lecture notes2

So ax ker conversely if i l is an ideal a l x i

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: a ∈ L then ϕ[ax] = [ϕ(a)ϕ(x)] = [ϕ(a)0] = 0. So, [ax] ∈ ker ϕ. Conversely, if I ⊂ L is an ideal, a ∈ L, x ∈ I then [(a + I )(x + I )] ⊆ [ax] + [aI ] + [Ix] + [II ] ⊆ [ax] + I So, the bracket is well-defined in L/I , [ϕ(a)ϕ(x)] = ϕ[ax] and I = ker ϕ. ￿ Proposition 2.1.5. Any epimorphism (surjective homomorphism) of Lie algebras ϕ : L → L￿ gives a 1-1 correspondence between ideals I ￿ in L￿ and ideals I of L containing ker ϕ. Proof. The correspondence is given by I ￿ = ϕ(I ). This is an ideal since [L￿ , I ￿ ] = [ϕ(L), ϕ(I )] = ϕ[LI ] ⊆ ϕ(I ) = I ￿ and I = ϕ−1 (I ￿ ) which is an ideal since it is the kernel of L → L￿ → L￿ /I ￿ . ￿ Example 2.1.6. The center of a Lie algebra L is defined to be Z (L) = {x ∈ L | [xL] = 0}. Then Z (L) is an ideal in L. Z (L) also the kernel of the adjoint representation ad : L → gl(L). If I ⊆ L is an ideal, then the adjoint representation restricts to a representation on I : ad : L ...
View Full Document

This document was uploaded on 03/09/2014 for the course MATH 223a at Brandeis.

Ask a homework question - tutors are online