TITLE: A four-dimensional counterexample?
QUESTION [13 upvotes]: Does anyone know an example of a smooth hyperbolic surface bundle over a hyperbolic surface (surface = compact two-manifold) 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?

REPLY [8 votes]: Igor, suppose the total space of a bundle is complex. Then the fibration can be realized
holomorphically. Such fibrations are classified by holomorphic maps of the base Riemann surface to the moduli space of the fiber. Thus, your question about a "procedure" essentially reduces to the problem of what homomorphisms from surface groups to the mapping class group are realized holomorphically. There are obstructions of Hodge-theoretic nature (since the total space of the bundle is Kahler), for instance, Betti numbers of the 4-manifold should be even, etc. If the fiber has genus 2 (and, thus, is hyperelliptic), then its moduli space admits a finite cover which is a domain in ${\mathbb C}^3$, so no non-constant holomorphic maps in this case. However, for higher genus of the fiber, there is nothing like a complete set of invariants, even conjecturally. 
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).