664 6. REDUCING L2(2°) TO n = 2 AND V ORTHOGONAL
In contrast to the previous chapter, we find now when n(H) = 1 for each
H E 7-{* (T, M) that weak-closure methods are frequently effective.
LEMMA 6.1.3. (1) If Vis a TI-set under M, then Hypothesis E.6.1 holds.
(2) Either(I) r(G, V) = 1, or
(II) J1(T) i Cr(V).
PROOF. Part (1) of Hypothesis E.6.1 follows from Hypothesis 6.1.1. We saw
Cr(V) = 02 (LT), so that M = !M(LT) = !M(NMv(Cr(V)), giving part (3) of
Hypothesis E.6.1. This establishes (1). Further n(H) = 1 for all HE H*(T,M)
by Hypothesis 6.1.1.2, so the hypotheses of E.6.26 are satisfied with "j" equal to
- Therefore (2) follows from E.6.26. D
6.1.1. Initial reductions. In this subsection, we establish various reductions
culminating in the two cases of Proposition 6.1.15; eliminating the first of those
cases is then the goal of the second subsection.
LEMMA 6.1.4. VL/CvL (L) is the natural module for L.PROOF. Assume that the lemma fails. This assumption excludes case (3) of
5.1.3, so by Remark 6.1.2 and 5.1.3, either
(A) n > 2 is even and V is the 04 (2n/^2 )-module, or
(B) Vis the sum of two copies of the natural module.
Similarly by Remark 6.1.2 and 5.1.2, J(T) ::::) LT and M = !M(NLr(J(T))).Thus [Z, H] = 1 for each H E H*(T, M) by 5.1.7. Further by Hypothesis 6.1.1.2,
n(H) = 1. Enlarging V if necessary, we may take V = R 2 (LT).
Assume that there is A E Ai(T) with A =f 1. Then by B.2.4.1,
m(V/Cv(A)):::; m(A) + 1:::; m2(L'f') + 1=n+1. (*)
But in case (B), m(V/Cv(A)) ~ 2n > n+l since n > 1, contrary to(*), so case (A)
holds. Hence by H.1.1.2 with n/2 in the role of "n", n = 4, and A is of rank 1 and
generated by an orthogonal transvection. Further fort E T-Cr(V), m(V/Cv(t)) ~2n in case (A), and m(V/Cv(t)) ~ n in case (B) by H.1.1.1. Therefore we have
shown that either:
(i) n = 4, Vis the 0.4(4)-module, and if J1(T) =f 1 then J 1 (T) is generated by
an orthogonal transvection, or(ii) m(LT, V) > 2 and J1(T):::; Cr(V) = 02 (LT), so that J1(T) ::::) LT.
Suppose first that case (ii) holds. Then r(G, V) = 1 by 6.1.3.2. Now if Hy-
pothesis E.6.1 is satisfied, then since m(LT, V) > 2 in case (ii), r(G, V) > 1 by
Theorem E.6.3, a contradiction. Thus Vis not a TI-set in M by 6.1.3.l. Thereforeas L ::::) M, L is not irreducible on V, so case (B) holds where V = V 1 EB Vi is
the sum of two natural modules V1 and Vi. Further we may choose V1 to be T-
invariant (cf. the proof of A.1.42.1). As Lis irreducible on Vi, Vi is a TI-set under
M. As r(G, V) = 1, there is a hyperplane W of V with Ca(W) i Na(V). Set
Ui := W n Vi. Then Ca(W) :::; Ca(Ui) and m(Vi/Ui) :::; m(V/W) = 1, so Ui =f 1as m(Vi) ~ 4. Thus if Ca(W) :::; M, then as V 1 and V 2 are TI-sets in M, Ca(W)
normalizes 'Vi EB Vi = V, contrary to our choice of W. Therefore Ca(W) i M, so
Ca(U1) i M. But as V1 ::::) LT, Na(V1) :SM, so r(G, V1) = 1. As Vi is a TI-set in