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Proof of existence of opposite roots in semisimple algebras?

  1. Jan 4, 2012 #1

    naima

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    Happy new year from France.

    I am reading books on elementary particle and i see that their
    gauge bosons may be neutral or have opposite charge. They live
    in semisimple Lie algebras. So I searched in math books how to prove
    that in a semisimple Lie algebra if α is a root so is -α.
    I found that it is related to the fact that the killing form is not degenerate.

    Could you comment this:

    If α is a root the space gα is not nul and orthogonal to Ʃλ ≠- α g λ. Since the Killing form is non degenerate g must be ≠ 0 then -α is a root.

    Here g λ = {x ∈ g |∃n ∈ N ∀h ∈ h , (π(h) − λ(h))n (x) = 0}.
     
  2. jcsd
  3. Jan 7, 2012 #2

    morphism

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    That is certainly a correct proof, but be sure that you're not using anything you're not allowed to (i.e. that you're not being circular).
     
  4. Jan 15, 2012 #3

    naima

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    I found it in a book.

    the main thing is to see that [gα , gβ ] ⊂ gα+β
    by definition the killing form K(x,y) = trace (ad x ad y) where ad y (z) = [y,z].

    We have ad x ad y (z) = [x,[y,z]]
    if x ∈ g α y ∈ g β and z ∈ g γ then [x,[y,z]] is in g α+β+γ
    so if α+β not null gγ is mapped on another space and does not participate to the trace (outside the diagonal)
    x is orthogonal to y (K =0) if β not = -α
    it cannot be orthogonal to all the vectors (K not degenerate) so -α is a root.

    Could you give me an example where gα (eigenvectors) is strictly included in gα
     
  5. Jan 15, 2012 #4

    morphism

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    For a complex semisimple g, what you denote by g^\alpha ("generalized" eigenvectors) is always equal to g_\alpha.
     
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