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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:

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?

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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2 Answers 2

up vote 12 down vote accepted

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.

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+1 for (1); I guess by algebra-topological I'm including things like the cohomology of the sheaf of holomorphic functions on $E$, which should be more rigid. Or I'd be OK with obstructions to deforming the complex structure on $E$ to one inducing a complex structure on $B$. –  Daniel Litt Jan 7 '11 at 20:18

This seems to be a question about holomorphicity of diffeomorphisms in a given complex structure. Replace your covering map $E \to B$ by its Galois closure (= frame bundle) $X \to B$. Now by construction $X \to B$ is a covering space which is Galois with Galois group $G$ (= groups of self bijections of a fixed fiber of $E \to B$). Since $X \to B$ factors through $E$, every complex structure on $E$ will induce a complex structure on $X$, and a complex structure on $B$ makes $E \to B$ holomorphic if and only if it makes $X \to B$ holomorphic. But the later question is just the question of whether all elements of $G$ which act as diffeomorphisms of $X$ will preserve the complex structure. Some of them preserve it automatically, e.g. the elements of the subgroup $H \subset G$ for which $E = X/H$. But for the rest it is an actual condition. If all those diffeomorphisms preserve your complex structure, then the quotient exists as a complex manifold. If one of them doesn't, then your are out of luck.

I don't think you can get more concrete obstructions.

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