Welcome to MathOverflow

Question

3

1

I have the following question:

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

Is there a way to classify all topologically distinct three-dimensional compact manifolds $M$ whose boundary is $C$, i.e., $\partial M =C$?
Is there always a three-dimensional compact manifold $M$ such that $\partial M =C$ and is contractible?
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!

flag	asked Dec 23 2011 at 19:41 D. S. Park 113●5

Igor Rivin · Accepted Answer · 2011-12-23 19:51:16Z UTC

5

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"

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

3.Vacuous, because of 2.

answered Dec 23 2011 at 19:51

Igor Rivin
31.2k●3●37●71

	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 2011 at 22:01
	There's only one simply connected compact 3-manifold, up to punctures... – Steve D Dec 24 2011 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 2011 at 22:27

Will Sawin · Answer 2 · 2011-12-23 19:58:42Z UTC

(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)

Chris Gerig · Answer 3 · 2011-12-24 04:38:53Z UTC

1

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.

edited Dec 24 2011 at 4:38

answered Dec 24 2011 at 3:55

Chris Gerig
5,646●1●12●40

However, note that neither is contractible. – Will Sawin Dec 24 2011 at 4:05

1

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 2011 at 4:44

Questions on 3-manifolds with a given boundary

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

3 Answers

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.