Is every holomorphic line bundle on the - say - punctured unit disc $\dot{\Delta} \subseteq \mathbf{C}$ trivial? Griffiths-Harris (p. 39) prove that $H^{p,q}_{\overline{\partial}}(\Delta) = 0$ (for $q \geq 1$), and mention that by replacing discs by annuli that one could prove also $H^{p,q}_{\overline{\partial}}(\dot{\Delta}^k \times \Delta^\ell) = 0$. They seem to imply that in particular $H^{p,q}_{\overline{\partial}}(\dot{\Delta}) = 0$. If this were true, using Dolbeault's theorem and the Kummer sequence one could conclude $H^1(\dot{\Delta},\mathcal{O}^\times_{\dot{\Delta}}) = 0$, hence that every holomorphic line bundle on $\dot{\Delta}$ is trivial.
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Yes, every holomorphic vector bundle of any rank is trivial on the punctured disk $\dot{\Delta}$ . Indeed, since $\dot{\Delta}$ is a Stein manifold ( like any non-compact Riemann surface ! ) the Oka meta-principle (here a Theorem of Grauert ) says that the classification of holomorphic vector bundles on that manifold is the same as that of topological vector bundles. But since the punctured disk is homotopically equivalent to a circle , all topological complex vector bundles are trivial, hence the triviality of all holomorphic vector bundles. (Do not confuse with the Möbius vector bundle, which is a non-trivial real vector bundle!)
Bibliography If you want to read complete proofs, you can consult Otto Forster's Lectures on Riemann Surfaces.
On page 229, Theorem 30.3 states that every holomorphic line bundle on a non-compact Riemann surface $X$ is trivial, and immediately below (on the same page) Theorem 30.4 proves that holomorphic vector bundles of any rank on $X$ are also trivial.
answered Apr 21, 2011 at 11:38
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$\begingroup$ Hey, thanks very much for your answer! Although I must say that I'm a bit disappointed, since for some reason I thought that there should be lots of nontrivial bundles. $\endgroup$ Commented Apr 21, 2011 at 13:05
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$\begingroup$ It you take C^2 \{0} you get an infinite family of topologically trivial line bundles because H^(0,1)_{\bar{\partial}} is infinite dimensional. $\endgroup$– CraigCommented Apr 21, 2011 at 15:12
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$\begingroup$ I'm a little confused: every smooth affine curve is a Stein manifold, and they all deformation-retract onto a graph (or a point). Isn't any complex vector bundle on a graph trivial? But there are definitely affine curves with nonvanishing Picard group, such as any once-punctured elliptic curve. What am I missing here? $\endgroup$ Commented Mar 23, 2016 at 16:45
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3$\begingroup$ @Xander Flood: An algebraic vector bundle may be bon-trivial and have a trivial underlying holomorphic vector bundle.That's exactly what happens with algebraic vector bundles on your punctured elliptic curve. $\endgroup$ Commented Mar 23, 2016 at 18:00
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$\begingroup$ Dear @Georges Elencwajg , can we prove it alternatively using Deligne's canonical extension theorem since the connection is always flat on the punctured disc (due to dimensional reason), it has an extension to the whole disc, which is simply connected and the extension must be a trivial bundle that is trivialized by global frame consists of flat sections, therefore the restriction of this vector bundle back to punctured disc is also trival? $\endgroup$– yi liCommented May 2, 2023 at 13:05