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I've recently come across some mid-sized 3-manifolds that I think are likely hyperbolic, but SnapPea has some trouble with them. This is related to my previous question

can you fool SnapPea?

but in this case I'm dealing with closed, orientable 3-manifolds instead of knot and link complements.

The 3-manifolds I've come across have 11, 12 and 13 tetrahedra in their triangulations. One of them SnapPea finds a solution to the gluing equations but it has "negatively oriented tetrahedra". Does this mean what I think it means -- that once you've put the geometric structure on the tetrahedra, you have a tetrahedron folded-over? If you view the gluing equations from the upper half-space model, they say that the sum of a bunch of angles should be $2\pi$. Is this the case where one of those angles is negative?

The other two triangulations SnapPea finds geometric structures with degenerate tetrahedra. In the upper half-space model this is where one of the angles is zero, I believe. Is there a way to fix this, so that I could get a Dirichlet domain, drill and fill, etc? This is my primary question.

Here are the triangulations, both in SnapPea format and Regina format.

Regina file, all triangulations

11-tet, SnapPea

12-tet, SnapPea

13-tet, SnapPea

With the 12-tetrahedron example, SnapPea can generate a Dirichlet Domain. This one has what looks almost like bevelled-edges. Is there a way to find a better basepoint to grow the Dirichlet domain from?

12-tet Dirichlet domain

edit: After working a bit with Nathan's answer, I've identified these three manifolds and got a little closer to understanding how to work SnapPea to maximal advantage.

tri11, as Nathan mentioned, is hyperbolic. It has a fairly pretty Dirichlet domain.

Tri11 Dirichlet Domain

Another common name for this manifold would be the 0-surgery on the 2-component link $7a_6$ (in the Thistlethwaite table). Similarly tri12 can be identified as Nathan says.

After playing around with Regina a bit I found an incompressible torus in tri13 that splits tri13 into the union of an orientable $I$-bundle over the Klein bottle and a figure-8 complement. So this answers the core of my question.

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(I hope my edit in the title is not too controversial!) –  Mariano Suárez-Alvarez Nov 14 '12 at 23:59
    
Looks good to me. –  Ryan Budney Nov 15 '12 at 0:08
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Only a Spanish-speaking person would make such an edit. Being one myself, I'm loving it! –  Alberto García-Raboso Nov 15 '12 at 4:04
    
T11 and T13 continue to have degenerate tetrahedra after using the simplify and randomize functions of SnapPy. It's my experience that this means the manifolds are not hyperbolic and I start looking for incompressible spheres, tori or a Seifert fibered structure. Experimenting in the same way with T12 seems to stably give a structure with negatively oriented tetrahedra. So it could be hyperbolic or not. These manifolds come from surgery on a hyperbolic manifold right? If so, which one? Have you checked the lengths of these curves and tried applying a 6-theorem argument? –  Neil Hoffman Nov 15 '12 at 10:33
    
T11 and T12 are prime, and have no incompressible tori. T13 might have incompressible tori... –  Ryan Budney Nov 15 '12 at 22:14
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2 Answers

up vote 9 down vote accepted

The first two of these manifolds are hyperbolic. The problem, I suspect, is that they are being specified as Dehn fillings on 1-cusped manifolds with essential tori. So one needs to pick a different knot in the closed manifold to actually see the hyperbolic structure. One way to force SnapPy to do this is to create a 1-vertex non-ideal triangulation and then use one of the edges there as the knot. For example:

sage: M = Manifold('tri11.txt')
sage: N = Manifold(M.filled_triangulation()._to_string())
sage: N.solution_type()
'all tetrahedra positively oriented'
sage: closed = snappy.OrientableClosedCensus()
sage: closed.identify(N)
m038(1,2)

The tri12 manifold is "m032(5,2)". I strongly suspect the tri13 manifold is not hyperbolic. Instead, I suspect it has a nontrivial JSJ splitting where one piece is the figure-8 complement, since sometimes the volume appears to be 2.0988...

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Thanks Nathan. I think I've found the incompressible tori in the 3rd manifold. So I should know the answer to that question soon. –  Ryan Budney Nov 16 '12 at 6:19
    
@Nathan: What does "N=Manifold(M.filled_triangulation()._to_string())" do that N.randomize() and N.simplify() don't? –  Neil Hoffman Nov 16 '12 at 11:56
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@Neil: The methods "randomize" and "simplify" preserve the underlying Dehn filling description of the closed manifold, ans so correspond to retriangulating the fixed cusped manifold involved. M.filled_triangulation is a non-ideal 1-vertex triangulation of M which SnapPy can't really work with directly, at least for finding geometric structures. The hackish code Manifold(blah._to_string()) reopens that manifold in SnapPy, which requires it to choose a Dehn filling description of the manifold; it does this by picking at random an edge of the finite triangulation. –  Nathan Dunfield Nov 16 '12 at 17:31
    
Hi Nathan, strangely if I attempt your filled_triangulation()._to_string() construction in SnapPy, I continue to get negatively-oriented tetrahedra in my solution_type(). I'm using version 1.7 of SnapPy in Ubuntu. But after randomizing the triangulation and repeating, I get your outcome on the nose. Thanks. –  Ryan Budney Nov 16 '12 at 22:22
    
I can also duplicate your outcome on tri12, identifying it as m032(5,2) but my solution_type always has negatively oriented tetrahedra. –  Ryan Budney Nov 16 '12 at 22:43
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That's just a long comment about tri13. I don't have Regina now, but you can get some informations by looking only at the 1-skeleton $G$ of the closed triangulation for tri13. Among the (experimentally) minimal triangulations for tri13, i.e. the ones that you find by trying to simplify it as much as you (or Regina) can, you probably find one where the 1-skeleton $G$ contains a pair of disconnecting edges, which split the $G$ into two parts $G_1$ and $G_2$. In any experimentally minimal triangulation such a pair of edges is always transverse to a torus $T$, which cut the manifold $M$ into two parts $M_1$ and $M_2$, containing $G_1$ and $G_2$ correspondingly.

According to Nathan's suggestion, the torus $T$ should separate a figure-8 knot complement (or its sibling) $M_1$ and a small Seifert space $M_2$. You can see this from the 1-skeleton $G$: the portion $G_1$ must have at least 8 vertices to be hyperbolic, and it has probably 8 or 9 (corresponding to the figure-8 knot sibling and the figure-8 knot complement), and the other portion $G_2$ has of course 5 or 4 vertices. If $G_2$ is layered, then the torus is compressible and you have a Dehn fillig of $M_1$. If (as we expect) $G_2$ is not layered, then it is certainly not a solid torus (minimal triangulations of solid tori with small number of tetrahedra must be layered, as far as I remember).

Summing up, I think that:

if a triangulation for $M$ is experimentally minimal with a reasonable number of tetrahedra and its 1-skeleton $G$ decomposes into two pieces $G_1$ and $G_2$, noone of which is layered, then you can conclude theoretically that it is not hyperbolic.

The numer of tetrahedra must be reasonable, because minimal triangulations of solid tori are classified only for small number of tetrahedra (I don't know how many...)

A fool-proof alternative is to use Matveev's Recognizer, which tells you exactly the JSJ decomposition of the manifold.

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Thanks Bruno. Yes, I've identified the 3rd triangulation now. There's a single torus in its JSJ-decomposition, which separates the manifold into a figure-8 complement union the orientable $I$-bundle over the Klein bottle. So this is one of those manifolds where the "geometric" decomposition isn't the same as, say, the Hatcher 3-manifold lecture notes version of the JSJ-decomposition. –  Ryan Budney Nov 16 '12 at 22:05
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