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Let $G$ be a connected algebraic group over an algebraically closed field $\overline{k}$ acting on an irreducible variety $X$. A geometric quotient is a morphism of varieties $\pi: X \rightarrow X/\sim$ which on closed points (that is, as a morphism of classical varieties) satisfy the following:

(i): $\pi$ is a surjective open map, and the fibres are exactly the $G$-orbits of $X$.

(ii): for any open set $U$ of $X/\sim$, the ring homomorphism $\pi^{\ast}: \overline{k}[U] \rightarrow \overline{k}[\pi^{-1}U]$ is an isomorphism onto the $G$-fixed points of codomain.

Rosenlicht's theorem says that there exists a $G$-stable open subset of $X$ for which the geometric quotient exists.

Is there any generalization of Rosenlicht's theorem for when $G$ and $X$ are defined over an arbitrary field? The case I'm interested in is when $G$ and $X$ are geometrically connected subgroups of upper triangular unipotent matrices over a $p$-adic field $F$ (so all orbits are closed), and $X$ is normalized by $G$.

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  • $\begingroup$ Assume that $G$ and $X$ are defined over a perfect field $k$. You can choose a $G$-stable open subset $U$ of $X$ (for which the geometric quotient exists) defined over $k$. It seems that your geometric quotient $U/\sim$ is has no automorphisms (compatible with the identity automorphism of $U$). If $U/\sim$ is quasi-projective, you can construct a $k$-model of it by Galois descent. (I hope all this is correct....) $\endgroup$
    – Mikhail Borovoi
    Commented Jan 18, 2018 at 9:24
  • $\begingroup$ @MikhailBorovoi. You need to make sure that the $G$-stable open subset $U$ descends, or equivalently, that its complement descends. Since there are finitely many coefficients of the finitely many defining equations of the closed complement, the open subset $U$ descends to some finite field extension $L/k$. Since $k$ is perfect, this field extension is Galois. Now you can replace $U$ by the intersection of the finitely many Galois conjugates of $U$. This $G$-stable open subset descends. $\endgroup$
    – Jason Starr
    Commented Jan 18, 2018 at 13:43
  • $\begingroup$ @JasonStarr: Yes, this is what I meant, the intersection of finitely many Galois conjugates of $U$. I was in hurry and could not write it down. $\endgroup$
    – Mikhail Borovoi
    Commented Jan 18, 2018 at 16:30
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    $\begingroup$ @D_S: It would help to give a reference for Rosenlicht's result. He wrote quite a few papers on structural questions for algebraic groups and their actions, often in language which is now getting archaic but also very often dealing with fields of definition. $\endgroup$
    – Jim Humphreys
    Commented Jan 18, 2018 at 17:06
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    $\begingroup$ @D_S: An unrelated question is why you've included the tag 'reductive-groups'? Probably 'group-actions' would be a better choice if you need another tag. $\endgroup$
    – Jim Humphreys
    Commented Jan 18, 2018 at 17:12

1 Answer 1

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Jim Humphreys was right: No generalization is necessary. In Theorem 2 of

Rosenlicht, Maxwell: Some basic theorems on algebraic groups, Amer. J. Math. 78 (1956) 401--443,

the result is already stated over arbitrary base fields.

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