# Can the commuting condition in Jordan-Chevalley decomposition be replaced with this global criterion?

Let $G$ be a reductive linear algebraic group defined over an algebraically closed field $k$ of arbitrary characteristic, and write $\mathfrak{g}$ for its Lie algebra. The Jordan-Chevalley decomposition of $X\in\mathfrak{g}$ is the unique decomposition $X=s+n$ with

1. $s\in \mathfrak{g}$ semisimple (i.e. in the Lie algebra of a torus in $G$),
2. $n\in\mathfrak{g}$ nilpotent (i.e. in the Lie algebra of a unipotent group in $G$), and
3. $[s,n]=0$.

The commutation relation (3) is equivalent to $[s,X]=0$, and hence to $X\in\mathfrak{c}_{\mathfrak{g}}(s)$.
In very good characteristic, for every $X$, the stabilizer $C_G(X)$ is smooth and satisfies $\text{Lie}\, C_G(X) =\mathfrak{c}_{\mathfrak{g}}(X)$. Therefore $X\in \text{Lie}\, C_G(X)$, and replacing (3) by

3'. $C_G(X)(k)\subset C_G(s)(k)$

still uniquely specifies the Jordan-Chevalley decomposition. Does replacing (3) with (3') uniquely specify the Jordan-Chevalley decomposition in any characteristic?

Extra (more vague) questions if this is too hard:

Would looking at the centralizer instead of just $k$-points help?

Do centralizers of nilpotent elements (known if I understand correctly) give centralizers for general $X$ in an explicit enough way, or would verifying the question through the classification require ideas in addition to brute force?

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This characterization might or might not be true in general, though it's certainly not just a formal consequence of standard general facts (about the nice fit between global and infinitesimal centralizers of semisimple elements in $\mathfrak{g}$, for example). Even if true, it's hard to visualize how the hypotheses in this version of Jordan-Chevalley decomposition would be verified in practice. Are there motivating examples? –  Jim Humphreys Aug 18 at 23:36
The motivation is seeing whether/how the "Jordan decomposition" of Kac-Vinberg differs from Jordan-Chevalley in positive characteristic. –  Jason Aug 18 at 23:58