twisted The weighted fundamental lemma for the transfer of automorphic forms from GSp(4) to GL(4)
Thesis by
David Whitehouse
In Partial Fulllment of the Requirements for the Degree of Doctor of Philosophy
California Institute of Technology Pasadena, California
2005 (Defended May 20, 2005)
ii
c
2005
David Whitehouse All Rights Reserved
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To Joan and Olive
iv
Acknowledgements
It is a pleasure to thank my advisor, Dinakar Ramakrishnan, for his guidance and support throughout my time in graduate school. I am grateful to Mladen Dimitrov, Matthias Flach and Hee Oh for serving on my defense committee. I also thank Jiu-Kang Yu for answering my questions. This work could not have been undertaken without nancial assistance. I thank Caltech for their support throughout the last ve years, Steve and Rosemary Johnson for support through a Scott Russell Johnson Prize and the Fields Institute, with funding from the National Science Foundation, for hosting me in 2003. The Caltech mathematics department has been a very friendly home. Many thanks are due to the administrative sta for all their help and my fellow graduate students from whom I have beneted both personally and professionally. I cannot forget the lads back home. I thank you all for ensuring that I havent forgotten where I come from. Finally, I thank my family for their love and encouragement.
v
Abstract
We prove the twisted weighted fundamental lemma for the group GL(4) GL(1) relative to a certain outer automorphism , which yields GSp(4) as a twisted endoscopic group. This version of the fundamental lemma is needed to stabilize the twisted trace formula for the pair (GL(4) GL(1), ). This stabilized twisted trace formula is required for Arthurs classication of the discrete spectrum of GSp(4) in terms of automorphic representations of GL(4).
vi
Contents
Acknowledgements Abstract 1 Introduction 1.1 1.2 1.3 The trace formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iv v 1 1 4 5 10 10 11 11 12 13 14 15 15 16 18 18 19 21 23 24 26 26
Stabilization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Transfer from GSp(4) to GL(4) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2 Preliminaries 2.1 2.2 2.3 2.4 2.5 2.6 Twisted conjugacy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Parabolic subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . GSp(4) and Sp(4) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Weighted orbital integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Endoscopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3 The twisted weighted fundamental lemma 3.1 3.2 The twisted weighted fundamental lemma . . . . . . . . . . . . . . . . . . . . . . . . Our case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4 Endoscopic groups 4.1 4.2 4.3 4.4 4.5 Twisted endoscopic groups for GL(4) GL(1) . . . . . . . . . . . . . . . . . . . . . . Twisted endoscopic groups for the (2,2) Levi . . . . . . . . . . . . . . . . . . . . . . Twisted endoscopic groups for the (1,2,1) Levi . . . . . . . . . . . . . . . . . . . . . Twisted endoscopic groups for the diagonal Levi . . . . . . . . . . . . . . . . . . . . Endoscopic groups for GSp(4) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5 Weight functions 5.1 Notations and denitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
vii 5.2 Twisted weight functions 5.2.1 5.2.2 5.2.3 5.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 28 32 35 43 43 45 48 50 51 53 57 57 59 61 61 62 62 63 65 70 78 83 86 87 90
The (2,2) Levi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The (1,2,1) Levi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The diagonal Levi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Weight functions for GSp(4) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3.1 5.3.2 5.3.3 The Siegel Levi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The Klingen Levi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The diagonal Levi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.4 5.5 5.6
Other groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Normalization of volumes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Weighted orbital integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6 The fundamental lemma for the (2,2) Levi 6.1 6.2 6.3 Twisted integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Explicit statement of the fundamental lemma . . . . . . . . . . . . . . . . . . . . . . Computation of P , P1 and P2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3.1 6.3.2 6.3.3 6.4 Calculation of P . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Calculation of P1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Calculation of P2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Reduction in case 1 Reduction in case 2 Reduction in case 3 Proof when M = N . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Proof of the fundamental lemma for split tori . . . . . . . . . . . . . . . . . . . . . . 6.4.1 6.4.2 6.4.3 6.4.4
6.5
Proof of the fundamental lemma for elliptic tori . . . . . . . . . . . . . . . . . . . . . 6.5.1 6.5.2 6.5.3 6.5.4 6.5.5 Proof when b is a unit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Reduction when |b| | det A 1| . . . . . . . . . . . . . . . . . . . . . . . . .
Proof when |b| = | det A 1| . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102 Reduction when | det A 1| |b2 D| . . . . . . . . . . . . . . . . . . . . . . . 108 Proof when |b2 D| | det A 1| < |b| . . . . . . . . . . . . . . . . . . . . . . . 115 120
7 The fundamental lemma for the (1,2,1) Levi I 7.1 7.2 7.3
Stable conjugacy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120 Statement of the fundamental lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 Proof of the fundamental lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 7.3.1 7.3.2 s equal to the identity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124 s central . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
viii 7.3.3 7.3.4 s diagonal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128 s elliptic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 143
8 The fundamental lemma for the (1,2,1) Levi II 8.1 8.2
Statement of the fundamental lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . 143 Proof of the fundamental lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144 8.2.1 8.2.2 s equals the identity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 s not equal to the identity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 152
9 The fundamental lemma for the diagonal Levi 9.1 9.2
Statement of the fundamental lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . 152 Proof of the fundamental lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153 9.2.1 s equal to the identity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153 9.2.1.1 9.2.1.2 9.2.1.3 9.2.1.4 9.2.2 Region 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156 Region 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158 Region 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167 Putting it altogether . . . . . . . . . . . . . . . . . . . . . . . . . . . 173
s not equal to the identity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174 181 189
10 Some p-adic integrals Bibliography
1
Chapter 1
Introduction
We give a brief outline of the trace formula and the problem of stabilization. Introductions to the stabilization of the trace formula, from which much of this section is taken, can be found in [Art97] and in the introduction to [Art02b].
1.1
The trace formula
We take G to be a connected reductive algebraic group dened over a number eld F . We let A denote the ring of adeles of F . Via the diagonal embedding G(F ) embeds discretely in G(A). The group G(A) of adelic points of G acts on the Hilbert space L2 (Z(A)G(F ) \ G(A)) by right translation; here Z denotes the center of G. Automorphic representation theory is concerned with the representations of G(A) that occur in this action. The trace formula is a powerful tool in this study. Suppose for the moment that G/F is anisotropic modulo the center, that is Z(A)G(F ) \ G(A) is compact, for example we could take G to be the group of units in a denite quaternion algebra over Q. In this case the representation of G(A) on L2 (Z(A)G(F ) \ G(A)) decomposes discretely. That is we have L2 (Z(A)G(F ) \ G(A)) = m
(G(A))
as a direct sum over irreducible representations of G(A) with nite multiplicities m Z0 . Let
f Cc (G(A)) be a smooth function on G(A) with compact support; we have a linear map
R(f ) : L2 (Z(A)G(F ) \ G(A)) L2 (Z(A)G(F ) \ G(A)) given by integrating f against the action of G(A), that is (R(f ))(x) =
G(A)
f (y)(xy) dy.
2 The map R(f ) is of trace class and the trace formula, due in this case to Selberg, gives two expansions for the trace of R(f ). On the one hand we can write the trace of R(f ) as a sum over representations of G(A), tr R(f ) =
(G(A))
m tr((f )).
On the other we have tr R(f ) =
(G(F ))
vol(Z(A)G (F ) \ G (A))
f (g 1 g) dg,
G (A)\G(A)
a sum over conjugacy classes in G(F ) of orbital integrals; here G denotes the connected component of the centralizer of . In the case that G is not anisotropic modulo the center the arguments above immediately break down since the action of G(A) on L2 (Z(A)G(F ) \ G(A)) does not decompose discretely. The cusp forms do, however, lie in the discrete spectrum and we have a trace formula due to Arthur that applies to any reductive group G; see [Art78] and [Art80]. The trace formula gives an identity of distributions, Jo (f ) =
oO X
J (f )
for f a smooth function on G(A). The sum on the left is over conjugacy classes in G(F ) while the sum on the right is over automorphic representations of G and its Levi subgroups. The rst renement of Arthur was to make the trace formula invariant, that is to write each of the expansions above as a sum of invariant distributions. We take the invariant trace formula from [Art88a]. The invariant trace formula gives two expansions for a certain linear form I(f ); we have a geometric expansion I(f ) =
M G M |W0 ||W0 |1
aM ()IM (, f )
(M (F ))
given as a sum over conjugacy classes of Levi subgroups M of G. The terms IM (, f ) are built out of weighted orbital integrals. And we have a spectral expansion I(f ) =
M G M |W0 ||W0 |1
aM ()IM (, f ) d
(M )
given in terms of data associated to representations of the Levi subgroups M . Some of the terms in the above expansions are entirely similar to the case that G/F is anisotropic.
3 On the geometric side we have, for G(F ) which is semisimple and elliptic, IG (, f ) =
G (A)\G(A)
f (g 1 g) dg.
On the spectral side we have, for an irreducible representation of G(A) which occurs discretely in the representation of G(A) on L2 , aG () equal to the multiplicity of in the discrete spectrum and IG (, f ) = tr((f )). The most powerful applications of the trace formula, for example to questions of functoriality, come about when the trace formula is not used in isolation. Identities between the geometric sides of the trace formula for dierent groups produce corresponding identities between the spectral sides out of which one can hope to prove relationships (functorialities) between the automorphic representations of the groups in question. Suppose now that we have two groups G and G . In order to match up the geometric sides of the trace formula for G and G one needs to be able to 1. transfer conjugacy classes between the two groups, and 2. transfer functions between the two groups given by identities between orbital integrals. We note that it is enough to treat 2 for functions of the form
v
fv given as a product over the local
groups, this turns the problem of transferring functions into a local one. This is precisely the strategy carried out by Jacquet and Langlands in [JL70] to prove functoriality for G = D , the group of units in a quaternion algebra D, and G = GL(2). The characteristic polynomial for elements of G and G gives an injection i : (G(F )) (G (F )). Jacquet and Langlands also dene a correspondence f=
from Cc (G(A)) to Cc (G (A)), such that
fv f =
fv
v
IG ( , f ) = IG (, f ) if = i() and is zero if (G (F )) does not lie in the image of i. Using the trace formula they then prove functoriality between G and G . There is a serious obstacle in applying this argument to prove functoriality for other groups; for example between a group and its quasi-split inner form. The transfer of conjugacy classes between G and G is only dened up to conjugacy over the algebraic closure of F . For the cases considered by Jacquet and Langlands this is not a problem; elements of GL(2, F ) which are conjugate in GL(2, F ) are already conjugate in GL(2, F ). However, in general, the notion of stable conjugacy (conjugacy
4 over F ) is a strictly coarser equivalence relation. This is already apparent in the case of SL(2); see [LL79]. In the same way one can only hope to transfer functions between groups up to their values on stable orbital integrals, that is, the sum of the orbital integrals over the conjugacy classes within a stable conjugacy class. One would therefore like to be able to write I(f ) as a sum of stable distributions parameterized by stable conjugacy classes in G(F ). The rst problem is that I(f ) is, in general, not stable. To see this let G(F ) be semisimple and elliptic. We take a nite set of places S of F . Suppose that for each v S we choose an element v G(Fv ) such that and v are conjugate in G(F v ). We can then dene = (v ) G(A) by setting v = G(Fv ) for v S. In order for the distribution I(f ) to be stable we need the orbital integral IG ( , f ) to appear in the geometric expansion of I(f ). But this only happens if is conjugate in G(A) to an element of G(F ), and in general one cannot guarantee that this will happen. Therefore the distribution I(f ) is not stable.
1.2
Stabilization
In [Lan83], Langlands suggested a stabilization of the geometric side of the trace formula, at least for the M = G terms, of the form I(f ) =
H
(G, H)S H (f H ).
Here the sum is over a certain family {H} of quasi-split groups, known as elliptic endoscopic groups, attached to G via dual group considerations. The coecients (G, H) are explicitly dened and the distributions S H are stable distributions, which depend only on the group H. The existence of the functions f H on H depend on certain local conjectures; see below. In the case that G is quasi-split one can write this as I(f ) = S G (f ) +
H=G
(G, H)S H (f H ).
Each of the groups H = G have dimension strictly smaller than G, and one can view this as writing I(f ) as a main stable term S G (f ) together with an explicit error term coming from the proper elliptic endoscopic groups of G. Such a decomposition for the M = G terms in the trace formula was achieved by Langlands [Lan83] and Kottwitz [Kot86] subject to these local conjectures which we now describe. In order to
5 obtain the existence of the function f H above it is sucient to consider functions of the form f=
v
fv
on G(A), where fv is a smooth compactly supported function on G(Fv ), and fv is the characteristic function of a xed hyperspecial maximal compact subgroup of G(Fv ) for almost all v. To obtain
H H the function f H we need to obtain functions fv on H(Fv ) for each v. The conjectural functions fv
need to satisfy an identity between their stable orbital integrals on H and a certain unstable linear combination of orbital integral of f . In the case that v is archimedean, this has been achieved by
H work of Shelstad [She82], for non-archimedean v the existence of fv is conjectural. This conjecture
is known in a limited number of cases. The fundamental lemma asserts that for almost all v we
H can take fv to be the characteristic function of a xed hyperspecial maximal compact subgroup of
H(Fv ), which guarantees that the function fH =
v H fv
is a function on H(A). Furthermore, Waldspurger in [Wal97] has shown that the analog of the fundamental lemma for the Lie algebra of G implies the corresponding transfer conjecture. Building on the work of Langlands and Kottwitz, Arthur [Art02b] has stabilized, subject to a fundamental lemma for weighted orbital integrals, the remaining terms in the geometric expansion for I(f ). In this thesis we will be interested in the stabilization of the twisted trace formula. This trace formula applies to a pair (G, ) of a connected reductive group G and an automorphism of G. Roughly speaking, the twisted trace formula for (G, ) gives two expansions for the trace of the operator R()R(f ) on L2 (G(F ) \ G(A)), where R() acts on L2 by composition with 1 . The stabilization of the elliptic part of the twisted trace formula has been completed, subject to appropriate local conjectures for twisted orbital integrals, by Kottwitz and Shelstad in [KS99] and Labesse in [Lab04]. The stabilization of the remaining terms in the twisted trace formula is work in progress; however, it relies on a fundamental lemma for twisted weighted orbital integrals given in [Art02a]; see also Chapter 3.
1.3
Transfer from GSp(4) to GL(4)
We now specialize to the case of interest in this thesis. In [Art04] Arthur gives, subject to the twisted weighted fundamental lemma proven in this thesis, a classication theorem for the automorphic discrete spectrum of GSp(4); this theorem applies to both generic and non-generic forms and
6 in particular applies to representations attached to holomorphic Siegel cuspforms. This classication includes a parameterization of the representations of the local groups GSp(4, Fv ) into packets together with a decomposition of the discrete spectrum of GSp(4) in terms of automorphic representations of GL(4). This also gives a multiplicity formula for representations which appear in the discrete spectrum of GSp(4). Much is known about the automorphic representation theory of GL(4). Thus a knowledge of this transfer allows one to obtain important information about automorphic representations of GSp(4) from the known facts about GL(4). From the analytic point of view one is interested in properties of the L-functions attached to cuspforms on GSp(4), for example the location of poles. On the algebraic side one is interested in properties of the -adic representations attached to holomorphic Siegel cuspforms. We hope to explore some of the consequences of this transfer for these associated objects. The results of [Art04] are achieved by a comparison of the stable trace formula for GSp(4) with a stable twisted trace formula for GL(4) GL(1); see also [KS02, Remark 9.3]. This work is part of a program of Arthurs to prove functoriality between classical groups and the appropriate general linear groups subject to the local conjectures required in the stabilization of the necessary trace formulas. For GSp(4) the fundamental lemma for invariant orbital integrals is proven in [Hal97]; see also [Wei94]. The weighted fundamental lemma in [Art02b] required for the stabilization of the full trace formula does not apply to GSp(4) since its proper Levi subgroups are products of general linear groups, and hence do not possess proper elliptic endoscopic groups. Therefore, all the local conjectures required for the stabilization of the trace formula for GSp(4) have been established. For the stabilization of the twisted trace formula for GL(4) GL(1) and the automorphism given in Section 3.2 below, the twisted fundamental lemma for invariant orbital integrals is proven in [Fli99]. Flickers proof is for elds of odd residual characteristic, however, this is sucient for global applications. A weighted variant of the twisted fundamental lemma, stated in [Art02a], is also needed for the stabilization of the full twisted trace formula. This is needed since there are Levi subgroups of GL(4) GL(1) that have elliptic twisted endoscopic groups. It is this fundamental lemma which we prove in this thesis, we again restrict ourselves to local elds of odd residual characteristic. This is the rst such twisted weighted fundamental lemma to be proven when the twisting is of a group theoretic nature, moreover the geometric methods used to prove the unweighted fundamental lemma, for example in [LN04], do not apply to the weighted fundamental lemma. In Chapter 2 we give some denitions and results used throughout this thesis, we also give enough details in order to give the statement of the twisted weighted fundamental lemma in Chapter 3. The
7 conjectural identity is given by the formula
G M,K ( , k)rM (k) = kGreg (M (F )) G EM (G)
M (G, G )sG ( ). M
The left hand side consists of a nite linear combination of twisted weighted orbital integrals on the group G with respect to the Levi subgroup M . We take M to be an elliptic twisted endoscopic group for M ; the right hand side is then a nite linear combination of stable weighted orbital integrals on certain groups G that contain M as a Levi subgroup. From Chapter 4 onwards we specialize to the twisted weighted fundamental lemma for G equal to GL(4) GL(1). We begin in Chapter 4 by determining all endoscopic groups that appear in the statement of the twisted weighted fundamental lemma, and in Chapter 5 we compute the necessary weight functions, which appear in our weighted orbital integrals. As above, the twisted weighted fundamental lemma applies to a pair (M, M ) of a Levi subgroup M of G and an unramied elliptic twisted endoscopic group M for M . There are four such pairs given in the table below, here E denotes the unramied quadratic extension of the local nonarchimedean eld F . M (GL(2) GL(2)) GL(1) (GL(1) GL(2) GL(1)) GL(1) (GL(1) GL(2) GL(1)) GL(1) GL(1)4 GL(1) M GL(2) GL(1) GL(2) GL(1) ResE/F (GL(1)) GL(1) GL(1)3
We now outline the proof of the fundamental lemma for each pair (M, M ). We take F to be a local eld of characteristic zero. We let R denote the ring of integers in F . We denote by q the cardinality of the residue eld of F that for now we assume is odd and greater than three. In Chapter 6 we prove the fundamental lemma for the rst pair. We begin by writing both sides of the fundamental lemma in this case as untwisted orbital integrals on GL(2, F ). The identity to be proven then takes the form F L(A) : L(A) = R(A) indexed by elements A GL(2, F ). Moreover, since both sides vanish if the conjugacy class of A in GL(2, F ) does not intersect GL(2, R) we may assume that A GL(2, R). We split the proof into two cases depending on whether A lies in a split or elliptic torus. In the former case we may assume that a d
is diagonal. We nd that both L(A) and R(A) depend only on |a 1|, |d 1|, |a d| and |ad 1|. Since we are assuming that F has odd residual characteristic we have the following three cases
A=
8 Case 1. q M = |ad 1| = |a d| = |d 1| |a 1| = q N Case 2. q M = |a 1| = |d 1| = |ad 1| |a d| = q N Case 3. q M = |a 1| = |d 1| = |a d| |ad 1| = q N . In each case we denote L(A) (resp. R(A)) by L(M, N ) (resp. R(M, N )). In cases 1 and 3 we prove that qL(M, N + 1) L(M, N ) = qR(M, N + 1) R(M, N ) and in case 2 we prove that L(M, N + 1) L(M, N ) = R(M, N + 1) R(M, N ). In each case we exploit cancellations between the integrals on either side of F L(A) allowing us to readily compute the dierences. Thus the proof of the identity F L(A), when A lies in a split torus, is reduced to proving the identity under the assumption |ad 1| = |a d| = |d 1| = |a 1|. We then compute both sides of F L(A) under this assumption and show that they are equal. In the case that A lies in an elliptic torus we again reduce the proof to certain cases, which we then prove, by following a similar strategy. The proofs of the fundamental lemma for the Levi subgroup M = (GL(1) GL(2) GL(1)) GL(1) and both its unramied elliptic twisted endoscopic groups are given in Chapters 7 and 8. The proof uses the twisted topological Jordan decomposition which is described in Section 5.6. We can write any element with M (R) uniquely as = us = su, where s has nite order prime to q and u is topologically unipotent, i.e., uq I as n . Using this decomposition allows us to write the twisted weighted orbital integral at as an (untwisted) weighted orbital integral at u on the group Gs , the centralizer of s in G. The main part of the proof of the fundamental lemma is when s is the identity. In this case the twisted weighted orbital integrals become untwisted weighted orbital integrals on Sp(4). These integrals are of a type that appear on the right hand side of the fundamental lemma treated in Chapter 6. We are then able to use the calculations from there to prove the fundamental lemma for both pairs (M, M ). When s
n
9 is not the identity, the groups Gs have dimension strictly smaller than Sp(4) and the fundamental lemma can be readily veried in these cases. In Chapter 9 we prove the fundamental lemma for the diagonal torus in GL(4) GL(1). We again use the twisted topological Jordan decomposition. The main part of the proof comes down to proving an identity between weighted orbital integrals on Sp(4) with respect to the diagonal torus. We establish this identity by exploiting cancellations between the relevant integrals on Sp(4). We delay to Chapter 10 the computation of certain p-adic integrals that are needed in the proof of the fundamental lemma.
10
Chapter 2
Preliminaries
In this chapter we give a few preliminary denitions and results that will be used throughout this thesis, often without reference. We also give enough details so that the statement of the twisted weighted fundamental lemma, given in Chapter 3, makes sense.
2.1
Twisted conjugacy
Let G0 be a connected reductive group dened over F , a eld of characteristic zero. We take to be a quasi-semisimple automorphism of G0 . By denition this means that there exists a pair (B, T ) (not necessarily dened over F ) with B a Borel subgroup of G0 and T a maximal torus in B such that (B) = B and (T ) = T . Throughout we allow to be trivial in which case we recover the untwisted denitions. We assume further that is of nite order and form the semidirect product G+ = G0 take G to be the connected component G
0 0
and
G is semisimple, i.e., the automorphism of G0 given by Int() is quasi-semisimple in the sense above. The twisted conjugacy class of G0 is g 1 (g) : g G0 . We note that for g G0 we have
. We say that G is -semisimple if the element
g 1 g = g 1 (g),
and so the notion of twisted conjugacy of is equivalent to conjugacy of by elements of G0 , we will use these notions interchangeably. The twisted centralizer of G0 is ZG0 () = g G0 : g 1 (g) = .
11 We say that is strongly regular if ZG0 () is abelian. We denote by G the connected component of ZG0 ().
2.2
Parabolic subgroups
Having xed F by a parabolic subgroup of G0 we mean a parabolic subgroup dened over F . Similarly by a Levi subgroup of G0 we mean a Levi subgroup M 0 of a parabolic subgroup P 0 such that both M 0 and P 0 are dened over F . If M 0 is stable under we denote by M the connected component M 0 of M 0 . We say that M is strongly G0 -regular if is strongly regular
as an element of M and G = M . We denote by G0 the dual group of G0 ; see [Bor79, Section 2]. For a quasi-split group G0 there is a bijection between the parabolic subgroups of G0 and the parabolic subgroups of G0 the dual group of G0 ; see [Bor79, Section 3].
2.3
GSp(4) and Sp(4)
1 1 1 ,
We let J denote the matrix
1
and we set
Sp(4) = {g GL(4) : J t g 1 J 1 = g}, and GSp(4) = {g GL(4) : J t g 1 J 1 = g}, . The intersection with GSp(4) of the upper triangular Borel subgroup of GL(4) is a Borel subgroup of GSp(4). The proper parabolics of GSp(4) that contain this Borel subgroup are the Siegel parabolic, which has Levi decomposition g 1 t 1 aw g w x 1 r 1 s : g GL(2), a GL(1) , 1 r
12 where 1
and the Klingen parabolic which has Levi decomposition a g 1 1 a det g x 1 r : g GL(2), a GL(1) . 1 x 1 r s
w= 1
The intersection of each of these parabolic subgroups with Sp(4) is a parabolic subgroup of Sp(4), we refer to their intersection with Sp(4) by the same name. The dual group of GSp(4) is GSp(4, C) and under the bijection between parabolic subgroups of G and G the Siegel and Klingen parabolics are switched.
2.4
Weighted orbital integrals
We now assume that F is a local nonarchimedean eld of characteristic zero. Let M be a Levi subset of G. We x K a maximal compact subgroup of G0 (F ), which is admissible relative to M in the sense of [Art81, Section 1]. Associated to (G, M ) is a weight function vM : G+ (F ) C dened in [Art88b, Section 1]; see also Chapter 5. We do not give the denition here but note that vM is left M + (F ) invariant and right K invariant. Let M (F ) be strongly G0 -regular. We dene DG () = det(1 Ad())g/g , where Ad() denotes the adjoint action of on the Lie algebra g of G0 and g is the Lie algebra of G . Let f be a smooth complex valued function of compact support on G(F ). The weighted orbital integral of f is dened by JM (, f ) = |DG ()| 2
1
f (g 1 g)vM (g) dg.
G (F )\G0 (F )
The integral depends on the choice of a Haar measure on G (F ), however we suppress this from our notation. We will usually take the measure on G (F ) that gives its maximal compact sub-
13 group volume one. We also note that since G M 0 the function vM descends to the quotient G (F ) \ G0 (F ).
2.5
Endoscopy
WF the L-group of G0 . By duality
We continue with F local nonarchimedean of characteristic zero and we let WF denote the Weil group of F . Let G0 denote the dual group of G0 and L G0 = G0 we obtain an automorphism of G0 ; see [KS99, Section 1.2]. split group over F , H is a split extension of H by WF and : H L G0 is an L-homomorphism The denition of endoscopic datum (H, H, s, ) is found in [KS99, Section 2.1]. Here H is a quasi-
which maps H isomorphically onto ZG0 (s )0 , the connected component of the centralizer of s in G0 . There are many issues that are dealt with by Kottwitz and Shelstad in the theory of endoscopy that do not appear in our case. For us it will be sucient to consider ZG0 (s )0 together with a Galois action on ZG0 (s )0 given by a homomorphism : Gal(F /F ) ZG0 (s ). We use this homomorphism to form the L-group ZG0 (s )0 group with this L-group. Let H be a twisted endoscopic group for G0 . Then by [KS99, Theorem 3.3.A] we have a canonical map AH/G : Clss (H) Clss (G0 , ) from semisimple conjugacy classes in H(F ) to semisimple twisted conjugacy classes in G0 (F ). This map respects the action of Gal(F /F ). If we take T to be a maximal torus in G0 , which is stable under and TH , a maximal torus in H, then the map AH/G is given by a norm map N : T TH which factors through T /(1 )T . We say that H(F ) is strongly G0 -regular if its image under AH/G is a strongly regular WF and take H to be the quasi-split
twisted conjugacy class in G0 (F ). The stable conjugacy class of a strongly G0 -regular is the
intersection of H(F ) with the conjugacy class of in H(F ), it is a nite union of conjugacy classes in H(F ). Similarly the stable twisted conjugacy class of G0 (F ), which is strongly regular, is the intersection of G0 (F ) with the twisted conjugacy class of in G0 (F ); it is a nite union of twisted conjugacy classes in G0 (F ). Kottwitz and Shelstad dene a transfer factor (, ) in [KS99, Chapter 4]. The function (, ) depends only on the stable conjugacy class of and the twisted conjugacy class of in G0 (F ). Moreover (, ) is non-zero if and only if lies in AH/G (). Thus it follows that if we x then
14 there are only nitely many twisted conjugacy classes {} in G0 (F ) for which (, ) is non-zero.
2.6
Notation
In addition to the notation introduced above we x the following notation throughout this thesis. Throughout F will be a local nonarchimedean eld of characteristic zero. We let R denote the ring of integers in F and we let denote a uniformizer in R. We x the Haar measure on F that gives R volume one. We let q denote the order of the residue eld of F , which we take to have characteristic p. We use v and | | to denote the additive and multiplicative valuations on F , which
m we normalized so that v() = 1 and || = q 1 . We let UF denote the group of units in R and UF
denotes the subgroups of UF dened by U , if m = 0; F = 1 + m R, if m > 0.
m UF
We x an algebraic closure F of F and denote again by | | the extension of | | to F . We will frequently denote by the Galois group Gal(F /F ). For a group G1 with an action of Gal(F /F ) we let G denote the elements of G1 that are xed by . 1 For an algebraic group H we let X(H) denote the group of characters of H and H 0 denotes the connected component of the identity in H. For a eld extension E/F and H an algebraic group dened over E we let ResE/F H denote the restriction of scalars of H to F . For a compact open subgroup of a p-adic group we use 1K or charK to denote the characteristic function of K. For ease of notation we frequently use blank entries in matrices to denote zeros. Given Ai GL(ni ), 1 i k we let diag(A1 , . . . , Ak ) denote the block diagonal matrix in GL(n1 + . . . + nk ) with block diagonal entries A1 , . . . , Ak .
15
Chapter 3
The twisted weighted fundamental lemma
In this chapter we give the statement of the twisted weighted fundamental lemma from [Art02a]. As remarked there it is stated in such a way that it includes the statement of the weighted fundamental lemma found in [Art02b, Section 5].
3.1
The twisted weighted fundamental lemma
Let G0 be a connected reductive algebraic group dened over F , a local eld of characteristic zero. We let be a quasi-semisimple automorphism of G0 . We form the semidirect product G+ = G0 and we take G to be the connected component G0 of G+ . We assume that G0 is unramied over
F . We x a hyperspecial maximal compact subgroup K of G0 (F ). Let G0 denote the dual group of G0 and let denote the automorphism of G0 dual to . Let M = M 0 be a Levi subset as dened in [Art88b, Section 1]. Equivalently M 0 is a Levi component of a parabolic subgroup P 0 (dened over F ) of G0 such that both M 0 and P 0 are stable under . We let M 0 denote the dual group of M 0 . Suppose now that M represents an unramied, elliptic, twisted endoscopic datum (M , M , sM , M ) for M 0 ; see [KS99, Section 2.1]. Here sM is a semisimple element in M = M 0 suppose that M is an L-subgroup of M = M endoscopic data for G0 of the form
L 0 0
Let Z(M ) = Z(M 0 ) denote the centralizer of M in M 0 . We dene EM (G) to be the set of twisted
WF and that M is the inclusion of M in L M 0 .
G0
. We
(G , G , s , ), where s sM Z(M ) , G is the connected centralizer of s in G and is the identity embedding of
16 G = M G into L G0 . The elements of EM (G) are taken up to translation of s by Z(G) . We now set M (G, G ) = and
G rM (k) = JM (k, u),
|Z(M ) /Z(M ) | |Z(G ) /Z(G) |
,
Hecke algebra of G0 (F ).
where k M (F ) is strongly G0 -regular, and u = uK is the stabilizer in G(F ) of the unit in the
Conjecture 3.1. (The twisted weighted fundamental lemma) Let conjugacy class in M (F ). Then
G M,K ( , k)rM (k) = kGreg (M (F )) G EM (G)
be a strongly G0 -regular, stable
M (G, G )sG ( ), M
where sG ( ) is the function dened uniquely for the unramied connected pair (G , M ) in [Art02b, M Section 5] and M,K is the twisted transfer factor for M 0 , normalized relative to the hyperspecial maximal compact subgroup K M 0 (F ) of M 0 (F ). The transfer factor is dened as the product of the terms I , II and III from [KS99, Chapter 4]. We also remark that as in [Art99, (3.2)], the coecients M (G, G ) are zero unless G is elliptic. The function sG ( ) is inductively dened by M sG ( ) = M
1
G rM ( 1 )
M (G , G1 )sM1 ( ),
G1
G
where the rst sum is over representatives conjugacy class
1
for the conjugacy classes in M (F ) within the stable
and the second sum is over G1 EM (G ) with G1 = G .
3.2
Our case
We now describe the situation we are considering in this thesis. We take G0 = GL(4) GL(1) and we take to be the automorphism of G0 given by
: (g, e) (J t g 1 J 1 , e det g),
17 where J is the matrix 1 1 1 1 .
We have G0 = GL(4, C) GL(1, C) and the automorphism is given by : (h, t) (tJ t h1 J 1 , t). When one takes M 0 = G0 in the statement of the twisted weighted fundamental lemma one recovers the statement of the twisted fundamental lemma proven in [Fli99]. Therefore we consider only proper parabolic subgroups of G0 . The proper standard parabolics P 0 of G0 , which are stable under are those whose projection onto GL(4) are of the form , , .
J =
We take M 0 to be the Levi component in each of these parabolic subgroups that contain the diagonal torus in G0 . We refer to these Levi subgroups as the (2,2) Levi, the (1,2,1) Levi and the diagonal Levi.
G The integrals rM () depend on the choice of a measure on G , there is a similar such depen-
dence in the denition of sG ( ). Within a stable conjugacy class these measures are chosen so M that stable conjugacy is measure preserving. Having done this, if we are now given M (F ) and M (F ), such that ( , ) = 0, we normalize the measures on M and M such that under this normalization the (unweighted) twisted fundamental lemma holds for the pair (M, M ). For the proof of the twisted weighted fundamental lemma for (GL(4) GL(1), ) we assume that the residual characteristic of F is odd.
18
Chapter 4
Endoscopic groups
Throughout this chapter we adopt the notation of Section 3.2. We now determine the unramied elliptic twisted endoscopic groups M for each of the Levi subgroups M 0 of G0 given in Section 3.2. We refer to [KS99, Section 2.1] for the denition of twisted endoscopic groups. For each such endoscopic group M we also compute the set of elliptic twisted endoscopic groups for G0 in EM (G), which contain M as a Levi subgroup; and for each group G in EM (G) we compute the coecient M (G, G ). The elliptic twisted endoscopic groups for G0 itself are computed in [Fli99, Section I.F]; these results are recalled in Section 4.1. We use these results below in computing the sets EM (G) and the norm maps from M to M .
4.1
Twisted endoscopic groups for GL(4) GL(1)
In this section we recall results from [Fli99, Section I.F] on the twisted endoscopic groups for G0 . First we note that given s G, assumed semisimple, the twisted centralizer ZG0 (s ) depends only on the component of s lying in GL(4, C). Moreover, after twisted conjugation, we can assume that we have s = (diag(1, 1, c, d), 1) . Furthermore the -conjugacy class of s does not change if c is replaced by c1 , d by d1 and (c, d) by (d, c). We recall that a twisted endoscopic group H is called elliptic if (Z(H) )0 is contained in Z(G0 ). The elliptic twisted endoscopic groups of G0 are given below. 1. c = d = 1: The twisted centralizer of s is isomorphic to GSp(4, C) and we get GSp(4) as a twisted endoscopic group. 2. c = d = 1: The connected component of the twisted centralizer of s is isomorphic to
GL(2, C)2 /C with C embedded via z (z, z 1 ). If we have a trivial Galois action then we obtain (GL(2) GL(2)) , where the prime denotes the subgroup of pairs (A, B) with
19 det A = det B, as a twisted endoscopic group. We can also have a non-trivial Galois action with acting through a quadratic extension E/F in which case we obtain ResE/F GL(2) , with the prime here denoting determinant in F , as a twisted endoscopic group. 3. c = 1, d = 1: The connected component of the twisted centralizer of s is isomorphic to ab. In this case we only obtain elliptic endoscopic datum if acts through a quadratic extension E/F ; in which case we obtain (GL(2) ResE/F GL(1))/ GL(1), with GL(1) embedded as (z, z 1 ), as a twisted endoscopic group.
(GL(2, C) GL(1, C)2 ) with the prime denoting the subgroup of triples (A, a, b) with det A =
As noted in [Fli99, Section I.F] none of these groups, with the exception of GSp(4) (see 4.5), have proper elliptic endoscopic groups. Let H be a twisted endoscopic group for G0 and let TH denote the diagonal torus in H. Let T denote the diagonal torus in G0 . The norm maps N : T TH are given below. 1. H = GSp(4): N : (diag(x, y, z, t), w) diag(xyw, xzw, tyw, ztw) 2. H = (GL(2) GL(2)) : N : (diag(x, y, z, t), w) (diag(xyw, ztw), diag(xzw, ytw)) 3. H = ResE/F GL(2) : N : (diag(x, y, z, t), w) (diag(xyw, ztw), diag(xzw, ytw)) 4. H = (GL(2) ResE/F GL(1))/ GL(1): N : (diag(x, y, z, t), w) (diag(xw, tw), y, z).
4.2
Twisted endoscopic groups for the (2,2) Levi
In this section we take M 0 to be the (2,2) Levi in G0 . We have M 0 = GL(2, C)GL(2, C)GL(1, C), which sits inside G0 as the (2,2) Levi. The restriction of to M 0 is given by (A, B, t) (twt B 1 w, twt A1 w, t), where 1 .
Lemma 4.1. The only elliptic twisted endoscopic group for M 0 is GL(2) GL(1). Proof. Let s M 0 be such that s is semisimple. We may assume that s is diagonal, and after
w= 1
twisted conjugacy in M 0 we can assume that it is of the form
s = (diag(1, 1, 1 , 2 ), s2 ).
20 We now compute ZM 0 (s ). We see that (A, B, t) ZM 0 (s ) if and only if we have Awt Bw = and 1 B t t .
2
Which is if and only if we have A = twt B 1 w and 1 2
t wt Aw = 1 1 =
t2
B So if 1 = 2 then we have ZM 0 (s ) = while if 1 = 2 then we have
B 1
2
.
A, twt A1 w, t M 0 : A GL(2, C), t C ,
Both of these centralizers are connected hence we can only have a trivial Galois action. Therefore only when we have 1 = 2 do we get elliptic twisted endoscopic data for M 0 . In this case we have = ZM 0 (s ) GL(2, C) GL(1, C) and hence we get GL(2) GL(1) as a twisted endoscopic group for M 0.
x ty 1 , ZM 0 (s ) = y
tx1
, t M 0 : x, y, t C
.
We now compute EM (G). Lemma 4.2. Let M represent the elliptic twisted endoscopic datum for M 0 . Then the elliptic twisted endoscopic groups for G0 in EM (G) are GSp(4) and (GL(2) GL(2)) , the prime denoting the subgroup of pairs (A, B) with det A = det B. Each group occurs with multiplicity one. Proof. We may as well take s = (I, I, 1) M 0 which gives rise to M = GL(2) GL(1). We need to look at the translations of s by elements in Z(M ) taken modulo Z(G). We have Z(M ) = {(diag(a, a, b, b), ab)} and Z(G) = diag(a, a, a, a), a2 .
21 Thus we need to look for elliptic endoscopic datum for G0 arising from elements of the form (diag(1, 1, , ), ) G. So we get endoscopic datum only when we have = 1 and we must have a trivial Galois action in both cases.
We note that M sits inside GSp(4, C) as the Siegel Levi, hence we have M = GL(2) GL(1) sitting inside GSp(4) as the Klingen Levi. We have M sitting inside (GL(2)GL(2)) as (T GL(2)) where T is the diagonal torus in GL(2) and the prime again denotes the subgroup of pairs with equal determinant. The coecients M (G, G ) are equal to 1 for G equal to GSp(4) and (GL(2) GL(2)) .
4.3
Twisted endoscopic groups for the (1,2,1) Levi
In this section we take M 0 to be the (1,2,1) Levi in G0 . We have M 0 = GL(1, C) GL(2, C) GL(1, C) GL(1, C), which sits inside G0 as the (1,2,1) Levi. The restriction of to M 0 is given by (a, g, b, t) (tb1 , t(det g)1 g, ta1 , t). Lemma 4.3. The unramied elliptic twisted endoscopic groups for M 0 are GL(2) GL(1) and GL(1) ResE/F GL(1), where E/F is the unramied quadratic extension. Proof. After twisted conjugacy in M 0 we can assume that we have 1 1
s = 1, Then (a, g, b, t) ZM 0 (s ) if and only if 1 ab, g Hence we need ab = t and 1 g
, 2 , s2 .
1
(det g)g 1 , ab2 = t,
t t1
, t2 .
1
g 1 =
det g 1 t det g 1 t1
.
22 Therefore if 1 = 1 then we have g is any element of GL(2, C), while if 1 = 1 then x , g w y z .
Thus we see that if 1 = 1 then
ZM 0 (s ) = (a, g, a1 det g, det g) M 0 : g GL(2, C), a C , while if 1 = 1 then x , a1 xy, xy ZM 0 (s ) = a, y a, y x , a1 xy, xy ,
and if 1 = 1 then
When 1 = 1 we have a connected centralizer and hence we have a trivial Galois action. In this case we have elliptic endoscopic data and we get GL(2) GL(1) as a twisted endoscopic group for M 0. When 1 = 1 to get elliptic endoscopic datum we need to have a non-trivial Galois action acting through a quadratic extension by a, x y y
x , a1 xy, xy . ZM 0 (s ) = a, y
, a1 xy, xy, xy a,
, a1 xy, xy . x
In order for our endoscopic data to be unramied we need this quadratic extension to be unramied. In this case we get GL(1) ResE/F GL(1) as a twisted endoscopic group for M 0 . Finally, when 1 = 1 the data is never elliptic. We now compute EM (G) for M = GL(2) GL(1). Lemma 4.4. Let M = GL(2) GL(1). Then the only elliptic twisted endoscopic group for G0 in EM (G) is GSp(4) with multiplicity two. Proof. Recall that M is given by the element s = (diag(1, 1, 1, 2 ), s2 ) M . We have Z(M ) = diag(a, c, c, a1 c2 ), c2
and so we need to look for elliptic twisted endoscopic groups for G0 given by translating s by
23 elements of the form diag(1, , , 2 ), 2 G0 . Thus we need to look at elements of the form
diag(1, 1, 1, 2 2 ), s2 . Since we must have a trivial Galois action we get elliptic endoscopic data if and only if 2 = 1 ; in which case we get GSp(4). 2 We have M sitting inside GSp(4, C) as the Klingen Levi and so we get M = GL(2) GL(1) sitting inside GSp(4) as the Siegel Levi. We also have M (G, GSp(4)) = 1. Lemma 4.5. Let M = GL(1) ResE/F GL(1). Then the elliptic twisted endoscopic groups for G0 in EM (G) consists of (GL(2) ResE/F GL(1))/ GL(1) and ResE/F GL(2) . Each group appears with multiplicity two. Proof. Recall that M is given by the element (diag(1, 1, 1, 2 ), s2 ) M . We need to look for elliptic twisted endoscopic groups for G0 given by translating s by elements of the form diag(1, , , 2 ), 2 . Thus we need to look at elements of the form diag(1, , , 2 2 ), 2 s2 G. After conju2 = 1 . When 2 = 1 we get (GL(2) ResE/F GL(1))/ GL(1), while if 2 = 1 we get 2 2 2 ResE/F GL(2) . In this case we have M sitting inside each group in EM (G) as the diagonal torus. And we have M (G, G ) = 1 for both G = (GL(2) ResE/F GL(1))/ GL(1) and G = ResE/F GL(2) .
diag(1, , , 2 2 ), 2 s2 G. After twisted conjugacy we can look at the elements of the form
gacy we can look at the elements diag(1, 1, 1, 2 2 ), s2 G. Thus we get elliptic data if
4.4
Twisted endoscopic groups for the diagonal Levi
We now take M 0 to be the diagonal torus in G0 . We have M 0 = GL(1, C)5 , which sits inside G0 as the diagonal torus. Lemma 4.6. The unramied elliptic twisted endoscopic group for M 0 is GL(1)3 . Proof. Since M 0 is abelian we see that for any s M 0 we have ZM 0 (s ) = {(x, y, z, w, t) M 0 : xw = yz = t}. = Hence we have ZM 0 (s ) (C )3 and we get GL(1)3 as the only twisted endoscopic group for M 0 . Furthermore, it is both elliptic and unramied.
We now compute EM (G).
24 Lemma 4.7. Let M = GL(1)3 . Then the elliptic twisted endoscopic groups for G0 in EM (G) are GSp(4) and (GL(2) GL(2)) ; each group appears with multiplicity two. Proof. We have Z(M ) = diag(x, y, ty 1 , tx1 ), t .
Thus we need to look for the elliptic twisted endoscopic groups for G0 given by elements of the form diag(1, y 1 , yw, w), w G. We can conjugate such an element to diag(1, 1, y 2 w, w), w . Since we must have a trivial Galois action we get elliptic data when we have w = 1 and y 2 = 1, in which case we get GSp(4), or when we have w = 1 and y 2 = 1, in which case we get (GL(2) GL(2)) . For G equal to both GSp(4) and (GL(2) GL(2)) we have M sitting inside as the diagonal torus and we have M (G, G ) = 1
4.5
Endoscopic groups for GSp(4)
We will also need to know the endoscopic groups for GSp(4). There is only one proper elliptic endoscopic group for GSp(4) namely (GL(2)GL(2))/ GL(1) with GL(1) embedded as a (a, a1 ), see [Fli99, Section 1.F]. It is given by the element diag(1, 1, 1, 1) GSp(4, C). The norm map is given by diag(a, b, cb1 , ca1 ) diag(1, (ab)1 c), diag(a, b) . For each proper Levi subgroup M of GSp(4) we also need to compute the elliptic endoscopic groups for GSp(4) in EM (GSp(4)). Since we are taking M as an endoscopic group for itself the elements of EM (GSp(4)) are given by elements s Z(M ) taken modulo translation by Z(GSp(4, C)), which equals {diag(x, x, x, x)}. Lemma 4.8. Let M be the Siegel Levi in GSp(4). Then the elliptic endoscopic groups in EM (GSp(4)) are GSp(4) and (GL(2) GL(2))/ GL(1) each with multiplicity one. Proof. We have M sitting inside GSp(4, C) as the Klingen Levi. So we have Z(M ) = diag(x, y, y, x1 y 2 ) . And we get that the elliptic endoscopic groups in EM (GSp(4)) are GSp(4) and (GL(2)GL(2))/ GL(1) each with multiplicity one.
25 We have M sitting inside (GL(2)GL(2))/ GL(1) as (T GL(2))/ GL(1) where T is the diagonal
1 torus in GL(2). And we have M (GSp(4), (GL(2) GL(2))/ GL(1)) = 2 .
Lemma 4.9. Let M be the Klingen Levi in GSp(4). Then the only elliptic endoscopic group in EM (GSp(4)) is GSp(4) with multiplicity one. Proof. We have M sitting inside GSp(4, C) as the Siegel Levi. So we have Z(M ) = {diag(x, x, y, y)} . The only elliptic endoscopic group given by such an element is GSp(4) itself which we obtain when x = y = 1.
Lemma 4.10. Let M be the diagonal Levi in GSp(4). Then the elliptic endoscopic groups in EM (GSp(4)) are GSp(4) and (GL(2) GL(2))/ GL(1), each with multiplicity one. Proof. We have M sitting inside GSp(4, C) as the diagonal torus. So we have Z(M ) = diag(x, y, y 1 z, x1 z) , and we get that the elliptic endoscopic groups in EM (GSp(4)) are GSp(4) and (GL(2)GL(2))/ GL(1), each with multiplicity one.
We have M sitting inside (GL(2) GL(2))/ GL(1) as the diagonal torus and we have M (GSp(4), (GL(2) GL(2))/ GL(1)) = 1 . 2
26
Chapter 5
Weight functions
In this chapter we compute all the weight functions needed in the proof of the fundamental lemma.
5.1
Notations and denitions
We recall the necessary notations and denitions from [Art88b, Section 1] needed to dene the weight functions. Let G0 be a connected reductive algebraic group over F . Let be a quasi-semisimple automorphism of G0 dened over F . We form the semidirect product G+ = G0 connected component G
0
and take G to be the
A parabolic subgroup of G+ is the normalizer in G+ of a parabolic subgroup of G0 . A parabolic of G+ dened over F . Let P = P + G be a parabolic subset of G, a Levi component of P will be a subset of G is by denition a non-empty set of the form P + G where P + is a parabolic subgroup
G .
+
set M = M + P where M + is the normalizer in G+ of some Levi component of P 0 = G0 P which P 0 in G0 .
is dened over F . We call such an M a Levi subset of G. We denote by NP the unipotent radical of
Let M be a Levi subset of G. We dene F(M ) to be the collection of parabolic subsets of G which contain M , and let L(M ) denote the collection of Levi subsets of G which contain M . Any P F(M ) has a unique Levi component MP in L(M ). We write P(M ) for the set of P F(M ) with MP = M . Let AM denote the split component of the centralizer of M in M 0 = M + G0 . Let X(M )F denote the group of characters of M + dened over F and set aM = Hom(X(M )F , R), a real vector space of dimension equal to the dimension of AM . Since AM AM 0 we get a canonical embedding aM aM 0 . We x a Weyl invariant Euclidean metric on a maximal such space, the restriction of this metric provides a Euclidean metric on any subspace.
27 By restriction we have a canonical identication aM = Hom(X(AM ), R). We set a = X(M )F Z R = X(AM ) Z R. M P a denote the simple roots of (P, AP ). Let Q be a parabolic subset of G containing P . Then P we have AQ AP and by restriction we have a map X(AP ) X(AQ ), which yields a surjection a P an injection a = X(MQ )F Z R X(MP )F Z R = a . This gives rise to canonical splittings Q P aP = aQ aQ P and a = a (aQ ) . P Q P We also have the set of coroots
0 aG0 : 0 P 0 , P
0
We now x a Levi subset M of G. Let P F(M ), then we write AP = AMP and aP = aMP . Let
a and hence an embedding aQ aP . On the other hand, since P Q we have by restriction Q
a minimal parabolic subgroup of G0 contained in P 0 . Each 0 P 0 is the restriction to aG0 of a P
unique root 1 P0 . We then dene 0 to be the projection onto aG0 of 1 aG0 . P P
0 0
where P 0 = P + G0 . We recall the denition of these coroots from [Art78, Section 1]. Let P0 be
0
We have a natural inclusion aP aP 0 and so for each P we can dene =
0 0 ,
where the sum if over those 0 P 0 which equal when restricted to aP . We dene aG to P P be the set of such and note that is a basis of aG . Let aP,C = aP R C, then for a we P P P,C dene P () = vol(aG /Z( ))1 P P
P
( ).
We have a homomorphism HM : M + (F ) aM
28 dened by HM (m), = log |(m)| for X(M )F . Let P P(M ) then using the Iwasawa decomposition G+ (F ) = NP (F )M + (F )K we can extend HM to a map HP : G+ (F ) aM by taking HP to be zero on NP (F ) and K. Then we set vP (, x) = exp((HP (x))) for a and x G+ (F ). We now set P,C vM (, x) =
P P(M )
vP (, x)P ()1 .
By [Art81, Lemma 6.2] for each x G+ (F ) this function extends to a smooth function on ia . We M dene vM (x) to be the value of the function vM (, x) at = 0. We note that vM (mxk) = vM (x) for all m M + (F ) and k K.
5.2
Twisted weight functions
In this section we adopt the notation of Section 3.2 and compute the weight functions for the relevant Levi subgroups of G0 = GL(4) GL(1). We will use the following basic fact in computing the weight functions below. Lemma 5.1. For v = (v1 , . . . , vn ) F n dene |v| = max{|v1 |, . . . , |vn |}. Then for all k GL(n, R)
and v F n we have |vk| = |v|.
Proof. We clearly have |vk| |v| and replacing v by vk 1 yields the result.
5.2.1
The (2,2) Levi
. Let P 0 (resp. Q0 ) be
In this section we take M 0 to be the (2,2) Levi in G0 . We have M = M 0
the upper (resp. lower) block triangular parabolic in G0 with M 0 as its Levi component. We have
29 M = M0 and if we set P = P 0 and Q = Q0 then we have P(M ) = {P, Q}. We let NP
(resp. NQ ) denote the unipotent radical of P 0 (resp. Q0 ). Let x G0 (F ) and write x = nP mP kP = nQ mQ kQ
with obvious notation. We write mP = (AP , BP , cP ) GL(2) GL(2) GL(1) and similarly we write mQ = (AQ , BQ , cQ ). Lemma 5.2. With notation as above we have vM (x) = vol(aG /Z( )) (log | det AQ | log | det AP |) . P P
Proof. For (A, B, c) M 0 we have : (A, B, c) (wt B 1 w, wt A1 w, c det AB), and hence AM = a = diag(a1 , a1 ), diag(a1 , a1 )), a2 1 1 We x the basis {1 , 2 } of X(AM ) given by i : a ai . We have AM 0 = {b = (diag(b1 , b1 ), diag(b2 , b2 )), b3 )} and we x the basis {1 , 2 , 3 } of X(AM 0 ) given by i : b bi . We have P 0 = {1 2 }. Kronecker delta symbol. To determine (1 2 ) we may as well work inside GL(4). We set P0 equal to the upper triangular Borel subgroup of GL(4) and we take M0 to be the diagonal torus in P0 . We have AM0 = M0 = {c = diag(c1 , c2 , c3 , c4 )} and we x the basis {1 , 2 , 3 , 4 } of X(M0 ) given by i : c ci . We dene i aM0 similarly. We now describe the splittings a 0 = a 0 (aP0 ) and aP0 = aP 0 aP0 . The map X(AP 0 ) P P P P 1 1 2 1 3 2 4 2
0 0
.
We now compute (1 2 ) . Let 1 , 2 , 3 denote the basis of a 0 given by i (j ) = ij , the M
X(M0 ) is given by
30 and the map aP 0 aP0 is given by
1 1 2 (1 + 2 ) 1 2 2 (3 + 4 ).
Thus we have aP0 = aP 0 aP0 = Span{1 + 2 , 3 + 4 } Span{1 2 , 3 4 } P and a 0 = a 0 (aP0 ) = Span{1 + 2 , 3 + 4 } Span{1 2 , 3 4 }. P P P Therefore we have 1 2 = 1 1 1 1 (1 + 2 ) (3 + 4 ) = 2 3 + (1 2 ) + (3 4 ) 2 2 2 2
0 0
equal to the projection of 2 3 onto a 0 . Now (2 3 ) = 2 3 whose projection onto aP 0 P is 1 1 ( + 2 ) (3 + 4 ). 2 1 2 Hence we have (1 2 ) = 1 (1 2 ). 2 The map X(AM 0 ) X(AM ) is given by 1 1 We have P = {21 }, Q = {21 } and (21 ) : 1 Hence for = a1 1 + a2 2 a M,C we have P () = and Q () = a1 . G /Z( )) 2 vol(aP P a1 G /Z( )) 2 vol(aP P
1 2
2 1
3 2 .
2 0 .
We now make explicit the isomorphism between X(AM ) Z R and X(M )F Z R. We have a basis for X(M )F given by the characters 1 : ((A, B), c) det A det B 1 2 : ((A, B), c) c
31 of M 0 . The restriction map X(AM ) X(M )F is given by 1 41 and 2 2 . Now we have HM (mP ) : Therefore, vP (, x) = exp a1 1 log | det AP BP | a2 log |cP | 4 1
1 4 1 log | det AP BP |
2 log |cP |.
and similarly for vQ (, x). Hence vM (, x) equals
2 vol(aG /Z( )) a1 a1 1 1 P P exp log | det AP BP | a2 log |eP | exp log | det AQ BQ | a2 log |eQ | a1 4 4 .
Taking the limit as = a1 1 + a2 2 0 we get vM (x) = vol(aG /Z( )) 1 1 P P log | det AP BP | log | det AQ BQ | . 2
But we have | det AP BP | = | det AQ BQ | and hence vM (x) = vol(aG /Z( )) (log | det AQ | log | det AP |) , P P which completes our computation. We now compute vM on the unipotent radical of P 0 . Lemma 5.3. We have vM 1 1 x1 x3 1 x2 x4 , 1 = vol(aG /Z( )) log max{1, |x1 |, |x2 |, |x3 |, |x4 |, |x1 x4 x2 x3 |}. P P 1
Proof. We write
1 1
x1 x3 1
x2
Applying the vector (1, 0, 0, 0) (0, 1, 0, 0) and using Lemma 5.1 allows us to deduce that log | det AQ |= log max{1, |x1 |, |x2 |, |x3 |, |x4 |, |x1 x4 x2 x3 |}
x4 = y1 y3 1
1
1 y2 y4 1 1
AQ
BQ
kQ .
32 and the result follows.
5.2.2
The (1,2,1) Levi
In this section we take M 0 to be the (1,2,1) Levi in G. Let P 0 (resp. Q0 ) be the upper (resp. lower) triangular parabolic in G0 with M 0 as its Levi component. We have M = M 0 P = P0 and Q = Q0
0
and if we set
unipotent radical of P (resp. Q0 ). Let x G0 (F ) and write
then we have P(M ) = {P, Q}. We let NP (resp. NQ ) denote the
x = nP mP kP = nQ mQ kQ with obvious notation. We write mP = (aP , BP , cP , dP ) GL(1) GL(2) GL(1) GL(1) and similarly we write mQ = (aQ , BQ , cQ , dQ ). Lemma 5.4. With notation as above we have vM (x) = vol(aG /Z( )) P P log |aQ c1 | log |aP c1 | . Q P 2
Proof. We have AM = a = a1 , I, a1 , a2 1 .
We x the basis {1 , 2 } of X(AM ) given by i : a ai . We have AM 0 = {b = (b1 , diag(b2 , b2 ), b3 , b4 )} and we x the basis {1 , 2 , 3 , 4 } of X(AM 0 ) given by i : b bi . We have P 0 = {1 2 , 2 3 }. We now compute (1 2 ) and (2 3 ) . Let 1 , 2 , 3 , 4 denote the basis of a 0 given by M i (j ) = ij . To determine (1 2 ) and (2 3 ) we may as well work inside GL(4). We set P0 equal to the upper triangular Borel subgroup of GL(4) and we take M0 to be the diagonal torus in P0 . We have AM0 = M0 = {c = diag(c1 , c2 , c3 , c4 )} and we x the basis {1 , 2 , 3 , 4 } of X(M0 ) given by i : c ci . We dene i aM0 similarly.
33 We now need to describe the splittings a 0 = a 0 (aP0 ) and aP0 = aP 0 aP0 . The map P P P P X(AP 0 ) X(M0 ) is given by 1 1 and the map aP 0 aP0 is given by 1 1 Thus we have aP0 = aP 0 aP0 = Span{1 , 2 + 3 , 4 } Span{2 3 }, P and a 0 = a 0 (aP0 ) = Span{1 , 2 + 3 , 4 } Span{2 3 }. P P P Therefore we have 1 1 1 2 = 1 (2 + 3 ) = 1 2 + (2 3 ) 2 2 equal to the projection of 1 2 onto a 0 . Now we have (1 2 ) = 1 2 whose projection P onto aP 0 is 1 1 (2 + 3 ). 2 Hence we have (1 2 ) = 1 1 2 . Now 2 2 3 = 1 1 (2 + 3 ) 4 = 3 4 + (2 3 ) 2 2
0 0 0 0
2 2
3 2
4 3 ,
1 2 2 (2 + 3 )
3 4 .
equals the projection of 3 4 onto a 0 . Now we have (3 4 ) = 3 4 whose projection P onto aP 0 is 1 ( + 3 ) 4 . 2 2
1 Hence we have (2 3 ) = 2 2 3 .
The map X(AM 0 ) X(AM ) is given by 1 1 2 0 3 1 4 2 .
34 We have P = {1 }, Q = {1 } and (1 ) : 1 1 Hence for = a1 1 + a2 2 a M,C we have P () = and Q () = a1 . vol(aG /Z( )) P P a1 , G /Z( )) vol(aP P 2 0.
We now make explicit the isomorphism between X(AM ) Z R and X(M )F Z R. We have a basis for X(M )F given by the characters 1 : (a, B, c, d) ac1 , 2 : (a, B, c, d) acd2 det B on M 0 . The restriction map X(AM ) X(M )F is given by 1 21 and 2 22 . Now we have HM (mP ) : 1 Therefore, vP (, x) = exp a2 a1 log |aP b1 | log |aP cP d2 det BP | P P 2 2
1 2
log |aP c1 | P
2
1 2
log |aP cP d2 det BP |. P
and similarly for vQ (, x). We can set a2 = 0 and take the limit as a1 0 to give vM (x) = as wished. We now compute vM on the unipotent radical of P 0 . Lemma 5.5. We have vol(aG /Z( )) P P log |aQ c1 | log |aP c1 | Q P 2
equal to
vM
1 x1 1
x2
1
x4 , 1 x5 1
x3
vol(aG /Z( )) P P (log max{1, |x1 |, |x2 |, |x3 |} + log max{1, |x4 |, |x5 |, |x1 x4 + x2 x5 x3 |}) . 2
35
Proof. We write 1 x1 1 1 x2 x3 1 1 y4 y4 1 a Q cQ
x4 y1 = x5 y2 1 y3
1
BQ
kQ .
Applying the vector (1, 0, 0, 0) allows us to deduce that log |aQ | = log max{1, |x1 |, |x2 |, |x3 |}. Taking the transpose inverse of the above matrix equation and applying the vector (0, 0, 0, 1) allows us to deduce that log |c1 | = log max{1, |x4 |, |x5 |, |x1 x4 + x2 x5 x3 |} Q and the result follows.
5.2.3
The diagonal Levi
Let M 0 be the diagonal Levi subgroup in G0 . For the proof of the fundamental lemma it is (essentially) sucient to compute vM on elements of G0 xed by , i.e., elements of the form g = (g1 , g2 ) Sp(4) GL(1). For now we show that for such a g vM (g) is, up to a scalar, equal to vM1 (g1 ) where M1 is the diagonal Levi in Sp(4). We will then compute vM1 on the unipotent radical of the upper triangular Borel subgroup in Sp(4). Let B 0 (resp. B1 ) denote the upper triangular Borel subgroup of G0 (resp. Sp(4)). Lemma 5.6. For g (g1 , g2 ) G0 (F ) with g1 Sp(4) we have vM (g) = vol(aG /Z( )) B B vol(aB1
Sp(4)
/Z( 1 )) B
vM1 (g1 ).
Proof. We have AM = a = (diag(a1 , a2 , a1 , a1 ), a3 ) , 2 1 and we x the basis {1 , 2 , 3 } of X(AM ) given by i : a ai . We have AM 0 = {b = (diag(b1 , b2 , b3 , b4 ), b5 )} ,
36 and we x the basis {1 , . . . , 5 } of X(AM 0 ) given by i : b bi . The map X(AM 0 ) X(AM ) given by restriction is given by 1 1 We have AM1 = a = diag(a1 , a2 , a1 , a1 ) . 2 1 We identify aM1 with the subspace of aM of elements which are zero on 3 and we identify a 1 with M the subspace {a1 1 + a2 2 } of a . M We now compute B () for = a1 1 + a2 2 + a3 3 a . We have M,C B 0 = {1 2 , 2 3 , 3 4 }, and B = {1 2 , 22 }. We have (1 2 ) : 1 2 and (22 ) : 1 0 On the other hand B1 = {1 2 , 22 }, and we have (1 2 ) : and (22 ) : 1 0 2 1. 1 1 2 1, 2 1 3 0. 2 2 3 0, 2 2 3 2 4 1 5 3 .
each Borel subgroup of G0 , which is stable and which contains M 0 is of the form w1 B 0 w with w = (w1 , 1) where w1 is an element of the Weyl group of Sp(4). Hence we deduce that for each Borel subgroup P 0 of G0 which contains M 0 we have vol(aG /Z( ))P () = vol(aP1 P P
Sp(4)
Hence we see that for = 1 + a3 3 a M,C with 1 aM1 ,C we have B () = B1 (1 ). Now
/Z(1 ))P1 (1 ), P
where P1 denotes the Borel subgroup of Sp(4) which is contained in P 0 .
37 Next we compute vP (, g) and vP1 (1 , g1 ). In order to compute vP (, g) we need to write g = nP mP kP with nP NP (F ), mP M 0 (F ) and kP K. But if we write g1 = nP1 mP1 kP1 with obvious notation then we have g = (g1 , g2 ) = (nP1 , 1)(mP1 , g2 )(kP1 , 1). Hence we have for = 1 + a3 3 that vP (, g) = vP1 (1 , g1 )|g2 |a3 . Thus we get vM (, g) =
P P(M )
vP (, g)P ()1 vol(aG /Z( )) B B vol(aB1
Sp(4)
=
/Z( 1 )) B
|g2 |a3
vP1 (1 , g1 )P1 (1 )1 .
P1 P(M1 )
And now taking the limit as 0 gives the result. We now compute vM1 on the unipotent radical of B1 . We set 1 x1 1 x2 + x1 x4 x4 1 x3
n= and then if we write
x2 NB1 (F ). x1 1
In order to do this we need to write n = n1 m1 k1 for each Borel subgroup of Sp(4) containing M1 m1 = diag(a, b, b1 , a1 ) we need to compute |a| and |b|. The Weyl group of Sp(4) is isomorphic to D8 with generators 1 1 , w2 = 1
1 w1 =
1 1
1
. 1
38 Explicitly the Weyl group is given by {e, (12)(34), (23), (14), (1243), (1342), (13)(24), (14)(23)}, where we have e=I (12)(34) = w1 (23) = w2 (14) = w1 w2 w1 (1243) = w2 w1 (1342) = w1 w2 (13)(24) = w2 w1 w2 (14)(23) = w1 w2 w1 w2 . w = e. In this case we have |a| = |b| = 1. w = (12)(34). In this case we have 1 y 1 1 w NB 1 w = For m > 0 and u UF we have 1 u m 1 = u1 m 1 m m u1 u y2 y1 y4 1 y3 1 y1 y4 y 2 . 1
Multiplying n on the left by such an element we can put n in the form 1 x1 1
. 1 x1 1
1
m
and hence we deduce that |a| = max{1, |x1 |}
39 and |b|1 = max{1, |x1 |}. w = (23). In this case we have 1 y 2 1 w1 NB1 w = y 4 1 And as above we deduce that |a| = 1 and |b| = max{1, |x4 |}. w = (14). In this case we have 1 y y y 4 1 2 w1 NB1 w = y1 y3 . 1 y1 y1 . y 2 y 1 y 4 1 y3
1
Multiplying n on the left by such an element we can put n in the form
1 x4 1
. 1
1
y4 1
y1
y2
Using the vector (1, 0, 0, 0) we deduce that
|a| = max{1, |x1 |, |x2 + x1 x4 |, |x3 |}. and using (1, 0, 0, 0) (0, 0, 1, 0) we deduce that |ab1 | = max{1, |x1 |2 , |x3 + x1 x2 + x2 x4 |}. 1
40 w = (1243). In this case we have 1 y 2 1 w NB 1 w = y y1 1 y3 1 y2 y1 y4 y1 . 1
4
Using the vector (0, 0, 1, 0) we deduce that
|b| = max{1, |x1 |}1 , and using (1, 0, 0, 0) (0, 0, 1, 0) we deduce that |ab1 | = max{1, |x1 |2 , |x3 + x1 x2 + x2 x4 |}. 1 w = (1342). In this case we have 1 1 w NB 1 w = y 1 y1 y4 y2 1 y3 y1 1 y4 . y 2 1
Using the vector (0, 1, 0, 0) we deduce that
|b| = max{1, |x4 |, |x4 |}, and using (0, 1, 0, 0) (0, 0, 0, 1) we deduce that |a1 b| = max{1, |x4 |}. w = (13)(24). In this case we have 1 1 w NB 1 w = y 2 y y1 1 y3 y1 y4 y2 . 1 y 1 1
4
41 Using the vector (0, 1, 0, 0) we deduce that |b| = max{1, |x2 |, |x4 |}, and using (1, 0, 0, 0) (0, 1, 0, 0) we deduce that |ab| = max{1, |x2 |, |x4 |, |x3 x1 x2 |, |x2 x3 x4 + x1 x2 x4 |}. 2 w = (14)(23). In this case we have 1 y 1 w1 NB1 w = y y y 1 4 2 y3 . 1
1 y4 y2 1 y1
Using the vector (1, 0, 0, 0) we deduce that
|a| = max{1, |x1 |, |x2 + x1 x4 |, |x3 |}, and using (1, 0, 0, 0) (0, 1, 0, 0) we deduce that |ab| = max{1, |x2 |, |x4 |, |x3 x1 x2 |, |x2 x3 x4 + x1 x2 x4 |}. 2 Lets set = a1 1 +a2 2 a 1 ,C . Where i is the character of M1 mapping diag(a1 , a2 , a1 , a1 ) 1 1 M to ai . Let x Sp(4, F ) and let P1 be a Borel subgroup containing M1 . We write x = nP1 mP1 kP1 with the usual notation where mP1 = diag(aP1 , bP1 , b1 , a1 ). Then we have P1 P1 HP1 (x) : 1 log |aP1 | 2 log |bP1 |.
Hence we have vP1 (, x) = |aP1 |a1 |bP1 |a2 and therefore for = a2 1 + a2 2 a 1 ,C we have M vP1 (, x) = (|aP1 | |bP1 |)a2 . Next we compute P1 for each of these Borel subgroups P1 = w1 B1 w and = a2 1 + a2 2
a . These functions are given in the table below. M,C
42 w e (12)(34) (23) (14) (1243) (1342) (13)(24) (14)(23) P1 22 , 1 2 21 , 2 1 22 , 1 + 2 22 , 1 2 21 , 1 + 2 21 , 1 2 21 , 1 2 22 , 1 + 2 P1 ()/a2 2 1 (1 ) ( + 1) ( + 1) ( + 1) ( + 1) (1 ) 1
For C we set P1 () = P ()/a2 . We have 2 vM1 (x, ) =
P1
vol(aB1 /Z( 1 )) B (|aP1 | |bP1 |)a2 . a2 P1 () 2
GSp(4)
The value at a2 = 0 of this expression is equal to vol(aB1
Sp(4)
vM1 (x) =
/Z( 1 )) B 2
P1
1 ( log |aP1 | + log |bP1 |)2 P1 ()
for any value of . The calculations above give the following. Lemma 5.7. We have vM1 (n) equal to
vol(aB
Sp(4) 1
/Z( 1 )) B 2
times
(A2 + 2B 2 + 2C 2 + D2 + 2E 2 + F 2 ) + 2(AB + AE + BD + CD + EF ) where A = log max{1, |x2 |, |x4 |, |x3 x1 x2 |, |x2 x3 x4 + x1 x2 x4 |} 2 B = log max{1, |x1 |, |x2 + x1 x4 |, |x3 |} C = log max{1, |x1 |} D = log max{1, |x1 |2 , |x3 + x1 x2 + x2 x4 |} 1 E = log max{1, |x2 |, |x4 |} F = log max{1, |x4 |}.
Combining Lemmas 5.6 and 5.7 we get the following.
43 Corollary 5.8. For 1 x1 1 x2 + x1 x4 x4 1 x3
n = we have vM (n) equal to
x2 , 1 NB (F ) x1 1
vol(aG /Z( )) B B 2
times
(A2 + 2B 2 + 2C 2 + D2 + 2E 2 + F 2 ) + 2(AB + AE + BD + CD + EF ). where A, . . . , F are as in Lemma 5.7.
5.3
Weight functions for GSp(4)
In this section we compute the weight functions for the Levi subgroups of GSp(4).
5.3.1
The Siegel Levi
In this section we take M to be the Siegel Levi in GSp(4). Let P (resp. Q) be the upper (resp. lower) triangular parabolic in GSp(4) with M as its Levi component. Then we have P(M ) = {P, Q}. We let NP (resp. NQ ) denote the unipotent radical in P (resp. Q). Let x GSp(4, F ) and write x = nP mP kP = nQ mQ kQ with obvious notation. We write AP mP = and similarly for mQ . Lemma 5.9. With notation as above we have vM (x) = vol(aP
GSp(4)
bP wt A1 w P
M (F ),
/Z( )) (log | det AQ | log | det AP |) . P
Proof. We have AM = a = diag(a1 , a1 , a1 a2 , a1 a2 ) . 1 1
44 We x the basis {1 , 2 } of X(AM ) given by i : a ai . We have P = {21 2 } and Let 1 , 2 denote the basis of a given by i (j ) = ij . We set P0 equal to the upper triangular M
Q = {21 + 2 }. We now compute (21 2 ) .
Borel subgroup of GSp(4) and we take M0 to be the diagonal torus in P0 . We have AM0 = M0 = c = diag(c1 , c2 , c1 c3 , c1 c3 ) 2 1 and we x the basis {1 , 2 , 3 } of X(M0 ) given by i : c ci . We dene i aM0 similarly. X(AP ) X(M0 ) is given by 1 1 and the map aP aP0 is given by
1 1 2 1 + 2
We now need to describe the splittings a 0 = a (aP0 ) and aP0 = aP aP0 . The map P P P P 2 1 3 2
2 3 .
Thus we have aP0 = aP aP0 = Span{1 + 2 , 3 } Span{1 2 }, P and a 0 = a (aP0 ) = Span{1 + 2 , 3 } Span{1 2 }. P P P Therefore we have 21 2 = 1 + 2 3 = 22 3 + (1 2 ) equal to the projection of 22 3 onto aP . Now we have (22 3 ) = 2 whose projection onto
1 1 a is 2 (1 + 2 ). Hence we have (21 2 ) = 2 1 . P
Hence for = a1 1 + a2 2 a M,C we have P () =
a1 , G /Z( )) 2 vol(aP P a1 . 2 vol(aG /Z( )) P P
and Q () =
We now make explicit the isomorphism between X(AM ) Z R and X(M )F Z R. We have a
45 basis for X(M )F given by the characters A 1 : and A 2 :
bw A
t
1
w
det A b.
bwt A1 w
The restriction map X(M )F X(AM ) is given by 1 21 and 2 2 . Therefore, HM (mP ) : and so vP (, x) = exp a1 log | det AP | a2 log |bP | . 2 1
1 2
log | det AP | 2 log |bP |,
We have a similar expression for vQ (, x). Taking a2 = 0 and letting a1 0 gives vM (x) = vol(aP as desired.
GSp(4)
/Z( )) (log | det AQ | log | det AP |) P
The computation of vM on the unipotent radical of P follows directly from the proof of Lemma 5.3. Lemma 5.10. We have vM 1 1 x r 1 s = vol(aGSp(4) /Z( ))) log max{1, |x|, |r|, |s|, |xs r2 |}. P P 1 r
5.3.2
The Klingen Levi
In this section we take M to be the Klingen Levi in GSp(4). Let P (resp. Q) be the upper (resp. lower) triangular parabolic in GSp(4) with M as its Levi component, then we have P(M ) = {P, Q}. We let NP (resp. NQ ) denote the unipotent radical in P (resp. Q). Let x GSp(4, F ) and write x = nP mP kP = nQ mQ kQ
46 with obvious notation. We write mP = and similarly for mQ . Lemma 5.11. With notation as above we have vM (x) = vol(aP
GSp(4)
aP BP a1 det BP P
M (F )
/Z( )) (log |aQ | log |aP |) . P
Proof. We have AM = a = diag(a1 , a2 , a2 , a1 a2 ) . 2 1 We x the basis {1 , 2 } of X(AM ) given by i : a ai . We have P = {1 2 } and P = {2 1 }. We now compute (1 2 ) . Let 1 , 2 denote the basis of a given by i (j ) = ij . We set P0 equal to the upper triangular M
Borel subgroup of GSp(4) and we take M0 to be the diagonal torus in P0 . We have AM0 = M0 = c = diag(c1 , c2 , c1 c3 , c1 c3 ) , 2 1 and we x the basis {1 , 2 , 3 } of X(M0 ) given by i : c ci . We dene i aM0 similarly. We now describe the splittings a 0 = a (aP0 ) and aP0 = aP aP0 . The map X(AP ) P P P P 1 1 and the map aP aP0 is given by 1 1 Thus we have aP0 = aP aP0 = Span{1 , 2 + 23 } Span{2 }, P and a 0 = a (aP0 ) = Span{1 , 3 } Span{22 3 }. P P P Therefore we have 1 1 1 2 = 1 3 = 1 2 + 2 3 2 2
1 2 2 3 .
X(M0 )
is given by 2 2 3 22 ,
47 equal to the projection of 1 2 onto a . Now we have (1 2 ) = 1 2 whose projection P onto aP is 1 . Hence we have (1 2 ) = 1 . Therefore for = a1 1 + a2 2 a M,C we have P () = and Q () = a1 . G /Z( )) vol(aP P a1 vol(aG /Z( )) P P
We now make explicit the isomorphism between X(AM ) Z R and X(M )F Z R. We have a basis for X(M )F given by the characters a B a1 det B a1 det B
1 : and
a
The restriction map X(M )F X(AM ) is given by 1 1 and 2 22 . So we have HM (mP ) : 1 log | det aP | and therefore, vP (, x) = exp a1 log |aP | a2 log | det BP | . 2 2
1 2
a 2 : B
det B.
log | det BP |,
We have a similar expression for vQ (, x). Setting a2 = 0 and taking the limit as a1 0 gives vM (x) = vol(aP as desired.
GSp(4)
/Z( )) (log |aQ | log |aP |) P
The computation of vM on the unipotent radical of P follows directly from the proof of Lemma 5.5.
48 Lemma 5.12. We have vM 1 x 1 1 r r = vol(aGSp(4) /Z( ))) log max{1, |x|, |r|, |s|}. P P x 1 s
5.3.3
The diagonal Levi
In this section we take M to be the diagonal Levi in GSp(4). We will compute vM on the unipotent radical of the upper triangular Borel subgroup of GSp(4). We follow the strategy in the twisted case; we rst relate the function vM to vM1 , where M1 is the diagonal torus in Sp(4) and then use Lemma 5.7. Let B denote the upper triangular Borel subgroup of GSp(4) and let B1 denote its intersection with Sp(4). Lemma 5.13. For g Sp(4, F ) we have vM (g) = vol(aB
GSp(4) Sp(4)
/Z( )) B
vol(aB1
/Z( 1 )) B
vM1 (g).
Proof. We have AM = a = (diag(a1 , a2 , a1 a3 , a1 a3 ) 2 1 and we x the basis {1 , 2 , 3 } of X(AM ) given by i : a ai . We have AM1 = a = diag(a1 , a2 , a1 , a1 ) . 2 1 We identify aM1 with the subspace of aM given by those elements which are zero on 3 and we identify a 1 with the subspace {a1 1 + a2 2 } of a . M M We now compute B () for a . We have M,C
B = {1 2 , 22 3 } and (1 2 ) : 1 1 2 1 3 0
49 and (22 3 ) : 1 0 We have B1 = {1 2 , 22 }, and (1 2 ) : and (22 ) : 1 0 2 1. 1 1 2 1, 2 1 3 0.
each Borel subgroup of GSp(4) is of the form w1 Bw with w an element of the Weyl group of Sp(4). Hence we deduce that for each Borel subgroup P of GSp(4) that contains M we have vol(aP where P1 = P Sp(4). Next we compute vP (, g) and vP1 (1 , g). In order to compute vP (, g) we need to write g = nP mP kP with nP NP (F ), mP M 0 (F ) and kP K. Since we are assuming that g Sp(4) we can do this inside Sp(4) and assume that mP M1 for each P . Hence we have for = 1 + a3 3 that vP (, g) = vP1 (1 , g). And we get vM (, g) =
P P(M ) GSp(4)
Hence we see that for = 1 + a3 3 a , with 1 a 1 ,C , we have B () = B1 (1 ). Now M,C M
/Z( ))P () = vol(aP1 P
Sp(4)
/Z(1 ))P1 (1 ), P
vP (, g)P ()1 vP1 (1 , g1 )P1 (1 )1 . Sp(4) )) vol(aB1 /Z(B1 P1 P(M1 ) vol(aG /Z( )) B B
=
Taking the limit as 0 gives the result. Since the unipotent radical of B lies inside Sp(4) we conclude the following Corollary of Lemmas 5.13 and 5.7.
50 Corollary 5.14. Let
Then vM (n) is equal to
vol(aB
GSp(4)
n=
2
1 x1 1
x2 + x1 x4 x4 1
/Z( )) B
x2 NB (F ) x1 1
x3
times
(A2 + 2B 2 + 2C 2 + D2 + 2E 2 + F 2 ) + 2(AB + AE + BD + CD + EF ), where A = log max{1, |x2 |, |x4 |, |x3 x1 x2 |, |x2 x3 x4 + x1 x2 x4 |} 2 B = log max{1, |x1 |, |x2 + x1 x4 |, |x3 |} C = log max{1, |x1 |} D = log max{1, |x1 |2 , |x3 + x1 x2 + x2 x4 |} 1 E = log max{1, |x2 |, |x4 |} F = log max{1, |x4 |}.
5.4
Other groups
We will also need to compute weighted orbital integrals on groups closely related to GL(2). We now compute vM for M the diagonal torus in GL(2). Lemma 5.15. Let M be the diagonal torus in GL(2) and B the upper triangular Borel subgroup containing M . Then we have 1 x = vol(aGL(2) /Z( )) log max{1, |x|}. vM B B 1 Proof. Let Q denote the lower triangular Borel subgroup of GL(2). Then we have P(M ) = {P, Q}. We have AM = {a = (a1 , a2 )}
51 and we let i X(M ) be given by i : a ai . We have P = {1 2 }, Q = {2 1 } and (1 2 ) : Let = a1 1 + a2 2 a M,C then Q () = We set a2 a1 . vol(aG /Z( )) P P 1 1 2 1.
If x R then we have n GL(2, R) and vM (n) = 0. Next we note that for m > 0 and u UF we have 1 u m 1 = u1 m 1 1 m m m u1 u NQ (F )M (F ) GL(2, R)
1 x . n= 1
Therefore, if x R then vM (a1 1 + a2 2 , n) = vol(aG /Z( )) P P exp (a1 log |x| + a2 log |x|) a2 a1
and taking the limit as 0 gives vM (n) = vol(aG /Z( )) log |x| as required. P P
5.5
Normalization of volumes
Let M 0 be one of our Levi subgroups of G0 and let M be a twisted endoscopic group for M 0 . We need to normalize vol(aG /Z( )) for P a parabolic subset of G with Levi component M with P P vol(aG /Z( )) where G EM (G) and P is a parabolic subgroup of G with levi component M . P P The norm map gives an isomorphism between aP and aP ; and restricts to give an isomorphism between aG and aG . We choose measures on these spaces, which are preserved by this isomorphism. P P First we take M 0 to be the (2,2) Levi in G0 and P 0 the upper triangular parabolic in G0 with M 0 as a Levi component. Then we have AM = a = ((diag(a1 , a1 ), diag(a1 , a1 )), a2 ) , 1 1 and N (a) = diag(a2 a2 , a2 , a2 , a2 a2 ) GSp(4), 1 1
52 and N (a) = diag(a2 a2 , a2 a2 ), diag(a2 , a2 ) (GL(2) GL(2)) . 1 1 Using this we see that we have vol(aG /Z( )) = vol(aP1 P P
GSp(4)
/Z( )) = vol(aP2 P
(GL(2)GL(2))
/Z(2 )). P
Next we take M 0 to be the (1,2,1) Levi in G0 and P 0 the upper triangular parabolic in G0 with M 0 as a Levi component. First we take M = GL(2) GL(1). Then we have AM = a = (diag(a1 , 1, 1, a1 ), a2 ) , 1 and N (a) = diag(a1 a2 , a1 a2 , a1 a2 , a1 a2 ) GSp(4). 1 1 Using this we see that we have vol(aG /Z( )) = 2 vol(aP1 P P
GSp(4)
/Z( )). P
Next we take M 0 to be the (1,2,1) Levi in G0 and P 0 the upper triangular parabolic in G0 with M 0 as a Levi component. We take M = GL(1) ResE/F GL(1). Then we have AM = a = (diag(a1 , 1, 1, a1 ), a2 ) , 1 and N (a) = (diag(a1 a2 , a1 a2 ), diag(a1 a2 , a1 a2 )) ResE/F GL(2) , 1 1 and N (a) = (diag(a1 a2 , a1 a2 ), 1, 1) (GL(2) ResE/F GL(1))/ GL(1). 1 Using this we see that we have vol(aG /Z( )) = vol(aG /Z( )) P P P P for each elliptic endoscopic group G EM (G). with M 0 as a Levi component. We have AM = a = (diag(a1 , a2 , a1 , a1 ), a3 ) , 2 1
Next we take M 0 equal to the diagonal Levi in G0 and P 0 the upper triangular parabolic in G0
53 and N (a) = diag(a1 a2 a3 , a1 a1 a3 , a1 a2 a3 , a1 a1 a3 ) GSp(4), 2 1 1 2 and N (a) = (diag(a1 a2 a3 , a1 a1 a3 ), diag(a1 a1 a3 , a1 a2 a3 )) (GL(2) GL(2)) . 1 2 2 1 Using this we see that we have vol(aG /Z( )) = 2 vol(aP1 P P and vol(aG /Z( )) = 2 vol(aP2 P P
(GL(2)GL(2)) GSp(4)
/Z(1 )), P
/Z(2 )). P
We also need to do the same for GSp(4) and its elliptic endoscopic group (GL(2)GL(2))/ GL(1). First we take M equal to the Siegel Levi in GSp(4). Then we have AM = a = diag(a1 , a1 , a1 a2 , a1 a2 ) , 1 1 and N (a) = (diag(1, a2 a2 ), diag(a1 , a1 )) (GL(2) GL(2))/ GL(1). 1 Using this we see that we have vol(aP
GSp(4)
/Z( )) = P
1 (GL(2)GL(2))/ GL(1) vol(aP /Z( )). P 2
Next we take M equal to the diagonal Levi in GSp(4). We have AM = diag(a1 , a2 , a1 a3 , a1 a3 ) , 2 1 and N (a) = (diag(1, a1 a1 a3 ), diag(a1 , a2 )) (GL(2) GL(2))/ GL(1). 1 2 Therefore we have vol(aP
GSp(4)
/Z( )) = P
1 (GL(2)GL(2))/ GL(1) vol(aP /Z( )). P 2
5.6
Weighted orbital integrals
In this section we prove a couple of lemmas that will be useful in the computation of our weighted orbital integrals. We begin with the following lemma, which allows us to write our weighted orbital
54 integrals as integrals over the Levi subgroup itself. We continue with the notation of Section 5.1. Lemma 5.16. Assume further that NP and K are stable under . Let KM = M 0 (F ) K. For N N : n a1 na(n) and dene
a M 0 (F ) for which a is strongly G0 -regular let a : N N denote the inverse of the bijection
P (a) =
N (F )K
vM (a (n)) dn,
where the Haar measure on N (F ) is normalized to give N (F ) K volume one. Let M (F ) be strongly G0 -regular then
G rM () = |DM ()| 2
1
M (F )\M 0 (F )
1KM (m1 (m))P (m1 (m)) dm,
where the Haar measure on M 0 (F ) gives KM volume one. Proof. By the Iwasawa decomposition we have G0 (F ) = M 0 (F )N (F )K and we can write the Haar measure on G0 (F ) as dg = dm dn dk. By denition we have
G rM () = |DG ()| 2
1
1K (g 1 (g))vM (g) dg
G (F )\G0 (F )
= |DG ()| 2 = |DG ()|
1 2
1
1K (k 1 n1 m1 (m)(n)(k))vM (mnk) dm dn dk
K N (F ) M (F )\M 0 (F )
1K (n1 m1 (m)(n))vM (n) dm dn.
N (F ) M (F )\M 0 (F )
If we set a = m1 (m) M 0 (F ) then we have n1 m1 (m)(n) = a(a1 n1 a(n)), which lies in K if and only if a KM and a1 n1 a(n) N (F ) K. Hence we have
G rM () = |DG ()| 2
1
M (F )\M 0 (F )
1KM (m1 (m))
N (F )
1N (F )K (a1 n1 a(n))vM (n) dn dm.
Let n = a1 n1 a(n) so that n = a (n ) then we have 1N (F )K (a1 n1 a(n))vM (n) dn =
N (F ) N (F )K
vM (a (n ))
n n
dn .
But we have n DM () = n DG ()
1 2
55 and hence
G rM () = |DM ()| 2
1
M
(F )\M 0 (F )
1KM (m1 (m))
N (F )K
vM (m1 (m) (n)) dn
dm
= |DM ()| 2 as wished.
1
M (F )\M 0 (F )
1KM (m1 (m))P (m1 (m)) dm
We now give a reduction for weighted orbital integrals using the topological Jordan decomposition; see [BWW02, Section 3]. We continue with the notation above and assume that G0 is dened over R and let K = G0 (R). Assume further that the automorphism has order prime to the residual characteristic of F and that K is stable under . For G0 (R) we can write G uniquely as = us = su with s absolutely semisimple (i.e., s has nite order prime to the residual characteristic of F ) and u topologically unipotent (i.e., uq 1, the identity in G0 , as n ). We now make the assumption of [BWW02, Lemma 5.5]. That is, we assume that if s1 and s2 for s1 , s2 K are residually semisimple and conjugate by an element of G0 (F ) then they are also conjugate by an element of K. This is automatic in the case that is trivial. In the [Fli99, Section I.H]. Under this assumption we have for g G0 (F ) that if g 1 (g) G0 (R) then g ZG0 (s)(F )K. For g ZG0 (s) we have g 1 us(g) = g 1 ugs. Hence g 1 us(g) K if and only if g 1 ug K. Furthermore, if we x s and set G1 = ZG0 (s) then we have ZG0 (us) = ZG1 (u). Assume now that M 0 (R). Then we have u, s M 0 (R) and, as in Lemma [BWW02, Lemma 5.5],
G rM (us) = |DG (us)| 2
1 n
case that G0 = GL(4) GL(1) and is as in Section 3.2 this is veried in [BWW02]; see also
1K1 (g 1 ug)vM (g) dg,
G1,u (F )\G1 (F )
where G1,u denotes the connected component of the centralizer of u in G1 and the measure on G1 (F ) is taken to give K1 = G1 (F ) K volume one. We now assume further that G1 is connected. We note that this is the case if G0 = GSp(4)
56 and is trivial or if G0 = GL(4) GL(1) and is as in Section 3.2. Then K1 is a hyperspecial maximal compact subgroup of G1 (F ) and P1 = ZP 0 (s) is a parabolic subgroup of G1 . Hence by the Iwasawa decomposition we again have G1 (F ) = P1 (F )K1 . Moreover, P1 has Levi decomposition M1 N1 where M1 = ZM 0 (s) and N1 = ZNP (s). We normalize the Haar measures on M1 (F ) and N1 (F ) to give M1 K1 and N1 K1 volume one. We can now mimic the proof of Lemma 5.16 to deduce the following. Lemma 5.17. For a M1 (F ) strongly G1 -regular let a : N1 N1 denote the inverse of the bijection N1 N1 : n a1 n1 an and dene P (a) =
N1 (F )K1
vM (a (n)) dn.
With the notations above we have
G rM (us) = |DM1 (u)| 2
1
M1,u (F )\M1 (F )
1KM1 (m1 um)P1 (m1 um) dm.
57
Chapter 6
The fundamental lemma for the (2,2) Levi
In this chapter we take M 0 to be the (2,2) Levi in G0 . We have A , c : A, B GL(2), c GL(1) M 0 = B
and we write such an element as a triple (A, B, c). The restriction of to M 0 is given by : (A, B, c) (wt B 1 w, wt A1 w, c det AB), where 1 .
We set M = GL(2) GL(1) the unramied elliptic twisted endoscopic group for M . In this chapter we prove the fundamental lemma for the pair (M, M ).
w= 1
6.1
Twisted integrals
In this section we concentrate on the calculation of the twisted integrals. Note that we have (I, B, 1)1 (A, B, c)(I, B, 1) = (Awt B 1 w, I, c det B), and hence every twisted conjugacy class in M 0 contains a representative of the form (A, I, c). We now determine the stable twisted conjugacy class of such an element. Lemma 6.1. Assume that M (F ) be semisimple. Let m M (F ) such that m1 (m)
58 M (F ). Then there exists m1 M (F ) such that m(m1 ) = m1 (m1 ). 1
m1 (A, I, c)(m) M (F ). We have
Proof. We may assume that = (A, I, c). We take m = (D, E, f ) M (F ) and assume that
m1 (A, I, c)(m) = (D1 Awt E 1 w, E 1 wt D1 w, c det DE). Hence we have E1 = E 1 wt D1 w GL(2, F ) and therefore, GL(2, F ) D1 Awt E 1 w = D1 ADwt E1 w Now there exists D1 GL(2, F ) such that
from which it follows that D1 AD GL(2, F ).
1 1 1 D1 AD1 = D1 AD. Then we can take m1 = (D1 , wt D1 wE1 , 1).
Thus the stable twisted conjugacy class of a strongly regular element is equal to the twisted conjugacy class of . We now show that the twisted orbital integrals on G0 can be written as untwisted orbital integrals on GL(2). Lemma 6.2. Let = (A, I, c) M (F ) be semisimple and strongly G0 -regular. Then if c UF
G we have rM () = 0. Otherwise, let T1 denote the centralizer of A in GL(2) then we have
G rM () = |DM ()| 2
1
1GL(2,R) (C 1 AC)P (C 1 AC, I, 1) dC.
T1 (F )\GL(2,F )
Proof. By Lemma 5.16 we have
G rM () = |DM ()| 2
1
T (F )\M 0 (F )
1KM (m1 (m))P (m1 (m)) dm.
But now let m = (C, D, e) M 0 (F ) then we have m1 (m) = (C 1 Awt D1 w, D1 wt C 1 w, c det CD). Thus we see that if m1 (m) KM then we have D1 wt C 1 w GL(2, R) from which it follows vanishes.
G that we must have det CD UF . But this then forces c UF and hence if c UF then rM ()
59 Now assume that c UF . Then we have that m1 (m) KM if and only if D1 wt C 1 w = C1 GL(2, R) and C 1 Awt D1 w = C 1 ACwt C1 w GL(2, R).
Which is if, and only if, C 1 AC GL(2, R) and D = wt C 1 wC1 with C1 GL(2, R). So we have m1 (m) KM if, and only if,
m = (C, wt C 1 w, e)(I, C1 , 1) with C 1 AC, C1 GL(2, R). Now we note that for k KM and n N (F ) we have k1 (k) (n) = k 1 ((k)n(k)1 )k, and hence P (k 1 (k)) =
N (F )K
vM (k1 (k) (n)) dn vM (k 1 ((k)n(k)1 )k) dn
N (F )K
= =
N (F )K
vM ( ((k)n(k)1 )) dn,
which equals P () after a suitable change of variables.
G Therefore the integrand in rM () is invariant under right multiplication of m by an element of
KM . Thus if we set T1 equal to the centralizer of A in GL(2) then we have
G rM () = |DM ()| 2
1
1GL(2,R) (C 1 AC)P (C 1 AC, I, 1) dC
T1 (F )\GL(2,F )
as wished.
6.2
Explicit statement of the fundamental lemma
We now give an explicit statement of the fundamental lemma for the pair (M, M ). Let = (A, I, c) M (F ) be semisimple. Under the norm maps we have N () = c det A cA c M (F ) GSp(4, F ),
60 and to c det A N () =
GL(2, F ) and c F for which (A, I, c) M is strongly G0 -regular we have
G rM ((A, I, c)) = rM GSp(4)
By Lemma 6.1 the fundamental lemma for the pair (M, M ) is the assertion that for all A
, cA M (F ) (GL(2, F ) GL(2, F )) . c
(diag(c det A, cA, c)) + rM
(GL(2)GL(2))
(diag(c det A, c), cA).
From Lemma 6.2 we know that the twisted integral vanishes if c UF . It is also clear from Lemma 5.16 that the integrals on GSp(4) and (GL(2) GL(2)) vanish if c UF . Thus the fundamental lemma is proven in this case. Moreover, if c UF then all integrals that appear in the statement of the fundamental lemma are independent of c and so we may assume that c = 1. Furthermore, we may as well assume that A K1 = GL(2, R). Having xed A we let T1 denote the centralizer of A in GL(2). Then we can write GL(2, F ) =
m0
T1 (F )zm K1
for an explicit set of representatives zm to be given below. Let P1 (resp. P2 ) denote the upper triangular parabolics in GSp(4) (resp. (GL(2) GL(2)) ) of which M is a Levi component. By abuse of notation we write B P (B) = P P1 (B) = P1
I
det B
, 1 B
det B P2 (B) = P2 for B GL(2, F ).
1
1 , B
Therefore the fundamental lemma we wish to prove is given by the following. Proposition 6.3. Let A GL(2, R) be such that = (A, I, 1) is strongly G0 -regular. Assume
1 that we have zm Azm GL(2, R) if and only if m N (A). Then N (A)
|DM ()| 2
1
m=0
1 1 vol(K1 zm T1 (F )zm \ K1 )P (zm Azm )
61 is equal to
N (A)
1 2
|DM (N ())|
m=0
1 1 1 vol(K1 zm T1 (F )zm \ K1 ) P1 (zm Azm ) + P2 (zm Azm ) .
We label the identity of this Proposition by F L(A). We now proceed to prove F L(A). We split the proof into two cases, in the rst we assume that A lies in a split torus, while in the second we assume that A lies in an elliptic torus.
6.3
equal to
Computation of P , P1 and P2
1 log q ,
In this section we give the expressions for P , P1 and P2 . For ease of notation we set vol(aG /Z( )) P P which has the eect of replacing log by logq below. We suppress the q from our notation and for the rest of this chapter take log to be log to the base q. We normalize the other volumes as in Section 5.5.
6.3.1
We have
Calculation of P
x1 1 1 x 3 NP = 1 a c x2 x4 , 1 . 1
If we identify NP (F ) with F 4 using x1 , . . . , x4 as our coordinates then for x = (A, I, 1) M 0 (F ) with A= b d x 1 b x2 . 0 x3 a x4
the map n x1 n1 x(n) is given by
x1 d 0 x2 0 det A d det A1 x3 c 0 x4 det A c Let B denote this matrix then we have
b 0 det A a 0
det A
det B = det A2 (det A 1)(det A tr A + 1);
62 and after a change of variables we have, for A GL(2, R), P (A) equal to |(det A1)(det Atr A+1)| charR4 (B t (x1 , x2 , x3 , x4 )) log max{1, |x1 |, |x2 |, |x3 |, |x4 |, |x1 x4 x2 x3 |}.
F4
6.3.2
We have
Calculation of P1
1 = x 1 1 r r . x 1 s
NP 1
We identify NP1 with F 3 using x, r and s as our coordinates. For det A y= A with A= a c b d
1
the map n y 1 n1 yn is given by
x (det A a)x cr f : r det A1 . bx + (det A d)r 2 2 s (det A 1)s + bx + (d a)xr cr Therefore after a change of variables we have
P1 (A) = |(det A 1)(det A tr A + 1)| charR3 (f (x, r, s)) log max{1, |x|, |r|, |s|}.
F3
6.3.3
Calculation of P2
1 x , I charR ((det A 1)x)vM 1 log max{1, |x|}.
In this case we have
P2 (A) = | det A 1| = | det A 1|
F
|x|| det A1|1
63
6.4
Proof of the fundamental lemma for split tori
In this section we prove Proposition 6.3 when A lies in a split torus. After conjugation we may assume that A lies in the diagonal torus T1 . We begin by giving a double coset decomposition for GL(2, F ). Lemma 6.4. For each m 0 let xm F be an element of valuation of m. Then we have GL(2, F ) =
m0
T1 (F )
1 xm 1
K1 .
Proof. By the Iwasawa decomposition we have GL(2, F ) = T1 (F )U (F )K1 , where U denotes the subgroup of GL(2) of upper triangular unipotent matrices. But for u UF and x F we have 1 ux u = 1 1 x u1 1 .
1
1
To check that the union of double cosets is disjoint we note that 1 xm 1 a b 1 xn a axn bxm = , 1 b
and for this matrix to lie in K1 we would need a, b UF and m = n. We now x a sequence of elements (xm ) as in Lemma 6.4 and we set 1 xm 1
zm = Note that we have
1 zm
. ,
a d
and therefore,
zm =
a (a d)xm d
We now set
1, if m = 0; 1 vol(K1 zm T1 (F )zm \ K1 ) = (q 1)q m1 , if m > 0. A= a d
then in the notation of Proposition 6.3 we have N (A) = v(a d). Using the action of the Weyl
64 group in GL(2) we can assume that we have |a 1| |d 1|. We recall that we are assuming that F has odd residual characteristic, so we can split the proof of Proposition 6.3 into the following three cases Case 1. |ad 1| = |a d| = |d 1| |a 1| Case 2. |a 1| = |d 1| = |ad 1| |a d| Case 3. |a 1| = |d 1| = |a d| |ad 1|. Our strategy will be to show that each case follows from proving the identity F L(A) when |ad 1| = |a d| = |d 1| = |a 1|. We then prove that the identity F L(A) holds in this case. In order to guarantee that, for any M 0, there exists a, d UF such that |ad 1| = |a d| = |d 1| = |a 1| = q M we need to make the additional assumption that q > 3. See Remark 6.9 below for the case that q = 3. We will need to compute P , P1 and P2 at elements of the form d 0 0 ad a b d
with a, d UF and 0 < |a d| |b| 1. For P the matrix B of Section 6.3.1 equals 0 d(a 1) 0 0 b 0 a(d 1) 0 b . 0 a ad
After suitable row operations, invertible over R, we can put B in the form 0 0 (a 1)d 0 0 ad 1 0 b . d1 0 0 1 b
0 0 d
Since the function vM is invariant under right multiplication by K we may assume that x1 = d1 x4 . After multiplying x2 by d1 we get that P (A) is given by |a 1||d 1||ad 1| times the integral of log max{1, |x2 |, |x3 |, |x4 |, |x2 x2 x3 |} 4
65 over the region in F 3 given by |x3 | |d 1|1 (ad 1)x4 + bx3 R (a 1)x2 + bx4 R. We have P1 at the element a b d
equal to |a 1||d 1||ad 1| times the integral of
log max{1, |x|, |r|, |s|} over the region in F 3 given by |x| |d 1|1 bx + d(a 1)r R (ad 1)s + x(bx (a d)r) R.
6.4.1
Reduction in case 1
We assume that we have N M and q M = |ad 1| = |a d| = |d 1| |a 1| = q N . We let L(M, N ) (resp. R(M, N )) denote the left (resp. right) hand side of the identity F L(A) in this case. We will see that L(M, N ) and R(M, N ) are well dened. In this section we prove the following Proposition. Proposition 6.5. For all N M we have qL(M, N + 1) L(M, N ) = 3q M 3 + (3M + N + 1)(q 1) = qR(M, N + 1) R(M, N ).
1 Proof. We begin by considering the twisted integrals P (zm Azm ). We need to integrate
log max{1, |x2 |, |x3 |, |x4 |, |x2 x2 x3 |} 4 over the region given by
66 |x3 | |d 1|1 bx3 + (ad 1)x4 R (a 1)x2 + bx4 R where b R with |a d| |b| 1.
We rst consider when |b|1 < |x3 | |d 1|1 . Then we have x4 = (ad 1)1 bx3 u1
with u1 UF
v(bx3 )
and |x4 | = |ad 1|1 |bx3 | > |b|1 . Therefore, x2 = (a 1)1 bx4 u2 = (a 1)1 (ad 1)1 b2 x3 u1 u2
with u2 UF
v(bx4 )
and |x2 | = |a 1|1 |ad 1|1 |b2 x3 |. Therefore,
x2 x2 x3 = b2 x2 u1 (ad 1)2 (a 1)1 ((a 1)u1 (ad 1)u2 ) 3 4 Since |(a 1)u1 (ad 1)u2 | = |d 1| for all such u1 and u2 we have |x2 x2 x3 | = |(ad 1)1 (a 1)1 ||bx3 |2 . 4 The contribution to the integral is |a 1|1 |ad 1|1 log |(ad 1)1 (a 1)1 ||bx3 |2 .
|b|1 <|x3 ||d1|1
We are now left with the region given by |x3 | |b|1 |x4 | |ad 1|1 (a 1)x2 + bx4 R. We now consider the case that |x4 | > |b|1 . Then we have x2 = (a 1)1 bx4 u
67 with u UF
v(bx4 )
and |x2 | = |a 1|1 |bx4 |. Now |x2 x2 x3 | = |x2 ||x2 x1 x3 |, 4 2 4
and |x2 x1 | = |a 1||b|1 |x4 | |b|1 . 4 2 Therefore making the change of variables x3 x3 x2 x1 gives the contribution to the integral as 4 2 |a 1|1 log max{|a 1|1 |bx4 |, |a 1|1 |bx4 ||x3 |},
|x3
||b|1
|b|1 <|x4 ||ad1|1
which we can write as the sum of |a 1|1 |b|1 and |a 1|1 (|ad 1|1 |b|1 )
1<|x3 ||b|1
|b|1 <|x4 ||ad1|1
log |a 1|1 |bx4 |,
log |x3 |.
Finally we are left with the remaining contribution, which is log max{1, |x2 |, |x3 |, |x4 |, |x2 x2 x3 |}. 4
|x3
||b|1
|x4
||b|1
|x2
||a1|1
qL(M, N + 1) L(M, N ). For b with |b| = q k where 0 k M we set P (M, N, k) = P the dierence is given by q M +1
q k <|x3 |q M
We note that the integrals above depend only on M , N and |b|. We now compute the dierence
a
b d
.
We need to compute qP (M, N + 1, k) P (M, N, k). From the rst contribution to the integral
(M + N + 1 2k + 2 log |x3 |)
minus q M
q k <|x3 |q M
(M + N 2k + 2 log |x3 |).
The dierence between the second contributions is given by q 2M +k+1
q k <|x4 |q M
(N + 1 k + log |x4 |)
68 plus q 2M +1 (q M q k ) minus q 2M +k
q k <|x4 |q M 1<|x3 |q k
log |x3 |.
(N k + log |x4 |)
minus q 2M (q M q k )
1<|x3 |q k
log |x3 |.
And the dierence between the third contributions is q 2M N
|x3 |q k |x4 |q k |x2 |=q N +1
log max{1, |x2 |, |x2 x2 x3 |}. 4
We note that |x1 x2 | q 2kN 1 < q k and so making the change of variables x3 x3 + x1 x2 in 4 4 2 2 this last integral gives q 2M N
|x3 |q k |x4 |q k |x2 |=q N +1
N + 1 + log max{1, |x3 |}.
Using Lemma 10.1 we get qP (M, N + 1, k) P (M, N, k) = (3M + N 2k + 1)(q 1) 1 + q M . Now we have qL(M, N + 1) L(M, N ) equal to q M times
M 1
(qP (M, N + 1, M ) P (M, N, M )) + (q 1) Using the fact that (1 q 1 ) for all m 1 we get
k=0
(qP (M, N + 1, k) P (M, N, k))q M k1 .
m i=0
iq i = mq m
qm 1 q1
qL(M, N + 1) L(M, N ) = 3q M + (3M + N + 1)(q 1) 3. We now consider the right hand side of the identity F L(A). First we consider the relevant integrals on GSp(4). Here we need to integrate log max{1, |x|, |r|, |s|} over the region in F 3 given by
69 |x| |d 1|1 bx + d(a 1)r R (ad 1)s + x(bx (a d)r) R. First we suppose that |b|1 < |x|. Then r = d1 (a 1)1 bxu with u UF
v(bx)
. We have
bx (a d)r = bx (a d)d1 (a 1)1 bxu = bx(a 1)1 d1 (d(a 1) (a d)u) and we note that |d(a 1) (a d)u| = |d 1| for all u UF is |a 1|1 |ad 1|1 We are now left with the region |x| |b|1 |r| |a 1|1 (ad 1)s + x(bx (a d)r) R to integrate over. Making the change of variables s s (ad 1)1 x(bx (a d)r) we see that the contribution to the integral is log max{1, |x|, |r|, |s (ad 1)1 x(bx (a d)r)|}.
|b|1 <|x||d1|1 v(bx)
. Hence we must have |s| = |a 1|1 |bx2 |. Thus the contribution to the integral log |a 1|1 |bx2 |.
|x||b|1
|r||a1|1
|s||ad1|1
Multiplying x, r and s by suitable units this integral equals log max{1, |x|, |r|, |s M x(bx M r)|}.
|x||b|1
|r||a1|1
|s||ad1|1
The integral on (GL(2) GL(2)) is given by
1 P2 (zm Azm ) = |ad 1|
1<|x||ad1|1
log |x|.
We note that the integrals above depend only on M , N and |b|. For |b| = q k , 0 k M , we dene P1 (M, N, k) and P2 (M, N, k) as we did for P (M, N, k). We now compute (qP1 (M, N + 1, k) P1 (M, N, k)) + (qP2 (M, N + 1, k) P2 (M, N, k)).
70 First we compute qP1 (M, N + 1, k) P1 (M, N, k). The rst part of the integral contributes q M +1
q k <|x|q M
N k + 1 + 2 log |x|
minus q M
q k <|x|q M
N k + 2 log |x|.
While the second part of the integral contributes q N 2M
|x|q k |r|=q N +1 |s|q M
log max{1, |x||r|, |s M x(bx M r)|},
which equals q N 2M
|x|q k |r|=q N +1 |s|q M
log max{|r|, |s xr|},
which equals q N M
|x|q k |r|=q N +1
N + 1 + log max{1, |x|},
since k M N . Putting this together and using Lemma 10.1 gives (qP1 (M, N + 1, k) P1 (M, N, k)) + (qP2 (M, N + 1, k) P2 (M, N, k)) equal to (3M + N k + 1)(q 1) 2 + 2q M . And we get qR(M, N + 1) R(M, N ) equal to 3q M 3 + (3M + N + 1)(q 1) as required.
6.4.2
Reduction in case 2
We assume that N M and q M = |a 1| = |d 1| = |ad 1| |a d| = q N . We let L(M, N ) (resp. R(M, N )) denote the left (resp. right) hand side of the identity F L(A) in this case. We will see that L(M, N ) and R(M, N ) are well dened. In this section we prove the
71 following Proposition. Proposition 6.6. For all N M we have L(M, N + 1) L(M, N ) = 0 = R(M, N + 1) R(M, N ). Proof. We begin by analyzing the twisted integrals P . For b with |b| = q k we write P (M, N, k) = P and we dene, for 0 k N , e(M, N, k) = P (M, N + 1, k) P (M, N, k). Now we have
N
a
b d
q N +1 L(M, N + 1) = P (M, N + 1, N + 1) + (q 1)
k=0 N
P (M, N + 1, k)q N k
N
= P (M, N + 1, N + 1) + (q 1)
k=0
P (M, N, k)q N k + (q 1)
k=0 N
e(M, N, k)q N k e(M, N, k)q N k
k=0
= P (M, N + 1, N + 1) P (M, N, N ) + q N +1 L(M, N ) + (q 1)
and therefore,
N
q N +1 (L(M, N + 1) L(M, N )) = P (M, N + 1, N + 1) P (M, N, N ) + (q 1)
e(M, N, k)q N k .
k=0
Thus we will be done with the left hand side if we can show that P (M, N +1, N +1) = P (M, N, N ) and e(M, N, k) = 0 for all k. Now recall that P (M, N, k) is given by q 3M times the integral of log max{1, |x2 |, |x3 |, |x4 |, |x2 x2 x3 } 4 over the region given by |x3 | q M bx3 + (ad 1)x4 R (a 1)x2 + bx4 R. We now consider the integral over this region for |b| = q k . First suppose that q k < |x3 | q M .
72 Then we have x4 = (ad 1)1 bx3 u1 with u1 UF
v(bx3 )
and x2 = (a 1)1 bx4 u2 = (a 1)1 (ad 1)1 b2 x3 u2 u1
with u2 UF
v(bx4 )
. Therefore,
x2 x2 x3 = (ad 1)2 b2 x2 u2 (a 1)1 (ad 1)1 b2 x2 u1 u2 3 3 1 4 = (ad 1)2 (a 1)1 b2 x2 u1 ((a 1)u1 (ad 1)u2 ). 3 We have |(a 1)u1 (ad 1)u2 | = |d 1| for all u1 and u2 and hence in the range q k < |x3 | q M we have log max{1, |x2 |, |x3 |, |x4 |, |x2 x2 x3 } = log |x2 x2 x3 | = 2M 2k + 2 log |x3 |. 4 4 We are now left to integrate over the region |x3 | min{q k , q M } |x4 | q M (a 1)x2 + bx4 R. Next we suppose that q k < |x4 | q M . Then we have x2 = (a 1)1 bx4 u with u UF
v(bx4 )
. Hence,
x2 x2 x3 = x2 + (a 1)1 bx4 ux3 = (a 1)1 bx4 u(u1 (a 1)b1 x4 + x3 ). 4 4 Now |u1 (a 1)b1 x4 | q M +k q M = q k . Hence making the change of variables x3 x3 u1 (a 1)b1 x4
73 gives the integral over this region as qM
|x3 |q k q k <|x4 |q M
(M k + log max{|x4 |, |x3 x4 |}) .
And nally we are left with the integral log max{1, |x2 |, |x3 |, |x4 |, |x2 x2 x3 |}. 4
|x3 |min{q k ,q M }
|x4 |min{q k ,q M }
|x2 |q M
Its clear from above that P (M, N, k) does not depend on N and hence we have e(M, N, k) = 0 for all k. Moreover, we see that P (M, N, N ) = q 3M
|x2 |,|x3 |,|x4 |q M
log max{1, |x2 |, |x3 |, |x4 |, |x2 x2 x3 |} 4
and hence we have P (M, N + 1, N + 1) = P (M, N, N ). Now we turn to the right hand side of the identity F L(A). Let R1 (M, N ) (resp. R2 (M, N )) denote the contribution to R(M, N ) from the sum over the P1 (resp. P2 ). First we consider the integral on (GL(2) GL(2)) . We have for 0 m N
1 P2 (zm Azm ) = q M
|x|q M
log max{1, |x|}
and its clear from this that we have R2 (M, N ) = R2 (M, N + 1). Now we consider the integral on GSp(4). For |b| = q k , 0 k N , we set a P1 (M, N, k) = P1 and dene e1 (M, N, k) = P1 (M, N + 1, k) P1 (M, N, k). As above we have
N
b d
,
q
N +1
(R(M, N + 1) R(M, N )) = P1 (M, N +1, N +1)P1 (M, N, N )+(q1)
e1 (M, N, k)q N k .
k=0
We now show that this expression is equal to zero. Having xed M we set, for m Z, I(m) = q 3M
|r|q M |s|q M
|r| log max{1, |r|, |s m r2 |}.
We note that I(m) is constant for m 2M . We will express P1 (M, N + 1, N + 1) P1 (M, N, N )
74 and e1 (M, N, k)q N k in terms of I(m). We begin by computing e1 (M, N, k). Recall that P1 (M, N, k) is equal to q 3M times the integral of log max{1, |x|, |r|, |s|} over the region |x| q M bx + d(a 1)r R (ad 1)s + x(bx (a d)r) R. First we suppose that q k < |x| q M . Then we have r = d1 (a 1)1 bxu with u UF
v(bx)
. Therefore, x(bx (a d)r) = bx2 d1 (a 1)1 (d(a 1) (a d)u)
and we have |d(a 1) (a d)u| = |d 1| for all such u. Hence over this region the integrand is equal to log |ad 1|1 |bx2 | and therefore the contribution to e1 (M, N, k) is zero. We are now left with the region |x| min{q k , q M } |r| q M (ad 1)s + x(bx (a d)r) R. So after scaling our variables by suitable units we can take this region to be |x| min{q k , q M } |r| q M M s + x( k x N r) R.
1 Making the change of variables x x + 2 N k r and r 2r, which doesnt change the integrand,
this region becomes
75 |x| min{q k , q M } |r| q M M s + k (x + N k r)(x N k r) R. Thus we see that if |x| > | N k r| then we have | k (x + N k r)(x N k r)| = | k x2 | = | k (x + N +1k r)(x N +1k r)| and the contribution to e1 (M, N, k) is zero. Therefore e1 (M, N, k) is equal to the dierence between the integral of q 3M log max{1, |r|, |s|} over the regions |r| q M |x| q kN |r| M s + k (x + N +1k r)(x N +1k r) R, and |r| q M |x| q kN |r| M s + k (x + N k r)(x N k r) R. Over the rst region the integral is equal the sum of q 3M q kN (1 q 1 )I(2N k), the contribution when |x| = q kN |r|, q 3M q kN 2 I(2N k + 2) the contribution when |x| q kN 2 |r|, and
q 3M q kN 1 (1 3q 1 )I(2N k + 2) + q 3M q kN 1 the contribution when |x| = q kN 1 |r|.
a=1
2q a (1 q 1 )I(2N k + 2 + a)
76 Over the second region the integral is equal to the sum of q 3M q kN 1 I(2N k) the contribution when |x| q kN 1 |r|, and
q 3M q kN (1 3q 1 )I(2N k) + q 3M q kN the contribution when |x| = q kN |r|.
a=1
2q a (1 q 1 )I(2N k + a)
Hence we have e1 (M, N, k)q N k equal to q 3M times
q 1 I(2N k) + q 1 (1 2q 1 )I(2N k + 2) + q 1
a=1
2q a (1 q 1 ) (I(2N k + 2 + a) I(2N k + a)) ,
which equals q 3M times the sum of q 1 I(2N k) q 1 I(2N k + 2), and
2q
1
(1 q
1
)
a=0
q
a
I(2N k + 2 + a) 2q
1
(1 q
1
)
a=0
q a I(2N k + 1 + a).
We now sum from k = 0 to N . By telescoping we have
N
k=0
q 1 I(2N k) q 1 I(2N k + 2) = q 1 I(N ) + q 1 I(N + 1) 2q 1 I(2M ).
While we have
N N
k=0 a=0
q a I(2N k + 2 + a)
k=0 a=0
q a I(2N k + 1 + a)
equal to
N +1 k=1 a=0 N
q a I(N + k + 1 + a)
q a I(N + k + 1 + a),
k=0 a=0
which equals
q
a=0
a
I(2N + 2 + a)
q a I(N + 1 + a),
a=0
which equals
1 I(2M ) q a I(N + 1 + a), 1 q 1 a=0
77 using the fact that I(m) is constant for m 2M . Putting this altogether we get
N
q 3M (q 1) equal to
e1 (M, N, k)q N k
k=0
(1 q 1 )I(N ) + (1 q 1 )I(N + 1) 2(1 q 1 )2
q a I(N + 1 + a).
a=0
equal to q 3M times the integral of
Next we compute P1 (M, N + 1, N + 1) P1 (M, N, N ) in terms of I(m). We have P1 (M, N, N )
log max{1, |x|, |r|, |s|} over the region |x|, |r| q M M s + N x(x r) R, which becomes, after the change of variables r x r that doesnt aect the integrand, |x|, |r| q M M s + N xr R. Since the region and integrand are symmetric in x and r we can compute this integral as twice the integral when |x| |r| minus the integral when |x| = |r|. The contribution from when |x| |r| is
a=0 |r|,|s|q M
(1 q 1 )| a r| log max{1, |r|, |s N +a r2 |},
which equals
a=0
q a (1 q 1 )I(N + a).
While the contribution when |x| = |r| is equal to (1 q 1 )I(N ). Hence we have P1 (M, N + 1, N + 1) P1 (M, N, N ) equal to q 3M times
2
a=0
q a (1 q 1 )I(N + 1 + a) (1 q 1 )I(N + 1)
78 minus 2
a=0
q a (1 q 1 )I(N + a) (1 q 1 )I(N ).
But we have 2
a=0
q a (1 q 1 )I(N + 1 + a) 2
a=0
q a (1 q 1 )I(N + a)
equal to 2(1 q 1 )2
a=0
q a I(N + 1 + a) 2(1 q 1 )I(N ),
and hence we have q 3M (P1 (M, N + 1, N + 1) P1 (M, N, N )) equal to
2(1 q 1 )2
a=0
q a I(N + 1 + a) (1 q 1 )I(N + 1) (1 q 1 )I(N ).
Thus R1 (M, N + 1) R1 (M, N ) = 0 as required.
6.4.3
Reduction in case 3
We assume that N M and q M = |a 1| = |d 1| = |a d| |ad 1| = q N . We let L(M, N ) (resp. R(M, N )) denote the left (resp. right) hand side of the identity F L(A) in this case. We will see that L(M, N ) and R(M, N ) are well dened. In this section we prove the following Proposition. Proposition 6.7. For all N M we have qL(M, N + 1) L(M, N ) = 2q M 2 + 2(M + N + 1)(q 1) = qR(M, N + 1) R(M, N ).
1 Proof. We begin by considering the twisted integrals P (zm Azm ). Again we need to integrate
log max{1, |x2 |, |x3 |, |x4 |, |x2 x2 x3 |} 4 over the region in F 3 given by |x3 | |d 1|1 bx3 + (ad 1)x4 R (a 1)x2 + bx4 R.
79 We rst consider the contribution when |b|1 < |x3 |. Then we have x4 = (ad 1)1 bx3 u1 with u1 UF
v(bx3 )
. Therefore |x4 | = |ad 1|1 |bx3 | > |b|1 and hence x2 = (a 1)1 bx4 u2 = (a 1)1 (ad 1)1 b2 x3 u1 u2
with u2 UF
v(bx4 )
. Thus,
x2 x2 x3 = (ad 1)2 (a 1)1 b2 x2 u1 (u1 (a 1) (ad 1)u2 ). 3 4 Since |u1 (a 1) (ad 1)u2 | = |d 1| for all u1 and u2 we have |x2 x2 x3 | = |(ad 1)2 b2 x2 |. 4 3 So the contribution when |b|1 < |x3 | is |ad 1|1 |a 1|1 We are now left to integrate over |x3 | |b|1 |x4 | |ad 1|1 (a 1)x2 + bx4 R. Suppose that |x4 | > |b|1 . Then we have x2 = (a 1)1 bx4 u with u UF
v(bx4 )
|b|1 <|x3 ||d1|1
log |(ad 1)2 b2 x2 |. 3
, and x2 x2 x3 = x2 + (a 1)1 bx4 ux3 = x4 (x4 + (a 1)1 bux3 ). 4 4
So after multiplying x3 by a suitable unit the contribution to the integral is |a 1|1 log max{| M bx4 |, |x4 (x4 + M bx3 )|}.
|x3
||b|1
|b|1 <|x4 ||ad1|1
80 Finally, when |x4 | |b|1 the contribution is log max{1, |x2 |, |x3 |, |x4 |, |x2 x2 x3 |}. 4
|x3
||b|1
|x4
||b|1
|x2
||a1|1
We dene P (M, N, k) as before and now compute qP (N + 1, M, k) P (N, M, k). From the rst contribution to the integral the dierence is given by q M +1
q k <|x3 |q M
(2N 2k + 2 + 2 log |x3 |)
minus q M
q k <|x3 |q M
(2N 2k + 2 log |x3 |).
The dierence between the second contributions is 2q M N +k
|x4 |=q N +1
log |x4 |,
and the dierence between the third contributions is zero. Using Lemma 10.1 we get qP (M, N + 1, k) P (M, N, k) = 2(M + N k + 1)(q 1), and we compute qL(M, N + 1) L(M, N ) = 2q M 2 + 2(M + N + 1)(q 1). We now turn our attention to the right hand side of the identity F L(A). First we look at
1 computing the integrals P1 (zm Azm ). We are integrating the function
log max{1, |x|, |r|, |s|} over the region |x| |d 1|1 bx + d(a 1)r R (ad 1)s + x(bx (a d)r) R. If |b|1 < |x| then we have
r = d1 (a 1)1 bxu
81 with u UF
v(bx)
. Then bx (a d)r = bxd1 (a 1)1 (d(a 1) (a d)u),
and we have |d(a 1) (a d)u| = |d 1| for all such u. Hence we have |s| = |ad 1|1 |bx2 |. Therefore, the contribution to the integral is |ad 1|1 |a 1|1 The region thats left is given by |x| |b|1 |r| |a 1|1 (ad 1)s + x(bx (a d)r) R. Making the change of variables s s (ad 1)1 x(bx (a d)r) gives the remaining integral as log max{1, |x|, |r|, |s (ad 1)1 x(bx (a d)r)|}. log |ad 1|1 |bx2 |.
|b|1 <|x||d1|1
|x||b|1
|r||a1|1
|s||ad1|1
And making the change of variables r r + (a d)1 bx gives this integral as log max{1, |x|, |r + (a d)1 bx|, |s (ad 1)1 (a d)xr|}.
|x||b|1
|r||a1|1
|s||ad1|1
We see that if |xr| > |a d|1 then the integrand equals log |ad 1|1 |a d||xr|, and so the contribution to the integral from this region is |ad 1|1 log |ad 1|1 |a d||xr|.
1<|x||b|1
|a1|1 |x|1 <|r||a1|1
Now we look at the contribution when |xr| |a d|1 . This is given, after suitable change of
82 variables in x and s, by log max{1, |x|, |r + M bx|, |s|}.
|x||b|1
|r||a1|1 ,|xr||a1|1
|s||ad1|1
We dene P1 (M, N, k) as before and we now compute qP1 (M, N + 1, k) P1 (M, N, k). The dierence between the rst contributions to the integrals gives q M +1
q k <|x|q M
(N + 1 k + 2 log |x|) q M
q k <|x|q M
(N k + 2 log |x|) .
The dierence between the second contributions is q 2M (q 1) (N M + log |y|)
1<|y|q k
q M <|y|q M +k
q M |y||x|q k
|x|1
= q M 1 (q 1)2 plus q 2M +1
q M <|y|q M +k q M |y||x|q k
(N + log |y|)(k + 1 log |y|)
|x|1 = q M (q 1)
1<|y|q k
k + 1 log |y|.
And the dierence between the third contributions is q N 2M
|x|q k |r|q M ,|xr|q M |s|=q N +1
log |s|.
Putting these altogether gives qP1 (M, N + 1, k) P1 (M, N, k) = (2M + N k + 1)(q 1) 1 + q M . We note that we have qP2 (M, N + 1, k) P2 (M, N, k) = (N + 1)(q 1) and hence q(P1 (M, N + 1, k) + P2 (M, N + 1, k)) (P1 (M, N + 1, k) + P2 (M, N + 1, k)) equals (2M + 2N k + 2)(q 1) 1 + q M . We now compute qR(M, N + 1) R(M, N ) = 2q M 2 + 2(M + N + 1)(q 1)
83 as desired.
6.4.4
Proof when M = N
We assume that we have |a 1| = |d 1| = |ad 1| = |d 1| = q M . We let L(M ) (resp. R(M )) denote the left (resp. right) hand side of the identity F L(A). We now prove the following Proposition which completes the proof of Proposition 6.3 in the case that A lies in a split torus. Proposition 6.8. For all M 0 we have L(M ) = 4M 4 1 q M = R(M ). q1
Proof. We begin by computing the left hand side of F L(A). For b with |b| = q k we set a P (M, k) = P As we have seen P (M, k) is equal to the sum of q M
q k <|x3 |q M
b 2M d
.
log(q 2k |x3 |2 ),
and q 2M +k
q k <|x4 |q M
log(q M k |x4 |),
and (q M q 2M +k ) and q 3M
|x3 |q k |x4 |q k |x2 |q M 1<|x3 |q k
log |x3 |,
log max{1, |x2 |, |x3 |, |x4 |, |x2 x2 x3 |}. 4
Putting this altogether gives P (M, k) = (4M 2k) + 1 q 3k3M q 3M (2 + q M + q 3k3M ) . q1 q3 1
84 And we get
M 1
L(M ) = P (M, M ) + (q 1)
k=0
P (M, k)q M k1 = 4M 4
1 q M . q1
We now compute R(M ). We dene P1 (M, k) and P2 (M, k) similarly. First we note that P2 (M, k) = M 1 q M . q1
We now compute P1 (M, k). As we have seen this is equal to the sum of q M
q k <|x|q M
log(q M k |x|2 ),
and q 3M
|x|q k |r|q M |s|q M
log max{1, |x|, |r|, |s (ad 1)1 x(bx (ad 1)r)|}.
We turn our attention to computing this latter integral. Its clear that if |x(bx (ad 1)r)| > 1 then the nal term dominates. We begin by computing the contribution to the integral in this case. We need to compute the volume of x and r such that |x(bx (a d)r)| = q m for m > 0. Making the change of variables 1 x x + (ad 1)b1 r, r 2r 2 turns this into |b|1 |bx (a d)r||bx (a d)r|. We now make the change of variables u = bx (ad 1)r and v = bx + (ad 1)r, which multiplies the integral by |b|1 |ad 1|1 . Given m with 0 m < k the volume of u and v such that |uv| = q m is
m n=0
vol(|u| = q n ) vol(|v| = q m+n ) = (m + 1)q m (1 q 1 )2 .
Thus the contribution to P1 (M, k) when |x(bx (ad 1)r)| > 1 is
k1
q kM
m=0
(m + 1)(M + k m)q m (1 q 1 )2 .
We are now left the range of integration |x| q k , |r| q M , x(bx (ad 1)r) R |s| q M
85 and after making of change of variables in s we can take our integrand to be log max{1, |x|, |r|, |s|}. We set l = k/2 , so that |bx2 | > 1 if and only if |x| > q l . We dene, for a 0, F (q a ) =
|s|q M
log max{q a , |s|} = M q M
qM qa . q1
need r = (ad 1)1 bxu with u UF contribution to the integral is
Let us rst consider the case that q l < |x| q k . Then in order that x(bx (ad 1)r) R we
v(bx2 )
. The volume of such r equals |(ad 1)1 x1 | and the
q l <|x|q k
|(ad 1)1 x1 |F (q M |bx|).
Now we consider the contribution when |x| q l . In this case we need to have that (ad1)xr R. When |x| 1 the contribution is F (1) +
1<|r|q M
F (|r|).
Finally we are left with the region 1 < |x| q l and |r| q M |x|1 . Lets set |x| = q i with 1 i l. Then |r| q M i . Note that for all such i we have q i q M i . If we split up the cases that |r| q i and q i < |r| q M then the contribution to the integral is
l
vol(|x| = q i ) q i F (q i ) +
i=1 q i <|r|q M i
F (|r|) .
Putting this altogether gives P1 (M, k) + P2 (M, k) equal to (4M k) + q M l q 3M q 3M +3l+2 q 3M +2 3 + 4q M q M k+l + . 21 q1 q (q + 1)(q 3 1)
And we compute the right hand side of F L(A) to be 4M q M 4 as required. qM 1 q1
Remark 6.9. We made the assumption that q > 3 in order to ensure that we could reduce to this M = N case. However, in the case that q > 3 the reductions made are still valid. The identity proven in the Proposition above is again valid, its just that it doesnt actually represent a case of
86 the fundamental lemma. Hence the fundamental lemma for the (2,2) Levi is proven in the case that q = 3 as well.
6.5
Proof of the fundamental lemma for elliptic tori
In this section we prove Proposition 6.3 in the case that A lies in an elliptic torus. In this case we may assume that a A= b bD a
with v(D) = 0 or 1 and ED = F ( D) a quadratic extension of F . We note that for = (A, I, 1) M 0 (F ) we have 1 1 |DM ()| 2 = |b D| = |DM (N ())| 2 .
GL(2, R)
We take the following from [Fli99, Section I.I]. Let T1 denote the torus in GL(2) with x T1 (F ) = y yD GL(2, F ) : x + y D E . D x
Let zm = diag(1, m ) then we have the double coset decomposition GL(2, F ) =
m0
T1 (F )zm K1 ,
where K1 = GL(2, R). We have a m bD a
1 zm Azm = m b 1 and so zm Azm K1 if and only if m v(b). We have
.
1 K1 zm T1 (F )zm
1 So if we set vol(D, m) = vol(K1 zm T1 (F )zm \ K1 ) then we have
x = m y
m yD x
K1
.
1, vol(D, m) = (q + 1)q m1 , m q ,
if ED /F unramied and m = 0; if ED /F unramied and m > 0; if ED /F ramied.
87 We set T (A) = det A tr A + 1. Then we have
1 P (zm Azm ) = | det A 1||T (A)| F4
charR4 (B t (x1 , x2 , x3 , x4 )) log max{1, |x1 |, |x2 |, |x3 |, |x4 |, |x1 x4 x2 x3 |},
where B is the matrix
1 We have P1 (zm Azm ) equal to | det A 1||T (A)| times the integral of
0 m b det A
a
0 det A a 0
m
b
m bD . det A a 0 0 a 0
m bD
det A
log max{1, |x|, |r|, |s|} over the region in F 3 given by (det A a)x m br R m bDx + (det A a)r R (det A 1)s m b(r2 2m Dx2 ) R. And we have
1 P2 (zm Azm ) = | det A 1| 1<|x|| det A1|1
log |x|.
As in the case that A lies in a split torus we will reduce the proof of F L(A) to certain cases. We nd, in the course of the proof, that the integrals in the identity F L(A) depend only on |b| and | det A 1|. We rst prove the equality in the case that b is a unit. Using similar reductions as above we reduce the proof of F L(A) when |b| | det A 1| to the case that |b| = | det A 1|; we the case that | det A 1| = |b2 D|; we then prove F L(A) in the case that |b2 D| | det A 1| < |b|. 6.9 allows us to deduce the fundamental lemma in the case that q = 3 as well. then prove F L(A) in this case. Similarly we reduce the proof of F L(A) when | det A 1| |b2 D| to We again need to make the assumption that q > 3. However the same argument as in Remark
6.5.1
Proof when b is a unit
We begin by proving Proposition 6.3 under the assumption that b UF . Proposition 6.10. Let A be as above with b UF . If we have |T (A)| = 1 then both sides of F L(A) are equal to 2|D| 2 | det A 1|
1
|x|| det A1|1
log max{1, |x|}.
88
1 Otherwise we must have v(D) = 1 and a UF , then if we set | det A 1| = q k we have both sides
of F L(A) equal to |D| 2
1
2k + 1 + q k1 2
1 q k1 q1
.
Proof. We rst compute the twisted integral. In this case after applying row operations invertible over R we get B in the form 0 0 a1 (det A 1)T (A) det A a det A(det A a) b
0 0 b 0 0 b2 Hence we have
0 . 0 ab
|x3 | |(det A 1)T (A)|1 and we can take bx4 = (a 1)x3 , bx1 = (det A a)x3 and b2 x2 = (det A2 a det A a2 + a)x3 . Then b2 (x1 x4 x2 x3 ) = det Ab2 T (A)x2 3 and hence |x1 x4 x2 x3 | = |T (A)x2 |. So we have 3 P (A) = | det A 1||T (A)| The integral on (GL(2) GL(2)) is | det A 1| log max{1, |x|}. log max{1, |x3 |, |T (A)x2 |}. 3
|x3 || det A1|1 |T (A)|1
|x|| det A1|1
In order to compute the integral on GSp(4) we need to integrate log max{1, |x|, |r|, |s|} over (x, r, s) F 3 such that det A a bD x R2 r det A a b
89 and (det A 1)s + b(Dx2 r2 ) R. Doing the row operation R2 bR2 + (det A a)R1 in the matrix above gives T (A) det A Therefore if |T (A)| = 1 we have P1 (A) = | det A 1| log max{1, |s|} det A a b 0 .
Hence we need |x| |T (A)|1 and (det A a)x br R.
|s|| det A1|1
1 (1 1)(2 1) and hence if |T (A)| < 1 we must have v(D) = 1 and a UF . It follows that
and the result follows. Let 1 = a + b D and 2 = a b D be the eigenvalues of A in ED . We have T (A) =
|T (A)| = q 1 . We now assume that this is the case and set | det A 1| = q k . The twisted integral is |D| 2 q k1 and the integral on (GL(2) GL(2)) is |D| 2 q k
1 1
|x3
|q k+1
log max{1, q 1 |x3 |2 }
|x|q k
log(max{1, |x|}).
For the integral on GSp(4) we rst note that b(r2 Dx2 ) R if and only if x and r are in R, and hence if and only if x R. The integral on GSp(4) is therefore the sum of |D| 2 q k1 the term contributing when |x| 1, and |D| 2 q 1 the term contributing when |x| = q. We compute the twisted integral to be |D| 2
1 1 1
|s|q k
log max{1, |s|},
k+1
|x|=q
2q k1 (k + 1)q k+1
q k+1 1 q1
q k1 (q k+1 1) .
90 The integral on (GL(2) GL(2)) equals |D| 2 q k kq k and the integral on GSp(4) equals |D| 2
1 1
qk 1 q1
q k1 kq k
qk 1 q1
+ (k + 1)(1 q 1 ) .
Hence we get both the left and right hand sides of the identity F L(A) equal to |D| 2 and we are done.
1
2k + 1 + q k1 2
1 q k1 q1
For the rest of this chapter we assume that |b| < 1.
6.5.2
Reduction when |b| | det A 1|
In this section we reduce the proof of Proposition 6.3 in the case that |b| | det A 1| to the case that |b| = | det A 1|. We note that if we have |b| < 1 and | det A 1| = 1 then we have |T (A)| = 1 and | det A 1| = 1. It follows that both sides of F L(A) vanish in this case. Thus we may as well assume that we also have | det A 1| < 1. Under the assumption |b| | det A 1| < 1 we have | det A a| = |a 1| = | det A 1| = q M and hence |T (A)| = |a 1|2 = q 2M . We set n = det A then n a(a 1)(n a)1 = (n a)1 (a(a 1)2 (a + 1) b2 D(n + a(a 1))). Hence if |b| < |a 1| we have |n a(a 1)(n a)1 | = |n 1|. On the other hand if |b| = |a 1| then, provided q > 3, given b we can choose a such that |a 1| = |b| and |n a(a 1)(n a)1 | = |n 1|, we make this further assumption in the case that |b| = |a 1|.
91 We now assume that N M and q N = |b| | det A 1| = q M . We let L(M, N ) (resp. R(M, N )) denote the left (resp. right) hand side of the identity F L(A) in this case. We now prove the following Proposition. Proposition 6.11. With the notations and assumptions above we have, for all N M 1, L(M, N + 1) L(M, N ) = f q N 1 |D| 2
1
2M
1 q M 1 q 3M q1 q3 1
= R(M, N + 1) R(M, N )
where f = f (ED /F ) is the degree of the residue eld extension.
1 Proof. We begin by seeing how to compute P (zm Azm ). Recall we have
0 B= m b det A
a
0 det A a 0 m b
m bD 0 det A a 0
m bD . 0 a
det A
We now do a series of row operations invertible over R to get E in a suitable form. The row operation R1 n1 (R1 ( m bD)(n a)1 R3) gives (a 1)(n a)1 0 m b n 0 na 0 m b 0 bD . na 0 0 a
m
0
1
Now we do R2 R2 ( m bD)R1 and R1 aR1 + R4 to give (a 1)(n a)1 m bD m b n Now n a(a 1)(n a)1 = (n a)1 (a(a 1)2 (a + 1) + b2 D(n + a(a 1))) n a(a 1)(n a)1 m b na 0 m b 0 . na 0 0 a 0 0 0
92 and therefore provided a 1 UF we have |n a(a 1)(n a)1 | = |n a|1 |a 1|2 = |n a| > | m bD|. Next we do R2 R2 (a 1) m bD(a a2 + n2 an)1 R1 to give (n a)1 (a a2 + n2 an) 0 m b n But now |(a 1)(a a2 + n2 an)1 b2 D| = |a 1|1 |b2 D|. After multiplying row 2 by a suitable unit and adding row 1 to row 4 and multiplying it by a1 we get (n a)1 (a a2 + n2 an) 0 m b (a 1)(n a)1
m
m b n a (a 1)(a a2 + n2 an)1 b2 D 0 m b
0 0 na 0
0 . 0 a
0
b
na 0 0
Therefore in order to compute the twisted integral we need to integrate the function log max{1, |x1 |, |x2 |, |x3 |, |x4 |, |x1 x4 x2 x3 |} over the region |x2 | |n 1|1 (n a)1 (a a2 + n2 an)x1 + m bx2 R m bx1 + (n a)x3 R (a 1)(n a)1 x1 x4 R. Note that we can set x4 = (a1)(na)1 x1 and make the change of variables x3 (a1)(na)1 x3 to give our integral as the integral of log max{1, |x1 |, |x2 |, |x3 |, |x2 x2 x3 |} 1 over the region |x2 | |n 1|1
0 . na 0 0 1 0
0
0
93 (n a)1 (a a2 + n2 an)x1 + m bx2 R m bx1 + (a 1)x3 R. First we note that for m such that | m b| |n 1| this region becomes |x2 | |n 1|1 |x1 | |n 1|1 |x3 | |n 1|1 . Now assume that | m b| > |n 1|. First suppose that | m b|1 < |x2 | |n 1|1 . Then we have x1 = (n a)(a a2 + n2 an)1 m bx2 u1 with u1 UF
v( m bx2 )
and |x1 | = |n a|1 | m bx2 | > | m b|1
hence x3 = (a 1)1 m bx1 u2 = (a 1)1 (n a)(a a2 + n2 an)1 2m b2 x2 u1 u2 with u2 UF
v( m bx1 )
; and therefore |x3 | = |(n 1)2 || m b|2 |x2 |. Now we have x2 x2 x3 equal to 1
x2 2m b2 u1 (a 1)1 (n a)(a a2 + n2 an)2 ((n a)(a 1)u1 (a a2 + n2 an)u2 ). 2 And since (n a)(a 1) (a a2 + n2 an) = nT (A) so |(n a)(a 1)u1 (a a2 + n2 an)u2 | = |n 1|2 for all u1 and u2 . Hence we deduce that |x2 x2 x3 | = | m b(n a)1 x2 |2 . 1 Thus the contribution to the integral is |n 1|2 2 log | m b(n 1)1 x2 |.
| m b|1 <|x2 ||n1|1
94 So we are now left with the region |x2 | | m b|1 |x1 | |n 1|1 m bx1 + (a 1)x3 R. We rst consider the case that | m b|1 < |x1 | |n 1|1 . Then we have x3 = (a 1)1 m bx1 u with u UF
v(bx1 )
. Hence |x3 | = |(n 1)1 m b||x1 |. Then x2 x2 x3 = x2 + (a 1)1 m bx2 x1 u 1 1 = (a 1)1 m bux1 ((a 1) m b1 u1 x1 + x2 ).
Now |(a 1) m b1 u1 x1 | | m b|1 and so making the change of variables x2 x2 (a 1) m b1 u1 x1 gives the integral as |n 1|1 log max{|(n 1)1 m bx1 |, |(n 1)1 m bx1 ||x2 |}.
| m b|1 <|x1 ||n1|1
|x2 || m b|1
Finally we are left with the region |x2 | | m b|1 |x1 | | m b|1 |x3 | |n 1|1 . We see that the integrals above depend only on |b|, |n 1| and m. For |a 1| = q M and
1 P (M, N, m) = P (zm Azm )
|b| = q N we set
then its clear from above that we have P (M, N + 1, m + 1) = P (M, N, m)
95 for all m with 0 m N . So we have
|D| 2 q N (qL(M, N + 1) L(M, N )) =
m=0
1
N +1
N
vol(D, m)P (M, N + 1, m)
m=0
vol(D, m)P (M, N, m),
which equals
N
vol(D, 0)P (M, N + 1, 0) +
m=0
(vol(D, m + 1) vol(D, m))P (M, N, m)).
In the case that |D| = q 1 we have vol(D, m) = q m for all m and hence we see that q N +1 |D| 2 (L(M, N + 1) L(M, N )) = P (M, N + 1, 0). In the case that |D| = 1 we have vol(D, 0) = 1 and vol(D, m) = (q + 1)q m1 . Hence vol(D, 1) vol(D, 0) = q and if m > 0 then vol(D, m + 1) vol(D, m) = (q + 1)q m (q + 1)q m1 = (q 1)(q + 1)q m1 . So we see that if |D| = 1 then |D| 2 q N +1 (L(M, N + 1) L(M, N )) = P (M, N + 1, 0) + P (M, N, 0). Now for N M we have P (M, N, 0) = q 3M
|x1 |q M |x2 |q M |x3 |q M
1 1
log max{1, |x1 |, |x2 |, |x3 |, |x2 x2 x3 |}. 1
Hence we get from Lemma 10.7 that q N +1 |D| 2 (L(M, N + 1) L(M, N )) = f
1
2M
1 q 3M 1 q M q1 q3 1
.
We now turn to computing the right hand side of F L(A). First we consider the integral on GSp(4). Recall we need to integrate log max{1, |x|, |r|, |s|} over the region in F 3 given by
96 (n a)x m br R m bDx + (n a)r R (n 1)s m b(r2 2m Dx2 ) R. Now consider na m b na
Doing the row operation R2 R2 + m bD(n a)1 R1 gives na m b . 0 (n a)1 nT (A) Note that |T (A)| = |n 1|2 and hence we need to integrate log max{1, |x|, |r|, |s|} over the region in F 3 given by |r| |n 1|1 (n a)x m br R (n 1)s m b(r2 2m Dx2 ) R. First suppose that | m b|1 < |r| |n 1|1 . Then we have x = (n a)1 m bru with u UF
v( m br)
m bD
.
. So
(n 1)s m b(r2 2m Dx2 ) = (n 1)s m b(r2 2m D((n a)1 m bru)2 ) = (n 1)s m br2 (1 Db2 (n a)2 u2 ). Hence the contribution to the integral is |n 1|2 log |n 1|1 | m br2 |.
| m b|1 <|r||n1|1
We are then left with the region |r| min{|n 1|1 , | m b|1 }
97 |x| |n 1|1 (n 1)s m b(r2 2m Dx2 ) R. The integrals above depend only on |b| = q N , |n 1| = q M and m. We set
1 P1 (M, N, m) = P1 (zm Azm )
and write P1 (M, N + 1, m + 1) = P1 (M, N, m) + e(M, N, m). Let R1 (M, N ) denote the contribution of the GSp(4) integral to the right hand side of the identity F L(A). Then we have |D| 2 q N +1 R1 (M, N + 1) q N R1 (M, N ) equal to
N N
1
P1 (M, N + 1, 0) +
m=0
(vol(D, m + 1) vol(D, m))P1 (M, N, m) +
1
vol(D, m + 1)e(M, N, m).
m=0
Thus when |D| = q 1 we have q N +1 |D| 2 (R1 (M, N + 1) R1 (M, N )) equal to
N
P1 (M, N + 1, 0) +
m=0
1
vol(D, m + 1)e(M, N, m)
and when |D| = 1 we have q N +1 |D| 2 (R1 (M, N + 1) R1 (M, N )) equal to
N
P1 (M, N + 1, 0) + P1 (M, N, 0) +
m=0
vol(D, m + 1)e(M, N, m).
We now set about computing e(M, N, m), which is given by the dierence between integrating q 3M log max{1, |x|, |r|, |s|} over the region |r| min{q M , q N m } |x| q M M s N m (r2 D(x m+1 )2 ) R, and over the region |r| min{q M , q N m } |x| q M
98 M s N m (r2 D(x m )2 ) R. When |r| | m x| we have |r2 D(x m )2 | = |r|2 = |r2 D(x m+1 )2 | and the integrals cancel. Hence e(M, N, m) is given by the dierence between integrating q 3M log max{1, |x|, |s|} over the regions |x| q M |r| q m1 |x| M s N m (r2 D(x m+1 )2 ) R, and |x| q M |r| q m1 |x| M s N m (r2 D(x m )2 ) R. Now note that when |r| q m2 |x| we have |r2 D(x m+1 )2 | = |D(x m+1 )2 | and |r2 D(x m )2 | = |D(x m )2 |. Hence e(M, N, m) is given as the dierence between integrating q 3M q m2 |x| log max{1, |x|, |s|} over the region |x| q M M s N +m+2 Dx2 R and |x| q M
99 M s N +m Dx2 R plus the dierence between integrating q 3M (1 q 1 )q m1 |x| log max{1, |x|, |s|} over the region |x| q M M s N +m+2 x2 R and |x| q M M s N +m Dx2 R. But adding all this together gives e(M, N, m) as the dierence between integrating q 3M q m2 |x| log max{1, |x|, |s|} over |x| q M M s N +m+2 Dx2 R and |x| q M M s N +m+2 x2 R plus the dierence between integrating q 3M q m1 |x| log max{1, |x|, |s|} over the region |x| q M M s N +m+2 x2 R and |x| q M
100 M s N +m Dx2 R. Having xed M , N and D we set I(k) equal to the integral of |x| log max{1, |x|, |s|} over the region |x| q M M s N +k x2 R. Then if |D| = 1 we have e(M, N, m) = q 3M m1 (I(m + 2) I(m)) and if |D| = q 1 we have e(M, N, m) = q 3M m2 (I(m + 3) I(m + 2)) + q 3M m1 (I(m + 2) I(m + 1)). We need to compute
N
vol(D, m + 1)e(M, N, m).
m=0
When |D| = 1 this sum is equal to
N
q 3M
m=0
(q + 1)q 1 (I(m + 2) I(m)) = (1 + q 1 )(I(N + 2) + I(N + 1) I(1) I(0))
while if |D| = q 1 this sum is equal to
N
q
3M
m=0
(q 1 (I(m + 3) I(m + 2)) + I(m + 2) I(m + 1)) = q 1 (I(N + 3) I(2)) + I(N + 2) I(1).
Finally we also need to compute P1 (M, N, 0), which equals q 3M times the integral of log max{1, |x|, |r|, |s|} over the region |r| q M |x| q M M s N (r2 Dx2 ) R.
101 This equals
|x||r|q M M s N (r 2 Dx2 )R
log max{1, |x|, |r|, |s|}
plus
|r|<|x|q M M s N (r 2 Dx2 )R
log max{1, |x|, |r|, |s|},
which equals the sum of |r| log max{1, |r|, |s|}
|r|q M
M s N r 2 R
and q 1
|x|q M M s N Dx2 R
|x| log max{1, |x|, |s|}.
Hence we get P1 (M, N, 0) = q 3M (1 + q 1 )I(0) and P1 (M, N + 1, 0) = q 3M (1 + q 1 )I(1) if |D| = 1. While if |D| = q 1 we get P1 (M, N + 1, 0) = I(1) + q 1 I(2). We note that when m N we have I(m) =
|x|q M |s|q M
|x| log max{1, |x|, |s|}
which by Lemma 10.4 is equal to q q+1 Therefore we have |D| 2 q N +1 (R1 (M, N + 1) R1 (M, N )) = f
1
M q 3M
q 3M 1 q3 1
.
M
1 q 3M q3 1
.
We now compute the integrals on (GL(2) GL(2)) . We have
1 P2 (zm Azm ) = q M
|x|q M
log max{1, |x|}
=M
1 q M . q1
1 Thus if we set P2 (M, N, m) = P2 (zm Azm ) then we have
P2 (M, N + 1, m + 1) = P2 (M, N, m). Hence if we let R2 (M, N ) equal the contribution to the right hand side of F L(A) from the integral
102 on (GL(2) GL(2)) then we have q N +1 |D| 2 (R2 (M, N + 1) R2 (M, N )) = f Putting these together gives q N +1 |D| 2 (R(M, N + 1) R(M, N )) = f q N 1 |D| 2 as required.
1 1 1
M
1 q M q1
.
2M
1 q M 1 q 3M q1 q3 1
6.5.3
Proof when |b| = | det A 1|
In this section we prove Proposition 6.3 under the assumption that |b| = | det A 1|. It follows
|b| = | det A 1| = q N . We let L(N ) (resp. R(N )) denote the left (resp. right) hand side of the identity F L(A). We now prove the following Proposition. Proposition 6.12. With the notations and assumptions above for all N 1 we have L(N ) and R(N ) equal to |D| 2
1
that we have |a 1| = | det A a| = |b| and |T (A)| = |b|2 . Let N 1 and assume that we have
4q + 3q N +1 + 2q N q 2N q N +3 q 4N 4N q 2N q N + + q1 (q 1)2 (q 1)(q 3 1)
if |D| = q 1 and equal to
(q + 1) 4(q + 1) + q N 1 (3q 2 + 6q + 1) 2q 2N 4N 2N q N 1 q N +3 q 4N + +2 (2N + 1)q N 1 2 q1 (q 1) (q 1)(q 3 1)
if |D| = 1.
1 Proof. We begin by computing the integrals P (zm Azm ). As we saw in the proof of Proposition
6.11 we can make row operations to put the matrix B in the form n a(a 1)(n a)1 m b na 0 m b 0 0
(a 1)(n a)1 m bD m b n
0 . na 0 0 a 0
103 Next we do R2 R2 ( m b)1 (n a)R1 and multiply the second row by a suitable unit to get n a(a 1)(n a)1 b
m 1
m b 0 0 m b
T (A) b
m
n
0 0 . na 0 0 a
0
0
Next we do R4 a1 (R4 R1) to give n a(a 1)(n a)1 m b1 T (A) m b (a 1)(n a)1 So we wish to integrate log max{1, |x1 |, |x2 |, |x3 |, |x4 |, |x1 x4 x2 x3 |} over the region given by |x1 | | m b1 T (A)|1 = | m b|1 m bx2 + (n a(a 1)(n a)1 )x1 R (n a)x3 + m bx1 R x4 + (a 1)(n a)1 x1 R. Thus we can take x4 = (a 1)(n a)1 x1 and make the change of variables x3 (a 1)(n a)1 x3 to give it as the integral of log max{1, |x1 |, |x2 |, |x3 |, |x2 x2 x3 |} 1 over the region |x1 | | m b|1 m bx2 + (n a(a 1)(n a)1 )x1 R (a 1)x3 + m bx1 R. Lets see how to compute this integral. Recall that |n a(a 1)(n a)1 | |b|. First suppose m b 0 0 0 0 . na 0 0 1 0 0 0
104 that |x1 | > |n a(a 1)(n a)1 |1 . Then we have x2 = m b1 (n a(a 1)(n a)1 )x1 u1 with u1 UF
v((na(a1)(na)1 )x1 )
and we note that |x2 | |x1 |. Moreover since
|n a(a 1)(n a)1 |1 |b|1 we also have |x1 | > | m b|1 and hence x3 = (a 1)1 m bx1 u2 with u2 UF
v( m bx1 )
and we have |x3 | = | m x1 | |x1 |. Now
x2 x2 x3 = x2 (n a(a 1)(n a)1 )(a 1)1 x2 u1 u2 1 1 1 = x2 (1 (n a(a 1)(n a)1 )(a 1)1 u1 u2 ) 1 and since 1 (n a(a 1)(n a)1 )(a 1)1 = (a 1)1 (n a)1 ((a 1)(n a) n(n a) + a(a 1)) = (a 1)1 (n a)1 (n)T (A) we have |1 (n a(a 1)(n a)1 )(a 1)1 u1 u2 | = 1 for all u1 and u2 . Hence when |x1 | > |n a(a 1)(n a)1 |1 the integrand is equal to 2 log |x1 |.
Now suppose we have |b|1 < |x1 | |n a(a 1)(n a)1 |1 . Then we have |x2 | | m b1 | and x3 = (a 1)1 m bx1 u
with u UF
v( m bx1 )
. Therefore |x3 | = | m x1 | |x1 |. Now x2 x2 x3 = x2 + (a 1)1 m bx1 x2 u 1 1 = x1 (x1 + (a 1)1 m bx2 u)
but |(a 1)1 m bx2 u| = | m x2 | |b|1 < |x1 | and hence when |b|1 < |x1 | the integrand is equal to 2 log |x1 |.
105 So the contribution to the integral when |b|1 < |x1 | | m b|1 is 2| m b2 |1 We are now left with the region |x1 | |b|1 |x2 | | m b|1 (a 1)x3 + m bx1 R. Next we suppose that |x1 | > | m b|1 . Then we have x3 = (a 1)1 m bx1 u with u UF
v( m bx1 )
|b|1 <|x1 || m b|1
log |x1 |.
and so |x3 | |x1 | > |x2 |. Now
x2 x2 x3 = x2 + (a 1)1 m bux1 x2 1 1 = u m (a 1)1 bx1 (u1 m (a 1)b1 x1 + x2 ) and |u1 m (a 1)b1 x1 | = | m x1 | | m b|1 so making the change of variables x2 x2 u1 m (a 1)b1 x1 gives the contribution when | m b|1 < |x1 | |b|1 as |b|1
| m b|1 <|x1 ||b|1
|x2 || m b|1
log max{| m x1 |, | m x1 ||x2 |},
which equals the sum of |b|1 | m b|1 and |b|1 (|b|1 | m b|1 ) Finally we are left with the region |x1 | | m b|1 |x2 | | m b|1 |x3 | |b|1 .
1<|x2 || m b|1
| m b|1 <|x1 ||b|1
log | m x1 |
log |x2 |.
106
1 With |b| = q N we have P (zm Azm ) equal to the sum of
2q N m
q N <|x1 |q N +m
log |x1 |
and q N m
q N m <|x1 |q N
(m + log |x1 |)
and (q N q N m ) and q 3N
|x1 |q N m |x2 |q N m |x3 |q N 1<|x2 |q N m
log |x2 |
log max{1, |x1 |, |x2 |, |x3 |, |x2 x2 x3 |}. 1
Putting these together we get
1 P (zm Azm ) = (2N + 2m) +
q N + q 3m 2 q 3m q 3N . q1 q3 1
Now we compute the left hand side of F L(A). When |D| = q 1 we get L(N ) = |D| 2
1
4q + 3q N +1 + 2q N q 2N q N +3 q 4N 4N q 2N q N + + q1 (q 1)2 (q 1)(q 3 1)
,
and when |D| = 1 we get L(N ) equal to
(q + 1) 4N 2N q N 1 (4(q + 1)) + q N 1 (3q 2 + 6q + 1) 2q 2N q N +3 q 4N + +2 (2N + 1)q N 1 . 2 q1 (q 1) (q 1)(q 3 1)
1 We now look to compute the right hand side of F L(A). We have P1 (zm Azm ) equal to q 3N
times the integral of log max{1, |x|, |r|, |s|} over the region in F 3 given by |x| | m b|1 m br (n a)x R (n 1)s m b(r2 2m Dx2 ) R. First suppose that |b|1 < |x| | m b|1 . Then we have |r| = | m x| and | m b(r2 2m Dx2 )| = |b m x2 | > 1.
107 Therefore |s| = | m x2 | and the contribution to the integral is | m b2 |1 which equals q 2N m
q N <|x|q N +m
|b|1 <|x|| m b|1
log | m x2 |,
(2 log |x| m) .
We are then left to compute, after multiplying s by a suitable unit, the integral of log max{1, |x|, |r|, |s|} over the region in F 3 given by |x| |b|1 |r| | m b|1 N s N m (r2 D( m x)2 ) R. The contribution when |r2 D( m x)2 | > q N m is q N +m
xED ,q N m <|x|ED q 2(N m) N m 2
log |x|ED + m.
Having xed N and m we set l =
if |r| q l and |x| q l+m and the contribution to the integral is
. If |D| = 1 then m b(r2 D( m x)2 ) R if and only
|x|q l+m
|r|q l
|s|q N
log max{1, |x|, |r|, |s|}.
If |D| = q 1 then we have m b(r2 D( m x)2 ) R if and only if |r| q l and |x| q l1 +m where l1 =
N m+1 2
and the contribution to the integral is log max{1, |x|, |r|, |s|}.
|x|q l1 +m
|r|q l
|s|q N
1 When |D| = q 1 , if N m = 2l we get P1 (zm Azm ) equal to q 3N times
(2N + m)q 3N
2q 3N q 3N m q 2m+3l+1 1 q 3l+2 + 2 + 3 + , q1 q 1 q 1 (q + 1)(q 3 1)
1 while if N m = 2l + 1 we have P1 (zm Azm ) equal to q 3N times
(2N + m)q 3N
q 2m+3l+3 1 q 3l+2 2q 3N q 3N m + 2 + 3 + . q1 q 1 q 1 (q + 1)(q 3 1)
108 We compute the contribution of the integral on GSp(4) to R(N ) to be |D| 2
1
3N q (N 2)q N 3q + 3q N q N +3 q 4N + + 2 q1 (q 1) (q 1)(q 3 1)
.
1 Now suppose that |D| = 1. Then we have P1 (zm Azm ) equal to q 3N times the sum of
(2N + m)q 3N + (N m 2l 2)q N +m+2l and 2
q N +m+2l 2q 3N 2q 3N m q1 q1
q 3N m q N +m+2l+2 q 3l+2m+1 1 q 3l+2 + 2 + 3 + . q2 1 q 1 q 1 (q + 1)(q 3 1)
We compute that the contribution of the integral on GSp(4) to R(N ) is equal to 2N q N 1 2 + 3N (q + 1) 2N q N 1 3q 3 + (2q 2 2q + 6)q N q N +3 + q N 2q 4N + + . q1 (q 1)2 (q 1)(q 3 1)
Now we compute the contribution of the integral on (GL(2) GL(2)) to R(N ). We have
1 P2 (zm Azm ) = q N
1<|x|q N
log |x| = N
1 q N . q1
And we compute that the contribution when |D| = q 1 is |D| 2 while when |D| = 1 it is q N + (q + 1)(1 q N ) q1 N 1 q N q1 .
1
q q N q1
N
1 q N q1
Putting these calculations together gives the computation of R(N ) and nishes the proof.
6.5.4
Reduction when | det A 1| |b2 D|
We now assume that we have | det A 1| |b2 D|. In this section we reduce the proof of Proposition 6.3 in the case that | det A 1| < |b2 D| to the case that | det A 1| = |b2 D|. So we assume that we have N M and q N = | det A 1| |b2 D| = q 2M |D|. We let L(M, N ) (resp. R(M, N )) denote the left (resp. right) hand side of the identity F L(A).
109 We note that under the assumption that | det A1| |b2 D| we have |a1|, | det Aa| |b2 D| and
so |T (A)| = |b2 D|. For ease of notation we set n = det A. We now prove the following Proposition. 2M + v(D),
Proposition 6.13. With the notations and assumptions above we have, for all M 1 and N
q q M q1
qL(M, N +1)L(M, N ) = |D| 2
1
(2N + 2M + 3)q (2N + 1)q M 2
= qR(M, N +1)R(M, N )
when |D| = q 1 and
qL(M, N +1)L(M, N ) = (2N +2M +2)(q+1)(4N +4)q M 2(q+1) 1 q M = qR(M, N +1)R(M, N ) q1
when |D| = 1.
1 Proof. We begin by computing the twisted integrals P (zm Azm ). As above we have
0 na B= m b 0 n m b
a
0
m bD 0 na 0
m bD . 0 a
n
We now do a series of row operations invertible over R to get B in a suitable form. First we do R2 R2 (n a)( m b)1 R4 and then divide by n to give (n a) m b1 m b n a 0 0 0
m
m bD 0 na b 0
(a 1) m b1 . 0 a
n
Next we do R3 aR3 + m bR1 and then divide by n to give (n a) m b1 0 n a 0 0 0 m b m bD 0 a1 0 n (a 1) b m b a
m 1
.
110 Next we do R2 aR2 (n a) m b1 R1 to give 0 0 n Now we note that a2 a n2 + an = a(a 1)2 (a + 1) + b2 D(n + a(a 1)) and since |a 1| |b| < 1 so |a2 a n2 + an| max{|a 1|2 , |b2 D|} |b2 |. Thus we can do R2 R2 (a2 a n2 + an) 2m b2 R3 to give a 0 0 n Thus we need to integrate log max{1, |x1 |, |x2 |, |x3 |, |x4 |, |x1 x4 x2 x3 |} over the region |x3 | |(n 1)D 2m |1 m bx4 + (a 1)x3 R ax1 + m bDx3 + nx4 R nx1 + m bx2 ax4 R. Therefore we can take x1 = a1 m bDx3 + a1 nx4 and then we need to integrate log max{1, |x2 |, |x3 |, |x4 |, |a1 nx2 + (a1 m bDx4 x2 )x3 |} 4 over the region |x3 | |(n 1)D 2m |1 0 0 0 m b m bD a(n 1)T (A) 2m b2 a1 0 0 . m b a n a 0 0 0 m b m bD (n a) a1 0
2m
n D (a a n + an) b m b a
2 2
m 1
.
111 m bx4 + (a 1)x3 R a1 n m bDx3 + a1 n2 x4 + m bx2 ax4 R. Now we make the change of variables x2 a1 nx2 + a1 m bDx4 to give our integral as the integral of log max{1, |x2 |, |x3 |, |x4 |, |x2 x2 x3 |} 4 over the region |x3 | |(n 1)D 2m |1 m bx4 + (a 1)x3 R m bx2 + m bDx3 + (n 1)x4 R
1 and we have P (zm Azm ) equal to q N 2M |D| times this integral.
First suppose that |x3 | > |a 1|1 . Then we have
|x4 | = | m b1 (a 1)x3 | < |x3 |. Now |(n 1)x4 | |bD m (a 1)x3 | < | m bDx3 |, hence | m bDx3 + (n 1)x4 | = | m bDx3 | > 1, and so |x2 | = | 2m Dx3 |. Therefore we have |x4 |2 = | 2m b2 (a 1)2 x2 | < |x2 x3 | = | 2m Dx2 | 3 3 and the integrand equals log | 2m Dx2 |. 3 We are now left with the region |x3 | min{|b2 D2 2m |1 , |a 1|1 } |x4 | | m b|1 m bx2 + m bDx3 R. If |x3 | > | m bD|1 then we have |x2 | = | 2m Dx3 |
and so |x2 x3 | = | 2m Dx2 | > |b2 D|1 | m b|2 |x4 |2 3
112 therefore the integrand is equal to log | 2m Dx2 | in this case as well. So the contribution to the 3 integral when | m bD|1 < |x3 | |(n 1)D 2m |1 is | 2m b2 |1 which equals q 2M 2m
q M +m |D|1 <|x3 |q N +2m |D|1
| m bD|1 <|x3 ||(n1)D 2m |1
log | 2m Dx2 |, 3
(2 log |x3 | 2m + log |D|) .
We are then left with the region |x3 | | m bD|1 = q M +m |D|1 |x4 | | m b|1 = q M m |x2 | | m b|1 = q M m to integrate over.
1 P (zm Azm ) and we see that 1 Thus we see that P (zm Azm ) depends only on m, |b| and |n 1|. We dene P (M, N, m) =
qP (M, N + 1, m) P (M, N, m) = (q 1)(2N + 2m + 2 log |D|). So we have
M
qL(M, N + 1) L(M, N ) = q M |D| 2 which equals |D| 2 if |D| = q 1 and
1
1
m=0
vol(D, m)(q 1)(2N + 2m + 2 log |D|),
(2N + 2M + 3)q (2N + 1)q M 2
q q M q1
(2N + 2M + 2)(q + 1) (4N + 4)q M 2(q + 1) if |D| = 1.
1 q M q1
We now turn to the computation of the right hand side of the identity F L(A). First we consider
1 the integrals P1 (zm Azm ), which are equal to |n 1||b2 D| times the integral of
log max{1, |x|, |r|, |s|} over the region in F 3 given by (n a)x m br R
113 m bDx + (n a)r R (n 1)s m b(r2 2m Dx2 ) R. We consider na m b na
then doing R2 R2 + (n a) m b1 R1 gives Now
m bD
m bD + (n a) m b1 (n a)
na
m b 0
.
m bD + (n a) m b1 (n a) = m b1 (b2 D + (n a)2 ) = m b1 nT (A), which has absolute value | m bD|. So after more row operations we get the matrix 0 m bD m b 0 .
1 Therefore P1 (zm Azm ) is equal to |(n 1)b2 D| times the integral of
log max{1, |x|, |r|, |s|} over the region in F 3 given by |x| | m bD|1 |r| | m b|1 (n 1)s m b(r2 D 2m x2 ) R.
1 We set P1 (M, N, m) = P1 (zm Azm ). Then we have
q 2M +N |D|1 (qP1 (M, N + 1, m) P1 (M, N, m)) equal to the sum of (N + 1)q N (q 1) vol({x, r : | m b(r2 D( m x)2 )| 1}),
114 the contribution when | m b(r2 D( m x)2 )| 1, and the sum of (q N +1 q N ) and q N +1 vol({x, r : | m b(r2 D( m x)2 )| 1}), which is the contribution when | m b(r2 D( m x)2 )| > 1. Putting these contributions together gives qP1 (M, N + 1, m) P1 (M, N, m) as the sum of (N + 1)(q 1) and q 2M +m |D|(q 1) and q 2M +m |D| vol{|x| q M |D|1 , |r| q M m : |r2 Dx2 | > q M m }. The integral above can be written as the sum of q M |D|(q 1) and q 2M +m |D|(q 1)
xED ,q M m <|x|ED q 2(M m) |x|q M |D|1 ,|r|q M m ,|r 2 Dx2 |>q M m
x,r:| m b(r 2 D( m x)2 )|>1
N + log | m b(r2 D( m x)2 )|
log |r2 Dx2 | M + m
q M m <|x|q M |D|1
log |Dx2 | M + m
log |x|ED M + m
And we have vol{|x| q M |D|1 , |r| q M m : |r2 Dx2 | > q M m } equal to q M m (q M |D|1 q M m ) + vol{x ED : q M m < |x|ED q 2(M m) }. Now we compute qP1 (M, N + 1, m) P1 (M, N, m) equal to (N + M + m + 2)q (N + M + m + 3) + q m if |D| = q 1 . And when |D| = 1 we have qP1 (M, N + 1, m) P1 (M, N, m) equal to (N + M + m + 1)q (N + M + m + 2) + 2q m 2q M +1 + q M 2 when M m is even and equal to (N + M + m + 1)q (N + M + m + 2) + 2q m q M 2 q m q M +1 q+1 q m q M +2 q+1
115 when M m is odd. With similar notation we have qP2 (M, N + 1, m) P2 (M, N, m) = (N + 1)(q 1). Using these computations we get qR(N + 1, M ) R(N, M ) = |D| 2 when |D| = q 1 and qR(N + 1, M ) R(N, M ) = (2N + 2M + 2)(q + 1) (4N + 4)q M 2(q + 1) when |D| = 1. 1 q M q1
1
(2N + 2M + 3)q (2N + 1)q M 2
q q M q1
6.5.5
Proof when |b2 D| | det A 1| < |b|
In this section we assume that we have |b2 D| | det A 1| < |b| and we prove Proposition 6.3 in this case. We set |b| = q M and | det A 1| = q N . We then have |T (A)| = |b2 D| = q 2M |D|. We let L(M, N ) (resp. R(M, N )) denote the left (resp. right) hand side of the identity F L(A). Again for ease of notation we set n = det A. We now prove the following Proposition. Proposition 6.14. Let M and N be such that M < N 2M + v(D). Then L(M, N ) and R(M, N ) are equal to
|D| 2
1
4q 2(q + 1)q M q N +1 + q N M (2N + 2M + 1)q (2N + 1)q M q N +2 q N 3M 1 + q1 (q 1)2 (q 1)(q 3 1)
if |D| = q 1 and are equal to
(4N + 2)q M 4(q + 1) 4q M (q + 1) 2q N +1 + 2q N M 2(N + M )(q + 1) q N q N 3M +2 2 q1 q1 (q 1) (q 1)(q 3 1)
if |D| = 1.
1 Proof. We begin by computing P (zm Azm ). As we saw in the proof of Proposition 6.13, we have 1 P (zm Azm ) equal to |(n 1)b2 D| times the integral of
log max{1, |x2 |, |x3 |, |x4 |, |x2 x2 x3 |} 4 over the region |x3 | |(n 1)D 2m |1
116 m bx4 + (a 1)x3 R m bx2 + m bDx3 + (n 1)x4 R. As we saw above the contribution to this integral when |x3 | > | m bD|1 is | 2m b2 |1 log | 2m Dx2 |. 3
| m bD|1 <|x3 ||(n1) 2m D|1
We are then left to integrate over the region |x3 | | m bD|1 m bx4 + (a 1)x3 R m bx2 + (n 1)x4 R. We note that if | m bD|1 |n 1|1 then this region becomes |x3 | | m bD|1 |x4 | | m b|1 |x2 | | m b|1 . On the other hand if | m bD|1 > |n 1|1 then when |n 1|1 < |x3 | | m bD|1 we have x4 = m b1 (a 1)x3 u with u UF
v((n1)x3 )
and |x2 | | m b|1 . The integrand in this case equals log max{|x3 |, |x2 x2 x3 |}. 4
But for x3 in this range we have |x2 ||x3 |1 | m b|1 and so after a change of variables in x2 the 4 integral over this range becomes | m b|1 log max{|x3 |, |x2 x3 |}.
|n1|1 <|x3 || m bD|1
|x2 || m b|1
We can write this integral as the sum of | m b|2 and | m b|1 (| m bD|1 |n 1|1 ) And nally we are left to integrate over
|x2 || m b|1
|n1|1 <|x3 || m bD|1
log |x3 |
log max{1, |x2 |}.
117 |x3 | |n 1|1 |x4 | | m b|1 |x2 | | m b|1 .
1 We let e {0, 1} be such that |D| = q e . Using the results of Chapter 10 we get P (zm Azm )
equal to q N 2M e times
(2N + 2m + e)q N +2M +e
2q N +2M +e q 3M 3m q 2M +e q 3M 3m 1 . q1 q3 1
And we compute L(M, N ) to be equal to
|D| 2
1
4q 2(q + 1)q M q N +1 + q N M (2N + 2M + 1)q (2N + 1)q M q N +2 q N 3M 1 + 2 q1 (q 1) (q 1)(q 3 1)
when |D| = q 1 and to be equal to
(4N + 2)q M 4(q + 1) 4q M (q + 1) 2q N +1 + 2q N M 2(N + M )(q + 1) q N q N 3M +2 2 q1 q1 (q 1) (q 1)(q 3 1)
when |D| = 1. We now turn to the computation of the right hand side of F L(A). We begin with the integrals
1 P1 (zm Azm ), which equal |(n 1)b2 D| times the integral of
log max{1, |x|, |r|, |s|} over the region |x| | m bD|1 m br (n a)x R (n 1)s m b(r2 D 2m x2 ) R. For |n 1|1 < |x| | m bD|1 we have |r| = | m b1 (n 1)x| and |s| = |(n 1)1 bD m x2 |. Hence the contribution to the integral is |n 1|1 | m b|1 log(|(n 1)1 bD m x2 |).
|n1|1 <|x|| m bD|1
118 We are then left with the region |x| min{|n 1|1 , | m bD|1 } |r| | m b|1 (n 1)s m b(r2 D 2m x2 ) R and we can compute this integral as in the proof of Proposition 6.12 when |n 1| = |b|. Having xed M , N and m we set l = to N + M + m + 1 + q m1
M m 2
1 . When |D| = q 1 we compute P1 (zm Azm ) equal
q N 2M 1 q N l q N 2M +3l+1 2 q m1 + + 2 + , 31 q1 q q 1 (q + 1)(q 3 1)
1 and when |D| = 1 we have P1 (zm Azm ) equal to the sum of
N +M +m and
2 q m+1 q N 2M +2 2 + 3 q1 q 1 q 1
(M 2l m 2)q 2M +2l+m + q 2M +2l+m
2q + 1 q N 2M +3l+2m+1 q N 2M +3l+2 + + . q+1 q2 1 (q + 1)(q 3 1)
We now assume that |D| = q 1 . We compute the contribution of the integral on GSp(4) to the right hand side of F L(A) to be equal to |D| 2 times 3q q M +1 2q M q N +2 q N 3M 1 (N + 2M + 1)q N q M + q1 (q 1)2 (q 1)(q 3 1) when |D| = q 1 . And when |D| = q 1 the integral on (GL(2) GL(2)) contributes |D| 2 times q q M q1 The sum of these expressions equals L(M, N ). We now assume that |D| = 1. The contribution of the integral on GSp(4) to R(M, N ) is equal to the sum of (N + M )q M and (q + 1) and 1 q M 2M q M q N 2M q N 3M (N + 2M ) (N + M )q M 3 + 2 + 2 q1 (q 1) q 1 (q 1)(q 3 1) q N (q 3 + 1) q N 2M (q + 1) q M + . q+1 (q 1)(q 3 1) 2q M q N 3M + 3 21 q q 1 N 1 q N q1 .
1 1
119 The integral on (GL(2) GL(2)) contributes N 1 q N q1 q M + (q + 1) 1 q M q1
to R(M, N ). Adding these together we nd they are equal to L(M, N ).
120
Chapter 7
The fundamental lemma for the (1,2,1) Levi I
In this chapter we take M 0 to be the (1,2,1) Levi in G0 . We have a 0 M = , e : A GL(2), a, b, e GL(1) A b
and we write such an element as a tuple (a, A, b, e). The restriction of to M 0 is given by : (a, A, b, e) (b1 , det A1 A, a1 , abe det A). We set M = GL(2) GL(1) an unramied elliptic twisted endoscopic group for M . In this chapter we prove the fundamental lemma for the pair (M, M ).
7.1
Stable conjugacy
We begin by determining the stable twisted conjugacy class of an -semisimple element = (a, A, b, e) M 0 (F ). For m = (a1 , A1 , b1 , e1 ) M 0 we have m1 (m) = ((a1 b1 )1 a, det A1 A1 AA1 , (a1 b1 )1 b, a1 b1 det A1 e). 1 1 Now if we assume that m1 m(m1 ) M 0 (F ) then its clear that we must have a1 b1 F and 1
det A1 F . Moreover, after twisted conjugation over F , we can assume that A is either diagonal or else lies in an elliptic torus of the form x Dy : x + y D E D y x
with v(D) {0, 1} and ED
121 = F ( D) a quadratic extension of F .
Lemma 7.1. Assume that A lies in the diagonal torus. Then the stable twisted conjugacy class of is equal to the twisted conjugacy class of . Proof. Let T denote the diagonal torus in GL(2). Then the question is given A1 GL(2, F ) with A1 AA1 GL(2, F ) and det A1 F does there exist B GL(2, F ) such that B 1 AB = A1 AA1 1 1 multiplying C on the left by an element of T (F ) we can insist that det C = det A1 . and det B = det A1 . We know there exists C GL(2, F ) such that C 1 AC = A1 AA1 ; and by 1
For = (a, A, b, e) with A diagonal we take the Haar measure on M (F ), which gives its maximal compact subgroup volume one. Lemma 7.2. Assume that A is non-central and lies in an elliptic torus as above. Then the stable twisted conjugacy class of is equal to the disjoint union of the twisted conjugacy classes of =
(a, A, b, e) and (a, c1 A, b, ce) with c F \ NED /F ED .
Proof. Let T denote the torus in GL(2) containing A. First its clear that (a, A, b, e) and (a, c1 A, b, ce) are not twisted conjugate over F . Its also clear that they are stably conjugate, since we can conjugate them by an element of the form (1, B, 1, 1) with B T (F ) such that det B = c. Next we show that every element of the stable twisted conjugacy class of is conjugate to one of these elements. Let 1 = m1 (m) = ((a1 b1 )1 a, det A1 A1 AA1 , (a1 b1 )1 b, a1 b1 det A1 e) 1 1 lie in the stable twisted conjugacy class of . Then we can nd B GL(2, F ) such that A1 AA1 = 1
hence change det B by an element of NED /F (ED ). Thus 1 is twisted conjugate over F to either
B 1 AB. We can change our choice of B by multiplying B on the left by an element of T (F ) and (a, A, b, e) or (a, c1 A, b, ce).
We continue with the assumption that A lies in an elliptic torus as above. First suppose that ED /F is ramied. Then we may take c UF . We note that the weighted orbital integral at the element (a, c1 A, b, ce) is the same as the weighted orbital integral at the element (ca, A, cb, c1 e), having multiplied by the element (c, diag(c, c), c, c2 ) which lies in Z(G0 ) K. But now conjugating this element by m = (c, I, 1, 1) gives (a, A, b, e). Thus the weighted orbital integral along the twisted conjugacy class of = (a, A, b, e) is equal to the weighted orbital integral along the twisted conjugacy class of (a, cA, b, ce). For such an A we take the measure on M (F ) that gives its maximal compact subgroup volume two.
122 Next we assume that ED /F unramied and we take c d Dd c
A= Conjugating this element by
with v(D) = 0. In this case (a, A, b, 1 e) is stably conjugate but not conjugate to = (a, A, b, e). 1, a, c 1 d 1
gives
, 1, 1 , b, e .
Dd c
If the stable twisted conjugacy class of = (a, A, b, e) intersects M 0 (R) then we can assume that we have a, b, e UF and A GL(2, R) with A as above. If we assume that (a, A, b, e) M 0 (R) then we see that the twisted conjugacy class of (a, A, b, 1 e) intersects M 0 (R) if and only if v(d) 1; this is clear from the double coset decomposition found in Section 6.5. For such an A we take the measure on M (F ) that gives its maximal compact subgroup volume one.
7.2
Statement of the fundamental lemma
In this section we give the statement of the fundamental lemma for the pair (M, M ). We recall that M sits inside GSp(4) as the Siegel Levi and we have EM (G) equal to GSp(4) with multiplicity two. Thus in this case the fundamental lemma states that for stable conjugacy class in M (F ) we have
G rM (k) = 2sM k GSp(4)
a strongly G-regular,
( )
where the sum on the left is over those twisted conjugacy classes in M 0 (F ) for which N (k) = . We now compute the function sM Lemma 4.8 we see that for = diag(g, awt g 1 w) a (stable) conjugacy class in M (F ) we have sG ( ) = rM M
GSp(4) GSp(4)
( ) whose denition is given in [Art02b, Section 5]. From
1 G (diag(g, awt g 1 w)) rM (diag(1, a det g 1 ), g) 2
where G = (GL(2) GL(2))/ GL(1). Therefore the fundamental lemma for the pair (M, M ) is
123 given by the following Proposition. Proposition 7.3. For = (a, g, b, e) M (F ) semisimple and strongly G0 -regular we have GSp(4) eag G rM ( ) = 2rM
G rM
eb det g w g
t 1
w
1
a1 b
, eag
where the sum on the left hand side is over representatives for the twisted conjugacy classes within the stable twisted conjugacy class of . For = diag(eag, eb det gwt g 1 w) M (F ) we take the Haar measure on M (F ) that gives its maximal compact subgroup volume one. For P 0 the upper triangular (1,2,1) parabolic in G0 we set vol(aG /Z( )) = P P
1 log q
and normalize
the other volumes as in Section 5.5. This has the eect of replacing log by logq below. We suppress the q from our notation and for the rest of this chapter take log to be log to the base q.
7.3
Proof of the fundamental lemma
stable twisted conjugacy class of does not intersect M 0 (R) if |a| = |b|. Its clear that the integrals on GSp(4) and (GL(2) GL(2))/ GL(1) also vanish in this case.
In this section we prove Proposition 7.3. We begin by noting that for = (a, g, b, e) M 0 (F ) the
twisted conjugacy class of intersects M 0 (R) if and only if eg is conjugate in GL(2) to an element in GL(2, R). Its also clear that if eg is not conjugate to an element in GL(2, R) then the integrals on GSp(4) and (GL(2) GL(2))/ GL(1) also vanish. We now assume that we have M 0 (R). We use the twisted topological Jordan decomposition
If |a| = |b| then we may, after twisted conjugation, assume that a, b UF . Then the stable
to prove the fundamental lemma. We can write M (R) uniquely as = us = su with u M 0 (R) topologically unipotent and s M (R) absolutely semisimple. The twisted weighted orbital integrals can now be computed using 5.17. We set N equal to the unipotent radical of the upper triangular parabolic of which M 0 is a Levi component, we dene N in GSp(4) similarly. Given s = (a1 , g1 , b1 , e1 ) we have ZM 0 (s) = {(a, g, a1 , e) M 0 : g 1 g1 g = g1 , det g = 1}. For u = (a, g, a1 , e) ZM 0 (s) topologically unipotent we have that the norm of in GSp(4)
124 is equal to the product of the absolutely semisimple element g1
1 a1 b1 det g1 wt g1 w 1
a1 e1 and topologically unipotent element
ae
g a
2
det gw g
t 1
w
.
We can then also use Lemma 5.17 to compute the weighted orbital integrals on GSp(4). We now proceed to prove the fundamental lemma by analyzing the possibilities for s.
7.3.1
s equal to the identity
We rst consider the case that s is the identity. In this case we have ZG0 () = Sp(4) GL(1) and we take = (u, e) Sp(4, R) UF topologically unipotent. Lemma 7.4. Suppose that s is the identity, then the fundamental lemma holds. Proof. We have a u= a1
g
topologically unipotent. By Lemma 5.17 we have
Sp(4, R)
G rM ((u, e)) = rKlingen (u)
Sp(4)
and hence for = (u, e) we have
G rM ( ) = u
rKlingen (u )
Sp(4)
where {u } is a set of representatives for the conjugacy classes within the stable conjugacy class of u. But now using Lemma 5.16 and the double coset decompositions for SL(2, F ) given in [Fli99, Lemma I.I.3] we have rKlingen (u ) = rKlingen (u).
u Sp(4) GSp(4)
From the fundamental lemma for the (2,2) Levi proven above we have
G rKlingen (u) = r(2,2) ((diag(ag, 1), 1)) r(T GL(2)) GSp(4) (GL(2)GL(2))
(diag(a2 , 1), ag).
125 Therefore to prove Proposition 7.3 we need to show that
G r(2,2) ((diag(ag, 1), 1))r(T GL(2)) (GL(2)GL(2))
(diag(a2 , 1), ag) = 2rM
GSp(4)
G (diag(ag, wt (ag)1 w))rM (diag(a, a1 ), g).
First we note that
G rM (diag(a, a1 ), g) = r(T GL(2)) (GL(2)GL(2))
(diag(a2 , 1), ag).
Next we note that the element ag wt (ag)1 w GSp(4)
lies in Sp(4) and by Lemma 5.16 we have 2rM
GSp(4)
(diag(ag, wt (ag)1 w)) = 2rM
Sp(4)
(diag(ag, wt (ag)1 w)).
Since this element is topologically unipotent we can apply Lemma 5.17 to get 2rM
GSp(4) G (diag(ag, wt (ag)1 w)) = r(2,2) ((diag(ag, wt (ag)1 w), 1)).
After twisted conjugation we have
G G r(2,2) ((diag(ag, wt (ag)1 w), 1)) = r(2,2) ((diag((ag)2 , I), 1))
and from the calculations of Chapter 6 we have
G G r(2,2) ((diag((ag)2 , I), 1)) = r(2,2) ((diag(ag, I), 1))
and we are done.
7.3.2
s central
We now assume that s = (a1 , g1 , b1 , e1 ) with g1 a scalar matrix. Therefore we have u = (a, g, a1 , e) with a, e GL(1) and g SL(2). In this section we prove Proposition 7.3 for = us either by reducing the proof to Lemma 7.4 or by showing that both sides of the identity in Proposition 7.3 vanish. We begin with the following Lemma. Lemma 7.5. Let = (a, g, b, e) M (F ) be semisimple and strongly G0 -regular. Then for
126 , UF we have
G G rM () = rM ((a, g, b, e)).
G Proof. Since we are free to scale by an element of Z(G0 )K without changing the value of rM ()
we have
G G rM ((a, g, b, e)) = rM ((1 a, g, 1 b, 1 e)).
But now for m = (, I, 1 , 1) we have m1 (1 a, g, 1 b, 1 e)(m) = (a, g, b, e) and we are done.
G G Now suppose that a1 = b1 . Then by Lemma 7.5 we have rM () = rM (u) and the fundamental
lemma in this case follows from Lemma 7.4. Proposition 7.3 in the case that a1 = b1 follows from the following. Lemma 7.6. With notation as above assume that we have a1 = b1 . Then both sides of the fundamental lemma vanish. Proof. We rst compute N ZG0 (s), by abuse of notation we work inside GL(4). For n= we have (n) = 1 x1 1 1 x2 x3 x4 N x5 1
1 x5 1
x4
x3 x1 x4 x2 x5 x2 x1 1
1
and
1 s1 ns =
a1 c1 x1 1 1
a1 c1 x2 1 1
a1 b1 x3 1
b1 c1 x4 1 . b1 c1 x5 1 1
127 Thus we need x1 = b1 c1 x5 1 x2 = b1 c1 x4 1 x4 = a1 c1 x2 1 x5 = a1 c1 x1 . 1 From which it follows that x1 = a1 b1 x1 and x2 = a1 b1 x2 . But since we are assuming that a1 = b1 , 1 1 it follows that x1 = x2 = x4 = x5 = 0. But now we need x3 = a1 b1 x3 , and hence x3 = 0 in this 1 case as well. Thus when a1 = b1 the twisted integral vanishes by Lemma 5.17. We now consider the right hand side of the fundamental lemma. First we consider the integral on GSp(4). The absolutely semisimple part of N () is g1
1 a1 b1 det g1 wt g1 w 1
s1 = a1 e1 We now compute ZGSp(4) (s1 ) N . For
.
1 n= we have 1 s1 ns1 = 1
x1 1 x3 1
x2
x1 N 1 a1 b1 x2 1 a1 b1 x1 1 1
a1 b1 x1 1 1 a1 b1 x3 1 1
from which it follows that ZGSp(4) (s) N = {I} if a1 = b1 and hence by Lemma 5.17 the integral on GSp(4) vanishes. Finally we consider the integral on (GL(2) GL(2))/ GL(1). The norm of the element in (GL(2) GL(2))/ GL(1) is equal to 1 , e1 a1 eag1 g (GL(2) GL(2))/ GL(1).
a2 a1 b1 1
1 1 And therefore if a1 = b1 then a1 b1 UF , and since u is topologically unipotent a2 UF . Hence 1 1 we have a2 a1 b1 UF and the integral on (GL(2) GL(2))/ GL(1) vanishes. 1
128
7.3.3
s diagonal
In this section we prove Proposition 7.3 in the case that s is diagonal but not central. So we take c1 d1
s = a1 ,
, b1 , e1
with c1 = d1 . After twisted conjugation we may assume that a1 = c1 = 1. We now compute N1 = N ZG (s); by abuse of notation we consider N GL(4). Lemma 7.7. Let s = (1, diag(1, d1 ), b1 , e1 ). Then we have the following possibilities for N1 . 1. If b1 = d1 = 1 then 1 x 1 1 N1 = x1 x2 x2 . x1 1 0
x2
1
2. If b1 = d1 = 1 then
3. If b1 = d1 = 1 then 1
4. If b1 = 1 and d1 = 1 then
1 N1 =
1 x 1 1 N1 =
0 . 1 x1 1 0 0
0 x2 1 1
5. In all other cases N1 = {I}.
1 N1 =
d1 x2 . 0 1 x3 0 . 0 1
0 0 1 1
129 Proof. For 1 n= 1 x5 1 1
x1 1
x2
x3
1
we have
and
(n) =
x4 x5 1
x4
x3 x1 x4 x2 x5 x2 x1 1
Hence we need
s1 ns =
1 x1 1
d1 x2
1
b1 x4 . b1 d1 x5 1 1
b1 x3
x1 = b1 d1 x5 1 x2 = b1 x4 x4 = d1 x2 x5 = x1 . Thus unless b1 = d1 we have x1 = x5 = 0. And unless b1 = d1 we have x2 = x4 = 0. And the only 1 way both can happen is if b1 = d1 = 1 (since we are assuming that d1 = 1). We also need to have b1 x3 = x3 x1 x4 x2 x5 and hence we need to have (1 b1 )x3 = x1 x4 + x2 x5 = (d1 1)x1 x2 . Putting this all together completes the proof.
We now compute the twisted integral in each of the above cases. We have c c
1
u = a,
, a1 , e
130 and so the stable twisted conjugacy class of = us is equal to the twisted conjugacy class of .
G Lemma 7.8. With notation as in Lemma 7.7 the twisted integral rM () is given by the following.
1. If b1 = d1 = 1 then
G rM () = |ac 1||ac1 1|
|x1 ||ac1 1|1
|x2 ||ac1|
log max{1, |x1 |, |x2 |, |x1 x2 |}.
2. If b1 = d1 = 1 then
G rM () = |ac1 1|
|x1 ||ac1 1|1
log max{1, |x1 |}.
3. If b1 = d1 = 1 then 1
G rM () = |ac 1|
|x2 ||ac1|1
log max{1, |x2 |}.
4. If b1 = 1 and d1 = 1 then
G rM () = |a 1|
|x3 ||a1|1
log max{1, |x3 |}.
G 5. In all other cases rM () = 0.
Proof. In each case we compute u1 n1 un for n N . In the rst case we have 1 1 u n un = 1 (1 a1 c)x1 1 1 (1 a1 c1 )x2 (1 a1 c)(1 a1 c1 )x1 x2 (1 a1 c1 )x2 (1 a1 c)x1 1 .
In the second case we have u1 n1 un = 1 (1 a1 c)x1 1 0 . 1 (1 a1 c)x1 1 0 0
131 In the third case we have 1 1 u n un = In the fourth case we have u1 n1 un = 1 0 1 1 0 (1 a2 )x3 0 0 1 . 1 0 (1 a1 c1 )x2 1 1 (1 a1 c1 )d1 x2 . 0 1 0
And of course in the fth case the integral vanishes.
Now we turn to the corresponding integrals on GSp(4). The absolutely semisimple part of N () is 1 s1 = e1 b1 b1 d 1
d1
Lemma 7.9. With notation as above we have the following possibilities for N1 = ZGSp(4) (s1 ) N . 1. If b1 = 1 then 1 x1 1 0 N1 = 1 x1 . 1 0
.
2. If b1 = d1 = 1 then
1 0 1 x 3 N1 = 1
x2 0 . 1
132 3. If b1 = d1 {1, 1} then 1 N1 = 0 0 . 1
0 1 x3 1
4. If b1 = d1 {1, 1} then 1
5. In all other cases N1 = {I}. Proof. For
1 N1 = 1 n=
0 1 0 1
x2 0 . 1
x1 1 x3 1
x2
we have
and the result follows.
s1 ns1 = 1
x1 N 1 d1 b1 x2
1
b1 x1 1 b1 d1 x3 1 1
b1 x1 1
We now need to compute the weighted integral on ZGSp(4) (s1 ) at the element ac ac1 u= These integrals are given in the following Lemma. Lemma 7.10. With notation as above the integral 2rM 1. If b1 = 1 then 2rM
GSp(4) GSp(4)
a1 c a1 c1
.
(N ()) is given by the following.
(N ()) = 2|a 1|
|x1 ||a1|1
log max{1, |x1 |}.
133 2. If b1 = d1 = 1 then 2rM
GSp(4)
(N ()) = |ac 1||ac1 1|
|x2 ||ac1|1
|x3 ||ac1 1|1
log max{1, |x2 |, |x3 |, |x2 x3 |}.
3. If b1 = d1 = 1 then 2rM
GSp(4)
(N ()) = |ac1 1|
|x3 ||ac1 1|1
log max{1, |x3 |, }.
4. If b1 = d1 = 1 then 1 2rM
GSp(4)
(N ()) = |ac 1| (N ()) = 0.
|x2 ||ac1|1
log max{1, |x2 |, }.
5. In all other cases 2rM
GSp(4)
Proof. We take n N1 . In the rst case we have 1 u1 n1 un = In the second case we have 1 1 u n un = In the third case we have u1 n1 un = 1 1 0 (1 a2 c2 )x3 1 0 0 . 1 1 1 0 (1 a2 c2 )x3 1 1 (1 a2 c2 )x2 0 . (1 a 1
2
)x1
0 1
(1 a2 )x1 . 1
0
134 In the fourth case we have 1 1 u n un = 1 1 0 (1 a2 c2 )x2 0 1 1 0 .
And in the fth case its clear that the integral vanishes.
For the integral on (GL(2) GL(2))/ GL(1) the norm of is 1 , e1 ea diag(c, d1 c1 ) (GL(2) GL(2))/ GL(1).
a2 b1
Thus we see that the integral vanishes unless b1 = 1 in which case it equals |a 1| log max{1, |x|}.
|x||a1|1
Combining the above lemmas proves Proposition 7.3 in this case.
7.3.4
s elliptic
We now assume that we have g1 GL(2, F ) which is non-central and lies in an elliptic torus. After stable twisted conjugation we can assume that we have c1 d1 Dd1 c1
with d1 = 0 and v(D) = {0, 1}. We let ED = F ( D). For s to be absolutely semisimple we need to have xk k g1 =
g1 =
GL(2, R)
xk
for some x F and k prime to the residual characteristic of F . But then, as an element of ED , we
have g1 = x for some k th root of unity . Since were assuming that g1 is non-central we must have F . Hence we must have ED /F unramied and v(D) = 0. After twisted conjugation we can take s = 1, c1 1 D c1 , b1 , e1 .
135 We now compute N1 = N ZG0 (s), which by abuse of notation we consider as a subgroup of GL(4). Lemma 7.11. With notation as above we have the following possibilities for N1 . 1. If b1 = 1 and c1 = 0 then 1 x 1 1 N1 = 1 N1 = x2 (Dx2 x2 )/2 1 2 Dx1 . x2 1 x3 0 . 0 1
1
2. If b1 = 1 then
0 0 1 1
3. In all other cases we have N1 = {I}. Proof. For
we have
and
s1 ns =
(n) =
1 n= 1 x5 1
x1 1
x2
x3
1
x4 x5 1
x4
x3 x1 x4 x2 x5 x2 x1 1
1
1 c1 x1 + x2 1
Dx1 + c1 x2
1
(c2 D)1 (b1 c1 x4 b1 Dx5 ) 1 . (c2 D)1 (b1 x4 + b1 c1 x5 ) 1 1
b1 x3
136 Hence we need x1 = (c2 D)1 (b1 x4 + b1 c1 x5 ) 1 x2 = (c2 D)1 (b1 c1 x4 b1 Dx5 ) 1 x4 = Dx1 + c1 x2 x5 = c1 x1 x2 . So we have (c2 D)x1 = b1 x4 b1 c1 x5 1 and from the third and fourth equations we get (c2 D)x1 = x4 c1 x5 . 1 Hence we have (1 + b1 )x4 + c1 (1 b1 )x5 = 0. We also have (c2 D)x2 = b1 c1 x4 b1 Dx5 1 and from the third and fourth equations we get (c2 D)x2 = c1 x4 + Dx5 . 1 So we have (1 + b1 )x4 + c1 (1 b1 )x5 = 0 c1 (b1 1)x4 D(1 + b1 )x5 = 0. Hence we deduce that (c2 (1 b1 )2 D(1 + b1 )2 )x4 = 0 1 and (c2 (1 b1 )2 D(1 + b1 )2 )x5 = 0. 1 Thus unless b1 = 1 and c1 = 0 we have x1 = x2 = x4 = x5 = 0. Now we also need to have x3 x1 x4 x2 x5 = b1 x3 .
137 Thus if b1 = 1 then we can take x3 to be anything we like. On the other hand if b1 = 1 and c1 = 0 we have x4 = Dx1 , x5 = x2 and x3 = We take
1 2
Dx2 x2 . 1 2
1 to be topologically unipotent, so c UF and d (). We have
u = a,
c d
Dd , a1 , e ZM 0 (s) c
cc1 + Dd D(c + dc1 ) , a1 b1 , ee1 . us = aa1 , c + dc1 cc1 + Dd M 0 (R), i.e., the other twisted conjugacy class within the stable twisted conjugacy class of us does not intersect M 0 (R). The twisted integrals at the element us are given by the following lemma.
G Lemma 7.12. With notation as above the twisted integrals rM () are given by the following.
Now c+dc1 UF and hence we deduce that it is only the twisted conjugacy class of us that intersects
1. If b1 = 1 and c1 = 0 then
G rM () = 2|DG ()| 2
1
log max{1, |x1 |, |x2 |}
over the region (1 a1 c)x1 a1 dx2 R a1 dDx1 + (1 a1 c)x2 R. 2. If b1 = 1 then
G rM () = |a 1| G 3. In all other cases rM () = 0. |x3 ||a1|1
log max{1, |x3 |}.
Proof. First suppose we have b1 = 1 and c1 = 0 then we have 1 u1 n1 un = (1 a1 c)x1 a1 dx2 1 1 a1 dDx1 + (1 a1 c)x2 . 1
138 If b1 = 1, then we have
And in all other cases its clear that the integral vanishes.
1 1 u n un =
1 0 1
0 (1 a2 )x3 0 1 0 1
.
Next we look at the integrals on GSp(4). The absolutely semisimple part of N () is c 1 1 s1 = e1 D c1 b1 c1 b1
. b1 D b1 c1
Lemma 7.13. With notation as above we have the following possibilities for N1 = ZGSp(4) (s1 ) N . 1. If b1 = 1 and c1 = 0 then 1 1 N1 = 1 N1 = x1 x3 1 Dx3 x1 . 1 Dx3 0 . 1
2. If b1 = 1 then
0 1 x3 1
3. In all other cases N1 = {I}. Proof. For
n=
1 1
x1 x3 1
x2
x1 1
139 we have 1 2 1 s ns = (c1 D) Hence we need (c2 D) x1 = b1 (c2 + D) 1 1 (c2 D) x2 = 1 (c2 1 2b1 c1 D 2b1 c1 D) x3 = x1 x1 x1 b1 c1 + b1 x2 x2 x2 b1 c1 D + + b1 D 2 c2 b1 1 x3 x3 x3 . 1 1 b1 (c2 + D)x1 b1 c1 x2 b1 c1 Dx3 1 2b1 c1 x1 + b1 x2 + c2 b1 x3 1 1 b1 (c2 + D)x1 b1 c1 x2 b1 c1 Dx3 1 . 1 2b1 c1 Dx1 + b1 c2 x2 + b1 D2 x3 1
+ b1 c2 1
That is (b1 c2 + b1 D + D c2 ) x1 1 1 2b1 c1 D 2b1 c1 x1 x1 + + b1 c1 x2 + + (c2 b1 1 b1 c1 D b1 D 2 +D c2 ) 1 x3 x3 x3 =0 =0 = 0.
(D + b1 c2 c2 ) x2 1 1 b1 x2
Equation 1 times D plus equation 2 times c1 gives D(D c2 )(1 + b1 )x1 + c1 (1 b1 )(D c2 )x2 = 0 1 1 and since D c2 = 0 we have 1 D(1 + b1 )x1 + c1 (1 b1 )x2 = 0 Next we do equation 2 times c2 b1 + D c2 minus equation 3 times b1 D2 to give 1 1 2b1 c1 D(1 b1 )(c2 D)x1 + (b1 1)(c2 D)((c2 D) + b1 (c2 + D))x2 = 0 1 1 1 1 and since D c2 = 0 we have 1 2b1 c1 D(1 b1 )x1 + (b1 1)((c2 D) + b1 (c2 + D))x2 = 0. 1 1 Thus we have D(1 + b1 )x1 + c1 (1 b1 )x2 = 0 2b1 c1 D(1 b1 )x1 + (b1 1)((c2 D) + b1 (c2 + D))x2 = 0, 1 1
140 which yields (D(b1 + 1)2 c2 (b1 1)2 )x1 = 0 1 and (b1 1)(c2 (b1 1)2 D(b1 + 1)2 )x2 = 0. 1 Therefore if c1 = 0 and b1 = 1 we can take x1 and x2 to be whatever we like; and then we have Dx3 = x2 . Now if b1 = 1 then we have x1 = 0 and x2 = Dx3 . In all other cases we have x1 = x2 = x3 = 0.
Now we compute the integrals on GSp(4). We need to compute the relevant integrals at the element ac adD ac a1 c a1 d
GSp(4)
ad u = e
. a1 dD a1 c
Lemma 7.14. With notation as above 2rM
(N ()) is given by the following. (N ()) equal to
1
1. If b1 = 1 and c1 = 0 then we have 2rM
GSp(4)
2|DGSp(4) (N ())| 2 over the region (a2 c2 Dd2 )x1 + 2cdDx3 R 2cdx1 + (a2 c2 Dd2 )x3 R. 2. If b1 = 1 then we have 2rM
GSp(4)
log{1, |x1 |, |x3 |}
(N ()) equal to log max{1, |x3 |}.
2|a 1| 3. In all other cases we have 2rM
GSp(4)
|x3 ||a1|1
(N ()) = 0.
Proof. Lets consider the rst case. We have 1 u1 n1 un = (a2 c2 Dd2 )x1 + 2cdDx3 1 2cdx1 + (a2 c2 Dd2 )x3 1 . 1
141 In the second case we have 1 u1 n1 un = 0 1 (1 a2 )x3 1 1 D(1 a2 )x3 0 .
And its clear that in the third case that the integral vanishes.
Again we recall that the integral on (GL(2) GL(2))/ GL(1) vanishes unless b1 = 1 in which case it equals |a 1|
|x||a1|1
log max{1, |x|}.
Thus its clear that the fundamental lemma holds in all cases except perhaps when b1 = 1 and of log max{1, |x|, |y|} over the regions in F 2 given by and ac dD d c1 = 0. We have |DG ()| = |DGSp(4) (N ())| and in this case we need to show that the integrals x R2 ac y x R2 a2 c2 Dd2 y 2cdD
a2 c2 Dd2 2cd
are equal. We have
and
det
ac dD
= max{|a c|2 , |d|2 } ac
d
a2 c2 Dd2 det 2cd
2cdD a c Dd
2 2 2
= max{|a2 c2 Dd2 |2 , |d|2 } = max{|a2 c2 |2 , |d|2 } = max{|a c|2 , |d|2 }
1 since a, c UF .
142 Thus if |d| |a c| both these matrices lie in GL(2, R) and if |d| < |a c| then both these matrices lie in GL(2, R) ac . ac d d
Hence the integrals above are equal and the proof of Proposition 7.3 is now complete.
143
Chapter 8
The fundamental lemma for the (1,2,1) Levi II
In this chapter we again take M 0 to be the (1,2,1) Levi in G0 . We have a 0 M = , e : A GL(2), a, b, e GL(1) A b
and we write such an element as a tuple (a, A, b, e). The restriction of to M 0 is given by : (a, A, b, e) (b1 , det A1 A, a1 , abe det A). We set M = GL(1) ResE/F GL(1) an unramied elliptic twisted endoscopic group for M 0 . In this chapter we prove the fundamental lemma for the pair (M, M ).
8.1
Statement of the fundamental lemma
the multiplicative valuation on E normalized such that ||E = q 2 . Given E we let denote its Galois conjugate. We x the Haar measure on E that gives RE volume one.
Let E denote the unramied quadratic extension of F . We x D F with v(D) = 0 such that E = F ( D). Let RE denote the ring of integers in E and UE the group of units. We let | |E denote
We recall from Lemma 4.5 that the elliptic twisted endoscopic groups for G0 in EM (G) are G1 = ResE/F GL(2) and G2 = (GL(2) ResE/F GL(1))/ GL(1). Moreover each group appears with multiplicity two and we have M sitting inside both of these groups as the diagonal torus. The stable twisted conjugacy classes in M 0 (F ), which transfer to M (F ), are those with repre-
144 sentatives of the form c a bD b a d
Moreover as we saw in Section 7.1 the stable twisted conjugacy class of is the disjoint union of the twisted conjugacy classes of and c a b b 1 D a d
=
, e .
=
, e .
And we have, using [KS99, Chapter 4], (N (), ) = (1)v(b) and (N (), ) = (1)v(b)+1 . We let = a + b D E . Then since neither G1 or G2 have proper elliptic endoscopic groups the fundamental lemma is given by the following Proposition. Proposition 8.1. Let and be as above then we have ce ce de
G G G rM () rM ( ) = (1)v(b) 2rM1
de
+ 2rG2
M
, .
1 log q
For P 0 the upper triangular (1,2,1) parabolic in G0 we set vol(aG /Z( )) = P P
and normalize
the other volumes as in Section 5.5. This has the eect of replacing log by logq below. We suppress the q from our notation and for the rest of this chapter take log to be log to the base q.
8.2
Proof of the fundamental lemma
We note that both sides of the identity in Proposition 8.1 vanish if the stable twisted conjugacy class of does not intersect M 0 (R). Thus we may assume that we have c a bD b a d
=
, e M 0 (R).
145
G G We now compute 2rM1 (N ()) and 2rM2 (N ()). We have
G 2rM1 (N ()) = |c d|E
|x||cd|1 E
log max{1, |x|E }
and
G 2rM2 (N ()) = |c d| |x||cd|1
log max{1, |x|}.
As in the previous chapter we use the twisted topological Jordan decomposition of to prove the fundamental lemma. So we write = us = su as a commuting product of an absolutely semisimple element s and a topological unipotent element u. We again analyze the possibilities for s and prove the fundamental lemma for each such s and every topologically unipotent element u that commutes with it.
8.2.1
s equals the identity
We now assume that s is equal to the identity. With a slight change in notation we take c ac1 bc1 bc1 D ac1 c1
1 1 with c, e UF , = a + b D UE and a2 Db2 = c2 .
=
, e Sp(4, R) UF
In order to use the calculations and reductions of Chapter 6 we make the further assumption that q > 3. However, arguing as in Remark 6.9 will give the fundamental lemma in the case q = 3 as well. We set |b| = q M and |c 1| = q N . Then we have q 3N M , if N M ; 1 |DG ()| 2 = q N 3M , if N M .
1
|DG2 (N ())| 2 = q N .
|DM ()| 2 = q M q 2N , if N M ; 1 |DG1 (N ())| 2 = q 2M , if N M .
1
Using Lemma 5.17 we note that the twisted weighted orbital integrals we need to compute on G0 are equal to the weighted orbital integrals on GSp(4) with respect to the Klingen Levi. Let M1 denote the Klingen Levi in GSp(4) and P1 the upper triangular parabolic of which M1 is a Levi
146 component. We also set N1 equal to the unipotent radical in P1 . We let P1 denote the function P1 (a) =
N1 (F )GSp(4,R)
vM1 (a (n)) dn
where a : N1 N1 the inverse of the map N1 N1 : n a1 n1 an. By abuse of notation we
G G identify with its component lying in Sp(4); then we have rM () rM ( ) equal to |DM ()| 2 M
1
times P1 () + (q + 1)
1 (1)m q m1 P1 (zm zm ) m=1
where
zm
And Proposition 8.1 says that it equals (1)M if N M and equals (1)M if N M . We now set about computing 2q 2M 2q 2N
1 =
1 m
. 1 qN 1 q1
N q 2N
q 2N 1 q2 1
+ q N
N qN
M q 2M
q 2M 1 q2 1
+ q N
N qN
qN 1 q1
M
P1 () + (q + 1)
m=1
1 (1)m q m1 P1 (zm zm ).
To put us in the same shape as Chapter 6 we scale our element by c to give c2 a bD b a
=
, 1
1 which of course doesnt change the value of P1 (zm zm ). And in the notation of Chapter 6 we have
a n = det b
bD a
= c2 .
147 We have |n 1| = |c 1| = q N and |b| = q M . Note also that we have | 1| = max{|a 1|, |b|}. But since a2 b2 D = c2 it follows that | 1| = max{|c 1|, |b|}. We begin by proving Proposition 8.1 under the assumption that |b|2 |n 1| |b|. So we have
1 M N . As we have seen in Section 6.5.5 P1 (zm zm ) for 0 m M equals the sum of
N +M +m and
q m+1 q N 2M 2 +2 2 + 3 q1 q 1 q 1
(M 2l m 2)q 2M +2l+m + q 2M +2l+m where l =
N m 2
2q + 1 q N 2M +3l+2m+1 q N 2M +3l+2 + + 21 q+1 q (q + 1)(q 3 1)
. Using this we compute that
M
P1 () + (q + 1)
m=1
1 (1)m q m1 P1 (zm zm )
equals (1)M 2q M M q 2M
q 2M 1 q2 1
+ q N +M
N qN
qN 1 q1
.
And since |DM ()| = q M Proposition 8.1 follows in this case. In proving the fundamental lemma for the (2,2) Levi in the case of an elliptic torus we reduced the proof to this case. We now follow these same reductions for the fundamental lemma here. First we assume that we have |c 1| |b|2 . We set
1 P1 (M, N, m) = P1 (zm zm )
and L(M, N ) = q M P1 (N, M, 0) + (q + 1)
M
(1)m q m1 P1 (N, M, m) .
m=1
We now compute qL(M, N + 1) L(M, N ). As we have seen in the proof of Proposition 6.13 we have qP1 (M, N + 1, m) P1 (M, N, m) equal to (N + M + m + 1)q (N + M + m + 2) + 2q m 2q M +1 + q M 2 if M m is even and equal to (N + M + m + 1)q (N + M + m + 2) + 2q m q M 2 if M m is odd. q m q M +1 q+1 q m q M +2 q+1
148 Using this we compute that qL(M, N + 1) L(M, N ) equals (1)M times the sum of q N +M and 2q M +1 M q 2M as required. Now assume that we have q N = |c 1| = |n 1| |b| = q M . Again we set
1 P1 (M, N, m) = P1 (zm zm )
(N + 1)q N +1
q N +1 1 q1
q N +M
N qN
qN 1 q1
q 2M 1 q2 1
2q M
M q 2M
q 2M 1 q2 1
and L(M, N ) = P1 (M, N, 0) + (q + 1)
M
(1)m q m1 P1 (M, N, m).
m=1
We denote, as in the case of the (2,2) Levi, e(M, N, m) = P1 (M + 1, N, m + 1) P1 (M, N, m), and we have
M
L(M + 1, N ) + L(M, N ) = P1 (M + 1, N, 0) P1 (M, N, 0) + (q + 1) And as we have seen in the proof of Proposition 6.11
(1)m+1 q m e(M, N, m).
m=0
e(M, N, m) = q 3N m1 (I(m + 2) I(m)) where I(m) is...