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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 $A, B \in \mathfrak{g}$?

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    $\begingroup$ More generally we have $$\kappa(h_1, h_2) = \sum_{\alpha \in R} \dim_K(\mathfrak g_\alpha) \alpha(h_1)\alpha(h_2),$$ which you can find here at MSE (Torsten's answers). $\endgroup$ Apr 7, 2022 at 8:11
  • $\begingroup$ @DietrichBurde Interesting! Could you care to link such an answer? I'm afraid that I'm still a bit too novice to formulate my question in terms of Lie algebra such that I'd find it myself. $\endgroup$ Apr 7, 2022 at 8:14
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    $\begingroup$ Perhaps it is more convenient to write some answer. I am sure that there are several related posts, like this one, but it is perhaps quicker to write the argument again. $\endgroup$ Apr 7, 2022 at 13:03
  • $\begingroup$ In my answer to math.stackexchange.com/q/280090/96384 I claim a more general version of Dietrich's formula "follows more or less from definitions", and I stand by that: $ad(h)$ is scalar multiplication with $\alpha(h)$ on $\mathfrak g_\alpha$, and with $0$ on $\mathfrak h$ itself, and this completely describes it via the root space decomposition. With that, take traces of $ad(h_1) \circ ad(h_2)$. --- Your third question, for general elements from $\mathfrak g$ (not just $\mathfrak h$) is more intricate. I don't think there is a good general formula. There are for the classical LAs. $\endgroup$ Apr 7, 2022 at 16:04

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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.

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  • $\begingroup$ Could you further elaborate how Lie's theorem yields the result? $\alpha(h)^2$ makes sense as the diagonal elements of the matrix product $\mathrm{ad}(h)\circ \mathrm{ad}(h)$ are multiplied by themselves. But where does the $\mathrm{dim}(\mathfrak{g}_\alpha)$ come from? Moreover, what does the matrix $\mathrm{ad}(h)$ look like? Is it some direct sum of matrices, or something else? $\endgroup$ Apr 8, 2022 at 6:05
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    $\begingroup$ Yes, each block has size $m=\dim \mathfrak{g}_{\alpha}$, with the $\alpha(h)$'s on the diagonal, summed up for the trace. You should do a small explicit example. $\endgroup$ Apr 8, 2022 at 7:56
  • $\begingroup$ Do you happen to know some source which discusses the form of the $\mathrm{ad}(h)$ matrix in terms of the root space blocks? $\endgroup$ Apr 9, 2022 at 6:04
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    $\begingroup$ In some books the example of $L=\mathfrak{sl}(2)$ is explicitly given. Try Erdmann and Wildon. The root space decomposition for $L$ can be written quite easily in explicit terms, so that one can write down all matrices. You can try to do this yourself. Start with the root space decomposition with $R=\{\alpha,-\alpha\}$. $\endgroup$ Apr 9, 2022 at 8:23

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