Proof of existence of opposite roots in semisimple algebras?

[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 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}.

 

[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).

 

[naima]

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

 

[morphism]

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

 

[blue_raver22]

Happy new year to you as well from a fellow scientist! Your question is very interesting and touches on a fundamental aspect of semisimple algebras.

Firstly, for those who may not be familiar with the terminology, a semisimple algebra is a type of algebra that is made up of simpler, irreducible pieces. In the context of physics, semisimple Lie algebras are used to describe the symmetries of elementary particles.

Now, to address your question about the existence of opposite roots in semisimple algebras, let me start by clarifying some terminology. In this context, a root is a vector in the algebra that, when acted on by certain elements of the algebra, produces a scalar multiple of itself. In other words, it is an eigenvector of the action of the algebra.

In semisimple algebras, the roots are important because they correspond to the possible charges of the gauge bosons. The fact that the gauge bosons may have opposite charges is related to the fact that the algebra has both positive and negative roots.

To prove that -α is also a root if α is a root, we can use the fact that the Killing form is non-degenerate. The Killing form is a bilinear form on the algebra that measures the commutativity of its elements. If the Killing form is non-degenerate, it means that there are no non-trivial elements in the algebra that commute with all other elements. This is a crucial property in semisimple algebras.

Now, let's consider the space gα, which is the set of elements in the algebra that, when acted on by α, produce a scalar multiple of themselves. If gα is not null, it means that there are elements in the algebra that have the same properties as α. This is because gα is the span of α and any other elements that satisfy the same property.

Since gα is not null, it must be orthogonal to the space Ʃλ ≠- α g λ, which is the span of all the other roots in the algebra except for -α. This is because if gα and Ʃλ ≠- α g λ were not orthogonal, it would mean that there exists an element in the algebra that commutes with all other elements, which is not possible in a semisimple algebra with non-degenerate Killing form.

Therefore, since gα is orthogonal to Ʃλ ≠