introduction - 京都大学ikeda/maass.pdf · 2005-05-05 · 5 for each q ⊂ qd, set χq = y q∈q...

ON MAASS LIFTS AND THE CENTRAL CRITICALVALUES OF TRIPLE PRODUCT L-FUNCTIONS

ATSUSHI ICHINO AND TAMOTSU IKEDA

To the memory of Prof. Tsuneo Arakawa

Abstract. We express certain period integrals of Maass lifts whichappear in the Gross-Prasad conjecture in terms of the central crit-ical values of triple product L-functions in the imbalanced case.

Introduction

After Garrett [10] discovered an integral representation of tripleproduct L-functions, Harris and Kudla [16] determined the transcen-dental part of the central critical values of triple product L-functions.More precisely, for i = 1, 2, 3, let fi ∈ Sκi

(Γ0(Ni), χi) be a primitiveform. We may assume that κ1 ≤ κ2 ≤ κ3. We also assume that χ1χ2χ3

is trivial, so that w = κ1 + κ2 + κ3 − 3 is odd. Recall that the tripleproduct L-function L(s, f1 × f2 × f3) satisfies a functional equationwhich replaces s with w + 1− s. Set

p(f1, f2, f3) =

〈f1, f1〉〈f2, f2〉〈f3, f3〉 if κ1 + κ2 > κ3,

〈f3, f3〉2 if κ1 + κ2 ≤ κ3,

where

〈fi, fi〉 =

∫

Γ0(Ni)\H|fi(τ)|2 Im(τ)κi−2 dτ

is the Petersson norm of fi. Then the main theorems of Harris andKudla [16] say that

L(w+12, f1 × f2 × f3)

p(f1, f2, f3)= C2 ·

∏

p|N1N2N3

Cp · C∞

with some constants C ∈ Q(f1, f2, f3), Cp ∈ Q(f1, f2, f3)×, and

C∞ ∈πw+2 ·Q× if κ1 + κ2 > κ3,

π2κ3 ·Q× if κ1 + κ2 ≤ κ3.

Date: May 5, 2005.2000 Mathematics Subject Classification. 11F67.

1

2 ATSUSHI ICHINO AND TAMOTSU IKEDA

Here Q(f1, f2, f3) is the field generated over Q by the Fourier coef-ficients of f1, f2, f3. Moreover, Cp (resp. C∞) depends only on thelocal components of π1, π2, π3 at p (resp. at ∞), where πi is the irre-ducible cuspidal automorphic representation of GL2(AQ) determinedby fi. We remark that in the case κ1 +κ2 > κ3, the critical values havealso been studied by Garrett [10], Orloff [31], Satoh [35], Garrett andHarris [11]. Moreover, Gross and Kudla [13], Bocherer and Schulze-Pillot [5] expressed certain height pairings as the algebraic part of thecentral critical values in the case κ1 + κ2 > κ3. By contrast, in thecase κ1 + κ2 ≤ κ3, there are no results on the critical values except[16] to our knowledge. In the present paper, we express certain periodintegrals of Maass lifts which appear in the Gross-Prasad conjecture[14], [15], as the algebraic part of the central critical values in the caseκ1 = κ2 = κ3/2.

Now we give a more precise description of our result. Let K =Q(√−D) be an imaginary quadratic field with discriminant −D < 0,

O the ring of integers of K, wK the number of roots of unity in K,and χ the primitive Dirichlet character associated to K/Q. Let QD

denote the set of all primes dividing D. For each q ∈ QD, let χq

be the

quadratic character of Q×q associated to Qq(√−D)/Qq by class field

theory.Let κ be an even positive integer such that wK | κ. Let g ∈

Sκ−1(Γ0(D), χ) be a primitive form. Then g determines a cusp form

g∗(τ) =∞∑

n=1

ag∗(n)qn ∈ Sκ−1(Γ0(D), χ)

such that ag∗(n) = 0 if aD(n) = 0, where q = exp(2π√−1τ) and

aD(n) =∏

q∈QDq-n

(1 + χq(−n)).

Note that g∗ is a certain sum of “twists” of g (see §1 for more de-

tails). Let G ∈ Sκ(Γ(2)K ) be the hermitian Maass lift of g, where

Γ(2)K = U(2, 2)(Q) ∩ SL4(O). Then G has a Fourier expansion of the

form

G(Z) =∑H

AG(H) exp(2π√−1 tr(HZ))

for Z ∈ H2. Here H2 is the hermitian upper half space of degree 2, Hruns over all positive definite semi-integral hermitian matrices of size

3

2, and

AG(H) =∑

d|ε(H)

dκ−1α∗G

(D det(H)

d2

),

where α∗G is a function on Z≥0 such that

ag∗(n) = aD(n)α∗G(n)

and ε(H) = maxn ∈ N |n−1H is semi-integral.Let f ∈ S2κ−2(SL2(Z)) be a normalized Hecke eigenform and

h(τ) =∞∑

n=1

c(n)qn ∈ S+κ−1/2(Γ0(4))

a Hecke eigenform associated to f by the Shimura correspondence.Note that c(n) = 0 unless −n ≡ 0, 1 mod 4. Let F ∈ Sκ(Sp2(Z)) bethe Saito-Kurokawa lift of f . Then F has a Fourier expansion of theform

F (Z) =∑B

AF (B) exp(2π√−1 tr(BZ))

for Z ∈ H2. Here H2 is the Siegel upper half space of degree 2, B runsover all positive definite half-integral symmetric matrices of size 2, and

AF

((n r/2r/2 m

))=

∑

d|(n,r,m)

dκ−1c

(4nm− r2

d2

).

Recall that H2 ⊂ H2. We consider the period integral 〈G|H2 , F 〉 givenby

〈G|H2 , F 〉 =

∫

Sp2(Z)\H2

G(Z)F (Z) det(Im(Z))κ−3 dZ.

Let Λ(s, g × g × f) be the completed triple product L-function givenby

Λ(s, g×g×f) = (2π)−4s+4κ−8Γ(s)Γ(s−2κ+4)Γ(s−κ+2)2L(s, g×g×f).

Our main result is as follows.

Theorem 3.1.

Λ(2κ− 3, g × g × f)

〈f, f〉2 = −24κ−6D−2κ+3c(D)2 〈G|H2 , F 〉2〈F, F 〉2 .

This paper is organized as follows. In §1 and §2, we review the theoryof hermitian Maass lifts and Saito-Kurokawa lifts, respectively. In §3,


we state our main result. In §4, we compute restrictions of hermitianMaass lifts to H2 and prove an identity for the seesaw

O(4, 2)

PPPPPPPPPPPPPP SL2 × SL2 O(2, 2)

nnnnnnnnnnnnnn

O(3, 2)×O(1)

nnnnnnnnnnnnSL2 O(2, 1)×O(1)

PPPPPPPPPPPP

.

To compute the central critical values of triple product L-functions, weuse the seesaw

Sp3

UUUUUUUUUUUUUUUUUUUUUUU O(2, 2)×O(2, 2)×O(2, 2)

iiiiiiiiiiiiiiiiii

SL2× SL2× SL2 O(2, 2)

.

In §5, we give an explicit formula for theta lifts from GL2 to GO(2, 2).In §6, we compute the local zeta integrals which arise in the integralrepresentation of triple product L-functions. Using these two seesawidentities, we prove Theorem 3.1 in §7. Finally, in §8, we interpret ourresult in terms of the Gross-Prasad conjecture.

The authors would like to thank Prof. Hiroshi Saito and Dr. KaoruHiraga for useful discussions.

Notation

Let K = Q(√−D) be an imaginary quadratic field with discriminant

−D < 0, O the ring of integers of K, O] = (√−D)−1O the inverse

different ideal of K/Q, wK the number of roots of unity in K, x 7→ xthe non-trivial Galois automorphism of K over Q, and χ the primitiveDirichlet character associated to K/Q. Let QD denote the set of allprimes dividing D. For each q ∈ QD, put Dq = qordq(D). Let χq be theprimitive Dirichlet character mod Dq defined by

χq(n) = χ(n′)

for n ∈ Z with q - n, where n′ is an integer such that

n′ ≡n mod Dq,

1 mod D−1q D.

Then

χ =∏

q∈QD

χq.

5

For each Q ⊂ QD, set

χQ =∏q∈Q

χq and χ′Q =∏

q∈QD−Q

χq.

Let χ = ⊗vχvbe the Hecke character of A×Q/Q× determined by χ.

Then χv

is the quadratic character of Q×v associated to Qv(√−D)/Qv

by class field theory. If q ∈ QD, then χq(n) = χq(n) for n ∈ Z with

q - n. One should not confuse χq

with χq.

A hermitian matrix H = (Hij) is called semi-integral if Hii ∈ Z andHij ∈ O] for all i, j. The unitary group U(n, n) and the symplecticgroup Spn are defined by

U(n, n)(R) =

g ∈ GL2n(R⊗K)

∣∣∣∣ g(

0 −1n

1n 0

)tg =

(0 −1n

1n 0

),

Spn(R) =

g ∈ GL2n(R)

∣∣∣∣ g(

0 −1n

1n 0

)tg =

(0 −1n

1n 0

),

for any Q-algebra R, respectively. The hermitian upper half space Hn

and the Siegel upper half space Hn are defined by

Hn =

Z ∈ Mn(C)

∣∣∣∣1

2√−1

(Z − tZ) > 0

,

Hn =Z ∈ Mn(C)

∣∣ tZ = Z, Im(Z) > 0,

respectively. Set Γ(n)K = U(n, n)(Q)∩ SL2n(O). When n ≥ 2, the space

Mκ(Γ(n)K ) of hermitian modular forms of degree n and weight κ consists

of all holomorphic functions G on Hn which satisfy

G((AZ +B)(CZ +D)−1) = det(CZ +D)κG(Z)

for all (A BC D

)∈ Γ

(n)K

and Z ∈ Hn.

1. Hermitian Maass lifts

In this section, we review the theory of hermitian Maass lifts [25],[39], [26].

Let κ be an even positive integer such that wK | κ. Krieg [26]introduced the space M∗

κ−1(Γ0(D), χ) of all modular forms

g∗(τ) =∞∑

n=0

ag∗(n)qn ∈Mκ−1(Γ0(D), χ)


such that ag∗(n) = 0 if aD(n) = 0, where

aD(n) =∏

q∈QD

(1 + χq(−n)).

Set

S∗κ−1(Γ0(D), χ) = Sκ−1(Γ0(D), χ) ∩M∗κ−1(Γ0(D), χ).

Let

g(τ) =∞∑

n=0

ag(n)qn ∈Mκ−1(Γ0(D), χ)

be a primitive form. By Theorem 4.6.16 of [30], there is a primitiveform

gQ(τ) =∞∑

n=0

agQ(n)qn ∈Mκ−1(Γ0(D), χ)

such that, for each prime p,

agQ(p) =

χQ(p)ag(p) if p /∈ Q,

χ′Q(p)ag(p) if p ∈ Q.

Put

(1.1) g∗ =∑

Q⊂QD

χQ(−1)gQ.

Then g∗ ∈M∗κ−1(Γ0(D), χ).

Lemma 1.1. Let g ∈ Sκ−1(Γ0(D), χ) be a primitive form. Then theFourier coefficients of g∗ are purely imaginary.

Proof. For each Q ⊂ QD, the Fourier coefficients of gQ − g(QD−Q) arepurely imaginary. This yields the lemma. ¤

Let

G(Z) =∑H

AG(H) exp(2π√−1 tr(HZ)) ∈Mκ(Γ

(2)K )

be a hermitian modular form of degree 2 and weight κ. Here H runsover all positive semi-definite semi-integral hermitian matrices of size2. We say that G satisfies the Maass relation if there exists a functionα∗G on Z≥0 such that

AG(H) =∑

d|ε(H)

dκ−1α∗G

(D det(H)

d2

),

7

where ε(H) = maxn ∈ N |n−1H is semi-integral. Let MMaassκ (Γ

(2)K )

denote the space of hermitian modular forms of degree 2 and weight κ

which satisfy the Maass relation. For G ∈MMaassκ (Γ

(2)K ), put

Ω(G)(τ) =∞∑

n=0

aD(n)α∗G(n)qn.

Then Ω(G) ∈M∗κ−1(Γ0(D), χ), and the linear map

Ω : MMaassκ (Γ

(2)K ) −→M∗

κ−1(Γ0(D), χ)

is an isomorphism. Note that we slightly modified Krieg’s definition ofΩ by a scalar.

2. Saito-Kurokawa lifts

In this section, we review the theory of Saito-Kurokawa lifts [28],[29], [1], [41].

Let κ be an even positive integer. Kohnen [22] introduced the spaceM+

κ−1/2(Γ0(4)) of all modular forms

h(τ) =∞∑

n=0

c(n)qn ∈Mκ−1/2(Γ0(4))

such that c(n) = 0 unless −n ≡ 0, 1 mod 4. Set

S+κ−1/2(Γ0(4)) = Sκ−1/2(Γ0(4)) ∩M+

κ−1/2(Γ0(4)).

Let

F (Z) =∑B

AF (B) exp(2π√−1 tr(BZ)) ∈Mκ(Sp2(Z))

be a Siegel modular form of degree 2 and weight κ. Here B runs overall positive semi-definite half-integral symmetric matrices of size 2. Wesay that F satisfies the Maass relation if there exists a function β∗F onZ≥0 such that β∗F (n) = 0 unless −n ≡ 0, 1 mod 4 and such that

AF

((n r/2r/2 m

))=

∑

d|(n,r,m)

dκ−1β∗F

(4nm− r2

d2

).

Let MMaassκ (Sp2(Z)) denote the space of Siegel modular forms of degree

2 and weight κ which satisfy the Maass relation. For F ∈MMaassκ (Sp2(Z)),

put

ΩSK(F )(τ) =∞∑

n=0

β∗F (n)qn.


Then ΩSK(F ) ∈M+κ−1/2(Γ0(4)), and the linear map

ΩSK : MMaassκ (Sp2(Z)) −→M+

κ−1/2(Γ0(4))

is an isomorphism.

3. Statement of the main theorem

Recall that K = Q(√−D) is an imaginary quadratic field with dis-

criminant −D < 0, wK is the number of roots of unity in K, and χ isthe primitive Dirichlet character associated to K/Q. Let κ be an evenpositive integer such that wK | κ. Let

g(τ) =∞∑

n=1

ag(n)qn ∈ Sκ−1(Γ0(D), χ)

be a primitive form. For each prime p /∈ QD, the Satake parameterαp, χ(p)α−1

p of g at p is defined by

1− ag(p)X + χ(p)pκ−2X2 = (1− p(κ−2)/2αpX)(1− p(κ−2)/2χ(p)α−1p X).

For each q ∈ QD, put αq = q−(κ−2)/2ag(q). Let G = Ω−1(g∗) ∈SMaass

κ (Γ(2)K ) be the hermitian Maass lift of g, where g∗ ∈ S∗κ−1(Γ0(D), χ)

is given by (1.1). Recall that H2 ⊂ H2. Let G|H2 denote the restrictionof G to H2. Then G|H2 ∈ SMaass

κ (Sp2(Z)) by [26].Let

f(τ) =∞∑

n=1

af (n)qn ∈ S2κ−2(SL2(Z))

be a normalized Hecke eigenform. For each prime p, the Satake param-eter βp, β

−1p of f at p is defined by

1− af (p)X + p2κ−3X2 = (1− p(2κ−3)/2βpX)(1− p(2κ−3)/2β−1p X).

Let

h(τ) =∞∑

n=1

c(n)qn ∈ S+κ−1/2(Γ0(4))

be a Hecke eigenform associated to f by the Shimura correspondence[37], [22]. Note that h is unique up to scalars. Let F = (ΩSK)−1(h) ∈SMaass

κ (Sp2(Z)) be the Saito-Kurokawa lift of f . The Petersson normsof f and F are defined by

〈f, f〉 =

∫

SL2(Z)\H|f(τ)|2 Im(τ)2κ−4 dτ,

〈F, F 〉 =

∫

Sp2(Z)\H2

|F (Z)|2 det(Im(Z))κ−3 dZ,

9

respectively.Put w = 4κ−7. We define the triple product L-function L(s, g×g×f)

by an Euler product

L(s, g × g × f) =∏v<∞

Lv(s, g × g × f)

for Re(s) À 0. Here, for v = p /∈ QD,

Lp

(s+

w

2, g × g × f

)= det(18 − Ap ⊗ Ap ⊗Bp · p−s)−1

with

Ap =

(αp 00 χ(p)α−1

p

), Bp =

(βp 00 β−1

p

),

and for v = q ∈ QD,

Lq

(s+

w

2, g × g × f

)

=[(1− α2

qβqq−s)(1− α2

qβ−1q q−s)(1− α−2

q βqq−s)(1− α−2

q β−1q q−s)

]−1.

Let Λ(s, g × g × f) be the completed triple product L-function givenby

Λ(s, g×g×f) = (2π)−4s+4κ−8Γ(s)Γ(s−2κ+4)Γ(s−κ+2)2L(s, g×g×f).

By [32], [19], Λ(s, g × g × f) has a holomorphic continuation to thewhole s-plane and satisfies a functional equation which replaces s withw + 1− s.

Our main result is as follows.

Theorem 3.1.

Λ(2κ− 3, g × g × f)

〈f, f〉2 = −24κ−6D−2κ+3c(D)2 〈G|H2 , F 〉2〈F, F 〉2 .

Remark 3.2. We may assume that c(n) ∈ Q(f) for all n ∈ N. In partic-ular, 〈G|H2 , F 〉 ∈

√−1R by Lemma 1.1. Since the Fourier coefficientsof G (resp. F ) are in Q(g) (resp. Q(f)), we have

〈G|H2 , F 〉〈F, F 〉 ∈ Q(g, f).

Remark 3.3. Fix a normalized Hecke eigenform f ∈ S2κ−2(SL2(Z)) andassume that c(D) 6= 0. By Theorem 1 of [7], the restriction map

SMaassκ (Γ

(2)K ) −→ SMaass

κ (Sp2(Z))

G 7−→ G|H2


is surjective. Hence there exists a primitive form g ∈ Sκ−1(Γ0(D), χ)such that

Λ(2κ− 3, g × g × f) 6= 0.

Remark 3.4. Using the Maass space defined by Sugano [39], one mightremove the assumption that wK | κ. See also [21].

Example 3.5. We discuss the caseD = 7, κ = 10. Let g ∈ S9(Γ0(7), χ)be the primitive form with

ag(3) = − 1

108290a3 − 27

15470a2 − 56557

54145a− 2508

245,

ag(5) = − 22

10829a3 +

359

3094a2 − 268563

10829a+

13716

49,

where a = 8 + 2√

46 + 2√−2148− 213

√46. Here we have used Stein’s

database [38]. Let f ∈ S18(SL2(Z)) be the normalized Hecke eigenform,and h ∈ S+

19/2(Γ0(4)) the Hecke eigenform associated to f with c(3) = 1,

c(7) = −16. Then

〈G|H2 , F 〉〈F, F 〉 = β∗G|H2

(3) = 2(ag(3) + ag(5)) = 24

√−6088 + 442

√46.

Using Dokchitser’s computer program [8], we obtain

〈f, f〉 = 0.0000045947361976392466101158732480223961 . . . ,

Λ(17, g × g × f) = 0.0000007104357360884738902046072703848041 . . . .

Hence

Λ(17, g × g × f)

〈f, f〉2 = 33651.438624311611653588297096717239 . . .

= −234 · 7−17 · c(7)2 · 〈G|H2 , F 〉2〈F, F 〉2 .

4. Restrictions of hermitian Maass lifts to H2

In this section, we compute restrictions of hermitian Maass lifts toH2 and prove the following seesaw identity.

Proposition 4.1. Let G = Ω−1(g∗) ∈ SMaassκ (Γ

(2)K ) be the hermitian

Maass lift of a primitive form g ∈ Sκ−1(Γ0(D), χ). Then

(4.1) S+−D(ΩSK(G|H2)) = ag(D)2 TrD

1 (g2).

11

Here

S+−D : S+

κ−1/2(Γ0(4)) −→ S2κ−2(SL2(Z))∞∑

n=1

c(n)qn 7−→∞∑

n=1

∑

d|nχ(d)dκ−2c

(Dn2

d2

)qn

is the linear map defined by Kohnen [22], and

TrD1 : S2κ−2(Γ0(D)) −→ S2κ−2(SL2(Z))

f 7−→∑

γ∈Γ0(D)\ SL2(Z)

f |γ

is the trace operator.

The rest of this section is devoted to the proof of Proposition 4.1.First, we compute the right-hand side of (4.1). For each Q ⊂ QD, letgQ ∈ Sκ−1(Γ0(D), χ) be the primitive form as in §1. For convenience,we write aQ(n) = agQ

(n) for n ∈ N and put aQ(n) = 0 for n ∈ Q− N.Put

DQ =∏q∈Q

Dq, DQ =

DQ if D is odd or 2 ∈ Q,

2DQ if D is even, D2 = 4, and 2 /∈ Q,

4DQ if D is even, D2 = 8, and 2 /∈ Q,

D′Q = D−1

Q D, and D′Q = D−1

Q D. Then D′Q is a square-free integer. For

m ∈ N and f ∈ S2κ−2(Γ0(D)), we define Um(f) by

Um(f)(τ) =1

m

∑

l mod m

f

(τ + l

m

)=

∞∑n=1

af (mn)qn.

Lemma 4.2.

ag(D)2 TrD1 (g2) =

∑Q⊂QD

χQ(−1)ag(D′Q)2UDQ

(g2Q).

Proof. For each q ∈ QD, we choose γq ∈ SL2(Z) such that

γq ≡

(0 −1

1 0

)mod D2

q ,

(1 0

0 1

)mod D−2

q D2.

For each Q = q1, . . . , qr ⊂ QD, put

γQ = γq1

(Dq1 00 1

). . . γqr

(Dqr 00 1

)(D−1

Q 00 1

)∈ SL2(Z).


We define a subset RQ of SL2(Z) by

RQ =

R′Q if D is odd or 2 ∈ Q,

R′Q ∪R′′

Q if D is even, D2 = 4, and 2 /∈ Q,

R′Q ∪R′′

Q ∪R′′′Q if D is even, D2 = 8, and 2 /∈ Q,

where

R′Q =

γQ

(1 l0 1

) ∣∣∣∣ l mod DQ

,

R′′Q =

γQ

(1 0

2−1D′Q 1

)(1 l0 1

) ∣∣∣∣ l mod DQ

,

R′′′Q =

γQ

(1 0

4−1D′Q 1

)(1 l + l′DQ

0 1

) ∣∣∣∣ l mod DQ, l′ mod 2

.

Then ⋃Q⊂QD

RQ

is a set of representatives for Γ0(D)\ SL2(Z). Thus, it suffices to showthat

ag(D)2∑

γ∈RQ

g2|γ = χQ(−1)ag(D′Q)2UDQ

(g2Q).

We only consider the case when D is even, D2 = 8, and 2 /∈ Q. Theother cases are similar. By Corollary 4.6.18 of [30],

g|γq

(Dq 00 1

)= χq(−1)

√χq(−1)D(κ−2)/2

q ag(Dq)−1gq

for q ∈ QD. Thus,

g2|γQ = χQ(−1)Dκ−2Q ag(DQ)−2g2

Q|(D−1

Q 00 1

).

Hence ag(D)2∑

γ∈RQg2|γ is equal to the product of χQ(−1)Dκ−2

Q ag(D′Q)2

and∑

l mod DQ

[g2

Q|(D−1

Q 00 1

)(1 l0 1

)+ g2

Q|(

1 02−1D 1

)(D−1

Q 00 1

)(1 l0 1

)

+∑

l′ mod 2

g2Q|

(1 0

4−1D 1

)(1 l′

0 1

)(D−1

Q 00 1

)(1 l0 1

)].

For m ∈ Z, put

gQ,m = gQ|(

4−1 00 1

)(1 m0 1

).

13

It is easy to check that

gQ,m|(

1 02−1D 1

)= χ(1− 2−1Dm)gQ,m,

gQ,m|(

1 04−1D 1

)(1 l′

0 1

)= χ(1 + 4−1Dm)gQ,−m+l′ ,

and

gQ = 4(κ−3)/2aQ(4)−1∑

m mod 4

gQ,m.

By a direct calculation,

g2Q + g2

Q|(

1 02−1D 1

)+

∑

l′ mod 2

g2Q|

(1 0

4−1D 1

)(1 l′

0 1

)

= 4κ−2aQ(4)−2∑

m mod 4

g2Q,m

= ag(4)−2U4(g2Q).

Hence ag(D)2∑

γ∈RQg2|γ is equal to

χQ(−1)D−κ+2Q ag(D

′Q)2ag(4)−2

∑

l mod DQ

U4(g2Q)|

(D−1

Q 00 1

)(1 l0 1

)

= χQ(−1)ag(4−1D′

Q)2U4DQ(g2

Q).

¤

Next, we compute the left-hand side of (4.1). For each Q ⊂ QD, set

O](Q) =

t ∈ O]

∣∣∣∣∣ NK/Q(t) ∈ D−1(QD−Q)Z−

( ⋃Q1)Q

D−1(QD−Q1)Z

).

For i = 0, 1 and n ∈ N, set

O]i,n = t ∈ O] | trK/Q(t) = i,NK/Q(t) ≤ n

and O]i,n(Q) = O]

i,n ∩ O](Q). Note that

O]i,n =

⋃Q⊂QD

O]i,n(Q).

Put

bi,n(Q) =∑

t∈O]i,n

NK/Q(t)∈D−1Q Z

aQ(D(n− NK/Q(t))).


Lemma 4.3. Let β∗G|H2be the function on Z≥0 attached to G|H2 as in

§2. Then

β∗G|H2(4n− i) =

∑Q⊂QD

χQ(−1)bi,n(Q)

for i = 0, 1 and n ∈ N.

Proof. By [26, p. 680],

β∗G|H2(4n− i) =

∑

t∈O]i,n

α∗G(D(n− NK/Q(t))),

where α∗G is the function on Z≥0 attached to G as in §1. If t ∈ O](Q),then α∗G(D(n− NK/Q(t))) is equal to

∑Q1⊂Q

∑Q2⊂QD−Q

2−](QD−Q)χ(Q1∪Q2)(−1)a(Q1∪Q2)(D(n− NK/Q(t)))

=∑

Q1⊂Q

χ(Q1∪(QD−Q))(−1)a(Q1∪(QD−Q))(D(n− NK/Q(t)))

=∑

Q1⊃QD−Q

χQ1(−1)aQ1(D(n− NK/Q(t))).

Hence β∗G|H2(4n− i) is equal to

∑Q⊂QD

∑

t∈O]i,n(Q)

∑Q1⊃QD−Q

χQ1(−1)aQ1(D(n− NK/Q(t)))

=∑

Q1⊂QD

∑Q⊃QD−Q1

∑

t∈O]i,n(Q)

χQ1(−1)aQ1(D(n− NK/Q(t)))

=∑

Q1⊂QD

∑

t∈O]i,n

NK/Q(t)∈D−1Q1Z

χQ1(−1)aQ1(D(n− NK/Q(t))).

¤Now we prove Proposition 4.1. Note that

b0,n(Q) =∑

m∈ZD′Qm2≤DQn

aQ(D′Q)aQ(DQn− D′

Qm2).

Also, b1,n(Q) is equal to

∑

m∈ZD′Qm2≤DQ(n−1/4)

aQ(D′Q)aQ

(DQ

(n− 1

4

)− D′

Qm2

)

15

if D2 = 8 or 2 ∈ Q, and is equal to

∑

m∈ZD′Q(m+1/2)2≤DQ(n−1/4)

aQ(D′Q)aQ

(DQ

(n− 1

4

)− D′

Q

(m+

1

2

)2)

otherwise. For n = Dl2 with l ∈ N,

∑

m∈ZD′Qm2≤DQn

aQ

(DQn− D′

Qm2

4

)=

∑

m1,m2∈Z≥0

m1+m2=DQl

aQ(D′Q)aQ(m1m2),

hence

β∗G|H2(n) =

∑Q⊂QD

∑

m1,m2∈Z≥0

m1+m2=DQl

χQ(−1)aQ(D′Q)aQ(m1m2)

by Lemma 4.3. As in the proof of Proposition 3 of [24], S+−D(ΩSK(G|H2))(τ)

is equal to

∞∑n=1

∑

d|nχ(d)dκ−2β∗G|H2

(Dn2

d2

)qn

=∞∑

n=1

∑

d|n

∑Q⊂QD

∑

m1,m2∈Z≥0

m1+m2=DQn/d

χ(d)dκ−2χQ(−1)aQ(D′Q)2aQ(m1m2)q

n

=∑

Q⊂QD

∞∑n=1

∑

m1,m2∈Z≥0

m1+m2=DQn

χQ(−1)ag(D′Q)2

∑

d|(m1,m2)

χ(d)dκ−2aQ

(m1m2

d2

)qn

=∑

Q⊂QD

∞∑n=1

∑

m1,m2∈Z≥0

m1+m2=DQn

χQ(−1)ag(D′Q)2aQ(m1)aQ(m2)q

n

=∑

Q⊂QD

χQ(−1)ag(D′Q)2UDQ

(g2Q)(τ).

Therefore Proposition 4.1 follows from Lemma 4.2.

5. Theta lifts from GL2 to GO(2, 2)

In this section, we give an explicit formula for theta lifts from GL2

to GO(2, 2).


5.1. Weil representations for similitudes. Let k be a number fieldand A = Ak the adele ring of k. Fix a non-trivial additive characterψ of A/k. Let GSpn denote the symplectic similitude group and ν :GSpn → Gm the scale map. Let V = M2(k) be the quadratic spacewith bilinear form (x, y) = tr(xyι). Here ι is the main involution on V ,that is,

xι =

(x4 −x2

−x3 x1

)for x =

(x1 x2

x3 x4

)∈ V

Let H = GO(V ) denote the orthogonal similitude group and ν : H →Gm the scale map. Set H = (GL2×GL2)o 〈t〉 and H0 = GL2×GL2,where t is the involution on GL2×GL2 defined by

t(h1, h2) = ((hι2)−1, (hι

1)−1)

for h1, h2 ∈ GL2. Recall that there is an exact sequence

1 −→ Gm −→ Hρ−→ H −→ 1,

where ρ(h1, h2)x = h1xh−12 , ρ(t)x = xι for h1, h2 ∈ GL2 and x ∈ V .

Let ω be the Weil representation of Spn(A)×O(V )(A) with respectto ψ on the Schwartz space S(V n(A)). Set

R = (g, h) ∈ GSpn×H | ν(g) = ν(h).Following [17, §5.1], we extend the Weil representation to a represen-tation ω of R(A). Then

ω(g, h)ϕ = ω

(g

(1n 00 ν(g)−11n

), 1

)L(h)ϕ

for (g, h) ∈ R(A) and ϕ ∈ S(V n(A)), where

L(h)ϕ(x) = |ν(h)|−nϕ(h−1x)

for x ∈ V n(A). Let R be the pullback of R to GSpn×H. By abuse ofnotation, we write ω for the pullback of ω to R(A).

When n = 1, we also use a different model of the Weil representation.Let ϕ ∈ S(V (A)) be the partial Fourier transform of ϕ ∈ S(V (A)) givenby

ϕ

((x1 x2

x3 x4

))=

∫

A2

ϕ

((x1 x′2x3 x′4

))ψ(x2x

′4 − x4x

′2) dx

′2 dx

′4.

Here dx′2, dx′4 are the self-dual measures on A with respect to ψ. We

define a representation ω of R(A) (and of R(A)) on S(V (A)) by

ω(g, h)ϕ = ω(g, h)ϕ.

Note that ω(g, 1)ϕ(x) = ϕ(xg) for g ∈ SL2(A).

17

5.2. Fourier coefficients of theta lifts. Let n = 1. The theta func-tion, defined for (g, h) ∈ R(A) and ϕ ∈ S(V (A)) by

Θ(g, h;ϕ) =∑

x∈V (k)

ω(g, h)ϕ(x),

is left R(k)-invariant. Let φ be a cusp form on GL2(A). For h ∈ H(A),choose g′ ∈ GL2(A) such that ν(g′) = ν(h), and put

θ(φ, ϕ)(h) =

∫

SL2(k)\ SL2(A)

φ(gg′)Θ(gg′, h;ϕ) dg.

Here dg is the Tamagawa measure on SL2(A). This integral does not de-pend on the choice of g′, and the theta lift θ(φ, ϕ) is left H(k)-invariant.By abuse of notation, we write θ(φ, ϕ) for the pullback of θ(φ, ϕ) toH(A). For an irreducible cuspidal automorphic representation π ofGL2(A), set

θ(π) = θ(φ, ϕ) |φ ∈ π, ϕ ∈ S(V (A)).

By [36], [16, §7],

(5.1) θ(π)|H0(A) = π £ π∨

as spaces of functions on H0(A). Here π∨ is the contragredient repre-sentation of π.

Define the Whittaker function Wφ of φ by

Wφ(g) =

∫

k\Aφ

((1 b0 1

)g

)ψ(b) db

for g ∈ GL2(A). Here db is the self-dual measure on A with respect toψ. Similarly, define the Whittaker function Wθ(φ,ϕ) of θ(φ, ϕ) by

Wθ(φ,ϕ)(h) =

∫

(k\A)2θ(φ, ϕ)

(((1 b10 1

),

(1 b20 1

))h

)ψ(b1)ψ(b2) db1 db2

for h ∈ H(A).

Lemma 5.1.

Wθ(φ,ϕ)(h) =

∫

SL2(A)

Wφ(gg′)ω(g′, h)ϕ(g) dg.


Proof. By the Poisson summation formula,∫

k\Aθ(φ, ϕ)

((1,

(1 b0 1

))h

)ψ(b) db

=

∫

k\A

∫

SL2(k)\ SL2(A)

φ(gg′)∑

x∈V (k)

ω(gg′, h)ϕ(x)ψ(b(det(x)− 1)) dg db

=

∫

SL2(k)\ SL2(A)

φ(gg′)∑

x∈SL2(k)

ω(gg′, h)ϕ(x) dg

=

∫

SL2(A)

φ(gg′)ω(g′, h)ϕ(g) dg.

Hence Wθ(φ,ϕ)(h) is equal to∫

k\A

∫

SL2(A)

φ(gg′)ω(g′,

((1 b0 1

), 1

)h

)ϕ(g)ψ(b) dg db

=

∫

k\A

∫

SL2(A)

φ(gg′)ω(g′, h)ϕ((

1 −b0 1

)g

)ψ(b) dg db

=

∫

k\A

∫

SL2(A)

φ

((1 b0 1

)gg′

)ω(g′, h)ϕ(g)ψ(b) dg db

=

∫

SL2(A)

Wφ(gg′)ω(g′, h)ϕ(g) dg.

¤

5.3. An explicit formula for theta lifts. Let k = Q. Let ψ =⊗vψv be the standard additive character of A/Q, so that ψ∞(x) =exp(2π

√−1x) for x ∈ R. Let N be a positive integer. We temporarilylet χ denote an arbitrary primitive Dirichlet character mod N andχ = ⊗vχv

the Hecke character of A×/Q× determined by χ.Let

f(τ) =∞∑

n=1

af (n)qn ∈ Sl(Γ0(N), χ)

be a primitive form. Then f determines a cusp form f on GL2(A) bythe formula

f(g) = det(g∞)l/2j(g∞,√−1)−lf(g∞(

√−1))χ(k)−1

for g = γg∞k ∈ GL2(A) with γ ∈ GL2(Q), g∞ ∈ GL+2 (R), and k ∈

K0(N ; Z). Here

K0(N ; Z) =

(a bc d

)∈ GL2(Z)

∣∣∣∣ c ≡ 0 mod N Z

19

and

χ

((a bc d

))= χ(d) for

(a bc d

)∈ K0(N ; Z).

By definition, f satisfies

f(gk) = χ(k)−1f(g),(5.2)

f(gkθ) = exp(√−1lθ)f(g),(5.3)

for all g ∈ GL2(A), k ∈ K0(N ; Z), and

kθ =

(cos θ sin θ− sin θ cos θ

)∈ SO(2).

When N = 1,

(5.4) f(g) = f(gJ )

for all g ∈ GL2(A), where

J =

(−1 00 1

)∈ GL2(R).

Let π = ⊗vπv be the irreducible cuspidal automorphic representationof GL2(A) generated by f . Then πp is a principal series representation

IndGL2(Qp)

B(Qp) (| |sp £χ−1p| |−sp) of GL2(Qp) for each prime p, where B is the

standard Borel subgroup of GL2 and sp ∈ C. Also, π∞ is the (limit of)discrete series representation of GL2(R) of weight l. In the space of π,the conditions (5.2), (5.3) characterize the cusp form f up to scalars.Let Wf = ⊗vWv be the Whittaker function of f . We may assume thatWp(1) = 1 for all primes p. Then

W∞

((a 00 1

)kθ

)=

al/2 exp(−2πa+

√−1lθ) if a > 0,

0 if a < 0.

We define ϕf = ⊗vϕv ∈ S(V (A)) as follows.

• For each prime p - N , ϕp is the characteristic function of M2(Zp).• For each prime q | N , the partial Fourier transform ϕq of ϕq is

given by

ϕq(x) =

χ

q(x4) if x1, x2 ∈ Zq, x3 ∈ NZq, x4 ∈ Z×q ,

0 otherwise.

• ϕ∞(x) = (x1 +√−1x2 +

√−1x3 − x4)l exp(−π tr(xtx)).

Set K = SO(2) SL2(Z) and K0(N) = K ∩ SO(2)K0(N ; Z). Recall that∫

SL2(A)

φ(g) dg =

∫

A×A××K

φ

((1 x0 1

)(a 00 a−1

)k

)|a|−2 dx d×a dk


for φ ∈ L1(SL2(A)). Here dg is the Tamagawa measure on SL2(A), dxis the Tamagawa measure on A, d×a is the Tamagawa measures on A×,and dk is the Haar measure on K such that vol(K) = πζ(2)−1.

Proposition 5.2. As functions on H0(A),

θ(f , ϕf ) = 2l vol(K0(N))(f ⊗ fχ(det)).

Proof. By (5.1), θ(f , ϕf ) ∈ π £ π∨. A routine calculation shows that

ω(g, h)ϕp = ϕp,

ω

((det(h1h

−12 ) 0

0 1

), (h1, h2)

)ϕq = χ

q(h1h2)

−1χq(det(h2))ϕq,

ω(kθ, (kθ1 , kθ2))ϕ∞ = exp(√−1l(−θ + θ1 + θ2))ϕ∞,

for (g, h) ∈ R(Zp), h1, h2 ∈ K0(N ;Zq), and kθ, kθ1 , kθ2 ∈ SO(2), wherep (resp. q) is a prime such that p - N (resp. q | N). Hence there existsa constant C such that

θ(f , ϕf ) = C(f ⊗ fχ(det)).

By Lemma 5.1,

C = Wf (1)−2Wθ(f ,ϕf )(1) = exp(4π) vol(K0(N))C∞,

where C∞ is equal to∫

R×R×W∞

((1 x0 1

)(a 00 a−1

))ϕ∞

((1 x0 1

)(a 00 a−1

))|a|−2 dx d×a

= 2

∫ ∞

0

∫ ∞

−∞exp(2π

√−1x)al exp(−2πa2)

× (a−√−1a−1x+ a−1)l exp(−π(a2 + a−2x2 + a−2)) · a−2 dx d×a

= 2l exp(−4π).

This completes the proof. ¤

6. Local zeta integrals

In this section, following Gross and Kudla [13], we compute the localzeta integrals of Garrett [10], Piatetski-Shapiro and Rallis [32].

6.1. Preliminaries. Let

P =

(A ∗0 νtA−1

)∈ GSp3(Qv)

∣∣∣∣ A ∈ GL3(Qv), ν ∈ Q×v

21

be the Siegel parabolic subgroup of GSp3(Qv). Let Z be the center ofGSp3(Qv). Set K = GSp3(Zp) if v = p, and

K =

(A B−B A

) ∣∣∣∣ A+√−1B ∈ U(3)

if v = ∞. Let

G = (g1, g2, g3) ∈ GL2(Qv)3 | det(g1) = det(g2) = det(g3).

We regard G as a subgroup of GSp3(Qv) via the embedding

((a1 b1c1 d1

),

(a2 b2c2 d2

),

(a3 b3c3 d3

))7−→

a1 0 0 b1 0 00 a2 0 0 b2 00 0 a3 0 0 b3c1 0 0 d1 0 00 c2 0 0 d2 00 0 c3 0 0 d3

.

Set

T =

t(a) =

((a1 00 a−1

1

),

(a2 00 a−1

2

),

(a3 00 a−1

3

)) ∣∣∣∣ ai ∈ Q×v,

U0 =

((1 x1

0 1

),

(1 x2

0 1

),

(1 x3

0 1

)) ∣∣∣∣ xi ∈ Qv, x1 + x2 + x3 = 0

,

U =

u(x) =

(1, 1,

(1 x0 1

)) ∣∣∣∣ x ∈ Qv

,

and KG = G ∩K.For s ∈ C, let I(s) = Ind

GSp3(Qv)P (δ

s/2P ) denote the degenerate princi-

pal series representation of GSp3(Qv). Here δP is the modulus characterof P . Let f (s) be a holomorphic section of I(s). Note that

f (s)

((A ∗0 νtA−1

)g

)= | det(A)|2s+2|ν|−3s−3f (s)(g).

For i = 1, 2, 3, let πi be an irreducible admissible generic representationof GL2(Qv) with central character ωi. We assume that ω1ω2ω3 is trivial.Let Wi be a Whittaker function of πi. Define a function W = W1 ⊗W2 ⊗W3 on G by

W (g) = W1(g1)W2(g2)W3(g3)

for g = (g1, g2, g3) ∈ G. Then the local zeta integral is given by

Z(f (s),W ) =

∫

ZU0\Gf (s)(δg)W (g) dg,


where

δ =

1 1 1 −1 0 00 1 0 −1 1 00 0 1 −1 0 11 1 1 0 0 00 0 0 −1 1 00 0 0 −1 0 1

∈ Sp3(Z).

As in §5.1, let V = M2(Qv) be the quadratic space with bilinearform (x, y) = tr(xyι). Let Φ ∈ S(V 3). For g ∈ GSp3(Qv), chooseh ∈ GO(V )(Qv) such that ν(h) = ν(g), and put

f(0)Φ (g) = ω(g, h)Φ(0).

Then f(0)Φ (g) does not depend on the choice of h, and defines an element

of I(0). We extend f(0)Φ to a holomorphic section f

(s)Φ of I(s) so that

the restriction of f(s)Φ to K does not depend on s. Using a Bruhat

decomposition of δ, we obtain the following.

Lemma 6.1.

f(0)Φ (δg) =

∫

V

ω(g, h)Φ(y, y, y) dy.

6.2. The non-archimedean case. Let v = q ∈ QD. Let ψ be theadditive character of Qq given by ψ(x) = exp(−2π

√−1x) for x ∈Z[q−1]. Let χ = χ

qbe the quadratic character of Q×q associated to

Qq(√−D)/Qq by class field theory. Put d = ordq(D). Note that

d =

1 if q 6= 2,

2 or 3 if q = 2.

Let π = IndGL2(Qq)

B(Qq) (| |s′ £ χ| |−s′) and σ = IndGL2(Qq)

B(Qq) (| |s′′ £ | |−s′′),

where B is the standard Borel subgroup of GL2. Put α = q−s′ andβ = q−s′′ . Let Wπ be the Whittaker function of π with respect to ψsuch that Wπ(1) = 1,

Wπ(gk) = χ(k)Wπ(g)

for all g ∈ GL2(Qq) and k ∈ K0(D;Zq). Here

K0(D;Zq) =

(a bc d

)∈ GL2(Zq)

∣∣∣∣ c ≡ 0 mod DZq

and

χ

((a bc d

))= χ(d) for

(a bc d

)∈ K0(D;Zq).

23

Similarly, let Wσ be the Whittaker function of σ with respect to ψ suchthat Wσ(1) = 1,

Wσ(gk) = Wσ(g)

for all g ∈ GL2(Qq) and k ∈ GL2(Zq). Note that Wπ and Wσ areuniquely determined. Define ϕχ ∈ S(V ) so that

ϕχ

((y1 y2

y3 y4

))=

χ(y4) if y1, y2 ∈ Zq, y3 ∈ DZq, y4 ∈ Z×q ,

0 otherwise,

where ϕχ ∈ S(V ) is the partial Fourier transform of ϕχ given by

ϕχ

((y1 y2

y3 y4

))=

∫

Q2q

ϕχ

((y1 y′2y3 y′4

))ψ(y2y

′4 − y4y

′2) dy

′2 dy

′4.

Let ϕ0 ∈ S(V ) be the characteristic function of M2(Zq).

Proposition 6.2. Set W = Wπ ⊗Wπ ⊗Wσ and Φ = ϕχ ⊗ ϕχ ⊗ ϕ0 ∈S(V 3). Then

Z(f(0)Φ ,W ) = χ(−1)q−3d(1+q−1)−2(1−q−1)2α−4d vol(KG)L

(1

2, π × π × σ

).

Here

L(s, π × π × σ)

=[(1− α2βq−s)(1− α2β−1q−s)(1− α−2βq−s)(1− α−2β−1q−s)

]−1.

First, we compute the function W . It is well-known that

Wσ

((qn 00 1

))=

q−n/2β

n+1 − β−n−1

β − β−1if n ≥ 0,

0 otherwise.

Set

w =

(0 1−1 0

), k1 =

(1 0

qd−1 1

), k2 =

(1 0

qd−2 1

).

Lemma 6.3. Put ε =√χ(−1) and ζ8 = exp(π

√−1/4).

(i)

Wπ

((qn 00 1

))=

q−n/2αn if n ≥ 0,

0 otherwise,

Wπ

((qn 00 1

)w

)=

χ(qn+d)ε−1q−(n+d)/2α−n−2d if n ≥ −d,0 otherwise.


(ii) When q = 2,

Wπ

((qn 00 1

)k1

)=

q1/2α−1 if n = −1,

0 otherwise.

(iii) When q = 2 and d = 3, for u ∈ 1 + 2Z,

Wπ

((qnu 00 1

)k2

)=

χ(u)εζ−u

8 q1/2α−2 if n = −2,

0 otherwise.

Proof. Define φ ∈ π so that supp(φ) ∩GL2(Zq) = K0(D;Zq) and

φ(k) = χ(k)

for k ∈ K0(D;Zq). Let Wφ be the Whittaker function of π defined by

Wφ(g) =

∫

Qq

φ

(w

(1 x0 1

)g

)ψ(x) dx

for g ∈ GL2(Qq). A standard calculation shows that

Wπ = χ(qd)εqd/2α−2d ·Wφ

and proves the formula for Wπ. ¤

Next, we compute the section f(s)Φ . For each n ∈ Z, let φn denote

the characteristic function of qnZq. We define φχ ∈ S(Qq) by

φχ(x) =

χ(x) if x ∈ Z×q ,

0 otherwise.

Let φχ ∈ S(Qq) be the Fourier transform of φχ given by

φχ(x) =

∫

Qq

φχ(x′)ψ(xx′) dx′,

where dx′ is the Haar measure on Qq such that vol(Zq) = 1. Similarly,let ξχ ∈ S(Qq) be the Fourier transform of

φχ · (the characteristic function of 1 + q2Zq).

It is easy to check that

ω

((1 x0 1

)(qn 00 q−n

), 1

)ϕ0

((y1 y2

y3 y4

))

= q−2nφ−n(y1)φ−n(y2)φ−n(y3)φ−n(y4)ψ(x(y1y4 − y2y3)).

Also, a routine calculation shows the following.

Lemma 6.4.

25

(i)

ω

((qn 00 q−n

), 1

)ϕχ

((y1 y2

y3 y4

))

= q−2nφ−n(y1)φχ(qny2)φ−n+d(y3)φ−n(y4).

(ii) For b ∈ Zq,

ω

((qn 00 q−n

)(1 b0 1

)w, 1

)ϕχ

((y1 y2

y3 y4

))

= q−2n−dφ−n(y1)φ−n−d(y2)φχ(qny3)φ−n(y4)ψ(q2nb(y1y4 − y2y3)).

(iii) When q = 2,

ω

((qn 00 q−n

)k1, 1

)ϕχ

((y1 y2

y3 y4

))

= q−2nφ−n(y1)φ−n(y4)φχ(−qn−d+1y3)φχ(−q2n−d+1y2y3).

(iv) When q = 2 and d = 3, for e ∈ ±1,

ω

((qn 00 q−n

)ke

2, 1

)ϕχ

((y1 y2

y3 y4

))

= q−2nφ−n(y1)φ−n(y4)φχ(−qn−d+2ey3)ξχ(−q2n−d+2ey2y3).

Set

t(n1, n2, n3) =

((qn1 00 q−n1

),

(qn2 00 q−n2

),

(qn3 00 q−n3

)).

Put K0 = 1 and

Kd =

(1 b0 1

)w

∣∣∣∣ b ∈ Zq/DZq

.

When q = 2, put K1 = k1. When q = 2 and d = 3, put K2 =k2, k

−12 . It is easy to check that

f(s)Φ (δg) = (qn1+n2+n3 max(q−2n1 , q−2n2 , q−2n3 , |x|))−2sf

(0)Φ (δg)

for g = u(x)t(n1, n2, n3)k with x ∈ Qq, n1, n2, n3 ∈ Z, k ∈ Ki×Kj×1.Lemma 6.5. Let g = u(x)t(n1, n2, n3)k with x ∈ Qq, n1, n2, n3 ∈ Z,and k ∈ Ki ×Kj × 1. If i 6= j, then

f(0)Φ (δg) = 0.

Proof. If i 6= j, then ω(g, 1)Φ(y, y, y) = 0 for all y ∈ V . This yields thelemma. ¤Lemma 6.6. Let g = u(x)t(n1, n2, n3)k with x ∈ Qq, n1, n2, n3 ∈ Z,and k = (h1, h2, 1) ∈ Ki ×Ki × 1.


(i) For i = 0 and h1 = h2 = 1, f(0)Φ (δg) is equal to

χ(−1)q−2n3−d(1− q−1) if n1 = n2 ≤ n3 − d, x ∈ q2n1Zq,

0 otherwise.

(ii) For i = d and

h1 =

(1 b10 1

)w, b2 =

(1 b20 1

)w

with b1, b2 ∈ Zq, f(0)Φ (δg) is equal to

q−n1−n3−2d(1− q−1) if n1 = n2 ≤ n3 < n1 + d

and q2n1b1 + q2n1b2 + x ∈ qn1+n3Zq,

q−2n3−d(1− q−1) if n1 = n2 ≤ n3 − d

and q2n1b1 + q2n1b2 + x ∈ q2n1+dZq,

0 otherwise.

(iii) When q = 2, for i = 1 and h1 = h2 = k1, f(0)Φ (δg) is equal to

χ(−1)q−2n3−d(1− q−1) if n1 = n2 ≤ n3 − d

and x ∈ q2n1+1Zq,

−χ(−1)q−2n3−d(1− q−1) if n1 = n2 ≤ n3 − d

and x ∈ q2n1 + q2n1+1Zq,

0 otherwise.

(iv) When q = 2 and d = 3, for i = 2 and h1 = ke12 , h2 = ke2

2 with

e1, e2 ∈ ±1, f (0)Φ (δg) = 0 unless n1 = n2 ≤ n3 − d, in which

case, f(0)Φ (δg) is equal to

q−2n3−d(1− q−1) if e1 = e2, x ∈ −q2n1e1 + q2n1+2Zq,

−q−2n3−d(1− q−1) if e1 = e2, x ∈ q2n1e1 + q2n1+2Zq,

χ(−1)q−2n3−d(1− q−1) if e1 6= e2, x ∈ q2n1+2Zq,

−χ(−1)q−2n3−d(1− q−1) if e1 6= e2, x ∈ q2n1+1 + q2n1+2Zq,

0 otherwise.

Proof. We only prove (ii). The other cases are similar. Put m =min(n1, n2, n3), m

′ = min(n1+d, n2+d, n3), and x′ = q2n1b1+q2n2b2+x.

Since

ω(g, 1)Φ(y, y, y) = q−2n1−2n2−2n3−2dφ−m(y1)φ−m′(y2)φ−m(y4)

× φχ(qn1y3)φχ(qn2y3)φ−n3(y3)ψ(x′(y1y4 − y2y3)),

27

f(0)Φ (δg) = 0 unless n1 = n2 ≤ n3, in which case, f

(0)Φ (δg) is equal to

∫

Q2q

q−3m+m′−2n3−2dφ−m(y4)φχ(qmy3)2φ−n3(y3)φm(x′y4)φm′(x′y3) dy3 dy4.

This integral is equal to

q−3m+m′−2n3−2d

∫

y3∈q−mZ×q

∫

y4∈q−mZq

dy4 dy3 = q−m+m′−2n3−2d(1− q−1)

if x′ ∈ qm+m′Zq, and vanishes otherwise. ¤

Now we compute the local zeta integral Z(f(s)Φ ,W ). Set

q =

((q 00 1

),

(q 00 1

),

(q 00 1

)).

Note that f(s)Φ (δqg) = q−s−1f

(s)Φ (δg). Since UTKG ∪ qUTKG is a fun-

damental domain of ZU0\G, Z(f(s)Φ ,W ) is equal to

∫

Qq×(Q×q )3×KG

f(s)Φ (δu(x)t(a)k)W (u(x)t(a)k)|a|−2 dx d×a dk

+

∫

Qq×(Q×q )3×KG

f(s)Φ (δqu(x)t(a)k)W (qu(x)t(a)k)q2|a|−2 dx d×a dk

as in [13, §4]. Here dx is the Haar measure on Qq such that vol(Zq) =1, d×ai is the Haar measure on Q×q such that vol(Z×q ) = 1, d×a =d×a1d

×a2d×a3, and |a| = |a1a2a3|. Set

K ′G = G ∩ (K0(D;Zq)×K0(D;Zq)×GL2(Zq)).

Then the function g 7→ f(s)Φ (δg)W (g) on G is right K ′

G-invariant. Since

⋃

0≤i,j≤d

Ki ×Kj × 1

is a set of representatives for KG/K′G, we have

Z(f(s)Φ ,W ) = vol(K ′

G)d∑

i=0

d∑j=0

1∑

l=0

Z(l)ij (s),


where

Z(0)ij (s) =

∑

k∈Ki×Kj×1

∑

n1,n2,n3∈Z

∫

Qq

× f(s)Φ (δu(x)t(n1, n2, n3)k)W (u(x)t(n1, n2, n3)k)q

2n1+2n2+2n3 dx,

Z(1)ij (s) =

∑

k∈Ki×Kj×1

∑

n1,n2,n3∈Z

∫

Qq

× f(s)Φ (δqu(x)t(n1, n2, n3)k)W (qu(x)t(n1, n2, n3)k)q

2n1+2n2+2n3+2 dx.

If i 6= j, then Z(l)ij (s) = 0 by Lemma 6.5.

Lemma 6.7.

(i)

Z(0)00 (s) = χ(−1)q−2ds−2d(1− q−1)(β − β−1)−1

× [β2d+1(1− β2q−2s−1)−1(1− α4β2q−2s−1)−1

− β−2d−1(1− β−2q−2s−1)−1(1− α4β−2q−2s−1)−1],

Z(1)00 (s) = χ(−1)q−2ds−s−2d−1/2(1− q−1)α2(β − β−1)−1

× [β2d+2(1− β2q−2s−1)−1(1− α4β2q−2s−1)−1

− β−2d−2(1− β−2q−2s−1)−1(1− α4β−2q−2s−1)−1].

(ii)

Z(0)dd (s) =

Y

(0)0 (s) if d = 1,

Y(0)0 (s) + Y

(0)1 (s) if d = 2 or 3,

Z(1)dd (s) =

Y

(1)0 (s) + Y

(1)1 (s) if d = 1 or 2,

Y(1)0 (s) + Y

(1)1 (s) + Y

(1)2 (s) if d = 3,

29

where

Y(0)0 (s) = χ(−1)q−d(1− q−1)α−4d(β − β−1)−1

× [β(1− β2q−2s−1)−1(1− α−4β2q−2s−1)−1

− β−1(1− β−2q−2s−1)−1(1− α−4β−2q−2s−1)−1],

Y(0)1 (s) = χ(−1)q−2s−d−1(1− q−1)α−4d+4(β − β−1)−1

× [β3(1− β2q−2s−1)−1 − β−3(1− β−2q−2s−1)−1

],

Y(1)0 (s) = χ(−1)q−s−d−1/2(1− q−1)α−4d−2(β − β−1)−1

× [β2(1− β2q−2s−1)−1(1− α−4β2q−2s−1)−1

− β−2(1− β−2q−2s−1)−1(1− α−4β−2q−2s−1)−1],

Y(1)1 (s) = χ(−1)q−s−d−1/2(1− q−1)α−4d+2(β − β−1)−1

× [β2(1− β2q−2s−1)−1 − β−2(1− β−2q−2s−1)−1

],

Y(1)2 (s) = χ(−1)q−3s−d−3/2(1− q−1)α−4d+6(β − β−1)−1

× [β4(1− β2q−2s−1)−1 − β−4(1− β−2q−2s−1)−1

].

(iii) When q = 2, Z(0)11 (s) = 0 and

Z(1)11 (s) = χ(−1)q−2ds+s−2d+1/2(1− q−1)α−2(β − β−1)−1

× [β2d(1− β2q−2s−1)−1 − β−2d(1− β−2q−2s−1)−1

].

(iv) When q = 2 and d = 3, Z(1)22 (s) = 0 and

Z(0)22 (s) = χ(−1)q−2ds+2s−2d+1(1− q−1)α−4(β − β−1)−1

× [β2d−1(1− β2q−2s−1)−1 − β−2d+1(1− β−2q−2s−1)−1

].

Proof. We only compute Z(0)dd (s). The other cases are similar. Let

x ∈ Qq, n1, n2, n3 ∈ Z, and

k =

((1 b10 1

)w,

(1 b20 1

)w, 1

)

with b1, b2 ∈ Zq. Then the integrand of Z(0)dd (s) vanishes unless n1 =

n2 ≤ n3. Put

Υ(m,n) = χ(−1)q−n(2s+1)−3d(1−q−1)α−4m−4d(β−β−1)−1(β2n+1−β−2n−1).

For n1 = n2 = m and n3 = n, the integral∫

Qq

f(s)Φ (δu(x)t(n1, n2, n3)k)W (u(x)t(n1, n2, n3)k)q

2n1+2n2+2n3 dx


is equal to Υ(m,n) if0 ≤ m ≤ n if d = 1,

0 ≤ m ≤ n, or m = −1 and n ≥ 1 if d = 2 or 3,

and vanishes otherwise. It is easy to check that

q2d

∞∑m=0

∞∑n=m

Υ(m,n) = Y(0)0 (s), q2d

∞∑n=1

Υ(−1, n) = Y(0)1 (s).

This proves the formula for Z(0)dd (s). ¤

By a direct calculation,

Z(f(s)Φ ,W ) = χ(−1)q−d(1−q−1)(1−q−2s−1)α−4d vol(K ′

G)L

(s+

1

2, π × π × σ

).

This completes the proof of Proposition 6.2.

6.3. The archimedean case. Let v = ∞. Let ψ be the additivecharacter of R given by ψ(x) = exp(2π

√−1x) for x ∈ R.For each l ∈ N, let σl denote the (limit of) discrete series represen-

tation of GL2(R) of weight l. Let Wl be the Whittaker function of σl

with respect to ψ given by

Wl

((a 00 1

)(cos θ sin θ− sin θ cos θ

))=

al/2 exp(−2πa+

√−1lθ) if a > 0,

0 if a < 0.

Define ϕl ∈ S(V ) by

ϕl

((y1 y2

y3 y4

))= (y1+

√−1y2+√−1y3−y4)

l exp(−π(y21+y

22+y

23+y

24)).

Proposition 6.8. Set W = Wl ⊗ Wl ⊗ W2l and Φ = ϕl ⊗ ϕl ⊗ω(1, (J ,J ))ϕ2l ∈ S(V 3), where

J =

(−1 00 1

).

ThenZ(f

(0)Φ ,W ) = 2−4l+1π−4l+2Γ(l)2Γ(2l − 1) vol(KG).

First, we compute the section f(s)Φ .

Lemma 6.9. Let r ∈ R×+ and x ∈ R. For each l ∈ Z≥0, put

Jl =

∫

R4

((y1 + y4)2 + (y2 + y3)

2)l exp(−πr(y21 + y2

2 + y23 + y2

4))

× exp(−2π√−1x(y1y4 + y2y3)) dy1 dy2 dy3 dy4.

31

Then

Jl = 2lπ−ll!(r +√−1x)−l−1(r −√−1x)−1.

Proof. Put

Fl(z) =

∫

R4

((y1 + y4)2 + (y2 + y3)

2)l exp(−πr(y21 + y2

2 + y23 + y2

4))

× exp(−2π√−1x(y1y4 − y2y3))

× exp(2π√−1(z1y1 + z2y2 + z3y3 + z4y4)) dy1 dy2 dy3 dy4.

Then

F0(z) = R−1 exp(−πrR−1(z21 +z2

2 +z23 +z2

4)+2π√−1R−1x(z1z4+z2z3))

and Fl(z) = ∇lF0(z), where R = r2 + x2 and

∇ =1

(2π√−1)2

((∂

∂z1

+∂

∂z4

)2

+

(∂

∂z2

+∂

∂z3

)2).

Hence

Jl = Fl(0) = 2lπ−ll!(r +√−1x)−lR−1.

¤

Lemma 6.10. For x ∈ R and a = (a1, a2, a3) ∈ (R×+)3, f(s)Φ (δu(x)t(a))

is equal to

22lπ−2l(2l)!(r +√−1x)−s−2l−1(r −√−1x)−s−1a2s+l+2

1 a2s+l+22 a2s+2l+2

3 ,

where r = a21 + a2

2 + a23.

Proof. It is easy to check that

f(s)Φ (δu(x)t(a)) = (a1a2a3(r

2 + x2)−1/2)2sf(0)Φ (δu(x)t(a))

and f(0)Φ (δu(x)t(a)) = al+2

1 al+22 a2l+2

3 J2l. This yields the lemma. ¤

Now we compute the local zeta integral Z(f(s)Φ ,W ). Note that supp(W ) =

Z SL2(R)3. Since UTKG is a fundamental domain of U0\ SL2(R)3,

Z(f(s)Φ ,W ) is equal to

1

2

∫

R×(R×)3×KG

f(s)Φ (δu(x)t(a)k)W (u(x)t(a)k)|a|−2 dx d×a dk

as in [13, §6]. Here dx, dai are the Lebesgue measures on R, d×ai =|ai|−1dai, d

×a = d×a1d×a2d

×a3, and |a| = |a1a2a3|. Since the function


g 7→ f(s)Φ (δg)W (g) on G is right KG-invariant, vol(KG)−1Z(f

(s)Φ ,W ) is

equal to

22l+2π−2l(2l)!

×∫

R×(R×+)3(a2

1 + a22 + a2

3 +√−1x)−s−2l−1(a2

1 + a22 + a2

3 −√−1x)−s−1

× a2(s+l)1 a

2(s+l)2 a

2(s+2l)3 exp(−2π(a2

1 + a22 + a2

3 −√−1x)) dx d×a

= 2−4s−4l+1π−s−4l+2(2l)!Γ(s+ 2l + 1)−1Γ(2s+ 2l)−1

× Γ(s+ l)2Γ(s+ 2l − 1)Γ(s+ 2l)

by Lemma 2.6 of [20]. This completes the proof of Proposition 6.8.

7. Proof of Theorem 3.1

Recall that g ∈ Sκ−1(Γ0(D), χ) is a primitive form and f ∈ S2κ−2(SL2(Z))is a normalized Hecke eigenform. As in §5.3, let g (resp. f) be thecusp form on GL2(A) determined by g (resp. f), π (resp. σ) the irre-ducible cuspidal automorphic representation of GL2(A) generated byg (resp. f), and Wg (resp. Wf ) the Whittaker function of g (resp. f).Let V = M2(Q) be the quadratic space as in §5.1. We define ϕg, ϕf ∈S(V (A)) as in §5.3. Set W = Wg ⊗Wg ⊗Wf and

Φ = ϕg ⊗ ϕg ⊗ ω(1, (J ,J ))ϕf ∈ S(V 3(A)),

where

J =

(−1 00 1

)∈ GL2(R).

Put S = ∞ ∪QD and

ZS(s) = vol(SO(2))−3∏

q∈QD

vol(SL2(Zq))−3 ·

∏v∈S

Z(f(s)Φv,Wv),

where Z(f(s)Φv,Wv) is the local zeta integral as in §6.1. Let

G = (g1, g2, g3) ∈ (GL2)3 | det(g1) = det(g2) = det(g3),

H = (h1, h2, h3) ∈ GO(V )3 | ν(h1) = ν(h2) = ν(h3).We regard F = g⊗ g⊗ f as a cusp form on G(A) and define the thetalift θ(F,Φ) to H(A) as in §5.2. Put

I(θ(F,Φ)) =

∫

ZH0 (A)H0(Q)\H0(A)

θ(F,Φ)(h, h, h) dh,

where ZH0 is the center of H0 = GL2×GL2 and dh is the Haar measure

on H0(A) such that vol(ZH0(A)H0(Q)\H0(A)) = 1.

33

By Main Identity 9.1 of [16],

(7.1) vol(K)3ZS(0)ζS(2)−2LS

(1

2, π × π × σ

)= 2I(θ(F,Φ)),

where K = SO(2) SL2(Z) and vol(K) = πζ(2)−1. By Propositions 6.2and 6.8,

ZS(0)ζS(2)−2LS

(1

2, π × π × σ

)

= 2−4κ+5π−4κ+6Γ(κ− 1)2Γ(2κ− 3)

×∏

q∈QD

χq(−1)D2κ−7

q (1 + q−1)−2(1− q−1)2ag(Dq)−4

× ζS(2)−2L(2κ− 3, g × g × f)

= −2π2D2κ−7ag(D)−4∏

q∈QD

(1 + q−1)−4

× ζ(2)−2Λ(2κ− 3, g × g × f).

By Proposition 5.2 and (5.4),

θ(F,Φ) = θ(g, ϕg)⊗ θ(g, ϕg)⊗ θ(f , ω(1, (J ,J ))ϕf )

= 24κ−4 vol(K0(D))2 vol(K)

× (g ⊗ gχ(det))⊗ (g ⊗ gχ(det))⊗ (f ⊗ f).

Thus,

I(θ(F,Φ)) = 24κ−6 vol(K0(D))4 vol(K)〈TrD1 (g2), f〉2.

Therefore

Λ(2κ− 3, g × g × f) = −24κ−6D−2κ+3ag(D)4〈TrD1 (g2), f〉2

= −24κ−6D−2κ+3〈S+−D(ΩSK(G|H2)), f〉2

by Proposition 4.1 and (7.1). If c(D) = 0, then Λ(2κ−3, g×g×f) = 0.If c(D) 6= 0, then f = c(D)−1S+

−D(h) and

〈S+−D(ΩSK(G|H2)), f〉

〈f, f〉 = c(D)〈ΩSK(G|H2), h〉

〈h, h〉 = c(D)〈G|H2 , F 〉〈F, F 〉 .

This completes the proof of Theorem 3.1.

8. The Gross-Prasad conjecture

In this section, we interpret our result in terms of the Gross-Prasadconjecture [14], [15].


Let H1 = SO(n + 1) and H0 = SO(n) be special orthogonal groupsover a number field k with embedding ι : H0 → H1. Let π1 ' ⊗vπ1,v

and π0 ' ⊗vπ0,v be irreducible cuspidal automorphic representationsof H1(Ak) and H0(Ak), respectively. We assume that

HomH0(kv)(π1,v, π0,v) 6= 0

for all places v of k. Gross and Prasad conjectured that, when π1 andπ0 are tempered, the period integral

〈F1|H0 , F0〉 =

∫

H0(k)\H0(Ak)

F1(ι(h0))F0(h0) dh0

does not vanish for some F1 ∈ π1 and some F0 ∈ π0 if and only if

L

(1

2, π1 × π0

)6= 0.

To relate our result to the Gross-Prasad conjecture, we must

• remove the assumption that π1 and π0 are tempered,• formulate an identity which relates the period integral to special

values of automorphic L-functions.

Following Ginzburg, Piatetski-Shapiro, and Rallis [12], we put

Pπ1,π0(s) =L(s, π1 × π0)

L(s+ 12, π1,Ad)L(s+ 1

2, π0,Ad)

,

where Ad is the adjoint representation of LHi on the Lie algebra ofLHi. Then the identity

(8.1)|〈F1|H0 , F0〉|2〈F1, F1〉〈F0, F0〉 = Pπ1,π0

(1

2

)

would hold up to an elementary constant. This conjectural identity iscompatible with the results of Waldspurger [40] for n = 2, Harris andKudla [16], [18] for n = 3, Bocherer, Furusawa, and Schulze-Pillot [4]for n = 4.

Now we discuss the case n = 5. We retain the notation of §3. Notethat H1 = SO(4, 2) ∼ SU(2, 2) and H0 = SO(3, 2) ∼ Sp2. We mayassume that g∗ 6= 0. Let π1 (resp. π0) be the irreducible cuspidalautomorphic representation of H1(AQ) (resp. H0(AQ)) determined byG (resp. F ). Then π1 and π0 are non-tempered. It is easy to checkthat

L(s, π1) = L(s, Sym2(π))ζ(s+ 1)ζ(s)ζ(s− 1),

L(s, π0) = L(s, σ)ζ

(s+

1

2

)ζ

(s− 1

2

),

35

and

Pπ1,π0

(1

2

)=

L(12, Sym2(π)× σ)L(3

2, σ)

L(2, Sym2(π))L(1, π,Ad)L(1, σ,Ad).

Here π (resp. σ) is the irreducible cuspidal automorphic representationof GL2(AQ) determined by g (resp. f). On the other hand, if c(D) 6= 0,then

〈f, f〉〈h, h〉|〈G|H2 , F 〉|2〈F, F 〉2 ∼ L

(1

2, Sym2(π)× σ

)

by Theorem 3.1 and the Kohnen-Zagier formula [24]. According to [21,§15], it is expected that

〈G,G〉〈g∗, g∗〉 ∼ L(2, Sym2(π))ζ(2).

Indeed, Raghavan and Sengupta [34] proved it for D = 4. By the resultof Kohnen and Skoruppa [23],

〈F, F 〉〈h, h〉 ∼ L

(3

2, σ

)ζ(2).

It is well-known that 〈g∗, g∗〉 ∼ L(1, π,Ad) and 〈f, f〉 ∼ L(1, σ,Ad).Therefore (8.1) is compatible with Theorem 3.1.

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Department of Mathematics, Graduate School of Science, OsakaCity University, 3-3-138 Sugimoto, Sumiyoshi-ku, Osaka 558-8585, Japan

E-mail address: [email protected]

Graduate school of mathematics, Kyoto University, Kitashirakawa,Kyoto 606-8502, Japan

E-mail address: [email protected]

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