# Varieties which become isomorphic to algebraic groups over an algebraic closure

My question is as follows:

Let $k$ be a field of characteristic zero and let $\overline{k}$ be an algebraic closure. Let $V$ be an algebraic variety over $k$ and let $\overline{V}=V \times_k \overline{k}$. Suppose that $\overline{V}$ admits the structure of an algebraic group. Then is $V$ itself a principal homogeneous space for some algebraic group?

Note that the answer to my question is yes when $V$ is projective; in this case it is well-known that such a $V$ is a principal homogeneous space for its Albanese variety. So I am really interested in the case where $V$ is affine. Here I am not aware of an analogue of the Albanese variety for linear algebraic groups.

Even if the answer is no in general, I would still be interested in some positive results for special cases, e.g. for reductive groups or semisimple groups.

-
Since the structure of an algebraic group can be defined by writing down finitely many polynomials, it will always show up over some finite field extension, so a twist will show up as a map $Gal \to Aut(G)$ with finite image. I think this will descend to a homogeneous space if and only if $Lie(G)$ inside $H^0(G,T_G)$ is an invariant subspace under the action of this group. We can check this in some basic cases like the torus, but I'm not sure in general. – Will Sawin May 24 '13 at 16:38
I should probably also mention that I have come across proofs of the fact that the answer is also yes when $\overline{V}$ is an algebraic torus, see e.g. Skorobogatov - Torsors and rational points - Lemma 2.4.4. The proof here however makes key use of the fact that $\overline{V}$ is an algebraic torus, namely it uses the fact that one can recover an algebraic torus from its character group. – Daniel Loughran May 24 '13 at 17:03
I don't think that restricting to simple groups will help. Some simple groups are isomorphic as a variety of a product of a vector space with another space. So if you had a strange form of a vector space, you could take a product to get a candidate strange form of the simple group. – Ben Wieland May 29 '13 at 21:15

Here is a counterexample with $V$ finite.

Suppose that $k'$ is a separable extension of degree $5$ of $k$, and set $V = \mathop{\rm Spec} k \sqcup \mathop{\rm Spec} k'$. Suppose that $V$ is a homogenous space; then it is a group scheme, since $V(k) \neq \emptyset$.

If $\overline k$ is the separable closure of $k$, then $V_{\overline k}$ is the disjoint union of six copies of $\mathop{\rm Spec} \overline k$, and these copies form a group of order $6$. The Galois group of $k$ acts on $V_{\overline k}$ by permuting the non-identity component transitively. But a group of order $6$ contains elements of order $2$ and $3$, so this is impossible.

I am convinced that one can get geometrically connected examples using forms of $\mathbb G_{\rm a}^n$.

-
This works whenever the Galois group, as a subgroup of $S_n$, does not normalize any simply transitive subgroups. – Will Sawin May 25 '13 at 0:01
Thanks Angelo. If anybody could say anything at all about geometrically connected examples (except for tori) I would be most interested. – Daniel Loughran May 25 '13 at 9:26

I think the answer should be "no" if $V$ has finite order automorphisms as a variety which are not conjugate to group automorphisms. Principal homogenous spaces for $G$ are described by $H^1(Gal(\bar{k}/k), G(\bar{k}))$. I believe groups over $k$ which be come isomorphic to $G$ over $\bar{k}$ are described by $H^1(Gal(\bar{k}/k), Out(G))$ and principal homogenous spaces for groups of this sort are described by $H^1(Gal(\bar{k}/k), Aut_{grp}(G))$. On the other hand, spaces isomorphic to $G$ as a variety are $H^1(Gal(\bar{k}/k), Aut_{var}(G))$.

So a good place to look should be algebraic groups $G$ whose automorpism group as varieties is much larger than their automorphism group as groups. The automorphism group of $\mathbb{G}_a^n$ is huge, but most of it is elements have infinite order and Galois cocycles have finite image. So I don't have an example to propose yet.

-
(For $A^1$ it is easy and for $A^2$ it follows (not easy, though!) from the amalgamated product structure of the automorphism group, that finite order automorphisms are linearizable. There is an example by Asanuma of a non-linerlizable automorphism in higher dimension, but in positive characteristic.) – Mariano Suárez-Alvarez May 24 '13 at 17:22
Abelian varieties have many finite-order automorphisms of the underlying variety that are not group automorphisms. Why does your argument fail for that case? (To be sure, I am quite convinced that what you say must be close to true. I just don't fully understand it yet.) – René May 24 '13 at 17:28
All automorphisms of an abelian variety are of the form $x \mapsto Ax + b$ where $A$ is an automorphism of the abelian variety and $b$ is a point of the abelian variety. If $A - \mathrm{Id}$ is invertible, at least, that's conjugate (in the group of variety automorphisms) to $x \mapsto Ax$. I haven't actually worked through the Galois cohomology I sketch above, but I suspect that things conjugate to group automorphisms won't do it. It isn't clear to me what happens if $A - \mathrm{Id}$ is not invertible. – David Speyer May 24 '13 at 19:46
I think any $Ax+b$ is clearly fine. In the nonabelian case, an automorphism composed with a left/right multiplication will be fine. You can mod out by the translation subgroup to get the action by automorphism, which gives a twist of the group. But then the action of that twist on your variety is Galois-equivariant, by construction, so your variety is a homogeneous space. – Will Sawin May 24 '13 at 21:16
I think forms of $G$ use not $Out(G)$, but $Aut(G)$; and homogeneous spaces over forms of $G$ use $G\rtimes Aut(G)$ (aka $(G\times G)/\Delta(Z(G))$, I think). As a check, $Out(SL_2)$ is trivial, but there are forms of $SL_2$. – Ben Wieland May 25 '13 at 21:34