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# This implies that l has no 2 dimensional ideal i

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Unformatted text preview: he standard basis for sl(2, F ) is given by ￿ ￿ ￿ ￿ ￿ ￿ 01 00 10 x= , y= , h= 00 10 0 −1 and computation gives: [hx] = 2x, [hy ] = −2y, [xy ] = h This proves that [LL] = L. This implies that L has no 2-dimensional ideal I since then L/I would be 1-dimensional and thus abelian which would imply that [LL] ⊆ I . This also implies that L = sl(2, F ) has no 1-dimensional ideal I since, in that case, L/I would be 2-dimensional and thus would have a 1-dimensional ideal by Proposition 2.1.3 and this would correspond to a 2-dimensional ideal in L by Proposition 2.1.5. Therefore, sl(n, F ) is simple. ￿ Proposition 2.1.11. Every simple Lie algebra is linear. Proof. The center of L is trivial. So, the adjoint representation gives an embedding ad : L ￿→ gl(L). ￿ 2.2. nilpotent elements and automorphisms. First, suppose that char F = 0 and δ is a nilpotent derivation of any nonassociative algebra A. I.e., δ k = 0 for some k . Then we claim that k −1 ￿ δi exp δ = i! i=0 is an automorphism of A. It is...
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## This document was uploaded on 03/09/2014 for the course MATH 223a at Brandeis.

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