Proof of existence of opposite roots in semisimple algebras?

1. Jan 4, 2012

naima

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. Jan 7, 2012

morphism

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).

3. Jan 15, 2012

naima

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

4. Jan 15, 2012

morphism

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