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Hecke Operators and L L L-Functions of Modular Forms

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Hecke Operators and L L L-Functions of Modular Forms HECKE OPERATORS AND L-FUNCTIONS OF MODULAR FORMS AMR UMERI, MUZE REN 1. Motivation We want to dene Hecke operators, which are linear maps, indexed by m 2 N and sending modular forms of a given weight k to modular forms of the same weight and cusp forms to cusp forms: Tm : Mk(SL(2; Z)) ! Mk(SL(2; Z)); Tm : Sk(SL(2; Z)) ! Sk(SL(2; Z)): Hecke operators enable us to apply methods from linear algebra into the theory of modular forms: Hecke operators are pairwise commuting and self-adjoint on , Sk(SL(2; Z)) where the hermitian form is taken to be the Petersson inner product. Hence by Spec- tral Theorem we have the existence of an orthonormal basis of consist- Sk(SL(2; Z)) ing of simultaneous eigenfunctions (called Hecke eigenforms) of all of the operators . Tm; m 2 N 2. Definition of Hecke Operators and their first properties Fix and consider the space: . We have m 2 N Mm = fA 2 Mat(2; Z) j det A = mg an action of on by left-multiplication: SL(2; Z) Mm SL(2; Z) × Mm ! Mm: Hence we have a decomposition of Mm into orbits. Hecke operators can be thought of as averaging operators on the space of modular forms, where the sum is taken over orbit representatives of the above action. Before dening the Hecke operators we want to study the above action and give explicitly a set of orbit representatives of the group action. This will be done in the following proposition: Proposition 2.1. A set of orbit representatives of the above action is given by the following matrices: a b M 0 = f j ad = m; 0 ≤ b ≤ d − 1; d > 0g: m 0 d 0 Moreover two dierent elements of Mm are inequivalent, meaning that each element gives rise to a distinct orbit. Proof. First we want to show that each element A 2 Mm appears in some orbit. a b a b Hence we want to nd γ 2 SL(2; ) and 2 M 0 such that γ−1A = : Z 0 d m 0 d x x First let 1 2 and set: x1 , x3 . Hence A = 2 Mm a = (x1; x3); γ1 = a γ3 = a x3 x4 and by lemma of Bézout we can nd such that . (γ1; γ3) = 1 γ2; γ4 2 Z γ1γ4 −γ2γ3 = 1 γ1 γ2 Hence 2 SL(2; Z). Now consider the following calculations: γ3 γ4 1 2 AMR UMERI, MUZE REN γ γ −1 x x γ x − γ x γ x − γ x a γ x − γ x 1 2 1 2 = 4 1 2 3 4 2 2 4 = 4 2 2 4 . γ3 γ4 x3 x4 γ1x3 − γ3x1 γ1x4 − γ3x2 0 γ1x4 − γ3x2 This follows from: γ4x1 − γ2x3 = γ4γ1a − γ2γ3a = a(γ1γ4 − γ2γ3) = a. γ1x3 − γ3x1 = γ1γ3a − γ3γ1a = 0. 0 Now set: b = γ4x2 − γ2x4; d = γ1x4 − γ3x2. a b0 Hence we have γ−1A = : We can ensure that d > 0 by multiplying from the 0 d −1 0 left by the matrix . By the division algorithm we can also nd b = b0 −kd, 0 −1 1 −k where k 2 and 0 ≤ b ≤ d − 1 then we just multiply from the left by Z 0 1 to obtain the matrix with the desired properties: a b . It remains to prove that 0 d a b ad = m. This follows by simple calculations or by observing that γ0A = , for 0 d some γ0 2 SL(2; Z) and that the determinant of A equals m. 0 Now we want to prove the second claim, namely that dierent elemens in Mm give a b a0 b0 dierent orbits. Let , 2 M 0. Assume there is γ 2 SL(2; ) such 0 d 0 d0 m Z that a b a0 b0 γ = : 0 d 0 d0 This means that 0 0 aγ1 bγ1 + dγ2 a b = 0 : aγ3 aγ3 + dγ4 0 d This implies that γ3 = 0, but then since γ1γ4 = 1, we have that: γ1; γ4 = 1 or− 1: 0 0 0 Since d; d > 0, we have that γ1 = γ4 = 1. But then d = d and 0 ≤ b; b ≤ d − 1 0 0 implies that −d + 1 ≤ b − b ≤ d − 1. But b − b = dγ2. Hence γ2 = 0. Hence γ a b a0 b0 is necessarily the identity matrix and = . The converse implication 0 d 0 d0 proves the claim. Before dening the Hecke operators we need to recollect some denitions: a b Denition 2.2. Let γ = 2 GL+( ) and z 2 . Dene j(γ; z) = cz + d. c d 2 R H 1 Now let f : H ! C and an integer k xed. Dene the function f[γ]k : H ! C; k=2 −k f[γ]k(z) = (det γ) j(γ; z) f(γz): This function has the following properties: + f[γ1γ2]k = f[γ1]k[γ2]k 8γ1; γ2 2 GL2 (R); is a linear operator + f 7! f[γ]k 8γ 2 GL2 (R): 1 In literature one often nds the following notation for this function: fjkγ. HECKE OPERATORS AND L-FUNCTIONS OF MODULAR FORMS 3 Remark 2.3. In general we cannot expect f[A]k to be a modular form for arbitrary + and . Hecke operators (which will be dened below) A 2 GL2 (R) f 2 Mk(SL(2; Z)) are special in that regard: By summing elements of the form f[A]k for a given modular form f we indeed get back a modular form! Now we are ready to dene the Hecke operators: Denition 2.4. Let . We dene the m-th Hecke operator as follows: m 2 N Tm For let f 2 Mk(SL(2; Z)) k=2−1 X (2.1) Tmf := m f[A]k: A2Mm=∼ The sum is meant to be taken over arbitrary orbit representatives.2 After showing that the above statement is well-dened, i.e independent of the choice of orbit represen- tatives, we can insert our distinguished set of orbit representatives from proposition 2.1 to obtain: d−1 X X a b T f = mk=2−1 f : m 0 d ad=m b=0 k d>0 For z 2 H we have: d−1 X X az + b T f(z) = mk−1 d−kf : m d ad=m b=0 d>0 Lemma 2.5. The Hecke operators are well-dened, i.e the denition does not depend on the choice of orbit representatives. Proof. We want to show now that the Hecke operators are well-dened. For this let be the (nite) number of orbits and Let 0 N fAi j i = 1;:::;Ng; fAi j i = 1;:::;Ng be two arbitrary orbit representatives. Then up to reordering of the indices we have that 0 for some Then the claim follows directly Ai = γiAi γi 2 SL(2; Z); i = 1; : : : ; N: from the weak modularity of f: N N N N X X 0 X 0 X 0 f[Ai]k = f[γiAi]k = f[γi]k[Ai]k = f[Ai]k: i=1 i=1 i=1 i=1 Theorem 2.6. Let , . Then is m 2 N f 2 Mk(SL(2; Z)) Tmf : H ! C (i) holomorphic on H, (ii) holomorphic at 1, (iii) weakly modular of weight k on SL(2; Z). Hence and is a linear Tmf 2 Mk(SL(2; Z)) Tm : Mk(SL(2; Z)) ! Mk(SL(2; Z)) operator on the space of modular forms of weight k. Proof. The rst claim is obviously true. The second claim will follow from the next lemma, where we will discuss the action of Tm on the Fourier expansion of f 2 . Let us prove the third claim. Given we must show Mk(SL(2; Z)) f 2 Mk(SL(2; Z)) that: Tmf[γ]k = Tmf 8γ 2 SL(2; Z): 2 This notation may seem ambiguous, since the elements in Mm= ∼ are orbits. However this should not lead to any confusion. 4 AMR UMERI, MUZE REN Consider the following: k=2−1 X k=2−1 X Tmf[γ]k = m f[A]k [γ]k = m f[Aγ]k = Tmf: A2Mm=∼ A2Mm=∼ The second equality follows from known properties. The last equality follows from the fact that is another set of orbit representatives. fAγ j A 2 Mm= ∼} To prove holomorphicity at 1, we need to know how the Hecke operator acts on the Fourier expansions of modular forms. For that let us consider the next lemma: Lemma 2.7. Let have the Fourier expansion: f 2 Mk(SL(2; Z)) 1 X 2πinz f(z) = fne : n=0 Then the Fourier expansion of Tmf is given by the following formula: 1 X X k−1 2πinz Tmf(z) = a f mn e : a2 n=0 ajn;m Hence Tmf is holomorphic at 1. Proof. We need the following easy result for the proof: d−1 ( X d; if d j n e2πinb=d = : 0; otherwise b=0 Let P1 2πinz be the Fourier expansion. Applying the Hecke operator we f(z) = n=0 fne obtain: d−1 1 d−1 1 k−1 X X −k X 2πin az+b 1 X X mk X 2πin az 2πin b T f(z) = m d f e d = f e d e d m n m d n ad=m b=0 n=0 ad=m b=0 n=0 d>0 d>0 1 1 X mk−1 X 2πin az X mk−1 X 2πinaz = f e d = f e d n d nd ad=m n=0 ad=m n=0 d>0 djn d>0 1 1 X k−1 X 2πinaz X X k−1 2πirz = a f nm e = a f rm e : a a2 ajm n=0 r=0 ajm ajr Corollary 2.8.
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