$\newcommand{\g}{\mathfrak{g}}$Let $G$ be a reductive algebraic group over field $k = \overline{k}$ and consider the characteristic polynomial $\g \to \g/\!/G := \operatorname{Spec} (k[\g]^G)$ induced by the adjoint action. Assume $|W|$ is invertible in $k$, so Chevalley-Todd-Shepard implies the quotient is affine space.

I expect the following basic geometric properties to generalize from the $\operatorname{GL}_n$ case.

  1. The $k$-points of $\g/\!/G$ are in bijection with Zariski-closed $k$-orbits.
  2. Every $k$-rational orbit has a unique closed orbit in its closure, and the map on $k$-points is this identification.
  3. The locus of regular elements (those $X \in \g$ with $\dim G_X = \operatorname{rank} G$, as small as possible) is exactly the smooth locus of the morphism.
  4. The semisimple $X \in \g$ (in the sense of additive Jordan-Chevalley) are exactly those with closed $k$-orbits.

Do these hold in the setting above? If so, how much more general are they? The first two at least seem like general facts about actions on an affine $k$-scheme.

Edit: After some more reading and Ben's helpful answer, I was able to resolve most of this, though I haven't found anything that ensures surjectivity of the quotient morphism in the positive characteristic case: in characteristic zero the argument uses that for any $A^G$-algebra $S$, we have the identity $S = (S \otimes_{A^G} A)^G$. This doesn't work in positive characteristic (and GIT never claims that $(\operatorname{Spec} A)/\!/G$ is a "universal categorical quotient" unless char 0).

I'll leave this open for a bit in case Ben or anyone else can chip in with a substitute argument or counterexample.


1 Answer 1


Yes, all these things are true, and 1. and 2. hold for any action of a reductive group $G$ on an affine variety $X$. The key point is that if I have two closed orbits $O_1,O_2$, then the functions on their disjoint union is a quotient of functions on $X$ (because $X$ is affine), so there's a function on $X$ that is 1 on $O_1$ and 0 on $O_2$. Since the group is reductive, you can assume this function is $G$ invariant by projecting it to the trivial rep.

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    $\begingroup$ Thanks so much, that's a big help. By the way, is the good-characteristic assumption enough to make this projection argument work in positive characteristic, where reductive doesn't imply complete reducibility of representations? $\endgroup$
    – C.D.
    Jan 16, 2023 at 2:55
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    $\begingroup$ Answering my own side-question: the argument works with no characteristic assumptions at all, see Corollary A.1.3 and the prior discussion on geometric reductivity in Appendix A to Mumford's GIT. $\endgroup$
    – C.D.
    Jan 16, 2023 at 4:26
  • $\begingroup$ Happy to accept this, though I was wondering whether you had any reference for (2) why/whether the quotient map is surjective in characteristic p -- it doesn't seem to follow from any of the general results I've found. $\endgroup$
    – C.D.
    Jan 17, 2023 at 19:33
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    $\begingroup$ It’s enough to check that a Cartan $\mathfrak{h}$ maps surjectively, and that’s true because the map is etale (and thus closed) on the Cartan. $\endgroup$
    – Ben Webster
    Jan 18, 2023 at 21:32
  • $\begingroup$ This argument works because finite maps are closed, but by the way $$\mathfrak{h} \to \mathfrak{g}/\!/G \cong \mathfrak{h}/\!/W$$ is not etale except on the regular locus, right? (e.g. ramified over zero) $\endgroup$
    – C.D.
    Jan 19, 2023 at 19:57

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