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Let $H$ be an abelian subalgebra, in a complex semisimple Lie algebra $L\subset {\rm gl}\ ({\bf C}^n)$, whose elements are semisimple.

Assume that $H$ is abelian Lie subalgebra.

Define $$H^\ast = \{ \alpha |\ \alpha : H\rightarrow {\bf C}\ is\ {\bf C}-linear\ \}$$

Then define $$ L_\alpha = \{ x\in L|\ [h,x]=\alpha(h)x \ \forall h\in H \ \} $$

Note that $$ H\subset L_0 $$

And we have $$ L = L_0\oplus \bigoplus L_\alpha $$

Question : $L \supseteq L_0\oplus \bigoplus L_\alpha$ is reasonable. But why these sets are equal ?

HK Lee
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    This seems like a strange way to formulate this. The set of semisimple elements will usually not form a subalgebra, so it seems pointless to draw conclusions from that assumption. Usually what one does is consider a subalgebra which is maximal among those subalgebras consisting of only semisimple elements. Since any subalgebra consisting of only semisimple elements will indeed be abelian, the conclusions then become meaningful. – Tobias Kildetoft Dec 21 '13 at 12:28
  • Thank you for your reply and correcting. I rewrite. – HK Lee Dec 21 '13 at 14:36

1 Answers1

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Since $H$ is abelian and consists of semisimple elements, the endomorphisms $ad(h) = [h,.]$ of $L$ are all semisimple (that is diagonlizable since you work on the complex numbers) and they all commute. Then it is a standard result that all the $ad(h)$ are diagonalizable in the same basis.

Fix some basis such that all $ad(h)$ are diagonal. Then if you look at the $i$-th eigenvalue, you show easily that it is a linear form, when considered as a function of $h$.

Finally you obtain your decomposition by assembling together the indices that give the same linear form (otherwise said, that give the same eigenvalue for all $ad(h)$).

Jeremy Daniel
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