Mike Freedman on SPC4 April 19, 2011Posted by Scott Morrison in low-dimensional topology, Poincaré conjecture.
Mike Freedman gave a talk last week at Berkeley titled “(Still) thinking about the smooth 4-dimensional Poincare conjecture”, and I’d like to try and relate the main idea.
Back in 2009 Mike, Bob Gompf, Kevin Walker and I wrote a paper “Man and machine thinking about the smooth 4-dimensional Poincare conjecture”, in which we discussed various equivalent and stronger conjectures with more of a “three manifold flavour”, as well as proposing a way to show that certain potential counterexamples, the Cappell-Shaneson spheres, really were counterexamples, using Khovanov homology.
As it turned out, for the Cappell-Shaneson spheres that our computers could cope with Khovanov homology didn’t provide an obstruction. Shortly thereafter, Selman Akbulut posted a short paper showing that a certain integer family of CS spheres (including the two examples we considered) were all in fact standard, and not long after that Bob Gompf killed off some more. In fact, there’s been a whole flurry of work on SPC4 recently: Nash proposing some more counterexamples, Akbulut killing these off too, Akbulut proving a very general class of 4-spheres are standard, along with ,  and . There’s also a paper on Property 2R by Gompf, Scharlemann and Thompson.
Even though recent progress means that there are far far fewer potential counterexamples to SPC4 available, Mike has come up with another idea for detecting counterexamples! Essentially, from each proposed counterexample to the Andrews-Curtis conjecture (below), one can produce a homotopy 4-spheres. This part of the story is old news, but I’ll go over it carefully below. Mike points out that this construction also provides a family of embedded homology 3-spheres. Now any 3-manifold embeds in , but not every 3-manifold embeds in (e.g. lens spaces, but see more below). The hope now is to show that one of these homology 3-spheres can not embed in the standard , and thus that the homotopy 4-sphere it sits in must be exotic. Mike gave a condition on a 3-manifold Y that exactly determines whether it embeds in , and talked about his ideas towards making this condition an effective test.
So, what is the Andrews-Curtis conjecture? A balanced presentation of a group is simply a presentation with equal numbers of generators and relations. The Andrews-Curtis moves on a balanced presentation are
- “1-handle slides”: replace a pair of generators by .
- “2-handle slides”: replace a pair of relations by (here is any word in the generators).
- “handle cancellation”: add a generator along with a new relation killing it
The Andrews-Curtis conjecture says that any balanced presentation of the trivial group can be transformed by Andrews-Curtis moves to the trivial presentation. (There’s a stronger version that says handle cancellation isn’t even needed.) Despite the name, it’s widely expected to be false.
Before continuing, I should explain the funny names I’ve given the moves. Given any presentation of a group, we can build a two-complex whose fundamental group is the group: just take some 1-handles for the generators, and attach 2-handles killing the relations. The names I’ve given the moves correspond to the appropriate geometric operations on this 2-complex.
As an example, consider the presentation . It’s easy to check that this is the trivial group: , so . Thus , and . Even though that was easy, no one has found a sequence of Andrews-Curtis move, even after having looked extremely hard! (e.g. this paper) This is just one case of the family of proposed counterexamples that appears in the paper “A potential smooth counterexample in dimension 4 to the Poincaré conjecture, the Schoenflies conjecture, and the Andrews-Curtis conjecture”, by Selman Akbulut and Rob Kirby (Topology 24 (1985), 375–390). So far the Andrews-Curtis conjecture is just a problem in group theory, but the title of that paper should clue you in that we’re about to construct some manifolds!
Given a balanced presentation , we can build a 5-manifold —just like we built the 2-complex above, start with a 5-ball, attach some 5-dimensional 1-handles for the generators, and then attach some 5-dimensional 2-handles for the relations. To attach the two handles, we actually need to pick links in the boundary of the 0- and 1-handles representing in the relation, but since these boundaries are 4-dimensional there are actually no choices to make. If we started with a presentation of the trivial group, then is a homotopy 4-sphere. Moreover, if the presentation is a counterexample to the Andrews-Curtis conjecture, then perhaps this homotopy 4-sphere is a good candidate counterexample to SPC4: you can’t make it standard just by handle slides corresponding to some Andrews-Curtis moves! Of course, you have more freedom to simplify presentations of 4-manifolds, so just having a counterexample to AC doesn’t ensure a counterexample to SPC4.
As it turns out, Bob Gompf subsequently showed that is in fact the standard 4-sphere (by introducing a cancelling pair of 2- and 3-handles!). Nevertheless all the higher values of n are still open.
We can alternatively build in a different way. Start with a 4-ball, attached some 4-dimensional 1-handles for the generators, and now pick some links realizing the relations, and attach 4-dimensional 2-handles along these. This produces a 4-manifold , which genuinely depends on , but whose double (the boundary of ) is just our homotopy 4-sphere . This way of building the homotopy 4-sphere gives us something extra; a 3-manifold sitting inside . In fact, this 3-manifold is a homology sphere.
The challenge now is to pick some counterexample to the AC conjecture, along with a corresponding link, and then to prove that the resulting homology sphere can not be embedded in the standard 4-sphere.
Now Mike wasn’t claiming he knew how to do this; but he did explain a relatively easy theorem giving a precise characterization of when a 3-manifold embeds in the standard 4-sphere, which has a very “3-manifold flavour”. This was
Theorem. A 3-manifold embeds in if and only if it has an Heegaard diagram and an embedding , such that is an unlink, and is an unlink.
Mike calls this a doubly unlinked embedded Heegaard diagram. The proof of this theorem is via ambient Morse theory, but I won’t try to give the details here. In fact, every homology 3-sphere has an embedded Heegaard diagram which is doubly null, meaning that the linking matrices for and vanish. Of course, actually using this theorem to show that doesn’t embed would involve a lot of work. It seems that we’d need to consider all possible embeddings of , as well as all possible systems of curves and . I think Mike has some ideas about this (if you go “far out in the mapping class group”, it should be impossible that the images of the curves are unlinked), but I don’t think I could reproduce those ideas usefully here. Perhaps some of our readers can comment on how plausible this seems, and maybe if we’re lucky Mike will turn up too and say some more.