Timeline for Are there lots of integer homology three-spheres?
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| Nov 23, 2012 at 17:40 | comment | added | Neil Hoffman | On the question's last point, I am not sure I believe the problem of bounding the number of 3-spheres is easier than bounding the number of homology 3 spheres. In fact, I would venture the opposite is probably true. For instance, describing all of the Dehn fillings on a knot complement $S^3-K$ that are integral homology spheres is a relatively easy problem, whereas showing that a non-trivial prime knot cannot admit two $S^3$ fillings is much harder. Also, a computer can recognize a finite presentation for a trivial abelian group more easily that a trivial 3-manifold group. | |
| Feb 3, 2012 at 12:04 | comment | added | Bruno Martelli | ... and each graph manifold can be triangulated with (say) $100N$ tetrahedra, where $N$ is the number of vertices of the graph. Therefore you get $C^{N\log N}$ distinct manifolds triangulated with $N$ tetrahedra. So a fortiori you get $C^{N\log N}$ distinct triangulations of closed manifolds. | |
| Feb 3, 2012 at 12:03 | comment | added | Bruno Martelli | @Igor: The number of distinct 4-valent graphs is of order $C^{N\log N}$. The number of triangulations with fixed dual graph is "only" exponential, of order $6^{2N}$. Therefore the number of distinct triangulations is at most $C^{N\log N}$. On the other hand, for every 4-valent graph you can construct a graph-manifold by putting a fixed manifold at each vertex (say, the sphere with four holes times $S^1$) and by gluing at each edge by the map which reverses longitudes and meridians. Distinct graphs yield distinct graph manifolds. | |
| Feb 3, 2012 at 3:19 | comment | added | Igor Rivin | Why are the "roughly order $C^{N \log N}$ combinatorial three-manifolds on $N$ simplices"? That does not seem obvious (I will believe that there is an upper bound of this sort). | |
| Feb 3, 2012 at 2:14 | comment | added | Ryan Budney | I think triangulations are fairly biased towards seeing homology spheres, especially homology spheres with simple JSJ-decompositions. Roughly speaking, it takes a lot of tetrahedra to triangulate non-trivial homology classes and incompressible tori. That's not a formal argument but it's what I see when I look at the census of triangulated manifolds. | |
| Feb 3, 2012 at 1:49 | history | asked | John Pardon | CC BY-SA 3.0 |