a differential introduction to modular forms and elliptic...

Hossein Movasati

A Differential Introduction toModular Forms and EllipticCurvesThe text is under construction and it is very messy. DO NOT PRINT IT

August 8, 2019

Publisher

Contents

1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.1 Fibonacci sequence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Fermat’s last theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 Aritmetic modularity theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.4 Beyond elliptic curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.5 Prereuisites . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2 Modular forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.1 Elliptic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.2 The modular group and its action . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.3 Slash operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.4 Weierstrass ℘-function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.5 Differential equation of ℘ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.6 Eisenstein series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.7 Fourier expansion of Eisenstein series . . . . . . . . . . . . . . . . . . . . . . . . . 192.8 The Eisenstein series E2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.9 The algebra of moduler forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232.10 Ramanujan relations between eisenstein series . . . . . . . . . . . . . . . . . . 242.11 The product formula for discriminant . . . . . . . . . . . . . . . . . . . . . . . . . . 252.12 The j function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262.13 Poincare metric . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272.14 The numbers e1, e2,e3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272.15 Growth of coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282.16 Dedekind eta function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302.17 Modular forms as k-fold differential . . . . . . . . . . . . . . . . . . . . . . . . . . . 302.18 Petersson scalar product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312.19 Computing E2 as a double sum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

3 Elliptic curves and integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353.2 Elliptic integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

v

vi Contents

3.3 Elliptic curves in Weierstrass format . . . . . . . . . . . . . . . . . . . . . . . . . . . 393.4 Picard-Lefschetz theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393.5 Weierstrass uniformization theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 403.6 Sketch of the proof of Theorem (2.9.1) . . . . . . . . . . . . . . . . . . . . . . . . . 413.7 Some identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 423.8 Schwarz function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 433.9 CM elliptic curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 443.10 Fourier expansions and elliptic integrals . . . . . . . . . . . . . . . . . . . . . . . . 44

4 Rudiments of Algebraic Geometry of curves . . . . . . . . . . . . . . . . . . . . . . 474.1 Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 474.2 Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 484.3 Schemes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 494.4 Singular and smooth curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 504.5 Resultant and discriminant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 504.6 Discriminant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 514.7 The main property of the discriminant . . . . . . . . . . . . . . . . . . . . . . . . . 534.8 Descriminant of projective curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 534.9 Curves of genus zero . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 544.10 Curves of genus bigger than one . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 544.11 Elliptic curves in Weierstrass form . . . . . . . . . . . . . . . . . . . . . . . . . . . . 554.12 Elliptic curves in Weierstrass form . . . . . . . . . . . . . . . . . . . . . . . . . . . . 554.13 Real geometry of elliptic curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 564.14 Complex geometry of elliptic curves . . . . . . . . . . . . . . . . . . . . . . . . . . . 564.15 The group law in elliptic curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 574.16 Divisors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 594.17 Riemann-Roch theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 604.18 Weierstrass form revised . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 604.19 Moduli of elliptic curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 614.20 The addition formula for ℘ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 624.21 Why Schemes? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

5 Mordell-Weil Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 655.1 Mordell-Weil theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 655.2 Weak Mordell-Weil theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

6 Torsions and isogeny . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 736.1 Torsion points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 736.2 Isogeny . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 746.3 Isogeny II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

7 Hecke operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 777.1 Hecke operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 777.2 Hecke and cusp forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 817.3 Proof of Theorem 6.3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

Contents vii

8 Riemann zeta function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 858.1 Riemann zeta function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 858.2 The big Oh notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 868.3 Gamma function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 878.4 Mellin transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 888.5 Analytic extension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 888.6 Functional equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 898.7 Second proof for functional equation . . . . . . . . . . . . . . . . . . . . . . . . . . 908.8 Zeta and primes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 918.9 Other zeta functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 928.10 Dedekind Zeta function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 938.11 L-function of cusp forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 938.12 Hecke’s L-functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 948.13 L-function of CY-modular forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

9 Congruence groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 999.1 Congruence groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 999.2 Weil pairing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1009.3 Moduli spaces of elliptic curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1019.4 Modular forms for congruence groups . . . . . . . . . . . . . . . . . . . . . . . . . 1019.5 q-expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1029.6 Transcendental degree of modular forms . . . . . . . . . . . . . . . . . . . . . . . 104

10 Elliptic curves as Diophantine equations . . . . . . . . . . . . . . . . . . . . . . . . . . 10710.1 Finite fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10710.2 Zeta functions of elliptic curves over finite fields . . . . . . . . . . . . . . . . 10710.3 Nagell-Lutz Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10910.4 Mazur theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10910.5 One dimensional algebraic groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11010.6 Reduction of elliptic curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11010.7 Zeta functions of curves over Q . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11110.8 Hasse-Weil conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11210.9 Birch Swinnerton-Dyer conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11310.10Congruent numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11310.11p-adic numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

11 Theta series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11911.1 Two-squares theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12211.2 Poisson summation formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

12 Picard curious example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

viii Contents

13 Online supplemental items . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12913.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12913.2 How to start? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12913.3 Ramanujan differential equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133

Preface

There are so many books on modular forms and elliptic curves that it might seemuseless to add another one. None of these books approach modular forms from thepoint of view of differential equations and this differentiate the present book fromothers. This has resulted in a tremendous generalization of modular forms whoseorigin partially comes from many q-expansion computations in theoretical physicsand in particular string theory.

Hossein MovasatiJanuary 2020

Rio de Janeiro, RJ, Brazil

1

Chapter 1Introduction

Our main interest on modular forms is the fact that they are generating functions formany unexpected countings in mathematics. Why generating functions are usefulmight be explained with the simple example of Fibonacci numbers.

1.1 Fibonacci sequence

The Fibonacci sequence is defined in the following way

Fn+2 = Fn+1 +Fn, n> 0, F0 = 1, F1 = 1 (1.1)

Few elements of this sequence are

1,1,2,3,5,8,13,21, . . .

Once you have a sequence of natural numbers in mathematics, it is recommandedto put it in a generating function:

F (q) : = q+q2 +2q3 +3q4 + . . .+Fnqn + · · ·

At the beginning this is just a formal power series, however, soon it will becomeclear that its a convergence and radius of convergence carries many information ofthe sequence Fn itself. For now, let us do the following manipulation:

F (q) =∞

∑n=0

Fnqn = q+∞

∑n=2

(Fn−1 +Fn−2)qn

= q+q ·F (q)+q2F (q)

which implies thatF (q) =

q1−q−q2 (1.2)

3

4 1 Introduction

Therefore, F (q) converges to a rational function. In order to find the radius of con-vergence of a rational function, we have to find the roots of its denominator:

F (q) =q

1−q−q2 =q

(1−α ·q)(1−β ·q)=

(α−β )−1

1−α ·q− (α−β )−1

1−β ·q

=∞

∑n=0

(αn−β n

α−β

)qn

where α = 12

(1+√

5), β = 1

2

(1+√

5)

. We conclude that

Fn =

(1+√

52

)n−(

1−√

52

)n

√5

Which at first glance looks strange because we have found a formula for the in-teger Fn in terms of square root of 5. Since the radius of convergence of F (q) ismin

1|α| ,

1|β |

= max|α| , |β |= |α|, we conclude that

limn→∞

Fn

Fn−1= lim

n→∞F

1n

n =12

(1+√

5)

This number is called the golden rati or the golden number.

Exercise 1 Show that (1 11 0

)n

=

(Fn+1 FnFn Fn−1

).

1.2 Fermat’s last theorem

Modular forms and elliptic curves are firmly rooted in the fertil grounds of numbertheory. As a proof of the mentioned fact and as an introduction to the present text wemention the followings: For p prime, the Fermat last theorem ask for a non-trivialinteger solution for the Diophantine equation

ap +bp + cp = 0

For a hypothetical solution (A,B,C) = (ap,bp,cp) of the Fermat equation withabc 6= 0, Gerhart Frey considered the elliptic curve

EA,B,C : y2 = x(x−A)(x+B)

From this one construct a modular form fA,B,C and a Galois representation withcertain properties and then one proves that such objects does not exist. During thispassage one encounters the Modularity conjecture which claims that every elliptic

1.3 Aritmetic modularity theorem 5

curve over Q is modular. Roughly speaking this means that every elliptic curve overQ appears in the Jacobian of of a modular curve of level N. Another formulation ofmodularity property is by using L functions which generalizes the famous Riemannzeta function

ζ (s) :=∞

∑n=1

1ns

Riemann hypothesis claims that all the non-trivial zeros of ζ lies on ℜ(s) = 12 and it

has strong consequences on the growth of prime number. For the L functions associ-ated to elliptic curves one has the Birch-Swinnerton Dyer conjecture which predictsthe rank of an elliptic curve to be the order of vanishing of the corresponding L-function at s = 1.

1.3 Aritmetic modularity theorem

Modular forms as generating functions have many fascinating and mysterious ap-plications. Aritmetic modularity theorem is one of these. In many books and articleswe find the expression

“Let E be an elliptic curve over Z” .This has an intrinsic definition, that for now, we don’t want to get into its details.We content ourselves with the example.

E : y2 + y = x3− x2

which the reader might consider it as a Diophantine equation, that is, we are inter-ested to find x and y in the ring of integers, the field of rational numbers, finite fields,etc. Let p be a prime number (don’t take the Grothendieck’s prime) i We count thenumber of solutions Np of E modulo the prime p.

P Solutions Np2 (0,0),(0,1), (1,0), (1,1) 43 (0,0),(0,2), (1,0), (1,2) 45 (0,0),(0,4), (1,0), (1,4) 47 (0,0),(0,6), (1,0), (1,6), ( , ),. . . 9

11 (0,0),(0,10), (1,0), (1,10),( , ),. . . 10

In total we have to substitute p2 pairs (x,y) ,x,y = 0,1, . . . , p−1 inside E and verifywhether modulo prime p it is zero or not. The first four solutions in the above tablehave to do with the fact that over integers E has already four solutions.

(0,0) ,(0,−1) ,(1,0) ,(1,−1)

i A. Grothendieck (1928-2014) is one of the founders of modern Arithmetic Algebraic Geometry.Once he was asked to give an example of a prime number and he answered: 57.

6 1 Introduction

A priori, if we have computed N2,N3,N5, . . . ,N11, this doesn’t give any clue how tofind the number N13. We have to check 132 cases again. In modern language, we saythat, we are counting the number of Fp-rational points of E and we write

Np : = #E (Fp)

Here Fp := 0,1,2, . . . , p−1 is the finite field with p elements.

Exercise 2 Find Np for all p6 20.

The theory of modular forms, and in particular arithmetic modularity theorem, saysthat there is a closed formula for the generating function of Np’s. This is as follows.Let

η (q) = q124

∞

∏n−1

(1−qn) (1.3)

be the Dedekind eta function. We consider it as a formal product. Let

F (q) = η (q)2η(q11)2

= q224+

2·1124

∞

∏n−1

(1−qn)2∞

∏n−1

(1−q11n)2

= q−2q2−q3 +2q4 +q5 +2q6−2q7−2q9−2q10 +q11−2q12 +4q13 + . . .

=∞

∑n=1

fnqn

The arithmetic modularity theorem tells us that

Np = p− fp (1.4)

and f is a modular form. More precisely,“ f is a weight 2 new form for Γ0 (11)”

One of the aims of the present text is to understand this statement. This phenomenais a part of a general theorem:

Theorem 1.3.1 (Arithmetic modularity theorem) (A. Wiles, R.Taylor, C. Breuil,B. Conrad, F. Diamond) For any elliptic curve E over Q, there is a modular formf = ∑

∞n=1 fnqn such that (1.4) holds for all except a finite number of primes.

A precise statement, together with other equivalent versions will be presented in thistext.

Exercise 3 Show that the radius of convergence of the Dedekind η function is 1.

1.5 Prereuisites 7

1.4 Beyond elliptic curves

There is a tremendous amount of effort to generalize the arithmetic modularity the-orem beyond elliptic curves. Here we give an example taken from [Schuett2013].Let us consider the Fermat quartic surface

X = X42 ⊆ P3 : x4

0 + x41 + x4

2 + x43 = 0

We count the number of solutions of this Diophantine equation over the field Fp, p 6=2

#X(Fp) : = #[x0 : x1 : x2 : x3]|x4

0 + x41 + x4

2 + x43 = 0

.

Here, [x0 : x1 : x2 : x3] is the equivalence class

(x0 : x1 : x2 : x3)∼ (y0,y1,y2,y3)⇔∃a ∈ Fp−0

such that xi = ayi,∀i = 0,1,2,3. It turns out that

#X(Fp) = 1+bp +h · p+ p2

where

η(4τ)6 = q∞

∏n=1

(1−q4n)6 =∞

∑n=1

bnqn

h = 5+3χ−1(p)+6 · (χ2(p)+χ−2(p))

and χa(p) : = ( ap ) is the Legendre symbol.

Exercise 4 Verify the above affirmation for p= 3,5, . . . what goes wrong for p= 2?

1.5 Prereuisites

It is assumed that the reader has a basic knowledge in algebraic geometry of curvesand complex analysis in one variable.

Chapter 2Modular forms

I have told the story before, but it is ironic that being at the same university, Artinhad discovered a new type of L-series and Hecke, in trying to figure out what kindof modular forms of weight one there were, said they should correspond to somekind of L-function. The L-functions Hecke sought were among those that Artin haddefined, but they never made contact-it took almost forty years until this connectionwas guessed and ten more before it was proved, by Langlands. Hecke was olderthan Artin by about ten years, but I think the main reason they did not make contactwas their difference in mathematical taste. Moral: Be open to all approaches to asubject, (J. Tate in [RS11] page 446).

2.1 Elliptic functions

Definition 2.1.1 A lattice Λ in C is a discrete subgroup of (C,+) which generate itas an R-vector space

It follows easily that

Λ = Zω1 +Zω2 = nω1 +mω2|n,m ∈ Z

where ω1,ω2 ∈C, ω1,ω2 6= 0, Im(ω2/ω1) 6= 0. By changing the order of ω1,ω2, ifnecessary, we can assume that

Im(τ)> 0, τ :=ω1

ω2(2.1)

A lattice in general is equipped with a bilinear map Λ ×Λ → Z. In our case it isskew-symmetric, that is,

〈a,b〉=−〈b,a〉 ∀a,b ∈Λ

and so 〈ω1,ω1〉= 0,〈ω2,ω2〉= 0. Therefore,

9

10 2 Modular forms

Fig. 2.1 Lattice

〈ω2,ω1〉 := 1 (2.2)

determines〈·, ·〉 uniquely. The choice of ω1,ω2 with (2.1) and hence with (2.2) isalso called an orientation of Λ . If we choose another basis ω ′1,ω

′2 with (2.2) then[

ω ′1ω ′2

]= A

[ω1ω2

], A ∈ SL(2,Z),

where

SL(2,Z) :=[

a bc d

]∣∣∣∣a,b,c,d,∈ Z, ad−bc = 1. (2.3)

Let P be the space of lattice in C. The group

Gm = C∗ = (C−0, ·) (2.4)

acts on P from the right

P×C∗→ P(Λ ,λ ) 7−→Λ ·λ := Zω1λ +Zω2λ

For a lattice Λ the associate complex tori is simply

E := C/Λ

this means that two points z1,z2 ∈ C are equivalent if z1− z2 ∈ Λ . The set C/Λ

is an example of a Riemann surface or complex manifold. It is called real torus of

2.1 Elliptic functions 11

Fig. 2.2 Torus

dimension two or complex torus of dimension one. It has the structure of an abeliangroup which inherits from (C,+). Still we do not call it an elliptic curve as thisname is reserved for a similar object in algebraic geometry. In mathematics whenwe have a space, then we start to study the set of its functions. In our case, we areinterested on meromorphic functions on C/Λ as we have:

Exercise 5 There is no holomorphic function on E.

The pull-back of a meromorphic function by the projection map C→ E correspondsto a meromorphic function f with

f : C→ C,f (z+ω) = f (z) ∀z ∈ C, ω ∈Λ .

Since Λ is generated by ω1,ω2, (2.5) is equivalent to

f (z+ω1) = f (z)f (z+ω2) = f (z)

∀z ∈ C ,

that is f is double periodic.We may also view f as a function in both z ∈ C andΛ ∈ P. In this case we write f (z) = f (z,Λ). Since P is equipped with a C∗-action,it is natural to characterize functions f with

f (λ z,λΛ) = λ−a f (z,Λ),∀λ ∈ C∗ (2.5)

for some fixed a ∈ Z.

12 2 Modular forms

Definition 2.1.2 A meromorphic function f with the property (2.5) is called anelliptic function (of weight a).

Let us consider an elliptic function f and write its Laurent series at z = 0,

f (z,Λ) =+∞

∑n=−∞

fn(Λ)zn

The coefficients fn(Λ) are functions of the lattice Λ and it is easy to see that (2.5)is equivalent to the following functional equations for fn(Λ)’s

fn(λΛ) = λ−a−n fn(Λ) ∀λ ∈ C∗.

This is as follows

f (λ z,λΛ) =+∞

∑n=−∞

fn(λΛ)(λ z)n

= λ−a

(+∞

∑n=−∞

fn(Λ)zn

)

Definition 2.1.3 A meromorphic function f on the space P of lattices is called ameromorphic modular form of weight n ∈ Z if

f (λΛ) = λ−n f (Λ) ∀λ ∈ C∗, Λ ∈ P.

Therefore, from a meromorphic elliptic function of weight a we get meromorphicmodular form fn of weight n+a. If we evaluate a meromorphic modular form f ofweight n on lattices Λ =Zτ +Z with τ in the upper half plane H and regard them asa function in τ , we get a meromorphic function f in H with the following functionalequation.

(cτ +d)−n f (aτ +bcτ +d

) = f (τ), ∀τ ∈H. (2.6)

2.2 The modular group and its action

The upper half plane is defined as

H := τ ∈ C| Im(τ)> 0.

The following group acts on H by Mobius transformation

2.3 Slash operator 13

SL(2,R) :=(

a bc d

)∣∣∣∣ad−bc = 1,a,b,c,d,∈ R

SL(2,R)×H−→H(A,τ) 7−→ Aτ := aτ+b

cτ+d

where A =

(a bc d

)∈ SL(2,R). This follows from

Im(Aτ) =Im(τ)det(A)|cτ +d|2

(2.7)

An element A ∈ SL(2,R) acts as identity on H if it is ±I, where

I =(

1 00 1

)is the indentity matrix. Therefore, it is usefull to define

PSL(2,R) = SL(2,R)/± I

The protagonist of the present text is the group SL(2,Z) defined in (2.3). Let τ ∈Hand assume that it has non-trivial stablizer under the action of SL(2,Z), that is, thereis A ∈ SL(2,Z), A 6=±I such that Aτ = τ . Since detA =+1 and Im(τ)> 0 we get

|a+d|< 2 and τ =a−d+i

√(a+d)2−4c .

Exercise 6 Show that the group SL(2,Z) is generated by

T :=(

1 10 1

), S :=

(0 −11 0

). (2.8)

2.3 Slash operator

For A ∈ GL(2,R) and a modular form f of weight k we define the slash operator

f |kA := (detA)k−1(cτ +d)−k f (Aτ), A =

(∗ ∗c d

).

In some books, the power k− 1 of detAis different. For instance, in Chapter 7 theslash operator is different. For A ∈ SL(2,R) this will not make any difference.

Proposition 2.3.1 If B ∈ GL(2,R)and f is a meromorphic modular form of weightk for some group Γ ∈ GL(2,R) then f |kB is a modular form of the same weight forB−1Γ B.

Proof. This follows from

14 2 Modular forms

( f |kB)|kB−1AB = ( f |kA)|kB = f |kB

ut

2.4 Weierstrass ℘-function

When a discrete group Γ acts on a space M and we want to construct functions onthe quotient space

Γ \M := M/∼ x∼ y⇔ x = Ay for some A ∈ Γ

the first recepie is to start with a function f on M and define the formal sum

f (τ) = ∑A∈Γ

f (Aτ) (2.9)

If we don’t care about the convergence of f , the we can check that it is invariantunder the action of Γ . For any B ∈ Γ we have

f (Bτ) = ∑A∈Γ

f (ABτ)

= ∑A∈Γ

f (Aτ) = f (τ)

Note that we have used the fact that the multiplication by B from the right induces abijection Γ → Γ . We get the function

f : Γ \M→ C, f ([τ]) = f (τ)

which we denot it again by f = f If Γ is finite then (2.9) is a finite sum an so f iswell-defined, however, in general such a sum might not be convergent. In Our case,the lattice Λ as an additive group acts on C

Λ ×C→ C,(λ ,z) 7−→ z+λ .

Theorem 2.4.1 Show that the number of zeros of a non-constant elliptic functioncounted in C/Λ is equal to the number of poles, counted with multiplicity, and it isbigger than or equal to 2.

Proof. See [Apostol 1989], page 5-6.

For our purpose we start with f (z) = z−a, a ∈ Z and define

2.4 Weierstrass ℘-function 15

fa(z) = fa(z,ω) : = ∑ω∈Λ

(z+ω)−a

= ∑(n,m)∈Z2

(z+nω1 +mω2)−a

Proposition 2.4.1 The infinite series fa(z) converges absolutely for a ∈N with a>3

Proof. (See lemma 2 page 8 of [Apostol]). We can assume that the sum is taken forall ω ∈ Λ , |ω| > R and |z| < R. There is a constant M depending on R and a suchthat

1|z−ω|a 6

M|ω|a

∀ω ∈Λ , |ω|> R∀z ∈ C, |z|< R

Therefore, it is enough to prove that the sum ∑ω∈Λ1|ω|a is convergent. Let r and R

be the minimum and maximum distanced of 0 from the parallelogram formed by±ω1 +±ω2. Then the parallelogram Pn formed by four vertices n(±ω1±ω2) hasthe minimum and maximum distances nr and nR, respectively, from 0. Moreover, ithas 8n points of the lattice. Therefore, the sum Sn corresponding to all parallelogralsP1,P2, . . . ,Pn satisfies

8Ra (1+2a−1 + · · ·+na−1)≤ Sn ≤

8ra (1+2a−1 + · · ·+na−1)

ut

Exercise 7 Show that f2 does not converge in any z ∈ C,z /∈Λ

Note that∂ fa

∂ z=−a · fa+1

Definition 2.4.1 The Weierstrass ℘ function (read P) is

℘(z,Λ) =℘(z) :=1z2 + ∑

ω 6=0

1(z−ω)2 −

1ω2

Proposition 2.4.2 ℘ is convergent

Proof. (See [Apostol] Theorem 1.10). We have∣∣∣ 1(z−ω)2 − 1

ω2

∣∣∣= ∣∣∣ z(2ω−z)(z−ω)2ω2

∣∣∣6 M·R·(2|ω|+R)|ω|2|ω|2

6 M·R·(2+R/|ω|)|ω|3 6 3MR

|ω|3

where we used the notation in the proof Proposition (2.4.2) ut

16 2 Modular forms

It is easy to see that℘(−z) =℘(z)

that is ℘ is an even function.

Exercise 8 Show that there is no function f (ω),ω ∈Λ such that

∑ω∈Λ

1z−ω

+ f (ω)

is convergent

The function ℘ has poles at the points of Λ . We write the Laurent expansion of ℘

at z = 0

Theorem 2.4.2 We have

℘(z) =1z2 +

∞

∑n=1

(2n+1)G2n+2 · zn

whereG2n+2 = ∑

ω 6=0

1ω2n+2 · (2.10)

and0 < |z|< r := min|ω| | ω 6= 0· (2.11)

Proof. For z in (2.11) we have∣∣ z

ω

∣∣< 1 and

1(z−ω)2 =

1ω2(1− z

ω)2 =

1ω2

(1+

∞

∑n=1

(n+1)( z

ω

)n)

and so1

(z−ω)2 −1

ω2 =∞

∑n=1

n+1ωn+2 · zn

Summing over ω ∈Λ ,ω 6= 0,we get the result. ut

The notationg2 = 60G4, g3 = 140G6 (2.12)

is frequently used.

2.5 Differential equation of ℘

Theorem 2.5.1 The function ℘ satisfies the differential equation

℘′(z)2 = 4℘(z)3−60G4x−140G6 (2.13)

2.6 Eisenstein series 17

Proof. Let f (z) be the difference of both sides of (2.13). This is clearly an ellipticfunction with possible poles at z ∈ Λ . We show that f is holomorphic at z = 0 andso f = 0. We have

℘′(z) = −2z3 +6G3 · z+20G6 · z3 + . . .

℘′(z)2 = 4z6 − 24G4

z2 −80G6 + . . .

4℘(z)3 = 4z6 +

36G4z2 +60G6 + . . .

hence℘′(z)2 = 4℘(z)3 =− 60G4

z2 −140G6 + . . .

℘′(z)2 = 4℘(z)3 =+60G4℘(z) =−140G6 + . . . ut

Definition 2.5.1 An elliptic function f is a meromorphic function in C such thatIt is double periodic, that is, there is two Z linearly independent complex numbersω1,ω2 ∈ Z such that

f (z+ω1) = f (z), f (z+ω2) = f (z)

If f is a non-constant elliptic function, we can easily see that Im(ω1,/ω2) 6= 0.

Exercise 9 ([Apostol] page 23, Exercise 5) Prove that every elliptic function f canbe written as

R1[℘(z)]+℘′(z)R2[℘(z)]

where R1,R2 are rational functions and ℘ has the same set of periods as f .

Exercise 10 ([Apostol] page 24, Exercise 9)

℘(2z) =(℘(z)2 + 1

4 g2)2 +2g3 ·℘(z)

4℘3(2)−g2℘(z)−g3

= −2℘(z)+14

(℘′′(z)℘′(z)

)2

Exercise 11 Show that℘′′(z) = 6℘(z)− 1

2g2

Can you compute ℘′′(z) in terms of ℘(z),℘′(z)

2.6 Eisenstein series

The series

Gn :=∞

∑ω∈Λ , ω 6=0

1ωn , n even, n> 4

are called Eisenstein series. From the convergence of ℘ we can derive the conver-gence of Gn’s. They satisfy

18 2 Modular forms

Gn(λΛ) = λ−nG(Λ), ∀λ ∈ C∗ (2.14)

we usually define

Gn(τ) = Gn(Zτ +Z) = ∑(a,b)∈Z2, (a,b)6=(0,0)

1(a+bτ)n , τ ∈H

Exercise 12 Show that G2 doesn’t converge at any point τ ∈ H. Show also thatGn ≡ 0, for n an odd number.

From the functional equation (2.14) we deduce the following: For all A ∈ SL(2,Z)

Gn(Aτ) = G(

aτ +bcτ +d

)= G

(aτ +bcτ +d

Z+Z)

= (cτ +d)+nG((aτ +b)Z+(cτ +d)Z)= (cτ +d)nG(τZ+Z)= (cτ +d)nG(τ)

Later we will see thatlim

Im(τ)→∞

Gn(τ) exist

This motivate us to define (holomorphic) modular forms.

Definition 2.6.1 Let k ∈ Z be an integer and f be a meromorphic function on theupper half plane. Then f is called a meromorphic modular form for the subgroupΓ ⊆ SL(2,R) if

(cτ +d)−k f(

aτ +bcτ +d

)= f (τ) ∀

(a bc d

)∈ Γ (2.15)

Let us assume that(

1 10 1

)∈ Γ . Then

f (τ +1) = f (τ)

Therefore, f defines a meromorphic function g in the punctured disc:

D∗ = z ∈ C| |z|< 1\0

which is defined byf (τ) = f (q), where q := e2πiτ

The map H→ D∗ is depicted in Figure (*)We write the laurent expansion of f at q = 0.

2.7 Fourier expansion of Eisenstein series 19

Fig. 2.3 q map

f (q) =n=+∞

∑n=−∞

fn ·qn , fn ∈ C (2.16)

This is also called the Fourier expansion of f .

Definition 2.6.2 A meromorphic modular form for Γ = SL(2,Z) apart from theproperty (2.15) is meromorphic at i∞, that is, in (2.16) we have some M ∈ Z suchthat fn = 0 for all n 6 M. It is called weakly modular form if it is holomorphic inH and possibly meromorphic at i∞. It is called (holomorphic) modular form if it isholomorphic in H∪i∞.

From now we use the letter f for f too, and write the q−expansion of a modularform as

f =∞

∑n=0

fn ·qn =∞

∑n=0

fn · e2πiτ

It will be clear from the text whether we consider f as a function of τ or q.

2.7 Fourier expansion of Eisenstein series

We know that the Eisenstein series are weakly holomorphic moduler forms. In thissection we show that they are holomorphic at i∞ and so they are moduler forms. Thecomputation in this section can be found in [koblitz] page 110, [Apostol] page 18,Serre page 91.Recall the Bernouli numbers

xex−1

=∞

∑k=0

Bk ·xk

k!

20 2 Modular forms

For instance, B0 = 1, B1 =−12 , B2 =

16 , B4 =

−130 , B6 =

142

It is easy to see that for any odd k > 3 we have Bx = 0

Theorem 2.7.1 The Eisenstein series Gk has the following q-expansion

Gk = 2ζ (k)

(1− 2k

Bk

∞

∑n=1

σk−1(n)qn

)

where ζ (k) := ∑∞n=1

1nk is the Riemann’s zeta function and

σa(n) := ∑d|n

da·

We follow [Koblitz] page 110. Let us first state the main ingredient of the proofof Theorem 2.7.1

Proposition 2.7.1 We have

ζ (k) =− (2πi)2

k Bk

k!k even > 2 (2.17)

∑n∈Z

1(a+n)k =

(−2πi)k

(k−1)!

∞

∑n=0

nk−1e2πina, k ∈ N, k ≥ 2, a ∈ C−Z (2.18)

Proof. We have the following product formula for sine function

sin(πa) = πa∞

∏n=1

(1− a2

n2

), a ∈ C (2.19)

Take the logarithmic derivative if (2.19) and we have

π · cot(πa) =1a+

∞

∑n=1

(1

a+n+

1a−n

), (2.20)

The left hand side of this is

π · cot(πa) = πieπia + e−πia

eπia− e−πia = πi+2πi

e2πia−1

= πi−2πi

(∞

∑n=0

e2πina

)(2.21)

In (2.20) multiply both side with a and set x := 2πia.

2.7 Fourier expansion of Eisenstein series 21

x2+

xex−1

= 1+∞

∑n=1

x(x+2πin)

+x

(x−2πin)

= 1+∞

∑n=1

x2πin

(∞

∑k=0

(−x

2πin

)k

−( x

2πin

)k)

= 1−2∞

∑n=1

∞

∑k=1

k odd

xk+1

(2πin)k+1

= 1−2∞

∑k=1

k odd

xk+1

(2πi)k+1ζ (k+1)

We get the well-know formula (2.17). We differentiate (2.20) and (2.21) with respectto a, k−1 times, and we get

(−1)(−2) . . .(−k+1) ∑n∈Z

1(a+n)k =−(2πi)k

∞

∑n=0

nk−1e2πina

which is (2.18).

Proof (of Theorem 2.7.1). Take a = mτ,m ∈ Z,m 6= 0, and we have

∑n∈Z

1(mτ +n)k =

(−2πi)k

(k−1)!

∞

∑n=1

nk−1qnm

The result follows immediately:

Gk = 2ζ (k)+2∞

∑m=1

+∞

∑n=−∞

1(mτ +n)k (2.22)

= 2ζ (k)

(1+

(−2πi)k

ζ (k)(k−1)!

∞

∑m,n=1

nk−1qnm

)ut (2.23)

We know that the Eisenstein series Gk for k ≥ and odd number is identically zeroand hence the equality (2.22) is valid for k ≥ 4 an even number. However, note thatthe equality (2.23) is valid also for k ≥ 3 an odd number. In this case the number

πk

ζ (k) is conjecturally a transcendetal number. For this reason we may take (2.22) asthe definition of Gk which works also for odd k ≥ 3.

Exercise 13 ∗ Describe the functional equation of Gk(τ),k≥ 3 under the action ofSL(2,Z). See the Master thesis of H. Bachmann.

Exercise 14 Prove the product formula for sine in (2.19). Hint: both side have thesame zero set.

We will use the following new notation

http://www.henrikbachmann.com/uploads/7/7/6/3/77634444/msc_henrik_bachmann.pdf

22 2 Modular forms

Ek = Gk/2ζ (k) = 1− 2kBk

∞

∑n=1

σk−1(n)qn (2.24)

E4 = 1+240

(∞

∑n=1

σ3(n)qn

)(2.25)

E6 = 1−504

(∞

∑n=1

σ5(n)qn

)(2.26)

E8 = 1+480

(∞

∑n=1

σ7(n)qn

)(2.27)

2.8 The Eisenstein series E2

We follow [Koblitz]p. 112. We define

G2(τ) =∞

∑m=−∞

∞

∑n=−∞

′ 1(mτ +n)2

Where ′ means that if m = 0 then n 6== 0. The arguments in §2.7 shows that theinner sums converge for any m and τ ∈ H and then the other sum converges. Here,the order of summation is important.

Exercise 15 Show that the double sum

∑(n,m)∈Z2\(0,0)

1(mτ +n)2

doesn’t converge.

In a similar way as in §2.7 we get

G2(τ) = 2ζ (2)E2(τ), E2(τ) = 1−24∞

∑n=1

σ1(n)qn


(cτ +d)−2E2

(aτ +bcτ +d

)−E2(τ) =

122πi

ccτ +d

(2.28)

for all(

a bc d

)∈ SL(2,Z)

Proof. First we note that if we define

f ‖2 A := (cτ +d)−2 f (Aτ)− c(cτ +d)−1

2.9 The algebra of moduler forms 23

for a holomorphic function f on H then

( f ‖2 A) ‖2 B = f ‖2 AB

Since PSL(2,Z) is generated by T and S and the (2.28) is trivial for T , it isenough to verify (2.28) for S, that is

τ−2E2

(−1τ

)= E2(τ)+

122πi

1τ

ut

2.9 The algebra of moduler forms

Theorem 2.9.1 The Eisenstein series E4 and E6 are algebraically independent overC, that is, there is no polynomial P(X ,Y ) with coefficients in C such that P(E4,E6)=0. Moreover any holomorphic moduler form f of weight k can be written uniquelyas

f = P(E4,E6)

where P is a homogeneous polynomial of degree k in the ring

C[X ,Y ], weight (X) = 4, weight (Y ) = 6 (2.29)

There is a clasical proof of Theorem (2.9.1) which can be found in almost all bookson modular forms. In §3.6 we will give a new proof which is inspired by the author’sstudy of the generalized period domain in [Movasati2008]. The proof is based onthe study of elliptic integrals and Gauss-Manin connection.

Note that a homogeneous polynomial P of degree k in the ring (2.29) is of theform

P(X ,Y ) = ∑4i+6 j=k

ai, jX iY j ai, j ∈ C, i, j ∈ N0

and so P(X ,Y ) is clearly a moduler form of weight k for SL(2,Z). Let

M = M(SL(2,Z)) :=⊕k∈ZMk

be the algebra of moduler forms. By Theorem (2.9.1), for k ∈ Z, k 6 2 or k odd wehave Mk = 0 and

M = C[E4,E6], Mk = C[E4,E6]kweight (E4) = 4, weight(E6) = 6

Exercise 16 Show that

E24 = E8, E4E6 = E10, E6 ·E8 = E14

24 2 Modular forms

and derive the corresponding equalities for 6k(n). For instance

67(n) = 63(n)+120n−1

∑m=1

63(n−m)

The dimension of the space of modulex forms Mk is listed below:

k 0 2 4 6 8 10 12 14 16 18 k k+12dimMk 1 0 1 1 1 1 2 1 2 2 d d +1

Note thatdimMk = ](x,y) ∈ N2/4x+6y = k

It is easy to see that∞

∑k=0

dimMkqk =1

(1−q4)(1−q6). (2.30)

2.10 Ramanujan relations between eisenstein series

The derivation of a modular form with respect to τ is no more a modular form.Instead, we have

Proposition 2.10.1 We have the following map

Mk→Mk+2, f 7→ ∂ f∂τ−2πi

k12

E2 · f . (2.31)

which is called the Serre derivative of f

Proof. Let g be the Serre derivative of f . We have to show that g ∈Mk+2. Only thefunctional equation of g is non-trivial:

g(Aτ) = f ′(Aτ)− (2πi)k

12f (Aτ)E2(Aτ)

= · · ·= (cτ +d)k+2 · f (τ)

for A =

(a bc d

)∈ SL(2,Z).

We have∂

∂τ:= 2πiq

∂

∂q.

and sometime it is useful to divide the Serre derivative over 2πi and redefine itf 7→ q ∂ f

∂q −k

12 E2 f .

Proposition 2.10.2 we have the following equalities between Eisenstein series andtheir derivatives

2.11 The product formula for discriminant 25q ∂E2

∂q = 112 (E

22 −E4)

q ∂E4∂q = 1

3 (E2E4−E6)

q ∂E6∂q = 1

2 (E2E6−E24 )

, (2.32)

Proof. The proof of the second and third equality follows from the Serre derivative.The proof of the first equality follows in a similar way. We need to prove that f (τ) :=− 12

2πi∂E2∂τ

+E22 is a modular form of weight 4 and its constant term is 1. Therefore,

by Theorem 2.9.1 it must be E4. ut

2.11 The product formula for discriminant

Recall that

ζ (4) =π4

90, ζ (6) =

π6

945,

We define

∆ := t32 −27t2

3 = (2ζ (4)60E4)3−27(2ζ (6)140)2E2

6 =(2π)12

1728(E3

4 −E26 )

and we have

(2π)−12∆ =

11728

(E34−E2

6 )= (∞

∑n=1

τ(n)qn = q−24q2+252q3−1472q4+4830q5+· · ·=

τ(n) is called the Ramanujan function of n.


11728

(E34 −E2

6 ) = q∞

∏n=1

(1−qn)24 (2.33)

Proof. This follows from Ramanujan relations between Eisenstein series. The log-arithmic derivative of both sides in (2.33) is E2. ut

Exercise 17 ([Koblitz] III, §2, 4)

1. Show that

E12−E26 =

(2πi)−12 ·26 ·35 ·72

691·∆

2. From this derive an expression for τ(n) in terms of σ11 and σ53. Show that

τ(n)≡ σ11(n)( mod 691)

26 2 Modular forms

The following was conjectured by Ramanujan and proved by Deligne in [Deligne]as a by-product of his proof of the Weil conjectures.

Theorem 2.11.1 We have|τ(n)|< n

1112 σ · (n),

where σ0(n) is the number of divisors of n.

Another interesting function is

F(q) :=1728 ·qE3

4 −E26=

∞

∏n=1

1(1−qn)

=∞

∑n=0

Pnqn

It can be easily checked that Pn is the unrestricted partition function, that is, Pn is thenumber of ways a positive integer n can be expressed as a sum of positive integers:

n = a1 +a2 . . .+ak , ak ∈ N.

There is no restriction on k, order of ai’s, and repetion of ai’s is allowed. For moreinformation see Apostol’s book [Apostol] Chapter 5.

2.12 The j function

The following

j(τ) = 1728t32

t32 −27t2

3=

∞

∑n=−1

cnqn

= 1728E3

4

E34 −E2

6=

1q+744+196884q+21493760q2 +864299970q3 + · · ·

is called the j-function, or Klein’s modular function. It is holomorphic in H and hasa pole of order one at i∞. From the functional equation of Eisenstein series it followsthat j is invariant under the action of SL(2,Z):

j(aτ +bcτ +d

) = j(τ), ∀(

a bc d

)∈ SL(2,Z). (2.34)

Theorem 2.12.1 The map j : SL(2,Z)\H→ C is one to one and surjective.

There is a beautiful history behind the j-function. According to [Apostol], Berwickin 1916 calculated the first seven coefficients of j, Zuckerman the first 24 in 1939,and Van Wijngaarden the first 100 in 1953. The only reason for computing suchnumbers, seems to be only the joy of playing with them and their mysteriousness.In [Apostol] we find also some divisibility properties of cn’s due to D.H. Lehmer in1942 and J. Lehner in 1949. An asymptototic formula due to Petersson in 1932 andRademacher in 1932 is also reported in [Apostol].

2.14 The numbers e1, e2,e3 27

In 1978 MacKay noticed that 196884 = 196883+ 1 and 196883 is the numberof dimensions in which the Monster group can be most simply represented. Basedon this observation J.H. Conway and S.P. Norton in 1979 formulated the Monstrousmoonshine conjecture which relates all the coefficients in the j-function to the rep-resentation dimensions of the Monster group. In 1992 R. Borcherds solved this con-jecture and got fields medal, see [Gan06] for more information on this conjecture.There proof does not give any clue why elliptic curves must have something to dowith the Monster group, and so the mystery involved around it still exists. For in-stance, in a private conversation J.H. Conway expressed the fact that the proof forhim is not satisfactory.

2.13 Poincare metric

In the upper half plane H one usually use the Poincare metric:

ds =dx⊗dx+dy⊗dy

y= Im(ω), ω :=

dz⊗dzIm(z)

ω is called the Hermitian form.

Exercise 18 The Poincare metric is invariant under the action of SL(2,R). Hint:Show that

d(Az)⊗d(Az)Im(Az)

=dz⊗dzIm(z)

The volume form of the Poincare metric is given by the imaginary part of ω:

dV = Im(ω) =−dx⊗dy+dy⊗dx

y=

dx∧dyy

=1−2i

dz∧dzIm(z)

We use the Poincare metric to prove the convergency of Eisenstein series.

2.14 The numbers e1, e2,e3

The following functions are defined in [Apostol]

e1 := ℘(Zω1 +Zω2,

ω12

)e2 := ℘

(Zω1 +Zω2,

ω22

)e3 := ℘

(Zω1 +Zω2,

ω1+ω22

)Proposition 2.14.1 The numbers e1,e2,e3 are distinct and we have

28 2 Modular forms

4℘3(z)−g2 ·℘(z)−g3 = 4

(℘(z)− e1

)(℘(z)− e2

)(℘(z)− e3

)(2.35)

Proof. Since ℘(z) is even, ℘′(z) is odd. Therefore,

−℘′(

12

ω

)=℘

′(−1

2ω

)=℘

′(

ω− 12

ω

)=℘

′(

12

ω

)∀ω ∈Λ

This implies that ω12 , ω2

2 , ω1+ω22 are roots of ℘′(z). The function ℘′(z) has a pole of

order 3 at z = 0∈ E and so the mentioned three points are simple roots of℘′(z). Thedifferential equation of ℘(z), implies that e1, e2, e3 are roots of the left hard side of2.35 Now e1, e2, e3 are distinct. We show that e1 6= e2. The elliptic function℘(z)−eihas a double root at ω1

2 , because ℘′( 1

2 ωi)= 0, all these for i = 1,2. If e1 = e2 then

this function must have pole order > 4 at z = 0, which is a contradiction. ut

Now let us regard ei’s as functions in τ ∈H and redefine

e1 = e1

(Zτ +Z,

τ

2

), e2 = e2

(Zτ +Z,

12

), e3 = e3

(Zτ +Z,

τ +12

)and consider them as holomorphic functions in τ . [PUT HERE t1, t2, t3 of Darboux-Halphen, We have ti = ei +g1 i = 1,2,3, See [Quasi-Moduler forms, I]].

2.15 Growth of coefficients

We follow [Serre].

Theorem 2.15.1 (Hecke) If f is a holomorphic cusp form for a group SL(2,Z) ofweight k then

fn = O(nk)

where f =∞

∑n=1

fnqn is the q-expansion of f .

Proof. Cauchy residue formula implies that

fn =1

2πi

∫δ

f (q)q−n dqq

Where δ is a small circle turning around q = 0 ∈ C anticlockwise. This is

fn =∫

f (τ)e−2πnτidτ = e2πyn∫ 1

0f (x+ iy)e−2πinxdx (2.36)

The function(

f (τ)(Imτ)k2

)is invariant under the action of SL(2,Z), and so, it

gives us a function in SL(2,Z)\H. Since

2.15 Growth of coefficients 29

| f (τ)| ∼ O(q)∼ O(e−2πy)

This function sis bounded when y→ ∞, and so, there exists a constant M such that

| f (τ)|6My−k2 ∀ ↑∈H

Putting this in 2.36 we have

| fn|6 e2πyn ·M · y−k2

Here, y can be any positive number, we put y = 1n and get the desired result.

As a corollary we get

Proposition 2.15.1 Let f be a holomorphic modular form of weight k for SL(2,Z).Then

fn = O(nk−1)

Proof. Let σk be the Einstein series of weight k, and λ ∈ C, be constant such thatλσk + f is a cusp form. The proposition follow from theorem 2.15.1 and the asymp-totic behaviour of Forier coefficients of σk:

σk−1(n) = ∑d|n

dk−1 = O(nk−1)

Because

σk−1(n)nk−1

= ∑d|n

(dn

)k−1

= ∑d|n

(1d

)k−1

6∞

∑d=1

1dk−1 = ξ (k−1)< ∞

Since n k2 compared to nk−1 is negligible, we get the result. Deligne in [Deligne

1973] has shown that for a cusp form f , we have

fn = O(

nk2−

12 σ0(n)

)(2.37)

Where σ0(n) is the number of positive divisors of n. This implies that

fn = O(

nk2−

12+ε

), ∀ε > 0

According to [Milne], Deligne in 1969 proved that 2.37 follows from the Weil con-jectures for varieties over finite fields and he proved the Weil conjectures in 1973.

30 2 Modular forms

2.16 Dedekind eta function

According to [Apostol] the Dedekind eta function

η(τ) := e2πiτ12

∞

∏n=1

(1− e2πinτ

)was introduced by Dedekind in 1877. We know that

∆(τ) = 2πi)12η

24

and the functional equation of ∆ with respect to the action of SL(2,Z). Taking 12throot of this we get

η

(aτ +bcτ +d

)= ε(cτ +d)

12 η(τ)

for som ε which is a 12th root of unity and depends only on A and τ . In fact

Theorem 2.16.1 (Dedekind functional equation) We have

η

(aτ +bcτ +d

)= ε(A)

(− i(cτ +d)

) 12η(τ)

For all A =

(a bc d

)∈ SL(2,Z) with c > 0, where

ε(A) : = exp(

πi(a+d

2c+S(−d,c)

))S(d,c) : =

k−1

∑γ=1

γ

k

(γ · h

r

− 1

2

)a : = a− [a]

For a proof see [Apostol] Theorem 3.4. For A =

(0 1−1 0

)we get

η

(−1τ

)= (−iτ)

12 η(τ)

2.17 Modular forms as k-fold differential

Let f be a meromorphic function in H. Show that the k-fold differential

f (τ)dτ⊗dτ⊗·· ·⊗dτ︸︷︷︸k−times

2.18 Petersson scalar product 31

is invariant under SL(2,Z),( and hence it induces a k-fold differential in SL(2,Z)\H)if and only if f is a meromorphic modular form of weight 2k for SL(2,Z)

2.18 Petersson scalar product

We follow [Lang] page 37. Our main theorem in this section is that the space ofcusps of weight k for SL(2,Z) has a basis of eigenforms. For this we have to put aHermitian form on Sk(SL(2,Z)). This will be the Petersson scalar product.

Let Γ ⊆ SL(2,Z) be a sugroup of finite index and let f ,g be two holomorphicfunctions on H. We define

〈 f ,g〉 :=1

[SL(2,Z) : Γ ]

∫Γ \H

f (τ) g(τ) yk dx∧dyy2

where τ = x+ iy. Ino order to digest this definition we make the following com-ments:

1. We havedx∧dy

y2 =i2

dτ ∧dτ

Im(τ)

and this is invariant under GL+(2,R) action. This follows from

d(Aτ) = (cτ +d)−2 det(A) ·dτ

Im(Aτ) = Im(τ) ·det(A)|cτ +d|2(2.38)

Let us define

f |kA = f (Aτ)(cτ +d)−k(detA)k2

Ω( f ,g) : = f (τ) g(τ) Im(τ)k i2· dτ ∧dτ

Im(τ)2

We haveΩ(

f |k A, g|k A)= Ω( f ,g)(Aτ) (2.39)

which follows from (2.38).2. We will take f ,g two modular form of weight k for a subgroup Γ ⊆ SL(2,Z) of

definite index. In this way for A ∈ Γ , Ω( f ,g) is invariant under the action of Γ

and so it gives a differential form on Γ \H. The integration is over this space. Theintegration can be also taken over a fundamental domain of the action Γ on H.

3. The factor [SL(2,Z) : Γ ] is inserted so that 〈 f ,g〉 becomes independent of thegroup Γ , that is if f ,g are modular forms of weight k for Γ1 ⊆ Γ2 then 〈 f ,g〉defined using the group Γ1 is the same as 〈 f ,g〉 defined by the group Γ2. For twogroups Γ1,Γ2 we repeat this argument twice for Γ1∩Γ2 ⊂ Γi, i = 1,2

4. We have to check that the integral is convergent. For this we have to assume thatf ,g are cusp forms. The fundamental domain for Γ is a union of finite number

32 2 Modular forms

of translates (under Γ ) of the classical fundamental domain F for SL(2,Z)

F =∪A∈Γ \SL(2,Z)AF

It is enough to show that th integral converges for Im(τ)→+∞. For a cusp formf we have

| f (τ)|<< c1e−c2Im(τ)

for sufficiently large Im(τ), and so, if one for g is a cusp form, then the integralconverges

For A ∈ GL+(2,R) Let us define

A′ := A−1 ·det(A).

Theorem 2.18.1 Let f ,g be cusp forms of weight k for Γ ⊂ SL(2,Z). Let also A ∈GL+(2,Q). We have

1. 〈 f |k A,g|k A〉= 〈 f ,g〉2. 〈 f |k A,g〉= 〈 f ,g|k A′〉3. The scalar product in item 2 above above depends only on double cosets

Γ AΓ

Proof.

〈 f |kA, g|k A〉= 1[SL(2,Z) : Γ ]

∫F

Ω(

f |k A, g|k A)

= ′′∫

FΩ( f ,g)(Aτ) = ′′

∫AF

Ω( f ,g)

Here, F is a fundamental domain for Γ , and so AF is a fundamental domain forAΓ A−1. We take smaller Γ such that

AΓ A−1 ⊆ SL(2,Z)

For instance, we replace Γ with A−1SL(2,Z)A∩Γ . This has still finite index inSL(2,Z). This gives us

[SL(2,Z) : Γ ] = [SL(2,Z) : AΓ A−1]

and the first part follows.The second part follows from the first part and

g|k A′ =(g|k A−1)

Note that det(A′) = det(A) and here we use the new definition of the slash operator.

2.19 Computing E2 as a double sum 33

Theorem 2.18.2 The Hecke operators acting on Sk(SL(2,Z)

)are Hermitian with

respect to the Petersson scalar product that is

〈Tn f , g〉= 〈 f , Tn g〉

Proof. For A ∈Matn(2,Z) we have

〈 f |k A, g〉= 〈 f , g|k A′〉

The mapSL(2,Z)Matn(2,Z)→ SL(2,Z)Matn(2,Z)

A 7→ A′

is a bijection

Proposition 2.18.1 The eigenvalues of T ′ns are totally real algebraic numbers 2.The space of cusp form has a basis of eigenforms.

2.19 Computing E2 as a double sum

In [Mov 12] we have considered the following family of elliptic curves

: y2 = 4x3− t2x− t3

α =dxy, ω = (x+ t1)

dxy

(t1, t2, t3) =(

a1 (τ),a2E4(τ),a3E6(τ)

)We have E→C/Zτ +Z and if δ1,δ2 are cycles in E corresponding to vectors τ,1

then (∫δ1

dxy ,

∫δ1(x+ t1) dx

y∫δ2

dxy

∫δ2(x+ t1) dx

y

)=

(τ −11 0

)Let us consider the equality corresponding to 1,2 and (2,2) entries.

t1 · τ =−1∫

δ1

xdxy

=−1−∫

τ

0℘(τ,z)dz (2.40)

t1 =−∫

δ2

xdxy

=−∫ 1

0℘(τ,z)dz (2.41)

The integration in 2.41 can be replaced with integration over ℘ bellow:

34 2 Modular forms

−ε ε 1− ε

Recall weierstrass zeta function

ξ (z) =1z+ ∑

ω∈Λω 6=0

1(z−ω)

+1ω

+z

ω2

It has the following properties

ddz

ξ (z) =−℘(z) (2.42)

ξ (z+ω)−ξ (z) = 2ξ

(12

ω

), ω /∈ 2Λ (2.43)

see [Silvermann] page 40. We continue the computation of t1

t1 +∫

℘

dξ (z) = xi(1− ε)−ξ (−ε) = 2ξ12

and so

a1E2(τ) = 4+ ∑d|(n,m)

6= (0,0)4

(1−2nτ2m)+

2(nτ +m)

+1

(nτ +m)2

In a similar way

t1 · τ =−1+2ξ

(τ

2

)( integration over δ1)

t1 · (τ +1) =−1+2ξ

(τ +1

2

)(integration over δ1 +δ2)

Problem 2.19.1 Write E2 as a convergent power series on the generalized perioddomain.

Chapter 3Elliptic curves and integrals

Although most of the seminars I couldn’t understand, after 10 times I started to getsomething and that something could be very useful for my development in mathe-matics or even to physics eventually, (S.-T. Yau in Kavli IPMU News No. 33 March2016).

3.1 Introduction

In this chapter we study elliptic curves over complex numbers and the correspondingelliptic integrals.

3.2 Elliptic integrals

We start with an elliptic integral of the form∫ b

a

dx√p(x)

, (3.1)

where p(x) is a polynomial of degree 3 and with three distinct real roots, and a,bare two consecutive elements among the roots of p and ±∞. For instance, the poly-nomial p(x) := 4x3 − t2x− t3, t2, t3 ∈ C has three distinct roots if and only if∆ := 27t3

2 − t23 6= 0. If p(x) has repeated roots one can compute it easily.

Exercise 19 Compute the indefinite integral∫dx√p(x)

, (3.2)

35

36 3 Elliptic curves and integrals

Fig. 3.1 The elliptic curve y2 = p(x) in the four dimensional space C2.

where p is a polynomial of degree 1 and 2. Compute it also when p is of degree 3 butit has double roots. These integrals are computable because y2 = P(x) is a rationalcurve! Let p be a polynomial of degree 3 and with three real roots t1 < t2 < t3. Showthat two of the four integrals∫ t1

−∞

dx√p(x)

,

∫ t2

t1

dx√p(x)

,

∫ t3

t2

dx√p(x)

,

∫ +∞

t3

dx√p(x)

,

can be computed in terms of the other two.

In many calculus books we find tables of integrals and there we never find aformula for elliptic integrals. Already in the 19th century, it was known that if wechoose p randomly (in other words for generic p) such integrals cannot be calculatedin terms of until then well-known functions. For particular examples of p we havesome formulas calculating elliptic integrals in terms of the values of the Gammafunction on rational numbers.

Exercise 20 For particular examples of polynomials p of degree 3, there are someformulas for elliptic integrals 3.2 in terms of the values of the Gamma function onrational numbers. For instance, verify the equality∫ +∞

7

dx√x3−35x−98

=Γ ( 1

7 )Γ ( 27 )Γ ( 4

7 )

2πi√−7

. (3.3)

In [Wal06] page 439 we find also the formulas

3.2 Elliptic integrals 37∫ 1

0

dx√1− x3

=Γ ( 1

3 )2

243 3

12 π

,∫ 1

0

dx√x− x3

=Γ ( 1

4 )2

232 π

12.

These formulas can be also derived using the software Mathematica. The Chowla-Selberg theorem, see for instance Gross’s articles [Gro78, Gro79], describes thisphenomenon in a complete way. The right hand side of (3.3) can be written in termsof the Beta function which is more natural when one deals with the periods of alge-braic differential forms.

The fact that we only need two of the integrals in (3.1) in order to calculate theothers, can be easily seen by considering the integration in the complex domainx ∈ C, in which we may discard the assumption that p has only real roots. Theintegration is done over a path γ in the x ∈ C domain which connects two pointsin the set of roots of p and ∞, and avoids other roots except at its start and endpoints. An amazing fact that we learn in a complex analysis course is that if the pathγ moves smoothly, without violating the properties as before, then the value of theintegral does not change. This is certainly the origin of homotopy theory, or at leastone of them. The next step in the study of elliptic integrals is the invention of the yvariable which is basically the square root of p(x):

E :=(x,y) ∈ C2 | y2 = p(x)

. (3.4)

This is called an elliptic curve in Weierstrass form.

Theorem 3.2.1 The set E as a toplogical space is a compact torus mines one point,see Figure 3.1.

Proof.

We add another point O to E and will call it the point at infinity. We write E =E∪O and sometimes by abuse of notation use the same letter E for E. If we writethe equation of E in homogeneous coordinates [x : y : z] ∈ P2 then

O = [0 : 1 : 0].

We define H1(E,Z) as the abelization of the fundamental group of E, that is, thequotient of the fundamental group of E by its subgroup generated by commutators:

H1(E,Z) := π1(E,b)/[π1(E,b),π1(E,b)], (3.5)

where for a group G, [G,G] is the subgroup of G generated by the commutatorsaba−1b−1, a,b ∈ G. It turns out that the integrals∫

δ

ω, δ ∈ H1(E,Z).


are well-defined. The following Proposition can be proved without the help of The-orem 3.2.1.

Proposition 3.2.1 The abelian group H1(E,Z) is free of rank 2, and hence, it isisomorphic to (Z2,+).

Proof (Sketch). We prove that the non-abelian group π1(E,b) is free and it is gen-erated by two elements. Let π : E→ C be the projection into x-coordinate and a bethe x-coordinate of b. The map π is 2 to 1, except at (t1,0),(t2,0),(t3,0), where ti’sare the root of p(x). An element δ of the homotopy group π1(E,b) can be identifiedwith δ := π(γ) ∈ π1(C\t1, t2, t3,a) which has this property that the multivaluedfunction y :=

√p(x) along γ is one valued. The closed paths γ1 and γ2 in Figure 3.1

are in the image of π , let us say π(δi) = γi, i = 1,2, and δ1,δ2 generate π1(E,b)freely. ut

Another important ingredient of H1(E,Z) is the skew symmetric intersection bilin-ear map

H1(E,Z)×H1(E,Z)→ Z

The cycles δ1 and δ2 can be choosen in such a way that 〈δ1,δ2〉=−1.

Proposition 3.2.2 Let δ1 and δ2 be generators of H1(E,Z) with 〈δ1,δ2〉=−1. Thenthe quotient

τ :=

∫δ1

dxy∫

δ2dxy

has positive imaginary part.

Proof (sketch). First we note that ω := dxy restricted to E is holomorphic even at the

infinity point O. The statement follows from

√−1(∫

δ1

ω

∫δ2

ω−∫

δ2

ω

∫δ1

ω

)=√−1∫

Eω ∧ ω > 0,

where ω = dxy . The equality follows from Stokes theorem for the complement of δ1

and δ2 in E. In local holomorphic coordinate system z = x1 +√−1x2 in E we have

ω = f (z)dz, f (z) holomorphic, and

ω ∧ ω = | f (z)|2dz∧dz =−√−1| f (z)|2dx1∧dx2.

ut

Definition 3.2.1 The lattice of elliptic integrals is∫H1(E,Z)

dxy

= Z∫

δ1

dxy+Z

∫δ

ω2,

where δ1,δ2 is a basis of H1(E,Z) with 〈δ1,δ2〉=−1.

3.4 Picard-Lefschetz theory 39

3.3 Elliptic curves in Weierstrass format

We consider again the familly of elliptic curves

Et = Et2,t3 =(x,y) ∈ C2∣∣y2 = 4x3− t2x− t3

, t ∈ T (3.6)

T := C2−(t2, t3) ∈ C2 | ∆ = 0, ∆ := t32 −27t2

3 . (3.7)

The curve Et is called an elliptic curve in the Weierstrass format.

3.4 Picard-Lefschetz theory

By Ehresmann’s theorem the fibration Et , t ∈ T is a C∞ bundle over T , i.e. it is lo-cally trivial. This is the basic stone for the Picard-Lefschetz theory (see for instance[Mov04] and the references there). It gives us the following linear action:

π1(T,b)×H1(Eb,Z)→ H1(Eb,Z)

where b ∈ T is a fixed point In order to calculate it we proceed as follows: Firstwe choose two cycles δ1,δ2 ∈ H1(Eb,Z). For the fixed parameter t2 6= 0, define thefunction f in the following way:

f : C2→ C, (x,y) 7→ −y2 +4x3− t2x.

The function f has two critical values given by t3, t3 =±√

t32

27 . In a regular fiber Et =

f−1(t3) of f one can take two cycles δ1 and δ2 such that 〈δ1,δ2〉 = 1 and δ1 (resp.δ2) vanishes along a straight line connecting t3 to t3 (resp. t3). The correspondinganti-clockwise monodromy around the critical value t3 (resp t3) can be computedusing the Picard-Lefschetz formula:

δ1 7→ δ1, δ2 7→ δ2 +δ1 ( resp. δ1 7→ δ1−δ2, δ2 7→ δ2).

It is not hard to see that the canonical map π1(C\t3, t3,b)→ π1(T, t) induced byinclusion is an isomorphism of groups and so the image of the monodromy groupwritten in the basis δ1 and δ2 is:

〈A1,A2〉= SL(2,Z), where A1 :=(

1 01 1

), A2 :=

(1 −10 1

).

Note that g1 := A−12 A−1

1 A−12 =

(0 1−1 0

), g2 := A−1

1 A−12 =

(1 1−1 0

)and SL(2,Z) =

〈g1,g2 | g21 = g3

2 =−I〉, where I is the identity 2×2 matrix.

Exercise 21 Discuss the Picard-Lefschetz theory as above for the Legendre familyof elliptic curves:


y2 = x(x−1)(x−λ )

3.5 Weierstrass uniformization theorem

Lett := (g2,g3) = (60G4(Λ),140G6(Λ)),

where G4 and G6 are complex numbers defined in (2.10). From Proposition 2.14.1it follows that g3

2−27g23 6= 0 and so t ∈ T, where T defined in (3.7) (do not use the

product formula for the discriminant in (2.33)). Let Et be the corresponding ellipticcurve in (3.6). From the differential equation of the Weierstrass ℘ function, seeTheorem 2.5.1, it follows that we have a map

f : C/Λ −→ Etz 7−→ (℘(Λ ,z),℘′(Λ ,z)), f (0) = O.

(3.8)

Theorem 3.5.1 The map (3.8) is an isomorphism of sets.

Actually, f is an isomorphism of Riemann surfaces. Moreover, we will see that Ethas a structure of a group and it is also a morphism of groups.

Proof. The heart of the proof is Theorem 2.4.1. Let (x,y) ∈ Et . The elliptic function℘(z)− x has a pole of order two at z = 0. Therefore, it must have two zeros z1,z2 inC\Λ . By the differential equation of ℘, we know that ℘′(z1),℘

′(z2) = y,−y.If y 6= 0 then there is exactly one of zi’s, let us say z1, such that ℘′(z1) = y. If y = 0then ℘ has a zero of multiplicity 2 at z1 and hence z1 = z2 in C/Λ . This argumentproves that f is one to one and surjective. ut

Let δ1 and δ2 be closed paths in C/Λ which are the images of the vectors ω1,ω2 ∈Cunder the canonical map C→ C/Λ . We also use the same notation for their imagesin Et under the map (3.8). We have∫

δi

dxy

= ωi, i = 1,2. (3.9)

In particular, the lattice of elliptic integrals for Et as above is Λ . The inverse of f in(3.8) can be given explicitly as follows:

f−1 : Et2,t3 → C/Λ (3.10)

f−1(P) :=∫ P

O

dxy

The integration can be interpreted in the following way. We take a path in the x-plane which connects the infity to the x-coordinate of P. We also choose a branch ofdxy = dx√

P(x). In geometric terms, this is to say, we connect P to the point at infinity

3.6 Sketch of the proof of Theorem (2.9.1) 41

O of E and we integrate dxy over this path. The map (3.10) is well-defined. We claim

that f−1 is the inverse of f .

f−1 f (z) =∫ f (z)

O

dxy

=∫ z

O

d℘(z)℘′(z)

=∫ z

Odz = z

Let P be the space of lattices.

Theorem 3.5.2 The map given by

p : C2\t32 −27t2

3 = 0→P

(t2, t3) 7→∫

H1(Et ,Z)

dxy

is well-defined and it is a biholomorphism which satisfies

p(t2λ−4, t3λ

−6) = λp(t2, t3), λ ∈ C, λ 6= 0.

Its inverse p−1 is given by the Eisenstein series:

Λ → (g2(Λ),g3(Λ)) = (60G4,140G6)

Proof. The fact that p is well-defined follows from Theorem 3.2.2. From Theorem3.8 we can easily derive pp−1 = Id which implies that p−1 is injective. The equal-ity p−1 p= Id is equivalent to the fact that p is injective. This in turn is equivalentto the injectivity of j : SL(2,Z)\H→ C.[Proof of this fact!!!!]

Exercise 22 Ex. 6.2, 6.4,6.6,6.7,6.14 of [Sil92].

3.6 Sketch of the proof of Theorem (2.9.1)

We regard modular forms as functions on the space of lattices P . Under the bijec-tion p any moduler form of weight k becomes a holomorphic function f in T withthe property

f (t2λ4, t3λ

6) = λk f (t2, t3) (3.11)

We use the growth condition of f and prove that f is a polynomial of degree k inthe weighted ring

C[t2, t3], weight (t2) = 4, weight (t3) = 6.

Since the lattices associated to τ and τ+1 are the same, we get a map D−0→P ,where D is the disc of radius 1 and center 0 in C. We compose it with p−1 and getthe map


Fig. 3.2 Discriminant

i : D→ C2, q 7→ (g2(q),g3(q))

Note that this map is even defined at 0 ∈ D and the image of 0 is in the discrim-inant locus ∆ = 0. The growth condition of f implies that f |Im(i) extends as aholomorphic function at i(0). Now, the C∗-action in C2 implies that f extends toa holomorphic function in C2\(0,0). By Hartogs theorem we conclude that f isholomorphic in C2. We write the Taylor series of f at (0,0) and (3.11) implies thedes

3.7 Some identities

Let Et , t = (t2, t3) be an elliptic curve in Weierstrass format and let P= (x(P),y(P))be a point of Et . In Theorem 3.5.1 in order to prove f f−1 = Id, we need to provethat:

x(P) =

(∫ P

O

dxy

)−2

+∑

(∫ P

O

dxy

)−2

−

(∫ P

O

dxy−∫ P

O

dxy

)−2

y(P) = (−2)∑

(∫ P

O

dxy

)−3

where the sum is taken over all, except one, non-homotopic paths in E wich connectO to P and

∫ PO means integration over this path. In the above formulas the integration

3.8 Schwarz function 43

over the exceptional path is denoted by∫ P

O . It is easy to see that this doesn’t dependon the choice of exceptional path.

The Eisenstein series can be written in the following way. Let δ0 ∈ H1(E,Z) bea primitive element, that is, it is not divisable by ab integer. We have

ζ (k) · ∑δ a monodromy of δ0

(∫δ

ω

)−k

= Gk(E,ω), k ≥ 4,k even (3.12)

where the sum runs in all monodromies δ ∈ H1(E,ω) of δ0. Since the monodromygroup of the Weierstrass family is SL(2,Z), we can also take the sum over all prim-itive elements of H1(E,Z). The sums

∑δ a monodromy of δ0, 〈δ ,δ0〉>0

(∫δ

ω

)−k

, k ≥ 3 (3.13)

are related to the discussion after Theorem 2.7.1. These functions seem to give anembedding of the universal cover of C2\27t3

2 − t23 = 0 inside some affine space.

The following sum might be also interesting:

∑δ a clockwise monodromy of δ0

(∫δ

ω

)−k

. (3.14)

3.8 Schwarz function

Let us take the Legendre family of elliptic curves and consider the Schwarz function

λ 7→∫

δ1dxy∫

δ2dxy

∈H

It is multivalued beacuse of the choice of δ1,δ2. In order to get a one valued functionwe restrict λ to H and choose cycles δi, i = 1,2 such that the projection of δ1 (resp.δ2) to the x-palne is a cycle around 0 (resp. 1 )and λ . In this way the Shwarz functionis a biholomorphisim. Its analytic continuations around λ = 0,1 corresponds to thePicard-Lefschetz transformation of δ1 and δ2.

Exercise 23 Show that the image of the Schwarz function is the region depicted in3.3i. The analytic continuations of the Schwarz map gives us the trianglization ofH.

i Reproduced from Wikipedia


Fig. 3.3 Fundamental domain

3.9 CM elliptic curves

Recall that P is a complex manifold whose points are lattices Λ = Zω1 +Zω2. Infact one can reinterpret it as follows:

Exercise 24 P is the moduli of triples (E,ω, p), where E is a Riemann surface ofgenus one, p ∈ E, and ω is a holomorphic differential form on E. The canonicalaction of a ∈ C∗ on P corresponds to the multiplication of ω with a−1.

Definition 3.9.1 The endomorphisim group of an elliptic curve E = EΛ is the set ofall holomorphic maps EΛ → EΛ which are in addition homomorphisims of groups.It is in one to one correspondance with

End(E) = α ∈ C | αΛ ⊂Λ

We have Z⊂ End(E) and we say that E is CM if the inclusion is strict.

Later, we will encounter two spcial CM elliptic curves as follows:

Exercise 25 Classify all elliptic curves E with α ∈ End(E) which is not multipli-cation by ±1 and is an isomorphism. More precisely show that we have only twosuch elliptic curve

E = E〈z,1〉, z = i,−1+ i

√3

2.

Exercise 26 Ex. 8,9,10,12 of Koblitz.

3.10 Fourier expansions and elliptic integrals

Let Ez,z ∈ P′ be a family of elliptic curves with a singular fiber at z = 0. Further,assume that the monodromy

H1(Ez,Z)→ H1(Ez,Z)

For z near 0 and in a basis δ1,δ2 ∈ H1(E,Z),〈δ1,δ2〉=−1 is given by

3.10 Fourier expansions and elliptic integrals 45(δ1δ2

)→(

1 10 0

)(δ1δ2

)Moreover, for a holomorphic differential form

I2 : =∫

δ2

ω = holomorphic at z = 0 and I2(0) 6= 0

I1 =lnz+ I1

2πiI1 = holomorphic at z = 0 and I1(0) 6= 0

For an explicit example see [Movasati 2012]. The mirror map is defined in the fol-lowing way

τ =I1

I2=

12πi

(lnz+

I1

I2

)For z near to O and taking the branch of lnz with 0 < Im(lnz)< 2π,τ is near i∞ andit is in the band

τ ∈ C|06 Re(τ)6 1, Im(τ)> 0

For a modular form f we use 2.37 and we get

fn =∫

γ

f(

I1(z)I2(z)

)e−2πin I1

I2(z) d

(I1

I2(z))

=∫

γ

f(

I1

I2

)−

∞

∑m=0

Z−n

m!

(−n

I1

I2

)m I2θ I1− I1 ·θ I2

I22

dzZ

=∞

∑m=0

(−n)m

m!Residue

(gn,m(z),z = 0

), (3.15)

gn,m(z) := f(

I1

I2

)I2 ·θ I1− Iiθ I2

I22

(I1

I2

)m

Z−n

In many concrete examples, f(

I1I2

)turns out to be a rational function in I2,θ I2,Z.

More over, for large m,gn,m(z) turns out to to be holomorphic at z and so the sumin 3.15 is a actually finite.

Chapter 4Rudiments of Algebraic Geometry of curves

Throughout the present text we work with a field k of arbitrary characteristic andnot necessarily algebraically closed. By k we mean the algebraic closure of k. Themain examples that we have in mind are

k=Q,R,C,Fp :=ZpZ

, k(t)

and a number field. A number field k is a field that contains Q and has finitedimension, when considered as a vector space over Q. A function field k(t) =k(t1, t2, . . . , ts) over a field k is the field of rational functions a(t1,t2,··· ,ts)

b(t1,t2,...,ts), where a and

b are polynomials in indeterminates t1, t2, . . . , ts and with coefficients in k. Later, wewill also use the field of p-adic numbers.

4.1 Curves

Let k be a field and k[x,y] be the space of polynomial in two variables x,y and withcoefficients in k. The n dimensional affine space over k is by definition

An(k) = k×k×·· ·×k, n times

and the projective n dimensional space is

Pn(k) := An+1(k)−(0,0, · · · ,0)/∼

a∼ b if and only if ∃λ ∈ k,a = λb

We will consider the following inclusion

An(k)→ Pn(k), (x1,x2, · · · ,xn) 7→ [x1;x2; · · · ;xn;1]

47

48 4 Rudiments of Algebraic Geometry of curves

and call Pn the compactification of An. The projective space at infinity is defined tobe

Pn−1∞ (k) = Pn(k)−An(k) = [x1;x2; · · · ;xn;xn+1] | xn+1 = 0.

For simplicity, in the case n = 1,2 and 3 we use x, (x,y) and (x,y,z) instead ofx1,x2, . . ..

Any polynomial f ∈ k[x,y] defines an affine curve

C(k) := (x,y) ∈ k2 | f (x,y) = 0.

The most famous Diophantine curve is give by f = xn +yn−1. We denote it by Fn.

Remark 4.1.1 The set C(k) may be empty, for instance take k=Q, f = x2+y2+1.This means that the identification of a curve with its points in some field is not agood treatment of curves. One of the starting points of the theory of schemes is thissimple observation.

For f ∈ k[x,y] we define the homogenization of f

F(x,y,z) = zd f (xz,

yz), d := deg( f ).

F defines a projective plane curve in P2(k):

C(k) := [x;y;z] ∈ P2(k) | F(x,y,z) = 0.

Note that∀c ∈ k,(x,y,z) ∈ k3, F(cx,cy,cz) = cdF(x,y,z).

One has the injectionC(k)→ C(k), (x,y) 7→ [x;y;1]

and for this reason one sometimes says that C(k) is the compactification of C(k).Let g be the las homogeneous piece of the polynomial f . By definition it is a homo-geneous polynomial of degree d. The points in

C(k)−C(k) = [x;y] ∈ P1∞ | g(x,y) = 0

are called the points at infinity of C(k). The set of points at infinity of Fn is empty ifn is even and it is [1;−1] if n is odd.

4.2 Charts

From now on we use the notation C to denote the curve C in the previous section. Wesimply say that the curve C in an affine chart is given by f (x,y) = 0. The projectivespace P2 is covered by three canonical charts:

αi : A2(k) → P2(k)

4.3 Schemes 49

α1(x,y) = [x;y;1], α2(x,z) = [x;1;z], α3(y,z) = [1;y;z].

and the curve in each chart is repectively given by

f1(x,y) := F(x,y,1) = 0, f2 := F(x,1,z) = 0, and f3 := F(1,y,z) = 0.

We are also going to use the notion of an arbitrary curve over k from algebraicgeometry of schemes. Roughly speaking, a curve C over k means C over k and theingredient polynomials of C are defined over k. The reader who is not familiar withthose general objects may follow the text for affine and projective curves as above.The set C(k) is now the set of k-rational points of C.

4.3 Schemes

We defined P2(k) and C(k) without defining P2 and C. In this section we fill thisgap and we explain the rough idea behind the definition of the schemes P2 and C.

By the affine scheme A2 we simply think of the ring k[x,y]. Open subsets of A2

are given by the localization of k[x,y]. We will need two open subsets of A2 givenrespectively by

k[x,y,1y] and k[x,y,

1x]

By the projective scheme P2 we mean three copies of A2, namely

k[x,y], k[x,z], k[y,z]

together with the isomorphism of affine subsets:

k[x,y,1y]∼= k[x,z,

1z], x 7→ x

z, y 7→ 1

z(4.1)

k[x,y,1x]∼= k[y,z,

1z], x 7→ 1

z, y 7→ y

z

k[x,z,1x]∼= k[y,z,

1y], x 7→ 1

y, z 7→ z

y

The best way to see these isomorphisims is, for instance: we look at an element ofk[x,y] as a function on the first chart A2(k) and for (a,b) in this chart we use theidentities

[a;b;1] = [ab

;1;1b] = [1;

ba

;1a].

We think of the the scheme C in the same way as P2, but replacing k[x,y] withk[x,y]/〈 f1〉 and so on. Here 〈 f1〉 is the ideal k[x,y] generated by a single elementf1. We can also think of C in the same way as P2 but with the following additionalrelations between variables:


f1(x,y) = 0 in k[x,y]

f2(x,z) = 0 in k[x,z]

andf3(y,z) = 0 in k[y,z].

Remark 4.3.1 The above discssion does not use the fact that k is a field. In fact, wecan use an arbitrary ring R instead of k. In this way, we say that we have an schemeC over the ring R.

The function field of the projective space P2 is defined to be

k(P2) := k(x,y)∼= k(x,z)∼= k(y,z).

where the isomorphisms are given by (4.1). The field of rational function on thecurve C is the field of fractions of the ring k[x,y]/〈 f1〉. Using the isomorphism (4.1),this definition does not depend on the chart with (x,y) coordinates. We can also thinkof k(C) as k(x,y) but with the relation f1(x,y) = 0 between the variables x,y. Anyf ∈ k(C) induces a map

C(k)→ k

that we denote it by the same letter f .

4.4 Singular and smooth curves

Definition 4.4.1 We say that an affine curve given by f (x,y) = 0 is singular if thereis a point (a,b) ∈ k2 such that

f (a,b) = fx(a,b) = fy(a,b) = 0.

where fx is the derivation of f with respect to x and so on. The point (a,b) is calleda singularity of the affine curve. A projective curve is singular if one of its affinecharts is singular. For a curve C, affine or projective, we denote by Sing(C)⊂C(k)the set of singular points of C.

4.5 Resultant and discriminant

Let f ,g ∈ Z[x]. In the order to compute the resultant of f ,g we proceed as follows.We take one of f and g, let us say f , and define

V := Q[x]/ f (x)·Q[x]

4.6 Discriminant 51

This is a vector space of dimension deg( f ) and it has the basis

1, x, x2, . . . ,xdeg f−1 (4.2)

Now consider the linear map

A : V →V

A(

P(x))

:= P(x) ·g(x)

We write this map in the basis 4.2 and compute its characteristic polynomial

F(s) := det(A− sI)

It might be better from computational point of view to find the minimal polynomialF(s) of A, that is, a polynomial of minimal degree and with coefficients in Q suchthat

F(A) = 0

It follows thatF(

g(x))= P(x) · f (x)

for some polynomial P(x) ∈ Q[x]. We have

P(x) · f (x)+Q(x) g(x) = F(0) (4.3)

where Q(x) =F(0)−F

(g(x)

)g(x)

∈ Q[x].

After multiplying 4.3 with a natural number we get an equality

P(x) · f (x)+Q(x) g(x) = ∆

where ∆ ∈Z, P,Q∈Z[x]. We call ∆ the resultant of f , g. The resultant of f (x), f ′(x)is called the discriminant of f .

4.6 Discriminant

Let R be a ring and k be the field of fractions of k.

Definition 4.6.1 Let us be given a polynomial f ∈ R[x,y, . . .], where (x,y, . . .) is amulti variable. The discriminant ideal of f contains all element ∆ ∈ R such that

∆ = f a1 + fxa2 + fya3 + · · · , for some a1,a2, . . . ∈ R[x,y, . . .].

When the discriminant ideal is principal we denote by ∆ its generator and we callit the discriminant of the polynomial f . It is defined up to units of the ring R, and


hence in the case R= Z it is defined up to sign. In this case we fix this ambiguity byassuming that ∆ is positive.

Exercise 27 Forf = y2− x3− t4x− t6, t4, t6 ∈ R; (4.4)

show that the discriminant ideal is generated by

∆ = 2(4t34 +27t2

6 ),

The corresponding a1,a2 and a3 in this case are given by

a1 = 2(27x3−27y2 +(27t4)x+(−27t6)),

a2 = 2(−9x4 +(−15t4)x2 +(−4t24 )), a3 =−54x3y+27y3 +(−54t4)xy.

In this case, ∆ is the resultant of the polynomials P(x) and Px(x), where P = x3 +t4x+ t6.

In all the case below, the discriminant ideal is principal and we have calculated itsgenerator.

f = y2− x3− t4x− t6− t2x2 + t1xy+ t3y. (4.5)∆ = (t61 t6 − t51 t3t4 + t41 t2t23 +12t41 t2t6 − t41 t24 −8t31 t2t3t4 − t31 t33 −36t31 t3t6 +8t21 t22 t23 +48t21 t22 t6 −8t21 t2t24

+30t21 t23 t4 −72t21 t4t6 −16t1t22 t3t4 −36t1t2t33 −144t1t2t3t6 +96t1t3t24 +16t32 t23 +64t32 t6 −16t22 t24

−72t2t23 t4 −288t2t4t6 +27t43 +216t23 t6 +64t34 +432t26 ).

a1 = 432x3 −432y2 +(−432t1)xy+(432t2)x2 +(−432t3)y+(432t4)x+(−t61 −12t41 t2 +36t31 t3 −48t21 t22

+72t21 t4 +144t1t2t3 −64t32 +288t2t4 −216t23 −432t6)

a2 =−144x4 +(−48t21 −192t2)x3 +(−t41 −8t21 t2 −120t1t3 −16t22 −240t4)x

2 +(−t51 )y+(6t41 t2 −20t31 t3+

24t21 t22 −40t21 t4 −80t1t2t3 +32t32 −160t2t4)x+(t41 t4 +4t31 t2t3 +8t21 t2t4 −16t21 t23

+8t1t22 t3 −64t1t3t4 +16t22 t4 −64t24 ).

a3 =−432x3y+216y3 +(−144t1)x4 +(324t1)xy2 +(54t21 −432t2)x

2y+(−3t31 −120t1t2 −216t3)x3+

(324t3)y2 +(108t1t3 −432t4)xy+(−t51 +4t31 t2 −21t21 t3 +8t1t22 −96t1t4 −216t2t3)x

2 +(t61 +6t41 t2 −18t31 t3 +24t21 t22

−36t21 t4 −72t1t2t3 +32t32 −144t2t4 +162t23 )y+(−t51 t2 + t41 t3 +2t31 t4 +4t21 t2t3 +8t1t2t4 −27t1t23 −216t3t4)x

+(−t51 t4 + t41 t2t3 −4t31 t2t4 − t31 t23 +8t21 t22 t3 +14t21 t3t4 −8t1t22 t4 −36t1t2t23 +32t1t24 +16t32 t3 −72t2t3t4 +27t33 );

This modulo 2 is:

∆ = t41 t2t2

3 + t51 t3t4 + t6

1 t6 + t31 t3

3 + t41 t2

4 + t43

a1 = t61 , a2 = t4

1 x2 + t51 y+ t4

1 t4

a3 = t51 x2 + t3

1 x3 + t61 y+ t5

1 t2x+ t41 t3x+ t2

1 t3x2 + t41 t2t3 + t5

1 t4 + t31 t2

3 + t1t23 x+ t3

3

For the casef = y2− x3− t4x− t6− t2x2. (4.6)

we have∆ = 2(4t3

2 t6− t22 t2

4 −18t2t4t6 +4t34 +27t2

6 );

4.8 Descriminant of projective curves 53

a1 = 2(27x3−27y2 +(27t2)x2 +(27t4)x+(−4t32 +18t2t4−27t6));

a2 = 2(−9x4 +(−12t2)x3 +(−t22 −15t4)x2 +(2t3

2 −10t2t4)x+(t22 t4−4t2

4 ));

a3 =−54x3y+27y3 +(−54t2)x2y+(−54t4)xy+(4t32 −18t2t4)y;

Modulo 3 this is:∆ = t2

2 t24 − t3

2 t6− t34

a1 = t32 , a2 = t2

2 x2 + t32 x+ t2t4x− t2

2 t4 + t24 , a3 = t3

2 y.

4.7 The main property of the discriminant

The main property of the singular ideal is

Proposition 4.7.1 Let I be any maximal ideal of R and so R/I is a field. The affinevariety f = 0 is singular over R/I if and only if the discriminant ideal is a subset ofI.

Let us describe our main examples for the above theorem.

1. R= k and so I = 0.2. R= Z. This is a principal ideal domain and so the discriminat ideal is generated

by some ∆ ∈ N and I is generated by some prime p ∈ N. In this case, f = 0 issingular over Fp if and only if p | ∆ .

3. R = k[t] and for a ∈ ks, I is the ideal of R generate by ti − ai, i = 1,2, . . . ,s.Let also assume that the discriminant ideal is generated by ∆ . In this case, thecurve f = 0 with the evaluation of the parameters t = a is singular if and only if∆(a) = 0.

Proof. We can assume that R= k is a field.

4.8 Descriminant of projective curves

Let C ⊂ P2 be a curve over the ring R. We define its discriminant ideal to be theideal generated by all discriminant ideals of C in some affine coordinates.

Exercise 28 Let us take the curve y2z− x3− 17z3 = 0 over Z. Its discriminant inthe affine coordinates z = 1, respectively y = 1, is 2 · 3 · 172, respectively 24. Itsdiscriminant is 24 ·3 ·72 (see the tex file of this text for the corresponding Singularcode).


4.9 Curves of genus zero

Let f ∈ k[x,y] and let C be the curve induced by f = 0 in P2. Let us assume that fis of degree 2 and it is irreducible over k. Further, assume that C(k) has at least onepoint P. Note that if we look C in an affine chart then this point can be a point atinfinity. The following procedure finds all the points of C(k). We fix a line L in P2,for instance take y = 0. For any point X ∈ L(k), we connect X to P by a line L′ andfind the second intersction f (X) of L′ with L. Since P ∈C(k), we have f (X) ∈C(k)and we get a bijection

f : L(k)→C(k)

Exercise 29 Use the above geometric argument and find all k-rational points of theDiophantine equation

tx2 + sy2 = t + s.

for some s, t ∈ k.

The argument discussed in the previous paragraph works if f ∈ k[x,y] is irreducibleover k and has degree 3 and the induced curve C in P2 is singular. In this case, we canprove that the singularity P of C is defined over k, that is P ∈C(k), and it is unique.This point serves us as the point with the same name in the previous paragraph.

Exercise 30 Find all the k-rationa points of the Diophantine equation

y2− x3− t4x− t6 = 0

for some t4, t6 ∈ k with 4t34 +27t2

6 = 0.

4.10 Curves of genus bigger than one

Let C/Q be a smooth projective curve of degree d in P2, i.e. its defining polynomialis of degree d. Its genus is by definition

g(C) :=(d−1)(d−2)

2

The main objective of the Diophantine theory is to describe the set C(Q) for thecurves defined over Q. The most famous example is the Fermat curve given by thepolynomial f = xn + yn− 1. The machinery of algebraic geometry is very usefulto distinguish between various types of Diophantine equations. For instance, onecan describe the rational points of genus zero curves. Genus one curves are calledelliptic curves and the study of their rational points is the objective of the presenttext. For higher genus we have a conjecture of Mordell around 1922 which is provedby Faltings in 1982:

4.12 Elliptic curves in Weierstrass form 55

A non-singular projective curve of genus> 1 and defined over Q has only finitelymany Q-rational points.

In fact, the above theorem is true even for number fields. For instance the abovetheorem says that the Fermat curve Fn has a finite number of Q-rational points.However, it does not say something about the nature of Fn(Q). Mordell’s conjecturefor function fields was proved by Y. Manin in 1963, see [Man63] .

4.11 Elliptic curves in Weierstrass form

In this chapter, we define what is an elliptic curve over a field k, we describe theWeierstrass format of an elliptic curve and we define the group structure of an ellip-tic curve. If the reader is not familiar with the notion of an a curve over a field, hecan use the curves in p2 which we worked out in the previous section.

We are ready to give the definition of an elliptic curve:

Definition 4.11.1 An elliptic curve over k is a pair (E,O), where E is a genus onecomplete smooth curve and O is a k-rational point of E, that is, O ∈C(k).

Therefore, by definition an elliptic curve over k has at least one k-rational point. Asmooth projective curve of degree 3 is therefore an elliptic curve if it has a k-rationalpoint. For instance, the Fermat curve

F3 : x3 + y3 = z3

is an elliptic curve over Q. It has Q-rational points [0,1,1] and [1,0,1]. However

E : 3x3 +4y3 +5z3 = 0

has not Q-rational points and so it is not an elliptic curve defined over Q. It is aninteresting fact to mention that E(Qp) for all prime p and E(R) are not empty. Thisexample is due to Selmer (see [Cas66, Sel51]).

4.12 Elliptic curves in Weierstrass form

An elliptic curve in the Weierstrass form E is the affine curve given by the polyno-mial

Et2,t3 : y2− x3− t2x− t3, t2, t3 ∈ k,

∆ := 2(4t32 +27t2

3 ) 6= 0,

and in particular k is not of charachteristic 2. In homogeneous coordinates it iswritten in the form

zy2− x3− t2xz2− t3z3 = 0.


It has only one point at infinity, namely [0;1;0], which is considered as the markedpoint in the definition of an elliptic curve. It is in fact a smooth point of E which istangent to the projective line at infinity of order 3 and [0;1;0] is the only intersectionpoint of the line at infinity with E. If char(k) = 2 then the curve Et2,t3 is alwayssingular. We have already seen in Proposition 4.7.1 that ∆ = 0 if and only if thecorresponding curve is singular.

4.13 Real geometry of elliptic curves

For a projective smooth curve C defined over R the set C(R) has many connectedcomponents, all of them topologically isomorphic to a circle. We call each of theman oval.

For an elliptic curve E defined over R we want to analyze the topology of E(R).For simplicity (in fact because of Proposition 4.18.1 which will be presented later)we assume that E = Et2,t3 is in the Weierstrass form. For (t2, t3)∈R2 let ∆ = 2(4t3

2 +27t2

3 ) be the discriminant of the elliptic curve E. We have:

1. If ∆ < 0 then E(R) has two connected components, one is a closed path in R2,which we call it an affine oval, and the other a closed path in P2(R). We call it aprojective oval.

2. If ∆ > 0 then E(R) has only one component which is a projective oval.3. If ∆ = 0 and t3 < 0 then E(R) is an α-shaped path in R2 (∞-shaped path in

P2(R)). In this case, we say that E has a real nodal singularity.4. If ∆ = 0 and t3 > 0 then E(R) is a union of a point and a projective oval. In this

case, we say that E has a complex nodal singularity.5. If t2 = t3 = 0 then E(R) look likes a broken line in R2. In this case, we say that

E has a cuspidal singularity.

Note that E(R) intersects the line at infinity only at [0;1;0]. To see/prove all thetopological statements above, it is enough to take an example in each class anddraw the corresponding E(R). Note that in the (t2, t3)-space each set defined by theabove items is connected and the topology of E(R) does not change in each item(see Figure 4.1 and 4.2, the correspondence between the values of t2, t3 and Et2,t3(R)are done by colours).

4.14 Complex geometry of elliptic curves

Let C be a smooth projective curve of genus g over a subfield k of C. It can beshown that C(C) is a compact (Riemann) surface with g(C) wholes (a sphere withg handles).

4.15 The group law in elliptic curves 57

-2.4 -1.6 -0.8 0 0.8 1.6 2.4

-1.6

-0.8

0.8

1.6

Fig. 4.1 Elliptic curves: y2− x3− t2x− t3 = 0

-5 -2.5 0 2.5 5

-2.5

2.5

Fig. 4.2 The discriminant curve 4t32 + t2

3 = 0

Exercise 31 Give a proof of the above statement using the followings. 2. Any com-pact Riemann surface is diffeomorphic to a sphere with some handles. 2. Riemann-Hurwitz formula.

In genus one case, therefore, the set C(C) is torus.

Exercise 32 For a smooth elliptic curve E over R and in the Weierstrass form de-scribe the real curves E(R) inside the torus (Hint: Use the Riemann-Hurwitz for-mula).

4.15 The group law in elliptic curves

Let C be a smooth cubic curve in P2. Let also P,Q ∈ C(k) and L be the line in P2

connecting two points P and Q. If P = Q then L is the tangent line to C at P. Theline L is defined over k and it is easy to verify that the third intersection R := PQ ofC(k) with L(k) is also in C(k). Fix a point O ∈C(k) and call it the zero element ofC(k). Define

P+Q = O(PQ)


For instance, for an elliptic curve in the weierstrass form take O = [0;1;0] the pointat infinity. By definition O+O = O.

Theorem 4.15.1 The above construction turns C(k) into a commutative group.

Proof. The only non-trivial piece of the proof is the associativity property of +:

(P+Q)+R = P+(Q+R)

The proof constitute of three pieces:1. Let Pi = [xi;yi;zi] be 8 points in P2(k) such that the vectors (x3

i , · · · ,z3i )∈ k10of

monomials of degree 3 in xi,yi,zi are linearly independent. A cubic polynomial Fpassing through all Pi’s corresponds to a vector a∈ k10 such that Pi ·a = 0 and so thespace of such cubic polynomials is two dimensional. This means that there is twocubic polynomial F and G such that any other cubic polynomial passing throughPi’s is of the form λF +µG and so it crosses a ninth point too.

2. We apply the first part to the eight points O,P,Q,R,PQ,QR,P + Q,Q + Rand conclude that (P+Q)R = P(Q+R). Note that from these 8 points it crossesthere cubic polynomials: C, the product of lines through (0,PQ,P+Q), (R,Q,QR),(P(Q+R),P,Q+R) and the product of the lines (0,QR,Q+R), (PQ,Q,P), (P+Q,R,(P+Q)R): P+Q PQ O

R Q QR(P+Q)R,P(Q+R) P Q+R

(each column or row corresponds to aline).

3. The morphisms C×C×C→C,(P,Q,R) 7→ (P+Q)+R,P+(Q+R) coincidesin a Zariski open subset and so they are equal.

Exercise 33 ([Sil92] p. 60) On the elliptic curve

E : y2 = x3 +17

over Q. We have points

P1 = (−2,3), P2 = (−1,4), P3 = (2,5), P4 = (4,9), P5 = (8,23)P6 = (43,282)

P7 = (52,375), P8 = (5234,378661)

verify:P5 = 2P1, P4 = P1−P3, 3P1−P3 = P7,

Prove that E(Q) is freely generated by P1 and P3 and there are only 16 integralpoints ±Pi, i = 1,2, . . . ,8 (see [Nag35]).

Exercise 34 [Kob93], p. 35, Problem 4b: For the elliptic curve En : y2 = x3− n2xfind an explicit formula for the x coordinates of inflection points.

Exercise 35 [Kob93], p. 36, Problem 7: How many elements of En(R) or of order2,3 and 4? Describe geometrically where these points are located.

4.16 Divisors 59

Exercise 36 [Kob93], p. 36, Problem 9: For an elliptic curve over R prove thatE(R) (as a group) is isomorphic to R/Z or R/Z×Z/2Z.

Exercise 37 [Kob93], p. 36, Problem 11:

4.16 Divisors

Let E/k be an elliptic curve over a field k. A divisor in E and defined over k is aformal finite sum D := ∑i ni pi, ni ∈ Z, pi ∈ E(k) such that it is invariant under theGalois group Gal(k/k), that is,

σ(D) = D, ∀σ ∈ Gal(k/k)

whereσ(∑

ini pi) := ∑

iniσ(pi)

The set of divisors over k, let us denote it by Div(E/k) form an abelian group in anatural way. For any rational function f ∈ k(E) we define

div( f ) := ∑i

ni pi

where f is of order ni at pi. It is a divisor defined over k. The set of such divisorsform an abelian group which we denote it by Div(k(E)). The Picard-Group of E isdefined to be

Pic(E) := Div(E)/Div(k(E)).

The Chern class map is defined in the following way

c : Pic(E)→ Z, c(∑i

ni pi) := ∑i

ni

we definePic0(E) := ker(Pic(E)→ Z)

We have a canonial map

E(k)→ Pic0(E), P 7→ P−O (4.7)

Proposition 4.16.1 The map 4.7 is an isomorphism of groups.

Proof. First of all we notice that it is a group morphism. Just for this proof we denoteby ⊕ the addition structure in E(k). Let L1, respectively L2, be the equation of theline in P2 passing through P,Q,PQ, respectively O,PQ,P⊕Q. We have L1

L2∈ k(E)

with the divisor


P+Q+PQ−O−PQ−P⊕Q

and so in Pic0(E) we have P−O+Q−O = P⊕Q−O.

4.17 Riemann-Roch theorem

Definition 4.17.1 We say that a divisor D = ∑i ni pi is positive and write D ≥ 0 ifall coefficients ni are non negative integers. In a similar way we define D≤ 0.

For a divisor D on a curve C/k define the linear system

P(D) = f ∈ k(C), f 6= 0 | div( f )+D≥ 0∪0

andl(D) = dimk(P(D)).

Theorem 4.17.1 (Riemann-Roch theorem) Let C be a smooth curve over k.

l(D)− l(K−D) = deg(D)−g+1,

where K is the canonical divisor and g is the genus of C.

We only need to know that the canonical divisor satisfies:

deg(K) = 2g−2

and so for deg(D)> 2g−2, equivalently deg(K−D)< 0, we have

l(D) = deg(D)−g+1. (4.8)

4.18 Weierstrass form revised

In this section we prove that any elliptic curve can be realized as a certain curve inP2. The following proposition is proved in [Sil92], III, Proposition 3.1.

Proposition 4.18.1 Let E be an elliptic curve over a field k. There exist functionsx,y ∈ k(E) such that the map

E→ P2, a 7→ [x(a);y(a);1]

give an isomorphism of E/k onto a curve given by

y2 +a1xy+a3y = x3 +a2x2 +a4x+a6, a1, · · · , . . . ,a6 ∈ k

4.19 Moduli of elliptic curves 61

sending O to [0;1;0]. If further char(k) 6= 2,3 we can assume that the image curveis given by

y2 = 4x3− t2x− t3, t2, t3 ∈ k, t32 −27t2

3 6= 0.

We call x and y the weierstrass coordinates of of E.

Proof. Using Riemann-Roch theorem and in particular (4.8) with g = 1 and D = nOwe get l(D) = n. For n = 2 we can choose x,y ∈ k(E) such that 1,x form a basisof P(2O) and 1,x,y form a basis of P(3O). The function x (resp. y) has a poleof order 2 (resp. 3) at O. Now P(6O) has dimension 6 and 1,x,y,x2,xy,y2,x3 ∈P(6O). It follows that there is a relation

ay2 +a1xy+a3y = bx3 +a2x2 +a4x+a6, a1, · · · , . . . ,a6,a,b ∈ k.

Note that ab 6= 0, otherwise every term would have a different pole order at O and soall the coefficients would vanish. Multiplying x,y with some constants and dividingthe whole equation with another constant, we get the desired equation. The mapinduced by x and y is the desired map (check the details).

If char(k) 6= 2,3 we make the change of variables x′ = x,y′ = y− a1x2 and we

eliminate xy term. A change of variables x′ = x− a23 , y′ = y− a3

2 will eliminate x2

and y terms.

Exercise 38 Write the following elliptic curves in the Weierstrass form:

y2 = x4−1, O = [0;1;0]

x3 + y3 = 1, O = [0;1;1]

4.19 Moduli of elliptic curves

Now we can state what is the moduli of elliptic curves.

Proposition 4.19.1 Assume that char(k) 6= 2,3. Two elliptic curves Et2,t3 and Et ′2,t′3

are isomorphic if and only if there exists λ ∈ k, λ 6= 0 such that

t ′2 = λ4t2, t ′3 = λ

6t3

The isomorphism is given by

(x,y) 7→ (λ 2x,λ 3y).

Proof. Let (x,y) and (x′,y′) be two sets of Weierstrass coordinate functions on anelliptic curve Et2,t3 . It follows that 1,x and 1,x′ are both bases of P(2O), andsimilarly 1,x,y and 1,x′,y′ are both bases for P(3O). Writing x′,y′ in termsof x,y and substituting in the equation of Et ′2,t

′3

we get the first affirmation of theproposition. The second affirmation is easy to check.


Combining Proposition 4.18.1 and Proposition 4.19.1 we conclude that the modulispace of elliptic curves over a field of characteristic 6= 2,3 is

M1(k) :=(A2(k)−(t2, t3) | 4t3

2 +27t23 = 0

)/∼

where

(t2, t3)∼ (t ′2, t′3) if and only if ∃λ ∈ k, λ 6= 0, (t ′2, t

′3) = (λ 4t2,λ 6t3).

If k is algebraically closed then this is the set of k-rational points of the weightedprojective space P2,3(k) mines a point induced in P2,3 by ∆ = 0. In this case thej-invariant of elliptic curves

j : M1(k)→ A(k), j[t2; t3] =1728 ·4t3

2

4t32 +27t2

3

is an isomorphism and so the moduli of elliptic curves over k is A1(k). However,note that if k is not algebraically closed then j has non-trivial fibers. For instance,all the elliptic curves

y2 = x3− t3, t3 ∈Q

are isomorphic over Q but not over Q.If j0 6= 0,1728 consider the elliptic curve:

E j0 : y2 + xy = x3− 36j0−1728

x− 1j0−1728

.

It satisfies j(E) = j0.

4.20 The addition formula for ℘

For a fixed z′ ∈ C, z′ 6= 0, let us consider ℘(z+ z′) which is double periodic in z,and hence, it is a rational function in ℘ and ℘′


℘(z+ y) =14

(℘′(z)−℘′(y)℘(z)−℘(y)

)2

−℘(z)−℘(y) (4.9)

Proof. Let f (z) be the difference between the left and the right hand sides of 4.9.Its only possible poles are in

z = 0, ±y

We examine the Lauerent expansion of f (z) at the point z = 0 and see that it isholomorphic at z = 0 and there it vanishs. In a similar way, it has no poles at z = y

4.20 The addition formula for ℘ 63

and so, at worst it has a simple pole at z =−y. Since f is double periodic we get theresult.

We let y goes to z and we get

℘(2z) =14

(℘′′(z)℘′(z)

)2

−2℘(z).

If we make derivation of 4.9 with respect to z we get a similar formula for℘′(x+y).A better organization of this formula is∣∣∣∣∣∣

℘(z), ℘′(z) |℘(y), ℘′(y) |−℘(x+ y) ℘′(z+ y), |

∣∣∣∣∣∣= 0

Note that we can state the same formulas in the algebraic context, that is, for anelliptic curve E over the field k with the Weierstrass coordinates x and y we have

x(P+Q) =14

(y(P)− y(Q)

x(P)− x(Q)

)2

− x(P)− y(Q) (4.10)

∣∣∣∣∣∣x(P), y(P) |x(Q), y(Q) |

−x(P+Q) y(P+Q), |

∣∣∣∣∣∣= 0 (4.11)

x(2P) =14

(6x(P)− 1

2 g2

y(P)

)2

−2x(P) (4.12)

Even though, we have proved 4.10, 4.11, 4.12 in the complex context, theyare valid for an elliptic curve over an arbitrary field. The argument for instance for4.11 is as follows: We consider g2,g3,x(P),y(P),x(Q),y(Q),x(P+Q),y(P+Q) asvariables. The equation 4.11 is in fact a collection of equations:

y(P)2 = 4x(P)2−g2x(P)−g3y(Q)2 = 4x(Q)2−g2 · x(Q)−g3The equation of line x passing through P,QThe equation of P+Q ∈ x

y(P+Q)2 = 4x(P+Q)3−g2 · x(P+Q)−g3

This is a variety defined over Z[

16

]


4.21 Why Schemes?

One of the fundamental observation in Grothendieck’s revolution of Algebraic Ge-ometry, replacing varieties with schemes, is that

Spec(Z) := prime ideals of Z ' 2,3,5, . . . , p, . . .

is like a parameter, for instance, the parameter λ in the legendry family of ellipticcurves:

Eλ : y2 = 4χ(2−1)(χ−λ ), λ ∈ C

We consider Eλ as a curve over the ring

C[λ ] := polynomials in λ with C− coe f f icients.

The prime ideals of C[λ ] are

Spec(C[λ ]) := (λ −λ0)C[λ ]∣∣∣λ0 ∈ C ' C

For a prime ideal P⊆ C[λ ], the residue field is naturally isomorphic to C

C[λ ]/(λ−λ0)C[λ ]−→C

p(λ ) 7−→ P(λ0)

The process of substituting λ with λ0, can be interpreted as considering Eλ overthe residue field. In a similar way for an elliptic curve over Z, for instance

E : y2 = x3 +1

The process of working modulo a prime number p is the same as considering E overthe residue field

Fp = Z/pZ

Another reason for using schemes is that defining ideal of affine varieties have moredata than the underlying variety. For instance, the underlying variety of the idealI = 〈x,xy〉 ⊂ k[x,y] is just x = 0, however, we have I = 〈x〉∩ 〈x,y〉 which meansthat we must look at the underlying variety as a union of x = y = 0 and x = 0.

Chapter 5Mordell-Weil Theorem

We have seen that for an elliptic curve over Q the set E(Q) is an abelian group andso

E(Q)/E(Q)tors

whereE(Q)tors := x ∈ E(Q) | nx = 0, for some n ∈ N,

is a freely generated Z-module.

Theorem 5.0.1 For an elliptic curve E over a number field k the group E(k) ofk-rational points is finitely generated abelian group.

The above theorem for k = Q was proved by Mordell. Its generalization for anarbitrary number field was proved by Andre Weil and it is known as the Mordell-Weil theorem. For the proof see [Lan78a, Hus04].

Let us take k = Q. The above theorem implies that the set of torsion pointsE(Q)tors of E(Q) is finite and there is a number r ∈ N such that

E(Q)∼= Zr⊕E(Q)tors.

The non-negative integer r is called the rank of E(Q).

5.1 Mordell-Weil theorem

The free part of E(k), even for k = Q, is mysterious. We do not know whetherthere exist an elliptic curve of arbitrary rank or not. Until 1977, only, elliptic curvesof rank 6 9 was known. This is somehow trivial. Then, Mestre rank 6 15, Nago1992-1994, rank=17, 20, 21, Fermigier, 1992-1997 rank 19, 20, Martin-McMiller,1998-2000, rank=23, 24, and finally in 2006, N.Elkies found an elliptic curve ofrank 28. This is the largest rank known until now. There are results saying that withprobability 1, an elliptic curve E/Q has rank 0 or 1.

65

66 5 Mordell-Weil Theorem

Proof of Mordell’s theorem consist of two stepsE(Q)/2E(Q) is finite (Weak Mordell)Existence of a hight function

Theorem 5.1.1 (descent theorem). Let Γ be a commutative group. Suppose thatthere is a function

h : Γ → [0,∞)

such that

1. For any real number M, the set P ∈ Γ |h(P)6M is finite.2. For every P0 ∈ Γ , there is a constant k0 such that

h(P+P0)6 2h(P)+ k0 ∀P ∈ Γ

3. There is a constant k such that

h(2P)6 4h(P)− k ∀P ∈ Γ

Then Γ is finitely generated.

Proof. Let Γ /2Γ = Q1, Q2, . . . , Qn. It means that for every P∈Γ ∃Qi1 , dependingon P such that

P−Qi1 = 2P2 ∈ 2Γ

We repeat this for P2. P−Qi1 = 2P1P1−Qi2 −2P2

...Pm−1−Qim = 2Pm

This implies that

P = Qi1 +2Qi2 +22Qi2 + . . .+2m−1Qim +2mPm

Now we use the height function . We have

h(P−Qi)6 2h(P)+ k′ ∀P ∈ Γ i = 1, . . . ,n

Where k′ is the maximum of ki’s attached to each Qi in property 2 of h. We useproperty 3 of h and we have

5.1 Mordell-Weil theorem 67

4h(Pj) 6 h(2Pj)+ k = h(Pj−1−Q j−1

)+ k

6 2h(Pj−1)+ k′+ k ⇒

h(Pj) 612

h(Pj−1

)+

k+ k′

4

=34

h(Pj−1)−14

(h(Pj−1)− (k+ k′)

)Therefore, if h(Pj−1)> k+ k′ then

h(Pj)634

h(Pj−1)

We do this process until for some m h(Pm) 6 k+ k′ and so by property 1 of h theset of such Pm is finite. We conclude that Γ is generated by

Q1, Q2, . . . , Qn, P ∈ Γ∣∣ h(P)6 k+ k′

Now, let us construct the heught function for E(Q). Let P = (x, y) ∈ E(Q). Write

x := mn , (m,n) = 1

H(P) := H(x) = max|m|, |n|h(P) := logH(P)

Let P = (x,y) =(

me2 ,

ne3

), m,n,e ∈ Z. We have

|n|2 6 |m|3 + |t2| e4|m|+ |t3| e6

6 k2 H(P)3 ⇒

|n|6 kH(P)32

Where k is a constant number which only depends on E. Let P=(x,y), P0 =(x0,y0).We would like to estimate h(P+P0).

h(P+P0) =

(y− y0

x− x0

)2

− x− x0

=(y− y0)

2− (x− x0)2(x+ x0)

(x− x0)2

=Ay+Bx2 +Cx+D

Ex2 +Fx+G

Here, we have used y2 = x3− t2x− t3 and A, B, . . . , G are constants depending onlyon E and P0. Now set P =

(me2 ,

ne3

), (m,e2) = 1 and (n,e3) = 1


H(P+P0)6max ∣∣Ane+Bm2 +Cme2 +De4∣∣, ∣∣Em2 +Fme2 +Ge4∣∣

But, we know that

|e|6 H(P)12 , |n|6 k ·H(P)

32 , |m|6 H(P)

Which implies thatH(P+P0)6 constant. H(P)2

Where the constant term depends only on E and P0. Taking the logarithm of this, wehave the property 2 of h.Let us prove property 3 of h. We have

x(2P) =f ′(x)

4 f (x)−2x

Where y2 = x3− t2x− t3 = f (x) is the elliptic curve E We assume that 2P 6= 0

x(2P) =P(x)Q(x)

, degP(x) = 4, degQ(x) = 3

Where the coefficients of P,Q only depends on the elliptic curve.Let 4(4 t3

2 +27 t23 ) This is the resultant of P(x)andQ(x). This means that

f1(x)P(x)+ f2(x)Q(x) = ∆

f1, f2 ∈ Z[x], deg f1, deg f2 6 3

We need alsog1(x)P(x)+g2(x)Q(x) = ∆ · x7

g1, g2 ∈ Z[x], deg g1, deg g2 6 3

This can be considered as the resultant of x4P( 1

x

), x4Q

( 1x

)Let x = a

b , (a,b) = 1 and

x(2P) =F(a,b)G(a,b)

, δ :=(

F(a,b), G(a,b))

Therefore,

f1(a,b) F(a,b)+ f2(a,b) G(a,b) = 4∆b7

g1(a,b) F(a,b)+ g2(a,b) G(a,b) = 4∆a7 (5.1)

This givesδ |4∆ and |δ |6 |4∆ |

and so

5.1 Mordell-Weil theorem 69

H(2P)>max |F(a,b)|, |G(a,b)| / |4∆ |

On the other hands

|4∆ b7|6 2 max | f1|, | f2| max |F |, |G| |4∆ a7|6 2 max |g1|, |g2| max | f |, |G|

(5.2)

Where f1, f2 . . .are evaluated at (a,b).Now f1, f2, g1, g2 are polynomials of degree 6 3.

max | f1|, | f2|, |g1|, |g2| 6C max |a|3, |b|3 (5.3)

Where C is a constant which only depends on E. Combining 5.2 and 5.3 we get

|4∆ |max |a7|, |b7| 6 2C ·max |a3|, |b3| max F(a,b), G(a,b)

and soH(2P) = max|F(a,b)|,|G(a,b)|

δ>

max|F(a,b)|,|G(a,b)|4∆

> (2C)−1max|a|, |b|= (2C)−1H(P)

For more on descent procedure see [Silvermann I] page 199 chapter VIII.

Proposition 5.1.1 Let E be an elliptic curve over Q. We have

h(P1 +P2)+h(P1−P2)6 2h(P1)+2h(P2)+ k (5.4)

Where k only depends on E.For a proof see [Silvermann I] page 216

Proposition 5.1.2 (Tate). Let E/Q be an elliptic curve

h(P) :=12

limN→∞

4−Nh(2NP)

exists.

Proof. Taken from [Silvermann I] page 288. We show that the sequence in lim isCauchy. In 5.4 we pu P = P1 = P2∣∣h(2P)−4h(P)

∣∣6 k

Where k is a constant depending only on k. For N >M > 0 integers, we have∣∣∣4−Nh(2NP)−4−Mh(2MP)∣∣∣= ∣∣∣∣N−1

∑n=M

4−n−1 h(2n+1P)−4−n h(2nP)∣∣∣∣

6N−1∑

n=M4−n−1

∣∣∣∣ h(2n+1P)−4h(2nP)∣∣∣∣6 N−1

∑n=M

4−n−1 C

6 C4M+1


Definition 5.1.1 h is called the canonical or Neron-Tate height of E.

Theorem 5.1.2 (Neron-Tate) The canonical height h satisfies

1. For all P,Q ∈ E(Q)

h(P+Q)+ h(P−Q) = 2h(P)+2h(Q)

2. For all P ∈ E(Q) and m ∈ Z

h(M,P) = m2h(P)

3. h is a quadratic form in E(Q), that is, h is even and the paring

〈· , ·〉 : E(Q)×E(Q)→ R〈P,Q〉= h(P,Q)− h(P)− h(Q)

is bilinear.4. For P ∈ E(Q) we have h(P)> 0 and

h(P) = 0 ⇔ P is a torsion point.

5. 2h−h is bounded on E(Q).

Proof. Taken from [Silvermann I] page 2295. We have ∣∣∣4−N h(2NP)−4−M h(2MP)

∣∣∣6 C4M+1

Take M = 0 and let N→ ∞. We get

|2h(P)−h(P)|6 C4

1. We haveh(P+Q)+h(P−Q) = 2h(P)+2h(Q)+O(1)

replace P,Q by 2NP, 2NQ and multiply it with 12 4−Nand then N→ ∞.

2. We know that h(mP)−m2h(P) is bounded. Again replace P by 2−NP, multiplywith 4−N and N→ ∞

The rest is left to the reader.Let P1, P2, . . .Pγ be a basis of the free part of E(Q) The regulator of E/Q is

definedRE/Q := det

∣∣〈Pi, Pj〉∣∣

5.2 Weak Mordell-Weil theorem 71

5.2 Weak Mordell-Weil theorem

Theorem 5.2.1 Let E be an elliptic curve over a number field k. Then E(k)/mE(k)is finite for all m ∈ N.

We give a proof of weak Mordell-Weil theorem for k = Q, the family

Ea,b : y2 = x3 + ax2 +bx

and for E(Q)/2E(Q). We follow Husemller, chapter 6 3, in which he use an explicit2-isogeny in chapter 4 5.

Chapter 6Torsions and isogeny

6.1 Torsion points

Using Weierstrass theorem we have seen a correspondence

The space of lattices −→ the space of pairs (E,ω) (6.1)

Where E is an elliptic curve over C and ω is a regular differential 1-form. Since wehave not defined these objects intrinsicly, then E must be taken in the Weierstrassformat an ω = dx

y . In the right hand side of 6.1 we can talk about pairs defined overan arbitrary field.

Definition 6.1.1 A pair E,ω is called an enhanced elliptic curve. There will beother enhancements, and in order to reduce confusion, we say that E is enhancedwith a regular differential 1-form.

Definition 6.1.2 For an elliptic curve E ′ over a field of characteristic zero k, the setof m-torsion in

E[m] = P ∈ E(k) | mP = O

This is a subgroup of E(k).

Proposition 6.1.1 We have an embedding of groups

E[m] → Z/mZ ×Z/mZ

and for k an algebraically closed field, this is an isomorphism.

Proof. We first prove the second part for k = C. We can assume that E = C/Λ inthis case

E[m]' 1m

Λ/Λ ' Z/mZ ×Z/mZ

Now, we prove the second part. Since k is a field of characteristic zero and E uses afinite numbers of elements of E we can assume that there is an embedding of fields

73

74 6 Torsions and isogeny

σ : k → C

Let Eσ be the elliptic curve over C obtained from E and regarding its coefficients ascomplex numbers. We have an embedding of groups

E[m] → Eσ [m]' Z/mZ ×Z/mZ

For k an algebraically closed field, the algebraic equation mP = O has m2 distinctsolution over C, and it is defined over k. This means that all these solutions aredefined over k and the result follows.

In particular, for an elliptic curve E over Q we have

E[m]' Z/mZ ×Z/mZ

LetE[2] = (t1,0),(t2,0),(t3,0),O

For families of elliptic curves with a 3-torsion point see [Husemller] Chapter 4, 2.

6.2 Isogeny

Let Λ ⊆Λ ⊆ C be two lattices and let

N := #Λ/Λ

This gives a map of torif : E→ E (6.2)

Which is induced by the identity map C→ C. Here, E = C/Λ and E = C/Λ

. Thisis actually a holomorphic map between two Riemann surfaces. It is as a morphismof groups. We have

f ∗ω = ω,

where ω and ω are the differential form dz induced in E, respectively. Furthermore

f−1([z])= [z]+Λ

Λ

which means that f is a N to 1 map with no ramification points.

Definition 6.2.1 f as in 6.2 is called an isogeny of degree N.

Definition 6.2.2 Let N := #Λ/Λ

. we have

NΛ ⊆ Λ ⊆Λ

6.2 Isogeny 75

and this gives us the maps

Eg−→ E

f−→ E[z] 7−→ [Nz]

[z] −→ [z]

In the level of differential forms

Nω ←[ ω ← [ ω

g is called the dual isogeny of f . Note that both F0g, g0 f are multiplication by Nmap.

Exercise 39 Let G be a finite abelian group generated by at most two elements.There are unique d1, d2 ∈ N such that

G' Z/d1Z⊕ Z/d1d2Z

Conclude that any isogeny E1 7→ E2 can be uniquely written as

E1α−→ E1

β−→ E2[z1]→ [d1z1]

Where α is a multiplication by d1 and ker(β ) is cyclic of order d2Let a ∈ C∗ and Λ ⊆ C be a lattice. Let also

E = C/aΛ ,E = C/Λ

We have the mapfa : E→ E [z] 7→ [az]

which is a bijection and its inverse is given by fa−1 . Moreover

f ∗a ω = aω

Under the mentioned isomorphism, the lattices aΛ and Λ corresponds to (E,aω),(E,ω)

Exercise 40 Show that the number of sublattices Λ ⊆Λ of a fixed lattice Λ with #Λ/Λ=

n isσ(n) = ∑

d|nd

Hinit: Take a basis of Λ ′ and Λ and show that

Λ ⊆Λ , #Λ/Λ= n−→SL(2,Z)\Matn(2,Z)

Λ ′ generated by(

a bc d

)(ω1ω2

)←−[

(a bc d

)This quotient has representatives

76 6 Torsions and isogeny(a bc d

)06 b6 d−1, a ·d = n

Note that for n = P prime the set of such lattices is

Λ = ZPω1 +Zω2, Z(ω1 +ω2)+ZPω206 b6 P−1

6.3 Isogeny II

Let E be an elliptic curve in the Weierstrass format and defined over Q, that is, t2, t3 ∈Q. We have

G := Λ/Λ⊆ E[n]

G⊆ E(Q)

We way think of E as the quotient E/G. The elliptic curve E is also defined over(Q) and the reason is as follows.The pull-back of ℘(z,Λ), ℘(z,Λ) by the map of isogeny is an elliptic functionwith respect to the lattice Λ . Therefore, we have

℘(z,Λ) = P(℘(z,Λ),℘′(z,Λ)

)℘′(z,Λ) = Q

(℘(z,Λ),℘′(z,Λ)

)where P and Q are rational functions in two variables and with coefficients in C

Theorem 6.3.1 If E is defined over Q then the isogeny

E→ E (x,y)→(

P(x,y),Q(x,y))

is defined over Q, that is E and P, Q are defined over Q.

This will be proved in §7.3 when we introduce Hecke operator.

Chapter 7Hecke operators

In this chapter we introduce one of the fundamental features of modular forms whichis responsible for many arithmetic properties. This is namely the Heck operatorsacting on the space of modular forms. There are many text books covering this topicperfectly, see for instance Apostol’s book [Apo90] Chapter 6. We will adopt a moregeometric approach suitable for the same topic in the context of algebraic geometryof elliptic curves. The first application of Hecke operators is the following.

Theorem 7.0.2 The numbers τ(n) are multiplicative, that is for all n,m ∈ N with(n,m) = 1 we have

τ(n ·m) = τ(n) · τ(m)

7.1 Hecke operators

So far, we have interpreted modular forms as functions in theree spaces:

1. The Poicare upper half plan H,2. the space P of lattices,3. and the affine space A2

C with the coordinate system (t2, t3).

In this section for each natural number n we want to define the Hecke operator

Tn : Mk→Mk

which is a linear map. It is given by one of the following equivalent definitions:

1. For f : H→ C a modular form of weight k we have

Tn( f ) =s

∑i=1

f |kAi,

where [A1], [A2], · · · [As]= SL(2,Z)/Matn(2,Z).

77

78 7 Hecke operators

2. For f : P → C a modular form of weight k we have

Tn( f )(Λ) = nk−1∑Λ ′

f (Λ ′),

where Λ ′ runs through all sublattices Λ ′ ⊂ Λ of index n. This means that#(Λ/Λ ′) = n.

3. For f a homogeneous polynomial of degree k in C[tt , t3], deg(t2)= 4, deg(t3)= 6we have

Tn( f )(t2, t3) = nk−1∑t ′

f (t ′),

where t ′ = (t ′2, t′3) runs through all parameters for which there is an isogeny α :

Et ′ → Et such that α∗( dxy ) =

dxy and deg(α) = n.

Exercise 41 Prove the equivalence of the above definitions.

Exercise 42 Prove that each equivalence class in SL(2,Z)/Matn(2,Z) is repre-sented exactly by one of the matrices(

d b0 n

d

), d|n, 0≤ b <

nd.

Using the above exercise we know that the action of the Hecke operator Tn on amodular form of weight k defined on the upper half plane is given by

Tn( f )(τ) = nk−1∑

a·d=n, 0≤b≤d−1d−k f (

aτ +bd

) (7.1)

Proposition 7.1.1 For two natural numbers n.m and Hecke operators Tn,Tm ∈Mk→Mk prove that

Tn Tm = ∑d|(n,m)

dk−1Tnmd2. (7.2)

In particular, for n and m coprime we have

Tn Tm = Tnm

and for p a prime number

Tp Tpe = Tpe+1 + pk−1Tpk−1 . (7.3)

Proof. We follow [Silvermann] page 69.The idea of the proof is more and less in [Silvermann I] page 62. Let n, m ∈

N and d|(n,m) be fixed. We prove that for pairs of isogenies.

E1α−−→ E2

β−−→ E, degα = n, degβ = m

There is a unique isogeny E1℘−−→ E, deg℘= nm

d2 such that

7.1 Hecke operators 79

℘ [d] = β α

and for h fixed we have d pairs of such isogenies α, β

This decomposition is inspired by the identity

σ(n) ·σ(m) = ∑d|(n,m)

d ·σ(nm

d2

)If this is the case, then

Tn Tm f (E,ω) = (n,m)k−1∑

E1 α−→E2 β−→

Ef(

E1,(β α)∗ω)

= ∑d|(n,m)

(nm)k−1 d ∑E1℘−→

Ef(

E1,(℘ [d])∗ω)

= ∑d|(n,m)

(nm)k−1 d−k f (E1, ℘∗ω)

= ∑d|(n,m)

dk−1 Tnmd2

we have used [d]∗ω = dω and f (E,d∗) = d−k f (E,∗)in the order to prove the affirmation on isogenies we prove the corresponding

affirmation on lattices. Let Λ ⊆ C be a fixed lattices and

Λ1 ⊆Λ2 ⊆Λ #Λ2

Λ1= n, #

Λ

Λ2= m

For (n,m) = 1, Λ2 is uniquely characterized by

Λ2 = x ∈Λ | nx ∈Λ1

and the affirmation is trivial. We fin a subgroup

Gd := Z/dZ×Z/dZ ⊆Λ/Λ1

and defineΛ3 = pull-back of Gd by Λ →Λ/Λ1

We havedΛ3 ⊆Λ1 ⊆Λ3

and the index in both inclusions dΛ3 ⊆Λ3 and Λ1 ⊆Λ3 is d2. Therefore Λ3 =Λ1.We have to show that there are d such Gd and hence Λ3.

ut

The formula (??) is summarized in the following formal equality:


∞

∑n=1

Tnn−s = ∏p(1−Tp p−s + pk−1−2s)−1

Let f be a modular form with the Fourier expansion:

f (z) =∞

∑n=0

anqn, q = e2iπz.

For m ∈ N, we have Tm f (z) =∞

∑n=0

bnqn, where

bn = ∑d|gcd(m,n)

dk−1amn/d2 .

Exercise 43 Can you show by algebraic geometric methods that for fixed t ∈C2\∆ = 0 the set of parameters t ′ with

α : Et ′ → Et ,α∗ dx

y=

dxy, deg(α) = n

is finite.

Let A be an element in the group generated by GL+(2,R) and (

λ 00 λ

)| λ ∈C∗ ∼=

C∗. Let also ω =

(ω1ω2

)∈P . Then

Aω ∈P.

Using the map p in Weierstrass uniformization theorem, we can translate the aboveprocess to the (t2, t3)-space. Namely, for each t ∈ C2 − ∆ = 0 and a basis ofthe homology δ1,δ2 ∈ H1(Et ,Z) with 〈δ1,δ2〉 = 1 and A as above we have a localholomorphic map t 7→ α(t). If we choose another basis of H1(Et ,Z) obtained by theprevious one by an element B ∈ SL(2,Z), then we have a new period matrix ABω .This is equal to Aω in P if and only if

ABA−1 ∈ SL(2,Z).

We conclude that if we choose representatives for the quotient

(A ·SL(2,Z) ·A−1∩SL(2,Z))\(A ·SL(2,Z) ·A−1) = [Ai] | i = 1,2, . . . ,s

then to each Ai we have associated a local map αi(t).

Exercise 44 Let Ai = ASiA−1, Si ∈ SL(2,Z). Show that

SL(2,Z)/Matn(2,Z) = [ASi] | i = 1,2, . . . ,s

and vice-versa.

7.2 Hecke and cusp forms 81

Exercise 45 Let n ∈ N. Are there polynomials pn,i(t2, t3), i = 2,3 such that pn :=(pn,2, pn,3) leaves ∆ = 0 invariant and

Tn( f )(t) = f (p−1n (t)), f ∈Mk,

where f (X) = ∑x∈X f (x).

7.2 Hecke and cusp forms

Theorem 7.2.1 Let f be a cusp form of weight k and suppose that f is an eigenform for all Hecke operators and fn = 1. Then

Tn f = fn · f

Proof. We have Tn f = λn · f and soλn fm = (Tn f )m = ∑

d|(n,m)dk−1 fnm

d2

We put m = 1 and get λn = fn

Definition 7.2.1 A normalized eigen function is a modular form f with1. f is a cusp form with f1 = 12. Tn f = fn · f

Exercise 46 (1.25 Silvermann II page 92) 1. Ek is an eigenform for all Heckeoperators. 2. if f is an eigen form for all Hecke operators and f is not a cuspform then f is a multiple of Ek

Theorem 7.2.2 Let f (τ) = ∑∞n=1 fn qn be a normalized eigen function of weigh k.

Thenfnm = fn fm (n,m) = 1fpn · fp = fpn+1 + pk−1 fpn−1 p prime

Proof. Tn f = fn · f and the theorem follows from [taken from Silvermann page 79]Our main example is

∆ = qπ(1−qn)24 = ∑n=1

τ(n)qn

We have

τ(n)τ(m) = τ(nm) (n,m) = 1

τ(P)τ(Pn) = τ(Pn+1) +P11τ(Pn−1)

(7.4)

This follows from the fact that S12

(SL(2,Z)

)is generated by ∆

The identities were conjectured by Ramanujan and proved by Mordell.

Exercise 47 Can you find a basis of Sn

(SL(2,Z)

), n = 14, 16, 18 wich are nor-

malized eigen forms


7.3 Proof of Theorem 6.3.1

First we prove that E is defined over Q.

In order to see the analogy between holomorphic and algegraic context we define(x,y) =

(℘(z,Λ), ℘′(z,Λ)

)(x, y) =

(℘(z,Λ), ℘′(z,Λ)

)f−1(O) and f−1(P) both contains N points and let

x(

f−1(O))= ∞,a2,a3, ...,as ⊆Q

x(

f−1(P1))= b1,b2, ...,br ⊆Q

(7.5)

Since E is defined over Q its torsion points are also defined over Q and the inclusionsQ above follows. There is no repetition among ai

′s (resp.bi′s). Note that the map

E→ E, X →−X

leaves both sets f−1(O) and f−1(P1) invariant. For f−1(P1) this follows from P1 =−P1. Moreover

x(x) = x(−x), X ∈ E

It follows that s = r. Analysing the set of poles and zeros of (x− e1) of we get

(x− e1)o f = a

r∏i=1

(x−bi)

r∏i=2

(x−ai)

For some a ∈ C. Since f sends torsions of E to E evaluating (x− e1)o f at any tor-sion points of E different from 7.5 we get a ∈Q.

In a similar way, we have

yo fy

=∏(x− ci)

∏(x−di)

Where ci, di′s are algebraic numbers obtained by evaluating x at f−1

(E[2]

).

Let f ∈Mk be modular form defined over Q.

P(x) =∏

(x− f (Λ)

)Λ ⊆Λ , #Λ/

Λ=n

7.3 Proof of Theorem 6.3.1 83

Fig. 7.1 Isogeny

The coefficients of P(x) are symmetric polynomials in N =σ(n) quantitiesN∑

i=1xi, ∑

i< jxi ji ∑

i< j<kxix jxk, ...

We can write these as polynomials with coefficients in Q of the quantities

N

∑i=1

xmi , m = 1,2,3, ...

These are Tn f m. it is enough to prove that Hecke operators send modular formsdefined over Q to modular forms defined over Q. This follows from the computationof q-expansion of (Tn f )′s.

We know that f E(Q)tors to E(Q)tors, therefore for an elliptic function P on Ewith poles and zeros on torsion points of E, the pull-back of P by f has poles andzeros in torsion points of E.

We write ℘′2(z,Λ)2 = 4

(℘(z,Λ)− e1

)(℘(z,Λ)− e2

)(℘(z,Λ)− e3

)℘(z,Λ)− e1 has a pole of order two at O and a zero of multiplicity 2 at a 2-torsionpoin. The pull-back of this function gives us the desired result.

Chapter 8Riemann zeta function

In this chapter we introduce the Riemann zeta function. We will follow mainly theRiemann’s original article ([Rie]) and the book [Edw01] which explain an historicalaccount on Riemann’s paper. Euler considered the zeta function

ζ (s) :=∞

∑n=1

1ns

for real s and Riemann introduced ζ (s) for complex s and its extension as a mero-morphic function to the whole s-plane. In particular, it was known before Riemannthat

ζ (2) =π2

6, ζ (4) =

π4

90,ζ (6) =

π6

945, ζ (8) =

π8

9450

8.1 Riemann zeta function

In this section we are going to study the first of all Zeta function, namely Riemannzeta function:

ζ (s) :=∞

∑n=1

1ns

We mainly use the Riemman’s original article [Rie].

Proposition 8.1.1 The series ζ (s) converges for all s ∈ C with ℜ(s)> 1 and

ζ (s) = ∏p

1(1− p−s)

.

where p runs over all primes.

Proof. We have |n−s| = nℜs and so it is enough to prove the proposition for s ∈R,s > 1. We have

85

86 8 Riemann zeta function

∞

∑n=2

1ns ≤

∫∞

1x−s =

x−s+1

−s+1|+∞

1 =1

s−1if s > 1

Again we assume that s is a real number bigger than 1. We have p−s < 1 and so

(1− p−s) =∞

∑m=0

p−ms.

By unique factorization theorem

∏p≤N

(1− p−s)−1 = ∑n≤N

n−s +RN(s).

Clearly

RN(s)≤∞

∑n=N+1

n−s.

Since ζ (s) converges we have RN(s)→ 0 as N→ ∞ and the result follows.

8.2 The big Oh notation

In this section for a ∈ R∪±∞ we define the interval Ia to be a small one sidedneighborhood of a. If a ∈ R this means that Ia = (a,a+ ε) or = (a− ε,a) for somesmall ε and for a =+∞ this means Ia = (b,+∞) for a big positive number b and fora =−∞ this means Ia = (−∞,−b) for a big positive number b.

Definition 8.2.1 Let f ,g be two complex valued function in Ia we write

f = O(g) or f ∼x→a g

to say that f (x)g(x) is bounded near a, that is there exists a constant M such that

| f (x) |≤M | g(x) |, ∀x ∈ Ia.

For three complex valued functions f ,g and h in Ia we write f (x) = h(x)+O(g(x))if f (x)−h(x) = O(g(x)).

We mainly use the following convergence criterion: Let f be a complex valuedcontinuous function in Ia,a ∈ R and

f ∼x→a (x−a)s, s ∈ R.

For a = ±∞ we assume f ∼x→a xs. We can extend this assumption to s = ±∞. Forinstance for a ∈ R the expression f ∼x→a (x−a)+∞ means

∀s ∈ R+, f ∼x→a (x−a)s

8.3 Gamma function 87

For a,s ∈ R the integral ∫Ia

f (x)dx

converges if s > −1. If a = ±∞, s ∈ R then the integral∫

Ia f (x)dx converges ifs <−1.

8.3 Gamma function

The gamma function is defined by

Γ (s) =∫

∞

0xs−1e−xdx, ℜ(s)> 0

It converges because near 0

xs−1e−x ∼x→0 xℜ(s)−1

and so for ℜ(s) > 0 the integral near zero converges. Near infinity it is alway con-vergent because

xs−1e−x ∼x→+∞ x−n−1, ∀n ∈ N.

We haveΓ (s) = (s−1)Γ (s−1) (8.1)

becauseΓ (s) =−

∫∞

0xs−1de−x = · · ·= (s−1)Γ (s−1)

Since Γ (1) = 1 this implies that

Γ (n) = (n−1)!, n ∈ N

and so the Γ -function is the interpolation of the factoriel function.The equalities

Γ (s) =Γ (s+1)

s= · · ·= Γ (s+n+1)

s(s+1) · · ·(s+n), n ∈ N

enables us to continue Γ analytically onto all of the complex s-palne with poles ofsimple order at s = 0,−1,−2, . . .. We have also Euler’s reflection formula

Γ (1− s)Γ (s) =π

sin(πs)(8.2)

which shows that Γ has not zeros. We have also the equality


Γ (s) Γ

(s+

12

)= 21−2s √

π Γ (2s).

Remark 8.3.1 Gauss introduced the notation Π(s) = Γ (s + 1) which is used inRiemann’s original article. The notation Γ is due to Legendre, see [Edw01] p. 8.

8.4 Mellin transform

The content of this section is taken from Wikipedia. The Mellin transform of afunction f defined on R+ is ∫

∞

0xs f (x)

dxx.

If f (x) is locally integrable along the positive real line, and

f (x)x→0+ = O(xu) and f (x)x→+∞ = O(xv)

then its Mellin transform converges in the fundamental strip [−u,−v].By definition the Γ function is the Mellin transform of e−x.

8.5 Analytic extension

From the definition of Γ -function it follows:

Γ (s)ns =

∫∞

0xs−1e−nxdx, ℜ(s)> 1 (8.3)

Taking sum for n = 1,2, . . . we obtain

Γ (s)ζ (s) =∫

∞

0

xs−1dxex−1

, ℜ(s)> 1.

One can see that near +∞ we have 1ex−1 = O(x−∞) and near 0 we have 1

ex−1 =

O(x−1). Therefore, the convergence strip for the above integral is ℜ(s) ∈ (1,+∞)(see the section on Mellin transform).

Now, we consider the integral

I(s) :=∫ +∞

+∞

(−x)s

ex−1dxx

Here, we have taken the branch of (−x)s = es ln(−x) in C\R+ such that ln(−x) fornegative x is a real number. The path of integration begins at +∞, moves to the leftup to the positive axis, circles the origin once in the counterclockwise direction, andreturns down to the positive real axis to +∞. Now the above integral is convergent

8.6 Functional equation 89

for all s and it gives an entire function in s. A simple calculation show that it is equalto

I(s) = (eπis− e−πis)∫

∞

0

xs

ex−1dxx, ℜ(s)> 1

In particular, this shows that I(s) vanishes in s = 2,3, . . .. Therefore, we get

2isin(πs)Γ (s)ζ (s) =∫ +∞

+∞

(−x)s

ex−1dxx

and by (8.2)

ζ (s) =Γ (1− s)

2πi

∫ +∞

+∞

(−x)s

ex−1dxx

This equality shows that

1. ζ (s) extends to a meromorphic function on the s palne2. it has a unique pole in s = 1. The point s = 1 is a simple pole of ζ .3. It vanishs in s =−2,−4,−6, . . ..

The third item follows from the following: The Bernoulli numbers Bk are given by

xex−1

=∞

∑k=1

Bkxk

k!= 1+

−121!

x+162!

x2 +−1304!

x4 +1426!

x6 +1328!

x8 +566

10!x10 + · · ·

For s ∈ Z,s≤ 0 we have

ζ (s) =−Γ (1− s)2πi

∫|x|=ε

(∞

∑m=0

(−1)s−1 Bmxm+s−2

m!)dx =−Γ (1− s)(−1)s−1 B1−s

(1− s)!

8.6 Functional equation

By Cauchy theorem for s with ℜ(s)< 0 we get

2sin(πs)Γ (s)ζ (s) = (2π)s((−i)s−1 + is−1)∞

∑n=1

ns−1

In other words the functionΓ (

s2)π−

s2 ζ (s)

remains invariant under s 7→ 1− s. By analytic continuation this holds in the wholes-plane.

Exercise 48 Find the values of ζ for even positive integers:

ζ (2n) =(2π)2n(−1)n+1B2n

2 · (2n)!


8.7 Second proof for functional equation

In the equality (8.3) we make the change of variables n→ n2π and s→ s2 and we

obtain:Γ ( s

2 )

ns π− 1

2 =∫

∞

0x

s2−1e−n2πxdx, ℜ(s)> 1

Define

ψ(x) :=∞

∑n=1

e−n2πx

and soζ (s)Γ (

s2)π−

12 =

∫∞

0x

s2−1

ψ(x)dx, ℜ(s)> 1 (8.4)

Exercise 49 Prove that ψ converges and

1+2ψ(x) = x−12 (1+2ψ(

1x). (8.5)

Moreover, the above integral converges for ℜ(s)> 1. We have the theta series

θ3(z) :=∞

∑n=−∞

q12 n2

, q = e2πiz, z ∈H

which is related to ψ byθ3(z) = 1+2ψ(−iz).

Using (8.5) one can see that

ζ (s)Γ (s2)π−

12 =

∫∞

1(x

s2 + x

1−s2 )ψ(x)

dxx− 1

s(1− s), s ∈ C (8.6)

for which the right hand side is convergent for all s ∈ C. We multiply the aboveequality by s(s−1)

2 and define

ξ (s) := Γ (s2+1)(s−1)π−

s2 ζ (s)

which is an entire function and satisfies ξ (s) = ξ (1− s).

Exercise 50 From 8.4 we have

is2 ζ (s)Γ (

s2)π−

12 =

∫ i∞

0z

s2−1

ψ(−iz)dz, ℜ(s)> 1

In the above formula, use the Schwarz map of the family of elliptic curves y2−x3 +3x− t = 0 and write ζ as an integral in the t-domain. For this you have to calculate

8.8 Zeta and primes 91

ψ(

∫δ1

dxy∫

δ2dxy) as a polynomial in elliptic integrals. In particulat, one may prove in this

way that ζ (s) for s integer is a period.

8.8 Zeta and primes

In this section we sketch the relation between primes and the zeros of ζ .

Theorem 8.8.1 The sum

∑ρ

(ln(1− sρ)+ ln(1− s

1−ρ))

where ρ runs over all roots of ξ with Im(ρ)> 0, converges absolutely. In particular,

ξ (s) = ξ (0)∏ρ

(1− sρ).

where the infinite product is taken in an order which pairs each root ρ with 1−ρ .

We havelnζ (s) = ∑

p∑n

1n

p−ns = s∫

∞

0J(x)x−s−1dx, ℜ(s)> 1 (8.7)

where

J(x) :=12( ∑

pn<x

1n+ ∑

pn≤x

1n).

The order of summation in 8.7 is unimportant because it is absolutely convergent.Now we use the Fourier inversion

J(x) =1

2πi

∫ a−i∞

a+i∞lnζ (s)xs ds

s, a > 1 (8.8)

we have also

ln(ζ (s)) = ln(ξ (0))+∑ρ

ln(1− sρ)− lnΓ (

s2+1)+

s2

lnπ− ln(s−1). (8.9)

The direct substitution of 8.9 in (8.8) leads to divergent integrals. For this reason wewrite

J(x) =1

2πi ln(x)

∫ a−i∞

a+i∞

dds

(lnζ (s)

s)xsds, a > 1 (8.10)

Now the substitution of (8.9) gives us convergent integrals and a formula for J(x):

J(x) = Li(x)− ∑Im(ρ)>0

(Li(xρ)+ ln(x1−ρ))+∫

∞

x

dtt(t2−1) ln(t)

+ ln(ξ (0)), x > 1,


whereLi(x) :=

∫ x

0

dtln(t)

.

Now we derive a formula forπ(x) = ∑

p<x1

using

π(x) =∞

∑n=1

µ(n)n

J(x1n ),

where µ(n) is zero if n is divisable by a prime square, 1 if n is a product of an evennumber of distinct primes, and −1 otherwise.

8.9 Other zeta functions

There are many generalizations of the Riemann Zeta function. One of them is al-ready used in §10.2. Bellow we explain how the zeta functions of a curve over afinite field is a generalization of the Riemann zeta function.

Consider a plane affine curve C : f (x,y) = 0, f ∈ Fp(x,y) defined over the fieldFp. In analogy with the Riemann zeta function we define

ζ (C,s) = ∏p

1(1− (Np)−s)

. (8.11)

where p runs over all non-zero prime ideals of Fp[C] := Fp[x,y]/〈 f (x,y)〉. Here Npis the order of the quotient Fp[C]/p. Since such a quotient is a finite integral domainit is a field and hence it has pn elements. We define deg(p) := n. This allows us toredefine the zeta function as follows:

Z(C,T ) = ∏p

1(1−T deg(p))

.

with ζ (C,s) = Z(C, p−s).

Proposition 8.9.1 Let C be a curve ove the finite field Fp. We have

Z(C,T ) = exp(∞

∑r=1

#C(Fpr)

rT r)

For a proof see [Mil96] p. 90.

Exercise 51 Calculate the zeta functions of A1 and P1.

Exercise 52 Calculate the zeta function of degenerated elliptic curves:

y2 = x3 +ax+b, 4a3 +27b2 = 0.

8.11 L-function of cusp forms 93

8.10 Dedekind Zeta function

In this section we give a summary of Dedekind Zeta functions. Let k be anumberfield and Ok be its ring of integers.

Theorem 8.10.1 The integer ring of a number field is a Dedekind domain, i.e everyideal a⊂Ok in a unique way can be written

a= pα11 pα2

2 · · ·pαss ,

where α1, . . . ,αs ∈ N0 and p1, . . . ,ps are prime ideals.

A character χ on Ok is a map from the set of non-zero ideals of Ok to C such that itis multiplicative:

χ(a1a2) = χ(a1)χ(a2).

We mainly use the Character χ ≡ 1. Formally, the Dedekind Zeta function is definedin the following way:

ζk(s) = ∑a

χ(a)

N(a)s = ∏p

11−χ(p)N(p)−s , N(a) = #(Ok/a),

where the sum is running in non-zero ideals of Ok and the product is running in theprime ideals of Ok.

Exercise 53 Discuss the convergence of the Dedekind Zeta function (put χ ≡ 1).

Exercise 54 Discuss the fact that the integer ring of a number field is not necessarilya unique factorization domain/principal ideal domain. Give examples of irreduciblebut not prime elements. Hint Q(

√−5)

2.3 = (1+√−5)(1−

√−5)

〈6〉= 〈2,1+√−5〉〈2,1−

√−5〉〈3,1+

√−5〉〈3,1−

√−5〉

Exercise 55 The ring of Gaussian integers is Z[i]. Prove that the prime ideals ofZ[i] are of two types:

p= 〈p〉, ifp≡ 3(4), = 〈a+ ib〉, if a2 = b2 = p≡ 1(4).

In the second case we say that p splits in Z[i]. Show that the only units of Z[i] are±1,±i. Show also that the only ideal which ramifies is p= 〈1+ i〉, i.e. p2 = 〈p〉.

8.11 L-function of cusp forms

We follows [Silvermann II] page 80-84We first define L-functions for the full modular group .


Let f =∞

∑n=1

fnqn f1 = 1, be a normalized eigenfunction of weight k. Then

fmn = fm fn , (n,m) = 1fpe · fp = fpe+1 + pk−1 fpe−1 e> 1


L( f ,s) :=∞

∑n=1

fn

ns = ∏P prime

11− fp p−s + pk−1−2s (8.12)

for Re(s)> k2 +1

Proof. The convergence follows from fn ∼ nk and the same convergence statementfor Riemanns zeta function

L( f ,s) = ∑ fn ·n−s = ∏P prime

∑e>

fpe · p−es

we have

(1− fp p−s + pk−1−2s)

(∑e>

fpe · p−es)= ∑

e>fpe p−es

︸︷︷︸A

−∑e>

fp · fpe p−s−es

︸︷︷︸B

+ ∑e>

fpe pk−1−2s−es

︸︷︷︸C

= A+C−∑e> 1( fpeH + pk−1 fpe−1 p−s−es− fp · p−s)

= A+C− (A−1− fp p−s)−C− fp−s = 1

The product in 8.12 is also called the Euler product of L-function.

Theorem 8.11.1 (Hecke): Let f be a cusp form of weight k for SL(2,Z) Then

1. L( f ,s) has an analytic extension to C2. R( f ,s) := (2π)−sΓ (s)( f ,s). Then R( f ,k− s) = (−1)

k2 R( f ,s)

Note that R is symmetric with respect to Re(s) = k2

We prove this theorem in a general context introduced by Hecke, see §8.12.

8.12 Hecke’s L-functions

The following is due to [Hecke1936]. Let us consider a series of the form

f = f0 + f1qλ−1+ f2 ·q2·λ−1

+ . . .+ fnqnλ−1+ . . .

8.12 Hecke’s L-functions 95

Where fi ∈C and λ ∈R+. we assume that f is convergent in the unit disk and henceif we set q = e2πiτ then it defines a holomorphic function

f : H→ C

We further assume that f satisfies

f(−1

τ

)= γ · (−iτ)k f (τ)

for some γ =±1 and k ∈QThe L-function attached to f is

L( f ,s) =∞

∑n=1

fn

ns

which converges in the region Re(s)> a if we assume that fn ∼ na. We have

Γ (s)ns =

∫∞

0xs−1 e−nx dx

which implies

L( f ,s)Γ (s) =∫

∞

0xs−1

( ∞

∑n=1

fn · e−nx)

dx

we make the change of variables x→ 2πx ·λ−1 and get

R( f ,s) :=L( f ,s) ·Γ (s)(2πλ−1)s =

∫∞

0xs−1 f (τ)dx , τ := ix

Where f = f0 + f . Note that the functional equation of f is

f(−1τ

)= f0 · (γxk −1)+ γxk f (τ)

We have

R( f ,s) =∫ 1

0+∫

∞

1=∫

∞

1x1−s f

(−1τ

)x−2 dx+ xs−1 f (τ)dx

=∫

∞

1γx−1−s+k f (τ)dx+ xs−1 f (τ)dx+ f0

∫∞

1x−1−s(γxk −1)dx

=∫

∞

1

(γx−1−s+k + xs−1

)f (τ)dx+ f0 ·

(1−s− γ

−s+ k

)We have used Re(s) > k, Re(s) > 0 in order to compute the last integral. the firstintegral is a holomorphic entire function in s ∈ C and so we conclude that R( f ,s)extends to a meromorphic function in s ∈ C with possible poles at s = 0 and s = k.Moreover, it satisfies


R( f , k− s) = γ R( f ,s)

Here is the place, where we use γ =±1Riemann has used the theta series

θ3 =+∞

∑n=−∞

q12 n2

= 1+2∞

∑n=1

q12 n2

In this case

θ3

(−1τ

)= (−iτ)

12 θ3(τ)

and so k = 12 , γ =+1, λ = 2. In this case

L(θ3,s) = 2∑1

(n2)s = 2ξ (2s)

and so we get the functional equation of ξ (s)

Note that we could also use Einstein series and define.

L(Ek ,s) = ∑n

σk−1(n)ns =

∞

∑d=1

∞

∑m=1

1ds−k ms

= ξ (s− k)ξ (s)

and so we will get again the functional equation of ξ (s)

Exercise 56 Use the functional equation of E2

E2

(−1τ

)= τ

2 E2(τ)+122πi

τ

and describe L(E2,s).

8.13 L-function of CY-modular forms

The theory of Calabi-Yaw forms developed in [Mov17, ?] still lives its infancy.The lack of applications is one the big obstacles for developing the theory further.Looking at the history of (elliptic) modular forms, the author feel himself like E.Hecke who is one of the main resposables for the arithmetic of modular forms,however , he didnt see the most amazing arithmetic applications of modular formsin his life time. In the present text, we would like to develope of L-functions attachedto CY-modular forms.Recall the classical L-function attached to a cusp form f .

8.13 L-function of CY-modular forms 97

L( f , s) =∫

∞

of (τ)τs · dτ

τ=∫

℘

We regard ℘ as path ℘1−℘2, where ℘1 connects i∞ to i and ℘2 is

℘2 = S ·℘1, S =

(0 1−1 0

)Which connects i to O. in some sense, L is attached to S and its fixed point i.Now, consider A =

(1 1−1 0

), with A3 =−I, AP = P.

Under the iteration of A we have

i∞→−1→ 0→ i∞τ → τ+1

−τ→ −1

τ+1 → τ

Furthermore, we have the functional equations

f(

τ +1−τ

)= (−τ)k f (τ) f

(−1

τ +1

)= (τ +1)k f (τ)

Let δ1 be the path Re(P) which connects i∞ to P, δ2 = A δ1, δ3 = Aδ2, see FigureLet also

℘1 = δ1−δ2, ℘2 = δ2−δ3, ℘3 = δ3−δ1

Li :=∫

℘i

f (τ)τS−1, i = 1,2,3(8.13)

We haveL1 +L2 +L3 = 0

The integrals in 8.13 are convergent at i∞,−1, 0 respectively.For τ = it we have

| f (it)|6M · e−2πt

andlim

t→+∞e−t · tm = 0 ∀m> 1

Therefore ∣∣∣∣∫℘1

f (τ)τS−1dτ

∣∣∣∣6 ∣∣∣∣∫ ∞

Pf (it)tS−1dt

∣∣∣∣6M∫ +∞

PtRe(s)−1+1


Fig. 8.1 L-function

Chapter 9Congruence groups

In this chapter we work with modular forms for a congruence subgroup of SL(2,Z).One of the most well-known applications of such modular forms is the so-calledarithmetic modularity theorem.

9.1 Congruence groups

We have seen that SL(2,Z) appears as the monodromy group of the Weierstrassfamilly of elliptic curves. If we take other families of elliptic curves and computethe corresponding mondromy group then we will get subgroups of SL(2,Z) of finiteindex. Congruence groups are the most well-known subgroups of SL(2,Z). Let Nbe a positive integer number. Define

Γ (N) :=

A ∈ SL(2,Z)|A≡(

1 00 1

)( mod N)

It is the kernel of the canonical homomorphism of groups SL(2,Z)→ SL(2,Z/NZ).

Definition 9.1.1 A subgroup Γ ⊂ SL(2,Z) is called a congruence subgroup of levelN if

1. Γ (N)⊂ Γ

2. A(

1 N0 1

)A−1 ∈ Γ ∀A ∈ SL(2,Z)

Our main examples are

Γ0(N) :=

A ∈ SL(2,Z)∣∣A≡ (∗ ∗0 ∗

)( mod N)

Γ1(N) :=

A ∈ SL(2,Z)∣∣A≡ (1 ∗

0 1

)( mod N)

99

100 9 Congruence groups

For a description of a fundamental domain for the action of Γ0(p), p a prime number,see Apostol’s book [Apo90] Theorem 4.2.

Definition 9.1.2 A holomorphic function f : H→ C is called a moduler form ofweight k for Γ if

1. f |kA = f ∀A ∈ Γ

2. For all A ∈ SL(2,Z) lim f |kA = exists and < ∞

Im(τ)→+∞

9.2 Weil pairing

Let Let Λ = Zω1 +Zω2 ⊂ C be a lattice with Im(ω1ω2)> 0 and let E = C/Λ be the

corresponding elliptic curve. For a natural number N ∈ N an N-torsion [z] ∈ E is anelement with Nz = 0 in E, or equivalently Nz ∈Λ . The subgroup of E of N-torsionsis

E[N] : = z ∈ E|nz = 0 '1N Λ

Λ

'

aω1 +bω2

N

∣∣∣∣a,b ∈ ZNZ

' Z

NZ× Z

NZ

The following definition of Weil pairing is taken from [Silverman II], page 89.

Definition 9.2.1 Let E be an elliptic curve. The Weil pairing is

eN E[N]×E[N]→ µN

eN

(aω1 +bω2

N,

cω1 +dω2

N

)= e2πi ad−bc

N

HereµN := e

2πikN | k ∈ Z.

Exercise 57 Prove that the above definition is well-defined.

Theorem 9.2.1 Let

Y0(N) := Γ0(N)\H, Y1(N) := Γ1(N)\H, Y (N) := Γ (N)\H.

1. The set Y0(N) is the moduli space of pairs (E,C), where E is a complex ellipticcurve and C is a cyclic subgroup of E of order N.

2. The set Y1(N) is the moduli space of pairs (E, p), where E is a complex ellipticcurve and p is a point of E of order N.

9.4 Modular forms for congruence groups 101

3. The set Y (N) is the moduli space of pairs (E,(p,q)), where E is a complex ellipticcurve and (p,q) is a pair of points of E that generates the N-torsion subgroup ofE with Weil pairing eN(p,q) = e

2πiN .

Proof. We only prove item 2. The others are left to the reader.

Exercise 58 Prove items 1 and 3 above.

9.3 Moduli spaces of elliptic curves

We consider the following moduli spaces, which we call them period domains. Thedifference between these moduli spaces and those in theorem (9.2.1) is the presenceof the differential form ω together with E. Recall that by integration the pair (E,ω)is identified with a lattice Λ ⊂ C.

P := moduli of elliptic curves (E,ω)

P1(N) := moduli of (E,ω,z), z ∈ E[N]

P(N) := moduli of (E,ω,z1,z2), z1,z2 ∈ E[N], eN(z1,z2) = ζN

P0(N) := moduli of (E,ω,C), C ⊆ E[N] cyclic group of order N.

Let f be an elliptic function of weigh k with poles at 0. For instance, we use f =℘,℘′ which are of weight 2 and 3, respectively. We know the following functionson period domains

f1,N : P1(N)→ C, f1,N(E,z) = f (Λ ,z)

f iN : P(N) → C, fN(E,z1,z2) = f (Λ ,zi), i = 1,2

f0,N : P0(N)→ C, f0,N(E,C) = f (Λ ,z) = ∑z∈C

f (Λ ,z)

Any function g as above has the following functional equation:

g(E,aω,∗) = a−kg(E,ω,∗), ∀a ∈ C∗. (9.1)

where k is the wight of the elliptic function f .

9.4 Modular forms for congruence groups

We consider the following maps


i : H→P1(N) τ 7→ (E,ω,1N)

i : H→P(N) τ 7→(

E,ω,(

τ

N,

1N

))i : H→P0(N) τ 7→

(E,ω,

1N,

2N, . . . ,

N−1N

)here E = C/Zτ +Z and ω = dz

Proposition 9.4.1 Let f be an elliptic function of weigh k with poles at 0. Thenthe pull-back of f1,N , fN , f0,N by i is a holomorphic modular form of weight k forΓ1(N),Γ (N), and Γ0(N) respectively.

Proof. We only prove the P1(N) case

f1,N

(aτ +bcτ +d

)= f

(Z

aτ +bcτ +d

+Z,1N

)= (cτ +d)+k f (Z(aτ +b)+Z(cτ +d),

cτ +dN

= (cτ +d)+k f (Zτ +Z,1N)

For all(

a bc d

)∈ Γ1(N). Now we have to show the growth condition. For this it is

enough to assume that f =℘ or ℘′. In these cases the affirmation follows from

limIm(τ)→∞

℘(Zτ +Z,z) = ∑m∈Z

1(z−m)2 −

1m2

limIm(τ)→∞

℘′(Zτ +Z,z) = −2 ∑

m∈Z

1(z−m)3

ut

Exercise 59 Show that

#[SL(2,Z) : Γ1(N)] = deg(P1(N)→P) = n2 Π

(1− 1

p2

)#[SL(2,Z) : Γ (N)] = deg(P(N)→P) = n3 Π

(1− 1

p2

)#[SL(2,Z) : Γ0(N)] = deg(P0(N)→P) = n Π

(1− 1

p

)

9.5 q-expansion

Let f be a modular form for a congruence group Γ of level N. It follows that for allA ∈ SL(2,Z), f |kA has a q-expansion. Let

9.5 q-expansion 103

qN := e2πiτ

N

We have

T =

(1 10 0

), T N =

(1 N0 1

)∈ Γ (N)

and so( f |kA)T N = f |k(AT NA−1)A = f |kA

This implies that

f |kA =∞

∑n=0

an qnN , an ∈ C.

Proposition 9.5.1 If f is a modular form of weight k for SL(2,Z) then g(τ) :=f (Nτ) is a modular form of weight k for Γ0(N).

Proof. First note that Nk−1 ·g = f |k(

N 00 1

)and so g(τ) is a priori a modular form

for (N 00 1

)−1

SL(2,Z)(

N 00 1

).

For A =

(a bc d

)∈ SL(2,Z) we have

(N−1 0

0 1

)(a bc d

)(N 00 1

)=

(a N−1b

N · c d

)which means that g is modular for Γ0(N).

Exercise 60 Compute the q-expansions of ℘1,N , ℘N , ℘0,N .

We can construct modular functions for congruence groups by division for in-stance:

ϕ(τ) :=∆(Nτ)

∆(τ)

is a modular function for the group Γ0(N). This follows from the functional equationof ∆(τ) is follows that ϕ is holomorphic in H It is also holomorphic at i∞:

ϕ(τ) =

qN∞

∏n−1

(1−qnN)24

q∞

∏n−1

(1−q)= qN−1

(1+

∞

∑n=1

bn qn

)

and actually it has a zero of order N−1 at q= 0. There are no non-constant holomor-phic functions on a compact Riemann surface, and so, ϕ as a function on Γ0(N)\His necessarily meromorphic. Since in Γ0(N)\H∪i∞ it has only one zero which is


of order N− 1 doing q-expansions in other cusps we must have poles. For this weuse

∑∗∈Γ0(N)\Q

ord∗( f ) = 0

Note that the Fourier coefficients bn of ϕ are integers. For more information on ϕ

see Apostol’s book [Apostol], section 4.7

9.6 Transcendental degree of modular forms

Let Γ ⊆ SL(2,Z) be a subgroup of finite index a. Let also f ∈Mk(Γ ). We define

a

∑i=0

ga−i ·X i = ∏A∈Γ \SL(2,Z)

(X− f |kA) , g0 := 1 (9.2)

Proposition 9.6.1 We have gi ∈Mk·i(SL(2,Z)).

Proof. Let P(X) be the right hand side of 9.2. Then for B ∈ SL(2,Z) we have

P(X)|kB = ∏A∈Γ \SL(2,Z)

(X− ( f |kA)|kB

)= P(X)

Therefore, the coefficients of P(X) has the correct functional equation. The finitegrowth of gi’s follow from the finite growth of f |kA’s for all A ∈ Γ \SL(2,Z). ut

Conversely, let us be given gi ∈Mk·i(SL(2,Z)

), i = 1,2, ..a, and define

P(X) =a

∑i=0

ga−i ·X i

The resultant of P(X) is a homogeneous polynomial of degree 2 · k ·a in

Q[g1,g2, . . . ,ga], weight(gi) = ki

This is a weight 2ka modular form for SL(2,Z). Assume that this resultant has nozeros. Since H is simply connected, we can find holomorphic functions f1, f2, .., fa :H→ C such that

P(X) = (X− f1)(X− f2) · · ·(X− fa)

We have the representation

χ : SL(2,Z)→ GL(a,Z)

whose image is isomorphic to the permutation group in a elements and such that

9.6 Transcendental degree of modular forms 105f1|kAf2|kA

...fa|A

= χ(A)

f1f2

fa

, ∀A ∈ SL(2,Z). (9.3)

Here χ(A) is just a permutation matrix in 1,2, ..,a. Let

Γi := A ∈ SL(2,Z)∣∣χ(A)ei = ei

where ei = [0,0, . . . , 1︸︷︷︸i−th place

,0, . . . ,0]tr.

Proposition 9.6.2 We have fi ∈Mk(Γi).

Proof. This is a direct consequence of 9.3 and the definition of Γi. ut

Exercise 61 Show that P(X) is irreducible over Mk(SL(2,Z)

)[X ] if and only if an

orbit of χ in 1,2, . . . ,a is the whole set. It follows that if P(X) is irreducible overMk(SL(2,Z)

)[X ]. Then

f1, f2, . . . , fa= fi|kA, A ∈ Γ \SL(2,Z)Γi := A−1 ·Γ1A A ∈ Γ \SL(2,Z)

The following question is natural to ask: under which conditions on gi’s, Γ1 is acongruence group?We have proved:

Theorem 9.6.1 The ring of modular forms for congruence groups is of transcen-dental degree 2. More, precisely any modular form for a congruence group is in thealgebraic closure of C(E4,E6).

Chapter 10Elliptic curves as Diophantine equations

10.1 Finite fields

A finite field, as its name indicates, is a field with finite cardinality. By definition ofa field and finiteness property, the characteristic of a finite field is a prime numberp > 1. Finite fields are completely classified as follows:

1. The order of a finite field of characteristic p is pn for some n ∈ N.2. There is a unique (up to isomorphism of fields) finite field with pn, n ∈ N ele-

ments.3. For a prime number the finite field with cardinality p is simply the quotient

Fp :=ZpZ

.

4. For q = pn, n ∈ N the finite field with cardinality pn is denoted by Fq. It is thespliting field of the polynomial xq− x over Fp.

5. Every finite integral domain is a field and in particular6. Let f (T ) be a monic irreducible polynomial of degree n in Fp[T ]. Then the quo-

tient Fq[T ]/〈 f 〉 is a finite field with pn elements.7. Let f (x,y) ∈ Fp[x,y] be a polynomial and I be a non zero prime ideal of R :=

Fp[x,y]/〈 f 〉. Then the quotient R/I is a finite field.

For more on finite fields the reader is referred to [Jac85].

10.2 Zeta functions of elliptic curves over finite fields

Let V be an affine or projective variety defined over Fq. The zeta function of V isdefined to be the formal power series in T :

107

108 10 Elliptic curves as Diophantine equations

Z(V,T ) = exp(∞

∑r=1

#V (Fqr)

rT r)

Theorem 10.2.1 Let E be an elliptic curve defined over Fp. Then

Z(E,T ) =1+2aET + pT 2

(1−T )(1− pT ). (10.1)

where aE is an integer depending only on E. Moreover, the Riemann hypothesisholds for E, i.e. the only zeros of

ζ (C,s) := Z(E,q−s)

are in the line ℜ(s) = 12 .

Let1−2aET +qT 2 = (1−αT )(1−βT )

and soα +β = 2aE , αβ = q (10.2)

Note that α and β are algebraic integers:

α,β = aE ±√

a2E −q.

We take the logarithmic derivative of both sides of (10.1) and one easily finds theequalities

#E(Fpr) = pr +1−αr−β

r, r = 1,2,3, . . .

For r = 1 we obtain#E(Fp) = p+1−2aE

We conclude that for elliptic curves over a finite field Fq the number of Fq-rationalpoints determine the number of Fqr -rational points.

Concerning the Riemann hypothesis, we note that it is equivalent to the inequal-ity:

|#E(Fp)− p−1|< 2√

p. (10.3)

The Riemann hypothesis holds if and only if |α| = |β | = p12 . If these equalities

happen then|#E(Fp)− p−1|= |2aE |= |α +β |< 2

√p.

(the equality cannot occur because p is prime). Conversely, if (10.3) happens thena2

E − p < 0 and so the roots of the polynomial 1−2aET +qT 2 are complex conju-gate, β = α and so |α|= |β |= p

12 .

The general reuslt as in (10.1) was conjectured by Andre Weil [Wei49] and wasproved by P. Deligne (see for instance [Kat76] for an exposition of Deligne results).

10.4 Mazur theorem 109

10.3 Nagell-Lutz Theorem

In this section we state Nagell-Lutz theorem which gives a finite set of of possibili-ties for a torsion point of an elliptic curve.

Theorem 10.3.1 (Nagell-Lutz Theorem) Let E be an elliptic curve with the Weier-strass equation:

y2 = x3 + t2x+ t3, t2, t3 ∈ Z,∆ := 4t32 +27t2

3 6= 0.

Then for all non-zero torsion points P = (a,b) ∈ E(Q) we have:

1. The coordinates of P are in Z, i.e. a,b ∈ Z.2. If P is of order greater than 2, then b2 divides ∆ .3. If P is of order 2 then b = 0 and a3 + t2a+ t3 = 0.

A proof can be found in [Sil92], p. 221 or in [ST92] p.56.

Exercise 62 [Mil96], Exercise 8.11. For four of the following elliptic curves com-pute the torsion subgroup.

y2 = x3 +2, · · ·

See the reference above for the list of elliptic curves.

10.4 Mazur theorem

Theorem 10.4.1 (Mazur, [Maz77, Maz78]) Let E be an elliptic curve over Q. Thenthe torsion subgroup E(Q)tors is one of the following fifteen groups:

Z/NZ, 1≤ N ≤ 10, or N = 12

Z/2Z×Z/2NZ, 1≤ N ≤ 4

Note that the above theorem implies that for an elliptic curve over Q we have always:

#(E(Q)tors)≤ 16.

It is natural to conjecture that: If E is an elliptic curve over a number field k, theorder of the torsion subgroup of E(k) is bounded by a constant which depends onlyon the degree of k over Q. This is known uniform boundedness conjecture (UBC).It is proved by S. Kamienny in [Kam92] for all quadratic fields and by L. Merel in[Mer96] for all number fields.

For the proof of all the statements above one needs the notion of modular curvesX0(N) and modular forms which will be introduced in the forthcoming chapters.


10.5 One dimensional algebraic groups

We follow [Mil96] p. 23. When elliptic curves degenerate we find the followingalgebraic groups:

1. The additive group Ga := (A1(k),+).2. The multiplicative group Gm := (A(k), ·).3. Twisted multiplicative group Gm[a].

Exercise 63Gm[a]∼=Gm[ac2], a,c ∈ k−0,

Exercise 64Gm[a](Fq) = q+1.

By Bezout theorem a cubic curve E in P2 has a unique singular point (if there aretwo singularities then the line connecting that points meets the curve in 4 pointscounted with multiplicities). The singular point is defined over k because it is fixedunder the action of the Galois group Gal(k/k). Let S be the singular point of of Eand

Ens(k) := E(k)\S.

The same definition of group law for elliptic curves applies for Ens and it turns outthat Ens as a group and:

Exercise 65 If the elliptic curve E is given by the Weierstrass form

y2 = x3 + t4x+ t6, t2, t3 ∈ k,∆ = 2(4t34 +27t2

6 ) = 0.

then Ens is isomorphic to the three one dimensional group described above:

Ens(k)∼=Gm(k) or Gm[c](k), or Ga(k)

mentioned in the lectures. Does we need char(k) 6= 2,3?

Exercise 66 For char(k) = 3 (resp. char(k) = 2) we have to consider the case (4.6)(resp. (4.4)). Discuss the reduction modulo 2 and 3 in such cases.

10.6 Reduction of elliptic curves

We take an elliptic curve in the Weierstrass form

y2 = x3 + t4x+ t6, t2, t3 ∈Q,∆ := 2(4t34 +27t2

6 ) 6= 0.

and by change of coordinates (x,y) 7→ (c2x,c3y), c ∈Q we assume that |∆ | is mini-mal. For p prime different from 2 and 3 we have the curve E/Fp and the reductionmap

10.7 Zeta functions of curves over Q 111

E(Q)→ E(Fp).

1. Good reduction. If p does not divide ∆ then E/Fp is an elliptic curve.2. Cuspidal reduction/additive reduction. The reduced curve E/Fp has a cusp as a

singularity and so its non-singular part is an additive group. If char(k) 6= 2,3 thiscase happens if and only if p | ∆ , and p | 2t4t6.

3. Nodal reduction/split multiplicative. The reduced curve Ens/Fp is a multiplica-tive group.

4. Nodal reduction/nonsplit multiplicative. The reduced curve Ens/Fp is a twistedmultiplicative group.

Exercise 67 Reduction moulo 3 of the above elliptic curve in Weierstrass form issingular if and only if t4 = 0. In the singular case it is always a cusp. In reductionmodulo 2 the elliptic curve E/F2 is always singular and its singular point is S =(t4, t6). Find the four groups Ens(F2) corresponding to the four choice of (t4, t6).

Exercise 68 Let E/Q : y2 + y = x3− x2 + 2x− 2. Show that 1. the primes of badreduction for E are p = 5 and 7. 2. The reduction at p = 5 is additive, while thereduction at p = 7 is multiplicative. 3. NE/Q = 175.

10.7 Zeta functions of curves over Q

We follow [Mil96] p. 102. The non-complete zeta function of a smooth curve E :f (x,y) = 0, f ∈ Z[x,y] is defined to be

ζS(E,s) = ∏p6∈S

ζ (E/Fp,s).

where S is a finite number of prime numbers such that E/Fp is singular.

Exercise 69 Can you justify the definition of the zeta function of a variety over Qby interpreting it as a Euler product, the one similar to (8.11).

In the case of elliptic curves it is natural to define

LS(E,s) := ∏p6∈S

11+(#(E(Fp))− p−1)ps + p1−2s

and so we have

ζS(E,s) =ζS(s)ζS(s−1)

LS(E,s)

Proposition 10.7.1 The product ζS(E,s) and hence LS(E,s) converges for ℜ(s)> 32

Proof. It is direct consequence of the Riemann hypothesis for elliptic curves overfinite fields and the convergence of the Riemann zeta function(see [Mil96] p.102).


We we want to define the complete L function by adding bad prime numbers p ∈ S.We define

Lp(T )=

1+(#(E(Fp))− p−1)T + pT 2 p good1−T modulo p we have split multiplicative reduction1+T modulo p we have non-split multiplicative reduction1 modulo p we have additive reduction

We have defined this in such a way that

Lp(p−1) =#Ens(Fp)

p

Now we define the L-function of an elliptic curve E over Q:

L(E,s) = ∏p

1Lp(p−s)

.

10.8 Hasse-Weil conjecture

The conductor of an elliptic curve over Q is defined to be

NE/Q = ∏p bad

p fp

where fp = 1 if E has multiplicative reduction at p, fp = 2 if p 6 |2,3 and E hasadditive reduction at p. For the case in which we have additive reduction modulop = 2,3 we have fp ≥ 2, fp ∈ N and fp depends on wild ramification in the actionof the inertia group at of Gal(Q/Q) on the Tate module of E.

Exercise 70 Discuss the case p= 2,3 in the above definition. [Mil96] is also talkingabout a formula of Ogg fp = ordp∆ +1−mp using Neron models. Can you obtainsome information on this.

DefineΛ(E,s) := N

s2

E/Q(2π)−sΓ (s)L(E,s)

Theorem 10.8.1 (Hasse-Weil conjecture for elliptic curves) The function Λ(E,s)can be analytically continued to a meromorphic function on the whole C and itsatisfies the functional equation

Λ(E,s) =±Λ(E,2− s).

This theorem was first proved for CM elliptic curves by Deuring 1951/1952. It isproved in its generality by the works of Eichler and Shimura, Wiles, Taylor, Dia-mond and others.

10.10 Congruent numbers 113

10.9 Birch Swinnerton-Dyer conjecture

For the functional equation of L the value s = 1 is in the middle, i.e. it is the fixedpoint of s 7→ 2− s.

Conjecture 10.9.1 (BSD conjecture) For an elliptic curve E over Q, the functionL(E,s) is holomorphic at s = 1 and its order of vanishing at s = 1 is the rank of theelliptic curve E.

A weak form of this conjecture is not also proved:

Conjecture 10.9.2 (weak BSD conjecture) L(E,1) = 0 if and only if E has in-finitely many rational points.

For papers on BSD conjecture see [CW77, BSD63, BSD65, Tun83, Lan78b][Ser89], [Mor69].

10.10 Congruent numbers

A natural number n is said to be congruent if it is the area of a right triangle whosesides have rational length. In other words we are looking for the Diophantine equa-tion:

Cn : x2 + y2 = z2, n =12

xy

in Q, where x,y and z are the sides of the triangle. Consider the affine curve Cn/Qin A3 defined by the above equations. It intersects the projective space at infinity in4 points:

[x;y;z;w] = [0;±1;1;0], [±1;0;1;0].

LetC : y2 = x4−n2, En : y2 = x3−n2x.

We have morphisms

Cn→C, (x,y,z) 7→ (z2,

x2− y2

4)

andC→ En, (x,y) 7→ (x2,xy)

defined over Q.

Proposition 10.10.1 A necessary and sufficient condition for the point (x,y) ∈En(Q) be in the image of Cn(Q)→ En(Q) is that

1. x to be a square and that2. its denominator be divisible by two3. and its numerator has no common factor with n.

The proof is simple and is left to the reader (see [Kob93]).


Exercise 71 Let Cn be the projectivization of Cn in P3. Is Cn smooth? If yes deter-mine its genus.

Exercise 72 Ex. 1,2,3,4 of Koblitz, page 5.

We want to analyze the torsion points of

En : y2 = x3−n2x

By definition of the group structure of En we know that

O,(0,0),(0,±n)

are 2-torsions of En. Following the lines of [Kob93] p. 44 Proposition 4, we want toprove:


En(Q)tors = O,(0,0),(0,±n)

and so #En(Q)tors = 4.

Proof. Let us first give the strategy of the proof. Let E/Q be an a elliptic curve inthe Weierstrass form and let p > 2 be a prime number which does not divide thediscriminant of E. By a linear change of variable (x,y) 7→ (a2x,a3y) we can assumethat the ingredient coefficients of E are in Z. Let E/Fp be the elliptic curve obtainedfrom E by considering the coefficients of E modulo p. The main ingredient of theproof is the reduction map

E(Q)→ E(Fp),

which is a group homomorphism. Note that by our assumption on p, E/Fp is notsingular. This is an injection of E(Q)tors inside E(Fp) for all but finitely many p andso for such primes m := #E(Q)tors divides #E(Fp). In fact, we have not yet provedthat E(Q)tors is finite (a corollary of Mordell-Weil theorem). Therefore, we take afinite subgroup G of #E(Q)tors and prove that the reduction map restricted to G isan injection and so m := #G divides #E(Fp). From another side, we prove that forE = En:

#En(Fp) = p+1, ∀p prime p≡−1 mod 4 (10.4)

Therefore, for all but finitely many primes p≡−1 mod 4 we have p≡−1 mod m.This implies that m = 4. Therefore, every finite subgroup of E(Q)tors is of order 4.Since all the elements of E(Q)tors are torsion, we conclude that #En(Q)tors = 4.

Now let us prove that the reduction map induces an injection in a finite subgroupG of E(Q)tors . Two points P = [x;y;z],Q = [x′;y′;z′] ∈ E(Q) are the same afterreduction if and only if

xy′− x′y,xz′− x′z,yz′− y′z (10.5)

are zero modulo p. For all pairs P,Q in G, the number of numbers (10.5) is finite andso there are finitely many primes dividing at least one of them. For all other primes

10.10 Congruent numbers 115

p, we have the injection of G in E(Fp) by the reduction map. The proof of (10.4) isdone in the next proposition.

Proposition 10.10.3 Let q = p f , p 6 |2n. Suppose that q ≡ −1 mod 4. Then thereare q+1 Fq points on the elliptic curve En : y2 = x3−n2x.

Proof. Consider the map

f : Fq→ Fq, f (x) = x3−n2x

f is an odd function, i.e. f (−x) = − f (x), and −1 is not in its image (this followsfrom the hypothesis on p). It follows that the index of the multiplicative group F2

q−0 in Fq−0 is two and so for all x ∈ Fq−0 exactly one of x or −x is squareand so for all x ∈ Fq−0,n,−n exactly one of f (x) or f (−x) is square. Each sucha pair (x,y), y = f (x) gives us two points (x,y), (x,−y) ∈ En(Fq) and so in total wehave 3+2 q−1

2 points in En(Fq).

Proposition 10.10.4 The natural number n is congruent if and only if En(Q) hasnon-zero rank.

Proof. If n is a congruent number then by Proposition 10.10.1, En has Q-rationalpoint with x-coordinate in (Q+)2. The x coordinates of 2-torsion points in the affinechart x,y are 0,±n. The fact that n is square free and Proposition 10.10.2 impliesthat such a rational point is of infinite order.

Conversely, suppose that P is a rational point of infinite order in En. Then byExercise 73, the One can

Exercise 73 ([Kob93], p. 35, Ex. 2c) If P is a point not of order 2 in En(Q), thenthe x-coordinate of 2P is a square of rational number having an even denominator.By Proposition 10.10.1, 2P comes from a point in Cn(Q) and hence n is a congruentnumber.

Exercise 74 ([Kob93], p. 49-50) Ex. 4,5,6, 7,9.

Let us now state the main result in §10.2 for the elliptic curve En related to thecongruent numbers. The Legendre symbol is defined for integers a and positive oddprimes p by (

ap

)=

0 if p divides a1 for some x ∈ Z, a≡ x2 mod p−1 otherwise

Proposition 10.10.5 In the zeta function of En : y2 = x3−n2x defined over Fp, p aprime p 6 |2n, we have:

α =

i√

p if p≡ 3( mod 4) in this case aEn = 0

2k+(

np

)+2ki if p≡ 1( mod 4) in this case aEn = 2k+

(np

)In the second case k is determined by the fact that αα = p


Exercise 75 ([Mil96], Ex. 19.12) Let E be the elliptic curve

E : y2 = x3−4x2 +16

Consider also the formal power series given by

F(q) = q∞

∏n=1

(1−qn)2(1−q11n)2 = q−2q2−q3 +2q4 + · · ·

1. Compute Np = #E(Fp) for all primes 3≤ p≤ 13.2. Calculate the coefficient of Mn of qn in F(q) for n≤ 13.3. Compute the sum Mp +Np for p prime p≤ 13.4. Formulate a conjecture on the sum Mp +np. Can you prove it?.

The bad prime numbers for the elliptic curve En : y2 = x3− nx are those whichdivide 2n. For p | 2n, p 6= 2 or p = 2, 2|n we have an additive reduction. For p = 2and p 6 |n we have apparently a multiplicative reduction: y2 = x3 + x. The singularpoint in this case is S = (1,0) and Ens(F2) = O,(0,0) which is isomorphic to(A(F2),+) and so it is additive.

The conductor of En is:

NEn/Q =

24n2 if n is even25n2 if n is odd

In Theorem 10.8.1 the root number ± is determined in the following way:+1 if n≡ 1,2,3 (8)−1 if n≡ 5,6,7 (8)

Reformulating Proposition 10.10.5 and using Exercise 55 we have:

(1−T )(1− pT )Z(En/Fp,T ) = ∏p|〈p〉

(1− (αpT )deg(p))

where

αp =

i√

p if p= 〈p〉a+ ib if p splits, where a+ ib is the unique generator of p

which is congruent to ( np ) mod 2+2i.

0 p | 2n

The L function of En is

L(En,s) = ∏p⊂Z[i] prime

(1− (αp)deg(p)(Np)−s)−1

Now Z[i] is a Dedekind domain and so we can define a unique map χ from the idealsof Z[i] to C such that χn(p) = α

deg(p)p . Therefore

10.11 p-adic numbers 117

L(En,s) = ∏p⊂Z[i] prime

(1−χ(p)(Np)−s)−1 = ∑a⊂Z[i]

χn(a)(Na)−s

where the sum is taken over all non-zero ideals.

10.11 p-adic numbers

By definition a p-adic integer is an element in the inverse limit of

· · · → Z/p3Z→ Z/p2Z→ Z/pZ.

One can show that a p-adic integer is identified with a formal series

a1 p+a2 p2 +a3 p3 + · · · , ai ∈ 0,1,2, . . . , p−1.

The set of p-adic integers is denoted by Zp:

Zp := lim∞←n

Z/Zpn.

Zp is a ring without zero divisor, i.e. if ab = 0,a,b ∈ Zp then either a = 0 or b = 0.The field Qp of p-adic numbers is the quotient field of Zp. The ring Z of integers isa subring of Zp in a natural way and so Qp is a field extension of Q, Q⊂Qp.

Exercise 76 Show that the Diphantine equation x3 + y3− 3 = 0 has not a solutionin Q3 and hence it has not a solution in Q.

There is another way to define p-adic numbers. Any non-zero rational number a canbe expressed in the form a = pr m

n with m,n ∈ Z and not divisable by p. We define

ordp(a) := r, |a|p :=1pr , |0|p := 0

We have

1.|a|p = 0 if and only if a = 0

2.|ab|p = |a|p|b|p, a,b ∈Q.

3.|a+b|p ≤max|a|p, |b|p and so ≤ |a|p + |b|p

Therefore,dp(a,b) := |a−b|p

is a metric on Q. The field Qp of p-adic numbers is the completion of Q withrespect to dp. We have a cononical matric, call it again dp, on Qp which extends the


previous one on Q ⊂ Qp (this inclusion is given by sending a ∈ Q to the constantCauchy sequence a,a, . . .). The same construction with the usual norm of Q, i.e.d(a,b) = |a−b| yields to the field of real numbers R.

Exercise 77 Prove that the two definitions of Qp presented above are equivalent.Prove also

Zp = a ∈Qp | |a|p ≤ 1= the closure of Z in Qp

Chapter 11Theta series

Jacobi’s theta function is the followingg infinite sum

θ(z,τ) =+∞

∑n=−∞

e2πinz+πin2τ

Proposition 11.0.1 The Jacobi’s theta function satisfies:

1. It is a holomorphic function in C×H2. θ(z+1,τ)−θ(z,τ)3. θ(z+ τ,τ)−θ(z,τ) e−πiτ e−2πiz

4. θ(z,τ)−0 for z = 12 +

τ

2 +n+mτ, n,m ∈ Z

Proof. 1. For |Z|< M and Im(τ)> to we have

∞

∑n=−∞

∣∣∣e2πinZ+πin2τ

∣∣∣6C∞

∑n=0

e2πnMe−πn2t0

for some C ∈ R+. This shows that θ is converges is C×H.2. This is immediate3.

θ(z+ τ,τ) =+∞

∑n=−∞

e2πinz eπi(n2+2n)τ

=+∞

∑n=−∞

e2πi(n+1)z eπi(n+1)2τ e−πiτ e−2πiz

= θ(z,τ) · e−πiτ e−2πiz

4.

θ

(12+

τ

2,τ

)=

∞

∑n=−∞

(−1)n eπi(n2+n)τ

For n> 0 the terms corresponding to n and −n−1 cancel each other.

119

120 11 Theta series

We will frequently use the followings:

θ3(τ) = θ(0,τ)∞

∑n=−∞

q12 n2

θ4(τ) = θ( 1

2 ,τ) ∞

∑n=−∞

(−1)n q12 n2

θ2(τ) = θ(

τ

2 ,τ)= q−

18

+∞

∑n=−∞

q12 (n+ 1

2 )2

Theorem 11.0.1 We have

θ(z,τ) =∞

∏n=1

(1−qn)(

1+qn− 12 e2πiz

)(1+qn− 1

2 e−2πiz)

(11.1)

Proof. Let π(z,τ) be the right hand side of 11.1. We prove that π(z,τ) is a holo-morphic function in C×H and satisfies the same properties as of θ(z,τ) in

1. For the convergence we use the criterion for convergence of infinite products.

(1−qn)(

1−qn− 12 e2πiz

)(1−qn− 1

2 e2πiz)= 1+O

(|9|n−1 e2π|z|

)and

∞

∑n=1|9|n converges.

2. This is immediate3.

π(z,τ) =∞

∏n=1

(1−qn)(

1−qn+ 12 e2πiz

)(1−qn− 3

2 e−2πiz)

=

(1−q−

12 e−2πiz

)(

1−q12 e2πiz

) π(z,τ)

We have 1+χ

1+χ−1 = χ, χ 6=−1 and 3 follows.4. The product vanishes at a point (z,τ) if

±Z +

(n− 1

2

)τ ∈ Z+

12

which gives us the result.

Now, let us prove the equality 11.1. Let F(z,τ) = θ(z,τ)π(z,τ) . This as a function in

z is double periodic and has no poles. Therefore, it is constant as a function in z,Therefore C(τ) = θ(z,τ)/π(z,τ).

We put z = 12

11 Theta series 121

C(τ) =

∞

∑n=−∞

(−1)n q12 n2

∞

∏n=1

(1−q

12 n)(

1−qn− 12

) (11.2)

We put z = 14

C(τ) =

+∞

∑n=−∞

(−1)n q2n2

∞

∏n=1

(1−q2n)(1−q4n−2)(11.3)

11.2 and 11.3 imply C(4τ) =C(τ) for all τ ∈H. Since q4k → 0 when k→∞ wecan conclude that C(τ) = 1

We have

θ3(τ) =η(τ)5

η( 1

2 τ)2

η(2τ)2(11.4)

θ4(τ) =η( 1

2 τ)2

η(τ)(11.5)

θ2(τ) =2η(2τ)2

η(τ)(11.6)

Taken from Wikipedia and [Oliver]. Note that θ2θ3θ4 = 2η(τ)3.

Theorem 11.0.2 For τ ∈H and z ∈ C we have

θ

(z,−1τ

)=

√τ

ieπiτz2

θ(zτ,τ)

Here we have chosen a branch of√

τ

i ,τ ∈H such that for imaginary τ , it is positive.

Proof. It is enough to prove the formula z=α ∈R and τ = it, t ∈R+. This is exactlythe Poisson summation formula.

we get the following functional equations for θ2, θ3, θ4

θ3

(−1τ

)=

√τ

iθ3(τ) (11.7)

θ4

(−1τ

)=

√τ

iξ8 ·θ2(τ)

η

(−1τ

)=

√τ

i· η(τ)

122 11 Theta series

Let f (τ) = θ3(8τ)8

l f (τ +1) = f (τ)

f(−14τ

)=(

τ

2

)4f (τ)

Which says that f is a modular form for the group⟨(1 10 1

),

(0 −1

212 0

)⟩⊆ SL(2,Q)

11.1 Two-squares theorem

For k ∈ N a = (a1, a2, . . .ak) ∈ Zk

γk, a(n) = #(x1, x2, . . . ,xk) ∈ Zk∣∣a1x21 +a2x2

2 + · · ·+akx2k = n

Then

θ(2a1τ) θ(2a2τ) · · ·θ(2akτ) =∞

∑n=0

γk, a(n)qn. (11.8)

Therefore, in order to find formulas for γk,a(n) we have to study the analyticfunction in the left hand side of 11.8.

Let d1(n) denotes the number of divisors of n of the form 4k+ 1, and d3(n) thenumber of divisors of n of the form 4k+3.

Theorem 11.1.1 For n> 1

γ2(n) = 4(

d1(n)−d3(n))

(11.9)

The generating function of the right hand side of 11.9 is

C(τ) : = 2+∞

∑n=−∞

1qn +q−n

= 1+4∞

∑n=1

qn

1+q2n

= 1+4∞

∑n=1

qn

1−q4n −q3n

1−q4n

Therefore, in order to prove 11.1.1, we have to prove that

θ23 (2τ) =C(τ)

11.2 Poisson summation formula 123

For the rest of the proof see [Stein] page 299.

11.2 Poisson summation formula

Let ϕ : R→ R be any continuous function which decreases rapidly, let us say

ϕ(x)∼ |x|−c c > 0

as x→+∞. Then the Fourier transform of ϕ is

ϕ(y) :=∫R

ϕ(x) e−2πixy dx

The Poisson summation formula says that

∑n∈Z

ϕ(x+n) = ∑γ∈Z

ϕ(x+ γ) (11.10)

Proof. (Taken from Zagier) The growth condition on ϕ(x) ensures that the left handside of 11.10 converges to a continuous function φ . This function satisfies

Φ(x+1) = Φ(x)

and so Φ has Fourier expansion

Φ(x) = ∑γ∈Z

cγ · e2πiγx cγ =∫ 1

0Φ(x) e−2πiγx dx

substituting cγ in Φ(x) we get

cγ =∫ 1

0

∞

∑n=−∞

ϕ(x+n) e−2πiγ(x+n) dx =∞

∑n=−∞

∫ n+1

nϕ(x) e−2πiγx dx

=∫

∞

−∞

ϕ(x) e−2πiγx dx

This gives us

∑n∈Z

ϕ(x+n) = ∑γ∈Z

(∫R

ϕ(t) e−2πiγt dt)

e2πiγx

which is the desired statement.

Chapter 12Picard curious example

In this section we reproduce an example of a foliation, see Theorem 12.2, which isessentially due to E. Picard in [Pic89] pages 298-299, see also Mazzocco’s article[Maz01]. Picard has taken an evaluation of the Weierstrass ℘ function to give a so-lution to Painlev VI equation with a particular parameter, see also Corollary 2 and3 in [Lor16]. This particular case of Painleve VI written as a vector field is bira-tional to the foliation in Theorem 12.2. We have derived Picard’s foliation using theGauss-Manin connection of half elliptic integrals and in this way its generalizationto families of curves of arbitrary genus is reachable.

Let T be the moduli of triples (E,ω,P), where E is an elliptic curve, ω is aholomorphic differential form on E and P is a point in E with P 6= O. Note that inthis text an elliptic curve comes with a point O which is the neutral point of thegroup structure of of E. This is almost the same moduli as F. Loray’s moduli spacein [Lor16]. We take the line bundle L = [P]− [O], which automatically comes witha meromorphic section s with the diviser P−O. The connection ∇ : L→Ω 1

E ⊗L isdetermined by ∇s = ω⊗ s. Loray in [Lor16] works with the Legendre family whichthe moduli elliptic curves with a 2-torsion point structure, whereas in out contextwe do not consider this.

The universal family of pairs (E,ω) is given by the Weierstrass family

Et2,t3 y2 = 4x3− t2x− t3, ω =dxy, (t2, t3) ∈ C2\27t2

3 − t32 = 0 (12.1)

and for a given (E,ω), its Weierstrass coordinates (x,y) are unique. We have fourfunctions

t2, t3, x := x(P), y := y(P)

on T which satisfy the relation (12.1). We may choose (x, y, t2) as coordinate systemon T. On T we have the following holomorphic multi-valued functions∫ P

Oα,∫

δ1

α,∫

δ2

α (12.2)

125

126 12 Picard curious example

where δ1,δ2 ∈ H1(E,Z) form a basis and varies continousy when E changes in themoduli space (flat sections of the Gauss-Manin connection after going to H1), andα = [ dx

y ,xdx

y ].

Theorem 12.0.1 There is a unique algebraic one dimensional foliation F in T suchthat

1. The three vector valued holomorphic functions (12.2) restricted to the leaves ofF are linearly dependent (with constant coefficients depending only on the leafL).

2. The foliation F has the first integral f := t32

t23

3. The foliation F has an enumerable set of algebraic leaves birationl to mod-ular curves, we denote them by X0(d),d ∈ N, d ≥ 2. There is a sequencec2,c3, · · · ,cd , · · · ∈C∪∞ with c2 = ∞ such that X0(d)⊂ f−1(cd) and the point(E,ω,P) of X0(d) is characterized by the fact that P is a torsion point of order din E.

4. The foliation F is given by the vector field

y(

23

t2∂

∂ t2+ t3

∂

∂ t3− y2(6x2t2− t2

2 +9xt3)∂

∂x

)+ (12.3)(

y2(12

t2−6x2)(6x2t2− t22 +9xt3)−

13

t2x− 12

t3

)∂

∂y(12.4)

Picard’s foliation has the following singular set

Sing(F) := t2 = y2−4x3 = 0∪x = y = 0∪y = 12t2− x2 = 0 (12.5)

Proof. In the following d refers to differential operator in T space. We have thefollowing differential equation d

(∫ P

O

dxy

)d(∫ P

O

xdxy

)=

− 112

d∆

∆, 3

2α

∆

− 18 t2 α

∆, 1

12d∆

∆

∫ P

O

dxy∫ P

O

xdxy

+

A1(x, y)+ dxy

A2(x, y)+ xdxy

(12.6)

where∆ := 27t2

3 − t32 , α := 3t3dt2−2t2dt3.

A1 = (9x2y− 32

t2y)dt2 +(9xy)dt3

A2 = (18x3y− 94

y3− 154

t2xy)dt2 +(9x2y− 32

t2y)dt3

The full periods∫

δα satisfy the above differential equations without the last term.

In this case the involving 2×2 matrix in t2, t3 variables is usually called the Gauss-Manin connection matrix of the family (12.1). The computation of this matrix can

12 Picard curious example 127

be found in [Sas74] p. 304, [Sai01]. For the algorithms which computes the Gauss-Manin connections in other cases, and in particular, the computation of the last termin the above equality see [Mov19].

We are looking for a foliation with the first property in Theorem 12.2. This isgiven by the two differential forms in the last matrix of (12.6). The vector field inthe theorem generates the common kernel of these 1-forms.

Chapter 13Online supplemental items

13.1 Introduction

In this chapter we explain many procedures of the library foliation.lib ofSingular, [GPS01], a computer programming language for polynomial computa-tions. We also explain few other softwares which are useful when one deal withcomputational aspects of modular forms and elliptic curves.

13.2 How to start?

One has to run Singular in the same directory, where foliation.lib lies. Thenin Singular’s command line one has to type:

LIB "foliation.lib";

In order to get an example and help of a command, for instance PeriodMatrix,one has to type respectively:

example PeriodMatrix;help PeriodMatrix;

In this chapter I will only sketch few procedures related to the topic of the presenttext. For more information the reader might consult the help and example of eachprocedure.

13.3 Ramanujan differential equation

129

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http://w3.impa.br/~hossein/myarticles/hodgetheory.pdf

Index

Λ , lattice, 9τ , coordinate in the upper half plane, 9

Dedekind eta function, 6

Elliptic curve, 37Elliptic function, 12

Lattice, 9

Lattice of elliptic integrals, 38

Modular form, 12

Point at infinity, 37

Serre derivative, 24

Weierstrass form, 37

133

a differential introduction to modular forms and elliptic...

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