# Vector bundles on $\mathbb{A}^n / G$

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

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.

• 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. Feb 11, 2020 at 21:27
• 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.
– ssx
Feb 16, 2020 at 17:04