Let $p: E\to B$ be a covering map of $C^\infty$ manifolds, where $E$ has a complex structure. There are many cases when we want to know whether $B$ has a complex structure (which is obviously unique) making $p$ an analytic map, for example in the construction of families of elliptic curves.


The difficulty is that given a small open set $U\subset B$, and pulling back the covering map to $$p|\_{p^{-1}(U)}: \bigsqcup\_\alpha U\_\alpha\to U,$$ the $U\_\alpha$ may induce incompatible complex structures on $U$. The easiest example of this phenomenon is the covering map $\mathbb{CP}^1\simeq S^2\to \mathbb{RP}^2$, where $\mathbb{RP}^2$ does not admit *any* complex structure, as it is not orientable. This case is not too badly behaved, however--in particular, given $U\_\alpha, U\_\beta\subset \mathbb{CP}^1$ over some $U\subset \mathbb{RP}^2$, the transition map $U\_\alpha\to U\to U\_\beta$ seems to me to be antiholomorphic.


So I have two questions about this general situation:



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> 1) Is there an example of a covering map $p: E\to B$ of $C^\infty$ manifolds with $E$ complex, such that $B$ admits *some* complex structure, but none making $p$ analytic?
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> 2) Given a covering map $p: E\to B$ with $E$ complex, is there an algebra-topological obstruction to the existence of a complex structure on $B$ making $p$ analytic?
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Pushing Complex Structure Forward
For 1): take a double covering $E\to B$, where $E$ and $B$ are compact oriented surface of genus 3 and 2 respectively, and give $E$ a structure of Riemann surface with trivial automorphism group.


About 2): well, in the example above you see that you can deform a complex structure with no compatible complex structure on $B$ into one with a compatible structure. An algebraic-topological obstruction should be discrete, so it seems to me that the example suggests that there isn't one.


