23
$\begingroup$

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?

$\endgroup$
8
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Evgeny Shinder Feb 11 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. $\endgroup$ – I. G. Noramus Feb 16 at 17:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.