Does every compact complex manifold of complex dimension greater than or equal two possess a nontrivial rank 2 holomorphic vector bundle?

$\begingroup$ Aren't complex tori Oka manifolds? $\endgroup$– Jason StarrAug 31 '15 at 9:33

$\begingroup$ Oops, complex tori don't work: $h^{0,1}$ is always nonzero. $\endgroup$– Jason StarrAug 31 '15 at 13:39

$\begingroup$ Consider the CalabiEckman manifold = $S^3 \times S^3$, I would guess that since $b_{2} = b_{4} = 0$, that there is not even nontrivial rank 2 complex bundle..? $\endgroup$– Nick LMay 18 '17 at 10:51

$\begingroup$ @Nick Lindsay But we may still have some nontrivial holomorphic structures in the trivial complex vector bundle over the CalabiEckman manifold you have mentioned. Am I right? $\endgroup$– HamedMay 25 '17 at 9:52

$\begingroup$ yeah, that is true. I don't know how to rule that out. $\endgroup$– Nick LMay 25 '17 at 12:14
Some partial answer:
On nonKahler surfaces we can give stronger version of your question and we have allways the following thorem
Theorem: Let $S$ be nonKahler surfaces, For any $n > 0$ there exists a Mumford stable rank2 vector bundle $E$ with trivial determinant, ($det(E) = \mathcal O_S$ ) and $c_2(E)=n$.
We have also the following theorem due to Banica  Le Potier
Theorem: If $S$ is a nonprojective surface and $N$ is enough big , then there exists rank2 vector bundles on $S$ with $c_2(E) = N$ which are nonfiltrable.
We have following theorem due to Ballico and Schwarzenberger
Theorem: If $S$ be a projective surface and $E$ is any topological rank2 vector bundle with $det(E) \in Pic(X)$, then $E$ has a holomorphic structure by applying elementary transformations starting from the trivial bundle.
We have CartanSerre theorem for construction of rank 2 vector bundles on projetive varieties.
Theorem: Let $X$ be a complex manifold, $L_1, L_2 ∈ Pic(X)$ are line bundles, $Z ⊂ X$ with $codim_X(Z) = 2$. Under some cohomological conditions, the sheaf $E$ sitting in
$$0 \to L_1 \to E \to L_2 \otimes J_Z → o$$ is locally free. Moreover, If $X$ is projective, any rank2 holomorphic vector bundle can be constructed this way

4$\begingroup$ Additional comment: Let $D$ be a smooth irreducible divisor on a smooth variety $X$. Then an element $\mu\in H^1(D,\mathcal N_{D/X}) $ defines a rank 2 vector bundle over $ D$. $\endgroup$– user21574Jul 31 '17 at 0:12