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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.

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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. –  Nicolas Schmidt Apr 21 '11 at 13:05
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. –  Craig Apr 21 '11 at 15:12

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