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Integers that can be written as the sum of two rationalcubes

Eugenia Rosu

Contents

1 Introduction 2

2 Background. 42.1 The L-function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.2 Ring class fields. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2.2.1 Idelic interpretation of ring class field . . . . . . . . . . . . . . . . . . . . 52.2.2 Ring class field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.2.3 Characterization of ideals in ring class fields . . . . . . . . . . . . . . . . . 8

2.3 The cubic character . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.3.1 Relating χD to the Galois conjugates of D13. . . . . . . . . . . . . . . . . 11

2.4 Hecke characters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.4.1 Converting the characters. . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

3 Computing the value LpED, 1q using Tate’s Zeta function 133.0.2 Schwartz-Bruhat functions. . . . . . . . . . . . . . . . . . . . . . . . . . . 133.0.3 Haar measure. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

3.1 Zeta functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143.1.1 Computing the finite part of Tate’s Zeta function Zf ps, χDϕ,Φq . . . . . 153.1.2 Adelic representatives for ClpO3Dq . . . . . . . . . . . . . . . . . . . . . . 173.1.3 Connection to the Eisenstein series . . . . . . . . . . . . . . . . . . . . . . 173.1.4 Fourier expansion of the Eisenstein series Eεps, zq at s “ 0. . . . . . . . . 203.1.5 Connection to the theta function ΘKpzq. . . . . . . . . . . . . . . . . . . 223.1.6 Final formula for Lp1, χDϕq . . . . . . . . . . . . . . . . . . . . . . . . . . 233.1.7 Turning the formula into a trace. . . . . . . . . . . . . . . . . . . . . . . . 243.1.8 SD is an integer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

4 Shimura reciprocity law in the classical setting. 264.1 Applying Shimura reciprocity law to K “ Qr

?´3s. . . . . . . . . . . . . . . . . . 29

4.1.1 fpωq is in the ring class field H3D. . . . . . . . . . . . . . . . . . . . . . . 294.1.2 Galois conjugates of fpωq. . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

1

5 Writing SD as a square. 325.1 Factorization Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325.2 Ratios of θr and θ0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 375.3 Applying the factorization lemma to get a square . . . . . . . . . . . . . . . . . . 40

5.3.1 Representatives of ClpO3Dq. . . . . . . . . . . . . . . . . . . . . . . . . . . 405.3.2 Using the factorization formula . . . . . . . . . . . . . . . . . . . . . . . . 41

5.4 Shimura reciprocity applied to θr . . . . . . . . . . . . . . . . . . . . . . . . . . . 445.4.1 θr as an automorphic form . . . . . . . . . . . . . . . . . . . . . . . . . . 455.4.2 Galois action on modular functions (Shimura reciprocity) . . . . . . . . . 46

5.5 The square is invariant under Galois action . . . . . . . . . . . . . . . . . . . . . 57

6 Appendix A: properties of ΘK 586.1 Properties of ΘKpp´b`

?3q6q. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

6.2 About ΘKpDp´3`?´3q6q. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

6.2.1 Traces of theta functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

Abstract

We are interested in finding for which positive integers D we have rational solutions forthe equation x3

` y3“ D. The aim of the paper is to compute the value of the L-function

LpED, 1q, for ED : x3` y3

“ D. For the case of p prime p ” 1 mod 9, two formulas havebeen computed by Rodriguez-Villegas and Zagier in [14]. We compute several formulasusing automorphic methods.

1 Introduction

In the current paper we are interested in finding an algorithm to decide which positive integerscan be written as the sum of two rational cubes:

x3 ` y3 “ D, x, y P Q (1)

After making the equation homogeneous, we get the equation x3 ` y3 “ Dz3 that has arational point at 8 “ r1,´1, 0s. Moreover, after a change of coordinates X “ 12D

z

x` y,

Y “ 36Dx´ y

x` ythe equation becomes

ED : Y 2 “ X3 ´ 432D2

which defines an elliptic curve over Q.We will assume D is cube free and D ‰ 1, 2 (trivial cases) throughout the paper. It is known

that EDpQq has trivial torsion for D ‰ 1, 2 (see [16]). Thus, (1) has a solution iff EDpQq haspositive rank. From the BSD conjecture, this is equivalent to the vanishing of LpED, 1q.

Without assuming BSD, from the work of Coates-Wiles [2], or more generally Gross-Zagier[6] and Kolyvagin [10], when LpED, 1q ‰ 0, we have rankEDpQq “ 0, thus no rational solutionsin p1q. We define an invariant SD of ED as follows:

SD “LpED, 1q

ΩD,8RED,

2

where:

• ΩD,8is the real period

• RED is the regulator

The definition is made such that in the case of LpED, sq ‰ 0 we expect to get from the fullBSD conjecture:

SD “ #XpEDqź

p|6D

cp, (2)

where #X is the order of the Tate-Shafarevich group and cp are the the Tamagawa numberscorresponding to the elliptic curve ED. From work of Cassels [1], using the Cassels-Tate pairing,we have that when X is finite, its order is going to be a square. Thus we expect that SD to bean integer square up to the Tamagawa numbers.

For the case of prime numbers, Sylvester conjectured that the answer is affirmative in thecase of p ” 4, 7, 8 mod 9. In the cases of p ” 2, 3, 5 mod 9 we have LpEp, 1q ‰ 0 and p isnot the sum of two cubes. This follows either from a 3-descent argument (Sylvester, Lucas andPepin) or from the theorem of Coates-Wiles [2].

In [14], Rodriguez-Villegas and Zagier computed formulas for LpEp, 1q in the case of primesp ” 1 mod 9. In this case it is predicted by BSD that the rank of EDpQq is either 0 or 2. Theycompute two formulas for SD. In the current paper, we are extending their results to all integersD. If we let K “ Qr

?´3s, we have:

Theorem 1.1. For all integers D, SD is an integer and we have the formula:

SD “ TrH3DK

ˆ

D13 ΘKpDωq

ΘKpωq

˙

, (3)

where:

• H3D is the ring class field associated to the order O3D “ Z` 3DOK ,

• ω “ ´1`?´3

2 is a third root of unity, and

• ΘKpzq “ÿ

a,bPZe2πizpa2`b2ábq is the theta function associated to the number field K.

A second result makes SD more easily computable. We also hope to extend this result toshow that SD is an integer square up to Tamagawa numbers:

Theorem 1.2. In the case of D “ś

pi”1 mod 3

peii , SD is an integer and we have:

SD “

ˇ

ˇ

ˇ

ˇ

TrHOH0

θ1pz0q

θ0pz0qD´13

ˇ

ˇ

ˇ

ˇ

2

(4)

where:

• θ1pzq “ř

nPZp´1qneπipn`1D´16q2z a 12-weight modular form

• z0 “´b`

?´3

2 a CM-point, with b2 ” ´3 mod 4D2,

3

• HO is the ray class field of modulus 3D and H0 is an intermediate field K Ă H0 Ă HOthat is the fixed field of a certain Galois subgroup G0 – ClpO3Dq.

Conjecture 1.1. We conjecture that the term I0 “ TrHOH3D

θ1pz0qθ0pz0D2q

D13 is an integer andI20 is the order of the Tate-Shafaravich group.

Using similar methods, we obtain a general formula for all integers N , for which Theorem1.2 is a particular case.

Theorem 1.3. Using the same notation as in Theorem 1.2, we have for all integers D:

SD “D´1ÿ

r“0

ˇ

ˇ

ˇ

ˇ

TrHOH3D

θrpDz0q

θ0pz0DqD´13

ˇ

ˇ

ˇ

ˇ

2

, (5)

where:

• θrpzq “ř

nPZp´1qneπipn`rD´16q2z a 12-weight modular form

• z0 “´b`

?´3

2 a CM-point

Conjecture 1.2. We conjecture that all terms Ir “ˇ

ˇ

ˇTrHOH3D

D´13 θrpDz0qθ0pz0Dq

ˇ

ˇ

ˇ

2

are equal for allr such that pr,Dq “ 1. This is indeed the case when D is a product of primes that split.

Further results not included in this draft:

In Appendix A we provide a second proof of Theorem 1.1 based on an idea of Xinyi Yuan. InAppendix B we compute a different formula for SD inspired by the Rallis inner product. Thisexpresses SN in the following way:

Theorem 1.4. For D a product of primes p ” 1 mod 4, we have

LpED, sq “ cDÿ

APClpO3Dq

E0ps, gA,ΦqχDpAqp´1qNmA´1

2 ,

where gA is the embedding of the generator of the ideal A “ pa` ba

´3q into GL2pAQq and

E0ps, gz,Φq “ÿ

m”0 mod D,pn,Dq“1

1

pmz ` nq|mz ` n|sis a sum of Eisenstein series defined by

Hecke in [7].

2 Background.

Let K “ Qr?´3s. Note that K is a PID and has the ring of integers OK “ Zrωs, where

ω “ ´1`?´3

2 is a fixed root of unity. We will denote Kv the localization of K at the place v.We will denote by Kp :“

ś

v|pKv – Qpr?´3s.

4

2.1 The L-function

Our goal is to compute several formulas for the special value of the L-function LpED, 1q of theelliptic curve ED : x3 ` y3 “ Dz3. The elliptic curve ED has complex multiplication (CM) byOK . Then LpED, sq is the L-function of a Hecke character that is computed explicitly in Irelandand Rosen [8]. We have:

LpED, sq “ Lps, χDϕq,

where χD and ϕ are classical Hecke characters such that ϕχD is the Hecke character correspond-ing to the elliptic curve ED. The Hecke character ϕ is the Hecke character corresponding toE1 and χD is the Hecke character corresponding to the cubic twist. More precisely, the Heckecharacters are defined to be:

• ϕ : Ip3q Ñ Kˆ is defined on the ideals prime to 3 by ϕpAq “ α, where α is the uniquegenerator of the ideal A such that α ” 1 mod 3.

• χD : Ip3Dq Ñ t1, ω, ω2u is the cubic character defined below in Section 2.3; it is definedon the space Ip3Dq of all fractional ideals of OK prime to 3D. Moreover, it is well-definedover ClpO3Dq the ring class group corresponding to the order O3D “ Z` 3DOK .

The L-function can be expanded:

LpED, sq “ÿ

APIp3Dq

χDpAqϕpAqpNmAqs

“ÿ

αPOK ,α”1pmod 3q

χDpαqα

Nαs.

2.2 Ring class fields.

Recall that an order O of K is a subring of OK that is a finitely generated Z-module and suchthat O bZ Q “ K. As K is a quadratic number field, each order is of the form O “ Z ` fOKand we call f “ rOK : Os the conductor of O. We can also write O using a Z-basis in the formO “ r1, fωsZ.

We define the class group ClpOq of the order O of conductor f is defined to be:

ClpOq :“ IOpfqPOpfq,

where IOpfq is the set of fractional O-ideals prime to the conductor f , and POpfq the subgroupof IOpfq of principal fractional O-ideals.

We define the ring class field to be the abelian extension HO of K corresponding to theGalois group ClpOq from class field theory, meaning:

GalpHOKq – ClpOq.

We denote by IpNq the group of fractional ideals in K prime to N . We denote the subgroupPZ,N “ tpαq: α P K such that α ” a mod N for some integer a such that gcdpa,Nq “ 1.Furthermore, let ON :“ Z `NOK be the order of K of conductor N . Then we can define thering class field of the order ON to be

ClpON q :“ IpNqPZ,N

5

Note that K has class number one and thus by the Strong Approximation theorem we have:

AˆK “ KˆCˆź

v-8OˆKv .

We would like to describe ClpON q adelically. We do this below:

Lemma 2.1. For N a positive integer, we can think of the ring class group adelically as:

ClpON q – UpNqzAˆK,fKˆ,

where UpNq “ś

ppZ`NZprωsqˆ.

Proof: From the Strong approximation theorem, as K is a PID, we have:

AˆK – KˆCˆź

v-8OˆKv .

Taking the quotient by KˆCˆ, we get:

AˆK,fKˆ –

ź

v-8OˆKv

˜

Kˆ Xź

v

OˆKv

¸

–ź

v-8OˆKv 〈´ω〉 ,

where 〈´ω〉 is the group of sixth roots of unity.Furthermore, note that UpNq –

ś

v-NOˆKv

ś

p|N

pZ`NZprωsqˆ. Moreover note that 〈´ω〉UpNq “

UpNq. Thus we have:

AˆK,fKÛpNq –

ź

v-8OˆKv 〈´ω〉UpNq –

ź

v|N

OˆKvź

p|N

pZ`NZprωsqˆ –ź

p|N

ź

v|p

OˆKvpZ`NZprωsqˆ

Finally, we need to show an isomorphism between ClpON q “ IpNqPZpNq andś

v|N OˆKvś

p|N

pZ`

NZprωsqˆ. We construct the map:

IpNq Ñź

v|N

OˆKv Ñź

v|p

OˆKvź

p|N

pZ`NZprωsqˆ

Let pk0q P IpNq be an ideal. Then we can map k0 Ñ pk0qv|N . After taking the projectionmap, we want to look at the kernel of the composition IpNq Ñ

ś

v|pOˆKvś

p|N

pZ `NZprωsqˆ.

This consists of ideals pk0q P IpNq such that k0 ” ap mod NZprωs, where ap P Z and pap, pq “ 1.By the Chinese remainder theorem, we can find a P Z such that a ” ap mod N for all p|N .

Then we have k0 ” a mod NZprωs for all a P Z. Thus pk0q P PZpNq and PZpNq is the kernelof the above map. Thus we get:

IpNqPZpNq –ź

v|p

OˆKvź

p|N

pZ`NZprωsqˆ,

which proves our claim.

6

Another easy result that we will use is the following straight forward application of theChinese remainder theorem. This map will be important in our proof:

Lemma 2.2. For any pl1,vqv|N Pś

v|N

OˆKv , we can find k1 P OK such that for all v|N we have:

l1,v ” k1 mod NOKv ,

Proof: For any v|N we can find a1,v P OK such that l1,v ” a1,v mod NOKv . We will pickfor N “

ś

v|N pevv , where pv is the prime corresponding to the place v:

k1 “ÿ

v|N

a1,vmvN

pevv,

where mv P OK , mvNpevv

” 1 mod pevv . We can find such an inverse since OK is a PID, thusOKNOK –

ś

v|N OKpevv OK .

2.2.1 Characterization of ideals in ring class fields

Recall that a primitive ideal is an ideal not divisible by any integral ideal. It is easy to prove:

Lemma 2.3. Any primitive ideal of OK can be be written in the form A “ ra, ´b`?´3

2 s asa Z-module, where b is an integer (determined modulo 2a) such that b2 ” ´3 mod 4a andNmA “ a. This implies that for A “ pαq, we have α “ a.

Conversely, given an integer satisfying the above congruence and A defined as above, we getthat A is an ideal in OK of norm a.

2.3 The cubic character

In the following we will define the cubic character χD and check that it is well defined on theclass group ClpO3Dq. Let ω “ ´1`

?´3

2 be a fixed cube root of unity. Then we can define thecubic residue character following Ireland and Rosen [8].

Definition 2.1. For α P Zrωs such that α is prime to 3, we define a cubic residue characterχα : Ip3αq Ñ t1, ω, ω2u on the fractional ideals of K prime to 3α. For every prime ideal p ofZrωs, the character is defined to be:

χαppq “ ωj ,

for j P t0, 1, 2u such that ωj is the unique third root of unity for which:

αpNm p´1q3 ” ωj mod p, for Nm p ‰ 3.

It is further defined multiplicatively on the fractional ideals of Ip3αq.

Notation: We will also denote χDp¨q “:`

D¨

˘

3.

First let us check that this definition makes sense. Since K is a PID, any prime ideal p has agenerator of the form π “ a` bω P Zprωs. Then the norm Np “ a2´ ab` b2 which is congruentto 0, 1 mod 3. Then, if p is prime to 3, we must have Np ” 1 mod 3, implying that 3 dividesNp´ 1.

7

Furthermore, the group pZrωspZrωsqˆ has Nm p ´ 1 elements, thus we have αNm p´1 ” 1

mod p. Then since Nm p´ 1 is divisible by 3, we can factor out:

p|pαpNm p´1q3 ´ 1qpαpNm p´1q3 ´ ωqpαpNm p´1q3 ´ ω2q

Finally sinceK “ Qr?´3s is an UFD, p divides exactly one of these terms, say pαpNm p´1q3´

ωiq. Thus we can take χαppq “ ωi and it is well-defined.

Following Ireland and Rosen, it is natural to look at the primary elements of K:

Definition 2.2. For a prime ideal p of K we call π primary if π generates p a prime idealand π ” 2 mod 3.

Lemma 2.4. For any ideal A prime to 3, we can find a generator α P Zrωs such that α ” 2

mod 3.

Proof: Since K is a PID, we can find a generator α0 “ a ` bω be a generator of A. Thennote that ˘α0,˘α0ω,˘α0ω

2 also generate the ideal A and exactly one of them is ” 2 mod 3.

Remark 2.1. Note that from the definition of χπ1 we have χπ1pπ2q “ χ´π1pπ2q, as πpNmπ2´1q31 “

p´π1qpNmπ2´1q3 when Nmπ2 is odd and πpNm 2´1q3

1 ” p´π1qpNm 2´1q3 ” 1 mod 2 when π2 “ 2.

Moreover χπ1pπ2q “ χπ1

p´π2q, as χπ1p´1q “ 1. Then we actually have for any choices of ˘:

χ˘π1p˘π2q “ χ˘π2p˘π1q

Theorem 2.1. (Cubic reciprocity law). For π1, π2 ” 2 mod 3 primary generators of primesp1, p2, Nπ1 ‰ Nπ2 and Nπ1, Nπ2 ‰ 3, then:

ˆ

π1

π2

˙

3

“

ˆ

π2

π1

˙

3

Corollary 2.1. For πi, π1i ” 2 mod 3, we have

χ˘π1...πnp˘π11 . . . π

1nq “ χ˘π11...π1np˘π1 . . . πnq

Proof: We will first show that χπ1...πnpπ1iq “ χπipπ1 . . . πnq. By definition, we have:

χπ1...πnpπ1iq ” pπ1 . . . πnq

pNmπ1i´1q3 mod π1i

Thus, we have:

χπ1...πnp

mź

i“1

π1iq “mź

i“1

χπ1...πnpπ1iq “

mź

i“1

nź

j“1

χπj pπ1iq

Using the cubic reciprocity, we have χπj pπ1iq “ χπ1ipπjq, thus we getśmi“1

śnj“1 χπj pπ

1iq “

śmi“1

śnj“1 χπ1ipπjq, which furthermore implies:

χπ1...πnp

mź

i“1

π1iq “ χπ11...π1mpnź

j“1

πiq.

Note that we can always write the elements of Zrωs that are congruent to ˘1 mod 3 as aproduct of primary elements up to sign. Using the above corollary for α and D, we get:

8

Corollary 2.2. If α ” ˘1 mod 3 and D an integer prime to 3, then we have:

χDpαq “ χαpDq

Proof: Since α,D ” ˘1 mod 3, we can write each of them in the form α “ ˘śni“1 πi and

D “ ˘śmj“1 π

1j .

Then using the previous Corollary and Remark 2.1, we have χ˘

nś

i“1πip˘

mś

j“1

πjq “ χ˘

mś

j“1πjp˘

nś

i“1

πiq.

Lemma 2.5. Let α be prime to 3 and p a prime ideal prime to 3. Then the cubic residue canalso be rewritten as:

χαppq ”

ˆ

α

α

˙pNmπ´1q3

mod π

Proof: We have by definition χαppq ” αpNmπ´1q3 ” ωi mod p. Taking the complexconjugate we have αpNmπ´1q3 ” ω2i mod p. Then by taking the ratio we get:

ˆ

α

α

˙pNmπ´1q3

”ω2i

ωimod p

Thus we have χαppq ” αpNmπ´1q3 ” ωi ”

ˆ

α

α

˙pNmπ´1q3

mod p which finishes the proof

of the lemma.

Corollary 2.3. Let D “mś

i“1

pi. For α P PZ,3D, we have χDpαq “ 1. Thus χD is well defined

on ClpO3Dq.

Proof: Recall from the previous Lemma that if α ” ˘1 mod 3, then we have:

χαppq ”

ˆ

α

α

˙pNm p´1q3

mod p

Let p|D. Since α P PZ,3D, we have α ” a mod 3D for some a P Z and pa, 3Dq “ 1. Thusα ” a mod p, which also α ” a mod p, which implies:

χαppq ”

ˆ

α

α

˙pNm p´1q3

”

ˆ

a

a

˙pNm p´1q3

” 1 mod p

Thus we get χαppq “ 1 for all p|D. Thus we have χαpDq “ 1. Moreover, using Corollary

2.2, we have χDpαq “mś

i“1

χpipαq “mś

i“1

χαppiq “ 1.

Remark 2.2. For any fractional ideal A of K, when we write χDpAq we will mean:

χDpAq :“ χDpαq,

where α is the unique generator of A such that α ” 1 mod 3.

9

2.3.1 Relating χD to the Galois conjugates of D13.

There is another way to look at the cubic character using the Galois conjugates of D13. Wehave the following lemma:

Lemma 2.6. Let D be an integer prime to 3. Then for a prime ideal p of K prime to 3D, wehave:

D13χDppq “ pD13qσp ,

where σp P GalpCKq is the Galois action corresponding to the ideal p in the Artin correspon-dence.

Proof: It is enough to prove the claim for σi P GalpF Kq, where L “ KrD13, D13ω,D13ω2s.Let σp “

´

LKp

¯

the Frobenius element corresponds to p the prime ideal of OK . Then using the

definition of the Frobenius element for D13 P L, we get:

pD13qσp ” pD13qNm p mod pOL

Furthermore, note that pD13qNm p “ D13DpNm p´1q3 ” D13χDppq mod pOL.Since theGalois conjugates ofD13 are the roots of x3´D, the Galois conjugate pD13qσp P tD13, D13ω,D13ω2u

and from the congruences above we get:

pD13qσp “ D13χDppq

Corollary 2.4. Let D be an integer prime to 3 and A an ideal of K prime to 3D. Moreover,let σA P GalpKabKq be the Galois action corresponding to the ideal A through the Artin map.Then for the cubic character χD, we have:

pD13qσA “ D13χDpAq. (6)

Proof: Let A “ś

j

pfjj the prime decomposition of A in K. Note that χDppiq P K, thus it

is preserved by the Galois action. Applying the above Lemma we get:

ppD13qσpi qσpj “ pD13χDppiqq

σpi “ D13χDppjqχDppjq

Using this step repeatedly, we get pD13qσA “ D13χDpAq “ D13χDpAq.

Remark 2.3. Note that for the complex conjugate character χD we have a similar result:

pD23qσA “ D23χDpAq. (7)

2.4 Hecke characters

There are two equivalent ways of defining a Hecke character: classically and adelically. Wedefine the classical Hecke character over K to be rχ : Ipfq Ñ Cˆ a character from the setof fractional ideals prime to f , where f is a nonzero ideal of OK . We further say that rχ has

10

infinity type rχ8 if it is characterized by the condition that on the set of principal ideals P pfqprime to f it satisfies the condition:

rχppαqq “ rεpαqrχ´18 pαq,

where:

• rε : pOKfOKqˆ Ñ T is called the pOKfOKqˆ-type character i.e. ε is a character takingvalues in a finite group T.

• rχ8 is an infinity type continuous character i.e. rχ8 : Cˆ Ñ Cˆ is a continuous character.

We define the idelic Hecke character to be a continuous character χ : AˆKˆ Ñ Cˆ.There is a unique correspondence between the idelic and the classical Hecke characters. The

correspondence can be explicitly constructed in the following way:

• χpOˆv $vq :“ χppvq, v - f

• χ8 is determined by χ8

• χv with v|f is determined by Weak Approximation Theorem.

2.4.1 Converting the characters.

We want to compute a formula for Lps, χq, where χ : AˆKKˆ Ñ Cˆ is the Hecke characterdefined by χ “ χ3Dϕ. Here χ3Dϕ are the adelic correspondent Hecke characters of the classicalHecke characters:

1. χ3D : Ip3Dq Ñ t1, ω, ω2u is the cubic character.

2. ϕ : Ip3q Ñ Cˆ is the Hecke character defined by χppαqq “ α for α ” 1 mod 3.

By abuse of notation, I will use ϕ, χ3D both for the classical and the adelic Hecke characters.This should be clear from the context. We can rewrite the two characters adelically:

1. ϕ : AˆK Ñ Cˆ such that:

$

’

’

’

’

&

’

’

’

’

%

ϕvppq “ ´p, ϕvpOˆKv q “ 1, if v|p, p ” 2 mod 3,

ϕvp$vq “ $v, ϕvpOˆKv q, where ωv uniformizer of OKv , $v ” 1 mod 3, if v|p, p ” 1 mod 3,

ϕ8px8q “ x´18 , v “ 8

ϕvp$vq can be determined from the Weak approximation theorem, if v “?´3|3

2. Note that χ3D is trivial on PZ,3D, thus χ3D is a character on ClpO3Dq. We will define thecharacter by making it trivial on Cˆ, Up3Dq and Kˆ. Then we can define using Lemma2.2:

χ3Dplq “ χ3Dpl1q “ χ3Dppk1qq.

11

More precisely, this will be:

$

’

’

&

’

’

%

χ3D,vp$vq “ χ3Dppωvqq, χ3D,vpOˆKv q “ 1, if v - 3D

χ3D,8px8q “ 1, v “ 8

χ3D,vp$vq can be determined from the Weak approximation theorem, if v|3D

We can generally compute χf plf q in the following way:

Lemma 2.7. If χ “ χ3Dϕ, let lf “ kl1, k P Kˆ, l1 Pś

v OˆKv

. Note that this decomposition isunique up to a unit of OˆK and pick k such that l1,3 ” 1 mod 3. Moreover take k1 P K

ˆ suchthat l1 ” k1 mod 3DOKv . Then:

χf plf q “ krχ3Dppk1qq

Proof: We start by writing:

χf plf q “ χf pkqχf pl1q “ χ8pkq´1χv|3Dpl1,vq

Moreover, from the Chinese remainder theorem, we can find k1 P Kˆ such that k1 ” l1,v

mod 3DOKv . As we have k´11 l1 P 1 mod 3DOKv and χ is trivial on pZ` 3DOKv qˆ for v|3D,

we get χvpk1q “ χpl1,vq. This implies:

χf plf q “ kχv|3Dpk1q “ kχv-3Dpk1q´1χ8pk1q

´1

Note that if we write k1 “ uś

v ωevv , where u P OˆKv , we get:

ź

v-3Dχvpk1q “

ź

v-3Dχvpωvq

ev “ź

v-3Dχppvq

ev “ χppk1qq

This moreover implies:

χf plf q “ kχppk1qq´1k1 “ kk´1

1 k1χ3Dppk1qq “ kχ3Dppk1qq

3 Computing the value LpED, 1q using Tate’s Zeta function

In this section we will compute the value of LpED, 1q “ Lp1, χDϕq, working with χD, ϕ asautomorphic Hecke characters. We will show the following result:

3.0.2 Schwartz-Bruhat functions.

We take V “ K a quadratic vector space over Q and VAK “ AQ bQ K. Then we can define theSchwartz-Bruhat functions Φ “

ś

vΦv, Φv P SpVAK q to be:

12

$

’

’

’

’

’

&

’

’

’

’

’

%

Φv “ charOKv , if v - 3D

Φp “ř

pa,Dq“1

charpa`DOKp q, if v|3D, v - 3,OKp “ś

v|pOKv

Φp “ charp1`3OKv q, if v “?´3

Φ8pzq “ ze´πqpdet zq, where z P C

Here qpzq “ |z|2 the usual absolute value on C.

Remark 3.1. charpa`pOKp qpmq “ś

v|p charpa`DOKv qpmq “ś

v|p charp1`pOKv qpa´1mq and each

charp1`pOKv q is a locally constant function with compact support. We are taking a linearcombination of these Schwartz-Bruhat functions, thus we do get a Schwartz-Bruhat function.

3.0.3 Haar measure.

We pick the usual additive Haar measure:#

dˆv xv “dxv|xv |v

, normalized such that volpOˆKv q “ 1, if v - 8dˆz “ dz

|z|C, dz usual Lebesgue measure, z P C, |z|C “ x2 ` y2, for z “ x` yi

We also define the multiplicative Haar measure:XXXXXX

3.1 Zeta functions

We recall Tate’s zeta function. For a Hecke character χv : Kˆv Ñ Cˆ and a Schwartz-Bruhatfunction Φv P SpKvq, it is defined locally to be:

Zvps, χv,Φvq “

ż

Kˆv

χvpαvq|αv|svΦvpαvqd

ˆαv,

where dˆαv is the multiplicative Haar measure defined above.We define globally Zps, χ,Φq “

ś

v Zvps, χv,Φvq. As a global integral, this is:

Zps, χ,Φq “

ż

AˆK

χpαq|α|sΦpαqdˆα,

Lemma 3.1. For all s and Φ Schwartz-Bruhat functions chosen as above, we have:

Lf ps, χDϕq “ Zf ps, χDϕ,ΦqV3D,

where V3D “ volp1` 3Z3rωsq volpZ`Dź

p|D

Zprωsqˆ “1

6

ź

p|D

1

pp´`

p3

˘

q

Proof: From Tate’s thesis, we have Lf ps, χDϕq “ Zf ps, χDϕq

ś

p|3D Lpps, χD,pϕpqś

p|3D Zf ps, χD,pϕp,Φpq.

Since χDϕ is ramified at 3D, we have Lpps, χD,pϕpq “ 1. We need to compute:

13

Zpps, χDϕ,Φpq “

ż

Qprωsˆ

χD,ppαpqϕppαpq|αp|spΦppαpqd

ˆαp

From the definition for p|D, we have Φp “ charpZ`3DZprωsqˆ , the integral reduces to Zpps, χDϕ,Φpq “ż

pZ`3DZprωsqˆ

χD,ppαpqϕppαpq|αp|spdˆαp. Note that for p ‰ 3, all the characters χD, ϕ and | ¨ |p

are unramified, thus we just get the volume vol´

pZ` 3DZprωsqˆ¯

.

For p “ 3, we have Φp “ charp1`3Z3rωsq. Similarly, we get vol´

p1` 3Z3rωsqˆ¯

.We compute the volumes. For D a product of primes, we have

vol´

pZ` 3DZprωsqˆ¯

“ vol´

pZ` pZprωsqˆ¯

“ pp´ 1q vol p1` pZprωsq “1

pp´`

p3

˘

q

Note that vol p1` pZprωsq “ 1p2´1 volpZprωsˆq when p is nonsplit and vol p1` pZprωsq “

1pp´1q2 volpZˆp q2 when p is split. This is computed by writing:

• p nonsplit: volpZprωsˆq “ř

volpa` bω ` pZprωsq, where the sum is taken over all a` bωprime to p and 0 ď a, b ď p ´ 1. We count p2 ´ 1 of them and we get volpZprωsˆq “pp2 ´ 1q volp1` pZprωsq.

• p split: volpZprωsˆq “ř

volpa` bω` pZprωsq. We count similarly p2´ 2p` 1 such terms,as p splits and we have to discard the divisors of p.

For p “ 3, we have vol p1` 3Z3rωsq “15 .

We compute:

• Z3rωs “ Z3r?´3s “ ta0 ` a1

?´3` a2p´3q ` . . . , 0 ď ai ď 2u

• volpZ3rωsqˆ “ 1

• pZ3rωsqˆ “

Ť

pa0 ` a1

?´3qp1` 3Z3rωsq, where a0 ` a1

?´3 is prime to 3. Then we have

6 possibilities and thus volp1` 3Z3rωsq “16 .

By plugging in s “ 1 in the above Lemma, we get:

Corollary 3.1. The finite part of the L-function at s “ 1 equals:

Lf p1, χDϕq “1

6

ź

p|D

1

pp´`

p3

˘

qZf p1, χDϕ,Φq,

3.1.1 Computing the finite part of Tate’s Zeta function Zf ps, χDϕ,Φq

In this section we will compute the value of Zf ps, χDϕ,Φq. We begin by rewriting Tate’s zetafunction Zf ps, χDϕ,Φq as a linear combination of Hecke characters:

14

Lemma 3.2. For all s P C and the Schwartz-Bruhat functions Φf P SpAK,f q, we have:

Zf ps, χDϕ,Φf q “ V3D

ÿ

αfPUp3DqzAˆK,f KÎps, αf ,Φf qχDpαqϕpαq,

where Ips, αf ,Φf q “ř

kPKˆ

k|k|2sC

Φf pkαf q.

Proof: By definition, we have Zf ps, χDϕ,Φf q “ż

AˆK,f

χDpαf qϕpαf q|αf |sfΦf pαf qd

ˆαf . We

rewrite the integral by taking a quotient by Kˆ:

Zf ps, χDϕ,Φf q “

ż

AˆK,f Kˆ

ÿ

kPKˆ

χD,f pkα1f qϕf pkα

1f q|kαf |

sfΦf pkα

1f qd

ˆα1f

Note that from the definition of Hecke characters, we have χD,f pkα1f q “ χ´1D,8pkqχD,f pα

1f q “

χD,f pα1f q, ϕf pkα

1f q “ ϕ´1

8 pkqϕf pα1f q “ kϕf pα

1f q and |kα

1f |sf “ |k|

´s8 |αf |

sf “ |k|

´2sC |α1f |

sf , where

| ¨ |C is the usual absolute value over C. Then the integral reduces to:

Zf ps, χDϕ,Φf q “

ż

AˆK,f Kˆ

˜

ÿ

kPKˆ

k

|k|2sCχD,f pα

1f qΦf pkα

1f q

¸

ϕf pα1f q|α

1f |sf d

ˆα1f

Furthermore, note that our choice of Schwartz-Bruhat functions Φf pkα1f q are invariant over

Up3Dq. Similarly:

• | ¨ |f is trivial on units, thus on Up3Dq

• χD is invariant on Up3Dq by definition

• ϕ is trivial on all the units at all the unramified places. At 3 it is invariant under 1`3Z3rωs,thus it is trivial on all of Up3Dq

Thus we can take the quotient by Up3Dq as well. Note that the integral is now a finite sum:

Zf ps, χDϕ,Φf q “ volpUp3Dqqÿ

α2fPUp3DqzAˆK,f K

ˆ

˜

ÿ

kPKˆ

k

|k|2sCχD,f pα

2f qΦf pkα

2f q

¸

ϕf pα2f q|α

2f |sf

Moreover, note that volpUp3Dqq “ volp1` 3Z3ωqś

p|D

volpZ`DZprωsq “ V3D.

By denoting Ips, αf ,Φf q “ÿ

kPKˆ

k

|k|sCΦf pkαf q, we get the conclusion of the Lemma.

Combining the Lemma ?? and Lemma ??, we get:

Corollary 3.2. For all s P C and the Schwartz-Bruhat functions Φf P SpAK,f q chosen above,we have:

Lf ps, χDϕq “ÿ


15

3.1.2 Adelic representatives for ClpO3Dq

From the Strong approximation theorem, we can write αf P AˆK “ CˆKˆś

v-8OˆKv in the form

αf “ γ8kαβf , where kα P Kˆ, γ8 P Cˆ and βf Pś

v-8OˆKv . Then we can take representatives

in αf P Up3DqzAˆK,f Kˆ such that αf Pś

v-8OˆK,v. Moreover, since we are taking the quotient

by the cube roots of six t˘1,˘ω,˘ω2u, we can pick αf such that α3 ” 1 mod 3. This can bedone by replacing αf by ˘αfωi for some i, 0 ď i ď 2.

Furthermore, note that representatives αf , α1f are in the same class in Up3Dq iff αfα´1f ” a

mod DZprωs, for some integer a such that pa,Dq “ 1.Moreover, we can define an ideal Aα that is generated by kα P OK such that

αp ” kα mod 3DZprωs.

Note that this ideal is unique only as a class in ClpO3Dq.

3.1.3 Connection to the Eisenstein series

Using the above representatives, note that ϕf and | ¨ |f are trivial for the representatives lf andthe Corollary ?? becomes:

Lf ps, χDϕq “ÿ

αfPUp3DqzAˆK,f KÎps, αf ,Φf qχDpαf q

We will now connect Ips, αf ,Φf q to an Eisenstein series. We define the following classicalEisenstein series of weight 1:

Eεps, zq “ÿ

m,n

εpnq

p3mz ` nq|3mz ` n|s,

where the sum is taken over all m,n P Z except for the pair p0, 0q, and ε “`

¨3

˘

is thequadratic character associated to the field extension KQ.

Note that the Eisenstein series does not converge absolutely. However, we can still computeits value at 0 using the Hecke trick in order for it to converge. We will compute its Fourierexpansion in the following section.

Recall that for αf Pś

v-8OˆKv , we have the corresponding ideal class rAαs in ClpO3Dq. Such

a representative is Aαf “ pkαq, where kα P OK is chosen such that kα ” αp mod 3DZprωs forp|3D. Note that we can pick Aα to be a primitive ideal.

We can further write Aα as a Z-lattice Aα “ ra, ´b`?´3

2 sZ, where a “ NmAα and b ischosen (not uniquely) such that b2 ” ´3 mod 4a. Then we can take the corresponding CMpoint zAα :“ ´b`

?´3

2a

Using this notation, we have the following equality:

Lemma 3.3. For αf Pś

v-8OˆKv and any choice of zAα as above, we have:

16

Ips, αf ,Φf q “1

2

pNmAαq1´s

kαEεps, zAαq

Remark 3.2. Note that the variable zAα on the left hand side is not uniquely defined. However,the function is going to be invariant on the class rAαs in ClpO3Dq.

Proof: Recall that Ips, αf ,Φf q “ÿ

kPKˆ

k

|k|2sCΦf pkαf q. We need to compute Φf pkαf q. Note

that for all finite places v we have Φvpkαvq ‰ 0 only for kαv P OKv , and since αv P OˆKv , wemust have k P OKv for all v - 8. This implies k P OK and for all v - 3D we get Φvpkαvq “ 1 fork P OK . Thus we can rewrite:

Ips, αf ,Φf q “ÿ

kPOK

k

|k|2sCΦ3Dpkα3Dq,

where Φ3D “ś

v|3D Φv and α3D “ pαvqv|3D.We can further compute Φvpkαvq for v|3D. Recall that for p|D we defined Φp “ charpZ`3DZprωsqˆ

and Φ3 “ charp1`3Z3rωsqˆ . Then we have Φ3Dpkα3Dq ‰ 0 iff kαp P a` 3DZprωs for some integerpa, pq “ 1 and for p “ 3 we need kα3 P 1` 3OK3

.Recall that we can define kα such that kα ” αp mod 3DZprωs for all p|3D. Then the we

have kkα P a`3DZprωs for pa, pq “ 1 and kkα P 1`3Z3rωs as well. Furthermore, for k P OK weactually have Φ3Dpkα3Dq “ Φ3Dpkkαq. Then we can rewrite Ips, αf ,Φf q using kα in the form:

Ips, αf ,Φf q “ÿ

kPOK

k

|k|2sCΦ3Dpkkαq,

We can rewrite this further:

Ips, αf ,Φf q “|kα|

2sC

kα

ÿ

kPOK

kkα|kkα|2sC

Φ3Dpkkαq,

Finally, we will make this explicit. Note that we must have kkα P Aα, where Aα “ pkαq,we well as kkα P ap ` DZprωs for some integer ap, pap, pq “ 1 as well as kkα P 1 ` 3Z3rωs.By the Chinese remainder theorem, we can find an integer a such that a ” ap mod D anda ” 1 mod 3. Then we have kkα P a ` D

ś

p|3D

Zprωs X OK , thus kkα P PZ,3D X P1,3. Here

PZ,3D “ tk P K : k ” a mod 3DOK for some integer a, pa, 3Dq “ 1u and P1,3 “ tk P K : k ” 1

mod 3u. We rewrite:

Ips, αf ,Φf q “|kα|

2sC

kα

ÿ

kPAαXPZ,DXP1,3

k

|k|2sC,

Finally, we want to write the elements of Aα X PZ,D X P1,3 explicitly.Recall that we can write Aα as a Z-lattice Aα “ ra, b`

?´3

2 s. Then all of the elementsof A are of the form ma ` n b`

?´3

2 for some integers m,n P Z. Moreover, note that theintersection of A and PZ,3D “ tk P OK : k ” n mod 3D, for some integer n, pn, 3Dq “ 1u istma` 3Dn b`

?´3

2 : m,n P Zu. Further taking the intersection with P1,3, we must have ma ” 1,thus we must have m ” 1 mod 3. Thus we can rewrite Ips, αf ,Φf q in the form:

17

Ips, αf ,Φf q “as

kα

ÿ

m,nPZ,m”1pmod 3q

ma` n´b`?´3

2

|ma` n´b`?´3

2 |2sC

,

Here we have also used the fact that |kα|C “ a. Note that we can further rewrite this as:

Ips, αf ,Φf q “ as´1kαÿ

m,nPZ,m”1pmod 3q

1

pma` n´b´?´3

2 q|ma` n´b`?´3

2 |2s´2C

,

Furthermore, by changing nÑ ń and taking out a factor of a1´2s, we have:

Ips, αf ,Φf q “ a´skαÿ

m,nPZ,m”1pmod 3q

1

pm` n b`?´3

2a q|m` n b`?´3

2a |2s´2C

,

Note that for Repsq ą 1 the integral converges absolutely, thus we can rewrite it in the form:


2a´skα

ÿ

m,nPZ,m”1pmod 3q

1

pm` n b`?´3

2a q|m` n b`?´3

2a |2s´2C

`1

2a´skα

ÿ

m,nPZ,m”2pmod 3q

1

p´m` n b`?´3

2a q| ´m` n b`?´3

2a |2s´2C

Changing nÑ ń in the second sum, we get:


2a´skα

ÿ

m,nPZ,m”1pmod 3q

1

pm` n b`?´3

2a q|m` n b`?´3

2a |2s´2C

´

´1

2a´skα

ÿ

m,nPZ,m”2pmod 3q

1

pm` n b`?´3

2a q|m` n b`?´3

2a |2s´2C

Thus we can write for Repsq ą 1 we can rewrite:


2a´skα

ÿ

m,nPZ,m”1pmod 3q

εpmq

pm` n b`?´3

2a q|m` n b`?´3

2a |2s´2C

On the right hand side we can recognize the Eisenstein series Eεp2s´ 2, zAαq, thus we get:


2a´skαEεp2s´ 2, zAαq “

1

2

a1´s

kαEεp2s´ 2, zAαq

By analytic continuation, we can extend the equality to all s P C.

Using this Lemma, we can rewrite the Corollary ?? in the form:

Corollary 3.3. For all s, we have:

Lf ps, χDϕq “1

2

ÿ

APClpO3Dq

Eεp2s´ 2, zAqχDpAqpNmAq1´s

kA,

Proof: Recall that in the Corollary ?? we got

Lf ps, χDϕq “ÿ


18

We can rewrite Ips, αf ,Φf q “ 12a1´s

kαEεp2s ´ 2, zAαq and ϕpαq “ 1, χDpαq “ χDpkαq “

χDpAαq. Then we get:

Lf ps, χDϕq “ÿ

αfPUp3DqzAˆK,f Kˆ

1

2

a1´s

kαEεp2s´ 2, zAαqχDpAαq

Finally, consider A as representatives of ClpO3Dq. Note that by changing A Ñ A we justinvert the classes of ClpO3Dq. Thus we get the result of the Corollary:

Lf ps, χDϕq “ÿ

APClpO3Dq

1

2

a1´s

kAEεp2s´ 2, zAqχDpAq.

3.1.4 Fourier expansion of the Eisenstein series Eεps, zq at s “ 0.

We want to connect the Eisenstein series Eεps, zq to the theta function ΘKpzq. In order to dothis, we will compute the Fourier expansion of Eεps, zq at s “ 0.

We will use the Hecke trick to compute the Fourier expansion of the Eisenstein series:

Eεps, zq “1ÿ

c,d

εpdq

p3cz ` dq|3cz ` d|2s

We will follow closely the proof of Pacetti [12]. This is also done by Hecke in [7]. We rederivethe formula:

E1pz, sq “1ÿ

d

εpdq

d1`2s` 2

8ÿ

c“1

2ÿ

r“0

ÿ

dPZ

εprq

p3cz ` p3d` rqq|3cz ` p3d` rq|2s

We divide by 32s`1 and get:

E1pz, sq “ 2Lpε, 1` 2sq ` 28ÿ

c“1

2ÿ

r“0

εprq

32s`1

ÿ

dPZ

εprq

p 3cz`r3 ` dq| 3cz`r3 ` d|2s

We define for z in the upper-half plane:

Hpz, sq “1ÿ

mPZ

1

pz `mq|z `m|2s

Following Shimura ??, for z “ x` yi and s ą 0 we have the Fourier expansion:

Hpz, sq “8ÿ

n“´8

τnpy, s` 1, sqe2πinx,

where τnpy, s` 1, sqiΓps` 1qΓpsq

p2πq2s`1“

$

’

’

&

’

’

%

n2se´2πnyσp4πny, s` 1, sq, if n ą 0

|n|2se´2π|n|yσp4π|n|y, s, s` 1q, if n ă 0

Γp2sqp4πyq´2s, if n “ 0,

where γpY, α, βq “8ż

0

pt` 1qα´1tβ´1eÝ tdt

19

For any s ą 0, Hpz, sq converges, thus we can compute the limits of each of its Fouriercoefficients:

• n “ 0: limsÑ0

p2πq2s`1

iΓps`1qΓp2sqΓpsq p4πyq

´2s “ ´2πi limsÑ0

Γp2sqΓpsq

• n ă 0: limsÑ0

p2πq2s`1

iΓps` 1qΓpsq|n|2se´2π|n|y

8ż

0

pt`1qs´1tse´4π|n|ytdt “ ´2πie´2π|n|y limsÑ0

1

Γpsq

8ż

0

pt`

1qs´1tse´4π|n|ytdt

• n ą 0: limsÑ0

p2πq2s`1

iΓps` 1qΓpsqn2se´2πny

8ż

0

pt` 1qsts´1e´4πnytdt

(COMPUTATION)

We should get:

limsÑ0

Hps, zq “ ´πi´ 2πi8ÿ

n“1

qn

Finally, note that:

E1ps, zq “ 2Lpε, sq ` 28ÿ

c“1

2ÿ

r“0

εprq

32s`1H

ˆ

3dz ` r

3, s

˙

Using the Fourier expansion of Hpz, sq, we get:


c“1

2ÿ

r“0

εprq

32s`1

ÿ

nPZτnpyn, s` 1, sqe2πin 3xc`r

3

Taking the limit as sÑ 0, and the Fourier expansion above, we get:


c“1

2ÿ

r“0

εprq

3

˜

´πi´ 2πi8ÿ

n“1

e2πinzcωnr

¸

We compute separately the inner sum:

2ÿ

r“0

εprq

3

˜

´πi`8ÿ

n“1

e2πinzcωnr

¸

“ ´2πi8ÿ

n“1

e2πinzcεpnq2ÿ

r“0

ωnrεprnq “ ´2πi

3Gpεq

8ÿ

n“1

e2πinzcεpnq,

where Gpεq “ř2r“0 εprqω

r “?´3 is the Gaussian quadratic sum corresponding to ε.

Then we get:

E1p0, zq “ 2Lpε, 1q ´4πi?´3

3

8ÿ

c“1

8ÿ

n“1

e2πinzcεpnq “ 2Lpε, 1q `4π?

3

3

8ÿ

N“1

¨

˝

ÿ

m|N

εpmq

˛

‚e2πiNz

20

Since ε is a quadratic character, we can compute Lp1, εq “ π?

39 (see Kowalski [?]). This

gives us the Fourier expansion:

E1p0, zq “2π?

3

9

¨

˝1` 68ÿ

N“1

¨

˝

ÿ

m|N

εpmq

˛

‚e2πiNz

˛

‚

3.1.5 Connection to the theta function ΘKpzq.

Recall the theta function ΘK associated to the number field K:

ΘKpzq “ÿ

a,bPZe2πipa2áb`b2qz.

Equivalently, we can rewrite the theta function in the form: ΘKpzq “ 1 ` 6ř

A e2πiNmAz,

where we sum over all ideals A. Thus we have the Fourier expansion for ΘK :

ΘKpzq “ 1` 6ÿ

ně1

cpnqqn,

where cpnq is the number of ideals of norm n. We will show the following version of Siegel-Weil theorem:

Theorem 3.1. For Eεps, zq defined in the previous section and ε the quadratic character cor-responding to to the extension KQ, we have:

Eεp0, zq “ 2Lp0, εqΘKpzq

The proof consists of comparing the Fourier expansions of the two sides. This is mainlygoing to be based on the lemma below:

Lemma 3.4. For n ě 1 then for the ideals in OK we have:ÿ

d|n

εpdq “ #ideals of norm n

Proof: We first show the result for powers of primes pe. We consider three cases:If p ” 1 mod 3, then there are two ideals of norm p: pa ` bωq and pa ´ bωq such that

a2 ´ ab ` b2 “ p. Then we have k ` 1 ideals of norm pk: pa ` bωqipa ` bωqkí for 0 ď i ď k.Moreover, since εppq “ 1, we have p1` εppq ` . . . εppkqq “ k ` 1.

If p ” 2 mod 3, then there are no ideals of norm p. Thus, if k is even, we have exacly oneideal of norm pk: A “ ppk2q. In this case p1 ` εppq ` . . . εppkqq “ 1 ´ 1 ` ¨ ¨ ¨ ` 1 “ 1. If k isodd, we have no ideals of norm p2k`1. Moreover p1` εppq ` . . . εppkqq “ 1´ 1` ¨ ¨ ¨ ´ 1 “ 0.

If p “ 3, then we have exactly one ideal of norm 3k, namely the ideal p?´3

kq. Moreover

εp3q “ 0, thus p1` εp3q ` . . . εp3kqq “ 1.It is easy to extend the result to all integers. As ε is a character, we have:

ÿ

d|n

εpdq “ź

pi|n

p1` εppiq ` ¨ ¨ ¨ ` εppiqciq,

21

where n “ś

i

pcii , ei ě 1 and pi are primes. If we have any ideal A of norm n, then

A “ś

pvpevv , and we must have n “

ś

vNm pevv . Moreover, we have #ideals of norm n “

ś

pi|n

#ideals of norm pNm piqci , which finishes the proof.

We are ready to state the proof of the theorem. Using the above Lemma we can rewrite theFourier expansion of ΘK as:

ΘKpzq “ 1` 68ÿ

N“1

¨

˝

ÿ

m|N

εpmq

˛

‚e2πiNz

Multiplying by a factor of 2π?

39 , we recognize the Eisenstein series Eεp0, zq. Thus it implies

Eεp0, zq “2π?

39 ΘKpzq. Note that this is the same as:

Eεp0, zq “ 2Lp1, εqΘKpzq

3.1.6 Final formula for Lp1, χDϕq

Applying Corollary ?? for s “ 1 we get:

Lf p1, χDϕq “1

2

ÿ

APClpO3Dq

1

kAEεp0, DzAqχ3DpAq

Furthermore, from Theorem 3.1 this is the same as:

Lf p1, χDϕq “π?

3

9

ÿ

APClpO3Dq

1

kAΘKpDzAqχ3DpAq (8)

We need one more step before rewriting the formula as a trace. This is going to be thefollowing lemma:

Lemma 3.5. For A “„

a,´b`

?´3

2

a primitive ideal of norm NmA “ a, with generator

A “ pkAq, where kA ” 1 mod 3, we have:

ΘK

ˆ

´b`?´3

2a

˙

kAa“ ΘK

ˆ

´1`?´3

2

˙

Proof: Since A “”

a, ´b`?´3

2

ı

Zas a Z-lattice, we can write its generator kAin the form

kA “ ma ` 3n´b`

?´3

2for some integers m,n. Moreover, kA “ m ´ 3n

b`?´3

2and

kAa“

m´3nb`

?´3

2a. Moreover, since kA is the generator of a primitive ideal, we have gcdpm, 3nq “ 1.

Then we can find through the Euclidean algorithm integers A,B such thatmA`3nB “ 1, which

makesˆ

A B

´3n m

˙

a matrix in Γ0p3q. Since Θ is a modular form of weight 1 for Γ0p3q, we have:

ΘK

˜

A´b`?´3

2a `B

´3n´b`?´3

2a `m

¸

“

ˆ

m´ 3n´b`

?´3

2a

˙

ΘK

ˆ

´b`?´3

2a

˙

22

Noting that´3n´b`?´3

2a `m “ kAa “ 1kA, we can compute A´b`?´3

2a `B

´3n´b`?´3

2a `m“pA´b`

?´3

2 `BaqkAa .

This is paB À´b`?´3

2 qpma` 3n b`?´3

2 qa. After expanding, we get:

´3nAb2 ` 3

4a` abB `

bp´mA` 3nBq

2`

?´3

2

Note that mA` 3nB “ 1 implies that mA and 3nB have different parities. Also we chose bodd, since b2 ` 3 ” 0 mod 4a. Then we note that ´3nA b2`3

4a ` abB ` bp´mA`3nBq`12 P Z and

thus using the period 1 of ΘK we get:

ΘK

˜

A´b`?´3

2a `B

´3n´b`?´3

2a `m

¸

“ ΘK

ˆ

´1`?´3

2

˙

This finishes the proof.

Note that the Lemma above is equivalent to ΘKpτAq “ kAΘKpωq, where τA “´b`

?´3

2a .Then we can rewrite (??) as:

Proposition 3.1.

Lf pED, 1q “π?

3

9ΘK pωq

ÿ

APClpO3Dq

ΘKpDτAq

ΘKpτAqχ3DpAq (9)

3.1.7 Turning the formula into a trace.

We will rewrite (9) as a trace. First, let fpzq “ΘKpDzq

ΘKpzq. This is a modular function for Γ0p3Dq.

We will prove in the following section ?? the following proposition:

Proposition 3.2. Take A representative ideals for ClpO3Dq. We can take all A to be primitiveand we can write them in the form A “ ra, ´b`

?´3

2 sZ. Then the Galois conjugates of fpωq are:

fpωqσ´1A “

Θ´

D´b`?´3

2a

¯

Θ´

´b`?´3

2a

¯

We will also rewrite the character χD to include a trace. In the Introduction we have alsoshowed in Lemma ?? that pD13qσ

´1A “ D13χDpAq.

Then the formula (9) becomes:

Lf pED, 1q “π?

3

9D´13ΘKpωq

ÿ

APClpO3Dq

ˆ

D13 ΘKpDωq

ΘKpωq

˙σA´1

(10)

Moreover, we also have D13 P H3D. See Cohn [3] for a proof. Thus we can rewrite the sumon the left hand side as TrH3DK

´

D13 ΘKpDωqΘKpωq

¯

. We can compute the extra terms as well.

23

• Rodriguez-Villegas and Zagier in [] cite ΘK

´

´9`?´3

18

¯

“ ´3Γ`

13

˘3p2πq2. We will

use several properties of ΘK proved in section ??. We can rewrite ΘK

´

´9`?´3

18

¯

as

ΘK

´

´3`?´3

18 ´ 13

¯

and using formula ??, we get:

ΘK

ˆ

´3`?´3

18´

1

3

˙

“ p1´ ω2qΘK

ˆ

´3`?´3

6

˙

` ω2ΘK

ˆ

´3`?´3

18

˙

Using ΘK

´

´3`?´3

6

¯

“ 0, we get ΘK

´

´9`?´3

18

¯

“ ω2ΘK

´

´3`?´3

18

¯

.

Furthermore, the functional equation Θp´13zq “ ´?´3zΘpzq for z “ 3`

?´3

2 , we get

´?´3 3`

?´3

2 Θ pωq “ ΘK

´

´3`?´3

18

¯

. Note that´?´3 3`

?´3

2 “ 3ω, thus we get ΘK

´

´9`?´3

18

¯

“

3Θ pωq.

This gives us the value Θ pωq “ Γ`

13

˘3p2πq2

• L8ps, χDϕq “ L8ps, ϕ8q, where ϕ8pzq “ z´1. Then we can compute:

L8ps, ϕ8q “ L8ps´ 12, | ¨ |128 ϕ8q “ 2p2πqsΓpsq.

This gives us L8p1, χDϕq “ 4π.

XXXX Should have 2 instead here.

• The real period ΩD of the elliptic curve ED. The real period of E1 isΓ`

13

˘3

9π(see []). Then

to compute the real period of ED we twist by a factor of D´13 (see []) and get:

ΩD “ D´13 Γ`

13

˘3

18π

Check this.

Multiplying all the terms, we get:

LpED, 1q “ 2π?

3

9D´13 Γ

`

13

˘3

p2πq2TrH3DK

ˆ

D13 ΘKpDωq

ΘKpωq

˙

This gives us the theorem:

Theorem 3.2.

LpED, 1q “

?3Γ

`

13

˘3

18πD´13 TrH3DK

ˆ

D13 ΘKpDωq

ΘKpωq

˙

3.1.8 SD is an integer

In the previous section we showed that SD P K. Note that D13ΘpDωqΘpωq “ D13Θp´D `

DωqΘp´1` ωq “ D13ΘpDωqΘpωq. Thus SD P Q. We would like to show that SD P Z.First we look at the Fourier expansion of fpzq “ ΘpDzqΘpzq:

24

Θpzq “ 1` 6ÿ

NPZě1

cpNqqN ,

where cpNq “ # ideals with norm N in K and, q “ e2πiz. Then we also have the Fourierexpansion of ΘpDzq:

ΘpDzq “ 1` 6ÿ

NPZě1

cpNqqDN ,

By taking their ratio we getΘpDzq

Θpzq“

ÿ

nPZanq

n, an P Z. This is easy to see just by straight

computation. The minimal polynomial of D13fpωq is:ź

APClpO3Dq

pX ´D13χDpAqpfpωqqσAq P Zrω,D13spX, qq

This implies that TrH3DK D13fpωq P Zrω,D13s. We already know that TrH3DK D

13fpωq P Q,thus TrH3DK D

13fpωq P Z.

4 Shimura reciprocity law in the classical setting.

Let F be the field of modular functions over Q. From CM theory (see [], for example), it isknown that if τ P KXH and f P F , then we have fpτq P Kab, where Kab is the maximal abelianextension of K. Shimura reciprocity law gives us a way to compute the Galois conjugates offpτqσ when acting with σ P GalpKabKq. We will follow the exposition of Stevenhagen []. Formore details see Gee [].

We recall that F “Ť

Ně1 FN , where FN is the space of modular functions of level N .Moreover, we can think of FN as the function field of the modular curve XpNq “ ΓpNqzH˚ overQpζN q, where ζN “ e2πiN and H˚ “ H Y P1pQq. We can compute explicitly FN “ Qpj, jN q,where j is the j-invariant and jN pzq “ jpNzq. In particular, we have F1 “ Qpjq.

When working over Q, one has an isomorphism:

GalpFNF1q – GL2pZNZqt˘1u.

More precisely, if we denote by γσ the Galois action corresponding to the matrix γ P

GL2pZNZq under the isomorphism above, it is enough to define the Galois action for SL2pZNZq

and for GN “!

ˆ

1 0

0 d

˙

, d P pZNZqû. We state explicitly the two actions below.

• Action of α P SL2pZNZq on FN .

pfpτqqσα “ fαpτq :“ fpατq,

where α is acting on the upper half plane via fractional linear transformations.

• Action ofˆ

1 0

0 d

˙

P pZNZqˆ on FN . Note that for f P FN we have a Fourier expansion

25

fpzq “ř

ně0anq

nN with coefficients an P QpζN q, q “ e2πiz. If we denote ud :“

ˆ

1 0

0 d

˙

,

then the action of σud is given by

pfpτqqσud “ fudpτq :“ÿ

ně0

aσdn qnN ,

where σd is the Galois action in GalpQpζN qQq that sends ζN Ñ ζdN .

As the restriction maps between the fields FN are in correspondence with the natural mapsbetween the groups GL2pZNZqt˘1u we can take the projective limit to get the isomorphism:

GalpFF1q – GL2ppZqt˘1u.

To further get all the automorphisms of F we need to consider the action of GL2pAQ,f q. Weget the exact sequence:

1 Ñ t˘1u Ñ GL2pAQ,f q Ñ AutpFq Ñ 1

For this to make sense, we need to extend the action from GL2ppZq to GL2pAQ,f q. We dothis by using the action of GL2pQq`:

• Action of α P GL2pQq` on F .

fαpτq “ fpατq,

where α acts by fractional linear transformations.

We extend the action of GL2ppZq to GL2pAQq by writing the elements γ P GL2pAQq in theform γ “ uα, where u P GL2ppZq and α P GL2pQq`. Note that this decomposition is not uniquelydetermined. However, by combining the two actions of u and α, a well defined action is givenby:

fuα “ pfuqα.

We want to look at the action of GalpKabKq inside AutpFq. From class field theory wehave the exact sequence:

1 Ñ Kˆ Ñ AˆK,fr¨,KsÝÝÝÑ GalpKabKq Ñ 1,

where r¨,Ks is the Artin map.We are going to embed AˆK,f into GL2pAQ,f q such that the Galois action of AˆK,f through

the Artin map and the action of the matrices in GL2pAQ,f q are compatible. We do this byconstructing a matrix gτ pxq for the idele x P AˆK,f .

Let O be the order of K generated by τ i.e. O “ Zrτ s. We define the matrix gτ pxq to be

the unique matrix in GL2pAQq such that xˆ

τ

1

˙

“ gτ pxq

ˆ

τ

1

˙

. We can compute it explicitly. To

do that, consider the minimal polynomial of τ :

ppXq “ X2 `BX ` C

26

Then if we write xp P Qˆp in the form xp “ spτ ` tp P Qˆp with sp, tp P Qp, we can compute:

gτ pxpq “

ˆ

tp ´ spB ´spC

sp tp

˙

Shimura reciprocity law is going to make the following diagram commute:

1 Kˆ AˆK,f GalpKabKq 1

1 t˘1u GL2pAQ,f q AutpFq 1

gτ

r¨,Ks

We make the statement explicit below:

Theorem 4.1. (Shimura reciprocity law) For f P F and x P AˆK,f , we have:

pfpτqqrx,Ks “ fgτ px´1qpτq,

where rx,Ks is the Galois action corresponding to the idele x via the Artin map, gτ is definedabove and the action of gτ pxq is the action in GL2pAQ,f q.

Remark 4.1. Note that the elements of Kˆ have trivial action. This can be easily seen byembedding Kˆ ãÑ GL2pQq` given by k ãÑ gτ pkq. Noting that τ is fixed by the action of thetorus Kˆ, we have:

fgτ pk´1qpτq “ fpgτ pk

´1qτq “ fpτq

Remark 4.2. We can also rewrite the theorem for ideals in K. Let f P FN and O “ Zrτ s ofconductor M . Going through the Artin map, we can restate Shimura reciprocity in this case inthe form:

fpτqσA “ fgτ pAq´1

pτq, (11)

where σA is the Galois action corresponding to the ideal A through the Artin map and

gτ pAq :“ gτ ppαqp|NmpAqq.

Note that gτ pAq is unique up to multiplication by roots of unity in K. However, these havetrivial action on f . This can be easily seen by multiplying by an element of p˘ωjqv P Kˆ andnoticing that we get trivial action at the unramified places p -MN .

Remark 4.3. Note that the action of gτ pAq is the same as the action of gτ ppαqp|MN q´1.

Remark 4.4. Note that the maps above are based on the map between the ideals A prime toMN and the ideles:

27

IpMNq Ñ AˆK,fKˆ

A “ź

v

pevv Ñ p$vqevv ,

where $v is the uniformizer of the ideal pv at the place v - 8.

4.1 Applying Shimura reciprocity law to K “ Qr?´3s.

Lemma 4.1. The function fpzq “ΘKpDzq

ΘKpzqis a modular function of level 3D with integer

Fourier coefficients at the cusp 8.

Proof: Since ΘKpzq is a modular form of weight 1 for Γ0p3q, it can be easily seen thatΘpDzq is a modular form of weight 1 for Γp3Dq. Furthermore, their ratio is modular function

for Γ0p3Dq. We check this below. For γ “ˆ

a b

c d

˙

P Γp3Dq, we have:

fpγzq “

Θ

ˆˆ

D 0

0 1

˙ˆ

a b

c d

˙

z

˙

Θ

ˆˆ

a b

c d

˙

z

˙“

Θ

ˆˆ

a bD

cD d

˙

pDzq

˙

Θ

ˆˆ

a b

c d

˙

z

˙“pcz ` dqΘpDzq

pcz ` dqΘpzq“ fpzq

To find the Fourier expansion of fpzq at 8, it is enough to write the Fourier expansions ofΘpDzq and Θpzq:

ΘpDzq

Θpzq“

1`ř

Ně1

cpNqqND

1`ř

Ně1

cpNqqN“

ÿ

Mě0

aMqM

We can compute the Fourier coefficients explicitly from the equality:

1`ÿ

Ně1

cpNqqND “ p1`ÿ

Ně1

cpNqqN qpÿ

Mě0

aMqM q

Note that we have a0 “ 1 and aM “ áM´1cp1q ´ aM´2cp2q ´ ¨ ¨ ¨ ´ a1cpM ´ 1q ´ a0cpMq

if D -M and aM “ cpMDq ´ aM´1cp1q ´ aM´2cp2q ´ ¨ ¨ ¨ ´ a1cpM ´ 1q ´ a0cpMq if D|M . Byinduction, since cpNq P Z, we get all the coefficients aM P Z.

4.1.1 fpωq is in the ring class field H3D.

From CM-theory, we have that if f P F3D and τ generating OK , we have fpτq P H3D,OK the rayclass field of conductor 3D. We claim that fpωq P H3D. Recall that we have GalpKabH3Dq –

Up3DqzAˆK,fKˆ. Thus in order to show that fpωq P H3D, we need to check that fpωq isinvariant under the action of Up3Dq.

Lemma 4.2. For ω “ ´1`?´3

2 and fpzq “ ΘKpDzqΘKpzq

we have fpωq P H3D.

28

Proof: In order for fpωq P H3D, we need to show that it is invariant under GalpKabH3Dq.Using Shimura reciprocity law, we need to show:

fpωq “ frωpsqpωq,

for all s P KÛp3Dq. From Remark 4.1, the action of Kˆ is trivial. Thus it is enough to showthe result for all elements l “ pAp ` Bpωqp P Up3Dq. By the definition of Up3Dq, this impliesthat Ap `Bpω P pZprωsqˆ for all p and A3 ” 1 mod 3, B3 ” 1 mod 3, Bp ” 0 mod D for allp|D. Since the action for p - 3D is trivial, s has the same action lD “ pAp `Bpωqp|3D P Up3Dq.Moreover, this has the same action as l0 “ pA ` Bωqp|3D, where A ` Bω P OK and A ” Apmod 3DZp and B ” Bp mod 3DZp for all p|3D.

Note further that we can pick A,B such that pA ` Bωq generates a primitive ideal A inOK . Moreover, from above we have 3D|B and A ” 1 mod 3. Recall that we can rewrite anyprimitive ideal in the form A “ ra, ´b`

?´3

2 sZ, where a “ NmA and b2 ” ´3 mod 4a. Thenthe generator is A`Bω “ ta` s´b`

?´3

2 for t, s P Z, 3D|s.Now observe that fpωq “ fpτq, where τ “ ´b`

?´3

2 , thus from Shimura reciprocity law, wehave:

pfpτqqσl´1 “ frτ plqpωq.

Here rτ plq “´

Ap´bBp ´BpcBp Ap

¯

pand rτ plq has the same action as rτ pl0q, where l0 “ pA `

Bωqp|3D and A`Bω “ ta` s´b`?´3

2 . Then we need to compute the action of:

pfpτqqσl´1 “ frτ pl0qpτq.

Note that rτ pl0q “`

ta´sb ´scs ta

˘

p|3D, where c “ b2`3

4 . Then we can rewrite the action ofrτ pl0q:

frτ pl0qpτq “ f

´

ta´sb ´scas t

¯

p|3Dp 1 0

0 a qp|3D pτq “ fp 1 0

0 a qp|3D p`

ta´sb ´scas t

˘

τq

Since a|c, the matrix`

ta´sb ´scas t

˘

P SL2pZq and we can rewrite:

fp`

ta´sb ´scas t

˘

zq “ΘK

`

pD 00 1 q

`

ta´sb ´scas t

˘

z˘

ΘK

``

ta´sb ´scas t

˘

z˘ . “

ΘK

´´

ta´sb ´scDasD t

¯

pDzq¯

ΘK

``

ta´sb ´scas t

˘

z˘ .

Note that since 3D|s, we actually have´

ta´sb ´scDasD t

¯

,`

ta´sb ´scas t

˘

P Γ0p3q and we canapply the properties of the modular form ΘK :

ΘK

´´

ta´sb ´scDasD t

¯

pDzq¯

ΘK

``

ta´sb ´scas t

˘

z˘ . “

psz ` tq´1ΘK pDzq

psz ` tq´1ΘK pzq. “ fpzq

Finally, note that since pa, 3Dq “ 1 and f has rational coefficients, the action of p 1 00 a qp|3D

is trivial. This finishes the proof that fpωq is invariant under the Galois action coming fromUp3Dq, thus fpωq P H3D.

29

Remark 4.5. A different proof is shown in the in Appendix ??, where we reinterpret theclassical Shimura reciprocity law in the setting of Shimura curves following Hida [].

4.1.2 Galois conjugates of fpωq.

Let A “”

a, ´b`?´3

2

ı

Zbe a primitive ideal prime to 3D. For τ1 “ ´b`

?´3

2 , let OD “ Z`DτZ.

Lemma 4.3. Let f P FN be a modular function of level N with rational Fourier coefficients inits Fourier expansion. Let τ1 “ ´b`

?´3

2 be a CM point and let A “”

a, ´b`?´3

2

ı

be a primitiveideal prime to N . Then we have the Galois action:

fpτqσ´1A “ fpτaq

Proof: From Shimura reciprocity (11), we have:

fpτqσ´1A “ fgτ pAqpτq.

Note that the minimum polynomial of τ is pτpXq “ X2 ` bX ` b2`34 . Now let α “ ta `

s´b`?´3

2 “ ta ` sτ be a generator of A. Then we have gτ pAq “´

ta´sb ´s b2`34

´s ta

¯

p|a. We can

rewrite the matrix in the form:

gτ pAq “´

ta´sb b2`34a

´s t

¯

p|ap 1 0

0 a qp|a

As´

ta´sb ´ b2`34a

´s t

¯

p|aP SL2pZpq for p - ND, it has a trivial action. Then:

fgτ pAqpτq “ fp 1 0

0 a qp|apτq

We rewrite the matrix p 1 00 a qp|a “ p 1 0

0 a qp-a p1 00 a qQ, where

`

1 00 1a

˘

pP GL2ppZq and p 1 0

0 a qQ P

GL2pQq`.Note that the action of

`

1 00 1a

˘

pis only given by

`

1 00 1a

˘

p|NM. However, since f has rational

Fourier coefficients in its Fourier expansion, this action is trivial. Thus we are left with:

fgτ pAqpτq “ fp 1 0

0 a qQpτq

This is just fgτ pAqpτq “ fpτaq.

Lemma 4.4. Take the primitive ideals A “”

a, ´b`?´3

2

ı

Zto be the representatives of the ring

class field H3D such that all norms NmA are relatively prime to each other and b2 ” ´3 mod 4a

for all the a “ NmA chosen.

Then the only Galois conjugates of fpωq “ΘKpDωq

ΘKpωqare the following:

ˆ

ΘKpDωq

ΘKpωq

˙σ´1A

“ΘK

´

D´b`?´3

2a

¯

ΘK

´

´b`?´3

2a

¯

30

Proof: Note thatΘKpDωq

ΘKpωq“

ΘK

´

D´b`?´3

2

¯

ΘK

´

´b`?´3

2

¯ and apply lemma 4.3 to τ “ ´b`?´3

2 and

fpzq “ΘKpDzq

ΘKpzq. These are the only Galois conjugates from Lemma ??.

5 Writing SD as a square.

In this section we will show the following result:

Theorem 5.1. For D “ś

pi”1 mod 3

peii , let τ “´b`

?´3

2 such that b2 ” ´3 mod 12D2. More-

over, let b˚ ” b´1 mod D.Let HO be the ray class field of conductor 3D and let H0 Ă HO be the subfield of HO that is

the fixed field of G0 “ tr P pZDZqˆ, r ” 1 mod 6 : A˝r “´

1` b˚p1´ rq´b`?´3

2

¯

u. Then wehave

SD “ |TrHOH0pf1pτqD

23q|2

and SD P Z.

The main tool in proving Theorem 5.1 is a Factorization Formula of Rodriguez-Villegasand Zagier [13]. We will apply the Factorization Formula 12 to the formula for the L-functionLpED, 1q in Theorem ??.

5.1 Factorization Formula

We recall the version of Factorization Formula ([13], Theorem, page 7) simplified to the case ofα “ p “ 0:

Theorem 5.2. (Factorization formula.) For a P Zą0, µ, ν P Q, z “ x ` yi P C andQzpm,nq “

|mzń|2

2y , we have:

ÿ

m,nPZe2πipmν`nµqeπpimn´

|mzń|2

2y qa“a

2ayθ

„

aµ

ν

pa´1zq ¨ θ

„

µ

áν

pázq, (12)

where θ„

µ

ν

pzq “ř

nPZ`µeπin

2z`2πiνn is a theta function of half integral weight.

Using the formula above, we will prove the following Proposition:

Proposition 5.1. For a ” 1 mod 6, D ” 1 mod 6 and b2 ” ´3 mod 4D2a2a1, b ” 1

mod 16, we have:

3

2Θ

ˆ

D´b`

?´3

2a

˙

´1

2Θ

ˆ

D´b`

?´3

6a

˙

“ÿ

rPZDZ

4?

3

D?a1eπipa´1q6θar

ˆ

´b`?´3

2a2a1

˙

θr

ˆ

b`?´3

2a1

˙

,

(13)

31

where θspzq “ř

nPZeπipn`sD´16q2zp´1qn is a theta function of weight 12 for s non-negative

integer. Here we use the notation r P ZDZ to mean any representatives r for the residuesmod D.

We start by applying the Factorization Formula (12) several times for µ :“ µ`rD , 0 ď r ď D´1

and z :“ zD. Summing up the formulas, we are going to get:

Lemma 5.1. We have the following factorization lemma:

ÿ

rPZDZ

?2ay?D

θ

„

apµ` rqD

ν

´

Dz

a

¯

θ

„

pµ` rqD

áν

páDzq “ÿ

m,nPZe2πipmν`nµqeπpmni´

|n´mz|2

2y qDa

Proof: Plugging in µ :“ µ`rD , z :“ zD in 12, we get:

a

2ayθ

«

a pµ`rqD

ν

ff

´z

a

¯

θ

«

pµ`rqD q

áν

ff

pázq “ÿ

m,nPZe2πipmν`npµ`rqDqeπpmni´

|n´mz|2

2y q 1a

We sum from r in ZDZ:

ÿ

rPZDZ

a

2ayθ

«

a pµ`rqD

ν

ff

´z

a

¯

θ

«

pµ`rqD q

áν

ff

pázq “ÿ

rPZDZ

ÿ


|n´mz|2

2y q 1a

We change the two sums on the RHS:

ÿ

rPZDZ

ÿ


|n´mz|2

2y q 1a “ÿ

m,nPZ

ÿ

rPZDZe2πipmν`npµ`rqDqeπpmni´

|n´mz|2

2y q 1a

Note that the LHS can be rewritten asÿ

m,nPZe2πipmν`npµqDqeπpmni´

|n´mz|2

2y q 1aÿ

rPZDZe2πinrD

and note further that:

ÿ

rPZDZe2πinrD “

D´1ÿ

r“0

e2πinrD “

#

0, for D - nD, for D|n

Thus we are only summing over the n’s that are multiples of D:

ÿ

rPZDZ

ÿ


|n´mz|2

2y q 1a “ Dÿ

m,n1PZe2πipmν`n1pµ`rqqeπpmn

1i´|n1´mpzDq|2

2pyDq qDa

Going back to our initial equality, we can replace z1 “ zD and get:

a

2aDy1ÿ

rPZDZθ

«

a pµ`rqD

ν

ff

ˆ

Dz1

a

˙

θ

«

pµ`rqD q

áν

ff

`

áDz1˘

“ Dÿ

m,n1PZe2πipmν`n1pµ`rqqe

πpmn1i´|n1´mz1|2

2y1qDa

32

Corollary 5.1. Another version of the factorization lemma above is:

ÿ

rPZDZ

?2ay?D

θ

„

aµ` arD

ν

´

Dz

a

¯

θ

„

µ` rD

áν

páDzq “ÿ

m,nPZe2πipmν`nDµqeπpmni´

|n´mz|2

2y qDa

(14)

Proof: We apply the previous factorization lemma for µ :“ Dµ.

We will apply Corollary 5.1 for µ “ ´16 and ν “ 12, D odd, z “ ´b`?´3

2aa1, where b2 ” ´3

mod 4a2a1D and b ” 1 mod 16. This gives us:

ÿ

rPZDZ

a

2ayDθ

„

á6 `

arD

12

ˆ

D´b`

?´3

2a2a1D

˙

θ

„

´ 16 `

rD

á2

ˆ

Db`

?´3

2Da1

˙

“

“ÿ

m,nPZe2πipm2ńD6qe

πpmni´|nDaa1´m

´b`?´3

2|2

Daa1?

3qDa

We will analyze firs the LHS of the equation. Note that from the definition of θ„

µ

ν

we have:

θ

„

´ 16 `

rD

á2

pzq “ÿ

nPZeπipn`

rD´

16 q

2zeáπipn`

rD´

16 q “ eáπirDeaπi6θrpzq.

Similarly, as a ” 1 mod 6, we have θ„

á6 `

arD

12

pzq “ θ

„

´ 16 `

arD

12

pzq. Then from the

definition of θr we get:

θ

„

á6 `

arD

12

pzq “ÿ

nPZeπipn`

arD ´

16 q

2zeπipn`

arD ´

16 q “ eπiarDe´πi6θarpzq

Also, since D ” 1 mod 6 we also have: e´2πinD6 “ e´2πin6. We can also compute:?

2ay?D

“

b

2a?

32aa1D?D

“

4?

3

D?a1

. Thus we can rewrite the formula:

4?

3

D?a1

ÿ

rPZDZeπipa´1q6θar

ˆ

´b`?´3

2a2a1

˙

θr

ˆ

b`?´3

2a1

˙

“ÿ

m,nPZe2πipm2ń6qe

πpmni´|nDaa1´m

´b`?´3

2|2

Daa1?

3qDa

(15)Now we are going to analyze below the RHS of the equation (15):

ÿ

m,nPZe2πipm2ń6qe

πpmni´|nDaa1´m

´b`?´3

2|2

Daa1?

3qDa

First note that we have the following lemma:

Lemma 5.2. For b ” 1 mod 16, b ” 0 mod 3, b2 ” ´3 mod 4a2a1D, we have:

e2πipm2`n2qeπpmni´

|naa1D´m´b`

?´3

2|2

aa1D?

3qDa “ e2πi

|naa1D´m´b`

?´3

2|2

aa1DD´b`

?´3

6a

33

Proof: We only need to show that:

2πi

ˆ

m

2`n

2`Dmn

2a

˙

” ´2πi|naa1D ´m

´b`?´3

2 |2

aa1DDb

6amod 2πiZ.

After dividing by 2πi, we compute the RHS of the identity:

|naa1D ´m´b`

?´3

2 |2

aa1DDb

6a“

ˆ

Dm2 bpb2 ` 3q

24a2a1´D

b2mn

6a`Dba1n

2

6

˙

Thus our claim turns into:ˆ

m

2`n

2`Dmn

2a

˙

”

ˆ

Dm2 bpb2 ` 3q

24a2a1´D

b2mn

6a`Dba1n

2

6

˙

mod Z

Equivalently:

m

2`n

2”

ˆ

Dm2

2

b

3

pb2 ` 3q

4a2a1´D

pb2 ` 3qmn

6a`n2

2

b

3Da1

˙

mod Z

We have b2 ” ´3 mod 4aa21, b ” 1 mod 16, b ” 0 mod 3. Note that this implies that b is

odd and that b2 ` 3 ” 4 mod 8, as well as b2 ` 3 ” 0 mod 3. Then, since a, a1, D are odd, weget:

• m2 ” m22 ” Dm2

2

b

3

pb2 ` 3q

4a2a1mod Z

• n2 ” n22 ”n2

2

b

3Da1 mod Z

• ´D pb2 ` 3qmn

6aP Z

This finishes the proof.

Lemma 5.3. Under the same conditions as above we have:

ÿ

m,nPZe2πin3e2πi

|m¨b`?´3

2`naa1|

2

aa1¨z“

3

2Θp3zq ´

1

2Θpzq,

where z P H, AA1 “ raa1,´b`

?´3

2 s and b ” 0 mod 3, b2 ” ´3 mod 4aa1.

Proof: Note first that by changing mÑ ´m and ´m ¨ ´b`?´3

2 ` naa1 to its conjugate, we

haveÿ

m,nPZe2πin3e2πi

|´m¨´b`

?´3

2`naa1|

2

aa1z“

ÿ

m,nPZe2πin3e2πi

|m¨b`?´3

2`naa1|

2

aa1z.

We can split the sum in three terms, depending on n mod 3:ř

m,3|nPZe

2πi|m¨b`?´3

2`naa1|

2

aa1¨z`ω

ř

m,nPZ,n”1 mod 3

e2πi|m¨

b`?´3

2`naa1|

2

aa1¨z`ω2

ř

m,nPZ,n”2 mod 3

e2πi|m¨

b`?´3

2`naa1|

2

aa1¨z

Note that the first term equalsř

m,nPZe2πi

|m¨b`?´3

2`n3aa1|

2

3aa1¨3z“ ΘKp3zq.

34

Also note that the two termsř

m,nPZ,n”1 mod 3

e2πi|m¨

b`?´3

2`naa1|

2

aa1¨z and

ř

m,nPZ,n”2 mod 3

e2πi|m¨

b`?´3

2`naa1|

2

aa1¨z

equal each other, by changing in the latter nÑ ń and mÑ ´m. Thus we got so far:

Θp3zq ` pω ` ω2qÿ

m,nPZ,n”1 mod 3

e2πi|m¨

b`?´3

2`naa1|

2

aa1¨z

Furthermore, we have:

ÿ

m,nPZ,n”1 mod 3

e2πi|m¨

b`?´3

2`naa1|

2

aa1¨z“

1

2

ÿ

m,nPZ,pn,3q“1

e2πi|m¨

b`?´3

2`naa1|

2

aa1¨z

Finally, this is just:

1

2

ÿ

m,nPZe2πi

|m¨b`?´3

2`naa1|

2

aa1¨z´

1

2

ÿ

m,nPZe2πi

|m¨b`?´3

2`3naa1|

2

3aa1¨3z“

1

2pΘpzq ´Θp3zqq

Finally, we getř

m,nPZe2πin3e2πi

|m¨b`?´3

2`naa1|

2

aa1¨z“ Θp3zq ´ 1

2 pΘpzq ´ Θp3zqq “ 32Θp3zq ´

12Θpzq.

From the previous two lemmas, we get the following corollary:

Corollary 5.2. Under the above conditions, we have:

ÿ

m,nPZe2πipm2ń6qe

πpmni´|naa1D´m

´b`?´3

2|2

aa1D?

3qDa “

3

2Θ

ˆ

D´b`

?´3

2a

˙

´1

2Θ

ˆ

D´b`

?´3

6a

˙

Proof: Note that we can rewrite the LHS in the form:

ÿ

m,nPZe2πipm2ń6qe

πpmni´|naa1D´m

´b`?´3

2|2

aa1D?

3qDa “

ÿ

m,nPZe2πipm2ń2`n3qe

πpmni´|naa1D´m

´b`?´3

2|2

aa1D?

3qDa

Then, from Lemma 5.2, we have:

ÿ

m,nPZe2πipm2ń6qe

πpmni´|naa1D´m

´b`?´3

2|2

aa1D?

3qDa “

ÿ

m,nPZe2πin3e

|naa1D´m´b`

?´3

2|2

aa1DqD´b`

?´3

6a

Now apply Lemma 5.3 for z “ D´b`?´3

6a , we get:

ÿ

m,nPZe2πin3e

|naa1D´m´b`

?´3

2|2

aa1DD´b`

?´3

6a “3

2Θ

ˆ

D´b`

?´3

2a

˙

´1

2Θ

ˆ

D´b`

?´3

6a

˙

Finally, from (15) and Corollary 5.2 we get the result of Proposition 5.1:

35

3

2Θ

ˆ

D´b`

?´3

2a

˙

´1

2Θ

ˆ

D´b`

?´3

6a

˙

“

4?

3

D?a1

ÿ

rPZDZeπipa´1q6θar

ˆ

´b`?´3

2a2a1

˙

θr

ˆ

b`?´3

2a1

˙

.

A particular case of Proposition 5.1 is going to be the following result:

Corollary 5.3. For b2 ” ´3 mod 12a2a1, b ” 1 mod 16, we have:

3

2Θ

ˆ

´b`?´3

2a

˙

“

4?

3?a1eπipa´1q 16 θ0

ˆ

´b`?´3

2a2a1

˙

θ0

ˆ

b`?´3

2a1

˙

,

where θ0pzq “ř

nPZeπipn´16q2zp´1qn.

Proof: Applying the Proposition 5.1 for D “ 1 we get:

3

2Θ

ˆ

´b`?´3

2a

˙

´1

2Θ

ˆ

´b`?´3

6a

˙

“

4?

3?a1eπipa´1q 16 θ0

ˆ

´b`?´3

2a2a1

˙

θ0

ˆ

b`?´3

2a1

˙

.

Furthermore, using the result from Appendix A, Lemma 6.3 that Θ´

´b`?´3

6a

¯

“ 0, we get theresult of the Corollary.

We can rewrite further the ratios:

Corollary 5.4. Under the same conditions as above, we have:

Θ´

D´b`?´3

2a

¯

Θ´

´b`?´3

2a

¯ ´1

3

Θ´

D´b`?´3

6a

¯

Θ´

´b`?´3

2a

¯ “ÿ

rPZDZ

θar

´

´b`?´3

2a2a1

¯

θr

´

b`?´3

2a1

¯

θ0

´

´b`?´3

2a2a1

¯

θ0

´

b`?´3

2a1

¯

Proof: We begin by writing the ratio of the formulas in Proposition 5.1 and Corollary 5.3:

Θ´

D´b`?´3

2a

¯

Θ´

´b`?´3

2a

¯ ´1

3

Θ´

D´b`?´3

6a

¯

Θ´

´b`?´3

2a

¯ “

4?3a1eπipa´1q6

ř

rPZDZθar

´

´b`?´3

2a2a1

¯

θr

´

b`?´3

2a1

¯

4?3a1eπipa´1q6θ0

´

´b`?´3

2a2a1

¯

θ0

´

b`?´3

2a1

¯

Simplifying, we get the result of the Corollary.

Remark 5.1. Note that θ0pzq “ ηpz3q, where η is the Dedekind eta function.

5.2 Ratios of θr and θ0

Now we will apply the Factorization Lemma once more to connect the theta functions θr to thetheta function θ0. We do this by applying the Factorization Formula (12) twice and comparingthe results.

Note first that any primitive ideal A in OK prime to 6 has a generator pnaa`ma´b`

?´3

2 q

such that a “ NmpAq, b2 ” ´3 mod 12a and na ” 1 mod 3. Moreover, note that a “n12a a

2 `m2ab2`3

4 ´manaab, thus manab ” 1 mod a, as a|pb2 ` 3q4.Using this notation, we have:

36

Lemma 5.4. For b ” 0 mod 3, b2 ” ´3 mod 4D2aa1, na1 ” 1 mod 3, we have:

θr

ˆ

´b`?´3

2aa1

˙

θ0

ˆ

b`?´3

2D2aa1

˙

“1?a1θna1r

ˆ

´b`?´3

2a

˙

θ0

ˆ

b`?´3

2D2a

˙

Proof: We write the generator of A1 in the form pna1a1 ` ma1

´b`?´3

2 q, where b2 ” ´3

mod 4aa1D2. Moreover, we can pick na1 ” 1 mod 3. Then, using the Factorization Formula(12) for µ “ ´ 1

6 `rD , ν “ 1

2 , a :“ D and z “ ´b`?´3

2aa1D , we have:

4?

3

2?aa1

θ

„

´ 16 `

rD

D2

ˆ

D´b`

?´3

2aa1D

˙

θ

„

´D6

´12

ˆ

b`?´3

2D2aa1

˙

“ÿ

m,n

e2πinrD e2πin3 e2πipmn2D `m2 `

n2 qe2πi

|m´b`

?´3

2`naa1D|2

aa1D

?´3

6D

Note that on the LHS we have θ„

´ 16 `

rD

D2

pzq “ e´πi6eπirθrpzq and θ„

´D6

´12

“ eπi6θrpzq.

Furthermore, using Lemma 5.2, the RHS equals:

ÿ

m,n

e2πinrD e2πin3 e2πi|m´b`

?´3

2`naa1D|2

aa1D

´b`?´3

6D .

Thus we got:

4?

3

2?aa1

eπirθr

ˆ

D´b`

?´3

2aa1D

˙

θ0

ˆ

b`?´3

2D2aa1

˙

“ÿ

m,n


?´3

2`naa1D|2

aa1D

´b`?´3

6D .

(16)Note that if we write any element of AA1D, we can write it as an element of AD multiplied

by the generator of A1. Thus if we write an element of AA1D, in the form m´b`?´3

2 `naa1D, itis going to equal an element m0

´b`?´3

2 ` n0aD P AD times the generator ma1´b`

?´3

2 ` na1a1

of A1:

m´b`

?´3

2` naa1D “ pm0

´b`?´3

2` n0aDqpma1

´b`?´3

2` na1a

1q

This gives us:#

m “ m0ma1 ` n0na1 ´m0ma1b

n “ n0na1 ´m0ma1b2`34aa1D

Since b2 ` 3 ” 0 mod 4D2, it implies that n ” n0na1 mod D. Then we have:

ÿ

m,n


?´3

2`naa1D|2

aa1D

´b`?´3

6D “ÿ

m0,n0

e2πin0na1

r

D e2πin0na1

3 e2πi|m0

´b`?´3

2`n0aD|

2

aD´b`

?´3

6D

Since we picked na1 ” 1 mod 3, this is the same asÿ

m0,n0

e2πin0na1

r

D e2πin03 e2πi

|m0´b`

?´3

2`n0aD|

2

aD´b`

?´3

6D .

Then applying the Factorization Formula (12) again for µ :“ ´ 16 `

n1arD , ν :“ 1

2 , a :“ D and

37

z :“ ´b`?´3

2aD , we get:

ÿ

m0,n0

e2πin0na1

r

D e2πin03 e2πi

|m0´b`

?´3

2`n0aD|

2

aD´b`

?´3

6D “

4?

3

2?aθ

«

´ 16 `

n1arD

D2

ff

ˆ

D´b`

?´3

2aD

˙

θ

„

´D6

12

ˆ

´b`?´3

2D2a.

˙

Moreover, on the RHS we have the theta functions θ

«

´ 16 `

n1arD

D2

ff

pzq “ e´πi6eπin1arθrpzq

and θ„

´D6

12

pzq “ eπi6θ0pzq. Thus we can rewrite the equality as:

ÿ

m0,n0

e2πin0na1

r

D e2πin03 e2πi

|m0´b`

?´3

2`n0aD|

2

aD´b`

?´3

6D “

4?

3

2?aeπin

1arθn1ar

ˆ

D´b`

?´3

2aD

˙

θr

ˆ

´b`?´3

2D2a.

˙

(17)Comparing the two relations (16) and (17), we get:

1?a1eπirθr

ˆ

´b`?´3

2aa1

˙

θ0

ˆ

b`?´3

2D2aa1

˙

“ eπin1arθna1r

ˆ

´b`?´3

2a

˙

θ0

ˆ

b`?´3

2D2a

˙

Lemma 5.5. Under the same conditions as above, we have:

eπirθr

´

´b`?´3

2aa1

¯

θ0

´

´b`?´3

2aa1D2

¯ “eπina1rθna1r

´

´b`?´3

2a

¯

θ0

´

´b`?´3

2aa1D2

¯

Proof: Note that from Corollary 5.3, we have 32Θ

´

´b`?´3

2

¯

“4?3

D?aa1θ0

´

´b`?´3

2aa1D2

¯

θ0

´

b`?´3

2aa1D2

¯

.

Moreover, we also have from the same corollary that 32Θ

´

´b`?´3

2

¯

“4?3D?aθ0

´

´b`?´3

2aD2

¯

θ0

´

b`?´3

2aD2

¯

,thus:

1?a1θ0

ˆ

´b`?´3

2aa1D2

˙

θ0

ˆ

b`?´3

2aa1D2

˙

“ θ0

ˆ

´b`?´3

2aD2

˙

θ0

ˆ

b`?´3

2aD2

˙

Recall from the previous Lemma that we also have:

1?a1eπirθr

ˆ

´b`?´3

2aa1

˙

θ0

ˆ

b`?´3

2aa1D2

˙

“ eπina1rθna1r

ˆ

´b`?´3

2a

˙

θ0

ˆ

b`?´3

2aD2

˙

Dividing the two relations, we get exactly:

eπirθr

´

´b`?´3

2aa1

¯

θ0

´

´b`?´3

2aa1D2

¯ “

eπina1rθna1r

´

´b`?´3

2a

¯

θ0

´

´b`?´3

2aD2

¯

38

5.3 Applying the factorization lemma to get a square

We would like to apply the factorization lemma for the formula in Theorem ?? for certain idealsthat are representatives of the ring class field ClpO3Dq. We will pick this ideals below.

5.3.1 Representatives of ClpO3Dq.

Recall that, using Cox [4], for the ideal class group of conductor 3D, we have:

ClpO3Dq “ pO3D3DOKqˆpZ3DZqˆpOˆKt˘1uq

Moreover,we can compute explicitly that for D “ś

pi”1 mod 3

pi we have ClpO3Dq – pZDZqˆ

which also gives us # ClpO3Dq “ φpDq, where φ is Euler’s totient function.Furthermore, we are claiming that we can take as representatives of ClpO3Dq ideals with

norm NmAk ” k mod D for k P pZDZqˆ. We construct these ideals in the following lemma:

Lemma 5.6. We can take as representatives of ClpO3Dq the ideals:

Ak “ˆ

nkak `mk´b`

?´3

2

˙

,

where NmAk “ ak ” k mod D for k P pZDZqˆ, ak ” 1 mod 6 and nk ” 1 mod D. Wecan pick such an ideal if we take mk ” b´1pk ` 1q mod D. We can further put the conditionsnk,mk ” 1 mod 3 to determine the ideal uniquely modulo 3D.

Proof: Note first that two ideals A,B are in the same class in ClpO3Dq if we can findgenerators α, β for A and B, respectively, such that αβ´1 ” m mod 3D, where m is an integerprime to 3D. Note that this implies αβ´1 ” ˘1 mod 3.

Let us assume that Ak and Al are in the same class in ClpO3Dq. Then we must have˘ωi

´

nkak `mk´b`

?´3

2

¯

” ˘ωjR´

nlal `ml´b`

?´3

2

¯

mod 3D for some i, j. Since we chose

nk,mk, nl,ml ” 1 mod 3 and b is odd we actually have nkak`mk´b`

?´3

2 ” nlal`ml´b`

?´3

2 ”

ω mod 3, which determines the choice of ˘ωi “ ˘ωj on both sides. We further need thecondition:

nkak `mk´b`

?´3

2” Rpnlal `ml

´b`?´3

2q mod D

Note that this is equivalent to:

k ` b´1pk ` 1q´b`

?´3

2” Rpl ` b´1pl ` 1q

´b`?´3

2q mod D

Furthermore, this can be rewritten as:

kb` pk ` 1q?´3

2” R

lb` pl ` 1q?´3

2mod D

This implies k ” lR mod D and k ` 1 ” lR ` R mod D, thus R ” 1 mod D and k ” l

mod D.Finally, we have #pZDZqˆ such ideals, all in different classes of ClpO3Dq, thus we have

representatives in every class of ClpO3Dq.

39

5.3.2 Using the factorization formula

We will pick representatives as in the above Lemma to rewrite the Proposition 5.1 and applyCorollary 5.4. We will denote by ts P pZDZqˆ, s ” 1 mod 6u the norms of the ideals chosenin Lemma 5.6. Furthermore, we are going to choose in Proposition 5.1 all r to be even. We willuse the notation tr P ZDZ, r evenu to express this.

Lemma 5.7. Picking representatives of s P pZDZqˆ such that s ” 1 mod 6 and r P ZDZalso such that r ” 0 mod 2, we have

ÿ

sPpZDZqˆs”1 mod 6

Θ´

D´b`?´3

2as

¯

Θ´

´b`?´3

2as

¯ χpAsq “ÿ


ÿ

rPZDZ,r even

θsr

´

´b`?´3

2

¯

θ0

´

´b`?´3

2D2

¯ χpArAsq¨θr

´

b`?´3

2

¯

θ0

´

b`?´3

2D2

¯χpArq

Proof: We fix φpDq ideals Ak as in Lemma 5.6. Recall that we pick Ak such that NmAk “ak ” k mod D for k P pZDZqˆ, ak ” 1 mod 6 and nk ” 1 mod D. We can pick such an idealif we take Ak “ pnkak `mk

´b`?´3

2 q with mk ” b´1pk ` 1q mod D. We will try to compute:

S “ÿ

kPpZDZqˆk”1 mod 6

Θ´

D´b`?´3

2ak

¯

Θ´

´b`?´3

2ak

¯ χpAkq

Recall that from Corollary 5.4, we have:

Θ´

D´b`?´3

2as

¯

Θ´

´b`?´3

2as

¯ ´1

3

Θ´

D´b`?´3

2as

¯

Θ´

´b`?´3

6as

¯ “ÿ

rPZDZ,r even

θasr

´

´b`?´3

2a2s

¯

θr

´

b`?´3

2

¯

θ0

´

´b`?´3

2D2a2s

¯

θ0

´

b`?´3

2D2

¯

Moreover since r, ars are both even, we have eπina2srs “ eπirs “ 1 and thus in Lemma 5.5 wehave:

θrs

´

´b`?´3

2a2s

¯

θ0

´

b`?´3

´2a2sD2

¯ “θrs

´

´b`?´3

2

¯

θ0

´

´b`?´3

2D2

¯

Then our sum can be written in the form:

Θ´

D´b`?´3

2as

¯

Θ´

´b`?´3

2as

¯ ´1

3

Θ´

D´b`?´3

6as

¯

Θ´

´b`?´3

2as

¯ “ÿ

rPZDZ,r even

θsr

´

´b`?´3

2

¯

θr

´

b`?´3

2

¯

θ0

´

´b`?´3

2D2

¯

θ0

´

b`?´3

2D2

¯ (18)

Now summing up for all s P pZDZqˆ, we get the result of the lemma:

ÿ


Θ´

D´b`?´3

2as

¯

Θ´

´b`?´3

2as

¯ χpAsq ´1

3

ÿ


Θ´

D´b`?´3

6as

¯

Θ´

´b`?´3

2as

¯ χpAsq “

“ÿ


ÿ

rPZDZ,r even

θrs

´

´b`?´3

2

¯

θ0

´

´b`?´3

2D2

¯ χpArAsq ¨θr

´

b`?´3

2

¯

θ0

´

b`?´3

2D2

¯χpArq

40

From Lemma 6.7 in Appendix A, we haveÿ


Θ´

D´b`?´3

6as

¯

Θ´

´b`?´3

2as

¯ χpAsq “ 0. This give us

the result of the Lemma.

Proposition 5.2. Under the conditions above, we have:

ÿ


Θ´

D´b`?´3

2as

¯

Θ´

´b`?´3

2as

¯ χpAsq “

ˇ

ˇ

ˇ

ˇ

ˇ

ˇ

ˇ

ˇ

ÿ


θs

´

´b`?´3

2

¯

θ0

´

´b`?´3

2D2

¯χpAsq

ˇ

ˇ

ˇ

ˇ

ˇ

ˇ

ˇ

ˇ

2

Proof: Only for the purpose of this proposition we will use the following notation for θr, toemphasize how it depends on D:

θrDpzq “ÿ

nPZeπipn`

rD´

16 q

2zp´1qn

Using the new notation, in the previous Lemma we have proved:

ÿ


Θ´

D´b`?´3

2as

¯

Θ´

´b`?´3

2as

¯ χpAsq “ÿ


ÿ

rPZDZ,r even

θsrD

´

´b`?´3

2

¯

θ0

´

´b`?´3

2D2

¯ ¨θrD

´

b`?´3

2

¯

θ0

´

b`?´3

2D2

¯ χpAsq.

Note that using Corollary 5.3 for a “ D2 can rewrite:

θ0

ˆ

´b`?´3

2D2

˙

θ0

ˆ

b`?´3

2D2

˙

“D4?

3Θ

ˆ

b`?´3

2

˙

Thus the equation becomes:

ÿ


Θ´

D´b`?´3

2as

¯

Θ´

´b`?´3

2as

¯ χpAsq “1

D4?3

Θ´

b`?´3

2

¯

ÿ


ÿ

rPZDZ,r even

θsrD

ˆ

´b`?´3

2

˙

θrD

ˆ

b`?´3

2

˙

χpAsq

(19)Let R ” R1 mod D, R even and S ” 1 mod 6. Then we have by definition:

θRSpz1qΘRpz2q “ÿ

nPZeπipn`RSD´16q2z1eπin

ÿ

mPZeπipm`RD´16q2z2eπim

By changing nÑ n`S and mÑ m` 1, we change RÑ D`R and R`D ” R1 mod 2D. Weget

θRSpz1qΘRpz2q “ÿ

nPZeπipn`R

1SD´16q2z1eπinp´1qSÿ

mPZeπipm`R

1D´16q2z2eπimp´1q “ θR1Spz1qΘR1pz

12q

41

Thus we can choose in the formulas above all r to be actually odd. Furthermore, by makinga change of r ˘ 2D we can also choose r ” 1 mod 3. Then we can rewrite the equation as:

ÿ


Θ´

D´b`?´3

2as

¯

Θ´

´b`?´3

2as

¯ χpAsq “1

D4?3

Θ´

b`?´3

2

¯

ÿ


ÿ

rPZDZ,r”1 mod 6

θsrD

ˆ

´b`?´3

2

˙

θrD

ˆ

b`?´3

2

˙

χpAsq

(20)Denote τD “

´b`?´3

2 . Note that we are summing over all residues r mod D. We canseparate the terms, depending on whether a prime divisor pi divides both D and r. We do thisby using the Inclusion-Exclusion principle and note that the sum gets rewritten as:

ÿ


ÿ

rPZDZ,r”1 mod 6

θsrDpτDqθrDp´τDqχpAsq “ÿ


ÿ

rPpZDZqˆr”1 mod 6

θsrDpτDqθrDp´τDqχpAsq

`ÿ

pi|D

ÿ


ÿ

rPZpDpiqZr”1 mod 6

θ srpDpiq

pτDqθ rpDpiq

p´τDqχpAsq

´ÿ

pipj |D

ÿ


ÿ

rPZpDpipjqZr”1 mod 6

θ srpDpipjq

pτDqθ rpDpipjq

p´τDqχpAsq

. . .

` p´1qn´1ÿ

p1...pn|D

ÿ


ÿ

rPZpDp1...pnqZr”1 mod 6

θ srpDp1...pnq

pτDqθ rpDp1...pnq

p´τDqχpAsq

Using Lemma 5.8 proved below, all of the terms except for the first one equal 0. Thus gettingback to the equation (19), we get:

ÿ


Θ´

D´b`?´3

2as

¯

Θ´

´b`?´3

2as

¯ χpAsq “ÿ

s,rPpZDZqˆs”r”1 mod 6

ÿ


θsrD

´

´b`?´3

2

¯

θ0

´

´b`?´3

6D2

¯ χpArAsq ¨θrD

´

b`?´3

2

¯

θ0

´

b`?´3

6D2

¯ χpArq

“ÿ

s,r sPpZDZqˆs”1 mod 6

θsrD

´

´b`?´3

2

¯

θ0

´

´b`?´3

2D2

¯ χpArAsq ¨θrD

´

´b`?´3

2

¯

θ0

´

´b`?´3

2D2

¯ χpArq

“

ˇ

ˇ

ˇ

ˇ

ˇ

ˇ

ˇ

ˇ

ÿ


θsD

´

´b`?´3

2

¯

θ0

´

´b`?´3

2D2

¯ χpAsq

ˇ

ˇ

ˇ

ˇ

ˇ

ˇ

ˇ

ˇ

2

Below we prove Lemma 5.8 used in the proof of Proposition 20:

42

Lemma 5.8. If D “ p1 . . . pn and D1 “ Dppi1 . . . pikq, then:

ÿ


ÿ

rPZD1Zr”1 mod 6

θsrD1

ˆ

´b`?´3

2

˙

θrD1

ˆ

b`?´3

2

˙

χpAsq “ 0

Proof: Note that first that we can rewrite the sum in the form:

ÿ


ÿ

rPZD1Zr”1 mod 6

θsrD1

ˆ

´b`?´3

2

˙

θrD1

ˆ

b`?´3

2

˙

χpAsq “

“ θ0

ˆ

´b`?´3

2D12

˙

θ0

ˆ

b`?´3

2D12

˙

ÿ


ÿ

rPZD1Zr”1 mod 6

θrsD1´

´b`?´3

2

¯

θ0

´

´b`?´3

2D12

¯ ¨

θrD1´

b`?´3

2

¯

θ0

´

b`?´3

2D12

¯ χpAsq

Using (18) for D :“ D1, we recognize the sum on the LHS to be:

ÿ


Θ´

D1 ´b`?´3

2as

¯

Θ´

´b`?´3

2as

¯ χpAsq

Denote m “ DD1 “ pi1 . . . pik . Moreover, recall that from the definition of the cubiccharacter we have:

D13χDpAsq “ pD13qσAs “ pD113qσAs pm13qσAs “ D113χD1pAsqm13χmpAsq

Then we can rewrite the sum as:

ÿ


Θ´

D1 ´b`?´3

2as

¯

Θ´

´b`?´3

2as

¯ χDpAsq “ÿ

s1PpZD1Zqˆ,s1”1 mod 6

Θ´

D1 ´b`?´3

2a1s

¯

Θ´

´b`?´3

2a1s

¯ χD1pAsqÿ

sPpZDZqˆ,s”s1”1 mod 6s”s1 mod D1

χmpAsq

Note that as D “ p1 ¨ ¨ ¨ pn, we have ts P pZDZqˆ, s ” s1 mod D1u – pZmZqˆ. Moreover,note that χmpAsq depends only on s mod m. Thus we are summing the character χmpAsq “χmpAs2q over s2 P pZmZqˆ.

Moreover, χmpAsq is a nontrivial character as a function of s, as m13χmpAsq “ pm13qσAs “

m13 for all As iff m13 P Qr?´3s. As we are summing a non-trivial character over a group, the

sum is just 0:ÿ

s2PpZmZqˆχmpAs2q “ 0,

thus the whole sum is zero.

We left out the case r ” 0 mod D. In this case we have:

43

ÿ


θ0

´

´b`?´3

2

¯

θ0

´

b`?´3

2

¯

D4?3

Θ´

b`?´3

2

¯ χpAsq “ÿ


1

DχpAsq “ 0

5.4 Shimura reciprocity applied to θr

We define:

frpzq “θrpzq

θ0pzq“

ř

nPZeπipn`

rD´

16 q

2zeπin

ř

nPZeπipn´

16 q

2zeπin

Then we can rewrite Proposition 5.2 as:

ÿ

sPpZDZqˆ

Θ´

D´b`?´3

2as

¯

Θ´

´b`?´3

2as

¯ χpAsq “

ˇ

ˇ

ˇ

ˇ

ˇ

ˇ

ÿ

sPpZDZqˆfspτqχpAsq

ˇ

ˇ

ˇ

ˇ

ˇ

ˇ

2 ˇˇ

ˇ

ˇ

ˇ

ˇ

θ0

´

´b`?´3

2

¯

θ0

´

´b`?´3

2D2

¯

ˇ

ˇ

ˇ

ˇ

ˇ

ˇ

2

Note that from the Corollary 5.3, we can compute 32Θ

´

´b`?´3

2

¯

“4?31 θ0

´

´b`?´3

2

¯

θ0

´

b`?´3

2

¯

as well as 32Θ

´

´b`?´3

2

¯

“4?3D θ0

´

´b`?´3

2D2

¯

θ0

´

b`?´3

2D2

¯

. Taking the ratio of the two relations,gives us:

ˇ

ˇ

ˇ

ˇ

ˇ

ˇ

θ0

´

´b`?´3

2

¯

θ0

´

´b`?´3

2D2

¯

ˇ

ˇ

ˇ

ˇ

ˇ

ˇ

2

“ D

Thus we get:

ÿ

sPpZDZqˆ

Θ´

D´b`?´3

2as

¯

Θ´

´b`?´3

2as

¯ χpAsq “ D

ˇ

ˇ

ˇ

ˇ

ˇ

ˇ

ÿ

sPpZDZqˆfspτqχpAsq

ˇ

ˇ

ˇ

ˇ

ˇ

ˇ

2

.

By further multiplying by D13, we have:

ÿ

sPpZDZqˆ

Θ´

D´b`?´3

2as

¯

Θ´

´b`?´3

2as

¯ D13χpAsq “

ˇ

ˇ

ˇ

ˇ

ˇ

ˇ

ÿ

sPpZDZqˆfspτqχpAsqD23

ˇ

ˇ

ˇ

ˇ

ˇ

ˇ

2

. (21)

Our goal in this section is to show that all the terms fspτqχpAsqD23 are Galois conjugatesof each other.

5.4.1 θr as an automorphic form

We will first look closer at the function θr. We will rewrite θr as an automorphic theta functionΘ : SL2pAQq Ñ C:

44

Θpgq “ÿ

mPQrpgqΦpmq,

where Φ P pAQq is a Schwartz-Bruhat function and r is the Weil representation defined by:

• rˆˆ

a 0

0 a´1

˙

Φ

˙

pxq “ χ0paq|a|12Φpaxq

• rˆˆ

1 b

0 1

˙

Φ

˙

pxq “ ψpbx2qΦpxq

• rˆˆ

0 1

´1 0

˙

Φ

˙

pxq “ γpΦpxq,

where ψppxq “ e´2πiFracppxq and ψ8pxq “ e2πix, γ is an 8th root of unity, and χ0 is acharacter.

XXX define χ0 and γ

We define the following Schwartz-Bruhat functions for θ. Let fµ “ś

vppµqă0

p. We define

Φµ “ś

vΦµ,v, where:

$

’

’

’

’

’

’

’

&

’

’

’

’

’

’

’

%

Φprqp “ charZp , if p - D

Φprqp “ charZp´ r

D, if p|D, p - 2, 3

Φprq3 “ charZ3`

13,

Φprq2 pnq “ eπiFrac2pnq charZ2`

12pnq,

Φprq8 pxq “ e´2πqpxq.

We define the theta function:

ΘΦprqpgq “ÿ

nPQrpgqΦprqpnq

Note that Φprqf pnq ‰ 0 for n P Q implies n ´ r

D `16 P Zp for all p. This implies n ´ r

D `

16 P Z, thus n P Z ` r

D ´16 Also note that for gz “

´

y12 y´12x

0 y´12

¯

, we have rpgzqΦ8pnq “

r´´

y12 0

0 y´12

¯´

1 xy´1

0 1

¯

Φ¯

pnq “ y12e2πipx`yiqn2

. Then we can compute:

ΘΦprqpgz, 1f q “ÿ

nPZ` rD´

16

e2πizn2

eπiFrac2pnq “ y12θrp2zq

Note that: θrp2zq “ y´12ΘΦprqpgz, 1f q and θ0p2zq “ y´12ΘΦp0qpgz, 1f q, which implies:

θrpzq

θ0pzq“

ΘΦprqpgz2, 1f q

ΘΦp0qpgz2, 1f q

45

5.4.2 Galois action on modular functions (Shimura reciprocity)

Recall the function fr:

frpzq “θrpzq

θ0pzq“

ΘΦprqpgz2, 1f q

ΘΦp3qpgz2, 1f q

Lemma 5.9. The theta function θrpzq is modular form of weight 12 for Γp72D2q.

Proof: Recall that θrpzq “ ΘΦprqpgz2q. We will compute θrˆˆ

a b

c d

˙

z

˙

, forˆ

a b

c d

˙

P

Γp72D2q. Note first that:

θr

ˆˆ

a b

c d

˙

z

˙

“ ΘΦprq

ˆˆ

1?

2 0

0?

2

˙ˆ

a b

c d

˙

gz

˙

“ ΘΦprq

ˆˆ

a b2

2c d

˙ˆ

1?

2 0

0?

2

˙

gz

˙

As ΘΦprq is invariant under SL2pQq, we can rewrite ΘΦprq as:

ΘΦprq

ˆˆ

a b2

2c d

˙ˆ

1?

2 0

0?

2

˙

gz, 1

˙

“ ΘΦprq

˜

ˆ

1?

2 0

0?

2

˙

gz,

ˆ

a b2

2c d

˙´1¸

We will compute separately the two terms, using the Weil representations. For the RHS,

note that we have to compute rˆ

a b2

2c d

˙´1

Φprqf “ r

ˆ

d ´b2

´2c a

˙

Φprqf . We will show:

ΘΦprq

˜

ˆ

1?

2 0

0?

2

˙

gz,

ˆ

a b2

2c d

˙´1¸

“ź

p|6D

γ2pΘΦprq

ˆˆ

1?

2 0

0?

2

˙

gz,

˙

We rewrite the matrix as:ˆ

d ´b2

´2c a

˙

“

ˆ

1a ´b2

0 a

˙ˆ

1 0

´2ca 1

˙

At p - 6D, the action ofˆ

d ´b2

´2c a

˙

is trivial, as it belongs to SL2pZpq and Φprqp is the

characteristic function of Zp. For p|6D, we compute:

First we compute rˆ

1 0

´2cd 1

˙

Φprqp pxq “ γ2

pΦprqp pxq. We rewrite the matrix as

ˆ

1 0

´2cd 1

˙

“

ˆ

´1 0

0 ´1

˙ˆ

0 ´1

1 0

˙ˆ

1 2cd

0 1

˙ˆ

0 ´1

1 0

˙

and compute the Weil representation action:

• rˆ

0 ´1

1 0

˙

Φprqp pxq “ γp

y

Φprqp pxq. Note that we can compute:

y

Φprqp pxq “

ż

Qp

e´2πiFracpp2xyq charZp` rDpyqdy “

ż

Zp

e´2πiFracpp2xpy`rDqqdy “ e´2πiFracpp2rxD charZppxq

46

• rˆ

1 2cd

0 1

˙

yΦprqppxq “ e´2πiFracpp2cdx2

e´2πiFracpp2rxD charZppxq. As vppcdq ě 0, we

have e´2πiFracpp2cdx2

“ 1, thus the action is trivial on y

Φprqp pxq

• rˆ

0 ´1

1 0

˙

y

Φprqp pxq “ γp

y

y

Φprqp pxq. By the choice of the self-dual Haar measure, this equals

γpΦprqp p´xq.

• rˆ

0 ´1

1 0

˙

Φprqp p´xq “ Φ

prqp pxq

Now we also want to compute the action of rˆ

1a ´b2

0 a

˙

Φprqp pxq. We rewrite the matrix

as:ˆ

1a ´b2

0 a

˙

“

ˆ

1a 0

0 a

˙ˆ

1 ´ba2

0 1

˙

and compute the action:

• rˆ

1 ´ba2

0 1

˙

Φprqp pxq “ e2πiFracppba2x

2q charZp`rDpxq. AsD

2|ba2, we have e2πiFracppba2x2q,

thus we have trivial action.

• rˆ

1a 0

0 a

˙

Φprqp pxq “ χ0paq|a|

12p Φ

prqp pxaq. As a ” 1 mod D, we get Φ

prqp pxaq “ Φ

prqp pxq,

as well as χ0paq “ |a|12p “ 1.

For p “ 3 the computation is similar. For p “ 2, we compute again the action ofrˆ

1 0

´2cd 1

˙

Φprqp pxq “

γ2pΦprqp pxq:

• rˆ

0 ´1

1 0

˙

Φprq2 pxq “ γ2

y

Φprqp pxq. Note that we can compute:

y

Φprq2 pxq “

ż

Q2

e´2πiFrac2p2xyq charZ2`12pyqeπiFrac2pyqdy “ eπi2

ż

Z2

e´2πiFrac2p2xpy`12qqeπiFrac2pyqdy “

eπi2e´2πiFrac2pxq

ż

Z2

e´2πiFrac2pp2x´12qyqdy “ eπi2e´2πiFrac2pxq char 12 pZ2`12qpxq

• rˆ

1 2cd

0 1

˙

yΦprq2pxq “ e´2πiFrac2p2cdx2yΦprq2pxq. As vpp2cdq ě 4, we have e´2πiFracpp2cdx

2

“

1, thus the action is trivial on y

Φprqp pxq

• rˆ

0 ´1

1 0

˙

y

Φp2qp pxq “ γ2

y

y

Φprq2 pxq. By the choice of the self-dual Haar measure, this equals

γ2Φprq2 p´xq.

• rˆ

0 ´1

1 0

˙

Φprq2 p´xq “ Φ

prq2 pxq

47

We compute similarly the action of rˆ

1a ´b2

0 a

˙

Φprq2 pxq:

• rˆ

1 ´ba2

0 1

˙

Φprq2 pxq “ e2πiFracppba2x

2qeπiFrac2pxq charZ2`12pxq. As 4|ba2, we have

e2πiFrac2pba2x2q “ 1, thus we have trivial action.

• rˆ

1a 0

0 a

˙

Φprq2 pxq “ χ0paq|a|

122 Φ

prq2 pxaq. As a ” 1 mod 8, we get Φ

prq2 pxaq “ Φ

prq2 pxq,

as well as χ0paq “ |a|122 “ 1.

This finishes the computation of the finite part. We got:

ΘΦprq

˜

ˆ

1?

2 0

0?

2

˙

gz,

ˆ

a b2

2c d

˙´1¸

“ 2´14y12ÿ

mPZZ` rD´

16

eπim2zp´1qm “ 2´14y12θrpzq

(22)We will compute now the infinite part. Note first that rpgzqΦ8pmq “ y14e2πiz|m|2 We

rewrite the matrix:ˆ

a b

c d

˙

“

ˆ

´1 0

0 ´1

˙ˆ

1 bd

0 1

˙ˆ

1d 0

0 d

˙ˆ

0 ´1

1 0

˙ˆ

1 ćd

0 1

˙ˆ

0 ´1

1 0

˙

We compute the Weil representation action:

• F1pmq :“

ˆ

0 ´1

1 0

˙

e2πizm2

“ γ81?ze´2πi 1z

• F2pmq :“ r

ˆ

1 ćd

0 1

˙

F1pmq “ e´2πi cdm2

F1pmq “ γ81?ze´2πi cz`ddz

• F3pmq :“ r

ˆ

0 ´1

1 0

˙

F3pmq “ γ8F3pmq “ γ28

1?ze2πi dz

cz`d p˚q

b

dzcz`d “ γ2

8p˚q

b

dcz`de

2πi dzcz`dm

2

• F4pmq :“ r

ˆ

1d 0

0 d

˙

F3pmq “ sgnpdqd12F3pmdq “ sgnpdqd12F3pmdqγ28p˚q

b

dcz`de

2πi zdpcz`dqm

2

“

sgnpdqF3pmdqγ28p˚q

b

1cz`de

2πi zdpcz`dqm

2

• F5pmq :“ r

ˆ

1 bd

0 1

˙

F4pmq “ e2πi bdm2

F4pmq “ sgnpdqγ28p˚q

b

1cz`de

2πip bd`z

dpcz`dq qm2

“

sgnpdqγ28p˚q

b

1cz`de

2πippbc`1qz`bddpcz`dq qm

2

“ sgnpdqγ28p˚q

b

1cz`de

2πip az`bcz`d qm2

• rˆ

´1 0

0 ´1

˙

F5pmq “ ´F5p´mq “ ´ sgnpdqγ28p˚q

b

1cz`de

2πip az`bcz`d qm2

We still have to compute the action of rˆ

1?

2 0

0?

2

˙

oný12 sgnpdqγ28p˚q

b

1cz`de

2πip az`bcz`d qm2

.

This gives us just:

48

2´14y12 sgnpdqγ28p˚q

c

1

cz ` deπip

az`bcz`d qm

2

Thus we have:

Θ

ˆˆ

1?

2 0

0?

2

˙ˆ

a b

c d

˙

z, 1

˙

“ 2´14y12 sgnpdqγ28p˚q

c

1

cz ` d

ÿ

mPZ` rD´

16

eπipaz`bcz`d qm

2

p´1qm

Note that this is exactly:

2´14y12 sgnpdqγ28p˚q

c

1

cz ` dθr

ˆ

az ` b

cz ` d

˙

(23)

From (21) and (22) we get that:

sgnpdqγ28p˚q

c

1

cz ` dθr

ˆ

az ` b

cz ` d

˙

“ γ2fθrpzq

XXXXXX γ should disappear here

Lemma 5.10. f P F is a modular function for Γp72D2q.

Proof: We need to check that for γ “ˆ

a b

c d

˙

P Γp3Dq we have:

frpγzq “ frpzq.

Proof: Using the previous lemma, for γ P Γp72D2q we have sgnpdqγ28p˚q

b

1cz`dθr pγzq “

θrpzq. Applying the same computation for r “ 0, we get sgnpdqγ2p˚q

b

1cz`dθ0 pγzq “ θ0pzq.

Thus we have:

θrpγzq

θ0pγzq“p˚q?cz ` dθrpzq

p˚q?cz ` dθ0pzq

“θrpzq

θ0pzq

Lemma 5.11. The modular function fr has rational Fourier coefficients.

Proof: Note that θrpzq “ qpD´rq272p1 `

ř

Mě1

aMqMp72D2

qq, where am P Z and θ0pzq “

q172p1`ř

Mě1

bMqM72q

Then we can compute frpzq “ qppD´rq2´1q72p1`

ř

mamq

m72D2

q with am P Z.

From CM-theory we have fpτq P HO, where HO is the ray class field of modulus 72D2.In order to compute its Galois conjugates over K we can use Shimura reciprocity law. In itsgenerality:

Shimura reciprocity law. For τ P K XH with minimal polynomial X2 ` BX ` C “ 0,we have its Galois conjugates:

49

frpτqσ´1x “ fgτ pxqr pτq,

for x P AˆKf , gτ pxq “ˆ

t´ sB ´sC

s t

˙

.

In our case, we want to compute the Galois conjugates of frpτq, where τ “ ´b`?´3

2 . Notethat it has the minimum polynomial X2 ` bX ` b2`3

4 . Thus we have to compute the action of

all gτ ppxpqpq “ź

p

˜

tp ´ spb ´spb2`3

4

sp tp

¸

p

.

We will compute all these actions. However, we claim that it is enough to compute the actionof the ideals A through the correspondence:

Ip3q Ñ AˆK,fKˆ

A “ pA`Bωq ÝÑ pA`Bωqp|6D,

where A`Bω ” 1 mod 3 is the generator of the ideal A.More precisely, in order to find the Galois conjugates over K, we will compute the action of

all Galois actions corresponding to pAp `Bpωqp P AˆK and we will prove that the Galois actionfrom Shimura reciprocity law is:

Proposition 5.3. For A “ pnaa `ma´b`

?´3

2 q, where b2 ” ´3 mod 4Db2 is an ideal primeto 6D, we have:

f1pτqσA “ fnapτq

and frpτq are all the Galois conjugates of fpτq, where r P pZDZqˆ. Moreover, this impliesthat f1pτq P H6D.

Proof: First we note that we do not have to consider the action of all pxpqp P AˆK . Byapplying the Strong Approximation Theorem for GL1 and the number field K that is a PID,we have:

AˆK “ Kˆ ˆź

v-8OˆKv ˆ Cˆ

This implies:

AˆKf “ Kˆ ˆź

v-8OˆKv

Then any x “ pxvq P AˆK,f can be written as x “ kplvq, where k P Kˆ, plvqv Pś

v-8OˆKv .

Since Nm k ą 0, we have the embedding:

k P Kˆ ãÑ GL2pQq`

50

We also have the embedding:

plvqv Pź

v-8OˆKv ãÑ

ź

p

GL2pZpq

Thus if we know the Galois action of Kˆ and of pOˆK , we will know the Galois action of AˆK,f .

We recall the way the action of gτ pxq is defined for. For α P GL2pQq`, fα is defined byfαpτq “ fpατq. In our case we only need to look at the action of Kˆ. Recall that k P Kˆ

embeds into GL2pQq` under the map:

k “ t` s´b`

?´3

2ãÑ gτ pkq “

ˆ

t´ sb ´sc

s t

˙

Then the Galois action from Shimura reciprocity is:

fpτqk´1

“ fgτ pkqpτq “ fpgτ pkqτq

Note that t` sτ Ñ`

t´sb ´scs t

˘

is the torus that preserves τ , thus we have:

fpτqk´1

“ fpgτ pkqτq “ fpτq

Now all we have left is to compute the action ofś

OˆKv . Note that for all v - 6D the actionis trivial. For v|6D we project the action of pgτ pxvqqv gτ px

1q P GL2pZ6D2Zq.

Remark 5.2. Note that we have for p˘ωiqp ãÑ AˆKf acting trivially. Thus we have for x P AˆKf :

pfrpτqqσ˘ωix “ ppfrpτqq

σ˘ωi qσx “ pfgτ p˘ω

iq

r pτqqσx “ frpτqσx

Lemma 5.12. For x Pś

v OˆKv

we can find ωi, i “ 0,˘1 such that:

px2 ˘ ωiq2 “ pt2 ` s2ωq

with v2pt2q “ 0, v2ps2q ě 1 and

px3 ˘ ωiq3 “ pt3 ` s3ωq

with t3 ` s3 ” 1 mod 3.

Proof: Note first that if v2psq ě 1, then we must have v2pt2q “ 0, as we need x2ωi P

pZ2rωsqˆ. Thus we must find xωi such that v2psq ě 1. We write x2 “ t12 ` s

12ω. Then:

x2ω “ t12ω ` s12ω

2 “ pt12 ´ s12qω ` s

12

x2ω2 “ t12ω

2 ` s12 “ p´t12qω ` ps

12 ´ t

12q

One of t12, s12, t12´s12 must have positive valuation. Assume this is not true: v2pt12q “ v2ps

12q “

0. Then s12, t12 ” 1 mod 2, thus s12 ´ t12 ” 0 mod 2 and has positive valuation. Thus we can

always pick xωi as claimed above at the place 2.

51

Now since take x3ωi “ s13ω ` t

13 “ s13

´3`?´3

2 ` pt13 ` s13q. Then, since x3 is a unit in Z3rωs,

we must have v3ps13 ` t13q “ 0, thus s13 ` t13 ” ˘1 mod 3. We pick x3ω or ´x3ω to get the

condition s13 ` t13 ” 1 mod 3.

Since from the remark above x and ˘ωix act the same, we can consider the Galois action ofσxωi as in the lemma above. We compute it below.

Let xp Pś

v OˆKv

chosen as above. Then:

xp “ tp ` sp´b`

?´3

2ãÑ gτ pxpq “

ˆ

tp ´ spb ´spc

sp tp

˙

Elements ofś

pGL2pZpq project to GL2pZ6D2Zq, which is the action we care about. From

Chinese remainder theorem, we can find k0 P K such that k0 ” xp mod 6D2Zp for all p|6D.Note that k0 is independent of the choice of τ .

Then we only need to compute the action of:

frpτqσ´1x “ fgτ pxqpτq “ f

gτ pxvqv|6Dr pτq “ fgτ pt`sτqv|6D pτq

We will now compute fgτ pxqp|6Dr pτq. Note that, for c1 “ b2`34 , we have the map :

k0 “ sτ ` tÑ gτ pk0q “

ˆ

t´ sb1 ´sc

s t

˙

Let Nmpk0q “ a. We write the action:

fpτqσx “ f

¨

˝

t´ sb ´sca

s ta

˛

‚

p|6D

¨

˝

1 0

0 a

˛

‚

p|6D

Note thatˆ

1 0

0 a

˙

p|6D

acts trivially on fr as both functions θ„

´ 16 `

rD

12

e´πiprD´16q and

θ

„

´ 16

12

eπi6q have rational Fourier coefficients.

Thus we need to compute the action:

f

¨

˝

t´ sb ´sca

s ta

˛

‚

p|6Dr pτq

Note thatˆ

t´ sb ´sca

s ta

˙ˆ

t´ sb ´sca˚

s ta˚

˙

P SL2pZ6D2Zq and we can lift it to an ele-

ment of SL2pZq.Lift from SL2pZ6D2q to SL2pZq.

Lemma 5.13. We can always lift a matrix inˆ

A B

C D

˙

P SL2pZNZq to SL2pZq.

Proof: Takeˆ

A B

C D

˙

P SL2pZNZq, A,B,C,D P Z. We can further assue pC,Dq “ 1. Let

AD ´BC “ k P Z. Then we can take:

52

A0 “ A`NA1

B0 “ B `NB1

C0 “ C `NC1

D0 “ D `ND1

We want to have the condition:1 “ A0D0 ´ B0C0 “ AB ´ CD ` NpAD1 ` A1D ´ BC1 ´ B1Cq ` N2pA1D1 ´ B1C1q “

1`Nk `NpAD1 À1D ´BC1 ´B1Cq `N2pA1D1 ´B1C1q

For example, pick D1 “ C1 “ 0. Then we only need:

pA1D ´B1Cq “ ´k

Note that since pC,Dq “ 1, we can find mC ` nD “ 1. Then p´knqD ´ kmC “ ´k, thuspick A1 “ ´kn and B1 “ km.

We look at such a matrixˆ

a b

c d

˙

P SL2pZq such that:

ˆ

a0 b0c0 d0

˙

”

˜

s´ tb ´s b2`34 a˚

s t

¸

mod 6D2

Conditions obtained:

• v2psq ě 0 and v2ptq “ 0 imply b0, c0 ” 0 mod 2, a0, d0 ” 1 mod 2.

• From the choice 3|b we also have a0 ” d0 mod 3 and b0 ” 0 mod 3. Since we pickedk0 “ t0 ` s0ω ” s´b`

?´3

2 ` t with s0 ` t0 ” 1 mod 3, we must have t ” t0 ` s0 mod 3,thus d0 ” t0 ” 1 mod 3.

• From the choice of t ` s´b`?´3

2 unit inś

v|6DOˆKv

, we have pt,Dq “ 1. Otherwise note

that the norm is t2 ´ tsb` s2 b2`34 is divisible by p|D, a contradiction.

We will find the action using the following lemma:

Lemma 5.14. Forˆ

a b

c d

˙

P SL2pZq such that vppdq “ 0 and d ” 1 mod 6, we have:

ΘΦprq

ˆˆ

1 0

0 2

˙ˆ

a b

c d

˙

z

˙

“ ΘΦpd´1rq pz2q

Here by d´1 we mean d´1 mod D.

Proof: We compute:

ΘΦprq

ˆˆ

1 0

0 2

˙ˆ

a b

c d

˙

z

˙

“ ΘΦprq

ˆˆ

a b2

2c d

˙ˆ

1 0

0 2

˙

z

˙

Moreover, it equals:

ΘΦprq

„

z2,

ˆ

d ´b2

´2c a

˙

Note that for p - 6D we haveˆ

d ´b2

´2c a

˙

p

in SL2pZpq, thus acts trivially.

53

For p|3D, we have Φr “ charZp´ 16`

rD. For now, we will call µr :“ ´ 1

6 `rD .

If vppdq “ 0, we rewrite:ˆ

d ´b2

´2c a

˙

“

ˆ

1 0

´2cd 1

˙ˆ

d ´b2

0 d´1

˙

We can further write it in the form:

ˆ

d ´b2

´2c a

˙

“

ˆ

´1 0

0 ´1

˙ˆ

0 1

´1 0

˙ˆ

1 0

2cd 1

˙ˆ

0 1

´1 0

˙ˆ

d 0

0 d´1

˙ˆ

1 ´bp2dq

0 1

˙

• rˆ

1 ´bp2dq

0 1

˙

Φppxq “ e´2πiFracpp´bp2dqx2qΦppxq “ Φppxq

• rˆ

d 0

0 d´1

˙

Φppxq “ |d|pχppdqΦppdxq “ Φpd´1rq

p pxq

Note that Φppdxq ‰ 0 iff dx P Zp ` µr iff x P d´1Zp ` d´1µr “ Zp ` d´1µr. Note thatd´1µr “ d´1rD ´ d´16. Since we picked d ” 1 mod 6, this is the same as µd´1r.

Note: We need to check that the character corresponding to Q is trivial on units (χppdq “1).

• rˆ

0 1

´1 0

˙

Φppd´1rqpxq “ e2πiFracpp2d

´1xrDq charZp`12pxq

• rˆ

1 2cd

0 1

˙

pe2πiFracpp2xd´1rDq charZp`12pxqq “ e2πiFracpp2cdx

2qpe2πiFracpp2xd

´1rDq charZppxq

“ pe2πiFracpp2xd´1rDq charZppxqq

• rˆ

0 1

´1 0

˙

pe2πiFracpp2xd´1rDq charZppxqq “ Φ

pd´1rqp p´xq

• rˆ

´1 0

0 ´1

˙

Φpd´1rqp p´xq “ Φ

pd´1rqp pxq

In here we have used the Fourier transform:

y

Φprq3 pxq “

ż

QpΦprqp pyqe

´2πiFracpp2xyqdy “

ż

Zp` rD

Φprqp pyqe2πi2xydy “

ż

Zpe´2πiFracpp2xpy`rDqqdy

“

ż

Zp

e´2πiFracpp2xyqqe´2πiFracppxrDqdy “ e´2πiFracpp2xrDq

ż

Zp

e´2πiFracpp2xyqdy

“ e´2πiFracpp2xrDq charZp´12pxq “ e´2πiFracpp2xrDq charZppxq

Similarly we get y

Φprq3 pxq “ e´2πiFrac3px3q charZppxq

54

Note that the only difference for p “ 3 in the action ofˆ

d ´b2

´2c a

˙

is that it does not

modify rD. Instead, it leaves Φprq unchanged.

At the place p “ 2, we have Φ2 “ eπiFrac2pxq charZ2´12pxq. We can compute:

• rˆ

1 ´bp2dq

0 1

˙

Φppxq “ e´2πiFrac2p´bp2dqx2qeπix charZ2´12pxq “ e2πib8deπix charZ2´12pxqΦppxq

Note that we picked 2|b. Then we have x P Z2 ´ 12 iff x “ n ´ 12 for n P Z2. Then´b2dpn´ 12q2 “ ´bp2dqn2 ` bp2dqn´ bp8dq P Z2 ´ b8d.

• rˆ

d 0

0 d´1

˙

e2πib8deπix charZ2´12pxq “ e2πib8deπidx charZ2´12pdxq “ e2πib8deπix charZ2´12pxq

Note that we have used above v2pdq “ 0.

• rˆ

0 1

´1 0

˙

Φp

2rqpxq “ e2πib8deπix charZ2´12pxq “ e2πib8de2πiFrac2px`14q char 12Z2´14pxq

Below we compute the Fourier transform:ż

Q2

eπiFrac2pyq charZ2´12pyqe´2πiFrac2p2xyqdy “

ż

Z2´12

e2πiFrac2py2`2xyq “ş

Z2e2πiFrac2py2`14`2xy`xqdy “

e2πiFrac2px`14qş

Z2e2πiFrac2pyp12`2xqqdy “ e2πiFrac2px`14q char 1

2Z2´14pxq

• rˆ

1 0

2cd 1

˙

e2πib8de2πiFrac2px`14q char 12Z2´14 “ e2πib8de2πiFrac2p2cdx

2qe2πiFrac2px`14q char 1

2Z2´14

Note that we have the assumptions 2|c and 2 - d. We have x ` 14 “ 12n, n P Z2. Notex2 “ pn´12q4 “ pn2ń`14q4 and then e2πiFrac2p2cdx

2q “ e2πiFrac2pc2dpn

2ń`14qq “

e2πiFrac2pc8dq. Here we have used the fact that c2dpn2 ´ nq P Z2. Thus we get:

e2πiFrac2ppc`bq8dqe2πiFrac2px`14q char 12Z2´14pxq

• rˆ

0 1

´1 0

˙

e2πiFrac2ppc`bq8dqe2πiFrac2px`14q char 12Z2´14pxq “ e2πiFrac2ppc`bq8dqe2πiFrac2p´xq charZ2´12p´xq

• rˆ

´1 0

0 ´1

˙

e2πiFrac2ppc`bq8dqe2πiFrac2p´xq charZ2´12p´xq “ e2πiFrac2ppc`bq8dqe2πiFrac2pxq charZ2´12pxq “

e2πiFrac2ppc`bq8dqΦ2pxq

Finally we are ready to prove the proposition. We have showed so far that:

frpτqσ´1x “ fgτ pxqr pτq “ f

pgτ pk0qqp|6Dr pτq “ f

¨

˝

t´ sb ´sc

s t

˛

‚

p|6D pτq “ f

¨

˝

a0 b0c0 d0

˛

‚

p|6D pτq

From the above lemma we get immediately:

ˆ

Θprqpτ2q

Θp0qpτ2q

˙σx

“ f

¨

˝

a0 b0c0 d0

˛

‚

r pτq “

ΘΦprq

ˆˆ

1 0

0 2

˙ˆ

a0 b0c0 d0

˙

τ

˙

ΘΦp0q

ˆˆ

1 0

0 2

˙ˆ

a0 b0c0 d0

˙

τ

˙“

ΘΦpd´1rq pτ2q

ΘΦp0q pτ2q“ fd´1rpτq

55

For A P ClpO3Dq, A “ pkAq “ pnaa`ma´b`

?´3

2 q, where a “ NmA, we take the map:

x “ pkAqp|6D Ø A

This gives us:

x´1 Ø A´1

Then we have:

frpτqσA´1 “ frpτq

σx´1 “ fgτ pxpqp|6Dr pτq “ f

gτ pkAqp|6Dr pτq “ fn´1

A rpτq

This implies for r ” nA mod D that we have fnApτqσA´1 “ f1pτq, or equivalently:

f1pτqσA “ fnApτq

Remark. This implies that for Ar “ p1 ¨ r ` b˚pr ´ 1q´b`?´3

2 q we have:

f1pτqσAr “ f1pτq

Also it implies that ar “ pr´1q “ pr ¨ r´2 ` 0´b`?´3

2 q we have:

f1pτqσar “ frpτq

5.5 The square is invariant under Galois action

We are finally ready to prove Theorem 5.1.

We define A˝r “´

1` b˚p1´ rq´b`?´3

2

¯

. Note nr “ r´1. Note that A˝r “ Arpr´1q, thus Arand A˝r are in the same class in ClpO3Dq. This implies:

χDpArq “ χDpA˝rq

Moreover, from the definition of χD we have: pD23qσA˝r “ D23χDpA˝rq

Moreover, from Proposition 5.3:

f1pτqσAr˝ “ fnA˝r

pτq “ frpτq

Then we can rewrite the term in Proposition 20:

κ :“ÿ

rPpZDZqˆfrpτqD

23χpArq “ÿ

rPpZDZqˆfrpτqD

23χpA˝rq “ÿ

rPpZDZqˆf1pτq

σAr˝ pD23qσA˝r

“ÿ

rPpZDZqˆpf1pτqD

23qσAr˝

We want to write κ as a Galois trace of a modular function at a CM-point. Note that theideals tA˝,prPZDZqˆ

r u for a group, as we have A˝rA˝s “ A˝rs. Then take G0 “ tr P pZDZqˆ :

56

A˝ru – pZDZqˆ that is a subgroup of GalpHOKq, where HO is the ray class field of conductor3D.

We define fixed field of G0 in H:

H0 “ th P HO : σphq “ h,@σ P G0u

From abelian Galois theory this implies GalpHOH0q – G0. Then we got:

κ “ TrHOH0pf1pτqD

23q (24)

Thus we have proved so far that:

SD “ |κ|2,

where κ P H0. We claim that actually |κ|2 P Q. To prove this, it is enough to show that|κ|2 P Kˆ, as

Lemma 5.15. We have κ3 P K.

Proof: We will show that the Galois conjugates of κ over K are κω and κω2.Take A P ClpOq. Then we have:

κσA “ÿ

rPpZDZqˆpf1pτqD

23qσAr˝A

We can write A “ A˝spmq. Then we have:

κσA “ÿ

rPpZDZqˆpf1pτqD

23qσArs˝pmq

Note that pmq acts trivially on D23, but acts as A˝m on f1pτq. Then we have:

κσA “ÿ

rPpZDZqˆpf1pτqq

σArsm˝D23χpA˝rsq “ χpA˝mqÿ

rPpZDZqˆpf1pτqq

σArsm˝D23χpA˝rsmq “ χpA˝mqκ

Remark 5.3. Recall that |κ|2 P Q. Let κ3 “ a` b?´3 P K. Then |κ|6 “ a2`3b2 and we must

have a2 ` 3b2 “ m3 for some m P Q. With this notation we have |κ|2 “ m “3?a2 ` 3b2.

Remark 5.4. If we try to apply κ, this implies:κσA˝r “ pκ

σpA˝rq

´1q “ κ

κσAprq “ pκσAr q “ χpA˝mqκκσAr “ κχpArq “ κχpArq

6 Appendix A: properties of ΘK

In this appendix we would like to present a few properties of ΘK . First, we have a functionalequation for the theta function (see [9]):

57

ΘKp´13zq “3

?´3

zΘKpzq (25)

Furthermore, we can compute the transformation of ΘKpz ˘ 13q in the lemma below:

Lemma 6.1. We have the following relations:

(i) Θ

ˆ

z `1

3

˙

“ p1´ ωqΘp3zq ` ωΘpzq

(ii) Θ

ˆ

z ´1

3

˙

“ p1´ ω2qΘp3zq ` ω2Θpzq

Proof: We will rewrite the Fourier expansion of Θpzq for z :“ z ` 13:

Θ

ˆ

z `1

3

˙

“ÿ

m,nPZe2πipm2

`n2´mnqpz` 1

3 q.

We split the sum in two parts, depending on whether the ideal pm`nωq is prime to p?´3q.

Then we have:

Θ

ˆ

z `1

3

˙

“ÿ

m,nPZ,p?´3q|pm`nωq

e2πipm2`n2

´mnqpz` 13 q`

ÿ

m,nPZ,p?´3q-pm`nωqq

e2πipm2`n2

´mnqpz` 13 q.

Note that on the RHS we can rewrite the first term as:

ÿ

m,nPZ,p?´3q|pm`nωq

e2πipm2`n2

´mnqpz` 13 q “

ÿ

m,nPZe2πipm2

`n2´mnqp3z`1q “ Θp3z ` 1q “ Θp3zq

Also note that when 3 - m2 ` n2 ´mn, then we have m2 ` n2 ´mn ” 1 mod 3. Then thesecond term on the RHS can be rewritten as:

ÿ


e2πipm2`n2

´mnqpz` 13 q “

ÿ


e2πipm2`n2

´mnqzω.

We rewrite this:

ÿ


e2πipm2`n2

´mnqpz` 13 q “ ω

ÿ

m,nPZe2πipm2

`n2´mnqz´ω

ÿ

m,nPZ,p?´3q|pm`nωqq

e2πipm2`n2

´mnqz

Finally we recognize the two terms as theta functions ΘK :

ÿ


e2πipm2`n2

´mnqpz` 13 q “ ωΘpzq ´ ωΘp3zq

Now going back to our initial computation, we get:

58

Θ

ˆ

z `1

3

˙

“ Θp3zq ` ωΘpzq ´ ωΘp3zq “ p1´ ωqΘp3zq ` ωΘpzq

This finishes the proof of the first formula. We get the second formula by applying the firstformula for z :“ z ´ 13. We get Θ pzq “ p1 ´ ωqΘp3z ´ 1q ` ωΘpz ´ 13q and this is easilyrewritten to give us the second formula.

6.1 Properties of ΘKpp´b`?

3q6q.

Lemma 6.2. ΘK

´

´3`?´3

6

¯

“ 0

Proof: We apply the functional equation 24 for z “ ´3`?´3

6 :

Θ

ˆ

´3`?´3

6

˙

“ p´?´3q

´3`?´3

6Θ

ˆ

3`?´3

6

˙

.

Since Θ´

´3`?´3

6

¯

“ Θ´

3`?´3

6

¯

, we get the result of the lemma.

Lemma 6.3. For the primitive ideal A “ ra, ´b`?´3

a sZ prime to 3, where a “ NmA, b ” 0

mod 3 and b2 ” ´3 mod 4a, we have:

ΘK

ˆ

´b`?´3

6a

˙

“ 0

Proof: The proof is similar to that of Lemma 3.5. We can write the generator of primitive

ideal A “

”

a, ´b`?´3

2

ı

in the form kA “ ma ` n´b`

?´3

2for some integers m,n. Note

that pm, 3q “ 1, thus we can find through the Euclidean algorithm integers A,B such that

mA ` 3nB “ 1, which makesˆ

A B

´3n m

˙

a matrix in Γ0p3q. Since Θ is a modular form of

weight 1 for Γ0p3q, we have:

ΘK

˜

A´b`?´3

6a `B

´3n´b`?´3

6a `m

¸

“

ˆ

m´ n´b`

?´3

2a

˙

ΘK

ˆ

´b`?´3

6a

˙

Noting that´3n´b`?´3

6a `m “ kAa “ 1kA, we can compute A´b`?´3

6a `B

ń´b`?´3

2a `m“pA´b`

?´3

2 `3BaqkA3a .

This is p3aB À´b`?´3

2 qpma` n b`?´3

2 qp3aq. After expanding, we get:

ńAb2 ` 3

4a` abB3`

bp´mA` 3nBq

6`

?´3

6

Note that mA` 3nB “ 1 implies that mA and 3nB have different parities. Also we chose bodd, since b2 ` 3 ” 0 mod 4a. Finally, recall 3|b and thus using the period 1 of ΘK we get:

ΘK

˜

A´b`?´3

6a `B

´3n´b`?´3

2a `m

¸

“ ΘK

ˆ

´3`?´3

6

˙

59

From the previous Lemma, we have ΘK

´

´3`?´3

6

¯

, thus ΘK

´

´b`?´3

6a

¯

“ 0 which finishesthe proof.

6.2 About ΘKpDp´3`?´3q6q.

In this section we will show that for D a product of split primes p ” 1 mod 3 and for therepresentative ideals A “ ra, ´b`

?´3

2 s of ClpO3Dqwith b ” 0 mod 3, we have:

ÿ

APClpO3Dq

Θ´

´b`?´3

6a

¯

Θ´

´b`?´3

2a

¯χDpAqD13 “ 0

We will first show that the LHS is equal to the trace ofΘK

´

D´b`?´3

6

¯

Θpωq D13 with b ” 0 mod 3.We will show this by using Shimura reciprocity law. Note first that:

Lemma 6.4. The modular function f0pzq “ΘpDz3q

Θpzqis a modular function for Γp3Dq and

f0pzq has rational Fourier coefficients at the cusp 8.

Proof: The proof that f0 is invariant under Γp3Dq is straightforward. The proof that theFourier coefficients are rational is also similar to the proof of Lemma ??

Lemma 6.5. For f0 as above and τ “ ´b0`?´3

2 , we have f0pτq P H3D.

Proof:To show that fpτq P H3D, we need to look at action of Up3Dq. We follow closely theproof of Lemma ??. We rewrite the primitive ideal A “ pA ` Bωq as A “ ra, ´b`

?´3

2 sZ withb ” b0 mod 3. The only difference is computing:

f0p`

ta´sb ´scas t

˘

zq “ΘK

`

pD 00 3 q

`

ta´sb ´scas t

˘

z˘

ΘK

``

ta´sb ´scas t

˘

z˘ “

ΘK

´´

ta´sb ´scDp3aq3sD t

¯

pDzq¯

ΘK

``

ta´sb ´scas t

˘

z˘ .

Note that we still have´

ta´sb ´scDp3aq3sD t

¯

,`

ta´sb ´scas t

˘

P Γ0p3q, thus we simply get f0pzq

and all the arguments from Lemma ?? follow.

Lemma 6.6. For A “

”

a, ´b`?´3

2

ı

a primitive ideal ideal with a “ NmA and b2 ” ´3

mod 4a, we have:

Θ´

D´b`?´3

6a

¯

Θ´

´b`?´3

2a

¯ “

¨

˝

Θ´

D´b`?´3

6

¯

Θpωq

˛

‚

σ´1A

Proof: Note that f0pzq satisfies the properties of Lemma ??, thus applying its result forf0

´

´b`?´3

2

¯

gives us the result.

From the previous two lemmas, we immediately get the following Corollary:

60

Corollary 6.1. For A “”

a, ´b`?´3

2

ı

primitive ideals that are representatives of ClpO3Dq asin Lemma ??, we have:

TrH3DK

ΘpD´b`?´3

6 q

Θ´

´b`?´3

2

¯ D13 “ÿ

APClpO3Dq

Θ´

´b`?´3

6a

¯

Θ´

´b`?´3

2a

¯χDpAqD13

6.2.1 Traces of theta functions

We will show the following lemma:

Lemma 6.7. For D ” 1 mod 3, b0 ” 0 mod 3 as before, we have:

TrH3DK

Θ´

D´3`?´3

6

¯

ΘpωqD13 “

ÿ

APClpO3Dq

ΘK

´

D´b0`?´3

6a

¯

ΘK

´

D´b0`?´3

6a

¯χDpAqD13 “ 0.

Proof: The method will be to apply Lemma 6.1 two times. We first apply Lemma ?? (i)for z “ 1´2D

6D to get:

Θ

ˆ

1`?´3

6D

˙

“ p1´ ωqΘ

ˆ

1`?´3

2D

˙

` ωΘ

ˆ

1´ 2D `?´3

6D

˙

This can be rewritten as:

Θ´

1`?´3

6D

¯

Θpωq“ p1´ ωq

Θ´

1`?´3

2D

¯

Θpωq` ω

Θ´

1´2D`?´3

6D

¯

Θpωq

By taking the inverses and denoting B1 :“ ´1` 2D, a1 :“ pB21 ` 3q4, we have:

3DΘ´

D´1`?´3

2

¯

Θpω3q“ p1´ ωq

Θ´

1`?´3

2D

¯

Θpωq` 3Dω

Θ´

DB1`?´3

2a

¯

Θ´

B1`?´3

6a

¯

Note that B1 ” 1 ´ 2D ” 1 mod 3. Furthermore, noting that Θpω3q “ p1 ´ ωqΘpωq andΘ´

B1`?´3

6a

¯

“ p1´ ω2qΘ´

B1`?´3

2a

¯

, we get:

3D

1´ ω

Θ´

D´1`?´3

2

¯

Θpωq“ p1´ ωq

Θ´

1`?´3

2D

¯

Θpωq`

3Dω

1´ ω2

Θ´

DB1`?´3

2a1

¯

Θ´

B1`?´3

2a

¯

Multiplying by D13 and rewriting the first term on the RHS, we have:

3D

1´ ω

Θ´

D´1`?´3

2

¯

ΘpωqD13 “ p1´ωqp1´ω2q

Θ´

1`?´3

2D

¯

p1´ ω2qΘpωqD13`

3Dω

1´ ω2χDpA1q

´1Θ´

DB1`?´3

2a1

¯

Θ´

B1`?´3

2a

¯ D13χDpA1q

61

By taking the trace from H3D to K and denoting by A1 :“´

B1`?´3

2

¯

, we have:

3D

1´ ωTrH3DK

Θ pDωq

ΘpωqD13 “ 3 TrH3DK

Θ`

´Dω2˘

Θp´ω23qD13`

3Dω

1´ ω2χDpA1q

´1TrH3DK

Θ´

DB1`?´3

2a1

¯

Θ´

B1`?´3

2a

¯ D13χDpA1q

Note that by definition we have χDpA1q “ χD

´

B1`?´3

2 ω¯

. We can compute the value ofthe character using Lemma ??. For each p|D, we have:

χp

ˆ

B1 `?´3

2ω

˙

“

ˆ

p1´ 2D ´?´3qω2

p1´ 2D `?´3

qω

˙pNm p´1q3

“

ˆ

´1

1

˙pNm p´1q3

“ 1.

Thus we get χDpA1q “ 1, and we can rewrite the equation above as:

3D

1´ ωTrH3DK

Θ pDωq

ΘpωqD13 “ 3 TrH3DK

Θ`

´Dω2˘

Θp´ω23qD13`

3Dω

1´ ω2TrH3DK

Θ´

DB1`?´3

2a1

¯

Θ´

B1`?´3

2a

¯ D13χDpA1q.

Furthermore, from Lemma ??, we haveΘ´

DB1`

?´3

2a1

¯

Θ´

B1`?´3

2a

¯ D13χDpA1q “

´

ΘpDωqΘpωq D

13χDpA1q

¯σ´1A,

thus:

TrH3DK

Θ´

DB1`?´3

2a1

¯

Θ´

B1`?´3

2a

¯ D13χDpA1q “ TrH3DKΘ pDωq

Θ pωqD13

Denoting S :“ TrH3DKΘpDωqΘpωq D

13, we get the relation:

3D

1´ ωS “ 3 TrH3DK

Θ´

1`?´3

2D

¯

Θp´ω23qD13 `

3Dω

1´ ω2S

This implies:

3D

1´ ω2S “ 3 TrH3DK

Θ´

1`?´3

2D

¯

Θp´ω23qD13

This is equivalent to:

D

1´ ω2S “ TrH3DK

Θ´

1`?´3

2D

¯

Θp´ω23qD13

Note that if we apply the transformation z Ñ ´13z given by the functional equation (24)to both theta functions on the RHS we get:

1

1´ ω2S “

1

3TrH3DK

Θ´

D´1`?´3

6

¯

ΘpωqD13

62

This is equivalent to:

p1´ ωqS “ TrH3DK

Θ´

D´1`?´3

6

¯

ΘpωqD13. (26)

We will apply now Lemma 6.1 (ii) for z “ D´b1`?´3

2a , where b1 ” 1 mod 3. We denote byb0 an integer b0 ” 0 mod 3 such that b0 ” b1 mod 4a. Then we have:

Θ

ˆ

D´b0 `

?´3

6a

˙

“ p1´ ω2qΘ

ˆ

D´1`

?´3

2a

˙

` ω2Θ

ˆ

D´b1 `

?´3

6a

˙

This can be rewritten as:

Θ´

D´b0`?´3

6a

¯

Θ´

´b0`?´3

2a

¯ D13χDpAq “ p1´ω2qΘ´

D´b0`?´3

2a

¯

Θ´

´b0`?´3

2a

¯ D13χDpAq`ω2Θ´

D´b1`?´3

6a

¯

Θ´

´b0`?´3

2a

¯ D13χDpAq

By taking the sums, we get:

M “ p1´ ω2qS ` ω2p1´ ωqS “ 0

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Integers that can be written as the sum of two rational cubes › ~rosu › main.pdf · H O is the ray class ﬁeld of modulus 3Dand H 0 is an intermediate ﬁeld K•H 0 •H O...

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