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Is there some general description of all homology 3-spheres?

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4 Answers 4

Certainly. There's a general description of all compact 3-manifolds now that geometrization is about.

So for homology 3-sphere's you have the essentially unique connect sum decomposition into primes.

A prime homology 3-sphere has unique splice decomposition (Larry Siebenmann's terminology). The splice decomposition is just a convienient way of encoding the JSJ-decomposition. The tori of the JSJ-decomposition cut the manifold into components that are atoroidal, so you form a graph corresponding to these components (as vertices) and the tori as edges.

The splice decomposition you can think of as tree where the vertices are decorated by pairs (M,L) where M is a homology 3-sphere and L is a link in M such that M \ L is an atoroidal manifold.

By geometrization there's not many candidates for pairs (M,L). The seifert-fibred homology spheres that come up this way are the Brieskorn spheres, in that case L will be a collection of fibres in the Seifert fibering. Or the pair (M,L) could be a hyperbolic link in a homology sphere. That's a pretty big class of manifolds for which there aren't quite as compact a description, compared to, say, Brieskorn spheres.

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An excellent example of how this decomposition can be used to say something about general homology 3-spheres: front.math.ucdavis.edu/0606.5220 –  Ryan Budney Nov 10 '09 at 2:10

On the other hand, there is no particular classification of hyperbolic homology 3-spheres, much less hyperbolic links in homology 3-spheres, other than in general terms that they all come from hyperbolic groups.

For instance, part of geometrization establishes that if a finite group acts freely on $S^3$, then it is equivalent to an action by isometries on a round $S^3$ and is a subgroup of $\mathrm{SO}(4)$. Before geometrization, Milnor and Lee established severe restrictions on how a finite group $G$ can act freely on any homology 3-sphere, with the case of $S^3$ particularly in mind. Either $G$ is a spherical group, or it is one other family that hasn't been excluded. For all we know, if $G$ acts freely on any homology 3-sphere, then it acts on $S^3$ too. I think that this is still an open problem, and geometrization by itself doesn't settle it.

The working description homology 3-spheres for many purposes, in particular quantum topological invariants, is rather different. In practice, a homology 3-sphere is often given by surgery on a link in $S^3$ (or in some other homology 3-sphere) whose matrix has determinant 1. The big drawback of course is that the description is far from unique.

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What geometrization does do for you in this case (as in the paper I included a link for) is reduce problems like this to questions about symmetry groups of hyperbolic links in homology spheres. –  Ryan Budney Nov 10 '09 at 2:41

A nice historical note - Dehn observed that if M and N are knot complements and if you glue M to N switching meridian and longitude then the result is a homology sphere. Of course this is a special case of what Ryan was saying.

Another nice fact: the Poincare homology sphere is the only one with finite fundamental group.

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Another way to represent homology spheres is to take a Heegaard splitting for $S^3$, cut and reglue by an element of the Torelli group. This is not canonical, but any two Heegaard splittings are equivalent after some number of stabilizations. If you wanted to enumerate every homology sphere, you could list elements of the Torelli group, and construct 3-manifolds, then throw away repeats by using some solution to the homeomorphism problem for 3-manifolds. This is not really feasible to carry out in practice, but is one way to give a "general description" of homology spheres at least in theory, by giving a recursive enumeration of them.

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