133(3), 2005, p. 331–348
KLOOSTERMAN-FOURIER INVERSION FOR SYMMETRIC MATRICES
by Omer Offen
Abstract. — We formulate a Kloosterman transform on the space of generalized Kloosterman integrals on symmetric matrices, and obtain an inversion formula. The formula is a step towards a fundamental lemma of the Jacquet type. At the same time it hints towards a conjectural relative trace formula identity, associated with the metaplectic correspondence.
R´esum´e(Inversion de Kloosterman-Fourier pour les matrices sym´etriques)
Nous d´efinissons une transformation de Kloosterman sur l’espace des int´egrales de Kloosterman g´en´eralis´ees sur les matrices sym´etriques et nous obtenons une formule d’inversion. Cette formule est une ´etape vers un lemme fondamental de type de Jacquet.
En mˆeme temps, elle indique une identit´e conjecturale de la formule des traces relative associ´ee `a la correspondance m´etaplectique.
1. Introduction
LetF be a non-archimedean local field,OF the ring of integers inF and℘ the maximal ideal ofOF. Let | · |denote the normalized absolute value onF so that for a uniformizer$ ofF we have|$|−1= #(OF/℘) is the size of the
Texte re¸cu le 23 avril 2003, r´evis´e le 12 f´evrier 2004, accept´e le 3 mars 2004
Omer Offen, Max-Planck-Institut f¨ur Mathematik, Vivatsgasse 7, D-53111 Bonn (Germany) E-mail :[email protected] • Url :http://www.math.columbia.edu/~omer/
2000 Mathematics Subject Classification. — 11F70, 11L05.
Key words and phrases. — Kloosterman integrals, Kloosterman transform, Jacquet’s relative trace formula.
BULLETIN DE LA SOCI´ET´E MATH ´EMATIQUE DE FRANCE 0037-9484/2005/331/$ 5.00 c
Soci´et´e Math´ematique de France
residual field. Let ψ be a non-trivial additive character of F. We recall the formula
(1)
Z fb(x)ψ(ax2)dx=|2a|−12γ(a, ψ) Z
f(x)ψ(−a−1x2)dx
which we use to define the Weil constant γ. Here a ∈ F×, f ∈ Cc∞(F) is a Schwartz function onF andfbis the Fourier transform off defined by
fb(x) = Z
f(y)ψ(−2xy)dy.
The measure dxis the self dual Haar measure onF with respect toψ. Thus, it satisfies
(2) f(0) =|2|
Z fb(x)dx.
LetN=Nnbe the subgroup of upper triangular unipotent matrices in GLn(F).
Define the non-degenerate characterθ=θn ofN by θ
u
=ψn−1X
i=1
xi,i+1
whereu= (xi,j)∈N. The Haar measure dxonF determines a Haar measure onN and a self dual Haar measure on any finite dimensional F-vector space.
We will use the measures determined by dxunless otherwise specified. Denote byMm×n(F) the set of allm×nmatrices with entries inF. Let
Mn(F) =Mn×n(F) and denote byS=Sn the space of symmetric matrices
S =
X ∈Mn(F); tX=X . We consider the action (u, s)7→tusuofN onS.
Definition 1.1. — An element s∈ S is calledrelevant if θ is trivial on the stabilizerNs ofsinN.
Our objects of interest are the generalized Kloosterman integrals
(3) ω[Φ, ψ;s] =
Z
Ns\N
Φ(tusu)θ(u2)du for a relevants∈ S,Φ∈Cc∞(S). Let
Sn=Sn∩GLn(F).
The orbits in Sn are fully described in [6]. To describe a set of representatives for all orbits in Sn we view the elements of the Weyl group as permutation matrices in GLn(F). Thus a complete set of representatives for the orbits inSn is the set of allwa, wherewis the longest element in the Weyl group of a
o
standard Levi subgroupM of GLn(F) andais in the center ofM. All relevant orbits inSnwith zero determinant contain an element of the form s0
, where s∈Sn−1. This is proved in [3] for Hermitian matrices. The proof for symmetric matrices is identical and we omit it here. Representatives for the relevant orbits of zero determinant are therefore given as above in terms of representatives of orbits in Sn−1. When w= 1 andais a diagonal matrix, the stabilizer ofain Nn is trivial.
In this sense the diagonal matrices are representatives of the largest orbits.
In a sense explained in [3] and [2], the Kloosterman integrals for smaller orbits, i.e. withw6= 1 are determined by Kloosterman integrals of the largest orbits.
For this reason, our main concern in this work is the space of Kloosterman integrals, restricted to the relevant diagonal matrices. These are of the form a = diag(a1, . . . , an) where a1, . . . , an−1 ∈ F× and an ∈ F. We will denote byω[Φ, ψ;a1, . . . , an] the Kloosterman integralω[Φ, ψ; diag(a1, . . . , an)].
To state our main theorem it will be convenient to introduce a normalization.
We introduce the normalizing factors
σn(a) =an−11 an−22 · · ·an−1, Γn(a, ψ) =γ(a1, ψ)n−1γ(a2, ψ)n−2· · ·γ(an−1, ψ).
The normalized Kloosterman integral is
(4) ωeψ[Φ, ψ;a1, . . . , an] = Γn(−a, ψ)σn(a)12ω[Φ, ψ;a1, . . . , an].
The purpose of the notationωeψis to emphasize the dependence of the normal- ization on the characterψ. Let Ωψ,nbe the space of functionsωon (F×)n−1×F of the form
ω(a1, . . . , an) =ωeψ[Φ, ψ;a1, . . . , an]
for some Φ ∈ Cc∞(S). Denote by [·,·] : F× ×F× 7→ {±1} the quadratic Hilbert symbol. It is defined by the condition [a, b] = 1 iffais representable by the quadratic forma=x2−by2. We define the Kloosterman transformKψ,n
on Ωψ,n by
Kψ,nω(a1, . . . , an) = Z
ω(p1, . . . , pn)ψ
− Xn i=1
pian+1−i+
n−1X
i=1
1 pian−i
(5)
×n−1Y
i=1 n−iY
j=1
[ai, pj]
dpndpn−1· · ·dp1
where the integral overpi∈F,i= 1, . . . , nis only iterated. Although the inte- grand isa priorionly defined forp1, . . . , pn−1∈F× andpn∈F we make sense of (5) in the proof of theorem 1.2. Denote by wn ∈GLn(F) the permutation matrix with a unit anti-diagonal. For any matrixX ∈Mn(F) we will denote by Tr(X) the trace ofX. For a function Φ∈Cc∞(S) let
(6) Φ(s) =ˆ
Z
S
Φ(t)ψ −Tr(st) dt
be the standard Fourier transform of Φ. We will consider the Fourier transform
(7)
ˇ
Φ(s) =ˆ
Φ(wnswn)of Φ. Our main theorem is
Theorem 1.2. — The integral (5) defining the Kloosterman transform on Ωψ,n is a convergent iterated integral. Moreover, let Φ∈Cc∞(S)then,
Kψ,nωeψ[Φ, ψ;.]
(a1, . . . , an) (8)
=|2|12n(n−1)γ(1,ψ)¯ 12n(n−1)eωψ¯[
ˇ
Φ,ψ;¯ a1, . . . , an].The theorem shows in particular thatKψ,nis a transform from Ωψ,nto Ωψ,n¯ . Combining the theorem with Fourier inversion onS we obtain the inversion of the Kloosterman transform.
Corollary 1.3. — The Kloosterman transform satisfies Kψ,n¯ ◦Kψ,n=|2|n(n−1)Id whereId is the identity map onΩψ,n.
The motivation to the problem lies in a conjectural trace formula identity of the Jacquet type. The identity is concerned with the metaplectic corre- spondence of [1]. It is a lifting of genuine automorphic representations of the metaplectic double coverGLfn of GLn to automorphic representations of GLn. Jacquet suggests the following characterization for the image of this lift: A cus- pidal automorphic representation of GLnwith trivial central character is a lift- ing from GLfn iff it is (H, χ)-distinguished for some subgroupH of orthogonal similitudes of GLn and some id´ele class quadratic characterχ.
For more detail and definitions we refer to [6]. This characterization of the image of metaplectic correspondence will follow from the relative trace formula identity
(9)
Z
KΦ(u)θ(u2)du= Z
Kf(u1, u2)θ(u1u2)du1du2.
Herekis a global field. The integration is overu, u1, u2∈Nn(k)\Nn(Ak), where on the right hand sideNnis viewed as its splitting inGLfn,KfandKΦare kernel functions depending on the quadratic characterχ, of operators corresponding to the smooth functions of compact supportf onGLfn(Ak) and Φ onSn(Ak).
Again for more details we refer to [6]. The fundamental lemma for this situation is a matching of Kloosterman integrals. If Φ0 is the characteristic function of S ∩KwhereK= GLn(OF) is the standard maximal compact of GLn(F), then ω[Φ0, ψ;a] matches in an appropriate way, an integral of the form
(10)
Z
N×N
Ψ0(tu1au2)θ(u1u2)du1du2
o
where Ψ0is the unit element of the genuine spherical Hecke algebra ofGLfn(F).
In [6], Mao proved the fundamental lemma for the case n = 3 by brute force computation. In [4], Jacquet developed a method to prove Kloosterman inte- gral identities of the above type. The method requires in both sides an inversion formula for a Fourier-Kloosterman transform on the space of Kloosterman inte- grals. For the case of a quadratic extension the inversion formulas are obtained in [3] and the method is carried out in [4] to prove the identity of Kloosterman integrals which serves as a fundamental lemma for a relative trace formula. In this work we provide a step towards the fundamental lemma associated with the trace formula (9). At the same time, the formula (5) and mainly the os- cillating factorQn−1
i=1
Qn−i
j=1[ai, pj] in it, hint to a relation with the metaplectic group. Ifσis the 2-cocycle that defines multiplication inGLfn(F) as defined in [5], then fora= diag(a1, . . . , an) andp= diag(p1, . . . , pn) we have
(11) σ(a, wnpwn) =
n−1Y
i=1 n−iY
j=1
[ai, pj].
The rest of this manuscript is organized as follows: The main tool we use to prove Theorem 1.2 is Weil’s formula. In Chapter 2 we write it in a form convenient for our needs. We then prove the theorem by induction. Chapter 3 provides an inversion formula for some intermediate integrals designed to use an inductive argument. In Chapter 4 the inductive step is carried out to finish the proof of the inversion. Chapter 5 provides a much simpler formula associated with the smallest orbits. We present it here, since once the analogous results for the metaplectic case will be obtained, the method of Jacquet requires this formula in order to prove smooth matching. The proof of the inversion formula closely follows the guidelines of [3], the new ingredient is the occurrence of the Hilbert symbol in the Kloosterman transform. The problem was suggested to me by Jacquet. For the project and for much help and support, I am most thankful to him. Most of this work was written during my visit at IH´ES.
I thank the IH´ES for a very pleasant and productive visit.
2. Weil’s formula Let
Vn,m=n0n X
tX 0m
; X ∈Mn×m(F)o . We viewVn,mas a self-dual space via the pairing
(12) 0n X
tX 0m
,0n tY Y 0m
7−→Trh
−0n X
tX 0m
0n tY Y 0m
i
whereX ∈Mn×m(F) andY ∈Mm×n(F). The Fourier transform of a function Φ∈Cc∞(Vn,m) is defined by
(13)
ˆ
Φ0n tX X 0m
= Z
Mn×m
Φ0n Y
tY 0m
ψ
−2 Tr(XY) dY,
X ∈Mm×n(F). We recall Weil’s formula for the Fourier transform of a char- acter of second order for the spaceVn,m. Taking our definition of the Fourier transform into acount, from [7] we have that for all A∈Sn, C∈Sm there is a constant γmn(A, C, ψ) such that for all Φ∈Cc∞(Vn,m)
Z
Mn×m(F)
Φ0n X
tX 0m
ψ
Tr(CtXAX) dX (14)
=|2|12mn· |detA|−12m· |detC|−12n
×γmn(A, C, ψ) Z
Mm×n(F)
ˆ
Φ0n tZ Z 0mψ
−Tr(C−1ZA−1tZ) dZ.
ForP ∈Mm×n(F) the Fourier transform of the function
(15) 0n X
tX 0m
7−→Φh0n X
tX 0m
iψ
Tr(P X) X ∈Mn×m(F), is the function
(16) 0n tZ
Z 0m
7−→
ˆ
Φ 0n t(Z−12P) Z−12P 0m
Z ∈Mm×n(F). Applying Weil’s formula (14) to this function, we get after the change of variablesZ7→Z+12P that
Z
Mn×m(F)
Φh0n X
tX 0m
iψ
Tr(P X) + Tr(CtXAX) dX (17)
=|2|12mn|detA|−12m· |detC|−12nγmn(A, C, ψ) Z
Mm×n(F)
ˆ
Φ0n tZ Z 0m
×ψ
−Tr C−1(Z+12P)A−1t(Z+12P) dZ.
We remark that in the case m = n = 1 the Weil constant is the 1-dimen- sional Weil constant defined in (1)
(18) γ11(A, C, ψ) =γ(AC, ψ).
For generalmandnwe can describe the Weil constant in terms of the 1-dimen- sional case.
Lemma 2.1. — Let A=tg1diag(a1, . . . , an)g1 andC=g2diag(c1, . . . , cm)tg2, whereA∈Sn, C ∈Sm, g1∈GLn(F)andg2∈GLm(F), then
(19) γmn(A, C, ψ) = Y
1≤i≤n 1≤j≤m
γ(aicj, ψ).
o
Proof. — We start with the case m = 1. SinceCc∞(Vn,1) 'Cc∞(F)⊗n, it is enough in the definition of γ1n to consider functions Φ of the form Φ(v) = Qn
i=1fi(xi) wherev∈Vn,1 is of the formv= 0tXnX0
,X =t(x1, . . . , xn)∈Fn and fi ∈ Cc∞(F). Note that in this case
ˆ
Φ(v) = Qni=1fˆi(xi), therefore if A= diag(a1, . . . , an) the factorization ofγn1(A,1, ψ) is straightforward from (1) and (2). Indeed
Z
Φ0n X
tX 0 ψ
Tr(tXAX dX =
Yn i=1
Z
fi(xi)ψ(aix2i)dxi
=|2|12n Yn i=1
|ai|−12γ(ai, ψ)
Z fˆi(xi)ψ(−a−1i x2i)dxi
=|2|12n|detA|−12Yn
i=1
γ(ai, ψ) Z
ˆ
Φ0n tZ Z 0
ψ
−Tr(ZA−1tZ) dZ, therefore in this case indeed we get
(20) γ1n(A,1, ψ) =
Yn i=1
γ(ai, ψ).
Let g ∈ GLn(F) be such that A = tgag with a = diag(a1, . . . , an), for any A∈Sn. Note that the Fourier transform of the function
0n X
tX 0m
7−→Φ 0n g−1X
tXtg−1 0 is the function
0n tZ Z 0m
7−→ |detg| ·
ˆ
Φ0n tgtZ Zg 0
and therefore using the changes of variables X 7→ g−1X, Z 7→ Zg−1 and applying (20) for awe get
Z
Φ0n X
tX 0 ψ
Tr(tXAX)
dX=|detg|−1 Z
Φ 0n g−1X
tXtg−1 0 ψ
Tr(tXaX) dX
=|2|12n|deta|−12Yn
i=1
γ(ai, ψ) Z
ˆ
Φ0n tgtZ Zg 0
ψ
−Tr(Za−1tZ) dZ
=|2|12n|detg|−1· |deta|−12Yn
i=1
γ(ai, ψ) Z
ˆ
Φ0ntZ Z 0
ψ[−Tr(ZA−1tZ) dZ.
Since detA = detg2deta we get (20) for any A ∈ Sn. For a general m, if C= diag(c1, . . . , cm) is diagonal then
Tr(CtXAX) = Xm j=1
Tr(tXcjAX).
Therefore, as in the casem= 1 the fact thatCc∞(Vn,m)'Cc∞(Vn,1)⊗mimplies that
(21) γmn(A, C, ψ) =
Ym j=1
γ1n(cjA,1, ψ)
and (19) then follows from the casem= 1. For anyAandCas in the statement of the lemma, since
Tr(CtXAX) = Tr(XCtXA)
we may compute as before, denotingc= diag(c1, . . . , cm) and using the changes of variablesX 7→Xg2−1andZ 7→g2−1Z that
Z
Φ0n X
tX 0 ψ
Tr(CtXAX) dX
=|detg2|−n Z
Φ 0n Xg−12
tg−12 tX 0 ψ
Tr(ctXAX) dX
=|2|12mn· |detA|−12m· |detc|−12n·γmn(A, c, ψ) Z
ˆ
Φ 0n tZtg2g2Z 0
ψ[−Tr(c−1ZA−1tZ) dZ
=|2|12mn· |detg2|−n· |detA|−12m· |detc|−12n·γmn(A, c, ψ) Z
ˆ
Φ0ntZZ 0
ψ[−Tr(C−1ZA−1tZ) dZ.
We then have
γmn(A, C, ψ) =γmn(A, c, ψ) and the lemma follows from the case when C is diagonal.
The symmetry on the right hand side of (19) implies that (22) γmn(A, C, ψ) =γnm(C, A, ψ).
From (1) it is easy to see that the one-dimensional Weil constant satisfies
(23) γ(a, ψ) =γ(−a,ψ).¯
Applying Weil’s formula twice we get that
(24) γ(a, ψ)γ(−a−1, ψ) = 1.
We also recall thatγ satisfies
(25) γ(ac, ψ) =γ(1, ψ)−1γ(a, ψ)γ(c, ψ)[a, c]
fora, c∈F×. Therefore, forAandC as in Lemma 2.1 (26) γmn(A, C, ψ) =γ(1, ψ)−mnYn
i=1
γ(ai, ψ)mYm
j=1
γ(cj, ψ)n Y
i,j
[ai, cj] .
o
In particular
(27) γmn(A,1m, ψ) = Yn i=1
γ(ai, ψ)m=γnm(1m, A, ψ) and
(28) γnm(1n, C, ψ) = Ym j=1
γ(cj, ψ)n =γmn(C,1n, ψ).
The expressionQ
i,j[ai, cj] is thus only dependent onAandCand we can and will denote
(29) [A, C] =Y
i,j
[ai, cj].
We finish this chapter, with an identity we will need later.
Lemma 2.2. — One has
(30) γmn(−A,1m, ψ)γmn(A−1, C, ψ) =γ(1,ψ)¯ mn[A, C]γnm(−C,1n,ψ).¯ Proof. — From (26), (27) and (28) we get that
γmn(−A,1m, ψ)γnm(A−1, C, ψ) (31)
=γ(1, ψ)−mnγmn(−A,1m, ψ)γmn(A−1,1m, ψ)γnm(C,1n, ψ)[A, C].
Using (24) and (27) we get that
γmn(−A,1m, ψ)γmn(A−1,1m, ψ) = 1.
From (23) and (28) we have γnm(C,1n, ψ) = γnm(−C,1n,ψ) and from (23)¯ and (24) that γ(1, ψ)−1 = γ(1,ψ).¯ Therefore, the right hand side of (31) is now equal toγ(1,ψ)¯ mn[A, C]γnm(−C,1n,ψ) and the lemma follows.¯
3. Intermediate orbital integrals
For n, m ≥ 1 and a function Φ ∈ Cc∞(Sm+n) we define the intermediate orbital integral
ωmnh
Φ, ψ;An
Bm
(32) i
= Z
Mn×m(F)
Φh1n tX 1m
An
Bm
1n X 1m
iθm+n
h1n 2X 1m
idX
= Z
Mn×m(F)
Φh An AnX
tXAn Bm+tXAnX iψ
2 Tr(X) dX
where An ∈ Sn, Bm ∈ Sm and =nm = (δ(i,j),(1,n)) ∈ Mm×n(F). We also define a normalized intermediate integral
e ωn,ψm h
Φ, ψ;An
Bm
i (33)
=γmn(−An,1m, ψ)· |detAn|12m·ωnmh
Φ, ψ;An
Bm
i.
Proposition 3.1. — Let Cm ∈ Sm, Dn ∈ Sn and Φ ∈ Cc∞(Sm+n). The function mapping An to
[An, Cm]ψ
Tr(wmCm−1wmnmA−1n tnm) (34)
× Z
Sm
e ωmn,ψh
Φ, ψ;An
Bm
i ψ
−Tr(BmwmCmwm) dBm
originally defined forAn ∈Sn, extends to a smooth function of compact support on Sn. The iterated integral
Z
Sn
n Z
Sm
e ωn,ψm h
Φ, ψ;An
Bm
iψh
−Tr(BmwmCmwm)i dBm
o· (35)
×[An, Cm]ψ
Tr(wmCm−1wmnmA−1n tnm) ψ
−Tr(AnwnDnwn) dAn
is therefore convergent. It is equal to (36) |2|12mn·γ(1,ψ)¯ mn·ωenm,ψ¯h
ˇ
Φ,ψ;¯ Cm
Dn
i .
Proof. — First we remark that from the right hand side of (32) it is easily observed that for a fixed An∈Sn the function
Bm7−→ωmnh
Φ, ψ;An
Bm
i
is smooth and of compact support onSm. Therefore, for a fixedCm∈Smthe function Θ defined by
(37) Θ(Dn) =|2|12mnγ(1,ψ)¯ mn·ωenm,ψ¯h
ˇ
Φ,ψ;¯ wmCmwm
wnDnwn
i , is in Cc∞(Sn). Consider the partial Fourier transform with respect to Bm
(38)
Z
Sm
e ωmnh
Φ, ψ;An
Bm
iψ
Tr(−CmBm) dBm.
Expanding along (32) it becomes after a change of variablesBm7→Bm−tXAnX γmn(−An,1m, ψ)· |detAn|12m·
(39)
× Z
Φh An AnX
tXAn Bm
iψ[Tr(2nmX) + Tr(CmtXAnX)]dX ψ
Tr(−CmBm) dBm
o
and after another change of variablesX7→A−1n X it becomes γmn(−An,1m, ψ)|detAn|−12m
(40) Z
ΦhAn X
tX Bm
iψ
Tr(2nmA−1n X) + Tr(CmtXA−1n X) dX ψ
−Tr(BmCm) dBm. Applying Weil’s formula (17) withP = 2nmA−1n , we see that this is equal to
|2|12mnγnm(A−1n , Cm, ψ)γmn(−An,1m, ψ)|detCm|−12n (41)
× Z
ΦAn Y
tY Bm
ψ
−2 Tr(XY) dY ψ
−Tr(Cm−1(X+nmA−1n )An(tX+A−1n tnm)) dX ψ
−Tr(BmCm) dBm.
We now use Lemma 2.2. After expanding (41) it becomes:
|2|12mn·γ(1,ψ)¯ mn·γmn(−Cm,1n,ψ)¯ · |detCm|−12n (42)
ψ
−Tr(Cm−1nmA−1n tnm)
[An, Cm]
× Z
ΦAn Y
tY Bm
ψ
−2 Tr(XY)−Tr(Cm−1XAntX)
−2 Tr(Cm−1Xtnm)
dYdX·ψ
−Tr(BmCm) dBm. We showed so far that for a fixedCm∈Sm, the function Ψ onSn defined by
Ψ(An) = [An, Cm]·ψ
Tr(Cm−1nmA−1n mn)
· (43)
× Z
Sm
e ωmn,ψh
Φ, ψ;An
Bm
i ψ
−Tr(BmCm) dBm
is equal to
|2|12mn·γ(1,ψ)¯ mn·γnm(−Cm,1n,ψ)¯ · |detCm|−12n
(44) Z
ΦAn Y
tY Bm
ψ
−2 Tr(XY)−Tr(Cm−1XAntX)
−2 Tr(Cm−1Xtnm)−Tr(CmBm)
dYdXdBm. Changing variablesX 7→CmX we get
Ψ(An) =|2|21mn·γ(1,ψ)¯ mn·γnm(−Cm,1n,ψ)¯ · |detCm|12n (45)
× Z
ΦAn Y
tY Bm
·ψ
−2 Tr(CmXY)−Tr(XAntXCm)
−2 Tr(Xtnm)−Tr(BmCm)
dYdXdBm.
Next we expand Θ(Dn):
Θ(Dn) =|2|12mn·γ(1,ψ)¯ mn·γmn(−Cm,1n,ψ)¯ · |detCm|12n (46)
× Z
Mm×n(F)
ˆ
Φh wm+n
wmCmwm wmCmwmX
tXwmCmwm wnDnwn+tXwmCmwmX
wm+n
i ψ
−2 Tr(mnX) dX
=|2|12mnγ(1,ψ)¯ mn·γnm(−Cm,1n,ψ)¯ · |detCm|12n
× Z
Mm×n(F)
ˆ
ΦhDn+wntXwmCmwmXwn wntXwmCm
CmwmXwn Cm
i
ψ[−2 Tr(mnX)]dX.
After a change of variablesX 7→wmXwn,Θ(Dn) becomes
|2|12mn·γ(1,ψ)¯ mn·γnm(−Cm,1n,ψ)¯ · |detCm|12n· (47)
× Z
Mm×n(F)
ˆ
ΦhDn+CtXCmX tXCmmX Cm
iψ
−2 Tr(Xtnm) dX where we use the fact that
Tr(mnwmXwn) =X1,n= Tr(Xtnm).
Expanding further we have
Θ(Dn) =|2|12mn·γ(1,ψ)¯ mn·γnm(−Cm,1n,ψ)¯ · |detCm|12n· (48)
× Z
Φhsn Y
tY Bm
i ψh
−Trsn Y
tY Bm
Dn+tXCmX tXCm
CmX Cm
i dYdsndBmψ
−2 Tr(Xtnm) dX
=|2|12mn·γ(1,ψ)¯ mn·γnm(−Cm,1n,ψ)¯ · |detCm|12n
× Z
Φhsn Y
tY Bm
iψ
−Tr(sntXCmX)−Tr(snDn) dsn
ψ
−Tr(BmCm)−2 Tr(CmXY)
dYdBmψ
−2 Tr(Xtnm) dX.
Using Fourier inversion we then see that Θ(−Ab n) =
Z
Sn
Θ(Dn)ψ
Tr(AnDn) dAn
(49)
=|2|12mnγ(1,ψ)¯ mnγnm(−Cm,1n,ψ)|¯ detCm|12n
× Z
ΦhAn Y
tY Bm
i ψh
−Tr(AntXCmX)i ψ
−Tr(BmCm)−2 Tr(CmXY)
dYdBm
ψ[−2 Tr(Xtnm)]dX.
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Comparing with (45) we get that
(50) Θ(−Ab n) = Ψ(An)
for all An ∈ Sn. This proves the first part of the proposition. Furthermore, regarding Ψ now as its extension to Sn and writing explicitly the equality Ψ(Db n) = Θ(Dn) we have for allCm∈Sm, Dn∈ Sn
Z
Sn
h Z
Smωen,ψm h
Φ, ψ;An
Bm
i ψ
−Tr(BmCm) dBm
i (51)
[An, Cm]ψ
Tr(Cm−1nmA−1n tnm) ψ
−Tr(AnDn) dAn
=|2|12mn·γ(1,ψ)¯ mn·ωem,n ψ¯h
ˇ
Φ,ψ;¯ wmCmwmwnDnwn
i.
Observing that [An, Cm] = [An, wmCmwm] we get the proposition by replacing (Cm, Dn) with (wmCmwm, wnDnwn).
We end this chapter with a reduction formula that we will need for the proof of the main theorem. Let Φ∈Cc∞(Sm+n) and define onSn× Smthe function (52) Ξ(An, Bm) =ωnmh
Φ, ψ;An
Bm
i .
Associated with the action ofNn×NmonSn× Sm, we consider the generalized Kloosterman integral
(53) ωh
Ξ, ψ; (An, Bm)i
= Z
Ξ(tu1Anu1,tu2Bmu2)θn(u1)θm(u2)du1du2. For a relevant elementx∈ Sm+n of the formx= xnxm
, wherexn∈Sn and xm∈ Smare relevant, we have
(54) ω[Φ, ψ;x] =ω
Ξ, ψ,(xn, xm) .
4. Proof of the main theorem
We prove the functional equation by induction on n, the casen= 1 being simply the definition of the Fourier transform. For a fixeda1∈F×, let
Ψ(An−1) = [a1, An−1]ψ
Tr(a−11 A−1n−1t) (55)
× Z
e ωn−1,ψ1 h
Φ, ψ;An−1
pn
i
ψ[−pna1]dpn, Θ(Dn−1) =|2|12(n−1)·γ(1,ψ)¯ n−1·ωe1,n−1ψ¯h
ˇ
Φ,ψ;¯ a1Dn−1
i . (56)
Applying Proposition 3.1, with (1, n−1) in the role of (m, n), we get that Ψ extends to Sn−1. In fact Ψ,Θ ∈ Cc∞(Sn−1) and Θ =
ˇ
Ψ. Let a(1) =diag(a2, . . . , an) be relevant inSn−1. By induction applied to Ψ we have Kψ,n−1ωeψ[Ψ, ψ;·]
(a(1)) (57)
=|2|12(n−1)(n−2)·γ(1,ψ)¯ 12(n−1)(n−2)·ωeψ¯
Θ,ψ;¯ a(1) . By (54) and (19) we have
ω[Θ,ψ;¯ a(1)] =|2|12(n−1)·γ(1,ψ)¯ n−1·γn−11 (−a1,1n−1,ψ)¯ · |a1|12(n−1) (58)
×ω
ωn−11 [
ˇ
Φ,ψ;¯ ·],ψ; (a¯ 1, a(1))=|2|12(n−1)·γ(1,ψ)¯ n−1·γ(−a1,ψ)¯ n−1· |a1|12(n−1)·ω[
ˇ
Φ,ψ;¯ a1, . . . , an] therefore the right hand side of (57) satisfies|2|12(n−1)(n−2)·γ(1,ψ)¯ 12(n−1)(n−2)·ωeψ¯[Θ,ψ;¯ a(1)] (59)
=|2|12n(n−1)·γ(1,ψ)¯ 12n(n−1)·Γn−1(−a(1),ψ)¯ ·σn−1(a(1))12
×γ(−a1,ψ)¯ n−1· |a1|12(n−1)·ω[
ˇ
Φ,ψ;¯ a1, . . . , an]=|2|12n(n−1)·γ(1,ψ)¯ 12n(n−1)·ωeψ¯[
ˇ
Φ,ψ;¯ a1, . . . , an].Next we treat the left hand side of (57). We start with the computation of the Kloosterman integral of Ψ atx= diag(p1, . . . , pn−1).Note that
[a1,tuxu]·ψ
Tr(a−11 u−1x−1tu−1t)
=n−1Y
j=1
[a1, pj] ψh 1
a1pn−1
i
for allu∈Nn−1is constant on the orbit ofx. So e
ωψ[Ψ, ψ;p1, . . . , pn−1] = Γn−1(−p1, . . . ,−pn−1, ψ) (60)
×σn−1(p1, . . . , pn−1)12n−1Y
j=1
[a1, pj] ψh 1
a1pn−1
i
× ZZ
e ω1n−1,ψh
Φ, ψ;tuxu pn
i
ψ[−pna1]dpnθ(u)du.
Since Φ is smooth and of compact support, it is easy to see from the right hand side of (32), that forφ∈Cc∞(F×) the function
(61) (An−1, pn)7−→φ(detAn−1)·ωe1n−1,ψh
Φ, ψ;An−1
pn
i
is smooth of compact support on GL(n−1, F)×F. Since the orbit tuxu, u ∈ Nn−1 has fixed determinant, the double integral in (60) is absolutely
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convergent and we may switch order of integration. From (54) we get Γn−1(−p(n), ψ)·σn−1(p1, . . . , pn−1)12
(62)
× Z
e ωn−1,ψ1 h
Φ, ψ;tuxu pn
i θ(u)du
= Γn−1(−p(n), ψ)·σn−1(p1, . . . , pn−1)12 ·γ1n−1(−x,1, ψ)
×|detx|12 Z
ωn−11 h
Φ, ψ;tuxu pn
i θ(u)du
=eωψ[Φ, ψ;p1, . . . , pn].
By switching order of integration in the right hand side of (60) we get that e
ωψ[Ψ, ψ;p1, . . . , pn−1] (63)
=n−1Y
j=1
[a1, pj] ψh 1
a1pn−1
i Z e ωψh
Φ, ψ;p1, . . . , pn
i
ψ[−pna1]dpn. Finally, we see that
Kψ,n−1ωeψ[Ψ, ψ;·] (a(1)) (64)
=Z n−1Y
j=1
[a1, pj] ψh 1
a1pn−1
i Z eωψ
Φ, ψ;p1, . . . , pn
ψ[−pna1]dpn
×ψh
−
n−1X
i=1
pian+1−i+
n−2X
i=1
1 pian−i
in−1Y
i=2 n−iY
j=1
[ai, pj]
dpn−1· · ·dp1
= Kψ,nωeψ[Φ, ψ;·]
(a1, . . . , an).
The theorem now follows from (57), (59) and (64).
Corollary 1.3 is now immediate sinceγ(1, ψ)·γ(1,ψ) = 1.¯ 5. A formula for the smallest orbits Letκ(n) = 12n(n+ 1)−1.
Proposition 5.1. — ForΦ ∈ Cc∞(Sn), the function φ(a) =ω[Φ, ψ, wna] is a smooth function of compact support on F×. Furthermore, it satisfies the functional equation
(65) |a|κ(n)·ω[Φ, ψ, wna] = Z
ωh
ˇ
Φ,ψ;¯ −wn−1a−1 b
i db.
Proof. — We can writeφas follows (66) ω[Φ, ψ, wna] =
Z
Φ(m0)ψXn
i=2
xi,n+2−i
⊗
i+j≥n+2dxi,j
where
m0i,j=
0 if i+j≤n, a if i+j=n+ 1, axi,j if i+j≥n+ 2, andxi,j=xj,i ifi+j≥n+ 2. Let
κ1(n) = 1
4n2 if n−even,
1
4(n2−1) if n−odd, κ2(n) =κ1(n−1) +n−1.
Thenκ(n) =κ1(n) +κ2(n).After a change of variables (66) can be writen as (67) ω[Φ, ψ, wna] =|a|−κ1(n)
Z
Φ(m)ψ a−1
Xn i=2
xi,n+2−i
⊗
i+j≥n+2dxi,j
where
mi,j=
0 ifi+j ≤n, a ifi+j =n+ 1, xi,j ifi+j ≥n+ 2,
and xi,j = xj,i if i+j ≥ n+ 2. The smoothness of φ follows from the fact that Φ is smooth. Also Φ(m) = 0 for large enough|a|and for allxi,j as above, since Φ is of compact support. Thereforeφ(a) vanishes when|a|is sufficiently large. Let X be the n×n symmetric matrix with 0 in the (i, j)-th entry wheneveri+j ≤n+ 1 and xi,j in the (i, j)-th entry whenever i+j ≥n+ 2.
LetZ be a similar variable matrix with entries zi,j, thus
(68) m=
a
· x2,n
· ... a x2,n · · · xn,n
, X=
0
· x2,n
· ... 0x2,n· · · xn,n
, Z=
0
· z2,n
· ... 0z2,n· · · zn,n
.
Since Φ is smooth and of compact support, there are integers` < ksuch that Φ(m) = 0 unlessxi,j ∈℘`and Φ(m+Z) = Φ(m) wheneverzi,j ∈℘k for alli, j.
Therefore, φ(a) = X
xi,j∈℘`/℘k
Φ(m) Z
zi,j∈℘k
ψ a−1
Xn i=2
(xi,n+2−i+zi,n+2−i)
⊗
i+j≥n+2dzi,j. This integral factors, for example, through the integral R
℘kψ(a−1z2,n)dz2,n
which vanishes whenever |a|is small enough. So we get also thatφ(a) = 0 for
|a| sufficiently small. We can write Z
ωh
ˇ
Φ,ψ;¯ −wn−1a−1 bidb=
Z
ˇ
Φ(p0)ψ− Xn i=1
yi,n+1−i
⊗
2n>i+j≥n+1dyi,jdb
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where
p0i,j =
0 if i+j≤n−1,
−a−1 if i+j=n,
−a−1yi,j if 2n > i+j ≥n+ 1, b if i+j= 2n,
andyi,j=yj,iifi+j≥n+ 1. After a change of variables this becomes Z
ωh
ˇ
Φ,ψ;¯ −wn−1a−1 b
i
db=|a|κ2(n)
Z
ˇ
Φ(p00)ψ aXn i=1
yi,n+1−i
i+j≥n+1⊗ dyi,j
where
p00i,j=
0 if i+j≤n−1,
−a−1 if i+j=n, yi,j if i+j≥n+ 1, andyi,j=yj,iifi+j≥n+ 1 which is the same as writing
Z ωh
ˇ
Φ,ψ;¯ −wn−1a−1 b
i db (69)
=|a|κ2(n)
Z
ˆ
Φ(p)ψ aXn i=1
yi,n+1−i
i+j≤n+1⊗ dyi,j
where
pi,j=
yi,j if i+j≤n+ 1,
−a−1 if i+j=n+ 2, 0 if i+j≥n+ 3,
andyi,j =yj,iifi+j≤n+ 1. Next letY be then×nsymmetric matrix with Yi,j=
yi,j if i+j≤n+ 1, 0 if i+j≥n+ 2.
LetA0 be defined byp=Y+A0 and letXandZ be matrix variables as defined in (68). Thus
p=
y11 · · · y1n
... −a−1
· y1n −a−1
, Y =
y11· · · y1n
... 0
· y1n 0
, A0=
0 · · · 0 ... −a−1
· 0−a−1
.
Using Fourier inversion formula for the functionΦ(Y
ˆ
+(·)) in the (i, j)-th entries for alli+j≥n+ 2, we obtain from (69) that|a|−κ2(n) Z
ωh
ˇ
Φ,ψ;¯ −wn−1a−1 b
idb
=
Z Φ(Yb +Z)ψ
−Tr(XZ) dZ ψ
Tr(XA0) dXψ
a Xn i=1
yi,n+1−i
⊗
i+j≤n+1dyi,j. LetA=wna, one easily observes now that Tr(ZA) = Tr(Y X) = 0 and there- fore Tr((Y +Z)(A−X)) = Tr(Y A−ZX) =aPn
i=1yi,n+1−i−Tr(XZ), so we get by the Fourier inversion formula applied to
ˆ
Φ that|a|−κ2(n) Z
ωh
ˇ
Φ,ψ;¯ −wn−1a−1 b
idb (70)
=
Z Φ(Yb +Z)ψ
Tr((Y +Z)(A−X))]dYdZ ψ[Tr(XA0) dX
= Z
Φ(A−X)ψ
Tr(XA0) dX=
Z
Φ(A+X)ψ
Tr(−XA0) dX.
Butm=A+X and Tr(−XA0) =a−1Pn
i=2xi,n+2−i, so comparing with (67), the right hand side of (70) is equal to |a|κ1(n)ω[Φ, ψ;wna].
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