# non-trivial locus of a holomorphic vector bundle

Let $X$ be a holomorphic vector bundle over $Y$ (where $Y$ is an arbitrary complex manifold, not necessary projective). Does there exist an analytic subset $Z$ of $Y$ such that the restriction of $X$ to $Y \setminus Z$ is a trivial vector bundle?

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If $Y$ is a projective variety, this follows from GAGA. –  Ben McKay Feb 26 '12 at 14:00
Thanks for the remark, I was wondering about the general case (which I have now made explicit). –  Dima Sustretov Feb 26 '12 at 14:08
This is also true for Stein manifolds.This follows from corollary 5.6.3 page 149 Hormander Introduction to several complex variables 3edn. –  Mohan Ramachandran Feb 26 '12 at 15:36
Mohan, this is very nice, thanks a lot. –  Dima Sustretov Feb 26 '12 at 17:03
@Dima:A sufficient condition is global generation at some point of Y upto a twist by a holomorphic line bundle i.e a version of Cartan_Serre theorem A . –  Mohan Ramachandran Feb 26 '12 at 18:53

If your manifold is complex projective, then the answer is yes. Otherwise it is no. You can take a $K3$ surface without complex curves and just consider its tangent bundle. Of curse it will stay holomorphically non-trivial, if you throw away finite number of points from $K3$, since any holomorphic vector field on a $K3$ surface defined outside a finite set is zero.
I don't think that the tangent bundle of a $K3$ surface is topologically trivial. Indeed the second Chern class of $X$ is never zero, otherwise $X$ would be a finite étale quotient of a complex torus. –  diverietti Feb 26 '12 at 15:09