A vector bundle over a complex manifold is said to be **holomorphic** if its trivialization maps are biholomorphic maps. What is a "natural" example example of a vector bundle over compact complex manifold which is not holomorphic? I guess by "natural" I mean that one would be interested in such examples for reasons besides them being a counter example.


TO CLARIFY: I am interested in 


i) COMPLEX VECTOR BUNDLES 


ii) NOT ADMITTING ANY HOLOMORPHIC STRUCTURE



Vector bundle over compact complex manifold which is not holomorphic?
I interpret the question as "Are there any complex bundles over the complex manifold which do not admit the holomorphic structure". The answer to this is "yes", but I'm not sure if it is known in full generality for what manifolds the answer is positive and for what it is negative.


**Line bundles case.** Suppose X is compact Kahler, L - complex line bundle over X. Then it has a holomorphic structure iff it's 1st Chern class is (1,1). See the comment by @abx


Proof: the "only if" part of the statement is the fact that holomorphic bundle admits so-called Chern connection, which has curvature of type (1,1). "If" part comes from the following consideration:


Complex line bundles are classified by $H^1(X, C^{\infty}(X, \mathbb{C}^\*))$. This sheaf is subject to the following exponential exact sequence:


$0 \rightarrow \mathbb{Z} \rightarrow C^{\infty}(X, \mathbb{C}) \rightarrow C^{\infty}(X, \mathbb{C}^\*) \rightarrow 0$


Similarly, holomorphic bundles are classified by $H^1(X, \mathcal{O}^\*)$. There is also an analogous exact sequence (which is also embedded in the previous one by the inclusion of the holomorphic functions into the smooth ones):


$0 \rightarrow \mathbb{Z} \rightarrow \mathcal{O} \rightarrow \mathcal{O}^\* \rightarrow 0$


Now, middle term in the former exact sequence is acyclic, so $H^1(X, C^{\infty}(X, \mathbb{C}^\*)) \simeq H^2(X, \mathbb{Z})$ (this delta homomorphism is one of the equivalent formulations of the 1st Chern class). The latter exact sequence, however, gives (after standard identification $H^k(X, \mathcal{O}) \simeq H^{0,k}(X)$ the following description for holomorphic bundles:


$0 \rightarrow H^{0,1}(X)/H^1(X, \mathbb{Z}) \rightarrow H^1(X, \mathcal{O}^\*) \rightarrow H^2(X, \mathbb{Z}) \cap (H^{1,1}(X) \oplus H^{2,0}(X)) \rightarrow 0$


The third term of this sequence is actually $H^2(X, \mathbb{Z}) \cap H^{1,1}(X)$ (because $H^2(X, \mathbb{Z})$ is real, and $H^{2,0}(X)$ is conjugate to $H^{0,2}(X)$.


This gives the following obstruction for line bundle - it's first Chern class should be (1,1), and provided this obstruction holds there is a complex torus $H^{0,1}(X)/H^1(X, \mathbb{Z})$ of different holomorphic structures.


**General case.** I think in the general case the similar classification is not known. The criterion that all Chern classes should be of (p,p) type clearly holds, but I think it is not enough. I'm not up to date with the current state of this field but you can see from here <http://www.numdam.org/item/?id=SB_1978-1979__21__80_0> that even for projective spaces when this criterion is empty the question is far from trivial (and if I've understood correctly has a negative answer). Also see @Angelo 's comment which has a reference to the recent survey of this problem <https://arxiv.org/pdf/1506.08111.pdf>


Maybe it is also worth noting that over a manifold of complex dimension 1 any complex bundle admits holomorphic structure: indeed, one can choose an almost holomorphic structure (take any connection and take it's (0,1) part), and then it is automatically integrable by dimensional reasons).


