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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?

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  • $\begingroup$ 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 differential-geometric 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 non-empty when $X$ is a curve. Higher dimensional cases should also be well known, I guess. $\endgroup$ – Xin Nie Nov 13 '14 at 15:46
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    $\begingroup$ @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 $c_1^2-4c_2\leq 0$. 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
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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), 601--622).

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  • $\begingroup$ 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

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