# Homology spheres bounding homology balls but not embedding into $S^4$

Are there any examples of integer homology spheres $$Y^3$$ that bound smooth integer homology balls but that do not smoothly embeded into $$S^4$$?

• This is a major open problem.
– mme
Commented Nov 6, 2018 at 22:50
• @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. Commented Nov 6, 2018 at 23:15
• @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.
– mme
Commented Nov 7, 2018 at 0:00
• @DavidSnyder I believe every 3-dimensional homology sphere bounds a topological contractible 4-manifold (But this is a hard theorem of Freedman). By doubling this we get a 4-sphere (again - a theorem of freedman) Commented Nov 14, 2018 at 5:16
• So, Kervaire's example isn't smoothable (though topologically flat) which was basically observed by Freedman in his paper. Commented Nov 15, 2018 at 17:07

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 4-sphere.

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 $$S^4$$.

• That doesn't sound to me like a smooth manifold. OP is asking about compact smooth 3-manifolds 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).
– mme
Commented Oct 31, 2022 at 9:30
• @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 homology-manifolds (spaces with same local homology as Rn). Perhaps some sewing of crumpled cubes bounds a homology-manifold but does not embed in S4?
– mme
Commented Oct 31, 2022 at 10:17
• I am a 3 or 4-dimensional topologist so let me give an example in that spirit. It is a remarkable fact that many topological 5- and 6-manifolds 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 4-manifold with $H^3(M;\Bbb Z/2) = 0$ with zero Kirby-Siebenmann invariant then $M \times \Bbb R^2$ is smoothable. (Note they identify the degree-4 obstruction as KS(M) at the beginning of p2.)
– mme
Commented Oct 31, 2022 at 10:36
• So one may take any closed topological 4-manifold M with KS(M) = 0. A celebrated example is Freedman's "E8 manifold", a closed topological manifold with negative-definite 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 E8-manifold and use those results to see $M \times \Bbb R^2$ is smoothable, in fact even $M \times D^2$.
– mme
Commented Oct 31, 2022 at 10:37
• @AruRay Thanks for correcting that comment!
– mme
Commented Nov 21, 2023 at 1:42