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It would be interesting to me obtain an answer to the following easy to state question:

Does there exist a (smooth) fibre bundle $\pi\colon E\rightarrow B$ with typical fibre $F$ such that $E$, $B$ and $F$ are hyperbolic (connected, closed) manifolds?

I am not familiar with hyperbolic manifolds. From a colleague I have learned that "Thurston gave a necessary and sufficient criterion for a surface bundle over the circle to be hyperbolic," which is a similar question (see http://en.wikipedia.org/wiki/Hyperbolic_3-manifold for the quoted part) and could be a sign that this is not an easy question. Nevertheless I would like to ask it.

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If you count S^1 as hyperbolic, then you just gave an example (surface bundles over a circle can be hyperbolic). –  John Pardon May 25 '11 at 21:54
Why would S^1 be hyperbolic? –  Mariano Suárez-Alvarez May 25 '11 at 22:20
$S^1$ is the quotient of 1-dimensional hyperbolic space modulo any infinite cyclic group of isometries. –  Ryan Budney May 25 '11 at 22:23
It's a silly kind of "hyperbolic". –  Ryan Budney May 25 '11 at 22:24
The case when dim F = dim B = 2 is a major open problem. –  Tom Church May 25 '11 at 23:24
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3 Answers

up vote 15 down vote accepted

As Ryan points out, the interesting case is when the fiber is 2-dimensional. As Igor points out, this is a difficult open problem when the fiber has dimension 2.

When the fiber is a surface $F$, the fundamental group of the base $B$ admits a representation into the mapping class group $\mathrm{Mod}(F)$ of $F$. A combination of theorems of Farb-Mosher and Hamenstaedt shows that the fundamental group of the bundle is $\delta$-hyperbolic if and only if the map $\pi_1(B) \to \mathrm{Mod}(F)$ has finite kernel and the image is ``convex cocompact." (A subgroup of $\mathrm{Mod}(F)$ is convex cocompact if it acts on the Teichmueller space of $F$ with quasiconvex orbits.)

The only known examples of convex cocompact subgroups of mapping class groups are all virtually free, and so we are pretty far from knowing if there is a bona fide hyperbolic example like you are asking about. As Igor says, we don't even know if there is a $\delta$-hyperbolic surface-by-surface group.

However, it is conjectured that there is no hyperbolic (meaning truly hyperbolic) surface bundle $E$ over a surface. The reasoning is as follows: by an argument of Thurston, $E$ is symplectic. Using this fact, you can show that $E$ has a finite cover with nonvanishing Seiberg-Witten invariants. On the other hand, it is conjectured that the Seiberg-Witten invariants of a hyperbolic 4-manifold vanish. (See the article "Surface subgroups of mapping class groups" by Alan Reid in "Problems on Mapping Class Groups and Related Topics" edited by Benson Farb.)

For an introduction to convex cocompactness, and more references, see the survey "Subgroups of mapping class groups from the geometrical viewpoint" by me and Leininger.

Added: From the coarse perspective, the case where $F$ and $B$ are surfaces is key. If there were a $\delta$-hyperbolic $E$ with $F$ a surface and $B$ a hyperbolic manifold, then there is a $\delta$-hyperbolic $E'$ with fiber $F$ and base $B'$ a surface. To see this, note that recent work of Kahn-Markovic shows that $\pi_1(B)$ contains a quasiconvex surface group (I believe their theorem holds in all dimensions), and pulling back the bundle to this surface gives an example.

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Am I correct that the true goal of this activity is to understand convex cocompact subgorups of $\mathrm{Mod}(F)$? If so, it is unclear why one should focus on the case when $E$ is real hyperbolic. This is surely an important test case, but once it is answered (e.g. with help of Seiberg-Witten invariants) how much more do we know about convex cocompact subgroups of $\mathrm{Mod}(F)$? –  Igor Belegradek May 26 '11 at 12:49
@Igor, yes, that seems to be the correct summary. I think of the question of $E$ being real hyperbolic as something that allows you to motivate things when giving a colloquium. Convex cocompactness can't tell that the bundle is real hyperbolic, only that its group is $\delta$-hypebolic. –  Richard Kent May 26 '11 at 14:49
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I can think of at least one case where the answer is clearly no.

For example, if $F$ is a 3-dimensional compact hyperbolic manifold and the base space is any compact manifold, $E$ can't be hyperbolic.

The idea that that diffeomorphism group of $F$ is homotopy-discrete (this is a combination of work of Hatcher, Waldhausen and Mostow), having the homotopy-type of $Isom(M)$. So in this case the bundle $F \to E \to B$ has to have structure group a group of isometries, making $E$ a product geometry. Product geometries aren't hyperbolic, by the Margulis lemma (take elements of infinite order in $\pi_1 F$ and $\pi_1 B$ and you can construct a $\mathbb Z^2$-subgroup of $\pi_1 E$ with little effort).

I suspect an argument like this should work in more generality.

edit: yes, this argument holds in greater generality. Since it's essentially a fundamental-group issue you can avoid the Hatcher + Waldhausen part of the argument above and appeal directly the Mostow. Even if the monodromy for your bundle isn't of finite order, it is up to homotopy by Mostow, provided the dimension of the fibre is $\geq 3$.

So I think that means the only case that has not been covered is when the fibre is 2-dimensional.

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Thank you very much. –  Daniel Pape May 26 '11 at 5:17
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Ryan addressed the (easy) case when the fiber has dimension $>2$. The case when the fiber is $2$-dimensional is a well-known (and I think still open) problem with quite a bit of recent activity by the mapping class group community. For example, I think it is unknown whether a surface-by-surface group can be word-hyperbolic (assuming the fiber and the base are hyperbolic). See e.g. the survey by Lee Mosher "Problems in the geometry of surface group extensions" in here. Also look at this paper where Misha Kapovich showed that the total space of a surface bundle cannot be complex hyperbolic.

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Thank you for the references, I'll have a look at them. –  Daniel Pape May 26 '11 at 5:17
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