# Can we ignore the holomorphic trivialisation'' in the definition of a holomorphic vector bundle?

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 $$$$\phi_U\colon \pi^{-1}(U)\to U\times\mathbb C^k$$$$ are biholomorphic maps. This is equivalent to requiring that transition functions $$$$t_{UV}\colon U\cap V\to > \text{GL}_k(\mathbb C)$$$$ 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 $$\phi_U$$.

• I think en.m.wikipedia.org/wiki/Hartogs%27s_theorem is your solution. – Mindlack Jan 10 at 18:38
• @Mindlack sorry but I can not see why they are related. – Display Name Jan 10 at 18:50
• 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. – user98602 Jan 10 at 18:57
• 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. – Mindlack Jan 10 at 18:57