
#1
Jan412, 09:02 AM

PF Gold
P: 310

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 g^{α} is not nul and orthogonal to Ʃ_{λ ≠ 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}. 



#2
Jan712, 07:53 PM

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).




#3
Jan1512, 10:04 AM

PF Gold
P: 310

I found it in a book.
the main thing is to see that [g^{α} , g^{β} ] ⊂ g^{α+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 g^{γ} 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 g_{α} (eigenvectors) is strictly included in g^{α} 



#4
Jan1512, 11:24 AM

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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