# Are all holomorphic vector bundles on a contractible complex manifold trivial?

It is true that over a contractible manifold all differentiable vector bundles are trivial. However the method of proof does not apply in the holomorphic category.

It is also true that a contractible one dimensional complex manifold has no non-trivial line bundles. (Shown using the exponential sequence and a bit of tinkering.)

Moreover, I have read that the topological and holomorphic classification of bundles are the same over Stein manifolds. (No adequate reference was given.)

Hence my question:

Is it true that all holomorphic vector bundles are trivial over a contractible complex manifold?

Probably this is false. In this case, could you give me counter-examples?

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For the Stein case (where the statement is true) a precise reference is here: mathoverflow.net/questions/131453/… – Francesco Polizzi May 5 '14 at 13:40

No, even for line bundles. We have the short exact sequence of sheaves $$0 \to \underline{\mathbb{Z}} \overset{2 \pi i}{\longrightarrow} \mathcal{O} \overset{\exp}{\longrightarrow} \mathcal{O}^{\ast} \to 0$$ where $\underline{\mathbb{Z}}$ is locally constant $\mathbb{Z}$ valued functions, $\mathcal{O}$ is holomorphic functions and $\mathcal{O}^{\ast}$ is nonzero holomorphic functions. If $X$ is contractible, then $H^1(X,\underline{\mathbb{Z}}) = H^2(X,\underline{\mathbb{Z}}) = 0$. So $H^1(X, \mathcal{O}) \cong H^1(X, \mathcal{O}^{\ast}) \cong Pic(X)$.
Now use the standard example of a contractible manifold with nonzero $H^1(X,\mathcal{O})$:
$$X = \{ (z,w) \in \mathbb{C}^2 : (|z|, |w|) \in [0,1) \times [0,2) \cup [0,2) \times [0,1) \}.$$
How do you show that $H^1(X, \mathcal{O}) \neq 0$? – Michael Albanese Nov 8 '14 at 7:10