ZEROS OF MAASS FORMS

Transcription

1 ZEROS OF MAASS FORMS OSCAR E. GONZÁLEZ Abstract. We stud the location of the zeros of the Maass form obtained b appling the Maass level raising operator R k = i x + k to E k. We find that this Maass form has the same number of zeros on the bottom arc of F as E k+, and conjecture that all of its zeros in F lie on this arc. We note that this seems to hold for the Maass form obtained b appling the level raising operator multiple times to the Eisenstein series. 1. Introduction and Statement of Results The zeros of modular forms have been well studied over the ears. For example, in a ver influential paper [5], Rankin and Swinnerton-Der showed that all of the zeros of the weight k Eisenstein series, E k, in the standard fundamental domain { F lie on the bottom arc B = e iθ π : θ π } of F. In stark contrast, it was shown b the work of Rudnick [6] and Holowinsk-Soundararajan [] that the zeros of Hecke eigenforms become equidistributed with respect to the hperbolic measure on the fundamental domain as the weight grows. In this paper we stud the location of the zeros of the Maass form obtained b appling the Maass raising operator R k to E k. The result is no longer a modular form, but instead a weight k + Maass form. We find that this Maass form has the same number of zeros on A as E k+. Define mk b { k/1, if k mod 1, 1 mk = k/1 1, if k mod 1. Theorem 1.1. The Maass form R k E k has mk + zeros along the arc A. Since R k E k is no longer holomorphic, we do not have a valence formula for it. Hence, this theorem does not exclude the possibilit of there being other zeros inside the fundamental domain. However, as can be seen in Figure [?], numerical experiments with Mathematica suggest that all the zeros are on A. Conjecture 1.. All of the zeros of R k E k on F lie on the bottom arc A.. Preliminaries A modular form of weight k is a complex-valued function f on the upper halfplane H = {z C, Imz > 0}, satisfing the following three conditions: Date: Jul 4, Mathematics Subject Classification. Primar?? 1

2 OSCAR E. GONZÁLEZ Figure 1. ReR 6 E 6 z = 0 and ImR 6 E 6 z = 0 in solid and dotted lines, respectivel. a b a For an z in H and an matrix γ = SL c d Z, f satisfies az + b fγz = f = cz + d k fz. cz + d b f is a complex analtic function on H. c f is required to be holomorphic as z. A Maass form is a generalisation of modular forms in which the last two conditions of modular forms are replaced b the following: b f is an eigenvector of the Laplacian operator k = x + + ik x + i. c f is of at most polnomial growth as z. Throughout this paper we will use the following standard notation Γ = SL Z, S =, T =, ρ = e πi Let E k z be the Eisenstein series of weight k, k even, defined b E k z = 1 cz + d k. Its Fourier expansion is given b E k z = 1 + gcdc,d=1 ζ1 k σ k 1 ne πinz, n=1

3 ZEROS OF MAASS FORMS where σ k 1 n is the divisor function. Let M k denote the complex vector space of modular forms of weight k for Γ. If f M k, the valence formula is given b 4 k 1 = 1 ord if + 1 ord ρf + ord f + τ Γ\H {i,ρ} For details and a proof of this formula see III. of []. Define mk b { k/1, if k mod 1, 5 mk = k/1 1, if k mod 1, ord τ f. and write k = 1mk + s. Note that s determines the residue class of k modulo 1. Using this and the valence formula, we obtain { 1, if k mod 4, 6 ord i f 0, if k 0 mod 4, and, if k mod 6, 7 ord ρ f 1, if k 4 mod 6, 0, if k 0 mod 6. Define the derivative of a modular form f to be Df = 1 df πi dz. Recall the Maass raising and lowering operators R k and L k on functions f : H C which are defined b R k = i z + k = i x + k, L k = i z + k = i x + With respect to the Petersson slash operator, these satisf the properties R k f k γ = R k f k+ γ and L k f k γ = L k f k γ, for an γ SL R. This is, if f transforms b, then R k f and L k f transform like modular forms of weight k + and weight k, respectivel. Here R k satisfies the relation R k = 4πD + k. If we appl R k to the weight k Eisenstein series, we obtain R k E k z = 4πDE k z + k E kz = ki gcdc,d=1 ccz + d k 1 + k 1. cz + d k. gcdc,d=1 We are interested in the properties of this form. In particular, we stud the number and the location of its zeros inside the fundamental domain F = {z H : 1/ Rez 1/, z 1}.

4 4 OSCAR E. GONZÁLEZ. Basic properties of zeros of R k f. Definition.1. R j k = R k+j R k+ R k. Lemma.1. Suppose f satisfies and k mod4. Then fi = 0. Proof. Note that Si = i. Since f transforms like a weight k modular form, we have that 8 fsi = i k fi. Since i is a fourth root of unit, we have that fi = 0 whenever k mod 4. Corollar.. Suppose f satisfies and k jmod4. Then R j k f has a zero at i. Lemma.. Suppose f satisfies and k, 4mod6. Then fρ = 0. Proof. Note that ST ρ = ρ. Since f transforms like a weight k modular form, we have that 9 fst ρ = ρ + 1 k fρ. Since ρ + 1 is a sixth root of unit, we have that fρ = 0 whenever k + j, 4 mod 6. Corollar.4. Suppose f satisfies and k jmod6 or k 4 jmod 6 Then R j kf has a zero at ρ. Lemma.5. Suppose f is a modular form of weight k with real Fourier coefficients. Then z k+ Rk f is real valued on A. Proof. It is known that if the coefficients a n of the Fourier expansion of gz = a n e πinx are real and g satisfies, then z k gz is real valued for z = 1. n=0 See, for example, Proposition.1 of [1]. Since f has real Fourier coefficients, then R k f also has real Fourier coefficients. Hence, since R k f transforms like a k + modular forms, we have that z k+ Rk f is real valued on A. 4. Proof of theorem 1.1 We will prove this theorem b showing that R k Ek e iθ k+ iθ e has mk+ zeros in A. Some of the calculations in this section were performed using Mathematica 10. If we restrict R k E k zz k+ to F and z = 1, then we have R k E k z z k+ = kie iθ/ gcdc,d=1 cce iθ/ + de iθ/ k 1 + keiθ 1 sinθ ce iθ/ + de iθ/ k, gcdc,d=1 where π θ π. Taking the terms with c + d = 1, we obtain kθ k cscθ cos + θ.

5 ZEROS OF MAASS FORMS 5 Taking the real part of the terms with c + d =, we get using Mathematica 1 10 k cscθ 1 k 1 k θ k θ k θ k sin csc csc. Adding these two and simplifing, we obtain kθ 11 k cscθ cos + θ + 1 k θ k θ k k sin csc. Lemma 4.1. R k E k e iθ k+ iθ e = k cscθ M 1 θ + M θ + E 1 θ + E θ, where kθ M 1 θ = cos + θ, M θ = 1 k θ k θ k, k sin csc E 1 θ = ie iθ/ cce iθ/ + de iθ/ k 1 E θ = eiθ 1 sinθ gcdc,d=1 c +d 5 ce iθ/ + de iθ/ k gcdc,d=1 c +d 5 Lemma 4.. The term M is never positive in the interval π θ π. Furthermore, M 1 Proof. Taking the derivative of M, we get k 1 k θ θ cos sin k 1 4 tan θ θ k csc, which is clearl negative on π θ π. Therefore, M is decreasing. We evaluate this at θ = π and at θ = π. At θ = π/, M θ = k 1 k 1 which is less than or equal to 0 and greater than or equal to 1 4. At θ = π/, M θ = than or equal to 1. Hence, M 1. Lemma 4.. The number of zeros of M 1 in 1 k 4 1 which is less than or equal to 5 π, π is mk +. and greater Proof. This argument is similar to the one used b Rankin and Swinnerton-Der in [5]. Write θ = mπ and k + = 1n + s, where s = 4, 6, 8, 10, 0 or 14. k + Then M 1 θ = cosmπ which is either or depending on whether m is odd or kθ π even. Hence the number of zeros of cos + θ in, π is equal to the

6 6 OSCAR E. GONZÁLEZ [ k + number of integers in the closed interval 4 verified to be mk +. π Lemma 4.4. M = M 1 + M has mk + zeros in, π, k + ] minus one, which can be Proof. Since cosmπ is either or, and M is less than or equal to 1 in absolute value, then M 1 + M has the same number of zeros as M 1. Therefore, it onl remains to show that E 1 θ + E θ is less than one. We will use the estimates provided in [5]. z k+ Since we know that R k E k z is real valued on the bottom arc of F b ] Lemma.5, then we have that R k E k z z k+ = Re [R k E k z z k+, which is equal to 1 k M + ReE 1 + ReE Since Re[z] z we will provide bounds for the absolute values of the remaining parts. We will show that the sum of these two absolute values is less than 1. Lemma 4.5. ReE 1 < 0.44 Proof. [ Re ie iθ/ cce iθ/ + de iθ/ k 1] ie iθ/ cce iθ/ + de iθ/ k 1 c +d 5. c +d 5 = cce iθ/ + de iθ/ k 1 c +d 5 5 k/ 4 + which is decreasing on k and at k = 8 is less than Lemma 4.6. ReE < 0.6 Proof. [ e iθ 1 Re c +d 5 ] ce iθ/ + de iθ/ k. = eiθ eiθ N=10 1 ce iθ/ + de iθ/ k. c +d 5 5N 1/ 1 N k/ 1 ce iθ/ + de iθ/ k c +d 5 5 k/ 4 + N=10 5N 1/ 1 N k/ which is decreasing on k and at k = 8 is less than 0.6. Lemma 4.7. ReE 1 + ReE < 1 Proof. B Lemma 4.5 and Lemma 4.6, we have that ReE 1 +ReE <.7 < 1

7 ZEROS OF MAASS FORMS Jacobian Let f be a modular form. Recall the Cauch-Riemann equations: Write f = u + iv. Then, u x = v u = v x R k fz = i f z + k f = f + k f = u + k u + i v + k v. Let a = u + k u and b = v + k v. Then R kf = a + ib. The aim of this section is to calculate the Jacobian of R k f. JR k f = a x b a b x. From the Cauch-Riemann equations, one can obtain the following useful identities: Anticipating a simplification, define u = u xx, v = v xx, u x = v xx, v x = c xx. A = v xx, B = u xx, C = v, D = u, 18 X = A + k C, Y = B + k D. Theorem 5.1. Let f M k. Then JacR k f = X + Y k uy + vx. We will separate the calculation in several parts: 5.1. a x b. We begin b calculating a x : a x = u + k x c = u x + k u x Now b. b = v + k v = v + k v + k v

8 8 OSCAR E. GONZÁLEZ Multipling these two: 5.. a b x. a x b = 4u x v + k u xv k u xv k v u x k u xv + k u xv = 4v xx v xx k v xxv + 4k v xxv k v v + k v v We begin b calculating a : a = u + k c = u + k u + k u Now b x. b x = v + k x v Multipling these two: = v x + k v x a b x = 4v x u + k v xu k v xu k u v x k v xu + k v xu 5.. a x b a b x. = 4u xx u xx + k u xxu 4k u xxu + k u u k u u Adding these, we obtain 4v xx v xx + u xx u xx k v xx v + u xx u + 4k v xx v + u xx u k v v + u u + k v v + u u Then JacR k f = X + Y k uy + vx X, Y, u and v, we get. Writing a and b in terms of 19 0 b = X k v, a = Y k u. Then, when a = b = 0 we have 1 JacR k f = b + v v + a + u u.

9 ZEROS OF MAASS FORMS 9 If f is the weight k Eisenstein series, then, v = k σ k 1 ne πn sinπnx B k n=1 u = 1 k σ k 1 ne πn cosπnx. B k n=1 Question 5.1. If f M k, does the Jacobian of R k f vanish? 6. Conjectures and Future Work Let R j k = R k+j R k+ R k and suppose f is a modular form of weight k. Then, j 4 R j j k + k f = 1 r 4 r rj r r j r D r f r=0 where a m = aa + 1 a + m 1 is the Pochhammer smbol. See 4.15 of [4] for details. Also, it can be shown that 5 E j k = 1j k j πi j gcdc,d=1 c j cz + d k+j Hence, in the case of Eisenstein series, we have 6 R j k E kz = k j j j j i r c r cz + d k+r r r=0 Conjecture 6.1. All of the zeros of R k E k z inside the fundamental domain lie on A. Conjecture 6.. R j k E k has the same amount of zeros as E k+j. Furthermore, all of the zeros of R j k E k inside the fundamental domain lie on A. Conjecture 6.. ReE k z > 0 on the interior of F. References [1] J. Getz. A generalization of a theorem of Rankin and Swinnerton-Der on zeros of modular forms. Proceedings of the American Mathematical Societ, 18:1 1, Januar 004. [] R. Holowinsk and K. Soundararajan. Mass equidistribution for hecke eigenforms. Annals of Mathematics, 17: , September 010. [] N. Koblitz. Introduction to Elliptic Curves and Modular Forms. Springer-Verlag, [4] J. Lewis and D. Zagier. Period functions for Maass wave forms. Annals of Mathematics, 151:191 58, Januar 001. [5] F. K. C. Rankin and H. P. F. Swinnerton-Der. On the zeros of Eisenstein series. Bulletin of the London Mathematical Societ, : , [6] Z. Rudnick. On the asmptotic distribution of zeros of modular forms. IMRN, pages , 005. Department of Mathematics, Universit of Puerto Rico, San Juan, PR address: oscar.gonzalez@upr.edu