Graded Lie algebras with classical reductive null component

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4 . General information on graded Lie algebras

In this section we assume that G = G_^©...©G_i®Go©Gi®...©G, is an irreducible, transitive, graded Lie algebra with q>l, which contains a simple graded ideal S = S_^® ...eS-ieSo©...®^, such that S^, = G., = [G_ 1, G_,+1] for all f >0. We prove several general lemmas about G and about the annihilât or, Ann^^N, of N in M for certain subspaces M, N of G. In this we use M"* to denote the set of all m-tuples of nonnegative integers. For each i= (ïi,...,ïJeN"*, we write adSi = adSij... adS^^ and use adS_i as a shorthand notation for adS_,j... adS_j-^.

Lemma 6. For any x in G\G^q, [G-i,x] = (0) implies x=0.

Proof Since G is the direct sum of its gradation spaces, it will suffice to prove the result for xeGp — ^ + 1 ^/gr. If O^i^r, the result holds by transitivity. When —q + i^i<0, recall that Si = Gi, and that S^^ generates the negative gradation spaces of S. It follows that [S^,x]=(0), for all;, -^^j^ -1. Then

I = I^>Q Z^g j^m ad Sj(x)

is an ideal of the simple algebra S. But I must be zero, since it does not contain S_^. Thus, X is central ш S and so x=0, as desired. П

Lemma 7. Let G and S be as above. Then S^ = (adS_i)*"^5s for allj, —qSjus. If q{t-l)us, then (adS_^)'S = (0) if and only if (adS_/S,=(0) for some i, q(t-l)

Proof Note first that since S is simple, we must have that S equals the ideal ^o^oi^^S- ifS,, so that Sj = (adS_ if'-'S,. For the second assertion the "only if' implication is obvious. To see the "if' implication, we assume that (adS_^)'S, = (0) for some i^q{t—l) and show that (adS_^ySj = (0) for all; by considering three cases:

j<q { t - l ) , (1) q(t-mj<i, (2) and i<jSr. (3)

In case (1), we have; —r^< -q, so that (adS_/Sj5S^_j^ = (0). In the second case, (adS_,yS^=(adS_/(adS_ J-^S,=(adS_iy"^(adS_/S.=(0). In the final case, (0)=(adS_,)^S,=(adS_/(adS_iy-'S,=(adS.iy-4adS_/S^ so by Lemma6, (adS_/Sj.=(0). We have shown that (adS_/Sj=(0) for all; which establishes the sufBciency. П

Lemma 8. lG^q,GJ+{0) for alli^O, .,,,r. In addition, Sj^isLdG^J'^G^ for all j, — ^g;^r—1, so that 5 = r—1 or r.

Proof Since G _ 1 = S _ 1 and G is transitive, [S _ i, G J Ф (0). The (nonzero) ideal R of S generated by [S_ i, GJ is Ä=(ad [S_ iSJjG^ + Z'^> o(adS_ if G,. By simplicity this must be S, and the second assertion is seen to be true. To see the first assertion note that if [G_^,GJ = (0), then [S_ç,S] = [S_^,R] = (0) as S.^ commutes with 5^1. On the other hand, if [G»^,GJ = (0) for some i<r, then (0) = [G_^,SJ = [S-^, Sf], and we have by Lemma 7 (with t=1) that again [S.^, S] =(0). But S.^ cannot lie in the center of S as S is simple. This contradiction shows that IG^^GJ cannot be zero for any r=0,..., r. П