Absolute Modular Forms

Shin-ya Koyama and Nobushige Kurokawa

Abstract. We describe examples of absolute modular forms coming from differentiations of multiple sine functions. We give an identity in weight 3 indicating the graded structure.

Key words: multiple sine function, Eisenstein series, absolute modular form AMS Classification: 11M06

1 Introduction

The absolute modular form is a new kind of modular forms. We say that a (holomorphic) function f on

D_{r} = {(u_{1}, ⋯, u_{r}) ∈ C^{r} |u_{1}, ⋯, u_{r} and 1 belong to one side with respect to a line crossing 0}

is an absolute modular form of weight k if it satisfies the following two conditions:

(1) f(u_{1}, ⋯, u_{r}) is symmetric,

(2) f(\frac{1}{u_{1}} , \frac{u_{2}}{u_{1}} , ⋯, \frac{u_{r}}{u_{1}} ) = u_{1}^{k}f(u_{1}, u_{1}, ⋯, u_{r}).

The Eisenstein series

e_{k}(u_{1}, ⋯, u_{r}) = \begin{matrix}∑ & ’ \\ n_{1},··· ,n_{r+1}≥0 \end{matrix}(n_{1}u_{1} + ⋯ + n_{r}u_{r} + n_{r+1})^{−k}  (k > r + 1)  (1.1)

is a typical example, which is analogous to the classical modular forms. A strict difference is that the sum is taken over a semi-lattice here, while the classical sum was taken over a whole lattice.

We notice the shapes of D_{r} for r = 1, 2 as

D_{1} = C − R_{≤0} = {u ∈ C | − π < arg(u) < π}

and

D_{2} = {(u, v) ∈ D_{1}^{2} | − π < arg(u) − arg(v) < π}.

The absolute modular form is also called the Stirling modular form, since it was originated in an old paper of Barnes [B] (p.397), where a function ρ_{r}(ω_{1}, ⋯, ω_{r}) was called an “absolute modular form” associated to the multiple gamma function (the name “absolute” is coming from the Stirling (asymptotic) formula for n! as n → ∞) (cf. [K3, K4, K5, KK3, Ko]): we refer to [KOW] and [CCM] for mathematics over F_{1}. The absolute modula form is a function of the semi-lattice

Z_{≥0}ω_{1} + ⋯ + Z_{≥0}ω_{r}.

The absolute modular group is identified as

GL_{r}(F_{1}) = S_{r} = Aut(Z_{≥0}ω_{1} + ⋯ + Z_{≥0}ω_{r}).

From this original viewpoint, we may call a function F on

D_{r} = {(ω_{1}, ⋯, ω_{r}) ∈ C^{r} | ω_{1}, ⋯, ω_{r} belong to one side with respect to a line crossing 0}

as an absolute modular form if it satisfies the following two conditions

(1) F (ω_{1}, ⋯, ω_{r}) is symmetric,

(2) F (ω_{1}, ⋯, ω_{r}) is homogeneous of degree −k: F (cω_{1}, ⋯, cω_{r}) = c^{−k}F (ω_{1}, ⋯, ω_{r}) for c ∈ C \ {0}.

We remark that ρ_{r}(ω_{1}, ⋯ ω_{r}) is a complicated function and it is rather difficult to make a general theory containing it. For example, ρ_{r}(ω_{1}, ⋯, ω_{r}) satisfies (1) above, but unfortunately we need a homogeneous function k = k(ω_{1}, ⋯, ω_{r}) for (2). So we do not treat ρ_{r}(ω_{1}, ⋯, ω_{r}) directly in this paper. (But, see Theorem 6 below.)

It is useful to compare the situation with the case of ordinary modular forms which are considered to be functions of the lattice Zω_{1} + Zω_{2}. In that case we usually regard such functions as functions in τ = ω_{2}/ω_{1} on the upper (or lower) half plane to simplify the treatment. Here the modular group is obtained as

GL_{2}(Z) = Aut(Zω_{1} + Zω_{2}).

Similarly we associate a function f : D_{r} → C to F : D_{r+1} → C under

f(u_{1}, ⋯, u_{r}) = F (u_{1}, ⋯, u_{r}, 1)

and

F (ω_{1}, ⋯, ω_{r+1}) = ω_{r+1}^{−k}f (\frac{ω_{1}}{ω_{r+1}} , ⋯, \frac{ω_{r}}{ω_{r+1}} ) .

It is easy to see that the conditions (1)(2) on f and F are equivalent under this correspondence.

Thus, hereafter we deal with functions f on D_{r} of weight k ≥ 0. (Of course it is advisable to keep in mind the functions F on D_{r+1} as well.) In this paper we pick up two constructions:

(a) 𝒮_{k}(u_{1}, ⋯, u_{r}) = S_{r+1}^{(k)}(0, (u_{1}, ⋯, u_{r}, 1)) for k ≥ 0.

(b) e_{k}(u_{1}, ⋯, u_{r}) = ζ_{r+1}(k, (u_{1}, ⋯, u_{r}, 1)) for k ≥ 0.

Here, S_{r}(x, (ω_{1}, ⋯, ω_{r})) is the multiple sine function defined as

S_{r}(x, (ω_{1}, ⋯, ω_{r})) = ∏ ∐_{n_{1},··· ,n_{r}≥0}(n_{1}ω_{1}+⋯+n_{r}ω_{r}+x) ( ∏ ∐_{m_{1},··· ,m_{r}≥1}(m_{1}ω_{1} + ⋯ + m_{r}ω_{r} − x))^{(−1)^{r−1}}

with the regularized product notation ∏ ∐ due to Deninger [D]:

∏ ∐_{λ}λ = exp (− \frac{d}{ds} ∑_{λ} λ^{−s}|_{s=0}) .

Using the multiple gamma function

Γ_{r}(x, (ω_{1}, ⋯, ω_{r})) = ( ∏ ∐_{n_{1},··· ,n_{r}≥0}(n_{1}ω_{1} + ⋯ + n_{r}ω_{r} + x))^{−1}

we can express the multiple sine function as

S_{r}(x, (ω_{1}, ⋯, ω_{r})) = Γ_{r}(x, (ω_{1}, ⋯, ω_{r}))^{−1}Γ_{r}(ω_{1} + ⋯ + ω_{r} − x, (ω_{1}, ⋯, ω_{r}))^{(−1)^{r}}.

We refer to [K1] [K2] [KK1] for a theory of multiple sine functions; see Manin [M] for an excellent survey.

The other notation ζ_{r}(s, (ω_{1}, ⋯, ω_{r})) is a kind of multiple version of the Riemann (or Hurwitz) zeta function

ζ_{r}(s, (ω_{1}, ⋯, ω_{r})) = \begin{matrix}∑ & ’ \\ n_{1},··· ,n_{r}≥0 \end{matrix}(n_{1}ω_{1} + ⋯ + n_{r}ω_{r})^{−s}.

We remark that e_{k}(u_{1}, ⋯, u_{r}), which is given by (1.1) for k > r + 1, is defined by the analytically continued ζ_{r}(k, (u_{1}, ⋯, u_{r}, 1)) for k ≤ r + 1. Hence, we can consider e_{0}(u_{1}, ⋯, u_{r}) for example.

We find a neat expression (not mentioned in [B]) for ρ_{r}(ω_{1}, ⋯, ω_{r}):

ρ_{r}(ω_{1}, ⋯, ω_{r}) = ∏’_{n_{1},··· ,n_{r}≥0}(n_{1}ω_{1} + ⋯ + n_{r}ω_{r})
= exp (−ζ′_{r}(0, (ω_{1}, ⋯, ω_{r}))) .

Hereafter, it would be suggestive to regard e_{k} and 𝒮_{k} as “(generalized) Eisenstein series” and “cusp forms,” respectively. The present state of our experience on absolute modular forms is primitive, so we must postpone developing the general theory to future papers.

We also remark that the Kronecker’s Jugendtraum gives a strong motivation for studying absolute modular forms and absolute modular functions. It is a famous problem to construct abelian extensions (class fields) of an algebraic number field via division values of a suitable function. The studies in the rational number case (Kronecker) and the real quadratic field case (Shintani [S]) suggest looking at the extension Q (S_{r} (\frac{ω_{1}+···+ω_{r}}{N} , (ω_{1}, ⋯, ω_{r})) , ω_{1}, ⋯, ω_{r}) over Q(ω_{1}, ⋯, ω_{r}). This function S_{r} (\frac{ω_{1}+···+ω_{r}}{N} , (ω_{1}, ⋯, ω_{r})) on D_{r} (or the function S_{r+1} (\frac{u_{1}+···+u_{r}+1}{N}, (u_{1}, ⋯, u_{r}, 1)) on D_{r}) is a typical absolute modular function (or an absolute modular form of weight 0).

In this paper we calculate 𝒮_{k} and e_{k} to some extent in the one-variable case with a remark made on the two-variable case. The following four theorems concern the one-variable case.

Theorem 1

𝒮_{1}(u) = \frac{2π}{\sqrt{u}}.

Theorem 2

𝒮_{2}(u) = \begin{cases}\frac{8π^{2}i}{\sqrt{u}} (\frac{1}{u}E_{1} (−\frac{1}{u}) − E_{1}(u)) & if Im(u) > 0, \\ −\frac{8π^{2}i}{\sqrt{u}} (\frac{1}{u}E_{1} (\frac{1}{u}) − E_{1}(−u)) & if Im(u) < 0, \end{cases}

where

E_{1}(τ) = −\frac{1}{4} + ∑_{n=1}^{∞} d(n)e^{2πinτ}

for Im(τ) > 0 with d(n) = ∑_{d|n} 1.

Theorem 3

e_{0}(u) = \frac{1}{12}(u + \frac{1}{u} − 9).

Theorem 4

𝒮_{3}(u) = \frac{3}{4}𝒮_{2}(u)^{2}𝒮_{1}(u)^{−1} − \frac{3}{8}(4e_{0}(u) + 3)𝒮_{1}(u)^{3}.

Among others, Theorem 2 shows an unexpected relation in weight 2 using to an Eisensteinlike series of weight 1; remark that 1/\sqrt{u} is of weight 1 as in Theorem 1. The identity in Theorem 4 is indicating a graded structure in weight 3. Our proof shows that this identity is coming from the (quasi-)modularity of

E_{2}(τ) = −\frac{1}{24} + ∑_{n=1}^{∞} σ(n)e^{2πinτ}

with

σ(n) = ∑_{d | n} d.

Here we are using the notation

E_{k}(τ) = \frac{ζ(1 − k)}{2} + ∑_{n=1}^{∞} σ_{k−1}(n)e^{2πinτ}

with

σ_{k−1}(n) = ∑_{d | n} d^{k−1}.

In this paper we do not treat the “cuspidality.” We think that the cuspidality of 𝒮_{k}(u) means that 𝒮_{k}(∞) = 0 at now. This vanishing is obvious for 𝒮_{1}(u) from Theorem 1. It is not so difficult to show 𝒮_{2}(±i∞) = 0 and 𝒮_{3}(±i∞) = 0 from Theorem 2 and Theorem 4, respectively. We hope to report on them in a future paper.

Our results have some applications. For example, from Theorem 2 and 𝒮_{2}(1) = −4π we see that:

Corollary 1

lim_{{τ→1 \atop Im(τ)>0}} (E_{1} (−\frac{1}{τ}) − τE_{1} (τ)) = −\frac{1}{2πi}.

This indicates a merit of absolute modular forms, where the domain D_{1} contains the positive real numbers R_{>0} (See Figure 1). We refer to [K3] [KK3] for related results.

Oy  x

Figure 1. The domain D_{1}

Lastly we notice calculations in the two-variable case: Theorem 5

e_{0}(u, v) = \frac{1}{24} (u + \frac{1}{u} + v + \frac{1}{v} + \frac{u}{v} + \frac{v}{u} − 21) .

Theorem 6

𝒮_{1}(u, v) = \frac{ρ_{3}(u, v, 1)^{2}ρ_{1}(u)ρ_{1}(v)ρ_{1}(1)}{ρ_{2}(u, v)ρ_{2}(u, 1)ρ_{2}(v, 1)}
= \frac{(2π)^{\frac{3}{2}}}{\sqrt{uv}} · \frac{ρ_{3}(u, v, 1)^{2}}{ρ_{2}(u, v)ρ_{2}(u, 1)ρ_{2}(v, 1)}.

2 Proofs of Theorems 1 and 2

Since we have seen these results in other contexts essentially (see [K2] for example), we give concise proofs here. Proof of Theorem 1:

It is sufficient to show that

S′_{2}(0, (ω_{1}, ω_{2})) = \frac{2π}{\sqrt{ω_{1}ω_{2}}} .

The periodicity proved in [KK1]

S_{2}(x, (ω_{1}, ω_{2})) = S_{2}(x + ω_{2}, (ω_{1}, ω_{2}))S_{1}(x, ω_{1})

with

S_{1}(x, ω) = 2 sin (\frac{πx}{ω})

implies

S′_{2}(0, (ω_{1}, ω_{2})) = S_{2}(ω_{2}, (ω_{1}, ω_{2}))\frac{2π}{ω_{1}} .

Here,

S_{2}(ω_{2}, (ω_{1}, ω_{2})) = \frac{Γ_{2}(ω_{1}, (ω_{1}, ω_{2}))}{Γ_{2}(ω_{2}, (ω_{1}, ω_{2}))}
= lim_{x→0} \frac{Γ_{2}(ω_{1} + x, (ω_{1}, ω_{2}))}{Γ_{2}(ω_{2} + x, (ω_{1}, ω_{2}))}
= lim_{x→0} \frac{Γ_{2}(x, (ω_{1}, ω_{2}))Γ_{1}(x, ω_{2})^{−1}}{Γ_{2}(x, (ω_{1}, ω_{2}))Γ_{1}(x, ω_{1})^{−1}}
= lim_{x→0} \frac{Γ_{1}(x, ω_{1})}{Γ_{1}(x, ω_{2})}

= lim_{x→0} \frac{\frac{Γ( \frac{x}{ω_{1}} )}{\sqrt{2π}}ω_{1}^{\frac{x}{ω_{1}} − \frac{1}{2}}}{\frac{Γ( \frac{x}{ω_{2}} )}{\sqrt{2π}}ω_{2}^{\frac{x}{ω_{2}} − \frac{1}{2}}}

= \sqrt{\frac{ω_{1}}{ω_{2}}} .

Thus

S′_{2}(0, (ω_{1}, ω_{2})) = \frac{2π}{\sqrt{ω_{1}ω_{2}}} .

Proof of Theorem 2:

A formula of Shintani [S] (Proposition 5) shows that

Cot_{2}(x, (1, τ)) = πi (\frac{x}{τ} − \frac{1}{2} (1 + \frac{1}{τ}))

−2πi ∑_{n=0}^{∞} ∑_{m=1}^{∞} e^{2πinmτ}e^{2πimx} + \frac{2πi}{τ} ∑_{n=1}^{∞} ∑_{m=1}^{∞} e^{− \frac{2πinm}{τ}} e^{\frac{2πimx}{τ}} ,

where we use the double cotangent function

Cot_{2}(x, (1, u)) = \frac{S′_{2}(x, (1, u))}{S_{2}(x, (1, u))}.

Hence

Cot_{2}(τ, (1, τ)) = \frac{πi}{2} (1 − \frac{1}{τ}) − 2πi ∑_{n=1}^{∞} ∑_{m=1}^{∞} e^{2πinmτ} + \frac{2πi}{τ} ∑_{n=1}^{∞} ∑_{m=1}^{∞} e^{− \frac{2πinm}{τ}}

= \frac{πi}{2} (1 − \frac{1}{τ}) − 2πi (E_{1}(τ) + \frac{1}{4}) + \frac{2πi}{τ} (E_{1} (−\frac{1}{τ}) + \frac{1}{4})
= \frac{2πi}{τ} (E_{1} (−\frac{1}{τ}) − τE_{1}(τ)) .

Now we prove that

S′′_{2}(0, (1, τ)) = \frac{4π}{\sqrt{τ}} Cot_{2}(τ, (1, τ)).

This is obtained from the periodicity

S_{2}(x, (1, τ)) = S_{2}(x + τ, (1, τ))S_{1}(x)

with S_{1}(x) = 2 sin(πx). In fact, differentiating this periodicity relation twice leads to S′′_{2}(x, (1, τ)) = S′′_{2}(x + τ, (1, τ))S_{1}(x) + 2S′_{2}(x + τ, (1, τ))S′_{1}(x) + S_{2}(x + τ, (1, τ))S′′_{1}(x). Hence, we obtain the desired relation

S′′_{2}(0, (1, τ)) = 2S′_{2}(τ, (1, τ))S′_{1}(0)
= 4πS′_{2}(τ, (1, τ))
= \frac{4π}{\sqrt{τ}} · \frac{S′_{2}(τ, (1, τ))}{S_{2}(τ, (1, τ))}
= \frac{4π}{\sqrt{τ}} Cot_{2}(τ, (1, τ)),

where we used the fact

S_{2}(τ, (1, τ)) = \frac{1}{\sqrt{τ}}

proved in [KK2]. Thus we have

𝒮_{2}(τ) = S′′_{2}(0, (1, τ)) = \frac{8π^{2}i}{τ\sqrt{τ}} (E_{1} (−\frac{1}{τ}) − τE_{1}(τ)) .

This proves Theorem 2 for Im(u) > 0. The case Im(u) < 0 is given by the reflection.

3 Proof of Theorem 3

We prove that

ζ_{2}(0, (ω_{1}, ω_{2})) = \frac{1}{12} (\frac{ω_{2}}{ω_{1}} + \frac{ω_{1}}{ω_{2}} − 9) .

For this we recall the Riemann-Mellin type integral expression for ζ_{2}(s, (ω_{1}, ω_{2})) (cf. Barnes [B]). It says that

ζ_{2}(s, (ω_{1}, ω_{2})) = \frac{1}{Γ(s)} ∫_{0}^{∞} Θ(t)t^{s−1}dt

in Re(s) > 2 with

Θ(t) = \begin{matrix}∑ & ’ \\ n_{1},n_{2}≥0 \end{matrix} e^{−(n_{1}ω_{1}+n_{2}ω_{2})t}

= \frac{1}{(1 − e^{−tω_{1}})(1 − e^{−tω_{2}})} − 1
= \frac{e^{tω_{1}} + e^{tω_{2}} − 1}{(e^{tω_{1}} − 1)(e^{tω_{2}} − 1)}.

Let

Θ(t) = \frac{a_{−2}}{t^{2}} + \frac{a_{−1}}{t} + a_{0} + a_{1}t + ⋯

be the Laurent expansion around t = 0. An easy calculation shows that

a_{−2} = \frac{1}{ω_{1}ω_{2}} ,  a_{−1} = \frac{1}{2} (\frac{1}{ω_{1}} + \frac{1}{ω_{2}} ) ,  a_{0} = \frac{1}{12} (\frac{ω_{2}}{ω_{1}} + \frac{ω_{1}}{ω_{2}} − 9) .

Now, to make an analytic continuation of ζ_{2}(s, (ω_{1}, ω_{2})) in Re(s) > −1 we split the integral into three parts:

ζ_{2}(s, (ω_{1}, ω_{2})) = \frac{1}{Γ(s)} ∫_{1}^{∞} Θ(t)t^{s−1}dt + \frac{1}{Γ(s)} ∫_{0}^{1} (Θ(t) − \frac{a_{−2}}{t^{2}} − \frac{a_{−1}}{t} − a_{0}) t^{s−1}dt

+\frac{1}{Γ(s)} ∫_{0}^{1} (\frac{a_{−2}}{t^{2}} + \frac{a_{−1}}{t} + a_{0}) t^{s−1}dt.

Here, the first term is holomorphic for all s ∈ C since the integral is absolutely convergent. The second term is holomorphic in Re(s) > −1 since

Θ(t) − \frac{a_{−2}}{t^{2}} − \frac{a_{−1}}{t} − a_{0} = O(t)

as t → 0. The third term is

\frac{1}{Γ(s)} (\frac{a_{−2}}{s − 2} + \frac{a_{−1}}{s − 1} + \frac{a_{0}}{s} ) ,

which is meromorphic in s ∈ C. Hence, the above expression gives an analytic continuation of ζ_{2}(s, (ω_{1}, ω_{2})) in Re(s) > −1. Moreover this calculation shows that ζ_{2}(s, (ω_{1}, ω_{2})) is holomorphic at s = 0 and its value is given by ζ_{2}(0, (ω_{1}, ω_{2})) = a_{0}. Thus

ζ_{2}(0, (ω_{1}, ω_{2})) = \frac{1}{12} (\frac{ω_{2}}{ω_{1}} + \frac{ω_{1}}{ω_{2}} − 9) .

4 Proof of Theorem 4

We prove the following facts (a)(b): (a)

S′′′_{2}(0, (1, τ)) = \frac{6π}{\sqrt{τ}} (Cot′_{2}(τ, (1, τ)) + Cot_{2}(τ, (1, τ))^{2} − \frac{π^{2}}{3} ) ,

(b)

Cot′_{2}(τ, (1, τ)) = \frac{π^{2}}{6} (1 − \frac{1}{τ^{2}} ) .

Before proving (a) and (b) we show that Theorem 4 follows from them. In fact (a) says that

𝒮_{3}(τ) = S′′′_{2}(0, (1, τ))
= \frac{6π}{\sqrt{τ}} Cot′_{2}(τ, (1, τ)) + \frac{6π}{\sqrt{τ}} Cot_{2}(τ, (1, τ))^{2} − \frac{2π^{3}}{\sqrt{τ}} ,

so using (b) and the formula

Cot_{2}(τ, (1, τ)) = \frac{\sqrt{τ}}{4π} S′′_{2}(0, (1, τ))
= \frac{\sqrt{τ}}{4π} 𝒮_{2}(τ)

proved in §2, we have

𝒮_{3}(τ) = \frac{π^{3}}{\sqrt{τ}} (1 − \frac{1}{τ^{2}} ) + \frac{3\sqrt{τ}}{8π} 𝒮_{2}(τ)^{2} − \frac{2π^{3}}{\sqrt{τ}}
= −\frac{1}{8} (\frac{2π}{\sqrt{τ}} )^{3} (τ + \frac{1}{τ}) + \frac{3}{4} (\frac{2π}{\sqrt{τ}} )^{−1} 𝒮_{2}(τ)^{2}
= −\frac{1}{8}𝒮_{1}(τ)^{3}(12e_{0}(τ) + 9) + \frac{3}{4}𝒮_{1}(τ)^{−1}𝒮_{2}(τ)^{2}.

This is Theorem 4. Proof of (a):

The definition

Cot_{2}(x, (1, τ)) = \frac{S′_{2}(x, (1, τ))}{S_{2}(x, (1, τ))}

implies

Cot′_{2}(x, (1, τ)) = \frac{S′′_{2}(x, (1, τ))}{S_{2}(x, (1, τ))} − \frac{S′_{2}(x, (1, τ))^{2}}{S_{2}(x, (1, τ))^{2}} .

Hence

Cot′_{2}(τ, (1, τ)) = \frac{S′′_{2}(τ, (1, τ))}{S_{2}(τ, (1, τ))} − (\frac{S′_{2}(τ, (1, τ))}{S_{2}(τ, (1, τ))})^{2}
= \frac{S′′_{2}(τ, (1, τ))}{S_{2}(τ, (1, τ))} − Cot_{2}(τ, (1, τ))^{2}.

Now we show that

S′′_{2}(τ, (1, τ)) = \frac{1}{6π} (S′′′_{2}(0, (1, τ)) + \frac{2π^{3}}{\sqrt{τ}} ) .

We obtain this by differentiating the periodicity relation

S_{2}(x, (1, τ)) = S_{2}(x + τ, (1, τ))S_{1}(x)

three times at x = 0. In fact the identity

S′′′_{2}(x, (1, τ)) = S′′′_{2}(x + τ, (1, τ))S_{1}(x) + 3S′′_{2}(x + τ, (1, τ))S′_{1}(x)
+3S′_{2}(x + τ, (1, τ))S′′_{1}(x) + S_{2}(x + τ, (1, τ))S′′′_{1}(x)

gives

S′′′_{2}(0, (1, τ)) = 6πS′′_{2}(τ, (1, τ)) − 2π^{3}S_{2}(τ, (1, τ))
= 6πS′′_{2}(τ, (1, τ)) − \frac{2π^{3}}{\sqrt{τ}}

since S_{1}(x) = 2 sin(πx) and S_{2}(τ, (1, τ)) = \frac{1}{\sqrt{τ}} . Thus

Cot′_{2}(τ, (1, τ)) = \frac{\sqrt{τ}}{6π} S′′′_{2}(0, (1, τ)) + \frac{π^{2}}{3} − Cot_{2}(τ, (1, τ))^{2}.

This gives (a). Proof of (b):

We use the method of the proof of Theorem 2. In this case we start from

Cot′_{2}(x, (1, τ)) = \frac{πi}{τ} + 4π^{2} ∑_{n=0}^{∞} ∑_{m=1}^{∞} me^{2πinmτ}e^{2πimx} − \frac{4π^{2}}{τ^{2}} ∑_{n=1}^{∞} ∑_{m=1}^{∞} me^{− \frac{2πinm}{τ}} e^{\frac{2πimx}{τ}} .

Then we have

Cot′_{2}(τ, (1, τ)) = \frac{πi}{τ} + 4π^{2} ∑_{n=1}^{∞} ∑_{m=1}^{∞} me^{2πinmτ} − \frac{4π^{2}}{τ^{2}} ∑_{n=1}^{∞} ∑_{m=1}^{∞} me^{− \frac{2πinm}{τ}}
= \frac{πi}{τ} + 4π^{2} (E_{2}(τ) + \frac{1}{24}) − \frac{4π^{2}}{τ^{2}} (E_{2} (−\frac{1}{τ}) + \frac{1}{24})
= 4π^{2} (E_{2}(τ) − \frac{1}{τ^{2}} E_{2} (−\frac{1}{τ})) + \frac{πi}{τ} + \frac{π^{2}}{6} (1 − \frac{1}{τ^{2}} )
= \frac{π^{2}}{6} (1 − \frac{1}{τ^{2}} ) ,

where we used the modularity of E_{2}(τ):

E_{2} (−\frac{1}{τ}) = τ^{2}E_{2}(τ) − \frac{τ}{4πi}.

This proves (b). Remark Similar calculations show that

𝒮_{5} = \frac{5}{2}𝒮_{4}𝒮_{2}𝒮_{1}^{−1} − \frac{15}{16}𝒮_{2}^{4}𝒮_{1}^{−3} + \frac{15}{16}(4e_{0} + 3)𝒮_{2}^{2}𝒮_{1} + \frac{1}{192}(144e_{0}^{2} + 216e_{0} + 89)𝒮_{1}^{5}.

5 Proof of Corollary 1

Theorem 2 implies

lim_{{τ→1 \atop Im(τ)>0}} (E_{1} (−\frac{1}{τ}) − τE_{1} (τ)) = \frac{1}{8π^{2}i}𝒮_{2}(1)
= \frac{1}{8π^{2}i}S′′_{2}(0, (1, 1)),

and Theorem 4(1) of [K2] says that

S′′_{2}(0, (1, 1)) = −4π.

Hence we get Corollary 1.

6 Proof of Theorem 5

We calculate ζ_{3}(0, (ω_{1}, ω_{2}, ω_{3})). Exactly similarly to the case of ζ_{2}(0, (ω_{1}, ω_{2})) treated in §3 we have

ζ_{3}(0, (ω_{1}, ω_{2}, ω_{3})) = a_{0} = a_{0}(ω_{1}, ω_{2}, ω_{3})

where a_{0} is the constant term of the Laurent expansion of

Θ(t) = \frac{1}{(1 − e^{−tω_{1}})(1 − e^{−tω_{2}})(1 − e^{−tω_{3}})} − 1

around t = 0:

Θ(t) = \frac{a_{−3}}{t^{3}} + \frac{a_{−2}}{t^{2}} + \frac{a_{−1}}{t} + a_{0} + a_{1}t + · · · .

The direct calculation shows that

a_{0} = \frac{ω_{1}ω_{2}^{2} + ω_{2}ω_{3}^{2} + ω_{3}ω_{1}^{2} + ω_{1}^{2}ω_{2} + ω_{2}^{2}ω_{3} + ω_{3}^{2}ω_{1} − 21ω_{1}ω_{2}ω_{3}}{24ω_{1}ω_{2}ω_{3}} .

Hence

e_{0}(u, v) = ζ_{3}(0, (u, v, 1))
= \frac{1}{24} (u + \frac{1}{u} + v + \frac{1}{v} + \frac{u}{v} + \frac{v}{u} − 21) .

7 Proof of Theorem 6

We calculated S′_{3}(0, (ω_{1}, ω_{2}, ω_{3})) in [K2] as

S′_{3}(0, (ω_{1}, ω_{2}, ω_{3})) = \frac{ρ_{3}(ω_{1}, ω_{2}, ω_{3})^{2}ρ_{1}(ω_{1})ρ_{2}(ω_{2})ρ_{3}(ω_{3})}{ρ_{2}(ω_{1}, ω_{2})ρ_{2}(ω_{2}, ω_{3})ρ_{2}(ω_{3}, ω_{1})}

with

ρ_{1}(ω) = \sqrt{\frac{2π}{ω}} .

Hence

𝒮_{1}(u, v) = S′_{3}(0, (u, v, 1))
= \frac{(2π)^{\frac{3}{2}}}{\sqrt{uv}} · \frac{ρ_{3}(u, v, 1)^{2}}{ρ_{2}(u, v)ρ_{2}(u, 1)ρ_{2}(v, 1)}.

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Shin-ya Koyama: The Institute of Mathematical Sciences, Ewha Womans University, Daehyun-dong 11-1, Sedaemoon-ku, 120-750, Seoul, South Korea. Department of Biomedical Engineering, Toyo University, 2100 Kujirai, Kawagoe-shi 350-8585, Japan. koyama@toyonet.toyo.ac.jp

Nobushige Kurokawa: Department of Mathematics, Tokyo Institute of Technology, 2-12-1, Oh-okayama, Meguro-ku, Tokyo 152-8551, Japan. kurokawa@math.titech.ac.jp