# Which compact 3-manifolds with boundary embed in $\mathbb{S}^3.$

This is a more sensible (IMHO) restatement of this question:

Which Compact $3$-manifolds with boundaries embed in $\mathbb{S}^3?$ Is there any hope of a characterization?

• The embedding problem is algorithmically decidable, arxiv.org/abs/1402.0815 – Slava Krushkal Jan 3 '16 at 18:15
• As far as I know, no characterisation exists and any would constitute original research. There is a well-known answer in the case your manifold's boundary is $S^2$. But as soon as you get to $S^1 \times S^1$ boundary, distinguishing between knot exteriors in $S^3$ vs homology spheres is fussy business. But it also depends on what you consider a worthwhile answer. If "the fundamental group is normally generated by a meridian" works for you, then you do have a satisfactory answer, but your comments below suggest you think this is not satisfactory. – Ryan Budney Apr 24 '17 at 18:14

Theorem: Every compact connected 3-submanifold $Y$ of the 3-sphere can be reimbedded in the 3-sphere so that the exterior of the image of $Y$ is a union of handlebodies, i.e. regular neighborhoods of embedded graphs.
• But I am not sure that really answers the question, since if I give you some hairy manifold with boundary, how do you know if it is the complement of a bunch of handlebodies in $\mathbb{S}^3?$ If it is, there is obviously a decision procedure (keep trying elements of the mapping class group, and checking the result), the other direction seems much harder (but apparently addressed in the Krushkal comment), Still, some human-understandble criterion would be good... – Igor Rivin Jan 5 '16 at 12:05