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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.

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1 Answer 1

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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!

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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

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