MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
For the Stein case (where the statement is true) a precise reference is here:… – Francesco Polizzi May 5 '14 at 13:40
up vote 16 down vote accepted

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

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

share|cite|improve this answer
How do you show that $H^1(X, \mathcal{O}) \neq 0$? – Michael Albanese Nov 8 '14 at 7:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.