April 12, 12:00 pm 1 Introduction The study of modular forms is typically reserved for graduate students, because the amount of background needed to fully appreciate many of the constructions and methods is rather large. However, it is possible to get a first look at modular forms without relying too heavily on the theory of complex analysis, harmonic analysis, or differential geometry. In some sense, we've been doing this all semester { using identities which can be understood more generally in the language of modular forms to prove classic problems in number theory that were only completely resolved after years of work beyond Ramanujan by a host of talented mathematicians. In the remainder of this course, we'll be exploring the theory of modular forms equipped with a large number of examples coming from the generating functions we've been analyzing all semester. 2 The Modular Group In our recent study of elliptic functions, we were led to consider the set of matrices a b GL(2; ) = j a; b; c; d 2 ; ad − bc = ±1 Z c d Z which related any two pairs of complex numbers (!1;!2) generating the period mod- ule M. Furthermore, we learned that to any lattice, we may choose a basis h!1;!2i with τ = !2 2 uniquely determined by the criteria !1 C −1 1 τ : Im(τ) > 0; jτj ≥ 1; < Re(τ) ≤ 2 2 (which, in turn, almost uniquely determined the pair !1;!2). In reviewing the definition for modular forms, we typically state definitions with respect to the group a b SL(2; ) = j a; b; c; d 2 ; ad − bc = 1 : R c d R There is an action of SL(2; R) on C given by a b az + b γ = 2 SL(2; ): γ(z) = c d R cz + d 1 It is sometimes convenient to extend this action to the set C [ f1g by letting a d γ(1) = c and γ( c ) = 1. Moreover, one checks (EXERCISE, if you haven't tried it before) that with notation for γ as above, Im(z) Im(γ(z)) = : jcz + dj2 Hence, we see that the action takes R [ 1 to itself, and stabilizes the upper and lower half planes. Let H denote the upper half plane fz 2 C j Im(z) > 0g. Moreover −1 0 the matrix −I = acts trivially on H, i.e. fixes all elements. 0 −1 Proposition 1 The group P SL(2; R) := SL(2; R)={±Ig acts faithfully on H. Recall that a faithful group action is one for which no non-indentity element of the group fixes all elements of the set. We leave the proof of this proposition as an EXERCISE. The \modular group" G is the subgroup SL(2; Z)={±Ig in P SL(2; R), consisting of matrices with coefficients in Z up to equivalence by ±I. Technically, we should denote elements of this quotient group as cosets, but typically no confusion will arise a b by continuing to use the matrix representation of γ = in SL(2; ) to denote c d R elements of G. (It is important that SL(2; Z) is a discrete subgroup of SL(2; R), that is a topological group with the discrete topology. The theory of modular forms can be presented for arbitrary discrete groups of SL(2; R) with some additional complications. For more facts about discrete subgroups of SL(2; R), see Shimura's book. 3 The Fundamental Domain for G In this section, we show that the domain D = fz 2 H j jzj ≥ 1; jRe(z)j ≤ 1=2g is a fundamental domain for the action of G on H (and explain what is meant by this term \fundamental domain"). To this end, consider the matrices 0 1 −1 1 1 S = : S(z) = T = : T (z) = z + 1 −1 0 z 0 1 which satisfy the relations S2 = I; (ST )3 = I: We will prove the following two main results in this section: 2 Theorem 1 D is a fundamental domain for the action of G on H. That is, by way of definition, 1. For every z 2 H there exists γ 2 G such that γ(z) 2 D. 0 0 1 0 2. Given two points z; z 2 D with z = γ(z ), then either Re(z) = ± 2 and z = z ±1 −1 OR jzj = 1 and z = z0 . 3. To each z 2 D, let Stab(z) = fγ 2 G j γ(z) = zg: Then Stab(z) = fIg for all z 2 D UNLESS • z = i, then Stab(z) = hSi of order 2. • z = e2πi=3, then Stab(z) = hST i of order 3. • z = eπi=3, then Stab(z) = hTSi of order 3. Theorem 2 The group G is generated by S and T , i.e. every element γ 2 G can be written as a word in S and T of form γ = T n1 ST n2 S ··· ST nk for some choice of integers ni (though the representation is clearly not unique ac- cording to the relations mentioned above). We will use Serre's proof of these facts, which proves both theorems at the same time. Before launching into the proof, a simple example will serve to illustrate an alternate proof of Theorem 2. Given any matrix, say 4 9 γ = 11 25 we seek to represent γ in terms of S and T . Note 4 9 1 n 4 4n + 9 4 9 0 −1 9 −4 γT n = = ; γS = = : 11 25 0 1 11 11n + 25 11 25 1 0 25 −11 Hence, we may choose n to reduce the size of coefficients in the matrix γT n. In particular, noting the way in which S switches the columns of the matrix, we may choose n (say n = −2 in our case) so that jdj < jcj. Then 4 1 4 1 4 1 0 −1 1 −4 γT −2 = : Now γT −2S = = = : 11 3 11 3 11 3 1 0 3 −11 3 And we may continue this process, now reducing 11 mod 3 by multiplying by say T 4 (though T 3 would work as well). Carrying this example to the bitter end, we may obtain γ = ST −3ST −4ST 2: EXERCISE: Turn the ideas used in this example into a rigorous general proof of Theorem 2. Proof (of Theorems 1 and 2): Let G0 be the subgroup of G generated by S and T . We will first show part 1 of Theorem 1 by demonstrating an element of γ0 2 G0 such that γ0(z) 2 D. There exists a γ 2 G0 such that Im(γ(z)) is maximal, because the number of pairs of integers (c; d) with jcz + dj < k for any given number k is finite, and Im(z) Im(γ(z)) = : (1) jcz + dj2 Now choose an n so that T n(γ(z)) is shifted into the vertical strip between −1=2 and 1=2. But then z0 = T n(γ(z)) 2 D, since if jz0j < 1, then S(z0) = −1=z0 would have a larger imaginary part than z0, contradicting the maximality of the imaginary part of γ(z). For parts 2 and 3 of Theorem 1, suppose that given γ 2 G and z 2 D, γ(z) 2 D as well. As the pairs (z; γ) and (γ(z); γ−1) play symmetric roles here, we may assume without loss of generality that Im(γ(z)) ≥ Im(z), which implies from (1) that jcz + dj ≤ 1. Remembering that z 2 D, then jcz + dj ≤ 1 for very few choices of c and d, in particular only if c = 0; 1; −1 and we separate into three cases accordingly. ±1 b If c = 0, then d = ±1 so γ = , i.e. γ(z) = z ± b. But if γ(z) is also 0 ±1 assumed in D which has width 1, then either b = 0 so γ = ±I or b = ±1, and then Re(z); Re(γ(z)) are −1=2 and 1=2 (or vice versa). The cases c = ±1 follow similarly. EXERCISE: Finish the remaining cases c = ±1 in the above proof to complete the proof of theorem 1. (Note: so far, the c = 0 case provided no non-identity matrices that stabilized a point in D. As written in the theorem, there are points with non-trivial stabilizers, so they must occur for c = ±1.) Finally, it remains to prove that G0, our group generated by S and T , is G. Pick 0 any point z0 in the interior of D. For any γ 2 G, we must show γ 2 G . If z = γ(z0), 0 0 0 0 then there exists a γ 2 G such that γ (z) = γ γ(z0) 2 D (by our above argument 0 for Theorem 1, part 1). But since z0 was chosen in the interior of D, and z0; γ γ(z0) 0 0 are both in D, then part 2 of Theorem 1 implies that γ γ = I and hence γ 2 G . 4 Serre also notes that one can prove the slightly stronger statement that the only relations on elements in G are those generated by S2 = 1 and (ST )3 = 1. 4 Modular Functions, Modular Forms Definition 1 For a given integer k, we say a function f is weakly modular of weight 2k if f(z) is meromorphic on H and satisfies az + b a b f(z) = (cz + d)−2kf for all 2 SL(2; ) cz + d c d Z Note that weakly modular functions of odd weight are all 0, since taking γ = −I in the above definition would give f(z) = (−1)2k+1f(z).