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Since char f 0 this implies xy 0 we can now show that

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Unformatted text preview: e must have Wk = Wk+1 for some k . This means that (x − λ(x))k = 0 on Wk . So, for any x ∈ J , the matrix of x as an endomorphism of Wk is upper triangular with λ(x) on the diagonal and we have Tr(x|Wk ) = λ(x) dim Wk Claim For any y ∈ L, y (Wk ) ⊆ Wk+1 = Wk . Suppose for a moment that this is true. Then, for any x ∈ J, y ∈ L we have Tr([xy ]|Wk ) = 0 = λ[xy ] dim Wk . Since char F = 0 this implies λ[xy ] = 0. We can now show that y (w) ∈ W for all y ∈ L, w ∈ W : x(y (w)) = yx(w) + [xy ](w) = λ(x)y (w) + λ[xy ](w) = λ(x)y (w) Thus it suffices to prove the claim. Proof of Claim: y (Wi ) ⊆ Wi+1 . To prove this we must show that, for any w ∈ Wi , x ∈ J we have (x − λ(x))y (w) ∈ Wi . This is a calculation similar to the one above: x(y (w)) − λ(x)y (w) = y (x − λ(x))(w) + [xy ](w) ∈ y (Wi−1 ) + Wi ⊆ Wi proving the claim by induction. (It is clear when i = 0.) ￿ 12 MATH 223A NOTES 2011 LIE ALGEBRAS Exercise 3.2.4. (1) Show that t(n, F ) is solvable. (2) Show that any solvable subalgebra of gl(n, C) is, up to isomorphism, isomorphic to a subalgebra of t(n, C). (3) If J is an ideal in L then show that L is solvable if and only if J and L/J are solvable. (4) Prove that the following are equivalent. (a) L is solvable. (b) L has a sequence of ideals L ⊃ J1 ⊃ J2 · · · ⊃ Jn = 0 so that Jk /Jk+1 is abelian for each k . (c) L has a sequence of subalgebras L ⊃ L1 ⊃ L2 · · · ⊃ Ln = 0 so that Lk+1 is an ideal in Lk and Lk /Lk+1 is abelian for all k ....
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This document was uploaded on 03/09/2014 for the course MATH 223a at Brandeis.

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