Johan Andersson Summation Formulae and Zeta Functions

Johan Andersson Summation formulae and zeta functions

Department of Mathematics

Stockholm University

2006

Johan Andersson Summation formulae and zeta functions 4

Doctoral Dissertation 2006 Department of mathematics Stockholm University SE-106 91 Stockholm

Typeset by LATEX2e c 2006 by Johan Andersson e-mail: [email protected] ISBN 91-7155-284-7 Printed by US-AB, Stockholm, 2006. ABSTRACT

In this thesis we develop the summation formula

Z ∞ X a b X 1 σ2ir(|m|)σ2ir(|n|)F (r; m, n)dr f = “main terms”+ c d π ir 2 ad−bc=1 m,n6=0 −∞ |nm| |ζ(1 + 2ir)| c>0 ∞ θ(k) ∞ X X X 1 X X + ρj,k(m)ρj,k(n)F 2 − k i; m, n + ρj(m)ρj(n)F (κj; m, n), k=1 j=1 m,n6=0 j=1 m,n6=0 where F (r; m, n) is a certain integral transform of f, ρj(n) denote the Fourier coeﬃcients for the Maass wave forms, and ρj,k(n) denote Fourier coeﬃcients of holomorphic cusp forms of weight k. We then give some generalisations and applications. We see how the Selberg trace formula and the Eichler-Selberg trace formula can be deduced. We prove some results on moments of the Hurwitz and Lerch zeta-function, such as

Z 1 ∗ 1 2 −5/6 ζ 2 + it, x dx = log t + γ − log 2π + O t , 0 and

Z 1 Z 1 ∗ 1 4 2 5/3 φ x, y, 2 + it dxdy = 2 log t + O (log t) , 0 0 where ζ∗(s, x) and φ∗(x, y, s) are modiﬁed versions of the Hurwitz and Lerch zeta functions that make the integrals convergent. We also prove some power sum results. An example of an inequality we prove is that

n √ √ X v n ≤ inf max zk ≤ n + 1 |z |≥1 v=1,...,n2 i k=1 if n + 1 is prime. We solve a problem posed by P. Erd˝oscompletely, and disprove some conjectures of P. Tur´anand K. Ramachandra.

PREFACE

I have been interested in number theory ever since high school when I read classics such as Hardy-Wright. My special interest in the zeta function started in the summer of 1990 when I studied Aleksandar Ivić’sbook on the Riemann zeta function and read selected excerpts from “Reviews in number theory”. I remember long hours from that summer trying to prove the Lindelöfhypothesis, and working on the zero density estimates. My first paper on the Hurwitz zeta function came from that interest (as well as some ideas I had about Bernoulli polynomials in high school). The spectral theory of automorphic forms, and its relationship to the Rie- mann zeta function has been an interest of mine since 1994, when I first visited Matti Jutila in Turku, and he showed a remarkable paper of Y¯oichi Motohashi to me; “The Riemann zeta function and the non-euclidean Laplacian”. Not only did it contain some beautiful formulae. It also had interesting and daring speculations on what the situation should be like for higher power moments. The next year I visited a conference in Cardiff, where Motohashi gave a highly enjoyable talk on the sixth power moment of the Riemann zeta function. One thing I remember - “All the spectral theory required is in Bump’s Springer lecture notes”. Anyway the sixth power moment might have turned out more difficult than originally thought, but it further sparked my interest, and I daringly dived into Motohashi’s interesting, but highly technical Acta paper. My first real break-through in this area came in 1999, when I discovered a new summation formula for the full modular group. Even if I have decided not to include much related to Motohashi’s original theory (the fourth power moment), the papers 7-13 are highly influenced by his work. Although somewhat delayed, I am pleased to finally present my results in this thesis.

Johan Andersson

ACKNOWLEDGMENTS

First I would like to thank the zeta function troika Aleksandar Ivić, Y¯oichi Mo- tohashi and Matti Jutila. Aleksandar Ivićfor his valuable comments on early versions of this thesis, as well as his text book which introduced me to the zeta function; Y¯oichi Motohashi for his work on the Riemann zeta function which has been a lot of inspiration, and for inviting me to conferences in Kyoto and Ober- wolfach; Matti Jutila for first introducing me to Motohashi’s work, and inviting me to conferences in Turku. Further I would like to thank Kohji Matsumoto and Masanori Katsurada, for being the first to express interest in my first work, and showing that someone out there cares about zeta functions; Dennis Hejhal for his support when I visited him in Minnesota; Jan-Erik Roos for his enthusiasm during my first years as a graduate student; My friends at the math department (Anders Olofsson, Jan Snellman + others) for mathematical discussions and a lot of good times; My family: my parents, my sister and especially Winnie and Kevin for giving meaning to life outside of the math department. And finally I would like to thank my advisor Mikael Passare, for his never ending patience, and for always believing in me.

Johan Andersson

CONTENTS

Introduction ...... 13

Part I - The power sum method 21

Introduction ...... 23

On some Power sum problems of Tur´anand Erd˝os ...... 29

Disproof of some conjectures of P Tur´an ...... 41

Disproof of some conjectures of K Ramachandra ...... 47

Part II - Moments of the Hurwitz and Lerch zeta functions 53

Introduction ...... 55

Mean value properties of the Hurwitz zeta function ...... 61

On the fourth power moment of the Hurwitz zeta-function ...... 67

On the fourth power moment of the Lerch zeta-function ...... 79

Part III - A new summation formula on the full modular group and Klooster- man sums 89

Introduction ...... 91

A note on some Kloosterman sum identities ...... 99

A note on some Kloosterman sum identities II ...... 107

A summation formula on the full modular group ...... 115 Contents 12

A summation formula over integer matrices ...... 133

A summation formula over integer matrices II ...... 145

The summation formula on the modular group implies the Kuznetsov summation formula ...... 159

The Selberg and Eichler-Selberg trace formulae ...... 169 INTRODUCTION

This thesis in analytic number theory consists of three parts and altogether thirteen papers. Each of the parts has a separate introduction. This thesis is by no means complete and is a result of a number compromises. There are additional papers I would have liked to include, some of which I refer to as forthcoming papers. Even if they are mostly complete they would still need some more work to be included in this thesis. However, even if they would have painted a fuller picture and provided further motivation of my main result, the current version does not depend upon them and it is entirely self contained. Part I and parts of part II of this thesis have also been included in my licentiate thesis [3], and three of the papers have been published in mathematical journals ([1], [2] and [4]).

The Riemann zeta-function

The Riemann zeta-function1 ∞ X ζ(s) = n−s (Re(s) > 1) (0.1) n=1 is perhaps the deepest function in all of mathematics. First of all it is arithmetic

Y 1 ζ(s) = , (Re(s) > 1) (0.2) 1 − p−s p prime and has an Euler product. It is symmetric π ζ(s) = χ(s)ζ(1 − s) χ(s) = 2sπs−1 sin s Γ(1 − s) (0.3) 2 and has a functional equation. Furthermore it is random. A theorem illustrating this is the Voronin universality theorem (see e.g [8]). If f is a non-vanishing analytic function on D for a compact subset D ⊂ {s ∈ C : 1/2 < Re(s) < 1}, there exists a sequence Tk = Tk(D, f) so that

lim max |ζ(s + iTk) − f(s)| = 0. (0.4) k→∞ s∈D 1 Introduced by Euler [6] and studied by Riemann [10]. General references: Ivi´c[7] and Titchmarsh [12]. Introduction 14

The depth of the function comes from its simple definition and these three properties. I particular The Euler product implies that the function contains all information about the primes, and therefore its study belongs in the realms of number theory. The most fundamental open problem about the Riemann zeta-function is the Riemann hypothesis2 that says that ζ(s) = 0 implies that either Re(s) = 1/2 or s is an even negative integer. Of near equal importance are the weaker conjectures 1. The Lindelöfhypothesis: |ζ(1/2 + it)| = O(|t|), ∀ > 0. 2. The density hypothesis: N(σ, T ) = O(T 2(1−σ)+), ∀ > 0, where N(σ, T ) denote the number of zeroes ρ with real part Re(ρ) ≥ σ, and imaginary part −T ≤ Im(ρ) ≤ T . It is obvious that the Riemann hypothesis implies the density hypothesis and with some complex analysis, it is possible to show that the Riemann hypothesis implies the Lindelöfhypothesis, and the Lindelöfhypothesis implies the density hypothesis. Many consequences of the Riemann hypothesis are in fact implied already by the density hypothesis.

Part I - The power sum method

In an attempt to prove the density hypothesis we were lead to study the Tur´an power sum method. The method allows us to obtain lower bounds for

n X v max 1 + zk , v=M(n),...,N(n) k=1 and in the first papers on his method Turánstated some conjectures on these bounds that would indeed imply the density hypothesis. However we prove that Turán’s conjectures were in fact false. We also prove some other results in the field, in particular solving a problem of Erd˝os. Although it does not strictly use the power sum method, we have also included a disproof of a certain conjecture of Ramachandra. This conjecture would also have had implications on the Riemann zeta function if true. The Voronin universality theorem (0.4) is a crucial part of the disproof. All of the above together constitutes part I of our thesis.

Part II - Moments of the Hurwitz and Lerch zeta functions

A problem closely related to the Lindel¨ofhypothesis is the moment problem, which consists in estimating

Z T 1 2k Zk(T ) = ζ 2 + it dt. (0.5) 0 2 See Riemann [10] or Bombieri [5] Introduction 15

The general conjecture is that

k2 Zk(T ) ∼ ckT log T (0.6) for all integers k ≥ 1. It should here be mentioned that by the method of Bala- subramanian and Ramachandra (see [9]), the lower bound

k2 Zk(T ) ≥ dkT log T for some dk > 0 has been proved. It is also known that the Lindel¨ofconjecture is equivalent to a weak form of the conjecture

1+ Zk(T ) T (0.7) for all integers k ≥ 1, and not even the Riemann hypothesis seems to imply the strong form of the conjecture (0.6), which has so far only been proved for k = 1, 2. A way of trying to understand the Riemann zeta-function is to see how it relates to other zeta functions, such as the closely related Hurwitz zeta-function

∞ X ζ(s, x) = (k + x)−s, (Re(s) > 1) k=0 and the Lerch zeta-function

∞ X φ(x, y, s) = e(kx)(k + y)−s. (Re(s) > 1) k=0 The Lindel¨ofhypothesis is expected to hold also for the Hurwitz and Lerch zeta- functions. However the Riemann hypothesis is false. This is due to the fact that while the Hurwitz and Lerch zeta-functions admit a functional equation, they do not have an Euler product. In more generality the Riemann hypothesis is assumed for a class of zeta-functions with Euler product and functional equation (the Selberg class - see Selberg [11]). It is thus interesting to study what diﬀers between the Riemann zeta function and the Hurwitz and Lerch zeta functions, in order to better understand the arithmetic nature of the Riemann zeta function. In particular, for the Hurwitz zeta function it is of interest to study the corresponding moments Z 1 Z T ∗ 1 2k ζ 2 + it, x dtdx, (0.8) 0 0 Z 1 ∗ 1 2k ζ 2 + it, x dx, (0.9) 0 Introduction 16 and

Z T ∗ 1 2k ζ 2 + it, x dt, (0.10) 0 where ζ∗(s, x) = ζ(s, x) − x−s. In Part II we show some new results on these problems.

Part III - A new summation formula on the full modular group and Kloosterman sums

In the ﬁnal part of our thesis we consider the following question. What is the appropriate generalization of the classical Poisson summation formula

∞ ∞ X X f(n) = fˆ(n) n=−∞ n=−∞ to the case of the full modular group? There are several generalizations in the literature. The pre-trace formula, or even the Selberg trace formula has been suggested. The pre-trace formula allows us to get an expansion of SO(2) bi- invariant functions f on SL(2, R) in terms of modular forms and Eisenstein series. The central result in Part III of the thesis is a proof of such a formula without assuming SO(2) bi-invariance. We will obtain a formula of the following type:

Z ∞ X a b X 1 σ2ir(|m|)σ2ir(|n|)F (r; m, n)dr f = “main terms”+ c d π ir 2 ad−bc=1 m,n6=0 −∞ |nm| |ζ(1 + 2ir)| c>0 ∞ θ(k) ∞ X X X 1 X X + ρj,k(m)ρj,k(n)F 2 − k i; m, n + ρj(m)ρj(n)F (κj; m, n), k=1 j=1 m,n6=0 j=1 m,n6=0 where F (r; m, n) is a certain integral transform of f, the ρj(n) denote the Fourier coeﬃcients for the Maass wave forms, and the ρj,k(n) denote Fourier coeﬃcients of holomorphic cusp forms of weight k. For an exact version of this formula we refer to Theorem 1 in “A summation formula on the full modular group”. Our method of proof will use the Kloosterman sums

X mh + nh¯ S(m, n; c) = e , c hh¯≡1 (mod c) 0≤h,h

D, or in other words, the image of SL(2, Z) under Hecke operators TD. We prove some related identities about Kloosterman sums. We show that the Kuznetsov summation formula as well as the Selberg trace formula and the Eichler-Selberg trace formula can be proved as consequences. We also indicate how these results are closely connected to the explicit formula given by Motohashi for the fourth power moment of the Riemann zeta function. BIBLIOGRAPHY

[1] J. Andersson, Mean value properties of the Hurwitz zeta-function, Math. Scand. 71 (1992), no. 2, 295–300.

[2] , On some power sum problems of Tur´anand Erd˝os, Acta Math. Hun- gar. 70 (1996), no. 4, 305–316.

[3] , Power sum methods and zeta-functions, Licentiate thesis, Stockholm University, 1998.

[4] , Disproof of some conjectures of K. Ramachandra, Hardy-Ramanujan J. 22 (1999), 2–7.

[5] E. Bombieri, Problems of the millenium: The Riemann hypothesis, http://www.maths.ex.ac.uk/˜mwatkins/zeta/riemann.pdf.

[6] L. Euler, Introductio in analysin inﬁnitorum, Bousquet, Lausanne, 1748.

[7] A. Ivi´c, The Riemann zeta-function, A Wiley-Interscience Publication, John Wiley & Sons Inc., New York, 1985, The theory of the Riemann zeta-function with applications.

[8] A. Laurinˇcikas, Limit theorems for the Riemann zeta-function, Kluwer Aca- demic Publishers Group, Dordrecht, 1996.

[9] K. Ramachandra, On the mean-value and omega-theorems for the Riemann zeta-function, Tata Institute of Fundamental Research Lectures on Mathe- matics and Physics, vol. 85, Published for the Tata Institute of Fundamental Research, Bombay, 1995.

[10] B. Riemann, Uer¨ die Anzahl der Primzahlen unter einer gegebenen Gr¨osse, Monatsberichte der Berliner Akademie (November 1859), 671–680, also avail- able at http://www.emis.de/classics/Riemann/.

[11] A. Selberg, Old and new conjectures and results about a class of Dirichlet series, Proceedings of the Amalﬁ Conference on Analytic Number Theory (Maiori, 1989) (Salerno), Univ. Salerno, 1992, pp. 367–385. Bibliography 19

[12] E. C. Titchmarsh, The theory of the Riemann zeta-function, second ed., The Clarendon Press Oxford University Press, New York, 1986, Edited and with a preface by D. R. Heath-Brown.

PART I - THE POWER SUM METHOD

INTRODUCTION

The density hypothesis and the power sum method

From the explicit formula

∞ X Z ∞ X Z ∞ Λ(n)f(n) = f(x)dx − xρ−1f(x)dx, (I.11) n=1 0 ζ(ρ)=0 0 where Λ(n) denotes the Van-Mangholt function, it is easy to see that

2 Z H ∞ Z 1 X −1/2−it 2σ−1+ Φ(n/T )Λ(n)n dt T (N(σ, 2H) + 1)dσ (I.12) 0 n=1 0

∞ for a positive test-function Φ ∈ C0 (]0, ∞[) and log T log H, where N(σ, T ) denotes the number of zeroes of the Riemann zeta function with real part greater or equal to σ, and with imaginary part between −T and T . (Note: the Riemann hypothesis states that N(σ, T ) = 0, for all σ > 1/2.) Can we invert this formula? More precisely, can we ﬁnd for some H = H(T ) so that log H ∼ log T and such that 2 Z 1 Z T ∞ 2σ−1 X −1/2−it T N(σ, T/2)dσ T Φ(n/H)Λ(n)n dt. (I.13) 1/2 0 n=1 If this were true we could use a standard mean value formula for Dirichlet-polynomials

2 Z T X it−1/2 X 2 X 2 akk dt T |ak| /k + |ak| (I.14)

0 k≤T k

This result is called the density hypothesis and can be can be used as a substitute for the Riemann zeta function in a lot of applications, such as local prime number Introduction 24 theorems. Both the density hypothesis and the Riemann hypothesis imply h π(x + h) − π(x) ∼ , (xθ < h < x) (I.16) log x for each θ > 1/2. Turánintroduced a new method in 1941 (see [10]) in an attempt to prove the density hypothesis with this idea. However he was never able to prove the density hypothesis. Instead he stated two conjectures that would imply the density hypothesis. The problems with inverting (I.12) comes from the fact that there is no way of knowing that locally the sum over the zeroes in the critical strip do not cancel each other. If we knew that the zeroes in the critical strip (but not on the critical line) were “isolated”, i.e. N(σ, T +1)−N(σ, T ) = o(log T ), for σ > 1/2, we could have used Turán’s method. However this has been shown to be equivalent to the even stronger Lindelöfhypothesis (see Titchmarsh [9], Chapter 13). In our paper “Disproof of some power sum problems of P. Turán” we show that the two conjectures of Turánare false3. Furthermore it can be shown that it is a futile attempt to try to prove the density hypothesis just by these methods. One would need at least something like

n X inf max eλkx > e−o(n), (C > 1) n λ∈ ,λ1=0 x∈[1,C] C k=1 to be true, and we manage to prove that there is no such estimate. Even though the density hypothesis was not proved by these methods, the Tur´anpower sum theory allowed us to obtain non-trivial estimates (and the best in its time) for the density of the zeroes (and a θ < 1) in (I.16) before more modern zero-detecting methods were found. This could be done because (I.13) can be proved for positive constants c1, c2 and for each T and some H so that c1 log T ≤ log H ≤ c2 log T . Even though the Tur´antheory currently does not give as good estimates as the state-of the art zero-detecting methods, we suspect that it will eventually lead to the same-strength estimates (neither better nor worse).

The power sum method

Turánpower sum theory has over the years found a number of other applications. In his book [11] P. Turánhas 15 chapters for the theory and 40(!) chapters of applications. Turánhimself thought of his method as his most important contribution to mathematics. His book also contains two chapters of open problems. In

3 This result was presented at the Halberstam conference in analytic theory in Urbana Cham- paign, 1995 and a slightly diﬀerent version of this paper also appeared in our Licentiate thesis [2]. Introduction 25 the paper “On some power sum problem of Tur´anand Erd˝os” we solve some of them4. One of the problems we solve completely is the following one of P. Erd˝os.

Problem. Does there exist for each integer 1 ≤ m ≤ n − 1 a c = c(m) so that

v v |z1 + ··· + zn| max v ≥ m? (I.17) 1≤v≤c(m)n,v integer mink=1,...,n |zk|

We show that in fact one can use c(m) = max(1021m, [(2π2m)m/2]). Another problem of P. Tur´anconcerns that of estimating

n X v inf max zk . (I.18) |z |≥1 v=1,...,n2 i k=1 √ √ We√ also show that if n + 1 is prime then n ≤ (I.18) ≤ n + 1, and that we have n ≤ (I.18) for all integers n. We would like to state here a similar open problem of Erd˝os-Newman (not power sum though) that Erd˝osdid send to us in 1995.

Problem. Is it true that there is an absolute constant c > 0 such that for every choice of ak = ±1, 0 ≤ k ≤ n one has

n X k √ max akz > (1 + c) n? (I.19) |z|=1 k=0

For many similar open problems, see Montgomery [6]. None of the results in this paper have any applications on zeta-functions. (However we use some results from that theory, e.g. the distribution of consecutive primes.) It should be noted that the constants 23/84 and 65/42 in Proposition 2 can be improved to 0.2675 and 1.535 thanks to improvements on the diﬀerence between consecutive primes (Harman-Baker [4]).

Omega results for the Riemann zeta-function in short intervals As well as considering order estimates of the Riemann zeta-function there is interest in ﬁnding omega-estimates (that is estimates from below) in short intervals. The method used here is the one by Ramachandra-Balasubramian (see Ramachan- dra’s book [7]), which gives omega-estimates for the 2k-th moment:

Z T +H 1 2k k2 ζ 2 + it dt ≥ ck(log H) , for H log log T. (I.20) T 4 This paper has also been published as [1] and appeared in our licentiate thesis [2] Introduction 26

In a series of papers Ramachandra and Balasubramian generalized these results to a wide class of Dirichlet series, called the Titchmarsh series. In a paper in Asterisque [8], Ramachandra states some conjectures which essentially would allow us to localize these results even further. Ramachandra-Balasubramian wrote another paper [5] which contained further conjectures. In our paper “Disproof of some conjectures of Ramachandra” we disprove these conjectures (2 diﬀerent disproofs) as well as an even weaker conjecture given by Ramachandra in a private communication5. An excerpt from this paper which essentially disprove all these conjectures, is the following result. Result. Suppose that H, > 0, 0 < δ < 1/2. Then there exist an N,a1 = 1, δ−1 |ak| ≤ k so that

N X it max ann < t∈[0,H] n=1 We ﬁnally remark that this introduction is also taken in a somewhat modiﬁed form from our Licentiate thesis [2].

5 This paper has been published as [3] and also appeared in our Licentiate thesis BIBLIOGRAPHY

[1] J. Andersson, On some power sum problems of Tur´anand Erd˝os, Acta Math. Hungar. 70 (1996), no. 4, 305–316.

[2] , Power sum methods and zeta-functions, Licentiate thesis, Stockholm University, 1998.

[3] , Disproof of some conjectures of K. Ramachandra, Hardy-Ramanujan J. 22 (1999), 2–7.

[4] R. C. Baker and G. Harman, The diﬀerence between consecutive primes, Proc. London Math. Soc. (3) 72 (1996), no. 2, 261–280.

[5] R. Balasubramanian and K. Ramachandra, On Riemann zeta-function and allied questions. II, Hardy-Ramanujan J. 18 (1995), 10–22.

[6] H. L. Montgomery, Ten lectures on the interface between analytic number theory and harmonic analysis, CBMS Regional Conference Series in Math- ematics, vol. 84, Published for the Conference Board of the Mathematical Sciences, Washington, DC, 1994.

[7] K. Ramachandra, On Riemann zeta-function and allied questions, Astérisque (1992), no. 209, 57–72, JournéesArithmétiques, 1991 (Geneva).

[8] , On the mean-value and omega-theorems for the Riemann zeta- function, Tata Institute of Fundamental Research Lectures on Mathematics and Physics, vol. 85, Published for the Tata Institute of Fundamental Re- search, Bombay, 1995.

[9] E. C. Titchmarsh, The theory of the Riemann zeta-function, second ed., The Clarendon Press Oxford University Press, New York, 1986, Edited and with a preface by D. R. Heath-Brown.

[10] P. Tur´an, Uber¨ die Verteilung der Primzahlen. I, Acta Univ. Szeged. Sect. Sci. Math. 10 (1941), 81–104.

[11] , On a new method of analysis and its applications, Pure and Applied Mathematics, John Wiley & Sons Inc., New York, 1984, With the assistance of Bibliography 28

G. Halászand J. Pintz, With a foreword by Vera T. Sós,A Wiley-Interscience Publication. On some Power Sum problems of Turánand Erd˝os

Johan Andersson∗

1 Introduction

In this paper I will show some elementary inequalities, and e.g. solve a problem given by Paul Erd˝os.I will also consider some problems of Paul Tur´an.Lemma 1 was given by Cassels [1], and its proof will follow his approach, although I will use it in a more general setting. In this paper e(x) will denote e2πix, bxc will denote the integral part of x, and {x} will denote the distance between x and its nearest integer. Furthermore v will always be integer valued.

2 Main theorem

Lemma 1. (Cassels) If zk are complex numbers we have

n ! X v max Re zk ≥ 0. v=1,...,2n+1 k=1

Proof. Let ak, k = 1,..., 2n be deﬁned by

n 2n Y X k (z − zk)(z − zk) = a2n−kz k=1 k=0

Pn v v Pn v 0 and σv = k=1 zk + zk, Clearly 2 Re( k=1 zk) = σv. By the construction the aks are real. Now suppose σv < 0 for v = 1,..., 2n + 1. We now apply the Newton identities

σv + a1σv−1 + ··· + av−1σ1 + vav = 0, v = 1, 2,..., 2n (1)

∗This paper has been published in Acta Math. Hungar. 70 (1996), no. 4, 305–316.

29 and successively get a1 > 0, a2 > 0, . . . , a2n > 0 and with the relation σ2n+1 + a1σ2n + ··· + a2nσ1 = 0 we get σ2n+1 > 0, and a contradiction, hence maxv=1,...,2n+1 σv ≥ 0.

Theorem 1. If zk are complex numbers and 1 ≤ m ≤ n we have

Pn v | k=1 zk| max q ≥ 1. v=1,...,2nm−m(m+1)+1 Pm 2v k=1 |zk|

Pn v Pb v Proof. Put sv = k=1 zk, and sa,b,v = k=a zk. We get

2 2 2 2 |sv| = |s1,m,v + sm+1,n,v| = |s1,m,v| + |sm+1,n,v| + 2 Re(s1,m,vsm+1,n,v) = m ! X 2v 2 X v v = |zk| + |sm+1,n,v| + 2 Re s1,m,vsm+1,n,v + zj zk = k=1 1≤j

m(m+1) where B(v) is a pure power sum with nm − 2 elements. i. e.

m(m+1) nm− 2 X v B(v) = wk. (2) k=1 From Lemma 1 we have an integer v such that 1 ≤ v ≤ 2nm − m(m + 1) + 1, 2 Pm 2v such that Re(B(v)) ≥ 0, and for that v we have |sv| = k=1 |zk| + something pPm 2v non-negative, and |sv| ≥ k=1 |zk| .

Corollary 1. If 1 ≤ m ≤ n and |zk| ≥ 1 then

n X v √ max zk ≥ m. v=1,...,2nm−m(m+1)+1 k=1 Proof. This follows from the theorem and the inequality v u m √ uX 2v v t |zk| ≥ m min |zk| . k=1,...,n k=1

From the two extreme cases in corollary 1 (m = 1 and m = n) we get

30 Corollary 2. If |zk| ≥ 1 then

n X v max zk ≥ 1 v=1,...,2n−1 k=1 which is the classic result by Cassels (see [1] or [8] section 3). We also get

Corollary 3. If |zk| ≥ 1 then

n X v √ max zk ≥ n v=1,...,n2−n+1 k=1 which is a rather interesting new result, which we will use in the following sections. Note that the only condition is that |zk| ≥ 1. Under the assumption |zk| = 1 similar results have been proved earlier by e.g. Cassels, Newman and Szalay who even proved v n u n X v u n − 1 X max bkzk ≥ t 1 − bk, (3) 1≤v≤bcnc bcnc k=1 k=1 when bk > 0 and c ≥ 1. (See Tur´an [8] page 80, or Leenmann-Tijdeman [5] for a weaker result.)

3 Problem 10 of Tur´an

In his classic book [8] Tur´an states a number of problems. One of them (Problem 10, page 190) is the following: If minj |zj| = 1 then what is

n X v inf max zk ? (4) zj 1≤v≤n2,v integer k=1 We see that Corollary 3 clearly gives a lower estimate for this expression. In the following we will show that this estimate is rather strong, in the sense that for inﬁnitely many integers, e.g. the primes the upper estimate is very close. We begin with two lemmas which give upper estimates for (4).

Lemma 2. (H. Montgomery) If p is prime then there exists numbers z1, . . . , zp−1 Pp−1 v √ 2 such that |zk| = 1 and | k=1 zk| ≤ p for all integers 1 ≤ v ≤ p − p − 1.

Proof. Take zk = χ(k)e(k/p), χ a character mod p of order p − 1, see [8] page 83. See Queﬀelec [6] for a fuller account.

31 Lemma 3. If q is a prime power then there exists numbers z1, . . . , zq+1 such that Pq+1 v √ 2 |zk| = 1 and | k=1 zk| ≤ q for all integers 1 ≤ v ≤ q + q. Proof. (Here I essentially follow Fabrykowski [3].) Suppose q is a prime power. By the Singer theorem (see [7]), we can ﬁnd numbers a1, . . . , aq+1, such that the 2 q(q + 1) numbers ai − aj, i 6= j form all nonzero residues modulo q + q + 1. 2 Pq+1 v 2 Pq+1 Consider zk = e(ak/(q + q + 1)). We get | k=1 zk| = q + 1 + k6=j e((ak − 2 Pq2+q 2 1−e(v) aj)v/(q + q + 1)) = q + k=0 e(kv/(q + q + 1)) = q + 1−e(v/(q2+q+1)) = q for all 2 Pq+1 v √ integers v such that 1 ≤ v ≤ q + q, and thus we even have | k=1 zk| = q for those v. It is now an easy task to state and prove the following proposition.

Proposition 1.

n √ X For n+1 prime we have: n ≤ inf max zv ≤ (i) 2 k zk∈ ,|zk|≥1 v=1,...,n −n+1 C k=1 n X √ ≤ inf max zv ≤ n + 1. 2 k zk∈ ,|zk|≥1 v=1,...,n +n−1 C k=1 n √ X For n-1 prime power we have: n − 2 ≤ inf max zv ≤ (ii) 2 k zk∈ ,|zk|≥1 v=1,...,n −n C k=1 n √ √ X ≤ n − 1 ≤ n ≤ inf max zv . 2 k zk∈ ,|zk|≥1 v=1,...,n −n+1 C k=1 Proof. The inequality

n √ X n − 2 ≤ inf max zv 2 k zk∈ ,|zk|≥1 v=1,...,n −n C k=1 in (ii) follows from Corollary 1 with m = n − 2. The inequality

n √ X n ≤ inf max zv 2 k zk∈ ,|zk|≥1 v=1,...,n −n+1 C k=1 in (i) and (ii) is exactly Corollary 3. The inequality

n X √ inf max zv ≤ n + 1 2 k zk∈ ,|zk|≥1 v=1,...,n +n−1 C k=1

32 in (i) follows from Lemma 2. The inequality

n X √ inf max zv ≤ n − 1 2 k zk∈ ,|zk|≥1 v=1,...,n −n C k=1 in (ii) follows from Lemma 3. From Proposition 1 (i) we immediately obtain the best interval estimate so far for (4) and inﬁnitely many integers. Corollary 4. If n + 1 is a prime we have

n √ X √ n ≤ inf max zv ≤ n + 1. 2 k zk∈ ,|zk|≥1 v=1,...,n C k=1 For arbitrary n we can’t use the methods of Corollary 4 which requires pri- mality. But since we know that the primes are rather densely distributed we can use local prime density results from Iwaniec and Pintz (see [4]). First, however we need a lemma.

Lemma 4. (Erd˝osand R´enyi)There exists numbers zk for k = 1, . . . , n such Pn v p that |zk| = 1 and maxv=1,...,m | k=1 zk| ≤ 6n log(m + 1). Proof. See Erd˝os and R´enyi [2].

Similar results with explicit zk’s have been proved by Leenmann-Tijdemann (see [5] or [8] page 82). Proposition 2.

n X v √ 23/84p inf max zk = n + O n log n zk∈ ,|zk|≥1 v=1,...,bn2−n65/42c C k=1

1 23/42 Proof. We know that there exists a prime p between n−1 and n− 2 n , ∀n ≥ N0 by Iwaniec-Pintz’s theorem (see [4]). Choose z1, . . . , zp+1 fulﬁlling the condition in Lemma 3, and zp+2, . . . , zn fulﬁlling the condition in Lemma 4. We get for 1 ≤ v ≤ p2 + p.

p+1 n n X v X v X v zk ≤ zk + zk ≤ k=1 k=1 k=p+2 √ ≤ p + p6(n − p − 1) log(p2 + p + 1) = √ = p + O(n23/84plog n)

33 2 65/42 1 23/42 2 2 It is enough since bn − n c ≤ (n − 2 n ) ≤ p + p. 65/84 In the other direction we put m = n−2bn c for n ≥ N0, and use Corollary 1.

n √ q √ X v 65/84 23/84 max zk ≥ m = n − 2bn c = n + O(n ) v=1,...,2nm−m(m+1)+1 k=1

We also get 2nm − m(m + 1) + 1 = 2n(n − 2bn65/84c) − (n − 2bn65/84c)(n − 2bn65/84c + 1) + 1 = n2 − 4bn65/84c2 − n + 2bn65/84c + 1 ≤ n2 − bn65/42c. Queﬀelec [6] has showed the ≤ part of the proposition with essentially the same method. See also Queﬀelec [6] for an application of these estimates in functional analysis. Note that we still have not considered (4) for all n’s. In the next section we will do that (and obtain much worse results than e.g. Corollary 4 or Proposition 2), but we will also generalize results in the previous sections.

4 A further result

With Lemma 4 of Erd˝osand R´enyi in mind we might consider the following generalization of (4) If minj |zj| = 1 then what is

n X v inf max zk ? (5) zj v=1,...,n2m,v integer k=1 For general integers m ﬁrst I prove an analogue of Corollary 3:

Theorem 2. If |zk| ≥ 1 then

n 1 2m X v n 2 max zk ≥ m! . v=1,...,n2m m k=1 Proof.

n 2m n !m n !m X X X zv = zv zv = A(v) + B(v) + B(v) k k k k=1 k=1 k=1 Where A(v) is a power sum consisting of only positive real elements. Since A(v) + B(v) + B(v) consists of exactly n2m elements and A(v) is non-empty 1 2m we must have that B(v) have less than 2 (n − 1) elements. We now ﬁnd a lower estimate of the number of terms of A(v). It is clear that for each m-tuple

34 2 k1, . . . , km of distinct integers ki we have m! number of terms contributing to A(v), since each permutation on both sides occur. By elementary combinatorics n we know that we can pick m such sets of integers from 1, . . . , n. and since n 2 |zk| ≥ 1 we have A(v) ≥ m m! , ∀v. We now apply Lemma 1 on B(v) and we obtain the theorem. We now consider a well known version of the Stirling formula, which we will use √ n+ 1 −n n! ≥ 2πn 2 e (6) see e.g. Abramowitz-Stegun, Handbook of mathematical functions, formula 6.1.38.

Proposition 3. If |zk| ≥ 1, and 1 ≤ m ≤ n then

n X v p −1 max zk ≥ (n − m)me v=1,...,n2m k=1 Proof. By Theorem 2 we have n X v max zk ≥ v=1,...,n2m k=1 1 n 2m 1 ≥ m!2 = (n(n − 1) ··· (n − m + 1)m!) 2m ≥ m m 1 √ 1 ≥ ((n − m) m!) 2m = n − m(m!) 2m ≥ √ 1 ≥ (According to (6)) ≥ p(n − m)me−1( 2πm) 2m ≥ p(n − m)me−1.

We now have the means to obtain an interval estimate for (5) Corollary 5. We have for 1 ≤ m ≤ n r n m 1 X 1 e−1(1 − ) ≤ √ inf max zv ≤ p12 log(n) 1 + 2m k n mn zk∈ ,|zk|≥1 v=1,...,n n C k=1 Proof. The ﬁrst inequality is exactly Proposition 3. The second inequality is seen by Lemma 4 and the fact that r 1 plog(n2m + 1) = 2m log n + log(1 + ) ≤ n2m r 1 1 2m log n + ≤ p2m log n(1 + ) n2m n

35 We note that the quotient between the upper and lower estimate is essentially independent of m for small m’s.

5 A problem of Erd˝os

In this section I will use Corollary 1, Proposition 3 and a classic result by Dirichlet in Diophantine approximation to solve the following problem of Erd˝os, given in Tur´an[8] as problem 47 on page 196. Does there exist for each integer 1 ≤ m ≤ n − 1 a c = c(m) so that

v v |z1 + ··· + zn| max v ≥ m? (7) 1≤v≤c(m)n,v integer mink=1,...,n |zk| I will start by proving an essential lemma.

Lemma 5. If |zk| ≥ 1, and n ≥ m + 1 then n X v max zk ≥ m n/2 “ 2π2n ” v=1,...,b n−m c k=1 Proof. We use the following version of Dirichlet’s theorem:

Given real numbers a1, a2, . . . , an, and an A ≥ 2 then we can ﬁnd 1 an integer k in the range 1 ≤ k ≤ An such that {ka } ≤ . (8) i A See e.g. Tur´an[8] section 15. q 2π2n n Now put A = n−m and let zk = |zk|e(θk). By (8) choose a v, 1 ≤ v ≤ A 1 such that {vθk} ≤ A . We get n n ! n !

X v X v X v zk ≥ Re zk = Re e(vθk)|zk| = k=1 k=1 k=1 n X x2 = cos(2π{vθ })|z |v ≥ (By the relation 1 − ≤ cos x) ≥ k k 2 k=1 n X 1 ≥ 1 − (2π{vθ })2 |z |v ≥ 2 k k k=1 n 2! X 1 2π v v n − m ≥ 1 − |zk| ≥ min |zk| n 1 − ≥ m. 2 A k=1,...,n n k=1

36 We now state the main result in this section.

Proposition 4. If |zk| ≥ 1, we have for n ≥ m + 1 that n X v max zk ≥ m v=1,...,bmax 1021m,(2π2m)m/2 nc ( ) k=1 Proof. We will show this result in three cases. Case 1: m + 1 ≤ n ≤ 32m

2π2x x/2 Proof. First we put f(x) = ( x−m ) . From Lemma 5 we obtain that h(m) = 1 m maxn=m+1,...,32m f(n) clearly is suﬃcient as c(m) in case 1. We get h(m) ≤ 1 m maxx∈[m+1,32m] f(x). We now notice that f(x)’s only critical point for x > m is a minimum point, hence the maximum must be obtained in the interval 1 1 end-points, and we get h(m) ≤ m max(f(m + 1), f(32m)). Since m f(32m) = 1 2 16m 21m m ((32/31)2π ) ≤ 10 , and

1 m/2p 2 1 1 2 m+1 2 m/2 (1 + m ) 2π (m + 1) f(m + 1) = (2π (m + 1)) 2 = (2π m) ≤ m m m ≤ (2π2m)m/2, for e.g. m > 1000 Since clearly 1021m is the dominating term for m < 1000 this is enough. We have now proved the proposition in Case 1. Case 2: 32m ≤ n ≤ m2 Proof. We see that it only makes sense for m ≥ 32. Then we have by Proposition 3 that

n X v p −1 max zk ≥ (n − bm/8c)bm/8ce ≥ v=1,...,n2bm/8c k=1 r 3 1 r 31 ≥ 31m me−1 ≥ m 3 e−1 ≥ m. 4 8 32 2bm/8c m 2 m/2 Since n ≤ m 2 ≤ (2π m) when n is in the above interval, we have proved the proposition in case 2. Case 3: m2 ≤ n Proof. By Corollary 1 we get

n X v max zk ≥ m v=1,...,2nm2 k=1 Which clearly is suﬃcient.

37 Since we have proved all three cases we have proved the proposition. We see that Proposition 4 solves the problem (7), of Erd˝osaﬃrmative with c(m) = max(1021m, (2π2m)m/2) We also note that we do not really need the powerful results of section 4. By using either Lemma 1, Corollary 1 and Dirichlet’s theorem or just Corollary 1 and Dirichlet’s theorem we can solve the problem, although the results one gets will be asymptotically worse than the ones proved 5m/2 m2 above(they will be of the order (C1m) , respectively C2 for constants C1 and 21m 2 m/2 2 m/2 C2). Note that for large m, 10 < (2π m) . hence it is the term (2π m) that will be asymptotically signiﬁcant. It seems that this is essentially the asymptotically best possible estimate obtainable with these methods, although for small m it is easy to obtain better estimates.

6 On Tur´an’s problem 17 and 18

I will ﬁnally present a short solution to two further power sum problems from Tur´an’sbook [8]. Problem 17 page 191 is the following: Show that for arbitrarily small ε > 0, there is an m0(ε, n), such that for every ∗ ∗ ∗ integer m > m0(ε, n) there exists a system (z1, . . . , zn) with maxj |zj | = 1 for which the inequality

n X ∗v n max zk ≤ ε (9) m+1≤v≤m+n,v integer k=1

∗ holds. In problem 18 he asks if problem 17 also holds if maxj |zj | = 1 is replaced ∗ by minj |zj | = 1. I will now show a proposition which implies that there in fact exists an m0(ε, n) such that the statements in both problems are true.

2 −n Proposition 5. For each m ≥ 4πn ε there exists complex numbers zk , with |zk| = 1 such that

n X v n max zk ≤ ε m+1≤v≤m+n,v integer k=1

38 2πi k Proof. Let zk = e mn . For m + 1 ≤ v ≤ m + n we get

n n n X v X v 2πi k X 2πi k n n zk ≤ zk − e + e = k=1 k=1 k=1 n n X 2πi v k 2πi k X 2πi v −1 k m n n ( m ) n = e − e ≤ 1 − e ≤ k=1 k=1 2 ix m + n 4πn n ≤ (By the relation |e − 1| ≤ 2x) ≤ n 4π − 1 ≤ ≤ ε m m

References

[1] J. W. S. Cassels, On the sums of powers of complex numbers, Acta Math. Acad. Sci. Hungar. 7 (1956), 283–289.

[2] P. Erd¨osand A. R´enyi, A probabilistic approach to problems of Diophantine approximation, Illinois J. Math. 1 (1957), 303–315.

[3] J. Fabrykowski, A note on sums of powers of complex numbers, Acta Math. Hungar. 62 (1993), no. 3-4, 209–210.

[4] H. Iwaniec and J. Pintz, Primes in short intervals, Monatsh. Math. 98 (1984), no. 2, 115–143.

[5] H. Leenman and R. Tijdeman, Bounds for the maximum modulus of the ﬁrst k power sums, Nederl. Akad. Wetensch. Proc. Ser. A 77=Indag. Math. 36 (1974), 387–391.

[6] H. Queffélec, Sur un théorèmede Gluskin-Meyer-Pajor, C. R. Acad. Sci. Paris Sér.I Math. 317 (1993), no. 2, 155–158.

[7] J. Singer, A theorem in ﬁnite projective geometry and some applications to number theory, Trans. Amer. Math. Soc. 43 (1938), 377–385.

[8] P. Tur´an, On a new method of analysis and its applications, Pure and Applied Mathematics, John Wiley & Sons Inc., New York, 1984, With the assistance of G. Hal´aszand J. Pintz.

Disproof of some conjectures of P. Tur´an

Johan Andersson∗

1 Introduction

In his ﬁrst paper on power sum theory [4], P. Tur´anstated a power sum conjecture which implies the density hypothesis of the Riemann zeta-function. In this paper we will show that a considerably weaker statement is still false. That is if Claim. For some C > 1 one has that n X inf max eλkx > e−o(n). (1) n λ∈C ,λ1=0 x∈[1,C] k=1 Then Theorem. Claim is false.

2 Some conjectures

We first state the two original conjectures of Turàn [4]. It should be noted that Halász in a note in [6], (page 2083) wrote that the conjectures were probably false, but that no disproof had been found.

Conjecture 1. If ω(n) is a positive increasing function with limn→∞ ω(n) = ∞ and for some n ≥ C1 there exists a M = M(n) such that the inequality n2 nω(n) ≤ M(n) ≤ , ω(n)2 then

n2 v v − Mω(n) max |z1 + ··· + zn| > e 1 M(1− ω(n) )≤v≤M v integer

∗This result was ﬁrst presented at the Halberstam conference in analytic number theory in Urbana-Champaign, 1995.

41 for all systems (z1, . . . , zn) satisfying the conditions

n2 z = 1, and 1 − ≤ |z | ≤ 1. (v = 2, 3, . . . , n) (2) 1 2M 2 v

Conjecture 2. For n ≥ c0, z1 = 1 and |z2| ≤ 1,..., |zn| ≤ 1 one has that

v v −n0.09 max |z1 + ··· + zk| > e . n3/2(1−n−0.42)≤v≤n3/2

In [4] Tur`an shows that Conjecture 2 implies the density hypothesis for the Riemann zeta-function. If we use a modiﬁcation of his method it is possible to show that the conjectures can be weakened. However we would still need something like the existence of a C > 1 so that Claim is true to prove the density hypothesis by this approach. We have

Proposition 1. Conjecture 1 and Conjecture 2 are false.

Proof. The falsity of the conjectures follows from Theorem and the fact that both conjectures implies Claim. To prove the ﬁrst two implications, pick a λ ∈ Cn. We can assume that maxk Re(λk) = 0, since otherwise if j is such that maxk Re(λk) = ˆ ˆ ˆ Re(λj), then consider λk = λk − λj, with λ1 and λj interchanged. 1. Conjecture 1 =⇒ Claim. 1 λk/(M(1− ω(n) )) n Choose zk = e . First divide {zk}k=1 into two disjoint sets n0 n00 n0 {zki }i=1 and {zmj }j=1 such that {zki }i=1 is exactly the set that fulﬁlls (6). n 2 2 Now considerz ˆ ∈ C , wherez ˆki = zki , andz ˆmi = 1 − n /(2M ). Then for e.g. ω(n) (log n)2 we have

e−o(n) e−n/ω(n)2 , e−n2/(Mω(n)) − 2n(1 − n2/(2M 2))M , v v v v v v max |zˆ1 + ··· +z ˆn| − |zˆ1 − z1 + ··· +z ˆn − zn|, 1 M(1− ω(n) )≤v≤M v integer n X ≤ max eλkx , x∈[1,1/(1−1/ω(n))] k=1 n X max eλkx , x∈[1,C] k=1 for some C > 0.

42 2. Conjecture 2 =⇒ Claim. 3/2 −0.42 λk/(n (1−n )) Choose zk = e . We obtain

e−o(n) e−n0.09 , v v max |z1 + ··· + zk|, n3/2(1−n−0.42)≤v≤n3/2 n X ≤ max eλkx , x∈[1,1/(1−n−0.42)] k=1 n X max eλkx , x∈[1,C] k=1 for some C > 0.

3 Proof of theorem

First we remark that in the next section we will only use the principal part of the logarithm, and αβ will mean eβ log α. We have the following lemma Lemma. One has for α and β > 1 real numbers that

∞ ∞ β−1 X −β 2π X 2πk (1 + αki) = e−2πk/α. αΓ(β) α k=−∞ k=1 Proof. By using the Poisson summation formula

∞ ∞ X Z ∞ X f(x)e−ikxdx = 2π f(2πk) k=−∞ −∞ k=−∞ on ( xβ−1e−x/α, when x > 0, f(x) = 0, when x ≤ 0; we get for β > 1 the identity

∞ ∞ X X Γ(β)(1/α + ik)−β = 2π (2πk)β−1e−2πk/α. k=−∞ k=1

By dividing both sides with Γ(β)αβ we obtain the lemma.

43 Note that the Lemma is a special case of the functional equation for the Lerch zeta-function. In fact a proof for the functional equation for the Lerch zeta-function along these lines has been given by Oberhettinger [3]. Proposition 2. One has that if 0 < a ≤ b < 2π, then n −nx X ik 1 + e−Cn n k=−n for each x ∈ [a, b], and some C = C(a, b) > 0. 1 Proof. By using β = nx, and α = n in the Lemma we obtain n −nx X ki 1 + = A + B, (3) n k=−n where −nx X ki A = 1 + , n |k|>n and ∞ 2πn X nx−1 B = (2πkn) e−2πkn. Γ(nx) k=0 It is easy to see that

A n2−na/2 (4) for x ≥ a. The dominating term will be the ﬁrst, which will have absolute value 2−x/2. To estimate B we use the Stirling formula √ Γ(λ) ∼ 2πλλ+1/2e−λ. We get ∞ X nx−1 B n(nx)−nx−1/2enx (2πkn) e−2πkn k=0 ∞ nx X 2πk pn/x e1−2πk/x x k=0 (5) 2π −nx pn/x e1−2π/x x 2π −nb pn/b e1−2πb , b

44 for x ≤ b, since the terms tends to zero fast enough. From equations (3), (4) and (5), we see that we can choose b 2π a log 2 C(a, b) = min b log + 1 − , − , (6) 2π b 2 which is a constant > 0 for some > 0, since the function 2π 2π f(x) = log + − 1 x x has the unique maximum 0 at x = 2π. Proof of Theorem. That Claim is false is a direct consequence of Proposition 2. Choose for λ ∈ C2n+1, 1 ki λ = − log 1 − , 2k+1 C n and 1 ki λ = − log 1 + . 2k C n

4 Summary

The results of our Theorem shed new light on Turán’sachievement. The power sum result n n X k n max bkz ≥ 1.007 min |b1 + ··· + bk| (7) m+1≤v≤m+n 4e(m + n) k=1,...,n v integer k=1 for |z1| = 1 ≥ |z2| ≥ · · · ≥ |zn|. This result, which is a refined version of Turán’s second main theorem (see Turán [5](with constant 4e instead of 8e), or Kolesnik- Straus [1]), seems more powerful than ever when we consider the much more special case when bk = 1 and the maximum with respect to v is considered for v in the interval [m, n + m] instead of just the integers m, . . . , n + m, for e.g. n = m and we still obtain results of decreasing exponential order). It should also be noted that already Makai [2] showed that 4e is in general the best possible constant in (7). Even though the situation might seem disappointing, note that conjectures 1 to 2 and the Claim deal with conditions on the maximum norm, maxk |zk| = 1. Claim should be true for the minimum norm (Re(λk) > 0) and it is quite likely that we have

45 Conjecture 3. If C > 1 one has that

n X inf max eλkx > e−o(n). (8) n λ∈ ,Re(λk)≥0 x∈[1,C] C k=1 It is possible to show that the corresponding non-pure power sum problem (i.e. when bk = 1 do not need to be true), then the analogue of Conjecture 3 is false. However in the pure power sum case this seems likely (to us). It would not (to our knowledge) have any consequences to the theory of the Riemann zeta- function. Nevertheless it would be of interest to investigate it further. Another problem is to see to what extent Claim fails. That is Problem. Given C > 1. Find the smallest constant B = B(C) so that

n X inf max eλkx e−(B+)n, (9) n λ∈C ,λ1=0 x∈[1,C] k=1 for all > 0. From (7) it is easy to see that B ≤ log(4eC/(C − 1)). We also see that equation (6) can give us a lower bound for B.

References

[1] G. Kolesnik and E. G. Straus, On the sum of powers of complex numbers, Studies in pure mathematics - To the memory of Paul Tur´an, Birkh¨auser, Basel-Boston, Mass., 1983, pp. 427–442.

[2] E. Makai, On a minimum problem II, Acta. Math. Hung. 15 (1964), no. 1–2, 63–66.

[3] F. Oberhettinger, Note on the Lerch zeta-function, Paciﬁc J. Math. 6 (1956), 117–120.

[4] P. Tur´an, Uber¨ die Verteilung der Primzahlen (1), Acta. Sci. Math. Szeged 10 (1941), 81–104.

[5] , On a new method of analysis and its applications, John Wiley & Sons, New York, 1984.

[6] , Collected papers of Paul Turán, Akadémiai Kiadó(Publishing House of the Hungarian Academy of Sciences), Budapest, 1990.

46 Disproof of some conjectures of K. Ramachandra

Johan Andersson∗

1 Introduction

In a recent paper [9] K. Ramachandra states some conjectures, and gives consequences in the theory of the Riemann zeta function. In this paper we will will present two different disproofs of them. The first will be an elementary application of the Szász-Müntz theorem. The second will depend on a version of the Voronin universality theorem, and is also slightly stronger in the sense that it disproves a weaker conjecture. An elementary (but more complicated) disproof has been given by Rusza-Lazkovich [11].

2 Disproof of some conjectures

2.1 The conjectures We will ﬁrst state the three conjectures as given by Ramachandra [9], and Ramachandra- Balasubramian [5]:

Conjecture 1. For all N ≥ H ≥ 1000 and all N−tuples a1 = 1, a2, . . . , aN of complex numbers we have

N 1 Z H X a nit dt ≥ 10−1000. H n 0 n=1

Conjecture 2. For all N ≥ H ≥ 1000 and all N−tuples a1 = 1, a2, . . . , aN of complex numbers we have when M = H(log H)−2 that

N 2 M 1 Z H X X a nit dt ≥ (log H)−1000 |a |2. H n n 0 n=1 n=1

∗This paper has been published in Hardy-Ramanujan J. 22 (1999), 2–7.

47 Conjecture 3. There exist a constant c > 0 such that

2 Z T N X it X 2 ann dt ≥ c |an| . 0 n=1 n≤cT

2.2 The Szász-Müntz theorem To disprove these conjectures, we first consider the following classic result of Szász

Lemma 1. (Sz´asz) If we have that

∞ X 1 + 2 Re(λn) = +∞, 1 + |λ |2 n=1 n

λn where Re(λn) ≥ 0 then the set of ﬁnite linear combinations of x is dense in L2(0, 1).

Proof. See Sz´asz [12], theorem A. We will now state a theorem that will eﬀectively disprove the above conjectures:

Theorem 1. For each D ≥ 0 and ε > 0 there exists an N ≥ 0 and complex numbers a2, . . . , aN , such that 2 Z D N X it 1 + ann dt ≤ ε. 0 n=2

Proof. Since −1 ∈ L2(0, 1) and

∞ X 1 + 2 Re(−i log n) = +∞ (1) 1 + | − i log n|2 n=2 we have by the lemma that for each δ > 0 there exists an N > 0 and complex numbers a2, . . . , aN such that 2 Z 1 N X −i log n 1 + anx dx < δ. 0 n=2

48 We obtain 2 2 Z 1 N Z 1 N X −i log n X −i log n δ > 1 + anx dx ≥ 1 + anx dx = −D 0 n=2 e n=2 = (Substituting t = − log x) = 2 2 Z D N Z D N −t X it −D X it e 1 + ann dt ≥ e 1 + ann dt. 0 n=2 0 n=2 By choosing δ = e−Dε we obtain the theorem. It is now an easy task to falsify the conjectures. Proposition. Conjectures 1, 2 and 3 are false Proof. For Conjecture 1, choose ε < 10−2000H and D = H in Theorem 1 and apply the Cauchy-Schwarz inequality v N u N 2 Z H X √ uZ H X it t it ann dt ≤ H ann dt. 0 n=1 0 n=1 For Conjecture 2, choose ε < (log H)−1000H, and D = H in Theorem. To disprove Conjecture 3, choose e.g. a1 = T = 1, and ε = c/2.

2.3 The Voronin theorem In private correspondence, Ramachandra asked whether the conjectures hold un- 100 der the additional growth assumption |ak| (Hk) . Ramachandra and Bal- asubramian have proved conjectures 1 and 2 under the this and the additional further assumption that N < exp(exp(cH)). This shows that N must be very large compared to H for the conjectures to be false. However, they are in fact still false, although their proof requires a deeper result. A version of the Voronin universality theorem for the Riemann zeta-function. We will state the theorem that shows that the conjectures are still false below:

1 δ−1 Theorem 2. Suppose that H, ε > 0, 0 < δ < 2 . Then there exists |ak| ≤ k such that

N X it max 1 + ann dt < ε. t∈[0,H] n=2 The idea goes as follows: We use the following version of the Voronin universality theorem

49 Theorem 3. (Voronin-Bagchi) For any compact subset K of the complex numbers 1 such that x ∈ K =⇒ 2 < Re(x) < 1, non vanishing analytic function f on K and ε > 0, we have a real t such that |ζ(z + it) − f(z)| ≤ , for all z ∈ K

Proof. See Bagchi [1]. and the following version of the approximate functional equation for the Rie- mann zeta-function

N X 1 ζ(σ + it) − k−σ−it ≤ CN σ−1, for e.g. t < N < 2t and σ ≥ . 2 k=1

Proof. (of Theorem 2) Choose T0 so that when T > T0 then

X ε ζ(1 − δ + it) − kδ−1−it < 3 k<2T when T ≤ t ≤ T + H. Now choose T > T0 so that ε |ζ(1 − δ + it)| < 3 for T ≤ t ≤ T + H (This is possible by applying Theorem 3 to f(z) = ε/3, and K = [1 − δ, 1 − δ + iH]). By using the triangle inequality we get

X δ−1−it k < ε k<2T

δ−1−iT for all T < t < T + H. Now we can choose N = T and an = n in the theorem.

3 Summary

In [7] K. Ramachandra stated a similar conjecture to conjectures 1 and 2. Al- though it was more general in the sense that it considered Dirichlet series of form P −s A(s) = 1 + anλn , for certain λn generalizing λn = n it was much weaker as it had three further conditions. H depended on N, |an| were bounded from above and A was bounded in certain regions in the complex plane. Under these additional assumptions, A(s) is called a Titschmarsh-series and for these, analogues of conjectures 1 and 2, and similar conjectures were proved in Ramachandra [8] , and Balasubramian and Ramachandra [4], [3] and [2]. It would certainly be

50 interesting to see if all these additional assumptions are needed, or if a certain subset of them implies the truth of conjectures 1 and 2. For a disproof of the Lp(0,D), p > 2 version of conjectures 1 and 2, or for considering much more sparse sequences than log n in (1), we need a stronger version of the Lemma, the reader is referred to the literature on the theory of completeness of complex exponentials, see e.g. the classic by Levinson [6] or Redheffer [10] for a more recent survey. It should be noticed that problems similar to conjectures 1 and 2 are also studied in Turán power sum theory (although there it is essentially the λn’s which vary, instead of the an’s), see Turán [13] for a thorough treatment.

References

[1] B. Bagchi, A joint universality theorem for dirichlet l-functions. (english), Math. Z. 181 (1982), no. 3, 319–334.

[2] R. Balasubramanian and K. Ramachandra, Proof of some conjectures on the mean-value of Titchmarsh series. I, Hardy-Ramanujan J. 13 (1990), 1–20.

[3] , Proof of some conjectures on the mean-value of Titchmarsh series. II, Hardy-Ramanujan J. 14 (1991), 1–20.

[4] , Proof of some conjectures on the mean-value of Titchmarsh series. III, Proc. Indian Acad. Sci. Math. Sci. 102 (1992), no. 2, 83–91.

[5] , On Riemann zeta-function and allied questions. II, Hardy- Ramanujan J. 18 (1995), 10–22.

[6] N. Levinson, Gap and Density Theorems, American Mathematical Society, New York, 1940, American Mathematical Society Colloquium Publications, v. 26.

[7] K. Ramachandra, Progress towards a conjecture on the mean-value of Titch- marsh series-i, Recent progress in Analytic number theory (H. Halberstam and C. Hooley, eds.), vol. 1, Academic Press, London, New York, Toronto, Sydney, San Fransisco, 1981, pp. 303–318.

[8] , Proof of some conjectures on the mean-value of Titchmarsh series with applications to Titchmarsh’s phenomenon, Hardy-Ramanujan J. 13 (1990), 21–27.

[9] , On Riemann zeta-function and allied questions, Astérisque(1992), no. 209, 57–72, JournéesArithmétiques,1991 (Geneva).

51 [10] R. M. Redheﬀer, Completeness of sets of complex exponentials, Advances in Math. 24 (1977), no. 1, 1–62.

[11] M. Rusza, I. Lazkovich, Sums of periodic functions and a problem of Ra- machandra, Tech. report, Mathematical institute of the Hungarian Academy of Sciences, 1996.

[12] 0. Sz´asz, Uber¨ die Approximation stetiger Funktionen durch lineare Aggregate von Potenzen, Math Ann. 77 (1916), 482–496.

[13] P. Turán, On a new method of analysis and its applications, Pure and Ap- plied Mathematics, John Wiley & Sons Inc., New York, 1984, With the assistance of G. Halász and J. Pintz, With a foreword by Vera T. Sós, A Wiley-Interscience Publication.

52 PART II - MOMENTS OF THE HURWITZ AND LERCH ZETA FUNCTIONS

INTRODUCTION

It is conjectured that for the Riemann zeta-function one has

Z T 1 2k k2 Zk(T ) = ζ 2 + it dt ∼ ckT log T. (II.21) 0 This has so far only been proved for k = 1, 2, although the lower bound has been proved for a general k ≥ 1 by the method of Balasubramanian and Ramachandra (see [5]). A way to understand the Riemann zeta-function better is to see how it relates to other zeta functions, such as the Hurwitz zeta-function deﬁned by

∞ X ζ(s, x) = (k + x)−s, k=0 and the Lerch zeta-function deﬁned by

∞ X φ(x, y, s) = e(kx)(k + y)−s, k=0 for Re(s) > 1 and by analytic continuation elsewhere. We may thus ask: What can be said about the moments Z T 1 2k ζ 2 + it, x dt, (II.22) 0 and

Z T 1 2k φ x, y, 2 + it dt? (II.23) 0 This problems turns out to be difficult, since the behavior will be different depending on whether x and y are rational or not. This is due to the fact that while the Hurwitz and Lerch zeta-functions admit a functional equation, they do not have an Euler product, and they are in general non-arithmetic. For the special case of rational x and y, we can express them in terms of finite sums of Dirichlet L-functions, and they will inherit arithmetical properties. Introduction 56

The mean square of the Hurwitz zeta function

As we have a parameter x in the Hurwitz zeta function we may consider moments with respect to x instead of t. We will start with the mean square. In a paper from 1956 [4], M Mikol´asused the functional equation (the Fourier-expansion)

∞ ∞ ! 2Γ(1 − s) πs X sin 2πkx πs X cos 2πkx ζ(s, x) = sin + cos , (II.24) (2π)1−s 2 k1−s 2 k1−s k=1 k=1 valid for Re(s) < 0 and x ∈ [0, 1], and the Parseval identity to prove the following formula for the inner product of two Hurwitz zeta-functions:

Z 1 π ζ(z, x)ζ(w, x)dx = 2(2π)z+w−2Γ(1 − z)Γ(1 − w) cos (z − w) ζ(2 − z − w), 0 2 (II.25) for Re(z), Re(w), Re(z +w) < 1. By then setting w =z ¯ he obtained a mean-square formula for Re(z) < 1/2. However on the critical line Re(z) = 1/2, the integral diverges. This lead Koksma and Lekkerkerker [3] to instead consider the integral

Z 1 ∗ 1 2 ζ 2 + it, x dx, (II.26) 0 where ζ∗(s, x) = ζ(s, x) − x−s, and by using the approximate functional equation X ζ∗(s, x) = (k + x)−s + O(1), (Re(s) > 0) (II.27) k<2 Re(s)

1 for s = 2 +it, squaring and using explicit estimates they obtained (II.26)=O(log T ). When we wrote our paper “Mean value properties of the Hurwitz zeta function” the best result we knew of at the time was an O(1) error term. However Wen Peng Zhang [6] had recently proved that (II.26) = log t + γ − log 2π + Ot−θ for θ = 7/36 − . Furthermore he had conjectured that one can choose θ = 1/4. In our paper we used analytic continuation of Mikol´asformula as well as analytic continuation with respect z of Z 1 ζ∗(s, x)x−zdx, (II.28) 0 to obtain Z 1 0 1 ∞ 1 ! ∗ 1 2 Γ 2 + it X ζ 2 + it + k − 1 ζ + it, x dx = γ − log 2π + Re − 2 . 2 Γ 1 + it 1 + it + k 0 2 k=1 2 (II.29) Introduction 57

This showed that we can could chose θ = 47/56, by the best order estimate of the Riemann zeta-function known at the time we wrote the paper. This especially proved Wen Peng Zhang’s conjecture. By an estimate of Huxley [1] this can be improved to θ = 481/570. Of course the Lindel¨ofhypothesis gives θ = 1 − , for each > 0. For a more detailed history of the problem as well as further developments see e.g. Katsurada-Matsumoto [2].

The Fourth power moment

In analogy with the Riemann zeta function case it might seem natural to assume that Z 1 ∗ 1 4 4 ζ 2 + it, x dx ∼ c log t. (II.30) 0 In our paper “On the fourth power moment of the Hurwitz zeta function” we show that a formula like (II.30) cannot be true. In fact the moment

Z 1 ∗ 1 4 ζ 2 + it, x dx 0 will be rather chaotic in the same sense as the absolute values of the Riemann zeta function in the critical strip. On one hand we will be able to prove that

Z 1 1/4 ∗ 1 4 c(log t) ζ 2 + it, x dx = Ω exp . 0 log log t On the other hand we prove that its mean value

Z T Z 1 1 ∗ 1 4 2 5/3 ζ 2 + it, x dxdt = 2 log T + O (log T ) , T 0 0 will grow just like a log power of t. We notice that the log power here is 2 and not 4 as in the case of the Riemann zeta function. For rational x = p/q with (p, q) = 1 we will have Z T 4 1 ∗ 1 p 4 ζ + it, dx ∼ cq log T T 0 2 q though. Similarly in our paper “On he fourth power moment of the Lerch zeta function” we show the corresponding results

Z T Z 1 1 ∗ 1 4 2 5/3 φ x, y, 2 + it dxdt = 2 log T + O (log T ) , (0 ≤ y ≤ 1), T 0 0 (II.31) Introduction 58 and Z T Z 1 1 ∗ 1 4 2 5/3 φ x, y, 2 + it dydt = 2 log T + O (log T ) . (0 ≤ x ≤ 1), T 0 0 (II.32) and indicate that for rational values of x, y Z T 4 1 ∗ p h 1 4 φ , , + it dt ∼ ch,k,p,q log T. (II.33) T 0 q k 2 Here we use the deﬁnition

φ∗(x, y, s) = φ(x, y, s) − y−s − (2π)s−1Γ(1 − s)× 1 − s s − 1 × e − yx xs−1 + e + y(1 − x) (1 − x)s−1 , (II.34) 4 4 which is chosen so that φ does not tend to inﬁnity when x, y → 0, 1. For the Lerch zeta function we are able to prove that Z 1 Z 1 ∗ 1 4 2 5/3 φ x, y, 2 + it dxdy = 2 log t + O (log t) , 0 0 which is a natural generalization to (II.29).

Higher power moments

In general we expect the following conjectures to be true Z T 2k 1 ∗ 1 p k2 ζ + it, dx ∼ cp,q,k log T, T 0 2 q Z T 2k 1 ∗ p m 1 k2 φ , , + it dt ∼ cm,n,p,q,k log T, T 0 q n 2 Z T Z 1 1 ∗ 1 2k k ζ 2 + it, x dxdt ∼ k! log T, T 0 0 Z T Z 1 1 ∗ 1 2k k φ x, y, 2 + it dxdt ∼ k! log T, (0 ≤ y ≤ 1), T 0 0 Z T Z 1 1 ∗ 1 2k k φ x, y, 2 + it dydt ∼ k! log T, (0 ≤ x ≤ 1), T 0 0 and Z 1 Z 1 ∗ 1 2k k φ x, y, 2 + it dxdy ∼ k! log t. 0 0 We will return to this topic in the forthcoming work. BIBLIOGRAPHY

[1] M. N. Huxley, Exponential sums and the Riemann zeta function. IV, Proc. London Math. Soc. (3) 66 (1993), no. 1, 1–40.

[2] M. Katsurada and K. Matsumoto, Explicit formulas and asymptotic expansions for certain mean square of Hurwitz zeta-functions. I, Math. Scand. 78 (1996), no. 2, 161–177.

[3] J. F. Koksma and C. G. Lekkerkerker, A mean-value theorem for ζ(s, w), Ned- erl. Akad. Wetensch. Proc. Ser. A. 55 = Indagationes Math. 14 (1952), 446– 452.

[4] M. Mikl´os, Transformation und orhogonalit¨at bei ζ(s, u). Verallgemeinung der Riemannschen Funktionalgleichung von ζ(s), Acta. Sci. Mat. Szeged. 17 (1956), 143–164.

[5] K. Ramachandra, On the mean-value and omega-theorems for the Riemann zeta-function, Tata Institute of Fundamental Research Lectures on Mathe- matics and Physics, vol. 85, Published for the Tata Institute of Fundamental Research, Bombay, 1995.

[6] W. P. Zhang, On the Hurwitz zeta-function, Illinois J. Math. 35 (1991), no. 4, 569–576.

Mean value properties of the Hurwitz zeta-function

Johan Andersson∗

1 Introduction

The Lindelöfhypothesis for the Hurwitz zeta-function states that ζ(1/2+it, x) = O(t) for each fixed x and > 0. When x = 1 we have the usual hypothesis. The hypothesis is far from being proved but in 1952 Koksma and Lekkerkerker proved the estimate Z 1 ∗ 1 2 |ζ ( 2 + it, x)| dx = O(log t), 0 where ζ∗(s, x) = ζ(s, x + 1) and since then the result has been sharpened. The result is very similar to Lindelöf’s hypothesis, but it states that the mean square is small, not the function for fixed x. In this paper I will continue in Lekkerkerker’s tradition and calculate some integrals of which

1 Z 47 ∗ 1 2 − 56 + |ζ ( 2 + it, x)| dx = log t + γ − log 2π + O(t )( > 0) 0 is an improvement on former estimates. Especially I will obtain a better error- term depending on the Lindel¨of hypothesis. Thus the hypothesis implies a better mean square formula.

2 Integrals involving the Hurwitz zeta-function

From Hurwitz formula for the Hurwitz zeta-function ∞ ∞ ! 2Γ(1 − s) πs X sin 2πkx πs X cos 2πkx ζ(s, x) = sin + cos , (2π)1−s 2 k1−s 2 k1−s k=1 k=1 Re(s) < 0 ∧ x ∈ [0, 1] (1)

∗This paper has been published in Math. Scand. 71 (1992), no. 2, 295–300.

61 we see directly with Parseval’s identity (see Mikl´osMikol´as [5], [4],[6]): Z 1 ζ(z, x)ζ(w, x)dx = (2) 0 π 2(2π)z+w−2Γ(1 − z)Γ(1 − w) cos( (z − w))ζ(2 − z − w) 2 max(Re(z), Re(w), Re(z + w)) < 1 The formula holds initially for max(Re(z), Re(w)) < 0, but I will show later that it holds for the extended region. We have that ζ(s, x) = ζ∗(s, x) + x−s, and ζ∗(s, x) is continuous w.r.t. x for x ∈ [0, 1] and thus Z 1 Z 1 ζ∗(z, x)ζ∗(w, x)dx = (ζ(z, x) − x−z)(ζ(w, x) − x−w)dx = (3) 0 0 Z 1 (ζ(z, x)ζ(w, x) + x−(z+w) − ζ(z, x)x−w − ζ(w, x)x−z)dx = 0 Z 1 (ζ(z, x)ζ(w, x) − x−(z+w) − x−wζ∗(z, x) − x−zζ∗(w, x))dx = 0 Z 1 Z 1 (ζ(z, x)ζ(w, x) − x−(z+w))dx − (ζ∗(z, x)x−w + ζ∗(w, x)x−z)dx 0 0 max(Re(w), Re(z)) < 1 I will now deduce an expression for the last two integrals

Z 1 ζ∗(z, x)x−wdx = (partial integration) = 0 ∂ζ∗ (we use (s, x) = −sζ∗(s + 1, x)) ∂x x1−w 1 z Z 1 ζ∗(z, x) + ζ∗(z + 1, x)x1−wdx = 1 − w 0 1 − w 0 ζ(z) − 1 z Z 1 + ζ∗(z + 1, x)x1−wdx 1 − w 1 − w 0 Consider

n X (z)k sn(z, w) = (ζ(z + k) − 1), (1 − w)k+1 k=0 Z 1 ∗ −w we see that ζ (z, x)x dx = sn(z, w) + Rn 0 Z 1 (z)n+1 ∗ n+1−w where Rn = ζ (z + n + 1, x)x dx (1 − w)n+1 0

62 R 1 ∗ n−w n after n + 1 partial integrations. We have 0 ζ (n + z)x dx = O((1/2) ), ∗ −z−n (1−w)n (z)n+1 since ζ (z+n, x) ∼ (x+1) when n → ∞. We also have limn→∞ = (z)n (1−w)n+1 1, hence Rn → 0, and {sn(z, w)} converges when n → ∞, and

Z 1 ∞ X (z)k ζ∗(z, x)x−wdx = (ζ(z + k) − 1) (4) (1 − w)k+1 0 k=0 − z 6∈ Z ∪ {0, 1} ∧ Re(w) < 1 We combine formula (3) and (4) and get

Z 1 ζ∗(z, x)ζ∗(w, x)dx = (5) 0 π 2(2π)z+w−2Γ(1 − z)Γ(1 − w) cos( (z − w))ζ(2 − z − w)+ 2 ∞ 1 X (z)k (w)k + − (ζ(z + k) − 1) + (ζ(w + k) − 1) 1 − z − w (1 − w)k+1 (1 − z)k+1 k=0 z∈ / Z and w∈ / Z ∧ z + w 6= 1 I will now show (by analytic continuation) that the equality really holds in the region stated. We know that the integral in (5) is analytic w.r.t. z and w for all z, w∈ / Z (Since the function under the integral-sign is analytic w.r.t. z and w and uniformly continuous, w.r.t x for z, w 6= 1. I will now show that the right hand side of the equality (5) is analytic w.r.t. z and w. The ﬁrst product is clearly 1 analytic since its factors are analytic. The expression 1−z−w is also analytic when z + w 6= 1. We therefore only have to consider the last sum. First we notice that by symmetry it is enough to prove that

∞ X (z)k (ζ(z + k) − 1) (1 − w)k+1 k=0 is analytic for w and z. We have

∞ M−1 X (z)k X (z)k (ζ(z + k) − 1) = (ζ(z + k) − 1)+ (1 − w)k+1 (1 − w)k+1 k=0 k=0 ∞ (z)M X (z + M)k (ζ(k + M + z) − 1) (1 − w)M+1 (2 − w + M)k k=0 Clearly the ﬁrst term in this expression is analytic since it is a ﬁnite sum of analytic functions (when z, w∈ / Z). We see that it is enough to prove that the last

63 sum is analytic, since it is multiplied by an analytic function. Let w ∈ {w : |w − w0| < 1} and z ∈ {z : |z − z0| < 1}. Choose M > | min(Re(1 − w0), Re(z0))| + 3. We see that

∞ ∞ X (z + M)k X (|z + M| + 1)k (ζ(k + M + z) − 1) ≤ (ζ(2 + k) − 1) (2 − w + M)k (|2 − w + M| − 1)k k=0 k=0 which converges. By Weierstrass M-test the sum converges uniformly with z and w in the given neighborhoods and is thus analytic. We already have the equality (5) for max(Re(z), Re(w)) < 0 and by uniqueness of analytic continuation we get (5) in the region stated. From (5) we see when z = σ + it and z = σ − it

Z 1 |ζ∗(σ + it, x)|2dx = (6) 0 2(2π)2σ−2|Γ(1 − σ − it)|2 cosh(πt)ζ(2 − 2σ)+ ∞ ! 1 X (σ + it)k + − 2 Re (ζ(σ + it + k) − 1) 2σ − 1 (1 − σ + it)k+1 k=0 ¬(σ ∈ Z ∧ t = 0) From (6) we see (with Stirling’s formula)

Z 1 1 |ζ∗(σ + it, x)|2dx = + (2π)2σ−1ζ(2 − 2σ)t1−2σ− (7) 0 2σ − 1 2 1 − Im(ζ(σ + it)) + O , σ > 0 t t

From (3), (4) and (5) we also see that

Z 1 (ζ(z, x)ζ(w, x) − x−z−w)dx = (8) 0 π 1 2(2π)(z+w−2)Γ(1 − z)Γ(1 − w) cos( (z − w))ζ(2 − z − w) − 2 1 − z − w max(Re(z), Re(w)) < 1 ∧ z + w 6= 1

From (8) we see that the convergence region in (2) holds. If we put w := 1−s−

64 and z := s − we get

Z 1 (ζ(s − ε, x)ζ(1 − s − ε, x) − x2ε−1)dx = (according to 3) = 0 π 1 2(2π)(−2ε−1)Γ(1 − s + ε)Γ(s + ε) cos( (1 − 2s))ζ(1 + 2ε) − = 2 2ε = (Laurent series development) = 1 ((1 − 2 log(2π)ε + O(ε2)))(Γ(1 − s) + Γ0(1 − s)ε + O(ε2)) π 1 1 (Γ(s) + Γ0(s)ε + O(ε2)) sin(πs)( + γ + O(ε)) − = 2ε 2ε 1 1 1 ( Γ(s)Γ(1 − s) sin(πs) − ) + 2π 2 ε sin(πs) + ((γ − log 2π)Γ(1 − s)Γ(s)+ π 1 0 0 + 2 (Γ(1 − s)Γ (s) + Γ (1 − s)Γ(s))) + O(ε) = = (according to the reﬂexion formula for the Gamma-function) = Γ0(s) Γ0(1 − s) γ − log 2π + 1 + + O(ε) = 2 Γ(s) Γ(1 − s) 1 γ − log 2π + 2 (Ψ(s) + Ψ(1 − s)) + O(ε) We let ε tend to zero and get

Z 1 (ζ(s, x)ζ(1 − s, x) − x−1)dx = 0 (9) 1 γ − log 2π + 2 (Ψ(s) + Ψ(1 − s)). Re(s) ∈ (0, 1)

We now consider a special case. First when w + z = 1. From formula (9), (3) and (4) we get

Z 1 ∗ ∗ 1 ζ (s, x)ζ (1 − s, x)dx = γ − log 2π + 2 (Ψ(1 − s) + Ψ(s)) − 0 ∞ X ζ(s + k) − 1 ζ(1 − s + k) − 1 + . s∈ / (10) s + k 1 − s + k Z k=0 First we only have the formula for Re(s) ∈ (0, 1), but by the same argument as in (5) we know that the formula holds in the region stated. (In fact the integral and the sum is just a special case of the sum and integral discussed in the proof of (5). We also need that Ψ(s) is an analytic function and then by analytic continuation

65 the equality is valid). When s = 1/2 + it we get

∞ ! Z 1 X ζ(1/2 + it + k) − 1 |ζ∗( 1 + it, x)|2dx = γ − log 2π + Re Ψ( 1 + it) − 2 . 2 2 1/2 + it + k 0 k=0 (11)

1 9 56 + Directly from (11) we see (since Ψ(s) = log(s)+O(| s |) and ζ(1/2+it) = O(t ) (see [2])) Z 1 ∗ 1 2 2 1 1 |ζ ( 2 + it, x)| dx = log t + γ − log 2π − Im(ζ( 2 + it)) + O( ) (12) 0 t t − 47 + = log t + γ − log 2π + O(t 56 ). ( > 0) We see that the truth of the Lindel¨of hypothesis would imply the error-term O(t−1) The estimate (12) is however far better than the previously best estimate: Z 1 ∗ 1 2 |ζ ( 2 + it, x)| dx = log t + O(1). 0 See [7] V.V. Rane, and also [3], and [1] for former estimates.

References

[1] R. Balasubramian, A note on Hurwitz zeta-function, Ann.Acad.Sci. Fenn. A I Math. 4 (1979), 41–44. [2] E. Bombieri and H. Iwaniec, On the order of ζ(1/2+it), Ann. Sc. Norm. Sup. Pisa. 13 (1986), 449–372. [3] J. F. Koksma and C. G. Lekkerkerker, A mean-value theorem for ζ(s, w), Nederl. Akad. Wetensch. Proc. Ser. A. 55 = Indagationes Math. 14 (1952), 446–452. [4] M. Mikl´os, Transformation und orhogonalit¨atbei ζ(s, u). Verallgemeinung der Riemannschen Funktionalgleichung von ζ(s), Acta. Sci. Mat. Szeged. 17 (1956), 143–164. [5] , Integral formulae of arithmetical characteristics relating to the zeta- function of Hurwitz, Publ. Math. Debrecen 5 (1957), 44–53. [6] , Uber¨ die Charakterisirung der Hurwitzschen Zetafunktion mittels Funktional-gleichungen, Acta. Sci. Math. Szeged. 19 (1958), 247–250. [7] V. V. Rane, On Hurwitz zeta-function, Math. Ann 264 (1983), 147 – 151.

66 On the fourth power moment of the Hurwitz zeta function

Johan Andersson∗

Abstract It is well known that Z 1 ∗ 1 2 −5/6 ζ 2 + it, x dx = log x + γ − log 2π + O t , 0 where ζ∗(s, x) = ζ(s, x) − x−s and ζ(s, x) denote the Hurwitz zeta function. See e.g. Andersson [3]. In this paper we prove the less known fact that

Z 1 1/4 !! ∗ 1 4 c(log t) ζ 2 + it, x dx = Ω exp , (1) 0 log log t

for some c > 0. In particular this means that there is no corresponding formula for higher power moments. This result was ﬁrst presented at the Kubelius conference in Palanga 1996, and ﬁrst appeared in print in the introduction of our licentiate thesis [2]. We also prove that on the average with respect to t we have

Z T Z 1 1 ∗ 1 4 2 5/3 ζ 2 + it, x dxdt = 2 log T + O (log T ) , T 0 0

whereas for rational parameters x = p/q we have

Z T 4 1 ∗ 1 p 4 3 ζ + it, dt = cq log T + O log T T 0 2 q whenever (p, q) = 1.

Contents

1 Introduction 68

2 Omega estimates for moments of the Hurwitz zeta function 69

3 The fourth power moment of the Hurwitz zeta function 71

∗Department of Mathematics, Stockholm University, [email protected]

67 1 Introduction

The moment problem is an important problem in the theory of the the Riemann zeta function. The mean square Z T 1 1 2 −1/2 ζ 2 + it dt = log T + 2γ − 1 + O T (2) T 0 for the Riemann zeta function has been used as a motivation to study the mean square of the Hurwitz zeta function. In our paper [3] we considered this problem, and we proved Z 1 ∗ 1 2 −1 1 −1 ζ 2 + it, x dx = log x + γ − log 2π − 2t Im ζ 2 + it + O t , (3) 0 where ζ∗(s, x) = ζ(s, x)−x−s, is deﬁned so that the integral is convergent. For the Riemann zeta function we have a good formula for the fourth power moment1: Z T 1 1 4 4 ζ 2 + it dt ∼ c4 log T. (4) T 0 It is natural to consider the corresponding problem for the Hurwitz zeta function. In analogy with the Riemann zeta function case it might seem natural to assume that Z 1 ∗ 1 4 4 ζ 2 + it, x dx ∼ c log t. (5) 0 In this paper we will show that a formula like (5) cannot be true. In fact the moment Z 1 ∗ 1 4 ζ 2 + it, x dx 0 will be rather chaotic in the same sense as the absolute values of the Riemann zeta function in the critical strip. On one hand we will be able to prove that Z 1 1/4 ∗ 1 4 c(log t) ζ 2 + it, x dx = Ω exp . 0 log log t On the other hand we prove that its mean value with respective to t Z T Z 1 1 ∗ 1 4 2 5/3 ζ 2 + it, x dxdt = 2 log T + O (log T ) , T 0 0 will grow just like a log power of t. We notice that the log power here is 2 and not 4 as in the case of the Riemann zeta function. This comes from the fact that the Riemann zeta function by its nature is arithmetic and has an Euler product, whereas the Hurwitz zeta function in general is not arithmetic. In the special case of x being rational it will have arithmetic properties though, since it can be expressed as sums of Dirichlet L-functions, which do have Euler products2. This will allow us to prove that Z T 4 1 ∗ 1 p 4 ζ + it, dt ∼ cq log T, T 0 2 q whenever (p, q) = 1.

1A result of Ingham [5]. 2For a further discussion of this phenomenon see our forthcoming paper [1].

68 2 Omega estimates for moments of the Hurwitz zeta function

Lemma 1. There exists constants C1,C2 > 0 such that if κ ≥ 1 and t ≥ 2 are real numbers, then

Z 1 2κ ! ∗ 1 2κ 1 1 −2κ−1 ζ 2 + it, x dx > C1 ζ + − it − C2 (log t) . 0 2 2κ

Proof. The case κ = 1 follows trivially from (3). Hence we can assume κ > 1. The approximate functional equation for the Hurwitz zeta function (see e.g. Rane [9]) says that

∗ 1 X −1/2−it ζ 2 + it, x = (n + x) 1≤n≤X 1 X −1/2+it −1/2 −1/2 + χ 2 + it k e(−kx) + O |X| + |Y | , (2πXY = t) (6) 1≤k≤Y where χ(s) is the factor in the functional equation for the Riemann zeta function (in particular we have that |χ(1/2 + it)| = 1). By choosing Y = t this gives us

∗ 1 1 X −1/2+it ζ 2 + it, x = χ 2 + it k e(−kx) + O(1). (7) 1≤k≤t Another estimate we will use is

X −λ λ−1 λ−1 1−λ e(kx)k min (1 − x) + x , t . (0 < x, λ < 1) 1≤k≤t

In particular this implies that

Z 1 X −λ e(kx)k dx ≤ Cλ, (0 < λ < 1) (8) 0 1≤k≤t as well as

1/(1−λ) Z 1 X −λ e(kx)k dx log t. (2 ≤ t, 0 < λ < 1) 0 1≤k≤t (9)

From the approximate functional equation for the Riemann zeta function (Titchmarsh [10], Theorem 4.11) we have that

1 1 X ζ + − it = k−1/2−1/(2κ)−it + O(1), (κ > 0) 2 2κ 1≤k≤t

69 which in view of the Parseval identity equals

Z 1 ! ! 1 X −1/(2κ) X −1/2−it χ 2 + it e(kx)k e(−kx)k dx + O(1). 0 1≤k≤t 1≤k≤t

Applying the identity (7) we can rewrite this as

Z 1 ! 1 X −1/(2κ) ∗ 1 χ 2 + it e(kx)k ζ 2 + it, x + O(1) dx + O(1). 0 1≤k≤t By using equation (8) we can remove the O(1) term within the integral and we obtain ! 1 1 Z 1 X ζ + − it = χ 1 + it e(kx)k−1/(2κ) ζ∗ 1 + it, xdx + O(1). 2 2κ 2 2 0 1≤k≤t By using the H¨olderinequality [4]

1/p 1/q Z b Z b Z b 1 1 |f(x)g(x)|dx ≤ |f(x)|p |g(x)|qdx , + = 1 a a a p q with 2κ a = 0, b = 1, p = , and q = 2κ, 2κ + 1 together with the fact that |χ(1/2 + it)| = 1, we get that 1 1 ζ + − it + O(1) 2 2κ  2κ/(2κ+1) (2κ+1)/(2κ) Z 1 Z 1 1/(2κ) X 2κ e(kx)k−1/2−1/(2κ) dx ζ∗ 1 + it, x dx .   2 0 1≤k≤t 0

With (9) this implies that

Z 1 1/(2κ) 1 1 (2κ+1)/(2κ) ∗ 1 2κ ζ + − it + O(1) (log t) ζ 2 + it, x dx . 2 2κ 0 The Lemma follows by raising this inequality to the 2κ’th power.

Theorem 1. There exist constants C1,C2 > 0 such that whenever

C1 log log T ≤ H ≤ T, and κ > 1, then

Z 1 (κ−1)/(2κ) ∗ 1 2κ C2(log H) max ζ 2 + it, x dx > exp . T ≤t≤T +H 0 log log H

70 Proof. This follows from Lemma 1 and Theorem 3.3.1 (for the special case of z = 1) in Ramachandra [6]. Theorem 1 should be compared to the Omega estimates of Ramachandra and Sankara- narayanan [7]

s !! ∗ 1 p log t ζ + it, = Ω exp Cp,q , 2 q log log t for the Hurwitz zeta function with a rational parameter. We also remark that from the theorem it follows that there exist no formula like (3) for any integer or fractional moment ∞ strictly greater than 2. In fact for any κ > 1 we can choose a sequence {tk}k=1 such that Z 1 ∗ 1 2κ N ζ 2 + itk, x dx > (log tk) (k > k0(N)) 0 will grow faster than any polynomial in log tk.

3 The fourth power moment of the Hurwitz zeta function

From Theorem 1 we have the following consequence in the special case of κ = 2 that

Corollary. There exist constants C1,C2 > 0 such that whenever

C1 log log T ≤ H ≤ T, then

Z 1 1/4 ∗ 1 4 C2(log H) max ζ 2 + it, x dx > exp . T ≤t≤T +H 0 log log H Since it is impossible to get a nice asymptotic formula for the fourth power moment when we integrate with respect to x we may ask whether it is possible to get a nice asymptotic formula when we integrate with respect to t. In the case of x being rational we show that

Theorem 2. Let (p, q) be relatively prime integers. Then

Z T 4 1 ∗ 1 p 4 3 ζ + it, dt = cq log T + O log T , T 0 2 q where

−1 Y −2 2 −1 cq = q (1 − 1/P ) (1 − 1/P ) . P |q P prime

71 Proof. We have that the Hurwitz zeta-function can be expressed in terms of Dirichlet L-functions p 1 X ζ s, = qsχ(p)L(s, χ). q φ(q) χ (mod q)

From this we obtain

Z T 4 2 1 ∗ 1 p X χ3(p)χ4(p)q χ1(p)χ2(p) ζ + it, dt = 4 × T 0 2 q T (φ(q)) χ1,χ2,χ3,χ4 (mod q) Z T 1 1 1 1 × L 2 + it, χ1 L 2 + it, χ1 L 2 − it, χ3 L 2 − it, χ4 dt (10) 0 By the fact that3

Z T 1 1 2 1 2 2 L 2 + it, χ1 L 2 + it, χ2 dt = O log T T 0 whenever χ1 6= χ2 or χ1 6= χ2, we see by the Cauchy-Schwarz inequality that the contribution coming from the non diagonal case is

Z T 1 1 1 1 1 3 L 2 + it, χ1 L 2 + it, χ1 L 2 − it, χ3 L 2 − it, χ4 dt = O log T T 0 unless χ1 = χ2 = χ3 = χ4, or χ1 = χ2 = χ3 = χ4. Hence we have that (10) equals

2 Z T X 2q 1 4 3 L + it, χ dt + O log T . T (φ(q))4 2 χ (mod q) 0

The result follows from the following Theorem of Rane [8].

Z T X 1 1 4 1 Y 1 − 1/P 4 L + it, χ dt = T log (T q) φ(q) 2 2π2 1 − 1/P 2 χ (mod q) 0 P |q P prime + Olog3(qT )(log log 3q)5, and the fact that Y φ(q) = q (1 − 1/P ). P |q P prime

3This is easily proved with e.g. the approximate functional equation of Dirichlet L-series and and a somewhat generalized version of the Ramanujan identity.

72 We see that the method depends heavily on arithmetic properties, and it is not possible to generalize it to arbitrary x. To see how the Hurwitz zeta-function behaves for arbitrary x we will instead consider the fourth power moment when we integrate with respect to both x and t. We will prove

Theorem 3. One has that Z T Z 1 1 ∗ 1 4 2 5/3 ζ 2 + it, x dxdt = 2(log T ) + O (log T ) . T 0 0

Theorem 3 can certainly be improved if we use more careful estimates.√ We believe that by using the approximate functional equation (6) with X Y t, and not just carelessly throwing away terms into the error term, we should be able to prove that

Z T +H Z 1 1 ∗ 1 4 1/2+ 1− ζ 2 + it, x dxdt = P2(log T ) + o(1), T < H < T H T 0 and by using more advanced estimates from the theory of exponential sums we might improve the result to an even shorter interval [T,T + H], for H = T θ and some θ < 1/2. For now we will be content with the present version of the theorem since we are at this point in time mainly concerned with the leading term.

Proof of Theorem 3 We will ﬁrst prove some Lemmas.

Lemma 2. Suppose that 0 < j, k < n are integers such that k, n − k 6= j. Then

Z B j(n − j) it n2

dt ≤ . A k(n − k) |n − j − k| |j − k| Proof. By symmetry we may assume that k, j ≤ n/2, since otherwise we can consider k → n − k and/or j → n − j. Likewise we may assume k < j since otherwise we can consider (j, k) → (k, j). Assume that 0 < y < x < 1/2. We deﬁne 1 − 2x f(x) = log(x(1 − x)), and obtain f 0(x) = . (1 − x)x

Since f(x) is increasing on the interval (0, 1/2] we have that

x + y f(x) − f(y) > f − f(y). (11) 2

We have by the mean value theorem that

x + y x + y f − f(y) = f 0(ξ) − y , 2 2

73 for some y < ξ < (x + y)/2. Since f 0 > 0 is decreasing on the interval (0, 1/2) this implies that

x + y x + y x + y f − f(y) ≥ f 0 − y . 2 2 2

By simplifying this and combining it with (11) we obtain

2(x − y)(1 − x − y) f(x) − f(y) ≥ . (x + y)(2 − (x + y))

Since the denominator is positive and less than or equal to 1 for 0 < x, y < 1 this gives us

f(x) − f(y) ≥ 2(x − y)(1 − x − y). (12)

We remark that the integral can be written as

Z B j(n − j) it Z B j k dt = exp it f − f dt, A k(n − k) A n n and by the inequality

Z B iαB iαA iαt e − e 2 e dt = ≤ , A iα |α| valid for real α we have that

Z B j(n − j) it 2 dt ≤ , k(n − k) j k A f n − f n and by (12) this gives us

n2 ≤ . |n − j − k| |j − k|

Lemma 3. Suppose that B > A. Then

M 4 M !2 M 1 Z B Z 1 X X X M log2 M n−1/2+ite(−nx) dxdt = 2 j−1 − j−2 + O . B − A B − A A 0 n=1 j=1 j=1

Proof. By squaring the Dirichlet polynomial

M X n−1/2+ite(−nx), n=1

74 we obtain that

M !2 2M ! X X X n−1/2+ite(−nx) = (j(n − j))−1/2+it e(−nx). n=1 n=2 1≤j,n−j≤M

With the Parseval identity we obtain

4 Z 1 M X −1/2+it n e(−nx) dx = 0 n=1 2M ! ! X X X = (j(n − j))−1/2+it (k(n − k))−1/2−it . (13) n=2 1≤j,n−j≤M 1≤k,n−k≤M

By identifying the diagonal term we get the main term and by using Lemma 2 on the non diagonal terms we get

M 4  M !2 M  Z B Z 1 X −1/2+it X −1 X −2 n e(−nx) dxdt = (B − A)2 j − j  A 0 n=1 j=1 j=1  

 2M 2  X X −1/2 n  + O (jk(n − j)(n − k)) .  |n − j − k||j − k| k=1 1≤j,n−j≤M  1≤k,n−k≤M n−k,k6=j

By summing the elements in the error terms this can be estimated by

 M !2 M  X −1 X −2 2 (B − A)2 j − j  + O M log M . j=1 j=1

Proof of Theorem 2. By dyadic division it is suﬃcient to prove the result for

Z 2T Z 1 1 ∗ 1 4 ζ 2 + it, x dxdt. T T 0 Let us ﬁrst assume that

T ≤ t ≤ 2T.

By using the approximate functional equation (6) with

T t Y √ , and X = p3 log T, 3 log T 2πY

75 we obtain

∗ 1 X −1/2−it 1 X −1/2+it −1/2 ζ 2 + it, x = (n + x) + χ 2 + it k e(−kx) + O X . 1≤n≤X 1≤k≤Y

By estimating√ the first sum trivially by absolute values, we find that the first term can be estimated by X, and we obtain

∗ 1 1 X −1/2+it p6 ζ 2 + it, x = χ 2 + it k e(−kx) + O log T . (14) 1≤k≤Y

From Lemma 3 we have that 4 1 Z 2T Z 1 X k−1/2+ite(−kx) dxdt = 2 log2 T + O(log T )5/3. (15) T T 0 1≤k≤Y

Combining the fact that |χ(1/2 + it)| = 1, with the equations (14) and (15) ﬁnally gives us Z 2T Z 1 1 ∗ 1 4 2 5/3 ζ 2 + it, x dxdt = 2 log T + O (log T ) . T T 0

References

[1] J. Andersson, Higher power moments of the Hurwitz and Lerch zeta functions, forthcoming.

[2] , Power sum methods and zeta-functions, Licentiate thesis, Stockholm Univer- sity, 1998.

[3] , Mean value properties of the Hurwitz zeta-function, Math. Scand. 71 (1992), no. 2, 295–300.

[4] O. H¨older, Uber¨ einen Mittelwertsatz., G¨ottingenNachr., 38-47, 1889.

[5] A. E. Ingham, Mean value theorems in the theory of the Riemann zeta-function, proc. lond. math. soc,(2), 27 (1926),273-300.

[6] K. Ramachandra, On the mean-value and omega-theorems for the Riemann zeta- function, Published for the Tata Institute of Fundamental Research, Bombay, 1995.

[7] K. Ramachandra and A. Sankaranarayanan, Omega-theorems for the Hurwitz zeta- function, Arch. Math. (Basel) 53 (1989), no. 5, 469–481.

[8] V. V. Rane, A note on the mean value of L-series, Proc. Indian Acad. Sci. Math. Sci. 90 (1981), no. 3, 273–286.

76 [9] , On Hurwitz zeta function, Math. Ann. 264 (1983), no. 2, 147–151.

[10] E. C. Titchmarsh, The theory of the Riemann zeta-function, second ed., The Claren- don Press Oxford University Press, New York, 1986, Edited and with a preface by D. R. Heath-Brown.

On the fourth power moment of the Lerch zeta function

Johan Andersson∗

Abstract It is well known that Z 1 ∗ 1 2 −5/6 ζ 2 + it, x dx = log t + γ − log 2π + O t , 0 where ζ∗(s, x) = ζ(s, x) − x−s and ζ(s, x) denotes the Hurwitz zeta function. In a recent paper [2] we proved that it is not possible to extend this result to higher power moments for the Hurwitz zeta function. In this paper we show that when we consider the Lerch zeta function we have an analogous result for the fourth power moment

Z 1 Z 1 ∗ 1 4 2 5/3 φ x, y, 2 + it dxdy = 2 log t + O (log t) , 0 0 where

φ∗(x, y, s) = φ(x, y, s) − y−s − (2π)s−1Γ(1 − s)× 1 − s s − 1 × e − yx xs−1 + e + y(1 − x) (1 − x)s−1 , 4 4

is defined so that the integral converges. This result was first presented at the Kubelius conference in Palanga, 1996, but this is the first time it appears in print.

1 Introduction

In our paper [4] we discussed the problem of estimating

Z 1 ∗ 1 2 ζ 2 + it, x dx, (1) 0 where ζ∗(s, x) = ζ(s, x) − x−s is deﬁned so that the integral converges. Here

∞ X ζ(s, x) = (k + x)−s, (Re(s) > 1) k=0

∗Department of Mathematics, Stockholm University, [email protected]

79 denotes the Hurwitz zeta function. This has been studied in several papers, starting with Koksma-Lekkerkerker [7]. For an excellent historic account of this problem, see Katsurada- Matsumoto [6]. Although it is very diﬃcult to estimate ζ(1/2 + it, a) for ﬁxed a, we have obtained very sharp estimates for (1). In our paper [4] we obtained

Z 1 ∗ 1 2 −1 1 −1 ζ 2 + it, x dx = log t + γ − log 2π + t Im ζ 2 + it + O(t ). 0 It would be natural to continue these studies with the integral

Z 1 ∗ 1 4 ζ 2 + it, x dx. 0 However, in our paper [2] we showed that there exists no similar formula for this expression1. In fact we showed that

Z 1 1/4 ∗ 1 4 c(log t) ζ 2 + it, x dx = Ω exp . (2) 0 log log t We believe that the right problem to study, if we want to obtain an analogous formula to (3), is that of estimating

Z 1 Z 1 ∗ 1 4 φ x, y, 2 + it dxdy. (3) 0 0 Here

∞ X φ(x, y, s) = e(nx)(n + y)−s (Re(s) > 1) n=0 is the Lerch zeta function (For its theory see Laurinˇcikas-Garunkˇstis[8]), and

φ∗(x, y, s) = φ(x, y, s) − y−s − (2π)s−1Γ(1 − s)× 1 − s s − 1 × e − yx xs−1 + e + y(1 − x) (1 − x)s−1 , (4) 4 4 is deﬁned so that the integral converges. Although we will not be able to get as sharp a result as (3) we will obtain the the quite satisfactory estimate

Z 1 Z 1 ∗ 1 4 2 5/3 φ x, y, 2 + it dxdy = 2 log t + O (log t) . (5) 0 0 We will also extend our result Z T Z 1 1 ∗ 1 4 2 2 5/3 ζ 2 + it, x dxdt = 2 log T + O (log T ) , T 0 0 1This was also presented at the Kubelius conference, Palanga 1996. It ﬁrst appeared in print in a less reﬁned form in the introduction our licentiate thesis [3].

80 from our paper [2] to the case of the Lerch zeta function. We will prove that Z T Z 1 1 ∗ 1 4 2 5/3 φ x, y, 2 + it dxdt = 2 log T + O (log T ) , (0 ≤ y ≤ 1), (6) T 0 0 as well as Z T Z 1 1 ∗ 1 4 2 5/3 φ x, y, 2 + it dydt = 2 log T + O (log T ) . (0 ≤ x ≤ 1), (7) T 0 0 Notice that the log power in these three problems (5), (6) and (7) is 2, and not 4 as is the case2 Z T 1 1 4 4 ζ 2 + it dt ∼ c4 log T (8) T 0 of the Riemann zeta function. This comes from the fact that the Riemann zeta function by its nature is arithmetic and has an Euler product, whereas neither the Hurwitz zeta function nor the Lerch zeta function in general are arithmetic3. For a further discussion of this phenomenon see our forthcoming paper [1].

2 Preliminaries

We will use the notation ( x, 1 ≤ x < N + 1, {x}N = (9) 0, otherwise. With this notation we have that ∗ X −1/2−it φN (x, y, s) = (k + x) e(ky), (10) 1≤k≤N ∞ X −s = {k + x}N e(ky). (0 ≤ x, y < 1) k=−∞ We obtain ∞ ∗ 2 X −s −s (φN (x, y, s)) = {k1 + x}N {k2 + x}N e((k1 + k2)y), k1,k2=−∞ and with the Parseval identity we get Z 1 ∗ 2 ∗ 2 (φN (x, y, s)) (φN (x, −y, z)) dy = 0 ∞ X −s −s −z −z {k1 + x}N {k2 + x}N {k3 + x}N {k4 + x}N . k1,k2,k3,k4=−∞ k1+k2=k3+k4 2A result of Ingham [5]. 3In the special case of x, y being rational, the Lerch zeta function will have arithmetic properties. In particular it can be expressed as sums of Dirichlet L-functions, which do have Euler products. Thus the corresponding fourth power moment will have a log T term of the fourth order.

81 By integrating this with respect to x we have

Z 1 Z 1 ∗ 2 ∗ 2 (φN (x, y, s)) (φN (x, −y, z)) dydx = 0 0 ∞ Z 1 X −s −s −z −z {k1 + x}N {k2 + x}N {k3 + x}N {k4 + x}N dx. 0 k1,k2,k3,k4=−∞ k1+k2=k3+k4 With

k1 = n + [x], k2 = m + [x], k3 = n + m + [x], and k4 = [x], where [x] as usual denote the integer part of x, we obtain a proof for the following Lemma.

Lemma 1. Let s, w be complex variables, and let ΦN be deﬁned by (10). One has the identity

Z 1 Z 1 ∗ 2 ∗ 2 (φN (x, y, s)) (φN (x, −y, z)) dydx = 0 0 ∞ Z N+1 X −s −s −z −z {n + x}N {m + x}N {n + m + x}N x dx. n,m=−∞ 1 By specializing this to s = σ + it, and z = σ − it, (11) we obtain

Lemma 2. Let ΦN be deﬁned by (10), and σ, t be real numbers. One has the identity Z 1 Z 1 ∗ 4 |φN (x, y, σ + it)| dydx = 0 0 ∞ Z N+1 X −σ−it −σ−it −σ+it −σ+it {n + x}N {m + x}N {n + m + x}N x dx. n,m=−∞ 1 The most important case is the critical line σ = 1/2. We will ﬁrst prove a result that allows us to estimate the terms arising from Lemma 2. Lemma 3. Suppose that m, n, n + m 6= 0 are integers, and t is a real number. Then there exist a constant C > 0 such that Z N+1 −1/2−it −1/2−it −1/2+it −1/2+it CN {n + x}N {m + x}N {n + m + x}N x dx ≤ . 1 (|t| + 1)|mn| Proof. By symmetry we may assume that m, n ≥ 1, since otherwise we will get an integral of such type with the substitution y = x + min(m, n). With the choice of functions (n + x)(m + x) f(x) = , x(n + m + x) nm = 1 + , (12) x(n + m + x)

82 and 1 g(x) = , x(n + m + x) we can now write the integral as

Z B (f(x))it−1/2g(x)dx, A for some 1 ≤ A ≤ B ≤ N + 1. By (12) we see that f(x) is decreasing and that f 0(x) < 0, hence we can write the integral as

Z B 0 it−1/2 g(x) f (x)(f(x)) 0 dx. A f (x) We do partial integration and we ﬁnd that the integral equals

" #B (f(x))it+1/2 g(x) Z B (f(x))it+1/2 d g(x) 0 − 0 dx. (13) it + 1/2 f (x) A it + 1/2 dx f (x) A We have that g(x) x(n + m + x) = − , (14) f 0(x) mn(m + n + 2x) and

d g(x) (m + n + x)2 + x2 = − . dx f 0(x) mn(m + n + 2x)2

We see that the last integral equals s Z B 2 2 1 it (n + x)(m + x) (m + n + x) + x (f(x)) 2 dx it + 1/2 A (m + n + x)x mn(m + n + 2x)

By estimating the integral by its absolute values, this term can be estimated by CN . (|t| + 1)|mn|

Finally, by (14) we see that the ﬁrst two terms in (13) can also be estimated by

CN . (|t| + 1)|mn|

83 Lemma 4. Let ΦN be deﬁned by (10). One has the estimate

Z 1 Z 1 ∗ 1 4 φN x, y, 2 + it dydx = 0 0 N X 4 n 3N N log2 N log(n + 1) + log 1 − − + O . n N + 1 N + 1 |t| + 1 n=1 Proof. By Lemma 2 we have that

Z 1 Z 1 ∗ 1 4 φN x, y, 2 + it dydx = 0 0 ∞ Z N+1 X −1/2−it −1/2−it −1/2+it −1/2+it {n + x}N {m + x}N {n + m + x}N x dx. (15) n,m=−∞ 1

From the diagonal terms, that is when

n = 0, or m = 0, we obtain the terms ∞ Z N+1 Z N+1 X −1 −2 −2 2 {n + x}N x dx − x dx. n=−∞ 1 1 which equals

N X Z N+1−n dx Z N+1 dx 4 − 3 . (16) (n + x)x x2 n=1 1 1 We have that 1 1 1 1 = − (17) (n + x)x n x n + x and we have that equation (16) equals

N X 4 3N (log(N + 1 − n) + log(n + 1) − log(N + 1)) − . (18) n N + 1 n=1 It is clear that the integral in (15) vanishes when |m| > N or |n| > N, hence it is suﬃcient to consider the case |m|, |n| ≤ N and from Lemma 3 we obtain the error term

N X N N log2 N O = O . (|t| + 1)|mn| |t| + 1 n,m=−N n,m6=0

84 3 The main results

We are now ready to combine our lemmas to obtain

Theorem 1. One has that Z 1 Z 1 ∗ 1 4 2 5/3 φ x, y, 2 + it dxdy = 2 log t + O (log t) . 0 0

∗ ∗ Proof. By using the deﬁnition of φN (x, y, s), and φ (x, y, s), equations (10) and (4), the approximate functional equation for the Lerch zeta function can be written as

∗ 1 ∗ 1 1 ∗ 1 φ x, y, 2 + it = φX x, y, 2 + it + χ 2 + it φY y, 1 − x, 2 − it + O |X|−1/2 + |Y |−1/2 , (2πXY = t > 0) where χ(s) is the factor in the functional equation for the Riemann zeta function (in particular we have that |χ(1/2 + it)| = 1). By using it with t t X √ , and Y = p3 log t, 3 log t 2πX and by estimating the second term by its absolute values, we ﬁnd that

∗ 1 ∗ 1 1/6 φ x, y, 2 + it = φX x, y, 2 + it + O (log t) . (19) Since

N X log(n) = 1 (log N)2 + O(log N), n 2 n=1 we get by Lemma 4 that

Z 1 Z 1 ∗ 1 4 5/3 φX x, y, 2 + it dydx = 2 log t + O (log t) . 0 0 From this and (19) the Theorem follows.

Theorem 2. One has the estimates Z T Z 1 1 ∗ 1 4 2 5/3 φ x, y, 2 + it dxdt = 2 log T + O (log T ) , (0 ≤ y ≤ 1) T 0 0 and

Z T Z 1 1 ∗ 1 4 2 5/3 φ x, y, 2 + it dydt = 2 log T + O (log T ) . (0 ≤ x ≤ 1) T 0 0

85 Proof. First we notice that by the approximate functional equation of the Lerch zeta- function it is suﬃcient to prove the ﬁrst of the estimates. The proof is then identical to the proof of the corresponding result for the Hurwitz zeta function, Theorem 2 in our paper [2], so we will skip the proof here. We will just remark that instead of Lemma 3 from that paper

Z B j(n − j) it n2 j, k, n, n − k > 0 dt ≤ (20) k, n − k 6= j A k(n − k) |n − j − k| |j − k| we will need the corresponding inequality   it j, k, n, n − k > 0 Z B (j + y)(n − j + y) (n + 2y)2 dt ≤ k, n − k 6= j (21) (k + y)(n − k + y) |n − j − k| |j − k|   A 0 ≤ y ≤ 1 This follows by the same proof though, since the assumption that n, j, k are integers is actually not needed in the proof of the Lemma, and the case (21) follows by letting (j, k, n) be (j + y, k + y, n + 2y) in (20).

4 Final discussion

In our talk at the Kubelius conference in Palanga, 1996 we were using the approximate functional equation with

r t X = Y = . (22) 2π This has the advantage that if done carefully, and we estimate all the occurring terms we should obtain a better error term, since Lemma 3 gives a better error term for shorter ranges of N. There are certain new terms appearing though, and the proof will necessarily be a bit more technical. At the moment we are content with the present version of the theorem. It should also be mentioned that we have the following Z T 4 1 ∗ p h 1 4 φ , , + it dt ∼ ch,k,p,q log T. (23) T 0 q k 2 In the special case of h = p = 0 this reduces to the fourth power moment of the Riemann zeta function (8). In the case of h = 0 we have that φ∗(h/k, p/q, 1/2 + it) = ζ∗(1/2+it, p/q) and this follows from Theorem 2 from our paper [2]. The proof for the general case can be constructed along the same lines as that proof by expressing the Lerch zeta functions in terms of Dirichlet L-functions. We may ﬁnally ask. Can we obtain a similar formula as the one in Theorem 1 for higher power moments Z 1 Z 1 ∗ 1 2k φ x, y, 2 + it dxdy? 0 0 A reason to doubt it is that the higher moment case fails for the Hurwitz zeta function. However our best bet is that it will not fail in this case.

86 Conjecture. One has that

Z 1 Z 1 ∗ 1 2k k φ x, y, 2 + it dxdy ∼ k! (log t) . 0 0 We will discuss more about this conjecture in [1]. The truth of the conjecture depends on just how dense the rational numbers are among the real numbers. By the method of Ramachandra-Balasubramanian [9] it is namely possible to show that for rational parameters of x, y and the integral with respect to t from 0 to T the integral can be estimated 2 from below by cT logk T , and not just cT log2k T . We remark that although we believe in the Conjecture it should indeed be a very difficult conjecture to prove. The Lindelöf hypothesis for the Riemann zeta function as well as the Lindelöfconjecture for the Lerch zeta function would follow from the Conjecture.

References

[1] J. Andersson, Higher power moments of the Hurwitz and Lerch zeta functions, forthcoming.

[2] , On the fourth power moment of the Hurwitz zeta function, 2006.

[3] , Power sum methods and zeta-functions, Licentiate thesis, Stockholm Univer- sity, 1998.

[4] , Mean value properties of the Hurwitz zeta function, Math. Scand. 71 (1992), no. 2, 295–300.

[5] A. E. Ingham, Mean value theorems in the theory of the Riemann zeta-function, proc. lond. math. soc,(2), 27 (1926),273-300.

[6] M. Katsurada and K. Matsumoto, Explicit formulas and asymptotic expansions for certain mean square of Hurwitz zeta-functions, Proc. Japan Acad. Ser. A Math. Sci. 69 (1993), no. 8, 303–307.

[7] J. F. Koksma and C. G. Lekkerkerker, A mean-value theorem for ζ(s, w), Nederl. Akad. Wetensch. Proc. Ser. A. 55 = Indagationes Math. 14 (1952), 446–452.

[8] A. Laurinˇcikas and R. Garunkˇstis, The Lerch zeta-function, Kluwer Academic Publish- ers, Dordrecht, 2002.

[9] K. Ramachandra, On the mean-value and omega-theorems for the Riemann zeta- function, Tata Institute of Fundamental Research Lectures on Mathematics and Physics, vol. 85, Published for the Tata Institute of Fundamental Research, Bombay, 1995.

PART III - A NEW SUMMATION FORMULA ON THE FULL MODULAR GROUP AND KLOOSTERMAN SUMS

INTRODUCTION

A summation formula for the modular group

The classical Poisson summation formula ∞ ∞ X X f(n) = fˆ(n). n=−∞ n=−∞ is a fundamental tool in analytic number theory The importance of this theorem suggests that one should study generalizations of this result. In particular we wish to consider analogues for discrete groups such as the full modular group. Can we ﬁnd some similar expansion of X f(γ)?

γ∈PSL(2,Z) When f is SO(2) bi-invariant an expansion in terms of spectral theory can obtained by the pre-trace formula, and indeed Terras ([11], p. 265) choose to call this the “Poisson summation formula for SL(2, Z)”. The condition that f is SO(2) bi-invariant seems too restrictive though, and indeed the condition can quite easily be reduced to SO(2) right (or left)-invariant (for details see our forthcoming paper [1]). The condition that f is SO(2) right (or left)- invariant still seems to restrictive. The summation makes sense whenever the sum is convergent. In particular ∞ f ∈ C0 (PSL(2, R)) should be a sufficient condition. In part III of our thesis we develop such a formula. We generalize it to matrices of fixed determinant, and show how it relates to the Kuznetsov summation formula as well as the Selberg and Eichler-Selberg trace formulae. It turns out that this question is closely related to the expansion of the fourth power moment of the Riemann zeta function Z ∞ 1 4 g(t) ζ 2 + it dt, −∞ given by Y¯oichi Motohashi in his important paper [8] in terms of spectral theory, as well as to the additive divisor problem given by Kuznetsov [6] and Motohashi [9]. This was indeed our first approach to this subject and we will describe this relation in the next section. Introduction 92

Summation formulae and functional equations

A general idea in the analytic theory of numbers is that whenever you have a summation formula, you should have a functional equation, and vice verse (see e.g Ferrar [4]). For example, the classical functional equation of the Riemann zeta function is proved by the classical Poisson summation formula (after some trick to solve the convergence problems). The existence of a Poisson summation formula and a functional equation are in fact equivalent. We can also derive the Poisson summation formula from the functional equation6. The following table provides some other examples:

summation formula Functional equation

P f(n) = P fˆ(n) ζ(s) = χ(s)ζ(1 − s) (Poisson summation formula)

P τ(χ)χ(−1) P ˆ n χ(s)τ(χ) χ(n)f(n) = N χ(n)f N L(s, χ) = N s L(1 − s, χ) (twisted Poisson summation formula)

P d(n)f(n) = P d(n)f˜(n) ζ2(s) = χ2(s)ζ2(1 − s) (Voronoi summation formula)

Selberg trace formula Selberg zeta

P f(n + α, m + β) = P fˆ(n, m)e(−nα − mβ) two Hurwitz zeta (shifted double Poisson summation formula)

The way of going back and forth between these is by using Mellin inversions

∞ ∞ X 1 Z c+∞i X a f(n) = Mf(s)A(s)ds, A(n) = a n−s, n 2πi n n=1 c−∞i n=1 where Z ∞ Mf(s) = xs−1f(x)dx 0 is the Mellin transform of f. We should note that even though the functional equation and the summation formula are equivalent, the summation formula sometimes gives better results. For example, sometimes it is better to use an approximate

6 Actually the functional equation for the Riemann zeta-function corresponds to the Poisson summation formula for the Fourier-cosine transform. Introduction 93 version of the functional equation rather than the full functional equation for the Riemann zeta-function. This is most simply proved by the Poisson summation formula directly. In Motohashi’s series of papers and book (see e.g [10], page 165), he essentially discovers, even though it is in a certain sense hidden, some formulae of the type

ζ(u)ζ(v)ζ(w)ζ(z) = “main term”+ X 1 1 αjHj( 2 (u + v + w + z − 1))Hj( 2 (u + z − v − w + 1))× j 1 × Hj( 2 (v + z − u − w + 1))F (u, z, v, w; κj) + “similar terms”, (III.35) where F is a certain hypergeometric function, and ∞ X −s Hj(s) = tj(n)n (III.36) n=1 are Dirichlet-series (Hecke L-series corresponding to Maass wave forms - see [10], page 104). In a strict sense (III.35) is false. The right hand side is divergent. However, if we integrate the left hand side over test functions of certain restricted class we have equality. Our general analogy suggests that this is a sort of functional equation that should have a corresponding summation formula. If we formally apply four dimensional Mellin-transformation a quadruple sum can be written as

∞ X a b f = c d a,b,c,d=1 1 Z 2+∞i Z 2+∞i Z 2+∞i Z 2+∞i u v 4 ζ(u)ζ(v)ζ(w)ζ(z)Mf dudvdwdz, (2πi) 2−∞i 2−∞i 2−∞i 2−∞i w z (III.37) u v where Mf( w z ) now denotes the four-dimensional Mellin transform. By using the expression (III.35) for the product of four zeta-function, and using the linear change of variables 1 s1 = 2 (u + v + w + z − 1), 1 s2 = 2 (u + z − v − w + 1), 1 s3 = 2 (v + z − u − w + 1), we ﬁnd that the right hand side in equation (III.37) can be re-written as

1 Z 1/2+∞i Z 1/2+∞i Z 7/2+∞i 3 “main term”+ (2πi) 1/2−∞i 1/2−∞i 7/2−∞i ∞ X + αjHj(s1)Hj(s2)Hj(s3)F (u, z, v, w; κj) + “similar terms” ds1ds2ds3, j=1 Introduction 94 and we should be able, by using (III.36) and once again using Mellin inversion, to recover ∞ ∞ ∞ X a b X X f = “main term”+ α t (n)t (m)t (D)f˜(κ ; m, n, D)+ c d j j j j j a,b,c,d=1 j=1 m,n,D=1 + “other similar terms” + “certain terms of less importance”, (III.38) where f˜ is some transform of f, and the kernel-function is some (possibly complicated) hypergeometric function. If we look at the papers of Motohashi [9] and Kuznetsov [6] on the additive divisor problem, we see that we can in fact get a similar formula for ﬁxed D for sums X a b f c d ad−bc=D where the summation over D in the right hand side of (III.38) should be sup- pressed. This of course corresponds to the change of summation order ∞ ∞ X a b X X a b f = f . c d c d a,b,c,d=−∞ D=−∞ ad−bc=D Note that Motohashi calls the dissection D < 0,D = 0, D > 0, the “Atkinson dissection” after Atkinson’s paper [2] on the mean square of the Riemann zeta function. We will emphasize that in this case the summation formula we will ﬁnd in Part III in our thesis will be new, but the corresponding functional equation is known.

The summation formula - A simpliﬁed approach

We originally thought of the idea to use the results coming from the Fourth power moment in the spring of 1999. We were looking at some quadruple exponential sums coming from a problem involving the Lerch zeta function, when we realized that we had a Mellin Barnes integral involving four zeta functions, which we could treat with Motohashi’s method. It was clear to us that there had to be simpler proofs. A first simplification was given at a seminar in Turku in the fall of 1999 and is presented in the paper “A summation formula over integer matrices II”. In the spring of 2000 we discovered a further simplification and we will here describe the main idea in the proof as given in the paper “A summation formula on the full modular group”. The idea relies on the Bruhat decomposition of the Big cell of PSL(2, Z), or in other words the fact that any element in PSL(2, Z), which is not the identity can be written in a unique way as 1 m h ∗ 1 n 0 1 c h¯ 0 1 Introduction 95 whenever m, n, h,¯ h, c are integers such that c > 0, and 0 ≤ h, h¯ < c. The idea is then to proceed by using the double Poisson summation formula, and then the classical Kloosterman sums X mh + nh¯ S(m, n; c) = e c hh¯≡1 (mod c) 0≤h,h0 ˆ for fc(m, n) a certain integral transform of f. The case when m, n 6= 0 can now be treated with the Kuznetsov summation formula [5]. This will allow us to obtain a general summation formula of the following type

Z ∞ X a b X 1 σ2ir(|m|)σ2ir(|n|)F (r; m, n)dr f = “main terms”+ c d π ir 2 ad−bc=1 m,n6=0 −∞ |nm| |ζ(1 + 2ir)| c>0 ∞ θ(k) ∞ X X X 1 X X + ρj,k(m)ρj,k(n)F 2 − k i; m, n + ρj(m)ρj(n)F (κj; m, n). k=1 j=1 m,n6=0 j=1 m,n6=0 (III.39)

Here F (r; m, n) is a certain integral transform of f, the ρj(n) denote the Fourier coeﬃcients for the Maass wave forms, and the ρj,k(n) denote Fourier coeﬃcients of holomorphic cusp forms of weight k. For an exact version of this formula we refer to Theorem 1 in “A summation formula on the full modular group”.

The summation formula and Hecke operators

In order to relate this formula to matrices of determinant D in order to apply it on quadruple exponential sums such as in (III.38), we need to apply the Hecke operators in some way or another. We will choose to do so directly on the Kloosterman sums and by doing so the generalized Kloosterman sums X mh + nk S(D, m, n; c) = e c hk≡D (mod c) 0≤h,k

X mD c S(D, m, n; c) = d S , n; , d2 d d|(D,m,c) which they attribute to Heath-Brown. However, a clear and simple proof of this identity seems to be lacking in the literature, and thus we chose to write the paper “A note on some Kloosterman sum identities” on this subject. We also remark that this identity gives a simple and elementary proof of the Selberg identity. This is a subject which has been of some interest to other researchers, e.g. Matthes [7] who gave a diﬀerent elementary proof.

Applications

There are several applications of the summation formula. As already indicated it is related to the additive divisor problem and the additive circle problem, as well as the Fourth power moment. We have chosen not to include this in our thesis but will instead return to this topic in forthcoming papers. Instead we will concentrate on showing that it can be used to prove the Selberg trace formula and the Eichler-Selberg trace formula in a uniﬁed way. We also see that it is in fact equivalent with the Kuznetsov summation formula, since the summation formula can be used to prove the Kuznetsov summation formula in a simple way. BIBLIOGRAPHY

[1] J. Andersson, A summation formula on the modular group and the pre-trace formula, forthcoming.

[2] F. V. Atkinson, The mean-value of the Riemann zeta function, Acta Math. 81 (1949), 353–376.

[3] V. Bykovsky, N. Kuznetsov, and A. Vinogradov, Generalized summation formula for inhomogeneous convolution, Automorphic functions and their applications (Khabarovsk, 1988), Acad. Sci. USSR Inst. Appl. Math., Khabarovsk, 1990, pp. 18–63.

[4] W. L. Ferrar, Summation formulae and their relation to Dirichlet series, Com- positio Math. (1935), no. 1, 344–360.

[5] N. V. Kuznetsov, The Petersson conjecture for cusp forms of weight zero and the Linnik conjecture. Sums of Kloosterman sums, Mat. Sb. (N.S.) 111(153) (1980), no. 3, 334–383, 479.

[6] , Convolution of Fourier coeﬃcients of Eisenstein-Maass series, Zap. Nauchn. Sem. Leningrad. Otdel. Mat. Inst. Steklov. (LOMI) 129 (1983), 43– 84, Automorphic functions and number theory. I. MR 86h:11039

[7] R. Matthes, An elementary proof of a formula of Kuznecov for Kloosterman sums, Resultate Math. 18 (1990), no. 1-2, 120–124.

[8] Y. Motohashi, An explicit formula for the fourth power mean of the Riemann zeta-function, Acta Math. 170 (1993), no. 2, 181–220.

[9] , The binary additive divisor problem, Ann. Sci. Ecole´ Norm. Sup. (4) 27 (1994), no. 5, 529–572.

[10] , Spectral theory of the Riemann zeta-function, Cambridge University Press, Cambridge, 1997.

[11] A. Terras, Harmonic analysis on symmetric spaces and applications. I, Springer-Verlag, New York, 1985.

A note on some Kloosterman sum identities

Johan Andersson∗

Abstract In this paper we will give an elementary proof of the identity

X mD c S(D, m, n; c) = d S , n; , d2 d d|(D,m,c)

where

X mh + nk S(D, m, n; c) = e c hk≡D (mod c) 0≤h,k

are certain generalized Kloosterman sums and S(m, n; c) = S(1, m, n; c) denote the classical Kloosterman sums. This identity was ﬁrst stated in Bykovsky-Kuznetsov-Vinogradov [4]. Heath-Brown [5] had previously proved a closely related result. In particular this will give us a new proof of the Selberg identity X mn c S(m, n; c) = d S , 1; . d2 d d|(m,n,c)

The identity was given by Selberg [12] without proof. Kuznetsov [9] published the ﬁrst proof using his summation formula. Matthes [10] has also provided an alternative elementary proof.

1 Kloosterman sums

Kloosterman sums X mh + nh¯ S(m, n; c) = e e(x) = e2πix (1) c hh¯≡1 (mod c) 0≤h,h

∗Department of Mathematics, Stockholm University, [email protected]

99 have had a lot of applications in analytic number theory ever since Kloosterman’s paper [8]. An important estimate is Weil’s [13] algebro-geometric estimate

|S(m, n; c)| ≤ d(c)pc · (m, n, c), (2) P where (m, n, c) is the greatest common divisor, and d(n) = d|n 1 denotes the divisor function. This estimate follows from his result on the Riemann hypothesis for curves. In the seventies Kuznetsov [9] developed a summation formula1 that has since then been of prime importance, and is an essential point in Motohashi’s work [11] on the Riemann zeta function. Kuznetsov obtained a summation formula of the type

∞ ∞ X X Φ(c)S(m, n; c) = ρj(n)ρj(m)Φm,n(κj) + “other similar terms”, (3) c=1 j=1 where Φm,n is a certain integral transform of Φ, ρj(n) are certain Fourier co- eﬃcients of Maass wave forms, and the “other similar terms” is the spectral contribution coming from the holomorphic cusp forms and the Eisenstein series. For an introduction to “Kloostermania”, the theory of Kloosterman sums and their applications, see Huxley [6].

2 Main results

Theorem 1. (Bykovsky-Kuznetsov-Vinogradov identity) One has the identity

X mD c S(D, m, n; c) = d S , n; , d2 d d|(D,m,c) where

X mh + nk S(D, m, n; c) = e c hk≡D (mod c) 0≤h,k

Proof. Whenever

hk ≡ D (mod c), 0 ≤ h, k < c,

1Kuznetsov [9] and Bruggeman [3] independently developed the closely related trace formula.

100 we have with

hk − D p = , c that

h p c k is an integer matrix of determinant D. From the theory of Hecke-operators (see e.g Motohashi [11] pages 97-98), we know that it can be written uniquely as a product

h p a b x q = · , c k 0 d l y where

ad = D, 0 ≤ b < d, and the second factor

x q l y is an element in SL(2, Z). We get that h p a b x q = , c k 0 d l y bl + ax aq + by = , dl dy which gives us

c k D l = , y = , and a = . d d d Let us also assume thatx ¯ is deﬁned so that 0 ≤ x¯ < l andx ¯ ≡ y (mod l). This implies that

mh + nk m(bl + ax) + ndx¯ e = e . c c

101 Summing over hk ≡ D (mod c) will now correspond to summing over d|(c, D), xx¯ ≡ 1 (mod c/d), 0 < x, x¯ < c/d and b = 0, . . . , d − 1. We have that

X mh + nk S(D, m, n; c) = e , c hk≡D (mod c) 0≤h,k

d−1 d−1 ( X mbl X mb d, d|m, c e = e = l = , c d 0, otherwise, d b=0 b=0 we deduce that X X max + ndx¯ S(D, m, n; c) = d e , c d|(c,D,m) xx¯≡1 (mod c/d) 0≤x,x

Theorem 1 was stated in Bykovsky-Kuznetsov-Vinogradov [4] (see page 48, equation (2.29)). It is not proved in their paper though. Instead they refer to Heath-Brown [5] for proof. Heath-Brown never proves Theorem 1, but the corresponding identity for X u(n; q, r) = S(r, a, b; q). ab=n See [5], equation (51) page 411. He does not explicitly deﬁne S(D, m, n; c). His proof applies in this setting as well though, and it is similar to our proof. It should also be mentioned that M.N. Huxley [7] found a similar argument independently of the author. Neither Heath-Brown [5] nor Huxley [7] mentions Hecke operators. We believe they are a key point if we would like to generalize the theorem to more general Kloosterman sums associated with arithmetical groups.

102 Bykovsky-Kuznetsov-Vinogradov [4] used Theorem 1 to prove a generalized Kuznetsov summation formula (compare with (3)): ∞ ∞ X X Φ(c)S(D, m, n; c) = tj(D)ρj(n)ρj(m)ΦD,m,n(κj) + “other similar terms” c=1 j=1 (4) where

ρj(D) tj(D) = ρj(1) are the eigenvalues of the Hecke operator TD acting on the Maass wave form. The identity in Theorem 1 will occur in the same setting in our forthcoming work [2], where we develop a summation formula over integer matrices with determinant D. In fact we needed a formula like (4) as a Lemma for our other work and had developed an independent proof of (4) when K. Matsumoto made an interesting remark at our seminar [1] at the Millennial conference of number theory. The result had previously been proved by Bykovsky-Kuznetsov-Vinogradov [4]! Theorem 2. One has the identity

S(n1, n2, n3; c) = S(nσ(1), nσ(2), nσ(3); c) where S(n1, n2, n3; c) is deﬁned as in Theorem 1 and σ ∈ S3 is any permutation acting on {1, 2, 3}. Proof. From the deﬁnition of the generalized Kloosterman sums X mh + nk S(D, m, n; c) = e c hk≡D (mod c) 0≤h,k

103 Corollary. (The Selberg identity) We have that X mn c S(m, n; c) = d S , 1; . d2 d d|(m,n,c)

Proof. We have

S(m, n; c) = S(1, m, n; c), which by using Theorem 2 equals

= S(n, m, 1; c), which by using Theorem 1 equals

X mn c = d S , 1; . d2 d d|(m,n,c)

This is the classical Selberg identity. It was given by Selberg [12] without proof, and Kuznetsov [9] published the ﬁrst proof using his summation formula. Matthes [10] has given an elementary proof, by division into cases and using multiplicative properties of the Kloosterman sums. It is also of interest to note that Kuznetsov’s method of proof used the Hecke operators and his summation formula. Matthes gave an elementary proof, without the summation formula and without Hecke operators. We give an elementary proof but we use the notion of Hecke operators.

3 Acknowledgments

We would like to thank Martin Huxley for his interest in the problem, and for encouraging us to publish our results.

References

[1] J. Andersson. A Poisson summation formula for SL(2,Z). Conference abstract, Millenial conference in number theory, Urbana Champaign, http://atlas-conferences.com/c/a/e/w/11.htm, 2000.

[2] A summation formula over integer matrices, 2006.

104 [3] R. W. Bruggeman. Fourier coeﬃcients of cusp forms. Invent. Math., 45(1):1– 18, 1978.

[4] V. Bykovsky, N. Kuznetsov, and A. Vinogradov. Generalized summation formula for inhomogeneous convolution. In Automorphic functions and their applications (Khabarovsk, 1988), pages 18–63. Acad. Sci. USSR Inst. Appl. Math., Khabarovsk, 1990.

[5] D. R. Heath-Brown. The fourth power moment of the Riemann zeta function. Proc. London Math. Soc. (3), 38(3):385–422, 1979.

[6] M. N. Huxley. Introduction to Kloostermania. In Elementary and analytic theory of numbers (Warsaw, 1982), pages 217–306. PWN, Warsaw, 1985.

[7] . Three parameter kloosterman sum, private communication. 2002.

[8] H. D. Kloosterman. On the representation of numbers in the form ax2 + by2 + cz2 + dt2. Acta Math., 49:407–464, 1926.

[9] N. V. Kuznetsov. The Petersson conjecture for cusp forms of weight zero and the Linnik conjecture. Sums of Kloosterman sums. Mat. Sb. (N.S.), 111(153)(3):334–383, 479, 1980.

[10] R. Matthes. An elementary proof of a formula of Kuznecov for Kloosterman sums. Resultate Math., 18(1-2):120–124, 1990.

[11] Y. Motohashi. Spectral theory of the Riemann zeta-function. Cambridge University Press, Cambridge, 1997.

[12] A. Selberg. Uber¨ die Fourierkoeﬃzienten elliptischer Modulformen negativer Dimension. In C.R. Neuvieme Congres Math Scandinaves, pages 320–322. Helsinki, 1938.

[13] A. Weil. On some exponential sums. Proc. Nat. Acad. Sci. U. S. A., 34:204– 207, 1948.

105

A note on some Kloosterman sum identities II

Johan Andersson∗

Abstract In a previous paper [1] we gave new proof of Kloosterman sum identities of Bykovsky-Kuznetsov-Vinogradov and Selberg. In this paper we will generalize the identities to Kloosterman sums and generalized Kloosterman sums over Hecke congruence subgroups Γ0(N) . Let c be such that N|c and χ a Dirichlet character modulo N. We will deﬁne the generalized Kloosterman sum associated to Γ0(N) with character χ to be

X ma + nd S (D, m, n; c) = χ(k)e . χ c ad≡D (mod c) 0≤a,d

Here Sχ(m, n; c) = Sχ(1, m, n; c) denotes the classical Kloosterman sum associated to Γ0(N) and character χ. We will prove the identities

X mD c S (D, m, n; c) = χ(d)dS , n; , χ χ d2 d d|(D,m,c)

and X mn c S (m, n; c) =χ ¯(n) χ(d)dS , 1; . ((n, N) = 1) χ χ d2 d d|(m,n,c)

The ﬁrst identity is a generalization of an identity by Bykovsky-Kuznetsov- Vinogradov [3]. The last identity is a generalization of an identity of Sel- berg [7].

Contents

1 Kloosterman sums 108 ∗Department of Mathematics, Stockholm University, [email protected]

107 2 Main results 109

1 Kloosterman sums

The classical Kloosterman sums

X mh + nh¯ S(m, n; c) = e , (e(x) = e2πix) (1) c hh¯≡1 (mod c) 0≤h,h

X ma + nd S(n, m; c) = e . (2) c a b ∈PSL(2, ) ( c d ) Z 0≤a,d

The Hecke congruence subgroups Γ0(N) is the subgroup of the full modular group PSL(2, Z) given by a b Γ (N) = ∈ PSL(2, ): c ≡ 0 (mod N) . (3) 0 c d Z

The Kloosterman sums associated to a character modulo N, and to the group Γ0(N), can now be deﬁned similarly to (2) as (see Venkov [8] pp. 32)

X ma + nd S (n, m; c) = χ(d)e . (4) χ c a b ∈Γ (N) ( c d ) 0 0≤a,d

Similarly as in our paper [1] we will also consider generalized Kloosterman sums where Γ0(N) is replaced by its image under Hecke operators. We deﬁne

X ma + nd S (D, n, m; c) = χ(d)e . (c|N) (5) χ c ad−bc=D 0≤a,d

We have that Sχ(1, n, m; c) = Sχ(n, m; c), and the Kloosterman sum deﬁned by (4) is thus a proper generalization of the classical Kloosterman sum associated with Γ0(N).

108 2 Main results

Theorem 1. (generalized Bykovsky-Kuznetsov-Vinogradov identity) We have that X mD c S (D, m, n; c) = χ(d) S , n; , χ χ d2 d d|(D,m,c) where Sχ(D, m, n; c) deﬁned by (5) are generalized Kloosterman sums. Proof. By the deﬁnition (5)

X ma + nd S (D, n, m; c) = χ(d)e . χ c ad−bc=D 0≤a,d

a b c d is an integer matrix of determinant D. From the theory of Hecke-operators (see e.g Motohashi [6] pages 97-98), we know that it can be written uniquely as a product a b α β x q = · , c d 0 δ l y where

αδ = D, 0 ≤ β < δ, and

x q l y is an element in SL(2, Z). We get that a b βl + αx αq + βy = , c d δl δy which gives us

D c α = , d = δy, and l = δ δ

109 Let us also assume thatx ¯ is deﬁned so that 0 ≤ x¯ < l andx ¯ ≡ y (mod l). This implies that

ma + nd m(βl + αx) + nδx¯ e = e . c c

Summing over ad − bc = D hence corresponds to summing over δ|(c, D), xx¯ ≡ 1 (mod c/δ), and β = 0, . . . , δ − 1. We get

X ma + nd S (D, n, m; c) = χ(d)e , χ c ad−bc=D 0≤a,d

d−1 X X mαx + nδx¯ X mβl = χ(δ) χ(¯x)e e . c c δ|(c,D) xx¯≡1 (mod c/δ) β=0 0≤x,x

By applying the identity

δ−1 δ−1 ( X mβl X mβ δ, δ|m, c e = e = l = , c δ 0, otherwise, δ β=0 β=0 this implies that

X X max + nδx¯ S (D, m, n; c) = χ(δ) δ e , χ c δ|(c,D,m) xx¯≡1 (mod c/δ) 0≤x,x

X mD c = χ(δ) δ S , n, . χ δ2 δ δ|(c,D,m)

110 Remark 1. The proper way of doing the proof would of course be to use the Hecke operators of the congruence subgroups, instead of referring to the Hecke operators of the full modular group. In this case the Hecke operators are the same (see e.g. Atkin-Lehner [2]), and that is why it still works. Similarly as in the case of classical Kloosterman sum we have a Kuznetsov summation formula (see e.g Venkov [8] chapter 5).

∞ ∞ X X Φ(c)Sχ(m, n; c) = ρj,χ(n)ρj,χ(m)Φm,n(κj) + “other terms”, c=1 j=1 where Φm,n is a certain integral transform of Φ, and ρj(n) are certain Fourier coeﬃcients of Maass wave forms. The “other terms” are the corresponding terms coming from the holomorphic cusp forms and the Eisenstein series. In the same way as in Bykovsky-Kuznetsov-Vinogradov [3], it will now be possible to develop a generalized Kuznetsov summation formula.

∞ ∞ X X Φ(c)S(D, m, n; c) = tj,χ(D)ρj,χ(n)ρj,χ(m)Φm,n(κj) + “other terms” c=1 j=1 where tj,χ(D) is the eigenvalue for the Hecke operator acting on the Maass wave form. We will return to this topic in a later paper. The identity in Theorem 1 will also allow us to develop the Selberg identity for these Kloosterman sums. Theorem 2. One has the following identities

(i) Sχ(D, m, n; c) = Sχ(m, D, n; c),

(ii) Sχ(D, m, n; c) = χ(D)Sχ¯(D, n, m; c) whenever (D,N) = 1. Proof. To prove (i) we notice that the right hand side of the identity in Theorem 1

X mD c S (D, m, n; c) = χ(d)dS , n; , (6) χ χ d2 d d|(D,m,c) is symmetric in D, m, hence so is the left hand side. To prove (ii) We use the deﬁnitions of the generalized Kloosterman sums

X mh + nk S (D, m, n; c) = χ(k)e , (7) χ c hk≡D (mod c)

Since (D,N) = 1, since N|c, and since hk ≡ D (mod N) we get that

χ(hk) = χ(D), and χ(k) = χ(D)¯χ(h),

111 and ﬁnally

X mh + nk S (D, m, n; c) = χ(D) χ¯(h)e . χ c hk≡D (mod c)

Corollary. (generalized Selberg identity) We have that if (n, N) = 1 1 then X mn c S (m, n; c) =χ ¯(n) χ(d)dS , 1; . (8) χ χ d2 d d|(m,n,c)

Proof. By the fact that Sχ(m, n, c) = Sχ(1, m, n; c) we have

Sχ(m, n; c) = Sχ(1, m, n; c), which by using Theorem 2 (ii) equals

= Sχ¯(1, n, m; c), which by using Theorem 2 (i) equals

= Sχ¯(n, 1, m; c), which by using Theorem 2 (ii) equals

=χ ¯(n)Sχ(n, m, 1; c). which by using Theorem 1 equals

X mn c =χ ¯(n) χ(d)dS , 1; . χ d2 d d|(m,n,c)

This is the analogue of the Selberg identity. In the case of the classical Kloost- erman sums it was given by Selberg [7] without proof, and proved by Kuznetsov [4] with alternative elementary proofs given in Matthes [5] and Andersson [1].

1can this condition be removed and we can still ﬁnd a similar formula? It can be replaced by (m, N) = 1. But in general?

112 References

[1] J. Andersson. A note on some Kloosterman sum identities, 2006.

[2] A. O. L. Atkin and J. Lehner. Hecke operators on Γ0(m). Math. Ann., 185:134–160, 1970.

[3] V. Bykovsky, N. Kuznetsov, and A. Vinogradov. Generalized summation formula for inhomogeneous convolution. In Automorphic functions and their applications (Khabarovsk, 1988), pages 18–63. Acad. Sci. USSR Inst. Appl. Math., Khabarovsk, 1990.

[4] N. V. Kuznetsov. The Petersson conjecture for cusp forms of weight zero and the Linnik conjecture. Sums of Kloosterman sums. Mat. Sb. (N.S.), 111(153)(3):334–383, 479, 1980.

[5] R. Matthes. An elementary proof of a formula of Kuznecov for Kloosterman sums. Resultate Math., 18(1-2):120–124, 1990.

[6] Y. Motohashi. Spectral theory of the Riemann zeta-function. Cambridge University Press, Cambridge, 1997.

[7] A. Selberg. Uber¨ die Fourierkoeﬃzienten elliptischer Modulformen negativer Dimension. In C.R. Neuvieme Congres Math Scandinaves, pages 320–322. Helsingfors, 1938.

[8] A. B. Venkov. Spectral theory of automorphic functions and its applications. Kluwer Academic Publishers Group, Dordrecht, 1990. Translated from the Russian by N. B. Lebedinskaya.

113

A summation formula on the full modular group

Johan Andersson∗

Abstract In this paper we will develop a new type of formula that will allow us to expand a sum over PSL(2, Z) for test functions f deﬁned on PSL(2, R) X a b f c d “ ” a b ∈PSL(2, ) c d Z into spectral objects such as Fourier coeﬃcients for Maass wave forms, holomorphic cusp forms and Eisenstein series. A similar result, the pre-trace formula has previously been known in the special case when f is SO(2) bi-invariant. Our proof is as follows: Use the Bruhat decomposition to express the sum as a sum of Kloosterman sums and then apply the Kuznetsov summation formula.

Contents

1 Introduction 115 1.1 The Poisson summation formula ...... 115 1.2 The theory of modular forms ...... 116 1.3 Kloosterman sums ...... 118

2 Main result 119

3 Some remarks 123

4 Appendix - Integral transforms 124

1 Introduction

1.1 The Poisson summation formula It is fairly trivial and straightforward to obtain an analogue of the classical Poisson summation formula X X f(n) = fˆ(n) n∈Z n∈Z ∗Department of Mathematics, Stockholm University, [email protected]

115 from Fourier analysis in the case of ﬁnite abelian groups, where we have similarly X X f(g) = fˆ(g), (1) g∈G g∈G◦ where the dual group G◦ denotes the group of the characters on G, that is functions ρ : G → C such that ρ(g1)ρ(g2) = ρ(g1 + g2). The case of general locally compact abelian groups can also be treated although it will be a bit more complicated since the dual group will be non compact. However in the case of non abelian groups the relevant representation theory is non trivial. The concept of a dual group does not exist anymore. In this paper we will consider the non abelian locally compact group PSL(2, Z), the full modular group, and develop an equation where the left hand side is X a b f . c d a b ∈PSL(2, ) ( c d ) Z What will the right hand side be? It will not have a group theoretical structure but we will see that it can be expressed in terms of the spectral theory of the full modular group. It will come from the classical holomorphic modular forms as well as the Eisenstein series and the more mysterious Maass wave forms [11].

1.2 The theory of modular forms We will start with some definitions, and a short review of the theory of modular forms1. The Poincaréupper half plane is defined as

+ H = {z = x + iy : x ∈ R, y ∈ R } An element g in PSL(2, R) acts on H by Moebius transformations a b az + b · z = , c d cz + d and maps the Poincaréupper half plane on itself. A function f defined on H is said to be modular if a b f · z = f(z). c d We have that F = {x + iy : −1 < x ≤ 1, x2 + y2 ≥ 1} is a fundamental domain of the full modular group. That means that for each element z ∈ H there exist a unique element γ ∈ PSL(2, Z) such that γz ∈ F . We define the Petersson inner product as Z dxdy hf, gi = f(z)g(z) 2 . F y 1See e.g. Iwaniec [8] or Motohashi [12].

116 A modular form is a modular function that furthermore is an eigenfunction of the hyper- bolic Laplacian

∂2 ∂2 ∆ = y2 + . ∂x2 ∂y2

It is clear that any modular function is periodic in z with period 1, since

1 n · z = z + n, 0 1 and hence every modular function f has a Fourier series in x

∞ X 2πix f(x + iy) = fn(y)e(nx). e(x) = e n=−∞ We say that f is a cusp form if it is a modular form and its constant in the Fourier series vanishes, that is f0(y) = 0. The cusp forms for the full modular group are called Maass wave forms, and they are enumerable. We will use the notation

∞ X 1 Ψ (x + iy) = ρ (n)e(nx)K (|2πny|), ∆Ψ = + κ2 Ψ (z) (2) j j iκj j 4 j j n=−∞

∞ where Kiκj (x) denote the K-Bessel functions and {Ψj}j=1 is an orthonormalized basis of Maass wave forms and the norm comes from the Petersson inner product. The Eisenstein series Γ(s − 1/2)ζ(2s − 1) E(z; s) =y1−s + Γ(s)ζ(2s) 2√ 2π y X s−1/2 + |n| σ (|n|)K (2π|n|y)e(nx) Γ(s)ζ(2s) 1−2s s−1/2 n6=0 are the only non cusp modular forms. A fundamental result (Maass [11], Selberg [15]) is that the modular forms span the space of modular functions. We have

∞ X 1 Z ∞ f(z) = hf, Ψ iΨ (z) + f, E·; 1 + ir Ez; 1 + irdr. j j π 2 2 j=1 −∞

Similarly we will let

θ(k) {Ψj,k}j=1 , (3) be an orthonormal basis of holomorphic cusp forms of weight k. We will let ρj,k(n) be the coeﬃcients in its power series

∞ X n Ψj,k(z) = ρj,k(n)z . n=1

117 The cusp forms of weight k fulﬁll the transformation equation a b f · z = (cz + d)2kf(z). c d and the Petersson inner product will be Z hf, gi = f(z)g(z)y2k−1dxdy. F 1.3 Kloosterman sums Kloosterman sums X mh + nh¯ S(m, n; c) = e (4) c hh¯≡1 (mod c) 0≤h,h

S(m, n; c) = S(n, m; c), (5) as well as the facts that

S(m, 0; c) = cc(m), (6) and

S(0, 0; c) = φ(c), (7) where

X mh c (m) = e (8) c c (h,c)=1 0≤h

|S(m, n; c)| ≤ d(c)pc · (m, n, c), (9) P where (m, n, c) is the greatest common divisor, and d(n) = d|n 1 denotes the divisor function. It follows from his result on the Riemann hypothesis for curves. In the seventies Kuznetsov [10] developed an important summation formula2 that has since then been a useful tool in the analytic theory of numbers3. He developed the summation formula

2Kuznetsov [10] and Bruggeman [5] independently developed the closely related trace formula. 3For example it is an essential point in Motohashi’s [12] work on the fourth power moment of the Riemann zeta function.

118 Lemma 1. (Kuznetsov summation formula) Suppose that Φ ∈ C3(0, ∞) is a function on R+, and that for some δ > 0 and v = 0, 1, 2, 3 it satisﬁes t−v Φ(v)(t) . (10) 3 +δ t 2 + t−δ Then

∞ ∞ Z ∞ X X 1 Φm,n(r)σ2ir(|m|)σ2ir(|n|)dr S(m, n; c)Φ(c) = ρj(m)ρj(n)Φm,n(κj) + π ir 2 c=1 j=1 −∞ |mn| |ζ(1 + 2ir)| ∞ θ(k) X X 1 + ρj,k(m)ρj,k(n)Φm,n 2 − ki , k=1 j=1 where Z ∞ mn Φm,n(r) = Mr 2 Φ(t)dt, 0 t and Mr is deﬁned by

( iπ √ √ 2 sinh πr (J2ri(4π x) − J−2ri(4π x)) , x ≥ 0, Mr(x) = √ (11) 2 cosh πrK2ri(4π −x), x < 0.

Proof. After the change of variables ! 1 p|mn| Φ(t) = φ 4π (12) t t it reduces to the classical Kuznetsov summation formula (This version of the formula is taken from Motohashi [12], Theorem 2.3 and Theorem 2.5).

2 Main result

We have now provided the necessary tools to state and prove our summation formula.

Theorem 1. Let f be a function on PSL(2, R), fulﬁlling the condition

n m v −v ∂ ∂ ∂ ξt ∗ t n, m, v = 0, 1, 2, 3, f , (13) ∂ξn ∂ηm ∂tv t ηt (t2+δ + t−1−δ)(ξ2 + η2 + 1)1+δ t > 0.

Let ∗ denote the number making the determinant of the matrix to be 1. One then has the

119 identity

∞ X a b X Z ∞ Z ∞ cξ ∗ f = − φ(c) f dξdη c d c cη ad−bc=1 c=1 −∞ −∞ c>0 ∞ ∞ X X Z ∞ cn ∗ cξ ∗ + c (n) f + f dξ c c cξ c cn n=−∞ c=1 −∞ ∞ Z ∞ X X X 1 σ2ir(|m|)σ2ir(|n|)F (r; m, n)dr + ρj(m)ρj(n)F (κj; m, n) + π ir 2 j=1 m,n6=0 m,n6=0 −∞ |nm| |ζ(1 + 2ir)| ∞ θ(k) X X X 1 + ρj,k(m)ρj,k(n)F 2 − k i; m, n k=1 j=1 m,n6=0 where Z ∞ Z ∞ Z ∞ mn tξ ∗ F (r; m, n) = e(mξ + nη)Mr 2 f dξdηdt, (14) 0 −∞ −∞ t t tη

Mr is defined by (11), K2ri(x) and J2ri(x) are the K-Bessel and J-Bessel-functions respectively, ρj(n) and ρj,k(n) are defined by (2), and (3) are Fourier coefficients of Maass wave forms and holomorphic modular forms respectively, cc(n) are the Ramanujan sums defined by (8) and φ is the Euler phi function. The integral transform can also be given in Iwasawa coordinates √ Z mny f(g) y F (r; m, n) = e(mx)e(−(my + n) cot θ)M dg, (15) r 2 sin θ g∈PSL(2,R) sin θ where √ 1 x y 0 cos θ sin θ g = · · , (16) √1 0 1 0 y − sin θ cos θ and the integral is with respect to the Haar measure of the group.

Remark 1. Even if we have not bothered to get the best possible test function class (13) in Theorem 1, it is in fact a rather wide class of test functions, and contains e.g. the class

∞ C0 (PSL(2, R)) of all smooth functions with compact support.

We will now give the simplest proof we have found4 (assuming the Kuznetsov summation formula) for our main result. We will use results on integral transforms and Bessel functions proved in the Appendix.

4This proof was presented at the millennial conference on number theory, Urbana-Champaign, May 2000 see conference abstract [1]. A diﬀerent and a bit more involved proof had been presented in Turku in October 1999. We will return to that in [3].

120 Proof. We will now use the Bruhat decomposition, a simple change of summation trick, and the Kuznetsov summation formula to prove our theorem. By letting a = mc + h, and d = nc + h¯, with 0 ≤ h, h¯ < c, we obtain ∞ ∞ X a b X X X mc + h ∗ f = f c d c nc + h¯ ad−bc=1 c=1 m,n=−∞ hh¯≡1 (mod c) c>0 0≤h,h

∞ ∞ X X X mh + nh¯ e fˇ(m, n), (18) c c c=1 hh¯≡1 (mod c) m,n=−∞ 0≤h,h

" ∞ X X ρj(m)ρj(n)F (κj; m, n) m,n6=0 j=1 1 Z ∞ σ (|m|)σ (|n|)F (r; m, n)dr + 2ir 2ir ir 2 π −∞ |mn| |ζ(1 + 2ir)| ∞ θ(k) # X X 1 + ρj,k(m)ρj,k(n)F 2 − k i; m, n . (21) k=1 j=1

121 From the estimate for the Fourier coeﬃcients of cusp forms

X 2 2 √ |ρj(m)| K + d3(m) log(2m) m, (22)

K/2<κj ≤K θ(k) X 2 √ |ρj,k(m)| kd3(m) m log(2m), (23) j=1 valid uniformly for k, K, m ≥ 1 (see Motohashi [12] Lemma 2.4, and equation (2.2.10)), and the fact that

− 9 − 5 |F (r; m, n)| |mn| 4 (1 + |r|) 2 , −2 7 1 − 3 F 2 + k i; m, n |mn| k , when f satisﬁes (13) (see the Appendix Lemma 12 and Lemma 13), we see that the summation and summation-integration in (21) is absolutely convergent, and hence we can change the summation order, and we thus obtain the spectral part of Theorem 1. We will now consider the contribution coming from m = 0 or n = 0. We get from (20) the contribution ∞ X X ˇ S(m, n; c)fc(m, n), (24) m=0 or n=0 c=1 By equations (5),(6) and (7) this equals

∞ ∞ ∞ X X ˇ ˇ X ˇ cc(n) fc(0, n) + fc(0, n) − φ(c)fc(0, 0). (25) n=−∞ c=1 c=1 By applying the Poisson summation formula on the ﬁrst term in (25) again, and using equations (17) and (19) it equals

∞ ∞ X X Z ∞ cn ∗ cξ ∗ c (n) f + f dξ c c cξ c cn n=−∞ c=1 −∞ ∞ X Z ∞ Z ∞ cξ ∗ − φ(c) f dξdη c cη c=1 −∞ −∞ which account for the second and ﬁrst term in the theorem. To show how to express the integral transform in Iwasawa coordinates we start with the integral transform (14) Z ∞ Z ∞ Z ∞ mn tξ ∗ F (r; m, n) = e(mξ + nη)Mr 2 f dξdηdt. (26) 0 −∞ −∞ t t tη By using the explicit expression of a group element g ∈ SL(2, R), by equation (16) into Iwasawa coordinates, we obtain √ √ √ ! x√sin θ x√cos θ 1 x y 0 cos θ sin θ y cos θ − y sin θ y + y g = · 1 · = . 0 √ sin√ θ cos√ θ 0 1 y − sin θ cos θ − y y

122 Using the change of variables √ √ ! ξt ∗ y cos θ − x√sin θ sin θ y + x√cos θ = y y , sin√ θ cos√ θ t ηt − y y in equation (26), we get the equations

sin θ ξ = −y cot θ + x, t = − √ , and η = − cot θ. y

We have the Jacobian

∂η ∂t ∂ξ 1 cos θ y − 2 − √ − 2 ∂θ ∂θ ∂θ sin θ y sin θ 1 ∂η ∂t ∂ξ = 0 sin θ − cot θ = − , ∂y ∂y ∂y 2y3/2 3/2 ∂η ∂t ∂ξ 2y sin θ ∂x ∂x ∂x 0 0 1 since the matrix is upper triangular and the determinant equals the product of the diagonal elements. We get

dθdxdy dξdηdt = − , 2y3/2 sin θ and we see that

Z ∞ Z ∞ Z 2π mny F (r; m, n) = e(m(−y cot θ + x) − n cot θ)Mr 2 × −∞ 0 0 sin θ √ √ ! y cos θ − x√sin θ sin θ y + x√cos θ dθdxdy × f y y . (27) sin√ θ cos√ θ 3/2 − y y 2y sin θ

Using the fact that the Haar measure on the group PSL(2, R) can be given by Z 1 Z 2π Z ∞ Z ∞ dydxdθ f(g)dg = f(g) , 2 y2 g∈PSL(2,R) 0 −∞ 0 we can write equation (27) as √ Z mny f(g) y F (r; m, n) = e(mx)e(−(my + n) cot θ)M dg, r 2 sin θ g∈PSL(2,R) sin θ where g is given in Iwasawa coordinates by equation (16).

3 Some remarks

The method used in this papers also generalizes to sums over congruence subgroups and characters, and the Kuznetsov summation formula has been generalized to that setting by

123 Proskurin ([13], [14]). It should also be noted that from the inversion formulae for the Titchmarsh transforms Z ∞ (J2it(y) − J−2it(y)) f(y)dy Lf (t) = 0 2i sinh πt y Z ∞ (J2it(y) − J−2it(y)) f(y) = Lf (t) tanh(πt)tdt, −∞ 2 sin πit and Kontorevich-Lebedev transforms

Z ∞ f(y)dy Lf (t) = Kit(y) 0 y Z ∞ sinh(πt) f(y) = Lf (t)Kit(y) 2 tdt, −∞ π see Iwaniec [8], page 228-231, our main summation formula can be inverted. When f is SO(2) invariant our formula will reduce to the spectral expansion of a modular function, and when f is SO(2) bi-invariant it will reduce to the pre-trace formula. Even if the proof is fairly straightforward, it tends to become a bit technical (a certain tricky integral) so we will defer the proof to a forthcoming paper [2].

4 Appendix - Integral transforms

In this section we will prove the estimates for integral transforms that are needed in the paper. We will ﬁrst calculate the Mellin transform and Mellin-Barnes integral of the kernel function Mr as deﬁned by (11). Lemma 2. One has the following Mellin-transform and Mellin-Barnes representations of the Kernel function Mr Z ∞ s−1 −2s cos πs ε = 1 (i) Mr(ε · x)x dx = (2π) Γ(s + ri)Γ(s − ri) , 0 cosh πr ε = −1 (|Im (r)| < Re(s) < 3/4) Z c+∞ 1 −2s cos πs x > 0 −s (ii) Mr(x) = (2π) Γ(s + ri)Γ(s − ri) |x| ds 2πi c−∞ cosh πr x < 0 (|Im(r)| < c < 1/2)

Proof. To calculate the Mellin-transform of Mr we will need the Mellin transforms for the J and K Bessel functions (see [6], 6.8 (1) and 6.8 (22))

Z ∞ s−1 s−2 s + v s − v Kv(x)x dx = 2 Γ Γ , |Re(v)| < Re(s) (28) 0 2 2 Z ∞ s−1 s−1 Γ((s + v)/2) Jv(x)x dx = 2 , − Re(v) < Re(s) < 3/2. 0 Γ((2 + v − s)/2)

124 By the functional equation for the gamma-function π Γ(s)Γ(1 − s) = , sin πs we have that Z ∞ Γ((s + v)/2) J (x)xs−1dx = 2s−1 v Γ((2 + v − s)/2) 0 (29) 1 s + v s − v π = 2s−1Γ Γ sin (s − v) . π 2 2 2 √ With the substitution y = 4π x we get

Z ∞ Z ∞ 2 s−1 −2s y 2s−1 Mr(±x)x dy = 2(2π) Mr ± 2 y dy, 0 0 (4π) and by the deﬁnition of Mr(x), (29) and (28) we get Z ∞ s−1 −2s i Mr(x)x dx =(2π) Γ(s + ri)Γ(s − ri)(sin(π(s − ri)) − sin(π(s + ri))), 0 2 sinh πr Z ∞ s−1 −2s cosh πr Mr(−x)x dx =(2π) Γ(s + ri)Γ(s − ri), 0 π so by the fact that cosh πr = cos πri the lemma is clear for Mr(−x). Finally, the fact that i(sin(π(s − ri)) − sin(π(s + ri))) 2i cos(πs) sin(−πri) = 2 sinh πr 2 sinh πr = cos(πs), concludes the calculation of the Mellin-transform. The Mellin-Barnes integral follows from Mellin-inversion, see Titchmarsh[16], Theorem 30. The restriction c < 1/2 in (ii) comes from the fact that the Mellin transform of the function should tend to 0 which is one of the conditions of that theorem. Lemma 3. Let F (x) be a real diﬀerentiable function such that F 0(x) is monotone and F 0(x) ≥ m > 0 or F 0(x) ≤ −m < 0 for 0 ≤ x ≤ b. Then

Z b iF (x) 4 e dx ≤ a m Proof. This is taken from Ivi´c[7], Lemma 2.1. Lemma 4. Let F (x) be a real, twice-diﬀerentiable function in [a, b] such that F 00(x) ≥ m > 0 or F 00(x) ≤ −m < 0. Then

Z b iF (x) 8 e dx ≤ √ a m Proof. This is taken from Ivi´c[7], Lemma 2.2.

125 ˆ Lemma 5. Suppose that f fulfills condition (13). Then Φ(t) = ft(m, n), where ft(m, n) is defined by (17) fulfills condition (10) Proof. We have that

Z ∞ Z ∞ tx ∗ Φ(t) = e(nx + my)f dxdy −∞ −∞ t ty and we get

Z ∞ Z ∞ v (v) ∂ tx ∗ Φ (t) = e(nx + my) v f dxdy, (v = 0, 1, 2, 3) −∞ −∞ ∂t t ty Using (13) this can be estimated by

Z ∞ Z ∞ t−v 1 t−v 2+δ −1−δ 2 2 1+δ dxdy ≤ Cδ 2+δ −1−δ , v = 0, 1, 2, 3, −∞ −∞ (t + t ) (x + y + 1) (t + t ) which fulﬁll condition (10). In Lemmas 6 and 9 we will ﬁrst prove the estimate x2 |Jn(x)| , |n|7/3 which will be used to provide estimates for certain integral transforms appearing in our main summation formula. Lemma 6. Suppose that n ≥ 1, and x is real. Then 18 |J (x)| ≤ n πn1/3 Proof. By the Bessel representation (see Guo–Wang [17] 7.5 (20) or Bessel [4]) 1 Z π Jn(z) = cos(z sin θ − nθ)dθ, (30) 2π −π

we have that

Z π Z π 1 1 iF (x) Jn(z) = cos(z sin θ − nθ)dθ = Re e dx , (31) 2π −π 2π −π where

F (x) = z sin x − nx. (32)

Suppose that n/2 > z. Then we have by the triangle inequality that n F 0(x) = z cos x − n ≤ − , 2

126 so we can use Lemma 3, and (31) to get

Z π 8 cos(z sin θ − nθ)dθ ≤ . (33) −π n If n/2 ≤ z we divide the integral

 1  Z π Z −π+n−1/3 Z −n− 3 Z n−1/3 Z π−n−1/3 Z π iF (x) iF (x) e dx =  + + + + e dx. −π −π −π+n−1/3 −n−1/3 n−1/3 π−n−1/3

For the case

n−1/3 ≤ x ≤ π − n−1/3, (34) we have by the inequality x π ≤ sin x, 0 ≤ x ≤ 2 2 π and the fact that sin x is an increasing function for 0 ≤ x ≤ 2 , that n2/3 F 00(x) = −z sin x ≤ − . (35) 4 From the identity

sin(π − x) = sin(x), we have the inequality (35) in the whole interval (34), and by Lemma 4 we get

−1/3 Z π−n 16 eiF (x)dx ≤ . (36) 1/3 n−1/3 n Similarly we also get

−1/3 Z −n 16 eiF (x)dx ≤ . (37) 1/3 −π+n−1/3 n Using (31), (36), (37) and the triangle inequality, and adding the terms we get

Z π

cos(z sin θ − nθ)dθ −π  1  Z π Z −π+n−1/3 Z −n− 3 Z n−1/3 Z π−n−1/3 Z π iF (x) iF (x) ≤ e dx =  + + + + e dx −1/3 −1/3 −1/3 −1/3 −π −π −π+n −n n π−n 1 16 2 16 1 36 ≤ + + + + = . n1/3 n1/3 n1/3 n1/3 n1/3 n1/3

127 Lemma 7. One has that uniformly for |n| ≥ 3 and real x,

x2 |Jn(x)| . |n|7/3

Lemma 8. One has that when k is an integer, then

k √ Mi(1/2−k)(x) = π(−1) J2k−1 4π x , (x > 0)

Proof. Since x > 0 it follows from (11) πi √ √ M (x) = J (4π x) − J (4π x) , i(1/2−k) 2 sinh π(i(1/2 − k)) 2k−1 1−2k by applying the identities (see Guo–Wang [17], 7.2 (8)),

n Jn(x) = (−1) J−n(x), and

sinh π(i(1/2 − k)) = sin π 1 − k i 2 = cos(πk) = (−1)k.

Lemma 9. One has that when k ≥ 2 is an integer, then

x2 Mi(1/2−k)(x) . |k|7/3

Proof. This follows from Lemma 7 and Lemma 8. Proof. By the identity (see Guo–Wang [17], 7.2 (8))

n Jn(x) = (−1) J−n(x), (38) it is suﬃcient to consider n ≥ 3. Applying the identity (see Guo–Wang [17], 7.23 (14)) x J (x) = (J (x) + J (x)) (39) n 2n n−1 n+1 two times on Jn(x) gives us

2 x2 X |J (x)| |J (x)|, (40) n n2 n+j j=−2 and the result then follows from Lemma 6.

128 Lemma 10. Suppose that f fulﬁlls condition (13). Then

∞ ∞ X X X ˇ fc(m, n) < ∞. c=1 hh¯≡1 (mod c) m,n=−∞ 0≤h,h

Z ∞ Z ∞ cξ ∗ fc(m, n) = e(mξ + nη)f dξdη −∞ −∞ c cη Z ∞ Z ∞ 2 2 −2 ∂ ∂ cξ ∗ = (mn) e(mξ + nη) 2 2 f dξdη −∞ −∞ ∂ξ ∂η c cη By (13) this gives us

−2−δ −2 |fc(m, n)| c (mn)

If either m or n equals zero, we similarly do partial integration with respect to one variable, and we get

−2−δ −2 −2−δ −2 |fc(m, 0)| c m , |fc(0, n)| c n .

When both m = 0 and n = 0, we immediately get

−2−δ |fc(0, 0)| c , and this concludes our proof. Lemma 11. One has that Z t nm 1/2 3/4 3/2 −3/2 yMr 2 dy t |mn| + t (1 + |r|) , 0 y Z x Z t nm 3/2 3/4 3/2 −5/2 yMr 2 dydx x |mn| + x (1 + |r|) . 0 0 y Proof. We will use the Mellin-Barnes representation, Lemma 2 with c = 1/4. We get

nm yM = r y2 1 Z c+∞i cos πs, nm > 0 Γ(s + ri)Γ(s − ri) (2π)−2s|nm|−sy2s+1ds. 2πi c−∞i cosh πr, nm < 0 Integrating this with respect to y gives us

Z t nm yMr 2 dy = 0 y 1 Z c+∞i Γ(s + ri)Γ(s − ri) cos πs, nm > 0 (2π)−2s|nm|−st2s+2ds. (41) 2πi c−∞i 2s + 2 cosh πr, nm < 0

129 Integrating this again with respect to t we get

Z x Z t nm yMr 2 dydx = 0 0 y 1 Z c+∞i Γ(s + ri)Γ(s − ri) cos πs, nm > 0 (2π)−2s|nm|−sx2s+3ds. (42) 2πi c−∞i (2s + 2)(2s + 3) cosh πr, nm < 0 Now moving the integration line to c = −3/4 we will pick up the residues coming from s = ±ir. In order to estimate these residue terms we will use the following variant of the Stirling formula(see e.g. Ivi´c[7], A.34)

σ−1/2 − π |t| |Γ(σ + it)| |t| e 2 σ0 ≤ σ ≤ σ1, 0 < t0 ≤ |t| (43) For (41), the residue part coming from s = ±ir can be estimated by Ot2(1 + |r|)−3/2. Using (43) again, the remaining integral can be estimated by O t1/2(1 + |r|)−3/2|mn|3/4 .

Similarly for (42) the residue part coming from s = ±ir can be estimated by Ox3(1 + |r|)−5/2, and the remaining integral can be estimated by O x3/2(1 + |r|)−5/2|mn|3/4 .

Lemma 12. One has that if f fulfills condition (13) then the integral transform F (r; m, n) defined by (14) fulfills

|F (r; m, n)| |mn|−9/4(1 + |r|)−5/2. Proof. We will use partial integration with respect to x, y, t and diﬀerentiate tx ∗ t−1f t ty three times with respect to x, y and two times with respect to t, and do the corresponding integration of nm e(nx + my)tM . r t2 Condition (13) enables us to do this and ensures that the “limit terms” will disappear. We get 1 Z ∞ Z ∞ Z ∞ F (r; m, n) = 3 e(nξ + mη)H(t)G(ξ, η, t)dξdηdt, (44) (nm) 0 −∞ −∞

130 where ∂8 tξ ∗ G(ξ, η, t) = t−1f , ∂ξ3∂η3∂t2 t tη and

Z t Z x nm H(t) = yMr 2 dydx. 0 0 y By Lemma 11 we get |H(t)| t3/2 |mn|3/4 + t3/2 (1 + |r|)−5/2. (45)

By the fact that f fulﬁll condition (13), we have that

t−3 1 |G(x, y, t)| . (46) (t2+δ + t−1−δ) (x2 + y2 + 1)1+δ

By combining (44), (45) and (46), the proof is completed.

Lemma 13. One has that if f fulfills condition (13) then the integral transform F (r; m, n) defined by (14) fulfills

1 −2 −7/3 F 2 − k i; m, n |mn| k . (47) Proof. We will use partial integration with respect to x, y and diﬀerentiate

tx ∗ f (48) t ty two times with respect to each variable and do the corresponding integration of

nmD e(nx + my)M . (49) i(1/2−k) t2

Condition (13) ensures that we can do this and that the “limit terms” disappear. By then applying Lemma 9 the result follows.

References

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131 [4] F. W. Bessel. Analytishe Auﬂ¨osung der Keplerschen Aufgabe. Berliner Abh. 1816-17, pages 49–55, 1819.

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[11] H. Maass. Uber¨ eine neue Art von nichtanalytischen automorphen Funktionen und die Bestimmung Dirichletscher Reihen durch Funktionalgleichungen. Math. Ann., 121:141–183, 1949.

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132 A summation formula over integer matrices

Johan Andersson∗

Abstract In a previous paper [4] we proved a summation formula for the full modular group. In this paper we will generalize the result to integer matrices of determinant D. We develop a formula for X a b f c d ad−bc=D c>0 for test functions f. We are able express the sum into spectral objects coming from the full modular group. The proof is the same as in [4], but instead of using the classical Kuznetsov summation formula we use a generalized summation formula of Bykovsky-Kuznetsov-Vinogradov [7] for the generalized Kloosterman sums X mh + nk S(D, m, n; c) = e . c hk≡D (mod c) 0≤h,k

Contents

1 Kloosterman sums 133 1.1 Elementary properties ...... 133 1.2 The Kuznetsov summation formula ...... 134

2 Main result 136

3 Some remarks 139

4 Appendix - Integral transforms 139

1 Kloosterman sums

1.1 Elementary properties Kloosterman sums X mh + nh¯ S(m, n; c) = e (1) c hh¯≡1 (mod c) 0≤h,h

∗Department of Mathematics, Stockholm University, [email protected]

133 is an important part of number theory. In our paper [1] we introduced the generalized Kloosterman sums X mh + nk S(D, m, n; c) = e , (2) c hk≡D (mod c) 0≤h,k

X mD c S(D, m, n; c) = d S , n; . (3) d2 d d|(D,m,c)

We proved the symmetry

S(n1, n2, n3; c) = S(nσ(1), nσ(2), nσ(3); c) (4) where σ ∈ S3 is any permutation acting on {1, 2, 3}. Furthermore, for the classical Kloost- erman sums we have the trivial identities

S(m, 0; c) = cc(m), (5) as well as

S(0, 0; c) = cc(0) = φ(c), (6) where

X mh c (m) = e (7) c c (h,c)=1 0≤h

1.2 The Kuznetsov summation formula We will now state a generalized version of the classical Kuznetsov summation formula

Lemma 1. (Bykovski-Kuznetsov-Vinogradov1 summation formula) Suppose that Φ ∈ C3(0, ∞) is a function on R+, which for some δ > 0 and v = 0, 1, 2, 3 fulﬁlls t−v Φ(v)(t) . (8) 3 +δ t 2 + t−δ 1The formula was initially developed independently by the author, but it was pointed out by K. Mat- sumoto at the millennial conference on number theory, Urbana-Champaign, May 2000, that the same version of the Kuznetsov summation formula, with the generalized Kloosterman sums S(D, m, n; c), had been previously proved by Bykovski-Kuznetsov-Vinogradov [7].

134 Then

∞ ∞ X X S(D, m, n; c)Φ(c) = αjtj(D)tj(m)tj(n)ΦD,m,n(κj) c=1 j=1 1 Z ∞ Φ (r)σ (|D|)σ (|m|)σ (|n|)dr + D,m,n 2ir 2ir 2ir ir 2 π −∞ |nmD| |ζ(1 + 2ir)| ∞ θ(k) X X 1 + αj,ktj,k(D)tj,k(m)tj,k(n)ΦD,m,n 2 − ki , k=1 j=1 where S(D, m, n; c) are the generalized Kloosterman sums deﬁned by equation (2). Z ∞ Dmn ΦD,m,n(r) = Mr 2 Φ(t)dt, 0 t with Mr deﬁned by

( iπ √ √ 2 sinh πr (J2ri(4π x) − J−2ri(4π x)) , x ≥ 0, Mr(x) = √ (9) 2 cosh πrK2ri(4π −x), x < 0;

K2ri(x) and J2ri(x) are the K-Bessel and J-Bessel-functions respectively; ρj(n) are the Fourier-coeﬃcients

∞ X Ψj(x + iy) = ρj(n)e(nx)Kiκj (|2πny|) n=−∞

∞ of an orthonormalized basis {Ψj}j=1 of Maass wave forms, and

2 |ρj(1)| ρj(n) αj = and tj(n) = , cosh πκj ρj(1)

θ(k) are the eigenvalues for the Hecke-operators Tn. Similarly {Ψj,k}j=1 is an orthonormal basis of holomorphic cusp forms of weight k,

∞ X ρj,k(n) Ψ (z) = ρ (n)e(nz), t (n) = , j,k j,k j,k ρ (1) n=1 j,k and

1−4k −2k−1 2 a(k) = 2 π Γ(2k), αj,k = a(k)|ρj,k(1)| .

Proof. For proof, see also Bykovski-Kuznetsov-Vinogradov [7]. The notation αj, αj,k, tj(n), tj,k(n) is taken from Motohashi [9]. When D = 1 the formula reduces after the change of variables ! 1 p|mnD| Φ(t) = φ 4π t t

135 to the usual formula (See Motohashi [9], Theorem 2.3 and Theorem 2.5.). For D 6= 1 we reduce it to D = 1 by using the identity X mD c S(D, m, n; c) = dS , n; , (10) d2 d d|(D,m,c) (see Theorem 1 of Andersson [1] for a proof ) and X mn t (m)t (n) = t , j j j d2 d|(m,n) together with the corresponding identities

−ir σ2ir(m) σ2ir(n) X mn mn X mn = σ , t (m)t (n) = t , mir nir 2ir d2 d2 j,k j,k j,k d2 d|(m,n) d|(m,n) for the divisor functions and the Fourier coeﬃcients for holomorphic modular forms (see Motohashi[9], formula (3.1.4), (3.1.14)).

2 Main result

We will now state the main result

Theorem 1. Let f be a function on {A ∈ GL(2, R) : det A = D}, where D is a nonzero integer, fulﬁlling the conditions

n m v −v ∂ ∂ ∂ xt ∗ t n, m, v = 0, 1, 2, 3, f , (11) ∂xn ∂ym ∂tv t yt (t2+δ + t−1−δ)(x2 + y2 + 1)1+δ t > 0, for some δ > 0, where ∗ denotes the number making the determinant of the matrix equal to D. One then has the identity

∞ X a b X X c Z ∞ Z ∞ cx ∗ f = − φ f dxdy c d d c cy ad−bc=D c=1 d|(D,c) −∞ −∞ c>0 ∞ ∞ X X X Z ∞ cn ∗ cx ∗ + c (n) f + f dx c/d c cx c cn n=−∞ c=1 d|(D,c) −∞

∞ X X + αjtj(D)tj(m)tj(n)F (κj; D, m, n) j=1 m,n6=0 Z ∞ X 1 σ2ir(|D|)σ2ir(|m|)σ2ir(|n|)F (r; D, m, n)dr + π ir 2 m,n6=0 −∞ |Dmn| |ζ(1 + 2ir)| ∞ θ(k) X X X 1 + αj,ktj,k(D)tj,k(m)tj,k(n)F 2 − k i; D, m, n , k=1 j=1 m,n6=0

136 where Z ∞ Z ∞ Z ∞ Dmn tx ∗ F (r; D, m, n) = e(mx + ny)Mr 2 f dxdydt, (12) 0 −∞ −∞ t t ty

Mr(x) is deﬁned by (9), and αj, αj,k, tj(n), tj,k(m) are deﬁned as in Lemma 1, φ(c) denotes the Euler phi function and cc(n) denotes the Ramanujan sums (7). Proof. The proof will closely follow our proof from [4]. We will now use the Bruhat decomposition, a simple change of summation trick, and the Kuznetsov summation formula to prove our theorem. By letting a = mc + h, and d = nc + k, with 0 ≤ h, k < c, we obtain

∞ ∞ X a b X X X mc + h ∗ f = f c d c nc + k ad−bc=D c=1 m,n=−∞ hk≡D (mod c) c>0 0≤h,k

Condition (11) for x and y ensures that we can use the shifted double Poisson summation formula (see e.g Zygmund [10] (13.3)), and we get

∞ ∞ X X X mh + nk e fˇ(m, n), c c c=1 hk≡D (mod c) m,n=−∞ 0≤h,k

137 version of the Kuznetsov summation formula, Lemma 1, and we ﬁnd that the contribution coming from mn 6= 0 is equal to

∞ X X αjtj(D)tj(m)tj(n)F (κj; D, m, n) m,n6=0 j=1 1 Z ∞ σ (|D|)σ (|m|)σ (|n|)F (r; D, m, n)dr + 2ir 2ir 2ir ir 2 π −∞ |Dmn| |ζ(1 + 2ir)| ∞ θ(k)  X X 1 + αj,ktj,k(D)tj,k(m)tj,k(n)F 2 − k i; D, m, n  . (16) k=1 j=1 From the estimate for the Fourier coeﬃcients of cusp forms

X 2 2 √ αj|tj(m)| K + d3(m) log(2m) m, (17)

K/2<κj ≤K θ(k) X 2 √ αj,k|tj,k(m)| kd3(m) m log(2m), (18) j=1 valid uniformly for k, K, m ≥ 1 (see Motohashi [9] Lemma 2.4, and equation (2.2.10)), together with the fact that

− 9 3 − 5 |F (r; D, m, n)| |mn| 4 |D| 4 (1 + |r|) 2 , −2 7 1 − 3 F 2 + k i; D, m, n |mn| |D|k , when f satisﬁes (11) (see the Appendix, Lemma 7 and Lemma 8), we see that the summation and summation-integration in (16) is absolutely convergent, and hence we can change the summation order, and we get the spectral part of Theorem 1. We now consider the contribution coming from m = 0 or n = 0. We get from (15) the contribution ∞ X X ˇ S(D, m, n; c)fc(m, n), (19) m=0 or n=0 c=1 By the symmetry (4) of S(D, m, n; c), this equals

∞ ∞ ∞ X X ˇ ˇ X ˇ S(D, 0, n; c) fc(0, n) + fc(0, n) − S(D, 0, 0)fc(0, 0). (20) n=−∞ c=1 c=1 By the Bykovsky-Kuznetsov-Vinogradov identity (3) we have that X c S(D, 0, n; c) = S 0, n; d d|D which by the deﬁnition of the Ramanujan sums (7) equals X cc/d(n). d|D

138 By applying the Poisson summation formula on the ﬁrst term in (20) again, and using cl(0) = φ(l) we ﬁnd that equation (20) equals

∞ ∞ X X X Z ∞ cn ∗ cx ∗ c (n) f + f dx c/d c cx c cn n=−∞ c=1 d|(D,c) −∞ ∞ X X c Z ∞ Z ∞ cx ∗ − φ f dxdy d c cy c=1 d|(D,c) −∞ −∞ which accounts for the second and ﬁrst term in the theorem.

3 Some remarks

We believe that our main summation formula will have plenty of applications. We shall present some of them in forthcoming papers, e.g. [3] and [2]. An obvious application is the following: Since

σ2ir(n) = 1, tj(n) = 1, and tj,k(n) = 1 the summation formula over the full modular group as proved in [4] will be the special case when D = 1 of this formula. About the proof: This was not our first proof. Our first sketch of a proof used Motohashi’s work [9] on the theory of the Riemann zeta-function. For a sketch of that proof idea, see the introduction of our thesis [6]. Another proof, that we presented in Turku in the fall of 1999, used the Kuznetsov summation formula in a different way. We did not need to use the Kloosterman sum identity (3). Instead we used a formula relating the Kloosterman sums and Ramanujan sums. We will present this proof in [5]. Another proof that we would like to develop and hope to return to later is to prove the summation formula over integer matrices of determinant D directly from the summation formula over the full modular group from [4], by means of Hecke operators. In the current proof we use Hecke operators indirectly as they are used to prove the identity (3) in [1]. However a more direct proof would be nice as well.

4 Appendix - Integral transforms

In this appendix we prove estimates of integral transforms that are needed in the rest of the paper. Lemma 2. One has the following Mellin-transform and Mellin-Barnes representations of the Kernel function Mr Z ∞ s−1 −2s cos πs ε = 1 (i) Mr(ε · x)x dx = (2π) Γ(s + ri)Γ(s − ri) , 0 cosh πr ε = −1 (|Im (r)| < Re(s) < 3/4) Z c+∞ 1 −2s cos πs x > 0 −s (ii) Mr(x) = (2π) Γ(s + ri)Γ(s − ri) |x| ds 2πi c−∞ cosh πr x < 0 (|Im(r)| < c < 1/2)

139 Proof. For proof, see Lemma 3 in Andersson [4]. Lemma 3. One has that when k ≥ 2 is an integer, then

x2 Mi(1/2−k)(x) . |k|7/3 Proof. This is Lemma 9 in Andersson [4]. ˆ Lemma 4. Suppose that f fulfills condition (11). Then the function Φ(t) = ft(m, n), where ft(m, n) is defined by (13), fulfills condition (8). Proof. For a proof, see Andersson [4], Lemma 5. Lemma 5. Suppose that f fulfills condition (11). Then

∞ ∞ X X X ˇ fc(m, n) < ∞ c=1 hh¯≡D (mod c) m,n=−∞

Proof. The proof of this result is identical to the one in Andersson [4], Lemma 10. Lemma 6. One has the estimates Z t nmD 1/2 3/4 3/2 −3/2 yMr 2 dy t |mnD| + t (1 + |r|) , 0 y Z x Z t nmD 3/2 3/4 3/2 −5/2 yMr 2 dydx x |mnD| + x (1 + |r|) . 0 0 y Proof. We will use the Mellin-Barnes representation, Lemma 2 with c = 1/4. We get

nmD yM = r y2 1 Z c+∞i cos πs, nmD > 0 Γ(s + ri)Γ(s − ri) (2π)−2s|nmD|−sy2s+1ds. 2πi c−∞i cosh πr, nmD < 0 Integrating this with respect to y gives us

Z t nmD yMr 2 dy = 0 y 1 Z c+∞i Γ(s + ri)Γ(s − ri) cos πs, nmD > 0 (2π)−2s|nmD|−st2s+2ds. (21) 2πi c−∞i 2s + 2 cosh πr, nmD < 0 Integrating this again with respect to t we get

Z x Z t nmD yMr 2 dydx = 0 0 y 1 Z c+∞i Γ(s + ri)Γ(s − ri) cos πs, nmD > 0 (2π)−2s|nmD|−sx2s+3ds. (22) 2πi c−∞i (2s + 2)(2s + 3) cosh πr, nmD < 0

140 By moving the integration line to c = −3/4 we will pick up the residues coming from s = ±ir. In order to estimate these residue terms we will use the following variant of the Stirling formula (see e.g. Ivi´c[8], A.34)

σ−1/2 − π |t| |Γ(σ + it)| |t| e 2 , σ0 ≤ σ ≤ σ1, 0 < t0 ≤ |t|. (23)

For (21), the residue part coming from s = ±ir can be estimated by

Ot2(1 + |r|)−3/2.

Using (23) again, the remaining integral can be estimated by O t1/2(1 + |r|)−3/2|mnD|3/4 .

Similarly for (22) the residue part coming from s = ±ir can be estimated by

Ox3(1 + |r|)−5/2, and the remaining integral can be estimated by O x3/2(1 + |r|)−5/2|mnD|3/4 .

Lemma 7. If f fulfills condition (11) then the integral transform F (r; D, m, n) defined by (12) satisfies

|F (r; D, m, n)| |mn|−9/4|D|3/4(1 + |r|)−5/2.

Proof. We will use partial integration with respect to x, y, t and diﬀerentiate

tx ∗ t−1f t ty three times with respect to x, y and two times with respect to t, and perform the corresponding integration of the function nmD e(nx + my)tM . r t2

Condition (11) enables us to do this and ensures that the “limit terms” will disappear. We get 1 Z ∞ Z ∞ Z ∞ F (r; D, m, n) = 3 e(nx + my)H(t)G(x, y, t)dxdydt, (24) (nm) 0 −∞ −∞ where ∂8 tx ∗ G(x, y, t) = t−1f , ∂x3∂y3∂t2 t ty

141 and Z t Z x nmD H(t) = yMr 2 dydx. 0 0 y By Lemma 6 we get |H(t)| t3/2 |mnD|3/4 + t3/2 (1 + |r|)−5/2. (25)

By the fact that f fulfills condition (11), we have that t−3 1 |G(x, y, t)| . (26) (t2+δ + t−1−δ) (x2 + y2 + 1)1+δ By combining (24), (25) and (26), the proof is completed. Lemma 8. If f fulfills condition (11) then the integral transform satisfies F (r; D, m, n) defined by (12) fulfills

1 −2 −7/3 F 2 − k i; D, m, n |mn| |D|k . (27) Proof. We will use partial integration with respect to x, y and diﬀerentiate tx ∗ f (28) t ty two times with respect to each variable and do the corresponding integration of nmD e(nx + my)M . (29) i(1/2−k) t2 Condition (11) ensures that we can do this and that the “limit terms” disappear. By applying Lemma 3 the result then follows.

References

[1] J. Andersson. A note on some Kloosterman sum identities, 2006. [2] On the additive circle problem, forthcoming. [3] The Selberg and Eichler-Selberg trace formulae, 2006. [4] A summation formula on the full modular group, 2006. [5] A summation formula over integer matrices II, 2006. [6] Summation formulae and zeta-functions, Thesis. [7] V. Bykovsky, N. Kuznetsov, and A. Vinogradov. Generalized summation formula for inhomogeneous convolution. In Automorphic functions and their applications (Khabarovsk, 1988), pages 18–63. Acad. Sci. USSR Inst. Appl. Math., Khabarovsk, 1990.

142 [8] A. Ivi´c. The Riemann zeta-function. John Wiley & Sons Inc., New York, 1985. The theory of the Riemann zeta-function with applications.

[9] Y. Motohashi. Spectral theory of the Riemann zeta-function. Cambridge University Press, Cambridge, 1997.

[10] A. Zygmund. Trigonometric series: Vols. I, II. Cambridge University Press, London, 1968.

143

A summation formula over integer matrices II

Johan Andersson∗

Abstract In a previous paper [3] we proved a summation formula over integer matrices with ﬁxed determinant D, that is we obtained a formula for

X a b f . c d ad−bc=D

In doing so we used an identity by Bykovsky-Kuznetsov-Vinogradov (see [1])

X mD c S(D, m, n; c) = dS , n; , (1) d2 d d|(D,m,c)

where

X mh + nk S(D, m, n; c) = e (2) c hk≡D (mod c) 0≤h,k

are generalized Kloosterman sums. In this paper we will give an alternative proof without using this identity.

Contents

1 Introduction 146

2 Kloosterman sums and Ramanujan sums 147

3 Spectral theory 152

4 Some concluding remarks 156

∗Department of Mathematics, Stockholm University, [email protected]

145 1 Introduction

In [2] we proved a new type of summation formula over the full modular group, and in [3] we proved the corresponding identity for sums over integer matrices of fixed determinant D. We first thought of ways to obtain an identity like this in the spring of 1999 when we were visiting Minneapolis. It occurred to us when we were studying a certain sixth power moment of the Lerch zeta-function1, that an identity in the paper of Motohashi [7] on the fourth power moment of the Riemann zeta-function, could be seen as a “functional equation” corresponding to a yet to be discovered “summation formula”2. In the summer of 1999 we presented our idea at a symposium in Turku. The method of proof was not a very practical one. Even if it was simple in principle, the integrals would turn out to be cumbersome, and we never proved an explicit version of our summation formula with this method. Starting with that idea we searched for a simpler method of proof, and we soon found a nice proof, using just the classical Poisson summation formula, the Kuznetsov summation formula and some simple ideas concerning the Ramanujan sums and the Kloosterman sums. We presented our results at a seminar in Turku, in October 1999 and in this paper we will present the proof as given at that seminar. Even if it is a bit more involved than the proof from [2] and [3], which we found later and presented at the Millennial conference, May 2000, Urbana-Champaign, we still believe it has some interest, and is worth publishing. We do not have to define the generalized Kloosterman sums

X mh + nk S(D, m, n; c) = e (3) c hk≡D (mod c) 0≤h,k

X mD c S(D, m, n; c) = dS , n; (4) d2 d d|(D,m,c) as we did in [3]. Instead we will use the identity

l 1 X −mp − nq c (nm − D) = S(pq, D; l)e , (5) l l l p,q=1 that relates the classical Kloosterman sums with the Ramanujan sums cl(m). Although probably less important than the identity (4), we have not (yet) found the identity (5) anywhere in the literature, and it might be of some independent interest.

1At the time we were hoping that the main summation formula, Theorem 1 in this paper would have applications on estimating certain exponential sums arising from this problem. We have not returned to that question, but our current belief is that the summation formula is better for estimating certain other exponential sums, not these particular ones. 2For a further discussion of this matter, see the introduction in our Thesis [4]. 3We originally believed that we might have found a new identity in (4). However it turned out that it had previously been stated by Bykovsky-Kuznetsov-Vinogradov [5], and in a less reﬁned form by Heath- Brown [6]. For the full story see our paper [1].

146 2 Kloosterman sums and Ramanujan sums

Kloosterman sums

X mh + nh¯ S(m, n; c) = e (6) c hh¯≡1 (mod c) 0≤h,h

S(m, n; c) = S(n, m; c), (7) S(n, m; c) = S(−n, −m, c), (8) as well as the facts that

S(m, 0; l) = cl(m), (9) and

S(0, 0; c) = φ(c), (10) where

X mh c (m) = e , (11) l l (h,l)=1 0≤h

cl(m) = cl(−m). (12)

We will ﬁrst show some results about the Ramanujan sums.

Lemma 1. Suppose that f is a function on R such that

∞ X x f , k < ∞. (13) k k=1 One then has the identity

∞ X X cl(m) m f(a, d) = f kl, , (14) kl kl ad=m k,l=1 a>0

147 Proof. The condition (13), the estimate |cl(m)| ≤ φ(l), which follows immediately from P the deﬁnition of the Ramanujan sums, and the identity n = d|n φ(d) ensures that the sum on the right hand side in the identity (14) is absolutely convergent. Hence we can change the summation order and Lemma (1) follows from

X X X hm c (m) = e , d d d|k d|k (h,d)=1 0≤h

Kloosterman sums and Ramanujan sums are closely related, as is seen by the elementary identity (9). We will now prove a somewhat more involved identity involving the Kloosterman sums as well as the Ramanujan sums, that we will need for our main argument.

Lemma 2. One has the following identity

l 1 X −mp − nq c (nm − D) = S(pq, D; l)e . l l l p,q=1

Proof. We have by the deﬁnition (11) of the Ramanujan sums, and the symmetry equation (12), that

X −(nm − D)h c (nm − D) = e , l l (h,l)=1 0≤h

We will now express e(−nmh/l) through the identity

l l −nmh 1 X −mq X −p(−qh¯ + n) e = e e (17) l l l l q=1 p=1

This identity can be proved as follows. By starting with the right hand side of equation

148 (17) we obtain

l l l l 1 X −mq X −p(−qh¯ + n) X 1 X −mq + pqh¯ − np e e = e , l l l l l q=1 p=1 p=1 q=1 l l X −np1 X (−m + ph¯)q = e e , l l l p=1 q=1 l ( X −np 1, −m + ph¯ ≡ 0 (mod l), = e l p=1 0, otherwise. Since the only contribution will come when −m + ph¯ ≡ 0 (mod l), or in other words p ≡ mh (mod l), this equals −nmh e . l By equations (16) and (17), and changing the summation order, we will now get that

l l X 1 X −mq X −p(−qh¯ + n) Dh c (nm − D) = e e e , l l l l l (h,l)=1 q=1 p=1 0≤h

  l 1 X  X pqh¯ + Dh  −mp − nq =  e e , l  l  l p,q=1 (h,l)=1 0≤h

l 1 X −mp − nq = S(pq, D; l)e . l l p,q=1

Lemma 3. Let l > 0 be an integer and f a continuously diﬀerentiable function on R2, such that |f(x, y)| ≤ (x2 + y2)−1−. ( > 0) One then has ∞ ∞ X 1 X m n f(m, n)c (mn − D) = fˆ , S(mn, D; l), (18) l l l l m,n=−∞ m,n=−∞ where Z ∞ Z ∞ fˆ(t, τ) = e(−τx − ty)f(x, y)dxdy (19) −∞ −∞ is the two dimensional Fourier transform of f, the Ramanujan sums cl(n) are deﬁned by (11) and the Kloosterman sums S(m, n; c) by (6).

149 Proof. We will start with the right hand side of equation (18). We have with m = al+p, n = bl + q that

∞ l ∞ 1 X m n X X p q fˆ , S(mn, D; l) = S((al + p)(bl + q), d; l)f a + , d + . l l l l l m,n=−∞ p,q=1 a,d=−∞

Since by their deﬁnition (6), the Kloosterman sums S(m, n; l) only depend on the residue class of m, n (mod l), we see that this equals

l ∞ X X p q S (pq, D; l) f a + , d + . (20) l l l p,q=1 a,d=−∞

By using the fact that if p q g(x, y) = f x + , y + l l then −ξp − ηq gˆ(ξ, η) = e fˆ(ξ, η), l we can sum over p and q (mod l) and the conditions on f ensure that we can use the two dimensional version of the Poisson summation formula (see e.g Zygmund [9] (13.3)) on the inner sum in (20). We obtain

l ∞ X X −pm − qn S(pq, D; l) e f(m, n). (21) l p,q=1 m,n=−∞

By changing the summation order this equals

∞ l ! X X −pm − qn S(pq, D; l)e f(m, n). (22) l m,n=−∞ p,q=1

By Lemma 2

l X 1 X −mp − nq c (nm − D) = S(pq, D, l)e , l l l (h,l)=1 p,q=1 0≤h

∞ X f(m, n)cl(nm − D). m,n=−∞

150 Lemma 4. Let f be a continuously diﬀerentiable test function on all real 2 × 2 matrices with determinant D, such that x ∗ 2 2 −1− f ≤ C() |t|(x + y ) . ( > 0) (23) t y One then has the following identity

∞ ∞ X a b X X f = kS(mn, D, l)fˆ (m, n), c d kl ad−bc=D m,n=−∞ k,l=1 c>0 where

Z ∞ Z ∞ ˆ xt ∗ ft(a, b) = e(−ax − by)f dxdy. −∞ −∞ t yt Proof. We have that

∞ X a b X X a b f = f , c d c d ad−bc=D b,c=−∞ ad=bc+D c>0 by expressing the inner sum as a sum of Ramanujan sums with Lemma 1, this equals

∞ ∞ ad−D X X cl(ad − D) a f kl , kl kl d a,d=−∞ k,l=1

By changing summation order and using the notation

ξ ∗ g (ξ, ν) = f , (24) t t ν we see that this equals

∞ ∞ X X cl(ad − D) g (a, d). kl kl k,l=1 a,d=−∞

We now use Lemma 3 on the inner sum and we obtain

∞ ∞ X 1 X m n gˆ , S(mn, D, l), (25) kl2 t kl kl k,l=1 m,n=−∞ where

Z ∞ Z ∞ ξ ∗ gˆt(µ, τ) = e(−µξ − τν)f dξdν −∞ −∞ t ν

151 With the substitutions

ξ = tx, and ν = ty (26) we obtain that a b gˆ , = t2fˆ(a, b), t t t t where Z ∞ Z ∞ ˆ 2 xt ∗ ft(a, b) = t e(−ax − by)f dxdy −∞ −∞ t yt is the Fourier transform of

xt ∗ f (x, y) = f . t t xy

Hence equation (25) equals

∞ ∞ X 1 X (kl)2fˆ(m, n)S(mn, D, l) kl2 t k,l=1 m,n=−∞ and the lemma follows from 1/(kl2) · (kl)2 = k and by changing the summation order.

3 Spectral theory

We have now expressed the sum in terms of Kloosterman sums and we are ready to prove our main summation formula by means of the Kuznetsov summation formula: Lemma 5. (Kuznetsov summation formula) Suppose that Φ ∈ C3(0, ∞) is a function on R+, and that for some δ > 0 and v = 0, 1, 2, 3 it satisﬁes t−v Φ(v)(t) . (27) 3 +δ t 2 + t−δ Then ∞ ∞ Z ∞ X X 1 Φm,n(r)σ2ir(|m|)σ2ir(|n|)dr S(m, n; c)Φ(c) = αjtj(m)tj(n)Φm,n(κj) + π ir 2 c=1 j=1 −∞ |mn| |ζ(1 + 2ir)| ∞ θ(k) X X 1 + tj,k(m)αj,ktj,k(n)Φm,n 2 − ki , k=1 j=1 where Z ∞ mn Φm,n(r) = Mr 2 Φ(t)dt, 0 t

152 and Mr is deﬁned by

( iπ √ √ 2 sinh πr (J2ri(4π x) − J−2ri(4π x)) , x ≥ 0, Mr(x) = √ (28) 2 cosh πrK2ri(4π −x), x < 0.

K2ri(x) and J2ri(x) are the K-Bessel and J-Bessel-functions respectively, ρj(n) are the Fourier-coeﬃcients

∞ X Ψj(x + iy) = ρj(n)e(nx)Kiκj (|2πny|) (29) n=−∞

∞ of an orthonormalized basis {Ψj}j=1 of Maass wave forms, and

2 |ρj(1)| ρj(n) αj = and tj(n) = , (30) cosh πκj ρj(1)

θ(k) are the eigenvalues for the Hecke-operators Tn. Similarly {Ψj,k}j=1 is an orthonormal basis of holomorphic cusp forms of weight k,

∞ X ρj,k(n) Ψ (z) = ρ (n)e(nz), t (n) = , (31) j,k j,k j,k ρ (1) n=1 j,k and

1−4k −2k−1 2 a(k) = 2 π Γ(2k), αj,k = a(k)|ρj,k(1)| . (32)

Here the automorphic forms Ψj(z) and Ψj,k(z) are also chosen to be eigenfunctions of the Hecke operators. (For more about the notation see Motohashi [8]).

Proof. After the change of variables ! 1 p|mn| Φ(t) = φ 4π (33) t t it reduces to the classical Kuznetsov summation formula.(This version of the formula is taken from Motohashi [8], Theorem 2.3 and Theorem 2.5). The same formula appears in [2], except that we now choose to use the notation αj, αj,k, tj(n), tj,k(n). We are now ready to prove our main theorem

Theorem 1. Let f be an even function on {A ∈ GL(2, R) : det A = D}, where D is a nonzero integer, fulﬁlling the conditions

n m v −v ∂ ∂ ∂ xt ∗ t n, m, v = 0, 1, 2, 3, f , (34) ∂xn ∂ym ∂tv t yt (t2+δ + t−1−δ)(x2 + y2 + 1)1+δ t > 0,

153 for some δ > 0, where ∗ denote the number so that the determinant of the matrix is D. One then has

∞ ∞ X a b X X X Z ∞ cn ∗ cx ∗ f = c (n) f + f dx c d c/d c cx c cn ad−bc=D n=−∞ c=1 d|(D,c) −∞ c>0 ∞ X X c Z ∞ Z ∞ cx ∗ − φ f dxdy d c cy c=1 d|(D,c) −∞ −∞ ∞ X X + αjtj(D)tj(m)tj(n)F (κj; D, m, n) j=1 m,n6=0 Z ∞ X 1 σ2ir(|D|)σ2ir(|m|)σ2ir(|n|)F (r; D, m, n)dr + π ir 2 m,n6=0 −∞ |Dmn| |ζ(1 + 2ir)| ∞ θ(k) X X X 1 + αj,ktj,k(D)tj,k(m)tj,k(n)F 2 − k i; D, m, n , k=1 j=1 m,n6=0 where Z ∞ Z ∞ Z ∞ Dmn tx ∗ F (r; D, m, n) = e(mx + ny)Mr 2 f dxdydt, (35) 0 −∞ −∞ t t ty

Mr(x) is deﬁned by (28), and αj, αj,k, tj(n), tj,k(m) are deﬁned as in Lemma 1, φ(c) denote the Euler phi function and cc(n) denote the Ramanujan sums(see (11)). Proof. By Lemma 4 we have that

∞ ∞ ∞ X a b X X X f = S(mn, D, l)kfˆ (m, n), (36) c d kl ad−bc=D m,n=−∞ k=1 l=1 c>0 where

Z ∞ Z ∞ xt ∗ ft(ξ, ν) = e(−ξx − νy)f dxdy. −∞ −∞ t yt ˆ By Lemma 7 in [3] we have that Φ(t) = ft(n, m) fulﬁlls the test function condition (27). Hence we can apply the Kuznetsov summation formula, Lemma 5 on the inner sum with ˆ Φ(l) = kfkl(m, n). We obtain

Z ∞ Z ∞ Z ∞ mn ktx ∗ Φm,n(r) = k Mr 2 e(−mnx − Dy)f dtdxdy, −∞ −∞ 0 t kt kty which with the substitution τ = tk equals

Z ∞ Z ∞ Z ∞ mnk2 τx ∗ = Mr 2 e(−mnx − Dy)f dτdxdy, −∞ −∞ 0 τ τ τy

154 which by deﬁnition of F (r; m, n, D) equals

= F (r; mn, D, k2). The contribution for ﬁxed m, n, k 6= 0 coming from the inner sum in (36) thus equals

∞ X 2 tj(D)tj(mn)F (κj; D, mn, k ) j=1 1 Z ∞ σ (|D|)σ (|mn|)F (r; D, mn, k2)dr + 2ir 2ir ir 2 π −∞ |Dmn| |ζ(1 + 2ir)| ∞ θ(k) X X 1 2 + αj,ktj,k(D)tj,k(mn)F 2 − k i; D, mn, k . (37) k=1 j=1 From the estimate for the Fourier coeﬃcients of cusp forms

X 2 2 √ αj|tj(m)| K + d3(m) log(2m) m, (38)

K/2<κj ≤K θ(k) X 2 √ αj,k|tj,k(m)| kd3(m) m log(2m), (39) j=1 valid uniformly for k, K, m ≥ 1 (see Motohashi [8] Lemma 2.4, and equation (2.2.10)), and the fact that

− 9 3 − 5 |F (r; D, m, n)| |mn| 4 |D| 4 (1 + |r|) 2 , −2 7 1 − 3 F 2 + k i; D, m, n |mn| |D|k , when f satisﬁes the test function condition of the theorem, equation(34) (see Andersson [3], Lemma 7 and Lemma 8), we see that when we sum over m, n, k in (37) the summation and summation-integration is absolutely convergent, and hence we can change the summation order. To obtain the spectral part of the theorem it is now suﬃcient to prove the identities

∞ ∞ ∞ X X 2 X tj(mn)F (κj; D, mn, k ) = tj(m)tj(n)F (κj; D, m, n) k=1 m,n=−∞ m,n=−∞ m,n6=0 m,n6=0 and the corresponding identities for the holomorphic cusp forms and Eisenstein series. They follow from the identities X mn t (m)t (n) = t , (40) j j j d2 d|(m,n) as well as the corresponding identities

−ir σ2ir(m) σ2ir(n) X mn mn X mn = σ , t (m)t (n) = t , mir nir 2ir d2 d2 j,k j,k j,k d2 d|(m,n) d|(m,n) (41)

155 for the divisor functions and the Fourier coeﬃcients for holomorphic modular forms (see Motohashi [8], formula (3.1.4), (3.1.14)). It remains to consider the case when at least one of m, n are zero. We then get the contribution

∞ ∞ ∞ ∞ ∞ X X X ˆ ˆ X X ˆ k S(0, D, l) + fkl(0, n) + fkl(n, 0) − k S(0, D, l)fkl(0, 0). n=−∞ k=1 l=1 k=1 l=1 Using the one dimensional version of the Poisson summation formula again as we did in [3] on the ﬁrst term, and using equation (9) to express the Kloosterman sums in terms of the Ramanujan sums we obtain the ﬁrst two terms in the theorem.

4 Some concluding remarks

The Bykovsky-Kuznetsov-Vinogradov identity, equation (4) can in fact be used to prove that the identity

∞ ∞ X a b X X f = kS(mn, D, l)fˆ (m, n), c d kl ad−bc=D m,n=−∞ k,l=1 c>0 of Lemma 4, in fact equals

∞ ∞ X X ˆ = S(D, m, n; c)fc(m, n). m,n=−∞ c=1

This appears as equation (15) in [3]. Hence the two methods of proofs are clearly related. It is interesting4 to see that whatever method we use in the proof, the Hecke operators are involved in one way or another. In [1] we used the Hecke operators, and the identities (40) and (41) are clearly a direct consequence of X mn T (m)T (n) = T , (42) d2 d|(m,n)

(see Motohashi [8] equation (3.1.4)). As we wrote in [3] a more direct proof of the summation formula over integer matrices of determinant D using the corresponding formula for the full modular group (from [2]) and Hecke operators would be interesting.

References

[1] J. Andersson. A note on some Kloosterman sum identities, 2006.

[2] A summation formula on the full modular group, 2006.

[3] A summation formula over integer matrices, 2006.

4By the nature of the problem it is not surprising though.

156 [4] Summation formulae and zeta-functions, Thesis.

[5] V. Bykovsky, N. Kuznetsov, and A. Vinogradov. Generalized summation formula for inhomogeneous convolution. In Automorphic functions and their applications (Khabarovsk, 1988), pages 18–63. Acad. Sci. USSR Inst. Appl. Math., Khabarovsk, 1990.

[6] D. R. Heath-Brown. The fourth power moment of the Riemann zeta function. Proc. London Math. Soc. (3), 38(3):385–422, 1979.

[7] Y. Motohashi. An explicit formula for the fourth power mean of the Riemann zeta- function. Acta Math., 170(2):181–220, 1993.

[8] Spectral theory of the Riemann zeta-function. Cambridge University Press, Cambridge, 1997.

[9] A. Zygmund. Trigonometric series: Vols. I, II. Cambridge University Press, London, 1968.

157

The summation formula on the modular group implies the Kuznetsov summation formula

Johan Andersson∗

Abstract In a previous paper [1] we showed that the Kuznetsov summation formula implies a new type of summation formula on the full modular group PSL(2, Z). For test functions on PSL(2, R) we obtained an expansion for

X a b f c d “ ” a b ∈PSL(2, ) c d Z

in terms of spectral objects such as Fourier coeﬃcients for Maass wave forms, holomorphic cusp forms and Eisenstein series. In this paper we shall show the converse. Our general summation formula with the choice of a suitable test function implies the Kuznetsov summation formula

Contents

1 Introduction 159

2 Our main summation formula 160

3 A smooth deﬁnition of the Kloosterman sum 161

4 The Kuznetsov summation formula 163

5 Some remarks 166

1 Introduction

The Kloosterman sum

X mh + nh¯ S(m, n; c) = e (1) c hh¯≡1 (mod c) 0≤h,h

∗Department of Mathematics, Stockholm University, [email protected]

159 is an important exponential sum, with wide ranging applications in number theory. Special cases of the Kloosterman sums include

S(m, 0; c) = cc(m), (2) S(0, 0; c) = φ(c), (3) where

X mh c (m) = e (4) c c (h,l)=c 0≤h

∞ ∞ Z ∞ X X 1 Φm,n(r)σ2ir(|m|)σ2ir(|n|)dr S(m, n; c)Φ(c) = ρj(m)ρj(n)Φm,n(κj) + π ir 2 c=1 j=1 −∞ |nm| |ζ(1 + 2ir)| ∞ θ(k) X X 1 + ρj,k(m)ρj,k(n)Φm,n 2 − ki , k=1 j=1 in terms of spectral objects coming from the full modular group. Here we have

∞ ∞ X X n Ψj(x + iy) = ρj(n)e(nx)Kiκj (|2πny|), and Ψj,k(z) = ρj,k(n)z , (5) n=−∞ n=1

∞ θ(k) where {Ψj}j=1 is an orthonormalized basis of Maass wave forms, and {Ψj,k}j=1 is an orthonormal basis of holomorphic cusp forms of weight k (see Andersson [1]). In [1] we used the Kuznetsov summation formula as the principal tool to prove a a new type of general summation formula on the full modular group. In this paper we will show the converse. Our summation formula on the full modular group will in a natural way imply the Kuznetsov summation formula. It is sorts of a circular proof, since we originally used the Kuznetsov summation formula to prove our main summation formula. However in our forthcoming paper [2] we will present an independent proof of our summation formula, and then the approach in this paper will become a natural way to prove the Kuznetsov summation formula.

2 Our main summation formula

We start by stating our summation formula which is Theorem 1 in [1]:

Theorem 1. Let f be a function on PSL(2, R), fulﬁlling the condition n m v −v ∂ ∂ ∂ ξt ∗ t n, m, v = 0, 1, 2, 3, f , (6) ∂ξn ∂ηm ∂tv t ηt (t2+δ + t−1−δ)(ξ2 + η2 + 1)1+δ t > 0.

160 Let ∗ denote the number making the determinant of the matrix to be 1. We have

Mr is deﬁned by

( iπ √ √ 2 sinh πr (J2ri(4π x) − J−2ri(4π x)) , x ≥ 0, Mr(x) = √ (8) 2 cosh πrK2ri(4π −x), x < 0.

K2ri(x) and J2ri(x) are the K-Bessel and J-Bessel-functions respectively, ρj(n) and ρj,k(n), defined by (5) are Fourier coefficients of Maass wave forms and holomorphic modular forms respectively, cc(n) are the Ramanujan sums defined by (4) and φ is the Euler phi function.

3 A smooth deﬁnition of the Kloosterman sum

The Kloosterman sums are deﬁned by

X mh + nh¯ S(m, n; c) = e . (9) c hh¯≡1 (mod c) 0≤h,h

Let m, n be ﬁxed non zero integers. It is clear that if we deﬁne ( a b Φ(c)e ma+nd , 0 ≤ a < c, 0 ≤ d < c, f = c (10) c d 0, otherwise. then we have that

∞ X a b X f = S(m, n; c)Φ(c). (11) c d ad−bc=1 c=1 c>0

161 We can not apply our main summation formula though, since the function f does not fulﬁll the conditions of the test function class (6). In fact f is neither diﬀerentiable nor continuous. We will now show how we can solve this problem and see how we can use a partition of unity to express the Kloosterman sums in a smooth way.

Lemma 1. Suppose that Ψ ∈ C∞(R) is a smooth function deﬁned so that ( 1, x ≥ C, Ψ(x) = (12) 0, x ≤ −C,

for some constant C > 0. Then we have that

X ma + nd S(m, n; c) = (Ψ(a + c) − Ψ(a))(Ψ(d + c) − Ψ(d))e . c ad≡1 (mod c) a,d∈Z Proof. We ﬁrst notice that the condition (12) implies

∞ X (Ψ(a + kc + c) − Ψ(a + kc)) = 1. (13) k=−∞

By deﬁning h, h,¯ k, l so that

a = kc + h and d = lc + h,¯ with 0 ≤ h, h¯ < c, we obtain

X ma + nd (Ψ(a + c) − Ψ(a))(Ψ(d + c) − Ψ(d))e = c ad≡1 (mod c) a,d∈Z ∞ X X m(ck + h) + n(cl + h¯) (Ψ(kc+h+c)−Ψ(kc+h))(Ψ(lc+h¯+c)−Ψ(lc+h¯))e . c hh¯≡1 (mod c) k,l=−∞ 0≤h,h

Since e(x) has period 1 we have

m(ck + h) + n(cl + h¯) mh + nh¯ e = e , c c and by changing the summation order, equation (14) equals

∞ ! ∞ ! X X mh + nh¯ (Ψ(kc + h + c) − Ψ(kc + h)) (Ψ(lc + c + h¯) − Ψ(lc + h¯)) e . c k=−∞ l=−∞

162 By equation (13) this equals

X mh + nh¯ e , c hh¯≡1 (mod c) 0≤h,h

S(m, n; c) the Kloosterman sum.

4 The Kuznetsov summation formula

By the following Lemma we will be able to use our summation formula: Lemma 2. Suppose that Ψ fulﬁlls the conditions of Lemma 1 and that that Φ ∈ C(0, ∞) is a continuous function on R+ such that for some δ > 0, one has for positive integers c > 0 that |Φ(c)| c−2−δ. (15)

Also suppose that

a b ma + nd f = (Ψ(a) − Ψ(a − c))(Ψ(d) − Ψ(c − d))Φ(c)e . c d c

Then

∞ X a b X f = Φ(c)S(m, n; c) c d ad−bc=D c=1 c>0 Proof. The condition (15) ensures absolute convergence, and by changing the summation order this follows from Lemma 1. Before stating the Kuznetsov summation formula we will prove yet another lemma that we will need.

Lemma 3. Suppose that Ψ ∈ C∞(R) is a function such that ( 1, x ≥ C, Ψ(x) = (16) 0, x ≤ −C, for some constant C > 0. Then for n being an integer one has that ( Z ∞ 1, n = 0, e(nx)(Ψ(xt) − Ψ(t(x − 1)))dx = −∞ 0, otherwise.

163 Proof. We get by using the substitution ξ = xt that Z ∞ Z ∞ e(nx)(Ψ(xt) − Ψ((x − 1)t))dx = e(nξ/t)(Ψ(ξ) − Ψ(ξ − t))dξ, −∞ −∞ which if n 6= 0 equals

Ψ(ˆ n/t) − e(nt/t)Ψ(ˆ n/t) = (1 − e(nt/t))Ψ(ˆ n/t), = 0, since n is an integer. In case n = 0 we get

Z ∞ Z M e(nx)(Ψ(xt) − Ψ((x − 1)t))dx = lim e(0 · x)(Ψ(xt) − Ψ((x − 1)t))dx −∞ M→∞ −M Z M Z −M = lim Ψ(xt)dx − Ψ(xt)dx , M→∞ M−1 −M−1 = 1 − 0, = 0, by equation (16). We are now ready to prove the Kuznetsov summation formula Theorem 2. (Kuznetsov summation formula) Suppose that Φ ∈ C3(0, ∞) is a function on R+, which for some δ > 0 and v = 0, 1, 2, 3 fulﬁlls −v (v) t Φ (t) . (17) t2+δ + t−1−δ Then ∞ ∞ Z ∞ X X 1 Φm,n(r)σ2ir(|m|)σ2ir(|n|)dr S(m, n; c)Φ(c) = ρj(m)ρj(n)Φm,n(κj) + π ir 2 c=1 j=1 −∞ |nm| |ζ(1 + 2ir)| ∞ θ(k) X X 1 + ρj,k(m)ρj,k(n)Φm,n 2 − ki , k=1 j=1 where Z ∞ mn Φm,n(r) = Mr 2 Φ(t)dt, 0 t

ρj(n), ρj,k(n) are deﬁned by (5) and Mr is deﬁned by Proof. By Lemma 2 we have that

∞ X X a b Φ(c)S(m, n; c) = f , c d c=1 ad−bc=D c>0

164 where f is defined by a b ma + nd f = (Ψ(a) − Ψ(a − c))(Ψ(d) − Ψ(c − d))Φ(c)e . c d c We have defined Ψ, Φ so that the test function f satisfies condition (6), which means that we can use Theorem 1. We have that Z ∞ Z ∞ cx ∗ f dxdy −∞ −∞ c cy Z ∞ Z ∞ = e(mx + ny)Φ(c)(Ψ(cy) − Ψ(c(y − 1)))(Ψ(cx) − Ψ(c(x − 1)))dxdy, −∞ −∞ Z ∞ Z ∞ =Φ(c) e(mx)(Ψ(cx) − Ψ(c(x − 1)))dx e(ny)(Ψ(cy) − Ψ(c(y − 1)))dy. −∞ −∞ Since m, n 6= 0 we have by Lemma 3 that this vanishes, and it follows that the first term

∞ X Z ∞ Z ∞ cx ∗ φ(c) f dxdy = 0 c cy c=1 −∞ −∞ also vanishes. Similarly by Lemma 3 the integral in the second and third terms can be written as Z ∞ cn ∗ cx ∗ f + f dx −∞ c cx c cn Z ∞ Z ∞ = Φ(c) e(mx)(Ψ(cx) − Ψ(c(x − 1)))dx + e(nx)(Ψ(cx) − Ψ(c(x − 1)))dx , −∞ −∞ = Φ(c) · (0 + 0), = 0, and hence the second and third terms ∞ ∞ X X Z ∞ cn ∗ cx ∗ c (n) f + f dx = 0, c c cx c cn n=−∞ c=1 −∞ also vanish. We now consider the integral transform and we have that Z ∞ Z ∞ Z ∞ µν tx ∗ F (r; µ, ν) = e(µx + νy)Mr 2 f dxdydt, 0 −∞ −∞ t t ty Z ∞ Z ∞ Z ∞ µν = e(µx + νy)Mr 2 Φ(t)(Ψ(xt) − Ψ((x − 1)t))× 0 −∞ −∞ t × (Ψ(yt) − Ψ(t(y − 1)))e(mx + ny)dxdydt.

By changing the order of integration, we see that this equals Z ∞ µν Z ∞ = Mr 2 Φ(t) e((m + µ)x)(Ψ(xt) − Ψ((x − 1)t))dx × 0 t −∞ Z ∞ × e((n + ν)y)(Ψ(yt) − Ψ((y − 1)t))dy dt. −∞

165 The integrals with respect to x and y can be treated with Lemma 3 and we get Z ∞ e((m + µ)x)(Ψ(xt) − Ψ((x − 1)t))dx = δm,−µ, −∞ and Z ∞ e((n + ν)y)(Ψ(yt) − Ψ((y − 1)t))dy = δn,−ν. −∞

We thus ﬁnd that Z ∞ µν F (r; µ, ν; r) = δµ,−mδν,−n Φ(t)Mr 2 dt. 0 t

The formula in Lemma 2 now follows from the fact that

ρj(−µ)ρj(−ν) = ρj(µ)ρj(ν), together with the corresponding identities for Fourier coeﬃcients of holomorphic cusp forms, and divisor functions.

5 Some remarks

Remark 1. The generalized Kuznetsov summation formula of Bykovsky-Kuznetsov- Vino- gradov [4]

X mh + nk S(D, m, n; c) = e . c hk≡D (mod c) 0≤h,k

Remark 2. We have that the test function condition t−v Φ(v)(t) . (v = 0, 1, 2, 3 and some δ > 0) 3 +δ t 2 + t−δ

166 for the Kuznetsov summation formula when we use it as the starting point in our proof of the main summation formula in [1]. When we use the main summation formula to prove the Kuznetsov summation formula we get the condition

−v (v) t Φ (t) . (v = 0, 1, 2, 3 and some δ > 0) t2+δ + t−1−δ for the functions we can use in the test function class. Hence we lose some information. It would be interesting to see if we can sharpen the proof of the Main summation formula so we can get back the same class as we started with.

References

[1] J. Andersson. A summation formula on the full modular group, 2006.

[2] A summation formula on the full modular group II, forthcoming.

[3] A summation formula over integer matrices, 2006.

[5] N. V. Kuznecov. The Petersson conjecture for cusp forms of weight zero and the Linnik conjecture. Sums of Kloosterman sums. Mat. Sb. (N.S.), 111(153)(3):334–383, 479, 1980.

[6] A. Weil. On some exponential sums. Proc. Nat. Acad. Sci. U. S. A., 34:204–207, 1948.

167

The Selberg and Eichler-Selberg trace formulae

Johan Andersson∗

Abstract In this paper we will show how we can use our summation formula from [2] to prove the Selberg and Eichler-Selberg trace formula in a uniﬁed way. Our version of the Selberg trace formula is for the full modular group and with respect to Hecke operators and it also distinguishes between odd and even forms. Related formulae have previously been proved by Selberg [21] (without Hecke operators), Hejhal [13] (with Hecke operators of prime orders), Str¨ombergsson [23] (arbitrary Hecke operators - all forms), Bogomolny-Georgeot-Giannoni-Schmit [3] (for Hecke operators and odd forms), but with other methods.

Contents

1 Introduction 169

2 Preliminaries 170 2.1 The summation formula on integer matrices ...... 170 2.2 A class number formula ...... 170 2.3 Some smoothing lemmas ...... 173 2.4 The Rankin-Selberg zeta function ...... 175

3 The trace formulae 177

4 Appendix 189

1 Introduction

The Selberg Trace Formula [21] is an important result that gives a connection between the spectral theory of the group Γ and the geometry of the Riemann surface Γ\H. In this paper we will show how our summation formula on the full modular group implies the Selberg trace formula for the full modular group as well as the Eichler-Selberg trace formula. We will give an arithmetic form of the formula that involves class numbers instead of lengths of closed geodesics. However, as shown by Sarnak [20] it is possible to translate between the arithmetic and geometric form of the trace formula.

∗Department of Mathematics, Stockholm University, [email protected]

169 2 Preliminaries

2.1 The summation formula on integer matrices We will start by stating our summation formula which is Theorem 1 in [2]:

Theorem 1. Let f be a function on {A ∈ GL(2, R) : det A = D}, where D is a nonzero integer, fulﬁlling the conditions

n m v −v ∂ ∂ ∂ xt ∗ t n, m, v = 0, 1, 2, 3, f , (1) ∂xn ∂ym ∂tv t yt (t2+δ + t−1−δ)(x2 + y2 + 1)1+δ t > 0, for some δ > 0, where ∗ denotes the number which makes the determinant equal to D. One then has the identity

∞ X a b X X c Z ∞ Z ∞ cx ∗ f = − φ f dxdy c d d c cy ad−bc=D c=1 d|(D,c) −∞ −∞ c>0 ∞ ∞ X X X Z ∞ cn ∗ cx ∗ + c (n) f + f dx c/d c cx c cn n=−∞ c=1 d|(D,c) −∞ ∞ X X + αjtj(D)tj(m)tj(n)F (κj; D, m, n) j=1 m,n6=0 Z ∞ X 1 σ2ir(|D|)σ2ir(|m|)σ2ir(|n|)F (r; D, m, n) + dr π ir 2 m,n6=0 −∞ |Dmn| |ζ(1 + 2ir)| ∞ θ(k) X X X 1 + αj,ktj,k(D)tj,k(m)tj,k(n)F 2 − k i; D, m, n , k=1 j=1 m,n6=0 where Z ∞ Z ∞ Z ∞ Dmn tx ∗ F (r; D, m, n) = e(mx + ny)Mr 2 f dxdydt, (2) 0 −∞ −∞ t t ty the function Mr(x) is deﬁned by

( iπ √ √ 2 sinh πr (J2ri(4π x) − J−2ri(4π x)) , x ≥ 0, Mr(x) = √ (3) 2 cosh πrK2ri(4π −x), x < 0, and αj, αj,k, tj(n), tj,k(m) are deﬁned as in Andersson [2] or Motohashi [18], φ(c) denotes the Euler phi function and the cc(n) denote the Ramanujan sums.

2.2 A class number formula We will ﬁrst prove an asymptotic formula involving the Hurwitz class number H(∆), that counts the number of reduced positive deﬁnite quadratic forms (a, b, c) of discriminant ∆ = 4ac − b2.

170 Lemma 1. Let ∆ be a non square integer. One has then that ( 1 X − c 6 H(−∆) 2 log εD, ∆ > 0, lim e N = × × N→∞ N π2 p|∆| π, ∆ < 0, b2−4ac=∆ 0

h(−D) X D H(∆) = f −1 µ(d)c, (∆ = Df 2,D square free) w(−D) d cd|f h(∆) is the narrow class number (see Gauss [10]), and √ x + Dy ε = d d D 2 √ the fundamental unit in the number field O( D). Remark 1. The number H(∆) as defined in Lemma 1 coincides with the Hurwitz class number for ∆ > 0 (see Zagier [25] page 132) and is a natural generalization for ∆ < 0, where the Hurwitz class number is usually defined to be zero. Proof. Following Zagier [25], equation (6), we introduce the zeta function

X X 1 ζ(s, ∆) = . (Re(s) > 1) φ(m, n)s 2 φ mod Γ (m,n)∈Z / Aut(Φ) |φ|=∆ φ(m,n)>0 where the summation is over quadratic forms φ of discriminant ∆. When ∆ = Df 2, and D 6= 1 is square free we have ([25] Proposition 3.1 (iii))  0, ∆ ≡ 2, 3 (mod 4),  ζ(s, ∆) = ζ(s)ζ(2s − 1), ∆ = 0,  P D −s f ζ(s)LD(s) d|f µ(d) d d σ1−2s d , ∆ ≡ 0, 1 (mod 4), ∆ 6= 0, where

∞ X D L (s) = n−s (Re(s) > 1) D n n=1

D is the Dirichlet L-series and n denotes the Kronecker symbol. We have that the product

√ ζ(s)LD(s) = ζQ( D)

171 √ is also the Dedekind zeta-function of the quadratic number ﬁeld Q( D) (see e.g the discussion in Cohn [4] Chapter X, section 7). The residue of the Dedekind zeta-function at s = 1 can be expressed (see e.g. Fr¨ohlich-Taylor [9], Theorem 61) by

s+t t 2 π RK hK lim ζK (s)(s − 1) = , s→1 p wK |dK |

where hK is the class number of K, RK is the regulator of K, s (resp 2t) is the number of real (resp. imaginary) embeddings of K, wK is the number of roots of unity in K, and√dK denotes the discriminant of K. In the case of a real quadratic number ﬁeld K = Q( D) we have that

s = 1, t = 0,RK = log(εD), hK = h(−D), wK = 1,

√ where εD is√ the fundamental unit in Q( D). In the case of an imaginary quadratic number ﬁeld Q = −D we have that

s = 0, t = 2,RK = 1, hK = h(D), wK = w(D), and we get when ∆ ≡ 0, 1 (mod D) that

( 2π h(−D) X D −1 f w(D) , D > 0, lim ζ(s, ∆)(1 − s) = µ(d) d σ−1 × (4) s→1 p d d log ε , D < 0. |d| d|f D

The factor X D f µ(d) d−1σ d −1 d d|f can be written as X D f −1 µ(d)c. d cd|f

With the notation

2 a∆(n) = #{x : x ≡ ∆ (mod n)}, (5) we have that ([25] Proposition 3.1 (i))

∞ X −s ζ(s, ∆) = ζ(2s) a∆(n)n . (Re(s) > 1) (6) n=1

172 From equations (4), (6), the fact that ζ(2) = π2/6, and the general fact that

∞ ∞ X −s C 1 X − n lim ann = =⇒ lim ane N = C, s→1+ s − 1 N→∞ N n=1 n=1 we get that

∞ ( 1 X − n 6 H(−∆) 2 log εD, ∆ > 0, lim a∆(n)e N = × N→∞ N π2 p n=1 |∆| π, ∆ < 0.

The lemma now follows from noticing that by the deﬁnition (5), and identifying n = c and b = x, we obtain

∞ X − n X − c a∆(n)e N = e N . n=1 b2−4ac=∆ 0

2.3 Some smoothing lemmas In this section we will prove some auxiliary results that we will need later.

Lemma 2. Suppose that Ψ ∈ C∞(R) is a function such that ( 1, x ≥ C, Ψ(x) = (7) 0, x ≤ −C,

for some constant C > 0. Let n be an integer and take t > 0. Then ( Z ∞ 1, n = 0, e(nx)(Ψ(xt) − Ψ(t(x − 1)))dx = −∞ 0, otherwise.

Proof. We get by using the substitution ξ = xt that Z ∞ Z ∞ e(nx)(Ψ(xt) − Ψ((x − 1)t))dx = e(nξ/t)(Ψ(ξ) − Ψ(ξ − t))dξ, −∞ −∞ which, if n 6= 0, equals

Ψ(ˆ n/t) − e(nt/t)Ψ(ˆ n/t) = (1 − e(nt/t))Ψ(ˆ n/t), and since n is an integer this is equal to

(1 − 1)Ψ(ˆ n/t) = 0.

173 In case n = 0 we get

Z ∞ Z M e(nx)(Ψ(xt) − Ψ((x − 1)t))dx = lim e(0 · x)(Ψ(xt) − Ψ((x − 1)t))dx −∞ M→∞ −M Z M Z −M = lim Ψ(xt)dx − Ψ(xt)dx , M→∞ M−1 −M−1 which by equation (7) equals

= 1 − 0 = 1.

Lemma 1 is not quite in a form that we can apply later, so we will now prove a variation of it.

∞ Lemma 3. Let ∆ be a non square integer. Suppose that θN ∈ C0 (R) is a sequence of test functions converging to the unit step function, with the further properties

0 ≤θN (x) ≤ 1, (x ∈ R) (8) ( 1, x ≥ 0, θN (x) = (9) 0, x ≤ −1/N 2.

Then that ( 1 X b b − c 6 H(−∆) 2 log εD ∆ > 0 lim θN − θN − 1 e N = × N→∞ N c c π2 p|∆| π ∆ < 0 b2−4ac=∆ where H(∆) is deﬁned as in Lemma 1.

Proof. This follows from Lemma 1 and X − c X b b − c e N = θ − θ − 1 e N + O(1). (10) N c N c b2−4ac=∆ b2−4ac=∆ 00

This can be seen as follows: If we choose not to exclude the case when b = c and c = 0 from the sum, the diﬀerence between the diﬀerent sides of the equality in (10) will be of order

∞ X ce−c/N = ON 2e−N . c=N 2 For the remaining case b = 0 or b = c there will be at most 2d(∆) diﬀerent choices of c which is bounded.

174 2.4 The Rankin-Selberg zeta function Before we proceed with our main theorems we will state some facts about the Rankin- Selberg zeta functions. We will consider the following versions.

1. The Rankin-Selberg zeta function

∞ X 2 −s Zj(s) = αj |tj(n)| n (Re(s) > 1) n=1 associated with Maass wave forms;

2. The Rankin-Selberg zeta function

∞ X 2 −s Zj,k(s) = αj,k |tj,k(n)| n (Re(s) > 1) n=1 associated with holomorphic cusp forms;

3. The Rankin-Selberg zeta function

∞ X 2 −s Z(r; s) = |σ2ir(n)| n (Re(s) > 1) n=1 associated with the Eisenstein series.

The Rankin-Selberg zeta function associated with the Eisenstein series is easiest to treat since the Ramanujan identity (see [11], theorem 305)

∞ X ζ(s)ζ(s − α)ζ(s − β)ζ(s − α − β) σ (n)σ (n)n−s = , Re(s−α),Re(s−β), , α β ζ(2s − α − β) Re(s−α−β),Re(s)>1 n=1 implies that

1 ζ(2ir + s)ζ(s − 2ir)ζ2(s) Z(s; r) = . (11) |ζ(1 + 2ir)|2 ζ(2s)

After logarithmic diﬀerentiation this implies the following result.

Lemma 4. The function Z(r; s) has the following Laurent expansion at s = 1:

c c (r) Z(s; r) = −2 + −1 + O(1), (s − 1)2 s − 1 where

6 ζ0(1 + 2ir) ζ0(2) c = , and c (r0) = Re − . −2 π2 −1 ζ(1 + 2ir) ζ(2)

175 The corresponding result for the Rankin-Selberg zeta functions of Maass wave forms can be found in Motohashi [18] Lemma 3.5, and for the Rankin-Selberg zeta functions of holomorphic cusp form in Rankin [19] and Selberg [22]:

Lemma 5. The functions Zj(s) and Zj,k(s) are analytic for Re(s) > 0 with the exception of a simple pole at s = 1 with residue 12/π2. Equation (11) together with the functional equation for the Riemann zeta function also implies that Z∗(r; s) = Z∗(r; 1 − s), (12) where

∗ 2 −s 2 1 1 Z (r; s) = (2π ) ζ(2s)Γ(s)Γ(s/2) Γ( 2 (s + ri))Γ( 2 (s − ri))Z(r; s), The corresponding functional equation can also be proved for the Rankin-Selberg zeta function ∗ ∗ Zj (s) = Zj (1 − s), (13) where

∗ 2 −s 2 1 1 Zj (s) = (2π ) ζ(2s)Γ(s)Γ(s/2) Γ( 2 (s + κji))Γ( 2 (s − κji))Zj(s), corresponding to Maass wave forms (Lemma 3.5 in Motohashi [18]), as well as the Rankin- Selberg zeta function ∗ ∗ Zj,k(s) = Zj,k(1 − s), (14) where

∗ 2 −s 2 1 1 1 1 Zj,k(s) = (2π ) ζ(2s)Γ(s)Γ(s/2) Γ( 2 (s + 2 − k))Γ( 2 (s − 2 + k))Zj,k(s), associated with holomorphic cusp forms (Rankin [19] or Selberg [22]). Lemma 6. Let > 0 be given. There exist a constant 1/2 < c < 1 such that |Zj(c + it)| ≤ (κj(1 + |t|)) , |Zj,k(c + it)| ≤ (k(1 + |t|)) , |Z(r; c + it)| ≤ ((1 + |r|)(1 + |t|)). Proof. This follows from the functional equations (12), (13) and (14), the Phragm´en- Lindel¨of principle and the fact that x X 2 |tj(n)| ≤ xκj, n=1 (see Iwaniec [15] Theorem 8.3) together with The Ramanujan-Petersson conjecture

|tj,k(n)| ≤ d(n), which was proved for holomorphic cusp forms by Deligne [5].1

1To refer to Deligne is really overkill and not necessary since a weaker version is suﬃcient.

176 Remark 2. Motohashi and Jutila [16] have recently proved sharp estimates on the corresponding Hecke L-functions Hj(1/2+it) and Hj,k(1/2+it) in both κj and t simultaneously (similarly to Lemma 6). Even though it is clearly an interesting problem to ﬁnd sharp estimates (in particular sub convexity estimates) for the Rankin-Selberg zeta-functions in both κj and t, for our purposes the weak estimates of Lemma 6 are suﬃcient.

3 The trace formulae

We are now ready to prove the trace formula

∞ Theorem 2. Suppose that g ∈ C0 (R) is a smooth function with compact support, such that d2 g = 0 whenever d2 − 4D is a square. (15) 4|D| One then has the identity

∞ 2 ( 2 ) 2 X H(4D − d ) 2 log εd2−4D 4D − d < 0 d × × g = p 2 π 4D − d2 > 0 4|D| d=1 |4D − d | Z ∞ 0 ∞ θ(k) π −ir ζ (1 + 2ir) X X σ (|D|)|D| Re g˜(r)dr + t (D)˜g( 1 − k)i 6 2ir ζ(1 + 2ir) j,k 2 −∞ k=1 j=1 ∞ X p Z ∞ g(y)dy 1 X Z ∞ g(x2) dx + tj(D)˜g(κj) + σ−2(|D|) |D| √ − , 0 y 2 −∞ p p j=1 d|D 2x − ( |D|/d + d/ |D|) where ε∆,H(∆), are deﬁned as in Lemma 1,

 √ 2ri R ∞ 2 2 −1/2 2  1 Re x + x − 1 (x − 1) g(x )dx, D > 0, g˜(r) = √ 2ri R ∞ 2 2 −1/2 2  0 Re 1 + 1 + x (1 + x ) g(x )dx, D < 0, and ( (−1)k R 1 cos((2k − 1) arcsin x)(1 − x2)−1/2g(x2)dx, D > 0, g˜ 1 − ki = 0 2 0, D < 0.

Proof. We have that b d b2 − 4ac d2 2 − 2 a det b d = − . c 2 + 2 4 4

Now −∆ = 4ac − b2 is the discriminant of the quadratic form. We see that if

( 1 (x−w)2 z 2 − N x y π N g 4|D| e , 0 ≤ x + w ≤ z, gN = × z w 6 0, otherwise,

177 then we get with b/2 − d/2 = e and b/2 + d/2 = f that (Since the function g has compact support there are only ﬁnitely many non zero elements in d. Hence the limit and summation can be interchanged.):

∞ 2 2 X e a X 1 π d X − c lim gN = lim g e N . (16) N→∞ c f N→∞ N 6 4|D| ef−ac=−D d=−∞ b,a,c c>0 b2−4ac−d2=−4D 0≤b≤c

Notice that by condition (15), g(d2/4|D|) vanishes when −4D + d2 is square, and by using Lemma 1 on the inner sum, we ﬁnd that this equals

∞ 2 ( 2 ) 2 X H(4D − d ) 2 log εd2−4D 4D − d < 0 d × × g . p 2 π 4D − d2 > 0 4|D| d=−∞ |4D − d |

The function gN tends to 0 nicely as the argument tends to inﬁnity. However it is not smooth at x + w = z. That is the reason why we proved the smoothed class number formula Lemma 3. By choosing instead

2 2 x y π 1 (x − w) − z f = g e N (θ (x + w) − θ (x + w − z)), N z w 6 N 4|D| N N where θN is a function converging to the unit step function that satisﬁes equations (8) and (9), we have that fN fulﬁll equation (1), the condition of Theorem 1. By Lemma 3 we have that (Again since function g has compact support, the summation and limit can be interchanged):

∞ 2 ( 2 ) 2 X H(4D − d ) 2 log εd2−4D, 4D − d < 0 d X a b × × g = lim fN , p 2 π, 4D − d2 > 0 4|D| N→∞ c d d=−∞ |4D − d | ad−bc=−D c>0 (17) and we can apply Theorem 1. We get that the ﬁrst term is

∞ X X c Z ∞ Z ∞ cx ∗ − φ f dxdy = d N c cy c=1 d|(D,c) −∞ −∞

∞ π2 X X c e−c/N Z ∞ Z ∞ c2(x − y)2 − φ g (θ (c(x + y)) − θ (c(x + y − 1)))dxdy. 6 d N 4|D| N N c=1 d|(D,c) −∞ −∞

With the substitution

cx − cy p|D|dµdν ν = , µ = y + x, we have dxdy = , 2p|D| c

178 the integral splits and we obtain

p ∞ ∞ ∞ π2 |D| X X c e−c/N Z Z − φ g(ν2)dν (θ (µ) − θ (µ − 1))dµ. 6 d cN N N c=1 d|(D,c) −∞ −∞

By Lemma 2 we get that the ﬁrst term equals

p ∞ ∞ π2 |D| X X c e−c/N Z − φ gν2dν. (18) 6 d cN c=1 d|(D,c) −∞

The second term is

∞ ∞ X X X Z ∞ cn ∗ cx ∗ c (n) f + f dx = c/d N c cx N c cn c=1 n=−∞ d|(c,D) −∞ 2 ∞ ∞ Z ∞ 2 2 π X X X 1 − c c (x − n) c (n) e N g (θ (c(x + n)) − θ (c(x + n − 1)))dx. 3 c/d N 4|D| N N c=1 n=−∞ d|(c,D) −∞

Since g has ﬁnite support this can be estimated by

2 ∞ ∞ Z ∞ 2 2 π X X X 1 − c c x c (n) e N g (θ (cx) − θ (c(x − 1)))dx 3 c/d N 4|D| N N c=1 n=−∞ d|(c,D) −∞     p ∞ Z ∞ 2 2 |D| X −1 X X − c 1 − c c (x − n) + O c  cc/d(n) e N e N g dx. N N 4|D| c=1 1≤|n|≤M d|(c,D) −∞ With the substitution ν = cx/2p|D| the error term can be estimated by O p|D|/N , and by ﬁrst using (2) and then the same substitution ν = cx/2p|D| on the main term we obtain

2p ∞ ∞ − c c p ! 2π |D| X X X c e N Z |D| φ g(ν2)dν + O . 3 d cN N c=1 n=−∞ d|(c,D) 0

Since g has compact support we see that if we integrate to inﬁnity instead of c we introduce an error O p|D|/N . Also since g(x2) is even we can write the second term as

2p ∞ ∞ − c ∞ p ! π |D| X X X c e N Z |D| φ g(ν2)dν + O . (19) 3 d cN N c=1 n=−∞ d|(c,D) −∞

We see that for ﬁxed D, the sum of the ﬁrst two terms equations (18) and (19) will be

2p ∞ ∞ − c ∞ p ! π |D| X X X c e N Z |D| φ g(ν2)dν + O . 6 d cN N c=1 n=−∞ d|(c,D) −∞

179 By letting k = c/d this equals

p ∞ ∞ π2 |D| X X φ(k)e−dk/N Z gν2dν. (20) 6 dkN d|D k=1 −∞

P∞ −s By the fact that ζ(s)/ζ(2s) = n=1 φ(n)n we can prove that

∞ X 6 √ φ(n)e−n/M = M + O M , π2 n=1 and by this result we ﬁnd that eq. (20) equals

Z ∞ r ! p 2 |D| σ−2(|D|) |D| g ν dν + O . −∞ N

After the substitution y = ν2 this equals r ! p Z ∞ g(y)dy |D| σ−2(|D|) |D| √ + O . (21) 0 y N

We will now consider the integral transform FN (r; −D, m, n). We get π2 Z ∞ Z ∞ Z ∞ −Dmn tx ∗ FN (r; −D, m, n) = e(mx + ny)Mr 2 fN dxdydt 6 0 −∞ −∞ t t ty 2 ∞ ∞ ∞ 2 2 π Z Z Z −Dmn 1 t t (x − y) − N = e(mx + ny)Mr 2 e g × 6 0 −∞ −∞ t N 4|D|

× (θN (t(x + y)) − θN (t(x + y − 1)))dxdydt.

By using the substitution

µ = x + y, ν = x − y, we have

1 1 1 dxdy = 2 dµdν, x = 2 (µ + ν), y = 2 (µ − ν), and we get that

2 Z ∞ Z ∞ Z ∞ π 1 1 −Dmn FN (r; −D, m, n) = e 2 (m + n)µ + 2 (m − n)ν Mr 2 × 12 0 −∞ −∞ t 2 2 t ν 1 − t × g e N (θ (µt) − θ (µ(t − 1)))dµdνdt. 4|D| N N N We ﬁrst consider the integral with respect to ν. We have

2 Z ∞ 2 2 π 1 t ν e 2 (m − n)ν g dν, 12 −∞ 4|D|

180 which with the change of variables

tν p|D| ξ = , dν = dξ, p|D| t becomes ! ! π2 p|D| Z ∞ (m − n)p|D|ξ ξ2 π2 p|D| (m − n)p|D| e g dξ = G , 12 t −∞ 2t 4 12 t 2t where Z ∞ x2 G(t) = e(xt)g dx, (22) −∞ 4 is the Fourier transform of g(ξ2/4). In particular we have that G belongs to the Schwartz class since g(x) and hence also g(ξ2/4) belongs to the Schwartz class, and the Fourier transform maps the Schwartz class onto the Schwartz class2. By Lemma 2 we get that Z ∞ 1 (θN (µt) − θN (µ(t − 1)))e 2 (m + n)µ dµ = δm,−n. −∞ We thus obtain

2 ∞ p ! π Z (m − n) |D| −Dmn 1 t dt p − N FN (r; −D, m, n) = δm,−n |D| G Mr 2 e . 12 0 2t t N t

With the substitution

p|Dn2| dt dx x = , = , t t x we then get √ |Dn2| 2 Z ∞ − π p 2e xN dx FN (r; −D, m, n) = δm,−n |D| G(x)Mr sign(D) · x . 12 0 N x We will ﬁrst consider the contribution coming from a single Maass wave form. The factor 1/2 disappears when we consider the pair (m, n) = (n, −n), and (m, n) = (−n, n) jointly. By further using the identity

2 tj(−D)tj(n)tj(−n) = tj(D)tj (n), we see that the contribution from a single Maass wave form will be √ |D|n 2 ∞ ∞ − π X Z e xN dx t (D) α t2(n) G(x)M sign(D) · x2 p|D| . 6 j j j κj N x n=1 0 2See any text book on Fourier analysis.

181 By using the Rankin-Selberg zeta function

∞ X 2 −s Zj(s) = αj tj (n)n , n=1 together with Mellin transforms and complex integration, we ﬁnd that this equals

√ |D|t 2 Z 2+∞i Z ∞ Z ∞ − π 1 s−1 2e xN p dx tj(D) Zj(s) t G(x)Miκj sign(D) · x |D| dtds. 6 2πi 2−∞i 0 0 N x

By changing the integration in the Mellin transform the inner integral becomes

√ |D|t ∞ − Z e xN dt ts . 0 N t

With the substitution

p|D|t dt dy y = , = , xN t y this equals

s−1 s−1 ! Z ∞ ! x xN s−1 −y x N s−1 p p y e dt = p p Γ(s)x , |D| |D| 0 |D| |D| and the contribution from a cusp form can be expressed as a Mellin-Barnes integral involving the Rankin-Selberg zeta-function

√ |D|n 2 ∞ ∞ − π X Z e xN α t (D) t2(n) G(x)M sign(D) · x2 dx = 6 j j j κj N n=1 0 !s−1 π2 1 Z 2+∞i N Z ∞ t (D) Z (s)Γ(s) xs−1G(x)M sign(D) · x2dxds. (23) j j p κj 6 2πi 2−∞i |D| 0

Since by Lemma 5 the Rankin-Selberg zeta-function is holomorphic for Re(s) > 0 with the exception of a simple pole at s = 1 with residue 12/π2 we can move the integration line to c as given by Lemma 7, and we see that equation (23) equals

Z ∞ 2 2tj(D) G(x)Mr sign(D) · x dx 0 !s−1 π2 1 Z c+∞i N Z ∞ + t (D) · Γ(s) xs−1G(x)M sign(D) · x2dxds. j p κj 6 2πi c−∞i |D| 0

182 Since G belongs to the Schwartz class it follows from Lemma 17 Z ∞ 2 c−1+it 2 (|t| + 1) x G(x)Mr sign(D) · x dx 5/2 , 0 (|r| + 1) the Stirling formula (see e.g. Ivi´c[14], A.34)

c−1/2 − π |t| |Γ(c + it)| |t| e 2 , the growth estimate for the Rankin-Selberg zeta function given by Lemma 7

|Zj(c + it)| ((|t| + 1)(|κj| + 1)) , and the estimate

!c+it−1 !c−1 N N , p|D| p|D| that !s−1 π2 1 Z c+∞i N Z ∞ t (D) Z (s)Γ(s) xs−1G(x)M sign(D) · x2dxds j j p κj 6 2πi c−∞i |D| 0 (1/2

Similarly by the corresponding identities for the Rankin-Selberg zeta-function of holomorphic cusp forms from Lemma 5 and Lemma 6

∞ X 2 −s 12 Zj,k(s) = αj,k tj,k(n)n , lim(s − 1)Zj,k(s) = ,Zj(c + it) |(1 + |t|)k| , s→1 π2 n=1 Lemma 18 Z ∞ c+it−1 2 1 x G(x)Mi(k−1/2) sign(D) · x dx 7/3 , 0 |k| and the Stirling formula, we get that the contribution from a holomorphic cusp form will be  c−1  Z ∞ ! 2 N −7/3 2tj,k(D) G(x)Mi(k−1/2) sign(D) · x dx + Otj,k(D) p k . (25) 0 |D|

183 The contribution coming from the Eisenstein series Z ∞ X 1 σ2ir(|D|)σ2ir(|m|)σ2ir(|n|)F (r; D, m, n)dr π ir 2 m,n6=0 −∞ |Dmn| |ζ(1 + 2ir)| can in a similar way be written as

s−1 Z ∞ Z 2+∞i ! π σ2ir(|D|) 1 N ir Z(r; s)Γ(s) p × 6 −∞ |D| 2πi 2−∞i |D| Z ∞ s−1 2 × x G(x)Mr sign(D) · x dxdsdr, (26) 0 where

∞ 1 X Z(r; s) = σ2 (|n|)n−2ir−s. |ζ(1 + 2ir)|2 2ir n=1 In the same way as in the Maass wave form case we can now move the integration path in (26) to c given by Lemma 7. We will pick up the residue coming from s = 1, and we see that (26) equals

s−1 Z ∞ " " ! π σ2ir(|D|) N ir × Ress=1 Z(r; s)Γ(s) p × 6 −∞ |D| |D|  c−1  Z ∞ # ! # s−1 2 σ2ir(|D|) N −5/2 × x G(x)Mr sign(D) · x dx + O ir p (|r| + 1)  dr. 0 |D| |D| (27)

By Lemma 5

c c (r) Z(s; r) = −2 + −1 + O(1) (s → 1) (s − 1)2 s − 1 and the fact that

!s−1 !! !s−1 d Nx Γ0(s) Nx Nx Γ(s) = + log , ds p|D| Γ(s) p|D| p|D| we obtain that

 s−1  ! Z ∞ N s−1 2 Ress=1 Z(r; s)Γ(s) p x G(x)Mr sign(D) · x dx = |D| 0 Z ∞ 0 2 ζ (1 + 2ir) G(x)Mr sign(D) · x Re + c0 + c1 log x dx 0 ζ(1 + 2ir)

184 where the constants !! ζ0(2) 6 N 6 c0 = − + γ + log , and c1 = ζ(2) π2 p|D| π2 come from the deﬁnition of c−2 and c−1(r) in Lemma 4. By integrating the error term in (27) we see that the contribution from the Eisenstein series equals

Z ∞ 0 Z ∞ π σ2ir(|D|) ζ (1 + 2ir) 2 ir Re G(x)Mr sign(D) · x dxdr 6 −∞ |D| ζ(1 + 2ir) 0 Z ∞ Z ∞ π σ2ir(|D|) 2 + ir (c0 + c1 log x)G(x)Mr sign(D) · x dxdr 6 −∞ |D| 0  !c−1 N + O . (28) p|D|

We will thus treat the integral

Z ∞ Z ∞ π σ2ir(|D|) 2 ir (c0 + c1 log x)G(x)Mr sign(D) · x dxdr. (29) 6 −∞ 0 |D|

By changing the integration order we ﬁnd that it equals

Z ∞ Z ∞ π σ2ir(|D|) 2 G(x)(c0 + c1 log x) ir Mr sign(D) · x drdx. 6 0 −∞ |D|

By using Lemma 14

Z ∞ p !! σ2ir(|D|) 2 X |D| d Mr(ε · x )dr = π cos 2πx + ε , |D|ir d p −∞ d|D |D| we see that equation (29) equals

∞ p !! π2 Z X |D| d G(x)(c + c log x) cos 2πx + sign(D) dx, 6 0 1 d p 0 d|D |D| which with the deﬁnition of G, equation (22), equals

 p !2 π2 X 1 |D| d c log p|D|N g · + sign(D) 6 0 4 d p  d|D |D| ∞ p !! Z X |D| d + G(x) cos 2πx + sign(D) log x dx. (30) d p 0 d|D |D|

185 But if ! D k p|D| d k = + d, then = + sign(D) , (31) d 2p|D| d p|D| and we have that  !2 1 p|D| d k2 g · + sign(D)  = g . 4 d p|D| 4|D|

We also have that

k2 − 4D = D2/d2 + 2D + d2 − 4D D 2 = − d d is a square, and by the assumption (15) we ﬁnd that

k2 g = 0. 4|D|

Hence the ﬁrst term in equation (30) vanishes, and eq. (29) equals

∞ p !! Z X |D| d G(x) cos 2πx + sign(D) log x dx. d p 0 d|D |D|

According to Lemma 7 in the Appendix and equation (31) this equals

1 X Z ∞ g(x2) dx − , 2 −∞ p d|D 2x − k/ |D| with k deﬁned by (31), and we get the contribution

1 X Z ∞ g(x2) dx − . 2 −∞ p p d|D 2x − ( |D|/d + d/ |D|) We thus ﬁnd that the contribution coming from the Eisenstein series equals

Z ∞ 0 Z ∞ π σ2ir(|D|) ζ (1 + 2ir) 2 ir Re G(x)Mr sign(D) · x dxdr 6 −∞ |D| ζ(1 + 2ir) 0  !c−1 1 X Z ∞ g(x2)dx N − + O p . (32) 2 −∞ p p |D| d|D 2x − ( |D|/d + d/ |D|)

186 The main term in Theorem 2 now comes from summing over the ﬁrst two terms in our summation formula (21), the Maass wave form parts (24), the holomorphic cusp form parts (25), and the Eisenstein series part (32). The error terms coming from the ﬁrst two terms in our main summation formula (21), Maass wave form parts (24), the holomorphic cusp form parts (25), and the Eisenstein series part (32), now sum up to   r ! ∞ !c−1 |D| X N O + O t (D)(|κ | + 1)−5/2 N  j j p  j=1 |D|

∞ θ(k)  !c−1   !c−1 X X N −7/3 N + Otj,k(D) p k  + O p . k=1 j=1 |D| |D| From the estimate for the Fourier coeﬃcients of cusp forms X 2 2 √ αj|tj(m)| K + d3(m) log(2m) m,

K/2<κj ≤K θ(k) X 2 √ αj,k|tj,k(m)| kd3(m) m log(2m), j=1 valid uniformly for k, K, m ≥ 1 (see Motohashi [18] Lemma 2.4, and equation (2.2.10)), we get that the error terms can be estimated by r !  !c−1 |D| N O + O . N p|D| Since c − 1 < 0 and D is a constant with respect to N we have that the error term tends to zero when N tends to inﬁnity. By the deﬁnition of G, equation (22), we have Z ∞ Z ∞ Z ∞ 2 2 t 2 2 G(x)Mr sign(D) · x dx = 2 e(xt)g dt Mr sign(D) · x dx, 0 0 −∞ 4 Z ∞ Z ∞ 2 2 t = 2 e(xt)Mr sign(D) · x dx g dt, −∞ 0 4 which under the substitution 2x = ξ, 2dx = dξ, transforms to Z ∞ Z ∞ ξt ξ2 t2 e Mr sign(D) · dξ g dt. −∞ 0 2 4 4 By Lemma 11 this equals

0 |t/2| ≤ 1, ε = 1    2ri  Z ∞  p 2 2 −1/2  t2 Re |t/2| + (t/2) − 1 ((t/2) − 1) |t/2| > 1, ε = 1 g dt, 4 −∞  2ri   p 2 2 −1/2  Re 1 + 1 + (t/2) (1 + (t/2) ) ε = −1 

187 for r real, and ( ) Z ∞ (−1)k cos((2k − 1) arcsin |t/2|)(1 − (t/2)2)−1/2, |t/2| < 1 t2 g dt, −∞ 0, |t/2| > 1 4 for r = i(k − 1/2). With the further substitution x = t/2, we ﬁnish our proof. We will see that both the Eichler-Selberg trace formula (see Eichler [6]), as well as the classical Selberg trace formula (see e.g. Selberg [21] or Hejhal ([12] and [13])) both with Hecke characters, and the factor εj that determines whether the form is odd or even will follow. This means that our proof gives a uniﬁed way of proving both formulae. Note that the Eichler-Selberg trace formula can be obtained in an elementary way (see e.g. Zagier [26]), so this proof is certainly not the simplest. We should also remark that Kuznetsov’s paper [17] might be relevant, but we have not yet gotten hold on it to verify this. Corollary 1. (Eichler-Selberg) Let D > 0 be a positive integer. Suppose that g ∈ C∞([0, 1]) and g(1) = 0. One then has the identity

∞ θ(k) X d2 √ X X H4D − d2g = D t (D)˜g(k) 4D j,k d2≤4D k=1 j=1 √ √ Z 1 r Z 1 2 2 2σ−2(D)D 1 − y D X g(x ) 1 − x dx + g(y)dy − √ √ , π y π D/d + d/ D − 2x 0 d|D −1 where H(∆) denotes the Hurwitz class number, and

(−1)k Z 1 g˜(k) = cos((2k − 1) arcsin x)gx2dx. π 0 Proof. This follows from Theorem 2 by specializing to D > 0, and to a function g with support on [−1, 1]. Corollary 2. (Selberg trace formula with Hecke operators) Let D > 0 be a positive integer. ∞ Suppose that g ∈ C0 (R) has support on [1, ∞[ and is such that d2 g = 0 whenever d2 − 4D is a square. 4|D| One then has the identity

∞ 2 X 2 d H 4D − d log ε 2 g = d −4D 4D d=1 ∞ π Z ∞ ζ0(1 + 2ir) X √ σ (D)D1/2−ir Re g˜(r)dr + t (D) Dg˜(κ ) 6 2ir ζ(1 + 2ir) j j −∞ j=1 √ √ ∞ 2 ∞ r D X Z g(x ) x2 − 1 dx Z y − 1 − √ √ + σ−2(D)D g(y)dy, 2 −∞ 1 y d|D 2x − 2( D/d + d/ D)

188 where ε∆,H(∆), are deﬁned as in Lemma 1, and

Z ∞ √ 2ri g˜(r) = g(x2) Re x + x2 − 1 dx. 1 Proof. This follows from Theorem 2 by specializing to D > 0, and to a function g with support on [1, ∞[.

Corollary 3. (Selberg trace formula with Hecke operators, and εj) Let D > 0 be a positive integer. Suppose that g ∈ S(R) belongs to the Schwartz class, such that d2 g = 0 whenever d2 + 4D is a square. 4|D| We have

∞ 2 X 2 d H −4D − d log ε 2 g = d +4D 4D d=1 ∞ π Z ∞ ζ0(1 + 2ir) X √ σ (D)D1/2−ir Re g˜(r)dr + ε t (D) Dg˜(κ ) 6 2ir ζ(1 + 2ir) j j j −∞ j=1 √ √ ∞ 2 ∞ r D X Z g(x ) 1 + x2 dx Z 1 + y − √ √ + σ−2(D)D g(y)dy, 2 −∞ 0 y d|D 2x − 2( D/d − d/ D)

3 where ε∆ , H(∆), are deﬁned as in Lemma 1, and

Z ∞ √ 2ri g˜(r) = Re 1 + 1 + x2 g(x2)dx. 0 Proof. This follows from Theorem 2 by specializing to D < 0 in that result and by letting D → −D. Remark 3. The classical Selberg trace formula is usually given in a more geometric form, rather than arithmetic form, using length of closed geodesics instead of class numbers. It is well known that these approaches are equivalent, see e.g. Sarnak [20]. Remark 4. In the sense of the common distinction between the Kuznetsov summation formula, and the Kuznetsov trace formula, our results should maybe have been called the Selberg and Eichler-Selberg summation formulae instead of trace formulae (the trace formula being the inversion of the summation formula).

4 Appendix

In the Appendix we will prove some results on integral transforms and special functions needed in the main text.

3 Notice that εd2+4D is a unit in a quadratic number ﬁeld, and εj determines weather the j’th Maass wave form is odd or even. Hence despite using the same notation, they are not the same.

189 ∞ Lemma 7. Suppose that g ∈ C0 (R), the function G is defined as Z ∞ x2 α2 G(t) = e(xt)g dx, and g = 0. (33) −∞ 4 4 One then has the identity Z ∞ 1 Z ∞ g(x2)dx G(t) cos(αt) log t dt = − . 0 2 −∞ |2x − α| Proof. Without loss of generality we will assume that g and hence also G are real functions. We will consider the Laplace transform Z ∞ −st LG(s) = e G(t) log t dt. (Re(s) > 0) 0 With the definition of G this equals Z ∞ Z ∞ y2 e(yt)g dy e−st log t dt 0 −∞ 4 This integral is absolutely convergent and hence we can change the integration order. We obtain Z ∞ Z ∞ y2 e(2πiy−s)t log t dt g dy. −∞ 0 4 By the the Laplace transform (see e.g. Erdélyi-Magnus-Oberhettinger-Tricomi [7] 4.6 (1)) Z ∞ log(γz) e−zt log t dt = − (Re(z) > 0) 0 z we obtain with z = s − 2πiy that Z ∞ log(γ(s − 2πiy)) y2 LG(s) = − g dy. (Re(s) > 0) (34) −∞ s − 2πiy 4 Since by the assumption g and G are real we have for a real > 0 that Z ∞ −t Re(LG( − 2πiα)) = e cos(2πα)G(t) log t dt. (35) 0 By the assumption that g(α2/4) = 0 this will converge when → 0+ and by the equations (34) and (35) we obtain Z ∞ Z ∞ log(γ(−2πi(α − y))) y2 cos(2πα)G(t) log t dt = Re − g dy 0 −∞ −2πi(α − y) 4 which simplifies to 1 Z ∞ y2 dy − g 4 −∞ 4 |α − y| The substitution x = y/2 finishes our proof.

190 Lemma 8. The Kernel function Mr has the following Mellin-transform and Mellin-Barnes representations: Z ∞ s−1 −2s cos πs ε = 1 (i) Mr(ε · x)x dx = (2π) Γ(s + ri)Γ(s − ri) , 0 cosh πr ε = −1 (|Im (r)| < Re(s) < 3/4) Z c+∞ 1 −2s cos πs x > 0 −s (ii) Mr(x) = (2π) Γ(s + ri)Γ(s − ri) |x| ds 2πi c−∞ cosh πr x < 0 (|Im(r)| < c < 1/2) Proof. For a proof, see Lemma 3 in Andersson [1]. Lemma 9. Suppose that k is an integer. Then

k √ Mi(1/2−k)(x) = π(−1) J2k−1 4π x . (x > 0) Proof. See Andersson [1] Lemma 8. Lemma 10. Suppose that n ≥ 1, and x is real. Then 18 |J (x)| ≤ . n πn1/3 Proof. See Andersson [1] Lemma 7.

2 Lemma 11. For real r the Fourier transform of Mr(ε · x /4)/x equals  0, |t| ≤ 1, ε = 1, Z ∞ x2  √ e(itx)M ε · dx = Re (|t| + t2 − 1)2ri(t2 − 1)−1/2, |t| > 1, ε = 1, r 4 √ −∞ Re (1 + 1 + t2)2ri(1 + t2)−1/2, ε = −1. Proof. By using the substitution dξ ξ = 2πx, = dx, 2π we obtain ! Z ∞ x2 1 Z ∞ x 2 e(itx)Mr ε · dx = cos(ξt)Mr ε · dξ. −∞ 4 π 0 4πξ

With the deﬁnition (3) this equals ( ) 1 Z ∞ iπ (J (ξ) − J (ξ)) , ε = 1 cos(ξt) × 2 sinh πr 2ri −2ri dξ. π 0 2 cosh πrK2ri(ξ), ε = −1

For ε = −1 the result follows from Erd´elyi-Magnus-Oberhettinger-Tricomi [7], 2.12 (40) √ √ Z ∞ π((1 + 1 + t2)ν + (1 + 1 + t2)−ν) cos(xt)Kν(x)dx = √ 2 0 4 cos(νπ/2) t + 1

191 when we put ν = 2ri. For ε = 1 it follows from the discontinuous integral (see Watson [24], 13.42 (5))

( cos (ν arcsin |t|) Z ∞ √ , |t| < 1, 1−t2 cos(xt)Jν(x)dx = sin(πν/2) √ ν (Re(ν) > −1) (36) √ |t| + t2 − 1 , t > 1. 0 t2−1

By putting ν = 2ri and noticing that by (36) the integral

Z ∞ cos(xt)J2ir(x)dx (37) 0 is odd in r for |t| ≤ 1 and thus we see from the deﬁnition of Mr that it vanishes when |t| < 1. For |t| > 1 we obtain √ √ √ (|t| + t2 − 1)2ri + (|t| + t2 − 1)−2ri Re (|t| + t2 − 1)2ri √ = √ . 2 t2 − 1 t2 − 1

2 Lemma 12. For integer k the Fourier transform of M(1/2−k)i(±x /4) is equal to ( Z ∞ x2 (−1)k cos((2k − 1) arcsin |t|)(1 − t2)−1/2, |t| < 1, e(itx)M(1/2−k)i dx = −∞ 4 0, |t| > 1.

Proof. By using Lemma 9 we have that

Z ∞ 2 Z ∞ x k e(itx)M(1/2−k)i dx = (−1) 2π cos(2πtx)J2k−1(2πx)dx. −∞ 4 0

With the substitution

ξ = 2πx, dξ = 2πdx, we ﬁnd that this equals

Z ∞ k (−1) cos(tξ)J2k−1(ξ)dξ, 0 and the result follows from (36). Lemma 13. Suppose that r and x > 0 are real numbers. One then has that Z ∞ −ir 2ir−1 x Mr(x) = |x| t cos 2π t + dt. (38) 0 t Proof. We will consider the Mellin transform (in z)

Z ∞ Z ∞ x x > 0 xz−1−ir t2ir−1 cos 2π t ± dtdx, 0 0 t 0 < Re(z) < 1/2

192 We change the integration order (That we can do this has to be justiﬁed. See Remark 5) Z ∞ Z ∞ x t2ir−1 xz−1−ir cos 2π t ± dxdt. (39) 0 0 t By the summation formula for the cosine function we have x 2πx 2πx cos 2π t ± = cos 2πt cos ∓ sin 2πt sin , (40) t t t and by the following expressions for the gamma function Z ∞ πs Z ∞ πs sin(xt)xs−1dx = Γ(s) sin t−s, cos(xt)xs−1dx = Γ(s) cos t−s, (41) 0 2 0 2 we see that the integral given by eq. (39) equals Z ∞ π(z + ir) π(z + ir) (2π)ir−zΓ(z − ir) tir+z−1 cos 2πt cos ∓ sin 2πt sin dt. 0 2 2 By using (40) and (41) again we see that it equals

(2π)ir−zΓ(z − ir)(2π)−z−irΓ(z + ir)× π(z − ir) π(z + ir) π(z + ir) π(z − ir) × cos cos ∓ sin sin , 2 2 2 2 which according to (40) equals π (2π)−2zΓ(z − ir)Γ(z + ir) cos ((z − ir) ± (z + ir)) . (42) 2 The result finally follows from the fact that by Lemma 8 this is the Mellin transform of Mr(x). Remark 5. In the proof of Lemma 13 we have to justify that we can change the integration order, since the integral is not absolutely convergent. Doing a limit argument/analytic continuation in r and z also present some difficulties since if the situation improves at ∞ it will get worse at 0. A way it can be done is as follows: Divide the integral Z ∞ Z ∞ x t2ir−1xz−1−ir cos 2π t ± dxdt = 0 0 t Z 1 Z ∞Z 1 Z ∞ x + + t2ir−1xz−1−ir cos 2π t ± dxdt, 0 1 0 1 t and do a limit argument/analytic continuation in r and z separately for all four integrals. An alternative way to prove Lemma 13 is to use to the following results (see Erdélyi- Magnus-Oberhettinger-Tricomi [7], 6.5 (35) and (36)) Z ∞ 2 −1 s−1 s cos[a(x + b x )]x dx = −πb [Js(2ab) sin(πs/2) + Ys(2ab) cos(πs/2)], (|Re(s)| < 1) 0 Z ∞ 2 −1 s−1 s cos[a(x − b x )]x dx = 2b Ks(2ab) cos(πs/2), (|Re(s)| < 1) 0

193 and (see Erd´elyi-Magnus-Oberhettinger-Tricomi [8], 7.2.1 (4))

1 Y (z) = (J (z) cos(νπ) − J (z)), ν sin νπ ν −ν together with the deﬁnition of Mr(x), equation (3). Lemma 14. Suppose that x 6= 0 is a real number, and D > 0. Then √ Z ∞ !! −ir 2 X D d σ2ir(D)D Mr(ε · x )dr = π cos 2πx + ε√ . d D −∞ d|D Proof. We use the expression from Lemma 13 to get

Z ∞ Z ∞ Z ∞ 2 −ir 2 −ir −ir 2ir−1 x σ2ir(D)D Mr(ε · x )dr = σ2ir(D)D |x| t cos 2π t + ε · dtdr, −∞ −∞ 0 t

With the substitution s = 2ri this equals

Z +∞i √ Z ∞ 2 1 −s s−1 x π · σs(D)(|x| D) t cos 2π t + ε · dtds. 2πi −∞i 0 t

We recognize this as a Mellin Barnes integral and by Mellin inversion this equals √ !! X D d π cos 2π|x| + ε√ . d D d|D

Lemma 15. When k ≥ 2 is an integer one has the estimate

x2 Mi(1/2−k)(x) . |k|7/3 Proof. This is Lemma 9 in Andersson [1]. Lemma 16. We have that Z t nmD 1/2 3/4 3/2 −3/2 yMr 2 dy t |mnD| + t (1 + |r|) , 0 y Z x Z t nmD 3/2 3/4 3/2 −5/2 yMr 2 dydx x |mnD| + x (1 + |r|) . 0 0 y Proof. This is Lemma 6 in Andersson [2].

Lemma 17. If f ∈ S(R) belongs to the Schwartz class, then for r real, and for each 0 < c1 < c2 one has the estimate Z ∞ 2 2 s−1 (|s| + 1) f(x)Mr(ε · x )x dx 5/2 . (c1 ≤ Re(s) ≤ c2) 0 (1 + |r|)

194 Proof. With the substitution 1 dx dt t = , = , x x t we get

Z ∞ Z ∞ 2 s−1 1 −s−2 ε f(x)Mr(ε · x )x dx = f t · Mr 2 dt. 0 0 t t Since f belongs to the Schwartz class, we get by doing partial integration twice that this equals

Z ∞ Z t Z x 00 1 0 1 1 2 −6−s ε f + f t2(3 + s) + f t (2 + s)(3 + s) t yMr 2 dydxdt. 0 t t t 0 0 y (43)

By Lemma 16 we have with D = ε, n = 1, m = 1

Z t Z x ε 3/2 3/2 −5/2 yMr 2 dydx t 1 + t (1 + |r|) , 0 0 y and it is clear that equation (43) can be estimated by

2 Z ∞ 2 (|s| + 1) X (v) 1 2 −4−Re(s) 3/2 3/2 f (t + 1)t t 1 + t dt. (c1 ≤ Re(s) ≤ c2) (1 + |r|)5/2 t 0 v=0 (44)

Since f belongs to the Schwartz class it is of the order

Z ∞ 2 2 −1−Re(s) −5/2 (|s| + 1) (|s| + 1) t dt (1 + |r|) 5/2 . (c > 0) (45) 1 (1 + |r|)

Lemma 18. If f ∈ S(R) belongs to the Schwartz class, then for k integer, and for each 0 < c1 < c2 one has the estimate Z ∞ 2 s−1 −7/3 f(x)Mi(1/2−k)(ε · x )x dx k , (c1 ≤ Re(s) ≤ c2) 0 Proof. By Lemma 15, we have for k ≥ 2 x M (x) = O , i(1/2−k) k7/3 and we get that

Z ∞ Z ∞ 2 s−1 Re(s)+1 −7/3 f(x)Mi(1/2−k)(ε · x )x dx x |f(x)|dx k , (c1 ≤ Re(s) ≤ c2) 0 0

195 whenever f belongs to the Schwartz class. The case k = 1 follows from Lemma 9 √ Mi(1/2−k)(x) = −πJ1 4π x , and Lemma 10 18 |J (x)| ≤ . 1 π

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