Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers 2

up vote 8 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.

share|improve this answer
1  
Great. I think the precise reference is Proposition 2 of this paper: jstor.org/stable/2372637 –  Anton Geraschenko Dec 14 '09 at 3:25
add comment

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

Theorem

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.

share|improve this answer
add comment

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.