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(xWk ) = 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 ....
View
Full
Document
This document was uploaded on 03/09/2014 for the course MATH 223a at Brandeis.
 Fall '11
 K.Igusa
 Algebra

Click to edit the document details