# Condition for a complex vector bundle to be holomorphic?

Suppose that $(E,\pi,M)$ is a complex vector bundle. I've seen it suggested in a few places that if $E$ and $M$ are complex manifolds, and $\pi$ is a holomorphic map, then $(E,\pi,M)$ is in fact a holomorphic vector bundle. In other words, there exist local trivializations which are biholomorphisms. Is this true?

Edit: As explained by Mike Miller in the comments, this cannot be true as stated, because the complex vector bundle might have an "anti-holomorphic" structure to being with. A better question would be whether there is some holomorphic vector bundle structure $(E,\pi',M)$, which is isomorphic as a complex vector bundles to $(E,\pi,M)$. Alternatively, one could drop the requirement that $(E,\pi,M)$ be a complex vector bundle, and instead ask what milder conditions (e.g. $\pi$ is a smooth constant rank submersion) would be sufficient to guarantee that $(E,\pi,M)$ is holomorphic (assuming still that $E$ and $M$ are complex manifolds and $\pi$ is holomorphic).

• RE: Edit. Yes. The proof is essentially identical to that of smooth vector bundles having smooth transition maps. – user98602 Oct 22 '15 at 6:08
• (If you'd like more explanation than the brief comment, ping me - I'm glad to expound on it tomorrow if nobody else has.) – user98602 Oct 22 '15 at 6:18
• Oh, I'm very sorry for my previous comments. This is just not true. Take $E = \Bbb C, M = \{pt\}$. Let the scalar multiplication be, like, $(\lambda, z) \mapsto \overline{\lambda} z$. A holomorphic vector bundle would have this map be holomorphic. If you demand in addition to the conditions on $(E,\pi,M)$ that the scalar multiplication $\Bbb C \times E \to E$ and $E \times_M E \to E$ are holomorphic, then it's true that you have holomorphic trivializations/transition functions. – user98602 Oct 22 '15 at 22:40
• What I'm not sure about is whether or not there's an example where you can't give it the structure of a holomorphic bundle isomorphic to the previous complex vector bundle. So, changing the vector space structure, but in a way that doesn't change the topological type of the complex vector bundle. – user98602 Oct 22 '15 at 23:11
• @MichaelWan: But a trivialization here is just an isomorphism with $\Bbb C$. There is one for two good reasons: $E \to \Bbb C$, $z \mapsto \bar z$, where $E$ is $\Bbb C$ equipped with my scalar multiplication, is a complex linear isomorphism. Alternatively: there has to be an iso, because there's only one complex vector space of dimension 1 up to iso! – user98602 Oct 25 '15 at 6:52