# can you fool SnapPea?

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/

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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.

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Are you the guy that's always trying to sell people hyperbolic real-estate? –  Ryan Budney Nov 10 '09 at 23:03
BTW it was your comment about choosing 30 vertices randomly in the plane to create a random knot that got me thinking about this again. In general a 30 vertex knot should have something like 28! crossings? There's got to be a decent probability of things like round-off error for SnapPea's approach to the gluing equations, let alone the problem of triangulating those monster diagrams. –  Ryan Budney Nov 10 '09 at 23:08
1. I will be very happy to sell you as much as you wish to buy. I also have some lightly used Seifert fibred spaces. 2. That wasn't my comment (I don't think) but the round-off error may not be where you think it is. A random knot (for various values of random) will have no short geodesics. So probably the triangulation, while having lots of tetrahedra, has no very flat or very close to degenerate tetrahedra. So you can find the volume with reasonable confidence. What you cannot do is compute the Dirichlet domain: once the volume is large you'll have too many vertices, too close together. –  Sam Nead Nov 10 '09 at 23:34
@Ryan Budney: I believe you're referring to Mel Slugbate; cf. euclid.colorado.edu/~jnc/MelSlugbate.html –  Robert Haraway Feb 13 '13 at 2:47

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.

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I'm asking about something similar to your 2nd case, though I'm interested in knot and link complements -- no Dehn fillings. Moreover I'd like concrete examples to look at. –  Ryan Budney Nov 10 '09 at 20:55

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.)

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I think what you've found is a feature of SnapPea, not a fault. It's giving you the sum of the volumes of the hyperbolic bits in the JSJ-decomposition. In your example the JSJ-decomposition is the union of a figure-8 complement and a connect sum of two Hopf links. –  Ryan Budney Nov 10 '09 at 21:43
In current versions of SnapPea if you submit your example it gives you the volume of the figure-8 knot, but when you request solution_type() it returns "contains degenerate tetrahedra". So SnapPea isn't messing up here. –  Ryan Budney Nov 10 '09 at 21:50
But what is SnapPea really thinking? Haha. –  Richard Kent Nov 10 '09 at 21:51

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.

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Yeah, that's essentially what Mr. Sam Nead and I were discussing in the "probabilistic knot theory" discussion, except instead of Whitehead doubling we were doing a more subtle Borromean splicing operation. I'll try a few more (similar) examples.. –  Ryan Budney Nov 12 '09 at 1:08
Oh, I missed his edit. For the hell of it, you could try Haken's unknots. Question edited to add link to them. –  Richard Kent Nov 12 '09 at 1:10
I just did this with 10_165. SnapPea instantly gives volume zero and fundamental group ZZ. (The procedure is less painful than you might think. First enter your knot. Then draw a push-off -- I used the blackboard framing -- and fix all the crossings using the original as a guide. Then pop open some vertices and make the unclasped clasps.) –  Sam Nead Nov 12 '09 at 1:14
When he says "my SnapPea can't handle it" I think he might be referring to link editor problems. Have you tried the new cross-platform SnapPy, Richard? math.uic.edu/~t3m/SnapPy/doc –  Ryan Budney Nov 12 '09 at 1:18
Yeah, I probably should have said "I" couldn't handle it. I wonder if SnapPea can "see" the band when its working. –  Richard Kent Nov 12 '09 at 1:21

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!

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It depends on what you mean by "gluing equations are satisfied" but if you mean it in the strict exact sense, then the answer is no. If the gluing equations are satisfied, the manifold is hyperbolic, end of story. What I'm interested in is manifolds for which the ideal triangulation is so complicated that SnapPea makes numerical errors due to things like the limited precision of floating-point numbers, and gives a false positive. –  Ryan Budney Nov 12 '09 at 0:28
Perhaps you were thinking of the following: An ideal triangulation is "geometric" if all of the volumes are positive and the gluing equations are satisfied. It is an open question whether all cusped hyperbolic 3-manifolds have geometric ideal triangulations. –  Sam Nead Feb 15 '10 at 19:13