Classical Modular Forms T.N. Venkataramana∗ School of Mathematics, Tata Institute of Fundamental Research, Colaba, Mumbai, India Lectures given at the School on Automorphic Forms on GL(n) Trieste, 31 July - 18 August 2000 LNS0821002 ∗[email protected] Contents 1 Introduction 43 2 A Fundamental Domain for SL(2, Z) 43 3 Modular Forms; Definition and Examples 49 4 Modular Forms and Representation Theory 54 5 Modular Forms and Hecke Operators 62 6 L-functions of Modular Forms 70 Classical Modular Forms 43 1 Introduction The simplest kind of automorphic forms (apart from Gr¨ossencharacters, which will also be discussed in this conference) are the “elliptic modular forms”. We will study modular forms and their connection with automor- phic forms on GL(2), in the sense of representation theory. Modular forms arise in many contexts in number theory, e.g. in questions involving representations of integers by quadratic forms, and in expressing elliptic curves over Q as quotients Jacobians of modular curves (this was the crucial step in the proof of Fermat’s Last Theorem by Andrew Wiles), etc. The simplest modular forms are those on the modular group SL(2, Z) and we will first define modular forms on SL(2, Z). To begin with, in section 2, we will describe a fundamental domain for the action of SL(2, Z) on the upper half plane h. The fundamental domain will be seen to parametrise isomorphism classes of elliptic curves. In section 3, we will define modular forms for SL(2, Z) and construct some modular forms, by using the functions E4 and E6 which we encounter already in the section on elliptic curves. In section 4, a representation theoretic interpretation of modular forms will be given, which will enable us to think of them as automorphic forms on GL(2, R). In section 5, we will give an adelic interpretation of modular forms. This will enable us to think of Hecke operators as convolution operators in the Hecke algebra; using this, we show the commutativity of the Hecke operators. We will also prove a special case of the Multiplicity One theorem. 2 A Fundamental Domain for SL(2, Z) Notation 2.1 Denote by h the “Poincar´eupper half-plane” i.e. the space of complex numbers whose imaginary part is positive: h = {z ∈ C; z = x + iy, x, y ∈ R, y> 0}. If z ∈ C, denote respectively by Re(z) and Im(z) the real and imaginary parts of z. On the upper half plane h, the group GL(2, R)+ of real 2 × 2 matrices a b with positive determinant operates as follows: let g = ∈ GL(2, R)+, c d and let z ∈ h. Set g(z) = (az + b)/(cz + d). Notice that if cz + d = 0 44 T.N. Venkataramana and c 6= 0 then, z = −d/c is real, which is impossible since z has positive imaginary part. Thus, the formula for g(z) makes sense. Observe that Im(g(z)) = Im(z)(det(g))/ | cz + d |2 . (1) The equation (1) shows that the map (g, z) 7→ g(z) takes GL(2, R)+ × h into h. One checks immediately that this map gives an action of GL(2, R)+ on the upper half plane h. Note also that | cz + d |2= c2y2 + (cx + d)2. (2) Therefore, | cz + d |2≥ y2 or 1 according as | c |= 0 or nonzero. Therefore, 2 Im(γ(z)) ≤ y/ min{1,y } ∀γ ∈ Γ0 ⊂ SL(2, Z), (3) 2 2 where min{1,y } denotes the minimum of 1 and y and Γ0 ⊂ SL(2, Z) is 1 1 0 1 the group generated by the elements T = ( 0 1 ) and S = ( −1 0 ). Later we will see that Γ0 is actually SL(2, Z). The element T acts on the upper half plane h by translation by 1: 1 1 T (z)= (z)= z + 1 ∀z ∈ h. (4) 0 1 Similarly, the element S acts by inversion: 0 1 S(z)= (z)= −1/z ∀z ∈ h. (5) −1 0 Consider the set F = {z ∈ h; −1/2 < Re(z) ≤ 1/2, | z |≥ 1, and 0 ≤ Re(x)} if | z |= 1. Theorem 2.2 Given z ∈ h there is a unique point z0 ∈ F and an element γ ∈ SL(2, Z) such that γ(z) = z0. Moreover, given γ ∈ SL(2, Z), we have γ(F ) ∩ F = φ unless γ lies in a finite set ( of elements of SL(2, Z) which fix the point ω = 1/2+ i31/2/2 ∈ h or i ∈ h). [one then says that F is a fundamental domain for the action of SL(2, Z) on the upper half plane h]. Proof We will first show that any point z on the upper half plane can be translated by an element of the subgroup Γ0 of SL(2, Z) (generated by 1 1 0 1 T ( 0 1 ) and S = ( −1 0 )) into a point in the “fundamental domain” F . Classical Modular Forms 45 Now, given a real number x, there exists an integer k such that −1/2 < x + k ≤ 1/2. Therefore, equation (4) shows that given z ∈ h there exists an integer k such that the real part x′ of T k(z) satisfies the inequalities −1/2 <x′ ≤ 1/2. Let y denote the imaginary part of z and denote by Sz the set Sz = {γ(z); γ ∈ Γ0, Im(γ(z)) ≥ y , −1/2 < Re(γ(z)) ≤ 1/2} . We will first show that Sz is nonempty and finite. Let k be as in the pre- vious paragraph. Then −1/2 < Re(T k(z)) ≤ 1/2 and Im(T k(z) = Im(z); k therefore, T (z) lies in Sz and Sz is nonempty. Now, equation (3) shows that the imaginary parts of elements of the set 2 Sz are all bounded from above by y/ min 1,y . By definition, the imaginary parts of points on Sz are bounded from below by y. The definition of Sz shows that Sz is a relatively compact subset of h. We get from (3) that 2 2 a b | cz + d | ≤ 1; now (2) shows that | c |≤ 1/y . Suppose γ ∈ γ0 = ( c d ) is 2 such that γ(z) ∈ Sz then, c is bounded by 1/y and is in a finite set. The fact that cz +d is bounded now shows that d also lies in a finite set. Since Sz is relatively compact in h, it follows that γ(z) = (az + b)/(cz + d) is bounded for all γ(z) ∈ Sz; therefore, az + b is bounded as well, and hence a and b run through a finite set. We have therefore proved that Sz is finite. Let y0 be the supremum of the imaginary parts of the elements of the ′ ′ finite set Sz; let S1 = {z ∈ Sz; Im(z )= y0} and let z0 ∈ S1 be an element whose real part is maximal among elements of S1. We claim that z0 ∈ F . ′ ′ ′ First observe that if z ∈ Sz then S(z ) = −1/z has imaginary part y0/ | 2 ′ ′ 2 ′ 2 z | = Im(z )/ | z | ≤ y0 whence | z | ≥ 1. If | z0 |> 1, then it is immediate from the definitions of F and Sz that z0 ∈ F . Suppose that | z0 |= 1. Then, S(z0)= −1/z0 also has absolute value 1, its imaginary part is y0 and its real part is the negative of Re(z0); hence S(z0) ∈ S1. The maximality of the real part of z0 among elements of S1 now implies that Re(z0) ≥ 0. Therefore, z0 ∈ F . We have proved that every element z0 may be translated by an element of Γ0 into a point in the fundamental domain F . −1 a b Suppose now that z ∈ γ (F )∩F for some γ ∈ SL(2, Z). Write γ = ( c d ) with a, b, c, d ∈ Z and ad − bc = 1. Suppose that Im(γ(z)) ≥ Im(z) = y (otherwise, replace z by γ(z)). Then, by (3) one gets (cx + d)2 + c2y2 ≤ 1. (6) Since z ∈ F , we have x2 + y2 ≥ 1 and 0 ≤ x ≤ 1/2. Therefore y2 ≥ 3/4 and (1 ≥)c2y2 ≥ c24/3. (7) 46 T.N. Venkataramana This shows that c2 ≤ 1 since c is an integer. Suppose c = 0. Then, ad = 1, a, d ∈ Z and we may assume (by multiplying by the matrix −Id [minus identity] if necessary) that d = 1. 1 b Hence γ = ( 0 1 ). Then, γ(z)= z + b ∈ D which means that 0 ≤ x + b ≤ 1/2 and 0 ≤ x ≤ 1/2. Thus, −1/2 ≤ b ≤ 1/2, i.e. b = 0 and γ is the identity matrix. The other possibility is c2 = 1, and by multiplying by the matrix −Id (minus identity) we may assume that c = 1. Suppose first that d = 0. Then, bc = −1 whence b = −1. Now, (7) shows that x2 + y2 ≤ 1. Moreover, γ(z)= az + b/z = a + bz/ | z |2= a − z whence its real part is a − x which lies between 0 and 1/2. Since 0 ≤ x ≤ 1, 0 −1 it follows that 0 ≤ a ≤ 1. If a = 0 then γ = ( 1 0 ) and lies in the isotropy 1 −1 of the point i ∈ h. If a = 1 then, γ = ( 1 0 ) which lies in the isotropy of the point ω = 1/2+ i31/2/2. We now examine the remaining case of c = 1 and d 6= 0. From (6) we get (x + d)2 + y2 ≤ 1. If d ≥ 1 then the inequality 0 ≤ x ≤ 1/2 shows that 1 ≤ d ≤ x + d which contradicts the inequality (x + d)2 + y2 ≤ 1, which is impossible.