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I have learnt two definitions about holomorphic vector bundles over a complex manifold $M$.

  1. $E\to M$ is a smooth complex vector bundle with a trivialisation such that the transition functions are holomorphic

  2. $E\to M$ is a smooth complex vector bundle and $E$ is a complex manifold such that the map $E\to M$ is holomorphic (from Wikipedia)

It is easy to see that the first definition implies the second and I wanted to show that the second also implies the first.

In Wikipedia, they say

Specifically, one requires that the trivialization maps \begin{equation} \phi_U\colon \pi^{-1}(U)\to U\times\mathbb C^k \end{equation} are biholomorphic maps. This is equivalent to requiring that transition functions \begin{equation} t_{UV}\colon U\cap V\to > \text{GL}_k(\mathbb C) \end{equation} are holomorphic maps.

The problem happens to say ``the trivialisation maps $\phi_U$'s are biholomorphic'', because I think if we only assume $E\to M$ satisfies the second definition, we can only get smooth trivialisations. Even $E$ has complex coordinate charts, I can not require the chart has that special form as $\phi_U$.

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  • $\begingroup$ I think en.m.wikipedia.org/wiki/Hartogs%27s_theorem is your solution. $\endgroup$ – Mindlack Jan 10 at 18:38
  • $\begingroup$ @Mindlack sorry but I can not see why they are related. $\endgroup$ – Display Name Jan 10 at 18:50
  • $\begingroup$ I am skeptical that (2) is the correct definition. Rather you should mimic the definition of smooth vector bundle, holomorphically: it is a complex manifold $E$ with a holomorphic submersion $\pi: E \to M$, a holomorphic section $0: M \to E$ and holomorphic maps $+: E \times_M E \to E$ and $\lambda: \Bbb C \times E \to E$, satisfying the vector space axioms fiberwise. $\endgroup$ – user98602 Jan 10 at 18:57
  • $\begingroup$ Basically, your trivialization $U \times \mathbb{C}^r \rightarrow E_{|U}$ is smooth, and holomorphic wrt to the linear part (it is, well, linear) and holomorphic wrt the space part (it is the inverse of a holomorphic function, but something somewhere must be written out carefully). At least, that was the point I was trying to make, but thinking of it again, it might be a bit light for the second holomorphy. $\endgroup$ – Mindlack Jan 10 at 18:57

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