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\centerline{\bf SELBERG'S TRACE-FORMULA}
\centerline{a generalization of the Poisson summation formula}
\smallskip
\centerline{\it a lecture of G\'abor Hal\'asz}
\centerline{in the Bolyai College of the E\"otv\"os Lor\'and
University, Budapest}
\medskip\noindent{\CAP First talk:}
 
\beginsection The Poisson summation formula
 
Let a function $k(x)$ be given, and define the function
$K(x)=\sum\limits_{n=-\infty}^\infty k(x+n)$. (We disregard convergence
problems in this heuristic discussion.) Clearly, $K(x+1)=K(x)$,
and its Fourier series can be written as
$K(x)=\sum\limits_{j=-\infty}^\infty a_je^{2\pi ijx}$ with
$$
\align
a_j&=\int_0^1 K(x)e^{-2\pi ijx}\,dx=\int_0^1\sum_{n=-\infty}^\infty
k(x+n)e^{-2\pi ijx}\,dx\\
&=\sum_{n=-\infty}^\infty\int_0^1 k(x+n)e^{-2\pi
ijx}\,dx=\int_{-\infty}^\infty k(x)e^{-2\pi ijx}\,dx=\kappa(2\pi j),
\endalign
$$
where, $\kappa(t)=\int_{-\infty}^\infty k(x)e^{-itx}\,dx$ is the Fourier
transform of the function $k(x)$.
 
In such a way one gets by expressing $K(0)$ in two different ways the
relation
$$
\sum_{n=-\infty}^\infty k(n)=\sum_{j=-\infty}^\infty\kappa(2\pi j)\;.
$$
 
The main benefit of the last formula: The left-hand side is an
approximating sum of the integral $\int_{-\infty}^\infty k(x)\,dx$ which
equals $\kappa(0)$. The sum of the terms $j\neq 0$ expresses the error
of this approximation. (In the two-dimensional version of this formula
--- in this case the vectors with integer coordinates replace the role
of the integers ---, if $k(x)$ is the indicator function of a large
circle, the left-hand side equals the number of lattice points in the
circle, while at the right-hand side the term corresponding to the
origin equals the area of the circle, and this gives a possibility to
handle the not completely solved circle problem about the order of the
error term. A version of this problem will be discussed later.
 
\noindent{\CAP The same subject in a high-brow style}\noindent\medskip
 
We shall prove the Poisson summation formula in the following more
complicated form:
$$
\sum_{n=-\infty}^\infty k(n)=\text{sum of the eigenvalues of an integral
operator,}
$$
where the integral operator is appropriately defined. The reason for
this more complicated approach is that it is simpler to generalize this
form of the result to more general spaces.
 
Consider the real line $\text{\BBB R}^1$, together with the usual metric
$|x-y|$ and the Lebesgue measure. Let $\bold T_u\: x\to x+u$ denote the
shift with $u$. These transformations form a group for which the
above introduced metric and measure are invariant. Let us denote the
operator $\bold T_uf(x)=f(x+u)$ acting
on the space of functions $f(x)$, $-\infty<x<\infty$, by the same
letter.
 
A general operator $L$ will be called invariant, if
$L_xf(x+u)=Lf(x)|_{x\to x+u}$, i.e.\ if $L\bold T_u= \bold T_uL$, (The
subscript $x$ in $L_x$
means that the operator $L$ maps the function as a function of
the variable $x$ to a function of the variable
$x$. The remaining variables of the function where $L$ is acting are
fixed parameters.)
 
\noindent
We consider the following question: When is the integral operator
$Lf\!=\!\int\limits_{-\infty}^\infty k(x,y)f(y)dy$ invariant?
 
We have $L\bold T_uf=\int^\infty_{-\infty} k(x,y)f(y+u)\,du$ and
$\bold T_u Lf=\int_{-\infty}^\infty k(x+y,y)f(y)\;dy
=\int_{-\infty}^\infty k(x+u,y+u)f(y+u)\,dy$, which means that the
integral
operator with kernel $k(x,y)$ is invariant if $k(x,y)=k(x+u,y+u)$, i.e.\
if $k(x,y)$ depends only on $x-y$; $k(x,y)=k(x-y)$,
$Lf=\int_{-\infty}^\infty k(x-y)f(y)\,dy$ is a convolution.
 
\noindent
When is the differential operator $Lf=\sum\limits_{\nu=0}^m
a_\nu(x)\dfrac{d^\nu f}{dx^\nu}$ invariant?
 
$L\bold T_uf=\sum\limits_{\nu=0}^m a_\nu(x)f^\nu(x+u)$,
$\bold T_u Lf=\sum\limits_{\nu=0}^m a_\nu(x+u)f^\nu(x+u)$, i.e.\ it is
invariant if $a_\nu(x)$ is constant, $L=\sum\limits_{\nu=0}^ma_\nu D^\nu$, where
$D=\dfrac d{dx}$. In such a way we described all invariant differential
operators.
 
The above defined invariant operators are exchangeable. For instance, if
$Lf=\int_{-\infty}^\infty k(x-y)f(y)\,dy$, then
$DLf=\left(\int_{-\infty}^\infty k(x-y)f(y)\,dy\right)'=$ (we do not have
to differentiate immediately) $=\left(\int_{-\infty}^\infty k(y)
f(x-y)\,dy\right)'=\int_{-\infty}^\infty k(x-y)f'(y)\,dy=LDf$.
 
In the non-trivial step we exploited the fact that $L$ is invariant: If
$L^0f\overset{\text{def}}\to = Lf|_{x=0}$, then because of the invariance
$L_xf(x)=L^0_y\bold T_xf(y)$. (This means in our case that
$\int_{\infty}^\infty k(x-y)f(y)\,dy=\int_{-\infty}^\infty
k(0-y)f(y+x)\,dy$.)
\medskip\noindent{\bf Claim.} {\it Any two invariant operators $L_1$
and $L_2$ are exchangeable.}\medskip\noindent
{\it Proof:}\/ By the previous argument $L^0$ determines an invariant
operator $L$, and since both $L_1L_2$ and $L_2L_1$ are invariant, it is
enough to show that $(L_1L_2)^0=(L_2L_1)^0$, i.e.\  that
$L_1^0L_2=L_2^0L_1$.
 
We know that $L_2f=L^0_{2y}f(y+x)$, hence $L^0_{1x}L^0_{2y}f(y+x)=$
(We exploit the fact that $L_1^0$ and $L_2^0$ are acting on different
arguments. In this proof we restrict our attention to differential and
integral operators, where the statement of this step can be simply
checked.) $=L^0_{2y}L^0_{1x}f(y+x)=$ (because $x+y=y+x$, the group of
shifts is commutative) $=L^0_{2y}L^0_{1x}f(x+y)=L^0_2L_1f$.  \medskip
 
The exchangeability is important, because this means that the
spectral decomposition of the operators is the same. For instance, if
$f(x)$ is an eigenfunction of the operator $D$ with an (arbitrary
complex) eigenvalue $\lambda$, $f(x)=e^{\lambda x}$ (all functions
$Ae^{\lambda x}$ are such eigenfunctions), then $DLf=LDf=L\lambda
f=\lambda Lf$, that is $Lf$ is also an eigenfunction of $D$ with the
same eigenvalue. Since these eigenfunctions constitute a one-dimensional
space $Lf=\Lambda e^{\lambda x}=\Lambda f(x)$, where $\Lambda$ depends
only on $\lambda$ and $L$.
 
Let us determine this $\Lambda$ for the operator
$Lf=\int_{-\infty}^\infty k(x-y)f(y)\,dy$.
 
Since $f(0)=1$, $\Lambda=L^0f=\int_{-\infty}^\infty k(-y)e^{\lambda
y}\,dy=\int_{-\infty}^\infty k(y)e^{-\lambda y}\,dy$.
 
A typical discrete subgroup of the group of shifts is the group $\bold
T_n$, $n=0,\pm1,\pm2,\dots$. If $\bold T_nf=f$, i.e.\ $f(x+n)=f(x)$ for
all $n$, then
the function $f$ is called ``automorph" (periodic) with respect to the
subgroup. If $L$ is invariant, then $\bold T_nLf=L\bold T_n f=Lf$,
i.e.\ $L$ maps periodic functions into periodic functions. If $L$ is an
integral operator, then
$$
Lf=\int_{-\infty}^\infty k(x-y)f(y)\,dy=
\sum_{n=-\infty}^\infty\int_0^1k(x-(y+n))f(y)\,dy
=\int_0^1K(x,y)f(y)\,dy)\;,
$$
where $K(x,y)=\sum\limits_{n=-\infty}^\infty k(x-(y+n))$ is a periodic
kernel function in both variables.
 
In the space of periodic functions the operator $iD$ is self-adjoint,
because $(Df,g)=\int_0^1f'(x)\bar g(x)\,dx=$ (integrating by parts)
$=-\int_0^1 f(x)\bar g'(x)\,dx=-(f, Dg)$, i.e.\
$(iDf,g)=i(Df,g)=-i(f,Dg)=(f,iDg)$. Hence in this space the eigenvalues
of $D$ are imaginary, (If $Df=\lambda f$, then $(Df,f)=\lambda(f,f)$,
$(Df,f)=-(f,Df)=-\bar\lambda(f,f)$ $\Rightarrow$
$\lambda=-\bar\lambda$), and the eigenfunctions corresponding to
different eigenvalues are orthogonal: ($(Df,g)=\lambda(f,g)$,
$(Df,g)=-(f,Dg)=-\bar\nu(f,g)$,
and if $\lambda\neq\nu$, then $(f,g)=0$.)
 
We know that the (normalized) eigenfunctions are  the
functions $f_j(x)=e^{\lambda_jx}=e^{2\pi ijx}$. If we know that they
constitute a complete orthonormal system (this follows from general
results) then we can express the function $K(x,y)$ as
$$
K(x,y)=\sum\limits_{j,l=-\infty}^\infty c_{j,l}f_j(x)\bar f_l(y)\;.
$$
If $\Lambda_j$ denotes the eigenvalue of the operator
$Lf=\int_{-\infty}^\infty
k(x-y)f(y)\,dy$ corresponding to the eigenvalue $\lambda_j$ of $D$
about which we know that
$$
\Lambda_j=\int_{-\infty}^\infty
k(y)e^{-\lambda_jy}\,dy=\int_{-\infty}^\infty k(y)e^{-2\pi
ijy}\,dy=\kappa(2\pi j)\;,
$$
and
$$
\align
\Lambda_jf_j(x)&=Lf_j=\int_0^1K(x,y)f_j(y)\,dy=\int_0^1
\sum_{i,l=-\infty}^\infty
c_{i,l}f_i(x)\bar f_l(y)f_j(y)\,dy\\
&=\sum_{i,l=-\infty}^\infty c_{i,l}f_i(x)
\int_0^1\bar f_l(y)f_j(y)\,dy=\sum_{i=-\infty}^\infty c_{i,j}f_i(x) \;.
\endalign
$$
From here
$$
c_{i,j}= \cases 0,&\text{if }i\neq 0\\
\Lambda_j,&\text{if }i=j
\endcases\;.
$$
Hence $K(x,y)=\sum\limits_{j=-\infty}^\infty \Lambda_jf_j(x)\bar
f_j(y)$.
 
Thus,
$$
K(0,0)=\sum\limits_{j=-\infty}^\infty\Lambda_j
=\sum\limits_{j=-\infty}^\infty\kappa(2\pi j)\;.
$$
On the other hand,
$$
K(x,x)=\sum_{n=-\infty}^\infty
k(x-(x+n))=\sum\limits_{n=-\infty}^\infty k(n)\;,
$$
or because the function $K(x,x)$ is constant, the trace of $L$ equals
$$
\int_0^1 K(x,x)\,dx=K(0,0)=\sum_{j=-\infty}^\infty \Lambda_j
=\sum_{j=-\infty}^\infty \kappa(2\pi j)\;,
$$
and
$$
\int_0^1 K(x,x)\,dx=K(0,0)=\sum_{n=-\infty}^\infty k(n)\;.
$$
A comparison of the last two relations yields the Poisson formula.
\bigskip
Our  program is the following: We want to find the analog of
the above formula in a general space with a given group of isometries
(shifts). We want to determine the invariant differential and
integral operators, to calculate the eigenfunctions of the
integral operators and the eigenvalues corresponding to them. Then we
restrict the operators to the functions automorph with respect to a
discrete subgroup of these isometries. We want to determine the
eigenfunctions and corresponding eigenvalues of the invariant
differential operators in this space, to calculate the kernel function
of the integral operator. In such a way we get an analog of the Poisson
summation formula.
\bigskip
In a special case we carry out the above program. We shall discuss
the hyperbolic plane, i.e.\ the
Poincar\'e representation of the Bolyai--Lobachevski geometry. We write
down the invariant differential and integral operators with respect to
the group of motions in this space and determine, by using an
argument suggested by the previous proof, the analog of the Poisson
summation formula in this space. One has to overcome certain technical
difficulties which are related to the fact that the group of motions
in this space is non-commutative.
\vfill\eject
 
\noindent{\CAP Second talk:}
 
\beginsection Invariant transformations in the hyperbolic plane
 
We recall the following two classical (equivalent) models of the
hyperbolic plane.
\smallskip
\noindent {\it The space:}
\smallskip
\halign{\hskip0.4truecm\vtop{\hsize=6.6truecm \parindent=0pt #}&
\hskip2truecm\vtop{\hsize=6.6truecm \parindent=0pt #}\hskip0.4truecm\cr
{The unit circle  of the complex plane
$\{z\: |z|<1\}$ }&
{The upper half plane of the complex plane
$H=\{z\: \text{Im}\,z>0\}$}\cr}
\smallskip
\noindent {\it The metric (length) in the space:}
\smallskip
\halign{\hskip0.4truecm\vtop{\hsize=6.6truecm \parindent=0pt #}&
\hskip2truecm\vtop{\hsize=6.6truecm \parindent=0pt #}\hskip0.4truecm\cr
{$\dfrac {|dz|}{1-|z|^2}$
$$I(\gamma)=\int_\gamma \frac {|dz|}{1-|z|^2}$$}&
{$\dfrac {|dz|}{y}$ if $z=x+iy$.
$$I(\gamma)=\int_\gamma \frac {|dz|}{y}$$
$$\align d(z,w)&=\min_{\overset{\frown}\to{z,w}} I(\overset{\frown}\to
{z,w})\\ &=\log\frac {1+\left|\frac{z-w}{z-\bar w}\right|}
{1-\left|\frac{z-w}{z-\bar w}\right|}\endalign $$
 
This expression is the function of
 
$\frac4{\left|\frac{z-\bar w}{z- w}\right|^2-1}
=\frac{|z-w|^2}{yv}$
 
if $z=x+iy$ and $w=u+iv$.}\cr}
\smallskip
\noindent {\it The measure (area) in the space:}
\smallskip
\halign{\hskip0.4truecm\vtop{\hsize=6.6truecm \parindent=0pt #}&
\hskip2truecm\vtop{\hsize=6.6truecm \parindent=0pt #}\hskip0.4truecm\cr
{$$\frac{dxdy}{(1-|z|^2)^2}$$}&
{$$d\sigma \overset\text{def}\to =\frac{dxdy}{y^2}$$}\cr}
\smallskip
\noindent {\it The length and measure preserving movements of the
space:} \smallskip
\halign{\hskip0.4truecm\vtop{\hsize=6.6truecm \parindent=0pt #}&
\hskip2truecm\vtop{\hsize=6.6truecm \parindent=0pt #}\hskip0.4truecm\cr
{$$\bold Tz=\rho\frac{z-\zeta}{1-z\bar\zeta},$$
$|\rho|=1$, $|\zeta|<1$}&
{$$\bold Tz=\frac{az+b}{cz+d}$$
$a$, $b$, $c$, $d$ are real numbers, $ad-bc>0$ $(=1)$
 
(This group of transformations can be identified with the quotient
group $SL_2(R)/\{\pm \text{Id}\}$, where $SL_2(R)$ denotes the group of
$2\times2$ matrices with real coefficients and determinant~1.)}\cr}
\smallskip\noindent
The group of all isometries: The above transformations together with their
compositions with a reflection through an axis.
 
The movements written down are the orientation preserving isometries,
their composition with a reflection are the orientation changing
transformations.
 
Given a function $f(z)$ and a movement $\bold T$ on the
hyperbolic plane, define $\bold T f$ as $\bold
Tf(z)\overset\text{def}\to=f(\bold T z)$. \medskip\noindent
{\bf Definition.} {\it An operator $L$ acting on the functions over a
hyperbolic space is called invariant, if\/ $\bold TL=L\bold T$ for all
movements $\bold T$ of the space.}\medskip
\noindent
{\it Question:}\/ Which integral operators $Lf=\int_H k(z,w)f(w)
\sigma(\,dw)$ are invariant?
 
$$
\bold TLf(z)=\int_H k(\bold Tz,w)f(w)\sigma(\,dw)=
\int_H k(\bold Tz,\bold Tw)f(\bold Tw)\sigma(\,dw)\;,
$$
and
$$
L\bold Tf(z)=\int_H k(z,w)f(\bold Tw)\sigma(\,dw)\;.
$$
Hence $L$ is invariant
if and only if $k(z,w)=k(\bold Tz,\bold Tw)$ for all $\bold T$. The
group of the movements $\bold T$ is strongly transitive, i.e.\ for any
pairs $(z_1,w_1)$ and $(z_2,w_2)$ such that $d(z_1,w_1)=d(z_2,w_2)$ there
exists a transformation $\bold T$ such that $z_2=\bold Tz_1$ and
$w_2=\bold Tw_1$. Hence the kernel function of an invariant integral
operator depends only on $d(z,w)$. In the positive half-plane
model of the hyperbolic plane, because of the special form of the
distance in this space, the kernel function of an invariant integral
operator can be given in the form
$$
k(z,w)=k\left(\frac{|z-w|^2}{yv}\right)\;.
$$
 
\noindent
Which differential operators $Lf=\sum a_{i,j}(z)\dfrac{\partial^{i,j}}
{\partial x^i \partial y^j}f(z)$ are invariant?
 
Here we follow an argument similar to that given in the Euclidean space.
We fix a point $z_0$, define the functional $L^0$, $L^0f=Lf(z_0)$
and want to reconstruct the invariant operator $L$ by means of $L^0$.
\medskip\noindent
{\it Remark:}\/ The functional $L^0$ cannot be arbitrary. Indeed, if the
transformation $\bold T$ is such that $\bold Tz_0=z_0$, then $L^0\bold
Tf=L\bold Tf(z_0)=\bold T Lf(z_0)=Lf(\bold Tz_0)=Lf(z_0)=L^0f$ for this
$\bold T$, i.e.\ the functional $L^0$ is rotation invariant. On the
other hand, if $L^0$ is such a functional, then it defines an invariant
operator $L$ in a  unique way by the formula $Lf(z)=L^0\bold
T_zf=L_w^0f(\bold T_zw)$. We formulate this statement in the following
lemma. \medskip \noindent
{\bf Lemma.} {\it Let us fix some number $z_0$ in the hyperbolic plane.
We call a functional $L^0$ on the space of functions in the hyperbolic
plane rotation invariant if $L^0f=L^0\bold Tf$ for all such movements
$\bold T$ of the hyperbolic plane for which $\bold Tz_0=z_0$. If $L_0$
is a rotation invariant functional, then we define the operator $L$ by
the formula $Lf(z)=L^0\bold T_zf=L_w^0f(\bold T_zw)$, where $\bold T_z$
is a movement such that $\bold T_zz_0=z$. This definition is meaningful,
i.e.\ the value of $Lf(z)$ does not depend on the special choice of
$\bold T_z$. This operator $L$ is invariant, and all invariant
operators $L$ can be obtained in such a way through the choice of
$L_0f=Lf(z_0)$.} \medskip\noindent
{\it Proof:}\/ The definition of $Lf$ is meaningful. Indeed, let
$\bold T_z$ and $\bar{\bold T}_z$ be two movements such that $\bold
T_zz_0=z$ and $\bar{\bold T}_zz_0=z$. We have to show that
$L^0\bar{\bold T}_zf=L^0\bold T_zf$, or with the notation $\bold
T_zf=g$, $L^0g=L^0\bar{\bold T}_z\bold T_z^{-1}g$
or $L^0g(w)=L^0_wg(\bold T_z^{-1}\bar{\bold T}_zw)$.
This formula follows from the rotational invariance of $L^0$ and the
relation $\bold T_z^{-1}\bar{\bold T}_zz_0=z_0$.
 
Let us consider an arbitrary motion $\bold T$. Then
$$
\bold  T Lf(z)=h(\bold T z)=L^0g
$$
with $h(u)=Lf(u)=L^0\bold T_uf$ ($h(\bold Tz)=L^0\bold T_{\bold Tz}f$)
and $g(u)=\bold T_{\bold Tz}f(u)=f(\bold T_{\bold Tz}u)$,
where $\bold T_{\bold Tz}$ is a motion of the hyperbolic plane such that
$\bold T_{\bold Tz}z_0=\bold Tz$.
$$
L\bold Tf(z)=L^0\bold T_z\bold Tf=L^0 p=L_0\bold T'g=L_0g
$$
with $p(u)=\bold T_z\bold Tf(u)=f(\bold T\bold T_zu)$ and $\bold
T'=\bold T^{-1}_{\bold Tz}\bold T\bold T_z$. The last but one identity
holds in this relation because
$p(u)=f(\bold T_{\bold Tz}\bold T^{-1}_{\bold Tz}\bold T\bold T_zu)
=g(\bold T'u)$, i.e.\ $p=\bold T'g$. Finally, the
last identity holds in this relation, because of the rotational
invariance of $L^0$ and the equality $\bold T'z_0=\bold T^{-1}_{\bold
Tz}\bold T\bold T_z z_0=\bold T^{-1}_{\bold Tz} \bold Tz=z_0$. A
comparison of the expressions obtained for $\bold TLf$ and $L\bold Tf$
shows that the operator $L$ defined in the lemma is invariant.
 
Finally, let $L$ be an invariant operator, and define $L_0f=Lf (z_0)$.
Then $L_0$ is a rotation invariant functional, and it determines the
original operator $L$, since $L^0\bold T_zf=L\bold T_zf(z_0)=\bold
T_zLf(z_0)=Lf(\bold T_zz_0)=Lf(z)$.
\medskip
Thus the invariant differential operators can be found by first
describing the rotation invariant differential functionals
$$
L^0f=\sum a_{i,j}\left. \frac{\partial^{i+j}}{\partial x^i\partial
y^j}f(z)\right|_{z=z_0}
$$
with coefficients $a_{i,j}=a_{i,j}(z_0)$.
 
The problem can be slightly simplified by working in the unit circle
model $\{z\: |z|<1\}$ of the hyperbolic space with $z_0=0$ and by
rewriting the differential operator with the replacement $\dfrac\partial
{\partial z}=\dfrac12\dfrac\partial{\partial x}-i\dfrac12\dfrac\partial
{\partial y}$ and $\dfrac\partial {\partial\bar z}=\dfrac12\dfrac\partial
{\partial x}+i\dfrac12\dfrac\partial{\partial y}$. In this (equivalent)
reformulation our problem is to find the rotational invariant
differential functionals
$$
L^0f=\sum b_{i,j}\left. \frac{\partial^{i+j}}{\partial z^i\partial
\bar z^j}f(z)\right|_{z=0}
$$
with some new coefficients $b_{i,j}$. The rotational invariance means in
this case that the expression
$$
L^0f=\sum b_{i,j}\rho^i\bar \rho^j\left. \frac{\partial^{i+j}}{\partial
z^i\partial \bar z^j}f(z)\right|_{z=0}
$$
has the same value for all $\rho$ if $|\rho|=1$. This relation holds if
$$
\sum_{i-j=h} b_{i,j}\rho^i\bar \rho^j\left.
\frac{\partial^{i+j}}{\partial z^i\partial \bar z^j}f(z)\right|_{z=0}=0
$$
for all functions $f$ and $h\neq0$. This statement holds if
$b_{i,j}=0$ for $i\neq j$. So
$$
L^0f=\sum_{i=0}^m b_{i,i}\left(\rho^i\bar \rho^i\left.
\frac{\partial^{2}}{\partial z\partial \bar
z}\right)^if(z)\right|_{z=0}=\sum_{i=0}^mc_i
\left.\Delta^if(z)\right|_{z=0}\;,\tag$*$
$$
because
$$
4\dfrac{\partial^2}{\partial z\partial\bar z}=
\dfrac{\partial^2}{\partial x^2}+\dfrac{\partial^2}{\partial y^2}
=\Delta\;.
$$
Thus for instance, if $L^0f=\Delta f|_{z=0}$, then the invariant
operator $L$ corresponding to it equals $Lf(z)=L_w^0f(\bold
T_zw)=\Delta f(\bold T_zw)|_{w=0}$.
The operator $\bold T_z$ can be chosen as $\bold
T_zw=\dfrac{w+z}{1+w\bar z}$, for which $\left.\dfrac{\partial\bold T_zw}
{\partial w}\right|_{w=0}=1-|z|^2$. This gives that
$$
\align
\Delta f(\bold T_zw)&=4\frac {\partial^2f(\bold T_zw)}{\partial w
\partial \bar w}=4\frac\partial {\partial \bar w}\left.\frac{\partial f}
{\partial w}\right|_{\bold T_z w}\frac{\partial \bold T_zw}{\partial w}\\
%+\frac{\partial f}{\partial\bar w}\right|_{\bold T_z w}
%\frac{\partial \bold T_zw}{\partial \bar w}\right)
&\left.=\frac{\partial^2}{\partial\bar w\partial w}f\right|_{\bold T_zw}
\left|\frac{\partial\bold T_zw}{\partial
w}\right|^2+\left.4\frac{\partial f}{\partial w}\right|_{\bold T_zw}
\frac{\partial}{\partial \bar w}\frac{\partial \bold T_z w}{\partial w}=
\Delta f|_{\bold T_zw}\left|\frac{\partial \bold T_zw}{\partial
w}\right|^2\;.
\endalign
$$
 
The above calculation yields with $w=0$ that the invariant differential
operator corresponding to $L^0f=\dfrac{\partial^2}{\partial z\partial
\bar z}f$ is $\Delta
f(\bold T_zw)|_{w=0}=\bigl(1-|z|^2\bigr)^2\Delta f(z)\overset\text{def}
\to{=}D^*f(z)$, which is the hyperbolic Laplace operator. Also the
polynomials
$$
Lf=\sum_{l=0}^mc_l{D^*}^lf \tag$**$
$$
are invariant differential operators. We claim that all invariant
differential operators are given by this formula. Indeed, if $L^0$ is a
rotational invariant functional of the form ($*$), then the invariant
differential operator $Lf$ is a polynomial of the derivatives of $f$. The
coefficient of the leading term $\frac{\partial^{2m}}{\partial
z^m\partial\bar z^m}$ is a non-vanishing function. (It is
$c_m(1-|z|^2)^{2m}$.) Hence the class of invariant differential
operators
of order $m$ with $0\le m\le 2N$ constitute an $N+1$-dimensional linear
subspace of the differential operators of order less than or equal to
$2N$. Since the operators of the form ($**$) with $0\le m\le N$ give an
$N+1$-dimensional subspace of it, all invariant differential operators
can be given in the form ($**$). This means that in the hyperbolic plane
the ring of invariant differential operators is generated by the single
element $D^*$. This result can be rewritten to the
half-plane model of the hyperbolic plane. In this case
the invariant differential operators are of the form $Lf=\sum\limits_
{l=0}^mc_lD^mf$, with $Df(z)=y^2\Delta f(z)$, where $z=x+iy$ and
$\Delta$ is the Laplace operator. \medskip\noindent
{\bf Theorem.} {\it Let $L_1$ and $L_2$ be two invariant differential or
integral operators on the hyperbolic plane. Then they are
exchangeable, i.e.\ $L_1L_2=L_2L_1$.}\medskip
{\it Proof:}\/ We work in the positive half-plane model of the
hyperbolic plane. Since all invariant differential operators are the
polynomials of the operator $D$, they are exchangeable. So we may assume
that one of the operators, say $L_2$ is an integral operator with a kernel
function $k_2(z,w)$, and $L_1$ is either the operator $D$ or an
integral operator with a kernel function $k_1(z,w)$.
 
In the proof we try to adapt the proof of the Claim in the first talk.
Since $L_1L_2$ and $L_2L_1$ are also
invariant operators, by the lemma it is enough to show that
$(L_1L_2)^0f=(L_2L_1)^0f$. We can write $(L_1L_2)^0f=L_1g(z_0)$ with
$$
g(z)=L_2f(z)=\bold T_zL_2f(z_0)=
L_2\bold T_zf(z_0)=\int k_2(z_0,w)f(\bold T_zw)\sigma(\,dw)
$$
and $(L_2L_1)^0f=L_2h(z_0)$ with $h(z)=L_1f(z)$. Here
$$
h(z)=\bold T_zL_1f(z_0)=
L_1\bold T_zf(z_0)=\int k_1(z_0,w)f(\bold T_zw)\sigma(\,dw)
$$
if $L_1$ is an integral operator, and
$$
h(z)=\bold T_zL_1f(z_0)= L_1\bold T_zf(z_0)=D_wf(\bold T_zw)|_{w=z_0}
$$
if $L_1=D$.
If both $L_1$ and $L_2$ are integral operators, then
$$
(L_1L_2)^0f=L_1g(z_0)=\iint k_1(z_0,z)k_2(z_0,w)f(\bold
T_zw)\sigma(\,dz)\sigma(\,dw)
$$
and
$$
\align
(L_2L_1)^0f&=\iint k_2(z_0,z)k_1(z_0,w)f(\bold
T_zw)\sigma(\,dz)\sigma(\,dw)\\
&=\iint k_1(z_0,z)k_2(z_0,w)f(\bold T_wz)\sigma(\,dz)\sigma(\,dw)\;.
\endalign
$$
If $L_1=D$, then
$$
\align
(L_1L_2)^0f&=L_1g(z_0)=D_z\left.g(z) \right|_{z=z_0}=
D\int k_2(z_0,w)f(\bold T_zw)\sigma(\,dw)\biggl|_{z=z_0}\\
&=\int k_2(z_0,w)\left.D_zf(\bold T_zw)\right|_{z=z_0}\sigma(\,dw)
\endalign
$$
and
$$
\align
(L_2L_1)^0f=L_2h(z_0)=\int k_2(z_0,z)h(z)
\sigma(\,dz)&=\int k_2(z_0,z)\left.D_wf(\bold
T_zw)\right|_{w=z_0}\sigma(\,dz)\\
&=\int k_2(z_0,w)\left.D_zf(\bold
T_wz)\right|_{z=z_0}\sigma(\,dw)\;.
\endalign
$$
 
We have certain freedom in the choice of the operators $\bold T_z$.
If we could define these operators in such a way that $\bold T_zw=\bold
T_z\bold T_w z_0$ and $\bold T_wz=\bold T_w\bold T_z z_0$ agreed, then
a comparison of the expressions $(L_1L_2)^0)$ and $(L_1L_2)^0)$ would
imply the Theorem just as it was done in the proof of the Claim in the
first talk. But we cannot define these operators in such a way because
of the lack of commutativity in the group of motions in this space.
Hence some new ideas are needed, and Selberg found such ideas. He
introduced a weak commutativity property defined below which is
satisfied by the hyperbolic plane, and worked out a more refined
argument which proves the result under this weaker property.
 
With an appropriate choice of the transformations $\bold
T_z$ the following weak commutativity property holds:
$$
d(z_0,\bold T_zw)=d(z_0,\bold T_wz)\quad\text{for all $w$ and $z$.}
$$
 
Indeed, by choosing the motions $\bold T_z$ in such a way that $\bold
T_zz_0=z$ and $\bold T_zz=z_0$ we get that
$$
d(z_0,\bold T_wz)=d(\bold T_ww,\bold T_wz)=d(w,z)=d(\bold T_zw,\bold
T_zz)=d(\bold T_zw,z_0)\;.
$$
Let us first consider rotation invariant functions $f$, i.e.\ such
functions for which $f(z)=f(\bold Tz)$ for any motion $T$ with the
property $\bold Tz_0=z_0$. Then the value of $f(z)$ depends only on
$d(z_0,z)$. The previous calculations yield that $(L_1L_2)^0f=
(L_2L_1)^0f$ for rotation invariant functions. To consider a general
function $f$ let us introduce the following ``averaging" operator
$$
f_0(z)=Mf(z)=\frac1{2\pi}\int_0^{2\pi}\bold T_{\vartheta}f(z)
\,d\vartheta\;,\tag+
$$
where $\bold T_\vartheta$, $0\le \vartheta<2\pi$, ``the rotation with
angle $\vartheta$", is defined in the following way: Let us consider an
analytic automorphism $\bold U$ of the unit circle to the positive
half-plane such that $\bold U$ maps the origin to $z_0$. If $\bold
S_\vartheta z=e^{i\vartheta}z$ denotes the rotation with angle
$\vartheta$ in the unit circle, then $\bold T_\vartheta$ is the
transformation corresponding to it through this automorphism, i.e.\
$\bold T_\vartheta z=\bold U\bold S_\vartheta\bold U^{-1}z$. The
function $f_0=Mf$ is rotation invariant, since for all $\vartheta'$
$$
f_0(\bold T_{\vartheta'}z)=\frac1{2\pi}\int_0^{2\pi}f(\bold
T_\vartheta\bold T_{\vartheta'}z)\,d\vartheta
=\int_0^{2\pi}f(\bold T_{\vartheta+\vartheta'}z)\,d\vartheta
=\int_0^{2\pi}f(\bold T_{\vartheta}z)\,d\vartheta=f_0(z_0)\;,
$$
and all movements of the hyperbolic plane with the fixed point $z_0$
equals some $\bold T_\vartheta$.
 
If $L$ is an invariant operator, then $LM=ML$, since $L$ is exchangeable
with all $\bold T_\vartheta$. Hence
$$
(L_1L_2)^0f=L_1L_2f(z_0)=ML_1L_2f(z_0)=L_1L_2Mf(z_0)=L_1L_2f_0(z_0)
$$
and
$$
(L_2L_1)^0f=L_2L_1f(z_0)=ML_2L_1f(z_0)=L_2L_1Mf(z_0)=L_2L_1f_0(z_0)\;.
$$
Since $L_1L_2f_0=L_2L_1f_0$ the above relations imply that
$(L_1L_2)^0f=(L_2L_1)^0f$, and $L_1L_2=L_2L_1$ by the lemma. The theorem
is proved.
\vfill\eject
 
\noindent
{\CAP Third talk:}
 
\beginsection The eigenvalues of invariant integral operators
 
We can find many eigenfunctions of the operator $D=y^2\Delta$ in the upper
half-plane model of the hyperbolic space. Thus for instance $y^s$,
$z=x+iy$, is an eigenfunction with the eigenvalue $\lambda=s(s-1)$,
since $Dy^s=y^2\Delta y^s=y^2s(s-1)y^{s-2}=s(s-1)y^s$. If $L$ is an
arbitrary invariant differential or integral operator, then by the
previous statement $DLy^s=LDy^s=\lambda Ly^s$, i.e.\ $Ly^s$ is an
eigenfunction of $D$ with the same eigenvalue. Following the argument
of the first talk we would like to deduce from this statement that
$Ly^s=\Lambda y^s$ with some $\Lambda$. However, the argument of the
first talk does not work in the present case. In the real line case,
considered in the first talk, it was exploited that the
eigenspace of the differential operator corresponding to an eigenvalue
$\lambda$ is one-dimensional. In the hyperbolic space $H$ the
analogous statement does not hold. The eigenspace of the operator $D$
corresponding to an eigenvalue $\lambda$ is very large. For
instance, the functions $\bold Ty^s=(\text{Im}\,\bold T z)^s$ are
eigenfunctions with eigenvalue $\lambda$ for all motions $\bold T$.
 
\noindent
What can be saved from the uniqueness of eigenfunctions?
 
If $M$ is the ``averaging operator" with some center $z_0$ defined in
formula (+) in the second talk, and $Df=\lambda f$, then for $f_0=Mf$
$$
Df_0=DMf=MDf=\lambda Mf=\lambda f_0\;,
$$
and $f_0$ is rotation invariant around the center $z_0$.
\medskip\noindent
{\bf Proposition.} {\it Let $g(z)=g(z,z_0)$ satisfy the following
properties: $Dg=\lambda g$, $g(z)$ depends only on the value of
the distance $d(z,z_0)$, (with fixed $z_0$ and $\lambda$), and
$g(z_0)=1$. Then the function $g(z)$ is defined uniquely by these
properties.} \medskip\noindent
{\it Proof:}\/ Here again it is simpler to work in the unit square model
$\{|z|<1\}$ with $z_0=0$. If $z=re^{i\vartheta}$, then the function $g$
can be written in the form $g(z)=g(r)$ with $r=|z|$. The Laplace operator
$\Delta$ can be written in polar coordinate system as
$\Delta=\dfrac{\partial^2}{\partial r^2}+\dfrac1{r^2}\dfrac{\partial^2}
{\partial\vartheta^2}$. Hence we can write because of the rotation
invariance of $g$
$$
D^*g=(1-r^2)^2g''(r)=\lambda g(r)
$$
with the boundary conditions $g(0)=1$ and $g'(0)=0$. (The last relation
holds, because $g(r)=g(-r)$.) Since the coefficient $(1-r^2)^2$ in this
differential equation is non-vanishing, it can be solved uniquely.
\medskip \noindent
How does the function $g(z,z_0)$, the unique rotation invariant
eigenfunction of $D$ with center $z_0$, with eigenvalue $\lambda$ and
normalization $g(z_0,z_0)=1$, depend on $z_0$?
 
If $\bold Tz_0=z'_0$,
then $g(\bold Tz,z'_0)$ depends only on $d(\bold Tz,z'_0)=d(z,\bold
T^{-1}z_0')=d(z,z_0)$, \ $g(\bold Tz_0,z'_0)=g(z'_0,z'_0)=1$, and
with the notation $h(u)=g(\bold Tu,z'_0)$, $p(u)=g(u,z'_0)$
$$
Dh(z)=D\bold T p(z)=\bold
TDp(z)=\lambda\bold Tp(z)=\lambda g(\bold Tz,z'_0)=\lambda h(z)\;.
$$
This means that $h(z)=g(\bold Tz,z'_0)$ is an eigenfunction of $D$
with the eigenvalue $\lambda$, it depends only on $d(z,z_0)$, i.e.\
$h(z)=h(z')$ if $d(z,z_0)=d(z',z_0)$, which means that it is rotation
invariant with center $z_0$, and $h(z_0)=1$. Hence the Proposition
implies that $h(z)=g(\bold Tz,z'_0)=g(z,z_0)$, or in other words
$g(\bold Tz,\bold Tz_0)=g(z,z_0)$ for all motions $\bold T$ of the
hyperbolic plane. This means that $g(z,z_0)$ depends
only  on $d(z,z_0)$, i.e.\ it has the same properties as the kernel
function of an invariant integral operator.
 
If $L$ is an arbitrary invariant differential or integral operator, then
$$
D_zL_zg(z,z_0)=L_zD_zg(z,z_0)=\lambda L_zg(z,z_0)\;.
$$
But $L_zg(z,z_0)$, as a function of $z$ and $z_0$, depends only on
$d(z,z_0)$, since
$$
L_zg(\bold Tz,\bold Tz_0)=Lp(\bold Tz)=\bold TLp(z)=L\bold T p(z)=
Lq(z)=L_zg(z,z_0)
$$
for all movements $\bold T$ with $p(u)=g(u,\bold Tz_0)$ and
$q(u)=g(\bold Tu,\bold Tz_0)=g(u,z_0)$. In particular, because of the
rotational symmetry $L_zg(z,z_0)=\Lambda g(z_,z_0)$ with some
$\Lambda=\Lambda(z_0)$. Moreover, $\Lambda(z_0)$ does not depend on
$z_0$, since $\Lambda(z_0)=L^0g(z_0,z_0)$, and this function has the
same value for all~$z$.
 
If $f$ is an arbitrary function such that $Df=\lambda f$, and we fix
some number $z_0$ as the center of the averaging operator $M$, then
$Df_0=\lambda f_0$ with $f_0=Mf$, and $f_0$ is rotation invariant
around the point $z_0$. Because of the uniqueness of rotational
invariant eigenfunctions $f_0(z)=f_0(z_0)g(z,z_0)$, and this implies
for all invariant integral or differential operators $L$ that
$$
Lf_0(z)=f_0(z_0)L_zg(z,z_0)=f_0(z_0)\Lambda g(z,z_0)=\Lambda f_0(z)
$$
with a $\Lambda$ independent of the function $f$ and $z_0$. From here
$$
Lf(z_0)=MLf(z_0)=LMf(z_0)=Lf_0(z_0)=\Lambda f_0(z_0)=\Lambda f(z_0)\;.
$$
Since $z_0$ can be chosen in an arbitrary way, and $\Lambda$ is
independent of $z_0$, $Lf=\Lambda f$.
This is a very important observation which we formulate in the following
Theorem.\medskip\noindent
{\bf Theorem.} {\it If $f$ is an eigenfunction of the operator $D$ with
eigenvalue $\lambda$, i.e.\ $Df=\lambda f$, and $L$ in an invariant
differential or integral operator, then $Lf=\Lambda f$ with a $\Lambda$
which depends only on the operator $L$ and the eigenvalue $\lambda$,
but not on the eigenfunction $f$.} \medskip
If $f$ is an eigenfunction of $D$, then we can calculate the eigenvalue
of $Lf$ by this theorem without the knowledge of $f$, if we know an
arbitrary eigenfunction of $D$ with the same eigenvalue $\lambda$.
We know such an eigenfunction, namely the function $y^s$ with the
eigenvalue $\lambda$. This eigenvalue $\lambda$ can be arbitrary with
the choice of $s=\dfrac12\pm \sqrt{\dfrac 14+\lambda}$.
 
Let us calculate for instance the eigenvalue $\Lambda$ of the
integral operator $Lf=\int_Hk(z,w)f(w)w(\,d\sigma)$ if
$k(z,w)=k\left(\dfrac{|z-w|^2}{yv}\right)$.
 
We know that
$$
\int_H k(z,w)v^s\sigma(\,dv)=\Lambda y^s\
$$
with $w=u+iv$. Taking the point $z=i$ ($y=1$), we get
$$
\align
\Lambda&=\int_H k(i,w)v^s\sigma(\,dv)=\int _0^\infty\int_{-\infty}^\infty
k\left(\frac{|i-w|^2}v\right)v^{s}\frac{\,du\,dv}{v^2}\\
&=\int _0^\infty\int_{-\infty}^\infty
k\left(\frac{u^2+(v-1)^2}v\right)v^{s-2}\,du\,dv\;.
\endalign
$$
Put $t\overset\text{def}\to = \dfrac{(v-1)^2}v=v+\dfrac1v-2$. Then
$$
\align
\int_{-\infty}^{\infty}k\left(\frac{u^2+(v-1)^2}v\right)\,du
&=\int_{-\infty}^\infty k\left(\dfrac{u^2}v+t\right)\,du=2\int_0^\infty
k(\tau+t)\dfrac{\sqrt v}{2\sqrt \tau}\,d\tau \\
&=\sqrt v\int_0^\infty
\frac{k(\tau+t)}{\sqrt \tau}\,d\tau\overset\text{def}\to =v g(\log v)\;,
\endalign
$$
$$
\Lambda=\int_0^\infty vg(\log v)v^{s-2}\,dv=\int_{-\infty}^\infty
g(y)e^{sy}\,dy\overset\text{def}\to= G(s)\;,
$$
where $\lambda=s(s-1)$, $s=\dfrac12\pm\sqrt{\dfrac14+\lambda}$.
\vfill\eject
 
\noindent{\CAP Fourth talk:}
 
\beginsection Discrete subgroups of the group of motions in the
hyperbolic plane
 
{\bf Definition.} {\it The group $\Gamma\subset\text{\rm group of
motions in }H$ is called discrete if the set
$\{\gamma z\:\gamma\in\Gamma\}$
has no point of density for a fixed $z\in H$.}
\medskip\noindent
{\bf Definition.}{\it The function $f(z)$, $z\in H$, is automorph with
respect to the group $\Gamma$, if $f(\gamma z)=f(z)$ for all
$\gamma\in\Gamma$.} \medskip
From now on $f(z)$ will always denote an automorph function. We also
introduce the notation $z_1\sim z_2$ (mod$\,\Gamma$) if there exists
a motion $\gamma\in \Gamma$, such that $\gamma z_1=z_2$.
 
Put $\Cal S\overset\text{def}\to=\text {equivalence classes of $z\in H$
with respect to $\sim$}$. A set $\Cal D\subset H$ is called a
fundamental domain if it contains exactly one point from all
equivalence classes. In other words, the sets $\{\gamma\Cal D\}$,
($\gamma\in \Gamma$) produce a disjoint cover of the space $H$. (In the
Euclidean plane the discrete subgroups of the group of motions have the
form $\Gamma=\Gamma_{w_1,w_2}=\{\gamma_{m,n}\:z\to z+nw_1+mw_2\}$, with
some non-colinear vectors $w_1$ and $w_2$. The automorph functions are
the doubly periodic functions, $\Cal D$ can be chosen as a lattice
parallelogram, $S$ is topologically a torus. In the one-dimensional
case $\Cal D=[0,1)$, $\Cal S$ is a circle.)
 
Let us also assume that $\Gamma$ is free of fixed points. This means
that if $\gamma\in \Gamma$ has a fixed point, ($\gamma z_0=z_0$), then
$\gamma$ is the identity. We also assume that $\Gamma$ has a ``compact"
fundamental domain $\Cal D$ whose closure $\bar{\Cal D}$ (in the
compactified complex sphere) is a subset  of $H$.
 
It is far from trivial that there are such subgroups $\Gamma$ of the
motions of $H$ which satisfy all above conditions. It follows from the theory
of Riemann surfaces that this is possible. The fundamental
domains corresponding to the group $\Gamma$ show some similarities
with the torus appearing as a fundamental domain in the Euclidean
case. If it has genus $k$, $k>1$, which heuristically means that the
domain $\Cal D$ is topologically a ``sphere with $k$~ears", then there
exists a discrete subgroup $\Gamma$ with a fundamental domain $\Cal
D$ which is a regular polygon with $4k$ edges and angles $\pi/2$. The
area of this fundamental domain is determined by its genus $k$. There
are such pairs of edges of the boundary which can be stuck together
in opposite direction, and the points stuck together are equivalent
with respect to $\Gamma$.
%paired in such a way that  boundary $\partial \Cal D$ has the following
%structure: It consists of $k$ four-tuples
%$\left(\ell^{(j)}_1,\ell^{(j)}_2,\ell^{(j)}_3,
%\ell^{(j)}_4\right)$, $j=1,\dots,k$, in such a way that the pairs
%$\left(\ell^{(j)}_1,\ell^{(j)}_2\right)$ and
%$\left(\ell^{(j)}_3,\ell^{(j)}_4\right)$ are ``stuck together", i.e.\
%$\gamma\ell^{(j)}_1=\ell^{(j)}_2$ andJ\
%$\gamma\ell^{(j)}_3=\ell^{(j)}_4)$ for all $\gamma\in\Gamma$ and
%$j=1,\dots,k$.
The explanation of these results is not the subject of the present
lecture.
 
Under these conditions $\Cal S$ is a compact (closed) Riemann surface,
topologically it is a sphere $\Cal G$ with ``ears", its genus $g>1$.
There exists a universal cover $H\to \Cal S$. This universal cover
together with the metric and measure on $H$ induce a finite metric and
measure on $\Cal S$, invariant with respect to $\Gamma$.
 
\beginsection
The hyperbolic Laplace operator in a fundamental domain
 
If $L$ is an invariant operator, $f(z)$ is an automorph function, then
$Lf$ is again an automorph function, since $Lf(\gamma z)=\bold
T_\gamma Lf(z)=L\bold T_\gamma f(z)=Lf(z)$ for all $z\in H$ and
$\gamma\in\Gamma$, where the operator $\bold T_\gamma$ is defined by the
relation $\bold T_\gamma f(z)=f(\gamma z)$. Hence the operator $L$ can
be considered as an operator acting in the space of automorph functions.
In particular, this restriction can be done for  the operator $D
=y^2\Delta$. The operator $D$ is a self-adjoint negative operator in the
space $L_2(\Cal S, d\sigma)$. Indeed, the Green formula in the domain
$\Cal D$ gives that the operator $D$ is negative, i.e.
$$
\align
(Df,f)&=\int_{\Cal D} Df(z)\overline{f(z)}\sigma(\,dz)=\iint_{\Cal
D}y^2\Delta f(z)\overline{f(z)}\frac{dx\,dy}{y^2}\\
&=\undersetbrace=0\to
{\int _{\partial D}\frac{\partial f}{\partial n}\bar
f\,ds} -\undersetbrace\ge0\to
{\iint_{\Cal D}(f_x \bar f_x+f_y\bar
f_y)\,dx\,dy} \le 0\;,
\endalign
$$
since, as it is proved  in the theory of the Riemann surfaces, the
boundary $\partial \Cal D$ consists of finitely many pairs of curves
$(\ell,\ell')$ such that an appropriate $\gamma\in\Gamma$ maps $\ell$ to
$\ell'=\gamma \ell$, and
$$
\left.\frac{\partial f}{\partial n}\right|_{z}
=\left.-\frac{\partial f}{\partial n}\right|_{\gamma
z}|\gamma'(z)|\;.
$$
This relation implies that
$$
\int_{\ell}\frac{\partial f}{\partial n}\bar f\,ds=-\int_{\ell}
\left.\frac{\partial f}{\partial n}\right|_{\gamma z}|\gamma'(z)|\bar
f\, ds=-\int_{\gamma\ell}\frac{\partial f}{\partial n}\bar f\, ds\;,
$$
and the first integral at the right-hand side of the expression for
$(Df,f)$ equals zero. It can be proved similarly that $(Df,g)=
{(f,Dg)}$, i.e.\ $D$ is a self-adjoint operator.
 
If $Df\equiv 0$, then $f$ is a harmonic function, and
$f\equiv\text{const.}$ by the maximum principle. It can be proved that
the set of functions $Df$ is everywhere dense
in the space $L^{2}_0(\Cal D,\sigma)$, consisting of the
functions $f$, $\int_{\Cal D}f^2(z)\sigma(\,dz)<\infty$ and
$\int_{\Cal D}f(z)\sigma(\,dz)=0$. It is proved in the functional
analysis that because of these properties the operator $D$ can
be inverted, and the inverse of $D$ is an integral operator of the form
$$
D^{-1}f(z)=\int_{\Cal D} G(z,w)f(w)\sigma(\,dw)
$$
with a kernel function $G(z,w)$, $G(z,w)=G(w,z)$, and $G(z,w)$ is a
automorph function in both variables. This statement is equivalent to
the classical result in the analysis by which the Laplace operator on
the domain $\Cal D$ (with periodic boundary conditions) in the space
$L^2_0(\Cal D,\text{Lebesgue measure})$ is
invertible, and the inverse is an integral operator with a kernel
function $G(z,w)$. The kernel function $G(z,w)$ is the same in these two
problems. Its symmetry properties follow from the properties of the
operator $D$ (or $\Delta$).
 
We only give a very sketchy explanation why the inverse operator
$D^{-1}$ can be expressed in such a way. If $G(z,w)$ is an automorph
function in the variable $w$ which is of the form $-\log|z-w|+\log
|z_0-w|+\text{analytic function in the variable }w$, ($z_0$ is a
fixed  point in $H$) then it is relatively simple to show by means of
the Green formula that it can be taken as the kernel function of the
integral operator which is the inverse of $D$. It follows from the
general theory that such a function $G$ exits.
 
The integral operator is again a negative operator, and since the kernel
function $G(z,w)$ in it is a nice function, (it has only logarithmical
singularities) the inverse of the operator $D$ is a Hilbert--Schmidt
operator. It follows from the general theory of such operators that
there exists a system of complete orthonormal system $\{f_i(z)\}_{i=1}
^\infty$ of eigenfunctions of the operator $D^{-1}$ with eigenvalue
$\lambda_i$ in the space $L^2_0(\Cal D,\sigma)$. Moreover,
$\lambda_i\to 0$, if $i\to\infty$, and only the eigenvalue zero can
have infinite multiplicity. But in our case the zero eigenvalue does
not appear, since $F\equiv0\Rightarrow \Delta F\equiv0$.
%in $L^2_0(\Cal D,\sigma)$.
In such a way we get a complete orthonormal system of
eigenfunctions $f_i(z)$ of $D$ in the space $L^2_0(\Cal D,\sigma)$,
$Df_i=\lambda_if_i$ with $\lambda_i<0$, and $\lambda_i\to-\infty$ as
$i\to\infty$. Let us adjust the number $\lambda_0=0$ and
$f_0(z)=\dfrac1{\sqrt{\sigma(\Cal D)}}$ to this system. Then
$\{f_i\}_{i=0}^\infty$ is a complete orthonormal system consisting of
eigenfunctions of $D$ in the $L^2(\Cal D,d\sigma)$ space.
 
\beginsection The trace of an invariant integral operator
 
For an invariant integral operator $L$ with kernel function $k(z,w)$
$$
\align
Lf(z)&=\int_H
k(z,w)f(w)\sigma(\,dw)=\sum_{\gamma\in\Gamma}\int_{\gamma\Cal
D} k(z,w)f(w) \sigma(\,dw)\\
&=\sum_{\gamma\in\Gamma}\int_{\Cal D}k(z,\gamma w)\,f(w)\sigma
(\,dw)=\int_{\Cal D}K(z,w)f(w)\,\sigma(\,dw)\;,
\endalign
$$
where
$$
K(z,w)=\sum_{\gamma\in\Gamma} k(z,\gamma w)\;.
$$
The function $K(z,w)$ is automorph in both variables, hence it can be
expanded by the complete orthonormal system
$\{f_i(z)\overline{f_j(w)}\}$ in $\Cal D\times\Cal D$,
just as in the real line model considered in the first talk. This
expansion has the form
$$
K(z,w)=\sum_{i=1}^\infty \Lambda_i f_i(z)\overline{f_i(w)}\;,
$$
where $Lf_i=\Lambda_i f_i$.
 
The number $\Lambda_i$ can be calculated, as a function of $\lambda_i$,
without the knowledge of $f_i$, and this was already done in the third
talk. If we take $z=w$, and integrate on the diagonal, i.e.\ we
calculate the trace of
the operator $L$ on $\Cal D$, then the functions $f_i$ disappear from
the formulas, and we get
$$
\int_{\Cal D}K(z,z)\sigma(\,dz)=\sum_{i=0}^\infty \Lambda_i\int_{\Cal
D}|f_i(z)|^2\sigma(\,dz)=\sum_{i=0}^\infty\Lambda_i\;,
$$
that is,
$$
\sum_{\gamma\in\Gamma}\int_{\Cal D}k(z,\gamma z)\sigma(\,dz) =
\int_{\Cal D}\sum_{\gamma\in\Gamma}k(z,\gamma z)\sigma(\,dz)=
\sum_{i=0}^\infty\Lambda_i\;. \tag++
$$
We want to rewrite the left-hand side of (++) in a more
appropriate form. In the real line case the kernel function $k(z,\gamma
z)$ ($=k(x-(x+n))$) of the corresponding integral operator was in the
variable $z$ ($x$) constant. Let us look the corresponding question in
the present case. When does the equation $k(w,\gamma_0 w)=k(z,\gamma
z)$ hold?
 
If $w=\bold T z$, then
$$
k(w,\gamma_0 w)=k(\bold Tz,\gamma_0\bold Tz)=k(z,\bold
T^{-1}\gamma_0\bold Tz)\;
$$
hence $k(w,\gamma_0 w)=k(z,\gamma z)$ if $\gamma=\bold
T^{-1}\gamma_0\bold T$. If
$\gamma=\gamma_0$, then this condition means that $\gamma_0$ and $\bold
T$ are exchangeable. (In the real line case they are always
exchangeable, but in the present case this property may be violated). In
the general
case this equality holds if $\gamma$ and $\gamma_0$ are conjugate.
Therefore, let us sum for different conjugate classes separately. Let
us denote the conjugate classes in the form $\{\gamma_0\}$.
 
Let us conjugate by an element be $g\: \gamma=g^{-1}\gamma_0g$. (This
conjugation would be meaningful also for $g\notin\Gamma$, but we shall
consider only conjugations with elements from $\Gamma$.)
Two such conjugates agree, i.e.\ $g_1^{-1}\gamma_0 g_1
=g^{-1}\gamma_0 g$, if and only if
$g_1g^{-1}\gamma_0=\gamma_0g_1g^{-1}$, which means that
$g_1g^{-1}\in [\gamma_0]$, or $g_1\in[\gamma_0]g$, where
$[\gamma_0]$ denotes the
group of element in $\Gamma$ exchangeable with $\gamma_0$. This group
$[\gamma_0]$ is called in algebra the centralizator group of
$\gamma_0$. Thus the sum $\sum\limits_{\gamma\in\{\gamma_0\}}$ can also
be written in the form $\sum\limits\Sb g\\ g\in\Gamma/[\gamma_0]\endSb
$, where $g$ are fixed representation elements from the right cosets of
the subgroup $[\gamma_0]$.
\bigskip
If $\gamma\in\{\gamma_0\}$, then  $\gamma=g^{-1}\gamma_0g$ with some
$g$, and
$$
\align
\int_{\Cal D}k(z,\gamma z)\sigma(\,dz)=
\int_{\Cal D}k(z,g^{-1}\gamma_0g z)\sigma(\,dz)&=
\int_{\Cal D}k(gz,\gamma_0g z)\sigma(\,dz)\\
&=\int_{g\Cal D}k(z,\gamma_0 z)\sigma(\,dz)\;.
\endalign
$$
$$
\sum_{\gamma\in\{\gamma_0\}}\int_{\Cal D}k(z,\gamma z)\sigma(\,dz)=
\sum\Sb g\\
g\in\Gamma/[\gamma_0]\endSb \int_{g\Cal
D}k(z,\gamma_0 z)\sigma(\,dz)
=\int\limits\Sb \cup g\Cal D\\
g\in\Gamma/[\gamma_0]\endSb
k(z,\gamma_0 z)\sigma(\,dz)\;.
$$
If the representation elements $g\in \Gamma/[\gamma_0]$ are multiplied
from the left with the
elements of $[\gamma_0]$, then we get the elements of $\Gamma$,  and all
elements exactly once. Thus $\Cal
D(\gamma_0)\overset\text{def}\to=\bigcup\limits_{g\in \Gamma/[\gamma_0]}
g\Cal D$ is such a set that the sets $\gamma\Cal D(\gamma_0)$,
$\gamma\in[\gamma_0]$, produce a disjoint cover of
$H$. In other words, $\Cal D(\gamma_0)$ is a fundamental
domain of the discrete subgroup $[\gamma_0]$. We know that
$k(z,\gamma_0z)$ is automorph with respect to $[\gamma_0]$,
($k(gz,\gamma_0gz)=k(z,g^{-1}\gamma_0gz)=k(z,\gamma_0z)$ for
$g\in[\gamma_0]$), hence it has
no importance how to choose the fundamental domain where we integrate.
We formulate the result we obtained in the following lemma.
\medskip\noindent
{\bf Lemma.} {\it For all $\gamma_0\in\Gamma$ the relation
$$
\sum_{\gamma\in\{\gamma_0\}}\int_{\Cal D}k(z,\gamma z)\sigma(\,dz)=
\int_{\Cal D(\gamma_0)}k(z,\gamma_0z)\sigma(\,dz)
$$
holds, where $\{\gamma_0\}=\{g^{-1}\gamma_0 g;\;g\in\Gamma\}$ denotes
the class of elements in $\Gamma$ which are conjugated to $\gamma_0$,
and $\Cal D(\gamma_0)$ is a fundamental domain of the centralizator group $[\gamma_0]\subset \Gamma$
of $\gamma_0$. The function $k(z,\gamma_0z)$ at the right-hand side of
this formula is $[\gamma_0]$ automorph.}
\medskip\noindent
{\it Remark:}\/ It is possible that $\{\gamma_0\}=\{\gamma_1\}$ with
some $\gamma_0\neq\gamma_1$ and even $[\gamma_0]\neq [\gamma_1]$. But
the right-hand side of the integral in the lemma agrees for such
$\gamma_0$ and $\gamma_1$. Indeed, if $\gamma_1=g^{-1}\gamma_0g$, then
$$
\int_{\Cal D(\gamma_1)}k(z,\gamma_1z)\sigma(\,dz)=
\int_{g\Cal D(\gamma_1)}k(z,\gamma_0z)\sigma(\,dz)\;,
$$
and $g\Cal D(\gamma_1)$ is a fundamental domain of $[\gamma_0]$. (If
$h_j\Cal D(\gamma_1)$, $j=1,\dots$, is a disjoint cover of $H$,
and the transformations $h_j$ are the elements of $[\gamma_1]$, then
the sets $gh_j g^{-1}(g\Cal D(\gamma_1))$, $j=1,\dots$, also give a
disjoint cover of $H$, and $gh_jg^{-1}$ are the elements of
$[\gamma_0]$.)
 
We shall write the expression in $(++)$ in a simpler form by means of
the Lemma and a further investigation which will show that in the group
$\Gamma$ the centralizators $[\gamma_0]$ have a very special form.
\vfill\eject
 
\noindent{\CAP Fifth talk:}
 
\beginsection The proof of Selberg's trace formula in the hyperbolic
plane
 
Let $\bold T$ be a motion of $H$. Its analytic extension to the complex
sphere, which is a linear rational function, has exactly two fixed
points (with multiplicity). If it is a double fixed point, which must
be because of the relation $\bold T\bar z=\overline {\bold Tz}$ either
a real number or infinity, then the transformation $\bold T$ is called
parabolic. If $\bold T$ has a fixed point in $H$ (its conjugate, which
is not in $H$, is also a fixed point), then it is called an elliptic
transformation. In the remaining case, the transformation $\bold T$ is
called hyperbolic. Hyperbolic transformations have two different real
or ($\infty$) fixed points.
 
A discrete subgroup $\Gamma$ of the motions in $H$ cannot contain an
elliptic transformation, since the non-unity elements of $\Gamma$
cannot have a fixed point in $H$. We also excluded the possibility of
$\gamma\in \Gamma$ for a parabolic transformation $\gamma$ by the
assumption that $\bar {\Cal D}$ is compact. Indeed, for a parabolic
transformation $d(z,\gamma z)$ can be taken arbitrary small with an
appropriate choice of $z\in H$, (by a conjugation which takes the
fixed point to infinity the proof of this statement can be reduced to
the case when the double fixed point of $\gamma$ is infinity, i.e.\
$\gamma z=z+c$  with some real number $c$, and in this case it is easy
to check this property) and we show that because of this property
$\Gamma$ cannot contain a parabolic transformation. Otherwise, by
mapping a sequence of points with the above properties
into the domain $\Cal D$ we could choose a sequence $z_n$ with $z_n\in
\Cal D$ and $\gamma_n\in \Gamma$ in such a way that $0<d(z_n,\gamma_n
z_n)\to0$, as $n\to\infty$. Then, because of the compactness of
$\bar{\Cal D}$ we may assume, by considering a subsequence of the
sequence $z_n$, that $z_n\to z_0\in H$. Since
$d(\gamma_nz_n,\gamma_nz_0)=d(z_0,z_n)\to 0$, $d(z_0,\gamma_n z_0)\le
d(z_0,z_n)+d(z_n,\gamma_nz_n)+d(\gamma_n z_n,\gamma_nz_0)\to0$,
as $n\to0$, and $\gamma_n\neq\text{Id}$. But this contradicts to the
fact, that since $\Gamma$ has no fixed point, and it is discreet,
$\min\limits \Sb\gamma\in \Gamma\\ \gamma\neq\text{Id}\endSb
d(z_0,\gamma z_0)>0$.  So $\Gamma$ contains only hyperbolic
transformations.
 
Let $\gamma$ be a hyperbolic transformation. If $\bold T$ is such a
motion of $H$ (not necessarily in the group $\Gamma$) which
maps the two fixed points of $\gamma$  into 0 and $\infty$, then the
two fixed points of $\bold T^{-1}\gamma\bold T$ are 0 and $\infty$,
hence $\bold T^{-1}\gamma\bold Tz=\rho z$ with some $\rho>0$,
$\rho\neq1$. Moreover,
we may assume with a possible conjugation with $\dfrac1z$ that $\rho>1$.
This number $\rho$ is uniquely determined by $\gamma$ (and even by
$\{\gamma\}$). We call it the norm of $\gamma$, and introduce the
notation $\rho=N(\gamma)=N(\{\gamma\})$.
 
Two motions of $H$ are exchangeable if and only if their fixed points
agree. For the motions exchangeable with $\gamma_0$ the same motion
$\bold T$ can be chosen for writing it as the conjugate of a linear
transformations, and we see that
this group is isomorphic to the multiplicative group of the positive
numbers $\rho>0$. A discrete subgroup of this group is an infinite
cyclic group, and we call a generator $\gamma^*$ of it a primitive
element. $[\gamma_0]=\{{\gamma^*}^m\}_{m=-\infty}^{\infty}$. Put
$\rho^*=N(\gamma^*)$.
$$
\align
\int_{\Cal D(\gamma_0)} \!k(z,\gamma_0z)\sigma(\,dz)&=
\int_{\Cal D(\gamma_0)}\!k(\bold T^{-1}z,\bold
T^{-1}\gamma_0z)\sigma(\,dz)
=\int_{\bold T^{-1}\!\Cal D(\gamma_0)}k(z,\bold
T^{-1}\gamma_0\bold Tz)\sigma(\,dz)\\  \allowdisplaybreak
&=\int_{\bold T^{-1}\Cal D(\gamma_0)}k(z,\rho_0z)\sigma(\,dz)\;,\quad
\rho_0=N(\gamma_0)\;,
\endalign
$$
$\bold T^{-1}\Cal D(\gamma_0)$ is a fundamental domain of the group
$\bold T^{-1}[\gamma_0]\bold T=\{ {\rho^*}^mz\}_{m=-\infty}^\infty$,
and since $k(z,\bold T^{-1}\gamma_0\bold Tz)$ is automorph for
$\bold T^{-1}[\gamma_0]\bold T$ we can choose an arbitrary
fundamental domain of $\bold T^{-1}[\gamma_0]\bold T$ to calculate the
last integral.
 
It is clear that the half-ring $\{1\le|z|<\rho^*\}\cap H$ is such a
domain. The function $k(z,\rho_0z)$ is invariant for  all
transformations $\bold Tz=\rho z$. Hence, it is constant along
the rays, and we have to integrate only with respect to the angle
$\vartheta$, where $z=re^{i\vartheta}$.
$$
\align
I&=\int_{\bold T^{-1}\Cal D(\gamma_0)}k(z,\rho_0z)\sigma(\,dz)=
\int_0^\pi\int_1^{\rho^*}k(e^{i\vartheta},\rho_0e^{i\vartheta})
\frac{r\,dr\,d\vartheta}{r^2\sin^2 \vartheta}\\
&=\int_0^\pi
k(e^{i\vartheta},\rho_0e^{i\vartheta})\frac{\log\rho^*}{\sin^2
\vartheta}\,d\vartheta
=2\int_0^{\pi/2}k\left(\frac{\left|e^{i\vartheta}
-\rho_0e^{i\vartheta}\right|^2}{\sin\vartheta\rho_0\sin\vartheta}
\right)\frac{\log\rho^*}{\sin^2\vartheta}\,d\vartheta\\
&=2\log\rho^*\int_0^{\pi/2}k\left(\frac{(\rho_0-1)^2}
{\rho_0\sin^2\vartheta}\right) \frac{d\vartheta}{\sin^2\vartheta}\;.
\endalign
$$
Put $t=\dfrac{(\rho_0-1)^2}{\rho_0}=\rho_0+\dfrac1{\rho_0}-2$ and
$\tau=\dfrac t{\sin^2\vartheta}$, that is
$\vartheta=\arcsin\sqrt{\dfrac t\tau}$. Then with this notation
$$
\align
I&=2\log\rho^*\int_t^\infty k(\theta)\frac\tau t\frac1{\sqrt{1-\frac
t\tau}}\frac{\sqrt t}{2\tau^{3/2}}\,d\tau\\
&=\frac{\log \rho^*}{\sqrt t}\int_t^\infty
\frac{k(\tau)}{\sqrt{\tau-t}}
\,d\tau=\frac{\log \rho^*}{\sqrt t}\int_0^\infty \frac{k(\tau+t)}{\sqrt
\tau}\,d\tau\\
&=\frac{\log \rho^*}{\sqrt t}\sqrt{\rho_0}g(\log \rho_0)=
\frac{\log \rho^*}{\frac{\rho_0-1}{\sqrt{\rho_0}}}
\sqrt{\rho_0}g(\log\rho_0)=\frac{\log \rho^*}{1-\frac1{\rho_0}}
g(\log\rho_0)
\endalign
$$
by the definition of $g(\cdot)$.
 
Putting these results together
$$
\sum_{\gamma\in\{\gamma_0\}}\int_{\Cal D}k(z,\gamma z)\sigma(\,dz)=
\frac{\log N(\gamma^*)}{1-\frac1{N(\gamma_0)}}g(\log N(\gamma_0))\;.
$$
This relation does not hold for the
one element conjugate class containing the origin. But
$$
\int_{\Cal D}k(z,z)\,\sigma(\,dz)=k(0)\sigma(\Cal D)=\frac{\sigma(\Cal
D)}{4\pi}\int_{-\infty}^\infty r\tanh(\pi
r)G\left(\frac12+ir\right)\,dr
$$
with the previous definition of $G(\cdot)$, what can be shown by a
long but for us not so interesting calculation.
\bigskip
 
So the starting trace formula
$$
\sum_{\gamma\in\Gamma}\int_{\Cal D}K(z,\gamma z)\sigma(\,dz)
=\sum_{i=0}^\infty\Lambda_i
$$
has the following form:
$$
\frac{\sigma(\Cal D)}{4\pi}\int_{-\infty}^\infty r\tanh(\pi r)
G\left(\frac12+ir\right)\,dr+\sum_{\{\gamma_0\}}\frac{\log N(\{\gamma^*
\})}{1-\frac1{N(\{\gamma_0\})}}g(\log N(\{\gamma_0\}))
=\sum_{i=0}^\infty G(s_i)\;.
$$
Here
$$
G(s)=\int_{-\infty}^\infty g(y)e^{sy}\,dy\;,
$$
(the functions $g(\cdot)$, $G(\cdot)$ and $k(\cdot)$ can be calculated
from each others by means of simple formulas),
$$
s_i=\frac12+\sqrt{\frac14+\lambda_i}\;,
$$
(one has to consider only one of the square roots), where $\lambda_0=0$,
$\lambda_i<0$, ($i=1,2,\dots$) are the eigenvalues of $D$ with
multiplicity,
$$
\sigma(\Cal D)=\text{the area of a fundamental domain}=4\pi (g-1).
$$
(The number $g$ is the genus of $\Cal S$.) In the sum the $\{\gamma_0\}$
runs through the conjugate
classes not containing the identity, $\{\gamma^*\}$ is the class of the
corresponding primitive element, i.e.\ $\gamma^*$ is defined by the
relations $\gamma^*\in\Gamma$, and ${\gamma^*}^m=\gamma_0$, with the
largest possible $m$, $N(\cdot)$ denotes  the norm of an
element $\gamma\in\Gamma$ introduced in this talk.
 
This is the famous Selberg formula in the simplest case.
\vfill\eject
 
\bye
 
