# Lie theory question

1. Jul 21, 2009

After a long break doing my day job, I am back working from Fulton & Harris.

So ley me pose the question

Take the case of a 3-dimensional Lie algebra $$\mathfrak{g}$$. Direct computation reveals there is a basis vector, say $$H \in \mathfrak{g}$$ such that, for any other $$X_i \in \mathfrak{g}$$ one has that $$ad(H)X_i \equiv [H,X_i] = \alpha X_j$$, where one need not assume $$i \ne j$$.

This looks like merely to state that $$\alpha_i$$ is an eigenvalue for the adjoint action of $$H$$ on each $$X_i \in\mathfrak{g}$$. However, it seems this doesn't quite generalize.

Take the $$n$$-dimensional case. The vector subspace for which $$[H_i,H_j] =0$$ is called a Cartan subalgebra iff it is maximal in $$\mathfrak{g}$$. Call this a subalgebra $$\mathfrak{h} \ni H_i,\,\,\, i < n$$.

Now it is elementary fact from operator theory that, if 2 operators commute, then they "share" an eigenvector, though the eigenvalues need not coincide on this "shared" eigenvector (Umm - that's a fudge, I hope you know what I am getting at!).

Write $$ad(H_1)(X) = \alpha_1 X, \quad ad(H_2)(X) = \alpha_2 X, \quad \alpha_i \in \mathbb{C}$$. (I remind myself that $$ad(X)(Y) = [X,Y]$$, by definition)

It seems this is wrong, rather one must have that $$\alpha_i \in \mathfrak{h^*}$$ as a functional $$\mathfrak{h} \to \mathbb{C}$$, so that $$ad(H_i)(X) = \alpha_i(H_i)X,\quad ad (H_j)(X) = \alpha_j(H_j)X$$ where, of course, $$\alpha(H) \in \mathbb{C}$$

I find this confusing. Here's my Micky Mouse thinking:

Since for each $$H \in \mathfrak{h}$$ there is a simultaneous eigenvector say $$X \in \mathfrak{g}$$ for the adjoint action of $$H$$ on all $$X,\,Y,\,Z,\,...$$, then the requirement that the eigenvalue, say $$\beta$$, is (possibly) unique for each $$ad(H)(X_i)$$, then this is only guaranteed by the linear functional $$\alpha_i(H_i) = \beta_i$$ for this. Which appears to be $$ad(H)(X_\alpha) \equiv [H,X_\alpha] = ad(H)(X_\beta) = \alpha(H)(X_\alpha)$$ for all $$X$$, and all $$H$$.

Any help out there?

2. Aug 6, 2009

### morphism

I don't really understand what the question is. Could you please be more explicit?