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I'm reading the proof of the Jacobson-Morozov theorem in the note of Ana Bălibanu (Theorem 5.1). There are a few lines where I can't understand the logic within.

Let $\mathfrak{g}$ be a semisimple Lie algebra over $\mathbb{C}$. For $x \in \mathfrak{g}$, let $\mathfrak{g}^x = \ker(\operatorname{ad}_x: \mathfrak{g} \to \mathfrak{g})$. During the inductive step in the first half of the proof, the following two implications are presented:

(1) Let $s \in \mathfrak{g}$ be a semisimple element. Then $\mathfrak{g}^s$ is a proper reductive Lie subalgebra of $\mathfrak{g}$.

(2) If $e \in \mathfrak{g}^s$ is nilpotent, then $e \in [\mathfrak{g}^s, \mathfrak{g}^s]$.

I'm trying to give a proof to these claims, but I don't find much success. Claim (1) is not clear to me to any extent. I think we are supposed to hook up a decomposition $$\mathfrak{g}^s = \mathfrak{g}_0^s \oplus Z(\mathfrak{g}^s)$$ with $\mathfrak{g}_0^s$ a semisimple ideal.

I have minimal progress on (2) as well. If we assume (1), then (as semisimple Lie algebra are stable under the bracket), we clearly have $$\mathfrak{g}_0^s = [\mathfrak{g}_0^s, \mathfrak{g}_0^s] = [\mathfrak{g}^s, \mathfrak{g}^s]$$ so to prove (2), it suffices to prove that $e$ has no contribution from $Z(\mathfrak{g}^s)$. But's as far as I can get.

Am I missing something subtle? It would be much appreciated if someone can point me to the correct direction.

Ray
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1 Answers1

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$(1)$ It is well known that the centraliser of a semisimple element is reductive, see for example here, where $H$ consists of semisimple elements.

$(2)$ The text says that because $e$ is nilpotent, $e\in \mathfrak{g}^s$ implies $e\in [\mathfrak{g}^s,\mathfrak{g}^s]$. Did you try to use the nilpotency yet?

For an alternative proof see Theorem $4.10$ here, on page $42$. The last line there is, in your notation, $e=\frac{1}{2}[h,f]\in [\mathfrak{g}^s,\mathfrak{g}^s]$.

Dietrich Burde
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