Triviality of vector bundles on affine open subsets of affine space

Question

Let $k$ be a field. By the Quillen-Suslin theorem, all vector bundles on $\mathbb{A}^n_k$ are trivial for all $n \geq 0$. If $U \subset \mathbb{A}^n_k$ is an affine open subset, then vector bundles on $U$ (of small rank) need not be trivial in general, but there do exist (non-trivial) such $U$, e.g., (split) algebraic tori, on which all vector bundles are indeed trivial.

Are there simple conditions on $U$ as above which imply that all vector bundles on $U$ are trivial?

My interest is in conditions for arbitrary $n$; I would also be interested in conjectural answers or results for special $k$, e.g., $k$ algebraically closed or $k = \mathbb{C}$.

I am particularly interested in the following special case:

Let $U$ be the complement of a finite union of (affine) hyperplanes in $\mathbb{A}^n_k$. Is every vector bundle on $U$ trivial? If not, is there a large class of such $U$ for which it is known (or conjectured) that all vector bundles on $U$ are trivial?

Answer 1

For your final question, the answer is that all vector bundles over $U$ are trivial.

Sketch of proof: Let $R=k[x_1,\ldots,x_n]$.

Let $L_1, L_2,\ldots, L_m$ be the equations of hyperplanes in $R$. Without loss of generality, we can assume $L_1=x_1$.

Set $A=R[L_1^{-1},\ldots, L_m^{-1}]$ and set $B=R[L_2^{-1},\ldots,L_m^{-1}]$. We want to show that all projective modules over $A$ are trivial. Of course if $m=0$, we are done by Quillen-Suslin. Otherwise, we can assume by induction that all projective modules over $B$ are trivial.

Now consider this diagram:

$$\matrix{B&\rightarrow&A\cr \downarrow&&\downarrow\cr B\otimes_{k[x_1]}k[x_1]_{(x_1)}&\rightarrow & B\otimes_{k[x_1]}k(x_1)\cr }$$

Check that this allows you to construct projective modules on $B$ by patching.

Now by induction all projective modules over the lower right corner are trivial. So any projective module over $A$ can be patched to a trivial module, and therefore lifts to a projective module over $B$, where we've already agreed that the statement is true.

I believe this argument is essentially due to Ofer Gabber.