There used to be many candidates for an exotic 4-sphere, but a lot of them are now known to be the standard smooth $S^4$. The ones of Cappell-Shaneson (maybe not all of them?) were described in terms of handlebody decompositions where there are no 3-handles, but I think the proofs that these were diffeomorphic to $S^4$ involved altering the decompositions with a cancellation trick by introducing some 3-handles.

I am not sure what is left over, but I seek the following:

1) Is there a potentially exotic 4-sphere which is prescribed by an explicit 4-handlebody decomposition with only 0,1,2-handles and a single 4-handle?

2) Of the standard Cappell-Shanelson spheres with a given handlebody decomposition involving no 3-handles, is it known that we can prove [edit: some of them] they're standard without the trick involving 3-handles?

The reason I ask is because I'd like to study certain differential 2-forms on these spheres which "play well" with the handles, and to understand the deformation of these 2-forms as we change the handlebody decompositions.


1 Answer 1


Not really an answer, but may be helpful...

First, I would not say that the problem gets easy by introducing a canceling 2/3-handle pair. Akbulut, Kirby and Gompf worked several years on this.

S. Akbulut, and R. Kirby, A potential smooth counterexample in dimension 4 to the Poincaré conjecture, the Schoenflies conjecture, and the Andrews-Curtis conjecture. Topology 24 (1985), 375–390.

R. Gompf, Killing the Akbulut-Kirby 4-sphere, with relevance to the Andrews-Curtis and Schoenflies problems. Topology 30 (1991), 97–115.

Even to just understand the results will take some time. See for example the introduction to

R. Gompf, More Cappell-Shaneson spheres are standard. Algebr. Geom. Topol. 10 (2010), 1665–1681.

for some explanation on the history.

In the same paper, it is also explained exactly which Cappell-Shaneson spheres are standard. The handle decompositions of the others should be an answer to your first question. However, it is conjectured by Gompf that in fact, all Cappell-Shaneson spheres are standard.

The difference between allowing to introduce canceling 2/3-handle pairs and not allowing is given by the Andrew-Curtis conjecture. If you have a handle decomposition without 3-handles of a homotopy sphere you get a balanced presentation of the trivial group. If you can prove that these handle decomposition is a decomposition of the standard sphere without introducing 3-handles then the corresponding presentation of the trivial group given by the handle decomposition is Andrews-Curtis trivial.

The Andrews-Curtis conjecture (which is open) says that every balanced presentation of the trivial group is Andrews-Curtis trivial. This conjecture is believed to be false, many potential counterexamples are known. Probably the presentations given by the handle decompositions of the Cappell-Shaneson spheres are counterexamples. The original work of Gompf contains more information.

R. Gompf, Killing the Akbulut-Kirby 4-sphere, with relevance to the Andrews-Curtis and Schoenflies problems. Topology 30 (1991), 97–115.

  • $\begingroup$ I'm aware of those examples in Gompf's paper (More Cappell-Shaneson spheres are standard), but I don't think all the Cappell-Shaneson spheres have been described in terms of handlebodies, only an infinite family of them. And I don't see explicit handlebody decompositions in this paper, nor does Gompf's method use handlebody manipulations here (only in previous work). Correct me if I'm wrong. $\endgroup$ Sep 16, 2018 at 14:57
  • $\begingroup$ So as for the handlebody moves without 3-handles, it's only hard because Andrews-Curtis is (currently) hard. Thanks for pointing that out, I glossed over that part of the literature. Perhaps the Cappell-Shaneson spheres are not counterexamples! $\endgroup$ Sep 16, 2018 at 14:59
  • 1
    $\begingroup$ @ Chris Gerig: You are right. Gompf does not use Kirby calculus in "More Cappell-Shaneson spheres are standard". And I do not know any source where diagrams of all Cappell-Shaneson spheres are written down. That is why I sad it is not an answer. And as I said it is also conjectured that all Cappell-Shaneson spheres are standard. So even if you could do the same as Akbulut and Kirby did to get a Kirby diagram for some Cappell-Shaneson sphere not covered by Gompf's result, it will be probably just a complicated Kirby diagram of the standard sphere. $\endgroup$
    – Marc Kegel
    Sep 19, 2018 at 11:48

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