Modular Forms and Elliptic Curves Sriram Gopalakrishnan February 6, 2019 Abstract We define what it means for a function to be a modular form, prove fundamental theorems about modular forms, and explore the relationship between lattice functions and modular forms. We conclude by giving a small glimpse of the relationship between modular forms and elliptic curves. 1 Introduction In this paper, we focus on the fundamentals of modular forms. We first give a definition of a modular form through a careful series of theorems and proofs. We follow with a discussion of lattices and their relations to modular forms. We give a common example of a lattice function that is also a modular form, namely Eisenstein series and we conclude by providing some motivations for the study of modular forms by exploring their relationship with elliptic curves. This paper follows Chapter 7 of J.P. Serre's text A Course in Arithmetic, while filling in details that Serre omits. The final proof follows that of Washington in his book Elliptic Curves, Number Theory, and Cryptography. 2 Preliminary 2.1 The Modular Group Recall the definition of the special linear group of 2 × 2 matrices over the integers: a b SL2(Z) := a; b; c; d 2 Z; ad − bc = 1 c d We call G = SL2(Z)= ± 1 where 1 is the identity element (the identity matrix) in SL2(Z) the modular group. The modular group forms the starting point for our discussion of modular forms, so it is beneficial to state and prove some facts about it. We define the complex upper half plane H as follows: H := fz 2 C j Im(z) > 0g What follows is a discussion and development of the G-action on H a b Let g 2 G be a matrix . Then, for every g 2 G, z 2 H, the action of g on z is given c d by az + b gz = cz + d 1 0 −1 1 1 Let S; T 2 G where S = ;T = The S and T actions on a complex variable 1 0 0 1 z 2 H are then 1 Sz = − T z = z + 1 z 2 3 1 Furthermore, S = 1 and (ST ) = 1. Now, let D := z 2 H jzj ≥ 1; jRe(z)j ≤ 2 . The following figure represents the transformations of D by the elements: f1;T; −T;T −1S;STS;ST;S;ST −1; ST −1S; T Sg 2 G -T 1 T T−1S S TS STS ST ST−1 ST−1S Theorem 2.1.1. The modular group G is generated by the elements S and T . Proof. We will give an explicit expression of an arbitrary element g 2 G in terms of S and T . For some matrix in G, we have the following multiplications on the left by the elements S and T . a b −c −d a b a + nc b + nd S = ;T n = (1) c d a b c d c d a b Now let g = be an arbitrary element in G. Suppose c 6= 0. If jaj > jcj, write a = cq + r c d (division algorithm), where 0 ≤ r < jcj. By (1), the upper left entry of the matrix T −qg is a − qc = r < jcj. We can apply S to T −qg to swap the rows of T −qg. Once more, if our new matrix has a nonzero entry in the bottom left spot, we can perform the division algorithm again to obtain a new power of T by which we can multiply T −qS by to again reduce the element in the bottom left entry. We can continue this process until we obtain a matrix with lower left entry 0. Since this matrix must still be in G (as it is a product of matrices in G), it must have integer entries and determinant 1. Thus, it is of the form ±1 m 0 ±1 2 where m 2 Z and the diagonal entries have the same sign. This matrix is either T m or −T −m, so there exists h 2 G that is the multiple of powers of T and copies of S such that hg = ±T n for some n 2 Z. As T n 2 G and S2 = I, we can write g = ±h−1T n where h−1 is generated by S and T and we are done. Theorem 2.1.2. Let D, G, and H be as above. (1) For every z 2 H, there exists g 2 G such that gz 2 D. (2) Suppose z; z0 2 H are congruent modulo the G-action. That is, suppose z and z0 have the same orbit under the G-action. Then, 1 0 (i) Re(z) = ± 2 and z = z ± 1 or 0 1 (ii) jzj = 1 and z = − z (3) Fix z 2 D and let Gz = fg 2 G j gz = zg be the stabilizer of z under the G-action. Gz = f1g except if: (i) z = i, in which case Gz is generated by S and is a group of order 2; 2πi (ii) z = e 3 , in which case Gz is generated by ST and is a group of order 3; πi (iii) z = e 3 , in which case Gz is generated by TS and is a group of order 3. a b Proof. Let G0 ≤ G be the subgroup of G generated by S and T . Let g 2 G0, g = . c d Suppose z 2 H, z = r + si. Then, az + b ar + asi + b (ar + b) + asi Im(gz) = Im = Im = Im cz + d cr + csi + d (cr + d) + csi (ar + b)(cr + d) − csi(ar + b) + asi(cr + d) (da − bc)si = Im = Im jcz + dj2 jcz + dj2 Im(z) = jcz + dj2 Fix M 2 Z. Then, jf(c; d) 2 Z2 j jcz + dj < Mgj < 1. Thus, there exists a pair (c; d) such that Im(z) 0 jcz + dj is minimal and consequently jcz+dj2 is maximal. In other words, we can find h 2 G such Im(z) n 1 that Im(hz) = jcz+dj2 is maximized. We now choose an integer n such that jRe(T hz)j ≤ 2 . Let z0 = T nhz. Then, −1 Im (z0) Im = z0 jz0j2 0 −1 0 0 0 n so if jz j < 1, Im z0 > Im (z ). As this is impossible, jz j ≥ 1. Therefore, g = T hz is in the a b fundamental domain D. Let z 2 D and choose g 2 G, g = such that gz 2 D. Suppose c d Im(gz) ≤ Im(z). Then, replace the pair (z; g) with (gz; g−1) to obtain Im(g−1gz) = Im(z) ≤ Im(gz). We can thus assume without loss of generality, that Im(gz) ≥ Im(z), so jcz + dj ≤ 1. This is only possible if jcj < 2, so we consider the cases c = −1; 0; 1. Suppose c = 0. Since g is an element of the modular group, its determinant must be 1. Therefore, 3 1 1 d = ±1. Therefore, gz = z ± b. Since z; gz 2 D, jRe(z)j ≤ 2 and jRe(gz)j ≤ 2 . Hence, we have ! 8 1 0 >0 if g = ; jRe(z)j = jRe(gz)j < 1 <> 0 1 2 b = −1 if Re(z) = 1 ; Re(gz) = − 1 > 2 2 :> 1 1 1 if Re(z) = − 2 ; Re(gz) = 2 If c = 1, jz + dj ≤ 1, so d = 0, unless ( 2πi e 3 =) d = 0 or 1 z = πi e 3 =) d = 0 or − 1 1 1 When d = 0, jzj = 1, and b = −1. Therefore, gz = a − z . Thus, a = 0 except when Re(z) = ± 2 2πi πi 2πi (equivalently when z = e 3 or z = e 3 ), in which case a = 0; −1 or a = 0; 1. When z = e 3 2πi a−1 3 and d = 1, a − b = 1 and 2πi = a + e , implying that a = 0; 1. Similarly, we find that when 1+e 3 πi z = e 3 and d = −1, a must be 0 or −1. Suppose c = −1. We can change the signs of all a; b; c; d in g, which does not change the g-action on an element in H, and follow the same argument as the case when c = 1. 2.2 Lattices We begin with the definition of a lattice. There are many equivalent definitions, but we will use the following one for the sake of this paper. Definition 2.2.1. Let V be a finite dimensional real vector space. A lattice Γ of V is a discrete subgroup of V (where V is viewed as a group) that generates V . Consider the vector space C over the field R. Let R be the set of lattices of C. Let 2 !1 M := (!1;!2) 2 C j Im > 0 !2 !1 The condition that Im > 0 is equivalent to saying that !1 and !2 cannot both be real. We !2 associate the lattice Γ(!1;!2) = Z!1 ⊕ Z!2 with the lattice basis f!1;!2g. Suppose a b g = 2 SL ( ); (! ;! ) 2 M c d 2 Z 1 2 0 0 0 0 Then, the pair f!1;!2g where !1 = a!1 + b!2 and !2 = c!1 + d!2 forms a different lattice basis 0 !1 0 !1 for Γ(!1;!2). Indeed, let z = and z = 0 . Then, !2 !2 a! + b! a !1 + b az + b z0 = 1 2 = !2 = = gz !1 c!1 + d!2 c + d cz + d !2 Theorem 2.2.1.