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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.

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    $\begingroup$ See Prop. 2.2 of arxiv.org/abs/2009.03053 $\endgroup$
    – Ian Agol
    Mar 4 at 20:14
  • $\begingroup$ @IanAgol Oh, this is somewhat surprising! Thanks for pointing out the reference. $\endgroup$
    – Qiuyu Ren
    Mar 4 at 20:34

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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.

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  • $\begingroup$ The universal cover of every 3-manifold embeds in $S^3$. $\endgroup$ Mar 4 at 20:40
  • $\begingroup$ Thanks Danny. The paper Ian pointed out to me in the comment essentially used this proof, together with the fact Moishe pointed out (which they refered to Lemma 2.11 in this paper: ams.org/journals/proc/2017-145-12/S0002-9939-2017-13667-3). I'll mark this question as solved. $\endgroup$
    – Qiuyu Ren
    Mar 4 at 21:47
  • $\begingroup$ @MoisheKohan--Thanks. I was pretty sure that was true but didn't have a chance to think it through. $\endgroup$ Mar 5 at 1:49
  • $\begingroup$ @MoisheKohan the universal covers of compact 3-manifolds embeds in S^3. There are simply-connected 3-manifolds that do not embed in S^3. doi.org/10.2307/1970374 $\endgroup$
    – Ian Agol
    Mar 5 at 20:20
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    $\begingroup$ @IanAgol: I meant universal covers of closed manifolds (in the context of the answer). $\endgroup$ Mar 5 at 20:22

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