# Proof of existence of opposite roots in semisimple algebras?

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 PF Gold P: 375 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-α must be ≠ 0 then -α is a root. Here g λ = {x ∈ g |∃n ∈ N ∀h ∈ h , (π(h) − λ(h))n (x) = 0}.
 Sci Advisor HW Helper P: 2,020 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).
 PF Gold P: 375 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
 Sci Advisor HW Helper P: 2,020 Proof of existence of opposite roots in semisimple algebras? For a complex semisimple g, what you denote by g^\alpha ("generalized" eigenvectors) is always equal to g_\alpha.

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