Suppose that $\mathfrak{g}$ is a complex Lie algebra with a root space decomposition $\mathfrak{g} = \mathfrak{h}_\mathbb{C}\oplus \bigoplus_{\alpha \in R}\mathfrak{g}_\alpha$ where $R$ is the set of roots of $\mathfrak{g}$, and each $\mathfrak{g}_\alpha$ is a root space. Furthermore, suppose that $H \in \mathfrak{h}_\mathbb{R}, \mathfrak{h}_\mathbb{R} = i\mathfrak{h}_\mathbb{C}$ is an element of the real Lie subalgebra of the complexified ($\mathfrak{h}_\mathbb{C})$ Cartan subalgebra $\mathfrak{h}\subset \mathfrak{g}$.
My reading material claims that it follows trivially from the root space decomposition, that for we have the equality $K(H, H) = \sum_{\alpha \in R}\alpha(H)^2\mathrm{dim}(\mathfrak{g}_\alpha)$, where $K(.,.)$ is the Killing form, defined as $K(A, B) \equiv \mathrm{tr}(\mathrm{ad}_A\circ \mathrm{ad}_B)$, i.e. the trace of the compositions of the adjoint actions.
I currently have only a vague understanding of why the equality holds: something along the lines that in the root space decomposition of $\mathrm{ad}_H\circ \mathrm{ad}_H$, each summand $\mathfrak{g}_\alpha$ can somehow be represented as a matrix which maps its elements to $\alpha(H)$, therefore giving us $\alpha(H)^2$ in the product. However it is not clear to me 1.) how you get the dimension in the sum 2.) how to prove the equality rigorously, 3.) what is $K(A, B)$ for general $A, B \in \mathfrak{g}$?
Best Answer
The formula follows by using Lie's theorem. So for all $\alpha\colon \mathfrak{h}\rightarrow \Bbb C$ there exists a basis of $\mathfrak{g}_{\alpha}$, such that the endomorphisms ${\rm ad} (h)\mid_{\mathfrak{g}_{\alpha}}$ are simultaneously represented by strictly upper-triangular matrices in $\mathfrak{gl}_m(\Bbb C)$, $m=\dim \mathfrak{g}_{\alpha}$, and with diagonal elements equal to $\alpha (h)$. This immediately yields the formula by taking traces.