Let $k$ be a characteristic zero field, $V \subset \mathbb{A}^n_k$ an open subscheme, $G$ a split reductive group over $k$ and $T$ a $G$-torsor over $V$ (in the etale, equivalently fppf topology). Suppose that $T$ extends along codimension $1$ points of $\mathbb{A}^n_k$, i.e. the torsor class $[T_{k(\mathbb{A}^n)}] \in \mathrm{H}^1(k(\mathbb{A}^n),G)$ lies in the image of $\mathrm{H}^1(\mathcal{O}_{\mathbb{A}^n,Z},G)\rightarrow \mathrm{H}^1(k(\mathbb{A}^n),G)$ for all closed irreducible $Z\subset \mathbb{A}^n$ of codimension $1$. (This is true, for example, if $\mathbb{A}^n\setminus V$ has codimension $\geq 2$ in $\mathbb{A}^n$.)

Question: does there exist a $G$-torsor $T'$ on $\mathbb{A}^n$ such that $T$ and $T'|_V$ are locally isomorphic in the Zariski topology?

Some remarks:

  1. By known cases of the Grothendieck-Serre conjecture [Panin], it suffices to show that there exists a $G$-torsor $T'$ on $\mathbb{A}^n$ with $[T_{k(\mathbb{A}^n)}]=[T'_{k(\mathbb{A}^n)}]$ in $\mathrm{H}^1(k(\mathbb{A}^n),G)$.
  2. We cannot require that $T$ literally extends to a $G$-torsor on $\mathbb{A}^n$; indeed there exists a $\mathrm{GL}_2$-torsor on $\mathbb{A}^3\setminus\{0\}$ that does not extend to $\mathbb{A}^3$.
  3. The question has a negative answer if we allow $V$ to be an open subset in an arbitrary smooth affine $k$-variety [Antieau--Williams].
  4. The question is trivial for $G = \mathrm{GL}_n$ or $\mathrm{Sp}_n$, since every torsor under such a group is Zariski locally trivial.

Panin, Ivan A., Proof of the Grothendieck-Serre conjecture on principal bundles over regular local rings containing a field, Izv. Math. 84, No. 4, 780-795 (2020); translation from Izv. Ross. Akad. Nauk, Ser. Mat. 84, No. 4, 169-186 (2020). ZBL1458.14024.

Antieau, Benjamin; Williams, Ben, Topology and purity for torsors, Doc. Math. 20, 333-355 (2015). ZBL1349.14068.

  • $\begingroup$ I'm probably missing something, but the answer would be yes if your $G$-torsor is already trivial, locally for the Zariski topology, right? So, if $G$ is special, the answer is yes. (This is just a bit more general than your 4.) $\endgroup$ Commented Jul 5, 2021 at 15:22
  • $\begingroup$ Maybe you can use 3 to cook up an example. Indeed, let $X$ be smooth affine, and let $U\subset X$ such that $T\to U$ is a counterexample. Now, choose $X\to\mathbb{A}^n$ finite surjective and consider $U\to V$. Assuming the latter is finite (i.e., proper), you could maybe use Weil restriction of scalars along $U\to V$. $\endgroup$ Commented Jul 5, 2021 at 15:24
  • $\begingroup$ @AriyanJavanpeykar: the question is indeed trivial if $G$ is special. About your suggestion: it sounds interesting, but filling in the details does not seem easy. Also if $T$ is a $G$-torsor over $U$, then the Weil restriction to $V$ would be a $G'$ torsor, where $G'$ is the Weil restriction of $G_U$ along $U\rightarrow V$. I think $G'$ will not be split unless $U\rightarrow V$ is of a very degenerate form. $\endgroup$
    – Jef
    Commented Jul 5, 2021 at 16:39
  • $\begingroup$ Ah yes, if you insist on $G$ being split, this probably won't work... $\endgroup$ Commented Jul 9, 2021 at 18:58

1 Answer 1


In case this might be useful to anyone, it turns out that results of Colliot--Thelene can be used to resolve a closely related question: see Theorem 6.1 in the following preprint (apologies for the self-promotion):

https://www.dpmms.cam.ac.uk/~jcsl5/ADEpaper.pdf (version 5 January 2022)

It seems that a very similar proof establishes the claim of the question.


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