Let $E$ be a real vector bundle on a smooth manifold $X$. Let $J : E \to E$ be a vector bundle morphism (i.e. $\pi \circ J = \pi$, where $\pi : E \to X$ is the projection map) with $J^2 = -\textrm{id}$. Then $E$ is a complex vector bundle ($E_x$ has a complex structure given by $(a + bi)v = av + bJ_x(v)$). I will call $J$ an almost complex structure on $E$ - for $E = TX$, this is standard.

Is there any relationship between an almost complex structure on a vector bundle $E$ and the existence of a complex structure on $E$ as a manifold? That is, is there anything we can say about $J$ that will help us to determine whether $E$ is a complex manifold?

Another way of viewing this question is the following:

If you have a smooth manifold, which happens to be a complex vector bundle over some other manifold, is it any easier to determine whether or not it has a complex structure?

  • $\begingroup$ The underlying manifold will at least have to be even dimensional. Also $X\times \Bbb C$ works for any manifold $X$, so I'm skeptical you can get anything out of it. Is there maybe something I'm misunderstanding in your question? $\endgroup$ Oct 3, 2012 at 22:02
  • $\begingroup$ I think you have interpreted my question as: Does the existence of a complex vector bundle on $X$ help you determine whether or not $X$ is a complex manifold? (Is this how you interpreted the question?) This is not the question I intended to ask. My question is: Does the fact that $E$ is a complex vector bundle make it any easier to determine whether or not $E$ is a complex manifold? $\endgroup$ Oct 3, 2012 at 22:27
  • $\begingroup$ I interpreted your question correctly, my comment stands. $\endgroup$ Oct 4, 2012 at 0:05
  • $\begingroup$ Sorry, I see what you mean now. $\endgroup$ Oct 4, 2012 at 0:13

1 Answer 1


I think the answer is simply "no." Let's consider a specific case. We will take the base space $X$ to be a complex manifold and $(E, \pi, J)$ to be a (complex) rank $k$ complex vector bundle over $X$. Let us call $\pi: E \longrightarrow X$ a holomorphic vector bundle if there exists a trivialization of $E$ such that the transition functions $$g_{\alpha\beta}: U_\alpha \cap U_\beta \longrightarrow \mathrm{GL}(k, \mathbb{C})$$ of $E$ are holomorphic. One can show that $E$ is a holomorphic vector bundle if and only if there exists a complex manifold structure on the total space $E$ such that the projection $\pi: E \longrightarrow X$ is a holomorphic map. Therefore in the specific case where the base space is a complex manifold and we require that $\pi: E \longrightarrow X$ be holomorphic, we see that the question is equivalent to whether or not a given complex vector bundle over $X$ can be given a trivialization with holomorphic transition functions.

With this point of view developed, I now bring your attention to a relevant discussion on MathOverflow. In particular, the paper

  • C. Bănică and M. Putinar. On complex vector bundles on projective threefolds. Invent. Math. 88 (1987), 427-438.

mentioned in the answer the the linked discussion proves that all complex vector bundles over a projective 3-fold are holomorphic. I think you will agree that with the work they need to do there, it is not particularly "easy" to prove that these bundles are holomorphic. Also mentioned in the answer is that whether all complex bundles on $\mathbb{C}P^n$ are holomorphic is an open question for $n \geq 4$. Therefore we can see that at least determining if a complex vector bundle is holomorphic is in general a difficult question.

If we don't require $\pi: E \longrightarrow X$ to be holomorphic, just that $E$ has some complex structure, I'm not sure how much easier this problem becomes.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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