I wonder if it is true that all vector bundles on the punctured affine plane are trivial, and I would like to have a reference in that case.

4$\begingroup$ Yes, use the Corollary after Theorem 4.1 in Horrocks' paper "Vector bundle on punctured..." plus the fact that any vector bundle on the whole affine plane is trivial. $\endgroup$ – Hailong Dao Feb 4 '13 at 15:30

1$\begingroup$ Or, one can quote algebraic Hartog's Lemma plus the fact mentioned above. $\endgroup$ – Hailong Dao Feb 4 '13 at 15:35

4$\begingroup$ Using Cech cohomology, the claim is equivalent to $\mathrm{GL}_n(k[X,Y]_{X \cdot Y}) = \mathrm{GL}_n(k[X,Y]_X) \cdot \mathrm{GL}_n(k[X,Y]_Y)$. I wonder, is there a purely algebraic proof (similar to Hazewinkels Short and elementary proof of Grothendieck's Theorem on algebraic vector bundles over the projective line)? $\endgroup$ – Martin Brandenburg Feb 4 '13 at 16:03

1$\begingroup$ The topological statement is that $\pi_3(BU(n)) = 1$. It seems interesting to note that $\pi_3(BO(n)) = Z_2$ for $n$ large enough. Is there some algebraic avatar of this nontrivial class, maybe a vector bundle with nondegenerate symmetric form? $\endgroup$ – Allen Knutson Feb 4 '13 at 16:59

1$\begingroup$ The conjecture of Serre: Is every algebraic vector bundle $E_{alg} \to\mathbb C^n$algebraically trivial , When $E $ is stabilized; see Bass, H.: Algebraic KTheory. New York: W.A.Benjamin Inc. 1968. Now my conjecture is that serre conjecture holds on special affine variety in thesense of Griffiths see page 82 link.springer.com/article/10.1007/BF01389905 $\endgroup$ – user21574 Nov 15 '17 at 10:31
Thanks Hailong Dao, for the answer! Algebraic Harthog's lemma says that if $A$ is a noetherian normal domain, then $A$ is the intersection of all localizations of $A$ at prime ideals of codimension 1. What should $A$ be in my situation?
For $n=1$, $H^1(X,GL(n,\mathcal O))$ classifies the vector bundles of rank $n$ on $X$ is this true also for $n>1$? Should I use $A=\{1$cocycles with values in $GL(n,\mathcal O)\}$, with respect to the covering by the open affines $D(x)$ and $D(y)\subset\mathbb A^2$? I suppose that all vector bundles over the affine pricipal open subsets $D(x)$ and $D(y)$ are trivial. Then a nontrivial vector bundle on $\mathbb A^2_*$ would give a nontrivial bundle also on $\mathbb A^2$?

$\begingroup$ Thanks for your comment Martin! All I need is to understand the result, but of course it would be a very nice bonus if it could be understood by means of a short and elementary algebraic proof! $\endgroup$ – anonymous Feb 4 '13 at 16:30

$\begingroup$ Yes this was my idea (except that I don't want to extend the vector bundle to $\mathbb{A}^2$ as in Hailong's comment (which is an answer)). For every $n$ there is a bijection between the set of isomorphism classes of rank $n$ vector bundles on $X$ and the Cech cohomology set $\check{H}^1(X,\mathrm{GL}_n(\mathcal{O}_X))$, even for an arbitrary ringed space $X$ (so this is just a fancy way of stating that vector bundles can be equivalently described by transition functions). $\endgroup$ – Martin Brandenburg Feb 5 '13 at 9:35

$\begingroup$ PS: Please read mathoverflow.net/faq, your "answer" should have been a comment. $\endgroup$ – Martin Brandenburg Feb 5 '13 at 9:41

1$\begingroup$ Even better! Suppose $\mathcal{F}^\prime$ is a vector bundle on the punctured affine plane. Then you can extend $\mathcal{F}^\prime$ to a coherent sheaf $\mathcal{F}$ on the affine plane itself (cf. EGA I, p. 174 here: archive.numdam.org/ARCHIVE/PMIHES/PMIHES_1960__4_/…) Now since we are on the affine space, $\mathcal{F}=\tilde{M}$, where $M$ is a finitely generated module over $\mathbb{C}[X,Y]$. $\endgroup$ – Mahdi MajidiZolbanin Feb 5 '13 at 18:32

1$\begingroup$ We already know that $M$ is free when localized at any prime ideal of $\mathbb{C}[X,Y]$ other than $(X,Y)$ because $\tilde{M}$ was the extension of $\mathcal{F}^\prime$. Now localize both $M$ and $\mathbb{C}[X,Y]$ at the ideal $(X,Y)$. You get a regular twodimensional local ring and a vector bundle $M_{(X,Y)}$ on its punctured spectrum and Corollary 4.1.1 says it's free. This shows $M$ is a vector bundle on $\mathbb{C}[X,Y]$. By Serre's Problem $M$ must be free then. $\endgroup$ – Mahdi MajidiZolbanin Feb 5 '13 at 18:32