## Are surface bundles over a surface with non-zero signature necessarily complex (or algebraic)?

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By "surface bundle over a surface" I mean a compact, oriented 4-manifold $X$ which is the total space of an oriented fiber bundle $X\to B$ over an oriented 2-manifold $B$. Assume that the signature of the 4-manifold $X$ is non-trivial.

Conjecture 1: $X$ and $B$ can be given complex structures such that the map $X \to B$ is holomorphic.

Conjecture 2: $X$ and $B$ can be given the structure of complex algebraic varieties such that the map $X \to B$ is algebraic.

Question: Are the above conjectures true? If not, can you provide a counter example?

Background: The bundle $X\to B$ induces a map $f:B \to M_g$ where $M_g$ is the moduli space of curves and $g$ is the genus of the fiber. The signature of $X$ is given by

$$\sigma(X) = 4\int _B f^*(\lambda)$$

where $\lambda$ is the first Chern class of the Hodge bundle. Since the signature of $X$ is non-zero, the map $f$ is non-trivial in homology and one might hope that an argument along the following lines is true: Find a representative $\tilde{f}$ in the same homotopy class as $f$ which has minimal energy. Using the fact that $M_g$ has a hyperbolic metric ($g>2$ since $\sigma(X)\neq 0$), prove this minimal energy map is holomorphic or even algebraic. $\tilde{f}$ provides $X$ with the desired structure.

This question is a variant on this recent question: http://mathoverflow.net/questions/69344/a-four-dimensional-counterexample the counter example given there has zero signature.

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## 2 Answers

I seem to have found an answer to my own question.

The first observation is that a holomorphic fibration is automatically algebraic. This is because $M_g$ is quasi-projective and so the map $f:B\to M_g$, which trivially extends to a map to $\overline{M}_g$ is algebraic by GAGA (perhaps one should first pass to some finite cover to take care of stack issues).

However, my conjectures are false --- here is a counterexample. Let $X \to B$ be a surface bundle having non-trivial signature $\sigma$, base genus $h$, and fiber genus $g$. Let $X_N$ be the surface bundle obtained by taking the fiber connect sum of $X$ with the trivial bundle of fiber genus $g$ and base genus $N-h$. So for each $N\ge h$, we get a surface bundle $X_N \to B_N$ of signature $\sigma$ and base genus $N$. If $X_N \to B_N$ were all holomorphic as in my conjecture, then we would obtain a family of complete smooth curves $B_N$ in $M_g$ of fixed degree and unbounded genus, a contradiction. (The degree here is taken with respect to $\lambda$ which is ample on $M_g$ and fixed by the signature formula).

What happens in my minimal energy argument is that as the map $B_N \to M_g$ becomes harmonic, it bubbles off a collapsing component.

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I have a question about the first posting of Jim Bryan.

what does it mean that f tilda has minimal energy?

And, how is it relevant to the nontriviality of f_{*} on the second homology?

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