WEIGHTED AVERAGES OF MODULAR L-VALUES

ANDREW KNIGHTLY AND CHARLES LI

Abstract. Using an explicit relative trace formula on GL(2), we derive a formula for averages of modular L-values in the critical strip, weighting by Fourier coefficients, Hecke eigenvalues, and Petersson norms. As an application we show that a GRH holds for these averages as the weight or the level goes to ∞. We also use the formula to give explicit zero-free regions of the form | Im(s)| ≤ τ_{0} for some particular modular L-functions.

July 31, 2009

1. Introduction

Let S_{k}(N, ω′) denote the space of cusp forms h on Γ_{0}(N) satisfying

h(\frac{az + b}{cz + d}) = ω′(d)^{−1}(cz + d)^{k} h(z) (( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ Γ_{0}(N)).

The Mellin transform of h is the analytic function

Λ(s, h) = ∫_{0}^{∞} h(iy)y^{s−1}dy,

which converges absolutely for all s ∈ C ([Sh], p. 94). Write h(z) = ∑_{r>0} a_{r}(h)e^{2πirz}.

When Re(s) > 1 + k/2, we have additionally

∫_{0}^{∞} ∑_{r>0} |a_{r}(h)e^{−2πry}y^{s−1}|dy < ∞.

Therefore for such s,

Λ(s, h) = ∑_{r>0} a_{r}(h) ∫_{0}^{∞} e^{−2πry}y^{s−1}dy = ∑_{r>0} a_{r}(h) ∫_{0}^{∞} e^{−t}t^{s−1}(2πr)^{−s}dt

= (2π)^{−s}Γ(s) ∑_{r>0} \frac{a_{r}(h)}{r^{s}} = (2π)^{−s}Γ(s)L(s, h),

where L(s, h) is the Dirichlet series attached to h. The completed L-function Λ(s, h) satisfies a functional equation relating values at s and k − s, which in the case of N = 1 is simply

(1)  Λ(s, h) = i^{k}Λ(k − s, h).

Hence the critical line of the L-function is Re(s) = k/2. If h is a newform determining the cuspidal representation π, then Λ(s, π) = Λ(s+ \frac{k−1}{2}, h), and Λ(s, π) satisfies a functional equation relating its values at s and 1 − s.

The central values of L-functions have deep arithmetic significance. If the Hecke eigenvalues are known, one can compute the central values of a particular L-function using the approximate functional equation (see e.g. [Mi], §1.3.2). We can also use the trace formula to get information about averages of L-values as h ranges through an orthogonal Hecke eigenbasis F for S_{k}(N, ω′). In this paper, we will explicitly compute such an average, with the L-values weighted by Hecke eigenvalues, Fourier coefficients and Petersson norms.

The asymptotics of such averages have been studied widely. Duke showed that when k = 2, N is prime, ω′ is trivial, and χ is a Dirichlet character unramified at N,

\frac{1}{ψ(N)} ∑_{h∈F} \frac{a_{1}(h)L(1, h ⊗ χ)}{∥h∥^{2}} = 4π + O(N^{−1/2} log N),

where ψ(N) = [SL_{2}(Z) : Γ_{0}(N)], [Du]. Here we have normalized the Petersson norm as in (2) below. With a more careful estimation, Ellenberg improved Duke’s error term to O(N^{−1+ε}), while at the same time allowing a_{r}(h) in place of a_{1}(h), [El]. Of the many other generalizations of Duke’s work, we mention two: Akbary extended it to weight k > 2 with an error term of O_{k}(N^{−1/2}(log N)^{k−1}) [Ak], and Kamiya further allowed composite N and L(1 + it, h ⊗ χ) with an error term of O_{t,k}(N^{−k/4}) [Ka]. The method of Duke uses the Petersson trace formula. Another approach, based on the Eichler-Selberg trace formula, was found by Royer (see §4.3 of [Ro]).

Here we consider the case k > 2. For the weighted averages we obtain an error term of O(N^{−k/2}) on the critical line. In fact, we give an explicit formula for the average (Theorem 1.1). At the same time, we allow s to vary through the whole critical strip. We will also give the asymptotic behavior of the average as k → ∞.

To state the main theorem, for h ∈ S_{k}(N, ω′), let h^{−} ∈ S_{k}(N, ω^{′−1}) denote the “complex conjugate” of h, given by h^{−}(z) = ∑ \overline{a_{n}(h)}q^{n}. If ω′ is trivial, then h^{−} = h, and in general Λ(s, h^{−}) = \overline{Λ(s̄, h)}.

Theorem 1.1. Let r, N, n, k ∈ Z^{+} with (n, N) = 1 and k > 2. Fix a Dirichlet character ω′ of conductor dividing N, and suppose S_{k}(N, ω′) ≠ {0}. Let F be an orthogonal basis for S_{k}(N, ω′) consisting of eigenfunctions for the Hecke operator T_{n}. Then for any s ∈ C with 1 < Re(s) < k − 1,

∑_{h∈F} \frac{λ_{n}(h)a_{r}(h)Λ(s, h^{−})}{∥h∥^{2}}

=\frac{ψ(N)2^{k−1}Γ(s)(2πrn)^{k−s−1}}{(k − 2)!} ∑_{m| gcd(n,r)} \frac{m^{2s−k+1}}{ω′(m)}

+ δ_{N,1} \frac{2^{k−1}Γ(k − s)(2πrn)^{s−1}}{(k − 2)! i^{k}} ∑_{m| gcd(n,r)} m^{k−2s+1}

+ \frac{ψ(N)(4πrn)^{k−1}}{N^{s}(k − 2)! e^{iπs/2}} ∑_{{a≠0,d>0 \atop gcd(a,Nd)| gcd(r,n)}}\frac{a^{−(k−s)}d^{−s} gcd(a, Nd)}{ω′(a) e^{2πirℓ_{0}/a}}_{1}f_{1}(s; k; \frac{2πirn}{Nad}),

where T_{n}h = λ_{n}(h)h, ℓ_{0} is any integer satisfying ℓ_{0}Nd ≡ n mod a, and

1f_{1}(s; k; w) = \frac{Γ(s)Γ(k − s)}{Γ(k)}_{1}F_{1}(s; k; w)

for the confluent hypergeometric function _{1}F_{1}(s; k; w) = 1 + \frac{s}{k}w + \frac{s(s+1)}{k(k+1)} \frac{w^{2}}{2!} + · · ·. When a < 0, we take a^{s} = e^{iπs}|a|^{s}. Throughout we use the convention that ∑_{m|n} is a sum over positive divisors of n.

This theorem generalizes a result of Kohnen, who derived the special case n = N = 1 using a Poincaré series-type argument ([Ko], p. 188). Our approach here is quite different.

From its integral representation (cf. (17) on page 16), it follows that

|_{1}f_{1}(s; k; 2πirn/Nad)| ≤ 1.

Thus the sum over a, d is bounded independently of N (see Prop. 4.2 for a precise bound), and we have the following.

Corollary 1.2. With notation as above and 1 < Re(s) < k − 1,

\frac{1}{ψ(N)} ∑_{h∈F} \frac{λ_{n}(h)a_{r}(h)Λ(s, h^{−})}{∥h∥^{2}}

= \frac{2^{k−1}Γ(s)(2πrn)^{k−s−1}}{(k − 2)!} ∑_{m| gcd(n,r)} \frac{m^{2s−k+1}}{ω′(m)} + O(N^{− Re(s)}).

The implied constant is effective and depends only on k, n, r and s, uniformly for s in compact subsets of the given strip.

According to the Grand Riemann Hypothesis, when h is a Hecke eigenform all zeros of Λ(s, h) inside the critical strip \frac{k−1}{2} < Re(s) < \frac{k+1}{2} lie on the critical line Re(s) = k/2. Using Theorem 1.1, we will show that a GRH holds for averages (see also [Ko] for the N = 1 case). Note that Corollary 1.2 implies nonvanishing of the average when N is large, at least when gcd(n, r) = 1. By the results of Section 4.1 in which we determine the asymptotic behavior as k → ∞, the average is also nonzero when k is large. To state the result, we shift the L-functions so that the critical strip becomes 0 ≤ Re(s) ≤ 1, independent of k.

Corollary 1.3. Assume N > 1, k > 3, gcd(n, r) = 1, and that S_{k}(N, ω′) ≠ {0}. For τ_{0} > 0, let R be the rectangle consisting of s with 0 ≤ Re(s) ≤ 1 and | Im(s)| ≤ τ_{0}. Then there exist constants C_{k}, C_{N} > 0 depending only on R, n and r, such that if either k > C_{k} or N > C_{N}, the sum

∑_{h∈F} \frac{λ_{n}(h)a_{r}(h)Λ(s + \frac{k−1}{2}, h^{−})}{∥h∥^{2}}

is nonzero for every s ∈ R. In particular, for any s ∈ R there exists an eigenform h ∈ S_{k}(N, ω′) such that λ_{n}(h), a_{r}(h) and Λ(s + \frac{k−1}{2}, h) are all nonzero.

Some of the hypotheses of Corollary 1.3 can be weakened with minor modifications. To allow gcd(n, r) > 1, we simply need to exclude the left edge of the strip. Thus the boundary of R should be shrunk to δ ≤ Re(s) ≤ 1 for any 0 < δ < 1/2. If in addition we exclude the right edge by considering δ ≤ Re(s) ≤ 1 − δ for such δ, then the statement is also valid for k = 3. When N = 1, the situation is a little more delicate because, if s lies on the critical line, the first two terms in the formula for the average may cancel each other out and we cannot say anything. Indeed if k ≡ 2 mod 4, the L-values themselves vanish at s = k/2 because of the functional equation (1). So when N = 1 we must assume that R is a compact region which does not meet the critical line Re(s) = \frac{1}{2}.

Suppose it happens that dim S_{k}(N, ω′) = 1. Then the theorem gives a computable formula for the values of the L-function of the cusp form. Using an effective version of Corollary 1.3, we obtain zero-free regions for several such L-functions in Section 4.2. As a final illustration, we show how to use the formula to compute some familiar data, namely values of Ramanujan’s τ-function. This is achieved by taking a quotient of two different averages. The resulting expression can be estimated to any desired precision using partial sum approximations, and since τ(r) is known to be an integer, we can pinpoint its value with just a few terms.

Theorem 1.1 is proven using a relative trace formula on GL(2). We start with a Hecke operator and integrate its associated kernel over the group N × M, where N is unipotent and M is diagonal. This is a hybrid of the techniques of the papers [Li], [KL1] (which used N ×N) and [RaRo] (which used M ×M). The paper [RaRo] of Ramakrishnan and Rogawski gives an asymptotic formula for certain averages of the form ∑_{h∈F} \frac{λ_{p^{n}} (h)Λ(k/2,h⊗χ)Λ(k/2,h)}{∥h∥^{2}}, which yields a weighted equidistribution result for the Hecke eigenvalues. They use a regularization procedure since they assert that the terms on their geometric side are not absolutely convergent. Thus the replacement here of just one factor of M by the unipotent group N (of compact quotient) is enough to give an absolutely convergent trace formula.

We mention that Feigon and Whitehouse refined the method of [RaRo] in many cases by using the Jacquet-Langlands correspondence to avoid the convergence issues inherent to GL(2), [FW]. They obtain closed formulas for the averages at the central point, over a totally real field.

A version of Theorem 1.1 involving twisted L-functions Λ(s, h ⊗ χ) should be obtainable by similar methods, using a test function as in [RaRo]. Of course, the presence of a nontrivial character χ will only help the convergence of the trace formula.

We would like to thank David Bradley and George Knightly for their helpful comments on the hypergeometric function. The numerical calculations in Section 4 were made using Mathematica. The first author was supported by the University of Maine Office of the Vice President for Research, and NSA grant H98230-06-1-0039.

2. Notation and preliminaries

We briefly recall the notation and test function of [KL2], which contains proofs of the various facts mentioned in this section. Let A, A_{fin} be the adeles and finite adeles of Q, and let G = GL(2). We write \overline{G} for G/Z, where Z is the center. Fix a level N ≥ 1 and a Dirichlet character ω′ of conductor dividing N. For a weight k > 2, let S_{k}(N, ω′) denote the space of cusp forms satisfying

h(γz) = ω′(γ)^{−1}j(γ, z)^{k}h(z) (γ ∈ Γ_{0}(N)).

Here ω′(( \begin{matrix}a & b \\ c & d \end{matrix} )) = ω′(d) and

j(( \begin{matrix}a & b \\ c & d \end{matrix} ), z) = (ad − bc)^{−1/2}(cz + d) (( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ G(R)^{+}).

Using A^{∗} = Q^{∗}(R_{+}^{∗} × Ẑ^{∗}), define

ω : A^{∗} → Ẑ^{∗} → (Z/NZ)^{∗} → C^{∗},

where the last arrow is ω′. For an idele x, let x_{N} denote the idele which agrees with x at the places p|N, and which is 1 at all other places. Then for any integer d prime to N,

ω(d_{N}) = ω′(d).

To each h ∈ S_{k}(N, ω′) we associate ϕ_{h} ∈ L_{0}^{2}(ω) = L_{0}^{2}(\overline{G}(Q)\\overline{G}(A), ω) by using strong approximation:

ϕ_{h}(γ(g_{∞} × k)) = j(g_{∞}, i)^{−k}h(g_{∞}(i))

for γ ∈ G(Q), g_{∞} ∈ G(R)^{+} and k ∈ K_{1}(N) = \{( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ G(Ẑ)| c, d − 1 ∈ NẐ\}.

We normalize the Petersson norm by

(2)  ∥h∥^{2} = \frac{1}{ψ(N)} ∫_{Γ_{0}(N)\H} |h(z)|^{2}y^{k}\frac{dx dy}{y^{2}} .

If we normalize Haar measure on \overline{G}(A) so that meas(\overline{G}(Q)\\overline{G}(A)) = π/3, then the Petersson norm corresponds to the L^{2}-norm and the map h ↦ ϕ_{h} is an isometry. We normalize Haar measure on A so that meas(Q\A) = 1. We take Lebesgue measure dx on R and d^{∗}y = \frac{dy}{|y|} on R^{∗}. On A_{fin}^{∗} we normalize so that meas(Ẑ^{∗}) = 1.

Fix n ∈ Z^{+} with gcd(n, N) = 1, and define a test function f = f_{∞} × f^{n} as follows. Define

M(n, N) = \{g = ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ M_{2}(Ẑ)| det g ∈ nẐ^{∗} and c ≡ 0 mod NẐ\}.

The support of f_{fin} = f^{n} is the set Z(A_{fin})M(n, N) = Z(Q^{+})M(n, N). By definition,

f^{n}(z_{Q}m) = \frac{ψ(N)}{ω(m)} (z_{Q} ∈ Z(Q^{+}), m ∈ M(n, N)),

where for m = ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ M(n, N) we define ω(m) = ω(d_{N}). We take f_{∞}(g) = \frac{1}{d_{k}} \overline{⟨π_{k}(g)v_{0}, v_{0}⟩}, where π_{k} is the weight k discrete series of GL_{2}(R) with formal degree d_{k} = \frac{4π}{k−1} and lowest weight unit vector v_{0}. Explicitly, if g = ( \begin{matrix}a & b \\ c & d \end{matrix} ), then

f_{∞}(g) = \{ \begin{matrix}\frac{(k−1)}{4π} \frac{det(g)^{k/2}(2i)^{k}}{(−b+c+(a+d)i)^{k}} & if det(g) > 0, \\ 0 & otherwise \end{matrix}

(see [KL2], Theorem 14.5). By construction, f(zg) = ω(z)^{−1}f(g) for z ∈ Z(A).

This function f is integrable precisely when k > 2. Hence for such k it defines an operator R(f) on L^{2}(ω) by

R(f)ϕ(x) = ∫_{\overline{G}(A)} f(g)ϕ(xg)dg.

Then as shown in [KL2], we have the following commutative diagram:

L^{2}(ω) \overline{\begin{matrix}n^{\frac{k}{2} −1}R(f) \\  \end{matrix}}L^{2}(ω)

S_{k}(N, ω′)

orthog. proj.

                                 T_{n} S_{k}(N, ω′)

where T_{n} is the classical Hecke operator. Letting F be any orthogonal basis for S_{k}(N, ω′), the kernel of R(f) is the function on G(A) × G(A) given by

(3)  K(g_{1}, g_{2}) = ∑_{γ∈\overline{G}(Q)} f(g_{1}^{−1}γg_{2}) = ∑_{h∈F} \frac{R(f)ϕ_{h}(g_{1})\overline{ϕ_{h}(g_{2})}}{∥ϕ_{h}∥^{2}}.

Lastly, we let θ : A → C^{∗} denote the standard character of A. It is defined by

θ_{∞}(x) = e^{−2πix}, x ∈ R,

and

θ_{p}(x) = e^{2πir_{p}(x)}, x ∈ Q_{p},

where r_{p}(x) ∈ Q is the principal part of x, a number with p-power denominator characterized (up to Z_{p}) by x ∈ r_{p}(x)+Z_{p}. Then θ is trivial on Q and θ_{fin} = ∏_{p} θ_{p} is trivial precisely on Ẑ. In particular, for any q ∈ Q, θ_{fin}(q) = θ_{∞}(q)^{−1} = e^{2πiq}. The characters of Q\A are parametrized by r ∈ Q via

θ_{r}(x) = θ(−rx).

3. Proof of the theorem

3.1. Spectral side. The theorem is proven by computing the following

(4)  ∫_{Q^{∗}\A^{∗}} ∫_{Q\A} K((\begin{matrix}1 & x \\ 0 & 1 \end{matrix}) , (\begin{matrix}y & 0 \\ 0 & 1 \end{matrix})) \overline{θ_{r}(x)}|y|^{s−k/2}dx d^{∗}y

using the two expressions for the kernel (3). We will see presently that the integral (4) is absolutely convergent for all s.

For the spectral side, choose F in (3) to be an orthogonal basis of eigenvectors of T_{n}. Then R(f)ϕ_{h} = n^{1−k/2}λ_{n}(h)ϕ_{h} for h ∈ F, so (4) is equal to

(5) ∑_{h∈F} \frac{n^{1−k/2}λ_{n}(h)}{∥ϕ_{h}∥^{2}} ∫_{Q\A} ϕ_{h}(( _{1}^{1 x} ))\overline{θ_{r}(x)}dx ∫_{Q^{∗}\A^{∗}} \overline{ϕ_{h}(( \begin{matrix}y \\ & 1 \end{matrix} ))}|y|^{s−k/2}d^{∗}y

= \frac{n^{1−k/2}}{e^{2πr}} ∑_{h∈F} \frac{λ_{n}(h)a_{r}(h)Λ(s, h^{−})}{∥h∥^{2}},

by the following lemma.

Lemma 3.1. For r ∈ Q,

∫_{Q\A} ϕ_{h}(( _{1}^{1 x} )) \overline{θ_{r}(x)}dx = \{ \begin{matrix}e^{−2πr}a_{r}(h) & if r ∈ Z^{+}, \\ 0 & otherwise \end{matrix}

and

∫_{Q^{∗}\A^{∗}} \overline{ϕ_{h}(( \begin{matrix}y \\ & 1 \end{matrix} ))} |y|^{s−k/2}d^{∗}y = Λ(s, h^{−}).

Proof. For a proof of the first statement, see [KL2], Corollary 12.4. For the second, note that ϕ_{h}(( \begin{matrix}y \\ & 1 \end{matrix} )) = y^{k/2}h(iy) when y ∈ R_{+}^{∗}. Furthermore, \overline{h(iy)} = ∑ \overline{a_{r}(h)}e^{−2πry} = h^{−}(iy). We can integrate over the fundamental domain R_{+}^{∗} × Ẑ^{∗}. The integrand is invariant under Ẑ^{∗}, which has measure 1. Thus

∫_{Q^{∗}\A^{∗}} \overline{ϕ_{h}(( \begin{matrix}y \\ & 1 \end{matrix} ))}|y|^{s−k/2}d^{∗}y = ∫_{0}^{∞} h^{−}(iy)y^{s−1}dy = Λ(s, h^{−}).

□

The two integrals in (5) are absolutely convergent for all s, so we have the following.

Proposition 3.2. The double integral (4) is absolutely convergent for all s ∈ C.

3.2. Geometric side. On the geometric side, we use the formalism of Jacquet’s relative trace formula. Let N be the upper triangular unipotent subgroup of G, and let M be the diagonal subgroup. Let \overline{M} = M/Z, where Z is the center. Setting H = N × \overline{M}, the integral (4) is taken over H(Q)\H(A). Using K(n, m) = ∑_{γ∈\overline{G}(Q)} f(n^{−1}γm), we would like to pull the sum out of (4); however the individual terms f(n^{−1}γm) are not well-defined modulo H(Q). We have to break \overline{G}(Q) into H(Q)-orbits and then sum over these orbits. The action of H is (n, m) · γ = n^{−1}γm. For δ ∈ \overline{G}(Q), its orbit is

[δ] = \{( _{1}^{1 −x} )δ( \begin{matrix}y \\ & 1 \end{matrix} )| x ∈ Q, y ∈ Q^{∗}\} = \{n^{−1}δm| (n, m) ∈ H_{δ}(Q)\H(Q)\},

where H_{δ} is the stabilizer of δ. It is easy to check that in fact H_{δ} = {1} for any δ. Thus the geometric expression for (4) is equal to

(6)  ∑_{[δ]} ∫_{A^{∗}} ∫_{A} f(( _{1}^{1 −x} )δ( \begin{matrix}y \\ & 1 \end{matrix} ))\overline{θ_{r}(x)}|y|^{s−k/2}dx d^{∗}y.

To justify this manipulation we have to show that (6) converges absolutely.

Proposition 3.3. Suppose 1 < Re(s) < k − 1. Then

∑_{[δ]} ∫_{A^{∗}} ∫_{A} |f(( _{1}^{1 −x} )δ( \begin{matrix}y \\ & 1 \end{matrix} ))\overline{θ_{r}(x)}|y|^{s−k/2}| dx d^{∗}y < ∞.

Thus for such s, the geometric side (6) converges absolutely and equals the spectral side (5).

We postpone the proof of the proposition until Section 3.3 below. Assuming it for now, let I_{δ}(f) denote the double integral attached to δ in (6). By the proposition, I_{δ}(f) is absolutely convergent on the given strip. We just need to determine the set of δ and compute each of these geometric integrals. We assume throughout that the hypothesis of the proposition is satisfied.

The set of orbits [δ] is in one-to-one correspondence with N(Q)\\overline{G}(Q)/\overline{M}(Q). By the Bruhat decomposition

G(Q) = N(Q)M(Q) ∪ N(Q)( \begin{matrix}& 1 \\ −1 \end{matrix} )N(Q)M(Q),

a set of representatives is given by

\{1\} ∪ \{( \begin{matrix}0 & 1 \\ −1 & t \end{matrix} )| t ∈ Q\}.

Proposition 3.4. When δ = 1, the integral

I_{1}(f) = ∫_{A^{∗}} ∫_{A} f((\begin{matrix}y & −x \\ 0 & 1 \end{matrix}))θ(rx)dx|y|^{s−k/2}d^{∗}y

converges absolutely on 0 < Re(s) < k − 1, and for such s it is

= \frac{n^{1−k/2}}{e^{2πr}} \frac{ψ(N)2^{k−1}Γ(s)(2πrn)^{k−s−1}}{(k − 2)!} ∑_{m| gcd(n,r)} \frac{m^{2s−k+1}}{ω′(m)} .

Proof. The absolute convergence will be proven in Prop. 3.10 below. For s as given, we factorize the integral as I_{1}(f)_{∞}I_{1}(f)_{fin}. To start with,

I_{1}(f)_{fin} = ∫_{A_{fin}^{∗}} ∫_{A_{fin}} f^{n}((\begin{matrix}y & −x \\ 0 & 1 \end{matrix}))θ_{fin}(rx)dx|y|_{fin}^{s−k/2}d^{∗}y.

The value of f^{n} is nonzero if and only if there exists m ∈ Q^{+} such that ( \begin{matrix}my & −mx \\ 0 & m \end{matrix} ) ∈ M(n, N). In particular, m ∈ Ẑ ∩ Q^{+} = Z^{+}. Furthermore,

(i) my ∈ Ẑ (ii) m^{2}y ∈ nẐ^{∗} (iii) mx ∈ Ẑ.

Together, the first two conditions imply that m|n. Conversely, if m|n, condition (ii) implies condition (i). Assuming that m|n and y satisfies (ii), we have

∫_{A_{fin}} f^{n}(( \begin{matrix}y & −x \\ 0 & 1 \end{matrix} ))θ_{fin}(rx)dx = \frac{ψ(N)}{ω(m_{N})} ∫_{_{\frac{1}{m}} Ẑ} θ_{fin}(rx)dx.

Because m|n, it follows that (m, N) = 1, so ω(m_{N}) = ω′(m). Hence the above is

= \begin{cases}mψ(N)/ω′(m) & if m|r \\ 0 & otherwise. \end{cases}

Thus

∫_{A_{fin}} f^{n}((\begin{matrix}y & −x \\ 0 & 1 \end{matrix}))θ_{fin}(rx)dx = \begin{cases}mψ(N)/ω′(m) & if y ∈ \frac{n}{m^{2}} Ẑ^{∗} for \\ & some m| gcd(n, r), \\ 0 & otherwise. \end{cases}

We note that if such m exists, it is uniquely determined by y. Now

I_{1}(f)_{fin} = ∑_{m| gcd(n,r)} \frac{mψ(N)}{ω′(m)} ∫_{_{\frac{n}{m^{2}}} Ẑ^{∗}} |y|_{fin}^{s−k/2}d^{∗}y = ψ(N) ∑_{m| gcd(n,r)} \frac{m(m^{2}/n)^{s−k/2}}{ω′(m)} .

For the infinite part, recall that f_{∞} vanishes on matrices with negative determinant. Thus

I_{1}(f)_{∞} = ∫_{0}^{∞} ∫_{R} f_{∞}((\begin{matrix}y & −x \\ 0 & 1 \end{matrix}))θ_{∞}(rx)dx |y|^{s−k/2}d^{∗}y.

We have

∫_{R} f_{∞}(( \begin{matrix}y & −x \\ 0 & 1 \end{matrix} ))θ_{∞}(rx)dx = \frac{k − 1}{4π}y^{k/2}(2i)^{k} ∫_{−∞}^{∞} \frac{e^{−2πirx}}{(x + (y + 1)i)^{k}} dx.

Use a clockwise semicircular contour integral in the lower complex half-plane. The integrand has a pole at x = −(y + 1)i inside the contour. By the residue theorem, the above is

= −\frac{k − 1}{4π}y^{k/2}(2i)^{k}\frac{2πi}{(k − 1)!} \frac{d^{k−1}}{dx^{k−1}} |_{x=−(y+1)i} e^{−2πirx}

= −\frac{k − 1}{4π}y^{k/2}(2i)^{k}\frac{2πi}{(k − 1)!}(−2πir)^{k−1}e^{−2πr(y+1)} = \frac{(4πr)^{k−1}}{(k − 2)! e^{2πr}} y^{k/2}e^{−2πry}.

Therefore using Re(s) > 0,

I_{1}(f)_{∞} = \frac{(4πr)^{k−1}}{(k − 2)!e^{2πr}} ∫_{0}^{∞} y^{s−1}e^{−2πry}dy = \frac{(4πr)^{k−1}}{(k − 2)!e^{2πr}} (2πr)^{−s}Γ(s).

All together we have

I_{1}(f) = \frac{ψ(N)2^{k−1}Γ(s)n^{k/2−s}(2πr)^{k−s−1}}{(k − 2)! e^{2πr}} ∑_{m| gcd(n,r)} \frac{m^{2s−k+1}}{ω′(m)} .

□

Next we need to compute I_{δ}(f) for δ = ( \begin{matrix}0 & 1 \\ −1 & t \end{matrix} ) with t ∈ Q. We begin with the special case t = 0.

Proposition 3.5. If δ = ( \begin{matrix}& 1 \\ −1 \end{matrix} ), then

(7)  I_{δ}(f) = ∫_{A^{∗}} ∫_{A} f((\begin{matrix}yx & 1 \\ −y & 0 \end{matrix})) θ(rx)dx|y|^{s−k/2}d^{∗}y

converges absolutely for 1 < Re(s) < k. For such s, I_{δ}(f) = 0 unless N = 1. When N = 1,

I_{δ}(f) = \frac{n^{1−k/2}}{e^{2πr}} \frac{2^{k−1}Γ(k − s)(2πrn)^{s−1}}{(k − 2)! i^{k}} ∑_{m| gcd(n,r)} m^{k−2s+1}.

Proof. For the absolute convergence, see Prop. 3.10 below. The value of f^{n} in I_{δ}(f)_{fin} = ∫_{A_{fin}^{∗}} ∫_{A_{fin}} f^{n}(( \begin{matrix}yx & 1 \\ −y & 0 \end{matrix} ))θ_{fin}(rx)dx|y|_{fin}^{s−k/2}d^{∗}y is nonzero if and only if there

exists m ∈ Q^{+} such that ( \begin{matrix}myx & m \\ −my & 0 \end{matrix} ) ∈ M(n, N). This means m ∈ Z^{+}, my ∈ NẐ and m^{2}y ∈ nẐ^{∗}. It follows that N|n, which is only possible if N = 1. Assuming N = 1, we have m|n. The last requirement for nonvanishing is x ∈ \frac{1}{my} Ẑ = \frac{m}{n}Ẑ, in which case f^{n}(( \begin{matrix}yx & 1 \\ −y & 0 \end{matrix} )) = 1. Hence for fixed m|n and y ∈ \frac{n}{m^{2}} Ẑ^{∗},

∫_{A_{fin}} f^{n}((\begin{matrix}yx & 1 \\ −y & 0 \end{matrix})) θ_{fin}(rx)dx = ∫_{_{\frac{m}{n}} Ẑ} θ_{fin}(rx)dx = \begin{cases}n/m & if \frac{rm}{n} ∈ Ẑ \\ 0 & otherwise. \end{cases}

Now we have

I_{δ}(f)_{fin} = ∑_{{m|n, \atop \frac{n}{m} |r}} \frac{n}{m} ∫_{_{\frac{n}{m^{2}}} Ẑ^{∗}} |y|_{fin}^{s−k/2}d^{∗}y = ∑_{{m|n, \atop \frac{n}{m} |r}} \frac{n}{m}(m^{2}/n)^{s−k/2}

= n^{k/2−s+1} ∑_{{m|n, \atop \frac{n}{m} |r}} m^{2s−k−1} = n^{k/2−s+1} ∑_{m| gcd(n,r)}(n/m)^{2s−k−1}

= n^{s−k/2} ∑_{m| gcd(n,r)} m^{k−2s+1}.

For the infinite part I_{δ}(f)_{∞} = ∫_{R^{∗}} ∫_{R} f_{∞}(( \begin{matrix}yx & 1 \\ −y & 0 \end{matrix} ))θ_{∞}(rx)dx|y|^{s−k/2}d^{∗}y, as before we can assume y > 0. We have

∫_{R} f_{∞}((\begin{matrix}yx & 1 \\ −y & 0 \end{matrix}))e^{−2πirx}dx = \frac{k − 1}{4π}y^{k/2}(2i)^{k} ∫_{−∞}^{∞} \frac{e^{−2πirx}}{(−1 − y + (yx)i)^{k}} dx

= \frac{(k − 1)(2i)^{k}}{4π} y^{k/2}(iy)^{−k} ∫_{−∞}^{∞} \frac{e^{−2πirx}}{(x + (\frac{1+y}{y})i)^{k}} dx.

Take a clockwise semicircular contour integral in the lower half-plane. The integrand has a pole at x = −i(1 + \frac{1}{y}). By the residue theorem the above is

= −\frac{(k − 1)2^{k}}{4π} y^{−k/2} \frac{2πi}{(k − 1)!} \frac{d^{k−1}}{dx^{k−1}} |_{x=−i(1+ \frac{1}{y})} e^{−2πirx}

= \frac{−i2^{k−1}}{(k − 2)!} y^{−k/2}(−2πir)^{k−1}e^{−2πr(1+1/y)}

= \frac{(4πr)^{k−1}e^{−2πr}}{(k − 2)! i^{k}} y^{−k/2}e^{−2πr/y}.

Therefore

I_{δ}(f)_{∞} = \frac{(4πr)^{k−1}e^{−2πr}}{(k − 2)! i^{k}} ∫_{0}^{∞} y^{s−k−1}e^{−2πr/y}dy.

For any α > 0, ∫_{0}^{∞} t^{w−1}e^{−α/t}dt = α^{w}Γ(−w) when Re(w) < 0, so we get

I_{δ}(f) = \frac{(4πr)^{k−1}e^{−2πr}}{(k − 2)! i^{k}} (2πr)^{s−k}Γ(k − s)n^{s−k/2} ∑_{m| gcd(n,r)} m^{k−2s+1}.

□

For the case of δ = ( \begin{matrix}0 & 1 \\ −1 & t \end{matrix} ) with t ∈ Q^{∗}, we use the following lemma, which is very easy to prove.

Lemma 3.6. For any n, m, r ∈ Ẑ,

rẐ ∩ (n + mẐ) = \begin{cases}rc_{0} + \frac{rm}{gcd(r,m)} Ẑ & if gcd(r, m)|n \\ ∅ & if gcd(r, m) ∤ n, \end{cases}

where c_{0} ∈ Z is any fixed solution to rc_{0} ≡ n mod mẐ.

We also need to recall the definition of the confluent hypergeometric function

1F_{1}(s; k; w) = ∑_{m=0}^{∞} \frac{(s)_{m}}{(k)_{m}} \frac{w^{m}}{m!}

where (s)_{0} = 1 and for m > 0, (s)_{m} = s(s + 1)(s + 2) ⋯ (s + m − 1). This is absolutely convergent for all s, k, w ∈ C, except when k is a nonpositive integer. We have the following useful integral representation:

(8) _{1}F_{1}(s; k; w) = \frac{Γ(k)}{Γ(k − s)Γ(s)} ∫_{0}^{1} e^{wt}t^{s−1}(1 − t)^{k−s−1}dt (Re(k) > Re(s) > 0)

(see [Sl], §3.1).

Proposition 3.7. If δ = ( \begin{matrix}0 & 1 \\ −1 & t \end{matrix} ) for t ∈ Q^{∗}, then I_{δ}(f) is absolutely convergent when 0 < Re(s) < k. It vanishes unless t ∈ \frac{N}{n} Z. For such t, write t = \frac{N}{n} b. Then

I_{δ}(f) = \frac{(4πr)^{k−1}ψ(N)n^{k/2}}{(k − 2)! e^{iπs/2}e^{2πr}N^{s}} b^{s−k}_{1}f_{1}(s; k; \frac{2πirn}{Nb}) ∑_{{d|b \atop gcd(b/d,Nd)| gcd(r,n)}} \frac{gcd(b/d, Nd)}{d^{2s−k}ω′(b/d)}e^{− \frac{2πirℓ_{0}}{b/d}} ,

where ℓ_{0} ∈ Z is any integer satisfying ℓ_{0}(Nd) ≡ n mod (b/d), and

1f_{1}(s; k; w) = \frac{Γ(s)Γ(k − s)}{Γ(k)}_{1}F_{1}(s; k; w).

When b < 0, we take b^{s−k} = |b|^{s−k}e^{iπ(s−k)}.

Proof. The absolute convergence will be proven in Prop. 3.9 below. We can factorize the integral as I_{δ}(f) = I_{δ}(f)_{∞}I_{δ}(f)_{fin}. First we compute

I_{δ}(f)_{fin} = ∫_{A_{fin}^{∗}} ∫_{A_{fin}} f^{n}((\begin{matrix}yx & 1 − tx \\ −y & t \end{matrix}))θ_{fin}(rx)dx|y|^{s−k/2}d^{∗}y.

Suppose f^{n}(( \begin{matrix}yx & 1−tx \\ −y & t \end{matrix} )) ≠ 0. Then there exists m ∈ Q^{+} such that

(\begin{matrix}myx & m − mtx \\ −my & mt \end{matrix}) ∈ M(n, N).

This means:

(i) my ∈ NẐ  (iv) mxy ∈ Ẑ
(ii) m^{2}y ∈ nẐ^{∗}  (v) m − mtx ∈ Ẑ.
(iii) mt ∈ Ẑ

The first two conditions imply that m = \frac{n}{Nd} for some integer d > 0, and that y ∈ \frac{N^{2}d^{2}}{n} Ẑ^{∗}. By the third condition, t ∈ \frac{Nd}{n}Ẑ, or equivalently, t ∈ \frac{N}{n} Z and d|\frac{n}{N} t. This proves the first assertion. Condition (iv) is now equivalent to x ∈ \frac{1}{Nd} Ẑ. Conversely, if m, y, t, x are given in this way, they will satisfy (i)-(iv). Thus we have

I_{δ}(f)_{fin} = ∑_{d| \frac{n}{N} t} \frac{n^{s−k/2}}{(Nd)^{2s−k}} ∫_{_{\frac{N^{2}d^{2}}{n}} Ẑ^{∗}} ∫_{_{\frac{1}{Nd}} Ẑ} f^{n}(( \begin{matrix}yx & 1−tx \\ −y & t \end{matrix} ))θ_{fin}(rx)dx d^{∗}y.

Write t = \frac{N}{n} b for nonzero b ∈ dZ. Then mt = b/d, so the fifth condition is equivalent to x ∈ \frac{n}{Nb} + \frac{d}{b} Ẑ. Thus the inner integral is taken over

x ∈ \frac{1}{Nd}Ẑ ∩ (\frac{n}{Nb} + \frac{d}{b}Ẑ).

By Lemma 3.6 (multiply the above through by Nb), this set is nonempty if and only if gcd(b/d, Nd)|n, in which case it is equal to \frac{1}{Nd}c_{0} + \frac{1}{gcd(b/d,Nd)} Ẑ, where c_{0} is any solution to (b/d)c_{0} ≡ n mod Nd.

Note that gcd(b/d, Nd)|n implies that b/d is prime to N. Therefore the value of f^{n} in the integrand is \frac{ψ(N)}{ω′(b/d)}. Thus

I_{δ}(f)_{fin} = ∑_{{d|b \atop gcd(b/d,Nd)|n}} \frac{n^{s−k/2}}{(Nd)^{2s−k}} \frac{ψ(N)}{ω′(b/d)} ∫_{\frac{1}{Nd} c_{0}+ \frac{1}{gcd(b/d,Nd)} Ẑ} θ_{fin}(rx)dx

(9)  = ∑_{{d|b \atop gcd(b/d,Nd)|n}}  \frac{n^{s−k/2}}{(Nd)^{2s−k}} \frac{ψ(N)}{ω′(b/d)}θ_{fin}(\frac{rc_{0}}{Nd}) ∫_{_{\frac{1}{gcd(b/d,Nd)}} Ẑ} θ_{fin}(rx)dx

= \frac{ψ(N)n^{s−k/2}}{N^{2s−k}} ∑_{{d|b \atop gcd(b/d,Nd)| gcd(r,n)}} \frac{gcd(b/d, Nd)}{d^{2s−k}ω′(b/d)}e^{2πirc_{0}/Nd}.

For the archimedean part, the inner integral is

∫_{R} f_{∞}(( \begin{matrix}yx & 1−tx \\ −y & t \end{matrix} ))e^{−2πirx}dx

= \frac{k − 1}{4π}(2i)^{k}y^{k/2} ∫_{−∞}^{∞} \frac{e^{−2πirx}}{(tx − 1 − y + (yx + t)i)^{k}} dx

= \frac{k − 1}{4π}(2i)^{k}y^{k/2}(t + iy)^{−k} ∫_{−∞}^{∞} \frac{e^{−2πirx}}{(x − \frac{1+y−it}{i(y−it)})^{k}} dx.

The integrand has a pole at x = −i(1 + \frac{1}{y−it}) in the lower half-plane. Using a clockwise lower semicircular contour integral, this is

= −\frac{k − 1}{4π}(2i)^{k}\frac{2πi}{(k − 1)!}(−2πir)^{k−1}y^{k/2}(i)^{−k}(y − it)^{−k}e^{−2πr(1+ \frac{1}{y−it})}.

Thus

I_{δ}(f)_{∞} = \frac{(4πr)^{k−1}}{(k − 2)! i^{k}e^{2πr}} ∫_{0}^{∞} y^{s−1}(y − it)^{−k}e^{−2πr/(y−it)}dy.

This has an essential singularity at y = it. We define y^{s−1} as a holomorphic function of y by taking the principal value of log y on the positive real axis, and making a branch cut along the positive imaginary axis if t > 0 or the negative imaginary axis if t < 0. Now pulling out t and making a change of variables, we get

I_{δ}(f)_{∞} = \frac{(4πr)^{k−1}t^{s−k}}{(k − 2)! i^{k}e^{2πr}} ∫_{0}^{±∞} y^{s−1}(y − i)^{−k}e^{−2πr/t(y−i)}dy,

where the sign in the upper limit is the sign of t, and by our choice of branch, t^{s−k} = |t|^{s−k}e^{iπ(s−k)} if t < 0. In the notation of the next lemma below, the integral is G(s, k, r/t). By the result of the lemma and setting t = Nb/n, this gives

I_{δ}(f)_{∞} = \frac{(4πr)^{k−1}N^{s−k}}{(k − 2)! e^{iπs/2}e^{2πr}n^{s−k}} \frac{b^{s−k}_{1}f_{1}(s; k; 2πirn/Nb)}{e^{2πirn/Nb}}.

When we multiply this by I_{δ}(f)_{fin}, we can combine the terms

e^{−2πirn/Nb}e^{2πirc_{0}/Nd} = e^{2πir(c_{0}(b/d)−n)/Nb}.

Writing c_{0}(b/d) − n = −Ndℓ_{0} for some ℓ_{0} ∈ Z, we have Ndℓ_{0} ≡ n mod (b/d), and the above is equal to e^{−2πirℓ_{0}/(b/d)}. The result now follows. □

Lemma 3.8. For s, w ∈ C and k ∈ Z^{+}, define

G(s, k, w) = ∫_{0}^{∞} y^{s−1}(y − i)^{−k}e^{−2πw/(y−i)}dy.

This function converges absolutely for 0 < Re(s) < k. On this strip we can represent G(s, k, w) in terms of the confluent hypergeometric function:

G(s, k, w) = i^{k}e^{−iπs/2}e^{−2πiw} \frac{Γ(s)Γ(k − s)}{Γ(k)} _{1}F_{1}(s; k; 2πiw).

Furthermore, the integral defining G is unchanged if we replace ∞ by −∞.

Proof. Let t = 1 + \frac{i}{y−i}, so that y − i = \frac{−i}{1−t}. This linear fractional transformation takes the positive real axis to the upper semicircle C of radius 1/2 centered at z = 1/2. Then dy = \frac{−i}{(1−t)^{2}} dt and

G(s, k, w) = ∫_{C} (\frac{−it}{1 − t})^{s−1} (\frac{−i}{1 − t})^{−k} e^{2πiw(t−1)} \frac{−i}{(1 − t)^{2}} dt.

We define y^{s−1} = e^{(s−1) log y} by taking the principal value of log y for y > 0, and making a cut along the positive imaginary axis in the y-plane. This cut corresponds in the t-plane to cuts on the real axis from 0 to −∞ and from 1 to ∞. We choose log(−i) = −iπ/2, and choose the principal branches of log(t) and log(1 − t). Then for t ∈ (0, 1), −3π/2 < arg(−it/(1 − t)) < π/2 and therefore these choices are compatible with the choice of log y. Now

G(s, k, w) = e^{−isπ/2}(−i)^{−k}e^{−2πiw} ∫_{C} e^{2πiwt}t^{s−1}(1 − t)^{k−s−1}dt.

The integrand is holomorphic in t and single-valued in the cut plane, and by Cauchy’s theorem, its integral around the following contour vanishes:

C ε  1 − ε

Using the fact that 0 < Re(s) < k, it is straightforward to show that the contribution along the small arcs goes to 0 as ε → 0. It follows that the integral along C can instead be taken along the real axis, so

G(s, k, w) = i^{k}e^{−iπs/2}e^{−2πiw} ∫_{0}^{1} e^{2πiwt}t^{s−1}(1 − t)^{k−s−1}dt.

= i^{k}e^{−iπs/2}e^{−2πiw} \frac{Γ(k − s)Γ(s)}{Γ(k)}_{1}F_{1}(s; k; 2πiw)

by (8). If the upper limit of G is replaced by −∞, then t will instead traverse the lower semicircle \overline{C} from 0 to 1, which can likewise be moved to the real axis. In fact a more general path independence property can be proven in a similar way. □

3.3. Proof of Proposition 3.3. For each δ, we set

I_{δ}^{abs}(f) = ∫_{A^{∗}} ∫_{A} |f(( _{1}^{1 −x} )δ( \begin{matrix}y \\ & 1 \end{matrix} ))\overline{θ_{r}(x)}|y|^{s−k/2}| dx d^{∗}y.

Because f^{n} is compactly supported modulo the center and bounded by ψ(N), the finite part I_{δ}^{abs}(f)_{fin} converges for all s to a value depending on δ. Thus we primarily need to consider the infinite part

I_{δ}^{abs}(f)_{∞} = ∫_{0}^{∞} ∫_{−∞}^{∞} |f_{∞}(( _{1}^{1 −x} )δ( \begin{matrix}y \\ & 1 \end{matrix} ))| dx y^{Re(s)−k/2−1}dy.

We will repeatedly use the fact that for g = ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ G(R)^{+},

(10)  |f_{∞}(g)| = \frac{k − 1}{4π} \frac{det(g)^{k/2}2^{k}}{(a^{2} + b^{2} + c^{2} + d^{2} + 2 det(g))^{k/2}} .

This follows easily from the explicit formula for f_{∞}.

Proposition 3.9. Let δ_{t} = ( \begin{matrix}0 & 1 \\ −1 & t \end{matrix} ) for t ∈ Q^{∗}. Then if 0 < Re(s) < k, (a) I_{δ_{t}}^{abs}(f) < ∞

(b) I_{δ_{t}}^{abs}(f)_{∞} ≪ |t|^{Re(s)−k}.

Furthermore, if 1 < Re(s) < k − 1, then

(c) ∑_{t∈Q^{∗}} I_{δ_{t}}^{abs}(f) < ∞.

Proof. We need to estimate the expression

∫_{0}^{∞} ∫_{−∞}^{∞} |f_{∞}((\begin{matrix}yx & 1 − tx \\ −y & t \end{matrix})| dx y^{Re(s)−k/2−1}dy.

By (10), the inner integral is

≪ y^{k/2} ∫_{−∞}^{∞} \frac{dx}{(y^{2}x^{2} + y^{2} + t^{2} + (1 − tx)^{2} + 2y)^{k/2}}

= y^{k/2}(t^{2} + y^{2})^{−k/2} ∫_{−∞}^{∞} \frac{dx}{(x^{2} − \frac{2t}{t^{2}+y^{2}} x + \frac{1+t^{2}+y^{2}+2y}{t^{2}+y^{2}} )^{k/2}}

We will show that the integral is bounded, independently of y and t. Completing the square, the integral is equal to

∫_{−∞}^{∞} \frac{dx}{((x − \frac{t}{t^{2}+y^{2}} )^{2} + \frac{(1+y)^{2}+t^{2}}{t^{2}+y^{2}} − \frac{t^{2}}{(t^{2}+y^{2})^{2}} )^{k/2}} = ∫_{−∞}^{∞} \frac{dx}{(x^{2} + \frac{(1+2y+y^{2}+t^{2})(t^{2}+y^{2})−t^{2}}{(t^{2}+y^{2})^{2}} )^{k/2}}

= ∫_{−∞}^{∞} \frac{dx}{(x^{2} + \frac{(t^{2}+y^{2}+y)^{2}}{(t^{2}+y^{2})^{2}} )^{k/2}} < ∫_{−∞}^{∞} \frac{dx}{(x^{2} + 1)^{k/2}} < ∞.

Therefore writing s = σ + iτ,

I_{δ_{t}}^{abs}(f)_{∞} ≪ ∫_{0}^{∞} y^{σ−1}(t^{2} + y^{2})^{−k/2}dy.

For convergence as y → 0, we need σ − 1 > −1, i.e. σ > 0. For convergence as y → ∞, we need σ − 1 − k < −1, i.e. σ < k. This proves the absolute convergence of I_{δ_{t}}(f) on the given strip.

In order to sum over t, we need to bound the above integral in terms of t. We have

I_{δ_{t}}^{abs}(f)_{∞} ≪ ∫_{0}^{∞} y^{σ−1}|t|^{−k}(1 + \frac{y^{2}}{t^{2}} )^{−k/2}dy

= |t|^{−k} ∫_{0}^{∞} (\frac{y^{2}}{t^{2}} )^{\frac{σ}{2}} |t|^{σ}(1 + \frac{y^{2}}{t^{2}} )^{−k/2}d^{∗}y.

Letting u = (y/t)^{2} so d^{∗}u = 2d^{∗}y, the above is

= \frac{1}{2}|t|^{σ−k} ∫_{0}^{∞} \frac{u^{\frac{σ}{2} −1}}{(1 + u)^{k/2}} du,

which proves the second assertion since k > σ > 0. (As an aside, this last integral equals B(\frac{σ}{2} , \frac{k−σ}{2} ) where B(n, m) = Γ(n)Γ(m)/Γ(n + m) is the Beta function.)

As in the proof of Proposition 3.7, I_{δ_{t}}^{abs}(f)_{fin} vanishes unless t = \frac{N}{n} b for some b ∈ Z − {0}. By (9), we see that

I_{δ_{t}}^{abs}(f)_{fin} ≪ \frac{n^{σ−k/2}ψ(N)}{N^{2σ−k}} ∑_{d|b} d^{−2σ+k}.

If σ > k/2, then d^{−2σ+k} ≤ 1. If σ ≤ k/2, then d^{−2σ+k} ≤ |b|^{−2σ+k}. The number of divisors of b is ≪ b^{ε} for any ε > 0. Since I_{δ_{t}}^{abs}(f)_{∞} contributes |b|^{σ−k}, we have

(11)  ∑_{t∈Q^{∗}} I_{δ_{t}}^{abs}(f) ≪ \begin{cases}∑_{b∈Z−\{0\}} |b|^{−σ+ε} & if σ ≤ k/2 \\ ∑_{b∈Z−\{0\}} |b|^{σ−k+ε} & if σ > k/2. \end{cases}

Hence ∑_{t} I_{δ_{t}}^{abs}(f) < ∞ as long as 1 < σ < k − 1. □

The following will complete the proof of Proposition 3.3.

Proposition 3.10. For δ = 1, I_{1}^{abs}(f) < ∞ provided

0 < Re(s) < k − 1.

For δ = ( \begin{matrix}& 1 \\ −1 \end{matrix} ), I_{δ}^{abs}(f) < ∞ provided

1 < Re(s) < k.

Proof. For any a > 0, a change of variables gives

(12)  ∫_{−∞}^{∞} \frac{dx}{(x^{2} + a^{2})^{k/2}} = a^{−k+1} ∫_{−∞}^{∞} \frac{du}{(u^{2} + 1)^{k/2}} .

We again write s = σ + iτ. When δ = 1, using (10) we have

I_{1}^{abs}(f) ≪ ∫_{0}^{∞} ∫_{−∞}^{∞} \frac{y^{σ−1}}{(x^{2} + y^{2} + 2y + 1)^{k/2}} dx dy.

By (12), this is

≪ ∫_{0}^{∞} y^{σ−1}(y + 1)^{−k+1}dy.

This converges precisely when 0 < σ < k − 1.

Similarly, for δ = ( \begin{matrix}& 1 \\ −1 \end{matrix} ),

I_{δ}^{abs}(f) ≪ ∫_{0}^{∞} ∫_{−∞}^{∞} \frac{y^{σ−1}}{(x^{2}y^{2} + y^{2} + 2y + 1)^{k/2}} dx dy

= ∫_{0}^{∞} y^{σ−1−k} ∫_{−∞}^{∞} \frac{dx}{(x^{2} + (1 + \frac{1}{y})^{2})^{k/2}} dy

≪ ∫_{0}^{∞} y^{σ−k−1}(1 + y^{−1})^{−k+1}dy.

As y → 0, we need σ − k − 1 + k − 1 > −1, i.e. σ > 1. As y → ∞, we need σ − k − 1 < −1, i.e. σ < k. This proves the proposition. □

3.4. Proof of Theorem 1.1. We have now proven that the geometric side converges absolutely when 1 < Re(s) < k − 1, and therefore it is equal to the spectral side on this strip. When we sum the contribution of Prop. 3.7 over all b ≠ 0, we set a = b/d so that b = ad. Then

∑_{b≠0} b^{s−k}_{1}f_{1}(s; k; \frac{2πirn}{Nb}) ∑_{{d|b \atop gcd(b/d,Nd)| gcd(r,n)}} \frac{gcd(b/d, Nd)}{d^{2s−k}ω′(b/d)}e^{− \frac{2πirℓ_{0}}{b/d}}

= ∑_{{a≠0,d>0 \atop gcd(a,Nd)| gcd(r,n)}} \frac{a^{s−k}d^{−s} gcd(a, Nd)}{ω′(a)e^{2πirℓ_{0}/a}} _{1}f_{1}(s; k; \frac{2πirn}{Nad}).

The theorem now follows immediately upon equating the two sides of the trace formula and dividing through by e^{−2πr}n^{1−k/2}.

4. Estimates and Examples

4.1. Asymptotic behavior. For two functions A, B, we write A ∼ B to mean that A/B → 1 in a limiting sense which will be clear from the context. For example, by Stirling’s approximation we have the following:

(13)  Γ(z + b) ∼ \sqrt{2π}e^{−z}z^{z+b−1/2}  (z → ∞, | arg z| < π)

([AS], 6.1.39). The ∼ notation here depends on b; i.e. given ε > 0 there is a constant N(b) > 0 such that the quotient is within ε of 1 whenever |z| > N(b).

We now estimate each term of Theorem 1.1 as k → ∞. It will turn out that the first two terms are dominant, provided their sum does not vanish. In order to ensure nonvanishing of ∑_{m| gcd(n,r)} m^{2s−k+1}/ω′(m), we will assume for simplicity that gcd(n, r) = 1. However in general one can prove that this sum can only vanish on the left edge of the critical strip, i.e. on the line Re(s) = \frac{k−1}{2}.

Proposition 4.1. Let s = k/2 + α + iτ, with 1 < k/2 + α < k − 1. Assume gcd(n, r) = 1. Then as k → ∞ the identity term in Theorem 1.1 satisfies

(14)  |\frac{ψ(N)2^{k−1}Γ(s)(2πrn)^{k−s−1}}{(k − 2)!} | ∼ \frac{2\sqrt{π} ψ(N)(4πrn)^{k/2−α−1}k^{k/2+α−1/2}}{(k − 2)! e^{k/2}}

∼ \sqrt{2}ψ(N)e^{α+1} (\frac{4πrne}{k})^{k/2−α−1} .

If N = 1, then as k → ∞ the second term in Theorem 1.1 satisfies

(15)  |\frac{2^{k−1}Γ(k − s)(2πrn)^{s−1}}{(k − 2)! i^{k}}| ∼ \frac{2\sqrt{π}(4πrn)^{k/2+α−1}k^{k/2−α−1/2}}{(k − 2)! e^{k/2}}

∼ \sqrt{2}e^{−α+1} (\frac{4πrne}{k})^{k/2+α−1} .

Remark: The ∼ notation here depends on α + iτ as discussed after (13).

Proof. Using (13), the lefthand side of (14) is

=\frac{ψ(N)2^{k−1}|Γ(k/2 + α + iτ)|(2πrn)^{k/2−α−1}}{(k − 2)!}

∼ \frac{ψ(N)2^{−1}2^{k}(2πrn)^{k/2−α−1}\sqrt{2π}e^{−k/2}(k/2)^{k/2+α−1/2}}{(k − 2)!} .

For the second line of (14) we substitute (k − 2)! = Γ(k − 1) ∼ \sqrt{2π}e^{−k}k^{k−1−1/2}. The second estimate is similar, as the lefthand side of (15) is

∼ \frac{2^{−1}2^{k}(2πrn)^{k/2+α−1}e^{−k/2}(k/2)^{k/2−α−1/2}}{(k − 2)!} .

□

We now show that the third term in Theorem 1.1 decays much more rapidly in comparison with the first terms as k → ∞. We can rewrite it as a sum over a, d > 0. Note that ω′(−a) = (−1)^{k}ω′(a). Thus the third term is equal to

\frac{ψ(N)(4πrn)^{k−1}}{N^{s}(k − 2)! e^{iπs/2}} ∑_{{a,d>0 \atop gcd(a,Nd)| gcd(r,n)}}[a^{s−k}d^{−s}_{1}f_{1}(s; k; \frac{2πirn}{Nad})e^{−2πirℓ_{0}/a} (16)

+ e^{iπs}a^{s−k}d^{−s}_{1}f_{1}(s; k; −\frac{2πirn}{Nad})e^{2πirℓ_{0}/a}] \frac{gcd(a, Nd)}{ω′(a)},

where ℓ_{0} is any integer satisfying ℓ_{0}Nd ≡ n mod a. Write s = σ + iτ. If w is real,

|_{1}f_{1}(s; k; 2πiw)| = |∫_{0}^{1} e^{2πiwt}t^{s−1}(1 − t)^{k−s−1}dt|

≤ ∫_{0}^{1} t^{σ−1}(1 − t)^{k−σ−1}dt = B(σ, k − σ) (17)

for the Beta function B. Furthermore, |e^{iπs/2}| = e^{−πτ/2}. Thus the absolute value of (16) is

≤ \frac{ψ(N)(4πrn)^{k−1} gcd(r, n)B(σ, k − σ)}{N^{σ}(k − 2)!}e^{πτ/2} ∑_{a,d>0} a^{−(k−σ)}d^{−σ} |1 + e^{iπs}| .

Note that |1 + e^{iπs}| ≤ (1 + e^{−πτ}). Pulling this out of the sum, we obtain (e^{πτ/2} + e^{−πτ/2}) = 2 cosh(τπ/2), and we immediately arrive at the following.

Proposition 4.2. Write s = σ + iτ for 1 < σ < k − 1. Then the absolute value of the last term (16) of Theorem 1.1 is

≤ \frac{ψ(N)(4πrn)^{k−1} gcd(r, n)B(σ, k − σ)}{N^{σ}(k − 2)!}2 cosh(τπ/2)ζ(k − σ)ζ(σ)

for the Beta function B and the Riemann zeta function ζ.

We remark that when 1 < Re(s) < k − 1 as is the case here, the integrand in (17) is smaller than 1 so 0 < B(σ, k − σ) < 1.

If we restrict s to the critical strip \frac{k−1}{2} < Re(s) < \frac{k+1}{2}, then both zeta values approach 1 as k → ∞. Therefore we see that if N > 1, the identity term is dominant as k → ∞. If N = 1, then I_{1}(f) is the main term when σ > k/2, while I_{( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} )}(f)

is the main term when σ < k/2.

Corollary 1.3 now follows easily. In fact we can make it effective. Assume N > 1, k > 3 and gcd(n, r) = 1. Let

F (s) = \frac{ψ(N)2^{k−1}Γ(s)(2πrn)^{k−s−1}}{(k − 2)!}

denote the first term of the geometric side of Theorem 1.1, and let T (s) denote the other term, given in (16). Clearly the average of L-values is nonzero whenever |T (s)| < |F (s)|. By Prop. 4.2, this holds whenever

\frac{ψ(N)(4πrn)^{k−1}B(σ, k − σ)}{N^{σ}(k − 2)!}2 cosh(πτ/2)ζ(k − σ)ζ(σ)

< \frac{ψ(N)2^{k−1}|Γ(s)|(2πrn)^{k−σ−1}}{(k − 2)!} .

Using B(σ, k − σ) = Γ(σ)Γ(k − σ)/(k − 1)!, the above is equivalent to

(18)  2 cosh(τπ/2) < (\frac{N}{2πrn})^{σ} \frac{(k − 1)! |Γ(s)|}{ζ(k − σ)ζ(σ)Γ(k − σ)Γ(σ)}.

Lemma 4.3. For any s = σ + iτ with σ > 1,

|\frac{Γ(s)}{Γ(σ)}| ≥ e^{−τ arg(s−1/2)} (\frac{σ − 1/2 − \frac{ln(2)}{π\sqrt{2πe}}}{σ − 1/2 + \frac{ln(2)}{π\sqrt{2πe}}} ) .

Proof. This follows immediately from the following approximation due to Spouge:

(19)  Γ(s) = \sqrt{2π}(s − 1/2)^{s−1/2}e^{−s+1/2}[1 + ε(s)]  (σ > 1),

where

|ε(s)| < \frac{\frac{ln(2)}{π\sqrt{2πe}}}{σ − 1/2}

([Sp], Theorem 1.3.2). We apply this to Γ(s) and Γ(σ), and use

|(s − 1/2)^{s−1/2}| = |s − 1/2|^{σ−1/2}e^{−τ arg(s−1/2)} ≥ (σ − 1/2)^{σ−1/2}e^{−τ arg(s−1/2)}.

□

By the lemma and (18), we see that the average of Theorem 1.1 is nonzero whenever

(20) 2 cosh(τπ/2)e^{τ arg(s−1/2)} < \frac{(\frac{N}{2πrn})^{σ} (k − 1)!}{ζ(k − σ)ζ(σ)Γ(k − σ)} (\frac{σ − 1/2 − \frac{ln(2)}{π\sqrt{2πe}}}{σ − 1/2 + \frac{ln(2)}{π\sqrt{2πe}}} ) .

We remark that since | arg(s − 1/2)| < π/2, the lefthand side is bounded above by 2 cosh(τπ/2)e^{|τ|π/2} = e^{π|τ|} + 1, which would simplify but weaken the inequality.

Since the lefthand side of (20) increases with |τ|, we obtain the following.

Proposition 4.4. Suppose N > 1, k > 3, and gcd(n, r) = 1. Fix τ_{0} > 0, and let R denote the set of s = σ + iτ with |τ| ≤ τ_{0} and \frac{k−1}{2} ≤ σ ≤ \frac{k+1}{2}. Then the average in Theorem 1.1 is nonzero at every point of R if

(21)  2 cosh(τ_{0}π/2)e^{τ_{0} tan^{−1}( \frac{τ_{0}}{k/2−1} )} < \frac{(\frac{N}{2πrn})^{\frac{k±1}{2}} (k − 1)!}{ζ(\frac{k−1}{2})^{2} Γ(\frac{k+1}{2})} ( \frac{\frac{k}{2} − 1 − \frac{ln(2)}{π\sqrt{2πe}}}{\frac{k}{2} + \frac{ln(2)}{π\sqrt{2πe}}} ) .

Here we choose \frac{k−1}{2} if N > 2πrn, and \frac{k+1}{2} otherwise.

Because the righthand side of (21) tends to ∞ as N + k → ∞, Corollary 1.3 follows immediately.

4.2. Zero-free regions. We can use Prop. 4.4 to find zero-free regions of certain modular L-functions. The idea is to apply the proposition with n = r = 1 when dim S_{k}(N, ω′) = 1, since the average then gives an actual L-value. The exponent of \frac{N}{2π} in (21) is \frac{k+1}{2} unless N ≥ 7.

Example 4.5. Let h denote the unique normalized cusp form in S_{10}(2). When n = r = 1, N = 2 and k = 10, the righthand side of (21) is 8.97346, and the inequality holds for τ_{0} = 1.169259. Hence the value of Λ(s, h) is nonzero for all s in the critical strip with | Im(s)| ≤ 1.169259.

Example 4.6. Let h denote the unique normalized cusp form in S_{8}(3). Then Λ(s, h) is nonzero for all s in the critical strip with | Im(s)| ≤ 1.119308.

Example 4.7. Let h denote the unique normalized cusp form in S_{6}(5). Then the value of Λ(s, h) is nonzero for all s in the critical strip with | Im(s)| ≤ 0.852608.

Example 4.8. According to Stein’s Modular Forms Database, there exists a Dirichlet character χ mod 7 (unique up to Galois conjugacy) for which dim S_{5}(7, χ) = 1. If h is the normalized cusp form, then Λ(s, h) is nonzero for all s in the critical strip with | Im(s)| ≤ 0.501352.

4.3. Approximation by partial sums. In order to estimate the geometric side, we can truncate the last term (16). Let A, D be positive integers. Define the partial sum

S_{A,D} = \frac{ψ(N)(4πrn)^{k−1}}{N^{s}(k − 2)! e^{iπs/2}} ∑_{{1≤a≤A,1≤d≤D \atop gcd(a,Nd)| gcd(r,n)}}[a^{s−k}d^{−s}_{1}f_{1}(s; k; \frac{2πirn}{Nad})e^{−2πirℓ_{0}/a}

+ e^{iπs}a^{s−k}d^{−s}_{1}f_{1}(s; k; −\frac{2πirn}{Nad})e^{2πirℓ_{0}/a}] \frac{gcd(a, Nd)}{ω′(a)}

where as usual ℓ_{0}Nd ≡ n mod a. The error is given by the tail of the series

Δ_{A,D} = \frac{ψ(N)(4πrn)^{k−1}}{N^{s}(k − 2)! e^{iπs/2}} ∑_{a,d>0 a>A or d>D gcd(a,Nd)| gcd(r,n)}[a^{s−k}d^{−s}_{1}f_{1}(s; k; \frac{2πirn}{Nad})e^{−2πirℓ_{0}/a}

+ e^{iπs}a^{s−k}d^{−s}_{1}f_{1}(s; k; −\frac{2πirn}{Nad})e^{2πirℓ_{0}/a}] \frac{gcd(a, Nd)}{ω′(a)}.

As in the proof of Prop. 4.2, we have the following bound for the error:

|Δ_{A,D}| ≤ \frac{ψ(N)(4πrn)^{k−1} gcd(r, n)B(σ, k − σ)}{N^{σ}(k − 2)!}2 cosh(πτ/2) ∑_{a>A or d>D gcd(a,Nd)| gcd(r,n)} a^{−(k−σ)}d^{−σ}.

We can estimate the error using the following easy lemma.

Lemma 4.9. For s = σ + iτ,

∑_{a>A or d>D gcd(a,Nd)| gcd(r,n)} a^{−(k−σ)}d^{−σ} ≤ ζ(k − σ)ζ(σ) − ∑_{a=1}^{A} a^{−(k−σ)} ∑_{d=1}^{D} d^{−σ}.

4.4. Computing the τ-function. As a simple example, consider Ramanujan’s Δ(z) = ∑_{n=1}^{∞} τ(n)e^{2πinz} ∈ S_{12}(1). Writing

(22)  τ(r) = \frac{τ(r)Λ(6, Δ)/∥Δ∥^{2}}{τ(1)Λ(6, Δ)/∥Δ∥^{2}} ,

we can use the geometric side of Theorem 1.1 to compute the top and bottom. Taking n = 1, let F (r) denote the sum of the first two terms of the formula for \frac{τ(r)Λ(6,Δ)}{∥Δ∥^{2}}. We find that

F (r) = \frac{2^{12}(2πr)^{5}5!}{10!}.

Let S_{A}(r) denote the A^{th} partial sum (taking A = D above) of the last term of the formula. Then \frac{τ(r)Λ(6,Δ)}{∥Δ∥^{2}} ≈ F (r) + S_{A}(r) with an error of

(23)  ≤ \frac{2(4πr)^{11}B(6, 6)}{10!} [ζ(6)^{2} − (∑_{a=1}^{A} \frac{1}{a^{6}} )^{2}]

by Lemma 4.9.

As an illustration, we will compute τ(2). To estimate the denominator of (22), take r = 1 and A = 1. This gives

\frac{Λ(6, Δ)}{∥Δ∥^{2}} ≈ \frac{2^{12}(2π)^{5}5!}{10!} − \frac{(4π)^{11}}{10!} [_{1}f_{1}(6; 12; 2πi) +_{1}f_{1}(6; 12; −2πi)] = 1492.55

with an error of ≤ 8.584. So the exact value is in the interval [1483, 1502]. For r = 2 we need to use A = 3 to get a reasonable approximation. We get

\frac{τ(2)Λ(6, Δ)}{∥Δ∥^{2}} ≈\frac{2^{12}(4π)^{5}5!}{10!} − \frac{(8π)^{11}}{10!} ∑_{{a,d∈\{1,2,3\} \atop gcd(a,d)=1}}(ad)^{−6}[_{1}f_{1}(6; 12; \frac{4πi}{ad})e^{−2πirℓ_{0}/a}

+_{1}f_{1}(6; 12; −\frac{4πi}{ad})e^{2πirℓ_{0}/a}]

= −35769.72.

By (23) the error here is

≤ \frac{2(8π)^{11}B(6, 6)}{(10)!} (ζ(6)^{2} − (1 + \frac{1}{2^{6}} + \frac{1}{3^{6}} )^{2}) = 354.008.

Thus the exact value is in the interval [−36124, −35415].

Taking the quotient of the estimates, we find that

\frac{−36124}{1483} ≤ τ(2) ≤ \frac{−35415}{1502},

i.e.

−24.359 ≤ τ(2) ≤ −23.578.

Because τ(2) is an integer, it must equal −24.

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Department of Mathematics & Statistics, University of Maine, Neville Hall, Orono, ME 04469-5752, USA

Department of Mathematics, The Chinese University of Hong Kong, Shatin, Hong Kong