A FORMULA FOR THE CENTRAL VALUE OF CERTAIN HECKE L-FUNCTIONS

ARIEL PACETTI

Abstract. Let N ≡ 1 mod 4 be the negative of a prime, K = ℚ(\sqrt{N}) and O_{K} its ring of integers. Let D be a prime ideal in O_{K} of prime norm congruent to 3 modulo 4. Under these assumptions, there exists Hecke characters ψ_{D} of K with conductor (D) and infinite type (1, 0). Their L-series L(ψ_{D}, s) are associated to a CM elliptic curve 𝔄(N, D) defined over the Hilbert class field of K. We will prove a Waldspurger-type formula for L(ψ_{D}, s) of the form L(ψ_{D}, 1) = Ω ∑_{[A],I} r(D, [A], I)m_{[A],I}([D]) where the sum is over class ideal representatives I of a maximal order in the quaternion algebra ramified at |N| and infinity and [A] are class group representatives of K. An application of this formula for the case N = −7 will allow us to prove the non-vanishing of a family of L-series of level 7|D| over K.

1. Introduction

Given an elliptic curve E over ℚ, and a fundamental discriminant D, a formula of Waldspurger formula relates the value of L(E ⊗ D, 1), the twist of E by D, with the coefficients of a 3/2 modular form (see [Wal]). The purpose of this work is to get a formula for quadratic twists of a family of elliptic curves with complex multiplication not defined over the rationals.

Given an imaginary quadratic field K the theory of complex multiplication (see [Sh2]) gives a relation between elliptic curves with CM given by an order of K and L-series associated to Hecke characters ψ on K. The simplest case is when K = ℚ(\sqrt{N}) with N ≡ 1 mod 4 the negative of a prime and ψ is a character of conductor \sqrt{N}. In this case the L-series corresponds to a CM elliptic curve 𝔄(N) studied by Gross in [Gr], defined over H, the Hilbert class field of K. A formula for the central value of L(ψ, 1) was given by Villegas in [Vi].

In this paper we will study the central value of the L-series corresponding to the CM elliptic curves 𝔄(N, D), given by twists of 𝔄(N) by ideals \sqrt{N}D where D is a prime ideal of K prime to \sqrt{N} and with prime norm congruent to 3 modulo 4. The ideal D has associated h Hecke characters ψ_{D} of K of conductor D, where h is the class number of K. The relation between the L-series of 𝔄(N, D) and L(ψ_{D}, s) is given explicitly by :

L(𝔄(N, D)/H, s) = ∏_{ψ_{D}} L(ψ_{D}, s)L(\overline{ψ_{D}}, s)

where H is the Hilbert class field of K and the product is over the h Hecke characters associated to D (see [Gr] formula (8.4.4) and Theorem 18.1.7). If we define 𝔅 be the Weil restriction of scalars of 𝔄(N, D) to K, then 𝔅 is a CM abelian variety, and L(𝔄(N, D)/H, s) = L(𝔅/K, s).

This paper is the author’s Ph.D. thesis presented at the University of Texas on August 2003.

He would like to thank Dr. Fernando Rodriguez-Villegas and the University of Texas.

Let B be the quaternion algebra ramified at |N| and infinity. To the ideal D we will associate a maximal order O_{[D]} in B depending only on the class of D . If {I} are representatives for left O_{[D]}-ideals, the main theorem (2.1) gives the formula L(ψ_{D}, 1) = Ω ∑_{[A],I} r(D, [A], I)m_{[A],I}([D]) where the sum is over the ideals {I} and ideal representatives of O_{K}, Ω is a period, r(D, [A], I) is a rational integer and the numbers m_{[A],I}([D]) are algebraic integers.

In the second chapter we prove a first formula for the value of the L-series at 1. In the third chapter we prove the fact that m_{[A],I} is an algebraic integer. Furthermore we define the field it belongs to. In the forth chapter we make the relation between L-series and ideals, giving a method to determine if two special points are equivalent. At last, in the last chapter we study in detail the case when the class number of K is one. In this case the elliptic curve 𝔄(N) is defined over ℚ and the numbers m_{I} turn out to be rational integers. In the case N = −7 using the fact that the quaternion algebra has class number 1 for maximal ideals, we will be able to prove that the CM elliptic curves 𝔄(N, D) defined over K have a non-vanishing L-series for all primes D. We finish this work with a remarkable relation between the numbers m_{I} and the coordinates of the eigenvector of the modular form associated to 𝔄(N) represented in the Brandt matrices of level N^{2}.

2. L-series

2.1. L-series definition. Given a number field K, we will denote O_{K} its ring of integers, Cl(O_{K}) its class group and h its class number. Let N ≡ 1 mod 4 be the negative of a prime and K := Q(\sqrt{N}). Let D ≡ 1 mod 4 be the negative of a prime such that the ideal generated by D splits completely in K, i.e. (D) = (D) ¯ (D). We will denote L := ℚ(\sqrt{D}). The ideal D induces a quadratic character from O_{K}/D to \{±1\} by extending the Kronecker symbol _{(\overline{|D|})} so as to make the following diagram commute:

[figure: O_{K}/DO_{K}; ±1; ℤ/|D|ℤ; _{(_{\overline{|D|}})}]

We will denote ε_{D} this character. It induces a Hecke character ψ_{D} on principal ideals by ψ_{D}(⟨α⟩) = ε_{D}(α)α.

Proposition 2.1. The character ψ_{D} on principal ideals is well defined.

Proof. Since 1 and −1 are the only units in K, we must check that ε_{D}(α)α = −ε_{D}(−α)α. This follows from the fact that ε_{D} is multiplicative and |D| ≡ 3 mod 4, hence ε_{D}(−1) = −1 □ The character actually depends of the choice of D (i.e. we have one character associated to D and another one associated to D̄). Abusing notation we will denote just by ψ the character associated to D. The character ψ defined on principal ideals extends to h Hecke characters on I(O_{K}) the set of ideals of O_{K}. We fix an extension once and for all and we call it ψ. Then ψ : I(O_{K}) ⟶ T_{ψ}, where T_{ψ} is a degree h field extension of K.

Definition. The L-series associated to ψ is

(1)  L(ψ, s) := ∑_{A} \frac{ψ(A)}{NA^{s}}

where the sum is over all ideals A of O_{K}.

By Hecke’s work we know that L(ψ, s) extends to an analytic function in the upper half plane, and satisfies the functional equation:

(\frac{2π}{\sqrt{ND}} )^{−s} Γ(s)L(ψ, s) = w_{ψ} (\frac{2π}{\sqrt{ND}} )^{s−2} Γ(2 − s)L(ψ̄, 2 − s)

where w_{ψ} is the root number. The character ψ defines a weight 2 modular, f_{ψ}(z) = ∑_{A} ψ(A)e^{2πizNA}, for z in the upper half plane, of level ND. The root number is given by w_{ψ} = f_{ψ}(\frac{i}{\sqrt{ND}})/\overline{f_{ψ}(\frac{i}{\sqrt{ND}})}. The characters ψ are associated to a CM elliptic curve 𝔄(N, D) defined over H, the Hilbert class field of K, by the formula:

L(𝔄(N, D)/H, s) = ∏_{ψ_{D}} L(ψ_{D}, s)L(\overline{ψ_{D}}, s)

See [Gr] formula (8.4.4) and Theorem 18.1.7.

2.2. Choosing characters in a consistent way. Given an ideal D we choose an extension of the Hecke character ψ defined on principal ideals to all the ideals and we got a field T_{ψ} depending on the extension chosen. Given another prime ideal D′, if we extend the character associated to D′ in an arbitrary way, the image of both characters will lie in different fields. There is a natural way of defining a Hecke character ψ_{D′} associated to D′ such that ψ_{D′}(I(O_{K})) ⊂ T_{ψ}. Since the class group has order h we know that any ideal raised to the h-power is principal, hence for all ideals A prime to DD′ we define:

(2)  ψ_{D′}(A) = ψ_{D}(A)\frac{ε_{D′}(A^{h})}{ε_{D}(A^{h})}

Proposition 2.2. there exists a Hecke character associated to D′ taking values in T_{ψ} and defined as above on ideals prime to DD′.

Proof. We start by proving that the character defined above is a Hecke character on ideals prime to DD′. If A is principal, say A = ⟨α⟩, then ψ_{D′}(α) = ε_{D}(α)α\frac{ε_{D′} (α)^{h}}{ε_{D}(α)^{h}}. Since h is odd, and ε takes the values ±1, we get that ψ_{D′}(α) = ε_{D′}(α)α, hence it is a Hecke character. Let 𝔮 be a prime ideal in the same class equivalence as D and prime to DD′ (there exists such an ideal by Tchebotarev density theorem), say 𝔮β = D. Then ψ_{D′}(D) = ψ_{D′}(𝔮β) = ψ_{D′}(𝔮)ψ_{D′}(β) = ψ_{D′}(𝔮)ε_{D′}(β)β. In this way we can extend the character to all ideals prime to D′ and clearly this is well defined, taking values in T_{ψ}. □ Given a prime ideal 𝔭 prime to D, we will denote ψ_{𝔭} the Hecke character associated to 𝔭. From now on given two different characters ψ_{D} and ψ_{D′} we will always assume that they are chosen in a consistent way.

Proposition 2.3. The root number in the functional equation is w_{ψ} = − (\frac{2}{|N|}) i\frac{α}{|α|},

where α = ±ψ_{N} (D) and the sign is chosen such that K(\sqrt{α\sqrt{N}}) is the quadratic extension of K associated to the character ψ, i.e. it is +1 if 2 is unramified in K(\sqrt{α\sqrt{N}}) and −1 if not.

Proof. See [Bu-Gr] proposition 10.6, page 20 □

Given a quadratic imaginary field L = ℚ[\sqrt{d}] we denote w_{d} the number of units in the ring of integers of L. Our main result is the following:

Theorem 2.1. Let B be the quaternion algebra over ℚ ramified at |N| and infinity. Let O be a maximal order in B, {I} a set of left O-ideal representatives and for each ideal D of K, ψ_{D} the character defined in (2). Then :

(3)  L(1, ψ_{D}) = \frac{2π}{w_{|D|}\sqrt{|D|}}η(D̄)η(O_{K}) (∑_{I} ∑_{[A]} r(D, [A], I)m_{[A],I}([D]))

with r([A], D) ∈ ℤ and m_{[D]}([A], I) an algebraic integer lying in a finite field extension of ℚ (see diagram 3.3).

The term Ω = _{\frac{2π}{w_{|D|}\sqrt{|D|}}}η(D̄)η(O_{K}) on (3) corresponds to a period of the abelian

variety 𝔅 and the number r([A], I) counts the number of special points on each class ([A], I) with a ±1 weight. The rest of this paper will be a constructive proof of Theorem 2.1.

2.3. Computing the L-series value at 1. Given A an ideal of K , we will denote [A] its class in the class group. We can decompose the L-series as

(4)  L(ψ, s) = ∑_{[A]} ∑_{B∼A} \frac{ψ(B)}{NB^{s}}

Proposition 2.4. All integral ideals equivalent to A are of the form cA for some c ∈ A^{−1}.

Proof. easy to check. □ Since the only units in O_{K} are 1 and −1,

∑_{B∼A} \frac{ψ(B)}{NB^{s}} = \frac{1}{2} ∑_{c∈Ā} \frac{ψ(c)ψ(A)}{ψ(NA)} \begin{matrix}NA & s \\ Nc & s \end{matrix} = \frac{1}{2}NA^{s} \frac{ψ(A)}{ψ(NA)} ∑_{c∈Ā} \frac{ψ(c)}{Nc^{s}}

Since ψ is multiplicative ψ(A)ψ(Ā) = ψ(NA), then \frac{ψ(A)}{ψ(NA)} = \frac{1}{ψ(Ā)}. Using the fact

that NA = NĀ it follows that ∑_{B∼A} \frac{ψ(B)}{NB^{s}} = \frac{1}{2} \frac{NĀ^{s}}{ψ(Ā)} ∑_{c∈Ā} \frac{ψ(c)}{Nc^{s}} and we can write the L-series as:

(5)  L(s, ψ) = \frac{1}{2} ∑_{[A]∈Cl(O_{K})} \frac{NA^{s}}{ψ(A)} ∑_{c∈A} \frac{cε_{D}(c)}{Nc^{s}}

Without loss of generality, we may assume that A = aℤ + \frac{b+\sqrt{N}}{2} ℤ and D = |D|ℤ + \frac{b+\sqrt{N}}{2} ℤ, hence AD = a|D|ℤ + \frac{b+\sqrt{N}}{2} ℤ (see [Vi] §2.3 page 552). If c ∈ A then c = ma + n\frac{b+\sqrt{N}}{2} , and ε_{D}(c) = ε_{D}(ma + n\frac{b+\sqrt{N}}{2} ). Since n\frac{b+\sqrt{N}}{2} ∈ D, ε_{D}(c) = ε_{D}(a)ε_{D}(m) = ε_{D}(NA)ε_{D}(m). We will denote z_{A} the point \frac{b+\sqrt{N}}{2a} (respectively z_{D} the point \frac{b+\sqrt{N}}{2|D|} and z_{AD} the point \frac{b+\sqrt{N}}{2a|D|} ). Also we denote by ∑′ the sum removing the zero element (or zero vector depending on the context). We have:

(6) L(s, ψ) = \frac{1}{2} ∑_{[A]∈Cl(O_{K})} \frac{NA^{1−s}ε_{D}(NA)}{ψ(A)} ∑_{m,n∈ℤ}′ \frac{ε_{D}(m)(m + z_{AD}|D|n)}{N(m + z_{AD}|D|n)^{s}}

If we change m by −m in the sum, since ε_{D}(−1) = −1, the term in the inner sum can be written as \frac{ε_{D}(m)}{(m+(−z̄_{AD})|D|n)|m+(−z̄_{AD})|D|n|^{2s−2}} , where the point −z̄_{AD} is in the upper half plane. This sum is related to Eisenstein series that we define below:

Definition. Let p be a prime integer and ε(m) := (\frac{m}{p}). We define the Eisenstein

series associated to ε by E_{1}(z, s) = ∑′_{m,n∈ℤ} \frac{ε(m)}{(m+zpn)|m+zpn|^{2s}}.

By (6) we get the relation:

(7)  L(s, ψ) = \frac{1}{2} ∑_{[A]∈Cl(O_{K})} \frac{NA^{1−s}ε_{D}(NA)}{ψ(A)}E_{1}(−z̄_{AD}, s − 1)

E_{1}(z, s) turns out to be a modular form of weight 1 with a character. We need to compute its value at s = 0 for a point z in the upper half plane. This series converges only for ℜ(s) > \frac{3}{2}, but it can be analytically continued to the whole plane and satisfy a functional equation. We will compute its value at s = 0 using Hecke’s trick. Since ε is a character of conductor p, we break the sum over m as:

(8) E_{1}(z, s) = ∑_{m∈ℤ}′ \frac{ε(m)}{m} + 2 ∑_{n=1}^{∞} ∑_{r mod p} ε(r) ∑_{m∈ℤ} \frac{1}{(zpn + r + mp)|zpn + r + mp|^{2s}}

and dividing the last sum by p^{2s+1} we get:

(9) E_{1}(z, s) = 2L(s, ε) + 2 ∑_{n=1}^{∞} ∑_{r mod p} \frac{ε(r)}{p^{2s+1}} ∑_{m∈ℤ} \frac{1}{(\frac{zpn+r}{p} + m)|\frac{zpn+r}{p} + m|^{2s}}

For z in the upper half plane we define:

H(z, s) = ∑_{m∈ℤ} \frac{1}{(z + m)|z + m|^{2s}}

Lemma 2.1. Let z = x + iy be a point in the upper half plane, then:

∑_{m=−∞}^{∞}(z + m)^{−(s+1)}(z̄ + x)^{−s} = ∑_{n=−∞}^{∞} τ_{n}(y, s + 1, s)e^{2πinx}

where τ_{n}(y, s + 1, s) is given by:

τ_{n}(y, s + 1, s)\frac{iΓ(s+1)Γ(s)}{(2π)^{2s+1}} = \{ \begin{matrix}n^{2s}e^{−2πny}σ(4πny, s + 1, s) & (n > 0) \\ |n|^{2s}e^{−2π|n|y}σ(4π|n|y, s, s + 1) & (n < 0) \\ Γ(2s)(4πy)^{−2s} & n = 0 \end{matrix}

and σ(y, α, β) = ∫_{0}^{∞}(t + 1)^{α−1}t^{β−1}e^{−yt}dt

Proof. This is Lemma 1 page 84 [Sh] □ The right side of lemma 2.1 equality converges for any s > 0, so we can compute the limit when s tends to 0 of τ_{n}(y, s + 1, s) in the different cases:

• Case n = 0: lim_{s→0} \frac{(2π)^{2s+1}}{iΓ(s+1)} \frac{Γ(2s)}{Γ(s)}(4πy)^{−2s} = −iπ • Case n < 0: lim_{s→0} \frac{(2π)^{2s+1}}{iΓ(s+1)Γ(s)}|n|^{2s}e^{2π|n|y} ∫_{0}^{∞} (t + 1)^{s−1}t^{s}e^{−4π|n|yt}dt = 0

• Case n > 0: lim_{s→0} \frac{(2π)^{2s+1}n^{2s}}{iΓ(s+1)} e^{−2πny} \frac{1}{Γ(s)} ∫_{0}^{∞} (t + 1)^{s}t^{s−1}e^{−4πnyt}dt.

We just need to compute lim_{s→0} \frac{1}{Γ(s)} ∫_{0}^{1} (t + 1)^{s}t^{s−1}e^{−4πnyt}dt. Doing integration by parts:

∫_{0}^{1} (t + 1)^{s}t^{s−1}e^{−4πnyt}dt = \frac{2^{s}e^{−4πny}}{s} − ∫_{0}^{1} t^{s}(t + 1)^{s−1}e^{−4πnyt}dt−

−\frac{1}{s} ∫_{0}^{1} t^{s}(t + 1)^{s}e^{−4Πnyt}(−4πnyt)dt

The function Γ(z) has a simple pole at z = 0 with residue 1. Dividing the integral by Γ(s) and taking the limit when s tends to zero we get:

(10)  lim_{s→0} τ_{n}(y, s + 1, s) = −2πie^{−2πny}

We just prove:

Lemma 2.2. lim_{s→0} H(s, z) = −πi − 2πi ∑_{n=1}^{∞} q^{n}

Equation (9) can be written as

E_{1}(z, s) = 2L(s, ε) + 2 ∑_{n=1}^{∞} ∑_{r mod p} \frac{ε(r)}{p^{2s+1}} H(\frac{zpn + r}{p} , s)

which by lemma 2.1 is the same as:

E_{1}(z, s) = 2L(s, ε) + 2 ∑_{n=1}^{∞} ∑_{r mod p} \frac{ε(r)}{p^{2s+1}} ∑_{k∈ℤ} τ_{k}(yn, s + 1, s)e^{2πik( \frac{xpn+r}{p})}

Let G(ε) := ∑_{r mod p} ε(r)ξ_{p}^{r} be the Gauss sum associated to the quadratic character

ε. Let ξ_{p} = e^{\frac{2πi}{p}}. If we take the limit as s tends to zero and use lemma (2.2) in the inner sum we get:

∑_{r mod p} \frac{ε(r)}{p}(−πi − 2πi ∑_{k=1}^{∞} q^{nk}ξ_{p}^{rk}) = −\frac{2πi}{p}G(ε) ∑_{k=1}^{∞} ε(k)q^{nk}

If p is congruent to 3 modulo 4 it is a well known result that G(ε) = i\sqrt{p} then:

(11)  lim_{s→0} E_{1}(z, s) = 2L(1, ε) + \frac{4π}{\sqrt{p}} ∑_{n=1}^{∞} (∑_{d|n} ε(d)) q^{n}

Knowing the value of E_{1}(z, 0), and using equation (7) we get the value of L(1, ψ). We will write this number in terms of theta functions so as to relate the value for different ideals D. Let A be any ideal of L. For z in the upper half plane, we define Θ_{A}(z) = ∑_{λ∈A} e^{2πiz \frac{Nλ}{NA}} = 1 + ∑_{n=1}^{∞} r_{A}(n)q^{n} where r_{A}(n) is the number of elements λ ∈ A of norm nNA.

Lemma 2.3. Let w be the number of roots of unity in L, and z a point in the upper half plane. Then \frac{w\sqrt{p}}{4π}E_{1}(z, 0) = ∑_{A∈Cl(O_{L})} Θ_{A}(z)

Proof. We need to check that the q-expansion on both sides is the same. The constant term first on the right side is h, the class number of ℚ(\sqrt{−p}). On the left side we have \frac{L(1,ε)w\sqrt{p}}{2π} which by the class number formula is h. Since the constant term is the same, we can apply the Mellin transform on both sides. Dividing by w we need to prove the equality:

(12)  ∑_{n=1}^{∞} \frac{∑_{d|n} ε(d)}{n^{s}} = \frac{1}{w} ∑_{A∈Cl(O_{L})} ∑_{n=1}^{∞} \frac{r_{A}(n)}{n^{s}}

Given a number field L the zeta function associated to it is:

ζ_{L}(s) = ∑_{A} \frac{1}{NA^{s}}

where the sum is over all integral ideals of L. It follows easily from the definition that ζ_{L}(s) = \frac{1}{w} ∑_{A∈Cl(O_{L})} ∑_{n=1}^{∞} \frac{r_{A}(n)}{n^{s}} which is the right hand side of (12).

It is a classical result that ζ_{L}(s) = ζ(s)L(s, ε) (see for example [Wa] Theorem 4.3, page 33). If we look at the Mellin transform of this product, we get (∑_{n=1}^{∞} \frac{1}{n^{s}} ) (∑_{m=1}^{∞} \frac{ε(m)}{m^{s}}) which is the right hand side of (12) □ Note that −z̄_{AD} = zĀD̄, hence by equation (7) and lemma 2.3 we get:

L(1, ψ) = \frac{2π}{w\sqrt{|D|}} ∑_{[A]∈Cl(O_{K})} \frac{ε_{D}(NA)}{ψ(A)} ∑_{[B]∈Cl(O_{L})} Θ_{B}(zĀD̄)

By equation (2) ψD̄(A) = ψ_{D}(A)εD̄(A^{h})ε_{D}(A^{h}) = ψ_{D}(A) (\frac{NA}{|D|})^{h}. Since h is odd

it follows that \frac{ε_{D}(NA)}{ψ_{D}(A)} = \frac{1}{ψD̄(A)}.

Theorem 2.2. The value at s = 1 of L(s, ψ) is given by:

L(1, ψ) = \frac{2π}{w\sqrt{|D|}} ∑_{[A]∈Cl(O_{K})} ∑_{[B]∈Cl(O_{L})} \frac{Θ_{B}(z_{AD̄})}{ψD̄(Ā)}

2.4. Theta functions in several variables. The goal now is to write the identity of theorem 2.2 in terms of theta functions in two variables so as to relate the Lfunction values for different primes D. Given an element in ℂ^{2}x𝔥_{2} (the Siegel space of dimension 2), the generalized theta function is defined by

θ(z⃗, Ω) = ∑_{n⃗∈ℤ^{2}} exp(πin^{t}Ωn⃗ + 2πin^{t}.z⃗)

It satisfies a functional equation for the group Γ_{12} (following Igusa notation), which

is defined to be: α = ( \begin{matrix}A & B \\ C & D \end{matrix} ) ∈ Sp_{2g}(ℤ) such that A^{t}C and B^{t}D have even

diagonal.We are interested in the cases:

(13)  θ(0, −(Qτ)^{−1}) = \sqrt{det(Q)} (−i)τθ(0, Qτ)
(14)  θ(z⃗, Ω + B) = ξ_{α}θ(z⃗, Ω)

where B is symmetric, entire and even diagonal. (see [Mu], §5, page 189).

L is an imaginary quadratic field, so given an ideal B of Cl(O_{L}) we can associate to it a quadratic form of discriminant D via the group isomorphism between Cl(O_{L}) and {quadratic forms of discriminant D}. More specifically, given a quadratic form of discriminant D, say [a, b, c] where b^{2} −4ac = D, we associate the ideal ⟨a, \frac{b+\sqrt{D}}{2} ⟩; and conversely given any primitive ideal (i.e. not divisible by any rational integer greater than 1) A, we can chose a pair of generators of the form A = ⟨a, \frac{b+\sqrt{N}}{2} ⟩, and associate to it the quadratic form [a, b, c] where c = (b^{2} − D)/(4a). We will

denote Q_{B} the matrix ( \begin{matrix}2a & b \\ b & 2c \end{matrix} ) associated to the quadratic form [a, b, c].

Given an ideal B in Cl(O_{L}), and a point z ∈ 𝔥, Θ_{B}(z) = ∑_{α∈B} e^{2πizN(α)/N(B)}. Let B = ⟨a, \frac{b+\sqrt{N}}{2} ⟩ with a = N(B). If α ∈ B then it can be written uniquely as

α = ma + n (\frac{b+\sqrt{N}}{2} ). Hence N(α) = a(am^{2} + mnb + n^{2} \frac{b^{2}−N}{4a}) and

(15)  Θ_{B}(z) = ∑_{(m,n)∈ℤ^{2}} exp [πiz(m, n) ( \begin{matrix}2a & b \\ b & 2c \end{matrix} ) ( \begin{matrix}m \\ n \end{matrix} )]

Since z ∈ 𝔥 and Q_{B} is symmetric, zQ_{B} ∈ 𝔥_{2}. Hence Θ_{B}(z) = θ(0, zQ_{B}). So we can rewrite the main formula of theorem 2.2 as:

(16)  L(1, ψ) = \frac{2π}{w\sqrt{|D|}} ∑_{[A]∈Cl(O_{K})} ∑_{[B]∈Cl(O_{L})} \frac{θ(0, z_{AD̄}Q_{B})}{ψD̄(Ā)}

3. Normalization of the Theta function

For z ∈ 𝔥, we recall the definition:

η(z) = e^{2πiz/24} ∏_{n=1}^{∞}(1 − e^{2πinz})

Without loss of generality any time we write a basis for an ideal we will assume that b is congruent to 3 modulo 48 to avoid keep track of roots of unity. Given a point z_{AD}, we define the normalizer:

Υ(z_{AD}) := η(D)η(O_{K})ψ_{D}(Ā)

Then the main formula (16) can be written as:

(17)  L(1, ψ) = \frac{2π}{w\sqrt{|D|}} ( ∑_{[A]∈Cl(O_{K})} ∑_{[B]∈Cl(O_{L})} \frac{θ(0, z_{AD̄}Q_{B})}{Υ(z_{AD̄})}) η(D̄)η(O_{K})

We are interested in studying the number: n_{A,B,D̄} = θ(0, z_{AD̄}Q_{B})/Υ(z_{AD̄}). The normalizer Υ is chosen so as to make n_{A,B,D̄} be an algebraic integer as we will see later.

3.1. Complex Multiplication. Let F_{M} be the field of all modular functions of level M whose q-expansion at every cusp has coefficients in ℚ(ζ_{M}). Let K(M) denote the ray class field of K mod M, and for a prime ideal 𝔭 in K relative prime to M (say of norm p), σ(𝔭) denotes the Frobenius automorphism of K(M)/K corresponding to 𝔭.

Theorem 3.1. let f(z) be in F_{M} and suppose that (p) = 𝔭𝔭̄ in K where p is a rational prime such that (p, NM) = 1. Suppose that A = [μ, ν] is a fractional ideal of K with ϑ = μ/ν in 𝔥 and let B\binom{μ}{ν} be a basis for 𝔭̄A. Then f(ϑ) is in K(M) and f(ϑ) ∘ σ(𝔭) = [f ∘ (pB^{−1})](Bϑ).

If in addition f is analytic in the interior of 𝔥 and has algebraic integer coefficients in its q-expansion at every cusp, then f(ϑ) is an algebraic integer.

Proof. This is Theorem 3 of [St] page 213. □

Proposition 3.1. Following the previous notation, θ(0, \frac{z}{a|D|}Q_{B})/η(\frac{z}{|D|})η(z) is in F_{24aD^{2}}.

For the proof we need the elementary result:

Lemma 3.1. if f(z) is a modular form of weight k and level N and D is a positive integer then f(\frac{z}{D}) is a modular form of weight k and level at most ND.

Proof of proposition 3.1. Let B be the ideal B := ℤa + ℤ\frac{b+\sqrt{d}}{2}. Then the quadratic form associated to B is [a, b, c] with b^{2} − 4ac = D and the matrix of the

bilinear form is ( \begin{matrix}2a & b \\ b & 2c \end{matrix} ) . The theta series θ_{B} is the theta series associated to

this quadratic form hence it has level |D|, weight 1 and a character ϵ(d) = (\frac{D}{d} ) (see [Ogg] Theorem 20, page VI-25). Using the previous lemma, we have that θ_{B}(\frac{z}{a|D|}) is a modular form of weight 1 and level aD^{2}.

The eta function is a modular form of weight 1/2 and level 24 , then η(\frac{z}{|D|}) has weight 1/2 and level 24|D| , so the product of the two eta functions has weight 1 and level 24|D|. Hence the quotient has weight 0 and level at most 24aD^{2}. We do not need a sharp estimate of the q-expansion, hence the real level is not important.

From the q-expansion of the functions θ_{B}, and η it is clear that the q-expansion at infinity of θ(0, \frac{z}{a|D|}Q_{B})/η(\frac{z}{|D|})η(z) is in ℚ(ξ_{24aD^{2}}), hence we just need to check this condition at the other cusps. For that purpose we will study the q-expansion of each form separately.

Since the theta function θ_{B} is a modular form for Γ_{0}(|D|), there are just two inequivalent cusps which may be taken to be 0 and ∞. One transformation that

send infinity to zero is given by the matrix S = ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) sending z to −1/z.

The functional equation (13) reads:

(18) θ (0, Q_{B}^{−1}(−1/z)) = det(Q_{B})^{1/2}(−i)zθ(0, Q_{B}z) = \sqrt{|D|}(−i)zθ(0, Q_{B}z)

Since Q_{B}^{−1} = Adj (Q_{B})/|D|, replacing z by z/|D| we get

(19)  θ (0, Adj (Q_{B})(−1/z)) = (−i)z/\sqrt{|D|} θ(0, Q_{B}z/|D|)

Replacing Q_{B} by its adjoint matrix, we see that the q-expansion at 0 includes a 4th root of unity and the square root of |D| (the z factor actually cancels out a factor coming from the eta function). Since \sqrt{D} ∈ ℚ(ξ_{D}), the q-expansion of θ(0, Q_{B}) has coefficients in ℚ(ξ_{8D}) at all cusps. Replacing z by z/a|D| we add at most (aD^{2})- th roots of unity to the q-expansions, hence the q-expansion of θ(0, \frac{z}{a|D|}Q_{B}) has coefficients in ℚ(ξ_{24aD^{2}}) at all cusps.

We will use the following explicit version of the transformation formula for η, which can be found in [Vi] page 560:

Lemma 3.2. Let ( \begin{matrix}α & β \\ γ & δ \end{matrix} ) ∈ Sl_{2}(ℤ) with γ even, δ positive (and odd), and τ ∈ 𝔥.

Then

(20)  η (\frac{ατ + β}{γτ + δ} ) = \binom{γ}{δ}e_{24}(κ)\sqrt{γτ + δ}η(τ)

where κ = 3(δ − 1) + δ(β − γ) − (δ^{2} − 1)γα.

For any matrix in Γ_{0}(2), the modular form η changes by a 24-th root of unity, hence its q-expansion at the cusps equivalent module Γ_{0}(2) have coefficients in ℚ(ξ_{24}) and the q-expansion of η(\frac{z}{|D|}) has coefficients in ℚ(ξ_{24aD^{2}}). But modulo Γ_{0}(2) there are just two not equivalent cusps which may be taken to be zero and infinity also. The eta function satisfies the functional equation η(−1/z) = \sqrt{z/i} η(z). Hence its q-expansion at zero has coefficients in ℚ(ξ_{8}) and η(\frac{z}{|D|}) certainly has a q-expansion with coefficients in ℚ(ξ_{24aD^{2}}) at zero. □

3.2. Field of definition.

Theorem 3.2. The number θ(0, z_{AD̄}Q_{B})/η(zD̄)η(O_{K}) is an algebraic integer in H, the Hilbert class field of K.

Proof. The eta function does not vanish in the upper half plane so we can apply Theorem 3.1 and θ(0, z_{a|D|}Q_{B})/η(z_{|D|})η(z) is an algebraic integer in F (some field extension of K containing H) for any z in the upper half plane.

Let g(z) := θ(0, \frac{z}{a|D|}Q_{B})/η(\frac{z}{|D|})η(z). Given an element σ of Gal(F/K) by complex multiplication theory there exists a prime ideal 𝔭 in K such that σ = σ_{𝔭}, where σ_{𝔭} is the element in Gal(F/K) corresponding to 𝔭 via the Artin map. We want to prove the quotient is in H hence we take 𝔭 to be principal and using Tchebotarev density theorem we may assume that 𝔭 is prime to A ,D̄ and the ideal (6). By theorem 3.1, g(z_{0}) ∘ σ(𝔭) = [g ∘ (pB^{−1})](Bz_{0}), where B is the matrix that sends O_{K} to 𝔭.

Since 𝔭, A and D̄ are prime to each other, we can choose b such that 𝔭̄ = ⟨\frac{b+\sqrt{N}}{2} , p⟩, A = ⟨\frac{b+\sqrt{N}}{2} , a⟩ ,D̄ = ⟨\frac{b+\sqrt{N}}{2} , |D|⟩ and O_{K} = ⟨\frac{b+\sqrt{N}}{2} , 1⟩. Let z_{0} denote the point \frac{b+\sqrt{N}}{2} . Then 𝔭̄AD̄ = ⟨\frac{b+\sqrt{N}}{2} , pa|D|⟩, and with this basis B is given by

( \begin{matrix}1 & 0 \\ 0 & p \end{matrix} ). Now Bz_{0} = \frac{z_{0}}{p} and pB^{−1} = ( \begin{matrix}p & 0 \\ 0 & 1 \end{matrix} ) = S^{−1}BS.

Let g^{⋆}(z) = g ∘ S(z) = g(−1/z) = θ(0, −1/(a|D|z)Q_{B})/η(\frac{−1}{|D|z})η(\frac{−1}{z}). If in (19) we replace z by za|D| and Q_{B} by Adj (Q_{B}), we get the equation:

(21)  θ(0, Q_{B}(−1/a|D|z)) = (−i)\sqrt{|D|}azθ(0, Adj (Q_{B})az)

The eta function satisfies the functional equation η(−1/z) = \sqrt{z/i} η(z). Replacing z by |D|z and multiplying both equations:

η(−1/z)η(−1/(|D|z)) = \sqrt{|D|}\frac{z}{i} η(z)η(|D|z)

Hence we get:

g(−1/z) = a\frac{θ(0, Adj (Q_{B})az)}{η(z)η(|D|z)}

The q-expansion of this quotient has rational coefficients hence it is fixed by the action of σ_{p}, i.e. σ_{p} ∘ g^{⋆} = g^{⋆}. Then [g ∘ (pB^{−1})] = g and (g(z_{0}))^{σ_{p}} = g(z_{0}/p).

Proposition 3.2. with the notation as above, if 𝔭 is principal, g(z_{0})^{σ_{𝔭}} = g(z_{0}).

Proof. The proposition reduces to prove that g(z_{0}/p) = g(z_{0}) if 𝔭 is principal of norm p which follows from the next two lemmas. This completes the proof of theorem 3.2 since it implies that g(z_{0})^{σ_{𝔭}} = g(z_{0}) for all principal ideals 𝔭. □

Lemma 3.3. Let 𝔭 = ⟨μ⟩ be a principal ideal prime to A and D̄. Then the theta function Θ_{B} satisfies the formula:

Θ_{B} (\frac{b + \sqrt{N}}{2ap|D|} ) = μ̄εD̄(μ) (\frac{p}{|D|}) Θ_{B} (\frac{b + \sqrt{N}}{2a|D|} )

Note: since εD̄(μ)εD̄(μ̄) = (\frac{p}{|D|}), the formula may be written as Θ_{B}(\frac{b+\sqrt{N}}{2ap|D|}) =

ψD̄(μ̄)Θ_{B}(\frac{b+\sqrt{N}}{2a|D|})

Proof. Θ_{B} is a modular form of weight 1 for Γ_{0}(|D|) with a quadratic character. The number b was chosen such that 𝔭AD̄ = ⟨\frac{b+\sqrt{N}}{2} , pa|D|⟩ = ⟨μ\frac{b+\sqrt{N}}{2} , μa|D|⟩.

Hence there exists a change of basis matrix M = ( \begin{matrix}α & β \\ γ & δ \end{matrix} ) in Sl_{2}(ℤ) such that

( \begin{matrix}α & β \\ γ & δ \end{matrix} ) ( ap|D|^{\frac{b+\sqrt{N}}{2}} ) = ( \begin{matrix}μ\frac{b+\sqrt{N}}{2} \\ μa|D| \end{matrix} ).

If μ = \frac{m+n\sqrt{N}}{2} , an easy computation shows that δ = \frac{m−nb}{2p} and γ = n|D|a. In particular M is in Γ_{0}(|D|) and by modularity of Θ_{B} we have:

Θ_{B} (\frac{b + \sqrt{N}}{2a|D|} ) = Θ_{B} (M.\frac{b + \sqrt{N}}{2ap|D|} ) = (γ \frac{b + \sqrt{N}}{2ap|D|} + δ) χ(δ)Θ_{B} (\frac{b + \sqrt{N}}{2ap|D|} )

And the formula:

(22)  Θ_{B} (\frac{b + \sqrt{N}}{2a|D|} ) = \frac{μ}{p}χ(δ)Θ_{B} (\frac{b + \sqrt{N}}{2ap|D|} )

where χ(d) = (\frac{D}{q} ) for any prime q which is sufficiently large and satisfies q ≡ d mod |D|. ([Ogg] Theorem 20, Chapter VI, page 25). Let q be a prime congruent to 1 modulo 4 and congruent to δ modulo |D|. Then χ(δ) = (\frac{D}{q} ) = (\frac{|D|}{q}) =

(\frac{q}{|D|}) = ( \frac{\frac{m−nb}{2p}}{|D|} ) = ( \frac{\frac{m−nb}{2}}{|D|} ) (\frac{p}{|D|}). Then the proof follows from the definition of

εD̄ and the fact that \frac{μ}{p} = (μ̄)^{−1}. □

Lemma 3.4. With the same assumptions as in the previous lemma, the eta function satisfies the equation η(\frac{b+\sqrt{N}}{2p|D|})η(\frac{b+\sqrt{N}}{2p}) = μ̄εD̄(μ) (\frac{p}{|D|}) η(\frac{b+\sqrt{N}}{2|D|})η(\frac{b+\sqrt{N}}{2} ). In term of ideals:

(23)  η(𝔭D̄)η(𝔭) = μ̄εD̄(μ) (\frac{p}{|D|}) η(D̄)η(O_{K})

Proof. Since we choose |N| ≡ 3 mod 4, and |N| ≠ 3, the number of units in H is 2 (see [Ha-Vi] tables 3 and 4 of page 507). Given μ ∈ O_{K}, define:

κ(μ) = χ_{4}(Nμ)\frac{1}{μ̄} \frac{η^{2}(u)}{η^{2}(O_{K})}

Since the number of units in H is 2, κ is a quadratic character (see [Ha-Vi], Lemma 14). We can write the left hand side of (23) as:

(24)  η(𝔭D̄)η(𝔭) = (\frac{η(𝔭D̄)}{η(D̄)} \frac{η(O_{K})}{η(𝔭)}) \frac{η^{2}(𝔭)}{η^{2}(O_{K})}η(O_{K})η(D̄)

If μ is a generator of 𝔭, \frac{η^{2}(𝔭)}{η^{2}(O_{K})} = κ(μ)μ̄χ_{4}(p). By proposition 10 of [Ha-Vi]

(\frac{η(𝔭)}{η(O_{K})})^{σ_{D}} = (\frac{p}{|D|}) \frac{η(𝔭D̄)}{η(D̄)}. Then we get:

(\frac{η(𝔭D̄)}{η(D̄)} \frac{η(O_{K})}{η(𝔭)}) = (\frac{p}{|D|}) (\frac{η(𝔭)}{η(O_{K})})^{σ_{D}−1} = (\frac{p}{|D|}) (\sqrt{κ(μ)μ̄χ_{4}(p)})^{σ_{D}−1}

By lemma 12 of [Ha-Vi], κ(−1) = −1. Since the right term of (23) remains unchanged replacing μ by −μ, without loss of generality we can choose μ such that κ(μ) = χ_{4}(p). Replacing each term on the right hand side of (24) we get:

η(𝔭D̄)η(𝔭) = (\frac{p}{|D|}) ε_{D}(μ̄) μ̄ η(O_{K})η(D̄)

And the result follows since ε_{D}(μ̄) = εD̄(μ). □

Theorem 3.3. The number n_{A,B,D̄} is in the field M = HT. It corresponds to the fields diagram:

[figure: M; h; h; H; M^{+}; T; H^{+}; h; K; 2; T^{+}; h; ℚ]

Diagram 1

Proof. By theorem 3.2 the number θ(0, z_{AD̄}Q_{B})/η(zD̄)η(O_{K}) is in H and T contains the image of ψD̄(ClO_{K}) hence n_{A,B,D̄} is in M. □

Proposition 3.3. The quotient θ_{Q_{B}}(z_{AD̄})/ψD̄(Ā) depends only on the class of B and the class of A.

Proof. Independence of B is clear since Θ_{B} depends only in the class of B. To prove independence of A, let α ∈ O_{K} be an element with prime norm q such that q ∤ 6a|D|. By definition Θ_{B}(z_{αAD}) = Θ_{B}(\frac{b+\sqrt{N}}{2aq|D|}). Then by lemma 3.3:

Θ_{B} (\frac{b + \sqrt{N}}{2an|D|} ) = ψD̄(ᾱ)Θ_{B} (\frac{b + \sqrt{N}}{2a|D|} ) □

We will denote by n_{[A],[B],D̄} the number n_{A,B,D̄}.

Proposition 3.4. The number n_{[A],[B],D̄} is an algebraic integer.

Proof. In theorem 3.2 we proved that θ_{Q_{B}}(z_{AD̄})/η(zD̄)/η(z_{O_{K}}) is an algebraic integer and the number ψD̄(Ā) has norm NA. Since the quotient depends on the class of the ideal A but not A itself, using Tchebotarev density theorem we can choose two prime ideals 𝔭_{1} and 𝔭_{2} in the same class of A of prime norms p_{1} and p_{2}. Looking at 𝔭_{1} we see that the minimal polynomial of n_{[𝔭_{1}],[B],D̄} has rational coefficients with only 1 or p_{1} in the denominator. Considering 𝔭_{2} we see that the minimal polynomial of n_{[𝔭_{2}],[B],D̄} only has 1 or p_{2} in the denominator. Since n_{[𝔭_{1}],[B],D̄} = n_{[𝔭_{1}],[B],D̄} its minimal polynomial must have integer coefficients.□

Proposition 3.5. n_{[A],[B̄],D̄} = n_{[A],[B],D̄}

Proof. It is easy to check that the theta function Θ_{B} associated to B is the same as the theta function Θ_{Adj B} associated to the adjoint matrix of B. Clearly the point z_{AD̄} and the number ψD̄(A) are independent of B. □

Lemma 3.5. The character ψD̄ satisfy: \overline{ψ_{D}(Ā)} = ψD̄(A)

Proof. \overline{ψ_{D}(Ā)}ψ_{D}(Ā) = NA, and NA = ψ_{D}(Ā)ψ_{D}(A)ε_{D}(NA) hence \overline{ψ_{D}(Ā)} = (\frac{NA}{|D|}) ψ_{D}(A). By (2), ψD̄(A) = ψ_{D}(A)εD̄(A^{h})ε_{D}(A^{h}) = ψ_{D}(A) (\frac{(NA)^{h}}{|D|} ). Since |N| is prime, h is odd. □

Proposition 3.6. \overline{n_{[A],[B],D̄}} = n_{[Ā],[B],D}

Proof. It is clear from their definition that \overline{Θ_{B}(z_{AD̄})} = Θ_{B}(−\overline{z_{AD̄}}) and \overline{η(z_{AD̄})} = η(−\overline{z_{AD̄}}) . Since −\overline{z_{AD̄}} = z_{ĀD} and \overline{ψ_{D}(Ā)} = ψD̄(A), the result follows. □.

Proposition 3.7. If the ideal D is principal in O_{K}, \overline{n_{[A],[B],D̄}} = n_{[Ā],[B],D̄}

Proof. The proof of this proposition involves the same kind of techniques used on the previous ones (a little more tedious) so we omit the proof. □

In particular this implies that if A and D are both principal then the number n_{[A],[B],D̄} lives in a subfield of M which we denote M^{+} (following [Bu-Gr] notation, see page 13) and corresponds to the previous field diagram (see theorem 3.3 Diagram 1). We will be need the next lemmas for the theorem relating the numbers n_{[A],[B],D̄} for different ideals D.

Lemma 3.6. Let D and D′ be two prime ideals of ℚ(\sqrt{N}) with norm |D| and |D′| respectively, and let μ ∈ ℚ(\sqrt{N}) be such that μD = D′. Then \frac{η^{2}(AD′)}{η^{2}(AD)} = μ̄κ(μ)χ_{4}(Nμ).

Proof. Note that although κ is defined on integer elements, since it is a character on (O_{K}/12O_{K})^{×}, we can extend it multiplicatively to all elements in ℚ(\sqrt{N}) with both numerator and denominator prime to 12. By definition κ(μ) = \frac{1}{μ̄}χ_{4}(Nμ)\frac{η^{2}(μ)}{η^{2}(O_{K})}

then we are led to prove that \frac{η^{2}(AD′)}{η^{2}(AD)} \frac{η^{2}(O_{K})}{η^{2}(μ)} = 1.

By Proposition 10 of [Ha-Vi] we can write the left hand side as (\frac{η^{2}(AD)}{η^{2}(O_{K})})^{σ_{(D̄′D̄^{−1})}−1}.

Since \frac{η^{2}(AD)}{η^{2}(O_{K})} is in H (by theorem 20 of [Ha-Vi]) then σ_{A} represents the classical Artin map from Cl(O_{K}) to Gal(H/K), and since D̄′D̄^{−1} is principal, σD̄′D̄^{−1} is the identity. □

Lemma 3.7. Let D and D′ be two prime ideals of ℚ(\sqrt{N}) such that D ∼ D′. Then \frac{η(AD′)η(D)}{η(AD)η(D′)} = ε_{D}(Ā^{h})ε_{D′}(Ā^{h})

Proof. By proposition 10 of [Ha-Vi] we have:

(25)  \frac{η(AD′)η(D)}{η(D′)η(AD)} = (\frac{η(A)}{η(O_{K})})^{σD̄′} (\frac{η(A)}{η(O_{K})})^{−σD̄} (\frac{a}{|D|}) (\frac{a}{|D′|})

Since the Artin-Frobenius map is an homomorphism:

(\frac{η(A)}{η(O_{K})})^{σD̄′ −σD̄} = ((\frac{η(A)}{η(O_{K})})^{σ_{(D̄′(D̄)^{−1})}−1})^{σD̄}

But (\frac{η(A)}{η(O_{K})})^{σ_{(D̄′(D̄)^{−1})}−1} = ±1 (see the proof of lemma 3.6), then σD̄ acts trivially

on it. Let μ ∈ ℚ(\sqrt{N}) be such that D̄′D̄^{−1} = \frac{μ̄}{|D|} then by theorem 19 of [Ha-Vi]:

(\frac{η(A)}{η(O_{K})})^{σ_{(D̄′(D̄)^{−1})}−1} = κ (\frac{μ}{|D|})^{\frac{a−1}{2}} (\frac{μ̄|D|}{Ā})

Since |D| is prime to 12, and κ is a multiplicative quadratic character, κ(\frac{μ}{|D|}) = κ(μ)κ(|D|). The character κ defined on (O_{K}/12O_{K})^{×} factors as a product of two characters, κ_{3} from (O_{K}/3O_{K})^{×} to the group of third roots of unity and κ_{4} from (O_{K}/4O_{K})^{×} to the group of fourth roots of unity (see lemma 14 of [Ha-Vi]). In our case κ_{3} = 1 and the character is completely determined from the congruence modulo 4. Then κ(|D|) = κ(−1) = −1. Using the quadratic reciprocity law,

(26)  (\frac{η(A)}{η(O_{K})})^{σ_{(D̄′(D̄)^{−1})}−1} = κ(μ)^{\frac{a−1}{2}} (\frac{μ̄}{Ā}) (\frac{a}{|D|})

Also since κ(μ)κ(μ̄) = κ(|D||D′|) = 1, κ(μ) = κ(μ̄) and we can write (25) as:

\frac{η(AD′)η(D)}{η(AD)η(D′)} = κ(μ̄)^{\frac{a−1}{2}} (\frac{μ}{A}) (\frac{a}{|D′|})

D̄D′ = μ and ε is a multiplicative quadratic character, then:

(27)  ε_{D}(Ā^{h})ε_{D′}(Ā^{h}) = ε_{D}(Ā^{h})εD̄(Ā^{h})ε_{D̄D′}(Ā^{h}) = (\frac{a}{|D|}) (\frac{Ā^{h}}{μ} )

Using the reciprocity law in ℚ(\sqrt{N}) (see for example theorem 21 of [Ha-Vi]):

(28) (\frac{Ā^{h}}{μ} ) = (\frac{μ}{Ā^{h}} ) κ(μ̄)^{\frac{a−1}{2}} = (\frac{μ}{Ā}) κ(μ̄)^{\frac{a−1}{2}} = κ(μ̄)^{\frac{a−1}{2}} (\frac{μ}{A}) (\frac{|D||D′|}{a})

And the lemma follows from (\frac{|D||D′|}{a}) = (\frac{a}{|D|}) (\frac{a}{|D′|}). □

Theorem 3.4. Let z_{AD}Q_{B} and z_{AD′}Q_{B′} be two points in 𝔥_{2} such that they are equivalent modulo Γ_{12} and D ∼ D′ in ℚ(\sqrt{N}). Then n_{[A],[B],D̄} = ±n_{[A],[B′],D̄′}

Proof. By proposition 3.6 and proposition 3.7 we may restrict to the case D′ ≠ D̄. For simplicity we will denote Ω_{D} := z_{AD}Q_{B} and Ω_{D′} := z_{AD′}Q_{B′}. Since Ω_{D}

is equivalent to Ω_{D′} there exists a matrix γ = ( \begin{matrix}A & B \\ C & D \end{matrix} ) in Sp_{4}(ℤ) such that

γ ⋆ (Ω_{D}) = Ω_{D′}. We can think points in the Siegel space like lattices with an extra data (see Lemma 4.1 of [Mu]), where the action is given as multiplication

on the left by 1/γ^{t}. Then γ ⋆ (\frac{I_{2}}{−Ω_{D}} ) = ( \begin{matrix}D & −C \\ −B & A \end{matrix} ) (\frac{I_{2}}{−Ω_{D}} ) = (\frac{CΩ_{D}+D}{−(AΩ_{D}+B)}) =

A FORMULA FOR THE CENTRAL VALUE OF CERTAIN HECKE L-FUNCTIONS 15

(\frac{I_{2}}{−Ω_{D′}} ) (CΩ_{D} + D)^{−1} By the coherent way we chose characters, \frac{ψ_{D}(A)}{ψ_{D′} (A)} = ε_{D}(A^{h})ε_{D′}(A^{h}). Hence:

\frac{n_{[A],[B],D̄}}{n_{[A],[B′],D̄′}} = \frac{θ(Ω_{D})}{θ(Ω_{D′})} \frac{η(D′)}{η(D)}ε_{D}(Ā^{h})ε_{D′}(Ā^{h}) = \frac{θ(Ω_{D})}{θ(Ω_{D′})} \frac{η(AD′)}{η(AD)}

The last equality follows from lemma 3.7. We claim that:

(29)  \frac{θ^{2}(Ω_{D})}{θ^{2}(Ω_{D′})} = Det(CΩ + D)^{−1} = \frac{η^{2}(AD)}{η^{2}(AD′)}

The first equality follows at once from the functional equation of the theta function. Since |D| is prime and Det(Q) = |D| there exists matrices U, V ∈ Sl_{2}(ℤ) such that UQV = ( \begin{matrix}1 & 0 \\ 0 & |D| \end{matrix} ) (respectively U′ and V′ for Q′). Then:

( \begin{matrix}V^{−1} & 0 \\ 0 & U \end{matrix} ) ( \begin{matrix}I_{2} \\ −Ω_{D} \end{matrix} ) V = ( \begin{matrix}I_{2} \\ −UQV z_{AD} \end{matrix} ) = ( \begin{matrix}1 & 0 \\ 0 & 1 \\ −z_{AD} & 0 \\ 0 & −z_{A} \end{matrix} )

Similarly:

( \begin{matrix}V^{′−1} & 0 \\ 0 & U′ \end{matrix} ) ( \begin{matrix}I_{2} \\ −Ω_{D′} \end{matrix} ) V′ = ( \begin{matrix}I_{2} \\ −U′Q′V′z_{AD′} \end{matrix} ) = ( \begin{matrix}1 & 0 \\ 0 & 1 \\ −z_{AD′} & 0 \\ 0 & −z_{A} \end{matrix} )

Since D′ ≠ D̄, we may choose basis D = ⟨|D|, \frac{b+\sqrt{N}}{2} ⟩, D′ = ⟨|D′|, \frac{b+\sqrt{N}}{2} ⟩ and A = ⟨a, \frac{b+\sqrt{N}}{2} ⟩. If μ is such that μD = D′, then AD′ = ⟨a|D′|, \frac{b+\sqrt{N}}{2} ⟩ = ⟨μa|D|, μ(\frac{b+\sqrt{N}}{2} )⟩ = μAD hence there exists a matrix M = ( \begin{matrix}α & β \\ γ & δ \end{matrix} ) in Sl_{2}(ℤ)

such that: M ( \begin{matrix}μ(\frac{b+\sqrt{N}}{2} ) \\ μa|D| \end{matrix} ) = ( a|D′|^{\frac{b+\sqrt{N}}{2}} )

Let N := ( \begin{matrix}δ & 0 & −γ & 0 \\ 0 & 1 & 0 & 0 \\ −β & 0 & α & 0 \\ 0 & 0 & 0 & 1 \end{matrix} ), then it is clear that:

N ( \begin{matrix}1 & 0 \\ 0 & 1 \\ −z_{AD} & 0 \\ 0 & −z_{A} \end{matrix} ) ( \begin{matrix}\frac{μ|D|}{|D′|} & 0 \\ 0 & 1 \end{matrix} ) = ( \begin{matrix}1 & 0 \\ 0 & 1 \\ −z_{AD′} & 0 \\ 0 & −z_{A} \end{matrix} )

Combining these results we get that:

( \begin{matrix}V^{′−1} & 0 \\ 0 & U′ \end{matrix} ) N ( \begin{matrix}V^{−1} & 0 \\ 0 & U \end{matrix} ) ( \begin{matrix}I_{2} \\ −Ω_{D} \end{matrix} ) ( \begin{matrix}\frac{μ|D|}{|D′|} & 0 \\ 0 & 1 \end{matrix} ) V V^{′−1} = ( \begin{matrix}I_{2} \\ −Ω_{D′} \end{matrix} )

and

( \begin{matrix}I_{2} \\ −Ω_{D′} \end{matrix} ) = ( \begin{matrix}D & −C \\ −B & A \end{matrix} ) ( \begin{matrix}I_{2} \\ −Ω_{D} \end{matrix} ) (CΩ_{D} + D)^{−1}

Since both lattices have the same volume then | Det(CΩ_{D} + D)|^{−1} = \frac{|μ||D|}{|D′|}.

By lemma 3.6, \frac{η^{2}(AD)}{η^{2}(AD′)} = \frac{1}{μ̄}κ(μ) = \frac{μ|D|}{|D′|}κ(μ). Now Det(CΩ_{D} + D)^{−1} and κ(μ)\frac{μ|D|}{|D′|} have the same absolute value and both lie in ℚ(\sqrt{N}) hence they differ by ±1. Then

(\frac{θ(Ω_{D})}{θ(Ω_{D′})} \frac{η(AD′)}{η(AD)})^{2} = Det(CΩ_{D} + D)^{−1}μ̄κ(μ) = ±1

Taking square roots:

\sqrt{±1} = \frac{θ(Ω_{D})}{θ(Ω_{D′})} \frac{η(AD′)}{η(AD)}

By theorem 3.3 we know that \frac{θ(Ω_{D})}{η(D)η(O_{K})} and \frac{θ(Ω′_{D})}{η(D′)η(O_{K})} are in H. Since \sqrt{−1} ∉ H the theorem follows. □

It is not clear how to determine the sign a priori, and we are not able to give any answer in this direction.

4. Equivalence of special points

The problem of determining whether two points in 𝔥_{2} are equivalent or not is complicated in general. For our case we will get this equivalence via ideals in quaternion algebras. A good reference for the basic definitions and some elementary facts about quaternion algebras is Pizer’s paper ([Pi]).

Proposition 4.1. Let B be a quaternion algebra over ℚ ramified at p_{1}, …, p_{n} and L be an ideal in B. Then O_{l}(L) is a maximal order if and only if disc(L)= (p_{1} … p_{n})^{2}N(L)^{4}

Proof. By definition disc(L) is the determinant of the bilinear associated to L on any basis. Since L is locally principal at all primes, given a finite prime q, L_{q} = O_{l}(L)_{q}α_{q}. Clearly disc(L_{q}) = N(α_{q})^{4} disc(O_{q}); then the statement follows from the fact that this proposition is true replacing L by an order O and N(L) by 1 (see [Pi] Proposition 1.1, page 344), and the fact that the norm of L is the product over all primes q of q^{v_{q}(Nα_{q})} where v_{q}(n) is the q-valuation. □

Lemma 4.1. Let O be a maximal order, {I_{1}, …, I_{h}} a set of left O-ideals representatives, and {R_{1}, …, R_{h}} be the right orders of {I_{1}, …, I_{h}} respectively. Then for a given i = 1, …, h the maximal order R_{i} appears twice on the list if and only if there is no embedding of ℤ[\sqrt{N}] into R_{i}.

Proof. Let P be the bilateral O-ideal of norm |N| . For a given left O-ideal I_{j}, the ideal PI_{j} is another left O-ideal. Note that if P_{j} is the ideal of norm |N| in R_{j}, then I_{j}^{−1}PI_{j} = P_{j} by the uniqueness of such a bilateral ideal. Then the ideals I_{j} and PI_{j} are equivalent if and only if there exists β ∈ R_{j}^{×} such that I_{j}β = PI_{j}. Multiplying on the left by I_{j}^{−1} we see that R_{j}β = i_{j}^{−1}PI_{j} = P_{j} hence P_{j} is principal, and the element β has norm |N|.

To see that this is the only way in which a maximal order R appears twice in the list of right orders, suppose that I and J are two nonequivalent left O-ideals with same right order R. Then I^{−1}J is a non-principal bilateral ideal for R. Let P_{R} be the ideal of norm |N| in R, then P_{R} is non-principal and J is equivalent to PI. □

4.1. Siegel Space and applications. Let A := ( \begin{matrix}0 & I_{n} \\ −I_{n} & 0 \end{matrix} ). A point in the

Siegel space 𝔥_{n} is equivalent to give a lattice L over ℤ of rank 2n together with: • A basis V_{0} of L.

• U ∈ ℝ^{2n×2n} such that U^{2} = −I_{2n}. (complex structure)

• A positive definite Hermitian form H for U on the vector space L⊗ℝ, with

the imaginary part of H having the matrix A on the basis V_{0}. See Lemma 4.1 of [Mu] for a proof. Clearly two points in the Siegel space will be equivalent if there is a map sending one basis to the other one, and preserving the Hermitian form.

Definition. Let L be a ℤ lattice of rank 2n and V the vector space L ⊗ ℝ. We call a triple (P, J, U) a Siegel point if:

• P is a positive definite real symmetric 2n × 2n form (that will correspond to the real part of H ).

• J is a real 2n × 2n non-degenerate skew symmetric matrix (that will correspond to the imaginary part of H).

• U ∈ ℝ^{2n×2n} is such that U^{2} = −I_{2g}. (complex structure) With the relation:

(30)  −JU = U^{t}J = P

Via the matrix U we can put a complex structure on the vector space V. Let H be the bilinear form H(x, y) := P (x, y) + iJ(x, y). The condition (30) implies that H(ix, y) = iH(x, y). Since J is skew symmetric and P symmetric, it follows that H(x, y) = \overline{H(y, x)}. Then H defined in this way is a positive definite Hermitian form. Each choice of a reduced basis for J will give a point in the Siegel space and different basis give equivalent points.

Given two lattices L and L′, a morphism γ : L → L′ is an ℤ-linear map from L to L′. Given γ : L′ → L an isomorphism of lattices, we define an action of γ on a Siegel point (P, J, U) as (γ^{∗}P, γ^{∗}J, γ^{∗}U) where given x, y ∈ L′, γ^{∗}P (x, y) = P (γ(x), γ(y)), γ^{∗}J(x, y) = J(γ(x), γ(y)) and γ^{∗}(x) = γ^{−1}(U(γ(x))). If we choose V_{0} to be a skew symmetric reduced base for J, i.e. a base where J is

of the form ( \begin{matrix}0 & I_{n} \\ −I_{n} & 0 \end{matrix} ), and γ is an automorphism sending a skew symmetric

reduced basis to another one, then γ ∈ Sp_{2n}(ℤ) and the action of γ on the Siegel point Ω associated to V_{0} is the usual action of Sp_{2n}(ℤ) on 𝔥_{n}.

4.2. Siegel Points from Quaternion algebras. Let N be the negative of prime congruent to 31 modulo 4, and B = (−1, N) the quaternion algebra ramified at N and infinity. Let O be a maximal order in B such that there exists an embedding (not necessarily optimal) of ℤ + ℤ\sqrt{N} into O. Let u ∈ O be the image of \sqrt{N}, i.e. u^{2} = N and T r(u) = 0. To a left O-ideal I we associate a Siegel point (P, J, U) as follows:

• We take V the real vector space V := B ⊗_{ℚ} ℝ.

• Define U acting on V as left multiplication by \frac{u}{\sqrt{|N|}}.

• We think of I as a full rank lattice in V.

• For x, y ∈ I define P (x, y) := \frac{1}{\sqrt{|N|}}T r(xȳ)/N(I).

• For x, y ∈ I define J(x, y) := T r(u^{−1}xȳ)/N(I).

Proposition 4.2. The triple (P, J, U) defined as before is a Siegel point.

Proof. We start checking the properties of the matrices P, J and U:

• P is a real form. Since T r(xȳ) is real, T r(xȳ) = T r(yx̄) which implies that P (x, y) is symmetric. Clearly P (x, x) = \frac{1}{\sqrt{|N|}}N(x)/N(I) is positive definite.

• J is a real form. Since u is pure imaginary, u^{−1} is also. Then J(x, x) = T r(u^{−1}N(x))/N(I) = 0. It is also clear that J(x, y) is non-degenerate, since for any non-zero x ∈ V, J(x, u^{−1}x) ≠ 0. So we are left to prove that J(x, y) = −J(y, x). We have the trace identity:

(31)  T r(xy) = T r(yx)

By definition, N(I)J(y, x) = T r(u^{−1}yx̄) = T r(xȳū^{−1}). Since u is pure imaginary, ū = −u. Then T r(xȳū^{−1}) = −T r(xȳu^{−1}) = −T r(u^{−1}xȳ) by (31) and J(y, x) = −J(x, y).

• Let x ∈ V, then U^{2}(x) = U(\frac{u}{\sqrt{|N|}}x) = \frac{u^{2}}{|N|}x = −x.

As for the relation, it is easy to check that J(\frac{u}{\sqrt{|N|}}x, y) = P (x, y) and that

J(x, \frac{u}{\sqrt{|N|}}y) = −P (x, y). □

Definition. Given a lattice I in B we define its dual by I^{#} := {b ∈ B : T r(bI) ⊂ ℤ}. Given an order R we define its different by R^{ι} := NR^{#}.

Proposition 4.3. If O is a maximal order, O^{ι} is a bilateral ideal for O of index N^{2}, and \frac{1}{N} O ⊂ O^{ι} ⊂ O.

Proof. See [Vig] Lemma 4.7, page 24.

Proposition 4.4. If x, y ∈ I then J(x, y) ∈ ℤ. Also the matrix of J on the basis given by I has determinant 1.

Proof. Since we are considering the reduced norm, if V is the matrix associated to multiplication (on the left or on the right) by v, then N(v) = \sqrt{det(V)}. Let W (x, y) := T r(xȳ) be the bilinear form of B. If we denote W the matrix of W (x, y) on the basis given by I, J = \frac{1}{N(I)}(U^{−1})^{t}W. Then det(J) = N(I)^{−4}N(u)^{−2} det(W). By definition det(W) = disc(I), which is an ideal for a maximal order, then by Proposition 4.1 disc(I) = N^{2}N(I)^{4} and det(J) = 1. Since the trace is linear, J(x, y) = T r(u^{−1}x\frac{ȳ}{N(I)}). For ideals I with maximal left order it is true that I^{−1} = Ī/N(I) and II^{−1} = O, hence J(x, y) ∈ ℤ for all x, y ∈ I if and only if T r(u^{−1}v) ∈ ℤ for all v ∈ O. By proposition 4.3 this is the same as u^{−1} ∈ O^{#}. But u^{−1} = −\frac{u}{N}, and since u ∈ O it follows that \frac{u}{N} ∈ \frac{1}{N} O ⊂ O^{#}. □

This gives a method for assigning to every left O-ideal a Siegel point. Note that choosing different skew symmetric reduced basis of I will give equivalent Siegel points. From now on we fixed a maximal order O with an embedding of ℤ[\sqrt{N}].

Proposition 4.5. Let u ∈ O with N(u) = |N| and T r(u) = 0, and denote by U the complex multiplication associated to u. If I, I′ are two equivalent left O-ideals, then the Siegel points (P, J, U)_{I} and (P, J, U)_{I′} are equivalent.

Proof. Since I ∼ I′ there exists α ∈ B^{×} such that I = I′α. Let W denote the isomorphism of B given by W (v) = vα. We claim that W is the isomorphism that makes the two Siegel points equivalent. Since W (I′) = I, we need to check that W^{∗}P = P′, W^{∗}J = J′ and W^{∗}U = U.

• If x, y ∈ I′ by definition (W^{∗}P)(x, y) := P (W (x), W (y)) = P (xα, yα) = \frac{T r(xαᾱȳ)}{N(I)} = \frac{Nα}{N(I)}T r(xȳ) = P′(x, y).

• The equality W^{∗}J = J′ follows from a similar argument.

• By definition U is given by multiplying on the left by u/\sqrt{|N|} while W is given by multiplying on the right by α then clearly this maps commute with each other and W^{∗}U := W^{−1} ∘ U ∘ W = U. □

Lemma 4.2. Let U be the complex multiplication given by u and α ∈ B an element such that αOα^{−1} = O. Define I′ = αIα^{−1} and u′ = αuα^{−1}, then (P, J, U) ∼ (P′, J′, U′).

Proof. Let W : B → B be the isomorphism defined by W (x) = αxα^{−1}. By hypothesis W (R) = R, W (I) = I′. It is easy to see that W^{∗}P = P′ and W^{∗}J = J′. If x ∈ B then W^{−1} ∘ U ∘ W (x) = W^{−1} ∘ U(αxα^{−1}) = W^{−1}(uαxα^{−1})/\sqrt{|N|} = α^{−1}uαx/\sqrt{|N|} = U′(x). □

This lemma suggests that we should consider not just elements u in O corresponding to \sqrt{N} (i.e. u^{2} = N and T r(u) = 0) but modulo conjugation by the normalizer of O. It is clear that N (O) = {h ∈ B | Oh is bilateral}. All bilateral ideals are principal, generated by u^{s}m where s = 0, 1 and m is a rational number (see [Ei2] proposition 1, page 92). The generator of an ideal is well defined up to units in O, then N (O) = {ζu^{s}m | s = 0 or 1 , m ∈ ℚ and ζ ∈ O is a unit}

Corollary. If I and I′ are left O-ideals with the same right order then the Siegel points (P, J, U)_{I} and (P, J, U)_{I′} are equivalent.

Proof. If I and I′ are equivalent this follows from proposition 4.5. If I and I′ are not equivalent, we know by lemma 4.1 that O_{r}(I) has no embedding of ℤ[\sqrt{N}]. Let u be the element in O giving the complex multiplication. Then uI has the same left and right order as I but they are not equivalent, hence uI ∼ I′ ∼ uIu^{−1}. By proposition 4.2 the Siegel points (P, J, U)_{I} and (P, J, U′)_{uI} are equivalent. Just note that U′ is given by u^{−1}uu = u. □

In particular we should index the Siegel points not by the class number of ideals, but by the type number of maximal orders. We still have equivalent Siegel points coming from conjugation by units of O and this are all the possibilities for N (O). For counting equivalent classes of Siegel points, fixed a maximal order O we have to count the number of embeddings of ℤ[\sqrt{N}] into O modulo conjugation by units of O.

Given a maximal ideal O, let V := {I_{1}, …, I_{h}} be a set of left O-ideal representatives and T := {R_{1}, …, R_{t}} the distinct right orders of the ideals in V. We index the Siegel points by pairs (ϕ, R_{i}) where ϕ is an embedding from ℤ[\sqrt{N}] to some R_{j} and R_{i} is an order in T. By this we mean the Siegel point obtained with the complex multiplication given by ϕ(\sqrt{N}), and an ideal I with left order R_{j} and right order R_{i}.

If d is a negative discriminant we denote by h(d) the class number of binary quadratic forms of discriminant d. Let u(d) = 1 unless d = −3, −4 when u(d) = 3, 2 respectively (half the number of units in the ring of integers of discriminant d). For D > 0 we define the Hurwitz’s class number H(D) by:

(32)  H(D) := ∑_{df^{2}=−D} \frac{h(d)}{u(d)}

Given D > 0 let L = ℚ[\sqrt{−D}] and O its ring of integers. Define H_{N}(D) be the modified invariant by:

(33) H_{N}(D) = \{ \begin{matrix}0 & if N splits in O \\ H(D) & if N is inert in O \\ \frac{1}{2}H(D) & if N is ramified in O but does not divide \\ & the conductor of O \\ H_{N}(D/N^{2}) & if N divides the conductor of O \end{matrix}

Then the number of embeddings of O into R_{i} (i = 1, …, n) modulo conjugation by R_{i}^{×}/{±1} is H_{N}(D) (see [Gr2] the proof of Proposition 1.9, page 122). In our case, N the negative of a prime and D = −4N, we get :

(34)  H_{N}(4N) = \{ \begin{matrix}\frac{1}{2}h(4N) & if N ≡ 1 mod 4 \\ h(N) & if N ≡ 7 mod 8 \\ 2h(N) & if N ≡ 3 mod 8 and N ≥ 11 \end{matrix}

Note that in the case D = 4N an order R_{i} on T appears twice as a right order if and only if it has no embedding of O_{4N}. In this case it does not contribute to the sum, and hence the number of embeddings of ℤ[\sqrt{N}] into the t orders in T is also H_{N}(4N). With this we proved:

Proposition 4.6. The number of non-equivalent Siegel points constructed is at most H_{N}(4N)t.

Proposition 4.7. Let B be a quaternion algebra over a commutative field K, and let B_{0} := {β ∈ B | T r(β) = 0}. If ψ : B_{0} → B_{0} is an isommetry of Kvector spaces then there exists an element β ∈ B^{⋆} such that σ(x) = βxβ^{−1} or σ(x) = −βxβ^{−1} = βx̄β^{−1}.

Proof. See [Vig] Theorem 3.3, page 12 □

Lemma 4.3. Let ψ : B → B be an isomorphism of ℚ-vector spaces (respectively σ : B_{q} → B_{q} an isomorphism of ℚ_{q}-vector spaces) such that σ(1) = 1 and σ is an isommetry. Then there exists an α ∈ B^{⋆} (respectively α ∈ B_{q}^{⋆}) such that σ(x) = αxα^{−1} or σ(x) = αx̄α^{−1}.

Proof. Since σ(1) = 1 and σ is a morphism, σ(ℚ) = ℚ. Denoting B_{0} the trace zero elements, σ(B_{0}) = B_{0} and σ|_{B_{0}} : B_{0} → B_{0} is an isommetry. By proposition 4.7 we get two different cases:

(1) σ_{B_{0}}(x) = αx̄α^{−1} for some α ∈ B^{⋆}. Then σ is the antiautomorphism given

by σ(x) = αx̄α^{−1}.

(2) σ_{B_{0}}(x) = αxα^{−1} for some α ∈ B^{⋆}. Then σ is an automorphism given by

σ(x) = αxα^{−1}. □

Theorem 4.1. The H_{N}(4N)t Siegel points {(ϕ, R_{i})} constructed above are nonequivalent.

Proof. The proof breaks in two steps. First we will prove that for a fixed embedding of ℤ[\sqrt{N}] into R (say u is the image of \sqrt{N}), the t left R-ideals give non-equivalent points (P, J, U) where U is multiplication by u/\sqrt{|N|}. Then we will prove that different embeddings give non-equivalent Siegel points.

Let I_{1}, I_{2} two left R-ideals. Abusing notation we will denote P_{i} the symmetric form P_{I_{i}} and analogously for J_{i}. Suppose there exists W : V → V an isomorphism making the Siegel points (P_{1}, J_{1}, U) and (P_{2}, J_{2}, U) equivalent. Let β = W (1), σ the map σ(v) = W (v)β^{−1} and V_{0} the space of elements in V with trace zero. We claim that σ is an isommetry. By hypothesis W^{∗}P_{1} = P_{2} then evaluating at (1, 1) we have

(W^{∗}P_{1})(1, 1) = P_{2}(1, 1) = \frac{2}{N(I_{2})}

By definition, (W^{∗}P_{1})(1, 1) = \frac{T r(W (1),\overline{W (1)})}{N(I_{1})} = 2\frac{N(β)}{N(I_{1})} hence N(β) = \frac{N(I_{1})}{N(I_{2})}. Then

∥x∥/\sqrt{N} = P_{2}(x, x)N(I_{2})/2 = W^{∗}(P_{1}(x, x))N(I_{2})/2 = \frac{∥W (x)∥}{N(I_{1})}N(I_{2}) = \frac{∥W (x)∥}{∥β∥} = ∥σ(x)∥/\sqrt{N}, i.e. σ is an isommetry. Since σ is an isommetry and σ(1) = 1, by lemma 4.3 we have two different cases:

(1) σ(x) = αx̄α^{−1} for some α ∈ B^{×}, i.e. σ is an antiautomorphism and

W (x) = αx̄α^{−1}β^{−1}.
(2) σ(x) = αxα^{−1} for some α ∈ B^{×} and W (x) = αxα^{−1}β^{−1}.

We know that W preserves the complex multiplication, i.e. W^{−1}∘U ∘W (x) = U(x). In the first case, W^{−1}(x) = α^{−1}β̄x̄α. Then W^{∗}U(x) = W^{−1}(uαx̄α^{−1}β^{−1}) = α^{−1}β̄β̄^{−1}ᾱ^{−1}xᾱūα = xα^{−1}ūα. It must be the case that ux = xα^{−1}ūα for all x ∈ B (which is the same as saying that uxα^{−1} = xα^{−1}ū) which would imply that u ∈ ℚ and is not the case. Then we must be in the second case.

Since W (I_{1}) = I_{2}, I_{2} = αI_{1}α^{−1}β^{−1}. In particular αRα^{−1} = R, i.e. α ∈ N (R). Then I_{1} and I_{2} have the same right order and represent the same class between the t left R-ideals we started with.

Assume that there is a left R-ideal I and a left R′-ideal I′ such that R and R′ are non-conjugate maximal orders and the Siegel points (P, J, U) and (P′, J′, U′) are equivalent. Then there exist an isomorphism W : V → V that sends one point to the other. Arguing as before we get the same two possible cases for W. In the first case, since W^{∗}U = U′ we would get that u′xα^{−1} = xα^{−1}ū for all x ∈ V. Taking x = α we would get that u′ = ū and it commutes with all elements of V, then it is rational which is not the case. Then W (x) = αxα^{−1}β^{−1} and I′ = αIα^{−1}β. In particular the orders R and R′ are conjugate which is a contradiction. □

4.3. Ideals associated to Siegel points. For finding relations between the numbers n_{[A],[B],D̄},we will assign a left O-ideal I in B (for some maximal order O) and an embedding of ℤ[\sqrt{N}] into O to each point z_{AD̄}Q_{B} on the Siegel space 𝔥_{2} such that the Siegel point (P, J, U) is z_{AD̄}Q_{B} (in the given basis). This will imply that there are at most H_{N}(4N)t different values (up to a sign) for n_{[A],[B],D̄}.

Proposition 4.8. There exists u and v in B such that:

• T r(uv̄) = 0, T r(u) = 0 and T r(v) = 0

• N(u) = |N|

• N(v) = |D|

• u and v are in a maximal order R of B

Proof. Since |N| ≡ 3 mod 4, we can assume B = (−1, N). Choosing u = j it is clear that T r(u) = 0 and N(u) = |N|, hence we are looking for v in B such that T r(uv) = 0 and T r(v) = 0 and N(v) = |D|. This conditions forces v to have the form v = xi + yk and we are looking for an integer solution of the quadratic equation:

(35)  x^{2} + |N|y^{2} − |D|z^{2} = 0

We can assume that the solution is primitive (i.e. gcd(x, y, z) = 1). If (x, y, z) is a solution, clearly gcd(z, N) = 1 = gcd(x, N) and gcd(x, D) = 1 = gcd(y, D). To prove the existence of such a solution we use the Hasse-Minkowski principle. Clearly (35) has a non-zero solution over ℝ, so we need to prove the existence of local non-zero solutions for all primes. We consider the different cases:

• For a prime p ≠ N and p ≠ D the quadratic form clearly has a local solution (see [Se] corollary 2, page 6).

• For the prime |N| by Hensel’s Lemma it is enough to look for solutions of (35) modulo |N|:

x^{2} − |D|z^{2} ≡ 0 mod |N| iff (xz^{−1})^{2} ≡ |D| mod |N|

This equation has solution if and only if (\frac{|D|}{|N|}) = 1. By quadratic reciprocity law and the fact that |N| ≡ 3 mod 4 this last condition is equivalent to ask that |D| splits in ℚ(\sqrt{N}) which is the case.

• For the prime |D|, looking at (35) modulo |D|:

x^{2} + |N|y^{2} ≡ 0 mod |D| iff N ≡ (xy^{−1})^{2} mod |D| iff (\frac{N}{|D|}) = 1

Which is the case since |D| splits in ℚ(\sqrt{N}). Given u and v as before, consider the rank 4 lattice R = ⟨1, u, v, uv⟩. It is easy to see that R is actually an order, hence contained in a maximal one. □ Remark: if we define R = ⟨1, \frac{1+j}{2} , v, (\frac{1+j}{2} ) v⟩ it is easy to see that this is also an order. The advantage of this order is that it contains an embedding of the ring of integers of ℚ(\sqrt{N}), but is still not maximal.

Let z_{AD̄}Q_{B} = (\frac{b_{1}+\sqrt{N}}{2a_{1}|D|}) ( \begin{matrix}2a & b \\ b & 2c \end{matrix} ), u and v as in proposition 4.8 (choosing u = j).

Define

(36)  I_{z} := ⟨(\frac{b_{1} − j}{2a_{1}|D|}) av, (\frac{b_{1} − j}{2a_{1}|D|}) (\frac{|D| + bv}{2} ) , \frac{v − b}{2}, a⟩

If we denote ϕ the embedding of ℚ(\sqrt{N}) into B and ψ the embedding of ℚ(\sqrt{D}) into B with ϕ(\sqrt{N}) = u and ψ(\sqrt{D}) = v and choosing the basis B = ⟨v_{1}, v_{2}⟩ (where in our notation v_{1} = \frac{b+\sqrt{D}}{2} and v_{2} = a) then the ideal I_{z} was defined by:

I_{z} = ⟨ϕ (\frac{b_{1} − \sqrt{N}}{2a_{1}|D|} ) ψ(\sqrt{D})ψ(v̄_{2}), ϕ (\frac{b_{1} − \sqrt{N}}{2a_{1}|D|} ) ψ(\sqrt{D})ψ(v̄_{1}), ψ(v̄_{1}), ψ(v̄_{2})⟩

Which can also be written as I_{z} = ⟨ϕ (\frac{b_{1}−\sqrt{N}}{2a_{1}|D|} ) ψ(\sqrt{D})ψ(B̄), ψ(B̄)⟩.

Proposition 4.9. The element \frac{1+j}{2} is in the left order of I_{z}.

Proof. This is an easy but tedious computation. We will just give the coordinates of the product of \frac{1+j}{2} with each element of the basis of I_{z} (given above) as a linear combination.

• (\frac{1+j}{2} ) a = [ba_{1}, −2aa_{1}, 0, \frac{b_{1}+1}{2}].

• (\frac{1+j}{2} ) (\frac{v−b}{2}) = [−2ca_{1}, ba_{1}, \frac{b_{1}+1}{2}, 0].

• (\frac{1+j}{2} ) (\frac{b_{1}−j}{2a_{1}|D|}) av = [\frac{1−b_{1}}{2} , 0, 2ac_{1}, bc_{1}].

• (\frac{1+j}{2} ) (\frac{b_{1}−j}{2a_{1}|D|}) (\frac{|D|+bv}{2} ) = [0, \frac{1−b_{1}}{2} , bc_{1}, 2cc_{1}]. □

Proposition 4.10. The element a_{1}v is in the left order of I_{z}.

Proof. Since B is an ideal, it is clear that v⟨w_{3}, w_{4}⟩ ⊂ ⟨w_{3}, w_{4}⟩. By the way we choose v, it satisfies vj = −jv, then

(37)  (a_{1}v) (\frac{b_{1} − j}{2a_{1}|D|}) = (\frac{b_{1} − j}{2a_{1}|D|}) (−a_{1}v) + \frac{b_{1}}{|D|}v

For the part corresponding to the first two elements of I_{z} note that they can be written as (\frac{b_{1}−j}{2a_{1}|D|}) v(a) and (\frac{b_{1}−j}{2a_{1}|D|}) v (\frac{v−b}{2}). Since B is an ideal, vB ⊂ B and the assertion follows from equation (37). □

Corollary. The order R = ⟨1, \frac{1+j}{2} , a_{1}v, \frac{1+j}{2} a_{1}v⟩ is contained in the left order of I_{z} and has discriminant (a_{1}^{2}ND)^{2} or index a_{1}^{2}|D| in a maximal order.

Proof. It is clear that R is in the left order of I_{z} by the previous two propositions. It is also clear that it is an order. To compute its discriminant, note that the bilinear matrix associated to it is:

( \begin{matrix}2 & 1 & 0 & 0 \\ 1 & \frac{1−N}{2} & 0 & 0 \\ 0 & 0 & 2a_{1}^{2}|D| & a_{1}^{2}|D| \\ 0 & 0 & a_{1}^{2}|D| & a_{1}^{2}|D|\frac{1−N}{2} \end{matrix} )

Then note that the index in a maximal order (which has discriminant N^{2}) is the square root of the discriminant. □

Theorem 4.2. Let U be the complex multiplication associated to \frac{−j}{\sqrt{|N|}}. Then the

Siegel point (P, J, U) associated to the ideal I_{z} in the given basis is z_{AD̄}Q_{B}.

Proof. This is a straight forward computation so we omit the details. Just check that the given basis of I_{z} is simplectic, i.e. that the matrix J(x, y) in the given basis

is a multiple of the matrix ( \begin{matrix}0 & I_{2} \\ −I_{2} & 0 \end{matrix} ) (since J(x, y) is skewsymmetric there are

half the conditions to check), and that the matrix U associated to the point z_{AD̄}Q_{B} is the same as the complex multiplication matrix on I_{z}. □

Theorem 4.3. The lattice I_{z} is an ideal for a maximal order.

Proof. The strategy is to prove that the quadratic form associated to the ideal I_{z} is locally equivalent to the maximal order for all primes. We need the next lemma:

Lemma 4.4. The quadratic form associated to the lattice I_{z} has discriminant N^{2}.

Proof. The bilinear form is the same as the Siegel point z_{AD̄}Q_{B} hence its bilinear

form matrix is: B_{I} = ( \begin{matrix}2c_{1}Q_{B} & b_{1}I_{2} \\ b_{1}I_{2} & 2a_{1}DQ_{B}^{−1} \end{matrix} )

Since Q_{B} has determinant D, it is an easy computation to prove that the determinant of this matrix is N^{2} (using that b_{1}^{2} − 4a_{1}c_{1}|D| = N). □

A maximal order for B = (−1, N) is given by O = ⟨\frac{1+j}{2} , \frac{i+k}{2} , j, k⟩ (see Proposition 5.2, page 369 of [Pi]), then it is easy to compute the matrix of the quadratic form trace and to check that it has discriminant N^{2}, and is an improperly primitive integral form. Since the discriminant of both forms is a unit for all primes p ≠ |N| then they are locally equivalent (see Corollary of Theorem 3.1 of [Cas], page 116). Hence (I_{z})_{p} is locally principal for all primes p ≠ |N|. For the ramified prime, D(I_{z}) = N^{2} hence it is locally principal. Locally principal ideals have the same discriminant as their left orders hence O_{l}(I_{z}) is maximal. □

4.4. Comparing Siegel Points. If I is an ideal for a maximal order, and U a complex multiplication, the Siegel point associated to (U, I) is the same as the one associated to the point (U, Iα) for any α ∈ B^{×} (with the same choice of basis). Suppose two Siegel points z and z′ have equivalent ideals I_{z} and I_{z′}, say I_{z} = I_{z′}α for some α ∈ B^{×}. Then since the complex multiplication is the same for all the ideals we construct, the two Siegel points are equivalent by proposition 4.5. Let M be the matrix in Sp_{4}(ℤ) making the change of basis between I_{z} and I_{z′}α.

Lemma 4.5. .The matrix M is in the subspace Γ_{1,2}.

Proof. Let M = ( \begin{matrix}A & B \\ C & D \end{matrix} ), z = (\frac{b_{1}+\sqrt{N}}{2a_{1}} ) Q and z′ = (\frac{b′_{1}+\sqrt{N}}{2a′_{1}} ) Q′ where Q and

Q′ have even diagonal. Since M sends the bilinear form associated to the ideal I_{z} to the bilinear form associated to the ideal I_{z′}α,

( \begin{matrix}A & B \\ C & D \end{matrix} )^{t} ( \begin{matrix}2c_{2}Q & b_{2}I_{2} \\ b_{2}I_{2} & 2a_{2}Q^{−1} \end{matrix} ) ( \begin{matrix}A & B \\ C & D \end{matrix} ) = ( \begin{matrix}2c′_{2}Q & b′_{2}I_{2} \\ b′_{2}I_{2} & 2a′_{2}Q^{′−1} \end{matrix} )

By the way we choose generators, b_{i} ≡ 1 mod 4, i = 1, 2 (also b′_{i} ≡ 1 mod 4, i = 1, 2)

hence 2Q ≡ ( \begin{matrix}0 & 2 \\ 2 & 0 \end{matrix} ) mod 4. Let J := ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ). Looking at the first 2×2 matrix

of the previous equality modulo 4 we get: 2c_{2}A^{t}JA + C^{t}A + A^{t}C + 2a_{2}C^{t}JC ≡ 2J mod 4. In particular 4 divides the diagonal.

If A := ( \begin{matrix}a & b \\ c & d \end{matrix} ) then A^{t}JA = ( \begin{matrix}2ac & ad + bc \\ ad + bc & 2bd \end{matrix} ) hence 4 divides the diagonal

of 2c_{2}A^{t}JA and 2a_{2}C^{t}JC. Also A^{t}C is symmetric hence A^{t}C + C^{t}A = 2A^{t}C and we get that 2 divides the diagonal of A^{t}C. The proof for B^{t}D is analogous looking at the last 2 × 2 matrix. □

Proposition 4.11. Fixed the ideals A and D, the left order of I_{z_{AD}Q_{B}} is independent of the ideal B.

Proof. We know I_{z} = ⟨ϕ (\frac{b_{1}−\sqrt{N}}{2a_{1}|D|} ) ψ(\sqrt{D})ψ(B̄), ψ(B̄)⟩. The ideal B_{q} := B ⊗ ℤ_{q} is

principal, hence there exists an element δ_{q} ∈ L_{q} := ℚ_{q}(\sqrt{D}) such that B_{q} = O_{L}δ_{q}. Then I_{z} ⊗ℤ_{q} = ⟨ϕ (\frac{b_{1}+\sqrt{N}}{2a_{1}|D|} ) ψ(\sqrt{D})ψ(O_{L}), ψ(O_{L})⟩ δ̄_{q}, hence its left order is clearly independent of B. □

Proposition 4.12. Let D and D′ be two split prime ideals of ℚ[\sqrt{N}] of norms |D| and |D′| respectively such that D′ = μD. Let B and B′ be ideals of ℚ[\sqrt{D}] and of ℚ[\sqrt{D′}] respectively. Then the ideals I_{z_{AD}Q_{B}} and I_{z_{AD′} Q_{B′}} have the same left order if following the notation of proposition 4.8 we take v′ = μv.

Proof. We are abusing notation while stating this theorem, since μ is an element of ℚ[\sqrt{N}]. We will denote indistinctly by μ the element in B or in ℚ[\sqrt{N}] via the identification \sqrt{N} ↦ j, and the case will be clear from the context. By proposition 4.11 it is enough to restrict to the case B and B′ principal. In this case we will prove that the ideals associated to them are slightly different and use this to prove the proposition. We can choose basis such that D = ⟨|D|, \frac{b_{1}+\sqrt{N}}{2} ⟩

and D′ = ⟨|D′|, \frac{b_{1}+\sqrt{N}}{2} ⟩. Let μ = \frac{α}{|D|} + \frac{β}{|D|} \sqrt{N}. Since μ (\frac{b_{1}+\sqrt{N}}{2} ) ∈ D′ and

μ^{−1} (\frac{b_{1}+\sqrt{N}}{2} ) ∈ D, \frac{α+βb_{1}}{|D|} ∈ ℤ and \frac{α−βb_{1}}{|D′|} ∈ ℤ. Since b = 1 the definition of the ideals is:

• I_{D} := I_{z_{AD}Q_{B}} = ⟨(\frac{b_{1}−j}{2a_{1}|D|}) v, (\frac{b_{1}−j}{2a_{1}|D|}) (\frac{v+|D|}{2}) , \frac{v−1}{2}, 1⟩

• I_{D′} := I_{z_{AD′} Q′_{B}} = ⟨(\frac{b_{1}−j}{2a_{1}|D′|}) v′, (\frac{b_{1}−j}{2a_{1}|D′|}) (\frac{v′+|D′|}{2}) , \frac{v′−1}{2}, 1⟩

where v and v′ are the elements of norm |D| and |D′| respectively as in proposition 4.8. We will write the elements of I_{D′} in the basis of I_{D}, the other case follows from symmetry.

• \frac{v′−1}{2} = [−a_{1}β, 0, \frac{α+b_{1}β}{|D|} , \frac{α+b_{1}β+D}{2|D|}]
• (\frac{b_{1}−j}{2a_{1}|D′|}) v′ = [\frac{α−βb_{1}}{|D′|} , 0, 4βc, 2βc] which has integer coefficients
• (\frac{b_{1}−j}{2a_{1}|D′|}) (\frac{v′+|D′|}{2}) = [\frac{α−βb_{1}−|D′|}{2|D′|}, 1, 2βc, βc]

We cannot say that the two ideals are the same, since the numbers α and β may have a 2 in the denominator, but (I_{D})_{p} = (I_{D′})_{p} for all primes p ≠ 2. In particular if we denote O_{D} and O_{D′} the left order of I_{D} and I_{D′} respectively, we get that (O_{D})_{p} = (O_{D′})_{p} for all p ≠ 2. Since the denominators are at most 2 it is easy to check that 4O_{D} + ℤ ⊂ O_{D′}, and has index at most 2^{8}. By corollary 4.3, the order R ⊂ O_{D′} with index a_{1}^{2}|D|, which is odd. Then 4O_{D} + R = O_{D′}. Also 4O_{D} + R = O_{D} hence both orders are the same. □

By theorem 3.4 we know that the numbers n_{[A],[B],D̄} depend on the equivalence class of A, the equivalence class of D and the class of z_{AD}Q_{B} modulo Γ_{12}. Fixed the class of A and the class of D we can associate ideals to the points z_{AD}Q_{B} such that they all have the same left order. Then we get at most h(B) different points in the Siegel space.

Theorem 4.4. The number of different n_{[A],[B],D̄} in T is at most h(O_{K})^{2}t(B), where t(B) is the type number.

Proof. This follows from corollary of lemma 4.12. □

Note that this number is independent of the class number of O_{L}. With this we finish the proof of theorem (2.1) □

Proposition 4.13. Let A be an ideal of ℚ(\sqrt{N}), then n_{[A],[B],D̄} and n_{[O_{K}],[B],D̄} differ by a unit in a quadratic extension of M.

Proof. Let σ_{A} be the automorphism of H corresponding to the ideal A via the Artin map. Then we proved that (\frac{θ(z_{O_{K} D}Q_{B})}{η(D)η(O_{K})})^{σ_{A}} = \frac{θ(z_{AD}Q_{B})}{η(AD)η(A)}. Hence n_{[A],[B],D̄} =

(\frac{η(A)η(AD)}{η(D)η(O_{K})ψD̄(A)}) (n_{[O_{K}],[B],D̄})^{σ_{A}}. Note that the quotient of etas squared is in H

while ψD̄(A) is in T, hence ζ := (\frac{η(A)η(AD)}{η(D)η(O_{K})ψD̄(A)}) is in a quadratic extension of M. Clearly N(ζ) = 1 as required. □

5. The class number one case

We study the situation in the case of imaginary quadratic fields of class number equal to one. In this case n_{[A],[B],D̄} is a rational integers for any choice of D. There are just six such cases (we exclude the case N = −3) so we can study all this cases by numerical computations. Here are some examples:

5.1. Case N = −7. This case is the simplest one, since the class number in the quaternion algebra is also one. Then the numbers n_{[A],[B],D̄} are integers and differ by a unit.

Theorem 5.1. Let N = −7 and D be any ideal of prime norm congruent to 3 modulo 4. Then L(1, ψ_{D}) ≠ 0.

Proof. By proposition 3.5 we know that the number associated to an ideal B is the same as the one associated to B̄. For a prime ideal D let Ω = η(D̄)η(O_{K})\frac{2π}{w\sqrt{|D|}}

where −D = N(D) and w is the number of units in ℚ[\sqrt{D}]. The formula 17 for L(1, ψ) reads:

(38) L(1, ψ) = ( ∑_{[B]∈Cl(O_{L})} n_{[O_{K}],[B],D̄}) Ω = (n_{[O_{K}],[O_{L}],D̄} + 2∑_{Φ}n_{[O_{K}],[B],D̄}) Ω

where the last sum is over non-principal ideals B, with any choice of ideal representatives modulo conjugacy (since |D| is prime, the class number of L is odd and we have such a representation).

Taking the maximal order O as left O-ideal representative, we see that the number associated to it is 1 up to a sign, then \frac{L(1,ψ)}{Ω} ≡ 1 mod 2. □

In the next table, we list some of the numbers n_{[O_{K}],[B],D̄} to show the behavior of the sign.

\begin{matrix}D & B & n_{[A],[B],D̄} \\ 11 & [1,-1,3] & 1 \\ 23 & [1, -1, 6] & 1 \\ 23 & [13, -17, 6] & -1 \\ 23 & [13, 17, 6] & -1 \\ 43 & [1, -1, 11] & -1 \\ 67 & [1, -1, 17] & 1 \\ 71 & [1, -1, 18] & -1 \\ 71 & [19, 9, 2] & -1 \\ 71 & [19, -9, 2] & -1 \\ 71 & [29, 33, 10] & 1 \\ 71 & [29, -33, 10] & 1 \\ 71 & [43, 141, 116] & -1 \\ 71 & [43, -141, 116] & -1 \end{matrix}

5.2. Case N = −11. In this case the quaternion algebra has type number 2 for maximal orders , so we get two different integers associated to different D’s. Each number n_{[O_{K}],[B],D̄} will be associated to an ideal class. Let B = (−1, −11) be the quaternion algebra ramified at 11 and infinity. Let O := ⟨\frac{1}{2} + \frac{j}{2}, \frac{i}{2} + \frac{k}{2} , j, k⟩ be a maximal order and I a non-principal ideal. Here is a table of n_{[O_{K}],[B],D̄} for different values of D and B, writing down the associated ideal also.

D  B  n_{[A],[B],D̄} Ideal
23 [1, -1, 6]  2  I_{1}
23 [13, -17, 6]  0  O
23 [13, 17, 6]  0  O
31 [1, -1, 8]  -2  I_{1}
31 [5, 17, 16]  0  O
31 [5, -17, 16]  0  O
47 [1, -1, 12]  0  O
47 [7, -17, 12]  2  I_{1}
47 [7, 17, 12]  2  I_{1}
47 [17, -53, 42]  0  O
47 [17, 53, 42]  0  O

Note that the number 0 is associated to the principal ideal, while the number 2 is associated to I_{1}. With the same reasoning as in theorem 5.1 we can get a partial result proving that the ideals D such that z_{D}Q_{O_{L}} is associated to the ideal I_{1} have a non-vanishing L-series. Following the method described in [Pa-Vi], taking {O, I_{1}} as representatives for the maximal order and constructing the Brandt matrices for level 11^{2} we get that the eigenvector associated to the modular form of weight 2 and level 11^{2} is [0, 0, 0, 1, −1, 0, 0, 0, 1, −1]. The first three zeros correspond to the principal ideal, and the ±1 to I_{1}. Then the number associated to each ideal is the same as the one associated to it via n_{[O_{K}],[B],D̄}, since the eigenvector is well defined up to a constant.

5.3. Case N = −163. Let B = (−1, −163) be the quaternion algebra ramified at 163 and infinity. In this case, the class number for maximal orders is 14 while the type number is 8. Consider the maximal order O := ⟨1, i, \frac{1}{2} + \frac{j}{2}, \frac{i}{2} + \frac{k}{2} ⟩ A set of representatives of left O-ideals is given by {I_{j}}_{j=1}^{14} with I_{1} = O and

• I_{2} := ⟨2, 2i, \frac{1}{2} + i + \frac{j}{2}, −1 + \frac{i}{2} + \frac{k}{2} ⟩
• I_{3} := ⟨3, 3i, \frac{1}{2} + i + \frac{j}{2}, −1 + \frac{i}{2} + \frac{k}{2} ⟩
• I_{4} := ⟨3, 3i, \frac{−1}{2} + i + \frac{j}{2}, −1 − \frac{i}{2} + \frac{k}{2} ⟩
• I_{5} := ⟨6, 6i, \frac{1}{2} + i + \frac{j}{2}, −1 + \frac{i}{2} + \frac{k}{2} ⟩
• I_{6} := ⟨6, 6i, \frac{−1}{2} + i + \frac{j}{2}, −1 − \frac{i}{2} + \frac{k}{2} ⟩
• I_{7} := ⟨4, 4i, \frac{3}{2} + i + \frac{j}{2}, −1 + \frac{3i}{2} + \frac{k}{2} ⟩
• I_{8} := ⟨4, 4i, \frac{−3}{2} + i + \frac{j}{2}, −1 − \frac{3i}{2} + \frac{k}{2} ⟩
• I_{9} := ⟨6, 6i, \frac{5}{2} + i + \frac{j}{2}, −1 + \frac{5i}{2} + \frac{k}{2} ⟩
• I_{10} := ⟨6, 6i, \frac{−5}{2} + i + \frac{j}{2}, −1 − \frac{5i}{2} + \frac{k}{2} ⟩
• I_{11} := 5, 5i, \frac{1}{3} + 2i + \frac{j}{2}, −2 + \frac{i}{2} + \frac{k}{2} ⟩
• I_{12} := ⟨5, 5i, \frac{−1}{2} + 2i + \frac{j}{2}, −2 − \frac{i}{2} + \frac{k}{2} ⟩
• I_{13} := ⟨7, 7i, \frac{5}{2} + 3i + \frac{j}{2}, −3 + \frac{5i}{2} + \frac{k}{2} ⟩
• I_{14} := ⟨7, 7i, \frac{−5}{2} + 3i + \frac{j}{2}, −3 − \frac{i}{2} + \frac{k}{2} ⟩

The pairs of ideals (I_{2j+1}, I_{2j+2}) with j = 1, …, 6, have the same right order, hence each pair will have the same integer associated. For the table we consider the range of primes between 150 and 200 so as to get all the ideals {I_{j}} associated to some number n_{[O_{K}],[B],D̄}. The table is:

\begin{matrix}D & B & n_{[A],[B],D̄} & Ideal \\ 151 & [1, -1, 38] & 20 & I_{2} \\ 151 & [29, 9, 2] & 14 & I_{8} \\ 151 & [29, -9, 2] & 14 & I_{8} \\ 151 & [11, -5, 4] & 8 & I_{13} \\ 151 & [11, 5, 4] & 8 & I_{14} \\ 151 & [43, 137, 110] & 4 & I_{12} \\ 151 & [43, -137, 110] & 4 & I_{12} \\ 167 & [1, -1, 42] & 0 & I_{1} \\ 167 & [157, 33, 2] & -20 & I_{2} \\ 167 & [157, -33, 2] & -20 & I_{2} \\ 167 & [61, 65, 18] & -2 & I_{4} \\ 167 & [61, -65, 18] & -2 & I_{3} \\ 167 & [29, 93, 76] & -10 & I_{6} \\ 167 & [29, -93, 76] & -10 & I_{5} \\ 167 & [127, -177, 62] & -14 & I_{7} \\ 167 & [127, 177, 62] & -14 & I_{8} \\ 167 & [19, -21, 8] & -12 & I_{9} \\ 167 & [19, 21, 8] & -12 & I_{10} \\ 179 & [1, -1, 45] & 0 & I_{1} \\ 179 & [19, 45, 29] & 2 & I_{3} \\ 179 & [19, -45, 29] & 2 & I_{4} \\ 179 & [13, 17, 9] & 4 & I_{12} \\ 179 & [13, -17, 9] & 4 & I_{11} \\ 199 & [1, -1, 50] & 0 & I_{1} \\ 199 & [31, -69, 40] & -20 & I_{2} \\ 199 & [31, 69, 40] & -20 & I_{2} \\ 199 & [43, -133, 104] & -4 & I_{12} \\ 199 & [43, 133, 104] & -4 & I_{11} \\ 199 & [13, 29, 20] & -14 & I_{8} \\ 199 & [13, -29, 20] & -14 & I_{7} \\ 199 & [131, 453, 392] & -8 & I_{14} \\ 199 & [131, -453, 392] & -8 & I_{13} \end{matrix}

The eigenvector for the Brandt matrices corresponding to the form of weight 2 and level 167^{2} is given by the vector [0, 10, 1, 1, 5, −5, 7, −7, −6, 6, 2, 2, −4, 4] with respect to the maximal order representatives {I_{j}}. Considering all the class number 1 imaginary quadratic fields (the computations being the same in all cases), we can prove:

Theorem 5.2. Let E be a CM elliptic curve over ℚ of level p^{2}. Then the coordinates of the eigenvector of the Brandt matrices associated to E are given up to a sign by n_{[O_{K}],[B],D̄}

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   E-mail address: apacetti@math.utexas.edu

University of Texas at Austin, TX 78712, USA