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.



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



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


