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Suppose we are given a compact orientable 3-manifold $M$ which is an integral homology $S^1\times S^2$. Then is there a way to determine whether $M$ bounds a smooth compact orientable 4-manifold which is an integral homology $S^1\times B^3$?

I think this may be a hard question, because I've heard that it is even not known that which integral homology $S^3$'s bound integral homology $B^4$'s.

So I want to consider the following special case. Let $M_K$ be obtained from $S^3$ by a 0-surgery on a knot $K$ in $S^3$. Then $M_K$ is an integral homology $S^1\times S^2$. Then is it known that for which knots $K$ the 3-manifold $M_K$ bounds an integral homology $S^1\times B^3$?

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    $\begingroup$ I don't think there is a closed-form answer that isn't tautological. But there are obstructions. There is an obstruction coming from Alexander polynomials, due to Kawauchi. I imagine there must be many other obstructions as well. $\endgroup$
    – Ryan Budney
    Commented Jun 13, 2022 at 1:49
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    $\begingroup$ Two tiny tangential notes: if you additionally ask for the $\pi_1$ of that 4-manifold to be normally generated by the meridian of $K$, then $K$ is necessarily topologically slice (see e.g. Proposition 3.1 of this paper arxiv.org/pdf/math/0505233.pdf). Disregarding the $\pi_1$ condition, as Ryan says, you might be interested in Kawauchi's "$\tilde{H}$-cobordism, I: The groups among among three dimensional homology handles". See also arxiv.org/abs/2004.05967 for a recent paper on the topic. $\endgroup$
    – Anthony Conway
    Commented Jun 13, 2022 at 6:31

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