630 5. THE GENERIC CASE: L2(2n) IN C.t AND n(H) > 1
an obstacle to applying this recognition theorem: the case where K ~ £* ( G, T),
leading to M 23. This case is dealt with in Theorem 5.2.10.
5.1. Preliminary analysis of the L 2 (2n) case
5.1.1. General analysis of V and H. Since this is the first case in the FSU
which we analyze, we begin with a lemma summarizing some of the basic tools
(developed in Volume I and earlier chapters of Volume II) to deal with the FSU.
We thank Ulrich Meierfrankenfeld for several improvements to the proofs in this
section.
LEMMA 5.1.1. (1) CT(V) = 02(LT).
(2) Each HE 'H*(T, M) is a minimal parabolic described in B.6.8, and in E.2.2
if n(H) > 1.
(3) For each H E 'H*(T, M), case (I) of Hypothesis 3.1.5 is satisfied with LTin the role of "Mo".
(4) LT is a minimal parabolic.
PROOF. Part (1) follows from 1.4.1.4, (2) follows from 3.3.2.4, (3) follows from
(1) and the fact that L :::) M, and (4) is well known and easy. D
We begin by discussing the possibilities for V:
LEMMA 5.1.2. One of the following holds:
(1) J(T) ::::; CM(V), so J(T) and Baum(T) are normal in LT and M
!M(Na(J(T))) = !M(Na(Baum(T))).(2) [V, J(T)] =f. 1 and V/Cv(L) is the natural module for L.
(3) [V, J(T)] =f. 1, n = 2, and V = Cv(LT) EEl [V, L] with [V, L] the 85 -module
for LT ~ 85.
PROOF. By 5.1.1.1, CT(V) = 02 (LT). Thus if J(T) ::::; CM(V), then J(T) =
J(02(LT)) and Baum(T)) = Baum(02(LT)) by B.2.3, so LT acts on J(T) and
Baum(T). However by 1.2.7.3, M = !M(LT), so (1) holds in this case. So assume
[V, J(T)] =f. 1. Then Vis an FF-module for LT by B.2.7, so by B.5.1.1, I:= [V, L] E
Irr +(L, V), and by B.5.1.5, V =I +Cv(L). By B.4.2, either I/Cr(L) is the natural
module, or n = 2 and I/Cr(L) is the A5-module. In the former case (2) holds as
V =I+ Cv(L), and in the latter (3) holds by B.5.1.4. D
LEMMA 5.1.3. One of the following holds:
(1) V is the direct sum of two natural modules for L.
(2) n = 2 and V is the direct sum of two 85 -modules for LT~ 85.
(3) [V, Ll/Crv,L] (L) is the natural module for L.
(4) n is even and Vis the 04(2nf^2 )-module for L.
(5) V = [V, L] EEl Cv(LT), and [V, L] is the 85 -module for LT~ 85.
REMARK 5.1.4. Recall that the A5-module and the 04_ (2)-module are the same.
Notice however that in case (4) we may have LT~ A 5 , which is not allowed in (5).
On the other hand in case (5) we may have Cv(L) =f. 1, which is not allowed in (4).
PROOF. If [V, J(T)] =f. 1 then (3) or (5) holds by 5.1.2. Thus we may assume
[V, J(T)] = 1, so that Cv(L) = 1 by 3.1.8.3.
· Next q(LT, V) ::::; 2 by 3.1.8.1. Hence in the language of Definition D.2.1, there
is A E Q(T, V). Recall that we are not yet assuming the FSU, so we will work with