MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $U$ be a $k$-scheme, where $k$ is a field. Let $G$ be a smooth affine $k$-group. Recall that a principal $G$-bundle over $U$ is a smooth surjective $U$-scheme $E$ with an action of $G$ on $E$ such that the action commutes with the projection to $U$ and the obvious morphism $G\times E\to E\times_UE$ is an isomorphism.

It is easy to see that every principal $G$-bundle is locally trivial in etale topology. However Serre in his Espaces fibres algebriques required a stronger condition called local isotriviality: for all $u\in U$ there is a a finite etale morphism $T'\to T$, where $T$ is a Zariski neighborhood of $u$ such that the pull-back of $E$ to $T'$ is trivial. I wonder if these definitions are known to coincide (I think I can prove it in some generality but the proof is not trivial).

Sketch of a proof. Let us embed $G$ into $GL(n)$. Consider the associated space $E'=E\times^GGL(n)$. It is a scheme because $GL(n)$ is affine and affine schemes can be glued in any reasonable topology. Moreover, it is a principal $GL(n)$-bundle, so passing to a Zariski cover we can assume it is trivial. The original bundle can be obtained from E' via reduction of the structure group, that is, it is a pull-back of the principal $G$-bundle $GL(n)\to GL(n)/G$. It remains to use the isotriviality of the latter bundle.

share|cite|improve this question
2  
I agree that the proof is not completely trivial, nor should it be. The obvious relative version is false: for a smooth, affine group scheme $G_U$ over $U$, there exist $G_U$-torsors over $U$ that are not "locally isotrivial" in the strong sense given. – Jason Starr Jun 20 '13 at 13:15
    
Thank you, Laurent, I edited the question. Jason, do you have a reference to such an example? – Roman Fedorov Jun 20 '13 at 14:20
1  
The conditions are in fact equivalent, at least when $k$ is infinite. I don't know a reference, but I know a proof, which I can post, if you'd like (although you seem to say that you have your own proof). – Angelo Jun 20 '13 at 15:23
    
@Angelo: Also the proof I know uses Bertini. Since we are allowed to make an (etale) field extension, I think probably we can get around the hypothesis that $k$ is infinite. – Jason Starr Jun 20 '13 at 15:56
    
@Roman: Oops! Upon closer inspection, the examples I thought I had produced are not truly torsors: they are generically torsors, and the bad fibers are "set theoretically" torsors. But the bad fibers are non-reduced, so these are not in fact torsors. I will keep thinking about this. – Jason Starr Jun 20 '13 at 16:57

This answer is coming late, but since I have also been struggling to find a reference, I hope this can be helpful to other people.

The answer is yes. Precisely, you can found a proof in

Raynaud, Michel Faisceaux amples sur les schémas en groupes et les espaces homogènes. (French) Lecture Notes in Mathematics, Vol. 119 Springer-Verlag, Berlin-New York 1970 ii+218 pp.

http://link.springer.com/book/10.1007%2FBFb0059504

Lemma XIV 1.4 Let $k$ be a field $G/k$ a smooth affine algebraic group $X/k$ a scheme $P$ a fpqc $G_X$-torsor. Then $P$ is representable and $P$ is locally isotrivial.

Remarks :

0) in fact "semi-locally isotrivial" in the original statement but this implies locally isotrivial,

1) the principle of the proof is the one you give,

2) this seems due to A.Grothendieck,

3) this is false is $G$ is not affine. There is a classical example also in Raynaud's book (XIII 3.1) where $X$ is a nodal curve and $G$ an abelian variety, see also

Brion, Michel Some basic results on actions of nonaffine algebraic groups. (English summary) Symmetry and spaces, 1–20, Progr. Math., 278, Birkhäuser Boston, Inc., Boston, MA, 2010.

and remark 3.1 in

Brion, Michel(F-GREN-F) On automorphism groups of fiber bundles. (English summary) Publ. Mat. Urug. 12 (2011), 39–66.

5) your definition of a torsor is a bit strange.

share|cite|improve this answer

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.