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Is there a compact, connected, smooth 3-manifold $M$ with boundary $S^2$ not diffeomorphic to $D^3$ (the closed unit ball)? If so, what is it?

The compactness condition rules out the complement of the open ball in $\mathbb{R}^3$, and the connectedness condition rules out the closed unit ball disjoint union with something having empty boundary.

My only idea is to consider things that have boundary $\mathbb{RP}^2 \cong S^2$, but this hasn't gotten me anywhere yet.

If you have an example, some explanation of how you thought of it would be great.

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$RP^2$ is not $\cong$ to $S^2$, for any sensible meaning of $\cong$. –  Mariano Suárez-Alvarez Oct 31 '12 at 19:20
    
could you believe i meant CP1? ;) –  AnonymousCoward Nov 12 '12 at 5:26

2 Answers 2

up vote 3 down vote accepted

I don't see why not.

Approach 1. You can probably perform a surgery on $D^3$ and attach a handle to it. That is, you remove two smaller open balls A and B from D^3. Now you have a manifold $X = D^3 \setminus A \setminus B$, and the boundary of X is diffeomorphic to $S^2 \sqcup S^2 \sqcup S^2$.

Now take $Y = S^2 \times D^1$. Its boundary is $S^2 \sqcup S^2$. Now you can glue $X$ to $Y$ along $S^2 \sqcup S^2$, and the result will have boundary $S^2$.

I guess there still remains some work to be done to prove that the resulting manifold is not diffesmorphic to $D^3$, and probably care should be taken to make everything smooth. Unfortunately, I'm not an expert in such things.

Approach 2. You also can take an arbitrary compact connected 3-manifold $Z$ without a boundary and remove an open ball $B^3$ from it. The resulting manifold will have $S^2$ as its boundary, and if $Z$ is complex enough, then $Z \setminus B^3$ will not be diffeomorphic to $D^3$. But again, I haven't done this in a while, so I can't describe this any more rigorously.

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It's definitely true for surfaces. Just take the torus and remove an open disk. Does it work with the 3-dimensional torus?

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