# extension of $G$-bundles

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

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.

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

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

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

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