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\def\leftheader{\centerline{S. KLIMEK and A. LESNIEWSKI}}
\def\rightheader{\centerline{QUANTUM RIEMANN SURFACES}}
\def\intsec{I}
\def\pionesec{II}
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{\baselineskip=12pt
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\line{\hfill \bf HUTMP 95/443}
\line{\hfill \today}
\vfill
\title{Quantum Riemann Surfaces}
\vskip 1cm
\title{for Arbitrary Planck's Constant}
\vskip 1in
\centerline{{\bf Slawomir Klimek}$^*$
and {\bf Andrzej Lesniewski}$^{**}$
}
\vskip 12pt
\centerline{$^*$Department of Mathematics}
\centerline{IUPUI}
\centerline{Indianapolis, IN 46205, USA}
\centerline{Supported in part by the National Science Foundation under
grant DMS--9500463}
\vskip 12pt
\centerline{$^{**}$Lyman Laboratory of Physics}
\centerline{Harvard University}
\centerline{Cambridge, MA 02138, USA}
\centerline{Supported in part by the National Science Foundation under grant
DMS-9424344}}
\vskip 1in
\centerline{\bf Abstract}
\vskip 3mm\noindent
We continue our study of quantum Riemann surfaces initiated in
[1, 2, 3]. We construct a one parameter family of deformations
of compact Riemann surfaces of genus $g\geq 2$. Our construction
does not require any discretness condition on the value of Planck's
constant. It coincides with the construction of [2] in the case
when Planck's constant assumes the discrete set of values dictated
by geometric quantization.
\vfill\eject
\section\intsec{Introduction}
In a series of papers [1], [2], [3], we studied non-perturbative
deformation quantization of Riemann surfaces. Our approach is based on
the ideas of [4] (for related developments, see also [5], [6], [7],
and references therein.) A satisfactory picture of uniformization of
exceptional quantum Riemann surfaces emerged from these investigations.
In the case of higher genus $(g\geq 2)$ Riemann surfaces, the
uniformization on the quantum level is a more complex issue. In fact,
if $M$ is a Riemann surface and $N$ is a covering of $M$, then the
quantization of $N$ is a covering of the quantization of $M$ in the
sense of [3] only if the fundamental group of the covering $N\to M$ is
abelian. Part of the problem is the presence of toplogical sectors
(similar to the $\theta$-vacua in gauge theory) in the quantum theory,
which is related to the non-simple connectedness of the classical phase
space. These sectors are classified by the characters of the fundamental
group of the phase space. This does not reflect the non-commutativity of
that group. On the other hand, whether the fundamental group is
commutative or not seems to be important for the quantum uniformization
in the sense of [3]. A similar phenomenon was discussed previously
in [8].
Quantization of Riemann surfaces in the framework of geometric
quantization (see e.g. [9]) requires a restriction on the allowed
values of the deformation parameter $r$ (``the quantization condition''.)
However, it is desirable to have a definition of quantum Riemann
surfaces for all values of $r$. A definition consistent with geometric
quantization was given in [2], for $r=n(2g-2)^{-1}$, where $n\in
\bN$. In principle, quantum uniformization allows for a construction
of quantum Riemann surfaces for all values of $r$, using the universal
covering, the Poincar\'e disk, as the point of departure. Since the
fundamental groups of higher genus Riemann surfaces are non-abelian,
it is likely, however, that the so defined algebra of quantized
functions does not reduce to the algebra of [2], when $r=n(2g-2)^{-1}$.
In this paper, we construct quantization of compact Riemann surfaces
for arbitrary values of $r>0$ in a manner consistent with [2].
Our starting point is a non-compact covering space $\hat M$ such
that the group of cover transformation is the abelian group $\bZ$.
Since $\hat M$ is a non-compact Riemann surface, all holomorphic
line bundles are holomorphically trivial [10], and geometric quantization
does not impose any restrictions on Planck's constant. It was
explained in [2] that geometric quantization of Riemann surfaces
leads to certain operator algebras on Hilbert spaces of automorphic
forms. We define the quantization of $M$ in terms of Toeplitz operators
with invariant symbols on a space of automorphic forms on $\hat M$,
and prove deformation estimates. The proof of the estimates is based
on the methods developed in [2] (see [11] for a different approach.)
The paper is organized as follows. In Section \pionesec , we study the
fundamental groups of $M$ and $\hat M$. We define and study suitable
spaces of automorphic forms on $M$ and $\hat M$ in Section \autformsec .
Section \toeplitzsec\ contains proofs of deformation estimates.
\section\pionesec{The fundamental groups}
In this section, we fix our notation and describe certain group
theoretic properties of the fundamental groups of Riemann surfaces
which will be useful in later sections. Let $M$ be a compact Riemann
surface of genus $g\geq 2$. We pick a basis $\{a_i,b_i\}$, $i=1,..,g,$
of one-cycles on $M$. The fundamental group $\Gamma$ of $M$ is a
finitely generated group with generators $\{a_i,b_i\}$, $i=1,\ldots ,
g,$ obeying the relation (see e.g [12])
$$
\prod_{i=1}^g a_ib_ia_i^{-1}b_i^{-1}=1\ .
$$
We denote $a:=a_g$ and $b:=b_g$. An element $\gamma\in\Gamma$ can be
represented as
$$
\gamma=\prod_{j=1}^\infty\prod_{i=1}^g a_i^{n_{i,j}}b_i^{m_{i,j}},
$$
where almost all integers $n_{i,j}, m_{i,j}$ are equal to zero.
Consider now the homomorphism
$$
\gamma\to\sum_j m_{g,j}\in\bZ\ .
$$
One can easily verify that the above map is well defined, i.e. it
does not depend on the way we represent $\gamma$, and is indeed a
homomorphism of groups. We denote the kernel of this homomorphism by
$\gz$. The group $\gz$ is no longer finitely generated. In fact, it is
not difficult to see that the following elements are its generators:
$$
a,\ b^na_ib^{-n},\ b^nb_ib^{-n},\ {\rm where}\ n\in \bN,\ i=0,1,\ldots ,
g-1\ ,
$$
and there are no relations among them, i.e. $\Gamma_0$ is the free
product of infinite cyclic groups generated by the above generators.
Let $\hat M$ be a covering of $M$ with no branching points and
such that $\pi_1(\hat M)=\gz$. The group of cover transformations
of the covering $\hat M\to M$ is then equal to $\bZ$. Intuitively,
$\hat M$ is obtained from $M$ by cutting along the $a$ cycle and then
continuing $M$ across the cut along the $b$ cycle. One can visualize
$\hat M$ as an infinite cylinder with infinite number of handle bodies
attached, each handle body having $g-1$ handles. $\hat M$ is a non-compact
Riemann surface.
The universal covering space of both $M$ and $\hat M$ is the unit disk
$\ud$. The groups $\Gamma$ and $\gz$ can be think of as Fuchsian groups
on $\ud$. For
$$
\gamma=\left[\matrix{a&b\cr
\bar a&\bar b\cr}\right]\in\Gamma,
$$
we denote
$$
\gamma(z)={az+b\over \bar b x+\bar a}\; ,\ \ z\in\ud.\ref{\gammaref}
$$
Since the derivative of \gammaref\ is $\gamma'(z)=(\bar bz+\bar a)^{-2}$,
we can set:
$$
\gamma'(z)^{1/2}:=(\bar bz+\bar a)^{-1}.
$$
Let $G$ be either $\Gamma$ or $\gz$, and let $r>0$. A multiplier
of weight $r$ for the group $G$ is a map $v:G\to \bC$ such that [12]
$$
|v(\gamma)|=1,
$$
and
$$
\gamma_1'(\gamma_2(z))^{-r/2}\gamma_2'(z)^{-r/2} v(\gamma_1)v(\gamma_2)=
(\gamma_1\gamma_2)'(z)^{-r/2} v(\gamma_1\gamma_2)\ .\ref{\multref}
$$
In \multref , the standard branch of the logarithm is taken
to define the $r$-th power. The existence of multipliers is a
cohomological question, see [13] for the modern treatment. Equation
\multref\ can be interpreted as the triviality of a 2-cocycle in the
group cohomology of $G$. A multiplier is essentially a 1-cochain whose
coboundary is that 2-cocycle. It is well-known that multipliers exist
for $\Gamma$ if and only if $r=n(2g-2)^{-1}$, $n=1,2,\ldots$. The
situation is different for $\gz$.
\prop\cohoprop{The second cohomology group $H^2(\gz,\bZ)$
of $\gz$ is trivial.}
\medskip\noindent{\it Proof.} Since $\gz=\pi_1(\hat M)$, and the
universal covering space of $\hat M$ is contractible, it follows from
Eilenberg-MacLane's theorem [14] that
$$
H^*(\gz,\bZ)\simeq H^*(\hat M,\bZ)\ .
$$
By Poincar\'e duality [15], $H^2(\hat M,\bZ)$ is isomorphic to the
compactly supported cohomology group of $\hat M$ in dimension zero.
The latter is trivial, as $\hat M$ is non-compact. $\square$
Consequently, multipliers for $\gz$ exist for arbitrary $r$. Let us also
remark that the ratio of two multipliers is a character of the group.
\section\autformsec{Automorphic forms}
As before, let $G$ be either $\Gamma$ or $\gz$, and let $v$ be a
multiplier for the group $G$. Recall that a holomorphic function
$\phi:\,\ud\to\bC$ is called an automorphic form for $G$ of
weight $r>0$ with multiplier $v$, if [12]:
$$
\phi(\gamma(z))=v(\gamma)\gamma '(z)^{-r/2}\phi(z)\ ,\ref{\autformref}
$$
for each $\gamma\in G$. Automorphic forms for infinitely generated
Fuchsian groups like $\gz$ have been studied less extensively than
those for finitely generated groups, but there is a fair amount of
information available, see [16], [17], and references therein.
Let $R$ be a fundamental polygon for $\Gamma$. Then
$R_0:=\bigcup_{n\in\bZ}b^nR$ is a fundamental polygon for $\gz$.
It has infinitely many sides. Here we use the same symbol $b$
to denote the group element of $\Gamma$ corresponding to the cycle
$b=b_g$. For $r=n(2g-2)^{-1}$, we define $\hg$ to be the Hilbert
space of automorphic forms for $\Gamma$ equipped with the following
scalar product:
$$
(\phi,\psi):=\int_R\overline{\phi(z)}\psi(z)\,d\mu^r(z),\ref{\scalprod}
$$
where the measure $d\mu^r(z)$ is
$$
d\mu^r(z):={r-1\over\pi}(1-|z|^2)^{r-2}\,d^2z\ .
$$
For arbitrary $r>0$, let $\hgz$ be the space of automorphic forms
for $\gz$ with the scalar product as in \scalprod\ but $R_0$
replacing $R$.
If $v$ is a multiplier for $\Gamma$ and $e^{i\theta}\in S^1$, we let
$v_\theta$ denote a new multiplier for $\Gamma$ defined by:
$$
v_\theta(b)=v(b)e^{i\theta},
$$
and $v_\theta=v$ on all other generators. In particular $v_0=v$.
All of the multipliers $v_\theta$ are equal when restricted to the
subgroup $\gz$. The restricted multiplier will be again denoted
by $v$.
\thm\autformthm{With the above definitions, there is a canonical
isomorphism
$$
\hgz\cong\int_{S^1}^\oplus\hgt\,d\theta,\hskip .5cm r=n(2g-2)^{-1},\
n=1,2,\ldots\ .
$$
}
\noindent{\it Proof.} For an automorphic form $\phi$ for $\gz$,
define
$$
U\phi(z)=b'(z)^{r/2}\phi(bz)\ .\ref{\udefref}
$$
We claim that \udefref\ is again an automorphic form, and in fact
$U$ is a unitary operator on $\hgz$. Let us first verify \autformref :
$$\eqalign{
U\phi(\gamma z)&=b'(\gamma z)^{r/2}\phi(b\gamma b^{-1}bz)\cr
&=v(b\gamma b^{-1}) b'(\gamma z)^{r/2}
(b\gamma b^{-1})'(bz)^{-r/2}\phi(bz).\cr
}$$
Using the chain rule and \multref , we obtain \autformref .
Unitarity of $U$ is a consequence of the following calculation:
$$\eqalign{
(U\phi,U\psi)&=\int\limits_{R_0}\overline{b'(z)^{r/2}\phi(bz)}
b'(z)^{r/2}\psi(bz)\,d\mu^r(z)\cr
&=\int\limits_{R_0}\overline{\phi(bz)}\psi(bz)\,d\mu^r(bz)=(\phi,\psi),\cr
}$$
where we have used the transformation properties of $d\mu^r(z)$ and
the fact that $R_0$ is invariant under $b$.
The isomorphism of Hilbert spaces that we want to establish is in essence
the spectral decomposition of $U$. Explicitly, we define
$$
P:\;\hgz\to\int_{S^1}^\oplus\hgt\,d\theta\
$$
by the following formula:
$$
P\phi(\zeta,\theta)=\sum_{n\in \bZ}v_\theta(b^{-n})
U^n\phi(z)\ .\ref{\isoref}
$$
We need to verify that the right hand side of \isoref\ is in $\hgt$.
If $\gamma\in\gz$, this follows from the fact that $U^n\phi(z)$ are
automorphic forms for $\gz$, and the fact that $v_\theta|_{\gz}=
v|_{\gz}$. If $\gamma=b$, we have
$$
U\left(\sum_{n\in \bZ}v_\theta(b^{-n}) U^n\phi(z)\right)=v_\theta(b)
\left(\sum_{n\in \bZ}v_\theta(b^{-n}) U^n\phi(z)\right)\ .
$$
The general case follows easily.
We verify that $P$ is an isometry:
$$
\eqalign{
\|P\phi\|^2&=\int\limits_{-\pi}^{\pi} \biggl(\int\limits_R
|P\phi(z,\theta)|^2\, d\mu^r(z)\biggr)\,{d\theta\over 2\pi}\cr
&=\int\limits_{-\pi}^{\pi} \biggl(\int\limits_R \sum_{n,m}
\overline{U^n\phi(z)} U^m\phi(z) v_\theta(b)^{n-m}
\, d\mu^r(z)\biggr)\,{d\theta\over 2\pi}\cr
&=\int\limits_R \sum_n |U^n\phi(z)|^2\,d\mu^r(z)\cr
&=\sum_n \int\limits_{b^nR} |\phi(z)|^2\,d\mu^r(z)\cr
&=\int\limits_{R_0} |\phi(z)|^2\,d\mu^r(z)\ .}
$$
Similar calculations show that the inverse of $P$ is given by:
$$
P^{-1}\psi(z)=\int\limits_{-\pi}^\pi \psi(z,\theta)\,d\theta\ .\quad
\square
$$
This result implies that the quantization of Riemann surfaces proposed
in this paper reduces, when $r=n(2g-2)^{-1}$, to the definition of [2].
\section\toeplitzsec{Toeplitz quantization}
In this section, we construct a quantization of the Riemann surface $M$
in terms of Toeplitz operators on $\hgt$. We first recall the relevant
definitions. The reproducing kernel $\kgz zw$ for $\hgz$ is given by the
following Poincar\'e series:
$$
\kgz zw =\sum_{\gamma\in\gz}v(\gamma)^{-1}\gamma '(z)^{r/2}
\k {\gamma(z)}w,\ref{\kernelref}
$$
where
$$
\k zw =(1-z\bar w)^{-1}\ .
$$
Let $C_\Gamma(\ud)$ be the $\bC^*$-algebra of bounded continuous
functions on $\ud$ which are invariant under $\Gamma$, so that
$C_\Gamma(\ud)\cong C(M)$. For $f\in C_\Gamma(\ud)$, we define
the Toeplitz operator $\tg f$ on $\hgz$ with symbol $f$ by:
$$
(\tg f \phi)(z)=\int_{\gz}\kgz zw f(w)\phi(w)\,d\mu^r(w)\ .
\ref{\toeplitzref}
$$
The goal of this section is to prove that the correspondence
$f\mapsto \tg f$ is a quantization of $M$. This means that for
$f\in C_\Gamma(\ud)$ we have the norm limit
$$
\lim_{r\to \infty} \| \tg f \| = \| f \|_{\infty},\ref{\normestref}
$$
where $||\cdot||$ denotes the operator norm, and where
$||\cdot||_\infty$ denotes the sup-norm. If, moreover, $f,g$
are smooth, then
$$
\lim_{r\to \infty}\|r\bigl[\tg f,\tg g\bigr]+\tg {i\{f,g\}}\|=0,
\ref{\comestref}
$$
where $\{f,g\}$ is the usual Poisson bracket,
$$
\{f,g\}(z)=i(1-|z|^2)^2[\partial f(z)\bar\partial g(z)-
\partial g(z)\bar\partial f(z)]\ .\ref{\bracketref}
$$
\medskip
\thm\defquantthm{With the above definitions, the correspondence
$$
f\mapsto \tg f
$$
is a quantization of $M$.
}
\medskip\noindent{\it Proof.} The details of analogous estimates were
explained in [1] and [2]. Here we follow [2], where the estimates
were proved for the quantization based on automorphic forms of $\Gamma$.
However, the compactness of the fundamental domain of $\Gamma$ was used
in an essential way in several places, so that the results cannot be
applied to the case of $\gz$ (as $R_0$ is not compact in $\ud$). The main
difference is that ``transfer of regularity" argument has to be done more
carefully in the present case.
To prove \normestref , we consider the vectors
$$
\phi_w(z):=\kgz ww^{-1/2}\kgz zw
$$
in $\hgz$, and verify that:
$$
\sup_{x\in R}|f(w)-(\phi_w,\tg f\phi_w)|\to 0,\ \ \ {\rm as}\ r\to\infty ,
\ref{\phiwestref}
$$
in a way analogous to [2]. The proof there was based on Lemmas 4.1
and 4.2 which are also valid for $\gz\subset\Gamma$. Since $f$ is
invariant under $\Gamma$, and not just $\gz$, and since the supremum
in \phiwestref\ is taken over a compact set, \phiwestref\ follows
exactly as in [2].
The estimate \comestref\ is a consequence of
$$
||r\left(\tg f\tg g-\tg {fg}\right)
+\tg {(1-|z|)^2\partial f\bar\partial g}||\to 0,\ \ \ {\rm as}\ r\to\infty,
\ref{\prodestref}
$$
for $f,g\in C_\Gamma^\infty$. To prove \prodestref , one expands
$(\phi,\tg f\tg g\psi)$ in a Taylor series as in [2] formula (5.6).
The first three terms in that formula combine to give the second and
third terms in \prodestref . The analog of the fourth term of
[2], formula (5.6), is $O(r^{-2})$ as in [1]. It remains to estimate
the remainder.
The technique developed in [1] for estimating the remainder terms can be
applied to our case with one modification. The integral
$\int_\ud |\psi(w)|^2\,d\mu^r(w)$ is infinite if $\psi\in\hgz$, and
one needs to transfer a power of $1-|w|^2$ to make it convergent. This
``transfer of regularity" trick was explained in detail on the last
two pages of [2] for $\Gamma$-automorphic forms. However, the
compactness of the fundamental domain of $\Gamma$ was used in an
essential way. We show below that a modification of the argument used
in [2] can be used to in our case.
\lemma\firstlemma{ Let $\psi\in\hgz$ and $s>1$. Then we have:
$$
\int_\ud|\psi(w)|^2(1-|w|^2)^s\,d\mu^r(w)=O(1)||\psi||^2\ .
$$
}
\lemma\secondlemma{Let $b\in SU(1,1)$ be hyperbolic, let $K$
be a compact set in $\ud$, and let $t>0$. Then we have:
$$
\sup_{z\in K}\;\sum_{n\in\bZ}\left|(b^n)'(z)\right|^t=O(1)\ .
$$
}
We will prove these lemmas after we have completed the main line of
the argument.
We now use the above lemmas to estimate the term in formula (5.16) of
[2]. This will conclude the proof of \defquantthm . That term reads:
$$
\int_{R_0\times\ud}|\phi(z)|(1-|z|^2)^{1-r/2}|\psi(\gamma_z(w))|
|\gamma_z'(w)|^{r/2}{|w|^2\over (1-|w|^2)^{11}}\,d\mu^r(z)\,d\mu^r(z),
\ref{\atermref}
$$
where $\gamma_z(w)=(w+z)/(\bar zw+1)$. Let $0<\epsilon<1/2$. We
multiply and divide the integrand by $(1-|\gamma_z(w)|^2)^{1-\epsilon}$,
and use the following elementary bound:
$$
{1\over (1-|\gamma_z(w)|^2)^{1-\epsilon}}\leq{O(1)\over
(1-|z|^2)^{1-\epsilon}(1-|w|^2)}\ .
$$
The integral in \atermref\ is consequently less then
$$
\eqalign{
O(1)&\int_{R_0\times\ud}|\phi(z)|(1-|z|^2)^{\epsilon-r/2}|
\psi(\gamma_z(w))|(1-|\gamma_z(w)|^2)^{1-\epsilon}|\gamma_z'(w)|^{r/2}\cr
&\times{|w|^2\over (1-|w|^2)^{12}}\,d\mu^r(z)\,d\mu^r(z)\ .\cr
}\ref{\btermref}
$$
\noindent Using the Schwarz inequality and changing variables in the
$\psi$ term, we get the following bound:
$$
\eqalign{
O(1)\;||\phi||\;&\left(\int_{R_0}(1-|z|^2)^{2\epsilon-r}\,
d\mu^r(z)\right)^{1/2}\left(\int_\ud|\psi(w)|^2(1-|w|^2)^{2-2\epsilon}\,
d\mu^r(w)\right)^{1/2}\cr
&\times\left(\int_\ud{|w|^4\over (1-|w|^2)^{24}}\,
d\mu^r(w)\right)^{1/2}\ .\cr}\ref{\ctermref}
$$
With our choice $0<\epsilon<1/2$, the exponent $2-2\epsilon$ in the
third factor is greater than $1$, and so we can apply \firstlemma\ to
it, and conclude that it is $O(1)||\psi||^2$. The fourth factor
is $O(r^{-1})$ by [2], formula (5.18). One can analyze the second
factor in \ctermref\ as follows:
$$
\eqalign{
\int_{R_0}(1-|z|^2)^{2\epsilon-r}\,d\mu^r(z)&=
O(r)\int_{R_0}(1-|z|^2)^{2\epsilon}\,d\mu_P(z)\cr
&=O(r)\sum_{n\in\bZ}\int_{b^{-n}R}(1-|z|^2)^{2\epsilon}\,d\mu_P(z)\cr
&=O(r)\sum_{n\in\bZ}\int_{R}(1-|b^nz|^2)^{2\epsilon}\,d\mu_P(z)\cr
&=O(r)\int_R\sum_{n\in\bZ}\left|(b^n)'(z)\right|^{2\epsilon}
(1-|z|^2)^{2\epsilon}\,d\mu_P(z)\cr
&\leq O(r)\sup_{z\in R}\;\sum_{n\in\bZ}\left|(b^n)'(z)
\right|^{2\epsilon}\ .\cr
}\ref{\dtermref}
$$
In \dtermref , $d\mu_P(z)$ is the Poincar\'e measure on $\ud$,
and we have used the fact that $R$ is compact. Since $2\epsilon>0$
it follows from \secondlemma\ that \dtermref\ is $O(r)$. This
concludes the transfer of regularity argument.
\noindent
{\it Proof of \firstlemma\ :}
We decompose $\ud$ into the translates of $R_0$:
$$
\eqalign{
&\int_\ud|\psi(w)|^2(1-|w|^2)^s\,d\mu^r(w)=
\sum_{\gamma\in\gz}\int_{\gamma^{-1}R_0}|\psi(w)|^2(1-|w|^2)^s
\,d\mu^r(w)\cr
&=\sum_{\gamma\in\gz}\int_{R_0}|\psi(w)|^2(1-|\gamma(w)|^2)^s
\,d\mu^r(w)\leq\left(\sup_{w\in\ud}\sum_{\gamma\in\gz}
(1-|\gamma(w)|^2)^s\right)||\psi||^2\ .\cr
}
$$
To estimate the supremum factor, we proceed as in the proof of
[2] lemma 4.2:
$$
\eqalign{
1&=(1-|w|^2)^s\int_\ud |\ks wz|^2\,d\mu^s(z)\cr
&=(1-|w|^2)^s\sum_{\gamma\in\gz}\int_{\gamma^{-1}R_0}
|\ks wz|^2\,d\mu^s(z)\cr
&=(1-|w|^2)^s\sum_{\gamma\in\gz}|\gamma '(w)|^s\int_{R_0}
|\ks {\gamma(w)}z|^2\,d\mu^s(z)\cr
&=\sum_{\gamma\in\gz}(1-|\gamma(w)|^2)^s\int_{R_0}
|\ks {\gamma(w)}z|^2\,d\mu^s(z)\cr
&\geq\sum_{\gamma\in\gz}(1-|\gamma(w)|^2)^s{1\over 2^{2s}}\int_{R_0}
\,d\mu^s(z)\ .
}
$$
Hence, $\sup_{w\in\ud}\sum_{\gamma\in\gz} (1-|\gamma(w)|^2)^s=O(1)$,
if $s>1$. This concludes the proof of \firstlemma .
\noindent
{\it Proof of \secondlemma\ :}
Since $b$ is a hyperbolic element of $SU(1,1)$, it has two real
eigenvalues $\lambda,\ 1/\lambda$ with $|\lambda|>1$. Letting
$$
b^n=\left[\matrix{\alpha_n&\beta_n\cr
\bar\beta_n&\bar\alpha_n\cr}\right],
$$
we have $\alpha_n=O(|\lambda|^{|n|})$. Furthermore, we have the
following bound:
$$
\eqalign{
|(b^n)'(z)|&=|\bar\beta_nz+\bar\alpha_n|^{-2}=
|\alpha_n|^{-2}|1+{\bar\beta_n\over\bar\alpha_N}z|^{-2}\cr
&\leq|\alpha_n|^{-2}(1-|z|)^{-2}.\cr
}
$$
Since $z$ varies over a compact set and $t>0$, it follows that the series
$\sum_{n\in\bZ}\left|(b^n)'(z)\right|^t$ is bounded, uniformly in $z$,
by a convergent geometric series. This concludes the proof of \secondlemma\
and \defquantthm\ . $\square$
\medskip\noindent
{\bf Acknowledgement.} We would like to thank Jerry Kaminker
for very helpful remarks.
\vfill\eject
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\baselineskip=12pt
\frenchspacing
\bigskip
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\end