can you fool SnapPea? - MathOverflow

can you fool SnapPea?

2009-11-10T20:39:04Z

A while back I thought I had some simple knots that "fooled" SnapPea. But I no longer remember those examples, if I ever had them to begin with.

What I'm looking for is a non-hyperbolizable knot or link in S^3 for which SnapPea thinks it finds a hyperbolic structure on the complement. Do you have such an example? I'm interested in the examples that work in SnapPea -- it's fine if Snap or the Harriet Moser criterion "knows" the gluing equations are not satisfied.

edit: to make my question more rigid, can you find an unknot (or a trivial link) for which SnapPea thinks there is a hyperbolic structure?

In case this is all jargon to you, SnapPea is software used primarily for finding and exploring hyperbolic structures on 3-manifolds: http://www.math.uic.edu/~t3m/SnapPy/doc/

Answer by Danny Calegari for can you fool SnapPea?

2009-11-10T20:51:56Z

If you are asking whether SnapPea rigorously certifies the existence of the hyperbolic structures that it "finds", then I think the answer is that it does not. Andrew Casson wrote a couple of programs to hyperbolize closed and cusped 3-manifolds, called "geo" and "cusp" respectively, and rigorously proved that they do not give "false positives" (this is, of course, relative to the correct functioning of the compiler, the computer hardware etc.)

If you are asking whether SnapPea in practice thinks that it has found a hyperbolic structure where none exists (or the structure it finds corresponds to a non-faithful or indiscrete representation), I think it can sometimes be fooled by analytically continuing a family of Dehn fillings, and where what is really being hyperbolized is the image of the manifold under a degree 1 map. Note that this information is from my experience playing with SnapPea ~15 years ago, so it may easily be out of date.

Answer by Richard Kent for can you fool SnapPea?

2009-11-10T21:40:20Z

I remember finding examples like this in graduate school.

Draw the figure eight knot and then draw a parallel pushoff of a meridian.

Snappea gives you a volume for this link. (It should tell you that there are degenerate tetrahedra, though.)

Answer by Sam Nead for can you fool SnapPea?

2009-11-10T23:01:24Z

Before you try to fool SnapPea, remember that you'll almost certainly have to go above 16 (17?) crossings to do so - see http://www.springerlink.com/content/y10185316280vpu4/ for the tale of the tabulation of knots by Hoste and Weeks and, independently, Thistlethwaite. Here is a nice quote: "...our methods for nonalternating knots are not algorithmic. Instead we simply employ a collection of methods that work for $N \leq 16$."

Edit: Ok, I googled "complicated unknot" and found a paper "Hard Unknots and Collapsing Tangles" by Louis H. Kauffman and Sofia Lambropoulou and a thesis "Interactive Topological Drawing" by Robert Glenn Scharein. I went through both and entered the unknots they give into SnapPea. In all cases SnapPea says that the volume is zero and, futhermore, reports that the fundamental group is $\mathbb{Z}$ (one generator, no relators). The unknots they discuss include the Goeritz unknot, Freedman's unknot, and several unknots that require increasing the complexity of the diagram before decreasing. (ie via Reidemeister moves). Another knot that SnapPea handled (~55 crossings) was the one on page 135 of the thesis, which is claimed to defeat KnotPlot.

SnapPea would report the results so quickly that I will conjecture that Newton's method, hyperbolic geometry, etc were not really involved. Instead, I think that SnapPea's retriangulation heuristic "detected" all of these unknots. That is: SnapPea takes the diagram you give it and produces a straightforward triangulation which is linear in terms of the crossing number. It then cleans this triangulation up, getting rid of material vertices and doing 4-1 and 3-2 moves wherever possible. I think that this first step must be getting rid of almost all of the tetrahedra.

Answer by Sam Lewallen for can you fool SnapPea?

2009-11-12T00:19:47Z

So I think this answers a (presumably non-hard) question I just had, but let me make sure: Is it true that there exist non-hyperbolic (non-compact) 3-manifolds, for which there is an ideal triangulation for which the gluing equations are satisfied? (Although NOT by solutions in the upper half plane, since that would of course allow you to glue up the triangulation into a bona-fide hyperbolic manifold).

I thought there was some long-standing technical open question regarding this business, but I can't recall it. It had to do with starting with an arbitrary ideal triangulation, of a non-compact hyperbolic 3-manifold, say, and trying to overcome the problem of degenerate and negatively oriented tetrahedra to use that triangulation, or a related one, to actually construct the hyperbolic structure. Does anyone know what I'm thinking of?

Thanks!

Answer by Richard Kent for can you fool SnapPea?

2009-11-12T00:50:22Z

Have you tried something like this (you probably have):

Take a diagram of a complicated hyperbolic knot and draw a diagram for the Whitehead double. Then change a crossing in the clasp.

I'd like to test this, but my SnapPea can't handle it.

Edit: You could also try Haken's unknots.