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Let $S$ be a smooth surface (let's say over an algebraically closed field) and let $D$ be a smooth divisor in $S$. Let also $G$ be a connected algebraic group. Assume that we are given a principal $G$-bundle ${\mathcal F}$ on $S\backslash D$. Under what conditions can we extend it to all of $S$? Do I understand correctly, that this is always the case when the derived group $[G,G]$ is simply connected?

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Sasha, there is a paper by Colliot-Thelene and Sansuc (I don't remember the title, but Keerthi M. does; hopefully he will see this) in the middle of which is buried of proof of the extension result for any connected reductive group when working over a regular 2-dimensional scheme with a torsor over complement of a codim-2 closed set (the proof views $G$ as subgroup of some GL$_n$ and affineness of GL$_n/G$ -- here is where reductivity is used -- to cleverly reduce the problem to GL$_n$). This reduces your question to the analogue for dvrs (i.e., just need to extend around generic pts of $D$). –  BCnrd Nov 25 '10 at 16:04
@BCnrd: I would expect that such extension result holds for any affine $G$: to prove, reduce to $GL(n)$ by Tannakian formalism. More explicitly, if $f:T\to X$ is a $G$-torsor, then $f$ is affine, so the torsor is determined by the sheaf of algeras $f_*O_T$ equipped with an action of $G$. But the sheaf $f_*O_T$ is a union of locally free $G$-invariant subsheaves of finite rank (corresponding to f. dimensional subrepresentations of the regular representation). So if you want to extend a $G$-torsor, the problem reduces to extending a bunch of vector bundles (+compatibilities). –  t3suji Nov 25 '10 at 17:51
The paper BCnrd refers to is: "Fibrés quadratiques et composantes connexes réelles", and the result is Thm 6.13. Link: springerlink.com/index/U5G225315W158721.pdf –  Keerthi Madapusi Pera Nov 25 '10 at 18:13
Dear t3suji: Your suggested exhaustion by $G$-stable subsheaves doesn't seem to encode the torsor property. Here are some examples. Consider $G = {\rm{PGL}}_2$ -- so twisted forms of $\mathbf{P}^1$ -- and char. not 2. Viewing these canonically as smooth conics in a $\mathbf{P}^2$-bundle, if the generic fiber as a conic over the function field has discriminant whose divisor on $S$ has odd multiplicity somewhere then we cannot hope to extend around the corresponding codimension-1 point of $S$. There are these "ramification" obstructions at generic pts of $D$ in general. –  BCnrd Nov 25 '10 at 19:21
Dear Brian: This is encoded in the "+compatibilities" part. That is, you need a bunch of vector bundles and some extra data (action of G and product on their union). This causes problems when you extend across codimension one, as in your example: although vector bundles extend, you can't extend them compatibly. However, across codimension two, the extension of vector bundles is unique, and compatibility is automatic. Perhaps I should have been more explicit: `such extension result' meant the result you were quoting (by Colliot-Thelene and Sansuc), not the original question. --Dima –  t3suji Nov 25 '10 at 19:40
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4 Answers

Nick Shepherd-Barron once asked me this question, and I think I can remember what was eventually concluded.

The short answer to the original question is negative for the additive group $\mathbb{G_a}$. Map $SL_2(\mathbb{C})$ to $X=\mathbb{C}^2-(0,0)$ by letting a matrix act on the vector $(1,0)^T$. This realizes the group as a torsor over $X$ for the group of unipotent matrices \[ \begin{pmatrix} 1 & a \\\\ 0& 1 \end{pmatrix}. \] We can trivialize it over the vectors $(v_1,v_2)^T$ with $v_1\neq 0$ with the section \[ \begin{pmatrix} v_1 & 0 \\\\ v_2& v_1^{-1} \end{pmatrix},\] while it trivializes on the set $v_2\neq 0$ via the section \[\begin{pmatrix} v_1 & -v_2^{-1} \\\\ v_2& 0 \end{pmatrix}.\] The transition function on the overlap is easliy computed to be \[\begin{pmatrix} 1 & (v_1v_2)^{-1}\\\\ 0& 1 \end{pmatrix}.\] This represents the standard non-trivial generator of $H^1(O_X)$, and hence, the bundle is non-trivial. On the other hand, if you could extend it to $\mathbb{C}^2$, it would trivialize, since there are no non-trivial $\mathbb{G}_a$-bundles on an affine variety.

There is indeed a correspondence between principal bundles and tensor functors from representations to vector bundles. But if I recall correctly, the functor is required to be exact. In the case at hand, the extension of vector bundles is the direct image with respect to the inclusion $X\hookrightarrow \mathbb{C}^2$, which fails this.

Added: OK, I see this was just an elaborate way to say: take any principal $\mathbb{G}_a$-bundle corresponding to a non-zero element of $H^1(O_X)$. Note, anyways, that the derived group is trivial in this example. Certainly the statement is false for general connected groups, contrary to some of the comments.

Added, 25, November, 2011:

This question came back to me today while I was thinking about something unrelated. It occurred to me then to point out that for the example above, if we work in the analytic category, we have $$H^1(X, \mathbb{G}_a)\simeq H^1(X, \mathbb{G}_m),$$ via the exponential sequence. On the other hand, $$H^1(\mathbb{C}^2, \mathbb{G}_a)=H^1(\mathbb{C}^2, \mathbb{G}_m)=0.$$

So the desired extension property is false on analytic spaces even for reductive structure groups.

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Thanks for the help, Alex! –  Minhyong Kim Nov 27 '10 at 13:47
No worries, you are welcome! –  Alex B. Nov 27 '10 at 13:53
Good point, thanks for posting this. –  t3suji Nov 27 '10 at 15:55
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Wait, are you assuming that G is a reductive group whose derived group is simply connected? In that case, I believe Serre's Conjecture II implies that the G-torsor is trivial over a dense Zariski open subset of S. So you can extend it across the codimension 1 points as a trivial torsor.

P.S. Sorry for writing this as a separate answer, as opposed to a reply to Sasha's comment, but I don't know how to make replies in MathOverflow (maybe you need to be registered).

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I wrote this before I saw Brian's comment. I am just saying exactly the same thing as Brian, sorry! Also I just figured out how to make comments. –  Jason Starr Nov 26 '10 at 14:36
Jason--You appear to have multiple accounts; that prevents the accumulation of reputation points. You need at least 50 points in your account to comment on other people's answers. –  Keerthi Madapusi Pera Nov 26 '10 at 16:05
Ah, yes, you also need to register. –  Keerthi Madapusi Pera Nov 26 '10 at 16:05
Jason, you don't need to register. Just keep using the same account. –  BCnrd Nov 26 '10 at 16:34
Jason, it is better to register. From the very beginning you can comment on your own answers. After you accumulate 50 points, you will be able to comment on other people's answers –  Mikhail Borovoi Nov 27 '10 at 14:38
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For extension across codimension 2 points, in the case of a reductive group, you can use Hartog's theorem / S_2 extension. As t3suji suggests, you embed G into GL_n and first extend the GL_n-bundle to a bundle E defined over all of S. Via its embedding in GL_n, G acts on E. Form the quotient space E/G. Then the original G-bundle structure away from D gives a rational section of E/G. Since G is reductive, E/G is affine over S. So if you extend your section of E/G at codim 1 points, then Hartog's theorem extends across codim 2 points. And the inverse image of this section of E/G in E is an extension of your original G-bundle.

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Jason, if I understand correctly, you're outlining how to geometrically make the construction if it can be done at the codim-1 pts of D. In general we may not be able to extend the section of $E/G$ around these points (as in the conic examples I mention above), unless I am misunderstanding something. The existence or not of this extended section seems to also depend on the choice of $E$, so it doesn't appear to show that the extension problem for the torsor at the generic pts of $D$ is etale-local on $S$. –  BCnrd Nov 25 '10 at 19:28
This is my understanding, too. (I guess my comment to the question was not quite clear on that, I tried to expand it.) –  t3suji Nov 25 '10 at 19:45
My question really was when can you extend to codimension 1 -- for some reason I thought that having $[G,G]$ simply connected guarantees this. Am I wrong? –  Alexander Braverman Nov 25 '10 at 22:12
Let $R$ be a dvr (allowing imperfect residue field, if char($k) > 0$ above), $K$ the frac. field, and $G$ a split conn'd reductive $R$-group with simply conn'd derived gp. Is ${\rm{H}}^1(R,G) \rightarrow {\rm{H}}^1(K,G)$ surjective, at least for $R$ excellent (e.g., local ring on surface)? Since $G/[G,G]$ is split torus, may assume $G$ is split semisimple and simply conn'd. Even if one has a version of Steinberg's vanishing result over a strict henselization, that seems too weak to help (since a finite sepble extn of $K$ unramified at one place over $R$ is usually ramified elsewhere). –  BCnrd Nov 26 '10 at 1:26
So, does it mean that you don't believe in this statement? Can you think of a counterexample? –  Alexander Braverman Nov 26 '10 at 12:37
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Just a quick idea on how I would proceed if the field was $\mathbb{C}$. Let see $S=S\setminus\mathcal{N}(D)\cup \mathcal{N}(D)$ where $\mathcal{N}(D)$ is a normal neighborhood of $D$. You have a $G$ principal bundle $\mathcal{F}$ over $S\setminus\mathcal{N}(D)$ and you want to extend it to all $S$.

Basically you could do so if there is a $G$ principal bundle, $P$, over $S$ such that $\pi^*(P)=\mathcal{F}\vert_{\partial\mathcal{N}(D)}$ where $\pi$ is the natural projection from $\partial\mathcal{N}(D)$ to $D$.

I have the feeling that the only obstruction is that $\mathcal{F}$ has to be trivial when restricted to the fibers of the unit normal neighborhood of $D$ in $S$.

As we are talking about surfaces then $S$ is of real dimension $4$, if $D$ is of real dimension $2$ then those fiber have dimension $1$. And since you assumed that $G$ was connected then it is necessarily trivial on the fibers...

This is just some thought maybe I'm wrong...

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Note that if you're still in the case of $\mathbb{C}$ and that dimension of $D$ is $0$ then the triviality of $\mathcal{F}$ on the fiber depends on $\pi_2(G)$ so if $G$ is a Lie group $\pi_2(G)$ is trivial, thus you can extend as well... –  Noz Nov 25 '10 at 14:03
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