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Let $G$ be a connected algebraic group. Is it true that every $G$-bundle on ${\mathbb A}^n$ is trivial? What is the reference? I am actually only interested in the case $n=2$.

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From the context, I assume $G$ is to be an affine algebraic group. The reductive case has been most studied, so it's also important to indicate if more generality is wanted. –  Jim Humphreys Nov 15 '10 at 18:39
Yes, I meant that $G$ is affine and $G$ reductive is sufficient for me (in fact, at least in characteristic $0$ it is clear that everything reduces to the reductive case). –  Alexander Braverman Nov 16 '10 at 14:32
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3 Answers

This is true for $G=GL(r)$, as shown by Quillen and Suslin.

For arbitrary $G$ there are counterexamples. Quite surprisingly, even $G$-bundles over $\textrm{Spec }k$ may not be trivial. See the paper

M. S. Raghunathan

"Principal bundles on affine space and bundles on the projective line",

Mathematische Annalen Volume 285, Number 2, 309-332

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This 1989 paper can be accessed freely at the German archive gdz.sub.uni-goettingen.de/ It would help to unpack Raghunathan's theorems (and his previous results) a little more, since he is mainly concerned with reductive groups over arbitrary fields. If I understand correctly, groups over algebraically closed fields are better behaved in the current context. –  Jim Humphreys Nov 15 '10 at 18:44
Thank you. I'll look at the paper. I am actually interested in the (split) reductive case but possibly over finite field. I wonder if there is a counterexample in that case. –  Alexander Braverman Nov 16 '10 at 14:40
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Over an algebraically closed field, for $G$ connected and reductive, every principal $G$-bundle on ${\Bbb A}^n$ is trivial, also by a theorem of Raghunathan:

"Principal bundles on affine space", in C. P. Ramanujam—a tribute, pp. 187–206, Tata Inst. Fund. Res. Studies in Math. 8 (1978).

(Unfortunately I can't find this reference free online.)

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This is reference [15] in Raghunathan's later paper mentioned by Francesco Polizzi. Is there a more accessible later reference for the algebraically closed case? –  Jim Humphreys Nov 15 '10 at 21:54
Thank you. Do you know if the same is true over finite field? –  Alexander Braverman Nov 16 '10 at 14:31
@Jim: Good question, I haven't been able to find a simpler reference. @Alexander: I'm no expert in this, but the strongest general results so far seem to be in an IHES paper from 1992 by Colliot-Thelene and Ojanguren. However, "infinite" is one of their hypotheses on the ground field -- but I don't think they give counter examples. –  Dave Anderson Nov 17 '10 at 4:38
@Alexander: I'm unaware of any explicit formulation for a reductive group defined over a finite field; but I'm not familiar enough with this literature. –  Jim Humphreys Nov 18 '10 at 14:52
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In characteristic $p$ you can make an easy counterexample with $n=1$, right? An exact sequence of commutative algebraic groups $0\to E\to X\to \mathbb A\to 0$ with $E$ an elliptic curve.

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Interesting -- but how does ${\Bbb A} = {\Bbb G}_a$ act on $E$? –  Dave Anderson Nov 15 '10 at 22:33
This is a naive question, but does this give us something for n=2, namely $E\times A \to X\times A \to A^2$? –  David Roberts Nov 15 '10 at 23:46
I meant: Choose an elliptic curve $E$ (over a given field $k$ of characteristic $p$) that has a subgroup of order $p$ (i.e. has a $k$-rational point of order $p$). In $\mathhbb G_a$ there is also a subgroup of order $p$, with quotient isomorphic to $\mathbb G_a$. In $E\times \mathbb G_a$ there is a 'diagonal' subgroup of order $p$. Divide out by it to get $X$. –  Tom Goodwillie Nov 16 '10 at 0:36
You can also take a supersingular elliptic curve as both it and $\mathbb G_a$ contains an $\alpha_p$. –  Torsten Ekedahl Nov 16 '10 at 5:13
I don't know. I'm just a topologist playing along. –  Tom Goodwillie Nov 17 '10 at 4:20
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