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lecture notes3 - MATH 223A NOTES 2011 LIE ALGEBRAS 9 3...

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MATH 223A NOTES 2011 LIE ALGEBRAS 9 3. Nilpotent and solvable Lie algebras I can’t find my book. The following is from Fulton and Harris. Definition 3.0.1. A Lie algebra is solvable if its iterated derived algebra is zero. In other words, D k L = 0 where DL = [ LL ] , D 2 L = [( DL )( DL )] = [[ LL ][ LL ]], etc. This is a recursive definition: The k -th derived algebra of L is the k 1st derived algebra of [ LL ]. Definition 3.0.2. A Lie algebra is nilpotent of class k if ad k L ( L ) = [ L · · · [ L [ L [ L k L ]]] · · · ] = 0 for some k . Note that if L is nilpotent of class k then ad k x = 0 for all x L since ad k x ( y ) = [ x [ x [ x · · · [ xy ] · · · ]]] [ L [ L [ L · · · [ LL ] · · · ]]] = 0. Every element of L is ad-nilpotent. 3.1. Engel’s Theorem. This is converse of the above statement. Theorem 3.1.1 (Engel) . If L is a finite dimensional Lie algebra in which every element is ad-nilpotent then L is nilpotent. We prove this theorem in a sequence of lemmas. The first lemma allows us to assume that L is a linear Lie algebra. Lemma 3.1.2. Suppose the image of the adjoint representation ad : L gl ( L ) is a nilpotent subalgebra of gl ( L ) of class k then L is nilpotent of class k + 1 . Proof. The kernel of the adjoint representation is the center Z = Z ( L ). If L/Z is nilpotent of class k then ad k L/Z ( L/Z ) = is the same as ad k L ( L ) Z . But [ LZ ] = 0 so ad k +1 L ( L ) = [ L ad k L ( L )] [ LZ ] = 0. 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 dimen- sional 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 .
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