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Suppose G is an algebraic group (over a field, say; maybe even over ℂ) and H⊆G is a closed subgroup. Does there necessarily exist an action of G on a scheme X and a point x∈X such that H=Stab(x)?

Before you jump out of your seat and say, "take X=G/H," let me point out that the question is basically equivalent to "Is G/H a scheme?" If G/H is a scheme, you can take X=G/H. On the other hand, if you have X and x∈X, then the orbit of x (which is G/H) is open in its closure, so it inherits a scheme structure (it's an open subscheme of a closed subscheme of X).

I say "basically equivalent" because in my argument, I assumed that the action of G on X is quasi-compact and quasi-separated so that the closure of the orbit (i.e. the scheme-theoretic closed image of G×{x}→X) makes sense. I'm also using Chevalley's theorem to say the the image is open in its closure, which requires that the action is locally finitely presented. I suppose it's possible that there's a bizarre example where this fails.

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up vote 10 down vote accepted

In his book "Linear algebraic groups", 6.8, p98, Borel shows that the quotient of an affine algebraic group over a field by an algebraic subgroup exists as an algebraic variety, and he notes p.105 that Weil proved a similar result for arbitrary algebraic groups.

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Great. I think the precise reference is Proposition 2 of this paper: – Anton Geraschenko Dec 14 '09 at 3:25

The representability theorem [Demazure-Gabriel, III.2.7.1, p. 318] implies the following.


Let $A$ be a local artinian ring, let $G$ be a group over $A$ locally of finite type, and let $H\hookrightarrow G$ be a closed subgroup which is flat over $A$. Then the quotient $G/H$ in the category of fppf sheaves is a scheme; and the canonical morphism $G\rightarrow G/H$ is faithfully flat and of finite presentation.

Note that the group $G$ in the above theorem need not be either affine or flat over $A$; also, Demazure-Gabriel write in comprehensible language, unlike Weil.

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