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The adjoint representation of a semisimple Lie algebra is completely reducible

  1. Oct 28, 2014 #1
    Hi,

    I am trying to work through a proof/argument to show that the adjoint representation of a semisimple Lie algebra is completely reducible.

    Suppose S denotes an invariant subspace of the Lie algebra, and we pick [itex]Y_i[/itex] in the invariant subspace S. The rest of the generators [itex]X_r[/itex] are such that the natural inner product is [itex](X_r, Y_i) = 0[/itex]. This can be done by some suitable Gram Schmidt orthogonalization if necessary.

    To begin with, I argue that the killing metric in this basis is block diagonal. If i denotes an index on Y and r denotes an index on X, then [itex]g_{ir} = 0[/itex] as the Killing form is the natural inner product or its negative depending on whether Y is chosen to be symmetric or antisymmetric. This is OK.

    But the following argument is unclear to me

    Since S is an invariant subspace, structure constants of the form [itex]{f_{ir}}^{s}[/itex] are zero.

    Is it reasonable to expect [X, Y] to be in S as well as its complement? The only way then that this would be possible is if [X, Y] = 0.

    The other argument (specious to me) is that Y and X live in different spaces so they must commute. This seems physically reasonable, but I don't see how to argue this mathematically.

    Any help would be greatly appreciated. Oh and I should point out, I am learning this from the standpoint of a theoretical physicist, so please feel free to point out mistakes/improvements in the reasoning (or holes in my understanding) from a purely mathematical perspective.

    Thanks!
     
    Last edited: Oct 28, 2014
  2. jcsd
  3. Nov 3, 2014 #2
    Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Nov 3, 2014 #3
    Hey Greg. I can't think of anything new, but I've kept it on my pending/to-do list. The argument used in the "physicist's proof" I had access to is already listed in my post. I don't have any more information.
     
  5. Nov 4, 2014 #4
    You have written [itex] \mathfrak{g}=S\oplus S^{\perp} [/itex] as vector spaces where we take the orthogonal complement with respect to the killing form. If we know that [itex] S^{\perp} [/itex] is an ideal, then we will have [itex] [S,S^\perp] \subseteq S\cap S^{\perp} [/itex] since in this case [itex] S [/itex] and [itex] S^\perp [/itex] are both ideals.

    To show [itex] S^{\perp} [/itex] is an ideal, let [itex] x\in S^\perp, g\in \mathfrak{g} [/itex]. Then for any [itex] s\in S [/itex], we know the Killing form is invariant so
    [tex] \kappa( s, [g,x])=\kappa([s,g],x)\in \kappa(S,x)=0. [/tex]
    where I used the fact that [itex] S [/itex] is an ideal in the second last step and the fact that [itex] x\in S^{\perp} [/itex] in the last step. Hence [itex] [g,x]\in S^\perp [/itex] for any [itex] g\in \mathfrak{g} [/itex] so [itex] S^{\perp} [/itex] is an ideal.

    Finally, you may now conclude that [itex] [X,Y]=0 [/itex] since it is equivalent to the fact that [itex] S\cap S^{\perp}=\varnothing[/itex] which follows using the nondegeneracy of the Killing form when restricted to S (which follows from the fact that [itex] \mathfrak{g} [/itex] is semisimple) as you mentioned.
     
    Last edited: Nov 4, 2014
  6. Nov 13, 2014 #5
    Thank you Terandol!
     
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