Here’s a problem I thought of while shoveling snow yesterday. (But it has nothing to do with snow.) In a comment to a previous post, Saul mentioned the placement space of a Heegaard surface (or of any subset of a manifold): Given a Heegaard surface *S* in a 3-manifold *M*, let *F* be a surface homeomorphic to *S* (but not necesarily embedded anywhere) and consider the space of all embeddings of *F* that are isotopic to *S*, modulo diffeomorphisms of *F*. You can think of this as all subsets of the ambient manifold that are isotopic to *S*. This definition is due to Hatcher for knots and links. The fundamental group of this space is roughly the ambiently trivial subgroup of the mapping class group of the Heegaard splitting. (By “roughly”, I mean that usually it is. Daryll McCullough and I have a theorem that determines the homotopy type of this space in all cases, but it’s currently being written up.)

One can also define the following space: Take all embeddings of *F* that are isotopic to *S*, but only quotient by isotopy trivial diffeomorphisms of *F*. Choose a connected component of this set and I will call that space the *marked placement space* of *S*. (This space covers the placement space and by the in-progress work with Daryll mentioned above, this space will often be contractible.)

If the ambient 3-manifold happens to have a Riemannian metric, then for each point in the marked placement space, there is an induced Riemannian metric on F (modulo isotopies of F). Such a metric determines a conformal structure on *F*, i.e. a point in the Teichmuller space. There is thus a canonical map from the marked placement space to the Teichmuller space of F. (There’s also a map from the (unmarked) placement space to moduli space.)

What I’d like to know is: Given a Heegaard splitting of a hyperbolic 3-manifold (i.e. if we can choose a nice metric on *M*), how “nice” is the map from the marked placement space to Teichmuller space? Is it onto? If not, is the image of the map contractible? Is the preimage of each point connected? Contractible? Is the map a homotopy equivalence?

The map can’t be surjective, for if is compact and the metric on is hyperbolic, then a curve in the Heegaard surface that’s essential in the manifold can’t be too short since the injectivity radius of is bounded below.

Comment by Richard Kent — December 21, 2008 @ 7:14 am |

Man, I shouldn’t have bothered with the latex, it looks terrible.

Comment by Richard Kent — December 21, 2008 @ 7:14 am |

I guess it’s conceivable that the map to the moduli space could be surjective, though.

Comment by Richard Kent — December 21, 2008 @ 7:15 am |

I think that Richard’s argument is bogus (happy new year, btw). Somehow, you can imagine creating a very very long cylinder with huge modulus in very little space (I don’t want to hurt anyone’s sensibilities, but the only image I have in mind is a condom). I bet that it is surjective and that if you put some nice topology it is something like a fiber bundle whose fibers can be disconnected (essentially, one for each element in some mapping class group of the Heegaard splitting or so) and where each connected component is contractible.

Comment by Juan Souto — January 2, 2009 @ 7:49 pm |

Oops. Juan’s right. You can make each essential curve have small extremal length that way.

Comment by Richard Kent — January 3, 2009 @ 8:23 am |

Juan — you could’ve avoided offending my delicate parts by talking instead about the bendy part of a bendy straw:

http://en.wikipedia.org/wiki/Drinking_straw

Jesse — does this mean that you and Darryl can decide finite generation of the Goeritz group of a splitting of S^3?

Comment by Saul Schleimer — January 7, 2009 @ 2:26 pm |

That’s a good question, Saul. I think the answer is “no” because all the spaces involved are infinite dimensional so there isn’t much chance of getting the compactness conditions that one would need.

Comment by Jesse Johnson — January 12, 2009 @ 12:10 pm |