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But lz 0 so adk1 l lz l l ladk l lz 0 l if

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Unformatted text preview: (L). If L/Z is nilpotent of class k then adk (L/Z ) = is the same as adk (L) ⊆ Z . But [LZ ] = 0 so adk+1 (L) = L/Z L L [Ladk (L)] ⊆ [LZ ] = 0. ￿ L If every element of L is ad-nilpotent then the image of the adjoint representation consists of nilpotent endomorphism of L considered as a vector space. Lemma 3.1.3. Suppose that L is a subalgebra of gl(V ) where V is a nonzero finite dimensional vector space over F . Suppose that every element of L is a nilpotent endomorphism of V . Then there exists a nonzero element v ∈ V so that x(v ) = 0 for all x ∈ L. Some people call this Engle’s Theorem since it is the key step in the proof of the theorem. To see that the lemma implies the theorem, let K be the subspace of V spanned by the vector v . Then the action of L on V induces an action on V /K which is nilpotent. So, the image of L in gl(V /K ) is a nilpotent Lie algebra of class, say k . This implies that Lk (V ) ⊆ K . So, Lk+1 (V ) = 0 making L nilpotent of class k + 1. Proof of key lemma. The proof is by induction on the dimension of L. If L is onedimensional, the lemma is clear. So, suppose dim L ≥ 2. Let J be a maximal proper subalgebra of L. Claim 1: J is an ideal of codimension 1 in L. 10 MATH 223A NOTES 2011 LIE ALGEBRAS Pf: J acts, by the adjoint action on L and this action leaves J invariant. Therefore, J act on the quo...
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This document was uploaded on 03/09/2014 for the course MATH 223a at Brandeis.

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