12.6. ELIMINATING As ON THE PERMUTATION MODULE S29(2) Vg E vca(v) for each v E V n V^9.
(3) r(G, V) :'.'.: 3, and r(G, V) :'.'.: 4 if Zv =I-l.
(4) If 1 =I-[V, V^9 ] s VnV^9 , then Zv = 1, VnV^9 is a totally singular 3-subspaceof V, and V^9 is the unipotent radical of an L 3 (2)/Es parabolic.
PROOF. Part (2) follows from A.1.7.1 in view of 12.6.19.1. If v EV is of weight
8 then Gv = M as we saw at the start of the section, while if v has weight 2 or
6, then Gv SM by Theorem 12.6.2. So by 12.6.19.1, we may apply A.l.7.2 to see
that each element of weight 2, 6, or 8 lies in a unique conjugate of V. Then (1)
follows, and (1) implies (3).
Assume the hypotheses of (4). Interchanging V and Vg if necessary, we may as-sume m(V^9 ) :'.'.: m(V/Cv(V9)). Then by B.l.4.4, Vg contains a member of P(T, V),
so the possibilities for vg are described in the discussion near the beginning of the
proofof12.6.l. As [V, Vg] s VnVg, (1) says [V, Vg]# ~ S, and the only possibility
satisfying this restriction is that given in (4). D
LEMMA 12.6.21. Ca(v) i. M for each v EV# of weight 4.
PROOF. As the groups in conclusions (2)-(4) of Theorem 12.2.13 do not have
a member L E £j(G,T) with L ~As and V/Cv(L) the permutation module,
conclusion (1) of 12.2.13 holds: Gv i. M for some v E V#. By 12.6.20, Gv S M
for v of weight 2, 6, or 8, so the lemma holds. D
LEMMA 12.6.22. If Zv =/-1, then
(1) Wo := Wo(T, V) centralizes V.
(2) If m(Vg /V^9 n M) S 1 for some g E G, then V^9 S Na(V).
(3) w(G, V) > 1.
PROOF. Notice (1) and (2) imply (3), so it remains to prove (1) and (2). As-
sume Zv =I-1. Then M = Na(V) by 12.2.2.3. Suppose A := V^9 n M with A =I-1.
Assume k := m(Vg /A) s 1, and if (1) fails, choose k = 0. Thus V 1:. Na(V9) if
k = 1: for otherwise by assumption (1) does not fail, so that VS Ca(V9), contra-
dicting A =I-1. Now by 12.6.20.3 and E.3.4, m(A) :'.'.: r(G, V) - k :'.'.: 4-k. Similarly
using 12.6.20.3 as in E.3.32,
U := (Cv(B): B s A and m(V^9 /B) s 3) s Na(V^9 ),
so [A, U] S V n V9. Now LT is As or Ss, and the maximal elementary abelian
2-subgroups of Ss are
(i) D 1 ~ Es regular on n.
(ii) D2 ~ E16 with two orbits of length 4.
(iii) D 3 ~ E 16 with one orbit of length 4, and two of length 2.
(iv) D4 ~ E16 with four orbits of length 2.
If k = 0, then m(.A) = 4, so A = Di for i = 2, 3, 4; while if k = }, then
m(A) ~ 3, so either A= Di for i = 1, 2, 3, 4 or A is of index 2 in Dj for j = 2, 3, 4.
In each case we find that [A, U] contains a vector of weight 2 or 8. This contradicts
12.6.20.1, as [A, U] s V n V9, so the proof is complete. D
LEMMA 12.6.23. Cv(L) = l. Thus V is the 6-dimensional orthogonal module