This version of the fundamental lemma is needed to stabilize the twisted trace formula for the pair (GL(4)×GL(1), α). This stabilized twisted trace formula is required for the Arthurian classification of the GSp(4) discrete spectrum in terms of GL(4) automorphic representations.
The trace formula
The most powerful uses of the trace formula, for example for questions of functionality, occur when the trace formula is not used in isolation. To match the geometric sides of the trace formula for GandG0 one must be able to.
Stabilization
In this thesis we will be interested in the stabilization of the twisted trace formula. Work is underway to stabilize the remaining terms in the rotated trace formula; however, it relies on a fundamental lemma for twisted weighted orbital integrals given in [Art02a]; see also chapter 3.
Transfer from GSp(4) to GL(4)
For the stabilization of the twisted trace formula for GL(4)×GL(1) and the automorphism α given in section 3.2 below, the twisted fundamental lemma for invariant orbital integrals has been proven in [Fli99]. In this case we start by writing both sides of the fundamental lemma as untwisted orbital integrals on GL(2, F).
Parabolic subgroups
GSp(4) and Sp(4)
The intersection of each of these parabolic subgroups with Sp(4) is a parabolic subgroup of Sp(4), we call their intersection with Sp(4) of the same name. The double group of GSp(4) is GSp(4,C) and under the bijection between parabolic subgroups of GandGb the Siegel and Klingen parabolas are switched.
Weighted orbital integrals
Endoscopy
Notation
In this chapter, we state the rotated weighted fundamental lemma from [Art02a]. As mentioned there, it is stated in such a way as to include the statement of the weighted fundamental lemma found in [Art02b, Section 5].
Our case
Elliptic twisted endoscopic groups for G0 itself are computed in [Fli99, Section I.F]; these results are recalled in Section 4.1. In this section we recall the results from [Fli99, Section I.F] on twisted endoscopic sets for G0.
Twisted endoscopic groups for the (2,2) Levi
In this case, we only obtain elliptical endoscopic datum if Γ acts through a quadratic extension E/F; in which case we obtain (GL(2) ×ResE/FGL(1))/GL(1), with GL(1) embedded as (z, z−1), as a twisted endoscopic group. Then the elliptic twisted endoscopic groups of G0 are iEM0(G) GSp(4) and (GL(2)×GL(2))0, where the prime denotes the subgroup of pairs (A, B) with detA= detB.
Twisted endoscopic groups for the (1,2,1) Levi
Therefore, we need to look for an elliptical endoscopic starting point for G0, which is derived from elements of the form (diag(1,1, λ, λ), λ)ˆα∈G. We need to look for elliptic twisted endoscopic groups for G0 given by translating sˆα over the form elements.
Twisted endoscopic groups for the diagonal Levi
Then the elliptical twisted endoscopic groups for G0 iEM0(G) consist of (GL(2)×ResE/FGL(1))/GL(1)andResE/FGL(2)0. Therefore, we need to look for the elliptic twisted endoscopic groups for G0 given by elements of the form.
Endoscopic groups for GSp(4)
Aparabolic subset of G is by definition a non-empty set of the form P+∩WhereP+ is a parabolic subgroup of G+ defined over F. Let P =P+∩G be a parabolic subset of G, a Levi component ofP will be a setM =M+∩P, where M+ is the normalizer inG+ of some Levi component ofP0=G0∩P which is defined over F.
Twisted weight functions
The (2,2) Levi
By [Art81, Lemma 6.2] for eachx∈G+(F) this function extends to a smooth function onia∗M. NQ) denotes the unipotent radical of P0 (resp. We set P0 equal to the upper triangular Borel subgroup of GL(4) and we take M0 as the diagonal torus inP0.
The (1,2,1) Levi
The diagonal Levi
Taking the inverse of the transpose of the above matrix equation and applying the vector (0,0,0,1) allows us to conclude. To do this we must write n=n1m1k1 for every Borel subgroup of Sp(4) containing M1.
Weight functions for GSp(4)
The Siegel Levi
We set P0 equal to the upper triangular Borel subgroup of GSp(4) and we take M0 to be the diagonal torus in P0.
The Klingen Levi
The calculation of vM on the impotent radical of P follows directly from the proof of Lemma 5.3. The calculation of vM on the impotent radical of P follows directly from the proof of Lemma 5.5.
The diagonal Levi
Now every Borel subgroup of GSp(4) is of the form −1bw With An of the Weyl group of Sp(4). Since we are assuming g∈Sp(4) we can do this inside Sp(4) and assume that P ∈M1 for every P .
Other groups
Normalization of volumes
Next, we take M0 to be the (1,2,1) Levi inG0 and P0 the upper triangular parabolic inG0 with M0 as a Levi component. Next, we take M0 equal to the diagonal Levi inG0 and P0 the upper triangular parabolic inG0 with M0 as a Levi component.
Weighted orbital integrals
We further assume that the automorphism α is prime with respect to the residual characteristic F and that K is stable under α. Thus, the stable twisted conjugation class of a strongly regular element γ is the same as the twisted conjugation class γ.
Explicit statement of the fundamental lemma
Moreover, ifc∈UF then all the integrals that appear in the statement of the fundamental lemma are independent of c and so we can assume that c = 1. We divide the proof into two cases, in the first we assume that A lies on a torus separated, while in the second we assume that Alies in an elliptic torus.
Calculation of σ P
Proof of the fundamental lemma for split tori
Reduction in case 1
By making the change of variables7→s−(ad−1)−1x(bx−(a−d)r), we see that the contribution to the integral is
Reduction in case 2
Over this region the integrand is therefore equal to log|ad−1|−1|bx2|and therefore the contribution toe1(M, N, k) is zero. So after scaling our variables with appropriate units we can assume that this region exists. Since the region and the integrand are symmetric, we can calculate this integral as twice the integral when|x| ≤ |r|minus the integral if|x|=|r|.
Reduction in case 3
We see that if|xr|>|a−d|−1 the integrand is equal to log|ad−1|−1|a−d||xr|, and so the contribution to the integral from this region is equal.
Proof when M = N
If we divide the cases that |r| ≤qi andqi<|r| ≤qM then the contribution to the integral is We made the assumption that >3 to ensure that we can reduce in this case M = N. The identity proved in the above Proposition is again valid, it just does not really represent a case of.
Therefore the basic lemma for Levi (2,2) is proved in the case that q= 3 as well.
Proof of the fundamental lemma for elliptic tori
Proof when b is a unit
For the integral on GSp(4) we first note that b(r2−Dx2)∈R if and only if xandr is inR, and therefore if and only ifx∈R.
Hence we take both the left and right sides of the identity F L(A) equal to. We have now decided to compute (M, N, m), which is given by the difference between the integration of q−3Mlog max{1,|x|,|r|,|s|}. Therefore, if we let R2(M, N) equal the contribution to the right-hand side of F L(A) from the integral.
Now we make the change of variablesx27→a−1nx2+a−1πmbDx4to give our integral as the integral of.
Then the stable class of the twisted conjugation γ is the same as the class of the twisted conjugation γ. Then the stable twisted conjugation class γ is equal to the disjoint union of the twisted conjugate classes γ = (a, A, b, e) and (a, c−1A, b, ce) with c∈F×\NED/ FED×. We then show that every element of the stable twisted conjugacy class γ is conjugate to one of these elements.
The weighted orbital integral along the twisted conjugation class of γ= (a, A, b, e) is therefore equal to the weighted orbital integral along the twisted conjugation class of (a, cA, b, ce).
Statement of the fundamental lemma
For such anA we take the measure onMγα(F) which gives the maximum compact subgroup volume one. Forγ0= diag(eag, ebdetgwtg−1w)∈M0(F) we include the Haar measureMγ00(F) which gives the maximum compact subgroup volume one. We drop theq from our notation and use log as log for the baseq in the remainder of this chapter.
Proof of the fundamental lemma
Then, according to Lemma 7.5, we have rMG(γα) =rMG(uα) and the fundamental lemma in this case follows from Lemma 7.4.
Nouc+dc1∈UF and therefore we deduce that it is only the rotated conjugation class that intersects M0(R), i.e. the other twisted conjugation class within the stable twisted conjugation class of ours does not intersect M0(R). It is therefore clear that the fundamental lemma holds in all cases, except perhaps when b1=−1 and c1 = 0. Furthermore, as we saw in Section 7.1, the stable twisted conjugation class of γ is the disjoint union of the twisted conjugation classes of γ and.
Since neither G1 nor G2 have proper elliptic endoscopic groups, the basic lemma is given by the following proposition.
Proof of the fundamental lemma
However, if you reason as in Remark 6.9, you also get the fundamental lemma in caseq= 3. Using Lemma 5.17, we note that the twisted weighted orbital integrals that we need to compute on G0 are equal to the weighted orbital integrals on GSp(4) regarding the Blades Levi. To get us into the same shape as Chapter 6, we scale up our elementγbyc to give.
When proving the fundamental lemma for the (2,2) Levi in the case of an elliptic torus we reduced the proof to this case.
Letting N0 denote the unipotent radical of the Borel subgroup containing M0, then we have N0∩ ZG1(N(sα)) ={I} unless. In this section, we prove the fundamental lemma for M0 equal to a diagonal torus in G0 and M0 equal to GL(1)3, the unique unbranched elliptic twisted endoscopic group for M0. This shows that the stable class of twisted conjugation γ is the same as the class of twisted conjugation γ.
Proof of the fundamental lemma
- Region 1
- Region 2
- Region 3
- Putting it altogether
Therefore, the contribution to I(M, N) from region 2 is given by q−N−2M times the integral of this function over the region. Considering the difference in these two regions, we need to calculate the integral. And we have I(M, N), which is given by q−N−3M times the integral of the same function over the domain.
Therefore, after removing the common region, we need to calculate the integral of our function over the region.
We take N0 equal to the unipotent radical of the upper triangular Borel in GSp(4) and we. In this chapter we calculate certain p-adic integrals that were necessary for the proof of the fundamental lemma. Since the integrand is symmetric inx2andx3, we might as well take twice the integral with|x3|<|x2| plus the integral with|x3|=|x2|.
