Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

There's an old problem in 4-manifold theory that, as far as I know, doesn't have a name associated with it and really deserves a name.

Let $M$ be a smooth 4-manifold with boundary. Let $S$ be a smoothly embedded 2-dimensional sphere in $\partial M$. Assume $S$ does not bound a ball in $\partial M$, but $S$ is null-homotopic in $M$. Does $S$ bound a smooth 3-ball in $M$? Perhaps you need to replace $S$ by another non-trivial $S'$ in $\partial M$ before you can find a 3-ball in $M$ bounding it?

You could think of this as the co-dimension one analogue to Dehn's lemma for 4-manifolds. Usually when people talk about a Dehn lemma for 4-manifolds they're interested in the co-dimension 2 analogue.

Does this problem / conjecture have a name? If not, do you have a good name for it? Do you know of anywhere in the literature where this issue is investigated?

Off the top of my head the only vaguely related things I know about in the literature is a 1975 paper of Swarup's.

share|improve this question

2 Answers 2

up vote 9 down vote accepted

I think that the conjecture is wrong. The following leads to counterexamples in the topological category and probably also smoothly: Take a closed oriented 4-manifold N with infinite cyclic fundamental group and remove an open neighborhood of a generating circle. Then you get a 4-manifold M with boundary $S^1 \times S^2$ where only the generators can be represented by embedded 2-spheres. If S denotes $pt \times S^2$ then the following hold:

  1. S is null homotopic in M via Poincare duality and the exact sequence of the pair.
  2. S bounds an embedded 3-ball in M if and only if N can be decomposed into $S^1 \times S^3$, connected sum with a simply connected manifold.

Hambleton and I found topological 4-manifolds N for which the intersection form is not extended from the integers, in particular 2 doesn't hold. With Friedl and Melvin we showed later that our examples don't have a smooth structure. But now I remember a discussion with Fintushel and Stern who mentioned that they constructed a smooth counterexample to 2 (and hence to the conjecture).

share|improve this answer
    
Thanks again Peter. I had some hope for this conjecture but it wouldn't crush me if it's false. I'll ask F. or S. about this the next time I bump into one of them. –  Ryan Budney Mar 29 '10 at 20:53

In dimension 3, you have the sphere theorem, the torus theorem, the annulus theorem, and the disk theorem (which is the loop theorem and Dehn's lemma put together).

So, if you didn't require the sphere to be embedded, and the problem already had an affirmative answer, I'd call it the Ball Theorem.

share|improve this answer
    
I'm still holding out for a historical name but perhaps this is something that's always been "hanging". So ignored, it does not have a name. :( –  Ryan Budney Dec 13 '09 at 2:44

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.