Does anyone know an example of a smooth hyperbolic surface bundle over a hyperbolic surface (surface = compact twomanifold) which does not have a complex structure? Is there any decision procedure to tell, given such a bundle, whether it has a complex structure, or is it more of a black art?

This paper by Hillmann addresses this question. He proves that a surface bundle over a surface which is a complex surface has a holomorphic fibration over the base, for some choice of complex structure on the base. He uses this to prove that when the fiber has genus 2, the bundle must be finitely covered by a product (see the last Corollary in the paper; this follows from a result of Kas). So for any fiber bundle with genus 2 fiber $S_2\to X\to S$, if the image of $\pi_1(S)\to Mod(S_2)$ induced by the fibration is infinite, then $X$ will not admit a complex structure. 


Igor, suppose the total space of a bundle is complex. Then the fibration can be realized There are interesting examples of complex structures on fibrations which fall into two classes: (a) Using the period map (or a similar construction) one can show that the moduli space $M_g$ (except for $g\le 2$) contains a lot of compact Riemann surfaces. For instance, S.Diaz ["Complete subvarieties of the moduli space of smooth curves", 1987] used this to show that one can construct a projective curve passing through any finite subset of $M_g$ for $g\ge 3$. (b) Kodaira's direct construction of complex surfaces which fiber: They are obtained as ramified covers over products of two Riemann surfaces, where the ramification locus is a (possibly disconnected) smooth Riemann surface in the product (graph of a, possibly multivalued, locally invertible holomorphic map between the surfaces). 


I'll try. Take a genus 2 Riemann surface $S$, and embed it (differentiably) in $\mathbb{R}^3$ so that it's symmetric with respect to a 2plane $\Pi$. Consider the intersection $S\cap \Pi$: choosing the embedding suitably, such intersection is the union of three circles $\alpha$, $\beta$ and $\gamma$. The complement of $\alpha \cup \beta \cup \gamma$ is a union of two noncompact surfaces $U_0$ and $V_0$, each homeomorphic to a sphere whith 3 holes, and each hole (within both $U_0$ and $V_0$) is bounded by the circles $\alpha$, $\beta$ and $\gamma$. If you thicken $U_0$ and $V_0$ you'll obtain an open cover {$U,V$} of $S$, such that $U\cap V$ is a union of thickenings of $\alpha$, $\beta$ and $\gamma$ which we call $W_{\alpha}$, $W_{\beta}$ and $W_{\gamma}$. Now, a bundle on $S$ with fiber $F$ is encoded (up to isomorphism) by (the homotopy class of) a map $\varphi:W_{\alpha}\cup W_{\beta} \cup W_{\gamma} \to \mathrm{Diff}(F)$. Take as $F$ your favourite hyperbolic surface, and suitably represent it in $\mathbb{R}^3$ as above, symmetrically with respect to a 2plane $\Pi$. Define $\varphi$ to be the identity on the connected components $W_{\alpha}$ and $W_{\beta}$ and the symmetry with respect to $\Pi$ on the connected component $W_{\gamma}$. Let $X$ be the total space of the resulting (differentiable) bundle. If I'm not mistaken, $X$ cannot have a complex structure because it's not orientable. As for the second question, orientability of the total space is of course a necessary condition for it to have a complex structure. Assuming my construction above is meaningful, if one knows $\pi_1(\mathrm{Diff}^+(F))$, he's able to tell apart the topological types of bundles on a given $S$ for which the orientability obstruction is not present. (I know I'm being quite vague, but) ...then one may try to look for an (existing?) topological classification of holomorphic (or maybe just algebraic) bundles, for various complex structures on $S$ and $F$, to see whether each topological class is in fact realized by a a holomorphic/agebraic example... 

