Elliptic Curves and Modular Forms

Home , Cusp form, Elliptic curve, Modular equation, Modular form, Riemann form

ALGANT Master Thesis - July 2018

Candidate Francesco Bruzzesi

Advisor Prof. Dr. Marc N. Levine

Universit´adegli Studi di Milano Universit¨atDuisburg–Essen

Introduction

The thesis has the aim to study the Eichler-Shimura construction associating elliptic curves to weight-2 modular forms for Γ0(N): this is the perfect topic to combine and develop further results from three courses I took in the first semester of the academic year 2017-18 at Universit¨atDuisburg-Essen (namely the courses of modular forms, abelian varieties and complex multiplication). Chapter 1 gives a brief overview on the algebraic geometry results and tools we will need along the whole thesis. Chapter 2 introduces and develops the theory of elliptic curves, firstly as an algebraic curve over a generic field, and then focusing on the fields of complex and rational numbers, in particular for the latter case we will be able to define an L-function associated to an elliptic curve. As we move to Chapter 3 we shift our focus to the theory of modular forms. We first treat elementary results and their consequences, then we seehow it is possible to define the canonical model of the modular curve X0(N) over Q, and the integrality property of the j-invariant. Chapter 4 concerns Hecke operators: Shimura’s book [Shi73 Chapter 3] introduces the Hecke ring and its properties in full generality, on the other hand the other two main references for the chapter [DS06 Chapter 5] and [Kna93 Chapters XIII & IX] do not introduce the Hecke ring at all, and its attributes (such as commutativity for the case of interests) are proved by explicit computation. We try to take an intermediate approach to the subject and rephrase everything just in terms of SL2(Z) and congruence subgroups. At the end of the chapter we will be able to associate an L-function to a cusp form for Γ0(N). Lastly in Chapter 5 we are ready to illustrate how to obtain an elliptic curve from a weight-2 cusp form for Γ0(N), making use of the theory of abelian varieties.

i ii

Notes and References

Chapter 1: The organization of Sections 1.1 and 1.2 is based on [Sil09, Chapter 1 & Chapter 2]. The first statement of Proposition 1.10 is taken from [DS06, Proposition 7.2.6]; Proposition 1.14 is from [Kna93, Proposition 11.43]; Proposition 1.22 comes from [Was03, Proposition C.2]. The main references for these two sections are [Har77], [Mir95] and [Sha77]. Section 1.3 treats some standard results on the Jacobian variety associated to a Riemann surface. The main references are [Kir92] and [Mir95].

Chapter 2: Section 2.1 is organized as [Sil09, Chapter 3]. Proposition 2.1 is taken from [Sil09, III.1.4(i)]; Proposition 2.3 and Proposition 2.5 are based on filling the detailsKna93 of[ , Theorem 11.57 and Theorem 11.58] (respectively); Proposition 2.12 comes from [Sil09, Theorem III.4.10]; Lemma 2.15 is from [Kna93, Lemma 11.63]; Theorem 2.16 and Theorem 2.17 try to fill the details of [Kna93, Theorem 11.64 and Theorem 11.66] (respectively). The main reference for Section 2.2 is [Shi73, Chapter 4]. For Section 2.3 we followed [Kna93, Chapter X].

Chapter 3: Some of the results arise as homeworks and/or are taken from the course of modular forms mentioned in the Introduction above and most of the results are standards: main references are [DS06], [Miy89] and [Lan76]. However Section 3.2.2 is taken from [Kna93, Chapter XI, Section 8].

Chapter 4: The structure of the Chapter is as in [Shi73, Chapter 3]. Propo- sition 4.5 is from [Shi73, Proposition 3.8]; Lemma 4.9 is from[Shi73, Lemma 3.12]; Proposition 4.10 is from [Shi73, Proposition 3.14]; Lemma 4.40 and 4.41 are respectively from [Shi73, Lemma 3.61 and 3.62]. Results from Theorem 4.33 to Proposition 4.39 are taken from [Kna93, Chapter XI, Section 5].

Chapter 5: The organization of the whole chapter is as in [Kna93, Chap- ter XI, Sections 10 & 11]. For Section 5.1 we used as references [Lan59] and [Swi74]. Section 5.2 is from the last part of [Kna93, Chapter XI, Section 10]; Proposition 5.15 is taken from [Kna93, Theorem 11.74]. iii

Notation

Z, Q, R, C integers, rationals, reals, complex numbers Fp the field with p elements H complex upper half plane Re(z), Im(z) real, imaginary part of z K algebraic closure of the field K Aut(K/L) automorphism group of K fixing the subfield L ⊂ K Mn(K) n × n matrices with coefficients in K GLn(K) invertible n × n matrices with coefficients in K SLn(K) n × n matrices with coefficients in K and determinant 1 [A : B] index of B in A or degree of A in B W ∨ dual space of the vector space W A× group of invertible elements of the ring A #S cardinality of the set S iv

Acknowledgements

First and foremost, I would like to express my gratitude to Professor Dr. Marc N. Levine, who accepted to supervise my study in this topic and patiently spent time to enlight me with his deep mathematical insights. He motivated me to do always better.

I want to thank all the people I met in the last two year in the Algant Master, both Professors and students. In particular Bob, John and Francesco: we shared every moment, all the laughs and all the dissapointments. We constantly helped each other, both in life and in university. We all know how frustating it feels to study math sometimes, and I am thankful that we overcome the difficulties of these two years together.

A special thanks goes to Federica, who had to bear me every single day of this stressful period. Even though she had to listen to my complaints, she made me smile in every situation. Thank you for having coped with all my struggles and problems.

Also I would like to mention my dear friend Antonio, the person who helped me to develop a taste for number theory, I couldn’t ask for a better mentor and friend. No matter how long we don’t see each other, we always have a great time and connection.

I am thankful to all my hometown friends and relatives. You are the reason why I keep coming back home and on all occasions it feels like time didn’t pass.

Last, but certainly most important, I want to thank my parents Lanfranco and Laura for providing me with unfailing support and continuous encourage- ment throughout my years of study. This accomplishment would not have been possible without them: words cannot really describe how grateful I am.

Francesco Contents

Introduction i Notes and References ...... ii Notation ...... iii Ackowledgements ...... iv

1 Algebraic Geometry 1 1.1 Algebraic Varieties ...... 1 1.2 Algebraic Curves ...... 6 1.2.1 Fields of positive characteristic ...... 10 1.2.2 Riemann-Roch theorem ...... 13 1.3 Riemann Surfaces ...... 16

2 Elliptic Curves 19 2.1 Elliptic curves over an arbitrary field ...... 19 2.1.1 Weierstrass form & abstract elliptic curves ...... 19 2.1.2 Isogenies ...... 24 2.2 Elliptic curves over C ...... 27 2.2.1 The Weierstrass ℘-function ...... 29 2.2.2 Isogenies over C ...... 33 2.2.3 Automorphisms of an elliptic curve ...... 35 2.3 Elliptic curves over Q ...... 36 2.3.1 L-function associated to an elliptic curve ...... 37 2.3.2 Hasse theorem ...... 38

3 Modular Forms 40 3.1 Modular forms for SL2(Z)...... 40 3.1.1 Functions of lattices ...... 40 3.1.2 The action of SL2(Z) on H ...... 42 3.1.3 Divisors of modular functions ...... 45 3.1.4 The space of modular forms ...... 49 3.1.5 The modular curve X0(1) ...... 52 3.2 Congruence subgroups ...... 58 3.2.1 Modular functions of higher level ...... 63 3.2.2 The canonical model of X0(N) over Q ...... 67 3.3 Integrality of the j-invariant ...... 69 3.3.1 j(z) is an algebraic number ...... 70 3.3.2 j(z) is integral ...... 71

v CONTENTS vi

4 Hecke Operators 74 4.1 The Hecke ring ...... 74 4.1.1 The structure of R(Γ, ∆) ...... 76 4.2 Action on modular functions ...... 83 4.2.1 Hecke operators on SL2(Z)...... 84 4.2.2 Hecke operators on congruence subgroups ...... 86 4.3 L-function of a cusp form ...... 93

5 Eichler-Shimura Theory 97 5.1 Complex abelian varieties and Jacobian varieties ...... 97 5.2 Technical results ...... 101 5.3 Elliptic curves associated to weight-2 cusp forms ...... 104 5.3.1 Perspective ...... 108

Bibliography 109 Chapter 1

Algebraic Geometry

Throughout this whole chapter, let K0 denote a field and K be an algebraically closed field containing K0

1.1 Algebraic Varieties

Definition. Define the affine n-dimensional space over K as

n n A = A (K) = {P = (x1, ..., xn) | xi ∈ K}

n Similarly, define the set of K0-points (or K0-rational points) of A as the set

n n A (K0) = {P = (x1, ..., xn) ∈ A | xi ∈ K0}

n Remark 1.1. We have an action of the Galois group Gal(K/K0) on A : let n σ σ σ σ ∈ Gal(K/K) and P ∈ A , then P = (x1 , ..., xn). It follows that

n n σ A (K0) = {P ∈ A | P = P ∀σ ∈ Gal(K/K0)} Recall that by the Hilbert basis theorem any polynomial ring over a field is a Noetherian ring, thus every ideal is finitely generated.

Definition. Consider the polynomial ring in n variables K[X1, .., Xn], to any n ideal J ⊂ K[X1, .., Xn] we can associate a subset of A , called affine algebraic set, n VJ = {P ∈ A | f(P ) = 0 ∀f ∈ J} Viceversa, to any algebraic set V ⊂ An we can associate an ideal of polynomials vanishing on V , called ideal of V ,

I(V ) = {f ∈ K[X1, .., Xn] | f(P ) = 0 ∀P ∈ V }

In particular we say that an algebraic set is defined over K0 if its ideal I(V ) can be generated by polynomials in K0[X1, .., Xn] and we write V/K0. Then if V is defined over K0, the set of K0-rational points of V is

n V (K0) = V ∩ A (K0)

1 1.1. ALGEBRAIC VARIETIES 2

Remark 1.2. Assume that V is an algebraic set defined over K0 and let def f1, ..., fr ∈ K0[X1, .., Xn] be the generators for I(V/K0) = I(V )∩K0[X1, .., Xn], then n V (K0) = {P = (x1, .., xn) ∈ A (K0) | f1(P ) = ... = fr(P ) = 0}

Noice that Gal(K/K0) acts on K[X1, .., Xn] by acting on the coefficients of an n element f ∈ K[X1, .., Xn]. Then for f ∈ K[X1, .., Xn], P ∈ A , σ ∈ Gal(K/K0) (f(P ))σ = f σ(P σ)

n Remark 1.3. If f ∈ K0[X1, .., Xn] and P ∈ A , then for σ ∈ Gal(K/K0) we σ σ have f(P ) = f(P ) . Therefore if V is defined over K0 we can characterize V (K0) as σ V (K0) = {P ∈ V | P = P ∀σ ∈ Gal(K/K0)} Definition. An affine algebraic set is irreducible if it is not union of two proper affine algebraic sets. An irreducible affine algebraic set iscalled affine variety Proposition 1.4. Let V ̸= ∅ be an affine algebraic set, then V is irreducible if and only if I(V ) is a prime ideal.

Proof. Assume V is reducible, so that V = V1 ∪ V2 for some proper algebraic subsets V1,V2 $ V . Since the containment is proper, there exist f ∈ I(V1)rI(V ) and g ∈ I(V2)rI(V ). Therefore fg vanishes on V1 ∪V2 = V and thus fg ∈ I(V ). We conclude that I(V ) is not a prime ideal. Viceversa, suppose that I(V ) is not prime, then there exist f, g ∈ K0[X1, .., Xn]r I(V ) but fg ∈ I(V ). Let us define J1 = (I(V ), f) and J2 = (I(V ), g), then V1 = V (J1),V2 = V (J2) are both strictly contained in V . On the other hand V ⊂ V1 ∪V2 since if P ∈ V then fg(P ) = 0, so that f(P ) = 0 or g(P ) = 0 which yields that P ∈ V1 or P ∈ V2. This shows that V is reducible. Definition. Given an affine variety V , define its affine coordinate ring as

K[V ] = K[X1, .., Xn]/I(V ) Since I(V ) is a prime ideal, K[V ] is an integral domain, thus we can form its quotient field K(V ) called function field of V . In case that V is defined over K0 we have K0[V ] = K0[X1, .., Xn]/I(V/K0). Since I(V ) is prime, so is I(V/K0): therefore K0[V ] is an integral domain as well. Remark 1.5. Recall that f is a polynomial function on V if there exists F ∈ K[X1, .., Xn] such that F (P ) = f(P ) ∀P ∈ V . Then K[V ] can be identified with the set {f : V −→ K | f is a polynomial function} Definition. The dimension dim(V ) of an affine variety V is the transcendence degree of K(V ) over K. In particular an affine curve is an affine variety of dimension one. n Definition. Let V ⊂ A be an affine variety and let I(V ) = ⟨f1, ..., fr⟩. A point P ∈ V is a nonsingular point if the matrix ⎛ ∂f1 (P ) ... ∂fr (P )⎞ ∂X1 ∂X1 ⎜ . . ⎟ ⎝ . . ⎠ (1.1.1) ∂f1 (P ) ... ∂fr (P ) ∂Xn ∂Xn 1.1. ALGEBRAIC VARIETIES 3 has rank n − dim(V ). V is nonsingular if each point P ∈ V is nonsingular. Remark 1.6. These definitions do not depend upon the choice of the generators f1, .., fr of I(V ). To see this, let P ∈ V and define

mP = {f ∈ K[V ] | f(P ) = 0} mP is maximal in K[V ] since evaluation at P is an isomorphism from K[V ]/mP 2 to K and mP /mP is a finite dimensional vector space over K. 2 Fact 1.7. P ∈ V is nonsingular if and only if dim(mP /mP ) = dim(V ) Proof. See [Har77] I.5.1 Definition. Define the local ring of V at P as

K[V ]P = {f/g ∈ K(V ) | g(P ) ̸= 0}

Its elements are said to be regular (or defined) at P .

K[V ]P is a local ring with maximal ideal

MP = {f/g ∈ K[V ]P | f(P ) = 0} = mP · K[V ]P

MP is maximal and consists of all non-invertible elements of K[V ]P , therefore it is the unique maximal ideal. 2 2 Let us now consider the natural map ι : mP −→ MP /MP given by f ↦→ f + MP 2 2 • ker(ι) = mP ∩ MP = mP

• ι is surjective since for all f/g ∈ MP we have f/g(P ) ∈ mP and

f f f · (g − g(P )) − = ∈ M 2 g(P ) g g · g(P ) P

2 which means that ι(f/g(P )) = f/g + MP .

2 ∼ 2 We can conclude that ι induces an isomorphism mP /mP −→ MP /MP Definition. The projective n-space over K is

n n n+1 P = P (K) = {0 ̸= P = (x0, .., xn) ∈ A }/ ∼

× where (x0, .., xn) ∼ (y0, .., yn) if there exists λ ∈ K such that yi = λxi for all i = 0, .., n. We denote [x0, .., xn] such equivalence class, thus we can write

n P (K) = {[x0, .., xn] | xi ∈ K, ∃i : xi ̸= 0}

n x0, .., xn are called homogeneous coordinates of the corresponding point in P n and the set of K0-rational points in P is

n n P (K0) = {[x0, .., xn] ∈ P | xi ∈ K0 ∀i} 1.1. ALGEBRAIC VARIETIES 4

n As for the affine case, the Galois group Gal(K/K0) acts on P by acting on homogeneous coordinates:

σ σ σ [x0, .., xn] = [x0 , .., xn]

Definition. A polynomial f ∈ K[X0, .., Xn] is homogeneous of degree d if d × f(λX0, .., λXn) = λ f(X0, .., Xn) for all λ ∈ K . An ideal J ⊂ K[X0, .., Xn] is homogeneous if it is generated by homogeneous polynomials.

Definition. To any homogeneous ideal J ⊂ K[X0, .., Xn] we cab associate its zero set in Pn

n WJ = {P = [x0, .., xn] ∈ P | f(P ) = 0 ∀f ∈ J} called projective algebraic set. Viceversa, if W ⊂ Pn is a projective algebraic set, its homogeneous ideal of polynomials vanishing on it is the ideal I(W ) generated by all homogeneous polynomials f ∈ K[X0, .., Xn] such that f(x0, .., xn) = 0 for all [x0, .., xn] ∈ W .

As for the affine case, W is defined over K0 if I(W ) can be generated by elements of K0[X0, .., Xn] and we write W/K0. The set of K0-rational points of n W is W (K0) = W ∩ P (K0). Definition. A projective algebraic set is irreducible if it is not union of two proper projective algebraic sets. A irreducible projective algebraic set W is called projective algebraic variety.

n n Now notice that P (K) = U0 ∪ · · · ∪ Un where Ui = {[x0, .., xn] ∈ P (K) | xi ̸= 0}. ϕi n Then for each i ∈ {0, .., n} the map Ui −→ A (K) given by

(x0 xi−1 xi+1 xn ) [x0, .., xn] ↦→ , .., , , .. xi xi xi xi is a bijection (actually it is a homeomorphism with respect to the Zariski topology). It follows that any projective variety W has a standard covering

n ⋃ W = Wi where Wi = W ∩ Ui i=0

And by mean of ϕi, each Wi can be regarded as an affine variety.

Remark 1.8. The map W ↦→ W0 = W ∩ U0 yields a bijection

{ projective varieties } { affine varieties } n ←→ n W ⊂ P | W * {x0 = 0} W0 ⊂ A which inverse map is given by the projective closure. Namely if V is an affine algebraic set with ideal I(V ), its projective closure is the projective set W = V whose homogeneous ideal is generated by {Xdf( X1 , .., Xn ) | f ∈ I(V )} 0 X0 X0 1.1. ALGEBRAIC VARIETIES 5

Definition. If W is a projective variety its function field is

K(W ) = {f/g | f, g ∈ K[X0, .., Xn], homogeneous of same degree d, g∈ / I(W )}/ ∼ where f/g ∼ f ′/g′ if and only if fg′ − gf ′ ∈ I(W ). The elements of K(W ) are called rational functions on W .

Lemma 1.9. Let W be a projective variety such that W * V (x0), with standard covering W = W0 ∩ .. ∩ Wn. We have an isomorphism

∼ K(W ) −→ K(W0)

Proof. We have the following maps are mutually inverse:

K(W ) −→ K(W0) given by f(X0, .., Xn)/g(X0, .., Xn) ↦→ f(1,X1, .., Xn)/g(1,X1, .., Xn)

(X1 Xn ) (X1 Xn ) K(W0) −→ K(W ) given by f(X1, .., Xn)/g(X1, .., Xn) ↦→ f , .., /g , .., X0 X0 X0 X0

Definition. The dimension dim(W ) of a projective variety W is the transcendence degree of K(W ). Definition. A rational function F ∈ K(W ) is regular at P ∈ W if there is a representation F = f/g such that g(P ) ̸= 0. The local ring of W at P ∈ W is

K[W ]P = {F ∈ K(W ) | F is regular at P } with maximal ideal MP = {F ∈ K[W ]P | F (P ) = 0}. 2 A point P ∈ W is a nonsingular point if dim(MP /MP ) = dim(W ); W is nonsingular if each point P is nonsingular.

m n Now let W1 ⊂ P and W2 ⊂ P be projective varieties in the respective projective spaces.

Definition. A rational map F : W1 −→ W2 is a tuple F = [f0, .., fn] of elements of K(W1) such that ∀P ∈ W1 where f0, .., fn are all defined, then F (P ) = [f0(P ), .., fn(P )] belongs to W2. F is defined over K0 if W1/K0, W2/K0 and f0, .., fn can be multiplied by the same invertible elements of K so that f0, .., fn ∈ K0(W1)

Definition. A rational map F = [f0, .., fn]: W1 −→ W2 is regular at P ∈ W1 if there exists g ∈ K(W1) such that gfi is regular at P and [gf0(P ), .., gfn(P )] is not the 0-tuple. A rational map regular at all points of W1 is called morphism.

Definition. We say that W1 and W2 are isomorphic if ∃F : W1 −→ W2 and ∃G : W2 −→ W1 morphisms such that F ◦ G and G ◦ F are the respective identity maps. W1/K0 and W2/K0 are isomorphic over K0 is F and G as above are defined over K0.

Definition. A rational map F : W1 −→ W2 is dominant if its image F (W1) is a Zariski dense subset of W2. 1.2. ALGEBRAIC CURVES 6

Now if F : W1 −→ W2 is a rational map and F is dominant, then we have a well-defined K-algebra map

∗ F : K(W2) −→ K(W1) such that F ∗(g/h) = F ∗(g)/F ∗(h) = g ◦ F/h ◦ F . In particular if F is an isomorphism then F ∗ is an isomorphism of function fields and an isomorphism of local rings at each point.

1.2 Algebraic Curves

Proposition 1.10. Let C be a projective curve and P be a nonsingular point ∞ 2 ⋂ h on the curve. Then MP = (t) for some t ∈ MP r MP and MP = {0} h=1

2 ∼ 2 Proof. P ∈ C nonsingular implies that mP /mP −→ MP /MP has dimension 1 2 as K-vector space, therefore mP /mP = ⟨t⟩ for some t ∈ K[C]. We claim that MP = t · K[C]P . def Let N = t · K[C]P ⊆ MP , and let us prove that the quotient MP /N is trivial, so that they coincide. Since MP is a K[C]P -module, so is the quotient MP /N, and notice that

2 • MP · (MP /N) = (N + MP )/N 2 2 • N + MP = (K · t + MP )K[C]P Therefore MP · (MP /N) = MP /N and by Nakayama’s lemma this can happen only if MP = N = (t). ∞ ∞ ⋂ h ⋂ h h Now let f ∈ MP = (t ) and write f = t · fh. Since fh = t · fh+1 we h=1 h=1 obtain a chain of inclusions

(f1) ⊆ (f2) ⊆ .. which has to stabilize due to the Noetherianity of K[C]P . Therefore there exists m > 0 such that (fm) = (fm+1), hence

fm+1 = afm = atfm+1 for some a ∈ K[C]P

Either at = 1 or fm+1 = 0. Since t ∈ MP , t is not invertible, and we can conclude that fm+1 = 0 =⇒ f = 0.

2 Definition. Any such t ∈ MP r MP is called uniformizer of C at P . Define ordP (·): K[C]P −→ Z ∪ {∞} as { +∞ if f = 0 ordP (f) = l l+1 (1.2.1) min{l ∈ Z>0 | f ∈ MP r MP } if f ̸= 0

l Remark 1.11. (i) If f ̸= 0 and l = ordP (f), then f = a · t for some a ∈ × K[C]P . 1.2. ALGEBRAIC CURVES 7

(ii) We can extend ordP to the whole K(C), namely if F = f/g with f, g ∈ K[C]P and g ̸= 0 then define

ordP (F ) = ordP (f) − ordP (g)

This is well defined since if l = ordP (F ) then l is the unique power for l which F/t is a unit in K[C]P .

Proposition 1.12. ordP : K(C) −→ Z ∪ {∞} satisfies the following properties:

(a) ordP (F · G) = ordP (F ) + ordP (G)

(b) ordP (F + G) ≥ min{ordP (F ), ordP (G)}

(d) MP = {F ∈ K(C) | ordP (F ) > 0}

(e) ordP (F ) = ∞ ⇐⇒ F = 0

In other words, K[C]P is a discrete valuation ring with valuation given by ordP . Theorem 1.13. Let C be a projective curve, P ∈ C be a nonsingular point and let W be a projective variety. If F : C −→ W is a rational map, then F is regular at P . In particular it follows that if C is nonsingular, then F is automatically a morphism.

Proof. Let F = [f0, .., fn] with fj ∈ K(C) and let t be a uniformizer at P . −m Define m = min {ordP (fj)}, then ordP (t · fj) ≥ 0 ∀j = 0, .., n and equality 0≤j≤n −m −m holding for some j = j0. This means that t · fj is regular at P and (t ·

fj0 )(P ) ̸= 0. Hence taking g = t−m exhibits that F is regular at P . Definition. A proper subring R ⊂ K(C) containing K is called discrete valuation ring of K(C) over K if there exists a function v : K(C) −→ Z ∪ {∞} called valuation, such that: (a) v(FG) = v(F ) + v(G); (b) v(F + G) ≥ min{v(F ), v(G)}; (c) v(F ) = ∞ ⇐⇒ F = 0; (d) R = {F ∈ K(C) | v(F ) ≥ 0}. In particular if R is a discrete valuation ring of K(C) over K, then R is a def local ring with maximal ideal MR = {F ∈ R | v(F ) > 0}. In particular, for a nonsingular projective curve C we have the following: Proposition 1.14. Let C be a nonsingular projective curve and let R be a discrete valuation ring of K(C) over K with valuation v. Then there exists P ∈ C such that v = c · ordP for some c ∈ Z>0, moreover for such P , R = K[C]P and MR = MP . 1.2. ALGEBRAIC CURVES 8

n Proof. Let [x0, .., xn] be the projective coordinates of P , then we can regard xj as a function from C to K, mapping a point P to its j-th coordinate. For what follows, do not consider the xj’s such that xj ∈ I(C) = {f ∈ K[X0, .., Xn] | f(P ) = 0 ∀P ∈ C}. For the remaining indices, regard xj/x0 as an element of K(C)

• If v(xj/x0) ≥ 0 ∀j pass to the affine curve C ∩ U0 = C0 (which is not empty since x0 ̸= 0 somewhere in C). Elements of the affine coordinate ring K[C0] are polynomials in the xj/x0’s, therefore v ≥ 0 on the whole ring. It follows that K[C0] ⊂ R.

• If v(xj/x0) < 0 for some j, then let j0 be the index for which v(xj0 /x0) is

the smallest possible. Then v(xj/xj0 ) = v(xj/x0) − v(xj0 /x0) > 0 for all j. Changing j0 with 0 we can assume C0 is not empty and K[C0] ⊆ R. def ∼ Therefore I = MR ∩ K[C0] is a proper ideal of K[C0] = K[X1, .., Xn]/I(C). Let J be the inverse image of I in K[X1, .., Xn] (so in particular I(C) ⊂ J). Then there are some points P ∈ An that annihilate all the elements of J, therefore those P belongs to C and for each f ∈ I we have ordP (f) ≥ 0. We claim that ∀F ∈ K(C)

v(F ) > 0 and ordP (F ) ≥ 0 =⇒ ordP (F ) > 0 (1.2.2)

In fact since C ∩ U0 ̸= ∅, by Lemma 1.9 we can identify K(C) with K(C0) and K[C]P with K[C0]P . Let F ∈ MR ∩ K[C0]P , F = f/g for f, g ∈ K[C0] and ordP (g) = 0; g ∈ K[C0] ⊂ R, implies that f = F · g belongs to I and therefore ordP (f) > 0. It follows that ordP (F ) = ordP (f) − ordP (g) > 0.

The contrapositive of (1.2.2) is ordP (F ) = 0 =⇒ v(F ) ≤ 0; which applied to 1/F yields

ordP (1/F ) = −ordP (F ) = 0 =⇒ 0 ≥ v(1/F ) = −v(F ) =⇒ v(F ) ≥ 0 and consequentely ordP (F ) = 0 =⇒ v(F ) = 0 (1.2.3)

Finally choose an uniformizer t ∈ K[C0] at P : since K[C0] ⊆ R we have that ℓ v(t) ≥ 0 and if F ∈ K(C), F = t · F0 with ℓ = ordP (F ) and ordP (F0) = 0 (so that by (1.2.3) v(F0) = 0), then

ℓ v(F ) = v(t ) + v(F0) = ℓ · v(t) = v(t) · ordP (F )

Since R is a proper subring of K(C), v cannot be identically zero and v(t) ≥ 0: therefore v(t) > 0. We obtained that v = v(t) · ordP and c := v(t) > 0 as we wanted. Theorem 1.15. Let L be a field of trancendental degree one and finitely generated over K and let CL be the set of discrete valuation rings R of L over K. ∼ Then there exists a nonsingular projective curve C such that K(C) = L and CL can be canonically identified with the set of points of C. Sketch of the proof. 1.2. ALGEBRAIC CURVES 9

Step 1 If B is an integral domain finitely generated as K-algebra then B is isomorphic to a coordinate ring of an affine variety. Namely, let x1, .., xn generate B over K, then the map Xj ↦→ xj yields a homomorphism K[X1, .., Xn] B which kernel I is prime since B is a domain. There- fore VI is the required affine variety.

Step 2 Let x ∈ L r K and consider the polynomial ring K[x]: then L is a finite extension of the quotient field K(x). Now if B is the integral closure of K[x] in L, the construction in the previous step generate a nonsingular affine curve (since in dimension 1 a ring is regular if and only if it is integrally closed)

Step 3 Let R ∈ CL, choose x ∈ R r K, then K[x] ⊆ R and since R is a discrete valuation ring, it is integrally closed in L. Hence B ⊆ R. Now let N = MR ∩ B, then N is maximal in B and therefore it corresponds to a unique point of the affine curve constructed from B.

Step 4 Finally maps CL diagonally into the product of finitely many projective closures (of affine curves). The image of CL will be the desired curve C.

Theorem 1.16. Let C1 be a projective nonsingular curve over K and let C2 be any curve over K. If F : C1 −→ C2 is a morphism, then either F (C1) is a point (i.e. F is constant) or F (C1) = C2 (i.e. F is surjective). In the latter case ∗ ∗ F is injective, K(C1) is a finite field extension of F (K(C2)) and F is a finite morphism. Proof. See [Har77] II.6.8

Theorem 1.17. Let C1 and C2 be projective nonsingular curves over K and let i : K(C2) −→ K(C1) be an injective map fixing K. Then there exists a unique ∗ non constant map F : C1 −→ C2 such that F = i

m Proof. Let C1 ⊂ P , and without loss of generality assume that C2 * V (x0). Then for each j = 0, .., m, let fj := xj/x0 ∈ K(C2), and define

F = [1, i(f1), .., i(fm)] : C1 −→ C2

∗ Such F satisfies F = i. If G = [g0, .., gm] is another map with the property ∗ ∗ ∗ G = i, then gj/g0 = G (fj) = F (fj) = i(fj) for all j = 0, .., m, therefore F = G.

Definition. Let F : C1 −→ C2 be a map between nonsingular projective curves over K. Let us define the degree of F as { 0 if F is constant deg(F ) = ∗ [K(C1): F (K(C2))] otherwise

Moreover if F is not constant, then F is separable, inseparable, or purely insep- ∗ arable if the field extension K(C1)/F (K(C2)) is respectively separable, inseparable or purely inseparable.

Remark 1.18. (i) If C1 and C2 are defined over K0, then

∗ ∗ [K(C1): F (K(C2))] = [K0(C1): F (K0(C2))] 1.2. ALGEBRAIC CURVES 10

(ii) Since any algebraic extension can be obtained as a separable extension followed by a purely inseparable one, we denote degsF the degree of the separable part and degiF the degree of the purely inseparable part of the ∗ extension K(C1)/F (K(C2)). Then deg(F ) = degs(F ) · degi(F )

Corollary 1.19. Let F : C1 −→ C2 be a non constant map between nonsingular projective curves over K. If deg(F ) = 1 then F is an isomorphism. Proof. See [Sil09] II.4.1

Definition. Let F : C1 −→ C2 be a non constant map between nonsingular projective curves. The ramification degree of F at a point P ∈ C1 is the positive integer ∗ eF (P ) = ordP (F (t)) = ordP (t ◦ F )

where t = tF (P ) is a uniformizer at F (P ). F is unramified at P if eF (P ) = 1, and it is unramified if it is unramified at each point of C1.

eP (F ) Notice that tF (P ) ◦F = tP g for tP uniformizer at P and some g in K(C1) such that ordP (g) = 0.

Proposition 1.20. Let F : C1 −→ C2 be a non constant map between nonsingular projective curves, then ∑ (a) For all Q ∈ C2 we have eF (P ) = deg(F ). P ∈F −1(Q)

−1 (b) For all but finitely many Q ∈ C2 we have |F (Q)| = degs(F ) Proof. (a) See [Sha77] III.2.1 (b) See [Har77] II.6.8

Corollary 1.21. A map F : C1 −→ C2 between nonsingular projective curves −1 is unramified if and only if |F (Q)| = deg(F ) ∀Q ∈ C2. Proof. By Proposition 1.20, we have that

−1 ∑ −1 |F (Q)| = deg(F ) ⇐⇒ eF (P ) = |F (Q)| P ∈F −1(Q)

Since eF (P ) ≥ 1, this happens if and only if eF (P ) = 1

1.2.1 Fields of positive characteristic We now want to take a look on what happens in the case of charK ̸= 0: the problem in such situation is that field extensions of fields with prime characteristic are not necessarely separable. Let p be a fixed rational prime and Fp be the field of p elements with algebraic r closure Fp. Recall that for any q = p , r ∈ Z>0, there exists a unique field Fq q of order q in Fp, namely the splitting field of X − X over Fp. In particular ⋃ Fp = Fpr . r≥1 1.2. ALGEBRAIC CURVES 11

q Definition. The q-th Frobenius map φq on Fp is φq : Fq −→ Fq , x ↦→ x r Proposition 1.22. Let q = p , r ∈ Z>0, then:

(a) Fq = Fp n (b) Let α ∈ Fq, then α ∈ Fqn if and only if φq (α) = α

φq(x + y) = φq(x) + φq(y) and φq(xy) = φq(x)φq(y)

Proof. (a) It is a more general fact that if M ⊂ L with L algebraic over M then M = L. In fact if α ∈ L and L is algebraic over M, then α is algebraic over M, thus α ∈ M. The other inclusion is trivial and equality follows.

q × (b) Let us first prove that Fq = {α ∈ Fq | α = α}. Since Fq ⊂ Fq is a group of q−1 × order q − 1 we have that α = 1 for all α ∈ Fq and therefore ∀α ∈ Fq we q q have α = α. This proves the containment Fq ⊂ {α ∈ Fq | α = α}. Now recall that a polynomial f(X) has multiple roots if and only if f(X) and its formal derivative f ′(X) have common roots. Since we are in characteristic p, d (Xq − X) = qXq−1 − 1 = −1 dX which means that Xq − X has no multiple roots and therefore we have q q q = def(X − X) dinstict α ∈ Fq such that α = α. Hence |Fq| = |{α ∈ Fq | q q α = α}| and Fq ⊆ {α ∈ Fq | α = α} therefore they must coincide. Finally, (b) follows by substituting q with qn.

(p) p! (c) Remark that k = k!(p−k)! and for all k = 1, .., p − 1 we have a factor p at its numerator which does not cancel by the denominator, thus (p) ∀k = 1, .., p − 1 = 0 mod p k

p p p It follows that φp(x + y) = (x + y) = x + y = φp(x) + φp(y). Now since r r if q = p then φq = φp, we have

φq(x + y) = φq(x) + φq(y) ∀x, y ∈ Fq

On the other hand the multiplicativity of φq follows by commutativity of Fq. Therefore φq is a homomorphism of fields, and this also gives us injectivity. n Now if α ∈ Fq, then α ∈ Fqn for some n ∈ Z>0 and therefore α = φq (α) = n−1 φq(φq (α)). Hence α ∈ im(φq), therefore φq is surjective.

r Fact 1.23. For q = p , Fq/Fp is a Galois extension with cyclic Galois group of order r generated by φp. n n n Definition. The Frobenius map on Fp is φp : Fp −→ Fp ,(x1, .., xn) ↦→ p p (x1, .., xn) 1.2. ALGEBRAIC CURVES 12

n n n Remark 1.24. φp : Fp −→ Fp is a bijection with fixed points Fp , which induces a bijection at level of projective classes

n n p p φp : P (Fp) −→ P (Fp)[x0, .., xn] ↦→ [x0, .., xn]

For brevity write X = (X0, .., Xn), e = (e0, .., en) and consider a homoge- ∑ e neous polynomial f(X) = aeX ∈ Fp[X0, .., Xn], then define e

(q) ∑ e f (X) = φq(ae)X ∈ Fp[X0, .., Xn] e and since we are in characteristic p if follows that

f (q)(Xq) = (f(X))q (1.2.4)

r Remark 1.25. Let q = p , and let C be a projective curve over Fp with ideal (q) of polynomials I(C) = ⟨f1, .., fr⟩. Then we can define a new curve C over Fp which ideal of polynomial I(C(q)) is the ideal generated by {f (q) | f ∈ I(C)}. We have that the Frobenius map on Pn induces a q-th Frobenius morphism

(q) q q φq : C −→ C given by [x0, .., xn] ↦→ [x0, .., xn]

In fact ∀[x0, .., xn] ∈ C

(q) (q) q q (1.2.4) ( )q f (φq(x0, .., xn)) = f (x0, .., xn) = f(x0, .., xn)    =0

(p) In particular notice that if C/Fp then C = C Theorem 1.26. Notation as above, we have:

∗ (q) q def q (a) φq (K(C )) = K(C) = {f | f ∈ K(C)}

(b) φq is purely inseparable;

(c) deg(φq) = q Proof. (a) Remark that every algebraically closed field is perfect, therefore each q q q element of K is a q-th power. Hence (K[X0, .., Xn]) = K[X0 , .., Xn]. Now let f/g ∈ K(C) with f, g homogeneous polynomial of the same degree. ∗ ∗ (q) Then for any φq (f/g) ∈ φq (K(C )) we have

∗ q q q q q q q q q q q φq (f/g) = f(X0 , .., Xn)/g(X0 , .., Xn) = f0 (X0 , .., Xn)/g0(X0 , .., Xn) ∈ K(C)

where f0 and g0 are polynomials which coefficients to the q-th power are the coefficients of f and g respectively. On the other hand K(C)q is the subfield of K(C) of elements

q q f(X0, .., Xn) /g(X0, .., Xn)

Therefore they coincide. (b) Follows from (a). 1.2. ALGEBRAIC CURVES 13

(c) deg(φq) = ordP (tφq (P ) · φq) = q where the first equality follows from Propo- sition 1.20 together with the fact that φq is a injective.

Corollary 1.27. Let C1,C2 be nonsingular projective curves, defined over a field of prime characteristic p. Then any map F : C1 −→ C2 factor as F = Fs ◦φq (q) where q = degi(F ), φq : C1 −→ C1 is the q-th Frobenius morphism and (q) Fs : C −→ C2 is a separable morphism.

∗ ∗ Proof. Let L = F (K(C2))sep be the maximal separable extension of F (K(C2)) in K(C1). Then K(C1)/L is purely inseparable of degree q = degi(F ), and in q particular we have that K(C1) ⊂ L. q ∗ (q) ∗ (q) From Theorem 1.26 K(C1) = φq (K(C1 )) and [K(C1): φq (K(C1 ))] = q. ∗ (q) By comparing the degrees, it follows that L = φq (K(C1 )). Thus we have the inclusions ∗ ∗ (q) F (K(C2)) ⊂ φq (K(C1 )) ⊂ K(C1) which correspond to φq (q) Fs C1 −→ C1 −→ C2

1.2.2 Riemann-Roch theorem Definition. The abelian group of divisors of a nonsingular curve C, Div(C), is the free abelian group generated by the points of C. In other words D ∈ Div(C) is a formal sum ∑ D = nP (P ) with nP ∈ Z, nP = 0 for almost all P P ∈C ∑ The degree of a divisor D is the (actual) sum deg(D) = nP ∈ Z. P ∈C The subgroup of degree-0 divisors is

Div0(C) = {D ∈ Div(C) | deg(D) = 0}

Remark that if C is defined over K0, then Aut(K/K0) acts on Div(C) by

σ ∑ σ D = nP (P ) P ∈C

σ We say that D is defined over K0 if D = D for all σ ∈ Aut(K/K0) The divisor of f ∈ K(C)× is ∑ ∑ ∑ div(f) = ordP (f) · (P ) = ordP (f) · (P ) − ordP (f) · (P ) P ∈C P ∈f −1(0) P ∈f −1(∞)

Divisors of the form div(f) are called principal divisors and its set is denoted by Divl(C). Notice that principal divisors have degree 0. Define the Picard group of C as the quotient Pic(C) = Div(C)/Divl(C) 1.2. ALGEBRAIC CURVES 14

Remark 1.28. If F : C1 −→ C2 is a dominant morphism between nonsingular projective curves, then F induces maps on divisors in both directions:

∗ F :Div(C2) −→ Div(C1) F∗ : Div(C1) −→ Div(C2) ∑ (Q) ↦→ eF (P ) · (P )(P ) ↦→ (F (P )) P ∈F −1(Q)

Definition. Let C be a nonsingular projective curve, the space of meromorphic differential 1-forms on C, denoted ΩC , is the K-vector space generated by the symbols dx for x ∈ K(C) such that ∀x, y ∈ K(C), ∀u ∈ K:

1. d(x + y) = dx + dy 2. d(xy) = xdy + ydx 3. du = 0

If F : C1 −→ C2 is a nonconstant map between nonsingular projective ∗ curves, then the map between function fields F : K(C2) −→ K(C1) induces a ∗ map on differentials F :ΩC2 −→ ΩC1 given by

∑ ∑ ∗ ∗ fidxi ↦→ F (fi)d(F xi)

Proposition 1.29. Let C be a curve, then

(a)Ω C is a 1-dimensional K(C)-vector space.

(b) Let x ∈ K(C), then dx is a basis for ΩC over K(C) if and only if K(C)/K(x) is a finite separable extension.

(c) Let F : C1 −→ C2 be a non constant map between curves, then F is ∗ separable if and only if F :ΩC2 −→ ΩC1 is injective (equivalently, non zero).

(d) Let P ∈ C and let t = tP be a uniformizer at P . Then for every ω ∈ ΩC there exists a unique function f ∈ K(C) (depending on ω and t), such that

ω = fdt

Proof. See [Sil09] II.4.2, II.4.3

Definition. Let C be a curve, let ω ∈ ΩC , let P ∈ C and t be a uniformizer at P . The order of ω at P is ordP (ω) = ordP (f) if ω = fdt. Then we can define the divisor associated to ω as ∑ div(ω) = ordP (ω) · (P ) ∈ Div(C) P ∈C

We say that the differential ω is regular (or holomorphic) if ordP (ω) ≥ 0 for all P ∈ C.

× Notice that if ω1, ω2 ∈ ΩC are nonzero, then there exists f ∈ K(C) such that ω2 = fω1 and consequently div(ω2) = div(f) + div(ω1). 1.2. ALGEBRAIC CURVES 15

Definition. Therefore define the canonical class of C as the image in Pic(C) of div(ω) for any non zero differential ω ∈ ΩC . Remark that for a curve C, there is a partial order in Div(C) as follows: ∑ we say that a divisor D = nP (P ) is positive (and write D ≥ 0) if nP ≥ P ∈C 0 ∀P ∈ C. Then for D1,D2 ∈ Div(C) the partial order is given by D1 ≥ D2 if D1 − D2 ≥ 0 Definition. Let D be a divisor on a curve C, the linear system associated to D is the (finite-dimensional) K-vector space L(D) = {f ∈ K | div(f) + D ≥ 0} Denote its dimension over K by ℓ(D) Definition. Let C be a curve defined over K, define the genus of C, denoted by g = g(C) as the dimension of L(κ) for any κ canonical divisor on C Let κ be a canonical divisor, say κ = div(ω), then L(κ) = {f ∈ K(C) | div(f) + κ ≥ 0}, or in other words f ∈ L(κ), hence 0 ≤ div(f) + κ = div(f) + div(ω) = div(fω). This means that fω is holomorphic. Viceversa, if fω is holomorphic, then f ∈ L(κ). It follows that ∼ 1 L(κ) = {ω ∈ ΩC | ω is holomorphic} =: Ω (C) Proposition 1.30. Let D be a divisor on a curve C, then (a) If deg(D) < 0, then L(D) = {0} and ℓ(D) = 0. (b) If D′ ∈ Div(C) is linearly equivalent to D i.e. there exists f ∈ K(C)× such that D′ = D + div(f), then L(D) =∼ L(D′) and consequently ℓ(D′) = ℓ(D).

Proof. (a) Let f ∈ L(D) r {0} then div(f) + D ≥ 0 then 0 ≤ deg(div(f)) + deg(D) = deg(D) ≤ 0 so we get a contradiction. Thus L(D) = {0}. (b) D′ = D + div(f) then the map L(D′) −→ L(D) such that g ↦→ g · f is an isomorphism of K-vector spaces.

Theorem 1.31. (Riemann-Roch) Let C be a nonsingular projective curve and let κ be a canonical divisor on C. Then for all divisors D ∈ Div(D) ℓ(D) − ℓ(κ − D) = deg(D) − g + 1 Proof. See [Har77] IV.1 Corollary 1.32. Let κ be a canonical divisor on C and let D be a divisor on C, then: (a) deg(κ) = 2g − 2 (b) If deg(D) > 2g − 2 then ℓ(D) = deg(D) − g + 1 Proof. (a) Apply Riemann-Roch theorem with D = κ, so that g − 1 = ℓ(κ) − ℓ(0) = deg(κ) − g + 1 thus deg(κ) = 2g − 2. 1.3. RIEMANN SURFACES 16

(b) By (a), deg(κ − D) < 0, hence by Proposition 1.30 ℓ(κ − D) = 0, and then the result follows by Riemann-Roch.

Proposition 1.33. Let C/K0 be a nonsingular projective curve and let D be a divisor defined over0 K . Then the K-vector space L(D) has a basis defined over K0 i.e. consisting of elements of K0(C). Proof. See [Sil09] II.5.8

Theorem 1.34. (Hurwitz) Let F : C1 −→ C2 be a nonconstant separable map between nonsingular projective curves. Let g1 = g(C1) and g2 = g(C2). Then ∑ 2g1 − 2 ≥ (deg(F ))(2g2 − 2) + (eF (P ) − 1)

P ∈C1 Moreover, we have equality if and only if either char(K) = 0 or char(K) = p > 0 and p does not divide eF (P ) for all P ∈ C1. Proof. See [Sil09] II.5.9

1.3 Riemann Surfaces

From now untill the end of the chapter, suppose that K = C. Then affine and projective spaces come with the complex topology, in addition to the Zariski topology. Then one can naturally give the points of a variety over C a topology inherited from the subspace topology. A little extra work (with the inverse function theorem and other analytic arguments) shows you that, if the variety is nonsingular, you have a nonsingular complex manifold. Nonetheless, in general the converse is false: there are many complex manifolds that do not arise from nonsingular algebraic varieties in this manner. However, in dimension 1, a miracle happens, and the converse is true: all compact one dimensional complex manifolds (i.e. compact Riemann Surfaces) are analytically isomorphic to the complex points of a nonsingular projective one dimensional variety (i.e. Curves), endowed with the complex topology instead of the Zariski topology. (See [Gun66] Chapter 10)

It follows that all the terminology and results we developed about curves over an arbitrary algebraically closed field hold for compact Riemann surfaces. In particular the genus g of a compact Riemann surface has the costumary topological meaning and one can show that the two definition coincide.

Let X be a Riemann surface of genus g ≥ 1, since X is a closed orientable surface of genus g, the ordinary homology group H1(X, Z) is free abelian on 2g generators. Let α1, .., αg, β1, .., βg be the standard basis over Z i.e. with the property that αiβj = δij and αiαj = 0 = βiβj ∀i ̸= j

∫ 1 ∨ def 1 Now consider the map : H1(X, Z) −→ Ω (X) = HomC(Ω (X), C) given by [ ∫ ] [γ] ↦→ ω ↦→ ω (1.3.1) γ 1.3. RIEMANN SURFACES 17

∫ 1 ∨ Since : H1(X, Z) −→ Ω (X) is injective, we can view the elements of the first homology groups as mapping holomorphic differentials on X to C, more explicitely: Proposition 1.35. Let X be a compact Riemann surface of genus g ≥ 1 and 1 let ω1, .., ωg be a basis over C for Ω (X) (which has dimension g since it is isomorphic to L(κ) for a canonical divisor κ). Then the 2g vectors ⎛∫ ⎞ ⎛∫ ⎞ ω1 ω1 α1 βg ⎜ . ⎟ ⎜ . ⎟ g λ1 = ⎝ . ⎠ , . . . , λ2g = ⎝ . ⎠ ∈ C ∫ ∫ ωg ωg α1 βg are linearly independent over R and therefore λ1, .., λ2g are a Z basis for a lattice Λ(X) in Cg. Proof. See [Swi74] I.1.7

1 In particular since dimC(Ω (X)) = g, that is also the complex dimension of its dual space. It follows that H1(X, Z) free abelian subgroup of rank 2g, generated by 2g indipendent elements over R, hence it is a lattice in Ω1(X)∨. Definition. The Jacobian variety of a compact Riemann surface X is 1 ∨ J(X) = Ω (X) /H1(X, Z) which is realized as the g-dimensional complex torus Cg/Λ(X)

Now fix a point x0 ∈ X and define the Abel-Jacobi map Φ: X −→ J(X) by (∫ x ∫ x ) x ↦→ ω1,..., ωg x0 x0 Remark 1.36. Since J(X) is a group, we can extend, by Z-linearity, Φ to a ∑ ∑ map Φ : Div(X) −→ J(X), D = ordx(D) · (x) ↦→ ordx(D) · (Φ(x)) x∈X x∈X

For a generic divisor D, this definition depends on the base point x0, however for a 0-degree divisor we have: Lemma 1.37. The Abel-Jacobi map Φ : Div(X) −→ J(X) restricted to 0 Div (X) is independent of the base point x0.

Proof. Let y0 be another base point and let γ be a path from x0 to y0. Then in the formula of Φ(x), if we change x0 to y0, we see that the image changes by (∫ ∫ ) λ = γ ω1,..., γ ωg mod Λ(X) ∑ ( ∑ Such element λ ∈ J(X) is indipendent of x, hence if nx = 0, then Φ nx · ) ∑ ∑ (x) changes by nx · (λ) = (λ) · nx = 0. Theorem 1.38. (Abel) Let X be a compact Riemann surface of genus g ≥ 1 and let D be a divisor on X. Then D is principal ⇐⇒ deg(D) = 0 and Φ(D) = 0 ∈ J(X) Or in other words the map from Pic0(X) to J(X) given by [ ] ∑ ∑ ∫ x nx · x ↦→ nx x x x0 is an isomorphism. 1.3. RIEMANN SURFACES 18

Proof. See [Mir95] VIII.2.2 Corollary 1.39. Let X be a compact Riemann surfact of genus g ≥ 1. Then the Abel-Jacobi map is injective Proof. Assume is not, then let Φ(x) = Φ(y) for some x ̸= y ∈ X. Since Φ is additive, Φ(x − y) = 0. Therefore by Abel’s theorem, x − y is a principal divisor, which means that there exists some meromorphic function f on X with a simple pole at y and no other poles. This implies that f is an isomorphism between X, of genus greater than 1, and P1(C), which has genus 0, hence a contradiction. Proposition 1.40. Let X be a compact Riemann surface of genus 1, then J(X) is isomorphic to X. Proof. Since X has genus one, J(X) is a one-dimensional complex torus, hence it is also a compact Riemann surface of genus one. Therefore for X of genus one, Φ is an injective holomorphic map between compact Riemann surfaces, therefore an isomorphism. Chapter 2

Elliptic Curves

2.1 Elliptic curves over an arbitrary field 2.1.1 Weierstrass form & abstract elliptic curves Definition. A Weierstrass equation over a field K is a cubic equation (in homogeneous coordinates) of the form

2 2 3 2 2 3 E : Y Z + a1XYZ + a3YZ = X + a2X Z + a4XZ + a6Z , a1, .., a6 ∈ K0 (2.1.1) If a1, .., a6 ∈ K0 then we say that E is defined over K0 .

We identify the Weierstrass equation E with the projective curve in P2 given def 2 2 3 by the points that annihilate F (X,Y,Z) = Y Z + a1XYZ + a3YZ − X − 2 2 3 a2X Z − a4XZ − a6Z i.e.

2 2 2 3 2 2 3 E = {[X,Y,Z] ∈ P | Y Z+a1XYZ+a3YZ −X −a2X Z−a4XZ −a6Z = 0} Remark that the only K-rational point on the line at infinity {Z = 0} is O = ∂F (X,Y,Z) [0, 1, 0], which is nonsingular since ∂Z (O) = 1 ̸= 0. Thus we can study the curve by working with the non-homogeneous coordinates x=X/Z, y=Y/Z:

2 3 2 E : y + a1xy + a3y = x + a2x + a4x + a6 (2.1.2) with associated curve

2 2 3 2 E = {(x, y) ∈ A | y + a1xy + a3y − x − a2x − a4x − a6 = 0} ∪ {∞ = [0, 1, 0]} Let us now simplify the equation in (2.1.2) under some assumption regarding the characteristic of K. First assume that Char(K) ̸= 2, then the variable change

(y − a x − a ) y ↦→ 1 3 2 completes the square and yields

2 3 2 y = 4x + b2x + 2b4x + b6 (2.1.3)

19 2.1. ELLIPTIC CURVES OVER AN ARBITRARY FIELD 20

2 2 where b2 = a1 + 4a2, b4 = 2a4 + a1a3 and b6 = a3 + 4a6. Moreover, if char(K) ̸= 3, then

(x − 3b y ) (x, y) ↦→ 2 , 36 108 produces the equation 2 3 y = x − 27c4x − 54c6 (2.1.4) 2 3 where c4 = b2 − 24b4 and c6 = −b2 + 36b2b4 − 216b6 2 2 2 Definition. Let b8 = a1a6 + 4a2a6 − a1a3a4 + a2a3 − a4, then define the discriminant ∆ of the curve given by the equation in (2.1.2) by

2 3 2 ∆ = ∆(E) = −b2b8 − 8b4 − 27b6 + 9b2b4b6 (2.1.5)

If ∆ ̸= 0 it makes sense to define the so-called j-invariant of E

3 j = jE = c4/∆ (2.1.6)

Lastly, define the invariant differential associated to E. dx dy ω = = 2 (2.1.7) 2y + a1x + a3 3x + 2a2x + a4 − a1y

Remark that for char(K) ̸= 2, 3 we have

c3 − c2 ∆ = 4 6 (2.1.8) 1728 Definition. An admissible change of variable in a Weierstrass equation is one of the form

X = u2X′ + r ; Y = u3Y ′ + su2X′ + t ; Z = Z′ for u, r, s, t ∈ K, u ̸= 0 (2.1.9) This change of variable fixes the point at infinity, [0 1, 0] and carries the line {Z = 0} to itself. Moreover one can compute all the new coefficients and see that u12∆′ = ∆ ; j′ = j and ω′ = uω (2.1.10) Proposition 2.1. The curve E given by a Weierstrass equation is nonsingular if and only if ∆ ̸= 0

2 2 3 2 Proof. Let E = {(x, y) ∈ A | f(x, y) = y +a1xy +a3y −x −a2x −a4x−a6 = 0} ∪ {∞}. We already saw that the point at infinity is never singular. Now suppose that E is singular at a some point P = (x0, y0) and consider the ′ ′ admissable change of variable given by x = x − x0 and y = y − y0 which leaves ∆ invariant. Therefore we can assume that E is singular at the point (0, 0) ∈ A2. Singularity means that

∂f(x, y) ∂f(x, y) 0 = f(0, 0) = −a ; 0 = (0, 0) = −a ; 0 = (0, 0) = a 6 ∂x 4 ∂y 3 2.1. ELLIPTIC CURVES OVER AN ARBITRARY FIELD 21

2 2 3 It follows that f(x, y) = y + a1xy − a2x − x and consequently 2 2 2 b2 = a1 + 4a2 and b4 = b6 = b8 = 0 =⇒ ∆ = −(a + 4a2) · 0 + 0 = 0 where ∆ is calculated using the formula in (2.1.5). For the other implication assume that char(K) ̸= 2 to simplify the computation. As in (2.1.3), 2 3 2 f(x, y) = y − 4x − b2x + 2b4x + b6 2 3 Then E is singular if and only if there is a point (x0, y0) such that y0 = 4x0 − 2 b2x0 + 2b4x0 + b6 and ∂f ∂f 0 = (x , y ) = 12x2 + 2b x + 2b ; 0 = (x , y ) = 2y ∂x 0 0 0 2 0 4 ∂y 0 0 0

The last equality forces y0 = 0 and thus (x0, 0) is singular if and only if it is a 3 2 double root of the polynomial q(x) = 4x − b2x + 2b4x + b6 if and only if the discriminant of q(x) is 0. Recall that for such polynomial its discriminant is

2 2 3 3 2 ∆q = b2 · (2b4) − 16(2b4) − 4b2b6 − 27 · 16b6 + 18 · 4b2(2b4)b6 = 2 2 3 3 2 = 4(b2b4 − 32b4 − b2b6 − 108b6 + 36b2b4b6) = (∗) 2 3 2 = 4(−4b2b8 − 32b4 − 108b6 + 36b2b4b6 = 16∆(E) 2 where the equality in (∗) follows from the fact that b4 − b2b6 = −4b8. This completes the proof for char(K) ̸= 2. The case of characteristic 2 is treated in [Sil09] A.1.2 Proposition 2.2. Let E be a singular curve given by a Weierstrass equation. Then E is birational to P1. Proof. As in the proof of Proposition 2.1, after an admissible linear change of variables we assume that E is singular at the point (0, 0) and thus E = {(x, y) ∈ 2 2 3 2 A | y + a1xy = x + a2x }. Then the rational maps

E → P1 P1 → E

( 2 3 2 2 ) ✤ ✤ y +a1xy−a2x y +a1xy −a2x y (x, y) → [x, y][x, y] → x2 , x3 are mutually inverse. Definition. An elliptic curve is a pair (E,O), where E us a nonsingular curve of genus 1 and O belongs to E. The elliptic curve is defined over K0 if E is defined over K0 and O ∈ E(K0).

Proposition 2.3. Let (E,O) be an elliptic curve defined over K0. Then there exist f, g ∈ K0(E) such that the morphism 2 F : E −→ P ,F = [f, g, 1] induces an isomorphism of E to a nonsingular curve given by a Weierstrass equation

2 2 3 2 2 3 Y Z + a1XYZ + a3YZ = X + a2X Z + a4XZ + a6Z for a1, .., a6 ∈ K0 such that F (0) = [0, 1, 0]. 2.1. ELLIPTIC CURVES OVER AN ARBITRARY FIELD 22

Proof. • Claim 1: There exists f, g ∈ K0(E) that satisfy a Weierstrass equation. By Riemann-Roch theorem we have that, since g = g(E) = 1, the dimension of the linear system L(n(O)) is n for any n ≥ 1. By Proposition 1.33 we can choose a basis of L(n(O)) defined over0 K . Therefore let f ∈ K0(E) be such that {1, f} is a basis over K0 for L(2(O)) and then choose any g so that {1, f, g} is a basis over K0 for L(3(O)). Notice that f must have a pole of order exactly 2 at O and g a pole of order exactly 3 at O. Therefore we have that

Space L(2(0)) L(3(0)) L(4(0)) L(5(0)) L(6(0)) Dimension 2 3 4 5 6 Generators {1, f}{1, f, g}{1, f, g, f 2}{1, f, g, f 2, fg}{1, f, g, f 2, fg, f 3, g2} Since seven functions are in L(6(O)), they must be linearly dependent. In particular the coefficients of f 3 and g2 cannot be 0, otherwise we would have a pole of order 6 which doesn’t cancel. After scaling coefficients we will have a linear relation 3 2 2 f + a2f + a4f + a6 = g + a1gf + a3g Therefore we have a map F : E −→ P2 given by F = [f, g, 1] whose image is in 3 2 2 3 2 2 the curve {X + a2X Z + a4XZ + a6Z = Y Z + a1XYZ + a3YZ } =: C. By Theorem 1.13 and Theorem 1.16 F is a surjective morphism. Moreover remark that since f (respectively g) has a pole of order 2 (resp. 3) at O, locally 1 1 we have f = z2 (1 + ...) (resp. g = z3 (1 + ...)), therefore F (O) = [f, g, 1]|O = 3 3 3 [z f, z g, z ]|O = [0, 1, 0].

• Claim 2: F has degree 1, so that it is an isomorphism by Corollary 1.19.

In fact by Proposition 1.20 the morphisms [f, 1] : E −→ P1 and [g, 1] : E −→ P1 have degree 2 and 3 respectively. Hence [K0(E):K0(f)] = 2 and [K0(E): K0(g)] = 3, and this implies that [K0(E):K0(f, g)] divides 2 and 3 therefore [K0(E):K0(f, g)] = 1.

• Claim 3: C is a nonsingular curve. Suppose not, then by Proposition 2.2 there is a degree 1 rational map h : C −→ P1 and therefore h◦F is a degree 1 morphism between nonsingular curves, hence an isomorphism by Corollary 1.19. This is a contradiction since E has genus 1 and P1 has genus 0 but the genus of a curve is a topological invariant. We can conclude that C is nonsingular and F is an isomorphism. Also the converse holds true: Proposition 2.4. If a curve E is given by a nonsingular Weierstrass equation (2.1.2) over K and O is taken as the usual point at infinity, thenE,O ( ) is an elliptic curve. Proof. To see this we only need to compute the genus of E. One first proves that for the invariant differential ω, div(ω) is a canonical divisor of degree 0 (See [Sil09] III.1.5). Then Riemann-Roch theorem with D = div(ω) yields g = dimL(div(ω)) = deg(div(ω)) + dimL(0) − g + 1 = 0 + 1 − g + 1 2.1. ELLIPTIC CURVES OVER AN ARBITRARY FIELD 23

Hence g = −g + 2 ⇒ g = 1. Proposition 2.5. Let (E,O) and (E′,O′) be two elliptic curves defined over ′ K0 given by Weierstrass equations over K0 with O and O corresponding to ′ ∞. If there exists an isomorphism F : E −→ E defines over K0 such that F (∞) = ∞, then E and E′ are related by an admissible change of variable (2.1.9) with coefficients in K0. Proof. Suppose that E, respectively E′, corresponds to the set of coordinates (f, g), resp. (x, y), as in Proposition 2.3. Then {1, f} and {1, x} are bases of L(2(O)) and L(2(O′)) respectively, while {1, f, g} and {1, x, y} are bases of L(3(O)) and L(3(O′)). Since F is an isomorphism of curves, it induces an iso- def morphism between function fields. Hence f ′ = F ∗(x) = x ◦ F ∈ L(2(O)) and ′ ∗ g = F (y) ∈ L(3(O)). In other words, there exist u1, u2, r, s1, t ∈ K0, u1, u2 ̸= 0 such that ′ ′ f = u1f + r and g = u2g + s1f + t ′ ′ 3 Since (f, g), (f , g ) satisfy the Weierstrass equation in E, we obtain that u1 = 2 2 u2. Letting u = u2/u1 and s = s1/u gives a change of variable as in (2.1.9). Let us now introduce a group operation on the elliptic curve. Definition. Let (E,O) be an elliptic curve, regarded as a nonsingular curve in P2. Let P,Q ∈ E and consider a line ℓ through these two points (we understand that the line is tangent to the curve at P if P = Q). By Bezout’s theorem there is a third point R in ℓ ∩ E, thus let ℓ′ be the line through R and O. Define the third point in ℓ′ ∩ E as P + Q. Proposition 2.6. This operation makes E into an abelian group with O as identity element. Moreover the inclusion E(K0) ⊂ E(K) is a group homomorphism. Proof. The core of the proof is to show associativity (See[Kna93] Theorem 3.8). The fact that the operation is commutative and that O is the identity element follows immediately from the above definition. Since by definition and associativity, three points P, Q, R on a line sum to O means that Q + R = −P i.e. we have an additive inverse.

Fact 2.7. One can prove that for an elliptic curve over K0 given by a Weierstrass equation over K0, addition and negative are morphisms defined over K0. Proof. See [Sil09] III.3.6 We can now reformulate Proposition 1.40 and obtain a description of the addition on E by mean of divisors. Theorem 2.8. Let (E,O) be an elliptic curve over K, then the map φ : ∑ ∑ Div(E) −→ E given by nP (P ) ↦→ [nP ]P induces an isomorphism

Pic0(E) = Div0(E)/Divl(E) −→∼ E

Therefore the principal divisors of E are characterized by

∑ l ∑ ∑ nP (P ) ∈ Div (E) ⇐⇒ nP = 0 and [nP ]P = O (2.1.11) 2.1. ELLIPTIC CURVES OVER AN ARBITRARY FIELD 24

2.1.2 Isogenies From now on we write E for an elliptic curve, and drop the reference to the group identity, which will be denoted O and for Weierstrass equations will be the point at ∞. Definition. Let E and E′ be elliptic curves over K. An isogeny between them is a non-constant morphism F : E −→ E′ such that F (O) = O. Notice that by Theorem 1.16 all isogenies are surjective morphisms. Proposition 2.9. Let F : E −→ E′ be a morphism such that F (0) = 0, then F is a group homomorphism. Proof. If F = 0 there is nothing to prove. Assume F ̸= 0, then F is a finite morphism and by Remark 1.28 we have a homomorphism F∗ : Div(E) −→ ′ 0 Div(E ) which induces a homomorphism, denoted again by F∗, Pic (E) −→ Pic0(E′). Therefore by Theorem 2.10, we can write F as the composition of three homomorphisms

F F : E =∼ Pic0(E) −→∗ Pic0(E′) =∼ E′

Example 2.10. Let E be an elliptic curve then 1. The multiplication by an integer, denoted by [m], is an isogeny from E to E.

2. If E is defined over Z/pZ then the Frobenius map is an isogeny. Let E and E′ be elliptic curves over K, denoted by Hom(E,E′) the set of all isogenies from E to E′. Remark 2.11. Hom(E,E′) is an abelian group with pointwise sum. Proof. Let F,G ∈ Hom(E,E′), then F + G is the composition of

E −→ E × E such that P ↦→ (P,P )

F × G : E × E −→ E′ × E′ such that (P,Q) ↦→ (F (P ),G(Q)) + : E′ × E′ −→ E′ such that (P,Q) ↦→ P + Q

Moreover, if E = E′, Hom(E,E) = End(E) is a ring, with multiplication given by composition of morphisms. ∗ For P ∈ E define a translation map TP : E −→ E by Q ↦→ Q + P . Then TP is an automorphism of K(E). Proposition 2.12. Let F : E −→ E′ be a nonzero isogeny between elliptic ∗ curves over K. Then ker(F ) is finite and the map P ↦→ TP induces an isomorphism ( ) ker(F ) =∼ Aut K(E)/F ∗(K(E′)) Furthermore, if F is separable then F is unramified, deg(F ) = #ker(F ) and K(E)/F ∗(K(E′)) is a Galois extension. 2.1. ELLIPTIC CURVES OVER AN ARBITRARY FIELD 25

Proof. Since F is non constant, it surjects E to E′ and it is a finite morphism, thus is particular has finite fibers, hence finite kernel. Now if P ∈ ker(F ) and f ∈ F ∗(K(E′)) then f = g ◦ F for some g ∈ K(E′). ∗ Hence TP (g ◦ F ) = g ◦ F ◦ TP = g ◦ F = f since F ◦ TP = F ∀P ∈ ker(F ). ∗ ∗ ′ Therefore TP fixes F (K(E )) for any P ∈ ker(F ). Moreover

∗ ∗ ∗ ∗ TP ◦ TQ(f) = f ◦ TQ ◦ TP = f ◦ TP +Q = TP +Q(f) = TQ+P (f)

∗ Hence P ↦→ TP is a homomorphism. −1 ′ By Corollary 1.20 (b), #F (Q) = degs(F ) for all but finitely many Q ∈ E . ′ ′ Now if F is an isogeny, the equality holds for all Q ∈ E , in fact ∀Q1,Q2 ∈ E there exists (by surjectivity of F ) R ∈ E such that F (R) = Q1 −Q2 and since F −1 −1 −1 is a homomorphism, the map TR : F (Q1) −→ F (Q2) sending P ∈ F (Q1) −1 −1 to P + R ∈ F (Q2) is a bijection. It follows that #F (Q) = degs(F ) for all ′ −1 Q ∈ E . In particular #ker(F ) = #F (O) = degs(F ) and by Galois theory

( ∗ ′ ) #Aut K(E)/F (K(E )) ≤ degs(F ) = #ker(F )

∗ It follows that it is enough to prove that the map P ↦→ TP is injective. But this ∗ ∗ is obvious since if TP = TQ then TP = TQ and therefore

∀ R ∈ E,R + P = TP (R) = TQ(R) = Q + R

Thus in particular P = O + P = O + Q = Q. Finally assume that F is separable, then

−1 ′ #F (Q) = degs(F ) = deg(F ) ∀Q ∈ E hence F is unramified by Corollary 1.21 and setting Q = O produces

#ker(F ) = #F −1(O) = deg(F )

∗ By the isomorphism P ↦→ TP it follows ( ) [K(E): F ∗(K(E′))] = deg(F ) = #ker(F ) = #Aut K(E)/F ∗(K(E′))

r Lemma 2.13. 1 − φq is a separable isogeny for any q = p , r ≥ 1.

Proof. Assume not, then 1 − φq factors as

r s 1 − φp = 1 − φq = f ◦ φp for some f : E −→ E separable morphism and s ≥ 1. Hence

{ s−r r−1 s r (f ◦ φp + 1) ◦ φp ◦ φp if s ≥ r 1 = f ◦ φp + φq = r−s s−1 (f + φp ) ◦ φp ◦ φp if r > s which contradicts the fact that 1 is an isomorphism, while the right hand side cannot be since φp is an inseparable map of degree p. 2.1. ELLIPTIC CURVES OVER AN ARBITRARY FIELD 26

Remark 2.14. Let E be an elliptic curve defined over Fp, then Proposition 2.12 gives us a way of computing #E(Fp). We saw that the Frobenius map φ fixes exactly those points of E that are in E(Fp), or in other words

Q ∈ E(Fp) ⇐⇒ φp(Q) = Q ⇐⇒ ([1]−φp)(Q) = O ⇐⇒ Q ∈ ker([1]−φp) hence #E(Fp) = #ker([1] − φp). Since 1 − φp is a separable isogeny, by Proposition 2.12

#E(Fp) = #ker([1] − φp) = deg([1] − φp) (2.1.12)

Lemma 2.15. Let F : E1 −→ E2 and G : E1 −→ E3 be nonzero isogenies between elliptic curves over K. Let F be separable and such that ker(F ) ⊆ ker(G). Then there exists a unique isogeny H : E2 −→ E3 such that G = H ◦ F . Moreover, if F and G are both defined over K0, so is H. Proof. By Proposition 2.12 we have

( ∗ ) ∼ ∼ ( ∗ ) Aut K(E1)/F (K(E2)) = ker(F ) ⊂ ker(G) = Aut K(E1)/G (K(E3))

∗ ∗ Hence is σ fixes F (K(E2)), then it also fixes G (K(E3)). Moreover since F ∗ ∗ is separable K(E1)/F (K(E2)) is a Galois extension, therefore G (K(E3)) ⊂ ∗ F (K(E2)) is an inclusion of fields. Theorem 1.17 yields a nonconstant morphism ∗ ∗ ∗ H : E2 −→ E3 such that F ◦ H = G . By uniqueness follows that G = H ◦ F , and this implies that H is an isogeny since O = G(O) = H(F (O)) = H(O) as required. ∗ If F and G are defined over K0 we can apply Theorem 1.17 to G (K0(E3)) ⊂ ∗ F (K0(E2)) to produce H defined over K0 aswell. Theorem 2.16. Let F : E −→ E′ be a nonzero isogeny of degree m, between elliptic curves over K. Then there exists a unique isogeny Fˆ : E′ −→ E such that Fˆ ◦ F = [m]. Such Fˆ is called dual isogeny of F .

Proof. Factor F as F = Fs ◦ φq for Fs a separable morphism and the q-th Frobenius map φq. Then it is enough to prove existence of dual isogeny for each ˆ factor. In fact, assume there exist Fˆs and φq, then ˆ ˆ ˆ ˆ (φq ◦ Fs) ◦ (Fs ◦ φq) = φq ◦ [degs(F )] ◦ φq = [degs(F )] ◦ φq ◦ φq = [m] ˆ Hence φq ◦ Fˆs = (Fˆs ◦ φ1).

• Let Fs be a separable morphism of degree m, then #ker(Fs) = m and therefore [m]: E −→ E annhilates ker(F ). It follows that ker(F ) ⊂ ker([m]) and we found ourselves in the assumption of Lemma 2.15 with ′ Fs : E −→ E and [m]: E −→ E. Therefore there exists a unique ′ H : E −→ E such that [m] = H ◦ Fs. Thus H = Fˆs.

r r • Let φq be the q-th Frobenius map for q = p , then φq = φp. Thus it ˆ it enough to prove existence of φp. Recall that deg(φp) = p, and since char(K) = p, multiplication by p is not separable1. Since [p] is not sepa- h ˆ h−1 rable, it has a factorization [p] = Gs ◦ φp , thus take φp = Gs ◦ φp .

1 It follows from Proposition 1.29 (c), since [p]∗ : ω ↦→ 0, where ω is the invariant differential 2.2. ELLIPTIC CURVES OVER C 27

Finally let us prove uniqueness: let F,ˆ Fˆ′ be two dual isogeny of F , then

(Fˆ − Fˆ′) ◦ F = [m] − [m] = 0

But by assumption F is nonzero, therefore Fˆ = Fˆ′. Definition. We say that E is isogenous to E′ if there exists a non zero isogeny from E to E′. The Theorem implies that ”being isogenous” is an equivalence relation.

The next result is a sort of converse of Proposition 2.12 Theorem 2.17. Let E be an elliptic curve over K and let S be a finite subgroup of E. Then there exist an elliptic curve E′ unique up to isomorphism and a separable isogeny F : E −→ E′ such that ker(F ) = S. Moreover if E is defined ′ over K0 and S is stable under Aut(K/K0), then E is also defined over0 K . Such elliptic curve is often denoted E/S. Proof. We give a proof in the case char(K) = 0, for a proof in positive characteristic see [DS06 Section 7.8]. ∗ To any P ∈ S associate the automorphism TP : f ↦→ f ◦ TP , ∀f ∈ K(E). Since ∗ ∗ ∗ TP +P ′ = TP ◦ TP ′ , S can be identified with a finite subgroup ofAut(K(E)), namely ∼ ∗ S = {TP | P ∈ S} Let K(E)S be the subfield of K(E) fixed by the elements in S, then K(E)/K(E)S is a Galois extension with Galois group Gal(K(E)/K(E)S) =∼ S. Moreover, K(E)S is a function field of dimension one over K, thus there exist a unique nonsingular projective curve E′ over K and a K-isomorphism F ∗ : K(E′) −→∼ K(E)S ↪→ K(E). The fields-curves correspondence in Theorem 1.17 yields a surjective morphism F : E −→ E′. We are left to show that such F is unramified and the genus of E′ is 1. For any X ∈ E, let Y = F (X) ∈ E′, then ∀P ∈ S, ∀f ∈ K(E)S we have

∗ f(X + P ) = f ◦ TP (X) = TP f(X) = f(X) =⇒ F (X + P ) = F (X) = Y

=⇒ X + S = {X + P | P ∈ S} ⊆ F −1(Y ) On the other hand we have

−1 ∑ ∗ ′ S #F (Y ) ≤ eX (F ) = deg(F ) = [K(E): F (K(E )] = [K(E): K(E) ] = #S X∈F −1(Y ) where the last equality follows by Galois theory. Hence we can conclude that F −1(Y ) = X + S and therefore F is unramified and deg(F ) = #S. Finally Theorem 1.34 in characteristic 0 yields that the genus of E′ is 1.

2.2 Elliptic curves over C

The theory of elliptic curves over C becomes much easier, since it is equivalent (in a categorial sense) to the theory of lattices in C or, equivalently again, to the one of complex tori. 2.2. ELLIPTIC CURVES OVER C 28

Definition. (i)A lattice Λ in C is a subgroup of (C, +) of the form Λ = Zw1 ⊕ Zw2 such that Λ ⊗Z R = Rw1 + Rw2 = C i.e. {w1, w2} is a R-basis for C. ( ) w1 (ii) An ordered basis of a lattice Λ is normalized if w1/w2 ∈ H w2 where H := {z ∈ C | Im(z) > 0}. (iii) Two lattices Λ and Λ′ are homothetic if there exists α ∈ C× such that Λ′ = αΛ.

× Remark 2.18. (a) (i) is equivalent to saying that w1, w2 ∈ C and w1/w2 ∈ C r R. (b) Λ is a lattice if and only if Λ is a free rank 2 discrete subgroup of (C, +). Let us denote by L the set of all lattices and by B the set of normalized bases. ( ) w1 Then we have a surjective map π : B → L given by ↦→ Zw1 + Zw2. w2

Lemma 2.19. The fibers of π are all of the form SL2(Z) · b for b ∈ B. Hence π induces a bijection π : SL2(Z)\B −→ L ( ) ( ′ ) w1 ′ w1 Proof. Notice that b = and b = ′ are the same basis for a lattice Λ w2 w2 (a b) if and only if there exists a matrix γ = ∈ GL ( ) such that c d 2 Z

( ′ ) ( )( ) ( ) w1 a b w1 aw1 + bw2 ′ = = w2 c d w2 cw1 + dw2

′ ′ ′ ′ ′ Setting τ = w1/w2 and τ = w1/w2 gives that τ, τ belong to C r R and τ = aτ+b ′ det(γ) ′ cτ+d . Noticing that Im(τ ) = |cτ+d|2 Im(τ) we deduce that b, b are normalized basis if and only if det(γ) > 0, therefore γ ∈ SL2(Z). Notice that there is an action of C× both on L and B. Namely, let α ∈ C×, (w ) (αw ) Λ ∈ L and 1 ∈ B, then αΛ ∈ L and 1 ∈ B. w2 αw2 [ ] × w1 In particular we have an identification B/C −→ H given by ↦→ w1/w2. w2 × The actions of C and SL2(Z) on B commute: therefore the action of SL2(Z) on B induces an action of SL2(Z) on H by fractional linear transformation

SL2(Z) × H −→ H ( (a b) ) aτ + b (2.2.1) , τ ↦→ c d cτ + d Corollary 2.20. The bijection π induces a natural bijection

× π˜ : SL2(Z)\H −→ L/C

In particular any lattice Λ is homothetic to a lattice of the form Λτ = Zτ +Z for some τ ∈ H. 2.2. ELLIPTIC CURVES OVER C 29

Remark 2.21. We will see that the set L/C× = {[Λ] = homothety class of the lattice Λ} is in bijection with the set of isomorphic classes of complex tori {[C/Λ] = isomorphism class of C/Λ} where isomorphism means holomorphic isomorphism. Moreover, the set of isomorphism classes of 1-dimensional complex tori is in bijection with the set of isomorphism classes of elliptic curves over C (with respect to algebraic isomorphism).

2.2.1 The Weierstrass ℘-function

Definition. Let Λ = Zw1 + Zw2 be a lattice in C. A meromorphic function f : C −→ C is an elliptic function of period Λ if f(z + w) = f(w) for all w ∈ Λ. This amount to saying that f(z + wi) = f(z) for i = 1, 2. 1 Definition. Let Λ = Zw1 + Zw2 be a lattice in C, the function ℘ : C −→ P given by 1 ∑ ( 1 1 ) ℘ (z) = ℘(z) = + − Λ z2 (z − w)2 w2 w∈Λ w̸=0 is called Weierstrass ℘-function for Λ.

Proposition 2.22. Let Λ = Zw1 + Zw2 be a lattice in C, and ℘ be the Weier- strass ℘-function for Λ, then: (a) ℘(z) is an even function whose singularities are double poles at lattice points; (b) ℘ is an elliptic function of period Λ; (c) deg(℘) = 2; (d) ℘ is surjective.

∑ ( 1 1 ) Proof. (a) One first proves that w∈Λ (z−w)2 − w2 converges absolutely and w̸=0 uniformely on compact subsets of C r Λ. Then to see that it has a double pole at lattice points, notice that the 1 function hv(z) = ℘(z)− (z−v)2 for some v ∈ Λ converges uniformely on (Cr 1 Λ)∪{v}. Hence in a neighbourhood of v, ℘(z) = (z−v)2 +holomorphic part. Finally absolute and uniform convergence on compact subsets allow us to reorder the summands and therefore since Λ = −Λ 1 ∑ ( 1 1 ) 1 ∑ ( 1 1 ) ℘(−z) = + − = + − = ℘(z) (−z)2 (−z − w)2 w2 z2 (z − (−w)2 (−w)2 w∈Λ w∈Λ w̸=0 w̸=0

(b) Since ℘ is meromorphic on C, its derivative ℘′ is also meromorphic. Moreover ′ 2 ∑ ( −2 ) ∑ ( 1 ) ′ ℘ (z) = − z3 + w∈Λ (z−w)3 = −2 w∈Λ (z−w)3 , hence ℘ (z + w) = w̸=0 w̸=0 ℘′(z) for all w ∈ Λ since w + Λ = Λ. Therefore ℘′ is an elliptic function, hence ℘(z + wi) = ℘(z) + ci for some ci ∈ C. Taking z = −wi/2 yields that wi wi ℘( 2 ) = ℘(− 2 ) + ci. But by (a), ℘ is even, thus ci = 0 for both i = 1, 2. We conclude that ℘ is an elliptic function. (c) Follows from Proposition 1.20 together with the fact that ℘ has poles of order 2 at any lattice point. 2.2. ELLIPTIC CURVES OVER C 30

(d) Since ℘ is an elliptic function of period Λ, it factors through the complect torus C/Λ, which is compact. Now ℘ is non constant therefore it must surject to P1(C).

Lemma 2.23. ℘′ has degree 3 and it has three distinct zeroes in the funda- def ∏ w1 w2 w3 mental domain = {t1w1 + t2w2 ∈ C | 0 ≤ t1, t2 < 1}, at 2 , 2 and 2 , where w3 = w1 + w2, each of multiplicity one. Proof. ℘′ has only poles of order 3 in w ∈ Λ, thus deg(℘′) = 3. Since ℘ is even, ℘′ is odd, and we saw that ℘′ is elliptic with period Λ. Therefore w w w w ℘′( i ) = −℘′( − i ) = −℘′( − i + w ) = −℘′( i ) 2 2 2 i 2

′( wi ) ′ Hence ℘ 2 = 0 for i = 1, 2, 3 and as deg(℘ ) = 3 these must be all the zeroes of ℘′ in ∏ and therefore have multiplicity one. Definition. For n ≥ 3, we call Eisenstein series of weight n the quantity

∑ 1 E (Λ) = n wn w∈Λ w̸=0

Theorem 2.24. Define g2 = g2(Λ) = 60E4(Λ) and g3 = g3(Λ) = 140E6(Λ), then for all z ∈ C: ′ 2 3 (℘ (z)) = 4℘(z) − g2℘(z) − g3 (2.2.2)

2 1 Proof. Recall the equality 1 + t + t + .. = 1−t for |t| < 1. Deriving we obtain

∞ 1 ∑ = 1 + 2t + 3t2 + ... = (k + 1)tk (1 − t)2 k=0 Hence for z near 0 we have ∞ 1 1 1 ( 1 ) 1 ∑ k + 1 − = −1 = (1+2(z/w)+...−1) = zk (z − w)2 w2 w2 (z/w − 1)2 w2 wk+2 k=1 Therefore we can write ℘(z) as

∞ ∞ 1 ∑ ( ∑ 1 ) 1 ∑ 1 ℘(z) = + (k+1) zk = + E (Λ)zk = +3E (Λ)z2+5E (Λ)z4+.. z2 wk+2 z2 k+2 z2 4 6 k=1 w∈Λ k=1 w̸=0

and 2 4 1 ℘′(z) = − + 6E (Λ)z + 20E (Λ)z3 + .. =⇒ (℘′(z))2 = − 24E (Λ) + ... z3 4 6 z6 4 z2

′ 2 3 Therefore setting h(z) = (℘ (z)) − 4℘(z) + g2℘(z) + g3 we see that h has no poles in 0 ∈ ∏ since all terms with zn for n < 0 cancel out, and h(0) = 0. As ′ ∏ ∏ ℘ and ℘ are holomorphic in r{0}, h is holomorphic in and Λ-periodic. Therefore h is constant and h(0) = 0 implies that h ≡ 0 as we wanted to show. 2.2. ELLIPTIC CURVES OVER C 31

Definition. For a lattice Λ ⊂ C, define its discriminant as 3 2 ∆(Λ) = g2(Λ) − 27g3(Λ) Remark 2.25. Recall that for a cubic polynomial of the form P (x) = 4x3 − 3 2 2 2 2 g2x − g3 its discriminant is ∆P = g2 − 27g3 = 16(e1 − e2) (e1 − e3) (e2 − e3) where e1, e2, e3 are the roots of P (x). It follows that for any lattice Λ in C,

∆P = ∆(Λ) ̸= 0 Proof. This is equivalent to prove that the roots of the polynomial are distinct. ′ w1 w2 w3 Since ℘ is even, ℘ is odd. Now let v ∈ { 2 , 2 , 2 } where Λ = Zw1 + Zw2 and ′ w3 = w1 +w2; then by Lemma 2.23 ℘ (v) = 0 and they are simple zeroes. Hence by the functional equation

′ 2 3 0 = ℘ (v) = 4℘(v) − g2℘(v) − g3

wi Therefore ℘( 2 ) is a root of P (x). We are reduced to show that ℘(wi) ̸= ℘(wj) for i ̸= j so that we have 3 distinct roots and nonzero discriminant. wi wi Consider the function ℘(z + 2 ) − ℘( 2 ) which is even since ℘ is and it is 0 at z = 0. Therefore it has a zero of order at least 2 at z = 0, but since it has degree 2, it has order exactly 2 at z = 0 and no other zeroes. Thus in particular wi+wj wj wi for z = 2 with i ̸= j, we obtain 0 ̸= ℘( 2 ) − ℘( 2 ) and this concludes the proof.

2 2 3 2 3 Therefore E = {[x, y, z] ∈ P | y z = 4x − g2(Λ)xz − g3(Λ)z } is a nonsingular projective curve for any lattice Λ. Moreover it makes sense to define the quantity g (Λ)3 j(Λ) = 1728 2 ∆(Λ) for any lattice Λ in C, called j-invariant.

Theorem 2.26. Let π : C −→ T = C/Λ be the canonical surjection on the quotient. Consider Φ : T −→ P2 given by { [℘(z), ℘′(z), 1] if z∈ / Λ t = π(z) ↦→ [0, 1, 0] if z ∈ Λ

Then Φ is holomorphic (as a map of complex manifolds) with image

def 2 2 3 2 3 Φ(T ) = E = {[x, y, z] ∈ P | y z = 4x − g2xz − g3z } Moreover, Φ is injective and has maximal rank (one) on T . Thus since T is compact, Φ : T −→ E is a biholomorphism, hence T =∼ E.

Proof. Φ is holomorphic on T r {0}: in fact let t ∈ T , t ̸= 0, then Φ(t) ∈ U2 = 2 2 {[x0, x1, x2] ∈ P | x2 ̸= 0} ⊂ P . We must check that for V ⊆ T r {0}, the map −1 −1 ˜ −1 F := ϕ2 ◦ Φ ◦ (πV ) is holomorphic on V = πV ⊆ C, where ϕ2[x0, x1, x2] ↦→ ( x0 , x1 ) ∈ 2. We have x2 x2 C ′ ′ F (z) = ϕ2(Φ(πV (z))) = ϕ2([℘(z), ℘ (z), 1]) = (℘(z), ℘ (z)) 2.2. ELLIPTIC CURVES OVER C 32 which are holomorphic in T r {0}. Moreover we can immidiatly see that Φ has maximal rank in T r {0}: in fact this is the case if and only if for all z ∈ C r Λ, (℘′(z)) (0) J (F ) = ̸= C ℘′′(z) 0

′ wi Suppose not, then ℘ (z) = 0, thus z = 2 for i = 1, 2, 3, but we saw that these ′′( wi ) are simple zeroes, therefore ℘ 2 ̸= 0. Let us now show that Φ is holomorphic at t = 0: we have Φ(0) = [0, 1, 0] ∈ U1. Let V˜ be a neighbourhood of 0 ∈ C and let z ̸= 0 in V˜ , then ℘(z) 1 F (z) = (ϕ ◦ Φ ◦ π)(z) = ( , ) 0 1 ℘′(z) ℘′(z)

(We can divide by ℘′(z) since we are near 0). Recall that 1 2 ℘(z) = (1 + a z + ...) ℘′(z) = − (1 + b z + ..) z2 1 z3 1 z z3 =⇒ F (z) = ( − (1 + c z + ...), − (1 + d z + ..)) 0 2 1 2 1

Therefore F0(z) is holomorphic at 0. Using the same expasion near 0 we see that

(( ′)′ ) ( 1 ) ( 1 ) ( ) ℘/℘ (z) − 2 + z(...) − 2 0 J (F ) = ′ = = ̸= C 0 (1/℘′) (z) − 3 z2 + z(...) 0 0 2 |z=0 |z=0 So we can conclude that Φ has maximal rank everywhere on T . To see that Φ(T ) = E , let P = [x, y, z] ∈ E: if z = 0 then x = 0, thus P = [0, 1, 0] = Φ(0) ∈ Φ(T ). Thus assume z ̸= 0 and write P = [x, y, 1], since ℘ is surjective, letx ˜ ∈ C be such that x = ℘(˜x). As P ∈ E we get

2 3 3 ′ 2 y = 4x − g2x − g3 = 4℘(˜x) − g2℘(˜x) − g3 = (℘ (˜x))

Hence y = ℘′(˜x) or y = −℘′(˜x). For the former case we have P = Φ(π(˜x)), for the latter notice that y = −℘′(˜x) = ℘′(−x˜) and x = ℘(˜x) = ℘(−x˜), therefore P = Φ(π(−x˜)). The other inclusion Φ(T ) ⊂ E is trivial by Theorem 2.27. ∏ We are left to prove that Φ is injective: let z1, z2 ∈ be such that Φ(π(z1)) = Φ(π(z2)). If z1 = 0 then Φ(π(0)) = [0, 1, 0] = Φ(π(z2)), which yields that z2 = 0. If z , z ̸= 0, then 1 2 { ℘(z1) = ℘(z2) (i) ′ ′ ℘ (z1) = ℘ (z2) (ii)

(i) implies that either z1 = z2 ( and we are done) or z1 = −z2 + w for some w ∈ Λ. Then ′ ′ ′ ′ ℘ (z2) = ℘ (z1) = ℘ (−z2 + w) = −℘ (z2)

′ wi wj hence ℘ (z2) = 0 so that z2 = 2 and z1 = 2 . Since ℘(z1) = ℘(z2), i = j and z1 = z2 as we wanted. 2.2. ELLIPTIC CURVES OVER C 33

2 3 Notice that the cubic equation y = 4x − g2x − g3 is in the form of (2.1.3) therefore E defined by such cubic is an elliptic curve over C. The Theorem shows how to associate to a lattice Λ in C an elliptic curve and implements this association by the biholomorphism Φ. However also the converse is true, for any elliptic curve E there exists a lattice Λ such that E is realized as C/Λ.

2.2.2 Isogenies over C Let E and E′ be elliptic curves isomorphic to C/Λ = T and C/Λ′ = T ′ respectively, for Λ, Λ′ lattices in C. Then every isogeny of E into E′ corresponds to a holomorphic homomorphism of T onto T ′, and viceversa. Proposition 2.27. With the same notation as above, we have that

′ ′ ′ Hom(E,E ) =∼ Hom(T,T ) = {α ∈ C | αΛ ⊂ Λ } Proof. Assume that f is non-constant and consider the following diagram:

C C

π1 π2 ↓ f ↓ T → T ′ Now by the Hurwitz theorem, f has no ramification points, therefore it is a ′ covering of T (as well as π2) and π1 is a covering of T . Since we have two ′ universal covering of T given by π2 and f ◦ π1, there exists F : C −→ C isomorphism of covering such that π2 ◦ F = f ◦ π1, and in particular F is ′ continuous. Fixing a point O = π1(0) ∈ T , we have that f(O) = O ∈ T −1 and ∀b ∈ π2 (O) there exists a unique F isomorphism of coverings such that F (0) = b. In particular since we require that F is a group homomorphism, there is a unique F so that F (0) = 0. Let us now show that F (z) = αz for some α ∈ C such that αΛ ⊂ Λ′: the fact that F passes to quotients means that for all w ∈ Λ there exists w′ ∈ Λ′ such that F (z +w) = F (z)+w′. Now notice that w′ = F (z +w)−F (z) is indipendent from z since the map F (∗ + w) − F (∗) is continuous, C is connected and Λ′ is discrete. Hence taking derivatives yields the equality F ′(z + w) = F ′(z) , or in other words F ′ is an elliptic function of period Λ. Therefore it defines a map ′ g : T −→ C holomorphic such that F = g◦π1. On the other hand T is compact, hence g is constant, and so F ′ is : say F ′(z) = α ̸= 0 ∀z ∈ C and consequently F (z) = αz since F (0) = 0. In particular for any w ∈ Λ, F (z + w) = F (z) + w′ implies aw = w′ ∈ Λ′ hence αΛ ⊂ Λ′. If f is constant, then F is constant, and in particular it is 0 (since F (0) = 0). Viceversa, let T = C/Λ,T ′ = C/Λ′ be complex tori and F (z) = αz with αΛ ⊂ Λ′. Then there exists f : T −→ T ′ holomorphic homomorphism, in fact such F passes to the quotient since π2 ◦ F (Λ) = π2(αΛ) = 0 = π1(Λ). Therefore there exists f continuous such that f ◦ π1 = π2 ◦ F . Finally f is holomorphic since −1 locally f|U = π2| ◦ F ◦ π1 is composition of holomorphic maps. V ′ |V In particular we have that

End(E) =∼ {α ∈ C | αΛ ⊂ Λ} 2.2. ELLIPTIC CURVES OVER C 34 and this always contains Z, since multiplication by integers maps Λ into Λ. Definition. We say that an elliptic curve E has complex multiplication if End(E) %Z. ′ ′ ′ Proposition 2.28. Let Λ = Zw1 + Zw2 and Λ = Zw1 + Zw2 be normalized lattices in C. Then E =∼ T = C/Λ and E′ =∼ T ′ = C/Λ′ are isogenous (resp. + isomorphic) if and only if there exists an element γ ∈ GL2 (Q) ∩ M2(Z) with + ′ GL2 (Q) = {γ ∈ GL2(Q) | det(γ) > 0} (resp. SL2(Z)) such that γ · τ = τ, ′ ′ ′ where τ = w1/w2, τ = w /w and the action of γ on H is by fractional linear ( )1 2 a b ′ transformation i.e. · τ ′ = aτ +b c d cτ ′+d

Proof. If 0 ̸= α ∈ Hom(E,E′) then αΛ ⊂ Λ′, therefore there exist a, b, c, d ∈ Z such that

′ ′ αw1 = aw1 + bw2 ′ ′ αw2 = cw1 + dw2

(a b) Thus we obtain γ = ∈ M ( ) ∩ GL ( ) such that τ = γ · τ ′. Moreover c d 2 Z 2 Q det(γ) > 0 since both bases are normalized. (a b) Conversely, if γ · τ ′ = τ for γ = ∈ M ∩ GL+( ) then for λ = cτ ′ + d c d 2 2 Q we have

( ) ( )( ′) ( ) ( )( ′ ) τ a b τ ′ w1 a b w1 λ = =⇒ (λw2/w2) = ′ 1 c d 1 w2 c d w2

′ ′ w2 ′ Hence αΛ ⊂ Λ for α = λ . In particular αΛ = Λ if and only if γ ∈ SL2( ). w2 Z

Proposition 2.29. Let Λ = Zw1 + Zw2 with τ = w1/w2 ∈ H. Then C/Λ has complex multiplication if and only if there exists a nonscalar element ξ ∈ + GL2 (Q) such that ξ · τ = τ. Proof. Let α ∈ End(E), then since αΛ ⊂ Λ, there exist a, b, c, d ∈ Z such that { ( ) ( )( ) ( ) ατ = aτ + b τ a b τ τ + =⇒ α = =: ξ for ξ ∈ M2(Z)∩GL2 (Q) α = cτ + d 1 c d 1 1 (2.2.3) Now notice that α is an eigenvalue for ξ for how ξ is defined, on the other hand α is an eigenvalue for ξ since

(τ) (τ) (τ τ)(α 0) (τ τ) α = ξ thus, in a compact way, we have = ξ 1 1 1 1 0 α 1 1

Therefore α ∈ Z ⇐⇒ b = 0 = c which yields a = d ⇐⇒ ξ is scalar.

Proposition 2.30. Let Λ = Zw1 + Zw2 with τ = w1/w2 ∈ H. Then C/Λ has complex multiplication if and only if Q(τ) is a quadratic imaginary field and End(E) is an order in Q(τ). 2.2. ELLIPTIC CURVES OVER C 35

Proof. C/Λ has complex multiplication if and only if α as in Proposition 2.32 is not in Z hence it is not real (since τ ∈ H) and ξ is not scalar. Therefore eliminating τ from (2.2.2) we obtain the quadratic equation α2 − (a + d)α + ad − bc = 0 (∗)

Since α = cτ + d, it follows that Q(τ) = Q(α) is a quadratic imaginary field. Moreover, (∗) tells us that α is integral over Z, thus contained in the ring of integers OQ(τ). Therefore End(E) ⊂ OQ(τ) and it is free of rank 2 because OQ(τ) is torsion free and End(E) contains 1 and an imaginary element: we conclude that End(E) is an order in Q(τ). From now on suppose that the elliptic curve E =∼ T = C/Λ has complex multiplication by the maximal order OK , for a quadratic imaginary field K and that Λ is of the form Zτ + Z. Under these assumptions, the condition End(E) = OK is equivalent to saying that OK Λ ⊂ Λ. Therefore Λ is a fractional OK -ideal. × Fact 2.31. If I is a fractional OK -ideal then there exists λ ∈ K such that λI ⊂ OK and therefore it is an OK -ideal.

In particular, we have m ∈ Z>0 such that mΛ ⊂ OK and consequently mΛ is a nonzero ideal of OK . Conversely, if a is a fractional OK -ideal, then the torus C/a satisfies End(C/a) = OK . In fact OK a ⊂ a, thus End(C/a) ⊃ OK and equality follows by maximality of OK in K. Recall that the class group of K is (multiplicative group of fractional ideals) Pic(O ) = = K (subgroup of principal fractional ideals) (multiplicative semigroup of nonzero ideals of O ) = K ∼ where I1 ∼ I2 if and only if there exist α1, α2 ∈ OK nonzero, such that α1I1 = α2I2. Therefore it follows that: Proposition 2.32. For a complex torus T , denote by [T ] its isomorphism class in the category of complex tori. After fixing one of the two possible embeddings ι : OK −→ C, the map

Pic(OK ) −→ {[T ] | End(T ) = OK } defined by [ι(a)] ↦→ [C/ι(a)] is a bijection.

Corollary 2.33. {[T ] | End(T ) = OK } is finite since Pic(OK ) is.

2.2.3 Automorphisms of an elliptic curve Let Aut(E) denote the group of all automorphisms of a given elliptic curve E defined over C. If E has no complex multiplication, Aut(E) = End(E)× = Z× = {±1}. ∼ Therefore assume that E has complex multiplication and let End(E) = O ⊆ OK × × be an order in an imaginary quadratic field K, so that Aut(E) = O ⊆ OK . × n Recall that OK = µ ∩ K where µ = {z ∈ C | z = 1 for some n ∈ N}. 2.3. ELLIPTIC CURVES OVER Q 36

× Proposition 2.34. OK = {±1} except for the following cases: × ∼ (a) if K = Q(i) then OK = {±1, ±i} = Z/4Z; 2πi/3 × 2 (b) if K = Q(ρ) with ρ = e , then OK = {±1, ±ρ, ±ρ }. × Proof. u ∈ OK if and only if u ∈ OK and Norm(u) = |u| = 1. Recall that for a quadratic imaginary field K, Gal(K/Q) = ⟨c√⟩ where c is the restriction of complex conjugation to K. Since u ∈ O , u = a+ db for a, b ∈ or a, b ∈ 1 + K √ Z 2 Z and d is a negative square-free integer such that K = Q( d). Now one has that √ √ √ Norm(a + db) = (a + db)(a − db) = a2 + |d|b2

2 |d| Therefore for |d| > 5, Norm(u) ≥ |d|b ≥ 4 > 1, for all b ̸= 0. It follows that × 2 u ∈ OK if and only if a = 1, b = 0 i.e. u = ±1. We are left with 4 cases:

2 2 • d = −1 ≡ 3 mod 4 =⇒ OK = Zi + Z. Hence Norm(u) = a + b = 1 if and only if a2 = 1, b = 0 or a = 0, b2 = 1, if and only if u = ±1 or u = ±i. √ 2 2 • d = −2 ≡ 2 mod 4 =⇒ OK = Zi 2+Z and then Norm(u) = a +2b = 1 if and only if a2 = 1, b = 0 if and only if u = ±1. √ 1+ 3 2 2 • d = −3 ≡ 1 mod 4 =⇒ OK = Z 2 + Z and Norm(u) = a + 3b = 1 with a, b ∈ 1 if and only if a2 = 1, b = 0 or a2 = 1 = b2 if and only if 2 Z √ 4 1 3 u = ±1 or u = ± 2 ± 2 . • d = −4 is not square-free.

In these 2 cases, O is the maximal order of K and the class number is 1, therefore by Proposition 2.35, there is exactly one elliptic curve up to isomor- ∼ phism over C such that End(E) = OK . 2 3 Now notice that the elliptic curve E : y = 4x − g3 has the 4 automorphisms

(x, y) ↦→ (x, ±y), (x, y) ↦→ (−x, ±iy)

2 3 And the elliptic curve E : y = 4x − g2x has the 6 automorphisms

(x, y) ↦→ (ρjx, ±y) for j = 0, 1, 2

2.3 Elliptic curves over Q

Let E be an elliptic curve defined over Q and we may assume that its Weierstrass equation has integer coefficients: in fact the change of variable 2 ′ 3 ′ ′ i (x, y) = (u x , u y ) gives a Weierstrass equation with coefficients ai = ai/u . Therefore a suitable choice of u makes this an integral equation. Then its discriminant ∆ will be an integer as well and which p-adic norm satisfies

|∆|p ≤ 1 with equality if and only if p - ∆ 2.3. ELLIPTIC CURVES OVER Q 37

Definition. A Weierstrass equation (2.1.2) is minimal for the prime p if for any admissible change of variable over Q, such that the new coefficients are p-integral, the power of p dividing ∆ cannot be decreased or, equivalently |∆|p cannot be increased. A Weierstrass equation (2.1.2) is a global minimal Weierstrass equation if it is minimal for all primes and its coefficients are integers. From now on we will assume that the elliptic curve E is given by a global minimal Weierstrass equation. This is in fact not a restriction, as one sees by the following results: Proposition 2.35. Let p be a fixed prime and let E be an elliptic curve over Q, then (a) There exists an admissible change of variable over Q such that the resulting equation is minimal for p. (b) If E has p-integral coefficients, then the change of variable in (a) issuch that u, r, s, t are p-integral. (c) Two equations that are minimal for p are related by an admissible change of varibale such that |u|p = 1 and r, s, t are p-integral. Proof. See [Kna93] Proposition 10.2

Theorem 2.36. (N´eron) Let E be an elliptic curve over Q, then there exists an admissible change of variable over Q such that the resulting equation is a global minimal Weierstrass equation. Moreover any two such global minimal Weierstrass equations are related by a change of variable with u = ±1 and r, s, t ∈ Z. Proof. See [Kna93] Theorem 10.3

2.3.1 L-function associated to an elliptic curve

Let E be an elliptic curve defined over Q and assume that E is given by a global minimal Weierstrass equation. Now for each prime p, write Ep for the reduction of E modulo p: such curve is defined over Fp and it is singular if and only if p | ∆ (For a complete discussion of singular Weierstrass equation see [Kna93] III.5 or [Sil09] III.1,III.2). For both singular and nonsingular cases define ap = p + 1 − #Ep(Fp) (2.3.1) Definition. The L-funtion associated to an elliptic curve E is

∏ ( −s)−1 ∏ ( −s 1−2s)−1 L(s, E) = 1 − app · 1 − app + p (2.3.2) p|∆ p-∆ Before giving a convergence result regarding L(s, E), we first need to prove the Hasse bound for elliptic curves over finite fields. 2.3. ELLIPTIC CURVES OVER Q 38

2.3.2 Hasse theorem r Notice that if E is defined over Fq with q = p , we have the obvious bound on 1 ≤ #E(Fp) ≤ 2q +1 since for any value of x there can be at most two values of y, and the point at ∞. In particular for q = p, it follows that |ap| ≤ p.

Definition. Let A be an abelian group, a map d : A −→ R is a quadratic form if:

(i) d(a) = d(−a) ∀a ∈ A;

(ii) B : A × A −→ R given by (a, b) ↦→ d(a + b) − d(a) − d(b) is bilinear. If additionally d(a) ≥ 0 ∀a ∈ A and d(a) = 0 ⇐⇒ a = 0 then d is said to be positive definite. ∼ Example 2.37. If E = C/Λ is an elliptic curve over K0 then deg : End(E) −→ Z is a positive definite quadratic form. In fact for α ∈ End(E), deg(α) = N(α) = α · α (see [Cox97 Theorem 10.14]) and therefore it is immidiate that N(α) = N(−α) and N(α) = 0 if and only if α = 0. Since B is symmetric it is enough to check bilinearity on the first variable:

B(aα + bβ, γ) = (aα + bβ + γ)(aα + bβ + γ) − (aα + bβ)(aα + bβ) − γγ =

= (aα + bβ)γ + γ(aα + bβ) = aαγ + aαγ + bβγ + bβγ = = (aα + γ)(aα + γ) − a2αα − γγ + (bβ + γ)(bβ + γ) − b2ββ − γγ =

= B(aα, γ) + B(bβ, γ) ∀a, b ∈ R, α, β, γ ∈ End(E)

Lemma 2.38. Let d : A −→ Z be a positive definite quadratic form, then |d(a − b) − d(a) − d(b)| ≤ 2√d(a)d(b) (2.3.3)

Proof. Let B : A × A −→ Z given by (a, b) ↦→ d(a − b) − d(a) − d(b): such B is bilinear. Notice that for all m, n ∈ Z, we have d(ma) = m2d(a) and

0 ≤ d(ma − nb) = m2d(a) + mnB(a, b) + n2d(b)

Setting m = −B(a, b) and n = 2d(a) to obtain

0 ≤ d(a)(4d(a)d(b) − B(a, b)2)

If d(a) = 0, then a = 0 and the both sides of (2.3.3) are zero and there is nothing to prove. Otherwise d(a) > 0 and dividing we obtain

4d(a)d(b) ≥ B(a, b)2 =⇒ |d(a − b) − d(a) − d(b)| ≤ 2√d(a)d(b)

r Theorem 2.39. (Hasse) Let E be an elliptic curve defined over Fq, q = p , then √ √ q + 1 − 2 q ≤ #E(Fq) ≤ q + 1 + 2 q (2.3.4) 2.3. ELLIPTIC CURVES OVER Q 39

Proof. Let φ = φq be the q-th Frobenius map and recall that Fq is the fixed field of φq, so that E(Fq) = ker(1 − φ) and by Lemma 2.13 follows that 1 − φq is a separable isogeny. Therefore #ker(1 − φ) = deg(1 − φ) and by the previous lemma √ √ |deg(1 − φ) − deg(1) − deg(φ)| = |#E(Fq) − 1 − q| ≤ 2 deg(1)deg(φ) = 2 q

Corollary 2.40. For an elliptic curve E defined over Q, the Euler product 3 L(s, E) as in (2.3.2) converges for Re(s) > 2 and is given there by an absolutely convergent Dirichlet series. Proof. For a proof that L(s, E) is given by a Dirichlet series see [Kna93] Corol- lary 10.6. ∏ ( −s We only need to prove convergence for the infinite product 1 − app + p-∆ −1 p1−2s) since L(s, E) is obtain by multiplicating this factor with a finite product. L(s, E) converges absolutely if and only is

∑ −s 1−2s | − app + p | ≤ ∞ p-∆ √ By Theorem 2.39 with q = p we get |ap| ≤ 2 p, thus

∑ −s 1−2s ∑ −s+ 1 1−2s |app + p | ≤ 2|p 2 | + p | ≤ p-∆ p-∆

∑ Re(−s+ 1 )( Re(−s+ 1 )) ∑ Re(−s+ 1 ) ∑ Re(−s+ 1 ) ≤ p 2 2 + p 2 ≤ p 2 ≤ n 2

p-∆ p n 3 Which converges for Re(s) ≥ 2 . Chapter 3

Modular Forms

3.1 Modular forms for SL2(Z) 3.1.1 Functions of lattices As in 2.2 denote by L the set of all lattices and by B the set of normalized basis.

Definition. A map F : L −→ P1(C) = C ∪ {∞} is a function of lattices of weight k ∈ Z if F (αΛ) = α−kF (Λ) for all α ∈ C and all Λ ∈ L. 1 Example 3.1. For k ≥ 3, consider the Eisestein series Ek : L −→ P (C) defined ∑ 1 as Λ ↦→ wk . One can prove that such series is absolute convergent, so 0̸=w∈Λ that Ek is a function of lattices of weight k. Note that Ek ≡ 0 for all k ≥ 3 odd. Let F : L −→ P1(C) be a function of lattices of weight k. Then we can view F as ˜ 1 F = F ◦ π : SL2(Z)\B −→ P (C) [τ] Then the map σ : H −→ SL ( )\B such that τ ↦→ yields, by further 2 Z 1 composing 1 f = F ◦ π ◦ σ : H −→ P (C) × Since the map H −→ SL2(Z)\B/C is surjective, f determines F . As F has weight k, we have [ ] [ ] ˜( w1 ) ˜( w1/w2 ) −k (w1 ) F = F w2 = w2 f w2 1 w2

We now wonder which functions f : H −→ P1(C) arise from F : L −→ P1(C). (a b) This means that f factors as f = σ◦F , or equivalently such that ∀γ = ∈ c d SL2(Z) (aτ + b) ( [ aτ+b ] ) f(γτ) = f = F cτ+d = cτ + d 1 ( [aτ + b] ) ( [τ] ) (cτ + d)kF = (cτ + d)kF = (cτ + d)kf(τ) cτ + d 1

40 3.1. MODULAR FORMS FOR SL2(Z) 41

Proposition 3.2. There is a one-to-one correspondence between ⎧ (aτ + b) ⎫ ⎪ f = (cτ + d)kf(τ)⎪ ⎨⎪ cτ + d ⎬⎪ { F : L −→ 1( ) of weight k } ←→ f : H −→ 1( ) P C P C (a b) ⎪ ∀ γ = ∈ SL ( )⎪ ⎩ c d 2 Z ⎭

Definition. We say that f : H −→ P1(C) is a weakly modular function of weight k ∈ Z for SL2(Z) if (a b) (i) f(γτ) = (cτ + d)kf(τ) for all γ = ∈ SL ( ) and all τ ∈ H; c d 2 Z

(ii) f is meromorphic on H. Remark 3.3. (a) Since f is meromorphic, f −1(∞) is a discrete subset of H. (−1 0 ) (b) If k is odd, then f ≡ 0 as one sees by taking γ = ∈ SL ( ) 0 −1 2 Z which acts on H as the identity, and by (1) we get f(τ) = −f(τ), thus f(τ) = 0.

(c) If f ̸= 0 is a weakly modular function of weight k, then 1/f is a weakly modular function of weight −k. To get to the definition of modular functions and modular forms we will require a growth condition at infinity. 2πiτ Consider the exponential function H −→ C defined by τ ↦→ qτ := e . 2πix −2πy If we write τ = x + iy ∈ H for x ∈ R, y ∈ R>0, then qτ = e e and in particular e2πix = cos(2πx) + isin(2πx) ∈ S1 = {z ∈ C : |z| = 1}. The mapping x ↦→ e2πix induces a real analytic isomorphism R/Z −→ S1 ; on the other hand the strictly decreasing real analytic map y ↦→ e−2πy yields a real analytic isomorphism R>0 −→]0, 1[. It follows that τ ↦→ qτ gives a holomorphic isomorphism

∗ H/Z −→ {z ∈ C | 0 < |z| < 1} =: D

where the action of Z on H is the translation (n, τ) ↦→ τ + n, ∀n ∈ Z, τ ∈ H. Now if f : H −→ P1(C) is a weakly modular function of weight k, it induces a map f : H/Z −→ P1(C) since the translation τ ↦→ τ + n, for any n ∈ Z, is (1 n) described by the action of γ = ∈ SL ( ) and f(τ +n) = f(γτ) = f(τ). 0 1 2 Z ∗ ∗ 1 ∗ Therefore f induces a function f : D −→ P (C) by f (qτ ) = f(τ) and the −2πy −2πIm(τ) relation |qτ | = e = e shows that q → 0 as Im(τ) → ∞. Consequently we say that f is meromorphic (resp. holomorphic) at ∞ is f ∗ is meromorphic (resp. holomorphic) at the center of D∗.

Definition. (i)A modular function of weight k ∈ Z for SL2(Z) is a weakly modular function of weight k for SL2(Z), which is meromorphic on H and at infinity.

(ii)A modular form of weight k ∈ Z for SL2(Z) is a weakly modular function of weight k for SL2(Z), which is holomorphic on H and at infinity. 3.1. MODULAR FORMS FOR SL2(Z) 42

(iii)A cusp form of weight k ∈ Z for SL2(Z) is a modular form of weight k for SL2(Z), that vanishes at infinity. Remark 3.4. (a) Note that the set of modular functions (resp. modular forms, resp. cusp forms) is equipped with a natural structure of C-vector space; (b) Modular functions of weight 0 form a field;

(c) If f is meromorphic at infinity, then in a neighbourhood of qτ = 0 (or equivalently for Im(τ) ≫ 0 ) we have a Laurant series expansion

∞ ∞ ∗ ∑ n ∑ 2πiτ f(τ) = f (qτ ) = anqτ = ane with an = an(f) ∈ C n=−N n=−N (3.1.1) This Fourier expansion is called the q-expansion of f. In particular, if f is a modular function, that it is also meromorphic on H, then its q-expansion converges on H.

3.1.2 The action of SL2(Z) on H

We now want to determine a set of representatives for the quotient SL2(Z)\H with respect the left action of SL2(Z) on H given by fractional linear transformation . Define Γ = SL2(Z)/{±1}: since −1 ∈ SL2(Z) acts trivially on H, then Γ acts on H. In particular the action of Γ on H is faithful, which means that γτ = τ ∀τ ∈ H implies γ = 1 ∈ Γ (0 −1) Let S ∈ Γ be an element represented by Sˆ = ∈ SL ( ), then Sτ = 1 0 2 Z (1 1) − 1 ; and let T ∈ Γ be represented by Tˆ = ∈ SL ( ), then T τ = τ + 1. τ 0 1 2 Z Remark that S2 = 1 and (ST )3 = 1 = (TS)3 in Γ; moreover since T nτ = τ + n for all n ∈ Z, the subgroup of Γ generated by T is infinite. Also define define D = {τ ∈ H | |τ| ≥ 1, |Re(τ)| ≤ 1/2}, then:

ρ i −ρ

−1 1 0 1 1 − 2 2 3.1. MODULAR FORMS FOR SL2(Z) 43

Theorem 3.5. (1) For all τ ∈ H there exists γ ∈ Γ such that γτ ∈ D.

′ ′ 1 (2) If τ ̸= τ ∈ D and γτ = τ for some γ ∈ Γ then either Re(τ) = ± 2 (and then τ ′ = τ ∓ 1) or |τ| = 1 (and then τ ′ = Sτ).

(3) For τ ∈ D, StabΓ(τ) = {γ ∈ Γ | γτ = τ} is trivial except for the following three cases: ∼ (a) τ = i, then StabΓ(i) = ⟨S⟩ = Z/2Z; 2πi/3 ∼ (b) τ = ρ = e , then StabΓ(ρ) = ⟨ST ⟩ = Z/3Z; 2πi/6 ∼ (c) τ = −ρ = e , then StabΓ(−ρ) = ⟨TS⟩ = Z/3Z Proof. (1): Define G = ⟨S, T ⟩ ⊆ Γ. We will prove that for any τ ∈ H there exists g ∈ G such that gτ ∈ D. (a b) If g ∈ G is represented byg ˆ = ∈ SL ( ) then Im(gτ) = Im(τ) . Hence c d 2 Z |cτ+d|2 there exists g ∈ G such that Im(gτ) is maximal since ∀k, the set {(c, d) ∈ Z2 : |cτ + d|2 < k2} is finite. n 1 Fix g ∈ G for which Im(gτ) is maximal, then ∃n ∈ Z such that |Re(T gτ)| ≤ 2 . We claim that T ngτ ∈ D, or in other words that |T ngτ| ≥ 1. Suppose it is not, then |T ngτ| < 1, therefore Im(T ngτ) Im(ST ngτ) = > Im(T ngτ) = Im(gτ) |T ngτ| which contradicts the maximality property with which we selected g. (2) & (3): If τ ′ = γτ with τ, τ ′ ∈ D, γ ∈ Γ, we can assume that Im(τ ′ = Im(γτ) ≥ Im(τ) i.e. |cτ + d| ≤ 1 (∗). It follows that Im(cτ + d) = |c|Im(τ) ≤ |cτ + d| ≤ 1. Since τ ∈ D, Im(τ) ≥ √ √ 3 3 Im(ρ) = 2 , hence |c| 2 ≤ 1 which forces c ∈ {0, ±1}. Also remark that the case c = −1 reduces to the case c = 1 by replacingγ ˆ with −γˆ. (±1 b ) • c = 0 yieldsγ ˆ = which means that γ is a translation. Since 0 ±1 1 |Re(τ)|, |Re(γτ| ≤ 2 the only possibilities are b = 0 if τ belongs to the 1 interior of D or b = ±1 and consequently |Re(τ)| = 2 , Re(γτ) = −Re(τ). • If c = 1, (∗) becomes |τ + d| ≤ 1. (a −1) If d = 0 then |τ| ≤ 1 hence |τ| = 1 and γ is represented byγ ˆ = 1 0 ′ 1 a hence τ = a − τ = T Sτ. If |Re(τ)| = 1/2 then a = 0 hence γ = S and Si = i, which also yields S ∈ StabΓ(i). If |Reτ| = 1/2 then τ = ρ, −ρ. For τ = ρ, γρ = −ρ hence a = 0 or a = −1, for τ = −ρ we have a = 0 or a = 1. If d ̸= 0 write τ = x+iy, then τ +d = (x+d)+iy and an easy computation from (∗) shows that d = ±1. For such values of d follows that τ = ρ if d = 1 and τ = −ρ for d = −1. The former yields that a − b = 1 and 1 γρ = a − 1+ρ = a + ρ hence a = 0 or a = 1. Similarly for the latter case we get a = 0 or a = −1. To conclude, it is enough to show that G = Γ: let τ belong to the interior of D and let γ ∈ Γ, then γτ ∈ H. The proof of part (1) shows that there exists g ∈ G = ⟨S, T ⟩ such that gγτ ∈ D. By parts (2) & (3), since τ is in the interior of −1 D, follows that gγτ = τ, hence gγ ∈ StabΓ(τ) = {1}, therefore γ = g ∈ G. 3.1. MODULAR FORMS FOR SL2(Z) 44

Definition. A subset A of H is a fundamental domain for Γ if : (i) A is open and connected;

(ii) The canonical map A −→ Γ\H is injective; (iii) The canonical map A −→ Γ\H is surjective. Corollary 3.6. The interior of D is a fundamental domain for Γ.

Notice that if A is a fundamental domain, then γA is also a fundamental domain for any γ ∈ Γ (Since γ is a holomorphic isomorphism of H). Remark 3.7. The theorem shows that Γ\H equipped with the quotient topology is homeomorphic to the complex plane A1(C). Corollary 3.8. There is a bijection between function of lattices of weight k ∈ Z and functions on H such that

f(T τ) = f(τ + 1) = f(τ) (3.1.2a) 1 f(Sτ) = f(− ) = τ kf(τ) (3.1.2b) τ Proposition 3.9. For τ ∈ H, k ≥ 3, the so-called Eisestain series of weight k

∑ 1 E (τ) = (3.1.3) k (mτ + n)k 2 (m,n)∈Z (m,n)̸=(0,0) defines a modular form of weight k. In particular ∀k ≥ 3 odd, Ek ≡ 0; for k ≥ 4 even, Ek does not vanish at infinity ∞ ∑ 1 and Ek(∞) = 2ζ(k), where ζ(s) = ns is the Riemann Zeta function. n=1 ∑ 1 Proof. For k ≥ 3 and for any lattice Λ ∈ L, the series Ek = wk converges 0̸=w∈Λ absolutely and Ek(τ) = Ek(Λτ ) for Λτ = Zτ + Z i.e. Ek is a function on H corresponding to Ek by the bijection in Proposition 3.2. In particular Ek ≡ 0 for all k odd and Ek satisfies the functional equation

k Ek(γτ) = (cτ + d) Ek(τ)

We are left to show that Ek is holomorphic on H and at infinity, and that it satisfies ∗ ∗ Ek(∞) = Ek (0) = lim Ek (q) = 2ζ(k) q→0 Assume at first that τ ∈ D, then

|mτ + n|2 = m2|τ|2 + 2mnRe(τ) + n2 ≥ m2 − mn + n2 = |mρ + n|2 where, as before, ρ = e2πi/3. Therefore

∑ 1 ∑ 1 ≤ = E (ρ) ≤ ∞ |mτ + n|2 |mρ + n|2 k (m,n)̸=(0,0) (m,n)̸=(0,0) 3.1. MODULAR FORMS FOR SL2(Z) 45

Hence for τ ∈ D we have absolute and uniform convergence on compact subsets, 1 therefore since (mτ+n)k is holomorphic on D, Ek(τ) is. Now write H = ⋃ γD and let τ ∈ γD, then γ−1τ ∈ D and therefore for γ∈Γ (a b) γ = c d

−1 −1 k −1 Ek(τ) = Ek(γγ τ) = (c(γ τ + d) Ek(γ τ)

The right hand side has both factors holomorphic, thus we conclude that Ek(τ) is holomorphic on each γD, hence on H. Due to uniform convergence we may compute lim E∗(q) term by term and q→0 k obtain: { 1 Im(τ)→∞ 0 if m ̸= 0 −−−−−−→ (mτ + n)k q→0 1 nk if m = 0 ∗ This calculation shows that Ek(τ) is holomorphic at infinity i.e. Ek (q) extends to the point q = 0 to a holomorphic function by defining E∗(0) = lim E∗(q) k q→0 k

3.1.3 Divisors of modular functions

Let f : H −→ P1(C) be a modular function of weight k ∈ Z, f ̸= 0. Since f is meromorphic, it makes sense to define for any τ ∈ H the order of f at τ (a b) ord (f) ∈ . Remark that for all γ = ∈ SL ( ) we have ord (f) = τ Z c d 2 Z γτ k ordτ (f) since f(γτ) = (cτ + d) f(τ) and τ ̸= −d/c since τ ∈ H while −d/c ∈ R. Definition. Let f be as above and let π : H −→ Γ\H be the canonical projection. Then define the divisor of f to be

∑ ( 1 ) div(f) = ordx(f) · x + (ord∞(f)) · ∞ = #StabΓ(x) x∈Γ\H (3.1.4) 1 1 ∑ = ord (f) · i + ord (f) · ρ + ord (f) · ∞ + ord (f) · x 2 i 3 ρ ∞ x x̸=i,ρ

def ∗ def where ord∞(f) = ord0(f ) and for x = π(τ), ordx(f) = ordτ (f). ∑ ( 1 ) Given a divisor div(f) = ordx(f) ·x+(ord∞(f))·∞, its degree #StabΓ(x) x∈Γ\H is 1 1 ∑ deg(div(f)) = ord (f) + ord (f) + ord (f) + ord (f) (3.1.5) 2 i 3 ρ ∞ x x̸=i,ρ

Theorem 3.10. Let f be a modular function of weight k ∈ Z, f ̸= 0. Then k deg(div(f)) = 12 Proof. First of all notice that te sum in (3.1.5) if finite. In fact since in a neighbourhood of ∞ there are no zeroes nor poles other than ∞ itself, we reduce to 3.1. MODULAR FORMS FOR SL2(Z) 46

consider singularities in a compact set of the form KM = D ∩ {τ ∈ H | Im(τ) ≤ M}: now since singularities form a discrete set and belong to a compact set, they are finite. We now want to deduce the theorem using the argument principle by perform- f ′ ing contour integration of dlog(f) = f dτ on the curve given by the boundary of KM . There is a problem on having singularities on such boundary, thus let us assume at first for simplicity that the only possible zeroes or poles onthe boundary are at ρ, −ρ, i and consider the contour integration given by

A E

C C′ B i D′ B′ D ρ −ρ

so that we avoid ρ, −ρ, i. Choose a parametrization for the curve C which is piecewise differentiable, makes it positively oriented and such that the winding number around any point in the interior is 1. The argument principle states that

1 ∫ f ′(τ) ∑ dτ = ord (f) 2πi f(τ) x C x∈Γ\H x̸=i,ρ

⌢ ⌢ ⌢ provided that the arcs BB′ , CC′ , DD′ and the segment EA are chosen such that all the singularities of Γ\H r {ρ, i} lie inside C. Since under these conditions ∑ ordx(f) x∈Γ\H x̸=i,ρ is constant, it must be equal to the quantity we obtain by taking the limits of the radius of those arcs going to zero (provided that the limits exist). Choosing symmetrical arcs, notice that the invariance of f under T : τ ↦→ τ + 1 implies ( ∫ B ∫ E ) (I) dlog(f) + dlog(f) = 0 A D′ ∫ A 2πiτ Consider E dlog(f) : the map q : τ ↦→ e transforms EA into a circle α around 0 of radius depending on M = Im(A) and which is negatively oriented. By change of variable we get

∫ A ∫ 1 1 ∗ ∗ (II) dlog(f) = f = −ord0(f ) = −ord∞(f) 2πi E 2πi α

∫ B′ Consider B dlog(f): we are integrating along an arc of radius ε > 0. Call Cε(ρ) the circle of radius ε centered at ρ negatively oriented. Then by the argument 3.1. MODULAR FORMS FOR SL2(Z) 47 principle 1 ∫ dlog(f) = −ordρ(f) 2πi Cε(ρ) ′ f ordρ(f) In a neighbourhood of ρ we have f = τ−ρ + (holom. part). Parametrize the ⌢ arc BB′ as t ↦→ ρ + ε−it for a ≤ t ≤ b, then

′ ′ ′ 1 ∫ B ord (f) ∫ B dτ 1 ∫ B dlog(f) = ρ + (holom)dτ = 2πi B 2πi B τ − ρ 2πi B ∫ b −it ∫ b ∫ b ordρ(f) −ie ordρ(f) = − −it dτ + (holom)dτ = − (b − a) + (holom)dτ 2πi a e a 2π a ′ ′ ∫ B As ε → 0, B → B and B (holom)dτ → 0. Moreover b − a → (angle of the arc), therefore

b − a 2π 1 ord (f) ord (f) −→ 6 = =⇒ − ρ (b − a) → − ρ (III) 2π 2π 6 2π 6 The same computation shows that

′ 1 ∫ D 1 1 (III) dlog(f) −→ ord−ρ(f) = − ordρ(f) 2πi D 6 6 Also similarly we obtain

′ 1 ∫ C 1 (III) dlog(f) −→ ordi(f) 2πi C 2 ( ∫ C ∫ D ) We are left to compute B′ dlog(f) + C′ dlog(f) .

1 ⌢′ ⌢′ Notice that S : τ ↦→ − τ takes B C to C D but with the opposite orientation ⌢′ ∫ D ∫ C (call it DC ). We relate C′ to B′ by the change of variable τ ↦→ Sτ. Now k df(Sτ) k−1 k ′ since f(Sτ) = τ f(τ) it follows that dτ = kτ f(τ) + τ f (τ). Therefore df(Sτ) k f ′τ = dτ + dτ which implies f(Sτ) τ f(τ)

1 ( ∫ C ∫ D ) 1 ∫ C (df(τ) df(Sτ)) (IV ) dlog(f) + dlog(f) = − = 2πi B′ C′ 2πi B′ f(τ) f(Sτ) 1 ∫ C k −k ∫ C dτ k π k = − dτ = −→ε→0 − ( − i) = 2πi B′ τ 2πi B′ τ 2πi 6 12 Combining together the results in (I), (II), (III) and (IV ) proves the theorem under our initial assumptions that the only singularities on C are on ρ, −ρ, i. 1 For the general situation, if τ0 is a zero or a pole in C then |Re(τ0)| = 2 and/or −1 |τ0| = 1, then also T τ0 (or T τ0) and/or Sτ0 is a singularity and we can take arcs around such that the two contributions cancel out: in other words one arc include the point in the interior of C while the arc for the other point leaves it outside. 3.1. MODULAR FORMS FOR SL2(Z) 48

A E C

τ0 T τ0

C C′ B i D′ B′ D ρ −ρ

We already defined the quatities g2 and g3 for lattices in Theorem 2.27, and by mean of the one-to-one correspondence given by Proposition 3.2, we have g2(τ) = 60E4(τ) and g3(τ) = 140E6(τ). Also define, in this new setting, the 3 2 modular discriminant as ∆(τ) = g2(τ) − 27g3(τ) and the modular j-invariant g2(τ) j(τ) = 1728 ∆(τ) . Proposition 3.11. (a) ∆(τ) is a cusp form of weight 12 which is nowhere vanishing on H and it has a simple zero at ∞. (b) j(τ) is a modular function of weight 0 with a simple pole at ∞.

Proof. Even though we could use the correspondence, we rather take a different approach.

(a) We already know that g2(τ) and g3(τ) are modular forms of weight 4 and 6 respectively, such that

∗ g2(∞) = g2 (0) = 60 · 2ζ(4) = 120ζ(4) ̸= 0 g3(∞) = 280ζ(6) ̸= 0

Moreover by Theorem 3.10 we have 4 1 div(deg(g )) = = =⇒ div(g ) = 1 · ρ 2 12 3 2 6 1 div(deg(g )) = = =⇒ div(g ) = 1 · i 3 12 2 3

where both implications follow by the fact that ordx(f) ∈ Z for all x ∈ H. π4 π6 4π4 Now since ζ(4) = 90 and ζ(6) = 945 it follows that g2(∞) = 3 and 8π6 g3(∞) = 27 . After these computation follows that ∆(τ) is a modular form of weight 12 3 2 since both g2(τ) and g3(τ) are modular forms of weight 12 and ∆ ̸= 0 on H since ∆(i), ∆(ρ) ̸= 0. Finally

( 4 3 8 2) ∆(∞) = π12 ( ) − 27( ) = 0 3 27 which means that ∆(τ) is a cusp form. By Theorem 3.10, deg(div(∆)) = 1, therefore we have a simple zero at infinity. 3.1. MODULAR FORMS FOR SL2(Z) 49

3 (b) j(τ) is a modular function of weight 0 since both ∆(τ) and g2(τ) are of weight 12. Moreover j(τ) is holomorphic on H since ∆(τ) ̸= 0 for all τ ∈ H. More precisely deg(div(j)) = 0, in fact from the computation above follows that div(j) = 1 · ρ − 1 · ∞.

Theorem 3.12. j induces a bijection j :Γ\H −→ C (denoted again by j). Proof. Since j has weight 0, it is Γ-invariant i.e. j(γτ) = j(τ) for all γ ∈ Γ, τ ∈ H. Hence it descends to a holomorphic map j :Γ\H −→ C. We must check that ∀λ ∈ C there exists a unique τ = τλ ∈ H such that j(τ) = C. 3 This is equivalent to saying that fλ(τ) = 1728g2(τ) − λ∆(τ) has a unique zero in Γ\H −→ C. Notice that such fλ(τ) is a modular form of weight 12, therefore 1 1 ∑ 1 = deg(div(f )) = ord (f ) + ord (f ) + ord (f ) + ord (f ) λ ∞ λ 2 i λ 3 ρ λ x λ x̸=i,ρ

Since ∆(∞) = 0 but g2(∞) ̸= 0, we have ord∞(fλ) = 0. Since fλ is a modular form, it is holomorphic on H, thus has no poles. Therefore we have to solve a b 1 = + + c for a, b, c ∈ 2 3 N

which yields (a, b, c) ∈ {(2, 0, 0), (0, 3, 0), (0, 0, 1)} and in each case fλ has only one zero as we wanted.

3.1.4 The space of modular forms

Let us denote by Mk (resp. Sk) the C-vector space of modular (resp. cusp) forms of weight k ∈ Z. ( ) Proposition 3.13. (a) dimC Mk/Sk ≤ 1 and equality holds if k ≥ 4.

(b) Mk = {0} for k < 0 and k = 2. (c) Multiplication by ∆ induces an isomorphism

·∆ Mk−12 −→ Sk

and Mk = Sk ⊕ CEk for any k ≥ 12.

(d) M0 = C, M4 = CE4, M6 = CE6, M8 = CE8 and M10 = CE10

Proof. (a) Since Sk is the kernel of the map Mk −→ C given by f ↦→ f(∞), ( ) we have dimC Mk/Sk ≤ 1. Now if k ≥ 4, then Ek ∈ Mk r Sk therefore Mk = S ⊕ CEk.

(b) If f ∈ Mk, f ̸= 0 then ordx(f) ≥ 0 for all x ∈ Γ\H∪{∞}. On the other hand k deg(div(f)) = 12 therefore there are no nonzero modular form of negative 1 weight, and for k = 2 we get deg(div(f)) = 6 but 1 a b = + + c + d has no solutions in 4. 6 2 3 N 3.1. MODULAR FORMS FOR SL2(Z) 50

(c) We know that div(∆) = 1 · ∞ and ∆(τ) ̸= 0 for all τ ∈ H. If f ∈ Sk then ord∞(f) ≥ 1 therefore f/∆ ∈ Mk−12. Since ord∞(∆) = 1, it follows that ord∞(f/∆) ≥ 0 . Therefore the C-linear maps Mk−12 −→ Sk, f ↦→ f · ∆ and Sk −→ Mk−12, f ↦→ f/∆ are mutually inverse. (d) We need to examine the cases k = 0, 4, 6, 8, 10. By (b) and (c), for these weights Sk = {0} (otherwise by (c) we would have nonzero modular forms of

negative weight, contradicting (b) ). On the other hand by (a), dimCMk ≤ 1 and just notice that 0 ̸= 1 ∈ M0 and for k ∈ {4, 6, 8, 10}, 0 ̸= Ek ∈ Mk.

def ∑ Consider the C-algebra structure on M∞ = Mk ⊂ {f : H −→ C | k≥0 f is holomorphic}. ⨁ Lemma 3.14. M∞ is actually a direct sum: M∞ = Mk. k≥0 ∑ Proof. Consider a relation of the form λifi = 0 for some finite set of indices i∈I I and fi ∈ Mi. We check that λi = 0 ∀i ∈ I: choose τ0 ∈ H such that fi(τ0) ̸= 0 ∑ ∑ for each i ∈ I. However λifi = 0 implies λifi(γτ0) = 0 ∀γ ∈ Γ. Let i∈I i∈I (a b) γ = then c d

∑ ∑ i 0 = λifi(γτ0) = λi(cτ0 + d) fi(τ0) i∈I i∈I

∑ i which means that the polynomial λifi(τ0)X has the (infinitely many) roots i∈I cτ0+d, thus it is the zero polynomial. Therefore λifi(τ0) = 0∀i ∈ I =⇒ λi = 0∀i since by our initial choice fi(τ0) ̸= 0 ∀i ∈ I.

Theorem 3.15. The map C[X,Y ] −→ M∞ given by P (X,Y ) ↦→ P (g2, g3) is an isomorphism of C-algebras. Proof. This map is clearly a C-algebra homomorphism. α β • Surjectivity: it is enough to check that the monomials in {g2 g3 | 4α+6β = k, α, β ≥ 0} generate Mk. We already know this for k = 0, 2, 4, 6 by Proposition 3.13. Therefore assume that k ≥ 8 and proceed by induction on k. α β Any k ≥ 8 even can be written as 4α + 6β with α, β ≥ 0; we have g2 g3 ∈ α β Mk and g2 g3 (∞) ̸= 0. Now let f belong to Mk, then there exists λ ∈ C α β α β such that (f −λg2 g3 )(∞) = 0, so that (f −λg2 g3 ) ∈ Sk and consequently α β by Proposition 3.13 (c), f − λg2 g3 = ∆ · g for some g ∈ Mk−12. By inductive hypothesis g = P (g2, g3) for some P (X,Y ) ∈ C[X,Y ] and ∆ = 3 2 g2 − 27g3, hence

3 2 α β ˜ f = P (g2, g3)(g2 − 27g3) + λg2 g3 =: P (g2, g3)

• Injectivity: let us prove that P (g2, g3) = 0 implies P (X,Y ) = 0 or equivalently that there are no non-trivial relations between g2 and g3. 3.1. MODULAR FORMS FOR SL2(Z) 51

By contradiction assume that there is a nontrivial algebraic relation among g2 and g3. We can assume that such relation is homogeneous in the sense α β that P (g2, g3) ∈ Mk for some k and P (X,Y ) is sum of monomials λX Y with 4α + 6β = k. Without loss of generality we can (and do) assume that k is minimal, then the relation is of the form

m n (I) g2 + g3P1(g2, g3) = 0 or (II) g3 + g2P2(g2, g3) = 0 1 1 Recall that div(g2) = 3 · ρ and div(g3) = 2 · i, then evaluating (I) at i and (II) at ρ we get a contradiction.

∼ Remark 3.16. (i) The isomorphism C[X,Y ] −→ M∞ is actually an isomorphism of graded C-algebras, where the grading in C[X,Y ] is defined by deg(XαY β) = 4α + 6β

·∆ ⨁ (ii) The isomorphism Mk−12 −→ Sk implies that S∞ = Sk is a principal k≥0 ideal of M∞ generated by ∆. We are now ready to describe all modular functions of weight 0: this amounts to characterize all meromorphic functions on Γ\H which are also meromorphic at infinity. Theorem 3.17. Let F be the set of a modular functions of weight 0. SL2(Z) Then F ∼ (j), or in other words f is a modular function of weight 0 if SL2(Z) = C and only if there exists P (t) ∈ C(t) such that f = P (j). Proof. If f = P (j) for some P (t) ∈ C(t) then f is a modular function of weight 0 since j is. Conversely, let f be a modular function of weight 0. Recall that j induces a bijection Γ\H −→ C and div(j) = 1 · ρ − 1 · ∞. Notice that for any τ0 ∈ H, j(τ) − j(τ0) is also a modular function of weight 0, with divisor div(j −j(τ0)) = 1·τ0 −1·∞. Therefore without loss of generality we can assume that f is holomorphic on H (otherwise multiply f by a function of the form ∏ ( )ki n j − j(τi) , with ki ≥ 0). Now by multiplying f for ∆ , n ≥ 0 we can also i n assume that g = f∆ is a modular form in M12n. Therefore by Theorem 3.15 we can write ∑ α β g = λαβg2 g3 α,β 4α+6β=12n α β Setting p = 3 and q = 2 , follows that p, q ≥ 0 are integers and p + q = n, hence g ∑ (g3)p(g2)q f = = λ 2 3 ∆n αβ ∆p · ∆q α,β 4α+6β=12n

3 g2 1 Thus we are left with weight 0 modular functions ∆ = 1728 j and (since ∆ = 2 3 3 2 g3 g2 −∆ 1 1 g2 − 27g3) ∆ = 27∆ = 27·1728 j − 27 which is linear is j, and this concludes the proof. Corollary 3.18. The modular function of weight 0 for Γ form a field isomorphic to C(t). 3.1. MODULAR FORMS FOR SL2(Z) 52

3.1.5 The modular curve X0(1)

def def As sets write Y0(1) = Γ\H and X0(1) = Y0(1) ∪ {∞}. Definition. The extended upper half plane is

∗ 1 H = H ∪ Q ∪ {∞} = H ∪ P (Q)

Remark that the action of Γ on H extends to H∗. Identifying P1(Q) with {[x] } ( 2 {0})/ × = | (x, y) ∈ 2 {0} modulo linearity Q r Q y Q r

2 then the action of SL2(Z) on Q r {0} given by (x) (a b)(x) ↦→ y c d y

yields an action of Γ on P1(Q). In fact P1(Q) is identified with Q ∪ {∞} by { [x] x if y ̸= 0 ↦→ ∈ Q y y ∞ if y = 0

And with such identification, Γ acts on Q ∪ {∞} by x x a y + b a γ · = x and γ · ∞ = (3.1.6) y c y + d c

(a b) for any γ = ∈ Γ. c d

1 Definition. The elements of Γ\P (Q) are called cusps of X0(1). Lemma 3.19. (a)Γ \P1(Q) contains a single element, which we will denote by ∞.

(b) StabΓ(∞) = ⟨T ⟩.

x Proof. (a) Given any rational number y we want to show that there exists γ ∈ Γ (a b) such that γ x = ∞. If γ = mod ± 1, then y c d

x x a y + b ax + by γ · = x = y c y + d cx + dy

Without loss of generality assume that x, y ∈ Z,(x, y) = 1, then by the Bezout identity there exist (a, b) ∈ Z2 such that ax + by = 1 therefore ( a b) ∈ SL ( ) and −y x 2 Z

( a b) x · = ∞. −y x y 3.1. MODULAR FORMS FOR SL2(Z) 53

(a b) (b) If γ · ∞ = ∞ then by (3.1.6) γ corresponds to some with c = 0. c d (±1 b ) This implies that ad = 1, thus a = d = ±1 and γ = = T ±b. We 0 ±1 conclude that StabΓ(∞) = ⟨T ⟩.

Let us now define a topology on H∗: • If τ ∈ H an open neighbourhood of τ will be an open neighbourhood of τ in C contained in H (i.e. equip H with the induced topology from C). • If τ = ∞ ∈ P1(Q) we define a base of open neighbourhoob to be the collection of sets of the form

{τ ∈ H | Im(τ) > A, A ∈ R>0} ∪ {∞} (3.1.7)

• If a ∈ Q we define a base of open neighbourhood of a to be the collection of sets of the form

{a} ∪ {open disc in H tangent to R at a} (3.1.8)

Remark 3.20. (i) Given a ∈ Q, let γ ∈ Γ be such that γ∞ = a (which exists by Lemma 3.19), then γ sends elements in the base of open neighbourhood of ∞ to elements of open neighbourhood of a.

(ii) Γ acts on the topological space H∗ via homeomorphisms. (iii) H∗ with the topology we just introduced is Hausdorff and connected.

Definition. Define the topology on X0(1) as the quotient topology induced by ∗ the canonical projection π : H −→ X0(1). In other words U ⊂ X0(1) is open if and only if π−1(U) is open in H∗. Lemma 3.21. π is an open map. Proof. If V ⊂ H∗ is open, then π−1(π(V)) = ⋃ γV is open since union of open γ∈Γ ∗ sets in H . Therefore π(V) is open in X0(1) by definition of its topology.

In the following we want to show that X0(1) is a compact Riemann surface. Before proving that we need a preparation lemma. Let us denote I(τ1, τ2) = {γ ∈ ∗ Γ | γτ1 = τ2} for any given τ1, τ2 ∈ H and let I(V1, V2) = {γ ∈ Γ | γV1 ∩V2 ̸= ∅} ∗ for any given open V1, V2 ⊂ H . Then:

∗ Lemma 3.22. For any pair τ1, τ2 ∈ H there exist open neighbourhoods V1 of τ1 and V2 of τ2 such that I(V1, V2) = I(τ1, τ2) (and such an action is called properly discontinuous). Proof. Due to the relations

−1 −1 I(γ1τ1, γ2τ2) = γ2I(τ1, τ2)γ1 and I(γ1V1, γ2V2) = γ2I(V1, V2)γ1

we may assume that τ1, τ2 ∈ D ∪ {∞} and we have 3 cases: 3.1. MODULAR FORMS FOR SL2(Z) 54

• Let τ1, τ2 ∈ D, and notice that I(D, D) is finite and contains the elements

−1 2 2 1Γ, T, S, T , ST, T S, (ST ) , (TS)

Define F = ⋃ γD ⊃ D and remark that also I(F, F) is finite since γ∈I(D,D) ⋃ F ⊆ I(γ1D, γ2D)

γ1,γ2∈I(D,D)

Consider the elements γ ∈ I(F, F) r I(τ1, τ2) =: I. For any such γ ∈ I holds that γτ1 ̸= τ2, thus we may fix Uγ ∋ γτ1 and Wγ ∋ τ2 open sets such that Uγ ∩ Wγ = ∅. Then define

⋂ −1 ⋂ V1 = F ∩ ( γ Uγ ) ∋ τ1 and V2 = F ∩ ( Wγ ) ∋ τ2 γ∈I γ∈I

V1 and V2 are open and I(V1, V2) = I(τ1, τ2): in fact if γ ∈ I(V1, V2) r −1 I(τ1, τ2) then γ ∈ I and by definition of V1 and V2 we have γ ∈ I(γ Uγ , Wγ ) = I(Uγ , Wγ )γ. Therefore 1Γ ∈ I(Uγ , Wγ ) which yields Uγ ∩ Wγ ̸= ∅ a contradiction. (a b) • Let τ = τ ∈ D and τ = ∞. Recall that for γ ∈ Γ given by 1 2 c d Im(τ) we have Im(γτ) = 2 . Now consider an open disc Dτ ∋ τ, then |cτ1+d| by the relation above we have Im(γτ) < M = M(Dτ ) for some M ∈ R>0 indipendent of τ. Let D∞ = {z ∈ H | Im(z) > M} ∪ {∞}, then I(Dτ ,D∞) = ∅ = I(τ, ∞).

• Let τ1 = τ2 = ∞, then I(∞, ∞) = StabΓ(∞) = ⟨T ⟩. Set V = {τ ∈ H | Im(τ) > 1} ∪ {∞} and let γ ∈ I(V, V) so that γV ∩ V ̸= ∅. Therefore ′ ′ ′ ′ ′ n ′ m γτ1 = τ2 for τ1, τ2 ∈ V r {∞}. Write τ1 = T z1 and τ2 = T z2 for some z1, z2 ∈ D. Then we have

n m −m n γT z1 = T z2 =⇒ z2 = T γT z1

−m n −1 Since Im(zi) > 1, by Theorem 3.5 follows that T γT ∈ {T,T , 1Γ} and consequently, γ ∈ ⟨T ⟩.

Theorem 3.23. X0(1) is a connected, compact and Hausdorff topological space.

∗ Proof. H is connected and π is continuous, hence X0(1) is connected. Now let −1 {Ui}i∈I be an open covering of X0(1), then {π (Ui)}i∈I is an open covering ∗ −1 of H . Say ∞ ∈ π (Ui0 ) for some i0 ∈ I, then some set of the form 3.1.7 is −1 −1 contained in π (Ui0 ). Since D r (π (Ui0 ) ∩ D) is compact because closed and bounded, there is a finite set of indeces i1, .., ik such that

k −1 ⋃ −1 D r (π (Ui0 ) ∩ D) ⊂ π (Uij ) j=1 3.1. MODULAR FORMS FOR SL2(Z) 55

Therefore X0(1) ⊂ Ui0 ∪ .. ∪ Uik , thus it is compact. We are left to prove that X0(1) is Hausdorff: let x1, x2 ∈ X0(1), x1 ̸= x2 and ∗ choose τi ∈ H such that π(τi) = xi, i = 1, 2. Since x1 ̸= x2 we have I(τ1, τ2) = ∅ therefore for i = 1, 2 by Lemma 3.22 we can select open neighbourhood Vi of τi ∗ in H such that I(V1, V2) = ∅ and this implies that π(V1) ∩ π(V2) = ∅. Since π is an open map, π(Vi) is an open neighbourhood of xi in X0(1). We conclude that X0(1) is an Hausdorff topological space.

We are now left to define a complex structure on X0(1) such that makes it into a compact Riemann surface. ∗ For each x ∈ X0(1) fix τ ∈ H such that π(τ) = x and fix a open neighbourhood ∗ Vx ⊂ H of τ such that I(Vx, Vx) = I(τ, τ) = StabΓ(τ). Then notice that

π(Vx) = StabΓ(τ)\Vx (3.1.9) and π(Vx) is a open neighbourhood of x. Let us now define a homeomorphism ψx : π(Vx) −→ Ux ⊆ C. • Let x∈ / {π(ρ), π(i), π(∞)} and as above let τ ∈ H∗ be such that π(τ) = x. Then StabΓ(τ) = {1Γ} and by (3.1.9) π(Vx) is homeomorphic to Vx.

Therefore choose Ux = Vx and ψx = idVx .

• Let x = π(∞) and choose τ = ∞. Recall that by Lemma 3.22, for V∞ we have I(V∞, V∞) = StabΓ(∞) = ⟨T ⟩. Hence π(V∞) = T Z\V∞ and the 2πiτ local parameter τ ↦→ qτ = e defines a homeomorphism

−2π ψ∞ : T Z\V∞ −→ {z ∈ C | |z| < e } ⊂ C such that [∞] ↦→ 0

• Let x = π(ρ), π(i), set τ = ρ, i respectively, write γ be a generator for StabΓ(τ) (as we found in Theorem 3.5) and let k be the order of γ (which is 3 and 2 respectively). Recall that we have a holomorphic isomorphism w − τ g : H −→ D = {z ∈ | |z| < 1} given by w ↦→ (3.1.10) x C w − τ Thus in particular τ ↦→ 0. Since γ is a holomorphic isomorphism of H of finite order k, then

∗ −1 γ = gx ◦ γ ◦ gx : D −→ D

is a holomorphic isomorphism of order k in D such that γ∗(0) = 0. By complex analysis one can prove that γ∗ is of the form z ↦→ ξz for some ξ ∈ S1 = {u ∈ C | |u| = 1} and the order of γ∗ is k. Therefore such ξ is a primitive k-th root of unity. Then define

k ψx : StabΓ\Vx = ⟨γ⟩\Vx −→ D by [z] ↦→ gx(z)

∗ Notice that ψx is well-defined in fact by definition of γ we have

∗ k ( ∗ )k ( )k k gx(γz) = γ (gx(z)) =⇒ gx(γz) = γ (gx(z)) = ξgx(z) = gx(z) 3.1. MODULAR FORMS FOR SL2(Z) 56

Moreover ψx is continuous since induced by the continuous map z ↦→ k k ′ k gx(z) . Let us now prove that ψx is also injective: let gx(z) = gx(z ) , then

r ′ ∗ r ′ r ′ gx(z) = ξ gx(z ) = (γ ) (gx(z )) = gx(γ z ) for some r ≥ 1

r ′ ′ By injectivity of gx z = γ z and therefore z ≡ z mod ⟨γ⟩. k Finally, since gx is a homeomorphism and w ↦→ w is an open map on D we get that ψx is an open map, hence a homeomorphism onto ψx(⟨γ⟩\Vx).

−1 A (boring) computation proves that the transition maps ψy ◦ ψx are holomorphic for all x, y ∈ X0(1). Therefore we can conclude that X0(1) is a compact Riemann surface, called modular curve of level 1.

1 Proposition 3.24. The modular j-function j : X0(1) −→ P (C) is a holomorphic isomorphism of compact Riemann surfaces, mapping [∞] to ∞. Proof. Let us proceed in multiple steps.

Claim 1: Let k be a positive integer, let νk : D −→ D be the holomorphic k 2π/k map given by z ↦→ z and let ξk = e . Now if f : D −→ C is a holomorphic function on D such that f(ξkz) = f(z) for all z ∈ D then f induces a map g : D −→ C such that f = g ◦ νk, ord0(f − f(0)) ≥ k and g is holomorphic. √ √ √ def k k i k In fact let g(z) = f(w) for w ∈ z, then since w ∈ z, also ξkw ∈ z for all i = 0, .., k − 1 and f(ξkz) = f(z) implies that g : D −→ C is well-defined and f = g ◦ νk. Now computing deriatives of f (which exists since f is holomorphic) one gets ′ ′ ′′ 2 ′′ f (z) = ξkf (ξkz); f (z) = ξkf (ξz) ; and so on.. Therefore f (m)(0) = 0 for all m such that n - m and consequently near z = 0 we have ∞ ∑ ni f(z) = f(0) + aniz (∗) i=1 which in particular implies ord0(f − f(0)) ≥ n. Finally using (∗) and the definition of g we get near z = 0

∞ n ∑ ni g(z ) = f(z) = f(0) + aniz i=1 and changing variable w = zn yields

∞ ∑ ni g(w) = g(0) + aniw i=1 Therefore g, having a power series expansion around 0 and radius of convergence at least 1 1 is holomorphic in D. Claim 2: j induces a holomorphic function on Y0(1).

1In fact we have that composition of two holomorphic functions H = F ◦ G only makes sense whenever G is analytic in a neighbourhood of z0 and F is analytic in a neighbourhood of G(z0). In our case we know that f is holomorphic on D and νk is analytic in all C, thus g has to be holomorphic in D. 3.1. MODULAR FORMS FOR SL2(Z) 57

Recall that j is holomorphic on H and is Γ-invariant therefore defines a map (which is a bijection by Theorem 3.12) ˆ : Y0(1) −→ C. Now consider the commutative diagram F

D → H j g−1 x ↘ ν π → k ˆ → ↗C ↓ π˜ ↓ D → Y0(1) G

−1 where νk is defined as above with k = #StabΓ(x), gx is the inverse of the biholomorphism in (3.1.10) centered at x, π is the canonical projection on the ∗ −1 quotient,π ˜ is such that π =π ˜ ◦exp with exp : H −→ D, and finally F = j ◦gx and G = ˆ◦ π˜. Due to commutativity we have F = G ◦ νk and F is holomorphic −1 since gx is a biholomorphism and j is holomorphic on H (by its definition). To show that j induces a holomorphic function on Y0(1) (as Riemann surfaces) is equivalent to show that G is holomorphic:

• For x ̸= i, ρ, ν1 : z ↦→ z therefore F = G and G is holomorphic. • For x = i we have k = 2 and if we prove that F (z) = F (−z) we can conclude by Claim 1. By the commutativity we in fact we have

2 F (z) = G(ν2(z)) = G(z ) = G(ν2(−z)) = F (−z)

• Similarly for x = ρ we get

3 F (z) = G(ν3(z)) = G(z ) = G(ν3(ξ3z)) = F (ξ3z)

Therefore ˆ is holomorphic. Claim 3: ˆ extends to a meromorphic function on X0(1) with a simple pole at the cusp. Recall that from the proof of Proposition 3.11 we have

div(j) = 1 · ρ − 1 · ∞

1 And since π(P (Q)) = [∞] ∈ X0(1), the claim follows. 1 Claim 4 j defines an isomorphism of Riemann surfaces from X0(1) to P (C). From Claim 3 we have that ˆ = j (with some abuse of notation) is a meromorphic function from X0(1) to C, now from the theory of Riemann surfaces we have that Mer(X) = Hol(X, P1(C)) r {c = ∞}. Therefore this is equivalent to say that j is a holomorphic function between the compact Riemann surfaces X0(1) and P1(C). Finally, since j has a single simple pole at ∞, deg(j) = 1, hence it is a biholomorphism.

Corollary 3.25. X0(1) has genus 0. Remark that in view of Theorem 3.17, Proposition 3.24 and the curves-fields correspondence, we have that the function field C(X0(1)) coincides with C(j). Moreover, since P1 can be defined over Q, the modular function j allows us to define the structure of non-singular projective curve over Q for X0(1). 3.2. CONGRUENCE SUBGROUPS 58

3.2 Congruence subgroups

Definition. Let N ∈ N, then define the principal congruence subgroup of level N to be {(a b) (a b) (1 0) } Γ(N) = ∈ SL ( ) ≡ mod N c d 2 Z c d 0 1

A subgroup Γ of SL2(Z) is a congruence subgroup of level N if Γ(N) ⊂ Γ for some N ∈ N. In particular define {(a b) (a b) (∗ ∗) } Γ (N) = ∈ SL ( ) ≡ mod N 0 c d 2 Z c d 0 ∗

{(a b) (a b) (1 ∗) } Γ (N) = ∈ SL ( ) ≡ mod N 1 c d 2 Z c d 0 1

Remark 3.26. (i) Since Γ(N) is the kernel of the natural homomorphism given by reduction modulo N, SL2(Z) −→ SL2(Z/NZ), thus Γ(N) is normal in SL2(Z). Moreover the homomorphism is actually surjective, hence it induces the isomorphism

∼ SL2(Z)/Γ(N) −→ SL2(Z/NZ) (3.2.1)

It follows that the index of Γ(N) in SL2(Z) is finite.

(ii)Γ( N) ⊂ Γ1(N) ⊂ Γ0(N) ⊂ SL2(Z): therefore also Γ1(N) and Γ0(N) have finite index in SL2(Z), as well as any congruence subgroup.

(iii) Notice that for N > 2, −1 ∈/ Γ(N), Γ1(N).

Proof. The only non-trivial statement is the surjectivity of SL2(Z) −→ SL2(Z/NZ), so that mod (N) 1 −→ Γ(N) −→ SL2(Z) −→ SL2(Z/NZ) −→ 1 is an exact sequence. (a b) Let γ = ∈ SL ( /N ) i.e. ad − bc = 1. c d 2 Z Z Write a, b, c, d for the representatives of a, b, c, d in Z ∩ {0, .., N − 1}, so in par- (a) ticular ad − bc = 1 + Nk for some k ∈ . We want to lift the 1st column Z c in a ”smart” way:

• If (a, c) = 1 then by the Bezout identity there exist β, δ ∈ Z such that aδ − cβ = 1 and we can choose β = b + Nr and δ = d + Ns for suitable r, s ∈ Z. In fact a(d + Ns) − c(b + Nr) = ad − bc + N(as − cr) = 1 + Nk + N(as − cr)

By coprimality of a and c, there exist r, s ∈ Z such that as − cr = −k. • If (a, c) ̸= 1, then notice that necessarely (a, c, N) = 1, otherwise f = (a, c, N) > 1 is not invertible mod (N) but f(a′d − bc′) = 1 and we get a contradiction. 3.2. CONGRUENCE SUBGROUPS 59

Established that, we are now looking for a lifting a′ of a such that a′ ≡ a mod (N) and (a′, c) = 1. Define l = ∏ p and let a′ = a + lN then we claim that (a′, c) = 1. p prime p|c,p-a In fact, ∀p | c we have a ̸≡ 0 mod (p) since

– If p | a then p - l, N hence lN ̸≡ 0 mod (p) ⇒ a′ ≡ lN ̸≡ 0 mod (p) – If p - a then p | l hence a′ ≡ a + 0 ̸≡ 0 mod (p) This implies that a′, c are coprime, and select them as lift for the 1st column. Using the same argument, there exist β = b+Nr, δ = d+Ns ∈ Z such that a′δ − cβ = 1 for suitable r, s ∈ Z. In fact a′(d+Ns)−c(b+Nr) = (a+lN)d−bc+N(a′s−rc) = 1+Nk+Nld+N(a′s−rc)

By the Bezout identity, we can choose r, s such that a′s − rc = −k − ld.

∗ For a modular subgroup Γ define the sets YΓ = Γ\H and XΓ = Γ\H = 1 YΓ ⊔ CΓ where CΓ = Γ\P (Q) is the set of cusps of Γ.

Lemma 3.27. CΓ is a finite set.

Proof. Consider the finite set decomposition SL2(Z) = Γγ1 ⊔ .. ⊔ Γγr, γi ∈ 1 SL2(Z) and r = [SL2(Z) : Γ]. We know that for all x ∈ P (Q) there exists ′ ′ γ ∈ SL2(Z) such that γ · ∞ = x. Write γ = γ · γi for some γ ∈ Γ and γi as above. Then

′ x = γ · ∞ = γ · γi · ∞ =⇒ [x] = [γi∞] in CΓ

Hence CΓ = {[γi · ∞], i = 1, .., r} which is finite.

Remark that from the proof we deduced more, namely that #CΓ ≤ [SL2(Z): Γ]. In general such inequality may be strict. For example we have that [SL2(Z):

Γ0(p)] = p + 1 and CΓ0(p) = {[∞], [0]} for any rational prime p. ∗ Theorem 3.28. The natural projection π = πΓ : H −→ XΓ equips XΓ with a natural structure of compact Riemann surface, XΓ is called modular curve of level Γ.

Proof. Let us prove the theorem for X0(N),X1(N) and X(N) which are the ∗ quotients of H by Γ0(N), Γ1(N) and Γ(N) respectively. ∗ def ∗ Recall that D = D ∪ {∞} surjects onto X0(1) and let us prove that D is ∗ ∗ compact in H . Let {Ui}i∈I be any open covering of D , then there exist some ∗ index i0 ∈ I such that ∞ ∈ Ui0 . Therefore there is a neighboorhood of ∞ in H , say VM = {z ∈ H : |z| > M} ∪ {∞} for some M ∈ R>0, such that Ui0 contains ∗ ∗ c VM ∩D . It follows that D ∩VM+1 is closed and bounded in H, hence compact. ∗ c ⋃ Since, trivially, D ∩ VM+1, ⊂ C ⊂ Ui, we get that {Ui}i∈I is a covering of i∈I ∗ c D ∩ VM+1 and by compactness, there exists a finite subcovering {Ui1 , .., Uik } k ∗ c ⋃ of it such that D ∩ VM+1 ⊂ Uij . We can conclude that the set {Ui0 , .., Uik } j=1 3.2. CONGRUENCE SUBGROUPS 60 is a finite subcovering the initial covering, therefore D∗ is compact in H∗. Step 1 X0(N),X1(N) and X(N) are compact. Since Γ(N) ⊂ Γ1(N) ⊂ Γ0(N) we have canonical (continuous) surjections π1 π0 X(N) −→ X1(N) −→ X0(N). Therefore it is enough to show that X(N) is compact, since if so, then the continuous image of a compact is compact. Let [SL2(Z) : Γ(N)] = r < ∞ and consider the right coset decomposition

SL2(Z) = Γ(N)γ1 ⊔ · · · ⊔ Γ(N)γr

n ⋃ ∗ for γ1, .., γr ∈ SL2(Z) cosets representatives. Define K = γi(D ), then K is i=1 ∗ compact in H since finite union of compact subsets (since the action of SL2(Z) ∗ is continuous, γiD is compact for all i’s), and such such K surjects to X(N). In fact let [τ] ∈ X(N) and let us prove that there exists ξ ∈ K such that π(ξ) = [τ], or equivalently that there exists γ ∈ Γ(N) for which γξ = τ. We ∗ ∗ already know that for every τ ∈ H there exist γ ∈ SL2(Z) and η ∈ D such that ′ γη = τ. Now using the coset decomposition we have that SL2(Z) ∋ γ = γ · γi ′ for some γ ∈ Γ(N) and γi ∈ SL2(Z) representative. Letting ξ := γiη, it follows that ξ ∈ K and π(ξ) = [τ]. Step 2 For any τ ∈ H the stabilizer of τ in Γ(N) is either {±1} or trivial. In fact remark that Stab (τ) ⊆ Stab (τ) for all τ ∈ H and Stab (τ) = Γ(N) SL2(Z) Γ(N) Stab (τ) ∩ Γ(N). Let us first analize the case for τ ∈ D: SL2(Z) • If τ ̸= i, ρ, −ρ, by Theorem 3.5 we already have Stab (τ) = {±1} and SL2(Z) just notice that ±1 ∈ Γ(2) but −1 ∈/ Γ(N) if N ≥ 3, hence StabΓ(2)(τ) = {±1} and for N ≥ 3, StabΓ(N)(τ) = {±1}. ( 0 1) • If τ = i we have Stab (i) = {±1, ±S}where S = . However SL2(Z) −1 0 ±S/∈ Γ(N), hence the stabilizer of i is the same of the generic τ above. • If τ = ρ, −ρ, similarly we obtain Stab (τ) = {±1, ±ST, ±(ST )2}; SL2(Z) however ±ST, ±(ST )2 ∈/ Γ(N) for any N. It follows that also the stabilizer of ρ and −ρ is as in the general case.

For any other point ξ ∈ H there exist γ ∈ SL2(Z) and τ ∈ D such that γ(τ) = ξ, and Stab (ξ) = γStab (τ)γ−1 SL2(Z) SL2(Z) In fact if η ∈ Stab (τ) then γηγ−1(ξ) = γη(τ) = γ(τ) = ξ. SL2(Z) Viceversa, if ν ∈ Stab (ξ) then γ−1νγ(τ) = γ−1ν(ξ) = τ. SL2(Z) Thus setting σ = γ−1νγ we obtain ν = γσγ−1. Therefore since the stabilizers are conjugated in SL2(Z) they have the same group structure (conjugation is an automorphism). It follows that we always are in one of the two cases:

∀τ ∈ H : StabΓ(2)(τ) = {±1}; or if N ≥ 3 StabΓ(N)(τ) = {1}

Step 3 Riemann surface structure on Y (N). Now recall that the action of SL2(Z)/{±1} is properly discontinuous, therefore also the actions of Γ(2)/{±1} and of Γ(N), for N ≥ 3, are. Let [τ] ∈ Y (N), and take Uγiτ a neighbourhood of γiτ in H (where γi is a coset representatives as 3.2. CONGRUENCE SUBGROUPS 61

above) such that ∀γ ∈ Γ(2) r {±1} or Γ(N) r {1} it holds γUγiτ ∩ Uγiτ = ∅.

Then we obtain that π(Uγiτ ) = StabΓ(N)(τ)\Uγiτ =: U[τ] is a neighbourhood of [τ] in Y (N). Taking as local coordinates for [τ], (U , π−1 ) yields a Riemann [τ] |U γiτ surface structure on Y (N). Step 4 Riemann surface structure on Y0(N) and Y1(N). Let us prove that for Γ = Γ0(N), Γ1(N) and for any τ ∈ H, ±StabΓ(τ)/ ± 1 is a cyclic group of order at most three. As above

Stab (τ) = Stab (τ) ∩ Γ ∼ γ−1Stab (ξ)γ ∩ Γ = Stab (ξ) Γ SL2(Z) = SL2(Z) Γ where ξ ∈ D and γ ∈ SL2(Z) are such that γ(ξ) = τ. Therefore it is enough to prove for elements in D, by Theorem 3.5 every point has cyclic stabilizer of order at most 6, and by quotienting for {±1} we get that such quotient is cyclic of order at most 3. Therefore define the charts as follows :

• If τ ∈ H has trivial stabilizer in then we can consider Uτ ⊂ H ⊂ C and the canonical projection π : U −→ π(U ) is already a homeomorphism, so |Uτ τ τ −1 take (π(Uτ ), π ) as local chart around π(τ). |Uτ • If τ has non-trivial stabilizer, say of order k, then fix a biholomorphism −1 −1 gτ : D −→ H with the property that gτ (0) = τ. Now since the stabilizer can be viewed as a cyclic subgroup of order k of the automorphisms of D such that they fix 0, then they operate on D as multiplication by powers of τ. This yields a commutative diagram

g−1 D τ → H

νk π ↓ π˜ ↓ D → Y

k where Y is either Y0(N) or Y1(N) and νk : w ↦→ w . Then there is a neighbourhood V of 0 in D such thatπ ˜(V) = π(U) = StabΓ(τ)\U. Therefore a complex chart around τ is given by (˜π(V), π˜−1). |V

This construction makes Y0(N) and Y1(N) into a Riemann surface. Step 5 Stabilizers of the cusps. We are left to study the cusps: let us show that for z ∈ P1(Q) we have Stab (z) = γ−1Stab (∞)γ ∩ Γ Γ SL2(Z) where γ ∈ SL2(Z) is such that γ(z) = ∞. Since Stab (z) = Stab (z) ∩ Γ it is enough to prove the equality Γ SL2(Z) Stab (z) = γ−1Stab (∞)γ SL2(Z) SL2(Z) (1 1) We already know that Stab (∞) = ⟨±T ⟩ where T = . Now let SL2(Z) 0 1 η ∈ Stab (∞), then SL2(Z) γ−1ηγ(z) = γ−1η(∞) = γ−1(∞) = z 3.2. CONGRUENCE SUBGROUPS 62 hence γ−1ηγ ∈ Stab (z) for any η ∈ Stab (∞). Viceversa, let ν ∈ SL2(Z) SL2(Z) Stab (z), then SL2(Z) γνγ−1(∞) = γν(z) = γ(z) = ∞ hence γνγ−1 =: η belongs to Stab (∞), thus ν = γ−1ηγ. SL2(Z) −1 We conclude that StabΓ(z) = γ ⟨±T ⟩γ ∩ Γ. Now consider two cases: • If z = ∞, we have

⎧⟨±T ⟩ if Γ = Γ (N), Γ (2) ⎪ 0 1 ⎪ −n n ⎨⟨T ⟩ if Γ = Γ1(N),N ≥ 3 StabΓ(∞) = T ⟨±T ⟩T ∩Γ = ⟨±T ⟩∩Γ = ( ) ⎪ N 1 N ⎪⟨T = ⟩ if Γ = Γ(N) ⎩ 0 1

α • If z ̸= ∞, then z ∈ Q so we may assume that z = β for some α, β ∈ Z coprime and β ̸= 0. (a b) First of all let us characterize γ = ∈ SL ( ) such that γ(z) = ∞: c d 2 Z

aα + bβ γ(z) = ∞ =⇒ = ∞ cα + dβ This forces cα + dβ = 0 (1) and aα + bβ ̸= 0 (2).

(1) =⇒ c = −kβ and d = kα for some k ∈ Z The condition on the determinant ad − bc = 1 implies that k = ±1 (otherwise we get a contradiction), so without loss of generality assume k = 1 , and also that a ̸= β, b ̸= −α (otherwise γ would be singular). Since α, β are comprime, by the Bezout identity, there exist suitable a, b ∈ Z such that aα + bβ = 1, thus choosing such pair we have

( a b) (α −b) γ = and γ−1 = −β α β a

Let us show ∀N ∈ N and ∀z ∈ Q we have StabΓ(N)(z) ̸= ∅ (and consequentely for all other Γ’s stabilizers are not empty). Consider the following:

(α −b)(1 N)( a b) (α −b)(a − Nβ b + Nα) = = β a 0 1 −β α β a −β α

(aα + bβ − Nαβ bα + Nα2 − bα) (1 − Nαβ Nα2 ) (1 0) = = ≡ mod (N) aβ − Nβ2 − aβ aα + bβ + Nαβ Nβ2 1 + Nαβ 0 1 Furthermore, each stabilizer is subgroup of a cyclic group, thus it is cyclic, (1 N) and it contains as a subgroup ⟨γ−1 γ⟩ which is infinite, hence it 0 1 is itself infinite. Step 6 Local parameter at the cusps. 1 ∗ Let [s] ∈ CΓ be a cusp, s ∈ P (Q). Consider the open subset Vs = H ∪ {s} of H , let γ ∈ SL2(Z) be such that γ(s) = ∞. Then we have the exponential map to D 3.2. CONGRUENCE SUBGROUPS 63

2πiγ(τ)/n given by τ ↦→ e . Since the action of SL2(Z) is continuous, γ(Vs) = V∞ and we just computed the stabilizer of ∞ for different Γ’s: StabΓ(∞) is generated (1 b) by for a suitable b ≥ 1. Choosing n = b is enough to have a bijection 0 1

2πiγ(τ)/n StabΓ(s)\Vs−→D given by τ ↦→ e Therefore consider the composition of maps

(left) action of γ exp π˜ ∗ Vs −→ V∞ −→ D −→ Γ\H

So s ↦→ 0 ∈ D and there exists U neighbourhood of 0 in D such thatπ ˜|U is a homeomorphism. Therefore take (StabΓ(s)\Vs, π˜|U ) as a local chart around the cusp [s].

3.2.1 Modular functions of higher level

Definition. Let Γ be a congruence subgroup of SL2(Z), (i)A weakly modular function of weight k ∈ Z for Γ is a meromorphic function (a b) f : H −→ 1( ) such that ∀γ = ∈ Γ,∀τ ∈ H it satisfies the P C c d functional equation f(γτ) = (cτ + d)kf(τ). (ii)A modular function of weight k for Γ is a weakly modular function for Γ which is also meromorphic at all the cusps in CΓ. (iii)A modular form of weight k for Γ is a modular function for Γ which is holomorphic at all the cusps in CΓ. (iv)A cusp form of weight k for Γ is a modular form for Γ which vanishes at all the cusps in CΓ. Let us explain more in details what this means. We use the local parameter at [s] ∈ CΓ, that we introduced in Step 6 of the proof of Theorem 3.28, to define what it means to be meromorphic, holomorphic, vanishing at [s]. Let f : H −→ P1(C) be a weakly modular function of weight k for Γ. For any (a b) σ = define j(σ, τ) = (cτ + d) and consider the meromorphic function c d on H given by −k f|[σ]k(τ) = f(σ(τ))j(σ, τ) (3.2.2)

Hence for σ ∈ Γ we have f|[σ]k(τ) = f(τ) due to the modularity property for Γ. Moreover a simple computation shows that f|[σ]k is a weakly modular function −1 −1 of weight k for σ Γσ. Now let γ ∈ SL2(Z) be such that γs = ∞, then f|[γ ]k ∗ is invariant for τ ↦→ τ + n, where n is as in Step 6. Define a function fs on some open disc of C centered at 0 with origin remuved, by ∗ 2πiτ/n −1 fs (e ) = f|[γ ](τ) (3.2.3) Then we say that f is meromorphic (resp. holomorphic, resp vanishes) at the ∗ cusp [s] if fs is meromorphic (resp. holomorphic, resp. vanishes) at 0. In other words −1 ∑ i/n 2πiτ f|[γ ]k(τ) = aiq = Φ(q) for q = e (3.2.4)

i≥n0 3.2. CONGRUENCE SUBGROUPS 64 called Puiseaux expansion. We can define divisors of modular functions for congruence subgroups in asim- ilar way as we did for SL2(Z). Let f be a modular function of weight k for Γ: let us define ordx(f) for all x ∈ XΓ.

• If x corresponds to a point z0 ∈ H, then fix a holomorphic isomorphism e g = gz0 : H −→ D such that g(z0) = 0. Let e = #StabΓ(z0), then t = g(z) ord(z−z0)(f) is the local normal form at x, therefore define ordx(f) = e .

• If x corresponds to a cusp s, let γ ∈ SL2(Z) be such that γ(s) = ∞. Then m { (1 h) } we saw that ±γStab (s)γ = ± m ∈ for some h ∈ . Γ 0 1 Z Z>0 2πiz/h −1 Let q = e , since f|[γ ]k(z) is invariant under z ↦→ z + h there exists −1 a meromorphic function Φ such that Φ(q) = f|[γ ]k(z), therefore define ordx(f) = ord0(Φ). ∑ Finally define div(f) = ordx(f) · x ∈ DivQ(XΓ) = Div(XΓ) ⊗Z Q. x∈XΓ We saw at the end of 3.1 that the analytic isomorphism given by the modular ∼ 1 function j of weight 0, j : X0(1) −→ P (C) allows us to define the structure of projective nonsingular curve over Q. Untill the end of these section and the next one, we want to adress the following question: Question: For a congruence subgroup Γ of SL2(Z), can we similarly define on XΓ the structure of nonsingular curve over Q (or at least over a finite extension of Q)? We will focus on Γ = Γ0(N) and imitating the case of X0(1) we start by describ- ing the field of modular functions FΓ0(N) = C(X0(N)) of weight 0 for Γ0(N). Remark 3.29. (i) Notice that j ∈ F = (X (N)) since j ∈ F = Γ0(N) C 0 SL2(Z) C(X0(1)).

(ii) Define jN by jN (τ) = j(Nτ) for any τ ∈ H, then jN ∈ C(X0(N)). In fact ( a b) (a Nb) let γ = ∈ Γ (N), then ∈ SL ( ) and Nc d 0 c d 2 Z

( aτ + b ) (a(Nτ) + Nb) ( (a Nb) ) j (γτ) = j N = j = j ·Nτ = j(Nτ) = j (τ) N Ncτ + d c(Nτ) + d c d N

Therefore jN is Γ0(N)-invariant and holomorphic on H since j is. Moreover j is meromorphic at [∞], hence jN is meromorphic at the cusps of Γ0(N). We first need to compute the q-expansion of j at ∞: Fact 3.30. The modular function j of weight 0 has a q-expansion of the form

1 ∑ j(τ) = + a(n)qn q n≥0 with a(n) ∈ Z and q = e2πiτ Proof. See [Kna93] Corollary 8.2 3.2. CONGRUENCE SUBGROUPS 65

Theorem 3.31. (The modular equation) ∼ (a) C(X0(N)) = C(j, jN );

(b) jN satisfies a non-zero polynomial in C(j)[Y ];

(c) If FN (j, Y ) ∈ C(j)[Y ] is the monic polynomial of minimal degree such that FN (j, jN ) = 0, then FN (X,Y ) ∈ Z[X,Y ].

Proof. Claim 1: Any f ∈ C(X0(N)) satisfies a polynomial in C(j)[Y ] of degree equal to r = [SL2(Z):Γ0(N)]. r ⨆ Consider the coset decomposition SL2(Z) = Γ0(N)γi for γi ∈ SL2(Z); for i=0 r ∏ any f ∈ C(X0(N)) define the polynomial Ff (Y ) = (Y − f(γiτ)). By its i=1 definition, degFf (Y ) = r and f(τ) is a root since there is an index for which γi ∈ Γ0(N). Notice that f(γiτ) only depends upon the right coset Γ0(N)γi since f(γiτ) = f(γγiτ) for any γ ∈ Γ0(N) as f is Γ0(N)-invariant. Moreover the coefficients of Ff (Y ) are symmetric polynomials in f(γiτ)’s and right multiplication of the above cosets by γ ∈ SL2(Z) induces a permutation σ = σγ of the cosets: Γ0(N)γiγ = Γ0(N)γσ(i). It follows that the coefficients of Ff (Y ) are SL2(Z)-invariant : if ci(τ) is a coefficient, then ci(τ) = S(f(γ1τ), .., f(γrτ)) = S(f(γσ(1)τ), .., f(γσ(r)τ)) = ci(γτ) for S = S(z1, .., zr) a symmetric polynomial. We can conclude that ci(τ) ∈ C(j), so that Ff (Y ) ∈ C(j)[Y ]. Claim 2 C(X0(N)) is a finite extension of C(j) of degree less or equal to r. Choose f ∈ C(X0(N)) such that its minimal polynomial Pf (Y ) ∈ C(j)[Y ] has maximal degree, then

(∗) C(X0(N)) = C(j, f)

In fact, suppose there exists g ∈ C(X0(N)) r C(j, f), then [C(j)(f, g): C(j)] > [C(j)(f): C(j)] and by the Primitive element theorem C(j)(f, g) = C(j)(h) and therefore degPh(Y ) > degPf (Y ) contradicting maximality for f. Therefore (∗) holds and by Claim 1 follows that [C(X0(N)) : C(j)] ≤ r. Claim 3 C(X0(N)) = C(j, jN ) and [C(X0(N)) : C(j)] = r.

By Claim 2, it is enough to prove that PjN (Y ) the minimal polynomial for jN over C(j) has exactly degree r. We actually reduce to check that

r ∏ FN (j, Y ) = FjN (Y ) = (Y − jN (γiτ)) (3.2.5) i=1

is the minimal polynomial of jN . Since FN (j, Y ) is satisfied by jN , it is enough to show that

(i) The roots jN (γiτ) of FN (j, Y ) are all conjugate over C(j); (ii) These roots are pairwise distinct.

Now if PjN (Y ) = PjN (j, Y ) is the minimal polynomial of jN over C(j), then

PjN (j(τ), jN (τ)) = 0 for all τ ∈ H. Therefore

0 = PjN (j(γiτ), jN (γiτ)) = PjN (j(τ), jN (γiτ)) = PjN (jN (γiτ)) 3.2. CONGRUENCE SUBGROUPS 66

where the second equality follows by the SL2(Z)-invariance of j. Hence jN (γiτ) are conugate. Suppose that jN (γiτ) = jN (γkτ) for all τ ∈ H, or more explicitly j(Nγiτ) = j(Nγkτ). This means that the holomorphic isomorphism (N 0) (N −1 0) η : H −→ H τ ↦→ γ γ−1 τ 0 1 k i 0 1 is such that j(τ) = j(ητ) for all τ ∈ H. Since j is a holomorphic isomorphism (a b) from SL ( )\H to , η is induced by a matrixγ ˜ = ∈ SL ( ), i.e. for 2 Z C c d 2 Z all τ ∈ H (N 0) (N −1 0) γτ˜ = γ γ−1 τ 0 1 k i 0 1 Therefore (N −1 0) (N 0) (N −1 0)(a b)(N 0) ( a N −1b) γ γ−1 = γ˜ = = k i 0 1 0 1 0 1 c d 0 1 Nc d Notice that the same computation from the second equality for a generic element of SL2(Z) yields (N −1 0) (N 0) SL ( ) ∩ SL ( ) ⊆ Γ (N) 0 1 2 Z 0 1 2 Z 0

−1 We can conclude that γkγi ∈ Γ0(N) and therefore Γ0(N)γi = Γ0(N)γk. Claim 4 FN = FN (j, Y ) ∈ C(j, Y ) actually belongs to Z[j, Y ]. The first step is to check that FN (j, Y ) ∈ C[j, Y ]: since the coefficients ci(τ)’s of FN are symmetric polynomials in holomorphic functions on H, they are them- selves holomorphic on H and belong to C(j). Therefore for any i, they are of Pi(j) the form λi for some λi ∈ and Pi(X),Qi(X) monic polynomials with Qi(j) C distinct roots. This acutally implies that Qi(X) = 1 for all indeces i, because if X − βi is a linear factor, then ci(τ) would have a pole at each τ ∈ H such that j(τ) = β (which exists by surjectivity of j) and this contradicts holomorphicity of ci(τ). It follows that FN (j, Y ) ∈ C[j, Y ]. ( (N 0) ) Now consider j (γτ) = j γ τ , then by linear algebra there exists N 0 1 k (N 0) (a b) γ ∈ SL such that γ γ = with ad = N, 0 ≤ b < d. Therefore 0 1 k 0 d ( (a b) ) aτ + b j (γ τ) = j = j( ) (3.2.6) N k 0 d d 1 ∑ n 2πiτ and recall that j(τ) = q + a(n)q with a(n) ∈ Z, q = e . n≥0 2πi aτ+b 2πib/d 2πiaτ/d Notice that e d = e e , thus since d | N, jN (γkτ) has a q- expansion in q1/N in the ring Z[e2πi/N ] for any k ∈ {1, .., r}. Therefore the same holds for the coefficients ck(τ) of FN (j, Y ) which are symmetric polynomials in jN (γkτ). We previously proved that ck(τ) = Pk(j) for some P (X) ∈ C[X], and we now claim that the coefficients of P (X) are algebraic integers in Z[e2πi/N ]. Write m ∑ n ∑ k P (j) = anj(τ) = bkq

n=0 k≥k0 3.2. CONGRUENCE SUBGROUPS 67

and notice that an ∈ Z[bk | k ≥ k0]. By comparison we have

am = b−m ; am−1 = b1−m − mama(1)

m(m − 1) a = b − (m + 1)a a(1) − a a(2); and so on.. m−2 2−m m−1 2 m ∑ m n Therefore FN (X,Y ) = am,nX Y with am,n algebraic integers and a0,r = 1 m,n since FN (j, Y ) is monic. Substituting the q-expansions of j and jN we obtain a system of linear equations in the indeterminates am,n corresponding to FN (j, jN ) = 0. Such system has a unique solution since FN arises from the unique monic minimal polynomial of jN over C(j). Moreover the coefficients of such system are rational, therefore by the theory of linear systems the unique solution (am,n)m,n has rational entries. Since they are also algebraic integers, they belong to Z.

Remark that Theorem 3.31 (b) tells us that jN is algebraic over C(j).

3.2.2 The canonical model of X0(N) over Q Let us state but not prove a technical lemma:

Lemma 3.32. Let C(x, y) be a field such that x is transcendental over C and y is algebraic over C(x). Then for a subfield K0 ⊂ C, the following are equivalent:

(a) y is algebraic over K0(x) and the minimal polynomial of y over K0(x) re- mains irreducible over C(x);

(b)[K 0(x, y):K0(x)] = [C(x, y): C(x)];

(d) K0 ∩ K0(x, y) = K0 Proof. See [Kna93] Theorem 11.36

Proposition 3.33. Let C(f, g) be a function field of dimension 1 over C, let K0 ⊂ C be a subfield such that0 K (f, g) ∩ C = K0. For x ∈ K0(f, g) r K0 let B be the integral closure of C[x] in C(f, g) and write V for the nonsingular affine curve associated (as in Step 1 of Theorem 1.15). Then V is defined over0 K and ∼ ∼ K0(V ) = K0(f, g), C(V ) = C(f, g). Proof. Suppose f is transcendental over C and g algebraic over C(f). By Lemma 3.32, f is transcendental over K0 and g algebraic over K0(f), therefore K0(f, g) has transcendence degree one over K0. Since K0(f, g) ∩ C = K0, such x ∈ K0(f, g) r K0 cannot belong to C and therefore x is transcendental over K0. It follows that {x} is a transcendental basis for K0(f, g) over K0 and in particular f, g are algebraic over K0(x). This means that K0(f, g)/K0(x) is a finite algebraic extension, thus by the Primitive element theorem there exists some y ∈ K0(f, g) such that K0(f, g) = K0(x, y) and consequently C(f, g) = C(x, y). Lemma 3.32 is indipendent of the generators, thus

n := [K0(f, g):K0(x)] = [C(f, g): C(x)] 3.2. CONGRUENCE SUBGROUPS 68

Since C[x] is a principal ideal domain, there exists a basis {x1, .., xn} of C(f, g) over C(x) consisting of elements of B, such that n ∑ B = C[x]xi (3.2.7) i=1

Thus we take x1, .., xn, x as generators of B over C so that the map

ϑ : C[X1, .., Xn+1] −→ B such that for i = 1, .., n, Xi ↦→ xi and Xn+1 ↦→ x shows that B is the affine coordinate ring of the curve V defined by I = kerϑ. We need to show that V is canonically defined over K0 or in other words that I is (finitely) generated by elements0 inK [X1, .., Xn+1]. To accomplish that, write B0 for the integral closure of K0[x] in K0(f, g), then there exists a basis {y1, .., yn} of K0(f, g) over K0(x) with yi ∈ B0 such that n ∑ B0 = K0[x]yi (3.2.8) i=1 and consider the map

ϑ0 :K0[X1, .., Xn+1] −→ B0 such that Xi ↦→ yi for i = 1, .., n and Xn+1 ↦→ x, with kernel I0 = kerϑ0. Since K0[X1, .., Xn+1] is a Noetherian ring, I0 is finitely generated. So let I0 =

⟨P1, .., Pℓ | Pj ∈ K0[X1, .., Xn+1]⟩. Now applying the exact functor · ⊗K0 C to the exact sequence

0 −→ I0 −→ K0[X1, .., Xn+1] −→ B0 −→ 0 it yields the short exact sequence

C ϑ0 0 −→ I0 ⊗K0 C −→ C[X1, .., Xn+1] −→ B0 ⊗K0 C −→ 0

C where ϑ0 = ϑ0 ⊗ idC. Now I0 ⊗K0 C is generated by P1, .., Pℓ over C and n ∑ B0 ⊗ C = C[x]yi. By Lemma 3.32 (a), yi’s are linearly indipendent over i=1 C(x) hence we have two bases for C(f, g) over C(x), namely {xi} and {yi}. Therefore we have a C(x)-linear map ϕ : C(f, g) −→ C(f, g) such that xi ↦→ yi for all i = 1, .., n. In particular ϕ is injective since sends a basis to a basis and C ϕ ◦ ϑ = ϑ0 as C-linear maps. We can conclude that

C I = kerϑ = kerϕ ◦ ϑ = kerϑ0 = I0 ⊗ C = ⟨P1, .., Pℓ⟩ where the second equality follows by injectiveness of ϕ. Moreover by construction, the affine coordinate ring of V is B and the function field is the quotient field of B, which by (3.2.7) is C(f, g). Finally if V is defined over K0, we have

I(V/K0) = I(V ) ∩ K0[X1, .., Xn+1] = I ∩ K0[X1, .., Xn+1] =

= (I0 ⊗ C) ∩ K0[X1, .., Xn+1] = I0

Thus by definition K0[V ] = K0[X1, .., Xn+1]/I0 = B0 and by (3.2.8) its quotient field is0 K (f, g). 3.3. INTEGRALITY OF THE J-INVARIANT 69

We can conclude that if K0 ⊂ C is a subfield and C is a nonsingular projective curve defined over0 K , then there exists f, g ∈ K0(C) such that:

K0(C) = K0(f, g); C(C) = C(f, g) and C ∩ K0(C) = K0

Viceversa, if C(f, g) is a function field of dimension 1 over C and K0 is a subfield of C such that K0(f, g)∩C = K0 then there exists a nonsingular projective curve C defined over K0 such that ∼ ∼ K0(C) = K0(f, g) and C(C) = C(f, g) Furthermore, such C is unique due to the field-curves correspondence we saw in Chapter 1. We can finally apply everything with f = j, g = jN and K0 = Q to obtain

Theorem 3.34. There exist a nonsingular projective curve C defined over Q and a biholomorphic map ϕ : X0(N) −→ C(C) such that

∗ ∗ ϕ (C(C)) = C(X0(N)) = C(j, jN ) and ϕ (Q(C)) = Q(j, jN )

Such curve is unique up to isomorphism defined over Q and ϕ is uniquely de- termined by the isomorphism of Q(C) with Q(j, jN ).

Definition. The pair (ϕ, C) is called canonical model for X0(N) over Q.

Remark 3.35. 1. Moreover one can prove that an element of C(j, jN ) is in Q(j, jN ) if and only if its q-expansion at ∞ has coefficients in Q. ( See [Kna93] Corollary 11.50 or [Shi73] Proposition 6.9)

2. An alternative way to construct the canonical model of X0(N) over Q is via

2 Y0(N) −→ V (FN ) = {(x, y) ∈ A | FN (x, y) = 0} ;[τ] ↦→ (j(τ), jN (τ))

The equation FN (X,Y ) = 0 defines a curve C over Q and by remuving singular points one obtains an affine nonsingular curve Cns over Q. Then ns C can be embedded in a projective nonsingular curve C such that (j, jN ) extends to an isomorphism X0(N) −→ C(C).

3. A similar discussion applies to FΓ for Γ = Γ1(N), Γ(N) and one obtains 2πi/N models for XΓ over Q or over Q(e ) respectively. However the method is less explicit and requires to study the automorphisms of C on FΓ (See [Shi73] Chapter 6).

3.3 Integrality of the j-invariant

This whole section will be dedicated to prove the following: Theorem 3.36. If z ∈ H belongs to an imaginary quadratic field then j(z) is an algebraic integer. The two steps of the proof will be to show that j(z) is an algebraic integer, and then its integrality. 3.3. INTEGRALITY OF THE J-INVARIANT 70

3.3.1 j(z) is an algebraic number

Claim 1: For any z ∈ C r Q there exists a transcendental basis Sz of C over Q containing z.

Let z ∈ C r Q and consider the inclusions Q ⊂ Q(z) ⊂ C. Obviously {z} is a trnscendental basis of Q(z) over Q. Now if {z} is a transcendental basis of C ′ over Q then set Sz = {z}. Otherwise there exists a transcendental basis S of C ′ over Q(z). So define Sz = S ∪ {z} and this is a transcendental basis of C over Q, in fact :

• C is algebraic over Q(Sz) by ”transitivity of being algebraic” (it is a more general fact that if F ⊂ E ⊂ K are such that E is algebraic over F and K is algebraic over E then K is algebraic over F ).

• The map Q[Xs : s ∈ Sz] −→ C such that Xs ↦−→ s is injective because we can view it as composition of two injective maps, namely

′ ′ Q[Xz][Xs : s ∈ S ] → Q(z)[Xs : s ∈ S ] → C

Xz ✤ → z ✤ → z

Xs ✤ → Xs ✤ → s

Step 2: Sz is uncountable.

If Sz was a countable transcendental basis {z1, z2, ..} of C over Q then C would be algebraic over Q(z1, z2, ..). Now since the polynomials over Q can be identified with finite sequences of rational numbers and countable product of a countable set is still countable, we would have |C| = |Q| contradicting uncountability of C.

Step 3: The orbit Aut(C) · z is uncountable.

Fix z ∈ C. Since Sz is uncountable, hence Sym(Sz) the set of permutations of Sz is an uncountable set. Since C is algebraic over Q(Sz) we can extend by linearity any such permutation to an automorphism of C. In particular

Aut(C) · z ⊃ Sym(Sz) · z = {σz | σ ∈ Sym(Sz)} = {s | s ∈ Sz} which is uncountable by Step 2.

Step 4: There is only a countable amount of isomorphism classes of complex tori with complex multiplication.

To begin with recall that every complex torus T is isomorphic to one of the form Tτ = C/(Zτ + Z) for some τ ∈ H and for a torus to have complex multiplication means that End(T ) = O for some order O in a quadratic imaginary field K. Also recall the bijection of Proposition 2.35

Pic(OK ) ←→ {[T ] : End(T ) = OK }

given by [a] ↦−→ [C/a], which implies that the latter set is finite since the first is. Moreover since any order O is of the form O = Z+cOK for some c ∈ Z>0, so that 3.3. INTEGRALITY OF THE J-INVARIANT 71

there are countably many orders in any fixed imaginary quadratic field√ extension K. Finally every quadratic imaginary field extension is of the form Q( −d) for some square-free positive integer d, in other words there is a countable number of quadratic imaginary field K/Q. Putting everything together, the claim follows.

Step 5: Let Λ = Zτ + Z, so that for E =∼ T = C/Λ we have j(τ) = j(E) = j(Λ). 2 3 If E is given by y = 4x − g2(Λ)x − g3(Λ), then for any σ ∈ Aut(C) let σ 2 3 −1 −1 σ E : y = 4x − σ g2(Λ))x − σ (g3(Λ)) and we have j(E) = σ(j(E )). Therefore if E has complex multiplication, then Aut(C) · j(E) ⊂ C is at most countable.

Simply by computation we have

( −1 3 ) σ σ (g2(Λ)) σ(j(E )) = σ 1728 −1 3 −1 2 = σ (g2(Λ)) − 27σ (g3(Λ))

3 (g2(Λ)) = 1728 3 2 = j(E) g2(Λ) − 27g3(Λ) σ Let ψσ :(x, y) ↦→ (σ(x), σ(y)) be the isomorphism from E to E, then −1 σ ψσ End(E)ψσ = End(E ). It follows that if E has complex multiplication, then σ also EΛ has complex multiplication (in fact conjugating is a group isomorphism). Now assume that E has complex multiplication and consider

−1 −1 σ −1 Aut(C) · j(Λ) = {σ j(Λ) | σ ∈ Aut(C)} = {j(E ) | σ ∈ Aut(C)} Since j is an invariant for the isomorphism class of a torus and by Step 4 we only have a countable number of classes with complex multiplication. Hence the orbit Aut(C) · j(Λ) is at most countable.

Step 6: If C/Λ has complex multiplication, then j(Λ) ∈ Q.

Contropositive of Step 3 is that Aut(C) · z is countable implies that z ∈ Q. And this is what we just proved about j(Λ).

3.3.2 j(z) is integral { ( ) } def a b Fix n ∈ and let A = α = ad = n, 0 ≤ b < d, (a, b, d) = 1 Z>1 0 d then we have a coset decomposition

(n 0) ⨆ SL ( ) SL ( ) = SL ( )α (3.3.1) 2 Z 0 1 2 Z 2 Z α∈A ∏ and consider the polynomial Fn(X, j) = (X − j ◦ α). Recall the q-expansion α∈A 1 ∑ n 2πiz of j, j(z) = q + a(n)q with a(n) ∈ Z, q = e . Hence for α ∈ A n≥0

( (a b) ) az + b ∑ j(αz) = j · z = j( ) = ξ−bq−a/d + a(n)ξnbqna/d (3.3.2) 0 d d d d n≥0 3.3. INTEGRALITY OF THE J-INVARIANT 72

2πi/d 1/d where ξd = e . Therefore j(αz) has a q-expansion in q with coefficients powers of ξd. Since d | n, these are algebraic integers of Q(ξn). In other words we 1/n have that Fn(X, j) ∈ Z[ξn]((q ))[X]. And we claim that actually Fn(X, j) ∈ Z[X, j]. ∼ Step 1: The coefficients of Fn(X, j) are invariant under the action of Gal(Q(ξn)/Q) = (Z/nZ)∗ =: G, hence they belong to Z((q1/n)). t Let σ ∈ G, then σ(ξn) = ξn for some t such that (n, t) = 1. Trasforming the (a b′) coefficients of j◦α by σ in (3.3.2) we obtain j◦β for some β = ∈ A with 0 d b′ ≡ bt mod d. Since α ↦→ β gives a permutation of the set A, the q-expansion in (3.3.2) has coefficients in Z. ∗ Step 2: Let f be a meromorphic function of SL2(Z)\H holomorphic on SL2(Z)\H whose q-expansion has integral coefficients. Then f ∈ Z[j]. ∗ Proceed by induction on n = −ord∞(f) = −ord0(f ). ∗ If n ≤ 0 then f is holomorphic at ∞ thus holomorphic on SL2(Z)\H which is compact. Therefore f is constant and its q-expansion has integral coefficients, hence f ∈ Z. −(n+1) ∑ k n+1 Assume true for n: let f = c−(n+1)q + ckq . Then ord∞(f−c−(n+1)j ) ≥ k≥n n+1 n+1 n, thus by inductive step g = f−c−(n+1)j ∈ Z[j]. Hence f = g+c−(n+1)j ∈ Z[j].

We can now conclude, in fact the coefficients of Fn(X, j) are meromorphic ∗ functions on SL2(Z)\H since they are SL2(Z)-invariant and holomorphic on SL2(Z)\H. By Step 1 every coefficient has q-expansion in Z, thus they lie in Z[j] by Step 2. def Lemma 3.37. If n is not a square, then the leading coefficients of Hn(X) = Fn(X,X) ∈ Z[X] is ±1. Proof. n not a square implies that in (3.3.2) a/d ̸= 1 (as ad = n). Therefore the leading coefficient of j − j ◦ α is a root of unity, and so is the leading coefficient of Hn(j). But since it is also rational, it has to be ±1.

Finally, let z ∈ H be such that Q(z) = K is a quadratic imaginay field, so ∼ that E = C/(Zz + Z) has complex multiplication by some order O ⊂ OK. Let us first assume that O = OK. Then there exists µ ∈ OK such that NormK/Q(µ) = n > 0 is square-free. In fact if√ K = Q(i), take µ = 1+i, to get Norm(µ) = µ·√µ = (1 + i)(1 − i) = 2; if K = Q( −m) for some square-free m > 1, let µ = i m, then Norm(µ) = µ · µ = m. In both cases, Norm(µ) is a square-free integer and 4 2 clearly µ ∈ OK since it is a root of the monic polynomial X +4 and X +NK/Q respectively. Moreover, since µ is imaginary, it actually belongs to OK r Z. Since µ ∈ OK r Z, µ ∈ End(E), and the multiplication by µ sends the lattice Λ = Zz + Z to a sublattice. Namely there are a, b, c, d ∈ Z such that µz = az + b and µ = cz + d (∗)

and we can define ξ ∈ M2(Z) by the relation (z) (a b)(z) (z) µ = =: ξ 1 c d 1 1 3.3. INTEGRALITY OF THE J-INVARIANT 73

Now notice that µ and µ are eigenvalues for ξ: in fact µ is an eigenvalue for how ξ is defined; µ is an eigenvalue since by (∗)

(z) (z) µ = ξ 1 1

Therefore (µ 0) det(ξ) = det = µµ = n = Norm(µ) 0 µ which, by the choice of µ, is square-free. (n 0) Since detξ = n, ξ ∈ SL ( ) SL ( ) Thus there exists α ∈ A such that 2 Z 0 1 2 Z

j(αz) = j(ξz) = j(z) and the second equality follows by the fact that ξ · z = z. So that

0 = Fn(j(z), j(αz)) = Fn(j(z), j(ξz)) = Fn(j(z), j(z)) = Hn(j(z)) (3.3.3)

Since n is square-free, by Lemma 3.37, the leading coefficient of Hn is ±1 and we conclude that j(z) is an algebraic integer.

+ Now if O $ OK, then by Proposition 2.31 there exists β ∈ GL2 (Q) ∩ M2(Z) ′ ′ such that for z = β(z) we have End(C/(Zz + Z)) = OK. Therefore j(z) is integral over Z[j(z′)] and by the previous argument j(z′) is integral. Therefore j(z) is. Chapter 4

Hecke Operators

In this chapter, we will introduce the action of Hecke operators on modular forms, modular curves and integral homology, and at the end of the chapter we will see how to associate a L-function to a cusp form.

4.1 The Hecke ring

Let Γ1 and Γ2 be congruence subgroups of SL2(Z). Then for any α ∈ + GL2 (Q) consider the double coset Γ1αΓ2 and denote R1,2 the Z-module generated by these double cosets. + Lemma 4.1. If Γ is a congruence subgroup of SL2(Z) and α ∈ GL2 (Q), then −1 the intersection αΓα ∩ SL2(Z) is a congruence subgroup of SL2(Z). Proof. Since αΓ(N)α−1 = (bα)Γ(N)(bα)−1 for all b ∈ Q∗, we can assume that α has integer coefficients, i.e. belongs to M2(Z). Now if d = det(α), then we claim that Γ(Nd) ⊆ αΓ(N)α−1 ⊆ αΓα−1 ′ −1 In fact, define α = dα ∈ M2(Z) and let γ ≡ 1 mod Nd (i.e. γ ∈ Γ(Nd)), then (d 0) α′γα ≡ α′α = mod (Nd) 0 d −1 −1 Hence α γα ≡ 1 mod N, which in particular implies that α γα ∈ M2(Z). Moreover det(α−1γα) = 1, so that α−1γα ∈ Γ(N) and we conclude that γ ∈ αΓα−1.

Proposition 4.2. If Γ1 and Γ2 are congruence subgroups of SL2(Z), then + Γ1\Γ1αΓ2 and Γ1αΓ2/Γ2 are finite for all α ∈ GL2 (Q). −1 Proof. Let Γ3 = α Γ1α ∩ Γ2 ⊂ G2. Then, the map given by [αγ] ↦→ [γ] is a bijection between Γ1\Γ1αΓ2 and Γ3\Γ2. Since Γ2 (by assumption) and Γ3 (by Lemma 4.1) are congruence subgroups, the quotient Γ3\Γ2 is finite. The ”dual” argument holds for Γ1αΓ2/Γ2. It follows that we can write n m ⨆ ⨆ ′ Γ1αΓ2 = Γ1αi = αjΓ2 (4.1.1) i=1 j=1

74 4.1. THE HECKE RING 75

Now let us consider the special case Γ1 = Γ2

Definition. Let Γ be a congruence subgroup of SL2(Z) and let ∆ be a semi- + group such that Γ ⊆ ∆ ⊆ M2 (Z) = {γ ∈ M2(Z) | det(γ) > 0}. Then define the Hecke ring with respect to Γ and ∆ to be { } ∑ R(Γ, ∆) = ck · ΓγkΓ | ck ∈ Z, γk ∈ ∆ k For every ΓγΓ ∈ R(Γ, ∆) define its degree deg(ΓγΓ) as the number of cosets Γξ contained in ΓγΓ, and extend the definition to the whole ring by Z-linearity. n ⨆ Let us define a composition law on R(Γ, ∆) as follows: consider ΓαΓ = Γαi i=1 m ⨆ and ΓβΓ = Γβj in R(Γ, ∆), then j=1

n m ∑ ∑ ∑ (ΓαΓ)(ΓβΓ) = Γαiβj = ckΓγkΓ (4.1.2) i=1 j=1 where the second sum is over all distinct double cosets R(Γ, ∆) ∋ ΓγkΓ ⊆ ΓαΓβΓ and

#{(i, j) | ΓγkΓ = ΓαiβjΓ} ck = #{(i, j) | Γγk = Γαiβj} = (4.1.3) #(Γ\ΓγkΓ) (For the second equality in (4.1.3) see [Shi73] Proposition 3.2) Proposition 4.3. The composition law defined in (4.1.2) is well-defined and associative.

Proof. To see that such law is well-defined we need to show that each ck depends only upon the double cosets ΓαΓ, ΓβΓ and ΓγkΓ and not on the choice of the representatives {αi}, {βj} and γk. Choosen γk, we have Γαiβj = Γγk if and only −1 if Γαi = Γγkβj and for a given index j, the last equality holds exactly for one i. It follows that

−1 #{(i, j) | Γγk = Γαiβj} = #{j | γkβj ∈ ΓαΓ} =

−1 −1 = #{j | βj ∈ Γα Γγk} = #{j | Γβj ⊂ Γα Γγk} −1 Since this last quantity coincides with the number of cosets Γξ in ΓβΓ∩Γα Γγk, ′ it is indipendent of the representatives {αi} and {βj}. Now if ΓγkΓ = ΓγkΓ, ′ −1 −1 ′ then γk = ϑγkδ for some ϑ, δ ∈ Γ, hence ΓβjΓ∩Γα Γγk = (ΓβjΓ∩Γα Γγk)δ, so that it is also indipendent of γk. For associativity see [Shi73] Proposition 3.4.

+ Lemma 4.4. Let Γ be a congruence subgroup of SL2(Z). Let α ∈ GL2 (Q) be such that #(Γ\ΓαΓ) = #(ΓαΓ/Γ), then there exists a common set of representatives for the two quotients, in other words there exists {αi} such that

n n ⨆ ⨆ ΓαΓ = Γαi = αiΓ i=1 i=1 4.1. THE HECKE RING 76

n n ⨆ ⨆ Proof. Let βi and δi be such that ΓαΓ = Γβi = δiΓ. It is enough to show i=1 i=1 that Γβi ∩ δiΓ ̸= ∅ for all indeces i, in fact if that is the case, then we can pick αi ∈ Γβi ∩ δiΓ for every i and obtain the result since Γβi = Γαi and δiΓ = αiΓ. ⨆ Now assume that Γβi ∩ δiΓ = ∅ for some i, then Γβi ⊂ δjΓ and therefore j̸=i ⨆ ΓαΓ = δjΓ which is a contradiction. j̸=i Proposition 4.5. Let Γ and ∆ be as above. If there exists an anti-involution ϕ : ∆ −→ ∆ such that ϕ(Γ) = Γ and ΓαΓ = Γϕ(α)Γ for all α ∈ ∆, then R(Γ, ∆) is a commutative ring with unit Γ1Γ = Γ.

n ⨆ Proof. Let α ∈ ∆ and ΓαΓ = Γβi. Since i=1

n n ⨆ ⨆ ΓαΓ = Γϕ(α)Γ = ϕ(ΓαΓ) = ϕ( Γβi) = ϕ(βi)Γ i=1 i=1 it follows that #(Γ\ΓαΓ) = #(ΓαΓ/Γ), so that we can apply Lemma 4.4. Now let α, β ∈ ∆, then we can write

n n m m ⨆ ⨆ ⨆ ⨆ ΓαΓ = Γαi = αiΓ and ΓβΓ = Γβj = βjΓ i=1 i=1 i=1 i=1 Therefore

∑ ∑ ′ (ΓαΓ)(ΓβΓ) = ckΓγkΓ and (ΓαΓ)(ΓβΓ) = ckΓγkΓ with the same components ΓγkΓ. Now using the characterization on the right ′ hand side of (4.1.3) we see that ck = ck.

4.1.1 The structure of R(Γ, ∆) In this section our goal will be to determine the structure of R(Γ, ∆). + Assume at first that Γ= SL2(Z) and ∆ = M2 (Z). By the theory of elementary ( ) a1 0 divisors, the diagonal matrices = diag(a1, a2), with ai ∈ Z>0 such 0 a2 that a1 divides a2, form a set of representatives for Γ\∆/Γ. In particular trans- position on matrices is an anti-automorphism which fixes every double coset ΓαΓ since we can assume α to be diagonal. Therefore, in view of Proposition 4.5, R(Γ, ∆) is a commutative ring and generated by the elements of the form T (a1, a2) = ΓαΓ with α = diag(a1, a2), ai ∈ Z>0, a1 | a2. To obtain informations regarding how elements multiply in the ring, our strat- egy will be to assign a lattice to each coset Γα and then to count the number of lattices instead of the number of cosets. 2 Consider Q as the space of row-vectors, then GL2(Q) acts on the right by multiplication. A submodule Λ of Q2 is a lattice in Q2 if finitely generated over Z 2 and Λ ⊗Z Q = Q . We can restate the fundamental theorem of elementary divisors in terms of lattices: 4.1. THE HECKE RING 77

2 2 Fact 4.6. Let Λ1, Λ2 be lattices in Q . Then there exist vectors u1, u2 ∈ Q and scalars b1, b2 ∈ Q such that Λ1 = Zu1 + Zu2,Λ2 = Zb1u1 + Zb2u2, with b2 ∈ b1Z.

The set {b1, b2} is called set of elementary divisors of Λ2 with respect to Λ1 and we write {Λ1 :Λ2} = {b1, b2}

In particular, Λ2 ⊆ Λ1 if and only if b1, b2 ∈ Z, and then [Λ1 :Λ2] = b1b2; ( ) b1 0 moreover if α = then {Λ1 :Λ1α} = {b1, b2}. 0 b2 Let us write Λ = Z2, then by Lemma 2.19,

Γ = SL2(Z) = {γ ∈ GL2(Q) | Λγ = Λ, det(γ) > 0} Remark 4.7. Notice for α, β ∈ ∆,

Λα = Λβ ⇐⇒ Λ = Λβα−1 ⇐⇒ βα−1 ∈ Γ (since det(βα−1) > 0)

⇐⇒ β ∈ Γα ⇐⇒ Γβ = Γα

2 Lemma 4.8. Let Λ1 and Λ2 be lattices in Q . Then {Λ:Λ1} = {Λ:Λ2} if and only if there exists α ∈ Γ such that Λ1α = Λ2.

2 Proof. Let {Λ:Λ1} = {Λ:Λ2} = {a1, a2}, then there exist u1, u2, v1, v2 ∈ Q such that Λ = Zu1 + Zu2 = Zv1 + Zv2,Λ1 = Za1u1 + Za2u2 and Λ2 = Za1v1 + Za2v2. Then define α ∈ GL2(Q) by uiα = vi for i = 1, 2. Then Λα = Λ and Λ1α = Λ2 however det(α) = ±1. If det(α) = 1 we are done, otherwise replace v1 with −v1. For the converse we have that

{Λ:Λ2} = {Λ:Λ1α} = {Λα :Λ1α} = {Λ:Λ1}

Lemma 4.9. Let ΓαΓ = T (a1, a2), then Γξ ↦→ Λξ is a one-to-one correspon- ′ ′ dence between cosets Γξ in T (a1, a2) and lattices Λ such that {Λ:Λ } = {a1, a2}

Proof. We can assume α = diag(a1, a2). Γξ ⊂ ΓαΓ, thus there exists some δ ∈ Γ such that Γξ = Γαδ, and consequentely:

{Λ:Λξ} = {Λ:Λαδ} = {Λ:Λα} = {a1, a2}

′ Viceversa, if {Λ:Λ } = {a1, a2} then by Lemma 4.8 there exists γ ∈ Γ such that Λ′ = Λαγ and Γαγ ⊂ ΓαΓ. The correspondence is one-to-one by Remark 4.7.

′ ′ In particular it follows that degT (a1, a2) = #{Λ | {Λ:Λ } = {a1, a2}}. ∑ Proposition 4.10. Let (ΓαΓ)(ΓβΓ) = ck(ΓγkΓ) with ck ∈ Z. Then ck corresponds to the number of lattices Λ′ such that

′ ′ {Λ:Λ } = {Λ:Λβ} and {Λ :Λγk} = {Λ:Λα} 4.1. THE HECKE RING 78

⨆ ⨆ Proof. Let ΓαΓ = Γαi and ΓβΓ = Γβj. Then by (4.1.3) and Lemma 4.9 i j

ck = #{(i, j) | Γαiβj = Γγk} = {(i, j) | Λαiβj = Λγk}

′ ′ Assume Λαiβj = Λγk and define Λ = Λβj, then {Λ:Λ } = {Λ:Λβ} and ′ ′ {Λ :Λγk} = {Λβj :Λαiβj} = {Λ:Λαi} = {Λ:Λα}. Viceversa, let Λ be a ′ ′ ′ lattice such that {Λ:Λ } = {Λ:Λβ} and {Λ :Λγk} = {Λ:Λα}, then Λ = Λβj −1 for one and only one index j, so that {Λ:Λγkβj } = {Λ:Λα}. By Lemma −1 ′ 4.9 Λγkβj = Λαi for some i. Therefore each Λ determines a pair (i, j) and viceversa. Remark 4.11. Notice that from the definition of the multiplicative law in R(Γ, ∆), it immidiatly follows that T (n, n)T (a1, a2) = T (na1, na2) for any positive integer n. In particular T (n, n) is not a zero-divisor in R(Γ, Λ). Definition. The elements T (n, n) ∈ R(Γ, ∆) are called diamond operators.

Proposition 4.12. For any T (a1, a2),T (b1, b2) ∈ R(Γ, ∆) such that (a2, b2) = 1 we have T (a1, a2)T (b1, b2) = T (a1b1, a2b2)

Proof. Since T (a1, a2) = T (a1, a1)T (1, a) for a = a2/a1 and T (b1, b2) = T (b1, b1)T (1, b) for b = b2/b1 and the ring R(Γ, ∆) is commutative, we reduce to check the equality T (1, a)T (1, b) = T (1, ab) for a, b coprime. (1 0) (1 0) Let α = and β = and suppose that T (1, a)T (1, b) = ∑ c (Γγ Γ). 0 a 0 b k k Claim 1: ck = 1 for any γk. Apply Proposition 4.10 and consider Λ1 and Λ2 such that

{Λ:Λ1} = {Λ:Λ2} = {Λ:Λβ} and {Λ1 :Λγk} = {Λ2 :Λγk} = {Λ:Λα}

Then [Λ1 + Λ2 :Λ1] = [Λ2 :Λ1 ∩ Λ2]. However the left hand side divides [Λ : Λ1] = b since Λ1 + Λ2 ⊂ Λ, and the right hand side divides [Λ : Λα] = a since Λγk ⊂ Λ1 ∩ Λ2. But a and b are coprime, therefore Λ1 + Λ2 = Λ1 and Λ1 ∩ Λ2 = Λ2, hence Λ1 = Λ2 and we conclude that ck = 1. Claim 2: The only double coset contained in ΓαΓβΓ is ΓαβΓ. s ⨆ Write ΓαΓβΓ = ΓγiΓ. We want to show that s = 1, which is equivalent to i=1 say that for any ξ1, ξ2 ∈ ΓαΓβΓ we have Γξ1Γ = Γξ2Γ. Let ξ1, ξ2 ∈ ΓαΓβΓ, then Λξi ⊂ Λi ⊂ Λ for some Λi as in Claim 1 and ∼ ∼ ∼ Λ/Λξ1 = Λ/Λ1 ⊕ Λ1/Λξ1 = Λ/Λα ⊕ Λ/Λβ = Λ/Λξ2 where the second isomorphism follows from coprimality of a and b. Therefore ξ1 and ξ2 have the same set of elementary divisors. Hence the equality Γξ1Γ = Γξ2Γ and s = 1. Since ΓαβΓ ⊆ ΓαΓβΓ, they have to coincide. As a consequence every element of R(Γ, ∆) is the product of elements of the form T (1, pr) and T (p, p) with p rational primes and r ≥ 1. Therefore understanding the structure of R(Γ, ∆) reduces to understand the one of Rp: the subring generated by T (1, pr),T (p, p), r ≥ 1. Lemma 4.13. T (1, p) and T (p, p) are algebraically indipendent. 4.1. THE HECKE RING 79

Proof. Assume notand let Q(X,Y ) ∈ Z[X,Y ] be a non zero polynomial such that Q(T (1, p),T (p, p)) = 0. Factoring the lowest power of T (p, p) we can write

Q(T (1, p),T (p, p)) = T (p, p)mQ˜(T (1, p),T (p, p))

Which implies Q˜(T (1, p),T (p, p)) = 0 since T (p, p) is not a zero divisors. Now write

n 0 = Q˜(T (1, p),T (p, p)) = a0Q0(T (1, p))+a1Q1(T (1, p))T (p, p)+..+anQn(T (1, p))T (p, p)

Therefore aiQi(T (1, p)) = 0 for all i = 0, .., n. Since T (1, p) is a transcendental element this implies that aiQi(X) = 0, and consequentely Q = 0 contradicting our initial assumption.

Theorem 4.14. The ring Rp is the polynomial ring Z[T (1, p),T (p, p)]. r Proof. Since Rp is generated by T (p, p) and T (1, p ), for any r ≥ 1, it is enough to show that T (1, pr) is a polynomial in T (p, p),T (1, p) for any r. Proceed by induction: (1 0) • T (1, p)2 = (Γ Γ)2 = ∑ c (Γγ Γ). Now notice that we are looking 0 p k k 2 for γk’s diagonal, with a1 | a2 and necessarely det(γk) = p . Therefore the { 2 } only possibilities are γk ∈ diag(1, p ), diag(p, p) . (1 0 ) By (4.1.3) let’s compute multiplicities and show that for γ = , k 0 p2 ck = 1: { Λ′ | {Λ:Λ′} = {Λ:Λ · diag(1, p)} } c = # = 1 k & {Λ′ :Λ · diag(1, p2)} = {Λ:Λ · diag(1, p)}

This is enough to prove that T (1, p2) is a polynomial in T (1, p) and T (p, p). • Assume that T (1, pk) is a polynomial is T (1, p) and T (p, p) for any k ≤ r and let us prove it for r+1. Similarly as above, we have that T (1, pr)T (1, p) = ∑ r+1 ck(ΓγkΓ) with γk diagonal and necessarely det(γk) = p . In particular γ = diag(1, pr+1) appears in the sum and its multiplicity is 1 by (4.1.3). i j All the remaining factors ΓγkΓ are of the form T (p , p ) with 1 ≤ i ≤ j, i + j = r + 1 and therefore we can factor them as T (p, p)T (pi, pj) with 0 ≤ i ≤ j, i + j = r − 1. Since by inductive hypotesis they are polynomials in T (p, p),T (1, p) as well as T (1, pr)T (1, p), the result follows.

Corollary 4.15. We can conclude that

R(Γ, ∆) = Z[T (1, p),T (p, p) | p rational prime] In particular R(Γ, ∆) is an integral domain. Definition. For any integer n ≥ 1, let ∆(n) = {α ∈ ∆ | det(α) = n} and define ∑ T (n) = ΓαΓ α∈∆(n) called Hecke operator. 4.1. THE HECKE RING 80

Lemma 4.16. T (n) belongs to R(Γ, ∆). Proof. We need to see that the sum is finite. We claim that there is a one-to- {(a b) } one correspondence between the set C(n) = ∈ ∆(n) 0 ≤ b < d and 0 d Γ\∆(n). (a b) • Surjectivity: Let A = ∈ ∆(n), we want to show that there exists c d (α β) γ = ∈ Γ such that γA ∈ C(n). Consider two different cases: θ δ – If c = 0 we have ad = n and in particular a, d ̸= 0. (α β)(a b) (aα bα + dβ) = θ δ 0 d aθ bθ + dδ To belong to C(n), take θ = 0 hence αδ = 1 and dδ > 0, thus let δ = sgn(d) = α. We are left to show that there exists β for which 0 ≤ sgn(d) · b + dβ < sgn(d) · d. Consider the Euclidian division of sgn(d) · b by sgn(d) · d with remainder 0 ≤ r < |d| so that sgn(d) · b = sgn(d) · d · q + r β = −sgn(d) · q satisfies 0 ≤ sgn(d) · b + dβ = r < sgn(d) · d = |d| as required a′ a ′ ′ – If c ̸= 0, let c′ = c with (a , c ) = 1. Then there exist α, β ∈ Z such that αa′ + βc′ = 1: therefore ( α β )(a b) (aα + cβ bα + dβ ) = =: T ∈ ∆(n) −c′ a′ c d 0 a′d − bc′ (z u) Let us now show that there exist γ′ = ∈ Γ such that γ′T ∈ v w (AB) C(n). Write T = to simplify notation, then 0 D (zA zB + uD) γ′T = vA vB + wD (|A| B + uD) Choosing v = 0 and z = w = sgn(D), gives γ′T = . 0 |D| We are left to show that there exists u ∈ Z such that 0 ≤ B + uD < |D|. Again by mean of the Euclidean division we have B = QD + R (a b) with 0 ≤ R < |D|. Thus take u = −Q and this yields that γ′γ c d belongs to C(n). This proves that C(n) surjects to Γ\M ∗(n). • Injectivity: Suppose that A, A′ ∈ C(n) are in the same class, so that there (α β) exists γ = for which θ δ (a b) (α β)(a′ b′ ) (a′α b′α + d′β) A = = = = γA′ 0 d θ δ 0 d′ a′θ b′θ + d′δ 4.1. THE HECKE RING 81

This equality forces θ = 0 hence αδ = 1. Since d, d′ > 0 from d = d′δ follows that δ has positive sign, thus δ = 1 = α. Therefore a = a′, d = d′. Now b = b′ + d′β and by hypothesis 0 ≤ b, b′ < d = d′ hence

′ ′ ′ |d β| = |b − b | < d =⇒ β = 0 (sinceβ ∈ Z) i.e. b = b′ which gives A = A′.

Proposition 4.17. (a) For any p prime we have T (p) = T (1, p). (b) If n, m are coprime then T (n)T (m) = T (nm). (c) For any n ≥ 1 we have ∑ T (n) = T (a, d) ad=n a|d

(d) For any r ≥ 2 we have T (1, pr) = T (pr) − T (p, p)T (pr−2).

r ∑ i i i (e) For any r ≤ s we have Tpr Tps = p T (p , p )Tpr+s−2i , and in particular i=0

T (p)T (pr) = T (pr+1) + pT (p, p)T (pr−1) (4.1.4)

Proof. (a) It is immidiate from the definition of T (p). (b) Follows from the definition and Proposition 4.12. (c) ∑ ∑ (a 0) ∑ T (n) = ΓαΓ = Γ Γ = T (a, d) 0 d α∈∆(n) ad=n ad=n a|d a|d where the second equality follows from the theory of elementary divisors. (1 0 ) (d) We have the three equalities: T (1, pr) = Γ Γ; 0 pr

∑ ∑ (pi 0 ) (1 0 ) (p 0 ) (ph 0 ) T (pr) = ΓαΓ = Γ Γ = Γ Γ+Γ Γ+..+Γ Γ 0 pj 0 pr 0 pr−1 0 pr−h α∈∆(pr ) i+j=r 0≤i≤j

where h ∈ Z>0 is such that h ≤ k − h and k − h − 1 < h + 1. ( (p 0) ) ( ∑ (pi 0 ) ) ∑ (pi 0 ) T (p, p)T (pr−2) = Γ Γ · Γ Γ = Γ Γ = 0 p 0 pj 0 pj i+j=r−2 i+j=r 0≤i≤j 1≤i≤j (p 0 ) (p 0 ) (ph 0 ) Γ Γ + Γ Γ + .. + Γ Γ 0 pr−1 0 pr−1 0 pr−h where the second equality follows from Remark 4.11. Putting these three formula together we obtain (b). 4.1. THE HECKE RING 82

(e) See [Shi73] Theorem 3.24 (4).

{ ( ) } def ∗ ∗ Now let ∆′ = ∆ = α ∈ M ( ) | det(α) > 0, α ≡ mod N and N 2 Z 0 ∗ (n 0) Γ′ = Γ (N). Let T ′(n, n) = Γ′ Γ′ and define the Hecke operators: 0 0 n Definition. For any integer n ≥ 1 define ∑ T ′(n) = Γ′αΓ′ ∈ R(Γ′, ∆′) α∈∆′(n)

{(a b) } Lemma 4.18. The C (n) = ad = n, 0 ≤ b < d, (a, N) = 1 is a N 0 d complete set of coset representatives for the right coset Γ0(N) on ∆N (n) = {α ∈ ∆N | det(α) = n} Proof. An analogous argument as in Lemma 4.16 gives the result. With some more work, one can obtain a structure theorem for R(Γ′, ∆′): namely

Theorem 4.19. R(Γ′, ∆′) is the polynomial ring generated over Z by the elements ′ ′ ′ ′ T (1, p),T (p, p) ∀p - N and T (p) = T (1, p) ∀p | N Moreover, these elements are algebraically indipendent. In other words R(Γ′, ∆′) is the homomorphic image of R(Γ, ∆) via the map ⎧ T (n) ↦→ T ′(n) ∀n ⎨⎪ T (p, p) ↦→ T ′(p, p) ∀p - N ⎩⎪T (p, p) ↦→ 0 ∀p | N

Proof. See [Shi73] Theorem 3.34.

Corollary 4.20. (a) For any prime p - N, and r ≥ 1: T (pr)T (p) = T (pr+1) + pT (p, p)T (pr−1)

(b) For any prime p | N, and r ≥ 1:

T (pr) = T (p)r

(c) T (n)T (m) = T (nm) if (n, m) = 1. Proof. All the relations immidiatly follows from Proposition 4.17 and Theorem 4.19. 4.2. ACTION ON MODULAR FUNCTIONS 83

4.2 Action on modular functions

+ Let k ∈ Z and recall the action of GL2 (C) on the space of modular functions for some congruence subgroup, as in (3.2.2), given by

k−1 −k (σ, f) ↦→ f|[σ]k(τ) = det(σ) f(σ(τ))j(σ, τ) (4.2.1)

Let Γ1 and Γ2 be congruence subgroups of SL2(Z). Then we have a map

Mk(Γ1) −→ Mk(Γ2) given by f ↦→ fρ for each ρ ∈ R1,2. + ⨆ Explicitly if ρ = Γ1αΓ2 for α ∈ GL2 (Q) such that Γ1αΓ2 = Γ1αj, we have

n k/2−1 ∑ f ↦→ f|[Γ1αΓ2] = det(α) f|[αj]k (4.2.2) j=1

Firstly, let us show that f|[Γ1αΓ2]k is a weakly modular form of weight k with n ⨆ respect to Γ2. Let γ ∈ Γ2, then det(γ) = 1 and Γ1αΓ2 = αjγ, therefore j=1

n n k/2−1 ∑ k/2−1 ∑ f|[Γ1αΓ2]k[γ]k = det(α) f|[αj]k[γ]k = det(α) f|[αjγ]k = f|[Γ1αΓ2]k j=1 j=1

We are left to see that f|[Γ1αΓ2] is holomorphic at the cusps of Γ2. For γ ∈ n k/2−1 ∑ SL2(Z) we have f|[Γ1αΓ2]k[γ]k = det(α) f|[αjγ]k, but since f is holo- j=1 morphic at the cusps, each f|[αjγ]k is holomorphic at ∞ therefore the whole sum is. Moreover by the last formula also follows that if f ∈ Sk(Γ1) then f|[Γ1αΓ2]k ∈ Sk(Γ2). Then (4.2.2) extends Z-linearly to R1,2.

In what follows we will focus on the case Γ1 = Γ2, so that we have an action of R(Γ, ∆) on Mk(Γ), and in particular on Sk(Γ). Before doing so, let us introduce an inner product in the space of cusp forms:

Definition. Let Γ be a congruence subgroup of SL2(Z), define the Petersson inner product ⟨· , ·⟩ : Sk(Γ) × Sk(Γ) −→ C 1 ∫ (f, g) ↦→ f(τ)g(τ)Im(τ)kdµ(τ) V (Γ) XΓ where dµ is the measure on XΓ induced by the SL2(Z)-invariant hyperbolic dxdy measure y2 on H (see [Kna93] 8.12) and V (Γ) is the (finite) volume of XΓ, Vol(XΓ) (see [Miy89] 1.9.1). Notice that even though f and g are not necessarely Γ-invariant, the integral is well-defined since f(τ)g(τ)Im(τ)k and dµ(τ) are. It follows from its definition that the Petersson inner product is Hermitian and positive definite. 4.2. ACTION ON MODULAR FUNCTIONS 84

4.2.1 Hecke operators on SL2(Z)

Let us consider the previous setting with Γ = SL2(Z) = Γ1 = Γ2, and focus µ on the action of the Hecke operators as in (4.2.2). Let {αi}i=1 be a complete set of representatives for the right cosets Γα of Γ on ∆(n) and let f ∈ Mk(Γ) then µ k/2−1 ∑ Tk(n)f = n f|[αi]k (4.2.3) i=1 {(a b) } We saw in Lemma 4.16 that C(n) = ad = n, 0 ≤ b < d is a complete 0 d set of coset representatives. Therefore explicitly, for every αi ∈ C(n), we have (aτ + b) f|[α ](τ) = nk/2f d−k (4.2.4) i d

∑ (aτ + b) Hence T (n)f(τ) = nk−1 f d−k (4.2.5) k d ad=n 0≤b

d−1 { ∑ d if d | n e2πinb/d = (4.2.6) 0 if d n b=0 -

Proof. If d | n then let a = n/d ∈ Z and therefore e2πia = 1. Since we have d summands the result follows. If d - n then e2πin/d ̸= 1 and

d−1 d−1 ( 2πin/d)d ∑ ∑ b 1 − e e2πinb/d = (e2πin/d) = = 0 1 − e2πin/d b=0 b=0

∞ ∑ m Proposition 4.22. Let f ∈ Mk(Γ) have q-expasion f(τ) = cmq . Then m=0 the q-expasion of Tk(n)f is

∞ ∑ ℓ Tk(n)f(τ) = bℓq ℓ=0

∑ k−1 ∑ k−1 where b0 = c0 d ; b1 = cn and bℓ = a cnℓ/a2 if ℓ > 1. 0

∞ ∑ (aτ + b) ∑ ∑ T (n)f(τ) = nk−1 f d−k = nk−1 c e2πin(aτ+b)/dd−k = k d m ad=n m=0 ad=n 0≤b

∞ ∞ ∞ k−1 ∑ ∑ ma −k+1 ∑ ∑ k−1 ma ∑ ℓ = n cmdq d = cmn/aa q = bℓq m=0 ad=n m=0 00 4.2. ACTION ON MODULAR FUNCTIONS 85

Let us compute the coefficient of powers of q on the left hand side: b0 is ∑ k−1 1 c0a ; the coefficient of q comes form a = m = 1 so that b1 = cn; 0

∑ m In particular for f(τ) = cmq and p prime, Tk(p)f(τ) has q-expansion m≥0 with coefficients { cpm if p - n bm = k−1 cpm + p cm/p if p | n

Remark that for f ∈ Mk(Γ), and T (n, n), (4.2.2) gives:

f|T (p, p) = pk−2f (4.2.7)

Therefore by (4.1.4), as an operator on Mk(Γ)

T (pr)T (p) = T (pr+1) + pk−1T (pr−1) ∀p prime, ∀r ≥ 1 k k k k (4.2.8) Tk(m)Tk(n) = Tk(mn) if (m, n) = 1

Theorem 4.23. (Petersson) The Hecke operators Tk(n) on Sk(Γ) are normal with respect to the Petersson inner product.

+ ∗ −1 Proof. One first proves that for all α ∈ GL2 (Q) holds [ΓαΓ]k = [Γ(detα)α Γ]k where ∗ denotes the adjoint operator. Then for every α in ∆(n)

ΓαΓ = Γ(detα)α−1Γ since α and (detα)α−1 have the same set of elementary divisors. Therefore the Hecke operators are self-adjoint. Since they also commute with each other, they are normal.

Theorem 4.24. There exists a basis of Sk(Γ) of simultaneous eigenvectors for the operators T (n). Proof. It follows immediately from the complex spectral theorem.

Proposition 4.25. Let f ∈ Sk(Γ) be a simultaneous eigenvector for Tk(n) such ∑ n that Tk(n)f = λ(n)f. If f has q-expansion at infinity f(τ) = cnq , then n≥1

cn = λ(n)c1

Proof. From Proposition 4.22 b1 = cn, on the other hand from being an eigenvector for Tk(n) follows that b1 = λ(n)c1. Since f is an eigenvector for every Hecke operator, the equality cn = λ(n)c1 holds for every n. Remark 4.26. As consequences we can see that

(i) f ̸= 0 implies c1 ̸= 0. (ii) The eigenvalues λ(n)’s determines f up to a scalar factor. 4.2. ACTION ON MODULAR FUNCTIONS 86

(iii) The previous point allows us to normalize f so that its q-expansion has c1 = 1, and therefore cn is the eigenvalue for Tk(n). (iv) In particular if f is normalized, from (4.2.8) follows that:

k−1 • For any p prime, cpr cp = cpr+1 + p cpr−1 .

• If (m, n) = 1 then cncm = cnm.

4.2.2 Hecke operators on congruence subgroups ′ ′ Fix a positive integer N and let Γ = Γ0(N),∆ = ∆N (n). Then for f ∈ ′ Mk(Γ ), (4.2.2) gives :

′ k/2−1 ∑ Tk(n)f = n f|[α]k (4.2.9)

α∈CN (n)

′ ′ ′ and recall that Tk(n) carries Mk(Γ ) (resp. Sk(Γ )) to itself. ∞ ′ ∑ n Proposition 4.27. Let f ∈ Mk(Γ ) have q-expansion at infinity f(τ) = cnq , m=0 ′ then Tk(n)f has q-expansion

∞ ′ ∑ ℓ Tk(n)f(τ) = bℓq ℓ=0

∑ k−1 ∑ k−1 where b0 = c0 a ; b1 = cn, and bℓ = a cnℓ/a2 if ℓ > 1. 0

′ r ′ ′ r+1 k−1 ′ r−1 Tk(p )Tk(p) = Tk(p ) + p Tk(p ) ∀p - N, ∀r ≥ 1 T ′(pr) = T ′(p)r ∀p | N, ∀r ≥ 1 (4.2.10) ′ ′ ′ Tk(m)Tk(n) = Tk(mn) if (m, n) = 1 We would now want to obtain a similar result as in Theorem 4.23, however: Theorem 4.28. (Petersson) For every positive integer n such that (n, N) = 1, the Hecke operators T ′(n) are self adjoint with respect to the Petersson inner ′ product in Sk(Γ ). In fact the same argument used to prove Theorem 4.23 only holds true if n ′ is coprime with N, otherwise not necessarely Tk(n) is self-adjoint. Distinguish- ing forms coming from the lower level would allow to remove the restriction (n, N) = 1 (see DS06 5.6) but this exceeds the purpuse of the thesis.

′ Nontheless since Hecke operators commute, Sk(Γ ) splits into orthogonal ′ sum of simultaneous eigenspaces for the operators Tk(n) with (n, N) = 1. 4.2. ACTION ON MODULAR FUNCTIONS 87

′ Definition. A cusp form f ∈ Sk(Γ) which is a eigenvector for all Tk(n), (n, N) = 1, is called eigenform.

′ ′ Proposition 4.29. Let f ∈ Sk(Γ ) be an eigenform such that Tk(n)f = λ(n)f ∑ n for all n coprime with N. If f has q-expansion at infinity f(τ) = cnq , then n≥1

cn = λ(n)c1

Proof. From Proposition 4.27 b1 = cn, on the other hand from being an eigen- ′ vector for Tk(n) follows that b1 = λ(n)c1. Since f is an eigenform, the equality cn = λ(n)c1 holds for every n,(n, N) = 1. Remark 4.30. As consequences we can see that

(i) f ̸= 0 implies c1 ̸= 0. (ii) The eigenvalues λ(n)’s determines f up to a scalar factor. (iii) The previous point allows us to normalize f so that its q-expansion has ′ c1 = 1, and therefore cn is the eigenvalue for Tk(n) with (n, N) = 1. (iv) In particular if f is normalized, from (4.2.8) follows that:

k−1 • For any p prime, p - N, cpr cp = cpr+1 + p cpr−1 . r • For any p prime, p | N, cpr = cp.

• If (m, n) = 1 then cncm = cnm.

Our next task is to deduce that T2(n)’s act on S2(Γ0(N)) as matrices with integer coefficients, so that their eigenvalues will be algebraic integers.

Proposition 4.31. Let x1, .., xr ∈ X0(N) correspond to all the elliptic point of ∗ Γ0(N) (i.e. the points in H with nontrivial stabilizer) of order e1, .., er respectively. Let s1, .., sm be the inequivalent cusps for Γ0(N). Let f be a modular k/2 function of weight k for Γ0(N) and for k even define ω = f(z)dz . Then ( r m ) k ∑ ei−1 ∑ (a) div(f) = div(ω) + ( ) · xi + sj if k is even. 2 ei i=1 j=1

r (b) deg(div(f)) = k (2g − 2 + m + ∑( ei−1 )) for any k. 2 ei i=1 Proof. (a) Assume that k is even.

• Recall that if x corresponds to a point z0 ∈ H then the local normal form is given by t = g(z)e where g : H −→ D is a holomorphic isomorphism such that g(z0) = 0 and e = #StabΓ(z0). Notice that dt i−1 ′ ( dt ) e−1 dz = eg(z) g (z) therefore ordt=0 dz = e , thus we obtain k k (∗) ord (ω) = ord (fdzk/2) = ord (f)+ ord (dz) = ord (f)+ (e−1−1) x x x 2 x x 2 4.2. ACTION ON MODULAR FUNCTIONS 88

∗ • Let s be a cusp corresponding tos ˜ ∈ H , let γ ∈ SL2(Z) be such that (1 h) γ(˜s) = ∞ and let q = e2πiz/h if generates ±γStab (˜s)γ−1. 0 1 Γ dq 2πi k/2 The local parameter is z = γ(t) and dz = h q, hence f(t)dt = −1 k/2 ( dz )k/2 k/2 f|[γ ]k(z)dz = Φ(q) dq (dq) . We obtain k k (∗∗) ord (ω) = ord (Φ) − = ord (f) − s 0 2 s 2 Putting together (∗) and (∗∗) we obtain (a).

k (b) By Riemann-Roch theorem we have deg(ω) = 2 (2g − 2) and (b) for k even immidiately follows. On the other hand if k is odd, apply the same 2 1 2 computation in (a) with f and notice that div(f) = 2 div(f ).

1 Corollary 4.32. S2(Γ0(N)) is isomorphic to the space Ω (X0(N)) of holomor-

phic 1-forms on X0(N) via the map f ↦→ f(z)dz. In particular dimCS2(Γ0(N)) = g.

1 It follows that the T (n)’s act on Ω (X0(N)), and by composing on the right 1 ∨ we obtain an action of the Hecke operators T (n)’s on Ω (X0(N)) and in particular an action on H1(X0(N), Z). If we prove that Hecke operators act as an endomorphism of H1(X0(N), Z) we will be able to conclude that T (n)’s act as matrices with integer coefficients.

Definition. We say that γ ∈ Γ0(N) is elliptic (resp. parabolic) if |Trγ| < 2 (resp. |Trγ| = 2).

Theorem 4.33. Let Γep be the normal subgroup of Γ0(N) generated by all the ∗ elliptic and parabolic elements. After fixing a point τ0 ∈ H , there is a canonical isomorphism ∼ ab ab H1(X0(N), Z) = Γ0(N) /Γep Proof. See [Kna93] Proposition 11.22

1 ∼ ∼ ab ab The correspondences Ω (X0(N)) = S2(Γ0(N)) and H1(X0(N), Z) = Γ0(N) /Γep respect integration, i.e. if ω ↔ f and c ↔ [γ] then

γτ def ∫ ∫ 0 ⟨c, ω⟩ = ω = f(τ)dτ (4.2.11) c τ0

Let us show that the right hand side of (4.2.11) is indipendent of τ0 and is 0 for elliptic and parabolic elements.

∗ • Let τ1 ∈ H then

∫ γτ1 ∫ τ0 ∫ γτ0 ∫ γτ1 f(τ)dτ = f(τ)dτ + f(τ)dτ + f(τ)dτ = τ1 τ1 τ0 γτ0

∫ τ0 ∫ γτ0 ∫ τ1 ∫ γτ0 = f(τ)dτ + f(τ)dτ + f(τ)dτ = f(τ)dτ τ1 τ0 τ0 τ0

where the secont equality follows from the fact that f is Γ0(N)-invariant. 4.2. ACTION ON MODULAR FUNCTIONS 89

∗ • Remark that given a cusp form f and fixed a point τ0 ∈ H , the association ∫ γτ0 γ ↦→ f(τ)dτ is a homomorphism from Γ0(N) to ( , +). τ0 C Therefore if γ is an elliptic element, then it has finite order in0 Γ (N), so does its image in ( , +). We conclude that ∫ γτ0 f(τ)dτ = 0. C τ0 (a b) • Now let γ = ∈ Γ (N) be a parabolic element and assume a+d = 2 c d 0 . Then az + b a − 1 γz = = z ⇐⇒ cz2 + (d − a)z − b = 0 ⇐⇒ z = cz + d c a+1 (for the case a + d = −2 one gets c ). a−1 This means that γz = z has a double root at z0 = c . Now let ρ ∈ SL2(Z) ′ −1 be such that ρ(z0) = ∞, then γ = ργρ ∈ StabSL2(Z)(∞). In particular (1 h) γ′ = ± for some h ∈ such that the width of the cusp z divides 0 1 Z 0 h. Therefore −1 ′ ∫ γτ0 ∫ ρ γ ρτ0 f(τ)dτ = f(τ)dτ = τ0 τ0 ′ ∫ γ ρτ0 ∫ ρτ0+h −1 −1 −2 −1 = f(ρ τ)j(ρ , τ) dτ = f|[ρ ]2(τ)dτ = 0 ρτ0 ρτ0 where the last equality follows from the fact that f is a cusp form, therefore it has zero residue. µ ⨆ Going back to the action of T (n) in H1(X0(N), Z): let ∆N (n) = Γ0(N)αi i=1 where #CN (n) = µ and αi ∈ CN (n). Then notice that for every element γ ∈ µ ⨆ Γ0(N) we have ∆N (n)γ ⊆ ∆N (n) = Γ0(N)αi. Therefore i=1

αiγ = γiαj(i) (4.2.12) for some γi ∈ Γ0(N) and i ↦→ j(i) a permutation of {1, .., µ}. Then the action of T (n) on H1(X0(N), Z) is µ ∑ T (n)[γ] = [γi] (4.2.13) i=1 with γi ∈ Γ0(N) are as in (4.2.12).

Proposition 4.34. The action of T (n) on H1(X0(N), Z) in (4.2.13) is well- defined, Z-linear and indipendent from the coset representatives αi ∈ CN (n). Proof. To verify that T (n) is well-defined and Z-linear we need to show that T (n)[γ1γ2] = T (n)[γ1] + T (n)[γ2] for every γ1, γ2 ∈ Γ0(N) and it is zero on Γep.

• Let αiγ1 = γiαj(i) and αjγ2 = δjαk(j). Then

αi(γ1γ2) = (γiαj(i))γ2 = γiδj(i)αk(j(i)) Hence since i ↦→ j(i) is a permutation of {1, .., µ} we get µ µ µ ∑ ∑ ∑ T (n)[γ1γ2] = [γiδj(i)] = [γi] + [δj(i)] = T (n)[γ1] + T (n)[γ2] i=1 i=1 i=1 4.2. ACTION ON MODULAR FUNCTIONS 90

r • Let γ ∈ Γ0(N) be elliptic so that there exists r ≥ 1 such that γ = 1, then µ ∑ rT (n)[γ] = T (n)[γr] = T (n)[1] = [1] = 0 i=1

since [1] = 0 (1 is parabolic). Now H1(X0(N), Z) is torsion free, hence T (n)[γ] = 0.

• Let γ be parabolic and let αiγ = γiαj(i). Then there exists some r ≥ 1 for which the r-th iteration of j is the identity, we obtain

r r−1 r−2 αiγ = γiαj(i)γ = γiγj(i)αj2(i)γ = ... = γiγj(i) · .. · γjr−1(i)αi

r −1 So we have that αiγ αi = γiγj(i) · .. · γjr−1(i) ∈ Γ0(N) and it is parabolic since parabolic elements form a subgroup and conjugating leaves them r r −1 stable. Therefore in (4.2.12) for γ we can use αiγ αi since they belong r r −1 to Γ0(N) for each index i and αiγ = (αiγ αi )αi. Hence µ r ∑ r −1 rT (n)[γ] = T (n)[γ ] = [αiγ αi ] = 0 i=1

r −1 because each αiγ αi is parabolic. As before the torsion free property of H1(X0(N), Z) yields that T (n)[γ] = 0. We are left to show that the action is indipendent from coset representatives: let {βi} be another set of representatives, then βi = δiαi for some δi ∈ Γ0(N). Now −1 if αiγ = γiαj(i), then βiγ = δi(γiαj(i)) = δiγiδj(i)βj(i). Therefore we conclude that µ µ µ µ µ ∑ −1 ∑ ∑ ∑ ∑ [δiγiδj(i)] = [δi] + [γi] − [δj(i)] = [γi] = T (n)[γ] i=1 i=1 i=1 i=1 i=1

1 Now we claim that if c ∈ H1(X0(N), Z) and ω ∈ Ω (X0(N)), with ω corresponding to f ∈ S2(Γ0(N)), then ∫ ∫ ⟨T (n)c, ω⟩ = ω = T (n)ω = ⟨c, T (n)ω⟩ (4.2.14) T (n)c c where T (n)ω is the holomorphic 1-form corresponding to T2(n)f.

µ ∑ Let γ correspond to c, if T (n)[γ] = [γi], then i=1 µ ∫ ∑ ∫ γiτ0 ω = f(τ)dτ T (n)c i=1 τ0 ∗ where τ0 ∈ H is such that the path [τ0, γτ0] in H has projection in X0(N) homologous to c. On the other hand µ ∫ ∫ γτ0 ∑ ∫ γτ0 T (n)ω = T2(n)f(τ)dτ = f|[αi]2(τ)dτ = c τ0 i=1 τ0 4.2. ACTION ON MODULAR FUNCTIONS 91

µ µ ∫ γτ0 ∫ αiγτ0 ∑ −2 (∗) ∑ ′ ′ = f(αiτ)j(αi, τ) dτ = f(τ )dτ = i=1 τ0 i=1 αiτ0 µ µ µ µ ∑ ∫ γiαj(i)τ0 ∑ ∫ αj(i)τ0 ∑ ∫ γiαj(i)τ0 ∑ ∫ γiτi f(τ)dτ = f(τ)dτ + f(τ)dτ = f(τ)dτ i=1 αiτ0 i=1 αiτ0 i=1 αj(i)τ0 i=1 τi    =0 where the first 3 equalities are tautological; (∗) follows by the change of vari- µ ′ ∑ ∫ αj(i)τ0 able τ = αiτ in each summand; and f(τ)dτ = 0 because i ↦→ j(i) αiτ0 i=1 is a permutation, thus the whole sum is a loop and f being a cusp form has 0-residue. Therefore the last equality follows by setting τi = αj(i)τ0, and this ∫ sum is equal to T (n)c ω since each integral is indipendent from the base point (as we proved in (4.2.11)).

Now extending by linearity ⟨c, ω⟩ we have it defined for c ∈ H1(X0(N), R) and H1(X0(N), C).

Remark 4.35. If ⟨c, ω⟩ = 0 for all ω ∈ H1(X0(N), R), then c = 0.

2g ∑ Proof. Let c1, .., c2g be a Z-basis for H1(X0(N), R), then c = rici for some i=1 1 1 ri ∈ R. Let ω1, .., ωg be a basis of Ω (X0(N)) over C, then any ω ∈ Ω (X0(N)) is g 2g ∑ ∑ of the form ω = sjωj. Since ⟨c, ω⟩ = 0 for all ω we must have ri⟨ci, ωj⟩ = 0 j=1 i=1 for all j = 1, .., g. By Proposition 1.35 this can only occur if ri = 0 for all i, hence c = 0. Consider the involution H −→ H given by τ ↦→ τ ∗ := −τ. This map in- (a b) duces an involution on X (N): let γ = ∈ Γ (N) and let τ = γτ , 0 c d 0 2 1 ( a −b) then τ ∗ = γ∗τ ∗ with γ∗ = ∈ Γ (N). Therefore by Theorem 4.33 we 2 1 −c d 0 have an involution on H1(X0(N), Z) which extends linearly to H1(X0(N), R) + and H1(X0(N), C). Since the eigenvalues of an involution are ±1, write H1 − and H1 for the eigenspace of +1 and −1 respectively. Then H1(X0(N), R) = + − + H1 (X0(N), R)⊕H1 (X0(N), R) and, similarly, H1(X0(N), C) = H1 (X0(N), C)⊕ − H1 (X0(N), C). ∗ Notice that τ ↦→ τ induces an involution also in the space S2(Γ0(N)), namely f ↦→ f ∗ given by f ∗(τ) = −f(τ ∗). Moreover if c ↦→ c∗ is the involution on the 1-cycle corresponding to γ ↦→ γ∗, then ∫ ∫ ∫ ⟨c∗, f⟩ = f(τ)dτ = −f(τ ∗)dτ = −f(τ ∗)dτ = ⟨c, f ∗⟩ c∗ c c Remark 4.36. If f = f ∗ then (if)∗ = −if and viceversa, therefore if we let + ∗ − ∗ S2 = {f ∈ S2(Γ0(N)) | f = f } and S2 = {f ∈ S2(Γ0(N)) | f = −f}, we obtain they have equal dimension g over R (since S2(Γ0(N)) has complex dimension g). 4.2. ACTION ON MODULAR FUNCTIONS 92

. + + Now if c ∈ H1 (X0(N), R) and f ∈ S2 , then

∗ ⟨c, f⟩ = ⟨c , f⟩ = ⟨c, f ∗⟩ = ⟨c, f⟩ =⇒ ⟨c, f⟩ ∈ R

+ − This implies that dimRH1 (X0(N), R) = g = dimRH1 (X0(N), R). In fact ar- + guing by contradiction, assume dimRH1 (X0(N), R) > g, then by rank-nullity theorem ( + + ) ∅= ̸ ker H1 (X0(N), R) −→ Hom(S2 , R) where the map is c ↦→ ⟨c, ·⟩. This means that there exists some 0 ̸= c0 ∈ + + H1 (X0(N), R) such that ⟨c0, f⟩ = 0 for all f ∈ S2 and by bilinearity also ⟨c0, if⟩ = 0. By Remark 4.36 ⟨c0, S2(Γ0(N))⟩ = 0 but by Remark 4.35 this can + only occour if c0 = 0. Therefore dimH1 (X0(N), R) ≤ g. The ”dual” argument − shows that dimH1 (X0(N), R) ≤ g and consequentely they are both equality since their sum is necessarely 2g.

+ Proposition 4.37. The linear extension of T (n) to H1(X0(N), C) maps H1 (X0(N), C) + to itself and the restriction of T (n) to H1 (X0(N), C) is the transpose of T2(n) 1 on Ω (X0(N)).

Proof. If c ∈ H1(X0(N), C) corresponds to the path [τ0, γτ0] and due to the independence of the base point we can freely select τ0 to be imaginary. With such ∗ ∗ ∗ ∗ choice of τ0, we obtain that c corresponds to [τ0, γ τ0], therefore [γ] = [γ ]. ∗ Now if αiγ = γiαj(i) and extend the involution to every matrix by

∗ (a b) ( a −b) = c d −c d

∗ ∗ ∗ ∗ ∗ then such map is multiplicative and therefore αi γ = γi αj(i). Since αi are coset representatives and the action is indipendent from those, it follows that

µ ∗ ∑ ∗ ( )∗ T (n)[γ ] = [γi ] = T (n)[γ] i=1

∗ ∗ + Hence T (n)c = (T (n)c) on H1(X0(N), C). Now if c ∈ H1 (X0(N), C), then T (n)c = T (n)c∗ = (T (n)c)∗

+ + i.e. T (n)c ∈ H1 (X0(N), C) and we conclude that T (n) maps H1 (X0(N), C) to itself. 1 The argument before the proposition prove that H1(X0(N), C) and Ω (X0(N)) are dual to one another, with perfect pairing ⟨c, ω⟩. Therefore

t ⟨c, T (n) f⟩ = ⟨T (n)c, f⟩ = ⟨c, T2(n)f⟩ where the second equality follows from (4.2.14).

+ + Lemma 4.38. H1 (X0(N), R) ∩ H1(X0(N), Z) is a lattice in H1 (X0(N), R) + and a Z-basis for the intersection is a basis for H1 (X0(N), C). + Proof. It is immidiate that H1 (X0(N), R)∩H1(X0(N), Z) is discrete and spans + H1 (X0(N), R). 4.3. L-FUNCTION OF A CUSP FORM 93

Proposition 4.39. There exists a basis of S2(Γ0(N)) for which the operators T2(n) act as matrices with integer coefficients.

+ + Proof. Let c1, .., cg ∈ H1 (X0(N), R)∩H1(X0(N), Z) be a C-basis for H1 (X0(N), C) + and let f1, .., fg be the dual basis of S2(Γ0(N)). Since T (n) maps H1 (X0(N), C) + to itself, it also maps H1 (X0(N), R) ∩ H1(X0(N), Z) to itself and therefore acting as an endomorphism of H1(X0(N), Z), its associated matrix has integer coefficients. Since by Proposition 4.37 T2(n) is its transpose, also T2(n) is acting as a matrix with integer coefficient.

We have finally reached our goal: since T2(n) is given by an integer matrix, its characteristic polynomial is monic with integer coefficients. Therefore we conclude that the eigenvalues of T2(n) are algebraic integers.

4.3 L-function of a cusp form

∑ n Definition. Let f ∈ Mk(Γ0(N)) have q-expansion at ∞ f(τ) = cnq . The n≥0 L-function associated to f is the Dirichlet series

∑ cn L(s, f) = ns n≥1 where s is a complex variable.

Our goal will be to determine convergence properties and a functional equation for the L-function associated to a cusp form.

Lemma 4.40. Let f ∈ Sk(Γ0(N)), then for all τ = x + iy ∈ H there exists a positive constant M indipendent from x such that |f(x + iy| ≤ My−k/2. Conversely, if f is a modular form for Γ0(N) of weight k such that |f(x + iy| ≤ −k/2 My for some positive constant M indipendent from x, then f ∈ Sk(Γ0(N)).

Proof. For any modular function f for Γ0(N) of weight k define a function hf = h : H −→ R by setting

h(τ) = h(x + iy) = |f(τ)|yk/2

−2 Recall the equality for all γ ∈ SL2(Z), Im(γτ) = Im(τ)|j(γ, τ)| , which implies that h is Γ0(N) invariant, and therefore factors through X0(N). Now suppose that f ∈ Sk(Γ0(N)), then in a neighbourhood of any cusp s we −1 q→0 have f|[γ ]k(τ) = Φ(q) for some γ ∈ SL2(Z) such that γs = ∞ and Φ(q) −→ 0. Hence h(γ−1(τ)) = |Φ(q)|Im(τ)k/2 −→ 0. This implies that h is continuous on X0(N). Since X0(N) is compact, h is bounded and the result follows. On the other hand if h(τ) is bounded, then Φ must be holomorphic at q = 0 and M Im(τ)→∞ |Φ(q)| = Im(τ)k/2 −→ 0. Therefore Φ(0) = 0, and thus f ∈ Sk(Γ0(N)).

∑ n 2πiτ/h Lemma 4.41. Let f ∈ Sk(Γ0(N)) and f(τ) = cnq with q = e where n≥1 h is the width of the cusp [∞]. Then there exists a constant C indipendent of n k/2 for which |cn| ≤ Cn for all n ≥ 1. 4.3. L-FUNCTION OF A CUSP FORM 94

∑ n Proof. Consider the function in the q variable F (q) = cnq ; by Cauchy n≥1 formula follows that ∫ 1 −n−1 cn = F (q)q dq 2πi |q|=r for some small r > 0. h −k/2 Now let τ have Im(τ) = y = 2πn , by Lemma 4.40, |F (q)| ≤ My = k/2 ( 2πn ) (2πi(x+iy))/h −2πy/h −1/n M h and |q| = |e | = e = e . Therefore taking r = e−1/n in the integral, yields

∫ [ −1/n 2πiϑ ] 1 −n−1 q = e e ⇒ |cn| ≤ |F (q)||q |dq = −1/n 2πiϑ = 2π r=e−1/n dq = 2πie e dϑ

2π ∫ 1 (2πn)k/2 (2πn)k/2 = M e(n+1)/ne−1/ndϑ = Me = Cnk/2 2π 0 h h

As a consequence we obtain a result about absolute convergence:

∑ cn Proposition 4.42. Let L(s, f) = ns be the L-function associated to the n≥1 ∑ n cusp form f = cnq ∈ Sk(Γ0(N)), then L(s, f) converges absolutely for n≥1 k Re(s) > 2 + 1 Proof. ∑ −s ∑ −Re(s) ∑ k/2−Re(s) |cnn | = |cn|n ≤ Bn n≥1 n≥1 n≥1 the inequality follows from Lemma 4.41 and the last sum converges for Re(s) − k 2 > 1.

Let L(s, f) be the L-function associated to some cusp form f for Γ0(N) of s/2 −s ∫ ∞ −x s−1 weight k, then define R(s, f) = N (2π) Γ(s)L(s, f) where Γ(s) = 0 e x dx is the Gamma function.

∑ n Theorem 4.43. Let f(τ) = cnq ∈ Sk(Γ0(N)). Then L(s, f) is absolutely n≥1 k convergent for Re(s) > 2 + 1 can be holomorphically continued to the whole complex plane and it satisfies the functional equation

k R(s, f) = i R(k − s, f|[ρ]k)

( 0 −1) with ρ = . N 0

∫ ∞ s−1 Proof. Define the Mellin transformation of f h(s) = 0 f(iy)y dy for y ∈ R>0 and proceed formally: ∫ ∞ ∫ ∞ ∫ ∞ s−1 ∑ −2πny s−1 ∑ −2πny s−1 h(s) = f(iy)y dy = cne y dy = cn e y dy = 0 0 n≥1 n≥1 0 4.3. L-FUNCTION OF A CUSP FORM 95

[ σ = 2πny ⇒ ] ∑ ∫ ∞ ( σ )s−1 dσ = (†) = c e−σ = dσ = 2πndy n 2πn 2πn n≥1 0 ∫ ∞ −s ∑ −s −σ s−1 −s −s/2 = (2π) cnn · e σ dσ = (2π) L(s, f)Γ(s) = R(s, f)N 0 n≥1       =Γ(s) =L(s,f) where the change of variable (†) is done for each n. k Claim: For Re(s) > 2 + 1 all the previous formal steps are actually rigorous. In fact: ∫ ε ∫ ε ∫ ε k f(iy)ys−1dy ≤ |f(iy)ys−1|dy ≤ M y−k/2yk/2dy −→ε→0 0 for Re(s) > +1 0 0 0 2 where the second inequality follows from Lemma 4.40. ∫ ∞ ∫ ∞ ∫ ∞ s−1 Re(s)−1 ∑ −2πny Re(s)−1 f(iy)y dy ≤ |f(iy)|y dy = | cne |y dy ≤ E E E n≥1

∫ ∞ ∫ ∞ ( ∑ ) −2πy Re(s)−1 2πy Re(s)−1 E→∞ ≤ |cn| e y dy ≤ A e y dy −→ 0 ∀s ∈ C E n≥1 E ∑ −2πny Now since f(iy) = cne is uniformely convergent for y ≥ ε n≥1

∫ E ∫ E s−1 ∑ −2πny s−1 f(iy)y dy = cn e y dy ε n≥1 ε

Then for any δ > 0 small there exists some B big enough such that

∫ E ∫ ∞ ∑ −2πny s−1 ∑ −2πny Re(s)−1 (†) cn e y dy ≤ |cn| e y dy = n≥B ε n≥B 0

−Re(s) ∑ −Re(s) = Γ(Re(s))(2π) |cn|n < δ n≥B Therefore we can conclude that

∫ ∞ B ∫ ∞ s−1 ∑ −2πny s−1 f(iy)y dy − cn e y dy = 0 n≥1 0

∫ E B ∫ E s−1 ∑ −2πny s−1 = lim f(iy)y dy − cn e y dy ≤ δ ε→0 E→∞ ε n≥1 ε Which proves the claim. 1 Similarly if g = f|[ρ]k we obtain ∫ ∞ g(iy)ys−1dy = Γ(s)(2π)−sL(s, g) = R(s, g)N −s/2 0

1 One can easily prove that g is a cusp form for Γ0(N) of the same weight k of f. 4.3. L-FUNCTION OF A CUSP FORM 96

Let us now prove the functional equation and deduce from that that L(s, f) can be holomorphically extended. Define A = N −1/2 and split the integral as

∫ ∞ ∫ A ∫ ∞ f(iy)ys−1dy = f(iy)ys−1dy + f(iy)ys−1dy 0 0 A By the computation we have done above, the first term converges for Re(s) > k 2 + 1 and the second converges for any s ∈ C. Now we have

k/2 −k k/2 −k g(iy) = f|[ρ]k(iy) = det(ρ) j(ρ, iy) f(ρ(iy)) = N (Niy) f(i/Ny)

Hence f(i/Ny) = g(iy)(iy)kN k/2. Therefore changing variable in the integral 1 1 ( dy ) y ↦→ Ny , dy ↦→ N − y2 yields:

∫ A ∫ A ∫ ∞ s−1 ( 1 )s−1 1 ( dy ) ( 1 )s−1 1 f(iy)y dy = f(i/Ny) − 2 = f(i/Ny) 2 dy = 0 ∞ Ny N y A Ny Ny ∫ ∞ ∫ ∞ = g(iy)(iy)kN k/2N −sy−s−1dy = ikN k/2−s g(iy)yk−s−1dy A A which is convergent for any s ∈ C. Similarly ∫ ∞ ∫ A f(iy)ys−1dy = ikN k/2−s g(iy)yk−s−1dy A 0

k which is convergent for Re(s) > 2 + 1. Therefore

R(s, f)N −s/2 = ikN k/2−sR(k − s, g)N −(k−s)/2 =⇒ R(s, f) = ikR(k − s, g) as we wanted to prove. and R(s, f)N −s/2 = Γ(s)(2π)−sL(s, f) can be holomorphically continued to the whole complex plane. Since Γ(s)−1 is an entire function, also L(s, f) can be extended holomorphically for any s ∈ C. Chapter 5

Eichler-Shimura Theory

5.1 Complex abelian varieties and Jacobian varieties

In this section we want to introduce the Jacobian variety of a nonsingular projective curve defined over Q. We will first give some results regarding abelian varieties over the complex numbers.

Definition. A lattice Λ in Cg is a discrete subgroup of maximal rank, or equivalently a free abelian group in Cg containing a R-basis of Cg. Let Λ be a lattice in Cg, the quotient Cg/Λ is called complex torus of dimension g. For a complex torus Cg/Λ, a positive definite Hermitian form H = S + iE on Cg such that H(Λ × Λ) ⊂ Z is called a Riemann form. Definition. A (complex) abelian variety is a pair (A, O) where A is a nonsingular projective variety over C, O ∈ A and A has an abelian group structure with identity O and such that the operations of addition and negative are morphisms. An abelian variety (A, O) is defined over Q if A is defined over Q, O ∈ A(Q) and addition and negative morphisms are defined over Q. The dimension of an abelian variety is the dimension as non-singular projective variety. Definition. Let A be an abelian variety and C be a subvariety. If C has a group structure compatible with that of A, then C is an abelian subvariety of A.

Remark 5.1. As for the case of elliptic curves over C, the theory of complex abelian varieties reduces to the one of complex tori admitting a Riemann form. (See [Swi74])

In what follows, if A is an abelian variety of dimension n, let us denote by V the n-dimensional complex vector space overlying A, in other words let A = V/Λ for a lattice Λ in V .

97 5.1. COMPLEX ABELIAN VARIETIES AND JACOBIAN VARIETIES 98

Theorem 5.2. Let (A1 = V1/Λ1,O1), (A2 = V2/Λ2,O2) be abelian varieties, write πi : Vi −→ Ai for the canonical projections, and let f : A1 −→ A2 be a holomorphic map such that f(O1) = O2. Then f is a group homomorphism and it is induced by a C-linear map F : V1 −→ V2 such that F (Λ1) ⊂ Λ2 and π2 ◦ F = f ◦ π1. Proof. See [Swi74] III.7 Theorem 32

Definition. A map f : A1 −→ A2 is a homomorphism between abelian varieties if it is a morphism of varieties and a homomorphism of groups. A homomorphism f is an isogeny if the map F : V1 −→ V2 as in Theorem 5.2 is an isomorphism.

Remark 5.3. Let f : A1 −→ A2 be a homomorphism of abelian varieties. The following are equivalent: (a) f is an isogeny;

(b) dim(A1) = dim(A2) and f has finite kernel;

As for the case of elliptic curves, given an isogeny f : A1 −→ A2, there exist an integer N and an isogeny g : A2 −→ A1 such that f ◦g and g◦f are respective multiplication by N, therefore being isogenous is an equivalence relation. We denote by Hom(A1,A2) the additive group (with respect to pointwise sum) of isogenies between the abelian varieties A1 and A2. One of the key results in the theory of complex abelian varieties is the so-called Poincar´e’scomplete reducibility theorem:

Theorem 5.4. (Poincar´ecomplete reducibility theorem) Let A1 = V1/Λ1 and A2 = V2/Λ2 be two abelian varieties and let f ∈ Hom(A1,A2). Then def A3 = im(f) is an abelian variety and there exist B1,B2 abelian varieties and α,˜ β˜ such that the following diagram is commutative:

f A1 → A2 ↑ α˜ β˜ ↓ p i A3 × B1 → A3 → A3 × B2

where p and i are the obvious projection and inclusion (respectively).

Proof. Step 1: A3 is an abelian variety. By Theorem 5.2 f : A1 −→ A2 is induced by F : V1 −→ V2 which is C-linear and F (Λ1) ⊂ Λ2. Hence F (Λ1) is discrete. Moreover RΛ1 = V1 thus RF (Λ1) = F (V1), and we can conclude that F (Λ1) is a lattice in F (V1). Therefore A3 = F (V1)/F (Λ1) =: V3/Λ3 is a complex torus. We are left to show that there exists a Riemann form on A3 with respect to Λ3. If H2 = S2 + iE2 is a Riemann form on A2 with respect to Λ2, then H2 is a Riemann form with respect to Λ3. |F (V1) Hence A3 is an abelian variety. Step 2: Construction of B1 andα ˜. Define W = kerF , we claim that W ∩ Λ1 is a lattice in W . It is discrete since contained in Λ and consider F :Λ −→ Λ . We can choose bases {λ , .., λ } 1 |Λ1 1 2 1 2n 5.1. COMPLEX ABELIAN VARIETIES AND JACOBIAN VARIETIES 99

1 and {µ1, .., µ2n} for Λ1 and Λ2 respectively, such that the matrix associated to the linear operator is of the form ⎛ ⎞ d1 ⎜ .. ⎟ ⎜ . ⎟ ⎜ ⎟ ⎜ dm ⎟ ⎜ ⎟ ⎜ 0 ⎟ ⎜ ⎟ ⎜ .. ⎟ ⎝ . ⎠ 0

Therefore ker(F ) = {λ , .., λ } and kerF = ⟨λ , .., λ ⟩ = W . Hence |Λ1 m+1 2n R n+1 2n W ∩ Λ1 is a lattice in W as we wanted. Now if H1 = S1 + iE1 one can prove (with some linear algebra) that ⊥ ⊥ ⊥ (i) W H1 = W E1 =: W ;

⊥ ⊥ (ii)Λ 1 ∩ W is a lattice in W ;

⊥ (iii)(Λ 1 ∩ W ) ⊕ (Λ1 ∩ W ) has finite index in Λ1. ⊥ ′ Now consider the projection p1 : V = W ⊕W −→ W such that v = w+w ↦→ w ˜ ˜ ˜ ˜ and let Λ = p1(Λ1). Since Λ1 ∩ W ⊂ Λ ⊂ W we have RΛ = W . Moreover Λ is ⊥ discrete: in fact by (iii) we have that (Λ1 ∩ W ) ⊕ (Λ1 ∩ W ) has finite index in Λ1, hence

⊥ Λ1 ∩ W = p1((Λ1 ∩ W ) ⊕ (Λ1 ∩ W )) ⊂ p1(Λ1) = Λ˜ has finite index in Λ.˜ We conclude that Λ˜ is a lattice in W and define B1 = W/Λ.˜ Let us prove that B1 is an abelian variety: let N = [Λ:Λ˜ 1 ∩ W ], then NΛ˜ ⊂ 2 Λ1 ∩ W ⊂ Λ1. Define H˜ = N H1, which is positive definite since N > 0 and H1 is positive definite by definition. Moreover H˜ (Λ˜ × Λ)˜ ⊂ Z because ˜ ′ 2 ′ ′ H(λ, λ ) = N H1(λ, λ ) = H1( Nλ , Nλ ) ∈ Z   ∈Λ1 ∈Λ1

Therefore B1 is an abelian variety. ⊥ α Now consider the map V1 = W ⊕ W −→ V3 ⊕ W given by ( ) f| 0 α = W ⊥ 0 idW and f is an isomorphism of -vector spaces since injective and dim(W ⊥) = |W ⊥ C dim(V3). Therefore dim(V1) = dim(V3 ⊕ W ) and thus α induces the isogenyα ˜

⊥ α V1 = W ⊕ W → V3 ⊕ W

↓ α˜ ↓ A1 → A3 × B1 ˜ Step 3: Construction of B2 and β.

′ ⊥H2 We have V3 ⊂ V2, let W = V3 . Similarly as before V3 ∩ Λ2 is a lattice in V3

1 Same n since f is an isogeny, therefore dim(V1) = dim(V2) 5.1. COMPLEX ABELIAN VARIETIES AND JACOBIAN VARIETIES 100

′ ′ ′ ′ and W ∩ Λ2 is a lattice in W . Define B2 = W /(W ∩ Λ2). Then we have an ′ ˜ isomorphism of C-vector spaces β : W ⊕ V3 −→ V2 which induces an isogeny β:

′ β W ⊕ V3 → V2

↓ β˜ ↓ B2 × A3 → A2

Remark 5.5. If A1,A2 and f : A1 −→ A2 are defined over Q in Theorem 5.4, then also A3 can be defined over Q.

Corollary 5.6. Let A1 be an abelian variety, C an abelian subvariety of A, then the quotient A1/C has an abelian variety structure in the sense that there exist a unique (up to isomorphism) pair (A2, f) where A2 is an abelian variety, f : A1 −→ A2 is a surjective homomorphism such that ker(f) = C and satisfying the following universal property: if g : A1 −→ B is a homomorphism of abelian ′ varieties such that ker(g) ⊃ C, then there exists f : A2 −→ B homomorphism of abelian varieties such that g = f ′ ◦ f. Moreover is if A1 and C are defined over Q then A2 and f can be defined over Q; if that is the case, and also B and g are defined over Q, then f ′ can be defined over Q. Definition. An abelian variety A is simple if it is not isogenous to the product of two non-trivial abelian subvarieties. This is equivalent to saying that A has no non-trivial subvarieties or no non-trivial quotient variety.

Corollary 5.7. Any abelian variety A is isogenous to the product of simple varieties, and factors are unique up to isogeny.

Remark 5.8. Let A1 and A2 be abelian subvarieties of an abelian variety B defined over Q. Then A1 + A2, which is the image of A1 × A2 ↪→ B × B −→ B by (a1, a2) ↦→ a1 + a2, is defined over Q. In fact A1 × A2 and + : B × B −→ B are both defined over Q. Moreover

dimA1 + dimA2 ≥ dim(A1 + A2) ≥ dimAi i = 1, 2

As in 1.3, if C is a nonsingular projective curve over C, then C has a complex manifold structure, and therefore it is a compact Riemann surface. Conse- quentely, we can define the Jacobian variety of C.

Theorem 5.9. Let C be a nonsingular projective curve over C, let J(C) the Jacobian variety of its underlying compact Riemann surface and let Φ : C −→ J(C) be the Abel-Jacobi map with base point x0 ∈ C. Then J(C) is a nonsigular projective variety such that: (a) it has a group structure which makes it into an abelian variety; (b) Φ is a morphism; 5.2. TECHNICAL RESULTS 101

(c) if F : C −→ A is a morphism into an abelian variety A, then F factors through J(C), i.e. there exists f : J(C) −→ A homomorphism of abelian varieties such that F = f ◦ Φ + F (x0)

Moreover, if C is defined over Q, then J(C) can be defined over Q in such a way that (a),(b) and (c) holds true with structures defined over Q. Proof. See [Lan59] II.2 Theorem 8 & Theorem 9.

As a consequence notice that J(X0(N)) is defined over Q since, by Theorem 3.34, X0(N) is defined over Q.

5.2 Technical results

In this section we want to see how to transform Hecke operators into elements of End(J(X0(N))) by applying Theorem 5.9 to X0(N) and further that these new elements are defined over Q. In order to do that, we first need a result due to Chow:

Theorem 5.10. (Chow) If V1,V2 are non-singular projective varieties over C and F : V1 −→ V2 is a holomorphic map between their underlying complex manifolds, then F is a morphism defined over C. Proof. See [Cho49] Theorem VII.

As a consequence the ring of holomorphic homomorphisms from J(X0(N)) into itself corresponds to End(J(X0(N))) the ring of homomorphisms of abelian varieties over C. Suppose that X0(N) has genus g ≥ 1 and fix a basis {ω1, .., ωg} of the space 1 Ω (X0(N)) and as in 1.3 define

1 ∨ J = J(X0(N)) = Ω (X0(N)) /H1(X0(N), Z)

∗ Write π : H −→ X0(N) for the canonical projection and, after fixing a point x0 ∈ X0(N), Φ : X0(N) −→ J for the Abel-Jacobi map. If we define fj(τ)dτ = ∗ ∗ π (ωj) for j = 1, .., g and π the pullback of π, then {f1, .., fg} is a basis of def ∗ S2(Γ0(N)) and the composition map Φ˜ = Φ ◦ π : H −→ J is given by ∫ τ ∫ τ ( ) −1 Φ(˜ τ) = f1(ξ)dξ, .., fg(ξ)dξ for some τ0 ∈ π (x0) (5.2.1) τ0 τ0

Definition. Let V ⊂ An(K) be an affine variety, the tangent space to V at P ∈ V is

{ n } ∑ ∂f T (V ) = (x , .., x ) ∈ n (P )(x − P ) = 0 ∀f ∈ I(V ) P 1 n A ∂X i i i=1 i where Pi is the i-th component of P . 5.2. TECHNICAL RESULTS 102

∼ ∨ 2 ∨ One can prove (See [Hul03] Theorem 3.14) that TP V = mP /(mP ) , therefore if P is a nonsingular point dimTP (V ) = dimV . Now if f : J −→ J is a holomorphic homomorphism, then by Theorem 5.2 it is 1 ∨ 1 ∨ induced by some F :Ω (X0(N)) −→ Ω (X0(N)) . Consequentely we have an ( −1) C-linear map between the tangent spaces: df = d(pr)◦F ◦d(pr) : T0(J) −→ ∼ g 1 ∨ T0(J)(= C ), where pr :Ω (X0(N)) −→ J and d(pr) is the isomorphism from 1 ∨ T0(Ω (X0(N)) ) to T0(J). Since any dinstict f yields a different df,we proved the existence of an injective ring homomorphism

End(J) −→ Mg(C) , f ↦→ [matrix associated to df] (5.2.2)

g 1 Let z1, .., zg be coordinates on C , then dz1, .., dzg form a basis for Ω (J), and let {e1, .., eg} be the dual basis for T0(J), in other words, such that:

1 ( ) Ω (J) × T0(J) −→ C, dzi, ej ↦→ dzi(ej) = ⟨dzi, ej⟩ = δij (5.2.3)

If we regard T0(J) as the space of invariant vector fields on J, we can extend and consider valid this formula at any point of J. Now let f ∈ End(J), we saw that f induces a map df : T0(J) −→ T0(J), therefore we have a map between dual spaces, namely δf :Ω1(J) −→ Ω1(J) such that

(( ) ) (( ) ) 1 δf (u) (v) = u df (v) ∀u ∈ Ω (J), ∀v ∈ T0(J) (5.2.4)

∗ Remark 5.11. We have that Φ˜ (dzi) = fi(τ)dτ.

∗ ∗ ∗ ∗ Proof. Since Φ˜ = π ◦ Φ , it is enough to show that Φ (dzi) = ωi for all (∫ x ∫ x ) i = 1, .., g, and this follows by its definition: xΦ( ) = ω1, .., ωg ∈ J. x0 x0

As a consequence we see that Φ˜ ∗ is a C-vector spaces isomorphism since 1 maps a basis to a basis. Hence we can define µ : S2(Γ0(N)) −→ Ω (J) as ( )−1 µ(f) = Φ˜ ∗ (f(τ)dτ) (5.2.5)

µ ⨆ Now recall the coset decomposition from Chapter 4: ∆N (n) = Γ0(N)αi i=1 ′ and write T (n) in place of T (n). T (n) acts on X0(N) by T (n): π(τ) ↦→ µ ∑ (π(αiτ)) ∈ Div(X0(N)). Also recall that since J is an additive group, we i=1 2 # can extend Φ : X0(N) −→ J to the surjective map Φ : Div(X0(N)) −→ J. Therefore composing we obtain

⎛∑ ∫ αiτ ⎞ f1(ξ)dξ τ0 µ i # ⎜ ⎟ # T (n) ∑ Φ ⎜ . ⎟ T (n): π(τ) ↦→ (π(αiτ)) ↦→ ⎜ . ⎟ (5.2.6) i=1 ⎝∑ ∫ αiτ ⎠ fg(ξ)dξ τ0 i

2Surjectivity follows from Theorem 1.38 5.2. TECHNICAL RESULTS 103 which shows that T #(n) is holomorphic, and consequentely by Theorem 5.10 it is a morphism of varieties over C. Let us now apply Theorem 5.9 (c) with A = J to obtain that # ( ) # ∗ T (n)(π(τ)) = t(n) Φ(τ) + T (n)(π(τ0)) ∀τ ∈ H (5.2.7) for some t(n) ∈ End(J). Let us be more explicit in order to compute dt(n); we have:

⎛∫ τ1 ⎞ f1(ξ)dξ τ0 ⎜ . ⎟ t(n) ⎝ . ⎠ = t(n) ◦ Φ(τ1) = ∫ τ1 fg(ξ)dξ τ0 ⎛∑ ∫ αiτ1 ⎞ f1(ξ)dξ τ0 ⎜ i ⎟ # # ⎜ . ⎟ = T (n)(π(τ1)) − T (n)(π(τ0) = ⎜ . ⎟    ⎝∑ ∫ αiτ1 ⎠ =0 fg(ξ)dξ τ0 i (5.2.8)

# where T (n)(π(τ0) = 0 since we integrate on a loop and each fi has zero-residue being a cusp form. ∗ Now since t(n) is additive, compute t(n) ◦ Φ(τ2) for some τ2 ∈ H , subtract and use (5.2.8) to obtain the equality

⎛∑ ∫ αiτ1 ⎞ ⎛∫ τ1 ⎞ f1(ξ)dξ f1(ξ)dξ τ2 τ2 ⎜ i ⎟ ⎜ . ⎟ ⎜ . ⎟ t(n) ⎝ . ⎠ = ⎜ . ⎟ (5.2.9) ∫ τ1 ⎝∑ ∫ αiτ1 ⎠ fg(ξ)dξ fg(ξ)dξ τ2 τ2 i

Finally, by differentiating (5.2.9) for τ2 → τ1 in every direction follows that for ∗ dt(n) ∫ τ1 all τ1 ∈ H , ∀j = 1, .., g, fj(ξ)dξ = dt(n)fj(τ1) and on the other hand dτ1 τ2

∫ αiτ1 ∫ αiτ1 d d(αiτ1) d d(αiτ1) fj(ξ)dξ = fj(ξ)dξ = f(αiτ1) dτ1 τ2 dτ1 d(αiτ1) τ2 dτ1 Therefore

⎛∑ dαiτ1 ⎞ ⎛∑ ⎞ ⎛ ⎞ f1(αiτ1) dτ f1|[αi]2(τ1) ⎛ ⎞ f1(τ1) 1 T2(n)f1(τ1) ⎜ i ⎟ ⎜ i ⎟ ⎜ . ⎟ ⎜ . ⎟ ⎜ . ⎟ ⎜ . ⎟ dt(n) ⎝ . ⎠ = ⎜ . ⎟ = ⎜ . ⎟ = ⎝ . ⎠ ⎝ ∑ αiτ1 ⎠ ⎝∑ ⎠ fg(τ1) fg(αiτ1) fg|[αi]2(τ1) T2(n)fg(τ1) dτ1 i i (5.2.10) a11 .. a1g ( .. ) By (5.2.2), let A = A(T2(n)) = .. be the matrix associated to dt(n), ag1 .. agg so that ⎛ ⎞ ⎛ ⎞ f1 T2(n)f1 ⎜ . ⎟ ⎜ . ⎟ A ⎝ . ⎠ = ⎝ . ⎠ fg T2(n)fg

Theorem 5.12. (Shimura-Tanyama) For any f ∈ S2(Γ0(N))

(δt(n))(µ(f)) = µ(T2(n)f) 5.3. ELLIPTIC CURVES ASSOCIATED TO WEIGHT-2 CUSP FORMS 104

Proof. Write aj forj-th column of A(T2(n)). Let {e1, .., eg} be the basis of T0(J) g ∑ as above and let f = ckfk. Then for all j = 1, .., g we have k=1 ∑ ⟨(δt(n))(µ(f), ej⟩ = ⟨(µ(f), dt(n)ej⟩ = ck⟨µ(fk), dt(n)ej⟩ = k ∑ ∑ ∑ = ck⟨µ(fk),A(T2(n))ej⟩ = ck⟨dzk, aj⟩ = ckakj k k k On the other hand: ∑ ∑ ⟨µ(T2(n)f), ej⟩ = ck⟨µ(T2(n)fk), ej⟩ = ck⟨µ(ak1f1 + .. + akgfg), ej⟩ = k k

g ∑ ∑ ∑ ∑ = ckaki⟨µ(fi), ej⟩ = ckaki ⟨dzi, ej⟩ = ckakj k i=1 k,i    k =0 if i̸=j

Lemma 5.13. The elements t(n): J −→ J are defined over Q. Proof. See [Kna93] Lemma 11.76. The last result we need before moving onto the next section is a structure theorem due to Wedderburn: Theorem 5.14. (Wedderburn) Let T be a finite-dimensional associative and commutative algebra with identity over a field K and let R be its nilradical. Then there exists a semisimple algebra S such that as a K-vector space T = S ⊕ R. Moreover S is direct sum of ideals, each of which is a simple algebra isomorphic to a finite algebraic field extension ofK. Proof. See [Jac43] Chapter 5, Theorem 37.

5.3 Elliptic curves associated to weight-2 cusp forms

We now have all the tools in order to prove the following:

Theorem 5.15. (Eichler-Shimura) Let f ∈ S2(Γ0(N)) be a normalized eigen- ∞ ∑ n 2πiτ form with q-expansion at ∞ f(τ) = cnq , q = e . If cn ∈ Z for all n ≥ 1, n=1 then there exists a pair (E, ν) such that:

(a)( E, ν) is the quotient of J by an abelian subvariety A ⊂ J defined over Q and E is an elliptic curve defined over Q.

(b) t(n) ∈ End(J) leaves A stable and acts on E as multiplication by cn. (c) µ(f) is a nonzero multiple of ν∗(ω) where ω is the invariant differential of E. 5.3. ELLIPTIC CURVES ASSOCIATED TO WEIGHT-2 CUSP FORMS 105

{ } def ∫ γτ0 ∼ (d)Λ f = f(ξ)dξ | γ ∈ Γ0(N) is a lattice in and E = /Λf over . τ0 C C C Proof. Step 1: Construction of A.

Let T be the commutative Q-subalgebra of EndQ(J) = End(J) ⊗Z Q generated by all the t(n) in End(J). By (5.2.2), t(n) can be identified with its differential dt(n), which is represented by the g × g matrix A. By Proposition 4.39, A has integer coefficients, therefore T is isomorphic to a subalgebra of Mg(Q), and in particular it is finite-dimensional over Q. We are under the hypothesis of Theorem 5.14, hence

T = S ⊕ R = (K1 ⊕ .. ⊕ Kr) ⊕ R

where R is the Nilradical of T and each Ki is an ideal of S isomorphic to a finite algebraic extension of Q. By Theorem 5.12 we have

(δt(n))(µ(f)) = µ(T2(n)f) = µ(cnf) = cnµ(f)

where the second equality follows from the fact that f is a normalized eigenform. It follows that we have a well-defined Q-algebra homomorphism

ρ : T −→ Q given by t(n) ↦→ cn

Notice that R ⊂ kerρ since Q is torsion free. Since t(1) = idT = (e1, .., er, 0) ↦→ 1, 2 ρ is surjective. Moreover each ei is idempotent i.e. ei = ei, therefore its image is either 1 or 0. Due to surjectivity there exists an index for which it is 1, and without loss of generality assume it is e1. Hence ρ(K1) = Q and since K1 is a field, it is isomorphic to Q. Now letρ ˜ : Q −→ K1 be the inverse map of ρ to K1, define U = (K2 ⊕ .. ⊕ Kr) ⊕ R ⊂ T and consider α ∈ U ∩ End(J). Since α ∈ U ⊂ T, it is a polynomial in t(n)’s with rational coefficients, thus there exists M ∈ Z such that Mα is a polynomial in t(n)’s with integer coefficients. By Lemma 5.13 t(n)’s are defined over Q, hence Mα is. Now since im(Mα) = im(α) ⊂ J, im(α) is an abelian subvariety of J defined over Q. By Remark 5.8 def ∑ A = im(α) α∈U∩End(J)

is an abelian subvariety of J defined over Q. Consequentely, J/A is an abelian variety in the sense of Corollary 5.6: there exists (E, ν) such that E is an abelian variety and ν : J −→ E is surjective with kerν = A, satisfying the following universal property: whenever ϕ : J −→ C is a homomorphism of abelian varieties such that A = kerν ⊆ kerϕ, then ϕ factors through ν, i.e. there exists ψ : E −→ C such that ϕ = ψ ◦ ν. (Notice that to prove (a) we are left to show that dim(E) = 1). Step 2: Action of t(n) on E. ∑ Let β ∈ T∩End(J) and a = αk(xk) ∈ A, where αk ∈ U ∩End(J) and xk ∈ J. Since U is an ideal of T, βαk ∈ U ∩ End(J) and therefore ∑ ∑ β(a) = βαk(xk) ∈ (βαk)(J) ⊂ A

In other words we proved that for all β ∈ T ∩ End(J), β(A) ⊂ A. In particular since t(n) belongs to T∩End(J), it leaves A stable. But further, since t(n)(A) ⊂ 5.3. ELLIPTIC CURVES ASSOCIATED TO WEIGHT-2 CUSP FORMS 106

A, we have ker(ν ◦ t(n)) ⊃ kerν = A: by the universal property of (E, ν) there exists t(n) ∈ End(E) such that

t(n) ◦ ν = ν ◦ t(n) (5.3.1)

Hence t(n) acts on E as t(n). By construction t(n)−ρ˜(cn) = t(n)−ρ˜(ρ(t(n))) ∈ U, and if [cn] denotes the multiplication by cn, thenρ ˜(cn) − [cn] ∈ U. Therefore t(n) − [cn] ∈ U ∩ End(J), and, by the construction of A follows that t(n) = [cn]. This proves (b). Step 3: dim(E) ≥ 1. This is equivalent to show that A $ J: since R is the Nilradical of T, there exists m m+1 m some m ≥ 0 such that K1R ̸= 0 but K1R = 0. Therefore let β ∈ K1R be a nonzero element, then, as above, there exists some integer M ≥ 1 for which m Mβ ∈ End(J), and notice that Mβ ∈ K1R , Mβ ̸= 0, hence ker(Mβ) $ J. On the other hand, for all α ∈ U we have (Mβ)α = 0 since by construction m ∑ K1 · Kj = 0 for all j = 2, .., r and RR = 0. Therefore if a = αk(xk) ∈ A, we get (Mβ)(a) = 0. In other words we have a nonzero element of End(J) which is identically zero on A. Hence A ⊂ ker(Mβ) $ J. Step 4: dim(E) ≤ 1. Since dim(E) ≥ 1, there exists 0 ̸= ω′ ∈ Ω1(E). Let ν∗ :Ω1(E) −→ Ω1(J) be the pullback map induced by ν. Since ν is a morphism of complex tori over C it is separable and therefore by Proposition 1.29 ν∗ is injective. Now taking the pullback of (5.3.1) yeilds

ν∗ ◦ δt(n) = δt(n) ◦ ν∗ (5.3.2)

Since t(n) = [cn], δt(n) = cn and therefore

( ∗ ′) ∗ ′ ∗ ′ δt(n) ν (ω = ν (cnω ) = cnν (ω ) (5.3.3)

Define f ′ = µ−1(ν∗(ω′)), then

′ ′ ∗ ′ ∗ ′ ′ µ(T2(n)f ) = (δt(n))(µ(f )) = (δt(n))(ν (ω )) = cnν (ω ) = cnµ(f ) (5.3.4)

′ ′ Since µ is an isomirphism we obtain that T2(n)f = cnf for all n. If dim(E) > 1, then there exist ω′, ω′′ ̸= 0 in Ω1(E) linearly independent. Since ν∗ is injective, also ν∗(ω′) and ν∗(ω′′) are linearly independent. However if f ′′ = µ−1(ν∗(ω′′)) ′′ ′′ then by the above argument, also T2(n)f = cnf for all n. Since µ is an isomorphism, f ′ and f ′′ are linearly independent, contradicting Remark 4.30 (ii). We conclude that dim(E) = 1 and we proved (a). In particular, taking ω′′ = ω the invariant differential of E, yields that f and f ′ are linearly dependent and (c) follows. Step 5:Λf is a lattice in C. g Recall from Section 1.3 that J is realized as the complex torus C /Λ0 with Λ0 = g ∑ Λ(X0(N)). Write f = rkfk for {f1, .., fk} basis of S2(Γ0(N)) and rk ∈ C, let k=1 ∫ ⎛ ω1⎞ cj ⎜ . ⎟ λj = ⎝ . ⎠ with {c1, .., c2g} basis of H1(X0(N), Z) and {ω1, .., ωg} basis of ∫ ωg cj 5.3. ELLIPTIC CURVES ASSOCIATED TO WEIGHT-2 CUSP FORMS 107

1 Ω (X0(N)). Let us compute µ(f) on Λ0: ∑ ∑ ∑ ∫ ∫ µ(f)(λj) = ⟨µ(f), λj⟩ = rk⟨µ(fk), λj⟩ = rk⟨dzk, λj⟩ = rk fk(ξ)dξ = f(ξ)dξ k k k cj cj ( ) ∑ ∫ =⇒ µ(f) Λ0 = Z f(ξ)dξ = Λf (5.3.5) j cj

g−1 ∼ ∼ g Let TO(A) be the tangent space to A at O: C = TO(A) ⊂ TO(J) = C . Since µ(f) is a nonzero multiple of ν∗(ω) we have

∗ kerµ(f) = {u ∈ TO(J) | ⟨ν (ω), u⟩ = 0} = {u ∈ TO(J) | ⟨ω, (dν)(u)⟩ = 0} =

= {u ∈ TO(J) | (dν)(u) = 0} = ker(dν) = TO(A) (5.3.6) where the third equality follows from the fact that ω is a basis of Ω1(E). There- fore ker(Φ| : TO(A) −→ A) = TO(A) ∩ Λ0 and this is a lattice in TO(A) of TO (A) rank 2g −2 since A is compact as it is an abelian variety over C (thus a complex torus). Let x1, .., x2g−2 be a Z-basis for TO(A) ∩ Λ0 and let x2g−1, x2g ∈ Λ0 be such ˜ def ˜ that Λ = Zx1 + .. + Zx2g has rank 2g. Then it follows that Λ has finite index m in Λ0, so that mΛ0 ⊂ Λ.˜ Therefore we have ( ) ( ) C = µ(f)(TO(J)) = µ(f) Rx1 + .. + Rx2g = µ(f) Rx2g−1 + Rx2g    =Λ˜⊗ ZR

Hence µ(f)(x2g−1) and µ(f)(x2g) are linearly independent over R. Moreover:

2g ( ) ( ∑ ) (˜) ( ) µ(f) Zx2g−1 + Zx2g = µ(f) Zxj = µ(f) Λ ⊂ µ(f) Λ0 ⊂ j=1    =Λf

( −1 ˜) ( −1 −1 ) ⊂ µ(f) m Λ = µ(f) m Zx2g−1 + m Zx2g

Since the first and last terms of the chain have rank 2, we deduce thatΛf if a free abelian subgroup of (C, +) of rank 2 and spans C over R. Therefore it is a lattice. Step 6: C/Λf is isomoprhic to E over C. ′ ′ ′ Let E = C/Λf be the elliptic curve associated to Λf . Now if π : C −→ E is ′ g ′ −1 ( ) the canonical projection, then π ◦ µ(f): C −→ E has kernel (µ(f) ) Λf = g Λ0 + kerµ(f) = Λ0 + TO(A). Since Φ : C −→ J is a universal coverining with ′ ′ kernel Λ0 ⊂ ker(π ◦ µ(f)), π ◦ µ(f) factors through Φ, namely there exists η : J −→ E′ such that π′ ◦ µ(f) = η ◦ Φ. Since η is a holomorphic map, by Theorem 5.10 it is a morphism over C and its kernel is Φ(TO(A) + Λ0) = A. Therefore ker(η) = ker(ν) and we can apply the universal property of ν to obtain ψ : E −→ E′ defined over C such that η = ψ ◦ ν. Finally since ker(ψ) = {0}, we have deg(ψ) = #ker(ψ) = 1 and by Corollary 1.19 follows that ψ is an isomorphism. Remark 5.16. Properties (a) and (b) of Theorem 5.15 characterize A uniquely and consequentely (E, ν) up to isomorphism over Q. 5.3. ELLIPTIC CURVES ASSOCIATED TO WEIGHT-2 CUSP FORMS 108

5.3.1 Perspective • Work of Igusa shows that the L-functions of E and f in Theorem 5.15 coincide as Euler products except for possibly finitely many primes (those dividing N). • With additional assumptions on f, the two L-functions L(s, E) and L(s, f) match exactly (See [Shi73, Chapter 7]).

• The modularity theorem states that any elliptic curve over Q can be obtained via a rational map with integer coefficients from the classical modular curve X0(N) for some integer N (See [DS06, Chapters 8 & 9]). • Andrew Wiles proved the modularity theorem for some class of elliptic curves. This was enough (together with Ribet’s Theorem in [Rib90]) to imply Fermat’s last theorem (See [Wil95]). Bibliography

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