Lifting from SL(2) to Gspin(1,4)

Home , Cuspidal representation

Lifting from SL(2) to GSpin(1,4)

DISSERTATION

Presented in Partial Fulﬁllment of the Requirement for

the Degree Doctor of Philosophy in the Graduate

School of The Ohio State Univesity

Ameya Pitale

*****

The Ohio State Univesity 2006

Dissertation Committee: Approved by

Professor Steve Rallis, Adviser

Professor James Cogdell Advisor

Professor Cary Rader Graduate Program in Mathematics

ABSTRACT

In this thesis we construct liftings of automorphic forms from the metaplectic group SLf 2 to GSpin(1, 4) using the Maaß Converse Theorem. In order to prove the non-vanishing of the lift we derive Waldspurger’s formula for Fourier coeﬃcients of half integer weight Maaß forms. We analyze the automorphic representation of the adelic spin group obtained from the lift and show that it is CAP to the Saito-

Kurokawa lift from SLf 2 to GSp4(A).

ii Dedicated to Swapna

iii ACKNOWLEDGMENTS

I would like to thank my thesis advisor Professor Steve Rallis for introducing me to this ﬁeld of Mathematics and this problem in particular. His guidance, patience and generosity were invaluable to me. I would like to thank James Cogdell and Cary

Rader for humoring my endless questions and giving me so much of their time and help. For their comments and suggestions I am grateful to Dinakar Ramakrishnan,

Tamotsu Ikeda, Winfred Kohnen, Dihua Jiang, Steve Kudla, William Duke, Peter

Sarnak and Eitan Sayag.

iv VITA

September 16, 1977...... Born - Nagpur, India 2000...... MSc Mathematics, Indian Institute of Technology, Kanpur, India 2000 − present ...... Graduate Teaching Assistant, The Ohio State University

PUBLICATIONS

Pitale A., “Lifting from SLf 2 to GSpin(1, 4)”, Intern. Math. Res. Not. 2005, No. 63, Pg 3919 - 3966

FIELDS OF STUDY Mathematics, Number Theory

v TABLE OF CONTENTS

Page

Abstract ...... ii

Dedication ...... iii

Acknowledgments ...... iv

Vita ...... v

Chapters:

1. Introduction ...... 1

2. Maaß Converse Theorem ...... 5

2.1 Cliﬀord algebras and Vahlen matrices ...... 5 2.2 Automorphic functions ...... 9

3. Deﬁnition and Automorphy of the Lift ...... 14

3.1 Maaß forms on SL2 of half-integral weight ...... 14 3.2 Deﬁnition of A(β) ...... 19 3.3 Proof of Theorem 3.3 ...... 22 3.3.1 Theta Function and Eisenstein Series ...... 25 3.3.2 Rankin Integral Formula ...... 29 3.3.3 The functional equation ...... 31

4. Non-Vanishing of the Lift ...... 36

4.1 Criteria for non-vanishing ...... 36 4.2 Waldspurger’s formula for Maaß forms ...... 38

vi 4.3 Non-vanishing of special values of L−functions ...... 44

5. Hecke Theory ...... 47

5.1 Hecke Algebra Hp for p odd...... 48 5.2 Hecke operator Tp ...... 63 5.3 Hecke operator Tp2 ...... 78

6. Automorphic representation corresponding to F ...... 89

6.1 Unramiﬁed Calculation ...... 91 6.2 CAP representation ...... 99

Bibliography ...... 102

vii CHAPTER 1

INTRODUCTION

In [16], Ikeda deﬁnes a lifting of automorphic forms from SLf 2, the metaplectic cover of SL2, to the symplectic group GSp4n for all n. Ikeda proves his result using the theory of Fourier Jacobi forms. For n = 1 he shows that his deﬁnition agrees with the classical Saito-Kurokawa lift. In [7], Duke and Imamo¯glu prove the automorphy of the classical Saito-Kurokawa lift from holomorphic half integer weight forms using the converse theorem for Sp4 due to Imai [17].

In this thesis we present the lifting of automorphic forms from SLf 2 to the spin group GSpin(1, 4) using the Maaß Converse Theorem [25]. The reason we choose to study these lifts is the following observation : For all primes p 6= 2 we have that

GSp4(Qp) ' GSpin(1, 4)(Qp) (l.c. Proposition 6.3). This relation between GSp4 and GSpin(1, 4) suggests the possibility of developing liftings for the spin group analogous to Ikeda’s liftings for the symplectic group. Another motivation for considering spin groups is to give applications for the Maaß Converse Theorem. As far as we know this theorem has been applied only once by Duke [6] to show that a non-cuspidal theta series is automorphic for the group GSpin(1, 5). In this paper we obtain a family of cuspidal automorphic forms for GSpin(1, 4) using the converse theorem.

1 To deﬁne the lifting we start with a weight 1/2 Maaß Hecke eigenform f on the complex upper half plane with respect to Γ0(4). The candidate function F on a symmetric space of GSpin(1, 4) is deﬁned in terms of a Fourier expansion with

Fourier coeﬃcients related to the Fourier coeﬃcients of f as in the formula (3.9).

The Maaß Converse Theorem 2.5 states that the function F is cuspidal automorphic with respect to the integer points of GSpin(1, 4) if and only if a family of Dirichlet series is “nice” (“nice” means analytic continuation, bounded on vertical strips and functional equation). To obtain the “nice” properties of the Dirichlet series we use the method of Duke and Imamo¯glu in [7] to rewrite the Dirichlet series in terms of a

Rankin triple integral of the function f, a certain Theta function and an Eisenstein series with respect to Γ0(4). Then essentially the “nice” properties of the Eisenstein series give us the corresponding “nice” properties of the Dirichlet series.

Another interesting result of the thesis is related to the non-vanishing of the lift. In [16] it is shown that the non-vanishing of the Ikeda lift is equivalent to the non-vanishing of certain Fourier coefficients of the holomorphic half-integer weight form, which follows from a straightforward calculation in [19]. In our case, the non- vanishing of the lift F is equivalent to the non-vanishing of certain negative Fourier coefficients of f. It seems that it is not possible to get non-vanishing of specifically negative Fourier coefficients of f using elementary methods as in [19]. To achieve this we use the work of Baruch and Mao [2] to prove a Waldspurger type (or Kohnen-

Zagier type) formula which relates the square of the negative Fourier coeﬃcient of f to special values of twisted L−functions (Theorem 4.3). Then we get the non- vanishing of the lift F by the results of Friedberg and Hoﬀstein [10] on non-vanishing of special values of twisted L−functions.

2 We also analyze the adelic automorphic representation πF of the group

GSpin(1, 4)(A) obtained from the automorphic function F. To do this, we first show that if f is a Hecke eigenfunction then so is F and explicitly calculate the eigenvalues of F in terms of the eigenvalues of f. Since the p−adic component (p an odd prime) of πF is an irreducible unramified representation of GSpin(1, 4)(Qp)(w GSp4(Qp)) we know that it is the unique spherical constituent of the representation obtained by induction from an unramified character on the Borel subgroup. In Theorem 6.5, we obtain the explicit description of this character in terms of the eigenvalue of f using the Hecke theory for the spin group.

Using the precise information about the local representations, we show that the representation πF is a CAP representation. Let us remind the reader that if G is a reductive algebraic group and P a parabolic subgroup then an irreducible cuspidal automorphic representation π of G(A) is called Cuspidal Associated to Parabolic (CAP) subgroup P if there is an irreducible cuspidal automorphic representation σ of the Levi subgroup of P such that π is nearly equivalent to an irreducible com-

G ponent of IndP (σ). The notion of CAP representations was ﬁrst introduced by I.

Piatetski-Shapiro [28] for the group Sp4. The CAP representations are very interesting because they provided counter-examples to the earlier versions of the generalized

Ramanujan conjecture. CAP representations on Sp4 have been extensively studied by Piatetski-Shapiro in [28] and Soudry in [35], [36]. In [11] and [29], families of CAP representations on the split exceptional group of type G2 have been constructed. In

[12] the authors give a criterion for an irreducible cuspidal automorphic representation of a split group of type D4 to be a CAP representation. In these papers the method used to construct CAP representations is theta lifting of various types.

3 The representation πF constructed here is interesting in this context because it is CAP to a representation of GSp4(A) instead of GSpin(1, 4)(A) (Theorem 6.7). If one considers Langlands’ philosophy then it is natural to extend the notion of CAP representations to the situation where the group G is replaced by two groups G1,G2 which satisfy G1,ν w G2,ν for almost all ν which is the case in our present setting.

4 CHAPTER 2

Maaß Converse Theorem

Maaß stated his converse theorem in terms of Clifford algebras and Vahlen matrices. In this chapter we give the definition and basic properties of Vahlen matrices and illustrate their relation to the spin group. We then define automorphic functions and state the Maaß Converse Theorem. We give a brief sketch of the proof of the converse theorem. In this chapter we perform all the calculations for the case of Spin(1, k + 2) for k ≥ 0. We will restrict to the case of k = 2 from Chapter 3. The main reference for this chapter is Maaß [25]. For details on Vahlen’s group of matrices we refer the reader to [8] and [9].

2.1 Cliﬀord algebras and Vahlen matrices

Let E be a vector space over R of dimension k. Let Q : E → R be a nondegenerate quadratic form and {i1, ··· , ik} a basis of E orthogonal with respect to Q. Deﬁne

2 iν = Q(iν), iνiµ = −iµiν for 1 ≤ µ, ν ≤ k and µ 6= ν. The Cliﬀord algebra Ck(Q)

k is generated by {i1, ··· , ik} with relations as above. Hence the 2 elements iν1 ··· iνp where 1 ≤ ν1 < ν2 < ··· < νp ≤ k, 0 ≤ p ≤ k form a basis of Ck(Q) over the reals

5 × (For p = 0 we mean the element 1) . Denote by Ck(Q) the invertible elements of

Ck(Q). Note that Ck−1(Q) is embedded in Ck(Q) by setting the coeﬃcient of all the above basis elements containing ik to be equal to zero. Ck(Q) consists of elements of the form

k X X a = a0 + aν1···νp iν1 iν2 ··· iνp

p=1 1≤ν1<···<νp≤k with aν1···νp , a0 ∈ R.

We deﬁne the following maps on Ck(Q):

k 0 X p X a 7→ a := a0 + (−1) aν1···νp iν1 iν2 ··· iνp ,

p=1 1≤ν1<···<νp≤k

k p(p+1) X 2 X a 7→ a := a0 + (−1) aν1···νp iν1 iν2 ··· iνp ,

p=1 1≤ν1<···<νp≤k

k p(p−1) ∗ 0 X 2 X a 7→ a := a = a0 + (−1) aν1···νp iν1 iν2 ··· iνp .

p=1 1≤ν1<···<νp≤k 0 ∗ These maps satisfy x = −x, x = −x, x = x for all x ∈ Ri1 ⊕ · · · ⊕ Rik and

0 0 0 ∗ ∗ ∗ (uv) = v u, (uv) = u v , (uv) = v u for all u, v ∈ Ck(Q). Let us denote Re(x0 +

··· + xkik) := x0. Let Vk(Q) ⊂ Ck(Q) be the (k + 1)−dimensional vector space over reals with basis {1, i1, i2, ··· , ik}.

Let Tk(Q) be deﬁned as the set of non-zero elements a in Ck(Q) with the property

0 that there exists a linear isomorphism φa : Vk(Q) → Vk(Q) such that ax = φa(x)a , for all x ∈ Vk(Q). Then, by [8] Pg 375, Tk(Q) is a group generated by the nonzero elements in Vk(Q). In [25], Maaß calls the elements of Tk(Q) as transformers.

6 The Vahlen group of matrices is deﬁned by   α β   SVk(Q) := {  ∈ M2(Ck(Q)) : α, β, γ, δ ∈ Tk(Q) ∪ {0}, γ δ ∗ ∗ ∗ ∗ αδ − βγ = 1, αβ , δγ ∈ Vk(Q)}. (2.1)     0 1 1 β     Lemma 2.1. SVk(Q) is generated by the matrices   ,   , with −1 0 0 1

β ∈ Vk(Q).

For the proof we refer the reader to [8] Pg 377.

+ We will now give the relationship between SVk(Q) and the Spin group. Let Ck (Q)

be the subalgebra of Ck(Q) with basis {iν1 ··· iνp : 1 ≤ ν1 < ··· < νp ≤ k, p ≡ 0

(mod 2)}. Then deﬁne

+ ∗ ∗ Spink(Q) := {s ∈ Ck (Q)|sEs ⊂ E, ss = 1}. (2.2)

Let E0 be a 3−dimensional vector space over reals with basis {f0, f1, f2}. Let Q0 be

2 2 2 the quadratic form on E0 deﬁned by Q0(y0f0 + y1f1 + y2f2) = y0 − y1 − y2. Equip ˜ ˜ + ˜ E := E0 ⊕ E with the quadratic form Q := Q0 ⊥ Q. The map · : E → Ck+3(Q) given byx ˙ = f0f1f2x extends to an injective R−algebra homomorphism · : Ck(Q) → + ˜ + ˜ Ck+3(Q). From [8] Pg 372 we get that the map ψ : M2(Ck(Q)) → Ck+3(Q) given by   α β   ˙ ˙ ψ   :=αu ˙ + βw1 +γw ˙ 0 + δv, γ δ

1 1 1 1 where u = 2 (1 + f0f1), w1 = 2 (f0f2 − f1f2), w0 = 2 (f0f2 + f1f2), and v = 2 (1 − f0f1), is an R−algebra isomorphism.

7 Proposition 2.2. The R−algebra isomorphism ψ deﬁned above restricts to a group isomorphism ˜ ψ : SVk(Q) → Spink+3(Q).

For the proof, we again refer the reader to [8] Pg 382.

2 2 k Let us now ﬁx Q = Qk(x1, ··· , xk) := −x1 − · · · − xk on R and consider Ck(Qk). Deﬁne the upper half space as

Hk+1 := {x0 + x1i1 + ··· + xk+1ik+1 ∈ Vk+1(Qk+1) such that (2.3)

xj ∈ R for j = 0, ··· , k + 1 and xk+1 > 0}.

2 2 2 dx0+···+dxk+1 The line element ds = 2 deﬁnes a Riemannian metric d on Hk+1. The xk+1 Laplace Beltrami operator is given by

k+1 k+1 X ∂ ∂ X ∂2 ∂ Ω := xk+2 (x−k ) = x2 − kx . (2.4) k+1 k+1 ∂x k+1 ∂x k+1 ∂x2 k+1 ∂x j=0 j j j=0 j k+1

We now state the result on the isometries of Hk+1 ([8] Pg 381 ).   α β   Proposition 2.3. If   ∈ SVk(Qk) and x ∈ Hk+1, then γx+δ ∈ Tk+1(Qk+1) γ δ and   α β   −1   · x := (αx + β)(γx + δ) ∈ Hk+1. (2.5) γ δ

Formula (2.5) deﬁnes an action of SVk(Qk) on Hk+1 by orientation preserving isometries. This action keeps the metric d and the Laplace Beltrami operator Ωk+1 invari-

+ ant. The resulting sequence 1 → {1, −1} → SVk(Qk) → Iso (Hk+1) → 1 is exact.

+ Here Iso (Hk+1) is the set of orientation preserving isometries of Hk+1.

8 2.2 Automorphic functions

Let us now deﬁne the automorphic functions for SVk. We ﬁx the quadratic form

Q = Qk given in the previous section and drop Q from the notation whenever there is no confusion.

Fix the subgroup ΓT of SVk deﬁned by     0 1 1 β     ΓT := h  ,   : β ∈ T i (2.6) −1 0 0 1 where T is a ﬁxed lattice in Vk.

∞ Deﬁnition 2.4 (Automorphic Function). A complex-valued C function F on Hk+1 is called automorphic with respect to ΓT if F satisﬁes the following conditions :

2 (k+1)2 1. Ωk+1F + (r + 4 )F = 0 for some r ∈ R,

κ1 2. For certain positive constants κ1 and κ2 we have F (x) = O(xk+1) for xk+1 →

−κ2 ∞,F (x) = O(xk+1) for xk+1 → 0 uniformly on x0, x1, ··· , xk and

3. F (γx) = F (x) for all γ ∈ ΓT , x ∈ Hk+1.

As given in [25], we have the following Fourier expansion of F (x):

X k+1 2πiRe(βx) F (x) = u0(xk+1) + A(β)xk+1 2 Kir(2π|β|xk+1)e (2.7) β∈S β6=0 where

 k+1 +ir k+1 −ir  2 2  a1xk+1 + a2xk+1 , if r 6= 0 u (x ) = 0 k+1 k+1 k+1  2 2  a1xk+1 + a2xk+1log(xk+1), if r = 0.

9 Here S is the lattice in Vk dual to the lattice T deﬁned by S = {β ∈ Vk :

Re(βT ) ⊂ Z}. For β = β0 + β1i1 + ··· + βkik we deﬁne the norm by the formula

2 2 2 2 |β| = β0 + β1 + ··· + βk. Also, Kir is the classical K−Bessel function satisfying

Kir(y) → 0 as y → ∞. If u0(xk+1) = 0 then we call F a cuspidal automorphic function. In the next chapter we will make a choice of the lattice T such that ΓT has only one cusp, hence the notation is justiﬁed.

For every integer l ﬁx a basis {Pl,ν} of spherical harmonic polynomials of degree l in k + 1 variables. In [25], Maaß proves the following Converse Theorem.

Theorem 2.5 (Maaß Converse Theorem). The following two statements are equivalent

k+1 P 2 2πiRe(βx) 1. F (x) = u0(xk+1) + A(β)xk+1Kir(2π|β|xk+1)e is an automorphic β∈S β6=0 function with respect to ΓT .

−2s ir ir P A(β)Pl,ν (β) 2. For all l, Pl,ν, the functions ξ(s, Pl,ν) := π Γ(s + 2 )Γ(s − 2 ) (|β|2)s β∈S β6=0 satisfy

k+1+ir k+1−ir (a) s − 2 s − 2 ξ(s, P0,ν) and ξ(s, Pl,ν), l > 0, have analyitic continuation to the complex plane,

(b) ξ(s, Pl,ν) are bounded on vertical strips and

k + 1 ξ( + l − s, P ) = (−1)lξ(s, P 0 ) (2.8) 2 l,ν l,ν

0 0 where Pl,ν(β) := Pl,ν(β ) for β ∈ Vk.

10 k+1+ir k+1−ir The residues of ξ(s, P0,ν) at s = 2 and 2 determine the constants a1 and a2 in the constant term u0(xk+1). Hence, if we want F to be a cuspidal automorphic function then we require ξ(s, P0,ν) to have analytic continuation to C. We give a sketch of the proof here. Note that the automorphy of a function F given by the Fourier expansion (2.7) is equivalent to the condition F (−x−1) = F (x), i.e. g(x) := F (−x−1) − F (x) is identically zero. Since F is an eigenfunction of

−1 Ωk+1 and x → −x preserves Ωk+1 we conclude that g is also an eigenfunction of

Ωk+1. The main ingredient of the proof of Theorem 2.5 is the following Proposition proved in [25] which gives a criterion for an eigenfunction of the operator Ωk+1 to be identically zero.

Proposition 2.6. Let g(x) be twice continuously diﬀerentiable function on Hk+1 sat-

2 (k+1)2 isfying Ωk+1g + (r + 4 )g = 0 for some r ∈ R. Since Ωk+1 is an elliptic operator, g(x) can be extended to a complex neighbourhood U about any point in Hk+1 such that g(x) represents a regular analytic function in complex variables x0, x1, ··· , xk+1 on U. g(x) vanishes identically in Hk+1 if and only if for any set of complex numbers

2 2 2 a0, a1, ··· , ak with a0 + a1 + ··· + ak = 0 and any real xk+1 > 0, the conditions

dl l g((a0 + a1i1 + ··· + akik)t + xk+1ik+1) = 0 dt t=0 are satisﬁed for l = 0, 1, 2, ··· .

For the proof of the above Proposition we refer the reader to [25]. Observe that in the case of a holomorphic function f on the complex upper half plane modularity with respect to SL2(Z) is equivalent to checking the periodicity (f(z +1) = f(z)) and the condition f(−z−1) = (−z)kf(z) only for z = iy. (The second condition is a direct

11 consequence of the holomorphy of f.) In our case, the above proposition replaces this condition. Note that l = 0 corresponds to g(ik+1xk+1) = 0.

2 2 2 Fix complex numbers a0, a1, ··· , ak with a0 +a1 +···+ak = 0 and real xk+1 > 0.

To prove Theorem 2.5 using Proposition 2.6 we need to show that for all l = 0, 1, 2, ··· we have dl l g((a0 + a1i1 + ··· + akik)t + xk+1ik+1) = 0. dt t=0 This gives us

dl F ((a + a i + ··· + a i )t + x i ) dtl 0 1 1 k k k+1 k+1 t=0 l d (−a0 + a1i1 + ··· + akik)t + xk+1ik+1 − l F 2 = 0. dt xk+1 t=0

−1 −2 2 2 Here we have used x = x|x| and |(a0 + a1i1 + ··· + akik)t + xk+1ik+1| = xk+1. Substituting for F (x) its Fourier expansion (2.7) we get

l X k+1 (2πi) [ul(xk+1) + A(β)xk+1 2 Kir(2π|β|xk+1) × (2.9) β∈S,β6=0

l l 1 X ×(β0a0 − β1a1 − · · · − βkak) ] − (−2πi) [ul( ) + A(β) × xk+1 β∈S,β6=0 k+1 − 2 −2l −1 l ×xk+1 Kir(2π|β|xk+1)(β0a0 + β1a1 + ··· + βkak) ] = 0 where

  u0(xk+1), if l = 0 ul(xk+1) =  0, if l 6= 0. Deﬁne

X k+1 +l Fl(y, Pl) := ul(y) + A(β)Pl(β)y 2 Kir(2π|β|y) (2.10) β∈S, β6=0

12 where y ∈ R, y > 0 and Pl is a spherical harmonic polynomial of degree l in k + 1

0 0 variables. Deﬁne Pl (β0 + β1i1 + ··· + βkik) = Pl(β0 − β1i1 − · · · − βkik) i.e. Pl (β) =

0 Pl(β ).

l l Then by multiplying (2.9) by xk+1 and setting Pl(β) := [β0a0 − β1a1 − · · · − βkak] we get 1 F (y, P ) − (−1)lF ( ,P 0) = 0. l l l y l

l We know that [β0a0 − β1a1 − · · · − βkak] , where a0, a1, ··· , ak range over the

2 2 2 complex numbers satisfying a0 +a1 +···+ak = 0, form a spanning set for spherical harmonic polynomials of degree l. Hence we conclude that 1 F (x) = F (−x−1) ⇐⇒ F (y, P ) = (−1)lF ( ,P 0 ) (2.11) l l,ν l y l,ν where for each l, {Pl,ν} ranges over a basis of spherical harmonic polynomials of degree l in k + 1 variables. Deﬁne ∞ Z 2s− k+1 −l dy ξ(s, P ) := 4 (F (y, P ) − u (y)) y 2 (2.12) l,ν l l,ν l y 0 essentially the Mellin Transform of Fl(y, Pl,ν)−ul(y). Condition (2) in Deﬁnition(2.4) guarantees the convergence of this integral in some right half plane.

Upon integration, we get

ir ir X A(β)Pl,ν(β) ξ(s, P ) = π−2sΓ(s + )Γ(s − ) . l,ν 2 2 (|β|2)s β∈S, β6=0 l 1 0 Using Mellin inversion formula, we now get that Fl(y, Pl,ν) = (−1) Fl( y ,Pl,ν) for every l and Pl,ν if and only if for every l, Pl,ν the ξ(s, Pl,ν) satisfy the following conditions :

k+1+ir k+1−ir s − 2 s − 2 ξ(s, P0.ν) and ξ(s, Pl,ν), l > 0 have analyitic continuation to the complex plane, ξ(s, Pl,ν) is bounded on vertical strips and ξ(s, Pl,ν) satisfy the

k+1 l 0 functional equation ξ( 2 + l − s, Pl,ν) = (−1) ξ(s, Pl,ν).

13 CHAPTER 3

Deﬁnition and Automorphy of the Lift

Let us restrict ourselves to the case k = 2. This corresponds to the group

Spin(1, 4). In this chapter we first give some details about half-integral weight Maaß forms in section 3.1. These will be the input data used to define a candidate for a cuspidal automorphic function F (x) on H3. We do this by giving an explicit formula (3.9) for the Fourier coefficients A(β) of F in section 3.2. In section 3.3 we prove the automorphy of the candidate function using the Maaß Converse Theorem 2.5. The main idea of the proof is to get a Rankin integral formula for the Dirichlet series which is achieved in Proposition 3.9.

3.1 Maaß forms on SL2 of half-integral weight

Automorphic forms for the metaplectic group SLf 2 can be realised as half integer   a b   weight forms on the upper half plane with respect to the group Γ0(4) := {  ∈ c d

SL2( ): c ≡ 0 (mod 4)}. Let S 1 (4), t ∈ be the space of functions f on the upper Z t+ 2 Z half plane H = {z = x + iy : x, y ∈ R, y > 0} satisfying :

14   a b   1. For all γ =   ∈ Γ0(4) and z ∈ H c d

1 !2t+1 c cz + d 2 f(γz) = −1 f(z) (3.1) d d |cz + d|

where   1, if d ≡ 1 (mod 4); d =  i, if d ≡ 3 (mod 4).

· and · is the Legendre symbol.

2. ∆ 1 f + λf = 0 for λ ∈ , where ∆ 1 is the Laplace Beltrami operator given t+ 2 C t+ 2 by 2 2 2 ∂ ∂ 1 ∂ ∆t+ 1 := y + − (t + )iy . 2 ∂x2 ∂y2 2 ∂x

1 3. f vanishes at the cusps of Γ0(4) (The cusps of Γ0(4) are given by ∞, 0 and 2 ).

f has the Fourier expansion

X 2πinx f(z) = c(n)W sign(n)(t+1/2) ir (4π|n|y)e , (3.2) 2 , 2 n∈Z n6=0

1 2 where λ = 4 + (r/2) and Wν,µ(y) is the classical Whittaker function. The numbers {c(n)} are called the Fourier coeﬀﬁcients of f. For more details on half-integral weight

Maaß forms we refer the reader to [21] and [18].

Deﬁne the plus space

+ t S 1 (4) := {f ∈ St+ 1 | c(n) = 0 whenever (−1) n ≡ 2, 3 (mod 4)}. t+ 2 2

This is analogous to the Kohnen plus space for holomorphic half integral weight

+ modular forms introduced in [19]. For t = 0,S 1 (4) is same as the plus space deﬁned 2

15 in [18] where it is shown to be nonempty. It then follows from Lemma 3.1 below that

+ S 1 (4) is nonempty for t ∈ Z. t+ 2 2 For t = 0 and every odd prime p deﬁne Hecke operators T (p ): S 1 (4) → S 1 (4), 2 2 2 P (p) 2πinx f → (T (p )f)(z) := c (n)W sign(n) , ir (4π|n|y)e by the formula n6=0 4 2

n n c(p)(n) = pc(np2) + p−1/2 c(n) + p−1c . (3.3) p p2

2 [18] gives us that the T (p ) commute with each other and commute with ∆ 1 . 2 From the deﬁnition above we can see that c(p)(n) = 0 if n ≡ 2, 3 (mod 4) if the same is true for c(n) since np2, n/p2 ≡ n (mod 4). This means that the operators T (p2)

+ keep S 1 (4) stable. 2

Deﬁne the following operators on St+1/2(4) :

∂ t+1/2 ∂ ∂ t+1/2 1. Lowering operator L 1 := (z − z) + = iy − y + : S 1 (4) −→ t+ 2 ∂z 2 ∂x ∂y 2 t+ 2

S 1 (4) satisﬁes ∆ 1 L 1 (f) = λL 1 (f) whenever ∆ 1 (f) = λf. t−2+ 2 t−2+ 2 t+ 2 t+ 2 t+ 2

∂ t+1/2 ∂ ∂ t+1/2 2. Raising operator R 1 := −(z−z) − = −iy −y − : S 1 (4) −→ t+ 2 ∂z 2 ∂x ∂y 2 t+ 2

S 1 (4) satisﬁes ∆ 1 R 1 (f) = λR 1 (f) whenever ∆ 1 (f) = λf. t+2+ 2 t+2+ 2 t+ 2 t+ 2 t+ 2

3. Inverting operator I 1 : S 1 (4) −→ S 1 (4) deﬁned by (I 1 f)(z) := f(z) t+ 2 t+ 2 −(t+ 2 ) t+ 2

satisﬁes ∆ 1 I 1 (f) = λI 1 (f) whenever ∆ 1 (f) = λf. −(t+ 2 ) t+ 2 t+ 2 t+ 2

For more on these operators see Duke [6]. Also, note that the raising and lowering operators above are the same as those deﬁned by Bump [3] Section (2.1) if we replace the half integers with integers. These operators along with Lemma 3.1 below give

+ + us the freedom to move f ∈ S1/2(4) to functions in St+1/2(4) for t ∈ Z. This will be crucial in the proof of the non-vanishing result (Theorem 4.5) of the next chapter.

16 The following recurrence relations ([26] Pg 302) for Whittaker functions Wν,µ(y)

0 1 1 y yW (y) = ( + µ − ν)( − µ − ν)W (y) − ( − ν)W (y), (3.4) ν,µ 2 2 ν−1,µ 2 ν,µ 0 y yW (y) = ( − ν)W (y) − W (y) (3.5) ν,µ 2 ν,µ ν+1,µ

give us

L R I Lemma 3.1. Let c(n), c (n), c (n) and c (n) be the Fourier coeﬃcients of f, L 1 (f), t+ 2

R 1 (f) and I 1 (f) respectively. Then we have t+ 2 t+ 2   c(n), if n < 0; cL(n) = (3.6)  ir 1 t+1/2 t+1/2 ir 1  2 + 2 − 2 2 + 2 − 2 c(n), if n > 0,  t+1/2 t+1/2  ir + 1 + − + ir − 1 c(n), if n < 0; cR(n) = 2 2 2 2 2 2 (3.7)  c(n), if n > 0, cI (n) = c(−n). (3.8)

Proof. (3.8) is clear from the deﬁnition of I 1 . We give the proof of (3.7) here. t+ 2 First let n > 0.

1 2πinx ∂ ∂ t + 2 2πinx R 1 W (t+ 1 ) (4πny)e = −iy − y − W (t+ 1 ) (4πny)e t+ 2 2 ir 2 ir 2 , 2 ∂x ∂y 2 2 , 2 2πinx = 2πnyW (t+1/2) ir (4πny)e 2 , 2

0 2πinx −4πnyW (t+1/2) ir (4πny)e 2 , 2 t + 1/2 2πinx − W (t+1/2) , ir (4πny)e . 2 2 2

17 (t+1/2) ir Using (3.5) with ν = 2 , µ = 2 and y replaced with 4πny we get

1 2πinx 2πinx (t + 2 ) 1 (t+1/2) ir 1 Rt+ W (4πny)e = 2πnyW (t+ ) ir (4πny)e − ((2πny − ) 2 2 , 2 2 2 , 2 2 2πinx ×W (t+1/2) ir (4πny) − W (t+1/2+2) ir (4πny))e 2 , 2 2 , 2

t + 1/2 2πinx − W (t+1/2) , ir (4πny)e 2 2 2 2πinx = W (t+1/2+2) ir (4πny)e . 2 , 2

This gives us cR(n) = c(n) for n > 0.

Now let −n > 0.

−2πinx −2πinx 1 1 −(t+1/2) ir Rt+ W −(t+ ) ir (4πny)e = −2πnyW (4πny)e − 2 2 2 , 2 2 , 2 0 −2πinx −4πnyW −(t+1/2) ir (4πny)e − 2 , 2 t + 1/2 −2πinx − W −(t+1/2) , ir (4πny)e . 2 2 2

(t+1/2) ir Using (3.4) with ν = − 2 , µ = 2 and y replaced with 4πny we get

−2πinx −2πinx Rt+ 1 W −(t+1/2) ir (4πny)e = −2πnyW −(t+1/2) ir (4πny)e − 2 2 , 2 2 , 2 1 ir (t + 1/2) 1 ir (t + 1/2) ( + + )( − + ) × 2 2 2 2 2 2 −2πinx W −(t+1/2+2) ir (4πny)e + 2 , 2

(t + 1/2) −2πinx (2πny + )W −(t+1/2) , ir (4πny)e 2 2 2 t + 1/2 −2πinx − W −(t+1/2) , ir (4πny)e 2 2 2 1 ir (t + 1/2) 1 ir (t + 1/2) = −( + + )( − + ) 2 2 2 2 2 2 −2πinx ×W −(t+1/2+2) ir (4πny)e . 2 , 2

R ir 1 t+1/2 t+1/2 ir 1 This gives us c (n) = 2 + 2 + 2 − 2 + 2 − 2 c(n) for n < 0. We get (3.6) in a similar way.

18 3.2 Deﬁnition of A(β)

Fix the lattice T := Z + Zi1 + Zi2 = V2(Z) ⊂ V2. This lattice is self dual, i.e. S = T.

Proposition 3.2. For T as above and ΓT as in (2.6), we get

ΓT = SV2(Z) := SV2 ∩ M2(C2(Z)) where C2(Z) is an order in C2 consisting of elements whose coeﬃcients are in Z, i.e. generated by {1, i1, i2, i1i2} over Z. In particular, ΓT is discrete, arithmetic and has ﬁnite co-volume.

Proof. It is clear that ΓT ⊂ SV2(Z). To show the other inclusion we will ﬁrst show   α β   that given a matrix M =   ∈ SV2(Z) we can ﬁnd an element g ∈ ΓT such γ δ   0 1   that gM is upper triangular. By multiplying on the left by the matrix   if −1 0 2 2 necessary, we can assume that |α| ≥ |γ| . From [8] Pg. 373 we know that αγ¯ ∈ V2,

−1 −1 2 which implies that αγ ∈ V2. Hence we can ﬁnd a u ∈ V2(Z) such that |αγ −u| < 1 ⇒ |α − uγ|2 < |γ|2. Now consider       0 1 1 −u γ δ           M =   . −1 0 0 1 −(α − uγ) −(β − uδ)

If α − uγ = 0 then we are done. If not, then repeat the same process (ﬁnitely many times since the norm of the lower left matrix entry is a non-negative integer and it decreases at each step) until you get an upper triangular matrix. So let g ∈ ΓT   α β   ∗ be such that gM =   . Since αδ = 1 we conclude that α is a unit, i.e. 0 δ

19   0 0   α = ±1, ±i1, ±i2 or ± i1i2 and δ = α . One can check easily that   ∈ ΓT 0 0 for any unit . Hence     α 0 1αβ ¯ −1     M = g     ∈ ΓT 0 α0 0 1 as required.

Notice that SV2(Z) is the stabilizer of the lattice C2(Z) × C2(Z) in the vector space C2 × C2. From [9] Pg. 261 − 262 we have that SV2(Z) and with it ΓT are discrete, arithmetic and have ﬁnite covolume. Remark : Let us note here that if we consider Spin(1, n) with n > 4 then the above Proposition is not true, i.e., ΓT (with T = Z + Zi1 + ··· + Zin−2) is not equal to the integer points of Spin(1, n), n > 4. In fact, the group ΓT does not have ﬁnite covolume and is not arithmetic. So, in a sense, Spin(1, 4) is the only interesting case to apply the Maaß Converse Theorem. For all higher values of n the converse theorem is an analog of Hecke’s converse theorem for triangle subgroups in SL2 [15].

+ Let f ∈ S 1 (4) be a non-zero Hecke eigenform with Fourier coeﬃcients {c(n)}. 2 u Let β = β0 + β1i1 + β2i2 ∈ V2(Z) and write gcd(β0, β1, β2) = 2 d, where u ≥ 0 and d is odd, positive.

Deﬁne u   X X −|β|2 A(β) := 23/4|β| c n−1/2 (−1)t2t/2. (3.9)  (2tn)2  t=0 n|d

20 The main result of this section is

Theorem 3.3. With A(β) as above,

X 3 2πiRe(βx) F (x) := A(β)x3 2 Kir(2π|β|x3)e β∈V2(Z), β6=0 is a cuspidal automorphic function on H3 with respect to SV2(Z).

1. We observe that the definition of A(β) above is similar to the Saito-Kurokawa ∞ P 2πnz lift for Sp4. In the Sp4 case, one starts with g(z) = b(n)e , a holomorphic n=1 1 k cusp form with respect to Γ0(4) of weight k + 2 with b(n) = 0 if (−1) n ≡ 2, 3 (mod 4). The Saito-Kurokawa lift is then given by G(Z) = P A(T )e2πiT r(TZ) : T >0 H2 → C (the sum is over all positive definite symmetric half-integral 2 × 2 matrices and H2 is the Siegel upper half space of genus 2) where   r m det(2T )  2  X k A(T ) := A = d b 2 .  r  d d|(m,n,r) 2 n The above formula differs from (3.9) in the sense that in (3.9) we have to treat

the prime 2 dividing gcd(β0, β1, β2) separately.

k0 2. Note that A(β) satisﬁes |A(β)| < C0|β| for some constants C0, k0 independent

of β which ensures the convergence of F (x) above. We can see this as follows :

+ k1 Since f ∈ S 1 (4) there exists positive C1, k1 ∈ R such that |c(n)| < C1|n| for 2 all nonzero integers n. This gives us   u 2 3 X X −|β| −1/2 t t/2 |A(β)| = 2 4 |β|| c n (−1) 2 |  (2tn)2  t=0 n|d u 3 t 3 X 2k1 2k1 < 2 4 |β| dC1|β| 2 2 ≤ 2 4 |β|udC1|β| |β| t=0 3 2k1+4 ≤ 2 4 C1|β| . (3.10)

21 3.3 Proof of Theorem 3.3

To prove Theorem 3.3, the Maaß Converse Theorem 2.5 says that it is enough to show that

ir ir X0 A(β)Pl,ν(β) ξ(s, P ) = π−2sΓ(s + )Γ(s − ) (3.11) l,ν 2 2 (|β|2)s β∈V2(Z) has analytic continuation, is bounded on vertical strips and satisﬁes the functional equation 3 ξ( + l − s, P ) = (−1)lξ(s, P 0 ) 2 l,ν l,ν for all integers l ≥ 0 and all spherical harmonic polynomials Pl,ν of degree l in 3 variables. Here the prime on the summation means that we exclude β = 0. Note 0 l X A(β)Pl,ν (β) that A(−β) = A(β) and Pl,ν(−β) = (−1) Pl,ν(β), which gives us (|β|2)s = β∈V2(Z) 0 l X A(β)Pl,ν (β) (−1) (|β|2)s . This implies that if l is odd then ξ(s, Pl,ν) = 0 and the above β∈V2(Z) conditions are trivially satisﬁed. Henceforth we will assume that l is even.

The key step in the proof of Theorem 3.3 is to get the Rankin integral formula

(3.29) for ξ(s, Pl,ν). For this we ﬁrst get another expression for the series ξ(s, Pl,ν) in

Proposition 3.5. We then introduce the theta function and Eisenstein series required 0 X A(β)Pl,ν (β) to formulate the Rankin integral. Let us set D(s, Pl,ν) := (|β|2)s . Observe β∈V2(Z) that from (3.10) we can conclude that D(s, Pl,ν) converges absolutely in some right half plane Re(s) ≫ 0.

Lemma 3.4.

−2s+l+1/2 ∞ X0 A(β)Pl,ν(β) 1 − 2 X c(−m)b(m) = 23/4 ζ(2s − l − 1/2) (3.12) (|β|2)s 1 + 2−2s+l+3/2 ms−1/2 β∈V2(Z) m=1 P where b(m) = Pl,ν(β). |β|2=m 3 β∈Z

22 u Proof. As in (3.9), for β = β0 +β1i1 +β2i2 ∈ V2(Z) we write gcd(β0, β1, β2) = 2 d with u ≥ 0 and d > 0 odd. Set ν2(β) := u.

Since D(s, Pl,ν) converges absolutely for Re(s) ≫ 0 we can rearrange the terms in the series without changing the value.

t t/2 P −|β|2 −1/2 ∞ u (−1) 2 c (2tn)2 n Pl,ν(β) X X X0 n|d D(s, P ) = 23/4 l,ν |β|2s−1 u=0 t=0 β∈V2(Z) ν2(β)=u t t/2 P −|β|2 −1/2 ∞ ∞ (−1) 2 c (2tn)2 n Pl,ν(β) X X X0 n|d = 23/4 . |β|2s−1 t=0 u=t β∈V2(Z) ν2(β)=u

Make the change of variable U = u − t and observe that {β|ν2(β) = U + t} =

t {2 β|ν2(β) = U}.

t t/2 P −|β|2 −1/2 β tl ∞ ∞ (−1) 2 c (2tn)2 n Pl,ν( 2t )2 0 3/4 X X X n|d D(s, Pl,ν) = 2 β 2s−1 t 2s−1 | t | (2 ) t=0 U=0 β∈V2(Z) 2 ν2(β)=U+t P −|β|2 −1/2 ∞ ∞ c n2 n Pl,ν(β) X X X0 n|d = 23/4 (−1)t2t/22tl(2t)1−2s |β|2s−1 t=0 U=0 β∈V2(Z) ν2(β)=U 3/4 P∞ −2s+l+3/2 t Set η(s) = 2 t=0(−2 ) to get

P −|β|2 −1/2 c n2 n Pl,ν(β) X0 n|d D(s, P ) = η(s) l,ν |β|2s−1 β∈V2(Z) P −|β|2 −1/2 c n2 n Pl,ν(β) ∞ n|gcd(β0,β1,β2) X X n odd = η(s) . |β|2s−1 v=1 β∈V2(Z) |β|2=v

23 Now rearranging terms we get

2 ∞ ∞ c −|β| n−1/2P (β) X X X n2 l,ν D(s, P ) = η(s) . l,ν |β|2s−1 n=1 v=1 β∈V2(Z) n odd |β|2=v n|gcd(β0,β1,β2) Notice that the third summation is nonempty only when v is divisible by n2. Hence we can replace v by mn2 to get

2 ∞ ∞ c −|β| n−1/2P (β) X X X n2 l,ν D(s, P ) = η(s) l,ν |β|2s−1 n=1 m=1 β∈V2(Z) n odd |β|2=mn2 n|gcd(β0,β1,β2) ∞ ∞ l X X X c(−m)Pl,ν(β/n)n = η(s) (m)s−1/2n2s−1/2 n=1 m=1 β∈V2(Z) n odd |β|2=mn2 n|gcd(β0,β1,β2)

∞ ∞ X X X c(−m)Pl,ν(β) D(s, P ) = η(s) n−2s+1/2+l l,ν ms−1/2 n=1 m=1 β∈V2(Z) n odd |β|2=m ∞ ∞ X X c(−m)b(m) = η(s) n−2s+1/2+l (3.13) ms−1/2 n=1 m=1 n odd P where b(m) = Pl,ν(β). Now for Re(s) ≫ 0 we have β∈V2(Z) |β|2=m

∞ X 1 η(s) = 23/4 (−2−2s+l+3/2)t = 23/4 . (3.14) 1 + 2−2s+l+3/2 t=0

∞ ∞ ∞ X X X n−2s+l+1/2 = n−2s+l+1/2 − (2n)−2s+l+1/2 n=1 n=1 n=1 n odd = ζ(2s − l − 1/2) − 2−2s+l+1/2ζ(2s − l − 1/2)

= (1 − 2−2s+l+1/2)ζ(2s − l − 1/2). (3.15)

(3.13), (3.14) and (3.15) completes the proof of the Lemma.

24 l 1 Substituting (3.12) in (3.11), replacing s by s + 2 + 4 and multiplying numerator and denominator by 2s we then have

Proposition 3.5.

s −s l 1 −2s−l− 1 3 2 − 2 l 1 ir l 1 ir ξ(s + + ,P ) = π 2 2 4 Γ(s + + + )Γ(s + + − ) 2 4 l,ν 2s + 21−s 2 4 2 2 4 2 ∞ X c(−m)b(m) ×ζ(2s) . (3.16) s+ l − 1 m=1 m 2 4

We will now get a Rankin integral formula for ξ(s, Pl,ν) of the form Z l 1 l+2 − 1 dxdy ξ(s + + ,Pl,ν) = q(s) f(z)Θl,ν(z)Ee∞(z, s)y 2 4 . (3.17) 2 4 y2 Γ0(4)\H

Here the integral involves a theta function Θl,ν, an Eisenstein series Ee∞(z, s) for Γ0(4) and q(s) is a rational function of 2s. We now give the deﬁnition and basic properties of the theta function and Eisenstein series.

3.3.1 Theta Function and Eisenstein Series

Deﬁnition 3.6.

∞ X 2πi|β|2z X 2πimz Θl,ν(z) := Pl,ν(β)e = b(m)e . (3.18) 3 β∈ Z m=0 3 Here we have identiﬁed the lattice V2(Z) deﬁned in Section 3.2 with Z . By [33]

3 Pg 456, Θl,ν(z) is a holomorphic modular form of weight l + 2 for the discrete group

Γ0(4) i.e. 2l+3 −1 c 1 Θ (γz) = (cz + d) 2 Θ (z) (3.19) l,ν d d l,ν   a b   for γ =   ∈ Γ0(4). It is a cusp form if l > 0 since b(0) = 0. Note that c d 0 the function Θl,ν(z) is not changed if we replace the polynomial Pl,ν(β) by Pl,ν (β) because {β : |β|2 = m} = {β0 : |β|2 = m}.

25 The following transformation laws for Θl,ν(z) will be used in Section 3.3.3 where we obtain the functional equation of ξ(s, Pl,ν) using the Rankin integral formula.

Proposition 3.7.

1 l 3 l+ 3 X 2πim z Θ (− ) = (−i) 2 2 (−iz) 2 b(m)e 4 , (3.20) l,ν 4z m≡0 (mod 4)

1 1 l 3 l+ 3 X 2πim z Θ (− + ) = (−i) 2 2 (−iz) 2 b(m)e 4 . (3.21) l,ν 4z 2 m≡3 (mod 4) Proof. Following Shimura [33], we deﬁne the following theta functions for each h ∈ Z3 :

X 2πi|β|2 z Θl,ν(z; h) := Pl,ν(β)e 4 . β≡h (mod 2) P It is clear from the deﬁnition that Θl,ν(z) = Θl,ν(4z; h). Shimura [33] Pg 454 h (mod 2) gives us the following formulae for Θl,ν(z; h):

t 1 l − 3 l+ 3 X 2πi kh Θ (− ; h) = (−i) 2 2 (−iz) 2 e 2 Θ (z; k) (3.22) l,ν z l,ν k (mod 2) t 2πi hh Θl,ν(z + 2; h) = e 2 Θl,ν(z; h) (3.23)

Here tk stands for the transpose. Using these we can now compute

1 X 1 Θ (− ) = Θ (− ; h). l,ν 4z l,ν z h (mod 2)

Using (3.22) we get

t 1 X l − 3 l+ 3 X 2πi kh Θ (− ) = (−i) 2 2 (−iz) 2 e 2 Θ (z; k) l,ν 4z l,ν h (mod 2) k (mod 2)   t l − 3 l+ 3 X X 2πi kh = (−i) 2 2 (−iz) 2  e 2  Θl,ν(z; k). k (mod 2) h (mod 2)

26 t 2πi kh 3 Use the fact that P e 2 = 2 if k ≡ 0 (mod 2) and 0 otherwise. h (mod 2)

1 l − 3 l+ 3 3 Θ (− ) = (−i) 2 2 (−iz) 2 2 Θ (z; 0) l,ν 4z l,ν l 3 l+ 3 X 2πi|β|2 z = (−i) 2 2 (−iz) 2 Pl,ν(β)e 4 β≡0 (mod 2)

 

1 l 3 l+ 3 X  X  2πim z Θl,ν(− ) = (−i) 2 2 (−iz) 2  Pl,ν(β) e 4 4z   m≡0 (mod 4) |β|2=m β≡0 (mod 2) l 3 l+ 3 X 2πim z = (−i) 2 2 (−iz) 2 b(m)e 4 . m≡0 (mod 4)

We get the last equality because in 3 variables m ≡ 0 (mod 4), |β|2 = m ⇒ β ≡ 0

(mod 2). Let us now consider

1 1 X 1 Θ (− + ) = Θ (− + 2; h). l,ν 4z 2 l,ν z h (mod 2)

Using (3.23) and (3.22) we get

t 1 1 X 2πi hh 1 Θ (− + ) = e 2 Θ (− ; h) l,ν 4z 2 l,ν z h (mod 2) t t X 2πi hh l − 3 l+ 3 X 2πi kh = e 2 (−i) 2 2 (−iz) 2 e 2 Θl,ν(z; k) h (mod 2) k (mod 2)   t l − 3 l+ 3 X X 2πi (h+k)h = (−i) 2 2 (−iz) 2  e 2  Θl,ν(z; k). k (mod 2) h (mod 2)

t 2πi (h+k)h 3 Use the fact that P e 2 = 2 if k ≡ (1, 1, 1) (mod 2) and 0 otherwise. h (mod 2)

1 1 l − 3 l+ 3 3 Θ (− + ) = (−i) 2 2 (−iz) 2 2 Θ (z; (1, 1, 1)) l,ν 4z 2 l,ν l 3 l+ 3 X 2πi|β|2 z = (−i) 2 2 (−iz) 2 Pl,ν(β)e 4 . β≡(1,1,1) (mod 2)

27 Hence  

1 1 l 3 l+ 3 X  X  2πim z Θl,ν(− + ) = (−i) 2 2 (−iz) 2  Pl,ν(β) e 4 4z 2   m≡3 (mod 4) |β|2=m β≡(1,1,1) (mod 2) l 3 l+ 3 X 2πim z = (−i) 2 2 (−iz) 2 b(m)e 4 . m≡3 (mod 4)

We get the last equality because in 3 variables m ≡ 3 (mod 4), |β|2 = m ⇒ β ≡

(1, 1, 1) (mod 2). Let us set

j X z 2πim 4 Θl,ν(z) := b(m)e (3.24) m≡j (mod 4) for j = 0, 3. As in [7], deﬁne the normalized Eisenstein series of weight −(l + 2) for

Γ0(4) by

l+2 l Γ(s + l/2 + 1) −s 1 X cz + d s Ee∞(z, s) := (4π) 2 π Γ(s)ζ(2s) Im(γz) . Γ(s) 2 |cz + d| γ∈Γ∞\Γ0(4) (3.25)   1 b   Here Γ∞ := {  | b ∈ Z}. The above series converges for Res ≫ 0. From 0 1

[7], Ee∞(z, s) has a meromorphic continuation to the whole complex plane with a possibility of 2 simple poles with constant residues. The Eisenstein series is also bounded on vertical strips, away from the poles.

The Eisenstein series satisﬁes

cz + d −(l+2) Ee∞(γz, s) = Ee∞(z, s) (3.26) |cz + d|   a b   for γ =   ∈ Γ0(4). c d

28 3.3.2 Rankin Integral Formula

Deﬁne the integral

Z l+2 − 1 dxdy I(s) := f(z)Θl,ν(z)Ee∞(z, s)y 2 4 . (3.27) y2 Γ0(4)\H

l+2 − 1 We ﬁrst check that it is well-deﬁned. Set h(z) := f(z)Θl,ν(z)Ee∞(z, s)y 2 4 . Then for γ ∈ Γ0(4) we have

˜ l+2 − 1 h(γz) = f(γz)Θl,ν(γz)E∞(γz, s)Im(γz) 2 4 " 1 # 2 2l+3 −1 c cz + d −1 c 1 = f(z) (cz + d) 2 Θ (z) × d d |cz + d| d d l,ν

" −(l+2) #" l+2 − 1 # cz + d y 2 4 × Ee∞(z, s) |cz + d| |cz + d|2

l+2 − 1 = f(z)Θl,ν(z)Ee∞(z, s)y 2 4

= h(z).

Here we have used (3.1), (3.19) and (3.26). Note that I(s) is convergent in some right ˜ half plane because f is a cusp form and Θl,ν and E∞ have moderate growth.

Proposition 3.8. For Re(s) ≫ 0, we have

∞ 1 1 l 1 ir l 1 ir X c(−m)b(m) −2s+ 4 −2s− 2 I(s) = π 2 Γ(s+ + + )Γ(s+ + − )ζ(2s) l 1 . (3.28) 2 4 2 2 4 2 s+ 2 − 4 m=1 m

Proof.

Z l+2 l−1 l −s X cz + d I(s) = 2 π 2 Γ(s + l/2 + 1)ζ(2s) f(z)Θ (z) × l,ν |cz + d| | {z } γ∈Γ∞\Γ0(4) u(s) Γ0(4)\H

s l+2 − 1 dxdy ×Im(γz) y 2 4 . y2

29 Here we have used (3.25). Now using (3.1), (3.19) we get

" 1 #−1 Z X c cz + d 2 I(s) = u(s) −1 f(γz) × d d |cz + d| γ∈Γ∞\Γ0(4) Γ0(4)\H 2l+3−1 l+2 −1 c 1 cz + d × (cz + d) 2 Θ (γz) × d d l,ν |cz + d| l+2 − 1 dxdy ×Im(γz)s Im(γz)|cz + d|2 2 4 y2 Z s+ l+2 − 1 dxdy = u(s) f(z)Θ (z)y 2 4 l,ν y2 Γ∞\H   Z ∞ Z 1 " ∞ # X 2πimx X 2πimz = u(s)  c(m)W sign(m) ir (4π|m|y)e  b(m)e ×  ,  0 0 4 2 m∈Z m=0 m6=0

s+ l+2 − 1 dxdy ×y 2 4 . y2

We have used (3.2), (3.18). Now we integrate with respect to x ﬁrst.

∞ Z ∞ X l 1 dy −2πmy s+ 2 − 4 I(s) = u(s) c(−m)b(m) W− 1 , ir (4πmy)e y . 4 2 y m=1 0

Change of variable y → (4πm)−1 y gives

∞ Z ∞ −(s+ l − 1 ) X c(−m)b(m) − y s+ l − 1 dy 2 4 1 ir 2 2 4 I(s) = u(s) (4π) l 1 W (y)e y − 4 , 2 s+ 2 − 4 y m=1 m 0 ∞ " l 1 ir l 1 ir # l 1 c(−m)b(m) Γ(s + + + )Γ(s + + − ) −(s+ 2 − 4 ) X 2 4 2 2 4 2 = u(s) (4π) l 1 s+ 2 − 4 Γ(s + l/2 + 1) m=1 m ∞ 1 1 l 1 ir l 1 ir X c(−m)b(m) −2s+ 4 −2s− 2 = π 2 Γ(s + + + )Γ(s + + − )ζ(2s) l 1 . 2 4 2 2 4 2 s+ 2 − 4 m=1 m

For the integral formula above involving the Whittaker function, we refer the reader to [26] Pg 316.

30 Comparing (3.8) with (3.16) we get

Proposition 3.9.

3s s Z l 1 −l− 3 5 2 − 2 l+2 − 1 dxdy ξ(s + + ,Pl,ν) = π 4 2 4 f(z)Θl,ν(z)Ee∞(z, s)y 2 4 2 4 2s + 21−s y2 Γ0(4)\H 3s s −l− 3 5 2 − 2 = π 4 2 4 I(s). (3.29) 2s + 21−s

Now using the meromorphic continuation of the Eisenstein series we get the analytic continuation of ξ(s, Pl,ν) to the entire complex plane. To see this, observe that at all points where the Eisenstein series is deﬁned, the integral I(s) converges since f is a cusp form and Θl,ν and the Eisenstein series have moderate growth. At the poles of the Eisenstein series its residue is a constant and hence the residue of I(s) is given

l+2 1 dxdy R 2 − 4 by a constant multiple of f(z)Θl,ν(z)y 2 . This integral is zero since f Γ0(4)\H y is a non-holomorphic cusp form and Θl,ν is a holomorphic modular form. Also, I(s) is bounded on vertical strips since the same is true of the Eisenstein series.

3.3.3 The functional equation

3 Observe that the functional equation from the Maaß Converse Theorem ξ( 2 + l −

l 0 l 1 l 1 0 s, Pl,ν) = (−1) ξ(s, Pl,ν) is equivalent to ξ(s + 2 + 4 ,Pl,ν) = ξ(1 − s + 2 + 4 ,Pl,ν). This implies that we have to show the functional equation

23s − 2s I(s) = 23(1−s) − 2(1−s) I(1 − s) (3.30) since the term 2s + 21−s in the denominator of (3.29) is already invariant under

0 s → 1 − s. (Note that I(s) is unchanged if Pl,ν is replaced by Pl,ν since the same is true of Θl,ν.)

31 We will use the functional equation for the Eisenstein series. For that, we need

1 to deﬁne two more Eisenstein series corresponding to the cusps 0 and 2 of Γ0(4) as in [7].

−z l+2 −1 Ee0(z, s) := Ee∞( , s), (3.31) |z| 4z −2z + 1 l+2 −1 Ee 1 (z, s) := Ee∞( , s). (3.32) 2 | − 2z + 1| 4z − 2

We have (refer [7] Pg 352)

Lemma 3.10.

24s−3 22s−2(1 − 22s−1) h i Ee∞(z, 1 − s) = Ee∞(z, s) + Ee0(z, s) + Ee 1 (z, s) . (3.33) 1 − 22s−2 1 − 22s−2 2 | {z } | {z } d1(s) d2(s)

l+2 1 dxdy 1 R 2 − 4 Let us deﬁne Ij(s) := f(z)Θl,ν(z)Eej(z, s)y 2 for j = 0, . (3.33) Γ0(4)\H y 2 above gives us h i I(1 − s) = d1(s)I(s) + d2(s) I0(s) + I 1 (s) . (3.34) 2

In order to get (3.30) we obtain a formula for I0(s) + I1/2(s) in terms of I(s) in the following proposition.

Proposition 3.11.

2s I0(s) + I 1 (s) = −2 I(s). (3.35) 2

Proof. We have (refer [18] or [7] Pg 354)

1 − 2 iπ z −1 √ e 4 f = 2f (z) (3.36) |z| 4z 0

1 − 2 iπ z −1 1 √ e 4 f + = 2f (z) (3.37) |z| 4z 2 1

y x P 2πim 4 where fj(z) := c(m)W sign(m) , ir (4π|m| 4 )e for j = 0, 1. m≡j (mod 4) 4 2

32 Let us now evaluate the integral I0(s).

Z l+2 − 1 dxdy I0(s) = f(z)Θl,ν(z)Ee0(z, s)y 2 4 . y2 Γ0(4)\H

−1 We make the change of variable z → 4z . This corresponds to action by the element   0 1  2    in SL(2, R) which normalizes Γ0(4). −2 0

Z −1 −1 −1 −1 l+2 − 1 dxdy I0(s) = f( )Θl,ν( )Ee0( , s)Im( ) 2 4 . 4z 4z 4z 4z y2 Γ0(4)\H Now using (3.20), (3.31) and (3.36) we get

" 1 # Z 2 √ − iπ z h l 3 l+ 3 0 i I (s) = e 4 2f (z) (−i) 2 2 (−iz) 2 Θ (z) × 0 |z| 0 l,ν Γ0(4)\H " −(l+2) #" l+2 − 1 # z y 2 4 dxdy × Ee∞(z, s) |z| | − 2z|2 y2 Z −l+ 1 0 l+2 − 1 dxdy = −2 2 f0(z)Θ (z)Ee∞(z, s)y 2 4 . l,ν y2 Γ0(4)\H

0 Θl,ν was deﬁned by (3.24). Proceeding as in the evaluation of I(s) we get

Z ∞ 1 X y y l 1 dy −l+ 2 −2πm 4 s+ 2 − 4 I0(s) = −2 u(s) c(−m)b(m) W− 1 , ir (4πm )e y . 4 2 4 y m≡0 (mod 4) 0 u(s) was deﬁned in Proposition 3.8. Change of variable y → (πm)−1 y gives

Z ∞ −l+ 1 −(s+ l − 1 ) X c(−m)b(m) 2 2 4 1 ir I0(s) = −2 u(s)(π) l 1 W (y) − 4 , 2 s+ 2 − 4 m≡0 (mod 4) m 0

− y s+ l − 1 dy ×e 2 y 2 4 y

1 −(s+ l − 1 ) X c(−m)b(m) −l+ 2 2 4 = −2 u(s)(π) l 1 × s+ 2 − 4 m≡0 (mod 4) m " # Γ(s + l + 1 + ir )Γ(s + l + 1 − ir ) × 2 4 2 2 4 2 Γ(s + l + 1)

33 Hence

− 1 −2s+ 1 l 1 ir l 1 ir I (s) = −2 2 π 4 Γ(s + + + )Γ(s + + − )ζ(2s) × 0 2 4 2 2 4 2 X c(−m)b(m) × l 1 . s+ 2 − 4 m≡0 (mod 4) m

−1 1 In the case of integral I 1 (s) we make a change of variable z → 4z + 2 corresponding 2   −1 1  2  to the action by the element   in SL(2, R). Since this change of variable −2 0 0 gives the same expression for f(z), Θl,ν(z), Ee 1 (z, s) and y with f0 and Θ replaced 2 l,ν 3 by f1 and Θ respectively the calculation for I 1 (s) proceeds in the exact same way l,ν 2 as above to yield

− 1 −2s+ 1 l 1 ir l 1 ir X c(−m)b(m) 1 2 4 I (s) = −2 π Γ(s + + + )Γ(s + + − )ζ(2s) l 1 . 2 2 4 2 2 4 2 s+ 2 − 4 m≡3 (mod 4) m (3.38)

This gives us

− 1 −2s+ 1 l 1 ir l 1 ir I0(s) + I 1 (s) = −2 2 π 4 Γ(s + + + )Γ(s + + − )ζ(2s) × 2 2 4 2 2 4 2 ∞ X c(−m)b(m) × s+ l − 1 m=1 m 2 4 = −22sI(s).

34 Lemma 3.12. 23s − 2s d (s) − 22sd (s) = (3.39) 1 2 23−3s21−s

Proof.

24s−3 22s−2(1 − 22s−1) d (s) − 22sd (s) = − 22s 1 2 1 − 22s−2 1 − 22s−2 24s−3 − 24s−2(1 − 22s−1) = 1 − 22s−2 −24s−3 + 26s−3 = 1 − 22s−2 22s − 1 22s 2s(2s − 2−s) = 24s−3 = 1 − 22s−2 22−2s (2 − 22s−1) 22s (2s − 2−s) 23s − 2s = = . 22−2s 2−s(2 − 22s−1) 23−3s − 21−s

Using (3.34), (3.35) and (3.39) we ﬁnally get

Proposition 3.13.

(23(1−s) − 2(1−s))I(1 − s) = (23s − 2s)I(s).

This gives us the required functional equation for ξ(s, Pl,ν) and completes the proof of our Main Theorem 3.3.

35 CHAPTER 4

Non-Vanishing of the Lift

We will ﬁrst show in Section 4.1 that A(β) is non-vanishing if and only if certain

Fourier coeﬃcients c(n) are non-zero. To get the non-vanishing of c(n), in Section

4.2, we prove an analogue of Waldspurger formula for coeﬃcients of non-holomorphic

Maaß forms for SL2(R) following the paper of Baruch and Mao [2]. This formula relates the value of |c(n)|2 with the central value of twisted L−functions. Finally, in

Section 4.3, we use the results of Friedberg and Hoﬀstein [10] to get the non-vanishing of the twisted L−values.

4.1 Criteria for non-vanishing

We have assumed that f(z) is a Hecke eigenfunction. So for odd prime p we have

n n pc(np2) + p−1/2 c(n) + p−1c = λ c(n) (4.1) p p2 p

2 where T (p )f = λpf.

Theorem 4.1. A(β) = 0 for all β ∈ V2(Z) if and only if c(−n) = 0 for all n > 0 and not of the form n = 4u(8k + 7).

36 Proof. We remind the reader of a fact in basic Number Theory : A positive integer n can be written as a sum of three squares if and only if n is not of the form

4u(8k + 7). (See Theorem 1, Pg. 38 in [14])

Let us now assume that A(β) = 0 for all β ∈ V2(Z). We want to show that if n = 4um, where m 6≡ 7 (mod 8) then c(−n) = 0. It is clear that we can take

4 6 | m. Let us ﬁrst consider the case when m is square-free. For a ﬁxed m we will use

2 2 2 induction on u. First consider the case u = 0. Write m = β0 + β1 + β2 for some integers β0, β1, β2. Observe that gcd(β0, β1, β2) = 1 since m is square-free. Choose

β = β0 + β1i1 + β2i2. Then from the deﬁnition (3.9) we get

A(β) = 23/4m1/2c(−m) =⇒ c(−m) = 0.

u−1 u Now let us assume that c(−m) = ··· = c(−4 m) = 0. Choose β = 2 (β0 + β1i1 +

β2i2) with β0, β1, β2 as above. Again from (3.9) we get

u X A(β) = 23/42um1/2 c(−4u−tm)(−1)t2t/2 t=0 u X =⇒ 23/42um1/2c(−4um) = A(β) − 23/42um1/2 c(−4u−tm)(−1)t2t/2 t=1 By the induction hypothesis we can conclude that c(−4um) = 0. Now let n = 4lj2m where j is odd and m is square-free, m 6≡ 7 (mod 8). Using (4.1) we can write

l 2 l c(−4 j m) = tjc(−4 m) where tj is a constant depending only on j, m and λp for p | j. Since c(−4lm) = 0 for m square-free, we get c(−4lj2m) = 0. This gives us one direction of the equivalence stated in the theorem.

To get the other direction let us assume that c(−n) = 0 unless n = 4l(8k+7). From the deﬁnition (3.9) we can immediately conclude that A(β) = 0 for all β ∈ V2(Z)

−|β|2 u since (2tn)2 is always of the form 4 m with m 6≡ 7 (mod 8).

37 4.2 Waldspurger’s formula for Maaß forms

In their paper [2], Baruch and Mao get an adelic Waldspurger’s formula and, as an application, use it to show the Kohnen-Zagier formula for holomorphic half-integral weight forms [20]. We will follow their paper and do the required calculation to get the Maaß forms case.

Let A denote the ring of adeles for the global ﬁeld Q. Let (π, Vπ) be an irreducible cuspidal automorphic representation of GL2(A) and ψ a non-trivial additive character

0 of A/Q. Then π ' ⊗νπν where (πν,Vπν ) is an irreducible cuspidal automorphic representation of GL2(Qν). Let {Wφ | φ ∈ Vπ} be the Whittaker model for π with respect to the character ψ. Fix a finite set S of places ν containing those where πν is not unramified. Fix unramified vectors φ0,ν for ν 6∈ S. Now choose local Whittaker functionals Lν for all ν such that Lν(φ0,ν) = 1 and kφ0,νkν = 1 for all ν 6∈ S. Here k . kν is defined by         Z a 0 a 0 da kφνkν = (φν, φν) := Lν πν   φν Lν πν   φν .        |a|ν ∗ 0 1 0 1 Qν

Let φ = ⊗ν∈Sφν ⊗ν6∈S φ0,ν ∈ Vπ such that Lν(φν) 6= 0 for ν ∈ S. Here we have ﬁxed

0 an identiﬁcation between Vπ and the restricted tensor product ⊗νVπν . Then deﬁne

|Wφ(1)| Y kφνk dπ(S, ψ) := . (4.2) kφk |Lν(φν)| ν∈S

It is shown in [2] that dπ(S, ψ) is independant of the choice of Lν and φ so long as the conditions above are satisﬁed.

D Let D ∈ Z and deﬁne ψ (x) := ψ(Dx). Let (˜π, Vπ˜) be an irreducible cuspidal automorphic representation of the adelic metaplectic group SLf 2(A). We have

38 0 π˜ ' ⊗νπ˜ν where (˜πν,Vπ˜ν ) is an irreducible cuspidal automorphic representation of SL ( ). Assume thatπ ˜ has a ψD−Whittaker model {W˜ D | φ˜ ∈ V }. Let S be finite f 2 Qν φ˜ π˜ set of places ν containing those whereπ ˜ν is not unramified. Fix unramified vectors

˜ ˜D ˜D ˜ φ0,ν for ν 6∈ S. Choose local Whittaker functionals Lν such that Lν (φ0,ν) = 1 and

˜ ˜ ˜D kφ0,νkν = 1 for all ν 6∈ S. Here kφνkν is deﬁned using Lν as in the linear case above.

˜ ˜ ˜ ˜D ˜ Let φ = ⊗ν∈Sφν ⊗ν6∈S φ0,ν ∈ Vπ˜ such that Lν (φν) 6= 0 for all ν ∈ S. Here we have ﬁxed

0 an identiﬁcation between Vπ˜ and the restricted tensor product ⊗νVπ˜ν . Then deﬁne

|W˜ D(1)| ˜ D φ˜ Y kφνk dπ˜(S, ψ ) := . (4.3) ˜ ˜D ˜ kφk ν∈S |Lν (φν|

D ˜D ˜ It is shown in [2] that dπ˜(S, ψ ) is independant of the choice of Lν and φ so long as the conditions above are satisﬁed.

We now state Theorem 4.3 in [2] for a representation π of GL2(A) with trivial central character which is unramiﬁed at all places. This is the case needed for our application.

Theorem 4.2. (Baruch-Mao) Let π be an irreducible cuspidal automorphic representation of GL2(A) with trivial central character which is unramiﬁed at all places.

Let π˜ = Θ(π, ψ) be the irreducible cuspidal automorphic representation of SLf 2(A) related to π by the theta correspondence. Let S be a ﬁnite set of places including ∞.

Then for a fundamental discriminant D,

1 Y |d (S, ψD)|2 = |d (S, ψ)|2LS , π ⊗ χ |D|−1 (4.4) π˜ π 2 D ν ν∈S

∗ ∗ where χD is the quadratic idele class character of A /Q corresponding to D and

S L (·, π ⊗ χD) is the partial L−function.

39 To apply this theorem, we ﬁrst give the dictionary for Maaß forms and automorphic representations. This diﬀers from [2] only at the place ν = ∞ since we are working with non-holomorphic forms.

Weight 2t Maaß forms

Let h(z) be a weight 2t Maaß Hecke eigenform with respect to SL2(Z) on the upper half plane. h determines a vector in the space of automorphic forms on GL2(A)   a b   by h → φh. The function φh is deﬁned as follows. For g =   ∈ GL2(R), c d −2t az+b cz+d + Q let h|g(z) := h cz+d |cz+d| . We have GL2(A) ' GL2(Q)GL2 (R) GL2(Zp). p + Write g ∈ GL2(A) as g = gQg∞k0, where gQ ∈ GL2(Q), g∞ ∈ GL2 (R) and k0 ∈ Q GL2(Zp). Now we deﬁne φh by the following formula : p

φh(g) = φh(gQg∞k0) := h|g∞ (i). (4.5)

Then φh is a vector in the space of an irreducible cuspidal representation π of GL2(A),

0 with trivial central character. The representation π ' ⊗ πν and the vector φh = ⊗φν can be described as follows :

1. When ν = ∞, π∞ is unitary principal series representation with even K-types

and φ∞ is a vector of K−type 2t.

2. When ν is p−adic, πν is an unramiﬁed representation and φν is an unramiﬁed

vector.

2πinx P ir ir As in [2] if h(z) = a(n)W (4π|n|y)e , then W (4π)a(1) = Wφh (1). sign(n)t, 2 sign(n)t, 2 n∈Z n6=0

Here Wφh is the function corresponding to φh in the Whittaker model of π.

40 1 Weight t + 2 Maaß forms ˜ ˜ ˜ Let A 1 be the space generated by vectors φ = ⊗φν in the space of cuspidal auto- t+ 2 morphic forms on SLf 2(A) satisfying :

˜ 1 1. When ν = ∞, φ∞ is the weight t + 2 vector in a principal series representation

of SLf 2(R).

˜ 2. When ν is p−adic, p 6= 2, φν is an unramiﬁed vector in an unramiﬁed repre-

sentation of SLf 2(Qp).

˜ 3. When ν is p−adic, p = 2, φ2 is a vector in the space of some representationπ ˜2   a b ˜ ˜   of SLf 2(Q2) such thatπ ˜2(σ, 1)φ2 = ˜2(σ)φ2 whenever σ =   ∈ SL2(Z2) c d

with |c|2 ≤ |4|2. Here

˜2(σ) =γ ˜(d)[c, d] when c 6= 0

=γ ˜(d)−1 when c = 0.

whereγ ˜ is the Weil constant and [ , ] is the Hilbert symbol.

˜ The existence of φ2 in condition (3) above puts a restriction onπ ˜2 forcing it to be a subrepresentation ofπ ˜(µ, ψ), a principal series representation of SLf 2(Q2), where µ

∗ is an unramiﬁed character of Z2. The space of vectors inπ ˜(µ, ψ) satisfying condition (3) is two dimensional and is spanned by the two vectors F [2, 1] and F [2, 22] as given in [37]. Deﬁne 1 + i F˜ := µ(22) F [2, 1] + F [2, 22]. (4.6) 2 4

˜n ˜ t ˜+ According to [2], we have L2 (F2) = 0 whenever (−1) n ≡ 2, 3 (mod 4). Deﬁne A 1 t+ 2 ˜ ˜ ˜ ˜ as the subspace of φ ∈ A 1 with φ2 ∈ F2. t+ 2 C

41 + ˜ ˜+ ˜ We then have a bijection given by g(z) ∈ S 1 (4) −→ φg ∈ A 1 , where φg is t+ 2 t+ 2 the unique function on SLf 2(A) that is continuous, left invariant under SL2(Q) and satisﬁes :  √ √     y x/ y cos θ sin θ (t+1/2) (t+1/2)iθ φ˜     , 1 = y 2 e g(x + iy), g  √     0 1/ y − sin θ cos θ where y > 0, x ∈ R and −π < θ ≤ π. Here g is taken to be a Hecke eigenform. The details of this bijection are similar to those in [2] where the holomorphic case is given.

P 2πinx If g(z) = b(n)W sign(n)(t+1/2) ir (4π|n|y)e , then W sign(n)(t+1/2) ir (4π|n|)b(n) = 2 , 2 2 , 2 n∈Z n6=0 ˜ n ˜ n ˜ n W (1). Here W is the function corresponding to φg in the ψ −Whittaker model of φ˜g φ˜g π.˜

Let us now return to the application of Theorem 4.2. We start with a Hecke

+ eigenform g(z) ∈ S 1 with Fourier coeﬃcients {b(n)} and let h(z) be the corre- t+ 2 sponding Maaß form of weight 2t with respect to SL2(Z) with Fourier coeﬃcients {a(n)} given by the Shimura correspondence for Maaß forms (l.c. [18], [21]). Let φ ˜ be the cuspidal automorphic form on GL2(A) obtained from h and φ be the cuspidal automorphic form on SLf 2(A) obtained from g as above. Then φ ∈ Vπ for some ˜ irreducible cuspidal automorphic representation π of GL2(A) and φ ∈ Vπ˜ for some irreducible cuspidal automorphic representationπ ˜ of SLf 2(A). Following [2], we can show thatπ ˜ = Θ(π, ψ). We can apply Theorem 4.2 to these π andπ ˜ with S = {∞, 2}

˜ t 2πix 2πixˆ and ψ(x) = ψ((−1) x) where ψ(x) = e if x ∈ R and for x ∈ Qp set ψ(x) = e where we choosex ˆ ∈ Q satisfying |x − xˆ|p ≤ 1.

t ˜ Choose D such that (−1) D > 0. Let us evaluate dπ(S, ψ)(= dπ(S, ψ) by [2]

˜D |D| Lemma (2.3)) and dπ˜(S, ψ )(= dπ˜(S, ψ )).

42 2 2 2 |Wφ(1)| Y kφνk |dπ(S, ψ)| = 2 2 kφk |Lν(φν)| ν∈{∞,2} 2 |W ir (4π)a(1)| 2 2 t, 2 kφ∞k kφ2k = 2 2 2 kφk |c1W ir (4π)| |L2(φ2)| t, 2 2 2 |a(1)| kφ∞k 3 −1−2ir −2 = 2 2 |1 − 2 | . (4.7) |c1| kφk 2

The calculation at ν = 2 is the same as in [2]. We have also used that L∞(φ∞) = c1W ir (4π) for some non-zero constant c1 due to the uniqueness of local Whittaker t, 2 function.

|D| |W˜ (1)|2 ˜ 2 |D| 2 φ˜ Y kφνk |dπ˜(S, ψ )| = kφ˜k2 ˜|D| ˜ 2 ν∈{∞,2} |Lν (φν)| 2 2 2 |W (t+1/2) , ir (4π|D|)b(|D|)| kφ˜ k kφ˜ k = 2 2 ∞ 2 2 2 |D| kφ˜k |cDW (t+1/2) ir (4π|D|)| |L˜ (φ˜ )|2 2 , 2 2 2 2 ˜ 2 |b(|D|)| kφ∞k = 2 2(D) (4.8) |cD| kφ˜k2 where again from [2], we have  3 − 1 −ir −2  2  4 |1 + 2 | , when D ≡ 1 (mod 4); 2(D) = (4.9)  3 −1−2ir −2 −1 D  4 |1 − 2 | |D|2 , when 4 ≡ 2, 3 (mod 4). ˜|D| ˜ We have also used that L∞ (φ∞) = cDW t+1/2 ir (4π|D|) for some non-zero constant 2 , 2 cD due to the uniqueness of local Whittaker function.

Note that the calculations above are valid only for choices of t, r and D such that

W t+1/2 ir (4π|D|) 6= 0 and Wt, ir (4π) 6= 0. To ensure this we look at the asymptotics 2 , 2 2 for the classical Whittaker functions. For ﬁxed r and ν ∈ R, |ν| large, from [26] Pg 319, we have

√ 1 1 h 1 1 π i 4 −ν ν− 4 2 2 Wν, ir (4π) ≈ 2(4π) e ν cos 2ν (4π) + (1 − 4ν) . 2 4

43 Hence for ﬁxed r, we can choose t a positive odd integer such that W ir (4π) 6= 0. To t, 2 get the condition on D we use the asymptotics in [26] Pg 317 where for ﬁxed ν, µ and y ∈ R, |y| large we have

∞ ( −1 )n − y ν 1 −1 1 −1 X 1 1 y W (y) ≈ e 2 y Γ( +µ−ν) Γ( −µ−ν) Γ( +µ−ν+n)Γ( −µ−ν+n) . ν,µ 2 2 2 2 n! n=0

Hence, for ﬁxed r and t there is a positive integer Mt,r such that whenever |D| > Mt,r

we have W t+1/2 ir (4π|D|) 6= 0. 2 , 2 Now, from (4.7), (4.8) and Theorem 4.2 we get

+ Theorem 4.3. Let g(z) ∈ S 1 with Fourier coeﬃcients {b(n)}. Let t, r and D be t+ 2 such that the above conditions for classical Whittaker functions are satisﬁed. Then

1 |b(|D|)|2 = CL{∞,2}( , π ⊗ χ ) (4.10) 2 D

2 2 ˜ 2 cD kφ∞k kφk 2 −1 3 −1−2ir −2 Q −1 where C = 2 2 ˜ 2 |a(1)| 2(D) |1 − 2 | ν∈{∞,2} |D|ν 6= 0 and D < c1 kφk kφ∞k 2

−Mt,r, a fundamental discriminant.

4.3 Non-vanishing of special values of L−functions

Let S be a ﬁnite set of places of Q and ξ a ﬁxed quadratic character of A∗/Q∗.

∗ ∗ Set Ψ(S, ξ) := {χ quadratic character of A /Q such that χν = ξν for all ν ∈ S}.

Theorem 4.4. (Freidberg-Hoﬀstein [10]) Let π be an irreducible cuspidal automorphic representation of GL2(A) which is self-contragradient. Assume that there exists

1 some χ ∈ Ψ(S, ξ) such that (π⊗χ, 2 ) = 1. Then there are inﬁnitely many χ ∈ Ψ(S, ξ)

1 such that L( 2 , π ⊗ χ) 6= 0.

44 ∗ ∗ ∗ Note that any quadratic character of A /Q is of the form χD for some D ∈ Q . We now state the result for non-vanishing of Fourier coeﬃcients of Maaß forms.

+ Theorem 4.5. Let f ∈ S 1 (4) be a Hecke eigenform with Fourier coeﬃcients {c(n)}. 2 Then for inﬁnitely many D < 0 with D = −n and n ≡ 3 (mod 8), we have c(D) 6= 0.

1 ir 2 Proof. Let λ = 4 + 2 be the eigenvalue of f with respect to the Laplace operator ∆ 1 . Choose an odd integer t such that W ir (4π) 6= 0. This is possible from the 2 t, 2 remark about the asymptotics of the classical Whittaker functions made before The-

+ orem 4.3. Now let g ∈ St+1/2(4) be deﬁned by g := Rt+1/2−2Rt+1/2−4 ··· R−1/2I1/2f. (R are the raising operators and I is the inverting operator) Let {b(n)} be the Fourier coeﬃcients of g. From (3.7) and (3.8), we know that if {c(n)} are the Fourier coef-

ﬁcients of f, then for n > 0 we have c(−n) = b(n). As in the previous section, let h be the Maaß form of weight 2t with respect to SL2(Z) corresponding to g and let φ be the adelic cuspidal automorphic form corresponding to h. Then φ lies in an irreducible cuspidal automorphic representation π of GL2(A) with trivial central character, hence π is self contragradient.

1 Since πp is unramiﬁed at all ﬁnite places p we obtain ( 2 , πp) = 1 for all p. π∞ is the unitary principal series representation with even K−type. If π∞ = π(µ1, µ2)

1 where µ1, µ2 are unitary unramiﬁed characters then from [13] we have ( 2 , π∞) =

1 1 1 ( 2 , µ1)( 2 , µ2) = 1 since ( 2 , µ) = 1 for an unramiﬁed character µ. From [2] (3.1),

1 1 ∗ 1 ∗ we have ( 2 , π ⊗ χD) = ( 2 , π) for all D ∈ Q . So ( 2 , π ⊗ χD) = 1 for all D ∈ Q and hence the hypothesis of Theorem 4.4 is satisﬁed by any choice of S and ξ. Let us choose S = {∞, 2} and ξ = χ−3. Then by deﬁnition χD ∈ Ψ(S, ξ) ⇐⇒ D ≡ 5

(mod 8) and D < 0. We can take D to be square-free since D and Dm2 give the same

45 1 character, i.e. χD = χDm2 . Theorem 4.4 then gives us L( 2 , π ⊗ χD) 6= 0 for inﬁnitely many D satisfying D ≡ 5 (mod 8) and D < 0.

The function g that we constructed above satisﬁes the hypothesis of Theorem 4.3 and hence we get that b(|D|) 6= 0 for inﬁnitely many D such that D < −Mt,r,D ≡ 5

(mod 8) and square-free. (Note that a number D satisfying the above conditions is a fundamental discriminant). This gives us the required result. Note that we can get similar non-vanishing result for diﬀerent D0s by choosing appropriate ξ = χD. Finally, from Theorem 4.1 and Theorem 4.5 we get the non- vanishing of A(β).

Theorem 4.6. A(β) 6= 0 for inﬁnitely many β ∈ V2(Z). In particular, the map f(z) 7−→ F (x) is injective.

46 CHAPTER 5

Hecke Theory

We now present the Hecke Theory for the Spin(1, 4) case. In Section 5.1 we get the generators for the Hecke algebra of the spin group. For this we will follow

Krieg’s work ([22], [23]) in which he obtains the generators of the Hecke algebra in a setting similar to ours. In Section 5.2, for an odd prime p, we deﬁne the Hecke operator Tp on automorphic functions on Spin(1, 4) with respect to Γ = SV2(Z) by

+ its action on the Fourier coeﬃcients A(β). We show that if f ∈ S1/2(Γ0(4)) is a Hecke eigenfunction then F deﬁned in Section 3.2 is an eigenfunction of Tp with eigenvalue

3/2 th p λp + p(p + 1), where λp is the p eigenvalue of f. In Section 5.3 we deﬁne the

Hecke operator Tp2 on automorphic functions on Spin(1, 4) with respect to Γ by its

+ action on the Fourier coeﬃcients A(β). We show that if f ∈ S1/2(Γ0(4)) is a Hecke eigenfunction then F deﬁned in Section 3.2 is an eigenfunction of Tp2 with eigenvalue

3/2 th (p + 1)p λp + (p − 1)(p + 1), where λp is the p eigenvalue of f.

47 5.1 Hecke Algebra Hp for p odd

As mentioned above, we will use the results of Krieg. In his papers ([22], [23]),

1 Krieg works with the Hurwitz order O := Z+Zi1 +Zi2 +Z 2 (1+i1 +i2 +i1i2) instead of the integer quaternions C2(Z) = Z + Zi1 + Zi2 + Zi1i2. We will use Krieg’s results whenever directly applicable or adopt his proofs in our case otherwise.   α β Deﬁne the similitude group GSV +( ) := {M =   ∈ Mat (C ( )) : 2 Q   2 2 Q γ δ ∗ ∗ + ∗ ∗ + αδ − βγ = µ(M) ∈ Q , αβ , δγ ∈ V2}. For M1,M2 ∈ GSV2 (Q) with single coset S decompositions ΓMiΓ = µ ΓMi,µ, i = 1, 2 deﬁne the product X (ΓM1Γ).(ΓM2Γ) := t(C)ΓCΓ with t(C) := ]{(µ, ν):ΓM1,µM2,ν = ΓC} + C∈Γ\GSV2 (Q)/Γ (5.1)

+ The Hecke algebra of GSV2 is the algebra over integers generated by the set of

+ double cosets {ΓMΓ: M ∈ GSV2 (Q)} with product as deﬁned in (5.1). Let

+ Hp be the subalgebra of double cosets generated by {ΓMΓ: M ∈ GSV2 (Q) ∩

−1 k −1 Mat2(C2(Z[p ])), µ(M) = p with k ∈ Z}. Here Z[p ] is the ring of rational numbers with only powers of p in the denominator. Note that for any ΓMΓ ∈ Hp we can

0 + ﬁnd M ∈ GSV2 (Q) ∩ Mat2(C2(Z)) and k ∈ Z such that    k p−1 0     0 ΓMΓ = Γ   Γ (ΓM Γ). (5.2) 0 p−1

Let Hep be the subalgebra of Hp generated by ΓMΓ with M ∈ Mat2(C2(Z)). Then   p−1 0   (5.2) implies that Hp is generated by Hep and the double coset Γ   Γ. 0 p−1 The main result of this section is to give the explicit calculation of the generators of

Hep.

48 + + Denote by GSV2 (Z) = GSV2 (Q) ∩ Mat2(C2(Z)). Fix anα ˆ ∈ C2(Z) such that

|αˆ|2 = p and p - αˆn for every n ≥ 1. (This condition is satisﬁed as long asα ˆ 6= −αˆ¯ and

2 2 2 |αˆ| = p. To see this use the fact that for any x ∈ C2(R) we have x = 2Re(x)x−|x| ).

The following Lemma shows that each double coset in Hep contains a diagonal matrix of a given type.

+ m Lemma 5.1. Let M ∈ GSV2 (Z) with µ(M) = p where p is an odd prime and   αˆlpk1 0   m ≥ 1. Then ΓMΓ contains an element of the form   with αˆ as 0 (ˆα0)lpk2 above and l, k1 ≥ 0, l + k1 + k2 = m and l + k1 ≤ k2.

Proof. Let n = min{|α|2 : α 6= 0 is a matrix entry of some matrix in ΓMΓ}.   α β   2 Let   ∈ ΓMΓ be such that |α| = n. ( Note that we can assume, without loss γ δ of generality, that α is the upper left matrix entry by multiplying on the left and right by elementary matrices corresponding to row and column operations since they belong to Γ.) We ﬁrst claim that γ = β = 0. Suppose γ 6= 0 then |γ|2 ≥ |α|2 by assumption.

+ −1 From the deﬁnition of GSV2 (Z) we have that γα¯ ∈ V2 ∩ C2(Z). Hence γα ∈ V2

−1 2 and so we can ﬁnd a v ∈ V2 ∩ C2(Z) which satisﬁes |γα − v| < 1. This implies that       −v 1 −v 1 α β 2 2       |γ−vα| < |α| . Consider   ∈ Γ and notice that     = −1 0 −1 0 γ δ   γ − vα β − vα   2   ∈ ΓMΓ which contradicts the minimality of |α| = n. Similarly −α −β   α 0   ∗ m we can also show that β = 0. So we have   ∈ ΓMΓ. Since αδ = p we have 0 δ α0 m 2 2m 2 2 2 m 2 t δ = |α|2 p and |α| |p . Since |α| ≤ |δ| we have |α| ∈ {1, p, ··· , p }. Let |α| = p

k1 and α = p τ with p - τ and set l = t−2k1. We claim that m−t ≥ l. First assume that

49       1 1 α 0 α α0pm−t 0       τ 6= ±τ . Since     =   ∈ ΓMΓ we conclude 0 1 0 α0pm−t 0 α0pm−t 0 m−t (by arguments as above) that α|α p i.e. there exists an element u ∈ C2(Z) ∩ V2

0 m−t 0 m−t τ¯ 0 m−t such that uα = α p and hence uτ = τ p =⇒ u = pl τ p ∈ C2(Z). This forces

0 0 m − t ≥ l. If τ = ±τ then (since τ 6= ±i1τ ) repeat the same procedure as above   1 i1   with   to get m − t ≥ l. Now we get the Lemma from : 0 1

Lemma 5.2. Let the notations be as above. Then there are two elements g1, g2 ∈ Γ     τ 0 αˆl 0     such that g1   g2 =   . 0 τ 0pm−t 0 (ˆα0)lpm−t

Proof. From the proof of Theorem 6(b), Pg. 194 in [22] we get an element

2 l ω ∈ C2(Z) ∩ V2 such that gcd(|ω| , p) = 1 and τω ∈ Oαˆ . Here O is the Hurwitz order

l mentioned above. Let µ ∈ O such that τω = µαˆ . We claim that µ lies in C2(Z).

τω(αˆ¯)l Observe that µ = pl . Since the numerator is in C2(Z) and the denominator is not divisible by 2 we can conclude that µ ∈ C2(Z). Choose a, b ∈ Z such that   aω bpm−t+l 2 2(m−t+l)   a|ω| − bp = 1. Taking g2 :=   ∈ Γ we get pm−t+l ω0

   −1 τ 0 αˆl 0       g2   ∈ Γ. 0 τ 0pm−t 0 (ˆα0)lpm−t

This completes the proof of Lemma 5.2 and hence the proof of Lemma 5.1. For the proof of the main theorem of this section we need the single coset decom-     1 0 αˆ 0     position of the two double cosets Γ   Γ and Γ   Γ. 0 p 0 pαˆ0

50 Proposition 5.3.           1 0 1 βˆ p 0 α δ   [      [   Γ   Γ =  Γ   ∪ Γ   ∪  Γ   (5.3) 0 0 p {βˆ} 0 p 0 1 {α,δ} 0 α

ˆ where β in the ﬁrst term runs through the set {β0 + β1i1 + β2i2 : 0 ≤ βj < p, j =

0, 1, 2} and {α, δ} in the third term runs through the set of equivalence classes of   α δ   the set C := {{α, δ} :=   : µ({α, δ}) = p} under the action of Γ by left 0 α0 multiplication. (We will give further precise information about this set in Proposition

5.4). Furthermore, the union above is a disjoint union.

Proof. It is clear that the right hand side of (5.3) is contained in the left hand   1 0 side since Γ   Γ = {M ∈ GSV +( ): µ(M) = p} by Lemma 5.1. Now we   2 Z 0 p + show the opposite inclusion. Let M ∈ GSV2 (Z) satisfy µ(M) = p. As in the proof of Proposition 3.2, we can ﬁnd an element g ∈ Γ such that gM is upper triangular.   α β   ∗ 2 2 2 Write gM =   . Now we have αδ = p. This gives us |α| |δ| = p and 0 δ α0 δ = |α|2 p. We have the following three cases.

2 0 Case I: |α| = 1. This implies that α ∈ {±1, ±i1, ±i2, ±i1i2} and δ = α p. Multiply   α¯ 0   gM on the left by   ∈ Γ to get 0 α∗

      α¯ 0 α β 1αβ ¯           =   . 0 α∗ 0 α0p 0 p

Writeαβ ¯ = x0 +x1i1 +x2i2 = p(y0 +y1i1 +y2i2)+(β0 +β1i1 +β2i2) with 0 ≤ βj ≤ p−1.

51 So       1 −(y0 + y1i1 + y2i2) 1αβ ¯ 1 β0 + β1i1 + β2i2           =   . 0 1 0 p 0 p   1 βˆ S   This gives us M ∈ Γ   . {βˆ≡0 (mod p)} 0 p 2 2 2 Case II: |α| = p . This implies that |δ| = 1 and hence δ ∈ {±1, ±i1, ±i2, ±i1i2}   δ∗ −δ∗βδ¯ 0   and α = δ p. Multiplying gM on the left by   ∈ Γ we get 0 δ¯

      δ∗ −δ∗βδ¯ pδ0 β p 0           =   . 0 δ¯ 0 δ 0 1

  p 0   This implies that M ∈ Γ   . 0 1   α β 2 0   Case III: Let |α| = p. This implies that δ = α and gM =   . Hence we 0 α0   α δ S   get that M ∈ Γ   where the summation runs over the set described in {α,δ} 0 α0 the statement of the Proposition.

This gives us (5.3). Finally, we show that the union on the right hand side of (5.3)   1 βˆ   ˆ is disjoint. It is clear that the cosets {Γ   : β ≡ 0 (mod p)} are mutually 0 p     ˆ ˆ 1 β1 1 β2     disjoint. (If for some g ∈ Γ we have g   =   then we must have 0 p 0 p   1 u   ˆ ˆ g =   for some u ∈ Z + Zi1 + Zi2. This implies that β2 = β1 + pu and then 0 1

52 ˆ ˆ ˆ the condition on β forces u = 0 and β1 = β2).       a c 1 β p 0       If for some g =   ∈ Γ we have g   =   then we get b d 0 p 0 1      1 βˆ p 0 c = 0 and d = 1 which is impossible. Hence  S Γ   ∩ Γ   p      {βˆ} 0 p 0 1     p 0 1 βˆ     is empty. Similarly we can show that the cosets Γ   and Γ   are 0 1 0 p   α δ   2 disjoint to Γ   with |α| = p. The precise description of the disjoint cosets 0 α0   α δ S   in Γ   is given in the next proposition. This completes the proof of {α,δ} 0 α0 proposition 5.3.

2 2 2 2 From [14] Pg. 30, we know that the number of solutions to α0 + α1 + α2 + α3 = p

2 for any odd prime p is exactly 8(p + 1). Let B := {α ∈ C2(Z): |α| = p} and

2 G := {µ ∈ C2(Z): |µ| = 1} be the group of 8 units in C2(Z).G acts on B by left multiplication and there are p + 1 orbits of this action with 8 elements each. Let us   α δ (1) (p+1)   ﬁx representatives {α , ··· , α } of these orbits. Now let C := {  ∈ 0 α0 + 2 GSV2 (Z): |α| = p}. Deﬁne an equivalence relation on C as follows : We say that g1 and g2 in C are equivalent (g1 ∼ g2) if there exists an element g ∈ Γ such that gg1 = g2 (i.e. we have Γg1 = Γg2). Then we have

Proposition 5.4. There are exactly p(p + 1) equivalence classes in C with respect to

∼ .

53       α δ1 a b αˆ δ2 2 2       Proof. Let |α| = |αˆ| = p. Suppose   =     for 0 α0 c d 0α ˆ0   a b   ∗ ∗ ∗ some element   ∈ Γ. Then we have c = 0. Also ad −bc = 1 implies ad = 1. c d   α δ   Hence a is a unit (i.e. a ∈ G) and α = aα.ˆ So we can assume that if   is a 0 α0 representative of an equivalence class of C then α = α(i) for some i = 1, ··· p+1. Now we claim that for each ﬁxed i = 1, ··· , p + 1 we get exactly p elements of the form   α(i) δ     that are not equivalent in C. Note that this claim completes the 0 (α(i))0 proof of Proposition 5.4. Let us ﬁx i and for simplifying notation let us write α for α(i)

¯0 0 whenever there is no confusion. We have δα ∈ V2 = R+Ri1 +Ri2. Hence δ = vα for     0 0 α v1α α v2α     some element v ∈ V2. Now if for some g ∈ Γ we have g   =   0 α0 0 α0   1 u   with v1, v2 ∈ V2 then g has to be of the form g =   for some u ∈ C2(Z) ∩ V2. 0 1 0 0 0 Hence v2α = v1α + uα . So for a ﬁxed α the distinct left cosets are given by the

0 choices of δ = vα where v = a + bi1 + ci2 satisﬁes

0 0 ≤ a, b, c < 1 and vα ∈ C2(Z). (5.4)

If α = α0 + α1i1 + α2i2 + α3i1i2 then

0 vα = (aα0+bα1+cα2)+(−aα1+bα0+cα3)i1+(−aα2−bα3+cα0)i2+(aα3−bα2+cα1)i1i2.

54 0 Now vα ∈ C2(Z) implies   t := aα + bα + cα ∈ ,  0 0 1 2 Z    t1 := −aα1 + bα0 + cα3 ∈ Z, (∗)  t := −aα − bα + cα ∈ ,  2 2 3 0 Z    t3 := aα3 − bα2 + cα1 ∈ Z.

2 This implies that α0t0 − α1t1 − α2t2 + α3t3 = a|α| = ap ∈ Z. Similarly bp, cp ∈ Z.

Denote by a1 := pa, b1 := pb, c1 := pc ∈ Z. We have a1, b1, c1 ∈ {0, 1, ··· , p − 1}. From the ﬁrst three conditions in (∗) above we have     α α α a  0 1 2   1          ≡ 0 (mod p). (5.5)  −α1 α0 α3   b1          −α2 −α3 α0 c1 | {z } A (The fourth condition in (∗) is linearly dependant on the ﬁrst three.) Consider A as a linear transformation on a 3−dimensional vector space over Z/pZ. Det(A) =

2 α0|α| = α0p ≡ 0 (mod p). Also A 6≡ 0 (mod p). One can check that rank of A is equal to 2. This implies that dim(Ker(A)) = 1. Hence the number of solutions to

(5.4) is exactly p. This completes the proof of Proposition 5.4.

Next we give the Peirce decomposition for quaternions. Fix r, s ∈ Z such that 1 + r2 + s2 ≡ 0 (mod p). The existence of r and s satisfying the above condition is proved in Lemma 5.11. Given α = α0 + α1i1 + α2i2 + α3i1i2 ∈ C2(Z) deﬁne   α0 − α2r − α3s α1 − α2s + α3r   ψp(α) :=   . (5.6) −α1 − α2s + α3r α0 + α2r + α3s

55 Lemma 5.5. For an odd prime p, the map ψp deﬁned above gives an isomorphism

2 of rings between C2(Z)/pC2(Z) and Mat2(Z/pZ). ψp satisﬁes |α| ≡ det(ψp(α)) (mod p).

This lemma follows directly from Lemma 1 on Pg. 329 in [23] when one observes that for an odd prime p we have C2(Z)/pC2(Z) ' O/pO where O is the Hurwitz order.

We can extend the map ψp naturally to a ring isomorphism between Mat4(Z/pZ) and Mat2(C2(Z)/pC2(Z)). For an element α ∈ C2(Z) (respectively, g ∈ Mat2(C2(Z))), deﬁne rk(α) (respectively, rk(g)), as the rank of the matrix ψp(α) as an element of Mat2(Z/pZ) (respectively, as the rank of the matrix ψp(g) as an element of

Mat4(Z/pZ)). Note that rk(g1gg2) = rk(g) for any g1, g2 ∈ Γ. Also for α ∈ C2(Z),

2 since |α| ≡ det(ψp(α)) (mod p) we have   0, if p|α;   rk(α) = 1, if p - α but |α|2 ≡ 0 (mod p); (5.7)   2  2, if p - |α| .

The notion of rank will play a crucial role in the proof of many results.

Proposition 5.6.         αˆ 0 pα 0 α δ p v   [   [   [   Γ   Γ = Γ   Γ   Γ   . (5.8) 0 pαˆ0 α 0 α0 α,δ 0 pα0 v 0 p

Here the union in the ﬁrst term runs over the orbits of the action of the group of

2 units in C2(Z) on the set {α ∈ C2(Z): |α| = p}. The union in the second term runs over the same set of equivalence classes for α and for each α the set δ is determined

56 in Lemma 5.7 below. The union in the third term is over the set {v = v0 +v1i1 +v2i2 :

2 v0, v1, v2 ∈ Z, 0 ≤ v0, v1, v2 ≤ p − 1, v 6= 0, |v| ≡ 0 (mod p)}.   αˆ 0 2   Proof. A matrix M with µ(M) = p lies in the double coset Γ   Γ 0 pαˆ0 if and only if rk(M) = 1. Take such a matrix M. By arguments similar to those in the proof of Proposition 5.3 we can ﬁnd an element g ∈ Γ such that gM =   µ δ   2 2 3 4 2 4   . We have |µ| = 1, p, p , p or p . Notice that |µ| = 1 or p implies µ0 2 0 |µ|2 p that one of the diagonal entries above is a unit and hence from (5.7) we get rk(M) ≥ 2, a contradiction. So, now we have to consider three cases.

2 3 µ0 2 Case I: |µ| = p . Since |µ|2 p ∈ C2(Z) we get that µ is divisible by p. Write µ = pα   pα δ 2   where |α| = p. Then we have gM =   . Notice that we can assume 0 α0 2 2 2 that |δ| < p. (Otherwise, if |δ| ≥ |α| = p then choose u ∈ C2(Z) ∩ V2 such that   1 u 0 2   |δ − uα | < p and replace gM by   gM). We get rk(gM) ≥ rk(δ) and from 0 1 (5.7) we have rk(δ) = 2 unless δ = 0 since 0 ≤ |δ|2 < p. Hence rk(M) = 1 forces   pα 0 S   δ = 0 and we can conclude that M ∈ Γ   . α 0 α0 Case II: |µ|2 = p2. We ﬁrst claim that µ = p for some unit . If not then rk(µ) = rk(µ0) = 1 which implies that rk(M) ≥ 2, a contradiction. So we get that gM =   p δ     . By multiplying on the left by a suitable element of Γ we get that δ = 0 p v = v0 + v1i1 + v2i2 with 0 ≤ v0, v1, v2 ≤ p − 1. v also satisﬁes rk(v) = rk(M) = 1.   p v 2 S   Hence we get |v| ≡ 0 (mod p) and v 6= 0. So, we have M ∈ Γ   . v 0 p

57   α δ 2   2 Case III: |µ| = p. Then gM is of the form   where |α| = p. The 0 pα0 possible values of δ are given by the following Lemma.

Lemma 5.7. With notations as above, δ has to satisfy the following two conditions.

0 1. δ ∈ C2(Z) has to be of the form uα where u = a + bi1 + ci2 ∈ V2 with 0 ≤ a, b, c < p and

2. α∗δ0 ≡ 0 (mod p).

The number of δ satisfying the above properties is p3.

Proof. The ﬁrst condition is clear by a similar argument as given in the proof of Proposition 5.4. To get the second condition we will use the fact that rk(M) = 1.

From (5.6) we have     α δ ψp(α) ψp(δ)     ψp   =   . 0 pα0 0 0

Let C1 and C2 (respectively, C3 and C4), be the columns of the matrices ψp(α)

(respectively, of ψp(δ)). The condition rk(M) = 1 implies that the determinant of every 2 × 2 matrix obtained from any two of the four columns C1,C2,C3 or C4 is equivalent to zero mod p. There are 6 such matrices and we get

2 2 ∗ 0 ∗ 0 ∗ 0 ∗ 0 |α| ≡ |δ| ≡ Re(α δ ) ≡ Re(i1α δ ) ≡ Re(i2α δ ) ≡ Re(i1i2α δ ) ≡ 0 (mod p).

This gives us the second condition in the statement of the Lemma.

Now we prove that the number of δ satisfying the above conditions is p3. We ﬁrst remind the reader that from Lemma 5.4 there are exactly p choices of δ ∈ C2(Z) of

58 the form uα0 where the coordinates of u lie between 0 and 1. Let us denote them by

0 0 0 0 = u1α , u2α , ··· , upα . (Note that again from Lemma 5.4 these p elements form a

0 line {tu2α : t = 0, 1, ··· , p − 1} mod p). Now we claim that all the possible choices

0 of δ satisfying the ﬁrst condition in the statement of the Lemma is {(u + uj)α : u = x+yi1 +zi2, x, y, z ∈ {0, 1, ··· , p−1}, j = 1, 2, ··· , p}. It is clear that when j = 1, or

0 equivalently uj = 0, the set {uα : u = x + yi1 + zi2, x, y, z ∈ {0, 1, ··· , p − 1}} does satisfy the ﬁrst condition. Now any elementu ˆ with coordinates between 0 and p can be written in the formu ˆ = u + v where u = x + yi1 + zi2, x, y, z ∈ {0, 1, ··· , p − 1}

0 0 and v has coordinates between 0 and 1. Hence ifuα ˆ ∈ C2(Z) then so does vα . But then from Lemma 5.4 we can conclude that v = uj for some j = 1, 2, ··· , p. This gives us the claim.

To show that the number of δ satisfying the conditions in the statement of the

3 Lemma is p , it is now enough to show that for every u = x + yi1 + zi2, x, y, z ∈

∗ 0 0 {0, 1, ··· , p − 1} there is a unique t = 0, 1, ··· , p − 1 such that α ((u + tu2)α ) ≡ 0

(mod p). To prove this we deﬁne a linear transformation Nα on the 3−dimensional

∗ 0 space U over Z/pZ by Nα(u) := α u α (mod p). Then the map Nα satisﬁes the following properties :

1. Rank of Nα is equal to 1,

2 2. The eigenvalues of Nα are given by 0 and 4α3, where α = α0 +α1i1 +α2i2 +α3i1i2

(Note that we can assume that α3 6= 0 by multiplying α by a suitable unit, if

necessary. We are allowed to do this since α is ﬁxed only upto a unit),

0 3. For every t = 0, 1, ··· , p − 1 the vectors tu2α are eigenvectors of Nα with

2 eigenvalue 4α3.

59 All of the above three properties can be proved by writing down Nα in the form of a 3×

3 matrix and doing the required calculations. From these properties we can conclude

0 that the image of the map Nα is precisely the line {tu2α : t = 0, 1, ··· , p − 1}. Hence given u = x+yi1+zi2, x, y, z ∈ {0, 1, ··· , p−1} choose the unique t0 ∈ {0, 1, ··· , p−1}

0 0 such that Nα(u) = −t0u2α. Then (u + t0)α satisﬁes the second condition in the statement of the Lemma. This proves Lemma 5.7 and with it Proposition 5.6. Now we state the main theorem on the generators of the Hecke algebra.     1 0 αˆ 0     Theorem 5.8. The double cosets X1 := Γ   Γ,X2 := Γ   Γ and 0 p 0α ˆ0p   p 0   X3 := Γ   Γ generate the algebra Hep and they are algebraically independent. 0 p 2 As in the remarks before Lemma 5.1, αˆ ∈ C2(Z) is chosen so that it satisﬁes |αˆ| = p and p - αˆn for all n ≥ 1.

Proof. We will prove that ΓMΓ, where µ(M) = pm, is given by an algebraic expression in terms of X1,X2 and X3 by induction on m. If m = 1 then the result follows from Lemma 5.1. If m = 2 then again from Lemma 5.1 we need to show that   1 0   Γ   Γ is given by an algebraic expression in terms of X1,X2 and X3. Write 0 p2         1 0 1 0 p 0 α 0   2       (Γ   Γ) = t0Γ   Γ + t1Γ   Γ + t2Γ   Γ 0 p 0 p2 0 p 0 α0p where t0, t1, t2 ∈ Z. We can do this because every double coset arising in the product above has similitude p2 and from Lemma 5.1 we know that every double coset with   1 0 2   similitude p is one of the three on the right hand side above. If Γ   Γ = 0 p

60 e S ΓMr is the decomposition into disjoint single cosets then by (5.1) we have t0 = r=1   1 0   card{(i, j):ΓMiMj = Γ  }. Now one can use Proposition 5.3 to check that 0 p2 t0 = 1. Hence we get the required result for m = 2.

Now let us assume that the theorem is true for all M with µ(M) = pk, k ≤ m − 1.   αˆlpk1 0 m   Let µ(M) = p and from Lemma 5.1 write ΓMΓ = Γ   Γ with 0 (ˆα0)lpk2 l, k1, k2 satisfying the conditions stated in Lemma 5.1. From (5.7) we get   0, if k > 0;  1  rk(M) = 1, if k1 = 0, l > 0; (5.9)    2, if k1 = l = 0.

We cannot have rk(M) = 3 or 4. For rk(M) = 0 the theorem follows from the   αˆlpk1−1 0   induction hypothesis and the relation ΓMΓ = pΓ   Γ. 0 (ˆα0)lpk2−1   αˆl−1 0   Now let rk(M) = 1. Denote by B the matrix   and consider 0 (ˆα0)l−1pk2−1 the product   αˆ 0   X (ΓBΓ)(Γ   Γ) = t(C)ΓCΓ (5.10) 0 0 pαˆ C∈Γ\Γ(pm)/Γ

(pm) + m where t(C) ∈ Z and Γ := {M ∈ GSV2 (Z): µ(M) = p }. Since the rank of every single coset representative of X2 is 1 the formula for t(C) in (5.1) gives us that t(C) 6=   αˆl1 0   0 implies that rk(C) ≤ 1. Fix C =   with l1+k3 = m, l1 > 0, k3 ≥ l1 0 (ˆα0)l1 pk3 S with rk(C) = 1 and t(C) 6= 0. Let ΓBΓ = ΓBi be the decomposition into disjoint

61   αˆ 0   single cosets where the Bi are chosen to be upper-triangular. Let Γ   Γ = 0 pαˆ0 S ΓAi be the decomposition into disjoint single cosets as in Proposition 5.6. Choose

Bi and Aj such that BiAj = gC for some g ∈ Γ. Since the left side is upper-   1 u   triangular we get that g has to be of the form   with u ∈ C2(Z) ∩ V2. (We 0 1 are allowed to have units as the diagonal entries but we can multiply both sides with suitable matrix from Γ and get g in the above form.) Again using Proposition   αˆ 0   5.6, we can conclude that Aj =   and hence Bi is forced to be equal 0 pαˆ0   αˆl1−1 0   to   . This is possible if and only if l1 = l and k3 = k2 i.e. 0 (ˆα0)l1−1pk3−1 C = M. Hence we can conclude that t(C) 6= 0 and rk(C) = 1 ⇐⇒ C = M and in this case t(M) = 1. Now (5.10) gives us   αˆ 0   X (ΓBΓ)(Γ   Γ) = ΓMΓ + t(C)ΓCΓ 0 m 0 pαˆ C∈Γ\Γ(p )/Γ, rk(C)=0 which gives us the result for rk(M) = 1 using the induction hypothesis and the result for rank 0.   1 0   Finally, if rk(M) = 2 we can assume that M =   . Using arguments as 0 pm above one can conclude that     1 0 1 0     X (Γ   Γ)(Γ   Γ) = ΓMΓ + t(C)ΓCΓ m−1 m 0 p 0 p C∈Γ\Γ(p )/Γ, rk(C)≤1 with t(C) ∈ Z. Since we have already shown that the theorem is valid for elements with rank ≤ 1, the above equation completes the proof of the theorem.

62 5.2 Hecke operator Tp

In this section, we deﬁne the Hecke operator Tp and get a formula for Tp in terms of the Fourier coeﬃcients A(β) in Proposition 5.10. In Theorem 5.12 we show

+ that if f ∈ S1/2(Γ0(4)) is a Hecke eigenfunction then F deﬁned in Section 3.2 is an

3/2 th eigenfunction of Tp with eigenvalue p λp + p(p + 1), where λp is the p eigenvalue of f.

Deﬁnition 5.9. Let F be a function on H3. For any odd prime p we deﬁne the Hecke operator Tp by ! X x + βˆ X (T F )(x) := F + F (px) + F ((αx + δ)(α0)−1). (5.11) p p {βˆ} {α,δ} ˆ Here {β} = {β0 + β1i1 + β2i2 : 0 ≤ β0, β1, β2 ≤ p − 1}, α, δ are as in Proposition 5.3 and x ∈ H3.

Note that the Hecke operator is deﬁned by evaluating F at the points obtained by acting by the single coset representatives of Proposition 5.3 on x and then adding them up. If we multiply the single-coset representatives by an element of Γ on the right, we again get a set of single-coset representatives. Hence if F is an automorphic form with respect to Γ, i.e. F (gx) = F (x) for all g ∈ Γ, then so is TpF.

(1) (p+1) 2 Let α , ··· , α be representatives for the orbits of {α ∈ C2(Z): |α| = p} under the left action by the units in C2(Z). We will identify the lattice T = V2(Z)

3 with Z and the multiplication of two elements will be as elements of V2(Z).

Proposition 5.10. 1. For any β ∈ Z3

p+1 0 ! X α(i) βα(i) (T F ) = p3/2A(pβ) + p3/2A(β/p) + p A . (5.12) p (β) p i=1

63 0 α(i) βα(i) Here (TpF )(β) is the β−th Fourier coeﬃcient of TpF. If β/p or p 6∈ C2(Z) then the corresponding terms are assumed to be zero.

2. For any prime q deﬁne νq(β) := the highest power of q dividing all the coordi-

nates of β.

0 α(i) βα(i) (a) If q 6= p then for every i = 1, ··· p + 1 we have νq(β) = νq( p ).

(b) Write β = gcd(β)(β0 + β1i1 + β2i2) with gcd(β0, β1, β2) = 1.

2 2 2 i. If −(β0 + β1 + β2 ) is not a square mod p, then for all i = 1, ··· , p + 1

0 α(i) βα(i) we have νp( p ) = νp(β) − 1.

2 2 2 2 2 2 2 ii. If β0 + β1 + β2 ≡ 0 (mod p) but β0 + β1 + β2 6≡ 0 (mod p ), then

0 α(i) βα(i) there is exactly one i such that νp( p ) = νp(β) and for all j 6= i 0 α(j) βα(j) we have νp( p ) = νp(β) − 1.

2 2 2 2 iii. If β0 + β1 + β2 ≡ 0 (mod p ), then there is exactly one i such that

0 0 α(i) βα(i) α(j) βα(j) νp( p ) = νp(β) + 1 and for all j 6= i we have νp( p ) =

νp(β) − 1.

2 2 2 iv. If −(β0 + β1 + β2 ) is a square mod p, then there are exactly two 0 α(iµ) βα(iµ) distinct iκ1 , iκ2 such that νp( p ) = νp(β) for µ = κ1, κ2 and for 0 α(j) βα(j) all j 6= iκ1 , iκ2 we have νp( p ) = νp(β) − 1.

Remarks : From the deﬁnition of A(β) in 3.9 we see that A(β) depends on the highest power of a prime that divides gcd(β).

1. (2)(a) above tells us that the highest power of prime q 6= p that divides gcd(β)

0 α(i) βα(i) does not change when β is replaced by p .

64 2. When we consider the Hecke relation (4.1) for SL2

n n pc(np2) + p−1/2 c(n) + p−1c = λ c(n) p p2 p

n we get four diﬀerent equations according to whether p = −1, n ≡ 0 (mod p)

2 2 n but n 6≡ 0 (mod p ), n ≡ 0 (mod p ) or p = +1. The four cases in (2)(b)

2 2 2 above correspond to these when we set n = −(β0 + β1 + β2 ).

Proof. We ﬁrst prove (2). Notice that we can assume without loss of generality that gcd(β) = 1, i.e. νq(β) = 0 for all primes q. Let β = β0 + β1i1 + β2i2 and

α = α0 + α1i1 + α2i2 + α3i1i2. (We will drop the superscript i for α whenever there is no confusion. We will also use the same notation for the element α and the orbit

0 ˆ ˆ ˆ corresponding to it whenever there is no confusion) Write α βα¯ = β0 + β1i1 + β2i2.

Writing in terms of coordinates of β and α we get   βˆ = −2α (β α − β α − β α ) + 2α (β α + β α + β α ) − pβ ,  0 3 2 1 1 2 0 3 0 0 0 1 1 2 2 0  ˆ (I) β1 = −2α2(β2α1 − β1α2 − β0α3) + 2α0(−β0α1 + β1α0 − β2α3) − pβ1,   ˆ  β2 = 2α1(β2α1 − β1α2 − β0α3) + 2α0(−β0α2 + β1α3 + β2α0) − pβ2.

Using these formulae we have

 ˆ ˆ ˆ 2 2 2 2  β0β0 + β1β1 + β2β2 = −p|β| + 2α0|β| + 2(β2α1 − β1α2 − β0α3) ,    ˆ ˆ ˆ 2 2 2 2  −β0β0 + β1β1 − β2β2 = −p|β| + 2α2|β| + 2(−β0α1 + β1α0 − β2α3) , (II)  β βˆ − β βˆ − β βˆ = −p|β|2 + 2α2|β|2 + 2(β α + β α + β α )2,  0 0 1 1 2 2 3 0 0 1 1 2 2   ˆ ˆ ˆ 2 2 2 2  −β0β0 − β1β1 + β2β2 = −p|β| + 2α1|β| + 2(−β0α2 + β1α3 + β2α0) . Step 1 : To prove (2)(b)(i) we have to show that for all the p + 1 values of α we have α0βα¯ 6≡ 0 (mod p) assuming that −|β|2 is not a square mod p. Suppose not, ˆ ˆ ˆ then β0, β1, β2 ≡ 0 (mod p). We can assume that α0 6= 0 (we can always multiply α

65 by a unit to ensure this). Then (II) above gives us

2 2 2 (β2α1 − β1α2 − β0α3) + α0|β| ≡ 0 (mod p)

−1 2 2 ⇒ (α0 (β2α1 − β1α2 − β0α3)) ≡ −|β| (mod p).

This contradicts the assumption that −|β|2 is not a square mod p. This proves

(2)(b)(i).

Step 2 : We will now prove (2)(b)(ii) and (iii). We have |β|2 ≡ 0 (mod p). In both cases we have to get a unique α with |α|2 = p such that α0βα¯ ≡ 0 (mod p) and show that the other values of α do not satisfy this condition. (In the case of (iii) we have to show the stronger condition α0βα¯ ≡ 0 (mod p2) holds for a unique α.) From (I) ˆ ˆ ˆ and (II) above we can deduce that β0, β1, β2 ≡ 0 (mod p) is equivalent to the matrix equation       β0 β1 β2 0 α0 0                    β1 −β0 0 −β2   α1   0      ≡   (mod p). (5.13)        β 0 −β β   α   0   2 0 1   2          0 β2 −β1 −β0 α3 0 | {z } Pβ (II) gives the forward direction (“ ⇒ ”) of the equivalence and (I) gives the reverse direction. Notice that this is also the same as βα¯ ≡ 0 (mod p). One can check that the determinant of the matrix Pβ is equivalent to 0 modulo p and its rank as a linear operator on the 4 dimensional vector space over Z/pZ is 2. The null space is given by ha(β0, β1, β2, 0) + b(β1, −β0, 0, −β2): a, b ∈ Z/pZi if β2 6≡ 0 (mod p) and is given by ha(β0, β1, 0, 0) + b(0, 0, −β1, −β0)i if β2 ≡ 0 (mod p). So any α ∈ Ker(Pβ) has to satisfy

66   (aβ0 + bβ1) + (aβ1 − bβ0)i1 + aβ2i2 − bβ2i1i2 (mod p), if p - β2; α ≡ (5.14)

 aβ0 + aβ1i1 − bβ1i2 − bβ0i1i2 (mod p), if p|β2.

Here a, b ∈ {0, 1, ··· , p−1}. We now have to distinguish between the two possibilities:

β2 divisible by p and β2 not divisible by p.

(1) First suppose that β2 is divisible by p. (Then β0, β1 are not divisible by p since p - gcd(β) and |β|2 ≡ 0 (mod p).) Note that this is possible only if p ≡ 1 (mod 4)

2 2 −1 2 because we have β0 + β1 ≡ 0 (mod p) ⇒ (β0β1 ) ≡ −1 (mod p), i.e. −1 is a square mod p. Let u, v be the unique integers (up to sign) such that u2 + v2 = p and let t be the unique element (up to sign) in Z/pZ satisfying t2 ≡ −1 (mod p). Choose the sign of u, v and t such that uv−1 ≡ t (mod p). Now given β as above we will

−1 −1 have β0β1 ≡ t or − t (mod p). If β0β1 ≡ t (mod p) then the only possibility for

2 −1 −1 α ∈ Ker(Pβ) with |α| = p is α = u + vi1 with a ≡ β0 u ≡ β1 v (mod p) and b = 0

−1 −1 in (5.14). (Note that one could take a = 0 and b ≡ β0 u ≡ β1 v (mod p) which would give usα ˆ = −vi2 − ui1i2. But we haveα ˆ = −i1i2α. Similar argument for other

−1 possible choices of a and b tell us that we get a unique orbit of α.) If β0β1 ≡ −t

2 (mod p) then the only possibility for α ∈ Ker(Pβ) with |α| = p is α = u − vi1 with

−1 −1 a ≡ β0 u ≡ −β1 v (mod p) and b = 0 in (5.14). (Arguing as above we can conclude that the orbit of α is unique.)

−1 2 (2) Now let us assume that β2 is not divisible by p. Then we have (β0β2 ) +

−1 2 (β1β2 ) ≡ −1 (mod p). We will obtain a bijection α 7→ (xα, yα) between the sets

2 2 2 S1 := { orbits of α with |α| = p, α2 + α3 6≡ 0 (mod p)} and S2 := {(x, y) ∈ Z/pZ ×

2 2 2 Z/pZ : x + y ≡ −1 (mod p)}. Then the unique α ∈ Ker(Pβ) with |α| = p will be

−1 −1 given by the pre-image of (β0β2 (mod p), β1β2 (mod p)) under the above bijection.

67 Given α ∈ S1 let (xα, yα) be the unique solution to       α2 −α3 xα α0           =   . α3 α2 yα α1

2 Then α0 ≡ xαα2 −yαα3 (mod p) and α1 ≡ xαα3 +yαα2 (mod p). Using |α| = p, this gives us that (xα, yα) ∈ S2. Let us show that the map α 7→ (xα, yα) is injective. If

2 (xη, yη) is a pair corresponding to η satisfying |η| = p and xα ≡ xη (mod p), yα ≡ yη

(mod p) then we can conclude that αη¯ ≡ 0 (mod p). Hence αη¯ = p, which implies

2 |α|2|η|2 η || = p2 = 1, i.e., is a unit. Hence α = p p = η, which means that η and α lie in the same coset. This gives us injectivity. To prove surjectivity it is enough to show that both sets S1 and S2 have the same number of elements. If p ≡ 1 (mod 4) then there are exactly p − 1 orbits in S1 and if p ≡ 3 (mod 4) then there are exactly p + 1 orbits in S1. We now obtain the surjectivity by the following Lemma.

2 2 Lemma 5.11. If N−1 is the number of solutions to x + y ≡ −1 (mod p) then   p + 1, if p ≡ 3 (mod 4); N−1 = (5.15)  p − 1, if p ≡ 1 (mod 4).

Proof. We ﬁrst claim that we get the same number of solutions to x2 + y2 ≡ c

(mod p) for any c 6≡ 0 (mod p). Notice that for any two c, d 6≡ 0 (mod p) we have that the number of solutions to x2 + y2 ≡ c2 (mod p) is the same as the number of solutions to x2 + y2 ≡ d2 (mod p). This is because each solution (x, y) to the ﬁrst equation gives a distinct solution (dc−1x, dc−1y) to the second equation and conversely each solution (x, y) to the second equation gives a distinct solution (cd−1x, cd−1y) to the ﬁrst equation. Also, if k is not a square mod p then by arguments similar to the one above the number of solutions to x2 + y2 ≡ k (mod p) is the same as the

68 number of solutions to x2 + y2 ≡ kc2 (mod p) for every c 6≡ 0 (mod p). We have that

{k ∈ Z/pZ not a square mod p} = {kcˆ 2 : c 6≡ 0 (mod p)} for any ﬁxed kˆ ∈ Z/pZ, which is not a square mod p.

Now to get the claim we have to show that for any ﬁxed k, which is not a square mod p, the number of solutions to x2 + y2 ≡ k (mod p) is the same as the number of solutions to x2 + y2 ≡ k2 (mod p). If (x, y) is a solution to the ﬁrst equation then

(±x, ±y) and (±y, ±x) are also solutions to the ﬁrst equation. (The second set is the same as the ﬁrst one if and only if x ≡ ±y (mod p).) Correspondingly one can check that (±(x2 − y2), ±2xy) and (±2xy, ±(x2 − y2) are solutions to the second equation.

(We get only four distinct solutions if and only if x ≡ ±y (mod p).) This shows us that the number of solutions to the first equation is less than or equal to the number of solutions to the second equation. Now fix (ˆx, yˆ) a solution to the first equation. Then for every solution (x1, y1) to the second equation (x2 := x1xˆ − y1y,ˆ y2 := x1yˆ + y1xˆ)

2 2 3 −1 is a solution to x + y ≡ k (mod p). Since x1 ≡ k (x2xˆ + y2yˆ) (mod p) and

−1 y1 ≡ k (−x2yˆ+y2xˆ) (mod p) we can conclude that the (x2, y2) above are all distinct.

This implies that the number of solutions to x2 + y2 ≡ k2 (mod p) is less than or equal to the number of solutions to x2 + y2 ≡ k3 (mod p) which is the same as the number of solutions to x2 + y2 ≡ k (mod p). This gives us the claim.

So we can ﬁnd an integer N which is the number of solutions to x2 + y2 ≡ c

(mod p) for any c 6≡ 0 (mod p). Now if p ≡ 3 (mod 4) then (0, 0) is the only solution to x2 + y2 ≡ 0 (mod p), whereas if p ≡ 1 (mod 4) then the number of solutions to x2 + y2 ≡ 0 (mod p) is 2p − 1. (The solutions are (0, 0) and (c, ±ct) for all c 6≡ 0

(mod p) where t satisﬁes t2 ≡ −1 (mod p).)

69 Since the total number of pairs (x, y) is p2 we have   1 + (p − 1)N, if p ≡ 3 (mod 4) p2 =  (2p − 1) + (p − 1)N, if p ≡ 1 (mod 4)   p + 1, if p ≡ 3 (mod 4); ⇒ N = N−1 =  p − 1, if p ≡ 1 (mod 4). as required.

−1 2 −1 2 Now given β = β0+β1i1+β2i2 with β2 6≡ 0 (mod p) we have (β0β2 ) +(β1β2 ) ≡

−1 −1 (mod p). Hence there is a unique orbit α such that β0β2 ≡ xα (mod p) and

−1 β1β2 ≡ yα (mod p). This is the unique α that satisﬁes βα¯ ≡ 0 (mod p) and hence α0βα¯ ≡ 0 (mod p).

Finally to complete the proof of (2)(b)(ii) and (iii) we have to show the following two claims.

1. Claim : If |β|2 ≡ 0 (mod p) and |β|2 6≡ 0 (mod p2) then α0βα¯ 6≡ 0 (mod p2)

for all cosets α.

0 2 2 ˆ ˆ ˆ Assume to the contrary that α βα¯ ≡ 0 (mod p ), i.e. p divides β0, β1, β2. We

already know that there is only one α for which α0βα¯ ≡ 0 (mod p) and that α

satisﬁes β2α1 −β1α2 −β0α3 ≡ 0 (mod p). Hence, from (II) above we are forced

to conclude that |β|2 ≡ 0 (mod p2), which is a contradiction.

2. Claim : If |β|2 ≡ 0 (mod p2) and α is the unique orbit which satisﬁes α0βα¯ ≡ 0

(mod p) then α0βα¯ ≡ 0 (mod p2) but α0βα¯ 6≡ 0 (mod p3).

From the assumption above we have that the expressions on the right hand side

of all the four equations in (II) are divisible by p2. Hence so are the expressions

on the left hand side. We know that at least 2 of β0, β1, β2 are not divisible by

70 p. Without loss of generality, assume that β0, β1 are not divisible by p. Adding

2 ˆ 2 the ﬁrst and third equation in (II) we get that p divides 2β0β0 and hence p ˆ divides β0. Similarly using the ﬁrst and second equation in (II) we can conclude

2 ˆ 4 2 2 0 2 ˆ2 ˆ2 ˆ2 that p divides β1. Since p divides p |β| = |α βα¯| = β0 + β1 + β2 we can

2 ˆ 0 2 conclude that p also divides β2. Hence we have shown that α βα¯ ≡ 0 (mod p ).

Now assume that we have α0βα¯ ≡ 0 (mod p3). Then p4 divides |β|2. Now (II)

implies that all of (β2α1 − β1α2 − β0α3), (β0α0 + β1α1 + β2α2), (−β0α1 + β1α0 −

2 β2α3), (−β0α2+β1α3+β2α0) are divisible by p . Hence from (I) we can conclude

that p divides β0, β1 and β2, which is a contradiction.

Step 3 : We will now prove (2)(b)(iv). Let β be given such that it satisﬁes −|β|2 ≡ x2

(mod p) for some x with x 6≡ 0 (mod p). We want to show that we get exactly 2 orbits of α such that α0βα¯ ≡ 0 (mod p). (We will get one orbit each corresponding to x and

−x.) Using (I) and (II) above and the fact that −|β|2 ≡ x2 (mod p) we conclude that α0βα¯ ≡ 0 (mod p) is equivalent to the matrix equation       β0 β1 β2 −x α0 0                    β1 −β0 −x −β2   α1   0      ≡   (mod p). (5.16)        β x −β β   α   0   2 0 1   2          −x β2 −β1 −β0 α3 0 | {z } Pβ,x

This is also the same as (β − xi1i2)¯α ≡ 0 (mod p). We have that atleast one of

2 2 2 2 2 2 2 2 x + β0 , x + β1 , x + β2 is not divisible by p. (Otherwise |β| + x ≡ 0 (mod p) ⇒

2 2 2 x ≡ 0 (mod p) which is not possible.) Let us ﬁrst assume that x +β2 6≡ 0 (mod p).

71 Then we have

((β0 + β1i1) + (β2 − xi1)i2)¯α ≡ 0 (mod p) (5.17)

⇔ (β2 + xi1)((β0 + β1i1) + (β2 − xi1)i2)¯α ≡ 0 (mod p).

Now (β2 +xi1)((β0 +β1i1)+(β2 −xi1)i2) = r0 +r1i1 +r2i2 where r0 := β2β0 −xβ1, r1 :=

2 2 2 2 2 β2β1 + xβ0 and r2 := β2 + x . One can easily check that r0 + r1 + r2 ≡ 0 (mod p). Hence from Step 2 above we can ﬁnd a unique orbit α satisfying α0βα¯ ≡ 0 (mod p)

2 2 for x and −x. We can proceed similarly in case β2 + x ≡ 0 (mod p) by replacing

2 2 β2 + xi1 in (5.17) by β1 + xi2 (respectively, β0 + xi1i2) in case β1 + x 6≡ 0 (mod p)

2 2 (respectively, β0 + x 6≡ 0 (mod p)). Note that it can be shown that we do not get diﬀerent orbits α for the three cases above. To see this, observe that (β −xi1i2)¯α ≡ 0

(mod p) and (β−xi1i2)¯η ≡ 0 (mod p) implies that αη¯ ≡ 0 (mod p). This implies that

α, η belong to the same orbit by arguments similar to those given before Lemma 5.11.

Finally, it is clear that p2 does not divide α0βα.¯ Otherwise p4 divides |α0βα¯|2 = p2|β|2 which implies that p2 divides |β|2 which is not possible. This completes the proof of

(2)(b)(iv).

Step 4 : We give the proof of (2)(a). Let q be any prime diﬀerent from p. If q = 2, ˆ ˆ ˆ then (I) above gives us β0, β1, β2 ≡ 0 (mod 2) ⇒ β0, β1, β2 ≡ 0 (mod 2) which is not

0 possible since we have assumed that ν2(β) = 0. Now let q 6= 2. If q divides α βα¯ then

2 q also divides |β| and hence (II) above implies that (β2α1 − β1α2 − β0α3), (β0α0 +

β1α1 +β2α2), (−β0α1 +β1α0 −β2α3), (−β0α2 +β1α3 +β2α0) are divisible by q. Finally

(I) above implies that β0, β1, β2 ≡ 0 (mod q) which is not possible since νq(β) = 0 by assumption. This completes the proof of (2)(a).

72 Step 5 : Now we will prove (1). We have

X 3/2 2πiRe(βx) F (x) = A(β)x3 Kir(2π|β|x3)e . 3 β∈Z , β6=0 To evaluate the ﬁrst term on the right-hand side of (5.11), we have to sum over the ˆ set {β} = {β0 + β1i1 + β2i2; 0 ≤ β0, β1, β2 ≤ p − 1} the following expression : ˆ! x + β X −1 ˆ F = A(β)(x p−1)3/2K (2π|β|p−1x )e2πiRe(βp x)e2πiRe(ββ/p). p 3 ir 3 3 β∈Z , β6=0

ˆ Since βˆ 7→ e2πiRe(ββ/p) is a character on (Z/pZ)3 we have   3 X ˆ  p , if p | β, e2πiRe(ββ/p) = (5.18) {βˆ}  0, otherwise.

We ﬁrst sum over βˆ and then (5.18) implies that we can replace β with pβ since the elements in the lattice Z3 that are not divisible by p vanish : ˆ! X x + β X 3/2 F = p3/2A(pβ)x K (2π|β|x )e2πiRe(βx). (5.19) p 3 ir 3 3 {βˆ} β∈Z ,β6=0 Next let us calculate

X 3/2 2πiRe(βpx) F (px) = A(β)(x3p) Kir(2π|β|px3)e . 3 β∈Z , β6=0 Replacing β by β/p we get

X 3/2 3/2 2πiRe(βx) F (px) = p A(β/p)x3 Kir(2π|β|x3)e . (5.20) 3 β∈Z , β6=0

β 3 The indexing set should consist of those β such that p ∈ Z . But since we assume

β 3 that A( p ) = 0 if p does not divide β, we can write the indexing set as β ∈ Z . Now let us consider

α∗ αxα∗ δα∗ F ((αx + δ)α0−1) = F ((αx + δ) ) = F + . p p p

73 αxα∗ Notice that the x3 component remains unchanged when x is replaced by p . We have

βαxα∗ βδα∗ 0−1 X 3/2 2πiRe p 2πiRe p F ((αx + δ)α ) = A(β)x3 Kir(2π|β|x3)e e . 3 β∈Z , β6=0

0 0 2 Replace β by α βα¯ . Using Re(ab) = Re(ba) and |β|2 = α βα¯ we get p p

0 0−1 X α βα¯ 3/2 2πiRe(βx) 2πiRe βαδ¯ F ((αx + δ)α ) = A x K (2π|β|x )e e ( p ). (5.21) p 3 ir 3 3 β∈Z , β6=0

α0βα¯ 0 We assume that A p = 0 if p does not divide α βα.¯ The indexing set in (5.21) can be written as β ∈ Z3 by the same argument as in (5.20). Claim : For every α such that α0βα¯ ≡ 0 (mod p) we have

X 2πiRe βαδ¯ e ( p ) = p. (5.22) δ

Proof of Claim According to Proposition 5.4 there are exactly p terms in the sum above. If p divides β then (5.22) is clear. If p does not divide β we have the following 3 cases :

1. If −|β|2 is not a square mod p then we have shown earlier that for all α we get

α0βα¯ 6≡ 0 (mod p). Hence the claim is true.

2. If |β|2 ≡ 0 (mod p) then we have shown earlier that α0βα¯ ≡ 0 (mod p) is true

2πiRe βαδ¯ if and only if βα¯ ≡ 0 (mod p). Hence e ( p ) = 1 for every δ which gives us

that the sum in (5.22) is equal to p.

3. If −|β|2 ≡ x2 (mod p) for some x 6≡ 0 (mod p) then we have shown earlier

0 that α βα¯ ≡ 0 (mod p) is true if and only if (β ± xi1i2)¯α ≡ 0 (mod p). Hence

74 βα¯ ≡ ±xi1i1α.¯ From the proof of Proposition 5.4 we know that δ is of the form

0 δ = vα where v ∈ V2. From this we get

βαδ¯ i i αvα¯ 0 α0i i αv¯ Re = ±xRe 1 2 = ±xRe 1 2 = 0 p p p

0 because α i1i2α¯ = −pi1i2 and Re(i1i2v) = 0.

This completes the proof of (5.22).

Finally, putting together (5.19), (5.20), (5.21) and (5.22) we get (5.12) as required.

This completes the proof of Proposition 5.10. We are now ready to state and prove the main theorem of this section. Let us remind the reader of the deﬁnition of the Fourier coeﬃcients A(β) of F given in

s u (3.9) : Let β = p 2 d(β0 + β1i1 + β2i2) with s, u ≥ 0, d odd such that p - d and gcd(β0, β1, β2) = 1. Then

s u   X X X −|β|2 A(β) := 23/4|β| c (prn)−1/2 (−1)t2t/2.  (2tprn)2  r=0 t=0 n|d

P 2πinx Theorem 5.12. Let f(z) = c(n)W sign(n) , ir (4π|n|y)e be a Hecke eigenform n6=0 4 2 + th in S 1 (4) with p eigenvalue given by λp for every odd prime p. Then F defined by 2 Fourier coefficients A(β) as above satisfies

3/2 TpF = (p λp + p(p + 1))F (5.23) for all odd prime numbers p.

3/2 Proof. For every β we have to show that (TpF )(β) = (p λp +p(p+1))A(β). From

Proposition 5.10 (2)(a) we know that the power of q 6= p dividing β does not change

α0βα when β is replaced by p . It will be clear from the proof that the computations only involve the prime p. Hence it is enough to show the result for the case νq(β) = 0

75 s s for all q 6= p. Hence let β = p (β0 + β1i1 + β2i2) =: p τ with gcb(β0, β1, β2) = 1. Then we have

s−1 X −p2s|τ|2p2 A(pβ) = 23/4|β|{ pc p−r/2 + (5.24) p2r r=0 +pc(−|τ|2p2)p−s/2 + pc(−|τ|2)p−s/2p−1/2},

s−1 X −p2s|τ|2 A(β/p) = 23/4|β| p−1c p−r/2, (5.25) p2rp2 r=0 and

 0  A(β), if ν α βα¯ = ν (β);  p p p   3/4 2 −s/2 α0βα¯ α0βα¯  A(β) − 2 |β|c(−|τ| )p , if νp = νp(β) − 1 ≥ 0; A = p 2 0 p  3/4 −|τ| −s/2 −1/2 α βα¯  A(β) + 2 |β|c 2 p p , if ν = ν (β) + 1;  p p p p   α0βα¯  0, if νp p < 0. (5.26)

0 Here we have used the fact that α βα¯ = |β|. We will use the following relation p satisﬁed by the coeﬃcients c(n) of f :

n n pc(np2) + p−1/2 c(n) + p−1c = λ c(n). (5.27) p p2 p

−p2s|τ|2 Let us ﬁrst assume that s ≥ 1. For r = 0, 1, ··· , s − 1 taking n = p2r in (5.27) we get −p2s|τ|2 −p2s|τ|2 1 −p2s|τ|2 pc .p2 + p−1c . = λ c . p2r p2r p2 p p2r Hence

s−1 ! X −p2s|τ|2 p3/2A(pβ) + p3/2A(β/p) = 23/4|β| p3/2λ c pr/2 + p p2r r=0   3/4  3/2 2 2 −s/2 3/2 2 −s/2 −1/2 +2 |β| p pc(−|τ| p )p + p pc(−|τ| )p p (5.28). | {z } (I) Next we consider the following four cases depending on |τ|2.

76 1. If −|τ|2 is not a square mod p then from (5.27) we have

2 2 −1/2 2 2 pc(−|τ| p ) − p c(−|τ| ) = λpc(−|τ| ).

Hence

(I) = p3/2 pc(−|τ|2p2) − p−1/2c(−|τ|2) + (p + 1)p−1/2c(−|τ|2) p−s/2

3/2 2 −s/2 2 −s/2 = p λpc(−|τ| )p + p(p + 1)c(−|τ| )p .

Also from Proposition 5.10 (2)(b)(i) and (5.26) we have

X α0βα¯ p A = p(p + 1)A(β) − p(p + 1)23/4|β|c(−|τ|2)p−s/2. p α

2. If |τ|2 ≡ 0 (mod p) but |τ|2 6≡ 0 (mod p2) then from (5.27) we have

2 2 2 pc(−|τ| p ) = λpc(−|τ| ).

Hence

3/2 2 −s/2 2 2 −s/2 (I) = p λpc(−|τ| )p + p c(−|τ| )p .

Also from Proposition 5.10 (2)(b)(ii) and (5.26) we have

X α0βα¯ p A = pA(β) + p2A(β) − p223/4|β|c(−|τ|2)p−s/2. p α

3. If |τ|2 ≡ 0 (mod p2) then from (5.27) we have

2 2 −1 2 2 2 pc(−|τ| p ) + p c(−|τ| /p ) = λpc(−|τ| ).

Hence

(I) = p3/2(pc(−|τ|2p2) + p−1c(−|τ|2/p2) + pc(−|τ|2)p−1/2

−p−1c(−|τ|2/p2))p−s/2

3/2 2 −s/2 2 2 −s/2 2 2 −s/2 −1/2 = p λpc(−|τ| )p + p c(−|τ| )p − pc(−|τ| /p )p p .

77 Also from Proposition 5.10 (2)(b)(iii) and (5.26) we have

X α0βα¯ p A = pA(β) + p23/4|β|c(−|τ|2/p2)p−s/2p−1/2 + p2A(β) p α −p223/4|β|c(−|τ|2)p−s/2.

4. If −|τ|2 ≡ x2 (mod p) for some x 6≡ 0 (mod p) then from (5.27)

2 2 −1/2 2 2 pc(−|τ| p ) + p c(−|τ| ) = λpc(−|τ| ).

Hence

(I) = p3/2 pc(−|τ|2p2) + p−1/2c(−|τ|2) + (p − 1)p−1/2c(−|τ|2) p−s/2

3/2 2 −s/2 2 −s/2 = p λpc(−|τ| )p + p(p − 1)c(−|τ| )p .

Also from Proposition 5.10 (2)(b)(iv) and (5.26) we have

X α0βα¯ p A = 2pA(β) + p(p − 1)A(β) − p(p − 1)23/4|β|c(−|τ|2)p−s/2. p α

We can do similar calculations for the case s = 0 as well. Putting together the above expressions with (5.24), (5.25) and (5.28) we ﬁnally get

X α0βα¯ p3/2A(pβ) + p3/2A(β/p) + p A = p3/2λ + p(p + 1) A(β) p p α as required. This completes the proof of Theorem 5.12.

5.3 Hecke operator Tp2

In this section we deﬁne the Hecke operator Tp2 and get a formula for Tp2 in terms of the Fourier coeﬃcients A(β) in Proposition 5.14. In Theorem 5.15 we show

78 + that if f ∈ S1/2(Γ0(4)) is a Hecke eigenfunction then F deﬁned in Section 3.2 is an

3/2 eigenfunction of Tp2 with eigenvalue (p + 1)p λp + (p − 1)(p + 1), where λp is the pth eigenvalue of f.

We will now give the deﬁnition of the Hecke operator Tp2 .

Deﬁnition 5.13. Let F be a function on H3. For any odd prime p we deﬁne the

Hecke operator Tp2 by

∗ ∗ ! X ∗ X αxα δα X v (T 2 F )(x) := F (αxα ) + F + + F x + . (5.29) p p2 p2 p α δ v

Here the sum over α, δ and v is the same as in Proposition 5.6 and x ∈ H3.

Note that just as in the case of the Hecke operator Tp in the previous section, Tp2 maps an automorphic function F to an automorphic function.

th Let (Tp2 F )(β) be the β Fourier coeﬃcient of Tp2 F. The following proposition describes (Tp2 F )(β) in terms of A(β).

Proposition 5.14. With notations as above, we have

0 ! 3/2 X 0 α βα¯ X 2πiRe( βv ) (T 2 F ) = p A(α βα¯) + A + e p A(β) (5.30) p (β) p2 α v where α runs through the orbits of the action of the group of units in C2(Z) on the set

2 {α ∈ C2(Z): |α| = p} and v runs through the set {v = v0 + v1i1 + v2i2 : v0, v1, v2 ∈

2 α0βα¯ Z, 0 ≤ v0, v1, v2 ≤ p − 1, v 6= 0, |v| ≡ 0 (mod p)}. Here we assume that A p2 = 0 if p2 does not divide α0βα.¯

Proof. We will evaluate (5.29) by substituting the Fourier expansion of F. Notice that the third term in the right hand side of (5.30) follows directly from the third

79 term on the right hand side of (5.29). Fix α such that |α|2 = p.

∗ X 3 2πiRe(βαxα∗) F (αxα ) = A(β)(px3) 2 Kir(2π|β|px3)e . 3 β∈Z , β6=0

∗ Here we have used that the coeﬃcient of i3 in αxα is equal to px3. Now replace β by (α0βα¯)/p2 to get

0 ∗ X 3/2 α βα¯ 3 2πiRe(βx) F (αxα ) = p A x 2 K (2π|β|x )e . (5.31) p2 3 ir 3 3 β∈Z , β6=0 Next, we have

∗ ∗ αxα δα X 3 ∗ 2 ∗ 2 F + = A(β)(x /p) 2 K (2π|β|x /p)e2πiRe(βαxα /p )e2πiRe(βδα /p ). p2 p2 3 ir 3 3 β∈Z , β6=0

∗ 2 Here we have used that the coeﬃcient of i3 in αxα /p is equal to x3/p. Now replace β by α0βα¯ to get

∗ ∗ αxα δα X −3/2 0 3 2πiRe(βx) F + = p A(α βα¯)x 2 K (2π|β|x )e . (5.32) p2 p2 3 ir 3 3 β∈Z , β6=0

We have used that e2πiRe(α0βαδα¯ ∗/p2) = e2πiRe(α∗α0βαδ/p¯ 2) = e2πiRe(βαδ/p¯ ) = 1 because

αδ¯ ≡ 0 (mod p) by the second condition in Lemma 5.7. The indexing set in (5.31) and (5.32) can be taken to be Z3 by arguments similar to those in the proof of Proposition 5.10. Since there are exactly p3 values of δ for each α, (5.31) and (5.32) complete the proof of Proposition 5.14. Now we state the main theorem of this section.

+ Theorem 5.15. Let f ∈ S1/2(Γ0(4)) be a Hecke eigenfunction with eigenvalue λp for every odd prime p. Let F be the function on H3 deﬁned in section 3.2. Then

3/2 Tp2 F = (p + 1)p λp + (p − 1)(p + 1) F. (5.33)

80 u s Proof. Let β = 2 dp τ with u, s ≥ 0, d odd not divisible by p and τ = τ0 + τ1i1 +

τ2i2 with gcd(τ0, τ1, τ2) = 1. From Proposition 5.10 (2)(a) we can assume without loss of generality as in the proof of Theorem 5.12 that u = 0 and d = 1. Hence β = psτ.

3/2 P 0 α0βα¯ We will ﬁrst evaluate Rβ := p A(α βα¯) + A p2 for diﬀerent possibilities α of s and |τ|2.

Case I : Let −|τ|2 not be a square mod p. Then from Proposition 5.10 (2)(b)(i) we

0 have νp(α βα¯) = νp(β) for all α. For s ≥ 2 we have

s−2 X p2s|τ|2 p2s|τ|2 1 R = (p + 1)p3/223/4|β| p−r/2 pc − p2 + p−1c − + β p2r p2r p2 r=0 +(p + 1)p3/223/4|β| pc(−p2|τ|2p2)p−(s−1)/2 + pc(−|τ|2p2)p−s/2 .

Use the relations

2 2 2 −1 2 2 2 pc(−p |τ| p ) + p c(−|τ| ) = λpc(−p |τ| ), (5.34)

2 2 −1/2 2 2 pc(−p |τ| ) − p c(−|τ| ) = λpc(−|τ| ) (5.35) and for r ≤ s − 2

p2s|τ|2 p2s|τ|2 1 p2s|τ|2 pc − p2 + p−1c − = λ c − (5.36) p2r p2r p2 p p2r to get

3/2 Rβ = (p + 1)p λpA(β). (5.37)

For s = 1 we have

3/2 3/4 2 2 2 2 2 −1/2 Rβ = (p + 1)p 2 |β| pc(−p |τ| p ) + pc(−p |τ| )p

3/2 3/4 2 2 −1 2 1/2 2 2 = (p + 1)p 2 |β| λpc(−|τ| p ) − p c(−|τ| ) + p c(−|τ| p )

3/2 3/4 2 2 −1/2 2 = (p + 1)p 2 |β|λp c(−|τ| p ) + p c(−|τ| )

3/2 = (p + 1)p λpA(β). (5.38)

81 Finally for s = 0 we have

3/2 3/4 2 2 Rβ = (p + 1)p 2 |β|pc(−p |τ| )

3/2 = (p + 1)p λpA(β) + (p + 1)pA(β). (5.39)

Case II : Let −|τ|2 be a square mod p. From Proposition 5.10 (2)(b)(iv) we get 2

0 values of α for which νp(α βα¯) = νp(β)+1 and the remaining p−1 values of α satisfy

0 νp(α βα¯) = νp(β).

For s ≥ 2 we have

s−1 X p2s|τ|2 p2s|τ|2 1 R = 2p3/223/4|β| p−r/2 pc − p2 + p−1c − + β p2r p2r p2 r=0 2p3/223/4|β| pc(−p2|τ|2)p−s/2 + pc(−|τ|2)p−(s+1)/2 + s−2 X p2s|τ|2 p2s|τ|2 1 (p − 1)p3/223/4|β| p−r/2 pc − p2 + p−1c − + p2r p2r p2 r=0 +(p − 1)p3/223/4|β| pc(−p2|τ|2p2)p−(s−1)/2 + pc(−|τ|2p2)p−s/2 .

In addition to (5.34) and (5.36), use

2 2 −1/2 2 2 pc(−p |τ| ) + p c(−|τ| ) = λpc(−|τ| ) (5.40) to get

3/2 3/2 3/4 −1/2 2 −s/2 Rβ = 2p λpA(β) + 2p 2 |β|(p − 1)p c(−|τ| )p +

3/2 3/2 3/4 −1/2 2 −s/2 +(p − 1)p λpA(β) − (p − 1)p 2 |β|2p c(−|τ| )p .

3/2 = (p + 1)p λpA(β). (5.41)

82 For s = 1 we have

3/2 3/4 2 2 2 2 2 −1/2 2 −1 −1 2 Rβ = 2p 2 |β| pc(−p |τ| p ) + pc(−|τ| p )p + pc(−|τ| )p + p c(−|τ| )

+(p − 1)p3/223/4|β|(pc(−p2|τ|2p2) + pc(−|τ|2p2)p−1/2)

3/2 3/2 3/4 2 −1 3/2 = 2p λpA(β) + 2p 2 |β|(p − 1)c(−|τ| )p + ((p − 1)p λpA(β) −

(p − 1)p3/223/4|β|2c(−|τ|2)p−1)

3/2 = (p + 1)p λpA(β). (5.42)

Finally, for s = 0 we have

3/2 3/4 2 2 2 −1/2 3/2 3/4 2 2 Rβ = 2p 2 |β| pc(−|τ| p ) + pc(−|τ| )p + (p − 1)p 2 |β|pc(−|τ| p )

3/2 3/4 2 2 3/4 2 = (p + 1)p λpA(β) − (p + 1)p2 |β|c(−|τ| ) + 2p 2 |β|c(−|τ| )

3/2 2 = (p + 1)p λpA(β) + (p − p)A(β). (5.43)

Case III : Let |τ|2 ≡ 0 (mod p) but |τ|2 6≡ 0 (mod p2). From Proposition 5.10

0 (2)(b)(ii) we know that there is exactly one value of α such that νp(α βα¯) = νp(β)+1

0 and the remaining p values of α satisfy νp(α βα¯) = νp(β).

For s ≥ 2 we have

s−1 X p2s|τ|2 p2s|τ|2 1 R = p3/223/4|β| p−r/2 pc − p2 + p−1c − + β p2r p2r p2 r=0 p3/223/4|β| pc(−p2|τ|2)p−s/2 + pc(−|τ|2)p−(s+1)/2 + s−2 X p2s|τ|2 p2s|τ|2 1 pp3/223/4|β| p−r/2 pc − p2 + p−1c − + p2r p2r p2 r=0 +pp3/223/4|β| pc(−p2|τ|2p2)p−(s−1)/2 + pc(−|τ|2p2)p−s/2 .

83 In addition to (5.34) and (5.36), use

2 2 2 pc(−|τ| p ) = λpc(|τ| ) (5.44) to get

3/2 3/2 3/4 2 −(s+1)/2 3/2 Rβ = p λpA(β) + p 2 |β|pc(−|τ| )p + pp λpA(β)

−pp3/223/4|β|p−1c(−|τ|2)p−(s−1)/2

3/2 = (p + 1)p λpA(β). (5.45)

For s = 1 we have

3/2 3/4 2 2 2 2 2 −1/2 2 −1 −1 2 Rβ = p 2 |β| pc(−p |τ| p ) + pc(−|τ| p )p + pc(−|τ| )p + p c(−|τ| )

+pp3/223/4|β|(pc(−p2|τ|2p2) + pc(−|τ|2p2)p−1/2)

3/2 3/2 3/4 2 3/2 3/2 3/4 −1 2 = p λpA(β) + p 2 |β|c(−|τ| ) + pp λpA(β) − pp 2 |β|p c(−|τ| )

3/2 = (p + 1)p λpA(β). (5.46)

Finally, for s = 0 we have

3/2 3/4 2 2 2 −1/2 3/2 3/4 2 2 Rβ = p 2 |β| pc(−|τ| p ) + pc(−|τ| )p + pp 2 |β|pc(−|τ| p )

3/2 3/2 3/4 2 −1/2 3/2 = p λpA(β) + p 2 |β|pc(−|τ| )p + pp λpA(β)

3/2 2 = (p + 1)p λpA(β) + p A(β). (5.47)

Case IV : Let |τ|2 ≡ 0 (mod p2). From Proposition 5.10 (2)(b)(iii) we know that

0 there is one value of α with νp(α βα¯) = νp(β) + 2 and the remaining p values of α

0 satisfy νp(α βα¯) = νp(β).

84 For s ≥ 2 we have

s X p2s|τ|2 p2s|τ|2 1 R = p3/223/4|β| p−r/2 pc − p2 + p−1c − + β p2r p2r p2 r=0 p3/223/4|β| pc(−|τ|2)p−(s+1)/2 + pc(−|τ|2/p2)p−(s+2)/2 + s−2 X p2s|τ|2 p2s|τ|2 1 pp3/223/4|β| p−r/2 pc − p2 + p−1c − + p2r p2r p2 r=0 +pp3/223/4|β| pc(−p2|τ|2p2)p−(s−1)/2 + pc(−|τ|2p2)p−s/2 .

In addition to (5.34) and (5.36), use

2 2 −1 2 2 2 pc(−|τ| p ) + p c(−|τ| /p ) = λpc(−|τ| ) (5.48) to get

3/2 3/2 3/4 2 −(s+1)/2 2 2 −(s+2)/2 Rβ = p λpA(β) + p 2 |β| pc(−|τ| )p + pc(−|τ| /p )p

3/2 3/2 3/4 −1 2 −(s−1)/2 −1 2 2 −s/2 +pp λpA(β) − pp 2 |β| p c(−|τ| )p + p c(−|τ| /p )p

3/2 = (p + 1)p λpA(β). (5.49)

For s = 1 we have

3/2 3/4 2 2 2 2 2 −1/2 2 −1 Rβ = p 2 |β|(pc(−p |τ| p ) + pc(−|τ| p )p + pc(−|τ| )p

+pc(−|τ|2/p2)p−3/2 + p−1c(−|τ|2) + p−1c(−|τ|2/p2)p−1/2)

+pp3/223/4|β|(pc(−p2|τ|2p2) + pc(−|τ|2p2)p−1/2)

3/2 = (p + 1)p λpA(β). (5.50)

Finally, for s = 0 we have

3/2 3/4 2 2 2 −1/2 2 2 −1 Rβ = p 2 |β|(pc(−|τ| p ) + pc(−|τ| )p + pc(−|τ| /p )p +

p−1c(−|τ|2/p2)) + pp3/223/4|β|pc(−|τ|2p2)

3/2 2 = (p + 1)p λpA(β) + p A(β). (5.51)

Now the theorem follows from the following lemma.

Here the sum on the left is over the set {v = v0 + v1i1 + v2i2 : v0, v1, v2 ∈ Z, 0 ≤

2 v0, v1, v2 ≤ p − 1, v 6= 0, |v| ≡ 0 (mod p)}.

Proof. The ﬁrst case is obvious since the number of elements in the set K :=

2 {v = v0 + v1i1 + v2i2 : v0, v1, v2 ∈ Z, 0 ≤ v0, v1, v2 ≤ p − 1, v 6= 0, |v| ≡ 0 (mod p)}

2 is p − 1. Denote by U = Z/pZ + Z/pZi1 + Z/pZi2 the 3−dimensional space over

Z/pZ. Since v ∈ K =⇒ tv ∈ K for t = 1, ··· , p − 1 we can see that the set K is the union of p + 1 lines without the origin. Let us ﬁx p + 1 vectors v1, ··· vp+1 such that

K = {tvj : t = 1, ··· , p − 1 and j = 1, ··· , p + 1}. Deﬁne a map Iβ from U to Z/pZ by Iβ(u) := Re(βu) (mod p) for u ∈ U. Since β 6= 0 the dimension of Ker(Iβ) is 2.

2 Case 1 : Let |β| ≡ 0 (mod p) but p - β. We claim that if u ∈ Ker(Iβ) ∩ K then u = tβ¯ (mod p) for some t = 1, ··· , p − 1. Suppose not, let u ∈ K, u 6= tβ¯ for any t

86 such that Re(βu) ≡ 0 (mod p). Hence |aβ + bu¯|2 = a2|β|2 + b2|u|2 + 2abRe(βu) ≡ 0

(mod p) for all a, b = 0, 1, ··· , p − 1. So we can conclude that K is the same as a

2−dimensional vector space U0 := hβ, ui without its origin. We will show that this ¯ forces β ≡ 0 (mod p). Let β = β0 + β1i1 + β2i2. Since β (mod p) and β (mod p)

¯ ¯ 2 2 are elements of K, we have β + β (mod p) ∈ U0. Hence we get |β + β| = 4β0 ≡ 0

(mod p). So β0 ≡ 0 (mod p). Now consider u := β2 +β3i1 (mod p) ∈ U0. By the same

2 argument as above we can conclude that β2 ≡ 0 (mod p). Since |β| ≡ 0 (mod p) we get that β3 ≡ 0 (mod p) and so β ≡ 0 (mod p), a contradiction.

From this we get

p−1 p−1 ¯ βv X 2πiRe( βv ) X 2πitRe( ββ ) X X 2πitRe( j ) e p = e p + e p

v t=1 vj 6=β¯ t=1 = (p − 1) + p(−1) = −1 (5.53) as required.

Case 2 : Let −|β|2 ≡ x2 (mod p) for some x 6≡ 0 (mod p). We claim that we can ﬁnd exactly two linearly independent elements vx and v−x in K which are in

2 2 2 2 2 2 Ker(Iβ). Since x is nonzero we know that atleast one of β0 + x , β1 + x and β2 + x

2 2 is not divisible by p. Let us ﬁrst assume that β2 + x 6≡ 0 (mod p). Deﬁne vx :=

2 2 (β2 − xi1)(β0 + β1i1 + (β2 + xi1)i2) = (β2β0 + xβ1) + (β2β1 − xβ0)i1 + (β2 + x )i2 and

2 2 v−x := (β2β0 − xβ1) + (β2β1 + xβ0)i1 + (β2 + x )i2. Now both vx and v−x are nonzero,

2 Re(vx, v−x) 6= 0 and |avx + bv−x| = 0 only if ab = 0. If vx = tv−x then t = 1 which gives us 2xβ0 ≡ 2xβ1 ≡ 0 (mod p). This is not possible since we have assumed that x and β are not zero. Hence vx and v−x are linearly independent. Using the formulae of vx and v−x above we can show that Re(βvx) ≡ Re(βv−x) ≡ 0 (mod p) and so vx and v−x form a basis for Ker(Iβ). Hence Ker(Iβ) ∩ K = {avx, bv−x : 1 ≤ a, b ≤ p − 1}.

87 2 2 If β2 + x ≡ 0 (mod p) then repeat the above argument with a suitably modiﬁed deﬁnition of vx. This gives us the claim.

Now we have

p−1 p−1 p−1 βv βv X 2πiRe( βv ) X 2πitRe( βvx ) X 2πitRe( −x ) X X 2πitRe( j ) e p = e p + e p + e p

v t=1 t=1 vj 6=vx,v−x t=1 = 2(p − 1) + (p − 1)(−1) = p − 1 (5.54) as required.

2 T Case 3 : Let us assume that −|β| is not a square mod p. We claim that Ker(Iβ) K is empty. We notice that there is a bijection between the set of all lines in U and the set of all two dimensional subspaces in U where a line generated by a nonzero vector u is related to Ker(Iu). An easy calculation shows us that the number of elements in the set {Uˆ a two dimesional subspace of U : Uˆ T K 6= ∅} is (p + 1)(p + 2)/2. We already know that the number of lines generated by vectors v such that |v|2 ≡ 0 (mod p) is p + 1. One can show that the number of lines generated by vectors v such that −|v|2 is a square mod p is equal to (p + 1)p/2. Since (p + 1) + (p + 1)p/2 = (p + 1)(p + 2)/2 we can conclude from Cases 1 and 2 above that {Uˆ a two dimesional subspace of U :

ˆ T 2 U K 6= ∅} = {Ker(Iv): v 6= 0, −|v| is a square (including zero) mod p}. Hence we get the claim.

Now we have

p+1 p−1 βv X 2πiRe( βv ) X X 2πitRe( j ) e p = e p v j=1 t=1 = (p + 1)(−1) = −(p + 1). (5.55)

This completes the proof of the Lemma 5.16 and hence the proof of Theorem 5.15.

88 CHAPTER 6

Automorphic representation corresponding to F

In this chapter we will give the classical to adelic calculation. Starting from the automorphic cuspidal function F deﬁned in (3.9), which is an eigenfunction for the

Hecke operators Tp and Tp2 for every odd prime p, we deﬁne a cuspidal automorphic form on the adelic group. This form gives an irreducible cuspidal automorphic representation and we will explicitly calculate it’s p−adic component for p 6= 2.

In Chapters 2−5 we have considered the function F : H3 −→ C as an automorphic function for the group SV2(R) ' Spin(1, 4)(R). Since H3 is also a symmetric space

+ + for GSV2 (R) ' GSpin (1, 4)(R), the similitude group, we can consider F to be an automorphic function for the group GSpin+(1, 4)(R) with trivial central character.

Let us denote by G := GSpin(1, 4) the similitude group. Let A be the ring of adeles for the global ﬁeld Q. We have the following decomposition for G(A) from the

× × Q × strong approximation for Spin(1, 4) and the decomposition GL1(A) ' Q R+ Zp : p<∞

+ Y G(A) ' G(Q)G (R)K0 where K0 := G(Zp). (6.1) p>∞

We refer the reader to Theorem 104 : 4 of [27] for details on the strong approximation for Spin(1, 4). Given a cuspidal automorphic Hecke eigenform F : H3 → C with

89 + respect to SV2(Z), write g = gQg∞k0 where g ∈ G(A), gQ ∈ G(Q), g∞ ∈ G (R) and k0 ∈ K0 and deﬁne ΦF : G(A) → C as ΦF (g) := F (g∞(i3)).

ΦF satisﬁes the following properties :

1. ΦF (ρg) = ΦF (g) for ρ ∈ G(Q),

2. ΦF (gk0) = ΦF (g) for k0 ∈ K0,

3. ΦF (gz) = ΦF (g) for z ∈ Z(A) ' GL1(A).

Since F is a cusp form we can show that ΦF is cuspidal as in Lemma 5 of [1]. Now we consider the representation of G(A) obtained from ΦF by right translations. Denote by (πF ,VF ) an irreducible component such that the projection of ΦF onto VF is non-zero. (We will drop the subscript F whenever there is no confusion).

0 Write π ' ⊗pπp where πp is a representation of Gp := G(Qp). Note that for p, an odd prime, πp is an irreducible unramiﬁed representation of Gp since G(Zp) is the maximal compact subgroup of Gp. From [5] we know that there exists an unramiﬁed character χ of the Borel subgroup of Gp, unique up to the Weyl group orbit, such that πp is isomorphic to the unique spherical constituent πχ of the normalized induced

Gp representation IndB (χ). Again from [5] we have a one to one correspondence between unramiﬁed representations of Gp and algebra homomorphisms of the p−adic Hecke algebra H(Gp,Kp) to C. Here Kp = G(Zp). We will use the calculation involving the classical Hecke algebra in Chapter 5 to obtain the Hecke algebra homomorphism corresponding to πp. Using this homomorphism we will get an explicit formula for the unramiﬁed character χ.

90 6.1 Unramiﬁed Calculation

For the rest of the chapter p is an odd prime. We have shown in Chapter

5 that the classical Hecke algebra Hp is generated by double cosets ΓMΓ where

M ∈ GSpin(1, 4)+(Z[p−1]) =: G+(Z[p−1]). Here Z[p−1] is the set of rational numbers with only powers of p in the denominator. In Theorem 5.8 we gave the four algebraically independent generators of this algebra. H(Gp,Kp) is the convolution algebra of Kp−biinvariant compactly supported functions on Gp. This is generated by the characteristic functions of double cosets KpgKp with g ∈ Gp.

Proposition 6.1. With notations as above

Hp ' H(Gp,Kp). (6.2)

Proof. We have the following bijections from the natural inclusions

+ −1 −1 Γ\G (Z[p ])/Γ ' G(Z)\G(Z[p ])/G(Z) ' Kp\Gp/Kp.   1 0   The ﬁrst bijection is obtained from the fact that G(Z) = Γ ∪   Γ and 0 −1   1 0 −1 + −1   + −1 G(Z[p ]) = G (Z[p ]) ∪   × G (Z[p ]). To get the second one, note 0 −1 −1 that we get injectivity from the fact that G(Z[p ]) ∩ Kp = G(Z). We get surjectivity

−1 from the isomorphism Gp ' G(Q)Kp ' G(Z[p ])Kp. This tells us that the two Hecke algebras are isomorphic as vector spaces. We get similar bijections for single cosets.

This is used to check that the classical double coset multiplication coincides with the convolution product on the p−adic Hecke algebra.

91 Hence H(Gp,Kp) is generated by the functions     1 αˆ 1 1 φ1 := char(Kp   Kp), φ2 := char(Kp   Kp), V ol(Kp)   V ol(Kp)   p pαˆ0     p p−1 1 1 φ3 := char(Kp   Kp), φ4 := char(Kp   Kp). V ol(Kp)   V ol(Kp)   p p−1

Since the representation πp is unramiﬁed we have a unique (up to a scalar) Kp−ﬁxed

0 unramiﬁed vector Fp ∈ Vπp . Now Proposition 6.1, Theorem 5.12 and Theorem 5.15 give us the following proposition.

0 3/2 0 Proposition 6.2. With notations as above we have φ1Fp = (p λp + p(p + 1))Fp ,

0 3/2 2 0 0 0 0 0 φ2Fp = ((p + 1)p λp + (p − 1))Fp , φ3Fp = Fp and φ4Fp = Fp .

Since the φi generate the Hecke algebra freely Proposition 6.2 gives us an algebra homomorphism of H(Gp,Kp). Our next step is to find the unramified character which corresponds to this homomorphism. For this it is convenient to work in the setting   AB   of the symplectic group. Define GSp4(Qp) := {M =   ∈ Mat4(Qp): CD t t × t t t t A D − B C = µ(M)I2, µ(M) ∈ Qp ,B D = D B,A C = C A}.

Proposition 6.3. Let p be an odd prime. Then with notations as above we have

Gp ' GSp4(Qp) and Kp ' GSp4(Zp).

2 2 Proof. Fix r, s ∈ Zp such that r + s = −1. (We can always choose r, s as above √ √ since p does not divide the discriminant of Zp[ −1] and hence every unit in Zp[ −1] is a norm.) For α = α0+α1i1+α2i2+α3i1i2 ∈ C2(Qp) deﬁne ψp : C2(Qp) → Mat2(Qp) by the formula

92   α0 − α1r − α2s −α3 − α1s + α2r   ψp(α) :=   . α3 − α1s + α2r α0 + α1r + α2s

ψp satisﬁes the following properties :

1. ψp is an isomorphism of Qp−algebras

2 2. det(ψp(α)) = |α|

0 t −1 ∗ t −1 3. ψp(α ) = p ψp(α) , ψp(α ) = ψp(α) and ψp(¯α) = p ψp(α) .

Extend this map to ψp : Gp → GSp4(Qp)     α β ψp(α) ψp(β)     M =   7→   . γ δ ψp(γ) ψp(δ)

∗ ∗ This is well-deﬁned since according to the deﬁnition of Gp we have αδ − βγ =

µ(M), αγ∗ and βδ∗ are vectors which gives us the precise conditions of the deﬁnition of GSp4(Qp) above. Hence we get the isomorphism Gp ' GSp4(Qp). One can also check that ψp maps Kp onto GSp4(Zp).

From now on, we will use the notation Gp for GSp4(Qp) and Kp for GSp4(Zp).

Following Asgari-Schmidt [1], Gp = BKp where B is the Borel subgroup. Let N :=   tD−1 X   {  : D upper triangular with 1 on the diagonal, X symmetric } be the 0 D   a1      a2    × unipotent radical and A := {a =   : a0, a1, a2 ∈ Qp } be    a−1a   1 0    −1 a2 a0 −3 2 4 the torus so that B = NA. Then δB(a) = |a0 a1a2|p is the modular function coming

93 × × from the Haar measure. Given unramiﬁed characters χ0, χ1, χ2 on Qp (trivial on Zp ) deﬁne the character χ on A by χ(a) := χ0(a0)χ1(a1)χ2(a2). Extend χ to a character

Gp of B = NA by setting it to be trivial on N. Deﬁne I(χ) := IndB (χ) = {f : Gp → C :

1/2 locally constant function such that f(nag) = δB (a)χ(a)f(g) for n ∈ N, a ∈ A, g ∈

Gp}.Gp acts on this space by right translation and we get the normalized induced representation I(χ).

We will now find the unramified character χ such that πp is isomorphic to the unique spherical constituent πχ of I(χ). The strategy is to apply the generators of the Hecke algebra H(Gp,Kp) to the unramified vector by convolution and evaluate at identity. The values will be polynomials in χ0(p), χ1(p), χ2(p) and then from

Proposition 6.2 we get a relation between the character values and the eigenvalues of our lift F. We solve these equations to get the character χ.

Note that χ is unique upto the action of the Weyl group W. The Weyl group of

Gp is of order 8 and is generated by the matrices       0 1 0 0 0 0 1 0 1 0 0 0                    1 0 0 0   0 1 0 0   0 0 0 1  w1 :=   , w2 :=   , w3 :=   .        0 0 0 1   −1 0 0 0   0 0 1 0              0 0 1 0 0 0 0 1 0 −1 0 0

The Weyl group acts on the character χ by the formula χw(a) := χ(w−1aw). If

χ(a) = χ0(a0)χ1(a1)χ2(a2) then we have

w1 χ (a) = χ0(a0)χ2(a1)χ1(a2),

w2 −1 χ (a) = (χ0χ1)(a0)χ1 (a1)χ2(a2)

w3 −1 χ (a) = (χ0χ2)(a0)χ1(a1)χ2 (a2).

94 0 0 Let Fp be the unramiﬁed vector in the space of πχ satisfying Fp (1) = 1. Then

0 1/2 0 Fp (nak) = δB (a)χ(a) where n ∈ N, a ∈ A, k ∈ Kp. Any φ ∈ H(Gp,Kp) acts on FP by convolution according to the formula

Z 0 0 (φ ∗ Fp )(h) := φ(hg)Fp (g)dg for h ∈ Gp.

0 We will use Proposition 6.2 and the action of H(Gp,Kp) on Fp above to get equations satisﬁed by χ. For this we need the following right coset decomposition :

Proposition 6.4.       1 p b1 b2 1                    1  [  p b2 b3  [  1  Kp   Kp =   Kp   Kp        p  bi (mod p)  1   p    i=1,2,3           p 1 p     1 ∗ ∗ p ∗ ∗         p−1     [  −b p ∗ ∗  [  1 ∗ ∗    Kp   Kp. (6.3)     b=0  p b   1          1 p

In the third and fourth term there are exactly p choices for the upper right hand corner.

95     1 p           ψ (ˆα) p−1  −b p   1  p S   S   Kp   Kp =   Kp   Kp       pψ (ˆα0) b=0  p2 bp   p  p         p p2       p2 ∗ ∗ p ∗ ∗ p ∗ ∗              p ∗ ∗  p−1  −bp p2 ∗ ∗   p ∗ ∗  S   S   S     Kp   Kp   Kp.        1  b=0  p b   p              p 1 p

In the third and fourth term there are exactly p3 choices for the upper right hand corner and in the ﬁfth term there are exactly p2 − 1 choices.

Proof. The proposition follows from Proposition 5.3, Proposition 5.6 and Propo- sition 6.3. S Now let φ = char(KpMKp) ∈ H(Gp,Kp) with KpMKp = MiKp. Let Mi = niai. i Then

Z Z Z 0 0 0 X 0 (φ ∗ FP )(e) = φ(g)Fp (g)dg = FP (g)dg = Fp (g)dg i Gp KpMKp MiKp X 0 X 1/2 = V ol(Kp) Fp (niai) = V ol(Kp) δB (ai)χ(ai). (6.4) i i   1      1  1   Apply (6.4) with φ = φ1 = char(Kp   Kp). Then by the single V ol(Kp)    p      p

96 coset decomposition obtained in (6.3) and Proposition 6.2 we get

3/2 3/2 p λp + p(p + 1) = p χ0(p)[χ1(p)χ2(p) + 1 + χ2(p) + χ1(p)] . (6.5)   ψp(α) 1 Now take φ = φ2 = char(Kp   Kp) in (6.4). Then by the V ol(Kp)   0 pψp(α ) single coset decomposition obtained in (6.4) and Proposition 6.2 we get

3/2 2 2 2 2 2 (p + 1)p λp + (p − 1) = p χ0(p) χ2(p) + χ1(p) + χ1(p)χ2(p) + χ1(p)χ2(p)

2 2 +(p − 1)χ0(p)χ1(p)χ2(p). (6.6)   p      p  1   Finally take φ = φ3 = char(Kp   Kp) in (6.4) to get V ol(Kp)    p      p

2 1 = χ0(p)χ1(p)χ2(p). (6.7)

This tells us that πχ has trivial central character. (6.7) implies

−1/2 χ0(p) = (χ1(p)χ2(p)) . Using this we get

3/2 3/2 1/2 1 p λp + p(p + 1) = p ((χ1(p)χ2(p)) + 1/2 (χ1(p)χ2(p)) χ (p)1/2 χ (p)1/2 + 1 + 2 ) (6.8) χ2(p) χ1(p) 3/2 2 1 1 (p + 1)p λp = p + + χ1(p) + χ2(p) . (6.9) χ1(p) χ2(p)

From the earlier remark on the Weyl group of Gp it is clear that the above equations are not changed if we replace the character χ with χw for any w ∈ W, the Weyl group. Hence the solutions to the above equations will be in the same Weyl group orbit as expected.

97 Theorem 6.5. Up to the action of the Weyl group the character χ is given by

p 2 p 2 λp + λ − 4 λp − λ − 4 χ (p) = p1/2 p , χ (p) = p1/2 p , χ (p) = p−1/2. (6.10) 1 2 2 2 0 1/2 1/2 −1/2 χ1(p) Proof. Denote a = (χ1(p)χ2(p)) + (χ1(p)χ2(p)) and b = + χ2(p) 1/2 χ2(p) −1 −1 . Then we have χ1(p) + χ2(p) + χ1(p) + χ2(p) = ab. Hence from (6.8) χ1(p) and (6.9) we get

3/2 3/2 p λp + p(p + 1) = p (a + b)

3/2 2 (p + 1)p λp = p ab.

Hence we get the equation

p2ab p2a − p3/2(a + b) + p(p + 1) = − p3/2 b − p−1/2(p + 1) = 0. p + 1 p + 1

−1/2 1/2 −1/2 1/2 ±1/2 If a = p (p+1) = p +p then b = λp and this implies (χ1(p)χ2(p)) = p √ 1/2 λ ± λ2−4 and χ1(p) = p p . This gives us four solutions χ2(p) 2

p 2 λp ± λ − 4 χ (p) = p±1/2 p 1 2 p 2 λp ∓ λ − 4 χ (p) = p±1/2 p . 2 2

−1/2 1/2 −1/2 1/2 If b = p (p + 1) = p + p then a = λp and this implies (χ1(p)χ2(p)) = √ λ ± λ2−4 1/2 p p and χ1(p) = p±1/2. This gives us four more solutions 2 χ2(p)

p 2 λp ± λ − 4 χ (p) = p±1/2 p 1 2 p 2 λp ± λ − 4 χ (p) = p∓1/2 p . 2 2

From the comments before Proposition 6.4 regarding the Weyl group action on characters, one can check that the 8 choices for the character χ obtained above are in the same Weyl group orbit. This completes the proof of Theorem 6.5.

98 −1 Since χ1χ2 = | | it follows from [30] Lemma (3.2) that the induced representation I(χ) obtained above is not irreducible. From the classiﬁcation of automorphic representations of GSp(4) given in [31] we can conclude that the representation πχ is a representation of type IIb.

6.2 CAP representation

Deﬁnition 6.6. Let G1 and G2 be two groups such that G1,ν ' G2,ν for almost all places ν and P2 be a parabolic subgroup of G2. Then we will call an irreducible cuspidal automorphic representation π of G1 a CAP representation associated to

P2 if there is an irreducible cuspidal automorphic representation σ of M2, the Levi

0 0 component of P2, such that πν ' πν for almost all places ν, where π is an irreducible component of IndG2 (σ). P2

To deﬁne the normalized induction, extend (σ, Vσ) to a representation of P2 =

M2N2 by setting it to be trivial on the unipotent radical N2 and let δP2 be the modular function obtained from the Haar measure. Then

IndG2 (σ) := {f : G → V |f smooth, f(pg) = δ1/2(p)σ(p)f(g) for p ∈ P , g ∈ G }. P2 2 σ P2 2 2 (6.11)

Let G1 := GSpin(1, 4) and G2 := GSp4. Then from Proposition 6.3 we have G1,p '

G2,p for every odd prime p. Let   g B   P := {  : g ∈ GL2,B symmetric matrix , µ the similitude } µtg−1 be the Siegel parabolic subgroup of G2. We will now construct an irreducible cuspidal

+ automorphic representation σ of the Levi subgroup of P. Consider f ∈ S1/2(Γ0(4))

99 which is a Hecke eigenform with eigenvalue λp for every odd prime p. Let h be the weight 0 Maaß form with respect to SL2(Z) associated to f by the Shimura correspondence given in [18]. If we deﬁne the Hecke operator T (p) on weight 0 Maaß forms by p−1 X z + j (T (p)h)(z) := h + h(pz) p j=0

1/2 then from [18], Pg 199 and 223 we know that T (p)h = p λp. Let σ be the irreducible cuspidal automorphic representation of GL2(A) obtained from h using the decompo-

+ Q 0 sition of GL2 given by : GL2(A) ' GL2(Q)GL2 (R) p GL2(Zp). Write σ ' ⊗pσp ˆ ˆ ˆ where σp is given by σp ' I(η1, η2) := {f : GL2(Qp) → C | f smooth, f(bg) =   a1 0 δ (b)1/2η(b)fˆ(g)∀ b ∈ B , g ∈ GL }. Here B := {b =  } is the stan- B2 2 2 2   a3 a2   a1 0 dard lower-triangular Borel subgroup of GL , δ   = |a−1a | is the 2 B2   1 2 p 0 a2   a1 0   modular function obtained from the Haar measure on GL2 and η   := a3 a2 ∗ η1(a1)η2(a2) where η1 and η2 are unramiﬁed unitary characters of Qp. (We consider the lower triangular Borel subgroup here instead of the upper triangular since we have used the notations of [1] for our calculations regarding the symplectic group.)

Following calculations similar to (6.4), (6.5), (6.6) we get η1(p) + η2(p) = λp and

η1(p)η2(p) = 1.

Theorem 6.7. Let πF be the representation of G1 from the previous section. Then

G2 −1/2 πF is CAP to an irreducible component of IndP (η0 × σ × |det| ) where σ is as

1/2 above and η0(µ) := |µ| is an unramiﬁed character that acts on the similitude.

100 G2,p G2,p Proof. We claim that for every odd prime p we have IndB (χ) ' IndP (η0 ×

−1/2 0 σp × |det| ) with χ as in Theorem 6.5. We can then take π to be the irreducible

G2 −1/2 automorphic constituent of IndP (η0 ×σ×|det| ) such that the p−adic component

0 G2,p −1/2 of π is the spherical constituent of IndP (η0 ×σp ×|det| ). It follows from Lemma

0 0 1 of [24] that we can always ﬁnd a π with the above property. Then we have πF,p ' πp for every odd prime p and hence we get the result of the theorem.

We get the claim above essentially by transitivity of induction. Deﬁne the map

G2,p −1/2 G2,p L : IndP (η0 × σp × |det| ) −→ IndB (χ) by (Lf)(g) := (f(g))(I2). Here f :

G2,p −1/2 G2,p → I(η1, η2) is a function in IndP (η0 × σp × |det| ) and I2 is the identity matrix in GL2. We have to show that this map is well-deﬁned and an isomorphism.

To show that L is well-deﬁned we have to prove that for any b ∈ B, Lf satisﬁes   s ∗ (Lf)(bg) = δ1/2(b)χ(b)(Lf)(g). Write b =   where s ∈ GL ( ) is lower B   2 Qp 0 µts−1 triangular with s = an, a = diag(a1, a2) and n in the unipotent radical of GL2. Since b ∈ P and s ∈ B2 we have

1/2 −1/2 (Lf)(bg) = δP (b)η0(µ)|det(s)| (σp(s)f(g))(I2)

1/2 −1/2 = δP (b)η0(µ)|det(s)| f(g)(s)

= δ1/2(b)η (µ)|det(s)|−1/2δ1/2(s)η(s)f(g)(I ). P 0 B2 2

Now we have

δ1/2(b)η (µ)|det(s)|−1/2δ1/2(s)η(s) = |µ−3a3a3|1/2|µ|1/2|a−1a |1/2|a a |−1/2η (a )η (a ). P 0 B2 1 2 1 2 1 2 1 1 2 2

1/2 This implies that (Lf)(bg) = δB (b)χ0(µ)χ1(a1)χ2(a2)(Lf)(g) since by Theorem 6.5

1/2 1/2 1/2 we have p η1(p) = χ1(p), p η2(p) = χ2(p) and χ0(µ) = |µ| .

101 G2,p It is easy to see that L is injective. If f1 and f2 are two functions in IndP (η0 ×

−1/2 σ × |det| ) then, by deﬁnition, Lf1 = Lf2 implies that f1(g)(I2) = f2(g)(I2) for all g. Now applying the right translation σ(s) for s ∈ GL2 we see that f1(g)(s) = f2(g)(s) and hence f1 = f2.

˜ G2,p To show that L is an isomorphism we define the inverse map L : IndB (χ) −→   s 0 IndG2,p (η × σ × |det|−1/2) by (Lf˜ )(g)(s) := |det(s)|−3/2f(  g). One can P 0 p   0 ts−1 show that L˜ is well-defined using a similar calculation as above. It is clear from the ˜ definition that L ◦ L is the identity map. We note that from [32] Lemma 2.2 and Theorem 6.7 above we can conclude that the representation πF,p is precisely the local Saito-Kurokawa lift of σ.

We want to point out that Definition 6.6 is not the same as the definition of CAP representation found in the literature in the sense that we allow two different groups

G1 and G2 satisfying G1,ν ' G2,ν for almost all places ν instead of considering just one group. To the best of our knowledge this is the ﬁrst example where such CAP representations are constructed.

One can explain why we get CAP representation involving two diﬀerent groups if we consider Langlands functoriality. The two groups GSpin(1, 4) and GSp4 are inner forms of each other. Hence they have the same L−groups. Langlands functoriality tells us that corresponding to the identity L−homomorphism we should get a lifting of automorphic representations from the inner form GSpin(1, 4) to the split group

GSp4. Locally, when GSpin(1, 4)ν ' GSp4,ν the lifting is given by an isomorphism which is the content of Theorem 6.7. Hence one can say that Theorem 6.7 is a special case of the Langlands functoriality expected in this situation.

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105