Timeline for Lower bound on number of tetrahedra needed to triangulate a knot complement
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Nov 20, 2010 at 15:49 | vote | accept | Mark Bell | ||
| Nov 16, 2010 at 6:50 | answer | added | Ian Agol | timeline score: 14 | |
| Nov 15, 2010 at 23:00 | history | edited | Mark Bell | CC BY-SA 2.5 |
rephrased definition of n
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| Nov 15, 2010 at 22:58 | comment | added | Mark Bell | Yes Ryan you are correct, by number of crossings of a knot K I mean the minimal number of crossings of ANY diagram of K. I'll change the question to reflect this. | |
| Nov 15, 2010 at 21:46 | comment | added | Ryan Budney | And you can't have a lower bound of the number of tetrahedra needed in terms of $n$ since you can always unneccessarily complicate your knot diagram. Perhaps you want $n$ to be the minimal number of crossings in a diagram for the knot? | |
| Nov 15, 2010 at 21:45 | comment | added | Ryan Budney | Point of clarification -- do you want your triangulation to be a hyperbolic ideal triangulation? I think it's an open problem as to whether or not cusped hyperbolic manifolds admit hyperbolic ideal triangulations, isn't it? They have cusped polyhedral (Epstein-Penner) decompositions but they're generally not triangulations. | |
| Nov 15, 2010 at 21:42 | comment | added | Ryan Budney | There's an upper bound coming from the number of vertices in a planar knot diagram -- this comes from the algorithm SnapPea uses to construct a topological ideal triangulation of the complement. You can then subdivide appropriately to construct a proper triangulation. I believe the number of tetrahedra needed should be linear in n although I haven't thought it through precisely. | |
| Nov 15, 2010 at 20:18 | history | asked | Mark Bell | CC BY-SA 2.5 |