- #1
QuarkHead
- 8
- 0
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?
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?