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Let $L$ be a semisimple Lie algebra, and $H$ be a maximal toral subalgebra. I don't understand why $L$ is a sum of the spaces $L_\alpha$. enter image description here

My confusion: $H$ is the maximal toral subalgebra. $ad_L H$ consists of endomorphisms of $L$. Each of these endomorphisms commute and therefore can be diagonalized together. This where I get confused: Then $End(L)$ has a basis of common eigenvectors of the endomorphisms in $ad_L H$? Then I don't see how we get the $L_\alpha$.

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Since $H$ is Abelian, $[x,y]=0$ for all $x,y\in H$. Hence, the same holds in $\operatorname{ad}_L(H)$, since this is a Lie algebra homomorphism. Now, this means that we have a family of commuting endomorphisms in $\operatorname{End}(L)$, and as a consequence if they can be diagonalized, they are simultaneously diagonalizable. Since the endomorphisms are semisimple this means that their Jordan form is diagonal (e.g. no nilpotent part). Hence, they are simultaneously diagonalizable.

Now, take an eigenbasis $x_1,\ldots, x_k$ of $L$, and then define $\alpha_i(h)$ to be the unique scalar such that $[h,x_i]=\alpha_i(h)x_i$. Then, defining $L_\alpha$ as above, it becomes clear that $L=\bigoplus_{i=1}^k L_{\alpha_i}$. Essentially, these functions $\alpha_i$ keep track of the eigenvalue that $h$ takes on $x_i$ as a function of $h$.

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  • $\begingroup$ can you explain why in general we take $L=\oplus L_{\alpha}$ where $\alpha$ is in the whole $h^*$ not only finite set as you write? $\endgroup$
    – Mary Maths
    May 6, 2023 at 4:56

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