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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2$\begingroup$ If $Y$ is a projective variety, this follows from GAGA. $\endgroup$– Ben McKayCommented Feb 26, 2012 at 14:00
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$\begingroup$ Thanks for the remark, I was wondering about the general case (which I have now made explicit). $\endgroup$– Dima SustretovCommented Feb 26, 2012 at 14:08
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3$\begingroup$ This is also true for Stein manifolds.This follows from corollary 5.6.3 page 149 Hormander Introduction to several complex variables 3edn. $\endgroup$– Mohan RamachandranCommented Feb 26, 2012 at 15:36
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$\begingroup$ Mohan, this is very nice, thanks a lot. $\endgroup$– Dima SustretovCommented Feb 26, 2012 at 17:03
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$\begingroup$ @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 . $\endgroup$– Mohan RamachandranCommented Feb 26, 2012 at 18:53
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1 Answer
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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.
On the other hand, I guess, in this example the bundle is topoligcally trivial, since its first Chern class is zero.
Added. Let me sketch the proof of the statement, everything holds for compact complex aglebraic manifold. First every such a manifold admits a blow up that is projective. Pull back the complex bundle to the blow up. Then we get a holomorphic bundle over a complex projective manifold and such a bundle is algebraic. Hence it has meromorphic sections. Moreover we can chose meromorphic sections that a linearly independent at one point. It is clear that they trivialise the bundle over a complement to a complex analytic manifold.
answered Feb 26, 2012 at 14:01
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1$\begingroup$ You can take any complex algebraic manifold, since such a manifold contains open affine submanifiold. But I would be surprised is something more general could be said. General complex manifolds are too wild (I would say)... $\endgroup$ Commented Feb 26, 2012 at 14:33
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1$\begingroup$ 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. $\endgroup$ Commented Feb 26, 2012 at 15:09
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1$\begingroup$ Simone, you are right, it is not topologically trivial on K3 but it becomes topologically trivial once we throw away points, I believe $\endgroup$ Commented Feb 26, 2012 at 15:19
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2$\begingroup$ Jörg Winkelmann proved that every compact complex manifold admits a non-trivial vector bundle <homepage.rub.de/Joerg.Winkelmann/publ/papers/banica.html>, so any complex manifold without codimension 1 analytic subspaces would give a counterexample. $\endgroup$– AngeloCommented Feb 26, 2012 at 16:05
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3$\begingroup$ I cannot find anymore a comment by Dmitri where he was asking me if holomorphic vector bundles on Stein manifolds are trivial. The answer is no in general. Anyway a famous theorem of Grauert says that the holomorphic and the topological classifications of complex vector bundles on a Stein manifold are the same. Quite often (for instance if the manifold is contractible or if it is an open Riemann Surface) this implies that every holomorphic vector bundle is holomorphically trivial. $\endgroup$ Commented Feb 27, 2012 at 8:16