Surface bundles over a surface

What can be used to distinguish two $\Sigma_g$-bundles over $\Sigma_h$ up to

(1) homotopy? (2) homeomorphism? (3) fiberwise homeomorphism? (4) bundle isomorphism?

And can these always be computed given 2 specific surface bundles over $\Sigma_h$?

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 For the homotopy type of a given $\Sigma_g$ bundle over $\Sigma_h$, since $\Sigma_h$ can be realized as a $2$-cell (thought of as a $2h$-gon) glued along the bouquet isn't it enough to consider the homotopy class of $f:\vee^{2h} S^1\to \textrm{Aut}\Sigma_g$ ? – Somnath Basu Sep 15 2010 at 22:04 Somnath, Romeo -- if nether the base nor the fiber are spheres of projective planes, by the long homotopy exact sequence each such bundle is a $K(\pi,1)$ i.e. it is classified by its fundamental group up to homotopy equivalence. – algori Sep 15 2010 at 22:13 I guess Hansen and Gottlieb has results where they compute $\pi_1(\textrm{Map}(X,Y);\varphi)$ for $Y$ an Eilenberg-MacLane space. The answer is just the centralizer of $\varphi_\ast(\pi_1(X))$. The higher homotopy groups (which aren't needed) are zero. You can use this for $X=Y=\Sigma_h$. – Somnath Basu Sep 15 2010 at 22:17

I'll make some comments, but I don't know the complete answer to your questions. (I'll assume $g,h>0$ too.)

As Thurston says, the bundle is determined up to isomorphism by the image $\pi_1(\Sigma_h)\to Mod(g)$. I think this answers questions (3) and (4). If two fiber bundles are fiberwise homeomorphic, then the space of fibers determines the base space, and therefore the fibration, so I think (3) and (4) are equivalent. The algorithmic question is open, as far as I know (of determining when the image of two surface groups in $Mod(g)$ are conjugate).

I think questions (1) and (2) are more difficult. One issue is that a 4-manifold that fibers over a surface might fiber many different ways. Because I'm assuming $g,h>0$, these manifolds are $K(\pi,1)$'s, and so the homotopy question reduces to determining if $\pi_1$ are isomorphic. The homeomorphism problem would follow from the the Borel conjecture, but this is wide open in the case of 4-manifold with fundamental group of exponential growth.

There is a well-known open question, whether there is a convex-cocompact map (in particular injective) $\pi_1(\Sigma_h)\to Mod(g)$. If this exists, the $\pi_1$ of the associated bundle is a word-hyperbolic group. In this case, there is a theorem of Sela which allows one to algorithmically distinguish the fundamental groups, and therefore determine the homotopy type of the associated manifolds. However, this could be a theory of the empty set, since no examples are known!

Addendum: There are some techniques for distinguishing the diffeomorphism types of these manifolds. If the fiber $\Sigma_g$ is homologically non-trivial in the manifold, then Thurston proved that $M$ admits a symplectic structure. Work of Taubes implies that the Seiberg-Witten invariants of $M$ may be computed from the Gromov invariants of $M$. This might give methods for detecting when two bundles are not diffeomorphic. I suppose that such manifolds might have many different smooth structures, but it's not clear to me that the natural smooth structures associated to different fiberings could be different.

One might be able to distinguish homotopy types of these manifolds using group invariants. For example, these groups are subgroups of mapping class groups, and are therefore residually finite. One could count homomorphisms from the fundamental group to a finite group to obtain invariants that may distinguish the bundles.

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Hi Ian. The reason why (3) and (4) are not equivalent is that a bundle isomorphism is usually assumed to induce the identity map on the base. So a pair of bundles is isomorphic exactly when the maps into Mod(g) are conjugate. But bundles could be fiberwise homeomorphic in many other different ways. – Daniel Groves Sep 16 2010 at 11:37
Good point, thanks. – Romeo Sep 16 2010 at 14:11
Oh, I see, I hadn't thought this through carefully - I didn't remember the proper definition of bundle isomorphism. So one may precompose the homomorphism with elements of $Aut(\pi_1(\Sigma_h))$ to get non-isomorphic bundles which are fiberwise homeomorphic. I'll delete this part since it is addresses in Thurston's answer. – Agol Sep 16 2010 at 15:35
What's a good example or two of bundles that are fiberwise homeomorphic but not isomorphic? – Romeo Sep 17 2010 at 16:21

I'm supposing you mean for $g, h > 0$. Associated with a surface bundle, there is a homomorphism of $\pi_1(\Sigma_h)$ to the outer automorphism group of $\pi_1(\Sigma_g)$. This is equivalent (with slight low-genus modifications) to a homotopy class of maps into the modular orbifold, Teichmüller space modulo the mapping class groups. The group of homeomorphisms of a surface homotopic to the identity is contractible, so these bundles are determined up isomorphism that acts as the identity on the base and on one fiber.

The conjugacy problem for the mapping class group is solved, using either the theory of pseudo-Anosov homeomophisms or automatic group theory, and either of those tools allows you to solve isomorphism up to bundle maps that are the identity on the base. Peter Brinkmann's program xtrain, which you can find online, computes the dilatation constant, which is typically enough to distinguish conjugacy classes in the mapping class group. Snappea, also available online, will usually distinguish homeomorphism classes of the 3-manifolds obtained by an element of the mapping class group (with exceptions that can be analyzed). This will also distinguish conjugacy classes, by looking for homoemorphism preserving a cohomology class.

The action of the mapping class group of the base on bundle maps seems trickier, and I don't think I know an immediate answer of classifying them. The troublesome cases would be where the image of the surface group in the mapping class group is not a quasi-isometric map of groups.

A classification of homeomorphism types would include the special case when the surface bundle is induced from a map of the base to a circle, so the bundle comes from a 3-manifold that fibers over a circle. 3-manifolds can fiber in many different ways, so not all homeomorphisms in these cases are fiber preserving, and the homeomorphism classification for these particular cases is solvable, but it gets into a complicated theory that won't usually work for 4-manifolds. I'm not sure what's known about surface fiber bundles over surfaces that fiber in multiple ways, apart from these.

One other point: the fundamental group of such a 4-manifold has an action on $S^1$, namely, the circle at infinity to the fibers. The action is faithful if the monodromy of the bundle is faithful. In these cases, the isomorphism class of the 4-manifold I believe is determined by the subgroup of homeomorphisms of the circle, up to conjugacy.

For $h > 1$, there is always some branched cover of the base surface so that when you pull the bundle back to the branched cover, there is a section of the bundle, the map to the outer automorphism group of the fiber lifts to the automorphism group, and the fundamental group of the 4-manifold is a semi-direct product.

I'm not an expert in these, and I'm sure there is more that is known.

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Dear Professor Thurston, A minor quibble and a correction. There are nontrivial bundles with finite monodromy. These will correspond (like the trivial bundle) to a constant map into the modular orbifold. So bundles are not quite classified by maps to this orbifold. In order to classify surface bundles up to bundle isomorphism (with trivial induced map on the base), you need to solve the simulataneous conjugacy in the mapping class group. I believe that Benson Farb knows how to do this algorithmically (but it's not yet published, and you need more than just solve the conjugacy problem. – Daniel Groves Sep 15 2010 at 23:44
I believe that Farb also knows how to prove that the fiberwise homeomorphism' problem is undecidable for surface bundles over a surface, and that this is supposed to be in the same (not-yet-existent) paper. (Normally, I wouldn't speak about somebody's unpublished work, but he talked about it in Michigan in public in April of either this year or last -- I can't check because my browser is not getting to the Michigan Math website at the moment...). – Daniel Groves Sep 15 2010 at 23:48
@Daniel Groves, Thanks for the clarification. I was thinking of maps to the modular space as in some orbifold sense (secretly lifting to the K(pi,1)) I realized that it takes more than the individual conjugacy problem, but I figured it was doable, just annoying, although I didn't actually think through details. Feel free to add your own answer with more informed information. – Bill Thurston Sep 15 2010 at 23:59
@Daniel Groves: the normalizer of the generic kind of element (pseudo-Anosov) is virtually cyclic and easily computable, which makes the joint conjugacy problem relatively easy for those --- it turns into a bounded search. The other cases don't seem to create obstacles either. – Bill Thurston Sep 16 2010 at 0:02
@Bill Thurston: I'm not sure how hard it was. (What I heard from Farb is both results, and maybe one's very easy.) The pseudo-Anosov case is indeed straightforward. In the general case, the centralizer of an element can quite large. If you have a pair of pairs of elements (g_1, g_2) and (h_1,h_2) with all elements reducible, and you know that g_1 is conjugate to g_2 by an element k, then you need to know if there's an element of the k-coset of the centraliser of g_1 that sends h_1 to h_2. Probably, as you say, doable, just annoying'. – Daniel Groves Sep 16 2010 at 0:12
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