I’ve mentioned in previous posts the idea of the mapping class group of a Heegaard surface: the group of automorphisms of the ambient 3-manifold that take the Heegaard surface to itself, modulo isotopies that leave the surface on itself. There’s also the subgroup of mapping class elements that are isotopy trivial as automorphisms of the ambient manifold. This group is the kernel of the induced map from the mapping class group of the Heegaard surface to the mapping class group of the ambient manifold.

There’s nothing in this definition that requires that the surface be a Heegaard surface; we can define the same group for any embedded surface, in particular for incompressible surfaces. For a separating incompressible surface, elements of this group come from automorphisms of the complementary components that induce matching automorphisms of their shared boundaries. For non-separating surfaces, non-trivial automorphisms come from automorphism of the complement that match up when the two boundary components making the incompressible surface are glued together.

As with Heegaard surfaces, one can ask whether non-trivial elements of incompressible surface’s mapping class group can be isotopy trivial in the ambient manifold. For Heegaard surfaces, there are lots of examples, but for incompressible surfaces, the only obvious examples are the leaves of surface bundles. Pushing a leaf around the monodromy brings it back to itself by a non-trivial automorphism, so this mapping class element is in the kernel.

The disparity in the number of examples comes from the fact that Heegaard surfaces are in many ways more floppy than incompressible surfaces. In particular, the map from the fundamental group of a Heegaard surface to the fundamental group of the ambient manifold is nowhere close to injective, while the same map for an incompressible surface is injective. Thus an automorphism of an incompressible surface coming from an isotopy of the manifold implies an element of the fundamental group of the 3-manifold such that conjugating the surface’s fundamental group by this element brings it back to itself. (You can find this element by following the base point in the surface during the isotopy.) In other words, the normalizer of the surface’s fundamental group, as a subgroup of the 3-manifold’s fundamental group, is non-trivial.

I wouldn’t be at all surprised if it turns out that the only incompressible surfaces with non-trivial automorphism that are trivial in the ambient manifold are leaves of surface bundles. In fact, it seems like the algebraic situation, a surface subgroup with non-trivial normalizer, should only happen in a surface bundle. However, my knowledge of results along these lines is very limited. Do any readers know if this is the case? Also, if you know of any paper where mapping class groups of incompressible surfaces are discussed, please put that in a comment too.

Let’s restrict our attention to hyperbolic 3-manifolds. It’s quite well known that any geometrically finite subgroup has finite index in its normalizer. On the other hand, it’s a consequence of tameness that the only (finitely generated) geometrically infinite subgroups are `virtual fibres’, ie fibres in a finite-sheeted covering. So, in summary, for hyperbolic 3-manifolds, the conjecture that you make in your final paragraph is true `up to finite index’, and follows from tameness.

I think the reference for this consequence of tameness is detailed in Corollary 8.3 of:

R. D. Canary, A covering theorem for hyperbolic 3- manifolds and its applications, Topology

35 (1996), pp. 751–778.

I’m 99.9% certain that you can’t remove the `up to finite index’ from any of these statements, though I don’t know an example off the top of my head.

Comment by Henry Wilton — August 13, 2008 @ 9:54 am |

Oh – maybe that you’re talking about

embeddedsurfaces might enable you to do away with the `up to finite index’ part. I’m not sure.Comment by Henry Wilton — August 13, 2008 @ 10:11 am |

Jesse, I think your intuition is correct. The condition of a surface

subgroup having a nontrivial normalizer is extremely strong, and

forces one of the possibilities listed in Theorem 11.1 of John

Hempel’s book. To show that you have a surface bundle, you would just

have to rule out the possibility that the surface separates the

manifold into two twisted I-bundles. This might take some work (or

some tool), since there does exist a homotopy that moves the surface

back to itself with a nontrivial automorphism, just not an isotopy.

Comment by Darryl McCullough — August 18, 2008 @ 2:47 pm |