On zeta functions associated to symmetric matrices (II): Functional equations and special values

Tomoyoshi Ibukiyama and Hiroshi Saito

In the previous paper [5], we gave an explicit formula for zeta functions associated to symmetric matrices. In this paper, we shall give functional equations of these zeta functions. We also give special values of the zeta functions at nonpositive integers , poles and residues. As an application, we give the contribution of “central unipotent” elements to the dimension of the space S_{k}(Γ_{n}(N)) of Siegel cusp forms belonging to the principal congruence subgroup Γ_{n}(N). This leads us to an explicit conjecture on dimensions of S_{k}(Γ_{n}(N)) for k > 2n and N ≥ 3 for arbitrary n.

Historically, functional equations for zeta functions associated with a large class of prehomogeneous vector spaces were investigated by M. Sato and T. Shintani (see [12]), and the special cases treated in this paper were studied by Shintani [16]. These are functional equations between n + 1 dimensional vectors each of whose components is the zeta function associated with each real orbit of the prehomogeneous vector spaces. Later, Satake and Faraut simplified these functional equations by giving a kind of diagonalization. (See [9], [8]) Since we have given the zeta functions explicitly in the previous paper, we can calculate everything explicitly and we can show that, in this diagonalization, only two, or three components are non-zero, if n is odd, or even, respectively. So, our functinal equations are much simpler than the previously known ones. Actually, we give a more direct proof of the functional equations independently of the arguments by Shintani or Satake-Faraut. In fact, when n is odd, the zeta functions are described by the Riemann zeta functions, so the functional equations are easily obtained. When n is even, our argument is based on the functional equations of the Mellin transform of real analytic Eisenstein series of half integral weights and we can get the results for n ≥ 4 by some standard arguments. The case n = 2 is the pathological case, since the volumes μ(x) in the definition of our zeta functions are not finite for some x in this case. Hence we need a modification of the definition of the zeta functions to obtain a good functional equation. Such modification has been done first by T. Shintani [16] and later by F. Sato [11] in more intrinsic way. Here we also give the third alternative proof of this functional equation for n = 2 from our standpoint. The prototype of the proof of this style is found in Sturm [18] for an easier half of our cases where n = 2. For all n ≥ 1, this direct approach allows us to evaluate special values, situation of poles, and their residues. This paper is organized as follows. In section 1, we state functional equations and poles and residues of our zeta functions for n ≥ 3. In section 2, we review the Fourier coefficients of the real analytic Eisenstein series of half integral weight, which is more or less known by [15], [2], [21], [18], [3]. In section 3, we introduce a Cohen type linear combination of Eisenstein series ([2]) for a general parameter. For even n ≥ 4, we use it to show the functional equation between Dirichlet series D_{n}^{∗}(s, δ) and D_{n}(s, δ) which appear in some part of our zeta functions. In section 4, we prove the functional equations for n ≥ 3. In section 5, we assume n = 2, and give the alternative proof described above. In section 3 and 5, we need some calculation on the Mellin transform of Eisenstein series with parameters. Goldfeld and Hoffstein [3] calculated this kind of Mellin transform for completely different purposes from ours. Our calculation is very similar to theirs in several points, but the details and aims are different. So, we write the proofs in detail here. The results in section 6 was announced before in [4], and we give here the proofs. We evaluate the values of our zeta functions at non positive integers. As a application, we give the contribution of central uniponent conjugacy classes to the dimension of Siegel cusp forms. This also leads to a conjecture for an explicit dimension formular for Siegel cusp forms, which will be stated there.

The first author would like to thank the Max Planck Institut für Mathematik in Bonn for its hospitality during his visit when the paper was revised.

1 Results on functional equations, poles and residues

In this section, we shall give results on functional equations between our zeta fnctions, their poles and residues. In this section, we mainly treat the case n ≥ 3. The proof will be given in section 4. The case n = 1 is the trivial case and will be omitted in this paper. The case n = 2 is somewhat special but has been known by Shintani [16] and F. Sato [11]. This case will be treated separately in section 5 from our stand point with a new proof.

First, we recall the results from the previous paper part I [5]. Let L_{n} be the set of n × n symmetric matrices and L_{n}^{∗} be the set of half integral symmetric matrices. For L = L_{n} or L_{n}^{∗}, our zeta function ζ_{i}(s, L) was originally defined for each i with 0 ≤ i ≤ n by a certain infinite sum over elements of L with signature (i, n − 1). We have shown in our previous paper that ζ_{i}(s, L) depends only on δ = (−1)^{n−i} and ϵ = (−1)^{(n−i)(n−i+1)/2} and denoted it by ζ(s, L, δ, ϵ). (As for the precise definition, see [5].) We review the explicit formula for this zeta function.

For the sake of simplicity, we put

b_{n} = \frac{| ∏_{i=1}^{[(n−1)/2]} B_{2i}|}{2^{n−1}([\frac{n−1}{2}])!}

where B_{k} is the k-th Bernoulli number defined by te^{t}/(e^{t} − 1) = ∑_{k=0}^{∞} B_{k}t^{k}/k!. When n is even, we also use the notation

Bn/2^{′} = 2(\frac{n}{2})!(2π)^{−n/2}ζ(\frac{n}{2})(−1)^{1+n/2}.

For odd n, we put

Q_{n}(s) = ζ(s − \frac{n − 1}{2}) ∏_{i=1}^{(n−1)/2} ζ(2s − (2i − 1)),

R_{n}(s) = ζ(s) ∏_{i=1}^{(n−1)/2} ζ(2s − 2i).

If n is odd with n ≥ 3, by [5] Theorem 1.2, we have

ζ(s, L_{n}^{∗}, δ, ϵ) = b_{n}2^{(n−1)s}(Q_{n}(s) + ϵδ^{(n+1)/2}(−1)^{(n^{2}−1)/8}R_{n}(s)),
ζ(s, L_{n}, δ, ϵ) = b_{n}(2^{(n−1)/2}Q_{n}(s) + ϵδ^{(n+1)/2}(−1)^{(n^{2}−1)/8}R_{n}(s)).

For even n, we put

A_{n}(s) = ∏_{i=1}^{n/2−1} ζ(2s − 2i), and B_{n}(s) = ∏_{i=1}^{n/2} ζ(2s − (2i − 1)).

If n is even with n ≥ 4 or (n, δ) = (2, 1), we also put

D_{n}^{∗}(s, δ) = (−1)^{[n/4]}2(2π)^{−n/2}Γ(n/2) ∑_{(−1)^{n/2}δd_{K}>0} |d_{K}|^{(n−1)/2−s}L(\frac{n}{2}, χ_{K})\frac{ζ(2s)ζ(2s − n + 1)}{L(2s − \frac{n}{2} + 1, χ_{K})},

where d_{K} runs through all the discriminant of quadratic fields K or Q ⊕ Q such that (−1)^{n/2}δD_{K} > 0 and χ_{K} is the Dirichlet character associated with K. Let D_{n}^{∗}(s, δ) = ∑_{d=1}^{∞} H(n/2, d, δ)d^{−s}. We define

D(s, δ) = ∑_{i=1}^{∞} H(\frac{n}{2}, 4d, δ)d^{−s}.

By [5] Theorem 1.3, we have

ζ(s, L_{n}^{∗}, δ, ϵ) = b_{n}2^{ns}((−1)^{[n/4]}D_{n}^{∗}(s, δ)A_{n}(s) + ϵδ_{n}(−1)^{n(n+2)/8} \frac{2|Bn/2^{′}|}{n}B_{n}(s))
ζ(s, L_{n}, δ, ϵ) = b_{n}((−1)^{[n/4]}D_{n}(s, δ)A_{n}(s) + ϵδ_{n}(−1)^{n(n+2)/8} \frac{2^{(n+2)/2}|Bn/2^{′}|}{n}B_{n}(s)),

where δ_{n} = 1 if (−1)^{n/2} ≡ δ mod 4, and δ_{n} = 0 otherwise.

Using the notation of Shintani [16], for any s ∈ C and a positive integer n, we put

γ_{n}(s) = ∏_{i=0}^{n−1} Γ(s + 1 + \frac{i}{2}).

When n is odd with n ≥ 3, then ζ(s, L, 1, ϵ) = ζ(s, L, −1, (−1)^{(n+1)/2}ϵ) for L = L_{n}, or L_{n}^{∗}. So, it is enough to give a functional equation between ζ(s, L, 1, 1) and ζ(s, L, 1, −1).

Theorem 1 When n is odd with n ≥ 3, then

( \begin{matrix}ζ(\frac{n+1}{2} − s, L_{n}^{∗}, 1, 1) \\ ζ(\frac{n+1}{2} − s, L_{n}^{∗}, 1, −1) \end{matrix} ) = 2^{−ns+(n^{2}−1)/2}π^{−ns+n(n−1)/4}γ_{n}(s − \frac{n + 1}{2})
( \begin{matrix}a(s) + b(s) & −a(s) + b(s) \\ a(s) − b(s) & −a(s) − b(s) \end{matrix} ) ( \begin{matrix}ζ(s, L_{n}, 1, 1) \\ ζ(s, L_{n}, 1, −1) \end{matrix} )

where

a(s) = (cos \frac{πs}{2})(cos πs)^{(n−1)/2}

and

b(s) = (−1)^{(n^{2}−1)/8}(sin πs)^{(n−1)/2} × \{ \begin{matrix}cos \frac{πs}{2}, & if n ≡ 1 mod 4, \\ sin \frac{πs}{2}, & if n ≡ 3 mod 4. \end{matrix}

Next, we give poles and residues of the above zeta functions. When s_{0} ∈ C is the possible simple pole of ζ(s, L_{n}^{∗}, δ, ϵ), or ζ(s, L_{n}, δ, ϵ), respectively, we denote by r_{n,δ,ϵ}(s_{0}), or r_{n,δ,ϵ}^{∗}(s_{0}) the residue of each zeta function at s_{0}, respectively. We understand that this value is 0, when the zeta function is holomorphic at s_{0}. When n is odd, we can determine the poles and residues exactly, as is shown in the following Theorem.

Theorem 2 We assume that n ≥ 3 is an odd integer. Then, the zeta function ζ(s, L_{n}^{∗}, δ, ϵ) or ζ(s, L_{n}, δ, ϵ) has possible poles only at s = l/2, where l is an integer with 2 ≤ l ≤ n + 1. The residues are given as follows.

(1) When n = 3, we get

\begin{aligned}r_{3,δ,ϵ}^{∗}(1) & = & (2ϵ − 1)b_{3}, \\ r_{3,δ,ϵ}(1) & = & (ϵ − 1)b_{3}/2. \end{aligned}

In particular, r_{3,δ,1}(1) = 0. When n > 3, we get r_{3,δ,ϵ}^{∗}(1) = r_{3,δ,ϵ}(1) = 0 , and the zeta functions are holomorphic at 1.

(2)For each natural integer j with 2 ≤ j ≤ (n − 1)/2, we get

r_{n,δ,ϵ}^{∗}(j) = 2^{(n−1)j−1}b_{n}ζ(j − (n − 1)/2) ∏_{i=1,i≠j}^{(n−1)/2} ζ(2j − 2i + 1),

r_{n,δ,ϵ}(j) = 2^{(n−3)/2}b_{n}ζ(j − (n − 1)/2) ∏_{i=1,i≠j}^{(n−1)/2} ζ(2j − 2i + 1).

In particular, if j < (n − 1)/2 and j ≡ (n − 1)/2 mod 2, then the zeta functions are holomorphic at s = j.

(3) We get

r_{n,δ,ϵ}^{∗}((n + 1)/2) = 2^{(n^{2}−1)/2}b_{n} ∏_{i=1}^{(n−1)/2} ζ(n − 2i + 2),

r_{n,δ,ϵ}((n + 1)/2) = 2^{(n−1)/2}b_{n} ∏_{i=1}^{(n−1)/2} ζ(n − 2i + 2).

(4)For each natural number j with 1 ≤ j ≤ (n − 1)/2, we get

r_{n,δ,ϵ}^{∗}(j + 1/2) = 2^{(n−1)(j+1/2)}δ^{(n+1)/2}ϵ(−1)^{(n^{2}−1)/8}b_{n}ζ(j + 1/2) ∏_{i=1}^{(n−1)/2} ζ(2j − 2i + 1),

r_{n,δ,ϵ}(j + 1/2) = δ^{(n+1)/2}ϵ(−1)^{(n^{2}−1)/8}b_{n}ζ(j + 1/2) ∏_{i=1}^{(n−1)/2} ζ(2j − 2i + 1).

When n is even with n ≥ 4, then ζ(s, L, 1, 1) = ζ(s, L, 1, −1) if n ≡ 2 mod 4, and ζ(s, L, −1, 1) = ζ(s, L, −1, −1) if n ≡ 0 mod 4. We give the functional equation for each case.

Theorem 3 We put c(s) = (cos πs)^{n/2−1} sin πs. When n ≡ 0 mod 4,

( \begin{matrix}ζ(\frac{n+1}{2} − s, L_{n}^{∗}, 1, 1) \\ ζ(\frac{n+1}{2} − s, L_{n}^{∗}, 1, −1) \\ ζ(\frac{n+1}{2} − s, L_{n}^{∗}, −1, 1) \end{matrix} ) =

2^{−ns+n^{2}/2−1}π^{−ns+n(n−1)/4}γ_{n}(s − \frac{n + 1}{2})

× ( \begin{matrix}(cos πs)^{n/2} + (−1)^{n(n−2)/8}(sin πs)^{n/2} & (cos πs)^{n/2} − (−1)^{n(n−2)/8}(sin πs)^{n/2} & 0 \\ (cos πs)^{n/2} − (−1)^{n(n−2)/8}(sin πs)^{n/2} & (cos πs)^{n/2} + (−1)^{n(n−2)/8}(sin πs)^{n/2} & 0 \\ (cos πs)^{n/2−1} & (cos πs)^{n/2−1} & −2c(s) \end{matrix} )

× ( \begin{matrix}ζ(s, L_{n}, 1, 1) \\ ζ(s, L_{n}, 1, −1) \\ ζ(s, L_{n}, −1, 1) \end{matrix} ) .

When n ≡ 2 mod 4 with n ≥ 6,

( \begin{matrix}ζ(\frac{n+1}{2} − s, L_{n}^{∗}, 1, 1) \\ ζ(\frac{n+1}{2} − s, L_{n}^{∗}, −1, 1) \\ ζ(\frac{n+1}{2} − s, L_{n}^{∗}, −1, −1) \end{matrix} ) =

2^{−ns+n^{2}/2−1}π^{−ns+n(n−1)/4}γ_{n}(s − \frac{n + 1}{2})

× ( \begin{matrix}2(cos πs)^{n/2} & 0 & 0 \\ 2(cos πs)^{n/2−1} & c(s) + (−1)^{n(n−2)/8}(sin πs)^{n/2} & c(s) − (−1)^{n(n−2)/8}(sin πs)^{n/2} \\ 2(cos πs)^{n/2−1} & c(s) − (−1)^{n(n−2)/8}(sin πs)^{n/2} & c(s) + (−1)^{n(n−2)/8}(sin πs)^{n/2} \end{matrix} )

× ( \begin{matrix}ζ(s, L_{n}, 1, 1) \\ ζ(s, L_{n}, −1, 1) \\ ζ(s, L_{n}, −1, −1) \end{matrix} ) .

Now, we shall give the location of poles and their residues.

Theorem 4 We assume that n ≥ 4 is an even integer. Then, the zeta function ζ(s, L_{n}^{∗}, δ, ϵ) or ζ(s, L_{n}, δ, ϵ) has possible poles only at s = l/2, where l is an integer with 1 ≤ l ≤ n + 1. The residues are given as follows.

(1) Unless (n, δ) = (4, −1), we get

r_{n,δ,ϵ}^{∗}(1) = 2^{n}b_{n}ϵδ_{n}(−1)^{(n(n+2)/8} \frac{|B_{n′/2}|}{n} ∏_{i=2}^{n/2} ζ(3 − 2i),
r_{n,δ,ϵ}(1) = 2^{−n/2}r_{n,δ,ϵ}^{∗}(1),

and if (n, δ) = (4, −1), we get

r_{4,−1,ϵ}^{∗}(1) = r_{4,−1,ϵ}(1) = 12 · π^{−4}ζ(3).

(2)For each natural integer j with 2 ≤ j ≤ n/2, we get

r_{n,δ,ϵ}^{∗}(j) = 2^{nj}b_{n}ϵδ_{n}(−1)^{n(n+2)/8} \frac{|B_{n′/2}|}{n} ∏_{i=1,i≠j}^{n/2} ζ(2j − 2i + 1),
r_{n,δ,ϵ}(j) = 2^{n(1/2−j)}r_{n,δ,ϵ}^{∗}(j).

In particular, if (−1)^{n/2} ≡ −δ mod 4, then the zeta functions are holomorphic at s = j.

(3)We get

r_{n,δ,ϵ}^{∗}((n + 1)/2) = 2^{n^{2}/2}b_{n}π^{−n/2}Γ(n/2)ζ(n) ∏_{i=1}^{n/2−1} ζ(n + 1 − 2i),
r_{n,δ,ϵ}((n + 1)/2) = 2^{n^{2}/2}r_{n,δ,ϵ}^{∗}((n + 1)/2).

(4)For each natural number j with 1 ≤ j ≤ n/2 − 1, we get

r_{n,δ,ϵ}^{∗}(j + 1/2) = 2^{n(j+1/2)−1}(−1)^{[n/4]}b_{n}D_{n}^{∗}(j + 1/2, δ) ∏_{i=1,i≠j}^{n/2−1} ζ(2i + 1 − 2j),

r_{n,δ,ϵ}(j + 1/2) = (−1)^{[n/4]}b_{n}D_{n}(j + 1/2, δ) ∏_{i=1,i≠j}^{n/2−1} ζ(2i + 1 − 2j).

Remark, We do not know whether D_{n}^{∗}(j + 1/2, δ) or D_{n}(j + 1/2, δ) vanishes or not.

2 Fourier coefficients of Eisenstein series

In this section, we review several results on Fourier coefficients of some real analytic or holomorphic Eisenstein series of half integral weight of one variable. These are investigated by Shimura [15], Siegel [17], Cohen [2], Zagier [21], Sturm [18], Goldfeld and Hoffstein [3] and so on. Although we need some modification of their results, the way of the proof is essentially due to them. We shall sometimes sketch the proof for the reader’s convenience. For any σ, z ∈ C with z ≠ 0, we put z^{σ} = e^{σ log(z)}, where arg(log(z)) is taken so that −π < arg(log(z)) ≤ π. We denote the upper half plane by H = {z ∈ C; Im(z) > 0}. For the sake of simplicity, we write e(z) = e^{2πiz}. For each s ∈ C, each odd (positive or negative) integer k, and z ∈ H, define the Eisenstein series E(k, σ, z) by

E(k, σ, z) = y^{\frac{σ}{2}} ∑_{d=1, odd}^{∞} ∑_{c=−∞}^{∞} (\frac{4c}{d}) ϵ_{d}^{−k}(4cz + d)^{\frac{k}{2}} |4cz + d|^{−σ},

where (\frac{∗}{∗}) is the quadratic residue symbol whose precise meaning is as was given in Shimura [13], p.442 and ϵ_{d} = 1, or i, if d ≡ 1, or 3 mod 4, respectively. When −k + 2σ − 4 > 0, this series converges absolutely and uniformly. We also define E^{∗}(k, σ, z) by

E^{∗}(k, σ, z) = E(−\frac{1}{4z})(−2iz)^{\frac{k}{2}}
= y^{\frac{σ}{2}} 2^{\frac{k}{2} −σ}e(−\frac{k}{8}) ∑_{d=1, odd}^{∞} ∑_{b=−∞}^{∞} (\frac{−b}{d}) ϵ_{d}^{−k}(dz + b)^{\frac{k}{2}} |dz + b|^{−σ}.

As usual, put

Γ_{0}(4) = \{g = ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ SL_{2}(Z); c ≡ 0 mod 4\}.

For any element γ = ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ Γ_{0}(4), put j(γ, z) = θ(γz)/θ(z), where θ(z) =

∑_{n∈Z} e^{2πin^{2}z}. It is known that

j(γ, z) = ϵ_{d}(\frac{c}{d})(cz + d)^{1/2}.

We get

E(k, σ, γz) = j(γ, z)^{−k/2}E(k, σ, z) and E^{∗}(k, σ, γz) = j(γ, z)^{−k/2}E^{∗}(k, σ, z),

where we put γz = (az + b)(cz + d)^{−1}. So, these functions are not holomorphic if σ ≠ 0, but behave like modular forms of weight −k/2. When k < −5 and σ = 0, two holomorphic functions E(−k, 0, z) and E^{∗}(−k, 0, z) gives a basis of the orthogonal complement of cusp forms of weight −k/2 in modular forms of weight −k/2 belonging to Γ_{0}(4) (cf. Shimura loc. cit.). After Siegel, define a confluent hypergeometric function W (y, α, β) by

W (y, α, β) = ∫_{0}^{∞}(u + 1)^{α−1}u^{β−1}e^{−yu}du,

where y > 0 and α, β ∈ C. This is convergent for Re(β) > 0 and meromorphically continued to the whole (α, β) ∈ C^{2}. For any integer d and α, β ∈ C with Re(α) > 0, Re(β) > 0, Re(α + β) > 1, we also define τ_{d}(y, α, β) by

i^{α−β}(2π)^{−α−β}Γ(α)Γ(β)τ_{d}(y, α, β) = \{ \begin{matrix}d^{α+β−1}e^{−2πdy}W (4πdy, α, β) & (d > 0) \\ |d|^{α+β−1}e^{−2π|d|y}W (4π|d|y, β, α) & (d < 0) \\ Γ(α + β − 1)(4πy)^{1−α−β} & (d = 0) \end{matrix}

This is also meromorphically continued. The Fourier expansions of E(k, σ, z) and E^{∗}(k, σ, z) are of the following form:

y^{−σ/2}E(k, σ, z) = 1 + ∑_{d=−∞}^{∞} β(d, σ, k)e(dx)τ_{d}(y, \frac{σ − k}{2} , \frac{σ}{2} ),
y^{−σ/2}2^{σ−k/2}e(\frac{k}{8})E^{∗}(k, σ, z) = ∑_{d=−∞}^{∞} α(d, σ, k)e(dx)τ_{d}(y, \frac{σ − k}{2} , \frac{σ}{2} ),

where

α(d, σ, k) = ∑_{c=1}^{∞}(\frac{−4}{c})ϵ_{c}^{−k}c^{k/2−σ} ∑_{m=1}^{c}(\frac{m}{c})e(dm/c),
β(d, σ, k) = ∑_{c=1}^{∞}(4c)^{k/2−σ} ∑_{m=1}^{4c} ϵ_{m}^{−k}(\frac{4c}{m})e^{2πimd/4c}.

To get Fourier coefficients suitable for our purpose, we put

F (k, σ, z) = E(k, σ, z) + 2^{k/2−σ}(e(k/8) + e(−k/8))E^{∗}(k, σ, z).

This kind of linear combination has been already considered by Cohen [2], Zagier [21] and Sturm [18] for σ = 0. We need the coefficients also for σ ≠ 0. which was calculated in Goldfeld and Hoffstein [3]. Since we need slightly different details here, we review this. We write the Fourier expansion of F (k, σ, z) as

y^{−σ/2}F (k, σ, z) = ∑_{d=−∞}^{∞} c_{d}(y)d(dx)
= 1 + ∑_{d=−∞}^{∞} C(d, σ, k)e(dx)τ_{d}(y, \frac{σ − k}{2} , \frac{σ}{2} ),

where each c_{d}(y) is the Fourier coefficients depending on y and d and each C(d, σ, k) is a constant which does not depend on z. Following the method of Shimura [15] and Sturm [18], we can calcultate C(d, σ, k) explicitly. For the convenience of the readers, we sketch here the calculation. By definition,

C(d, σ, k) = β(d, σ, k) + 2^{k−2σ}(1 + i^{−k})α(d, σ, k).

By Sturm [18] p.227 Lemma 2 and the arguments in p.228, we have

C(d, σ, k) = α(d, σ, k)\{2^{k−2σ}(1 + i^{−k}) + ∑_{r=2}^{∞} ∑_{l=1}^{2^{r}} ϵ_{l}^{−k}(\frac{2^{r}}{l})e(\frac{dl}{2^{r}})2^{(k/2−σ)r}\}
= α(d, σ, k)A(d, σ, k),

where

A(d, σ, k) = 2^{k−2σ}(1 + i^{−k}) + ∑_{r=2}^{∞} \frac{1}{2}((1 + i^{−k})U(r) + (1 − i^{−k})V (r))2^{(k/2−σ)r}
U(r) = ∑_{l=1}^{2^{r}}(\frac{2^{r}}{l})e(\frac{dl}{2^{r}}),

V (r) = ∑_{l=1}^{2^{r}}(\frac{−2^{r}}{l})e(\frac{dl}{2^{r}}).

Assume that d ≠ 0 and put d = 2^{t}d_{0}, where d_{0} is an odd (positive or negative) integer and t is a non-negative integer. If t is even, then we get that U(2i) = 2^{2i−1} for 1 ≤ i ≤ t/2, U(t + 3) = (\frac{2}{d_{0}} )2^{t+3/2}, and U(r) = 0 for the other r, and that V (t + 2) = 2^{t+1}i(\frac{−1}{d_{0}}), V (t + 3) = 2^{t+3/2}(\frac{−2}{d_{0}}) and V (r) = 0 for the other r. If t is odd, then V (r) = 0 for all r and U(2i) = 2^{2i−1} for 1 ≤ i ≤ (t − 1)/2, U(t + 1) = −2^{t} and U(r) = 0 for the other r. Hence, if t is even, then we get

∑_{r=2}^{∞} \frac{1}{2}(1 + i^{−k})U(r)2^{(k/2−σ)r} =

(1 + i^{−k})(\frac{2^{k−2σ}(1 − 2^{(k/2−σ+1)t})}{1 − 2^{k−2σ+2}} + 2^{3(k/2−σ)+1/2+(k/2−σ+1)t}(\frac{2}{d_{0}} )),

∑_{r=2}^{∞} \frac{1}{2}(1 − i^{−k})V (r)2^{(k/2−σ)r} =

(−1)^{(k+1)/2}(1 + i^{−k})(2^{3(k/2−σ)+1/2} · 2^{(k/2−σ+1)t}(\frac{−2}{d_{0}}) + 2^{k−2σ+(k/2−σ+1)t}(\frac{−1}{d_{0}})),

and if t is odd, then

∑_{r=2}^{∞} \frac{1}{2}(1 + i^{−k})U(r)2^{(k/2−σ)r} =

(1 + i^{−k})(−2^{k/2−σ−1} · 2^{(k/2−σ+1)t} + \frac{2^{k−2σ}(1 − 2^{(k−2σ+2)(t−1)/2})}{1 − 2^{k−2σ+2}}),

∑_{r=2}^{∞} \frac{1}{2}(1 − i^{−k})V (r)2^{(k/2−σ)r} = 0

We get (−1)^{(k+1)/2}(\frac{−1}{d_{0}}) = 1, or −1, if d_{0} ≡ (−1)^{(k+1)/2} mod 4, or not, respectively.

Lemma 2.1 (1) If d_{0} ≡ (−1)^{(k+1)/2} mod 4, then

∑_{t=0}^{∞} A(d_{0}2^{2t}, σ, k)2^{2t(σ−k/2−1−s)} = (1 + i^{−k})2^{k−2σ+1}

×\frac{(1 + 2^{k/2−σ+1/2}(\frac{2}{d_{0}} ))(1 − 2^{σ−k/2−3/2−2s}(\frac{2}{d_{0}} ))}{(1 − 2^{−2s})(1 − 2^{2σ−k−2−2s})}.

(2) If d_{0} ≢ (−1)^{(k+1)/2} mod 4, then

∑_{t=0}^{∞} A(d_{0}2^{2t}, σ, k)2^{2t(σ−k/2−1−s)} = (1 + i^{−k})\frac{2^{−1−2s}(1 − 2^{k−2σ+1})}{(1 − 2^{−2s})(1 − 2^{2σ−k−2−2s})}.

(3) For any odd d_{0}, we get

∑_{t=0}^{∞} A(d_{0}2^{2t+1}, σ, k)2^{(2t+1)(σ−k/2−1−s)} = (1 + i^{−k})\frac{2^{σ−k/2−2−3s}(1 − 2^{k−2σ+1})}{(1 − 2^{−2s})(1 − 2^{2σ−k−2−2s})}.

By Shimura [15] Proposition 2, for any square free integer m, we get

∑_{f=1}^{∞} α(mf^{2}, σ, k)f^{2σ−k−2−2s} = \frac{L_{2}(σ − (k + 1)/2, ω_{1})ζ_{2}(2s)ζ(2s − 2σ + k + 2)}{ζ_{2}(2σ − k − 1)L_{2}(2s − σ + (k + 3)/2, ω_{1})},

where we denote by ζ_{2} or L_{2} the usual zeta or L functions such that their Euler 2-factors are omitted, and ω_{1} is the primitive character defined by

ω_{1}(a) = (\frac{−1}{a})^{(k+1)/2} (\frac{4m}{a})

for (a, 4m) = 1. Here (*/*) is the quadratic residue symbol, the precise meaning of which is in Shimura [13]. Now, for any square free integer m, we calculate the Dirichlet series

∑_{f=1}^{∞} e(\frac{k}{8})C(mf^{2}, σ, k)(|m|f^{2})^{σ−k/2−1−s}.

By the above formula (2) and (3) in Lemma 2.1, if f is odd, then we get C(mf^{2}, σ, k) = 0, unless m is odd and m ≡ (−1)^{(k+1)/2} mod 4. Hence, we can assume that d_{K}f_{1}^{2} = (−1)^{(k+1)/2}mf^{2} for some integer f_{1} and the fundamental discriminant d_{K} of the quadratic field K = Q((−1)^{(k+1)/2}m) or d_{K} = 1 when m = (−1)^{(k+1)/2}.

Proposition 2.2

∑_{f=1}^{∞} e(\frac{k}{8})C((−1)^{(k+1)/2}d_{K}f^{2}, σ, k)(|d_{K}|f^{2})^{σ−k/2−1−s} =

(−1)^{(k^{2}−1)/8}2^{k−2σ+3/2} \frac{|d_{K}|^{σ−k/2−1}}{|d_{K}|^{s}} × \frac{L(σ − (k + 1)/2, χ_{K})ζ(2s)ζ(2s − 2σ + k + 2)}{ζ(2σ − k − 1)L(2s − σ + (k + 3)/2, χ_{K})},

where χ_{K} is the Kroneker character attached to K, and regarded as the trivial character, if d_{K} = 1.

Proof. This can be shown by calculationg a kind of convolution product of the Dirichelte series in Lemma 2.1 and Shimura’s results, noting that e(\frac{k}{8})(1 + i^{−k}) = e(\frac{k}{8}) + e(−\frac{k}{8}) = \sqrt{2}(−1)^{(k^{2}−1)/8}. q.e.d.

Proposition 2.3 If σ > k/2 + 1, then we have

C(0, σ, k) = 2^{k−2σ+1}(1 + i^{−k})\frac{ζ(2σ − k − 2)}{ζ(2σ − k − 1)}.

Proof. When d = 0, we get V (r) = 0 for all r, and U(r) = 2^{r−1}, or 0 for r even, or odd, respectively. Since

∑_{r=1}^{∞} U(2r)2^{(k/2−σ)2r} = \frac{2^{k−2σ+1}}{1 − 2^{k−2σ+2}} ,

we get

A(0, σ, k) = 2^{k−2σ+1}(1 + i^{−k})\frac{1 − 2^{k−2σ+1}}{1 − 2^{k−2σ+2}} .

On the other hand, by Shimura [15] p.89 (3.9), we have

α(0, σ, k) = \frac{ζ_{2}(2σ − k − 2)}{ζ_{2}(2σ − k − 1)}.

Hence, we get our results.

Corollary 2.4 We get

c_{0}(y) = 1 + 2^{3k/2−3σ+7/2}(−1)^{(k^{2}−1)/8}π \frac{Γ(σ − k/2 − 1)ζ(2σ − k − 2)}{Γ((σ − k)/2)Γ(σ/2)ζ(2σ − k − 1)}y^{1−σ+k/2}.

3 Dirichlet series associated to Eisenstein series

3.1

First, we prove functional equations between D_{n}^{∗}(\frac{n+1}{2} − s, δ) and D_{n}(s, δ), where δ = 1, or −1. The case (n, δ) = (2, −1) is exceptional, and we need some special treatment in this case. So, we exclude this case for a moment, and treat it in section 5. Our proof is based on the usual argument by Hecke on functional equations of Mellin transforms of automorphic forms. This type of functional equation has been more or less calculated by Goldfeld and Hoffstein [3] for a different purpose, but the modular forms in question are slightly different and we need some new calculation here. For any function f(z) on H, we define two operators f|_{k}W_{4} and f|U_{4} by

(f|_{k}W_{4})(z) = f(−1/4z)(−2iz)^{k/2},
(f|U_{4})(z) = \frac{1}{4} ∑_{ν=0}^{3} f(\frac{z + ν}{4} ).

We define F (z) = F (k, σ, z) as in the last section.

Lemma 3.1 We have the following relation.

F |_{k}W_{4} = 2^{(k+1)/2}(−1)^{(k^{2}−1)/8}F |U_{4}.

Proof. This can be proved in the same way as in Kohnen [6], and we sketch the outline. Put

g_{1}(z) = \frac{1}{4}(F (\frac{z + 1}{4}) + F (\frac{z + 3}{4})),

and

g_{2}(z) = \frac{1}{4}(F (\frac{z}{4}) + F (\frac{z + 2}{4})).

Then, F |U_{4} = g_{1} + g_{2}. As was shown in the last section, the coefficients of F (z) of e(dx) are 0, unless d ≡ 0, or (−1)^{(k+1)/2} mod 4. Since e(d/2) + 1 = e(d/4) + e(3d/4), if d ≡ 0 mod 4, or d ≡ 1 mod 2, we get g_{1} = g_{2}. Hence, using the same argument as in Kohnen [6] p.255, we get

F |U_{4}|_{k}W_{4} = 2g_{1}|_{k}W_{4}
= 2^{−k/2−1}(F (z − \frac{1}{4})e(−k/8) + F (z + \frac{1}{4})e(k/8)).

It is easy to see that e(−d/4)e(−k/8) + e(d/4)e(k/8) = \sqrt{2}(−1)^{(k^{2}−1)/8}, if d ≡ 0, or (−1)^{(k+1)/2} mod 4. Hence, F |U_{4}|_{k}W_{4} = 2^{−(k+1)/2}(−1)^{(k^{2}−1)/8}F, and since W_{4}^{2} acts trivially on F by Shimura [14] p.448, we prove our lemma. q.e.d.

Now, we put g(z) = y^{−σ/2}F (k, σ, z) and h(z) = (y^{−σ/2}F (k, σ, z))|U_{4}. We denote the Fourier expansion of g(z) as

g(z) = ∑_{d=−∞}^{∞} c_{d}(y)e(dx).

Then, we get

h(z) = ∑_{d=−∞}^{∞} c_{4d}(y/4)e(dx).

Put G(y) = g(iy) and H(y) = h(iy). Then, by the above lemma, we get G(1/4y) = (−1)^{(k^{2}−1)/8}\sqrt{2}y^{σ−k/2}H(y), and G(1/2y) = C_{k}y^{σ−k/2}H(y/2), where C_{k} = 2^{(k+1)/2−σ}(−1)^{(k^{2}−1)/8}. Define two Dirichlet series by

Φ_{σ,k}(s) = ∫_{0}^{∞}(G(y) − c_{0}(y))y^{s−1}dy,
Ψ_{σ,k}(s) = ∫_{0}^{∞}(H(y) − c_{0}(y/4))y^{s−1}dy.

Then, by the usual argument of Hecke, we get

2^{s}Φ_{σ,k}(s) = ∫_{0}^{∞}(G(y/2) − c_{0}(y/2))y^{s−1}dy
= ∫_{1}^{∞}(G(y/2) − c_{0}(y/2))y^{s−1}dy + ∫_{0}^{1}(G(y/2) − c_{0}(y/2))y^{s−1}dy
= ∫_{1}^{∞}(G(y/2) − c_{0}(y/2))y^{s−1}dy + ∫_{1}^{∞}(C_{k}H(y/2)y^{σ−k/2} − c_{0}(1/2y))y^{−s−1}dy
= ∫_{1}^{∞}(G(y/2) − c_{0}(y/2))y^{s−1}dy + C_{k} ∫_{1}^{∞}(H(y/2) − c_{0}(y/8))y^{σ−k/2−s−1}dy
− ∫_{1}^{∞} c_{0}(1/2y)y^{−s−1}dy + C_{k} ∫_{1}^{∞} c_{0}(y/8)y^{σ−k/2−s−1}dy.

We have c_{0}(y) = 1 + c(σ)y^{1−σ+k/2} for a certain function of c(σ) of σ which does not depend on y. (The exact value is given in section 2, Corollary 2.4.) So, if s > σ − k/2 and s > 1, then

2^{s}Φ_{σ,k}(s) + ∫_{1}^{∞} c_{0}(1/2y)y^{−s−1}dy − C_{k} ∫_{1}^{∞} c_{0}(y/8)y^{σ−k/2−s−1}dy
= 2^{s}Φ_{σ,k}(s) + \frac{c(σ)2^{σ−k/2−1}}{s − σ + k/2 + 1} + C_{k} \frac{c(σ)8^{σ−k/2−1}}{1 − s} + \frac{1}{s} + C_{k} \frac{1}{σ − k/2 − s}
= ∫_{1}^{∞}(G(y/2) − c_{0}(y/2))y^{s−1}dy + C_{k} ∫_{1}^{∞}(H(y/2) − c_{0}(y/8))y^{σ−k/2−s−1}dy.

Now, denote the last line above by J_{k}(σ, s). Then, it can be proved that (σ − (k + 3)/2)(σ − (k + 2)/2)ζ(2σ − k − 1)J_{k}(σ, s) is holomorphic at whole σ, s ∈ C. Hence, Φ_{σ,k}(s) is meromorphically continued to the whole (σ, s) ∈ C^{2} and

ζ(2σ−k−1)(σ−(k+2)/2)(σ−(k+3)/2)s(s−1)(s−σ+k/2)(s−σ+k/2+1)Φ_{σ,k}(s) is holomorphic on (σ, s) ∈ C^{2}. In the same way, we get

2^{s}Ψ_{σ,k}(s) = ∫_{1}^{∞}(H(y/2) − c_{0}(y/8))y^{s−1}dy + C_{k}^{−1} ∫_{1}^{∞}(G(y/2) − c_{0}(y/2))y^{σ−k/2−s−1}dy
− ∫_{1}^{∞} c_{0}(1/8y)y^{−s−1}dy + C_{k}^{−1} ∫_{1}^{∞} c_{0}(y/2)y^{σ−k/2−s−1}dy.
= C_{k}^{−1}J_{k}(σ, σ − k/2 − s) − \frac{1}{s} − \frac{c(σ)8^{σ−k/2−1}}{s − σ + k/2 + 1} − \frac{C_{k}^{−1}}{σ − k/2 − s} − \frac{C_{k}^{−1}c(σ)2^{σ−k/2−1}}{1 − s} .

Hence, we get the following functional equation (cf. [3]).

Φ_{σ,k}(σ − k/2 − s) = 2^{2s−σ+k/2}C_{k}Ψ_{σ,k}(s) = 2^{2s−2σ+k+1/2}(−1)^{(k^{2}−1)/8}Ψ_{σ,k}(s).

3.2

Now, we shall apply functional equations between Φ_{σ,k}(s) and Ψ_{σ,k}(s) to those between D^{∗}(s, δ) and D(s, δ). This can be done by calculating the Mellin transforms term by term and by specializing σ and k. The Mellin transform of τ_{d}(y, (σ − k)/2, σ/2) has been given in [3]. For each α, β ∈ R and s ∈ C, we consider the following integral:

I(s, α, β) = ∫_{0}^{∞} \frac{(1 + u)^{α−1}u^{β−1}}{(1 + 2u)^{s}} du.

This integral converges absolutely when Re(β) > 0 and Re(s) > Re(α)+Re(β)−1. This is related with the usual hypergeometric functions. This can be shown as follows. For any non negative integer n and a variable a, we put (a)_{n} = Γ(a + n)/Γ(a). The Gauss hypergeometric function F (a, b, c; z) is by definition

F (a, b, c; z) = ∑_{n=0}^{∞} \frac{(a)_{n}(b)_{n}}{(c)_{n}n!} z^{n},

and has an integral expression

F (a, b, c; z) = \frac{Γ(c)}{Γ(b)Γ(c − b)} ∫_{0}^{1} t^{b−1}(1 − t)^{c−b−1}(1 − zt)^{−a} dt

for Re(c) > Re(b) > 0 and |z| < 1. Now, in our integral, substituting the variable u by u = v/(2 − 2v), we get

I(s, α, β) = \frac{1}{2} ∫_{0}^{1} v^{β−1}(1 − v)^{s−α−β}(1 − \frac{v}{2})^{α−1} dv
= \frac{Γ(β)Γ(s − α − β + 1)}{Γ(s − α + 1)}F (1 − α, β, s − α + 1; 1/2).

Now, we assume that Re(σ) > 0, Re(σ) > k, and Re(s) > Re(σ)−k/2−1. Then, the Mellin transforms of the Whittaker functions are given by

∫_{0}^{∞} τ_{d}(y, \frac{σ − k}{2} , \frac{σ}{2} )y^{s−1}dy =

\frac{e(k/8)(2π)^{σ−k/2}}{Γ((σ − k)/2))Γ(σ/2)} × \{ \begin{matrix}d^{σ−k/2−1}(2πd)^{−s}Γ(s)I(s, \frac{σ−k}{2}, \frac{σ}{2}), & if d > 0, \\ |d|^{σ−k/2−1}(2πd)^{−s}Γ(s)I(s, \frac{σ}{2}, \frac{σ−k}{2}), & if d < 0. \end{matrix}

Hence, if d = (−1)^{(k+1)/2}d_{K}f^{2} ≠ 0, we get

∫_{0}^{∞} c_{d}(y)y^{s−1}dy = e(\frac{k}{8})C((−1)^{(k+1)/2}d_{K}f^{2}, σ, k)(|d_{K}|f^{2})^{σ−k/2−1−s}

×\frac{(2π)^{σ−k/2−s}Γ(s)}{Γ((σ − k)/2))Γ(σ/2)} × \{ \begin{matrix}I(s, (σ − k)/2, σ/2), & if d > 0 \\ I(s, σ/2, (σ − k)/2), & if d < 0 \end{matrix}

If Re(s) > −k/2 − 1, the right hand side can be continued holomorphically to σ = 0. Indeed, under this condition on s and k, I(s, σ/2, (σ − k)/2)) is holomorphic at σ = 0, and also

\frac{1}{Γ(σ/2)}I(s, \frac{σ − k}{2} , \frac{σ}{2} ) = \frac{1}{Γ(σ/2)(e^{πiσ} − 1)} ∫_{∞}^{(0+)} \frac{(u + 1)^{(σ−k)/2−1}u^{σ/2−1}}{(2u + 1)^{s}} du,

and this function is 1 at σ = 0.

For the sake of simplicity, we put

Z^{∗}(k, σ, s, d_{K}) = |d_{K}|^{σ−k/2−1−s} \frac{L(σ − (k + 1)/2, χ_{K})ζ(2s)ζ(2s − 2σ + k + 2)}{ζ(2σ − k − 1)L(2s − σ + (k + 3)/2, χ_{K})},

and

C_{σ,k} = (−1)^{(k^{2}−1)/8} \frac{2^{k/2−σ+3/2}π^{σ−k/2}}{Γ((σ − k)/2)Γ(σ/2)}.

We write the above Dirichlet series as Z^{∗}(k, σ, s, d_{K}) = ∑_{n=1}^{∞} a(n)n^{−s} and put

Z(k, σ, s, d_{K}) = ∑_{n=1}^{∞} \frac{a(4n)}{n^{s}}.

Then, we get

Φ_{σ,k}(s) = C_{σ,k}(2π)^{−s}Γ(s)

×(  ∑_{(−1)^{(k+1)/2}d_{K}>0} Z^{∗}(k, σ, s, d_{K})I(s, \frac{σ − k}{2} , \frac{σ}{2} ) + ∑_{(−1)^{(k+1)/2}d_{K}<0} Z^{∗}(k, σ, s, d_{K})I(s, \frac{σ}{2} , \frac{σ − k}{2} ))

and Ψ_{σ,k}(s) =

C_{σ,k}(2π)^{−s}Γ(s)

×(  ∑_{(−1)^{(k+1)/2}d_{K}>0} Z(k, σ, s, d_{K})I(s, \frac{σ − k}{2} , \frac{σ}{2} ) + ∑_{(−1)^{(k+1)/2}d_{K}<0} Z(k, σ, s, d_{K})I(s, \frac{σ}{2} , \frac{σ − k}{2} ))

Case (I): We put k = −(n + 1) and σ = 0. We assume that n is even and n ≥ 4. By definition, we have

∑_{(−1)^{n/2}d_{K}>0} Z^{∗}(−(n + 1), 0, s, d_{K}) = ∑_{(−1)^{n/2}d_{K}>0} |d_{K}|^{(n−1)/2−s} \frac{L(n/2, χ_{K})ζ(2s)ζ(2s − n + 1)}{ζ(n)L(2s − n/2 + 1, χ_{K})}
= 2^{−1}(2π)^{n/2}(−1)^{[n/4]}Γ(n/2)^{−1}ζ(n)^{−1}D_{n}^{∗}(s, 1).

Hence, we get Φ_{0,−n−1}(s) = (−1)^{n(n+2)/8+[n/4]}π^{n+1/2}Γ((n+1)/2)^{−1}Γ(n/2)^{−1}ζ(n)^{−1}(2π)^{−s}Γ(s)D_{n}^{∗}(s, 1). In the same way, we get Ψ_{0,−n−1}(s) = (−1)^{n(n+2)/8+[n/4]}π^{n+1/2}Γ((n+1)/2)^{−1}Γ(n/2)^{−1}ζ(n)^{−1}(2π)^{−s}Γ(s)D_{n}(s, 1). Hence, we get

(2π)^{−(n+1)/2+s}Γ(\frac{n + 1}{2} − s)D_{n}^{∗}(\frac{n + 1}{2} − s, 1) = 2^{2s−(n+1)/2}(2π)^{−s}Γ(s)D_{n}(s, 1)C_{k},

where C_{k} = 2^{−n/2}(−1)^{n(n+2)/8}. Since

Γ((n + 1)/2 − s)^{−1} = (−1)^{n/2}π^{−1}(cos πs)Γ(s − (n − 1)/2),

we get Proposition 3.2 For any even positive integer n ≥ 4, we have

D_{n}^{∗}(\frac{n + 1}{2}−s, 1) = (−1)^{n(n−2)/8}2^{−n/2}π^{−2s+(n−1)/2}Γ(s)Γ(s−\frac{n − 1}{2})(cos πs)D_{n}(s, 1).

When n = 2 and δ = −1, we need some modification of functional equation, since L(1, χ_{K}) = ∞, if K = Q ⊕ Q. But if δ = 1, the definition of D_{2} itself is the same and we have Proposition 3.3

D_{2}^{∗}(3/2 − s, 1) = 2^{−1}π^{1/2−2s}Γ(s)Γ(s − 1/2)(cos(πs)D_{2}(s, 1) − ζ(2s − 1)).

This proposition was first proved by Siegel [17] and Shintani [16] without using Eisenstein series. (Incidentally, in Shintani’s notation, D_{2}(s, 1) = 2^{2s}ξ_{−}^{∗}(s) and D_{2}^{∗}(s, 1) = ξ_{−}(s).) Later, an alternative proof in the case n = 2 is obtained by Sturm [18]. His argument is to use the Eisenstein series above, noting that, in the above calculation, I(s, 0, 3/2) is holomorphic for s > 1/2 and that ζ(σ+1)/Γ(σ/2) is holomorphic at σ = 0. As for the case (n, δ) = (2, −1), we need a modification of the definition, which will be explained in section 5. Case(II): We put k = −n + 3 and σ = 2 with even n ≥ 4. By changing the variable u by v = 2u(2u + 1)^{−1}, we get

I(s, 1, \frac{n − 1}{2}) = 2^{−(n−1)/2} ∫_{0}^{1} v^{(n−3)/2}(1−v)^{s−(n+1)/2}dv = 2^{−(n−1)/2}Γ(s)^{−1}Γ(\frac{n − 1}{2})Γ(s−\frac{n − 1}{2}).

On the other hand, the function I(s, \frac{n−1}{2}, 1) can be continued meromorphically to the whole s plane and satisfies

I(s, \frac{n − 1}{2}, 1)+I(\frac{n + 1}{2}−s, \frac{n − 1}{2}, 1) = 2^{−(n−1)/2}Γ((3−n)/2)^{−1}Γ(1−s)Γ(s−\frac{n − 1}{2}).

This can be proved easily by using the Pochhammer integral formula for the beta function. Indeed, put v = (1 + 2u)/(2 + 2u). Then,

I(s, \frac{n − 1}{2}, 1) = 2^{−(n−1)/2} ∫_{1/2}^{1} v^{−s}(1 − v)^{s−(n+1)/2}dv.

Now, for α, β > 0, put f(x) = x^{α−1}(1 − x)^{β−1}. Take a positive branch at x = 1/2 and take an analytic continuation along the path starting at 1/2, encircling 1 and 0 in the positive direction, then encircling 1 and 0 in the negative direction. Then,

∫_{1/2}^{(1+,0+,1−,0−)} x^{α}(1 − x)^{β}dx = \frac{1}{(1 − e^{2πiα})(1 − e^{2πiβ})}B(α, β),

where B(α, β) = Γ(α)Γ(β)Γ(α + β)^{−1} is the usual beta function. We also have

∫_{1/2}^{1} v^{α}(1 − v)^{β}dv = \frac{1}{(1 − e^{2πiα})(1 − e^{2πiβ})}(∫_{1/2}^{(1+)} f(x)dx + ∫_{1/2}^{(1−)} f(x)dx),

and

∫_{0}^{1/2} v^{α}(1 − v)^{β}dv = \frac{1}{(1 − e^{2πiα})(1 − e^{2πiβ})}(∫_{1/2}^{(0+)} f(x)dx + ∫_{1/2}^{(0−)} f(x)dx),

and we get the above result. By virtue of the above integral expression, it is easy to see that Γ(s−(n−1)/2)^{−1}I(s, \frac{n−1}{2}, 1) is an entire function and that I(s, \frac{n−1}{2}, 1) has simple poles only at s = (n + 1)/2 − m (m = 1, 2, …). The residue for each m is (−1)^{m−1}2^{−(n−1)/2}Γ(m − (n − 1)/2)Γ((3 − n)/2)^{−1}Γ(m)^{−1}.

Now, we get

Φ_{2,−n+3}(s) = (−1)^{n(n+2)/8+1+[n/4]}π^{n+1/2}Γ(n/2)^{−1}Γ((n − 1)/2)^{−1}ζ(n)^{−1}(2π)^{−s}Γ(s)

×(D_{n}^{∗}(s, 1)I(s, \frac{n − 1}{2}, 1) + D_{n}^{∗}(s, −1)I(s, 1, \frac{n − 1}{2})),

and

Ψ_{2,−n+3}(s) = (−1)^{n(n+2)/8+1+[n/4]}π^{n+1/2}Γ(n/2)^{−1}Γ((n − 1)/2)^{−1}ζ(n)^{−1}(2π)^{−s}Γ(s)

×(D_{n}(s, 1)I(s, \frac{n − 1}{2}, 1) + D_{n}(s, −1)I(s, 1, \frac{n − 1}{2})).

Hence, we get

(2π)^{s−(n+1)/2}Γ(\frac{n + 1}{2})

×(D_{n}^{∗}(\frac{n + 1}{2} − s, 1)I(\frac{n + 1}{2} − s, \frac{n − 1}{2}, 1) + D_{n}^{∗}(\frac{n + 1}{2} − s, −1)I(\frac{n + 1}{2} − s, 1, \frac{n − 1}{2}))

= 2^{2s−n−1/2}(−1)^{n(n+2)/8+1}(2π)^{−s}Γ(s)

×(D_{n}(s, 1)I(s, \frac{n − 1}{2}, 1) + D_{n}(s, −1)I(s, 1, \frac{n − 1}{2})).

Here, using the functional equation between D_{n}^{∗}(s, 1) and D_{n}(s, 1), and the formulas for I(s, 1, \frac{n−1}{2}) and I(s, \frac{n−1}{2}, 1), we get

(2π)^{s−(n+1)/2}2^{−(n−1)/2}Γ(\frac{n − 1}{2})Γ(1 − s)D_{n}^{∗}(\frac{n + 1}{2} − s, −1)

= 2^{2s−n−1/2}(−1)^{n(n+2)/8+1}(2π)^{−s}Γ(s)2^{−(n−1)/2}

×(Γ((3 − n)/2)^{−1}D_{n}(s, 1)Γ(1 − s)Γ(s − \frac{n − 1}{2}) + Γ(\frac{n − 1}{2})D_{n}(s, −1)Γ(s)^{−1}Γ(s − \frac{n − 1}{2})).

Since Γ(s)Γ(1 − s) = π/ sin πs, and Γ(\frac{n−1}{2})Γ(\frac{3−n}{2}) = π/ sin(\frac{n−1}{2}π) = π(−1)^{n/2+1}, we get

Proposition 3.4 For even n ≥ 4, we get

D_{n}^{∗}(\frac{n + 1}{2} − s, −1) = 2^{−n/2}π^{−2s+(n−1)/2}(−1)^{n(n+2)/8+1}Γ(s)Γ(s − \frac{n − 1}{2})
×((sin πs)D_{n}(s, −1) + (−1)^{n/2+1}D_{n}(s, 1)).

As for the necessary modification in case n = 2, see section 5.

For later use, we shall give poles and residues of D_{n}^{∗}(s, δ) and D_{n}(s, δ) (δ = ±1.) If σ = 0 and k = −(n + 1), then c_{0}(y) = 1. So, Φ_{0,k}(s), or Ψ_{0,k}(s) has simple poles only at s = 0 and (n + 1)/2.

Proposition 3.5 The functions D_{n}^{∗}(s, 1), or D_{n}(s, 1) is holomorphic except for a simple pole at s = (n + 1)/2 with the residue (−1)^{[n/4]}(2π)^{−n/2}Γ(n/2)ζ(n) , or (−1)^{[n/4]}2^{n/2}π^{−n/2}Γ(n/2)ζ(n), respectively. The functions D_{n}^{∗}(s, −1), or D_{n}(s, −1) is holomorphic except for simple pole at s = 1 and s = (n + 1)/2. The residue at s = 1 is π^{−n−1/2}Γ(n/2)Γ((n−1)/2)ζ(n−1), or 2π^{−n−1/2}Γ(n/2)Γ((n−1)/2)ζ(n−1) , and one at s = (n+1)/2 is (−1)^{[n/4]}(2π)^{−n/2}Γ(n/2)ζ(n) or (−1)^{[n/4]}2^{n/2}π^{−n/2}Γ(n/2)ζ(n), respectively.

Proof. Since Φ_{0,−n−1}(s), or Ψ_{0,−n−1}(s) has a simple pole at s = 0, Φ_{0,−n−1}(s)/Γ(s), or Ψ_{0,−n−1}(s)/Γ(s) is holomorphic at s = 0. Hence, D_{n}^{∗}(s, 1), or D_{n}(s, 1) is holomorphic for s ≠ (n + 1)/2. At s = (n + 1)/2, the residue of Φ_{0,−n−1}(s), or Ψ_{0,−n−1}(s) is 2^{−n−1/2}(−1)^{n(n+2)/8} or 2^{−1/2}(−1)^{n(n+2)/8}. Hence we get our results for D_{n}^{∗}(s, δ) and D_{n}(s, δ) for δ = 1. Next, we treat the case δ = −1. By the results in section 3.1, for even n ≥ 4, we see that Φ_{2,−n+3}(s) and Ψ_{2,−n+3}(s) are holomorphic except for s = 0, 1, (n−1)/2, and (n+1)/2 and Γ(s)D_{n}^{∗}(s, 1) or Γ(s)D_{n}(s, 1) has simple poles only at s = 0, (n+1)/2. Since Γ(s−(n−1))/2)^{−1}I(s, (n−1)/2, 1) is entire, it is easy to see that D_{n}^{∗}(s, −1) and D_{n}(s, −1) are holomorphic at s ≠ 0, 1, and (n + 1)/2. Now, by Proposition 3.2 and the above result, we get

D_{n}^{∗}(0, 1) = D_{n}(0, 1) = (−1)^{n(n+2)/8+1+[n/4]}π^{−n−1/2}Γ(n/2)Γ((n + 1)/2))ζ(n).

Since the residue of Φ_{2,−n+3}(s) and Ψ_{2,−n+3}(s) at s = 0 is −1 and I(0, \frac{n−1}{2}, 1) = −2/(n − 1), we can see that D_{n}^{∗}(s, −1) and D_{n}(s, −1) are holomorphic at s = 0. The residue of Φ_{2,−n+3}(s), or Ψ_{2,−n+3}(s) at s = (n+1)/2 is 2^{−n−1/2}(−1)^{n(n+2)/8+1} , or 2^{−1/2}(−1)^{(n(n+2)/8+1}, and at s = 1, it is 2^{−(n+1)/2}ζ(n−1)ζ(n)^{−1}, or 2^{(1−n)/2}ζ(n− 1)ζ(n)^{−1}, respectively. Using I((n + 1)/2, (n − 1)/2, 1) = 2(1 − 2^{(1−n)/2})/(n − 1) and the estimates of residues of D_{n}(s, 1) and D_{n}^{∗}(s, 1) at s = (n + 1)/2 just obtained, we can easily get the residues for δ = −1. q.e.d.

4 Proof of functional equations for n ≥ 3

The functional equations for the case where n ≥ 3, as well as the residues are easily obtained just from the well known facts on the Riemann zeta function and results on D^{∗}(s, δ) and D(s, δ) in the last section. We carry out this calculation in this section. By using the following well known functional equation

ζ(1 − s) = π^{−s−1/2}(cos \frac{πs}{2})Γ(\frac{s}{2})Γ(\frac{s + 1}{2})ζ(s) = π^{−s}2^{1−s}(cos \frac{πs}{2})Γ(s)ζ(s),

we get

ζ(2(\frac{n + 1}{2} − s) − (2i − 1)) = π^{−2s+n−2i+1/2}Γ(s + i − \frac{n}{2})Γ(s + i − \frac{n + 1}{2})ζ(2s + 2i − n − 1)

× \{ \begin{matrix}(−1)^{i−(n+1)/2} cos πs & if n is odd, \\ (−1)^{i−n/2} sin πs & if n is even, \end{matrix}

ζ(2(\frac{n + 1}{2} − s) − 2i) = π^{−2s+n−2i−1/2}Γ(s + i − \frac{n}{2})Γ(s + i − \frac{n − 1}{2})ζ(2s + 2i − n)

× \{ \begin{matrix}(−1)^{i−(n−1)/2} sin πs & if n is odd, \\ (−1)^{i−n/2} cos πs & if n is even. \end{matrix}

Hence, for even n ≥ 4, we get

A_{n}(\frac{n + 1}{2} − s) = (−1)^{n(n−2)/8}π^{−(n−2)s+ \frac{n(n−3)}{4} + \frac{1}{2}} (cos πs)^{\frac{n}{2} −1} \frac{γ_{n}(s − \frac{n+1}{2})}{Γ(s)Γ(s − \frac{n−1}{2})}A_{n}(s)
B_{n}(\frac{n + 1}{2} − s) = (−1)^{n(n−2)/8}π^{−ns+ \frac{n(n−1)}{4}} γ_{n}(s − \frac{n + 1}{2})(sin πs)^{\frac{n}{2}} B_{n}(s).

Using Proposition 3.2 and 3.3, we get

D_{n}^{∗}(\frac{n + 1}{2} − s, 1)A_{n}(\frac{n + 1}{2} − s) = 2^{−n/2}π^{−ns+ \frac{n(n−1)}{4}} (cos πs)^{n/2}

×γ_{n}(s − \frac{n + 1}{2})A_{n}(s)D_{n}(s, 1),

D_{n}^{∗}(\frac{n + 1}{2} − s, −1)A_{n}(\frac{n + 1}{2} − s) = 2^{−n/2}π^{−ns+ \frac{n(n−1)}{4}} γ_{n}(s − \frac{n + 1}{2})A_{n}(s)

×((−1)^{n^{2}/4+1}(cos πs)^{n/2−1}(sin πs)D_{n}(s, −1) +(cos πs)^{n/2−1}D_{n}(s, 1)).

For odd n ≥ 3, we get

Q_{n}(\frac{n + 1}{2} − s) = π^{−ns+n(n−1)/4}2^{1−s}(−1)^{(n^{2}−1)/8}γ_{n}(s − \frac{n + 1}{2})(cos \frac{πs}{2})(cos πs)^{(n−1)/2}R_{n}(s),
R_{n}(\frac{n + 1}{2} − s) = π^{−ns+n(n−1)/4}2^{(n+1)/2−s}γ_{n}(s − \frac{n + 1}{2})(sin πs)^{(n−1)/2}Q_{n}(s)

× \{ \begin{matrix}cos \frac{πs}{2} & if n ≡ 1 mod 4, \\ sin \frac{πs}{2} & if n ≡ 3 mod 4. \end{matrix}

Now, the functional equations in section 1 are proved as follows. We put ϵ_{0} = ϵδ^{(n+1)/2}.

When n is odd, then for L = L_{n}^{∗} or L_{n}, the zeta function ζ(s, L, ϵ, δ) depends only on ϵ_{0}. For the sake of simplicity, we put ζ(s, L, ϵ, δ) = ζ^{+}(s, L), or ζ^{−}(s, L), if ϵ = 1, or −1, respectively. We have

ζ(\frac{n + 1}{2} − s, L_{n}^{∗}, δ, ϵ) = b_{n}2^{(n−1)((n+1)/2−s)}(Q_{n}(\frac{n + 1}{2} − s) + ϵ_{0}(−1)^{(n^{2}−1)/8}R_{n}(\frac{n + 1}{2} − s))
= π^{−ns+(n(n−1)/4}2^{−ns+(n^{2}+1)/2}γ(s − \frac{n + 1}{2})

×(cos \frac{πs}{2}(cos πs)^{(n−1)/2}R_{n}(s)

+(−1)^{(n^{2}−1)/8}2^{(n−1)/2}ϵ_{0}(sin πs)^{(n−1)/2}f(s)Q_{n}(s)), where we denote f(s) = cos \frac{πs}{2}, or sin \frac{πs}{2}, if n ≡ 1 mod 4, or 3 mod 4, respectively. We get the following equalities easily from the definitions.

Q_{n}(s) = b_{n}^{−1}2^{−(n+1)/2}(ζ^{+}(s, L_{n}) + ζ^{−}(s, L_{n})),
R_{n}(s) = b_{n}^{−1}2^{−1}(ζ^{+}(s, L_{n}) − ζ^{−}(s, L_{n})).

Hence, we get

ζ^{±}(\frac{n + 1}{2} − s, L_{n}^{∗}) = (2π)^{−ns}π^{n(n−1)/4}2^{(n^{2}−1)/2}γ_{n}(s − \frac{n + 1}{2})

×(cos \frac{πs}{2}(cos πs)^{(n−1)/2} ± (−1)^{(n^{2}−1)/8}(sin πs)^{(n−1)/2}f(s))ζ^{+}(s, L_{n})

+(− cos \frac{πs}{2}(cos πs)^{(n−1)/2} ± (−1)^{(n^{2}−1)/8}(sin πs)^{(n−1)/2}f(s))ζ^{−}(s, L_{n}).

Thus, we prove Theorem 1.

Next, we treat the case of even n. We prove the functional equation for each δ and n. Case(I): We take δ = 1 and n ≡ 0 mod 4 with n ≥ 4. Then, δ_{n} = 1 and

b_{n}(−1)^{[n/4]}D_{n}(s, 1)A_{n}(s) = 2^{−1}(ζ(s, L_{n}, 1, 1) + ζ(s, L_{n}, 1, −1)),
b_{n}(−1)^{n(n+2)/8} \frac{2|Bn/2^{′}|}{n}B_{n}(s) = 2^{−n/2−1}(ζ(s, L_{n}, 1, 1) − ζ(s, L_{n}, 1, −1)).

So, we get

ζ(\frac{n + 1}{2} − s, L_{n}^{∗}, 1, ϵ) =

2^{−ns+ \frac{n^{2}}{2} −1}π^{−ns+ \frac{n(n−1)}{4}} γ_{n}(s − \frac{n + 1}{2}) ×

((cos πs)^{n/2} × (ζ(s, L_{n}, 1, 1) + ζ(s, L_{n}, 1, −1))

+ϵ(−1)^{n(n−2)/8}(sin πs)^{n/2} × (ζ(s, L_{n}, 1, 1) − ζ(s, L_{n}, 1, −1))).

Case(II): We take δ = −1 and n ≡ 0 mod 4. Then δ_{n} = 0 and the zeta functions do not depend on ϵ.

ζ(\frac{n + 1}{2} − s, L_{n}^{∗}, −1, ϵ) =

2^{−ns−n^{2}/2}π^{−ns+ \frac{n(n−1)}{4}} γ_{n}(s − \frac{n + 1}{2})

×((−1)^{n^{2}/4+1}(cos πs)^{n/2−1}(sin πs)ζ(s, L_{n}, −1, ϵ) +(cos πs)^{n/2−1}2^{−1}(ζ(s, L_{n}, 1, 1) − ζ(s, L_{n}, 1, −1)).

Case(III): We take δ = 1, n ≡ 2 mod 4. Then, the zeta functions do not depend on ϵ. In this case, we have

ζ(\frac{n + 1}{2} − s, L_{n}^{∗}, 1, ϵ) = 2^{n( \frac{n+1}{2} −s)}2^{−n/2}π^{−ns+ \frac{n(n−1)}{4}}

×(cos πs)^{n/2}γ_{n}(s − \frac{n + 1}{2})ζ(s, L_{n}, 1, ϵ).

Case(IV): We put δ = −1 and n ≡ 2 mod 4. In this case, δ_{n} = 1, and

ζ(\frac{n + 1}{2} − s, L_{n}^{∗}, −1, ϵ) =

2^{−ns−n^{2}/2}π^{−ns+ \frac{n(n−1)}{4}} γ_{n}(s − \frac{n + 1}{2})

×((−1)^{n^{2}/4+1}2^{−1}(cos πs)^{n/2−1}(sin πs)(ζ(s, L_{n}, −1, 1) + ζ(s, L_{n}, −1, −1) +(−1)^{n(n−2)/8}2^{−1}ϵ(sin πs)^{n/2}(ζ(s, L_{n}, −1, 1) − ζ(s, L_{n}, −1, −1)) +(cos πs)^{n/2−1}ζ(s, L_{n}, 1, ϵ)).

Hence, we proved Theorem 3.

The results on residues are easily obtained as follows. As is well known, ζ(s) has a simple pole at s = 1 with residue 1 and is holomorphic for the other s. Also we have ζ(−2m) = 0 for all natural integers m. Theorem 2 follows very easily by these facts and the explicit formula of the zeta functions. To prove Theorem 4, we use Proposition 3.5.

5 Functional equation for n = 2

When n = 2, we need some modification of the definition of our zeta functions, since the volume μ(x) in the definition of zeta functions for n ≥ 3 is infinite for some x ∈ V when n = 2. Shintani has given a modified definition of the zeta function and has shown the functional equations for such zeta functions. Later, F. Sato [11] gave a more theoretical definition of the above Shintani’s modification by using a kind of Eisenstein series of O(2, 1) (a kind of zeta function of two variables) and gave an alternative proof of the functional equation. Here we give the third alternative proof of Shintani’s results above by using real analytic Eisenstein series of half integral weight. That is, we give a natural modification of the definition from our stand point, show the functional equation of our zeta functions, and give the location of poles and the principal parts of the Laurent expansion there. As for the zeta function of definite symmetric matrices (i.e. ζ_{2}(s, L, 1, 1)), this approach has been already used by J. Sturm [18]. So the main point is the case for indefinite symmetric matrices (i.e. ζ(s, L, −1, −1).) Our idea is as follows. When we fix an odd integer k, the functional equation of the Mellin transforms of Eisenstein series in the previous section is valid as a meromorphic function of s and σ for whole C^{2} plane. Now, we fix k = 1 from now on throughout this section. Then each side of the functional equation have a pole at σ = 2 and this is the reason that our original definition of our zeta function cannot be applied in case n = 2. That is, the most part of the coefficients of the above Dirichlet series with respect to s have definite values at σ = 2, which form the main part of our zeta function, but the modified terms come from the coefficients which have pole at σ = 2. That is, we can compare the constant terms of the Laurent expansion of both sides of the functional equation associated with Eisenstein series at σ = 2, and get the modification of zeta functions and the functional equation at the same time. By the way, it is easy to see that the residue of both sides at σ = 2 is

C_{2,1}ζ(2)^{−1}2^{s−2}π^{1/2−s}Γ(s − 1/2)ζ(2s − 1).

and this gives you nothing new. We define the main part of our zeta functions. Define four Dirichlet series

ξ_{+}^{M}(s) = ζ(2s) ∑_{d=1}^{∞} h(d) log(ϵ_{d})d^{−s},
ξ_{+}^{∗,M}(s) = ζ(2s){∑_{d=1}^{∞} h(4d) log(ϵ_{4d})(4d)^{−s} + 2^{−2s} ∑_{d=1}^{∞} h(4d + 1) log(ϵ_{4d+1})(4d + 1)^{−s}},
ξ_{−}(s) = 2ζ(2s) ∑_{d=1}^{∞} h(−d)w_{−d}^{−1}|d|^{−s},
ξ_{−}^{∗}(s) = 2ζ(2s){∑_{d=1}^{∞} h(−4d)w_{−4d}^{−1}|4d|^{−s} + 2^{−2s} ∑_{d=1}^{∞} h(−4d + 1)w_{−4d+1}^{−1}(4d − 1)^{−s}},

where h(d) is the class number in the narrow sense of the order O_{d} of conductor d in a quadratic field and ϵ_{d} is the fundamental unit of norm 1 of O_{d} and w_{d} is the number of units of O_{d}.

Then, by the classical class number formula, we can show that

ξ_{+}^{M}(s) = ∑_{d_{K}>0,d_{K}≠1} ζ(2)Z^{∗}(1, 2, s, d_{K}),
πξ_{−}(s) = ∑_{d_{K}<0} ζ(2)Z^{∗}(1, 2, s, d_{K}).

Now, put

h(σ, s) = \frac{ζ(2s)ζ(2s − 2σ + 3)}{ζ(2σ − 2)ζ(2s − σ + 2)}I(s, σ/2, (σ − 1)/2)
h^{∗}(σ, s) = \frac{ζ(2s)ζ(2s − 2σ + 3)}{ζ(2σ − 2)ζ(2s − σ + 2)} \frac{(1 − 2^{σ−2} + 2^{2σ−3} − 2^{2s+2σ−3})}{(1 − 2^{2s+σ−2})} × I(s, σ/2, (σ − 1)/2).

We define two functions g(σ, s) and g^{∗}(σ, s) so that the following equalities hold.

Φ_{σ,1}(s) = C_{σ,1}(2π)^{−s}Γ(s){g(σ, s) + ζ(σ − 1)h(σ, s)}
Ψ_{σ,1}(s) = C_{σ,1}(2π)^{−s}Γ(s){g^{∗}(σ, s) + ζ(σ − 1)h^{∗}(σ, s)}.

Then, we get

g(2, s) = ζ(2)^{−1}(ξ_{+}^{M}(s)I(s, 1, 1/2) + πξ_{−}(s)I(s, 1/2, 1))
g^{∗}(2, s) = 2^{2s}ζ(2)^{−1}(ξ_{+}^{∗,M}(s)I(s, 1, 1/2) + πξ_{−}^{∗}(s)I(s, 1/2, 1)).

To obtain the functional equation which coincides with Shintani’s original version, it is convenient to multiply by

ϵ(σ, s) = C_{σ,1}^{−1}(2π)^{σ−s−1/2}2^{(σ−1)/2}ζ(2σ − 2)Γ((σ − 1)/2)^{−1}Γ(1 − s)^{−1}

both sides of the functional equation between Φ_{σ,1}(s) and Ψ_{σ,1}(s) in the previous section. To rewrite the functional equation in this form, we put

A(σ, s) = 2^{(σ−1)/2}ζ(2σ − 2)Γ((σ − 1)/2)^{−1}

×\frac{Γ(σ − s − 1/2)}{Γ(1 − s)}g(σ, σ − s − 1/2)

A^{∗}(σ, s) = 2^{(σ−1)/2}ζ(2σ − 2)Γ((σ − 1)/2)^{−1}

×(2π)^{σ−2s−1/2}2^{2s−2σ+3/2} \frac{Γ(s)}{Γ(1 − s)}g^{∗}(σ, s)

B(σ, s) = \frac{Γ(σ − s − 1/2)}{Γ((σ + 1)/2 − s)} \frac{ζ(2σ − 1 − 2s)ζ(2 − 2s)}{ζ(σ + 1 − 2s)}

×F_{2}(σ, s)

B^{∗}(σ, s) = 2^{−σ+1}π^{σ−2s−1/2} \frac{Γ(s)Γ(s − σ + 3/2)}{Γ(1 − s)Γ(s − σ/2 + 1)}

×\frac{ζ(2s)ζ(2s − 2σ + 3)}{ζ(2s − σ + 2)} × \frac{1 − 2^{σ−2} + 2^{2σ−3} − 2^{−2s+2σ−3}}{1 − 2^{−2s+σ−2}}

×F_{1}(σ, s),

where, for the sake of simplicity, we put

F_{1}(σ, s) = F (1 − \frac{σ}{2} , \frac{σ − 1}{2}; s − \frac{σ}{2} + 1; \frac{1}{2}),
F_{2}(σ, s) = F (1 − \frac{σ}{2} , \frac{σ − 1}{2}, \frac{σ + 1}{2} − s; \frac{1}{2}).

Then, the functional equation reads as

A(σ, s) + B(σ, s)ζ(σ − 1) = A^{∗}(σ, s) + ζ(σ − 1)B^{∗}(σ, s).

Comparing the constant terms of the both sides at σ = 2, we get

A(2, s) + \frac{d}{dσ} B(σ, s)|_{σ=2}

= A^{∗}(2, s) + \frac{d}{dσ} B^{∗}(σ, s)|_{σ=2}.

As is easily shown, we get

A(2, s) = ξ_{+}^{M}(3/2 − s)

+\sqrt{2π} \frac{Γ(3/2 − s)}{Γ(1 − s)}ξ_{−}(3/2 − s)I(3/2 − s, 1/2, 1),

A^{∗}(2, s) = 2^{2s−1}π^{1/2−2s}Γ(s)Γ(s − 1/2)

×\{sin(πs)ξ_{+}^{∗,M}(s) + π^{3/2}\sqrt{2}\frac{ξ_{−}^{∗}(s)}{Γ(s − 1/2)Γ(1 − s)}I(s, 1/2, 1)\}.

It has been already proved in the previous section that

ξ_{−}(3/2 − s) = 2^{2s−1}π^{−2s+1/2}Γ(s)Γ(s − 1/2) cos(πs)ξ_{−}^{∗}(s)
−2^{−1}π^{−2s+1/2}Γ(s)Γ(s − 1/2)ζ(2s − 1).

Using this functional equation and the well known fact

cos(πs) = −π/Γ(s − 1/2)Γ(3/2 − s),

we get

A(2, s) = ξ_{+}^{M}(3/2 − s)
−2^{2s−1/2}π^{2−2s}Γ(s)Γ(1 − s)^{−1}ξ_{−}^{∗}(s)I(3/2 − s, 1/2, 1)
−2^{−1/2}π^{1−2s}Γ(s)Γ(s − 1/2)Γ(3/2 − s)Γ(1 − s)^{−1}
×ζ(2s − 1)I(3/2 − s, 1/2, 1).

Hence, by using the formula

I(s, 1/2, 1) + I(3/2 − s, 1/2, 1) = \frac{Γ(1 − s)Γ(s − 1/2)}{\sqrt{2π}},

and applying the functional equation, we get

ξ_{+}^{M}(3/2 − s) + \frac{d}{dσ} B(σ, s)|_{σ=2}

= 2^{2s−1}π^{−2s+1/2}Γ(s)Γ(s − 1/2){sin(πs)ξ_{+}^{∗,M}(s) + πξ_{−}^{∗}(s)}

+\frac{π^{1−2s}}{\sqrt{2}} \frac{Γ(s)Γ(s − 1/2)Γ(3/2 − s)}{Γ(1 − s)}ζ(2s − 1)I(\frac{3}{2} − s, \frac{1}{2}, 1) + \frac{d}{dσ} B^{∗}(σ, s)|_{σ=2}.

Now, the problem is to calculate the difference

\frac{d}{dσ} B^{∗}(σ, s)|_{σ=2} − \frac{d}{dσ} B(σ, s)|_{σ=2}.

When we differentiate B(σ, s) and B^{∗}(σ, s), roughly we regard these as products of functions of σ which are powers of two, the shifted Riemann zeta functions, gamma factors, and hypergeometric functions. We get

\frac{d}{dσ} B(σ, s)|_{σ=2} = ζ(2 − 2s)\{\frac{ζ′(3 − 2s)}{ζ(3 − 2s)} − \frac{ζ′(2 − 2s)}{ζ(2 − 2s)}\} + ζ′(2 − 2s)

+\frac{1}{2}ζ(2 − 2s)\frac{Γ′(3/2 − s)}{Γ(3/2 − s)}

+ζ(2 − 2s)\frac{d}{dσ} F_{2}(σ, s)|_{σ=2}.

and

\frac{d}{dσ} B^{∗}(σ, s)|_{σ=2} = π^{3/2−2s} \frac{Γ(s − 1/2)}{Γ(1 − s)}\{ζ(2s − 1)\{\frac{ζ′(2s)}{ζ(2s)} − \frac{ζ′(2s − 1)}{ζ(2s − 1)}\} − ζ′(2s − 1)\}

+(log π)π^{3/2−2s}ζ(2s − 1)\frac{Γ(s − 1/2)}{Γ(1 − s)}

+(log 2)π^{3/2−2s} \frac{Γ(s − 1/2)}{Γ(1 − s)} \frac{ζ(2s − 1)}{2(1 − 2^{−2s})}

+2^{−1}π^{1/2−2s}ζ(2s − 1)Γ(s)Γ(s − 1/2)(sin πs)\{\frac{Γ′(s)}{Γ(s)} − \frac{Γ′(s − 1/2)}{Γ(s − 1/2)}\}

−2^{−1}π^{1/2−2s}ζ(2s − 1)Γ(s)Γ(s − 1/2)(sin πs)\frac{Γ′(s − 1/2)}{Γ(s − 1/2)}

+π^{3/2−2s}ζ(2s − 1)\frac{Γ(s − 1/2)}{Γ(1 − s)} \frac{d}{dσ} F_{1}(σ, s)|_{σ=2}.

By the functional equation of the Riemann zeta function and the well known formula Γ(s)Γ(1 − s) = π/ sin(πs), we get

ζ(2 − 2s) = π^{3/2−2s}ζ(2s − 1)\frac{Γ(s − 1/2)}{Γ(1 − s)}
= π^{1/2−2s}(sin πs)ζ(2s − 1)Γ(s)Γ(s − 1/2).

Now we would like to calculate

ζ(2 − 2s)\frac{d}{dσ} (F_{1}(σ, s) − F_{2}(σ, s))|_{σ=2}.

The following formula of the hypergeometric functions is well known.

F (a, b, c, z) = \frac{Γ(c)Γ(c − a − b)}{Γ(c − a)Γ(c − b)}F (a, b; a + b − c + 1; 1 − z)

+(1 − z)^{c−a−b} \frac{Γ(c)Γ(a + b − c)}{Γ(a)Γ(b)}F (c − a, c − b; c − a − b + 1; 1 − z).

So, we get

F_{1}(σ, s) = \frac{Γ(s − σ/2 + 1)Γ(s − (σ − 1)/2)}{Γ(s)Γ(s − σ + 3/2)}F_{2}(σ, s)

+2^{−s+(σ−1)/2} \frac{Γ(s − σ/2 + 1)Γ(−s + (σ − 1)/2)}{Γ(1 − σ/2)Γ((σ − 1)/2)}

×F (s, s − σ + 3/2, s − (σ − 3)/2; \frac{1}{2}).

Now, we differentiate at σ = 2 the both sides of the above formula. Since Γ(1 − σ/2)^{−1} has a simple zero at σ = 2 and \frac{d}{dσ}Γ(1−σ/2)^{−1}|_{σ=2} = −1/2, the differential coefficient at σ = 2 of the second term of the right hand side is

−2^{−1/2−s} \frac{Γ(s)Γ(1/2 − s)}{Γ(1/2)}F (s, s − 1/2, s + 1/2; 1/2),

but we see

F (s, s − \frac{1}{2}, s + \frac{1}{2} : \frac{1}{2}) = \frac{Γ(s + \frac{1}{2})}{Γ(s − \frac{1}{2})} ∫_{0}^{1} t^{s− \frac{3}{2}} (1 − \frac{1}{2}t)^{−s}dt
= 2^{s− \frac{1}{2}} \frac{Γ(s + \frac{1}{2})}{Γ(s − \frac{1}{2})} ∫_{1/2}^{1}(1 − v)^{s− \frac{3}{2}} v^{−s}dt
= 2^{s} \frac{Γ(s + \frac{1}{2})}{Γ(s − \frac{1}{2})}I(s, \frac{1}{2}, 1).

Hence we get

\frac{d}{dσ} (F_{1} − F_{2})|_{σ=2} = −\frac{1}{2}(\frac{Γ′(s)}{Γ(s)} − \frac{Γ′(s − 1/2)}{Γ(s − 1/2)})

−(2π)^{−1/2} \frac{Γ(s)Γ(1/2 − s)Γ(s + 1/2)}{Γ(s − 1/2)}I(s, 1/2, 1).

Using the notation of Shintani, we define

ξ_{+}(s) = ξ_{+}^{M}(s) + ζ(2s − 1)(\frac{ζ′(2s)}{ζ(2s)} − \frac{ζ′(2s − 1)}{ζ(2s − 1)}),
ξ_{+}^{∗}(s) = ξ_{+}^{∗,M}(s) + 2^{1−2s}ζ(2s − 1)(\frac{ζ′(2s)}{ζ(2s)} − \frac{ζ′(2s − 1)}{ζ(2s − 1)}),
+2^{−2s}(log 2)(1 − 2^{−2s})^{−1}ζ(2s − 1).

We prepare two more relations. By differentiating the usual functional equation of the Rieman zeta function, we get

\frac{ζ′(2 − 2s)}{ζ(2 − 2s)} = log π − \frac{1}{2} \frac{Γ′(s − 1/2)}{Γ(s − 1/2)} − \frac{1}{2} \frac{Γ′(1 − s)}{Γ(1 − s)} − \frac{ζ′(2s − 1)}{ζ(2s − 1)}.

On the other hand, differentiating the relations Γ(s)Γ(1 − s) = π/ sin πs and Γ(s − 1/2)Γ(3/2 − s) = −π/ cos πs, we get

\frac{Γ′(s − 1/2)}{Γ(s − 1/2)} + \frac{Γ′(1 − s)}{Γ(1 − s)} = \frac{π}{(cos πs)(sin πs)} + \frac{Γ′(s)}{Γ(s)} + \frac{Γ′(3/2 − s)}{Γ(3/2 − s)}.

By comparing all the equalities we obtained above, we can show that

ξ_{+}(3/2 − s) = 2^{2s−1}π^{−2s+1/2}Γ(s)Γ(s − 1/2)((sin πs)ξ_{+}^{∗}(s) + πξ_{−}^{∗}(s))

+2^{−1}π^{−2s+1/2}Γ(s)Γ(s − 1/2)ζ(2s − 1)(sin πs)(\frac{Γ′(s)}{Γ(s)} − \frac{Γ′(s − 1/2)}{Γ(s − 1/2)}).

This is completely the same as Shintani’s functional equation.

Next, for the sake of completeness, we shall calculate principal part of the Laurent expansion at poles of ξ_{+}(s) and ξ_{+}^{∗}(s) by our method. For that purpose, we need more precise information on the location of poles and their residues of the functions which appear in Φ_{σ,1}(s) and Ψ_{σ,1}(s).

We define

I_{2}(s) = \frac{Γ(3/2 − s)}{2\sqrt{π}Γ(1 − s)} ∫_{0}^{1} u^{−1/2}(1 − u)^{−s} log(1 − u/2) du.

This function converges when Re(s) < 1. Differentiating the integral expression of the beta function

B(t + \frac{1}{2}, 1 − s) = \frac{Γ(t + 1/2)Γ(1 − s)}{Γ(t + 3/2 − s)} = ∫_{0}^{1} u^{t−1/2}(1 − u)^{−s} du

at t = 0 under the integral sign, we get

\frac{Γ(3/2 − s)}{\sqrt{π}Γ(1 − s)} ∫_{0}^{1} u^{−1/2}(1−u)^{−s} log(u) du = \frac{Γ′(1/2)Γ(3/2 − s) − Γ(1/2)Γ′(3/2 − s)}{\sqrt{π}Γ(3/2 − s)}.

Using this relation, and differentiating under the integral sign, we get

\frac{d}{dσ} (F_{1}(σ, s))|_{σ=2} = I_{2}(s) and
\frac{d}{dσ} (F_{2}(σ, s))|_{σ=2} = I_{2}(\frac{3}{2} − s).

As before, for a non negative integer n and a variable number a, we put (a)_{n} = Γ(a + n)/Γ(a). We define the Barnes generalized hypergeometric function 3F_{2} as usual by

3F_{2}(α_{1}, α_{2}, α_{3}; β_{1}, β_{2}; z) = ∑_{n=0}^{∞} \frac{(α_{1})_{n}(α_{2})_{n}(α_{3})_{n}}{(β_{1})_{n}(β_{2})_{n}} \frac{z^{n}}{n!} .

Then, we also get

I_{2}(s) = − ∑_{n=1}^{∞} \frac{1}{2^{n+1}n} \frac{Γ(n + 1/2)Γ(3/2 − s)}{\sqrt{π}Γ(n + 3/2 − s)}
= −\frac{Γ(3/2 − s)}{2^{3}Γ(5/2 − s)}_{3}F_{2}(1, 1, 3/2; 2, 5/2 − s; 1/2).

Since I_{2}(s) is holomorphic for Re(s) < 1 and I_{2}(3/2 − s) for Re(s) > 1/2, I_{2}(s) is meromorphically continued to the whole s plane by virtue of the formula for I_{2}(3/2−s)−I_{2}(s) we gave before. It is also easy to show that I_{2}(s) has simple poles at s = 1/2 + m (m = 1,2,…) with residues

(−1)^{m+1}2^{−1/2−2m}(2m)!(m!)^{−2}mI(1/2 + m, 1/2, 1),

and holomorphic for the other s. We get

\frac{d}{dσ} B(σ, s)|_{σ=2} = ζ(2 − 2s)(\frac{ζ′(3 − 2s)}{ζ(3 − 2s)} − \frac{ζ′(2 − 2s)}{ζ(2 − 2s)}) + ζ′(2 − 2s)

+\frac{1}{2}ζ(2 − 2s)\frac{Γ′(3/2 − s)}{Γ(3/2 − s)} + ζ(2 − 2s)I_{2}(s).

It was shown in section 3 that the function (σ − 2)s(s − σ + 1/2)(s − σ + 3/2)ϵ(σ, s)Φ_{σ,1}(σ −1/2−s) is holomorphic on (σ, s) ∈ U_{3/2}(2)×C, where U_{r}(a) = {σ ∈ C; |σ − a| < r}. We write the Taylor expansion at σ = 2 of this function as ∑_{n=0}^{∞} f_{i}(s)(σ − 2)^{n}. The first two terms are calculated explicitly as follows.

f_{0}(s) = s(s − 1/2)(s − 3/2)ζ(2 − 2s).
f_{1}(s) = s(s − 1/2)(s − 3/2)(A(2, s) + \frac{d}{dσ} B(σ, s)|_{σ=2} + γB(2, s))

−2s(s − 1)B(2, s)

= s(s − 1/2)(s − 3/2) × (ξ_{+}(3/2 − s)

+\sqrt{2π} \frac{Γ(3/2 − s)}{Γ(1 − s)}ξ_{−}(3/2 − s)I(3/2 − s, 1/2, 1) + ζ′(2 − 2s)

+\frac{1}{2}ζ(2 − 2s)\frac{Γ′(3/2 − s)}{Γ(3/2 − s)} + \frac{1}{2}ζ(2 − 2s)I_{2}(s) + γζ(2 − 2s))

−2s(s − 1)ζ(2 − 2s).

Now, notation being as in section 3, we put

J̃(σ, s) = 2^{s+1/2−σ}ϵ(σ, s)

×(∫_{1}^{∞}(G_{1}(y/2) − c_{0}(y/2))y^{σ−3/2−s} dy + C_{1} ∫_{1}^{∞}(H_{1}(y/2) − c_{0}(y/8))y^{s−1} dy).

Then, (σ − 2)Γ(1 − s)J̃(σ, s) is holomorphic on (σ, s) ∈ U_{3/2}(2) × C. By the formula in section 3, we get

A(σ, s) + ζ(σ − 1)B(σ, s) = J̃(σ, s) − π^{1−s} \frac{Γ(σ − 3/2)ζ(2σ − 3)}{Γ((σ − 1)/2)Γ(1 − s)} × (\frac{2^{σ/2−1}}{s − σ + 3/2} + \frac{2^{1−σ/2}}{1 − s})

−π^{−s} \frac{Γ(σ/2)ζ(2σ − 2)}{Γ(1 − s)} × (\frac{2^{(σ−3)/2}}{s} + \frac{2^{(3σ−5)/2}}{σ − 1/2 − s}).

We write the Laurent expansion of J̃(σ, s) at σ = 2 as J̃(σ, s) = J_{−1}(s)/(σ −

2) + J_{0}(s) + · · ·. Calculating the first few terms of the Taylor expansion at σ = 2 of the product of the right hand side of the above equality and (σ − 2)s(s − σ + 1/2)(s − σ + 3/2), we get the following expression of f_{i}(s) (i = 0, 1).

f_{0}(s) = −2^{−2}π^{1−s} \frac{s(s − 3/2)}{(1 − s)Γ(1 − s)} + s(s − 1/2)(s − 3/2)J_{−1}(s).
f_{1}(s) = −2s(s − 1)J_{−1}(s) + s(s − 1/2)(s − 3/2)J_{0}(s)

+\frac{\sqrt{2}π^{2−s}}{12} \frac{(s − 1/2)(s + 3/2)}{Γ(1 − s)}

−\frac{π^{1−s}}{8(1 − s)Γ(1 − s)}(3γs(s − 3/2) + s(s − 3/2)(1 − 4s) log(2) + 4s(1 − s)),

where γ is the Euler constant and γ = −Γ′(1). First, we calculate the Laurent expansion of ξ_{+}(s) at s = 1. It is clear that the order of the pole of ξ_{+}(s) at s = 1 is at most two. We write ξ_{+}(s) = a_{−2}(s − 1)^{−2} + a_{−1}(s − 1)^{−1} + O(1). Since ζ′(2 − 2s) = −(1 − 2s)^{−2} + O(1)) and −2s(s − 1)ζ(2 − 2s) = 2^{−1}(1 − 2s)^{−1} + O(s − 1/2), and since f_{1}(s) is holomorphic, we get −2^{−1}(a_{−2} − 4^{−1}) − 4^{−1} = 0, hence a_{−2} = −1/4.

Comparing the values at s = 1/2 of the both expression of f_{1}(s), we can calculate a_{−1} as follows. From the first equality above, we get

J_{−1}(s) = ζ(2 − 2s) + \frac{2^{−2}π^{1−s}}{(1 − s)(s − 1/2)Γ(1 − s)}
= \frac{1}{2}(γ − log(π)) + 1 − log(2) + O(s − 1/2).

We easily get

f_{1}(\frac{1}{2}) = \frac{5}{8}γ − \frac{1}{4} log(π) + \frac{1}{4} − \frac{5}{8} log(2).

On the other hand, we have

s(s − 1/2)(s − 3/2)ξ_{+}(3/2 − s) = +8^{−1}(s − 1/2)^{−1} + (4^{−1} + 2^{−1}a_{−1}) + o(s − 1/2), I(1, 1/2, 1) = \sqrt{2} log(\sqrt{2}+1), I_{2}(1/2) = −2^{−1}3 log(2)+log(\sqrt{2}+1), and ξ_{−}(3/2− s) = 4^{−1}(s − 2/1)^{−1} + O(1). Hence, we also get

f_{1}(1/2) = \frac{1}{4} + \frac{1}{2}a_{−1} + \frac{5}{8}γ − \frac{3}{8} log(2).

Hence, we get a_{−1} = −2^{−1} log(2π). Now, f_{1}(0) = −2^{−4}\sqrt{2}π^{2}. Since I(3/2, 1/2, 1) = 2^{−1} ∫_{0}^{1}(1 − 2^{−1}t)^{−1/2}dt = 2 − \sqrt{2} and the residue of ξ_{−}(3/2 − s) at s = 0 is −π/12. Hence, if we denote the residue of ξ_{+}(s) at s = 3/2 by d, then f_{1}(0) = −2^{−2}3d − 2^{−5}π^{2}(2 − \sqrt{2}). Hence d = π^{2}/12. Now, we shall show that ξ_{+}(s) is holomorphic for all s ≠ 1, 3/2. Since Γ(1 − s)^{−1}I(3/2 − s, 1/2, 1) is entire and I_{2}(s) has poles only at s = 1/2 + m (m = 1,2,…), we see ξ_{+}(3/2 − s) is holomorphic for s ≠ 0,1/2, 1/2+m. Since the residue of Γ(1/2−s) at s = m+1/2 is (−1)^{m}/Γ(m + 1), we see that the residue at s = m + 1/2 of I_{2}(s)/2 is

(−1)^{m}2^{−2}π^{−1}Γ(m)^{−1}Γ(m + 1/2)I(1/2 + m, 1/2, 1).

The residue of Γ′(3/2 − s)/Γ(3/2 − s) at m + 1/2 is −1/2. By the functional equation for ξ_{−}(s), we see that ξ_{−}(1 − m) = −2^{−1}(−1)^{m}ζ(1 − 2m) for natural numbers m ≥ 2 and ξ_{−}(0) = 2^{−1}ζ(−1) + 1/12. We get I(1 − m, 1/2, 1) + I(m + 1/2, 1/2, 1) = (2π)^{−1}Γ(1/2 − m)Γ(m) and I(3/2, 1/2, 1) = 2 − \sqrt{2}, I(0, 1/2, 1) = −2. So, comparing the Taylor expansion at s = m + 1/2 of the two expression of f_{1}(s), it is easy to see that ξ_{+}(3/2 − s) is holomorphic also at s = m + 1/2 for natural numbers m. Thus we obtained all the location of poles of ξ_{+}(s) and its principal part of the Laurent expansion at each pole. As for ξ_{+}^{∗}(s), we have

2^{1−2s}π^{2s−3/2}Γ(1 − s)Γ(s − 1/2)^{−1}f_{1}(s)

= s(s − 1/2)(s − 3/2)(ξ_{+}^{∗}(s) + (2π)^{1/2} \frac{Γ(s)}{Γ(s − 1/2)}ξ_{−}^{∗}(s)I(s, 1/2, 1)

−2^{1−2s}ζ′(2s − 1) + 2^{1−2s} log(π)ζ(2s − 1) + 2^{1−2s}γζ(2s − 1)

+2^{−2s}ζ(2s − 1)(\frac{Γ′(s)}{Γ(s)} − 2\frac{Γ′(s − 1/2)}{Γ(s − 1/2)}) + 2^{1−2s}ζ(2s − 1) × I_{2}(3/2 − s))

−2s(s − 1)2^{1−2s}ζ(2s − 1).

By the functional equation, we get ξ_{−}^{∗}(1 − m) = (−1)^{m+1}2^{2m−2}ζ(1 − 2m) for m = 2, 3, … and ξ_{−}^{∗}(0) = 0. Using this, we get the location of poles and the expansion of ξ_{+}^{∗}(s) at poles almost in the same way as before. We omit the details here. Thus, we get the alternative proof of the following results first obtained by Shintani [16] in somewhat more complicated method.

Proposition 5.1 (Shintani [16], F. Sato [11]) Two zeta functions ξ_{+}(s) and ξ_{+}^{∗}(s) are meromorphically continued to the whole s plane and satisfies the following functional equation.

ξ_{+}(3/2 − s) = 2^{2s−1}π^{−2s+1/2}Γ(s)Γ(s − 1/2)((sin πs)ξ_{+}^{∗}(s) + πξ_{−}^{∗}(s))

+2^{−1}π^{−2s+1/2}Γ(s)Γ(s − 1/2)ζ(2s − 1)(sin πs)(\frac{Γ′(s)}{Γ(s)} − \frac{Γ′(s − 1/2)}{Γ(s − 1/2)}).

Both zeta functions are holomorphic except for s = 1 and s = 3/2. The principal part of the Laurent expansions are given as follows. At s = 1, we get ξ_{+}(s) = −4^{−1}(s − 1)^{−2} − 2^{−1} log(2π)(s − 1)^{−1} + ⋯ and ξ_{+}^{∗}(s) = −8^{−1}(s − 1)^{−2} − 4^{−1} log(2π)(s − 1)^{−1} + · · · . At s = 3/2, we get ξ_{+}(s) = 12^{−1}π^{2}(s − 3/2) + ⋯, and ξ_{+}^{∗}(s) = 24^{−1}π^{2}(s − 3/2)^{−1} + · · ·.

.

6 Special values of zeta functions and dimensions of Siegel cusp forms

The values ζ^{+}(1 − m, L_{n}^{∗}, 1, 1) for natural numbers m are related with dimension formulas of Siegel cusp forms and important. In this section, first we shall give the formulas for the values of ζ^{+}(s, L, 1, 1) at non positive integers for L = L_{n}^{∗}, or L_{n}. These are rational, and described by Bernoulli numbers explicitly. Secondly we apply these results to the dimensions of Siegel cusp forms. The results in this section have been announced in [4] before, but we reproduce them with more detailed explanation and the proof.

For each non-negative integer m, we denote by B_{m} the Bernoulli number defined by

\frac{te^{t}}{e^{t} − 1} = ∑_{m=0}^{∞} B_{m} \frac{t^{m}}{m!}.

Next, we shall give special values of the above zeta functions.

Theorem 5 When n is odd (including the case n = 1), for each natural number m,

ζ(1 − m, L_{n}^{∗}, 1, 1) = \frac{(−1)^{(n+1)/2}|B_{2}B_{4} ⋯ B_{n−1}|B_{m+ \frac{n−1}{2}} B_{2m}B_{2m+2} ⋯ B_{2m+n−3}}{2^{(2m+1)(n−1)/2}(\frac{n−1}{2})!m(m + 1) ⋯ (m + \frac{n−1}{2})} ,

ζ(1 − m, L_{n}, 1, 1) = 2^{(2m−1)(n−1)/2}ζ(1 − m, L_{n}^{∗}, 1, 1).

Theorem 6 When n is even, for each natural number m with nm ≠ 2,

ζ(1 − m, L_{n}^{∗}, 1, 1) = \frac{(−1)^{[ \frac{n}{4} ]+ \frac{nm}{2}} |B_{2}B_{4} ⋯ B_{n−2}B_{n/2}|B_{2m}B_{2m+2} ⋯ B_{2m+n−2}}{2^{nm+ \frac{n−2}{2}} (\frac{n}{2})!m(m + 1) ⋯ (m + \frac{n−2}{2})} ,

ζ(1 − m, L_{n}, 1, 1) = 2^{nm− \frac{n}{2}} ζ(1 − m, L_{n}^{∗}, 1, 1),

and ζ(0, L_{2}^{∗}, 1, 1) = 1/48, ζ(0, L_{2}, 1, 1) = 0. In particular, if n ≡ 2 mod 4 and n > 2, then ζ(1 − m, L_{n}^{∗}, 1, 1) = ζ(1 − m, L_{n}, 1, 1) = 0.

Proof of Theorem 5, 6. The case when n is odd is obvious, noting the well known relation ζ(1 − m) = −B_{m}/m for the Riemann zeta function. Now, assume n is even. By Proposition 3.2 and 3.3, D_{n}^{∗}(s, 1) and D_{n}(s, 1) are holomorphic at 1 − m (m = 1, 2, …). When n = 2, we have ζ(s, L_{2}^{∗}, 1, 1) = 2^{2s−1}D_{2}^{∗}(s, 1) and ζ(s, L_{2}, 1, 1) = 2^{−1}D_{2}(s, 1). So, our results easily follow from Proposition 3.3. When n ≥ 4, we have A_{n}(1 − m) = 0 and

B_{n}(1 − m) = (−1)^{n/2} ∏_{i=1}^{n/2} B_{2m+2i−2}/(2m + 2i − 2),

and we get the special values. q.e.d.

Remark. When n = 2, these values have been essentially known by the functional equation. When n = 3, as a special case of the above results, we get ζ(0, L_{3}^{∗}) = \frac{1}{3456}. This value has been obtained by an indirect method by Hashimoto and Tsushima. They compared two expressions of the dimension of Siegel cusp forms of degree 3, one by the Selberg trace formula and the other by Riemann Roch Theorem.

Next, we shall give an explicit value of the contribution of ‘central’ unipotent elements to the dimension of Siegel cusp forms. For each natural numbers n and N, we denote by Γ_{n}(1) = Sp(n, Z) the full Siegel modular group of size 2n and Γ_{n}(N) the principal congruence subgroup of level N of degree n:

Γ_{n}(N) = {g ∈ Sp(n, Z); g ≡ 1_{2n} mod N}.

For a natural number k, we denote by S_{k}(Γ_{n}(N)) the space of Siegel cusp forms of weight k belonging to Γ_{n}(N). First we review the Godement dimension formula.

Theorem 7 (Godement) For wny integer k > 2n, we get

dim S_{k}(Γ_{n}(N)) = \frac{a(k)}{#Z(Γ_{n})} ∫_{Γ_{n}\H_{n}} ∑_{γ∈Γ_{n}} H_{γ}(Z)dZ,

where dZ = (det(Y))^{−n−1}dXdY,

H_{γ}(Z) = det(\frac{Z − γZ̄}{2i})^{−k} det(CZ̄ + D)^{−k}(det Y)^{k},

and

a_{n}(k) = \frac{1}{2^{n}(2π)^{n(n+1)/2}} ∏_{j=0}^{n−1} \frac{Γ(k − \frac{n−1}{2} + \frac{j}{2})}{Γ(k − n + \frac{j}{2})}
= \frac{1}{2^{n^{2}+2n}π^{n(n+1)/2}} ∏_{1≤i≤j≤n}(2k − 2n + i + j − 2).

Now, for each natural number r with 0 ≤ r ≤ n, let Π_{r} be the set of elements

u of Γ_{n}(N) such that u is Sp(n, Z)-conjugate to a matrix of the form ( \begin{matrix}1_{n} & x \\ 0 & 1_{n} \end{matrix} )

where x ∈ L_{n} and x is of rank r. When r = 0, this is nothing but 1. When r ≥ 1, we call such unipotent elements ‘central’, because they are Sp(n, Q)-conjugate to some elements of the center of the unipotent radical of a maximal parabolic subgroup of Sp(n, Q). Using Shintani’s notation, we denote by I_{n}(Π_{r}, N, k) the contribution of Π_{r} to dim S_{k}(Γ_{n}(N)). That is, we put

I_{n}(Π_{r}, N, k) = \frac{a(k)}{#Z(Γ_{n})} ∫_{Γ_{n}\H_{n}} ∑_{γ∈Π_{r}} H_{γ}(Z)dZ.

Shintani [16] has given a formula to express each I_{n}(Π_{r}, N, k) as a product of some elementary factors and a special value of ζ(s, L_{n}^{∗}) as follows. For each natural number l, we put

ω = ∏_{i=1}^{l} ζ(2i),

and

U_{l} = ∏_{i=1}^{l} \frac{2π^{i}}{Γ(i)}.

Theorem 8 (Shintani [16]) For each natural number n and each integer k > 2n, we get

I_{n}(Π_{0}, N, k) = [Γ_{n}(1) : Γ_{n}(N)]\frac{ω_{n}}{U_{n}(4π)^{n(n+1)/2}} ∏_{i=1}^{n} ∏_{j=0}^{i−1}(2k − n − i + 2j),

and for each r with 1 ≤ r ≤ n − 1, we get

I_{n}(Π_{r}, N, k) = [Γ_{n}(1) : Γ_{n}(N)]\frac{ω_{n−r}}{U_{n}(4π)^{(n−r)(n−r+1)/2}} ∏_{i=1}^{n−r} ∏_{j=0}^{i−1}(2k − n − i + 2j)ζ(r − n, L_{r}^{∗}, 1, 1).

Remark. In Shintani’s original theorem, he assumed that k ≥ 2n + 3, but it is easy to show that this also holds for k ≥ 2n + 1 by applying carefully the estimation he used. Incidentally, ζ(s, L_{n}^{∗}, 1, 1) = ζ_{r}^{∗}(s) in his notation.

Using this theorem, we get an explicit formula for I_{n}(Π_{r}, N, k) as a corollary to our Theorem 5 and 6. We state this below.

Before stating this result, we prepare more notations. For each non-negative integer r with 0 ≤ r ≤ n, we set

C_{n−r} = C_{n−r}(k, N) = [Sp(n, Z) : Γ_{n}(N)]N^{−r(n− \frac{r−1}{2} )} ∏_{t=1}^{n−r} \frac{t!}{(2t − 1)!!} ( \begin{matrix}k − 1 − \frac{n}{2} + \frac{t}{2} \\ t \end{matrix} ) ,

where, for any natural number t, we denote by ( \begin{matrix}α \\ t \end{matrix} ) the binomial coefficient

( \begin{matrix}α \\ t \end{matrix} ) = \frac{α(α − 1) ⋯ (α − t + 1)}{t!},

and (2t − 1)!! = 1 · 3 ⋯ (2t − 1) = 2^{−t}t! ( \begin{matrix}2t \\ t \end{matrix} ). When r = n, the product part

in the above definition of C_{n−r} is regarded as 1.

Theorem 9 For each natural numbers n, N, k with k > 2n, the contribution I_{n}(Π_{r}, k, N) is given as follows.

(1) For odd r with 1 ≤ r ≤ n,

I_{n}(Π_{r}, k, N) =

C_{n−r} × \frac{(−1)^{[ \frac{n+1}{2} ]}|B_{n− \frac{r−1}{2}} B_{2}B_{4} ⋯ B_{r−1}B_{2}B_{4} ⋯ B_{2n−2r}B_{2n−2r+2}B_{2n−2r+4} ⋯ B_{2n−r−1}|}{2^{n+ \frac{r−1}{2}} (n − \frac{r−1}{2})!(\frac{r−1}{2})!}.

In particular, if n − \frac{r−1}{2} is an odd number greater than 1, then I_{n}(Π_{r}, k, N) = 0.

(2) For even r with 0 ≤ r ≤ n,

I_{n}(Π_{r}, k, N) =

C_{n−r} × \frac{(−1)^{r(1+δ_{nr})/2}|B_{2}B_{4} ⋯ B_{r−2}B\frac{r}{2} | · |B_{2}B_{4} ⋯ B_{2n−2r}B_{2n−2r+2}B_{2n−2r+4} ⋯ B_{2n−r}|}{2^{2n− \frac{r}{2}} (\frac{r}{2})!(n − \frac{r}{2})!},

where δ_{nr} is the Kronecker symbol. In particular, for r such that r ≡ 2 mod 4 with r ≠ 2, we get I_{n}(Π_{r}, k, N) = 0.

The proof is straightforward, if we note π^{−i}Γ(i)ζ(2i) = \frac{2^{i−1}}{(2i−1)!!}π^{i}|B_{2i}|.

When N ≥ 3, it is conjectured, independently by several mathematicians, that any contribution to dim S_{k}(Γ_{n}(N)) of conjugacy classes other than central unipotent conjugacy classes and the identity element would vanish. So, we are naturally led to the following conjecture. The point of our conjecture is that everything is explicit.

Conjecture When N ≥ 3 and k > 2n,

dim S_{k}(Γ_{n}(N)) = ∑_{r=0}^{n} I_{n}(Π_{r}, k, N),

where each I_{n}(Π_{r}, k, N) is as given explicitly in Theorem 5 above.

This conjecture is known to be true for n ≤ 3 by Christian [1], Morita [7], Yamazaki [20] and Tsushima [19].

Numerical examples of the conjecture: We assume N ≥ 3 hereafter.

(1) When n = 4 and k > 8, the conjecture reads as

dim S_{k}(Γ_{4}(N)) = [Sp(4, Z) : Γ_{4}(N)]×

\{\frac{1}{2^{25} · 3^{8} · 5^{4} · 7^{2}} (2k − 2)(2k − 3)(2k − 4)^{2}(2k − 5)^{2}(2k − 6)^{2}(2k − 7)(2k − 8)

+\frac{1}{N^{4}} × \frac{1}{2^{17} · 3^{7} · 5^{3} · 7}(2k − 3)(2k − 4)(2k − 5)^{2}(2k − 6)(2k − 7)

−\frac{1}{N^{7}} × \frac{1}{2^{15} · 3^{5} · 5 · 7}(2k − 4)(2k − 5)(2k − 6) + \frac{1}{N^{10}} × \frac{1}{2^{12} · 3^{4} · 5}\}.

(2) When n = 5 and k > 10, the conjecture reads as

dim S_{k}(Γ_{5}(N)) = [Sp(5, Z) : Γ_{5}(N)]×

\{\frac{(2k − 2)(2k − 3)(2k − 4)^{2}(2k − 5)^{2}(2k − 6)^{3}(2k − 7)^{2}(2k − 8)^{2}(2k − 9)(2k − 10)}{2^{33} · 3^{12} · 5^{5} · 7^{3} · 11}

−\frac{1}{N^{9}} × \frac{1}{2^{23} · 3^{7} · 5^{3} · 7}(2k − 4)(2k − 5)(2k − 6)^{2}(2k − 7)(2k − 8)

−\frac{1}{N^{12}} × \frac{1}{2^{17} · 3^{7} · 5^{2} · 7}(2k − 5)(2k − 6)(2k − 7)

+\frac{1}{N^{14}} × \frac{1}{2^{16} · 3^{6} · 5 · 7}(2k − 6)\}.

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Department of Mathematics, Graduate School of Science, Osaka University, Machikaneyama 1-16, Toyonaka, Osaka, 560 Japan. ibukiyam@math.wani.osaka-u.ac.jp

Department of Mathematics, Graduate School of Human and Environmental Studies, Kyoto University, Kyoto, 606-01 Japan. saito@platon.kula.kyoto-u.ac.jp