can you fool SnapPea?

Question

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/

score 13 · Answer 1 · edited Apr 17 '22 at 08:25

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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 https://doi.org/10.1007/BF03025227 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.

edited Apr 17 '22 at 08:25

The Amplitwist

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answered Nov 10 '09 at 23:01

Sam Nead

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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
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I will be very happy to sell you as much as you wish to buy. I also have some lightly used Seifert fibred spaces.
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

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@Ryan Budney: I believe you're referring to Mel Slugbate; cf. http://euclid.colorado.edu/~jnc/MelSlugbate.html – Robert Haraway Feb 13 '13 at 02:47

@RobertHaraway: right, that's who I was thinking of. I contacted Mel a few years ago and confirmed he is not Mr. Nead. – Ryan Budney Feb 13 '16 at 18:03

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I have vivid memories from attending a real estate marketing seminar given by the great Mr Slugbate. One pearl of wisdom: if you live in the Gieseking manifold (very affordable, by the way), and if you stroll through a face of the two-skeleton every time you drink a beer, then you can figure out the parity of the number of beers by checking if your heart is on the left or on the right. – Sam Nead Feb 14 '16 at 11:10

score 5 · Answer 2 · answered Nov 10 '09 at 20:51

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.

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

score 4 · Answer 3 · answered Nov 10 '09 at 21:40

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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.)

answered Nov 10 '09 at 21:40

Autumn Kent

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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
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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
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But what is SnapPea really thinking? Haha. – Autumn Kent Nov 10 '09 at 21:51

score 3 · Answer 4 · edited Apr 17 '22 at 09:27

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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. (Wayback Machine)

edited Apr 17 '22 at 09:27

Martin Sleziak

4,608

answered Nov 12 '09 at 00:50

Autumn Kent

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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 01:08
Oh, I missed his edit. For the hell of it, you could try Haken's unknots. Question edited to add link to them. – Autumn Kent Nov 12 '09 at 01:10
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 01:14
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? http://www.math.uic.edu/~t3m/SnapPy/doc/ – Ryan Budney Nov 12 '09 at 01:18
Yeah, I probably should have said "I" couldn't handle it. I wonder if SnapPea can "see" the band when its working. – Autumn Kent Nov 12 '09 at 01:21
Yeah, I did really mean the knew link editor. It's much worse than the one for OS 9. I haven't tried out SnapPy yet. – Autumn Kent Nov 12 '09 at 01:21
I've thought a bit about how to enter the Haken examples without going blind. The best idea I can come up with is to get two projectors and project them both onto the same screen. One projector shows the unknot you are interested in and the other is a blank SnapPea window. Then you "trace" the knot and Bob's your uncle. – Sam Nead Nov 12 '09 at 01:29
SnapPea's triangulation simplification code is pretty brilliant. It's one thing I've never really looked into in much detail. But it's far more effective than you'd think such code could ever be. – Ryan Budney Nov 12 '09 at 01:29
You could also print Haken's example on tracing paper and then tape it to your monitor. :) – Autumn Kent Nov 12 '09 at 01:36
Richard+Ryan: By the way, supposing you enter one of those huge diagrams into SnapPea, and SnapPea says it has positive volume. Will you conclude (A) SnapPea is broken or (B) Haken was wrong or (C) "Crud, I made a mistake entering the knot." ?? – Sam Nead Nov 12 '09 at 01:38
(C) is a pretty common mistake on my end. – Ryan Budney Nov 12 '09 at 01:48
Ok, I entered the third of the knots you linked to. It is a nice construction: the knot is basically is a ribbon and the ribbon consistently twists right. (Except in one place!) Anyway, SnapPea took slightly more time than usual but still less than a second. – Sam Nead Nov 12 '09 at 02:16

score 0 · Answer 5 · answered Nov 12 '09 at 00:19

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!

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 00:28
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

can you fool SnapPea?

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