Jamie Beacom - Part C Dissertation HT 2015

Modular Forms of Weight Oneand Quadratic Number Fields

Candidate Number: 473929

Submitted as a Part C Dissertationfor the Final Honour School of Mathematics

University of Oxford

Hilary Term 2015Word Count: 9997

Contents

Introduction 5

1 L-Series and Converse Theorems 71.1 Preliminary Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.2 A Functional Equation for Sk(N,χ) . . . . . . . . . . . . . . . . . . . . . . . . . 111.3 Weil’s Converse Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2 Modular Forms from Quadratic Number Fields 252.1 Großencharakters, Hecke Characters and L-series . . . . . . . . . . . . . . . . . 252.2 Modular Forms from Quadratic Number Fields . . . . . . . . . . . . . . . . . . . 31

3 Representations from Quadratic Number Fields 393.1 Representations of Galois Groups and Artin L-functions . . . . . . . . . . . . . 393.2 Selected Results from Class Field Theory . . . . . . . . . . . . . . . . . . . . . . 413.3 Representations of Galois Groups and Modular Forms . . . . . . . . . . . . . . . 43

4 Examples of Weight One Modular Forms 474.1 A Level 23 Modular Form from Q(

√−23) . . . . . . . . . . . . . . . . . . . . . 47

4.2 A Level 39 Modular Form from Q(√

13), Q(√−3) and Q(

√−39) . . . . . . . . . 49

4.3 A Level 145 Modular Form from Q(√

5) . . . . . . . . . . . . . . . . . . . . . . 52

A Induced Representations 55

Acknowledgements 57

Bibliography 59

3

4 CONTENTS

Introduction

Classical modular forms are an important class of objects to number theorists. They areclosely related to elliptic curves, Diophantine equations and to representations of the absoluteGalois group of Q.

It is, therefore, important to understand modular forms and how to construct them. Thisinvolves understanding the spaces Mk(Γ) of weight k modular forms for a congruence subgroupΓ and, where we can, constructing bases of these spaces. If the weight k is greater than 1,there are geometric methods we can use to achieve this goal. For example, Bryan Birch andYuri Manin independently developed the idea of a modular symbol, which allows for simplercalculations of cohomology in certain smooth projective curves (modular curves) over Q.

The case where the weight is one is unfortunately not directly amenable to these geometricmethods, and the construction of weight one modular forms is correspondingly harder. Thanksto a result due to Deligne and Serre in [3], we know that there is a correspondence betweenweight one normalised cuspidal eigenforms of level N and nebentypus χ and odd 2-dimensionalirreducible complex representations of the absolute Galois group over Q.

Our main aim is to prove Theorem 3.3.1 :

Theorem 3.3.1. Let K/Q be a real or imaginary quadratic field with discriminant dK. Let ξbe a character of a ray class group modulo n, an integral ideal. If K is real, then assume ξadditionally satisfies the condition that

ξ((a)) = 1 ∀a ≡ 1 mod n

Then

• there exists a weight one modular form f(z) of level |dK |NK/Q(n) and nebentypus χ,where χ(m) = χdK (m)ξ((m))

and

• there exists a Galois extension L/Q and a 2-dimensional complex Galois representationρ : Gal(L/Q)→ GL2(C) such that L(s, f) = L(s, ρ)

This says that starting off with a quadratic number field K and a character of some ray classgroup for K, we may construct a weight one modular form and a Galois representation whichhave the same L-series.

In Chapter 1, we introduce and prove Weil’s Converse Theorem (Theorem 1.3.1). This saysthat when a holomorphic function f with a q-expansion satisfies a certain infinite family offunctional equations, then f is modular. This theorem can also determine the weight and level,

5

6 CONTENTS

and whether or not f is a cusp form. The focus of this chapter will be to prove this theorem,and will largely be a re-ordering of material from [2] and [10].

Chapter 2 is a drawing together of material primarily from [11] and [12], followed by materialin [10]. We define the Großencharakters first introduced by Hecke, and use these to describethe characters of ray class groups. When these come from quadratic number fields, we usetheir functional equation and Theorem 1.3.1 to show they define modular forms. For reasons ofspace we do not prove that the L-function of a Hecke Großencharacter satisfies a functionalequation (Theorem 2.1.11).

In Chapter 3 we will explore the ray class groups in more detail. The material here is largelybased on on that from [2] and [11]. By introducing some ideas from Class Field Theory, namelythe Artin Reciprocity Theorem (Theorem 3.2.7) and the Existence Theorem (Theorem 3.2.2),we will be able to find an abelian extension M/K with corresponding Galois group isomorphicto the ray class group via the Artin map. By taking a minimal Galois extension L/Q whichcontains M , we lift our Hecke character to a character of the Galois group Gal(L/K). We canthen induce a complex Galois representation with the same L-series as our modular form.

In Chapter 4, we provide some examples of modular forms constructed in this way. These areoriginal calculations by the author. The first comes from an imaginary quadratic field, thesecond from both real and imaginary quadratic fields and the third from a real quadratic fieldonly. I would like to thank Kevin Buzzard and Steven Donnelly for the Magma code used inmaking these calculations.

I shall assume that the reader is familiar with the B3.1 Polynomial Rings and Galois Theory[13] course given by Xenia de la Ossa in Michaelmas 2012, B3.4 Algebraic Number Theory [6]course given by Victor Flynn in Hilary Term 2013, and, C3.6 Modular Forms [8] course givenby Alan Lauder in Michaelmas Term 2014. For brevity, any results taken from these courseswill not be proved and any definitions will not be stated.

Chapter 1

L-Series and Converse Theorems

Given a quadratic number field, we would like to show how we can construct a modular formof weight one. In order to demonstrate that the functions we construct are modular, we willneed to develop some analytic tools.We shall show how under certain conditions, thecoefficients of an L-series define the coefficients of the q-expansion of a modular form. Theseconditions are given by Weil’s Converse Theorem, and this is the major result of this chapter.Section 1 introduces some background ideas and tools, while Section 2 provides somemotivation for the statement of the theorem. In Section 3, we provide a complete proof ofWeil’s Converse Theorem. Throughout this chapter, we closely follow Bump in [2, §1.5] andMiyake in [10, §4.3].

1.1 Preliminary Concepts

Let Γ be a congruence modular group. Then there is a positive integer N such that Γ ⊃ Γ(N)by definition. Hence

Mk(Γ) ⊂Mk(Γ(N))

If f(z) ∈Mk(Γ(N)), then f(Nz) ∈Mk(Γ1(N2)), and the Fourier coefficients of f(Nz) and f(z)

coincide. Therefore the study of modular forms with respect to congruence subgroups isreduced to the study of of Mk(Γ1(N).

Given a Dirichlet character mod N we may define a character χ of Γ0(N) by

χ(γ) = χ(d), γ =

(a bc d

)∈ Γ0(N) (1.1.1)

We say that f(z) ∈Mk(Γ0(N), χ) if f |kγ = χ(γ)f for every γ ∈Mk(Γ0(N), χ).

We have the following lemma:

Lemma 1.1.1. For a positive integer N, we have

Mk(Γ1(N) =⊕χ

Mk(Γ0(N), χ), Sk(Γ1(N) =⊕χ

Sk(Γ0(N), χ)

where χ runs over all Dirichlet characters mod N .

Proof. This is a standard result. See [10, §4.3 Lemma 4.3.1] for a proof.

7

8 CHAPTER 1. L-SERIES AND CONVERSE THEOREMS

Thus we have reduced our study of congruence subgroups of the modular group to studyingMk(Γ0(N), χ) and Sk(Γ0(N), χ).

For a modular form f(z) =∑∞

n=0 anqn ∈Mk(Γ0(N), χ), we will write

fρ(z) =∞∑n=0

anqn

We also put for any positive integer N ,

ωN =

(0 −1N 0

)Lemma 1.1.2. Let χ be a Dirichlet character mod N . Then

1. If χ(−1) 6= (−1)k then Mk(Γ0(N), χ) = {0}.

2. The map “f 7→ f |kωN” gives an isomorphism:

Mk(Γ0(N), χ) ∼= Mk(Γ0(N), χ)

3. If f(z) ∈Mk(Γ0(N), χ) then

fρ(z) = f(−z)

and fρ(z) ∈Mk(Γ0(N), χ).

Proof. 1. This is obvious since −I ∈ Γ0(N).

2. Note that

ωNΓ0(N)ω−1N = Γ0(N)

Moreover, if γ =

(a bc d

)∈ Γ0(N) then N |c and ad− bc = 1 so that χ(d) = χ(a).

Hence, setting g = f |kωN , we have

g|kγ = f |kωNγω−1N |kωN = χ(a)g = χ(d)g

The remainder is then straightforward.

3. The first part is straightforward. For an element γ =

(a bc d

)∈ Γ(1) we let

γ′ =

(a −b−c d

)Then we observe that

fρ|kγ = (f |kγ′)ρ (1.1.2)

So in particular if γ ∈ Γ0(N) then fρ|kγ = χ(γ)fρ. Identity (1.1.2) verifies conditions atthe cusps.

1.1. PRELIMINARY CONCEPTS 9

Remark 1.1.3. Let H be the upper half plane. Consider the set of holomorphic functions f(z)on H with a Fourier expansion

f(z) =∞∑n=0

anqn (1.1.3)

converging absolutely and uniformly on any compact subset of H, and for which there is av > 0 such that

f(z) = O(=(z)−v) (=(z)→ 0)

uniformly on <(z). This set has a bijective correspondence to all sequences of complexnumbers {an}∞n=0 such that an = O(nv) for all n for some v > 0.

To determine the behaviour of weakly modular functions near cusps we need the followinglemma:

Lemma 1.1.4. Let f(z) be a weakly modular holomorphic function of weight k with respect toa subgroup Γ of Γ(1) of finite index. If there exists a positive real number v such that

f(z) = O(=(z)−v) (=(z)→ 0)

uniformly with respect to <(z), then f belongs to Mk(Γ). Moreover, if we can take v so thatv < k then f belongs to Sk(Γ).

Proof. Let x be a cusp of Γ. If x ∈ Q, take σ ∈ SL2(Z) so that σx =∞. Let h be a positiveinteger satisfying

σΓxσ−1.{±I2} = {±

(1 h0 1

)m: m ∈ Z}

where Γx is the stabiliser of x in Γ. Then f |kσ−1 has an expansion of the form:

(f |kσ−1)(z) =∞∑

n=−∞

ane2πinz/h

which is uniformly convergent on any compact subset of H. The coefficient an can be expressedas

an =1

h

∫ z0+h

z0

(f |kσ−1)(z)e−2πinz/hdz

for any fixed z0 ∈ H. Write σ−1 =

(a bc d

). Then c 6= 0 since σ−1∞ ∈ Q. Now =(σ−1z)

==(z)/|cz + d|2 = O(1/=(z)) as =(z)→∞ uniformly on |<(z)| ≤ h/2. By assumption thenwe have

f |k(σ−1)(z) = f(σ−1z)(cz + d)−k

= O(=(z)v−k) (=(z)→∞)

uniformly on |<(z) ≤ h/2. Taking z0 = −h/2 + iy in our expression for an we get

|an| = O(yv−ke2πny/h)

If n < 0 then an = 0. Moreover, if v < k then a0 = 0. In other words f(z) is holomorphic at x,and if v < k then f(z) has a zero at x. Now let x =∞. Then we note that Γ 6= Γ∞. So for anyγ ∈ Γ \ Γ∞, a real point γ∞ is a cusp of Γ equivalent to ∞. Therefore f(z) is holomorphic atall cusps of Γ, and if v < k then f(z) has a zero at every cusp.


As a preliminary to tackling Weil’s Converse theorem, we now introduce two important tools:

Lemma 1.1.5 (Phragmen-Lindelof Principle). For two real numbers v1 and v2 (v1 < v2) put

F = {s ∈ C|v1 ≤ <(s) ≤ v2}

Let φ be a holomorphic function on a domain containing F and satisfying

|φ(s)| = O(e|τ |δ

) (|τ | → ∞), s = σ + iτ

uniformly on F with δ > 0. For a real number b if

|φ(s)| = O(|τ |b) (|τ | → ∞), <(s) = v1 or v2

then

|φ(s)| = O(|τ |b) (|τ | → ∞), uniformly on F

Proof. By assumption there exists L > 0 such that |φ(s)| ≤ Le|τ |δ. Suppose now that b = 0,

then there is an M > 0 such that |φ(s)| ≤M on the lines <(s) = v1 and <(s) = v2. If we nowtake a positive integer m such that m ≡ 2 mod 4 then putting s = σ + iτ , <(sm) is apolynomial in σ and τ , with highest term of τ being −τm. So

<(sm) = −τm +O(|τ |m−1) (|τ | → ∞)

uniformly on F so that <(sm) is bounded above on F . Taking m and N so that m < δ and<(sm) ≤ N we have for any ε > 0

|φ(s)eεsm| ≤MeεN on <(s) = v1 and <(s) = v2

and

|φ(s)eεsm | = O(e|τ |

δ−ετm)→ 0 (|τ | → ∞)

By the maximum-modulus principle, then we must have that

|φ(s)eεsm| ≤MeεN ∀ s ∈ F

If we now let ε→ 0 then we get |φ(s)| = O(|τ |0).

If we now assume that b 6= 0, we may define a holomorphic function ψ(s) by

ψ(s) = eb log(s−v1+1)

where log takes the principal value. Now as <(log(s− v1 + 1) = log |s− v1 + 1|, we haveuniformly on F that

|ψ(s)| = |s− v1 + 1|b ∼ |τ |b (|τ | → ∞)

Now set φ1(s) = φ(s)/ψ(s). Then φ1 satisfies the same assumptions as φ with b = 0, and so φ1

is bounded on F . So we obtain our result.

We will also need to make use of the following transformation in Section 3:

1.2. A FUNCTIONAL EQUATION FOR SK(N,χ) 11

Definition 1.1.6. Let φ be a continuous function on the open interval (0,∞). We define theMellin transform of φ to be

Φ(s) =

∫ ∞0

φ(y)ysdy

y

wherever this integral is absolutely convergent.

We note that if ∫ 1

0

φ(y)ysdy

y

is absolutely convergent for some s0, it is absolutely convergent on <(s) > <(s0), and if∫ ∞1

φ(y)ysdy

y

is absolutely convergent for some s1, it is absolutely convergent on <(s) < <(s1). So there areσ1, σ2 ∈ [−∞,∞] such that Φ(s) is defined for <(s) ∈ (σ1, σ2) and is divergent when <(s) liesoutside this interval. So, if Φ(s) has a non-empty domain, then its domain is a vertical strip.

Lemma 1.1.7 (Mellin Inversion Formula). Suppose that Φ(s) is the Mellin transform of somefunction φ(y) defined on some non-empty vertical strip σ1 < <(s) < σ2. For σ ∈ (σ1, σ2) and0 < y <∞ we have:

φ(y) =1

2πi

∫ σ+i∞

σ−i∞Φ(s)y−sds

Proof. This is an application of the inversion formula for the Fourier transform. See [2, §1.5]just after the statement of Proposition 1.5.1 for a proof.

1.2 A Functional Equation for Sk(N,χ)

In this section we will demonstrate a functional equation for elements of Sk(Γ0(N), χ) twistedby Dirichlet characters, motivating Weil’s Converse Theorem. Hecke believed this reasoningshould be reversible. In the more simple case of Sk(Γ0(N)), Hecke did find a converse theorem,which is stated below.

Here will set out some notation:

Mk(N,χ) := Mk(Γ0(N), χ)

and

Sk(N,χ) := Sk(Γ0(N), χ)

First, we turn our attention to holomorphic functions on the upper-half plane and theirassociated Dirichlet series.

Definition 1.2.1. For a holomorphic function f(z) on H with a Fourier expansion

f(z) =∞∑n=0

anqn


converging absolutely and uniformly on any compact subset of H, we put

L(s, f) =∞∑n=1

ann−s

This is the Dirichlet series associated to f .

Now, by Remark 1.1.3, we must have that for some v > 0, an = O(nv) and so L(s, f) convergesabsolutely and uniformly on any compact subset of <(s) > 1 + v, and so is holomorphic on<(s) > 1 + v.

Definition 1.2.2. For N > 0 we put

ΛN(s, f) = (2π/√N)−sΓ(s)L(s, f)

We call this the completed Dirichlet series associated to f .

We now state the first converse theorem due to Hecke. It is a simpler cousin to Weil’s ConverseTheorem, but will be important in motivating the statement and proof:

Theorem 1.2.3 (Hecke). Let f(z) =∑∞

n=0 anqn and g(z) =

∑∞n=0 bnq

n be holomorphicfunctions on H satisfying the conditions in Remark 1.1.3. For positive numbers k and N , thefollowing conditions (A) and (B) are equivalent:

(A) g(z) = (−i√Nz)−kf(−1/Nz)

(B) Both ΛN(s, f) and ΛN(s, f) can be analytically continued to the whole s-plane, andsatisfy the functional equation

ΛN(s, f) = ikΛN(k − s, g)

and

ΛN(s, f) +a0s

+ikb0k − s

is holomorphic on the whole s-plane and bounded in any vertical strip.

Proof. A proof of this is can be found in [10, §4.3, Theorem 4.3.5].

Since any element f(z) of Sk(N,χ) satisfies the conditions of Remark 1.1.3, we obtain thefollowing:

Corollary 1.2.4. For any element f(z) of Sk(N,χ), ΛN(s, f) is holomorphic on the wholes-plane and satisfies the functional equation

ΛN(s, f) = ikΛN(k − s, f |kωN)

Proof. This is clear by applying Theorem 1.2.3 to f and f |kωN , and noting that f is a cuspform.

We may show (with extra work) that the functional equation in Corollary 1.2.4 is also satisfied

by elements of Mk(N,χ). As a consequence of this, we note that since ω1 =

(0 −11 0

)and

T =

(1 10 1

), we may obtain the following result:


Theorem 1.2.5. Let k ≥ 2 be an even integer. Assume that a holomorphic function f(z) on Hsatisfies the conditions of Remark 1.1.3. Then f(z) belongs to Mk(Γ(1)) if and only ifΛ(s, f) = (2π)−sΓ(s)L(s, f) can be analytically continued to the whole s-plane,

Λ(s, f) +a0s

+(−1)k/2a0k − s

is holomorphic on H and bounded on any vertical strip, and Λ(s, f) satisfies the functionalequation:

Λ(s, f) = (−1)k/2Λ(k − s, f)

Moreover if a0 = 0 then f(z) is a cusp form.

For N > 1 we will need to introduce twists of the L-functions since Γ(N) will have morecomplicated generators than those described above:

Definition 1.2.6. For a holomorphic function f(z) satisfying the conditions of Remark 1.1.3and a Dirichlet character ψ, we put

fψ(z) =∞∑n=0

ψ(n)anqn (1.2.1)

and

L(s, f, ψ) =∞∑n=0

ψ(n)ann−s (1.2.2)

Clearly fψ also satisfies the conditions of Remark 1.1.3. So the sum in (1.2.1) convergesabsolutely and uniformly on compact subsets of H, and fψ is holomorphic on the upper-halfplane.

Definition 1.2.7. Let m = mψ be the conductor of ψ, and put

ΛN(s, f, ψ) = (2π/m√N)−sΓ(s)L(s, f, ψ)

Thus, by definition

L(s, fψ) = L(s, f, ψ) (1.2.3)

ΛNm2(s, fψ) = ΛN(s, f, ψ) (1.2.4)

Lemma 1.2.8. Let f(z) and g(z) be two holomorphic functions H satisfying the conditions ofRemark 1.1.3, and ψ a primitive Dirichlet character of conductor m(> 1). The the followingconditions (Aψ) and (Bψ) are equivalent:

(Aψ)fψ|kωNm2 = Cψgψ

(Bψ) ΛN(s, f, ψ) can be holomorphically continued to the whole s-plane, bounded on anyvertical strip, and satisfies the functional equation

ΛN(s, f, ψ) = ikCψΛN(k − s, g, ψ)

for a constant Cψ.


Proof. Apply Theorem 1.2.3 for fψ, Cψgψ and Nm2 in place of f , g and N .

Now it is useful to define for a ∈ R

α(a) =

(1 a0 1

)For a primitive Dirichlet character ψ of conductor m, let W (ψ) :=

∑m−1a=0 ψ(a)e2πia/m be the

Gauss sum of ψ. The Gauss sum has the following properties:

• For any integer b, ψ(b)W (ψ) =∑m−1

a=0 ψ(a)e2πiab/m

• W (ψ) = ψ(−1)W (ψ)

• |W (ψ)|2 = m

For a proof, of these properties, see [4, §4.3] or [10, §3.1].

Lemma 1.2.9. 1. Let f be a holomorphic function on H satisfying the conditions ofRemark 1.1.3, and ψ a primitive Dirichlet character of conductor m. Then for an integerk > 0, we have

W (ψ)fψ =m∑u=1

ψ(u)(f |kα(u/m))

2. Let f(z) be an element of Mk(N,χ), mχ the conductor of χ, ψ a primitive Dirichletcharacter of conductor m = mψ and M the least common multiple of N , m2 and mmχ.Then fψ ∈Mk(M,χψ2).

Proof. 1. For any integer u

(f |kα(u/m))(z) =∞∑n=0

ane2πinu/mqn

so by using the properties of W (ψ) listed above

m∑u=1

ψ(u)(f |kα(u/m))(z) =∞∑n=0

(m∑u=1

ψ(u)e2πinu/m

)anq

n

=∞∑n=0

W (ψ)ψ(n)anqn

= W (ψ)fψ(z)

2. Assume f ∈Mk(N,χ) and set m = mψ. As

α(u/m)−1Γ0(N)α(u/m) ⊃ Γ(Nm2)

f |kα(u/m) belongs to Mk(Γ(Nm2)). From 1. we see that fψ must also be inMk(Γ(Nm2)) as a finite linear combination of elements of this space.

Now let γ =

(a bcM d

)∈ Γ0(M). We let


γ′ := α(u/m)γα(d2u/m)−1

which we can easily check lies in Γ0(M). Also, f ∈Mk(M,χ) since Γ0(M) ⊂ Γ0(N). If

we let γ′ =

(a′ b′

c′ d′

)∈ Γ0(M), we have

d′ ≡ d− cd2uM/m ≡ d mod mχ

and so

f |kα(u/m)γ = f |kγ′α(d2u/m) = χ(d′)f |kα(d2u/m) = χ(d)f |kα(d2u/m)

Hence by 1.:

fψ|kγ = W (ψ)−1

(m∑u=1

ψ(u)(f |kα(u/m))

)|kγ

= W (ψ)−1m∑u=1

ψ(u)(f |kα(u/m))|kγ

= W (ψ)−1m∑u=1

ψ(u)(f |kα(u/m)γ)

= W (ψ)−1m∑u=1

ψ(u)χ(d)(f |kα(d2u/m))

= χ(d)ψ(d2)

(W (ψ)−1

m∑u=1

ψ(u)ψ(d2)(f |kα(d2u/m))

)

= χ(d)ψ(d2)

(W (ψ)−1

∞∑n=0

(m∑u=1

ψ(u)ψ(d2)e2πid2nu/m

)anq

n

)

= χ(d)ψ(d2)

(W (ψ)−1

∞∑n=0

W (ψ)ψ(n)ψ(d2)ψ(d2)anqn

)(by properties of W (ψ))

= χ(d)ψ2(d)fψ = (χψ2)(d)fψ.

Theorem 1.2.10. Let f(z) be an element of Mk(N,χ), and ψ a primitive Dirichlet characterof conductor m. If (m,N) = 1, then

fψ|kωNm2 = Cψgψ

where g = f |kωN and

Cψ = χ(m)ψ(−N)W (ψ)

W (ψ)= χ(m)ψ(N)

W (ψ)2

m

Proof. For u an integer coprime to m, we take integers n, v so that nm−Nuv = 1. Note thenthat ψ(u) = ψ(−Nv). Then

α(u/m)ωNm2 = mωN

(m −v−uN n

)α(v/M) (1.2.5)


As g = f |kωN is in Mk(N,χ) by Lemma 1.1.2, this implies

f |kα(u/m)ωNm2 = χ(n)g|kα(v/m) = χ(m)g|kα(v/m)

so that by Lemma 1.2.9

W (ψ)fψ|kωNm2 =m∑u=1

ψ(u)f |kα(u/m)ωNm2

= χ(m)m∑v=1

ψ(−Nv)g|kα(v/m)

= χ(m)ψ(−N)m∑v=1

ψ(v)g|kα(v/m)

= χ(m)ψ(−N)W (ψ)gψ

on noting that the contribution from terms in the sum where gcd(u,m) 6= 1 is 0.

Applying this theorem together with Lemma 1.2.8 we obtain the following:

Theorem 1.2.11. Let f(z) be an element of Sk(N,χ), and ψ a primitive Dirichlet characterof conductor m. If (m,N) = 1, then ΛN(s, f, ψ) can be holomorphically continued to the wholes-plane, is bounded on any vertical strip, and satisfies the functional equation:

ΛN(s, f, ψ) = ikχ(m)ψ(N)W (ψ)2

mΛN(k − s, f |kωN , ψ)

.

1.3 Weil’s Converse Theorem

We can now give a statement of Weil’s Converse Theorem, motivated by Theorem 1.2.11. Theproof will be the focus of the remainder of this chapter. The main idea will be that nonon-zero holomorphic function on H can be fixed by an infinite order elliptic transformation.

For coprime integers a and b put

A(a, b) = {a+ nb : n ∈ Z}

and let MN be the set of odd prime numbers or 4 satisfying the conditions

1. Any element of MN is prime to N

2. MN ∩ A(a, b) 6= ∅ for all A(a, b)

By Dirichlet’s theorem on primes in arithmetic progressions, MN to be the set of all oddrational primes coprime to N.

Theorem 1.3.1 (Weil). Let k and N be two positive integers, and χ a Dirichlet charactermod N such that χ(−1) = (−1)k. For two sequences of complex numbers {an}∞n=0 and {bn}∞n=0

such that an = O(nv), bn = O(nv)(v > 0) put

1.3. WEIL’S CONVERSE THEOREM 17

f(z) =∞∑n=0

ane2πinz

and

g(z) =∞∑n=0

bne2πinz

for z ∈ H.

Then f ∈Mk(N,χ), g ∈Mk(N,χ) and g = f |kωN , if the following conditions are satisfied:

1. ΛN(s, f) and ΛN(s, g) satisfy condition (B) in Theorem 1.2.3;

2. for any primitive Dirichlet character ψ whose conductor mψ is a member of MN ,ΛN(s, f, ψ) and Λ(s, g, ψ) satisfy condition (Bψ) in Lemma 1.2.8 with the constant

Cψ = χ(m)ψ(N)W (ψ)2/m

Moreover, if L(s, f) is absolutely convergent at s = k − δ for δ > 0, then f and g are cuspforms.

As previously stated, we need to know how infinite order elliptic transformations affectholomorphic functions on H. This is given by the following lemma:

Lemma 1.3.2. Let f be a holomorphic function on H, and let M ∈ SL2(R) be an ellipticelement of infinite order such that f |kM = f . Then f ≡ 0.

Proof. Let a ∈ H be the fixed point of the elliptic element M . Consider the Cayley transform

C =

(1 −a1 −a

)which maps H onto D the unit disk, and takes a to the origin. Then

CMC−1 =

(α 00 α−1

)where α and α−1 are eigenvalues of M. They have absolute value 1 as M is elliptic, but are notroots of unity as M has infinite order. Let f1 = f |kC−1. Then

f1 = f1|k(α 00 α−1

), f1(z) = akf1(a

2z) (1.3.1)

Now we know that f1 has a power series expansion f1(z) =∑λnz

n. By (1.3.1) we haveλn = α2n+kλn, and as α is not a root of unity we must have that λn = 0 for all n. The resultthen follows.

If we could construct an elliptic transformation of infinite order fixing f |k(χ(m)− γ) forγ ∈ Γ0(N) then we would be able to demonstrate that f is invariant under the action of Γ0(N)by the slash operator. Now for integers m, v with (m, vN) = 1 we take integers n, u so thatmn− uvN = 1 and set

γ(m, v) =

(m −v−uN n

)∈ Γ0(N)


Clearly γ(m, v) is not uniquely determined, but u mod m is and

α(u/m)ωNm2 = m.wNγ(m, v)α(v/m) (1.3.2)

Now, we extend the weight k action of GL+2 (R) on functions on H linearly to the group algebra

C[GL+2 (R)], setting for a function f(z) on the upper-half plane

f |kβ =∑α

aαf |kα for β =∑α

aαα ∈ C[GL+2 (R)] (1.3.3)

We will use the fact that matrix γ(m, v) is not uniquely determined to construct thetransformation we want. To do this, however, we need to show that we can use it in a welldefined way:

Lemma 1.3.3. Let k and N be positive integers, χ a Dirichlet character mod N satisfyingχ(−1) = (−1)k, and f and g are holomorphic functions on H satisfying the conditions ofRemark 1.1.3 and conditions (A) from Theorem 1.2.3. Let m be an odd prime number or 4prime to N . If f and g satisfy (Aψ) in Lemma 1.2.8 for all primitive Dirichlet characters ψmodulo m with the constant

Cψ = χ(m)ψ(N)W (ψ)2

m

then for integers u, v prime to m

g|k(χ(m)− γ(m,u))α(u/m) = g|k(χ(m)− γ(m, v))α(v/m)

Proof. By assumption (Aψ) and Lemma 1.2.9:

m∑u=1

ψ(u)f |kα(u/m)ωNm2 = χ(m)ψ(−N)m∑u=1

ψ(u)g|kα(u/m) (1.3.4)

on noting that∑m

u=1 ψ(u)f |kα(u/m)ωNm2 =(∑m

u=1 ψ(u)f |kα(u/m))|kωNm2

For each integer u prime to m, take an integer v so that −uvN ≡ 1 mod m. By (1.3.2) thenwe have

f |kα(u/m)ωNm2 = g|kγ(m, v)α(v/m) (1.3.5)

by using assumption (A). As the left-hand side is independent of the choice of a representativeof u mod m, the right-hand side independent of the choice of a representative of γ(m, v). Using(1.3.5) , we may rewrite (1.3.4) as∑

v

ψ(v)g|k(χ(m)− γ(m, v))α(v/m) = 0 (1.3.6)

where v runs over a complete set of representatives for Z/mZ in Z using the identity ψ(u) =ψ(v)ψ(−N). Note that (1.3.6) is independent of the choice of such representatives. Let v1, v2be two integers prime to m. Multiply both sides of (1.3.6) by ψ(v1)− ψ(v2) and take sum withrespect to all non-trivial Dirichlet characters ψ mod m. Note that for the trivial characterψ(v1)− ψ(v2) = 0.

Dirichlet characters satisfy the following orthogonality relations:

∑ψ

ψ(v)ψ(v1) =

{ϕ(m) if v = v1

0 otherwise


and

∑ψ

ψ(v)ψ(v2) =

{ϕ(m) if v = v2

0 otherwise

A proof of these can be found in [4, §4.3].

Hence

0 =∑ψ

(ψ(v1)− ψ(v2))∑v

ψ(v)g|k(χ(m)− γ(m, v))α(v/m)

=∑v

∑ψ

(ψ(v1)− ψ(v2))ψ(v)g|k(χ(m)− γ(m, v))α(v/m)

=∑v

g|k(χ(m)− γ(m, v))α(v/m)∑ψ

ψ(v1)ψ(v)−∑v

g|k(χ(m)− γ(m, v))α(v/m)∑ψ

ψ(v2)ψ(v)

= g|k(χ(m)− γ(m, v1))α(v1/m)ϕ(m)− g|k(χ(m)− γ(m, v2))α(v2/m)ϕ(m)

= (g|k(χ(m)− γ(m, v1))α(v1/m)− g|k(χ(m)− γ(m, v2))α(v2/m))ϕ(m)

Noting that ϕ(m) 6= 0 completes the proof.

This identity tells us that for our purposes the choice of γ(m, v) doesn’t matter provided thatconditions (A) and (Aψ) hold. We will now use this to our advantage to show that a specificsubset of Γ0(N) leaves g fixed under the weight k action.

Lemma 1.3.4. Let m and n be odd prime numbers or 4. Assume both m and n are prime toN . If f(z) and g(z) satisfy condition (Aψ) in Lemma 1.2.8 with the constantCψ = χ(m)ψ(−N)W (ψ)/W (ψ) for any primitive Dirichlet character ψ whose conductormψ = m or n, then

g|kγ = χ(γ)g

for all γ ∈ Γ0(N) of the form γ =

(m −v−uN n

).

Proof. Set γ′ =

(m vuN n

). Since we may take γ and γ′ for γ(m, v) and γ(m,−v) in Lemma

1.3.3 respectively, we have

g|k(χ(m)− γ′)α(−v/m) = g|k(χ(m)− γ)α(v/m)

so that

g|k(χ(m)− γ′)α(−2v/m) = g|k(χ(m)− γ) (1.3.7)

We may also take γ′−1 =

(n −v−uN m

)and γ−1 =

(n vuN m

)for γ(n, v) and γ(n,−v) and

obtain:

g|k(χ(n)− γ−1)α(−2v/n) = g|k(χ(n)− γ′−1) (1.3.8)

As χ(n)χ(m) = 1, we see that


χ(n)− γ′−1 = −χ(n)(χ(m)− γ′)γ′−1

(χ(n)− γ−1)α(−2v/n) = −χ(n)(χ(m)− γ)γ−1α(−2v/n)(1.3.9)

as elements of C[GL+2 (R)]. Rewriting (1.3.8) using (1.3.9), we have

g|k(χ(m)− γ′) = g|k(χ(m)− γ)γ−1α(−2v/n)γ′ (1.3.10)

and so by (1.3.7)

g|k(χ(m)− γ)(1− γ−1α(−2v/n)γ′α(−2v/m)) = 0 (1.3.11)

If we now put

h = g|k(χ(m)− γ)

and

β = γ−1α(−2v/n)γ′α(−2v/m) =

(1 −2v/m

2uN/m 4/mn− 3

)observe that h(z) is holomorphic on H and

h|kβ = β (1.3.12)

Observe that |tr(β)| = | − 2 + 4/mn| < 2, and that |tr(β)| 6= 0, 1 by the conditions imposed onm and n. Hence β is an elliptic transformation. Note that as the entries of β are all rationalnumbers, the eigenvalues of β belong to a quadratic field. Hence, if an eigenvalue of β were tobe a root of unity, it would have to be one of ±1,±i,±eπi/3 or ±e2πi/3, so that tr(β) = 0,±1 or±2, a contradiction. If β has finite order, then necessarily its eigenvalues are roots of unity. Soβ has infinite order.

So β is an elliptic transformation of infinite order. By Lemma 1.3.2 then h ≡ 0 so thatg|kγ = χ(m)g.

We are now ready to prove Theorem 1.3.1:

Proof of Theorem 1.3.1. First we will prove here that condition (B) implies that g = f |kωN .The Mellin inversion formula implies that

e−t =1

2πi

∫ α+i∞

α−i∞Γ(s)t−sds (α > 0)

and hence that for any α > 0

f(iy) = a0 +1

2πi

∞∑n=1

an

∫ α+i∞

α−i∞Γ(s)(2πny)−sds

If α > v + 1, then L(s, f) is uniformly convergent and bounded on <(s) = α, so that bySterling’s estimate

Γ(s) ∼√

2πτα−1/2e−π|τ |/2 (s = α + iτ, |τ | → ∞)


we may check that ΛN(s, f) = (2π/√N)−sΓ(s)L(s, f) is absolutely integrable (that is to say∫ α+i∞

α−i∞ |ΛN(s, f)|ds <∞). Hence we may change the order of integration and summation, andwe get that

f(iy) = a0 +1

2πi

∫ α+i∞

α−i∞(√Ny)−sΛN(s, f)ds

Sterling’s estimate gives us that

|ΛN(s, f)| = O(|=(s)|−µ) (|=(s)| → ∞)

for any µ > 0 on <(s) = α. Now choose β so that k − β > v + 1. A similar argument impliesthat for any µ > 0

|ΛN(s, f)| = |ΛN(k − s, g)| = O(|=(s)|−µ) (|=(s)| → ∞)

on <(s) = β. By assumption

ΛN(s, f) +a0s

+b0

k − sis bounded and holomorphic on the domain β ≤ <(s) ≤ α. So for any µ > 0 we may apply thePhragmen-Lindelof Principle and we find that

|ΛN(s, f)| = O(|=(s)|−µ) (|=(s)| → ∞)

holds uniformly on β ≤ <(s) ≤ α. Furthermore, we may assume that α > k and β < 0. Since(√Ny)−sΛN(s, f) has simple poles at s = 0 and s = k with residues a0 and (

√Ny/i)−kb0

respectively, we may change the path of integration from <(s) = α to <(s) = β – consider thepath around the rectangle with vertices β ± iR and α± iR for R > 0 and then let R→∞. Sowe obtain

f(iy) = (√Ny/i)−kb0 +

1

2πi

∫ β+i∞

β−i∞(√Ny)−sΛN(s, f)ds

The functional equation in condition (B) then gives

f(iy) = (√Ny/i)−kb0 +

1

2πi

∫ β+i∞

β−i∞(√Ny)−sikΛN(k − s, g)ds

= (√Ny/i)−kb0 +

1

2πi

∫ k−β+i∞

k−β−i∞(√Ny)s−kikΛN(s, g)ds

= ik(

(√Ny)−kb0 +

1

2πi

∫ k−β+i∞

k−β−i∞(√Ny)s−kΛN(s, g)ds

)= ik(

√Ny)−kg(−1/Niy)

Since f and g are holomorphic functions on H, then the differencef(z)− (

√Nz)−ki2kg(−1/Nz) is holomorphic on H. The difference also vanishes on the positive

imaginary axis, and since the upper-half plane is connected and open, we have by the IdentityTheorem that the difference is identically 0 on the upper-half plane.

Hence, we must have that

f(z) = (−1)k(√Nz)−kg(−1/Nz)


or

f(−1/Nz) = (−1)k(−√N/Nz)−kg(z)⇒ g(z) = (

√Nz)−kf(−1/NZ)

and so g = f |kωN . Condition (2) implies that condition (Aψ) in Lemma 1.2.8 holds for f and gby a proof identical to the one above.

We will now prove that g|kγ = χ(γ)g for γ =

(a bcN d

)∈ Γ0(N). If c = 0 then a = d = ±1,

so that g|kγ = χ(d)g = χ(γ)g as χ(−1) = (−1)k and e2πib = 1. Now assume c 6= 0. As(a, cN) = (d, cN) = 1, there exist integers s, t such that a+ tcN ∈MN and d+ scN ∈MN bycondition (2) on MN . Set

m = a+ tcN n = d+ scNu = −c v = −(b+ sm+ stuN + nt)

Then (a bc d

)=

(1 −t0 1

)(m −v−uN n

)(1 −s0 1

)Hence, Lemma 1.3.4

g|kγ = χ(n)g = χ(d)g = χ(γ)g

Now by Remark 1.1.3, g(z) = O(=(z)−v), so that by Lemma 1.1.4 we have g ∈Mk(N,χ). Sincef = (−1)kg|kωN by condition (1), we have f ∈Mk(N,χ) by Lemma 1.1.2 (2). So we have thefirst half of the assertion.

Now assume that L(s, f) is absolutely convergent at s = k − δ for a δ > 0. Put

c0 = 0

cn =n∑

m=1

|am| (n ≥ 1)

Then for n ≥ 1

cn =n∑

m=1

|am| =n∑

m=1

|am|msm−s

≤ ns(n∑

m=1

|am|m−s)

≤ nk−δ(∞∑m=1

|am|m−k+δ)

so that cn = O(nk−δ).

It is a well-known result due to Euler and Gauss that for l > 0 the Gamma function satisfies

limn→∞

nl

(−1)n(−l−1

n

) = Γ(l + 1)

So there is an L ≥ 0 such that


cn ≤ L(−1)n(−k + δ − 1

n

)for all n ≥ 0. Setting z = x+ iy we have

∞∑n=0

cn|qn| = L

(∞∑n=0

(−1)n(−k + δ − 1

n

)e−2πny

)= L(1− e−2πy)−k+δ−1

Since (1− e−2πy) = O(y) as y → 0, we see that∑∞

n=0 cne−2πy is convergent and∑∞

n=0 cne−2πy = O(y−k+δ−1)(y → 0). As |an| = cn − cn−1, we see that

|f(z)− a0| ≤ (1− e−2πy)

(∞∑n=1

cne−2πy

)so that f(z) = O(yk−δ). So by Lemma 1.1.4 we must have that f is a cusp form, and hence sois g.

Chapter 2

Modular Forms from QuadraticNumber Fields

This chapter comes in two sections. In the first, we introduce the Großencharakters. These arecharacters of the group of fractional ideals of a number field, introduced by Hecke as ageneralisation to the Dirchlet characters for the rational integers. With these, we can definethe Hecke characters. These are characters of ray class groups which are generalisations ofclass groups.

We will then show how a Hecke character coming from a quadratic number field may be usedto construct a modular form. We will use the functional equation for the L-series of the Heckecharacter and Weil’s Converse Theorem (Theorem 1.3.1) to show modularity.

2.1 Großencharakters, Hecke Characters and L-series

We will follow Neukirch in [11, Chapter VII, §6 - §7] and Snyder in [12, §1.4.5-§1.4.10],providing the proof to Lemma 2.1.13. The definition of the Gauss sum we use is that of Miyakein [10, §3.3].

Let n be an integral ideal of a field extension K/Q and let IK be the set of fractional ideals ofOK , and PK the principal ideals of OK . We say that a ∈ K× is congruent to 1 modulo n ifa = b/c for algebraic integers b, c relatively prime to n and b ≡ c mod n.

Suppose that K/Q is of degree n = r1 + 2r2 where r1 is the number of real embeddings of K/Qand r2 is the number of pairs of complex embeddings of K/Q. Let ρ1, .., ρr1 , τ1, τ1, .., τr2 , τ r2 bethe real and complex embeddings of K. The let σ : K → (R×)r1 × (C×)r2 be the embeddingsending x 7→ (ρ1(x), .., ρr1(x), τ1(x), .., τr2(x)).

Definition 2.1.1. Let InK be the set of fractional ideals of K relatively prime to n. AGroßencharakter modulo n is a character ξ : InK → S1, where S1 is the unit circle in C, forwhich there are a pair of characters

ξf : (OK/n)× → S1 and ξ∞ : (R×)r1 × (C×)r2 → S1

such that

ξ((a)) = ξf (a)ξ∞(σ(a))

for every algebraic integer a ∈ OK coprime to n Note that we usually write ξ∞(a) for ξ∞(σ(a)).We may extend the definition of ξ to IK by letting ξ(a) = 0 for any a ∈ IK not coprime to n.

25

26 CHAPTER 2. MODULAR FORMS FROM QUADRATIC NUMBER FIELDS

If we consider any ε ∈ Un := {a ∈ O×K | a ≡ 1 mod n}, then as ε is both a unit and congruentto 1 mod n, we have 1 = ξ((ε)) = ξf (ε)ξ∞(ε) = ξ∞(ε). Hence ξ∞ factors through(R×)r1 × (C×)r2/Un, where Un = σ(Un).

We may explicitly describe the characters of (R×)r1 × (C×)r2 :

Proposition 2.1.2. The characters λ : (R×)r1 × (C×)r2 → S1 are given explicitly by

λ(x1, ...xr1 , z1, ...zr2) =

r1∏l=1

(xl|xl|

)pl|xl|iql

r2∏k=1

(zk|zk|

)p′k|zk|iq

′k

where the pl’s are 0 or 1, the p′k’s are integers and the q’s and q′’s are real numbers.

Proof. Let x ∈ (R×)r1 × (C×)r2 . Then we note that we may write this as x =x

|x||x|, and so we

obtain then a decomposition of (R×)r1 × (C×)r2 :

(R×)r1 × (C×)r2 = U.Rr1+r2+

where U = {x ∈ (R×)r1 × (C×)r2 : |x| = 1}. We determine the characters of U and Rr1+r2+

separately.

U =

r1∏l=1

{±1} ×r2∏k=1

S1

The characters of {±1} clearly correspond to exponentiation by 0 or 1. The characters of S1

correspond to exponentiation by an integer power. To find the characters of Rr+s+ ,we consider

the log map:

log : Rr1+r2+ → Rr1+r2

±

A character of R± corresponds to a map

y 7→ eiqy

for some real number q. Using this and the log isomorphism, we get that the characters ofRr1+r2

+ are such that

x 7→r1∏l=1

xiqll

r2∏k=1

xiq′kk

for some real numbers ql and q′k. Putting these three pieces together finishes the proof.

Now, to the integral ideal n we may associate the following group, which generalises the notionof the class group:

Definition 2.1.3. Let K be a number field and let n be an integral ideal of K.

Let

P nK = {(α) ∈ PK | α ≡ 1 mod n, α is totally positive}

where totally positive means that σ(a) > 0 for all embeddings σ : K ↪→ R. Then we call

ClnK :=InKP nK

the ray class group modulo n and P nK is a ray class modulo n.

2.1. GROSSENCHARAKTERS, HECKE CHARACTERS AND L-SERIES 27

We can now make our definition of a Hecke character:

Definition 2.1.4. A Hecke character modulo n is a character

ξ : ClnK → S1

Note that Neukirch defines a Hecke character to be a character of the idele class group in [11,Chapter VII, Section 6, Definition 6.11]. He shows in [11, Chapter VII, Section 6, Corollary6.14] that these are in a 1-1 correspondence with the characters we have defined above.

The following result will allow us to describe the Hecke characters in terms of theGroßcharakters we have already introduced:

Proposition 2.1.5. The Hecke characters modulo n are precisely the Großcharakters modulo nsuch that with pl, ql, p

′k and q′k defined as in Proposition 2.1.2

• ql = q′k = 0 for all l, k

• p′k = 0 for all k

In other words, for every a ∈ OK relatively prime to m we have

ξ((a)) = ξf (a)

r1∏l=1

(ρl(a)

|ρl(a)|

)plfor some character ξf of (OK/n)×.

Proof. Let χ be a Großencharakter modulo n, with corresponding characters χf and χ∞ of(OK/n)× and (R×)r1 × (C×)r2/Un respectively with pl, ql, p

′k and q′k satisfying the conditions of

the proposition.

So for any a ∈ OK totally positive such that a ≡ 1 mod n then necessarily we must have thatχf (a) = 1, and by choice of the p’s and q’s, χ∞(a) = 1. Hence, χ((a)) = 1, and so χ factorsthrough P n

K .

Now let ξ be a Hecke character modulo n. That is to say, ξ is a character of InK such thatξ(P n

K) = 1. Let Kn := {a ∈ K× | a ≡ 1 mod n}, Kn+ := {a ∈ Kn | a totally positive} and

G := {(x1, .., xr1 , z1, .., zr2) ∈ (R×)r1 × (C×)r2 | ∀i xi > 0}. We get an isomorphism then

Kn/Kn+ → (R×)r1 × (C×)r2/G ∼=

r1∏i=1

{±1}

Hence, the composite map

Kn/Kn+

( )→ ClnKξ→ S1

gives a character of (R×)r1 × (C×)r2 which sends G to 1. So by Proposition 2.1.2, then it has

the form ξ∞(a) =∏r1

l=1

(ρl(a)|ρl(a)|

)plfor some pl’s either 0 or 1. Hence, for a ∈ Kn, we have

ξ((a)) = ξ∞(a). So then ξf (a) := ξ((a))ξ∞(a)−1 defines a character of (OK/n)×.

With this result, therefore, we may describe all the characters of ray class groups in terms ofGroßencharakters. Just as Dirichlet characters may factor through a smaller modulus, so mayHecke characters. We can describe the conductor of the Hecke character by the conductor ofits associated Großencharakter:


Definition 2.1.6. The conductor of a Großencharakter ξ is the least integral ideal f(ξ) suchthat the finite part of ξ, ξf , factors through (O/f(ξ))×.

Thus, a Hecke character ξ modulo n is primitive if and only if n = f(ξ).

We will now introduce the L-series of Hecke characters. We focus on the Großencharakters tostart off with:

Definition 2.1.7. The Hecke L-seriesfor ξ a primitive Großencharakter character is

L(s, ξ) =∑a

ξ(a)NK/Q(a)−s (s ∈ C)

where a runs over all integral ideals of OK and NK/Q(a) is the usual norm of ideals a. For aHecke character ξ modulo n we define its Hecke L-series to be the Hecke L-series of thecorresponding primitive character modulo f(ξ).

The Hecke L-series for a Hecke character is the Hecke L-series of the Großencharakterassociated to it.

By the convergence of the Riemann zeta function, it is easily seen that this L-series isabsolutely and uniformly convergent for <(s) ≥ 1 + ε for any ε > 0:

Proposition 2.1.8. The L-series L(s, ξ) converges absolutely and uniformly in the domain<(s) ≥ 1 + ε for all ε > 0 and one has the Euler product

L(s, ξ) =∏p

1

1− ξ(p)NK/Q(p)−s

where p varies over the prime ideals of OK.

Proof. Consider the product

E(s) =∏p

1

1− ξ(p)NK/Q(p)−s

We take the logarithm of the product to obtain

logE(s) =∑p

∞∑m=1

ξ(p)

mNK/Q(p)ms

This converges absolutely and uniformly for <(s) = σ ≥ 1 + ε. As |ξ(p)| ≤ 1 and |NK/Q(p)s|= |NK/Q(p)σ| ≥ pfp(1+ε) ≥ p1+ε for some rational prime p, and since |{p|pOK}| ≤ [K : Q] = n,we obtain the following bound on this logarithm independent of s:∑

p,m

n

mpm(1+ε)= n log(1 + ε)

This shows that the product E(s) which is exp(∑

p

∑∞m=1

ξ(p)

mNK/Q(p)ms) is absolutely and

uniformly convergent for <(s) ≥ 1 + ε. Develop the factors in this product as

1

1− ξ(p)NK/Q(p)−s= 1 +

ξ(p)

NK/Q(p)s+

ξ(p)2

NK/Q(p)2s+ ...

for the finitely many prime ideals p1, p2, ..., pr such that NK/Q(pi) ≤ N . This yields that

2.1. GROSSENCHARAKTERS, HECKE CHARACTERS AND L-SERIES 29

r∏i=1

1

1− ξ(pi)NK/Q(pi)−s=

∞∑v1,v2,...,vr=0

ξ(p1)v1 ...ξ(pr)

vr

(NK/Q(p1)v1..NK/Q(pr)vr)s

=′∑a

ξ(a)

NK/Q(a)s

where we take the sum over all ideals a which are divisible by at most the prime idealsp1, p2, ..., pr. Any such sum can be written as

r∏i=1

1

1− ξ(pi)NK/Q(pi)−s=

∑NK/Q(a)≤N

ξ(a)

NK/Q(a)s+

′∑NK/Q(a)>N

ξ(a)

NK/Q(a)s

Thus we get

∣∣∣∣∣r∏i=1

1

1− ξ(pi)NK/Q(pi)−s− L(s, ξ)

∣∣∣∣∣ ≤∣∣∣∣∣∣∣∣∑

NK/Q(a)>Npi-a

ξ(a)

NK/Q(a)s

∣∣∣∣∣∣∣∣≤

∑NK/Q(a)>N

1

NK/Q(a)1+ε

As N →∞ the right hand side will tend to 0 by the convergence of E(s) on <(s) ≥ 1 + ε.

We now wish to talk about the completed L-function of a Großencharakter ξ. This is becauseit will provide a convenient way to normalise the L-series so that it has a “nice” functionalequation.

Definition 2.1.9. Using the same notation as in Proposition 2.4.2, if ξ is a Großencharakterthen the Γ-factor associated to the Hecke L-series of ξ is

L∞(s, ξ) =

r1∏l=1

π−s+pl−iql

2 Γ

(s+ pl − iql

2

) r2∏k=1

2(2π)−(s+p′k−iq

′k)Γ(s+ p′k − iq′k)

Now we can define the completed L-series of a Hecke character ξ:

Definition 2.1.10. Suppose that ξ is a Großencharakter of K/Q modulo n. The completedL-series associated to ξ is defined to be:

Λ(s, ξ) = (|dK/Q|NK/Q(f(ξ)))s2L(s, ξ)L∞(s, ξ)

where dK/Q is the discriminant of K/Q.

Of significant importance to us will be the functional equation for Λ(s, ξ) – it will allow us toshow in the next section that the conditions of Weil’s Converse Theorem hold.

Theorem 2.1.11. Let ξ be a primitive Großencharakter of K/Q of conductor n. Thecompleted Hecke L-series attached to ξ has a meromorphic continuation to the entire s-plane,which is holomorphic everywhere (aside from simple zeroes at s = 0 and s = 1 in the case thatξ is trivial). Further, it is bounded on any set of the form


{s ∈ C|a ≤ <(s) ≤ b, |=(s)| ≥ 0} (a < b, 0 < c)

and satisfies the functional equation

Λ(s, ξ) = T (ε)Λ(1− s, ξ)

where T (ξ) ∈ S1 and ξ(a) = ξ(a).

Proof. See the proof provided by Neukirch in [11, Chapter VIII, §8].

We now restrict our attention to Hecke characters. To describe the factor T (ε) appearing inthe functional equation for a Hecke character, we will need to define its Gauss sum (which isanalogous to the same for a Dirichlet character). For an extension K/Q consider the set

D−1 = {a ∈ K|TrK/Q(ab) ∈ Z ∀b ∈ OK}

Then D−1 is a fractional ideal of K and its inverse D is an integral ideal called the different ofK. Now let ξ be a primitive Hecke character modulo n the type described before. Let c be anintegral ideal of K such that Dnc is principal and gcd(n, c) = 1. We let b ∈ OK be a generatorof the ideal Dnc.

Then the sum

ξ∞(b)

ξ(c)

∑a

ξf (a)e2πiTrK/Q(a/b),

where a runs over a complete set of coset representatives for c/mc, is independent of the choiceof c, b and a set of representatives.

Definition 2.1.12. For a primitive Hecke character ξ modulo n, we define the Gauss sumW (ξ) to be

W (ξ) =ξ∞(b)

ξ(c)

∑a

ξf (a)e2πiTrK/Q(a/b)

where the sum is taken over a set of coset representatives of c/mc. This definition isindependent of the choice of b and c.

Like Dirichlet characters, we have:

|W (ξ)|2 = NK/Q(n)

Lemma 2.1.13. Let ξ and η be two primitive Hecke characters of conductors n and mrespectively. If gcd(n,m) = 1, then

W (ξη) = ξ(m)η(n)W (ξ)W (η)

Proof. Since gcd(n,m) = 1, we may find a c /OK such that Dmnc = (b) for some b ∈ OK . Sothen we may write

W (ξ) =ξ∞(b)

ξ(mc)

∑a

ξf (a)e2πiTrK/Q(a/b)

summing over representatives of mc/nmc and

2.2. MODULAR FORMS FROM QUADRATIC NUMBER FIELDS 31

W (η) =η∞(b)

η(nc)

∑a′

ηf (a′)e2πiTrK/Q(a

′/b)

summing over representatives of c/nmc. Now, note that taking all sums a+ a′ gives us acomplete set of representatives of c/nmc since n and m are coprime. Also, note that a ∈ m anda′ ∈ n.

So then

ξ(m)η(n)W (ξ)W (η) = ξ(m)η(n)ξ∞(b)η∞(b)

ξ(mc)η(nc)

∑a,a′

ξf (a)ηf (a′)e2πiTrK/Q(a+a

′/b)

=(ξη)∞(b)

(ξη)(c)

∑a,a′

(ξη)f (a+ a′)e2πiTrK/Q(a+a′/b)

= W (ξη)

We now use the Gauss sum to give a description of the constant T (ξ) in Theorem 2.1.11

Proposition 2.1.14. Let ξ be a Hecke character with pl as in Proposition 2.1.5. Then withT (ξ) defined as in Theorem 2.1.11 we have

T (ξ) =W (ξ)

ip√NK/Q(n)

where

p =

r1∑l=1

pl

2.2 Modular Forms from Quadratic Number Fields

We will now explore how Hecke characters connected to quadratic number fields can give riseto modular forms. We will follow the outline given by Miyake in [10, §4.8] , providing themissing details in Lemma 2.2.7 and proving Theorem 2.2.9.

We need to introduce the Kronecker symbol:

For p an odd prime, let(np

)be the Legendre symbol. We extend the definition of the Legendre

symbol to all pairs of integers (a, b) ∈ Z2 \ {(0, 0)}. We do this in the following way:

(a2

)=

1 if a ≡ 1 mod 8

−1 if a ≡ 5 mod 8

0 otherwise(a

−1

)=

{1 if a ≥ 0

−1 if a < 0(a1

)= 1


(a0

)=

{1 if a = 1

0 otherwise

Definition 2.2.1. Let b be a non-zero integer. We may express b as the product of ±1 andprime numbers p1, p2, .., pr:

b = εr∏i=1

paii (ε = ±1, pi, prime ai ∈ Z)

Then put

(ab

)=(aε

) r∏i=1

(a

pi

)aiWe can use the Kronecker symbol to define a Dirichlet character now from a quadratic numberfield:

Definition 2.2.2. For any d 6= 0 the mapping χd : n 7→(d

n

)is a Dirichlet character.

Remark 2.2.3. If d is the discriminant of a quadratic field then χd is a primitive Dirichletcharacter of conductor |d|.

Lemma 2.2.4. If d 6= 0 is the discriminant of the quadratic number field K = Q(√d)/Q, then

for any prime number p:

1. If χd(p) = 1 then p splits in K

2. If χd(p) = −1 then p is prime in K

3. If χd(p) = 0 then p ramifies in K

Proof. This is a simple consequence of the fact that χd(p) =(dp

), and if p is an odd prime this

is the Legendre symbol. The p = 2 case is then trivial by the choice of definition of χd.

The following theorem will be important:

Theorem 2.2.5. Let K = Q(√d) be a quadratic number field with discriminant d. Then the

Dedekind zeta function of K

ζK(s) =∑a

NK/Q(a)−s

is equal to

ζK(s) = ζ(s)L(s, χd)

As a special consequence of the functional equation of Hecke L-series (Theorem 2.1.11), wehave the following:


Theorem 2.2.6. Suppose χ is a primitive Dirichlet character of conductor m, and δ is 0 if χis even and 1 if χ is odd. Put

Λ(s, χ) = (π/m)−s/2Γ

(s+ δ

2

)L(s, χ)

Then Λ(s, χ) is meromorphic on the whole s-plane and is bounded on any set of the form

{s ∈ C|a ≤ <(s) ≤ b, |=(s)| ≥ c} (a < b, 0 < c)

and satisfies the functional equation

Λ(1− s, χ) =W (χ)

iδ√m

Λ(s, χ)

Consider now the quadratic number field K = Q(√d) with discriminant d, p a rational prime

with gcd(p, d) = 1 and ψ a primitive Dirichlet character of conductor p. Then

ψ ◦NK/Q(a) = ψ(Nk/Q(a)) a ∈ IKdefines a primitive Hecke character of conductor (p). The following lemma will be useful:

Lemma 2.2.7. 1. W (χd) =

{√d d > 0

i√|d| d < 0

2. If gcd(d, p) = 1 then W (ψ ◦NK/Q) = χd(p)ψ(|d|)W (ψ)2

Proof. Now since ζK(s) = ζ(s)L(s, χd), from the functional equation of ζK(s) we get that

W (χd)

iδ√|d|

= 1

where δ is 0 if χd is even and 1 otherwise. Now by definition of the Kronecker symbol,χd(−1) = 1 if and only if d > 0. This gives part 1.

Now, we consider the Euler products associated to the L-series L(s, ψ ◦NK/Q) andL(s, ψ), (s, ψχd).

If p is a prime ideal lying above p, then consider the factors

(1− ψ(NK/Q(p))NK/Q(p)−s)−1

and

(1− ψ(p)p−s)−1(1− ψ(p)χd(p)p−s)−1 = (1− ψ(p)(1 + χd(p))p

−s + ψ(p2)χd(p)p−2s)−1

Using Lemma 2.1.8, we see that these factors are equal by considering the cases when pramifies, splits and is inert in K, so that

L(s, ψ ◦NK/Q) = L(s, ψ)L(s, ψχd)

and hence T (ψ ◦NK/Q) = T (ψ)T (ψχd). Therefore

W (ψ ◦NK/Q)

NK/Q((p))1/2=W (ψ)W (ψχd)

iδ√p√p|d|

So by part 1 and Lemma 2.1.13 we are done.


We are now ready to prove the existence of weight 1 modular forms coming from quadraticnumber fields. We need to consider the real and imaginary cases separately, but the argumentswill be essentially the same. First of all, we note that in the imaginary case, any Heckecharacter must send P n

K to 1. As there are no real embeddings of K, ξ((a)) = 1 for all a ≡ 1mod n.

Theorem 2.2.8. Let K = Q(√d) be an imaginary quadratic field with discriminant d, and ξ a

Hecke character mod n Let

f(z) = f(z, ξ) =∑a

ξ(a)e2πiNK/Q(a)z

where a runs over all integral ideals of K. Then f(z) ∈M1(N,χ), where N = |d|NK/Q(n) andχ is a Dirichlet character defined by

χ(m) = χd(m)ξ((m)) (m ∈ Z)

Proof. Let p be a prime ideal. If ξ′ is the modulo pn character induced by ξ then we note that

f(z, ξ′) = f(z, ξ)− ξ(p)f(NK/Q(p)z, ξ)

since ξ′(pa) = 0 for integral a. So we only need to consider primitive Hecke characters.

Now let ψ be a primitive Dirichlet character of conductor p a prime which is coprime to N .Then

L(s, f, ψ) =∑a

ξ(a)ψ(NK/Q(a))

NK/Q(a)s

= L(s, ξ(ψ ◦NK/Q))

Note that the completed Hecke L-series for ξ(ψ ◦NK/Q) is

Λ(s, ξ(ψ ◦NK/Q)) =

(p2N

(2π)2

)s/2Γ (s)L(s, ξ(ψ ◦NK/Q)) (2.2.1)

since the infinite part of ξ(ψ ◦NK/Q) has the form

a 7→ 1

which gives the corresponding Γ-factor as(p2N(2π)2

)s/2Γ (s) by taking pl, ql, p

′l and q′l equal to 0 in

Definition 2.1.9 for all l. So we get that

ΛN(s, f, ψ) =

(p√N

2π

)s

Γ(s)L(s, f, ψ)

=

(p√N

2π

)s

Γ(s)L(s, ξ(ψ ◦NK/Q))

=

(p√N

2π

)s

Γ(s)

(p√N

2π

)−sΓ(s)−1Λ(s, ξ(ψ ◦NK/Q))

= Λ(s, ξ(ψ ◦NK/Q))


by (2.2.1). Now we use the functional equation in Theorem 2.1.11 and note that by Lemmas2.1.13 and 2.2.7

T (ξ(ψ ◦NK/Q)) =W (ξ(ψ ◦NK/Q))

NK/Q((p)n)1/2

=ξ((p))ψ(NK/Q(n))W (ξ)χd(p)ψ(|d|)W (ψ)2

pNK/Q(n)1/2

=W (ξ)χ(p)ψ(N)W (ψ)2

pNK/Q(n)1/2

If we let

Cψ =χ(p)ψ(N)W (ψ)2

p

g(z) = i−1W (ξ)

NK/Q(n)1/2

∑a

ξ(a)e2πiNK/Q(a)z

then

ΛN(s, f, ψ) = Λ(s, ξ(ψ ◦NK/Q))

= T (ξ(ψ ◦NK/Q))Λ(1− s, ξ(ψ ◦NK/Q))

= iCψΛN(1− s, g, ψ)

Since |ξ(a)| = 1 for any integral ideal a in K, and by Theorem 1.2.11 the completed L-seriesΛN(s, f, ψ) is bounded in any vertical strip, we may apply Theorem 1.3.1. Thus we deducethat f(z) ∈M1(N,χ).

We will now show the analogous result for the real case. It is slightly more complicated due tothe appearance of 2 gamma functions in the completed Hecke L-series for the Hecke character.In order to deal these we will need to make use of the Duplication Theorem

Γ(s)Γ(s+1

2) =√πΓ(2s)21−2s ∀ s ∈ C (2.2.2)

In the case where K is a real quadratic field, if ξ is a Hecke character whose L-seriescorresponds to that of a modular form, then by comparing the functional equation for themodular form with that of ξ we see that p := p1 + p2 in Proposition 2.1.14 is 1. Asp1, p2 ∈ {0, 1}, we must have that either p1 = 1, p2 = 0 or p1 = 0, p2 = 1.

Theorem 2.2.9. Let K = Q(√d) be a real quadratic field with discriminant d, and ξ a Hecke

character mod n such that we have either that

ξ((a)) = a/|a| for all a ≡ 1 mod n

or

ξ((a)) = a′/|a′| for all a ≡ 1 mod n

where a′ is the K/Q-conjugate of a. We put

f(z) = f(z, ξ) =∑a

ξ(a)e2πiNK/Q(a)z


where a runs over all integral ideals of K. Then f(z) ∈ S1(N,χ) where N = dNK/Q(n) and χis a Dirichlet character defined by

χ(m) = χd(m)ξ((m)) (m ∈ Z)

Proof. As in the imaginary quadratic case we only need to consider primitive Hecke characters.Now let ψ be a primitive Dirichlet character of conductor p a prime which is coprime to N .Then

L(s, f, ψ) = L(s, ξ(ψ ◦NK/Q))

and that

Λ(s, ξ(ψ ◦NK/Q)) =

(2p√N

2π

)s

Γ(s

2

)Γ

(s+ 1

2

)L(s, ξ(ψ ◦NK/Q))

by noting that r1 = 2, r2 = 0 and that {p1, p2} = {0, 1} and using the definition of theL∞-factor from Definition 2.1.9. By the Duplication Theorem (2.2.2) then

Λ(s, ξ(ψ ◦NK/Q)) =√π21−s

(2p√N

2π

)s

Γ(s)L(s, ξ(ψ ◦NK/Q)) (2.2.3)

Hence we get that

ΛN(s, f, ψ) =

(p√N

2π

)s

Γ(s)√π−1

2s−1

(2p√N

2π

)−sΓ(s)−1Λ(s, ξ(ψ ◦NK/Q))

=1

2√π

Λ(s, ξ(ψ ◦NK/Q))

Note that

T (ξ(ψ ◦NK/Q)) =i−1χd(p)ξ((p))W (ξ)W (ψ)2

pNK/Q(n)1/2

=−iW (ξ)χ(p)ψ(N)W (ψ)2

pNK/Q(n)1/2

So we can use the functional equation from Theorem 2.1.11 for Λ(s, ξ(ψ ◦NK/Q)) to get that

ΛN(s, f, ψ) =1

2√π

Λ(s, ξ(ψ ◦NK/Q))

=1

2√πT (ξ(ψ ◦NK/Q))Λ(1− s, ξ(ψ ◦NK/Q))

=1

2√πT (ξ(ψ ◦NK/Q))

(2p√N

2π

)1−s

Γ

(1− s

2

)Γ

(2− s

2

)L(1− s, ξ(ψ ◦NK/Q))

= T (ξ(ψ ◦NK/Q))

(p√N

2π

)1−s

Γ(1− s)L(1− s, ξ(ψ ◦NK/Q))

where in the last line we have again used the Duplication Theorem (2.2.2). Now setting


Cψ =χ(p)ψ(N)W (ψ)2

p

and

g(z) =−W (ξ)

NK/Q(n)1/2

∑a

ξ(a)e2πiNK/Q(a)z

we get

ΛN(s, f, ψ) = iCψΛN(1− s, g, ψ)

and the conditions of Weil’s Converse Theorem are satisfied as in the imaginary case so thatwe get that f(z) ∈M1(N,χ).

Chapter 3

Representations from QuadraticNumber Fields

This chapter will cover the main ideas characterising extensions of the Hecke charactersdiscussed in the previous chapter to Galois representations. Some of the results from ClassField Theory that are involved are quite technical and so we refer the reader to Janusz in [7]and Neukirch in [11, Chapter VI] for a more complete exposition of these. Finally, we drawtogether the results from Chapter 2 and this chapter.

3.1 Representations of Galois Groups and Artin

L-functions

In this section we will introduce the Artin L-functions, which are L-functions attached torepresentations of Galois groups. We will follow the path outlined by Bump in [2, §1.8] in thissection. His outline of Artin L-functions is slightly less rigid than some other sources (in thathe ignores ramified primes), but this helps to cement some broader understanding of whatthese functions are like.

Let L/K be a finite Galois extension of number fields and let ρ : Gal(L/K)→ GL(V ) be ann-dimensional representation. We will show how to associate an L-series L(s, ρ) to this.

Let p be a prime ideal of K that does not ramify in L and let P be a prime ideal of L lyingabove p. Let ϕP/p be the Frobenius automorphism. This is the element of Gal(L/K) such that

ϕP/p(x) = xNK/Q(p) mod P ∀x ∈ OKFor more details on the Frobenius automorphism, see [7, Chapter III, Section 2]. Note thatthen that since the primes in L lying above p are conjugate under the action of Gal(L/K),then the Frobenius automorphism ϕp is uniquely defined up to conjugation by elements of theGalois group.

Definition 3.1.1. We define the local factor of the Artin L-series of ρ at a prime p of Kwhich is unramified in L to be

Lp(s, ρ) = det(In − ρ(ϕp)NK/Q(p)−s)−1

Since ϕp is unique up to conjugacy, the local factor at an unramified prime is unique. This alsomeans that the eigenvalues of ρ(ϕp) are unique as well. Let these be α1(p), α2(p), ..., αn(p). Sowe may rewrite the local factor at p as

39

40 CHAPTER 3. REPRESENTATIONS FROM QUADRATIC NUMBER FIELDS

Lp(s, ρ) =n∏r=1

(1− αr(p)NK/Q(p)−s)−1

We may now define the Artin L-function:

Definition 3.1.2. The Artin L-function of a representation ρ : Gal(L/K)→ GL(V ) of theGalois group of a finite Galois extension L/K is

L(s, ρ) =∏p

Lp(s, ρ)

where the product is taken over all prime ideals of K that do not ramify in L.

If we wish to consider the primes which ramify, then we also need to take into account the factthat the Frobenius Automorphism will not be unique, as the inertia group will have sizegreater than 1. In that case, the action of In − ρ(ϕp)NK/Q(p)−s must be restricted to thesubspace of invariants:

V IP = {v ∈ V | ρ(g)(v) = v ∀g ∈ IP}

See Neukirch in [11, Chapter VII, §10] for further details.

The most important property of Artin L-functions for us will be how they behave on inducedrepresentations. See Appendix A for the definition of the induced representation.

Theorem 3.1.3. Let L/M/K be a sequence of finite Galois extensions with L/K Galois. LetG := Gal(L/K) and H := Gal(L/M) Let ρ : H → GLn(C) be a complex representation ofdegree n. Then the induced representation ρ′ := IndGH(ρ) is of degree [M : K]n and

L(s, ρ) = L(s, ρ′)

Proof. We will only consider the case where M/K is quadratic. See [12, Section 2.2, Theorem2.2.1] for a proof for the general case.

Assuming that M/K is quadratic, then Gal(L/M) is of index 2 in Gal(L/K). Let γ be arepresentative of the non-identity coset of Gal(L/M) in Gal(L/K). We can then realise theinduced representation ρ′ by the matrices

ρ′(σ) =

(ρ(σ) 0

0 ρ(γσγ−1)

)if σ ∈ Gal(L/M)(

0 ρ(σγ−1)

ρ(γσ) 0

)otherwise

using the definition from A.2.

Suppose now that p is prime of K which splits in M , so that pOM = P1P2. We have bydefinition that ϕp, ϕP1 and ϕP2 in Gal(L/K) are all conjugate in Gal(L/K) (though notnecessarily in Gal(L/M)).

Note that ϕp depends only on the choice of prime Q in L dividing p. So Q divides either P1 orP2. We may assume without loss of generality that Q divides P1. Hence ϕp is conjugate inGal(L/M) to ϕP1 . Since γ will interchange P1 and P2, then γϕpγ

−1 is conjugate in Gal(L/M)

3.2. SELECTED RESULTS FROM CLASS FIELD THEORY 41

to ϕP2 .

Say now that ρ(ϕP1) has eigenvalues α1, ..αn, and that ρ(ϕP2) has eigenvalues β1, ..., βn. Then

ρ′(ϕp) =

(ρ(ϕP1) 0

0 ρ(ϕP2)

)and so ρ′(ϕp) has eigenvalues α1, ..., αn, β1, ...βn. Hence

Lp(s, ρ′) = LP1(s, ρ)LP2(s, ρ)

Say now that p is inert. Hence, pOM = P. Now since NL/M(P) = NL/K(p)2, we observe thatϕ2p = ϕP. Now, suppose that ρ(ϕP) has eigenvalues α1, α2, ..., αn. We now let β1, β2, ...β2n be

the eigenvalues of ρ′(ϕp). So∑βri = tr(ρ′(ϕrp)). If r is odd, then ϕrp 6∈ Gal(L/M) and so∑

βri = 0. Otherwise, it is

tr

(ρ(ϕ

r/2P ) 0

0 ρ(γϕr/2P γ−1)

)= 2

∑αr/2i

and hence we see that the βi are given by

±√α1, ...,±

√αn

Hence

Lp(s, ρ′) =

n∏i=1

(1−√αiNL/K(p)−s)−1(1 +

√αiNL/K(p)−s)−1

=n∏i=1

(1− αiNL/K(p)−2s)−1

=n∏i=1

(1− αiNL/M(P)−s)−1

Since we assume our factors cover only the unramified primes, we are done.

3.2 Selected Results from Class Field Theory

In this section we will outline two important results which will help us to show how to extendthe Hecke characters from Theorems 2.2.8 and 2.2.9 to Galois representations. We will followNeukirch here in [11, Chapter VI, §6 - §7]. The proofs of these results are beyond the scope ofthis dissertation, and so we refer the reader where appropriate to other sources, namely [7] and[11].

Class Field Theory concerns itself with the classification of all abelian extensions of a numberfield K by properties of K. By an abelian extension, we mean that the Galois group of theextension is abelian. The most important theorems encapsulating this theory are the ArtinReciprocity Theorem and the Existence Theorem.

We simplify the discussion here by considering only modulii that are integral ideals. Moregenerally, a modulus of a number field K is a product of primes or places. These are


equivalence classes of valuations of K. The finite primes are the equivalence classes ofnonarchimedean valuations, and the infinite primes are the equivalence classes of archimedeanprimes. The infinite primes are obtained from real and complex embeddings of K, whilst thefinite primes are in direct correspondence with the prime integral ideals. In general, however,all of what follows can be extended to include the case where the infinite part of a modulus isnot trivial. This is explained further in Neukirch [11, Chapter III, §1].

To properly talk about class fields, we need to introduce congruence subgroups:

Definition 3.2.1. A subgroup H of IK is called a congruence subgroup if there is an integralideal n such that

P nK ⊆ H ⊆ InK

It is then a congruence subgroup modulo n. If L is an abelian extension of K, an ideal groupdefined mod n is a group Hn = (NL/K(InL))P n

K.

We can now state the Existence Theorem:

Theorem 3.2.2 (Existence Theorem). The map

L 7→ Hn

is a 1-1 inclusion reversing correspondence between the finite abelian extensions L/K andcongruence subgroups H modulo n for any n. The field L/K corresponding to the congruencesubgroup H modulo n is called the class field for H. It satisfies

Gal(L/K) ∼= InK/H

Proof. See Neukirch in [11, Chapter VI, §6, Theorem 6.1] for the proof. The proof there is inthe idele theoretic formulation. In the discussion immediately after the statement of Corollary7.2 Chapter VI, §7, this result is given an ideal theoretic formulation.

So, now that we have the Existence Theorem, we know that there is a class field for P nK , Kn/K.

We call this the ray class field. Let L/K be an abelian extension. We have the following result:

Proposition 3.2.3. Every finite abelian extension L/K is contained in a ray class field Kn/K.

Proof. This is clear from the Existence Theorem.

Definition 3.2.4. Let L/K be a finite abelian extension. Then we call the conductor f(L/K)of L/K the greatest common divisor of all the integral ideals n such that L ⊆ Kn.

The following result indicates how primes behave in abelian extensions:

Proposition 3.2.5. Let L/K be a finite abelian extension and let f be its conductor. Then fora prime ideal p in K

p ramifies in L⇔ p | n

Proof. See Neukirch in [11, Chapter VI, §10, Corollary 6.6] for a proof.

3.3. REPRESENTATIONS OF GALOIS GROUPS AND MODULAR FORMS 43

Now, let L/K be a finite abelian extension, and let n be any integral ideal such that L iscontained in ray class field modulo n. We call such an ideal a module of definition. As for anyprime ideal p not dividing n we get p is unramified in L by Proposition 3.2.5, then we maydefine the Frobenius Automorphism for p. We let(

L/K

p

):= ϕp

We now define the Artin Map:

Definition 3.2.6. Let L/K be a finite abelian extension and n a module of definition for it.Say a ∈ InK has factorisation a =

∏ri=1 p

vii and define(

L/K

a

)=

r∏i=1

(L/K

pi

)viThis defines a map (

L/K)

: InK → Gal(L/K)

which we call the Artin map.

Note that since n is a module of definition for L/K, and hence any ideal in InK is coprime to n,the Artin map is well defined. We are now ready to state the next main theorem, the ArtinReciprocity Theorem:

Theorem 3.2.7 (Artin Reciprocity Theorem). Let L/K be an abelian extension, and let n bea module of definition for it. Then the Artin map induces a surjective homomorphism:(

L/K)

: ClnK → Gal(L/K)

with kernel Hn/P nK.

Proof. See Neukirch in [11, Chapter VI, §7, Theorem 7.1] for a proof. Alternatively, you mightwish to look at the proof in [7, Chapter 5, §5] provided by Janusz.

Finally, we state what will be the most useful formulation of this result for us:

Corollary 3.2.8. The Artin map depends on classes modulo P nK and so defines an

isomorphism (L/K

): InK/H

n ∼→ Gal(L/K)

3.3 Representations of Galois Groups and Modular

Forms

In this section we will construct a 2-dimensional complex Galois representation correspondingto a weight one modular form. The outline for this will follow the discussion by Miyake at theend of [10, §4.8].

Let K be a real or imaginary quadratic number field, and let ξ be a Hecke character for Kmodulo n as in the statements of Theorems 2.2.8 and 2.2.9. We may extend our definition of ξto all of IK by defining ξ(a) = 0 for a ∈ IK not coprime to n. Then let


J := {a ∈ IK | ξ(a) = 1}

Clearly, by definition of ξ, J is a congruence subgroup modulo n. So by the Existence Theorem(Theorem 3.2.2), there is an abelian extension M/K with Galois group

Gal(M/K) ' InK/J

and this isomorphism is given by the Artin Map by the corollary to the Artin ReciprocityTheorem (Corollary 3.2.8). Note that the primes of K ramifying in M are, therefore, containedin the set of primes which divide n by Proposition 3.2.5.

Let us take a minimal Galois extension L/Q containing M . Then by an application of GaloisTheory we get

Gal(L/K)/Gal(L/M) ' Gal(M/K) ' InK/J

Now, by construction, ξ is a character of the group InK/J , and so the isomorphismGal(M/K) ' InK/J induces a character ξ′ of Gal(M/K) in the following way:

ξ′(σ) := ξ(

(M/K

)−1(σ)) ∀σ ∈ Gal(M/K)

In particular, then, we note that ξ′(

(M/K

p

)) = ξ(p) for a prime ideal p of K.

The isomorphism Gal(L/K)/Gal(L/M) ' Gal(M/K) is given by restriction of the action ofelements of Gal(L/K) to the subfield M . So we can induce a character ξ of Gal(L/K) byletting ξ(σ) = ξ′(σ|M).

So we have a character of the group Gal(L/K), which is a 1-dimensional complexrepresentation:

ξ : Gal(L/K)→ GL1(C)

Additionally, note that

(L/K

p

)|M =

(M/K

p

)(see [7, Chapter III, Section 2, Property 2.3]

for a proof), and so we see that the Artin L-series for the representation ξ has the form:

L(s, ξ) =∏p

(1− ξ((L/K

p

))NL/K(p)−s)−1

=∏p

(1− ξ((M/K

)−1(

(L/K

p

)|M))NL/K(p)−s)−1

=∏p

(1− ξ((M/K

)−1(

(M/K

p

)))NL/K(p)−s)−1

=∏p

(1− ξ(p)NL/K(p)−s)−1

= L(s, ξ)

So, ξ has Artin L-series equal to the Hecke L-series of ξ. Now, note that Gal(L/K) is asubgroup of Gal(L/Q) of index 2, and so we may induce ξ to a 2-dimensional complex

3.3. REPRESENTATIONS OF GALOIS GROUPS AND MODULAR FORMS 45

representation ρ of Gal(L/Q). By Theorem 3.1.3 ρ and ξ have the same L-series . Note alsothat the modular forms we found in Theorems 2.2.8 and 2.2.9 have the same L-series as ξ. Wemay now summarise all of this in the following theorem:

Theorem 3.3.1. Let K/Q be a real or imaginary quadratic field with discriminant dK. Let ξbe a Hecke character modulo n, an integral ideal. If K is real, then assume ξ additionallysatisfies the condition that

ξ((a)) = 1 ∀a ≡ 1 mod n

Then

• there exists a weight one modular form f(z) of level |dK |NK/Q(n) and nebentypus χ,where χ(m) = χdK (m)ξ((m))

and

• there exists a Galois extension L/Q and a 2-dimensional complex Galois representationρ : Gal(L/Q)→ GL2(C) such that L(s, f) = L(s, ρ)

We may further consider ρ as a projective representation, by composing it with the naturalquotient map π : GL2(C)→ GL2(C)/C× = PGL2(C). The image of the projectiverepresentation, ρ = π ◦ ρ, is a a finite subgroup of PGL2(C), and these are Cn, Dn, A4, S4 andA5. If ρ is the representation in Theorem 3.3.1, then Im(ρ) is isomorphic to a dihedral group,and we call ρ dihedral.

We call a Galois representation ρ odd if det(ρ(σ)) = −1 for a σ ∈ Gal(L/Q) which is complexconjugation. When K is imaginary we may take a σ ∈ Gal(L/Q) with σ|K 6= idK . Bydefinition then for some coset representative γ of Gal(L/K) in Gal(L/Q)

ρ(σ) =

(0 ξ(σγ−1)

ξ(γσ) 0

)and so det(ρ(σ)) = −1 and ρ is odd since σ2 = idL. When K is real and ξ is a Hecke characteras in Theorem 3.3.1 then we can also show that ρ is odd. For a complete proof, see [5,Proposition 4.13]. It is worth emphasising that this proof uses tools which are more involvedthan those currently at our disposal.

Chapter 4

Examples of Weight One ModularForms

We will now demonstrate the existence of three weight one modular forms, one coming from animaginary quadratic field, one coming from both a real and two imaginary quadratic fields andone coming from a real quadratic field only.

4.1 A Level 23 Modular Form from Q(√−23)

In this section we construct a modular form of weight one, level 23 and character type χ−23from the imaginary quadratic number field K = Q(

√−23).

Note that the class group of K, CK , has order hK = 3. Any character on it will send agenerator of the group to a primitive cube root of unity ω. Let ξ be such a character. Since ξis a character of

CK = ClOKK = IOKK /POKK

ξ sends all principal fractional ideals to 1 as a character of IOKK . The discriminant of K is -23since −23 ≡ 1 mod 4. As the conditions of Theorem 3.3.1 are satisfied, there is a weight onemodular form of level | − 23|NK/Q(OK) = 23 and nebentypus χ−23 = χ, where

χ(m) =

(−23

m

)ξ((m)) =

(−23

m

)= χ−23(m) ∀m ∈ Z

We would like to calculate what this modular form looks like. By Theorem 2.2.8, we know ithas a q-expansion like

f(z) =∑a

ξ(a)e2πiNK/Q(a)

The most obvious approach will be to consider the coefficient ap of qp from a prime p. We cancompute these easily, by considering the cases when p is inert, ramifies and splits separately:

p is inert: When p is inert, we get that NK/Q(pOK) = p2, so that there are no terms in thesum contributing to the coefficient of qp. Hence, ap = 0.

p ramifies: Then p = 23. There is a prime ideal P in OK such that 23OK = P2. Hence, wemust have that ξ(P)2 = 1 and so the only possibility is that ξ(P) = 1. Hence, as this is theonly ideal which can contribute to the coefficient of q23, we get that a23 = 1.

47

48 CHAPTER 4. EXAMPLES OF WEIGHT ONE MODULAR FORMS

p splits: When p splits, there are prime ideals P1 and P2 such that pOK = P1P2. So, we musthave that ξ(P1)ξ(P2) = 1. Hence the only possibilities are that either ξ(P1) = ξ(P2) = 1 andso ap = ξ(P1) + ξ(P2) = 2, or ξ(P1) and ξ(P2) are different cube roots of unity and soap = ξ(P1) + ξ(P2) = −1.

Suppose that ξ(P1) = ξ(P2) = 1. Motivated by the example on page 43 in [1], we want to finda necessary and sufficient condition for this to be true. We must have that the classes of P1

and P2 are the class of OK , and hence both of these are principal.

Say that P1 = (a+ b(1+√−232

)). The norm of this ideal is therefore

p = a2 + ab+ 6b2

If p can be written like this, then the ideal P generated by a+ b(1+√−232

) has norm p and hencedivides pOK . As it has norm p, a rational prime, it must be a prime ideal. Since ξ(pOK) = 1and ξ(P) = 1, the other prime factor of p must also be principal. Hence, if p splits in K, thenthe prime ideals lying above p are both principal if and only if p may be written in the formp = a2 + ab+ 6b2 for some rational integers a and b. Computation shows the first such prime is59.

Therefore, we have proved the following about f :

ap =

0 if p is inert

1 if p = 23

−1 if p splits and cannot be expressed as a2 + ab+ 6b2

2 if p splits and can be expressed as a2 + ab+ 6b2

where a and b are rational integers.

For m =∏r

i=1 pvii an integer with given prime factorisation, we can compute am in the

following way. The contributions to am will come from all the terms like

ξ(a)e2πiNK/Q(a)

where a is an ideal of norm m. Summing over all ideals of norm m gives us am.

We can compute the first few terms of the q-expansion for f (with the aid of Magma) as:

f(z) = q − q2 − q3 + q6 + q8 − q13 − q16 + q23 − q24 + q25 + q26 + q27 − q29 +O(q30)

There is a 2-dimensional complex Galois representation associated to this modular formcoming from the character and the ray class field M of CK . This is just the ray class field Kmodulo OK , which is simply the Hilbert class field of K. This is the maximal abelianunramified extension of K. Here, this means that none of the prime ideals of K ramify in M .

With the aid of Magma,we can compute that this extension is the cubic extension of K foundby adjoining a root of the polynomial x3 − 3x− 1. The extension M/Q is Galois, and as K/Qis of degree 2 and M/K is of degree 3, the Galois group of M/Q is of order 6 and must beisomorphic to D6. So, ξ induces a 1-dimensional complex representation of Gal(M/K), byletting the generator of Gal(M/K) map to the same cube root of unity that the generator ofthe class group maps to. This induces a 2-dimensional complex dihedral representationρ : Gal(M/Q)→ GL2(C) therefore, and ρ has Artin L-series equal to the L-series of f .

4.2. A LEVEL 39 MODULAR FORM FROM Q(√

13), Q(√−3) AND Q(

√−39) 49

4.2 A Level 39 Modular Form from Q(√13), Q(

√−3) and

Q(√−39)

In this example we get a modular form from both a real and two imaginary quadratic field. LetK1 = Q(

√13) and K2 = Q(

√−3). We will factorise 3OK1 in K1 and 13OK2 in K2 using

Dedekind’s Theorem.

Up until now, we have only dealt with ray class groups modulo some prime ideal. However, asnoted at the start of Chapter 3 Section 2, it is possible to talk about infinite primes of K aswell.

Suppose that we have a modulus n which is a product of an integral ideal n0 and infinite primesn∞. Then we define InK as In0K and P n

K as the group of all principal ideals (a) ∈ InK for which

a ≡ 1 mod n0 and σ(a) > 0 for all σ : L ↪→ R in n∞.

Although the results we have given in Chapters 2 and 3 hold for any integral ideal, they canalso be shown to hold for any modulus with non-trivial infinite prime factors as defined above.We take the norm of n to be n0.

By Dedekind’s Theorem, in OK1 we get that 3OK1 = P3P′3, where P3 = (3, 1+

√13

2) and

P′3 = (3, −1+√13

2). Since NK1/Q(±1+

√13

2) = −3, then we get that P3 = (1+

√13

2) and

P′3 = (−1+√13

2). Similarly, in OK2 we find that 13OK2 = P13P

′13, where P13 = (3 + 1+

√−3

2) and

P′13 = (4− 1+√−3

2).

For K1, let n1 := P3.P∞, where P∞ is the non-trivial infinite prime in K1. For K2, letn2 := P13. Using Magma we compute that

Cln1K1

∼= Z/2Z and Cln2K2

∼= Z/2Z

Hence, the non-trivial character on each of these sends the generator of the group to -1. Letthese characters be ξ1 and ξ2 respectively.

We will consider K1 and ξ1 first. By Theorem 2.2.9, we get that ξ1 induces a modular form g1of weight one, level |13|NK1/Q(n1) = 39 and nebentypus

χ(m) = χ13(m)ξ1((m)) ∀m ∈ Z

This modular form has a q-expansion of the form:

g1(z) =∑a

ξ1(a)e2πiNK1/Q(a)

Let p be a rational prime and ap the coefficient of qp. We consider the cases when p ramifies, isinert and splits in OK1 separately. Since P3 lies above 3, we will need to consider a3 separately

also. Note that we cannot have P′3 ∈ Pn1K1

or else −1+√13

2≡ 1 mod P3, and so 1 is divisible by

9. Hence, we can take P′3 as a generator for the ray class group and so it gets sent to -1 by ξ1.

p ramifies : Then p is necessarily 13. The prime lying above 13 is (√

13), and sinceNK1/Q(−1 +

√13) = −12, we get (

√13) 6∈ P n1

K1or else 9 divides 12. Hence, we get that

ξ1(√

13) = −1 and so a13 = −1.


p splits : We consider the cases when p = 3 and p 6= 3 separately:

• p = 3: We know that ξ1(P3) = 0 and ξ1(P′3) = −1 so that a3 = −1

• p 6= 3: Suppose that pOK1 = P1P2. If pOK1 ∈ P n1K1

, then p ≡ 1 mod P3. Hence, there is

an α ∈ OK1 such that p− 1 = α 1+√13

2. Say α = a+ b1+

√13

2. Then the only possibility is if

a = −b and p = −3b+ 1. It is also clear that if this is the case, then p ≡ 1 mod P3.

Since p splits then we also have that(

13p

)= 1, and so by Quadratic Reciprocity, p is a

quadratic residue modulo 13. This means that p is also congruent to ±1,±3 or ±4modulo 13. We find then that p splits and pOK1 ∈ P n1

K1if and only if p is congruent to

1, 4, 10, 16, 22 or 25 modulo 39. In this case, ξ1(pOK1) = 1 = ξ1(P1)ξ1(P2) and soap = ±2.

If pOK1 6∈ P n1K1

, then ξ1(pOK1) = −1 = ξ1(P1)ξ1(P2) and so ap = 0.

p is inert : As in the first example we find that ap = 0. This is true for any primes congruent to±2,±5 or ±6 modulo 13. As p is also a quadratic non-residue modulo 3, it is congruent to -1modulo 3. Hence, we calculate that if p is congruent to 2, 5, 11, 20, 32 or 38 modulo 39, thenap = 0.

Hence for p prime

ap =

−1 if p = 3 or p = 13

±2 if p ≡ 1, 4, 10, 16, 22 or 25 mod 39

0 otherwise

Using Magma, we may compute the first few terms of the q-expansion of g1(z):

g1(z) = q − q3 − q4 + q9 + q12 − q13 + q16 − q25 − q27 +O(q30)

The above calculation for the split primes shows in fact that if p is an odd prime, then

ξ1((p)) =(p3

). By Quadratic Reciprocity then ξ1((p)) =

(−3p

). We also check that

ξ1((2)) =(−3

2

), and hence χ = χ−39 as χ13 and ξ1 are multiplicative.

We now consider K2 and ξ2. By Theorem 2.2.8, we get that ξ2 induces a modular form g2 ofweight one, level | − 3|NK2/Q(n2) = 39 and nebentypus

χ′(m) = χ−3(m)ξ2((m)) ∀m ∈ Z

This modular form has a q-expansion of the form:

g2(z) =∑a

ξ2(a)e2πiNK2/Q(a)

An explicit calculation similar to the above shows that if bp is the coefficient of qp in theq-expansion of g2, then ap = bp, and g1(z) = g2(z). For a prime p, the character ξ2((m)) is

simply(

13p

)and we find once again that χ′ = χ−39.

Starting from these two quadratic number fields, we have induced the same weight onemodular form g = g1 = g2 in two different but analogous ways. As this modular form has level39 and nebentypus χ−39, this suggests that we might also be able to construct it from the class


13), Q(√−3) AND Q(

√−39) 51

group of K3 = Q(√−39) as well.

The class group CK3 of K3 is cyclic of order 4. Any non-trivial character of it sends a generatorof the group to either i or −i. Let ξ3 be such a character. This character will induce a weightone modular form g3 of level | − 39|NK3/Q(OK3) = 39 and nebentypus χ′′ = χ−39:

χ′′(m) =

(−39

m

)ξ((m)) =

(−39

m

)= χ−39(m) ∀m ∈ Z

Once again, g3(z) has a q-expansion of the form:

g3(z) =∑a

ξ3(a)e2πiNK3/Q(a)

To find the coefficients cp in the q-expansion we will again consider the cases when p is inert,ramifies, or is prime separately:

p is inert : Then once again, cp = 0.

p ramifies : Then p must divide the discriminant of K3 which is 39, and so p = 3 or 13. Saythat Q3 and Q13 are prime ideals lying above 3 and 13 respectively. Now, neither of these canbe principal, since the equations

x2 + xy + 10y2 = 3 and u2 + uv + 10v2 = 13

have no solutions in Z. As 3OK3 and 13OK3 are principal, however, ξ3(Q23) = ξ3(Q

213) = 1.

Therefore, since we have just observed that Q3 and Q13 are not principal, ξ3 maps them bothto −1 and so c3 = −1 and c13 = −1.

p splits : Suppose p splits, so that there are prime ideals Q1 and Q2 such that pOK3 = Q1Q2.Hence, ξ3(pOK3) = 1 = ξ3(Q1)ξ3(Q2). So we have 3 cases to consider:

• ξ3(Q1) = ξ3(Q2) = 1: Then cp = 2. This happens if and only if Q1 and Q2 are principal.In this case, there are integers a and b such that a2 + ab+ 10b2 = p. Calculation showsthat if this is possible then p ≡ 1, 4, 10, 16, 22 or 25 mod 39.

• ξ3(Q1) = ξ3(Q2) = −1: Then cp = −2. In this case, Q1 and Q2 are not principal but Q21

and Q22 are. This can only happen if there are integers a and b such that

a2 + ab+ 10b2 = p2, but no integers c and d such that c2 + cd+ 10d2 = p. Using Magma,we can again check that if this is possible then p ≡ 1, 4, 10, 16, 22 or 25 mod 39.

• ξ3(Q1) = ±i, ξ3(Q2) = ∓i: Then cp = 0. In this case, Q1 and Q2 are generators for theclass group and hence have order 4. It is easily calculated that the primes dividing 2generate the class group CK3 . Let these be P2 and P′2. One of Q1 and Q2 lie in the sameclass as P2. We may assume it is Q1. So by definition, there exists α, β ∈ OK3 such thatP2(α) = Q1(β). Taking norms then, we find that there are integers a, b, c and d such that

2(a2 + ab+ 10b2) = p(c2 + cd+ 10d2) (4.2.1)

Since a2 + ab+ 10b2, c2 + cd+ 10d2 ≡ 1, 4, 10, 16, 22 or 25 mod 39, on taking inversesmodulo 39 in (4.2.1) we get that p ≡ 2, 5, 8, 11, 20 or 32 mod 39. Say (4.2.1) holds, butQ1 does not generate the class group. It cannot be principal as p ≡ 2, 5, 8, 11, 20 or 32mod 39. Without loss of generality Q1 and P2

2 are in the same class. Factorising (4.2.1)


then P2P′2 and P2

2Q2 are in the same class. Therefore, after cancelling and takingnorms, we get that there are integers a′, b′, c′ and d′ such that

2(a′2 + a′b′ + 10b′2) = 2p(c′2 + c′d′ + 10d′2)⇒ a′2 + a′b′ + 10b′2 = p(c′2 + c′d′ + 10d′2)

The right-hand side is either 2, 5, 8, 11, 20 or 32 mod 39 but the left-hand side is1, 4, 10, 16, 22 or 25 mod 39. Hence, Q1 must generate the class group if and only ifthere are integers a, b, c and d such that

2(a2 + ab+ 10b2) = p(c2 + cd+ 10d2)

So we have found that

cp =

−1 if p = 3 or p = 13

±2 if p ≡ 1, 4, 10, 16, 22 or 25 mod 39

0 otherwise

This suggests that g and g3 will have the same q-expansions, and using Magma we cancalculate that

g3(z) = q − q3 − q4 + q9 + q12 − q13 + q16 − q25 − q27 +O(q30)

and in fact that g = g3. We have seen, therefore, that there is a modular form h which can beconstructed from two imaginary and one real quadratic number field.

To the ray class groups Cln1K1and Cln2K2

, we associate the ray class fields M1 and M2. These areboth extensions of Q obtained by adjoining a root of x4 + x3 − x2 − x+ 1. Taking the minimalGalois extension M of Q containing M1 and M2, we get the splitting field of the polynomialx8 + 6x7 −x6 − 54x5 − 41x4 + 138x3 + 153x2 − 72x+ 9 over Q. This has Galois groupisomorphic to D8. Here, ξ1 and ξ2 induce a representation ρ1 : Gal(M/Q)→ GL2(C), whichhas the same L-series as g.

Similarly, we find that the Hilbert class field N associated to K3 is obtained by adjoining aroot of x8 + 164x6 + 9358x4 + 248436x2 + 1734489 to Q. N is also the splitting field of thispolynomial, and Gal(N/Q) is isomorphic to D8. ξ3 induces a representation ρ2 : Gal(N/Q)→GL2(C) which has the same L-series as h, and hence ρ1. It can be shown that this implies thatρ1 and ρ2 are isomorphic representations of D8. The Galois group Gal(N/Q) has a centre oforder 2, coming from complex conjugation. Considering the composition of ρ2 followed by thenatural quotient map π : GL2(C)→ GL2(C)/C×, we observe ρ2 is a projective representationof D4.

4.3 A Level 145 Modular Form from Q(√5)

This example is the most interesting of the examples, as it comes from a real quadratic fieldbut no imaginary quadratic fields (as there are no suitable ray class groups). In this we willconsider the quadratic number field K = Q(

√5), and a ray class group for K with a non-trivial

module of definition.

Now, let us consider the factorisation of 29 in OK = Z[1, 1+√5

2]. By Dedekind’s Theorem, we

find that 29OK = P29P′29, where P29 = (29, 5 + 1+

√5

2) and P′29 = (29, 23 + 1+

√5

2). The class


5) 53

number of L is 1, so both of these ideals must be principal. Noting that NK/Q(5 + 1+√5

2) = 29,

then we must have that the ideal generated by 5 + 1+√5

2is prime and is divisible by P29 and so

is equal to P29. Similarly, note that 6− 1+√5

2has norm 29 and lies in P′29, so similarly then

P′29 = (6− 1+√5

2).

Take the modulus n to be P29.P∞ where P∞ corresponds to the non-trivial real embedding ofK. We can compute with Magma that

ClnK∼= Z/4Z

So the ray class group associated to n is cyclic of order 4. Note that 6− 1+√5

2− 1,

(6− 1+√5

2)2 − 1 6∈ P29 or else 29 divides 100 and 91 respectively. Hence neither P′29 nor P′29

2

are in P nK , and so the class of P′29 has order 4 and hence is a generator of the ray class group.

We may take a non-trivial character ξ of the ray class group then which sends the class of P′29to i. We extend ξ to a character of IK in the usual way. Hence ξ induces a level|5|NK/Q(P29) = 145 modular form of weight 1, and nebentypus

χ(m) = χ5(m)ξ((m)) ∀m ∈ Z

by Theorem 2.2.9. This theorem tells us that this modular form has a q-expansion like:

g(z) =∑a

ξ(a)e2πiNK/Q(a)

where the sum is over integral ideals as usual. To calculate what this modular form will looklike, we want to know what the coefficient ap of qp is for p prime. We need to consider theprimes which ramify, split and are inert in OK , and since P29 lies above 29, we will need toconsider it separately. Let p be a rational prime.

p is inert : As in the previous example, we again get that ap = 0. This is true for any primecongruent to ±2 modulo 5

p ramifies : Then p is necessarily 5. Note that as 5− 1 = 4 6∈ P1 (or else 29 divides 16), thenξ(5OK) 6= 1. Hence, ξ(5OK) is either −1, i or −i. Yet, since 5 ramifies, there is a prime idealP5 such that P2

5 = 5OK . So, as ξ(P5) must also be a fourth root of unity, we get thatξ(P5) = ±i. We can check that P5 and P′29 are in different class in ClnK , and so ξ(P5) = −i.Hence, a5 = −i.

p splits : We consider the cases when p = 29 and p 6= 29 separately:

• p = 29: We know that ξ(P29) = 0 and ξ(P′29) = i, so that a29 = i.

• p 6= 29: This case is more complicated to deal with. Let us consider pOK = P1P2.

Say that ξ(pOK) = 1. So p ≡ 1 mod P29. So 29 divides p− 1. In this case,ξ(P1) = ξ(P2) = ±1 or ξ(P1) = ±i, ξ(P2) = ∓i and so ap = ±2 or 0.

Similarly, suppose ξ(pOK) = −1. Then p2 ≡ 1 mod P29 and so 29 divides p2 − 1. Henceξ(P1) = ξ(P2) = ±i or ξ(P1) = ±1, ξ(P2) = ∓1 and so ap = ±2i or 0.


Similarly, we can deduce that if ξ(pOK) = ı then ap = ±(i+ 1), and if ξ(pOK) = −i thenap = ±(−i+ 1).

So we find that

ap =

0 p ≡ ±2 mod 5

−i p = 5

i p = 29

0,±2,±2i,±(1 + i),±(1− i) p splits, p 6= 29

Using Magma, we may compute the first few terms of the q-expansion of g as:

g(z) = q + iq4 − iq5 + iq9 + (−i− 1)q11 − q16 + (−i− 1)q19 + q20 − q25 + iq29 +O(q30)

The 2-dimensional complex Galois representation associated to this modular form comes fromthe character ξ and the ray class field N of ClnK , as per Chapter 3 Section 3. Using Magma, wecompute that N is the degree 8 extension of Q obtained by adjoining a root of the polynomialx8 − x7 + 3x6 + 3x5 + 3x4 − 6x3 − 2x2 − 3x+ 1. We can also check that N/Q is a splitting fieldof this polynomial and hence is Galois, with Galois group isomorphic to D8 since N/K hasGalois group isomorphic to the cyclic group of order 4. Thus ξ induces a 2-dimensionalcomplex dihedral representation ρ : Gal(N/Q)→ GL2(C) which corresponds to the weight onemodular form g, and ρ and g have equal L-series.

Appendix A

Induced Representations

Here we outline the concept of an induced representation to be used in Chapter 3. The basicmaterial closely follows that set out by Martin in his online notes [9].

Definition A.1. We let K ⊂ L be a field extension, and define G := Gal(L/K). Let V be afinite dimensional complex vector space. A representation of G on V is a homomorphismρ : G→ GL(V ). The degree of (V, ρ) is dimC V .

Given a representation (V, ρ) of a group G and a subgroup H ⊂ G, there is a natural way toconstruct a representation of H, the restricted representation.There is also a way to constructa representation of a group G given a representation (V, ρ) of a subgroup H ⊂ G.

Definition A.2. Let W :=⊕n

i=1 γiV where γ1, γ2, ..., γn are a set of coset representatives of Hin G. We say that W = IndGH V .

Now, for each g ∈ G and γi there is a unique j and h ∈ H such that gγi = γjh. We then define

ρ′(g)(γiv) = ρ(γj)(ρ(hi)(v))

and extend this linearly to W . Then (IndGH V, ρ′) is the representation of G induced from the

representation (V, ρ) of H.

It is easily shown by direct calculation that IndGH V is indeed a representation of G.

Remark A.3. Note that, dim IndGH V = dimV|G||H|

.

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56 APPENDIX A. INDUCED REPRESENTATIONS

Acknowledgements

I would like to thank my supervisor, Prof. Alan Lauder, both for suggesting such aninteresting dissertation topic to me and for his time and help. I would also like to thank hisD.Phil. student Mr Jan Vonk for the interesting conversation during my presentation. Thanksalso go to my housemates Harriet, Tom, Tomas and Shiv for putting up with my endless talkabout modular forms and L-series over the past few months and to Billy and Rachel for theiruseful comments and considerable skill in the art of proofreading. I am also grateful to Will formaking sure I took time off not to think about modular forms in the final weeks of work.Finally, all my love and thanks go to my sisters, Jessica and Katie, and my parents, Laura andRodney, for their help and support as I wrote this.

57

Bibliography

[1] Jan Hendrik Bruinier, Gerard van der Greer, Gunter Harder, Don Zagier, The 1-2-3 ofModular Forms, Springer-Verlag Berlin and Heidelberg GmbH & Co. K., Berlin, 2008.

[2] Daniel Bump, Automorphic Forms and Representations, Cambridge University Press,Cambridge, 1996

[3] Pierre Deligne and Jean-Pierre Serre, ‘Formes modulaires de poids 1’, Ann. Sc. de l’ Ec.Norm. Sup. 7, 1974, 507-530

[4] Fred Diamond and Jerry Shurman, A First Course in Modular Forms, Springer-VerlagNew York Inc, New York, 2005.

[5] Andrea Ferraguti, ‘Galois representations attached to type (1, χ) modular forms’, Master’sthesis, web page at http://algant.eu/documents/theses/ferraguti.pdf

[6] Victor Flynn, ‘B3.4 Algebraic Number Theory’, lecture notes for the B3.4 course given inHilary Term 2013, available from the Mathematical Insitute, Univeristy of Oxford

[7] Gerald J. Janusz, Algebraic Number Fields, 2nd edition, American Mathematical Society,Providence, 1996.

[8] Alan Lauder, ‘Part C Modular Forms MT 2014’, lecture notes for the C3.6 course given inMichaelmas Term 2015, available from the Mathematical Insitute, Univeristy of Oxford

[9] S. Martin, ‘Representation Theory’ lecture notes, web page athttp://www.maths.bris.ac.uk/∼matyd/repthy/martin.pdf

[10] Toshitsune Miyake, Modular Forms, Springer-Verlag Berlin and Heidelberg GmbH & Co.K., Berlin, 1989.

[11] Jurgen Neukirch, Algebraic Number Theory, 1st edition, Springer-Verlag Berlin andHeidelberg GmbH & Co. K., Berlin, 1999.

[12] Noah Snyder, ‘Artin’s L-functions: An Historical Approach’, web page athttp://math.columbia.edu/∼nsnyder/thesismain.pdf

[13] Balasz Szendroi, ‘Polynomial Rings and Galois Theory’, lecture notes for the B3.1 coursegiven in Michaelmas Term 2012, available from the Mathematical Insitute, Univeristy ofOxford

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