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?


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 $H^1(X,\mathcal{O})$:

$$X = \{ (z,w) \in \mathbb{C}^2 : (|z|, |w|) \in [0,1) \times [0,2) \cup [0,2) \times [0,1) \}.$$

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    $\begingroup$ How do you show that $H^1(X, \mathcal{O}) \neq 0$? $\endgroup$ – Michael Albanese Nov 8 '14 at 7:10
  • $\begingroup$ @MichaelAlbanese I just noticed this comment. In case you still care: Let $U_1 = \{ (z,w) : (|z|, |w|) \in [0,1) \times [0,2) \}$ and $U_2 = \{ (z,w) : (|z|, |w|) \in [0,2) \times [0,1) \}$. Then $U_1$, $U_2$ and $U_1 \cap U_2$ are polydiscs, so their $H^1(\mathcal{O})$ vanishes, and we can compute $H^1(X, \mathcal{O})$ using the Cech complex on $U_1$ and $U_2$. So $H^1(X, \mathcal{O})$ is the cokernel of $\mathcal{O}(U_1) \oplus \mathcal{O}(U_2) \to \mathcal{O}(U_1 \cap U_2)$. $\endgroup$ – David E Speyer Feb 19 at 14:12
  • $\begingroup$ An explicit nonzero class in the cokernel is $\sum z^i w^i$, since the coefficient of $z^i w^i$ is $O(2^{-i})$ in any function which extends to $U_1$ or $U_2$. $\endgroup$ – David E Speyer Feb 19 at 14:15
  • $\begingroup$ Thanks! I asked a question on MSE about this and it remains unanswered. Would you mind writing an answer for it? $\endgroup$ – Michael Albanese Feb 19 at 16:01

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