A DIRICHLET SERIES RELATED TO EIGENVALUES OF THE LAPLACIAN FOR CONGRUENCE SUBGROUPS

David A. Cardon and Xian-Jin Li

Abstract. For any congruence subgroup Γ_{0}(N), an explicit Dirichlet series is given which represents an analytic function of s in the half-plane Re s > 1/2 except for having simple poles at s = 1, 1/2 + \sqrt{1/4 − λ_{j}}, j = 1, 2, ⋯, S, where λ_{j}, j = 1, 2, ⋯, S, are the exceptional eigenvalues of the non-Euclidean Laplacian for the congruence subgroup.

1. Introduction

Let N be a positive integer greater than one. Denote by Γ_{0}(N) the Hecke congruence subgroup of level N. The Laplacian Δ on the upper half-plane H is given by

Δ = −y^{2} (\frac{∂^{2}}{∂x^{2}} + \frac{∂^{2}}{∂y^{2}} ) .

Let D be the fundamental domain of Γ_{0}(N). Eigenfunctions of the discrete spectrum of Δ are nonzero real-analytic solutions of the equation

Δψ = λψ

such that ψ(γz) = ψ(z) for all γ in Γ_{0}(N) and such that

∫_{D} |ψ(z)|^{2}dz < ∞

where dz represents the Poincaré measure of the upper half-plane.

Let 𝔞 be a cusp of Γ_{0}(N). Its stabilizer is denoted by Γ_{𝔞}. An element σ_{𝔞} ∈ P SL(2, ℝ) exists such that σ_{𝔞}∞ = 𝔞 and σ_{𝔞}^{−1}Γ_{𝔞}σ_{𝔞} = Γ_{∞}. Let f

Typeset by AMS-TEX

1991 Mathematics Subject Classification. Primary 11F37, 11F72, 42A16.

Key words and phrases. Hecke congruence subgroups, Hecke operators, Laplace-Beltrami operator, Maass wave forms.

be a Γ_{0}(N)-invariant function. If 𝔞 is a cusp of Γ_{0}(N), then f(σ_{𝔞}z) is Γ_{∞}-invariant, and hence it admits a formal Fourier expansion

f(σ_{𝔞}z) = ∑_{n∈ℤ} c_{𝔞n}(y)e^{2nπix}.

A Γ_{0}(N)-invariant function is said to be a Maass cusp form if it is squareintegrable and is an eigenvector of Δ, such that the Fourier coefficient c_{𝔞0}(y) = 0 for every cusp 𝔞 of Γ_{0}(N). If ψ is a cusp form associated with a positive discrete eigenvalue λ, then it has the Fourier expansion [8]

ψ(σ_{𝔞}z) = \sqrt{y} ∑_{m≠0} ρ_{𝔞}(m)K_{iκ}(2π|m|y)e^{2mπix},

where κ = \sqrt{λ − 1/4} and K_{ν}(y) is given by the formula §6.32, [23]

(1.1)  K_{ν}(y) = \frac{2^{ν}Γ(ν + \frac{1}{2})}{y^{ν} \sqrt{π}} ∫_{0}^{∞} \frac{cos(yt)}{(1 + t^{2})^{ν+ \frac{1}{2}}} dt.

The complex numbers ρ_{𝔞}(m), m(≠ 0) ∈ ℤ, are called the Fourier coefficients of ψ around the cusp 𝔞.

Let h_{d} be the class number of indefinite rational quadratic forms with discriminant d. Define

ϵ_{d} = \frac{v_{0} + u_{0}\sqrt{d}}{2}

where the pair (v_{0}, u_{0}) is the fundamental solution [13] of Pell’s equation v^{2} − du^{2} = 4. Denote by Ω the set of all the positive integers d such that d ≡ 0 or 1 (mod 4) and such that d is not a square of an integer. The Möbius function μ(n) is defined to be one if n = 1, to be (−1)^{k} if n is the product of k distinct primes, and to be zero otherwise.

In this paper we obtain the following theorem, which is a generalization of the main theorem of Conrey and Li [1] in the case when n = 1. The result is related to that of Hejhal [6] in spirit, and some of our computation is implicit in Hejhal [5]. Let λ_{j}, j = 1, 2, ⋯, be an enumeration in nondecreasing order of all positive discrete eigenvalues of the Laplacian for Γ_{0}(N) with an eigenvalue of multiplicity m appearing m times, and let κ_{j} = \sqrt{λ_{j} − 1/4}.

Theorem 1. Let

L_{N}(s) = ∑_{m|N_{m square-free}} ∑_{k|N} k^{1−2s} \frac{μ((m, k))}{(m, k)} ∑_{d∈Ω} ∑_{u} (\frac{d}{m}) \frac{h_{d} ln ϵ_{d}}{(du^{2})^{s}} ∏_{p^{2l}|(d, \frac{N}{k})} p^{l},

where the summation on u is taken over all the positive integers u such that \sqrt{4 + dk^{2}u^{2}} ∈ ℤ and p^{2l} is the greatest even prime p-power dividing (d, N/k). Then L_{N}(s) is analytic for Re s > 1 and can be extended by analytic continuation to the half-plane Re s > 0 except for having a possible pole at s = 1/2 and for having simple poles at s = 1, \frac{1}{2} ± iκ_{j}, j = 1, 2, · · ·.

Discrete eigenvalues λ of the Laplacian for congruence subgroups are said to be exceptional if 0 < λ < 1/4. In particular, by using Theorem 1 we obtain the following result for the case when N is a prime number p.

Corollary 2. Let λ_{j}, j = 1, 2, ⋯, S, be the exceptional eigenvalues of the Laplacian for Γ_{0}(p). If

\begin{aligned}L_{p}(s) = ∑_{u=1}^{∞} \frac{1}{u^{2s}} & \{ ∑_{d∈Ω_{v^{2}−du^{2}=4}} (\frac{v^{2} − 4}{p}) d^{\frac{1}{2} −s} ∑_{k=1}^{d} \frac{1}{k} (\frac{d}{k} ) \\ & + p^{1−2s} ∑_{d∈Ω_{v^{2}−dp^{2}u^{2}=4}} d^{\frac{1}{2} −s} ∑_{k=1}^{d} \frac{1}{k} (\frac{d}{k} )\}, \end{aligned}

then L_{p}(s) represents an analytic function of s in the half-plane Re s > 1/2 except for having simple poles at s = \frac{1}{2} + \sqrt{1/4 − λ_{j}}, j = 1, 2, ⋯, S.

The paper is organized as follows. In section 2, we recall Selberg’s trace formula (cf. Murty [12] and Selberg [15]). Elements of Γ_{0}(N) can be divided into four types, the identity, hyperbolic, elliptic and parabolic elements. By using the Maass-Selberg relation (cf. Theorem 2.3.1 of [9]) and some computations of Kubota [9], we compute contributions of the identity, elliptic, hyperbolic and parabolic elements to the trace formula in section 3. From contributions of the identity, elliptic, and parabolic elements, we obtain analyticity information about a series formed by contributions of hyperbolic elements, and a precise statement is given in Theorem 3.4. In section 4, we compute explicitly the total contribution of hyperbolic elements by using results of Sarnak [13]. The result is stated in Theorem 4.4. Then Theorem 1 follows from Theorem 3.4 and Theorem 4.4. Finally we prove Corollary 2 in section 5 by using Dirichlet’s class number formula [7] and the Pólya-Vinogradov inequality [2].

The authors wish to thank the referee for his/her valuable comments. The second author also wishes to thank Ozlem Imamoglu for her helpful suggestions during preparation of the paper.

2. Trace formulas

Let s be a complex number with Re s > 1. Define

k(t) = (1 + \frac{t}{4})^{−s}

and

k(z, z′) = k (\frac{|z − z′|^{2}}{yy′} ) ,

for z = x+iy and z′ = x′ +iy′ in the upper half-plane. Then k(mz, mz′) = k(z, z′) for every 2 × 2 matrix m of determinant one with real entries. The kernel k(z, z′) is of (a)-(b) type in the sense of Selberg [15], p.60. Let

g(u) = ∫_{w}^{∞} k(t)\frac{dt}{\sqrt{t − w}}

with w = e^{u} + e^{−u} − 2. Write

h(r) = ∫_{−∞}^{∞} g(u)e^{iru}du.

Then

(2.1)  g(u) = \sqrt{w} ∫_{0}^{1}(t + \frac{w}{4} )^{−s}t^{s− \frac{3}{2}} \frac{dt}{\sqrt{1 − t}} = c(1 + \frac{w}{4} )^{\frac{1}{2} −s}

where c = 2\sqrt{π}Γ(s − \frac{1}{2})Γ^{−1}(s). Since

h(r) = c4^{s− \frac{1}{2}} ∫_{0}^{∞}(u + \frac{1}{u} + 2)^{\frac{1}{2} −s}u^{ir−1}du

= c4^{s− \frac{1}{2}} ∫_{1}^{∞}(u + \frac{1}{u} + 2)^{\frac{1}{2} −s} (u^{ir} + u^{−ir}) \frac{du}{u}

= c\frac{4^{s}(s − \frac{1}{2})}{(s − \frac{1}{2})^{2} + r^{2}} + A(r, s),

where A(r, s) is finite for |Im r| ⩽ 1/2 and for Re s > 0, we obtain that

(2.2) lim_{s→1/2+iκ}(s − \frac{1}{2} − iκ)h(r) = \{ \begin{matrix}4^{1/2+iκ}\sqrt{π} \frac{Γ(iκ)}{Γ(1/2+iκ)}, & for r = ±κ; \\ 0, & for r ≠ ±κ. \end{matrix}

Let ν be the number of inequivalent cusps of Γ_{0}(N), and let 𝔞_{1}, 𝔞_{2}, ⋯, 𝔞_{ν} be a complete set of inequivalent cusps of Γ_{0}(N). We choose an element σ_{𝔞_{i}} ∈ P SL(2, ℝ) such that σ_{𝔞_{i}}∞ = 𝔞_{i} and σ_{𝔞_{i}}^{−1}Γ_{𝔞_{i}}σ_{𝔞_{i}} = Γ_{∞} for i = 1, 2, ⋯, ν. The Eisenstein series E_{i}(z, s) for the cusp 𝔞_{i} is defined by

E_{i}(z, s) = ∑_{γ∈Γ_{𝔞_{i}} \Γ_{0}(N)} (Im(σ_{𝔞_{i}}^{−1}γz))^{s}

for Re s > 1 when z is in the upper half-plane. Define

K(z, z′) = ∑_{T ∈Γ_{0}(N)} k(z, T z′)

and

H(z, z′) = ∑_{i=1}^{ν} \frac{1}{4π} ∫_{−∞}^{∞} h(r)E_{i}(z, \frac{1}{2} + ir)E_{i}(z′, \frac{1}{2} − ir)dr.

It follows from (2.14) of [15], Theorem 5.3.3 of [9], and the spectral decomposition formula (5.3.12) of [9] that

(2.3)  h(−\frac{i}{2}) + ∑_{j=1}^{∞} h(κ_{j}) = ∫_{D}\{K(z, z) − H(z, z)\}dz

for Re s > 1.

3. Evaluation of components of the trace

For every element T of Γ_{0}(N), we denote by Γ_{T} the set of all the elements of Γ_{0}(N) commuting with T. Put D_{T} = Γ_{T} \H. Elements of Γ_{0}(N) can be divided into four types, of which the first consists of the identity element, while the others are respectively the hyperbolic, the elliptic and the parabolic elements. If T is not a parabolic element, put

c(T) = ∫_{D_{T}} k(z, T z)dz.

Every cusp of Γ_{0}(N) is equivalent to one of the following inequivalent cusps

(3.1)  \frac{u}{w} with u, w > 0, (u, w) = 1, w|N.

Two such cusps u/w and u_{1}/w_{1} are Γ_{0}(N)-equivalent if and only if w = w_{1} and u ≡ u_{1} modulo (w, N/w). Let 𝔞 = u/w be given as in (3.1). By (2.2) and (2.3) of [3], we have

Γ_{𝔞} = \{( \begin{matrix}1 + cu/w & −cu^{2}/w^{2} \\ c & 1 − cu/w \end{matrix} ) : c ≡ 0 (mod [w^{2}, N])\}

and

σ_{𝔞}∞ = 𝔞 and σ_{𝔞}^{−1}Γ_{𝔞}σ_{𝔞} = Γ_{∞}

where

(3.2)  σ_{𝔞} = ( \begin{matrix}𝔞\sqrt{[w^{2}, N]} & 0 \\ \sqrt{[w^{2}, N]} & 1/𝔞\sqrt{[w^{2}, N]} \end{matrix} ) .

3.1. The identity component.

We have

c(I) = ∫_{Γ_{0}(N)\H} dz.

3.2. Elliptic components.

There are only a finite number of elliptic conjugacy classes.

Lemma 3.1. Let R be an elliptic element of Γ_{0}(N). Then

c(R) = \frac{π}{2m sin θ} ∫_{0}^{∞} \frac{k(t)}{\sqrt{t + 4 sin^{2} θ}} dt,

where m is the order of a primitive element of Γ_{R} and where θ is defined by the formula trace(R) = 2 cos θ.

Proof. Since R is an elliptic element of Γ_{0}(N), an element σ ∈ P SL(2, ℝ) exists such that

σRσ^{−1} = ( \begin{matrix}cos θ & − sin θ \\ sin θ & cos θ \end{matrix} ) = R̃

for some real number 0 < θ < π. Denote by (σΓ_{0}(N)σ^{−1})R̃ the set of all the elements of σΓ_{0}(N)σ^{−1} which commute with R̃. We have

c(R) = ∫_{D_{R}} k(z, R̃z)dz

where DR̃ = (σΓ_{0}(N)σ^{−1})R̃\H.

Let γ = ( \begin{matrix}α & β \\ γ & δ \end{matrix} ) be an element of Γ_{0}(N) which has the same fixed points

as R = ( \begin{matrix}a & b \\ c & d \end{matrix} ). Then (α − δ)c = γ(a − d) and βc = γb. It follows that γ commutes with R. By Proposition 1.16 of [17], a primitive elliptic element γ_{0} of Γ_{0}(N) exists such that (ηΓη^{−1})R̃ is generated by ηγ_{0}η^{−1}. Since ηγ_{0}η^{−1} commutes with R̃, it is of the form

( \begin{matrix}cos θ_{0} & − sin θ_{0} \\ sin θ_{0} & cos θ_{0} \end{matrix} )

for some real number θ_{0}. By Proposition 1.16 of [17], θ_{0} = π/m for some positive integer m. It follows from the argument of [9], p.99 that

c(R) = \frac{1}{m} ∫_{0}^{∞} ∫_{−∞}^{∞} k (\frac{|z^{2} + 1|^{2}}{y^{2}} sin^{2} θ) dz.

By the argument of [9], p.100 we have

c(R) = \frac{π}{2m sin θ} ∫_{0}^{∞} \frac{k(t)}{\sqrt{t + 4 sin^{2} θ}} dt. □

3.3. Hyperbolic components.

Let P be a hyperbolic element of Γ_{0}(N). Then an element ρ ∈ SL_{2}(ℝ) exists such that

ρP ρ^{−1} = ( \begin{matrix}λ_{P} & 0 \\ 0 & λ_{P}^{−1} \end{matrix} ) = P̃

with λ_{P} > 1. The number λ_{P}^{2} is called the norm of P, and is denoted by NP. It follows that

c(P) = ∫_{D_{P}} k(z, NP z)dz

where DP̃ = (ρΓ_{0}(N)ρ^{−1})P̃\H. Let P_{0} be a primitive hyperbolic element of SL_{2}(ℤ), which generates the group of all elements of SL_{2}(ℤ) commuting with P. Then there exists a hyperbolic element P_{1} ∈ Γ_{0}(N), which generates Γ_{P}, such that P_{1} is the smallest positive integer power of P_{0} among all the generators of Γ_{P} in Γ_{0}(N). Detail discussions about the “primitive” hyperbolic element P_{1} are given in the proof of Lemma 4.1.

Theorem 3.2. Let P be a hyperbolic element of Γ_{0}(N). If P_{1} is a “primitive” hyperbolic element of Γ_{0}(N) which generates Γ_{P}, then

c(P) = \frac{ln NP_{1}}{(NP)^{1/2} − (NP)^{−1/2}} g(ln NP).

Proof. An argument similar to that made for elliptic elements shows that every element of Γ_{0}(N), which has the same fixed points as P, commutes with P. Because ρP_{1}ρ^{−1} commutes with P̃, it is of the form

( \begin{matrix}λ_{P_{1}} & 0 \\ 0 & λ_{P_{1}}^{−1} \end{matrix} )

for some real number λ_{P_{1}} > 1. Then

c(P) = ∫_{1}^{NP_{1}} \frac{dy}{y^{2}} ∫_{−∞}^{∞} k (\frac{(NP − 1)^{2}}{NP} \frac{|z|^{2}}{y^{2}} ) dx.

The stated identity follows. □

3.4. Parabolic components.

Let S be a parabolic element of Γ_{0}(N). An argument similar to that made for the elliptic elements shows that every element of Γ_{0}(N), which has the same fixed point as S, commutes with S. If 𝔞 = u/w is the fixed point of S, then Γ_{S} = Γ_{𝔞}, and hence we have σ_{𝔞}^{−1}Γ_{S}σ_{𝔞} = Γ_{∞} where σ_{𝔞} is given as in (3.2). It follows that

σ_{𝔞}^{−1}Sσ_{𝔞} = ( \begin{matrix}1 & b \\ 0 & 1 \end{matrix} )

for some integer b. Furthermore, elements of the form

S = σ_{𝔞} ( \begin{matrix}1 & b \\ 0 & 1 \end{matrix} ) σ_{𝔞}^{−1}, 0 ≠ b ∈ ℤ

constitute a complete set of representatives for the conjugacy classes of parabolic elements of Γ_{0}(N) having 𝔞 as its fixed point. For a large positive number Y, let

D_{Y} = {z ∈ D : Imσ_{𝔞_{i}}^{−1}z < Y, i = 1, 2, ⋯, h}.

Then we have

∑_{{S}} ∫_{D_{Y}} k(z, Sz)dz = ∫_{0}^{Y} ∫_{0}^{1} ∑_{0≠b∈ℤ} k(z, z + b)dz + o(1),

where the summation on {S} is taken over all parabolic classes represented by parabolic elements whose fixed point is 𝔞 = u/w and o(1) tends to zero as Y → ∞.

THE LAPLACIAN FOR CONGRUENCE SUBGROUPS 9

Theorem 3.3. Let

c(∞)_{Y} = ν ∫_{0}^{Y} ∫_{0}^{1} ∑_{0≠b∈ℤ} k(z, z + b)dz − ∫_{D_{Y}} H(z, z)dz.

Then

\begin{aligned}c(∞)_{Y} & = −νg(0) ln 2 + ∑_{i,j=1}^{ν} \frac{1}{4π} ∫_{−∞}^{∞} h(r)Re φ′_{ij}(\frac{1}{2} + ir)φ_{ij}(\frac{1}{2} − ir)dr \\ & − \frac{ν}{2π} ∫_{−∞}^{∞} h(r)\frac{Γ′}{Γ} (1 + ir)dr + \frac{1}{4}h(0)\{ν − ∑_{i=1}^{ν} φ_{ii}(\frac{1}{2})\} + o(1) \end{aligned}

where o(1) tends to zero as Y → ∞. Proof. By the argument of [9], pp.102–106 we have

\begin{matrix}∫_{0}^{Y} ∫_{0}^{1} ∑_{0≠b∈ℤ} k(z, z + b)dz \\ = g(0) ln Y − \frac{1}{2π} ∫_{−∞}^{∞} h(r)\frac{Γ′}{Γ} (1 + ir)dr − g(0) ln 2 + \frac{1}{4}h(0) + o(1). \end{matrix}

Let

φ_{ij,m}(s) = ∑_{c} \frac{1}{|c|^{2s}} (∑_{d} e(md/c))

where summations are taken over c > 0, d modulo c with ( \begin{matrix}∗ & ∗ \\ c & d \end{matrix} ) belonging to σ_{𝔞_{i}}^{−1}Γ_{0}(N)σ_{𝔞_{j}}. Then we have

\begin{aligned}E_{i}(σ_{𝔞_{j}}z, s) & = δ_{ij}y^{s} + φ_{ij}(s)y^{1−s} \\ & + \frac{2π^{s}\sqrt{y}}{Γ(s)} ∑_{m≠0} |m|^{s− \frac{1}{2}} K_{s− \frac{1}{2}} (2|m|πy)φ_{ij,m}(s)e(mx) \end{aligned}

where

φ_{ij}(s) = \frac{\sqrt{π}Γ(s − \frac{1}{2})}{Γ(s)}φ_{ij,0}(s).

By (1.1) and the Maass-Selberg relation (cf. Theorem 2.3.1 of [9]), we obtain that

(3.3) \begin{aligned}∫_{D_{Y}} E_{i}(z, s)E_{i}(z, s̄)dz & = \frac{Y^{s+s̄−1} − ∑_{j=1}^{ν} φ_{ij}(s)φ_{ij}(s̄)Y^{1−s−s̄}}{s + s̄ − 1} \\ & + \frac{φ_{ii}(s̄)Y^{s−s̄} − φ_{ii}(s)Y^{s̄−s}}{s − s̄} + o(1) \end{aligned}

for nonreal s with Re s > 1/2, where o(1) tends to zero as Y → ∞. By partial integration, we obtain

(3.4)  h(r) = \frac{1}{r^{4}} ∫_{0}^{∞} g^{(4)}(ln u)u^{ir−1}du

for nonzero r. It follows that

(3.5)  \begin{matrix}lim_{S→ \frac{1}{2}^{+}} ∫_{−∞}^{∞} h(r)\{∫_{D_{Y}} E_{i}(z, S + ir)E_{i}(z, S − ir)dz\}dr \\ = 4πg(0) ln Y − ∑_{j=1}^{ν} ∫_{−∞}^{∞} h(r)Re φ′_{ij}(\frac{1}{2} + ir)φ_{ij}(\frac{1}{2} − ir)dr \\ + ∫_{−∞}^{∞} h(r)\frac{φ_{ii}(\frac{1}{2} − ir)Y^{2ir}}{ir} dr + o(1). \end{matrix}

By the Riemann-Lebesgue theorem (cf. §1.8 of [21]), we have

lim_{Y →∞} ∫_{−∞}^{∞} h(r)\frac{φ_{ii}(\frac{1}{2} − ir)Y^{2ir}}{ir} dr

= lim_{Y →∞} ∫_{−∞}^{∞} h(r)Re (φ_{ii}(\frac{1}{2} − ir)) \frac{sin(2r ln Y)}{r}dr = πh(0)φ_{ii}(\frac{1}{2}).

Then it follows from (3.5) that

\begin{aligned}∫_{D_{Y}} H(z, z)dz & = ∑_{i=1}^{ν}\{g(0) ln Y + \frac{1}{4}h(0)φ_{ii}(\frac{1}{2}) \\ & − ∑_{j=1}^{ν} \frac{1}{4π} ∫_{−∞}^{∞} h(r)Re φ′_{ij}(\frac{1}{2} + ir)φ_{ij}(\frac{1}{2} − ir)dr\} + o(1). \end{aligned}

The stated identity then follows. □

It follows Theorem 3.3 that (3.6)

\begin{matrix}lim_{Y →∞} c(∞)_{Y} = −νg(0) ln 2 + \frac{1}{4}h(0) (ν − ∑_{i=1}^{ν} φ_{ii}(\frac{1}{2})) \\ − \frac{ν}{2π} ∫_{−∞}^{∞} h(r)\frac{Γ′}{Γ} (1 + ir)dr + ∑_{i,j=1}^{ν} \frac{1}{4π} ∫_{−∞}^{∞} h(r)φ′_{ij}(\frac{1}{2} + ir)φ_{ij}(\frac{1}{2} − ir)dr. \end{matrix}

Denote by c(∞) the right side of the identity (3.6). We conclude that the trace formula (2.3) can be written as

(3.7)  h(−\frac{i}{2}) + ∑_{j=1}^{∞} h(κ^{j}) = c(I) + ∑_{\{R\}} c(R) + ∑_{\{P \}} c(P) + c(∞)

for Re s > 1, where the summations on the right side of the identity are taken over conjugacy classes.

Theorem 3.4. Let c(P) = ∫_{Γ_{P} \H} k(z, P z)dz for hyperbolic elements P ∈ Γ_{0}(N). Then the series

∑_{{P }} c(P)

represents an analytic function in the half-plane Res > 0 except for having a possible pole at s = 1/2 and for having simple poles at s = 1, \frac{1}{2} ± iκ_{j}, j = 1, 2, · · ·.

Proof. We have

g^{(4)}(log u) = A(s)u^{\frac{1}{2} −s} + O_{s}(u^{− \frac{1}{2}} ),

where A(s) is an analytic function of s for Re s > 0 and where O_{s}(u^{− \frac{1}{2}} ) means that, for every complex number s with Re s > 0, there exists a finite constant B(s) depending only on s such that

|O_{s}(u^{− \frac{1}{2}} )| ⩽ B(s)u^{− \frac{1}{2}} .

Moreover, for every fixed value of u, the term O_{s}(u^{− \frac{1}{2}} ) also represents an analytic function of s for Re s > 0. Since

h(r) = \frac{1}{r^{4}} ∫_{0}^{∞} g^{(4)}(ln u)u^{ir−1}du

for nonzero r, we have

h(r) = \frac{1}{r^{4}} ∫_{1}^{∞} g^{(4)}(ln u)(u^{ir} + u^{−ir})\frac{du}{u}

= \frac{A(s)}{r^{4}} ∫_{1}^{∞} u^{− \frac{1}{2} −s}(u^{ir} + u^{−ir})du + O_{s} (\frac{1}{r^{4}} ∫_{1}^{∞} u^{−1−ϵ}du)

= \frac{A(s)}{r^{4}} (\frac{1}{s − \frac{1}{2} − ir} + \frac{1}{s − \frac{1}{2} + ir} ) + O_{s}(r^{−4})

for Re s > 1 and for nonzero r with |Im r| < \frac{1}{2} −ϵ. By analytic continuation, we obtain that

(3.8)  h(r) = \frac{2A(s)(s − \frac{1}{2})}{r^{4}[(s − \frac{1}{2})^{2} + r^{2}]} + O_{s}(r^{−4})

for Re s > 0 and for nonzero r with |Imr| < \frac{1}{2} − ϵ. It follows from results of [22] and (2.2) that the left side of (3.7) is an analytic function of s for Re s > 0 except for having simple poles at s = 1, \frac{1}{2} ± iκ_{j}, j = 1, 2, · · ·. Then the right side of (3.7) can be interpreted as an analytic function of s in the same region by analytic continuation.

Since k(t) = (1 + t/4)^{−s}, by Lemma 3.1 we have that c(R) is analytic for Re s > 0 except for a simple pole at s = 1/2. There are only a finite number of elliptic conjugacy classes {R}. The term c(I) is a constant.

Since g(0) = 2\sqrt{π}Γ(s − \frac{1}{2})Γ(s)^{−1},

h(0) = 2\sqrt{π}4^{s} \frac{Γ(s − \frac{1}{2})}{Γ(s)} ∫_{1}^{∞}(u + \frac{1}{u} + 2)^{\frac{1}{2} −s} \frac{du}{u}

and

g(ln \frac{a}{d}) = 2\sqrt{π}4^{s− \frac{1}{2}} \frac{Γ(s − \frac{1}{2})}{Γ(s)} (\frac{a}{d} + \frac{d}{a} + 2)^{\frac{1}{2} −s} ,

the sum of first two terms on the right side of the identity (3.6) is analytic for Re s > 0 except for a pole at s = 1/2.

By Stirling’s formula the identity

(3.9)  \frac{Γ′(z)}{Γ(z)} = ln z + O(1)

holds uniformly when | arg z| ≤ π − δ for a small positive number δ. It follows from (3.8) and (3.9) that the third term on the right side of the identity (3.6) is analytic for Re s > 0 except for a possible pole at s = 1/2.

By Theorem 4.4.1 of Kubota [9], each Eisenstein series E_{i}(z, s) has a meromorphic continuation to the whole s-plane, and the identity

∑_{j=1}^{ν} φ_{ij} (\frac{1}{2} + ir) φ_{ij} (\frac{1}{2} − ir) = 1

hold for all real r and for i = 1, 2, ⋯, ν. It follows that functions φ_{ij}(s), i, j = 1, 2, ⋯, ν, are analytic on the line Re s = 1/2. Let Y be a fixed

THE LAPLACIAN FOR CONGRUENCE SUBGROUPS 13

large positive number. By (3.3) we have

(3.10)  \begin{matrix}Re ∑_{j=1}^{ν} φ′_{ij} (\frac{1}{2} + ir) φ_{ij} (\frac{1}{2} − ir) \\ = 2 ln Y + \frac{φ_{ii}(1/2 − ir)Y^{2ir} − φ_{ii}(1/2 + ir)Y^{−2ir}}{2ir} \\ − ∫_{D_{Y}} E_{i} (z, \frac{1}{2} + ir) E_{i} (z, \frac{1}{2} − ir) dz + o(1). \end{matrix}

Let 𝔞_{i} = u_{i}/w_{i} be a cusp given in (3.1), and let η = 1/2 + ir. By the argument following Lemma 3.6 of [3], we have

(3.11) \begin{aligned}E_{i}(z, η) & = δ_{𝔞_{i}∞}y^{η} + \frac{\sqrt{π}Γ(η − \frac{1}{2})}{Γ(η)}φ_{𝔞_{i}∞,0}(η)y^{1−η} \\ & + \frac{2π^{η}\sqrt{y}}{Γ(η)} ∑_{m≠0} |m|^{η− \frac{1}{2}} K_{η− \frac{1}{2}} (2|m|πy)φ_{𝔞_{i}∞,m}(η)e(mx) \end{aligned}

where

\begin{matrix}φ_{𝔞_{i}∞,0}(η) \\ = \frac{φ(w_{i})}{φ((w_{i}, N/w_{i}))} (\frac{(w_{i}, N/w_{i})}{w_{i}N})^{η} ∏_{p|N} \frac{1}{1 − p^{−2η}} ∏_{p| \frac{N}{w_{i}}} (1 − p^{1−2η})\frac{ζ2η − 1}{ζ(2η)} \end{matrix}

and

φ_{𝔞_{i}∞,m}(η) = (\frac{(w_{i}, N/w_{i})}{w_{i}N})^{η} ∑_{(c,N/w_{i})=1} \frac{1}{c^{2η}} ∑_{d mod(cw_{i}), (d,cw_{i})=1_{cd≡u_{i} mod(w_{i},N/w_{i})}} e (−\frac{md}{cw_{i}} ) .

It follows from the functional identity of the Riemann zeta-function that

\begin{matrix}\frac{Γ(η − \frac{1}{2})}{Γ(η)}φ_{𝔞_{i}∞,0}(η) = & π^{2η− \frac{3}{2}} \frac{Γ(1 − η)}{Γ(η)} \frac{ζ(2 − 2η)}{ζ(2η)} \frac{φ(w_{i})}{φ((w_{i}, N/w_{i}))} \\ × (\frac{(w_{i}, N/w_{i})}{w_{i}N})^{η} ∏_{p|N} \frac{1}{1 − p^{−2η}} ∏_{p| \frac{N}{w_{i}}} (1 − p^{1−2η}). \end{matrix}

By using Stirling’s formula |Γ(σ + it)| ∼ \sqrt{2π}e^{−π|t|/2}|t|^{σ−1/2} for any fixed real value of σ as t → ∞, we obtain that

(3.12)  \frac{Γ(η − \frac{1}{2})}{Γ(η)}φ_{𝔞_{i}∞,0}(η)y^{1−η} ≪ \sqrt{y} ln^{2}(|r| + 1)).

By (1.1) and by partial integration, we find that (3.13)

\frac{K_{η−1/2}(2|m|πy)}{Γ(η)} = \frac{2^{η−1/2}}{(2|m|πy)^{η−1/2}\sqrt{π}} ∫_{0}^{∞} \frac{cos(2|m|πyt)}{(1 + t^{2})^{η}}dt ≪ \frac{1 + |r|^{3}}{|m|^{3}y^{3}} .

Then it follows from (3.8), (3.10)-(3.13), Lemma 4.7 of [3] together with the proof of Theorem 1 in [10], and Cauchy’s inequality that the fourth term on the right side of the identity (3.6) is analytic for Re s > 0 except for a possible pole at s = 1/2.

Also, it follows from (3.4) and Lemma 8.2 of Strömbergsson [20] that the fourth term on the right side of the identity (3.6) is analytic for Re s > 0 except for a possible pole at s = 1/2.

Therefore, by (3.7) we have proved that the series

∑_{{P }} c(P)

represents an analytic function of s in the half-plane Res > 0 except for having a possible pole at s = 1/2 and for having simple poles at s = 1, \frac{1}{2} ± iκ_{j}, j = 1, 2, · · ·. □

4. Proof of the Main Theorem

A quadratic form ax^{2} + bxy + cy^{2}, which is denoted by [a, b, c], is said to be primitive if (a, b, c) = 1 and b^{2} − 4ac = d ∈ Ω. Two quadratic forms [a, b, c] and [a′, b′, c′] are equivalent if an element γ ∈ SL_{2}(ℤ) exists such that

( \begin{matrix}a′ & b′/2 \\ b′/2 & c′ \end{matrix} ) = γ^{t} ( \begin{matrix}a & b/2 \\ b/2 & c \end{matrix} ) γ,

where γ^{t} is the transpose of γ. This relation partitions quadratic forms into equivalence classes, and two such forms from the same class have the same discriminant. The number of classes h_{d} of primitive indefinite quadratic forms of a given discriminant d is finite, and is called the class number of indefinite quadratic forms. Two quadratic forms [a, b, c] and [a′, b′, c′] are Γ_{0}(N)-conjugate if an element γ ∈ Γ_{0}(N) exists such that

( \begin{matrix}a′ & b′/2 \\ b′/2 & c′ \end{matrix} ) = γ^{t} ( \begin{matrix}a & b/2 \\ b/2 & c \end{matrix} ) γ.

Lemma 4.1. Let P be a hyperbolic element of Γ_{0}(N). Then there exists a primitive indefinite quadratic form [a, b, c] of discriminant d such that

P = ( \begin{matrix}\frac{v−bNu(a,N)^{−1}}{2} & −cNu(a, N)^{−1} \\ aNu(a, N)^{−1} & \frac{v+bNu(a,N)^{−1}}{2} \end{matrix} )

with

v^{2} − {dN^{2}(a, N)^{−2}}u^{2} = 4.

If λ_{P} is an eigenvalue of P, then

λ_{P} − \frac{1}{λ_{P}} = ±\frac{Nu}{(a, N)}\sqrt{d}.

Let (v_{1}, u_{1}) with v_{1}, u_{1} > 0 be the fundamental solution of Pell’s equation v^{2} − d_{1}u^{2} = 4, where d_{1} = dN^{2}/(a, N)^{2}. Then Γ_{P} is generated by the hyperbolic element

P_{1} = ( \begin{matrix}\frac{1}{2} (v_{1} − b\frac{Nu_{1}}{(a,N)}) & −c\frac{Nu_{1}}{(a,N)} \\ a\frac{Nu_{1}}{(a,N)} & \frac{1}{2} (v_{1} + b\frac{Nu_{1}}{(a,N)}) \end{matrix} )

of Γ_{0}(N). Let

P_{0} = ( \begin{matrix}\frac{v_{0}−bu_{0}}{2} & −cu_{0} \\ au_{0} & \frac{v_{0}+bu_{0}}{2} \end{matrix} ) ,

where the pair (v_{0}, u_{0}) is the fundamental solution of Pell’s equation v^{2} − du^{2} = 4. If P is Γ_{0}(N)-conjugate to a hyperbolic element P′ ∈ Γ_{0}(N), then P_{0} is Γ_{0}(N)-conjugate to P′_{0}, where P′_{0} is associated with P′ similarly as P_{0} is associated with P.

Proof. Let

P = ( \begin{matrix}A & B \\ C & D \end{matrix} )

be a hyperbolic element of Γ_{0}(N). By Lemma 3.3 of [1], fixed points r_{1}, r_{2} of P are not rational numbers, and they satisfy the equation Cr^{2} + (D − A)r − B = 0. This implies that Γ_{P} is the subgroup of elements in Γ_{0}(N) having r_{1}, r_{2} as fixed points. Let a = C/μ, b = (D − A)/μ and c = −B/μ, where μ = (C, D −A, −B). Then [a, b, c] is a primitive quadratic form with r_{1}, r_{2} being the roots of the equation ar^{2} + br + c = 0. By Sarnak [13], the subgroup of elements in SL_{2}(ℤ) having r_{1}, r_{2} as fixed points consists of matrices of the form

( \begin{matrix}\frac{v−bu}{2} & −cu \\ au & \frac{v+bu}{2} \end{matrix} )

with v^{2} − du^{2} = 4 where d = b^{2} − 4ac, and it is generated by the primitive hyperbolic element

P_{0} = ( \begin{matrix}\frac{v_{0}−bu_{0}}{2} & −cu_{0} \\ au_{0} & \frac{v_{0}+bu_{0}}{2} \end{matrix} )

where the pair (v_{0}, u_{0}) is the fundamental solution of Pell’s equation v^{2} − du^{2} = 4.

Since P and P_{0} have the same fixed points, we have A = D − bC/a and B = −cC/a. Since P belongs to Γ_{0}(N) and AD − BC = 1, C satisfies

(4.2)  \{ \begin{matrix}aD^{2} − bDC + cC^{2} = a \\ a|C, N|C. \end{matrix}

Let λ_{P} be an eigenvalue of P. Then it is a solution of the equation λ^{2} − (A + D)λ + 1 = 0. By using A = D − bC/a and B = −cC/a, we obtain that

(4.3)  λ_{P} − \frac{1}{λ_{P}} = ±\frac{C}{a} \sqrt{d}

and

(4.4)  λ_{P} + \frac{1}{λ_{P}} = 2D − \frac{b}{a}C.

Conversely, let a pair (C, D) be a solution of the equation (4.2). Define A = D − bC/a and B = −cC/a. Then the matrix

P = ( \begin{matrix}A & B \\ C & D \end{matrix} )

is a hyperbolic element of Γ_{0}(N) having the same fixed points as P_{0}, and its eigenvalues satisfy (4.3) and (4.4).

Next, let v = 2D − b\frac{C}{a} and u = \frac{C}{a}. Then the equation (4.2) becomes v^{2} − du^{2} = 4 with N|au. Since N|au, this equation can be written as

v^{2} − \frac{dN^{2}}{(a, N)^{2}} u^{2} = 4.

We also have

(4.5)  ( \begin{matrix}A & B \\ C & D \end{matrix} ) = ( \begin{matrix}\frac{v−bNu(a,N)^{−1}}{2} & −cNu(a, N)^{−1} \\ aNu(a, N)^{−1} & \frac{v+bNu(a,N)^{−1}}{2} \end{matrix} ) .

Since Γ_{P} is cyclic, a solution v_{1}, u′_{1} > 0 of Pell’s equation v^{2} − du^{2} = 4 exists such that Γ_{P} is generated by

P_{1} = ( \begin{matrix}\frac{v_{1}−bu′_{1}}{2} & −cu′_{1} \\ au′_{1} & \frac{v_{1}+bu′_{1}}{2} \end{matrix} )

and such that P_{1} is the smallest positive integer power of P_{0} among all powers of P_{0} belonging to Γ_{0}(N). Note that the eigenvalues of P_{1} are

\frac{v_{1} ± \sqrt{d}u′_{1}}{2} .

Since Γ_{P} is generated by P_{1}, (v_{1}, u′_{1}) is the minimal solution of the equation v^{2} − du^{2} = 4 with N|au′_{1} in the sense that (v_{1} + \sqrt{d}u′_{1})/2 is of the smallest value among all such solutions. Since N|au′_{1}, we have \frac{N}{(a,N)}|u′_{1}. Write

u′_{1} = \frac{Nu_{1}}{(a, N)}.

Then the pair (v_{1}, u_{1}) with v_{1}, u_{1} > 0 must be the fundamental solution of the Pell equation v^{2} − d_{1}u^{2} = 4, where d_{1} = dN^{2}(a, N)^{−2}.

Assume that P is Γ_{0}(N)-conjugate to P′. Let P′_{0} be the primitive hyperbolic element of SL_{2}(ℤ) corresponding to [a′, b′, c′] of discriminant d′, which is associated with P′ similarly as P_{0} is associated with P. Note that eigenvalues of P and P′, which are greater than one, are

\frac{v + \frac{Nu}{(a,N)} \sqrt{d}}{2}

and

\frac{v′ + \frac{Nu′}{(a′,N)} \sqrt{d′}}{2} ,

respectively. Since P and P′ are conjugate, the two eigenvalues must be equal, and hence we must have v = v′. If P′ = γ^{−1}P γ for some element γ ∈ Γ_{0}(N), then by using the expression (4.5) for P and P′ we find that

\frac{Nu}{(a, N)}γ^{t} ( \begin{matrix}a & b/2 \\ b/2 & c \end{matrix} ) γ = \frac{Nu′}{(a′, N)} ( \begin{matrix}a′ & b′/2 \\ b′/2 & c′ \end{matrix} ) .

Since [a, b, c] and [a′, b′, c′] are primitive, we have

\frac{Nu}{(a, N)} = \frac{Nu′}{(a′, N)},

and hence we must have d = d′. Thus, [a, b, c] and [a′, b′, c′] are two Γ_{0}(N)- conjugate primitive indefinite quadratic forms. Since two forms [a, b, c] and [a′, b′, c′] of the same discriminant are equivalent in Γ_{0}(N) if and only if an element γ ∈ Γ_{0}(N) exists such that γ^{−1}P_{0}γ = P′_{0}, P_{0} is Γ_{0}(N)-conjugate to P′_{0}.

This completes the proof of the lemma.

Corollary. If [a, b, c] and [a′, b′, c′] are two Γ_{0}(N)-conjugate primitive indefinite quadratic forms, then we have (a, N) = (a′, N).

Proof. Assume that an element γ ∈ Γ_{0}(N) exists such that

( \begin{matrix}a & b/2 \\ b/2 & c \end{matrix} ) γ = ( \begin{matrix}a′ & b′/2 \\ b′/2 & c′ \end{matrix} ) .

Then we have

a′ = aα^{2} + bαγ + cγ^{2}.

Since N|γ, this identity implies that (a, N)|(a′, N). Similarly, we can obtain that (a′, N)|(a, N), and therefore we have (a, N) = (a′, N). □

We recall that two quadratic forms [a, b, c] and [a′, b′, c′] are equivalent in Γ_{0}(N) if an element γ ∈ Γ_{0}(N) exists such that

( \begin{matrix}a′ & b′/2 \\ b′/2 & c′ \end{matrix} ) = γ^{t} ( \begin{matrix}a & b/2 \\ b/2 & c \end{matrix} ) γ.

This relation partitions the quadratic forms into equivalence classes, and two such forms from the same class have the same discriminant. The number of such classes of a given discriminant d is finite, and is denoted by H_{d}.

Lemma 4.2. Let [a_{j}, b_{j}, c_{j}], j = 1, 2, ⋯, H_{d}, be a set of representatives for classes of primitive indefinite quadratic forms of discriminant d, which are not equivalent under Γ_{0}(N). Then we have

∑_{{P }: hyperbolic P ∈Γ_{0}(N)} c(P)

= 4\sqrt{π} \frac{Γ(s − \frac{1}{2})}{NΓ(s)} ∑_{d∈Ω} ∑_{j=1}^{H_{d}} ∑_{u} \frac{(a_{j}, N) ln ϵ_{d_{1}}}{u\sqrt{d}} (1 + \frac{d(Nu)^{2}}{4(a_{j}, N)^{2}} )^{\frac{1}{2} −s}

for Re s > 1, where d_{1} = dN^{2}(a_{j}, N)^{−2} and where the summation on u is taken over all the positive integers u such that 4 + dN^{2}(a_{j}, N)^{−2}u^{2} is the square of an integer.

Proof. It follows from (2.1) and Theorem 3.2 that

∑_{\{P \}} c(P) = 2\sqrt{π} \frac{Γ(s − \frac{1}{2})}{Γ(s)} ∑_{\{P \}} \frac{ln NP_{1}}{λ_{P} − 1/λ_{P}} (1 + \frac{(λ_{P} − 1/λ_{P})^{2}}{4} )^{\frac{1}{2} −s}

for Re s > 1, where λ_{P} > 1 is an eigenvalue of P and P_{1} is given in Lemma 4.1. Let P be associated with a primitive indefinite quadratic form [a_{j}, b_{j}, c_{j}] as in Lemma 4.1. Then by Lemma 4.1 we have

λ_{P} − \frac{1}{λ_{P}} = \frac{Nu}{(a_{j}, N)}\sqrt{d}.

By Lemma 4.1, we also have

\sqrt{NP_{1}} = \frac{v_{1} + \sqrt{d_{1}}u_{1}}{2} = ϵ_{d_{1}}.

If P′ is a hyperbolic of Γ_{0}(N), and is associated with a primitive indefinite quadratic form [a_{j′}, b_{j′}, c_{j′}] as in Lemma 4.1, then P and P′ are Γ_{0}(N)-conjugate only if [a_{j}, b_{j}, c_{j}] and [a_{j′}, b_{j′}, c_{j′}] are Γ_{0}(N)-conjugate by Lemma 4.1. Next, let T be a hyperbolic element of Γ_{0}(N). Assume that T is associated with a primitive indefinite quadratic form [a, b, c] as in Lemma 4.1. If the discriminant of [a, b, c] is not equal to the discriminant of [a_{j}, b_{j}, c_{j}] which is associated with P, then P and T are not Γ_{0}(N)- conjugate by the argument in the last paragraph of the proof of Lemma 4.1. The stated identity then follows. □

Lemma 4.3. Let k be a divisor of N. Then the number of indefinite primitive quadratic forms [a, b, c] with (a, N) = k of discriminant d, which are not equivalent under Γ_{0}(N), is equal to

h_{d_{1}} ∏_{p^{2l}|(d,k)} p^{l} · ∏_{p|k} (1 + (\frac{d}{p}))

where d_{1} = dN^{2}/k^{2}, where the first product runs over all distinct primes p with p^{2l} being the greatest even p-power factor of (d, k) and the second product is taken over all distinct primes p dividing k.

Proof. Let [a, b, c] and [a′, b′, c′] be two indefinite primitive quadratic forms of discriminant d with (a, N) = k = (a′, N). If they are equivalent under Γ_{0}(N), then an element ( \begin{matrix}α & β \\ γ & δ \end{matrix} ) ∈ Γ_{0}(N) exists such that

\frac{b′}{2} = α(aβ + \frac{b}{2}δ) + γ(\frac{b}{2}β + cδ).

This implies that b′ ≡ b (mod 2k). In particular, if b′ ≢ b (mod 2k), then [a, b, c] and [a′, b′, c′] are not Γ_{0}(N)-equivalent.

Assume that ϱ is an integer with 1 ⩽ ϱ ⩽ 2k. Denote by Λ_{k,d,ϱ} the set of representatives of indefinite primitive quadratic forms [a, b, c] of discriminant d with (a, N) = k and b ≡ ϱ (mod 2k), which are not equivalent under Γ_{0}(N). Let L_{N,d_{1},ϱN/k}^{0} be the set of representatives of indefinite quadratic forms [aN, b, c] of discriminant d_{1} with (a, b, c) = 1, (N, b, c) = N/k and b ≡ ϱN/k (mod 2N), which are not equivalent under Γ_{0}(N).

A map from Λ_{k,d,ϱ} to L_{N,d_{1},ϱN/k}^{0} is defined by

T : [a, b, c] → [aN/k, bN/k, cN/k].

We claim that T is bijective. By the definition of Λ_{k,d,ϱ}, we see that T is injective. Conversely, if [a_{1}N, b_{1}, c_{1}] is an element of L_{N,d_{1},ϱN/k}^{0}, then we have (a_{1}, b_{1}, c_{1}) = 1, (N, b_{1}, c_{1}) = N/k, b_{1} ≡ ϱN/k (mod 2N) and b_{1}^{2} − 4Na_{1}c_{1} = d_{1}. Since (N, b_{1}, c_{1}) = N/k, we have N/k|c_{1}. Let a = a_{1}k, b = b_{1}k/N and c = c_{1}k/N. Then we have b ≡ ϱ (mod 2k) and d = b^{2} − 4ac. Since (a_{1}, N/k) = 1, we have (a, N) = k. We claim that (a, b, c) = 1, that is, (a_{1}N, b_{1}, c_{1}) = N/k. Since (a_{1}, b_{1}, c_{1}) = 1 and (N, b_{1}, c_{1}) = N/k, we have (a_{1}N, b_{1}, c_{1}) = N/k. Therefore, we have (a, b, c) = 1. Thus, [a, b, c] is an element of Λ_{k,d,ϱ}, and T maps it into the element [a_{1}N, b_{1}, c_{1}] of L_{N,d_{1},ϱN/k}^{0}. Therefore, T is surjective. Thus, we have proved that T is a bijection if the set Λ_{k,d,ϱ} is not empty. By Proposition, p. 505 of Gross, Kohnen and Zagier [4], the number of elements contained in L_{N,d_{1},ϱN/k}^{0} is h_{d_{1}}, and hence the set Λ_{k,d,ϱ} contains h_{d_{1}} elements if it is not empty.

If ϱ_{1} ≢ ϱ_{2} (mod 2k), then the set of Γ_{0}(N)-equivalence classes represented by elements in Λ_{k,d,ϱ_{1}} is disjoint from the set of Γ_{0}(N)-equivalence classes represented by elements in Λ_{k,d,ϱ_{2}}. Now, we want to count the number of non-empty sets Λ_{k,d,ϱ}. That is, we want to count the number of solutions ϱ of the equation

(4.6)  ϱ^{2} ≡ d (mod 4k), 1 ⩽ ϱ ⩽ 2k.

If (d, k) = 1, by Theorem 3.4 of Chapter 12, Hua [7] the number of solution of the equation (4.6) is equal to

∏_{p|k} (1 + (\frac{d}{p})) .

If there exists a prime number q satisfying q|k, q^{2} ∤ k and q^{2}|d, then we have q|ϱ, and the equation (4.6) can be written as

(4.7)  (\frac{ϱ}{q})^{2} ≡ \frac{d}{q^{2}} (mod \frac{4k}{q}), 1 ⩽ \frac{ϱ}{q} ⩽ \frac{2k}{q} .

Dividing out ϱ, d and k all such prime numbers q as in (4.7), we can reduce the second case to the first case when (d, k) = 1. Then, by using properties of the Legendre symbol we obtain that the number of solution of the equation (4.6) in the second case is still equal to

∏_{p|k} (1 + (\frac{d}{p})) .

Next, we consider the case when there exists a prime number q satisfying q^{2}|k and q^{2}|d. Then we have q|ϱ, and the equation (4.6) can be written as

(\frac{ϱ}{q})^{2} ≡ \frac{d}{q^{2}} (mod \frac{4k}{q^{2}} ), 1 ⩽ \frac{ϱ}{q} ⩽ \frac{2k}{q} .

Dividing out ϱ, d and k all such prime numbers q as in (4.7), we can reduce this case to the first case when (d, k) = 1. Then, by using properties of the Legendre symbol we obtain that the number of solution of the equation (4.6) in this case is still equal to

∏_{p^{2l}|(d,k)} p^{l} · ∏_{p|k} (1 + (\frac{d}{p}))

where p^{2l} is the greatest even power of p dividing (d, k).

Finally, we consider the case when there exists a prime number q satisfying q|k, q|d and q^{2} ∤ d. We have again q|ϱ. The equation (4.6) can be written as

(4.8)  q (\frac{ϱ}{q})^{2} ≡ \frac{d}{q} (mod \frac{4k}{q}), 1 ⩽ \frac{ϱ}{q} ⩽ \frac{2k}{q} .

We can assume that q ≠ 2. Otherwise, if q = 2 then we must have q^{2}|d. If (4.8) is solvable, then we have (q, 4k/q) = 1. Otherwise, we must have q^{2}|d. Hence a number x_{q} exists such that qx_{q} ≡ 1 (mod 4k/q). Then the equation (4.8) can be written as

(4.9)  (\frac{ϱ}{q})^{2} ≡ x_{q} \frac{d}{q} (mod \frac{4k}{q}), 1 ⩽ \frac{ϱ}{q} ⩽ \frac{2k}{q} .

Note that we have

(4.10)  (\frac{x_{q}d/q}{p} ) = (\frac{q}{p}) (\frac{d/q}{p} ) = (\frac{d}{p})

for any prime number p|(k/q). Dividing out all such primes q as in (4.9) and using (4.10), we obtain that the number of solution of the equation (4.6) in the final case is equal to

∏_{p|k} (1 + (\frac{d}{p})) .

This completes the proof of the lemma. □

Theorem 4.4. We have

∑_{\{P \}: hyperbolic P ∈Γ_{0}(N)} c(P) = 4\sqrt{π} \frac{Γ(s − \frac{1}{2})}{Γ(s)} ∑_{m|N_{m square-free}} ∑_{k|N} \frac{μ((m, k))}{(m, k)}

× ∑_{d∈Ω} ∑_{u} (\frac{d}{m}) ∏_{p^{2l}|(d,N/k)} p^{l} · \frac{h_{d} ln ϵ_{d}}{u\sqrt{d}} (1 + \frac{dk^{2}u^{2}}{4} )^{\frac{1}{2} −s}

for Re s > 1, where the summation on u is taken over all the positive integers u such that \sqrt{4 + dk^{2}u^{2}} ∈ ℤ.

Proof. Let [a_{j}, b_{j}, c_{j}], j = 1, 2, ⋯, H_{d}, be a set of representatives for classes of primitive indefinite quadratic forms of discriminant d, which are not equivalent under Γ_{0}(N). By Lemma 4.2, we have

∑_{\{P \}} c(P) = 4\sqrt{π} \frac{Γ(s − \frac{1}{2})}{Γ(s)} ∑_{d∈Ω} ∑_{j=1}^{H_{d}} ∑_{u} \frac{ln ϵ_{d_{1}}}{u\sqrt{d_{1}}} (1 + \frac{d_{1}u^{2}}{4} )^{\frac{1}{2} −s}

for Re s > 1, where d_{1} = dN^{2}(a_{j}, N)^{−2} and the summation on u is taken over all the positive integers u such that \sqrt{4 + d_{1}u^{2}} ∈ ℤ. Let k = N/(a_{j}, N). By using Lemma 4.3, we can write the above identity as (4.11)

\begin{aligned}∑_{\{P \}} c(P) = 4\sqrt{π} \frac{Γ(s − \frac{1}{2})}{Γ(s)} ∑_{k|N} ∑_{d∈Ω} ∑_{u} & ∏_{p^{2l}|(d,N/k)} p^{l} · ∏_{p| \frac{N}{k}} (1 + (\frac{d}{p})) \\ & × \frac{h_{d_{1}} ln ϵ_{d_{1}}}{u\sqrt{d_{1}}} (1 + \frac{d_{1}u^{2}}{4} )^{\frac{1}{2} −s} \end{aligned}

for Re s > 1, where d_{1} = dk^{2}. By using Dirichlet’s class number formula

h_{d_{1}} ln ϵ_{d_{1}} = \sqrt{d_{1}}L(1, χ_{d_{1}})

and by using the identity (See Theorem 11.2 of Chapter 12, Hua [7])

L(1, χ_{d_{1}}) = L(1, χ_{d}) ∏_{p|k} (1 − (\frac{d}{p}) p^{−1}) ,

we can write (4.11) as

\begin{matrix}∑_{\{P \}} c(P) = & 4\sqrt{π} \frac{Γ(s − \frac{1}{2})}{Γ(s)} ∑_{k|N} ∑_{d∈Ω} ∑_{u} ∏_{p^{2l}|(d,N/k)} p^{l} · ∏_{p| \frac{N}{k}} (1 + (\frac{d}{p})) \\ × ∏_{p|k} (1 − (\frac{d}{p}) p^{−1}) \frac{h_{d} ln ϵ_{d}}{u\sqrt{d}} (1 + \frac{dk^{2}u^{2}}{4} )^{\frac{1}{2} −s} \end{matrix}

for Re s > 1, where the summation on u is taken over all the positive integers u such that \sqrt{4 + dk^{2}u^{2}} ∈ ℤ. Since

∏_{p| \frac{N}{k}} (1 + (\frac{d}{p})) ∏_{p|k} (1 − (\frac{d}{p}) p^{−1}) = ∑_{m|N_{m square-free}} (\frac{d}{m}) \frac{μ((m, k))}{(m, k)}

where μ is the Möbius function, we have (4.12)

∑_{\{P \}} c(P) =4\sqrt{π} \frac{Γ(s − \frac{1}{2})}{Γ(s)} ∑_{m|N_{m square-free}} ∑_{k|N} \frac{μ((m, k))}{(m, k)}

× ∑_{d∈Ω} ∑_{u} (\frac{d}{m}) ∏_{p^{2l}|(d,N/k)} p^{l} · \frac{h_{d} ln ϵ_{d}}{u\sqrt{d}} (1 + \frac{dk^{2}u^{2}}{4} )^{\frac{1}{2} −s} .

Next, we show that

∑_{d∈Ω} ∑_{u} (\frac{d}{m}) ∏_{p^{2l}|(d,N/k)} p^{l} · \frac{h_{d} ln ϵ_{d}}{u\sqrt{d}} (1 + \frac{dk^{2}u^{2}}{4} )^{\frac{1}{2} −s}

is absolutely convergent for σ = Re s > 1. Since

(4.13)  \begin{matrix}| ∑_{d∈Ω} ∑_{u} (\frac{d}{m}) ∏_{p^{2l}|(d,N/k)} p^{l} · \frac{h_{d} ln ϵ_{d}}{u\sqrt{d}} (1 + \frac{dk^{2}u^{2}}{4} )^{\frac{1}{2} −s} | \\ ≪ ∑_{d∈Ω} ∑_{u} \frac{h_{d} ln ϵ_{d}}{u\sqrt{d}} (1 + \frac{dk^{2}u^{2}}{4} )^{\frac{1}{2} −σ} . \end{matrix}

It is proved in Li [11] that the right side of (4.13) is convergent for σ > 1, and hence, the right side of the stated identity is absolutely convergent for Re s > 1.

This completes the proof of the theorem. □

Proof of Theorem 1. By the argument given at the end of Li [11], we see that

\begin{matrix}∑_{d∈Ω,u} & |\frac{h_{d} ln ϵ_{d}}{\sqrt{d}u} (1 + \frac{dk^{2}u^{2}}{4} )^{\frac{1}{2} −s} − k^{1−2s} \frac{h_{d} ln ϵ_{d}}{(du^{2})^{s}} | \\ ≪ ∑_{d∈Ω,u} \frac{(du^{2})^{\frac{1}{2} +ϵ−1−σ}}{u^{1+2ϵ}} < ∞ \end{matrix}

for σ = Re s > 0. Then it follows from Theorem 4.4 and Theorem 3.4 that the function L_{N}(s) represents an analytic function in the half-plane Res > 0 except for having a possible pole at s = 1/2 and for having simple poles at s = 1, \frac{1}{2} ± iκ_{j}, j = 1, 2, · · ·.

This completes the proof of the theorem.

5. Proof of Corollary 2

Lemma 5.1. Let

L(s) = ∑_{d∈Ω} ∑_{u>0_{v^{2}−du^{2}=4}} (1 + (\frac{d}{p})) \frac{h_{d} ln ϵ_{d}}{(du^{2})^{s}}

+ ∑_{d∈Ω} ∑_{u>0_{v^{2}−dp^{2}u^{2}=4}} \frac{h_{dp^{2}} ln ϵ_{dp^{2}}}{(dp^{2}u^{2})^{s}} − ∑_{d∈Ω} ∑_{u>0_{v^{2}−du^{2}=4}} \frac{h_{d} ln ϵ_{d}}{(du^{2})^{s}}

for Re s > 1. Then L(s) is analytic for Re s > 1 and has an analytic continuation to the half-plane Re s > 1/2 except for having simple poles at s = \frac{1}{2} + \sqrt{1/4 − λ_{j}}, j = 1, 2, ⋯, S.

Proof. It is proved in Li [11] that the series

D(s) = ∑_{d∈Ω} \frac{h_{d} ln ϵ_{d}}{d^{s}} ∑_{u>0_{v^{2}−du^{2}=4}} \frac{1}{u^{2s}} ,

represents an analytic function of s in the half-plane Re s > 1/2 except for having a simple poles at s = 1. Let

h(r) = 2\sqrt{π}Γ(s − \frac{1}{2})Γ^{−1}(s)4^{s− \frac{1}{2}} ∫_{0}^{∞}(u + \frac{1}{u} + 2)^{\frac{1}{2} −s}u^{ir−1}du

for Re s > \frac{1}{2}. By (2.3), (3.4), (3.5), Lemma 3.5, Lemma 4.1, and Lemma 4.2 of Li [11], we have that

D(s) − h(−i/2)

is analytic in the half-plane Re s > 1/2. Let L_{N}(s) be given as in Theorem 1 with N = p. Then, by (2.3), (3.13), (3.14), Theorem 3.12, Lemma 4.1, and Theorem 4.6 of Conrey and Li [1], we have that

L_{N}(s) − h(−i/2)

is an analytic function of s in the half-plane Re s > 1/2 except for simple poles at s = \frac{1}{2} + \sqrt{1/4 − λ_{j}}, j = 1, 2, ⋯, S. It follows that

L(s) = L_{N}(s) − D(s)

is an analytic function of s in the half-plane Re s > 1/2 except for simple poles at s = \frac{1}{2} + \sqrt{1/4 − λ_{j}}, j = 1, 2, ⋯, S.

This completes the proof of the lemma. □

By using the formula (See Theorem 11.2 of Chapter 12, Hua [7])

h_{dp^{2}} ln ϵ_{dp^{2}} = (p − (\frac{d}{p})) h_{d} ln ϵ_{d},

we find that

L(s) = ∑_{d∈Ω} ∑_{u>0_{v^{2}−du^{2}=4}} (\frac{d}{p}) \frac{h_{d} ln ϵ_{d}}{(du^{2})^{s}} + ∑_{d∈Ω} ∑_{u>0_{v^{2}−dp^{2}u^{2}=4}} \frac{h_{dp^{2}} ln ϵ_{dp^{2}}}{(dp^{2}u^{2})^{s}}

= ∑_{d∈Ω} ∑_{u>0, p∤u_{v^{2}−du^{2}=4}} (\frac{d}{p}) \frac{h_{d} ln ϵ_{d}}{(du^{2})^{s}} + p^{1−2s} ∑_{d∈Ω} ∑_{u>0_{v^{2}−dp^{2}u^{2}=4}} \frac{h_{d} ln ϵ_{d}}{(du^{2})^{s}} .

Therefore, we can write (5.1)

L(s) = ∑_{u=1,p∤u}^{∞} \frac{1}{u^{2s}} ∑_{d∈Ω_{v^{2}−du^{2}=4}} (\frac{d}{p}) \frac{h_{d} ln ϵ_{d}}{d^{s}} + p^{1−2s} ∑_{u=1}^{∞} \frac{1}{u^{2s}} ∑_{d∈Ω_{v^{2}−dp^{2}u^{2}=4}} \frac{h_{d} ln ϵ_{d}}{d^{s}}

for Re s > 1.

Lemma 5.2. Let u be the product of a power of 2 and a power of an odd prime. Then all integers d ∈ Ω, which satisfy v^{2}−du^{2} = 4 for some integers v, are given by

d = \{ \begin{matrix}n^{2}u^{2} ± 4n, & if 2 ∤ u; \\ n^{2}u^{2} ± 4n, & if 2|u and 4 ∤ u; \\ n^{2}u^{2} ± 4n or (n + \frac{1}{4})^{2}u^{2} − (4n + 1), \\ or (n + \frac{3}{4})^{2}u^{2} + (4n + 3), & if 4|u and 8 ∤ u; \\ n^{2}u^{2} ± 4n or (n + \frac{1}{4})^{2}u^{2} + (4n + 1), & if 8|u \end{matrix}

for n = 1, 2, 3, · · ·.

Proof. The idea is to solve Pell’s equation (v − 2)(v + 2) = du^{2} for integers v.

If 2 ∤ u, then we must have u^{2}|v−2 or u^{2}|v+2. Otherwise, if u contains a prime factor q dividing both v−2 and v+2, then q|4, and hence q = 2. This is a contradiction. Therefore, we have v = ±2 + nu^{2} for n = 1, 2, 3, · · ·. From Pell’s equation v^{2} − 4 = du^{2}, we deduce that d = n^{2}u^{2} ± 4n for n = 1, 2, 3, ⋯, which belong to the set Ω.

If 2|u and 4 ∤ u, then we have v = ±2 + \frac{1}{2}mu^{2} for m = 1, 2, · · ·. Since we want d to be in the set Ω, m has to be an even integer. Therefore, we have d = n^{2}u^{2} ± 4n for n = 1, 2, 3, · · ·.

If 4|u, then we have v = ±2 + \frac{1}{4}mu^{2} for m = 1, 2, · · ·. Note that d = \frac{1}{16}m^{2}u^{2} ± m in this case. First, we consider the case when 8 ∤ u. Then d ≡ m^{2} ± m (mod 4). If m is even, then d ≡ ±m (mod 4), and hence m ≡ 0 (mod 4). Thus, we have d = n^{2}u^{2} ± 4n for n = 1, 2, 3, · · ·. If m is odd, then d ≡ 1 ± m (mod 4). When m = 4n + 1 for n = 1, 2, 3, ⋯, we have d = (n + \frac{1}{4})^{2}u^{2} − (4n + 1). When m = 4n + 3 for n = 1, 2, 3, ⋯, we have d = (n + \frac{3}{4})^{2}u^{2} + (4n + 3). Next, we consider the case when 8|u. Then d ≡ ±m (mod 4). This implies that m = 4n or m = 4n + 1 with d = \frac{1}{16}m^{2}u^{2} + m, that is, d = n^{2}u^{2} ± 4n or d = (n + \frac{1}{4})^{2}u^{2} + (4n + 1) n = 1, 2, 3, · · ·.

This completes the proof of the lemma. □

By using the Pólya-Vinogradov inequality [2], we obtain that

| ∑_{k=d+1}^{∞} (\frac{d}{k} ) k^{−1}| ⩽ c\frac{ln d}{\sqrt{d}}

where c is an absolute constant (cf. Siegel [19]). ¿From Dirichlet’s class number formula

h_{d} ln ϵ_{d} = \sqrt{d} ∑_{k=1}^{∞} \frac{1}{k} (\frac{d}{k} ) ,

we deduce that

(5.2)  h_{d} ln ϵ_{d} = \sqrt{d} ∑_{k=1}^{d} \frac{1}{k} (\frac{d}{k} ) + O(ln d)

where the constant implied by the symbol O is an absolute constant.

Proof of Corollary 2. By an argument similar to that made in the proof of Lemma 5.2, we find that all integers d ∈ Ω satisfying v^{2} − du^{2} = 4 are among the integers d’s given by (5.3)

du_{1}^{2} = \{ \begin{matrix}n^{2}u_{2}^{2} ± 4n, & if 2 ∤ u_{2}; \\ n^{2}u_{2}^{2} ± 4n, & if 2|u_{2} and 4 ∤ u_{2}; \\ n^{2}u_{2}^{2} ± 4n or (n + \frac{1}{4})^{2}u_{2}^{2} − (4n + 1), \\ or (n + \frac{3}{4})^{2}u_{2}^{2} + (4n + 3), & if 4|u_{2} and 8 ∤ u_{2}; \\ n^{2}u_{2}^{2} ± 4n or (n + \frac{1}{4})^{2}u_{2}^{2} + (4n + 1), & if 8|u_{2} \end{matrix}

for n = 1, 2, 3, ⋯ and for all decompositions u = u_{1}u_{2} with u_{2} ⩾ u_{1}. Similarly, we can find all integers d ∈ Ω satisfying v^{2} − dp^{2}u^{2} = 4. Let σ = Re s, and let d(u) denote the number of positive divisors of u. Then by (5.1)-(5.3) we have

L(s) − L_{p}(s) ≪ ∑_{u=1}^{∞} \frac{d(u)}{u^{2σ}} ∑_{n=1}^{∞} \frac{ln n}{n^{2σ}} < ∞

for σ > 1/2. Note that (\frac{d}{p}) = (\frac{v^{2}−4}{p}) if v^{2} − du^{2} = 4 and p ∤ u. Therefore, the stated result follows from Lemma 5.1.

This completes the proof of the corollary. □

References

1. J. B. Conrey and Xian-Jin Li, On the trace of Hecke operators for Maass forms for congruence subgroups, Forum Math. 11 (1999).

2. H. Davenport, Multiplicative Number Theory, Second Edition, Revised by Hugh L. Montgomery, Springer-Verlag, New York, 1980.

3. J. -M. Deshouillers and H. Iwaniec, Kloosterman sums and Fourier coefficients of cusp forms, Invent. Math. 70, 219–288 (1982).

4. B. Gross, W. Kohnen, and D. Zagier, Heegner points and derivatives of L-series. II, Math. Ann. 278 (1987), 497–562.

5. D. A. Hejhal, The Selberg Trace Formula for P SL(2, ℝ), II, Lecture Notes in Math., no. 1001, Springer-Verlag, New York, 1983.

6. D. A. Hejhal, Roots of quadratic congruences and eigenvalues of the non-Euclidean Laplacian, in “The Selberg Trace Formula and Related Topics,” Edited by D. Hejhal, P. Sarnak, and A. Terras, Contemp. Math., 53, Amer. Math. Soc., Providence, R.I., 1986, 277–339.

7. L. K. Hua, Introduction to Number Theory, Springer-Verlag, New York, 1982.

8. H. Iwaniec, Introduction to the Spectral Theory of Automorphic Forms, Revista Matemática Iberoamericana, 1995.

9. T. Kubota, Elementary theory of Eisenstein series, Halsted Press, New York, 1973.

10. N. V. Kuznecov, Petersson’s conjecture for cusp forms of weight zero and Linnik’s conjecture, Math. USSR Sbornik 39 (1981), 299–342.

11. Xian-Jin Li, On the trace of Hecke operators for Maass forms, CRM Proc. and Lecture Notes, Vol. 19, Amer. Math. Soc., Providence (1999), 215–229.

12. M. Ram Murty, An introduction to Selberg’s trace formula, J. Indian Math. Soc. 52 (1987), 91–126.

13. P. Sarnak, Class numbers of indefinite binary quadratic forms, J. Number Theory 15 (1982), 229–247.

14. P. Sarnak, Arithmetic quantum chaos, The Schur Lectures (1992, Tel Aviv), Israel Math. Conf. Proc., 8, Bar-Ilan Univ., Ramat Gan (1995), 183–236.

15. A. Selberg, Harmonic analysis and discontinuous groups on weakly symmetric Riemannian spaces with applications to Dirichlet series, J. Indian Math. Soc. 20 (1956), 47–87.

16. A. Selberg, On the estimation of Fourier coefficients of modular forms, Proc. Symp. Pure Math. VII, Amer. Math. Soc., Providence, R. I. (1965), 1–15.

17. G. Shimura, Introduction to the Arithmetic Theory of Automorphic Functions, Princeton Univ. Press, 1971.

18. C. L. Siegel, Über die Classenzahl quadratischer Zahlkörper, Acta Arith. 1 (1935), 83–86.

19. C. L. Siegel, The average measure of quadratic forms with given determinant and signature, Ann. of Math. 45 (1944), 667–685.

20. A. Strömbergsson, The Selberg trace formula for modular correspondences, Preprint 1998.

21. E. C. Titchmarsh, Introduction to the Theory of Fourier Integrals, Second Edition, Oxford, 1948.

22. A. B. Venkov, Remainder term in the Weyl-Selberg asymptotic formula, J. Soviet Math. 17 (1981), 2083–2097.

23. G. N. Watson, A Treatise on the Theory of Bessel Functions, Cambridge Univ. Press, 1952.

Department of Mathematics, Brigham Young University, Provo, Utah 84602 USA

   E-mail address: cardon@math.byu.edu,  xianjin@math.byu.edu