Quantum Variance of Maass-Hecke Cusp Forms

Home , Kuznetsov trace formula, Zeev Rudnick

QUANTUM VARIANCE OF MAASS-HECKE CUSP FORMS

DISSERTATION

Presented in Partial Fulﬁllment of the Requirements for

the Degree Doctor of Philosophy in the Graduate

School of the Ohio State University

Peng Zhao

*****

The Ohio State University 2009

Dissertation Committee: Approved by Professor Wenzhi Luo, Advisor

Professor James Cogdell Advisor Professor Manfred Einsiedler Graduate Program in Mathematics

ABSTRACT

In this thesis we study quantum variance for the modular surface X = Γ\H where

Γ = SL2(Z) is the full modular group. This is an important problem in mathematical physics and number theory concerning the mass equidistribution of Maass-Hecke cusp forms on the arithmetic hyperbolic surfaces. We evaluate asymptotically the quantum variance, which is introduced by S. Zelditch and describes the ﬂuctuations of a quantum observable. We show that the quantum variance is equal to the classical variance of the geodesic ﬂow on S∗X, the unit cotangent bundle of X, but twisted by the central value of the Maass-Hecke L-functions.

Our approach is via Poincare series and Kuznetsov trace formula, which transfer the spectral sum into the sum of Kloosterman sums. The treatment of the non- diagonal terms contributions is subtle and forms the core of this thesis. It turns out that the continuous spectrum part is not negligible and contributes to the main term, but in general, it is small in the cuspidal subspace. We then make use of Watson’s explicit triple product formula to determine the leading term in the asymptotic formula for the quantum variance and analyze its structure, which leads to our main result.

ii Dedicated to my parents

iii ACKNOWLEDGMENTS

I express my deepest thanks and gratitude to my advisor, Professor Wenzhi Luo, for his patience, guidance and constant encouragement which made this work possible.

I am very grateful for his enlightening discussions and suggestions as well as the intellectual support he has provided.

I would like to thank the members of my dissertation committee, Professor James

Cogdell and Professor Manfred Einsiedler for their valuable time and generous help.

I would like to acknowledge my special thanks to Professor Peter Sarnak and

Professor Zeev Rudnick for their warmful encouragement and their interests to this work.

Last but not least, I wish to thank my parents, my wife and my daughter for their constant support.

iv VITA

1978.7 ...... Born in Linyi, China

1999.6 ...... B.Sc. in Mathematics, Shandong University, China

2002.6 ...... M.Sc. in Mathematics, Shandong University, China

2009.6 ...... (expected) Ph.D. in Mathematics, The Ohio State University

FIELDS OF STUDY

Major Field: Mathematics

Specialization: Number Theory

v TABLE OF CONTENTS

Abstract ...... ii

Dedication ...... ii

Acknowledgments ...... iv

Vita...... v

CHAPTER PAGE

1 Introduction ...... 1

2 Poincar´eSeries ...... 16

2.1 Preliminary ...... 16 2.2 The Kuznetsov Formula ...... 20 2.2.1 The Diagonal Term (2.5) ...... 22 2.2.2 The Non-diagonal Term (2.7) ...... 23 2.2.3 The Continuous Spectrum Part (2.6) ...... 28 2.3 The Asymptotic Formula for the Variance ...... 33

3 Symmetry Properties of B ...... 37

3.1 Self-adjointness of Laplacian for B ...... 37 3.2 Self-adjointness of Hecke Operators for B ...... 41

4 Extension and Diagonalization of B ...... 49

5 Eigenvalues of B ...... 54

5.1 Watson’s Formula ...... 54 5.2 The Kuznetsov Formula ...... 55 5.2.1 The Diagonal Term ...... 57 5.2.2 The Non-diagonal Term ...... 58 5.2.3 The Continuous Spectrum Part ...... 60 5.3 Conclusion ...... 63

vi Bibliography ...... 66

vii CHAPTER 1

INTRODUCTION

Let Γ = SL(2, Z) be the full modular group and X = Γ\H be the corresponding modular surface, where H denotes the upper half plane. H is equipped with the Poincar´emetric,

ds2 = y−2(dx2 + dy2) with constant negative curvature −1. On X we have the normalized invariant hyperbolic measure 1 dxdy dµ = , Area(X) y2 and the Beltrami-Laplace operator ∂2 ∂2 ∆ = y2( + ). ∂x2 ∂y2 Deﬁne A(X) to be the space of automorphic functions, i.e.

A(X) = {f : H → C|f(γz) = f(z), for all γ ∈ Γ} and As(X) be the subspace of automorphic forms (i.e. Maass forms consistings of all f ∈ A(X) which have continuous second derivatives and satisfy 1 (∆ + s(1 − s))f = 0 with s = + it. 2 Let L(X) be the Hilbert space of square integrable automorphic functions with the inner product Z hf, gi = f(z)¯g(z)dµ. X 1 Deﬁne C(X) to be the space of smooth, bounded automorphic functions whose zero-th terms vanish in their Fourier expansion at ∞. Automorphic forms in the space C(X) are called cusp forms.

Deﬁne E(X) to be the space of incomplete Eisenstein series

X E(z|ψ) = ψ(Im(γz))

γ∈Γ∞\Γ with ψ any smooth compactly supported function on R+ and

( 1 ∗ ) Γ∞ = γ = ∈ Γ . 0 1

It is well known that the space L(X) has the following orthogonal decomposition

[17],

L(X) = C˜(X) ⊕ E˜(X), where the tilde means the closure in the Hilbert space L(X).

In addition to the Laplace operator, we have the commuting family of Hecke operators Tn, n ≥ 1 acting on A(X) by

1 X T f(z) = √ f(γz), n n γ∈Γ\Gn where ( a b ) Gn = ∈ M2(Z), ad − bc = n c d and it is well known that all Tn’s commute with ∆. Let λf (n) be the eigenvalue of the

Hecke operator Tn for an automorphic function f. The eigenvalues are multiplicative and satisfy X mn λ (m)λ (n) = λ ( ). f f f d2 d|(m,n)

2 Recall that Hecke operators commute with each other and are self-adjoint, thus we can choose an orthonormal basis {φj(z)} for the space of cusp forms C(X), such that for any n,

Tnφj(z) = λj(n)φj(z).

Functions in As(X) which are simultaneous eigenfunctions for all Tn are called Maass- Hecke eigenforms, i.e., the Maass form φ satisﬁes

∆φ + λφ = 0,Tnφ = λφ(n)φ,

1 2 The Maass-Hecke eigenform φ with Laplacian eigenvalue 4 +t has a Fourier expansion of the type √ X φ(z) = y ρφ(n)Kit(2π|n|y)e(nx) n6=0 where Kit is the K-Bessel function and ρφ(n) is proportional to the n-th Hecke eigenvalue λφ(n), i.e.

ρφ(n) = λφ(n)ρφ(1).

We are mainly interested in the distribution of the probability measures on X

2 dµj := |φj(z)| dµ, as j → ∞. The quantum unique ergodicity conjecture (QUE) is a deep problem concerning the mass equidistribution of automorphic forms for X. The goal of QUE is to understand the limiting behavior of Laplacian eigenfunctions. This is an important problem in mathematical physics and number theory and it has origin from Quan- tum Chaos, which attempts to understand the relation between classical physics and quantum physics. In the number theory setting, we consider the case of Maass cusp forms associated with large Laplace eigenvalues for X.

Schnirelman [44], Colin de Verdiere [4] and Zelditch [51] have shown in general that if the geodesic ﬂow on the unit cotangent bundle of a manifold Y is ergodic,

3 (by the similar deﬁnition for X, we form dµj and dµ on Y ) there exist a full density

subsequence {φjk } of {φj} become equidistributed as jk → ∞, i.e. Z Z

lim ψdµjk = ψdµ. jk→∞ Y Y

∞ for any Schwartz function ψ ∈ C0 (Y ). The geodesic flow being ergodic means almost all orbits of the flow become equidistributed. The above result can be viewed as the quantum analogue of the geodesic flow being geodesic. Works of Hejhal-Rackner [14] and Rudnick-Sarnak [40] suggest that there are no exceptional subsequences such that dµj 9 dµ as j → ∞, that is called quantum unique ergodicity. More precisely, the quantum unique ergodicity conjecture for X states that:

Conjecture (Rudnick and Sarnak). For any Jordan measurable compact region A ⊂

X, we have Z Z lim dµj = dµ. j→∞ A A This is equivalent to Z Z lim ψdµj = ψdµ. j→∞ X X ∞ for any Schwartz function ψ ∈ C0 (X).

We will mention several important approaches to this conjecture shortly.

Next we describe the quantization introduced by Zelditch [51].

We realize the upper half plane H as the homogeneous space SL2(R)/SO2(R) and

Y = SL2(Z) \ SL2(R) as the unit cotangent bundle for X and functions on X can be viewed as SO2(R) invariant functions on Y . The motion along geodesics on X gives rise to a Hamiltonian ﬂow Gt on Y   t e 2 0 Γg → Γg   , t ∈ R  − t  0 e 2

4 Recall that the geodesic ﬂow on a negative curvature manifold is ergodic [15], see [1] for X, we can consider this Hamiltonian ﬂow as a classically chaotic system. From this system, we construct a quantum model as follows.

∞ Let C0 (Y ) be the space of smooth functions on Y which decay rapidly at cusps

∞ ∞ ∞ and similarly we deﬁne the space C0 (X). Let C0,0(X) be the subspace of C0 (X) Z 1 consisting of smooth functions whose zeroth Fourier coeﬃcient ψdx is zero for 0 R y large enough and mean zero, i.e. X ψdµ = 0. We can view any function ψ ∈ ∞ C0 (Y ) as a classical observable which includes information about the system. The quantization of Zelditch relates the classical observable ψ to the quantum observable, a self-adjoint operator Op(ψ) on the Hilbert space of square integrable automorphic forms L(X). To be precise,

Z hOp(ψ)φj, φji = Op(ψ)dµj X

∞ gives the value of this quantum observable at state φj. For ψ ∈ C0 (X), Op(ψ) is the multiplication operator (Op(ψ)h)(z) = ψ(z)h(z) [34] and therefore the inner product R becomes hOp(ψ)φj, φji = X ψdµj. Returning to the above quantum unique ergodicity, there are several diﬀerent approaches to this conjecture so far.

From Watson’s explicit triple product formula [49], let f be a Maass-Hecke eigenform, Z 1 2 1 2 Λ( 2 , f ⊗ sym φj)Λ( 2 , f) | f(z)dµj| = 2 2 2 , X Λ(1, sym f)Λ(1, sym φj) where Λ is the completed L-function with the infinite part included. The special values L(1, sym2·) have not much effect since we have the following effective bounds due to Iwaniec and Hoffstein-Lockhart, for any > 0:

− 2 λj L(1, sym φj) λj.

5 Thus, the subconvexity bound of the degree 6 triple product L-function

1 1 L( , f ⊗ sym2ϕ ) = o(λ 2 ) 2 j j would imply the above conjecture. Currently the best bound is

1 1 + L( , f ⊗ sym2ϕ ) = O (λ 2 ) 2 j ,f j which follows from the Phragmen-Lindelof convexity principle. The General Riemann hypothesis would imply the Lindelof Hypothesis

1 L( , f ⊗ sym2ϕ ) = O (λ), 2 j ,f j hence QUE holds under GRH. Moreover, it predicts the rate of convergence for the equidistribution

Z 1 − 4 + f(z)dµj = O,f (λj ). X In [29], Luo and Sarnak proved the quantum unique ergodicity with the above rate on the average, i.e.

∞ Theorem 1.1. Let f ∈ C0,0(X), then Z X 2 1 + | fdµj| λ 2 X λj ≤λ where the implied constant depends on and f.

For the continuous spectrum of X, they showed that, if A, B are two compact

Jordan measurable subsets of X, then

µ (A) Area(A) lim t = , t→∞ µt(B) Area(B) where Z 1 2 dxdy µt(A) = |E(z, + it)| 2 . A 2 y 6 Actually they proved that as t → ∞,

48 µ (A) ∼ Area(A) log t. t π

∗ Jakobson [20] has extended Theorem 1.1 to the microlocalizations of µt to S X, the unit cotangent bundle of X (the Wigner distributions on PSL2(Z) \ PSL2(R)). For CM forms, forms with complex multiplication, QUE also holds by the work of Sarnak [43]. By the above Watson’s triple-product formula, one can deduce the quantum unique ergodicity conjecture from the subconvexity bound for the triple product L-function at the central point. For CM forms, the advantage is the degree

6 triple product L-function factorizes into L-functions for which the subconvexity bound can be established (one is degree 4 Rankin-Selberg L-function and the other one L(s, f) is degree 2).

Lindenstrauss [26] has established quantum unique ergodicity for compact arithmetic quotients of H by ergodic methods.

Theorem 1.2. Let M = Γ\H with Γ a congruence lattice over Q. Then for compact M, the only arithmetic quantum limit is the normalized invariant measure dµ. For

M non-compact, any arithmetic quantum limit is of the form cdµ with 0 ≤ c ≤ 1.

Therefore, in the case of X = Γ\H, quantum unique ergodicity conjecture is true up to a constant c, where 0 ≤ c ≤ 1.

Holowinsky and Soundararajan recently proved the holomorphic version of QUE for X [13].

Theorem 1.3. Let f be a Hecke cusp eigenform of weight k.

(i) Let φ be a Hecke-Maass cusp form, then

− 1 + |µf (φ)| φ, (log k) 30 .

7 (ii) Let ψ be a ﬁxed smooth function compactly supported in (0, ∞). Then

3 − 2 + | (µ (E(·|ψ)) − µ(E(·|ψ))) | (log k) 15 . f π ψ, where Z k 2 dxdy µf (ψ) = ψ(z)y |f(z)| 2 X y Combined with the work of Lindenstrauss, Soundararajan [45] has proven the original version of the QUE conjecture, i.e., for noncompact arithmetic surfaces in

Theorem 1.2, c = 1.

Therefore, for QUE in the case of arithmetic surfaces, the case of holomorphic forms on compact arithmetic surfaces is still unsolved.

We will study this equidistribution problem by calculating the quantum variance.

This variance sum

X 2 Sψ(λ) = |µj(ψ)|

λj ≤λ is ﬁrst introduced by Zelditch [51]. It describes the ﬂuctuations of a quantum observ-

∞ ∞ able hOp(ψ)µj, µji = µj(ψ) for ψ ∈ C0,0(X). For the classical observable ψ ∈ C0,0(Y ), where Y = Γ \ SL2(R), as studied by Ratner [39], the ﬂuctuations along a generic orbit of the geodesic ﬂow obey the central limit theorem, i.e. the distribution of 1 Z T √ ψ(Gt(g))dt T 0 become Gaussian with mean 0 and the variance given by the non-negative Hermitian

∞ form on C0,0(Y ):    t Z ∞ Z e 2 0 VC(φ, ψ) = φ g   ψ(g)dgdt.   − t  −∞ Γ\SL(2,R) 0 e 2

∞ Restricting VC to C0,0(X), it can be diagonalized by the Maass cusp forms φ with

1 2 Laplacian eigenvalue λ = 4 + t and the eigenvalue of VC on φ equals |Γ( 1 − it )|4 V (φ, φ) = 4 2 C 1 2 2π|Γ( 2 − it)| 8 It is shown that the means of the classical observable and the quantum observable are the same by the work of Zelditch [51]. There should be a deep relation between the quantum variance and the classical variance. In [9] and [10], the physicists conjectured that for generic quantum chaotic system, one expects a central limit theorem for the statistical ﬂuctuations of the quantum observable. For the quantum variance, it should correspond to the classical variance VC (ψ, ψ). In [34], Luo and Sarnak examined the analogous quantum variance for holomorhic

Hecke eigenforms. The automorphic forms they considered are the holomorphic cusp forms in Sk(Γ) of even integral weight k for Γ. Sk(Γ) has a Hilbert space structure with respect with the Petersson inner product. Let Hk be the orthonormal basis of

∞ Hecke eigenforms for Sk(Γ). For ψ ∈ C0,0(X), they studied the variance of µf (ψ) as f runs through Hk. They considered a weighted version of the quantum variance sum by inserting arithmetic weights L(1, Sym2f) associated with the symmetric-square

L-function (these appear naturally in the Petersson formula), they proved that for

∞ φ, ψ ∈ C0,0(X), as K → ∞,

X X 2 L(1, Sym f)µf (φ)µf (ψ) ∼ B(φ, ψ)K,

k≤K f∈Hk ∞ where B(φ, ψ) is a non-negative Hermitian form on C0,0(X). It turns out that B(φ, ψ) has some important symmetric properties. Thus B is diagonalized by the orthonormal

π 1 basis of Maass-Hecke cusp forms and the eigenvalues of B at ψj is 2 L( 2 , ψj). We will turn to these similar properties for our case in Chapter 3.

In this thesis, we generalize the result in [34] to Maass cusp forms, i.e., we consider the quantum variance of Masse-Hecke cusp forms. In contrast to the case of the holomorphic cusp forms, the real analytic case is more challenging and complicated.

New diﬃculties arise when one applies the Kuznetsov formula instead of the Petersson formula, in view of the occurrence of the continuous spectrum contribution, the subtle analysis of the hypergeometric functions and the sum of Kloosterman sum weighted 9 by more involved Hankel transform. It turns out that the quantum variance is equal to the classical variance of the geodesic ﬂow on S∗X, the unit cotangent bundle of

X, but twisted by the central value of the Maass-Hecke L-functions. More precisely, we obtain the following:

Theorem 1.4. Let ϕj(z) be the j-th Maass-Hecke eigenform, with the Laplacian

1 2 eigenvalues λj = 4 + tj . For φ, ψ ∈ C(X), We have

X 1 µj(φ)µj(ψ) ∼ λ 2 V (φ, ψ). (1.1)

λj ≤λ

V is diagonalized by the orthonormal basis {φj} of Maass-Hecke cusp forms and the eigenvalue of V at a Maass-Hecke cusp form φ is

1 L( , φ)V (φ, φ). 2 C

In fact, we prove a weighted version of (1.1). The weights are mildly varying and have no eﬀect on this asymptotic. To prove this theorem, we ﬁrst establish a preliminary theorem for the incomplete Poincare series with the test functions satisfying some analytic conditions in view of the Kuznetsov formula. We then remove the analyticity condition on the test functions by using Fourier transform to obtain the general asymptotic formula for the quantum variance sum. For the incomplete

Poincar´eseries, we can take the advantage of unfolding the integral and represent the integral as expressions involving the Fourier coeﬃcients in quadratic polynomials which allows us to apply the Kuznetsov formula and transfer the spectral sum into the sum of Kloosterman sums. The treatment of the non-diagonal terms contributions is subtle and forms the core of my thesis. It turns out that the continuous spectrum part is not negligible and contributes to the main term. The preliminary theorem is as follows.

10 Let h(t) be a smooth function with compact support. Deﬁne the Poincar´eseries

(m 6= 0): X Ph,m(z) = h(y(γz))e(mx(γz)).

γ∈Γ∞\Γ For the continuous spectrum part, we consider the Eisenstein series E(z, s), and deﬁne Z 1 2 dxdy µt(ψ) = ψ(z)|E(z, + it)| 2 . X 2 y

Theorem 1.5. Let ϕj denote the eigenfunctions of Laplacian, with the corresponding

1 2 eigenvalues λj = 4 + tj . For m1, m2 6= 0, h1, h2 be smooth functions with compact support for i = 1, 2, ﬁx > 0, we have

t −T t +T X −( j )2 −( j )2 2 T 1− T 1− (e + e )L(1, sym ϕj)µj(Ph1,m1 )µj(Ph2,m2 )

tj Z ∞ −( t−T )2 −( t+T )2 2 T 1− T 1− + (e + e )|ζ(1 + 2it)| µt(Ph1,m1 )µt(Ph2,m2 )dt 0 1 1− 2 − = T B(Ph1,m1 ,Ph2,m2 ) + O(T ), (1.2)

The Laplacian ∆ and the Hecke operators Tn are self-adjoint with respect to B.

t −T t +T −( j )2 −( j )2 Remark 1.1. (i) The weight e T 1− + e T 1− is in eﬀect localizing the spectrum sum to the range

1− |tj − T | T .

(ii) To prove (1.2), we apply the Kuznetsov trace formula, and analyze the contributions of the diagonal, non-diagonal and the continuous spectrum parts. The integral term in Theorem 1.5 comes from the continuous spectrum. It turns out the continuous spectrum part is not negligible and contributes to the main term. How- ever, if we restrict to Lcusp(X), the continuous spectrum part is small in view of the following result [L-S1],

− 1 + |µt(φ)| φ, |t| 6 ,

11 for a Maass cusp form φ.

(iii) To determine B(φ, ψ) in Theorem 1.4 explicitly, we apply Watson’s formula, converting the problem to study the triple product L-values, by means of approximate functional equation and Kuznetsov formula. Here we appeal to Jutila’s subconvexity bound [23]

1 1/3+ L( 2 + it, ψj) (tj + t) .

(iv) We call the eigenvalue of V at φ as the arithmatic variance or quantum variance of the Maass-Hecke cusp form φ. It is consistent with the Feingold-Peres

[10] prediction for the variance of generic chaotic systems, i.e., it recovers the classical variance of φ 1 itφ 4 |Γ( 4 − 2 )| 1 2 2π|Γ( 2 − itφ)| 1 given by Ratner [38] multiplied by the central value L( 2 , φ). As a by-product, we 2 obtain that, for Maass-Hecke form ψ, L(1/2, ψ ⊗ sym (φj)) is non-vanishing for some

1 φj and the central value L( 2 , ψ) is non-negative. (v) In order to apply the Kuznetsov trace formula, we consider the weighted

2 version sums of the quantum variance in (1.2). The weights L(1, sym ϕj) are mildly varying. For them, we have the following bounds due to Iwaniec and Hoﬀstein-

Lockhart:

− 2 tj L(1, sym ϕj) tj.

Thus, we can remove these weights using the methods in [19], see also [18].

(vi) If we consider the cotangent bundle S∗X = Γ \ G to X and lift the measures

∗ ∗ dµj on X to the Wigner distribution dωj on S X, we could extend our result to S X, but we do not pursue them in this thesis.

(vii) The quantum variance V (ψ) is the same as the variance for the ﬂuctuations of the arithmetic measures on the modular surface obtained by collecting all closed geodesics. See the work of Luo, Rudnick and Sarnak [28]. 12 (viii) In [24], Kurlberg and Rudnick formulated an interesting analogous variance for the quantum cat map, a linear hyperbolic map of the torus R2/Z2.

We end the introduction with an outline of this thesis.

In Chapter 2, we ﬁrst establish a preliminary version of Theorem 1.5 for the incomplete Poincare series Pm,h with h satisﬁng some special analytic conditions in order to apply the Kutznesov formula. Now these Poincare series form a dense

∞ subspace of C0,0(X) and unfolding method allows us to analyze µj(Pm,h) in terms of sums over Fourier coeﬃcients. We average over the orthonormal basis for the cuspidal subspace by Kutznesov formula, and deal with the diagonal, non-diagonal terms separately. Here the subtle test functions arise due to the complicated hypergeometric functions. For non-diagonal terms, we apply the Fourier transform in (Bateman, volume 1, p. 59) ! J (x)\− J (x) 2it −2it (y) = −i cos(x cosh(πy)). sinh(πt)

The continuous spectrum parts is not negligible and contributes to the main term.

We then remove the analyticity condition on h by using the Fourier transform to obtain the general asymptotic formula for the quantum variance sum Sψ(λ). Chapter 3 and Chapter 4 are devoted to analyze the structure of the quardratic form B. In Chapter 3, we obtain the self-adjointness of B with respect to the Lapla- cian ∆ and the Hecke operators using the series expression as obtained in the Theorem

1.4. To be precise, we analyze the bilinear form B and using the series expression as obtained in Chapter 2, we prove it is self-adjoint with respect to the Laplacian ∆ and the Hecke operators Tn, n ≥ 1, i.e., for Poincar´eseries ψ1, ψ2, B satisﬁes the symmetries

B(∆ψ1, ψ2) = B(ψ1, ∆ψ2), (1.3)

13 and for n ≥ 1,

B(Tnψ1, ψ2) = B(ψ1,Tnψ2). (1.4)

This part is more complicated than the holomorphic case in [34] due to the subtle series expression of B.

∞ In Chapter 4, we then diagonalize and extend B to C0,0(X). We restrict B to the Maass-Hecke cusp forms. The continuous spectrum integral is small enough so we can obtain the ﬁrst part of main Theorem 1.4. If we restrict B to the Maass-

Hecke cusp forms, it can be shown that B is diagonalized by the orthonormal basis of

Hecke-Maass cusp forms from the symmetry properties of B obtained in Chapter 3.

The idea of the extension is choosing a partition of unity subordinate to a covering

∞ of X, then any ψ ∈ C0,0(X) can be written as a linear combination of Ph,m, where h is smooth with compact support functions.

In Chapter 5, we establish the second part of Theorem 1.4. We compute the value of B on φ(z), an even Maass-Hecke cuspidal eigenform. After applying Watson’s for-

2 mula, the quantum variance sum over φj boils down to averaging L(1/2, φ⊗sym (φj)) with respect to the spectrum. By Rankin-Selberg theory, we can express this sum in a suitable series to which the Kuznetsov formula is applied. The non-diagonal terms has no contribution to the main term as in the holomorphic case. For the continuous spectrum, it can be shown that it is small enough by the Jutila’s subconvexity bound

[23]. The resulting asymptotics is used to determine explicitly the quantum variance.

Therefore, we complete the proof of Theorem 1.4.

We sketch the logic graph of this thesis as follows.

14 The Logic Graph

Reduce problems to incomplete Poincar´e

series to get the preliminary theorem P PP PP PP PP PP

Diagonal terms Nondiagonal terms Continuous terms P PP PP PP PP PP

Symmetry properties of the quantum variance B P PP PP PP PP PP

Self-adjointness with Laplacian Self-adjointness with Hecke operators P PP PP PP PP PP ∞ Extend B to C0,0(X) and diagonalize B on the o.n.b of Maass-Hecke cusp forms

The eigenvalue of B on a Maass-Hecke eigenform Q Q Q Q QQ

Watson formula Kuznetsov formula

Quantum variance equals classical variance twisted by the central L-values

15 CHAPTER 2

POINCARE´ SERIES

In this chapter we state the basic facts and prove auxiliary results that are needed for the main Theorem 1.3. We ﬁrst establish a preliminary version of part (A) of

Theorem 1.3 for Poincar´eseries Pm,h with h satisfying some special analytic conditions in order to apply the Kutznesov formula. Now Poincar´eseries form a dense subspace

∞ of C0,0(X) and the unfolding method allows us to analyze µj(Pm,h) in terms of sums over Fourier coeﬃcients. We average over the orthonormal basis for the cuspidal subspace by the Kutznesov formula, and deal with the diagonal, non-diagonal terms separately, see details in Section 2.2. It turns out the continuous spectrum parts is not negligible and contributes to the main term. We then remove analyticity condition on h by using Fourier transform to obtain the general asymptotic formula for the quantum variance sum Sψ(λ) in Section 2.3.

2.1 Preliminary

Let ϕj denote the j-th Hecke-Maass eigenform with the corresponding Laplacian

1 2 eigenvalue λj = 4 + tj , Hecke eigenvalues λj(n) and we normalize kϕjk2 = 1. Let 1 sj = 2 + itj, then ϕj has the Fourier expansion √ X ϕj(z) = y ρφ(n)Kitj (2π|n|y)e(nx) n6=0 where Ks(y) is the K-Bessel function. 16 Let h(t) be an even real analytic function on R satisfying h(n)(t) (1 + |t|)−N for any n > 0 and any large N, and h(t) t10 when t → 0. Deﬁne the Poincar´eseries

(m ∈ Z, m 6= 0): X Ph,m(z) = h(y(γz))e(mx(γz)).

γ∈Γ∞\Γ 2 Deﬁne the probability measures dµj(z) = |ϕj(z)| dµ(z). We normalize

1/2 ρj(n) = cosh(πtj) aj(n), we have [31],

2 2 |aj(1)| = 2 , aj(n) = aj(1)λj(n) L(1, sym ϕj) From the well-known facts

1 1 π Γ( + itj)Γ( − itj) = , 2 2 cosh(πtj) we have Z 2 2 X < Ph,m, |ϕj| > = |ϕj(z)| ( h(y(γz))e(mx(γz)))dµ(z) Γ\H γ∈Γ∞\Γ X Z ∞ m r dr = ρ (k)ρ (k + m) K (r)K (| 1 + |r) h j j itj itj k 2π|k| r k6=0,−m 0 Z 2 cosh(πtj) X 1 G(s) −3 = 2 λj(k)λj(k + m) s 2 L(1, sym ϕj) 2πi (πk) k6=0,−m (σ) s s Γ + it Γ( − it )f(k, t , s)ds 2 j 2 j j Z 2π X 1 G(s) −3 = 2 λj(k)λj(k + m) s 2 ds L(1, sym ϕj) 2πi (πk) k6=0,−m (σ) s s Γ 2 + itj Γ( 2 − itj) 1 1 f(k, tj, s) (2.1) Γ( 2 + itj)Γ( 2 − itj) where

Z 1 2τm −s/2−itj itj s/2−1 s/2−1 2 f(k, tj, s) = (1 + m/k) τ (1 − τ) 1 + + τ(m/k) dτ. 0 k

17 Here we also applied the formulas (6.576) and (9.111) in [11]

Z ∞ s−1 m r Kitj (r)Kitj (1 + )r dr 0 k Γ( s )2 s s s s m = 2−3−s(1 + m/k)itj 2 Γ( + it )Γ( − it )F ( + it , , s, 1 − (1 + )2) Γ(s) 2 j 2 j 2 j 2 k s s Z 1 −3−s itj s/2−1 s/2−1 = 2 (1 + m/k) Γ( + itj)Γ( − itj) τ (1 − τ) 2 2 0 2τm −s/2−itj 1 + + τ(m/k)2 dτ. k

We can show f(k, tj, s) is even in tj by the integral identities of p.959 9.131 in [11] since

m s s m (1 + )itj F ( + it , , s, 1 − (1 + )2) k 2 j 2 k m Γ(s)Γ(−it ) s s m itj j 2 = (1 + ) s s F ( + itj, , itj + 1, (1 + ) ) k Γ( 2 − itj)Γ( 2 ) 2 2 k m Γ(s)Γ(it ) s s m −itj j 2 + (1 + ) s s F ( − itj, , −itj + 1, (1 + ) ) k Γ( 2 + itj)Γ( 2 ) 2 2 k We also applied unfolding integral, Mellin transform:

Z ∞ dy G(s) = h(y)y−s . 0 y

For the Gamma functions in (2.1) we can get an asymptotic formula so that we can use the Mellin inversion:

1 Z σ+i∞ h(y) = G(s)ysds 2πi σ−i∞ for any σ > 0.

By Stirling Formula, for any vertical strip 0 < a ≤ Re(s) ≤ b, as tj → ∞, we have

s s s +it − 1 − s Γ( + it ) ( + it ) 2 j 2 e 2 2 j ∼ 2 j 1 1 − 1 Γ( + itj) itj 2 2 ( 2 + itj) e 1−s +( s +it − 1 ) log( s +it )−it log( 1 +it ) = e 2 2 j 2 2 j j 2 j s s 1 1 s 1 2 +itj 1−s ( 2 − 2 ) log( 2 +itj )+( 2 +itj − 2 ) log 1 + 2 = e 2 +itj (2.2)

18 Next we show s +it ( s +it − 1 ) log 2 j + 1−s 2 j 2 1 2 −1 2 +itj e = 1 + Os(tj ), i. e. s s 1 2 + itj 1 − s −1 + itj − log 1 + = Os(tj ). 2 2 2 + itj 2 We have

s s 1 2 + itj 1 − s + itj − log 1 + 2 2 2 + itj 2 s−1 s 1 2 −2 1 − s = + itj − 1 + Os(tj ) + 2 2 2 + itj 2 s 1 s − 1 2 + itj − 2 −1 1 − s = 1 + Os(tj ) + 2 2 + itj 2 s s − 1 2 − 1 −1 1 − s = 1 + 1 + Os(tj ) + 2 2 + itj 2 −1 = Os(tj ). (2.3)

From (2.2) and (2.3), we obtain

s−1 s 2 Γ 2 + itj 1 −1 1 = + itj 1 + Os(tj ) . Γ( 2 + itj) 2

Similarly, we have

s−1 s 2 Γ( 2 − itj) 1 −1 1 = − itj 1 + Os(tj ) . Γ( 2 − itj) 2

Thus,

s s Γ 2 + itj Γ( 2 − itj) 1 1 Γ( 2 + itj)Γ( 2 − itj) s−1 1 2 = + t2 1 + O (t−1) = (t )s−1 1 + O (t−1) . (2.4) 4 j s j j s j as tj → ∞.

19 2.2 The Kuznetsov Formula

The strategy to obtain the asymptotic for the variance

X 2 Sψ(λ) = |µj(ψ)|

λj ≤λ is to take a smoothed version of the above sum and apply the Kuznetsov trace formula.

From (2.1), (2.4) and Mellin inversion, we obtain

2 π X −1− < Ph,m, |µj| >= 2 λj(k)λj(k + m)H(tj, k, m) + O(tj ), 4tjL(1, sym ϕj) k6=0,−m where

!itj   Z 1 1 + m 1 t pτ(1 − τ) H(t , k, m) = k h j dτ j 2τm τm2   τ(1 − τ) q 2τm τm2 0 1 + k + k2 πk 1 + k + k2 Note: In the holomorphic case (for weight k modular form f) [34], we have

2π2 X k − 1 < P , µ >= λ (r)λ (r+m)h +O(k−1−) h,m f (k − 1)L(1, sym2(f)) f f 4π(r + m/2) r≥1 Thus, by the multiplicativity of Hecke eigenvalues, we have

2 2 < Ph1,m1 , |µj| > < Ph2,m2 , |µj| > 2 π X X m1 m2 = λ (k (k + ))λ (k (k + )) 2 2 2 j 1 1 j 2 2 16tj L (1, sym ϕj) d1 d2 d1|m1,d2|m2 k1,k2 −2− H1(tj, d1k1, m1)H2(tj, d2k2, m2) + O(tj ) 2 π X X m1 m2 = 2 2 aj(k1(k1 + ))aj(k2(k2 + )) 32tj L(1, sym ϕj) d1 d2 d1|m1,d2|m2 k1,k2 −2− H1(tj, d1k1, m1)H2(tj, d2k2, m2) + O(tj ).

Consider

1−ε 2 1−ε 2 X −((tj −T )/T ) −((tj +T )/T ) 2 (e + e )L(1, sym ϕj) j≥1 2 2 < Ph1,m1 , |µj| > < Ph2,m2 , |µj| > 2 π X X X m1 m2 −1− = aj(k1(k1 + ))aj(k2(k2 + ))eh(tj) + O(tj ), 32 d1 d2 d1|m1,d2|m2 k1,k2 tj 20 where

1 1−ε 2 1−ε 2 −((tj −T )/T ) −((tj +T )/T ) eh(tj) = 2 H1(tj, d1k1, m1)H2(tj, d2k2, m2)(e + e ). tj

Applying Kuznetsov’s formula [25] to the inner sum, we obtain

X m1 m2 aj(k1(k1 + ))aj(k2(k2 + ))eh(tj) d1 d2 tj

m1 m2 ∞ ∞ 2 δk1(k1+ ),k2(k2+ ) Z Z d1 d2 2 eh(t)dit(k1 + k1m1/d1) = 2 t tanh(πt)eh(t)dt − 2 π −∞ π 0 |ζ(1 + 2it)| 2i X d (k2 + k m /d )dt + c−1S(k2 + k m /d , k2 + k m /d ; c) it 2 2 2 2 π 1 1 1 1 2 2 2 2 c Z ∞ p 2 2 4π (k1 + k1m1/d1)(k2 + k2m2/d2) eh(t) J2it( )t dt. −∞ c cosh(πt) Here X dm + an S(m, n; c) = e( ) c ad≡1( mod c) is the Kloosterman sum and

it X d1 dit(n) = . d2 d1d2=n Thus, we have

1−ε 2 1−ε 2 X −((tj −T )/T ) −((tj +T )/T ) 2 (e + e )L(1, sym ϕj)

tj 2 2 < Ph1,m1 , |µj| > < Ph2,m2 , |µj| >

2 m1 m2 ∞ δk1(k1+ ),k2(k2+ ) Z π X d1 d2 = ( 2 t tanh(πt)eh(t)dt (2.5) 32 π −∞ d1,d2,k1,k2 Z ∞ 2 eh(t) 2 2 − 2 dit(k1 + k1m1/d1)dit(k2 + k2m2/d2)dt (2.6) π 0 |ζ(1 + 2it)| 2i X + c−1S(k2 + k m /d , k2 + k m /d ; c) π 1 1 1 1 2 2 2 2 c Z ∞ p 2 2 4π (k1 + k1m1/d1)(k2 + k2m2/d2) eh(t) J2it( )t dt), (2.7) −∞ c cosh(πt) Next, we will estimate these three terms respectively. 21 2.2.1 The Diagonal Term (2.5)

m1 m2 Since k1(k1 + ) = k2(k2 + ) has at most ﬁnitely many solutions if m1/d1 6= m2/d2, d1 d2

m1 m2 m1 m2 and the integer solutions to k1(k1 + ) = k2(k2 + ) are only k1 = k2 if = . d1 d2 d1 d2 Thus, the diagonal terms are

1−ε 2 1−ε 2 π2 Z ∞ (e−((tj −T )/T ) + e−((tj +T )/T ) ) X X H1(t, d1k, m1)H2(t, d2k, m2)dt+O(1) 32 −∞ t m1/d1=m2/d2 k≥1 where

H1(t, d1k, m1)H2(t, d2k, m2) Z 1 Z 1 1 m m = cos( 1 t(2τ − 1)) cos( 2 t(2η − 1)) 0 0 τη(1 − τ)(1 − η) d1k d2k     tpτ(1 − τ) tpη(1 − η) h h dτdη 1  q 2  2  q 2  2τm1 τm1 2ηm2 ηm2 πd1k 1 + + 2 2 πd2k 1 + + 2 2 d1k d1k d2k d2k

(n) −N For i = 1, 2; restricting hi on R and hi satisfy hi (t) (1 + |t|) for any n > 0

10 suﬃciently large N and hi(t) t when t → 0. Thus, hi are continuous uniformly on R. For the sum over k, we estimate it as

X H1(t, d1k, m1)H2(t, d2k, m2) k≥1   Z 1 Z 1 Z ∞ m m tpτ(1 − τ) = cos( 1 t(2τ − 1)) cos( 2 t(2η − 1))h 1  q 2  0 0 0 d1k d2k 2τm1 τm1 πd1k 1 + + 2 2 d1k d1k   tpη(1 − η) 1 h dk dτdη + O(1) 2  q 2  2ηm2 ηm2 τη(1 − τ)(1 − η) πd2k 1 + + 2 2 d2k d2k Z 1 Z 1 Z ∞ p ! m1 m2 t τ(1 − τ) = cos( t(2τ − 1)) cos( t(2η − 1))h1 0 0 0 d1k d2k πd1k ! tpη(1 − η) 1 h2 dk dτdη + O(1) πd2k τη(1 − τ)(1 − η) ! Z 1 Z 1 Z ∞ cos( πm1 ξ(2τ − 1)) cos( πm2 ξ(2η − 1)) p t d1 d2 ξ τ(1 − τ) = h1 π 0 0 0 τη(1 − τ)(1 − η) d1

22 ! ξpη(1 − η) dξ h2 2 dτdη + O(1) d2 ξ

Therefore, we obtain the main term of the diagonal term is

3/2 Z 1 Z 1 Z ∞ πm1 πm2 π X cos( d ξ(2τ − 1)) cos( d ξ(2η − 1)) T (1−) 1 2 64 m m 0 0 0 τη(1 − τ)(1 − η) 1 = 2 d1 d2 ! ! ξpτ(1 − τ) ξpη(1 − η) dξ h1 h2 2 dτdη (2.8) d1 d2 ξ i.e. √ ξ τ(1−τ) πm1 cos( ξ(2τ − 1))h1 π3/2 X Z ∞ Z 1 d1 d1 T (1−) dτ 64 0 0 τ(1 − τ) m1/d1=m2/d2 √ ξ η(1−η) πm2 cos( ξ(2η − 1))h2 Z 1 d2 d2 dξ dη 2 . (2.9) 0 η(1 − η) ξ

For the holomorphic case [34], it is

Kπ Z ∞ X 1 Z ∞ dη u(ξ)dξ h1(d2η)h2(d1η) 2 . 4 0 d1d2 0 η m1/d1=m2/d2

2.2.2 The Non-diagonal Term (2.7)

The non-diagonal term is:

 q m1 m2  m1 m2 Z 4π k1k2(k1 + )(k2 + ) ˜ X X X S(k1(k1 + d ), k2(k2 + d ); c) d1 d2 h(t)t 1 2 J dt c 2it  c  cosh(πt) d1|m1 k1,k2 c≥1 R d2|m2 where

1 1−ε 2 1−ε 2 h˜(t) = H (t, d k , m )H (t, d k , m )(e−((t−T )/T ) + e−((t+T )/T ) ), t2 1 1 1 1 2 2 2 2

!it   Z 1 1 + m 1 tpτ(1 − τ) H (t, k, m) = k h dτ; j 2τm τm2 j   τ(1 − τ) q 2τm τm2 0 1 + k + k2 πk 1 + k + k2 for j = 1, 2.

23 q m1 m2 4π k1k2(k1+ )(k2+ ) d1 d2 Let x = c , the inner integral in the non-diagonal terms is

1 Z J (x) − J (x) I (x) = 2it −2it h˜(t) tanh πtdt T 2 sinh(πt) R Since tanh(πt) =sgn(t) + O(e−π|t|) for large |t|, we can remove tanh(πt) by getting a negligible term O(T −N ) for any N > 0.

Next we apply the Parseval identity and the Fourier transform in (Bateman, volume 1, p. 59) ! J (x)\− J (x) 2it −2it (y) = −i cos(x cosh(πy)). sinh(πt)

Remark : This transform is very useful in the analysis of the automorphic forms recently, e.g. Liu and Ye [27] applied this in their proof for a subconvexity bound for

Rankin-Selberg L-functions.

By the evaluation of the Fresnel integrals, we have ! 1 Z J (x)\− J (x) I (x) = 2it −2it (y)(\h˜(t)t)(y)dy + O(T −N ) T 2 sinh(πt) R −i Z = cos(x cosh(πy))(\h˜(t)t)(y)dy + O(T −N ) 2 R −i Z 1 = cos(x + π2xy2)(\h˜(t)t)(y)dy + O(T −N ) 2 2 R −i Z ∞ rxy rxy πy dy = h˜( ) cos(x − y + )√ + O(T −N ) 2 0 2 2 4 πy Z ∞ √ √ r Z 1 Z 1 −i −(( xy −T )/T 1−)2 −(( xy +T )/T 1−)2 2 = (e 2 + e 2 ) 2 0 xy 0 0   cos( m1 p xy (2τ − 1)) cos( m2 p xy (2η − 1)) p xy pτ(1 − τ) d1k 2 d2k 2 h 2 1  q 2  τη(1 − τ)(1 − η) 2τm1 τm1 πd1k 1 + + 2 2 d1k d1k   xy p p η(1 − η) π dy h 2 dτdη cos(x − y + )√ 2  q 2  2ηm2 ηm2 4 πy πd2k 1 + + 2 2 d2k d2k Thus, the non-diagonal terms are equal to

24 m1 m2 Z ∞ √ −i X X X S(k1(k1 + d ), k2(k2 + d ); c) −(( xy −T )/T 1−)2 1 2 (e 2 + 2 c 0 d1|m1 k1,k2 c≥1 d2|m2 √ r Z 1 Z 1 m1 p xy m2 p xy −(( xy +T )/T 1−)2 2 cos( d k 2 (2τ − 1)) cos( d k 2 (2η − 1)) e 2 ) 1 2 xy 0 0 τη(1 − τ)(1 − η) p xy p ! p xy p ! 2 τ(1 − τ) 2 η(1 − η) π dy h1 h2 dτdη cos(x − y + )√ πd1k1 πd2k2 4 πy +O(1) (2.10)

(n) −N Since both h1(t) and h2(t) satisfy hi (1 + |t|) for any n > 0 and suﬃciently

10 large N, and hi(t) t when t → 0, the above sum is concentrated on p xy 2 − T | | T 2 T 1−

1 xyτ(1 − τ) − 10 T 2 T k1

1 xyη(1 − η) − 10 T 2 T k2 Thus we can get the following range

rxy ∼ T. 2

−1 Note that here x ∼ k1k2c , the ranges for k1, k2, c are as follows

1− p 21 +p T 2 τ(1 − τ) k1 T 20 τ(1 − τ),

1− p 21 +p T 2 η(1 − η) k2 T 20 η(1 − η),

1 c yT 10

(In the holomorphic case [34], one could get k1 ∼ k2 ∼ T and y c, so by partial integration, the terms with c T contribute O(1).)

Here by the above relations and partial integration suﬃciently many times, we will get suﬃciently large power of y, k1 and k2 occurring in the denominator, so we

1 + get the terms with c T 10 contribute O(1). 25 Denote the sum (2.10) as

m1 m2 X X X S(k1(k1 + d ), k2(k2 + d ); c) 1 2 J + O(1). c k1,k2,c d1|m1 k1,k2 c≥1 d2|m2

√ xy 2 Making the change of variable t = T , we get Jk1,k2,c is

3 ∞ 2 1 1 m1 2 Z (t−1) 2 (t+1) 2 Z Z cos( tT (2τ − 1)) 2 −( ) −( ) 1 2(tT ) π d1k √ (e T − + e T − ) sin(−x + − ) πx 0 t x 4 0 0 τ(1 − τ) ! ! cos( m2 tT (2η − 1)) p p d2k tT τ(1 − τ) tT η(1 − η) h1 h2 dτdηdt η(1 − η) πd1k1 πd2k2

By Taylor expansion, r 4πi m1 m2 xi = k1k2(k1 + )(k2 + ) c d1 d2 2πi m2k1 m1k2 = (2k1k2 + + + ··· ) c d2 d1

So we can write

m2k1 m1k2 Jk1,k2,c = Im(ec(−(2k1k2 + + ))fc(k1, k2)), d2 d1 where

2 2 3 Z ∞ m1m2 m1k2 m2k1 2 2 −( (t−1) )2 −( (t+1) )2 1 T − T − fc(k1, k2) = ec( − 2 − 2 + ··· )√ (e + e ) 2d1d2 4d1k1 4d2k2 πx 0 t 2 1 1 m1 m2 2(tT ) π Z Z cos( tT (2τ − 1)) cos( tT (2η − 1)) i( − ) d1k d2k e x 4 0 0 τη(1 − τ)(1 − η) ! ! tT pτ(1 − τ) tT pη(1 − η) h1 h2 dτdηdt πd1k1 πd2k2

2πiz and we use the notation ec(z) = e c .

Reducing the summation over k1, k2 into congruence classes mod c, we have,

X m1 m2 m2k1 m1k2 S(k1(k1 + ), k2(k2 + ); c)ec(−(2k1k2 + + ))fc(k1, k2) d1 d2 d2 d1 k1,k2≥1

26 X m1 m2 m2a m1b = S(a(a + ), b(b + ); c)ec(−(2ab + + )) d1 d2 d2 d1 a,b mod c X fc(k1, k2)

k1≡a,k2≡b mod c 1 X X m1 m2 = 2 S(a(a + ), b(b + ); c) c d1 d2 u,v mod c a,b mod c m2 m1 X ec(−(2ab + ( + u)a + ( + v)b))( fc(k1, k2)ec(−uk1 − vk2)). d2 d1 k1,k2

Apply the Poisson summation for the sum in k1, k2 and obtain,

X X ZZ u v fc(k1, k2)ec(−uk1 − vk2) = fc(k1, k2)e((l1 − )k1 + (l2 − )k2)dk1dk2. R2 c c k1,k2 l1,l2

c c We can assume |u| ≤ 2 , |v| ≤ 2 , by partial integration suﬃciently many times, we get

ZZ X u v −A fc(k1, k2)ec(−uk1 − vk2) = fc(k1, k2)e(− k1 − k2)dk1dk2 + O(T ) R2 c c k1,k2 for any A > 1. For (u, v) 6= (0, 0), by partial integration suﬃciently many times, we also obtain ZZ u v −A fc(k1, k2)e(− k1 − k2)dk1dk2 T , R2 c c 1 + for any A > 0. Thus only (u, v) = (0, 0) contributes. We can allow c T 10 in the

T 2c c-summation, notice that we have the term in fc(k1, k2), so by partial integration k1k2 suﬃciently many times,

ZZ −A 2 fc(k1, k2)dk1dk2 c T , R2 for any A > 0. For ﬁxed di, mi (i = 1, 2), denote

X m1 m2 m2a m1b Sc = S(a(a + ), b(b + ); c)ec(−(2ab + + )) d1 d2 d2 d1 a,b mod c

27 Thus, the non-diagonal contribution is ZZ X X Sc Im( 2 fc(k1, k2)dk1dk2) + O(1) c R2 d1|m1 c≥1 d2|m2 ZZ 2 2 3 Z ∞ X X Sc m1m2 m1k2 m2k1 2 2 −( (t−1) )2 −( (t+1) )2 1 √ T − T − = Im( 2 ec( − 2 − 2 ) (e + e ) c R2 2d1d2 4d1k1 4d2k2 πx 0 t d1|m1 c≥1 d2|m2

2 1 1 m1 m2 p ! 2(tT ) π Z Z cos( tT (2τ − 1)) cos( tT (2η − 1)) i( − ) d1k1 d2k2 tT τ(1 − τ) e x 4 h1 0 0 τη(1 − τ)(1 − η) πd1k1 ! tT pη(1 − η) h2 dτdηdtdk1dk2) + O(1) πd2k2 ZZ 2 2 3 Z ∞ X X Scζ8 m1m2 m1φ m2ξ 2 2 −( (t−1) )2 T − = T Im( 3 ec( − 2 − 2 )√ 3 (e + c 2 R2 2d1d2 4d1ξ 4d2φ π(ξφ) 2 0 d1|m1 c≥1 d2|m2 1 1 m1ξ m2φ p ! (t+1) 2 Z Z cos( (2τ − 1)) cos( (2η − 1)) ξ τ(1 − τ) −( − ) d1 d2 e T )e(ξφc) h1 0 0 τη(1 − τ)(1 − η) πd1 ! φpη(1 − η) h2 dτdηdtdξdφ) + O(1) πd2

ZZ 2 2 3 1− X X Scζ8 m1m2 m1φ m2ξ 2 2 = T Im( 3 ec( − 2 − 2 ) 3 c 2 R2 2d1d2 4d1ξ 4d2φ (ξφ) 2 d1|m1 c≥1 d2|m2 m ξ m φ ! Z 1 Z 1 cos( 1 (2τ − 1)) cos( 2 (2η − 1)) p d1 d2 ξ τ(1 − τ) e(ξφc) h1 0 0 τη(1 − τ)(1 − η) πd1 ! φpη(1 − η) h2 dτdηdξdφ) + O(1). πd2

2.2.3 The Continuous Spectrum Part (2.6)

Next, we study the continuous spectrum integrals, which is

Z ∞ ˜ X X h(t) 2 2 2 dit(k1 + k1m1/d1)dit(k2 + k2m2/d2)dt. 0 |ζ(1 + 2it)| d1|m1 k1,k2≥1 d2|m2

28 Recall that |ζ(1 + 2it)| ≥ c log−2/3(2 + |t|) for some c > 0 [47],

it X d1 dit(n) = = O(n ), d2 d1d2=n and

1 −( t−T )2 −( t+T )2 h˜(t) = H (t, d k , m )H (t, d k , m )(e T 1− + e T 1− ), t2 1 1 1 1 2 2 2 2 !it   Z 1 1 + m 1 tpτ(1 − τ) H (t, k, m) = k h dτ; j 2τm τm2 j   τ(1 − τ) q 2τm τm2 0 1 + k + k2 πk 1 + k + k2 for j = 1, 2.

1 The Fourier expansion of E(z, 2 + it) is

1 ∞ 1 1 +it 1 1 −it 2y 2 X E(z, + it) = y 2 + ϕ( + it)y 2 + d (n)K (2πny) cos(2πnx) 2 2 ξ(1 + 2it) it it n=1 where 1 ξ(2it) ϕ( + it) = 2 ξ(1 + 2it)

− s s ξ(s) = π 2 Γ( )ζ(s). 2 Thus, we have Z 2 1 2 X < Ph,m, |µt| > = |E(z, + it)| ( h(y(γz))e(mx(γz)))dµ(z) Γ\H 2 γ∈Γ∞\Γ X Z ∞ m r dr = d (k)d (k + m) K (r)K (| 1 + |r) h it it itj itj k 2π|k| r k6=0,−m 0 Then, by the similar calculation in Section 2.1, we obtain the continuous spectrum integral is

Z ∞ −( t−T )2 −( t+T )2 2 2 T 1− T 1− 2 (e + e )|ζ(1 + 2it)| < Ph1,m1 , |µt| > < Ph2,m2 , |µt| >dt. 0

Remark 2.1. We notice that the continuous spectrum part is not negligible and contributes to the main term, but in general, it is small in the cuspidal subspace.

Z ∞ H(s) = h(y)y−s−1dy. 0

1 2 Since |ϕ( 2 + it)| = 1, the ﬁrst term of (2.8) contributes Z ∞ dy 2 h(y) 0 y and a rapidly decreasing function of t as t → ∞. We denote the second term of (2.8) as

Z ∞ 2 Z ∞ 1 X |σ−2it(n)| I(t) = H(s)( ) |K (2πy)|2)ys−1dy. πi|ξ(1 + 2it)|2 ns it R(s)=2 n=1 0 The series inside can be evaluated as was the work of Ramanujan [37],

∞ 2 2 X |σ−2it(n)| ζ (s)ζ(s − 2it)ζ(s + 2it) = . ns ζ(2s) n=1 Hence, combined with the evaluation of the K-Bessel functions [11],

Z ∞ 2 s s s 2 s−1 Γ ( 2 )Γ( 2 − it)Γ( 2 + it) |Kit(2πy)| )y dy = , 0 Γ(s) we have

Z −s 2 2 s s s 2 π H(s)ζ (s)ζ(s − 2it)ζ(s + 2it)Γ ( 2 )Γ( 2 − it)Γ( 2 + it) I(t) = 2 ds. πi|ξ(1 + 2it)| R(s)=2 ζ(2s)Γ(s) 30 1 Denote the above integrand as B(s) and shift the above integral to R(s) = 2 , we have Z 4πi 2 −20 I(t) = 2 Ress=1B(s) + 2 B(s)ds + O(t ). πi|ξ(1 + 2it)| πi|ξ(1 + 2it)| 1 R(s)= 2 The error term comes from the contribution of poles at s = 1 ± 2it of ζ(s ∓ 2it) and the fact that H(s) is rapidly decreasing in Im(s). Applying Weyl’s bound [50]

1 1 + ζ( + it) t 6 , 2 and Stirling’s formula

1/2 σ− 1 −π|t|/2 |Γ(σ + it)| ∼ (2π) t 2 e we have

2 Z 1 − 6 + 2 B(s)ds t . πi|ξ(1 + 2it)| 1 R(s)= 2 For the residue term, we can write B(s) as ζ2(s)G(s) where

π−sH(s)ζ(s − 2it)ζ(s + 2it)Γ2( s )Γ( s − it)Γ( s + it) G(s) = 2 2 2 ζ(2s)Γ(s)

0 is holomorphic at s = 1. Thus Ress=1B(s) = 2γG(1) + G (1). We can calculate that

24 24 Z ∞ dy Z G(1) = H(1) = h(y) 2 = Ph,0dµ(z) = 0, π π 0 y Γ\H π−1ζ(1 − 2it)ζ(1 + 2it)Γ2( 1 )Γ( 1 − it)Γ( 1 + it) G0(1) = H0(1) 2 2 2 . ζ(2) From Stirling’s formula 1 Γ( + 2it) ∼ (2π)1/2e−π|t| 2 and the arithmetic estimate [47]

(log t)−1 ζ(1 + it) log t, the residue term

Z ∞ 4πi 24 0 24 dy 2 Ress=1B(s) = H (1) = h(y) log y 2 . πi|ξ(1 + 2it)| π π 0 y 31 We obtain

∞ ∞ Z dy 24 Z dy 1 2 − 6 < Ph,0, |µt| >= 2 h(y) + h(y) log y 2 + O(t ), 0 y π 0 y Z ∞ dy ∞ if h(y) 2 = 0 or Ph,0 ∈ C0,0(X). 0 y We can easily choose h(y) ∈ C∞(R+), such that the above main term is nonzero Z ∞ dy 2 and h(y) 2 = 0. For example, let h(y) = y C(y) where C(y) is a step function 0 y Z ∞ Z ∞ such that C(y)dy = 0 and C(y)(y + log y)dy 6= 0, then apply approximation 0 0 argument to obtain a smooth function which satisﬁes the same condition.

Therefore, we conclude that the above continuous spectrum integral also contributes cT 1−ε.

In Chapter 5, if we restrict B to the Maass-Hecke cusp forms, we can show the continuous spectrum integral is small enough so we can obtain the ﬁrst part of main

Theorem 1.4.

32 2.3 The Asymptotic Formula for the Variance

From last section’s results, we obtain the following:

Proposition 2.1. Let ϕj denote the j-th Maass-Hecke eigenform with Laplacian

10 suﬃciently large N, hi(t) t when t → 0, i = 1, 2, ﬁx > 0, we have

t −T t +T X −( j )2 −( j )2 2 2 T 1− T 1− 2 (e + e )L(1, sym ϕj) < Ph1,m1 , |µj| > < Ph2,m2 , |µj| >

tj Z ∞ −( t−T )2 −( t+T )2 2 2 T 1− T 1− 2 + (e + e )|ζ(1 + 2it)| < Ph1,m1 , |µt| > < Ph2,m2 , |µt| >dt 0 1 1− 2 − = T B(Ph1,m1 ,Ph2,m2 ) + O(T ), where

B(Ph1,m1 ,Ph2,m2 ) √ ξ τ(1−τ) πm1 cos( ξ(2τ − 1))h1 π3/2 X Z ∞ Z 1 d1 d1 = dτ 64 0 0 τ(1 − τ) m1/d1=m2/d2 √ ξ η(1−η) πm2 cos( ξ(2η − 1))h2 Z 1 d2 d2 dξ dη 2 + (2.12) 0 η(1 − η) ξ ZZ 2 2 3 X X Scζ8 m1m2 m1ξ m2φ 2 2 Im( 3 ec( − 2 − 2 ) 3 c 2 R2 2d1d2 4d1φ 4d2ξ (ξφ) 2 d1|m1 c≥1 d2|m2 Z 1 Z 1 2 cos(πm1d2ξ(2τ − 1)) cos(πm2d1φ(2η − 1)) e((d1d2) ξφc) 0 0 τη(1 − τ)(1 − η) p p h1(ξd2 τ(1 − τ))h2(φd1 η(1 − η))dτdηdξdφ) (2.13)

Next, we remove the restriction on h1 and h2 by using Fourier transformation so that we can obtain a more general asymptotic formula for the quantum variance sum

Sψ(λ).

33 For any ϕ(x) smooth function with compact support on R+, write ϕ(x) = e−x2 xAψ(x), where A is a suﬃciently large even integer, thus, by Fourier inversion, Z ∞ Z ∞ ϕ(x) = e−x2 xA ψˆ(y)e(xy)dy = ψˆ(y)(e(xy)e−x2 xA)dy −∞ −∞ where Z ∞ ψˆ(y) = ψ(x)e(−xy)dx −∞ ∞ ˆ −x2 A since ψ(x) ∈ C0 (0, ∞), ψ(y) is analytic. ht(x) = e(xt)e x satisfy the condition of the Proposition 2.1 for each t, so we have

t −T t +T X −( j )2 −( j )2 2 2 T 1− T 1− 2 (e + e )L(1, sym ϕj)hPh,m1 , |µj| ihPh,m2 , |µj| i

tj Z ∞ −( t−T )2 −( t+T )2 2 2 T 1− T 1− 2 + (e + e )|ζ(1 + 2it)| hPh1,m1 , |µt| ihPh2,m2 , |µt| idt 0 1 1− 2 − = T B(Ph,m1 ,Ph,m2 ) + O(T ).

We also have X Pϕ,m(z) = ϕ(y(γz))e(mx(γz))

γ∈Γ∞\Γ Z ∞ X 2 = ψˆ(t)e(ty(γz))e−(y(γz)) (y(γz))Adt e(mx(γz)) −∞ γ∈Γ∞\Γ Z ∞ X 2 = ψˆ(t) e(ty(γz))e−(y(γz)) (y(γz))A e(mx(γz))dt −∞ γ∈Γ∞\Γ Z ∞ ˆ = ψ(t)Pe(xt)e−x2 xA,mdt (2.14) −∞ Thus, we have Z ∞ 2 ˆ 2 hPϕ,m, |µj| i = ψ(t)hPe(xt)e−x2 xA,m, |µj| idt (2.15) −∞

So, for any ϕ1, ϕ2 smooth with compact support, by (2.14) and (2.15),

t −T t +T X −( j )2 −( j )2 2 2 T 1− T 1− 2 (e + e )L(1, sym ϕj)hPϕ1,m1 , |µj| ihPϕ2,m2 , |µj| i

tj ∞ ∞ Z Z tj −T 2 tj +T 2 ˆ ˆ X −( 1− ) −( 1− ) 2 = ψ1(t1)ψ2(t2) (e T + e T )L(1, sym ϕj) −∞ −∞ tj 2 2 hP −x2 A , |µ | ihP −x2 A , |µ | idt dt e(xt1)e x ,m1 j e(xt2)e x ,m2 j 1 2 34 Z ∞ Z ∞ ˆ ˆ 1− = ψ (t )ψ (t )T B(P −x2 A ,P −x2 A )dt dt 1 1 2 2 e(xt1)e x ,m1 e(xt2)e x ,m2 1 2 −∞ −∞ 1 − +O(T 2 )

1 1− 2 − = T B(Pϕ1,m1 ,Pϕ2,m2 ) + O(T ).

For the last equality, we applied the expression of B in Proposition 2.1.

Therefore, we obtain the preliminary version for ﬁrst part of Theorem 1.5:

Theorem 2.2. Let ϕj denote the eigenfunctions of Laplacian, with the corresponding

1 2 eigenvalues λj = 4 + tj . For m1, m2 6= 0, h1, h2 be smooth functions with compact support for i = 1, 2, ﬁx > 0, we have

t −T t +T X −( j )2 −( j )2 2 2 T 1− T 1− 2 (e + e )L(1, sym ϕj)hPh1,m1 , |µj| ihPh2,m2 , |µj| i

tj Z ∞ −( t−T )2 −( t+T )2 2 2 T 1− T 1− 2 + (e + e )|ζ(1 + 2it)| < Ph1,m1 , |µt| > < Ph2,m2 , |µt| >dt 0 1 1− 2 − = T B(Ph1,m1 ,Ph2,m2 ) + O(T ), where B satisﬁes the expression in Proposition 2.1.

Remark 2.2. If any incomplete Poincar´eseries in the above theorem is replaced by incomplete Eisenstein series, i.e. mi = 0 with zero mean, the theorem is still valid.

For the case m1 = m2 = 0, there is a slight change for B: √ √ ξ τ(1−τ) ξ η(1−η) h1 h2 X Z ∞ Z 1 d1 Z 1 d2 dξ dτ dη 2 0 0 τ(1 − τ) 0 η(1 − η) ξ d1,d2≥1 √ √ P ξ τ(1−τ) P ξ η(1−η) h1 h2 Z ∞ Z 1 d1≥1 d1 Z 1 d2≥1 d2 dξ = dτ dη 2 0 0 τ(1 − τ) 0 η(1 − η) ξ By Euler-MacLaurin summation formula, we have

p ! ∞ p ! X ξ τ(1 − τ) Z ξ τ(1 − τ) dα h1 = − b2(α)H1 2 , d1 0 α α d1≥1 35 0 2 0 where b2(α) is the Bernoulli polynomial of degree 2, H1(x) = (h1(x)x ) . For the sum over d2, we have the similar expression.

36 CHAPTER 3

SYMMETRY PROPERTIES OF B

In this chapter we analyze the bilinear form B and using the series expression as obtained in last chapter, we prove it is self-adjoint with respect to the Laplacian ∆ and the Hecke operators Tn, n ≥ 1, i.e., for Poincar´eseries ψ1, ψ2, B satisﬁes the symmetries

B(∆ψ1, ψ2) = B(ψ1, ∆ψ2), (3.1) and for n ≥ 1,

B(Tnψ1, ψ2) = B(ψ1,Tnψ2). (3.2)

3.1 Self-adjointness of Laplacian for B

x ∞ Let Lm = Lm be the diﬀerential operator on C0 (0, ∞) given by

d2 L h(x) = (x2 − 4π2m2x2)h(x). m dx2

∞ If we deﬁne the inner product on C0 (0, ∞) by

Z ∞ dx (h1, h2) = h1(x)h2(x) 2 , 0 x then Lm is symmetric with respect to (, ), i.e.

(Lmh1, h2) = (h1,Lmh2).

37 It is easy to check

∆(h(y)e(mx)) = (Lmh)(y)e(mx), hence

∆Ph,m = PLmh,m. (3.3)

To prove (3.1), we ﬁrst deal with the case of m1m2 6= 0. We want to show

B(P ,P ) = B(P ,P ) Lm1 h1,m1 h2,m2 h1,m1 Lm2 h2,m2

For the diagonal part, it is suﬃce to prove, √ √ ξ τ(1−τ) ξ η(1−η) πm1 πm2 cos( ξ(2τ − 1))Lm h1 cos( ξ(2η − 1))h2 Z ∞ Z 1 d1 1 d1 Z 1 d2 d2 dξ dτ dη 2 0 0 τ(1 − τ) 0 η(1 − η) ξ is symmetric with respect to h1, h2. By integration by part, √ √ 00 ξ τ(1−τ) ξ η(1−η) πm1 πm2 cos( ξ(2τ − 1))h cos( ξ(2η − 1))h2 Z ∞ Z 1 d1 1 d1 Z 1 d2 d2 2 dτ dηdξ 0 0 d1 0 η(1 − η) √ √ 0 ξ τ(1−τ) 0 ξ η(1−η) cos( πm1 ξ(2τ − 1))h cos( πm2 ξ(2η − 1))h Z ∞ Z 1 Z 1 d1 1 d1 d2 2 d2 = − p p dτdηdξ 0 0 0 d1 τ(1 − τ) d2 η(1 − η) √ √ 0 ξ τ(1−τ) ξ η(1−η) h h2 Z ∞ Z 1 Z 1 1 d1 d2 πm1 πm2 + p (sin( ξ(2τ − 1)) cos( ξ(2η − 1)) 0 0 0 d τ(1 − τ) η(1 − η) d1 d2 πm πm πm πm 1 (2τ − 1) + cos( 1 ξ(2τ − 1)) sin( 2 ξ(2η − 1)) 2 (2η − 1))dτdηdξ d1 d1 d2 d2

The ﬁrst integral in the above sum is symmetric with respect to h1 and h2. For the second integral, by integration by part in terms of τ, it is easy to check the ﬁrst term

38 including sin( πm1 ξ(2τ − 1)) cos( πm2 ξ(2η − 1)) can be cancelled with the other term d1 d2 in the laplacian operator Lm1 , √ √ ξ τ(1−τ) ξ η(1−η) 2 πm1 πm2 m cos( ξ(2τ − 1))h1 cos( ξ(2η − 1))h2 Z ∞ Z 1 1 d1 d1 Z 1 d2 d2 2 4π 2 dτ dηdξ. 0 0 d1 0 η(1 − η)

For the second term including cos( πm1 ξ(2τ − 1)) sin( πm2 ξ(2η − 1)), by integration by d1 d2 parts in terms of ξ and η, it equals √ √ ξ τ(1−τ) ξ η(1−η) πm1 πm2 cos( ξ(2τ − 1))h1 cos( ξ(2η − 1))h2 2 Z ∞ Z 1 d1 d1 Z 1 d2 d2 2 m2 −π 2 dτ dηdξ. d2 0 0 τ(1 − τ) 0 η(1 − η)

It is symmetric since m1 = m2 . d1 d2 Note: In holomorphic case, the integral in the diagonal term is

Z ∞ ξ ξ dξ h1( )h2( ) 2 , 0 d1 d2 ξ which is self-adjoint with respect to ∆.

For the non-diagonal part, by change of variables

m m ξ → 2 ξ, φ → 1 φ, d2 d1 we symmetrize the integral kernel to

sin A cos B cos C 3 τη(1 − τ)(1 − η)(ξφ) 2 where

π πm1m2ξ πm1m2φ A = − − − + 2πd1d2m1m2cξφ, 4 2cd1d2φ 2cd1d2ξ

B = πm1m2ξ(2τ − 1) and C = πm1m2φ(2η − 1).

We can verify the following symmetry

d2 d2 (ξ2 − 4π2m2m2ξ2τ(1 − τ))K(ξ, φ) = (φ2 − 4π2m2m2φ2η(1 − η))K(ξ, φ), dξ2 1 2 dφ2 1 2 39 where

K(ξ, φ) = pξφ sin A cos B cos C, and A, B, C as above.

To show this, we check the symmetry in the following terms of

d2 (ξ2 − 4π2m2m2ξ2τ(1 − τ))K(ξ, η) dξ2 1 2 2 p πm1m2 πm1m2η 2 = −ξ sin A cos B cos C ξη(− + 2 + 2πm1m2d1d2ηc) (3.4) 2cd1d2η 2cd1d2ξ 2 p πm1m2 πm1m2η −2ξ cos A sin B cos C ξη(− + 2 + 2πm1m2d1d2ηc) 2cd1d2η 2cd1d2ξ

πm1m2(2τ − 1) (3.5) r 2 η πm1m2 πm1m2η +ξ cos A cos B cos C (− + 2 + 2πm1m2d1d2ηc) (3.6) ξ 2cd1d2η 2cd1d2ξ 2 p πm1m2η +ξ cos A cos B cos C ηξ(− 3 ) (3.7) cd1d2ξ 2 p 2 −ξ sin A cos B cos C ηξ(2πm1m2(2τ − 1)) (3.8) rη −ξ2 sin A sin B cos C (2πm m (2τ − 1)) (3.9) ξ 1 2 2 2 2 2 p −4π m1m2ξ sin A cos B cos C ηξτ(1 − τ) (3.10)

− sin A cos B cos Cpξη (3.11)

By some careful calculation, it can be seen the above terms are symmetric with respect to ξ and η, for example, the sum of (3.4), (3.8) and (3.10) can be canceled and only some symmetric terms left. More precisely, these three parts are

2 2 2 2 p 2 p −2π m1m2ξ sin A cos B cos C ηξ − ξ sin A cos B cos C ξη

πm1m2 πm1m2η 2 (− + 2 + 2πm1m2d1d2ηc) 2cd1d2η 2cd1d2ξ 2 2 2 2 p 2 p = −2π m1m2ξ sin A cos B cos C ηξ − ξ sin A cos B cos C ξη 2 2 2 πm1m2 2 πm1m2η 2 2 π m1m2 (( ) + ( 2 ) + (2πm1m2d1d2ηc) − 2 2 2 2 2cd1d2η 2cd1d2ξ 2c d1d2ξ 2π2m2m2η2 −2π2m2m2 + 1 2 1 2 ξ2

40 2 2 2 2 2 p 2 2 2 2 p = −2π m1m2(ξ + η ) sin A cos B cos C ηξ − π m1m2ξ sin A cos B cos C ξη

1 2 η 2 2 1 (( ) + ( 2 ) + (2d1d2ηc) − 2 2 2 2 ) 2cd1d2η 2cd1d2ξ 2c d1d2ξ 2 2 2 2 2 p 2 2 2 = −2π m1m2(ξ + η ) sin A cos B cos C ηξ − π m1m2 sin A cos B cos C 5/2 3/2 1 2 ξ 1 2 η 2 5/2 1 (( ) 3/2 + ( ) 5/2 + (2d1d2c) (ξη) − 2 2 2 ) 2cd1d2 η 2cd1d2 ξ 2c d1d2 Thus, combine with (3.3) and the expression of B in Theorem 2.2, we obtain (3.1) when m1m2 6= 0.

If m1m2 = 0, it can be shown by using the continuity argument for the desym- metrized integral kernel.

3.2 Self-adjointness of Hecke Operators for B

In order to prove (3.2), we can check it for each Hecke operator Tp, where p is a prime, i.e.

B(TpPh1,m1 ,Ph2,m2 ) = B(Ph1,m1 ,TpPh2,m2 )

We will use the fact (see the proof of Theorem 6.9 in [16]

2 X d 1 TnPh,m(z) = ( ) 2 Ph( ny ), mn (z), (3.12) n d2 d2 d|(m,n) and the explicit evaluation of Sc, m1 , m2 (γ) to verify it. m2 d2 Write

B(Ph1,m1 ,Ph2,m2 ) = B∞(Ph1,m1 ,Ph2,m2 ) + Bf (Ph1,m1 ,Ph2,m2 ), where √ √ ξ τ(1−τ) ξ η(1−η) h1 h2 X Z ∞ Z 1 d1 Z 1 d2 dξ B∞(Ph1,m1 ,Ph2,m2 ) = dτ dη 2 0 0 τ(1 − τ) 0 η(1 − η) ξ d1|m1,d2|m2 m1/d1=m2/d2

41 and

Bf (Ph1,m1 ,Ph2,m2 ) ZZ 2 2 3 X X Scζ8 m1m2 m1φ m2ξ 2 2 = Im( 3 ec( − 2 − 2 ) 3 e(ξφc) c 2 R2 2d1d2 4d1ξ 4d2φ (ξφ) 2 d1|m1 c≥1 d2|m2 2iφηm 2iξτm 1 1 2 − 1 p ! p ! Z Z e d2 d1 ξ τ(1 − τ) φ η(1 − η) h1 h2 dτdηdξdφ) 0 0 τη(1 − τ)(1 − η) 2πd1 2πd2

We consider the following 4 cases:

(1) If p - m1m2, B∞(TpPh1,m1 ,Ph2,m2 ) = B∞(Ph1,m1 ,TpPh2,m2 );

(2) If p - m1m2, Bf (TpPh1,m1 ,Ph2,m2 ) = Bf (Ph1,m1 ,TpPh2,m2 );

a (3) If p k (m1, m2), B∞(TpPh1,m1 ,Ph2,m2 ) = B∞(Ph1,m1 ,TpPh2,m2 );

a (4) If p k (m1, m2), Bf (TpPh1,m1 ,Ph2,m2 ) = Bf (Ph1,m1 ,TpPh2,m2 ). To prove (1), we use the fact

− 1 TpPh,m(z) = p 2 Ph(p·),pm(z)

pm1 m2 from (3.4). Also, from the conditions d1|pm1, d2|m2 and = we have p|d1. d1 d2

Thus, making the change of variable d1 → pd1, we have

B∞(TpPh1,m1 ,Ph2,m2 )

1 − 2 = p B∞(Ph1(p·),pm1 ,Ph2,m2 ) √ √ ξ τ(1−τ) ξ η(1−η) h1 h2 Z ∞ Z 1 d1 Z 1 d2 1 X dξ − 2 = p dτ dη 2 0 0 τ(1 − τ) 0 η(1 − η) ξ d1|m1,d2|m2 m1/d1=m2/d2 1 − 2 = p B∞(Ph1,m1 ,Ph2(p·),pm2 )

= B∞(Ph1,m1 ,TpPh2,m2 ).

For (2), we have

42 Bf (TpPh1,m1 ,Ph2,m2 )

1 − 2 = p Bf (Ph1(p·),pm1 ,Ph2,m2 ) ZZ 2 2 2 3 Z 1 Z 1 1 X X Scζ8 pm1m2 p m φ m ξ 2 2 − 2 1 2 = p Im( 3 ec( − 2 − 2 ) 3 e(ξφc) c 2 R2 2d1d2 4d1ξ 4d2φ (ξφ) 2 0 0 d1|pm1 c≥1 d2|m2 2iφηm2 − 2iξτpm1 p ! p ! e d2 d1 pξ τ(1 − τ) φ η(1 − η) h1 h2 dτdηdξdφ) τη(1 − τ)(1 − η) 2πd1 2πd2 ZZ 2 2 2 3 Z 1 Z 1 1 X X Scζ8 pm1m2 p m φ m ξ 2 2 − 2 1 2 = p Im( 3 ec( − 2 − 2 ) 3 e(ξφc) c 2 R2 2d1d2 4d1ξ 4d2φ (ξφ) 2 0 0 d1|m1 c≥1 d2|m2 2iφηm2 − 2iξτpm1 p ! p ! e d2 d1 pξ τ(1 − τ) φ η(1 − η) h1 h2 dτdηdξdφ) (3.13) τη(1 − τ)(1 − η) 2πd1 2πd2 ZZ 2 2 3 Z 1 Z 1 1 X X Scζ8 m1m2 m φ m ξ 2 2 − 2 1 2 +p Im( 3 ec( − 2 − 2 ) 3 e(ξφc) c 2 R2 2d1d2 4d1ξ 4d2φ (ξφ) 2 0 0 d1|m1 c≥1 d2|m2 2iφηm2 − 2iξτm1 p ! p ! e d2 d1 ξ τ(1 − τ) φ η(1 − η) h1 h2 dτdηdξdφ) (3.14) τη(1 − τ)(1 − η) 2πd1 2πd2

The ﬁrst sums in (3.13) and (3.14) correspond to the conditions p - d1, and p|d1 respectively.

Similarly, we have

Bf (Ph1,m1 ,TpPh2,m2 ) ZZ 2 2 2 3 Z 1 Z 1 1 X X Scζ8 pm1m2 m φ p m ξ 2 2 − 2 1 2 = p Im( 3 ec( − 2 − 2 ) 3 e(ξφc) c 2 R2 2d1d2 4d1ξ 4d2φ (ξφ) 2 0 0 d1|m1 c≥1 d2|m2 2iφηpm2 − 2iξτm1 p ! p ! e d2 d1 ξ τ(1 − τ) pφ η(1 − η) h1 h2 dτdηdξdφ) (3.15) τη(1 − τ)(1 − η) 2πd1 2πd2 ZZ 2 2 3 Z 1 Z 1 1 X X Scζ8 m1m2 m φ m ξ 2 2 − 2 1 2 +p Im( 3 ec( − 2 − 2 ) 3 e(ξφc) c 2 R2 2d1d2 4d1ξ 4d2φ (ξφ) 2 0 0 d1|m1 c≥1 d2|m2 2iφηm2 − 2iξτm1 p ! p ! e d2 d1 ξ τ(1 − τ) φ η(1 − η) h1 h2 dτdηdξdφ) (3.16) τη(1 − τ)(1 − η) 2πd1 2πd2 43 ξ (3.14) is same as (3.16). Making the change of variables ξ → p , φ → pφ, we have (3.13) is equal to (3.15), i.e., we have proved (2).

In the cases (3) and (4), we use the fact

− 1 1 T P (z) = p 2 P (z) + p 2 P · m (z). p h,m h(p·),pm h( p ), p where if p m, we understand that P · m (z) = 0. - h( p ), p Thus, for the case (3), we have

B∞(TpPh1,m1 ,Ph2,m2 )

− 1 1 2 2 · m = p B∞(Ph (p·),pm , ph2,m2 ) + p B∞(P 1 ,Ph2,m2 ) 1 1 h1( p ), p = A + B

Similarly,

B∞(Ph1,m1 ,TpPh2,m2 )

− 1 1 2 2 m · m = p B∞(Ph1,m1 , ph (p·),pm ) + p B∞(P 1 ,P 2 ) 2 2 h1, p h2( p ), p

= A1 + B1

Next, we check that

A(p|d1) = A1(p|d2),

A(p - d1) = B1(p - d1),

B(p - d2) = A1(p - d2),

B(p|d2) = B1(p|d1).

Hence, we get (3).

The proof of (4) is the most tedious one and we will use the induction to prove that. We have

44 Bf (TpPh1,m1 ,Ph2,m2 )

− 1 1 2 2 · m = p Bf (Ph (p·),pm ,Ph2,m2 ) + p Bf (P 1 ,Ph2,m2 ) 1 1 h1( p ), p

From the expression of (2.10), it equals

ZZ 2 2 2 3 Z 1 Z 1 1 X X Scζ8 pm1m2 p m φ m ξ 2 2 − 2 1 2 p Im( 3 ec( − 2 − 2 ) 3 e(ξφc) c 2 R2 2d1d2 4d1ξ 4d2φ (ξφ) 2 0 0 d1|pm1 c≥1 d2|m2 2iφηm2 − 2iξτpm1 p ! p ! e d2 d1 pξ τ(1 − τ) φ η(1 − η) h1 h2 dτdηdξdφ) τη(1 − τ)(1 − η) 2πd1 2πd2 ZZ 2 2 2 3 Z 1 Z 1 1 X X Scζ8 pm1m2 p m φ m ξ 2 2 2 1 2 +p Im( 3 ec( − 2 − 2 ) 3 e(ξφc) c 2 R2 2d1d2 4d1ξ 4d2φ (ξφ) 2 0 0 d1|pm1 c≥1 d2|m2 2iφηm2 − 2iξτpm1 p ! p ! e d2 d1 pξ τ(1 − τ) φ η(1 − η) h1 h2 dτdηdξdφ) τη(1 − τ)(1 − η) 2πd1 2πd2 ZZ 2 2 2 3 Z 1 Z 1 1 X X Scζ8 pm1m2 p m φ m ξ 2 2 − 2 1 2 = p Im( 3 ec( − 2 − 2 ) 3 e(ξφc) c 2 R2 2d1d2 4d1ξ 4d2φ (ξφ) 2 0 0 d1|m1 c≥1 d2|m2 2iφηm2 − 2iξτpm1 p ! p ! e d2 d1 pξ τ(1 − τ) φ η(1 − η) h1 h2 dτdηdξdφ) (3.17) τη(1 − τ)(1 − η) 2πd1 2πd2 ZZ 2 2 3 Z 1 Z 1 1 X X Scζ8 m1m2 m φ m ξ 2 2 2 1 2 +p Im( 3 ec( − 2 − 2 ) 3 e(ξφc) c 2 R2 2d1d2 4d1ξ 4d2φ (ξφ) 2 0 0 d1|m1 c≥1 d2|m2 2iφηm2 − 2iξτm1 p ! p ! e d2 d1 ξ τ(1 − τ) φ η(1 − η) h1 h2 dτdηdξdφ) (3.18) τη(1 − τ)(1 − η) 2πd1 2πd2

We denote the above sum as I1 + I2. Similarly,

Bf (Ph1,m1 ,TpPh2,m2 )

− 1 1 2 2 · m = p Bf (Ph1,m1 ,Ph (p·),pm ) + p Bf (Ph1,m1 ,P 2 ) 2 2 h2( p ), p

From the expression of (2.10), it equals

45 ZZ 2 2 2 3 Z 1 Z 1 1 X X Scζ8 pm1m2 p m φ m ξ 2 2 − 2 1 2 p Im( 3 ec( − 2 − 2 ) 3 e(ξφc) c 2 R2 2d1d2 4d1ξ 4d2φ (ξφ) 2 0 0 d1|pm1 c≥1 d2|m2 2iφηm2 − 2iξτpm1 p ! p ! e d2 d1 pξ τ(1 − τ) φ η(1 − η) h1 h2 dτdηdξdφ) τη(1 − τ)(1 − η) 2πd1 2πd2 ZZ 2 2 2 3 Z 1 Z 1 1 X X Scζ8 pm1m2 p m φ m ξ 2 2 2 1 2 +p Im( 3 ec( − 2 − 2 ) 3 e(ξφc) c 2 R2 2d1d2 4d1ξ 4d2φ (ξφ) 2 0 0 d1|pm1 c≥1 d2|m2 2iφηm2 − 2iξτpm1 p ! p ! e d2 d1 pξ τ(1 − τ) φ η(1 − η) h1 h2 dτdηdξdφ) τη(1 − τ)(1 − η) 2πd1 2πd2 ZZ 2 2 2 3 Z 1 Z 1 1 X X Scζ8 pm1m2 p m φ m ξ 2 2 − 2 1 2 = p Im( 3 ec( − 2 − 2 ) 3 e(ξφc) c 2 R2 2d1d2 4d1ξ 4d2φ (ξφ) 2 0 0 d1|m1 c≥1 d2|m2 2iφηm2 − 2iξτpm1 p ! p ! e d2 d1 pξ τ(1 − τ) φ η(1 − η) h1 h2 dτdηdξdφ) τη(1 − τ)(1 − η) 2πd1 2πd2 ZZ 2 2 3 Z 1 Z 1 1 X X Scζ8 m1m2 m φ m ξ 2 2 2 1 2 +p Im( 3 ec( − 2 − 2 ) 3 e(ξφc) c 2 R2 2d1d2 4d1ξ 4d2φ (ξφ) 2 0 0 d1|m1 c≥1 d2|m2 2iφηm2 − 2iξτm1 p ! p ! e d2 d1 ξ τ(1 − τ) φ η(1 − η) h1 h2 dτdηdξdφ) τη(1 − τ)(1 − η) 2πd1 2πd2

= II1 + II2

According to whether or not p|(c, ∗, ∗) in Sc,∗,∗, we can decompose the above sums

I1, I2, II1, II2 into the following 8 sums

I1 = I11 + I12,I2 = I21 + I22,II1 = II11 + II12,II2 = II21 + II22.

2 2 Note if p|(c, ∗, ∗), Sc,∗,∗ = 0 unless p |c. Let c = p c1, we have

2 δ(p, c1) S |m1p| |m2| = S |m1| |m2| p (1 − ), c, , c1, , d1 d2 d1 pd2 p

0 00 where δ(p, c1) = 0 if p|c1; δ(p, c1) = 1 if p - c1. Hence we can write I11 = I11 − I11 correspondingly. 46 Similarly we have

2 δ(p, c1) S |m1| |m2| = S |m1| |m2| p (1 − ), c, , c1, 2 , pd1 d2 p d1 pd2 p

0 00 and write I21 = I21 − I21,

2 δ(p, c1) S |m1| |m2p| = S |m1| |m2| p (1 − ), c, , c1, , d1 d2 pd1 d2 p

0 00 and write II11 = II11 − II11,

2 δ(p, c1) S |m1| |m2| = S |m1| |m2| p (1 − ), c, , c1, , 2 d1 pd2 pd1 p d2 p

0 00 and write II21 = II21 − II21 corresponding p|c1 or not.

m1 m2 0 0 0 0 By the induction hypothesis on ( p , p ), we have I11 + I21 = II11 + II21. 2 We have Scp,a,b = p Sc,a,b and Stp2,ap,b = 0 if p - bc. Using this and the evaluation of Sc,a,b we can verify that

I12(p|d1) = II12(p|d2), where I12(p|d1) means the partial sum of I12 in which p|d1). Similarly, we have

I12(p - d1, p - d2, p - c) = II12(p - d2, p - d1, p - c),

I12(p - d1, p k d2, p - c) = II12(p - d2, p k d1, p - c),

2 00 2 I12(p - d1, p |d2, p - c) = I11(p - d1, p |m2/d1),

2 00 I12(p - d1, p |d2, p - c) = I11(p - d1, p k m2/d2),

00 2 2 II11(p - d2, p |m1/d1) = II12(p - d2, p |d1, p - c),

00 II11(p - d2, p k m1/d1) = II12(p - d2, p - c),

00 00 I11(p|d1) = II11(p|d2),

I22(p|d2) = II22(p|d1),

I22(p - d2, p - d1, p - c) = II22(p - d1, p - d2, p - c), 47 I22(p - d2, p k d1, p - c) = II22(p - d1, p k d2, p - c),

2 00 3 I22(p - d2, p |d1, p - c) = I21(p - d2, p |m1/d1),

00 2 I22(p - d2, p|c) = I21(p - d2, p k m1/d1),

00 00 II21(p|d1) = I21(p|d2),

2 00 3 II22(p - d1, p |d2, p - c) = II21(p - d1, p |m2/d2),

00 2 II22(p - d1, p|c) = II21(p - d1, p k m2/d2).

Hence we deduce from the above identities that

Bf (TpPh1,m1 ,Ph2,m2 ) = Bf (Ph1,m1 ,TpPh2,m2 ).

This completes the proof of

B(TpPh1,m1 ,Ph2,m2 ) = B(Ph1,m1 ,TpPh2,m2 )

for each Tp, p is a prime. Thus, the bilinear form B(·, ·) deﬁned on the space spanned by Ph,m’s is self-adjoint with respect to the Hecke operators Tn, n ≥ 1.

48 CHAPTER 4

EXTENSION AND DIAGONALIZATION OF B

∞ In this chapter, we apply the method in [34] to extend B to C0,0(X), where

X = Γ \ H. If we restrict B to the Maass-Hecke cusp forms, it can be shown that B is diagonalized by the orthonormal basis of Hecke-Maass cusp forms. The idea of the extension is choose a partition of unity subordinate to a covering of X, then

∞ any ψ ∈ C0,0(X) can be written as a linear combination of Ph,m, where h is smooth with compact support functions. From last chapter’s symmetry properties of B, it is easy to see that B can be diagonalized by the orthonormal basis of the subspace of

Hecke-Maass cusp forms.

Let P be the projection map from H to X, and construct the following open sets with compact closures in H whose projections to X form a locally ﬁnite open covering of X: D00 is a neighborhood of i and the restriction of P on D00 is two to one. D01

2πi is a neighborhood of e 3 and the restriction of P on D01 is three to one.

1 D = {z|I(z) < 2, |R(z)| < , |z| > 1}. 02 2 1 1 D = {z|I(z) < 2, − ≤ |R(z)| < 0, |z| > 1}∪{z|I(z) < 2, −1 < |R(z)| ≤ − , |z+1| > 1}. 03 2 2 For k ≥ 1, let 3k D = {z| < I(z) < 3k+1, −1 < |R(z)| < 0}. k1 2

3k 1 1 D = {z| < I(z) < 3k+1, − < |R(z)| < }. k2 2 2 2 49 Let {fkj} be the partition of unity subordinate to the above covering of X, then each fkj can be viewed as an automorphic function with respect to Γ. We can extend fk,j ˜ ˜ to a smooth Γ∞ periodic function fk,j on H. There exists y0 > 0 such that fk,j are all ˜ supported in the half-plane y ≥ y0, where fk,j(z) = fk,j(z), except when k = 0 and j = 2, 3.

∞ Let ψ ∈ C0,0(X), we have

X ψ(z) = fk,j(z)ψ(z), k,j and 1 X ˜ fk,j(z)ψ(z) = fk,j(γz)ψ(γz), nk,j γ∈Γ∞\Γ where nk,j = 2 if k = 0, and j = 0; nk,j = 3 if k = 0, and j = 1; nk,j = 1, if otherwise. ˜ Expanding fk,j(z)ψ(z) into the Fourier series in x, we have

˜ X fk,j(z)ψ(z) = hk,j,m(y)e(mx), m∈Z where hk,j,m(y) are smooth with compact support and satisfy the rapid decay condition

−A −A hk,j,m(y) A y |m| for any A > 0. Thus, we have the following decomposition:

X X 1 ψ(z) = Phk,j,m,m(z) nk,j k,j m∈Z X X 1 X 1 2 = Phk,j,m,m(z) + Phk,j,0,0(z) − Ph0,0,0,0(z) − Ph0,1,0,0(z) nk,j 2 3 k,j m6=0 k,j X X 1 1 2 = Phk,j,m,m(z) + PH,0(z) − Ph0,0,0,0(z) − Ph0,1,0,0(z), nk,j 2 3 k,j m6=0

∞ where H(y) ∈ Cc (0, ∞) is deﬁned as

X H(y) = hk,j,0(y). k,j 50 P ˜ This follows from the fact that k,j fk,j(z) = 1, and fk,j(z) are all supported in the half-plane y ≥ y0 where we have

Z 1 X X ˜ hk,j,0(y) = ( fk,j(z))ψ(z)dx k,j 0 k,j Z 1 X ˜ ˜ = ( fk,j(z) + (f0,2(z) − f0,2(z)) + (f0,3(z) − f0,3(z)))ψ(z)dx 0 k,j Z 1 ˜ ˜ = ((f0,2(z) − f0,2(z)) + (f0,3(z) − f0,3(z)))ψ(z)dx, 0 R 1 since 0 ψ(z)dx = 0. Moreover we have

Z 1 2 (PH,0(z) − Ph0,0,0,0(z) − Ph0,1,0,0(z))dµ = 0. X 2 3

Let 1 2 P (z) − P (z) − P (z) = P (z), H,0 2 h0,0,0,0 3 h0,1,0,0 h,0 with 1 2 h = H − h − h . 2 0,0,0 3 0,1,0 We have X X 1 ψ(z) = Phk,j,m,m(z) + Ph,0(z), nk,j k,j m6=0 with Z Ph,0(z)dµ = 0. X ∞ Thus, it follows from Theorem 2.2 that for ψ and φ in C0,0(X), we have

tj −T 2 tj +T 2 X −( 1− ) −( 1− ) 2 2 2 (e T + e T )L(1, sym ϕj)hψ, |µj| ihφ, |µj| i

tj ∞ Z t−T 2 t+T 2 −( 1− ) −( 1− ) 2 2 2 + (e T + e T )|ζ(1 + 2it)| < ψ, |µt| > < φ, |µt| >dt 0 1− 1 − = T B(ψ, φ) + O(T 2 ),

51 where

X 1 B(ψ, φ) = B(P ,P ) hk1,j1,m1 ,m1 hk2,j2,m2 ,m2 nk1,j1 nk2,j2 k1,j1,m16=0;k2,j2,m26=0 X 1 + B(P ,P ) hk1,j1,m1 ,m1 hφ,0 nk1,j1 k1,j1,m16=0 X 1 + B(P ,P ) hψ,0 hk2,j2,m2 ,m2 nk2,j2 k2,j2,m26=0

+B(Phψ,0,Phφ,0). (4.1)

Since for φ and ψ we have the rapid decay condition

−A −A hk,j,m(y) A y |m| for any A > 0, we can see the above series converges abosolutely.

From last chapter’s results of (3.1) and (3.2) it follows that the bilinear form

∞ ∞ B(ψ, φ) is deﬁned on C0,0(X) × C0,0(X) and satisﬁes

B(∆ψ, φ) = B(ψ, ∆φ), and for n ≥ 1,

B(Tnψ, φ) = B(ψ, Tnφ).

If we restrict B to the Maass-Hecke cusp forms, we can show the continuous spectrum integral is small enough, see Section 5.2.3. It can be shown that the contin-

1 + uous spectrum integral contributes O(T 2 ). Hence, we obtain the ﬁrst part of main

Theorem 1.4.

Now restrict B to the Maass-Hecke cusp forms, it is easy to see B is diagonalized by the orthonormal basis of this subspace from the symmetry properties of B. If ψ,

φ are two distinct Maass-Hecke cusp forms, then for n ≥ 1,

B(Tnψ, φ) = B(ψ, Tnφ).

52 We have

λn(ψ)B(ψ, φ) = λn(φ)B(ψ, φ).

Since there must be an n such that λn(ψ) 6= λn(φ) for ψ and φ, we obtain

B(ψ, φ) = 0 for distinct Maass-Hecke cusp forms. Thus, we have shown that B is diagonalized by the orthonormal basis of the subspace of the Maass-Hecke cusp forms.

We are going to calculate the eigenvalue of B on such a Maass-Hecke cusp form in next chapter.

53 CHAPTER 5

EIGENVALUES OF B

In this chapter, we compute the value of B on φ(z), an even Maass-Hecke cuspidal

1 2 eigenform for the modular group Γ, with the Laplacian eigenvalue λφ = 4 + tφ. To do so we return to the original asymptotic of

t −T t +T X −( j )2 −( j )2 2 2 T 1− T 1− (e + e )L(1, sym φj)|hµφj , φi| . j≥1

2 After applying Watson’s formula on |hµφj , φi| , the quantum variance sum over φj

2 boils down to averaging L(1/2, φ ⊗ sym (φj)). By Rankin-Selberg theory, we can express this sum in a suitable series to which the Kuznetsov formula is applied. The estimation of diagonal and non-diagonal terms is similar to the way we did before.

However, for the continuous spectrum part, we apply Jutila’s subconvexity bound to show it is negligible in this case.

5.1 Watson’s Formula

Let L(s, φ) be the associated standard L-function, which admits analytic continuation to the whole complex plane and satisﬁes the functional equation:

s + it s − it Λ (s) := π−sΓ φ Γ φ L(s, φ) = Λ (1 − s). φ 2 2 φ

Moreover, we have

−3s/2 s s s 2 Λ 2 (s) = π Γ Γ + it Γ − it L(s, sym (φ)), sym (φ) 2 2 φ 2 φ 54 −3s s + itψ s + itψ s + itψ Λ 2 (s) = π Γ + it Γ − it Γ ψ⊗sym (φ) 2 φ 2 φ 2 s − it s − it s − it Γ ψ + it Γ ψ − it Γ ψ L(s, ψ ⊗ sym2(φ)), 2 φ 2 φ 2 here ψ(z) is also a Maass-Hecke cuspidal eigenform and we normalize ψ(z) so that its ﬁrst Fourier coeﬃcient aψ(1) = 1. By Watson’s formula [49] and Stirling formua, we have

1 1 2 Λψ⊗sym2(φ)( 2 )Λψ( 2 ) |hµφ, ψi| = 2 Λsym2(φ)(1) Λsym2(ψ)(1) 1 1 2 1 itψ 4 2 L( 2 , ψ)L( 2 , ψ ⊗ sym (φ)) cosh(πtψ)|Γ( 4 + 2 )| |aφ(1)| −1 = 2 2 (1 + O(tφ )) tφL(1, sym (φ))L(1, sym (ψ)) 1 1 itψ 4 2 L( 2 , ψ ⊗ φ ⊗ φ)|Γ( 4 + 2 )| |aφ(1)| −1 = 2 (1 + O(tφ )) tφL(1, sym (φ))

5.2 The Kuznetsov Formula

Next, we compute

t −T t +T X −( j )2 −( j )2 2 2 T 1− T 1− (e + e )L(1, sym φj)|hµφj , ψi| j≥1

Let Φ be the cuspidal automorphic form on GL(3) which is the Gelbart-Jacquet lift of the cusp form φ, with the Fourier coeﬃcients aΦ(m1, m2) [2], where

X m1 m2 a (m , m ) = λ ( , 1)λ ( , 1)µ(d), Φ 1 2 Φ d Φ d d|(m1,m2) and

X 2 λΦ(r, 1) = λφ(t ). s2t=r The Rankin-Selberg convolution L(s, ψ ⊗ sym2(φ)) is represented by the Dirichlet series [3],

2 X 2 −s L(s, ψ ⊗ sym (φ)) = λψ(m1)aΦ(m1, m2)(m1m2) , m1,m2≥1 55 where λψ(r) is the r-th Hecke eigenvalue of ψ. We have

2 Z ds Λψ⊗sym2(φ)(1/2) = Λψ⊗sym2(φ)(s + 1/2) . 2πi (2) s

So,

2 2 X 2 −1/2 m1m2 L(1/2, ψ ⊗ sym (φj)) = 2 λψ(m1)aΦj (m1, m2)(m1m2) V ( 2 ) tj m1,m2≥1 where Z 2 1 −s γ(1/2 + s, ψ ⊗ sym (φj)) ds V (y) = y 2 , 2πi (2) γ(1/2, ψ ⊗ sym (φj)) s and

s + it s + it s + it γ(s, ψ ⊗ sym2(φ )) = π−3sΓ ψ + it Γ ψ − it Γ ψ j 2 j 2 j 2 s − it s − it s − it Γ ψ + it Γ ψ − it Γ ψ . 2 j 2 j 2

Thus, we have

t −T t +T X −( j )2 −( j )2 2 2 T 1− T 1− (e + e )L(1, sym φj)|hµφj , ψi| j≥1

tj −T 2 tj +T 2 −( 1− ) −( 1− ) 1 1 1 4 X e T + e T 2 2 = L( , ψ)|Γ( + it)| |aj(1)| L(1/2, ψ ⊗ sym (φj)) 2 4 2 tj tj ≥1

tj −T 2 tj +T 2 −( 1− ) −( 1− ) 1 1 1 4 X e T + e T 2 = L( , ψ)|Γ( + it)| |aj(1)| 2 4 2 tj tj ≥1 2 X 2 −1/2 m1m2 λψ(m1)aΦj (m1, m2)(m1m2) V ( 2 ) tj m1,m2≥1 3 2 1 1 1 4 X X µ(d) X d n1n2 2 −1/2 = L( , ψ)|Γ( + it)| 3 λψ(dn1)V ( 2 )(n1n2) 2 4 2 d 2 tj tj ≥1 d≥1 n1,n2≥1

tj −T 2 tj +T 2 −( 1− ) −( 1− ) e T + e T 2 |aj(1)| λΦj (n1, 1)λΦj (n2, 1) tj 3 2 4 2 1 1 1 4 X X µ(d) X 2 d s1t1s2t2 2 4 2 −1/2 = L( , ψ)|Γ( + it)| 3 λψ(ds1t1)V ( 2 )(s1t1s2t2) 2 4 2 d 2 tj tj ≥1 d≥1 s1,s2,n1,n2≥1

tj −T 2 tj +T 2 −( 1− ) −( 1− ) e T + e T 2 2 2 |aj(1)| λj(t1)λj(t2) tj 56 1 1 1 4 X µ(d) X 2 2 4 2 −1/2 = L( , ψ)|Γ( + it)| 3 λψ(ds1t1)(s1t1s2t2) 2 4 2 d 2 d≥1 s1,s2,n1,n2≥1

tj −T 2 tj +T 2 3 2 4 2 −( 1− ) −( 1− ) X d s1t1s2t2 e T + e T 2 2 2 V ( 2 ) |aj(1)| λj(t1)λj(t2) tj tj tj ≥1 For the inner sum, by the Kuznetsov formula [25], we have

tj −T 2 tj +T 2 3 2 4 2 −( 1− ) −( 1− ) X d s1t1s2t2 e T + e T 2 2 2 V ( 2 ) |aj(1)| λj(t1)λj(t2) tj tj tj ≥1 Z ∞ 3 2 4 2 δ(t1, t2) d s1s2t1t2 −( t−T )2 −( t+T )2 T 1− T 1− = 2 V ( 2 )(e + e ) tanh(πt)dt (5.1) π −∞ t

t−T 2 t+T 2 Z ∞ 3 2 4 2 −( 1− ) −( 1− ) 2 d s1t1s2t2 e T + e T 2 2 − V ( 2 ) 2 dit(t1)dit(t2)dt (5.2) π 0 t t|ζ(1 + 2it)| 2 2 Z ∞ 3 2 4 2 X S(t , t ; c) 4πt1t2 d s s t1t + 1 2 J ( )V ( 1 2 2 ) c 2it c t2 c≥1 −∞

−( t−T )2 −( t+T )2 dt (e T 1− + e T 1− ) (5.3) cosh(πt)

Next, we will estimate the above three sums respectively.

5.2.1 The Diagonal Term

The diagonal term coming from (5.1) is

∞ 1 1 1 Z t−T 2 t+T 2 4 −( 1− ) −( 1− ) X −2 L( , ψ)|Γ( + itψ)| (e T + e T ) s2 2 4 2 −∞ s2≥1 X s2s4 s−1λ (s2)V ( 1 2 ) tanh(πt)dt 1 ψ 1 T 2 s1≥1

For the sum over s1, we have

2 4 Z 2 4 X −1 2 s1s2 1 X λψ(s1) s2 −s ds s1 λψ(s1)V ( 2 ) = 2s+1 Ut(s)( 2 ) , T 2πi (2) s T s s1≥1 s1≥1 1 where s+1/2−itψ s+1/2+itψ Γ 2 Γ 2 Ut(s) = (1 + Pt(s)) , 1/2−itψ 1/2+itψ Γ 2 Γ 2 57 and N+2 X pr+1(s) |s| P (s) = + O( ) t tr tN+1 1≤r≤N is an analytic function in Rs ≥ −2. pr+1(s) is a polynomial of degree at most r + 1. Also, we have [18] 2 X λψ(s1) 1 2 s = L(s, sym ψ). s1 ζ(2s) s1≥1

Thus, moving the line of integration in the sum over s1 to R(s) = −1/4 + , we get

X s2s4 1 s−1λ (s2)V ( 1 2 ) = L(1, sym2ψ) + O(T −1/2+) 1 ψ 1 T 2 ζ(2) s1≥1 Therefore, we get the diagonal terms contribute

π 1 1 1 T 1−L(1, sym2ψ)L( , ψ)|Γ( + it )|4 + O(T 1/2+). 4 2 4 2 ψ

5.2.2 The Non-diagonal Term

Next, we estimate the non-diagonal term from (5.3)

2 2 X µ(d) X 2 2 4 2 −1/2 X S(t1, t2; c) 3 λψ(ds1t1)(s1t1s2t2) d 2 c d≥1 s1,s2,t1,t2≥1 c≥1 Z ∞ 3 2 4 2 4πt1t2 d s1t1s2t2 −( t−T )2 −( t+T )2 dt T 1− T 1− J2it( )V ( 2 )(e + e ) −∞ c t cosh(πt)

4πt1t2 Let x = c , the inner integral in the non-diagonal terms is

Z ∞ 3 2 4 2 1 J2it(x) − J−2it(x) d s1t1s2t2 −( t−T )2 −( t+T )2 T 1− T 1− V ( 2 )(e + e ) tanh(πt)dt. 2 −∞ sinh πt t

Since tanh(πt) =sgn(t) + O(e−π|t|) for large |t|, we can remove tanh(πt) by getting a negligible term O(T −N ) for any N > 0. Applying the Parseval identity, the Fourier transform in [27], (see also Bateman, volume 1, p. 59) ! J (x)\− J (x) 2it −2it (y) = −i cos(x cosh(πy)). sinh(πt) and the evaluation of the Fresnel integrals, the integral is 58 Z ∞ 3 2 4 2 1 J2it(x) − J−2it(x) d s1t1s2t2 −( t−T )2 −( t+T )2 T 1− T 1− V ( 2 )(e + e ) tanh(πt)dt 2 −∞ sinh πt t Z ∞ 3 2 4 2 1 J2it(x) − J−2it(x) ∧ d s1t1s2t2 −( t−T )2 −( t+T )2 ∧ T 1− T 1− = ( ) (y)(V ( 2 )(e + e )) (y)dy 2 −∞ sinh πt t +O(T −N ) Z ∞ 3 2 4 2 −i d s1t1s2t2 −( t−T )2 −( t+T )2 ∧ −N T 1− T 1− = (cos(x cosh(πy)))(V ( 2 )(e + e )) (y)dy + O(T ) 2 −∞ t Z ∞ 3 2 4 2 −i 1 2 2 d s1t1s2t2 −( t−T )2 −( t+T )2 ∧ −N T 1− T 1− = (cos(x + π xy ))(V ( 2 )(e + e )) (y)dy + O(T ) 2 −∞ 2 t Z ∞ 3 2 4 2 r −i π d s1t1s2t2 −( t−T )2 −( t+T )2 xy dy T 1− T 1− = (cos(x − y + ))(V ( 2 )(e + e ))( )√ 2 0 4 t 2 πy +O(T −N ) √ √ xy xy Z ∞ 3 2 4 2 −T +T −i π 4d s1t1s2t2 −( 2 )2 −( 2 )2 dy = (cos(x − y + ))(V ( )(e T 1− + e T 1− ))√ 2 0 4 xy πy +O(T −N ) r 4πt t c−1y Z ∞ 3 2 4 2 1 2 −T −i −1 π 4d s1t1s2t2 −( 2 )2 T 1− = (cos(4πt1t2c − y + ))(V ( −1 )(e 2 0 4 4πt1t2c y r 4πt t c−1y 1 2 +T −( 2 )2 dy −N +e T 1− ))√ + O(T ) πy

Thus, by partial integration in terms of y, the non-diagonal terms is concentrated on

2 2− −1 2 T − T t1t2c y T .

3 2 4 2 2+ So, we can assume d s1t1s2t2 T since V (ξ) has exponential decay as ξ → ∞.

2−4 By partial integration, the terms with c T and t1t2 T contribute O(1). So we can assume c T and t t T 2−4, therefore we have t T 5, also we have 1 2 √ 2 −1 2πt1t2c y the sum over s1 and s2 converges. Let t = T , the inner integral is √ Z ∞ 3 2 4 2 T c −((t−1)/T −)2 −((t+1)/T −)2 −1 2 π 4d s1t1s2t2 (e −e )(cos(4πt1t2c −(tT ) c/(2πt1t2)+ ))V ( 2 2 )dt 2πt1t2 0 4 t T From the bound ([17] Theorem 8.1),

X −1/2 λψ(r)r R , r≤R 59 and the Hecke relation

X λψ(r1r2) = µ(d)λψ(r1/d)(r2/d);

d|(r1,r2) and partial summation, we get the non-diagonal terms contribute O(T 5).

5.2.3 The Continuous Spectrum Part

To evaluate the continuous part from (5.2), we need rewrite

X µ(d) X 2 2 4 2 −1/2 3 λψ(ds1t1)(s1t1s2t2) d 2 d≥1 s1,s2,t1,t2≥1 t−T 2 t+T 2 Z ∞ 3 2 4 2 −( 1− ) −( 1− ) d s1t1s2t2 e T + e T 2 2 V ( 2 ) 2 dit(t1)dit(t2)dt 0 t t|ζ(1 + 2it)| with respect to L-function, then we can use some mean value bounds of L-function to show the continuous part is also small. By Theorem 3 in [21], we have

Z T 1 2 2 + |L( + it, ϕj)| dt = TP (log T ) + O(T 3 ), 0 2 here P (t) is a polynomial of degree 4. So

2 Z T +T 3 1 2 2 + |L( + it, ϕj)| dt = O(T 3 ), T 2

The Eisenstein series E(z, s) deﬁned by

1 X ys E(z, s) = 2 (cz + d)2s (c,d)≥1 has Fourier expansion

s 1−s X E(z, s) = y + φ(s)y + φ(n, s)Ws(nz), n6=0 where Ws(z) is the Whittaker function given by

1 Ws(z) = 2|y| 2 K 1 (2π|y|)e(x) s− 2 60 and Ks(y) is the K-Bessel function,

√ Γ(s − 1 )ζ(2s − 1) φ(s) = π 2 Γ(s)ζ(2s) with ζ(s) be the Riemann zeta function and

s −1 −1 − 1 φ(n, s) = π Γ(s) ζ(2s) |n| 2 η(n, s) with s− 1 X a 2 η(n, s) = . d ad=n 1 For E(z, 2 + it), we associate the L−function

Y L(s, E) = (1 − p−s+it)−1(1 − p−s−it)−1, p and denote Y L(s, sym2E) = ζ(s) (1 − p−s+2it)−1(1 − p−s−2it)−1, p If

Y −s −1 −s −1 L(s, ψ) = (1 − αpp ) (1 − βpp ) , p we have

L(s, ψ ⊗ sym2E) = L(s − 2it, ψ)L(s + 2it, ψ)L(s, ψ)

Thus 1 1 1 L( , ψ ⊗ sym2E) = |L( + 2it, ψ)|2L( , ψ). 2 2 2 For Eisenstein series, we can take the advantage of unfolding integral,

Z 1 dxdy Ij(s) = ϕj(z)E(z, + it)E(z, s) 2 X 2 y Z ∞ Z 1 1 s dxdy = ϕj(z)E(z, + it)y 2 0 0 2 y

61 Z ∞ Z 1 ∞ 1 X 1 +it 1 1 −it = (y 2 λ (n)K (2πny) cos(2πnx))(y 2 + ϕ( + it)y 2 + j itj 2 0 0 n=1 1 ∞ 2y 2 X dxdy nitσ (n)K (2πny) cos(2πnx)ys) ξ(1 + 2it) −2it it y2 n=1 ∞ 1 Z ∞ X dxdy = ( λ (n)K (2πny)nitσ (n)K (2πny))ys ξ(1 + 2it) j itj −2it it y 0 n=1 ∞ it Z ∞ 1 X λj(n)n σ−2it(n) dxdy = ( ) K (2πy)K (2πy)ys ξ(1 + 2it) ns itj it y n=1 0 L(ϕ , s − it)L(ϕ , s + it) 2π−s Γ( s+itj +it )Γ( s+itj −it )Γ( s−itj +it )Γ( s−itj −it ) = j j 2 2 2 2 ζ(2s) ξ(1 + 2it) Γ(s)

Thus,

|Γ( 1 + itj )|2 Γ( 1 − itj − it)Γ( 1 + itj − it) 1 1 hϕ , µ i = 2π−2it 4 2 4 2 4 2 L(ϕ , )L(ϕ , − 2it) j t 2 1 2 j j |ζ(1 + 2it)| |Γ( 2 + it)| 2 2

We have 1 1 1 L( , ψ ⊗ sym2E) = |L( + 2it, ψ)|2L( , ψ). 2 2 2 On the other hand

2 1 X m1m L( , ψ ⊗ sym2E) = 2 λ (m )a (m , m )(m m2)−1/2V ( 2 ) 2 ψ 1 E 1 2 1 2 t2 m1,m2≥1 where Z 2 1 −s γ(1/2 + s, ψ ⊗ sym (E)) ds V (y) = y 2 , 2πi (2) γ(1/2, ψ ⊗ sym (E)) s Thus, we have

1 2 1/3+ L( 2 + it, ψj) (tj + t) ,

1 + we obtain the continuous part contributes O(T 2 ).

5.3 Conclusion

Combining last section’s results, we conclude that

t −T t +T X −( j )2 −( j )2 2 2 T 1− T 1− (e + e )L(1, sym φj)|hµφj , ψi| j≥1 π 1 1 1 = T 1−L(1, sym2ψ)L( , ψ)|Γ( + it )|4 + O(T 1/2+). 4 2 4 2 ψ 63 We have

2 L(1, sym ψ) = 2 < ψ, ψ > cosh(πtψ), since ψ(z) is normalized so that its first Fourier coefficient is 1 [31]. Thus, we obtain the eigenvalue of B at ψ is 1 |Γ( 1 − itψ )|4 L( , ψ) 4 2 . 1 2 2 2π|Γ( 2 − itψ)| Next, let us recall the classical variance of the geodesic flow which is studied by

Ratner.

We realize the upper half plane H as the homogeneous space SL2(R)/SO2(R) and Y = SL2(Z) \ SL2(R) as the unit cotangent bundle for X. The motion along geodesics on X gives rise to a Hamiltonian ﬂow Gt on Y   t e 2 0 Γg → Γg   , t ∈ R  − t  0 e 2

Since the geodesic ﬂow on a negative curvature manifold is ergodic [15], we can consider this Hamiltonian ﬂow as a classically chaotic system. Ratner [39] has shown

∞ that, for a classical observable ψ ∈ C0,0(Y ), the ﬂuctuations along a generic orbit of the geodesic ﬂow obey the central limit theorem, i.e. the distribution of 1 Z T √ ψ(Gt(g))dt T 0 become Gaussian with mean 0 and the variance given by the Hermitian form on

∞ C0,0(Y ),    t Z ∞ Z e 2 0 VC(φ, ψ) = φ g   ψ(g)dgdt,   − t  −∞ Γ\SL(2,R) 0 e 2 and VC is diagonalized by the Maass forms on SL2(R)/SO2(R). Let ψ be a Maass

1 2 form on SL2(R)/SO2(R) with Laplacian engenvalue 4 + r , we have explicitly [46]    t Z e 2 0 ψ g   ψ(g)dg = P− 1 +ir(cosh t),   − t  2 Γ\SL(2,R) 0 e 2

64 where P 1 (z) is the generalized Legendre function and we have the following inte- − 2 +ir gral representation [17]

π Z √ 1 1 2 − +ir P 1 (z) = (z + z − 1 cos x) 2 dx. − 2 +ir π 0

Thus, Z ∞ VC (ψ, ψ) = P 1 (cosh t)dt, − 2 +ir ∞ which can be explicitly calculated using the integral table [11] and we obtain

|Γ( 1 − ir )|4 V (ψ, ψ) = 4 2 . C 1 2 2π|Γ( 2 − ir)|

This completes the proof of Theorem 1.4.

Remark 5.1. As a by-product, we obtain that, for Maass-Hecke form ψ, L(1/2, ψ ⊗

2 1 sym (φj)) is non-vanishing for some φj and the central value L( 2 , ψ) is non-negative.

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