MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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 '14 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 '14 at 23:58
Is it obvious that $C_G(s)$ is even defined over $k$? (Of course this is true if $s$ is rational, but may fail if $k$ is imperfect.) EDIT: Oops, never mind; I guess that $C_G(s) = C_G(s^{[p]^N}) = C_G(X^{[p]^N})$ for sufficiently large $N$. – L Spice Jan 14 at 15:00
Also, I'm not sure that your equivalent definition of nilpotence in (2) remains equivalent (to the usual definition of having nilpotent image in some faithful representation) in bad characteristic. ((1) does remain equivalent, but that's a theorem.) EDIT: OK, sorry; it turns out that I'm just re-treading ground already trod by @user54268 in an answer to your previous question. – L Spice Jan 14 at 15:04

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.