On The Theory of Modular Formswr07/book.pdf · These notes serve as an explicit introduction to...

On The Theory of Modular Forms

Wissam Raji

Tobias Muhlenbruch

Department of Mathematics, American University of Beirut, Beirut,Lebanon

E-mail address: [email protected]

Department of Mathematics and Computer Science, FernUniver-sitat in Hagen, 58084 Hagen, Germany

E-mail address: [email protected]

http://www.aub.edu.lb/fas/math/

http://www.aub.edu.lb/

mailto:[email protected]

http://www.fernuni-hagen.de

http://www.fernuni-hagen.de

mailto:[email protected]

Preface

These notes were developed from course notes after years of teaching graduatecourses on the theory of elliptic modular forms. Several chapters have been adaptedas well from the excellent lectures of Professor Marvin Knopp, my PhD advisor. Itintends to shed light on the importance of the theory of modular forms in differentareas of mathematics and mathematical physics.

It is adapted to meet the expectations of all those who want to learn about thetheory of modular forms from an analytic approach with an elementary approachto discontinuous and discrete groups. Topics range from introductory to researchlevel topics depending on the depth of the chapters and on the importance of thetopic being discussed. Readers of these notes are required to have basic knowledgein complex analysis, real analysis and abstract algebra. Although all elementaryconcepts are being introduced in every chapter but still knowledge about certaintopics in analysis are being taken for granted. We also include an appendix thatcontains all the needed theorems that are being used repetitively.

These notes serve as an explicit introduction to theory of modular forms. Itis intended to give a precise introduction about the most important concepts inthe theory of modular forms and the recent developments on the theory of periodpolynomials emerging from modular forms on the full modular group. Period poly-nomials have been the center of attention for many researchers in the field andmany important results have been contributed to the deep understanding to therole of those polynomials.

We start by introducing discrete and discontinuous groups and as a result con-gruence subgroups and the study of modular forms on subgroups of finite index inthe full modular group. We proceed with detailed treatment of Hecke operators andtheir properties. Period polynomials are at the heart of the classical modular formspart, connections with L-functions are represented and the proof of Eichler-Shimuratheorem is proved explicitly. Actions of Hecke operators on period polynomials arealso provided in full details. Moreover, recent developments about the zeroes ofthose polynomials are mentioned.

iii

Contents

Preface iii0.1. Motivation 1

Chapter 1. Discreteness And Discontinuity Of Subgroups 31.1. Notations And Some Simple Results 31.2. Properties Of Linear Fractional Transformations 61.3. Discrete And Discontinuous Groups 101.4. Groups Fixing The Unit Disc 19

Chapter 2. The Modular Group And The Congruence Subgroups 252.1. Introduction 252.2. Fundamental Regions For The Full Modular Group 292.3. Fundamental Domains For Subgroups Of Γ(1). 332.4. Cuspidal Points 37

Chapter 3. Modular Forms 433.1. Introduction – A Toy Example 433.2. Introduction – Modular forms 443.3. Multiplier Systems 463.4. Modular Forms 493.5. Petersson scalar product and the Hilbert space of cusp forms 62

Chapter 4. Existence of Modular Forms 674.1. Eisenstein Series 674.2. The Discriminant Function ∆(z) 794.3. Applications To Mk,1

(Γ(1)

)And Sk,1

(Γ(1)

)81

4.4. Holomorphic Poincare series 874.5. Theta series 1024.5.1. Jacobi Theta function 1024.5.2. Theta series and sums of squares 113

Chapter 5. Hecke Operators 1275.1. The Slash Operator 1275.2. Two Principles 1285.2.1. Decent to a subgroup 1285.2.2. The subgroup method 1285.2.3. A combination of both principles 1305.3. Hecke Operators Of Index n ∈ N 1325.4. Transformations of Order n ∈ N 1355.5. The Action of Hecke Operators on Modular Forms 1375.6. Multiplicative Properties Of Hecke Operators 1395.6.1. Properties of T (n) and the proof of Theorem 5.30 1415.7. Eigenforms 1515.8. Hecke Operators, Eisenstein and Poincare Series 154

v

vi CONTENTS

5.9. Hecke Operators And The Petersson Inner Product 1595.10. L-functions Associated to Cusp Forms 1615.10.1. The Mellin transform 1615.10.2. L-series associated for cusp forms on Γ(1) 1645.10.3. L-functions of Hecke eigenforms 170

Chapter 6. Period Polynomials 1756.1. Period polynomials via integral transformations 1756.2. Eichler Integrals And Period Polynomials 1786.2.1. The (k − 1)-fold antiderivative 1816.2.2. An integral representation of the Eichler integral for cusp forms 1826.2.3. An integral representation of the Eichler integral for entire forms 1846.2.4. The Niebur integral representation for entire forms 1866.3. The Eichler Cohomology Group 1886.4. The Eichler Cohomology Theorem 1936.4.1. Injectivity of the map µ 1946.4.2. Surjectivity of the map µ 1966.5. Hecke Operator And Period Polynomials 2026.6. Roots Of Period Polynomials Associated To Hecke Eigenforms 212

Part 1. Appendix I

Appendix A. Background Material IIIA.1. Open Mapping Theorem IIIA.2. Weierstrass M -Test IIIA.3. Chinese Remainder Theorem – Special Case IIIA.4. Maximum Modulus Principle IVA.5. Phragmen-Lindelof principle IVA.6. Stirlings estimate of the Gamma-function IVA.7. Cauchy-Riemann equations IVA.8. Cauchy’s integral theorem VA.9. Green’s theorem VA.10. Rouche theorem VA.11. Beta function and Gamma function VA.12. A result on normal operators VIA.13. Unique continuation principle for holomorphic functions VIA.14. Whittaker functions VIA.15. Generalized hypergeometric functions VIIA.16. Poisson Summation Formula VIIA.17. Liouville’s theorem VIIIA.18. Residue theorem VIIIA.19. Geometric series IXA.20. Continued fractions IXA.21. Hurwitz zeta function XIA.22. Incomplete Gamma function XI

Appendix B. Solutions of Selected Problems XIIIB.1. Problems of Chapter 1 XIIIB.2. Problems of Chapter 2 XVIB.3. Problems of Chapter 3 XXB.4. Problems of Chapter 4 XXVIB.5. Problems of Chapter 5 XXXIIB.6. Problems of Chapter 6 XXXIV

CONTENTS vii

Appendix. Acronyms XLIX

Appendix. Index LI

Appendix. Bibliography LIX

0.1. MOTIVATION 1

0.1. Motivation

Let us examine the following old question in number theory for the moment:Is it possible to write any natural number into a sum of four squares of integers?A direct follow-up question would be to count the number of ways. how to write anatural number as different sums of four squares of integers. The first question wasanswered positively by Joseph Louis Lagrange in 1770. Carl Gustav Jacob Jacobiis attributed with answering the second question.

To formulate the questions more precisely, consider the counting function

r4 : N→ Z≥0; n 7→ r(n) := ]

(x1, x2, x3, x4) ∈ Z4; x21 + x2

2 + x23 + x2

4 = n.

We have the following results:

Lagrange’s four-square theorem:: r4(n) ≥ 1 for all n ∈ N.

Jacobi’s theorem:: r4(n) =

8σ1(n), 4 - n8σ1(n)− 32σ1

(n4

), 4 | n

for all n ∈ N =

Z≥1. Here σ1(n) denotes the “sum of all divisors” function.

Jacobi’s proof of his result leads already to the concept of theta-functions andmodular forms by forming the generating function of r(n). (We refer to §4.5 inChapter 4 for more details.) Such generating functions turned out to appear in theFourier expansions of certain functions that have nice transformation propertiesunder some discrete groups. These functions are known as modular forms. Thedimension of the space of modular forms with a given weight turned out to befinite dimensional and thus interesting arithmetical properties for many interestingnumber theoretical functions emerged.

In Chapter 1 we introduce and discuss the concepts of discrete and discontin-uous (sub)-groups of Mobius transformations action on C ∪ ∞. In Chapter 2 wefocus the discussion on the modular group and its congruence subgroups. We alsodiscuss the action of those groups on the complex upper half-plane. After preparingthe geometric aspects of modular group acting on the upper complex half-plane,we consider functions on the upper half-plane which are compatible to the actionof the modular group or its subgroups in Chapter 3. This leads to the concept ofmodular forms. Chapter 4 discusses some classical examples of modular forms. Wecontinue in Chapter 5 and Chapter 6 to present operators acting in the spaces onmodular forms, the Hecke operators, and associated polynomials, the period poly-nomials. Moreover, recent developments about the zeroes of those polynomials arementioned.

CHAPTER 1

Discreteness And Discontinuity Of Subgroups

In this chapter, we introduce discrete groups and discontinuous groups. Weshow that discrete and discontinuous groups are the same when dealing with reallinear fractional transformations. After classifying real linear transformations intothree: elliptic, parabolic, and hyperbolic, we investigate how the stabilizer groupsassociated to these transformations are generated.In the last section, we discuss discrete groups that fix the unit disc and prove resultsrelated to the limit set of these groups.

1.1. Notations And Some Simple Results

We denote the set of natural numbers and the set of integers by

N = 1, 2, 3, . . . and Z = . . . ,−2,−1, 0, 1, 2, . . ..

The set of rational numbers, real numbers, and complex numbers are denoted by Q,R, and C respectively. Let z = x+ iy ∈ C denote a complex number with x, y ∈ R.Its real and imaginary parts are denoted by < (z) = x and = (z) = y, respectively.

Let

(1.1) H =z ∈ C; = (z) > 0

and H− =

z ∈ C; = (z) < 0

,

denote the upper half plane and the lower half plane, respectively.In addition we define the group of linear fractional transformations or general

linear group, denoted by GL2 (C), to be the set of all 2× 2 invertible matrices withcomplex entries:

(1.2) GL2 (C) :=

(a bc d

); a, b, c, d ∈ C and ad− bc 6= 0

.

Its subgroup, GL2 (R), is given by

(1.3) GL2 (R) :=

(a bc d

); a, b, c, d ∈ R and ad− bc 6= 0

⊂ GL2 (C) .

A subgroup of GL2 (R), known as the special linear group SL2 (R), is given by:

(1.4) SL2 (R) :=

(a bc d

); a, b, c, d ∈ R and ad− bc = 1

.

The next subgroup, a special subgroup with nice properties, is SL2 (Z), the set ofall 2× 2 matrices in SL2 (R) with integer entries:

(1.5) SL2 (Z) :=

(a bc d

)∈ SL2 (R) ; a, b, c, d ∈ Z

.

We call this group the full modular group. We also use the notation Γ(1) = SL2 (Z)later on, see e.g. §2. Two important elements of each of the above matrix groupsare the matrices:

(1.6) 1 :=

(1 00 1

)and − 1 :=

(−1 00 −1

),

3

4 1. DISCRETENESS AND DISCONTINUITY OF SUBGROUPS

where 1 is of course called the identity matrix. We also need sometimes the set of2× 2 matrices with integer respectively real entries:

Mat2 (Z) : =

(a bc d

); a, b, c, d ∈ Z

and

Mat2 (R) : =

(a bc d

); a, b, c, d ∈ R

.

(1.7)

Definition 1.1. Let V =

(a bc d

)be any matrix (with complex entries). We

define the determinant of V :

detV = ad− bc.

Throughout the book, in particular in the first part, we often consider theRiemann sphere

(1.8) S := C ∪ ∞.Basically S forms the space on which the above groups will act. To do so, we haveto introduce the group action.

Definition 1.2. For z ∈ C ∪ ∞, we define the action of the group GL2 (C)on S = C ∪ ∞ by

(1.9)

(a bc d

)z :=

az+bcz+d if z ∈ C, z 6= −dcac if z =∞ and

∞ if z = −dc .

We call this group action the linear fractional transformation or Mobius transfor-mation.

Lemma 1.3. The Mobius transformation is a well defined group action. Wehave

(1.10)(MV

)z = M

(V z)

for all z ∈ S and M,V ∈ GL2 (C).

Proof. Consider invertible matrices

(a bc d

)and

(α βγ δ

)with complex en-

tries and z ∈ S = C ∪ ∞. We have[(a bc d

)(α βγ δ

))z =

(aα+ bγ aβ + bδcα+ dγ cβ + dδ

)z =

(aα+ bγ)z + aβ + bδ

(cα+ dγ)z + cβ + dδ

=

aαz+aβ+bγz+bδγz+δ

cαz+cβ+dγz+dδγz+δ

=aαz+βγz+δ + b

cαz+βγz+δ + d=

(a bc d

)αz + β

γz + δ

=

(a bc d

)[(α βγ δ

)z

].

The next lemma answers the question about the kernel of the Mobius transfor-mations, that is the set of matrices which act trivially on S.

Lemma 1.4. The kernel of the group GL2 (C) action on S by Mobius transfor-mation is given by (

λ 00 λ

); λ ∈ C 6=0

.

Proof. The proof is left for the reader as an exercise.

1.1. NOTATIONS AND SOME SIMPLE RESULTS 5

The statement of the lemma means that each element of the kernel set acts asan identity operation:(

λ 00 λ

)z = z for all z ∈ S and λ ∈ C 6=0.

Multiplying this equation with any matrix

(a bc d

)∈ GL2 (C) gives(

λa λbλc λd

)z =

(a bc d

)z

This allows us to identify the matrix element

(a bc d

)∈ SL2 (C) with the factional

linear transformation az+bcz+d up to multiplication of a unit in the entries. In particu-

lar, the kernel of the SL2 (R) action on S by Mobius transformation is given by thematrices

1,−1

.

Now, we consider the group SL2 (R) and its action on S. It is easy to see thatSL2 (R) preserves the upper half plane H, the compactified real line R ∪ ∞, andthe lower half plane H−. This is due to the fact that

= (V z) = =(az + b

cz + d

)= =

((az + b)

(cz + d

)|cz + d|2

)

==((az + b)

(cz + d

))|cz + d|2

=ad= (z) + bc= (z)

|cz + d|2

=(ad− bc

) = (z)

|cz + d|2= detV

= (z)

|cz + d|2

(1.11)

for all V =

(a bc d

)∈ GL2 (R) using detV = ad − bc. Summarizing, we get the

following result.

Lemma 1.5. We have

SL2 (R) H ⊂ H, SL2 (R) H− ⊂ H− and SL2 (R)(R ∪ ∞

)⊂ R ∪ ∞.

Proof. The lemma follows from the above discussion.

Exercise 1.6. Let V =

(a bc d

)∈ SL2 (R). Similar to the identity = (V z) =

=(z)

|cz+d|2 in (1.11) show

(1.12) d(V z) =dz

(cz + d)2.

We have also the following existence result.

Lemma 1.7. Let z1, z2 ∈ S = C ∪ ∞ be two distinct numbers. There alwaysexists a matrix A ∈ GL2 (C) such that

Az1 =∞ and Az2 = 0

holds.If z1, z2 ∈ R ∪ ∞ holds then we may choose A ∈ SL2 (R).

Proof. We can write z1 = ac and z2 = b

d as fractions, using∞ = 10 if necessary,

with a, b, c, d ∈ C. Put

(1.13) B :=

(a bc d

).


Obviously,

B∞ =a

c= z1 and B 0 =

b

d= z2

holds.Our goal is to invert B and define A = B−1. Hence, we have to show detB 6= 0.

Since z1 and z2 are distinct we have

0 6= z1 − z2 =ad− bccd

.

This shows detB = ad− bc 6= 0 if and only if cd 6= 0 holds.If c = 0 we know that z1 = ∞ was chosen. This implies also that z2, which is

distinct from z1, is a complex number and hence can be written as bd with d 6= 0.

Hence B =

(1 b0 d

)and by construction detB = d 6= 0.

The case d = 0 can be treated analogously. (Note that c = d = 0 can nothappen since this would imply z1 = z2 =∞).

Now assume that z1, z2 ∈ R ∪ ∞ with z1 6= z2. Since the points are eitherreal (and written as quotients of real numbers) or ∞ = 1

0 , we see from the aboveconstruction that B in (1.13) has only real entries. In other words, we have B ∈GL2 (R). Put φ := detB and define

A :=

( aφ bcφ d

)−1

=

(d −b− cφ

aφ

).

By construction we have

detA = da

φ− (−b)

(− cφ

)=ad− bcφ

= 1.

Hence A ∈ SL2 (R).

Throughout the chapter and for the sake of abbreviation, we will denote se-quences of matrices

(Vn)n∈N by Vn.

1.2. Properties Of Linear Fractional Transformations

In the following definition, we divide linear fractional transformations into threekinds in terms of their trace. This classification will actually lead to another de-scription that can be used as a definition as well in terms of the number of fixedpoints. Some other properties of the linear fractional transformations are also given.

Definition 1.8. Let V =

(a bc d

)be any matrix (with complex entries). We

call the sum of the diagonal elements the trace of V :

trace (V ) = a+ d.

Definition 1.9. Let V ∈ SL2 (R) which acts non-trivially. This means thatV /∈

1,−1

. We say:

(1) V is elliptic if |trace (V )| < 2.(2) V is hyperbolic if |trace (V )| > 2.(3) V is parabolic if |trace (V )| = 2.

Definition 1.10. Let V ∈ GL2 (C) be a matrix. A point z ∈ C∪∞ is calleda fixed point of V , if V z = z holds.

For nontrivial elements of SL2 (R) the notion of a fixed point is related to theclassification above. We have the following

1.2. PROPERTIES OF LINEAR FRACTIONAL TRANSFORMATIONS 7

Proposition 1.11. Let V ∈ SL2 (R), V /∈1,−1

be a matrix. We have the

following equivalences:

(1) V is elliptic ⇐⇒ V has two distinct fixed points in H ∪ H− which areconjugates of each other.

(2) V is hyperbolic ⇐⇒ V has two distinct fixed points in R ∪ ∞.(3) V is parabolic ⇐⇒ V has one fixed point in R ∪ ∞.

Proof. We have to show that the fixed point conditions stated in the proposi-tion correspond to |trace (V )| < 2, |trace (V )| > 2, and |trace (V )| = 2 respectively.

First, we discuss, what it means that z is a fixed point of V . Let z be a fixed

point of V =

(a bc d

)∈ SL2 (R). We have

V z = z ⇐⇒ az + b

cz + d= z.

This is equivalent to z solving the quadratic equation

(1.14) cz2 + (d− a)z − b = 0.

We assume c 6= 0 for the moment. Solving the quadratic equation (1.14) withthe usual quadratic formula gives us the solutions

z1 =(a− d)−

√(d− a)2 + 4bc

2cand z2 =

(a− d) +√

(d− a)2 + 4bc

2c.

Using the fact that ad− bc = 1 allows us to rewrite the term (d− a)2 + 4bc underthe square-root as

(d− a)2 + 4bc = (a+ d)2 − 4ad+ 4bc = (a+ d)2 − 4.

This gives us the solutions

z1 =(a− d)−

√(a+ d)2 − 4

2cand z2 =

(a− d) +√

(a+ d)2 − 4

2c.

Depending on whether the term under the square-root is strictly positive, zero ornegative, we have the three cases

trace (V )2

= (a+ d)2

> 4,

= 4 and

< 4.

These correspond to the three cases V is hyperbolic, parabolic, and elliptic byDefinition 1.9.

Now, consider the case c = 0 in (1.14). The matrix V has the form V =(a b0 d

)∈ SL2 (R) and its fractional linear transformation is V z = az+b

d . This

implies directly that ∞ is a fixed point of V , V∞ =∞. Also, if ∞ is a fixed point

of V then V has to be of the form V =

(a b0 d

).

As a consequence, any other fixed point of V needs to be of the form bd−a if

this fraction is defined, i.e., if d 6= a. This leads to the following two cases:

a 6= d: Now, V has two distinct fixed points ∞ and bd−a ∈ R. Using the

determinant condition ad = detV = 1 we get the relation d = 1a . Then

a 6= 1a implies a2 6= 1. We may assume a > 0. Then we have

(a− 1)2 > 0

since a 6= 1. This implies

0 < (a− 1)2 = a2 + 1− 2a


⇐⇒|trace (V )| = a+1

a=a2 + 1

a>

2a

a= 2.

Hence, V is hyperbolic. A similar calculation holds for the case a < 0.Conversely, if V is hyperbolic, then 2 < |trace (V )| =

∣∣a+ 1a

∣∣ holds.

This implies a 6= ±1 and d = 1a 6= a. Hence, V has two fixed points in

R ∪ ∞.a = d: Since detV = a2 = 1 holds, we have a = ±1. Hence V is of the form(±1 b0 ±1

). The linear fraction transform V z = z ± b only admits one

fixed point ∞. Hence, V is parabolic which corresponds to the fact that|trace (V )| = 2.

Definition 1.12. For obvious reasons, we call a matrix V of the form V =(±1 b0 ±1

)with b 6= 0 a translation.

Proposition 1.13. Let V ∈ SL2 (R), V 6= ±1. A parabolic transformation Vthat fixes ∞ is a translation and conversely.


Proposition 1.14. Let V,A ∈ GL2 (C) be linear fractional transformations,then

trace (V ) = trace(AV A−1

).


Proposition 1.15. A transformation V ∈ GL2 (C) that fixes both 0 and ∞ isof the form

V z = κ z

for some κ 6= 0.

Proof. Write V =

(a bc d

). Since V fixes 0, we have

0 = V 0 =b

d.

This implies b = 0.Since V fixes ∞, we have c = 0 by a similar calculation. Thus,

V z =a

dz

holds for all z ∈ C ∪ ∞.Put κ = a

d to complete this direction of the proof. (κ does not vanish sincead = detV 6= 0 holds).

The converse direction follows easily by choosing non-vanishing a and d in Z,R, or C respectively (depending on weather V is supposed to be in SL2 (Z), SL2 (R)or GL2 (C) )such that κ = a

d holds.

Exercise 1.16. Let V ∈ SL2 (R) satisfy: There are two distinct points α, β ∈ Ssatisfying

V α = β and V β = α.

Then V is elliptic.

In the following proposition, we show that a necessary and sufficient conditionfor real fractional transformations to commute is to fix the same set of points.

1.2. PROPERTIES OF LINEAR FRACTIONAL TRANSFORMATIONS 9

Proposition 1.17. Suppose V1 and V2 are non-trivial real linear fractionaltransformations (i.e., V1, V2 ∈ SL2 (R) and V1, V2 /∈

1,−1

), then V1V2 = V2V1 if

and only if V1 and V2 have the same set of fixed points.

Remark 1.18. We can denote the commuting property in Proposition 1.17above also as a vanishing commutator relation:

V1V2 = V2V1 ⇐⇒ [V1, V2] = 0.

As usual, the commutator brackets are defined as

(1.15) [V1, V2] := V1V2 − V2V1.

Proof of Proposition 1.17. Assume that V1 and V2 have the same set offixed points. Proposition 1.11 shows that both transformations V1 and V2 areboth either parabolic, hyperbolic, or elliptic. We consider the following two casesseparately.

(1) Both transformations V1 and V2 are parabolic. Let A ∈ SL2 (R) be a realfractional transformation with

Aξ =∞,

where ξ is the common single fixed point of both V1 and V2. Note that itis always possible to find such a matrix A, see Lemma 1.7. Let

V ′1 = AV1A−1 and V ′2 = AV2A

−1.

V ′1 and V ′2 are both parabolic, since the trace is preserved by Proposi-tion 1.14, and they both fix ∞. Thus by Proposition 1.13, we get

V ′1 z = AV1A−1 z = z + λ1 and V ′2 z = AV2A

−1 z = z + λ2.

Thus, V ′1 and V ′2 are translations and obviously they commute. This doesnot change under conjugation:

V ′1 V′2 = V ′2 V

′1

⇐⇒ AV1A−1AV2A

−1 = AV2A−1AV1A

−1

⇐⇒ AV1V2A−1 = AV2V1A

−1

⇐⇒ V1 V2 = V2 V1.

(2) Now, assume that V1 and V2 are either both elliptic or both hyperbolic.Let ξ1 and ξ2 be the two common fixed points, i.e.

V1ξ1 = ξ1, V1ξ2 = ξ2 and V2ξ1 = ξ1, V2ξ2 = ξ2.

By Lemma 1.7 let A be a nonsingular transformation satisfying

Aξ1 = 0 and Aξ2 =∞.

As above put

V ′1 = AV1A−1 and V ′2 = AV2A

−1.

Notice that both V ′1 and V ′2 fix both 0 and ∞ and by Proposition 1.15,we get

V ′1z = AV1A−1z = κ1z and V ′2z = AV2A

−1z = κ2z.

As a result, V ′1 V′2 = V ′2 V

′1 , and similar to the parabolic case, we get

V1 V2 = V2 V1.

Conversely, suppose that V1 and V2 commute. We distinguish the followingcases:


(1) Assume that V1 is parabolic with a fixed point α. Since

V1 V2 = V2 V1

and V1 fixes α we have

(1.16) V1(V2α) = V2(V1α) = V2α.

Thus, V2α is a fixed point of V1 as well. This directly implies V2α = α, aswe assumed that V1 fixes α only. Thus, V2 fixes α.

We shall prove now that V2 fixes nothing else. Assume V2β = β, forsome β 6= α. Then, V2 fixes V1β by the same argument as in (1.16). SinceV2 has at most two fixed points, we have V1β = β or V1β = α. Notethat the last case is impossible since V1α = α, and V1 induces a one-to-one mapping. Thus, β has to be another fixed point of V1, which is acontradiction to our assumption.

(2) Assume that V1 has two distinct fixed points α and β. Then, V2α andV2β are also fixed points of V1 by the same argument used in (1.16). Soeither V2 fixes both α and β, or interchanges them. In the first case, weare done.

In the second case, assume V2α = β, and V2β = α. Then it is clearthat V 2

2 has two fixed points α and β. Suppose V2 fixes an additionalthird point γ, where γ /∈ α, β. Thus, V 2

2 has (at least) three fixedpoints, implying V 2

2 = 1 by Proposition 1.11. Hence V 22 has the two fixed

points α and β whereas V2 interchanges these values. Thus, V2 is ellipticby Example 1.16, and has two complex fixed points in H ∪H− say γ ∈ Hand γ ∈ H−. So, V1 fixes α and β, while V2 fixes γ and γ with γ ∈ H.Using the commutativity of V1 and V2 similar to (1.16), we have

V2(V1γ) = V1(V2γ) = V1γ.

Hence, we get V1γ is a fixed point of V2 and similarly, V1γ is also a fixedpoint of V2. Since V1γ must be either γ or γ and V1 as an element ofSL2 (R) satisfies V1H ⊂ H, see Proposition 1.11, we conclude that γ isfixed by V1. Then, γ must be either α or β, since V1 has exactly the twodistinct fixed points α and β. The same arguments for V1γ conclude thatγ, γ

=α, β

as set identity. Hence, V1 and V2 have the same set of

fixed points.

1.3. Discrete And Discontinuous Groups

Two important and related properties of subgroups of the SL2 (C) are “dis-crete” and “discontinuous”. We first introduce the discrete subgroups and givea characterization. Next, we introduce the concept of discontinuous groups. Weconclude the section by relating both properties.

Definition 1.19. A group Γ ⊂ SL2 (C) is called discrete if the setV ; V ∈ Γ

is a discrete subset of C4. This means that for any given V ∈ SL2 (C) there doesnot exist a sequence of distinct elements Vn in Γ such that Vn converges to V .

To prove discreteness of a group, one has to find a sequence of matrices in thisgroup that converge to some matrix. However, the following proposition gives anecessary and sufficient condition for discreteness that will be practically easier touse.

Proposition 1.20. A group Γ is discrete if and only if there does not exist asequence of distinct elements Vn in Γ such that Vn converges to 1 where 1 is theidentity matrix.

1.3. DISCRETE AND DISCONTINUOUS GROUPS 11

Proof. Assume Γ is discrete. By Definition 1.19 there does not exist a se-quence of distinct elements Vn ∈ Γ such that Vn converges to 1 ∈ SL2 (C).

To prove sufficiency, suppose that Γ is not discrete. Take any V ∈ SL2 (C).Then there exists a sequence of distinct elements Vn in Γ which converges to V . PutWn = Vn+1V

−1n . Γ being a group implies immediately Wn ∈ Γ. Also, since Vn+1

converges to V we have that V −1n converges to V −1 as well. Hence, Wn = Vn+1V

−1n

converges to V V −1 = 1.What remains to prove is that Wn contains infinitely many distinct elements of

Γ. Assume that it does not, then there are at most finitely many distinct elementsof Wn. Since Wn converges to 1, there exists an N ∈ N such that Wn = 1 for alln ≥ N . Going back to the sequence Vn, this means that Vn+1 = Vn for all n ≥ N .Thus, the equalities VN = VN+1 = VN+2 = . . . must hold. This contradicts the factthat Vn are distinct.

We continue to define discontinuous groups. This concept needs neighborhoodsaround points in S. In addition to the usual neighborhood around a point z0 ∈ C,we define a neighborhood of ∞ by

(1.17)

z ∈ C; |z| > 1

ε

∪ ∞.

Definition 1.21. A point α ∈ S is called a limit point of a group Γ ⊂ GL2 (C)if there exists z ∈ S and Vn a sequence of distinct elements in Γ such that Vnzconverges to α.

Example 1.22. Let S ∈ Γ(1), with S =

(1 10 1

). Then Sn =

(1 n0 1

)∈ Γ(1)

for all n ∈ Z. Since obviously Sn∞ =∞ holds, Definition 1.21 implies that ∞ is alimit point of Γ(1).

Definition 1.23. The set of limit points of Γ ⊂ GL2 (C) is denoted by Lim (Γ).The set of ordinary points of Γ is denoted by Ord (Γ) = S r Lim (Γ).

We now give the definition of discontinuous groups. Notice that discontinuityand discreteness are not the same in every case.

Definition 1.24. A group Γ is called discontinuous if Ord (Γ) 6= ∅, or equiva-lently, if Lim (Γ) 6= S.

Definition 1.25. Let ∅ 6= Σ ⊂ Ord (Γ). We say that Γ acts discontinuous onΣ.

Remark 1.26. How do limit sets of related groups relate to each other? Inparticular, what happens if we have two groups where one is the subgroup of theother? Here is a list of some of the more obvious facts:

(1) For each V ∈ Γ we know that V(

Lim (Γ))

= Lim (Γ) and V(

Ord (Γ))

=Ord (Γ) holds.

Proof of this statement. Let α ∈ Lim (Γ). There exists a se-quence Vn ∈ Γ and a point z ∈ S such that Vn z converges to α. Hence,V Vn z converges to V α. Definition 1.21 shows that V α is a limit pointof Γ. Therefore V α ∈ Lim (Γ). This shows the inclusion V

(Lim (Γ)

)⊂

Lim (Γ).Now consider V −1. The same argumentation as above shows

V −1(

Lim (Γ))⊂ Lim (Γ) .

By multiplying V from the left we get Lim (Γ) ⊂ V(

Lim (Γ)). Therefore

equality holds: Lim (Γ) = V(

Lim (Γ)).


(2) If Γ1 ⊂ Γ2, then we have Ord (Γ2) ⊂ Ord (Γ1), or equivalently, Lim (Γ1) ⊂Lim (Γ2).

Proof of this statement. Let α ∈ Lim (Γ1). There exists a se-quence Vn ∈ Γ1 and a point z ∈ S such that Vn z converges to α. We haveobviously Vn ∈ Γ1 ⊂ Γ2, and thus also Vn ∈ Γ2 for all n. This impliesdirectly α ∈ Lim (Γ2). We just showed the relation Lim (Γ1) ⊂ Lim (Γ2).Using Definition 1.23 gives us directly

Ord (Γ2) = S r Lim (Γ2) ⊂ S r Lim (Γ1) = Ord (Γ1) .

(3) If Γ1 ⊂ Γ2 with finite index [Γ2 : Γ1] <∞, then Ord (Γ1) = Ord (Γ2) andLim (Γ1) = Lim (Γ2).

Proof of this statement. We decompose Γ2 in its right coset de-composition:

Γ2 = Γ1A1 ∪ Γ1A2 ∪ ... ∪ Γ1Aµ

with µ = [Γ2 : Γ1]. Using the second statement above shows

Ord (Γ2) ⊂ Ord (Γ1) and Lim (Γ1) ⊂ Lim (Γ2) .

Take an α ∈ Lim (Γ2). There exists Vn ∈ Γ2 and z ∈ S such that Vn zconverges to α. Write Vn = DnAin with 1 ≤ in ≤ µ and Dn ∈ Γ1 inits coset decomposition. Thus, there exists an i ∈ 1, . . . , µ that occursinfinitely often. We use this Ai and consider the subsequence Vnm =DnmAi characterized by inm = i. It is clear that Dnm are distinct becausethe Vnm ’s are distinct. Also, we have that (DnmAi) z converges to α asm approaches ∞. Hence, Dnm (Aiz) converges to α as m → ∞. Henceα ∈ Lim (Γ1) by Definition 1.21. This implies Lim (Γ1) = Lim (Γ2).

The following theorem shows that the concepts of discontinuity and discretenessare the same for subgroups of SL2(R).

Theorem 1.27. Let Γ ⊂ SL2 (R). Γ is discontinuous if and only if it is discrete.

Proof. Assume Γ is not discrete, then there exists a sequence Vn ∈ Γ suchthat Vn converges to 1, see Proposition 1.20. Hence, Vnz converges to z for allz ∈ S. By Definition 1.21 this implies Lim (Γ) = S and Ord (Γ) = ∅. Therefore,Definition 1.24 implies that Γ is not discontinuous.

Conversely, assume Γ is real and discrete, this means Γ ⊂ SL2 (R)). Supposez0 = x0 + iy0 ∈ H such that z0 ∈ Lim (Γ) is a limit point of Γ. Then, there existsdisjunct Vn ∈ Γ and a z ∈ S such that Vnz converges to z0 ∈ H, see Definition 1.21.

Writing Vn =

(an bncn dn

)and using (1.11) gives

= (Vn z) == (z)

|cnz + dn|2.

Therefore,= (z)

|cnz + dn|2→ y0 > 0

holds. This shows that |cnz + dn|2 is bounded, which in turn implies that the cn’sand dn’s are both bounded. Also, we have

|Vn z|2 =

∣∣∣∣anz + bncnz + dn

∣∣∣∣2 → |z0|2 > 0.


As above, this implies that the an’s and bn’s are bounded. By Bolzano-Weierstrass,we can find a convergent subsequence of Vn which we call Vnm . With other words, wejust found a converging subsequence Vnm (with convergence in R4) which containsonly distinct matrices. This contradicts the assumption that Γ is discrete, as wesee in Definition 1.19. Thus, no point of the upper half plane can be a limit point:We have H ⊂ Ord (Γ) and hence Lim (Γ) 6= S.

We would like to point out that the proof of Theorem 1.27 also showed that limitpoints of real discrete groups do not lie in the upper half plane. The same argumentcan be used to show that limit points also do not lie in the lower half plane. Hence,the limit points of any discontinuous subgroup of SL2 (R) are contained in R∪∞.This leads to the following:

Corollary 1.28. If Γ ⊂ SL2 (R) is a group of real linear transformations,then either Lim (Γ) = S or Lim (Γ) ⊂ R ∪ ∞.

Proof. If Γ is not discontinuous, then Lim (Γ) = S. If Γ is discontinuous,then H ⊂ Ord (Γ) as we showed already in the proof of Theorem 1.27. Similarly,H− ⊂ Ord (Γ) by basically the same argument. This leads to

H ∪H− ⊂ Ord (Γ)

and, using Definition 1.23,

Lim (Γ) ⊂ S r(H ∪H−

)= R ∪ ∞.

Another direct consequence is the following

Corollary 1.29. A discrete group Γ ⊂ SL2 (R) contains no elliptic elementsof infinite order.

Proof. Let Γ be a discrete group and assume that V ∈ Γ is elliptic of infiniteorder. Let z0 ∈ H be a fixed point of V (which has to exist). Then V nz0 = z0

converges obviously to z0. Notice that V n are all distinct because if V n1 = V n2

then V n2−n1 = 1 which would imply that V is of finite order. Thus, z0 ∈ Lim (Γ).We just showed Lim (Γ) ∩ H 6= ∅. Corollary 1.28 now shows that Lim (Γ) = Shas to hold. Definition 1.24 implies that Γ is not discontinuous and hence, usingTheorem 1.27, Γ is not discrete. This is a contradiction.

We continue our discussion of discontinuous groups by introducing the stabi-lizer group. We also give a characterization of stabilizer groups at the end of thissubsection.

Definition 1.30. Let z ∈ S and Γ a group. The stabilizer subgroup of z in ΓisV ∈ Γ; V z = z

which we denote by Γz.

Remark 1.31. Let Γ ⊂ SL2 (R) and pick one A ∈ SL2 (R). It is straightforwardto check that ΓAz = AΓzA

−1 holds.

Lemma 1.32. Let Γ ⊂ SL2 (R) be a discrete group. Any abelian subgroup of Γmust be cyclic.

Proof. Let Γ′ be an abelian subgroup of Γ. This means that all elements ofΓ′ commute with each other. Proposition 1.17 implies that they all have the sameset of fixed points, except the trivial element 1. In particular, except for 1, theyare all parabolic, or all hyperbolic, or all elliptic.


(1) Assume all elements of Γ′ (except 1) are parabolic and that they fix anx0 ∈ R ∪ ∞. By Lemma 1.7 there exists an A ∈ SL2 (R) satisfyingAx0 = ∞. Using this A and Remark 1.31 we may consider the groupΓ := AΓ′A−1. It is obvious that Γ is cyclic is equivalent to Γ′ is cyclic.

Hence, we may assume without loss of generality, that x0 =∞. SinceV∞ = ∞ for all V ∈ Γ′ holds, Proposition 1.13 shows that each V is atranslation. Thus Γ′ can be written as

Γ′ ⊂(±1 a0 ±1

); a ∈ R

.

Put

Ω :=

λ;

(1 λ0 1

)∈ Γ′

.

Now, notice that Ω has a smallest positive element λ0. (If not, then thereexists a sequence of λn’s in Ω satisfying λn → 0. This is a contradictionto Γ being discrete).

We show next that all elements of Ω are generated by λ0. Take aλ ∈ Ω and write it as λ = qλ0 + r with 0 ≤ r < λ0, q ∈ Z. Obviously wehave r = λ− qλ0 ∈ Ω. Since λ0 is the smallest positive integer, we knowthat r = 0 must hold. Hence, we can write λ as λ = qλ0 for some q ∈ Z.Therefore we can deduce that Γ′ is generated by(

1 λ0

0 1

).

(2) Assume Γ′ consists of only hyperbolic elements and 1. The hyperbolicelements have all the same fixed point set

x1, x2

with x1 6= x2 and

x1, x2 ∈ R ∪ ∞. Since for any pair x1 and x2 there exists non-singularmatrix A such that

Ax1 = 0 and Ax2 =∞

we can assume without loss of generality that the set

0,∞

are thosefixed points of Γ′. (We may conjugate Γ′ with A otherwise). Let V ∈ Γ′,V 6= 1, and put

V =

(a bc d

).

V (0) = 0 implies b = 0 and V (∞) = ∞ implies c = 0. Therefore, V is ofthe form

V =

(a 00 1

a

).

It would not satisfy detV = 1 otherwise. Hence the action of V on z isgiven by

V z =az + 0

0z + 1a

= a2 z =: κz

for some 1 6= κ > 0. Therefore, V is of the form

Vκ :=

(κ

12 0

0 κ−12

).

Let

Ω :=κ; Vκ ∈ Γ′

.

Clearly, Ω is a multiplicative group. Taking the logarithm we consider theset

K :=

log κ; Vκ ∈ Γ′.


This is a additive group. Using similar arguments as in the case above wesee that K is also discrete. As in the case before, we can prove that Kand hence Ω is cyclic. This in turn proves that Γ′ is cyclic.

(3) Assume now that all elements of Γ′ are elliptic (or the identity 1). Again,using Lemma 1.7 we may assume without loss of generality that the set ofcommon fixed points of the elements in Γ′ are again

0,∞

. Let V ∈ Γ′,

V 6= 1 and write

V =

(a bc d

).

Again, V (0) = 0 and V (∞) =∞ implies that V has the form

V = Vκ :=

(κ

12 0

0 κ−12

)for some complex κ with |κ| > 0 and κ 6= 1. The square root κ

12 is defined

as

κ12 =√ρ ei

θ2

where κ = ρeiθ with ρ > 0 and −π < θ ≤ π. The trace of Vκ is obviouslyreal since Vκ is conjugated to an element in Γ′ ⊂ SL2 (R). On the otherhand, we have

trace (V ) = κ12 + κ−

12

=√ρei

θ2 +

1√ρe−i

θ2

=

(√ρ+

1√ρ

)cos θ + i

(√ρ− 1√ρ

)sin θ.

Using trace (V ) ∈ R implies then that either sin θ = 0 or ρ = 1. Hence,either κ−1/2 is real or |κ| = 1. The first case is impossible, since then V willbe a matrix with real entries, i.e. Vκ ∈ SL2 (R), and hence is hyperbolic.This contradicts the assumption of V is elliptic. Hence we have |κ| = 1.Therefore, ρ = 1 and we have

V = Vκ = Vθ :=

(eiθ2 0

0 e−iθ2

).

Consider the set θ; Vθ ∈ Γ′

which is a discrete additive subgroup of the reals. Following the argumentsfor the set Ω in the parabolic case, we know that this group is cyclic. Itfollows that Γ′ is cyclic.

Lemma 1.33. Let Γ ⊂ SL2 (R) be a subgroup and assume that M ∈ Γ is hyper-bolic. If there exists a V ∈ Γ such that V has precisely one fixed point in commonwith M , then Γ is not discrete.

Proof. Without loss of generality, let us assume that M has fixed points 0and ∞, and V has ∞ as a fixed point but not 0. Following the argumentation inthe proof above we have Mz = α2z for some α ∈ R with α2 /∈ 0, 1. Similarly, wehave V z = a2z + ab for some a, b ∈ R with ab 6= 0 and V (0) 6= 0. The matrices Mand V have the form

M =

(α 00 α−1

)and V =

(a b0 a−1

).


Put

Cn := V Mn V −1M−n for all n ∈ Z.Then,

Cn =

(1 ab(1− α2n)0 1

).

Obviously, Cn ∈ Γ and all the elements Cn, n ∈ Z are distinct. Also, we have thethe following limits:

• If |α| < 1 we have limn→+∞ Cn =

(1 ab0 1

)= C0 ∈ Γ.

• If |α| > 1 we have limn→−∞ Cn =

(1 ab0 1

)= C0 ∈ Γ.

Hence, Γ is not discrete by Definition 1.19.

The following theorem shows that the group of stabilizers is cyclic. It is im-portant to keep in mind that the group of stabilizers of a point z ∈ H is the sameas the group of stabilizers of z, the complex conjugate of z.

Theorem 1.34. Let Γ ⊂ SL2 (R) be a discrete group of linear fractional trans-formations. If z ∈ H, then Γz is a finite cyclic group. If z ∈ R ∪ ∞, then Γz isan infinite cyclic group or the identity.

Proof. Γz is discrete since it is a subgroup of a discrete group.Now, consider the case z ∈ H. Then all elements of Γz fix z and, by conjugation,

z. Hence, all elements of Γz are elliptic, see Proposition 1.11. Corollary 1.29 impliesthat all of them are of finite order. This and the discreteness of Γz implies thatΓz is finite. Moreover, Γz is an abelian subgroup of Γ as the following argumentshows:

A,B ∈ Γz =⇒ [A,B]z = ABz −BAz = z − z = 0.

Lemma 1.32 implies that Γz is cyclic.Now, consider the remaining case z ∈ R ∪ ∞. We have to show that Γz is

infinite cyclic or contains only the identity element.If z is fixed by a parabolic element, using Lemma 1.33, z can not be fixed by

any hyperbolic element or by any elliptic element. Therefore, Γz consists entirelyof parabolic elements with fixed point z. Using Proposition 1.17 and Lemma 1.32,Γz is abelian and cyclic. Hence, Γz is infinite cyclic or 1.

If z is fixed by a hyperbolic element in Γz, we follow the same argument.

Definition 1.35. Let Γ be a group. A subgroup Γ′ ⊂ Γ is normal if Γ′ isinvariant under conjugation. This means that V ∈ Γ′ and M ∈ Γ which meansthat M−1VM ∈ Γ′.

Exercise 1.36. Suppose Γ is an infinite group of linear functional transforma-tions, prove that Lim (Γ) 6= ∅.

Definition 1.37. A discrete group Γ of linear fractional transformations withlimit set Lim (Γ) = R ∪ ∞ is called a horocyclic group.

Remark 1.38. The terminology “horocyclic group” was introduced by Ranking[100] in 1954. Another name for the same object is Fuchsian group of the first kindintroduced by Poincare. Older German literature sometimes uses the name Grenz-kreisgruppe.

Example 1.39. We show later in Theorem 2.8 that SL2 (Z) is a horocyclicgroup.


Lemma 1.40. Let Γ be a group of real linear fractional transformations such thatΓ is non-abelian and contains only hyperbolic elements and 1. Moreover, assumethat the points 0 and ∞ are fixed points of at least one hyperbolic element in thegroup.

If Γ contains a sequence of distinct elements Vn → 1, then there exists a N ∈ Nsuch that

Vn =

(ρn 00 ρ−1

n

)with ρn 6= 0,±1 for all n ≥ N .

Proof. Consider a sequence of distinct elements Vn ∈ Γ satisfying

Vn =

(an bncn dn

)→ 1.

Recall that, there exists A ∈ Γ such that A is hyperbolic fixing 0 and ∞. Hence

A =

(λ 00 λ−1

)with λ > 0 and λ 6= 1.

Put

(1.18) Cn = AVnA−1 V −1

n

for all n ∈ N. Multiplying the matrices, we get

Cn =

(1− bncn(λ2 − 1) anbn(λ2 − 1)cndn(λ−2 − 1) 1− bncn(λ−2 − 1)

).

Next, put Dn = ACnA−1C−1

n for all n ∈ N. Again, after multiplying the matrices,we get

Dn =

(1+anbncndn(λ2−1)(λ−2−1)(1−λ2) ?

? 1+anbncndn(λ2−1)(λ−2−1)(1−λ2)

).

Taking the trace of Cn and Dn gives

trace (Cn) = 2− bncn(λ2 − 2 + λ−2

)= 2− bncn

(λ− λ−1

)2(1.19)

and

trace (Dn) = 2 + anbncndn(λ2 − 1)(λ−2 − 1)(2− λ2 − λ−2)

= 2 + anbncndn(λ− λ−1)4.

Since Cn ∈ Γ, we know that either Cn = 1 or Cn is hyperbolic. Hence the tracesatisfies

|trace (Cn)| ≥ 2.

Moreover, the assumption Vn → 1 implies that Cn → 1 and in particular

trace (Cn)→ 2.

Hence, we find that

bncn → 0 as n→∞since

(λ− λ−1

)2is strictly positive. Therefore, there exists an N1 ∈ N such that

trace (Cn) ≥ 2 for n > N1.

In particular, we have

(1.20) bncn ≤ 0 for n > N1.

As in the argument above we have Dn ∈ Γ. Hence,

|trace (Dn)| ≥ 2.


A similar argumentation as done for Cn above gives that there exists an N2 ∈ Nsuch that

trace (Dn) ≥ 2 for all n > N2

holds. Therefore, we have

(1.21) anbncndn ≥ 0 for n > N2.

Assume that bncn < 0 holds for all n > N (which sharpens (1.20)). Combiningthe assumption with (1.21) implies

andn ≤ 0 for all n > maxN1, N2.Hence, an and dn have different signs. But this contradicts Vn → 1 since this limitimplies necessarily

an → 1 and dn → 1 as n→∞.Hence, there exists an N3 ∈ N such that

andn > 0 for all n > N3.

Since our assumption bncn < 0 is wrong we have bncn = 0. Plugging this into(1.19) gives

trace (Cn) = 2 for n > maxN1, N2, N3,i.e.,

Cn = 1 for n > maxN1, N2, N3.Therefore A and Vn, n > maxN1, N2, N3, commute, as can be seen from (1.18).Thus, A and Vn have the same fixed point sets, see Proposition 1.17. Hence, Vnhas fixed points 0 and ∞. Then

bn = cn = 0 for all n > maxN1, N2, N3.Therefore, Vn has the form

Vn =

(an 00 dn

).

Using the determinant condition andn = det(Vn) = 1 gives dn = a−1n . Hence Vn

has the form

Vn =

(an 00 a−1

n

)for all n > maxN1, N2, N3.

Lemma 1.41. Suppose Γ ⊂ SL2 (R) is non-abelian and contains only hyperbolicelements and 1. Then Γ is discrete.

Proof. Assume Γ is not discrete.

Case 1: Γ contains an element A fixing the two points 0 and∞. Lemma 1.40shows that there exists a sequence Vn ∈ Γ with Vn → 1 and a N ∈ N suchthat Vn is given by

Vn =

(ρn 00 ρ−1

n

)with ρn /∈ 0,±1 for n > N.

Hence, the common fixed points of the hyperbolic matrices Vn, n ≥ N ,are 0 and ∞.

Γ is non abelian. There exists therefore an element 1 6= B ∈ Γ suchthat the fixed points of B are not 0 and ∞, see Proposition 1.17. Say, B

has the matrix entries B =

(α βγ δ

). The conditions on Γ imply that B

is hyperbolic, i.e., |trace (B)| > 2. Hence, we have

(1.22) |trace (B)| = |α+ δ| > 2.

1.4. GROUPS FIXING THE UNIT DISC 19

Also, we deduce that the off diagonal terms satisfy

(1.23) β2 + γ2 > 0,

since B does not fix the elements of 0,∞ (and hence must have at leastone non-vanishing off-diagonal element).

Write

Xn = VnBV−1n B−1 for all n ≥ N.

Multiplying all matrices gives

Xn =

(? (ρ2

n − 1)αβ(ρ−2n − 1)γδ ?

).

Since Vn → 1 by assumption, we have obviously also Xn → 1. ByLemma 1.40 we find that Xn must be of the form

Xn =

(? 00 ?

)for all n ≥ N1 for some N1 ∈ N.

Recall that ρn /∈ 0,±1. Hence ρ−2n 6= 1. Since the off-diagonal

terms of Xn vanish for n ≥ N1, we conclude αβ = γδ = 0.Suppose α = 0. Then γ 6= 0 and δ 6= 0 because of the determinant

condition detB = 1. This gives α + δ = 0, which contradicts equation1.22.

Now, suppose α 6= 0. Then β = 0 must vanish. This leads to δ 6= 0and γ = 0. Hence, we just showed β2 +γ2 = 0 which contradicts equation1.23.

This shows that our starting assumption “Γ is not discrete” is wrong.Hence, Γ is discrete.

Case 2: Γ contains no non-trivial element fixing the two points 0 and ∞.Let 1 6= B ∈ Γ. Since B is hyperbolic, it has exactly two distinct fixedpoints z1, z2 ∈ R∪ ∞. By Lemma 1.7 there exists an C ∈ SL2 (R) suchthat Cz1 =∞ and Cz2 = 0 holds.

Conjugating Γ with C gives Γ′ := CΓC−1 which is non-abelian andcontains only hyperbolic elements and 1. Moreover, it contains an el-ement A := CBC−1 fixing exactly the points 0 and ∞. B hyperbolicimplies A hyperbolic, see Proposition 1.14. Since Γ is by assumption notdiscrete, there exists a sequence of distinct elements Vn satisfying Vn → 1,see Proposition 1.20. Conjugation gives CVnC

−1 → C1C−1 = 1. Thisimplies that Γ′ is not discrete, see Proposition 1.20. Now, we may applyCase 1 to Γ′ and conclude that Γ′ has to be discrete. Similar, as abovewe now show that Γ is discrete.

1.4. Groups Fixing The Unit Disc

An essential indigent of this section is the notion of limit sets Lim (Γ) of groupsΓ ⊂ GL2 (C), introduced in Definitions 1.21 and 1.23.

Denote the closed unit disc in C by

(1.24) U =z ∈ C; |z| ≤ 1

.

Also, define

(1.25) Lim∞ :=α ∈ S; ∃ sequence of distinct Vn ∈ Γ with α = lim

n→∞Vn∞

.

Note that Lim∞ ⊂ Lim (Γ).


Remark 1.42. Put

(1.26) A :=

(1 1i −i

)=⇒ z 7→ Az =

z + 1

iz − i=

1

i

z + 1

z − 1.

The matrix A maps the unit disc U to H: AU = H. In particular we have theimages A0 = i and A1 =∞.

We have the following equivalence: Γ ⊂ GL2 (C) satisfies ΓH ⊂ H if and onlyif its conjugation A−1ΓA satisfies A−1ΓAU ⊂ U.

Exercise 1.43. If V ∈ SL2 (R), then A−1V A satisfies

A−1V AU = U

and has the form

V =

(α ββ α

)for some α, β ∈ C, |α|2 − |β|2 = 1.

Lemma 1.44. Let Γ be a discrete group. We have

(1) V Lim∞ = Lim∞ for all V ∈ Γ and(2) the set Lim∞ is closed.

Proof. (1) Let α ∈ Lim∞. Then there exists distinct elements Vn suchthat Vn∞ converges to α, see (1.25). Hence, V

(Vn∞

)converges to V α.

This implies V α ∈ Lim∞.(2) Let λ be an accumulation point of Lim∞. So, there exists distinct λk ∈

Lim∞, k ∈ N, such that

|λk − λ| <1

kfor all k ∈ N.

We can also choose Vk ∈ Γ such that

|Vk∞− λk| <1

k.

Hence, the triangle inequality gives

|Vk∞− λ| ≤ |Vk∞− λk|+ |λk − λ| <2

k

for all k ∈ N. Therefore, we have shown λ ∈ Lim∞.

We already indicated in Exercise 1.42 that some elements of SL2 (C) which mapU onto itself have a special form. In fact, we have the following characterization ofelements in SL2 (C) which map U onto itself.

Proposition 1.45. Let V ∈ SL2 (C). The following statements are equivalent:

(1) V maps the unit disc U onto itself: V U = U.(2) V is of the form

(1.27) V =

(α ββ α

)for some α, β ∈ C, |α|2 − |β|2 = 1.

Exercise 1.46. Define the unit circle

(1.28) T := z ∈ S; |z| = 1.

For V ∈ SL2 (C) satisfying V U = U we have V T = T.


Proof of Proposition 1.45. We define the inversion at T on S by

(1.29) I : S→ S; z 7→ 1

z.

The inversion I is best explained in polar coordinates: If z = reiφ ∈ S, r ∈R≥0 ∪ ∞ and −π < φ ≤ π, we have

Iz = I(reiφ

)=

1

reiφ=

1

re−iφ=

1

reiφ =

z

|z|2.

Notice that I keeps the unit circle T invariant and maps the unit disc U ontoz ∈ S; |z| ≥ 1:

IT = T and IU = z ∈ S; |z| ≥ 1.We consider the two implications separately.

“(1) =⇒ (2)”:: Write V ∈ SL2 (C) as

V =

(a bc d

)with ad− bc = 1

and assume that V maps the unit disc U onto itself: V U = U. UsingExercise 1.46, V maps the unit circle T onto itself:

V T = T.

Consider IV I. We find that IV I maps T onto T since I and V map Tonto itself individually. Similar to the argument for I, we have that IV Imaps the unit disc U either onto itself or on z ∈ S; |z| ≥ 1. We have

IV Iz =

(az−1 + b

cz−1 + d

)−1

=dz + c

bz + a

for all z ∈ S. Since IV I and V coincide on the unit circle T, that meansIV Iz = V z for all z ∈ T, we find that IV I and V must be the sameMobius transformation on T. Hence,

V =

(a bc d

)!=

(εd εc

εb εa

)= IV I

for a suitable unit ε ∈ C such that ε2 = 1. Hence, ε ∈ +1,−1.If ε = −1 holds, we have(

a bc d

)!=

(−d −c−b −a

),

in particular a = −d and c = −b. Then the determinant condition implies

1 = ad− bc = −aa+ bb = −(|a|2 − |b|2

).

This case cannot happen since(a −bb −a

)0 =

b

a=

1 + a

a> 1

which implies (a −bb −a

)U = z ∈ S; |z| ≥ 1.

If ε = 1 holds, we have(a bc d

)!=

(d c

b a

),


in particular a = d and c = b. Hence, V has to have the form in (1.27).The determinant condition

1 = ad− bc = aa− bb = |a|2 − |b|2

in (1.27) follows also immediately. Also, we see(a bb a

)0 =

b

a=

b

b+ 1< 1

which implies (a bb a

)U = U.

“(1) ⇐= (2)”:: Let V be of the form given in (1.27) and recall that 1z = z

holds for all z ∈ T. Then we have

|V z|2 =

∣∣∣∣αz + β

βz + α

∣∣∣∣2 =αz + β

βz + α

αz + β

βz + α

=αz + β

βz + α

αz−1 + β

βz−1 + α=αz + β

βz + α

α+ βz

β + αz

= 1

for all z ∈ T. Hence V maps U either onto itself or onto z ∈ S; |z| ≥ 1.Using the condition |α|2 − |β|2 = 1 we have

|V 0|2 =β

α

β

α=|β|2

|α|2=

|β|2

1 + |β|2< 1.

This implies that V 0 ∈ U.Together, the two arguments show that V maps U onto U.

Lemma 1.47. Let Γ ⊂ SL2 (C) be a group and consider the sets

A =

x;

(x ?? ?

)∈ Γ

and

B =

y;

(? y? ?

)∈ Γ

.

If Γ is a discrete group satisfying ΓU ⊂ U then the sets A and B are discretein the complex plane C.

Proof. To prove discreteness of B we suppose that b is a finite accumula-tion point of B. That means, that there exists a sequence of distinct elementsbn ∈ B such that bn converges to b. Thus there exists distinct elements Vn ∈ Γ.Proposition 1.45 shows that these Vn have the form

Vn =

(an bnbn an

)with |a|2 − |b|2 = 1.

From the condition bn → b it follows

|an|2 → 1 + |b|2 .

As a result, an is a bounded sequence. So there exists a converging subsequence

anm with limit a. So, the subsequence Vnm of Vn converges to

(a bb a

). We just

constructed a contradiction to the discreteness of Γ. Hence, B is discrete.The discreteness of A follows similarly.


Theorem 1.48. Suppose that Γ ⊂ SL2 (C) is a discrete group fixing U. Thenthe set of limit points satisfies

Lim∞ = Lim (Γ) .

Proof. We know that Lim∞ ⊂ Lim (Γ), see below (1.25). It is sufficient toprove the inclusion Lim∞ ⊃ Lim (Γ).

Let λ ∈ Lim (Γ). There exists a point z ∈ S and a sequence of distinct Vn ∈ Γwith λ = limn→∞ Vnz. Proposition 1.45 guarantees that all Vn can be written as

Vn =

(an bnbn an

)∈ Γ.

Assume without loss of generality, that all bn 6= 0. Consider |bnz + an|. Herewe have two cases:

(1) There exists an N ∈ N with |bnz + an| ≥ 1 for all n > N .(2) |bmz + am| < 1 for some subsequence (nm)m.

In the second case, we have ∣∣∣∣z +apbp

∣∣∣∣ < 1

|bp|.

So,

(1.30)∣∣z − V −1

p ∞∣∣ < 1

|bp|.

Using Lemma 1.47 we find

|bp| → ∞ as p→∞.

Thus using (1.30), we get

V −1p ∞→ z as p→∞.

Then, z ∈ Lim∞. Since

λ = limn→∞

Vnz,

Lemma 1.44 implies that Lim∞ is closed. Hence, λ ∈ Lim∞.In the first case,

|Vnz − Vn∞| =∣∣∣∣anz + bnbnz + an

− anbn

∣∣∣∣ =

∣∣∣∣ 1

bn(bnz + an)

∣∣∣∣ ≤ 1

|bn|for all n > N . Using Lemma 1.47, the right hand side converges to 0 as n → ∞.Now

λ− Vn∞ =(λ− Vnz

)+(Vnz − Vn∞

)→ 0.

Thus, λ ∈ Lim∞.Summarizing we have Lim (Γ) = Lim∞.

Corollary 1.49. Let Γ be a discrete group of SL2 (C) that keeps either U orH invariant. Then Lim (Γ) is closed and Ord (Γ) is open.

Proof. Let Γ keep U invariant. Then Lim (Γ) is closed and Ord (Γ), as its setcomplement, is open.

Now, let Γ keep H invariant. Remark 1.42 shows that AΓA−1 keeps U invariant(with A given in (1.26)). Moreover,

Lim(AΓA−1

)= ALim (Γ) :=

Aλ; λ ∈ Lim (Γ)

and

Ord(AΓA−1

)= AOrd (Γ) :=

Az; z ∈ Ord (Γ)

.


Now, by Lemma 1.44 and by Theorem 1.48 we have Lim(AΓA−1

)is a closed set.

Thus, we getLim (Γ) = A−1 Lim

(AΓA−1

).

Thus, Lim (Γ) is closed.

Exercise 1.50. If Γ is a discrete group with ΓU = U, then Lim (Γ) ⊂ T.

CHAPTER 2

The Modular Group And The CongruenceSubgroups

In this chapter, we introduce the full modular group and its congruence sub-group. We also define the fundamental region of the full group and show how todetermine the fundamental regions of congruence subgroups and their cusps fromcoset copies of the fundamental region of the full modular groups.

2.1. Introduction

The previous chapter discussed and classified properties of (elements of) sub-groups Γ ⊂ GL2 (C). In particular, it identified three different classes of elementsbased on the number and position of their fixed points and introduced and dis-cussed discontinuous groups. One of the prime examples of discontinuous groupsare the full modular group and its subgroups of finite index.

In this chapter, we use the full modular group Γ(1) and its discontinuous actionon H to determine an associated (standard) fundamental region. We also determinethe fundamental region of some special subgroups of finite index and the set ofinequivalent cusps corresponding to those domains. It is important to mention thatthe theory of modular forms discussed in this text deals with congruence subgroupsof finite index in the full modular group. However, properties of modular forms onnon-congruence groups has been a recent area of interest, see e.g. [109] and [119].

Recall that the full modular group is given by

(2.1) Γ(1) = SL2 (Z)(1.5)=

(a bc d

); ad− bc = 1, a, b, c, d ∈ Z

,

see (1.5) on page 3.An interesting normal subgroup of SL2 (Z) is known as the principle subgroup

of level N . It is given by

(2.2) Γ(N) :=

(a bc d

)∈ SL2 (Z) ;

(a bc d

)≡(

1 00 1

)(mod N)

.

Notice that since Γ(1)/Γ(N) ∼= SL2 (Z/NZ), Γ(N) is of finite index in Γ(1). Werefer to [79, Theorem 4.2.5] for an explicit calculation of the index.

Definition 2.1. A congruence subgroup is a subgroup of Γ(1) that containsΓ(N) for some N ∈ N.

Two very useful congruence subgroups of finite index in Γ(1) are

Γ0(N) : =

(a bc d

)∈ SL2 (Z) ; c ≡ 0 (mod N)

(2.3)

=

V ∈ SL2 (Z) ; V ≡

(? ?0 ?

)(mod N)

and

Γ1(N) : =

(a bc d

)∈ SL2 (Z) ; c ≡ 0 and a ≡ d ≡ 1 (mod N)

(2.4)

25

26 2. THE MODULAR GROUP AND THE CONGRUENCE SUBGROUPS

=

V ∈ SL2 (Z) ; V ≡

(1 ?0 1

)(mod N)

.

The first one is sometimes called the Hecke congruence subgroup of level N .The companion groups to Γ0(N) and Γ1(N) are the groups

Γ0(N) : =

(a bc d

)∈ SL2 (Z) ; b ≡ 0 (mod N)

(2.5)

=

V ∈ SL2 (Z) ; V ≡

(? 0? ?

)(mod N)

and

Γ1(N) : =

(a bc d

)∈ SL2 (Z) ; b ≡ 0 and a ≡ d ≡ 1 (mod N)

(2.6)

=

V ∈ SL2 (Z) ; V ≡

(1 0? 1

)(mod N)

.

Notice that we have the inclusions

Γ(N) ⊂ Γ1(N) ⊂ Γ0(N) ⊂ Γ(1) and

Γ(N) ⊂ Γ1(N) ⊂ Γ0(N) ⊂ Γ(1)(2.7)

which follow directly from the definitions.Consider the two matrices

(2.8) S =

(1 10 1

)and T =

(0 −11 0

).

Obviously, they are elements of the full modular group Γ(1). They also generatethe group, as the next theorem will show. This will allow us to write any elementof Γ(1) as a word in S and T .

Remark 2.2. We would like to emphasize that using the letters S and T for thematrices in (2.8) is quite common in the mathematical literature, (even though Q

rarely replaces S). However, there is no consent whether S indeed denotes

(1 10 1

)and T the matrix

(0 −11 0

)or if the letters are used the other way around.

Definition 2.3. Let (c, d) ∈ Z26=(0,0) be a pair of integers, which do not vanish

simultaneously. The greatest common divisor gcd(c, d) of the pair is defined as

gcd(c, d) = maxn ∈ N; n | c and n | d

.

Here, the notation n | c means that n divides c, i.e., there exists an j ∈ Z such thatnj = c.

Note that the above definition implies gcd(0, d) = |d| for all d ∈ Z6=0.

Theorem 2.4. Every A ∈ Γ(1) can be expressed in the form

A = ±Sn1 T Sn2 T . . . T Snk ,

where the ni’s are integers with n1, nk ∈ Z and ni ∈ Z6=0 for all i ∈ 2, . . . , k − 1.Notice that this representation is not unique.

Proof. First notice that only the first power of T appears in the expressionsince T 2 = −1. This allows us to reduce any power of T to either 0 or 1 bymultiplying suitably the scalar sign −1.

Consider

A =

(a bc d

)∈ Γ(1).

We will prove the theorem by induction on c.

2.1. INTRODUCTION 27

If c = 0, then the determinant condition ad − bc = 1 implies ad = 1 and soa = d = ±1. Hence A has the form

A = ±(

1 b0 1

)= ±Sb.

Now assume c = 1. Again the determinant condition ad − bc = 1 impliesb = ad− 1. Hence,

A =

(a ad− 11 d

)=

(1 a0 1

)(0 −11 0

)(1 d0 1

)= Sa T Sd.

This proves the theorem for c = 0 and c = 1.Next, assume that the theorem holds for all A with lower left entry c ≥ 1. The

determinant ad− bc = 1 implies that the lower row entries are coprime: gcd(c, d) =1. Dividing d by c, we get

d = qc+ r for some 0 < r < c and q ∈ Z6=0.

Note that the remainder r = 0 cannot vanish since c and d are coprime. Thisimplies

AS−q =

(a bc d

)(1 −q0 1

)=

(a −aq + bc r

).

Multiplying T we get

AS−qT =

(−aq + b −a

r −c

).

Since 0 < r < c, our induction hypothesis applies on the last matrix. So, the lastmatrix is a product of powers of T and S. Hence, AS−qT can be written as

AS−q T = Sn1 T Sn2 T . . . T Snk .

Multiplying with the inverses of T and S−1, we get

A = Sn1 T Sn2 T . . . T Snk T Sq

which is of the required form.The remaining case c < 0 easily follows since we may multiply the matrix with

the scalar −1. This shows

−1A = −1

(a bc d

)=

(−a −b−c −d

).

We may now apply the induction above. (Note that T 2 = −1 · 1 = −1).To show that this expression is not unique, we consider the induction process

above in an example. Consider the matrix

A =

(11 69 5

).

Multiplying with Sn gives

ASn =

(11 11n+ 69 9n+ 5

).

Choose n ∈ Z such that |9n+ 5| < |9| holds. For example, for n = −1 we get

AS−1 =

(11 −59 −4

).


Consider the product

AS−1 T Sn =

(−5 −5n− 11−4 −4n− 9

).

Again, choose n ∈ Z such that |−4n− 9| < |−4| holds. For example, we may choosen = −2. Continuing in this manner, we find that

AS−1 T S−2 T S4T = S.

Rewritten, we get

A = S T S−4 T S2 T S.

Notice that at every stage, we were choosing n to obtain “|d| < |c|”, i.e., such thatthe lower left entry of our matrix bounds the lower right entry in absolute values.The choice of the n may not be unique; there may be more than one such n. Thisis the reason why the process is not unique.

Example 2.5. The generators S and T of the full modular group satisfy theidentities

(2.9) T 2 = −1 = (TS)3.

Indeed, we have

T 2 =

(0 −11 0

)2

=

(−1 00 −1

)= −1

and

(TS)3 =

((0 −11 0

)(1 10 1

))3

=

((0 −11 1

))3

=

(−1 00 −1

)= −1.

Corollary 2.6. Every A ∈ Γ(1) can be expressed in the form

A = T ε1 Sn1 T Sn2 T . . . T Snk

where ε1 and ni’s are integers with n1, nk ∈ Z and ni ∈ Z6=0 for all i ∈ 2, . . . , k−1and ε1 ∈ 0, 2.

Proof. Theorem 2.4 implies that each A ∈ Γ(1) has a representation

A = ±Sn1 T Sn2 T . . . T Snk

where the ni’s are integers with n1, nk ∈ Z and ni ∈ Z6=0 for all i ∈ 2, . . . , k − 1.We only need to encode the leading scalar multiple ±1.

Recall that scalar multiplication can be encoded as multiplication by a diagonalmatrix:

±1M =

(±1 00 ±1

)M for any matrix M ∈ GL2 (C) .

Moreover, the matrix T in (2.8) satisfies

T 0 = 1 and T 2 = −1 =

(−1 00 1

).

Combining these arguments, we see that each A ∈ Γ(1) can be written as a wordof the form

A = T ε1 Sn1 T Sn2 T . . . T Snk

where ε1 and ni’s are integers with n1, nk ∈ Z and ni ∈ Z6=0 for all i ∈ 2, . . . , k−1and ε1 ∈ 0, 2.

2.2. FUNDAMENTAL REGIONS FOR THE FULL MODULAR GROUP 29

Definition 2.7. For each M ∈ Γ(1) we define the Eichler length l(M) of thematrix M by

l(M) := minε1/2 + |n1|+ . . .+ |nk|+ k; M = T ε1 Sn1 T Sn2 T . . . T Snk

.

The minimum is taken over the sum of the absolute values of the exponents of eachword in S and T which expresses M .

We put

l(1) := 0.

The existence of the minimum of above set is shown in Corollary 2.6; eachelement M ∈ Γ(1) has at least one such representation as word in S and T and theelements of the set are non-negative integers.

Recall horocyclic groups in Definition 1.37.

Theorem 2.8. The full modular group Γ(1) = SL2 (Z) is horocyclic.


2.2. Fundamental Regions For The Full Modular Group

We first define the notion of fundamental regions.

Definition 2.9. Let Γ ⊂ SL2 (R) be a group of real linear functional transfor-mations which acts discontinuously. A fundamental region F = FΓ of Γ is an opensubset F = FΓ ⊂ H such that:

(1) No two distinct points of F are equivalent under Γ. This means that fortwo distinct points z1, z2 ∈ F , z1 6= z2, there is no matrix V ∈ Γ suchthat V z1 = z2 holds.

(2) Every point in H is equivalent to some point in the closure F of F underΓ. This means that for every point z ∈ H there exists an V ∈ Γ such thatV z ∈ F .

We will need the following technical result later.

Lemma 2.10. Let V ∈ Γ(1) and FΓ(1) be a fundamental region of the fullmodular group.

If there exists z0 ∈ FΓ(1) satisfying V z0 = z0, then V is trivial in the senseV ∈ 1,−1.

Proof. Consider the map

H→ H; z 7→ V z

induced by the linear fractional transformation in (1.9). Since this map is obviouslycontinuous, bijective, and has a fixed point V z0 = z0, we find that the open set

D := FΓ(1) ∩ V −1 FΓ(1) ⊃ z0

does not vanish. (FΓ(1) is open by definition and its pre-image under V is alsoopen). Hence there exists neighborhoods N1, N2 of z0 satisfying

N1 ⊂ N2 ⊂ FΓ(1)

and

V N1 ⊂ N2 ⊂ FΓ(1).

Since FΓ(1) is a fundamental region of Γ(1), it follows that V z = z for all z ∈ N1

using the first condition in Definition 2.9. However, a non-trivial linear functionaltransformation must have at most two fixed points, see Proposition 1.11. Therefore,V must be trivial: V = ±1.


It is important to bear in mind that the fundamental region is not unique asthe lemma and as the example below illustrate.

Lemma 2.11. Let F be a fundamental region for the group Γ ⊂ SL2 (R) in thesense of Definition 2.9. (This means that Γ acts discontinuously.) Then, for eachV ∈ Γ the set

V F :=V z; z ∈ F

is also a fundamental region of Γ. It satisfies

V F ∩ F = ∅ ⇐⇒ V 6= ±1.


Example 2.12. We consider the group

Γ =Sb; b ∈ Z

which is generated by the translation matrix S given in (2.8). Since the group isobviously discrete in the sense of Definition 1.19 we know by Theorem 1.27 that Γacts discontinuous on H.

Take the open set

F =z ∈ H; 0 < < (z) < 1

.

This set is a fundamental region of Γ since it satisfies the following two conditions:

(1) Assume that z1, z2 ∈ F are equivalent in Γ. Hence there exists an b ∈ Zsuch that

z2 = Sb z1 ⇐⇒ z2 = z1 + b

holds. Since z1, z2 ∈ F we have |< (z1 − z2)| < 1. This implies b = 0 andhence z1 = z2.

(2) Take an z ∈ H. Then, there exists an b ∈ Z such that 0 ≤ < (z + b) ≤1 holds. (Note that there may exist two such b’s if < (z) ∈ Z). Byconstruction we have Sb z ∈ F . Indeed, <

(Sb z

)∈ [0, 1] and the closure

F of F is obviously given by

F =z ∈ H; 0 ≤ < (z) ≤ 1

.

Another fundamental region of Γ is the diagonal strip

F1 :=z ∈ H; = (z) < < (z) < = (z) + 1

which can be proven using the same arguments above.

Exercise 2.13. Let λ > 1 be a positive number. We consider the group Γgiven by

Γ =

(λn2 0

0 λ−n2

); n ∈ Z

.

Prove the following statements:

(1) Γ is indeed a group consisting of all powers of the hyperbolic matrix(√λ 0

0 1√λ

).

(2) Γ is a discrete group and acts discontinuously on H.(3) Show that F =

z ∈ H; 1 < |z| < λ

is a fundamental region of Γ.

(4) Show that F1 =z ∈ H; 1 < = (z) < λ

is another fundamental region

of Γ.

2.2. FUNDAMENTAL REGIONS FOR THE FULL MODULAR GROUP 31

Figure 1. The left diagram illustrates the standard fundamentalregion FΓ(1) of the full modular group Γ(1), the right diagramillustrates the standard fundamental region FΓ0(2) of the Heckecongruence subgroup Γ0(2) with coset representatives 1, T , andTS.

Figure 2. The diagram illustrates the standard fundamental re-gion FΓ(1) of the full modular group Γ(1), indicating which part ofthe boundary gets mapped onto each other by S and T , respec-tively.

Definition 2.14. The Ford region is given by:(2.10)

F? =

z ∈ H; |< (z)| < 1

2, |cz + d| > 1 for all c, d ∈ Z, gcd(c, d) = 1, c 6= 0

.

In the following theorem we present the so-called standard fundamental regionF of the full modular group Γ(1) and show that this region F is the same as Ford’s


region. Illustrations of the standard fundamental region F are given in Figures 1and 2.

Theorem 2.15. (1) The open set

(2.11) FΓ(1) =

z ∈ H; |z| > 1, |< (z)| < 1

2

is a fundamental region for Γ(1).

(2) We have FΓ(1) = F?.

Proof. We show first that FΓ(1) = F?. It is clear that F? ⊂ FΓ(1), which canbe seen by taking c = 1, d = 0.

Conversely, take an z = x+ iy ∈ F and let c, d ∈ Z satisfying gcd(c, d) = 1 andc 6= 0. We get

|cz + d|2 =(c(x+ iy) + d

)(c(x+ iy) + d

)= c2(x2 + y2) + 2cdx+ d2

> c2 − |cd|+ d2 ≥(|c| − |d|

)2+ |cd| ≥ 1,

using the fact that |z| > 1 and |< (x)| < 12 . Thus, we showed F? ⊃ F . Together

we get F? = F .Next, we show the second condition of fundamental regions in Definition 2.9:

We show that every point in H is equivalent under Γ(1) to some point in the closureof F . Take any z0 ∈ H and find an m0 ∈ Z such that

|< (z0 +m0)| ≤ 1/2.

This can be done since all powers of the translation S with S z = z + 1 belong toΓ(1). Put

z′0 := z0 +m0 = Sm0 z0.

If |z′0| ≥ 1, we are done. In the case of |z0| < 1, put

z1 := − 1

z′0= T z′0 = T Sm0 z0.

Now choose m1 ∈ Z such that

|< (z1 +m1)| ≤ 1/2.

Considerz′1 := z1 +m1 = Sm1 z1.

Note that of course y1 = = (z1) = = (z′1) holds. Also, the definition of z′1 aboveimplies

(2.12) y1 = = (z′1) = = (z1)z1=Tz′0=

= (z′0)

|z′0|2 =

y0

|z′0|2

|z′0|<1

> y0.

If |z′1| ≥ 1, we are done. Otherwise continue this procedure.If the algorithm above stops in finite many steps, we are fine. But can it happen

that it does not stop? We assume, for the moment, that the algorithm does notstop in finite many steps for some initial point z0 ∈ H. We get a sequence ofΓ(1) equivalent points

(z′n)n∈N satisfying z′n = Vn z0 with Vn ∈ Γ(1) all distinct,

|< (z′n)| ≤ 12 and = (z′n) < =

(z′n+1

)for all n ∈ N. Iterating the same argument as

used in the inequality (2.12) we get

=(z′n+1

) zn+1=Tz′n== (z′n)

|z′n|2 = . . . =

= (z′0)

|z′0 · · · z′n|2 .

We know that all z′n are within the compact setz; |< (z)| ≤ 1

2 ,= (z0) ≤ |z| ≤ 1

and have increasing imaginary part. This implies that the sequence has a convergingsubsequence and the limit point of the subsequence lies in H. In other words, we just

2.3. FUNDAMENTAL DOMAINS FOR SUBGROUPS OF Γ(1). 33

showed that Lim (Γ(1))∩H 6= ∅. This is a contradiction to the result of Theorem 2.8where we show that Γ(1) is horocyclic and hence satisfies Lim (Γ(1)) = R∪∞, seeDefinition 1.37. This shows that our assumption is false and the algorithm abovestops after finite many steps.

To show that no two points in F are equivalent under Γ(1), we take an z ∈ F?and an element

(2.13) M =

(a bc d

)∈ Γ(1).

Recall that detM = 1 implies gcd(c, d) = 1.Consider the case c 6= 0. The definition of the Ford region F? in Definition 2.14

implies |cz + d| > 1. Consider the transformation

w := M z =az + b

cz + d.

Then,

|−cw + a| =∣∣∣∣−caz + b

cz + d+ a

∣∣∣∣ detM=1= |cz + d|−1

< 1.

Also, −c and a are coprime, (−c, a) = 1, since detM = ad− bc = 1. The definitionof the Ford region now implies w /∈ F?.

Now, consider the case c = 0. The determinant condition detM = 1 togetherwith a, b, d ∈ Z imply a = d = ±1. Then, w = M z is given by

w =az + b

d= z ± b.

Hence, the point w is in the Ford region only if b = 0. (We have |< (w)| > 12 unless

b = 0.) We just showed c = 0 = b and a = d = ±1. Hence M in (2.13) is M = ±1and the points z and w are equal.

Exercise 2.16. Let ∂FΓ(1) denote the set-boundary of the fundamental regionFΓ(1). Show that ∂FΓ(1) can be decomposed into 4 sets C1, . . . , C4 such that Ci∩Cjcontains at most one point and C1 ∪ · · · ∪C4 = ∂FΓ(1) holds. Moreover, C1, . . . , C4

can be chosen such that

S C1 = C2 and T C3 = C4

holds.

2.3. Fundamental Domains For Subgroups Of Γ(1).

Throughout this section, we assume Γ to be a subgroup of finite index in Γ(1),in symbols [Γ(1) : Γ] <∞. We also denote the fundamental region of a group Γ byFΓ. Also D means the closure and D means the interior of a set D.

In this section, we will present the standard fundamental region of any subgroupof finite index in Γ(1) in terms of the fundamental region of Γ(1).

Theorem 2.17. Let Γ be be a subgroup of Γ(1) with finite index µ. We denoterepresentatives of its right cosets in Γ with A1, . . . , Aµ (which means that Γ(1) =⋃µi=1 ΓAi holds). Then

(2.14) FΓ =

µ⋃i=1

Ai FΓ(1)

is a fundamental region for Γ.


Proof. First, we need to show that no two distinct points are equivalent withrespect to Γ in FΓ. Let z1, z2 ∈ FΓ be equivalent with respect to Γ. Then

z2 = M z1 with M ∈ Γ.

We have to show M ∈ 1,−1 and z2 = z1.Since Γ is a subgroup of Γ(1) there exists 1 ≤ i, j ≤ µ and u,w ∈ FΓ(1) with

z1 = Ai u and z2 = Aj w.

Hence,

Aj w = z2 = M z1 = MAi u.

This implies the identity

w = A−1j

(MAi u

)=(A−1j MAi

)u ∈ FΓ(1).

Applying Lemma 2.10 we get A−1j MAi = ±1. Thus, we have ±M = AjA

−1i which

implies

z2 = AjA−1i z1 and AjA

−1i ∈ Γ.

Since the two matrices Ai and Aj are representatives of right cosets of Γ in Γ(1)

we find that AjA−1i ∈ Γ holds only for i = j. Therefore, Ai = Aj and z1 = z2.

Now, let z ∈ H. We have to show that there exists an element V ∈ Γ such thatV z ∈ FΓ.

We know that there exists V1 ∈ Γ(1) such that

V1 z ∈ FΓ(1).

On the other hand, there exists M ∈ Γ and a representative Ai, 1 ≤ i ≤ µ satisfyingV1 = MAi. Hence, we have

V1 z =(M Ai

)z = M

(Ai z

)∈ FΓ(1).

Multiplying M−1 from the left gives

M−1V1 z = Ai z ∈ Ai FΓ(1) ⊂ Ai FΓ(1) ⊂ FΓ.

Since a fundamental region is not unique, we define what is known as standardfundamental region for subgroups of the full modular group with finite index basedon the above theorem.

Definition 2.18. Let Γ be subgroup of Γ(1) with finite index µ and right cosetrepresentatives A1, . . . Aµ. The set

FΓ =

µ⋃i=1

Ai F?

is called the standard fundamental region of Γ.

Remark 2.19. (1) The definition above implies that the standard funda-mental region of the full modular group is the Ford region F? since 1 isthe only right coset representative of Γ(1) in Γ(1).

(2) The standard fundamental region as defined above depends on the cosetrepresentation of the subgroup in the full modular group. If we haveanother set of representatives of the right cosets of Γ in Γ(1), say we havealso the representatives A′1, . . . A

′µ, then the standard fundamental region

becomes

F ′Γ =

µ⋃i=1

A′i FΓ(1).

2.3. FUNDAMENTAL DOMAINS FOR SUBGROUPS OF Γ(1). 35

We now give an example of a subgroup of finite index in Γ(1) and determine itscosets. We then use the cosets to determine the fundamental region of the subgroupusing Theorem 2.17.

Theorem 2.20. Let p be any prime. Then, for every V ∈ Γ(1), V /∈ Γ0(p),there exists P ∈ Γ0(p) and an integer k with 0 ≤ k < p satisfying

V = P T Sk.

Proof. Consider the matrix

V =

(α βγ δ

)with γ 6≡ 0 (mod p).

We need to find a matrix

P =

(a bc d

)with c ≡ 0 (mod p)

and an integer k with 0 ≤ k < p such that(α βγ δ

)= V = P T Sk =

(a bc d

)(0 −11 0

)(1 k0 1

)=

(a bc d

)(0 −11 k

).

Since all these matrices are invertible, we get(a bc d

)=

(α βγ δ

)(0 −11 k

)−1

=

(α βγ δ

)(k 1−1 0

)=

(kA−B AkC −D C

).

By the Chinese remainder theorem, see Appendix A.3, we can choose k to solve

kC ≡ D (mod p) with 0 ≤ k < p.

This is possible since C 6= 0 (mod p). Clearly, we have the identities

c = kC −D, a = kA−B, b = A, and d = C.

Now, let us check the determinant and the modulus:

detP = ad− bc = (kA−B)C −A(kC −D) = AD −BC = detV = 1,

andc ≡ 0 (mod p).

This shows P ∈ Γ0(p).

Theorem 2.21. For any prime p, the fundamental region of Γ0(p) is given bythe set

FΓ0(p) =

[p−1⋃k=0

TSk FΓ(1)

]∪ FΓ(1).

Proof. Put

D :=

[p−1⋃k=0

TSk FΓ(1)

]∪ FΓ(1).

We have to check the two conditions of Definition 2.9 to show that D is indeed afundamental region of Γ0(p).

First we show that for any z ∈ H there exists a V ∈ Γ0(p) such that V z ∈ D.Take a z ∈ H. Then, there exists a point z1 ∈ FΓ(1) and an element A ∈ Γ(1)

such that Az = z1. By Theorem 2.20 we can write

A−1 = P W

where P ∈ Γ0(p) and W is either W = 1 or W = TSk for some k with 0 ≤ k < p.Then,

P = A−1W−1


and

P−1 = W A.

Let V = P−1, then V ∈ Γ0(p) since P ∈ Γ0(p) and Γ0(p) is a group. So,

V z = P−1 z = WAz = W z1

where W is either W = 1 or W = TSk. Hence, we have shown

V z ∈W FΓ(1) ⊂ D.

Next, we have to show that no two distinct points of D are equivalent underΓ0(p).

Suppose z1, z2 ∈ D and V z1 = z2 for some V ∈ Γ0(p). We need to show thatz1 = z2. There are three cases:

(1) Assume that both z1 ∈ FΓ(1) and z2 ∈ FΓ(1). Then, V ∈ Γ0(p) ⊂ Γ(1)implies that z1 and z2 are also Γ(1) equivalent. This shows z1 = z2 sinceFΓ(1) is a fundamental region of Γ(1).

(2) Assume that z1 ∈ FΓ(1) and z2 ∈ TSk FΓ(1). The assumption on z2 canbe written as

z2 = TSk z3 for some z3 ∈ FΓ(1).

Then, V z1 = z2 implies

V z1 = TSk z3 ⇐⇒ z1 = V −1TSk z3

with both z1 and z3 in FΓ(1) and V −1TSk z3 ∈ Γ(1). This shows thatz1 = z3 since FΓ(1) is a fundamental region of Γ(1). Lemma 2.10 implies

V −1TSk = ±1. Hence, we have

±V = TSk =

(0 −11 k

).

This contradicts the fact that V ∈ Γ0(p).(3) Assume z1 ∈ TSk1 FΓ(1) and z2 ∈ TSk2 FΓ(1) for some 1 ≤ k1, k2 < p.

There exists z′1, z′2 ∈ FΓ(1) such that

z1 = TSk1 z′1 and z2 = TSk2 z′2.

Inserting this relations into V z1 = z2 we get

V TSk1 z′1 = TSk2 z′2.

Again, applying Lemma 2.10 we get

V TSk1 = TSk2

and

V = TSk2T−k1T−1 =

(1 0

k1 − k2 1

).

This matrix V has also to be an element of Γ0(p). Hence, k1 − k2 ≡ 0(mod p) must hold. Since k1 and k2 also satisfy the inequality 1 ≤ k1, k2 <p, we have k2 = k1. Summarizing, we just showed

V = TS0S = S2 = −1

which implies directly z1 = z2.

Remark 2.22. An illustration of the standard fundamental region FΓ0(2) isgiven in Figure 1. Some recent algorithms to compute fundamental regions ofsubgroups of Γ(1) of finite index can be found in [58] and [116].

2.4. CUSPIDAL POINTS 37

2.4. Cuspidal Points

We devote this section to cusps in the closure of the fundamental region ofsubgroups of finite index in the full modular group. Following the definition ofthose points we show that cusps are rational points. The cusps are used in thenext chapter where we determine the expansion of modular forms at those cuspidalpoints.

Recall that S denotes the Riemann sphere C ∪ ∞ as introduced in §1.1. We

denote the closure of a set D ⊂ S in S by DS.

Definition 2.23. Let Γ be a subgroup of SL2 (R) and FΓ a fundamental region.The elements of the set

CΓ,FΓ := FΓS ∩(R ∪ ∞

)are called the cusps of Γ in FΓ. The set is called the set of cusps of Γ in FΓ.

Remark 2.24. We point out again that the above closure of the fundamentalregion FΓ is taken in the Riemann sphere S and not in H. To illustrate the pointwe give both closures for the standard fundamental region FΓ(1) = F?:

FΓ(1) =

z ∈ H; |< (z)| ≤ 1

2, |z| ≥ 1

and

FΓ(1)S

=

z ∈ H; |< (z)| ≤ 1

2, |z| ≥ 1

∪ ∞ = FΓ(1) ∪ ∞.

The only cusp of Γ(1) in FΓ(1) is∞. The set of cusps of Γ(1) in FΓ(1) is CΓ(1),FΓ(1)=

∞.

Definition 2.25. Let Γ be a subgroup of SL2 (R).

(1) Two points x and w are equivalent under the group Γ if there exists V ∈ Γsuch that V x = w.

(2) The set

CΓ : =x ∈ R ∪ ∞; x is equivalent to a cusp of Γ in FΓ

=x ∈ R ∪ ∞; ∃V ∈ Γ and w is a cusp of Γ in FΓ satisfying x = V w

(2.15)

is called the set of cusps of Γ. Each element is called a cusp of Γ.

Exercise 2.26. Prove that any rational point is equivalent to∞ under the fullmodular group Γ(1).

Example 2.27. It is easy to see that the only cusp of Γ(1) in FΓ(1) is∞. Usingthe above exercise we easily deduce: The set of cusps of Γ(1) is

CΓ(1) =x ∈ R ∪ ∞; ∃V ∈ Γ(1) such that x = V ∞

= Q ∪ ∞.

Lemma 2.28. Let Γ ⊂ Γ(1) be a subgroup with finite index µ and representativesA1, . . . , Aµ of the right cosets. Its standard fundamental region is denoted by FΓ.

The number of cusps of Γ in FΓ satisfies

]CΓ,FΓ ≤ µ

and we have

CΓ,FΓ = A1∞, . . . , Aµ∞.


Proof. Recall that the standard fundamental region of Γ is given by

FΓ =

µ⋃i=1

Ai FΓ(1),

see Definition 2.18. From this and the definition of a cusp in FΓ, see Definition 2.23,we deduce

CΓ,FΓ= FΓ

S ∩(R ∪ ∞

)=

∞⋃i=1

AiF?S

∩(R ∪ ∞

)(Definition 2.18)

=

∞⋃i=1

AiF?S ∩(R ∪ ∞

)=

µ⋃i=1

Ai CΓ(1),FΓ(1)(Remark 2.24)

= A1∞, A2∞, . . . , Aµ∞

form a set of cusps for Γ in FΓ. We see in particular that the cardinaity of the setCΓ,FΓ

bounded by µ:

]CΓ,FΓ≤ µ

Remark 2.29. The matrices Ai, i ∈ 1, . . . , µ, in Lemma 2.28 are sometimescalled the scaling matrix of the cusp qi for the group Γ ⊂ Γ(1).

Recall that an element

(a bc d

)of SL2 (R) is called parabolic if |a+ d| = 2

holds, see Definition 1.9.

Definition 2.30. Let Γ ⊂ SL2 (R) and x ∈ R ∪ ∞.x is called a parabolic point of Γ, if there exists a parabolic matrix M ∈ Γ

satisfying M x = x.

It is easy to see in the proof of Proposition 1.11 that parabolic matrices can onlyfix rational points and thus parabolic points are all rational. The other directiontakes a bit more work.

Lemma 2.31. Let ab ∈ Q∪∞ with gcd(a, b) = 1 where we understand 1

0 :=∞.Put

(2.16) M ab

:=

(1− ab a2

−b2 1 + ab

).

We have:

(1) M ab∈ Γ(1).

(2) M ab

is parabolic and M ab

ab = a

b .

(3) There exists an A ∈ Γ(1) such that A∞ = ab . We have Mn

ab

= ASnA−1

for any n ∈ Z.(4) Suppose V ∈ Γ(1) and V fixes a

b . Then V = ±Mnab

for some n ∈ Z.

In particular, we may choose A =

(a ?b ?

)and have

Mnab

= ASnA−1 =

(1− nab na2

−nb2 1 + nab

).


Proof. The first and second part follow by direct calculation: All entries areintegers and we have

detM ab

= (1− ab)(1 + ab) + a2b2 = 1.

Hence M ab∈ Γ(1). We have

trace(M a

b

)= 1− ab+ 1 + ab = 2

which implies that M ab

is parabolic. Also ab is a fixed point of M a

b:

M ab

a

b=

(1− ab)ab + a2

−b2 ab + 1 + ab=

a−a2b+a2bb

−ab+ 1 + ab=a

b.

Now, we show the third part. Since a and b are coprime integers there existsx, y ∈ Z such that

ay − bx = 1.

(This follows from the Chinese remainder theorem, see Appendix A.3) Put

A :=

(a xb y

)∈ Γ(1)

which is obviously in the full modular group. The matrix A satisfies

A∞ =a

b.

Moreover, we have

ASnA−1 =

(a xb y

)(1 10 1

)n(a xb y

)−1

=

(a xb y

)(1 n0 1

)(y −x−b a

)=

(1− nab na2

−nb2 1 + nab

)for any positive integer n ∈ N. Using this identity for n = 1 we get immediately

M ab

= ASA−1 =⇒ Mnab

= ASA−1 · · · ASA−1︸︷︷︸n times

= ASnA−1.

Inverting the above relation, if necessary, gives

Mnab

= AStA−1

for all n ∈ Z.For the remaining fourth part, consider a matrix V ∈ Γ(1) satisfying

Va

b=a

b.

Using the properties of the matrix A above, we have

A−1V A∞ =∞ and A−1V A ∈ Γ(1).

Using (1.9), we see that A−1V A has to be of the form

A−1V A =

(u ?0 ?

).

Since detA−1V A = 1 as element of the full modular group, we know that u = ±1.Moreover, the only way to complete the entries of the matrix above is as follows:

A−1V A =

(±1 n0 ±1

)for some n ∈ Z. But this just shows

A−1V A = ±S∓n ⇐⇒ V = ±AS∓nA−1.


Proposition 2.32. Let Γ ⊂ Γ(1) of finite index µ, let q ∈ Q ∪ ∞, and letΓq be the stabilizer of q in Γ. Then, there exists M ∈ Γ, such that M is parabolicand Γq =

⟨M,−1

⟩. Also, every element in Γq r

± 1

is parabolic.We may take M = Mq defined in Lemma 2.31.

Proof. We first discuss the case

q =1

0=∞.

Consider the powers of the translation matrix S, S2, . . . , Sµ. Since Γ has finiteindex µ in Γ(1) we may apply Lemma 2.28. We find that there exists an n ∈ Z,0 ≤ n ≤ µ such that

Sn ∈ Γ and Sm /∈ Γ for all 0 ≤ m < n.

We claim that

(2.17) Γ∞ =⟨Sn,−1

⟩.

Recall

V ∈ Γ∞ =⇒ V = ±Sl for some l ∈ IZ.If we have l = 0, then V is of the form V = ±1. Obviously such a V satisfies

V ∈ Γ∞ =⟨Sn,−1

⟩.

Next, assume l < 0. Then consider V −1 = ±S−l instead of V .Now, assume 0 < l < n. Such an element V cannot exist since we assumed n

to be the smallest positive power satisfying Sm ∈ Γ∞.If l > 0 we write l as

l = l0 +mn

for suitable integers 0 ≤ l0 < n and m ≥ 0. Such a remainder expression alwaysexist for integers. Put

V ′ := V S−mn ∈⟨Sn,−1

⟩.

The new V ′ has now the form

V ′ = V S−mn = Sl0+mnS−mn = Sl0

for some 0 ≤ l0 < n. But this implies that l0 = 0 since all elements in⟨Sn,−1

⟩are

of the form ±Sm′m. Hence, V must also be of the form

V = ±Sm′n = ±(Sn)m

′∈⟨Sn,−1

⟩.

We just showed

Γ∞ =⟨Sn,−1

⟩.

Now, consider the case

q =a

b6=∞

with (a, b) = 1. Applying Lemma 2.31 above we find that each element of Γ ab

is of

the form ±M tab

for some t ∈ Z. M ab

is known to be parabolic.

The same reduction argument as in the previous case shows that there existsan integer 0 < t0 ≤ µ such that

Γ ab

=⟨M t0

ab,−1

⟩.

Remark 2.33. Notice that we already used the following fact several times:Assume that two elements A,A′ ∈ Γ(1) satisfy A∞ = a

b = A′∞. Then we haveA = A′St for some integer t.


Figure 3. The figure illustrates the fundamental domain FΓθ ofthe theta group Γθ discussed in Exercise 2.36. Clearly visible arethe two cusps ∞ of length 2 and 1 of length 1.

Definition 2.34. Let Γ ⊂ Γ(1) such that Γ(1) =⋃µi=1AiΓ with A1 = 1. This

means that the Ai’s are representatives of the right cosets of Γ in Γ(1) and we fixedthe first representative A1 = 1.

The width of the cusp ∞ is the smallest λ1 ∈ N such that Sλ1 ∈ Γ.For the other cusps Aj∞, we define the width of the cusp Aj∞ to be the

smallest λj ∈ N satisfying MλjajbJ

∈ Γ. The fraction Aj∞ =:ajbj

satisfies gcd(aj , bj) =

1 and bj 6= 0.

Example 2.35. By Example 2.27 we know that CΓ(1),FΓ(1)= ∞ where FΓ(1)

denote the standard fundamental region of the full modular group, see Theorem 2.15and Definition 2.18. Since S∞ = ∞ and S ∈ Γ(1), we see immediately that thewidth of the cusp ∞ is λ∞ = 1.

Example 2.36. We consider the theta group Γθ. Its index is[Γ(1) : Γθ

]= 3

and the group is generated by

Γθ =

⟨(1 20 1

),

(0 −11 0

)⟩=⟨S2, T

⟩.

The representatives of the right cosets of Γθ in Γ(1) are 1, S, and ST given by:

1 =

(1 00 1

), S =

(1 10 1

)and ST =

(1 −11 0

).

As a result, we can see that the inequivalent cusps are ∞ and 1.For the cusp 1 = 1

1 = ST ∞, we have that

M 11

(2.16)=

(0 1−1 2

)=

(0 1−1 0

)(1 −20 1

)= T 3 S−2.

Hence, M 11

is already in Γθ. This implies that the width of the cusp 1 is 1.

The cusp∞ has width 2, since it is already clear from the generators of Γθ thatS 6∈ Γθ but S2 ∈ Γθ.

A fundamental domain FΓθ of the theta group Γθ is given by

FΓθ = FΓ(1) ∪ S FΓ(1) ∪ ST FΓ(1) = FΓ(1) ∪(

1 10 1

)FΓ(1) ∪

(1 −11 0

)FΓ(1),

illustrated in Figure 3.


The next lemma presents a consistency condition for the width of cusps. Whathappens if you take two sets of representatives for the right cosets? Obviously, thestandard fundamental region changes and location of the cusps as well. But is thewidth of the cusps still the same?

Lemma 2.37. Let Γ be a subgroup of Γ(1) with finite index µ with two setsAi and A′i of representatives of the right cosets of Γ in Γ(1). Suppose that therepresentatives are arranged so that

A′i ∈ AiΓ for all 1 ≤ i ≤ µ.Assume that each cusp qi := Ai∞ has width λi and each cusp q′i := A′i∞ has widthλ′i.

Then, λi = λ′i for all i.

Proof. The proof of this lemma is left as an exercise.

CHAPTER 3

Modular Forms

This chapter is devoted to modular forms which we will introduce on congruencesubgroups with arbitrary multiplier systems. We prove some basic properties ofmodular forms and determine growth conditions for entire forms and cusp forms inaddition to growth conditions of their Fourier coefficients.Throughout this chapter, we assume that Γ is a subgroup of finite index in Γ(1).

3.1. Introduction – A Toy Example

Let us start with a toy example for the moment. We argue a bit relaxed withsome informality.

For the moment, we consider the full modular group Γ(1) and look at invariantfunctions under Γ = Γ(1): That means, we consider holomorphic functions

F : H→ H; z 7→ F (z)

satisfying

F (Mz) = F (z) for all M ∈ Γ and z ∈ H.Such functions are called modular functions. Theorem 2.4 and Equation (2.8) implyin particular that F satisfies

(3.1) F (z + 1) = F (z) = F

(−1

z

)for all z ∈ H.

Moreover, we assume that F is bounded at the cusp ∞:

(3.2) ∃K ∈ R>0 such that |F (z)| ≤ K for all z ∈ Uand U is a neighborhood of ∞, see (1.17).

The periodicity in (3.1) allows us to expand F into its Fourier expansion:

F (z) =∑n∈Z

ane2πinz for all z ∈ H.

What does the boundedness of F at ∞ imply?Since each Fourier term of the above expansion satisfies

e2πin(x+iy) →

∞ if n < 0,

1 if n = 0 and

0 if n > 0,

we may argue that boundedness of F at ∞ implies

limy→∞

F (x+ iy) = a0

and

a−n = 0 for all n ∈ N.Recall T in (2.8). Its Mobius transformation satisfies

T (x+ iy) =−1

x+ iy=−x+ iy

|x+ iy|2= O

(1

y

)43

44 3. MODULAR FORMS

for fixed x and y →∞. Using the right equality in (3.1) gives

limy↓0

F (x+ iy)(3.1)= lim

y↓0F(T (x+ iy)

)= lim

y↓0F

(−x+ iy

|x+ iy|2

)= limy→∞

F (x+ iy) = a0

andan = 0 for all n ∈ N.

Summarizing, any holomorphic function F on H which is invariant under Γ(1)and bounded at a neighborhood of∞ has to be a constant function. Such a functionspace is not very rich. We solve this problem by relaxing the “invariant condition”(3.1) a bit. Roughly speaking, instead of modular functions we consider “modularforms”.

3.2. Introduction – Modular forms

Recall that we assume that Γ ⊂ Γ(1) has finite index. In principle, we wouldlike to study invariant functions under the group action. However, as we indicatedin §3.1, the space of invariant holomorphic and bounded functions may not be thatinteresting.

Informally speaking: In order to get a more interesting function space, we con-sider meromorphic functions F on H which satisfy a slightly different transformationlaw

(3.3) F (Mz) = v(M) j(M, z)k F (z) for all z ∈ H and for all M ∈ Γ.

The terms v(M) and j(M, z) denote a suitable multiplier system and the auto-morphic factor introduced below. Functions F satisfying (3.3) are called modularforms.

The automorphic factor is defined by

(3.4) j : Mat2 (R)× C→ C (M, z) 7→ j(M, z) := cz + d

where we write M =

(a bc d

). The real parameter k ∈ R in (3.3) is the weight of

F .The multiplier system or short multiplier is a function v : Γ → C 6=0 such that

(3.3) allows nontrivial solutions. We only consider the case of unitary multipliers:|v(M)| = 1 for all M ∈ Γ.

The correct definition of multiplier systems and modular forms will be givenbelow in §3.3 and §3.4, respectively.

An immediate problem of the informal definition above is the following: Whydo we know that the multiplier v does not depend on z?

Lemma 3.1. The multiplier v associated to non-constant functions respectingthe transformation law (3.3) are independent of the complex variable z.

Proof. Let M ∈ Γ be a matrix and assume that v(M) depends on z. Thetransformation law (3.3) gives the identity

v(M) =F (M z)

j(M, z)k F (z)

for all z ∈ H with j(M, z)k F (z) 6= 0. We consider the map

Hrz ∈ H; j(M, z)k F (z) 6= 0

→ C; z 7→ v(M) =

F (M z)

j(M, z)k F (z).

By the open mapping theorem in §A.1, a non-constant holomorphic function on aconnected open set in the complex plane is an open mapping. But since the imageunder v is a closed set, we have that v(M) is independent of z.

3.2. INTRODUCTION – MODULAR FORMS 45

The next question is about real powers of complex numbers. We have a realweight k and look at expressions of the form j(M, z)k for M ∈ Γ and z ∈ H. Wehave to specify a branch cut to make this notion precise.

For a non-vanishing complex number w ∈ C 6=0 we put

(3.5) wk := |w|k eik arg(w) with − π ≤ arg(w) < π.

Notice that we adhere to the convention arg(−1) = −π.Now, we assume that there exists a non-zero function F satisfying the trans-

formation law (3.3). Given two matrices M1,M2 ∈ Γ and z ∈ H we consider

F(M1M2z

).

Using (1.10) we see that

v(M1)v(M2) j(M1,M2z)kj(M2, z)

k F (z) = F(M1M2z

)= v(M1M2) j(M1M2, z)

k F (z)

holds. Since F does not vanish everywhere we just showed the following conditionfor multipliers:

Lemma 3.2. If v is a multiplier associated to non-constant functions respectingthe transformation law (3.3), then v satisfies

(3.6) v(M1M2) j(M1M2, z)k = v(M1)v(M2) j(M1,M2z)

kj(M2, z)k

for all M1,M2 ∈ Γ and z ∈ H.

Proof. Repeating the arguments above Lemma 3.2, consider

F((M1M2) z

)= F

(M1(M2z)

)= v(M1) j(M1,M2 z)

k F (M2 z)

= v(M1)v(M2) j(M1,M2 z)kj(M2, z)

k F (z).

On the other hand, we have

F((M1M2) z

)= v(M1M2) j(M1M2, z)

k F (z).

Setting both expressions equal and dividing out the common term F (z), assumingof course that F (z) does not vanish everywhere, gives (3.6).

Lemma 3.3. If k ∈ Z is an integer, then there are no branching problems whenwe take powers of the automorphic factors. We have

(3.7) k ∈ Z =⇒ j(M1M2, z)k = j(M1,M2z)

kj(M2, z)k

for all M1,M2 ∈ Mat2 (R), and z ∈ H.If M1,M2 ∈ Γ, we have a multiplicative multiplier

(3.8) v(M1M2) = v(M1) v(M2).

Proof. To see this assume k = 1 for the moment. Write M1 =

(a1 b1c1 d1

)and

M2 =

(a2 b2c2 d2

)with

M1M2 =

(? ?

c1a2 + d1c2 c1b2 + d1d2

).

We find immediately

j(M1,M2 z)j(M2, z) =(c1(M2 z) + d1

)(c2z + d2

)=

(c1

(a2z + b2c2z + d2

)+ d1

)(c2z + d2)

46 3. MODULAR FORMS

= c1(a2z + b2) + d1(c2z + d2)

= (c1a2 + d1c2)z + (c1b2 + d1d2)

= j(M1M2, z).

Since integer powers of complex numbers have no branching problems, we im-mediately get

j(M1M2, z)k = j(M1,M2 z)

kj(M2, z)k

for any integer k.The multiplier condition 3.8 follows directly from (3.7) plugged into (3.6).

Definition 3.4. Property (3.6) is known as consistency condition of the mul-tiplier v.

Exercise 3.5. Consider k ∈ R. Show that the absolut value of the automorphicfactor satisfies

(3.9)∣∣j(M1M2, z)

k∣∣ =

∣∣j(M1,M2z)k∣∣ ∣∣j(M2, z)

k∣∣

for all z ∈ H and M1,M2 ∈ Mat2 (R).

3.3. Multiplier Systems

We like to introduce the term “multiplier system” correctly after using it in-formally in the introduction above.

Definition 3.6. Let Γ ⊂ Γ(1) be a subgroup with finite index and k ∈ R. Afunction

v : Γ→ C 6=0

is called a multiplier system or short multiplier with respect to (Γ, k) provided:

(1) |v(M)| = 1 for all M ∈ Γ and(2) v satisfies the consistency condition (3.6), for any M1,M2 in Γ.

Remark 3.7. If k ∈ Z, then (3.6) reduces to (3.8).

Next, we determine the value of the multiplier system at the identity elementand minus identity element of the group.

Example 3.8. First, we take M1 = M2 = 1 ∈ Γ and use (3.6). Since 12 = 1,we have

v(12) (1)k = v(1)v(1) 12k

which impliesv(1) = v(1)2.

Since v is a multiplier system and takes only values with |v(·)| = 1, we get

v(1) = 1.

Now, let M1 = M2 = −1 where we assume −1 ∈ Γ. The same calculation with(−1)2 = 1 leads to

v(1) (1)k = v(−1)2 (−1)k(−1)k.

Since we already know v(1) = 1, we have

1 = v(−1)2 (−1)2k ⇐⇒ v(−1)2 = (−1)−2k (3.5)= e2πik.

Hence, we havev(−1) ∈

±eπik

.

Now, consider a non-zero function F satisfying (3.3). Using the transformationlaw (3.3) for a function F , we get

F (z) = F((−1) z

) (3.3)= v(−1) (−1)k F (z)

3.3. MULTIPLIER SYSTEMS 47

for all z ∈ H. Dividing F (z) on both sides gives

1 = v(−1) (−1)k = v(−1) e−πik.

As a result, we calculated the value of v(−1):

(3.10) v(−1) = eπik.

Definition 3.9. We assume −1 ∈ Γ. Condition (3.10) is called the non-triviality condition for multipliers.

Remark 3.10. Pasles discusses multipliers in a slightly more general settingin [91]. He also proves the consistency condition and non-triviality condition inDefinitions (3.4) and 3.9.

Exercise 3.11. Let Γ ⊂ Γ(1) be a subgroup with finite index, k ∈ R andv : Γ → C 6=0 a multiplier in the sense of Definition 3.6 which satisfies the non-

triviality condition (3.10). Assume T ∈ Γ with T =

(0 −11 0

)as defined in (2.8).

Show that the multiplier v satisfies

(3.11) v(T ) = e−πi2 k,

respecting the argument convention (3.5).

Exercise 3.12. Consider the full modular group Γ(1), k ∈ R and v : Γ→ C 6=0

a multiplier in the sense of Definition 3.6 which satisfies the non-triviality condition

(3.10). Recall that S, T ∈ Γ(1) with S =

(1 10 1

)and T =

(0 −11 0

)as defined in

(2.8).Using (ST )3 = −1 show that the multiplier v satisfies

v(S) = eπi6 k and

v(T ) = e−πi2 k,

(3.12)

respecting the argument convention (3.5).

Example 3.13. Define the function

∆ : H→ C; z 7→ ∆(z) := e2πiz∏n∈N

(1− e2πinz

)24.

This function is also known as the discriminant function. We will show in §4.2 that∆ satisfies the transformation law (3.3) trivial multiplier v ≡ 1, weight 12 and thefull modular group Γ(1).

The Dedekind η-function, given by

η : H→ C; z 7→ η(z) := e2πiz24

∏n∈N

(1− e2πinz

),

satisfies

∆(z) = (η(z))24

(z ∈ H).

Moreover, η satisfies the transformation laws

η(Tz) = e−πi4 z

12 η(z) and η(Sz) = e

πi12 η(z).

More details about the Dedekind η-function can be found for example in [2, Chap-ter 3] or in [61, Chapter IX, §1].

Let r ∈ R> 0 be a positive real number. Then, the (24 · r)th power of theη-function, η24r, satisfies in particular

(3.13)(η(Tz)

)24r= e−6πir z12r

(η(z)

)24rand

(η(Sz)

)24r= e2πir

(η(z)

)24r

48 3. MODULAR FORMS

for all z ∈ H. Comparing these transformation laws to (3.3), we are inspired todefine the missing factor as a value of a suitable multiplier v12r.

Let k > 0 be positive weight. We define the η-multiplier vk by its values onthe generators S and T of Γ(1):

(3.14) vk(T ) := e−πik2 and vk(S) := e

πik6 .

Then, we extend vk to Γ using the consistency condition in Definition (3.4).In particular, vk is a multiplier system with respect to

(Γ(1), k

), k > 0.

Example 3.14. Another typical example of a multiplier system is the Dirichletcharacter χ which is defined as follows.

We call a function χ : Z → C 6=0 a Dirichlet character of modulus n ∈ N if itsatisfies the following conditions:

(1) χ(l) = χ(l + n) for all l ∈ Z,(2) if gcd(l, n) > 1 then χ(l) = 0 and if gcd(l, n) = 1 then χ(l) 6= 0,(3) χ(lm) = χ(l)χ(m) for all l,m ∈ Z.

We refer to [80] and [33, Dirichlet character] for more details about the Dirichletcharacters.

Let k ∈ 2N for the moment and consider the congruence subgroup Γ0(n) andlet χ denote a Dirichlet character of modulus n. We construct a multiplier v onΓ0(n) compatible with k by

v

((a bc d

)):= χ(d)

for all

(a bc d

)∈ Γ0(n). The above properties of a Dirichlet character ensure that

v is a multiplier system for(Γ0(n), k

)in the sense of Definition 3.6.

We need another technical definition relating cusps and values of multipliers.

Definition 3.15. Let Γ ⊂ Γ(1) be of finite index, v be a multiplier system ofweight k on Γ, and let q be a cusp of Γ in the sense of Definition 2.30. Denote byλq the width of the cusp q in Γ introduced in Definition 2.34.

We define the real number κq such that it satisfies

(3.15) v(Mλqq

)= e2πiκq ,

where κq can be chosen satisfying 0 ≤ κq < 1. Here, the matrix Mq is given in(2.16).

Example 3.16. Recall the η-multiplier vk, k > 0, defined in Example 3.13.It is a multiplier with respect to

(Γ(1), k

). In particular, we stated in (3.14) that

vk(S) has the value

vk(S) = eπik6 .

Recall also that M∞ defined in (2.16) satisfies M∞ =

(1 10 1

)= S, that the cusp

∞ of the full modular group Γ(1), and that the width of the cusp∞ is λ∞ = 1, seeExample 2.35.

Then, κ∞ introduced in Definition 3.15 above satisfies

e2πiκ∞ (3.15)= v

(Mλ∞∞) 2.35

= v(S) (3.14)

= eπik6 .

Hence, we have

κ∞ ≡k

12mod 1.

3.4. MODULAR FORMS 49

3.4. Modular Forms

In the following section, we give the definition of a modular form on subgroupsof finite index in the full modular group. We show that such functions have Fourierexpansions at all the cusps of the fundamental region of the subgroup. We alsogive several important properties of modular forms that will be of use in the laterchapters.

However, we do not present explicit examples of modular forms. Such examplesare given in Chapter 4.

Before we proceed with the definition of modular forms we present a lemmathat will be useful in giving the expansion of functions satisfying the transformationlaw (3.3) and admit certain growth conditions.

Lemma 3.17. Let q = ab ∈ Q with coprime integers gcd(a, b) = 1. Recall Mq in

(2.16) and assumeM tq ∈ Γ(1)

for some t ∈ Z.For any k ∈ R, we have

(3.16)(M tq z − q

)k= j(M tq , z)−k

(z − q)k.

Proof. Write q as q = ab with a ∈ Z, b ∈ Z6=0, gcd(a, b) = 1. Lemma 2.31

shows that M tq can be written as

M tq =

(1− tab ta2

−tb2 1 + tab

).

We shall prove the lemma first for the weight k = 1. As a second step, wethen take integer powers of both sides of the equation. The last step is to extendeverything to real k.

k = 1: Thus, we need to show the identity(M tq z − q

)j(M tq , z) ?

= z − q.Direct calculation of the left hand side yields to(

(1− tab)z + ta2

tb2z + 1 + tab− q)

(−tb2z + 1 + tab)

= (1− tab)z + ta2 − q(−tb2z + 1 + tab)

=((1− tab) + qtb2

)z −

(− ta2 + q(1 + tab)

)=((1− tab) + qtb2

)(z − −ta

2 + q(1 + tab)

(1− tab) + qtb2

)=((1− tab) + qtb2

)(z −M−tq q

).

Next, we insert the fixed point property M tq q = q and get(

M tq z − q

)j(M tq , z)

=

((1− tab)z + ta2

tb2z + 1 + tab− q)

(−tb2z + 1 + tab)

=((1− tab) + qtb2

)(z −M−tq q

)=((1− tab) + qtb2

)(z − q).

Using q = ab we get

(1− tab) + qtb2 = 1− tab+a

btb2 = 1.

Hence, we have proven the identity

(3.17) j(M, z)(M z − q

)= z − q.

50 3. MODULAR FORMS

k ∈ Z: We can extend the result directly to all integer powers, i.e., k ∈ Z.Hence, we have

j(M, z)k(M z − q

)k= (z − q)k

for all k ∈ Z.k ∈ R: Now, we consider real weight k. We have to show(

M tq z − q

)k ?= j(M tq , z)−k

(z − q)k.

Using the already proven identity (3.17) we have

(M tq z − q

)k=

(z − q

j(M tq , z))k .

We plug in the explicit entries of M tq into the right hand side and get(

z − qj(M tq , z))k =

(z − q

−tb2z + (1 + tab)

)k.

To establish the desired identity we have to show(z − q

−tb2z + (1 + tab)

)k?=

(z − q)k(− tb2z +−(1 + tab)

)k⇐⇒

(z − q

−tb2z + (1 + tab)

)k(−tb2z + (1 + tab))k

(z − q)k?= 1

(3.18)

for real k ∈ R. Since real powers of complex numbers are defined by

wr = |w|r eir argw for w ∈ C 6=0 and r ∈ R,

we check the moduli and the arguments separately. We need also theformula

eik arg(a/b) = eik(

arg(a)−arg(b))

which allows us to factor out k when we compare the arguments.We obviously have for the moduli∣∣∣∣∣(

z − q−tb2z + (1 + tab)

)k∣∣∣∣∣ =

∣∣(z − q)k∣∣∣∣∣(− tb2z + (1 + tab))k∣∣∣ .

It remains to check the arguments. We have to show

arg

(z − q

−tb2z + (1 + tab)

)?= arg(z − q)− arg

(− tb2z + (1 + tab)

).

As usual, we know that this identity holds up to integer multiples of 2π:

(3.19) arg

(z − q

−tb2z + (1 + tab)

)= arg(z − q)− arg

(− tb2z + (1 + tab)

)+ 2πn

for suitable integer n ∈ Z.Recall the proven relation (3.17). The argument convention (3.5)

implies

0 < arg(z − q) < π

since z−q ∈ H. Also, z ∈ H and b 6= 0 implies −tb2z+(1+tab) ∈ CrR≤0.Written as an argument, we have

−π < arg(− tb2z + (1 + tab)

)< π.


As a result, we have

2π >

∣∣∣∣arg

(z − q

−tb2z + (1 + tab)

)− arg(z − q) + arg

(− tb2z + (1 + tab)

)∣∣∣∣(3.19)

= 2π |n| .

The only integer n solving the inequality is n = 0. Plugging n = 0 into(3.19) proves (3.18).

We just showed that

j(M tq , z)k (

M tq z − q

)k= (z − q)k

holds for all k ∈ R.

We now present a theorem that will be needed in later chapters. It simplystates that if a function satisfies a specific transformation law on the generators ofthe full modular group, then it satisfies the same transformation law on every otherelement of that group.

Theorem 3.18. Let k ∈ 2Z be an even integer and let F : H→ C be a functionsatisfying the two relations

(3.20) F (z + 1) = F (z) and F

(−1

z

)= zk F (z)

for all z ∈ H. Then, F satisfies

(3.21) F(M z

)= j(M, z)k F (z)

for all z ∈ H and all M ∈ Γ(1).

Remark 3.19. Theorem 3.18 shows that if F satisfies (3.20) then F satisfiesthe transformation law (3.3) for even integral weight k ∈ 2Z, trivial multiplier v ≡ 1and the full modular group Γ(1).

Proof of Theorem 3.18. Each Element M ∈ Γ(1) can be expressed as wordin the generators S and T of Γ(1), see Corollary 2.6, and has an Eichler length l(M),see Definition 2.7. We make an induction argument in the Eichler length to provethe theorem.

Consider the case l(M) = 0 as starting point. This implies immediately M = 1and (3.21) simplifies to the trivial identity

F (z) = F(1 z) !

= j(1, z)k F (z) = 1k F (z).

Suppose that (3.21) holds for all elements M ′ ∈ Γ(1) with Eichler lengthl(M ′) < m. We consider an element M ∈ Γ(1) with Eichler length l(M) = m.Since Γ(1) is generated by S and T , we know that M can be written as (at least)one of the three cases

M = M ′ S, M = M ′ S−1 , or M = M ′ T

and m′ has a smaller Eichler length l(M ′) < m.Assume M = M ′S for some M ′ ∈ Γ(1) with l(M ′) ≤ m− 1. Hence we have

F(M z

)= F

(M ′ S z

)= j(M ′, S z)k F

(S z)

= j(M ′, S z)kj(S, z)k F (z)

(3.7)= j(M ′S, z)k F (z)

= j(M, z)k F (z).

52 3. MODULAR FORMS

The other two cases M = M ′ S−1 and M = M ′ T follow by the same argument,based on (3.7).

We are now getting closer to the definition of modular forms. In the followingdefinition, we give an informal and incomplete definition of modular forms that willbe completed later in the chapter.

Definition 3.20. Let k ∈ R, and let Γ be a subgroup of finite index in Γ(1). Amodular form of weight k and multiplier system v on Γ is a meromorphic functionF : H→ C satisfying (3.3) and such that F (z) satisfies “certain growth condition”at the cusps of the standard fundamental region FΓ of Γ.

Remark 3.21. The wording “certain growth condition” indicates that thisdefinition is not yet complete. We still need to specify what these words mean.

We now present a theorem that give the Fourier expansion of functions satis-fying (3.3) in addition to other conditions. If we impose certain growth conditionson those functions, those functions will be modular forms.

Theorem 3.22. Let Γ ⊂ Γ(1) be a subgroup of finite index, k ∈ R be a realweight, and v : Γ→ C 6=0 be a multiplier respecting (Γ, k). Suppose

F : H→ C; z 7→ F (z)

is a meromorphic function which satisfies the transformation law (3.3). Moreover,F has at most finitely many poles in the closure of the standard fundamental regionFΓ ∩ H of Γ. Let q1 = ∞, q2, . . . , qµ ∈ Q ∪ ∞ be the cusps of Γ in the stan-dard fundamental region FΓ and A1, A2, . . . , Aµ be representatives of the associatedright cosets of Γ in Γ(1) satisfying qj = Aj∞, as introduced in Lemma 2.28. Letλ1, λ2, . . . , λµ be the width of the cusp qi and κ1, κ2, . . . , κµ be the correspondingreal numbers associated to qj by (3.15).

Then for each j, j ∈ 1, . . . , µ, there exists a non-negative real yj such that Fhas an expansion at the cusp qj given by(3.22)

F (z) = σj(z)

∞∑n=−∞

aj(n) e2πi(n+κj)A

−1j

zλj valid for all z ∈ H, =

(A−1j z

)> yj

where σj(z) is given by

(3.23) σk,j(z) =

1 if qj =∞ and

(z − qj)−k if qj 6=∞.

Proof. In a first step, we determine first the expansion of F at the cuspq1 =∞.

F is meromorphic in H with only finite many poles in FΓ ∩H by assumption.Then, there exists y1 ∈ R>0 such that F (z) has no poles in z ∈ H; = (z) > y1 ⊂H. The cusp width λ1 and the value κ1 satisfy v

(Sλ1

)= e2πiκ1 , see Definitions 2.34

and 3.15.We apply (3.3) to get

F(Sλ1z

)= F (z + λ1)

(3.3)= e2πiκ1 F (z)

for all z ∈ H with = (z) > y1. We define the function g : z ∈ H; = (z) > y1 → Cby

g1(z) := e−2πiκ1zλ1 F (z).

The function g is holomorphic on z ∈ H; = (z) > y1 ⊂ H, which it just inheritsform F , and it is periodic:

g1(z + λ1) = g1(z).


Now put

w1 := e2πi zλ1 and ρ1 := e−2πy1λ1

which is a coordinate transformation, mapping z ∈ H; = (z) > y1 onto thepunctured disc 0 < |w1| < ρ1 such that z and z + λ1 map onto the same point w1.We define the next function h1 implicitly on the punctured disc 0 < |w1| < ρ1 usingthe above coordinate transformation:

h(w1) = g(z).

It can be easily shown that h is well defined due to the periodicity of g and thath is holomorphic in 0 < |w| < ρ1. As a result, h has a Laurent expansion, valid in0 < |w1| < ρ1, which is given by

h(w1) =

∞∑n=−∞

a1(n)wn1 .

Hence g has the expansion

g(z) =∞∑

n=−∞a1(n)

(e2πi zλ1

)n=

∞∑n=−∞

a1(n) e2πin zλ1

and F admits

F (z) =

∞∑n=−∞

a1(n) e2πi(n+κ1) zλ1

for all z ∈ H with = (z) > y1.The next step determines the expansions for the other cusps qj 6=∞. We map

the cusp qj = Aj∞ back to the cusp ∞ (by applying z 7→ A−1j z) and follow the

calculation above for the cusp q1 =∞.Let q ∈ Q be a parabolic cusp. (Note that all cusps are in Q ∪ ∞ since

they satisfy qj = Aj∞, see Definition 2.23. Recall that Γqj = 〈Mλjqj ,−1〉, see

Proposition 2.32. Put

Fj(z) :=(z − qj

)kF (z) (z ∈ H).

Using (3.3) and Lemma 3.17, we get

Fj(Mλjqj z

)=(Mλjqj z − qj

)kF(Mλjqj z

)=

(z − qj)k

j(Mλjqj , z)

kv(Mλjqj

) (j(Mλj

qj , z)k F (z)

)= v(Mλjqj

)Fj(z)

for all z ∈ H. By Definition 3.15 we have v(M

λjqj

)= e2πiκqj and hence

Fj(Mλjqj z

)= e2πiκqj , Fj(z) for all z ∈ H.

Recall as well that Mqj = Aj S A−1j Γ, where Aj∞ = qj with Aj ∈ Γ(1) is the

representative of the right coset associated to the cusp qj . Thus we get

Fj(Aj S

λj A−1j z

)= e2πiκj Fj(z).

Substituting z → Aj z we get

Fj(Aj S

λj z)

= e2πiκj Fj(Aj z

).

We now follow the proof above for the expansion for the cusp ∞: Let

gj(z) := e−2πiκj

zλj Fj

(Aj z

).

54 3. MODULAR FORMS

It is easy to see that gj is λj-periodic:

gj(Sλj z

)= gj(z + λj)

= e−2πiκj

z+λjλj Fj

(AjS

λj z)

= e−2πiκj

zλj e−2πiκj e2πiκj Fj

(Aj z

)= gj(z).

Of course, there exists a yj > 0 such that gj is holomorphic on z ∈ H; = (z) > yjsince F and so Fj only have finite many poles within the closure of the standard

fundamental region FΓ ∩H. We now put

wj := e2πi zλj and ρj := e

−2πyjλj

and define implicitly

hj(w) := gj(z)

(= gj

(λj logwj

2πi

) ).

Notice that h is well defined and holomorphic on 0 < |w| < ρj . Hence, h has aLaurent expansion valid in the punctured disc 0 < |wj | < ρj and we get

gj(z) =

∞∑n=−∞

aj(n) e2πin z

λj

for all z with = (z) > y1. This implies

e−2πiκj

zλj(Aj z − qj

)kF(Aj z

)=

∞∑n=−∞

aj(n) e2πin z

λj .

Substituting z 7→ A−1j z, we get the required expansion.

Exercise 3.23. Assume that FΓ and F ′Γ are two standard fundamental regionsof Γ (i.e. both are generated using different sets of representatives for the right cosetsof Γ in Γ(1)) and assume that F (z) is a meromorphic function on H satisfying (3.3).F (z) has a finite number of poles in FΓ ∩H if and only if F (z) has a finite numberof poles in F ′γ ∩H.

Exercise 3.24. If F : H→ C is a meromorphic function and has an expansionof the form (3.22) at each cusp qj of the standard fundamental region FΓ, then F

has at most a finite number of poles in F ′γ ∩H.

Definition 3.25. Let F be as in Theorem 3.22 and consider the expansions ofF at the cusp’s qj ∈ CΓ,FΓ

.

(1) If only a finite number of terms with n < 0 appear in (3.22) at any cuspqj , we say that F is meromorphic at qj .

(2) If the first non-zero aj(n) occurs for n = −nj < 0 for some nj ∈ N, wesay F has a pole at qj of order nj − κj .

(3) If the first non-zero aj(n) occurs for n = n0 ≥ 0 for some nj ∈ N, we sayF is regular at qj with a zero at qj of order n0 + κj .

We now give the formal definition of modular forms.

Definition 3.26. Let k ∈ R and let v be a multiplier system with respect to(Γ, k). A meromorphic function F : H→ C is called a weakly holomorphic modularform of weight k and multiplier system v on Γ provided:

(1) F (z) satisfies (3.3) for all elements of Γ.(2) F (z) has at worst a finite number of poles in FΓ ∩H.


(3) F (z) is meromorphic at each cusp qj ∈ CΓ,FΓwhich is a parabolic point.

We will denote the vector space of modular forms of weight k and multipliersystem v on a group Γ by M !

k,v(Γ). A useful subspace is the space S!k,v(Γ) ⊂

M !k,v(Γ) where the Fourier expansion 3.22 of each weakly holomorphic modular

form f ∈ S!k,v(Γ) has vanishing zero-terms aj(0) = 0.

We call F (z) an entire modular form of weight k with multiplier v on Γ ifF ∈ M !

k,v(Γ) is holomorphic in H and F (z) is regular at each cusp qj . The vector

space of entire modular forms will be denoted Mk,v(Γ).We call F ∈ Mk,v(Γ) a cusp form if F (z) is a modular form with a zero of

positive order at each cusp qj . The vector space of cusp forms will be denoted bySk,v(Γ).

In addition, we call F (z) a modular function if F (z) is a modular form of weight0 and trivial multiplier, i.e., v(M) = 1 for all M ∈ Γ.

Remark 3.27. The added part, “which is a parabolic point” in the third con-dition of the definition above, is redundant if Γ is a subgroup of Γ(1) of finite index.In this case, each cusp q in CΓ,FΓ is already a parabolic point as can be seen inLemma 2.31.

Remark 3.28. If we look at modular forms with trivial multiplier v ≡ 1, thatmeans that 1(M) = 1 for all M ∈ Γ, we denote the respective modular form spacesby M !

k(Γ), Mk(Γ) and Sk(Γ) instead of M !k,1(Γ), Mk,1(Γ) and Mk,1(Γ).

Remark 3.29. The notations Mk,v(Γ) and Sk,v(Γ) for entire modular and cuspforms are nowadays standardised (up to small variations). Also the space M !

k,v(Γ)is also known as weakly holomorphic modular forms in modern literature. However,you find sometimes different notations, mostly in some older sources like [54]. Forexample, there the spaceM !

k,v(Γ) is denoted by Γ, k, v, the space of entire modular

forms Mk,v(Γ) is denoted by C+(Γ, k, v), and the space of cusp forms Sk,v(Γ) isdenoted by C0(Γ, k, v).

In what follows, we will give important facts about modular forms concerningthe growth of these forms as one approaches the parabolic points in addition to thegrowth of their Fourier coefficients.

Lemma 3.30. Let Γ be a subgroup of Γ(1) with finite index and let f ∈M0(Γ) bea modular function on Γ. (That means that f(z) is holomorphic in H and regular atall cusps of the standard fundamental region FΓ). Then, for each cusp qj ∈ CΓ,FΓ

,we have

f(z)→ aj(0) as z → qj , z ∈ FΓ ∩H.

Furthermore, f(z) is bounded on H.

Proof. The expansion (3.22) of f given by Theorem 3.22 together with fbeing a modular function implies

f(z) =

∞∑n=0

aj(n) e2πinA−1

jzλj

for each cusp qj ∈ CΓ,FΓ, j ∈ 1, . . . , µ. (f modular function implies trivial weight

k = 0 and κj = 0 for all cusps).

Also, we can rewrite the limit process z ∈ FΓ∩H with z → qj using qj = Aj∞.We get

A−1j z →∞

56 3. MODULAR FORMS

1

Figure 1. We indicate the sets Di = Di(y0) in the standard fun-damental regions illustrated in Figure 1. The set D = D(y) in(3.24) is the remaining part of the standard fundamental region.

and < (z) is bounded. This limit process shows that all coefficients in the expansionof f above vanish except the first term aj(0). Hence, we have

f(z)→ aj(0) as z → qj with z ∈ FΓ.

To prove that f is bounded in H, what remains to prove is that f(z) is boundedin the fundamental region excluding a disc around the parabolic points from withinthe fundamental region. Let y0 > 0 be a large real number and consider D1 :=D1(y0) := z ∈ H; = (z) ≥ y0. Take all images Dj := Dj(y0) := Aj D1 :=Aj z; = (z) > y0 and remove the union of all Dj from the fundamental region.This means that we consider the set D given by:

D1 := D1(y0) := z ∈ H; = (z) ≥ y0Dj := Dj(y0) := Aj D1 := Aj z; = (z) > y0 and

D := D(y0) :=(FΓ ∩H

)r

µ⋂j=1

Dj

.

(3.24)

(See Figure 1 for an illustration of the sets Dj and D). Notice that D is a compactset since we removed all parts close to a cusp. Since z → |f(z)| is continuouson H we have that z → |f(z)| is bounded on the compact set D. Together withthe boundedness of the limit process above we see that f bounded on FΓ. Thetransformation property (3.17) extends the boundedness of f to the whole upperhalf plane H since H =

⋃M∈ΓM FΓ.

Exercise 3.31. Let Γ and f be as in Lemma 3.30. We define

G(z) :=

µ∏j=1

(f(z)− aj(0)

)(z ∈ H).

Show that G(z) is invariant under Γ.

Theorem 3.32. Let Γ and f be as in Lemma 3.30. Then f is constant.


Proof. Put

F (z) =

µ∏j=1

(f(z)− aj(0)

)(z ∈ H).

It is clear that F (z) is invariant under Γ, see Exercise 3.31. Using Lemma 3.30, weknow that F is bounded on H. Also, observe that

F (z)→ 0 as z → qj

within FΓ for each cusp qj . Now, consider

C := supz∈H|F (z)| = sup

z∈FΓ∩H|F (z)| .

The reduction to the fundamental region is valid since F is invariant under Γ.We claim C = 0 which would imply that the range of F is just 0. Using thecontinuity of F , this implies that F vanishes everywhere. Hence, at least one of themultiplicative factors f(z)− aj(0) vanishes everywhere. This implies f(z) = aj forall z ∈ H for the vanishing factor with index j, showing that f is constant.

We still have to show that C = supz∈FΓ∩H |F (z)| is zero. Assume the contraryand consider the set D(y0) constructed in the proof of Lemma 3.30. Since

(3.25) F (z)→ 0 as z → qj , z ∈ FΓ ∩H,

there exists a (large) y0 > 0 such that

|F (z)| < C

2for all z ∈

µ⋃i=1

Dj ∩ FΓ

where the Dj ’s are defined in the proof of Lemma 3.30. As a result, the maximumof F has to occur in the compact set D = D(y0), also constructed in the proof ofLemma 3.30, since

FΓ ∩H =

(D(y0) ∪

µ⋃i=1

Dj

)∩ FΓ.

By the maximum modulus principle, see Appendix A.4, we find that F has to beconstant F ≡ C which is nonzero by assumption. This contradicts (3.25). HenceC = 0 must be true.

Corollary 3.33. Let Γ be a subgroup of Γ(1) with finite index. If f is amodular function on Γ and f is bounded in H, then f is entire and hence constant.

Proof. If f has the expansion

f(z) =

∞∑n=−n0

aj(n)e2πinA−1

jzλj , z ∈ H,

given by Theorem 3.22. Then n0 can not be positive or else f can not be boundedas z → qj . As a result, f has to be regular at the cusps leading to the fact that fis entire and hence by Theorem 3.32, f is constant.

We now determine what is widely known as Hecke bound for the Fourier coef-ficients of modular cusp forms of a given weight k. However, to do that, we needto determine a growth condition for modular cusp forms.

Lemma 3.34. Let Γ be a subgroup of Γ(1) with finite index and take a cuspform F ∈ Sk,v(Γ). There exists a positive constant K > 0 such that

|F (z)| ≤ K = (z)− k2 for all z ∈ H.

58 3. MODULAR FORMS

Proof. Define the function φ : H→ C by

φ(z) := = (z)k2 |F (z)| for all z ∈ H.

We need to show that φ(z) is bounded for all z ∈ H. Using the transformation law(3.3) for F (Mz) and (1.11) for = (Mz) we have

φ(M z

)== (z)

k2

|j(M, z)|k∣∣v(M) j(M, z)k F (z)

∣∣ = φ(z) for all M ∈ Γ.

We used the property |v(M)| = 1 for multipliers and (1.11). Hence we see that φis invariant on Γ.

Next we want to show that φ is bounded on H. To do so, we need to show thatφ(z)→ 0 as z → qj within FΓ ∩H for all cusps qj ∈ CΓ,FΓ

, 1 ≤ j ≤ µ. Recall thatthe expansion of F at the cusp qj is given by

F (z) = σk,j(z)

∞∑n=nj

aj(n)e2πi(n+κj)A−1j z/λj

in Theorem 3.22 and Definition 3.26 with nj ∈ N and

σk,j(z) =

1 if qj =∞ and(z − qj

)−kif qj ∈ Q

in (3.23).Recall that we have qj = Aj∞, see Lemma 2.28. Writing qj =

ajbj∈ Q ∪ ∞

with the understanding that gcd(aj , bj) = 1 and aj = 1, bj = 0 if qj = ∞(= 1

0

),

we have

Aj =

(a ?b ?

)and A−1

j =

(? ?−b a

).

Thus, we have

=(A−1j z

)=

= (z) if qj =∞ and=(z)

|bjz−aj |2if qj ∈ Q.

This leads to

=(A−1j z

) k2 |σj(z)|−1

=

= (z)k2 if qj =∞ and

=(z)k2

|bj |kif qj ∈ Q.


= (z)k2 |σj(z)| =

=(A−1j z

) k2 if qj =∞ and

=(A−1j z

) k2 |bj |k if qj ∈ Q.

First, we discuss the cases qj =∞. We have

φ(z) = = (z)k2

∣∣∣∣∣∣∞∑

n=nj

aj(n) e2πi(n+κj)

zλj

∣∣∣∣∣∣= = (z)

k2 e−2π(nj+κj)

=(z)λj

∣∣∣∣∣∣∞∑

n=nj

aj(n) e2πi(n−nj) zλ1

∣∣∣∣∣∣→ 0 as z →∞.


For any finite cusp qj ∈ Q, we have

φ(z) = = (z)k2 |σj(z)|

∣∣∣∣∣∣∞∑

n=nj

aj(n) e2πi(n+κj)A

−1j

zλj

∣∣∣∣∣∣= |bj |k =

(A−1j z

) k2 e−2π(nj+κj

)=(A−1j

zλj

) ∣∣∣∣∣∣∞∑

n=nj

aj(n) e2πi(n−nj)A−1

jzλj

∣∣∣∣∣∣→ 0 as z → qj .

As a result, we have that φ(z) approaches 0 within the fundamental region FΓ asz approaches the cusps. By continuity, there exists a K > 0 such that φ(z) ≤ Kholds for all z ∈ FΓ ∩H. By the invariance of φ(z), we get φ(z) ≤ K for all z ∈ H.

Plugging in the definition of φ we get for F :

|F (z)| ≤ K= (z)− k2 for all z ∈ H.

Exercise 3.35. Let Γ be a subgroup of Γ(1) with finite index and take a cuspform F ∈ Sk,v(Γ). Show that there exists a δ > 0 such that F satisfies the growthestimate

F (z) = O(e−2πδ=(z)

)as = (z)→∞.

We now show that the Fourier coefficients of the expansion of a cusp form atinfinity have polynomial growth.

Theorem 3.36. Let F ∈ Sk,v(Γ) with Γ ⊂ Γ(1) being of finite index. Assumethat F has an expansion at ∞ given by

F (z) =

∞∑n=n0

a(n) e2πi(n+κ) zλ

with n0 + κ > 0.Then, the Fourier coefficients a(n) satisfy

a(n) = O(nk2

)as n→∞.

Proof. Consider the function F (z) e−2πi(n+κ) zλ . Using the expansion at ∞we have

F (z) e−2πi(n+κ) zλ =

( ∞∑m=m0

a(m) e2πi(m+κ) zλ

)e−2πi(n+κ) zλ

=

∞∑m=m0

a(m)(e2πi(m+κ) zλ e−2πi(n+κ) zλ

)=

∞∑m=m0

a(m) e2πi(m−n) zλ .

Since F as cusp form is holomorphic on H, it is bounded on any compact subset ofH. Using this bound as constant integrable function on the compact set, we may usethe Lebesgue’s theorem on dominated convergence to interchange the integrationand summation: We have∫ z+λ

z

F (ξ) e−2πi(n+κ) ξλ dξ =

∫ z+λ

z

∞∑m=m0

a(m) e2πi(m−n) ξλ dξ

60 3. MODULAR FORMS

=

∞∑m=m0

a(m)

∫ z+λ

z

e2πi(m−n) ξλ dξ

= λ a(n),

since ∫ z+λ

z

e2πi(m−n) ξλ dξ =

0 if m− n 6= 0 and

λ if m− n = 0.

Thus we get

a(n) =1

λ

∫ z+λ

z

F (ξ)e−2πi(n+κ) ξλ dξ

for all n ≥ n0. Using the bound for F (z) in Lemma 3.34, we find

(3.26) |a(n)| ≤ 1

λλK= (z)

− k2 e2π(n+κ)=(z)λ

for some K > 0. Now choosing = (z) = 1n , we see that

|a(n)| ≤ K ′ n k2

for some K ′ > 0.

It is worth mentioning that when k < 0, we can do much better. This isillustrated in the following corollary.

Corollary 3.37. Let k < 0 be a negative weight and Γ be a subgroup of Γ(1)with finite index. If F ∈ Sk,v(Γ) then F ≡ 0.

Proof. In (3.26) of the proof of Theorem 3.36 we have shown

|a(n)| ≤ K = (z)−k/2

e2π(n+κ)=(z)λ .

Now, fix n and let = (z) approach 0. We find that a(n) = 0 for all n. Hence, F ≡ 0vanishes everywhere.

We now determine the growth of an entire modular form in order to determinethe polynomial growth of the Fourier coefficient of the expansion of the form at theinfinite cusp.

Lemma 3.38. Let Γ be a subgroup of Γ(1) with finite index. If F ∈Mk,v(Γ) isan entire modular form which has an expansion (3.22), then

|F (z)| ≤ K(

1 + = (z)−k)

(z ∈ H)

for some K > 0.

Proof. Following the proof of Lemma 3.34, we define

φ(z) := = (z)k2 |F (z)|

for all z ∈ H.At the infinite cusp∞, we follow the same steps as in the proof of Lemma 3.34

to determine a bound for φ(z). We observe that the 0th Fourier coefficient a1(0) isnot necessarily 0 since F (z) is only regular at∞. Following the same arguments asin the proof of Lemma 3.34, but allowing that the Fourier expansion starts at the0th term, i.e., n1 = 0, we get an upper estimate for some suitable K1 > 0

|φ(z)| ≤ K1= (z)k2


for all z ∈ D1, where D1 is defined in (3.24). Again, as in the proof of Lemma 3.34,we use the same bound for φ(z) at the finite cusps q and we get for some positiveconstant Kq > 0

|φ(z)| ≤ Kq= (z)

k2

|z − q|k

for all z ∈ Dq, where Dq is defined in (3.24) (identifying the cusp q with the indexj of the cusp q). Notice that |z − q| ≥ = (z) for all z ∈ H. Hence, we get

|φ(z)| ≤ Kq = (z)− k2

for all z ∈ Dq. Also, F as entire function is bounded within compact subsets of thefundamental region: there exists an K such that

|φ(z)| ≤ K

for all z ∈ D, where D is defined in (3.24).As a result, we get the bound

|φ(z)| ≤ K?(= (z)

k2 + = (z)

− k2 + 1)

≤ K ′(= (z)

k2 + = (z)

− k2)

for suitable 0 < Kq,K ≤ K? ≤ K ′ and for all z ∈ FΓ ∩H.Recall that φ is invariant under Γ:

φ(M z

)= φ(z)

for all M ∈ Γ and z ∈ H. (We have shown this in the proof of Lemma 3.34). Wecombine this with the second property of fundamental regions in Definition 2.9.Take any w ∈ H. Then, there exists an M ∈ Γ and z ∈ FΓ ∩H such that w = M zholds. Hence, we have

|φ(w)| =∣∣φ(M z

)∣∣= |φ(z)|

≤ K ′(= (z)

k2 + = (z)

− k2).

Using the definition of the auxiliary function φ, we finally have

|F (z)| ≤ K ′(= (z)

−k+ 1)

for all z ∈ H.

Using the above lemma, we will be able to determine the growth of the Fouriercoefficients of entire modular forms at the infinite cusps.

Theorem 3.39. Let Γ be a subgroup of Γ(1) with finite index and let F ∈Mk,v(Γ) be a modular form of non-negative weight k ≥ 0. Assume that F has anexpansion at infinity given by

F (z) =

∞∑n=n0

a(n) e2πi(n+κ) zλ

with n0 + κ ≥ 0. Here, λ ∈ N and 0 ≤ κ < 1 denote the width of the cusp ∞ andits associated κ-value, see Definitions 2.34 and 3.15 respectively.

Then we have a(n) = O(nk)

as n→∞.

62 3. MODULAR FORMS

Proof. We follow the same steps as the proof of Theorem 3.36. The onlydifference is that we use the estimate in Lemma 3.38 instead of the one for cuspforms in Lemma 3.34. Hence, we get from the integral representation of a(n) in theproof of Theorem 3.36 the following estimate:

|a(n)| ≤ K(

1 + = (z)−k)e2π(n+κ)=(z)

λ (z ∈ H).

Taking y = 1n , we get our result.

In what follows, we will show that if we consider an entire modular form ofnegative weight on a subgroup of finite index in the full group, then this form willbe constant. In order to do that, we need the following theorem.

Lemma 3.40. If F ∈ Mk,v(Γ) with negative weight k < 0, then there exists apositive constant K > 0 such that

|F (z)| ≤ K = (z)−k/2

for all z ∈ H.Proof. The proof of this theorem will be left as an exercise for the reader.

Once we established the above lemma we may proceed as before in the cuspidalcase. The same arguments that lead to Theorem 3.36 and Corollary 3.37 prove:

Proposition 3.41. Let F (z) ∈Mk,v(Γ) with negative weight k < 0. Then, themodular form F ≡ 0 vanishes everywhere.

Proof. Using Lemma 3.40 we can literally copy the proof of Theorem 3.36and Corollary 3.37.

Exercise 3.42. Let F ∈ M !k,1(Γ) and G ∈ M !

l,1(Γ). Then F · G ∈ M !k+l,1(Γ),

where the function product is defined as(F ·G

)(z) := F (z) ·G(z).

3.5. Petersson scalar product and the Hilbert space of cusp forms

Let Γ ⊂ Γ(1) be a subgroup of finite index µ and let k ∈ R a weight and v amultiplier with respect to (Γ, k). We consider the map 〈·, ·〉Γ,k given by

〈·, ·〉Γ,k : Mk,v(Γ)× Sk,v(Γ)→ C

(f, g) 7→ 〈f, g〉Γ,k :=1

µ

∫Ff(z) g(z)= (z)

k = (z)−2

dλ2(z)(3.27)

where F is a standard fundamental region of the group Γ, see e.g. §2.3. A typicalchoice of a fundamental region of Γ is derived in Theorem 2.17.

The integration measure λ2(z) denotes the usual Euclidean surface Lebesguemeasure usually given by

(3.28) dλ2(z) = dxdy,

if z = x+iy ∈ C are the usual real and imaginary coordinates. The derived measure

(3.29) = (z)−2

dλ2(z) := y−2 dxdy

for z = x+ iy ∈ H is called the hyperbolic surface measure.

Remark 3.43. The hyperbolic distance function d : H×H→ R≥0 is given by

(3.30) cosh d(z, z′) = 1 +

1

2

|z − z′|= (z) = (z′)

(z, z′ ∈ H).

The induced metric ds satisfies

(3.31) (ds)2 =dx2 + dy2

y2(z = x+ iy ∈ H).

It is conformal which means that the angles measured in the hupper half-plane arethe same as in the actual hyperbolic plane.

3.5. PETERSSON SCALAR PRODUCT AND THE HILBERT SPACE OF CUSP FORMS 63

Some questions naturally arise from the definition above:

(1) How does the measure = (z)−2

dλ2(z) behave under Mobius transforma-tions?

(2) Is 〈·, ·〉Γ,k well defined? In other words, is 〈f, g〉Γ,k finite for any pair(f, g) ∈Mk,v(Γ)× Sk,v

(Γ(1)

)?

(3) Does 〈·, ·〉Γ,k depend on the choice of the fundamental region F?(4) How does 〈·, ·〉Γ,k relate to transformations under Γ or even under SL2 (R)?(5) For f, g are modular forms of a group Γ1 with Γ ⊂ Γ1 ⊂ Γ(1), how does〈f, g〉Γ,k and 〈f, g〉Γ1,k relate?

We answer the questions in the following lemmas.

Lemma 3.44. The hyperbolic surface measure = (z)−2

dλ2(z) is invariant undercoordinate changes by Mobius transformations:

(3.32) = (M z)−2

dλ2(M z) = = (z)−2

dλ2(z)

for all z ∈ H and M ∈ SL2 (R).

Proof. Let

(a bc d

)= M ∈ SL2 (R). Recalling (1.11) we already have

= (Mz) == (z)

|cz + d|2.

Now, we need to calculate the transformation behavior of dλ2(z). Consider thecoordinate transformation induced by x+ iy = z 7→ V z =: u+ iv: We have

x+ iy = z 7−→ V z =az + b

cz + d

=ac(< (z)

2+ = (z)

2 )+ < (z) (ad+ bc) + bd

|cz + d|2+ i

= (z)

|cz + d|2

=: u+ iv.

(3.33)

In particular, we have

u(x, y) =ac(x2 + y2

)+ x(ad+ bc) + bd

(cx+ d)2 + (cy)2and v(x, y) =

y

(cx+ d)2 + (cy)2

with partial derivatives

(3.34)∂u

∂x=

(cx+ d)2 − (cy)2((cx+ d)2 + (cy)2

)2 =(cx+ d)2 − (cy)2

|cz + d|4

and

(3.35)∂v

∂x= − 2c(cx+ d)y(

(cx+ d)2 + (cy)2)2 = −2c(cx+ d)y

|cz + d|4.

To calculate the Jacobian det

(∂u∂x

∂u∂y

∂v∂x

∂v∂y

)we use the Cauchy-Riemann equations,

see §A.7:∂u

∂x=∂v

∂yand

∂u

∂y= −∂v

∂x.

We get

det

(∂u∂x

∂u∂y

∂v∂x

∂v∂y

)=∂u

∂x

∂v

∂y− ∂u

∂y

∂v

∂x

64 3. MODULAR FORMS

=

(∂u

∂x

)2

+

(∂v

∂x

)2

(3.34),(3.35)=

1

|cz + d|4.(3.36)

Now we can calculate the left hand side of (3.32). We have

= (M z)−2

dλ2(M z)(1.11)

== (z)

−2

|cz + d|−4 dλ2(M z)

(3.29),(3.33)=

= (z)−2

|cz + d|−4

(dudv

)== (z)

−2

|cz + d|−4 det

(∂u∂x

∂u∂y

∂v∂x

∂v∂y

) (dxdy

)(3.36)

== (z)

−2

|cz + d|−4

1

|cz + d|4(dxdy

)(3.29)

= = (z)−2

dλ2(z).

This proves (3.32).

Exercise 3.45. Show∫FΓ(1)

= (z)−2

dλ2(z) = π3 .

Lemma 3.46. For (f, g) ∈Mk,v(Γ)×Sk,v(Γ) the evaluation of 〈·, ·〉Γ,k in (f, g)is finite: 〈f, g〉Γ,k ∈ C.

Proof. We consider 〈f, g〉Γ,k. Using (3.27) we have

〈f, g〉Γ,k =1

µ

∫Ff(z) g(z)= (z)

k = (z)−2

dλ2(z).

According to Definition 3.26 f and g are continuous (since they have no poles inH), and bounded (Lemma 3.30). Moreover, Exercise 3.35 shows that g in Sk,v(Γ)implies g(z) vanishes exponentially fast as z → i∞. Similar exponential decayhappens as z goes to any cusp of the fundamental region (as can be seen from theproof of Lemma 3.34). This shows that the integrant

f(z) g(z)= (z)k

is bounded on FΓ and vanishes exponentially fast as we tend to a cusp of thefundamental region.

For the full modular group Γ(1) we have the estimate

〈f, g〉Γ(1),k = O

(∫FΓ(1)

= (z)−2

dλ2(z)

)3.45= O (1) .

Using the linearity of the integral and FΓ given as in (2.14) as disjoint union oftransformations of FΓ(1), we calculate the hyperbolic area of the fundamental regionFΓ as

1

µ

∫FΓ

= (z)−2

dλ2(z)(2.14)

=

µ∑i=1

∫FAiΓ(1)

= (z)−2

dλ2(z)

(3.32)=

µ∑i=1

∫FΓ(1)

= (z)−2

dλ2(z)

= µ

∫FΓ(1)

= (z)−2

dλ2(z)

3.5. PETERSSON SCALAR PRODUCT AND THE HILBERT SPACE OF CUSP FORMS 65

= µπ

3.

Hence we have the estimate

〈f, g〉Γ,k = O (1) .

This proves the lemma.

Lemma 3.47. For (f, g) ∈Mk,v(Γ)× Sk,v(Γ) and M ∈ Γ we have(3.37)

1

µ

∫MF

f(z) g(z)= (z)k = (z)

−2dλ2(z) =

1

µ

∫Ff(z) g(z)= (z)

k = (z)−2

dλ2(z).

Proof. We have

1

µ

∫MF

f(z) g(z)= (z)k = (z)

−2dλ2(z)

=1

µ

∫Ff(Mz) g(Mz)= (Mz)

k = (Mz)−2

dλ2(Mz)

(3.32)=

1

µ

∫Ff(Mz) g(Mz)= (Mz)

k = (z)−2

dλ2(z)

(3.21)=

1

µ

∫Fv(M)j(M, z)k f(z) v(M)j(M, z)k g(z)

= (z)k

|j(M, z)|2k= (z)

−2dλ2(z)

=1

µ

∫Ff(z) g(z)= (z)

k = (z)−2

dλ2(z)

using the properties |j(M, z)|2k = j(M, z)k j(M, z)k and v(M) v(M) = |v(M)| = 1.(The last identity is due to the Definition 3.6 of multiplier systems). This proves(5.23).

Remark 3.48. Let FΓ be a fundamental region of Γ of the form given in (2.14):

FΓ =

µ⋃i=1

Ai FΓ(1)

with A1, . . . , Aµ a right coset representation of Γ in Γ(1). Let M ∈ Γ. Then thematrices A1M, . . . , AµM are also a right coset representation of γ in Γ(1). Thefundamental region F ′Γ associated to the new right coset representation is

F ′Γ =

µ⋃i=1

AiM FΓ(1).

Lemma 3.47 shows that the Petersson scalar product 〈·, ·〉Γ,k defined in (3.27) doesnot depend on the choice of the standard fundamental region FΓ as derived in(2.14).

Theorem 3.49. Let Γ ⊂ Γ(1) be a subgroup of finite index µ and let k ∈ Ra weight and v a multiplier with respect to (Γ, k). The Petersson scalar product〈·, ·〉Γ,k defined in (3.27) satisfies:

(1) 〈·, ·〉Γ,k is a scalar product: It is well defined and satisfies

〈αf, g〉Γ,k = α〈f, g〉Γ,k = 〈f, αg〉Γ,k(f ∈Mk,v(Γ), Sk,v(Γ), α ∈ C

),

〈f1 + f2, g〉Γ,k = 〈f1, g〉Γ,k + 〈f2, g〉Γ,k

and

〈f, g1 + g2〉Γ,k = 〈f, g1〉Γ,k + 〈f, g2〉Γ,kfor f, f1, f2 ∈Mk,v(Γ), g, g1, g2 ∈ Sk,v(Γ) and α ∈ C.

66 3. MODULAR FORMS

(2) We have

〈f, g〉Γ,k = 〈g, f〉Γ,kfor f, g ∈ Sk,v(Γ) both cusp forms.

Proof. The well-definiteness of the Petersson scalar product follows directlyfrom Lemma 3.46. It is independent of the choice of the fundamental region, as canbe seen from Remark 3.48. The linearity in the first component and anti-linearity inthe second follows directly from the linearity of integrals. The hermitian propertyalso follows directly from the definition in (3.27), assuming that both forms f, g arecusp forms.

Definition 3.50. A (complex) Hilbert space H is a vector space with a com-patible scalar product 〈·, ·〉 : H×H → C.

Corollary 3.51. Let Γ ⊂ Γ(1) be a subgroup of finite index µ and let k ∈ Ra weight and v a multiplier with respect to (Γ, k). The space Sk,v(Γ) is a Hilbertspace with the Petersson scalar product 〈·, ·〉Γ,k as scalar product.

Proof. The proof of Corollary 3.51 is left to the reader as an exercise.

CHAPTER 4

Existence of Modular Forms

In this chapter, we give examples of modular forms. We start by describing theEisenstein series (in §4.1) and the discriminant function (in §4.2). In §4.3 we showthat these two types of modular forms generate the space of entire modular formsfor the full modular group Γ(1). We also give some results about the dimensionof the space Mk

(Γ(1)

). The next section, §4.4, contains the construction of the

holomorphic Poincare series for the full modular group Γ(1). We also show thesefunctions form a basis for the space of cusp forms. In the last section, §4.5, weintroduce the Jacobi theta function and relate it to counting functions similar to r4

mentioned in the introduction of Part?? on page ?? and relate it to modular formsfor the theta group Γθ.

4.1. Eisenstein Series

Here, we describe Eisenstein series and prove that those series are actually entiremodular forms. We also give the Fourier expansion of these forms and observe thattheir coefficients are divisor functions. Using properties of Eisenstein series, wewill show in later sections that we will be able to determine arithmetic relationsbetween the divisor functions of given integers.

Definition 4.1. We define the prime summation∑′

(m,n) as follows:

(4.1)∑

(m,n)

′:=

∑(m,n)∈Z2

(m,n) 6=(0,0)

:=∑m∈Z

∑n∈Z

(m,n)6=(0,0)

which is a double summation over the non-vanishing pairs (m,n) ∈ Z26=(0,0).

Remark 4.2. The prime notation is a typical notation for Eisenstein seriesgenerated by summing a lattice. Typically, we sum over some expressions whichare not defined for a finite number of lattice points. The prime notation indicatesthat those lattice points are omitted in the summation.

Lemma 4.3. Let k > 2 be a weight. The double series∑(m,n)

′(mz + n)−k

converges absolutely for all z ∈ H.Moreover, for each α > 0 the convergence is uniform on the subset

(4.2) Eα :=

z ∈ H; |< (z)| ≤ 1

αand = (z) ≥ α

.

Remark 4.4. The second part of Lemma 4.3 implies that the convergence isuniform on all compact subsets of H.

Proof of Lemma 4.3. Let z ∈ H. We consider the lattice

Lz :=mz + n; m,n ∈ Z

⊂ C.

67

68 4. EXISTENCE OF MODULAR FORMS

Let ν ∈ Z≥0, and put

(4.3) πν(z) :=± νz + µ, µz ± ν; µ ∈ Z with |µ| ≤ ν

⊂ Lz.

Example 4.5. For example, it is easy to see that

π0(z) =

0,π1(z) =

z + 1, z − 1,−z + 1,−z − 1, z,−z, 1,−1

and(4.4)

π2(z) =

2z + 2, 2z + 1, 2z, 2z − 1, 2z − 2,−2z + 2,−2z + 1,−2z,−2z − 1,

− 2z − 2, z + 2, 2,−z + 2, z − 2,−2,−z − 2.

Note that πν(z) is a finite set. We have

#πν(z) ≤

8ν if ν > 0 and

1 if ν = 0.

The sets πν(z) form a tessellation of the lattice Lz: We have

Lz =⋃

ν∈Z≥0

πν(z)

andπν(z) ∩ πν′(z) = ∅ for all ν, ν′ ∈ Z≥0, ν 6= ν′.

Consider the (Euclidian) distance function

d(H,w) := infd(v, w); v ∈ H

,

where d(v, w) := |v − w| denotes the Euclidian distance between v and w in C andH ⊂ C is a set of points. Form (4.4) follows immediately d

(π1(z), 0

)> 0 since

z ∈ H always contains an imaginary part.

Example 4.6. Let z ∈ H. The construction of πν(z) implies in particular

d(π1(z), 0

)= inf

d(w, 0); w ∈ π1(z)

(4.4)= min

|z + 1| , |z − 1| , |z| , 1

> 0

since z ∈ H implies 0 < = (z) < min|z| , |z + 1| , |z − 1|

is strictly positive.

Exercise 4.7. Let z ∈ H be a element of the upper half-plane and ν ∈ N anatural number. Denote by par(ν, z) the parallelogram spanned by πν(z). Showthe following relations of the distance to 0 ∈ C:

(4.5) d(πν(z), 0

)≥ d(par(ν, z), 0

)> 0

and

(4.6) d(par(ν, z), 0

)= ν d

(par(1, z), 0

).

Exercise 4.7 leads immediately to the estimate

(4.7) d(πν(z), 0)(4.5)

≥ d(par(ν, z), 0

) (4.6)= ν d

(par(1, z), 0

).

Applying it to an element mz + n ∈ Lz ∩ πν(z), we have

|mz + n| ≥ d(πν(z), 0) ≥ ν d(par(1, z), 0

)= C ν

with C := d(par(1, z), 0

)being a positive constant depending only on z.

Combining the above arguments, we get∑(m,n)

′|nz +m|−k =

∞∑ν=1

∑mz+n∈πν(z)

|mz + n|−k(4.8)

≤∞∑ν=1

ν−k∑

mz+n∈πν(z)

C−k

4.1. EISENSTEIN SERIES 69

= C−k∞∑ν=1

ν−k(#πν(z)

)= C−k

( ∞∑ν=1

ν−k 8ν

)= 8C−k ζ(k − 1),

where ζ(s) :=∑∞n=1 n

−s denotes the Riemann zeta function and C is the constantabove. Since the series representation of the Riemann zeta function converges

absolutely for < (s) > 1, we find that∑′m,n |mz + n|−k converges absolutely for

k > 2.

Now, we show that the series converge uniformly on the sets Eα. It will besufficient to show that

(4.9) infz∈Eα

d(π1(z), 0

)=: ρα > 0

holds uniformly on Eα for some strictly positive constant ρα depending only on α.

Then, we have∑′

(m,n) |nz +m|−k uniformly bounded on Eα by (4.8): Using (4.7)we get

infz∈Eα

d(πν(z), 0

) (4.7)

≥ ν infz∈Eα

d(par(1, z), 0

) (4.9)= ν ρα.

Thus for z ∈ Eα, we have∑mz+n∈πν(z)

|mz + n|−k ≤ 8ν

(νρ)k= 8ρ−k ν1−k.

Plugging this into (4.8), we have∑(m,n)

′|mz + n|−k ≤ 8ρ−kα

∞∑ν=1

ν1−k = 8ρ−kα ζ(k − 1),

Using the Weierstrass M -test, see e.g. Appendix A.2, we get uniform convergencefor z ∈ Eα.

It remains to prove (4.9) which we leave to the reader as an exercise.

Exercise 4.8. Prove the estimate (4.9).

We now define Eisenstein series on the full modular group for weights k > 2and we show that those series are actually entire modular forms.

Definition 4.9. Let 2 < k ∈ 2N be a positive even integer weight.We define the Eisenstein series Gk of weight k by

(4.10) Gk : H→ C; z 7→ Gk(z) :=∑′

(m,n)

(mz + n)−k.

Remark 4.10. (1) Obviously, Lemma 4.3 implies that the series in thedefinition above converges absolutely for every fixed z and uniformly onall compact subsets of H.

(2) It is important to mention that Gk ≡ 0 for all odd k since the sum runsover all integers. (The terms mz + n and (−m)z + (−n) appear in theseries and cancel each other).

(3) Also, we have

limz→i∞

Gk(z) = 2ζ(k) 6= 0


for even integer weights k. Here, the limit z → i∞ is understood as= (z)→∞ and < (z) is bounded. This is because of the uniform conver-gence of the series on the set Eα, see Lemma 4.3. Also, the only terms inthe series that survive the limit z →∞ are the ones with m = 0.

Exercise 4.11. For 2 < k ∈ 2N being a positive even integer weight show

limz→i∞

Gk(z) = 2ζ(k)

where the limit z → i∞ is understood as = (z)→∞ and < (z) is bounded.

Theorem 4.12. Let k ∈ 2Z, k > 2. We have Gk ∈ Mk,1

(Γ(1)

)where the

multiplier 1 is the trivial multiplier 1(M) = 1 for all M ∈ Γ(1).

Proof. Using Lemma 4.3 and the fact that k > 2, we see that Gk(z) is welldefined and holomorphic in H.

We now show the transformation law of Gk(z). For M =

(α βγ δ

)∈ Γ(1) we

have the identity

m(Mz)+n = mαz + β

γz + δ+n =

mαz +mβ + nγz + nδ

γz + δ=

(mα+ nγ)z + (mβ + nδ)

γz + δ

for all z ∈ H. Hence, we have

Gk(M z

)=∑′

(m,n)

(m(Mz) + n)−k = (γz + δ)k

∑′

(m′,n′)

(m′z + n′)−k,

where m′ = mα + nγ and n′ = mβ + nδ. As (m,n) runs through all pairs inZ2 r

(0, 0)

so does (m′, n′).

We now determine the Fourier expansion of the Eisenstein series and we ob-serve that their Fourier coefficients are actually arithmetic functions that captureimportant arithmetic identities.

In order to do that, we need to present some basic preliminary results. Westart with the partial fraction decomposition of the cot-function.

Lemma 4.13. For z ∈ Cr R, we have the partial fraction decomposition

(4.11) π cot(πz)

=1

z+

∞∑m=1

(1

z +m+

1

z −m

).

Proof. Denote by

R(z) :=1

z+

∞∑m=1

(1

z +m+

1

z −m

)the right hand side of (4.11) and let

g(z) := π cot(πz)−R(z)

be the difference between the left hand side and the right hand side. Our aim is toprove that g(z) is an bounded entire function on C. Once we established that g isbounded on C, we can conclude by Liouville’s theorem that g is a constant function.Then, using the fact that cot and R are odd functions, we see immediately thatg(z) is an odd function, too. Hence, g satisfies g(z) = −g(−z) for all z ∈ C and wehave in particular g(0) = 0. This implies immediately g(z) ≡ 0.

To show that g is entire on C, observe that∞∑m=1

1

|z2 −m2|(z ∈ Z)


converges uniformly in any set of the form

Sρ,ε :=z ∈ C; |z| ≤ ρ, |z −m| ≥ ε for all m ∈ Z

.

Thus, R(z) is meromorphic in C and holomorphic in C r Z. Also, R has simplepoles at the integers with residue 1.

Now, since π cot(πz) is periodic of period 1, we need to show as well thatR(z + 1) = R(z). To do so, write

R(z + 1) = limk→∞

1

z + 1+

k∑m=1

(1

z + 1 +m+

1

z + 1−m

)

= limk→∞

1

z + 1+

(1

z + 2+

1

z

)+

(1

z + 3+

1

z + 1

)+ . . .

. . .+

(1

z + k + 1+

1

z − k + 1

)= limk→∞

1

z+

(1

z + 1+

1

z − 1

)+ . . .

. . .+

(1

z + k − 1+

1

z − k + 1

)+

(1

z + k + 1+

1

z + k

)= R(z) + lim

k→∞

(1

z + k + 1+

1

z + k

)= R(z).

Observe that this follows from reordering of some terms in the series definition ofR(z).

Recall that π cot(πz) has a pole of order 1 at all integers z ∈ Z with residue 1.As a result, we conclude that g(z) is an entire and periodic function on C.

What remains to show is that g(z) is bounded. To do that, we first showthat π cot(πz) is bounded in 0 ≤ < (z) ≤ 1 as = (z) approaches infinity. Writingπ cot(πz) in exponential terms, we have

π cot(πz) = iπe2πiz + 1

e2πiz − 1.

Notice that as = (z) → +∞, then π cot(πz) approaches −iπ and it approaches iπfor = (z)→ −∞. We still need to show that R(z) is also bounded. To do so, it willbe enough to show that

|z|∞∑m=2

1

|z2 −m2|

is bounded for 0 ≤ < (z) ≤ 1 and |= (z)| ≥ 1. We have that

|z|∞∑m=2

1

|z2 −m2|= |z|

∞∑m=2

1

|z −m| |z +m|

≤ |z|∞∑m=2

1

|z −m|2

= |z|∞∑m=2

1

(< (z)−m)2 + = (z)2

≤ |z|∞∑m=2

1

(1−m)2 + = (z)2


= |z|∞∑m=1

1

m2 + = (z)2

≤ |z|∫ ∞

0

1

t2 + = (z)2 dt

=|z|= (z)

[arctan

(t

= (z)

)]∞0

=π

2

|z|= (z)

,

which is bounded for all z with 0 ≤ < (z) ≤ 1 and |= (z)| ≥ 1. As a result, weget from the periodicity of g that g(z) is bounded for all z with |= (z)| ≥ 1. Thisextends to all z ∈ C by continuity of g.

Lemma 4.14. For z ∈ H we have

limd→∞

π cot(dπz)− limd→∞

π cot(−dπz) = −2πi

and ∑d∈Zr0,1

(π cot

(π(d− 1)z)− π cot(πdz)

)= 2π cot(πz) + 2πi

Proof. Recall the definition of the cot-function:

cot z =cos(z)

sin(z)= i

eiz + e−iz

eiz − e−iz.

This implies immediately that for z ∈ H ∪H− we have the limit

limd→∞

cot(dz) =

−i if = (z) > 0 and

i if = (z) < 0.

Hence, we have

limd→∞

π cot(dπz)− limd→∞

π cot(−dπz) = −πi− πi = −2πi.

Moreover, we have∑d∈Zr0,1

(π cot


)=∑d∈Z≥2

(π cot


)+

∑d∈Z≤−1

(π cot


)=

((π cot

(π1z

)− π cot(π2z)

)+(π cot

(π2z

)− π cot(π3z)

)+ . . .

. . .+(π cot

(π(−2)z)− π cot

(π(−1)z

))+

+(π cot

(π(−3)z

)− π cot

(π(−2)z

))+ . . .

)= π cot

(πz)

+ 0 + 0 + . . .+ 0− limd→∞

π cot(πdz)

− π cot(− πz

)+ 0 + 0 + . . .+ 0 + lim

d→∞π cot(−πdz)

= 2π cot(πz) + 2πi

using a telescope sum argument and the limit calculated above.


We now give the the expansion of a certain sum that will be used in determiningthe Fourier expansion of the Eisenstein series for weight k > 2.

Lemma 4.15. For k ∈ 2Z with k ≥ 2, we have∞∑

m=−∞

1

(z +m)k=

(−2πi)k

(k − 1)!

∞∑n=1

nk−1 e2πinz (z ∈ H).

Proof. For z ∈ H we have∣∣e2πiz

∣∣ < 1. Hence,

π cotπz = iπ

(1− 2

1− e2πiz

)= iπ − 2πi

∞∑n=0

e2πinz.

Using Lemma 4.13, we get

1

z+

∞∑m=1

(1

z +m+

1

z −m

)= −πi− 2πi

∞∑n=1

e2πinz.

Differentiating this expression we get

− 1

z2−∞∑m=1

(1

(z +m)2+

1

(z −m)2

)= −(2πi)2

∞∑n=0

ne2πinz.

So we get∞∑

m=−∞

1

(z +m)2= (2πi)2

∞∑n=0

ne2πinz

which proves the lemma for k = 2. If we differentiate k − 1 times instead of onlyonce, we get

∞∑m=−∞

1

(z +m)k=

(−2πi)k

(k − 1)!

∞∑n=0

nk−1e2πinz

proving the lemma to all integer k ≥ 2.

Theorem 3.39 in §3.4 shows the form of the Fourier expansions (at the cusp∞) for modular forms in Mk

(Γ(1)

). Together with Theorem 4.12, which shows

Gk ∈ Mk

(Γ(1)

), we find that the Eisenstein series Gk has a Fourier expansion at

the cusp ∞ given by

Gk(z) =

∞∑n=0

a(n) e2πinz (z ∈ H).

Using the factlimz→i∞

Gk(z) = 2ζ(k),

see Exercise 4.11, we find

Gk(z) = 2ζ(k) +

∞∑n=1

a(n)e2πinz.

In the following theorem, we determine the Fourier coefficients of Eisensteinseries of weight k > 2 in terms of the arithmetic divisor functions.

Theorem 4.16. The Fourier expansion of Gk(z) for 2 < k ∈ 2N is given by

Gk(z) = 2ζ(k) + 2(2πi)k

(k − 1)!

∞∑n=1

σk−1(n)e2πinz (z ∈ H)

where

(4.12) σk(n) =∑d|nd>0

dk


is the kth arithmetic divisor function.

Proof. For k > 2 being an even integer, we have

Gk(z) = 2ζ(k) + 2

∞∑m=1

∑n∈Z

(mz + n)−k

by simply rearranging the summation terms, see e.g. the calculation done in Exer-cise 4.11. Applying Lemma 4.15 to the inner sum with z′ = mz, we get

Gk(z) = 2ζ(k) + 2

∞∑m=1

(2πi)k

(k − 1)!

∞∑s=1

sk−1 e2πismz.

Let l = ms. We use absolute convergence and the equivalence d | l ⇐⇒ ld | l to

get

Gk(z) = 2ζ(k) + 2(2πi)k

(k − 1)!

∞∑n=1

∑m|n

( nm

)k−1

e2πinz

= 2ζ(k) + 2(2πi)k

(k − 1)!

∞∑n=1

σk−1(n) e2πinz.

Remark 4.17. It is important to note that, being inspired by Theorem 4.16,we might define a weight 2 Eisenstein series by

G2(z) = 2ζ(2) + 2(2πi)2∞∑n=1

σ1(n)e2πinz (z ∈ H).

However, we will show later that G2(z) is not a modular form by showing that theonly entire modular form of weight 2 on the full modular group is the constantfunction 0. In the following, we determine the transformation law of G2(z) underthe inversion z 7→ Tz = −1

z and conclude that G2(z) is not a modular form. Itsbehavior is known as being a rational period.

Definition 4.18. We define the function G2 as

(4.13) G2(z) := 2ζ(2) + 2(2πi)2∞∑n=1

σ1(n) e2πinz (z ∈ H).

Lemma 4.19. G2 is well defined on H (in the sense that the series expansion ofG2 in (4.13) converges absolutely for all z ∈ H) and forms a holomorphic functionon H.

Proof. We know that the series∑n∈N

(n+ 1)2 e2πinz

converges absolutely for all z ∈ H since∑n∈N

∣∣(n+ 1)2 e2πinz∣∣ =

∑n∈N

(n+ 1)2 e−2πn=(z) <∞

holds. Using the Gauß formula, we find the (crude) estimate

σ1(n) =∑

0<d|n

d ≤n∑d=1

d =n(n+ 1)

2≤ (n+ 1)2


for the divisor sum σ1(n). Combining this and the triangle inequality, we find forG2 in (4.18) the estimate

|G2(z)| ≤ |2ζ(2)|+ 8π2∞∑n=1

∣∣σ1(n) e2πinz∣∣

≤ 2ζ(2) + 8π2∞∑n=1

(n+ 1)2 e−2πn=(z) <∞

since = (z) > 0 for all z ∈ H and ζ(2) = π2

6 is positive, see e.g. [89, A013661].This implies that G2 in (4.18) is well defined (and its series definition converges

absolutely) on H; the function G2 : H→ C is holomorph.

Theorem 4.20. Consider G2 as defined above.

(1) G2 satisfies

(4.14) G2(z) =

( ∞∑m=−∞

∞∑n=−∞

)′1

(mz + n)2

which converges conditionally. Here,(∑∞

m=−∞∑∞n=−∞

)′means that we

calculate first the inner sum, which runs over n, and then the outer sum,which runs over m; the prime indicating that that the term m = n = 0 isskipped.

(2) We have the identity

(4.15) G2(z) =2πi

z+

( ∞∑n=−∞

∞∑m=−∞

)′1

(mz + n)2

similar to Theorem 4.16 for Eisenstein series Gk with k > 2. The series

in the right hand term converges conditionally. Here,(∑∞

n=−∞∑∞m=−∞

)′means that we calculate first the inner sum, which runs over m, and thenthe outer sum, which runs over n; the prime indicating that that the termm = n = 0 is skipped.

(3) G2 satisfies the transformation properties

(4.16) z−2G2

(−1

z

)= G2(z)− 2πi

z⇐⇒ G2

(T z)

= j(T, z)2G2(z)− 2πiz

and

(4.17) G2(z + 1) = G2(z)

for all z ∈ H.

Proof. First, we consider the definition of G2. Lemma 4.15 gives( ∞∑m=−∞

∞∑n=−∞

)′1

(mz + n)2=∑n∈Z6=0

1

n2+

∑m∈Z6=0

∑n∈Z

1

(mz + n)2

= 2ζ(2) +∑

m∈Z6=0

(−2πi)2∑s∈N

se2πismz

= 2ζ(2) + 2∑m∈N

(2πi)2∑s∈N

se2πismz.

As in the proof of Theorem 4.16, put l = ms and recall d | l ⇐⇒ ld | l. This gives

us ( ∞∑m=−∞

∞∑n=−∞

)′1

(mz + n)2= 2ζ(2) + 2

∑n∈N

(2πi)2∑s∈N

se2πismz


= 2ζ(2) + 2(2πi)2∑l∈N

∑d|l

d e2πilz

= 2ζ(2) + 2(2πi)2∑l∈N

σ1(l) e2πilz

= G2(z).

We just proved (4.14) of the theorem.

Define the auxiliary series H2 as follows

H2(z) :=

( ∞∑c=−∞

∞∑d=−∞

)]1

(cz + d)(cz + d− 1)(z ∈ H),

where the ] indicates that the terms (c, d) ∈

(0, 0), (0, 1)

are skipped. (Thismeans (c, d) 6= (0, 0) and (c, d) 6= (0, 1)). It is easy to see that the telescoping seriesvanishes for c 6= 0:

∞∑d=−∞

1

(cz + d)(cz + d− 1)=

∞∑d=−∞

1

cz + d− 1− 1

cz + d= 0 if c 6= 0.

The case c = 0 gives ∑d∈Zr0,1

1

d(d− 1)= 2

∞∑d=2

1

d(d− 1)= 2.

Hence, we have

(4.18) H2(z) = 2 (z ∈ H).

Next, define the second auxiliary series

H2(z) :=

( ∞∑d=−∞

∞∑c=−∞

)]1

(cz + d)(cz + d− 1)(z ∈ H)

where we interchanged the order of summation compared to H2. Again, the ] meansthat the terms (c, d) ∈

(0, 0), (0, 1)

are skipped.

We consider j(T, z)−2 H2

(T z):

z−2H2

(−1

z

)= z−2

( ∞∑d=−∞

∞∑c=−∞

)]1(

c−1z + d

) (c−1z + d− 1

)=

( ∞∑d=−∞

∞∑c=−∞

)]1(

dz − c)(

(d− 1)z − c)

=1

z

( ∞∑d=−∞

∞∑c=−∞

)](1

(d− 1)z − c− 1

dz − c

)for all z ∈ H. We calculate the sum at the right hand side above in three steps:

(1) For d /∈ 0, 1, we use Lemma 4.13 and find

∞∑c=−∞

1

dz − c= π cot(π dz).

Using Lemma 4.14 and a telescope summation argument we have

1

z

( ∞∑d=−∞

∞∑c=−∞

)](1

(d− 1)z − c− 1

dz − c

)


=1

z

∑d∈Zr0,1

(π cot

(π(d− 1)z

)− π cot(πdz)

)

=1

z

(π cot(πz) +

(π cot(−πz)

)+

1

zlimd→∞

π cot(−πdz)− 1

zlimd→∞

π cot(πdz)

=2

zπ cot

(πz)

+2πi

z.

(2) For d = 0, we have to evaluate 1z

∑c∈Z6=0

(1

−z−c −1−c

). Using symmetric

summation we find

1

z

∑c∈Z 6=0

(1

−z − c− 1

−c

)=

1

z

∑c∈N

(1

−z − c+

1

−z − (−c)+

1

−c+

1

c︸︷︷︸=0

)

=1

z

∑c∈N

(1

−z − c+

1

−z + c

).

Thus, Lemma 4.13 implies

1

z

∑c∈Z6=0

(1

−z − c− 1

−c

)=

1

z

(π cot(−πz)− 1

−z

)

= −π cot(πz)

z+

1

z2

for all z ∈ H.(3) For d = 1, we find with the same summation trick as above

1

z

∑c∈Z6=0

(1

−c− 1

z − c

)= −1

z

∑c∈Z 6=0

1

z − c.

Again, Lemma 4.13 implies

1

z

∑c∈Z 6=0

(1

c− 1

z − c

)=

1

z2− π cot(πz)

z.

Putting all the cases together, we get

z−2H2

(−1

z

)=

1

z

( ∞∑d=−∞

∞∑c=−∞

)]1(

(d− 1)z − c)(dz − c)

=2

zπ cot

(πz)

+2πi

z+ 2

(1

z2− π cot(πz)

z

)=

2

z2+

2πi

z

for all z ∈ H. Multiplying z2 and substituting z 7→ −−1z , we get

(4.19) H2

(−1

z

)= 2 + 2πiz ⇐⇒ H2(z) = 2− 2πi

z

for all z ∈ H.

Now, back to the function G2. Recall that we have already established (4.14).This implies that the difference between H2 and G2 can be calculated by

H2(z)−G2(z) =

( ∞∑c=−∞

∞∑d=−∞

)]1

(cz + d)(cz + d− 1)−

( ∞∑c=−∞

∞∑d=−∞

)′1

(cz + d)2


= −1 +

( ∞∑c=−∞

∞∑d=−∞

)](1

(cz + d)(cz + d− 1)− 1

(cz + d)2

)

= −1 +

( ∞∑c=−∞

∞∑d=−∞

)′1

(cz + d)2(cz + d− 1),

where the series on the right hand side converges absolutely.Similar to H2, define the function

G2(z) :=

( ∞∑n=−∞

∞∑m=−∞

)′1

(mz + n)2(z ∈ H)

which hast he summation order reversed compared to G2 in (4.14). Again, we have

H2(z)− G2(z) =

( ∞∑d=−∞

∞∑c=−∞

)]1

(cz + d)(cz + d− 1)−

( ∞∑d=−∞

∞∑c=−∞

)′1

(cz + d)2

= −1 +

( ∞∑d=−∞

∞∑c=−∞

)](1

(cz + d)(cz + d− 1)− 1

(cz + d)2

)

= −1 +

( ∞∑d=−∞

∞∑c=−∞

)′1

(cz + d)2(cz + d− 1).

Comparing both differences, we find the identity

H2(z)−G2(z) = H2(z)− G2(z)

for all z ∈ H. On the other hand, (4.18) and (4.19) imply

H2(z)− H2(z) = 2−(

2− 2πi

z

)=

2πi

z

for all z ∈ H. Combining both relations gives

G2(z)− G2(z) = H2(z)− H2(z) =2πi

z.

This shows relation (4.15).

To complete the proof we look at z−2G2

(− 1z

)for z ∈ H: we have

z−2G2

(−1

z

)= z−2

( ∞∑m=−∞

∞∑n=−∞

)′1(

−mz + n)2

=

( ∞∑m=−∞

∞∑n=−∞

)′1

(nz −m)2

=

( ∞∑n=−∞

∞∑m=−∞

)′1

(mz + n)2

= G2(z) = G2(z)− 2πi

z.

This proves (4.16).The second transformation property follows immediately from the Definition

of G2: Since e2πin = 1 for all n ∈ N, we have

G2(z + 1)(4.13)

= 2ζ(2) + 2(2πi)2∞∑n=1

σ1(n) e2πin(z+1)

4.2. THE DISCRIMINANT FUNCTION ∆(z) 79

= 2ζ(2) + 2(2πi)2∞∑n=1

σ1(n) e2πinz eeπin︸︷︷︸=1

= 2ζ(2) + 2(2πi)2∞∑n=1

σ1(n) e2πinz

(4.13)= G2(z).

Exercise 4.21. Show that G2 satisfies the limit

(4.20) lim=(z)→∞

G2(z) = 2ζ(2) =π2

3.

Moreover, there exists δ > 0 such that

(4.21) |G2(z)− 2ζ(2)| = O(e−2πδ=(z)

)holds as = (z)→∞.

We conclude the section on Eisenstein series’ by defining the normalized version.

Definition 4.22. Let k ∈ 2Z, k > 2 be a positive even weight. We define thenormalized Eisenstein series

(4.22) Ek(z) =∑′

c∈Z≥0, d∈Zgcd(c,d)=1c=0⇒d=1

(cz + d)−k

for all z ∈ H.

In fact, we have that

(4.23) Ek(z) =1

2ζ(k)Gk(z).

Exercise 4.23. Show (4.23).

It follows from Theorem 4.16 for k ∈ 2N, k ≥ 4 and from Definition 4.18 fork = 2, that Ek(z) has a Fourier expansion at infinity of the form

(4.24) Ek(z) = 1 +(2πi)k

(k − 1)! ζ(k)

∞∑n=1

σk−1(n) e2πinz

for all z ∈ H. The Fourier expansion at infinity in (4.24) is valid for all k ∈ 2N.

4.2. The Discriminant Function ∆(z)

In this section, we define a famous function called the discriminant function ∆which is a cusp form of weight 12 for the full modular group.

Definition 4.24. Define the discriminant function ∆ : H→ C by

(4.25) ∆(z) := e2πiz∞∏n=1

(1− e2πinz

)24(z ∈ H).

It is clear that this product converges uniformly on compact subsets of H. Theproduct expansion also implies immediately that ∆ is translation invariant:

∆(z + 1) = ∆(z) for all z ∈ H.The following theorem shows that ∆(z) is a cusp form of weight 12 and of trivialmultiplier system 1 on the full modular group Γ(1).


Theorem 4.25. We have ∆ ∈ S12,1

(Γ(1)

).

Proof. Clearly, the leading factor e2πiz in (4.25) implies

limz→i∞

∆(z) = 0

where the limit z → i∞ is understood as = (z)→∞ and < (z) is bounded.Since ∆ is 1-periodic and vanishes at the cusp ∞, we have that ∆ admits a

Fourier expansion of the form

(4.26) ∆(z) =

∞∑n=1

τ(n) e2πinz (z ∈ H)

with Fourier coefficients τ(n). (Use the same arguments as in the proof of Theo-rem 3.22 establishing the expansion at the cusp ∞).

To show that ∆ satisfies the transformation law (3.3) for weight 12, trivialmultiplier 1 and the full modular group, we like to apply Theorem 3.18. In orderto do so, we have to show that ∆ satisfies

∆

(−1

z

)= z12 ∆(z) ⇐⇒ ∆

(T z)

= j(T, z)12 ∆(z) (z ∈ H).

We consider the logarithmic derivative of ∆ given by

d

dz

(log ∆(z)

)=

∆′(z)

∆(z).

Using the Taylor expansion

− log(1− x) =

∞∑k=1

xk

k

of the logarithm, we have

log ∆(z) = 2πiz + 24

∞∑n=1

log(1− e2πinz

)= 2πiz − 24

∞∑n=1

∞∑k=1

1

ke2πinkz

= 2πiz − 24

∞∑k=1

1

k

∞∑n=1

e2πinkz.

As a result, we get that

∆′(z)

∆(z)= 2πi− 24

∞∑k=1

1

k

∞∑n=1

(2πink) e2πinkz

= 2πi

(1− 24

∞∑l=1

σ1(l) e2πilz

).

Recalling the value ζ(2) = π2

6 , see e.g. [89, A013661], we have from Definition 4.18

G2(z) =π2

3+ 2(2πi)2

∞∑n=1

σ1(n) e2πinz.

Comparing this to the above result, we get

∆′(z)

∆(z)=

6i

πG2(z).

4.3. APPLICATIONS TO Mk,1

(Γ(1)

)AND Sk,1

(Γ(1)

)81

Using Theorem 4.20, we get for the logarithmic derivative of ∆(Tz)

∆′(− 1z

)∆(− 1z

) ==6i

πG2

(−1

z

)=

6i

π

(z2G2(z)− 2πiz

)= z2 ∆′(z)

∆(z)+ 12z.

This implies

z−2 ∆′(− 1z

)∆(− 1z

) =∆′(z)

∆(z)+

12

z.

Integrating both sides, we get

log ∆

(−1

z

)= log ∆(z) + 12 log z + k

for some integration constant k ∈ C. Thus,

∆

(−1

z

)= ek z12∆(z) ⇐⇒ ∆

(T z)

= ek j(T, z)12 ∆(z)

for all z ∈ H. Inserting the special value z = i on both sides yields ek = 1 and weget the desired result.

Remark 4.26. The coefficients τ(n) of the Fourier expansion of the discrimi-nant function ∆ in (4.26) form a function

τ : N→ Z; n 7→ τ(n).

This function is known as the Ramanujan τ function, which is an importantnumber-theoretic function. By construction, we have τ(1) = 1 since only one terme2πiz appears in the product definition of δ in (4.25). Some of its values are listedin [89, A000594].

4.3. Applications To Mk,1

(Γ(1)

)And Sk,1

(Γ(1)

)Throughout this section, and for practical purposes, we will denote Sk,1(Γ) and

Mk,1(Γ) by Sk(Γ) and Mk(Γ)) respectively, see e.g. Remark 3.28.The plan for this section is to determine the dimensions of the spaces of cusp

forms and entire forms of integer weights on the full group with trivial multiplierv ≡ 1. We start by giving the dimension of the spaces of cusp forms and entireforms for small weights.

Using Theorem 4.16, we have that the Eisenstein series Gk, with even integerk ≥ 4, satisfies

Gk ∈Mk

(Γ(1)

)r Sk

(Γ(1)

).

As a result, we get that

dimMk

(Γ(1)

)≥ 1.

Lemma 4.27. Assume that Sk(Γ(1)

)is finite dimensional. For even integer

k ∈ 2Z with k ≥ 4 we have

(4.27) dimMk

(Γ(1)

)= 1 + dim Sk

(Γ(1)

).

Proof. Let S = g1, g2, . . . , gm be a basis for Sk(Γ(1)

). We have to show

that

S′ := g1, g2, . . . , gm, Gkis a basis for Mk

(Γ(1)

).


To show that elements of S′ are linearly independent, we let α1, α2, . . . , αm, βbe complex numbers such that

α1g1 + α2g2 + . . .+ αmgm + βGk ≡ 0.

Since each of the gi’s is a cusp form, we get

βGk(z) = −α1g1(z)− α2g2(z)− . . .− αmgm(z)

for all z ∈ H. Now, take the limit z → i∞ in the sense of = (z) → ∞ and < (z)bounded. Using Exercise 4.11 and Definition 3.26 of cusp forms we get

0 = − limz→i∞

α1g1(z) + . . .+ αmgm(z) = limz→i∞

βGk(z) = 2βζ(k).

Since ζ(k) 6= 0 for positive even integer k we get β = 0. Recall that the elements ofthe set S are a basis of Sk

(Γ(1)

)which implies that the gl’s are linearly independent.

Hence,α1 = α2 = . . . = αm = 0 = β.

We still need to show that S′ spans Mk

(Γ(1)

). So let f ∈ Mk

(Γ(1)

). By

Definition 3.26 f has a Fourier expansion

f(z) = c(0) +

∞∑n=1

c(n)e2πinz.

This implies that the limit

limz→∞z∈FΓ(1)

f(z)− c(0)

2ζ(k)Gk(z) = 0.

Hence, the difference function f(z)− c(0)2ζ(k)Gk(z) is a cusp form which is given by a

linear combination of the elements in the basis S:

f(z)− c(0)

2ζ(k)Gk(z) =

m∑i=1

αigi(z) (z ∈ H)

for a suitable choice of coefficients α1, . . . , αm ∈ C, which do not all vanish. Thisshows that any function f ∈ Mk

(Γ(1)

)is given as a linear combination of the

elements of S′.

We now determine the dimension of the space of entire and cusp forms forcertain integer weights.

Proposition 4.28. For k ∈ 2, 4, 6, 8, 10 we have

dim Sk(Γ(1)

)= 0.

For k = 12, we havedim Sk

(Γ(1)

)= 1.

Proof. Let f(z) ∈ Sk(Γ(1)

)be a cusp form. Since f satisfies the transforma-

tion law for weight k and ∆ in (4.25) for weight 12, we find

f(M z

)∆(M z

) =j(M, z)k f(z)

j(M, z)12 ∆(z)= j(M, z)k−12 f(z)

∆(z)

for all M ∈ Γ(1) and z ∈ H. Moreover,z → f(z)∆(z) is holomorphic in H and regular

at the infinite cusp. The first follows from the product definition of ∆ in (4.25)since ∆(z) 6= 0 for all z ∈ H. The second follows from the Fourier expansions ofboth functions at ∞, the constant terms of f and ∆ vanish and the order of thefirst non-vanishing term of f and ∆ is ≤ 1 respectively 1. We just showed that

f(z)

∆(z)∈Mk−12

(Γ(1)

)


(Γ(1)

)AND Sk,1

(Γ(1)

)83

holds.Consider the case k ∈ 2, 4, 6, 8, 10 and observe that k − 12 < 0 is negative.

Using Proposition 3.41, we get that

f(z)

∆(z)= 0

for all z ∈ H. Hence, f ≡ 0 and Sk(Γ(1)

)= 0. This shows that the dimension of

the space of cusp forms is zero:

dim Sk(Γ(1)

)= dim 0 = 0.

Consider k = 12. Similarly as above, we show

f(z)

∆(z)∈M0

(Γ(1)

).

Using Corollary 3.33 we have that f(z) and ∆(z) are proportional for all z ∈ H.Hence, there exists an c ∈ C with f = c∆:

S12

(Γ(1)

)= 〈∆〉.

This shows that the dimension of the space of cusp forms is one:

dim S12

(Γ(1)

)= dim 〈∆〉 = 1.

Exercise 4.29. Show that 20G34 − 49G2

6 ∈ S12

(Γ(1)

).

We will show that there are no entire modular forms of weight 2 on the fullgroup with trivial multiplier system. In order to do so, we need to introduce whatis known as abelian integrals.

Definition 4.30. Let F : H → C be a meromorphic function in the upperhalf plane and meromorphic at the infinite cusp. Then, F (z) is called an abelianintegral if there exists a sequence of numbers pM ∈ C, M ∈ Γ(1), such that

(4.28) F(M z) = F (z) + pM

holds for all M ∈ Γ(1).

We now show that the antiderivative of a modular form of weight 2 on the fullmodular group is an abelian integral.

Definition 4.31. Let f : H→ C have a Fourier expansion at ∞ of the form

f(z) =

∞∑n=−µ

a(n)e2πinz (z ∈ H).

The antiderivative of f is defined as

F (z) =a(−µ)

(−2πiµ)e−2πiµz + . . .+

a(−1)

(−2πi)e−2πiz + a(0)z +

∞∑n=1

a(n)

(2πin)e2πinz

for all z ∈ H.

Lemma 4.32. Let f ∈M !2

(Γ(1)

)and assume additionally that f(z) is holomor-

phic in H. Write its Fourier expansion at the cusp ∞ as

f(z) =

∞∑n=−µ


The antiderivative F of f given in Definition 4.31 is an abelian integral.


Proof. Since f ∈M !2

(Γ(1)

)by assumption, we get

j(M, z)−2 f(M z

)− f(z) = 0

for all M ∈ Γ(1) and z ∈ H. Denote by F the antiderivative of f . The definitiondirectly implies

d

dzF (z) = f(z).

Using Chain ruled

dzF(M z

)= j(M, z)−2 F ′

(M z

)where F ′(z) = d

dzF (z) gives

(4.29)d

dzF(M z

)− d

dzF (z) = 0.

Integrating both sides of (4.29), we get

F(M z

)− F (z) = pM (z ∈ H)

for suitable constants pM ∈ C and for all M ∈ Γ(1).

Finally, using the above lemma, we can show that the dimension of the spaceof entire forms of weight 2 on the full group and of multiplier system is 0.

Proposition 4.33. We have dim M2

(Γ(1)

)= 0.

Proof. Let f ∈ M2

(Γ(1)

)be a modular form of weight 2. Consider the

antiderivative F (z) of f(z). Lemma 4.32 shows that we have

F(T z)

= F (z) + pT

for all z ∈ H. Now apply T again; we use on one hand T 2 = −1 and on the otherhand above relation for TZ and z respectively to get

F (z)TT z=z

= F(TT z

)(4.28)

= F(T z)

+ pT

(4.28)= F (z) + 2pT

for all z ∈ H. As a result, we get pT = 0.Apply the same arguments using the matrix ST three times – since (ST )3 = −1

– we get also pST = 0.Consider now F (z+ 1) = F (Sz). By inserting the identity −1 = TT and using

the above cases we get

F(S z)

= F(STT z

)= F

((ST ) (T z)

)= F

(T z)

= F (z).

Hence, F is periodic. This periodicity combined with the expansion of the abelianintegral F (z), see Lemma 4.32, and the expansion of the antiderivative F in Def-inition 4.31, shows that the term a(0) must vanish; the function F would not beperiodic otherwise.

Using again that F is invariant under S and T together with its vanishingat the cusps, we see that F is a modular function (which vanishes at the cusps).Corollary 3.33 implies that F is a constant function which has to be constant zero:

F (z) = 0

for all z ∈ H. Hence, f was already the zero-function. This shows

M2

(Γ(1)

)= 0

and proves the proposition.


(Γ(1)

)AND Sk,1

(Γ(1)

)85

Corollary 4.34. For k = 14 we have

dim Sk(Γ(1)

)= 0.

Proof. Recall the inclusion S14

(Γ(1)

)⊂ M14

(Γ(1)

). Using Proposition 4.33

we find

0 ≤ dim S14

(Γ(1)

)≤ dim M14

(Γ(1)

)= 0.

This proves the corollary.

Definition 4.35. We define the Gauß brackets [·] : R → Z as the functionassigning each real number largest not-larger integer:

(4.30) [x] := maxn ∈ Z; n ≤ x

(x ∈ R).

Remark 4.36. The Gauss brackets [·] can be characterized by

[x] =n ∈ Z; x− 1 < n ≤ x

.

It is also known as the floor -function.

Theorem 4.37. Let k ∈ Z≥0. The dimension of the space of entire modularforms for the full modular group and weight k is

(4.31) dimMk

(Γ(1)

)=

[k12

]for k ≡ 2 mod 12 and[

k12

]+ 1 for k 6≡ 2 mod 12.

Proof. For 0 ≤ k ≤ 14 the result follows from Propositions 4.28, 4.33 andCorollary 4.34.

Now for k ≥ 14, we prove the theorem by induction. Assume that (4.31) holdsfor all weights less than k. Consider the map

(4.32) H : Sk(Γ(1)

)→Mk−12

(Γ(1)

); F (z) 7→

(H(F )

)(z) :=

F (z)

∆(z)(z ∈ H).

It is straightforward to show that H is a vector space isomorphism. Therefore, wehave

dim Sk(Γ(1)

)= dim Mk−12

(Γ(1)

)=

[k12

]− 1 for k ≡ 2 mod 12 and[

k12

]for k 6≡ 2 mod 12,

where the first equality is due to the vector space isomorphism H and the secondequality is due the induction assumption. Using Lemma 4.27, we get from theabove line

dim Mk

(Γ(1)

)= 1 + dim Sk

(Γ(1)

)= 1 +

[k12

]− 1 for k ≡ 2 mod 12 and[

k12

]for k 6≡ 2 mod 12,

=

[k12


k12

]+ 1 for k 6≡ 2 mod 12.

Theorem 4.38. The space of modular forms of weight k and trivial multiplier1 for the full modular group with k ∈ 2N is spanned by the modular forms Gm4 G

n6

with 4m+ 6n = k:

Mk

(Γ(1)

)=⟨Gm4 Gn6 ; 4m+ 6n = k, n,m ∈ Z≥0

⟩.


Proof. We use an induction in weight k. We already know that

dimMk

(Γ(1)

)=

0 if k < 0 or k = 2,

1 if k = 0,

see Propositions 3.41, 4.33 and Corollary 3.33. The last mentioned corollary implies,in particular, M0

(Γ(1)

)= 〈1〉 contains only constant functions.

Suppose that k > 2 and F ∈ Mk

(Γ(1)

). First, observe that any even integer

k > 2 may be written as 4m + 6n for m,n ∈ Z≥0. Theorem 4.16 shows that G4

and G6 have the non-vanishing constant term 2ζ(4) respectively 2ζ(6) as 0th termin their Fourier expansion.

Consider the expansion of F at the cusp ∞ by Theorem 3.39 and denote the0th coefficient with a(0). The difference function

h := F − a(0)Em4 En6

has no constant term in its expansion at ∞. We use the normalized Eisensteinseries Ek which has a 1 as constant term in its Fourier expansion, see (4.24) and(4.23) for its relation to Gk. Hence, h ∈ Sk

(Γ(1)

)is a cusp form by Definition 3.26.

Using the vector space isomorphism H in (4.32) in the proof of Theorem 4.37 shows

g :=h

∆=F − a(0)Em4 En6

∆∈Mk−12

(Γ(1)

)is an entire modular form of weight k− 12. By induction, g is a linear combinationof the modular forms Ej4 E

l6 with 4j+6l = k−12. Since 20E3

4−49E26 ∈ S12

(Γ(1)

),

see Exercise 4.29, ∆ ∈ S12

(Γ(1)

), see Theorem 4.25 and dimS12

(Γ(1)

)= 1, see

Proposition 4.28, we conclude that ∆ is given by a linear combination of E34 and

E26 . Using h = g∆, we see that h is a sum of monomials of the form Ep4 E

q6 with

4p+ 6q = k. Hence, F is of the same form. The theorem follows since G4 and G6

are scalar multiples of E4 and E6, see (4.23).

To conclude the section, we will give a bound on the Fourier coefficients of entiremodular forms which improves the bond given in Theorem 3.39. The key step is thatthe following proposition determines a bound for the k-divisor arithmetic functions.

Proposition 4.39. For k ∈ Z, k ≥ 3, we have

nk−1 ≤ σk−1(n) ≤ ζ(k − 1)nk−1

for all n ∈ N.

Proof. Recall

σk−1(n) =∑d|n

dk−1 ≥ nk−1.

On the other hand,

σk−1(n)

nk−1=∑d|n

(d

n

)k−1

=∑δ|n

1

δk−1≤ ζ(k − 1).

Corollary 4.40. For even k ∈ Z≥4, let f(z) ∈ Mk

(Γ(1)

)be a modular form

of non-negative weight k ≥ 4. Assume that f has an expansion at infinity given by

F (z) =

∞∑n=0

a(n) e2πinz.

Then, we have

a(n) = O(nk−1

)(n ∈ N).

4.4. HOLOMORPHIC POINCARE SERIES 87

Proof. For even k ∈ Z≥4, we know by Lemma 4.27 that the space of entireforms is spanned by the Eisenstein series Gk and cusp forms. Hence, f ∈Mk

(Γ(1)

)can be written as

f(z) = βGk(z) + g(z) (z ∈ H)

with β ∈ C and g ∈ Sk(Γ(1)

).

Proposition 4.39 shows that the Fourier coefficients σk−1(n) of Gk satisfy

σk−1(n) = O(nk−1

)(n ∈ N),

whereas Theorem 3.36 implies that the Fourier coefficients b(n) of the expansion ofg at ∞

g(z) =

∞∑n=1

b(n) e2πinz (z ∈ H)

satisfy

b(n) = O(nk2

)(n ∈ N).

Hence, the coefficients a(n) of f satisfy

a(n) = O(

maxnk−1, n

k2

)= O

(nk−1

)(n ∈ N).

The dimension formula of the cusp spaces Sk(Γ(1)

)for small k gives rise to

interesting identities on the level of Eisenstein series.Now, Proposition 4.28 and Corollary 4.34 immediately imply the identities

given in the following example.

Example 4.41. (1) E24(z)− E8(z) ∈ S8

(Γ(1)

)implies

(4.33) E24(z) = E8(z) (z ∈ H).

(2) E4(z)E6(z)− E10(z) ∈ S10

(Γ(1)

)implies

(4.34) E4(z)E6(z) = E10(z) (z ∈ H).

(3) E4(z)E10(z)− E14(z) ∈ S14

(Γ(1)

)implies

(4.35) E4(z)E10(z) = E14(z) (z ∈ H).

4.4. Holomorphic Poincare series

Recall the normalized Eisenstein series Ek in Definition 4.22, with k ≥ 4 aneven integer. Part of the defining formula involves a sum over all coprime pairs(c, d) ∈ Z2

6=(0,0), gcd(c, d) = 1. Using Proposition 2.31 we might write the sum over

all coprime pairs as a sum over the quotient set Γ∞\Γ(1).

Lemma 4.42. (1) Let c ∈ Z≥0, d ∈ Z such that gcd(c, d) = 1. Assumethat we have two matrices M(c,d),M

′(c,d) ∈ SL2 (Z) with the same given

lower row (c, d). Then both matrices differ by a power of S: M ′(c,d) =

SnM(c,d) for some n ∈ Z.(2) A complete set of representatives of the right cosets

Γ∞\Γ(1) =

[(? ?c d

)]Γ(1)∞

; c ∈ Z≥0, d ∈ Z, (c, d) 6= (0, 0) and gcd(c, d) = 1

is given by

(4.36)

(? ?0 1

),

(? ?c d

); c ∈ N, d ∈ Z and gcd(c, d) = 1

.


(3) Another complete set of representatives of the right cosets Γ∞\Γ(1) isgiven by

(4.37)

(? ?c d

)∈ Γ(1); c ∈ Z, −d ∈ N, and gcd(b, d) = 1

∪(

? ?1 0

).

Proof. (1) First, we show that for each coprime (c, d) ∈ Z6=(0,0) thereexists a, b ∈ Z such that(

a bc d

)∈ Γ(1).

This follows immediately from the Chinese remainder theorem, see e.g.Appendix A.3. Since left-multiplication by S given in (2.8) does notchange the lower row,

S

(a bc d

)=

(a+ c b+ dc d

),

we have Sn(a bc d

); n ∈ Z

⊂ Γ(1).

Moreover, the values a and b of the upper row are only determined moduloc respectively d. This shows the first part of the lemma.

(2) Using Proposition 2.32 to calculate the stabilizer Γ∞ = 〈S,−1〉 of thecusp ∞ of Γ(1), we find

Γ∞

(? ?c d

)⊂ Γ(1)

and (Γ∞

(? ?c d

))∩(

Γ∞

(? ?c′ d′

))= ∅

for all (c, d), (c′, d′) ∈ Z6=(0,0), (c, d) 6= ±(c′, d′).On the other hand, each element of Γ(1) has coprime entries (c, d) in

its lower row. Hence, it belongs uniquely to the left coset Γ∞

(? ?c d

)up

to a common sign.This shows that Γ(1) can be written as the disjoint union

Γ(1) = Γ∞

(? ?0 1

)∪

⋃(c,d)

c∈N, d∈Zgcd(c,d)=1

Γ∞

(? ?c d

)

of left cosets and proves the second part of the lemma.(3) The third part of the lemma is left to the reader as an exercise.

Using the above lemma, we can rewrite the normalized Eisenstein series Ek,k ≥ 4, in Definition 4.22 as

(4.38) Ek(z)(4.22)

=∑

c∈Z≥0, d∈Zgcd(c,d)=1c=0⇒d=1

′(cz + d)−k =

∑M∈Γ∞\Γ(1)

j(M, z)−k

for all z ∈ H. This simple calculation shows that Ek is in fact given by summingover a set of representatives of the left cosets Γ∞\Γ(1).

This leads directly to the principle of defining Poincare series below by replacingthe automorphic factor j(M, z) with other suitable functions.


Definition 4.43. Let m ∈ Z≥0 and k ∈ N even, k > 2. The series

(4.39) Pm,k(z) :=∑

M∈Γ∞\Γ(1)

j(M, z)−k e2πimMz (z ∈ H)

is called the mth Poincare series of weight k.

The definition of Pm,k above raises two immediate questions: What freedomdo we have in the choice of M? Is Pm,k well defined? The answers are given in thefollowing two lemmas.

Lemma 4.44. Let c, d ∈ Z be coprime: gcd(c, d) = 1. Then, Vc,d =

(? ?c d

)∈

Γ(1) exists and is uniquely defined up to left multiplication by powers of S.

Proof. By the Chinese remainder theorem, see A.3, there exists a, b ∈ Z such

that ad − bc = 1. This means that

(a bc d

)∈ SL2 (Z). Putting Vc,d :=

(a bc d

)proves the existence of Vc,d.

Assume that there exists another element

(a′ b′

c d

)∈ SL2 (Z). The determi-

nant conditions det

(a bc d

)= 1 =

(a′ b′

c d

)implies(

a bc d

)(a′ b′

c d

)−1

=

(a bc d

)(d −b′−c a′

)=

(1 a′b− ab′0 1

)= Sa

′b−ab′

with a′b − ab′ ∈ Z. Both matrices

(a bc d

)and

(a′ b′

c d

)are equal up to left

multiplication by a power of S.

Lemma 4.45. For m ∈ Z≥0, k ∈ 2N, k ≥ 4, the function Pm,k in (4.39) is welldefined.

Proof. Since Lemma 4.3 shows that the double series∑m,n′ |mz + n|−k con-

verges absolutely, we have

|Pm,k(z)| =

∣∣∣∣∣∣∑

M∈Γ∞\Γ(1)

j(M, z)−k e2πimMz

∣∣∣∣∣∣4.42,4.44

=

∣∣∣∣∣∣∣∣∣∣∑

c∈Z≥0,d∈Zgcd(c,d)=1c=0⇒d=1

′ e2πim(Vc,dz

)(cz + d)k

∣∣∣∣∣∣∣∣∣∣≤

∑c∈Z≥0,d∈Zgcd(c,d)=1c=0⇒d=1

′

∣∣∣∣∣∣e2πim

(Vc,dz

)(cz + d)k

∣∣∣∣∣∣≤


′ ∣∣(cz + d)−k∣∣ ≤ ∑

c,d∈Z

′|cz + d|−k <∞

for all z ∈ H.Lemma 4.44 implies that Vc,d is unique up to left multiplication by powers of

S. Plugging in SlVc,d, l ∈ Z into (4.39) instead of Vc,d gives


′ e2πim(SlVc,dz

)(cz + d)k

=∑


′ e2πim(Vc,dz+l

)(cz + d)k


=∑


′ e2πim(Vc,dz

)e2πiml

(cz + d)k

=∑


′ e2πim(Vc,dz

)(cz + d)k

(using e2πiml = 1)

=∑

M∈Γ∞\Γ(1)

j(M, z)−k e2πimMz (using 4.42 and 4.44)

= Pm,k(z, ν).

Hence, Pm,k is well defined and independent of the particular choice of thecoset representatives Vc,d of Γ∞\Γ(1).

We saw in the proof above that we can express Pm,k as a series over the pairs(c, d); c ∈ Z≥0, d ∈ Z, gcd(c, d) = 1, c = 0 ⇒ d = 1

. This is summarized in the

next lemma.

Lemma 4.46. For m ∈ Z≥0 and k ∈ 2N, k ≥ 4 the Poincare series Pm,k hasthe expansion

(4.40) Pm,k(z) =∑


′ e2πimVc,dz

(cz + d)k

for all z ∈ H.

Proof. It follows directly from rewriting the index set M ∈ Γ∞\Γ(1) used inthe defining Equation (4.39) using Lemma 4.42 and Vc,d defined in Lemma 4.44.

Lemma 4.47. Let m ∈ Z≥0 and k ∈ 2N, k ≥ 4. The Poincare series Pm,k(z)satisfies modular transformation law (3.3) for weight k and trivial multiplier v = 1:

Pm,k∣∣kM = Pm,k for all M ∈ Γ(1).

Proof. Using Lemma 4.46, we rewrite the definition of Pm,k in (4.39) and get

Pm,k(z) =∑


′(cz + d)−k e2πimVc,dz (z ∈ H)

where Vc,d =

(a bc d

)∈ SL2 (Z) is a matrix with given second row (c, d), see

Lemma 4.42. Lemma 4.45 shows that Pm,k is well defined. Moreover, the aboveseries expression shows immediately that Pm,k(z) is invariant under z 7→ z + 1:

Pm,k(z + 1) =∑


′(c(z + 1) + d

)−ke2πimVc,d(Sz)

=∑


′(cz + c+ d)−k e

2πim

a a+ bc c+ d

z


=∑


′(cz + c+ d)−k e2πimVc,c+dz

(c,c+d)7→(c,d)=


′(cz + d)−k e2πimVc,dz

= Pm,k(z),

using Vc,dS =

(a bc d

)(1 10 1

)=

(a a+ bc c+ d

)= SlVc,c+d for some l ∈ Z, which

implies the equality e2πimSlVc,c+dz = e2πimVc,c+dz. Also, we used the set identity(c, c+ d); c ∈ Z≥0, d ∈ Z, gcd(c, d) = 1, c = 0⇒ d = 1

=

(c, d); c ∈ Z≥0, d ∈ Z, gcd(c, d) = 1, c = 0⇒ d = 1

for the last step.

Now, consider Pm,k(Tz). We have

Pm,k(Tz) =∑


′(cTz + d

)−ke2πimVc,d(Tz)

(3.7)= j(T, z)k


′j(Vc,dT, z)

−k e2πimVc,dT z,

where we used Lemma 3.3 with(cTz+d

)−k= j(Vc,d, T z

)−k (3.7)=

j(Vc,dT,z)−k

j(T,z)−k. Since

Vc,dT =

(a bc d

)(0 −11 0

)=

(b −ad −c

),

we have

j(T, z)−k Pm,k(Tz) =∑


′j(Vc,dT, z

)−ke2πimVc,dT z

=∑


′j

((b −ad −c

), z

)−ke

2πim

b −ad −c

z.

Using the third statement of Lemma 4.42, wee see that the series on the left handside above runs again over a complete set of representatives of the cosets in Γ∞\Γ(1).We have

j(T, z)−k Pm,k(Tz) =∑

c∈Z≥0,d∈Zgcd(c,d)=1

′j

((b −ad −c

), z

)−ke

2πim

b −ad −c

z

4.42=

∑M∈Γ∞\Γ(1)


(4.39)= Pm,k(z).


We just showed that Pm,k satisfies

j(S, z)−k Pm,k(Sz) = Pm,k(z) and j(T, z)−k Pm,k(Tz) = Pm,k(z)

for all z ∈ H. Since k ∈ 2N is even, Theorem 3.18 implies

Pm,k∣∣kM = Pm,k

for all M ∈ Γ(1).

Proposition 4.48. Let m ∈ Z≥0 and k ∈ 2N, k ≥ 4. The Poincare seriesPm,k(z) is a modular form of weight k and trivial multiplier for the full modulargroup: Pm,k ∈Mk

(Γ(1)

).

If m ∈ N, then Pm,k ∈ Sk(Γ(1)

)is a cusp form.

Proof. Lemma 4.47 shows that Pm,k satisfies the transformation law (3.3) forweight k and trivial multiplier. Since Pm,k is a holomorphic function by construc-tion, Theorem 3.22 shows that Pm,k admits a Fourier expansion of the form

Pm,k(z) =∑n∈Z

a(n) e2πinz for all z ∈ H.

We will not perform this calculation here but refer to Theorem 4.59 at the end ofthis section where we calculate the a(n)s explicitly. One can notice that the Fouriercoefficients a(n) vanish for all n < 0 if m ≥ 0 and for n ≤ 0 if m > 0. Definition 3.26implies that Pm,k is a modular form, respectively a cusp form.

What happens if we take the Petersson scalar product between Poincare seriesand a cusp form?

Theorem 4.49. Let m ∈ N, k ∈ 2N, k ≥ 4 a weight and f ∈ Sk(Γ(1)

)a cusp

form whose expansion at the cusp ∞ is given by f(z) =∑∞n=0 a(n) e2πinz for all

z ∈ H with a(0) = 0.The Petersson scalar product between the Poincare series Pm,k and the cusp

form f is given by

(4.41)⟨Pm,k, f

⟩=a(m)

µ

Γ(k − 1)

(4πm)k−1

for all m ∈ N. Here, µ denotes the hyperbolic surface area of the fundamentalregion FΓ(1), see e.g. Exercise 3.45.

Proof. First, we consider an auxiliary identity. We have

j(M, z)−k f(z)= (z)k (3.3),(1.11)

= j(M, z)−k j(M, z)−k f(Mz)(|j(M, z)|2 = (Mz)

)k=

|j(M, z)|2k

j(M, z)k j(M, z)kf(Mz)= (Mz)

k

= f(Mz)= (Mz)k.

Using the Definition of the Poincare series and the above identity gives⟨Pm,k, f

⟩(3.27)

=1

µ

∫FΓ(1)

Pm,k(z) f(z)= (z)k = (z)

−2dλ2(z)

(4.39)=

1

µ

∫FΓ(1)

∑M∈Γ∞\Γ(1)


f(z)= (z)k = (z)

−2dλ2(z)


(3.3),(1.11)=

1

µ

∫FΓ(1)

∑M∈Γ∞\Γ(1)

e2πimMz = (Mz)kf(Mz) = (z)

−2dλ2(z).

Since FΓ(1) is the fundamental domain of Γ(1) in Handwe sum over all cosetsΓ∞\Γ(1), we see that the integral can be viewed as⟨Pm,k, f

⟩=

1

µ

∫FΓ(1)

∑M∈Γ∞\Γ(1)

e2πimMz = (Mz)kf(Mz) = (z)

−2dλ2(z)

(3.32)=

1

µ

∫FΓ(1)

∑M∈Γ∞\Γ(1)

e2πimMz = (Mz)kf(Mz) = (Mz)

−2dλ2(Mz)

=1

µ

∑M∈Γ∞\Γ(1)

∫FΓ(1)

e2πimMz = (Mz)kf(Mz) = (Mz)

−2dλ2(Mz)

=1

µ

∑M∈Γ∞\Γ(1)

∫MFΓ(1)

e2πimz = (z)kf(z) = (z)

−2dλ2(z)

=1

µ

∫[− 1

2 ,12 ]×(0,∞)

e2πimz f(z) = (z)k−2

dλ2(z)

where we integrate over vertical stripz ∈ H; |< (z)| ≤ 1

2

in the last line. Using

Fubini’s theorem, we find⟨Pm,k, f

⟩=

1

µ

∫[− 1

2 ,12 ]×(0,∞)

e2πimz f(z) = (z)k−2

dλ2(z)

=1

µ

∫ ∞0

(∫ 12

− 12

e2πim (x+iy) f(x+ iy) dx

)yk−2dy.

Since f is given by the Fourier expansion f(z) =∑∞n=0 a(n) e2πinz, we can rewrite

the above expression as⟨Pm,k, f

⟩=

1

µ

∫ ∞0

(∫ 12

− 12

e2πim (x+iy) f(x+ iy) dx

)yk−2dy

=1

µ

∫ ∞0

(∑n=0

a(n)

∫ 12

− 12

e2π(m−n) x dx

)e−2πi(m+n)yyk−2dy.

Now, it is known that the inner integral can be calculated. We have∫ 12

− 12

e2π(m−n) x =

1 if m = n and

0 if m 6= n.

Hence, we have ⟨Pm,k, f

⟩=a(m)

µ

∫ ∞0

e−4πimyyk−2dy.

Using the integral representation of the Gamma function, see Appendix A.11 orExample 5.61, we find⟨

Pm,k, f⟩

=a(m)

µ

∫ ∞0

e−4πimyyk−2dy

4πmy 7→y=

a(m)

µ

∫∞0e−yyk−2dy

(4πm)k−1

(5.58)=

a(m)

µ

Γ(k − 1)

(4πm)k−1.


Corollary 4.50. Let k ∈ 2N, k ≥ 4 be a weight. The space of cusp formsSk(Γ(1)

)is spanned by Poincare series Pm,k, m ≥ 0.

Proof. To show that Sk(Γ(1)

)is spanned by Poincare series, we have to show

that each cusp form can be expressed as a linear combination of Poincare series’.Assume that f ∈ Sk

(Γ(1)

)is orthogonal to all Poincare series:

〈Pm,z, f〉 = 0 for all m ∈ Z≥0.

Theorem 4.49 implies that all Fourier coefficients of f vanish. Hence, f must bethe 0-function: f(z) = 0 for all z ∈ H. This means that all cusp forms lie in thespan of the Poincare series.

Similar to Theorem 4.16, we can calculate the Fourier expansion of the Poincareseries. However, we first need to introduce the Kloosterman sums S(m,n; c) andthe J-Bessel functions.

Definition 4.51. For m,n, c ∈ N the Kloosterman sum K(m,n; c) is given by

S(m,n; c) :=∑

0≤l<cgcd(l,c)=1

e2πi(ml+nl?)

c

where l? is defined as a solution of l l? ≡ 1 mod c.

Remark 4.52. The Kloosterman sums are usually denoted by the letter Swhich should not be confused with the matrix S in (2.8). The notation goes backto Kloostermans original paper [53].

The following result was already shown in [42, Theorem 292].

Lemma 4.53. For m, s ∈ N with s > 1 we have the identity

∞∑c=1

c−s

∑0≤l<c

gcd(l,c)=1

e2πi lmc

=σs−1(m)

ms−1ζ(s)

where ζ(s) =∑∞n=1 n

−s denotes the Riemann ζ-function (for < (s) > 1) andσs−1(m) the arithmetic divisor function introduced in Theorem 4.16.

Proof. We follow the proof of [42, Theorem 292]. Define the Mobius functionas

µ : N→ −1, 0, 1; n 7→ µ(n)

with µ(n) given by

µ(n) :=

1 if n is square-free with an even number of prime factors,

−1 if n is square-free with an odd number of prime factors, and

0 if n has at least one squared prime factor.

The Mobius function is multiplicative

µ(mn) = µ(m)µ(n) for all m,n ∈ N with gcd(m,n) = 1

and satisfies ∑d|n

µ(n) =

1 if n = 1 and

0 if n > 1..


Now, we can write∑

0≤l<cgcd(l,c)=1

e2πi lmc in terms of the Mobius function. We have

[42, Theorem 271] ∑0≤l<c

gcd(l,c)=1

e2πi lmc =∑

d|gcd(c,m)

µ( cd

)d.

Rewriting the summation of the right hand side, we have∑d|gcd(c,m)

µ( cd

)d =

∑d,d′

d|mdd′=c

µ(d′)d.

This implies

∞∑c=1

c−s

∑0≤l<c

gcd(l,c)=1

e2πi lmc

=

∞∑c=1

∑d,d′

d|mdd′=c

µ(d′)d

(dd′)s

=

∞∑d′=1

µ(d′)(

d′)s ∑

d|m

1

ds−1.

Since∑d|m

1ds−1 = m1−s∑

d|m ds−1 = m1−s σs−1(m), we find

∞∑c=1

c−s

∑0≤l<c

gcd(l,c)=1

e2πi lmc

=

∞∑d′=1

µ(d′)(

d′)s ∑

d|m

1

ds−1

=σs−1(m)

ms−1

∞∑d′=1

µ(d′)(

d′)s

=σs−1(m)

ms−1

1

ζ(s).

For the last step we used the identity [42, Theorem 287]

1

ζ(s)=

∞∑l=1

µ(l)

ls(< (s) > 1).

Following the standard definitions for Bessel functions, see e.g. [33, Besselfunctions], [30, (10.2.2)], we define the J-Bessel function by its series expansion.

Definition 4.54. Let k ∈ Z be an integer. The Bessel function Jk of the firstkind is given by

Jk(z) =(

12z)k ∞∑

l=0

(−1)l(

14z

2)l

l! Γ(k + l + 1),

where Γ(x) denotes the Γ-function.

Remark 4.55. Let k ∈ Z be an integer. The Bessel function Jk defined aboveis an entire function and solves the Bessel differential equation [30, (10.2.1)]

(4.42) z2 dJ(z)

dz2+ z

dJ(z)

dz+(z2 − k2

)J(z) = 0 (z ∈ C).


Figure 1. The path C∞ of integration in (4.44).

Lemma 4.56. For c, k ∈ N, µ1 ∈ R, , µ2 > 0 and y > 0, we have(4.43)∫

R+iy

(cw)−ke−2πiµ1w−2πiµ2c2w dw =

2πikc

(µ1

µ2

) k−12

Jk−1

(4π√µ1µ2

c

)if µ1, µ2 > 0,

0 if µ1 ≤ 0.

The integration runs along the horizontal line with imaginary part y from the leftto the right.

Proof. Consider the integral on the left hand side in (4.43). We notice thatthe integrant is analytic on H. In Cauchy’s integral theorem, see Appendix A.8,the integral is independent of the actual choice of y > 0. We consider the two casesseparately.

µ1 ≤ 0: We consider the integral on the right hand side. Since the path ofintegration runs along a horizontal line with imaginary part y from the leftto the right, we see that the integrand is evaluated in w ∈ x+ iy; x ∈ Rand y > 0 fix. Hence, we have the lower estimate

|w| ≥ yfor all w along the path of integration. The integrand then satisfies∣∣∣(cw)−ke−2πiµ1w−2πi

µ2c2w

∣∣∣ =∣∣(cw)−k

∣∣ ∣∣e−2πiµ1w∣∣ ∣∣∣−e2πi

µ2c2w

∣∣∣≤ (cy)−k e−2π|µ1|y e

2π|µ2|y|cw|2

≤ (cy)−k e−2π|µ1|y e2π|µ2|y(cy)2

= (cy)−k e−2π|µ1|y e2π|µ2|c2y

for all w on the path of integration.Now, take y → ∞. Then, the above right hand side of the estimate

vanishes since (cy)−k → 0, e−2π|µ1|y → 0 and e2π|µ2|c2y stays bounded.

Hence, the integral vanishes in the limit y →∞:

limy→∞

∫R+iy

(cw)−ke−2πiµ1w−2πiµ2c2w dw = 0.

Since the integral is independent of y, see the arguments above, we find∫R+iy

(cw)−ke−2πiµ1w−2πiµ2c2w dw = 0

for all y > 0.


µ1, µ2 > 0:: Recall the identity [39, (8.412.2]

(4.44) Jν(z) =zν

2ν+1πi

∫C1

t−ν−1et−z2

4t dt

where we integrate over the contour C1, illustrated in Figure 1. Cauchy’sintegral theorem, see Appendix A.8, allows us to deform the contour C1

to a vertical line y + iR with positive real part y. Hence, we have

Jν(z)(4.44)

=zν

2ν+1πi

∫C1

t−ν−1et−z2

4t dt

=zν

2ν+1πi

∫ y+i∞

y−i∞t−ν−1et−

z2

4t dt

v=it=

zν

2ν+1πi

∫ −∞+iy

∞+iy

(−iv)−ν−1e−iv−iz2

4v (−i)dv

=zν

(−2i)ν+1π

∫ ∞+iy

−∞+iy

v−ν−1e−iv−iz2

4v dv.

Plugging in z =4π√µ1µ2

c and ν = k − 1, we have

Jk−1

(4π√µ1µ2

c

)

=

(4π√µ1µ2

)k−1

(−2i)kck−1π

∫ ∞+iy

−∞+iy

v−ke−iv−i

(4π√µ1µ2

)2

4c2v dv

w= v2πµ1=

(4π√µ1µ2

)k−1

(−2i)kck−1π

∫ ∞+iy′

−∞+iy′

(2πµ1w

)−ke−2πiµ1w−

i

(4π√µ1µ2

)2

4c2

(2πµ1w

)2πµ1dw

=ikc

2π

(√µ2

µ1

)k−1 ∫ ∞+iy′

−∞+iy′(cw)−k e−2πiµ1w− 2πiµ2

c2w dw

with y′ = y2πµ1

> 0. We move the path of integration back to the hori-

zontal line with imaginary part y and get

2π

ikc

(µ1

µ2

) k−12

Jk−1

(4π√µ1µ2

c

)=

∫ ∞+iy′

−∞+iy′(cw)−k e−2πiµ1w− 2πiµ2

c2w dw.

This proves the case of (4.43).

Lemma 4.57. For µ1, y > 0 we have

(4.45)

∫R+iy

w−k e−2πiµ1w dw =(2πi)k µk−1

1

Γ(k).

The integration runs along the horizontal line with imaginary part y from the leftto the right.

Proof. Consider the integral on the left hand side in (4.45). We notice that theintegrant is analytic on H. Then by Cauchy’s integral theorem, see Appendix A.8,the integral is independent of the actual choice of y > 0.

Recall Hankel’s loop integral [39, (5.9.2]

(4.46)1

Γ(k)=

1

2πi

∫C1

t−k et dt


where we integrate over the contour C1 illustrated in Figure 1. Cauchy’s integraltheorem, see Appendix A.8, allows us to deform the contour C1 to a vertical liney + iR with positive real part y. Hence, we have

1

Γ(k)=

1

2πi

∫C1

t−k et dt

=1

2πi

∫ y+i∞

y−i∞t−k et dt

v=it=

1

2πi

∫ ∞+iy

−∞+iy

(−iv)−k e−iv (−i)dv

=(−i)1−k

2πi

∫ ∞+iy

−∞+iy

v−k e−iv dv

=ik−2

2π

∫ ∞+iy

−∞+iy

v−k e−iv dv.

Plugging in v = 2πµ1w, we have

1

Γ(k)=ik−2

2π

∫ ∞+iy

−∞+iy

v−k e−iv dv

v=2πµ1w=

ik−2

2π

∫ ∞+iy′

−∞+iy′(2πµ1w)−k e−2πiµ1w 2πµ1dw

=ik−2

2π(2πµ1)k−1

∫ ∞+iy′

−∞+iy′w−k e−2πiµ1w dw

=(2πiµ1)1−k

2πi

∫ ∞+iy′

−∞+iy′w−k e−2πiµ1w dw

with y′ = y2πµ1

> 0.

Definition 4.58. The Kronecker δ-function δn,m is defined by

(4.47) δn,m :=

1 if n = m and

0 if n 6= m.

Theorem 4.59. For m ∈ Z≥0 and k ∈ 2N, k ≥ 4, the Poincare series Pm,khas the Fourier expansion

Pm,k(z) =∑n∈Z≥0

ane2πinz (z ∈ H)

at the cusp ∞. The coefficients an are given by:

• If m = 0, we have

an =

1 if n = 0 and(2πi)k σk−1(n)

(k−1)! ζ(k) if n > 0.

• If m > 0, we have

an = δn,m +∑c∈N

S(m,n; c)(2πi)k

c

( nm

) k−12

Jk−1

(4π√mn

c

).

Proof. Using Lemma 4.46 we have

Pm,k(z) =∑

c∈Z≥0,d∈Zgcd(c,d)=1

′(cz + d)−k e2πimVc,dz (z ∈ H)


where Vc,d =

(a bc d

)∈ SL2 (Z) is a matrix with given second row (c, d), see

Lemma 4.44.To calculate the nth Fourier coefficient of Pm,k(z) we start with the integral

formula for Fourier coefficients of periodic functions with period 1. We have

an =

∫ 1+it

it

Pm,k(z) dz

for some arbitrary t > 0. The Fourier coefficient an might depend on t at this step;we will show its independence later.

Using the above summation formula for the Poincare series Pm,k we get

an =

∫ 1+it

it

Pm,k(z) e−2πinz dz

=

∫ 1+it

it

∑c∈Z≥0,d∈Zgcd(c,d)=1


dz.

Using Lebesgue dominated convergence theorem, we may interchange integration

and summation. As dominating series, we use the integral of∑′c,d∈Z |cz + d|−k.

This is the same absolutely converging series appearing in the proof of Lemma 4.45.Hence, we have for the Fourier coefficient

an =

∫ 1+it

it

∑



e−2πinz dz

=∑


′∫ 1+it

it

(cz + d)−k e2πimVc,dz e−2πinz dz

=∑c=0d=1

∫ 1+it

it

(d)−k e2πimV0,dz e−2πinz dz+

+∑c∈N

∑d∈Z

gcd(c,d)=1

∫ 1+it

it


The first part gives∑c=0d=1

∫ 1+it

it

(d)−k e2πimV0,dz e−2πinz dz =

∫ 1+it

it

e2πi(m−n) z dz

=

1 if m = n and

0 if m 6= n.

= δn,m

since V0,1 =

(1 ?0 1

). Note that the integral

∫ 1+it

ite2πi(m−n) z dz is independent of

t > 0.


For the second part, write d ∈ Z as d = lc+ d′ for unique 0 ≤ d′ < c and l ∈ Z.Hence, the series can be expressed as∑

c∈N

∑d∈Z

gcd(c,d)=1

∫ 1+it

it


=∑c∈N

∑0≤d′<c

gcd(c,d′)=1

∑l∈Z

∫ 1+it

it

(c(z + l) + d′

)−ke2πimVc,d′+clz e−2πinz dz

=∑c∈N

∑0≤d′<c

gcd(c,d′)=1

∑l∈Z

∫ 1+it

it

(c(Slz) + d′

)−ke2πimVc,d′S

lz e−2πinz dz

We used that only the lower row of Vc,d′ is fixed; the upper row can be choosenarbitrarly provided Vc,d ∈ Γ(1), see Lemma 4.44.

=∑c∈N

∑0≤d′<c

gcd(c,d′)=1

∫R+it

(cz + d′

)−ke2πimVc,d′z e−2πinz dz.

Write Vc,d′ =

(a bc d′

), where of course a and b depend on the given c and d′. We

have the identities (cz + d′

)−k= c−k

(z +

d′

c

)−kand

Vc,d′z =az + b

cz + d′=

acz + bc

c2(z + d′

c

) =acz + ad− 1

c2(z + d′

c

) =a

c− 1

c2(z + d′

c

)using det

(a bc d

)= ad− bc = 1 and c 6= 0. Using these identities we get

∑c∈N

∑d∈Z

gcd(c,d)=1

∫ 1+it

it


=∑c∈N

∑0≤d′<c

gcd(c,d′)=1

∫R+it

(cz + d′

)−ke2πimVc,d′z e−2πinz dz

=∑c∈N

∑0≤d′<c

gcd(c,d′)=1

∫R+it

(c

(z +

d′

c

))−ke

2πim ac−2πi m

c2(z+ d′c ) e−2πinz dz

=∑c∈N

∑0≤d′<c

gcd(c,d′)=1

e2πim ac

∫R+it

(c

(z +

d′

c

))−ke−2πinz−2πi m

c2(z+ d′c ) dz.

Applying finally the substitution z + d′

c 7→ z, we get∑c∈N

∑d∈Z

gcd(c,d)=1

∫ 1+it

it


=∑c∈N

∑0≤d′<c

gcd(c,d′)=1

e2πim ac

∫R+it

(c

(z +

d′

c

))−ke−2πinz−2πi m

c2(z+ d′c ) dz


=∑c∈N

∑0≤d′<c

gcd(c,d′)=1

e2πim ac

∫R+it

(cz)−k e−2πin

(z− d′c

)−2πi m

c2z dz

=∑c∈N

∑0≤d′<c

gcd(c,d′)=1

e2πim ac+2πin d

′c

∫R+it

(cz)−k e−2πinz−2πi mc2z dz.

Next, we have to consider the cases m > 0 and m = 0 separately.

• Consider the case m > 0. Using Lemma 4.56 for the inner integral andDefinition 4.51 for the inner summation, we find∑

c∈N

∑d∈Z

gcd(c,d)=1

∫ 1+it

it


=∑c∈N

∑0≤d′<c

gcd(c,d′)=1

e2πim ac+2πin d

′c

∫R+it

(cz)−k e−2πinz−2πi mc2z dz

=

∑c∈N S(m,n; c) 2π

ikc

(nm

) k−12 Jk−1

(4π√nmc

)if n > 0 and

0 if n ≤ 0.

This concludes the proof in this case.• Consider the case m = 0 and n > 0. We get∑c∈N

∑d∈Z

gcd(c,d)=1

∫ 1+it

it


=∑c∈N

∑0≤d′<c

gcd(c,d′)=1

e2πim ac+2πin d

′c

∫R+it


m=0=∑c∈N

c−k∑

0≤d′<cgcd(c,d′)=1

e2πin d′c

∫R+it

z−k e−2πinz dz.

With Lemma 4.57 and using Γ(k) = (k − 1)!, the inner integral evaluatesto ∫

R+it

z−k e−2πinz dz =(2πi)k nk−1

Γ(k)=

(2πi)k nk−1

(k − 1)!

since n > 0 is positive. Using Lemma 4.53 for the outer two summationswe find∑

c∈N

∑d∈Z

gcd(c,d)=1

∫ 1+it

it


=∑c∈N

c−k∑


e2πin d′c

∫R+it

z−k e−2πinz dz

=(2πi)k nk−1

(k − 1)!

∑c∈N

c−k∑


e2πin d′c

=(2πi)k nk−1

(k − 1)!

σk−1

nk−1 ζ(k)=

(2πi)k σk−1(n)

(k − 1)! ζ(k).


This concludes the proof in this case.• Consider the remaining case m = n = 0. We get∑c∈N

∑d∈Z

gcd(c,d)=1

∫ 1+it

it


=∑c∈N

∑0≤d′<c

gcd(c,d′)=1

e2πim ac+2πin d

′c

∫R+it


m=n=0=

∑c∈N

c−k∑


∫R+it

z−k dz.

With Cauchys integral theorem, see Appendix A.8, we get∫R+it

z−k dz = 0.

Hence, we have∑c∈N

∑d∈Z

gcd(c,d)=1

∫ 1+it

it


m=n=0=

∑c∈N

c−k∑


∫R+it

z−k dz

=∑c∈N

c−k∑


0

= 0.

Putting both parts together, we get

an =∑c=0d=±1

∫ 1+it

it

(d)−k e2πimV0,dz e−2πinz dz +

+∑c∈N

∑d∈Z

gcd(c,d)=1

∫ 1+it

it


= 2δn,m +

+

0 if m = n = 0,(2πi)k σk−1(n)

(k−1)! ζ(k) if m = 0, n > 0 and∑c∈N S(m,n; c) (2πi)k

c

(nm

) k−12 Jk−1

(4π√mnc

)if m > 0.

Exercise 4.60. For k ∈ 2N, k ≥ 4, show that the 0th Poincare series P0,k andthe normalized Eisenstein series Ek satisfy

(4.48) P0,k = Ek.

4.5. Theta series

4.5.1. Jacobi Theta function.

4.5. THETA SERIES 103

Definition 4.61. The Jacobi theta function Θ : C×H→ C is defined by theseries

(4.49) Θ(w, z) :=∑n∈Z

eπin2z e2πinw

for all w ∈ C and z ∈ H.

Lemma 4.62. The Jacobi theta function Θ(w, z) is an entire function in thefirst argument for fixed z ∈ H. It is a holomorphic function in the second argumentfor fixed w ∈ C.

Proof. Let t0 > 0 and M > 0 be arbitrary positive constants, fixed through-out this proof. We consider the sets

Ht :=z ∈ H; = (z) ≥ t

⊂ H

and

UM :=w ∈ C; |w| ≤M

⊂ C.

For all z ∈ Ht and w ∈ UM , we have the series estimate

|Θ(w, z)| ≤∑n∈Z

∣∣∣eπin2z e2πinz∣∣∣

=(z)≥t0≤

∑n∈Z

e−πin2t0∣∣e2πinz

∣∣|w|≤M≤

∑n∈Z

e−πin2t0 e2π|n|M

<∞.

Hence, Θ(·, z) is entire on C in the first argument for each fixed z ∈ H. Similarly,Θ(w, ·) is holomorphic in the second argument for each fixed w ∈ C.

We have the following transformation behavior of the Jacobi theta function inthe first argument:

Proposition 4.63. We have:

(1) Θ(w + 1, z) = Θ(w, z) for all w ∈ C and z ∈ H.(2) Θ(w + z, z) = e−πiz e−2πiw Θ(w, z) for all w ∈ C and z ∈ H.(3) If w ∈ C and z ∈ H satisfy w = 1+z

2 +m+nz for some m,n ∈ Z, we haveΘ(w, z) = 0.

Proof. The first property follows directly from the expansion of Θ. We have,

Θ(w + 1, z)(4.49)

=∑n∈Z

eπin2z e2πin(w+1)

=∑n∈Z

eπin2z e2πinw e2πin︸︷︷︸

=1

=∑n∈Z

eπin2z e2πinw

(4.49)= Θ(w, z)

for all w ∈ C and z ∈ H.The second property is based on a completion of a square argument: we have

(4.50) n2 + 2n = (n+ 1)2 − 1.


Applying this to the expansion of Θ, we find

Θ(w + z, z)(4.49)

=∑n∈Z

eπin2z e2πin(w+z)

=∑n∈Z

eπi(n2+2n)z e2πinw)

(4.50)= e−πiz

∑n∈Z

eπi(n+1)z e2πinw)

= e−πiz e−2πiw∑n∈Z

eπi(n+1)z e2πi(n+1)w)

(4.49)= e−πiz e−2πiw Θ(w, z)

for all w ∈ C and z ∈ H.To prove the last property we consider Θ

(1+z

2 , z)

for some z ∈ H. We find

Θ

(1 + z

2, z

)(4.49)

=∑n∈Z

eπin2z e2πin 1+z

2

=∑n∈Z

eπi(n2+n)z eπin︸︷︷︸

=(−1)n

=∑n∈N

eπi(n2+n)z −

∑n∈Z≤0

eπi(n2+n)z

n=−m−1=

∑n∈N

eπi(n2+n)z −

∑m∈N

eπi(

(−m−1)2+(−m−1))z.

Using

(−m− 1)2 + (−m− 1) = m2 + 2m+ 1−m− 1 = m2 +m

we find

Θ

(1 + z

2, z

)=∑n∈N

eπi(n2+n)z −

∑m∈N

eπi(

(−m−1)2+(−m−1))z

=∑n∈N

eπi(n2+n)z −

∑m∈N

eπi(m2+m)z

=∑n∈N

eπi(n2+n)z − eπi(n

2+n)z︸︷︷︸=0

= 0.

Using the first two transformation relations we extend the above result to

0 = Θ

(1 + z

2, z

)= Θ

(1 + z

2+ n+mz, z

)for all m,n ∈ Z. This concludes the proof of Proposition 4.63.

Now, we consider the transformation behavior in the second argument. A firsttrivial result is given by the following:

Exercise 4.64. Show that we have Θ(w, z + 2) = Θ(w, z) for all w ∈ C andz ∈ H.

To show the transformation behavior under z 7→ −1z in the second argument of

the Jacobi theta function, consider first the following auxiliary result.


Lemma 4.65 ([111, Chapter 4, Equation (3) on Page 120]). For fixed t > 0and a ∈ R we have

(4.51)∑n∈Z

e−πt(n+a)2

=1√t

∑n∈Z

e−πn2

t e2πina.

Proof. Consider the function f : R→ R given by

f(x) = e−πt(x+a)2

.

Its Fourier transform is given by

f(k) :=

∫ ∞−∞

f(x) e−2πikx dx =1√te−π

k2

t e2πika.

Using The Poisson summation formula, see Appendix A.16, we find immediatelyRelation 4.51.

Proposition 4.66. For z ∈ H and w ∈ C, we have

(4.52) Θ

(w,−1

z

)=

√z

ieπizw

2

Θ(zw, z).

Here,√

zi denotes the branch of the square root defined on H which is positive if

z ∈ iR>0 is purely imaginary.

Remark 4.67. The square root in the proposition above can be defined asfollows: If z = r eiφ with r > 0 and φ = arg z ∈ (0, π) in polar coordinates, then√

zi is given by √

z

i=

√r ei(φ−

π2 ) !

=√r ei

φ−π2

2 =√r ei(

φ2−

π4 ).

Proof of Proposition 4.66. Since Θ(·, ·) is entire in the first argument andholomorphic in the second argument, it suffices to check Relation (4.51) for realand w = x + i0 ∈ R and purely imaginary z = it ∈ iR>0. If Relation (4.51) holdsfor these values, it holds everywhere, due to being entire in Θ’s first argument forfixed second argument and being holomorphic in Θ’s second argument for fixed firstargument, respectively. This leads to the question whether

Θ

(x,−1

it

)= Θ

(x,i

t

)(4.49)

=∑n∈Z

e−πn2

t e2πinx

?=√t e−πtx

2 ∑n∈Z

e−πn2t e−2πntx

(4.49)=

√z

ieπizw

2

Θ(zw, z)

holds. Reformulating the identity in question we have to show

(4.53)1√t

∑n∈Z

e−πn2

t e2πinx ?= e−πtx

2 ∑n∈Z

e−πn2t e−2πntx.

Using (4.51) of Lemma 4.65, we express the left hand side of (4.53) as

1√t

∑n∈Z

e−πn2

t e2πinx (4.51)=

∑n∈Z

e−πt(n+x)2

=∑n∈Z

e−πt(n2+2nx+x2)


= e−πtx2 ∑n∈Z

e−πn2t e−2πntx.

This is exactly the right hand side of (4.53), proving the validity of this identity.

Summarizing, we have the following transformation behaviour of the Jacobitheta function.

Theorem 4.68. For z ∈ H and w ∈ C we have

(4.54) Θ(w + 1, z) = Θ(w, z) and Θ(w + z, z) = e−πiz e−2πiw Θ(w, z)

in the first coordinate and

(4.55) Θ(w, z + 2) = Θ(w, z) and Θ

(w,−1

z

)=

√z

ieπizw

2

Θ(zw, z)

in the second coordinate. Here√

zi denotes the branch of the square root defined

on H which is positive if z ∈ iR>0 is purely imaginary.

Proof. The identities in the first coordinate follows from Proposition 4.63.The identities in the second coordinate follow from Exercise 4.64 and Proposi-tion 4.66.

Corollary 4.69. The function θ : H→ C given by

(4.56) θ(z) := Θ(0, z)

satisfies

(4.57) θ(z + 2) = θ(z) and θ

(−1

z

)=

√z

iθ(z).

The next corollaries give the growth behavior of θ(z) for z → i∞ and z → 1.

Corollary 4.70. The function θ satisfies the limit

(4.58) lim=(z)→∞

θ(z) = 1.

Proof. Using the Series definition of Θ, we get

θ(z)(4.56)

= Θ(0, z)

(4.49)=

∑n∈Z

eπin2z.

For n ∈ Z each individual summand satisfies

lim=(z)→∞

eπin2z = lim

=(z)→∞

(eπiz

)n2

=

0 if n2 6= 0 and

1 if n2 = 0.

We used the fact that 0 <∣∣eπiz∣∣ < 1 and

(eπiz

)0= 1 for all z ∈ H and that

lim=(z)→∞

eπiz = 0.

Hence, we have

θ(z) =∑n∈Z

eπin2z

= 1 +∑n∈Z6=0

eπin2z︸︷︷︸

→0

−→ 1

as = (z)→∞.


Corollary 4.71. The function θ satisfies

(4.59) θ

(1− 1

z

)=

√z

i

∑n∈Z

eπi(n+ 12 )

2z

for all z ∈ H.In particular, we have the growth estimate

(4.60) θ

(1− 1

z

)= O

(√z

ieπiz2

)as = (z)→∞.

Proof. Using (4.56) together with the expansion of Θ in (4.49) we find

θ(1 + z)(4.56)

= Θ(0, 1 + z)

(4.49)=

∑n∈Z

eπin2(1+z)

=∑n∈Z

eπin2︸︷︷︸

=(−1)n2

eπin2z

=∑n∈Z

(−1)n eπin2z,

using the fact that n ≡ n2 mod 2 for all n ∈ Z. (In words: We use that n and n2

have the parity). Using again that (−1)n = eπin we find

θ(1 + z) =∑n∈Z

(−1)n eπin2z

=∑n∈Z

eπin2z eπin

=∑n∈Z

eπin2z e2πin 1

2

(4.49)= Θ

(1

2, z

).

Now, using the transformation property (4.52) of Θ in Proposition 4.66 gives

θ

(1 +

1

z

)= Θ

(1

2,−1

z

)(4.52)

=

√z

ieπiz4 Θ

(z2, z)

(4.49)=

√z

ieπiz4

∑n∈Z

eπin2z eπinz

=

√z

i

∑n∈Z

eπi(n2+n+ 1

4 )z

=

√z

i

∑n∈Z

eπi(n+ 12 )

2z.

This shows expansion (4.59) of the corollary.The growth estimate now follows from the growth estimate of the individual

terms in the series expansion. The dominating terms are eπi(12 )

2z and eπi(−

12 )

2z

corresponding to n = 0 and n = −1 respectively. They both give a growth estimate

of O(√

zi e

πiz2

)as = (z)→∞. The other terms contribution is of lower order.


Next we consider a product expansion of the Jacobi theta function, followingthe arguments presented in [111, Chapter 4, §1].

To do so, we temporarily define a product Θ(w, z) which has the same trans-formational properties as the Jacobi theta function Θ. Then we show that the twofunctions Θ and Θ agree.

Definition 4.72. The triple product function Θ : C×H→ C is defined by theproduct

(4.61) Θ(w, z) :=∏n∈N

(1− e2πinz

) (1 + eπi(2n−1)z e2πiw

)(1 + eπi(2n−1)z e−2πiw

)for all w ∈ C and z ∈ H.

Remark 4.73. Using the notation q = eπiz the triple product function can bewritten as

Θ(w, z) =∏n∈N

(1− q2n

) (1 + q2n−1 e2πiw

) (1 + q2n−1 e−2πiw

).

Proposition 4.74. We have:

(1) The Jacobi theta function Θ(w, z) is an entire function in the first argu-ment for fixed z ∈ H. It is a holomorphic function in the second argumentfor fixed w ∈ C.

(2) Θ(w + 1, z) = Θ(w, z) for all w ∈ C and z ∈ H.

(3) Θ(w + z, z) = e−πiz e−2πiw Θ(w, z) for all w ∈ C and z ∈ H.(4) We have the following bijection:

Θ(w, z) = 0 ⇐⇒ w =1 + z

2+m+ nz for some m,n ∈ Z.

(5) The zeros of Θ(w, z) described above are simple. (This means that

Θ

(1 + z

2+m+ nz, z

)= 0 for all z ∈ H and m,n ∈ Z

and they are all simple zeros).

Proof. (1) Let t0 > 0. We have the estimate∣∣eπiz∣∣ ≤ e−πt0 < 1

for all z ∈ H with = (z) ≥ t0. This implies that the series∑n∈N

∣∣∣eπi(2n−1)z∣∣∣

converges absolutely. Hence, the product Θ(w, z) in (4.61) of Defini-tion 4.72 defines an entire function in w ∈ C for fixed z ∈ H and viceversa defines a holomorphic function in z ∈ H for fixed w ∈ C, see e.g.[111, Chapter V].

(2) The product expansion of Θ(w, z) in (4.61) implies immediately

Θ(w + 1, z)

(4.61)=

∏n∈N

(1− e2πinz

) (1 + eπi(2n−1)z e2πi(w+1)

)(1 + eπi(2n−1)z e−2πi(w+1)

)

=∏n∈N

(1− e2πinz

)1 + eπi(2n−1)z e2πiw e2πi︸︷︷︸=1

1 + eπi(2n−1)z e−2πiw e−2πi︸︷︷︸=1

=∏n∈N

(1− e2πinz

) (1 + eπi(2n−1)z e2πiw

)(1 + eπi(2n−1)z e−2πiw

)


(4.61)= Θ(w, z).

(3) We have

Θ(w + z, z)

(4.61)=

∏n∈N

(1− e2πinz

) (1 + eπi(2n−1)z e2πi(w+z)

)(1 + eπi(2n−1)z e−2πi(w+z)

)=∏n∈N

(1− e2πinz

) (1 + eπi(2n+1)z e2πiw

)(1 + eπi(2n−3)z e−2πiw

)=

(1 + e−πiz e−2πiw

1 + eπiz e2πiw

)·

·∏n∈N

(1− e2πinz

) (1 + eπi(2n−1)z e2πiw

)(1 + eπi(2n−1)z e−2πiw

)(4.61)

=


1 + eπiz e2πiw

)Θ(w, z).

Using the identity1 + a

1 + 1a

= a

with a = e−πiz e−2πiw, we find

Θ(w + z, z) =


1 + eπiz e2πiw

)Θ(w, z)

= e−πiz e−2πiw Θ(w, z).

(4) Recall that a product vanishes if at least one of its factors vanishes. Since∣∣e2πinz∣∣ < 1 for all z ∈ H we see that the factors 1−e2πinz = 1−

(e2πiz

)n 6=0 never vanishe for all z ∈ H and n ∈ N.

The second and third factors

1 + eπi(2n−1)z e2πiw and 1 + eπi(2n−1)z e−2πiw

in the product representation (4.61) vanish if

eπi(2n−1)z e2πiw != −1 = e−πi and eπi(2n−1)z e−2πiw !

= −1 = e−πi

holds respectively. This is the case if either

(2n− 1)z + 2w = −1 mod 2

or

(2n− 1)z − 2w = −1 mod 2

holds for some n ∈ N. Equivalently, this is the case if either

nz − z

2+

1

2= −w mod 1

or

nz − z

2+

1

2= w mod 2

holds for some n ∈ N. We can unify the above two condition to requiring

w =1

2+z

2+m+ nz

for some m,n ∈ Z. In other words, we have the following equivalenceshown:

Θ(w, z) = 0 ⇐⇒ w =1 + z

2+m+ nz for some m,n ∈ Z.


(5) The previous arguments, where we calculate the location of the zeros of

Θ(w, z), also show the following: For any fixed w ∈ C and z ∈ H satisfyingw = 1

2 + z2 + m + nz for some m,n ∈ Z, exactly one factor of either

1 + eπi(2n−1)z e2πiw or 1 + eπi(2n−1)z e−2πiw in the product representation(4.61) vanishes. Hence, the zero in this w and z is simple.

Exercise 4.75. Show that for every z ∈ H there exists a constant c(z) suchthat

(4.62) Θ(w, z) = c(z) Θ(w, z)

holds for all w ∈ C.

Theorem 4.76. We have

(4.63) Θ(w, z) = Θ(w, z)

for all w ∈ C and z ∈ H.

Proof. Exercise 4.75 shows that for every z ∈ H there exists a constant c(z)

such that Θ(w, z) = c(z) Θ(w, z) holds for all w ∈ C. To establish the identity, wehave to show that c(z) = 1 for all z ∈ H holds.

Step 1: Calculate c(z) for w = 12 .

Put w = 12 . Then (4.62) together with the defining equations (4.49)

and (4.61) imply∑n∈Z

eπin2z e2πin 1

2︸︷︷︸=(−1)n

=∑n∈Z

(−1)n eπin2z

(4.62)= c(z)

∏n∈N

(1− e2πinz

)1 + eπi(2n−1)z e2πi 12︸︷︷︸

=−1

1 + eπi(2n−1)z e−2πi 12︸︷︷︸

=−1

= c(z)

∏n∈N

(1− e2πinz

) (1− eπi(2n−1)z

)(1− eπi(2n−1)z

)= c(z)

∏n∈N

(1−

(eπiz

)2n)(1−

(eπiz

)2n−1)

︸︷︷︸(?)

(1−

(eπiz

)2n−1)

= c(z)∏n∈N

(1−

(eπiz

)n)︸︷︷︸(?)

(1−

(eπiz

)2n−1).

We use the fact that factors in (?) run through all even and odd positivepowers of eπiz so that we can re-index all factors:

(?)∏n∈N

(1−

(eπiz

)2n)(1−

(eπiz

)2n−1)

=∏n∈N

(1−

(eπiz

)n).

Hence, we have

(4.64) c(z) =

∑n∈Z(−1)n

(eπiz

)n2

∏n∈N

(1− (eπiz)

n) (1− (eπiz)

2n−1)

for all z ∈ H.


Step 2: Calculate c(z) for w = 14 .

Let us calculate Θ(

14 , z)

on the left hand of (4.62). We have

Θ

(1

4, z

)(4.49)

=∑n∈Z

eπin2z e2πin 1

4

=∑n∈Z

eπin2z

eπi2︸︷︷︸

=i

n

=∑n∈Z

in eπin2z

=∑n∈2Z

in eπin2z +

∑n∈1+2Z

in eπin2z

︸︷︷︸=0

=∑n∈2Z

in eπin2z

2n=m=

∑m∈Z

(−1)m e4πim2z.

We used the fact that the set 1+2Z contains all odd integers. In particular,if n ∈ 1 + 2Z then −n ∈ 1 + 2Z, too. Hence the terms

in eπin2z and i−n eπi(−n)2z = −in eπin

2z

cancel each other.Now, we calculate Θ

(14 , z)

on the right hand of (4.62). We have

Θ

(1

4, z

)(4.61)

=∏n∈N

(1− e2πinz

)1 + eπi(2n−1)z e2πi 14︸︷︷︸

=eπi2 =i

1 + eπi(2n−1)z e−2πi 1

4︸︷︷︸=e−

πi2 =−i

=∏n∈N

(1− e2πinz

) (1 + i eπi(2n−1)z

)(1− i eπi(2n−1)z

)︸︷︷︸

=1+eπi(4n−2)z

=∏n∈N

(1− e2πinz

) (1 + eπi(4n−2)z

),

using the binomial formula (1− a)(1 + a) = 1− a2 with a = i eπi(2n−1)z.Re-indexing the factors we get

Θ

(1

4, z

)=∏n∈N

(1− e2πinz

) (1 + eπi(4n−2)z

)=∏n∈N

(1− e2πinz

) ∏n∈N

(1 + eπi(4n−2)z

)2n=4m−2

2n=4m=∏m∈N

(1− eπi(4m−2)z

)(1− eπi(4m)z

) ∏n∈N

(1 + eπi(4n−2)z

),

using the fact that we can split the set of even natural numbers in twosets of m = 0 mod 4 and m = 2 mod 4. Next we apply again the binomial


formula (1− a)(1 + a) = 1− a2 with a = eπi(4m−2)z and get

Θ

(1

4, z

)=∏m∈N

(1− eπi(4m−2)z

)(1− eπi(4m)z

) ∏n∈N

(1 + eπi(4n−2)z

)=∏m∈N

(1− eπi(4m)z

) (1 + eπi(8m−4)z

).

Summarizing, we have

c(z)(4.62)

=Θ(

14 , z)

Θ(

14 , z)

=

∑m∈Z(−1)m

(eπiz

)4m2

∏m∈N

(1− (eπiz)

4m) (

1 + (eπiz)8m−4

)(4.65)

for all z ∈ H.Step 3: Show that c(4z) = c(z) holds.

Comparing the expressions of the right hand sides in (4.64) and (4.65)we get

c(4z)(4.64)

=

∑n∈Z(−1)n

(eπi(4z)

)n2

∏n∈N

(1−

(eπi(4z)

)n) (1−

(eπi(4z)

)2n−1)

=

∑n∈Z(−1)n

(eπiz

)4n2

∏n∈N

(1− (eπiz)

4n)(

1− (eπiz)8n−4

)(4.65)

= c(z)

for all z ∈ H.Step 4: Conclude c(z) = 1.

Iterating the identity c(4z) = c(z), we get

c(z) = c(4z) = c(42z)

= . . . = c(4kz)

for any k ∈ Z≥0 and all z ∈ H.

Now, consider the term(eπi(4

kz))m

which appears in c(4kz). We see

that for given z ∈ H, we have

limk→∞

(eπi(4

kz))m

=

1 if m = 0 and

0 if m 6= 0.

This allows us to argue as follows

c(z) = limk→∞

c(4kz)

= limk→∞

1∏n∈N

(1−

(eπi(4kz)

)n) (1−

(eπi(4kz)

)2n−1) +

+

∑n∈Z6=0

(−1)n[limk→∞

(eπi(4

kz))n2]

∏n∈N

(1−

[limk→∞

(eπi(4kz)

)n]) (1−

([limk→∞ eπi(4kz)

)2n−1])

= 1 +

∑n∈Z6=0

0∏n∈N 1

= 1.


n r2(n) r4(n) n r2(n) r4(n)0 1 1 11 0 961 4 8 12 0 962 4 24 13 8 1123 0 32 14 0 1924 4 24 15 0 1925 8 48 16 4 246 0 96 17 8 1447 0 64 18 4 3128 4 24 19 0 1609 4 104 20 8 14410 8 144 21 0 256

Table 1. Some values of the functions r2(n) and r4(n) defined in(4.67) and (4.68) respectively.

Corollary 4.77. For z ∈ H, we have

(4.66) θ(z) =∏n∈N

(1− e2πinz

) (1 + eπi(2n−1)z

)2

.

Proof. To show the corollary consider Theorem 4.76 and the product expan-sion of Θ(w, t) in Definition 4.72. Using θ(z) = Θ(0, z), see (4.56), we immediatelyget (4.66).

4.5.2. Theta series and sums of squares. Recall the introduction of PartI on page ??. We are interested in the following question; how often we can writea given integer n ∈ Z≥0 as a sum of four squares?

Lets start first with the question, whether we can write an integer n ∈ Z≥0 assum of two squares and how many ways of writing are there. This means that weare interested in the value of the function

(4.67) r2 : Z≥0 → Z≥0; n 7→ r2(n) := ]

(a, b) ∈ Z2; a2 + b2 = n.

Similarly, we define the function r4 counting the number of ways how an integern ∈ N can be written as a sum of four squares:

(4.68) r4 : Z≥0 → Z≥0; n 7→ r4(n) := ]

(a, b, c, d) ∈ Z4; a2+b2+c2+d2 = n.

Example 4.78. Some values of r2 are given in Table 1. For example we haver2(0) = 1 since 02 = 02 + 02 and r2(1) = 4 since 1 = (±1)2 + 02 = 02 + (±1)2.

In order to describe the behaviour of the function r2, we use generating function,following mostly the argumentation presented in [111, Chapter 10, §3].

Exercise 4.79. Show the identities

θ(z)2 = 1 +∑n∈N

r2(n) eπinz

=

∞∑n=0

r2(n) eπinz.

(4.69)

and

θ(z)4 = 1 +∑n∈N

r4(n) eπinz

=

∞∑n=0

r4(n) eπinz.

(4.70)


Consider the function

(4.71) F2 : H→ C; z 7→ f(z) := 2∑n∈Z

1

eπinz + e−πinz.

Of course, we have the trivial identity

F2(z)(4.71)

= 2∑n∈Z

1

eπinz + e−πinz

=∑n∈Z

1

cos(nπz)

(4.72)

for all z ∈ H using the trigonometric identity

cos(φ) =1

2

(eiφ + e−iφ

).

As a first step, we show that f in (4.72) is well defined.

Lemma 4.80. For any z ∈ H the series

2∑n∈Z

1

eπinz + e−πinzand 1 + 4

∑n∈N

eπinz

1 + e2πinz

converge absolutely and are equal.This implies that F1 is well defined on H and we have

(4.73) F1(z) = 2∑n∈Z

1

eπinz + e−πinz= 1 + 4

∑n∈N

eπinz

1 + e2πinz

for all z ∈ H. Moreover, the function f is holomorphic on H.

Proof. Let us first show that the two series are equal. Using

1

eπinz + e−πinz=

1

e−πinz1

e2πinz + 1=

eπinz

e2πinz + 1

we find

2∑n∈Z

1

eπinz + e−πinz= 1 + 2

∑n∈N

1

eπinz + e−πinz+ 2

∑n∈N

1

eπi(−n)z + e−πi(−n)z

= 1 + 2∑n∈N

eπinz

1 + e2πinz+ 2

∑n∈N

eπinz

1 + e2πinz

= 1 + 4∑n∈N

eπinz

1 + e2πinz.

This shows the right identity in (4.73).Now take a (fixed) z ∈ H. Since

∣∣eπinz∣∣ ≤ ∣∣eπiz∣∣ < 1 for all n ∈ N we have∣∣1 + e2πinz∣∣ ≥ ε > 0

with ε := 1−∣∣eπiz∣∣ = 1− e−π=(z) > 0 independent of n. Hence we can estimate∣∣∣∣∣∑n∈N

eπinz

1 + e2πinz

∣∣∣∣∣ ≤∑n∈N

∣∣∣∣ eπinz

1 + e2πinz

∣∣∣∣=∑n∈N

∣∣eπinz∣∣|1 + e2πinz|

≤∑n∈N

∣∣eπiz∣∣nε


=1

ε

∑n∈N

(e−π=(z)

)n<∞,

using that the geometric series∑∞n=0 q

n = 11−q converges absolutely for all |q| < 1.

The same trick shows that the series 2∑n∈Z

1eπinz+e−πinz converges absolutely

for all z ∈ H. This proves that F1 in (4.71) is well defined and holomorphic onH.

Before we show the transformation behaviour of F1(z) under z 7→ −1z , consider

first the following auxiliary result.

Lemma 4.81 ([111, Chapter 4, Equation (4) on Page 120]). For fixed t > 0and a ∈ R, we have

(4.74)∑n∈Z

e−2πian

cosh(πnt

) =∑n∈Z

t

cosh(π(n+ a)t)

.

Proof. Recall that the trigonometric function cosh(φ), the cosinus hyperboli-cus, is defined by

(4.75) cosh(φ) :=1

2

(eφ + e−φ

)and satisfies

(4.76) cosh(iφ) = cos(φ)

for all φ ∈ C.Consider the function g : R→ R given by

g(x) =e−πiax

cosh(πxt

) .Its Fourier transform is given by

g(k) :=

∫ ∞−∞

g(x) e−2πikx dx =t

cosh(π(k + a)t

) .Using The Poisson summation formula, see Appendix A.16, we find immediatelyRelation 4.74.

The above auxiliary lemma implies one part of the following transformationbehaviour of f .

Proposition 4.82. The function F1 satisfies

(4.77) F1

(−1

z

)=z

if(z)

and

(4.78) F1(z + 2) = F1(z)

for all z ∈ H.

Proof. The transformation (4.77) follows for z = it, t > 0, directly from therepresentation (4.72) together with (4.75) (with a = 0) from Lemma 4.81. Byholomorphic continuation we extend (4.77) to all z ∈ H.

For the second transformation property consider again the representation (4.72)of F1. Since n ∈ N and the trigonometric function cos is 2π-periodic, we conclude

F1(z + 2)(4.72)

=∑n∈Z

1

cos(nπz + 2nπ)


=∑n∈Z

1

cos(nπz)

(4.72)= F1(z).

Lemma 4.83. The function f defined in (4.71) satisfies

(4.79) lim=(z)→∞

F1(z) = 1.

Proof. Consider f as defined in (4.71). The exponentials appearing in thedenominators satisfy

lim=(z)→∞

∣∣e±πinz∣∣ = lim=(z)→∞

e∓πn=(z) =

0 if ∓n < 0,

1 if n = 0 and

∞ if ∓n > 0.

Hence the expression 1eπinz+e−πinz satisfies

lim=(z)→∞

1

eπinz + e−πinz=

0 if n 6= 0 and

2 if n = 0.

We conclude that f defined in (4.71) satisfies

lim=(z)→∞

F1(z)(4.71)

= lim=(z)→∞

2∑n∈Z

1

eπinz + e−πinz

= 1 +∑n∈Z6=0

lim=(z)→∞

1

eπinz + e−πinz

= 1.

Lemma 4.84. The function F1 defined in (4.71) satisfies

(4.80) F1

(1 +

1

z

)= 4

z

ieπiz2 +O

(|z| e

−3π=(z)2

).

as = (z)→∞.

Proof. We use Lemma 4.81 with a = 12 and get (for z = it, t > 0)

F1

(1 +

1

it

)(4.72)

=∑n∈Z

1

cos(πn+ πn

it

)=∑n∈Z

1

(−1)n cos(iπnt)

(4.75)=

∑n∈Z

(−1)n

cosh(πnt

)(4.74)

=∑n∈Z

t

cosh(π(n+ 1

2

)t)

=it

i

∑n∈Z

1

cos(π(n+ 1

2

)it) ,

using cos(−x) = cos(x) in the last step. Using analytic continuation in z, we extendthe growth estimate from z = it to all z ∈ H .


Splitting the right series into terms indexed by n ∈ −1, 0 and n 6∈ −1, 0we get

F1

(1 +

1

z

)=z

i

∑n∈Z

1

cos(π(n+ 1

2

)z)

=z

i

2

cos(πz2

) +z

i

∑n∈Zr−1,

1

cos(π(n+ 1

2

)z)

we get as estimate

F1

(1 +

1

z

)=z

i

2

cos(πz2

) + O

zi

∑n∈Z6=0

1

cos(π(n+ 1

2

)z)

=z

i

4

eπiz2 + e−

πiz2

+ O(t e−

3π=(z)2

).

Since4

eπiz2 + e−

πiz2

≈ 4 eπiz2

as = (z)→∞, we conclude the growth estimate

F1

(1 +

1

z

)= 4

z

ieπiz2 + O

(t e−

3π=(z)2

)as = (z)→∞.

Comparing now F1 and θ2 we get

Theorem 4.85. The functions F1 in (4.71) and θ2 are equal:

(4.81) F1(z) = θ2(z) for all z ∈ H.

Proof. Collecting the transformation properties of F1 in Proposition 4.82,together with its holomorphicity Lemma 4.80 and its growth estimate in Lem-mata 4.83 and 4.67 and similar properties for θ2 deduced from Corollaries 4.69,4.70 and 4.71, we see that the quotient function

g(z) :=F1(z)

θ2(z)for all z ∈ H

is holomorphic on H (since θ does not vanish – this follows from Corollary 4.77since none of the factors in the product expansion vanishes on H). The function gsatisfies the invariants

g(z + 2) = g(z) and g

(−1

2

)= g(z).

Moreover, the function is bounded at the cusps∞ and 1 since the growth estimatesfor both F1 and θ cancel there. Recalling the Mbius transformation of S2 and T ,see (1.9) and (2.8), and that the theta group Γθ is generated by S2 and T , seeExample 2.36, we conclude analogously as in the proof of Theorem 3.18, that gsatisfies

g(M z) = g(z) for all M ∈ Γθ and z ∈ H.Since Γθ ⊂ Γ(1) is a subgroup of finite index – the index is calculated to be 3 in

Exercise 2.36 – we may apply Theorem 3.32 and we conclude that g is a constantfunction. Hence, there exists a c ∈ C (independent of z ∈ H) satisfying

F1(z) = c θ2(z)

for all z ∈ H.


The constant c has to be 1: Using

lim=(z)→∞

θ2(z) = 1

in (4.58) andlim

=(z)→∞F1(z) = 1

in (4.79) we conclude

c =F1(z)


= lim=(z)→∞

F1(z)

θ2(z)

=lim=(z)→∞ F1(z)

lim=(z)→∞ θ2(z)

=1

1= 1.

One may wonder how the above result may help us with our motivating questionof understanding the counting function r2(n) in (4.67). Since r2(n) appears in theFourier expansion of θ2 as coefficient, see (4.69) in Exercise 4.79, we have

Corollary 4.86. We have

(4.82)∑n∈Z≥0

r2(n) eπinz = 1 + 4∑n∈N

eπinz

1 + e2πinz

for all z ∈ H.

Proof. Using (4.81) of Theorem 4.85 and substituting the expansions of θ2 in(4.69) and F1 in (4.73) we get immediately (4.82).


(4.83) r2(n) = 4(d1(n)− d3(n)

)for all n ∈ N.

Here, d1 :IN → Z and d3 : N→ Z denote the counting functions given by

d1(n) : = ]k ∈ N; 4k + 1 | n

and

d3(n) : = ]k ∈ N; 4k + 3 | n

.

(4.84)

Remark 4.88. The counting functions d1(n) and f3(n) in (4.84) count thenumber of divisors of n of the form 4k + 1 and 4k + 3, respectively.

Before we prove Corollary 4.87, we need a combinatorial result.

Lemma 4.89. We have

(4.85)eπinz

1 + e2πinz=(d1(n)− d3(n)

)eπinz

for all n ∈ N and z ∈ H.

Proof. Using the identity

1

1 + a2=

1− a2

1− a4=

1

1 + a4− a2

1− a4

with a = eπinz, we find

eπinz

1 + e2πinz=

eπinz

1− e4πinz− e3πinz

1− e4πinz.


Using the geometric series expansions

bn

1− b4n=

∑m∈Z≥0

bn(4m+1) n(4m+1)=k=

∑k∈N

d1(k) bk

andb3n

1− b4n=

∑m∈Z≥0

bn(4m+3) n(4m+3)=k=

∑k∈N

d3(k) bk

with b = eπiz, we find

eπinz

1 + e2πinz=

eπinz

1− e4πinz− e3πinz

1− e4πinz

=∑k∈N

d1(k) eπikz −∑k∈N

d3(k) eπikz.

The series on the right and side above converge since d1/3(k) ≤ k holds for allk ∈ N.

proof of Corollary 4.87. Since the Identity (4.82) holds for all z ∈ H weconclude ∑

n∈Z≥0

r2(n) eπinz(4.82)

= 1 + 4∑n∈N

eπinz

1 + e2πinz

(4.85)= 1 + 4

∑n∈N

(d1(n)− d3(n)

)eπinz

for all z ∈ H. Hence, the coefficients of the Fourier expansion agree and we get

r2(n) = 4(d1(n)− d3(n)

)for all n ∈ N.

We considered the question about the behaviour of r2(n), n ∈ N, which weanswered in Corollary 4.87. The key idea was that we consider the function θ2(z)which contains the values r2(n) as coefficients of the Fourier expansion of θ2(z).

We now discuss r4(n) given in (4.68). First, we recall the relation betweenr4(n) and the Fourier expansion of θ4(z) in (4.70) of Exercise 4.79. This relates theunderstanding of the counting function r4(n) to understanding θ4(z). As above forθ2, see Theorem 4.85, we would like to relate θ4(z) to another function which wecall F2.

Definition 4.90. The function F2 : H→ C is defined by the series

(4.86) F2(z) :=∑m∈Z

∑n∈Z

′ 1(m2 z + n

)2 − ∑m∈Z

∑n∈Z

′ 1(mz + n

2

)2 .Here the notation

∑m∈Z

∑n∈Z′

means that we calculate first the inner series (overn) and then the outer series (over m); the prime ′ indicates that we skip over theterm with (m,n) = (0, 0).

Remark 4.91. Recall that the above series are conditionally convergent andnot absolutely convergent. This means that the order of the summation playsa crucial role. This is similar to the two expansions of the Eisenstein series G2

presented in Theorem 4.20, where we also employ the prime notation. We refer toLemma 4.92 where we make the relation between F2 and G2 explicit.


Lemma 4.92. Recall the Eisenstein series G2 in Definition4.18. We have

(4.87) F2(z) = G2

(z2

)− 4G2(2z)

for all z ∈ H.

Remark 4.93. Lemma 4.92 shows in particular that F2 is well defined since itis represensted in (4.87) in terms of expressions of the Eisenstein series G2.

Proof of Lemma 4.92. Recall the definitions of the functions G2 and F2 in(4.18) and (4.86), respectively. The Eisenstein series has a description as condi-tionally convergent series in (4.14) of Theorem 4.20. Putting everything togetherwe get

F2(z)(4.86)

=∑m∈Z

∑n∈Z

′ 1(m2 z + n

)2 − ∑m∈Z

∑n∈Z

′ 1(mz + n

2

)2=∑m∈Z

∑n∈Z

′ 1(m z

2 + n)2 − ∑

m∈Z

∑n∈Z

′ 1(m2 (2z) + n

2

)2(4.14)

= G2

(z2

)− 4G2

(z2

)for all z ∈ H.

Next, we use the properties of the Eisenstein series G2 given in Theorem 4.20to derive properties of the function F2.

Exercise 4.94. Show that the function F2 is 2-periodic:

(4.88) F2(z + 1) = F (z)

for all z ∈ H.

Proposition 4.95. The function F2 satisfies the transformation property

(4.89) − z−2 F2

(−1

z

)= F2(z)

for all z ∈ H.It satisfies the growth estimates

(4.90) lim=(z)→∞

F2(z) = −π2

and

(4.91) F2

(1− 1

z

)= O

(|z|2 e−2πδ=(z)

)as = (z)→∞

for some (small) δ > 0.

Proof. We derive (4.89) from its companion property of G2 in (4.16). UsingLemma 4.92, we have

−z−2 F2

(−1

z

)(4.87)

= −z−2G2

(1

2

−1

z

)+ 4z−2G2

(2−1

z

)= −4

[(2z)−2G2

(−1

2z

)]+(z

2

)−2

G2

(−1z2

)(4.16)

= −4

[G2(2z)− 2πi

2z

]+ G2

(z2

)− 2πi

z2

= −4G2(2z) +4πi

z+G2

(z2

)− 4πi

z(4.87)

= F2(z).


To prove the growth estimate (4.90), we use again the corresponding result forG2. We have

lim=(z)→∞

F2(z)(4.87)

= lim=(z)→∞

(G2

(z2

)− 4G2(2z)

)= lim=(z)→∞

G2

(z2

)− 4 lim

=(z)→∞G2(2z)

(4.20)= 2ζ(2)− 4

(2ζ(2)

)= −6ζ(2).

Using ζ(2) = π2

6 , see e.g. [30, (25.6.1)], we conclude (4.90).To prove (4.91), we use again (4.87) and apply the transformation laws (4.16)

and (4.17) of G2 several times. We have


F2

(1− 1

z

)(4.87)

= G2

(1

2− 1

2z

)− 4 G2

(2− 2

z

)︸︷︷︸(4.17)

= G2(− 2z )

(4.17)= G2

(z − 1

2z

)− 4G2

(−1z2

)(4.16)

=

[(z − 1

2z

)−2

G2

(2z

1− z

)+

2πiz−12z

]− 4

(z2

)2[G2

(z2

)− 2πi

z2

]

=

(z − 1

2z

)−2

G2

(2z

1− z

)− 4

(z2

)2

G2

(z2

)+ 4πi

z

z − 1+ 4πiz

using 2z1−z = 2

1−z − 2

=

(z − 1

2z

)−2

G2

(2

1− z− 2

)︸︷︷︸

(4.17)= G2( 2

1−z )

−4(z

2

)2

G2

(z2

)+ 4πi

z

z − 1+ 4πiz

(4.17)=

(z − 1

2z

)−2

G2

(2

1− z

)− 4

(z2

)2

G2

(z2

)+ 4πi

z

z − 1+ 4πiz

(4.16)=

(z − 1

2z

)−2[(

2

1− z

)−2

G2

(z − 1

2

)+

2πi2

1−z

]−

− 4(z

2

)2

G2

(z2

)+ 4πi

z

z − 1+ 4πiz

= z2G2

(z − 1

2

)− z2G2

(z2

)−4πi

z2

z − 1+ 4πi

z

z − 1+ 4πiz︸︷︷︸

=0

= z2

[G2

(z − 1

2

)−G2

(z2

)]for all z ∈ H. Applying (4.21), we find

F2

(1− 1

z

)= O

(z2

[G2

(z − 1

2

)−G2

(z2

)])= O

(z2

[(G2

(z − 1

2

)− 2ζ(2)

)−(G2

(z2

)− 2ζ(2)

)])= O

(z2

[(G2

(z − 1

2

)− 2ζ(2)

)−(G2

(z2

)− 2ζ(2)

)])(4.21)

= O(|z|2 e−2πδ=(z)

)as = (z)→∞ for some (small) δ > 0. This shows (4.91)

Proposition 4.96. Define the modified divisor function σ(4-)1 : N→ N by

(4.92) σ(4-)1 (n) :=

∑0<d|n

4-d

d.

We have

(4.93) F2(z) = −π2 − 8π2∑n∈N

σ(4-)1 (n) eπinz


for all z ∈ H.

Proof. Recall the expansion (4.13) of G2 in Definition 4.18. Together with(4.87), we have

F2(z)

(4.87)= G2

(z2

)− 4G2(2z)

(4.13)= 2ζ(2) + 2(2πi)2

∞∑n=1

σ1(n) e2πin z2 −

− 4

(2ζ(2) + 2(2πi)2

∞∑n=1

σ1(n) e2πin(2z)

)= −6ζ(2) + 2(2πi)2

∑n∈N

σ1(n)eπinz − 8(2πi)2∑n∈N

σ1(n) e2πin(2z)

4n=m= −6ζ(2) + 2(2πi)2

∑n∈N

σ1(n)eπinz − 2(2πi)2∑m∈N4|m

4σ1

(m4

)eπimz

= −6ζ(2) + 2(2πi)2∑n∈N4-n

σ1(n) e−πinz +

+ 2(2πi)2∑n∈N4|n

(σ1(n)− 4σ1

(n4

))e−πinz

(4.92)= −6ζ(2) + 2(2πi)2

∑n∈N

σ(4-)1 (n) e−πinz.

Using ζ(2) = π2

6 , see e.g. [30, (25.6.1)], we conclude

F2(z) = −π2 − 8π2∑n∈N

σ(4-)1 (n) e−πinz.

Exercise 4.97. Show that the divisor functions σ1(n) defined in (4.12) and

the modified divisor function σ(4-)1 (n) in (4.92) satisfiy

(4.94) σ(4-)1 (n) =

σ1(n) if 4 - n and

σ1(n)− 4σ1

(n4

)if 4 | n

for all n ∈ N.

Matching θ4 and F2 we have

Theorem 4.98. The function F2 and θ4 satisfy

(4.95) F2(z) = −π2 θ4(z)

for all z ∈ H.

Exercise 4.99. Prove Theorem 4.98 by adapting the proof of Theorem 4.85.


(4.96) θ4(z) = 1 +∑n∈N

8σ(4-)1 (n) eπinz

for all z ∈ H.

Proof. Combine (4.93) in Proposition 4.96 with (4.95) in Theorem 4.98.



(4.97) r4(n) = 8σ(4-)1 (n)

for all n ∈ N.

Proof. We use (4.96) of the above Corollary 4.100 on one hand and (4.70) ofExercise 4.79 on the other hand. This gives

1 +∑n∈N

r4(n) eπinz(4.70)

= θ4(z)(4.96)

= 1 +∑n∈N

8σ(4-)1 (n) eπinz

for all z ∈ H. Hence, the coefficients have to agree and we conclude (4.97).

Remark 4.102. We just proved Jacobi’s theorem, which we mentioned in theintroduction of part 1, see page ??. As one can see, the study of arithmetic prop-erties of counting function of the type r2 in (4.67) or r4 in (4.68) leads to modularforms of its generating function θ2 or θ4.1

Now we know that the theta function is associated to the counting functions r2

and r4. What about other rk’s? We discuss this only for k ∈ N which are divisibleby 8.

Definition 4.103. Let 8 | k ∈ N. The counting function rk : Z≥0 → Z≥0 isdefined by

(4.98) rk(n) = ]

(x1, . . . , xk) ∈ Zk; x21 + . . .+ x2

k = n

for all n ∈ Z≥0.

Lemma 4.104. Let 8 | k ∈ N. The generating function of rk satisfies

θk(z) =∑

n1,...,nk∈Zeπi(n

21+...+n2

k)z

= 1 +∑n∈N

rk(n) eπinz(4.99)

for all z ∈ H.

Proof. Similar to Exercise 4.79 we have

θ(z)kE6.1.15a

= Θ(0, z)k

(4.49)=

(∑n∈Z

eπin2z

)k=

∑n1,...,nk∈Z

eπi(n21+...+n2

k)z

(4.98)=

∑m∈Z≥0

rk(m) eπimz.

Corollary 4.105. Let 8 | k ∈ N. We have

(4.100) θk(z + 2) = θk(z)

for all z ∈ H.

Proof. This follows immediately from θ(z+2)(4.57)

= θ(z) in Corollary 4.69.

1To do: Please find a good reverence for an in-depth discussion of theta functions!


To show the transformation property under z 7→ −1z consider first the following

axillary result.

Lemma 4.106. Let 8 | k ∈ N. We have

(4.101)∑

n1,...,nk∈Ze−π(n2

1+...+n2k)t =

1

yk2

∑n1,...,nk∈Z

e−πn2

1+...+n2k

t

for all t > 0.

Proof. Using Lemma 4.65, we find∑n1,...,nk∈Z

e−π(n21+...+n2

k)t =∑

n1,...,nk∈Ze−πn

21t · · · e−πn

2kt

(4.51)=

∑n1,...,nk∈Z

1√te−π

n21 · · · 1√

te−πn

2kt

=1

tk2

∑n1,...,nk∈Z

e−πn2

1+...+n2k

t .

Proposition 4.107. Let 8 | k ∈ N. We have

(4.102) z−k2 θk

(−1

z

)= θk(z)

for all z ∈ H.

Proof. Consider z = it with t > 0. We have

θk(it)(4.99)

=∑

n1,...,nk∈Ze−π(n2

1+...+n2k)t

(4.101)=

1

tk2

∑n1,...,nk∈Z

e−πn2

1+...+n2k

t

(4.99)=

1

tk2

θk(−1

it

).

Using the holomorphy of θ allows us to extend the above identity to all z ∈ H. Thisproves (4.102).

Proposition 4.108. Let 8 | k ∈ N. The function θk satisfies

(4.103) j(M, z)−k2 θk

(M z

)= θk(z)

fo all z ∈ H and M ∈ Γθ.

Proof. Recalling the Mbius transformation of S2 and T , see (1.9) and (2.8),we see that θk satisfies (4.103) for the elements S2 an T . Since the theta group Γθis generated by S2 and T , see Example 2.36, we can conclude analogously to theproof of Theorem 3.18, that θk satisfies (4.103).

Remark 4.109. The above proposition shows that θk (for 8 | k ∈ N) satisfiesthe transformation behaviour of a modular form of weight k

2 . We have to check the

growth in the cusps ∞ and 1 of FΓθ to check whether θk is a modular form in thesense of Definition 3.26.

Lemma 4.110. Let 8 | k ∈ N. The function θk satisfies

(4.104) lim=(z)→∞

θk(z) = 1


and

(4.105) θk(

1− 1

z

)= O

(|z| k2 e− kπ2 =(z)

)as = (z)→∞.

Proof. The first identity is a direct consequence of Corollary 4.70. The secondfollows from Corollary 4.71.

Theorem 4.111. Let 8 | k ∈ N. The function θk is a modular form of weightk2 for the theta group Γθ:

(4.106) θk ∈M k2

(Γθ).

Proof. θk satisfies the transformation property of modular forms, see Proposi-tion 4.108. The growth conditions in Lemma 4.110 imply that the Fourier expansionof θk is regular at the cusps∞ and 1. We also know that θk is holomorphic. Hence,θk ∈M k

2

(Γθ)

holds according to Definition 3.26.

CHAPTER 5

Hecke Operators

In this chapter, we define Hecke operators on the space of modular forms andgive their basic properties. We demonstrate how Hecke operators are self adjointunder Petersson inner product and we use this to show that the space of cusp formsis a Hilbert space. We then discuss the product representation of the L-functionsassociated to eigenforms and give the analytic continuation of these L-functions.

5.1. The Slash Operator

Definition 5.1. Let f be a meromorphic function on H and k be a real number.We define the slash operator of weight k by

(5.1)(f∣∣kM)(z) := j(M, z)k f(Mz) (z ∈ H)

where M ∈ Mat2 (R) is a 2 × 2 matrix and the automorphic factor j is defined in(3.4).

For example, the slash operator allows us to write (3.3) as

f∣∣kM = v(M) f for M ∈ Γ.

We sometimes extend the slash operator to work with formal sums of matrices inthe following sense:

(5.2) f∣∣k(M + V ) := f

∣∣kM + f

∣∣kV for M ∈ Γ.

Lemma 5.2. Let k ∈ 2Z and f : H→ C . We have

(5.3)(f∣∣kM)∣∣kV = f

∣∣kMV

for all matrices M,V ∈ GL2 (R).

Proof. Using (1.10) and Lemma 3.3, we find that((f∣∣kM)∣∣kV

)(z) = j(V, z)−k

(f∣∣kM

)(V z)

= j(V, z)−k j(M,V z)−k f(M(V z)

)= j(MV, z)−k f(MV z)

=(f∣∣kMV

)(z)

for all z ∈ H.

We can extend the slash operator to work with formal linear combinationsof matrices. To do so, we first define formal linear combinations of matrices inMat2 (Z). (Here, Mat2 (Z) is given in (1.1).)

Definition 5.3. Let n ∈ N. The set R? of finite linear combinations of ele-ments Mat2 (Z) with integer coefficients is defined as

(5.4) R? := Z[Mat2 (Z)

].

127

128 5. HECKE OPERATORS

The set Rn of finite linear combinations of elements Mat2 (Z) with fixed determi-nant n with integer coefficients is defined as

(5.5) Rn := Z[M ∈ Mat2 (Z) ; detM = n

].

We think of the elements of R? respectively Rn as formal finite linear combi-nations of matrices in Mat2 (Z) with or without fixed determinant. Obviously, R?respectively Rn is an additive ring: If

∑i aiMi,

∑j bjVj ∈ R?, then

∑i aiMi +∑

j bjVj ∈ R?. The other relations follow analogously.We can extend the definition of the slash operator to elements in R?.

Definition 5.4. Let f be a meromorphic function on H and k be a real number.We define the slash operator of weight k on R? by

(5.6)(f∣∣k

∑i

aiMi

)(z) :=

∑i

ai(j(M, z)k f(Miz)

)(z ∈ H,

∑i

aiMi ∈ R?).

The definition on Rn with (formal) linear combinations of matrices of fixed deter-minant n ∈ N is analog.

5.2. Two Principles

5.2.1. Decent to a subgroup. Let f(z) be a modular form of even integerweight k ∈ 2Z and trivial multiplier v0 ≡ 1 on the full modular group Γ(1). We fixa natural number n ∈ N and consider the following function

(5.7) g(z) := f(nz) (z ∈ H).

Our claim is that g is a modular form on Γ0(n) for weight k and trivial multiplier.

To prove this, let M =

(a bnc d

)∈ Γ0(n). We have

(5.8)

g(M z) = f(n (M z)

)= f

(naz + b

ncz + d

)= f

(a(nz) + nb

c(nz) + d

)= f

((a nbc d

)(nz)

).

Since

det

(a bnc d

)= 1 = det

(a nbc d

),

we have

(a nbc d

)∈ Γ0(n) ⊂ Γ(1) where Γ0(n) is defined in (2.5). Using that f is

a modular form of weight k and trivial multiplier allows us to apply (3.3). We find

(5.9) g(M z) = f

((a nbc d

)(nz)

)=(c(nz) + d

)kf(nz) =

(c(nz) + d

)kg(z).

Hence g satisfies the transformation property (3.3) for all M ∈ Γ0(n) with weightk and trivial multiplier. The function g inherits the other properties of a modularform from f , see Definition 3.26.

Remark 5.5. The calculuation above reveals that it is sufficient for g to be amodular form on Γ0(n) if f is modular on Γ0(n).

5.2.2. The subgroup method. Suppose that we have two groups Γ1 ⊂ Γ2

of finite index in Γ(1), and let Aii denote representatives of the right cosets ofΓ1 in Γ2. This means that we can write

Γ2 =

µ⋃i=1

Γ1Ai and Γ1Ai ∩ Γ1Aj = ∅ for all i 6= j.

Suppose f ∈M !k

(Γ1

)for some k ∈ 2Z and trivial multiplier, see Remark 3.28.

5.2. TWO PRINCIPLES 129

We define

(5.10) F (z) :=

µ∑i=1

(f∣∣kAi)(z) = f

∣∣k

(µ∑i=1

Ai

)for all z ∈ H using the slash operator introduced in (5.1) respectively its extensionto R? defined in (5.4).

Proposition 5.6. Under the above assumptions we have F ∈M !k

(Γ2

).

Before we prove the proposition, we need few intermediate steps which wepresent in the following lemmas.

Lemma 5.7. Let M ∈ Γ2. Then there exists an bijective map

1, . . . , µ → 1, . . . , µ; i 7→ i′,

and Mi ∈ Γ1 depending on M , such that

AiM = MiAi′

holds for all i.

Proof. In order to prove the lemma, we show that Ai′ runs through thecomplete set of right representatives ifAi does the same, by using a similar argumentas presented in the proof of Theorem 2.17.

Assume we have i, j ∈ 1, . . . , µ such that i′ = j′ (or in other words Ai′ = Aj′).This means that we have matrices Mi,Mj ∈ Γ1 such that

AiM = MiAi′ and AjM = Mj Aj′

holds. We have

AiA−1j =

(AiM

)(AjM

)−1

=(MiAi′

)(Mj Aj′

)−1

= Mi Ai′A−1j′︸︷︷︸

=1

M−1j

= MiM−1j ∈ Γ1.

Hence Ai ∈ Γ1Aj which implies

Ai = Aj ,

since the Ai’s are representatives of the right cosets of Γ1 in Γ2. The argumentabove shows that the map

1, . . . , µ → 1, . . . , µ; i 7→ i′

is injective. Hence the map is bijective since the cardinality of the domain is finiteand equal to that of the codomain. This shows that if i runs through 1, . . . , µthen i′ does it, too.

Lemma 5.8. Let i ∈ 1, . . . , µ. The function f∣∣kAi is independent of the

choice of the representative Ai.

This lemma shows that F in (5.10) is well defined.

Proof of Lemma 5.8. We want to show that f∣∣kAi is independent of the

choice of the representative Ai.Assume that Ai and Bi are two representatives of the same right coset of Γ1

in Γ2. Then, they differ by an element M ∈ Γ1:

Bi = M Ai for some M ∈ Γ1.


Using Lemma 5.1, we have

f∣∣kBi = f

∣∣k(MAi) =

(f∣∣kM)∣∣kAi.

We use the assumption f ∈M !k

(Γ1

)and get with (3.3)(

f∣∣kM)∣∣kAi = f

∣∣kAi.

This shows f∣∣kBi = f

∣∣kAi.

Proof of Proposition 5.6. First, we verify the transformation law (3.3) forF under the group Γ2. Let M ∈ Γ2 be a matrix. Using Lemma 5.2, we obtain

F∣∣kM =

µ∑i=1

f∣∣k(AiM).

Using the map i→ i′ and the elements Mi ∈ Γ1 introduced in the Lemma 5.7, wehave

F∣∣kM =

µ∑i=1

f∣∣k(AiM)

=

µ∑i=1

(f∣∣kMi

)∣∣kAi′

(3.3)=

µ∑i=1

f∣∣kAi′

=

µ∑i=1

f∣∣kAi = F.

The last identity is due to the fact that the map i 7→ i′ is in fact a permutation.

5.2.3. A combination of both principles. We combine the two principlesgiven in §5.2.1 and §5.2.2 above.

We start with f ∈M !k

(Γ1

), k ∈ 2Z, and define

g(z) := f(pz) (z ∈ H)

for some prime p as discussed in §5.2.1. For the sake of simplicity, we restrict ourdiscussion to prime p first and then generalize for any integer n ∈ N.

Next, we would like to apply the principle presented in §5.2.2 for Γ0(p) ⊂ Γ(1).Using Theorem 2.20, we have

(5.11) Γ(1) =

p−1⋃j=0

Γ0(p)TSj

∪ Γ0(p).

This gives us the following right coset representation of Γ0(p) in Γ(1):1, T, TS, . . . TSp−1

.

As in (5.10), we define

F := g∣∣k1 +

p−1∑j=0

g∣∣kTSj = g

∣∣k

1 +

p−1∑j=0

TSj

.

Combining both steps gives

F (z) =

g∣∣k1 +

p−1∑j=0

g∣∣kTSj

(z)

5.2. TWO PRINCIPLES 131

= f(pz) +

p−1∑j=0

(f(p·)

)∣∣kTSj(z)

for any z ∈ H. Using (2.8), we get

TSj =

(0 −11 0

) (1 j0 1

)=

(0 −11 j

).

Hence, F can be written as

F (z) = g(z) +

p−1∑j=0

(z + j)−k g

(−1

z + j

)

= f(pz) +

p−1∑j=0

(z + j)−k f

(−pz + j

).

With the slash operator defined in (5.1), the above formula can be written as

(5.12) F := f∣∣k

(p 00 1

)+

p−1∑j=0

f∣∣k

(0 −p1 j

)= f

∣∣k

(p 00 1

)+

p−1∑j=0

(0 −p1 j

) .

However, the formula in (5.12) is not yet exactly what we want. We use thefact that F is a modular form, and in particular satisfies (3.3), to rearrange thesum in (5.12). Using F

∣∣kT−1 = F , we get

F(5.12)

= f∣∣k

(p 00 1

)+

p−1∑j=0

f∣∣k

(0 −p1 j

)(3.3)= f

∣∣k

(p 00 1

)+

p−1∑j=0

(f∣∣kT−1

)∣∣k

(0 −p1 j

)(5.3)= f

∣∣k

(p 00 1

)+

p−1∑j=0

f∣∣kT−1

(0 −p1 j

)

= f∣∣k

(p 00 1

)+

p−1∑j=0

f∣∣k

(1 j0 p

),

where we use

T−1

(0 −p1 j

)=

(0 1−1 0

)(0 −p1 j

)=

(1 j0 p

).

This leads to the following definition

Definition 5.9. Let p ∈ N be a prime and k ∈ 2Z. We define the mapTp : M !

k

(Γ(1)

)→M !

k

(Γ(1)

)given by

(5.13) Tpf := f∣∣k

(p 00 1

)+

p−1∑j=0

f∣∣k

(1 j0 p

)= f

∣∣k

(p 00 1

)+

p−1∑j=0

(1 j0 p

) .

Remark 5.10. It is known that the operator Tp defined above preserves thespaces M !

k

(Γ(1)

), Mk

(Γ(1)

)and Sk

(Γ(1)

).


5.3. Hecke Operators Of Index n ∈ N

Definition 5.11. Let k be an integer and let f : H→ C. For n ∈ N and z ∈ H,we define(5.14)(

Tnf)(z) := nk−1

∑d|n

d−kd−1∑b=0

f

(nz + bd

d2

)= nk−1 f

∣∣k

∑d|n

d−1∑b=0

(nd b0 d

) (z)

the Hecke operator of index n.

Remark 5.12. Suppose n = p is a prime. Then Tp defined above and Tp in(5.13) are closely related. For n = p prime, we have d ∈ 1, p. Then Tp in (5.14)can be written as(

Tpf)(z)

(5.14)= pk−1

∑d|p

d−kd−1∑b=0

f

(pz + bd

d2

)

= pk−1

0∑b=0

f

(pz + b

12

)︸︷︷︸

case d=1

+ d−kd−1∑b=0

f

(pz + bd

d2

)︸︷︷︸

case d=p

= pk−1

[f(pz) +

p−1∑b=0

f

(z + b

p

)]= pk−1

(Tpf

)(z)

for z ∈ H.

Proposition 5.13. Let k be an integer and n ∈ N. If a function f : H→ C isperiodic, then Tnf is also periodic. In other words, we have

(5.15) f(z + 1) = f(z) =⇒(Tnf

)(z + 1) =

(Tnf)

)(z)

for all z ∈ H.

Proof. We calculate(Tnf

)(z + 1). Inserting z + 1 in (5.14), we have

(Tnf

)(z + 1) = nk−1

∑d|n

d−kd−1∑b=0

f

(nz + bd+ n

d2

)

= nk−1∑d|n

d−kd−1∑b=0

f

(nz +

(b+ n

d

)d

d2

).

Writing n = ad, we have for any t ∈ Z the identity(a b0 d

)(1 t0 1

)=

(1 t′

0 1

)(a b′

c d

)with b′ = b + at − dt′. We choose t′ ∈ Z such that b′ is between 0 and d − 1:0 ≤ t′ ≤ b− 1. For t = 1, we find in particular(

a b0 d

)(1 10 1

)=

(1 t′

0 1

)(a b′

c d

)and 0 ≤ b′ ≤ d − 1 for suitably chosen t′ ∈ −1, 0. Applying this to the abovecalculation, we get(

Tnf)(z + 1) = nk−1

∑d|n

d−kd−1∑b=0

f

(nz +

(b+ n

d

)d

d2

)

5.3. HECKE OPERATORS OF INDEX n ∈ N 133

= nk−1∑d|n

d−kd−1∑b=0

f

((a b0 d

)(1 10 1

)z

)

= nk−1∑d|n

d−kd−1∑b=0

f

((1 t′

0 1

)(a b′

0 d

)z

).

Using f(z + t′) = f(z) (since f is periodic) we get

f

((1 t′

0 1

)(a b′

0 d

)z

)= f

((a b′

0 d

)z

).

Hence, we have(Tnf

)(z + 1) = nk−1

∑d|n

d−kd−1∑b=0

f

((1 t′

0 1

)(a b′

0 d

)z

)

= nk−1∑d|n

d−kd−1∑b=0

f

((a b′

0 d

)z

)

= nk−1∑d|n

d−kd−1∑b=0

f

((a b0 d

)z

).

The last identity is due to the fact that b′ runs through all integers between 0 andd−1 if b does it. In fact, the map b 7→ b′ is a permutation on the set 0, 1, . . . , d−1.Summarizing, we just showed(

Tnf)(z + 1) =

(Tnf

)(z).

Theorem 5.14. Let k ∈ 2Z and n ∈ N. Suppose that the function f : H → Chas a Fourier expansion of the form

f(z) =∑m∈Z

c(m) e2πimz (z ∈ H).

Then the function Tnf admits the Fourier expansion

(5.16)(Tnf

)(z) =

∑m∈Z

γn(m) e2πimz

with

(5.17) γn(m) :=∑

0<d|gcd(n,m)

dk−1 c(mnd2

).

A direct consequence is the following corollary.

Corollary 5.15. In the setting of Theorem 5.14, consider the coefficientsγn(m) as defined in (5.17). These satisfy

(5.18) γn(m) = γm(n)

for all m,n ∈ N.

Proof of Corollary 5.15. Theorem 5.14 implies

γn(m)(5.17)

=∑

0<d|gcd(n,m)

dk−1 c(mnd2

)(5.17)

= γm(n)

for all m,n ∈ N.

Before we prove Theorem 5.14, we need a short lemma about certain Dirichletsums.


Lemma 5.16. For d ∈ N and m ∈ Z, we have

(5.19)

d−1∑b=0

e2πim bd =

d if d | m and

0 if d - m.

We leave the proof as an exercise to the reader.

Proof of Theorem 5.14. Plugging the Fourier expansion into (5.14), we get

(Tnf

)(z) = nk−1

∑d|nd−k

d−1∑b=0

∑m∈Z

c(m) e2πimnz+bd

d2

=∑m∈Z

∑d|n

(nd

)k−1

c(m) e2πimnzd2

1

de2πim b

d .

Using Lemma 5.16, we find

(Tnf

)(z) =

∑m∈Z

∑d|n

(nd

)k−1

c(m) e2πimnzd2

1

de2πim b

d

(5.19)=

∑m∈Z

∑d|nd|m

(nd

)k−1

c(m) e2πimnzd2

m=qd=

∑g∈Z

∑d|n

(nd

)k−1

c(gd) e2πigdnzd2

=∑g∈Z

∑d|n

(nd

)k−1

c(gd) e2πig nzd .

Observe that if d runs through all divisors of n then so does nd . Hence, we may

substitute d by nd in the inner sum. This gives

(Tnf

)(z) =

∑g∈Z

∑d|n

(nd

)k−1

c(gd) e2πig nzd

=∑g∈Z

∑d|n

(nnd

)k−1

c(gn

d

)e

2πig nznd

=∑g∈Z

∑d|n

dk−1 c(gnd

)e2πigdz

=∑g∈Z

∑d|n

dk−1 c(gnd

)e2πigdz

m=qd=

∑m∈Z

∑d|nd|m

dk−1 c(mnd2

)e2πimz.

Summarizing, we just showed that Tnf has the Fourier expansion(Tnf

)(z) =

∑m∈Z

∑d|gcd(m,n)

dk−1 c(mnd2

)e2πimz.

This proves Theorem 5.14.

5.4. TRANSFORMATIONS OF ORDER n ∈ N 135

5.4. Transformations of Order n ∈ N

We want to prove that Tn preserves the space of modular forms. We rewritethe Tn’s for this purpose.

Consider a typical term of(Tnf

)(z) as given in (5.14):

d−k f

(nz + bd

d2

)= d−k f

((nd

)z + b

0z + d

)for z ∈ H and integer 0 ≤ b < d and d | n.

Put a = nd which is a natural number since d divides n. Then, the above term

can be written as

d−k f

(az + b

d

)with ad = n.

Recalling the definition of the slash operator in (5.1) we write it as

d−k f

(az + b

d

)=(f∣∣kA)(z)

with A =

(a b0 d

)a 2 × 2 matrix with nonnegative integer coefficients and deter-

minant detA = ad = n.This easy calculation motivates the following rewriting of Tn. Let f : H → C

be a function on the upper half plane and n ∈ N. We have

(Tnf

)(z)

(5.14)= nk−1

∑d|n

d−kd−1∑b=0

f

(nz + bd

d2

)

= nk−1∑ad=n1≤a

0≤b<d

(f∣∣k

(a b0 d

))(z)(5.20)

= n−1∑ad=n1≤a

0≤b<d

ak f

((a b0 d

)z

)

for any z ∈ H.

Recall that

Mat2 (Z)(1.1)=

(a bc d

); a, b, c, d ∈ Z

denotes the set of 2× 2 matrices with integer coefficients.

Definition 5.17. The Mobius transformation Az of an element

(5.21) A ∈ Mat2 (Z) with detA = n

with positive n ∈ N is called a transformation of order n: We denote the set of suchmatrices by H(n):

(5.22) H(n) := A ∈ Mat2 (Z) ; detA = n.

Example 5.18. It is obvious from the definitions (2.1) and (5.22) that the fullmodular group Γ(1) and the set of transformation of order 1 are equal: Γ(1) = H(1).

Exercise 5.19. For A ∈ H(n) and B ∈ Γ(1), we have AB,BA ∈ H(n).

This exercise shows that Γ(1) forms an action on H(n) by multiplication.


Definition 5.20. Let A1, A2 ∈ H(n) for some n ∈ N. We call A1 and A2

equivalent or Γ(1)-equivalent denote A1 ∼ A2 if

∃V ∈ Γ(1) : A1 = V A2.

The relation ∼ introduced above is indeed an equivalence relation as the namealready suggests. This follows from the characterization

(5.23) A1, A2 ∈ H(n) are equivalent ⇐⇒ A1A−12 ∈ Γ(1).

Definition 5.21. Let n ∈ N and Γ ⊂ Γ(1). The right Γ-equivalence classes ofA ∈ H(n) are denoted by

(5.24) [A]Γ :=M A; M ∈ Γ

.

The set of right Γ-equivalence classes of H(n) is given by

(5.25) Γ\H(n) :=

[A]Γ; A ∈ H(n).

Lemma 5.22. Let n ∈ N. Each equivalence class in H(n) contains an element

of the form

(a b0 d

)or

(−a −b0 −d

)with a, b, d ∈ Z≥0, ad = n and 0 ≤ b < d.

Proof. Let A =

(a bc d

)∈ H(n) be a transformation of order n. We consider

two cases, c = 0 and c 6= 0:

c = 0:: If c = 0, then A has already the form

(a b0 d

). We may assume that

d > 0 is positive. Otherwise, we consider the matrix −1A =

(−a −b0 −d

)which is equivalent to A and satisfies d > 0. In particular, this ensuresalso that a > 0 is positive and detA = ad = n.

Let k ∈ Z such that b+kd satisfies 0 ≤ b+kd < d. In other words we

have b ≡ b+kd mod d. Consider the matrix Sk A with S =

(1 10 1

)∈ Γ(1)

given in (2.8). We have

Sk A =

(1 k0 1

) (a b0 d

)=

(a b+ kd0 d

).

This shows that A is equivalent to a matrix of the required form.c 6= 0:: Write the fraction − c

a = rs in normal form. This means that r ∈ Z,

s ∈ N are coprime. By the Chinese remainder theorem, see §A.3, we find

p, q ∈ Z such that V :=

(p qr s

)∈ Γ(1) holds.

Consider V A. We find

V A =

(p qr s

) (a bc d

)=

(? ?

ra+ sc ?

)=

(? ?0 ?

)since by construction − c

a = rs holds. Hence, A is equivalent to a matrix

V A ∈ H(n) with a vanishing lower left entry. We continue as in the casec = 0 above.

We just proved the lemma.

Theorem 5.23. A complete system of representatives of equivalence classes ofΓ(1)\H(n) is given by

(5.26) T (n) :=

(a b0 d

); 0 < d | n, ad = n, 0 ≤ b < d

.

5.5. THE ACTION OF HECKE OPERATORS ON MODULAR FORMS 137

Proof. Lemma 5.22 shows that each element A of H(n) is equivalent to anelement of the set T (n).

What remains to be done in order to prove the theorem is to show that two dis-

tinct elements of T (n) are not equivalent. Let A1 =

(a1 b10 d1

), A2 =

(a2 b20 d2

)∈

T (n). We have to show that

A1 ∼ A2 ⇐⇒ a1 = a2, d1 = d2 and b1 = b2.

The implication “⇐=” is trivial.To show the implication “=⇒” assume A1 ∼ A2. Hence, there exists a V =(

p qr s

)∈ Γ(1) such that A1 = V A2. We get

A1 = V A2 ⇐⇒(a1 b10 d1

)=

(p qr s

) (a2 b20 d2

)=

(pa2 pb2 + qd2

ra2 rb2 + sd2

).

Comparing the entries, we immediately get 0 = ra2 and hence r = 0 since a2 > 0.Now, the determinant condition detV = 1 implies

1 = ps− qr r=0= ps.

Since p and r are integers, we find that either p, r = 1 or p, r = −1. Comparingthe upper left entry of the matrix identity above, we see a1 = pa2. Hence, p = 1 ispositive since both a1 and a2 are positive. Moreover, this also shows d1 = d2 sincedetA1 = a1d1 = a2d2 = detA2 = n.

The above arguments show that V has the form

V =

(1 q0 1

)= Sq.

However, the upper right entry of the the matrix identity above implies

0 ≤ b1 = b2 + qd2 < d1 = d2.

This inequality is solved only by q = 0. Hence V = 1 is the identity.Summarizing, we have shown that A1, A2 ∈ T (n) are equivalent if and only if

they are already equal.

5.5. The Action of Hecke Operators on Modular Forms

Theorem 5.24. Let n ∈ N, A1 ∈ H(n) a transformation of order n andV1 inΓ(1) a matrix.

Then there exist a triangle matrix A2 ∈ T (n) and V2 ∈ Γ(1) such that

(5.27) A1 V1 = V2A2

holds. Moreover, we have the identity

j(A1, V1 z) j(V1, z) = j(V2, A2 z) j(A2, z)

for all z ∈ H where the automorphic factor j(·, ·) is defined in (3.4).

Proof. The first statement follows easily from Theorem 5.23. For A1 ∈ H(n)and V1 ∈ Γ(1) we have A1 V1 ∈ H(n), see Exercise 5.19. Theorem 5.23 shows thatthere exists exactly one triangle matrix A2 ∈ T (n) which is equivalent to A1 V1. Inother words, there exists exactly one A2 ∈ T (n) and a V2 ∈ Γ(1) such that

A1 V1 = V2A2

holds, see Definition 5.20.


The second part of the theorem follows from Lemma 3.3 (for weight k = 1)using (5.27). We have

j(A1, V1 z) j(V1, z)(3.7)= j(A1 V1, z)

(5.27)= j(V2A2, z)

(3.7)= j(V2, A2 z) j(A2, z)

for all z ∈ H.

Theorem 5.25. Suppose that k ∈ 2Z and n ∈ N.If f : H→ C satisfies

f(V z) = j(V, z)k f(z)

for all V ∈ Γ(1) and z ∈ H then Tn f defined in (5.14) satisfies(Tn f

)(V z) = j(V, z)k

(Tn f

)(z)

for all V ∈ Γ(1) and z ∈ H.

Proof. We apply Theorem 3.18. Proposition 5.13 implies(Tn f

)(S z) =

j(S, z)k(Tn f

)(z) for all z ∈ H, which is the left condition in (3.20).

Now, consider the element T =

(0 −11 0

)∈ Γ(1). Using the set T (n) in (5.26),

we rewrite Tn f in (5.14) and (5.20) as(Tn f

)(z)

(5.20)= nk−1

∑ad=n1≤a

0≤b≤d

(f∣∣k

(a b0 d

))(z)

(5.26)= nk−1

∑A∈T (n)

(f∣∣kA)(z)

for all z ∈ H. Now, consider(Tn f

)∣∣kT (z):

(Tn f

)∣∣kT (z) = nk−1

∑A∈T (n)

(f∣∣kA) ∣∣

kT (z)

(5.3)= nk−1

∑A∈T (n)

(f∣∣kAT)(z).

Using Theorem5.24, there exist VA ∈ Γ(1) such that AS = VAA′ for all A ∈ T (n).

Hence, we have(Tn f

)∣∣kT (z) = nk−1

∑A∈T (n)

(f∣∣kAT)(z)

(5.27)= nk−1

∑A∈T (n)

(f∣∣kVAA

′)(z) with AS = VAA′

(5.3)= nk−1

∑A∈T (n)

((f∣∣kVA)∣∣kA′)(z)

= nk−1∑

A∈T (n)

(f∣∣kA′)(z) since f

∣∣kVA = f.

The map T (n) → T (n): A 7→ A′ with AS = VAA′ is bijective by Theorem 5.23.

Hence, the A′ run through all elements of T (n) if A does the same. This meansthat we may substitute A′ in the sum above by A. We get(

Tn f)∣∣kT (z) = nk−1

∑A∈T (n)

(f∣∣kA′)(z)

5.6. MULTIPLICATIVE PROPERTIES OF HECKE OPERATORS 139

= nk−1∑

A∈T (n)

(f∣∣kA)(z)

=(Tnf

)(z).

We just showed that Tn f satisfies the conditions of Theorem 3.18. Applyingthis theorem, we get that Tnf satisfies the transformation property(

Tn f)(M z) = j(V, z)k

(Tn f

)(z)

for all V ∈ Γ(1) and z ∈ H.

Theorem 5.25 together with Theorem 5.14 immediately give the following

Corollary 5.26. Let n ∈ N and k ∈ 2Z. The operator Tn satisfies:

(1) If f ∈M !k

(Γ(1)

)then Tnf ∈M !

k

(Γ(1)

).

(2) If f ∈Mk

(Γ(1)

)then Tnf ∈Mk

(Γ(1)

).

(3) If f ∈ Sk(Γ(1)

)then Tnf ∈ Sk

(Γ(1)

).

Remark 5.27. Let f be function with Fourier expansion

f(z) =∑m∈IZ

c(m) e2πimz (z ∈ H)

and recall

γn(m)(5.17)

=∑

0<d|gcd(n,m)

dk−1 c(mnd2

)for even weigh k ∈ N in Theorem 5.14. In particular, Theorem5.14 shows therelations

γn(0) = σk−1(n) c(0) and γn(1) = c(n).

This simply reflects that γn takes Fourier coefficients of cusp forms to cusp forms.

5.6. Multiplicative Properties Of Hecke Operators

Lemma 5.28. For m,n ∈ N coprime let d | m and α | n. The elements αb+βdfor 0 ≤ b < d and 0 ≤ β < n

α are all distinct modulo dnα .

Proof. We assume that there exist 0 ≤ b, b′ < d and 0 ≤ β, β′ < nα such that

(5.28) αb+ βd ≡ αb′ + β′d mod dn

α.

In other words, the two sides differ by a multiple of nα :

αb+ βd− αb′ − β′d ∈ dnαZ.

Since gcd(nα , d) = 1 and dnαZ ⊂ dZ, we have

α(b− b′) = αb− αb′ ∈ dZ.In other words, we have d | α(b− b′). Since α and d are coprime, gcd(α, d) = 1, wesee that d has to divide b− b′:

d | b− b′.Using the assumption 0 ≤ b, b′ < d, we see that the difference b − b′ satisfies|b− b′| < d. Hence, the only value of b− b′ which is divided by d is 0. This showsb = b′. Plugging this identity int (5.28), we get

βd ≡ β′d mod dn

α⇐⇒ d(β − β′) ∈ dn

αZ.

Since d and nα are coprime, we may divide both sides by d and get

β − β′ ∈ n

αZ.


Again, this can be written asn

α| β − β′.

Using the assumption 0 ≤ β, β′ < nα shows, as above, that β = β′.

Summarizing, we have that (5.28) implies b = b′ and β = β′.

Theorem 5.29. Let n,m ∈ N be coprime and suppose that f : H → C isperiodic (i.e. f(z + 1) = f(z) for all z ∈ H). Then, we have

Tm Tn f = Tmn f.

Proof. RecallTn f = nk−1

∑A∈T (n)

f∣∣kA,

see (5.20). We now calculate Tm Tn f

Tm Tn f = Tm(Tn f

)= mk−1

∑A∈T (m)

(Tn f

)∣∣kA

= mk−1∑

A∈T (m)

nk−1∑

B∈T (n)

f∣∣kB

∣∣kA

= (mn)k−1∑

A∈T (m)B∈T (n)

f∣∣kBA.

For every A =

(a b0 d

)∈ T (m) and B =

(α β0 δ

)∈ T (n), we have

BA =

(α β0 δ

)(a b0 d

)=

(αa αb+ βd0 δd

).

Since the determinant is multiplicative, we have detBA = detB detA = mn.Then, Lemma 5.28 shows that these upper triangular matrices BA are all distinctif A and B run through the sets T (m) and T (n) respectively. By Theorem 5.23,the upper triangular matrix BA is equivalent to an element in T (mn), i.e., thereexists an k ∈ Z depending on a and B such that T kBA ∈ T (mn) (and givedistinct elements if A and B run through the respective sets). Also, each elementof C ∈ T (mn) is reached by the product since the cardinality of T (mn) is equal tothe product of the cardinalities ofT (n) and T (m):

]T (mn) = ]T (m) · ]T (n).

Using the fact that f is periodic, we have f∣∣T k = f . Hence, Tm Tn f can be written

as

Tm Tn f = (mn)k−1∑

A∈T (m)B∈T (n)

f∣∣kBA

= (mn)k−1∑

C∈T (mn)

f∣∣kC

= Tmn f.

The main idea of the proof above is to study properties of the triangle set T (n)and apply it to periodic f with the slash operator.

The same idea can be used to prove the result below: one of the main resultson Hecke operators. The proof will be presented in §5.6.1.


Theorem 5.30. For m,n ∈ N and f : H→ C being periodic, we have

(5.29) Tm Tn f =∑

d|gcd(m,n)

dk−1 Tmnd2f.

Of course, the role of n and m above can be reversed. This leads to

Corollary 5.31. For m,n ∈ N and f : H→ C being periodic we have

Tm Tn f = Tn Tm f.

Proof. Using (5.29), we have

Tm Tn f =∑

d|gcd(m,n)

dk−1 Tmnd2f = Tn Tm f

since the expression in the middle is symmetric in n and m.

5.6.1. Properties of T (n) and the proof of Theorem 5.30. In this sec-tion, we consider certain sets of matrices which will later represent Hecke operators.In order to deal with multiplication of two Hecke operators, we need to introducea set multiplication and some other terminologies.

Definition 5.32. Let A,B ⊂ GL2 (C) be two sets of matrices and C ∈ GL2 (C)a matrix.

We define the dot-multiplication of two sets A · B by

(5.30) A · B := AB; A ∈ A, B ∈ B.

For C ∈ SL2 (Z), we say A and B are equal modulo C and write A ≡ B mod Cif the sets satisfy(5.31)A ≡ B mod C :⇐⇒ ∃ bijective map φ : A× B; A 7→ CkA for some k ∈ Z.

The integer k = kA depends only on A.We introduce the multiplicity of a set in the following way: The set A has

multiplicity l, denoted by A×l, if each element of A is counted l-times:

A ∈ A×l =: A, . . . , A︸︷︷︸l times

;A ∈ A.

We understand the l copies of A are distinct elements of A ∈ A×l.

Example 5.33. To illustrate the new notations, we have(1 23 4

)·(

a bc d

),

(α βγ δ

)=

(1a+ 2c 1b+ 2d3a+ 4c 3b+ 4d

),

(1α+ 2γ 1β + 2δ3α+ 4γ 3β + 4δ

).

Examples of equivalent sets areT, S

≡ST =

(1 −11 0

), S101 =

(1 1010 1

)mod S and

T, S≡ 1, TS mod T.

The multiplicity of a set is illustrated by

S, T, ST×2 = S, S, T, T, ST, ST

where the two copies of each element in S, T, ST are understood as distinctelements.


Lemma 5.34. Let p, i ∈ N with p being a prime. The set T (·) satisfies themultiplication rule

(5.32) T (p) · T (pi) = T (pi+1) ∪(

p 00 p

)·(T (pi−1)

)×pmod S.

Proof. Recall the set T (n) in (5.26). For n = pi, we have

T (pi) =

(a b0 d

); ad = pi, 0 ≤ b < d

=

⋃0≤t≤i

(pi−t bt

0 pt

); 0 ≤ bt < pt

(5.33)

where the sets involved in the union are disjoint. Similarly, applying the formulaabove to n = p, we get

T (p) =

(a b0 d

); ad = p, 0 ≤ b < d

=

(p 00 1

)∪(

1 b0 p

); 0 ≤ b < p

.

Thus,

T (p) · T (pi) =

((p 00 1

)∪(

1 b0 p

); 0 ≤ b < p

)·

(5.34)

·

⋃0≤t≤i

(pi−t bt

0 pt

); 0 ≤ bt < pt

=

⋃0≤t≤i

(pi−t+1 pbt

0 pt

); 0 ≤ bt < pt

∪⋃

0≤t≤i

(pi−t bt + bpt

0 pt+1

); 0 ≤ bt < pt, 0 ≤ b < p

=⋃

1≤t≤i

(pi−t+1 pbt

0 pt

); 0 ≤ bt < pt

∪

((pi+1 0

0 1

)∪

∪⋃

0≤t≤i

(pi−t bt + bpt

0 pt+1

); 0 ≤ bt < pt, 0 ≤ b < p

)=: Σ1 ∪ Σ0.

The last line defines the two disjoint sets Σ1 and Σ0 which we consider separately.We have

Σ0 =

(pi+1 0

0 1

)∪⋃

0≤t≤i

(pi−t bt + bpt

0 pt+1

); 0 ≤ bt < pt, 0 ≤ b < p

which we like to compare to

T (pi+1) =⋃

0≤t≤i+1

(pi−t bt

0 pt

); 0 ≤ bt < pt

.


We have

Σ0 =

(pi+1 0

0 1

)∪⋃

0≤t≤i

(pi−t bt + bpt

0 pt+1

); 0 ≤ bt < pt, 0 ≤ b < p

=

(pi+1 0

0 1

)∪⋃

0≤t≤i

(pi−t b′t

0 pt+1

); 0 ≤ bt < pt, 0 ≤ b < p

where we put b′t := bt + bpt.

We think of t as fixed. Then bt runs through the integers in [0, pt − 1] and bruns through the integers in [0, p− 1]. This implies that b′t = bt + bpt runs throughthe integers in [0, pt+1 − 1] since on one hand we have the estimate

0 ≤ bt + bpt ≤ (pt − 1) + (p− 1)pt = pt − 1 + pt+1 − pt = pt+1 − 1

and on the other hand b′t = bt + bpt takes pt · p = pt+1 different values.This shows that Σ0 can be written as

Σ0 =

(pi+1 0

0 1

)∪⋃

0≤t≤i

(pi−t b′t

0 pt+1

); 0 ≤ bt < pt, 0 ≤ b < p

=

(pi+1 0

0 1

)∪⋃

0≤t≤i

(pi−t b′t

0 pt+1

); 0 ≤ b′t < pt+1

=

(pi+1 0

0 1

)∪

⋃1≤t≤i+1

(pi+1−t b′t

0 pt

); 0 ≤ b′t < pt

=⋃

0≤t≤i+1

(pi+1−t b′t

0 pt

); 0 ≤ b′t < pt

= T (pi+1).

Next, consider the set Σ1 given by

Σ1 =⋃

1≤t≤i

(pi−t+1 pbt

0 pt

); 0 ≤ bt < pt

.

We compare it to(p 00 p

)· T (pi−1) =

(p 00 p

)·⋃

0≤t≤i−1

(pi−1−t bt

0 pt

); 0 ≤ bt < pt

=⋃

0≤t≤i−1

(pi−t pbt

0 pt+1

); 0 ≤ bt < pt

=⋃

1≤t≤i

(pi−t+1 pbt

0 pt

); 0 ≤ bt < pt−1

.

We immediately see that both sets satisfy(p 00 p

)· T (pi−1) ⊂ Σ1.

Moreover, it seems that Σ1 contains p-times as much elements as

(p 00 p

)·

T (pi−1): each element of

(p 00 p

)· T (pi−1) has p S-equivalent copies in Σ1:(

pi−t+1 pbt0 pt

)∈(

p 00 p

)· T (pi−1)


⇐⇒ Sk(pi−t+1 pbt

0 pt

)=

(pi−t+1 pbt + kpt

0 pt

)∈ Σ1 ∀0 ≤ k ≤ p− 1.

In other words, we have

Σ1 ≡(

p 00 p

)·(T (pi−1)

)×pmod S.

Plugging both results into (5.34), we get

T (p) · T (pi)(5.34)

= Σ1 ∪ Σ2

≡(

p 00 p

)·(T (pi−1)

)×p ∪ T (pi+1) mod S.

This proves the lemma.

The above result can be generalized to products of the form T (pi) · T (pj).

Lemma 5.35. Let p ∈ N be a prime number and i, j ∈ N two natural numbers.We have

(5.35) T (pi) · T (pj) ≡mini,j⋃l=0

(pl 00 pl

)·(T (pi+j−2l)

)×plmod S.

Proof. If i = 1, then Lemma 5.34 gives

T (p1) · T (pj)(5.32)

=

(p 00 p

)· T (pi−1) ∪ T (pi+1)

=

(p1 00 p1

)· T (pi+1−2) ∪

(p0 00 p0

)· T (pi+1−0).

Hence, (5.35) holds in this case.Now, we consider the general case i ∈ N. Similarly as above, we have

T (pi) · T (pj)(5.33)

=

⋃0≤t≤i

(pi−t bt

0 pt

); 0 ≤ bt < pt

··

⋃0≤s≤j

(pj−s bs

0 ps

); 0 ≤ bs < ps

=

⋃0≤t≤i0≤s≤j

(pi−t bt

0 pt

)(pj−s bs

0 ps

); 0 ≤ bt < pt, 0 ≤ bs < ps

=⋃

0≤t≤i0≤s≤j

(pi−tpj−s pi−tbs + btp

s

0 ptps

); 0 ≤ bt < pt, 0 ≤ bs < ps

=⋃

0≤t≤i0≤s≤j

(pi+j−t−s pi−tbs + btp

s

0 pt+s

); 0 ≤ bt < pt, 0 ≤ bs < ps

=⋃

0≤t≤i0≤s≤j

(pl 00 pl

)· with l := mini− t, s

·(

pi+j−t−s−l pi−t−lbs + btps−l

0 pt+s−l

); 0 ≤ bt < pt, 0 ≤ bs < ps

.


We point out that pl is the greatest common divisor of the matrix entries of(pi+j−t−s pi−tbs + btp

s

0 pt+s

)(excluding the vanishing entry in the lower left):

pl = gcd(pi+j−t−s, pi−tbs + btp

s, pt+s)⇐⇒ l = mini− t, s.

The next step is to reorder the above union of sets. Instead of using the divisors ofpi and pj , we order the union in terms of l. For this, we need to run over all possiblel values. Since i − t and s are bounded by 0 from below and i and j respectively,we can write

T (pi) · T (pj)

=⋃

0≤t≤i0≤s≤j

(pl 00 pl

)· with l = mini− t, s

(5.36)

·(

pi+j−t−s−l pi−t−lbs + btps−l

0 pt+s−l

); 0 ≤ bt < pt, 0 ≤ bs < ps

=

mini,j⋃l=0

(pl 00 pl

)·

⋃l≤t+s

l+t+s≤i+jl=mini−t,s

0≤t≤i−ll≤s≤j

(pi+j−t−s−l pi−t−lbs + btp

s−l

0 pt+s−l

); 0 ≤ bt < pt, 0 ≤ bs < ps

=

mini,j⋃l=0

(pl 00 pl

)·

·⋃

0≤t+s−l≤i+j−2ll=mini−t,s

0≤t≤i−ll≤s≤j

(pi+j−2l−(t+s−l) pi−t−lbs+btp

s−l

0 pt+s−l

); 0 ≤ bt < pt, 0 ≤ bs < ps

=

mini,j⋃l=0

(pl 00 pl

)·

·

( ⋃0≤t+s−l≤i+j−2l

l=s0≤t≤i−l

(pi+j−2l−(t+s−l) pi−t−lbs+bt

0 pt+s−l

); 0 ≤ bt < pt, 0 ≤ bs < ps

∪

∪⋃

0≤t+s−l≤i+j−2ll=i−tl<s≤j

(pi+j−2l−(t+s−l) bs+btp

s−l

0 pt+s−l

); 0 ≤ bt < pt, 0 ≤ bs < ps

)

=

mini,j⋃l=0

(pl 00 pl

)·

(5.37)

( ⋃0≤t≤i+j−2l

0≤t≤i−l

(pi+j−2l−t pi−t−lbs + bt

0 pt

); 0 ≤ bt < pt, 0 ≤ bs < pl


∪⋃

2l≤i+s≤i+jl<s≤j

(pj−s bs + btp

s−l

0 pi+s−2l

); 0 ≤ bt < pi−l, 0 ≤ bs < ps

).

Now, we substitute i+ s− 2l =: t ⇐⇒ s = t+ 2l− i in the last row. This leads to⋃2l≤i+s≤i+j

l<s≤j

(pj−s bs + btp

s−l

0 pi+s−2l

); 0 ≤ bt < pi−l, 0 ≤ bs < ps

i+s−2l=t=

⋃0≤t≤i+j−2li−l<t≤j+i−2l

(pj+i−2l−t b+ βpt+l−i

0 pt

); 0 ≤ β < pi−l, 0 ≤ b < pt+2l−i

.

Hence, T (pi) · T (pj) in (5.37) can be written as

T (pi) · T (pj)

=

mini,j⋃l=0

(pl 00 pl

)·( ⋃

0≤t≤i+j−2l0≤t≤i−l

(pi+j−2l−t pi−t−lbs+bt

0 pt

); 0 ≤ bt < pt, 0 ≤ bs < pl

∪⋃

2l≤i+s≤i+jl<s≤j

(pj−s bs+btp

s−l

0 pi+s−2l

); 0 ≤ bt < pi−l, 0 ≤ bs < ps

)

=

mini,j⋃l=0

(pl 00 pl

)·

(5.38)

( ⋃0≤t≤i+j−2l

0≤t≤i−l

(pi+j−2l−t pi−t−lβ+b

0 pt

); 0 ≤ b < pt, 0 ≤ β < pl

∪⋃

0≤t≤i+j−2li−l<t≤i+j−2l

(pj+i−2l−t βpt+l−i+b

0 pt

); 0 ≤ β < pi−l, 0 ≤ b < pt+2l−i

.

The cardinality of the set above can be determined as follows: For fixed 0 ≤ l ≤mini, j, we have

]

(pi+j−2l−t pi−t−lβ + b

0 pt

); 0 ≤ b < pt, 0 ≤ β < pl

= ]pi−t−lβ + b; 0 ≤ b < pt, 0 ≤ β < pl

= pt pl = pt+l if 0 ≤ t ≤ i− l

and

]

(pj+i−2l−t βpt+l−i + b

0 pt

); 0 ≤ β < pi−l, 0 ≤ b < pt+2l−i

= ]βpt+l−i + b; 0 ≤ β < pi−l, 0 ≤ b < pt+2l−i

= pi−l pt+2l−i = pt+l if i− l < t ≤ i+ j − 2l.


On the other hand, consider the right hand side of (5.35):

mini,j⋃l=0

(pl 00 pl

)·(T (pi+j−2l)

)×pl

=

mini,j⋃l=0

(pl 00 pl

)·

⋃0≤t≤i+j−2l

(pi+j−2l−t bt

0 pt

); 0 ≤ bt < pt

×pl

.

(5.39)

The cardinality of the set above can be calculated as follows: For fixed 0 ≤ l ≤mini, j we have

]

(pi+j−2l−t bt

0 pt

); 0 ≤ bt < pt

= pt for 0 ≤ t ≤ i+ j − 2l.

We easily see that for fixed t = t, we have the inclusion(pi+j−2l−t bt

0 pt

); 0 ≤ bt < pt

⊂(

pi+j−2l−t pi−t−lβ + b0 pt

); 0 ≤ b < pt, 0 ≤ β < pl

if t = t ≤ i− l and(

pi+j−2l−t bt0 pt

); 0 ≤ bt < pt

⊂(

pj+i−2l−t βpt+l−i + b0 pt

); 0 ≤ β < pi−l, 0 ≤ b < pt+2l−i

if t = t > i− l.

Similar to the proof of Lemma 5.34, see the calculation of the set Σ1, wecompare the sets in (5.38) and (5.39). We find( ⋃

0≤t≤i+j−2l0≤t≤i−l

(pi+j−2l−t pi−t−lβ + b

0 pt

); 0 ≤ b < pt, 0 ≤ β < pl

∪⋃

0≤t≤i+j−2li−l<t≤i+j−2l

(pj+i−2l−t βpt+l−i + b

0 pt

); 0 ≤ β < pi−l, 0 ≤ b < pt+2l−i

≡(

pl 00 pl

)·

⋃0≤t≤i+j−2l

(pi+j−2l−t bt

0 pt

); 0 ≤ bt < pt

×pl

mod S

for fixed 0 ≤ l ≤ mini, j. This shows that the identity (5.35) holds and concludedthe proof.

The next lemma can be derived from Theorem 5.29. However, we also give adirect proof on the matrix level.

Lemma 5.36. Let p, q ∈ N to distinct primes and i, j ∈ N. Then

(5.40) T (pi) · T (qj) ≡ T (piqj) mod S.

Proof. Recall T (pi) as calculated in (5.33):

T (pi) =⋃

0≤t≤i

(pi−t bt

0 pt

); 0 ≤ bt < pt

.


Hence T (pi) · T (qj) can be written as

T (pi) · T (qj) =

( ⋃0≤t≤i

(pi−t bt

0 pt

); 0 ≤ bt < pt

)

·

( ⋃0≤s≤j

(qj−s βs

0 qs

); 0 ≤ βs < qs

)

=⋃

0≤t≤i0≤s≤j

(pi−t bt

0 pt

)(qj−s βs

0 qs

); 0 ≤ bt < pt, 0 ≤ βs < qs

=⋃

0≤t≤i0≤s≤j

(pi−tqj−s βsp

i−t + btqs

0 ptqs

); 0 ≤ bt < pt, 0 ≤ βs < qs

.


T (piqj) =⋃

0≤t≤i0≤s≤j

(pi−tqj−s b

0 ptqs

); 0 ≤ b < ptqs

.

We easily see that both sets have the same number of elements.Since we want to show that both sets are equal modulo S, we only have to

show each element

(pi−tqj−s βsp

i−t + btqs

0 ptqs

)of T (pi) · T (qj) satisfies

(5.41)

(1 k0 1

) (pi−tqj−s βsp

i−t + btqs

0 ptqs

)∈ T (piqj)

for some k and that these elements are distinct. The converse relation follows fromthe fact that both sets have the same number of elements.

To show (5.41), it is enough to prove that the βspi−t + btq

s’s are all distinctmodulo ptqs. But this follows directly from Lemma 5.28. Hence (5.40) holds.

Repeating the same argument, we can show the following generalization of(5.40) above.

Lemma 5.37. Let m,n ∈ N such that m and n are coprime. Then

(5.42) T (m) · T (n) ≡ T (mn) mod S.

Proof. We follow the proof of Lemma 5.36.Using

T (m)(5.26)

=

(md b0 d

); 0 < d | m, 0 ≤ b < d

,

we have

T (m) · T (n) =

(md b0 d

); 0 < d | m, 0 ≤ b < d

·(

nδ β0 δ

); 0 < δ | n, 0 ≤ β < δ

=

⋃0<d|m0<δ|n

(md b0 d

)(nδ β0 δ

); 0 ≤ b < d, 0 ≤ β < δ

=⋃

0<d|m0<δ|n

(mndδ

md β + bδ

0 dδ

); 0 ≤ b < d, 0 ≤ β < δ

.



T (mn)(5.26)

=

(mnd

b

0 d

); 0 < d | mn, 0 ≤ b < d

.

We easily see that both sets have the same number of elements.Since we want to show that both sets are equal modulo S, we only have to

show each element

(mndδ

md β + bδ

0 dδ

)of T (m) · T (n) satisfies(

1 k0 1

) (mndδ

md β + bδ

0 dδ

)∈ T (pin)

for some k ∈ Z; and that these elements are distinct. The converse inclusion thenfollows from the fact that both sets have the same number of elements.

To show the identity above, it is enough to prove that the md β + bδ’s are all

distinct modulo dδ. But this follows directly from Lemma 5.28 since d | m and δ | nare coprime.

Combining Lemmas 5.34 and 5.37, we find

Proposition 5.38. Let n,m ∈ N be two natural numbers. We have

(5.43) T (m) · T (n) ≡⋃

0<d|gcd(n,m)

(d 00 d

)·(T(mnd2

))×dmod S.

Proof. We write m and n in its prime-factor decomposition

m =

U∏u=1

piuu and n =

V∏v=1

qjvv

for suitable U, V, iu, jv ∈ N and prime numbers pu and qv. Using Lemma 5.37 wecan decompose T (m) and T (n) into its prime power factors:

T (m) ≡U∏u=1

T(piuu)

mod S and T (n) ≡V∏v=1

T(qjvv)

mod S

where we sort the primes pu and qv such that there exists a w ∈ N with

p1 = q1, . . . , pw−1 = qw−1 and pw, . . . , pU ∩ qw, . . . , qV = ∅.

This means in particular that we can decompose the dot product

T (m) · T (n)

≡ T(pi11)· · · T

(piUU)· T(qj11

)· · · T

(qjVV)

mod S

≡ T(pi11)· T(qj11

)· · · T

(piw−1

w−1

)· T(qjw−1

w−1

)·

· T(piww)· · · T

(piUU)· T(qjww)· · · T

(qjVV)

mod S

(5.43)≡ T

(pi11)· T(qj11

)· · · T

(piw−1

w−1

)· T(qjw−1

w−1

)· T(piww · · · p

iUU · q

jww · · · q

jVV

)mod S.

The first factor T(pi11)· T(qj11

)· · · T

(piw−1

w−1

)· T(qjw−1

w−1

)can be rewritten using

Lemma 5.35. For pu = qu, 1 ≤ u ≤ w − 1 we get

T(piuu)· T(pjuu) (5.35)≡

miniu,ju⋃lu=0

(plu 00 plu

)·(T (pi+j−2lu)

)×plumod S.

This implies

T(pi11)· T(qj11

)· · · T

(piw−1

w−1

)· T(qjw−1

w−1

)


=

w−1∏u=1

T(piuu)· T(qjuu)

mod S

(5.35)≡

w−1∏u=1

miniu,ju⋃lu=0

(plu 00 plu

)·(T (pi+j−2lu)

)×plu.

Both parts of T (m) · T (n) together give

T (m) · T (n) ≡

w−1∏u=1

miniu,ju⋃lu=0

(plu 00 plu

)·(cT (pi+j−2lu)

)×plu ·· T(piww · · · p

iUU · q

jww · · · q

jVV

)mod S

=∏

d|gcd(m,n)

(d 00 d

)·(cT(mnd2

))×d.

This concludes the proof.

Now we are able to prove Theorem 5.30, using the above results.

Proof of Theorem 5.30. Let f be a periodic function, i.e., f∣∣kS = f and

consider Tnf . Recalling (5.20), we have

Tnf(5.20)

= nk−1∑

A∈T (n)

(f∣∣kA)

Considering Tm Tn f we have

Tm (Tn f) = mk−1∑

A∈T (m)

(Tn f

∣∣kA)

= (mn)k−1∑

A∈T (m)

∑B∈T (n)

f∣∣kAB

= (mn)k−1∑

C∈T (n)·T (m)

f∣∣kC.

Using Proposition 5.38, we get

Tm (Tn f) = (mn)k−1∑

C∈T (n)·T (m)

(5.43)= (mn)k−1

∑C∈

⋃0<d|gcd(n,m)

d 0

0 d

·(T(mnd2

))×d f∣∣kC

= (mn)k−1∑

0<d gcd(n,m)

C ∈(

d 00 d

)·(T(mnd2

))×df∣∣kC

= (mn)k−1∑

0<d gcd(n,m)

d−k∑

C∈T(mnd2

)×d f∣∣kC

= (mn)k−1∑

0<d gcd(n,m)

d1−k∑

C∈T(mnd2

) f ∣∣kC=

∑0<d gcd(n,m)

(mnd

)k−1 ∑C∈T

(mnd2

) f ∣∣kC

5.7. EIGENFORMS 151

=∑

0<d gcd(n,m)

dk−1

(mnd2

)k−1 ∑C∈T

(mnd2

) f ∣∣kC

(5.20)=

∑0<d gcd(n,m)

dk−1(Tmnd2f).

This proves the relation

Tm (Tn f) =∑

0<d gcd(n,m)

dk−1(Tmnd2f).

5.7. Eigenforms

Let k ∈ 2N and f ∈ M !k

(Γ(1)

). Suppose that f admits a Fourier expansion

f(z) =∑m∈Z c(m) e2πimz for all z ∈ H and satisfies

Tnf = c(n) f for all n ∈ N.

Then, the Fourier coefficients satisfies

(5.44) c(m) c(n) =∑

0<d|gcd(m,n)

dk−1 c(mnd2

)using Theorem 5.14. For example, it is known that the arithmetic divisor functionσk−1(n), defined in (4.12), satisfies this relation, see Example 5.42 below.

Definition 5.39. For k ∈ 2N, let 0 6= f ∈M !k

(Γ(1)

)satisfy

(5.45) Tn f = λ(n) f for some λ(n).

We then call f an eigenfunction of the Hecke operator Tn with eigenvalue λ(n).If f is an eigenfunction of Tn for all n ∈ N, we call f a simultaneous eigenfunc-

tion of all Hecke operators.We call a simultaneous eigenfunction f normalized if the first Fourier coefficient

of f(z) =∑n c(n) e2πinz satisfies c(1) = 1.

Example 5.40. Consider ∆ ∈ M12

(Γ(1)

)as defined in (4.25) with Fourier

expansion in (4.26). Since dimM12

(Γ(1)

)= 1, see Proposition 4.28, ∆ must be an

eigenfunction of Tn for every n ∈ N:

Tn ∆ = λ(n) ∆ for all n ∈ N.

We calculate the first Fourier coefficient on both sides of the identity. For the lefthand side, we use Theorem 5.14 and get

γn(1) =∑

0<d|gcd(1,n)

d11 τ(nd2

)= τ(n).

And for the left hand side, we get immediately

λ(n) τ(1) = λ(n)

since τ(1) = 1 as can easily be seen from the product expansion in (5.14). We haveshown that the Fourier coefficients equal the eigenvalues:

τ(n) = λ(n) for all n ∈ N.

Hence, the Fourier coefficients τ(n) satisfy the multiplication rule (5.44).


Example 5.41. Lemma 4.27 and Theorem 4.37 imply that

dimMk

(Γ(1)

)= 1 for weight k ∈ 12, 16, 18, 20, 22, 26.

Corresponding cusp forms of weight 12, 16, 18, 20, 22 and 26 are

∆,∆E4,∆E6,∆E8,∆E10 and ∆E14.

Using the Fourier expansion of Ek in (4.24) shows that the leading Fourier coef-ficient of the cusp form ∆Ek is 1. This allows us to repeat the argument usedin Example 5.39 for weight 12. We get that ∆Ek, k ∈ 12, 16, 18, 20, 22, 26, isa simultaneous eigenfunction and that the Fourier coefficient c(n) ist the eigen-value for the Hecke operator Tn. Again, the Fourier coefficients of ∆Ek, k ∈12, 16, 18, 20, 22, 26, satisfy the multiplication rule (5.44).

Example 5.42. Theorem 4.37 implies

dimMk

(Γ(1)

)= 1 for weight k ∈ 4, 6, 8, 10, 14.

Moreover, we have Ek ∈Mk

(Γ(1)

).

Take k ∈ 4, 6, 8, 10, 14. Using the dimension argument, we see immediatelythat Ek has to be a simultaneous eigenfunction with eigenvalue λ(n) (which maydepend on k)

TnEk = λ(n)Ek for all n ∈ N.Calculating the first Fourier coefficients of above left and right hand sides, we find

λ(n) c(0) = λ(n)

for the right hand side and

γn(0) = c(0)∑

0<d|gcd(n,1)

dk−1 = σk−1(n)

for the left hand side. We used that the 0th Fourier coefficient of Ek is 1, see e.g.(4.24), and the others are given by

c(n) =(2πi)k

(k − 1)! ζ(k)σk−1(n) for all n ∈ N.

Comparing again the Fourier terms of the above identity gives

γn(m) = σk−1(n) c(m)

⇐⇒∑

0<d|gcd(m,n)

dk−1 c(nd2

)= σk−1(n) c(m)

⇐⇒ (2πi)k

(k − 1)! ζ(k)

∑0<d|gcd(m,n)

dk−1 σk−1

(nd2

)= σk−1(n)

(2πi)k

(k − 1)! ζ(k)σk−1(m)

⇐⇒∑

0<d|gcd(m,n)

dk−1 σk−1

(nd2

)= σk−1(n)σk−1(m)

(5.46)

for all m,n ∈ N.

Remark 5.43. There also exist elementary proofs for the identity (5.46) of thearithmetic divisor function. Using the Hecke operators, one can also establish theidentities for k ∈ 4, 6, 8, 10, 14.

The above examples demonstrate that simultaneous eigenfunctions of the Heckeoperators have some interesting multiplicative properties which we discuss next.

5.7. EIGENFORMS 153

Theorem 5.44. Let k ∈ 2N. If f ∈Mk

(Γ(1)

)is a simultaneous eigenfunction

with nonvanishing eigenvalues λ(n) 6= 0, n ∈ N. If f admits the Fourier expansionf(z) =

∑n∈Z≥0

c(n) e2πinz, then its first Fourier coefficient c(1) 6= 0 does not

vanish.

Proof. Theorem 5.14 and f being an simultaneous eigenfunction imply that

Tnf(z) =

∞∑m=0

∑0<d|gcd(m,n)

dk−1 c(mnd2

) e2πimz = λ(n)

∞∑m=0

c(m) e2πimz

for all z ∈ H and n ∈ Z≥0. This implies for the Fourier coefficients

c(n) = λ(n) c(1) for all n ∈ N

with λ(n) 6= 0 by assumption.Now, if c(1) = 0, then c(n) = 0 for all n ∈ N and f(z) = c0 has to be constant.

Using that f ∈ Mk

(Γ(1)

)is modular with weight k and trivial multiplier, we see

that f satisfies

f∣∣kT = f ⇐⇒ z−k f

(−1

z

)= f(z) for all z ∈ H.

But this identity can only hold for constant functions f if f = 0 vanishes every-where. This is a contradiction because f needs to be nonvanishing in order to be aeigenfunction, see Definition 5.39. This shows that c(1) = 0 cannot happen. Hencec(1) 6= 0 holds, concluding the proof.

Theorem 5.45. Let k ∈ 2N and let 0 6= f ∈ Sk(Γ(1)

)admit the Fourier

expansion

f(z) =∑n∈N

c(n) e2πinz.

The following statements are equivalent:

(1) f is a normalized simultaneous Hecke eigenform.(2) c(m) c(n) =

∑0≤d|gcd(m,n) d

k−1 c(mnd2

)for all m,n ∈ N.

(3) Tnf = c(n) f for all n ∈ N.

Proof. (1)⇒ (3):: Suppose that f is a normalized simultaneous eigen-function,i.e., c(1) = 1 and Tnf = λ(n) f . Again, we compare the firstFourier coefficients of the left and right hand side. Using Theorem 5.14gives

c(n) = γn(1) = λ(n) c(1) = λ(n).

Hence

Tnf = c(n) f

holds for all n ∈ N.(3)⇒ (2):: Again, we use Theorem 5.14 to calculate the Fourier coefficients

on both sides of the identity Tnf = c(n) f . This gives∑0≤d|gcd(m,n)

dk−1 c(mnd2

) (5.17)= γn(m)

!= c(n) c(m).

(2)⇒ (1):: Assume that

c(m) c(n) =∑

0≤d|gcd(m,n)

dk−1 c(mnd2

)holds for all m,n ∈ N. The left hand side can be interpreted as the mth

Fourier coefficient of c(n) f . Theorem 5.14 allows us to interpret the right


hand side as mth Fourier coefficient of Tnf . Since the identity holds forall m ∈ N, we see that

c(n) f = Tnf

holds for all n ∈ N. This shows that f is a simultaneous eigenfunction ofthe Hecke operators. Plugging m = 1 into the identity, we get

c(1) c(n) = c(n) for all n ∈ N.

Plugging n = 1 into this identity implies the condition c(1)2 = c(1). Theonly solution to this condition is either c(1) = 0 or c(1) = 1. Assumingc(1) = 0, we see that the above identity implies c(n) = 0 for all n ∈ N. So,f = 0 vanishes everywhere. But this contradicts the assumption. Hencec(1) 6= 0 holds.

5.8. Hecke Operators, Eisenstein and Poincare Series

Recall the normalized Eisenstein series Ek which we introduced in (4.22). Fork ∈ 2N, k ≥ 4, the Eisenstein series Ek is an entire modular form of weight k andtrivial multiplier for the full modular group Γ(1) with Fourier expansion given in(4.24).

The action of the Hecke operators on Ek are described in the following

Theorem 5.46. Let k ∈ 2N, k ≥ 4 and n ∈ N. We have

(5.47) TnEk = σk−1(n)Ek

where σk(n) is the kth arithmetic divisor function defined in (4.12).

Corollary 5.47. Let k ∈ 2N and suppose f ∈ Mk

(Γ(1)

)r Sk

(Γ(1)

). Then

the first of the following statements implies the second.

(1) f is a normalized simultaneous Hecke eigenform.

(2) f = αk Ek with αk = (k−1! ζ(k)(2πi)k

and Ek the normalized Eisenstein series

given in Definition 4.22.

Proof. Write the Fourier expansion of f as

f(z) =

∞∑n=0


Since 0 6= f ∈Mk

(Γ(1)

)rSk

(Γ(1)

), we know that a(0) does not vanish. (Otherwise

f be a cusp form contradicting the assumption).

(1)⇒ (2):: Let f be a normalized Hecke eigenform. Then a(1) = 1 byDefinition 5.39 and Tnf = a(n)f holds for all n ∈ N, see Theorem 5.45.

a(n)a(0) =∑

0<d|gcd(n,0)

dk−1 a(n0d2

)=

∑0<d|gcd(n,0)

dk−1 a(0).

Since a(0) 6= 0 we may divide by a(0) and get

a(n)∑

0<d|gcd(n,0)

dk−1 = σk−1(n) for all n ∈ N

where we used the arithmetic divisor function σk−1(n) defined in (4.12).This shows that f satisfies

f(z) = a(0) +∑n∈N

σk−1(n) e2πinz with z ∈ H

5.8. HECKE OPERATORS, EISENSTEIN AND POINCARE SERIES 155

with non-vanishing constant term c(0) 6= 0. Comparing the Fourier ex-pansion of f with the expansion of the normalized Eisenstein series Ek in(4.24), we find that the difference function

(2πi)k

(k − 1! ζ(k)f(z)− Ek(z) =

(2πi)k

(k − 1! ζ(k)c(0)− 1

is constant. Using the transformation property of the difference functionfor T (since f and Ek are modular of the same weight k > 0 and trivialmultiplier) we find, as in the proof of Theorem 5.44, that the differencevanishes everywhere. Hence

(2πi)k

(k − 1! ζ(k)f(z) = Ek(z)

holds for all z ∈ H.(2)⇒ (1):: The normalized Eisenstein series Ek is a simultaneous Hecke

eigenform with eigenvalues σk−1(n) of Tn, n ∈ N, see Theorem 5.46.Hence, f as scalar multiple of Ek is a simultaneous Hecke eigenform.

It remains to check whether f is normalized in the sense of Defini-tion 5.39. Using the Fourier expansion of Ek in (4.24) we have

f(z) =(k − 1! ζ(k)

(2πi)kEk(z)

=(k − 1! ζ(k)

(2πi)k

(1 +

(2πi)k

(k − 1)! ζ(k)

∞∑n=1

σk−1(n) e2πinz

)

=(k − 1! ζ(k)

(2πi)k+

∞∑n=1

σk−1(n) e2πinz

for all z ∈ H. In particular, the first Fourier coefficient σk−1(1) satisfies

σk−1(1)(4.12)

=∑

0<d|1

dk−1 = 1.

Hence, f is a normalized simultaneous Hecke eigenform.

The remainder of this section is used to prove the theorem above.

For k ∈ 2N, k ≥ 4, recall the Poincare series Pν,k : H × Z≥0 → C given in(4.39):

Pν,k(z) :=∑c,d∈Z

gcd(c,d)=1

e2πiν(Vc,dz

)(cz + d)k

with Vc,d =

(? ?c d

)∈ Γ(1) run through a set of representatives of the right cosets

Γ(1)∞\Γ(1).Consider Tn Pν,k for given 4 ≤ k ∈ 2N and ν ∈ Z≥0. We have

Tn Pν,k(z) = nk−1∑

A∈T (n)

Pν,k∣∣kA(z)

= nk−1∑

A∈T (n)

∑c,d∈Z

gcd(c,d)=1

e2πiν(Vc,dAz

)(cAz + d)k


= nk−1∑

A∈Γ(1)\H(n)

∑Vc,d∈Γ(1)∞\Γ(1)

e2πiν(Vc,dAz

)(cAz + d)k

= nk−1∑

A∈Γ(1)\H(n)

∑Vc,d∈Γ(1)∞\Γ(1)

(e2πiν ·) ∣∣

k

(Vc,dA

)(z),(5.48)

using the representations of Γ(1)∞\Γ(1) given in Lemma 4.42 and of T (n) =Γ(1)\H(n) given in Theorem 5.23. We would like to “commute” Vc,d = V ∈Γ(1)∞\Γ(1) and A in the above formula. However, these matrices do not commutein general.

Lemma 5.48. Let n ∈ N.

(1)V A; V ∈ Γ(1)∞\Γ(1), A ∈ Γ(1)\H(n)

forms a complete set of right

coset representatives of Γ(1)∞\H(n).(2) The set

AV ; V ∈ (4.37), A ∈ T (n)

forms also a complete set of right

coset representatives of Γ(1)∞\H(n).

Proof. Theorem 5.23 shows that T (n) is a complete set of representatives ofthe right cosets of Γ(1)\H(n). Hence, each B ∈ H can be written as B = V A withV ∈ Γ(1) and A ∈ T (n). Using Lemma 4.42 and (2.17), we write V ∈ Γ(1) asVc,d representing a right coset in Γ(1)∞\Γ(1) for suitable coprime c, d ∈ Z and asuitable power St ∈ Γ(1)∞, t ∈ Z. Summarizing, we have

B = St Vc,dA

Assume that we have A1, A2 ∈ T (n) and V1, V2 in Γ(1)∞\Γ(1) such that

V1A1 = StV2A2

holds for some t ∈ Z. Hence, A1 = V −11 StV2A2 ∈ T (n). Since the set T (n) form

a complete set of representatives of the right cosets in Γ(1), \H(n) and V −11 StV2 ∈

Γ(1) we have A1 = A2 ∈ T (n). Since A1 = A2, we conclude V1 = St V2. Hence, V1

and V2 have the same entries in the lower row. This implies with Lemma 4.42 thatV1 and V2 represent the same right coset in Γ(1)∞\Γ(1). This proves the first partof the lemma.

For the second part of the lemma consider an

(α βγ δ

)∈ H(n) and assume

that δ ≥ is nonnegative and also that γ > 0 if δ = 0.Now, assume the case δ ≥ 0. Put x = δ

gcd(γ,δ) and y = γgcd(γ,δ) . By construction

we know x ∈ Z≥0, y ∈ Z, gcd(x, y) = 1,

xγ = xy gcd(γ, δ) = δy and aδ − bγ = (ax− by) gcd(γ, δ) = gcd(γ, δ) ∈ N.

By the Chinese reminder theorem, see §A.3, there exists a matrix

(a by x

)∈ Γ(1).

We may assume that

(a by x

)is in the set (4.37) of representatives of the left

cosets in Γ(1)∞\Γ(1); in particular the matrix satisfies either 0 ≤ b < x if x > 0 or(a by x

)= T if x = 0. We find(

α βγ δ

) (a by x

)−1

=

(α βγ δ

) (x −b−y a

)=

(xα− yβ aβ − bαxγ − δy aδ − bγ

)=

(xα− yβ aβ − bα

0 gcd(γ, δ)

).

5.8. HECKE OPERATORS, EISENSTEIN AND POINCARE SERIES 157

Since by construction gcd(γ, δ) ∈ N is positive, we see that the determinant con-

dition det

(α βγ δ

)= n and det

(a by x

)= 1 implies the upper left entry of the

triangular matrix is positive, too. We have

det


0 gcd(γ, δ)

)= (xα− yβ) gcd(γ, δ) = n.

Hence,

(α βγ δ

)can be written as(

α βγ δ

)=


0 gcd(γ, δ)

) (a by x

)=

(1 t0 1

) (1 −t0 1

) (xα− yβ aβ − bα

0 gcd(γ, δ)

) (a by x

)=

(1 t0 1

) (xα− yβ aβ − bα− t gcd(γ, δ)

0 gcd(γ, δ)

) (a by x

)where we chose t ∈ Z so that the inequality

0 ≤ aβ − bα− t gcd(γ, δ) < gcd(γ, δ)

holds. Hence, we have written

(α βγ δ

)as

H(n) 3(α βγ δ

)= StAVy,x

with A =

(xα− yβ aβ − bα− t gcd(γ, δ)

0 ∈

)T (n) and Vy,x =

(a by x

)representing

a coset in Γ(1)∞\Γ(1).

Next, assume δ < 0. Then, we multiply

(α βγ δ

)∈ H(n) by −1 ∈ Γ(1)∞ and

consider

(−α −β−γ −δ

)∈ H(n). Hence, we reduced this case to the case above.

To prove uniqueness of this product representation, assume

St1 A1 Vy1,x1=

(α βγ δ

)= St2 A2 Vy2,x2

with A1, A2 ∈ T (n) and Vy1,x1, Vy2,x2

∈ Γ(1)∞\Γ(1) are elements in the set (4.37)of representatives. This implies

A1 = St2−t1 A2 Vy2,x2V−1y1,x1

.

Since A1, A2, St2−t1 are upper triangular matrices, we know that its lower left entry

vanishes. This implies that Vy2,x2V −1y1,x1

= St1−t2 A−12 A1 must also have a vanishing

lower left entry. Now, using the determinant condition Vy2,x2V −1y1,x1

∈ Γ(1) shows

that the diagonal elements of of Vy2,x2V −1y1,x1

are either both +1 or both −1.

Vy2,x2V −1y1,x1

=

(1 b0 1

)or Vy2,x2

V −1y1,x1

=

(−1 b0 −1

).

The second case cannot happen since

Vy2,x2V−1y1,x1

= St1−t2 A−12 A1

and all matrices on the right hand side are triangular matrices with positive entrieson the diagonal. Hence, we have

Vy2,x2V −1y1,x1

=

(1 b0 1

)


for some b ∈ Z. This shows that Vy1,x1and Vy2,x2

have the same lower row.Using Lemma 4.42, Vy1,x1

and Vy2,x2both represent the same equivalence class in

Γ(1)∞\Γ(1). Hence, they must be equal, using the description of the elements inthe set (4.37). Summarizing, we have Vy1,x1 = Vy2,x2 and

A1 = St2−t1 A2.

Comparing A1 and A2 shows that they coincide, using the fact that they both areelements in T (n), see (5.26).

Applying Lemma 5.48 to the expression of Tn Pν,k(z) in (5.48), we have

Tn Pν,k(z) = nk−1∑

A∈Γ(1)\H(n)

∑Vc,d∈Γ(1)∞\Γ(1)

(e2πiν ·) ∣∣

k

(Vc,dA

)(z)

= nk−1∑

A∈Γ(1)\H(n)

∑Vc,d∈Γ(1)∞\Γ(1)

(e2πiν ·) ∣∣

k

(AVc,d

)(z)

= nk−1∑

A∈Γ(1)\H(n)

∑Vc,d∈Γ(1)∞\Γ(1)

((e2πiν ·) ∣∣

kA) ∣∣kVc,d(z)

= nk−1∑

A∈T (n)

∑Vc,d∈Γ(1)∞\Γ(1)

((e2πiν ·) ∣∣

kA) ∣∣kVc,d(z)

= nk−1∑ad=na,d>00≤b<d

∑Vc,d∈Γ(1)∞\Γ(1)

(d−k e2πiν a·+bd

) ∣∣kVc,d(z)

= nk−1∑ad=na,d>0

∑Vc,d∈Γ(1)∞\Γ(1)

(d−k e2πiν a·d

(d−1∑b=0

e2πiν bd

))∣∣kVc,d(z).(5.49)

Using (5.19), we can simplify (5.49). We have

Tn Pν,k(z) = nk−1∑ad=na,d>0

∑Vc,d∈Γ(1)∞\Γ(1)

(d−k e2πiν a·d

(d−1∑b=0

e2πiν bd

))∣∣kVc,d(z)

= nk−1∑ad=na,d>0d|ν

∑Vc,d∈Γ(1)∞\Γ(1)

(d1−k e2πiν a·d

) ∣∣kVc,d(z)

= nk−1∑

d|gcd(n,ν)

d1−k∑

Vc,d∈Γ(1)∞\Γ(1)

(e2πiνn ·

d2

) ∣∣kVc,d(z)

= nk−1∑

0<d|gcd(n,ν)

d1−k∑c,d∈Z

gcd(c,d)=1

(e2πiνn ·

d2

) ∣∣kVc,d(z)

= nk−1∑

0<d|gcd(n,ν)

d1−k P νnd2 ,k(z).

We just proved the following

Proposition 5.49. Let d ∈ N and ν ∈ Z≥0. We have

(5.50) Tn Pν,k(z) = nk−1∑

0<d|gcd(n,ν)

d1−k P νnd2 ,k(z)

for all z ∈ H.

Using this proposition, we can prove Theorem 5.46.

5.9. HECKE OPERATORS AND THE PETERSSON INNER PRODUCT 159

Proof of Theorem 5.46. We have

TnEk(z) =1

2

(Tn P0,k

)(z)

(5.50)=

1

2nk−1

∑0<d|n

d1−k P0,k(z)

=1

2

∑0<d|n

(nd

)1−k P0,k(z)

(4.12)=

1

2σk−1(n)P0,k(z)

= σk−1(n)Ek(z)

for all z ∈ H.

Exercise 5.50. Let k ∈ 2N, k ≥ 4 and n ∈ N. Show

(5.51) Tn P1,k = nk−1 Pn,k.

Corollary 5.51. For m,n ∈ N we have

(5.52) mk−1 TnPm,k = nk−1 TmPn,k.

Proof. Using Proposition 5.49, we find

mk−1(TnPm,k

)(z)

(5.50)= (mn)k−1

∑0<d|gcd(m,n)

d1−k Pmnd2 ,k(z)

(5.50)= nk−1

(TmPn,k

)(z)

for all z ∈ H.

5.9. Hecke Operators And The Petersson Inner Product

Next, we consider Hecke operators and describe how these operators interactwith the Petersson scalar product.

Lemma 5.52. Let k ∈ 2N. For m,n ∈ N and f ∈ Sk(Γ(1)

), we have

(5.53) mk−1⟨Pm, Tnf

⟩= nk−1

⟨Pn, Tmf

⟩.

Proof. Let us denote the expansion of f at ∞ by

f(z) =∑m∈N

c(m) e2πimz (z ∈ H).

Theorem 5.14 shows that Tnf has the expansion(Tnf

)(z)

(5.16)=

∑m∈N

γn(m) e2πimz

at ∞ with γn(m) given by

γn(m)(5.17)

=∑

0<d|gcd(n,m)

dk−1 c(mnd2

).

Using Theorem 4.49 gives

mk−1⟨Pm, Tnf

⟩ (4.41)= mk−1 γn(m)

µ

Γ(k − 1)

(4πm)k−1

=γm(n)

µ

Γ(k − 1)

(4π)k−1


(5.18)=

γn(m)

µ

Γ(k − 1)

(4π)k−1

= nk−1 γn(m)

µ

Γ(k − 1)

(4πn)k−1

(4.41)= nk−1

⟨Pn, Tmf

⟩.

Definition 5.53. Let H be a complex Hilbert space. We call an operatorT : H → H self-adjoint if T satisfies

(5.54) 〈Tf, g〉 = 〈f, Tg〉for all f, g ∈ H.

Proposition 5.54. For n ∈ N and 4 ≤ k ∈ 2N a positive even weight. TheHecke operator Tn is a self-adjoint operator on the Hilbert space of cusp formsSk(Γ(1)

).

Proof. Let g ∈ Sk(Γ(1)

)be a cusp form and denote its expansion of g at

∞ by∑∞m=0 a(m) e2πimz and the expansion of Tng at ∞ by

∑∞m=0 γn(m) e2πimz.

We calculate first the case 〈TnPm,k, g〉 where Pm,k denotes the Poincare series, asdefined in (4.39). We have

〈TnPm,k, g〉(5.50)

= nk−1∑

0<d|gcd(m,n)

d1−k 〈Pmnd2, g〉

(4.41)= nk−1

∑0<d|gcd(m,n)

d1−k a(mnd2

)µ

Γ(k − 1)(4π(mnd2

))k−1

=Γ(k − 1)

µ (4πm)k−1

∑0<d|gcd(m,n)

dk−1 a(mnd2

)(5.17)

=Γ(k − 1)

µ (4πm)k−1γn(m)

(4.41)= 〈Pm,k, Tng〉.

This shows

(5.55) 〈TnPm,k, g〉 = 〈Pm,k, Tng〉for all n ∈ N, m ∈ Z≥0 and all cusp forms g ∈ Sk

(Γ(1)

).

To show 〈Tnf, g〉 = 〈f, Tng〉 for all f, g ∈ Sk(Γ(1)

)we recall that Sk

(Γ(1)

)is

spanned by Poincare series, see Corollary 4.50. This implies that we may replacef by a (linear combination of) Poncare series Pm,k. Using the linearity of thePetersson scalar product in its first component, see Theorem 3.49, we reduce thisto the already established identity (5.53).

Corollary 5.55. For 4 ≤ k ∈ 2N a positive even weight and n ∈ N. Iff ∈ Sk

(Γ(1)

)is an eigenfunction of the Hecke operator Tn then its eigenvalue is

real.

Proof. Consider first f = Ek being an Eisenstein series. Theorem 5.46 showsthat Ek is an eigenfunction of Tn with eigenvalue σk−1(n), which is a real numberby its definition in (4.12).

Now, consider 0 6= f ∈Mk

(Γ(1)

)being a cusp form satisfying Tnf = λn f with

eigenvalue λn. Proposition 5.54 and Theorem 3.49 imply that

λn〈f, f〉 = x〈λn f, f〉 = 〈Tnf, f〉 = 〈f, Tnf〉 = 〈f, λn f〉 = λn〈f, f〉.

5.10. L-FUNCTIONS ASSOCIATED TO CUSP FORMS 161

Dividing through 〈f, f〉 on both sides – this is allowed since f 6= 0 implies 〈f, f〉 6= 0– gives λn = λn. Hence λn has to be a real value.

Lemma 4.27 shows that each element in g ∈ Mk

(Γ(1)

)can be written as a

linear combinationcEk + f

for suitable c ∈ C with f ∈ Sk(Γ(1)

)being a cusp form. This proves the corollary.

Now we can state a result by Hecke.

Theorem 5.56. Let 4 ≤ k ∈ 2N. Then, the space Sk(Γ(1)

)of cusp forms

admits an orthonormal basis consisting only on simultaneous eigenfunctions of allHecke operators in the sense of Definition 5.39.

Proof. Proposition 5.54 shows that all Hecke operators are self-adjoint andTheorem 4.37 shows that Sk

(Γ(1)

)is finite dimensional. Hence, we may apply the

result stated in Appendix A.12.

5.10. L-functions Associated to Cusp Forms

First, we introduce the notion of an L-series.

Definition 5.57. Let C ∈ R be a constant and(a(n)

)n∈N be a sequence of

complex numbers satisfying the growth estimate

a(n) = O(nC)

(n ∈ Z≥0).

For given 0 ≤ κ < 1, the function

(5.56) L(s) :=

∞∑n∈N

a(n)n−s (< (s) > C + 1)

is called the L-series of(a(n)

)n.

Exercise 5.58. Show that the L-series defined above is indeed well-defined forall < (s) > C + 1.

Remark 5.59. (1) If the sequence(a(n)

)n

contains the Fourier coeffi-cients of a cusp form F for the full modular group, we call the L-series in(5.56) the L-series of F .

(2) The L-series in (5.56) are also known as Dirichlet series.(3) Another area of interest are the domains of conditional convergence or

uniform convergence of Dirichlet series. We refer to [70] and referencestherein for more information on this subject.

5.10.1. The Mellin transform.

Definition 5.60. Let f : (0,∞) → C be a Lebesgue-integrable function. ItsMellin transform is defined as

(5.57) M(f)(s) :=

∫ ∞0

f(t)ts−1 dt

for all s ∈ C if the integral exists.

Example 5.61. A typical example is the Gamma-function and its integralrepresentation

(5.58) Γ(s) :=

∫ ∞0

e−yys−1 dy =M(e−t)

(s) (< (s) > 0).

The next lemma summarizes some obvious existence results.


Lemma 5.62. Let f : (0,∞)→ C be continuous with asymptotic estimates

f(t) =

O (t−a) as t→ 0 and

O(t−b)

as t→∞

for some real a < b. Then M(f)(s) exists for any s ∈ C with a < < (s) < b.

If f satisfies additionally the stronger estimate

f(t) = O(e−ct

)as t→∞

for some c > 0, then M(f)(s) exists for all < (s) > a.

Proof. Assume that f satisfies the first estimates. Splitting the integral at 1,∫ ∞0

f(t)ts−1 dt =

∫ 1

0

f(t)ts−1 dt+

∫ ∞1

f(t)ts−1 dt

= O(∫ 1

0

ts−a−1 dt

)+O

(∫ ∞1

ts−b−1 dt

).

We see from the first and second terms respectively that the the integral exists if< (s− a) > 0 and < (s− b) < 0 holds. Hence, the Mellin transform is well definedfor a < < (s) < b.

The second part follows from the rough estimate

f(t) = O(e−ct

)= O

(t−b)

as t→∞

for any b ∈ R.

Definition 5.63. For real a < b let ϕ : s ∈ C; a < < (s) < b → C beholomorphic satisfying the asymptotic estimate

ϕ(s)→ 0 uniformly as |= (s)| → ∞.

We also assume that the path-integral∫ c+i∞c−i∞ |ϕ(s)|ds exists for any c with a+ δ <

c < b− δ (for arbitrary but fixed δ > 0).The inverse Mellin transform is defined by

(5.59) M−1(ϕ)(x) :=

1

2πi

∫ c+i∞

c−i∞ϕ(s)x−s ds (x > 0)

for any a < c < b.

Lemma 5.64. In the setting of Definition 5.63, the inverse Mellin transformM−1(ϕ) exists for any a < c < b and is independent of c.

Proof. Since by assumption ϕ(s)→ 0 uniformly as |= (s)| → ∞ on the stripa + δ < < (s) < b − δ, we may move the path of integration in (5.59) to anya < c < b without changing the integral. This shows that M

(f)

does not dependon the choice of c.

Proposition 5.65 (Mellin inversion formula). For real a < b, let ϕ : s ∈C; a < < (s) < b → C be holomorphic satisfying the asymptotic

ϕ(s)→ 0 uniformly as |= (s)| → ∞.

We also assume that the path-integral∫ c+i∞c−i∞ |ϕ(s)|ds exists for any c with a+ δ <

c < b− δ and arbitrary but fixed δ > 0.In this situation the function f :=M−1(ϕ) satisfies the inversion formula

ϕ(s) =M(f)(s)

for any s in the strip a < < (s) < b.


Conversely, suppose that f : (0,∞)→∞ is piecewise continuous, and supposethat the integral

∫∞0|f(t)| t<(s)−1 dt exists for all a < < (s) < b. Then, we have

f(x) =M−1

(M(f)

)(x) (x > 0).

Proof. We follow the arguments in [27, pp.103–105]. Fix a c with a < c < b,i.e., fix a path of integration in (5.59). Now, choose a < c1 < c < c2 < b. We have∫ ∞

0

f(t) ts−1 dt

=

∫ 1

0

f(t) ts−1 dt+

∫ ∞1

f(t) ts−1 dt

=1

2πi

∫ 1

0

∫ c+i∞

c−i∞ϕ(s)t−s ds ts−1 dt+

1

2πi

∫ ∞1

∫ c+i∞

c−i∞ϕ(s)t−s ds ts−1 dt.

The first integral can be estimated by∣∣∣∣ 1

2πi

∫ 1

0

∫ c+i∞

c−i∞ϕ(s)t−s ds ts−1 dt

∣∣∣∣ ≤ 1

2π

∫ ∞−∞|ϕ(c1 + iy)| dy

∫ 1

0

tc−c1−1 dt <∞,

and the second by∣∣∣∣ 1

2πi

∫ 1

0

∫ c+i∞


∣∣∣∣ ≤ 1

2π

∫ ∞−∞|ϕ(c1 + iy)| dy

∫ ∞1

tc−c2−1 dt <∞,

We may apply Fubini and interchange the integrals, we get∫ ∞0

f(t) ts−1 dt

=1

2πi

∫ 1

0

∫ c+i∞

c−i∞ϕ(s)t−s ds ts−1 dt+

1

2πi

∫ ∞1

∫ c+i∞


=1

2πi

∫ c2+i∞

c2−i∞

ϕ(s)

s− sds− 1

2πi

∫ c1+i∞

c1−i∞

ϕ(s)

s− sds.

Applying Cauchy’s integral formula, we see that the difference at the right handside equals ϕ(s) since the integrant vanishes as |= (s)| → ∞.

For the second part, use the substitution x = eu and put g :=M(f). We get

M−1(g)(x) =

1

2πi

∫ c+i∞

c−i∞g(s)x−s ds

x 7→eu=

1

2π

∫ ∞−∞

g(c+ it) e−u(c+it) dt

=1

2π

∫ ∞−∞

(∫ ∞0

f(y) yc+it−1 dy

)e−u(c+iy) dy

y 7→ev=

1

2π

∫ ∞−∞

(∫ ∞−∞

f(ev)ev(c+it−1) evdv

)e−u(c+iy) dy

=e−uc

2π

∫ ∞−∞

∫ ∞−∞

f(ev)eit(v−u)) evc dvdy.

Now, the Fourier integral theorem shows that the right hand side is equal to

e−uceuc f(eu)

= f(x).

We conclude

M−1(M(f)

)= f

and finish the proof.


5.10.2. L-series associated for cusp forms on Γ(1). We consider a mod-ular cusp forms F on the full modular group Γ(1), see (2.1), for weight k ∈ 2N andtrivial multiplier system. Let F ∈Mk

(Γ(1)

), see Definition 3.26 and Remark 3.28.

As shown in Theorem 3.22, the cusp form F has a Fourier expansion of the form

(5.60) F (z) =

∞∑n=1


The Fourier coefficients a(n) satisfy the growth estimate

(5.61) a(n) = O(nk2

)as n→∞,

see Theorem 3.36. We associate to F the L-series of F given in (5.56):

L(s) =∑n∈N

a(n)n−s (< (s) >k

2+ 1).

How are L and F related?

Definition 5.66. Let F be a cusp form of weight k ∈ 2N with trivial multiplierfor Γ(1) and admits the Fourier expansion (5.60). Define the L-function L?(s) bytaking the Mellin transform of F along the upper imaginary axis

(5.62) L?(s) :=

∫ ∞0

F (iy) ys−1 dy

for all complex s.

Remark 5.67. The L-function defined in (5.62) is sometimes called the com-pleted L-function.

Lemma 5.68. In the situation of Definition 5.66, we have

(1) L?(s) is well defined for all s ∈ C.(2) The L-series L(s) associated to the cusp form F exists for < (s) ≥ k

2 +1 >0.

(3) We have

(5.63)

∫ ∞0

F (iy) ys−1 dy = L?(s) = (2π)−sΓ(s)L(s)

for < (s) ≥ k2 + 1 > 0 where Γ(s) denotes the Gamma function, see (5.58)

or Appendix A.11.

Proof. (1) The existence can be shown as follows: Consider the definingintegral ∫ ∞

0

ys−1 F (iy) dy.

Splitting the path of integration at 1 and applying the substitution y 7→ 1y ,

we get∫ ∞0

ys−1 F (iy) dy =

∫ 1

0

ys−1 F (iy) dy +

∫ ∞1

ys−1 F (iy) dy

y 7→ 1y

=

∫ ∞1

y−s−1 F

(−1

iy

)dy +

∫ ∞1

ys−1 F (iy) dy

(3.3)=

∫ ∞1

yk−s−1 F (iy) dy +

∫ ∞1

ys−1 F (iy) dy

=

∫ ∞1

(yk−s−1 + yl

)F (iy) dy.(5.64)


The last integral exists since F (iy) = O(e−δy

)as y →∞ for some δ > 0,

see Exercise 3.35. This implies that the L-function

L?(s) =

∫ ∞0

ys−1 F (iy) dy

exists for all s.(2) By estimate (5.61), there exists a constant C > 0 such that

|a(n)| ≤ C n k2

holds for all n ∈ N. Exercise 5.58 implies this statement.(3) Define G : (0,∞)→ R by

G(y) := C

∞∑n=1

nk2 e−2πny (y > 0).

We can easily see that G(y) dominates F (x + iy) for all x ∈ R as thefollowing calculation shows

|F (x+ iy)| ≤∞∑n=1

∣∣∣a(n) e2πin(x+iy)∣∣∣

=

∞∑n=1

|a(n)| e−2πny

≤ C∞∑n=1

nk2 e−2πny

= G(y)

for all y > 0 and x ∈ R.Now, for < (s) > k

2 + 1, Fubini’s theorem allows us to interchangeintegration and summation in∫ ∞

0

G(y) ys−1 dy = C

∫ ∞0

∞∑n=1

nk2 e−2πny ys−1 dy

= C∞∑n=1

nk2

∫ ∞0

e−2πnyys−1 dy

= C (2π)−s∞∑n=1

nk2−s

∫ ∞0

e−yys−1 dy

= C (2π)−sΓ(s)

∞∑n=1

nk2−s

where we use the integral representation of the Gamma-function

Γ(s) =

∫ ∞0

e−yys−1 dy (< (s) > 0),

see e.g. (5.58) or Appendix A.11. The domain < (s) > k2 + 1 of absolute

convergence of the series∑∞n=1 n

k2−s ensures that the Mellin transform of

G exists.Now, applying the dominated convergence theorem, we find for L?

L?(s) =

∫ ∞0

F (y) ys−1 dy

=

∫ ∞0

∞∑n=1

a(n) e−2πny ys−1 dy


=

∞∑n=1

a(n)

∫ ∞0

e−2πnyys−1 dy

=

∞∑n=1

a(n) (2πn)−s∫ ∞

0

e−yys−1 dy

= (2π)−sΓ(s)

∞∑n=1

a(n)n−s

for all s with < (s) > k2 + 1.

Lemma 5.69. In the situation of Definition 5.66, the L-function L? of the cuspform F extends holomorphically to the complex plane and satisfies the functionalrelation

(5.65) L?(k − s) = (−1)k2 L?(s) (s ∈ C).

Proof. Recall that F satisfies (3.3) for the trivial multiplier system. In par-ticular, F satisfies the identity

zk F (z) = j(T, z)k F (z)(3.3)= F

(Tz)

= F

(−1

z

)(z ∈ H)

applying the element T ∈ Γ(1) defined in (2.8). Also notice that the substitutionz 7→ −1

z keeps the upper imaginary axis invariant. It acts on the upper imaginary

axis iy; y > 0 as y 7→ 1y . Applying this substitution to the defining integral of

L? in (5.62), we get

L?(s)(5.62)

=

∫ ∞0

F (iy) ys−1 dyy 7→ 1

y= −

∫ 0

∞F

(−1

iy

)y1−s y−2dy

(3.3)=

∫ ∞0

(iy)k F (iy) y−1−sdy

= ik∫ ∞

0

F (iy) yk−1−s dy(5.62)

= ik L?(k − s).

Using 1ik

= (−i)k = (−1)k2 , this proves (5.65) assuming that L? is a holomorphic

function on C.Next, we show that L? extends holomorphically to C. Using the first part of

Lemma 5.68, we know that L?(s) exists for all s ∈ C. Since L?(s) is holomorphicfor < (s) > k

2 + 1, see (5.63) and recall that L(s) and Γ(s) are holomorphic on asuitable right half-plane, the L-function extends holomorphically to C.

Exercise 5.70. Use the expression of L?(s) in (5.64) to show the functionalequation (5.65).

Exercise 5.71. For k ∈ 2N and F ∈ Sk(Γ(1)

), consider the associated L-series

L(s) defined in (5.56). Show the following statements:

(1) L(s) extends to a holomorphic function on C and(2) L(s) satisfies the functional equation

(5.66)(2π)k−s

Γ(k − s)L(s) = ik

(2π)s

Γ(s)L(k − s)

for all s ∈ C.

Combining Lemma 5.68 and Lemma 5.69, we have proven the following


Proposition 5.72. Let F ∈ Sk(Γ(1)

)be a cusp form of weight k ∈ 2N (and

trivial multiplier). Then, there exists the L-series L(s) of F defined in (5.56) whichconverges absolutely for < (s) > k

2 + 1. Its L-function L?(s) given in (5.62) extendsto a holomorphic function on C and satisfies the functional equation (5.65).

An obvious question is whether the following holds: Given L and L?, doesthere exist a cusp form F such that L is the L-series of F? Before we answer thisquestion, we recall the inverse Mellin transform, see e.g. AppendixA.5.

Definition 5.73. For a < b let g(s) be an analytic function on the stripa < < (s) < b such that g(s) → 0 uniformly as |= (s)| → ∞. The inverse Mellintransform is defined by

(5.67) M−1(ϕ)

=1

2πi

∫ c+i∞

c−i∞x−sϕ(s) ds (x > 0).

Lemma 5.74. Let L(s) =∑n∈N ann

−s be an L-series which is absolutely con-

vergent on the half-plane < (s) > k2 + 1. Assume that its associated L-function

L?(s) := (2π)−sΓ(s)L(s) is a holomorphic function on the same half-plane < (s) >k2 + 1.

Then, the function f : i(0,∞) → C on the upper imaginary axis defined byf :=M−1

(L?)

satisfies

f(iy) =

∞∑n=1

ane−2πny (y > 0).

Proof. Using Stirling’s estimate of the Gamma-function in Appendix A.6, wefind the upper bound

|L?(s)| =∣∣(2π)−sΓ(s)L(s)

∣∣= (2π)−σ |L(s)| |γ(s)|

≤ C2 (2π)−σ

( ∞∑n=1

|an|n−σ)e−

π|t|2 |t|σ−

12

for s = σ + it with σ > k2 + 1 bounded, |t| > 2 and C2 > 0 a constant defined

in Appendix A.6. We used an = O(nk2

), which follows from the fact that L(s)

converges absolutely for < (s) > k2 + 1.

This bound allows us to apply the inverse Mellin transform defined in (5.59) toL?. Taking c > k

2 + 1, we get

M−1(L?)(y) =

1

2πi

∫ c+i∞

c−i∞L?(s)y−s ds

=1

2πi

∫ c+i∞

c−i∞(2π)−sΓ(s)L(s) y−s ds

for all y > 0. Let us assume for the moment that we may interchange the pathintegration with the summation hidden in the L-series L(s), we get

M−1(L?)(y) =

1

2πi

∫ c+i∞

c−i∞(2π)−sΓ(s)L(s) y−s ds

=

∞∑n=1

an1

2πi

∫ c+i∞

c−i∞Γ(s) (2πny)−s ds

=

∞∑n=1

an1

2πi

∫ c+i∞

c−i∞Γ(s) (2πny)−s ds.


Using (5.58), together with the first part of proposition 5.65, we get

1

2πi

∫ c+i∞

c−i∞Γ(s) y−s ds = e−y (y > 0).

Hence, the above inverse Mellin transform gives

M−1(L?)(y) =

∞∑n=1

an1

2πi

∫ c+i∞

c−i∞Γ(s) (2πny)−s ds

=

∞∑n=1

an e−2πny.

We still have to show that we may interchange integration and summation.Using the upper bound we calculated for |L?(s)| above, we may use the dominatedconvergence theorem to show this.

Lemma 5.75. For k ∈ 2N, let L(s) = be an L-series which is absolutely con-vergent on the half-plane < (s) > k

2 + 1. Assume that its associated L-functionL?(s) := (2π)−sΓ(s)L(s) extends to a holomorphic function on C and satisfies the

functional equation L?(s) = (−1)k2 L?(k − s).

Then the function f : i(0,∞) → C on the upper imaginary axis defined byf :=M−1

(L?)

satisfies

(iy)k f(iy) = f

(−1

iy

)(y > 0).

Proof. We have already established in the proof of Lemma 5.74 that theinverse Mellin transform M−1

(L?)

is well defined. For c > k2 + 1, we have

f(iy) =M−1(L?)(y)

=1

2πi

∫ c+i∞

c−i∞L?(s)y−s ds

for all y > 0. Applying the functional equation L?(s) = (−1)k2 L?(k − s), we get

f(iy) =1

2πi

∫ c+i∞

c−i∞L?(s) y−s ds

(5.65)=

(−1)k2

2πi

∫ c+i∞

c−i∞L?(k − s)y−s ds

=−(−1)

k2

2πi

∫ k−c−i∞

k−c+i∞L?(s)ys−k ds

= y−k(−1)

k2

2πi

∫ k−c+i∞

k−c−i∞L?(s)

(1

y

)−sds

for all y > 0. The integral on the right hand side is again interpreted as the inverseMellin transform, integrating along the path (k− c) + iR instead of c+ iR. Hence,we find

f(iy) = y−k(−1)

k2

2πi

∫ k−c+i∞

k−c−i∞L?(s)

(1

y

)−sds

= (−1)k2 y−kM−1

(L?)(1

y

)= (−1)

k2 y−k f

(i1

y

)


for all y > 0. Since k is even, we write −1 = i−2 and get

f(iy) = (iy)−k f

(−1

iy

)(y > 0).

Lemma 5.76. Let f : i(0,∞)→ C be a function given by the Fourier expansion

f(iy) =

∞∑n=1

an e−2πny (y > 0)

such that the series∑∞n=1 an e

−2πny converges absolutely for all y > 0.

Then f extends holomorpically to a function f : H→ C given by

f(z) =

∞∑n=1

an e2πinz (z ∈ H).

The function satisfies f(z + 1) = f(z) for all z ∈ H.

Proof. Consider the following estimate based on the triangular inequality∣∣∣∣∣∞∑n=1

an e2πinz

∣∣∣∣∣ =

∞∑n=1

|an|∣∣e2πinz

∣∣=

∞∑n=1

|an|∣∣e2πinx

∣∣ e−2πny

=

∞∑n=1

|an| e−2πny

where z = x + iy ∈ H with real x and y > 0. Since∑∞n=1 an e

−2πny convergesabsolutely for any y > 0, we see that the right hand side and hence the left handside of the above inequality converge. Hence,

f : H→ C; z 7→ f(z) :=

∞∑n=1

an e2πinz

is well defined and forms a holomorphic function on H (since no dependency of

z appears). The periodicity of f is obvious from the Fourier expansion (usinge2πin = 1).

Collecting all results from the previous three lemmas we get the following con-verse statement.

Proposition 5.77. For k ∈ 2N, let L(s) =∑n∈N ann

−s be an L-series which

is absolutely convergent on the half-plane < (s) > k2 + 1. Assume that its associated

L-function L?(s) := (2π)−sΓ(s)L(s) extends to a holomorphic function on C and

satisfies the functional equation L?(s) = (−1)k2L?(k − s).

Then, the function f : H→ C defined by

f(z) =

∞∑n=1

an e2πinz

is a cusp form f ∈ Mk

(Γ(1)

)of weight k, trivial multiplier for the full modular

group.


Proof. Lemma 5.74 shows that f = M−1(L?) is well defined on the upperimaginary axis and has the required Fourier expansion. Lemma 5.76 shows f ex-tends holomorphically to H and satisfies f(z+ 1) = f(z) for all z ∈ H. Lemma 5.75shows the relation j(T, z)k f(z) = f(T z) for all z ∈ i(0,∞) on the the upper imag-inary axis. This relation holds for all z ∈ H since f is holomorphic (by the uniquecontinuation principle, see A.13). Applying Theorem 3.18 shows that f satisfies re-quired transformation property (3.21). Hence, f ∈ Sk

(Γ(1)

)by Definition 3.26.

Collecting Propositions 5.72 and 5.77, we immediately have the following

Corollary 5.78 (Hecke’s converse theorem). Let k ∈ 2N. The following twospaces are equivalent:

(1) Sk(Γ(1)

)and

(2) the space of L-series L(s) which converges absolutely on the half-plane< (s) > k

2 + 1 such that its associated L-function

L?(s) := (2π)−sΓ(s)L(s)

extends to a holomorphic function on C and satisfies the functional equa-tion

L?(s) = (−1)k2 L?(k − s).

The Mellin transform and inverse Mellin transforms are isomorphisms between thetwo spaces.

Remark 5.79. Hecke alredy published his result in [43] in 1936.

Remark 5.80. An interesting question is whether a type of Hecke’s conversetheorem exists for subgroups Γ ⊂ Γ(1) of finite index. Only partial results existon this topic. For example, regarding Hecke congruence subgroups, we have Weil’sconverse theorem [22, Section 1.5].

5.10.3. L-functions of Hecke eigenforms. We assume the following settingthroughout this section. Let k ∈ 2N and consider a normalized Hecke eigenformf ∈Mk

(Γ(1)

). This means, in particular, that f satisfies

Tnf = a(n) f (n ∈ N)

where a(n) is the nth Fourier coefficient in the expansion (5.60).Using Theorem 5.30, we get the following

Corollary 5.81. For k ∈ 2N and let f ∈ Mk

(Γ(1)

)be a normalized Hecke

eigenform with Fourier coefficients a(n). Then, the Fourier coefficients satisfy

(5.68) a(m)a(n) =∑

0<d|gcd(m,n)

dk−1 a(mnd2

)for all m,n ∈ N.

Proof. We just apply Theorem 5.30:

a(m) a(n) f = TmTnf(z)

(5.29)=

∑0<d|gcd(m,n)

dk−1 Tmnd2f

=∑

0<d|gcd(m,n)

dk−1 a(mnd2

)f.

for all m,n ∈ N.


Corollary 5.82. For k ∈ 2N, let f ∈Mk

(Γ(1)

)be a normalized Hecke eigen-

form with Fourier coefficients c(n). Let p be a prime. Then, the Fourier coefficientssatisfy

a(pnm

)= a

(pn)a(m) for all m,n ∈ N with gcd(m, p) = 1 and

a(pn+1

)= a(p) a

(pn)− pk−1a

(pn−1

)for all n ∈ N.

(5.69)

Proof. Using Corollary 5.81, we calculate the coefficients. For all m,n ∈ Nwith gcd(m, p) = 1, we get

a(pn)a(m)

(5.68)=

∑0<d|gcd(pn,m)

dk−1 a

(pnm

d2

)p-m= a

(pnm

).

Next consider a(p) a(pn)

for n ∈ N. We have

a(p) a(pn) (5.68)

=∑

0<d|gcd(p,pn)

dk−1 a

(ppn

d2

)= a

(ppn)

+ pk−1a(pn−1

).

This shows the second identity in (5.69).

Lemma 5.83 ([61, Lemma, p.21]). Let a(n), n ∈ N, be a sequence of numbersand let p be a prime. The power series identity

(5.70)1

1− a(p)X + pk−1X2=

∞∑i=0

a(pi)Xi

is true if and only if a(n) satisfies

(5.71) a(1) = 1 and a(pi+1

)= a(p) a

(pi)− pk−1a

(pi−1

)for all i ∈ N.

Proof. Assume that a(n) satisfies (5.71). Then,(1− a(p)X + pk−1X2

) ∞∑i=0

a(pi)Xi

=

∞∑i=0

a(pi)Xi −

∞∑i=0

a(p)a(pi)Xi+1 +

∞∑i=0

pk−1a(pi)Xi+2

= a(1) + a(p)X +

( ∞∑i=2

a(pi)Xi

)− a(p)a(1)X−

−

( ∞∑i=1

a(p)a(pi)Xi+1

)+

∞∑i=0

pk−1a(pi)Xi+2

a(1)=1= 1 +

( ∞∑i=2

a(pi)Xi

)−

( ∞∑i=1

(a(p)a

(pi)− pk−1a

(pi−1

))Xi+1

)(5.70)

= 1 +

( ∞∑i=2

a(pi)Xi

)−

( ∞∑i=1

a(pi+1

)Xi+1

)= 1.

Hence, (5.70) holds.For the converse, assume (5.70) holds. Hence,

1 =(1− a(p)X + pk−1X2

) ∞∑i=0

a(pi)Xi

=

∞∑i=0

a(pi)Xi −

∞∑i=0

a(p)a(pi)Xi+1 +

∞∑i=0

pk−1a(pi)Xi+2


=

(a(0) + a(p)X +

∞∑i=2

a(pi)Xi

)−

(a(p)a(1)X +

∞∑i=1

a(p)a(pi)Xi+1

)+

+

∞∑i=0

pk−1a(pi)Xi+2

= a(0) +(a(p)− a(p)a(1)

)X +

∞∑i=2

(a(pi)− a(p)a

(pi−1

)+ pk−1a

(pi−2

))Xi.

Comparing the coefficients of the formal power series on the left hand side and theright hand side gives a(0) = 1 in the constant X0 term, a(p) = a(p)a(1) in the termX and a

(pi)

= a(p)a(pi−1

)+ pk−1a

(pi−2

)for all i ≥ 2. Hence the a(n)’s satisfy

(5.71).

Now consider the L-series associated to normalized Hecke eigenform f . Wehave the following result.

Proposition 5.84. For k ∈ 2N and let f ∈ Sk(Γ(1)

)be a cusp form with

associated L-series L(s) defined in (5.56).If f is a normalized Hecke eigenform, then the associated L-series L(s) admits

the Euler product

(5.72) L(s) =∏

p prime

(1− a(p)p−s + pk−1−2s

)−1(< (s) >

k

2+ 1)

where the product runs over the set of all prime numbers p.

Proof. Each n ∈ N can be uniquely written as

n = pi11 · · · pill

for some l ∈ Z≥0, i1, . . . il ∈ N and p1, . . . , pl distinct primes, where the case l = 0is understood as the empty product 1.

Let f ∈ Sk(Γ(1)

)be a normalized Hecke eigenform and consider the series∑∞

i=0

a(pi)psi .

Using the identities derived in Corollary 5.82, we get

L(s) =∑n∈N

a(n)

ns

= a(1) +∑l∈N

∑i1,...,il∈N

p1,...,pldistinct primes

a(pi11 · · · p

ill

)(pi11 · · · p

ill

)s(5.69)

= a(1) +∑l∈N

∑i1,...,il∈N

p1,...,pl distinct primes

a(pi11)(

pi11)s · · · a (pill )(

pill)s

=∏

p prime

∑i∈Z≥0

a(pi)

(pi)s

where we used that a(1) = 1 holds for normalized Hecke eigenforms. ApplyingLemma 5.83 we find

L(s) =∏

p prime

∑i∈Z≥0

a(pi)

pis(5.70)

=∏

p prime

1

1− a(p)p−s + pk−1p−2s

for all s with < (s) > k2 + 1. This proves (5.72).

Does the converse of Proposition 5.84 also hold? We will derive a partial result.


Definition 5.85. A sequence a(n), n ∈ N of (complex) numbers is calledmultiplicative if a(mn) = a(m) a(n) holds for all coprime m,n ∈ N. It is calledcompletely multiplicative if a(mn) = a(m) a(n) holds for all m,n ∈ N.

Proposition 5.86. Let k ∈ 2N and assume that a(n), n ∈ N, is a nonvanishingmultiplicative sequence of complex numbers such that the associated L-series L(s)defined in (5.56) belongs to the second space described in Corollary 5.78. (Nonva-nishing means that a(n) 6= 0 for at least one n ∈ N).

If L(s) additionally admits the Euler product (5.72), then

f : H→ C; f(z) :=∑n∈N

a(n) e2πnz

is a normalized Hecke eigenform with f ∈ Sk(Γ(1)

).

Proof. Corollary 5.78 shows f ∈ Sk(Γ(1)

). It remains to show that f is

indeed a normalized Hecke eigenform.Since a(n) is multiplicative, we have

a(n) = a(1n) = a(1) a(n) for all n ∈ N.

Hence, either a(1) = 1 or a(n) = 0 for all n ∈ N. The second option contradictsthe nonvanishing condition of a(n).

We may expand the L-series as in the first step of the proof of Proposition 5.84:

L(s) =∑n∈N

a(n)

ns

= a(1) +∑l∈N

∑i1,...,il∈N

p1,...,pldistinct primes

a(pi11 · · · p

ill

)(pi11 · · · p

ill

)s= a(1) +

∑l∈N

∑i1,...,il∈N

p1,...,pl distinct primes

a(pi11)(

pi11)s · · · a (pill )(

pill)s

(a(n) is multiplicative)

=∏

p prime

∑i∈Z≥0

a(pi)

(pi)s

where we used in the last step that a(1) = 1 holds.On the other hand, we know that the L-series admits an Euler product (5.72).

Hence we have∏p prime

∑i∈Z≥0

a(pi)

(pi)s

!=

∏p prime

(1− a(p)p−s + pk−1−2s

)−1(< (s) >

k

2+ 1).

Replacing ps with the formal variable X shows that the sequence a(n) satisfiesthe identity (5.70). Applying Lemma 5.83 now implies that (a(n) satisfies (5.71).Hence a(n) satisfies

a(m) a(n) =∑

0<d|gcd(m,n)

dk−1 a(mnd2

)for all m,n ∈ N

expanding each natural number in its unique prime number decomposition. Apply-ing Theorem 5.45 shows that f is indeed a normalized simultaneous Hecke eigen-form.


Remark 5.87. In [26, Lemma 2.4] Conrey, Farmer and Imamoglu obtainedthe following estimates:

For k ∈ 2N, let F ∈ Sk(Γ(1)

)be a normalized Hecke eigenform and let L(s) be

its associated L-series. Then, for < (s) ≥ 3k4 , we have

(5.73) |L(s)1| ≤ 5k4

and, for < (s) ≥ k2 , we have

(5.74) L(s) ≤ 1 + 2√k log(2k).

CHAPTER 6

Period Polynomials

In this chapter, we discuss period polynomials that emerge from Eichler In-tegrals on the full modular group. We give all basic properties of these periodpolynomials and prove the Eichler-Shimura Isomorphism theorem. We continue todescribe the action of Hecke operator on period polynomial and we show that theroots of period polynomials associated to Hecke eigenforms all lie on the unit circle.

6.1. Period polynomials via integral transformations

Definition 6.1. For a cusp form F ∈ Sk(Γ(1)

), we call the function P = PF

defined by the integral transformation

(6.1) P (X) := PF (X) :=

∫ i∞

0

(z −X)k−2 F (z) dz (X ∈ C).

the period polynomial P of F (for Γ(1)). The path of integration is the upperimaginary axis iR>0 connecting the cusp 0 to the cusp ∞.

Let C[X]k−2 denote the set of polynomials in X of degree at most k − 2.

Lemma 6.2. The integral transformation in (6.1) is well defined and the result-ing period polynomial P is indeed a polynomial. The degree of P is smaller than orequal to k − 2. Thus P (X) ∈ C[X]≤k−2.

Proof. Recall that F ∈ Sk(Γ(1)

). To show that the integral exists, we split

the path of integration at i and consider the two integrals separately.

Consider∫ i∞i

(z−X)k−2 F (z) dz. Since h satisfies the growth condition derivedin Exercise 3.35, we see that F decays exponentially at the cusp i∞. Hence theintegrand decays exponentially fast and the integral exists.

Consider the integral∫ i

0(z−X)k−2 F (z) dz. Since F satisfies the transformation

property (3.3), we have in particular

zk F (z) = h

(−1

z

).

Using the substitution z 7→ −1z , we get∫ i

0

(z −X)k−2 F (z) dzz 7→Tz

= −∫ i

i∞

(−1

z−X

)k−2

F

(−1

z

)z−2dz

(3.3)=

∫ i∞

i

(−1

z−X

)k−2

zk−2 F (z) dz

=

∫ i∞

i

(−1− zX)k−2

F (z) dz.

Again, the growth condition derived in Exercise 3.35 ensures the existence of theintegral for all X ∈ C.

Together, we see that the integral∫ i∞

0

(z −X)k−2 F (z) dz =

∫ i

0

(z −X)k−2 F (z) dz +

∫ i∞

i

(z −X)k−2 F (z) dz

175

176 6. PERIOD POLYNOMIALS

exists for all X ∈ C.Since (z −X)k−2 is a polynomial in X of degree k − 2 we also see that∫ i∞

0

(z −X)k−2 F (z) dz ∈ C[X]k−2

is a polynomial of degree at most k − 2.

Remark 6.3. Recall the binomial formula

(z −X)k−2 =

k−2∑l=0

(k − 2

l

)zlXk−l−2.

Applying this identity to the integrand in (6.1) and compare the result with (5.62)we see directly that period polynomials and L-functions are related. We discussthis connection later.

We already used in the proof above that the term (z − X)k−2 has certaintransformation properties under z 7→ Tz with T given in (2.8). The next lemmashows that these properties essentially extend to all M ∈ SL2 (R).

Lemma 6.4. Let k ∈ 2Z and X denote a complex variable. We have

(6.2)(Mz −MX

)k−2=

(z −X)k−2

j(M, z)k−2 j(M,X)k−2

for all M ∈ SL2 (R), z ∈ H and X ∈ C provided that j(M,X) 6= 0.

Proof. We leave the proof to the reader as an exercise.

A direct application is the following transformation behavior for period poly-nomials.

Lemma 6.5. Let k ∈ 2N, F ∈ Sk(Γ(1)

)and M ∈ Γ(1). We have

(6.3) PF∣∣2−kM =

∫ M−1i∞

M−10

(z −X)k−2 F (z) dz (X ∈ C).

The path of integration is the geodesic line M−1iR>0 connecting the cusp M−1 0 tothe cusp M−1 i∞.

Proof. Consider F ∈Mk

(Γ(1)

)for k ∈ 2N and trivial multiplier. Using (6.1)

and a suitable substitution we have

PF∣∣2−kM(X)

(5.1)= j(M,X)k−2 PF (M X)

(6.1)= j(M,X)k−2

∫ i∞

0

(z −MX)k−2 F (z) dz

z 7→Mz= j(M,X)k−2

∫ M−1i∞

M−10

(Mz −MX)k−2 F (Mz) j(M, z)−2 dz

(6.2)=

∫ M−1i∞

M−10

(z −X)k−2 j(M, z)−kF (Mz) dz

(3.3)∫ M−1i∞

M−10

(z −X)k−2 F (z) dz

for all X ∈ C.

The above lemma allows us to connect the group action of Γ(1) on the cuspforms to transformations on the path of integration between cusp.

6.1. PERIOD POLYNOMIALS VIA INTEGRAL TRANSFORMATIONS 177

Lemma 6.6. For k ∈ 2N, consider a cusp form F ∈ Sk(Γ(1)

). The associated

period polynomial PF satisfies

(6.4) PF + PF∣∣2−kT = 0 and PF + PF

∣∣2−k(TS) + PF

∣∣2−k(TS)2 = 0

where S and T are given in (2.8).

Proof. Recall the relations T 2 = −1 = (TS)3 as shown in Exercise 2.5. We

consider the relation T 2 = −1 first. Since T =

(0 −11 0

)and T 2 =

(−1 00 −1

)=

−1, we have T−10 = i∞ and T−1i∞ = 0. The cusps 0 and i∞ are mapped ontoeach other and the path of integration is reversed. Hence, we have

PF (X)(6.1)=

∫ i∞

0

(z −X)k−2 F (z) dz

= −∫ 0

i∞(z −X)k−2 F (z) dz

= −∫ T−1i∞

T−10

(z −X)k−2 F (z) dz

(6.3)= −

(PF∣∣2−kT

)(X)

for all X ∈ C.

Now consider the matrices TS =

(0 −11 1

)and (TS)2 =

(−1 −11 0

). Applying

the Mobius transformations gives

(TS)−1 0 = i∞, (TS)−2 0 = −1, (TS)−1 i∞ = −1 and (TS)−2 i∞ = 0.

Hence the path of integration iR>0 connecting the cusp 0 to i∞ can be transformedto

iR>0 = (TS)−1 iR>0 ∪ (TS)−2 iR>0.

Here, we use the fact that the integrand is holomorphic in z and vanishes at thecusps. Hence the integral can be written as

PF (X)(6.1)=

∫ i∞

0

(z −X)k−2 F (z) dz

=

∫ −1

0

(z −X)k−2 F (z) dz +

∫ i∞

−1

(z −X)k−2 F (z) dz

= −∫ (TS)−2 i∞

(TS)−2 0

(z −X)k−2 F (z) dz −∫ (TS)−1 i∞

(TS)−1 0

(z −X)k−2 F (z) dz

(6.3)= −

(PF∣∣2−k(TS)2

)(X)−

(PF∣∣2−kTS

)(X)

for all X ∈ C.

The above lemma inspirers the following definition, extending the notion ofperiod polynomials for Γ(1) a bit.

Definition 6.7. Let k ∈ 2N and P ∈ C[X]k−2 be a polynomial of degreeat most k − 2. We call P a period polynomial P (for Γ(1)) if it satisfies the twofunctional equations

(6.5) P + P∣∣2−kT = 0 and P + P

∣∣2−k(TS) + P

∣∣2−k(TS)2 = 0.

We denote the space of such period polynomials by Pk−2.

Exercise 6.8. Let k ∈ 2N. Show that the polynomial P (X) := Xk−2 − 1 is aperiod polynomial in the sense of Definition 6.7.


The next proposition explains the relation between L-functions and periodpolynomials, which we already have hinted at in Remark 6.3.

Proposition 6.9. Let k ∈ 2N and F ∈ Sk(Γ(1)

). The L-function L? and

period function P defined in (5.62) and (6.1) respectively that are associated to Fsatisfy

PF (X) =

k−2∑l=0

(k − 2

l

)(−X)k−l−2 il+1 L?(l + 1)

= −k−2∑l=0

(k − 2

l

)X l (−i)k−l−1 L?(k − l − 1) (X ∈ C).

(6.6)

Proof. Using the binomial formula

(z −X)k−2 =

k−2∑l=0

(k − 2

l

)zl (−X)k−l−2,

we have

P (X)(6.1)=

∫ i∞

0

(z −X)k−2 F (z) dz

=

∫ i∞

0

k−2∑l=0

(k − 2

l

)zl(−X)k−l−2 F (z) dz

=

k−2∑l=0

(k − 2

l

)(−X)k−l−2

∫ i∞

0

zl F (z) dz

z 7→iy=

k−2∑l=0

(k − 2

l

)il+1(−X)k−l−2

∫ ∞0

yl F (iy) dy

(5.62)=

k−2∑l=0

(k − 2

l

)il+1(−X)k−l−2 L?(l + 1)

assuming that the integrals on the right hand side exists. This was shown inLemma 5.68.

The second identity in (6.6) follows from a similar calculation, starting fromthe binomial formula

(z −X)k−2 = (X − z)k−2 =

k−2∑l=0

(k − 2

l

)X l (−z)k−l−2

since k ∈ 2N.

Exercise 6.10. Let k ∈ 2N and F ∈ Sk(Γ(1)

). Show that the L-series and the

period function P defined in (5.56) and (6.1) respectively that are associated to Fsatisfy

(6.7) PF (X) = −k−2∑l=0

(k − 2)!

l!

L(k − l − 1)

(2πi)k−l−1X l (X ∈ C).

6.2. Eichler Integrals And Period Polynomials

There is another connection between period polynomials and Eichler integrals,these generalizations of abelian integrals introduced in Definition 4.30.

6.2. EICHLER INTEGRALS AND PERIOD POLYNOMIALS 179

Definition 6.11. Let k ∈ 2N and let F : H → C be a holomorphic function.The function F is called an Eichler integral of weight 2−k if there exists a sequenceof polynomials PM (z) ∈ C[z]k−2 of degree at most k − 2 such that F satisfies thetransformation property

(6.8) j(M, z)k−2 F(M z

)= F (z) + PM (z)

for all M ∈ Γ(1) and z ∈ H.

Remark 6.12. Each abelian integral F defined in Definition 4.30 is an Eichlerintegral for weight 0 in the sense of Definition 6.11.

Lemma 6.13. Let k ∈ 2N and let F be an Eichler integral of weight 2− k. Theassociated sequence of polynomials

(PM)M∈Γ(1)

(see Definition 6.11) satisfies

(6.9) P1 = P−1 = 0

and

(6.10) PMV = PM∣∣2−kV + PV

for all M,V ∈ Γ(1).

Proof. Consider the transformation property (6.8) of Eichler integrals. Wehave obviously

F (z) = j(1, z)2−k F (1z) and F (z) = j(−1, z)2−k F((−1)z

)since 1z = z, j(1, z) = 1 and (−1)z

(1.6)=

(−1 00 −1

)z

(1.9)= z, j(−1, z) = −1

with j(−1, z)2−k = 1 utilizing that k is an even integer. Hence the transformationproperty (6.8) the following identities

F (z) = F (z) + P1(z) and F (z) = F (z) + P−1(z).

This implies P1 = P−1 = 0.For the last part, we start with the transformation property (3.7) of the auto-

morphic factor j in Lemma 3.3 (since 2− k ∈ 2Z). Applying (3.7) to M,V ∈ Γ(1)we find

j(MV, z)2−k = j(M,vz)2−k j(V, z)2−k (z ∈ H).

Combining it with the transformation property (6.8) of Eichler integrals, we find

F (z) + PMV (z)(6.8)= j(MV, z)2−k F (MV z)

(3.7)= j(V, z)2−k

(j(M,V z)2−k F (M V z)

)(6.8)= j(V, z)2−k

(F (V z) + PM (V z)

)= j(V, z)2−k F (V z) + j(V, z)2−k PM (V z)

(6.8)= F (z) + PV (z) + j(V, z)2−k PM (V z)

for all z ∈ H. Subtracting F (z) on both sides, we get

PMV (z) = PV (z) + j(V, z)2−k PM (V z) (z ∈ H).

The above equation can be written as

PMV = PV + PM∣∣k−2

V

with the slash-notation given in Definition 5.1.


Proposition 6.14. Let k ∈ 2N. There exists a map from the space of Eich-ler integrals of weight 2 − k which vanishes under S to the space Pk−2 of periodpolynomials of weight 2− k (in the sense of Definition 6.7):

(6.11) F 7→ P := PT(6.8)= F

∣∣2−k(1− T ).

The kernel of the map is given by modular forms of weight 2− k for Γ(1).

Proof. We prove the lemma in two steps. First, we show that F −F∣∣2−kT is

a period polynomial of weight 2 − k provided F is an Eichler integral of the sameweight. Then we conclude by calculating the kernel.

Let F be an Eichler integral of weight 2−k. Then the transformation property(6.8) of Eichler integrals implies that P := PT given by

F∣∣2−kT = F + PT

is a polynomial of degree at most k − 2. Using the matrix identity T 2 = −1 andrepeatedly (6.9) of Lemma 6.13 we find

0 = P−1 = PT 2

(6.10)= PT

∣∣2−kT + PT .

Hence P = PT satisfies the left relation in (6.5) of Definition 6.7.Similarly, using the identity (TS)3 = −1 and repeatedly (6.10) of Lemma 6.13,

we find

0 = P−1 = P(TS)3

(6.10)= PTSTST

∣∣k−2

S + PS

(6.10)=

(PTSTS

∣∣k−2

T + PT

)∣∣k−2

S + PS

(5.3)= PTSTS

∣∣k−2

TS + PT∣∣k−2

S + PS

= . . .

= PT∣∣k−2

STSTS + PS∣∣k−2

TSTS + PT∣∣k−2

STS + PS∣∣k−2

TS + PT∣∣k−2

S + PS .

Since PS = F∣∣2−k(1− S) vanishes, as we assumed in the proposition, we find

0 = PT∣∣k−2

STSTS + PT∣∣k−2

STS + PT∣∣k−2

S

= PT∣∣k−2

S∣∣2−k

((TS)2 + TS + 1

).

Hence PT∣∣k−2

S vanishes under (TS)2 +TS+1. Using the matrix identity (TS)3 =

−1 again, we see

S(TS)2 = (−1)T, S(TS) = TS(−1)T and S = (TS)2(−1)T.

Hence PT satisfies

0 = PT∣∣k−2

S∣∣2−k

((TS)2 + TS + 1

)= PT

∣∣2−k

(1 + TS + (TS)2

)∣∣2−k(−1)T.

Since (−1)T =

(−1 −10 −1

), we conclude that PT satisfies the identity(

PT∣∣k−2

STST + PT∣∣k−2

ST + PT

)∣∣∣∣2−k

T = 0.

Using j(S,X)k−2 = 1 and P = −PT the above identity reads as(P∣∣k−2

STST + P∣∣k−2

ST + P

)(−X − 1) = 0.


This proves the right relation in (6.5) of Definition 6.7. Summarizing, we justshowed that P = PT is a period polynomial of weight 2− k in the sense of Defini-tion 6.7.

Next, we calculate the kernel of the map F 7→ P := F − F∣∣2−kT . An Eichler

integral F is in the kernel, if F − F∣∣2−kM vanishes. Since PS = 0 vanishes by

assumption, we find that F in the kernel satisfies the identities

F = F∣∣2−kT and F = F

∣∣2−kS.

Theorem 3.18 implies

F = F∣∣2−kM for all M ∈ Γ(1).

Also, F is holomorphic on H. Hence F satisfies all criteria of being a modular formof weight 2 − k for Γ(1) as introduced in Definition 3.26. In other words, if theEichler integral F is in the kernel of the map (6.11) then F ∈M !

2−k(Γ(1)

).

The remaining part of this section introduces different realizations of Eichler

integrals, the (k − 1)-fold antiderivative, the integral∫ i∞z

f(w) (z − w)k−2 dw for

f ∈ Sk(Γ(1)

), and the integral

∫ izf(w) (z − w)k−2 dw for f ∈Mk

(Γ(1)

).

6.2.1. The (k − 1)-fold antiderivative. Similar to the notion of the anti-derivative, as introduced in Definition 4.31, we define the (k−1)-fold antiderivativeas follows:

Definition 6.15. Let f : H → C be a holomorphic function admitting anFourier expansion of the form

f(z) =

∞∑n=−µ

ane2πinz (z ∈ H).

For k ∈ 2N define the (k − 1)-fold antiderivative of f as

F (z) =a(−µ)

(−2πiµ)k−1e−2πiµz + . . .

. . .+a(−1)

(−2πi)k−1e−2πiz +

a(0)

(k − 1)!zk−1 +

∞∑n=1

a(n)

(2πin)k−1e2πinz

for all z ∈ H.

Exercise 6.16. Let F : H → C be a holomorphic function, k ∈ 2N andM ∈ Γ(1). Prove Bol’s identity

(6.12)dk−1

dzk−1

(j(M, z)k−2 F

(M z

))= j(M, z)−k F (k−1)

(M z

)(z ∈ H)

where F (k−1) denotes the (k − 1)th derivative F (k−1)(z) = dk−1

dzk−1F (z) of F .

Remark 6.17. Bol’s identity (6.12) goes back to G. Bol’s paper [5, page 28].

Lemma 6.18. Let k ∈ 2N and f ∈M !k

(Γ(1)

). We assume in addition that f(z)

is holomorphic on H and that its Fourier expansion at the cusp ∞ is given by

f(z) =

∞∑n=−µ


The (k−1)-fold antiderivative F of f given in Definition 6.15 is an Eichler integralof weight 2− k.


Proof. Since f ∈M !k

(Γ(1)

)by assumption, we get

j(M, z)−k f(M z

)− f(z) = 0

for all M ∈ Γ(1) and z ∈ H. Denote by F the (k − 1)-fold antiderivative of f . Thedefinition directly implies that

(6.13)dk−1

dzk−1F (z) = f(z)

by interchanging summation and differentiation. Using Bol’s identity in (6.12)Chain rule we find

dk−1

dzk−1

(j(M, z)k−2 F

(M z

)− F (z)

) (6.12)= j(M, z)−k F (k−1)

(M z

)− F (k−1)(z)

= j(M, z)−k f(M z

)− f(z)

= 0.

(6.14)

Integrating k − 1 times the left hand and right hand side of (6.14), we get

j(M, z)k−2 F(M z

)− F (z) = pM (z) (z ∈ H)

for suitable polynomials pM (z) ∈ C≤k−2[z] of degree at most k − 2 and for allM ∈ Γ(1). Hence F satisfies condition (6.8) of Definition 6.11. This shows that Fis an Eichler integral of weight 2− k.

Remark 6.19. The authors of [8] considered an slight variant of the Eichlerintegral F given as (k − 1)-fold derivative in Definition 6.15. Assuming that f ∈M !k,1

(Γ(1)

)with Fourier expansion

f(z) =

∞∑n=−ν

an e2πinz

they associate to f the formal Eichler integral

(6.15) Ef (z) :=

∞∑06=n=−ν

annk−1

e2πinz

and use this definition to develop their theory.A vector valued version of the (k− 1)-fold derivative and its relation to Eichler

integrals was also discussed in [38].

6.2.2. An integral representation of the Eichler integral for cuspforms.

Definition 6.20. Let k ∈ 2N and f ∈ Sk(Γ(1)

). We define the function

F : H→ C by the integral transform

(6.16) F (z) :=

∫ i∞

z

f(w) (w − z)k−2 dw (z ∈ H).

The integral in (6.16) is well defined since the cusp form f ∈Mk

(Γ(1)

)satisfies

the exponential growth condition given Exercise (3.35).

Lemma 6.21. Let k ∈ 2N and f ∈ Sk(Γ(1)

).

• The function F : H → C given in (6.16) is an Eichler integral of weight2− k.

• If F denotes the (k − 1)-fold antiderivative of f defined in Lemma 6.18

then F and F satisfy

(6.17) − (k − 2)!F (z) = F (z) (z ∈ H).


Proof. We leave the proof of the first part as an exercise to the reader.For the second part, we first consider the derivative of F .

d

dzF (z) =

d

dz

∫ i∞

z

f(w) (w − z)k−2 dw

= limε0

1

ε

(∫ i∞

z+ε

f(w) (w − z − ε)k−2 dw −∫ i∞

z

f(w) (w − z)k−2 dw

)= limε0

1

ε

(∫ i∞

z

f(w + ε) (w − z)k−2 dw −∫ i∞

z

f(w) (w − z)k−2 dw

)=

∫ i∞

z

limε0

1

ε

(f(w + ε)− f(w)

)(w − z)k−2 dw

=

∫ i∞

z

f ′(w) (w − z)k−2 dw

= f(w) (w − z)k−2

∣∣∣∣w=i∞

w=z

− (k − 2)

∫ i∞

z

f(w) (w − z)k−3 dw.

Since f as cusp form satisfies the growth condition in Exercise 3.35, we see that thefirst term vanishes. Hence, we have

d

dzF (z) = −(k − 2)

∫ i∞

z

f(w) (w − z)k−3 dw.

Iterating the argument, we find

dk−2

dzk−2F (z) = (−1)k−2 (k − 2)!

∫ i∞

z

f(w) dw

and applying it once again we get,

dk−1

dzk−1F (z) = (−1)k−2 (k − 2)!

d

dz

∫ i∞

z

f(w) dw

= −(−1)k−2 (k − 2)! f(z).

The last step uses ddz

∫ i∞z

f(w) dw = −f(z). Hence

dk−1

dzk−1F (z) = −(k − 2)! f(z)

holds since k ∈ 2N. This equation together with (6.13) implies that

dk−1

dzk−1F (z) = −(k − 2)!

dk−1

dzk−1F (z)

where F is the (k − 1)-fold antiderivative of f .It remains to show that (6.17) holds given above identity for the derivatives.

By (k − 1) times indefinite integration, we get

(6.18) F (z) = −(k − 2)! F (z) + P (z)

where P (z) ∈ C[z]k−2 is a polynomial of degree at most k − 2 coming from theindefinite integration. To calculate the polynomial P (z) we consider the limit asz → i∞. On the left hand side, we have

limz→i∞

F (z) = limz→i∞

∫ i∞

z

f(w) (w − z)k−2 dw = 0.


To calculate limz→i∞ F (z), recall that f as cusp form has an Fourier expansion atthe cusp ∞ of the form

f(z) =

∞∑n=1

a(n) e2πinz (z ∈ H),

see Definitions 3.25 and 3.26. Then the (k − 1)-fold antderivative of f is given by

F (z) =

∞∑n=1

a(n)

(2πin)k−1e2πinz (z ∈ H),

see Definition 6.15. We see that the Fourier coefficients of F satisfy the estimate∣∣∣∣ a(n)

(2πin)k−1

∣∣∣∣ ≤ |a(n)|

for all n ∈ N. This implies that F can be estimated by f :

|F (z)| ≤∞∑n=1

∣∣∣∣ a(n)

(2πin)k−1

∣∣∣∣ e−2πn=(z) = O(e−2πδ=(z)

)as = (z) → ∞ for some δ > 0. The last estimate follows by the same argumentswhich were used to show Exercise 3.35. Hence,

limz→i∞

F (z) = 0

holds, too. Together, both limits show that the additive polynomial P (z) in (6.18)must be identical 0. This shows (6.17) and concludes the proof.

Exercise 6.22. For k ∈ 2N, let f ∈ Sk(Γ(1)

)and F the Eichler integral defined

by the integral (6.16). Show, using only the integral representation (6.16), that Fassociated to the period polynomial P in (6.11) admits the integral representation(6.1) and is a period polynomial in the sense of Definition 3.27. In particular, show

(6.19) F∣∣2−k

(T − 1

)= −Pf and F

∣∣2−k

(S − 1

)= 0

where Pf is given in (6.1).

Remark 6.23. For k ∈ 2N let f ∈ Sk(Γ(1)

)be a cusp form. Since we have

the inclusions Sk(Γ(1)

)⊂ S!

k

(Γ(1)

)⊂M !

k

(Γ(1)

), we may associate to f the formal

Eichler integral Ef (z) given in (6.24) of Remark 6.19. Then, Ef (z) can be interpreted

as the Eichler integral F (z) given in (6.16). We have

Ef (z) = − (2πi)k−1

Γ(k − 1)

∫ i∞

z

f(z) (w − z)k−2 dw

= − (2πi)k−1

Γ(k − 1)F (z).

(6.20)

This relation already appears in [8, (1.8)].

Exercise 6.24. Prove (6.20).

6.2.3. An integral representation of the Eichler integral for entireforms.

Definition 6.25. Let k ∈ 2N and f ∈ Mk

(Γ(1)


F : H→ C by the integral transform

(6.21) F (z) :=

∫ i

z

f(w) (w − z)k−2 dw (z ∈ H).

Obviously the integral in (6.21) is well defined for all z ∈ H.


Exercise 6.26. Let k ∈ 2N and f ∈ Mk

(Γ(1)

). F defined in (6.21) is an

Eichler integral of weight 2− k.

Lemma 6.27. Let k ∈ 2N and f ∈ Mk

(Γ(1)

). If F denotes the (k − 1)-fold

antiderivative of f defined in Lemma 6.18 then F and F satisfy

(6.22) F (z) = −(k − 2)!

(F (z) +

k − 2

(k − 1)!f(i)zk−1

)+ P (z) (z ∈ H)

for some polynomial P (z) ∈ C[z]k−2 of degree at most k− 2 which depends only onf .

Proof. Similar to the proof of the second part of Lemma 6.21, we considerthe derivative of F :

d

dzF (z) =

d

dz

∫ i

z

f(w) (w − z)k−2 dw

= limε0

1

ε

(∫ i

z+ε

f(w) (w − z − ε)k−2 dw −∫ i

z

f(w) (w − z)k−2 dw

)= limε0

1

ε

(∫ i

z

f(w + ε) (w − z)k−2 dw −∫ i

z

f(w) (w − z)k−2 dw

)=

∫ i

z

limε0

1

ε

(f(w + ε)− f(w)

)(w − z)k−2 dw

=

∫ i

z

f ′(w) (w − z)k−2 dw

= f(w) (w − z)k−2

∣∣∣∣w=i

w=z

− (k − 2)

∫ i∞

z

f(w) (w − z)k−3 dw

= f(i) (i− z)k−2 − (k − 2)

∫ i∞

z

f(w) (w − z)k−3 dw.

Iterating the argument, we find that

dk−2

dzk−2F (z) =

dk−3

dzk−3

(d

dzF (z)

)=

dk−3

dzk−3

(f(i) (i− z)k−2 − (k − 2)

∫ i∞

z

f(w) (w − z)k−3 dw

)= (−1)k−3 (k − 2)! f(i) (i− z) −

− (k − 2)dk−4

dzk−4

(d

dz

∫ i∞

z

f(w) (w − z)k−3 dw

)= (−1)k−3 (k − 2)! f(i) (i− z)

− (k − 2)dk−4

dzk−4

(f(i) (i− z)k−3 − (k − 3)

∫ i∞

z

f(w) (w − z)k−4 dw

)= (−1)k−3 (k − 2)! f(i) (i− z) + (−1)k−3 (k − 2)!f(i) (i− z)

+ (k − 2)(k − 3)dk−4

dzk−4

∫ i∞

z

f(w) (w − z)k−4 dw

= . . .

= (k − 2)

((−1)k−3(k − 2)! f(i) (i− z)

)+ (−1)k−2 (k − 2)!

∫ i

z

f(w) dw

= (−1)k−2 (k − 2)!

(∫ i

z

f(w) dw − (k − 2) f(i) (i− z)).


Using that k is even gives

dk−2

dzk−2F (z) = (k − 2)!

(∫ i

z

f(w) dw − (k − 2) f(i) (i− z)).

Applying the derivative once again, we find

dk−1

dzk−1F (z) = (k − 2)! (k − 2) f(i) + (k − 2)!

d

dz

∫ i

z

f(w) dw

= (k − 2)! (k − 2) f(i) − (k − 2)! f(z)

= (k − 2)! (k − 1) f(i) + (k − 2)!f(i) − (k − 2)! f(z)(6.23)

= (k − 1)! f(i)− (k − 2)!

(f(z)− f(i)

).

Equation (6.23) together with (6.13) implies

(6.24)dk−1

dzk−1F (z) = (k − 2)!(k − 2) f(i) − (k − 2)!

dk−1

dzk−1F (z)

where F is the (k − 1)-fold antiderivative of f .It remains to show that (6.22) holds given above identity for the derivatives.

By (k − 1) times indefinite integration, we get

F (z) = −(k − 2)!

(F (z) +

k − 2

(k − 1)!f(i)zk−1

)+Q(z)

where Q(z) ∈ C[z]k−2 is a polynomial of degree at most k − 2 coming from theindefinite integration. This shows (6.22) and concludes the proof.

Exercise 6.28. Let k ∈ 2N and f ∈ Sk(Γ(1)

). Show that F and F defined in

(6.16) and (6.21) respectively satisfy

(6.25) F (z) = F (z) + P (z)

where P ∈ C[z]k−2 is the polynomial of degree at most k − 2 given by

(6.26) P (z) =

∫ i∞

i

f(w) (w − z)k−2 dw

for all z ∈ H.

Remark 6.29. Let z0 ∈ H be a fixed reference point on H. Analogous to theabove exercise we easily see that any integral

∫ z0zf(w) (w − z)k−2 dw is also an

Eichler integral. The difference of the two terms is the polynomial∫ z0zf(w) (w −

z)k−2 dw ∈ C[z]k−2 of degree at most k − 2.

6.2.4. The Niebur integral representation for entire forms.

Definition 6.30. Let k ∈ 2N and g ∈ Mk

(Γ(1)


G− : H→ C by the integral transform

(6.27) G−(z) :=

∫ i

z

g(w)(w − z

)k−2dw (z ∈ H)

where · denotes complex conjugation of the overlined expression. We call G− theNiebur integral of g.

Obviously the integral in (6.21) is well defined for all z ∈ H.


Exercise 6.31. Let k ∈ 2N and g ∈Mk

(Γ(1)

). Show that G− defined in (6.27)

satisfies

(6.28) G−∣∣2−kM = G− + PM

for all M ∈ Γ(1) with polynomials PM ∈ C[z]k−2 given by

(6.29) PM (z) :=

∫ M−1i

i

g(w)(w − z

)k−2dw.

Lemma 6.32. Let k ∈ 2N and g ∈ Mk

(Γ(1)

). The family of polynomials

PM ∈ C[z]k−2, M ∈ Γ(1), associated to f by the Niebur integral G− and its trans-formation formula shown in Exercise 6.32 satisfies (6.9) and the transformationrelation (6.10).

Proof. This lemma can be proved exactly by the same arguments used in theproof of Lemma 6.13. The only modification necessary is referring to the transfor-mation property shown in Exercise 6.32 instead the one in Definition 3.17.

Exercise 6.33. Recall the usual definitions of partial derivatives

(6.30)∂

∂zH =

1

2

(∂

∂xH − i ∂

∂yH

)and

∂

∂zH =

1

2

(∂

∂xH + i

∂

∂yH

)for z = x + iy and H : H → C a function (not necessarily holomorphic) whichare derived from the Cauchy-Riemann equations for analytic functions, see Appen-dix A.7. If G− is a Niebur integral associated to the modular form g ∈ Mk

(Γ(1)

)in the sense of Definition 6.30, then G− satisfies

(6.31)∂

∂zG−(z) = g(z)

((z − z

)k−2).

Exercise 6.34. Let k ∈ 2N and g ∈ Sk(Γ(1)

). We define the function G− :

H→ C by the integral transform

(6.32) G−(z) :=

∫ i∞

z

g(w)(w − z

)k−2dw. (z ∈ H)

(1) Then G− satisfies

(6.33) G−∣∣2−kM = G− + PM

for all M ∈ Γ(1) with polynomials PM (z) ∈ C[z]k−2 given by

(6.34) PM (z) :=

∫ M−1i∞

i∞g(w)

(w − z

)k−2dw.

(2) We have

(6.35) G−(z) = G−(z) + P (z)

with P ∈ C[z]k−2 is the polynomial of degree at most k − 2 given by

(6.36) P (z) =

∫ i∞

i

g(w)(w − z

)k−2dw

for all z ∈ H.


6.3. The Eichler Cohomology Group

To introduce the Eichler cohomology group, we consider a collection of poly-nomials

(PM)M∈Γ(1)

.

Definition 6.35. Let k ∈ 2N and(PM)M∈Γ(1)

a collection of polynomials

PM ∈ C[X]k−2 of degree at most k − 2.

(1) We call the collection(PM)M∈Γ(1)

a cocycle if its elements satisfy

(6.37) PMV = PM∣∣2−kV + PV for all M,V ∈ Γ(1).

(2) We call the collection(PM)M∈Γ(1)

a coboundary if there exists a fixed

polynomial P ∈ C[X]k−2 such that PM = P∣∣2−kM − P holds for all

M ∈ Γ(1).

The set of cocycles and the set of coboundaries are denoted by Z1(Γ(1),C[X]k−2

)and B1

(Γ(1),C[X]k−2

), respectively.

It is obvious that the set of cocycles and the set of coboundaries are additivegroups.

Exercise 6.36. Show that every coboundary is a cocycle.

Exercise 6.37. Let k ∈ 2N. Show that every cocycle(PM)M∈Γ(1)

∈ Z1(Γ(1),C[X]k−2

)satisfies

(6.38) P±1 = 0 and P±T∣∣2−k(±T ) + P±T = 0.

Exercise 6.38. Let k ∈ 2N. Show that every cocycle(PM)M∈Γ(1)

∈ Z1(Γ(1),C[X]k−2

)satisfies

(6.39) PM∣∣2−kM

−1 + PM−1 = 0

for all M ∈ Γ(1).

Definition 6.39. We call the quotient group

(6.40) H1(Γ(1),C[X]k−2

):=

Z1(Γ(1),C[X]k−2

)B1(Γ(1),C[X]k−2

)the Eichler cohomology group. Elements of H1

(Γ(1),C[X]k−2

)are denoted by

PM, indicating the family of polynomials(PM)M∈Γ(1)

which represent a corre-

sponding cocycle.

We next introduce a natural and important subspace of H1(Γ(1),C[X]k−2),

the space of parabolic cohomology.

Definition 6.40. Let k ∈ 2N. A cocycle(PM)M∈Γ(1)

∈ Z1(Γ(1),C[X]k−2

)is

parabolic if there exists a polynomial Q ∈ C[X]k−2 such that PS indexed by theparabolic generator S of Γ(1) is given by

(6.41) PS = Q∣∣2−kS −Q.

The additive subgroup of parabolic cocycles is denoted by Z1par

(Γ(1),C[X]k−2

)⊂

Z1(Γ(1),C[X]k−2

). We call the quotient group

(6.42) H1par

(Γ(1),C[X]k−2

):=

Z1par

(Γ(1),C[X]k−2

)B1(Γ(1),C[X]k−2

)the parabolic Eichler cohomology group.

6.3. THE EICHLER COHOMOLOGY GROUP 189

Exercise 6.41. Show that a parabolic cocycle(PM)M∈Γ(1)

∈ Z1par

(Γ(1),C[X]k−2

)satisfies the following: For each parabolic element V ∈ Γ(1) there exists a polyno-mial Q?V ∈ C[X]k−2 such that PV can be expressed as

(6.43) PV = Q?V∣∣2−kV −Q

?V .

For f ∈Mk

(Γ(1)

), consider the Eichler integral F as defined in (6.21). Sine F

is an Eichler integral, it satisfies its defining relation (6.8), see Definition 6.11. Theassociated family of polynomials PM ∈ C[X]k−2, M ∈ Γ(1), can be expressed as

PM (z) =

∫ M−1i

i

f(w) (w − z)k−2 dw,

see Exercise 6.26. The family(PM)M∈Γ(1)

is a cocycle in the sense of Definition 6.35

since Lemma 6.13 shows that the PM ’s satisfy (6.37). This argument establishesthe following map

Mk

(Γ(1)

)→ Z1

(Γ(1),C[X]k−2

); f

(6.21)7→ F(6.8)7→

(PM)M∈Γ(1)

.

Using the fact that the cocycles are unique up to a coboundary, the above mapinduces an map α into the Eichler cohomology group.

Definition 6.42. Let k ∈ 2N. We define the map α as follows:

α :Mk

(Γ(1)

)−→ H1

(Γ(1),C[X]k−2

)f

(6.21),(6.8)7−→(PM)M∈Γ(1)

mod B1(Γ(1),C[X]k−2

).

(6.44)

Remark 6.43. If the realization of the Eichler integral in the Definition 6.42is replaced by another realization, for example by the (k − 1)-fold antiderivativeor the integral transformation for cusp forms, then the cocycle

(PM)M∈Γ(1)

might

change, but the associated cohomology class does not.

Using the same reasoning as above, we can define a map

Mk

(Γ(1)

)→ Z1

(Γ(1),C[X]k−2

); g

(6.27)7→ G−(6.28)7→

(PM)M∈Γ(1)

using the Niebur integral instead of the Eichler integral. This leads again to thefollowing

Definition 6.44. Let k ∈ 2N. We define the map β as

β :Mk

(Γ(1)

)−→ H1

(Γ(1),C[X]k−2

)f

(6.27),(6.28)7−→(PM)M∈Γ(1)

mod B1(Γ(1),C[X]k−2

).

(6.45)

Definition 6.45. Let k ∈ 2N. The combination of the maps α and β leads to

µ :Mk

(Γ(1)

)⊕ Sk

(Γ(1)

)−→ H1

(Γ(1),C[X]k−2

)(f, g)

(6.43),(6.45)7−→ α(f) + β(g).(6.46)

By construction, we have that the linear map µ maps the direct product spaceMk

(Γ(1)

)⊕ Sk

(Γ(1)

)into the Eichler cohomology group. In what follows, we will

discuss whether this map is an isomorphism or not.

Lemma 6.46. Let k ∈ 2N.

(1) For f ∈ Sk(Γ(1)

), recall that the image of α(f) =

(PM)M∈Γ(1)

is given

by transformation property PM = F∣∣2−kM − F of the Eichler integral F


defined in (6.21). We can express the values of this cocylce(PM)M∈Γ(1)

as

(6.47) PM =

(F∣∣2−kM − F

)+

((−P )

∣∣2−kM − (−P )

)where P ∈ C[z]k−2 is the polynomial given in (6.26). In particular wehave

PS =

(P∣∣2−kS − P

)which implies that the cocycle

(PM)M∈Γ(1)

is parabolic and

α : Sk(Γ(1)

)→ H1

par

(Γ(1),C[X]k−2

)maps into the parabolic cohomology.

(2) For g ∈ Sk(Γ(1)

), recall that the image of β(g) =

(QM

)M∈Γ(1)

is given by

transformation property QM = G−∣∣2−kM−G

− of the Niebour integral G−

defined in (6.27). We can express the values of this cocylce(QM

)M∈Γ(1)

as

(6.48) QM =

(G−∣∣2−kM − F

−)

+

(P∣∣2−kM − P

)where P ∈ C[z]k−2 is the polynomial given in (6.35). In particular wehave

PS =

(P∣∣2−kS − P

)which implies that the cocycle

(QM

)M∈Γ(1)

is parabolic and

β : Sk(Γ(1)

)→ H1

par

(Γ(1),C[X]k−2

)maps into the parabolic cohomology.

Proof. For f ∈ Sk(Γ(1)

)write the image of α(f) =

(PM)M∈Γ(1)

as given by

transformation property PM = F∣∣2−kM − F of the Eichler integral F defined in

(6.21). Using Exercise 6.28, we see

PM(6.8)= F

∣∣2−kM − F

(6.25)=

(F − P

)∣∣2−kM −

(F − P

)=

(F∣∣2−kM − F

)+

((−P )

∣∣2−kM − (−P )

)with P ∈ C[z]k−2 given by (6.26).

Using the fixed point condition S i∞ = i∞ we have

F∣∣2−kS(z) = F (z) +

∫ S−1i∞

i∞f(w) (w − z)k−2 dw

= F (z) +

∫ i∞

i∞f(w) (w − z)k−2 dw︸︷︷︸

=0

= F (z),

following the proof of the first part of Lemma 6.21 for the parabolic element S =(1 10 1

). This implies that the cocycle

(PM)M∈Γ(1)

is in fact parabolic cocycle.

We have (PM)M∈Γ(1)

∈ Z1par

(Γ(1),C[X]k−2

).

6.3. THE EICHLER COHOMOLOGY GROUP 191

The map α maps Sk(Γ(1)

)into the parabolic cohomology H1

par

(Γ(1),C[X]k−2

).

The second part is left to the reader as an exercise.

Example 6.47. For k ∈ 2N, let f ∈ Sk(Γ(1)

)be a cusp form. The period

polynomial P (X) associated to f by (6.1) and the parabolic cocycle(PM)M∈Γ(1)

attached to f by the Eichler integral F given in (6.16) are related as follows: Con-sider the cocycle given by its values PT and PS of the generators T and S of Γ(1).Using the transformation property (6.8) of Eichler integrals, we have

PS(X)(6.8)= F

∣∣2−kS(X)− F (X)

=

∫ S−1i∞


using the proof of the first part of Lemma 6.21

S i∞=i∞=

∫ i∞


= 0

and

PT (X)(6.8)= F

∣∣2−kT (X)− F (X)

=

∫ T−1i∞


using the proof of the first part of Lemma 6.21

T−1 i∞=0=

∫ 0


(6.1)= −P (z)

where P (z) is to f associated period function.Hence the cocycle

(PM)M∈Γ(1)

satisfies

(6.49) PS = 0 and PT = −P.We see in particular that period functions give a representative of the paraboliccocycle associated to cusp forms by the Eichler integral.

Lemma 6.48. Let k ∈ 2N.

(1) Each element of parabolic cohomology H1par

(Γ(1), k

)can be uniquely rep-

resented (up to scalar multiplication) by a parabolic cocycle(PM)M∈Γ(1)

∈Z1par

(Γ(1), k

)chosen such that PS = 0 vanishes.

(2) The map

H1par

(Γ(1), k

)→ Pk−2;

(PM)M∈Γ(1)

with PS = 0

7→ PT

has codimension 1. The one-dimensional subspace in Pk−2 is spanned byXk−2 − 1 ∈ C[X]k−2.

Proof. (1) Recall Definition 6.40: each element ofH1par

(Γ(1), k

)is repre-

sented by a parabolic cocycle(PM)M∈Γ(1)

∈ Z1par

(Γ(1), k

)which is unique

up to a coboundary in B1(Γ(1), k

). If

(QM

)M∈Γ(1)

∈ B1(Γ(1), k

)is a

coboundary then the cocycles(PM)M∈Γ(1)

and(PM +QM

)M∈Γ(1)

repre-

sents the same cohomology class.


Let(PM)M∈Γ(1)

∈ Z1par

(Γ(1), k

)be a parabolic cocycle. Defini-

tion 6.40 implies that there exists a Q ∈ C[X]k−2 such that

PS = Q∣∣2−k(S − 1).

Define the coboundary(QM

)M∈Γ(1)

and a new cocycle(P ′M)M∈Γ(1)

by

QM := −Q∣∣2−k(M − 1) and P ′M := PM +QM = PM −Q

∣∣2−k(M − 1)

for all M ∈ Γ(1). Hence(PM)M∈Γ(1)

and(P ′M)M∈Γ(1)

represent the same

cohomology element in H1par

(Γ(1), k

)since they differ by a coboundary.

By construction the cocycle(P ′M)M∈Γ(1)

is parabolic and vanishes. We

have

P ′S = PS +QS =

(Q∣∣2−k(S − 1)

)−(Q∣∣2−k(S − 1)

)= 0 = 0

∣∣2−k(S − 1).

(2) Using the first part of the lemma, we can represent each element ofH1

par

(Γ(1), k

)with a parabolic cocycle

(PM)M∈Γ(1)

∈ Z1par

(Γ(1), k

)satis-

fying PS = 0. We see as in the proof of Proposition 6.14 that the polyno-mial PT is a period function in the sense of Definition 6.7: PT ∈ Pk−2.

Now, let P ∈ Pk−2 be a period polynomial. We define

PS := 0 and PT := P.

Since the matrices S and T generate Γ(1), we see that the values of PS andPT determine the whole cocycle

(PM)M∈Γ(1)

, utilizing (6.43). Moreover,

the cocycle is parabolic (since PS = 0 = 0∣∣2−k(S − 1)). Hence, we just

constructed a map

Pk−2 → Z1par

(Γ(1), k

); P 7→

(PM)M∈Γ(1)

given by PS = 0 and PT = P

which induces a map

Pk−2 → H1par

(Γ(1), k

).

Next, we calculate the kernel of the above map. A P ∈ Pk−2 liesin the kernel of above map, if P is derived from a coboundary. Thismeans that there exists a Q ∈ C[X]k−2 such that P = Q

∣∣2−k(T − 1) and

0 = Q∣∣2−k(S − 1). Since the second condition on Q implies in fact that

Q is invariant under translation X 7→ S X = X + 1; hence Q must bea constant function. (The constant functions are the only polynomialsinvariant under translation.) Put Q := 1 since C[X]k−2 is a C-vectorspace. Hence, we have

PT (X) : = 1∣∣2−k

(T − 1

)(X) = Xk−2 − 1 and

PS(X) : = 1∣∣2−k

(S − 1

)(X) = 1− 1 = 0.

Since Xk−2 − 1 ∈ Pk−2, see Exercise 6.8, we conclude that the kernelof the map above is one dimensional and is spanned by Xk−2 − 1. Thisimplies

dimPk−2 = 1 + dimH1par

(Γ(1), k

).

6.4. THE EICHLER COHOMOLOGY THEOREM 193

6.4. The Eichler Cohomology Theorem

Theorem 6.49 (Eichler cohomology theorem). Let k ∈ 2N. The spaces

Mk

(Γ(1)

)⊕ Sk

(Γ(1)

)and H1

(Γ(1),C[X]k−2

)are isomorphic under the map µ defined in (6.46). Here, ⊕ denotes the direct sumof two spaces.

Remark 6.50. The isomorphism µ between the spaces Mk

(Γ(1)

)× Sk

(Γ(1)

)and H1

(Γ(1),C[X]k−2

), which was defined in (6.46), is also known as the Eichler-

Shimura isomorphism.

Corollary 6.51. Let k ∈ 2N. The spaces Sk(Γ(1)

)⊕Sk

(Γ(1)

)and H1

par

(Γ(1),C[X]k−2

)are isomorphic under the map µ defined in (6.46).

Proof. Consider (f, g) ∈ Sk(Γ(1)

)⊕Sk

(Γ(1)

). Then µ(f, g) ∈ H1

(Γ(1),C[X]k−2

)is represented by a cocycle, see Theorem 6.49. Using Lemma 6.46 we see that thecocycle is in fact parabolic. Hence µ maps into the parabolic cocycle space

µ : Sk(Γ(1)

)⊕ Sk

(Γ(1)

)→ H1

par

(Γ(1),C[X]k−2

)The injectivity of the restricted map follows directly from the injectivity of theoriginal map. Since Sk

(Γ(1)

)⊕ Sk

(Γ(1)

)is a subspace of Mk

(Γ(1)

)⊕ Sk

(Γ(1)

)of

codimension 1 and also H1par

(Γ(1),C[X]k−2

)is a subspace of H1

(Γ(1),C[X]k−2

)of codimension 1. This follows since Γ(1) = 〈S, T 〉 and only S is parabolic. Wesee that the restricted map remains surjective. Hence the restricted map is also anisomorphism.

Corollary 6.52. Let k ∈ 2N. The map

Sk(Γ(1)

)⊕ Sk

(Γ(1)

)→ Pk2

(f, g) 7→∫ i∞

0

f(w) (w −X)k−2 dw +

∫ i∞

0

g(w) (w −X)k−2 dw(6.50)

has codimension 1. The subspace in Pk−2 is spanned by Xk−2 − 1.

Proof. Corollary 6.51 implies that the map

Sk(Γ(1)

)⊕ Sk

(Γ(1)

)→ H1

par

(Γ(1),C[X]k−2

)induced by µ is an isomorphism. We may replace the Eichler and Niebur integralsF and G− in the construction of µ in (6.46) by the respective integrals F and G−,see Lemma 6.46, since we assumed the spaces of cusp forms as the initial spaces.With this replacement, define the generating values PS and PT of the paraboliccocycle µ(f, g) =

(PM)M∈Γ(1)

by

PS = −

(∫ S−1 i∞

i∞f(w) (w −X)k−2 dw +

∫ S−1 i∞

i∞f(w) (w −X)k−2 dw

)= 0

and

PT = −

(∫ T−1 i∞

i∞f(w) (w −X)k−2 dw +

∫ T−1 i∞

i∞f(w) (w −X)k−2 dw

)

=

∫ i∞

0

f(w) (w −X)k−2 dw +

∫ i∞

0

g(w) (w −X)k−2 dw.

Applying Lemma 6.48 concludes the proof.


In what remains in this section, we present the proof of the Eichler cohomologytheorem.

6.4.1. Injectivity of the map µ. We have the following variant of Green’sTheorem:

Lemma 6.53. Let H : H→ C be a non-holomorphic function satisfying the growestimate H(z) = O

(e−2πδ=(z)

)as = (z)→∞ for some δ > 0. Then H satisfies the

identity

(6.51)

∫∫FΓ(1)

∂

∂zH(z) dxdy = − i

2

∫∂FΓ(1)

H(z) dz

where the the path of integration on the right hand side is along the positively(anticlockwise) oriented set-boundary ∂FΓ(1) of the standard fundamental regionFΓ(1).

Proof. To do!1

Observe that the usual stated Green’s theorem does not apply directly to thefundamental region FΓ(1), since the later contains a vertical half-strip. This meansthat FΓ(1) is not a bounded subset of H. However, we can extend the theoremto cover this case by applying a simple limit argument and using the fact thatg ∈ Sk

(Γ(1)

)as cusp form vanishes at the cusp i∞.

Lemma 6.54. In the situation of Lemma 6.53, assume additionally that Hsatisfies the transformation property (3.3) with weight 2 (and trivial multiplier 1):j(M, z)−kH(M z) = H(z) for all M ∈ Γ(1) and z ∈ H.

The integral∫∂FΓ(1)

H(z) dz on the right had side of (6.51) vanishes:

(6.52)

∫∂FΓ(1)

H(z) dz = 0..

Proof. Using Exercise 2.16, we can decompose ∂FΓ(1) into the sets C1, . . . , C4,as defined in the proof of Exercise 2.16. Hence the left hand side of (6.52) can bewritten as ∫

∂FΓ(1)

H(z) dz =

∫C1

H(z) dz +

∫C2

H(z) dz

+

∫C3

H(z) dz +

∫C4

H(z) dz.

(6.53)

Using the set-identities S C1 = C2 and T C3 = C4 from Exercise 2.16 and noticingthat the orientation changes under the Mobius transformation, we conclude∫

C2

H(z) dz = −∫SC1

H(z) dz

= −∫C1

H(S−1z

)d(S−1z)

= −∫C1

H∣∣2S−1(z) dz

= −∫C1

H(z) dz

using the transformation property of the integrand. Similarly, we can show∫C4

H(z) dz = −∫C3

H∣∣2S−1(z) dz.

1To do!


This shows that the right hand side of (6.53) vanishes and this concludes the proof.

Proposition 6.55. The map µ : Mk

(Γ(1)

)⊕Sk

(Γ(1)

)−→ H1

(Γ(1),C[X]k−2

)defined in (6.46) is injective.

We follow [55, §4] for the proof of this proposition.

Proof. Since µ is a linear map, it suffices to show that the kernel of µ istrivial. Assume that (f, g) ∈Mk

(Γ(1)

)⊕ Sk

(Γ(1)

)satisfies

µ(f, g) = 0

in H1(Γ(1),C[X]k−2

). This means that the cocycle image PM + PM induced by µ

must be a coboundary. There exists a P ∈ C[z]k−2 such that

PM + PM = P∣∣2−kM − P

for all M ∈ Γ(1). Using Definitions 6.11 and 6.30 of Eichler and Niebur integrals

respectively we see that the above condition implies that the function F + G− −Psatisfies the transformation law (3.3) with weight 2− k(

F + G− − P)∣∣

2−kM = F∣∣2−kM + G−

∣∣2−kM − P

∣∣2−kM

(6.8),(6.28)= F + PM + G− + PM − P

∣∣2−kM

= F + G− + PM + PM︸︷︷︸=P∣∣2−k

M−P

−P∣∣−kM

= F + G− − P

for all M ∈ Γ(1). (Note that F + G− − P is not a modular form in the sense

of Definition 3.26 since G− and hence the whole function is not a holomorphic(or meromorphic) function. This means that ∂

∂z G−(z) does not vanish, see Exer-

cise 6.33.)Recall that g ∈ Mk

(Γ(1)

)is holomorphic. With the help of Exercise 6.33 and

the product rule we find

∂

∂z

(g(z) G−(z)

)= g(z)

∂

∂zG−(z)

(6.31)= g(z)g(z)

(z − z

)k−2

= |g(z)|(z − z

)k−2.(6.54)

Also, the function g(z)(F (z)− P (z)

)is holomorphic. Hence, we have

(6.55)∂

∂z

(g(z)

(F (z)− P (z)

))= 0.

Recall that the fundamental region FΓ(1) is defined in (2.11). Its boundary isdenoted by ∂FΓ(1), oriented in the positive (anticlockwise) sense. Using (6.54) and(6.55) we have∫∫

FΓ(1)

|g(z)|(z − z

)k−2dxdy

(6.54)=

∫∫FΓ(1)

∂

∂z

(g(z) G−(z)

)dxdy

(6.55)=

∫∫FΓ(1)

∂

∂z

(g(z)

(F (z)− P (z)

))+

∂

∂z

(g(z) G−(z)

)dxdy


=

∫∫FΓ(1)

∂

∂z

(g(z)

(F (z) + G−(z)− P (z)

))dxdy.

Recall that g appearing as multiplicative factor in the integrant is a cusp form.Exercise 3.35 shows that g itself satisfies the growth condition of the integrantrequired in Lemma 6.53. The second multiplicative factor, composed of Eichlerrespectivly Niebur integrals of entire modular forms and a polynomial, grows atmost polynomial at the cusp ∞: there exists an m ∈ Z≥0 such that

F (z) + G−(z)− P (z) = O(zk)

as = (z)→∞.

Hence, Lemma 6.53 applies to the integrant g(z)

(F (z) + G−(z)− P (z)

). We find∫∫

FΓ(1)

|g(z)|(z − z

)k−2dxdy =

∫∫FΓ(1)

∂

∂z

(g(z)

(F (z) + G−(z)− P (z)

))dxdy

(6.51)= − i

2

∫∂FΓ(1)

g(z)

(F (z) + G−(z)− P (z)

))dz.(6.56)

Next, we use the modular transformation property (3.3) for g with weight k

and F + G− − P with weight 2− k. Hence

g

(F + G− − P

))∣∣∣∣2

M =

(g∣∣kM

) ((F + G− − P

))∣∣∣∣2

M

)(3.3)= g

(F + G− − P

))holds for all M ∈ Γ(1). This shows that the integrand in (6.56) satisfies the trans-formation law (3.3) with weight 2 (and trivial multiplier). Applying Lemma 6.54to (6.56) we find∫∫

FΓ(1)

|g(z)|(z − z

)k−2dxdy

(6.56)= − i

2

∫∂FΓ(1)

g(z)(F (z) + G−(z)− P (z)

)dz

(6.52)= 0.

This can only happen if g vanishes.To complete the proof, it suffices to show that f = 0 assuming g = 0. Since

g = 0, we conclude G− = 0, see (6.27). Thus the weight 2 − k transformation

property of F + G− − P can be written as(F − P

)∣∣2−k = F − P.

Since f is an entire form we know that F is an entire function satisfying the modulartransformation property for weight 2 − k ≤ 0. Hence F − P ∈ M2−k

(Γ(1)

)and,

using Corollary 3.37 for negative 2 − k < 0 and §3.1 for 2 − k = 0, F − P is aconstant function. Using (6.13) and recalling that P (z) ∈ C[z]k−2, we concludethat f indeed vanishes.

Summarizing, we just showed that µ(f, g) = 0 implies f = g = 0. Hence themap µ is injective.

6.4.2. Surjectivity of the map µ. We now calculate the dimension of thecocycle space H1

(Γ(1),C[X]k−2

).

Lemma 6.56. Consider the operator

C[X]1 → C[X]1; Q 7→ Q∣∣−1TS.


acting on polynomials of degree at most 1 and put ρ = 1+√

3i2 . This operator has

two eigenspaces spanned by the polynomials

ρX + 1 and ρX + 1

with eigenvalues ρ and ρ respectively.

Proof. Since TS(2.8)=

(0 −11 0

)(1 10 1

)=

(0 −11 1

), we have

Q∣∣1

(TS)(X) = (X + 1)

(−ρ

X + 1+ 1

)= X + 1− ρ =

1

ρ

(ρX + (1− ρ)ρ

)for Q(X) = ρX + 1 ∈ C[X]1 with ρ 6= 0. Putting ρ := 1+

√3i

2 , we easily calculate

|ρ| = 1, r = 1r and (1 − r)r = 1. Hence the above calculation shows that Q(x) =

ρX + 1 is an eigenpolynomial of degree 1 of the operator with eigenvalue r = 1r :

Q∣∣1

(TS)(X) = r Q(X).

The analogous calculation shows that the polynomial ρX + 1 is also an eigenpoly-nomial with eigenvalue ρ of the operator. We just have seen that C[X]1 is spannedby two eigenspaces generated by ρX + 1 and ρX + 1 with eigenvalues ρ and ρrespectively.

Lemma 6.57 ([61, p.2, Lemma 2]). For k ∈ 2N, we define the subspaces E,F ⊂C[X]k−2 by

E =P ∈ C[X]k−2; P + P

∣∣2−kT = 0

and

F =P ∈ C[X]k−2; P + P

∣∣2−kTS + P

∣∣2−k(TS)2 = 0

.

(6.57)

The following statements hold

(1) For k ≥ 4, we have E +F = C[X]k−2 such that dim(E +F ) = k− 1, andfor k = 2, we have E + F = C[X]0 = C such that dim(E + F ) = 0.

(2) dimE =

k2 if k−2

2 is odd andk2 − 1 if k−2

2 is even,

(3) dimF = k − 1−

1 if k = 2,[k−2

3

]+ 1 if 4 ≤ k ≡ 1, 2 (mod 3) and[

k−23

]if 4 ≤ k ≡ 0 (mod 3).

Here [·] denote the Gauß brackets given in Definition 4.35.

Proof. Recall the matrix identities

(1− T )(1 + T ) = 1− (−1) and (1− TS)(1 + TS + (TS)2

)= 1− (−1).

Applying 1− (1) with the slash operator to any P ∈ C[X]k−2 gives

P∣∣2−k

(1− (−1)

)(X)

(5.2)=

(P∣∣2−k1

)(X)−

(P∣∣2−k(−1)

)(X)

(5.1)= P (X)− (−1)k−2P (X)

= 0

since k is an even integer. Hence, the above matrix identities imply

P∣∣2−k(1− T )(1 + T ) = 0 and P

∣∣2−k(1− TS)

(1 + TS + (TS)2

)= 0

for any P ∈ C[X]k−2. Using (5.3), we rewrite this to(P∣∣2−k(1−T )

)∣∣2−k(1 +T ) = 0 and

(P∣∣2−k(1−TS)

)∣∣2−k

(1 +TS+ (TS)2

)= 0


for any P ∈ C[X]k−2. Hence the polynomials P∣∣2−k(1 − T ) and P

∣∣2−k(1 − TS)

satisfy

P∣∣2−k(1− T ) ∈ E and P

∣∣2−k(1− TS) ∈ F.

Even more, we get all elements of the spaces E and F respectively in the same way.Summarizing, we have shown the alternative representation of the spaces E and Fas(6.58)E =

P∣∣2−k(1− T ); P ∈ C[X]k−2

and F =

P∣∣2−k(1− TS); P ∈ C[X]k−2

.

(1) Using (6.58), we can express any element of E as P∣∣2−k(1 − T ) and any

element of F as Q∣∣2−k(1− TS) for P,Q ∈ C[X]k−2. Then we have(

P∣∣2−k(1− T ) +Q

∣∣2−k(1− TS)

)∣∣∣∣2−k

T

= P∣∣2−k

(T − (−1)

)+Q

∣∣2−k(T − TST )

= P∣∣2−k

(T − (−1)

)+(Q∣∣2−kTS

)∣∣2−k

((−1)S − T

).

Since P∣∣2−k(−1) = P , we have(

P∣∣2−k(1−T ) +Q

∣∣2−k(1−TS)

)∣∣∣∣2−k

T = P∣∣2−k(T −1) +

(Q∣∣2−kTS

)∣∣2−k(S−T ).

This implies the following space inclusion

C[X]k−2 ⊃ E + F ⊃P∣∣2−k(S − 1); P ∈ C[X]k−2

.

Since the only polynomials invariant under the translation S are the con-stant functions, we see that

P∣∣2−k(S − 1); P ∈ C[X]k−2

has codi-

mension 1 in C[X]k−2. This shows that either E + F = C[X]k−2 orE + F =

P∣∣2−k(S − 1); P ∈ C[X]k−2

.

Let us calculate the spaceP∣∣2−k(S − 1); P ∈ C[X]k−2

.

For k ≥ 4 and starting from any polynomial

P (X) = aXk−2 +

k−3∑l=0

blXl ∈ C[X]k−2.

We have

P∣∣2−k(S − 1) = P (X + 1)− P (X)

= a (X + 1)k−2 − aXk−2 +

k−3∑l=0

bl((X + 1)l −X l

)= a

(k−3∑m=0

(k − 2

m

)Xm

)+

k−3∑l=0

bl((X + 1)l −X l

)∈ C[X]k−3.

This shows that P∣∣2−k(S − 1) ∈ C[X]k−3 does not contain any term

of order Xk−2. Since we already have shown that the space above hascodimension 1 in C[X]k−2, we conclude that

P∣∣2−k(S − 1); P ∈ C[X]k−2

= C[X]k−3.


The case k = 2 is easy. Since C[X]k−2 = C is the space of constant polyno-mials which is translation invariant by construction, we see P

∣∣2−k(S−1) =

0 for any P (X) ∈ C[X]0 = C. HenceP∣∣2−k(S − 1); P ∈ C[X]k−2

= 0

is the set containing only the zero function. Summarizing, we haveP∣∣2−k(S − 1); P ∈ C[X]k−2

=

0 if k = 2 and

C[X]k−3 if 4 ≤ k ∈ 2N.

Now consider the space E again. Since Q(X) := Xk−2 − 1 satisfies

Q∣∣2−k

(1 + T

)(X) =

(Xk−2 − 1

)+Xk−2

(1

Xk−2− 1

)= 0

we see thatXk−2 − 1 ∈ E.

Hence the space E and therefore also E+F contains polynomials of degreek − 2. Therefore

E + F = C[X]k−2

since we just showed that E+F is strictly larger thanP∣∣2−k(S−1); P ∈

C[X]k−2

. This concludes the first part of the lemma since dimC[X]k−2 =

k − 1.(2) Assume that a polynomial

Q(X) :=

k−2∑l=0

alXl ∈ C[X]k−2

satisfiesQ∣∣k−2

(1 + T ) = 0

which is the defining property of the space E in (6.57). This means thatQ satisfies

0 =

k−2∑l=0

alXl +

k−2∑l=0

(−1)lalXk−2−l =

k−2∑l=0

(al + (−1)k−2−lak−2−l

)X l.

Comparing the coefficients of the monomial factors X l shows that theysatisfy the following condition

al + (−1)k−2−lak−2−l = 0.

This means that E has the dimension stated in the lemma.(3) First notice that the set identity

F(6.58)

=P∣∣k−2

(1− TS); P ∈ C[X]k−2

= C[X]k−2 r

P ∈ C[X]k−2; P

∣∣k−2

(1− TS) = 0

holds. Hence,

dimF = k − 1− dimP ∈ C[X]k−2; P

∣∣k−2

(1− TS) = 0

and we only need to calculate the dimension of the right hand space.For k = 2, recall that constant functions are invariant under Γ(1), seeProposition 3.41. Hence,

dimP ∈ C[X]0; P

∣∣k−2

(1− TS) = 0

= dimC[X]0 = 1

and hencedimF = 2− 1− 1 = 0.


To calculate the dimension of the spaceP ∈ C[X]0; P

∣∣k−2

(1−TS) =

0

for 4 ≤ k ∈ 2N, we consider first the operator

C[X]1 → C[X]1; Q 7→ Q∣∣−1TS.

Lemma 6.56 shows that C[X]1 can be decomposed into two linearly inde-pendent eigenspaces of the operator spanned by ρX + 1 and ρX + 1 witheigenvalues ρ and ρ respectively. Hence the operator

C[X]k−2 → C[X]k−2; Q 7→ Q∣∣2−kTS

viewed as product operator admits an eigenspace decomposition

C[X]k−2 =

k−2⊕l=0

C(ρX + 1

)l (ρX + 1

)k−2−l

with eigenvalues ρl ρk−2−l, respectively. This allows us to consider thesolution space of

P∣∣2−k

(1− TS

).

It consists of the eigenspaces of the operator Q 7→ Q∣∣2−kTS discussed

above provided that the eigenvalues satisfy

ρl ρk−2−l = 1 for suitable 0 ≤ l ≤ k − 2.

Since ρ = 1+√

3i2 we have ρ = ρ5. This implies

ρl ρ5(k−2−l) = 1

for suitable 0 ≤ l ≤ k − 2. On the level of the index l, this means that0 ≤ l ≤ k − 2 has to satisfy

l + 5(k − 2)− 5l ≡ 0 mod 6 ⇐⇒ 4l ≡ 5(k − 2) mod 6 ⇐⇒ l ≡ k − 2 mod 3.

Counting solutions of above equality, we find

]

0 ≤ l ≤ k − 2; l ≡ k − 2 mod 3

=

[k−2

3

]+ 1 if k − 2 = 0, 2 (mod 3) and[

k−23

]if k − 2 = 1 (mod 3).

Summarizing, we have shown

dimP ∈ C[X]k−2; P

∣∣k−2

(1− TS) = 0

=

[k−2

3

]+ 1 if k − 2 = 0, 2 (mod 3) and[

k−23

]if k − 2 = 1 (mod 3).

and hence

dimF = k − 1− dimP ∈ C[X]k−2; P

∣∣k−2

(1− TS) = 0

= k − 1−

[k−2

3

]+ 1 if k − 2 = 0, 2 (mod 3) and[

k−23

]if k − 2 = 1 (mod 3).

This proves the third statement of the lemma.

Proposition 6.58. The map µ : Mk

(Γ(1)

)⊕ Sk

(Γ(1)

)→ H1

(Γ(1),C[X]k−2

)defined in (6.46) is surjective.


Proof. Following the arguments in [48, §4] to prove the surjectivity of thelinear map µ, we compare the dimensions of the spaces Mk

(Γ(1)

)⊕ Sk

(Γ(1)

)and

H1(Γ(1),C[X]k−2

). In particular we will show

dimMk

(Γ(1)

)+ dimSk

(Γ(1)

)= dimH1

(Γ(1),C[X]k−2

).

The left hand side is easily calculated by Theorem 4.37, Lemma 4.27, Proposi-tion 4.28 and Corollary 4.34. We have

dimMk

(Γ(1)

) (4.31)=

[k12


k12

]+ 1 for k 6≡ 2 mod 12

and (with Lemmata 4.27 and 4.28)

dimSk(Γ(1)

)=

0 for 2 ≤ k ≤ 10,[k12

]− 1 for 12 ≤ k ≡ 2 mod 12 and[

k12

]for 12 ≤ k 6≡ 2 mod 12.

Combining both results we get

dim

(Mk

(Γ(1)

)⊕ Sk

(Γ(1)

))= dimMk

(Γ(1)

)+ dimSk

(Γ(1)

)=

0 for k = 2,

2[k12

]− 1 for 4 ≤ k ≡ 2 mod 12 and

2[k12

]+ 1 for 4 ≤ k 6≡ 2 mod 12.

(6.59)

To calculate the dimension of H1(Γ(1),C[X]k−2

), we put D3 equal to the

dimension of the space of cocycles and D3 equal to the space of coboundaries

D3 := dimZ1(Γ(1),C[X]k−2

)and D4 := dimB1

(Γ(1),C[X]k−2

).

Definition 6.35 implies that the space of coboundaries is given by taking any poly-nomial P ∈ C[X]k−2 and form the coboundary

(P∣∣2−kV − P

)V ∈Γ(1)

. For such a

coboundary to vanish means that P satisfies (3.3) for all V ∈ Γ(1). In other words,we have P ∈M !

2−k(Γ(1)

). The toy example discussed in §3.1 and Proposition 3.41

respectively implies

dimM !2−k(Γ(1)

)=

1 for k = 2 and

0 for k > 2.

Hence, we have

D4 = dimB1(Γ(1),C[X]k−2

)= dimC[X]k−2 − dimM !

2−k(Γ(1)

)= k − 1−

1 for k = 2 and

0 for k > 2

=

0 for k = 2 and

k − 1 for k > 2.(6.60)

To calculate D3, we first observe the following: (6.37) implies that each cocycle(PM)M∈Γ(1)

is uniquely determined by its values on the generators S and T of

the full modular group Γ(1). Hence, we only need to assign polynomials PS , PT ∈C[X]k−2 in such a way, that the associated cocycle satisfies the relations related toT 2 = −1 = (TS)3 (see Exercise 2.5):

0(6.38)

= P−1 = PT 2

(6.37)= PT

∣∣2−kT + PT = PT

∣∣2−k

(T + 1)


and

0(6.38)

= P−1 = P(TS)3

(6.37)= PT

∣∣2−kS(TS)2 + PS(TS)2

(6.37)= . . .

(6.37)= PT

∣∣2−kS(TS)2 + PS

∣∣2−k(TS)2 + PT

∣∣2−kS(TS) + PS

∣∣2−k(TS) + PT

∣∣2−kS + PS

= PT∣∣2−kS

((TS)2 + TS + 1

)+ PS

∣∣2−k

((TS)2 + TS + 1

)=(PT∣∣2−kS + PS

)∣∣2−k

((TS)2 + TS + 1

).

By (6.57), this implies PT ∈ E and PT∣∣2−kS + PS ∈ F . In particular, we see that

the space D3 is isomorphic to the direct sum of the spaces E⊕F and the dimensionof D3 can be calculated by

dimD3 = dimE + dimF.

The terms appearing on the right hand side are already calculated in Lemma 6.57.We have

dimD3 = dimE + dimF

=

k2 if k−2

2 is odd andk2 − 1 if k−2

2 is even,

+ k − 1−

1 if k = 2,[k−2

3

]+ 1 if 4 ≤ k ≡ 1, 2 (mod 3) and[

k−23

]if 4 ≤ k ≡ 0 (mod 3).

= k − 2 +

0 if k = 2,

2[k12

]if 4 ≤ k ≡ 2 (mod 1)2 and

2[k12

]+ 2 if 4 ≤ k 6≡ 2 (mod 1)2.

(6.61)

Summarizing, we see that the dimension of H1(Γ(1),C[X]k−2

)is given by

dimH1(Γ(1),C[X]k−2

)= dimZ1

(Γ(1),C[X]k−2

)− dimB1

(Γ(1),C[X]k−2

)= dimD3 − dimD4

(6.60),(6.61)=

0 if k = 2,

2[k12

]− 1 if 4 ≤ k ≡ 2 (mod 1)2 and

2[k12

]+ 1 if 4 ≤ k 6≡ 2 (mod 1)2.

The dimension agrees with dim

(Mk

(Γ(1)

)⊕ Sk

(Γ(1)

))calculated in (6.59). We

have

dimH1(Γ(1),C[X]k−2

)= dim

(Mk

(Γ(1)

)⊕ Sk

(Γ(1)

)).

Hence the map µ defined in (6.46) is surjective.

6.5. Hecke Operator And Period Polynomials

In §6.5, we follow the approach presented in [24] which is based on matrixidentities of the formal polynomial ring of matrices Z

[Γ(1)

]modulo the relations

imposed by the period polynomials.Recall the sets R? and Rn, n ∈ N, as introduced in Definition 5.3. Note that

R? is a non-commutative ring with unit and is ”multiplicatively graded“ in thesense that RmRn ⊂ Rmn. In particular, Rm is a left and right module over thegroup ringR1 = Z

[Γ(1)

]. Also, we recall the slash operator introduced in (5.1) such

that for each k ∈ 2Z,∑l αlMl ∈ Rn and f : H → C, we have f

∣∣k

(∑l αlMl

)(z) =∑

l αl j(Ml, z)k f(Ml z) for z ∈ H.

6.5. HECKE OPERATOR AND PERIOD POLYNOMIALS 203

Recall the Hecke operator Tn, n ∈ N, as given in Definition 5.11. The actionof Tn on a function f : H → C can be described by the action of the element∑d|n∑d−1b=0

(nd b0 d

)∈ Rn:

Tn f(5.14)

= nk−1∑d|n

d−1∑b=0

f

∣∣∣∣k

(nd b0 d

).

We identify the Hecke operator Tn with its matrix representation. Hence, we denotethe matrix representation also with Tn:

(6.62) Tn :=∑d|n

d−1∑b=0

(nd b0 d

)=∑ad=n

0≤b<d

(a b0 d

)=

∑A∈T (n)

A ∈ Rn

where the set T (n) is defined in (5.26). The operator Tn is obtained by

Tnf = nk−1 f∣∣kTn.

However, some elements of Rn act trivially on functions using the slash op-

erator. For example let M =

(a bc d

)be a matrix in GL2 (Z) with determinant

n. Obviously the element (−M) =

(−a −b−c −d

), which we calculate from M by

exchanging each matrix entry with its negative value, is also an element in GL2 (Z)with determinant n. The formal difference M − (−M) ∈ Rn lies in Rn and theaction of this element on functions gives

f∣∣k

(M − (−M)

)(z) =

(f∣∣kM)(z)−

(f∣∣k(−M)

)(z)

= (cz + d)−k f

(az + b

cz + d

)− (−cz − d)−k f

(−az − b−cz − d

)= 0

since k ∈ 2Z. This shows that some elements of Rn lie in the kernel of its action onfunctions by the slash operator. To work around this issue, we make the followingassumption for this section: Within §6.5 we write identities of elements in anyRn up to trivial actions by the slash operator. To mark this identification we write.= instead of = for equality. This means for example that we identify (−1) with1, and M − (−M) with 0:

(−1).= 1 and M − (−M)

.= 0.

In [69], Manin gave the representation(6.63)

T ∗n =∑

ad−bc=na>c>0d>−b>0

[(a bc d

)+

(a −b−c d

)]+

∑ad=n

− 12d<b≤

12d

(a b0 d

)+

∑ad=n

− 12a<c≤

12a

c6=0

(a 0c d

)

for the Hecke operators in Rm acting on period polynomials in obvious sense. ForP a period polynomial in the sense of Definition 6.7, the element Tn ∈ Rn acts onP by

P 7→ nk−1 P∣∣2−kT

∗n .

Choie and Zagier presented in [24] a simple proof that T ∗n indeed representsthe Hecke operator of level n on period polynomials. To formulate this criterionconsider the matrices

(2.8) S =

(1 10 1

)and T =

(0 −11 0

)


and the right ideal J of R1 with

(6.64) J = (1 + T )R1 +(1 + TS + (TS)2

)R1.

Choie and Zagier proved

Theorem 6.59 ([24, Theorem 2]). For each integer n ≥ 1, the matrix repre-sentation Tn of the nth Hecke operator fulfills the relations

(6.65) Tn(S − 1).= 0, Tn(T − 1)

.= (T − 1)Tn (mod (S − 1)Rn)

for a certain element Tm ∈ Rm, which is unique modulo JRn and satisfies TnJ ⊂JRn. The elements Tm, Tn ∈ R∗ satisfy the product formula

(6.66) Tm Tn =∑

d|gcd(n,m)

d−1

(d 00 d

)Tnmd2

(mod JRnm).

Theorem 6.60 ([24, Theorem 3]). The set T ∗n satisfies the conditions put on

Tn in Theorem 6.59.

Corollary 6.61. Let k ∈ 2N. If P ∈ Pk−2 is a period polynomial in the senseof Definition 6.7 and n ∈ N, then the element T ∗n ∈ Rn given in (6.63) induces an

operator on period polynomials denoted also by Tn given by

(6.67) TnP := P∣∣2−kT

∗n .

In particular TnP is a period polynomial in the sense of Definition 6.7. Moreoverthe operators satisfy

(6.68) TmTnP = TnTmP =∑

d|gcd(m,n)

d2k−1 T ∗mnd2P

for all n,m ∈ N.

Proof. Let P be a period polynomial in the sense of Definition 6.7 and n ∈N. Using (6.65) of Theorem 6.59, we see that there exists a Tn ∈ Rn such that

P∣∣2−kTn corresponds to taking the Hecke operator Tn on the corresponding cusp

form. Theorem 6.60 shows that such an Tn ∈ Rn is realized by T ∗n .

The remainder of this section is used to prove Theorems 6.59 and 6.60 wherewe follow the arguments and reasonings presented in [24, §3] closely.

Before we explain Choi and Zagier’s argument, we need some preliminary re-sults. First, what we need is a generalization of the identities (1.12) and (6.2).

Exercise 6.62. For A ∈ Mat2 (Z) with detA 6= 0 and k ∈ 2Z, prove theidentities

d(Az) = det(A)dz

j(A, z)2(6.69)

and (Aw −Az

)k= det(A)k

(w − z)k

j(A,w)k j(A, z)k(6.70)

for all z, w ∈ H.

Let f ∈ Sk(Γ(1)

). We have seen (e.g. in Example 6.47 and Lemma 6.48) that a

period polynomial P can be written in terms of the Eichler integral F (see (6.16))of the cusp form f :

P = F∣∣2−k(T − 1) and 0 = F

∣∣2−k(S − 1).


Denoting the matrix representation inRn of the Hecke operator Tn of level n definedin (5.14) also by Tn, we have

(6.71) Tn =∑ad=n

0≤b<d

(a b0 d

).

Using this realization of the Hecke operator on cusp forms, we have for A =

(a b0 d

)with ad = n and 0 ≤ b < d:

F∣∣2−kA(z) = j(A, z)k−2 F (Az)

(6.16)= j(A, z)k−2

∫ i∞

Az

f(w)(w −Az

)k−2dw

w 7→Aw= j(A, z)k−2

∫ A−1i∞

z

f(Aw) (Aw −Az)k−2 d(Aw)

(6.69)(6.70)

= det(A)k−1

∫ A−1i∞

z

j(A,w)−kf(Aw) (w − z)k−2 dw

= det(A)k−1

∫ A−1i∞

z

(f∣∣kA)(w) (w − z)k−2 dw

A−1i∞=i∞=

∫ i∞

z

det(A)k−1(f∣∣kA)(w) (w − z)k−2 dw.

The last integral on the right hand side is the Eichler integral of the cusp formf∣∣kA. This shows that the action of the Hecke operator Tn on a cusp form f and

the action can be lifted to an action of the associated Eichler Integral F . The samearguments hold for the variant G− of the Niebur integral, see (6.32). We just haveproven the following

Lemma 6.63. For k ∈ 2N and f, g ∈ Sk(Γ(1)

)denote by F the associated

Eichler integral given in (6.16). For n ∈ N taken as the level, we have

f 7→ Tnf = nk−1 f∣∣kTn ⇐⇒ F 7→ F

∣∣2−kTn

andg 7→ Tnf = nk−1 g

∣∣kTn ⇐⇒ G− 7→ G−

∣∣2−kTn.

In other words, Lemma 6.63 shows that applying the Hecke operator of level non a cusp form is the same as applying the map F 7→ F

∣∣2−kTn on the Eichler integral

F and the corresponding expression for the Niebur integral. Since this Eichler andNiebur integrals also satisfies F

∣∣2−kS − 1) = 0, we get the following algebraic

description: Suppose that there are elements Xn, Tn and Yn in Rn satisfying

Tn (S − 1).= (S − 1)Xn and Tn (T − 1)

.= (T − 1) Tn + (S − 1)Yn.

Then, these elements satisfy(F∣∣2−kTn

)∣∣2−k(S − 1) =

(F∣∣2−k(S − 1)

)∣∣2−kXn

= 0∣∣2−kXn = 0

and(F∣∣2−kTn

)∣∣2−k(T − 1) =

(F∣∣2−k(T − 1)

)∣∣2−kTn +

(F∣∣2−k(S − 1)

)∣∣2−kYn

=(F∣∣2−k(T − 1)

)∣∣2−kTn + 0

= P∣∣2−kTn.


This shows that F∣∣2−kTn and its corresponding Niebur integral is again a mod-

ular integral with associated period polynomial∣∣2−kTn.

Lemma 6.64. We have

(6.72) M − 1 ∈(T − 1

)R1 +

(S − 1

)R1

for any M ∈ Γ(1).

Proof. Assume that there exists M ∈ Γ(1) satisfying (6.72). We could takeM = 1 for example, since 1 − 1 = 0 is the neutral element of R1. Then, considerthe elements

TM − 1 =(T − 1

)M +

(M − 1

),

SM − 1 =(S − 1

)M +

(M − 1

),

T−1M − 1 =(T − 1

)(− T−1M

)+(M − 1

)and

S−1M − 1 =(S − 1

)(− S−1M

)+(M − 1

).

This shows that the elements TM −1, TM −1, TM −1 and TM −1 are elementsin(T −1

)R1 +

(S−1

)R1. Using the fact that S and T generate Γ(1), we conclude

that all elements M ∈ Γ(1) satisfy (6.72).

Lemma 6.65. The Tn defined in (6.62) satisfies Tn(T −1

)∈(T −1

)Rn+

(S−

1)Rn.

Proof. Consider an element of the formal sum of matrices Tn. Each element

has the form A :=

(a b0 d

)with ad = n and 0 ≤ b < d. Consider

AT =

(a b0 d

)T =

(b −ad 0

)∈ H(n)

with H(n) defined in (5.22). Theorem 5.23 shows that AT can be written asAT = MAA

′ with some unique A′ ∈ T (n) and MA ∈ Γ(1), both depending on A.Hence Tn T can be written as

Tn T =∑

A∈cT(n)

AT =∑

A∈cT(n)

MAA′.

This shows that we can write Tn(T − 1

)as

Tn(T − 1

)=

∑A∈cT(n)

MAA′ −

∑A∈cT(n)

A =∑

A∈cT(n)

(MA − 1

)A′

since A′ runs through all representatives of T (n) if A does. Using Lemma 6.64, wefind

Tn(T − 1

)=

∑A∈cT(n)

(MA − 1

)A′ ∈

(T − 1

)Rn +

(S − 1

)Rn.

Suppose that V is an abelian group on which Γ(1) acts. Then V is a leftR1-module. For X ∈ R1 write

Ker(X) := v ∈ V ; Xv = 0and

Im(X) := Xv; v ∈ V for kernel and image of the map

V 7→ V ; v 7→ Xv.

We have the following


Definition 6.66. We call V acyclic if

• Ker(1 + T ) ∩Ker(1 + TS + (TS)2

) .= 0 (i.e., acts trivially by the slash

operator) and• Ker(1− T ) = Image(1 + T ) and Ker(1− TS) = Image

(1 + TS + (TS)2

)holds.

Remark 6.67. (1) The first condition that Ker(1 + T ) ∩ Ker(1 + TS +

(TS)2)

acts trivially by the slash operator, is required since we discussΓ(1) = SL2 (Z). Choie and Zagier discuss the quotient group PSL2 (Z)which divides out the trivial actions of Γ(1) = SL2 (Z) by Mobius trans-formations. This means that in PSL2 (Z) the elements(

a bc d

)and

(−a −b−c −d

)are identified and treated as one element.

(2) The second defining property of acyclicity is listed for the sake of com-pleteness since it is mentioned in [24, p.13]. We do not necessarily needthis extra assumption if we work over Q since

(1 − T

)v = 0 implies

v =(1 + T

)v2 and

(1− TS

)v = 0 implies v =

(1 + TS + (TS)2

)v3 . This

was already used in §6.4.2.

Example 6.68. For example the space of period functions Pk−2 is given by theintersection of

Pk−2 = Ker(1 + T ) ∩Ker(1 + TS + (TS)2

)with V = C[z]k−2, see Definition 6.7.

Choie and Zagier prove the following lemma using rational period functions.Since we did not introduce these functions, we just state this lemma as fact withoutproof.

Lemma 6.69 ([24, Lemma 2]). Rn is an acyclic R1-module for all n ∈ N.

Using the acyclicity of Rn, we are able to show

Lemma 6.70 ([24, Lemma 3]). If V is an acyclic R1-module and v, v′ ∈ V withv.= v′ then we have

(6.73)(1− T

)v ∈

(1− S

)V ⇐⇒ v′ ∈

(1 + T

)V +

(1 + TS + (TS)2

)V = J V

where the right ideal J of R1 is defined in (6.64).

Proof. We consider the two directions separately.

=⇒: Consider v ∈ J V . Then the form of J given in (6.64) implies thatthere exists u,w ∈ V such that v can be written as

v.= v′ =

(1 + T

)u+

(1 + TS + (TS)2

)w

Applying(1− T

)from the left gives(

1− T)v.=(1− T

)(1 + T

)u+

(1− T

)(1 + TS + (TS)2

)w

=(1− T + T − (−1)︸︷︷︸

=1−(−1)which acts trivially

)u+

(1 + TS + (TS)2 − T − T 2S − T 2STS

)w

.=(1 + TS + (TS)2 − T − T 2S − T 2STS

)w

=(1− S

)(1 + TS + (TS)2

)w +

+(S + STS + S(TS)2 − T − T 2S − T 2STS

)w


using the identities

T 2 = (−1) ⇐⇒ (−T ) = T−1 and (TS)3 = (−1).

(Recall that the − sign in front of the matrix elements is not a “minus”in the sense of Rn; it denotes the scalar multiplication of the matrix by

−1. For example we have (−1) =

(−1 00 −0

)and (−T ) =

(0 1−1 1

)0).

Calculating the right hand term we find

S + STS + S(TS)2 − T − T 2S − T 2STS

=

(1 10 1

)+

(1 01 1

)+

(0 −11 0

)−(

0 −11 0

)−(−1 −10 −1

)−(−1 0−1 −1

)= S + STS + T − T − (−S)− (−STS).= 0.

(6.74)

Hence,(1− T

)v can be written as

(1− T

)v.=(1− S

)(1 + TS + (TS)2

)w +

+(S + STS + S(TS)2 − T − T 2S − T 2STS

)w

.=(1− S

)(1 + TS + (TS)2

)w.

This shows that v satisfies(1− T

)v ∈

(1− S

)V.

⇐=: Assume that V is acyclic and v satisfies(1− T

)v =

(1− S

)w

for some w ∈ V . Hence, we have the identity(1− T

)(v − w) =

(1− T

)v −

(1− T

)w

=(1− S

)w −

(1− T

)w

=(T − S

)w.

Multiplying both sides with T from the left gives(T − 1

)(v − w)

.=(T − T 2

)(v − w)

=(T 2 − TS

)w

.=(1− TS

)w.

Hence, we have

X :=(1− T

)(v − w) = −

(1− TS

)w.

Since(1− T

)(v−w) ∈ Image

(1− T

)and −

(1− TS

)winImage

(1− TS

),

we see X lies in

X ∈ Image(1− T

)∩ Image

(1− TS

)= Ker

(1− T

)∩Ker

(1− TS

).=

0,

using the second and successively the first property of Definition 6.66.Hence X =

(1−T

)(v−w)

.= 0 which implies that the element v−w is in


the kernel of 1− T and similarly, −w in the kernel of 1− TS. This givesthe relation

v − w ∈ Image(1 + T

)and − w ∈ Image

(1 + TS + (TS)2

).

Hence v satisfies

v = v − w + w

∈ Image(1 + T

)∩ Image

(1 + TS + (TS)2

)=(1 + T

)V +

(1 + TS + (TS)2

)V = J V.

Next, we proceed with the proof of Theorem 6.59.

Proof of Theorem 6.59. Lemmas 6.69 and 6.70 show the following charac-terization of JRn: We have

JRn.=X ∈ Rn; (1− T )X ∈ (1− S)Rn

.

The conditions in (6.65) on Tn ∈ Rn imply

(1− T ) Tn(1 + T ).=

(Tn (1− T ) + (1− S)w

)(1 + T )

= Tn (1− T )(1 + T ) + (1− S)w(1 + T )

= Tn(1− (−1)︸︷︷︸

.=0

)+ (1− S)w(1 + T )

= (1− S)w(1 + T ).= 0 (mod

(S − 1

)Rn)

for some w ∈ Rn. Similarly, we conclude

(1− T ) Tn(1 + TS + (TS)2

).=

(Tn (1− T ) + (1− S)w

)(1 + TS + (TS)2

)= Tn (1− T )

(1 + TS + (TS)2

)+ (1− S)w

(1 + TS + (TS)2

)= Tn

(1 + TS + (TS)2 − T − T 2S − T 2STS

)+ (1− S)w

(1 + TS + (TS)2

).= Tn

(1− S

)(1 + TS + (TS)2

)+ (1− S)w

(1 + TS + (TS)2

).= 0 (mod

(S − 1

)Rn)

for some w ∈ Rn where we used the identity(1 + TS + (TS)2 − T − T 2S − T 2STS

)=(1− S

)(1 + TS + (TS)2

)+(S + STS + S(TS)2 − T − T 2S − T 2STS

)(6.67).

=(1− S

)(1 + TS + (TS)2

).

Property (6.66) follows from the following argument. We have

(T − 1)

Tm Tn − ∑d|gcd(n,m)

d−1

(d 00 d

)Tnmd2

.= Tm(T − 1)Tn −

∑d|gcd(n,m)

d−1

(d 00 d

)(T − 1)Tnm

d2


.= Tm

[Tn(T − 1)− (S − 1)Yn

]−

∑d|gcd(n,m)

d−1

(d 00 d

)Tnmd2

(T − 1)

.= (Tm Tn−) (T − 1)

∑d|gcd(n,m)

d−1

(d 00 d

)Tnmd2

(mod JRnm).

Since we have

Tm Tn −∑

d|gcd(n,m)

d−1

(d 00 d

)Tnmd2

.= 0 (mod (S − 1)Rn)

by (5.29), we conclude the proof of Theorem 6.59.

Continuing towards the proof of Theorem 6.60, consider the sets

An : =

(a bc d

)∈ H(n); a > c > 0, d > −b ≥ 0, and if b = 0 then a ≥ 2c

(6.75)

and

Bn : =

(a bc d

)∈ H(n); a > −c ≥ 0, d > b > 0, and if c = 0 then d ≥ 2b

.

(6.76)

Exercise 6.71. Show that the maps

α : A =

(a bc d

)7−→ TS−[ ac ]A =

(c d

−(a− c

[ac

])−b+ d

[ac

])and

β : B =

(a bc d

)7−→ S[ db ]T B =

(−c+ a

[db

]−(d− b

[db

])a b

)give inverse bijections between the sets An and Bn. Here, [·] denote the Gaußbrackets defined in (4.30).

Following the arguments in [24, §3.2] closely, we now proceed with the

Proof of Theorem 6.60. Recall T ∗n as defined in (6.63). We have to showthat T ∗n satisfies the right identity in (6.65):

Tn(T − 1).= (T − 1)T ∗n (mod (S − 1)Rn).

To shorten notation, we do not write (mod (S − 1)Rn) into each congruencerelation. We only indicate these by using the “

.=” symbol.

Since the congruence relation (mod (S−1)Rn) implies SrA.= A for any r ∈ Z

and A ∈ H(n), we have by Exercise 6.71∑A∈An

A.= T

∑B∈Bn

B

⇐⇒∑a>c>0d>−b>0ad−bc=n

[(a bc d

)− T

(a −b−c d

)].=

∑12d≥b>0ad=n

(0 −da b

)−

∑12a≥c>0ad=0

(a 0c d

).

Conjugating this identity with

(−1 00 1

)changes the sign of all the off-diagonal

coefficients and preserves the property “.=”. Adding this result to the original


identity, we get(6.77)

(1− T )∑a>c>0d>−b>0ad−bc=n

[(a bc d

)+

(a −b−c d

)].=

∑0<|b|≤ 1

2dad=n

[(0 −da b

)−(d 0b a

)].

(Recall that within §6.5, we write identities of elements in any Rn up to trivialactions by the slash operator). Identifying the sum on the left hand side as part ofT ∗n , see (6.63), we have

(1− T )T ∗n

.= (1− T )

∑

ad−bc=na>c>0d>−b>0

[(a bc d

)+

(a −b−c d

)]+

∑ad=n

− 12d<b≤

12d

(a b0 d

)+

∑ad=n

− 12a<c≤

12a

c 6=0

(a 0c d

)

.= (1− T )

∑ad−bc=na>c>0d>−b>0

[(a bc d

)+

(a −b−c d

)]

+ (1− T )

∑ad=n

− 12d<b≤

12d

(a b0 d

)+

∑ad=n

− 12a<c≤

12a

c6=0

(a 0c d

)(6.77).

=∑

0<|b|≤ 12d

ad=n

[(0 −da b

)−(d 0b a

)].+ (1− T )

∑ad=n

− 12d<b≤

12d

(a b0 d

)+

∑ad=n

− 12a<c≤

12a

c 6=0

(a 0c d

).=

∑− 1

2d<b≤12d

ad=n

(a b0 d

)(1− T ) +

∑a,d>0ad=nd even

[(0 −da − 1

2d

)−(

d 0− 1

2d a

)].

The first sum on the left hand side satisfies∑− 1

2d<b≤12d

ad=n

(a b0 d

)(1− T )

.= Tn(1− T )

and the second evaluates as∑a,d>0ad=nd even

[(0 −da − 1

2d

)−(

d 0− 1

2d a

)].=∑a,d>0ad=nd even

(S2 − 1

)( d 0− 1

2d a

)

.=(S − 1

) ∑a,d>0ad=nd even

(S + 1

)( d 0− 1

2d a

).= 0 (mod (S − 1)Rn),


recalling again that within §6.5, we write identities of elements in any Rn up to

trivial actions by the slash operator. This means in particular that matrices

(a bc d

)and

(−a −b−c −d

)are identified.

Remark 6.72. There exists an alternative approach using the integral repre-sentation (6.1) of period polynomials. This approach was originally used by Manin[69]. Later is was adapted by [85] to work for period functions of Maass cusp formstoo. The same author also extended Choie and Zagier’s algebraic approach in [84]to the case of period functions.

6.6. Roots Of Period Polynomials Associated To Hecke Eigenforms

As an application of the theory of period polynomials, we present a unimodular-ity result for roots of period polynomials associated to Hecke eigenforms followingclosely [34].

Definition 6.73. Let d ∈ N be a given natural number. A complex polynomial

P (z) =∑di=0 ci z

i ∈ C[z]d of degree d is said to be self-inversive if it satisfies

(6.78) P (z) = ε zd P

(1

z

)for some constant ε (necessarily of modulus 1), where P (z) :=

∑di=0 ci z

i and thebar above the ci denotes complex conjugation.

Remark 6.74. The class of self-reciprocal polynomials is the special case of(6.78) with ε = 1 and P = P .

Cohn [25] proved that a polynomial P (z) has all its zeros on the unit circle ifand only if it is self-inversive and its derivative P ′(z) has all its zeros in the closedunit disc |z| ≤ 1. Lalın and Smyth [59] extended Cohn’s result as follows.

Theorem 6.75 ([59, Theorem 1]). Let h(z) ∈ C[z]n be a nonzero complexpolynomial of degree n with all its zeros in |z| ≤ 1. Then, for d > n and any λ with|λ| = 1, the self-inversive polynomial

(6.79) P λ := zd−n h(z) + λ znh

(1

z

)has all its zeros on the unit circle.

In what follows, we relax the condition d > n to d ≥ n. We thus obtain thefollowing slightly stronger version.

Theorem 6.76. Let h(z) ∈ C[z]n be a nonzero polynomial of degree n with allits zeros in |z| ≤ 1. For d ≥ n and λ ∈ C set P λ as in (6.79). Then, for any|λ| = 1 for which P λ 6= 0, P λ has all its zeros on the unit circle.

Proof. Write h?(z) := zn h(

1z

), and temporarily assume that all n zeros of

h(z) are in the open disc |z| < 1. It follows easily that zd−n h(z) has all of its dzeros inside |z| < 1, whereas h?(z) has all of its n zeros in |z| > 1. For |z| = 1, wehave z = 1

z and thus

|h?(z)| =∣∣h(z

∣∣ =∣∣∣h(z)

∣∣∣ = |h(z)| =∣∣zdn h(z)|

∣∣ .Recall Rouche’s theorem, see Appendix A.10. If |λ| < 1, then we set g(z) =zd−n h(z) and f(z) = λh?(z) and we deduce with Rouche’s theorem that P λ(z)has all of its d zeros in |z| < 1. If |λ| > 1, then we can switch the choice of f and

6.6. ROOTS OF PERIOD POLYNOMIALS ASSOCIATED TO HECKE EIGENFORMS 213

g to deduce that P λ(z) has no zeros in |z| < 1, and hence all d of its zeros mustbe in |z| > 1. As the location of the zeros of P λ(z) are continuous functions ofλ, we see that, for |λ| = 1, P λ(z) must have all its zeros on the unit circle. Theresult under the weaker assumption that h has all its zeros in the closed unit disc|z| ≤ 1 follows from continuity of the zeros of P λ(z) as functions of the zeros ofh(z).

Lemma 6.77. Let k ∈ 2N and F ∈Mk

(Γ(1)

)a cuspidal Hecke eigenform. We

denote the period polynomial associated to F by the integral in (6.1) by PF (X) ∈C[X]k−2. Then the polynomial R = RF ∈ C[X]k−2 given by

(6.80) RF (X) :=

k−2∑l=0

L(k − l − 1)(2πX)l

l!= − (2πi)k−1

(k − 2)!PF (−iX)

is self-inversive and we have

(6.81) RF (X) = ikXk−2RF

(1

X

).

Proof. Consider the polynomial RF as defined in (6.80). We have

RF (X) : =

k−2∑l=0

L(k − l − 1)(2πX)l

l!

=(2πi)k−1

(k − 2)!

k−2∑l=0

(k − 2)!

l!

L(k − l − 1)

(2πi)k−l−1(−iX)l

(6.7)= − (2πi)k−1

(k − 2)!PF (−iX)

This shows the identity on the right hand in (6.80).Since PF is a period polynomial and hence satisfies PF

∣∣2−k(1 + T ) = 0, see

Lemma 6.6, we find

ikXk−2RF

(1

X

)(6.80)

= −ikXk−2 (2πi)k−1

(k − 2)!PF

(1

iX

)= ikXk−2 (2πi)k−1

(k − 2)!j

(T,

1

iX

)k−2

PF (−iX) using PF = −PF∣∣2−k(T )

= ikXk−2 (2πi)k−1

(k − 2)!(iX)2−k PF (−iX)

= i2(2πi)k−1

(k − 2)!PF (−iX)

(6.80)= RF (X).

Hence RF is self-inversive with constant ik, if the coefficients of RF are real.Recall that we assumed F to be a normalized simultaneous eigenfunction of

all Hecke operators. Corollary 5.55 implies that the Hecke eigenvalues are real inthis case. Then, applying Theorem 5.45 shows that the Hecke eigenvalues are infact the coefficients of the expansion of f at the cusp ∞. Hence, the values of theL-series in (6.80) are also real. This shows that

RF (X)(6.80)

=

k−2∑l=0

L(k − l − 1)(2πX)l

l!


has only real power series coefficients. In other words, we conclude RF = RF whereRF is defined in Definition 6.73.

Lemma 6.78. Set

(6.82) QF (X) :=1

2L

(k

2

)(2πX)

k2−1(

k2 − 1

)!

+

k2−2∑l=0

L(k − l − 1)(2πX)l

l!.

Then QF satisfies

(6.83) ik RF (X) = QF (X) + ikXk−2QF

(1

X

).

Proof. We have

RF (X)(6.80)

=

k−2∑l=0

L(k − l − 1)(2πX)l

l!

=1

2L

(k

2

)(2πX)

k2−1(

k2 − 1

)!

+

k2−2∑l=0

L(k − l − 1)(2πX)l

l!

+1

2L

(k

2

)(2πX)

k2−1(

k2 − 1

)!

+

k−2∑l= k

2

L(k − l − 1)(2πX)l

l!

(6.82)= QF (X) +

1

2L

(k

2

)(2πX)

k2−1(

k2 − 1

)!

+

k−2∑l= k

2

L(k − l − 1)(2πX)l

l!.

Now, we apply the functional equation (5.74) to the terms on the right hand side.Recall that we have l! = Γ(l + 1), see Appendix A.11. This gives

1

2L

(k

2

)(2πX)

k2−1(

k2 − 1

)!

+

k−2∑l= k

2

L(k − l − 1)(2πX)l

l!

=1

2L

(k

2

)(2π)

k2

Γ(k2

) X k2−1

2π+

k−2∑l= k

2

L(k − l − 1)(2π)l+1

Γ(l + 1)

X l

2π

(5.66)=

1

2L

(k

2

)ik

(2π)k2

Γ(k2

) X k2−1

2π+

k−2∑l= k

2

L(l + 1) ik(2π)k−l−1

Γ(k − l − 1)

X l

2π

l=k−2−n= ikXk−2

1

2L

(k

2

) (2πX

) k2−1

Γ(k2

)+ ik

k2−2∑n=0

L(k − n− 1)(2π)n

Γ(n+ 1)Xk−2−n

= ikXk−2

1

2L

(k

2

) (2πX

) k2−1

Γ(k2

) +

k2−2∑n=

L(k − n− 1)

(2πX

)nΓ(n+ 1)

(6.82)

= ikXk−2QF

(1

X

).

Summarizing, we just proved (6.83).

Remark 6.79. It is obvious that PF (X) would have all its zeros on |z| = 1 ifand only if the same is true for ik RF (X). By Theorem 6.76, the latter is true ifQF (X) has all its zeros in |z| ≤ 1. In the remaining part of this section, we willprove the latter, obtaining along the way some similar results for truncations ofcertain exponential series.

6.6. ROOTS OF PERIOD POLYNOMIALS ASSOCIATED TO HECKE EIGENFORMS 215

Before we continue, we prove that certain self-reciprocal polynomials built fromtruncations of e2πX are eventually unimodular. To that end, set

Tm(z) : =

m∑n=0

(2π)n

n!zn and

Hm(z) : = zm Tm(z) =

m∑n=0

(2π)n

n!zm−n

(6.84)

for m ∈ N. We consider the polynomial

(6.85) P λm (z) := zmHm(z) + λTm(z).

We will show next that the zeros of Pλm (z) are unimodular for m ≥ 20. (Note

that the assertion is in fact not true for m below that range!)

Theorem 6.80 ([34, Theorem 3.1]). For m ≥ 20, Hm has all of its zeros in

|z| < 1. Hence Pλm (z) has all of its zeros on the unit circle |z| = 1.

Proof. We start by finding a lower bound for

min|z|=1

|Hm(z)| = min|z|=1

|Tm(z)| .

Write the Taylor expansion of e2πz as

e2πz = Tm(z) +Rm(z).

By the well-known Taylor inequality, we get for the remainder term Rm(z), for|z| = 1,

|Rm(z)| ≤ (2π)m+1

(m+ 1)!e2π ≤ 0.0007513 for m ≥ 25.

For z = eiφ on the unit circle, we easily estimate∣∣e2πz∣∣ = e2π cosφ ≥ e−2π ≥ 0.001867.

It thus follows with the triangle inequality that

|Tm(z)| =∣∣e2πz −Rm(z)

∣∣ ≥ 0.001867− 0.0007513 = 0.0011157.

holds for all |z| = 1 and m ≥ 25. For M ≥ 25, we can write Hm(z) as

Hm(z) = zm−25H25(z) + gm(z)

with

gm(z) :=

m∑n=26

(2π)n

n!zm−n.

For |z| = 1, we find the estimate

|gm(z)| ≤m∑

n=26

(2π)n

n!≤

∞∑n=26

(2π)n

n!≤ e2π −H25(1) ≤ 0.000001823 < |H25(z)| .

It thus follows by Rouche’s theorem, see Appenix A.10, that the functions Hm(z)and zm−25H25(z) have the same number of zeros inside the unit circle. Numericalverification using PARI [90] gives that H25(z) has all 25 of its zeros inside the unitcircle (the largest of which has modulus less than 0.8); thus, it follows that Hm(z)has all m of its zeros inside the unit circle as well. For 20 ≤ m ≤ 24, the result canbe verified directly using PARI as well.

Theorem 6.81 ([34, Theorem 3.4]). For 4 ≤ k ∈ 2N let F ∈ Sk(Γ(1)

)be a

simultaneous Hecke eigenform (as defined in Definition 5.39). The by (6.1) asso-ciated period polynomial PF (X) has all of its zeros on the unit circle.


Proof. Note that, for any complex number µ, we have PµF (X) = µPF (X); sowe might assume, without loss of generality, that F is normalized. By Theorem 6.76,see also Remark 6.79, it suffices to prove that all the zeros of QF (X) are inside theunit circle.

Let m = k2 − 1. Using [26, Lemma 2.4] stated in Remark 5.87, we find

|Hm(X)−QF (X)| ≤m−1∑n=0

|1− L(k − n− 1)| (2π)n

n!+

∣∣∣∣1− 1

2L

(k

2

)∣∣∣∣ (2π)m

m!

≤[ k4 ]−1∑n=0

5 · 2− k4 (2π)n

n!+

m∑n=[ k4 ]

|1 + L(k − n− 1)| (2π)n

n!

≤ 5 e2π 2−k4 +

(2 + 2

√k log(2k)

)R[ k4 ](1).

Here, [·] denotes the Gauß brackets defined in (4.30) of Definition 4.35. ApplyingTaylor’s inequality as in the proof of Theorem 6.80, we see that

R[ k4 ](1) ≤ e2π (2π)[k4 ][

k4

]!.

It is not hard to show that, for k ≥ 124, we have

e2π (2π)[k4 ][

k4

]!≤(

2 + 2√k log(2k)

)e2π (2π)[

k4 ][

k4

]!≤ 0.000045

and5 e2π 2−

k4 ≤ 0.0000025.

It thus follows that, for k ≥ 124 and |X| = 1, we have |Hm(X)QF (X)| < |Hm(X)|.And it follows from Rouche’s theorem, see Appendix A.10, that QF (X) has thesame number of zeros as Hm(X) inside the unit circle. Theorem 6.80 shows thatthis are m zeros.

For cusp forms with 12 ≤ k ≤ 122, it can be verified directly (using PARI [90],for instance) that QF (X) has all its zeros in |z| < 1, and thus the result follows forall k ≥ 12. By Proposition 4.28 there are no cusp forms for k ≤ 10.

Part 1

Appendix

APPENDIX A

Background Material

A.1. Open Mapping Theorem

[101, Theorem 2.11]: If U ⊂ C is a connected open subset and

f : U → C

is a non-constant holomorphic function, then f is an open map. (Open map meansthat it sends open subsets of U to open subsets of C.)

A.2. Weierstrass M-Test

[118, §3.34]: Let ∅ 6= A ⊂ C be a set. We consider a sequence (fn)n∈N offunctions fn : A→ C satisfying

∃Mn ∈ R>0 such that |fn(z)| ≤Mn for all z ∈ A and n ∈ N.

Moreover, assume that∞∑n=1

Mn <∞

converges.Then the series

∑∞n=1 fn(z), z ∈ A, converges uniformly on A.

A.3. Chinese Remainder Theorem – Special Case

[33, Chinese remainder theorem]: Suppose n1, n2 ∈ N are coprime, that meansgcd(n1, n2) = 1. Then, for any given sequence of integers a1, a2, there exists anx ∈ Z solving the system of simultaneous congruences

x ≡ a1 (mod n1) and x ≡ a2 (mod n2).

Furthermore, all solutions x of this system are congruent modulo the product,N = n1n2. The solution is given by

x = a1b1N

n1+ a2b2

N

n2mod N

= a1b1n2 + a2b2n1 mod n1n2

for suitable b1, b2 ∈ Z.

In particular, we find the following special case: Let c, d ∈ Z be coprime, i.e.,gcd(c, d) = 1 holds. Then by the Chinese remainder theorem stated above thereexist an solution x ∈ Z satisfying

x ≡ 1 mod c and x ≡ 0 mod d.

Hence, there exists a, b ∈ Z satisfying

x = 1 + bc and x = 0 + ad.

In other words, there exists a, b ∈ Z satisfying

ad = 1 + bc ⇐⇒ ad− bc = 1 ⇐⇒(a bc d

)∈ SL2 (Z) .

III

IV A. BACKGROUND MATERIAL

A.4. Maximum Modulus Principle

[33, Maximum-modulus principle]: The modulus (or absolute value) of a non-constant holomorphic function defined on an open connected set does not attain alocal maximum.

A.5. Phragmen-Lindelof principle

[60, Section XIII.5]: Let φ(s) be a holomomorphic function on the upper partof a vertical strip

s ∈ C; σ1 ≤ < (s) ≤ σ2, = (s) > c

for some σ1 < σ2 and c ∈ R such that

φ(s) = O(e=(s)α

)for some α > 0 holds on the strip. If the function φ satisfies

φ(s) = O(= (s)

M)

for all s in the strip with < (s) ∈ σ1, σ2

on the vertical boundary of the strip for some M ∈ R then

φ(s) = O(= (s)

M)

for all s in the strip.

A.6. Stirlings estimate of the Gamma-function

Let Γ(s) denote the Gamma-function defined in (5.58). Uniformly for |σ| ≤ 2and |t| ≥ 2 we have have [30, (5.11.1)]

log Γ(σ + it) =−π |t|

2+

(σ − 1

2

)log |t|+ it log |t| − it+O (1) .

In particular, in this range there exist positive constants 0 < C1 < C2 such that

C1e−π|t|2 |t|σ−

12 ≤ |γ(s)| ≤ C2e

−π|t|2 |t|σ−12 .

Moreover we have the asymptotic estimate [30, (5.11.3)]

Γ(s) ∼√

2π e(s− 12 ) log s−s

uniformly for s→∞ and |arg(s)| < π − δ for some δ > 0, also known as Stirling’sformula.

A.7. Cauchy-Riemann equations

[33, Cauchy-Riemann equations]: Let f : U → C, U ⊂ C open, be a holomor-phic function and write f as real and imaginary part depending on real variables xand y:

z = x+ iy and f(x+ iy) =: u(x, y) + iv(x, y)

with real valued functions u(x, y) and v(x, y) of two real valued variables x and y.Then u and v satisfy

∂u

∂x=∂v

∂yand

∂u

∂y= −∂v

∂x.

Using the complex coordinates z = x+iy and z = x−iy the above Cauchy-Riemannequation can be written as

∂

∂zf(z) = 0.

A.11. BETA FUNCTION AND GAMMA FUNCTION V

A.8. Cauchy’s integral theorem

[33, Cauchy integral theorem] If D ⊂ C is a simply connected open set andf : D → C a holomorphic function, then the integral of

∫f(z) dz along any closed

rectifiable curve γ ⊂ D vanishes: ∫γ

f(z) dz = 0.

An equivalent version states that (under the same assumptions as above), givenany (rectifiable) path η : [0, 1]→ D the integral∫

η

f(z) dz

depends only upon the two endpoints η(0) and η(1), and hence it is independent ofthe choice of the path of integration η.

A.9. Green’s theorem

[30, (1.6.44)] Let F1, F2 : R× R → R are continuously differentiable functionsand assume that S be the closed and bounded set in the (x, y)-plane having a simpleclosed curve ∂S as boundary. If ∂S is oriented in the positive (anticlockwise) sense,then ∫∫

S

(∂

∂xF2 −

∂

∂yF1)

)dxdy =

∫∂S

F1 dx+ F2 dy.

A.10. Rouche theorem

[33, Rouche theorem] Let f(z) and g(z) be holomorphic functions of a complexvariable z in a domain D, let a simple closed piecewise-smooth curve Γ together withthe domain G bounded by it belong to D and let everywhere on Γ the inequality|f(z)| > |g(z)| be valid; then in the domain G the sum f(z) + g(z) has the samenumber of zeros as f(z).

A.11. Beta function and Gamma function

[30, §5.2–§5.5]: The Gamma function Γ(s) is defined for < (s) > 0 by

Γ(s) =

∫ ∞0

e−tts−1 ddt

and for < (s) ≤ 0 by analytic continuation. It is a function with no zeros and withsimple poles at s ∈ Z≤0. The Gamma function generalized the faculty symbol inthe following sense:

Γ(n) = (n− 1)! for all n ∈ N.The function satisfy the functional equations

Γ(s+ 1) = sΓ(s) and Γ(s)Γ(1− s) =π

sin(πs).

A variant of the Gamma function is the beta function defined by

B(x, y) =

∫ 1

0

tx−1(1− t)y−1,

see [39, 40, ((8.380.1))]. It is related to the Gamma function by

B(x, y) =Γ(x) Γ(y)

Γ(x+ y).

VI A. BACKGROUND MATERIAL

Moreover, B(x, y) admits the integral representation

B(x, y) = 2

∫ ∞0

t2x−1

(1 + t2)x+ydt,

see [39, 40, (8.380.3)].

A.12. A result on normal operators

[50, Proposition 6.14]: Let H denote a finite dimensional Hilbert space over Cwith scalar product 〈,〉 and let T : H → H denote a linear operator.

The adjoint operator T ? : H → H is defined by 〈Tf, g〉 = 〈f, T ?g〉. T is callednormal if T commutes with T ?. It is called self-adjoint if T = T ? holds.

We have the following result: If T denotes a commuting family of normaloperators T : H → H then there exists an orthonormal basis F of H which consistsof common eigenfunctions of all operators T in the family T .

A.13. Unique continuation principle for holomorphic functions

[33, Analytic function]: Let f, g : D → C be two analytic functions in a domainD ⊂ C. If they coincide on some set with an accumulation point in D then theyare identical everywhere.

Some important consequence is the following corollary: If D is a connectedopen set and f : D → C an holomorphic function which is not identically zero,then each zero z0 of f is isolated. In addition, for some neighborhood U of z0, thereare an holomorphic function g : U → C which never vanishes and a natural numbern, (called the multiplicity of the zero z0, or order of vanishing of f at z0) such thatf(z) = (z − z0)ng(z) for every z ∈ U .

A.14. Whittaker functions

[30, Chapter 13 Confluent Hypergeometric Functions]: We collect a few factsfrom the Digital Library of Mathematical functions [30] and some other referencesabout Whittaker functions which are used in this book.

Whittaker’s normalized differential equation is the second order ordinary dif-ferential equation

∂2

∂y2G(y) +

(−1

4+k

y+

14 − ν

2

y2

)G(y) = 0

for smooth functions G : (0,∞) → C and ν /∈ − 12N. According to [30, (13.14.2),

(13.14.3)], see also [68, Chapter 7], we have two solutions Mk,ν(y) and Wk,ν(y)with different behavior as y →∞:

Mk,ν(y) ∼ Γ(1 + 2ν)

Γ(

12 + ν + k

) e 12y y−k and

Wk,ν(y) ∼ e− 12y yk.

The asymptotic behavior is valid for k − ν /∈

12 ,

32 ,

52 , . . .

, see [30, (13.14.20),

(13.14.21)]. These functions satisfy also the recurrence relations [30, (13.15.1),(13.15.11)] and differentiation relations [30, (13.15.17), (13.15.20) and (13.15.23),(13.15.26)] and [68, §7.2.1, page 302].

We have the following identities and relations with other functions:

• [30, (13.18.2)]

Wk,k− 12(y) = Wk, 12−k

(y) = e−y2 yk (k ∈ R, y > 0)

A.16. POISSON SUMMATION FORMULA VII

• [30, (13.18.9)]

W0,ν(2y) =

√2y

πKν(y) (ν ∈ Cr−1

2N, y > 0)

• [110, (3.5.12)] Assume <(

12 − ν −

k2

)> 0. We have

Γ

(1

2− ν − k

2

)W k

2 ,−ν(y) = y

12−ν22ν

∫ ∞1

e−12yt(t− 1)−

12−ν−

k2 (t+ 1)−

12−ν+ k

2 dt

for y > 0.(Slater states the formula only under the condition <

(12 − ν ± k

)> 0;

but it is obvious that the integral converges also under the slightly relaxedcondition given here.)

A.15. Generalized hypergeometric functions

[30, Chapter 16], [110, §1.1.1]: We collect a few facts from the Digital Libraryof Mathematical functions [30] and some other references about the generalizedhypergeometric function 2F1 which are used in this book.

For p, q ∈ Z≥0 the generalized hypergeometric function pFq is defined by thePower series expansion

pFq

[a1, . . . , apb1, . . . , bq

∣∣∣∣∣z]

=

∞∑n=0

(a1

)n· · ·(ap)nzn(

b1)n· · ·(bq)nn!

where the Pochammer symbol(a)n

is defined by(a)

0:= 1,

(a)

1:= a, and

(a)n

:= a(a+ 1)(a+ 2) · · · (a+ n− 1)

for a ∈ C and n ∈ Z≥0.Several families of special functions can be derived by generalized hypergeo-

metric functions if we choose the parameters accordingly. For example we have thefollowing connection to the Whittaker Wk,ν-function:

Mk,ν(y) = 1F1

[12 − k + ν

1 + 2ν

∣∣∣y] e− y2 y 12 +ν

and

Wk,ν(y) = Γ(−2ν)1F1

[12 − k + ν

1 + 2ν

∣∣∣y]Γ( 1

2 − k − ν)e−

y2 y

12 +ν +

+ Γ(2ν)1F1

[12 − k − ν

1− 2ν

∣∣∣y]Γ( 1

2 − k + ν)e−

y2 y

12−ν

for k ∈ R, ν ∈ C and y > 0, see [110, (1.9.7), (1.9.10)].

A.16. Poisson Summation Formula

[111, Chapter 4, §1 and Theorem 2.4] For each a > 0 denote by Fa the class ofall functions f that satisfy the following two conditions:

(1) The function f is holomorphic in the horizontal strip

Sa =z ∈ C; |= (z)| < a

.

(2) There exists a constant A > 0 such that

|f(x+ iy)| ≤ A

1 + x2for all x+ iy ∈ Sa.

VIII A. BACKGROUND MATERIAL

We denote by F the class of all functions that belong to Fa for some a.

For f ∈ F the Poisson summation formula is the relation∑n∈Z

f(n) =∑n∈Z

f(n),

where f denotes the Fourier transformation

f(k) =

∫ ∞−∞

f(x) e−2πikx dx

of f .

A.17. Liouville’s theorem

[33, Liouville theorems] Liouville’s theorem states that every bounded entirefunction must be constant.

In other words: Let f : C→ C be a holomorphic function and M > 0 a positiveconstant such that

|f(z)| ≤M

holds for all z ∈ C. Then f is constant.

A.18. Residue theorem

[33, Residue of an analytic function] The theory of residues is based on theCauchy integral formula.

A simple form of the Cauchy integral formulais the following: Suppose U ⊂ Cis open, f : U → C is holomorphic and γ is a positively oriented simple closed curvecompletely contained in U. W have

f(a) =1

2πi

∮γ

f(z)

z − adz

where the contour integral is taken counter-clockwise.The residue theorem is the following statement: Suppose U is a simply con-

nected open subset of the complex plane, and a1, . . . , an are finitely many pointsin U and let f be a meromorphic function on U with poles in a1, . . . , an. If γ is aclosed rectifiable curve in U which does not meet any of the ak, we have∮

γ

f(z) dz = 2πi

n∑k=1

I(γ, ak) resz=ak

(f).

where I(γ, ak) is the winding number of γ around ak.It slightly simplified form reads as follows: If γ is a positively oriented simple

closed curve and I(γ, ak) =

1 if ak is in the interior of γ

0 elsewhere, then we have

∮γ

f(z) dz = 2πi∑

resz=ak

(f)

where the sum runs over those k for which ak is inside γ.

A.20. CONTINUED FRACTIONS IX

A.19. Geometric series

[33, Geometric progression], [1] The geometric series is given by∞∑n=0

qn.

The series converges absolutely for all |q| < 1 and has the value∞∑n=0

qn =1

1− q

Its variants∞∑n=0

nk qn

converge also absolutely for |q| < 1. Due to interchanging differentiation and sum-mation we can calculate the series and get for example

∞∑n=0

n qn = qd

dq

∞∑n=0

qn

= qd

dq

1

1− q=

q

(1− q)2

and∞∑n=0

n2qn = qd

dqq

d

dq

∞∑n=0

qn

= qd

dqq

d

dq

1

1− q

= qd

dq

q

(1− q)2

=q(1 + q)

(1− q)3.

Similar arguments show the growth estimate∞∑n=0

nkqn = O((1− q)−C

)as q → 1

for some C > 0.

A.20. Continued fractions

[33, Continued fraction] A continued fraction is a finite or infinite expressionof the form

a0 +b1

a1 +b2

a2 + · · ·+ bnan + · · ·

where (an)n and (bn)n are finite or infinite sequences of complex numbers.The regular Gauss continued fraction is defined as the expression

a0 +1

a1 +1

a2 + · · ·+ 1

an + · · ·

X A. BACKGROUND MATERIAL

with a0 ∈ Z and all following an ∈ N (if they exist).For every continued fraction the recurrence equations

Pn = anPn−1 + bnPn−2 and

Qn = anQn−1 + bnQn−2,

with the initial conditions

b0 = 1, P−2 = 0, P−1 = 1, Q−2 = 1, and Q−1 = 0

determine two sequences(Pn)n

and(Qn)n

of complex numbers. As a rule, it isassumed that these sequences are such that Qn 6= 0 for all n ∈ Z≥0 in the sequence.The above recurrence relation terminates if Qn+1 = 0.

The fraction PnQn

is called the nth convergent of the continued fraction. We have

P0

Q0= a0,

P1

Q1= a0 +

b1a1,

P2

Q2= a0 +

b1

a1 + b2a2

, · · · ,

moreover,

PnQn− Pn−1

Qn−1=

(−1)n−1 b1 · · · bnQnQn−1

.

It is convenient to denote the nth convergent of the regular Gauss continuedfraction by

[a0; a1, . . . , an] := a0 +1

a1 +1

a2 + · · ·+ 1

an

.

Its convergents satisfy the equalities

[an; . . . , a1] =QnQn−1

for n ≥ 1

and

[an; . . . , a0] =PnPn−1

for a0 6= 0 and n ≥ 0.

If the continued fraction is an infinite expression and the sequence of conver-gents Pn

Qnconverges to some limit l, then the continued fraction is called convergent

and the number l is its value. If the continued fraction is a finite expression, thenits value is defined as the last of its convergents.

The convergents of a regular Gauss continued fraction satisfy

a0 =P0

Q0<P2

Q2<P4

Q4< . . . < l < . . . <

P5

Q5<P3

Q3<P1

Q1= a0 +

1

a1.

Regular continued fractions are a very convenient tool for the approximationof real numbers by rational numbers. The following propositions hold:

(1) If PnQn and Pn+1

Qn+1are neighbouring convergents of the expansion of a number

r in a regular Gauss continued fraction, then∣∣∣∣r − PnQn

∣∣∣∣ ≥ ∣∣∣∣r − Pn+1

Qn+1

∣∣∣∣ and

∣∣∣∣r − PnQn

∣∣∣∣ ≤ 1

QnQn+1

holds. Equality holds in the estimate on the right side only if r = Pn+1

Qn+1.

(2) For two neighbouring convergents of the expansion of a number r in a reg-ular Gauss continued fraction, at least one of them satisfies the inequality∣∣∣∣r − Pn

Qn

∣∣∣∣ =1

2Q2n

.

A.22. INCOMPLETE GAMMA FUNCTION XI

(3) Let a and b be integers, b ≥ 1 and let r is a real number. If∣∣∣r − a

b

∣∣∣ ≤ 1

2b2

holds then ab is a convergent of the expansion of r in a regular continued

fraction.(4) Let Pn

Qnbe a convergent of the expansion of a number r into a regular

Gauss continued fraction. If the integers a and b with b > 0, PnQn6= a

b

satisfy ∣∣∣r − a

b

∣∣∣ ≤ ∣∣∣∣r − PnQ− n

∣∣∣∣then we have b ≥ Qn.

A.21. Hurwitz zeta function

[33, Hurwitz zeta function], [68] The Hurwitz zeta function ζ(s, q) is definedfor < (q) > 0 and < (s) > 1 as

ζ(s, q) =

∞∑n=0

(n+ q)−s.

The series converges absolutely, and defines for < (s) > 1 an analytic function. Thefunction possesses an analytic continuation to the whole s-plane except for a simplepole of residue 1 at s = 1.

For example ζ(s, q) admits the integral representation

ζ(s, q) =1

Γ(s)

∫ ∞0

ts−1e−qt

1− e−tdt

for < (s) > 1 and < (q) > 0.We have

lims→1

[ζ(s, q)− 1

s− 1

]= −Γ′(q)

Γ(q)

where Γ(q) denotes the Gamma-function, see Appendix A.11, and Γ′(q) its deriva-tive.

A.22. Incomplete Gamma function

[30, §8]: The incomplete Gamma function Γ(s, x) is defined for < (s) > 0 andx ≥ 0 by

Γ(s, x) =

∫ ∞x

e−tts−1 ddt

and for < (s) ≤ 0 by analytic continuation.If x 6= 0 then the function s 7→ Γ(s, x) is analytic, see [30, §8.2(ii)].It satisfies the growth estimate

Γ(s, x) = O(xs−1 e−x

)as x→∞, see [30, (8.11.2)].

APPENDIX B

Solutions of Selected Problems

B.1. Problems of Chapter 1

1.4 Assume that

(a bc d

)∈ GL2 (C) which acts trivially on S. This means that(a bc d

)z = z for all z ∈ S.

We may assume for the moment z ∈ S r−dc ,∞

. Using Definition 1.2 we get

az + b

cz + d= z ⇐⇒ az + b = cz2 + dz

for all z ∈ S. Comparing coefficients we get a = d ∈ C and b = c = 0.

For z = ∞ the defining equation (1.9) gives a∞+bc∞+d

(1.9)= a

c . Hence matrices of

the form

(λ 00 λ

), λ 6= 0, satisfy the required identity since λ

0 =∞.

For the third case z = −dc we get directly d = λ and c = 0 using the matrix(a bc d

)=

(λ 00 λ

), λ 6= 0. Hence z = −λ

0 =∞ and

(λ 00 λ

)is in the kernel of the

Mobius transformation.

1.6 Consider V =

(a bc d

)∈ SL2 (R). Similar to the calculation in (1.11) we have

d(V z)

= d

(az + b

cz + d

)=∂(az+bcz+d

)∂z

dz

=(cz + d)a− (az + b)c

(cz + d)2dz =

ad− bc(cz + d)2

dz

ad−bc=1=

dz

(cz + d)2.

1.13 Let V =

(a bc d

)be parabolic with fixed point ∞. Since

V z =

ac if c 6= 0 anda∞+bd =∞ if c = 0

holds by (1.9) we see immediately that c = 0 must hold. Hence V is of the form

V =

(a b0 d

)∈ SL2 (R) .

Moreover detV = 1 implies d = a−1. Since V is parabolic, we have 2 = |trace (V )| =∣∣a+ a−1∣∣. This equality can only hold for a = ±1. (a 6= ±1 implies

∣∣a+ a−1∣∣ > 2.)

We just proved that V is of the form

V =

(±1 b0 ±1

)XIII

XIV B. SOLUTIONS OF SELECTED PROBLEMS

and hence a translation by Definition 1.12.Now, Assume that V is a translation. Definition 1.12 implies that V is of the

form

V =

(±1 b0 ±1

).

Obviously trace (V ) = ±2 and hence V is parabolic by Proposition 1.11. Thematrix also fixes ∞ as the following calulation shows:

V∞ =∞± b =∞.

1.14 The trace is cyclic, i.e., trace (ABC) = trace (BCA). Hence

trace(AV A−1

)= trace

(V A−1A

)= trace (V )

holds.

1.16 Let V =

(a bc d

)∈ SL2 (R) such that V α = β and V β = α holds. We have

aα+ b

cα+ d= β and

aβ + b

cβ + d= α

which implies

aα+ b = cαβ + dβ and aβ + b = cαβ + dα.

Subtracting the right equation from the left one, we get

a(α− β) = −d(α− β).

Since by assumption we have α 6= β we may divide by α− β and get

a = −d.

Hence the trace of V vanishes:

trace (V ) = a+ d = 0.

Definition 1.9 implies that V is elliptic.1.36 Let Vn denote a sequence of distinct elements in Γ and take any z ∈ S. Thisis possible since Γ has infinite many elements.

Now, we have two cases. Either the set of all points

W := Vnz; n ∈ N

is bounded or it is unbounded.

W is bounded:: This means that there exists an w ∈ R+ such that |Vnz| ≤w for all n ∈ N. The BolzanoWeierstrass theorem implies that the se-quence Vn has a convergent subsequence

(Vnm

)m

. Hence we have founda sequence Vnm ∈ Γ such that the limit

limm→∞

Vnmz := α

exists. Since S is closed we conclude that α ∈ S. Definition 1.21 impliesα ∈ Lim (Γ).

W is unbounded:: This means that there exists a subsequence Vnm , m ∈ Nsatisfying |Vnmz| > w for all m large enough.

Given a neighborhood U ⊂ S of ∞, see above Definition 1.21, thisimplies in particular that the points Vnmz lay inside the neighborhood Ufor m large enough. Hence,

limm→∞

Vnmz =∞

holds. Definition 1.21 implies ∞ ∈ Lim (Γ).

B.1. PROBLEMS OF CHAPTER 1 XV

This concludes the exercise since we have shown in both cases that Lim (Γ) is notempty.1.43 Lemma 1.5 implies V H ⊂ H. Using Remark 1.42 we get

A−1V AU ⊂ Uwith A given in (1.26).

Moreover, we have det(A−1V A

)= detV = 1. This shows that A−1V A ∈

SL2 (C) since each matrix entry can be expressed as complex linear combination ofthe complex entries of A and V .

Say V is of the form V =

(a bc d

)with real entries a, b, c, d and determinant

ad− bc = 1. Matrix multiplication gives

A−1V A =

(1 1i −i

)−1(a bc d

)(1 1i −i

)=

1

2i

(i 1i −1

)(a+ ib a− ibc+ id c− id

)=

1

2i

(i(a+ d)− (b− c) i(a− d) + (c+ b)i(a− d)− (c+ b) i(a+ d) + (b− c)

)=

((a+d)+i(b−c)

2(a−d)−i(b+c)

2(a−d)+i(b+c)

2(a+d)−i(b−c)

2

).

Writing α = (a + d) + i(b − c) and β = (a − d) + i(b + c) we see that A−1V A hasthe form

A−1V A =

(α ββ α

).

Calculating the determinant, we get

1 = detV = det(A−1V A

)= αα− ββ = |α|2 − |β|2 .

1.50 A−1ΓA satisfies A−1ΓAH = H. This implies in particular that the set of limitpoints Lim

(A−1ΓA

)has to be contained in the closure of H in S:

Lim(A−1ΓA

)⊂ H ∪ R ∪ ∞.

ThusLim

(A−1ΓA

)⊂ R ∪ ∞

by Corollary 1.28. Hence we have

A−1 Lim (Γ) ⊂ R ∪ ∞.This implies

Lim (Γ) ⊂ A(R ∪ ∞

)= T.

1.46 Let V ∈ SL2 (C) such that z 7→ V z maps the closed unit disc U onto itself.Definition 1.2 of the Mobius transformation implies that the function z 7→ V z

is a rational function. The additional assumption V U = U implies that z 7→ V zdoes not have any poles in U. Hence V z is holomorphic on the open unit disc

(B.1) U := z ∈ C; |z| < 1and extends continuously to the clousure U.

The assumption V U = U implies

|V z| ≤ 1 for all z ∈ U.Now, assume that there exits an z0 ∈ U, i.e., |z0| < 1, such that Vz0 = 1 holds.Then,

|V z| ≤ |Vz0 | for all z ∈ U

XVI B. SOLUTIONS OF SELECTED PROBLEMS

holds. The maximum modulus principle, see Appendix A.4, implies that the mapz 7→ V z is constant on U. However, this cannot happen: Since V ∈ SL2 (C) andin particular detV = 1 implies that the top and bottom row of V are linearlyindependent we find that the function V z = az+b

cz+d cannot be constant.Hence we conclude that our assumption was wrong. We have have

|V z| < 1 for all z ∈ U.

This means that the maximum modulus of the map V z must be obtained on theunit circle T:

(B.2) |V z| = 1 =⇒ |z| = 1 for all z ∈ U.

Next, we need to show the converse implication. To do so, consider the inverseV −1 ∈ SL2 (C). Since V U = U, Lemma 1.3 implies

V −1U V U=U= V −1

(V U) (1.10)

=(V −1V

)U V −1V=1

= U.

Then, (B.2) shows the implication∣∣V −1z∣∣ = 1 =⇒ |z| = 1 for all z ∈ U.

Replacing z with V z, which is also in U, we find

|z| =∣∣V −1V z

∣∣ = 1 =⇒ |V z| = 1 for all z ∈ U

using (B.2).Summarizing, we have shown that the maximum modulus 1 of the function

z 7→ V z, which maps U onto itself, lies on the unit circle T. Moreover, z ∈ Timplies V z ∈ T. In other words, we have shown that the unit circle T is invariantunder the map V .


2.8 The group Γ(1) = SL2 (Z) is obviously a discrete group (since SL2 (Z) ⊂ Z4

is a discrete set in R4), see Definition 1.19. Theorem 1.27 shows that Γ(1) actsdiscontinuous. It remains to show that the set of limit points Lim (Γ(1)) is equalto R ∪ ∞.

Corollary 1.28 already implies

Lim (Γ(1)) ⊂ R ∪ ∞.To show the reverse inclusion, we start with Example 1.22 and conclude that ∞ isa limit point of Γ. By Definition 1.21 there exists a point z0 ∈ H ∪ R ∪ ∞ and asequence Vn ∈ Γ(1) such that

limn→∞

Vn z0 =∞

holds. The relation T∞ = 0 implies directly

limn→∞

TVn z0 = 0 and0 ∈ Lim (Γ(1)) .

Let q = ab ∈ Q be a rational number and a, b ∈ Z chosen such that gcd(a, b) = 1

holds. Since a and b are coprime integers there exists x, y ∈ Z such that

ay − bx = 1.

(This follows from the Chinese remainder theorem, see Appendix A.3.) Put

Aq :=

(a xb y

)∈ Γ(1)

which is obviously in the full modular group. The matrix A satisfies

Aq∞ = q.

B.2. PROBLEMS OF CHAPTER 2 XVII

As above, we see immediately limn→∞AqVn z0 = q implying q ∈ Lim (Γ(1)). Sum-marizing, we have shown

Q ∪ ∞ ⊂ Lim (Γ(1)) .

To complete the proof take any element z ∈ R. Then there exists a sequence ofrational qm approximation z: limm→∞ qm = z. Using a triangle argument on theindex sets we see that

z = limm→∞

qm = limm→∞

limn→∞

AqmVn z0.

The triangle argument is possible since we have countable index sets N2: We mayreorder all elements of N2 as

N2 =

(1, 1)∪ (2, 1), (2, 2)

∪ . . . ∪ (n, 1), . . . , (n, n)

∪ . . .

=

∞⋃n=1

n⋃m=1

(n,m)

without changing the limit above. This shows

R ∩ ∞ ⊂ Lim (Γ(1)) .

2.11 We take an element V ∈ Γ and consider V F . It is open since F is an openset as fundamental domain, see Definition 2.9, and V induces a bijection on H, seee.g. Lemma 1.5 together with V is invertible.

Now consider two points z1, z2 ∈ V F which are Γ equivalent. Hence thereexists an element M ∈ Γ such that z1 = M z2 holds. Multiplying with V −1 implies

V −1 z1 = V −1M z2

and also V −1 z1, V −1 z2 ∈ F . Using that F is a fundamental domain of Γ, weknow by the first property of Definition 2.9 that V −1 z1 and V −1 z2 are equal. Thisimplies directly z1 = z2 and shows that the first property of Definition 2.9 is holdsfor V F .

To check the second condition, take any point z ∈ H. Since F is a fundamentaldomain of Γ there exists an element M ∈ Γ with M z ∈ F . Hence we have VM z ∈V F . This completes the proof that V F is a fundamental domain.

Let’s proceed to the proof of the second part of the lemma and assume thatthe set V F ∩ F 6= ∅ is non-empty. Take an z ∈ V F ∩ F . By construction of V F ,the point w := V −1 z ∈ F satisfies w ∈ F and V w = z ∈ V F . Hence v and w areΓ equivalent points in the fundamental domain F . Therefore they must coincide:z = w.

We just showed thatz = V −1 z

holds for any point z in the open set V F ∩ F . Hence V −1 has infinite many fixedpoints, implying the same for V . Applying Proposition 1.11 this can only happenif V is of the form

V = ±1.

2.13 This example appeared already in [21, Part I, Example 3.2 on page 25].Let λ > 1.

(1) We first show that M :=

(√λ 0

0 1√λ

)is a hyperbolic matrix. This follows

directly from Definition 1.9 since

trace (V ) =√λ+

1√λ

=λ+ 1√λ

> 2

holds. (This can be seen by recognizing that this expression has the value2 for λ = 1 and has a positive derivative in λ.)

XVIII B. SOLUTIONS OF SELECTED PROBLEMS

Due to the diagonal matrix form of M we also easily calculate Mn forall n ∈ Z. We have

Mn =

(λn2 0

0 λ−n2

).

Hence we have

Γ =

(λn2 0

0 λ−n2

); n ∈ Z

=Mn; n ∈ Z

.

This shows directly that Γ is generated by M .That Γ is a group is also easy to verify since MnMm = Mn+m ∈ Γ

for all n,m ∈ Z. The neutral element 1 = M0 is also present.(2) We use Proposition 1.20. Assume that we have a sequence

(Vn)n∈N in Γ

such that Vn → 1 for n→∞. Since any Vn is a power of M , there existsmn ∈ Z such that Vn = Mmn . M being a diagonal matrix implies thatthe above limit can be written as

1 = limn→∞

Vn = limn→∞

Mmn =

(limn→∞ λ

mn2 0

0 limn→∞ λ−mn2

).

This implies

limn→∞

λmn2 = 1 and lim

n→∞λ−

mn2 = 1

simultaneously. Recall λ > 1. Hence above limits can only hold if themn’s satisfy

limn→∞

mn = 0.

Due to mn ∈ Z and Z has no accumulation points in R (i.e. Z is discretein R) we find mn = 0 for all n ≥ n0 and n0 sufficiently large. This showsVn = 1 for all n ≥ n0. Proposition 1.20 implies that Γ is discrete. Thediscontinuous action of Γ follows from Theorem 1.27.

(3) Consider two points z1, z2 ∈ F which are Γ equivalent. This means thatthere exists an n ∈ Z such that z1 = Mn z2 holds. Since M has thediagonal form given above and both points lie in the fundamental domainF , we find that

z1 = λn z2 and |z1| , |z2| ∈ (1, λ)

holds. Both conditions can only hold for n = 0 which shows that bothpoints coincite. They are not distinct.

We check the second condition in Definition 2.9. Take a z ∈ H. Thereexists an n ∈ Z such that

1 ≤ |z|λn≤ λ

holds. This implies directly

M−n z =z

λn∈ F .

(4) The arguments in this case are nearly identical to previous case.

2.26 Let V :=

(a bc d

)∈ Γ(1). The image of ∞ under V is given by

V∞ =a

c,

which is easily deduced from

V −1 a

c=

(d −b−c a

)a

c=

dac − b−cac + a

=ad−bcc

0=

1

0=∞.

B.2. PROBLEMS OF CHAPTER 2 XIX

Hence V∞; V ∈ Γ(1) ⊂ Q ∪ ∞.To show the converse direction pick any rational number a

c . We may assumegcd(a, c) = 1. By the Chinese remainder theorem, see Appendix A.3, there existsb, d ∈ Z such that

ad− bc = 1

holds. Put

M :=

(a bc d

).

We just have shown M ∈ SL2 (Z) = Γ(1). By the same calculation as above wehave M∞ = a

c . Hence ac ∈ V∞; V ∈ Γ(1) for any rational ac .

This shows that any rational point is equivalent to ∞ under Γ(1).2.16 Put

C1 =

−1

2+ it; t ≥

√3

2

,

C2 =

+

1

2+ it; t ≥

√3

2

,

C3 =

z; |z| = 1, −1

2≤ < (z) ≤ 0

and

C4 =

z; |z| = 1, +

1

2≥ < (z) ≥ 0

.

Since we have ∣∣∣∣±1

2+ it

∣∣∣∣ =

√1

4+ t2 ≥ 1 ⇐⇒ t ≥

√3

2,

we see easily that

C1 ∩ C3 =

−1

2+ i

, C2 ∩ C4 =

−1

2+ i

and C3 ∩ C4 = 0.

The remaining joints

C1 ∩ C2 = C1 ∩ C4 = C2 ∩ C3 = ∅are empty. Using (2.11) as description of the fundamental union FΓ(1) we seedirectly

C1 ∪ C2 ∪ C3 ∪ C4 = ∂FΓ(1).

The last statement follows easily since the matrix elements S and T given in(2.8) together with the definition of the Mobius transformation in (1.9) implies

S C1 =S z; z ∈ C1

!= C2 and T C3 = C4.

2.37 We fix an 1 ≤ i ≤ µ. Using the definition of the cusps qi and q′i in theassumptions of Lemma 2.37 we have

qi = Ai∞ and q′i = A′i∞which leads to the identity

q′i = A′iA−1i qi ⇐⇒ qi = Ai

(A′i)−1

q′i.

By the definition of the width of a cusp in Definition 2.34 and the characterizationof the matrix Mq in part c of Lemma 2.31, we have

Mλiqi = AiS

λiA−1i and M

λ′iq′i

= A′iSλ′i(A′i)−1

with

qi = Mλiqi qi and q′i = M

λ′iq′iq′i.

XX B. SOLUTIONS OF SELECTED PROBLEMS

Since λi respectivley λ′i are the width of the cusps qi respectively q′i we have

λi := mint ∈ N; qi = M t

qi qi

and λ′i := mint ∈ N; q′i = M t

q′iq′i

.

Combining the identities q′i = Mλ′iq′iq′i and q′i = A′iA

−1i qi we find

A′iA−1i qi = M

λ′iq′iA′iA

−1i qi

⇐⇒ qi = Ai(A′i)−1

Mλ′iq′iA′iA

−1i qi

⇐⇒ qi = Mλ′iqi qi

using the matrix identity

Ai(A′i)−1

,Mλ′iq′iA′iA

−1i

= Ai(A′i)−1

A′iSλ′i(A′i)−1

A′iA−1i

= AiSλ′iA−1

i = Mλ′iqi

for the last step. The definition of the width of a cusp implies now the inequality.

λ′i ≥ λi.An analogous calculation, starting width the identities

qi = Mλiq′iqi and qi = Ai

(A′i)−1

q′i,

shows the inequalityλ′i ≤ λi.

Hence we haveλ′i = λi

which concludes the proof of Lemma 2.37.


3.5 Recall that the absolute value function has no branshing problems with realpowers. This means that we have ∣∣zk∣∣ = |z|k

for all z ∈ C and k ∈ R.Using this identity and Lemma 3.3 we find∣∣j(M1M2, z)

k∣∣ = |j(M1M2, z)|k

(3.7)= |j(M1,M2 z) j(M2, z)|k

= |j(M1,M2 z)|k |j(M2, z)|k

for all M1,M2 ∈ Mat2 (R) and z ∈ H. This proves the identity (3.9).3.11 Inspired by Example 3.8 we use the consistency condition (3.6) for T ∈ Γ.Since T 2 = −1 we have

v(−1) j(−1, z)k =(v(T )

)2j(T, Tz)k j(T, z)k

for all z ∈ H. The left hand side simplifies directly to v(−1) since j(−1, z)(3.4)= −1

for all z ∈ H.Plugging in z = iy for some y > 0 gives

v(−1) (−1)k = v(−1) j(−1, z)k

(3.6)

T 2=−1=

(v(T )

)2j

(T,

i

y

)kj(T, iy)k

B.3. PROBLEMS OF CHAPTER 3 XXI

(3.4)=(v(T )

)2 ( iy

)k(iy)k

=(v(T )

)2i2k.

Using the argument convention (3.5) we have −1 = e−πi and i = eπi2 . Hence

e−πikv(−1) = eπi2 2k(v(T )

)2⇐⇒

(v(T )

)2= e−2πik v(−1)

holds. Using the non-triviality condition (3.10), v(−1) = eπik we find

v(T ) =(e−2πik v(−1)

) 12 (3.10)

= e−πik eπi2 k = e−

πi2 k.

3.12 We already calculated v(T ) = e−πi2 k, see (3.11) in Exercise 3.11. Using the

consistency condition (3.6), the non-triviality condition (3.10) and (TS)3 = −1,see Example 2.5 we find

eπik(3.10)

= v(−1) j(− 1, iy

)k(2.9)= v(TSTSTS) j

(TSTSTS, iy

)k(3.6)= v(T ) v(STSTS) j

(T, STSTS iy

)kj(STSTS, iy

)k(3.6)= . . .

(3.6)= v(T ) v(S) v(T ) v(S) v(T ) v(S) j

(T, STSTS iy

)kj(S, TSTS iy

)k ·· j(T, STS iy

)kj(S, TS iy

)kj(T, S iy

)kj(S, iy

)k(3.11)

= v(S)3(e−

πik2

)3

j(T, STSTS iy

)kj(S, TSTS iy

)k ·· j(T, STS iy

)kj(S, TS iy

)kj(T, S iy

)kj(S, iy

)k.

Calculating the matrices

S(2.8)=

(1 10 1

),

T S(2.8)=

(0 −11 0

)(1 10 1

)=

(0 −11 1

),

S TS =

(1 10 1

)(0 −11 1

)=

(1 01 1

),

T STS =

(0 −11 0

)(1 01 1

)=

(−1 −11 0

),

S TSTS =

(1 10 1

)(−1 −11 0

)=

(0 −11 0

)(2.8)= S, and

T STSTS(2.9)=

(0 −11 0

)(0 −11 0

)=

(−1 00 −1

)= −1.

The associated automorphic factors are

j(S, iy

)k= 1,

j(T, S iy

)k=(iy + 1

)k,

j(S, TS iy

)k= 1,

j(T, STS iy

)k=

(iy

iy + 1

)k= e

πik2 yk

(iy + 1

)−k,

XXII B. SOLUTIONS OF SELECTED PROBLEMS

j(S, TSTS iy

)k= 1, and

j(T, STSTS iy

)k=

(−1

iy

)k=

(i

y

)k(3.5)= e

πik2 y−k.

Hence we have

eπik = v(S)3(e−

πik2

)3

j(T, STSTS iy

)kj(S, TSTS iy

)k ·· j(T, STS iy

)kj(S, TS iy

)kj(T, S iy

)kj(S, iy

)k= v(S)3

(e−

πik2

)3

· eπik2 y−k · 1 · eπik2 yk(iy + 1

)−k · 1 ·(iy + 1

)k · 1

= v(S)3 e−πik2 .

Rewriting the above equation, we have

v(S)3 = eπik e−πik2 = e

πik2

⇐⇒ v(S) = eπik6 .

This proves (3.12).3.23 Let µ denotes the index µ = [Γ(1) : Γ]. Let Ai; i = 1, . . . , µ and A′i; i =1, . . . , µ be two sets of right coset representatives of Γ in Γ(1) such that the twostandard fundamental domains FΓ and F ′Γ are given by

FΓ =

µ⋃i=1

AiF? and F ′Γ =

µ⋃i=1

A′iF?,

see e.g. Definition 2.18.We assume that the representatives Ai and A′i are sorted such that Ai and A′i

belong to the same right coset of Γ: A′i ∈ AiΓ for all i = 1, . . . , µ and A′j 6∈ AΓi for

all j 6= i.Recall that A′i lies in the same right coset as Ai. Hence there exists an element

Mi ∈ Γ such that A′i = MiAi holds for all i = 1, . . . , µ. This allows us to rewriteF ′Γ:

F ′Γ ∩H =

(µ⋃i=1

A′iF?)∪H

=

(µ⋃i=1

MiAiF?)∪H.

Now we use that F satisfies the transformation law (3.3). Assume that F (z)

has a pole in z0 ∈ F ′Γ ∩ H. The pole lies in at most finite many of the segments

A′iF? ∩H. (Usually z0 lies only in one of those segments. But if z0 lies exactly onthe boundary of the segment, then it might also lay in another segment.) Applyingthe transformation law (3.3) we find that

F(M−1i z

)= v(M−1i

)j(M−1i , z

)kF (z)

has a pole in M−1i z0 ∈ AiF?.

Redoing this argument for each pole of F in F ′Γ ∩H we see that each pole of F

in F ′Γ ∩H corresponds to at most finite many poles of F in FΓ ∩H.

Summarizing, we just showed that finite many poles of F in F ′Γ∩H corresponds

to finite many in FΓ ∩H.3.24 We assume that F has an expansion of the form (3.22) at each cusp qj of thestandard fundamental region FΓ. These expansions converge on some neighborhood

B.3. PROBLEMS OF CHAPTER 3 XXIII

of each cusp qj in H. As in the proof of Lemma 3.30 consider the sets

D1 := D1(y0) := z ∈ H; = (z) ≥ y0Dj := Dj(y0) := Aj D1 := Aj z; = (z) > y0 and

D := D(y0) :=(FΓ ∩H

)r

µ⋂j=1

Dj

where we choose y0 > 0 so large, that each expansion of F of the form (3.22)converge for all z ∈ Dj(y0). Hence there are no poles of F in

⋃j Dj(y0) ∩ FΓ.

The remaining set D in the closure of the fundamental domain is compact.Since F is by assumption a meromorphic function, it is meromorphic on D. Then Fcan have at most finitely many poles in D. (Otherwise, F cannot be meromorphic.)

We just showed that F has at most finite many poles in D and no poles in⋃j Dj(y0) ∩ FΓ. Hence F has at most finite many poles in

D ∪⋃j

Dj(y0) ∩ FΓ = FΓ ∩H

3.31 Choose a M ∈ Γ. Since f is a modular function we have

f(M z

)= f(z)

see Definition 3.26.Hence we have

G(M z

)=

µ∏j=1

(f(M z

)− aj(0)

)=

µ∏j=1

(f(z)− aj(0)

)= G(z).

3.35 Let F ∈ C0(Γ, k, v) as given in the exercise. According to the definition ofcusp forms in Definition 3.25 there exists an 0 ≤ κ′ < 1 and a λ such that F hasan expansion at the cusp ∞ of the form

F (z) =∑

n+κ>0

an e2πinκλ z (z ∈ H).

Put δ := min n∈Zn+κ>0

nκλ which is strictly positive (since F is a cusp form). Then we

have

(B.3) F (z) = e2πiεz∑

n+κ>0

an e2πi(nκλ −δ)z

for all z ∈ H. Using the definition ofκ we can rewrite the expansion on the righthand side as ∑

n+κ>0

an e2πi(nκλ −δ)z =

∑nκλ −δ≥0


which is a one-sided expansion with constant term. Hence, the expansion is boundedon the upper half-plane = (z) ≥ 1. It satisfies∑

n+κ>0

an e2πi(nκλ −δ)z = O (1) as = (z)→∞.

Plugging this estimate into (B.3) we get

|F (z)| =∣∣e2πiδz

∣∣ ∣∣∣∣∣ ∑n+κ>0


∣∣∣∣∣

XXIV B. SOLUTIONS OF SELECTED PROBLEMS

=∣∣e2πiδz

∣∣ O (1)

= O(e2πiδz

)as = (z)→∞.

3.40 As in the proof of Lemma 3.34 we define the function φ : H→ C by

φ(z) := = (z)k2 |F (z)| for all z ∈ H.

Using the transformation law (3.3) for F (Mz) and (1.11) for = (Mz) we have

φ(M z

)== (z)

k2

|j(M, z)|k∣∣v(M) j(M, z)k F (z)

∣∣ = φ(z) for all M ∈ Γ

where we used |v(M)| = 1 for multipliers and (1.11). We just showed that φ isinvariant on Γ.

Next we want to show that φ is bounded on H. To do so, we need to show thatφ(z)→ 0 as z → qj within FΓ ∩H for all cusps qj ∈ CΓ,FΓ

, 1 ≤ j ≤ µ. Recall thatthe expansion of F at the cusp qj is given by

F (z) = σj(z)

∞∑n=nj

aj(n)e2πi(n+κj)A−1j z/λj

in Theorem 3.22 and Definition 3.26 with nj ∈ N and

σj(z) =

1 if qj =∞ and(z − qj

)−kif qj ∈ Q.

Recall that we have qj = Aj∞, see Lemma 2.28. Writing qj =ajbj∈ Q ∪ ∞

with the understanding that gcd(aj , bj) = 1 and aj = 1, bj = 0 if qj = ∞(= 1

0

),

we have

Aj =

(a ?b ?

)and A−1

j =

(? ?−b a

).

Thus we have

=(A−1j z

)=

= (z) if qj =∞ and=(z)

|bjz−a|2if qj ∈ Q.

This leads to

=(A−1j z

) k2 |σj(z)|−1

=

= (z)k2 if qj =∞ and

=(z)k2

|bj |kif qj ∈ Q.


= (z)k2 |σj(z)| =

=(A−1j z

) k2 if qj =∞ and

=(A−1j z

) k2 |bj |k if qj ∈ Q.

First we discuss the cases qj =∞. We have

φ(z) = = (z)k2

∣∣∣∣∣∣∞∑

n=nj

aj(n) e2πi(n+κj)

zλj

∣∣∣∣∣∣= = (z)

k2 e−2π(nj+κj)

=(z)λj

∣∣∣∣∣∣∞∑

n=nj

aj(n) e2πi(n−nj) zλ1

∣∣∣∣∣∣= O

(= (z)

k2

)as z →∞,

B.3. PROBLEMS OF CHAPTER 3 XXV

since nj + κj ≥ 0 may vanish. (F is regular at the cusp qj = ∞.) However, theassumption of negative weight k < 0 implies

φ(z) = O(= (z)

k2

)→ 0 as z →∞.

For any finite cusp qj ∈ Q, we have

φ(z) = = (z)k2 |σj(z)|

∣∣∣∣∣∣∞∑

n=nj

aj(n) e2πi(n+κj)A

−1j

zλj

∣∣∣∣∣∣= |bj |k =

(A−1j z

) k2 e−2π(nj+κj

)=(A−1j

zλj

) ∣∣∣∣∣∣∞∑

n=nj

aj(n) e2πi(n−nj)A−1

jzλj

∣∣∣∣∣∣O(|bj |k =

(A−1j z

) k2

)as z → qj

since nj + κj ≥ 0 may vanish. (F is regular at the cusp qj .) Again, the assumptionof negative weight k < 0 implies

φ(z) = O(|bj |k =

(A−1j z

) k2

)→ 0 as z → qj .

As a result, we have that φ(z) approaches 0 within the fundamental region FΓ asz approaches the cusps. By continuity, there exists a K > 0 such that φ(z) ≤ Kholds for all z ∈ FΓ ∩H. By the invariance of φ(z), we get φ(z) ≤ K for all z ∈ H.

Plugging in the definition of φ we get for F :

|F (z)| ≤ K= (z)− k2 for all z ∈ H.

3.42 Since F and G satisfy the transformation laws F (Mz) = j(M, z)k F (z) andG(Mz) = j(M, z)lG(z) for all M ∈ Γ and z ∈ H we have(

F ·G)(Mz) = F (Mz) · G(Mz)

= j(M, z)k F (z) · G(Mz)

= j(M, z)k F (z) · j(M, z)lG(z)

= j(M, z)k+l(F ·G

)(z).

We just showed F ·G ∈M !k+l

(Γ).

3.45 Using Fubini and z = x+ iy we have∫FΓ(1)

= (z)2

dλ2(z) =

∫ 12

− 12

(∫ ∞√

1−x2

y−2 dy

)dx

=

∫ 12

− 12

[− y−1

]∞√

1−x2

dx

=

∫ 12

− 12

1√1− x2

dx

=

[arcsin(x)

] 12

− 12

=π

6+π

6=π

3.

3.51 Theorem 3.49 shows that 〈·, ·〉Γ,k : Sk,v(Γ)×Sk,v

(Γ)→ C is an inner product

with respect to the space Sk,v(Γ). Hence Sk,v

(Γ)

is a Hilbert space with thePetersson inner product as inner product.

XXVI B. SOLUTIONS OF SELECTED PROBLEMS

Figure 1. The parallelogram par(1, z) spanned by π1(z). Thecircle indicates the Euclidean distance from the parallelogram tothe origin. Its radius is d

(par(1, z), 0

).


4.7 Let z ∈ H. We consider the set πν(z) as defined in (4.3). This set of pointsin C spans the parallelogram par(ν, z), see Figure 1 as an illustration (for the caseν = 1). We see that the parallelogram is always non-degenerate in the sense thatit does not collapse to an line segment on the real line. (z ∈ H implies thatthe horizontal line segments of par(ν, z) – these connect the points ±nuz + µ,µ ∈ −ν, 1 − ν, . . . , ν − 1, ν respectively – do not lie on the real line.) Since 0 /∈par(ν, z) and 0 is not an limit point of the parallelogram by construction we seethat d

(par(ν, z), 0

)> 0 is strictly positive. This shows the right part of (4.5). The

left inequality sign in (4.5) follows from the fact that πν(z) ⊂ par(ν, z) is a subset.Now consider the corner points of the parallelogram par(ν, z). By construction

of πν(z) they are given byνz + ν, νz − ν, −νz + ν and − νz − ν

= ν

z + 1, z − 1, −z + 1 and − z − 1

= ν π1(z).

Hence the spanned parallelograms satisfy

par(ν, z) = ν par(1, z).

This proves (4.6).4.8 Consider an z ∈ Eα. The definition of the set Eα in (4.2) implies

z ∈ Eα ⇐⇒ |z|≥=(z)≥α<(z)≤ 1

α.

This means in particular that the angle β between the real axis and the diagonalline segments of the parallelogram par(1, z) given by

cosβ =< (z)

|z|,

see Figure 1 for an illustration, satisfies

|cosβ| =∣∣∣∣< (z)

|z|

∣∣∣∣ ≤ 1− δ < 1

B.4. PROBLEMS OF CHAPTER 4 XXVII

for some small δ > 0 depending only on α. In other words, the parallelogrampar(1, z) is never degenerate for all z ∈ Eα and it admits at least minimal positivedistance to the origin. This means that

infz∈Eα

d(par(1, z), 0

)=: ρα > 0

holds. Combining this estimate with (4.5) gives

infz∈Eα

d(π(1, z), 0

)≥ infz∈Eα

d(par(1, z), 0

)=: ρα > 0

and proves Estimate (4.9).4.11 The Eisenstein series Gk is given by

Gk(z) =∑

(m,n)

′(mz + n)−k

see Definition 4.9. Since the series converge absolutely, we may reorder the sum-mation terms. We have

Gk(z) =∑

(m,n)

′(mz + n)−k

=∑

(m,n)m=0

′(mz + n)−k +

∑(m,n)m 6=0

′(mz + n)−k

=∑n∈Z6=0

n−k +∑

m∈Z6=0

∑n∈Z

(mz + n)−k.

Taking z → i∞, i.e. = (y)→∞, we see by interchanging limit and series that

limz→i∞

Gk(z) =∑n∈Z6=0

n−k +∑

m∈Z6=0

∑n∈Z

limz→i∞

(mz + n)−k

=∑n∈Z6=0

n−k +∑

m∈Z6=0

∑n∈Z

0

=∑n∈Z6=0

n−k

=

2ζ(k) for even k and

0 for odd k.

Since the series is dominated, see Lemma 4.3, interchanging series and limit isallowed.4.21 We start with the defining series expansion (4.13) of G2. This series expansionconverges absolutely as shown in Lemma 4.19.

Recalling

lim=(z)→∞

eπinz = lim=(z)→∞

e−πn=(z) = 0

for all n ∈ N, we conclude

lim=(z)→∞

G2(z)(4.13)

= lim=(z)→∞

2ζ(2) + 2(2πi)2∞∑n=1

σ1(n) e2πinz

= 2ζ(2) + 2(2πi)2∞∑n=1

σ1(n) lim=(z)→∞

e2πinz︸︷︷︸=0

2ζ(2) + 0.

Using ζ(2) = π2

6 , see e.g. [30, (25.6.1)], we derive (4.20) of the exercise.

XXVIII B. SOLUTIONS OF SELECTED PROBLEMS

To solve the second part, we look at G2(z)−−ζ(2). We have

G2(z)− 2ζ(2)(4.13)

= 2(2πi)2∞∑n=1

σ1(n) e2πinz

Since the above right hand side converges absolutely, we conclude as in Exercise 3.35that there exists a 1 > δ > 0 such that

2(2πi)2∞∑n=1

σ1(n) e2πi(n−δ)z.

still converges absolutely We have

2(2πi)2∞∑n=1

σ1(n) e2πi(n−δ)z = O (1)

as = (z)→∞. This implies

2(2πi)2∞∑n=1

σ1(n) e2πinz = O(e2πiδz

)= O

(e−2πδ=(z)

)as = (z)→∞.4.23 This can be easily seen by going back to Definition 4.9:

Gk(z) =∑

(m,n)

′(mz + n)−k.

Let δ = gcd(m,n) be the greatest common divisor of (m,n) ∈ Z26=(0,0). Note that

Definition 2.3 implies gcd(c, d) ∈ N is positive. We get

Gk(z) = Gk(z) =∑

(m,n)

′(mz + n)−k = 2

∑c∈Z≥0,d∈Zc=0⇒d>0

(cz + d)−k

= 2

∞∑δ=1


δ−k (cz + d)−k

= 2ζ(k)Ek(z).

4.29 Recall the values

ζ(2) =π2

6, ζ(4) =

π4

90and ζ(6) =

π6

945=

π6

33 · 5 · 7,

see e.g. [89, A013661, A013662, A013664]. We calculate easily(90 2ζ(4)

)3 − 2(945 2ζ(6)

)2=

(90 2

π4

90

)3

− 2

(945 · 2 π

6

945

)2

= 0.

Dividing by gcd(903, 2 · 9452) = 36450 we get

20(2ζ(4)

)3 − 49(2ζ(6)

)2= 0

Using Theorem 4.16 shows that the 0th term of expansion of 20G34−49G2

6 vanishes.Combining this with Exercise 3.42 we just showed

20G34 − 49G2

6 ∈ C0(Γ(1), 12, 1

).

4.42 The set of representatives of Γ∞\Γ(1) given in (4.37) can be similarly derivedas the second part of Lemma 4.42: Using Proposition 2.32 to calculate the stabilizerΓ∞ = 〈S,−1〉 of the cusp ∞ of Γ(1) we find

Γ∞

(? ?c d

)⊂ Γ(1)

B.4. PROBLEMS OF CHAPTER 4 XXIX

and (Γ∞

(? ?c d

))∩(

Γ∞

(? ?c′ d′

))= ∅

for all (c, d), (c′, d′) ∈ Z6=(0,0), (c, d) 6= ±(c′, d′).On the other hand, each element of Γ(1), has coprime entries (c, d) in its lower

row. Hence it belongs uniquely to the left coset Γ∞

(? ?c d

)up to a common sign.

This shows that Γ(1) can be written as the disjoint union

Γ(1) = Γ∞

(? ?1 0

)∪

⋃(c,d)

c∈Z, d∈Ngcd(c,d)=1

Γ∞

(? ?c d

)

of left cosets and proves the second part of the lemma.4.60 We can show identity (4.48) easily in two ways: We can compare the definitionsand we can compare the Fourier expansion.

(1) Comparing Definitions 4.22 and 4.43 for Ek and P0,k respectively, we find

P0,k(z)(4.39)

=∑

M∈Γ∞\Γ(1)

j(M, z)−k

(4.40)=

∑(c,d)∈Z2

6=(0,0)

gcd(c,d)=1c=0⇒d=1

(cz + d)−k

(4.1)=

′∑(c,d)∈Z2

gcd(c,d)=1c=0⇒d=1

(cz + d)−k

(4.22)= Ek(z)

for all z ∈ H.(2) Comparing the Fourier expansions of Ek and P0,k in (4.24) and Theo-

rem 4.59 respectively, we get

P0,k(z)4.59= 1 +

∞∑n=1

(2πi)k σk−1(n)

(k − 1)! ζ(k)e2πinz

(4.24)= Ek(z)

for all z ∈ H.

4.64 Using the expansion (4.49) of the Jacobi theta function we have

θ (w, z + 2)(4.49)

=∑n∈Z

eπin2(z+2) e2πinw

=∑n∈Z

eπin2z e2πin︸︷︷︸

=1

e2πinw

=∑n∈Z

eπin2z e2πinw

(4.49)= θ (w, z) .

4.75 Fix a z ∈ H and consider the quotient

f(w) :=θ(w, z)

θ(z, w)

XXX B. SOLUTIONS OF SELECTED PROBLEMS

for all w ∈ C. This quotient is entire since θ(·, z) as well as θ(·, z) are both entireand each simple zero of the denominator is cancelled by a zero of the numerator,see Propositions 4.79 and 4.84.

Next, recall that θ(·, z) as well as θ(·, z) are periodic with periods 1 and z ∈H, see again Propositions 4.79 and 4.84. Hence the the quotient f(w) os alsodouble periodic with periods 1 and z. This implies in particular that f is bounded.Applying Liouville’s theorem, see Appendix A.17, we conclude that f is a constantfunction. Hence there exists a c = c(z) ∈ C such that f(w) = c holds for all w ∈ C.Reformulating the above conclusion we have

θ(w, z) = c(z) θ(z, w)

for all w ∈ C.4.79 We have

θ(z)2 E6.1.15a= Θ(0, z)2

(4.49)=

(∑n∈Z

eπin2z

)2

=

(∑n1∈Z

eπin21z

)(∑n2∈Z

eπin22z

)=

∑n1,n2∈Z

eπi(n21+n2

2)z

=∑

m∈Z≥0

∑n1,n2∈Zn2

1+n22=m

e

πi

n21 + n2

2︸︷︷︸=m

z

=∑

m∈Z≥0

eπimz

∑

n1,n2∈Zn2

1+n22=m

1

︸︷︷︸=r2(m)

(4.67)

=∑

m∈Z≥0

r2(m) eπimz.

This shows (4.69).Analogously, we consider

θ(z)4 E6.1.15a= Θ(0, z)4

(4.49)=

(∑n∈Z

eπin2z

)4

=∑

n1,n2,n3,n4∈Zeπi(n

21+n2

2+n23+n2

4)z

(4.67)=

∑m∈Z≥0

r4(m) eπimz

to show (4.70).

B.4. PROBLEMS OF CHAPTER 4 XXXI

Using r2(0) = r4(0) = 1 and e0 = 1 we also have alternative expressions in(4.69) and (4.70).4.94 Using (4.87) of Lemma 4.92 we find

F2(z + 2)(4.87)

= G2

(z + 2

2

)− 4G2

(2(z + 2)

)= G2

(z2

+ 1)− 4G2

(2z + 4

)(4.17)

= G2

(z2

)− 4G2

(2z)

(4.87)= F2(z)

for all z ∈ H.4.99 Lets adapt the proof of Theorem 4.85 to our situation.

Collecting the transformation properties of F2 in Exercise 4.94 and Proposi-tion 4.95 and similar properties for θ4 deduced from Corollaries 4.69, 4.70 and 4.71,we see that the quotient function

g(z) :=F2(z)


is holomorphic on H (since θ does not vanish – this follows from Corollary 4.77since non of the factors in the product expansion vanishes on H). The function gsatisfies the invariants

g(z + 2) = g(z) and g

(−1

2

)= g(z).

Moreover, the function is bounded at the cusps∞ and 1 since the growth estimatesfor both F2 and θ cancel there. Recalling the Mbius transformation of S2 and T ,see (1.9) and (2.8), and that the theta group Γθ is generated by S2 and T , seeExample 2.36, we can conclude analogously to (the proof of) Theorem 3.18, that gsatisfies

g(M z) = g(z) for all M ∈ Γθ and z ∈ H.Since Γθ ⊂ Γ(1) is a subgroup of finite index – the index is calculated to be 3 in

Exercise 2.36 – we may apply Theorem 3.32 and we conclude that g is a constantfunction. Hence there exists a c ∈ C (independent of z ∈ H) satisfying

F1(z) = c θ2(z)

for all z ∈ H.The constant c has to be −π2: Using lim=(z)→∞ θ4(z) = 1 in (4.58) and

lim=(z)→∞ F2(z) = −π2 in (4.79) we conclude

c =F2(z)


= lim=(z)to∞

F2(z)

θ4(z)

=lim=(z)to∞ F2(z)

lim=(z)to∞ θ4(z)

=π2

1= −π2.

4.97 Consider first the case 4 - n. We have

σ(4-)1 (n)

(4.92)=

∑0<d|n

4-d

d

XXXII B. SOLUTIONS OF SELECTED PROBLEMS

=∑

0<d|n

d

(4.12)= σ1(n)

since 4 | n implies 4 does not divide any divisor of n.Now consider the case 4 | n. We have

σ(4-)1 (n)

(4.92)=

∑0<d|n

4-d

d

=∑

0<d|n

d −∑

0<d|n4|d

d

(4.12)= σ1(n) −

∑0<d|n

4|d

d

4l=d= σ1(n) −

∑0<l|n4

4l

(4.12)= σ1(n) − 4σ1

(n4

).


5.16 Consider the case d | m. Then m bd ∈ Z and hence e2πim b

d = 1. Hence thesummation reduces to

d−1∑b=0

e2πim bd =

d−1∑b=0

1 = d.

Now, consider the case d - m. We consider the expression

e2πimd

d−1∑b=0

e2πim bd =

d−1∑b=0

e2πim b+1d

Since the exponential x 7→ e2πix is periodic, we see that we only have to considerb+ 1 mod d. The map

0, 1, . . . , d− 1 → 0, 1, . . . , d− 1; b 7→ b+ 1 mod d

is bijective. In fact, this map is an permutation on the set of integers between 0and d− 1. This implies that the above sum is just a reordering; we have

d−1∑b=0

e2πim b+1d =

d−1∑b=0

e2πim bd .

This proves the identity

e2πimd

d−1∑b=0

e2πim bd =

d−1∑b=0

e2πim bd .

Since d - m we know e2πimd 6∈ 0, 1. We have now in an equation of the formA · B = B with nontrivial A. This identity can only be solved by B = 0. Hence∑d−1b=0 e

2πim bd = 0 must vanish.

5.19 Let A =

(a bc d

)∈ H(n) and B =

(l mn o

)∈ Γ(1). The definitions in (5.22)

and (2.1) imply that all matrix entries are integers

a, b, c, d, l,m, n, o ∈ Z

B.5. PROBLEMS OF CHAPTER 5 XXXIII

and that the determinants are

detA = and detB = 1.

We immediately see by matrix multiplication

AB =

(a bc d

) (l mn o

)=

(al + bn an+ bocl + bn cm+ do

)∈ Mat2 (Z) .

Since the Determinant is multiplicative, we get

detAB = detA detB = n 1 = n.

Hence AB ∈ H(n).Analogously we show BA ∈ H(n).

5.50 Apply (5.50) with ν = 1.5.58 Under the assumptions of Definition 5.57 we have∣∣a(n)n−s

∣∣ ≤ c nC−<(s) (n ∈ N)

for some constant c > 0. Assuming < (s) > C + 1 there exists a (small) ε > 0 suchthat < (s) ≥ C + ε+ 1 > C + 1 holds. Hence we have∣∣a(n)n−s

∣∣ ≤ c nC−<(s) ≤ n−(1+ε) (n ∈ N).

Hence the series

L(s) =∑n∈N

a(n)n−s

converges absolutely for < (s) > C + 1.5.70 Consider (5.64):

L?(s) =

∫ ∞1

F (iy)(ys−1 − yk−1−s) dy (s ∈ C).

On the other hand L?(k − s) gives

L?(k − s) =

∫ ∞1

F (iy)(yk−s−1 − ys−1

)dy (s ∈ C).

Since both right hand sides differ by a “−”-sign we see immediately the functionalequation

L?(s) = −L?(k − s) (s ∈ C).

5.71

(1) Starting from (5.63) we have

L(s) = (2π)sL?(s)

Γ(s).

Since the Gamma-function does not vanish anywhere we conclude that1

Γ(s) is holomorphic on C. Using that L?(s) is holomorphic, see Lemma 5.68,

we conclude that L(s) extends to a holomorphic function on C.(2) Starting from the functional equation (5.65) and applying (5.63) we find

(2π)k−s

Γ(k − s)L(s)

(5.63)=

(2π)k−s

Γ(k − s)(2π)s

Γ(s)L?(s)

(5.65)=

(2π)k−s

Γ(k − s)(2π)s

Γ(s)(−1)

k2 L?(k − s)

= (−1)k2

(2π)s

Γ(s)

(2π)k−s

Γ(k − s)L?(k − s)

(5.63)= (−1)

k2

(2π)s

Γ(s)L(k − s).

XXXIV B. SOLUTIONS OF SELECTED PROBLEMS

Since k is even we have ik = (−1)k2 . This shows (5.66).


6.4 For M =

(a bc d

)∈ SL2 (R) and z ∈ H and X ∈ C such that j(M,X) 6= 0 we

have

MX −Mz =aX + b

cX + d− az + b

cz + d

=(aX + b)(cz + d)− (az + b)(cX + d)

(cX + d)(cz + d)

=acXz + adX + bcz + bd− acXz − adz − bcX − bd

(cX + d)(cz + d)

=(ad− bc)X − (ad− bc)z

(cX + d)(cz + d)

ad−bc=1=

X − z(cX + d)(cz + d)

.

Using the Definition of the automorphic factor j in (3.4) we find

MX −Mz =X − z

j(M,X) j(M, z).

Taking both sides to the 2− kth power gives (6.2).6.8 For k ∈ 2N consider P (X) := Xk−2 − 1. Obviously, P is an polynomial ofdegree k − 2. By direct calculation we find

P (X) + j(T,X)k−2 P (TX) = Xk−2 − 1 +Xk−2

((−1

X

)k−2

− 1

)= Xk−2 − 1 + 1−Xk−2

= 0

and

P (X) + j(TS,X)k−2 P (TSX) + j((TS)2, X

)k−2P((TS)2X

)= Xk−2 − 1 + (X + 1)k−2

((−1

X + 1

)k−2

− 1

)+Xk−2

((−X − 1

X

)k−2

− 1

)= Xk−2 − 1 + 1− (X + 1)k−2 + (X + 1)k−2 −Xk−2

= 0

using (2.8) with TS =

(0 −11 1

), (TS)2 =

(−1 −11 0

)and (3.4). Hence P satisfies

the defining conditions for a period polynomial in Definition 6.7.6.10 Using the second identity in (6.6) of Proposition 6.9 gives

PF (X) = −k−2∑l=0

(k − 2

l

)X l (−i)k−l−1 L?(k − l − 1).

Now, rewriting the L-function n terms of the L-series, see (5.63) of Lemma 5.68,gives

PF (X)(6.6)= −

k−2∑l=0

(k − 2

l

)X l (−i)k−l−1 L?(k − l − 1)

B.6. PROBLEMS OF CHAPTER 6 XXXV

(5.63)= −

k−2∑l=0

(k − 2

l

)X l (−i)k−l−1 (2π)l+1−k Γ(k − l − 1)L(k − l − 1)

= −k−2∑l=0

(k − 2)!

l! (k − 2− l)!X l (−i)k−l−1 (2π)l+1−k (k − l − 2)!L(k − l − 1)

= −k−2∑l=0

(k − 2)!

l!

L(k − l − 1)

(2πi)k−l−1X l

using(ab

)= a!

b! (a−b)! and and the identity Γ(n) = (n − 1)! for all n ∈ N of the

Gamma function, see Appendix A.11.6.2.1 We follow the same calculation as in the proof of the first part of Lemma 6.21.Let M ∈ Γ(1). Recall that f ∈ C+

(Γ(1), k

)satisfies

j(M, z)−k f(Mw) = f(w) (w ∈ H),

see (3.3).For z ∈ H and using identities (1.12), (3.3) and (6.2) we have

j(M, z)k−2F (Mz)(6.2.2)

= j(M, z)k−2

∫ i

Mz

f(w) (w −Mz)k−2 dw

w 7→Mw= j(M, z)k−2

∫ M−1i

z

f(Mw) (Mw −Mz)k−2 d(Mw)

(1.12)= j(M, z)k−2

∫ M−1i

z

j(M,w)−2f(Mw) (Mw −Mz)k−2 dw

(6.2)=

∫ M−1i

z

j(M,w)−kf(Mw) (w − z)k−2 dw

(3.3)=

∫ M−1i

z

f(w) (w − z)k−2 dw

=

∫ i

z

f(w) (w − z)k−2 dw +

∫ M−1i

i

f(w) (w − z)k−2 dw

(6.16)= F (z) +

∫ M−1i

i

f(w) (w − z)k−2 dw.

Since pM (z) :=∫M−1i

if(w) (w− z)k−2 dw is obviously a polynomial in the variable

z of degree at most k − 2 we have

j(M, z)k−2F (Mz) = F (z) + pM (z).

Hence F satisfies (6.8) and is a Eichler integral by Definition 6.11.

6.16 Let F : H→ C be a holomorphic function, M =

(a bc d

)∈ Γ(1), and k ∈ 2N.

The chain rule gives

d

dz

(j(M, z)k−2 F (Mz)

)=

d

dz

((cz + d)k−2 F

(az + b

cz + d

))= (k − 2)(cz + d)k−3c F

(az + b

cz + d

)+

+ (cz + d)k−2 F ′(az + b

cz + d

)a(cz + d)− c(az + b)

(cz + d)2

XXXVI B. SOLUTIONS OF SELECTED PROBLEMS

= (k − 2)(cz + d)k−3c F

(az + b

cz + d

)+ (cz + d)k−4 F ′

(az + b

cz + d

)using detM = ad− bc = 1. We see that the first term contains the factor (k − 2),which would vanish for k = 2.

Iterating this argument once more, we get

d2

dz2

(j(M, z)k−2 F (Mz)

)=

d

dz

((k − 2)(cz + d)k−3c F

(az + b

cz + d

)+ (cz + d)k−4 F ′

(az + b

cz + d

))= (k − 2)(k − 3)(cz + d)k−4c2 F

(az + b

cz + d

)+

+ (k − 2)(cz + d)k−5c F ′(az + b

cz + d

)+

+ (k − 4)(cz + d)k−5c F ′(az + b

cz + d

)+ (cz + d)k−6 F (2)

(az + b

cz + d

)= (k − 2)(k − 3)(cz + d)k−4c2 F

(az + b

cz + d

)+

+ (2k − 6)(cz + d)k−5c F ′(az + b

cz + d

)+

+ (cz + d)k−6 F (2)

(az + b

cz + d

)We see that the first two terms contain the factor (k − 3), which would vanish fork = 3.

Let’s do one more iterations. We have

d3

dz3

(j(M, z)k−2 F (Mz)

)=

d

dz

((k − 2)(k − 3)(cz + d)k−4c2 F

(az + b

cz + d

)+

+ (2k − 6)(cz + d)k−5c F ′(az + b

cz + d

)+

+ (cz + d)k−6 F (2)

(az + b

cz + d

))

= (k − 2)(k − 3)(k − 4)(cz + d)k−5c3 F

(az + b

cz + d

)+

+ (k − 2)(k − 3)(cz + d)k−6c2 F ′(az + b

cz + d

)+

+ (2k − 6)(k − 5)(cz + d)k−6c2 F ′(az + b

cz + d

)+

+ (2k − 6)(cz + d)k−7c F (2)

(az + b

cz + d

)+

+ (k − 6)(cz + d)k−7c F (2)

(az + b

cz + d

)+ (cz + d)k−8 F (3)

(az + b

cz + d

)= (k − 2)(k − 3)(k − 4)(cz + d)k−5c3 F

(az + b

cz + d

)+

B.6. PROBLEMS OF CHAPTER 6 XXXVII

+ 3(k − 3)(k − 4)(cz + d)k−6c2 F ′(az + b

cz + d

)+ 3(k − 4)(cz + d)k−7c F (2)

(az + b

cz + d

)+

+ (cz + d)k−8 F (3)

(az + b

cz + d

).

We see that each term except the last contains the factor (k − 4), which wouldvanish for k = 4. This illustrates the mechanism.

To prove it, we use induction in the index n ∈ Z≥0 using the function

Hn(z) :=

n∑i=0

(n

i

) n+1∏j=2+i

(k − j)

cn−i (cz + d)k−2−n−i F (i)

(az + b

cz + d

),

where the empty product (with n = i = 0) is defined as∏1j=2(k − j) = 1 and the

empty binomial(

00

)= 1. The notation F i denotes the ith derivative of F ; special

cases are F (0) = F and F (1) = F ′.We see directly that

H0(z) = (cz + d)k−2 F

(az + b

cz + d

)holds. Comparing Hn(z) with above calculated examples we also have

Hn(z) =dn

dzn(j(M, z)k−2 F (Mz)

)for n ∈ 1, 2, 3.

To show the identity for all n ∈ Z≥0 by induction we need the induction stepn→ n+ 1. Consider

d

dzHn(z)

=d

dz

(n∑i=0

(n

i

) 1+n∏j=2+i

(k − j)


(az + b

cz + d

))

=n∑i=0

d

dz

((n

i

) 1+n∏j=2+i

(k − j)


(az + b

cz + d

))

=

n∑i=0

((n

i

) 1+n∏j=2+i

(k − j)

cn−i (k − 2− n− i)(cz + d)k−3−n−ic F (i)

(az + b

cz + d

)+

+

(n

i

) 1+n∏j=2+i

(k − j)

cn−i (cz + d)k−4−n−i F (i+1)

(az + b

cz + d

)),

using the chain rule in the last step. Rewriting the summation index of the lastsummand, we get

d

dzHn(z)

=

n∑i=0

(n

i

) 1+n∏j=2+i

(k − j)


(az + b

cz + d

)+

+

n∑i=0

(n

i

) 1+n∏j=2+i

(k − j)

cn−i (cz + d)k−4−n−i F (i+1)

(az + b

cz + d

)

XXXVIII B. SOLUTIONS OF SELECTED PROBLEMS

=

n∑i=0

(n

i

) 1+n∏j=2+i

(k − j)


(az + b

cz + d

)+

+

n+1∑i=1

(n

i− 1

) 1+n∏j=1+i

(k − j)

cn+1−i (cz + d)k−3−n−i F (i)

(az + b

cz + d

).

Now, we join the summations for the indices 1 ≤ i ≤ n and write the remainingtwo cases i = 0 and i = n+ 1 separately. We get

d

dzHn(z)

=

n∑i=0

(n

i

) 1+n∏j=2+i

(k − j)


(az + b

cz + d

)+

+

n+1∑i=1

(n

i− 1

) 1+n∏j=1+i

(k − j)

cn+1−i (cz + d)k−3−n−i F (i)

(az + b

cz + d

)

=

1+n∏j=2

(k − j)

cn (k − 2− n)(cz + d)k−3−nc F (0)

(az + b

cz + d

)+

+

n∑i=1

(n

i

) 1+n∏j=2+i

(k − j)


(az + b

cz + d

)+

+

n∑i=1

(n

i− 1

) 1+n∏j=1+i

(k − j)

cn+1−i (cz + d)k−n−3−i F (i)

(az + b

cz + d

)+

+ (cz + d)k−3−n−(n+1) F (n+1)

(az + b

cz + d

)

=

2+n∏j=2

(k − j)

cn+1 (cz + d)k−3−n F (0)

(az + b

cz + d

)+

+

n∑i=1

[(n

i

)(k − 2− n− i) +

(n

i− 1

)(k − 1− i)

]·

·

1+n∏j=2+i

(k − j)

cn+1−i (cz + d)k−3−n−i F (i)

(az + b

cz + d

)+

+ (cz + d)k−4−2n F (n+1)

(az + b

cz + d

).

We consider the middle term separately. Using the relations(n

i

)=

n!

i! (n− i)!and

(n

i

)+

(n

i− 1

)=

(n+ 1

i

)for binomial coefficients we get(

n

i

)(k − 2− n− i) +

(n

i− 1

)(k − 1− i)

=

((n

i

)+

(n

i− 1

))(k − 2− n)−

(n

i

)i+

(n

i− 1

)(n+ 1− i)

B.6. PROBLEMS OF CHAPTER 6 XXXIX

=

(n+ 1

i

)(k − 2− n)− n!

i! (n− i)!i+

n!

(i− 1)! (n+ 1− i)!(n+ 1− i)

=

(n+ 1

i

)(k − 2− n)− n!

(i− 1)! (n− i)!i+

n!

(i− 1)! (n− i)!

=

(n+ 1

i

)(k − 2− n).

With this identity we conclude

d

dzHn(z)

=

2+n∏j=2

(k − j)

cn+1 (cz + d)k−3−n F (0)

(az + b

cz + d

)+

+

n∑i=1

[(n

i

)(k − 2− n− i) +

(n

i− 1

)(k − 1− i)

]·

·

1+n∏j=2+i

(k − j)

cn+1−i (cz + d)k−3−n−i F (i)

(az + b

cz + d

)+

+ (cz + d)k−4−2n F (n+1)

(az + b

cz + d

)

=

2+n∏j=2

(k − j)

cn+1 (cz + d)k−3−n F (0)

(az + b

cz + d

)+

+

n∑i=1

2+n∏j=2+i

(k − j)

cn+1−i (cz + d)k−3−n−i F (i)

(az + b

cz + d

)+

+ (cz + d)k−4−2n F (n+1)

(az + b

cz + d

)= Hn+1(z).

Hence we have shown the relation

d

dzHn(z) = Hn+1(z)

which concludes the induction step n→ n+ 1. In particular, we have shown

dn+1

dzn+1H0(z) =

dn

dzn

(d

dzH0(z)

)=

dn

dznH1(z)

= . . . = Hn+1(z)

for all n ∈ Z≥0.

To finally prove Bel’s identity, consider n = k − 1. We have

dk−1

dzk−1H0(z) = Hk − 1(z).

Plugging in the definition of Hk−1(z) we have

dk−1

dzk−1

((cz + d)k−2 F

(az + b

cz + d

))

=dk−1

dzk−1H0(z)

= Hk−1(z)

XL B. SOLUTIONS OF SELECTED PROBLEMS

=

k−1∑i=0

(k − 1

i

) k∏j=2+i

(k − j)

ck−1−i (cz + d)−1−i F (i)

(az + b

cz + d

)

We observe that most terms in the right hand sum contain the multiplicative factor(k−j) with j runs from 2+ i to k. This terms vanish since k−k = 0 appears in thismultiplicative factor. The only surviving term is the term with i = k−1, since then

the product∏kj=2+i(k − j) =

∏kj=k+1(k − j) is the empty product and the empty

product was defined as 1, see the line below the definition of Hn(z). Summarizingthis arguments, we have

dk−1

dzk−1

((cz + d)k−2 F

(az + b

cz + d

))

=

k−1∑i=0

(k − 1

i

) k∏j=2+i

(k − j)

ck−1−i (cz + d)−1−i F (i)

(az + b

cz + d

)

=

k−2∑i=0

(k − 1

i

)k∏

j=2+i

(k − j)︸︷︷︸=0

ck−1−i (cz + d)−1−i F (i)

(az + b

cz + d

)+

+ (cz + d)−k F (k−1)

(az + b

cz + d

)= (cz + d)−k F (k−1)

(az + b

cz + d

).

This proves Bel’s identity (6.12).6.21 Let M ∈ Γ(1). Recall that f ∈ C0

(Γ(1), k

)satisfies

j(M, z)−k f(Mw) = f(w) (w ∈ H),


j(M, z)k−2F (Mz)(6.16)

= j(M, z)k−2

∫ i∞

Mz


w 7→Mw= j(M, z)k−2

∫ M−1i∞

z


(1.12)= j(M, z)k−2

∫ M−1i∞

z


(6.2)=

∫ M−1i∞

z


(3.3)=

∫ M−1i∞

z

f(w) (w − z)k−2 dw

=

∫ i∞

z

f(w) (w − z)k−2 dw +

∫ M−1i∞


(6.16)= F (z) +

∫ M−1i∞

i∞f(w) (w − z)k−2 dw.

B.6. PROBLEMS OF CHAPTER 6 XLI

Since pM (z) :=∫M−1i∞i∞ f(w) (w−z)k−2 dw is obviously a polynomial in the variable


(B.4) j(M, z)k−2F (Mz) = F (z) + pM (z).

Hence F satisfies (6.8) and is a Eichler integral by Definition 6.11.

6.22 Let f ∈ C0(Γ(1), k

). Then F is given by (6.16). Applying T we get, as in the

proof of the first part of Lemma 6.21,

F∣∣2−kT (z)

(6.16)= j(T, z)k−2

∫ i∞

Tz

f(w) (w − Tz)k−2 dw

w 7→Tw= j(T, z)k−2

∫ T−1i∞

z

f(Tw) (Tw − Tz)k−2 d(Tw)

(1.12),(6.2)=

∫ T−1i∞

z

j(T,w)−kf(Tw) (w − z)k−2 dw

(3.3)=

∫ T−1i∞

z

f(w) (w − z)k−2 dw

=

∫ i∞

z

f(w) (w − z)k−2 dw +

∫ T−1i∞


(6.16),T−1i∞=0= F (z) +

∫ 0


(6.1)= F (z)− Pf (z).

Similarly, we find for applying the translation S

F∣∣2−kS(z)

(6.16)= j(S, z)k−2

∫ i∞

Sz

f(w) (w − Sz)k−2 dw

w 7→Sw= j(S, z)k−2

∫ S−1i∞

z

f(Sw) (Sw − Sz)k−2 d(Sw)

(1.12),(6.2)=

∫ S−1i∞

z

j(S,w)−kf(Sw) (w − z)k−2 dw

(3.3)=

∫ S−1i∞

z

f(w) (w − z)k−2 dw

=

∫ i∞

z

f(w) (w − z)k−2 dw +

∫ S−1i∞


(6.16),S−1i∞=i∞= F (z) +

∫ i∞


= F (z).

6.24 Consider a cusp form f ∈ Sk(Γ(1)

)with Fourier expansion

f(z) =

∞∑n=1

a(n) e2πinz

for all z ∈ H, see Theorem 3.36 (with v = 1 and κ = 0). Using the set inclusion

f ∈ Sk(Γ(1)

)⊂ S!

k

(Γ(1)

)⊂M !

k

(Γ(1)

),

we find that the formal Eichler integral Ef of f in (6.23) can be written as

Ef (z) :=

∞∑n=1

a(n)n1−k e2πinz

XLII B. SOLUTIONS OF SELECTED PROBLEMS

for all z ∈ H.On the the hand, we consider the (k − 1)-fold antiderivative F of f in Defini-

tion 6.15. We have

F (z) = (2πi)1−k∞∑n=1

a(n)n1−k e2πinz

for all z ∈ H. This shows that Ef can be expressed in terms of the (k − 1)-foldantiderivative F of f :

Ef (z) = (2πi)k−1 F (z).

Next, apply (6.17) of Lemma 6.21. We have

Ef (z) = (2πi)k−1 F (z)

(6.17)= − (2πi)k−1

(k − 2)!F (z)

= − (2πi)k−1

Γ(k − 1)F (z),

using Γ(n) = (n− 1)!, see Appendix A.11. With Definition 6.20 we find

Ef (z) = − (2πi)k−1

Γ(k − 1)F (z)

(6.16)= − (2πi)k−1

Γ(k − 1)

∫ i∞

z

f(w) (w − z)k−2 dw.

This proves (6.20).6.26 We use the same arguments as in the proof of the first part of Lemma 6.21.

Let M ∈ Γ(1). Recall that f ∈ C+(Γ(1), k

)satisfies

j(M, z)−k f(Mw) = f(w) (w ∈ H),


j(M, z)k−2F (Mz)(6.21)

= j(M, z)k−2

∫ i

Mz


w 7→Mw= j(M, z)k−2

∫ M−1i

z


(1.12)= j(M, z)k−2

∫ M−1i

z


(6.2)=

∫ M−1i

z


(3.3)=

∫ M−1i

z

f(w) (w − z)k−2 dw

=

∫ i

z

f(w) (w − z)k−2 dw +

∫ M−1i

i

f(w) (w − z)k−2 dw

(6.21)= F (z) +

∫ M−1i

i

f(w) (w − z)k−2 dw.

Since pM (z) :=∫M−1i

if(w) (w− z)k−2 dw is obviously a polynomial in the variable


j(M, z)k−2F (Mz) = F (z) + pM (z).

B.6. PROBLEMS OF CHAPTER 6 XLIII

Hence F satisfies (6.8) and is a Eichler integral by Definition 6.11.6.28 We have

F (z)

(6.16)∫ i∞

z

f(w) (w − z)k−2 dw

=

∫ i

z

f(w) (w − z)k−2 dw +

∫ i∞

i

f(w) (w − z)k−2 dw

(6.21)= F (z) + +

∫ i∞

i

f(w) (w − z)k−2 dw.

The term∫ i∞i

f(w) (w− z)k−2 dw ∈ C[z]k−2 is a polynomial in z of degree at mostk − 2 by construction.6.31 To show the transformation property of G− consider a M ∈ Γ(1). We have

G−∣∣2−kM(z) = j(M, z)k−2 G−(Mz)

(6.27)= j(M, z)k−2

∫ i

Mz

g(w)(w −Mz

)k−2dw

= j(M, z)k−2

∫ M−1i

z

g(Mw)(Mw −Mz

)k−2d(Mw)

(6.2),(1.12)=

∫ M−1i

z

j(M,w)−k g(Mw)(w − z

)k−2dw

(3.3)=

∫ M−1i

z

g(w)(w − z

)k−2dw

=

∫ i

z

g(w)(w − z

)k−2dw +

∫ M−1i

i

g(w)(w − z

)k−2dw

(6.27),(6.29)= G−(z) + PM (z)

for all z ∈ H. Expanding the binomial in the definition of PM (z) we find

PM (z)(6.29)

=

∫ M−1i

i

g(w)(w − z

)k−2dw

=

k−2∑l=0

(k − 2

l

)(−1)k−2−l

(∫ M−1i

i

g(w)wl zk−2−l dw

)

=

k−2∑l=0

(k − 2

l

)(−z)k−2−l

∫ M−1i

i

g(w)wl dw.

Hence PM (z) ∈ C[z]k−2 is a polynomial in variable z of degree at most k − 2.6.33 We have

∂

∂zG−(z)

(6.27)=

∂

∂z

∫ i

z

g(w)(w − z

)k−2dw

(6.30)=

∂

∂z

∫ i

z

g(w)(w − z

)k−2dw

= −g(z)(z − z

)k−2

= g(z)((z − z

)k−2).

6.34

XLIV B. SOLUTIONS OF SELECTED PROBLEMS

(1) We follow similar arguments as in the solution of Exercise 6.31. To show

the transformation property of G− consider a M ∈ Γ(1). We have

G− |2−kM(z) = j(M, z)k−2 G−(Mz)

(6.32)= j(M, z)k−2

∫ i∞

Mz

g(w)(w −Mz

)k−2dw

= j(M, z)k−2

∫ M−1i∞

z

g(Mw)(Mw −Mz

)k−2d(Mw)

(6.2),(1.12)=

∫ M−1i∞

z

j(M,w)−k g(Mw)(w − z

)k−2dw

(3.3)=

∫ M−1i∞

z

g(w)(w − z

)k−2dw

=

∫ i∞

z

g(w)(w − z

)k−2dw +

∫ M−1i

i∞g(w)

(w − z

)k−2dw

(6.32),(6.36)= G−(z) + PM (z)

for all z ∈ H. Expanding the binomial in the definition of PM (z) we find

PM (z)(6.36)

=

∫ M−1i∞

i∞g(w)

(w − z

)k−2dw

=

k−2∑l=0

(k − 2

l

)(−1)k−2−l

(∫ M−1i∞

i∞g(w)wl zk−2−l dw

)

=

k−2∑l=0

(k − 2

l

)(−z)k−2−l

∫ M−1i∞

i∞g(w)wl dw.

Hence PM (z) ∈ C[z]k−2 is a polynomial in variable z of degree at mostk − 2.

(2) Similar to the solution of Exercise 6.28 we have

G−(z)(6.32)

=

∫ i∞

z

g(w)(w − z

)k−2dw

=

∫ i

z

g(w)(w − z

)k−2dw +

∫ i∞

i

g(w)(w − z

)k−2dw

(6.32)= G−(z) +

∫ i∞

i

f(w)(w − z

)k−2dw.

The term∫ i∞

i

f(w)(w − z

)k−2dw =

k−2∑l=0

(k − 2

l

)(−z)k−2−l

∫ i∞

i

f(w)wl dw ∈ C[z]k−2

is a polynomial in z of degree at most k−2 by construction. This concludesthe second part.

6.36 Let(PM)M∈Γ(1)

∈ B1(Γ(1),C[X]k−2

)be a coboundary. By Defininition 6.35

there exists a fixed polynomial P ∈ C[X]k−2 such that each PM can be written as

PM = P∣∣2−kM − P for all M ∈ Γ(1).

Let M,V ∈ Γ(1) be two matrices. Using the above relation, we have

PMV = P∣∣2−kMV − P

B.6. PROBLEMS OF CHAPTER 6 XLV

= P∣∣2−kMV − P

∣∣2−kV + P

∣∣2−kV − P

=(P∣∣2−kM − P

)∣∣2−kV + P

∣∣2−kV − P

= PM∣∣2−kV + PV .

This shows that the coboundary(PM)M∈Γ(1)

satieties (6.37). Hence(PM)M∈Γ(1)

∈Z1(Γ(1),C[X]k−2

)is a cocycle.

6.37 Since 1 = 11 and 1 = (−1)(−1) we conclude

P1 = P11 = P1

∣∣2−k1 + P1

and

P1 = P(−1)(−1) = P−1∣∣2−k(−1) + P−1

using (6.37). The first identity implies P1 = 2P1 which in turn implies P1 = 0.Now, the second identity implies

0 = P−1∣∣2−k(−1) + P−1 = 2P−1.

Again, we conclude P−1 = 0.The same argumentation, now applied to the element −1 = (±T )2 gives

0 = P−1(6.37)

= P±T∣∣2−k(±T ) + P±T .

6.38 Using the cocycle relation (6.37) and the left identity of (6.38) we get

0 = P1(6.38)

= PMM−1

(6.37)= PM

∣∣2−kM

−1 + PM−1 .

6.41 Let(PM)M∈Γ(1)

∈ Z1par

(Γ(1),C[X]k−2

)be a parabolic cocycle. If V ∈ Γ(1)

is parabolic then trace (V ) = 2, see Definition 1.11. Following the arguments ofthe parabolic case in the proof of Proposition 1.17 we see that V is conjugatedto a parabolic element fixing the cusp ∞: There exists an M ∈ Γ(1) such thatMVM−1 ∈ Γ(1)∞ (where the stabilizer subgroup Γ(1)∞ is given in Definition 1.30).Following Proposition 2.32 we have Γ(1)∞ = 〈S,−1〉. Hence we have

V = (±1)MSnM−1 for suitable M ∈ Γ(1) and n ∈ N.

Applying (6.37) to the polynomial PV we get

PV = P(±1)MSnM−1

(6.37)= P±1

∣∣2−kMSnM−1 + PMSnM−1

(6.9)= PMSnM−1 ,

since Exercise 6.37 implies P±1 = 0.Hence we may assume that V can be written as

V = MSnM−1.

(Because if we show the result for PMSnM−1 then we have it also for P(±1)MSnM−1 =PMSnM−1 .) We apply again the cocycle relation (6.37) and find

PV = PMSnM−1

(6.37)= PMSn

∣∣2−kM

−1 + PM−1

= . . .

(6.37)= PM

∣∣2−kS

nM−1 + PSn∣∣2−kM

−1 + PM−1 .

Using again (6.37) we reduce PSn to a sum over PS∣∣2−kS

·. We get

PV = PM∣∣2−kS

nM−1 + PM−1 + PSn∣∣2−kM

−1

(6.37)= PM

∣∣2−kS

nM−1 + PM−1 + PS∣∣2−kS

n−1M−1 + . . .+ PS∣∣2−kSM

−1 + PS∣∣2−kM

−1

XLVI B. SOLUTIONS OF SELECTED PROBLEMS

= PM∣∣2−kS

nM−1 + PM−1 +

n−1∑i=0

PS∣∣2−kS

iM−1.

Now, we use that(PM)M∈Γ(1)

is a parabolic cocycle. Definition 6.40 implies that

there exists a polynomial Q ∈ C[X]k−2 such that PS is given by

PS(6.41)

= Q∣∣2−kS −Q.

Hence PV can be written as

PV = PM∣∣2−kS

nM−1 + PM−1 +

n−1∑i=0

PS∣∣2−kS

iM−1

= PM∣∣2−kS

nM−1 + PM−1 +

n−1∑i=0

(Q∣∣2−kS −Q

)∣∣2−kS

iM−1

= PM∣∣2−kS

nM−1 + PM−1 +

n−1∑i=0

(Q∣∣2−kS

i+1M−1 −Q∣∣2−kS

iM−1).

Expanding the telescoping sum, we find

PV = PM∣∣2−kS

nM−1 + PM−1 +

n−1∑i=0

(Q∣∣2−kS

i+1M−1 −Q∣∣2−kS

iM−1)

= PM∣∣2−kS

nM−1 + PM−1 +Q∣∣2−kS

nM−1 −Q∣∣2−kM

−1

= (PM +Q)∣∣2−kS

nM−1 + PM−1 −Q∣∣2−kM

−1.

Now, we use Exercise 6.38 and write PM−1 as

PM−1 = −PM∣∣2−kM

−1.

Expanding the telescoping sum, we find

PV = (PM +Q)∣∣2−kS

nM−1 + PM−1 −Q∣∣2−kM

−1

(6.39)= (PM +Q)

∣∣2−kS

nM−1 − PM∣∣2−kM

−1 −Q∣∣2−kM

−1

= (PM +Q)∣∣2−kS

nM−1 − (PM +Q)∣∣2−kM

−1

= (PM +Q)∣∣2−kM

−1MSnM−1 − (PM +Q)∣∣2−kM

−1

= (PM +Q)∣∣2−kM

−1∣∣2−kMSnM−1 − (PM +Q)

∣∣2−kM

−1

= (PM +Q)∣∣2−kM

−1∣∣2−kV − (PM +Q)

∣∣2−kM

−1.

If we defineQV := (PM +Q)

∣∣2−kM

−1,

which is a polynomial in C[X]k−2, we can write above relation as

PV = QV∣∣2−kV −QV .

6.46 For g ∈ C0(Γ(1), k

)write the image of β(f) =

(PM)M∈Γ(1)

as given by

transformation property QM(6.28)

= F∣∣2−kM − F of the Niebour integral G− defined

in (6.27). Using Exercise 6.34 we see

QM(6.28)

= G−∣∣2−kM − G

−

(6.35)=

(G− − P

)∣∣2−kM −

(G− − P

)=

(F∣∣2−kM − F

)+

((−P )

∣∣2−kM − (−P )

)

B.6. PROBLEMS OF CHAPTER 6 XLVII

with P ∈ C[z]k−2 given by (6.36).Using the fixed point condition S i∞ = i∞ we have

G−∣∣2−kS(z) = G−(z) +

∫ S−1i∞

i∞f(w)

(w − z

)k−2dw

= G−(z) +

∫ i∞

i∞f(w)

(w − z

)k−2dw︸︷︷︸

=0

= G−(z).

following the proof of the first part of Lemma 6.21 for the parabolic element S =(1 10 1

). This implies that the sequence

(QM

)M∈Γ(1)

is a parabolic cocycle in

Z1par

(Γ(1),C[X]k−2

). The map β maps the cuspidal space C0

(Γ(1), k

)into the

parabolic cohomology H1par

(Γ(1),C[X]k−2

).

6.62 Consider A =

(a bc d

).

Similar to the calculation in Exercise 1.6 we have

d(Az) = d

(az + b

cz + d

)=

(cz + d)a− (az + b)c

(cz + d)2dz

=ad− bc

(cz + d)2dz = det(A)

dz

j(A, z)2

and (Aw −Az

)=

(aw + b

cw + d− az + b

cz + d

)=

(aw + b)(cz + d)− (cw + d)(az + b)

(cw + d) (cz + d)

=(ad− bc)(w − z)j(A,w) j(A, z)

= det(A)w − z

j(A,w) j(A, z).

Taking the last identity to the kth power concludes the proof.6.71 We show the statement of the exercise in three steps, α(An) ⊂ Bn, β(Bn) ⊂ Anand that the maps are inverse to each other.

α(An) ⊂ Bn: Now, start again with an A =

(a bc d

)∈ An. We have

α(A) =

(c d

−(a− c

[ac

])−b+ d

[ac

])Since A ∈ An we know that a > c > 0. Using the definition of [·] in (4.30)we have

a

c− 1 ≤

[ac

]≤ a

c

⇐⇒ c = a− c(ac− 1)≥ a− c

[ac

]≥ a− ca

c= 0.

Again, since A ∈ An we know that d > −b ≥ 0. In particular thisshows that the upper right entry and lower right entry of α(A) is strictlypositive.

XLVIII B. SOLUTIONS OF SELECTED PROBLEMS

To show that the lower right entry is strictly larger than the upperright entry of α(A) we consider

−b+ d[ac

]> d ⇐⇒ d

[ac

]> d+ b.

Since a > c implies[ac

]≥ 1 and −b ≥ 0 implies d+ b ≤ d we conclude

−b+ d[ac

]≥ d

holds. If b 6= 0 the the above inequality is strict (i.e., “>” holds). Forb = 0 the conditions on the set An require a ≥ 2c. This implies

[ac

]≥ 2

and the above inequality is again strict.To show the remaining condition on elements of the set cBn assume

that the lower left entry of α(A) vanishes. This means, we have

−(a− c

[ac

])= 0 ⇐⇒

[ac

]=a

c,

i.e., a is a multiple of c. Together with the assumption a > c we concludethat

[ac

]≥ 2. Hence the lower right entry of α(A) satisfies

−b+ d[ac

]≥ −b+ 2d ≥ 2d

since −b is nonnegative by assumption on A ∈ cAn. This shows thevalidity remaining condition of Bn on the image α(A). Hence we haveshown

α(A) ∈ Bn for all A ∈ cAn.β(Bn) ⊂ An: This case can be shown by similar arguments as the previous

case.

α and β are inverse to each other: Consider an element A =

(a bc d

).

Applying the map α on A 7→ TS−[ ac ]A gives

α(A) = TS−[ ac ]A =

(c d

−(a− c

[ac

])−b+ d

[ac

]) .Then applying β on α(A) gives

β(α(A)

)= β

((c d

−(a− c

[ac

])−b+ d

[ac

]))

=

(a− c [ac ])+ c

[−b+d[ ac ]

d

]−(−b+ d

[ac

])− d

[−b+d[ ac ]

d

]c d

=

(a− c

[ac

]+ c

[ac

]+ c

[−bd

]b+ d

[ac

]− d

[−bd

]− d

[ac

]c d

)=

(a+ c

[−bd

]b− d

[−bd

]c d

).

If we assume A ∈ An, then 0 ≤ −b < d by definition of An. This implies[−bd

]= 0. Hence A 7→ β

(α(A)

)is the identity map on An. A similar

calculation shows that the map B 7→ α(β(B)

)is the identity map on Bn.

We just have shown that α : An → Bn and β : Bn → An are inverse to each otherwhich implies that the sets An and cBn are bijective.

Acronyms

N: Set of natural numbers N = 1, 2, 3, . . .Z: Set of integers Z = . . . ,−2,−1, 0, 1, 2, . . .Q: Set of rational numbersR: Set of real numbersC: Set of complex numbersC[X]k: Set of complex polynomials in formal variable X of degree ≤ kSL2 (·): Set/group of 2× 2 matrices with determinant 1 and entries in ·GL2 (·): Set/group of 2× 2 invertible matrices with entries in ·Mat2 (·): Group of 2× 2 matrices with entries in ·trace (V ): The trace of matrix Vxt: Transpose of a matrix or vector x< (z): Real part of a complex variable z= (z): Imaginary part of a complex variable zsign (a): Sign function of real argument a|a|: Absolute value of complex argument a

XLIX

Index

(k − 1)-antiderivative, 181

−1, 4

A ∼ B, 136

A, 391

B, 391

B1(Γ(1),C[X]k−2

), 188

EMaassν , 252

Eα, 67

Ek, 79

Fn, 374

G2, 74

Gk, 69

H(n), 135

H+k , 434

H1(Γ(1),C[X]k−2

), 188

H1par

(Γ(1),C[X]k−2

), 188

Hk, 434

K(m,n; c), 94

L-function, 164

completed, 164

L-series, 161, 441

Dirichlet series, 161

Euler product, 172

of(a(n)

)n

, 161

of cusp forms, 161

of weakly holomorphic modular forms,

441

L?(s), 164

Lz0,z1 , 378

M !k,v(Γ), 55

M !k(Γ), 55

Mk,v(Γ), 55

MMaassk,v

(Γ, ν

), 242

Mk(Γ), 55

MMaassk

(Γ, ν

), 242

Pf , 434

Pm,k(·), 89

S, 26, 203

S!k,v(Γ), 55

S′, 359

Sk,v(Γ), 55

Hilbert space, 66

SMaassk,v (Γ, ν)

Hilbert space, 259

SMaassk,v

(Γ, ν

), 242

Sk(Γ), 55

SMaassk

(Γ, ν

), 242

T , 26, 203

TnHecke operator, 132

matrix representation, 203

Z1(Γ(1),C[X]k−2

), 188

Z1par

(Γ(1),C[X]k−2

), 188

ZS, 431

[A]Γ, 136

[·, ·], 9

[·], 85

∆, 79

∆0, 227

Friedrichs extension, 266

∆k, 227

E±k , 227

GL2 (C), 3

GL2 (R), 3

GL2 (Z), 391

Γ\H(n), 136

Γ(1), 25

Γ(N), 25

Γ(s), 165

Γ0(N), 26

Γ1(N), 26

Γ0(N), 26

Γ1(N), 26

Γθ, 41

Γz , 13

C, 3

C[X]k−2, 175

C′, 334, 359

H, 3

H−, 3

H?, 378

Mk,ν(y), 234

N, 3

Pk−2, 177

Q, 3

R, 3

S, 4

T, 20

U, 19

U, XV

Wk,ν(y), 234

Z, 3

Mat2 (R), 4

Mat2 (Z), 4, 135

SL2 (R), 3

SL2 (Z), 3, 25

LI

LII INDEX

f, g±

, 351

A · B, 141

A ≡ B mod C, 141

A, 408

A×l, 141

CF , 218

CG, 220

CΓ, 37

CΓ,FΓ, 37

Ef , 182

F?, 31

FΓ, 29

F , 29

FGL2(Z), 394

FΓ(1), 32

Ls, 412, 413

M(f), 161

M−1(ϕ), 162, 167

R?, 127

Rn, 128

T (n), 136

χ, 48

cosh, 115

detV , 4

η, 47, 254

η-multiplier, 48

ηk, 352

gcd(c, d), 26

F , 184

G−, 186

1, 4

= (·), 3

〈·, ·〉Γ,k, 62

d·e, 401

b·c, 388

Lim (Γ), 11

Mk,ν(y), XCI

µ-function, 94

⊕, 193

Ord (Γ), 11

ψk,ν(y), 277

< (·), 3

σ(4-)1 , 122

σk(n), 74

sign (·), 235∑′(m,n), 67

τ(n), 80, 81

LNS(q), 376

LN, 375

lev, 374

G−, 187

Tnmatrix representation, 203

Tp, 131

∆k, 433

C′, 421

Ls, 418

trace (V ), 6

ζ(·, ·), 415

ζ(s), 69, 94

ζR, 431

fG, 388

j(M, z), 44

ju(M, z), 225

l(M), 29

r(f ; ·), 444

r(y), 411

r2, 113

r4, 113

rk, 124

v(M), 44, 46

vk, 48

C, 391

abelian integral, 83

antiderivative, 83

(k − 1)-fold, 181

arithmetic divisor function, 74

Artin billiard, 396

automorphic factor, 44

unitary , 225

base points, 399

Bol’s identity, 181, 438

ceiling function, 401

closed geodesics, 408

coboundary, 188

cocycle, 188

parabolic, 188

commutator brackets, 9

completely multiplicative sequence, 173

congruence subgroup, 25

consistency condition, 46

continued fraction, 385

convergent, 385

finite, 385

infinite, 385

period length, 386

periodic, 386

prime period length, 386

prime periodic, 386

purely periodic, 386

regular, 385

convergent, 385

cosinus hyperbolicus, 115

curve, 218, 220

elliptic —

preliminary definition, 219

in H, 396

smooth, 218, 220

smooth at a point, 218, 220

cusp, 37

of Γ in FΓ, 37

meromorphic at a cusp, 54

pole at a cusp, 54

regular at a cusp, 54

width of, 41

cusp form, 55

cut-plane, 334, 359, 421

Dedekind η-function, 47, 254

differential formsf, g±

, 351

ηk, 352

Maass-Selberg form, 352

INDEX LIII

differential operator

Lk, 433

Rk, 433

Laplace operator

(weak harmonic Maass wave forms),

433

Maass operator, 227


433

raising an lowering —


433

yperbolic Laplace operator, 227

of real weight, 227

Dirichlet character, 48

discontinuous, 11

discrete group, 10

discriminant function, 47, 79

dynamical zeta function, 431

Eichler cohomology

B1(Γ(1),C[X]k−2

), 188

H1(Γ(1),C[X]k−2

), 188

H1par

(Γ(1),C[X]k−2

), 188

Z1(Γ(1),C[X]k−2

), 188

Z1par

(Γ(1),C[X]k−2

), 188

— Shimura isomorphism, 193

coboundary, 188

cocycle, 188

group, 188

parabolic cocycle, 188

theorem, 193

Eichler cohomology group, 188

parabolic, 188

Eichler integral, 179

Eichler length, 29

Eichler Shimura isomorphism, 193

Eisenstein series, 69

nonholomorphic, 252

normalized, 79

weight 2, 74

elliptic curve

preliminary definition, 219

elliptic matrix, 6

equation

Pell’s —, 390

equivalent points, 37

Euler product, 172

family of Maass cusp forms, 274

admissible, 274

range of admissible weights, 274

reference weight, 275

trivial, 274

Farey sequence, 374

level function, 374

fixed point, 6

floor function, 388

Ford region, 31

form

cusp form, 55

discriminant function, 79

entire modular form, 55

Maass wave —

weakly harmonic, 434

Maass wave form, 242

mock modular, 435

modular —, 44, 52

modular function, 55

weakly holomorphic modular form, 54

formal Eichler integral, 182

Fourier transformation, VIII

Fourier-Whittaker expansion, 236, 238, 239


Fuchsion group of the first kind, 16

full modular group, 25

function

(k − 1)-antiderivative, 181

Mk,ν(y), VI, 233

Wk,ν(y), VI, 233

Mk,ν(y), 234

Wk,ν(y), 234

floor, 388

Mk,ν(y), 234

µ-function, 94

ψk,ν(y), 277

ζ(·, ·), 415

fG, 388

r2, 113

r4, 113

rk, 124

abelian integral, 83

antiderivative, 83

arithmetic divisor —, 74

automorphic factor, 44

unitary, 225

Buchholts function, 234

ceiling —, 401

cosinus hyperbolicus, 115

counting function, 124

cusp form, 55

Dedekind η-function, 47, 254


discriminant function, 47, 79

divisor –

arithmetic, 74

divisor —

modified, 122

Eichler integral, 179

formal, 182


entire modular form, 55

Eucledian distance, 68

floor, 388

Gamma function, 165

Gauss bracket, 388

Gauß brackets, 85

Gau —, 388

generalized hypergeometric –, VII

Hurwitz ζ- —, 415

hyperbolic distance, 62

Jacobi theta function, 103

Kloosterman sum, 94

Kronecker δ, 98

left neighbor map, 375

level —, 374

LIV INDEX

logarithmic derivative, 80

Mobius function, 94

Maass cusp form, 242

Maass wave form, 242



mock modular form, 435

modified divisor function, 122

modular form, 44

modular function, 43, 55

multiplier, 44, 46

multiplier system, 44, 46

nearly periodic, 323

nonholomorphic Eisenstein series, 252

normalized Eisenstein series, 79

period —, 359, 444

Poincare series, 89

pole at a cusp, 54

polynomial

cubic, 218

homogeneous cubic, 219

Ramanujan τ function, 81

real-analytic, 233

recurrence time, 411


Riemann ζ-function, 69, 94

sign, 235

theta function

Jacobi, 103

unitary automorphic factor, 225

weakly holomorphic modular form, 54

Whittaker function, VI, 233

functional equation

three term —, 359, 365, 420

fundamental region, 29, 394

FΓ, 29

Artin billiard, 396

Ford region, 31

standard, 32

Gamma function, 165

Gauss bracket, 388

Gauss function, 388

recurrence time, 411

Gauß brackets, 85

generalized hypergeometric function, VII

geodesic, 396

GL2 (Z)-equivalence class, 401

GL2 (Z)-equivalent, 401

SL2 (Z)-equivalence class, 407

SL2 (Z)-equivalent, 407

— segment, 396

base points, 399

oriented, 396

geodesics

A, 408

closed, 408

greatest common divisor, 26

group

GL2 (R), 3

GL2 (Z), 391

Γ(1), 25

Γ(N), 25

Γ0(N), 26

Γ1(N), 26

Γ0(N), 26

Γ1(N), 26

SL2 (R), 3

SL2 (Z), 3

of linear fractional transformations, 3

acts discontinuous, 11

congruence subgroup, 25

discontinuous, 11

discrete, 10

Eichler cohomology, 188

Fuchsian group of the first kind, 16

full modular group, 3, 25

general linear group, 3

Hecke congruence subgroup, 26

horocyclic, 16

limit point, 11

normal subgroup, 16

parabolic Eichler cohomology, 188

principle subgroup, 25

special linear group, 3

stabilizer, 13

theta group, 41

half plane

lower, 3

upper, 3

Hankel’s loop integral, 97

Hecke congruence subgroup, 26

Hecke operator, 132

Tn, 203

T ∗n , 203

eigenfunction, 151

eigenvalue, 151

matrix representation, 203

normaized eigenfunction, 151

on period functions, 203

self-adjoint, 160

simultaneous eigenfunction, 151

Hilbert space, 66

Sk,v(Γ), 66

SMaassk,v (Γ, ν), 259

〈·, ·〉Γ,k, 66

〈·, ·〉Γ, 259

Petersson scalar product, 66

scalar product, 66, 259

self-adjoint operator, 160

horocyclic group, 16

Hurwitz ζ-function, 415

hyperbolic

distance function, 62

Laplace operator, 227


433


of real weight, 227

matrix, 6

surface measure, 62

identity

Bol, 181, 438

integral

INDEX LV

Ef , 182

F , 184

G−, 186

G−, 187

Eichler —, 179

formal, 182

Hankel’s loop integral, 97

Niebur, 186

regularized, 440

inverse Mellin transform, 162, 167

Jacobi theta function, 103

Kloosterman sum, 94

Kronecker δ function, 98

Laplace operator

hyperbolic —, 227

(weak harmonic Maass wave forms),433

of real weight, 227

left neighbor

map, 375

sequence, 376

left neighbor map, 375

left neighbor sequence, 376

level function, 374

limit point, 11

linear fractional transformation, 4, 393

on sets, 364

logarithmic derivative, 80

Mobius function, 94

Mobius transform

on geodesics, 398

Maass form

nonholomorphic Eisenstein series, 252

Maass operator, 227

(weak harmonic Maass wave forms), 433

Maass wave form

cusp form, 242

entire, 242

family of cusp forms, 274

admissible, 274



trivial, 274

mock modular —, 435

space of –, 242

spectral parameter, 242


principal part, 434

Maass-Selberg form, 352

matrix

−1, 4

A, 391

B, 391

C, 388

S, 26, 203, 388

S′, 359

T , 26, 203

Γ(1)-equivalence, 136

1, 4

C, 391

Eichler length, 29

elliptic, 6

hyperbolic, 6

parabolic, 6

scaling —, 38

trace, 4, 6

Mayer’s transfer operator, 418

Mellin transform, 161, 164

inverse, 162, 167

inversion formula, 162


mock mocular form, 435

modified divisor function, 122

modular form, 44, 52

cusp form, 55


entire, 55

modular function, 55

normalized Eisenstein series, 79

Poincare series, 89

weakly holomorphic, 54

modular function, 43, 55

Mobius transformation, 4, 393

on sets, 364

transformation of order n, 135

multiplicative sequence, 173

multiplier, 44, 46

η-multiplier, 48



non-triviality condition, 47

multiplier system, 44, 46

η-multiplier, 48


multiplier sytsem


nearly periodic function, 323

neighborhood of ∞, 11

Niebur integral, 186

non-triviality condition, 47

normal subgroup, 16

operator

Ls, 412, 413

Ls, 418

hyperbolic Laplace, 227


433

of real weight, 227

Maass —, 227

Mayer’s transfer —, 418

transfer —, 412, 413

parabolic matrix, 6

partition, 376

length, 376

minimal, 376

path, 378

lies in the quadrant, 378

simple —, 378

standard —, 378

Pell’s equation, 390

period function, 359, 444

LVI INDEX

period polynomial, 177

Pk−2, 177

of F , 175

space, 177

Petersson scalar product, 66

Pochammer symbol, VII, 284

Poincare series, 89

point

equivalent under Γ, 37

fixed point, 6

limit point, 11

ordinary point, 11

Poisson summation formula, VIII

pole at a cusp, 54

Polynomial

self-inversive, 212

polynomial

cubic, 218

homogeneous cubic, 219

period polynomial, 177

of F , 175

prime summation, 67

principle subgroup, 25

Ramanujan τ function, 81


recurrence time function, 411



regularized integral, 440

Riemann ζ-function, 69, 94

Riemann sphere, 4

scalar product, 66, 259

scaling matrix, 38

Selberg zeta function, 431

sequence

completely multiplicative, 173

Farey —, 374

left neighbor —, 376

multiplicative, 173

partition, 376

set

C′, 334, 359

H?, 378

A · B, 141

A, 408

F?, 31

FΓ, 29

FGL2(Z), 394

FΓ(1), 32

C′, 421

of limit points, 11

of transformation of order n, 135

Artin billiard, 396

curve, 218, 220

in H, 396

dot-multiplication, 141

equal modulo C, 141

Ford region, 31

fundamental region, 29, 394

standard, 32

geodesic, 396

GL2 (Z)-equivalence class, 401

GL2 (Z)-equivalent, 401

SL2 (Z)-equivalence class, 407

SL2 (Z)-equivalent, 407

— segment, 396

base points, 399

oriented, 396

geodesics

A, 408

lower half plane, 3

multiplicity l, 141

of cusps, 37

of ordinary points, 11

Riemann sphere, 4

unit circle, 20

unit disc

closed, 19

open, XV

upper half plane, 3

set of cusps, 37

set of limit points, 11

set of ordinary points, 11

simple path, 378

slash operator, 127, 226

on R?, 128

on differential forms, 351

with multiplier, 359

spectral parameter, 242

stabilizer, 13

standard path, 378

Stirling’s formula, IV

theorem

Cauchy’s integral –, V

Chinese remainder –, III

Eichler cohomology –, 193

Geeen’s –, V

open mapping –, III

Rouche –, V

theta function

Jacobi, 103

theta group, 41

three-term equation, 359, 365, 420

transfer operator, 412, 413

three-term equation, 420

transformation of order n, 135

set, 135

unit circle, 20

unit disc

closed, 19

open, XV

unitary automorphic factor, 225

Weierstraß

equation, 221

short, 221

form, 221

short, 221

weight, 44, 225

Whittaker

differential equation, VI, 236, 238

normalized, 233

INDEX LVII

Fourier-Whittaker expansion, 236, 238,

239

function, VI, 233normalized differential equation, 233

width of cusp, 41

zeta function

dynamical —, 431

Selberg —, 431

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