Logarithms of theta functions on the upper half space

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General Section

Logarithms of theta functions on the upper half space Hiroshi Ito Department of Mathematics, Faculty of Science, Kanagawa University, Hiratsuka, 259-1293, Japan

a r t i c l e

i n f o

Article history: Received 23 March 2019 Received in revised form 5 September 2019 Accepted 6 September 2019 Available online xxxx Communicated by A. Pal Keywords: Eisenstein cohomology Theta functions Imaginary quadratic fields Dedekind sums Gauss sums

a b s t r a c t Let K be an imaginary quadratic field whose discriminant is congruent to one modulo 8 and O be the ring of integers of K. Let Γ denote the group SL(2, O) which acts discontinuously on the upper half space H. In this paper, we study a homomorphism ϕ : Γ → Z obtained from a branch of the logarithm of a theta function on H which is automorphic with respect to Γ and does not vanish on H. In particular, we determine explicitly the decomposition ϕ = ϕc + ϕe of ϕ into the cusp part ϕc and the Eisenstein part ϕe , and prove a congruence conjectured by Sczech [14] between ϕ and ϕe modulo 8 under an assumption on the 2-divisibility of a certain L-value. © 2019 Elsevier Inc. All rights reserved.

1. Introduction Let H be the upper half space which consists of all quaternionnumbers  τ = z + jv a b (j 2 = −1, ij = −ji) with z ∈ C and v > 0. Every element A = of SL(2, C) c d acts on H by

E-mail address: [email protected]. https://doi.org/10.1016/j.jnt.2019.09.006 0022-314X/© 2019 Elsevier Inc. All rights reserved.

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Aτ = (aτ + b)(cτ + d)−1 , where the right hand side is taken in the skew-field of quaternions. Let K be an imaginary quadratic field embedded in the complex number field C, D its discriminant and O its ring of integers. The group Γ = SL(2, O) acts discontinuously on H. We shall consider the following theta function on H: ϑ(τ ) =

√

v

   2 2π|μ|2 v μ z+μ √ . exp − √ + πitr | D| D μ∈O+1/2 

Assume D ≡ 1 (mod 8) for simplicity. Then, it is known that ϑ(Aτ ) = ρϑ(τ ) (ρ8 = 1) for every A ∈ Γ, cf. Sczech [13], Theorem 8. Sczech [14] has studied the logarithm of ϑ(τ ) and stated several theorems and conjectures. He first states, as a theorem with a brief outline of proof, that ϑ(τ ) does not vanish on H. This allows him to define a continuous branch log ϑ(τ ) of the logarithm of ϑ(τ ) and, setting ϕ(A) =

4 {log ϕ(Aτ ) − log ϕ(τ )} πi

(A ∈ Γ),

he gets a homomorphism ϕ from Γ to Z, namely an element of H1 (Γ, Z). By a general theory on H1 (Γ, C) due to Harder, the homomorphism ϕ is decomposed into the cusp part ϕc and the Eisenstein part ϕe as ϕ = ϕc + ϕe , where ϕc and ϕe are in general Q-valued. Sczech obtained an expression (which is not explicitly described in [14]) for ϕe in terms of his Dedekind sums ([12]), calculated some examples of values of ϕ and ϕe , and conjectured that the congruence ϕ(A) ≡ ϕe (A) (mod 8Z2 ) hold for every A in Γ, where Z2 is the ring of 2-adic integers. The purpose of the present paper is to supply detailed accounts for the above mentioned two results due to Sczech (the non-vanishing of ϑ(τ ) on H and the expression of ϕe in terms of Dedekind sums), together with an explicit formula for the latter result, and to prove his conjecture under a hypothesis concerning a value of an L-function. Some of our previous results in [8] will be used. Now we shall state our results more explicitly. A general theory due to Harder gives a decomposition H1 (Γ, C) = H1cusp (Γ, C) ⊕ H1eis (Γ, C).

(1.1)

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Here, H1cusp (Γ, C) consists of χ ∈ H1 (Γ, C) such that χ(A) = 0 for every parabolic element A in Γ and H1eis (Γ, C) is the subspace of H1 (Γ, C) defined as follows. Let L be a lattice in C with complex multiplication by O. Let, for z ∈ C and n ∈ Z (n ≥ 0), 

En (z) = En (z, L) =

−n

(w + z)



−s

|w + z|

w∈L, w+z=0

, s=0

where the value at s = 0 is to be considered in the sense of analytic continuation. Setting I(z) = z − z, define ΦL : Γ → C by the following:  ΦL

 a b c d

⎧    ar  r 1  a+d ⎪ ⎪ − (0)I E E , E ⎪ 2 1 1 ⎨ c c c c r∈L/cL =   ⎪ b ⎪ ⎪ , ⎩ E2 (0)I d

c = 0, c = 0.

Then, by Sczech [12], ΦL is an element of H1 (Γ, C). The space H1eis (Γ, C) is defined as the subspace of H1 (Γ, C) generated by all ΦL with lattices L having complex multiplication by O. It is known that dim H1 (Γ, C) = h (the class number of K). Now, we can write ϕ = ϕc + ϕe according to the decomposition (1.1). First, we shall write ϕe as a linear combination of ΦL by comparing, as is indicated in [14], the both sides of the above equation on parabolic elements of Γ. Denote by η(z) the Dedekind η-function and set L = Oω with π ω= √ · 2 3|D|1/4

η

 √  4 1 + D /2  √ 2 . η 1+ D

(1.2)

The image of the homomorphism ΦL is contained in the absolute class field H of K and the set {ΦL σ | σ ∈ Gal(H/K)} is a basis of H1eis (Γ, C) over C (Section 4). Let G = Gal(H/K) = {σ1 , · · · , σh } and write ϕe =



dσ ΦL σ

(dσ ∈ C).

(1.3)

σ∈G

Denote by C1 , · · · , Ch the ideal classes of K. We choose, for each j (1 ≤ j ≤ h), an integral ideal aj ∈ Cj prime to 2 whose norm is minimum among the integral ideals prime to 2 belonging to Cj and put

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  σi T = E2 (0, a−1 , j L) 1≤i,j≤h cf. also Lemma 13. As we shall see in Section 5, detT is non-zero. Theorem 1. We have, for dσ (σ ∈ G) in (1.3), that √

D(dσ1 , · · · , dσh )T = (N a21 , · · · , N a2h ).

Next, by the use of the above theorem, we shall study the conjecture in Sczech [14] on a congruence relation between ϕ and ϕe . Take a grössencharacter ψ of K with conductor O satisfying ψ((γ)) = γ 2

(γ ∈ K × )

and put LH (ψ ◦ NH/K , s) =

 ψ(NH/K (a)) · N a−s

(Re(s) > 2).

a

Here, a runs over all integral ideals of H. We will see that the main part of LH (ψ ◦ NH/K , 2) is detT and prove the following theorem (Section 5) utilizing results in a previous work [8]. Theorem 2. The number 8(2ω)−2h LH (ψ ◦ NH/K , 2) is an integer in H and if it is prime to 2, the congruence ϕ(A) ≡ ϕe (A) (mod 8Z2 )

(1.4)

holds for every A in Γ. The congruence (1.4) can be seen as an analogue of a well-known congruence between the classical Dedekind sum and the quadratic residue symbol of Q (cf., for example, [8], §3). We do not have any value of D for which the number 8(2ω)−2h LH (ψ ◦ NH/K , 2) is not prime to 2. Since Theorem 3 (the non-vanishing of ϑ(τ )) holds also in the case D ≡ 5 (mod 8), it will not be difficult to extend our results here to this case applying the results in [9]. Although we will be concerned in this paper mainly with congruence relations modulo 8 between ϕ and ϕe , in view of works related to the Eisenstein cohomology (cf., for example, Berger [1]), congruence relations between the cusp part of ϕ and the Eisenstein part of ϕ will also deserve to be studied. The homomorphisms ϕc and ϕe are Q-valued 1 (Lemma 11) and we may take the smallest positive integer l such that Z contains the l values of ϕc and ϕe . If l > 1, we have a non-trivial congruence lϕc (A) ≡ lϕe (A) (mod l) for every A ∈ Γ (Sczech [14] contains a list of values of D for which l > 1). Various number theoretic consequences are expected from this kind of congruences, cf. [1]. The

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√ integer l is a divisor of D(a1 · · · ah )2 2h ω −2h LH (ψ ◦ NH/K , 2) as is seen from Theorem 1 and Lemma 14, and it is also a divisor of the index [ΦL (Γ) : ΦL (Γunip )] as can be seen from the proof of Lemma 11, where Γunip is the subgroup of Γ generated by unipotent elements of Γ. The discussion of this paper proceeds as follows. In §2, we consider an embedding of H into the Siegel upper half plane H2 of degree 2 and see that the theta function ϑ(τ ) is the pull-back of a well-studied theta function on H2 . Then, utilizing a known fact on the zeros of the theta function on H2 , we show that ϑ(τ ) does not vanish on H (Theorem 3). As has been explained, this enables us to define the homomorphism ϕ : Γ → Z. Next, in §3, as a preparation for §4, we calculate the Fourier expansion of ϑ(τ ) at each cusp of Γ. In §4, by the use of this expansion, we determine the value of ϕ on parabolic elements of Γ and prove Theorem 1. Finally, in §5, applying a result in [8] to an expression for ϕe obtained from Theorem 1, we prove Theorem 2. As has been mentioned, Theorem 1 and Theorem 3 are essentially due to Sczech [14]. 2. Theta functions In this section we only assume that D ≡ 1 (mod 4). Let      a θ exp πi t (n + a)Ω(n + a) + 2πi t (n + a)(w + b) (w, Ω) = b 2 n∈Z

for a, b in R2 , w in C2 and Ω in H2 . Also, for τ = z + jv in H, set 1 Ω(τ ) = √ t A D

 z −v

−v −z

 A

with  A=

 1 α 1 α

√ 1+ D . α= 2

,

The map τ → Ω(τ ) gives an embedding of H into H2 . It is observed in Kubota [10] that this embedding is induced by a suitably defined injective homomorphism from SL(2, R) to Sp(2, R). By some calculation, we see that

ϑ(τ ) = for τ ∈ H.

√

 c1 vθ d1



 (0, Ω(τ ))

 c1 =



 1/2 0

, d1 =

 0 1/2

(2.1)

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Now, set, for Ω ∈ H2 , Δ5 (Ω) =



  c θ (0, Ω). d

(c,d):even

Here, (c, d) runs over the 10 pairs of column vectors with entries 0, 12 which satisfy 4 t cd ≡ 0 (mod 2). The function Δ5 (Ω) is a cusp form of weight 5 with respect to Sp(2, Z) and the set of zeros of Δ5 (Ω) is well understood. From this knowledge, we can show the following theorem. Theorem 3 (Sczech [14]). We have Δ5 (Ω(τ )) = 0 for every τ in H. In particular, ϑ(τ ) = 0 for every τ in H. Proof. It is known that Δ5 (Ω) (Ω ∈ H2 ) is equal to zero if and only if there exists an element g of Sp(2, Z) such that g · Ω is a diagonal matrix (cf. Freitag [4], Kapitel III, Hilfssatz 1.3). We shall show that g · Ω(τ ) is not diagonal for every g in Sp(2, Z) and every τ in H, borrowing some ideas from Gritsenko and Nikulin [5]. First, we quote some facts from [5]. Consider a free Z-module of rank 4 M4 = Ze1 ⊕ Ze2 ⊕ Ze3 ⊕ Ze4 and define a symmetric bilinear form (u, v) on the free Z-module of rank 6 M4 ∧ M4 = ⊕i
(u, v ∈ M4 ∧ M4 ).

A homomorphism g : M4 → M4 induces a homomorphism g ∧ g : M4 ∧ M4 → M4 ∧ M4 and if g is in SL(M4 ) (if g is an automorphism with determinant one), then the bilinear form (u, v) is invariant under g ∧ g. Considering the matrix representation with respect to the basis e1 , e2 , e3 , e4 of M4 , we get an isomorphism Sp(2, Z) ∼ = {g : M4 → M4 | (g ∧ g)(e1 ∧ e3 + e2 ∧ e4 ) = e1 ∧ e3 + e2 ∧ e4 }. Hence, setting M = (e1 ∧ e3 + e2 ∧ e4 )⊥ = {u ∈ M4 ∧ M4 | (u, e1 ∧ e3 + e2 ∧ e4 ) = 0},

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we get a map ∧2 : Sp(2, Z) → O(M )

(g → g ∧ g).

The lattice M is a free Z-module of rank 5 with a basis f1 = e1 ∧ e2 , f2 = e2 ∧ e3 , f3 = e1 ∧ e3 − e2 ∧ e4 , f−2 = e4 ∧ e1 , f−1 = e4 ∧ e3 . Via this basis of M , we shall identify an element of O(M ) with its representation matrix, the projective space P(M ⊗ C) with P4 (C), respectively. Now, the group O(M ) acts on the space HIV = {Z ∈ P(M ⊗ C)|(Z, Z) = 0, (Z, Z) < 0} and we have HIV = H+ ∪ H+ if we put H+ = {Z = t (z22 − z1 z3 , z3 , z2 , z1 , 1)z0 ∈ P(M ⊗ C) | y1 > 0, y1 y3 − y22 > 0}. Here, yi = Im(zi ) (i = 1, 2, 3). Also, every element of ∧2 (Sp(2, Z)) leaves H+ invariant and hence Sp(2, Z) acts on H+ . Furthermore, the map 

 z1 z2

ρ : H2 → H+

z2 z3

 →

t

(z22

− z1 z3 , z3 , z2 , z1 , 1)

is one to one and preserves the action of Sp(2, Z), i.e., for Ω ∈ H2 and g ∈ Sp(2, Z), we have ρ(g · Ω) = (∧2 g) · ρ(Ω), cf. [5], Lemma 1.1. Therefore, it suffices for our purpose to show that h · ρ(Ω(τ )) = t (∗, ∗, 0, ∗, ∗) for every h ∈ O(M ) and every τ ∈ H. Let (hij )1≤i,j≤5 be the representation matrix of h with respect to f1 , f2 , f3 , f−2 , f−1 . We first note that h33 ≡ 1 (mod 2). In fact, we have h(f3 ) = h13 f1 + h23 f2 + h33 f3 + h43 f−2 + h53 f−1 and if f−1 = h(u) (u ∈ M ), then −h13 = (h(f3 ), f−1 ) = (f3 , u) ≡ 0 (mod 2) since

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(f3 , fi ) ≡ 0 (mod 2) (i = ±1, ±2, 3). Similarly, we see that hi3 ≡ 0 (mod 2) for i = 1, 2, 4, 5. It follows from det(hij ) = ±1 that h33 ≡ 1 (mod 2).   z1 z2 ∈ H2 , the condition h · ρ(Ω) = (∗, ∗, 0, ∗, ∗) For h = (hij ) ∈ O(M ) and Ω = z2 z3 is equivalent to the condition −h31 det Ω + h32 z3 + h33 z2 + h34 z1 + h35 = 0. If Ω = Ω(τ ) (τ = z + jv ∈ H), we have det Ω = −|z|2 − v 2 and 

⎛ ⎜ Ω=⎜ ⎝

tr  tr

√z D

√

1+ D 2



·



+ 2i |√vD|

√z D



+ i |√vD|

tr  tr

√ 1+ D 2 √

1+ D 2

2

·

√z D





√z D

⎞

+ i |√vD| + i 1−D 2 ·

√v | D|

⎟ ⎟. ⎠

Hence, the imaginary part of the equation h · ρ(Ω) = (∗, ∗, 0, ∗, ∗) implies that 0 = Im(h32 z3 + h33 z2 + h34 z1 )   1−D v + h33 + h34 · 2 √ . = h32 · 2 | D| Since D ≡ 1 (mod 4) and h33 ≡ 1 (mod 2), this is impossible. We have proved Theorem 3. 2 The author does not know whether Theorem 3 can be proved directly, namely without any reference to the identification ρ : H2 → H+ . 3. Fourier expansions of theta functions at cusps We shall assume that D ≡ 1 (mod 8). In this section, we prepare some expansions of theta functions in order to calculate in the next section the values of ϕ on parabolic elements of Γ. For an integral ideal a of K which is prime to 2, let ϑa (τ ) =

 2   √  μ z + μN 2π|μ|2 v + πitr v exp − |Δ| Δ N μ∈a+

√ with N = N a and Δ = N D.

2

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 a c

Lemma 4. For a matrix x =

9

b d

with a, b ∈ a, c, d ∈ a and det x = N , we have

ϑa (xτ ) = (a constant) · ϑ(τ ). We shall deduce the above lemma from a special case of it, namely from the following lemma. Lemma 5. The assertion of Lemma 4 holds if a is a prime ideal of degree one which is prime to 2. First, we prove Lemma 4 assuming Lemma 5. Take a prime ideal p of degree one prime to 2 which belongs to the ideal class of a. Let √ a = λp (λ ∈ K × ), p = N p, Δ = p D. We have N = |λ|2 p, Δ = |λ|2 Δ and ϑa (τ ) =

√

v

μ∈λp+

=

√

v

  2  2π|μ|2 v μ z + μ|λ|2 p exp − 2  + πitr |λ| |Δ | |λ|2 Δ |λ|2 p



2

  ˜2 2  2π|μ|2 v λ μ z + λμp . + πitr exp − |Δ | Δ λp

 μ∈p+

2

˜ = λ/|λ|. Note that λp is in λpp = ap and is prime to 2. Then, we see that Here, λ λp − p ≡ 0 (mod 2),

λp − p ≡ 0 (mod p)

and that λp − p ≡ 0 (mod 2p),

p+

p λp =p+ . 2 2

Furthermore, from (λ − 1)p ≡ 0 (mod 2p), we see that (λ − 1)p ⊂ (2) and (λ − 1)p ⊂ (2). p Hence, for μ ∈ p + , we have 2 λp p λp p ≡μ− + (mod 2) λμ ≡ λ(μ − ) + 2 2 2 2 and  tr It follows that

λμp Δ



 = tr

λμ √ D



 ≡ tr

μ √ D



 + tr

λp √ 2 D

 (mod 2).

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ϑa (τ ) =

√

v



     ˜2 2  2π|μ|2 v λp λ μ z + μp √ · exp πitr + πitr exp − |Δ | Δ 2 D p

μ∈p+ 2

   λp ˜ 2 z + jv) √ = exp πitr · ϑp (λ 2 D       λp λ 0 √ · ϑp τ . = exp πitr 0 λ 2 D 

 a b c d

Now, for a matrix x =

satisfying the condition of Lemma 4, let x =

  −1 λ a b = x. Then, we see that a , b ∈ p, c , d ∈ p and det x = p. It c d λ follows from Lemma 5 that     λ 0 λ 0  ϑa (xτ ) = (a constant) · ϑp xτ 0 λ 0 λ



= (a constant) · ϑp (x τ ) = (a constant) · ϑ (τ ) , which proves Lemma 4. Now we shall prove Lemma 5. Let √ a = p be a prime ideal of K which is of degree one and prime to 2. Let p = N p, Δ = p D. First of all, it is enough to show ϑp (xτ ) = (a constant) · ϑ(τ )   a b for a special matrix x = with a, b ∈ p, c, d ∈ p and det x = p, since then, for c d any such matrix x , we have x−1 x ∈ Γ and ϑp (x τ ) = ϑp (xx−1 x τ ) = (a constant) · ϑ(x−1 x τ ) = (a constant) · ϑ(τ ). Therefore, we shall calculate ϑp (xτ ) for  x=

 a b c d

⎛ =⎝

pa p

⎞ b√ m + D ⎠, 2 2

√ m+ D + Z, b is a number in O satisfying where m is an integer such that p = Z 2 √ ⎧ m+ D ⎨ ) −1 (mod 2 b≡ 2 ⎩ 0 (mod p)

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√ m+ D   a (a ∈ O). Note that det x = p. and a is defined by b + 1 = 2 2 Now, since 

 1 0

x=



a 1

0 −1 p 0



 1 0

d/c 1

,

we may write τ1 = xτ = τ2 + a ,

τ2 = −p−1 τ3−1 ,

d τ3 = τ + . c

Let τi = zi + jvi (zi ∈ C, vi > 0) for i = 1, 2, 3. First, we have ϑp (τ1 ) = ϑp (τ2 + a )  2    √ 2π|μ|2 v2 μ (z2 + a ) + μp = v2 + πitr exp − |Δ| Δ p μ∈p+ 2

√

=

v2

  ν∈p/2pp μ∈p



+πitr

 exp −

2π|ν + 2pμ + p2 |2 v2 |Δ|

(ν + 2pμ + p2 )2 (z2 + a ) + (ν + 2pμ + p2 )p Δ

 .

Here, note that  p 2   2 2 p   p 2   a − ν+ a = 4p μ + 4pμ ν + ν + 2pμ + a ≡ 0 (mod 2p), 2 2 2      2 2 ν + p2 a ν + 2pμ + p2 a = tr tr Δ Δ



and       νp  ν + 2pμ + p2 p (ν + 2pμ) p (mod 2). = tr ≡ tr tr Δ Δ Δ Then, we see that ϑp (τ1 ) =

√

v2





exp πitr

ν∈p/2pp

(ν + p2 )2 a + νp Δ



   (ν + 2pμ + p2 )2 z2 2π|ν + 2pμ + p2 |2 v2 + πitr exp − |Δ| Δ μ∈p 

×

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=

√





v2

exp πitr

ν∈p/2pp

×

(ν + p2 )2 a + νp Δ







p ν+ μ∈p+ 2p2

2π|2pμ|2 v2 + πitr exp − |Δ|



(2pμ)2 z2 Δ

 .

We apply the Poisson summation formula to the inner sum. For u > 0 and w, β ∈ C, let   f (t) = exp −2π|t + β|2 u + πitr{(t + β)2 w}u2

(t = x + iy).

Then,   fˆ(ξ) =

f (t)e2πitr(ξt) dxdy

C    1 2π|ξ|2 u z 2  = exp − 2 − πitr · ξ + 2βξ |w| + u2 |w|2 + u2 2 |w|2 + u2

and the Poisson summation formula implies that 

f (t) = κ ·



fˆ(ξ),

ξ∈p∗

t∈p

√ −1 where p∗ = Δ−1 p = p−1 D p and κ is a constant depending only on p. We shall consider the case where, in the definition of f (t), w=

ν + p2 |2p|2 v2 (2p)2 z2 ,u= ,β= . Δ |Δ| 2p

Let |τ |2 = |z|2 + v 2 for τ = z + jv ∈ H and note that |τ2 | =

1 , p|τ3 |

z2 = −

z3 , p|τ3 |2

v2 =

v3 . p|τ3 |2

Then, we have 16p2 16 (|z2 |2 + v22 ) = , |D| |D||τ2 |2 √ √ | D| | D| u 2 · pv2 |τ3 | = v3 , = |w|2 + u2 4 4 √ √ w D D 2 · pz2 |τ3 | = − z3 , = |w|2 + u2 4 4

|w|2 + u2 =

and

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 μ∈p+

p ν+ 2 2p

2π|2pμ|2 v2 + πitr exp − |Δ|



(2pμ)2 z2 Δ

13



  √    √ √ ν + p2 D | D||τ3 |  2 | D|v3 2 − πitr ξ − z3 + ξ exp −2π|ξ| =κ 8 4 4 p ∗ ξ∈p

√   2  | D||τ3 |  2π|ξ|2 v3 ξ z3 2ν + p √ √ − √ ξ =κ exp − + πitr . 8 p2 | D| p2 D p2 D ξ∈ 1 p 2

Hence, if we put κ =

√ | D| √ κ, 8 p

we see that √

ϑp (τ1 ) = κ





2π|ξ|2 v3 √ v3 exp − + πitr 2 | D| p 1 ξ∈ p



ξ 2 z3 √ p2 D

 · R(ξ),

2

with 



(ν + p2 )2 a + νp 2ν + p − √ ξ Δ p2 D ν∈p/2pp     p2 ξ 1 a ν 2 + ((a + 1)p − 2 )ν + a − ξ . exp πi tr = Δ p 4

R(ξ) =

exp πi tr

ν∈p/2pp

We want to see that ξ ∈ p(O + 12 ) if R(ξ) = 0. For this, note that, if ν  = ν + pμ (ν ∈ p, μ ∈ O),  ξ 2 a ν  + ((a + 1)p − 2 )ν  p

 ξ − a ν 2 + ((a + 1)p − 2 )ν p ξ  2 2  = a (2pνμ + p μ ) + ((a + 1)p − 2 )pμ p (mod 2p) ≡ a p2 μ(μ + 1) + (p2 − 2ξ)μ 2 (mod 2p). ≡ (p − 2ξ)μ Here, we used the fact that 2 decomposes in K. Hence,  R(ξ) = exp πi and, if R(ξ) = 0, we must have

1 2 (p − 2ξ)μ · R(ξ) (μ ∈ O) Δ

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14

 tr

1 2 (p − 2ξ)μ Δ

∈ 2Z (μ ∈ O),

and therefore 1 2 (p − 2ξ) ∈ 2O, Δ

  1 . ξ∈p O+ 2

The above calculation also shows that the summand for R(ξ) depends only on the class of ν modulo pO if ξ ∈ p(O + 12 ), and hence we have, for ξ ∈ O + 12 ,   p2  1  2  a ν + ((a + 1)p − 2ξ)ν + a − pξ R(pξ) = 4p exp πitr Δ 4 ν∈p/pO     p  1 · S(ξ) a −ξ = 4p exp πitr √ D 4 



with 

 " 1 !  2  S(ξ) = exp πitr √ a ν + ((a + 1)p − 2ξ)ν . p D ν∈p/pO 

We next calculate the sum S(ξ) (ξ ∈ O + 12 ). By the one-to-one correspondence O/p  μ → 2dμ ∈ p/pO and the congruence (a + 1)p · 2dμ ≡ 0 (mod 2pO), we see that   1   2 2 √ 4a d μ − 4dμξ p D μ∈O/p     4d(b + 1)  2 √ exp πitr μ − μξ . = p D μ∈O/p 

S(ξ) =



exp πitr

Here, we have also used the equality a d = b + 1 and the congruence 4dξ ≡ 4d(b + 1)ξ If we write q =

p+1 2 ,ξ

=η+

p 2

(mod 2pO).

(η ∈ O), then

2q ≡ 1, 2ξ ≡ 2η

(mod pO),

and, for μ ∈ O, μ2 − μξ ≡ μ2 − 2qμξ ≡ μ2 − 2qμη (mod pO). ≡ (μ − qη)2 − q 2 η 2 Noting also that d ≡ 0 (mod 2), we see

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 4d(b + 1)  2 √ μ − q2 η2 p D μ∈O/p   d(b + 1) 2 η = W · exp πitr − √ p D

S(ξ) =

15



exp πitr

with  4d(b + 1) 2 √ W = exp πitr μ . p D μ∈O/p 



Therefore,  T (pξ) = 4pW · exp πitr

pa √ 4 D



  ξ d(b + 1) 2 · exp πitr − √ − √ η . D p D

By the above calculation, setting κ = 4κ p W exp πitr



 pa √ , 4 D

we see that   2   2π|ξ|2 v3 ξ z3 √ ϑp (τ1 ) = κ v3 exp − √ + πitr √ | D| D ξ∈O+ 12   ξ d(b + 1) 2 ·exp πitr − √ − √ η D p D    2  2π|ξ|2 v3 ξ z−ξ  √ √ √ = κ v3 exp − + πitr | D| D ξ∈O+ 12  2  dξ d(b + 1) 2 ·exp πitr √ − √ η . c D p D If we note that c = p, bd ≡ 0 (mod 2p) and   dp2 p2 ≡ dξ 2 − d(b + 1)η 2 ≡ d(ξ 2 − η 2 ) ≡ d pη + 4 4 we see that ϑp (xτ ) = κ exp πitr This proves Lemma 5, and hence Lemma 4.



dp √ 4 D

 · ϑ(τ ).

(mod 2p),

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4. The Eisenstein part of the theta homomorphism As we have stated in §1, Theorem 3 allows us to get an element ϕ of H1 (Γ, Z) by defining ϕ(A) =

4 [log ϑ(Aτ ) − log ϑ(τ )] (A ∈ Γ), πi

and according to the decomposition (1.1), it is decomposed as ϕ = ϕc + ϕe . In this section, we prove that the Eisenstein part ϕe is determined as in Theorem 1. For this, we first calculate the values of ϕ on parabolic elements. We begin with the following lemma. Lemma 6. For an integral ideal a prime to 2, there exist exactly two elements in a + 12 N a whose norm gives the minimum value of the norms of elements of a + 12 N a. Proof. We note that the assertion holds for every imaginary quadratic field K with D ≡ 1 (mod 4). Let a0 be an integral  ideal with minimum norm in the absolute ideal class of a. It is known that N a0 < |D|. Since a0 is not divisible by any integer different from ±1, we can write √ s+ D , r = N a0 , |s| < r (s ∈ Z) a0 = Zr + Z 2 and the following inequalities for m, n in Z can be verified from |s| < r <

 |D|:

 √  1 c + D a a+n if (m, n) = (0, 0), (1, 0), m− > 2 2 2 √ √   1 c + D c + D if (m, n) = (0, 0), (0, 1), ma + n − > 2 2 4  √ √    1 1 c + D a c + D a+ n− m− > − if (m, n) = (0, 1), (1, 0). 2 2 2 2 4 Hence, if γ ∈ 12 ao , γ ∈ / a0 , the function |x − γ| of x ∈ a0 takes the minimum value at exactly two elements of a0 . Since we can write a = λa0 , (λ ∈ K × ), the same assertion holds also for a. Therefore, the function |x + 12 N a| of x ∈ a takes the minimum value at exactly two elements of a. This concludes the proof. 2  Lemma 7. Let a be an integral ideal of K and let x = a, c, d ∈ a and det x = N a.

 a c

b d

be a matrix with a, b ∈

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 (1) The matrix U (t) = x

 1 0

−1

17

t 1

x belongs to Γ if and only if t ∈ aa−1 .

(2) Assume a is prime with 2 and let β be an element of a + 12 N a with minimum norm. Then, we have for t ∈ aa−1 ,  ϕ(U (t)) = 4tr

β2t √ D · Na

 .

Proof. (1) Put N = N a. Since ad − bc = N , we see ac + ad = ON = aa and a = Oc + Od. By calculation, ⎛

t ⎜ N U (t) = ⎝ 2 t −c N 1 + cd

d2

t N

⎞

⎟ . t ⎠ 1 − cd N

t t t t This belongs to Γ if and only if c2 , cd , d2 ∈ O, which is equivalent to a2 ⊂ O N N N N and also to t ∈ a−2 N = aa−1 . (2) By Lemma 4, we may take a continuous function log ϑa (τ ) on H for an integral ideal a of K prime with 2 and we have log ϑ(τ ) = log ϑa (xτ ) + (a constant). Hence, for t ∈ aa−1 , 4 [log ϑ(U (t)τ ) − log ϑ(τ )] πi 4 = [log ϑa (xU (t)τ ) − log ϑa (xτ )] πi 4 [log ϑa (xτ + t) − log ϑa (xτ )] . = πi

ϕ(U (t)) =

with minimum norm are β  and −β.  Also, the summands β ∈ Z. Therefore, if we with μ = β, −β in the definition of ϑa (τ ) coincide since tr √ D put Now, the elements of a +

N 2

   β k = 2 exp πitr √ D

(= 2, −2),

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18

we have

 2   2 1 β z √ ϑa (τ ) − k exp − 2π|β| v + πitr √ v |Δ| Δ 

≤

μ∈a+ N 2 , μ=±β

 2π|μ|2 v exp − |Δ|

 2π(|β|2 + n4 )v exp −  |Δ| n=1 ∞ 

 2π(|β|2 + 14 )v  exp − |Δ|

(if v  1).

Here, the implied constant depends only on the ideal a and we have used the fact that the number of elements of a + N2 with norm |β|2 + n4 , (n ∈ N) is O(n). Hence, for v  1,  2       β z −πv 2π|β|2 v + πitr × 1 + O exp ϑa (τ ) = k v exp − |Δ| Δ 2|Δ| √

and  2  β z 2π|β|2 v + πitr log ϑa (τ ) = log k + log v − |Δ|    Δ πv . +(a constant in 2πiZ) + O exp − 2|Δ| 1 2

This gives  log ϑa (τ + t) − log ϑa (τ ) = πitr

β2t Δ



   πv + O exp − 2|Δ|

and 4 lim {log ϑa (τ + t) − log ϑa (τ )} = 4tr ϕ(U (t)) = πi v→∞ This proves Lemma 7.



β2t Δ

 .

2

Let L be a lattice in C with complex multiplication by O. We define the homomorphism ΦL : Γ → C as has been described in Section 1. Lemma 8 (Sczech [12], Sats 3). Let a, x and U (t) be the same as in Lemma 7 and let t ∈ aa−1 . Then, we have ΦL (U (t)) =

" 1 ! tE2 (0, a−1 L) − tE2 (0, a−1 L) . Na

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19

We shall deduce Theorem 1 from the above two lemmas. First, we study the algebraicity of the values of ΦL . Define ω by (1.2) and put L = Oω. This lattice has been used in Sczech [13] and Ito [8]. Lemma 9. The lattice L being as above, let ℘(z) be the Weierstraß ℘-function with respect to L, φ be a primitive 4-th division point of C/L. Let F = Q(j(L)) and denote by OF the integer ring of F . (1) (℘(φ) − ℘(2φ)2 = 144|D|. 1 (2) √ ΦL (A) ∈ OF (A ∈ Γ) D 2 (3) The coefficients g2 and g3 of the differential equation ℘ = 4℘3 − g2 ℘ − g3 satisfied by ℘(z) are algebraic integers. Proof. By Ito [8], Theorem 2, 12 ΦL (A) ∈ OF i(℘(φ) − ℘(2φ))

(A ∈ Γ).

Hence the assertion (2) follows from (1). For a lattice Λ in C, let ℘(z) = ℘Λ (z) be the Weierstraß ℘-function with respect to Λ and let Δ(Λ) = g2 3 − 27g3 2 with g2 = g2,Λ , g3 = g3,Λ being the coefficients of 2 the differential equation ℘ = 4℘3 − g2 ℘ − g3 . Also, we write Δ(w) for Δ(Λ) if Λ = Z + Zw (Im(w) > 0). It is known that Δ(w) = (2π)12 η(w)24 . Put 212 η(2α)24 212 Δ(2α) u=− =− 24 η(α) Δ(α)



√  1+ D α= . 2

Then,  √ 2 3|D|1/4 η(2α)2 · Δ(L) = ω Δ(α) = π η(α)4 12 24 2 η(2α) = 126 |D|3 · η(α)24 6 3 = 12 D u.

12

−12

· (2π)12 η(α)24

On the other hand, let γ = ℘(φ) − ℘(2φ), δ = ℘(2φ), t =

12δ γ

with ℘(z) = ℘L (z) and T (z) =

γ d , T1 (z) = γ −1/2 T (z). ℘(z) − ℘(2φ) dz

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20

Then, T1 2 = T (4T 2 + tT + 4), cf., for example, Cassou-Naguès and Taylor [3], p. 54. It follows from this that g2 = g2,L = 12δ 2 − 4γ 2 , g3 = g3,L = 4γ 2 δ − 8δ 3 and Δ(L) = g2 3 − 27g3 2 = 144γ 4 δ 2 − 64γ 6 . Hence, 144γ 4 δ 2 − 64γ 6 = 126 D3 u. Furthermore, from [3], p. 110, we have t2 − 64 = −u,

144δ 2 = (64 − u)γ 2

and hence, (64 − u)γ 6 − 64γ 6 = 126 D3 u,

γ 6 = −126 D3 .

Here, note that ω ∈ R, ℘(z) = ℘(z) and that t2 and γ 2 do not depend on the choice of φ (cf. [3], p. 109, Lemma 1.1). Then, we see that γ 2 ∈ R and γ 2 = −122 D = 144|D|. This proves (1). From the above, we see furthermore, δ 2 = (u − 64)D, g2 = 12δ 2 − 4γ 2 = 12D(u − 16), j g3 2

(u − 16)3 (12g2 )3 = , Δ(L) u  1 1  3 3 (g2 3 − Δ(L)) = 12 D ju − 126 D3 u = 27 27

=

= 26 D3 u(j − 1728). Since j is an algebraic integer, u, g2 and g3 are also algebraic integers. This completes the proof of Lemma 9. 2

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21

By the above lemma and Sczech [12], Satz 4, for an arbitrary lattice L with OL = L , we can take λ ∈ C and σ ∈ Gal(H/K) such that ΦλL = ΦL σ . Since ΦλL = λ−2 ΦL , we see that H1eis (Γ, C) =  ΦL σ | σ ∈ G C , where G = Gal(H/K). Hence, we may express ϕe as a linear combination of ΦL σ (σ ∈ G) as in (1.3). To prove Theorem 1, we shall prepare a lemma. Lemma 10. Let a be an integral ideal prime to 2 whose norm is minimum among integral ideas prime to 2 in the absolute ideal class of a. Then, the elements in a + 12 N a with minimum norm are ± 12 N a. Proof. By Lemma 6, it suffices to show that 21 N a is an element with minimum norm in a + 12 N a. For any λ in a + 12 N a, 2λ is in a and prime to 2. Hence, we can write (2λ) = ab with an integral ideal b prime to 2. Since b belongs to the absolute ideal class of a, we have N b = N b ≥ N a = N a and N (λ) = This proves Lemma 10.



1 1 N (ab) ≥ N a2 = N 4 4

 1 Na . 2

2

Now, we take, as in Section 1, an integral ideal aj prime to 2 with minimum norm   aj bj from each absolute ideal class Cj of K (j = 1, · · · , h). Let, for each j, xj = cj dj be a matrix with aj , bj ∈ aj , cj , dj ∈ aj and det xj = N aj , and  Uj (t) = xj

−1

 1 t 0 1

xj

(t ∈ aj aj −1 ).

By Lemma 10 and Lemma 7, (2), we have 

αj 2 t ϕ(Uj (t)) = 4tr √ D · N aj   t . = N aj tr √ D



 αj =

1 N aj 2



Since ϕc (Uj (t)) = 0, comparing the both sides of (1.3), we see by Lemma 8,

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22

   " t 1 ! = N aj tr √ dσ tE2 (0, aj −1 L)σ − tE2 (0, aj −1 L)σ . N aj D σ∈G Here, we note that E2 (0, aj −1 L) ∈ H by Lemma 8 and Lemma 9, (2). Considering the coefficients of t and t, we see that N aj 2 =

√

D



dσ E2 (0, aj −1 L)σ =

√

D

σ∈G



dσ E2 (0, aj −1 L)σ

(1 ≤ j ≤ h).

(4.1)

σ∈G

Hence, if we write G = {σ1 , · · · , σh } and   T = E2 (0, aj −1 L)σi 1≤i,j≤h , then, (N a1 2 , · · · , N ah 2 ) =

√

D(dσ1 , · · · , dσh )T.

This proves Theorem 1. We conclude this section by the following lemma. Lemma 11. For any A ∈ Γ, we have ϕe (A) ∈ Q. Proof. By Sczech [12], Satz 6 and Ito [7], we have n := [ΦL (Γ) : ΦL (Γunip )] < ∞, where Γunip is the subgroup of Γ generated by unipotent elements in Γ. Hence, for A ∈ Γ, there exists a matrix B ∈ Γunip such that ΦL (An ) = nΦL (A) = ΦL (B). Then, from (1.3) follows that nϕe (A) = ϕe (An ) =

 σ∈G

dσ ΦL (An )σ =



dσ ΦL (B)σ = ϕe (B).

σ∈G

Since ϕc (B) = 0, we have ϕe (A) =

1 1 1 ϕe (B) = ϕ(B) ∈ Z ⊂ Q. n n n

2

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23

5. Congruence between the theta homomorphism ϕ and its Eisenstein part ϕe The purpose of this section is to prove Theorem 2. We begin with several lemmas. As in Section 1, we let ψ be a grössencharacter of K with conductor (1) = O which satisfies ψ((γ)) = γ 2

(γ ∈ K × )

and let LH (ψ ◦ NH/K , s) =



ψ(NH/K (a))N a−s

(Re(s) > 2).

a⊂OH

Here, OH is the ring of integers of H and a runs over all integral ideals of H. Also, NH/K denotes the relative norm with respect to H/K and N denotes the absolute norm of ideals of H. Put L = Oω with ω defined by (1.2). Lemma 12. Let b1 , · · · , bh be a complete representative system for absolute ideal classes of K and set   T  = E2 (0, bj −1 bi ) 1≤i,j≤h . Then, the following assertions hold. (1) det T  = 2h LH (ψ ◦ NH/K , 2). (2) det T  = 0. (3) E2 (0, L) = 0. Lemma 13. For an integral ideal a of K, the number E2 (0, a−1 L) is an integer of H and E2 (0, a−1 L) ∈ a2 · OH . Moreover, if a is prime to 2, we have E2 (0, a−1 L) ≡ E2 (0, L)

(mod 8OH ).

Lemma 14. The number ω −2h LH (ψ ◦ NH/K , 2) belongs to H and we have (det T )OH = (a1 · · · ah )2 2h ω −2h LH (ψ ◦ NH/K , 2) · OH . Furthermore, we have det T = 0. We prove these three lemmas at the end of the paper. Note that dσ ∈ H (σ ∈ G) from Theorem 1.

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Lemma 15. If every dσ (σ ∈ G) is 2-integral, we have ϕ(A) ≡ ϕe (A)

(mod 8Z2 )

for every A in Γ. Proof. By Lemma 9, (1) and Ito [8], Theorem 2, 1 √ ΦL (A) ≡ E2 (0, L)ϕ(A) (mod 8OH ). D Then, from (1.3) and the assumption on dσ , we see that ϕe (A) ≡



√ dσ D E2 (0, L)σ ϕ(A) (mod 8),

σ∈G

μ−ν is a 2-integral algebraic number. Now, we have 8 aj = O for some j (1 ≤ j ≤ h) and hence from (4.1) follows that where μ ≡ ν (mod 8) means that √

D



dσ E2 (0, L)σ = 1.

σ∈G

Therefore, ϕe (A) ≡ ϕ(A) (mod 8). Since ϕe (A), ϕ(A) ∈ Q, we have ϕe (A) − ϕ(A) ∈ 8Z2 .

2

We shall now prove Theorem 2. By Lemma 13, all of the numbers in j-th column   of T = E2 (0, aj −1 L)σi 1≤i,j≤h are in aj 2 · OH , and hence det T is in (a1 · · · ah )2 · OH . Also, since det T is equal to the determinant of the matrix obtained from T by replacing E2 (0, aj −1 L)σi by E2 (0, aj −1 L)σi − E2 (0, a1 −1 L)σi (≡ 0 (mod 8OH )) for j = 2, · · · , h, it is contained in 8h−1 OH . Hence, det T ∈ 8h−1 (a1 · · · ah )2 · OH . It follows from Lemma 14 that 8(2ω)−2h L(ϕ ◦ NH/K , 2) ∈ OH . Let Ti be the matrix obtained from T by replacing the i-th row by (N a1 2 , · · · , N ah 2 ). Then, we have det Ti = N a1 2 · t˜i1 + · · · + N ah 2 · t˜ih

(1 ≤ i ≤ h),

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25

where t˜ij is the (i, j)-th cofactor of T . Furthermore, since N aj 2 ∈ aj 2 OH and N aj 2 ≡ N a1 2 ≡ 1 (mod 8), the determinant det Ti is also in 8h−1 (a1 · · · ah )2 OH . Hence, if T˜ denotes the cofactor matrix of T (the (i, j)-element of T˜ is t˜ji ), then every element of (N a1 2 , · · · , N ah 2 )T˜ belongs to 8h−1 (a1 · · · ah )2 OH . Since we see from Theorem 1 and Lemma 14 that (dσ1 , · · · , dσh ) = √

1 (N a1 2 , · · · , N ah 2 )T˜, D · det T

all of dσi (1 ≤ i ≤ h) are 2-integral if 81−h · det T is prime to 2 and we have for A ∈ Γ ϕ(A) ≡ ϕe (A)

(mod 8Z2 )

by Lemma 15. Finally, Lemma 14 tells that 81−h · det T is prime to 2 if and only if 8(2ω)−2h LH (ψ ◦ NH/K , 2) is prime to 2. This proves Theorem 2. To conclude the paper, we shall give proofs of Lemma 12, 13 and 14. Proof of Lemma 12. For a character χ of the absolute ideal class of K, let 

L(χψ, s) =

(χψ)(a)N a−s

(Re(s) > 1),

a⊂O

where a runs over all integral ideals of K. By class field theory, LH (ψ ◦ NH/K , s) =



L(χψ, s).

χ

For an absolute ideal class C, let L(C, ψ, s) =



ψ(a)N a−s

(Re(s) > 1),

a∈C

where a runs over all integral ideals belonging to the class C. If we take an ideal a ∈ C −1 , then since O× = {±1}, L(C, χ, s) =

1  ψ(wa−1 )N (wa−1 )−s 2 0=w∈a

=

 1 ψ(a)−1 N as w−2 |w|4−2s . 2 0=w∈a

Because |ψ(a)| = N a, we see that L(C, ψ, 2) =

1 ψ(a)E2 (0, a). 2

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Denoting the ideal class of bi −1 by Ci (1 ≤ i ≤ h),  

det T = det 2ψ(bj

−1

−1

bi )

L(Cj

−1

 Ci , ψ, 2)

  = 2h det L(Cj −1 Ci , ψ, 2) . By a fact on the group determinant (cf., for example, Borevich and Schafarevich [2], p. 421), 

det T = 2

h

 h   χ

= 2h



 χ(ci )L(Ci , ψ, 2)

i=1

L(χψ, 2)

χ

= 2h LH (ψ ◦ NH/K , 2). This proves (1). It is known by Hecke ([6], p. 215) that LH (ψ ◦ NH/K , 2) = 0. The assertion (2) follows from this. As is mentioned after the proof of Lemma 9, for any lattice L with complex multiplication by O, we have ΦλL = ΦL σ for some λ ∈ C× and σ ∈ Gal(H/K). Since  ΦL

 1 b 0 1

= E2 (0, L )I(b) (b ∈ O),

this implies that E2 (0, L ) = λ2 E2 (0, λL ) = λ2 E2 (0, L)σ . Now, if E2 (0, L) were zero, then E2 (0, L ) would be zero for every lattice L with OL = L and T  would be zero matrix, which is impossible by (2). This completes the proof of Lemma 12. 2 Proof of Lemma 13. Take a complete representative system b1 , · · · , bh of the absolute ideal class group of K which satisfies N bj ≡ 1 (mod 4),

bj ∩ Z = ZN bj

for every j (1 ≤ j ≤ h). We assume b1 = O. For example, take a prime ideal pj of degree √ one which is prime to 2D from j-th ideal class (2 ≤ j ≤ h) and put bj = pj or bj = Dpj according to N pj ≡ 1 (mod 4) or N pj ≡ 3 (mod 4). We can write √ bj + D bj = Zaj + Z 2

(aj = N bj , bj ∈ Z)

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and we have bj 2 − D ≡ 0 (mod 8aj ). We assume a1 = b1 = 1. Put, for each j, τj =

√ 4 bj + D π η(τj ) , ωj = √ · , 2aj 2 3|D|1/4 η(2τj )2

Lj = (Zτj + Z)ωj = bj

−1

ωj

√ 1+ D . The number ω1 coincides with ω in (1.2) and we have L1 = and τ = τ1 = 2 Oω = L. For each j, there exists σ ∈ G such that g2,Lj = g2 σ , g3,Lj = g3 σ , cf. [13], p. 103–p. 104. Then, we have ΦLj = ΦL σ by [12], Satz 4. Also, we see from Lemma 8 and Lemma 9 that E2 (0, a−1 L) ∈ H and E2 (0, a−1 Li ) = E2 (0, a−1 L)σ

(1 ≤ i ≤ h).

(5.1)

Now, we have  ω 12 j

ω

=

Δ(τj )2 Δ(2τ ) . Δ(2τj )Δ(τ )2

Here, Δ(2τj )/Δ(τj ) and Δ(2τ )/Δ(τ ) are units in H by a property of the automorphic function Δ(2z)/Δ(z) (cf. [3], VIII, §1 and IX, §5). Also, Δ(τj )/Δ(τ ) belongs to H and Δ(τj ) 12 · OH = bj · OH Δ(τ ) by the theory of complex multiplication (cf., for example, Robert [11], Proposition 3). Hence, we see that  ω 12 j

ω

· OH = bj

12

· OH .

If we take the number j such that bj belongs to the same ideal class as a and put a = λbj then we see

(λ ∈ K × ),

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ωj L, ω  2 ω 2 −1 2 E2 (0, a Lj ) = λ N bj E2 (0, L). ωj a−1 Lj = λ−1 N bj −1

Note that E2 (0, L) = 0 by Lemma 12. Then it follows from E2 (0, a−1 Lj ), E2 (0, L) ∈ H  2 ω that also belongs to H and ωj  ω 2 j

ω

2

· OH = bj · OH

(1 ≤ j ≤ h).

Since E2 (0, L) is in OF , we see that E2 (0, a−1 Lj ) ∈ λ2 bj 2 · OH = a2 OH and, by (5.1), E2 (0, a−1 L) ∈ a2 OH . Assume now that a is prime to 2. By the definition of E2 (z, L), we see 

E2 (0, a−1 L) =

E2 (r, L)

r∈a−1 L/L

and hence, 

E2 (0, a−1 L) − E2 (0, L) = 2

E2 (r, L).

0=r∈(a−1 L/L)/±1

Here, as is noted in [12], (7), we have E2 (r, L) = ℘(r) + E2 (0, L)

(r ∈ / L)

and, by [8], Lemma 2, ℘(2φ)−1 E2 (r, L) ≡ 0 (mod 4), where φ is taken as in Lemma 9. It follows that   ℘(2φ)−1 E2 (0, a−1 L) − E2 (0, L) ≡ 0 (mod 8). Finally, we have by Lemma 9 t=

℘(2φ) 12℘(2φ) = ± ℘(φ) − ℘(2φ) |D|

(5.2)

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and t2 − 26 is known to be a unit (cf. [3], IX, Theorem 5.10). Hence t and ℘(2φ) are prime to 2. This concludes the proof of Lemma 13. 2 Proof of Lemma 14. By Lemma 8 and Lemma 9, the elements of T and det T belong to H. Keeping the notation introduced in the proof of Lemma 13, write the elements of G = Gal(H/K) as σ1 , · · · , σh so that g2,Li = g2 σi , g3,Li = g3 σi

(1 ≤ i ≤ h).

Then, σ1 is the identity map and E2 (0, bj −1 L)σi = E2 (0, bj −1 Li ) = E2 (0, bj −1 bi

−1

ωi ) = N bi 2 · ωi −2 E2 (0, bj −1 bi ).

It follows that det T  = (N b1 · · · N bh )2 (ω1 · · · ωh )−2 · det T  with   T  = E2 (0, bj −1 L)σi 1≤i,j≤h . By (5.2) and Lemma 12, we see that ω −2h LH (ψ ◦ NH/K , 2) belongs to H and (det T  )OH = (b1 · · · bh )2 ω −2h 2h LH (ψ ◦ NH/K , 2)OH . The identity for (det T )OH in Lemma 14 follows from this. Since LH (ψ ◦ NH/K , 2) = 0, we have det T = 0. This concludes the proof. 2 References [1] T. Berger, On the Eisenstein ideal for imaginary quadratic fields, Compos. Math. 145 (2009) 603–632. [2] Z.I. Borevich, I.R. Schafarevich, Number Theory, Academic Press, 1966. [3] Ph. Cassou-Nogues, M.J. Taylor, Elliptic Functions and Rings of Integers, Progress in Mathematics, vol. 66, Birkhauser, Boston, MA, 1987. [4] E. Freitag, Siegelsche Modulfunktionen, Grundlehren der Mathematischen Wissenschaften, vol. 254, Springer Verlag, 1983. [5] V.A. Gritsenko, V.V. Nikulin, Siegel automorphic form corrections of some Lorentzian Kac-Moody Lie algebras, Amer. J. Math. 119 (1997) 181–224. [6] E. Hecke, Mathematische Werke, Vandenhoeck & Ruprecht, Göttingen, 1970. [7] H. Ito, On a property of elliptic Dedekind sums, J. Number Theory 27 (1987) 17–21. [8] H. Ito, Dedekind sums and quadratic residue symbols, Nagoya Math. J. 118 (1990) 35–43. [9] H. Ito, Dedekind sums and quadratic residue symbols of imaginary quadratic fields, J. Math. Soc. Japan 43 (1991) 447–456. [10] T. Kubota, Notes on Analytic Theory of Numbers, Lecture Note, University of Chicago, 1963. [11] G. Robert, Unites elliptiques et formules pour nenombre de classes des extensions abeliennes d’un corps quadratique imaginaire, Bull. Soc. Math. France, Memoire 36 (1973) 5–77.

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[12] R. Sczech, Dedekindsummen mit elliptischen Funktionen, Invent. Math. 76 (1984) 523–551. [13] R. Sczech, Dedekind sums and power residue symbols, Compos. Math. 59 (1986) 89–112. [14] R. Sczech, Theta functions on the hyperbolic three space, in: Kokyuroku RIMS, vol. 603, Kyoto Univ., 1987, pp. 9–20.