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Feb 21, 2014 at 16:51 comment added Ilya Kofman We showed a universal bound for the one-cusped case (i.e. knots) in "The 500 simplest hyperbolic knots," arXiv:1307.4439.
Jan 6, 2014 at 10:43 vote accept shestipalov
Dec 31, 2013 at 13:56 vote accept shestipalov
Jan 6, 2014 at 10:43
Dec 31, 2013 at 11:30 comment added Bruno Martelli I don't think you can recover a bound only from the volume $V$, you need to know the manifold. The cusp shapes tell you easily which curves have norm bigger than 6. With SnapPy simply use the function get_cusp_shape().
Dec 30, 2013 at 12:51 comment added shestipalov So perhaps a better question is: given a hyperbolic link of $n$ components which has volume $V$, can we explicitly find a constant $c(n,V)$ (so depending on the concrete manifold only minimally), such that filling $n-1$ components with slopes $p_i/q_i$ s.t. the norm of $(p_i,q_i)$ is bigger than this constant, we get a hyperbolic manifold with one cusp? Sorry for this long comment-question.
Dec 30, 2013 at 12:50 comment added shestipalov My problem is (before I do learn hyperbolic geometry) that I have infinite families of knots, and all I can say at the moment, that infinitely many of them are hyperbolic. But what I would really like to say is given that parameters (that come from slopes) are bigger than a concrete constant, then the knots are hyperbolic. SnapPy can just verify this for a finite number of them.
Dec 30, 2013 at 12:50 comment added shestipalov Thanks a lot for your answer! Somehow I got an incorrect idea that denominators were more important - thanks for clarifying that. Yes, I have SnapPy and seen references to Thurston's and Agol-Lackenby's theorems, but unfortunately have not yet had time to study this properly. So unfortunately I can't really understand what you are saying about the cusp shapes - that's something for me to learn, thanks for pointing this out.
Dec 29, 2013 at 13:23 history edited Bruno Martelli CC BY-SA 3.0
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Dec 29, 2013 at 12:55 history answered Bruno Martelli CC BY-SA 3.0