# Slice knots in 3-manifolds

Is there a nonslice knot $$K\subset S^3$$ that is slice in some closed oriented $$3$$-manifold $$Y$$? Here, when we say $$K$$ is slice in $$Y$$, it means that when regarded as a local knot in $$Y\times\{1\}$$, $$K$$ bounds an embedded disk in $$Y\times[0,1]$$. One can ask this question in both topological and smooth categories.

When $$Y$$ is irreducible, by taking the universal cover one sees that being slice in $$Y$$ is equivalent to being slice (in $$S^3$$). However, for $$Y$$ reducible I am not sure the same conclusion still holds.

In one dimension lower there is an analogous question, namely whether there is a nontrivial knot in $$S^3$$ that becomes trivial in some closed $$3$$-manifold $$Y$$. The answer is no, because surgering along essential spheres reduces the question to $$Y$$ irreducible where one can again take the universal cover to conclude. However, the same argument doesn't seem to carry directly to the $$4$$-dimensional case.

• See Prop. 2.2 of arxiv.org/abs/2009.03053 Mar 4 at 20:14
• @IanAgol Oh, this is somewhat surprising! Thanks for pointing out the reference. Mar 4 at 20:34

Suppose you know that the universal cover of $$Y$$ embeds in $$S^3$$, i.e. is $$S^3-A$$ for some $$A$$. For example, this happens when $$Y$$ is a connected sum of lens spaces. (I'm thinking this is always true but feel like I might be missing something.) Then by covering space theory, a disk in $$Y \times I$$ lifts to a disk in the universal cover. Adding $$A \times I$$ back in gives a slice in $$S^3 \times I$$, so your knot was slice in the first place.
• The universal cover of every 3-manifold embeds in $S^3$. Mar 4 at 20:40