An Exact Average Formula for the Symmetric Square L-Function at the Center

Transcription

1 An Exact Average Formula for the Symmetric Square L-Function at the Center Thesis by Aliekber Gürel In Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy California Institute of Technology Pasadena, California 008 Defended November 16, 007

2 ii c 008 Aliekber Gürel All Rights Reserved

3 iii In memory of my grandfather Hacı Paşa

4 iv Acknowledgements First, I want to thank my advisor, Dinakar Ramakrishnan, for getting me excited about the problem and for his kindness and endless support. Without his guidance and enlightenment, I would have been searching for a black cat in a dark room without knowing why. I thank Matthias Flach, Eric Wambach, and Jim Brown for graciously agreeing to be on my committee and for their useful comments. Special thanks to Solomon Friedberg, Peter Sarnak, Philippe Michel, David Whitehouse, and Kimball Martin for their time and helpful comments. Peter Sarnak also taught me that there is always time for L-functions, even when you are about to catch a flight. My fellow graduate students made life more enjoyable at Caltech. I thank Bahattin Yıldız, Roberto Carlos Pelayo, Orestis Raptis, Tonči Crmarić, Jerome Pinzuti, and many others for their friendship and support. Great thanks go to the department staff for their assistance whenever needed: Stacey Croomes, Kathy Carreon, Pamela Fong and, Seraj Muhammed. Last but not least, I would like to thank my parents and brothers for their support and love, which helped me get through toughest times in my journey.

5 v Abstract In this thesis, we find an exact formula for the weighted average of the symmetric square L-values at the center. The average is taken over a Hecke eigen basis of cusp forms of SL Z with a fixed weight k. The weights are the n-th Fourier Coefficients of these functions. The terms in the formula involve quadratic Dirichlet L-values at the center, Confluent Hypergeometric functions, and some arithmetic functions. The main ingredient, and the starting point, is a formula due Shimura, which relates the symmetric square L-function of a Hecke eigen form f to the inner product of f with the product of the theta function, θ; and a real analytic Eisenstein series of half integral weight, E. We apply Michel-Ramakrishnan s averaging technique on Shimura s formula to write the weighted average of symmetric square L-values in terms of the Fourier coefficients of the Eisenstein series. There are two complications. First, the levels of θ E and f are different. Second, E is not holomorphic. That is why we first take trace of θ E, and then we take the holomorphic projection. Computing the Fourier coefficients of the resulting function gives us the exact formula desired. Finally, we deduce the asymptotic behavior of these formulas as k.

6 vi Contents Acknowledgements iv Abstract v 1 Introduction Definitions and Notation Shimura s Formula Exact Average Principle Main Theorem: The Exact Average Formula at the Center Some Comments Computation of Trace Definition of Trace A set of representatives for Γ 0 4\Γ Transformation formulas for θz Transformation formulas for Ez, s Fourier Coefficients of T r1θzez, 4 s Fourier Coefficients of E and F Holomorphic Cuspidal Projection. 3.1 Definition of holomorphic cuspidal projection Growth Condition Holomorphic Cuspidal Projection of T rθzez, s

7 vii 4 Calculations for the Main Formula Calculations for b n, c n, and P n m, s Calculations for P n m, s Calculations for b n s Calculations for c n s Calculations for g ns Simplifying A n,k Main Theorem: The Exact Average Formula at the Center Asymptotic Behavior as k Asymptotic of A n,k Asymptotic of B n,k Asymptotic of the Main Formula Bibliography 53

8 1 Chapter 1 Introduction. 1.1 Definitions and Notation Let k > 1 be a positive integer and H k be the Hecke eigen-basis of S k SL Z, normalized so that the first Fourier coefficients are all 1. For f H k, we will denote its Fourier coefficients by c n f and normalized Fourier coefficients by λ n f: fz = c n fenz = λ n fn k 1/ enz weight k, level 1 n=1 n=1 We will let θz be the usual theta function: θz = en z weight 1/, level 4 n= For a positive integer N and a complex number s, Ez, s, 4N will denote a nonholomorphic real analytic Eisenstein series of half integral weight, which transforms like a modular form of weight k 1/ and level 4N: Ez, s, 4N = y s/ g Γ \Γ 0 4N jg, z 1 4k jg, z s weight k 1/, level 4N By abuse of notation, when N = 1 we will simply write Ez, s for Ez, s, 4, since we will mainly work with level 4. We will also apply similar abuse of notations later on with the dual Eisenstein series.

9 We use the standard notation Γ = SL Z, Γ = a b Γ 0 d, and Γ 0N = a b Nc d Γ c Z. z = x + iy is a point in the upper half plane H, ez = e πiz, and jg, z = 4c ε 1 d d 4cz + d1/, for g = a 4c b d Γ 0 4, where 4c d is the Kronecker s symbol, εd = 1 or i depending on d 1 or 3 mod 4, and the branch of square root is chosen so that 1 = +1. For a real analytic modular form h 1 and a real analytic cusp form h, both having the same weight k and level R, let < h 1, h > R := Γ 0 R\H k dxdy h 1 zh zy, y where H is the upper half plane, and z = x + iy. When R = 1, we will ignore the index and simply write < h 1, h >. δx will be the integer indicator function, i.e. δx = 1 or 0 depending on x is an integer or not. For a non-zero integer n, < n > will denote its square-free part with the sign. χ n will denote the character defined by χ n d = n d. For a character χ, and a

10 3 positive integer N, we let Ls, χ = n=1 L N s, χ = n,n=1 χn n s, χn n s = p N 1 χp 1. p s F 1 is the confluent hypergeometric function defined as F 1 a, b; c; z = n=0 a n b n c n We use the following conventions: z n n!, where r n = rr r + n 1. fk fk gk if lim = 1, and k gk fk << gk if there is a positive constant A s.t. fk < A gk, k. Finally, we let e k = k! 4π k 1 1. Shimura s Formula By using the Euler product Lf, s := c n fn s = p n=1 [1 α p p s 1 β p p s ] 1 product over primes p, one defines the symmetric square L-function as Lsym f, s := p [1 α pp s 1 α p β p p s 1 β pp s ] 1. By Theorem 1 in Shimura [8], Lsym f, s is a meromorphic function of s, with possible simple poles at only k and k 1. It also satisfies the following functional

11 4 equation. The function Λsym f, s = π s ΓsΓs k + 1ζs k + 1Lsym f, s is invariant under the transformation s 4k 1 s. symmetry, s = k 1/, the central point. We will call the center of An analogue of this result is known over arbitrary number fields by Gelbart and Jacquet [1]. If we put Λ u sym f, s = Λsym f, s + k 1, then we have the unitary functional equation Λ u sym f, s = Λ u sym f, 1 s. Moreover, by the formula 1.5 in Shimura [8], we have the following equation relating the symmetric square L-function to the Eisenstein series and theta function: Γs/Lsym f, s 4π s/ ζs 4k + = k dxdy fzθzez, s + 4ky 1.1 Γ 0 4\H y When s R, all the terms on the LHS of Shimura s equation 1.1 are real, and so, taking complex conjugate of RHS will not affect the identity. We can rewrite it as Γs/4π s/ ζs 4k + Lsym f, s = < θe., s + 4k, f > Exact Average Principle Starting with Shimura s Integral representation formula 1., we will apply the averaging method, developed and used by Michel-Ramakrishnan [5], to represent the average of symmetric square L-values in terms of the function θzez, s. However, there are a few complications. One is that the levels do not match. Another is that

12 5 the function Ez, s is real analytic but not holomorphic. To resolve these problems, we will first take trace and then holomorphic cuspidal projection of θzez, s. Let T r be the trace map sending real analytic modular forms of level 4 down to real analytic modular forms of level 1, and Pr 0 hol be the holomorphic cuspidal projection of a real analytic modular form. Define the following function: Then, G s z := Pr 0 holt rθzez, s = g n senz. n=1 Γs/4π s/ ζs 4k + Lsym f, s = < θe., s + 4k, f > 4 = < T rθe., s + 4k, f > = < G s+ 4k, f > 1.3 Averaging these equations over the orthogonal H k, we get Γs/4π s/ ζs 4k + Lsym f, s < f, f > f = G s+ 4k. 1.4 f H k We can rewrite this equation in terms of the Fourier coefficients as Γs/4π s/ n k 1/ ζs 4k + Lsym f, sλ n f < f, f > f H k = g n s + 4k 1.5 Note that, ζs 4k + has a simple pole at the central value s = k 1/. On the other hand, Lsym f, s is regular by Theorem 1 in Shimura [8]. We deduce that g n s + 4k has a zero at this point, and we can rewrite the equation as 4Γs/4π s/ n k 1/ s 4k + 1ζs 4k + Lsym f, sλ n f < f, f > f H k = g ns + 4k s k + 1/

13 6 As s k 1/, we get 4Γk 1/4n k 1/ Lsym f, k 1/λ n f 4π k 1/4 < f, f > f H k = g n3/ k 1.6 To find these coefficients {g n }, we need to understand T rθzez, s first. In Chapter we will find the Fourier coefficients of T rθzez, s in terms of the Fourier coefficients of the Eisenstein series. Then, in Chapter 3 we will see how these Fourier coefficients are affected by taking the holomorphic cuspidal projection. We will spend Chapter 4 with calculations to find {g ns}, which finally gives rise to the main formula. Before stating the Main Theorem, recall the following notations from Section 1.1: For a non-zero integer n, < n > denotes its square-free part with the sign. δx = 1 or 0 depending on x is an integer or not, and e k = k! 4π k 1

14 7 1.4 Main Theorem: The Exact Average Formula at the Center Main Theorem. For integers k > 1 and n > 0, we have the following formula at the center of symmetry s = k 1/: f H k L sym f, k 1 ek λ n f < f, f > = A n,k + m Z, m 4n 1 L, χ 4<m 4n> ξ n,k m. A n,k = δ n n 1/4 Γ k 1/ Γk 1/ + A lnn, where A = π + 6γ 6 ln π 10 ln 4 The weights ξ n,k m are positive for all but finitely many m. In fact, ξ n,k m > 0, whenever m > n. Explicitly, ξ n,k m = β4n m γ4n m F n,k m. For n 0, we write n = d n 1 with n 1 odd. Then βn = γn = µa 4 < n > a a a,b>0, ab n 1 d/ d even, n 1 1 mod 4, d/ + d even, n 1 3 mod 8, d/ + + d even, n 1 7 mod 8, d/ 1/ d odd. Finally, F n,k m = π 1 k Γk 1/4 n 1/4 Γk+1/4 F 1 1 4n k 1/4 m 4n n 1/4 m Ik m k 1, 3 k; 1; m 4 4 4n if m < 4n, if m > 4n, where I k c = 0 u k 3/4 du 1 + u k+1/4 1 + cu k 1/4

15 8 In Chapter 5, we will investigate the asymptotic and deduce two Corollaries: Corollary 1. For a fixed square integer n, letting k, f H k L sym f, k 1 ek λ n f < f, f > = 1 Γ k 1/ n 1/4 Γk 1/ + A lnn 1 +O k In particular, when n = 1, we have f H k L sym f, k 1 e k < f, f > = Γ k 1/ 1 Γk 1/ + A + O k Assuming that Lsym f, k 1 0 for all f H k, we get f H k L sym f, k 1 << kln k 4. An analogous result for the case n = 1 was recently proved by R. Khan [4] with an error term of Ok 1/0+ɛ. Corollary 1 improves the bound on the error term to Ok 1/ and extends the result to arbitrary n. More importantly, ours is an exact formula, and the error term is also given explicitly by the Main Formula. Corollary. For a fixed non-square integer n, we have f H k L As k, sym f, k 1 ek λ n f < f, f > = 1 L, χ 4<m 4n> ξ n,k m m Z f H k L sym f, k 1 ek λ n f 1 < f, f > = O k

16 1.5 Some Comments 9 A By Shimura s formula 1., Lsym f, s or Λsym f, s is essentially given by the scalar product < θe, s + 4k, f >. On the other hand, J. Sturm s expression for the Fourier coefficients of the series Ez, s involves a quadratic Dirichlet L-function, Ls + k 1, χ 4<n> see Section.6. So the nth Fourier coefficient of Ez, s + 4k involves Ls + 4k + k 1, χ 4<n> = Ls k + 1, χ 4<n> which becomes L1/, χ 4<n> at the central point s = k 1/. This explains the occurrence of the central values of Dirichlet L functions in the Main Theorem. B Perhaps, it is most interesting when n is non-square. Because, in this case our formula relates a finite weighted sum of Symmetric Square L-values at the center to an infinite weighted sum of quadratic Dirichlet L-values at the center. Both L-values are expected, by the Generalized Riemann Hypothesis, to be non-negative. Even though we are far from proving either, we hope to understand, in a sequel, the compatibility aspect of these two L series. C Assuming the non-negativity condition in Corollary 1, the bound we get for an individual Lsym f, k 1/ is k 1+ɛ which is weaker than the convexity bound k 1/. D For k = 6, S k Γ is one dimensional and H 1 = { } where z = q n 11 q n 4 = τ n q n n=1 q = ez. In this case, the Main Theorem relates the product of τ n and L sym, 11 infinite weighted average of quadratic Dirichlet L-values at the center. to an

17 10 Chapter Computation of Trace..1 Definition of Trace In this chapter we will find the Fourier coefficients of the trace of θzez, s in terms of the Fourier coefficients of the Eisenstein series. First, let us explain what we mean by this trace map. Let M k Γ 0 N be the set of modular forms of level N and weight k, which are real analytic but not necessarily holomorphic. For N M, we define the trace map T r M N : M k Γ 0 M M k Γ 0 N as follows: where for any γ = a c T rn M hz := h γ z b SL Z, d γ Γ 0 M\Γ 0 N h γ z = cz + d k hγz. Remark.1. Note that 1 h γ1 γ = h γ1 γ h γ doesn t depend on the right coset representative γ.

18 11 3 For any element g Γ 0 N, as γ runs through a list of right coset representatives, γg runs through a list of right coset representatives as well. Hence, the image T rn Mh is a modular form of level N. 4 It is not difficult to show that for a cusp form f with weight k and level N and a real analytic modular form g of weight k and level M where N M, one has: < f, T r M N g > N =< f, g > M. To proceed we need to find a set of right coset representatives of Γ 0 4\Γ first.. A set of representatives for Γ 0 4\Γ Lemma.. The set of matrices g 0 = I, g 1 = 1 0 1, g +j = j, j = 0, 1,, 3 is a set of representatives for the right cosets of Γ 0 4 in Γ = SL Z, i.e., 5 Γ = Γ 0 4g j. j=0 Proof. We need to show that any g Γ is in exactly one of these cosets. Let g = a Γ. Since detg = 1, we know that gcda, c = 1. So t = gcd4, c = c gcd4a, c is a divisor of 4. Hence, t = 1,, or 4. Also, in any case, there exists relatively prime integers m and n such that t = 4am + cn. Case 1: t = 4. In this case, c is a multiple of 4, and g is already in Γ 0 4 = Γ 0 4g 0. So, there is only one right coset in this form, and a representative is g 0 =

19 1 Case : t =. In this case, c is even but not divisible by 4. So = 4am + cn implies that n is odd, and since it is relatively prime with m, we conclude that gcd4m, n = 1. So, there exists a matrix h = Γ 0 4. Note that 4m n hg = 4m n a c = Γ 0 4g. Replacing g with hg in the same right coset, we may assume that c =, i.e., g = a. Given two matrices g and g in Γ, Γ 0 4g = Γ 0 4g g g 1 Γ 0 4. Let g = a b d and g = a b d. Observe that both d and d are odd because detg = detg = 1. So, g g 1 = a b d d b a = d d Γ 0 4, meaning that there is only one right coset in this form, and a coset representative,

20 13 for instance, is g 1 = Case 3: t = 1. In this case, c is odd, and 1 = 4am + cn. Letting h = c a 4m n Γ 0 4, we see that hg = c a 4m n a c = 0 1 Γ 0 4g. Hence, we may assume that a = 0, c = 1. Then, detg = 1 implies that g = Let g = d, g = f Γ. g g 1 = f d = d f Γ 0 4 d f mod 4. So, Γ 0 4g = Γ 0 4g if and only if d f mod 4..1 Hence, there are exactly four right cosets in this case, one for each number modulo

21 14 4, and we introduce the following four coset representatives: g = , g 3 = , g 4 = 0 1 1, and g 5 = So far, we have proved that any element g Γ is in one of the 6 right cosets, with representatives given as above. To prove the claim, it remains to check that these 6 cosets are different. Given h = 4m n Γ 0 4, and g = a c Γ, we have hg = 4m n a c = 4ma + nc. Note that n is odd because deth = 1. So, gcd4ma + nc, 4 = gcdnc, 4 = gcdc, 4. Observe that, gcd4, c depends only on the right coset Γ 0 4g, but not on the coset representative g. This shows that the right cosets we have found in different cases above are different. Finally, we have already seen that the right cosets of g, g 3, g 4 and, g 5 are all different; thus, the lemma is proved.3 Transformation formulas for θz Recall that Let θz = en z weight 1/, level 4. n= θ g z = cz + d 1/ θgz for any g = a b SL Z. c d

22 In this section we will find transformation formulas for θ gj z, j = 0,..., Lemma.3. We have the following transformation formulas: θ g1 z = θz/4 θz, and θ gj+ z = 1 i z + j θ for j = 0, 1,, 3. 4 Proof. We will use the following properties of the theta function repeatedly in the proof: θz + 1 = θz, θ 1 z = θz, and θ z + 1 = θ4z θz. 4z i The first two are famous formulas, whereas the third one can be obtained easily by using the Fourier expansion. Transformation formula at g 1 : 1 z 1 z 1 z 1 θ g1 z = z + 1 θ = θ z + 1 z + 1 iz 4z 1 1 = iz θ = 4z iz θ + 1 4z z z z = θ θ = iz z 4z iz i θ 4 i θz 1 z z 1 z = iz i θ θz = θz/4 θz. 4 iz i Transformation formulas at g, g 3, g 4, and g 5. For j = 0, 1,, 3 we have: θ gj+ z = = z + j z + j z + j θ = θ z + j z + j i 4 i z + j θ = 1 i z + j θ. 4 4

23 .4 Transformation formulas for Ez, s Recall that for a positive integer N, 16 Ez, s, 4N = y s/ g Γ \Γ 0 4N jg, z 1 4k jg, z s, weight k 1/, level 4N. We also define the dual Eisenstein series F as F z, s, 4N = 4N z 1/ k E 1, s, 4N. 4Nz In particular, For any g = a c b d F z, s = z 1/ k E Γ, let 1 4z, s. E g z, s, 4N = cz + d 1/ k Egz, s, 4N. In this section, we will find E gj z, s for j = 0,..., 5 in terms of E and F. Proposition.4. We have the following transformation formulas: E g1 z, s = s Ez/4, s, 8 Ez, s, and z + j E gj+ z, s = 1/ k F 4, s, for j = 0, 1,, 3. Proof. Transformation formula at g 1 : Let R 1 be a set of representatives for Γ \Γ 0 4, W 1 = {c, d Z gcd4c, d = 1, c 1 = c + d > 0, d 1 = d}, and V 1 = {c 1, d 1 Z gcd4c 1, d 1 = 1, c 1 > 0, c 1 odd}.

24 Note that there is a one to one correspondence among R 1, W 1, and V 1 : 17 a 4c b d R 1 c, d W 1 c + d, d V 1. On one hand, E g1 z, s = z z + 1 1/ k E z + 1, s = s/ z z + 1 1/ k Im z + 1 On the other hand, = z + 1 1/ k y s/ z + 1 s = y s/ ε 1 d c,d W 1 = y s/ = y s/ c 1,d 1 V 1 ε 1 4c d g R 1 j ε 1 d c,d W 1 g, z z+1 jg, z s z+1 4c d 4c + dz + d 1/ k 4c + dz + d s 1 4k c1 d 1 c1 z + d 1 1/ k d 1 c 1 z + d 1 s d 1 ε 1 d 1 c 1 odd>0 gcd4c 1,d 1 =1 Ez, s = y s/ 1 + c>0 gcd4c,d=1 = y s/ 1 + ε 1 d c even>0 gcd4c,d=1 c1 d 1 4c d ε 1 d 4cz z+1 + d1/ k 4cz z+1 + d s c1 z + d 1 1/ k c 1 z + d 1 s 4cz + d 1/ k 4cz + d s c cz + d 1/ k. d cz + d s Adding these two equations for E g1 z, s and Ez, s we get: E g1 z, s + Ez, s = s Ez/4, s, 8

25 18 Transformation formulas at g, g 3, g 4, and g 5. For j = 0, 1,, and 3 we have: 1 E gj+ z, s = z + j 1/ k E z + j, s k 1/ z + j = z + j 1/ k F z + j = 1/ k F 4, s. z + j 4, s.5 Fourier Coefficients of T r 4 1θzEz, s From the definition of trace and the transformation formulas derived earlier, we get G s z : = T r 4 1θzEz, s = 5 θzez, s gj = j=0 5 θ gj ze gj z, s = θzez, s + θz/4 θz s Ez/4, s, 8 Ez, s + 1 i 3 z + j z + j θ F 1/+k 4 4, s. j=0 j=0 Remark.5. Gs z is z z + 1 invariant. So are 3 j=0 θ z+j 4 F z+j, s and 4 θzez, s. We conclude that θ g1 ze g1 z, s is also z z + 1 invariant. This is not a surprise and could be derived by noticing that the right coset Γ 0 4g 1 is invariant when we multiply with 1 1 on the right. Using this with 0 1 z + 1 θ 4 z iθ = 1 iθz, 4 which can be proved easily using the Fourier expansion of θ, we conclude that E z + 1 z 4, s, 8 + ie 4, s, 8 = 1 + i s Ez, s.

26 19 Let the Fourier expansion of G s be: G s z = n Z g n y, senx. Also, let the Fourier expansions of E and F be: Ez, s, 4N = n Z A n y, s, 4N enx, and F z, s, 4N = n Z B n y, s, 4N enx. By considering the contribution of each term θ gj ze gj z, s towards g n, we get g n y, s = e πly A n l y, s, 4 + s l Z l odd 1 i 4 1/+k l Z e πl y 4 A 4n l y/4, s, 8 e πl y 4 B 4n l y/4, s, 4 Note that the term θz/4 θzez, s has no contribution in g n, and the contributions of θ gj+ ze gj+ z, s are equal for j = 0, 1,, 3. Next, let us see what these Fourier coefficients A n and B n are. +.6 Fourier Coefficients of E and F The formulas in this section are due Shimura [8] Proposition 1 and Sturm [9] Lemma. Note that our definition of E and F is slightly different than their definition of E and E. To be more precise, we have Ez, s, 4N = y s/ E JS z, s, λ, ω 0 and F z, s, 4N = 4N λ+s y s/ E JSz, s, λ, ω 0, where ω 0 is the trivial character, λ = k 1/, and to avoid confusion we used E JS and E JS for Sturm s definition of E and E.

27 0 Fourier Coefficients of F z, s, 4N. First we need to define b n and τ n. We start with b n : For a non-zero n: b 0 s, 4N = b 0s, 4N = L4N s + 4k 3 L 4N s + 4k b n s, 4N = L4N s + k 1, 4<n>N. β L 4N n s, 4N, s + 4k and b ns, 4N = L4N s + k 1, 4<n>. β L 4N s + 4k ns, 4N, with < n > denoting the square-free part of n when n 0: β n s, 4N = 4 < n > N µa a a 1 k s b 3 4k s, and βns, 4N = 4 < n > µa a 1 k s b 3 4k s, a the sums being over positive integers a, b satisfying ab n and gcdab, 4N = 1. Next, τ n is defined using the equation z + n α z + n β = τ n y, α, βenx. n= n= It can be written in terms of the Whittaker function as: n α+β 1 e πny Γα 1 W 4πny, α, β if n > 0, i α β π α β τ n y, α, β = n α+β 1 e π n y Γβ 1 W 4π n y, β, α if n < 0, Γα 1 Γβ 1 Γα + β 14πy 1 α β if n = 0,

28 1 where the Whittaker function is W y, α, β := Γβ u α 1 u β 1 e yu du 0 It is first defined for Reβ > 0 and then extended to all β analytically with a functional equation W y, α, β = y 1 α β W y, 1 β, 1 α. With these definitions, F z, s, 4N = n Z B n y, s, 4Nenx, where B n y, s, 4N = 4N 1/4 k s/ y s/ b n s, 4Nτ n y, s + 4k 1, s Fourier Coefficients of Ez, s, 4N. For A n, we need one more definition. Let c n s, 4N = 4N M 4N M p=1 M ε 1 p enp/mm 1/ k M s. p Then Ez, s, 4N = n Z A n y, s, 4Nenx, where A n y, s, 4N = y s/ b ns, 4Nc n s, 4Nτ n y, s + 4k 1, s if n 0, A 0 y, s, 4N = y s/ 1 + b 0s, 4Nc 0 s, 4Nτ 0 y, s + 4k 1, s Now that we have a Fourier expansion for the trace, and the levels are set, we are ready to take the holomorphic projection.

29 Chapter 3 Holomorphic Cuspidal Projection. 3.1 Definition of holomorphic cuspidal projection Gross-Zagier [] Section IV.5 summarizes the definition of holomorphic projection in the following proposition: Proposition 3.1. Let F be a complex valued C function on H which transforms like a modular form of weight k for Γ with the Fourier expansion: F z = ã n yenx. n= If F z = Oy ε as y for some ε > 0, then F z = Pr 0 hol F z = a n enz is a holomorphic cusp form of the same weight k for Γ, where a n = proj n ã n y = 4πnk 1 k! n=1 Moreover < g, F > = < g, F > for all g S k Γ. 0 ã n ye πny y k dy. To apply the proposition to T rθzez, s, first we need to check the growth condition required by the proposition.

30 3. Growth Condition 3 We would like to take the holomorphic projection of T rθzez, s when s > s 0 = 3/ k is a real number in some small neighborhood of s 0. The reason for considering not a single value but infinitely many s values near s 0 is to be able to take the derivative of the Fourier coefficients later on. Lemma 3.. When s > s 0 is a real number close to s 0, T rθzez, s satisfies the growth condition stated in the proposition above. More precisely, there exists some r > 0 such that for any s s 0, s 0 + r, T rθzez, s = Oy u, as y, where u = s s 0 > 0. Proof. We will prove the lemma by checking the growth condition for each summand of trace separately, starting with θzez, s itself. To simplify the notation, let u = s s 0, α = s/ + k 1/, β = s/. θzez, s: Ez, s = y s/ + n Z b n s, 4c n s, 4τ n y, α, βenx. The first term in the product is b n s, 4 = L4 s + k 1, χ 4<n> β L 4 n s, 4, s + 4k where β n s, 4 = 4 < n > a u b u µa, a a the sum being over positive odd integers a, b such that ab divides n. Since each term in absolute value is less than or equal to 1, and the number of terms is no more than n, we deduce that β n s, 4 n. Also, it is not difficult to prove that L 4 s + k 1, χ 4<n> 4 < n > 4n.

31 4 On the other hand, as s s 0, s+4k 1, so the denominator L 4 s+4k 1 goes to infinity. Hence, its reciprocal is bounded, say by 1. We conclude that, in some s-neighborhood of s 0, b n s, 4 n 3 for non-zero n. When n = 0, note that b 0 s, 4 = L4 u L 4 u + 1 = 1 u ζu 1 1 u ζu + 1 has a double-zero at u = 0 because of the 1 u and ζu + 1 terms. b 0 s, 4 << u. So, The second term, c n s, 4, is given as an infinite summation: c n s, 4 = 4 M 4 M p=1 which by letting M = j can be written as j= j p=1 M ε 1 p enp/mm 1/ k M s, p p j ε 1 p enp/ j ju. j For non-zero n, let n = d n 1 where n 1 is odd. When j > d + 3, the summation over 1 p j can be split into four parts, one for each odd p-congruence class j modulo 8. The terms ε 1 p, depend only on p modulo 8, and in each congruence p class we are summing the roots enp/ j, which form powers of a nontrivial root of unity, ep/ j, each power counted with the same multiplicity. We conclude that the sum is zero, and the only contribution is when j d + 3. Hence, d+3 c n s, 4 ju d + << n. j= For n = 0, we have c 0 s, 4 = j= j p=1 p j j ε 1 p ju ju = j=0 1 1 u << 1 u

32 5 Next, we bound the τ n terms. By Lemma 4 in Shimura [8], y β W y, α, β and y α W y, β, α are both bounded when y > 1 as s s 0. Since α > k and β > k, we have W y, α, β << y k and W y, β, α << y k. Using the relation between the Whittaker function and τ n we get the following bound for n 0: τ n y, α, β << n k+1 y k e π n y. For n = 0, τ 0 y, α, β has a simple pole at u = 0, and from i α β π α β τ 0 y, α, β = Γα + β 1 4πy 1 α β, ΓαΓβ we deduce that τ 0 y, α, β << y u u Combining these bounds and summing over n Z, we get Ez, s = Oy u as y. Also, it is easy to see that θz = 1 + Oe y. These two bounds combine to give θzez, s = Oy u. θz/4 θz s Ez/4, s, 8 Ez, s: Ez, s, 8 = y s/ + n Z b ns, 8c n s, 8τ n y, α, βenx. Applying the same ideas as before, we note that b ns, 8 = b n s, 4 is bounded by n 3 and c n s, 8 is bounded by a multiple of n. So, the Eisenstein terms are bounded

33 6 by y u as before. In θz/4 θz, the constant terms cancel each other. Hence, it decays exponentially as y. We conclude that θz/4 θz s Ez/4, s, 8 Ez, s = Oe y/4 as y. 3 j=0 θ z+j 4 F z+j, s : 4 Let us check θzf z, s first. The rest follows by the substitution z z + j/4. F z, s = y s/ n Z 4 s/ k+1/4 b n s, 4τ n y, α, βenx. Using the same bounds from before for b n s, 4 and τ n y, α, β, we also get that F z, s is bounded by a constant times y u, so is the product θz/4f z/4, s. Similarly, the other terms are bounded by y u as well, and we conclude that 3 z + j z + j θ F 4 4, s = Oy u as y j=0 This finishes checking the growth condition, since u = s s 0 > 0 when s > s 0. Although we are restricted to take holomorphic projection only for s s 0, s 0 + r, it is not a problem for our purposes. Our goal is to find the derivative of the Fourier coefficients of the holomorphic projection at s = s 0 and for this, knowing these coefficients for an infinite sequence of s values converging to s 0 suffices. Now, we are ready to take holomorphic cuspidal projection of T rθzez, s for s values larger and close to s Holomorphic Cuspidal Projection of T rθzez, s Recall that T rθzez, s = n Z g n y, senx M k Γ.

34 7 So far, we have computed these coefficients g n in terms of the fourier coefficients of E and F : g n y, s, 4 = e πly A n l y, s, 4 + s e πl y 4 A 4n l y/4, s, 8 l Z l odd 1 i +4 e πl y 1/+k 4 B 4n l y/4, s, 4 l Z Let the holomorphic cuspidal projection of T rθzez, s be g n senz S k Γ. n=1 For a positive integer n, we write these coefficients g n holomorphic cuspidal projection: using the definition of g n s = proj n g n y, s = 4πnk 1 k! 0 g n y, se πny y k dy Using the formula we derived earlier for g n y, s, we get g n s = δ n proj n e πny y s/ + l Z b n l s, 4c n l s, 4proj n + l odd b 4n l s, 8c 4n l s, 8proj n + 4k s 1 i l Z e πly y s/ τ n l y, α, β e πl y 4 y s/ τ 4n l y/4, α, β b 4n l s, 4proj n e πl y 4 y s/ τ 4n l y/4, α, β From the definition of proj n, it is immediate that proj n ãy/4 = proj 4n ãy.

35 8 Using this formula in the above equation for g n and letting P n m, s = proj n e πmy y s/ τ n m y, α, β for integers m 0, n > 0, we get g n s = δ n proj n e πny y s/ + b n l s, 4c n l s, 4P nl, s l Z + s b 4n l s, 8c 4n l s, 8P 4nl, s l odd + 4k s 1 i b 4n l s, 4P 4n l, s. l Z We will spend the next chapter with calculations to evaluate g ns 0.

36 9 Chapter 4 Calculations for the Main Formula. 4.1 Calculations for b n, c n, and P n m, s In this section, we will make calculations for the terms b n, c n and P n m, s at the point s 0 = 3/ k. Let us start with P n m, s Calculations for P n m, s. Recall that P n m, s = proj n e πmy y s/ τ n m y, s + L, s for integers m 0, n > 0, where τ n and proj n are as defined in the previous chapters. Let s 0 = 3/ k, u = s s 0, α = s/ + k 1/, α 0 = s 0 / + k 1 = 1/4 + k, β = s/, β 0 = s 0 / = 3/4 k.

37 30 For 0 m < n, let c = n m Then we have n P n m, s 0 = proj n e πmy y 3/4 k τ n m y, α 0, β 0 where = 4πnk 1 k! = π i 1 k 4πn k 1 Γk + 1/4k! 0 e πm+ny y k 1/4 τ n m y, α 0, β 0 dy y = π i 1 k 4πn k 3/4 Γk + 1/4k! J kc, J k c = 0 0 e 4πny y k 1/4 W 4πn my, α 0, β 0 dy y e y y k 1/4 W cy, α 0, β 0 dy y There are a few definitions of confluent hypergeometric functions which are closely related: W y, α, β = Ua, b, z = The relations being 1 Γβ 1 Γa u α 1 u β 1 e uy du, t a t b a 1 e zt dt, M k,µ z = z 1/+µ e z/ 1F 1 1/ + µ k, µ + 1; z, W k,µ z = Γ µm k,µz Γ1/ µ k + ΓµM k, µz Γ1/ + µ k W y, α, β = Uβ, α + β, y and W k,m z = e z/ z m+1/ U1/ + m k, 1 + m; z. Combining them, we get W y, α 0, β 0 = W y, 1 β 0, β 0 = Uβ 0, 1, y = y 1/ e y/ W 1/ β0,0y.

38 31 Let us rewrite J k c using this relation: J k c = 1 c 0 e 1 c/y y k 3/4 W k 1/4,0 cy dy y The following formula Bateman [7], page 16 relates the Laplace transform of the Whittaker function W k,m to the Hypergeometric function F 1 : 0 e pt t v W k,µ at dt t = Γµ + v + 1/Γv µ + 1/aµ+1/ Γv k + 1p + a/ µ+v+1/ F 1 µ + v + 1/, µ k + 1/; v k + 1; p a/ p + a/ With the following substitutions: p 1 c/, a c, k k 1/4, v k 3/4, µ 0, the formula reduces to J k c = Γk 1/4 F 1 k 1 π 4, 3 4 k; 1 ; 1 c Hence, P n m, s 0 = 1 + i π 1 k 4πn k 3/4 Γk 1/4 Γk + 1/4k! F 1 k 14, 34 k; 1 ; 1 c

39 3 For 0 < m = n we have P n n, s = proj n e πny y β τ 0 y, α, β = proj n e πny y β i β α α+β Γα + β 1 π 4πy 1 α β ΓαΓβ = i 1/ k α β Γα + β 1 π proj n y 1 α e πny ΓαΓβ = i 1/ k α β Γα + β 1 4πn k 1 π y k α 4πny dy e ΓαΓβ k! 0 y = i 1/ k α β Γα + β 1Γ1/ β 4πn α 1 π ΓαΓβ k! = 1 + i 1k π4πn k 3/4 k! πn u/ ΓuΓk 1/4 u/ Γk + 1/4 + u/γ3/4 k + u/ For 0 < n < m, let c = m n/m. We have P n m, s = proj n e πmy y β τ n m y, α, β = proj n e πmy y β i β α π α+β m n α+β 1 e πm ny Γβ 1 W 4πm ny, β, α = i1/ k π α+β m n α+β 1 proj n y β e πm ny W 4πm ny, β, α Γβ = i1/ k π α+β m n α+β 1 proj n y β e πm ny 1 + u β 1 u α ΓαΓβ e 4πm nyu = i1/ k π α+β m n α+β 1 ΓαΓβ = i1/ k k 1/ π α m n α+β 1 m k 1+β 4πn k 1 k! At the center, s 0 = 3/ k, we get n k 1 k! 0 0 Γk 1 + β ΓαΓβ P n m, s 0 = 1 + i4πk 3/4 Γk 1/4 k! 0 y k 1+β dy e 4πm1+cuy y 0 n k 1 m k 1/4 I kc, du u 1 + u β 1 u α du u 1 + u β 1 u α 1 + cu k 1+β du u where I k c = 0 u k 3/4 du 1 + u k+1/4 1 + cu k 1/4

40 Remark 4.1. Note that, in all the cases, we have the relation 33 P n m, s = 4 1 α P 4n 4m, s. We summarize these calculations in the following lemma. Lemma 4.. We have the following values for P n m, s 0, where n > 0 and m 0: 1 If 0 m < n, letting c = n m n, P n m, s 0 = 1+i π 1 k 4πn k 3/4 Γk 1/4 Γk + 1/4k! F 1 When 0 < m = n, P n n, s has a simple pole at s = s 0 and, P n n, s = 1 + i 1k π4πn k 3/4 k! 3 If 0 < n < m, letting c = m n m, where for c 0, 1, P n m, s 0 = 1 + i4πk 3/4 Γk 1/4 k! I k c = 0 k 14, 34 k; 1 ; 1 c πn u/ ΓuΓk 1/4 u/ Γk + 1/4 + u/γ3/4 k + u/ n k 1 u k 3/4 du 1 + u k+1/4 1 + cu k 1/4 m k 1/4 I kc, 4.1. Calculations for b n s Recall that for non-zero n, b n s, 4N = L4N s + k 1, 4<n>N β L 4N n s, 4N, s + 4k and b ns, 4N = L4N s + k 1, 4<n> β L 4N s + 4k ns, 4N,

41 34 with < n > denoting the square-free part of n, β n s, 4N = 4 < n > N µa a a 1 k s b 3 4k s, and βns, 4N = 4 < n > µa a 1 k s b 3 4k s a The sums are over positive integers a, b satisfying ab n and gcdab, 4N = 1. Also, b 0 s, 4N = b 0s, 4N = L4N s + 4k 3 L 4N s + 4k. To simplify the notation, let us introduce b n s := b n s, 4, β n s := β n s, 4. Note that b n s, 4 = b ns, 4 = b ns, 8, and that they all vanish at the center, s 0 = 3/ k, because the denominators, L 4N s + 4k, have a pole at this point. For our computations we still need to know their derivatives. For a non-zero n =< n > m, where < n > is a square-free integer, we have L 4 b s + k 1, 4<n>. ns 0 = β L 4 n s s + 4k s=s 0 L 4 s + k 1, 4<n>. = lim s s0 s + k 3/L 4 s + 4k β ns = L 4 1/, 4<n>. lim s s0 s s 0 L s s 0 β ns 0 = L4 1/, 4<n>. lim v 1 v 1L 4 v β ns 0 v = 1 + s s 0 L 4 1/, 4<n>. = 1 1 lim v 1 v 1ζv β ns 0 = 4 L 1/, χ 4<n> β n s 0 using Res s=1 ζs = 1

42 35 Here, β n s 0 = 4 < n > µa, a a the sum being over positive odd integers a, b satisfying ab m. Next, let us see what happens when n = 0: b 0 s = L4 s + 4k 3 L 4 s + 4k = 1 s 4k+3 ζs + 4k 3 1 s 4k+ ζs + 4k Note that, as s s 0 = 3/ k, s + 4k 3. The numerator has a simple zero, and the denominator has a simple pole. Hence, b 0 s has a double zero at s = s Calculations for c n s Let u = s s 0 = s + k 3/. Recall that c n s, 4N = 4N M 4N M M p=1 p ε 1 p M 1+u enp/m, where M 4N means that all prime divisors of M are also divisors of 4N, and ε p = 1 or i depending on p 1 or 3 mod 4. We will find the values of c n s, 4 and c n s, 8. From the definition, note that they only differ by M = 4 terms. This simple observation gives us the following relation: c n s, 4 = c n s, nn+1 1 i 4 1+u We will first calculate c n s, 8 and then use this relation to find c n s, 4.

43 36 When n = 0, c 0 s, 8 = = = c 0 s, 4 = 8 M 8 M M p=1 ε 1 p p = M 1+u j=3 j p=1 p j1+u j ε 1 p j i + 1 j + i j 3 j=3 j=3 j1+u j 1 i = 1 i 3+ju 4 1 i 44 u 1 j 4, even 1 ju = 1 i 4 u+1 4 u 1 and Note that c 0 s, 8 and c 0 s, 4 both have a simple pole at s = s 0 with residue: Next, for non-zero n, lim s s 0 c n s, 8 = lim s s 0 c n s, 4 = 1 i s s 0 s s0 4 ln 4 c n s 0, 8 = 8 M 8 M M p=1 p M ε 1 p e np M = j=3 S n j j, where j j np S n j = ε 1 p e. p j p=1 To compute S n j, let us first investigate the product +1, if p 1, 7 mod 8 = p 1, if p 3, 5 mod 8 p j ε 1 p. Using

44 37 and the definition of ɛ p, we get j ε 1 p = p 1, if p 1 mod 8, 1 j i, if p 3 mod 8, 1 j, if p 5 mod 8, i, if p 7 mod 8. By grouping p terms ranging from 1 to j that are congruent modulo 8, we get S n j = np e + 1 j i np np e + 1 j e + i j j j p 18 p 38 p 58 n 3n 5n 7n = e + 1 j ie + 1 j e + ie j j j j 8n j 8n 1 + e e j j n n 4n = e j ie j e j j j 8n j 8n 1 + e e j j p 78 np e j Note that the last factor 8n j 8n 1 + e e = j j j 3, if j 8n, 0, if j 8n. So, we have j 3 e n j ie n j j e 4n j, if j 8n j S n j = 0, if j 8n. This last condition encourages us to write n as d n 1, where n 1 is an odd integer. We will consider cases separately depending on how large j is compared to d.

45 38 For 3 j d; en/ j = en/ j = e4n/ j = 1 and S n j = 0, for odd j j 1 i, for even j. When j = d + 1; en/ j = 1, en/ j = e4n/ j = 1 and S n d + 1 = 0, for even d d 1 i 1, for odd d. When j = d + ; en/ j = i n 1, en/ j = 1, e4n/ j = 1, and S n d + = 0, for odd d d i n i, for even d. When j = d + 3 ; en/ j = i n 1, en/ j = i n 1, e4n/ j = 1, and S n d + 3 = 0, for odd d d+1 i n i n 1+1, for even d. Finally, for j > d + 3 as we have seen earlier S n j = 0. We now have all the information we need to calculate c n s 0, 8 = j=3 S n j j We will look case by case as d varies.

46 39 For d = 0, c n s 0, 8 = S n3 8 = i n 1 + i n+1 4 = 0, if n 1 modulo 4 i n / if n 3 modulo 4. For d = 1, c n s 0, 8 = 0. For d even, c n s 0, 8 = = d And for d 3 odd, 3 j even d 1 i 4 1 i iin iin i n i n i n i n1+1 4 c n s 0, 8 = 3 j even d 1 i + i = d 5 1 i 4 Finally, to find the values c n s 0, 4, we use the formula c n s 0, 4 = c n s 0, nn+1 1 i 4 Let us summarize these formulas in the following lemma.

47 Lemma 4.3. For a non-zero integer n = d n 1 where n 1 is odd, we have 40 c n s 0, 8 = 0 d = 0, n i d = 0, n i d = 0, n d = 1 d 4 1 i d even, n d 1 i d even, n d+ 1 i d even, n d 5 1 i d 3 odd 4 and c n s 0, 4 = d 1 i 4 d+ 1 i 4 d++ 1 i 4 d 3 1 i 4 d even, n d even, n d even, n d odd Also, letting γn = 1 + ic n s 0, 4 + 1, we have γn = d/ d even, n d/ + d even, n d/ + + d even, n d/ 1/ d odd. 4. Calculations for g ns 0 Recall that g n s = δ nproj n e πny y s/ + m Z b n m sc n m s, 4P n m, s + s m odd b 4n m sc 4n m s, 8P 4n m, s + 4k s 1 i m Z b 4n m sp 4n m, s.

48 41 Taking the derivatives at s = s 0, we get g ns 0 = δ n proj n e πny y s/ s=s0 + δ n b 0 sc 0 s, 4P n n, s s=s0 +δ n 4k s 1 ib 0 sp 4n 4n, s s=s0 + b n m s 0c n m s 0, 4P n m, s 0 + s 0 m Z, m n m odd + s i b 4n m s 0c 4n m s 0, 8P 4n m, s 0 b 4n m s 0P 4n m, s 0. m Z, m 4n Recall the following equations we derived in the previous section: P n m, s = 4 1 α P 4n 4m, s b n s = b 4n s c n s 0, 8 = c n s 0, 4 1 nn+1 c n s 0, 4 = c 4n s 0, 4 1 i 4 b ns 0 = 4L1/, χ 4<n> β n s 0 1 i 4 Using these equations above, and letting βn = β n s 0, we get g ns 0 = δ n proj n e πny y s/ s=s0 +δ n b 0 s c 0 s, i P n n, s s=s0 + s0 1 i b 4n m s 0γ4n m P 4n m, s 0 m Z, m 4n = δ n proj n e πny y s/ s=s0 +δ n b 0 s c 0 s, i P n n, s s=s0 + s 0 1 i L1/, χ 4<4n m >β4n m γ4n m P 4n m, s 0. m Z, m 4n

49 4 Rewriting formula 1.6 after scaling, we get f H k L sym f, k 1 ek λ n f < f, f > = k! 44π k 3/4 n k 1/ Γk 1/4 g ns 0 Now we plug in g ns 0 above with the values of P n m, s 0 we found earlier and deduce the following proposition. Proposition 4.4. where f H k L sym f, k 1 ek λ n f < f, f > = A n,k + B n,k, A n,k = B n,k = k! 44π k 3/4 n k 1/ Γk 1/4 δ n proj n e πny y s/ s=s0 k! + 44π k 3/4 n k 1/ Γk 1/4 δ n b 0 s c 0 s, i P n n, s s=s0 m Z, m 4n L 1/, χ 4<m 4n> ξ n,k m. The coefficients ξ n,k m are given by ξ n,k m = β4n m γ4n m F n,k m with F n,k m = π 1 k Γk 1/4 n 1/4 Γk+1/4 F 1 1 4n k 1/4 m 4n n 1/4 m Ik m k 1, 3 k; 1; m 4 4 4n if m < 4n if m > 4n. Lemma 4.5. ξ n,k m > 0 whenever m > 4n. Proof. First, note that F n,k m > 0 for m > 4n. Also, from Lemma 4.3, it is easy to see that γn 0 for all n. Moreover γ4n m > 0 for all m, n.

50 Then 43 Finally, for positivity of βn, let n = 4 r m t where m is odd and t is square-free. βn = µa 4t a a ab m Note that we are summing over the terms µa 4t a a function. Hence, it suffices to find the last sum for a prime power: ab p α µa 4t a a 1/ = 1 + α α 4t > 0. p p which is a multiplicative We conclude that βn, being a product of these over p m, is positive as well 4.3 Simplifying A n,k proj n e πny y s/ s=s0 proj n e πny y s/ = 4πnk 1 k! Taking the derivative at u = 0, we get 0 e 4πny k 1/4+u/ dy y y = 4πnk 3/4 Γk 1/4 + u/ 4πn u/ k! projn e πny y s/ s=s0 = 4πnk 3/4 Γk 1/4 k! b 0 s c 0 s, i Pn n, s s=s0 Γ k 1/4 Γk 1/4 ln4πn First, we will deal with each term, finding their first two coefficients of the Laurant expansions, around the point s = s 0 using the standard little-o notation. b 0 s = L4 s + 4k 3 L 4 s + 4k = 1 3 s 4k ζs + 4k 3 1 s 4k ζs + 4k = 1 u ζu 1 1 u ζ1 + u

51 44 We have already computed c 0 s, 4 earlier as c 0 s, 4 = 1 i 4 4 So, u+1 c 0 s, i = 1 i 4 u+1 1 u 1 Multiplying these two, we get As u 0, b 0 s c 0 s, i = 1 i ζu ζ1 + u ζu = lnπu + ou, using ζ0 = 1/, ζ 0 = lnπ/ ζ1 + u = 1 u + γ + o1 = γu + ou using lim ζs 1 = γ, u s 1 s 1 and so b 0 s c 0 s, i = u1 i 1 + lnπ γu + ou. Next, we use the calculations for P n n, s with n = 1 to get P n n, s = 1 + i 1k π4πn k 3/4 k! As u 0, we have nπ u/ ΓuΓk 1 4 u Γk u Γ 3 4 k + u = = Γk 1 4 u Γk uγ 3 k + u 4 4 Γk Γ k 1 u + ou 4 Γk Γ k + 1u + ou Γ 3 k Γ 3 ku + ou 4 Γk 1/4 Γk + 1/4Γ3/4 k 1 Γ k 1/4 Γk 1/4 + Γ k + 1/4 Γk + 1/4 + Γ 3/4 k Γ3/4 k u + ou. Gamma function has the property that ΓxΓ1 x = π/ sinπx Taking logarithmic derivatives of both sides, we get Γ 1 x Γ1 x = Γ x Γx + π cotπx.

52 45 In particular, letting x = k + 1/4, we have the two relations Γk + 1/4Γ3/4 k = 1 k π and Γ 3/4 k Γ3/4 k = Γ k + 1/4 Γk + 1/4 + π Also, Γz has the following expansion at 0: Γu = 1 1 γu + ou u Combining these, we get P n n, s = 1 + i 1k π4πn k 3/4 k! = 1 u 1 + i4πnk 3/4 Γk 1/4 k! lnπn 1 + Finally, we add the pieces together and get b 0 s 1 + nπ u/ ΓuΓk 1 4 u Γk u Γ 3 4 k + u γ Γ k 1/4 Γk 1/4 Γ k + 1/4 Γk + 1/4 π P n n, s = 4πnk 3/4 Γk 1/4 k! c 0 s, i lnπ 3γ + lnπn Γ k 1/4 Γk 1/4 Γ k + 1/4 Γk + 1/4 π After simplifications, we end up with b 0 s c 0 s, i P n n, s s=s0 = 4πnk 3/4 Γk 1/4 k! lnπ + 3γ lnπn + Γ k 1/4 Γk 1/4 + Γ k + 1/4 Γk + 1/4 + π Combining these two calculations gives A n,k = δ n n 1/4 u + ou. u + ou. Γ k 1/4 Γk 1/4 + Γ k + 1/4 Γk + 1/4 + π + 3γ 3 lnπ lnn.

53 46 Now, starting with the identity ΓxΓx + 1/ = π 1/ x Γx, and taking logarithmic derivatives, we derive the relation: Γ x Γx + Γ x + 1/ Γx + 1/ = Γ x ln. Γx Using this identity with x = k 1/4 above, we simplify A n,k to Proposition 4.6. A n,k = δ n n 1/4 Γ k 1/ Γk 1/ + π + 6γ 6 ln π 10 ln 4 lnn. By combining Propositions 4.4, 4.6, and Lemma 4.5 we arrive at our destination.

54 Main Theorem: The Exact Average Formula at the Center Main Theorem. For integers k > 1 and n > 0, we have the following formula at the center of symmetry s = k 1/: f H k L sym f, k 1 ek λ n f < f, f > = A n,k + m Z, m 4n 1 L, χ 4<m 4n> ξ n,k m. A n,k = δ n n 1/4 Γ k 1/ Γk 1/ + A lnn, where A = π + 6γ 6 ln π 10 ln 4 The weights ξ n,k m are positive for all but finitely many m. In fact, ξ n,k m > 0, whenever m > n. Explicitly, ξ n,k m = β4n m γ4n m F n,k m. For n 0, we write n = d n 1 with n 1 odd. Then βn = γn = µa 4 < n > a a a,b>0, ab n 1 d/ d even, n 1 1 mod 4, d/ + d even, n 1 3 mod 8, d/ + + d even, n 1 7 mod 8, d/ 1/ d odd. Finally, F n,k m = π 1 k Γk 1/4 n 1/4 Γk+1/4 F 1 1 4n k 1/4 m 4n n 1/4 m Ik m k 1, 3 k; 1; m 4 4 4n if m < 4n, if m > 4n, where I k c = 0 u k 3/4 du 1 + u k+1/4 1 + cu k 1/4

55 48 Chapter 5 Asymptotic Behavior as k. From the main formula we have f H k L sym f, k 1 ek λ n f < f, f > = A n,k + B n,k We will find the asymptotic of this formula as k by finding the asymptotic of the terms A n,k, B n,k. 5.1 Asymptotic of A n,k Recall that A n,k = δ n n 1/4 Γ k 1/ Γk 1/ + A lnn By taking logarithmic derivatives of the identity Γx + 1 = xγx, we get Γ x + 1 Γx + 1 = 1 x + Γ x Γx It follows that Γ k 1/ Γk 1/ ln k, and we conclude Lemma 5.1. As k, we have A n,k δ n n 1/4 ln k.

56 49 5. Asymptotic of B n,k We will first bound the F n,k m terms appearing in B n,k. Recall that B n,k = L 1/, χ 4<m 4n> β4n m γ4n m F n,k m m Z, m 4n Let m 0 > 0 be the smallest integer such that m 0 > 4n. Case 1: m < m 0. In this case, m < 4n as well, and we have F n,k m = π 1 k Γk 1/4 F n 1/4 1 k 1 Γk + 1/4 4, 3 4 k; 1 ; m. 4n From Stirling s formula, we have Γk 1/4 1 Γk + 1/4 = O k It remains to bound the Gauss hypergeometric function F 1, which is defined as F a, b; c; z = F 1 a, b; c; z = where r n = rr r + n 1. n=0 a n b n c n The series converge when z < 1, assuming that c is not a non-positive integer. For 0 < x < 1, we can write F k 1 4, 3 4 k; 1 ; x in terms of the Legendre functions P µ ν formula in Handbook of Mathematical Functions [6]. z n n!, F a, b; 1 ; x = π 1 a+b 3 Γ a + 1 Γ b x 1 1 a b x + P 1 a b x P 1 a b a b 1 a b 1 Letting a = k 1/4, b = 3/4 k gives F k 14, 34 k; 1 ; x = Γ k + 1/4 Γ 5/4 k π P 0 k 3/ x + P 0 k 3/ x

57 50 we get Using the relation F k 14, 34 k; 1 ; x Γ5/4 kγk 1/4 = π sinπk π/4 = π 1 k, = 1k+1 π Γ k + 1/4 P 0 Γk 1/4 k 3/ x + Pk 3/ 0 x By formula of Handbook of Mathematical Functions[6], we have the following asymptotic for the Legendre function. As ν, we have P µ ν cos θ = Γν + µ + 1 Γν + 3/ π sin θ 1/ cos ν + 1 θ π 4 + µπ + O 1 ν Letting ν = k 3/, µ = 0, we get as k P 0 k 3/ cos θ = Γk 1/ Γk 3π 4 1/ cos k 1θ π + O = O k k Hence, F k 14, 34 k; 1 ; x = O1 We deduce that 1 F n,k m = O k Since we have finitely many m such that m < 4n, the L, β, γ terms are all bounded by some constant, depending on n but not k. Thus we proved that for any fixed n, letting k, we have m Z, m <4n 1 L 1/, χ 4<m 4n> β4n m γ4n m F n,k m = O k Case, m > m 0: In this case, m > 4n as well, and we have F n,k m = 1 n 1/4 k 1/4 4n m 4n I k m m,

58 where I k c = 0 51 u k 3/4 du 1 + u k+1/4 1 + cu k 1/4 Note that, for m > 4n we have the following inequalities: 0 < γ4n m < 4m 4n < 4m L1/, χ 4<m 4n> < 4m 4n < 4m 0 < β4n m < m 4n < m 4. On the other hand, m 4n I k = 0 Then we get m Case 3, m = m 0: We have 0 u k 3/4 du 1 + u k+1/4 1 + m 4n du 1 + u1 + m 4nu m m 4n m m 0 u k 1/4 du 1 + u = m m 4n m L 1/, χ 4<m 4n> β4n m γ4n m F n,k m m Z, m >m 0 << k n k 1 m < k 1/ k n k dx m>m m 0 0 x k 1/ 4n k = << 1 k 4n k 3/m k 3/ k m 0 0 F n,k m = 1 n 1/4 L 1/, χ 4<m 0 4n> << F n,k m 0 < 1 n 1/4 k 1/4 4n m I 0 4n k m 0 m 0 m 0 β4n m 0γ4n m 0F n,k m 0 k 1/4 k 4n 4n m 0 << m 0

59 5 Note that 4n < 1, so the bounds in the last two cases decay exponentially with m 0 k. Thus, by adding the three bounds we got above, we have shown 1 Lemma 5.. For fixed n, as k, we have B n,k = O k 5.3 Asymptotic of the Main Formula The Main Theorem has two immediate corollaries. When n is square, the dominant term is A n,k, and we deduce Corollary 1. For a fixed square integer n, letting k, f H k L sym f, k 1 ek λ n f < f, f > = 1 Γ k 1/ n 1/4 Γk 1/ + A lnn 1 +O k In particular, when n = 1, we have f H k L sym f, k 1 e k < f, f > = Γ k 1/ 1 Γk 1/ + A + O k Assuming that Lsym f, k 1 0 for all f, we get f H k L sym f, k 1 << kln k 4. In the last statement, we used the following bound for < f, f > from [3],.3: < f, f > << 4π k k 1!ln k 3. In particular, each L-value, L sym f, k 1, is bounded by k 1+ɛ ; whereas the convexity bound is k. On the other hand, when n is not a square, A n,k vanishes because δ n = 0. Corollary. For a fixed non-square integer n, we have f H k L sym f, k 1 ek λ n f < f, f > = 1 1 L, χ 4<m 4n> ξ n,k m = O k m Z

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