Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If $\alpha$ is a root of a simple lie algebra, prove that $\langle \alpha,\alpha \rangle$ not equal to $0$. From this, I want to prove that the $\langle,\rangle$ could be used as a scalar product.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

Let $t$ be the Lie algebra of a maximal torus of a compact form. Then $\alpha \in (it)^*$ so that $i\alpha \in t^*$. But the Killing form is definite on $t^*$ so that $\langle \alpha,\alpha \rangle = - \langle i \alpha, i\alpha\rangle \ne 0$.

I'm not sure what you mean by trying to prove that $\langle,\rangle$ can be used as a scalar product-- it is a scalar product!

share|improve this answer
    
we define <α, β> ≡ (hα , hβ ) where ρ(k) = (hρ , k) where ρ is root. I wanted to prove that <,> satisfies the properties of a scalar product i.e. <α,α> is positive definite. –  ramanujan_dirac Mar 5 '12 at 13:03
    
It is because the Killing form is negative definite on $t^*$ and therefore will be positive definite on $it^*$. –  Eric O. Korman Mar 5 '12 at 13:32

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.