# Orthonormal basis of Cartan subalgebra relative to Killing form

I'm trying to understand a step in a proof:

Let $\mathfrak{g}$ be semi-simple (finite dimensional) Lie-algebra over $\mathbb{C}$, $\mathfrak{h}\subset\mathfrak{g}$ a Cartan subalgebra and let $\kappa:\mathfrak{g}\times\mathfrak{g}\to\mathbb{C}$ be the Killing form.

In this setting, the author of the proof chooses an orthonormal basis $h_1,\dots,h_n$ of $\mathfrak{h}$ relative to the Killing form, which is - to my understanding - a basis satisfying $\kappa(h_i,h_j)=\delta_{ij}$.

Why is it always possible to find such an orthonormal basis?

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This is because over an algebraically closed field, one can always find an orthonormal basis with respect to any symmetric bilinear form, as long as no non-zero vector is orthogonal to itself (so you need to know that the Killing form has this property when restricted to the Cartan subalgebra, which is the case because the elements of the Cartan subalgebra act via scalars when one looks at the adjoint action).

The Gram-Schmidt process (taught in most courses on linear algebra) then carries over to work in this general setting.

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Thus the algebraic closure of $\mathbb{C}$ is absolutely crucial to the classification of (semi-?)simple complex Lie Algebras. –  Jp McCarthy Jan 16 '13 at 15:12
Thank you! I did not know that fact about symmetric bilinear forms. –  Sh4pe Jan 16 '13 at 15:32
The Killing form is symmetric and non-degenerate(Cartan's criterion). For such bilinear forms you can always diagonalize it via a proper basis. So in particular over $\mathbb{C}$ you should be able to find an orthonormal basis.