lecture notes3

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## This document was uploaded on 03/09/2014 for the course MATH 223a at Brandeis.

Ask a homework question - tutors are online