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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 2dimensional ideal I since then
L/I would be 1dimensional and thus abelian which would imply that [LL] ⊆ I .
This also implies that L = sl(2, F ) has no 1dimensional ideal I since, in that case, L/I
would be 2dimensional and thus would have a 1dimensional ideal by Proposition 2.1.3
and this would correspond to a 2dimensional 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.
 Fall '11
 K.Igusa
 Algebra

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