Let $E\to X$ be a complex vector bundle over a compact Kahler surface $X$. Assume $c_{i}(E)\in H^{i,i}(X)$ for all i. Does the bundle $E$ admit a holomorphic structure?

$\begingroup$ The moduli space $\mathcal{M}(E)$ of holomorphic structures on $E$ is a classic subject in gauge theory and algebraic geometry (and involves some stability issue). $\mathcal{M}(E)$ can be interpreted by differentialgeometric means as the space of all integrable "partial connections" (a.k.a. pseudo connections, Dolbeault operators...) modulo gauge equivalence, see the book of Donaldson and Kronheimer Section 2.1.5. However, I can only ensure that $\mathcal{M}(E)$ is nonempty when $X$ is a curve. Higher dimensional cases should also be well known, I guess. $\endgroup$ – Xin Nie Nov 13 '14 at 15:46

1$\begingroup$ @XinNie: unfortunately, there are situations where there is a big gap between moduli spaces and the classification of vector bundles. For instance, if $X=\mathbb{P}^2$, semistable holomorphic rank 2 bundles on $X$ satisfy $c_1^24c_2\leq 0$. However, every complex vector bundle has a holomorphic structure, independent of what the Chern classes are. In this case, I would say that the moduli space is empty although there is an enormous amount of holomorphic structures... $\endgroup$ – Matthias Wendt Nov 13 '14 at 18:59
For line bundles, this is the Lefschetz (1,1)theorem which says that the map $H^1(X,\mathcal{O}^\times)\to H^2(X,\mathbb{Z})$ from the exponential sequence surjects onto the $H^{1,1}(X)$part.
For higher rank bundles, the answer is positive whenever $X$ is additionally assumed to be projective. In this case, a theorem of Schwarzenberger states that a complex vector bundle is algebraizable if and only if its determinant is algebraizable, see Theorem 9 in (R.L.E. Schwarzenberger: Vector bundles on algebraic surfaces. Proc. London Math. Soc. (3) 11 (1961), 601622).

$\begingroup$ Thank you for your answer. By the way, I think I should modify the condition in the question to just $c_{1}(E)\in H^{1,1}(X)$ as it is alway true for compact complex surface that $H^{4}(X,\mathbb{C})=H^{2,2}(X)$. $\endgroup$ – Jiang Nov 14 '14 at 3:52