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Lie theory question

  1. Jul 21, 2009 #1
    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 [tex]\mathfrak{g}[/tex]. Direct computation reveals there is a basis vector, say [tex]H \in \mathfrak{g}[/tex] such that, for any other [tex]X_i \in \mathfrak{g}[/tex] one has that [tex]ad(H)X_i \equiv [H,X_i] = \alpha X_j[/tex], where one need not assume [tex]i \ne j[/tex].

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

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

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

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

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

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

    Any help out there?
     
  2. jcsd
  3. Aug 6, 2009 #2

    morphism

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