# Is every homogeneous G-variety of the form G/H?

Let $G$ be an algebraic group over an algebraically closed field $k$. Then G/H is a quasi-projective homogeneous G-variety for any closed subgroup $H$. Now, several times I have seen something like "Let $X$ be a homogeneous $G$-variety, i.e. $X = G/H$ for a closed subgroup $H$ of $G$" and I wonder if this "i.e." is correct. This would imply that any homogeneous $G$-variety is already quasi-projective. I think this is true, when $\mathrm{char}(k) = 0$, because then the canonical abstract isomorphism $\pi:X \rightarrow G/G _x$ is separable and thus an isomorphism of varieties for any $x \in X$ (is this correct?). But what about $\mathrm{char}(k) > 0$? Are there counter-examples or is any (quasi-projective) homogeneous $G$-variety up to isomorphism of the form $G/H$?

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For those of us in pedants' corner, could someone edit the title to read "Is every homogeneous G-variety of the form G/H"? Thank you. – Joel Fine Apr 2 '10 at 12:27
Must assume $G$ and $X$ are smooth (or reduced; leads to the same since $k$ alg closed). Otherwise can have things like $G$ infinitesimal and $X = {\rm{Spec}}(k)$. Role of smoothness in proof is that for $x \in X(k)$, orbit map $G \rightarrow X$ is a surjective with all fibers equidimensional of the same dim., necessarily the "right" value (difference of pure dim's of $G$ and $X$), and so smoothness allow to apply "miracle flatness theorem" (23.1, Matsumura CRT) to deduce orbit map is (faithfully) flat. Thus, $G/G_x \rightarrow X$ is isom due to faithfully flat descent theory. – BCnrd Apr 2 '10 at 12:38
@Joel: your suggested edit has been done. – José Figueroa-O'Farrill Apr 2 '10 at 12:44

## 2 Answers

It depends on what you mean by "closed subgroup". If you mean a Zariski closed subset which forms a subgroup then the answer is no. If you mean a closed subgroup scheme, then the answer is yes. An example where you need to use the second definition is the Frobenius map $F\colon G \to G^{(p)}$. If we let $G$ act on $G^{(p)}$ through $F$ then the action is transitive and indeed $G^{(p)}$ is isomorphic to $G/Ker F$. However, unless $G$ is zero-dimensional $Ker F$ is a non-trivial finite group scheme whose $k$-points consist of just the identity.

Note however that $G/H$ is always quasi-projective even when $H$ is a subgroup scheme so all homogeneous $G$-spaces are quasi-projective.

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You definitely have to be more careful when working with possibly nonreduced group schemes. Standard references include the book by Demazure and Gabriel; see Part I, Chapter 5 in Jantzen's Representations of Algebraic Groups for a development in this spirit taking into account Frobenius kernels, etc. For reduced groups, the characteristic of the field doesn't really matter: see Chapter II in Borel's Linear Algebraic Groups (or similar material in books with the same title by Springer and me). – Jim Humphreys Apr 2 '10 at 13:42

An aside: BCnrd's argument above shows that, over any perfect field $k$, $X$ is isomorphic to $G/H$ if and only if $X$ has a $k$-point. Otherwise, consider, e.g., conics with no point: they are homogeneous under a twist of $G=SO_3$ but are not $G/H$. I don't know what happens for more general base schemes.

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Another example along the same lines is a principal homogeneous space over an elliptic curve over non-algebraically closed field (the algebraic group is not linear, but that wasn't required in the question). – Victor Protsak Aug 25 '10 at 19:17