Let $G$ be a finite group acting linearly on $\mathbb{A}^n$. Do we expect algebraic vector bundles on $X := \mathbb{A}^n/G$ to be trivial? Here by the quotient I mean the singular scheme, not the stack quotient.

What is known:

(1) For finite abelian groups $G$, $X$ is an affine toric variety, and vector bundles on $X$ are trivial by a Theorem of Gubeladze: https://iopscience.iop.org/article/10.1070/SM1989v063n01ABEH003266. In particular, when $G$ is the trivial group, triviality of vector bundles on $\mathbb{A}^n$ is an older result by Quillen-Suslin.

(2) Line bundles on $X$ are trivial. This can be shown by lifting a line bundle on $X$ to a $G$-line bundle on $\mathbb{A}^n$; then $\mathrm{Pic}^G(\mathbb{A}^n)$ is the group of characters of $G$, and since the linear $G$-action has a fixed point $0$, this character will be trivial, hence coming from a trivial line bundle on $X$.

(2') For vector bundles of higher rank the argument in (2) does not work. This has to do in particular with $G$-actions on $\mathbb{A}^N$ not being linearizable in general: https://link.springer.com/chapter/10.1007%2F978-94-015-8555-2_3

(3) The Grothendieck group of vector bundles is $\mathrm{K}_0(X) = \mathbb{Z}$. We prove it in https://arxiv.org/pdf/1809.10919.pdf, Prop. 2.1 indirectly, using comparison with cdh topology of differential forms.

(3') By homotopy invariance of K-groups in the smooth case, $\mathrm{K}^G_0(\mathbb{A}^n) \simeq \mathrm{K}^G_0(\mathrm{Spec}(k))$ which is the Grothendieck ring of $G$-representations; however this does not seem to help.

Is there any more evidence for/against the triviality of vector bundles on $X$? Is this question mentioned anywhere in the literature?


1 Answer 1


In the paper Affine varieties dominated by $\mathbf{C}^2$ Gurjar considers a slightly more general situation, namely an affine normal variety $\mathrm{X}$ with a proper surjective morphism $\mathbf{A}^2\rightarrow\mathrm{X}$. He shows that every line bundle on $\mathrm{X}$ is trivial; together with a result of Anderson (every vector bundle on $\mathrm{X}$ is the direct sum of a trivial bundle and a line bundle) this shows in particular that every vector bundle on $\mathbf{A}^2/\mathrm{G}$ is trivial.

  • 2
    $\begingroup$ Thanks! Looking at the proof, it seems that the key nontrivial result inside is that of Anderson, that in this setting vector bundles always split into line bundles: ams.org/journals/tran/1978-240-00/S0002-9947-1978-0485827-5/…, e.g. Prop. 5.1. $\endgroup$ Feb 11, 2020 at 21:27
  • $\begingroup$ You are right, one should give credit where credit is due. I think the point of Gurjar's paper is to consider line bundles in the more general situation; but of course in your case the triviality of line bundles is easy to see as you mentioned. $\endgroup$
    – Samuel
    Feb 16, 2020 at 17:04

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