Modular Forms and Related Topics

U.U.D.M. Project Report 2021:13

Linnea Rousu

Examensarbete i matematik, 30 hp Handledare: Wolfgang Staubach Examinator: Magnus Jacobsson Juni 2021

Department of Mathematics Uppsala University

ABSTRACT

This thesis is a study of modular forms and related topics. It begins with some necessary background information on the modular group, SL2(Z), and the fundamental domain. Following that, modular functions are presented. These functions are invariant under the action of the modular group and are then generalized to modular forms. As an example of modular forms, the Eisenstein series is derived. Thereafter, Hecke operators are discussed. They are deﬁned as averaging operators over a suitable collection of double cosets with respect to a group. This chapter contains a section about Hecke operators on modular forms. The ﬁnal part of the thesis covers Dirichlet series, which are series of arithmetic functions. An interesting connection to modular forms is discovered. Acknowledgements

First and foremost I would like to thank my supervisor, Wulf Staubach for all his help and guidance. I am very grateful that he has taken the time to supervise me. I would also like to thank Alexander Söderberg for his support and the discussions we have had throughout the process. Additionally I would like to thank my family for all the support throughout the years. Furthermore I am grateful to all the teachers who have inspired me to study mathematics. I would also like to thank my classmates for making the years at university memorable. 3

Contents

1 Introduction4

2 Modular group5

3 Fundamental domain7

4 Modular functions9

5 Modular forms 12 5.1 Eisenstein series...... 17

6 Hecke operators 23 6.1 Introduction...... 23 6.2 Hecke operators...... 23 6.3 Hecke operators on periodic functions...... 25 6.4 Hecke operators on modular forms...... 28

7 Dirichlet series 32 7.1 Introduction...... 32 7.2 Dirichlet series and modular forms...... 33 7.3 The half plane of absolute convergence...... 35 7.4 The function defined by a Dirichlet series...... 36 7.5 Multiplication of Dirichlet series...... 38 7.6 Euler products...... 40 7.7 The half-plane of convergence of a Dirichlet series.... 42 7.8 Analytic properties of Dirichlet series...... 44 7.9 Dirichlet series with non-negative coefficients...... 45 7.10 Analytic continuation...... 46 7.11 Mean value formulas for Dirichlet series...... 48 7.12 An integral formula for the coefficients of a Dirichlet series 49 4

1 Introduction

Historically the study of modular functions (Swe: "modulära funktioner") started in 19th century where they first appeared in the theory of elliptic functions, more specifically as elements of the function field of an elliptic curve. The term goes back to Peter Gustav Lejeune Dirichlet, although the functions also occurred in the works of Carl Friedrich Gauss, Niels Henrik Abel and Carl Gustav Jacobi (in connection to his work on theta functions). Later, they also played a significant role in the works of Leopold Kronecker, Gotthold Eisenstein and Karl Weierstrass. Towards the end of 19th centry, modular functions became a crucial source of inspiration for Felix Klein and Henri Poincaré in the development of the theory of automorphic functions. Indeed the theory of Riemann surfaces became an important tool in this context, and it was Klein who used the term "Modulform" for the first time. A sys- tematic study of modular forms on SL2(Z) and its congruence subgroups, was made by Erich Hecke in 1925, and this established the modular forms as an independent discipline within function theory and analytic number theory.

Modular forms are used in several mathematical topics due to their geometrical, arithmetical and topological properties. For example topological modular forms is currently a topic of big interest in research. We will now present some applications of modular forms, based on [8]. In the proof of Fermat’s last theorem, Andrew Wiles uses modular forms exten- sively. Additionally, from this proof new techniques were developed to solve certain diophantine equations. These developments relied on having access to tables or software for computing modular forms. Modular forms are actually used in cryptography and coding theory. More specifically, to construct elliptic curve cryptosystems one wants to count the number of points on the elliptic curves. In order to do so there are point counting algorithms which use modular forms. Furthermore, algebraic forms associated to modular forms are used in certain error-correcting codes. This essay is a compendium of modular forms and related topics. Our goal is to make it as self-contained as possible and to include the details which explain the theory. In order to read this essay one should be familiar with abstract algebra, Fourier series, functional analysis and theory of integration on a basic level, as well as complex analysis on an advanced level. We will begin this essay with section2, covering the modular group, which is the integer subgroup of the special linear group. Most of the mathematics in the following will be done on this group, or on some subgroup. In section3 we present the fundamental domain; given a topological space and a group Γ0 acting on it, a fundamental domain is a subspace (of the topological space) containing exactly one point of each Γ0−equivalence class. Section4 covers modular functions which are functions invariant under the modular group and meromorphic on H ∪ {∞}. The 5 modular functions are then generalized to modular forms in section5. These forms are, in the most basic case, modular functions holomorphic on H ∪ {∞}. Following that, we derive a famous example of modular forms, called Eisenstein series in 5.1. In section6 we discuss the properties of the so called Hecke operators. These act as averaging operators over a certain collection of double cosets with respect to a group. They arose from Hecke’s theory on classifying the modular forms having multiplicative Fourier coeﬃcients. In this chapter we also discuss the properties of Hecke operators on modular forms. Section7 covers the Dirichlet series, which are series of arithmetic functions. These series are central in the theory of analytic number theory. For example, the famous Riemann zeta function is actually a Dirichlet series. The chapter begins with an interesting connection between modular forms and Dirichlet series. We show that there is a one-to-one correspondence between certain modular forms and Dirichlet series satisfying a speciﬁc functional equation.

2 Modular group

The following chapter is based on [5] and [7]. The special linear group SL2(R) = a b : a, b, c, d ∈ , ad − bc = 1 acts on the complex upper half plane = c d R H az+b {z ∈ C : Im(z) > 0} by linear fractional transformations as γz = cz+d for a b γ = ∈ SL2( ). The action is indeed a group action. Firstly, since if c d R a b γ = ∈ SL2( ) and z ∈ , then c d R H

az + b (az + b)(cz + d) γz = = cz + d (cz + d)(cz + d) (bcz + adz) + (aczz + bd) = (1) (cz + d)(cz + d) ac|z|2 + bd + 2bc Re(z) + z = . |cz + d|2

Therefore the imaginary part of γz is

ad − bc Im (z) Im (γz) = Im (z) = > 0. (2) |cz + d|2 |cz + d|2

Thus SL2(R) × H → H. Secondly, for the identity element I we have that 1 · z Iz = = z. 1 6

Thirdly, the action also satisﬁes the compatibility condition, i.e.

(aa˜ + bc˜)z + (a˜b + bd˜) (γγ˜)z = = γ(˜γ(z)) (˜ac + dc˜)z + (˜bc + dd˜)

a b a˜ ˜b for all γ = , γ˜ = ∈ SL2( ) and z ∈ . The transformations de- c d c˜ d˜ R H scribed above are also called Möbius transformations, which one may recognize from complex analysis. A particular discrete subgroup of SL2(R) plays a fundamental role in various branches of mathematics and physics and is called the modular group. Deﬁnition 2.1. The modular group Γ(Swe: "den modulära gruppen") is the group a b of all matrices for a, b, c, d ∈ such that ad − bc = 1. c d Z We also note that changing the sign of γ ∈ Γ does not change the action. Since, a b if γ = ∈ Γ, then we have c d

−az − b az + b −γz = = = γz. (3) −cz − d cz + d

0 1 Theorem 2.2. The modular group Γ is generated by S = and T = −1 0 1 1 , i.e every A ∈ Γ can be expressed as A = T n1 ST n2 S...ST nk for n ∈ 0 1 j Z, j = 1, . . . , k. Proof. Since S,T ∈ Γ we have that hS,T i ⊆ Γ where hS,T i is the span of S and T . 1 x It remains to show that Γ ⊆ hS,T i. We have T x = ∈ hS,T i for x ∈ . 0 1 Z 1 x 1 0 So N := , x ∈ } ⊆ hS,T i. Since S−1T −yS = we also have 0 1 Z y 1 1 0 a b that N := , y ∈ ⊆ hS,T i. Now let be an arbitrary element of y 1 Z c d a b a b Γ. We want to show that ∈ hS,T i as well. First note that S−1 S = c d c d d −c , and so without loss of generality we may assume that |a| |d| . −b a 6 a b 1 x a ax + b Next observe that = . We may therefore assume c d 0 1 c cx + d that 0 6 b < |a| . 1 0 a b a b We also have that = . Hence we may assume y 1 c d ay + c by + d that 0 6 c < |a| . 7

a b Since ∈ Γ we have that ad and bc are integers such that ad − bc = 1. c d 2 2 By the assumptions on the matrix elements we have that |ad| > a and |bc| < a . a b 1 0 Therefore we must have ad = 1 and bc = 0, which means that = ± ∈ c d 0 1 N ⊆ hS,T i. Thus an arbitrary element of Γ is in hS,T i, so Γ ⊆ hS,T i.

Now we would like to study the induced maps of the generators of Γ. For 0 1 1 1 S = and T = we have that −1 0 0 1

−1 Sz = and T z = z + 1 for all z ∈ . z H Since S and T generate the modular group, the induced maps generate the group of transformations deﬁned by the modular group. An important class of subgroups of the modular group is called congruence subgroups, which we will now deﬁne.

Deﬁnition 2.3. A congruence subgroup (Swe: "kongruens delgrupp") of Γ is any 0 subgroup of Γ = SL2(Z) containing Γ (N) = ker (SL2(Z) → SL2(Z \ NZ)). The smallest such N is the level of Γ0.

In this essay we will mainly make use of the congruence subgroup Γ0(N) deﬁned below

Deﬁnition 2.4. Let N ∈ N, the modular group (Swe: "den modulära gruppen") Γ0(N) of level N is

a b Γ (N) = ∈ Γ: c ≡ 0 mod N . 0 c d

Note that Γ0(1) = Γ so the modular group is actually the modular group of level 1.

3 Fundamental domain

The following is based on [5].

Deﬁnition 3.1. Let F ∈ H be a closed set with connected interior and let Γ0 ⊆ Γ be a subgroup of the modular group. F is called a fundamental domain (Swe: "grundläggande domän") for Γ0 if

(i) Any z ∈ H is Γ0−equivalent to some point in F.

(ii) No two boundary points of F are Γ0−equivalent.

(iii) The boundary ∂F of F is a ﬁnite union of smooth curves in H ∩ F. 8

So given a topological space and a group acting on it, the fundamental domain is a subset of the space containing exactly one point from each equivalence class. For the modular group Γ the fundamental domain is given by the following proposition

Proposition 3.2. The set

1 = z ∈ : |Re (z)| , |z| 1 F H 6 2 > is a fundamental domain for Γ.

a b Proof. (i) We want to show that for all z ∈ there exists γ = ∈ Γ such H c d that γz ∈ F. Recall that by equality (2) we have Im (z) Im (γz) = |cz + d|2

2 for c, d ∈ Z. So c and d cannot both be zero. Consequently |cz + d| attains a minimum as γ ranges over Γ. Therefore we may choose γ such that Im (γz) 1 1 1 n is maximal. To this end let T = , then we have T n = . Now 0 1 0 1 n 1 0 n choose n ∈ Z such that |Re ((T γ)z)| 6 2 and deﬁne γ = T γ for such n. We would like to show that |γz| > 1. Towards contradiction, suppose not, and consider the point w = γ0z with w = u + iv. Here we observe that for 0 −1 S = we have 1 0

−1 −u + iv Sw = = . u + iv u2 + v2

If |w| < 1 then

Im (w) Im (Sw) = > Im (w), |w|2

i.e. Im (Sγ0z) > Im (γ0z) which implies

Im (Sγz) > Im (γz),

contradicting maximality of Im (γz).

(ii) Suppose z and w are distinct points in the interior of F and γz = w for some a b γ = ∈ Γ. We may assume that Im (w) > Im (z) (otherwise, just switch c d 9

z and w and replace γ with γ−1). Again, as in (2) we have

Im (z) Im (w) = , |cz + d|2

and therefore |cz + d| < 1. In particular

|c| Im (z) = |Im (cz + d)| < 1.

√ 3 But since z ∈ F we have that Im (z) > 2 , which yields 1 2 |c| < < √ . Im (z) 3

Thus since c ∈ Z we must have c ∈ {0, ±1}. If c = 0, then the determinant condition gives ad = 1. Multiplying γ with ±1 if necessary (recalling (3), this does not change γ in Γ) we may assume a = d = 1. Hence we must have γ = T n for some n ∈ Z, however T nz 6∈ F. Next suppose c = ±1. By multiplying γ with ±1 if necessary we may assume c = 1. Note that 3 1 > |cz + d|2 = Re (z + d)2 + Im (z)2 > Re (z + d)2 + , 4

1 1 which implies |Re (z + d)| < 2 . Since |Re (z)| < 2 we must have d = 0. Hence a −1 γ = , thus γz = a − 1 which cannot be in . To summarise, no two 1 0 z F distinct points in F are in the same equivalence class of Γ. (iii) The boundary

1 ∂ = z ∈ : |Re (z)| = , |z| = 1 F H 2

is obviously a union of smooth curves.

4 Modular functions

A modular function is a function that is invariant with respect to the modular group, and is meromorphic on H ∪ {∞}. The following section is based on [5] and [8]. Deﬁnition 4.1. A weakly modular function (Swe: "svagt modulär funktion) of weight k is a meromorphic function f on H such that

f(γz) = (cz + d)kf(z) (4) 10

a b for all γ = ∈ Γ and all z ∈ . c d H −k a b We will sometimes use the notation f|γ,k(z) = (cz + d) f(γz), for γ = . c d

Remark 4.2. Using the notation above, it is clear that the function f satisﬁes the weak modularity condition (4) if f|γ,k(z) = f(z) for all γ ∈ Γ and z ∈ H.

Example 4.1. The zero function is weakly modular of any weight k ∈ Z.

Example 4.2. The constant functions on H are weakly modular of weight 0. Lemma 4.3. Suppose f and g are weakly modular functions of weight k and l respectively for the modular group Γ. Then

(i) For k = l and λ ∈ C, λf + g is a weakly modular function of weight k for Γ. Hence weakly modular functions of weight k form a complex vector space,

which is denoted by Mk.

(ii) The product fg is a weakly modular function of weight k + l for Γ.

f (iii) The fraction g is a weakly modular function of weight k − l for Γ provided g 6≡ 0.

Proof. (i) We have

λf(z) + g(z) = (cz + d)−k(λf(γ(z)) + g(γ(z))).

Let Mk be the set of weakly modular functions of weight k. This is a subset of the vector space of meromorphic functions. Since 0 ∈ Mk (recall example 4.1) and λf +g ∈ Mk, Mk it is in fact a subspace. Hence Mk is itself a vector space.

(ii) We have

f(z)g(z) = (cz + d)−(k+l)f(γ(z))g(γ(z)).

(iii) We have

f(z) f(γz) = (cz + d)−(k−l) . g(z) g(γz)

Proposition 4.4. There is no non-zero weakly modular function of odd weight. 11

Proof. Assume f is weakly modular of odd weight k. Apply −I, then

−z f(z) = (−1)−kf = −f(z). −1

Hence f must be 0.

Remark 4.5. When k is even, the condition (4) gives

− k d(γz) 2 f(γz) = (cz + d)kf(z) = f(z). dz

Hence (4) is equivalent to

f(γz)(d(γ(z)))k/2 = f(z)(dz)k/2.

This means that the diﬀerential form f(z)(dz)k/2 of weight k is ﬁxed under the action of every element of the modular group Γ.

Theorem 4.6. Let f be a meromorphic function on H. Then f is a weakly modular function of weight 2k for k ∈ Z if and only if

f(z + 1) = f(z), −1 f = z2kf(z) z for all z ∈ H.

Proof. Let f be a meromorphic function. First assume that f is a modular function of degree 2k for k ∈ Z. We have 1 · z + 1 f(z + 1) = f = (0 · z + 1)2kf(z) = f(z). 0 · z + 1

We also have −1 0 · z − 1 f = f = (1 · z + 0)2kf(z) = z2kf(z), z 1 · z + 0

These yield

−1 f(z + 1) = f(z) and f = z2kf(z). z

Next assume that f satisﬁes

−1 f(z + 1) = f(z) and f = z2kf(z). z 12

0 1 1 1 If we can show that formula (4) holds for S = and T = then since −1 0 0 1 S and T generates Γ (theorem 2.2), formula (4) holds for all γ ∈ Γ. Consequently f is a weakly modular function of weight 2k. The condition f(z + 1) = f(z) gives

1 · z + 1 f(z + 1) = f = (0 · z + 1)2kf(z). 0 · z + 1

1 1 −1 2k Thus formula (4) holds for T = . Using f = z f(z) we have 0 1 z

−1 0 · z + 1 f = f = ((−1) · z + 0)2kf(z) = z2kf(z). z −1 · z + 0

0 1 Thus formula (4) holds for S = . Hence f is weakly modular of weight −1 0 2k. Deﬁnition 4.7. A modular function (Swe: "modulär funktion) of weight k is a weakly modular function of weight k that is also meromorphic at ∞.

Example 4.3. The zero function is a modular function of every weight k ∈ Z.

Example 4.4. Every constant function on H is a modular function of weight 0. Later on, in section 5.1, we shall see a less trivial example of modular functions.

5 Modular forms

A modular form is a holomorphic function on H satisfying a certain kind of functional equation with respect to the group action of the modular group. Modular forms appear in many areas of mathematics such as number theory and algebraic topology, and also in physics, in particular string theory [8]. The following chapter is based on [5] and [8]. Deﬁnition 5.1. A modular form (Swe:"modulär form") of weight k and level 1 is a modular function of weight k that is holomorphic on H ∪ {∞}. Compared to modular functions of weight k we demand holomorphicity instead of meromorphicity on H ∪ {∞}.

Example 5.1. The zero function is a modular form of every weight k ∈ Z and level 1.

Example 5.2. Every constant function on H is a modular form of weight 0 and level 1. 13

These are of course trivial examples and in Section 5.1 we shall see an example of a non-trivial modular form represented as an Eisenstein series. We may now generalise the modular forms to any level. In order to do so we need the following deﬁnitions. Deﬁnition 5.2. Let Γ0 ⊆ Γ. The cusps (Swe: "spetsarna") of Γ0 are the set of Γ0−equivalence classes of Q ∪ {i∞}. a b Example 5.3. Let ∈ Γ (2) then c ≡ 0 mod 2 and a ≡ b ≡ d ≡ 1 mod 2. c d 0 We have

a b a i∞ = . c d c

a So i∞ is Γ0(2)−equivalent to a nonzero rational number c if and only if c is even and a is odd. Similarly,

a b b 0 = . c d d

b Thus 0 is Γ0(2)−equivalent to a nonzero rational number d if and only if d is odd. Hence Γ0(2) has exactly two cusps, represented by 0 and i∞.

Remark 5.3. Note that we can generalise the above example so that, for any prime p, Γ0(p) has exactly the two cusps represented by 0 and i∞.

Lemma 5.4. For all α, β ∈ Q ∪ {i∞} there exists γ ∈ Γ such that γα = β.

Proof. • For α = β = i∞, choose γ = I, then I(i∞) = i∞. p b p b • For α = i∞, β = p ∈ , choose γ = ∈ Γ. Then i∞ = p . q Q q d q d q

p • For α = q ∈ Q, β = i∞, choose a, b, c, d ∈ Z such that cp + dq = 0 and a b ad − bc = 1, then p = ap+bq = i∞. c d q cp+dq

p r • For α = q , β = s ∈ Q, choose a, b, c, d ∈ Z such that ap + bq = r, cp + dq = s a b and ad − bc = 1. Then p = ap+dq = r . c d q cp+dq s

Proposition 5.5. For any congruence subgroup Γ0(N) the set of cusps is ﬁnite.

p p Proof. Consider any q ∈ Q in reduced form. We may assume that q is not Γ0(N)− equivalent to i∞. There exists c0, d ∈ Z relative prime such that c0pN + dq = gcd(pN, q) which equals gcd(N, q) since gcd(p, q) = 1 . Furthermore c0pN + dq = 14 gcd(N, q) implies gcd(c, d) = 1 for c = c0N. Hence, there exists a, b such that a b ∈ Γ0(N). Therefore, c d

a b p ap + bq ap + bq = = . c d q cp + dq gcd(N, q)

p Thus, replacing q by something Γ0(N)−equivalent we may assume q|N. Since T = 1 1 n p 1 n p p+nq p ∈ Γ0(N) we have T = = is also Γ0(N)− equivalent to so 0 1 q 0 1 q q q we may assume 0 6 p < q. This gives only a ﬁnite number of possible non-equivalent p choices of q . In order to make sense of holomorphicity of a weakly modular function f for an arbitrary congruence subgroup Γ0 at any α ∈ Q we present the following lemma.

Lemma 5.6. Let f : H → C be a weakly modular function of weight k for a 0 −1 congruence subgroup Γ . If δ ∈ Γ then f|δ,k is a weakly modular function for δ Γδ.

Proof. For s = δ−1γδ ∈ δ−1Γδ we have

(f|δ,k)|s,k = f|δs,k = f|γδ,k = f|δ,k.

Hence f|δ,k is a weakly modular function if f is.

Now ﬁx a weakly modular function f of weight k for a congruence subgroup Γ0 and suppose α ∈ Q. 1 1 If T = ∈ Γ0 then we would have f(z + 1) = f(z), but this is not the 0 1 1 N case for any congruence subgroup Γ0. Fortunately T N = ∈ Γ(N), so any 0 1 congruence subgroup of level N contains T N . Hence we have f(z + j) = f(z) for 1 j some positive j ∈ . Choose the minimal j > 0 such that ∈ δ−1Γ0δ where Z 0 1 δ(i∞) = α. If the function f is holomoprhic at i∞ we obtain a Fourier expansion P∞ n/j 1/j 2πiz/j f(z) = n=0 anq where q = e . For the other cusp α ∈ Q, lemma 5.4 gives some γ ∈ Γ such that γ(i∞) = α. So we consider f to be holomorphic at the cusp α if the weakly modular function f|γ,k is holomorphic at i∞. Proposition 5.7. If a weakly modular function f is holomorphic at a set of representative elements for the cusps of Γ0(N), then it is holomorphic at every element of Q ∪ {i∞}.

Proof. Let c1, . . . , cn ∈ Q ∪ {i∞} be representatives for the set of cusps of Γ0(N) such that f is holomorphic at c1, . . . , cn ∈ Q ∪ {i∞}. If α ∈ Q ∪ {i∞}, then by lemma 5.4 there exists a γ ∈ Γ0(N) such that α = γci for some i. If δ ∈ Γ is 15

such that δ(i∞) = ci then f|δ,k is holomorphic at i∞. Since f is a weakly modular function for Γ0(N) we have

f|δ,k = (f|γ,k)|δ,k = f|γδ,k. (5)

But γ(δ(i∞)) = γ(ci) = α so (5) makes f holomorphic at α.

Remark 5.8. Using proposition 5.7 we may generalize the discussion above. So that a weakly modular function f(z) is considered to be holomorphic at the cusps for a congruence subgroup Γ0(N) if f|γ,k(z) is holomorphic at z = i∞ for all γ ∈ Γ. To see this, let γ ∈ Γ be arbitrary. Assume the weakly modular function f|γ,k is holomorphic at i∞, then by proposition 5.7, f|γ,k(z) is holomorphic at every z ∈ Q ∪ {i∞}. Consider an arbitrary representative ci of the cusps. By proposition 0 5.5 we have i ∈ {1, . . . , n}. The equivalence class of ci is {γci : γ ∈ Γ }. We have

˜ −k f|γ,k(ci) = (˜cci + d) f(γci)

a˜ ˜b for all γ = ∈ Γ0. Thus, for f| holomorphic at c , f is also holomorphic c˜ d˜ γ,k i 0 at γci for all γ ∈ Γ .

Remark 5.9. There is a connection between a function being holomorphic at i∞ and it being bounded, which we will now explain. If f is a meromorphic periodic function of period 1 then f has a Fourier expansion X f(z) = a qn n (6) n∈Z

2πiz −2πy 2πx for some constants an and q = e . Consider the map z = x + iy 7→ e e , it is 1 1 an holomorphic bijection from the vertical strip z ∈ H : 2 6 Im z < 2 in H to the punctured disk Dx = {z ∈ C : 0 < |z| < 1}. So we might think of f(z) as a function F (q) on Dx. As z → i∞ we recieve q → 0. Thus, for f(z) to be holomorphic at i∞ we need F (q) to be holomorphic at q = 0. If F (q) is holomorphic at q = 0 it must have a Laurent expansion which must be equal to the Fourier-expansion (6), meaning that an = 0 for all n < 0. This is equivalent to the condition that f(z) grows less than exponentially. i.e. there exists m such that |f(z)| < emy for y = Im z 0 (see [5] page 20 for the corresponding equivalence for f meromorphic at i∞).

Deﬁnition 5.10. Let Γ0 be a congruence subgroup of Γ of level N. A modular form f of weight k and level N for Γ0 is a weakly modular function f : H → C of weight k, holomorphic on H and at the cusps of Γ0.

We denote the space of modular forms of weight k for Γ0(N) by Mk(N).

Proposition 5.11. Mk(N) is a complex vector space for even k > 0. 16

Proof. We will show that Mk(N) is a subspace of the vector space of holomorphic functions on H and hence a vector space.

• For f ∈ Mk(N) and c ∈ C we have cf ∈ Mk(N).

• For f, g ∈ Mk(N) we have

a b f(z) + g(z) = (cz + d)−k(f(γz) + g(γz)) for all γ = ∈ Γ. c d

• If f and g are holomorphic at i∞ then f|γ,k and g|γ,k are bounded at i∞ for a b all γ = ∈ Γ. Hence c d

(f + g)|γ,k = f|γ,k + g|γ,k

a b is bounded at i∞ for all γ = ∈ Γ, which means that f + g is holomor- c d phic at the cusps.

Lemma 5.12. If f ∈ Mk(N) and g ∈ Ml(N) then fg ∈ Mk+l(N). Proof. By lemma 4.3, fg is a weakly modular function of weight k + l. Since both f and g are holomorphic at H, so is fg. As in the proof of proposition 5.11, if f|γ,k, g|γ,k are bounded at i∞ for all a b γ = ∈ Γ then c d

(fg)γ,k+l = (f|γ,k)(g|γ,l)

a b is bounded at i∞ for all γ = ∈ Γ. Hence fg is holomorphic at the cusps. c d Another interesting feature of modular forms is their speciﬁc Fourier expansion in the upper half-plane. More precisely, if f is a modular form there are constants an such that

∞ X n f(z) = anq n=0 for all z ∈ H and q = e2iπz. This series converges for all z ∈ H since f(q) is holomorphic on the unit disk D so its Taylor series converges absolutely in D. This feature of modular forms is sometimes included in their deﬁnition see e.g. [8].

Now we deﬁne an important class of modular forms, the so-called cusp forms. 17

0 0 Definition 5.13. Let Γ be a congruence subgroup and let f ∈ Mk(Γ ). We define f as a cusp form (Swe: "spetsform") if f vanishes at the cusps, i.e. f(z) → 0 as z → i∞. Alternatively, we can view cusp forms as modular forms with the constant 0 Fourier coefficient vanishing, a0 = 0. The space of cusps is denoted by Sk(Γ ) or 0 Sk(N) for Γ = Γ0(N). As Mk(N), this is also a vector space.

In Section 5.1 we define the Eisenstein series and in Section 6.4 we define an inner product on the space of modular forms, called the Petersson product, which is defined when one of the forms is a cusp form (since cusp forms vanish at the cusps). It turns out that the Eisenstein series are the orthogonal complement of cusp forms with respect to the Petersson product.

5.1 Eisenstein series

We will now construct a non-trivial example of a modular form, called the Eisen- stein series, which were ﬁrst studied by Gotthold Eisenstein. It is actually easier to construct a modular function by considering its derivatives. If f is a modular function, then

f (k)(γz) = (cz + d)2kf (k)(z)

a b for γ = ∈ Γ0 where Γ0 is a congruence subgroup of Γ. Therefore we would c d like to construct a function that satisﬁes the above transformation properties. The idea is now to take a weighted average of some function g : H → Cˆ over a congruence subgroup Γ0, where Cˆ is the Riemann sphere C ∪ {∞}. To do so, given γ ∈ Γ deﬁne the automorphy factor as

a b j(γ, z):Γ × → ˆ, j(γ, z) = cz + d for γ = . H C c d

For even k ∈ Z consider the average X f(z) = j(γ, z)−kg(γz). (7) γ∈Γ

Then we have X f(γ z) = j(γ, γ z)−kg(γγ z) 0 0 0 (8) γ∈Γ for γ, γ0 ∈ Γ. Now we need the following lemma for the automorphy factor j. Lemma 5.14. Let Γ be the modular group. For γ, γ0 ∈ Γ we have

j(γγ0, z) = ±j(γ0, z)j(γ, γ0z). 18

Proof. We observe that (2) yields

Im (z) Im (γz) = |j(γ, z)|2 and hence

0 Im (z) 0 Im (γ z) Im (z) = Im (γγ z) = = . (9) |j(γγ0, z)|2 |j(γ, γ0z)|2 |j(γ, γ0z)j(γ0, z)|2

Thus the lemma holds up to absolute values. The result will now follow from the following claim: Suppose f and g are two meromorphic functions such that

|f(z)|2 = |g(z)|2 . (10)

Then f(z) = ζg(z) for ζ such that |ζ| = 1 when g 6≡ 0. Note that the claim says that f f¯ ff¯ = gg¯ and = . g g¯

The image of f/g is contained in

1 z ∈ : z = = z ∈ : |z|2 = 1 . C z¯ C

By the open mapping theorem (restrict to an open set where f/g is holomorphic) this is impossible unless f/g is constant. Then the assumption (10) implies f(z) = ζg(z) for |ζ| = 1. Hence the denominators of (9) gives

j(γγ0, z) = ζj(γ0, z)j(γ, γ0z) for |ζ| = 1.

Finally the fact that j(γ, z) ∈ Z for z ∈ Z implies ζ = ±1. By equality (8) and this lemma we have that

k X −k f(γ0z) = j(γ0, z) j(γγ0, z) g(γγ0z) γ∈Γ k X −k = j(γ0, z) j(γ, z) g(γz) (11) γ∈Γ k = j(γ0, z) f(z),

−1 where we changed γγ0 to γ in the middle step.

Remark 5.15. The equality (11) means that if f converges and is meromorphic on H then f is a weakly modular function of weight k. 19

Taking the average over the whole congruence subgroup Γ0 will be too much in order for f(z) to converge (see [5] page 48). Instead we want to start with a periodic function and average it over just what we need in order for f to satisfy the weakly modular condition for Γ. Thus, consider the modular group Γ0(N) of level N which 1 1 contains T = . Since T ∈ Γ0(N), for a weakly modular function f we have 0 1 f(T z) = f(z +1) = f(z) so f is periodic. Let P be the subgroup of Γ0(N) generated 1 1 by T = , i.e. 0 1

1 n P = hT i = : n ∈ ⊆ Γ (N). 0 1 Z 0

Remark 5.16. Note that for γ ∈ P we have j(γ, z) = 1 so any periodic function with period 1 satisﬁes the weak modularity condition (4) for T . Thus averaging over P \ Γ0(N) should yield something weakly modular.

Let us now consider the coset space P \ Γ0(N).

Lemma 5.17. A complete set of representatives for P \ Γ0(N) is parametrized by the set

2 (c, d) ∈ Z : c ≡ 0 mod N, gcd (c, d) = 1 \ {±1} where a coset representative with parameter (c, d) can be taken uniquely in the form a b a b 0 1 ∈ Γ (N) for 0 b < |d| or = if (c, d) = ±(1, 0) and c d 0 6 c d −1 0 N = 1. a b Proof. Let γ = ∈ Γ0(N). We can replace γ with c d

1 n a b a + cn b + dn = 0 1 c d c d which is in the same right P -coset of Γ0(N) as γ. First suppose that d = 0, then −bc = 1 so b = −c = 1 which implies N = 1. Choosing n = −a shows that 0 1 γ ∈ P . −1 0 Secondly, suppose d 6= 0. Then we may uniquely choose n such that 0 6 b+dn < |d| . Therefore we may assume 0 6 b < |d| . On the other hand, the determinant condition gives 1 + bc a = ∈ i.e. bc ≡ −1 mod d. d Z

This determines b and consequently also a uniquely. 20

We will now choose our modular form to be a periodic function over the coset space P \ Γ0(N). Let k > 4 be an even positive integer. For z ∈ H deﬁne the normalized Eisenstein series of weight k and level N as

X 1 1 X 1 E (z) = = . k,N j(γ, z)k 2 (cz + d)k 2 γ∈P \Γ0(N) (c,d)∈Z gcd (c,d)=1 c≡0 mod N

1 The change of summands comes from lemma 5.17 and the 2 −factor comes from that we allow both (c, d) and (−c, −d) in the rightmost series. We want to show that this series is a modular form.

Proposition 5.18. The Eisenstein series Ek,N is a modular form of weight k > 4 and level N.

Proof. Firstly, lemma 5.14 gives us

−k k −k j(γ, γ0z) = j(γ0, z) j(γγ0, z) , for γ ∈ Γ and γ0 ∈ Γ0(N) since k is even. Hence X X E (γ z) = j(γ, γ z)−k = j(γ , z)k j(γγ , z)−k. k,N 0 0 0 0 (12) γ∈P \Γ0(N) γ∈P \Γ0(N)

Since right multiplying P \ Γ0(N) by γ0 simply permutes the cosets of P \ Γ0(N) we −1 can replace γ by γγ0 in the rightmost sum of equality (12), which gives

k Ek,N (γ0z) = j(γ0, z) Ek,N (z). (13)

We will now show absolute convergence of Ek,N . Since Ek,N is a sum of strictly smaller sets of terms than Ek,1 it suﬃces to show convergence of Ek,1. Consider the fundamental domain for Γ;

1 = z ∈ : |Re z| , |z| 1 . F H 6 2 >

(2πi)/3 Now let z ∈ F and let ζ3 = e . We have

|cz + d|2 = (cz + d)(cz¯ + d) = c2 |z|2 + 2cdRe z + d2. (14)

1 Since z ∈ F, we have |z| > 1 and Re z 6 2 . Hence (14) becomes

2 2 2 2 |cζ + d| for cd > 0 |cz + d| > c − |cd| + d = 2 . |cζ3 − d| for cd < 0 21

So z ∈ F gives X 1 X 1 |Ek,1(z)| 6 k + k |cζ3 + d| |cζ3 − d| c,d>0 c>0, d60 gcd (c,d)=1 gcd (c,d)=1 X 1 = 2 k . |cζ3 + d| c,d>0 gcd (c,d)=1

2 We want to check how many (c, d) ∈ Z there are such that |cζ3 + d| < r for some r ∈ R or equivalently, such that cζ3 + d ∈ Dr when Dr is a disk centered at 0 with radius r. Firstly, consider the case when |c| > 2r. Then we have √ 3 |Im (cζ + d)| = |Im (cζ )| = |c| > r. 3 3 2

k |Ek,1(γz)| = j(γ, z) Ek,1(z) . (16)

Hence (15) gives us absolute convergence of Ek,1 at any point z ∈ H. However, using (15) we also see that Ek,1 is a bounded function on F, which by Weierstrass M-test implies uniform convergence of Ek,1(z) and hence also of Ek,N (z) on F. For Ek,1 this implies compact convergence on H by the property (16) since, for ﬁxed γ ∈ Γ, j(γ, z) is uniformly bounded on compact sets. Since the tail of a series for Ek,N (z) can be bounded by the tail of a series for Ek,1(z), Ek,N is also uniformly bounded on compact subsets of H and hence compactly convergent on H. So Ek,N is a series of holomorphic functions which converges absolutely and com- 22 pactly on H and is therefore, by Montel’s theorem holomorphic on H. Furthermore, by remark 5.15 it is a weakly modular function. Now it remains to show that the series is holomorphic at the cusps. Let τ = r s ∈ Γ. Then we have t u

1 X 1 Ek,n|τ,k(z) = 2(tz + u)k rz+u k 2 c + d (c,d)∈Z tz+u gcd (c,d)=1 c≡0 mod N 1 X 1 = 2 k 2 (c(rz + s) + d(tz + u)) (c,d)∈Z gcd (c,d)=1 c≡0 mod N 1 X 1 = 2 ((cr + dt)z + (cs + du))k 2 (c,d)∈Z gcd (c,d)=1 c≡0 mod N 1 X 1 = 2 (c0z + d0)k 2 (c,d)∈Z gcd (c,d)=1 c≡0 mod N 0 0 r t for c = cr + dt and d = cs + du. Now consider Tτ = ∈ GL2( ), it acts on s u Z c c0 × and gives Tτ = . Hence the ﬁnal sum runs over a certain subset of Z Z d d0 pairs (c0, d0) ∈ Z × Z \{(0, 0)} . Consequently we have 1 X 1 |E | (z)| k,N τ,k 6 2 |c0z + d0|k (c0,d0)∈Z2\{(0,0)} which we have shown is bounded for all z ∈ F. As z → i∞ we recieve ∞ 1 X 1 1 X 1 X 1 → = = ζ(k). 2 |c0z + d0|k 2 |d|k dk (c0,d0)∈Z2\{(0,0)} d∈Z\{0} d=1

This means that Ek,N |τ,k(z) is bounded at i∞, and hence holomorphic at i∞. Thus Ek,N (z) is holomorphic at the cusps. 23

6 Hecke operators

6.1 Introduction

Erich Hecke developed a general theory in order to classify the modular forms that ∞ have multiplicative Fourier coefficients {an}n=1 (i.e. amn = aman when gcd (m, n) = 1). In this theory he defined the Hecke operators, which we will define later on in this section. There are several different ways to define these operators. One can define them as functions acting on lattices, as operators acting on cusp forms or as double coset operators. In this thesis we will do the latter. Generally, Hecke operators are averaging operators over a suitable collection of double cosets with respect to a group. This means that much of the Hecke theory belongs to linear algebra. But when studying the spectral analysis of these operators, things become more interesting. There are many results known only for arithmetic groups and therefore we will focus on the modular group. Hecke operators commute and are self-adjoint and hence the modular forms which are eigenvectors with respect to all Hecke operators form a basis of the space of cusp forms. The following section is based on [4].

6.2 Hecke operators

Consider the set a b G = : a, b, c, d ∈ , ad − bc = n . n c d Z

Note that G1 = Γ.

Lemma 6.1. A complete set of right coset representations of Gn modulo Γ is

a b ∆ = : ad = n, 0 b < d . n 0 d 6

This means there is a disjoint partition [ Gn = Γρ.

ρ∈∆n

a ∗ Proof. Let ρ = ∈ Gn. There are integers γ, δ such that γa + δc = 0 c ∗ c a and gcd (a, c) = 1 (Let γ = gcd (a,c) , δ = − gcd (a,c) ). So there exists a matrix ∗ ∗ ∗ ∗ τ = ∈ Γ giving τρ = . γ δ 0 ∗ 24

In order for τρ to be in ∆n one may need to change the sign of τ, then we have a b τρ = for ad = n, d > 0. 0 d 1 n Multiplying on the left by for n ∈ gives 0 b + nd < d (since the 0 1 Z 6 matrix product should also be in ∆n) so we may assume 0 6 b < d. As ρ ranges over ∆n the cosets Γρ are disjoint, since

α β a b a0 b0 α β = for ∈ Γ γ δ 0 d 0 d0 γ δ gives

α β 1 0 = . γ δ 0 1

P Remark 6.2. The number of elements of ∆n is σ(n) = d|n d.

Lemma 6.3. There exists a one-to-one correspondance between ∆n ×Γ and Γ×∆n. 0 0 Namely for any ρ ∈ ∆n and τ ∈ Γ there exist unique τ ∈ Γ and ρ ∈ ∆n such that

ρτ = τ 0ρ0. (17)

Proof. We begin by showing uniquness. Let

a b a0 b0 α ∗ α0 ∗ ρ = , ρ0 = , τ = , τ 0 = . 0 d 0 d0 γ δ γ0 δ0

Then (17) becomes

αa + γb ∗ α0a0 ∗ = . (18) γd δd γ0a0 γ0b0 + δ0d0

The determinant condition gives gcd(α0, γ0) = 1 and hence

a0 = gcd(αa + γb, γd), α0 = (αa + γb)/ gcd(αa + γb, γd), γ0 = γd/ gcd(αa + γb, γd), d0 = n/ gcd(αa + γb, γd) because a0d0 = n. Moreover, the determinant condition gives α0δ0 ≡ 1 mod |γ0| so δ0 is given modulo |γ0|. Then, by equating the lower right entries of (18) we ﬁnd b0 0 0 0 0 0 mod d. Restricting b to 0 6 b < d determines b uniquely, and hence also δ . 25

Note that the argument for uniqueness also proves existence of ρ0, τ 0 satisfying (17).

Let χ : Z → C be any function, this yields a function χ on GL2(Z) deﬁned as

a ∗ χ(ρ) = χ(a) for ρ = ∈ GL2( ). (19) ∗ ∗ Z

Deﬁnition 6.4. Given χ and k ∈ Z we deﬁne the Hecke operator of f : H → C as

k −1 X Tnf = n 2 χ(ρ)f|ρ,k.

ρ∈∆n

By lemma 6.1 we, more explicitly, have

1 X X a b T = χ(a)ak n n 0 d (20) ad=n 06b

6.3 Hecke operators on periodic functions

In the following we assume that

χ(1) = 1 and χ(−1) = (−1)k. (22)

Consider the group

1 n Γ = ± : n ∈ . ∞ 0 1 Z

From now on we consider functions automorphic to the group Γ∞, which means

f|τ,k = χ(τ)f for all τ ∈ Γ∞. (23)

1 1 Taking τ = ∈ Γ∞ shows that f is periodic of period 1. 0 1

Theorem 6.6. The operator Tn maps periodic functions to periodic functions. 26

Proof. Consider

a b a b0 1 u 1 v ρ = , ρ0 = , τ = , τ 0 = 0 d 0 d 0 1 0 1

0 0 0 for any ρ, ρ ∈ ∆n and τ, τ ∈ Γ∞. Then (19) and (22) yields

χ(ρ) = χ(ρ0) = χ(a) and χ(τ 0) = χ(τ) = 1.

Consequently we have

χ(a) = χ(ρ)χ(τ 0) = χ(ρ0)χ(τ). (24)

Furthermore the relation ρτ = τ 0ρ0 (lemma 6.3) gives

0 f|ρτ = f|τ 0ρ0 = f|ρ0 = χ(τ )f|ρ0 . (25)

Combining (24) and (25) gives

k −1 X (Tnf)|τ = n 2 χ(ρ)f|ρτ

ρ∈∆n k −1 X 0 = n 2 χ(τ)χ(ρ )f|ρ0 = χ(τ)Tnf. 0 ρ ∈∆n

Thus by (23), Tnf is periodic of period 1.

Now we want to study the action of the Hecke operator on the coeﬃcients of a Fourier series.

2πiz P∞ Proposition 6.7. Let p(z) = e and consider f(z) = m=0 a(m)p(mz), which converges absolutely in H. Then

∞ X Tnf(z) = an(m)p(mz) m=0 with coeﬃcients

X k−1 −2 an(m) = χ(d)d a(mnd ). d| gcd (m,n)

Proof. By (21) we have

∞ d−1 1 X X X az + b T f(z) = a(m) χ(a)ak p m . n n d m=0 ad=n b=0 27

Let

∞ d−1 1 X X az + b S = a(m) p m . n d m=0 b=0

We have

d−1 d−1 X az + b az X b p m = p m p m . d d d b=0 b=0

Note that

d−1 X b p m 6= 0 if and only if m ≡ 0 mod d. d b=0

Further use that n = ad, this gives

1 X az S = a(m)p m d. ad d m=0 mod d

Hence

∞ X X k−1 Tnf(z) = χ(a)a a(dl)p(alz) l=0 ad=n   ∞ X  X k−1  =  χ(a)a a(dl) p(mz), m=0 ad=n al=m which proves the theorem.

Corollary 6.8. For m, n > 1 we have an(m) = am(n).

From now on we assume that χ is completely multiplicative,

χ(mn) = χ(m)χ(n).

We will also show that the operator Tn has a multiplicativity property.

Theorem 6.9. For m, n > 1, we have

X k−1 TmTn = χ(d)d Tmnd−2 . d| gcd (m,n) 28

Proof. By (20) we have X k X a1 b1 a2 b2 mnTmTn = χ(a1a2)(a1a2) . 0 d1 0 d2 a1d1=m b1 mod d1 a2d2=n b2 mod d2 P a1 b1 a2 b2 The proof is done by reducing the matrix b1 mod d1 and then b2 mod d2 0 d1 0 d2 rearranging the terms of the sum, see [4] page 96.

Corollary 6.10. The Hecke operators commute as TmTn = TnTm.

6.4 Hecke operators on modular forms

We observe that since by Lemma 6.1 the collection ∆n forms a complete set of right cosets of Gn modulo Γ, then by restricting Tn to Mk(N) one can exchange every ρ ∈ ∆n with anything in the coset Γρ. Thus the Hecke operator on a modular form is

k −1 X T f = n 2 f| n ρ,k (26) ρ∈Γ\Gn since

f|τρ,k = f = f|ρ,k for τ ∈ Γ and f ∈ Mk(N). (27)

Since a modular form is a periodic function (by theorem 4.6) all results from section 6.3 hold for modular forms. So for example, modular forms are commutative.

Theorem 6.11. The Hecke operator Tn maps a modular form to a modular form and a cusp form to a cusp form. I.e.

Tn : Mk(N) → Mk(N)

Tn : Sk(N) → Sk(N)

Proof. For f ∈ Mk(N) meromorphic on H, Tnf is meromorphic on H. The correla- tion ρτ = τ 0ρ0 and (27) gives

f|ρτ,k = f|τ 0ρ0,k = f|ρ0,k and hence

k −1 X k −1 X (Tnf)|τ,k = n 2 f|ρτ,k = n 2 f|ρ0,k = Tnf 0 ρ∈∆n ρ ∈∆n for all τ ∈ Γ so Tnf is a modular function (by remark 4.2). 29

If f|ρ,k is holomorphic on H for ρ ∈ ∆n then Tnf is holomorphic on H. Now we shall prove holomorphicity at the cusps. Let τ ∈ Γ and consider a b (f|ρ,k)|τ,k = f|ρτ,k for ρτ = . As Im z → ∞ we have c d

1 a f|ρτ,k → lim f Im z→∞ (cz)k c which is ﬁnite since f is holomorphic at the cusp a/c ∈ Q. Consequently, each f|ρτ,k is holomorphic at i∞ and hence Tnf is holomorphic at the cusps. Applying the argument of holomorphicity on cusps on cusp forms shows that if f vanishes at the cusps, then so does Tnf.

Corollary 6.12. The space of cusp forms Sk(N) is spanned by the Poincaré series X P (z) = j(τ, z)−kp(mτz). m (28) τ∈Γ∞\Γ

Proof. For proof see [4] page 53.

Hence TnPm is a linear combination of various Poincaré series.

Theorem 6.13. For k > 2, m > 0, n > 1 we have X nk−1 T P = P −2 . n m d mnd d| gcd (m,n)

Proof. Combining expressions for Tn (26) and Pm (28) gives

k−1 X −k TnPm(z) = n j(g, z) p(mgz).

g∈Γ∞\Gn

Let H and G be any sets of right cosets representatives of Γ∞ \ Γ and Γ \ Gn respectively. Then both HG and GH are sets of right coset representatives. Hence 30

k−1 X X −k TnPm(z) = n j(ρτ, z) p(mτz) ρ∈G τ∈H X X X aτz + b = nk−1 d−k j(τ, z)−kp d ad=n b mod d τ∈H X X am = nk−1 d1−k j(τ, z)−kp τz d ad=n τ∈H d|m X nk−1 X = j(τ, z)−kp mnd−2τz d d| gcd (m,n) τ∈Γ∞\Γ X nk−1 = P −2 (z) d mnd d| gcd (m,n)

For m = 0 theorem 6.13 gives us the Eisenstein series,

X nk−1 X X T P = j(τ, z)−kp(0) = j(τ, z)−k n 0 d d|n τ∈Γ∞\Γ τ∈Γ∞\Γ 1 X = (cz + d)−k = E (z). 2 k gcd (m,n)=1

Actually, theorem 6.13 implies that the Eisenstein series Ek is an eigenfunction of all the Hecke operators Tn with eigenvalue σk−1(n). I.e.

TnEk = σk−1(n)Ek.

Corollary 6.14. For m, n > 1 we have

k−1 k−1 m TnPm = n TmPn.

A similar symmetry occurs for the inner product hTnf, Pmi. To see this, we shall ﬁrst deﬁne the inner product on Mk(N) × Sk(N).

Deﬁnition 6.15. We deﬁne the Petersson inner product h·, ·i : Mk(N)×Sk(N) → C (after Hans Petersson) as Z hf, gi = f(z)g(z)(Im z)kdν(z), F

dxdy where F is the fundamental domain for Γ and ν(z) = y2 for z = x + iy is the hyperbolic volume form. 31

Proposition 6.16. For m, n > 1 and f ∈ Mk(N) we have

k−1 k−1 m hTnf, Pmi = n hTmf, Pni .

Proof. For veriﬁcation, see [4] page 99.

Theorem 6.17. The Hecke operators Tn acting on Sk(N) are self adjoint, meaning that

hTnf, gi = hf, Tngi for all f, g ∈ Sk(N). (29)

Proof. Since Sk(N) is spanned by the Poincare series (28), it suﬃces to check (29) for the Poincare series. Corollary 6.14 and proposition 6.16 gives

hTnPm,Pli = hTnPl,Pmi = hPm,TnPli since the Fourier coeﬃcients of the Poincare series, and hence the inner products are real.

Now we will use a well known result from linear algebra.

Proposition 6.18. Let A ⊆ Mν(C) be a commuting family of normal matrices. −1 There exists a unitary matrix U ∈ Mν(C) such that U AU is diagonal for every A ∈ A.

This proposition can be stated in terms of linear operators.

Proposition 6.19. Let S be a ﬁnite dimensional Hilbert space over C and T a commuting family of normal operators T : S → S. Then there exits an orthonormal basis F of S consisting of common eigenfunctions of all the operators in T .

Proof. T = {f } S Let i i6ν be an orthonormal basis of . Then X T fi = λij(τ)fj j with some λij(τ) ∈ C. In this way T is represented by a matrix Λ(T ) = (λij(T )) which commutes with its adjoint Λ(T )∗ = Λ(T )0. This means that Λ(T ) is a normal matrix. These matrices commute, so by a suitable linear transformation with a unitary matrix the system T can be diagonalized.

Now we will state an important theorem on Hecke eigenfunctions.

Theorem 6.20 (Hecke). In Sk(N) there exists an orthonormal basis F consisting of eigenfunctions of all the Hecke operators Tn.

Proof. Apply proposition 6.19 on the Hecke algebra. 32

7 Dirichlet series

7.1 Introduction

In order to prove Euclid’s theorem on the existence of inﬁnitely many primes Leon- hard Euler (1737) showed that P p−1, over all primes, diverges. To show that he used that the zeta function ζ(s), deﬁned as

∞ X 1 ζ(s) = for s > 1 (30) ns n=1 diverges as s → 1. Further on Bernhard Riemann developed the function by allowing s to be complex. The complex version is known as the Riemann zeta function. The distribution of the zeros of this function is one of the biggest questions in modern mathematics. In 1837 Peter Gustav Lejeune Dirichlet deﬁned the Dirichlet L-functions,

∞ X χ(n) L(s, χ) = (31) ns n=1 where χ is called a Dirichlet character and s > 1, in order to prove his famous theorem on primes. The series (30) and (31) can be generalized as

∞ X f(n) ns n=1 where f(n) is an arithmetic function (a function from Z>0 to some subset of C). These series are called Dirichlet series with coeﬃcients f(n). The series are very useful in analytic number theory. Following Riemann, we let s be a complex variable as s = σ + it for σ, t ∈ R. Then we have that ns = nσeit log n, which gives |ns| = nσ since eit log n = 1 because t log n ∈ R. We consider the half plane of points s = σ + it such that σ > a for some a ∈ R. In the following section we shall show that there is a half plane σ > σc where the Dirichlet series converges and another half plane σ > σa where the series converges absolutely. Additionally we will show that in the half plane of convergence, the series represent a holomorphic function of the complex variable s. The ﬁrst subsection 7.2 is based on [6] and the following sections, 7.3-7.12 are based on [2]. 33

7.2 Dirichlet series and modular forms

Recall from theorem 4.6 that a modular function f of weight 2k satisﬁes the conditions −1 f(z + 1) = f(z) and f = z2kf(z). z

Now let f be a modular form, the statement does still hold. The periodicity of f P n 2πiz implies that f has a Fourier expansionf(z) = anq for q = e , and hence gives P −s a Dirichlet series ϕ(s) = ann . As we will see in this section, Erich Hecke showed that the second condition implies that the Dirichlet series satisfies a functional equation and conversely, every Dirichlet series satisfying a functional equation and some holomorphicity condition arises from a modular form. This section is based on [6]. M In the following we will define the Dirichlet coefficients as an = O(n ) for some P∞ −s M. This gives absolute convergence of the Dirichlet series ϕ(s) = n=1 ann for Re s > M + 1.

Deﬁnition 7.1. The gamma function Γ(s) is deﬁned as

Z ∞ Γ(s) = e−xxs−1dx for Re s > 0. 0

The gamma function extends to a holomophic function on the entire complex (−1)n plane except at the simple poles at s = −n where it has a residue n! for n ∈ Z>0. Proposition 7.2. (Mellin inversion formula) For every real c > 0,

1 Z c+i∞ e−x = Γ(s)x−sds for x > 0. 2πi c−i∞

Proof. We have

Z c+i∞ ∞ −s X −s Γ(s)x ds = 2πi Ress=−nx Γ(s) c−i∞ n=0 ∞ ∞ X (−1)n X (−x)n = xn = 2πi = 2πi · e−x. n! n! n=0 n=0

Theorem 7.3. Let a1, a2,... be a sequence of complex numbers such that an = M P∞ −nx P∞ −s O(n ) for some M. Let f(x) = n=1 ane and ϕ(s) = n=1 ann . Then Z ∞ Γ(s)ϕ(s) = f(x)xs−1dx (32) 0 34 for Re s > max (0,M + 1) and

1 Z c+i∞ f(x) = ϕ(s)Γ(s)x−sds (33) 2πi c−i∞ for c > max(0,M + 1) and Re (s) > 0.

Proof. First consider (32), we have

Z ∞ Z ∞ ∞ s−1 X −nx s−1 f(x)x dx = ane x dx 0 0 n=1 ∞ Z ∞ X −nx s−1 = ane x dx n=1 0 ∞ X −s = anΓ(s)n n=1 = Γ(s)ϕ(s).

The change of integration and summation sign is a consequence of the monotone P∞ −nx M convergence theorem since n=1 ane converges for an = O(n ). In the last integral we change the variable nx to x. The equation (33) follows from proposition 7.2.

Deﬁnition 7.4. Let ϕ˜(s) be a function on C. We say that ϕ˜(s) is bounded on vertical strips, if for all real numbers a < b, ϕ˜(s) is bounded on the strip a 6 Re s 6 b as Im s → ±∞.

Theorem 7.5. (Hecke, 1936) Let a0, a1, a2,... be a sequence of complex numbers M such that an = O(n ) for some M. Given λ > 0 and k > 1 deﬁne

P −s • ϕ˜(s) = ann (converges for Re s > M + 1)

2π −s • φ(s) = λ Γ(s)ϕ ˜(s) f˜(z) = P a e2πinz/λ (converges for Im z > 0). • n>0 n Then the following conditions are equivalent

(i) The function a a φ(s) + 0 ± 0 s k − s can be analytically continued to a holomorphic function on the entire complex plane which is bounded on vertical strips and satisﬁes the functional equation

φ(k − s) = ±φ(s). 35

(ii) For z ∈ H we have −1 f = ±zkf(z). z

Proof. See [6], page 103. The proof uses theorem 7.3. Remark 7.6. Let Γ0(λ) be the subgroup of Γ generated by the maps z 7→ z + λ and z 7→ −1/z.A modular form of even weight k, level 1 and multiplier C for Γ0(λ) is a holomorphic function f on H ∪ {∞} such that −1 f(z + λ) = f(z) and f = Czkf(z). z

This is slightly more general than the deﬁnition in section5, but for λ = 1 and C = 1 the deﬁnitions agree.

The theorem gives a one-to-one correspondance between modular forms of weight 0 M k, level 1 and multiplier C for Γ (λ) whose Fourier coeﬃcients satisfy an = O(n ) for some M and Dirichlet series satisfying (i). P∞ −s Example 7.1. The Riemann zeta function ζ(s) = n=1 n corresponds to a modular form of weight 1/2 and multiplier 1 for Γ0(2).

7.3 The half plane of absolute convergence

s σ a Note that if σ > a we have |n | = n > n and hence

f(n) |f(n)| . ns 6 na

This means that if a Dirichlet series converges absolutely for s = a + ib then it converges absolutely for every s such that Re s = σ > a. This observation implies the following theorem. P∞ −s Theorem 7.7. Suppose that the series n=1 f(n)n does not converge nor diverge for every s. Then there exists a real number, called the abscissa of absolute convergence (Swe "abskissan av absolutkonvergens), such that the series converges absolutely for all σ > σa but not for σ < σa. P∞ −s Proof. Let D be the set of all real σ such that n=1 |f(n)n | diverges. By assumption of convergence D is not empty and it is bounded above. Therefore D has a least upper bound, denote it by σa. If σ < σa then σ ∈ D since otherwise, σ would be an upper bound for D smaller than the least upper bound. If σ > σa then σ 6∈ D since σa is an upper bound for D. This proves the theorem.

P∞ −s If the series n=1 |f(n)n | converges everywhere we deﬁne σa = −∞ and if it diverges everywhere we deﬁne σa = ∞. 36

Example 7.2. The Riemann zeta function

∞ X 1 ζ(s) = ns n=1 converges absolutely for Re s = σ > 1 but diverges for s = 1, so σa = 1.

7.4 The function deﬁned by a Dirichlet series

P∞ −s Assume that n=1 f(n)n converges for every s with Re s = σ > σa and let

∞ X f(n) F (s) = for σ > σa. (34) ns n=1

In this section we will discuss some properties of the sum function.

Lemma 7.8. For N > 1 and σ > c > σa we have

∞ ∞ X f(n) 1 X |f(n)| . ns 6 N (σ−c) nc n=N n=N

Proof. We have

∞ ∞ ∞ ∞ X f(n) X |f(n)| X |f(n)| 1 X |f(n)| = . ns 6 nσ ncnσ−c 6 N σ−c nc n=N n=N n=N n=N

Theorem 7.9. Let F (s) be deﬁned as in (34), then

lim F (σ + it) = f(1) σ→∞ uniformly for t ∈ (−∞, ∞).

Proof. We have

∞ X f(n) F (s) = f(1) + ns n=2 so we want to show that the second term tends to 0 as σ → ∞. Let c > σa, then for σ > c lemma 7.8 gives

∞ ∞ ! X f(n) 1 1 X |f(n)| , ns 6 2σ 2c nc n=2 n=2 which tends to 0 as σ → ∞. 37

Theorem 7.10. If f is an arithmetic function with f(1) 6= 0 then there exists a unique arithmetic function f −1 called the Dirichlet inverse of f such that

f ∗ f −1 = f −1 ∗ f = I.

Moreover f −1 is given by the recursion formula

1 1 X n f −1(1) = and f −1(n) = − f f −1(d) for n > 1. f(1) n d d|n d

Proof. See [2] page 30.

We will now present a theorem stating that all the coeﬃcients are uniquely determined by the sum function.

Theorem 7.11. Let

∞ ∞ X f(n) X g(n) F (s) = and G(s) = ns ns n=1 n=1 be two Dirichlet series, both absolutely convergent for σ > σa. If F (s) = G(s) for each s in an inﬁnite sequence {sk} such that σk → ∞ as k → ∞, then f(n) = g(n) for every n.

Proof. Let h(n) = f(n) − g(n) and let H(s) = F (s) − G(s). Towards contradiction, assume that h(n) 6= 0 for some n. Let N be the smallest integer such that h(n) 6= 0. Then we have

∞ ∞ X h(n) h(N) X h(n) H(s) = = + . ns N s ns n=N n=N+1

Rearranging the terms gives

∞ X h(n) h(N) = N sH(s) − N s . ns n=N+1

For s = sk we have H(sk) = 0 and hence

∞ X h(n) h(N) = −N sk . nsk n=N+1

Now choose k so that σk > c > σa. Then lemma 7.8 gives

σ c ∞ N k 1 X |h(n)| |h(N)| 6 N + 1 N + 1 nc n=N+1 38

which tends to 0 as k → ∞ (and hence also σk → ∞). Therefore h(N) = 0, which is a contradiction.

The uniqueness theorem implies the existence of a half-plane in which the Dirich- let series does not vanish (unless it vanishes identically).

P∞ −s Theorem 7.12. Let F (s) = n=1 f(n)n and assume that F (s) 6= 0 for s with Re s = σ > σa. Then there is a half-plane σ > c > σa in which F (s) is never zero.

Proof. Assume no such half-plane exists. Then for every k > 1 there is a point sk with σk > k such that F (sk) = 0. Since σk → ∞ as k → ∞ theorem 7.11 shows that f(n) = 0 for all n, contradicting the hypothesis that F (s) 6= 0 for some s.

7.5 Multiplication of Dirichlet series

The following theorem states that the product of two Dirichlet series corresponds to their convolution. Theorem 7.13. Let s = σ + it and consider the Dirichlet series

∞ ∞ X f(n) X g(n) F (s) = for σ > a and G(s) = for σ > b. ns ns n=1 n=1

Then in the half plane of convergence of both series we have

∞ X h(n) F (s)G(s) = ns n=1 where h = f ∗ g is the Dirichlet convolution of f and g, deﬁned as X n h(n) = f(d)g . d d|n

P∞ −s Conversely, if F (s)G(s) = n=1 α(n)n for all s in a sequence {sk} with σk → ∞ as k → ∞, then α = f ∗ g.

Proof. For all s such that both the series converge absolutely we have

∞ ∞ ∞ ∞ X f(n) X g(m) X X f(n)g(m) F (s)G(s) = = ns ms (mn)s n=1 m=1 n=1 m=1 ∞ ∞ X (P f(n)g(m)) X h(k) = mn=k = ks ks k=1 k=1 for X h(k) = f(n)g(m) = (f ∗ g)(k). mn=k 39

The other statement follows from theorem 7.11.

Example 7.3. We will give an example on multiplication of the Riemann zeta function and the Dirichlet series with the Möbius function µ(n) as coeﬃcients, deﬁned as

 1 if n is a square-free positive integer with an even number of prime factors  µ(n) = −1 if n is a square-free positive integer with an odd number of prime factors  0 if n has a squared prime factor

Theorem 7.14. For n > 1 we have

X 1 1 if n = 1, µ(d) = = n 0 if n > 1. d|n

Proof. See [2] page 25.

Now consider the Dirichlet series

∞ ∞ X 1 X µ(n) and . ns ns n=1 n=1

Both these series converge absolutely for s with Re s = σ > 1. By theorem 7.13 and 7.14 we have

∞ X µ(n) ζ(s) = 1 if σ > 1. ns n=1

This shows that ζ(s) 6= 0 for σ > 1 and that

∞ 1 X µ(n) = for σ > 1. ζ(s) ns n=1

P∞ −s Example 7.4. Let f(1) 6= 0 and consider F (s) = n=1 f(n)n and G(s) = P∞ −s −1 n=1 g(n)n for g = f . Then in any half-plane of absolute convergence for both the series, we have

1 F (s) 6= 0 and G(s) = . F (s) 40

7.6 Euler products

The following theorem was proved by Leonhard Euler in 1737 and is sometimes called the fundamental theorem of arithmetic. Theorem 7.15. Let f be a multiplicative arithmetic function such that the series P f(n) is absolutely convergent. Then we have

∞ X Y 2 f(n) = 1 + f(p) + f(p ) + ··· (35) n=1 p extended over all primes. If f is completely multiplicative, the product simpliﬁes as

∞ X Y 1 f(n) = . 1 − f(p) n=1 p

In each of these cases the product is called the Euler product of the series.

Proof. Let Y P (x) = 1 + f(p) + f(p2) + ··· p6x be a ﬁnite product over all primes p 6 x. Since it is a product of a ﬁnite number of absolutely convergent series we can multiply the series and rearrange the terms in any fashion without altering the sum. Using that f is a multiplicative function and the unique factorization theorem we can rewrite the product as X P (x) = f(n) n∈A where A is the set of integers having prime factors less than x. This gives

∞ X X f(n) − P (x) = f(n) n=1 n∈B where B is the set of integers having at least one prime factor greater than x. Therefore we have

∞ X X X f(n) − P (x) 6 |f(n)| 6 |f(n)| . n=1 n∈B n>x

Since the series P |f(n)n−s| is absolutely convergent, the rightmost term tends to 0 as x → ∞. Hence

∞ X P (x) → f(n) as x → ∞. n=1 41

Q Recall that an inﬁnite product (1 + an) converges absolutely whenever the corre- P sponding series an does. In this case we have

X 2 X 2 f(p) + f(p ) + ··· 6 |f(p)| + |f(p )| + ··· p6x p6x ∞ X 6 |f(n)| < ∞. n=2

P |f(p) + f(p2) + · · · | Since all partial sums p6x are bounded, the series P 2 p |f(p) + f(p ) + · · · | of all primes is absolutely convergent. Hence the product (35) is absolutely convergent. When f is completely multiplicative we have f(pn) = f(p)n and each series on the right of (35) is a convergent series with sum 1 . 1 − f(p)

We will now apply the theorem on Dirichlet series.

P −s Theorem 7.16. Assume f(n)n converges absolutely for s with Re s = σ > σa. If f is multiplicative we have

∞ X f(n) Y f(p) f(p2) = 1 + + + ··· if σ > σ ns ps p2s a n=1 p extended over all primes. If f is completely multiplicative we have

∞ X f(n) Y 1 = if σ > σ . ns 1 − f(p)p−s a n=1 p

Proof. Apply theorem 7.15 on absolutely convergent Dirichlet series.

Example 7.5. For the Riemann zeta function we have

∞ X 1 Y 1 ζ(s) = = if Re s = σ > 1. ns 1 − p−s n=1 p

Example 7.6. Following example 7.3 we have

∞ 1 X µ(n) Y = = 1 − p−s . ζ(s) ns n=1 p 42

7.7 The half-plane of convergence of a Dirichlet series

In order to prove existence of a half-plane of convergence we will use the following lemma

P −s0 Lemma 7.17. Let s0 = σ0 +it and let f(n)n be a Dirichlet series with bounded partial sums, say

X f(n) 6 M ns0 n6x for all x > 1. Then for each s such that Re s = σ > σ0 we have X f(n) |s − s0| σ0−σ s 6 2M 1 + . n σ − σ0 a

Proof. a(n) = f(n)n−s0 A(x) = P a(n) Let and let n6x . Then f(n) a(n) = . ns ns0−s

Hence we can apply Abel’s identity (Theorem 4.2 in [2]) with f(x) = xs0−s, which gives

f(n) Z b X s0−s s0−s s0−s−1 s = A(b)b − A(a)a + (s − s0) A(t)t dt. n a a

σ0−σ σ0−σ Since a < b and σ0 −σ < 0 one has that b < a , and so the triangle inequality yields σ −σ σ −σ σ −σ σ −σ σ −σ |b 0 − a 0 | 6 b 0 + a 0 < 2a 0 . Therefore since |A(x)| 6 M we obtain

X f(n) Z b Mbσ0−σ + Maσ0−σ + |s − s |M tσ0−σ−1dt s 6 0 n a a

The following theorem will be used only to prove the next theorem.

Theorem 7.18. (Cauchy condition for uniform convergence of series) Let fn be a P∞ sequence of functions. The series n=1 fn(x) converges uniformly on S if and only 43 if, for all ε > 0 there exists a N such that n > N gives

n+p

X fk(x) < ε k=n for all p = 1, 2,... and all x ∈ S.

Proof. See [1] page 223.

P −s Theorem 7.19. If the series f(n)n converges for s = σ0 + it0 then it also converges for all s with Re (s) = σ > σ0. If it instead diverges for s = σ0 + it0, it diverges for all s with Re (s) = σ < σ0.

Proof. To prove the ﬁrst statement we choose any s with σ > σ0. Lemma 7.17 shows that

X f(n) Kaσ0−σ ns 6 a

Theorem 7.20. If the series P f(n)n−s does not converge everywhere or diverge everywhere, then there exists a real number σc called the abscissa of convergence, such that the series converges for all s in the half-plane σ > σc and diverges for all s in the half-plane σ < σc.

The proof of this theorem is very similar to that of theorem 7.7

P∞ −s Proof. Let D be the set of all real σ such that n=1 f(n)n diverges. By assumption of convergence D is not empty and it is bounded above. Therefore D has a least upper bound, denote it by σc. If σ < σc then σ ∈ D. If σ > σc then σ 6∈ D since σa is an upper bound for D. This proves the theorem.

Remark 7.21. If the series converges everywhere we deﬁne σc = −∞ and if it converges nowhere we deﬁne σc = ∞.

Absolute convergence implies convergence and hence σa > σc. In the case that σa > σc, there is an inﬁnite strip σc < σ < σa where the series converges condition- ally. We will show that the width of this strip is bounded.

Theorem 7.22. For any Dirichlet series with σc ﬁnite we have

0 6 σa − σc 6 1. 44

P −s0 Proof. It is suﬃcient to show that if f(n)n converges for some s0, then it converges absolutely for all s with Re (s) = σ > σ0 + 1. Let A be an upper bound for the numbers |f(n)n−s0 |. Then we have

f(n) f(n) 1 A = 6 ns ns0 ns−s0 nσ−σ0

P∞ n −s Example 7.7. The Dirichlet series n=1(−1) n converges for s with Re s = σ > 0 and converges absolutely for σ > 1. So we have σc = 0 and σa = 1. One can compare the convergence properties of a Dirichlet series with those of a power series. A power series has a disk of convergence whereas a Dirichlet series has a half-plane of convergence. For a power series, the interior of the disk is the domain of absolute convergence, but for a Dirichlet series the domain of absolute convergence may be a proper subset of the half-plane of convergence. A power series represents a holomorphic function inside its disk of convergence. Similarly, we will show that a Dirichlet series represents a holomorphic function inside its half-plane of convergence.

7.8 Analytic properties of Dirichlet series

In order to deduce the analytic properties of Dirichlet series we will use the following well-known result from complex analysis.

Theorem 7.23 (Weierstrass, 1894). Let {fn} be a sequence of functions holomorphic on an open set Ω ⊆ C and assume that {fn} converge compactly on Ω to a 0 limit function f. Then f is holomorphic on Ω and the sequence of derivatives {fn} converge compactly on Ω to the derivative f 0. Proof. See [2] page 234. P∞ −s Theorem 7.24. A Dirichlet series n=1 f(n)n converges compactly in the interior of the half-plane of convergence σ > σc. Proof. It suﬃces to show that P f(n)n−s converges uniformly on every compact rectangle R = [α, β] × [γ, δ] with α > σc. Let s0 = σ0 for σc < σ0 < α and let s be any point with Re s = σ > σ0. On the rectangle R we have

X f(n) 6 M ns0 n6x for some M. Hence lemma 7.17 gives X f(n) |s − s0| σ0−σ s 6 2M 1 + . (36) n σ − σ0 a

For s ∈ R we have σ−σ0 > α−σ0 and |s − s0| < C where C is a constant depending on s0 and R but not on s. Then the inequality (36) gives

X f(n) C σ0−α s 6 2Mα 1 + . n α − σ0 a

Since ασ0−α → 0 as α → ∞ theorem 7.18 for uniform convergence is satisﬁed. Theorem 7.25. The sum function F (s) = P f(n)n−s of a Dirichlet series is holomorphic in its half-plane of convergence σ > σc and its derivative in σ > σc is given by

∞ X f(n) log (n) F 0(s) = − ns n=1 obtained by diﬀerentiating term by term.

Proof. Apply theorem 7.23 and 7.24 to the sequence of partial sums. Remark 7.26. The series

∞ X f(n) log (n) F 0(s) = − ns n=1 has the same abscissa of convergence and absolute convergence as the series for F (s).

Remark 7.27. Applying theorem 7.25 repeatedly gives

∞ X f(n)(log n)k F (k)(s) = (−1)k for s with Re (s) = σ > σ . ns c n=1

7.9 Dirichlet series with non-negative coeﬃcients

Some functions deﬁned by a Dirichlet series on their half-plane of convergence σ > σc can be continued analytically beyond the line σ = σc. A famous example of this phenomena is that the Riemann zeta function ζ(s) can be continued analytically beyond the line σ = 1 to a function holomorphic at all s except at the simple pole s = 1. Theorem 7.28. Let F (s) be represented in its half-plane of convergence σ > c by the Dirichlet series

∞ X f(n) F (s) = ns n=1 where c is ﬁnite and assume that f(n) > 0 for all n > n0. If F is holomorphic in some disk about the point s = c, then the Dirichlet series converges in the half- 46 plane σ > c − ε for some ε > 0. Consequently if the series has a ﬁnite abscissa of convergence σc, then F (s) has a singularity on the real axis at the point s = σc.

Proof. Let a = 1 + c. Since F is holomorphic at c it can be represented by the absolutely convergent power series

∞ X F (k)(a) F (s) = (s − a)k. k! (37) k=0

The radius of convergence of this series exceeds 1 because F is holomorphic at c. By applying theorem 7.25 repeatedly we obtain

∞ X f(n)(log n)k F (k)(a) = (−1)k nσ n=1 so (37) can be rewritten as

∞ ∞ X X (a − s)k f(n)(log n)k F (s) = · . k! nσ (38) k=0 n=1

Since the radius of convergence is greater than 1, the formula is valid for a real number s = c − ε for some ε > 0. For this s we have a − s = 1 + ε and hence the series in (38) has non-negative coeﬃcients for n > n0. So we can exchange the order of summation in (38) to obtain

∞ ∞ ∞ ∞ X f(n) X ((1 + ε) log n)k X f(n) X f(n) F (c − ε) = = e(1+ε) log n = . na k! na nc−ε n=1 k=0 n=1 n=1

This means that the Dirichlet series P f(n)n−s converges for s = c − ε and hence also for the half plane σ > c − ε.

7.10 Analytic continuation

P −s Theorem 7.29. Let F (s) = f(n)n be absolutely convergent for σ > σa and such that f(1) 6= 0. If F (s) 6= 0 for σ > σ0 > σa then for s such that Re s = σ > σ0 we have

F (s) = eG(s) for

∞ X (f 0 ∗ f −1)(n) 1 G(s) = log f(1) + · log n ns n=2 where f −1 is the Dirichlet inverse of f (deﬁned as in (7.10)) and f 0(n) = f(n) log n. 47

For complex z 6= 0, log z denotes the branch of the logarithm.

Proof. Since F (s) 6= 0 we have F (s) = eG(s) for some function G(s) holomorphic for s with Re s = σ > σ0. Diﬀerentiation gives

F 0(s) = eG(s)G0(s) = F (s)G0(s).

Hence G0(s) = F 0(s)/F (s). But by deﬁnition of F (s) we have

∞ ∞ X f(n) log n X f 0(n) F 0(s) = − = − and ns ns n=1 n=1

∞ 1 X f −1(n) = . F (s) ns n=1 Hence

∞ X (f 0 ∗ f −1)(n) G0(s) = − . ns n=2

Integration gives

∞ X (f 0 ∗ f −1)(n) 1 G(s) = C + · log n ns n=2 where C is a constant. By theorem 7.9 we have

lim F (σ + it) = eC σ→∞ so C = log f(1). Hence

∞ X (f 0 ∗ f −1)(n) 1 G(s) = log f(1) + · log n ns n=2 which converges absolutely for s with Re s = σ > σ0.

This theorem can be used to give an analytic continuation of the Riemann zeta function, which we will now show.

P∞ −s 0 Example 7.8. Consider ζ(s) = n=1 n . The coeﬃcients f(n) = 1 give f (n) = log n and f −1(n) = µ(n), hence

X n (f 0 ∗ f −1)(n) = log (d)µ . d d|n 48

So for σ > 1 we have ζ(s) = eG(s) for

∞ P n X d|n log (d)µ d 1 G(s) = · . log n ns n=2

Since G(s) is a sum of holomorphic terms in σ > 1, ζ(s) = eG(s) is a holomorphic function in σ > 1 and hence it is an analytic continuation of ζ(s).

7.11 Mean value formulas for Dirichlet series Theorem 7.30. Let F (s) = P f(n)n−s and G(s) = P g(n)n−s be two Dirichlet series of abscissae of absolute convergence σ1 and σ2 respectively. Then for a > σ1 and b > σ2 we have

∞ Z T X f(n)g(n) lim F (a + it)G(b − it)dt = . T →∞ na+b −T n=1

Proof. We have

∞ ! ∞ ! X f(m) X g(n) F (a + it)G(b − it) = ma+it nb−it m=1 n=1 ∞ ∞ X X f(m)g(n) n it = manb m m=1 n=1 ∞ ∞ X f(n)g(n) X f(m)g(n) n it = + . na+b manb m n=1 m,n=1 m6=n

Now we have the inequality

∞ ∞ ∞ X f(m)g(n) n it X |f(m)| X |g(n)| manb m 6 ma nb m,n=1 m=1 n=1 so the series converges absolutely and uniformly for all t. Hence we can integrate term by term and divide by 2T to obtain

1 Z T F (a + it)G(b − it)dt 2T −T ∞ X f(n)g(n) X f(m)g(n) 1 Z T = + · eit log (n/m)dt na+b manb 2T n=1 m,n=1 −T m6=n ∞ ∞ n X f(n)g(n) X f(m)g(n) sin T log = + · m . na+b manb T log n n=1 m,n=1 m m6=n 49

The double series converges uniformly with respect to T since sin x/x is bounded for every x, and hence the double sum vanishes as T → ∞. P∞ Corollary 7.31. If F (s) = n=1 converges absolutely for σ > σa we have

∞ ∞ X X |f(n)|2 F (s) = |F (σ + it)|2dt = . n2σ n=1 n=1

Proof. The formula follows by theorem 7.30 with g(n) = f(n).

7.12 An integral formula for the coeﬃcients of a Dirichlet series

Given a sum function F (s) = P f(n)n−s with unknown coefficients, we want to find the coefficients f(n). The following theorem gives a formula to do so.

Theorem 7.32. Assume the series

∞ X f(n) F (s) = ns n=1 converges absolutely for s with Re s = σ > σa. Then for σ > σa and x > 0, we have

1 Z T f(n) if x = n lim F (σ + it)xσ+itdt = . T →∞ 2T −T 0 otherwise

Proof. For σ > σa we have

∞ 1 Z T xσ Z T X f(n) xit F (σ + it)xσ+itdt = dt 2T 2T nσ n −T −T n=1 ∞ (39) xσ X f(n) Z T = eit log (x/n)dt 2T nσ n=1 −T since the series is uniformly convergent for any t in the interval [−T,T ] for any T . We will look at two cases. Firstly, when x is not an integer we have x/n 6= 1 for all n and

∞ T ∞ x xσ X f(n) Z xσ X f(n) sin T log eit log (x/n) = · n 2T nσ T nσ log x n=1 −T n=1 n which tends to 0 as T → ∞. Secondly when x is an integer, say x = k then the term in (39) with n = k becomes

kσ f(k) Z T k it f(k) · σ = · 2T = f(k) 2T k −T k 2T 50 and hence (39) becomes

∞ ∞ it xσ X f(n) Z T xit kσ X Z T k dt = f(k) + dt. 2T nσ n 2T n n=1 −T n=1 −T n6=k

The second term tends to 0 as T → ∞ and hence the statement is proved. 51

References

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