Why is there always a basis for the Cartan orthonormal relative to the Killing form?

Question

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?

Thank you for your help!

Answer 1

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 the entire space (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.

Answer 2

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.

Answer 3

This answer is just to show how much of the result generalizes to arbitrary ground fields of characteristic $0$, and along the way gives the result mentioned by Paul Garrett (in a comment to Tobias Kildetoft's canonical answer) that over the real numbers, the Killing form restricted to a maximal split toral subalgebra is actually positive definite i.e. has an orthonormal basis in the strongest sense of the word.

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Let $\mathfrak g$ be a semisimple Lie algebra over a field $K$ of characteristic $0$. Let $\mathfrak s \subset \mathfrak g$ be a maximal split toral subalgebra (which in general is not quite a Cartan subalgebra).

Then with respect to such $\mathfrak s$, generalizing the usual split case we have a decomposition

$$\mathfrak g = \mathfrak g_0 \oplus \bigoplus_{\alpha \in R} \mathfrak g_\alpha$$

where $\mathfrak g_\alpha = \{x \in \mathfrak g: [s,x]=\alpha(s)x \text{ for all } s \in \mathfrak s\}$, and $R$ is a finite subset of the dual $\mathfrak s^*$. Indeed one shows as in the split case that $R$ spans $\mathfrak s^*$. (One also shows that $R$ is (either empty or) a root system, just possibly non-reduced; but one might show this after what comes next here; it is not needed in the following.)

Now elementary considerations show that with respect to the Killing form $\kappa$ (which is non-degenerate on $\mathfrak g$ by Cartan's criterion), for all $\alpha, \beta \in R \cup \{0\}$ we have $\mathfrak g_\alpha \perp \mathfrak g_\beta$ unless $\beta =-\alpha$. This implies that the restriction of $\kappa$ to $\mathfrak g_0 \times \mathfrak g_0$ is non-degenerate as well.

In the split case, where $\mathfrak s$ is a Cartan subalgebra and hence equals its own centralizer $\mathfrak g_0 = C_\mathfrak g(\mathfrak s)$, we have now shown the Killing form restricted to the Cartan is non-degenerate. This is the case e.g. if $K$ is algebraically closed, so in that case now go to Tobias' answer.

However, in general $\mathfrak g_0 = C_\mathfrak g(\mathfrak s)$ will be bigger than $\mathfrak s$, and as further deviation from the well-known split case, the root spaces $\mathfrak g_\alpha$ can have dimension $\ge 2$. But it still follows more or less from definitions that the restriction of $\kappa$ to $\mathfrak s \times \mathfrak s$ is given by

$$\kappa(s_1, s_2) = \sum_{\alpha \in R} \dim_K(\mathfrak g_\alpha) \alpha(s_1)\alpha(s_2)$$

From this we conclude:

1. If $K =\mathbb R$ or a similar (namely, ordered) field, we are done again. Because for any $s \in \mathfrak s \setminus \{0\}$, by the above formula $\kappa(s,s)$ is a sum of squares times positive integers, at least one summand being non-zero, so it is positive, and we have shown a stronger result: that the restriction of $\kappa$ to $\mathfrak s \times \mathfrak s$ is positive definite. For $K=\mathbb R$, Gram-Schmidt will do the rest. (Note that e.g. over $K=\mathbb Q$, we cannot expect an orthonormal basis because scaling might not work.)

2. In general, the restriction of $\kappa$ to $\mathfrak s \times \mathfrak s$ is still non-degenerate. I would like to see a more direct proof for that. The one I have (which is spread out in lemma 3.1.9, 3.1.10 and proposition 3.1.11 in my thesis) needs a bit more input: Namely, first via fine-tuning the Jacobson-Morozov Theorem we construct for each root $\alpha$ an element $H_\alpha \in [\mathfrak g_\alpha, \mathfrak g_{-\alpha}] \cap \mathfrak s$ such that $\alpha(H_\alpha)=2$; one then shows that the $(H_\alpha)_{\alpha \in R}$ generate $\mathfrak s$, that $\kappa(H_\alpha, H_\alpha)$ is a positive integer (follows from the formula above), and with some root string considerations and $\mathfrak{sl}_2$-representations, that $\alpha(s) = 0 \implies \kappa(s, H_\alpha)=0$. This last fact immediately gives the seemingly much stronger $$ \alpha(s)= 2 \dfrac{\kappa(H_\alpha, s)}{\kappa(H_\alpha, H_\alpha)} $$ for all $s \in \mathfrak s, \alpha \in R$. And this, finally, implies that $\kappa$ is non-degenerate, because otherwise there would be an $s \in \mathfrak s \setminus \{0\}$ such that $\alpha(s)=0$ for all $\alpha \in R$, contradicting the fact that those span $\mathfrak s^*$.