Homology 4-balls with boundary $S^3$

Question

Are there interesting homology 4-balls with boundary $S^3$? Going the other way, must any homology 4-ball with boundary $S^3$ be homotopy equivalent/homeomorphic/diffeomorphic to $B^4$?

Answer 1

They are absolutely not all $B^4$. Given a smooth 4-manifold with boundary $S^3$, there is a unique way to cap that boundary component off with a copy of $B^4$, thus giving us a smooth closed 4-manifold. (Hidden in this statement is Cerf's quite nontrivial theorem that every diffeomorphism of $S^3$ extends over the 4-ball.) If the original manifold was a homology ball, then the new manifold is a homology sphere. So your question is entirely equivalent to "Are all homology spheres equivalent to $S^4$?"

The first reason that you should not believe this is that already in 3 dimensions the study of homology spheres is active, lively, and quite confusing. (As two pieces of support for this statement: first, that there's an entire book by Saveliev on invariants of homology spheres; second, they have a strong relationship with higher-dimensional topology: Manolescu's recent proof that there are no homology 3-spheres with Rokhlin invariant 1 and $\Sigma \# \Sigma$ bounding a homology ball implies that there exist high-dimensional non-triangulable manifolds!)

There are indeed vastly many homology 4-spheres. Here's a couple ways of constructing some.

1) Take a homology 3-sphere that bounds a homology ball and take the double of the homology ball. One easy case is eg $\Sigma \# \overline{\Sigma}$, which bounds the manifold you get when you tunnel through from one side to the other of $\Sigma \times [0,1]$. Taking the double of this gets you a homology sphere with fundamental group $\pi_1(\Sigma)$.

2) Take a homology 3-sphere and cross it with $S^1$, then do surgery on the circle $\{*\} \times S^1$. The resulting manifold is a homology sphere with fundamental group $\pi_1(\Sigma)$. (I wonder if this is actually the same construction as above? I dunno.) You could also do this but furl up instead a homology cobordism and surgery out some path from one side to another, which gives you a much more general class of examples.

One recentish advance in the study of homology 4-spheres (solving a problem on Kirby's list) is the existence of aspherical homology 4-spheres; these are universal in dimension 3 but seem much harder to come by in higher dimensions. See Ratcliffe and Tschantz's paper here.

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Of course if you also demand that the manifold be simply connected, the Hurewicz and Whitehead theorems imply that $M$ is homotopy equivalent to $S^4$, and Freedman's classification of simply connected closed topological 4-manifolds implies it's homeomorphic to $S^4$. This reduces us to the completely wide-open question of understanding smooth structures on $S^4$...

Thinking about simply connected homology 4-balls is one popular approach to the Smooth Poincare Conjecture in 4 dimensions. See this lovely paper by Freedman-Gompf-Morrison-Walker for a particular attempt at showing certain homotopy 4-spheres aren't the standard one by showing that there are knots in $S^3$ which are slice in an exotic 4-sphere but not in the standard one. Unfortunately, at the time it was thought that their approach doesn't work for two reasons:

1) the spheres they were testing it on turned out to be standard (proved 3 days later by Akbulut, then some more a month later by Gompf);

2) the invariant (Rasmussen's $s$-invariant) they were using to try to show that knots aren't smoothly slice in $S^4$ was thought to vanish for knots that are smoothly slice in an exotic 4-sphere - this was claimed by Kronheimer and Mrowka, but this claim has since been retracted; this is an interesting open question. In fact, I don't know a single example of a knot $K$ and a concordance invariant $c$ so that $K$ is slice in a homology ball, but $c(K)$ is nonzero.

This approach might be salvageable if one can find a good invariant that (2) doesn't kill, but finding such a thing sounds tough, and I'm sure more-or-less every researcher in 4-manifold topology would be very excited if they (or you!) found one and an example of a slice knot it doesn't vanish on!