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
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 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. Thus, d ≤ −1; then the inequality 0 ≤ x ≤ 1/2 implies that x + d ≤ 1/2 + (−1) = −1/2 whence (x + d)2 ≥ 1/4. Since y2 ≥ 3/4 the inequality (x+d)2+y2 ≤ 1 implies that equalities hold everywhere: y2 = 3/4, x = 1/2 and d = −1. Thus, z = ω and z −1= z2. Since 1 = ad−bc = −a−b (d = −1 and c = 1), and γ(z) = (az+b)/(z−1) = (az+b)/z2 = −(az+b)z = a+(−a−b)z = a+z ∈ D, the real part of γ(z) is a + x = a + 1/2 and is between 0 and 1/2, i.e. 0 −1 −1/2 ≤ a ≤ 0 i.e. a = 0 and b = −1. Therefore, γ = ( 1 −1 ) lies in the isotropy of ω. This completes the proof of Theorem (2.2). 1 1 Corollary 2.3 The group SL(2, Z) is generated by the matrices T = ( 0 1 ) 0 1 and S = ( −1 0 ). Proof In the proof of Theorem (2.2), a point on the upper half plane is brought into the fundamental domain F by applying only the transforma- tions generated by S and T . The fact that the points on the fundamen- tal domain are inequivalent under the action of SL(2, Z) now implies that SL(2, Z) is generated by S and T . Classical Modular Forms 47 (The Corollary can also be proved directly by observing that ST −1S−1 = 1 0 ( 1 1 ). Now, the usual row-column reduction of matrices with integral entries implies that T and STS−1 generate SL(2, Z)). Notation 2.4 Elliptic Functions. We recall briefly some facts on elliptic functions (for a reference to this subsection, see Ahlfors’ book on Complex Analysis). Given a point τ on the upper half plane h, the space Γτ = Z ⊕ Zτ of integral linear combinations of 1 and τ forms a discrete subgroup of C with compact quotient. The quotient Eτ = C/Γτ may be realised as the curve in P2(C) whose intersection with the comple- ment of the plane at infinity is given by 2 3 y = 4x − g2x − g3 (8) The curve Eτ = C/Γτ is called an “elliptic curve”. 2 ′ The map of C/Γτ to P is given by z 7→ (℘ (z),℘(z), 1) for z ∈ C. Recall the definition of ℘: if z ∈ C and does not lie in the lattice Γτ , then write ′ ℘(z) = 1/z2 + (1/(z + w)2 − 1/w2), ′ X where is the sum over all the non-zero points w in the lattice Γτ . The derivative ℘′(z) of ℘(z) is then given by P ℘′(z)= 1/(z + w)3, X where the sum is over all the points of the lattice Γτ . One has the equation (cf. equation (8)) ′ 2 3 ℘ (z) = 4℘(z) − g2(τ)℘(z) − g3(τ). (9) a b If γ = ∈ SL(2, Z) and τ ∈ h, then the elliptic curve E is c d γ(τ) isomorphic as an algebraic group (which is also a projective variety) to the elliptic curve Eτ . The explicit isomorphism on C is given by z 7→ z/(cτ + d). It is also possible to show that if Eτ and Eτ ′ are isomorphic elliptic curves, then τ ′ is a translate of τ by an element of SL(2, Z). Thus the fundamental domain F which was constructed in The- orem (2.2) parametrises isomorphism classes of elliptic curves. 48 T.N. Venkataramana In equation (9), recall that the coefficients g2 and g3 are given by ′ −4 g2(τ) = 60G4(τ) = 60 (mτ + n) X and ′ −6 g3(τ) = 140G6(τ) = 140 (mτ + n) ′ X where is the sum over all the pairs of integers (m,n) such that not both m and n are zero. The discriminant of the cubic equation in (9) is given by P 1/(16)∆(τ) where 3 2 ∆(τ)= g2 − 27g3 . (10) It is well known and easily proved that ℘′(z) has a simple zero at all the 2-division points 1/2,τ/2 and (1+τ)/2 and that ℘(1/2),℘(τ/2) and ℘((1+ τ)/2) are all distinct. Thus equation (9) transforms to ℘′(z)2 = 4(℘(z) − ℘(1/2))(℘(z) − ℘(τ/2))(℘(z) − ℘((1 + τ)/2)) (11) Thus the discriminant of the (nonsingular) cubic in equation (9) is non-zero and so we obtain that ∆(τ) 6= 0 (12) for all τ ∈ h. Notation 2.5 On the upper half plane h, there is a measure denoted y−2dxdy, as z = x + iy varies in h. This measure is easily seen to be invariant under the action of elements of the group GL(2, R)+ of nonsingular matrices with positive determinant. Lemma 2.6 With respect to this measure, the fundamental domain F has finite volume. Proof We compute the volume of F . Note that if z = x + iy lies in F , then, −1/2 ≤ x ≤ 1/2 and 1/(1 − x2)1/2 ≤ y < ∞. Thus the volume of F is the integral 1/2 ∞ dxdy/y2 = dx( dy/y2) 2 1/2 ZF Z−1/2 Z(1−x ) which is easily seen to be π/3. In particular, F has finite volume. Classical Modular Forms 49 Notation 2.7 Let S denote the inverse image of the fundamental domain F ⊂ h under the quotient map GL(2, R) → GL(2, R)/O(2)Z = h. Then, we have proved that GL(2, Z)S = GL(2, R). The set S is called a Siegel Fundamental Domain. 3 Modular Forms; Definition and Examples Notation 3.1 Given z ∈ h (h is the upper half plane) and an element a b g = ( c d ), write j(g, z)= cz + d. Note that if j(g, z) = 0, then by comparing the real parts and imaginary parts we get c = 0 and d = 0 which is impossible since ad − bc 6= 0. Thus, j(g, z) is never zero. Definition 3.2 A function f : h → C is weakly modular of weight w if the following two conditions hold. (1) f is holomorphic on the upper half plane. a b (2) for all γ ∈ SL(2, Z), with γ = ( c d ), we have the equation f((az + b)/(cz + d)) = (cz + d)wf(z). (13) a b Given g = ( c d ) and a function f on the upper half plane h, define g−1 ∗ f(z) = (cz + d)−wf(g(z)) ∀z ∈ h. Then, it is easily checked that the map (g,f) → g−1 ∗ f defines an action of GL(2, R) on the space of functions on h. Thus, the condition (2) above is that the function f there is invariant under this action by SL(2, Z). Now 0 1 by Corollary (2.3), SL(2, Z) is generated by the matrices S = ( −1 0 ) and 1 1 −1 T = ( 0 1 ). Thus condition (2) is equivalent to saying that γ ∗ f = f for γ = S, T . This amounts to saying that f(−1/z)= zwf(z) (14) and f(z +1) = f(z). (15) Note that the invariance of f under the action of −1 where 1 is the identity matrix in SL(2, Z) implies that f is zero of w is odd: f(z) = (−1)wf(z). Therefore, we assume from now on (while considering modular forms for the group SL(2, Z) ) that w = 2k where k is an integer. 50 T.N. Venkataramana Definition 3.3 The map exp : h → D∗ given by z 7→ e2πiz = q is easily seen to be a covering map of the upper half plane h onto the set D∗ of non-zero complex numbers of modulus less than one. The covering transformations are generated by T (z) = z + 1. A weakly modular function f is invariant under T and therefore yields a holomorphic map f ∗ : D∗ → C given by f ∗(q)= f(z) for all z ∈ h. We say that a weakly modular function of weight w is a modular function of weight w if f ∗ extends to a holomorphic function of D (the set of complex numbers of modulus less than one) i.e. f ∗ extends to 0 ∈ D. Let f be a weakly modular function on h. Then, f is a modular func- tion if and only if the function f ∗ has the “Fourier expansion” (or the “q- expansion”) ∗ n f (q)= anq , (16) nX≥0 where an are complex numbers and the summation is over all non-negative integers n. Observe that a weakly modular function is modular if and only if it is bounded in the fundamental domain F . We will say that a modular form is a cusp form if the constant term of its q-expansion is zero: i.e. a0 = 0 in the notation of equation (15). Notation 3.4 Examples of modular forms. First we note that if f and g are modular forms of weights w and w′ then, the product function fg is a modular form of weight ww′. We will first prove that for the modular group SL(2, Z), there are no non-constant “weight zero” modular forms. First note that if f is a weight zero modular form, then the function f ∗ extends to 0 and hence is bounded in a disc of radius r < 1. Its inverse image under exp : F → D∗ is precisely the set A = {z = x + iy ∈ F ; y > − log r} and f is bounded on the set A. The complement of the set A in the fundamental domain F is compact, and f is bounded there as well, whence f is bounded on all of the fundamental domain F as well as at “infinity”. By the maximum principle, f is constant. We will now show that there are no modular forms of weight two on SL(2, Z). Suppose f is one and let F (z) be its integral from z0 to z for some fixed z0 ∈ h. The modularity of f shows that γ 7→ F (γ(z0)) gives a homomorphism from SL(2, Z) to C. But, SL(2, Z) is generated by the finite order elements S and ST whence, this homomorphism is identically zero. This and the modularity of f shows that the integral F is invariant under SL(2, Z). It is easy to show that F ∗ is holomorphic at 0 (integrate both Classical Modular Forms 51 sides of equation (15)), and use the invariance of F under T ). Hence F is a modular form of weight zero. By the foregoing paragraph, f is a constant, i.e. f = 0. Fix an even positive integer 2k, with k ≥ 2. We will construct a modular form of degree k as follows. Let τ ∈ h and write (compare the definition of g2 and g4 in section (2.4)) ′ −2k G2k(τ)= (mτ + n) , (17) ′ X where is the sum over all the pairs of integers (m,n) not both of which are zero. Then, G is easily shown to be a weakly modular function of weight P 2k 2k on the upper-half plane. If τ is varying in the fundamental domain and its imaginary part tends to infinity, then it is clear from the formula for G ′ 2k −2k that G2k(τ) tends to n = 2ζ(2k) where P ζ(s)= n−s is the Riemann zeta function (the sumX is over all the positive integers n and in the sum, the real part of s exceeds 1). Consequently, G2k is a modular form of weight 2k. We will now outline a derivation of the q-expansion of G2k. Start with the partial fraction expansion πcot(πz)= z−1 + (z + n)−1 + (z − n)−1 (18) where the sum is over all positive integersX n. This series converges uniformly on compact subsets of the complement of Z in C. Write q = e2πiz (where i ∈ h and i2 = −1). Then one has the q-expansion πcot(πz)= πi(q + 1)/(q − 1) = −πi − 2πi qn (19) nX≥1 Differentiate 2k- times, the right-hand sides of equations (17) and (18) with respect to z. We then get the equality (z + n)−2k = ((2k − 1)!)−1(2πi)2k n2k−1qn (20) Z nX∈ nX≥1 Fix m and in equation (19) take for z the complex number mτ. Then sum over all m. We obtain by equations (16) and (18), the q-expansion −1 2k n G2k(τ) = 2ζ(2k) + ((2k − 1)!) (2πi) σ2k−1q (21) nX≥1 52 T.N. Venkataramana r where for an integer r and n ≥ 1, σr(n) is defined to be the sum d where d runs over the positive divisors of n. P By using the power series expansion (1 + x)−2 = nxn−1 nX≥1 and equation (17) one has the power series identity πcot(πz)= z−1 + 2 n2jz2j−1 = z−1 + 2 ζ(2j)z2j−1 (22) nX≥1 Xj≥1 Xj≥1 −1 2m By comparing the power series expansions cos(x)= m≥0((2m)!) x and sin(x)= ((2m+1)!)−1x2m+1 with the right-hand side of equation m≥0 P (21) one obtains P ζ(2) = π2/6, ζ(4) = π4/90 andζ(6) = π6/(33.5.7). (23) Using (20) and (22) we get 4 4 g2 = 60G4 = (4/3)π + 160π (q + · · · ) (24) where the expression q + · · · is a power series in q with integral coefficients with the coefficient of q being 1. Similarly, we get (again from (20) and (22)) 6 5 6 g3 = 140G6 = (8/27)π − 2 .7π /3(q + · · · ) (25) Therefore, we get, after some calculation, that for all z ∈ h, 3 2 11 12 n ∆(z)= g2(z) − 27g3(z) = 2 π (q + τ(n)q ) (26) nX≥2 where the τ(n) are integers. We recall that ∆(z) is never zero on the upper- half plane (section (2.4)). The equation (26) shows that the coefficient of q in q-expansion of ∆ is non-zero, (and that its constant term is zero). Lemma 3.5 There are no modular forms of negative weight. Proof Suppose that f is a modular form of weight −l with l > 0. Form the product g = f 12∆l. Since f and ∆ are modular forms, so is the product. Since its weight is zero, g is a constant (see the beginning of this subsection). But, (26) shows that the q-expansion of g has no constant term. Hence g = 0 whence, f = 0. Classical Modular Forms 53 Lemma 3.6 Suppose that f is a cusp form. Then ∆ divides f i.e., there is a modular form g such that f = ∆g. In particular, the weight of f is at least 12. Proof Consider the quotient g = f/∆. Since ∆ has no zero in h, it follows that g is holomorphic in h. Clearly, g is weakly modular of weight =weight of f -12. Now the q-expansion of f (and also ∆), has no constant term; and the coefficient of q in the q-expansion of ∆ is non-zero. Therefore, g∗extends to a holomorphic function in a neighbourhood of 0. That is, g is a modular function. Since the weight of g is non-negative (by Lemma (3.5)), it follows that the weight of f must be at least that of ∆, namely, 12. Corollary 3.7 The space of cusp forms of weight 12 (for SL(2, Z)), is one dimensional. Proof If f is a cusp form of weight 12, then f/∆ is a modular form of weight zero, hence is a constant. That is, the space of cusp forms of weight 12 is spanned by ∆. Theorem 3.8 The space of modular forms of weight 2k with k ≥ 0 is m n spanned by the modular forms G4 G6 with 4m + 6n = 2k. Proof Argue by induction on k. We have already excluded the possibilities k< 0 and k = 0 and k = 1. Suppose that k ≥ 2 and that f is modular of weight 2k. First observe that any integer k ≥ 2 may be written as 2m + 3n for non-negative integers m and n. Now, the q-expansion of G4 and G6 have non-zero constant term. m n Hence h = f −λG4 G6 ) for a suitable constant λ, has no constant term in its q-expansion, and is a cusp form. Now, Lemma (3.6) shows that g = h/∆ is a modular form of weight 2k−12. By induction, g is a linear combination of the a b modular forms G4G6 with k − 6 = 2a + 3b whence, h is a sum of monomials p q 3 2 of the form G4G6 with 2p + 3q = k (recall that ∆ is (60G4) − 27(140G6) ). Therefore, so is f. Notation 3.9 Define E2k(z) = G2k/2ζ(2k). Then, it follows from the Fourier expansion of G2 and G6 that the modular forms E4 and E6 have in- n tegral Fourier coefficients. One sometimes writes ∆(z)= q + n≥2 τ(n)q . Then, ∆ has integral Fourier coefficients as well. We now consider the Z- m n P module spanned by E4 E6 with 4m + 6n = 2k. We get an integral lattice 54 T.N. Venkataramana of modular forms of weight 2k. We will see later that this integral lattice is stable under the Hecke operators. 4 Modular Forms and Representation Theory Notation 4.1 We will begin with some calculations on the Lie algebra g of the group GL(2, R). Write, 0 1 1 0 1 0 0 1 X = , Y = , Z = , and H = . (27) 0 0 0 0 0 1 −1 0 The complexified Lie algebra of GL(2, R) is M2(C) the space of 2 × 2 matrices with complex entries; the Lie algebra structure is given by (a, b) 7→ [a, b]= ab−ba; M2(C) is spanned by X,Y,Z and A. Write A = −iH (where i ∈ h is the unique element whose square is -1). Then, A acts semisimply (under the adjoint action) on g with real eigenvalues. Write g = CE+ ⊕ CE− ⊕ CZ ⊕ CA (28) where E− = X + iY − (i/2)A − (i/2)Z and E+ = X − iY − (i/2)A + (i/2)Z. (29) Then E− and E+ are eigenvectors for A with eigenvalues −2 and 2 respec- tively. Of course, on A and Z, A acts by 0. Thus, the complex Lie algebra + − spanned by E , E and A is isomorphic to sl2(C). Definition 4.2 Fix the subgroup K∞ = O(2) of GL(2, R). This is the group generated by cosθ sinθ SO(2) = R = : θ ∈ R (30) θ −sinθ cosθ and −1 0 ι = . (31) 0 1 Then, O(2) is a maximal compact subgroup of GL(2, R). Suppose that (π, V ) is a module for g as well as for O(2) such that the module structures are compatible. That is, suppose that v ∈ V and ξ ∈ g, and σ ∈ O(2). Then, π(σ)π(ξ)(v)= π(σ(ξ))(v) Classical Modular Forms 55 where σ(ξ) is the inner conjugation action of O(2) on the Lie algebra g. One then says that (π, V ) is a (g, K∞)-module. If, as a K∞-module, (π, V ) is a direct sum of irreducible representations of K∞ with each irreducible repre- sentation occurring only finitely many times, then one says that the (g, K∞)- module is admissible. One then sees at once that a (g, K∞)-submodule (or a quotient module) of an admissible module is also admissible. One says that a vector v ∈ V generates (π, V ) as a (g, O(2))-module, if the smallest submodule of V containing v is all of V . Notation 4.3 The Tensor Algebra. Given a (complex) vector space V , denote by T n(V ) = V ⊗n the n-th tensor power of V . This is of course, spanned by the pure tensors, i.e. vectors of the form v1 ⊗···⊗ vn 0 1 with vi ∈ V . By definition, T (V ) = C and T (V ) = V . Denote by T (V )= ⊕T n(V ) where the direct sum is over all the non-negative integers n. Given non-negative integers m and n, there exists a linear map T m(V ) ⊗ T n(V ) → T m+n(V ) which on pure tensors is the map (v1 ⊗···⊗ vm) ⊗ (w1 ⊗···⊗ wn) 7→ (v1 ⊗···⊗ vm ⊗ w1 ⊗··· wn). This extends by linearity to all of T m(V ) ⊗ T n(V ) and thence to all of the direct sum T (V ). Under this “multiplication”, T (V ) becomes an associa- tive algebra, and is called the tensor algebra of the vector space V . The subspace T 0(V )= C acts simply by scalar multiplication. Notation 4.4 The Universal Enveloping Algebra. Given now a Lie algebra g, let u(g) denote the quotient of the tensor algebra T (g) of g, by the two sided ideal generated by the elements x⊗y−y⊗x−[x,y], as x and y vary over the elements of the Lie algebra g. Here, ⊗ denotes the multiplication in the tensor algebra T (g) and the bracket [x,y] denotes the Lie bracket in g. The algebra u(g) is called the universal enveloping algebra of g. Note that g is a subspace of u(g), with [x,y]= xy − yx. Here, x,y are the images of x,y ∈ g = T 1(g) under the quotient map T (g) → u(g). Suppose that u is some algebra over C and f : g → u a linear map such that f([x,y]) = f(x)f(y) − f(y)f(x) for all elements x,y ∈ g. Here f(x)f(y) refers to the product of the two elements in the algebra u. Then, there exists 56 T.N. Venkataramana a unique algebra map F : u(g) → u which extends f. This is why u(g) is called the universal enveloping algebra of g. In particular, if V is a module over g, we have a map f : g → End(V ) of Lie algebras, and we have f([x,y]) = [f(x),f(y)] where the bracket structure on End(V ) is simply the commutator in the algebra End(V ). Therefore, by the last paragraph, we get a unique extension F : u(g) → End(V ) , with F an algebra map. In other words, the g module V is naturally a module over u(g) as well. Theorem 4.5 The Poincar´e-Birkhoff-Witt Theorem: Let a, b and c be subalgebras of a Lie algebra g such that g = a ⊕ b ⊕ c. Then, one has the decomposition u(g)= u(a) ⊗ u(b) ⊗ u(c) Proof We must prove that every element of u(g) lies in the subspace of the right-hand side of the above equation. Argue by an induction on the degree of an element ξ ∈ T n(g). If we have an element yx for example, with y ∈ c and x ∈ c, then, we may write it as xy − [x,y]. Now xy is in the above subspace, and since g is by assumption a direct sum of a, b and c, the element [x,y] also lies in the relevant subspace. We omit the details, since this would be rather technical, and the reader can easily supply the details. We now return to the group G = GL(2, R) and its Lie algebra g. We will now prove the basic fact from representation theory which we will use. Theorem 4.6 Let (π, V ) be a (g, K∞)-module. Suppose that v ∈ V has the following properties: (1) v generates V . (2) The connected component SO(2) of O(2) acts on v by the character 2πiθm determined by Rθ(v)= e v, for some positive integer m (i.e. v is an eigenvector for A with eigenvalue m). (3) E−(v) = 0 and Z(v) = 0. Then the (π, V ) is admissible and irreducible. Proof Let u(g) denote the universal enveloping algebra of the Lie-algebra g. One has the decomposition (the Poincar´e-Birkhoff-Witt Theorem) u(g)= u(g).[E−]+ u(g).[Z] ⊕ C[E+] ⊗ C[A] (32) Classical Modular Forms 57 where C[ξ] denotes the algebra generated by the operator ξ. Therefore, if (as in (31)) −1 0 ι = 0 1 then by assumptions (1) and (2) of the Theorem, V = C[E+](v) ⊕ ιC[E+](v). (33) On E+ the element A acts by the eigenvalue 2. Therefore, for an integer p ≥ 0, the element (E+)p(v) is an eigenvector for A with eigenvalue (2p + m), and ι(E+)p(v) is an eigenvector with eigenvalue (−2p − m) (note that under the conjugation action of ι, the element A goes to −A, hence ι takes an r-eigenspace for A into the −r-eigenspace). Since all these weights are different, equation (33) shows that V is admissible as an SO(2) module (A generates the complexified Lie algebra of SO(2)). In fact, equation (33) shows that the multiplicity of an irreducible representation of SO(2) in V is at most one, i.e. V is admissible. Suppose that W ⊂ V is a submodule. In the last paragraph, we saw that the action of A on V is completely reducible; hence the same holds for W . Suppose that w is a weight vector in W of weight j, say. By replacing w by ι(w) if necessary, we may assume that j > 0. The last paragraph shows that j = 2p + m for some p ≥ 0 and also that (E+)p(v) = w (up to scalar multiples). We may assume that p is the smallest non-negative integer such that W contains the eigenvector (E+)p(v)= w with eigenvalue 2p + m. The minimality of p implies that E−(w) = 0. Let W ′ be the submodule of W generated by the vector w. To prove the irreducibility, it is enough to show that W ′ = V . We may assume then that W = W ′. Since v generates V and Z annihilates v, it follows that Z acts by zero on all of V . Therefore, the vector w satisfies all the properties that v does in the assumptions of the Theorem (except that in (2) the eigencharacter is 2p + m). Therefore, cf. equation (33), we have W = C[E+](w) ⊕ ιC[E+](w)= C[E+](E+)p(v) ⊕ ιC[E+](E+)p(v). (34) Now the equations (33) and (34) show that the codimension of W in V is finite: dim(V/W ) < ∞. Hence V/W also satisfies the assumptions of the Theorem (with v ∈ V replaced by its image v ∈ V/W ), but is finite dimensional. This is impossible by the finite dimensional representation theory of sl(2, C): a lowest weight vector (i.e. one killed by E− of sl(2)) 58 T.N. Venkataramana cannot have positive weight for A. But v has exactly this property in V/W . This shows that V/W = 0 i.e. W = V . Proposition 4.7 Given m > 0, there is a unique irreducible (g, K∞)- module ρm which satisfies the properties of 4.6. Proof The uniqueness follows easily from the above proof of Theorem (4.6). Let χm denote the one dimensional complex vector space on which the 2πimθ group SO(2) acts by the character Rθ 7→ e (where Rθ is, as in (30), the rotation by θ in 2-space). Consider the space u(g) ⊗ χm. This is a representation for SO(2) (as well as for the universal enveloping algebra u(g)). Let ρm be the O(2)-module induced from this SO(2)-module. Then, ρm satisfies the properties of Theorem (4.6) and is therefore irre- ducible. Moreover, it is clear that any module V of the type considered in 4.6 is a quotient of ρm. By irreducibility, V = ρm. Remark 4.8 The modules ρ2k are called the discrete series representa- tions of weight 2k of (g, O(2)). This means the following. Suppose there exists an irreducible unitary representation of the group GL(2, R), call it ρ. Suppose that this occurs discretely (i.e. is a closed subspace of ) in a space of functions L2(G, ω) which transform according to the unitary character ω of the centre Z of GL(2, R) and which are square summable with respect to the Haar measure on the quotient GL(2, R)/Z. Given such a unitary module ρ, consider the space of vectors whose translates under the compact group O(2) form a finite dimensional vector space. This is the Harish-Chandra module of the unitary representation ρ and is a (g, O(2))-module. The representations ρ2k are the Harish-Chandra modules of discrete series repre- sentations of even weight. We will show in the next section, that these are closely related to modular forms of weight 2k. There are also the discrete series representations of odd weight, which we will not discuss, since we are dealing with the group SL(2, Z) and it has no modular forms of odd weight. We are now in a position to state the precise relationship of modular forms with representation theory. Classical Modular Forms 59 Notation 4.9 Let f be a modular form of weight 2k with k > 0. We will now construct a function on the group G+ = GL(2, R)+ as follows. Set −2k k Ff (g)= j(g, i) f(g(i))det(g) where i ∈ h is the point whose isotropy is the group SO(2) as in equation (30). As before, j(g, i)= cz + d, where a b g = ( c d ). By using the modularity of f and the equation j(gh, z) = j(g, h(z))j(h, z) for the “automorphy factor” j(g, z), it is easy to see that Ff is invariant under left translation by elements of SL(2, Z) and also under the centre Z∞ of GL(2, R). We will now check that the (g, O(2))-module generated by Ff is isomor- phic to ρ2k, with ρ2k as in 4.7. Note that Ff is contained in the space ∞ C (Z∞GL(2, Z)\GL(2, R), the space of smooth functions on the relevant space and that the latter is a (g, O(2))-module under right translation by elements of GL(2, R). Moreover, for all y > 0 and x ∈ R we have y x F = ykf(x + iy) (35) f 0 1 The function g 7→ f(g(i)) is right invariant under the action of SO(2) −iθ since i is the isotropy of SO(2). Using the fact that j(Rθ, i) = e (where −2iθ Rθ is as in (30)) one checks that j(gRθ, i) = j(g, i)(e ). Therefore, it follows that 2ikθ Ff (gRθ)= Ff (g)e . (36) 0 1 This equation implies that under the action of the element A = i( −1 0 ) (A generates the Lie algebra of SO(2)), Ff is an eigenvector with eigenvalue 2k. − − Compute the action of E (E as in (29)) on Ff . Using the invariance of of Ff under Z∞ and that it is an eigenvector of A with eigenvalue 4k, one sees that − E (Ff ) = (X + iY )Ff − ikFf . − 2k Now use equation (35) to conclude that E Ff = y (∂f/∂z). Since f is − holomorphic, one obtains that E Ff = 0. The (g, O(2)) module generated by Ff satisfies the conditions of 4.6. 60 T.N. Venkataramana Notation 4.10 Growth Properties of Ff . Consider now the growth properties of Ff . We have the quotient map GL(2, R)+ → h(⊃ F ) where F is the fundamental domain constructed in section 2. Let S be the pre-image of F under this quotient map. Then, we have the inclusion y x S ⊂ Z O(2) : y2 > 3/4 and − 1/2 y x F = ykf(x + iy). f 0 1 From the modularity property of f, it follows that f is “bounded at infinity”, which means that there exists a constant C > 0 such that on the fundamental domain F of SL(2, Z), the function z 7→ f(z)= f(x + iy) is bounded by C: | f(z) |≤ C ∀z ∈ F. Therefore, on the Siegel set S, we have y x F ≤ Cyk, f 0 1 i.e. Ff has moderate growth on the Siegel Set. Suppose now that f is a cusp form. Then, the Fourier expansion at infinity of f is of the form n=∞ f(z)= anexp(2πinz) n=1 X where a(n) are the Fourier coefficients. The function n−1 anq X is clearly bounded in a neighbourhood of infinity in the fundamental domain F and the complement of a neighbourhood of infinity being compact, it is bounded on all of F too, by a constant C. Classical Modular Forms 61 This shows the existence of a constant C > 0 such that for all z ∈ F , one has f(x + iy) ≤ Cexp(−y). The Haar measure on the group GL(2, R) is the product of the Haar measure on the group O(2) and the invariant measure dxdy/y2 on the upper half plane h = GL(2, R)/O(2) constructed at the end of section 2. This is an easy exercise. −1 Thus, the square of the absolute value of Ff integrated on S = π (F ) (where π is the quotient map GL(2, R) → h ) is simply the integral of the square of the absolute value of f(z) on the domain F . The above estimate for f shows that this integral over F is finite: 1/2 ∞ dx (dy/y2)exp(−y) < ∞ 2 1/2 Z−1/2 Z(1−x ) ! Definition 4.11 Automorphic Forms. Recall the definition of automor- phic forms on GL(2, Z). These are smooth functions φ on the quotient GL(2, Z)\GL(2, R), which are (1) K-finite. That is, the space of right translates of φ under the compact group K = O(2) forms a finite dimensional vector space. (2) The function φ has moderate growth on the Siegel set St,1/2, i.e. St,1/2 is a set of the form N1/2AtKZ where, Z is the centre of GL(2, R), 1 x N1/2 is the set of matrices of the form n = ( 0 1 ) with −1/2 ≤ x ≤ 1/2, and y 0 At is the set of diagonal matrices of the form a = ( 0 1 ) with (0 <)t φ(nakz) ≤ CyN for some integer N and for all elements nakz ∈ St,1/2 in the Siegel set. (3) There is an ideal I of finite codimension in the centre of the universal enveloping algebra U(g) which annihilates the smooth function φ. The last few paragraphs imply the following Theorem 4.12 Let f be a modular form of weight 2k. Let Ff be the associ- ated function on GL(2, Z)\GL(2, R). Then, Ff is an automorphic form. Moreover, the (g, O(2)) module generated by Ff is isomorphic to ρ2k with ρ2k as in 4.7 62 T.N. Venkataramana Moreover, if f is a cusp form, then, the associated function Ff is rapidly decreasing on the Siegel domain, and is therefore square summable on the quotient space Z∞GL(2, Z)\GL(2, R). Proof We need only check that an ideal I of finite codimension in the centre z of the universal enveloping algebra of g annihilates Ff . But, the module generated by Ff is ρ2k by 4.6 (and 4.7). Now, the 2k eigenspace of the operator A in the representation ρ2k is one dimensional (and is generated by Ff ), and z commutes with the action of A (and in fact with all of u(g) as well). Therefore, the annihilator of Ff in z is an ideal I of codimension one. Theorem 4.13 The space M2k of modular forms of weight 2k for the group SL(2, Z) may be identified with the isotypical subspace of the irreducible (g, O(2)) module ρ2k in the space ∞ C (Z∞GL(2, Z)\GL(2, R)). The isomorphism is obtained by sending a modular form f to the span of the function Ff under the action of (g, O(2)) (the latter (g, O(2))-module is isomorphic to ρ2k). 5 Modular Forms and Hecke Operators Notation 5.1 Let Af be the ring of finite adeles over Q. Recall that this is the direct limit (the maps are inclusion maps) as the finite set S of primes varies, of the product AS = Qp × Zp. pY∈S p/Y∈S ∗ The group of units of Af is the group Af of ideles and is the direct limit as S varies, of ∗ ∗ ∗ AS = Qp × Zp, pY∈S p/Y∈S (where ∗ denotes the group of units of the ring under consideration). ∗ ∗ There is a natural inclusion of Q in Af (and hence of Q in Af and of GL(2, Q) in GL(2, Af )). Denote by P the set of primes. The Strong Approximation Theorem (Chinese Remainder Theorem) implies that Af = Q + Zp. (37) pY∈P Classical Modular Forms 63 This, and the fact that Z is a principal ideal domain imply that ∗ ∗ ∗ Af = Q . Zp. (38) pY∈P From this it is not difficult to deduce that GL(2, Af )= GL(2, Q). GL(2, Zp) (39) pY∈P Note that the intersection of GL(2, Q) with Kf = GL(2, Zp) is precisely GL(2, Z). Q Let A = R × Af be the ring of adeles over Q. Then, Q is diago- nally imbedded in A. Hence there is a diagonal imbedding of GL(2, Q) in GL(2, A)= GL(2, R) × GL(2, Af ). Then, GL(2, Q) is a discrete subgroup of GL(2, A). Now equation (39) (a consequence of strong approximation) implies that GL(2, A)= GL(2, Q)(GL(2, R) × GL(2, Zp)). (40) pY∈P Now, equation (40) and the last sentence of the previous paragraph imply that the quotient GL(2, Q)\GL(2, A) = GL(2, Z)\(GL(2, R) × GL(2, Zp)). (41) pY∈P Note that GL(2, A) acts by right translations on the left-hand side of the equation (41). Notation 5.2 A representation (π,W ) of GL(2, Af ) is said to be smooth if the isotropy of any vector in W is an open subgroup of GL(2, Af ). Define the “Hecke algebra” H of GL(2, Af ) as the space of compactly supported locally constant functions on GL(2, Af ). If W is a smooth representation of GL(2, Af ), then the Hecke Algebra H also operates on W by “convolutions”: if µ is a Haar measure on GL(2, Af ), φ ∈H, and w ∈ W is a vector, then the W valued function g 7→ φ(g)π(g)w is a locally constant compactly supported function and hence can be integrated with respect to the Haar measure µ. Define φ ∗ w = π(φ)(w)= φ(g)π(g)(w)dµ(g) (42) Z 64 T.N. Venkataramana This gives the GL(2, Af )-module π, the structure of an H-module. As is well known, the category of smooth representations of GL(2, Af ) is isomorphic to the category of representations of the Hecke algebra H, the isomorphism arising from the foregoing action of the Hecke algebra on the smooth module π. Notation 5.3 The group K0 = GL(2, Z) is an open compact subgroup of GL(2, Af ) and is the product over all primes p of the groups GL(2, Zp). Given g ∈ GL(2, Af ), consider the characteristicb function χg of the double coset set K0gK0. Then χg is an element of the Hecke algebra and elements of H which are bi-invariant under H are finite linear combinations of the functions χg as g varies. We will refer to the subalgebra generated by these elements as the ‘unramified Hecke algebra and denote it by H0. Under convolution, H is an algebra and H0 is a commutative subalgebra. Fix a prime p. Let H0(p) be the subalgebra generated by the elements χMp p 0 p 0 and χNp where Mp = ( 0 1 ) and Np = ( 0 p ). It is easily proved that for varying p, the algebras H0(p) generate the unramified Hecke algebra H0. Notation 5.4 The equation (41) implies that the space of smooth functions + on Z∞SL(2, Z)\GL(2, R) is isomorphic to the space V0 of K0-invariant smooth functions on the quotient GL(2, Q)Z(A)\GL(2, A). On V0 the un- ramified Hecke algebra operates. Suppose S denotes the image of F × K0 in Z(A)\GL(2, A). Then, S is contained in a Siegel set S0 whose elements are of the form y x z × k ∞ 0 1 0 2 where z∞ ∈ Z∞, k0 ∈ K0, | x |< 1/2 and y > 3/4. Suppose that f is a cusp form for SL(2, Z) and Ff be as in section (4.5). Given g ∈ GL(2, A)= GL(2, R) × GL(2, Af ), write g = (g∞, gf ) accordingly. Define the function Φf on GL(2, Q)\GL(2, A) as follows. Set Φf (g∞, gf ) = Ff (g∞) if gf ∈ K0 and extend to G(A) by demanding that Φf be GL(2, Q)-invariant. The SL(2, Z)-invariance of Ff implies that Φf is well defined. Now, 4.12 shows that Φf is an automorphic form on GL(2, A). By 4.13, Ff is rapidly decreasing on S0; moreover, Ff is a cuspidal automorphic form in the sense that for all g ∈ G(A), the following holds. Φf (ng)dn = 0 (43) ZU(Q)\U(A) Classical Modular Forms 65 where U is the group of unipotent upper triangular matrices in GL(2) with ones on the diagonal and dn is the Haar measure on U(A). To prove this, we note that the vanishing of the integral is unaltered by changing g on the (1) right by an element of Z∞O(2) × K0, since Φf is an eigenvector for the right translation action by K∞Z∞ × K0. We may hence assume that g = (g∞, gf ). Now, up to elements of K0 = GL(2, Z) an element of G(Af ) is upper triangular; (2) left by an element of B(Q) since G(Q) normalisesb U(A) and U(Q), and preserves the Haar measure on U(A). Note that (the Iwasawa decom- position) the double coset B(Q)\G(Af )/K0Z(A) is a singleton. Hence, we y x may assume that g = g∞ = ( 0 1 ). Then the above integral is the same as f(x + n + iy)dn ZU(Z)\U(R) which is nothing but the zero-th Fourier coefficient of f, and by the cuspi- dality of f, this is zero. On the Siegel domain, the modular function f satisfies an estimate of the form |f(x + iy)| < Cexp(−y) where C is some constant. This implies that on the Siegel set S, the function Φf satisfies an estimate of the form −N Φf (g)= O(|g| ) for some positive integer N. This can be shown to imply that the function Φf is square summable on the quotient Z(A)GL(2, Q)\GL(2, A) with respect to the Haar measure. Further, one has the L2-metric <, > on the space of cuspidal automorphic forms which translates to the “Petersson” metric Notation 5.5 From now on, we will fix our attention on cusp forms. We have the natural inclusion of GL(2, Q) in GL(2, Af ). Let p > 0 be a prime p 0 A and let gp = ( 0 1 ) be thought of as an element in GL(2, f ) under the 66 T.N. Venkataramana foregoing inclusion. Let Xp denote the characteristic function of the double coset of K0 through the element gp. If χp denotes the characteristic function Xp, and f is a cuspidal mod- ′ ular form of weight 2k, then Φ = Φf ∗ R(χp) (where R(φ) denotes the right convolution by the function φ) is a smooth function on the quotient GL(2, Q)Z(A)\GL(2, A) whose “infinite” component is still ρ2k (since χp commutes with the right action of GL(2, R) on the above quotient). Since ′ χp is K0 invariant, it follows that Φ is also right K0 invariant. Therefore, ′ it corresponds to a modular form g, i.e. Φ = Φg. It is easy to show that Φ′ is cuspidal (the space of cusp forms is stable under right convolutions). Therefore, g is a cusp form of weight 2k as well. Denote g = T (p)(f). Then, T (p) is called the Hecke operator corresponding to the prime p. By noting 2 that convolution by χp is self-adjoint for the L metric on cuspidal automor- phic functions on GL(2) one immediately sees that the operators T (p) are self-adjoint for the Petersson metric on the space of cusp forms of weight 2k. The commutativity of the unramified Hecke algebra implies that the operators T (p) (as p varies) commute as well. Definition 5.6 Now a commuting family of self-adjoint operators on a fi- nite dimensional complex vector space can be simultaneously diagonalised. Consequently, there exists a basis of cusp forms of weight 2k which are si- multaneous eigenfunctions for all the Hecke operators T (p); these are called Hecke eigenforms. If f is a Hecke eigenform for SL(2, Z) and has constant term 1, then it is called a normalised Hecke eigenform. Theorem 5.7 The Iwasawa Decomposition: Any matrix in GL(2, Af ) may be written as a product bk with b ∈ B(Af ) (the group of upper triangular matrices), and k ∈ K0 = GL(2, Z). Proof This is an easy applicationb of the elementary divisors theorem. By identifying B\G with the projective line P1, we see that the Iwasawa decom- 1 position amounts to the transitivity of the action of GL(2, Z) on P (Af ). 1 2 But, any element of P (Af ) may be written as a vector (x,y) ∈ Af where for every prime p, the p-th components (xp,yp) are not bothb zero. ∗ By changing (x,y) by an element of Af if necessary, (x,y) may be as- sumed to be in Z2. Further, x,y may be assumed to be coprime, in the sense that for every prime p, the p-adic components xp,yp of x,y are coprime. Now, by writingb everything in the notation of row vectors, we want to solve Classical Modular Forms 67 for g ∈ GL(2, Z) the equation b (x,y) = (0, 1)g (note that the isotropy at (0, 1) is precisely B(Af )). This amounts to finding (z,t) ∈ Z2 to a basis of Z2, which can be done precisely because x,y are coprime. b b Notation 5.8 This implies that for a prime p, we have p x 1 0 Xp = ∪( 0 1 )K0 ∪ ( 0 p )K0 where the union is a disjoint union, and 0 ≤ x ≤ p−1. Here K0 = GL(2, Z), as before. b Notation 5.9 We will now state without proof the computation of T (p) for a prime p. Note that by strong approximation, the K0 invariant function Φf on the quotient Z(A)GL(2, Q)\GL(2, A) is completely determined by y x its values on elements of the form ( 0 1 ) with y > 0, in the quotient. We compute (using the description of Xp in the previous section) y x Φf ∗ R(χp)( 0 1 ) and find that this is equal, to 2k−1 y x p Φg( 0 1 ) where g(z) = (1/p) f((z + m)/p)+ p2k−1f(pz)= T (p)(f)(z). 0≤mX≤p−1 The Fourier coefficients of g at infinity are given by g(m)= a(mp) if m is coprime to p and g(m)= a(mp)+ p2k−1a(m/p) if p divides m. 68 T.N. Venkataramana Notation 5.10 In particular, if f is an eigenfunction for all the T (p) with eigenvalue λ(p) say, the equation T (p)f = λ(p)f implies, by comparing the p-th Fourier coefficients, that a(p)= λ(p)a(1) for each p. Remark 5.11 In particular, we get from the last two paragraphs, that if f is an eigenform whose first Fourier coefficient a(1) is zero, then, a(m)=0for all positive m. This easily follows from induction and the formula a(mp)= λ(p)a(p) if m and p are coprime, and a(m)λ(p)= a(mp)+ p2k−1a(m/p) if p divides m. Hence f = 0. Thus, we have proved that every Hecke eigenform is a nonzero multiple of a normalised Hecke eigenform. Theorem 5.12 (The Multiplicity 1 Theorem): Let f1 and f2 be two normalised Hecke eigenforms for the action of the Hecke operators T (p) with the same eigenvalues λ(p) for every prime p. Then, f1 = f2. Proof We will prove this by showing that the Fourier coefficients of f1 and f2 are the same. This will imply, by the Fourier expansion for modular forms, that f1 = f2. Write f = f1 − f2. Now, the first Fourier coefficient of f is zero, since f1 and f2 are nor- malised. Further, f is also a Hecke eigenform, since f1 and f2 are so, and with the same eigenvalues. Therefore, by the previous remark, f = 0. Recall that we have identified [representations π of GL(2, A) whose infi- nite component π∞ is ρ2k and whose finite component πf contains a non-zero GL(2, Af ) invariant vector], with [normalised eigenforms f of weight 2k for the group GL(2, Z)]. Therefore, we have proved that the multiplicity of such a π in the space of cusp forms on GL(2, A) is one. Remark 5.13 Later, Cogdell will prove that the multiplicity of a cuspidal automorphic representation of GL(n) is always 1. This is the famous mul- tiplicity 1 theorem due to Jacquet-Langlands for GL(2) and to Piatetskii- Shapiro and Shalika in general. What we have proved is therefore a very special case when n = 2, the infinite component is ρ2k, and the representation is unramified at all the local places. Classical Modular Forms 69 Definition 5.14 We now define inductively, the operators T (n) as follows. If m and n are coprime, we define T (mn)= T (m)T (n). This reduces us to defining T (pm) where p is a prime. Define recursively the operators T (pm) by the formula T (p)T (pm)= T (pm+1)+ pT (pm−1). This now implies (by the similarity of the recursive formulae for T (n) and the Fourier coefficients a(n)), that if f = a(n)qn is a normalised Hecke eigenform, then T (n)f = a(n)f for all n. These T (n)’s are the classi- P cal Hecke operators. By construction, they commute, one has T (mn) = T (m)T (n) if m,n are coprime, and they are self-adjoint for the Petersson inner product on modular forms of weight 2k. Theorem 5.15 If f is a normalised Hecke eigenform for SL(2, Z), then, all its Fourier coefficients are algebraic integers. Proof We consider the action of the Hecke operators T (n) on the space 0 M2k of cups forms. Note that the space of cusp forms contains the (ad- ditive) subgroup L of those cusp forms whose Fourier coefficients are ra- tional integers. This subgroup is stable under the action of the operators T (n). To see this, first suppose that n = p is a prime. By the formula 2k−1 aT (p)f (m) = a(pm)+ p a(m/p) if p divides m and aT (p)f (m) = a(pm) otherwise, we see that T (p) stabilises the subgroup L. Since the T (p) gen- erate T (n), the operator T (n) also stabilises L. By induction on k, we see that the space M2k of modular forms of weight 2k has a basis whose Fourier coefficients are integral: M2k = CE2k ⊕ M2k−12∆, and ∆ and E2k have integral Fourier coefficients. This shows that the subgroup L of the last paragraph contains a basis of the space of cusp 0 forms M2k. Since the operator T (n) is self adjoint with respect to a suitable metric 0 on the space M2k, it follows that the eigenvalues of T (n) are all real and are all algebraic integers. Now the Fourier coefficients a(n) of the eigenform f are nothing but the eigenvalue λ(n) of T (n) corresponding to the eigenvector f, by the last paragraph of the previous section. Consequently, the Fourier coefficients of f are all real algebraic integers. 70 T.N. Venkataramana 6 L-functions of Modular Forms In this section, we define the L-function of a cusp form for SL(2, Z) and prove that it has analytic continuation to the entire plane and has a nice functional equation. Later, Cogdell will prove analogous statements for cus- pidal automorphic representations for GL(n). To begin with, we prove an estimate –due to Hecke– for the n-th Fourier coefficient of a cusp form of weight 2k. n=∞ n Lemma 6.1 (Hecke) Let f = n=1 a(n)q be a cusp form of weight 2k. Then, there exists a constant C > 0 such that P a(n) ≤ Cnk/2 ∀n ≥ 1. Proof Consider the function φ defined and continuous on the upper half plane h, given by φ(z)= yk/2|f(z)|. a b ′ ′ ′ ′ If γ = ( c d ) ∈ SL(2, Z) and z = γ(z) = x + iy , then recall that y = y/(|cz + d|2). Therefore, we obtain from the modularity property of f, that φ(γ(z)) = φ(z), i.e. φ is invariant under SL(2, Z). Hence, φ is determined by its restriction to the fundamental domain F . As z tends to infinity in F , the cuspidality condition of f shows that f(z)= O(exp(−2πy)). Therefore, φ is bounded on F and hence on all of the upper half plane h. Thus, there exists a constant C1 such that −k/2 |f(z)|≤ C1y ∀z ∈ h. Now consider the n-th Fourier coefficient a(n) of f. Clearly, 1 a(n)= e2πny f(x + iy)e−2iπnxdx. Z0 By applying the foregoing estimate for f to this equation, we obtain 2πny k/2 |a(n)|≤ C1e y for all y > 0. Take y = 1/n. We then get 2π k/2 a(n) ≤ (C1e )n . Classical Modular Forms 71 Remark 6.2 Deligne has proved that for all primes p, |a(p)| = O(p((k−1)/2)), by linking these estimates with the “Weil Conjectures” for the number of rational points of algebraic varieties over finite fields. Definition 6.3 If f(z) = a(n)qn is a cusp form of weight 2k, define for a complex variable s, the Dirichlet Series L(f,s) by the formula P ∞ L(f,s)= a(n)/ns. n=1 X From Lemma (6.1), it follows that the series converges and is holomorphic in the region Re(s) > 1 + (k/2). The function L(f,s) is called the L-function of the cusp form f. Notation 6.4 an integral expression for L(f,s)). We will now write an integral formula for the L-function of f. First consider the integral ∞ f(iy)ys(dy/y). Z0 Since f(iy)= O(exp(−2πy)) for all y > 0 (cf. the proof of Lemma (6.1)), it follows that if Re(s) > 0, then the integral converges. Let σ be the real part of s. Then, for each n ≥ 1 the integral ∞ |a(n)|e−2πnyyσ(dy/y) Z0 converges, and is equal to (|a(n)|/nσ)(2π)−σΓ(σ) where Γ is the classical Γ-function: ∞ Γ(z)= e−ttz(dt/t). Z0 From the Hecke estimate a(n)= O(nk/2) of Lemma (6.1), it follows that the infinite sum of these integrals also converges, provided σ >k/2 + 1. Thus, by the Dominated Convergence Theorem (to justify the interchange of sum and integral), we obtain the equation ∞ ∞ f(iy)ys(dy/y) = (2π)−sΓ(s) a(n)n−s. 0 Z nX=1 We finally obtain the integral expression: ∞ f(iy)ys(dy/y) = (2π)−sΓ(s)L(f,s). Z0 72 T.N. Venkataramana Remark 6.5 Notice that the integral expression says that the function L(f,s) can be analytically continued to the region Re(s) > 0, since the left-hand side is analytic there, and the function (2π)−sΓ(s) has no zero’s on the complex plane. Notation 6.6 The functional Equation. The L-Function and the in- tegral expression could have been defined for any holomorphic function f(z) = a(n)qn (q = e2πiz), provided a(n) satisfy a Hecke estimate. We will now prove a functional equation for L(f,s), by using the modularity P property, especially that f(−1/z) = (−1)kz2kf(z). ∞ s Consider now the integral I(s)= 0 f(iy)y (dy/y), which converges for Re(s) > 0. We write the integral as a sum of the integral from 1 to ∞ and R the integral from from 0 to 1. By making a change of variable y 7→ 1/y we get, 1 ∞ f(iy)ys(dy/y)= f(−1/(iy))y−s(dy/y). Z0 Z1 Note that as f(1/(iy)) is bounded on the interval [1, ∞], and Re(s) > 0, the integral on the right side converges. Therefore, we get, using the functional equation f(−1/iy) = (−1)ky2kf(iy), that 1 ∞ f(iy)ys(dy/y) = (−1)k f(iy)y2k−s(dy/y). Z0 Z1 We then get ∞ I(s)= f(iy)(ys + (−1)ky2k−s)(dy/y). Z1 This holds for all s, with Re(s) > 0. We now make the change of variable s 7→ 2k − s, for s in the region 0 < Re(s) < 2k. Then the above expression for I(s) shows that I(2k − s) = (−1)kI(s). This is the functional equation for L(f,s): (2π)−sΓ(s)L(f,s) = (−1)k(2π)2k−sΓ(2k − s)L(f, 2k − s) for all s in the region 0 < Re(s) < 2k. The left side of this equation is analytic in the region Re(s) > 0 and the right side is analytic in the region Re(s) < 2k. Using the functional equation (and the fact the (2π)−sΓ(s) never vanishes on the complex plane), we now see that L(f,s) has an analytic continuation over the entire complex plane. Classical Modular Forms 73 Notation 6.7 Euler Factors. In the last few sections, we derived the functional equation and analytic continuation of a cusp form for SL(2, Z). We now derive an Euler product, for the L-function of a normalised Hecke eigenform f. ∞ n Let then f = n=1 a(n)q be a normalised Hecke eigenform of weight 2k. Fix a prime p and consider the infinite sum P ∞ m ms Lp(f,s)= a(p )/p mX=0 where Re(s) > k/2+ 1. It converges, by the Hecke estimate a(n)= O(nk/2). We now use the relations a(pm)a(p)= a(pm+1)+p2k−1a(pm−1). Multiplying these equations by 1/p(m+1)s and then summing over all m ≥ 0 we get s 2k−1 2s a(p)Lp(f,s) = (Lp(f,s) − 1)p + (p /p )Lp(f,s). That is, −1 s 2k−1 2s Lp(f,s) = 1 − a(p)/p + p /p . We now use the fact that a(mn) = a(m)a(n) if m,n are coprime, since f is a normalised Hecke eigenform. Form the product over all primes p of these Lp(f,s). We then get (by using the Dominated Convergence Theorem s to justify interchanges) the equation Lp(f,s) = a(n)/n = L(f,s). Thus we have the infinite product expansion (the product being over all Q P primes p) L(f,s)= 1/(1 − a(p)/ps + p2k−1/p2s). p Y From now on we consider the function L∗(f,s) = (2π)−sΓ(s)L(f,s) and refer to this as the L-function of f. We have thus proved the following Theorem. ∞ n Theorem 6.8 Let f = n=1 a(n)q be a normalised Hecke eigenform of weight 2k for SL(2, Z). Then, the L-function L∗(f,s) = (2π)−sΓ(s) a(n)/ns P converges for Re(s) > k/2, has an analytic continuation to the entire com- P plex plane and satisfies the functional equation L∗(f,s) = (−1)kL∗(f, 2k − s). Moreover, in the region Re(s) > k/2, one has the Euler product L∗(f,s) = (2π)−sΓ(s) 1/(1 − a(p)/ps + p2k−1/p2s). p Y 74 T.N. Venkataramana Remark 6.9 Recall that to each Hecke eigenform f of weight 2k, there corresponds an irreducible cuspidal automorphic representation π(f)= π = π∞ ⊗p πp of the (restricted direct product) group GL(2, A) = GL(2, R) × p GL(2, Qp) such that π∞ is the discrete series representation ρ2k and each π is an irreducible unramified representation of GL(2, Q ). Qp p In the lectures of Cogdell, you will see that each cuspidal automorphic representation π of GL(n, A) has an L-function attached to it –denoted L(π,s)– which satisfies a functional equation, has an analytic continuation to the entire plane, and has an Euler product comprising of terms which are monic polynomials in p−s of degree n. It turns out that for the representation π(f)= π attached to the Hecke eigenform f of weight 2k, the L-function is nothing but L(π,s)= L∗(f,s + (k − 1/2)), which can easily be seen to satisfy the equation L(π,s) = (−1)kL(π, 1 − s). Moreover, the local factors are of the form −1 k−1/2 s 2s L(πp,s) = (1 − (a(p)/p )/p + 1/p , a monic polynomial in in p−s of degree two.