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I have the following question:

For a given two-dimensional Riemann surface $C$,

  1. Is there a way to classify all topologically distinct three-dimensional compact manifolds $M$ whose boundary is $C$, i.e., $\partial M =C$?

  2. Is there always a three-dimensional compact manifold $M$ such that $\partial M =C$ and is contractible?

  3. If there exists $M$ that satisfies condition 2, is it unique? If not unique, can the set of such manifolds be classified in a nice way?

Thank you in advance!

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3 Answers 3

up vote 5 down vote accepted
  1. When you say "Riemann surface", do you mean "topological surface"? Does the Riemann surface structure have any significance?

I assume below that you mean "two-manifold"

  1. Well, any three manifold contains a handlebody of your favorite genus, so this question is at least as hard as classifying three-manifolds (which is possible, by Thurston-Perelman).

  2. No, by the half-lives half-dies theorem.

3.Vacuous, because of 2.

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Thank you! The half-lives half-dies theorem---which I didn't know about until now---seems to do the job just fine. Just a comment and an additional question. (1) I do not care about the structure of the Riemann surface for now, so you are correct in assuming that it is a topological surface. (2) What happens if I weaken the assumption and say that the three-manifold M is simply connected? Thanks again for your response. –  D. S. Park Dec 23 '11 at 22:01
    
There's only one simply connected compact 3-manifold, up to punctures... –  Steve D Dec 24 '11 at 22:25
    
But even without the Poincare conjecture, "half lives, half dies" shows a manifold with at least one Riemann surface boundary component will have infinite first homology. –  Steve D Dec 24 '11 at 22:27
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(1) Consider the case where C is a sphere. For any such manifold, we can just add a 3-dimensional ball to get a closed compact 3-fold. So in this case it's just the classification of closed compact 3-folds.

For the torus, it's slightly more compact. One can again add a donut, so the problem includes the classification of closed compact 3-folds. But suppose this 3-fold turns out to be a sphere. We still don't know how the donut embeds - in particular, it could surround any knot. So it also includes the classification of knots.

(3) This is not entirely vacuous because it is a reasonable question to ask if C is a sphere. In this case, M, a closed ball, is unique up to homeomorphism, by the Poincare conjecture. (Glue a closed ball to it, get a homotopy 3-sphere, which is therefore homeomorphic to a 3-sphere)

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Good point, re (3). –  Igor Rivin Dec 23 '11 at 20:14
    
Thanks for the nice explanation. :) –  D. S. Park Dec 23 '11 at 22:03
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For (3), here is a nice counterexample if we ignore contractibility (taken from an article from The American Mathematical Monthly):

http://www.jstor.org/stable/pdfplus/2695643.pdf

Hopefully someone can embed the image (Figure 13). It is two topologically distinct manifolds with the torus as a boundary... one being a solid [knot] torus, and the other being a 'cylinder' with a knotted inner-hole.

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However, note that neither is contractible. –  Will Sawin Dec 24 '11 at 4:05
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Every knot complement is an example of a 3-manifold with torus boundary (and there are tons more), so I am not sure that this example is so exciting. –  Tom Church Dec 24 '11 at 4:44
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