Why are Killing form, Cartan ${\frak h}$, and roots $\alpha$, related by $\kappa(h,[x,y])=\alpha(h)\kappa(x,y)$?

Question

I am having some trouble with the following proof of a formula for the killing form. Let $\mathfrak{g}$ be a semisimple Lie algebra and $\mathfrak{h}$ a Cartan subalgebra.

I know that if $\mathfrak{h}$ is any nilpotent Lie algebra and $(V, \rho)$ is any representation. Then
$$ V = \bigoplus V_\lambda $$

Where $\lambda \in (\mathfrak{h}/D\mathfrak{h})^*$ are one dimensional representations.

And $V_\lambda = \{ v \in V: \forall x \in \mathfrak{h}, \exists n>0 \text{ such that } (\rho(x)-\lambda(x))^n(v) = 0$

Then a Cartan subalgebra $\mathfrak{h}$ is a nilpotent self-normalising subalgebra. (One can check that for any Lie algebra and any element $x$ then the generalised 0 eigenspace of $ad(x)$ is such an object.)

Then the restriction of $ad$ to $\mathfrak{h}$ makes $\mathfrak{g}$ into an $\mathfrak{h}$ representation:

$$ \mathfrak{g} = \mathfrak{h} \oplus \bigoplus_{\lambda \in \Phi} \mathfrak{g}_\lambda $$
is what I know as the Cartan decomposition of $\mathfrak{g}$.

It is claimed that, if $x \in \mathfrak{g}_\alpha, y \in \mathfrak{g} \text{ and } h \in \mathfrak{h}$ then $$ \kappa(h, [x, y]) = \alpha(h)\kappa(x,y) $$

$\textbf{Proof:}$

$$\kappa(h, [x, y]) = \kappa([h,x],y) =\kappa(\alpha(h)x,y)= \alpha(h)\kappa(x,y) $$

Where has the second equality come from. That is, why is it the case that: $\kappa([h,x],y) =\kappa(\alpha(h)x,y)$?

It surely is not the case that $[h,x] = \alpha(h)x$? Since $x$ is a generalised eigenvector of $ad(h)$, not necessarily an actual eigenvector, or have I majorly misunderstood something here??

Is it to do with the fact that the trace only looks at the diagonal elements of the action, and the diagonal elements of $ad(h)$ on $x$ will all be $\alpha(h)$ when in e.g Jordan Normal form?

Answer 1

The existing comment and answer both misunderstood your notation. By the decomposition you meant "decomposition to generalized eigenspaces", since a priori we don't know if the $ad(h)$ are simultaneously diagonalizable.

However, the point is that they are (this should kills your worry)! If you are following Humphreys book, read section 8.1. It is quite clear and short. You basically need the fact that elements in a toral algebra (Cartan subalgebra in your case) are simultaneously diagonalizable, which follows from commutativity (slightly nontrivial).

Note: I think you want your Lie algebra to be semisimple, and I suppose the underlying field is algebraically closed with characteristic $0$ (e.g. $\mathbb{C}$).

EDIT: Unfortunately, there are lots of equivalent definitions of a Cartan subalgebra, and the proof you want should depend on where you start. So, what's the definition of a CSA you are using?

Answer 2

We have $[h,x]=\alpha(h)x$, because we have $\mathfrak{g}_{\alpha}=\{x\in \mathfrak{g}\mid [h,x]=\alpha(h)x \; \forall h\in \mathfrak{h} \}$. This is not a "misunderstanding" of the notation, but simply true for both notations, generalized eigenspaces and root spaces, because $\mathfrak{g}$ is semisimple. Then it follows that $$\kappa(h, [x, y]) = \kappa([h,x],y) =\kappa(\alpha(h)x,y)= \alpha(h)\kappa(x,y) $$