Are there any examples of integer homology spheres $Y^3$ that bound smooth integer homology balls but that do not smoothly embeded into $S^4$?
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1$\begingroup$ This is a major open problem. $\endgroup$– mmeCommented Nov 6, 2018 at 22:50

1$\begingroup$ @MikeMiller: do you know who might have first formulated that problem, and where? If I were to guess I'd say perhaps Fintushel and Stern, perhaps a long ways back, but I'm not certain which paper to look in. $\endgroup$– Ryan BudneyCommented Nov 6, 2018 at 23:15

1$\begingroup$ @RyanBudney That's a good question. I looked for a while but had no luck; I'll ask around (or maybe Danny Ruberman will see this question and know). If it helps anyone, my standard reference for existence results and discussion of this problem is your and Burton's "Census..." As a side comment, one natural place to try to look for counterexamples is $\Sigma \# \overline \Sigma$. But it is a theorem of Gompf, nicely written up in Kyle Larson's thesis, that if $\Sigma$ is $\pm 1/n$surgery on a knot, that manifold does embed. $\endgroup$– mmeCommented Nov 7, 2018 at 0:00

1$\begingroup$ @DavidSnyder I believe every 3dimensional homology sphere bounds a topological contractible 4manifold (But this is a hard theorem of Freedman). By doubling this we get a 4sphere (again  a theorem of freedman) $\endgroup$– user101010Commented Nov 14, 2018 at 5:16

1$\begingroup$ So, Kervaire's example isn't smoothable (though topologically flat) which was basically observed by Freedman in his paper. $\endgroup$– David SnyderCommented Nov 15, 2018 at 17:07

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The first examples (infinitely many of them) are given in Clayton McDonald's recent note. In particular, the double branched cover of the knot depicted in Figure 1 bounds a homology ball but does not embed into any homotopy 4sphere.
Of course, the existence of a homology sphere which embeds into a homotopy sphere but not $S^4$ is still open, because it's still open whether homotopy spheres are $S^4$.

1$\begingroup$ That doesn't sound to me like a smooth manifold. OP is asking about compact smooth 3manifolds without boundary and with the same homology as a sphere, and their smooth embedding properties into $S^4$, whereas your examples are not even obviously topological manifolds (when they are known to be it is often the case the result is $S^3$, such as for the Alexander cell A). $\endgroup$– mmeCommented Oct 31, 2022 at 9:30

1$\begingroup$ @WlodAA No worries, that makes sense. Your question is interesting in its own right. McDonald answers the smooth question and Freedman answers the question for topological manifolds, and you pose the question for homologymanifolds (spaces with same local homology as Rn). Perhaps some sewing of crumpled cubes bounds a homologymanifold but does not embed in S4? $\endgroup$– mmeCommented Oct 31, 2022 at 10:17

1$\begingroup$ I am a 3 or 4dimensional topologist so let me give an example in that spirit. It is a remarkable fact that many topological 5 and 6manifolds admit smooth structures. Combining Theorem 1 (which implicitly assumes CLOSED manifolds, but works for the manifold with boundary $M^4 \times D^2$, as remarked below the theorem) and Theorem 2, we see that if $M$ is any 4manifold with $H^3(M;\Bbb Z/2) = 0$ with zero KirbySiebenmann invariant then $M \times \Bbb R^2$ is smoothable. (Note they identify the degree4 obstruction as KS(M) at the beginning of p2.) $\endgroup$– mmeCommented Oct 31, 2022 at 10:36

1$\begingroup$ So one may take any closed topological 4manifold M with KS(M) = 0. A celebrated example is Freedman's "E8 manifold", a closed topological manifold with negativedefinite intersection form E8. This form is not diagonalizable over Z so Donaldson's diagonalizability theorem asserts M is not smoothable. Further, KS(M) = 0 whenever the intersection form of M is even. So I propose to take M to be Freedman's E8manifold and use those results to see $M \times \Bbb R^2$ is smoothable, in fact even $M \times D^2$. $\endgroup$– mmeCommented Oct 31, 2022 at 10:37

1$\begingroup$ @AruRay Thanks for correcting that comment! $\endgroup$– mmeCommented Nov 21, 2023 at 1:42