its Borel sets.

As well as the usual examples of Polish spaces, such as $\bbR^n, \bbC^n,
2^{\bN},$ stripped down to their Borel structure, we should also mention
that given any Polish space $X$ the collection of all closed subsets, 
$\f(X)$, is a standard Borel space, when equipped with the $\sigma$-algebra generated by
sets of the form $\{ F\in\f(X):F\cap U\ne\emptyset\}$ for $U\subseteq X$
open; this more subtle example is known as the Effros standard Borel
space and is discussed in $\S$12.C of Kechris [95].  It follows that the
space of all {\it open} subsets of $X$ is a standard Borel space in
the $\sigma$-algebra generated by sets of the form 
$\{ O\subseteq X\,\,\text{open}:O\cap U=\emptyset\}$.

The collection of standard Borel spaces is closed under taking
countable disjoint unions and countable products.  In consequence of
$\S$13.A in Kechris [95], any Borel subset of a standard Borel space is
again standard Borel in the induced subspace Borel structure.

We can then proceed to define for each $n\ge 1$, a standard Borel space
$$\m^n$$
and a map
$$p\in\m^n\mapsto M_p$$
assigning to each $p\in\m^n$ an $n$-dimensional complex manifold $M_p$ 
such that for every $n$-dimensional complex manifold $M$ there is at
least one $p\in\m^n$ with $M\cong M_p$.  We call $\m^n$ the
{\it parameter space of $n$-dimensional complex manifolds}.  The
construction of $\m^n$ and the verification that it has a number of
reasonable properties that we will need in various parts of this
paper is technically cumbersome, although mathematically rather shallow.
We will thus postpone the precise definition and verification of these
facts until $\S$7.  In our proofs between now and then we will simply
state as lemmas the various facts about this coding that we need and
return to their proofs in $\S$7.

To start off, the following fact gives an upper bound for the
complexity of biholomorphic equivalence.

\medskip
\noindent {\bf Proposition 3.3.}~~{\it For each} $n\ge 1$, {\it the
relation}
$$p\cong_nq\Leftrightarrow p,q\in\m^n\,\,\,and\,\,\,M_p\cong M_q$$
{\it is} ${\bSi}^1_1$. 

\medskip
For $n=1$ we will denote $\m^1$ by $\r$ and $\cong_1$ by $\cong_R$ (for
Riemann surfaces) and we will see in $\S$4 that actually $\cong_R$ is 
Borel, i.e., the relation of conformal equivalence of Riemann surfaces is
Borel (in the parameters).  However we doubt if this is true in
higher dimensions.

\medskip
\noindent {\bf Conjecture 3.4.}~~{\it For  $n\ge 2$, the biholomorphic
equivalence relation $\cong_n$ of $n$-dimensional complex manifolds (in
the parameters) is not Borel.}

\medskip
We will also discuss in this paper a particular class of Riemann surfaces,
namely {\it domains} (open connected sets) in $\bbC$.  Again we will need
to parametrize these by elements of a standard Borel space.  We will thus 
define in $\S$7 a standard Borel space $\d$ and a map $d\in\d\mapsto D_d$ assigning to
each $d\in \d$ a domain $D_d\subseteq\bbC$ such that for each
domain $D\subseteq\bbC$ there is at least one $d\in\d$ with
$D=D_d$.  We call $\d$ the {\it parameter space of domains in $\bbC$}.
Of course for every domain $D$ there is also a $p\in \r$ such that
$M_p\cong D$.  The next fact asserts that such a $p$ can be
computed in a Borel way from any $d\in\d$ such that $D=D_d$.

\medskip
\noindent {\bf Proposition 3.5.}~~{\it There is a Borel function 
$f:\d\rightarrow\r$ such that $D_d\cong\m_{f(d)}$.}

\medskip
In particular if we denote by $\cong_D$ the relation of conformal
equivalence (in the parameters), i.e.,
$$d\cong_D e\Leftrightarrow d, e\in\d\,\,\,\&\,\,\,D_d\cong D_e$$
then we have that 
$$(\cong_D) \leq_B (\cong_R)  . $$

\section{Riemann surfaces}

Our main goal in this section is to compute the precise complexity of
conformal equivalence of Riemann surfaces and planar domains in the
hierarchy of Borel equivalence relations.
\medskip

\noindent{\bf Theorem 4.1.}  {\it The conformal equivalence relations
$\cong_R$, $\cong_D$ (in the parameters) of Riemann surfaces and
planar domains are Borel and}
$$(\cong_R)\sim_B(\cong_D)\sim_BE_\infty .$$

We will split the proof of this theorem in two parts:
\medskip

The {\it upper bound}, i.e., showing that $\cong_R$ is Borel and
$(\cong_R)\leq_B E_\infty .$

The {\it lower bound}, i.e., showing that $E_\infty\leq_B(\cong_D)$.
\medskip

Since, as we already pointed out in $\S$3, $(\cong_D)\leq_B(\cong_R)$
this will complete the proof of 4.1. 
\medskip

\noindent{\bf 4.A The upper bound}
\medskip

We will need here some basic facts from the Uniformization Theory of 
Riemann Surfaces (see, e.g. Forster [81]).

Given a Riemann surface $M$, we will devote by $\widehat M$ its {\it universal
covering} Riemann surface and by $\pi :\widehat M\rightarrow M$ the 
{\it covering
map}.  Thus $(\widehat M , M, \pi)$ has the following properties:

\begin{enumerate}
\item[(i)] $\widehat M$ is simply connected and $\pi$ is a holomorphic surjection
of $\widehat M$ onto $M$;

\item[(ii)] $\pi$ {\it evenly covers} $M$, i.e., for each $x\in M$ there
is open $U$ containing $x$, so that $\pi^{-1}[U]$ is the disjoint union
of open sets $\{ V_i\}$, where each $\pi\vert V_i:V_i\rightarrow U$ is a 
biholomorphism.

\item[(iii)] $\widehat M$ is uniquely, up to biholomorphism, determined
by (i), (ii) above, i.e., if $f:M\rightarrow N$ is a biholomorphism
and $(\widehat N, N, \rho )$ satisfy (i), (ii) above, then there is a
biholomorphism $\widehat f:\widehat M\rightarrow\widehat N$ such that $\rho\circ
\widehat f=f\circ\pi$.  Moreover, for each $\widehat x_0\in\widehat M, 
\widehat y_0\in\widehat N$ with $f(\pi (\widehat x_0))=\rho (\widehat y_0)$,
there is a unique $\widehat f$ as above with $\widehat 
f(\widehat x_0)=\widehat y_0$.
\end{enumerate}
\medskip

The following fundamental result is variously called the 
{\it Uniformization
Theorem or the Riemann Mapping Theorem for Riemann Surfaces}.
\medskip

\noindent{\bf Theorem 4.2.}  {\it Every simply connected Riemann 
surface is conformally equivalent to exactly one of the following:}

\begin{enumerate}
\item[(i)] $\bbC_\infty$\ =\ {\it the Riemann Sphere} $(=\bbC\cup\{
\infty\} )$;
\item[(ii)] $\bbC$ = {\it the complex plane};
\item[(iii)] $\bbH =\{ x+iy: y>0\}$ = {\it the upper half plane}.
\end{enumerate}

The group of automorphisms (i.e., biholomorphic correspondences) of a Riemann surface $M$ will be denoted by 
$Aut(M)$.  It turns out that the automorphism groups of the simply
connected Riemann surfaces can be explicitly described as follows
(see, e.g., Beardon [84], Bedford et al. [91]).

\begin{enumerate}
\item[(i)] $Aut(\bbC_\infty)=PSL_2(\bbC )$
\medskip

Here $PSL_2(\bbC )$ is the quotient of the group $SL_2(\bbC )$ of all
complex matrices $\left (
\begin{smallmatrix}
a&b\\ c&d
\end{smallmatrix}\right )$
with $ad-bc=1$ by its center
$\{ +I, -I\}$, where I is the identity matrix.  This acts on 
$\bbC_\infty$ as a {\it M$\ddot o$bius transformation}, i.e., for
$\left (\begin{smallmatrix}a&b\\ c&d\end{smallmatrix}\right )\in
PSL_2(\bbC ),\ z\in\widehat\bbC$

$$\left (\begin{matrix}a & b\\c & d\end{matrix}\right )(z)=
{{az+b}\over {cz+d}}$$

\item[(ii)] $Aut(\bbC )$ = the ``$az+b$'' group.

This is the group of all pairs $(a, b)\in\bbC^*\times\bbC$, when
$\bbC^*=\bbC -\{0\}$, with multiplication $(a, b)\cdot (a^\prime ,
b^\prime )=(aa^\prime , ab^\prime + b)$.  It acts on $\bbC$ by
$(a, b)(z)=az+b$.

\item[(iii)] $Aut(\bbH )=PSL_2(\bbR)$

Here $PSL_2(\bbR)$ is the quotient of the group $SL_2(\bbR)$ of all 
real matrices 
$\left (\begin{smallmatrix}a&b\\ c&d\end{smallmatrix}\right )$
with $ad-bc=1$ by its center
$\{ +I, -I\}$.  This again acts on $\bbH$ by M$\ddot {\rm o}$bius
transformations.
\end{enumerate}

In particular, these automorphism groups are Lie groups, therefore
Polish locally compact, and their actions on the underlying spaces
are continuous.

Now consider a Riemann surface $M$ and a universal covering
$\pi :\widehat M\rightarrow M$, where we now take $M$ to be one
of $\bbC_\infty,\ \bbC,\ \bbH $.  Let
$$F_\pi =\{ (x,y)\in\widehat\m :\pi (x)=\pi (y)\} .$$

\noindent
Then $F_\pi$ is a closed subset of $\widehat M^2$, i.e., $F_\pi\in\f (\widehat
M^2)$.  Now let $f: M\rightarrow N$ be a biholomophic map and
$\rho :\widehat N\rightarrow N$ a universal covering.  Then $\widehat N=\widehat M$
and there is $\sigma\in Aut(\widehat M)$ such that 
$\pi=\rho\circ\sigma$.  Thus
\begin{eqnarray*}
(x,y)\in F_\rho &\Leftrightarrow &\rho (x)=\rho (y)\\
&\Leftrightarrow &\pi \left (\sigma^{-1}(x)\right )=\pi\left 
(\sigma^{-1}(y)\right )\\
&\Leftrightarrow &\left (\sigma^{-1}(x), \sigma^{-1}(y)\right )
\in F_\pi\\
&\Leftrightarrow &(x,y)\in\sigma\cdot F_\pi ,
\end{eqnarray*}

\noindent where $Aut(\widehat M)$ acts on $\f (\widehat M^2)$ by
$$\sigma\cdot F=\{ (x,y): \left (\sigma^{-1}(x), \sigma^{-1}(y)
\right )\in F\}$$

\noindent Thus $F_\rho,\ F_\pi$ belong to the same orbit of this
action.

Conversely, suppose $\pi :\widehat M\rightarrow M$, $\rho :\widehat M
\rightarrow N$ are two universal coverings and for some $\sigma\in
Aut(\widehat\m)$, $F_\rho =\sigma\cdot F_\pi$.  Then we can define
$f :M\rightarrow N$ by
$$f(x)=\rho \left (\sigma (y)\right ),\ {\rm where\ }\pi (y)=x.$$

\noindent Notice that this is well-defined, since if $\pi (y)=\pi
(y^\prime )=x$, i.e., $(y, y^\prime)\in F_\pi$, then
$\left (\sigma (y), \sigma (y^\prime)\right )\in F_\rho$,
so $\rho \left (\sigma (y)\right )=\rho \left (\sigma (y^\prime )
\right )$.  It is now clear that $f$ is a biholomorphism between
$M , N$.

Thus we have seen that fixing $\widehat M\in \{ \bbC_\infty ,\ \bbC,\;\bbH
\}$ there is an injection from the set of conformal equivalence
classes of Riemann surfaces with universal covering surface $\widehat M$
into the orbits of the action of $Aut(\widehat M)$ on $\f (\widehat M^2)$.
We have thus essentially reduced our problem to the study of this
action.  But then we can use the following general
result:
\medskip

\noindent{\bf Theorem 4.3.} (Kechris [92])  {\it Let $G$ be
a Polish locally compact group, $X$ a standard Borel space,
$(g, x)\mapsto g\cdot x$ a Borel action of $G$ on
$X$.  Then there is a Borel set  $S\subseteq X$ which
meets every orbit in a countable nonempty set.}
\medskip

This implies the following, concerning the orbit equivalence 
relation $E^X_G$:
$$xE^X_Gy\Leftrightarrow\exists g\in G (g.x=y). $$

\noindent
First, it is well-known that $E^X_G$ is Borel (see Kechris 
[95, 35.49]).  Thus, $E^X_G\vert S$ is countable Borel and
satisfies $(E^X_G\vert S)\leq_B E_\infty$.  Also there is a Borel
function $f: X\rightarrow S$ with $f(x)E^X_Gx$, for all $x\in X$ (see Kechris 
[95, 18C]).  Thus $E^X_G\leq_BE^X_G\vert S$, so
$$E^X_G\leq_B E_\infty $$

In particular, this applies to the action of $Aut(\widehat M)$
on $\f (\widehat M^2)$, where $\widehat M\in\{ \bbC_\infty , \bbC,
\bbH\}$,  since this is clearly a Borel action.

One way to proceed then is to show the following:

\begin{enumerate}
\item[(i)] For each $\widehat M\in\{ \bbC_\infty , \bbC , \bbH\}$, the
set $U_{\widehat M} = \{ z\in\r : M_z$ has universal covering surface
$\widehat M\}$ is Borel.
\item[(ii)] There is a Borel function $F: U_{\widehat M}\rightarrow
\f (\widehat M^2)$ so that $F(z)$ is of the form $F_\pi$, for some
covering $\pi :\widehat M\rightarrow M$.
\end{enumerate}

Then letting $E_{\widehat M}$ be the orbit equivalence relation
induced by the action of $Aut(\widehat M)$ on $f (\widehat M^2)$, we have 
that for $z, w\in U_{\widehat M}$,
\begin{eqnarray*}
z\cong_Rw&\Leftrightarrow & M_z\cong M_w\\
&\Leftrightarrow & F(z)E_{\widehat M}F(w),
\end{eqnarray*} 

\noindent i.e., $(\cong_R)\leq_BE_{\widehat M}$ and since 
$E_{\widehat M}\leq_BE_\infty$ we have that $(\cong_R)\leq_BE_\infty$.

The drawback of this approach is that the construction of the function 
$F$ in (ii) above, although intuitively rather clear, involves some
messy computations, which we don't want to commit to print.
So we will follow an alternative, somewhat indirect, approach that will
minimize the technicalities.

We will use the following criterion (a proof of which can be found
in Hjorth [9?a, 5.2]).
\medskip

\noindent {\bf Proposition 4.4.} (Kechris)  {\it For a Borel 
equivalence relation E on a standard Borel space $X$, the following
are equivalent:}
\begin{itemize}
\item [(i)] $E\leq_BE_\infty$,
\item [(ii)] {\it There is a Borel function} $f: X\rightarrow Y$,
{\it with} $Y$ {\it a standard Borel space, such that for any} $x, y\in X$:

\hskip 2em (a) $f([x]_E)$ {\it is countable;}

\hskip 2em (b) $\neg xEy \Rightarrow f([x]_E)\cap f([y]_E=\emptyset $.
\end{itemize}

Put $\widehat M_1=\bbC_\infty$, $\widehat M_2=\bbC$, $\widehat M_3=\bbH$.
Consider the action of $Aut(\widehat M_i)$ on $\f (\widehat M_i^2)$
and let $E_i$ be the corresponding orbit equivalence relation. 
Let also $S_i$ be Borel sets that meet every $E_i$-class
in a countable nonempty set.  Then there are Borel functions
$f_{i, j},\ i=1,2,3,\ j=1,2,\cdots ,\ f_{i, j}: \f (\widehat M_i^2)
\rightarrow\f (\widehat M_i^2)$ such that $\{f_{i, 1}(F): j=1,2,\cdots\}=
S_i\cap [F]_{E_i}$ (see Kechris [95, 18.C]).  We can of course assume
that $f_{i, 1}(F)=F$ for $F\in S_i$.  Thus if we put $f_i=f_{i, 1}$,
$f_i$ satisfies (ii) of 4.4 for $E_i$.  Now define the following relation 
$P\subseteq\r\times Y$, 
where $Y=S_1\cup S_2\cup S_3$, and we of course consider $S_1, S_2, 
S_3$ as pairwise disjoint:
$$P(z, y)\Leftrightarrow\exists i\in\{1,2,3\}\exists\pi: \widehat M_i
\rightarrow M_z\ (\pi\ {\rm is\ a\ covering\ map\ and\ }
y=f_i(F_\pi)).$$

\noindent Denote by $P_z$ the section of $P$ determined by $z\in\r$.  
Then if $z_1\cong_Rz_2$ and $y_1\in P_{z_1}$, $y_2\in P_{z_2}$, say
$y_1=f_i(F_{\pi_1}),\ y_2=f_i(F_{\pi_2})$ (for some $i\in 
\{1, 2, 3,\})$, then clearly
$F_{\pi_1} E_i F_{\pi_2}$, so $y_1\in f_i([F_{\pi_1}]_{E_i})=
f_i([F_{\pi_2}]_{E_i})\ni y_2$.  Thus $P_{z_1}, P_{z_2}\subseteq
f_i([F_{\pi_1}]_{E_i})$.  It follows that for each $z\in \r ,\
\bigcup_{w\cong_Rz}P_w=\bar P_z$ is countable.

If now $\bar P_{z_1}\cap\bar P_{z_2}\not=\emptyset$ and $y\in\bar 
P_{z_1}
\cap\bar P_{z_2}$, then $y=f_i(F_{\pi_1})=f_i(F_{\pi_2})$,
where $i\in\{ 1,2,3\}$ and $\pi_1:\widehat M_i\rightarrow M_{z^\prime_1}$,
$\pi_2:\widehat M_i\rightarrow M_{z_2^\prime}$ are covering maps with
$z^\prime_1\cong_Rz_1,\ z^\prime_2\cong_Rz_2$.  
Then $F_{\pi_1}E_iF_{\pi_2}$, so
$z^\prime_1\cong_Rz^\prime_2$, and thus $z_1\cong_Rz_2$.

So we have:
\begin{enumerate}
\item[(i)] $z\in\r\Rightarrow\bar P_z$ is countable,
\item[(ii)] $z_1\not\cong_Rz_2\Rightarrow\bar P_{Z_1}\cap
\bar P_{Z_2}=\emptyset$.
\end{enumerate}

We can now complete the proof by showing the following two facts:
\begin{enumerate}
\item[(a)] $\cong_R$ is Borel;
\item[(b)] $P$ is Borel.
\end{enumerate}

Indeed, from (b) and the fact that every section of $P$ is 
countable, we 
can find a Borel function $f: \r\rightarrow Y$ such that $f$ uniformizes
$P$, i.e.
$$P(z, f(z))$$

\noindent for any $z\in\r$.  Then clearly $f$ satisfies condition (ii)
of 4.4 and so by (a) above we can apply 4.4 to conclude that 
$(\cong_R)\leq_BE_\infty$.
\medskip

\noindent {\bf Proof of (a):}  We need the following lemma concerning
our parameterization, which will be proved in \S 7.
\medskip

\noindent {\bf Lemma 4.5.}  {\it Let} $T_i\subseteq \r\times \f(\widehat
M_i^2)$ {\it be defined by}
$$(z, F)\in T_i\Leftrightarrow \exists\pi : \widehat M_i\rightarrow M_z\ 
(\pi\ is\ a\ covering\ map\ and\ F=F_\pi).$$

\noindent {\it Then} $T_i\ is\ \bSi^1_1$.
\medskip

It follows that if $C_i\subseteq\r$ is defined by
$$z\in C_i \Leftrightarrow
{\rm the\ universal\ cover\ of\ }M_z {\rm\ is\ }\widehat M_i,$$

\noindent then $C_i={\rm proj}_\r (T_i)$, so $C_i\in \bSi^1_1$ as well.
Since $\{C_1, C_2, C_3\}$ is a partition of $\r$, it follows that 
actually each $C_i$ is Borel.

We have already seen in \S 2 that $\cong_R$ is $\bSi^1_1$, so to show 
that it is Borel it is enough to show that it is $\bP^1_1$.
Recall
that the equivalence relations $E_i$ induced by the action of
$Aut(\widehat M_i)$ on $\f (\widehat M_i^2)$ are Borel.  Then we have:
$$
z_1\cong_R z_2\Leftrightarrow \exists i\in\{ 1,2,3\} [z_1, z_2\in
C_i\ \&\ \forall F_1\forall F_2[(z_1, F_1)\in T_i \ \&\ 
(z_2, F_2)\in T_i\Rightarrow$$
$$ F_1E_iF_2]$$

\noindent which shows that $\cong_R$ is $\bP^1_1$.
\medskip

\noindent {\bf Proof of (b):}  We have that
$$P(z,y)\Leftrightarrow \exists i\in\{ 1,2,3\} [(z, F)\in T_i\ \&\ 
f_i(F)=y],$$

\noindent so clearly $P$ is $\bSi^1_1$.  To see that it is also 
$\bP^1_1$, notice that
$$P(z, y)\Leftrightarrow \exists i\in\{ 1,2,3\} [z\in C_i\ \&\ 
\forall F ((z, F)\in T_i\Rightarrow\exists j (f_i(f_{i,j}(F))=y))].$$

\noindent {\bf 4.B.  The lower bound}
\medskip

We will use the following realization of $E_\infty$:
\medskip

Recall that we let $F_2$ be the free group with 2 generators and let $p(F_2)$
be the set of all subsets of $F_2$ with the topology given by its
identification with $2^{F_2}$, the latter having the product
topology.  Consider the left-translation action of $F_2$ on $p(F_2)$
$$g\cdot A=gA,$$

\noindent which is clearly continuous.  Remember
that we denote by $E(F_2, 2)$ the
associated orbit equivalence relation and we identify $E(F_2, 2)$ with
$E_\infty$.
It is therefore enough to show that
$$E(F_2, 2)\leq_B (\cong_D).$$

To do this we will associate to each $A\subseteq F_2$ a discrete subset
$S_A\subseteq \bbH $ and consider the domain $D_A=\bbH \setminus S_A$.
We will show that
$$A\ E(F_2, 2)B\Leftrightarrow D_A\cong D_B$$

\noindent (where $D_A\cong D_B$ means of course that $D_A,\ D_B$ are
conformally equivalent).  Moreover the construction is very explicit,
so that there is a Borel function $f: p(F_2)\rightarrow \bbH ^{\bbN}$
such that for any $A\subseteq F_2$, $f(A)$ enumerates $S_A$.  We will
prove in \S 7 the following easy fact about our parameterization of 
domains.
\medskip

\noindent {\bf Proposition 4.6.}  {\it There is a Borel function} 
$g: \bbH ^{\bbN}\rightarrow \d$ {\it so that for any} $x\in \bbH^{\bbN}$
{\it with} $\{x_n: n\in\bbN\}$ {\it discrete we have that}
$D_{g(x)}=\bbH \setminus \{ x_n: n\in\bbN\}$.
\medskip

Then if $h=g\circ f$ we have that
$$A\ E(F_2, 2)B\Leftrightarrow h(A)\cong_D h(B)$$

\noindent i.e., $E(F_2, 2)\leq_B(\cong_D)$.
\medskip

We will now proceed to the construction of $A\mapsto S_A$.

Consider the group $PSL (2, \bbZ )$ of all integer matrices in 
$PSL(2, \bbR )=Aut(\bbH )$.  This acts properly discontinuously on 
$\bbH$,
thus each orbit of this action is a discrete subset of $\bbH $ (see,
e.g., Katok [92]).  Moreover, there are subgroups of $PSL (2, \bbZ )$
isomorphic to $F_2$ which have no fixed points under this action.  An
example is the group generated by the two generators

$$\sigma =\left (\begin{matrix}1 & 2\\0 & 1 \end{matrix}\right ),
\ \ \tau =\left (\begin{matrix}1 & 0\\2 & 1\end{matrix}
\right )$$

\noindent (see Wagon [93, p. 61, 7.1]).  We will identify from now
on $F_2$ with the subgroup $\langle\sigma , \tau\rangle$ of
$PSL(2, \bbZ )$ generated by $\sigma, \tau$.  Then for any fixed
$x\in\bbH ,\ \{ g(x): g\in F_2\}$ is discrete and $g(x)=x$ iff $g=1$ 
(the identity of $F_2$).

So fix from now on $x_1\in\bbH$ and put $x_g=g(x)$.  
Then $g(x_h)=x_{gh}$.

Next we consider the hyperbolic metric $\rho$ on $\bbH$ given by

$$\rho (z, w)=\ln {{\vert z-\bar w\vert+\vert z-w\vert}\over
{\vert z-\bar w\vert -\vert z-w\vert}}.$$

\noindent It of course also induces the usual topology on $\bbH$.
The main fact that we will use is that each element of 
$Aut(\bbH )=PSL(2, \bbR)$ acts by isometries on $(\bbH ,\rho)$ 
(see again Katok [92]).  Since $\{ x_g: g\in F_2\}$ is discrete, there is
an $\epsilon > 0$ such that $\{ y\in\bbH :  \rho (x_1, y)<\epsilon \}
\cap \{ x_g: g\in F_2\} =\{x_1\}$.  It follows that for any
$g\in F_2$, $\{y\in\bbH : \rho (x_g, y)<\epsilon\}\cap \{ x_g: g\in F_2\}
=\{ x_g\}$.

Next put $x^{(0)}_1=x_1$ and choose three points $x^{(1)}_1,\ x^{(2)}_1,
\ x^{(3)}_1$ in $\bbH$, distinct from each other and from $x^{(0)}_1$, 
such
that $\rho (x^{(0)}_1, x^{(i)}_1)<{\epsilon\over 5}$ and the hyperbolic
distances $\epsilon_{ij}=\rho (x^{(i)}_1, x^{(j)}_1),\ 0\leq i\neq
j\leq 3$,
are all different from each other, except for $\epsilon_{ij}=\epsilon
_{ji}$.  Then define $x^{(i)}_g$, for $i=0, 1, 2, 3,$ by

$$x^{(i)}_g=g(x^{(i)}_1),$$

\noindent (so that $x^{(0)}_g=x_g$).  Then $\rho (x^{(i)}_g, x^{(j)}_g)
=\rho (x^{(i)}_1, x^{(j)}_1)$.  It follows that $\{y\in\bbH : \rho (y,
x^{(0)}_g)<{\epsilon\over 2}\}\cap\bigcup_{h\in F_2}\{ x^{(i)}_h: 0\leq i
\leq 3\}=\{x^{(i)}_g: 0\leq i\leq 3\}$ and that the set
$\bigcup_{g\in F_2}\{ x^{(i)}_g: 0\leq i\leq 3\}$ is also discrete.
Notice also that $g(x^{(i)}_h)=x^{(i)}_{gh}$.

Put for any $A\in F_2$

$$S_A=\{ x^{(i)}_g: g\in F_2, 0\leq i\leq 2\}\cup \{x^{(3)}_h:
h\in A\}.$$

\noindent We will show that this works.

First assume that $A\in E(F_2, 2)B$ and let $g\in F_2$ be such that
$gA=B$.  Then clearly $g(S_A)=S_B$ and so $g(D_A)=D_B$, thus
$D_A\cong D_B$.

Conversely, assume that $D_A\cong D_B$ via the biholomorphism $\pi$.
If $\Omega$ is a domain, $\bbD$ the unit disk, $a\in\Omega$ and a 
function
$f:\Omega\rightarrow\bbD$ is holomorphic in $\Omega\setminus \{ a\}$,
then by a classical theorem of Riemann (see, e.g., Rudin [66, 10.20])
$a$ is a removable singularity, so $f$ can be defined at $a$ so that it
remains holomorphic on $\Omega$.  Since $\bbH, \bbD$ are conformally 
equivalent it follows that there are holomorphic extensions $\pi^+:
\bbH\rightarrow \bbH $, $\pi^-: \bbH\rightarrow\bbH$ of $\pi , 
\pi^{-1}$, respectively.  Then $\pi^-\circ\pi^+$ is the identity
on $D_A$, and vice versa, so $(\pi^+)^{-1}=\pi^{-}$.
Thus there is a biholomorphism $\pi^+\in Aut (\bbH )$ extending $\pi$.
For convenience we will just write $\pi$ instead of $\pi^+$ from 
now on.
Clearly $\pi (S_A)=\pi (S_B)$.  We will next find $g\in F_2$ such that
$\pi = g$.  To find this $g$ consider $\pi (x^{(0)}_1)$.  For some
$g$, $\pi (x^{(0)}_1)=x^{(i)}_g$, $i\in \{ 0, 1, 2, 3 \}$.  We 
want to argue that $i=0$, i.e., $\pi (x^{(0}_1)=x^{(0)}_g$.
So assume $i\neq 0$, towards a contradiction.  Let us look at $\pi
(x^{(i)}_1)$.  We have that $\rho (\pi (x^{(i)}_1), x^{(i)}_g)
=\rho (\pi (x^{(i)}_1), \pi (x^{(0)}_1))=\rho (x^{(i)}_1, x^{(0)}_1)<
{\epsilon\over 5}$, so as $\rho (x^{(i)}_g, x^{(0)}_g)<{\epsilon\over
5}$, clearly $\rho (\pi (x^{(i)}_1), x^{(0)}_g)< {\epsilon\over 2}$
so that $\pi (x^{(1)}_1)$ must be one of $x^{(j)}_g$, $j\in \{
0, 1, 2, 3\} ,\ j\neq i$.  Thus, $\rho ( x^{(i)}_g, x^{(j)}_g)=
\epsilon_{ij}=\rho (x^{(0)}_1, x^{(i)}_1)=\epsilon_{0i}$, so, by the
choice of the $\epsilon_{ij}$, we must have $j=0$, i.e., 
$\pi (x^{(0)}_1)=x^{(i)}_g,
\pi (x^{(1)}_1)=x^{(0)}_g$.  Then let $j\in \{ 1, 2\}, j\neq i$, and
consider $\pi (x^{(j)}_1)$.  Again it must be one of the $x^{(k)}_g$,
$k\neq i, 0$.  But then $\epsilon_{k0}=\rho (x^{(k)}_g, x^{(0)}_g)
=\rho (\pi (x^{(j)}_1),\ \pi (x^{(i)}_1))=\rho (x^{(j)}_1, x^{(i)}_1)
=\epsilon_{ji}$, a contradiction.

So $\pi (x^{(0)}_1)=x^{(0)}_g$.  Then since $\rho (\pi (x^{(1)}_1), 
x^{(0)}_g)=\rho (x^{(1)}_1, x^{(0)}_1)=\epsilon_{10}$ we must have that
$\pi (x^{(1)}_1)=x^{(1)}_g$ and similarly $\pi (x^{(2)}_1)=x^{(2)}_g$.
Thus $\pi (x^{(i)}_1)=g(x^{(i)}_1),\ i\in \{ 0, 1, 2\}$.  Since both
$\pi, g$ are M$\ddot {\rm o}$bius transformations this implies that 
$\pi =g$.  Finally we need to show that $gA=B$, and by symmetry it is
enough to see that $gA\subseteq B$.  If $h\in A$, then $x^{(3)}_H\in
S_A$.  So $g (x^{(3)}_h)=\pi (x^{(3)}_h)=x^{(3)}_{gh}\subseteq S_B$, so
$gh\in B$.

It is obvious from our construction that there is a Borel function
$f: p(F_2)\rightarrow \bbH^{\bbN}$ with $f(A)$ enumerating $S_A$, so
the proof is complete.
\hfill$\dashv$

\section{Some consequences}

\noindent {\bf 5.A}
We first notice that the result in \S 4 answers the following question
in Becker-Henson-Rubel [80]:
\medskip

{\bf Q10.}  {\it Is it consistent with the ZFC axioms of set theory
to assume that there is no complete system of invariants $\Phi$, definable by a formula in set theory, where
$\Phi (G)$ is a countable (unordered) subset of $\bbC$ for each domain
G?}
\medskip

Using 4.1 and the fact that any two uncountable standard Borel spaces
are Borel isomorphic, it follows that there is a Borel function
$f: \d\rightarrow\bbC$ and a countable Borel equivalence relation
$E$ on $\bbC$ such that
$$d\cong_Re\Leftrightarrow f(d)Ef(e).$$

In particular, if to any planar domain $D$ we assign the invariant $\Phi (D)
=f([d]_{\cong_R})$, for any $d\in {\cal D}$ with $D_d=D$, we obtain complete
conformal invariants for planar domains which have the form of countable subsets of
$\bbC$.  Moreover, since the work in \S 4 is quite effective, one 
has that both $f$ and $E$ are actually effectively Borel, or $\Delta^1_1$,
and thus the invariant $\Phi (D)$ can be defined by an explicit simple
formula in set theory.  So Q10 is answered negatively in a strong form.
\medskip

\noindent {\bf 5.B}  By loosening-up somewhat a definition given in
Hjorth-Kechris [95] let us say that an equivalence relation $E$ on a 
standard Borel space $X$ is {\it Ulm-classifiable} if there is a map
$U:  X\rightarrow Y^{<\omega_1}$ where $Y$ is a standard Borel space and 
$Y^{<\omega_1}=\{ f: \alpha\rightarrow Y : \alpha <\omega_1\}$, 
with $xEy\Leftrightarrow
U(x)=U(y)$, and such that $U$ satisfies the following technical condition
expressing the ``niceness'' of $U$:

\begin{enumerate}
\item[(i)] If $\alpha (x) = \alpha^U (x)$ = domain of $U(x)< \omega_1$, 
then for each $\alpha < \omega_1$,
$$A_\alpha =A^U_\alpha =\{ x\in X:  \alpha (x)=\alpha \}$$

\noindent has the universal Baire property and the map $U | A_{\alpha}:
X\rightarrow Y^\alpha$ is universally Baire measurable.

\item[(ii)] The set
$$R_U=\{ (x, y)\in X^2: \alpha (x)\leq\alpha (y)\}$$

\noindent has the universal Baire property in $X^2$.
\end{enumerate}
\medskip

Here a set has the {\it universal Baire property} if its preimage by any
Borel function (on any Polish space) has the Baire property and a function 
is {\it universally Baire measurable} if its pre-composition with any such
Borel function is Baire measurable.

Recall that $E_0$ be the equivalence relation on $2^{\bbN}$ given by $x E_0y
\Leftrightarrow\exists n\forall m\geq n(x(m)=y(m))$.  
Then it is a folklore fact that $E_0$ is not
Ulm-classifiable.  To see this recall first that $E_0$ is generically ergodic
in the sense that any $E_0$-invariant set with the property of Baire is
either meager or comeager.  Assume now that $U: 2^{\bbN}\rightarrow 
2^{<\omega_1}$
verifies that $E_0$ is Ulm-classifiable.  By the Kuratowski-Ulam Theorem
(see Kechris [95]) applied to $R_U$ there is some $\alpha_0< \omega_1$, with
$A_{\alpha_0}$ non-meager.  Thus, $A_{\alpha_0}$ is comeager.  Since there
is a Borel embedding from $Y$ into $2^{\bbN}$ and so a Borel embedding of
$Y^{\alpha_0}$ into $(2^{\bbN})^{\alpha_0}$ we can assume, without loss of
generality, that $Y =2$.  For $\alpha <\alpha_0$, there is then a unique
$i_\alpha \in\{ 0,1\}$ so that $\{ x\in 2^{\bbN}: U(x)(\alpha )=i_\alpha \}
=B_\alpha$ is comeager, so $B=\bigcap_{\alpha <\alpha_0}B_\alpha$ is comeager,
which is absurd, since $B$ is a single $E_0$-equivalence class, thus
countable.

Since $E_0\leq_BE (F_2, 2)\sim_B(\cong_D)\sim_B(\cong_R)$, it follows
that $\cong_R,\ \cong_D$ are not Ulm-classifiable either, i.e., one
cannot find complete conformal invariants for planar domains which take the
form of countable transfinite sequences of reals, complex numbers, or members
of any standard Borel space, and which can be defined in a ``reasonable''
way.

In Becker-Henson-Rubel [80], the authors show that in the generic model
$V[G]$, obtained by adding a Cohen real $G$ to the universe $V$, there is no
$U: \{ domains\}\rightarrow\bbC^{<\omega_1}$ which is a complete system
of invariants and can be defined in $V[G]$ by a formula in set theory
with parameters in $V$.  Their proof can be recast in our language as follows: If such a $U$ existed, since $E_0\leq_B(\cong_D)$ by a $\Delta^1_1$
and thus explicitly defined function, we would have (in $V[G]$) a
$\Psi: 2^{\bbN}\rightarrow 2^{<\omega_1}$ definable by a formula of set theory
with parameters in $V$, such that $xE_0 y\Leftrightarrow
\Psi (x)=\Psi (y)$.  By standard
homogeneity properties of the Cohen poset it follows that $\Psi (H_1)=\Psi
(H_2)$ for any two Cohen generics $H_1, H_2\in V[G]$, which is absurd if we
take $H_1, H_2$ to be the odd, even parts of $G$.

Also since Shelah [84] showed that it is consistent relative to ZF+DC 
that every set in a Polish space has the property of Baire, it follows 
that it is
consistent with ZF that there is no map (definable or not) from planar
domains into $Y^{<\omega_1}$, $Y$ a standard Borel space, which gives complete
invariants for conformal equivalence.
\medskip

\noindent {\bf 5.C}  We have seen in \S 4 that the conformal equivalence
relation on domains of the form $\bbH\setminus S$, $S$ discrete, has exactly
the complexity of $E_\infty$.  Of course the same holds for domains of the
form $\bbD\setminus S$, $S$ discrete in $\bbD$ ($\bbD$ the disc).
In some sense these are the simplest Riemann surfaces for which
conformal equivalence can be so complex.

First if we restrict our attention to the compact case, a concrete
classification of $\cong_R$ for compact Riemann surfaces can be achieved,
and we refer the reader to the classical theory of moduli spaces (see, e.g.,
Imayoshi-Taniguchi [92]).  A similar result holds for the Riemann surfaces
obtained by subtracting finitely many points from a compact Riemann surface
(in particular all domains of the form $\bbC\setminus 
\{ x_1,\cdots x_n\}$) and domains of the form $\bbH\setminus \{ x_1, \cdots
, x_n\}$.  Next we consider domains of the form $\bbC\setminus S$, with
$S$ a discrete subset of $\bbC$.  Using the Picard Great Theorem,
it follows that if $\bbC\setminus S$, $\bbC\setminus S^\prime$
are conformally equivalent, then there is $\pi\in Aut (\bbC )$ with
$\pi (S)=S^\prime$ and of course the converse is true as well.  Thus, up
to $\sim_B$, conformal equivalence on domains of the form
$\bbC\setminus S$, $S$ a discrete subset of $\bbC$, is the same
as that of the orbit equivalence relation induced by the action
$$(g, S)\mapsto g\cdot S=g(S)$$

\noindent of $Aut(\bbC)$ on the standard Borel space of discrete subsets
of $\bbC$ (a Borel subset of ${\cal F}(\bbC )$).  We will use this fact
to show that conformal equivalence on the domains of the form 
$\bbC\setminus S$ is actually $<_BE_\infty$.

To see this, we will use the theory of amenability of countable Borel
equivalence relations, see, e.g., Kechris [91].

As in that paper, we call a countable Borel equivalence relation E
{\it amenable} if there is a map $C\mapsto\Phi_C$
associating to each $E$-equivalence class $C=[x]_E$ a mean $\Phi_C$ on $C$
(i.e., a continuous functional $\Phi_C$ on $l^\infty (C)$, with $inf (f)
\leq \Phi_C(f)\leq sup (f))$ such that $C\mapsto\Phi_C$
is universally measurable in the following sense:  if $F : X^2\rightarrow
[-1, 1]$ is Borel, the function $G: X\rightarrow [-1, 1]$ given by
$G(x)=\Phi_{[x]_E} (F_x)$ (where $F_x: [x]_E\rightarrow\bbR$ is defined by
$F_x(y)=F(x, y))$ is universally measurable.  (A function is {\it universally
measurable} if it is $\mu$-measurable for any probability Borel measure
$\mu$).

Since $E(F_2, 2)$ is not amenable (see Kechris [91, 2.3]) and amenability
is preserved downwards under $\leq_B$, it follows that if $E$ is amenable,
then $E<_BE_\infty$.

Denote the space of discrete subsets of $\bbC$ by ${\cal F}_d (\bbC )$
and the orbit equivalence relation induced by the action of $Aut (\bbC )$
on ${\cal F}_d (\bbC )$ by $E_d$.  Then, by 4.3, there is a Borel set
$T\subseteq{\cal F}_d (\bbC )$ meeting every $E_d$-equivalence class
in a countable nonempty set.  Then $E_d\sim_BE_d\vert T$ so in order to show
that $E_d<_B E_\infty$, it is enough to show that $E_d\vert T=E$ is
amenable.  Since the statement ``$E_d<_BE_\infty$'' is clearly $\Pi^1_2$,
so absolute under generic extensions, we can assume that the {\it Continuum
Hypothesis}, CH, holds.

By a theorem of Mokobodzki (see Dellacherie-Meyer [83, pp. 102--108]),
CH implies that there is a universally measurable shift-invariant mean
$\Phi_{\bbN}$ on $\bbN$.  This means that $\Phi_{\bbN}\vert [-1, 1]^{\bbN}:
[-1, 1]^{\bbN}\rightarrow [-1, 1]$ is universally 
measurable and $\Phi_{\bbN} ((x_0, x_1,\cdots ))
=\Phi_{\bbN}(x_1, x_2,\cdots )$.  Next we will use that $Aut (\bbC )=$ the
``$az+b$'' group, is solvable, thus amenable, and so it has a 
{\it F\" olner sequence} $\{ K_n\}$.  Thus $\{ K_n\}$ is a sequence of compact 
subsets of $Aut(\bbC )$ which have the following properties, where 
$\lambda$ is the left-invariant Haar measure on $Aut(\bbC )$:

\begin{enumerate}
\item[(i)] $\lambda (K_n)>0$

\item[(ii)] ${{\lambda (hK_n\Delta K_n)}\over{\lambda (K_n)}}\rightarrow 0$,
\end{enumerate}

\noindent
for each $h\in Aut(\bbC )$ (see Patterson [88, 4.16]).  We now define
a mean on $L^\infty (Aut (\bbC ), \lambda$), i.e., a continuous linear
functional $\Lambda$ on this Banach space with $essinf (f)\leq\Lambda
(f)\leq esssup (f)$, as follows

$$\Lambda (f)=\Phi_\bbN \left (n\mapsto {{\int_{K_n} f(g) d\lambda (q)}\over
{\lambda (K_n)}}\right ).$$

\noindent Then property (ii) of the F\" olner sequence and the shift-invariance
of $\Phi_\bbN$ imply that $\Lambda$ is left-invariant, i.e., for any
$f$ as above if $f_h(g)=f(hg)$, then

$$\Lambda (f)=\Lambda (f_h),\ \ \  \forall h\in Aut (\bbC ).$$

Now let $\pi: \f_d(\bbC )\rightarrow T$ be Borel such that $\pi (x)E_d x$,
$\forall x\in \f_d(\bbC )$.  Then for each $E \; (=E_d | T)$-equivalence class
$C=[x]_{E_d}\cap T$ define a mean $\Phi_C$ on $l^\infty (C)$ by the 
formula
$$\Phi_C(p)=\Lambda \left (g\mapsto p (\pi (g\cdot x))\right )$$

\noindent for any $p\in l^\infty (C)$, where $g\cdot x$ is the action of
$Aut (\bbC )$ on $\f_d(\bbC )$.

The left-invariance of $\Lambda$ shows that this is well-defined 
independently of the choice of $x$ in the same orbit of the action of
$Aut (\bbC )$ on $\f_d (\bbC )$.  Next we check that $C\mapsto\Phi_C$ is
universally measurable, and thus $E$ is amenable.  Fix Borel $F: T^2\rightarrow
[-1, 1]$.  We have to verify that $G: T\rightarrow [-1, 1]$ given by
$G(x)=\Phi_{[x]_E}(F_x)=\Lambda (g\mapsto F_x (\pi (g\cdot x)))=\Lambda (g
\mapsto F(x, \pi (g\cdot x)))$ is universally measurable.  From the definition
of $\Lambda$ we have
$$G(x)=\Phi_\bbN\left (n\mapsto {{\int_{K_n}F(x, \pi (g\cdot x))d\lambda 
(g)}\over{\lambda (K_n)}}\right ).$$

\noindent Put
$$\varphi (n, x)={{\int_{K_n}F(x, \pi (g\cdot x)d\lambda (g)}\over
{\lambda (K_n)}}$$

\noindent Then $\varphi : \bbN\times T\rightarrow [-1, 1]$ is Borel, so
$$\psi (x)=(n\mapsto B(n, x))$$

\noindent is a Borel function from $T$ into $[-1, 1]^\bbN $.  Since
$G=\Phi_\bbN \circ\psi$, and universally measurable functions are closed
under composition, $G$ is universally measurable.

On the other hand, let $E (\bbZ , 2)$ be the equivalence relation induced by
the shift on $p(\bbZ )$ and consider the subset $X\subseteq p(\bbZ )$ 
defined by
$$X =\{ A\subseteq \bbZ: A\ {\rm contains\ two\ consecutive\ integers.}\}$$

\noindent Then for $A, B\in X$
$$AE (\bbZ , 2)B\Leftrightarrow \bbC\setminus A, \bbC\setminus B\ 
{\rm are\ conformally\ equivalent.}$$

\noindent Now the map
$$A\in p (\bbZ\setminus \{\emptyset\})\mapsto A^\prime\in X$$

\noindent defined by
$$A^\prime = A\cup\{ n+1: n\in A\}$$

\noindent clearly has the property that
$$AE (\bbZ , 2)B\Leftrightarrow A^\prime E (\bbZ , 2)B,$$

\noindent thus $E(\bbZ , 2)\vert p(\bbZ )\setminus\{\emptyset\}\leq_B
E(\bbZ , 2)\vert X$.  But it is known, see Dougherty-Jackson-Kechris [94],
that
$$E_0 \sim_B E (\bbZ , 2)\vert p (\bbZ )\setminus\{\emptyset\},$$

\noindent from which it follows that $E_0$ is $\leq_B$ the conformal
equivalence of planar domains of the form $\bbC\setminus S$, $S$ a discrete
subset of $\bbC$.  We in fact conjecture the following.

\medskip
\noindent {\bf Conjecture 5.1.} {\it The conformal equivalence relation
on planar domains of the form $\bbC\setminus S$, $S$ discrete in
$\bbC$, {\it is } $\sim_BE_0$.}
\medskip

Notice also that, as in 5.B above, the conformal equivalence relation on
planar domains of the form $\bbC\setminus S$, $S$ discrete, is not 
Ulm-classifiable.
\medskip

\noindent {\bf 5.D}  As in Becker-Henson-Rubel [80, 6.8], we want to
discuss some implications of our result to the conjugacy action on
$PSL_2(\bbR )=Aut (\bbH )$ on its discrete subgroups.

Put
$$\s_d =\{F\subseteq PSL_2(\bbR ): F\ {\rm is\ a\ discrete\ subgroup}\}.$$

\noindent Then $\s_d$ is Borel in $\f (PSL_2 (\bbR ))$, so it is a standard
Borel space.  Consider the conjugacy action of $PSL_2 (\bbR )$ on $\s_d$,
which is clearly Borel and denote by $E_c$ the corresponding orbit
equivalence relation.  Thus,
$$FE_cG\Leftrightarrow F, G\ {\rm are\ conjugate\ discrete\ subgroups\ of\ } 
PSL_2(\bbR ).$$

\noindent It is essentially shown in 6.8 of Becker-Henson-Rubel [80] that $E_0\leq_B E_c$. We compute again the exact complexity of $E_c$.
\medskip

\noindent {\bf Theorem 5.2.} $E_c\sim_BE_\infty .$
\medskip

\noindent {\bf Proof.}  By 4.3, $E_c\leq_BE_\infty$ and, by 4.B, $E_\infty
\leq_B(\cong_R\vert C_3)$, where $C_3$ is the set of all $z\in\r$ with $M_z$
having universal covering surface $\bbH$, so it is enough to show that
$(\cong_R\vert C_3)\leq_BE_c$.  (Recall here that the universal covering
space of any domain of the form $\bbH\setminus S$, $S$ discrete, must be
$\bbH$.)

If $\pi: \bbH\rightarrow M$ is the universal covering of a Riemann surface
$M$, let $G_\pi$ be the {\it covering group}, i.e., the group of all
$g\in Aut (\bbH )=PSL_2(\bbR )$ with $\pi\circ g=g$.  The group $G_\pi
\subseteq PSL_2 (\bbR )$ is discrete and acts freely on $\bbH$.  Moreover
with the usual definition of $\bbH /G_\pi$, $\bbH /G_\pi$ 
is conformally equivalent to $M$.  Finally, if $G, F\subseteq PSL_2(\bbR )$ are
discrete subgroups acting freely on $\bbH$, then $\bbH /G$ is conformally
equivalent to $\bbH /F$ iff $G, F$ are conjugate in $PSL_2 (\bbR )$ (see
Beardon [84]).

We will next need the following technical lemma about our parametrization,
that we will prove in \S 7.
\medskip

\noindent {\bf Lemma 5.3.}  {\it There is  Borel map  $\varphi :\s_d\rightarrow
\r$  such that if $G\in \s_d$ acts freely on $\bbH$, then
$\bbH /G$ is conformally equivalent to $M_{\varphi (G)}$.}
\medskip

Let now $T\subseteq X=\s_d \cap \{G: G$ acts freely on $\bbH\}$ be
a Borel set meeting every $E_c$-equivalence class in $X$ at a 
countable nonempty set.  Define $Q\subseteq C_3\times T$ by
$$ (z, G)\in Q\Leftrightarrow\varphi (G)\cong_R z.$$

\noindent Thus $Q$ is Borel and each section $Q_z$ is countable, nonempty,
so let $F: C_3\rightarrow T$ be Borel with $(z, F(z))\in Q$.  Then
$$z\cong_R w\Leftrightarrow F(z)E_cF(w),$$

\noindent so $\cong_R\vert C_3\leq_BE_c$, and we are done.
\hfill$\dashv$

\section{Complex manifolds}

We will see here that for $n\geq 2$ the biholomorphic equivalence relation
on $n$-dimensional complex manifolds is much more complicated than that
of Riemann surfaces.  We have the following result:
\medskip

\noindent {\bf Theorem 6.1.} {\it For $n\geq 2$ the relation of
biholomorphic equivalence $\cong_n$ of $n$-dimensional complex manifolds
(in the parameters) does not admit classification by countable structures.}
\medskip

\noindent {\bf Proof.}  It is enough to consider the case $n=2$.  The
proof will be an application of the theory of turbulence, a consequence of
which is the following result.
\medskip

\noindent {\bf Proposition 6.2.} (Hjorth [9?, 3.3.3])  
{\it Consider the Polish
group $G=(\bbR^\bbN, +)$ and let $H\subseteq G$ be a proper Polishable subgroup.  If $H$ is strongly dense (i.e., for every $(x_0 , x_1, \cdots ,x_n)\in
\bbR^{n+1}$ there is $x_{n+1}, x_{n+2}\cdots$ with $(x_0 , x_1,\cdots ,
x_n, x_{n+1}, \cdots )\in H$), then the equivalence relation
$$xE_Hy\Leftrightarrow x-y\in H$$

\noindent does not admit classification by countable structures.}
\medskip

It will thus be enough to find such an $H$ with
$$E_H\leq_B (\cong_2).$$

Consider the unit disk $\bbD$ and the group $H_\bbD$ of all holomorphic
functions $f :\bbD\rightarrow\bbC$ with pointwise addition.  It is easy
(using, e.g., Rudin [66, 10.27])
to see that with the topology of uniform convergence on compacts this is a 
Polish group.  Consider the closed subgroup $H^*_\bbD$ of all $f\in H_\bbD$
such that $f(1/n)\in\bbR$ for all $n\in\bbN,\ n\geq 2$.  Define
$p: \bbH^*_\bbD\rightarrow\bbR^\bbN$ by $p(f)=\left (f\left ({1\over{n+2}}
\right )\right )$.  Then $p$ is a continuous homomorphism from $H^*_\bbD$ into
$\bbR^\bbN$ and so $p(H^*_\bbD) =H$
is a Polishable subgroup of $\bbR^\bbN$ (see Becker-Kechris [96, 1.6])
and is clearly proper (as the sequence $(1, 0, \cdots )\not\in H$).  We next
verify that $H$ is strongly dense in $\bbR^\bbN:$ if $a_0 ,\cdots , a_{m-1}, \in\bbR$,
then there is a polynomial $f$ with real coefficients for which 
$f({1\over{i+2}})=a_i$ for $i< m$, so
$$(f({1\over 2}),\cdots , f({1\over{m+1}}), f({1\over{m+2}}),\cdots )=p(f)\in 
H.$$

So it is enough to show that

$$E_H\leq_B (\cong_2).$$

For $x=(x_n)\in\bbR^\bbN$ let $M(x)$ be the following 2-dimensional complex
manifold:
$$M(x)=(\bbD\times\bbC )\backslash\bigcup_{n\in\bbN}\{{1\over{n+2}}\}\times
\{x_n+(n+2), x_n+(n+3), x_n+2(n+2)\}$$

\noindent  with the obvious chart.  The following
easy fact about our parametrization will be checked in \S 7.
\medskip

\noindent {\bf Lemma 6.3.}  {\it There is a Borel function $F: 
\bbR^\bbN\rightarrow M^2$ such that}
$$M(x)=M_{F(x)}$$

It is then enough to show that 
$$x-y\in H\Leftrightarrow M(x)\cong M(y).$$

\noindent If $x-y\in H$, so that $y-x\in H$ let $f\in H^*_\bbD$
be such that $p(f)=y-x$, i.e., $f$ is
holomorphic on $\bbD$ and $f\left ({1\over{n+2}}\right )=y_n-x_n$.
Then the map $\sigma: M(x)\rightarrow M(y)$ given by 
$$\sigma (z, w)=(z, w+f(z))$$

\noindent is a biholomorphism between $M(x), M(y)$, i.e., $M(x)\cong M(y)$.

Conversely assume that $M(x)\cong M(y)$ and $\sigma : M(x)\rightarrow M(y)$
is a biholomorphism.  Fix $z\in\bbD$.  The map $w\in\bbC\mapsto\pi_1 (\sigma
(z, w))\in\bbD$, where $\pi_1: \bbD\times\bbC\rightarrow\bbD$ is the first
projection function, is holomorphic and bounded, so by Liouville's
Theorem (see Rudin [66, 10.23]) it is constant, say $C(z)$.  (Strictly 
speaking, when $z={1\over{n+2}}$ this map is only defined in the complement of
a finite set, but it clearly has a holomorphic extension to all of $\bbC$.)

It is obvious that for each $n,\; C\left ({1\over{n+2}}\right )={1\over{m+2}}$
for some $m$.  Thus the map $w\in\bbC\backslash \{ x_n+(n+2), x_n+(n+3),
x_n+2(n+2)\}\mapsto\pi_2(\sigma (w))\in\bbC\setminus\{x_m+(m+2), x_m+(m+3),
x_m+2(m+2)\}$, where $\pi_2: \bbD\times\bbC\rightarrow\bbC$ in the second
projection function, is a biholomorphism, so clearly $\bbC\backslash\{n+2,
n+3, 2(n+2)\}$, $\bbC\backslash\{ m+2, m+3, 2(m+2)\}$ are conformally
equivalent.  Thus, as in 5.C, a transformation of the form $z\mapsto az+b$
of $\{n+2, n+3, 2(n+2)\}$, gives us $\{m+2, m+3, 2(m+2)\}$ by considering 
distances we see that $\vert a\vert =1$ and therefore $a=1$ and $m=n$, i.e.
$C \left ({1\over{n+2}}\right )={1\over{n+2}}$.

Since $C$ is holomorphic (being given by $C(z)=\pi_1 (\sigma (z, i)$)), it
follows that $C(z)=z$ for all $z\in\bbD$.

Thus $\sigma$ has the form

$$\sigma (z,w)=(z, \rho_z(w))$$

\noindent where $\rho_z\in Aut(\widehat\bbC )$, i.e., $\rho_z (w)=a(z)w + b(z)$.
As $b(z)=\rho_z(0)=\pi_2(\sigma (z,0))$ and $a(z)=-b(z)+{{\pi_2(\sigma (z, i))}
\over i}$, it follows that $a, b$ are holomorphic.  Since the map
$a\left ({1\over{n+2}}\right )w+b\left ({1\over{n+2}}\right )$ maps $\{ x_n
+(n+2), x_n+(n+3), x_n+2(n+2)\}$ onto $\{y_n+(n+2), y_n+(n+3), y_n+2(n+2)\}$
it follows that $a\left ({1\over{n+2}}\right )=1$ and $x_n+(n+2)=b\left (
{1\over{n+2}}\right )+y_n+(n+2)$, or $x_n-y_n=b\left ( {1\over{n+2}}\right )$,
so $b\in H^*_\bbD$ and $x-y=p(b)\in H$, and the proof is complete.
\hfill$\dashv$
\medskip

It is clear that the complex manifold used above are not compact.  So the
following is open:
\medskip

\noindent {\bf Problem 6.4.}  {\it What is the complexity of the biholomophic
equivalence relation on 2-dimensional compact complex manifolds?}
\medskip

We do not even know if it is Borel or even concretely classifiable.

\section{Technicalities of the parametrization}

We start with some definitions.
\medskip

\noindent {\bf Definition 7.1.} Let $Y^n_0$ be the set of all triples ($O_0,
O_1, \varphi$) such that $O_0$ and $O_1$ are open sets in $\bbC^n$ and
$\varphi : O_0\rightarrow O_1$ is holomorphic.  We provide $Y^n_0$ with the
Borel structure generated by sets of the form
$$\{ (O_0 , O_1 , \varphi ) : U\subseteq O_0 \}$$
$$\{ (O_0 , O_1 , \varphi ) : U\subseteq O_1 \}$$
$$\{ (O_0 , O_1 , \varphi ) : \varphi (V)\subseteq U \}$$

\noindent for $U, V\subseteq\bbC^n$ open.
\medskip

Fixing in advance for each $n$ some enumeration $(q^n_i)_{i\in \bbN}$ (or just
$(q_i)_{i\in \bbN}$ when context makes $n$ clear) of a dense subset of
$\bbC^n$, let $Y^n_1$ the collection of 4-tuples
$$(O_0, O_1, (c_i), (d_i)),$$

\noindent such that

\begin{enumerate}
\item[(i)] $O_0, O_1$ are open sets in $\bbC^n$; each $c_i$ is in $\bbC^n$;
each $d_i$ is in $M_n(\bbC )$ (the space of all $n\times n$ matrices over
$\bbC $);

\item[(ii)] for $q_i$ not in $O_0$ the values of $c_i$ and $d_i$ are both
$(0, 0,\cdots )$ (or any other fixed default value in $\bbC^n$);

\item[(iii)] there exists $\varphi :O_0\rightarrow O_1$ analytic such
that for each $q_i\in O_0$
$$\varphi (q_i)=c_i,$$
$$\varphi^\prime (q_i)=d_i.$$
\end{enumerate}

Letting $\o (\bbC^n)$ be the standard Borel space of open subsets of $\bbC^n$, we have then that $Y^n_1$ is a subspace of the standard Borel space $\o 
(\bbC^n )\times \o (\bbC^n)\times (\bbC^n)^\bbN\times M_n(\bbC )^\bbN$.
Here $\o (\bbC^n)$ is understood to have the Borel structure generated
by sets of the form $\{O\in \o (\bbC^n):U\subseteq O\}$ for $U$ open in
$\bbC^n$.  To see that this is a standard Borel space define
$$\pi :\f (\bbC^n)\rightarrow \o (\bbC^n)$$

\noindent by
$$F\mapsto \bbC^n\setminus F;$$

\noindent this is obviously a Borel bijection with Borel inverse between
$\f (\bbC^n)$ equipped with the Effros standard Borel structure and
$\o (\bbC^n)$.
\medskip

\noindent {\bf Lemma 7.2.}  $Y^n_1$ is a standard Borel space.
\medskip

\noindent {\bf Proof.}  By Kechris [95, 13.A] it suffices to show that $Y^n_1$
is a Borel subset of $\o (\bbC^n)\times\o (\bbC^n)\times (\bbC^n)^\bbN\times
M_n(\bbC )^\bbN$.

Let $(C_p)_{p\in\bbN}$ be the family of compact sets in $\bbC^n$ consisting
of all closures of basic open precompact sets.  Note then that ($O_0,
O_1, (c_i), (d_i))\in Y^n_1$ if and only if for all $C_p$ with $C_p
\subseteq O_0$ we have

\begin{enumerate}
\item[(a)] for all $\epsilon\in\bbQ^+$ there exists $\delta\in\bbQ^+$ such that
$$\forall q_i,q_j\in C_p(\vert q_i-q_j\vert<\delta
\Rightarrow\vert d_i-d_j\vert ,
\vert c_i-c_j\vert <\epsilon );$$
\item[(b)]for all $\epsilon\in\bbQ^+$ there exists $i_0, i_1,\cdots ,i_l$, a
finite sequence in $\bbN$, and $\delta_0, \delta_1, \cdots ,\delta_l$ in
$\bbQ^+$ such that
$$C_p\subseteq \bigcup_{j\leq l}B_{\delta_j}(q_{i_j})$$
and for all $j\leq l$ and $q_m, q_{m^\prime}\in B_{\delta_j}(q_{i_j})$ (the
open ball of radius $\delta_j$ around $q_{i_j}$) we have that if
$$(\zeta_1,\cdots ,\zeta_n)=(q_m-q_{m^\prime})\in\bbC^n,$$
$$(\zeta_1^\prime ,\cdots ,\zeta_n^\prime )=(c_m-c_{m^\prime})\in\bbC^n,$$
$[ \xi_{i, i^\prime} ]_{i,i^\prime\leq n}=d_{i_j}$, then
$$\vert (\zeta_1 ,\cdots , \zeta_n)[\xi_{i,i^\prime}]-
(\zeta^\prime_1,\cdots ,\zeta^\prime_n)\vert /\vert q_m-q_{m^\prime}\vert$$
$$=_{df}\vert(\sum_{i\leq n}\zeta_1\xi_{i,1},\sum_{i\leq n}\zeta_i
\xi_{i,2} ,\cdots ,\sum_{i\leq n}\zeta_i
\xi_{i, n})-(\zeta^\prime_1,\cdots ,
\zeta^\prime_n)\vert /\vert q_m-q_{m^\prime}\vert <\epsilon .
$$
\end{enumerate}

Since all continuous functions on compact sets are uniformly continuous, (a)
asserts that the assignments
$$q_i\mapsto c_i ,$$
$$q_i\mapsto d_i$$

\noindent extend to continuous functions on $C_p$ while (b) 
asserts that $q_i\mapsto d_i$ indicates the derivative of
$q_i\mapsto c_i$.  We leave to the reader the routine verifications that
(a) and (b) describe Borel sets.\hfill$\dashv$
\medskip

\noindent {\bf Corollary 7.3.}  {\it For each $n$, $Y^n_0$ is a standard
Borel space.}
\medskip

\noindent {\bf Proof.}  The natural map from $Y^n_1$ to $Y^n_0$ that associates
to each $(O_0,O_1, (c_i), (d_i))$ the unique holomorphic function
$\varphi$ with
$$\varphi (q_i)=c_i$$
$$\varphi^{\prime} (q_i)=d_i$$

\noindent for $q_i\in O_0$ is clearly one to one and an isomorphism
with respect to the relevant Borel structures.\hfill$\dashv$
\medskip

\noindent {\bf Definition 7.4.}  Let $X^\prime_0$ be the space of all
$(y_{i,i^\prime})_{i,i^\prime\in\bbN}\in\bbR^{\bbN\times\bbN}$ such that
there exists a complete metric space with dense subset $\{ a_i:i\in\bbN\}$,
where for each $i, i^\prime$
$$d(a_i, a_{i^\prime})=y_{i, i^\prime}.$$

Note that $X^\prime_0$ is a Borel subspace of 
$\bbR^{\bbN\times\bbN}$, since the only restrictions on the 
$(y_{i, i^\prime})_{i, i\in\bbN}$ is that they are $\geq 0$ and accord
with the triangle inequality, and that $y_{i, i^\prime}=0$ exactly when
$i=i^\prime$.  For each such $(y_{i, i^\prime})$ we let $X(y_{i, i^\prime})$
be a complete metric space with dense subset $\{a_i:i\in\bbN\}$ as
indicated; the exact choice of $X(y_{i, i^\prime})$ will be irrelevant in
what follows.

We then let $X_0$ be the collection of all $(y_{i, i^\prime})\in X^\prime_0$
such that for each $i\in\bbN$ the open ball
$$B_1(a_i)$$

\noindent of radius 1 around $a_i$ in $X(y_{i, i^\prime})$ is precompact.

Since precompactness can be phrased in terms of being $\epsilon$-bounded
for every rational $\epsilon >0$, this is a Borel subset of $X^\prime_0$, and
thus $X_0$ is a standard Borel space.  It is also seen that 
$X(y_{i, i^\prime})$ will be locally compact for each $(y_{i, i^\prime})$
in $X_0$.  Conversely we have:
\medskip

\noindent {\bf Lemma 7.5.}  {\it If $X$ is a locally compact separable
metric space, then there is some $(y_{i, i^\prime})$ in $X_0$ such that
$X(y_{i, i^\prime})$ is homeomorphic to $X$.}
\medskip

\noindent {\bf Proof.}  Starting with $X$ we may fix a complete $d$ and an
increasing union $(O_m)_{m\in\bbN}$ of precompact open sets so that for
each $m$ the closure of $O_m, \bar O_m$, is included in $O_{m+1}$ and such
that
$$X=\bigcup_{m\in\bbN}O_m.$$

\noindent Then we may choose continuous $f_m:X\rightarrow [m, m+2]$ such that
$f_m$ is constantly $m$ on $\bar O_m$ and constantly $m+2$ on $X\setminus
O_{m+1}$.  Then we may define $d^\prime$ on $X$ by
$$d^\prime (a, a^\prime)=d(a, a^\prime )+\sum_{m\in\bbN}\vert f_m (a)-
f_m(a^\prime)\vert .$$

\noindent Letting $\{ a_i:i\in\bbN\}$ be any dense subset of $X$, we obtain
$(y_{i, i^\prime})$ in $X^\prime_0$ with

$$(y_{i, i^\prime})=d^\prime (a_i,a_{i^\prime}).$$
\hfill$\dashv$

The next technical definition is rather cumbersome.  In any case, we ultimately
parametrize the complex manifolds in a Borel manner, and this is the only
real concern.  By a {\it basic open} set in $\bbC^n$ we mean a ball with
rational center and radius.
\medskip

\noindent {\bf Definition 7.6.}  Let $\m^n$ be the space of sequences
$((y_{i, i^\prime}), (A_i), (\xi_{i,i^{\prime}} ))_{i,i'\in\bbN}$ (or just
$((y_{i, i^\prime}), (A_i), (\xi_{i, i^\prime}))$ for short) such that
\begin{enumerate}
\item[(i)] $(y_{i, i^\prime})\in X_0$;
\item[(ii)] $A_i\in 2^\bbN$ (which we equip with the product topology and
identify with the set of all subsets of $\bbN$);
\item[(iii)] $\xi_{i, i^\prime}\in\bbC^n$;
\item[(iv)] $A_j$ corresponds to a regular open set in $X(y_{i, i^\prime})$,
in the sense that $i\in A_j$ if and only if $\exists\delta\in\bbQ^+\forall
i^\prime\in\bbN\forall q\in\bbQ^+(y_{i, i^\prime}<\delta\Rightarrow\exists
i^{\prime\prime}\in A_j(y_{i', i^{\prime\prime}}<q))$;
\item[(v)] $\{ a_i:i\in A_j\}$ is guaranteed to be precompact by having
diameter less than or equal to 1, in the sense that for all $i, i^\prime\in
A_j,$
$$y_{i, i^\prime}<1;$$

\item[(vi)] if we define $U_j=_{df}(\overline{\{ a_i:i\in A_j\}})^\circ$
for each $j$ to be the interior of the closure of $\{ a_i:i\in A_j\}$
and $V_j=_{df}(\overline{\{\xi_{j, i}:i\in A_j\}})^\circ$ for each $j$
to be the interior of the closure of $\{ \xi_{j, i}:i\in A_j\}$, then the
assignment
$$a_i\mapsto \xi_{j, i}$$

\noindent extends to a homeomorphism
$$\varphi_j:U_j\rightarrow V_j$$

\noindent and the sets $V_j$ are basic open and connected in $\bbC^n$;
\item[(vii)] the overlap maps
$$\varphi_j\circ\varphi^{-1}_i:\varphi_i (U_i\cap U_j)\rightarrow\varphi_j
(U_i\cap V_j)$$

\noindent are biholomorphic;
\item[(viii)] the topological space $X(Y_{i, i^\prime})$ is connected;
\item[(ix)] for each $i$ the closed set $\overline{B_1(a_i)}$ is covered
by finitely many open sets of the form $U_j$ as described in (vi).
\end{enumerate}

\noindent For $p\in\m^n$ we will use $M_p$ to denote the complex manifold
of dimension $n,\; X(y_{i, i^\prime}),$ with chart $\{ U_i,\varphi_i\}$ as
indicated above.
\medskip

\noindent {\bf Lemma 7.7.}  {\it $\m^n$ is a standard Borel space.}
\medskip

\noindent {\bf Proof.}  First let $M^n_0$ be the collection of all $((y_{i, 
i^\prime}), (A_i), (\xi_{i, i^\prime}), (V_{i, j}), (\varphi_{i, j}))$ such
that:

\begin{enumerate}
\item[(i)] $((y_{i, i^\prime}), (A_i), (\xi_{i, i^\prime}))$ satisfy (i)-(vi)
and (ix) of 7.6,
\item[(ii)] for each $i, j$ we have $(V_{i,j}, V_{j,i},\varphi_{i, j})$ in
$Y^n_0$,
\item[(iii)] each $V_{i,j}$ is the interior of the closure of $\{\xi_{i, l}:
l\in A_j\cap A_i\},$
\item[(iv)] for each $l\in A_j\cap A_i$; we have
$$\varphi_{i, j}(\xi_{i, l})=\xi_{j,l},$$

\item[(v)] for all $i, i^\prime$ there is a finite sequence $j_1=i, j_2,\cdots
, j_k=i^\prime$ such that each $A_{j_n}\cap A_{j_{n+1}}\neq\emptyset$.
\end{enumerate}

\noindent Note that (v) expresses that $M_p$ is connected.  Thus $((y_{i, i^\prime}),
(A_i), (\xi_{i, i^\prime}))$ is in $\m^n$ if and only if there is some
$(V_{i,j}),(\varphi_{i,j})$ with $((y_{i,i^\prime}), (A_i), (\xi_{i, i^\prime}), (V_{i,j}),
(\varphi_{i,j}))$ in $M^n_0$ and the functions
$$\varphi_j\circ\varphi^{-1}_i:\varphi_i(U_i\cap U_j)\rightarrow\varphi_j
(U_i\cap U_j)$$

\noindent are equal to $\varphi_{i,j}$.

$M^n_0$ is a subset of the standard Borel space $X_0\times (2^\bbN )^\bbN\times
(\bbC^n)^{\bbN\times\bbN}\times (Y^n_0)^{\bbN\times\bbN}$ and so we endow
it with the relative Borel structure.  To see it is a standard Borel space
we only need check that it is a Borel subset, and the first issue here is
whether 7.6 (vi) corresponds to a Borel condition; this in turn follows
using the precompactness of $\{ a_l: l\in A_j\}$, as in the proof of 7.2.
We also need to be concerned with showing 7.6 (ix) is Borel; but this
amounts to the assertion that there are finite sequences $i_1, i_2,\cdots
, i_n, j_1,\cdots j_n$ in $\bbN$ and $\delta_1, \cdots ,\delta_n\in\bbQ^+,
\delta^\prime_1,\cdots ,\delta^\prime_n\in\bbQ^+$ such that

\begin{enumerate}
\item[(a)] each $\delta_l<\delta^\prime_l$,
\item[(b)] for all $i^\prime\in\bbN$ and $l\leq n$, if $y_{i_l,i^\prime}<
\delta^\prime_l$ (i.e., $a_{i^\prime}\in B_{\delta^\prime_l}(a_{i_l}))$, then $i^\prime\in A_{j_l}$ (i.e. $a_{i^\prime}\in U_{j_l})$; and so, by $\delta_l
<\delta^\prime_l$, we have $B_{\delta_l}(a_{i_l})\subseteq U_{j_l}$;
\item[(c)] for all $i^\prime$ with $y_{i^\prime , i}<1$ (i.e. $a_{i^\prime}\in
B_1(a_i))$, we have some $l$ with $y_{i^\prime , i_l}<\delta_l$ (i.e. 
$a_{i^\prime}\in B_{\delta_l}(a_{i_l}))$.
\end{enumerate}

Granting all this, the projection function

$$((y_{i, i^\prime}), (A_i), (\xi_{i, i^\prime}), V_{i, j}, (\varphi_{i, j}))
\mapsto ((y_{i, i^\prime}), (A_i), (\xi_{i, i^\prime}))$$

$$M^n_0\rightarrow X_0\times (2^\bbN )^\bbN \times (\bbC^n)^{\bbN\times\bbN}$$

\noindent is one-to-one, Borel, and has $\m^n$ as its image.  Thus, by
Kechris [95, 15.1], $\m^n$ is a Borel subset of $X_0\times (2^\bbN )^\bbN\times
(\bbC^n)^{\bbN\times\bbN}$ and therefore standard Borel.

\hfill$\dashv$
\medskip

It is easily seen that for each $M$, a complex manifold of dimension $n$,
there is some $p\in\m^n$ with $M_p$ and $M$ isomorphic.  The only restriction
on our charts is that the sets $V_j$ from 7.6 (vi) are basic open and
connected, but refining a chart as needed we may always make this assumption.

Recall that for $p,q \in \m^n$ we set $p\cong_n q$ if $M_p$ and $M_q$ are biholomorphic. We are finally in a position to verify the lemmas in \S 3-\S 6, whose proofs
we postponed until now.
\medskip

\noindent {\bf Proposition 3.3.} {\it $\cong_n$ is 
$\bSi^1_1$.}
\medskip

\noindent {\bf Proof}.  Fix $p=((y_{i, i^\prime}), (A_i), (\xi_{i, i^\prime}))$
and $q=((y^\prime_{i, i^\prime}), (A^\prime_i), (\xi^\prime_{i, i^\prime}))$
in $\m^n$, let $V_i, U_i$ and $V^\prime_i, U^\prime_i$ be the respective
sets from 7.6 (vi) for $p$ and $q$, and let $\varphi_i$ and $\varphi^\prime_i$
be the respective functions.  Note that we may parametrize the holomorphic
functions from $M_p$ to $M_q$ by sequences of 4-tuples $(V_{i,j}, W_{i,j},
\psi_{i,j}, k(i, j))$ with $(V_{ij}, W_{ij}, \psi_{ij})$ in $Y^n_0$ and
$k(i, j)\in\bbN$, such that

\begin{enumerate}
\item[(i)] $V_i$ is the union of $\{V_{i,j}:j\in\bbN\}$;
\item[(ii)] $W_{i,j}\subseteq V^\prime_{k(i, j)}$;
\item[(iii)] for $l\in A_i\cap A_{i^*},\varphi_i(a_l)=_{df}\xi_{i,l},
\varphi_{i^*}(a_l)=_{df}\xi_{i^*, l}, \xi_{i,l}\in V_{i,j},\xi_{i^*,l}\in
V_{i^*,j^*}$, we have 
$$(\varphi^\prime_{k(i,j)})^{-1}(\psi_{i,j}(\xi_{i,l}))=(\varphi^\prime_{k(i^*,
j^*)})^{-1}(\psi_{i^*,j^*}(\xi_{i^*, l})).$$
\end{enumerate}

(ii) and (iii) are clearly Borel; local compactness of the space gives the
same conclusion for (i).  The significance of (iii) is to ensure that the
partial functions knit together in a well defined fashion and yield a
holomorphic function from $M_p$ to $M_q$.  So given $(V_{i,j}, W_{i,j},
\psi_{i,j}, k(i, j)))_{i,j\in\bbN\times\bbN}$ as above we define
$$\psi :M_p\rightarrow  M_q$$

\noindent extending the assignment
$$\psi (a_l)=(\varphi^{\prime}_{k(i,j)})^{-1}(\psi_{i,j}(\xi_{i,l}))$$

\noindent for $l\in V_{i,j}$.

Thus we have a Borel subset, $Z^n_0$, of $\m^n\times\m^n\times (Y_0^n\times
\bbN )^{\bbN\times\bbN}$ that provides parameters for the collection of all pairs of
complex $n$-dimensional manifolds and holomorphic functions between those
spaces.  For $(p, q,\vec u)\in Z^n_0$, let $\varphi_{\vec u}: M_p\rightarrow
M_q$ be the resulting holomorphic function.  Since it is easily seen that
$$Z^n_1=\{ (p, q,\vec u ,\vec v):(p,q, \vec u), (q, p, \vec v)\in Z^n_0,
\varphi_{\vec u}=(\varphi _{\vec v})^{-1}\}$$

\noindent is a Borel subset of $\m^n\times \m^n \times (Y^n_0\times\bbN )^{\bbN\times
\bbN} \times (Y^n_0\times\bbN )^{\bbN\times
\bbN}$, we then have that $\cong_n$, being the projection of $Z_1^n$ (over $\vec u, \vec v)$, is  $\bSi^1_1$.
\hfill$\dashv$\medskip

\noindent {\bf Definition 7.8.}  We put $\d =\{O\in\o (\bbC ):O$ is
connected$\}$.  It is not hard to see that $\d$ is a Borel subset of $\o
(\bbC )$.  For $d\in\d$, we let $D_d=d$.
\medskip

Recall that $\r$ is just $\m^1$, the space of parameters of Riemann surfaces.
\medskip

\noindent {\bf Proposition 3.5.}  {\it There is a Borel function $f: \d\rightarrow
\r$ such that for each $d\in\d ,\;  D_d\cong M_{f(d)}$.}
\medskip

\noindent {\bf Proof.}  For $d=U$ an open subset of $\bbC$, we can let $(a_i)$
enumerate $U\cap (\bbQ +i\bbQ)$ and let $(V_i)$ enumerate the basic open
sets of diameter $<1$ included in $U$.  We may then let $y_{i,i^\prime}=
d(a_i, a_{i^\prime}), A_j=\{i:a_i\in V_j\}$ and $\xi_{j,i}=a_i$ for $a_i\in
V_j$.  By local compactness of the spaces this can be done in a Borel in
$U$ fashion.  (i)-(ix) of 7.6 then follow easily.
\hfill$\dashv$
\medskip

\noindent {\bf Lemma 4.5.}  {\it Let $\widehat M_1=\bbC_\infty, \widehat M_2=\bbC , \widehat M_3=\bbH$.
For each $i=1, 2, 3,$ let $T_i\subseteq\r\times\f (\widehat\m^2_i)$ be the set
of $(p, F)$ for which there exists
$$\pi :\widehat M_i\rightarrow M_p$$
with $\pi$ a covering map and $F=F_\pi=\{ (x, y):\pi (x)=\pi (y)\}$.
Then $T$ is $\bSi^1_1$.}
\medskip

\noindent {\bf Proof.}  Following the proof of 3.1, we can obtain a
standard Borel space $Z_{2, i}$ of pairs $(\vec u, p)$ such that 
$\vec u$ appropriately represents a holomorphic $\pi_{\vec u}: \widehat M_i
\rightarrow M_p$.  The construction of 3.1 gives that for each
$\zeta, \zeta '\in D$, the fixed countable subset of $\widehat M_i$ and
$q, r\in\bbQ^+$ the set of pairs $(\vec u, p)\in Z_{2,i}$ with
$d(\pi_{\vec u}(\zeta ),\pi_{\vec u}(\zeta '))<q$ and $d(\pi_{\vec u}
(\zeta ), a_n)<r$ is Borel, where $d$ is the metric for $M_p$ and $(a_n)$
the fixed dense subset of $M_p$.

Now the statement that $\pi_{\vec u}$ be a covering map amounts to the
requirement that each unit ball around each $a_n$ in $M_p$ can be
covered by finitely many basic open sets, $B_{\delta_1}(a_{n_1}),\cdots
, B_{\delta_k}(a_{n_k})$, so that for all $j\leq k$ we have, letting
$W_j=\pi^{-1}_{\vec u}(B_{\delta_j}(a_{n_j}))$:

\begin{enumerate}
\item[(i)] any connected component of $W_j$ is precompact and any two
distinct connected components of $W_j$ have disjoint closures;
\item[(ii)] for any connected component $U$ of $W_j, \pi_{\vec u}|U:
U\rightarrow B_{\delta_j}(a_{n_j})$ is biholomorphic.
\end{enumerate}

We fix $j$, and we verify that (i) and (ii) are Borel conditions.
Put $D_j=D\cap W_j$.  Then if we define the equivalence relation:
$$\zeta\sim\zeta '\Leftrightarrow\zeta ,\zeta '\text{are in the same connected
component of }W_j,$$
for $\zeta ,\zeta '\in D_j$, we see that the connected components of $W_j$
are exactly the sets of the form
$$U_C=\text{the interior of the closure of} \;C,$$
where $C$ is a $\sim$-equivalence class.

Thus (i), (ii) above are equivalent to (i)$'$, (ii)$'$ below, where
$C$ varies over all the $\sim$-equivalence classes:

\begin{enumerate}
\item[(i)$'$] Every $U_C$ is precompact and for any two distinct
$C_1, C_2, \bar U_{C_1}\cap\bar U_{C_2}=\emptyset$
\item[(ii)$'$] $\pi_{\vec u}|U_C$ is biholomorphic.
\end{enumerate}

By the methods of 7.2 and the proof of 3.1 we see that these conditions
are Borel and thus the set of $(\vec u, p)$ in $Z_{2,i}$ such that
$\pi_{\vec u}$ is a covering map of $M_p$ is Borel.

It is left to show that $F=F_{\pi_{\vec u}}$ can be expressed in a
Borel manner.  However, this amounts only to requiring that for each
pair of precompact basic open $U_0 , U_1\subseteq\widehat M_i$, we
have that $F\cap (\bar U_0 \times\bar U_i)\neq\emptyset$ iff there
exists $V_0 , V_1$ basic open with $\bar V_0 \subseteq U_0 ,
\bar V_1\subseteq U_1$ and $\forall q\in\bbQ^+\exists\zeta_0 \in D
\cap V_0 \exists\zeta_1\in D\cap V_1(d(\pi_{\vec u}(\zeta_0 ),
\pi_{\vec u}(\zeta_1))<q)$.

\hfill$\dashv$\medskip

\noindent {\bf Proposition 4.6.} {\it There is a Borel function
$$g:\bbH^\bbN\rightarrow \d ,$$

\noindent so that whenever $x=(x_n)\in\bbH^\bbN$ enumerates a discrete
set, we have that $D_{g(x)}=g(x)=\bbH\setminus\{ x_n:n\in\bbN\} $.}
\medskip

\noindent {\bf Proof.}  First let $A$ be the set of sequences $(x_n)\in
\bbH^\bbN$ enumerating a discrete subset of $\bbH$.  This is Borel since
$(x_n)\in B$ if and only if
$$\forall n\exists\delta\in\bbQ^+\forall m(m\neq n\Rightarrow d(x_n,
x_m)>\delta ).$$

\noindent For $x\not\in A$, we can just let $g(x)=\bbH$.  For $x\in A$,
we let $g(x)=\bbH\setminus \{ x_n:n\in\bbN\} .$  This is a Borel
function, since for any basic open $U\subset\bbC$ we have that 
$U\subseteq g(x)$ if and only $U\subseteq\bbH$ and no $x_n$ is in $U$.
\hfill$\dashv$
\medskip

Recall that $S_d$ is the space of discrete subgroups of $PSL_2(\bbR )$.
This is a Borel subset of $\f (PSL_2(\bbR ))$ in the Effros
Borel structure, and hence a standard Borel space.
\medskip

\noindent {\bf Lemma 5.3.} {\it There is a Borel map $\varphi :S_d
\rightarrow\r$ such that $\bbH /G$ is conformally equivalent to
$M_{\varphi (G)}$ for all $G\in S_d$ acting freely on $\bbH$.}
\medskip

\noindent {\bf Proof.}  Note that we can indeed verify in a Borel
manner whether $G\in S_d$ acts freely on $\bbH$:  This amounts to
the claim that for all basic open $U\subseteq\bbH$ we may find a
finite sequence $V_0, V_1,\cdots , V_n$ of basic open sets covering
$\bar U$ and such that for all $i\leq n$,
\medskip

(*) for all $g\in G$ with $g\neq 1_G, g\cdot V_i\cap V_i=\emptyset$.
\medskip

\noindent However (*) is Borel, since it amounts to the assertion
that for all $W\subseteq PSL_2(\bbR )$ basic open not
containing the identity, if $G\cap W\neq\emptyset$ then there
exists $h\in W$ such that $h\cdot V_i\cap V_i=\emptyset$.

So let us just fix $G\in S_d$ acting freely on $\bbH $ and describe 
$\varphi (G)$.

First we let $(V_i)$ enumerate the basic open sets which are
$G$-discrete, in the sense of meeting each $G$-orbit in at
most one point, and have diameter $<1$ in the hyperbolic metric
$\rho$ (see 4.B).  As in the proof of 3.3, this can be used to
give a chart for a representative of $\bbH /G$ in $\r$.  The
one further problem is in uniformly obtaining a metric.

For any $\zeta , \xi\in\bbH$ and $g\in G$ we have
$$\text{inf}_{h\in G}\rho (h\cdot\zeta ,\xi )=\text{inf}_{h\in G}\rho
(h\cdot\zeta , g\cdot\xi ),$$

\noindent and moreover this quantity is greater than zero if and only
if $G\cdot\zeta\neq G\cdot\xi $.  In particular
$$\text{inf}_{h\in G}\rho (h\cdot\zeta ,\xi )=\text{inf}_{h, g\in G}\rho
(h\cdot\zeta , g\cdot\xi ).$$

\noindent Therefore
$$\tilde d (G\cdot\zeta , G\cdot\xi )=\text{inf}_{h, g\in G}\rho
(h\cdot\zeta , g\cdot\xi )$$

\noindent provides the needed metric on $\bbH /G$.

Thus if we let $(a_i)$ enumerate a maximal $G$-discrete subset of
$\bbH\cap (\bbQ +i\bbQ )$ (in the sense that $G\cdot a_i\cap G\cdot
a_j =\emptyset$ for all $i\neq j)$, $y_{i, i'}=
\tilde d (G\cdot a_i, G\cdot a_i'), A_i=\{j:G\cdot a_j\cap V_i\neq
\emptyset \}$, and $\zeta_{i,j}$ to be the unique element in $V_i
\cap G\cdot a_j$, if it exists, we obtain from $p=((y_{i, i'}), 
(A_i), (\zeta_{i, i'}))$ an element in $\r^n$ with $M_p$ conformally
equivalent to $\bbH /G$.

There is the further concern that all these steps can be performed
in the Borel context, but this is routine and resembles earlier
calculations.
\hfill$\dashv$
\medskip

Recall that for $x\in\bbR^\bbN$ we let $M(x)$ be the complex manifold
$$(\bbD\times\bbC )\setminus \bigcup_{n\in\bbN}\{ 1/(n+2)\}\times
\{ x_n+(n+2), x_n+(n+3), x_n+2(n+2)\}$$

\noindent endowed with the inherited complex structure.
\medskip

\noindent {\bf Lemma 6.3.} {\it There is Borel map

$$F:\bbR^\bbN\rightarrow \m^2$$

\noindent such that $M(x)$ and $M_{F(x)}$ are biholomorphic for all
$x\in\bbR^\bbN$.}
\medskip

\noindent {\bf Proof.}  This follows the method of 3.3.\hfill$\dashv$

\newpage

\begin{center}
\section*{References}
\end{center}
\bigskip

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\vskip 1in
\noindent Department of Mathematics, 
UCLA, Los Angeles, CA 90095;
greg@@math.ucla.edu
\medskip
\noindent Department of Mathematics,
Caltech,
Pasadena, CA 91125;
kechris@@caltech.edu

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