23
$\begingroup$

The problem of counting combinatorial three-spheres with $N$ simplices has implications for some partition functions in physics (see a paper by Benedetti and Ziegler for more background and references). To add some context, let's first make the observation that there are roughly order $C^{N\log N}$ combinatorial three-manifolds on $N$ simplices. Now there is a partition function in quantum gravity whose convergence depends on knowing that the number of combinatorial three-spheres on $N$-simplices is bounded by $C^N$ for some $C<\infty$. According to the paper by Benedetti and Ziegler, providing such a bound is an open problem.

One could ask whether a stronger property is true, namely whether the number of combinatorial integer homology three-spheres on $N$ simplices is bounded by $C^N$ for some $C<\infty$. Is this strengthened conjecture known to be false? (certainly it can't be known to be true, since the weaker statement about bounding the number of genuine three-spheres is still open).

Improve this question
$\endgroup$
5
  • 3
    $\begingroup$ 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. $\endgroup$
    – Ryan Budney
    Commented Feb 3, 2012 at 2:14
  • 1
    $\begingroup$ 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). $\endgroup$
    – Igor Rivin
    Commented Feb 3, 2012 at 3:19
  • 6
    $\begingroup$ @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. $\endgroup$
    – Bruno Martelli
    Commented Feb 3, 2012 at 12:03
  • 1
    $\begingroup$ ... 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. $\endgroup$
    – Bruno Martelli
    Commented Feb 3, 2012 at 12:04
  • 2
    $\begingroup$ 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. $\endgroup$
    – Neil Hoffman
    Commented Nov 23, 2012 at 17:40

0

Reset to default

You must log in to answer this question.