I'm quoting a question from p. 753 of Gromov's recent paper Singularities, Expanders and Topology of Maps:

Does there exist, for every closed oriented $n$-manifold $X_0$, a closed oriented $n$-manifold $X$ that admits a map $X \to X_0$ of positive degree and, at the same time, can be smoothly fibered over some $Y$ with $dim(Y) = n − 2$ ? (Bogomolov’s original question concerns parametrization of complex algebraic manifolds $X_0$ by algebraic manifolds fibred by surfaces.)

I'm wondering where Bogomolov's question came from, and for what cases it is known and why it may be of interest? Gromov doesn't give a reference, and searching for "Bogomolov conjecture" seems to bring up a different conjecture. Maybe an expert can point me quickly to a reference.

Also, if anyone has partial answers to Gromov's question, I would be interested. I think I can prove it for 3-manifolds, and there may be a 4-manifold topologist who might know something about the 4-dimensional case.

Addendum: I'll add my argument in the 3-dimensional case, then ask a question which would answer Gromov's question in dimension 4.

Any orientable 3-manifold is a cover of $S^3$ branched over the figure 8 knot $K$. The 3-fold cyclic cover branched over $K$ is a Euclidean manifold, which has a finite-sheeted cover which is a 3-torus. The preimage of the branched locus is geodesic, so we may find a product fibering of the 3-torus by tori transverse to the preimage of the branched locus. Thus, any covering of this cover will also fiber (by taking the preimage of the fibering). For a 3-manifold, we may take the common branched cover over $K$ with the 3-torus branched cover. This fibers since it covers the 3-torus, and has a non-zero degree (in fact bounded degree) map to the 3-manifold.

As mentioned in an answer to Ramified cover of 4-sphere, any closed orientable PL 4-manifold is a branched cover over $S^4$ with branched locus a surface. One could try to generalize my 3-dimensional argument to this context. The key fact is that there is a cover of $S^3$ branched over the figure 8 knot which has a fibration transverse to the branched locus. So I ask the question: for any surface in $S^4$, is there a cover of $S^4$ branched over the surface which has a fibration by surfaces which are transverse to the preimage of the branched locus? I have no evidence for or against this question, it just seems like a natural generalization of the 3-D case.

One could also ask the analogous question in higher dimensions and in the algebraic category. But one would need to know the corresponding branched covering result first. For projective algebraic varieties, maybe it would be natural to consider branched covers over projective space, analogous to Belyi's theorem.

  • $\begingroup$ I don't know the answer to the question about Bogomolov, but certainly for hyperbolic 3-manifolds M it's a much-thought-about question whether M necessarily has a finite cover smoothly fibered over the circle. $\endgroup$
    – JSE
    Dec 13, 2009 at 1:26
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    $\begingroup$ JSE - some of the most profound recent work on this is by, er, Agol (see eg arxiv.org/abs/0707.4522). $\endgroup$
    – HJRW
    Dec 13, 2009 at 2:05
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    $\begingroup$ It would be good to have group-theoretic version of your question. Here is one. Given a finitely presented group $G$, does there exists a finite index subgroup $G_0$ and and a surjective homomorphism $H\to G_0$ inducing a surjection in homology and such that $H$ is finitely presented and contains a surface subgroup that is normal. Note that degree one maps are surjective in homology (in all degrees). $\endgroup$ Dec 13, 2009 at 15:05
  • $\begingroup$ Igor, here's a silly answer to your question as stated. Let F be a free group that surjects G, let S be a surface group and let H=SxF. Now the obvious map H->F->G is a surjection in homology! $\endgroup$
    – HJRW
    Dec 13, 2009 at 16:09
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    $\begingroup$ Two who have thought about the topological version (one at least being a 4-manifold topologist) are Kotschick and Loeh, arXiv:0806.4540. Kotschick gave a pretty talk about it at Columbia, with Gromov in the audience. DK: "For the next five minutes I'm going to explain something standard. Misha, I expect you know this as Preismann's theorem?" MG: "It's just obvious." $\endgroup$
    – Tim Perutz
    Dec 13, 2009 at 16:22

1 Answer 1


There is strong circumstantial evidence that the question came from a personal conversation with Bogomolov. First, Gromov's paper has more than 50 references, but he doesn't cite Bogomolov at all. Second, he does later in the paper thank Bogomolov for a helpful conversation, referring to him in nickname form as Fedya. They are about the same age, they both have appointments at NYU, and Gromov has thanked Bogomolov in papers before. Third, when I skimmed over Bogomolov's papers (okay, the abstracts and some introductions), I couldn't find anything directly bearing on the question.

On the other hand, I did find a paper by Carlson and Toledo that said that 30 years ago, Gromov introduced the partial ordering on $n$-manifolds given by the existence of a map $M \to N$ with non-zero degree. The broad question is what this partial ordering looks like. Obviously Gromov norm is an important invariant for this question.

Unless you ask Gromov or Bogomolov directly, it may make more sense to make an educated guess based on Gromov's summary. I would say, if $X$ is a closed, smooth, $n$-dimensional algebraic variety over $\mathbb{C}$, is there a closed, smooth $n$-fold $Y$ with a non-constant morphism $Y \to X$, such that $Y$ also fibers smoothly over a complex curve? Unless there is some clear reason that this question is misstated, it surely works as motivation for Gromov's question.


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