ELLIPTIC APOSTOL SUMS AND THEIR RECIPROCITY LAWS

SHINJI FUKUHARA AND NORIKO YUI

Abstract. We introduce an elliptic analogue of the Apostol sums, which we call elliptic Apostol sums. These sums are defined by means of certain elliptic functions with a complex parameter τ with positive imaginary part. When τ → i∞, these elliptic Apostol sums represent the well-known Apostol generalized Dedekind sums. Also these elliptic Apostol sums are modular forms in the variable τ. We obtain a reciprocity law for these sums, which gives rise to new relations between certain modular forms (of one variable).

1. Introduction and statement of results

Let p and q be coprime positive integers. First let us recall the definition of Apostol’s generalized Dedekind sums s_{k}(q, p) from [1], which we will call the Apostol sums: for a positive integer k,

s_{k}(q, p) := ∑_{μ=1}^{p−1} \frac{μ}{p}B̄_{k}(\frac{μq}{p}).

Here B̄_{k}(x) denotes the k-th Bernoulli function. That is, B̄_{k}(x) is given by the Fourier expansion

B̄_{k}(x) := −k! ∑_{\begin{matrix}m=−∞ \\ m≠0 \end{matrix}}^{+∞} \frac{e^{2πimx}}{(2πim)^{k}} .

It is well-known that for 0 ≤ x < 1, B̄_{k}(x) reduces to the k-th Bernoulli polynomial. It is also easy to show that s_{k}(q, p) = 0 if k is even.

The fundamental property of the Apostol sums is the following formula:

p^{2n−2}s_{2n−1}(q, p) + q^{2n−2}s_{2n−1}(p, q)

=\frac{1}{pq} \{∑_{j=0}^{n} \frac{(2n − 1)!B_{2j}B_{2n−2j}}{(2j)!(2n − 2j)!} p^{2j}q^{2n−2j} + \frac{(2n − 1)B_{2n}}{2n} \} (n > 1)

(A reciprocity law for the Apostol sums [1, P.149])

(1.1) where B_{k} denotes k-th Bernoulli number.

1991 Mathematics Subject Classification. Primary 11F20; Secondary 11L03, 11F11.

Key words and phrases. Generalized Dedekind sums (Apostol sums), elliptic functions, elliptic Apostol sums, modular forms, reciprocity laws.

There are many different proofs for the Apostol’s reciprocity law (1.1) (refer, for example, [1, 2, 6, 14]). However, our approach [12] that makes use of the following trigonometric identity (1.2) appears to provide the most elementary proof.

For coprime positive integers p and q, and for z ∈ ℂ, it holds that ([7, Theorem 2.4], [11, 12]):

\frac{1}{p} ∑_{μ=1}^{p−1} cot(\frac{μqπ}{p}) cot(z + \frac{μπ}{p}) + \frac{1}{q} ∑_{μ=1}^{q−1} cot(\frac{μpπ}{q}) cot(z + \frac{μπ}{q})

= − cot(pz) cot(qz) + \frac{1}{pq} csc^{2}(z) − 1.

(1.2)

After expanding both sides of (1.2) as Laurent series, we compare the coefficients of z^{2n−2} of both sides. This leads us to the reciprocity law for the Apostol sums (1.1) for n > 1. For n = 1, it rediscovers the reciprocity law for the classical Dedekind sums [17, P.18]:

\frac{1}{p} ∑_{μ=1}^{p−1} cot(\frac{μqπ}{p}) cot(\frac{μπ}{p}) + \frac{1}{q} ∑_{μ=1}^{q−1} cot(\frac{μpπ}{q}) cot(\frac{μπ}{q}) = \frac{p^{2} + q^{2} + 1 − 3pq}{3pq} . (1.3)

In view of the fact that (1.2) is giving rise to the Apostol reciprocity law (1.1), the identity (1.2) may be regarded as a “generating function” of the reciprocity law (1.1).

The purpose of this article is firstly to obtain an elliptic analogue of the identity (1.2), and secondly, to investigate what kind of sums and reciprocity law will be satisfied by the elliptic analogue.

Elliptic analogues of classical Dedekind sums (so-called elliptic Dedekind sums) are studied in [18, 15]. Higher dimensional (or multiple) Dedekind sums are studied in Zagier [19]. As an elliptic analogue of Zagier’s multiple Dedekind sums, higher dimensional (or multiple) elliptic Dedekind sums are defined and studied in Egami [8], and also by Bayad [3]. In this paper we introduce elliptic Apostol sums (see Definition 1.1). They may be regarded as “higher order version” of elliptic Dedekind sums in parallel to the fact that the Apostol sums are “higher order version” of classical Dedekind sums.

We now recall three elliptic functions. Fix a complex number τ with positive imaginary part. The Weierstrass ℘-function, introduced by Weierstrass, is a meromorphic function given as follows:

℘(τ, z) := \frac{1}{z^{2}} + ∑_{\begin{matrix}γ∈2πi(ℤτ+ℤ) \\ γ≠0 \end{matrix}} (\frac{1}{(z − γ)^{2}} − \frac{1}{γ^{2}} ) .

Associated to ℘(τ, z), the function e_{1}(τ) is defined by

e_{1}(τ) := ℘(τ, πi).

It is known that the ℘-function has the following expansion at z = 0:

℘(τ, z) = \frac{1}{z^{2}} + ∑_{n=2}^{∞}(2n − 1)G_{2n}(τ)z^{2n−2}

where G_{2n}(τ) is the Eisenstein series of weight 2n, namely,

G_{2n}(τ) := ∑_{\begin{matrix}γ∈2πi(ℤτ+ℤ) \\ γ≠0 \end{matrix}} \frac{1}{γ^{2n}} .

Let Γ(n) and Γ_{0}(n) be the congruence subgroups of SL_{2}(ℤ) defined respectively by

Γ(n) := \{( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ SL_{2}(ℤ) | a ≡ d ≡ 1 (n), b ≡ c ≡ 0 (n)\} ,

Γ_{0}(n) := \{( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ SL_{2}(ℤ) | c ≡ 0 (n)\} .

It is well-known that G_{2n}(τ) (n > 1) and e_{1}(τ) are modular forms for SL_{2}(ℤ) and Γ_{0}(2) of weight 2n and 2 respectively.

The Weierstrass ζ-function, also introduced by Weierstrass, is a meromorphic function given as follows:

ζ(τ, z) := \frac{1}{z} + ∑_{\begin{matrix}γ∈2πi(ℤτ+ℤ) \\ γ≠0 \end{matrix}} (\frac{1}{z − γ} + \frac{1}{γ} + \frac{z}{γ^{2}} ) .

It holds that ∂ζ(τ, z)/∂z = −℘(τ, z). The function ℘(τ, z) is periodic with respect to 2πi(ℤτ + ℤ). On the other hand the function ζ(τ, z) satisfies

ζ(τ, z + 2πi) = ζ(τ, z) + η_{1}(τ), ζ(τ, z + 2πiτ) = ζ(τ, z) + η_{2}(τ)

where η_{1}(τ) and η_{2}(τ) are independent of z. That is, the function ζ(τ, z) is not periodic but quasi-periodic with respect to 2πi(ℤτ + ℤ).

The function φ(τ, z) is defined by

\begin{aligned}φ(τ, z) & = \sqrt{℘(τ, z) − e_{1}(τ)}, \\ φ(τ, z) & = \frac{1}{z} + O(1) (z → 0). \end{aligned} ; (1.4)

The function φ(τ, z) is periodic with respect to 2πi(2ℤτ + ℤ) and has the following Laurent expansion at z = 0:

(1.5)  φ(τ, z) = \frac{1}{z} ∑_{n=0}^{∞} H_{2n}(τ)z^{2n}

where, by (1.4), H_{2n}(τ) is recursively determined by

H_{0}(τ) = 1, H_{2}(τ) = −\frac{1}{2}e_{1}(τ),

H_{2n}(τ) = −\frac{1}{2} ∑_{j=1}^{n−1} H_{2j}(τ)H_{2n−2j}(τ) + \frac{2n − 1}{2}G_{2n}(τ) (n > 1).

It is obvious that H_{2n}(τ) is a polynomial in e_{1}(τ), G_{4}(τ), G_{6}(τ), …, G_{2n}(τ) with rational coefficients, and then H_{2n}(τ) is a modular form for Γ_{0}(2) with weight 2n. Also note that φ(τ, z) is an odd function, namely φ(τ, −z) = −φ(τ, z).

For a fixed τ, we will simply write ℘(z), ζ(z), φ(z), e_{1}, η_{1}, η_{2}, G_{2n} and H_{2n} dropping the variable τ. By f′(z), we denote the derivative of a function f(z) with respect to z. More generally, by f^{(k)}(z), we denote the k-th derivative of a function f(z).

Retaining these notations we can now formulate our results:

Theorem 1.1. Let p and q be coprime positive integers, and z ∈ ℂ.

(1) Suppose that p + q ≡ 1 (2). Then we have

\begin{aligned}\frac{1}{p} & ∑_{\begin{matrix}λ,μ=0 \\ (λ,μ)≠(0,0) \end{matrix}}^{p−1}(−1)^{λ}φ(\frac{2πiq(λτ + μ)}{p})φ(z + \frac{2πi(λτ + μ)}{p}) \\ & + \frac{1}{q} ∑_{\begin{matrix}λ,μ=0 \\ (λ,μ)≠(0,0) \end{matrix}}^{q−1}(−1)^{λ}φ(\frac{2πip(λτ + μ)}{q})φ(z + \frac{2πi(λτ + μ)}{q}) = −φ(pz)φ(qz) − \frac{1}{pq} φ′(z). \end{aligned} (1.6)

(2) Suppose that p + q ≡ 0 (2). Then we have

\begin{aligned}\frac{1}{p} & ∑_{\begin{matrix}λ,μ=0 \\ (λ,μ)≠(0,0) \end{matrix}}^{p−1}(−1)^{λ}φ(\frac{2πiq(λτ + μ)}{p})ζ(z + \frac{2πi(λτ + μ)}{p}) \\ & + \frac{1}{q} ∑_{\begin{matrix}λ,μ=0 \\ (λ,μ)≠(0,0) \end{matrix}}^{q−1}(−1)^{λ}φ(\frac{2πip(λτ + μ)}{q})ζ(z + \frac{2πi(λτ + μ)}{q}) \\ & = −φ(pz)φ(qz) − \frac{1}{pq} ζ′(z) + C(p, q; τ) \end{aligned} (1.7)

where C(p, q; τ) is a constant with respect to z.

Next we define the so-called elliptic Apostol sums which are elliptic analogues of the Apostol sums.

Definition 1.1. Let p and q be coprime positive integers, and let k be a positive integer. Define the elliptic Apostol sum s̃_{k}(q, p; τ) as follows:

(1) When p + q ≡ 1 (2),

s̃_{k}(q, p; τ) := −\frac{k}{p^{k}} ∑_{\begin{matrix}λ,μ=0 \\ (λ,μ)≠(0,0) \end{matrix}}^{p−1}(−1)^{λ}φ(\frac{2πiq(λτ + μ)}{p})φ^{(k−1)}(\frac{2πi(λτ + μ)}{p}).

(2) When p + q ≡ 0 (2),

\begin{aligned}s̃_{k}(q, p; τ) : & = −\frac{k}{p^{k}} ∑_{\begin{matrix}λ,μ=0 \\ (λ,μ)≠(0,0) \end{matrix}}^{p−1}(−1)^{λ}φ(\frac{2πiq(λτ + μ)}{p})ζ^{(k−1)}(\frac{2πi(λτ + μ)}{p}) \\ & (= \frac{k}{p^{k}} ∑_{\begin{matrix}λ,μ=0 \\ (λ,μ)≠(0,0) \end{matrix}}^{p−1}(−1)^{λ}φ(\frac{2πiq(λτ + μ)}{p})℘^{(k−2)}(\frac{2πi(λτ + μ)}{p}) (k ≥ 2)) . \end{aligned}

Remark 1.1. As we will discuss in Section 5 below, the functions φ(\frac{2πiq(λτ+μ)}{p}),

φ^{(k−1)}(\frac{2πi(λτ+μ)}{p}) and ζ^{(k−1)}(\frac{2πi(λτ+μ)}{p}) are modular forms for the principal congruence subgroup Γ(p) (or Γ(2p)) of weight 1, k and k, respectively. (In fact, we see that

ζ^{(k−1)}(\frac{2πi(λτ + μ)}{p}) = −℘^{(k−2)}(\frac{2πi(λτ + μ)}{p})

is a “level p Eisenstein series”, cf. [16, PP.131-135]). Hence s̃_{k}(q, p; τ) is also a modular form for the same congruence subgroup Γ(p) (or Γ(2p)) of weight k + 1. Moreover we can show that

(1.8)  lim_{τ→i∞} s̃_{2n−1}(q, p; τ) = s_{2n−1}(q, p)

for any n > 1. This is a reason why we may call s̃_{k}(q, p; τ) an elliptic Apostol sum. Another fundamental property of Apostol sums is that they satisfy

s_{2n−1}(q + p, p) = s_{2n−1}(q, p).

This implies that Apostol sums is a (generalized) Dedekind symbol (refer to [10]).

As for elliptic Apostol sums, we see that sthey satisfy

s̃_{2n−1}(q + 2p, p; τ) = s̃_{2n−1}(q, p; τ).

(This follows easily from the fact that φ(z +2·2πiτ) = φ(z).) A implication of this formula is that elliptic Apostol sum is not a Dedekind symbol but may be regarded as a “pre”-Dedekind symbol.

Now we are ready to formulate the reciprocity law for the elliptic Apostol sums. Theorem 1.2. Let p and q be coprime positive integers, and n be an integer with n > 1.

(1) Suppose that p + q ≡ 1 (2). Then we have

\begin{aligned}p^{2n−2} & s̃_{2n−1}(q, p; τ) + q^{2n−2}s̃_{2n−1}(p, q; τ) \\ & = \frac{(2n − 1)!}{pq} \{∑_{j=0}^{n} H_{2j}H_{2n−2j}p^{2j}q^{2n−2j} + (2n − 1)H_{2n}\} . \end{aligned} ; (1.9)

(2) Suppose that p + q ≡ 0 (2). Then we have

\begin{aligned}p^{2n−2} & s̃_{2n−1}(q, p; τ) + q^{2n−2}s̃_{2n−1}(p, q; τ) \\ & = \frac{(2n − 1)!}{pq} \{∑_{j=0}^{n} H_{2j}H_{2n−2j}p^{2j}q^{2n−2j} − (2n − 1)G_{2n}\} . \end{aligned} ; (1.10)

Remark 1.2. We may regard the formulae (1.9) and (1.10) as the identities between the modular forms for Γ(p) (or Γ(2p) if p is odd), Γ(q) (or Γ(2q) if q is odd) and Γ_{0}(2). In fact, the left hand sides of the identities represent modular forms for Γ(p) (or Γ(2p)) and Γ(q) (or Γ(2q)), while the right hand sides modular forms for Γ_{0}(2), all of weight 2n. This kind of reciprocity formulae between modular forms seem quite new to us. One intriguing question arises: Is there any geometric interpretation of these formulae?

2. Proof of Theorem 1.1

In this section we give proof of Theorem 1.1.

Proof of Theorem 1.1 (1). We will prove the identity (1.6). We let f(z) and g(z) denote the left and right hand side of (1.6), respectively. Namely

\begin{aligned}f(z) = & \frac{1}{p} ∑_{\begin{matrix}λ,μ=0 \\ (λ,μ)≠(0,0) \end{matrix}}^{p−1}(−1)^{λ}φ(\frac{2πiq(λτ + μ)}{p})φ(z + \frac{2πi(λτ + μ)}{p}) \\ & + \frac{1}{q} ∑_{\begin{matrix}λ,μ=0 \\ (λ,μ)≠(0,0) \end{matrix}}^{q−1}(−1)^{λ}φ(\frac{2πip(λτ + μ)}{q})φ(z + \frac{2πi(λτ + μ)}{q}) \end{aligned}

and

g(z) = −φ(pz)φ(qz) − \frac{1}{pq} φ′(z).

We claim that

f(z) = g(z).

First note that both f(z) and g(z) are meromorphic functions, and both have simple poles at the points z = −2πi(λτ +μ)/p+2πi(mτ +n) with λ, μ = 0, 1, …, p−1; (λ, μ) ≠ (0, 0); m, n ∈ ℤ and z = −2πi(λτ+μ)/q+2πi(mτ+n) with λ, μ = 0, 1, …, q − 1; (λ, μ) ≠ (0, 0); m, n ∈ ℤ. Furthermore, the residues at these points are equal, that is,

\begin{aligned}Res_{z=−2πi(λτ+μ)/p+2πi(mτ+n)}(f) & = \frac{(−1)^{λ}}{p} φ(2πiq(λτ + μ)/p) \\ & = Res_{z=−2πi(λτ+μ)/p+2πi(mτ+n)}(g), \end{aligned}

and

\begin{aligned}Res_{z=−2πi(λτ+μ)/q+2πi(mτ+n)}(f) & = \frac{(−1)^{λ}}{q} φ(2πip(λτ + μ)/q) \\ & = Res_{z=−2πi(λτ+μ)/q+2πi(mτ+n)}(g). \end{aligned}

Next we investigate other poles. We may assume without loss of generality that g(z) has no pole of order greater than 1. Indeed, the principal parts of −φ(pz)φ(qz) and −(1/pq)φ′(z) at z = 0 are −1/(pqz^{2}) and 1/(pqz^{2}), respectively, and therefore, they cancel out in g(z). Furthermore, the principal part of −φ(pz)φ(qz) at z = 2πiτ is 1/pq(z −2πiτ)^{2} (note that p+q ≡ 1 (2)), and the principal part of −(1/pq)φ′(z) at z = 2πiτ is −1/pq(z − 2πiτ)^{2}, respectively. Therefore, again these principal parts at z = 2πiτ cancel out in g(z). Hence any poles at z = 2πi(mτ + n) (m, n ∈ ℤ) cancel out in g(z) since g(z) is periodic with period 2πi(2ℤτ + ℤ). Clearly f(z) has no pole of order greater than 1. Thus we know that the principal parts of f(z) and g(z) coincide at all of their poles. Therefore f(z)−g(z) is an entire function.

Thus we can conclude that f(z)−g(z) is a doubly periodic entire function on ℂ, and then it must be a constant by the well-known Liouville Theorem. Hence

f(z) − g(z) = A(p, q; τ)

where A(p, q; τ) is a constant with respect to z.

Next we will determine the constant function A(p, q; τ). First note that (1.5) implies that the function g(z) = −φ(pz)φ(qz) − (1/pq)φ′(z) has the following expansion:

g(z) = −\frac{1}{pqz^{2}} (1 + H_{2}p^{2}z^{2} + ⋯)(1 + H_{2}q^{2}z^{2} + ⋯) + \frac{1}{pqz^{2}} (1 − H_{2}z^{2} − ⋯).

Tending z → 0, we obtain

lim_{z→0} g(z) = −\frac{1}{pq} (p^{2} + q^{2})H_{2} − \frac{1}{pq} H_{2}.

Observe also that lim_{z→0} f(z) = f(0). Since A(p, q; τ) is independent of z, we then have

A(p, q; τ) = lim_{z→0}{f(z) − g(z)}

=\frac{1}{p} ∑_{\begin{matrix}λ,μ=0 \\ (λ,μ)≠(0,0) \end{matrix}}^{p−1}(−1)^{λ}φ(\frac{2πiq(λτ + μ)}{p})φ(\frac{2πi(λτ + μ)}{p})

+ \frac{1}{q} ∑_{\begin{matrix}λ,μ=0 \\ (λ,μ)≠(0,0) \end{matrix}}^{q−1}(−1)^{λ}φ(\frac{2πip(λτ + μ)}{q})φ(\frac{2πi(λτ + μ)}{q}) + \frac{p^{2} + q^{2} + 1}{pq}H_{2}.

Our proof will be complete if we show that A(p, q; τ) is indeed equal to 0, and this will be done in Lemma 2.1 below. □

Lemma 2.1. Let p and q be coprime positive integers such that p+q ≡ 1 (2). Then we have

\begin{aligned}\frac{1}{p} & ∑_{\begin{matrix}λ,μ=0 \\ (λ,μ)≠(0,0) \end{matrix}}^{p−1}(−1)^{λ}φ(\frac{2πiq(λτ + μ)}{p})φ(\frac{2πi(λτ + μ)}{p}) \\ & + \frac{1}{q} ∑_{\begin{matrix}λ,μ=0 \\ (λ,μ)≠(0,0) \end{matrix}}^{q−1}(−1)^{λ}φ(\frac{2πip(λτ + μ)}{q})φ(\frac{2πi(λτ + μ)}{q}) + \frac{p^{2} + q^{2} + 1}{pq}H_{2} = 0. \end{aligned} (2.1)

Proof. Let h(z) be an elliptic function defined by

h(z) = φ(pz)φ(qz)φ(z).

Since p + q ≡ 1 (2), h(z) is periodic with period 2πi(ℤτ + ℤ). Then we know that the sum of the residues of h(z) in any fundamental domain is zero. For instance, we take the following fundamental domain {2πi(sτ + t) | \frac{1}{2pq} − 1 ≤ s, t ≤ \frac{1}{2pq} \}. In this fundamental domain, h(z) has simple poles at z = −2πi(λτ + μ)/p with λ, μ = 0, 1, …, p − 1; (λ, μ) ≠ (0, 0) and z = −2πi(λτ + μ)/q with λ, μ = 0, 1, …, q − 1; (λ, μ) ≠ (0, 0). Further the residues of h(z) at these poles are computed as follows:

Res_{z=−2πi(λτ+μ)/p}(h) = \frac{(−1)^{λ}}{p} φ(\frac{2πiq(λτ + μ)}{p})φ(\frac{2πi(λτ + μ)}{p})

and

Res_{z=−2πi(λτ+μ)/q}(h) = \frac{(−1)^{λ}}{q} φ(\frac{2πip(λτ + μ)}{q})φ(\frac{2πi(λτ + μ)}{q}).

Finally we calculate the residue of h(z) at z = 0. For this we note that h(z) has the following expansion at z = 0:

h(z) = φ(pz)φ(qz)φ(z) = \frac{1}{pqz^{3}} (1+H_{2}p^{2}z^{2}+⋯)(1+H_{2}q^{2}z^{2}+⋯)(1+H_{2}z^{2}+⋯).

Then we can compute the residue of h(z) at z = 0:

Res_{z=0}(h) = \frac{p^{2} + q^{2} + 1}{pq}H_{2}.

Adding the residues for h(z) at all poles, we obtain (2.1). □

Remark 2.1. The formula in Lemma 2.1 is a special case of Egami’s result [8, Theorem 1]. Our proof is also essentially the same with that of [8]. Unfortunately Egami’s original formula as well as its proof contained errors (e.g., the right hand side of his formula [8, Eq.5] should be divided by a_{0} ⋯ a_{r}). In the above Lemma and its proof, we reproduced Egami’s result correcting all errors.

We will use the following lemma in the proof of Theorem 1.1 (2). Lemma 2.2. Let p and q be coprime positive integers such that p+q ≡ 0 (2). Then we have (2.2)

\frac{1}{p} ∑_{\begin{matrix}λ,μ=0 \\ (λ,μ)≠(0,0) \end{matrix}}^{p−1}(−1)^{λ}φ(\frac{2πiq(λτ + μ)}{p}) + \frac{1}{q} ∑_{\begin{matrix}λ,μ=0 \\ (λ,μ)≠(0,0) \end{matrix}}^{q−1}(−1)^{λ}φ(\frac{2πip(λτ + μ)}{q}) = 0.

Proof. Let h(z) be an elliptic function defined by

h(z) = φ(pz)φ(qz).

Since p+q ≡ 0 (2), h(z) is periodic with period 2πi(ℤτ +ℤ). Hence the sum of the residues of h(z) in the fundamental domain \{2πi(sτ + t) | \frac{1}{2pq} − 1 ≤ s, t ≤ \frac{1}{2pq} \} is zero. In this fundamental domain, h(z) has simple poles at z = −2πi(λτ + μ)/p with λ, μ = 0, 1, …, p − 1; (λ, μ) ≠ (0, 0) and z = −2πi(λτ + μ)/q with λ, μ = 0, 1, …, q − 1; (λ, μ) ≠ (0, 0). Further the residues of h(z) at these poles are computed as follows:

Res_{z=−2πi(λτ+μ)/p}(h) = \frac{(−1)^{λ}}{p} φ(\frac{2πiq(λτ + μ)}{p})

and

Res_{z=−2πi(λτ+μ)/q}(h) = \frac{(−1)^{λ}}{q} φ(\frac{2πip(λτ + μ)}{q}).

To calculate the residue of h(z) at z = 0, we expand h(z) at z = 0:

h(z) = φ(pz)φ(qz) = \frac{1}{pqz^{2}} (1 + H_{2}p^{2}z^{2} + ⋯)(1 + H_{2}q^{2}z^{2} + ⋯).

Then we can compute the residue of h(z) at z = 0: Res_{z=0}(h) = 0. Adding the residues for h(z) at all poles, we arrive at the equation (2.2). □

Next we give a proof of Theorem 1.1 (2).

Proof of Theorem 1.1 (2). We will prove the identity (1.7). Our proof is along the same line as for Theorem 1.1 (1). Let f(z) and g(z) be defined as follows:

\begin{aligned}f(z) = & \frac{1}{p} ∑_{\begin{matrix}λ,μ=0 \\ (λ,μ)≠(0,0) \end{matrix}}^{p−1}(−1)^{λ}φ(\frac{2πiq(λτ + μ)}{p})ζ(z + \frac{2πi(λτ + μ)}{p}) \\ & + \frac{1}{q} ∑_{\begin{matrix}λ,μ=0 \\ (λ,μ)≠(0,0) \end{matrix}}^{q−1}(−1)^{λ}φ(\frac{2πip(λτ + μ)}{q})ζ(z + \frac{2πi(λτ + μ)}{q}) \end{aligned}

and

g(z) = −φ(pz)φ(qz) − \frac{1}{pq} ζ′(z).

We claim that f(z) − g(z) is a constant.

First note that both f(z) and g(z) are meromorphic functions, and both have simple poles at the points z = −2πi(λτ +μ)/p+2πi(mτ +n) with λ, μ = 0, 1, …, p−1; (λ, μ) ≠ (0, 0); m, n ∈ ℤ and z = −2πi(λτ+μ)/q+2πi(mτ+n) with λ, μ = 0, 1, …, q − 1; (λ, μ) ≠ (0, 0); m, n ∈ ℤ. Furthermore, the residues at these points are equal, that is,

\begin{aligned}Res_{z=−2πi(λτ+μ)/p+2πi(mτ+n)}(f) & = \frac{(−1)^{λ}}{p} φ(2πiq(λτ + μ)/p) \\ & = Res_{z=−2πi(λτ+μ)/p+2πi(mτ+n)}(g) \end{aligned}

and

\begin{aligned}Res_{z=−2πi(λτ+μ)/q+2πi(mτ+n)}(f) & = \frac{(−1)^{λ}}{q} φ(2πip(λτ + μ)/q) \\ & = Res_{z=−2πi(λτ+μ)/q+2πi(mτ+n)}(g). \end{aligned}

Next we investigate other poles. We may assume without loss of generality that g(z) has no pole of order greater than 1. Indeed, the principal parts of −φ(pz)φ(qz) and −(1/pq)ζ′(z) at z = 2πi(mτ +n) (m, n ∈ ℤ) are −1/pq(z− (mτ + n))^{2} and 1/pq(z − (mτ + n))^{2}, respectively, and furthermore, they cancel out in g(z). (This can be seen easily at z = 0. At z ≠ 0, this follows from noticing that g(z) is periodic with period 2πi(ℤτ + ℤ) from the hypothesis that p+q is even). Clearly f(z) has no pole of order greater than 1. Thus we know that the principal parts of f(z) and g(z) coincide at all of their poles. Therefore f(z) − g(z) is an entire function.

Now note that ζ′(z) is periodic with the period 2πi(ℤτ + ℤ) though ζ(z) is not. Take the derivative f′(z) − g′(z) of f(z) − g(z). Then f′(z) − g′(z) is periodic with period 2πi(ℤτ + ℤ). Thus we can conclude that f′(z) − g′(z) is a doubly periodic entire function on ℂ, and then it must be a constant by the well-known Liouville Theorem. Hence we may write

f(z) − g(z) = B(p, q; τ)z + C(p, q; τ)

where B(p, q; τ) and C(p, q; τ) are constants with respect to z.

Next we will show that B(p, q; τ) = 0.

By periodicity of φ and quasi-periodicity of ζ, we have

(2.3) ζ(z + 2πi) = ζ(z) + η_{1}, φ(z + 2πi) = φ(z) and ζ′(z + 2πi) = ζ′(z),

we obtain

2πiB(p, q; τ) = {f(z + 2πi) − g(z + 2πi)} − {f(z) − g(z)}

= η_{1} \{\frac{1}{p} ∑_{\begin{matrix}λ,μ=0 \\ (λ,μ)≠(0,0) \end{matrix}}^{p−1}(−1)^{λ}φ(\frac{2πiq(λτ + μ)}{p})

+ \frac{1}{q} ∑_{\begin{matrix}λ,μ=0 \\ (λ,μ)≠(0,0) \end{matrix}}^{q−1}(−1)^{λ}φ(\frac{2πip(λτ + μ)}{q})\}

(from (2.3)) = 0 (by Lemma 2.2).

Hence we can deduce that B(p, q; τ) = 0 completing our proof. □

Remark 2.2. At moment we do not know how to determine the exact value of the constant C(p, q; τ). The difficulty stems from the facts that ζ(z) is only quasi-periodic and that we do not have the reciprocity law for

\frac{1}{p} ∑_{\begin{matrix}λ,μ=0 \\ (λ,μ)≠(0,0) \end{matrix}}^{p−1}(−1)^{λ}φ(\frac{2πiq(λτ + μ)}{p})ζ(\frac{2πi(λτ + μ)}{p}).

3. Proofs of Theorems 1.2

In this section we give proof of Theorem 1.2.

Let f(z) be a meromorphic function on ℂ (in fact we consider the case where f(z) = φ(z) or ζ(z)). If f(z) has not a pole at z = α, then f(z) is expanded at z = α:

(3.1)  f(z + α) = f(α) + ∑_{k=1}^{∞} \frac{f^{(k)}(α)}{k!}z^{k}.

We now give proof of Theorem 1.2 (1).

Proof of Theorem 1.2 (1). We expand both sides of (1.6) and compare the coefficients of z^{2n−2} to obtain the claimed identity (1.9).

The first term of the left hand side of (1.6) is expanded as follows:

\frac{1}{p} ∑_{\begin{matrix}λ,μ=0 \\ (λ,μ)≠(0,0) \end{matrix}}^{p−1}(−1)^{λ}φ(\frac{2πiq(λτ + μ)}{p})φ(z + \frac{2πi(λτ + μ)}{p})

= \frac{1}{p} ∑_{\begin{matrix}λ,μ=0 \\ (λ,μ)≠(0,0) \end{matrix}}^{p−1}(−1)^{λ}φ(\frac{2πiq(λτ + μ)}{p})φ(\frac{2πi(λτ + μ)}{p})

+ ∑_{k=1}^{∞} \frac{1}{k!} \{\frac{1}{p} ∑_{\begin{matrix}λ,μ=0 \\ (λ,μ)≠(0,0) \end{matrix}}^{p−1}(−1)^{λ}φ(\frac{2πiq(λτ + μ)}{p})φ^{(k)}(\frac{2πi(λτ + μ)}{p})\} z^{k}.

We obtain a similar expansion for the second term of the left hand side of (1.6). Then the coefficient of z^{2n−2} in the left hand side of (1.6) is

\begin{aligned}\frac{1}{(2n − 2)!} & \{\frac{1}{p} ∑_{\begin{matrix}λ,μ=0 \\ (λ,μ)≠(0,0) \end{matrix}}^{p−1}(−1)^{λ}φ(\frac{2πiq(λτ + μ)}{p})φ^{(2n−2)}(\frac{2πi(λτ + μ)}{p}) \\ & + \frac{1}{q} ∑_{\begin{matrix}λ,μ=0 \\ (λ,μ)≠(0,0) \end{matrix}}^{q−1}(−1)^{λ}φ(\frac{2πip(λτ + μ)}{q})φ^{(2n−2)}(\frac{2πi(λτ + μ)}{q})\} . \end{aligned}

And this is nothing but equal to

(3.2)  −\frac{1}{(2n − 1)!} \{p^{2n−2}s̃_{2n−1}(q, p; τ) + q^{2n−2}s̃_{2n−1}(p, q; τ)\} .

Next we expand the right hand side of (1.6) as follows:

−φ(pz)φ(qz) − \frac{1}{pq} φ′(z)

= −\frac{1}{pqz^{2}} (1 + H_{2}p^{2}z^{2} + H_{4}p^{4}z^{4} + ⋯)(1 + H_{2}q^{2}z^{2} + H_{4}q^{4}z^{4} + ⋯)

+ \frac{1}{pqz^{2}} (1 − H_{2}z^{2} − 3H_{4}z^{4} − 5H_{6}z^{6} − ⋯).

Then the coefficient of z^{2n−2} in the right hand side of (1.6) is

(3.3)  −\frac{1}{pq} ∑_{j=0}^{n} H_{2j}H_{2n−2j}p^{2j}q^{2n−2j} − \frac{1}{pq} (2n − 1)H_{2n}.

Finally comparing the coefficients of z^{2n−2} in both sides of the equation (1.6), (3.2) and (3.3) are equal. This gives rise to the identity 1.2 (1). □

Proof of Theorem 1.2 (2). The proof is similar to that of Theorem 1.2 (1). The detail is left to the reader. □

4. Limits of elliptic reciprocity law

In this section we consider limits of elliptic reciprocity law (1.6) and (1.7) as τ → i∞.

First we recall a few formulae for elliptic functions. Let

q = e^{2πiτ} and ξ = e^{z}.

Then the functions φ(z) and ζ(z) have the following expansion (see [13, PP.144,147]):

(4.1)  φ(z) = \frac{1}{2} \frac{ξ^{1/2} + ξ^{−1/2}}{ξ^{1/2} − ξ^{−1/2}} ∏_{n=1}^{∞} \frac{(1 + q^{n}ξ)(1 + q^{n}ξ^{−1})(1 − q^{n})^{2}}{(1 − q^{n}ξ)(1 − q^{n}ξ^{−1})(1 + q^{n})^{2}} ,

(4.2)  ζ(z) = 2G_{2}z + \frac{1}{2} \frac{ξ^{1/2} + ξ^{−1/2}}{ξ^{1/2} − ξ^{−1/2}} + ∑_{n=1}^{∞} (\frac{−q^{n}ξ}{1 − q^{n}ξ} + \frac{q^{n}ξ^{−1}}{1 − q^{n}ξ^{−1}} ) .

Let us define ω(z) by (4.3)

ω(z) := ζ(z) − 2G_{2}z (= \frac{1}{2} \frac{ξ^{1/2} + ξ^{−1/2}}{ξ^{1/2} − ξ^{−1/2}} + ∑_{n=1}^{∞} (\frac{−q^{n}ξ}{1 − q^{n}ξ} + \frac{q^{n}ξ^{−1}}{1 − q^{n}ξ^{−1}} )) .

Lemma 4.1. (1) Suppose that x ≥ 0. Then

lim_{τ→i∞} φ(2πi(xτ + y)) = \begin{cases}−\frac{1}{2}(−1)^{[x]} & x ∉ ℤ \\ \frac{1}{2i} cot(πy) & x ∈ ℤ. \end{cases}

Here [x] denotes the largest integer not exceeding x.

(2) Suppose that 0 ≤ x < 1. Then

lim_{τ→i∞} ω(2πi(xτ + y)) = \begin{cases}−\frac{1}{2} & 0 < x < 1 \\ \frac{1}{2i} cot(πy) & x = 0. \end{cases}

Proof. The first identity (1) is proved in Egami [8, P.102] using the expansion (4.1).

The second identity (2) is proved similarly. In fact, the terms

∑_{n=1}^{∞} (\frac{−q^{n}ξ}{1 − q^{n}ξ} + \frac{q^{n}ξ^{−1}}{1 − q^{n}ξ^{−1}} )

in (4.3) tend to 0 as τ → i∞. Furthermore, for z = 2πi(xτ + y), we know

\frac{1}{2} \frac{ξ^{1/2} + ξ^{−1/2}}{ξ^{1/2} − ξ^{−1/2}} = \frac{1}{2} \frac{e^{2πi(xτ+y)} + 1}{e^{2πi(xτ+y)} − 1} → \begin{cases}−\frac{1}{2} & 0 < x < 1 \\ \frac{1}{2} \frac{e^{2πiy}+1}{e^{2πiy}−1} = \frac{1}{2i} cot(πy) & x = 0 \end{cases}

as τ → i∞. Thus we obtain the claimed formula (2). □

Now we apply Lemma 4.1 to obtain limits of elliptic reciprocity law (1.6) and (1.7) as τ → i∞.

Suppose that p + q ≡ 1 (2). Then the left hand side of (1.6) tends to the following limit when τ → i∞:

\begin{aligned}\frac{1}{p} & ∑_{\begin{matrix}λ,μ=0 \\ (λ,μ)≠(0,0) \end{matrix}}^{p−1}(−1)^{λ}φ(\frac{2πiq(λτ + μ)}{p})φ(z + \frac{2πi(λτ + μ)}{p}) \\ & = \frac{1}{p} ∑_{μ=0}^{p−1} ∑_{λ=1}^{p−1}(−1)^{λ}φ(\frac{2πiq(λτ + μ)}{p})φ(z + \frac{2πi(λτ + μ)}{p}) + \frac{1}{p} ∑_{μ=1}^{p−1} φ(\frac{2πiqμ}{p})φ(z + \frac{2πiμ}{p}) \\ & → −\frac{1}{4} ∑_{λ=1}^{p−1}(−1)^{λ+1+[qλ/p]} + \frac{1}{p} ∑_{μ=1}^{p−1} \frac{1}{2i} cot(\frac{πqμ}{p})\frac{1}{2i} cot(\frac{z}{2i} + \frac{πμ}{p}). \end{aligned} (4.4)

On the other hand, we see that the right hand side of (1.6) tends to the following limit when τ → i∞:

−φ(pz)φ(qz) − \frac{1}{pq} φ′(z) → −\frac{1}{2i} cot(\frac{pz}{2i})\frac{1}{2i} cot(\frac{qz}{2i}) + \frac{1}{pq} \frac{1}{(2i)^{2}} csc^{2}(\frac{z}{2i}) (4.5)

noting that cot′(z) = − csc^{2}(z).

We also recall the fact (see Berndt-Evans [5, P.3]) that the term ∑_{λ=1}^{p−1}(−1)^{λ+1+[qλ/p]} in (4.4) satisfies the following reciprocity law:

(4.6)  ∑_{λ=1}^{p−1}(−1)^{λ+1+[qλ/p]} + ∑_{λ=1}^{q−1}(−1)^{λ+1+[pλ/q]} = 1

where p and q coprime positive integers with p + q ≡ 1 (2).

Combining (4.4), (4.5) and (4.6) and substituting the variable z with z/2i, we recover the original reciprocity law (1.2).

Therefore this establishes our claim that Theorem 1.1 (1) is an elliptic analogue of (1.2) in the case p + q ≡ 1 (2).

Furthermore, expanding the last term in (4.4) and then showing that its coefficient of z^{2n−2} is equal to (−p^{2n−2}/(2n − 1)!)s_{2n−1}(q, p) (refer to [12, 11, 2] for the detailed calculation), we obtain the limit (1.8):

lim_{τ→i∞} s̃_{2n−1}(q, p; τ) = s_{2n−1}(q, p).

This fact also is in support of our claim on elliptic Apostol sum.

Now consider the case that p+q ≡ 0 (2). We reformulate (1.7) as follows:

\begin{matrix}\frac{1}{p} ∑_{\begin{matrix}λ,μ=0 \\ (λ,μ)≠(0,0) \end{matrix}}^{p−1}(−1)^{λ}φ(\frac{2πiq(λτ + μ)}{p}) \{ζ(z + \frac{2πi(λτ + μ)}{p}) − 2G_{2}(z + \frac{2πi(λτ + μ)}{p})\} \\ + \frac{1}{q} ∑_{\begin{matrix}λ,μ=0 \\ (λ,μ)≠(0,0) \end{matrix}}^{q−1}(−1)^{λ}φ(\frac{2πip(λτ + μ)}{q}) \{ζ(z + \frac{2πi(λτ + μ)}{q}) − 2G_{2}(z + \frac{2πi(λτ + μ)}{q})\} \\ = −φ(pz)φ(qz) − \frac{1}{pq} ζ′(z) + D(p, q; τ). \end{matrix} (4.7)

Here D(p, q; τ) is a constant with respect to z. This formula follows from (1.7) and the fact that

\begin{matrix}\{\frac{1}{p} ∑_{\begin{matrix}λ,μ=0 \\ (λ,μ)≠(0,0) \end{matrix}}^{p−1}(−1)^{λ}φ(\frac{2πiq(λτ + μ)}{p}) + \frac{1}{q} ∑_{\begin{matrix}λ,μ=0 \\ (λ,μ)≠(0,0) \end{matrix}}^{q−1}(−1)^{λ}φ(\frac{2πip(λτ + μ)}{q})\} G_{2} = 0 \\ (from (2.3)) \end{matrix}

where we set

\begin{aligned}D(p, q; τ) & = C(p, q; τ) − 2 \{\frac{1}{p} ∑_{\begin{matrix}λ,μ=0 \\ (λ,μ)≠(0,0) \end{matrix}}^{p−1}(−1)^{λ}φ(\frac{2πiq(λτ + μ)}{p})\frac{2πi(λτ + μ)}{p} \\ & +\frac{1}{q} ∑_{\begin{matrix}λ,μ=0 \\ (λ,μ)≠(0,0) \end{matrix}}^{q−1}(−1)^{λ}φ(\frac{2πip(λτ + μ)}{q})\frac{2πi(λτ + μ)}{q}\} G_{2}. \end{aligned}

Now we are ready to show that (4.7) is an elliptic analogue of (1.2) (up to addition of constant) in the case p+q ≡ 0 (2). Note that ζ(z)−2G_{2}z = ω(z), and apply Lemma 4.1 (2) to (4.7). Then we can see the left hand side of (4.7) tends to the following limit when τ → i∞:

−\frac{1}{4} ∑_{λ=1}^{p−1}(−1)^{λ+1+[qλ/p]} + \frac{1}{p} ∑_{μ=1}^{p−1} \frac{1}{2i} cot(\frac{πqμ}{p})\frac{1}{2i} cot(\frac{z}{2i} + \frac{πμ}{p})

− \frac{1}{4} ∑_{λ=1}^{q−1}(−1)^{λ+1+[pλ/q]} + \frac{1}{q} ∑_{μ=1}^{q−1} \frac{1}{2i} cot(\frac{πpμ}{q})\frac{1}{2i} cot(\frac{z}{2i} + \frac{πμ}{q}).

(4.8)

On the other hand, we know that the right hand side of (4.7) tends to the following limit when τ → i∞:

(4.9)  −\frac{1}{2i} cot(\frac{pz}{2i})\frac{1}{2i} cot(\frac{qz}{2i}) + \frac{1}{pq} \frac{1}{(2i)^{2}} csc^{2}(\frac{z}{2i}) + c(p, q)

where c(p, q) is a constant with respect to z. To get (4.9) we make use of the fact that

lim_{τ→i∞} −ζ′(z) = lim_{τ→i∞} ℘(z) = \frac{1}{(2i)^{2}} csc^{2}(\frac{z}{2i}).

Combining (4.8), (4.9) and (4.6) and substituting the variable z with z/2i, we recover the original reciprocity law (1.2) in the case p + q ≡ 0 (2).

We summarize the above discussion in the following Proposition.

Proposition 4.2. (a) Suppose that p + q ≡ 1(2). Then the limit of (1.6) as τ → i∞ yields the identity (1.2) on cotangent sums.

(b) Suppose that p + q ≡ 0(2). Then the limit of (4.7) as τ → i∞ yields the identity (1.2) on cotangent sums.

5. New relations between modular forms

We can read off some new relations between modular forms from the reciprocity formulae established in Theorem 1.2. Here we give some comments on our results from the point of view of modular forms. Let ℍ be the upper half complex plane. A function f : ℍ × ℂ → ℂ is called a Jacobi form of weight k and index 0 for the congruence subgroup Γ of SL_{2}(ℤ) if it satisfies the following two conditions:

(5.1)  f(\frac{aτ + b}{cτ + d}, \frac{z}{cτ + d}) = (cτ + d)^{k}f(τ, z) (( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ Γ) ,

(5.2)  f(τ, z + 2πi(mτ + n)) = f(τ, z) (m, n ∈ ℤ)

together with some regularity conditions. The reader can refer to [9, 13] for more detail on Jacobi forms.

It is known that the function φ is a Jacobi form of weight 1 and index 0 for Γ_{0}(2) while the Weierstrass ℘-function is a Jacobi form of weight 2 and index 0 for SL_{2}(ℤ) (see [13] and [9]). It is easily shown that if f is a Jacobi form of weight w and index 0, then f^{(k)} is a Jacobi form of weight w +k and index 0. Hence φ^{(k)} and ζ^{(k)} = −℘^{(k−1)} are Jacobi forms of weight k + 1 and index 0.

Lemma 5.1. Suppose that f(τ, z) is a Jacobi form of weight k and index 0 for SL_{2}(ℤ), and suppose that it has poles only at 2πi(ℤτ + ℤ). Then, for λ, μ = 0, 1, …, p − 1; (λ, μ) ≠ (0, 0), f(τ, 2πi(λτ + μ)/p) is a modular form for Γ(p) of weight k.

Proof. This follows from the following identities for ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ Γ(p):

f(\frac{aτ + b}{cτ + d}, \frac{2πi(λ\frac{aτ+b}{cτ+d} + μ)}{p}) = f(\frac{aτ + b}{cτ + d}, \frac{2πi}{cτ + d} \frac{λ(aτ + b) + μ(cτ + d)}{p})

= (cτ + d)^{k}f(τ, \frac{2πi\{λ(aτ + b) + μ(cτ + d)\}}{p})

(by (5.1))

= (cτ + d)^{k}f(τ, \frac{2πi(λτ + μ)}{p})

(because (5.2) and ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ Γ(p)).

□

Similarly we know that, if f(τ, z) is a Jacobi form of weight k and index 0 for Γ_{0}(2), f(τ, 2πi(λτ + μ)/p) is a modular form of weight k for Γ(p) or Γ(2p) according as p is even or odd.

Now we apply the above argument to f = φ, or φ^{(k)} or ζ^{(k)}. Proposition 5.2. The elliptic Apostol sum s̃_{2n−1}(q, p; τ) is a modular form of weight 2n for Γ(p) or Γ(2p) according as p is even or odd.

Since H_{2n} is a modular form for Γ_{0}(2), the reciprocity law for elliptic Apostol sums (Theorem 1.2) give us new relations between modular forms for Γ(p) (or Γ(2p) if p is odd), Γ(q) (or Γ(2q) if q is odd) and Γ_{0}(2).

The classical Dedekind sums s(q, p) first appeared in the transformation formula of log η(τ). The Apostol sums s_{k}(q, p) also appear in the transformation formula of a sort of Lambert series.

Question: Is there any geometrical interpretation of the elliptic Apostol sums? Is there a generalization of Lambert series whose transformation is described in terms of elliptic Apostol sums? Remark 5.1. Recently, Bayad [3] also addressed the similar problem that was discussed in this paper. Namely, Bayad’s paper considers elliptic analogues of the multiple Dedekind sums and reciprocity law satisfied by the new sums. The reciprocity law was proved in Theorem 2.2. His approach is an extensive use of Jacobi forms, and d-division points on a complex torus ℂ/L of dimension 1. His result generalizes that of Egami [8], which corresponds to 2–division points on ℂ/L.

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Department of Mathematics, Tsuda College, Tsuda-machi 2-1-1, Kodairashi, Tokyo 187-8577, Japan

E-mail address: fukuhara@tsuda.ac.jp

Department of Mathematics and Statistics, Queen’s University, Kingston, Ontario Canada K7L 3N6

E-mail address: yui@mast.queensu.ca