Multiple Zeta Functions I

Shin-ya Koyama (Keio University)

Nobushige Kurokawa (Tokyo Institute of Technology)

Running title. Multiple Zeta Functions Abstract. We compute the absolute tensor product of the Hasse zeta functions for finite fields.

1 Introduction

Let

Z_{j}(s) = ∏ ; ∐_{ρ∈ℂ}(s − ρ)^{m_{j}(ρ)}

be “zeta functions” expressed as regularized product, where

m_{j} : ℂ → ℤ

denotes the multiplicity function for j = 1, ..., r. (Later we will specify “zeta functions” to be treated.) As in the previous paper [K2] we define the absolute tensor product (Z_{1}⊗⋯⊗Z_{r})(s) as

(Z_{1} ⊗ ⋯ ⊗ Z_{r})(s) = ∏ ∐_{ρ_{1},··· ,ρ_{r}∈ℂ}(s − (ρ_{1} + ⋯ + ρ_{r}))^{m(ρ_{1},··· ,ρ_{r})}

with

m(ρ_{1}, ⋯, ρ_{r}) = m_{1}(ρ_{1}) ⋯ m_{r}(ρ_{r}) × \{ \begin{matrix}1 & Im(ρ_{j}) ≥ 0, & (j = 1, ..., r) \\ (−1)^{r−1} & Im(ρ_{j}) < 0 & (j = 1, ..., r) \\ 0 & otherwise. \end{matrix}

We refer to the excellent survey of Manin [M]. We are especially interested in the case of Hasse zeta functions Z_{j}(s) = ζ(s, A_{j}) for rings A_{1}, ..., A_{r}. We recall that the Hasse zeta function ζ(s, A) of a ring A is defined to be

ζ(s, A) = ∏_{m}(1 − N(m)^{−s})^{−1}

where m runs over maximal left ideals of A up to the following equivalence:

m_{1} ∼ m_{2} ⟺ A/m_{1} and A/m_{2} are isomorhpic as left A-modules,

and N(m) = #End_{A−mod}(A/m). See [K3] and [F]. (For a commutative ring A, the above ζ(s, A) coincides with the usual Hasse zeta function

ζ(s, A) = ∏_{m}(1 − N(m)^{−s})^{−1},

when m runs over maximal ideals of A and N(m) = #(A/m).)

For simplicity we write

ζ(s, A_{1} ⊗ ⋯ ⊗ A_{r}) = ζ(s, A_{1}) ⊗ ⋯ ⊗ ζ(s, A_{r}).

Actually, as was explained by Manin [M], we expect that our multiple zeta function would be the zeta function of the “absolute tensor product”

A_{1} ⊗_{F_{1}} ⋯ ⊗_{F_{1}} A_{r}

that is the tensor product over the (virtual) “one element field” F_{1}. In any way, we notice that ζ(s, A_{1}⊗⋯⊗A_{r}) has the following additive structure on zeros and poles: if ζ(s, A_{j}) = 0 (resp. ∞) and Im(s_{j}) (j = 1, ..., r) have the same signature, then ζ(s_{1} + ⋯ + s_{r}, A_{1} ⊗ ⋯ ⊗ A_{r}) = 0 (resp. ∞).

Such an additive structure was crucial in the study of Hasse zeta functions of positive characteristic (congruence zeta functions) pursued by Grothendieck [G] and Deligne [D].

In this Part I, we investigate

ζ(s, F_{p} ⊗ F_{q}) = ζ(s, F_{p}) ⊗ ζ(s, F_{q})

for primes p and q. We prove that it has a kind of Euler product expression in terms of the polylogarithm:

Li_{k}(x) := ∑_{n=1}^{∞} \frac{x^{n}}{n^{k}} .

Theorem 1.1 The following expressions hold in Re(s) > 0 with some polynomial Q(s) of degree at most two:

(1) When p ≠ q, we have

ζ(s, F_{p}) ⊗ ζ(s, F_{q}) = e^{Q_{p,q}(s)} (1 − p^{−s})^{\frac{1}{2}} (1 − q^{−s})^{\frac{1}{2}}
× exp (\frac{1}{2i} ∑_{k=1}^{∞} \frac{cot (πk \frac{log p}{log q})}{k}p^{−ks} + \frac{1}{2i} ∑_{n=1}^{∞} \frac{cot (πn\frac{log q}{log p})}{n}q^{−ns}) .

(2) When p = q, we have

ζ(s, F_{p}) ⊗ ζ(s, F_{p}) = e^{Q_{p,p}(s)} (1 − p^{−s})^{1− \frac{is log p}{2π}} exp (−\frac{Li_{2}(p^{−s})}{2πi}) .

In Part II we study

ζ(s, ℤ ⊗ ℤ) = ζ(s, ℤ) ⊗ ζ(s, ℤ)

where ζ(s, ℤ) = ζ(s) is the Riemann zeta function.

2 Double Poisson Summation Formula with Signature

Let H(t) be an odd function in L^{1}(ℝ) with H(t) = O(t^{−2}) as |t| → ∞. We put

H̃(u) = ∫_{−∞}^{∞} H(t)e^{itu}dt.

Definition A real number α is called generic if and only if

lim_{m→∞} ∥mα∥^{\frac{1}{m}} = 1,

where we put ∥x∥ := min{|x − n| : n ∈ ℤ} for x ∈ ℝ. Examples 2.1 (1) If α ∈ (\overline{ℚ} ∩ ℝ) \ ℚ, then α is generic.

(2) Let x, y ∈ \overline{ℚ} ∩ ℝ. If α = \frac{log x}{log y} ∉ ℚ, then α is generic (Baker [B, Theorem 3.1]). Lemma 2.2 Assume α is generic, then the power series

∑_{n=1}^{∞} cot(πnα)x^{n}  (2.1)

absolutely converges in |x| < 1. Proof. As α is generic, we have ∥nα∥^{−1} = O(e^{εn}) as n → ∞ for any ε > 0. Since cot(πx) ∼ 1/(πx) as x → 0, we have cot(πnα) = O(e^{εn}) for any ε > 0. Theorem 2.3 Assume a/b is generic and that the test function H(t) satisfies

H̃(x) = O(μ^{x})  (2.2)

as x → ∞ for some 0 < μ < 1, then we have

∑_{k,n>0} H (2π (\frac{k}{a} + \frac{n}{b})) + \frac{1}{2} (∑_{k>0} H (2π \frac{k}{a}) + ∑_{n>0} H (2π \frac{n}{b}))

= −\frac{ia}{4π} ∑_{k>0} cot (π \frac{ka}{b}) H̃(ka) − \frac{ib}{4π} ∑_{n>0} cot (π \frac{nb}{a}) H̃(nb) − \frac{iab}{8π^{2}} H̃′(0). (2.3)

Proof. Put Z_{a}(s) = sinh (\frac{as}{2}) and Z_{b}(s) = sinh (\frac{bs}{2}). Let D_{T} be the region defined by

D_{T} = {s ∈ ℂ | |s| > α, |Re(s)| < α, 0 < Im(s) < T }

with 0 < α < min \{\frac{2π}{a} , \frac{2π}{b} \}. By Cauchy’s theorem we have for an odd function h which is

regular in D_{T}

∑_{0<Im(ρ_{a}),Im(ρ_{b})<T} h(ρ_{a} + ρ_{b}) = \frac{1}{(2πi)^{2}} ∫_{∂D_{T}} ∫_{∂D_{T}} h(s_{1} + s_{2})\frac{Z′_{a}}{Z_{a}} (s_{1})\frac{Z′_{b}}{Z_{b}} (s_{2})ds_{1}ds_{2},  (2.4)

where ρ_{a} and ρ_{b} denote the zeros of Z_{a}(s) and Z_{b}(s), respectively, and the integrals along ∂D_{T} are taken counter clockwise. Considering the limits as T → ∞ in the both sides of (2.4), we have

∑_{Im(ρ_{a}),Im(ρ_{b})>0} h(ρ_{a} + ρ_{b}) = \frac{1}{(2πi)^{2}} ∫_{∂D} ∫_{∂D} h(s_{1} + s_{2})\frac{Z′_{a}}{Z_{a}} (s_{1})\frac{Z′_{b}}{Z_{b}} (s_{2})ds_{1}ds_{2},  (2.5)

where

D = {s ∈ ℂ | |Re(s)| < α, |s| > α, Im(s) > 0}.

We decompose ∂D = C_{1} ∪ C_{2} ∪ C_{3} with

C_{1} = {s ∈ ∂D | Re(s) = −α},
C_{2} = {s ∈ ∂D | |s| = α},
C_{3} = {s ∈ ∂D | Re(s) = α}.

We compute each double integral I_{ij} = \frac{1}{(2πi)^{2}} ∫_{C_{i}} ∫_{C_{j}} in (2.5).

First we treat the integral along the vertical lines.

I_{33} = \frac{1}{(2π)^{2}} ∫_{0}^{∞} ∫_{0}^{∞} h(2α + i(t_{1} + t_{2}))\frac{Z′_{a}}{Z_{a}} (α + it_{1})\frac{Z′_{b}}{Z_{b}} (α + it_{2})dt_{1}dt_{2}.  (2.6)

Since

\frac{Z′_{a}}{Z_{a}} (α + it_{1}) = \frac{a}{2} + a ∑_{k=1}^{∞} e^{−ka(α+it_{1})}

and

\frac{Z′_{b}}{Z_{b}} (α + it_{2}) = \frac{b}{2} + b ∑_{n=1}^{∞} e^{−nb(α+it_{2})},

by putting H_{α}(t) = h(2α + i(t_{1} + t_{2})) with t = t_{1} + t_{2}, (2.6) turns to

I_{33} = \frac{1}{4π^{2}} ∑_{k,n≥0}ε_{k,n}ab ∫_{0}^{∞} ∫_{0}^{t} H_{α}(t)e^{−ka(α+it_{1})}e^{−nb(α+i(t−t_{1}))}dt_{1}dt,

where we put

ε_{k,n} = \{ \begin{matrix}1/4 & (k = n = 0) \\ 1/2 & (k = 0, n ≠ 0 or k ≠ 0, n = 0) \\ 1 & (otherwise) \end{matrix} .

Thus

I_{33} = \frac{ab}{4π^{2}} ∑_{{k,n≥0 \atop ka≠nb}} ε_{k,n}e^{−(ka+nb)α} ∫_{0}^{∞} \frac{H_{α}(t)(e^{−ikat} − e^{−inbt})}{−i(ka − nb)}dt

+ \frac{ab}{4π^{2}} ∑_{{k,n≥0 \atop ka=nb}} ε_{k,n}e^{−2kaα} ∫_{0}^{∞} tH_{α}(t)e^{−ikat}dt.  (2.7)

We similarly compute that

I_{11} = −\frac{ab}{4π^{2}} ∑_{{k,n≥0 \atop ka≠nb}} ε_{k,n}e^{−(ka+nb)α} ∫_{0}^{∞} \frac{H_{α}(−t)(e^{ikat} − e^{inbt})}{i(ka − nb)}dt

− \frac{ab}{4π^{2}} ∑_{{k,n≥0 \atop ka=nb}} ε_{k,n}e^{−2kaα} ∫_{0}^{∞} tH_{α}(−t)e^{ikat}dt.  (2.8)

By (2.7) and (2.8) we have

I_{11} + I_{33} = \frac{ab}{4π^{2}} ∑_{{k,n≥0 \atop ka≠nb}} ε_{k,n}e^{−(ka+nb)α} ∫_{−∞}^{∞} \frac{H_{α}(t)(e^{−ikat} − e^{−inbt})}{−i(ka − nb)}dt

+ \frac{ab}{4π^{2}} ∑_{{k,n≥0 \atop ka=nb}} ε_{k,n}e^{−2kaα} ∫_{−∞}^{∞} tH_{α}(t)e^{−ikat}dt.

= \frac{iab}{4π^{2}} ∑_{{k,n≥0 \atop ka≠nb}} ε_{k,n} \frac{e^{−(ka+nb)α}}{ka − nb} (H̃_{α}(−ka) − H̃_{α}(−nb))

+ \frac{ab}{4π^{2}} ∑_{{k,n≥0 \atop ka=nb}} ε_{k,n}e^{−2kaα}{ \atop tH_{α}(t)}(−ka).

The assumption that a/b is generic implies that the second sum consists only of the term k = n = 0.

Next we calculate I_{13}. Since h(i(t_{1} + t_{2})) = H_{0}(t_{1} + t_{2}) and \frac{Z′_{a}}{Z_{a}} is an odd function, we have

I_{13} = \frac{−1}{(2πi)^{2}} ∫_{∞}^{0} ∫_{0}^{∞} h(i(t_{1} + t_{2}))\frac{Z′_{a}}{Z_{a}} (−α + it_{1})\frac{Z′_{b}}{Z_{b}} (α + it_{2})dt_{1}dt_{2}
= \frac{ab}{4π^{2}} ∑_{(k,n)≠(0,0)} ε_{k,n}e^{−(ka+nb)α} ∫_{0}^{∞} \frac{H_{0}(t)(e^{ikat} − e^{−inbt})}{i(ka + nb)}dt + \frac{ab}{16π^{2}} ∫_{0}^{∞} tH_{0}(t)dt.

(2.9)

Similarly

I_{31} = \frac{ab}{4π^{2}} ∑_{(k,n)≠(0,0)} ε_{k,n}e^{−(ka+nb)α} ∫_{0}^{∞} \frac{H_{0}(−t)(e^{−ikat} − e^{inbt})}{i(ka + nb)}dt + \frac{ab}{16π^{2}} ∫_{0}^{∞} tH_{0}(t)dt. (2.10)

Therefore (2.9) and (2.10) lead to

I_{13} + I_{31} = −\frac{iab}{4π^{2}} ∑_{(k,n)≠(0,0)} ε_{k,n} \frac{e^{−(ka+nb)α}}{(ka + nb)} (H̃_{0}(ka) − H̃_{0}(−nb)) − \frac{iab}{16π^{2}} H̃_{0}′(0).

Letting α → 0 gives

lim_{α→0}(I_{11} + I_{33} + I_{13} + I_{31}) = −\frac{iab}{2π^{2}} ∑_{(k,n)≠(0,0)} ε_{k,n} \frac{H̃_{0}(ka)ka − H̃_{0}(nb)nb}{k^{2}a^{2} − n^{2}b^{2}} − \frac{iab}{8π^{2}} H̃_{0}′(0), (2.11)

since H̃_{0}′ = i{ \atop tH_{0}(t)}.

Next we treat I_{2} := I_{21} + I_{22} + I_{23}. We compute

I_{2} = \frac{1}{2πi} ∫_{C_{2}} (\frac{1}{2πi} ∫_{∂D} h(s_{1} + s_{2})\frac{Z′_{a}}{Z_{a}} (s_{1})ds_{1}) \frac{Z′_{b}}{Z_{b}} (s_{2})ds_{2}
= \frac{1}{2πi} ∫_{C_{2}} ∑_{ρ_{a}} h(ρ_{a} + s_{2})\frac{Z′_{b}}{Z_{b}} (s_{2})ds_{2},

where ρ_{a} runs through the zeros of Z_{a}(s) with Im(ρ) > 0. Putting s_{2} = αe^{iθ}, we reach

lim_{α→0} I_{2} = \frac{1}{2π} ∫_{π}^{0} ∑_{ρ_{a}} h(ρ_{a})dθ = −\frac{1}{2} ∑_{ρ_{a}} h(ρ_{a}).  (2.12)

We similarly deal with I′_{2} := I_{12} + I_{22} + I_{32} to get

lim_{α→0} I′_{2} = −\frac{1}{2} ∑_{ρ_{b}} h(ρ_{b}).  (2.13)

The integral I_{22}, which appears in both (2.12) and (2.13), tends to 0 as α → 0. Thus taking (2.11), (2.12) and (2.13) into account, (2.5) equals

−\frac{iab}{2π^{2}} ∑_{(k,n)≠(0,0)} ε_{k,n} \frac{H̃_{0}(ka)ka − H̃_{0}(nb)nb}{k^{2}a^{2} − n^{2}b^{2}} − \frac{iab}{8π^{2}} H̃_{0}′(0)− \frac{1}{2} ∑_{k>0} H_{0} (2π \frac{k}{a})− \frac{1}{2} ∑_{n>0} H_{0} (2π \frac{n}{b}) .

Theorem follows from the formulas

∑_{n>0} \frac{2ka}{k^{2}a^{2} − n^{2}b^{2}} + \frac{1}{ka} = \frac{π}{b} cot (π \frac{ka}{b}) ,

∑_{k>0} \frac{2nb}{n^{2}b^{2} − k^{2}a^{2}} + \frac{1}{nb} = \frac{π}{a} cot (π \frac{nb}{a}) .

3 Expression of Double Sine

We use the multiple Hurwitz zeta function due to Barnes

ζ_{r}(s, z, \underline{ω}) = ∑_{n_{1},...,n_{r}=0}^{∞}(n_{1}ω_{1} + ⋯ + n_{r}ω_{r} + z)^{−s}

for \underline{ω} = (ω_{1}, ..., ω_{r}) and the definitions of the multiple gamma and the multiple sine:

Γ_{r}(z, \underline{ω}) = exp ( \frac{∂}{∂s}ζ_{r}(s, z, \underline{ω})|_{s=0}) ,

S_{r}(z, \underline{ω}) = Γ_{r}(z, \underline{ω})^{−1}Γ_{r}(ω_{1} + ⋯ + ω_{r} − z, \underline{ω})^{(−1)^{r}}.

When r = 2, we have \underline{ω} = (ω_{1}, ω_{2}) and

S_{2}(z, ω_{1}, ω_{2}) = Γ_{2}(z, ω_{1}, ω_{2})^{−1}Γ_{2}(ω_{1} + ω_{2} − z, ω_{1}, ω_{2}).

The double gamma function has an expression

Γ_{2}(z, ω_{1}, ω_{2})^{−1} = e^{Q_{1}(z)}z\begin{matrix}∏ & ′ \\ n_{1},n_{2}≥0 \end{matrix} P_{2} (−\frac{z}{n_{1}ω_{1} + n_{2}ω_{2}} )

and

Γ_{2}(ω_{1} + ω_{2} − z, ω_{1}, ω_{2})^{−1} = e^{Q_{2}(z)} ∏_{n_{1},n_{2}≥1} P_{2} (\frac{z}{n_{1}ω_{1} + n_{2}ω_{2}} )

where Q_{1}(z) and Q_{2}(z) are polynomials of degree 2 and P_{2}(u) := (1 − u) exp(u + \frac{u^{2}}{2} ). We then have

S_{2}(z, ω_{1}, ω_{2}) = e^{c_{0}+c_{1}z+c_{2}z^{2}} \frac{z\begin{matrix}∏ & ′ \\ n_{1},n_{2}≥0 \end{matrix} P_{2} (−\frac{z}{n_{1}ω_{1}+n_{2}ω_{2}} )}{∏_{n_{1},n_{2}≥1} P_{2} (\frac{z}{n_{1}ω_{1}+n_{2}ω_{2}} )}  (3.1)

where we put Q_{1}(z) − Q_{2}(z) = c_{0} + c_{1}z + c_{2}z^{2}.

Lemma 3.1

\frac{d^{2}}{dz^{2}} log(1 − e^{iaz}) = − ∑_{n=−∞}^{∞} \frac{1}{(z − \frac{2πn}{a})^{2}}

Proof. Since

log(1 − e^{iaz}) = −\frac{πi}{2} + log (2 sin \frac{az}{2} )

and

2 sin \frac{az}{2} = az ∏_{n=1}^{∞} (1 − (\frac{az}{2πn})^{2}) ,

we have

\frac{d^{2}}{dz^{2}} log(1 − e^{iaz}) = −\frac{1}{z^{2}} − ∑_{n=1}^{∞} (\frac{1}{(z − \frac{2πn}{a})^{2}} + \frac{1}{(z + \frac{2πn}{a})^{2}} ) = − ∑_{n=−∞}^{∞} \frac{1}{(z − \frac{2πn}{a})^{2}} .

Theorem 3.2 Assume ω_{1}/ω_{2} is generic, then the double sine function has the following expression in Im(z) > 0:

S_{2}(z, (ω_{1}, ω_{2})) = exp (\frac{1}{2i} ∑_{k=1}^{∞} \frac{1}{k} cot (πk \frac{ω_{2}}{ω_{1}} ) e^{2πik \frac{z}{ω_{1}}} + \frac{1}{2i} ∑_{n=1}^{∞} \frac{1}{n} cot (πn\frac{ω_{1}}{ω_{2}} ) e^{2πin \frac{z}{ω_{2}}}
+ \frac{1}{2} log (1 − e^{2πi \frac{z}{ω}_{1}}) + \frac{1}{2} log (1 − e^{2πi \frac{z}{ω}_{2}})
+ \frac{πiz^{2}}{2ω_{1}ω_{2}} − \frac{πi}{2} (\frac{1}{ω_{1}} + \frac{1}{ω_{2}} ) z + \frac{πi}{12} (\frac{ω_{2}}{ω_{1}} + \frac{ω_{1}}{ω_{2}} + 3))

Proof. Apply the odd function

H(t) = \frac{1}{(z − t)^{2}} − \frac{1}{(z + t)^{2}}

with z ∈ ℂ, Im(z) > 0 to our summation formula (2.3). As we have

H̃(x) = ∫_{−∞}^{∞} H(t)e^{ixt}dt = 2πiRes_{t=z} (H(t)e^{ixt}) = −2πxe^{ixz},

the condition (2.2) is satisfied. As H̃′(0) = −2π, by putting

F (z) = ∑_{k,n≥1} (\frac{1}{(z − 2π (\frac{k}{a} + \frac{n}{b}))^{2}} − \frac{1}{(z + 2π (\frac{k}{a} + \frac{n}{b}))^{2}} )

+ \frac{1}{2} ∑_{k>0} (\frac{1}{(z − 2π \frac{k}{a})^{2}} − \frac{1}{(z + 2π \frac{k}{a})^{2}} ) + \frac{1}{2} ∑_{n>0} (\frac{1}{(z − 2π \frac{n}{b})^{2}} − \frac{1}{(z + 2π \frac{n}{b})^{2}} ) ,

the summation formula (2.3) shows

F (z) = \frac{i}{2} ∑_{k>0} cot (π \frac{ka}{b}) ka^{2}e^{ikaz} + \frac{i}{2} ∑_{n>0} cot (π \frac{nb}{a}) nb^{2}e^{inbz} + \frac{iab}{4π}

= \frac{d^{2}}{dz^{2}} (\frac{1}{2i} ∑_{k>0} \frac{1}{k} cot (π \frac{ka}{b}) e^{ikaz} + \frac{1}{2i} ∑_{n>0} \frac{1}{n} cot (π \frac{nb}{a}) e^{inbz}) + \frac{iab}{4π} (3.2)

By (3.1) with n_{1} = k, n_{2} = n, ω_{1} = \frac{2π}{a} and ω_{2} = \frac{2π}{b}, we have

\frac{d^{2}}{dz^{2}} log S_{2}(z, ω_{1}, ω_{2}) = −\frac{1}{z^{2}} − ∑_{n_{1},n_{2}≥1} (\frac{1}{(z + n_{1}ω_{1} + n_{2}ω_{2})^{2}} − \frac{1}{(z − (n_{1}ω_{1} + n_{2}ω_{2}))^{2}} )

− ∑_{n_{1}≥1} \frac{1}{(z + n_{1}ω_{1})^{2}} − ∑_{n_{2}≥1} \frac{1}{(z + n_{2}ω_{2})^{2}} + 2c_{2}

= F (z) − \frac{1}{2} ∑_{k=−∞}^{∞} \frac{1}{(z − 2π \frac{k}{a})^{2}} − \frac{1}{2} ∑_{n=−∞}^{∞} \frac{1}{(z − 2π \frac{n}{b})^{2}} + 2c_{2}

= \frac{d^{2}}{dz^{2}} (\frac{1}{2i} ∑_{k>0} \frac{1}{k} cot (π \frac{ka}{b}) e^{ikaz} + \frac{1}{2i} ∑_{n>0} \frac{1}{n} cot (π \frac{nb}{a}) e^{inbz}

+ \frac{1}{2} log(1 − e^{iaz}) + \frac{1}{2} log(1 − e^{ibz})) + 2c_{2} + \frac{iab}{4π},  (3.3)

where we used (3.2) and Lemma 3.1. So if we put

E(z) := log S_{2}(z, ω_{1}, ω_{2}) − (\frac{1}{2i} ∑_{k>0} \frac{1}{k} cot (π \frac{kω_{2}}{ω_{1}} ) e^{\frac{2πikz}{ω_{1}}} + \frac{1}{2i} ∑_{n>0} \frac{1}{n} cot (π \frac{nω_{1}}{ω_{2}} ) e^{\frac{2πinz}{ω_{2}}}

+ \frac{1}{2} log (1 − e^{\frac{2πi}{ω_{1}} z}) + \frac{1}{2} log (1 − e^{\frac{2πi}{ω_{2}} z})) ,  (3.4)

it holds \frac{d^{2}}{dz^{2}} E(z) is constant and that E(z) is a polynomial of degree 2.

Thus we put E(z) = α + βz + γz^{2} and will compute α, β and γ. We first calculate β and γ by considering

E(z + ω_{1}) − E(z) = (βω_{1} + γω_{1}^{2}) + 2γω_{1}z.  (3.5)

It follows from (3.4) that (3.5) equals

log \frac{S_{2}(z + ω_{1}, ω_{1}, ω_{2})}{S_{2}(z, ω_{1}, ω_{2})} − \frac{1}{2i} ∑_{n>0} \frac{1}{n} cot (π \frac{nω_{1}}{ω_{2}} ) (e^{\frac{2πinω_{1}}{ω_{2}}} − 1) e^{\frac{2πinz}{ω_{2}}}

−\frac{1}{2} log (1 − e^{\frac{2πi}{ω_{2}} (z+ω_{1})}) + \frac{1}{2} log (1 − e^{\frac{2πi}{ω_{2}} z}) .

The sum over n is computed as

−\frac{1}{2i} ∑_{n>0} \frac{1}{n} cot (π \frac{nω_{1}}{ω_{2}} ) (e^{\frac{2πinω_{1}}{ω_{2}}} − 1) e^{\frac{2πinz}{ω_{2}}} = −\frac{1}{2} ∑_{n>0} \frac{1}{n} (1 + e^{\frac{2πinω_{1}}{ω_{2}}} ) e^{\frac{2πinz}{ω_{2}}}
= \frac{1}{2} log (1 − e^{\frac{2πi}{ω_{2}} (z+ω_{1})}) + \frac{1}{2} log (1 − e^{\frac{2πi}{ω_{2}} z}) .

We appeal to the formula [KK, (2.4)] to get

\frac{S_{2}(z + ω_{1}, ω_{1}, ω_{2})}{S_{2}(z, ω_{1}, ω_{2})} = S_{1}(z, ω_{2})^{−1} = (2 sin \frac{πz}{ω_{2}} )^{−1} .

Hence (3.5) is equal to

− log (2 sin \frac{πz}{ω_{2}} ) + log (1 − e^{\frac{2πi}{ω_{2}} z}) = − log (2 sin \frac{πz}{ω_{2}} ) + log (−2ie^{\frac{πi}{ω_{2}} z} sin \frac{πz}{ω_{2}} )
= −\frac{πi}{2} + \frac{πi}{ω_{2}} z.

Therefore we have

βω_{1} + γω_{1}^{2} = −\frac{πi}{2}

and

2γω_{1} = \frac{πi}{ω_{2}} .

We thus obtain

β = −\frac{πi}{2} (\frac{1}{ω_{1}} + \frac{1}{ω_{2}} )

and

γ = \frac{πi}{2ω_{1}ω_{2}} .

Next we deal with α by considering

E(z) + E (z + \frac{ω_{1}}{2} ) + E (z + \frac{ω_{2}}{2} ) + E (z + \frac{ω_{1} + ω_{2}}{2} ) − E(2z).  (3.6)

The constant term of (3.6) is

3α + β(ω_{1} + ω_{2}) + γ (\frac{ω_{1}}{4} + \frac{ω_{2}}{4} + \frac{(ω_{1} + ω_{2})^{2}}{4} ) = 3α − \frac{πi}{4} (\frac{ω_{2}}{ω_{1}} + \frac{ω_{1}}{ω_{2}} + 3) .  (3.7)

On the other hand we will compute (3.6) by using (3.4). We write (3.6) as ∑_{j=0}^{4} A_{j}, where

A_{0} = log \frac{S_{2}(z)S_{2} (z + \frac{ω_{1}}{2} ) S_{2} (z + \frac{ω_{2}}{2} ) S_{2} (z + \frac{ω_{1}+ω_{2}}{2} )}{S_{2}(2z)},

A_{1} = −\frac{1}{2i} ∑_{k>0} \frac{cot (π \frac{kω_{2}}{ω_{1}} )}{k} (e^{\frac{2πikz}{ω_{1}}} + e^{\frac{2πik}{ω_{1}} (z+ \frac{ω_{1}}{2} )} + e^{\frac{2πik}{ω_{1}} (z+ \frac{ω_{2}}{2} )} + e^{\frac{2πik}{ω_{1}} (z+ \frac{ω_{1}+ω_{2}}{2} )} − e^{\frac{4πikz}{ω_{1}}} ) ,

A_{2} = −\frac{1}{2i} ∑_{n>0} \frac{cot (π \frac{nω_{1}}{ω_{2}} )}{n} (e^{\frac{2πinz}{ω_{2}}} + e^{\frac{2πin}{ω_{2}} (z+ \frac{ω_{2}}{2} )} + e^{\frac{2πin}{ω_{2}} (z+ \frac{ω_{1}}{2} )} + e^{\frac{2πin}{ω_{2}} (z+ \frac{ω_{2}+ω_{1}}{2} )} − e^{\frac{4πinz}{ω_{2}}} ) ,

A_{3} = −\frac{1}{2} log \frac{(1 − e^{\frac{2πi}{ω_{1}} z}) (1 − e^{\frac{2πi}{ω_{1}} (z+ \frac{ω_{1}}{2} )}) (1 − e^{\frac{2πi}{ω_{1}} (z+ \frac{ω_{2}}{2} )}) (1 − e^{\frac{2πi}{ω_{1}} (z+ \frac{ω_{1}+ω_{2}}{2} )})}{1 − e^{\frac{4πi}{ω_{1}} z}},

A_{4} = −\frac{1}{2} log \frac{(1 − e^{\frac{2πi}{ω_{2}} z}) (1 − e^{\frac{2πi}{ω_{2}} (z+ \frac{ω_{1}}{2} )}) (1 − e^{\frac{2πi}{ω_{2}} (z+ \frac{ω_{2}}{2} )}) (1 − e^{\frac{2πi}{ω_{2}} (z+ \frac{ω_{1}+ω_{2}}{2} )})}{1 − e^{\frac{4πi}{ω_{2}} z}}.

The formula [KK, (2.5)] gives A_{0} = 0. Next A_{1} is computed as follows:

A_{1} = −\frac{1}{2i} ∑_{{k>0 \atop even}} \frac{1}{k} cot (π \frac{kω_{2}}{ω_{1}} ) (2e^{\frac{2πikz}{ω_{1}}} + 2e^{\frac{2πik}{ω_{1}} (z+ \frac{ω_{2}}{2} )}) + \frac{1}{2i} ∑_{k>0} \frac{1}{k} cot (π \frac{kω_{2}}{ω_{1}} ) e^{\frac{4πikz}{ω_{1}}}

= −\frac{1}{2i} ∑_{k>0} \frac{1}{k} (cot (π \frac{2kω_{2}}{ω_{1}} ) (1 + e^{\frac{2πikω_{2}}{ω_{1}}} ) − cot (π \frac{kω_{2}}{ω_{1}} )) e^{\frac{4πikz}{ω_{1}}}
= −\frac{1}{2} ∑_{k>0} \frac{1}{k}e^{\frac{2πikω_{2}}{ω_{1}}} e^{\frac{4πikz}{ω_{1}}}
= \frac{1}{2} log (1 − e^{\frac{4πi}{ω_{1}} (z+ \frac{ω_{2}}{2} )}) ,

where we used an identity

cot 2θ(1 + e^{2iθ}) − cot θ = ie^{2iθ}

with θ = π \frac{kω_{2}}{ω_{1}} . Similarly A_{2} is calculated as

A_{2} = \frac{1}{2} log (1 − e^{\frac{4πi}{ω_{2}} (z+ \frac{ω_{1}}{2} )}) .

The remaining terms are easily computed as

A_{3} = −\frac{1}{2} log (1 − e^{\frac{4πi}{ω_{1}} (z+ \frac{ω_{2}}{2} )}) ,

A_{4} = −\frac{1}{2} log (1 − e^{\frac{4πi}{ω_{2}} (z+ \frac{ω_{1}}{2} )}) .

Hence we deduced that (3.6)= ∑_{j=0}^{4} A_{j} = 0. Therefore its constant term (3.7) vanishes, which

leads to

α = \frac{πi}{12} (\frac{ω_{1}}{ω_{2}} + \frac{ω_{2}}{ω_{1}} + 3) .

4 Proof of Theorem 1.1

We first describe a more precise definition of the absolute tensor product of meromorphic functions. Let Z_{j} (j = 1, 2) be meromorphic functions of order μ_{j}. We put the Hadamard product as

Z_{j}(s) = s^{k_{j}}e^{Q_{j}(s)} \begin{matrix}∏ & ′ \\ ρ∈ℂ \end{matrix} P_{μ_{j}} (\frac{s}{ρ})^{m_{j}(ρ)} ,  (4.1)

where P_{r}(u) := (1 − u) exp(u + \frac{u^{2}}{2} + ⋯ + \frac{u^{r}}{r}), m_{j} denotes the multiplicity function with k_{j} := m_{j}(0), and Q_{j} is a polynomial with deg Q_{j} ≤ μ_{j}. Here the product over ρ ∈ ℂ means

lim_{R→∞} ∏_{0<|ρ|<R} P_{μ_{j}} (\frac{s}{ρ})^{m_{j}(ρ)}. The absolute tensor product is defined by

(Z_{1} ⊗ Z_{2})(s) := s^{k_{1}k_{2}}e^{Q(s)}\begin{matrix}∏ & ′ \\ ρ_{1},ρ_{2}∈ℂ \end{matrix}P_{μ_{1}+μ_{2}} (\frac{s}{ρ_{1} + ρ_{2}} )^{m(ρ_{1},ρ_{2})} ,  (4.2)

where Q(s) is a polynomial with deg Q ≤ μ_{1} + μ_{2} and

m(ρ_{1}, ρ_{2}) := m_{1}(ρ_{1})m_{2}(ρ_{2}) × \{ \begin{matrix}1 & if Im(ρ_{1}), Im(ρ_{2}) ≥ 0, \\ −1 & if Im(ρ_{1}), Im(ρ_{2}) < 0, \\ 0 & otherwise. \end{matrix}

Here we do not give the precise definition of the polynomial Q(s), since it is not necessary for our purpose.

In this section we will compute this absolute tensor product for the Hasse zeta functions for finite fields:

Z_{1}(s) = ζ(s, F_{p}) = (1 − p^{−s})^{−1},
Z_{2}(s) = ζ(s, F_{q}) = (1 − q^{−s})^{−1},

with p, q primes.

Proposition 4.1 The absolute tensor product of the Hasse zeta functions for finite fields is given as follows:

ζ(s, F_{p}) ⊗ ζ(s, F_{q}) = e^{Q(s)}S_{2} (is, (\frac{2π}{log p}, \frac{2π}{log q} )) ,

where Q(s) is a polynomial of degree at most two, which depends on p and q.

Proof. We easily compute that the Hadamard product (4.1) for the Hasse zeta function is given by

ζ(s, F_{p}) = s^{−1}e^{Q̃_{p}(s)}\begin{matrix}∞ \\ ∏ & ′ \\ n=−∞ \end{matrix} P_{1} ( \frac{s}{\frac{2πi}{log p}n})^{−1}

with Q̃_{p}(s) a linear polynomial depending on p. Thus by the definition (4.2) of the absolute tensor product,

ζ(s, F_{p}) ⊗ ζ(s, F_{q}) = se^{Q̃_{p,q}(s)}\begin{matrix}∏ & ′ \\ k,n∈ℤ \end{matrix}P_{2} ( \frac{s}{\frac{2πi}{log p}k + \frac{2πi}{log q}n})^{m_{k,n}} ,

where Q̃_{p,q}(s) is a polynomial of degree at most two and

m_{k,n} := m (\frac{2πi}{log p}k, \frac{2πi}{log q}n) = \{ \begin{matrix}1 & if k, n ≥ 0 \\ −1 & if k, n < 0 \\ 0 & otherwise. \end{matrix}

Hence

ζ(s, F_{p}) ⊗ ζ(s, F_{q}) = se^{Q̃_{p,q}(s)} \frac{\begin{matrix}∞ \\ ∏ & ′ \\ k,n=0 \end{matrix} P_{2} ( \frac{s}{\frac{2πi}{log p}k + \frac{2πi}{log q}n})}{∏_{k,n=1}^{∞} P_{2} (− \frac{s}{\frac{2πi}{log p}k + \frac{2πi}{log q}n})}.

We appeal to the r = 2 case of the formula [KK, Proposition 2.4]:

S_{2}(z, (ω_{1}, ω_{2})) = e^{Q\underline{ω}(z)}z\frac{\begin{matrix}∞ \\ ∏ & ′ \\ k,n=0 \end{matrix} P_{2} (−\frac{z}{ω_{1}k + ω_{2}n})}{∏_{k,n=1}^{∞} P_{2} (\frac{z}{ω_{1}k + ω_{2}n})}

where Q\underline{ω}(z) a polynomial with deg Q\underline{ω} ≤ 2. Putting z = is and \underline{ω} = (ω_{1}, ω_{2}) = (\frac{2π}{log p}, \frac{2π}{log q}), we reach the proposition.

Proof of Theorem 1.1. We take (ω_{1}, ω_{2}) = (\frac{2π}{log p}, \frac{2π}{log q}) in Theorem 3.2. Example 2.1 (2) tells that ω_{1}/ω_{2} is generic, if p ≠ q. Thus Proposition 4.1 gives the assertion (1). For proving (2), we put p = q in Proposition 4.1. We recall the formulas of the double sine function:

S_{2}(z, (ω, ω)) = S_{2} (\frac{z}{ω} , (1, 1))  ([KK, Theorem 2.1(c)])
= S_{2} (\frac{z}{ω} )^{−1} S_{1} (\frac{z}{ω} ) ,  ([KK, Example 3.6])  (4.3)

where S_{r}(z) (r = 1, 2) are the primitive multiple sine functions [KK]. We have by definition

S_{1}(z) = 2 sin πz

and the expression [KK, Theorem 2.8 (2.12)]:

S_{2}(z) = exp (\frac{1}{2πi}Li_{2}(e^{2πiz}) + z log(1 − e^{2πiz}) − \frac{πi}{2}z^{2} − \frac{ζ(2)}{2πi})

for Im(z) > 0. Thus putting z = is and ω = \frac{2π}{log p} in (4.3), we have the conclusion.

Remark 4.2 The assertion (2) can also be proved in the same manner as in §2 and §3.

References

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[D] P. Deligne: La Conjecture de Weil. I, Inst. Hautes Êtudes Sci. Publ. Math. 43 (1974) 273-307.

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[G] A. Grothendiecke: Séminaire de Géométrie Algébrique du Bois-Marie 1965-66 (SGA 5), Lecture Notes in Math. 589, Springer.

[K1] N. Kurokawa: Lectures on multiple sine functions (notes taken by Shin-ya Koyama), pp. 1-119, University of Tokyo, April-July, 1991.

[K2] N. Kurokawa: Multiple zeta functions: an example. Adv. Stud. in Pure Math. 21 (1992) 219-226.

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