Triviality of vector bundles on affine open subsets of affine space 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?

 A: 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.
