# Homology 4-balls with boundary $S^3$

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 $B^4$?

## 1 Answer

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.

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!

• I love the capitalization you made to distinguish it from the case solved by Freedman. Did not realize it at first. – Bombyx mori Jun 22 '16 at 6:41
• @Bombyxmori I actually intended to use the common acronym SPC4 later but didn't have enough to say to need to. Oops! – user98602 Jun 22 '16 at 7:27
• In July 2018 Kronheimer and Mrowka published a correction to their paper, in particular it is again open whether Rasmussen's s-invariant could detect an exotic 4-ball: arxiv.org/abs/1110.1297v2 – ben300694 Feb 14 at 16:03
• @ben300694 Thanks for bringing this outdated answer to my attention. – user98602 Feb 14 at 16:12