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?

  • $\begingroup$ I'm deleting a duplicate of this question (mathoverflow.net/questions/86580). The following was posted by Igor Riven as an answer: Such examples are abundant, and you could not do better than check out Kotschick's paper (arxiv.org/pdf/math/9911255.pdf) One implication (pointed out by Kotschick) is that a bundle with fiber of genus two is complex if (and only if) the monodromy representation of the base has finite image in the mapping class group of the fiber. $\endgroup$ – Anton Geraschenko Jan 26 '12 at 2:57

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.

  • $\begingroup$ I had just a quick look at the first article, and it seems to me that Hillman already starts from complex surfaces (not Riemann surfaces) hence the total space the bundles he deals with admit a complex structure by default. Hillman deals with fiber bundle structures on complex 2-dim manifolds, whereas the OP is looking for a $\mathcal{C}^{\infty}$ (real-)surface bundle that does not admit a complex structure. So, unless I'm missing something, the answer doesn't address the OP's question. $\endgroup$ – Qfwfq Jul 2 '11 at 23:43
  • $\begingroup$ Moreover, genus 0 and genus 1 compact Riemann surfaces are not hyperbolic. $\endgroup$ – Qfwfq Jul 2 '11 at 23:43
  • $\begingroup$ @unknowngoogle: I think if you apply the contrapositive to Hillman's theorem, you get that a surface bundle over a surface which cannot be realized by a holomorphic fiber bundle cannot be a complex surface. The corollary in the paper says that the base had genus 0 or 1, or the fiber has genus 2. In the second case (when the fiber has genus 2), no restriction is made on the base. The Corollary is not very precisely worded, so maybe there is some unstated assumption I am missing (since it is a corollary of Theorem 2, I was assuming the hypotheses of Theorem 2 hold). I'll have a closer look. $\endgroup$ – Ian Agol Jul 3 '11 at 0:07
  • $\begingroup$ @Agol: thanks for clarifying. I had a second look to the article now, and it seems he gives a purely topological (on the $\pi_1$) characterization for when such holomorphic submersion is there. +1 to the answer then! $\endgroup$ – Qfwfq Jul 3 '11 at 0:51
  • $\begingroup$ Thanks, that is exactly the sort of thing I was looking for! $\endgroup$ – Igor Rivin Jul 3 '11 at 17:33

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).

  • $\begingroup$ Thanks! You should take a look at Jim Bryan's slightly more recent question. But do I understand correctly that any complex bundle is Kahler? My understanding was that this was not necessarily true... $\endgroup$ – Igor Rivin Jul 4 '11 at 21:26
  • $\begingroup$ Also, for genus 2, what does the argument you describe give? $\endgroup$ – Igor Rivin Jul 4 '11 at 21:42

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 2-plane $\Pi$. 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 2-plane $\Pi$.

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...

  • 1
    $\begingroup$ Thanks. This is perfectly valid, but I should have added orientability to the original question (I won't edit it, so that your answer is an answer to the question). $\endgroup$ – Igor Rivin Jul 3 '11 at 17:32

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