SELBERG’S TRACE-FORMULA a generalization of the Poisson summation formula a lecture of Gábor Halász in the Bolyai College of the Eötvös Loránd University, Budapest

First talk:

The Poisson summation formula

Let a function k(x) be given, and define the function K(x) = ∑_{n=−∞}^{∞} 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) = ∑_{j=−∞}^{∞} a_{j}e^{2πijx} with

a_{j} = ∫_{0}^{1} K(x)e^{−2πijx} dx = ∫_{0}^{1} ∑_{n=−∞}^{∞} k(x + n)e^{−2πijx} dx

= ∑_{n=−∞}^{∞} ∫_{0}^{1} k(x + n)e^{−2πijx} dx = ∫_{−∞}^{∞} k(x)e^{−2πijx} dx = κ(2πj),

where, κ(t) = ∫_{−∞}^{∞} 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

∑_{n=−∞}^{∞} k(n) = ∑_{j=−∞}^{∞} κ(2πj) .

The main benefit of the last formula: The left-hand side is an approximating sum of the integral ∫_{−∞}^{∞} k(x) dx which equals κ(0). The sum of the terms j ≠ 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. The same subject in a high-brow style

We shall prove the Poisson summation formula in the following more complicated form:

∑_{n=−∞}^{∞} k(n) = 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 ℝ^{1}, together with the usual metric |x − y| and the Lebesgue measure. Let T_{u}: x → 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 T_{u}f(x) = f(x + u) acting on the space of functions f(x), −∞ < x < ∞, by the same letter.

A general operator L will be called invariant, if L_{x}f(x + u) = Lf(x)|_{x→x+u}, i.e. if LT_{u} = T_{u}L, (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.)

We consider the following question: When is the integral operator Lf = ∫_{−∞}^{∞} k(x, y)f(y)dy

invariant?

We have LT_{u}f = ∫_{−∞}^{∞} k(x, y)f(y + u) du and T_{u}Lf = ∫_{−∞}^{∞} k(x + y, y)f(y) dy = ∫_{−∞}^{∞} 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 = ∫_{−∞}^{∞} k(x − y)f(y) dy is a convolution.

When is the differential operator Lf = ∑_{ν=0}^{m} a_{ν}(x)\frac{d^{ν}f}{dx^{ν}} invariant?

LT_{u}f = ∑_{ν=0}^{m} a_{ν}(x)f^{ν}(x + u), T_{u}Lf = ∑_{ν=0}^{m} a_{ν}(x + u)f^{ν}(x + u), i.e. it is invariant

if a_{ν}(x) is constant, L = ∑_{ν=0}^{m} a_{ν}D^{ν}, where D = \frac{d}{dx}. In such a way we described all

invariant differential operators.

The above defined invariant operators are exchangeable. For instance, if Lf =

∫_{−∞}^{∞} k(x − y)f(y) dy, then DLf = (∫_{−∞}^{∞} k(x − y)f(y) dy)′ = (we do not have to differ-

entiate immediately) = (∫_{−∞}^{∞} k(y)f(x − y) dy)′ = ∫_{−∞}^{∞} k(x − y)f′(y) dy = LDf.

In the non-trivial step we exploited the fact that L is invariant: If L^{0}f =^{def} Lf|_{x=0}, then because of the invariance L_{x}f(x) = L_{y}^{0}T_{x}f(y). (This means in our case that ∫_{∞}^{∞} k(x − y)f(y) dy = ∫_{−∞}^{∞} k(0 − y)f(y + x) dy.)

Claim. Any two invariant operators L_{1} and L_{2} are exchangeable.

Proof: By the previous argument L^{0} determines an invariant operator L, and since both L_{1}L_{2} and L_{2}L_{1} are invariant, it is enough to show that (L_{1}L_{2})^{0} = (L_{2}L_{1})^{0}, i.e. that L_{1}^{0}L_{2} = L_{2}^{0}L_{1}.

We know that L_{2}f = L_{2y}^{0}f(y + x), hence L_{1x}^{0}L_{2y}^{0}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_{2y}^{0}L_{1x}^{0}f(y+x) = (because x+y = y+x, the group of shifts is commutative) = L_{2y}^{0}L_{1x}^{0}f(x + y) = L_{2}^{0}L_{1}f.

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 λ, f(x) = e^{λx} (all functions Ae^{λx} are such eigenfunctions), then DLf = LDf = Lλf = λLf, that is Lf is also an eigenfunction of D with the same eigenvalue. Since these eigenfunctions constitute a one-dimensional space Lf = Λe^{λx} = Λf(x), where Λ depends only on λ and L.

Let us determine this Λ for the operator Lf = ∫_{−∞}^{∞} k(x − y)f(y) dy.

Since f(0) = 1, Λ = L^{0}f = ∫_{−∞}^{∞} k(−y)e^{λy} dy = ∫_{−∞}^{∞} k(y)e^{−λy} dy.

A typical discrete subgroup of the group of shifts is the group T_{n}, n = 0, ±1, ±2, . . .. If T_{n}f = 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 T_{n}Lf = LT_{n}f = Lf, i.e. L maps periodic functions into periodic functions. If L is an integral operator, then

Lf = ∫_{−∞}^{∞} k(x − y)f(y) dy = ∑_{n=−∞}^{∞} ∫_{0}^{1} k(x − (y + n))f(y) dy = ∫_{0}^{1} K(x, y)f(y) dy) ,

where K(x, y) = ∑_{n=−∞}^{∞} 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) = ∫_{0}^{1} f′(x)ḡ(x) dx = (integrating by parts) = − ∫_{0}^{1} f(x)ḡ′(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 = λf, then (Df, f) = λ(f, f), (Df, f) = −(f, Df) = −λ̄(f, f) ⇒ λ = −λ̄), and the eigenfunctions corresponding to different eigenvalues are orthogonal: ((Df, g) = λ(f, g), (Df, g) = −(f, Dg) = −ν̄(f, g), and if λ ≠ ν, then (f, g) = 0.)

We know that the (normalized) eigenfunctions are the functions f_{j}(x) = e^{λ_{j}x} = e^{2πijx}. 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) = ∑_{j,l=−∞}^{∞} c_{j,l}f_{j}(x)f̄_{l}(y) .

If Λ_{j} denotes the eigenvalue of the operator Lf = ∫_{−∞}^{∞} k(x − y)f(y) dy corresponding to the eigenvalue λ_{j} of D about which we know that

Λ_{j} = ∫_{−∞}^{∞} k(y)e^{−λ_{j}y} dy = ∫_{−∞}^{∞} k(y)e^{−2πijy} dy = κ(2πj) ,

and

Λ_{j}f_{j}(x) = Lf_{j} = ∫_{0}^{1} K(x, y)f_{j}(y) dy = ∫_{0}^{1} ∑_{i,l=−∞}^{∞} c_{i,l}f_{i}(x)f̄_{l}(y)f_{j}(y) dy

= ∑_{i,l=−∞}^{∞} c_{i,l}f_{i}(x) ∫_{0}^{1} f̄_{l}(y)f_{j}(y) dy = ∑_{i=−∞}^{∞} c_{i,j}f_{i}(x) .

From here

c_{i,j} = \{ \begin{matrix}0, & if i ≠ 0 \\ Λ_{j}, & if i = j \end{matrix} .

Hence K(x, y) = ∑_{j=−∞}^{∞} Λ_{j}f_{j}(x)f̄_{j}(y).

Thus,

K(0, 0) = ∑_{j=−∞}^{∞} Λ_{j} = ∑_{j=−∞}^{∞} κ(2πj) .

On the other hand,

K(x, x) = ∑_{n=−∞}^{∞} k(x − (x + n)) = ∑_{n=−∞}^{∞} k(n) ,

or because the function K(x, x) is constant, the trace of L equals

∫_{0}^{1} K(x, x) dx = K(0, 0) = ∑_{j=−∞}^{∞} Λ_{j} = ∑_{j=−∞}^{∞} κ(2πj) ,

and

∫_{0}^{1} K(x, x) dx = K(0, 0) = ∑_{n=−∞}^{∞} k(n) .

A comparison of the last two relations yields the Poisson formula.

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.

In a special case we carry out the above program. We shall discuss the hyperbolic plane, i.e. the Poincaré 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.

Second talk:

Invariant transformations in the hyperbolic plane

We recall the following two classical (equivalent) models of the hyperbolic plane.

The space:

The unit circle of the complex plane {z: |z| < 1} The upper half plane of the complex plane H = {z: Im z > 0}

The metric (length) in the space:

\frac{|dz|}{1 − |z|^{2}} ; I(γ) = ∫_{γ} \frac{|dz|}{1 − |z|^{2}} \frac{|dz|}{y} if z = x + iy.

I(γ) = ∫_{γ} \frac{|dz|}{y}

d(z, w) = min_{z,w^{⌢}} I(z, w^{⌢})

= log \frac{1 + |\frac{z−w}{z−w̄} |}{1 − |\frac{z−w}{z−w̄} |}

This expression is the function of

\frac{4}{|\frac{z−w̄}{z−w} |^{2}−1} = \frac{|z−w|^{2}}{yv}

if z = x + iy and w = u + iv.

The measure (area) in the space:

\frac{dxdy}{(1 − |z|^{2})^{2}}  dσ =^{def} \frac{dxdy}{y^{2}}

The length and measure preserving movements of the space:

{Tz = ρ\frac{z − ζ}{1 − zζ̄}, \atop |ρ| = 1, |ζ| < 1} 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)/{±Id}, where SL_{2}(R) denotes the group of 2 × 2 matrices with real coefficients and determinant 1.)

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 T on the hyperbolic plane, define Tf as

Tf(z) =^{def} f(Tz).

Definition. An operator L acting on the functions over a hyperbolic space is called invariant, if TL = LT for all movements T of the space.

Question: Which integral operators Lf = ∫_{H} k(z, w)f(w)σ( dw) are invariant?

TLf(z) = ∫_{H} k(Tz, w)f(w)σ( dw) = ∫_{H} k(Tz, Tw)f(Tw)σ( dw) ,

and

LTf(z) = ∫_{H} k(z, w)f(Tw)σ( dw) .

Hence L is invariant if and only if k(z, w) = k(Tz, Tw) for all T. The group of the movements 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 T such that z_{2} = Tz_{1} and w_{2} = 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 (\frac{|z − w|^{2}}{yv} ) .

Which differential operators Lf = ∑ a_{i,j}(z)\frac{∂^{i,j}}{∂x^{i}∂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^{0}f = Lf(z_{0}) and want to reconstruct the invariant operator L by means of L^{0}.

Remark: The functional L^{0} cannot be arbitrary. Indeed, if the transformation T is such that Tz_{0} = z_{0}, then L^{0}Tf = LTf(z_{0}) = TLf(z_{0}) = Lf(Tz_{0}) = Lf(z_{0}) = L^{0}f for this 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}T_{z}f = L_{w}^{0}f(T_{z}w). We formulate this statement in the following lemma.

Lemma. 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^{0}f = L^{0}Tf for all such movements T of the hyperbolic plane for which 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}T_{z}f = L_{w}^{0}f(T_{z}w), where T_{z} is a movement such that T_{z}z_{0} = z. This definition is meaningful, i.e. the value of Lf(z) does not depend on the special choice of T_{z}. This operator L is invariant, and all invariant operators L can be obtained in such a way through the choice of L_{0}f = Lf(z_{0}).

Proof: The definition of Lf is meaningful. Indeed, let T_{z} and T̄_{z} be two movements such that T_{z}z_{0} = z and T̄_{z}z_{0} = z. We have to show that L^{0}T̄_{z}f = L^{0}T_{z}f, or with the notation T_{z}f = g, L^{0}g = L^{0}T̄_{z}T_{z}^{−1}g or L^{0}g(w) = L_{w}^{0}g(T_{z}^{−1}T̄_{z}w). This formula follows from the rotational invariance of L^{0} and the relation T_{z}^{−1}T̄_{z}z_{0} = z_{0}.

Let us consider an arbitrary motion T. Then

TLf(z) = h(Tz) = L^{0}g

with h(u) = Lf(u) = L^{0}T_{u}f (h(Tz) = L^{0}T_{Tz}f) and g(u) = T_{Tz}f(u) = f(T_{Tz}u), where T_{Tz} is a motion of the hyperbolic plane such that T_{Tz}z_{0} = Tz.

LTf(z) = L^{0}T_{z}Tf = L^{0}p = L_{0}T′g = L_{0}g

with p(u) = T_{z}Tf(u) = f(TT_{z}u) and T′ = T_{Tz}^{−1}TT_{z}. The last but one identity holds in this relation because p(u) = f(T_{Tz}T_{Tz}^{−1}TT_{z}u) = g(T′u), i.e. p = T′g. Finally, the last identity holds in this relation, because of the rotational invariance of L^{0} and the equality T′z_{0} = T_{Tz}^{−1}TT_{z}z_{0} = T_{Tz}^{−1}Tz = z_{0}. A comparison of the expressions obtained for TLf and LTf shows that the operator L defined in the lemma is invariant.

Finally, let L be an invariant operator, and define L_{0}f = Lf(z_{0}). Then L_{0} is a rotation invariant functional, and it determines the original operator L, since L^{0}T_{z}f = LT_{z}f(z_{0}) = T_{z}Lf(z_{0}) = Lf(T_{z}z_{0}) = Lf(z).

Thus the invariant differential operators can be found by first describing the rotation invariant differential functionals

L^{0}f = ∑ a_{i,j} \frac{∂^{i+j}}{∂x^{i}∂y^{j}} f(z)|_{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 \frac{∂}{∂z} = \frac{1}{2} \frac{∂}{∂x} − i\frac{1}{2} \frac{∂}{∂y} and \frac{∂}{∂z̄} = \frac{1}{2} \frac{∂}{∂x} + i\frac{1}{2} \frac{∂}{∂y}. In this (equivalent)

reformulation our problem is to find the rotational invariant differential functionals

L^{0}f = ∑ b_{i,j} \frac{∂^{i+j}}{∂z^{i}∂z̄^{j}} f(z)|_{z=0}

with some new coefficients b_{i,j}. The rotational invariance means in this case that the expression

L^{0}f = ∑ b_{i,j}ρ^{i}ρ̄^{j} \frac{∂^{i+j}}{∂z^{i}∂z̄^{j}} f(z)|_{z=0}

has the same value for all ρ if |ρ| = 1. This relation holds if

∑_{i−j=h} b_{i,j}ρ^{i}ρ̄^{j} \frac{∂^{i+j}}{∂z^{i}∂z̄^{j}} f(z)|_{z=0} = 0

for all functions f and h ≠ 0. This statement holds if b_{i,j} = 0 for i ≠ j. So

L^{0}f = ∑_{i=0}^{m} b_{i,i} (ρ^{i}ρ̄^{i} \frac{∂^{2}}{∂z∂z̄})^{i} f(z)|_{z=0} = ∑_{i=0}^{m} c_{i} Δ^{i}f(z)|_{z=0} ,  (∗)

because

4\frac{∂^{2}}{∂z∂z̄} = \frac{∂^{2}}{∂x^{2}} + \frac{∂^{2}}{∂y^{2}} = Δ .

Thus for instance, if L^{0}f = Δf|_{z=0}, then the invariant operator L corresponding to it equals Lf(z) = L_{w}^{0}f(T_{z}w) = Δf(T_{z}w)|_{w=0}. The operator T_{z} can be chosen as

T_{z}w = \frac{w + z}{1 + wz̄}, for which \frac{∂T_{z}w}{∂w} |_{w=0} = 1 − |z|^{2}. This gives that

Δf(T_{z}w) = 4\frac{∂^{2}f(T_{z}w)}{∂w∂w̄} = 4\frac{∂}{∂w̄} \frac{∂f}{∂w} |_{T_{z}w} \frac{∂T_{z}w}{∂w}

= \frac{∂^{2}}{∂w̄∂w} f|_{T_{z}w} |\frac{∂T_{z}w}{∂w} |^{2} + 4\frac{∂f}{∂w} |_{T_{z}w} \frac{∂}{∂w̄} \frac{∂T_{z}w}{∂w} = Δf|_{T_{z}w} |\frac{∂T_{z}w}{∂w} |^{2} .

The above calculation yields with w = 0 that the invariant differential operator

corresponding to L^{0}f = \frac{∂^{2}}{∂z∂z̄}f is Δf(T_{z}w)|_{w=0} = (1 − |z|^{2})^{2}Δf(z) =^{def} D^{∗}f(z), which is the hyperbolic Laplace operator. Also the polynomials

Lf = ∑_{l=0}^{m} c_{l}D^{∗}^{l}f  (∗∗)

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{∂^{2m}}{∂z^{m}∂z̄^{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 ≤ m ≤ 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 ≤ m ≤ N give an N + 1dimensional 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 = ∑_{l=0}^{m} c_{l}D^{m}f, with Df(z) = y^{2}Δf(z), where z = x + iy and Δ is the Laplace

operator.

Theorem. Let L_{1} and L_{2} be two invariant differential or integral operators on the hyperbolic plane. Then they are exchangeable, i.e. L_{1}L_{2} = L_{2}L_{1}.

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_{1}L_{2} and L_{2}L_{1} are also invariant operators, by the lemma it is enough to show that (L_{1}L_{2})^{0}f = (L_{2}L_{1})^{0}f. We can write (L_{1}L_{2})^{0}f = L_{1}g(z_{0}) with

g(z) = L_{2}f(z) = T_{z}L_{2}f(z_{0}) = L_{2}T_{z}f(z_{0}) = ∫ k_{2}(z_{0}, w)f(T_{z}w)σ( dw)

and (L_{2}L_{1})^{0}f = L_{2}h(z_{0}) with h(z) = L_{1}f(z). Here

h(z) = T_{z}L_{1}f(z_{0}) = L_{1}T_{z}f(z_{0}) = ∫ k_{1}(z_{0}, w)f(T_{z}w)σ( dw)

if L_{1} is an integral operator, and

h(z) = T_{z}L_{1}f(z_{0}) = L_{1}T_{z}f(z_{0}) = D_{w}f(T_{z}w)|_{w=z_{0}}

if L_{1} = D. If both L_{1} and L_{2} are integral operators, then

(L_{1}L_{2})^{0}f = L_{1}g(z_{0}) = ∫∫ k_{1}(z_{0}, z)k_{2}(z_{0}, w)f(T_{z}w)σ( dz)σ( dw)

and

(L_{2}L_{1})^{0}f = ∫∫ k_{2}(z_{0}, z)k_{1}(z_{0}, w)f(T_{z}w)σ( dz)σ( dw)
= ∫∫ k_{1}(z_{0}, z)k_{2}(z_{0}, w)f(T_{w}z)σ( dz)σ( dw) .

If L_{1} = D, then

(L_{1}L_{2})^{0}f = L_{1}g(z_{0}) = D_{z} g(z)|_{z=z_{0}} = D ∫ k_{2}(z_{0}, w)f(T_{z}w)σ( dw)|_{z=z_{0}}
= ∫ k_{2}(z_{0}, w) D_{z}f(T_{z}w)|_{z=z_{0}} σ( dw)

and

(L_{2}L_{1})^{0}f = L_{2}h(z_{0}) = ∫ k_{2}(z_{0}, z)h(z)σ( dz) = ∫ k_{2}(z_{0}, z) D_{w}f(T_{z}w)|_{w=z_{0}} σ( dz)
= ∫ k_{2}(z_{0}, w) D_{z}f(T_{w}z)|_{z=z_{0}} σ( dw) .

We have certain freedom in the choice of the operators T_{z}. If we could define these operators in such a way that T_{z}w = T_{z}T_{w}z_{0} and T_{w}z = T_{w}T_{z}z_{0} agreed, then a comparison of the expressions (L_{1}L_{2})^{0}) and (L_{1}L_{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 T_{z} the following weak commutativity property holds:

d(z_{0}, T_{z}w) = d(z_{0}, T_{w}z) for all w and z.

Indeed, by choosing the motions T_{z} in such a way that T_{z}z_{0} = z and T_{z}z = z_{0} we get that

d(z_{0}, T_{w}z) = d(T_{w}w, T_{w}z) = d(w, z) = d(T_{z}w, T_{z}z) = d(T_{z}w, z_{0}) .

Let us first consider rotation invariant functions f, i.e. such functions for which f(z) = f(Tz) for any motion T with the property Tz_{0} = z_{0}. Then the value of f(z) depends only on d(z_{0}, z). The previous calculations yield that (L_{1}L_{2})^{0}f = (L_{2}L_{1})^{0}f for rotation invariant functions. To consider a general function f let us introduce the following “averaging” operator

f_{0}(z) = Mf(z) = \frac{1}{2π} ∫_{0}^{2π} T_{ϑ}f(z) dϑ ,  (+)

where T_{ϑ}, 0 ≤ ϑ < 2π, “the rotation with angle ϑ”, is defined in the following way: Let us consider an analytic automorphism U of the unit circle to the positive half-plane such that U maps the origin to z_{0}. If S_{ϑ}z = e^{iϑ}z denotes the rotation with angle ϑ in the unit circle, then T_{ϑ} is the transformation corresponding to it through this automorphism, i.e. T_{ϑ}z = US_{ϑ}U^{−1}z. The function f_{0} = Mf is rotation invariant, since for all ϑ′

f_{0}(T_{ϑ′}z) = \frac{1}{2π} ∫_{0}^{2π} f(T_{ϑ}T_{ϑ′}z) dϑ = ∫_{0}^{2π} f(T_{ϑ+ϑ′}z) dϑ = ∫_{0}^{2π} f(T_{ϑ}z) dϑ = f_{0}(z_{0}) ,

and all movements of the hyperbolic plane with the fixed point z_{0} equals some T_{ϑ}.

If L is an invariant operator, then LM = ML, since L is exchangeable with all T_{ϑ}. Hence

(L_{1}L_{2})^{0}f = L_{1}L_{2}f(z_{0}) = ML_{1}L_{2}f(z_{0}) = L_{1}L_{2}Mf(z_{0}) = L_{1}L_{2}f_{0}(z_{0})

and

(L_{2}L_{1})^{0}f = L_{2}L_{1}f(z_{0}) = ML_{2}L_{1}f(z_{0}) = L_{2}L_{1}Mf(z_{0}) = L_{2}L_{1}f_{0}(z_{0}) .

Since L_{1}L_{2}f_{0} = L_{2}L_{1}f_{0} the above relations imply that (L_{1}L_{2})^{0}f = (L_{2}L_{1})^{0}f, and L_{1}L_{2} = L_{2}L_{1} by the lemma. The theorem is proved.

Third talk:

The eigenvalues of invariant integral operators

We can find many eigenfunctions of the operator D = y^{2}Δ in the upper half-plane model of the hyperbolic space. Thus for instance y^{s}, z = x + iy, is an eigenfunction with the eigenvalue λ = s(s − 1), since Dy^{s} = y^{2}Δy^{s} = y^{2}s(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} = λLy^{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} = Λy^{s} with some Λ. 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 λ is one-dimensional. In the hyperbolic space H the analogous statement does not hold. The eigenspace of the operator D corresponding to an eigenvalue λ is very large. For instance, the functions Ty^{s} = (Im Tz)^{s} are eigenfunctions with eigenvalue λ for all motions T. 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 = λf, then for f_{0} = Mf

Df_{0} = DMf = MDf = λMf = λf_{0} ,

and f_{0} is rotation invariant around the center z_{0}.

Proposition. Let g(z) = g(z, z_{0}) satisfy the following properties: Dg = λg, g(z) depends only on the value of the distance d(z, z_{0}), (with fixed z_{0} and λ), and g(z_{0}) = 1. Then the function g(z) is defined uniquely by these properties.

Proof: Here again it is simpler to work in the unit square model {|z| < 1} with z_{0} = 0. If z = re^{iϑ}, then the function g can be written in the form g(z) = g(r) with r = |z|. The

Laplace operator Δ can be written in polar coordinate system as Δ = \frac{∂^{2}}{∂r^{2}} + \frac{1}{r^{2}} \frac{∂^{2}}{∂ϑ^{2}} . Hence we can write because of the rotation invariance of g

D^{∗}g = (1 − r^{2})^{2}g′′(r) = λg(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.

How does the function g(z, z_{0}), the unique rotation invariant eigenfunction of D with center z_{0}, with eigenvalue λ and normalization g(z_{0}, z_{0}) = 1, depend on z_{0}?

If Tz_{0} = z′_{0}, then g(Tz, z′_{0}) depends only on d(Tz, z′_{0}) = d(z, T^{−1}z′_{0}) = d(z, z_{0}), g(Tz_{0}, z′_{0}) = g(z′_{0}, z′_{0}) = 1, and with the notation h(u) = g(Tu, z′_{0}), p(u) = g(u, z′_{0})

Dh(z) = DTp(z) = TDp(z) = λTp(z) = λg(Tz, z′_{0}) = λh(z) .

This means that h(z) = g(Tz, z′_{0}) is an eigenfunction of D with the eigenvalue λ, 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(Tz, z′_{0}) = g(z, z_{0}), or in other words g(Tz, Tz_{0}) = g(z, z_{0}) for all motions 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_{z}L_{z}g(z, z_{0}) = L_{z}D_{z}g(z, z_{0}) = λL_{z}g(z, z_{0}) .

But L_{z}g(z, z_{0}), as a function of z and z_{0}, depends only on d(z, z_{0}), since

L_{z}g(Tz, Tz_{0}) = Lp(Tz) = TLp(z) = LTp(z) = Lq(z) = L_{z}g(z, z_{0})

for all movements T with p(u) = g(u, Tz_{0}) and q(u) = g(Tu, Tz_{0}) = g(u, z_{0}). In particular, because of the rotational symmetry L_{z}g(z, z_{0}) = Λg(z_{,}z_{0}) with some Λ = Λ(z_{0}). Moreover, Λ(z_{0}) does not depend on z_{0}, since Λ(z_{0}) = L^{0}g(z_{0}, z_{0}), and this function has the same value for all z.

If f is an arbitrary function such that Df = λf, and we fix some number z_{0} as the center of the averaging operator M, then Df_{0} = λf_{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_{z}g(z, z_{0}) = f_{0}(z_{0})Λg(z, z_{0}) = Λf_{0}(z)

with a Λ independent of the function f and z_{0}. From here

Lf(z_{0}) = MLf(z_{0}) = LMf(z_{0}) = Lf_{0}(z_{0}) = Λf_{0}(z_{0}) = Λf(z_{0}) .

Since z_{0} can be chosen in an arbitrary way, and Λ is independent of z_{0}, Lf = Λf. This is a very important observation which we formulate in the following Theorem.

Theorem. If f is an eigenfunction of the operator D with eigenvalue λ, i.e. Df = λf, and L in an invariant differential or integral operator, then Lf = Λf with a Λ which depends only on the operator L and the eigenvalue λ, but not on the eigenfunction f.

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 λ. We know such an eigenfunction, namely the function y^{s} with the

eigenvalue λ. This eigenvalue λ can be arbitrary with the choice of s = \frac{1}{2} ± \sqrt{\frac{1}{4} + λ}.

Let us calculate for instance the eigenvalue Λ of the integral operator Lf =

∫_{H} k(z, w)f(w)w( dσ) if k(z, w) = k (\frac{|z − w|^{2}}{yv} ).

We know that

∫_{H} k(z, w)v^{s}σ( dv) = Λy^{s}

with w = u + iv. Taking the point z = i (y = 1), we get

Λ = ∫_{H} k(i, w)v^{s}σ( dv) = ∫_{0}^{∞} ∫_{−∞}^{∞} k (\frac{|i − w|^{2}}{v} ) v^{s} \frac{du dv}{v^{2}}

= ∫_{0}^{∞} ∫_{−∞}^{∞} k (\frac{u^{2} + (v − 1)^{2}}{v} ) v^{s−2} du dv .

Put t =^{def} \frac{(v − 1)^{2}}{v} = v + \frac{1}{v} − 2. Then

∫_{−∞}^{∞} k (\frac{u^{2} + (v − 1)^{2}}{v} ) du = ∫_{−∞}^{∞} k (\frac{u^{2}}{v} + t) du = 2 ∫_{0}^{∞} k(τ + t)\frac{\sqrt{v}}{2\sqrt{τ}} dτ

= \sqrt{v} ∫_{0}^{∞} \frac{k(τ + t)}{\sqrt{τ}} dτ =^{def} vg(log v) ,

Λ = ∫_{0}^{∞} vg(log v)v^{s−2} dv = ∫_{−∞}^{∞} g(y)e^{sy} dy =^{def} G(s) ,

where λ = s(s − 1), s = \frac{1}{2} ± \sqrt{\frac{1}{4} + λ}.

Fourth talk:

Discrete subgroups of the group of motions in the hyperbolic plane

Definition. The group Γ ⊂ group of motions in H is called discrete if the set {γz: γ ∈ Γ} has no point of density for a fixed z ∈ H.

Definition.The function f(z), z ∈ H, is automorph with respect to the group Γ, if f(γz) = f(z) for all γ ∈ Γ.

From now on f(z) will always denote an automorph function. We also introduce the notation z_{1} ∼ z_{2} (mod Γ) if there exists a motion γ ∈ Γ, such that γz_{1} = z_{2}.

Put S =^{def} equivalence classes of z ∈ H with respect to ∼. A set D ⊂ H is called a fundamental domain if it contains exactly one point from all equivalence classes. In other words, the sets {γD}, (γ ∈ Γ) produce a disjoint cover of the space H. (In the Euclidean plane the discrete subgroups of the group of motions have the form Γ = Γ_{w_{1},w_{2}} = {γ_{m,n}: z → z + nw_{1} + mw_{2}}, with some non-colinear vectors w_{1} and w_{2}. The automorph functions are the doubly periodic functions, D can be chosen as a lattice parallelogram, S is topologically a torus. In the one-dimensional case D = [0, 1), S is a circle.)

Let us also assume that Γ is free of fixed points. This means that if γ ∈ Γ has a fixed point, (γz_{0} = z_{0}), then γ is the identity. We also assume that Γ has a “compact” fundamental domain D whose closure D̄ (in the compactified complex sphere) is a subset of H.

It is far from trivial that there are such subgroups Γ 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 Γ 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 D is topologically a “sphere with k ears”, then there exists a discrete subgroup Γ with a fundamental domain D which is a regular polygon with 4k edges and angles π/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 Γ. The explanation of these results is not the subject of the present lecture.

Under these conditions S is a compact (closed) Riemann surface, topologically it is a sphere G with “ears”, its genus g > 1. There exists a universal cover H → S. This universal cover together with the metric and measure on H induce a finite metric and measure on S, invariant with respect to Γ.

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(γz) = T_{γ}Lf(z) = LT_{γ}f(z) = Lf(z) for all z ∈ H and γ ∈ Γ, where the operator T_{γ} is defined by the relation T_{γ}f(z) = f(γz). 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}Δ. The operator D is a self-adjoint negative operator in the space L_{2}(S, dσ). Indeed, the Green formula in the domain D gives that the operator D is negative, i.e.

(Df, f) = ∫_{D} Df(z)\overline{f(z)}σ( dz) = ∫∫_{D} y^{2}Δf(z)\overline{f(z)}\frac{dx dy}{y^{2}}

= \underbrace{∫_{∂D} \frac{∂f}{∂n}f̄ds}_{=0} − \underbrace{∫∫_{D}(f_{x}f̄_{x} + f_{y}f̄_{y}) dx dy}_{≥0} ≤ 0 ,

since, as it is proved in the theory of the Riemann surfaces, the boundary ∂D consists of finitely many pairs of curves (ℓ, ℓ′) such that an appropriate γ ∈ Γ maps ℓ to ℓ′ = γℓ, and

\frac{∂f}{∂n}|_{z} = −\frac{∂f}{∂n}|_{γz} |γ′(z)| .

This relation implies that

∫_{ℓ} \frac{∂f}{∂n}f̄ds = − ∫_{ℓ} \frac{∂f}{∂n}|_{γz} |γ′(z)|f̄ds = − ∫_{γℓ} \frac{∂f}{∂n}f̄ds ,

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 ≡ 0, then f is a harmonic function, and f ≡ const. by the maximum principle. It can be proved that the set of functions Df is everywhere dense in the space L_{0}^{2}(D, σ), consisting of the functions f, ∫_{D} f^{2}(z)σ( dz) < ∞ and ∫_{D} f(z)σ( 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) = ∫_{D} G(z, w)f(w)σ( 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 D (with periodic boundary conditions) in the space L_{0}^{2}(D, 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 Δ).

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| + 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}^{∞} of eigenfunctions of the operator D^{−1} with eigenvalue λ_{i} in the space L_{0}^{2}(D, σ). Moreover, λ_{i} → 0, if i → ∞, and only the eigenvalue zero can have infinite multiplicity. But in our case the zero eigenvalue does not appear, since F ≡ 0 ⇒ ΔF ≡ 0. In such a way we get a complete orthonormal system of eigenfunctions f_{i}(z) of D in the space L_{0}^{2}(D, σ), Df_{i} = λ_{i}f_{i} with λ_{i} < 0, and λ_{i} → −∞ as i → ∞. Let us adjust the number λ_{0} = 0

and f_{0}(z) = \frac{1}{\sqrt{σ(D)}} to this system. Then \{f_{i}\}_{i=0}^{∞} is a complete orthonormal system

consisting of eigenfunctions of D in the L^{2}(D, dσ) space.

The trace of an invariant integral operator

For an invariant integral operator L with kernel function k(z, w)

Lf(z) = ∫_{H} k(z, w)f(w)σ( dw) = ∑_{γ∈Γ} ∫_{γD} k(z, w)f(w)σ( dw)

= ∑_{γ∈Γ} ∫_{D} k(z, γw) f(w)σ( dw) = ∫_{D} K(z, w)f(w) σ( dw) ,

where

K(z, w) = ∑_{γ∈Γ} k(z, γw) .

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 D × D, just as in the real line model considered in the first talk. This expansion has the form

K(z, w) = ∑_{i=1}^{∞} Λ_{i}f_{i}(z)\overline{f_{i}(w)} ,

where Lf_{i} = Λ_{i}f_{i}.

The number Λ_{i} can be calculated, as a function of λ_{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 D, then the functions f_{i} disappear from the formulas, and we get

∫_{D} K(z, z)σ( dz) = ∑_{i=0}^{∞} Λ_{i} ∫_{D} |f_{i}(z)|^{2}σ( dz) = ∑_{i=0}^{∞} Λ_{i} ,

that is,

∑_{γ∈Γ} ∫_{D} k(z, γz)σ( dz) = ∫_{D} ∑_{γ∈Γ} k(z, γz)σ( dz) = ∑_{i=0}^{∞} Λ_{i} .  (++)

We want to rewrite the left-hand side of (++) in a more appropriate form. In the real line case the kernel function k(z, γz) (= 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, γ_{0}w) = k(z, γz) hold?

If w = Tz, then

k(w, γ_{0}w) = k(Tz, γ_{0}Tz) = k(z, T^{−1}γ_{0}Tz)

hence k(w, γ_{0}w) = k(z, γz) if γ = T^{−1}γ_{0}T. If γ = γ_{0}, then this condition means that γ_{0} and 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 γ and γ_{0} are conjugate. Therefore, let us sum for different conjugate classes separately. Let us denote the conjugate classes in the form {γ_{0}}.

Let us conjugate by an element be g: γ = g^{−1}γ_{0}g. (This conjugation would be meaningful also for g ∉ Γ, but we shall consider only conjugations with elements from Γ.) Two such conjugates agree, i.e. g_{1}^{−1}γ_{0}g_{1} = g^{−1}γ_{0}g, if and only if g_{1}g^{−1}γ_{0} = γ_{0}g_{1}g^{−1}, which means that g_{1}g^{−1} ∈ [γ_{0}], or g_{1} ∈ [γ_{0}]g, where [γ_{0}] denotes the group of element in Γ exchangeable with γ_{0}. This group [γ_{0}] is called in algebra the centralizator group of γ_{0}. Thus the sum ∑_{γ∈\{γ_{0}\}} can also be written in the form ∑_{\begin{matrix}g \\ g∈Γ/[γ_{0}] \end{matrix}}, where g are fixed

representation elements from the right cosets of the subgroup [γ_{0}].

If γ ∈ {γ_{0}}, then γ = g^{−1}γ_{0}g with some g, and

∫_{D} k(z, γz)σ( dz) = ∫_{D} k(z, g^{−1}γ_{0}gz)σ( dz) = ∫_{D} k(gz, γ_{0}gz)σ( dz)
= ∫_{gD} k(z, γ_{0}z)σ( dz) .

∑_{γ∈\{γ_{0}\}} ∫_{D} k(z, γz)σ( dz) = ∑_{\begin{matrix}g \\ g∈Γ/[γ_{0}] \end{matrix}} ∫_{gD} k(z, γ_{0}z)σ( dz) = ∫_{\begin{matrix}∪gD \\ g∈Γ/[γ_{0}] \end{matrix}} k(z, γ_{0}z)σ( dz) .

If the representation elements g ∈ Γ/[γ_{0}] are multiplied from the left with the elements of [γ_{0}], then we get the elements of Γ, and all elements exactly once. Thus D(γ_{0}) =^{def} ⋃_{g∈Γ/[γ_{0}]} gD is such a set that the sets γD(γ_{0}), γ ∈ [γ_{0}], produce a disjoint cover of H. In

other words, D(γ_{0}) is a fundamental domain of the discrete subgroup [γ_{0}]. We know that k(z, γ_{0}z) is automorph with respect to [γ_{0}], (k(gz, γ_{0}gz) = k(z, g^{−1}γ_{0}gz) = k(z, γ_{0}z) for g ∈ [γ_{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.

Lemma. For all γ_{0} ∈ Γ the relation

∑_{γ∈{γ_{0}}} ∫_{D} k(z, γz)σ( dz) = ∫_{D(γ_{0})} k(z, γ_{0}z)σ( dz)

holds, where {γ_{0}} = {g^{−1}γ_{0}g; g ∈ Γ} denotes the class of elements in Γ which are conjugated to γ_{0}, and D(γ_{0}) is a fundamental domain of the centralizator group [γ_{0}] ⊂ Γ of γ_{0}. The function k(z, γ_{0}z) at the right-hand side of this formula is [γ_{0}] automorph.

Remark: It is possible that {γ_{0}} = {γ_{1}} with some γ_{0} ≠ γ_{1} and even [γ_{0}] ≠ [γ_{1}]. But the right-hand side of the integral in the lemma agrees for such γ_{0} and γ_{1}. Indeed, if γ_{1} = g^{−1}γ_{0}g, then

∫_{D(γ_{1})} k(z, γ_{1}z)σ( dz) = ∫_{gD(γ_{1})} k(z, γ_{0}z)σ( dz) ,

and gD(γ_{1}) is a fundamental domain of [γ_{0}]. (If h_{j}D(γ_{1}), j = 1, …, is a disjoint cover of H, and the transformations h_{j} are the elements of [γ_{1}], then the sets gh_{j}g^{−1}(gD(γ_{1})), j = 1, …, also give a disjoint cover of H, and gh_{j}g^{−1} are the elements of [γ_{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 Γ the centralizators [γ_{0}] have a very special form.

Fifth talk:

The proof of Selberg’s trace formula in the hyperbolic plane

Let 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 Tz̄ = \overline{Tz} either a real number or infinity, then the transformation T is called parabolic. If 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 T is called hyperbolic. Hyperbolic transformations have two different real or (∞) fixed points.

A discrete subgroup Γ of the motions in H cannot contain an elliptic transformation, since the non-unity elements of Γ cannot have a fixed point in H. We also excluded the possibility of γ ∈ Γ for a parabolic transformation γ by the assumption that D̄ is compact. Indeed, for a parabolic transformation d(z, γz) can be taken arbitrary small with an appropriate choice of z ∈ 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 γ is infinity, i.e. γz = 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 Γ cannot contain a parabolic transformation. Otherwise, by mapping a sequence of points with the above properties into the domain D we could choose a sequence z_{n} with z_{n} ∈ D and γ_{n} ∈ Γ in such a way that 0 < d(z_{n}, γ_{n}z_{n}) → 0, as n → ∞. Then, because of the compactness of D̄ we may assume, by considering a subsequence of the sequence z_{n}, that z_{n} → z_{0} ∈ H. Since d(γ_{n}z_{n}, γ_{n}z_{0}) = d(z_{0}, z_{n}) → 0, d(z_{0}, γ_{n}z_{0}) ≤ d(z_{0}, z_{n}) + d(z_{n}, γ_{n}z_{n}) + d(γ_{n}z_{n}, γ_{n}z_{0}) → 0, as n → 0, and γ_{n} ≠ Id. But this contradicts to the fact, that since Γ has no fixed point, and it is discreet, min_{\begin{matrix}γ∈Γ \\ γ≠Id \end{matrix}} d(z_{0}, γz_{0}) > 0. So Γ contains

only hyperbolic transformations.

Let γ be a hyperbolic transformation. If T is such a motion of H (not necessarily in the group Γ) which maps the two fixed points of γ into 0 and ∞, then the two fixed points of T^{−1}γT are 0 and ∞, hence T^{−1}γTz = ρz with some ρ > 0, ρ ≠ 1. Moreover,

we may assume with a possible conjugation with \frac{1}{z} that ρ > 1. This number ρ is uniquely determined by γ (and even by {γ}). We call it the norm of γ, and introduce the notation ρ = N(γ) = N({γ}).

Two motions of H are exchangeable if and only if their fixed points agree. For the motions exchangeable with γ_{0} the same motion 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 ρ > 0. A discrete subgroup of this group is an infinite cyclic group, and we call a generator γ^{∗} of it a primitive element. [γ_{0}] = {γ^{∗}^{m}}_{m=−∞}^{∞}. Put ρ^{∗} = N(γ^{∗}).

∫_{D(γ_{0})}k(z, γ_{0}z)σ( dz) = ∫_{D(γ_{0})}k(T^{−1}z, T^{−1}γ_{0}z)σ( dz) = ∫_{T^{−1}D(γ_{0})} k(z, T^{−1}γ_{0}Tz)σ( dz)

= ∫_{T^{−1}D(γ_{0})} k(z, ρ_{0}z)σ( dz) , ρ_{0} = N(γ_{0}) ,

T^{−1}D(γ_{0}) is a fundamental domain of the group T^{−1}[γ_{0}]T = {ρ^{∗}^{m}z}_{m=−∞}^{∞}, and since k(z, T^{−1}γ_{0}Tz) is automorph for T^{−1}[γ_{0}]T we can choose an arbitrary fundamental domain of T^{−1}[γ_{0}]T to calculate the last integral.

It is clear that the half-ring {1 ≤ |z| < ρ^{∗}} ∩ H is such a domain. The function k(z, ρ_{0}z) is invariant for all transformations Tz = ρz. Hence, it is constant along the rays, and we have to integrate only with respect to the angle ϑ, where z = re^{iϑ}.

I = ∫_{T^{−1}D(γ_{0})} k(z, ρ_{0}z)σ( dz) = ∫_{0}^{π} ∫_{1}^{ρ^{∗}} k(e^{iϑ}, ρ_{0}e^{iϑ})\frac{r dr dϑ}{r^{2} sin^{2} ϑ}

= ∫_{0}^{π} k(e^{iϑ}, ρ_{0}e^{iϑ})\frac{log ρ^{∗}}{sin^{2} ϑ} dϑ = 2 ∫_{0}^{π/2} k (\frac{|e^{iϑ} − ρ_{0}e^{iϑ}|^{2}}{sin ϑρ_{0} sin ϑ} ) \frac{log ρ^{∗}}{sin^{2} ϑ} dϑ

= 2 log ρ^{∗} ∫_{0}^{π/2} k (\frac{(ρ_{0} − 1)^{2}}{ρ_{0} sin^{2} ϑ} ) \frac{dϑ}{sin^{2} ϑ} .

Put t = \frac{(ρ_{0} − 1)^{2}}{ρ_{0}} = ρ_{0} + \frac{1}{ρ_{0}} − 2 and τ = \frac{t}{sin^{2} ϑ}, that is ϑ = arcsin \sqrt{\frac{t}{τ}}. Then with

this notation

I = 2 log ρ^{∗} ∫_{t}^{∞} k(θ)\frac{τ}{t} \frac{1}{\sqrt{1 − \frac{t}{τ}}} \frac{\sqrt{t}}{2τ^{3/2}} dτ

= \frac{log ρ^{∗}}{\sqrt{t}} ∫_{t}^{∞} \frac{k(τ)}{\sqrt{τ − t}} dτ = \frac{log ρ^{∗}}{\sqrt{t}} ∫_{0}^{∞} \frac{k(τ + t)}{\sqrt{τ}} dτ

= \frac{log ρ^{∗}}{\sqrt{t}} \sqrt{ρ_{0}}g(log ρ_{0}) = \frac{log ρ^{∗}}{\frac{ρ_{0}−1}{\sqrt{ρ_{0}}}} \sqrt{ρ_{0}}g(log ρ_{0}) = \frac{log ρ^{∗}}{1 − \frac{1}{ρ_{0}}} g(log ρ_{0})

by the definition of g(·).

Putting these results together

∑_{γ∈\{γ_{0}\}} ∫_{D} k(z, γz)σ( dz) = \frac{log N(γ^{∗})}{1 − \frac{1}{N(γ_{0})}}g(log N(γ_{0})) .

This relation does not hold for the one element conjugate class containing the origin. But

∫_{D} k(z, z) σ( dz) = k(0)σ(D) = \frac{σ(D)}{4π} ∫_{−∞}^{∞} r tanh(πr)G (\frac{1}{2} + ir) dr

with the previous definition of G(·), what can be shown by a long but for us not so interesting calculation.

So the starting trace formula

∑_{γ∈Γ} ∫_{D} K(z, γz)σ( dz) = ∑_{i=0}^{∞} Λ_{i}

has the following form:

\frac{σ(D)}{4π} ∫_{−∞}^{∞} r tanh(πr)G (\frac{1}{2} + ir) dr + ∑_{\{γ_{0}\}} \frac{log N(\{γ^{∗}\})}{1 − \frac{1}{N(\{γ_{0}\})}}g(log N(\{γ_{0}\})) = ∑_{i=0}^{∞} G(s_{i}) .

Here

G(s) = ∫_{−∞}^{∞} g(y)e^{sy} dy ,

(the functions g(·), G(·) and k(·) can be calculated from each others by means of simple formulas),

s_{i} = \frac{1}{2} + \sqrt{\frac{1}{4} + λ_{i}} ,

(one has to consider only one of the square roots), where λ_{0} = 0, λ_{i} < 0, (i = 1, 2, …) are the eigenvalues of D with multiplicity,

σ(D) = the area of a fundamental domain = 4π(g − 1).

(The number g is the genus of S.) In the sum the {γ_{0}} runs through the conjugate classes not containing the identity, {γ^{∗}} is the class of the corresponding primitive element, i.e. γ^{∗} is defined by the relations γ^{∗} ∈ Γ, and γ^{∗}^{m} = γ_{0}, with the largest possible m, N(·) denotes the norm of an element γ ∈ Γ introduced in this talk.

This is the famous Selberg formula in the simplest case.