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?
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asked Jan 3, 2016 at 15:19
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6$\begingroup$ The embedding problem is algorithmically decidable, arxiv.org/abs/1402.0815 $\endgroup$– Slava KrushkalJan 3, 2016 at 18:15
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$\begingroup$ 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. $\endgroup$– Ryan BudneyApr 24, 2017 at 18:14
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1 Answer
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There is a theorem of Fox that more or less deals with this. Any such manifold is a complement of (possibly knotted) handlebodies.
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.
R. H. Fox, On the imbedding of polyhedra in 3-space, Ann. of Math. (2) 49 (1948), 462–470.
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2$\begingroup$ A relevant paper is Kei Nakamura's arxiv.org/abs/1202.4062 $\endgroup$ Jan 5, 2016 at 12:02
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$\begingroup$ 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... $\endgroup$ Jan 5, 2016 at 12:05
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$\begingroup$ This is the initial step in Sedgwick et. al.'s algorithm. $\endgroup$– Ian AgolApr 24, 2017 at 17:50