Timeline for What is known about the distribution of average edge-degrees for 3-manifold triangulations (with the number of 3-simplices less than a fixed constant)
Current License: CC BY-SA 4.0
13 events
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| Dec 15, 2022 at 17:12 | history | edited | Glorfindel | CC BY-SA 4.0 |
broken link fixed, cf. https://meta.mathoverflow.net/q/5301/70594
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| Jul 23, 2012 at 3:29 | history | edited | Aaron Trout | CC BY-SA 3.0 |
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| Jul 23, 2012 at 2:23 | vote | accept | Aaron Trout | ||
| Jul 22, 2012 at 19:29 | answer | added | Henry Segerman | timeline score: 6 | |
| Jul 22, 2012 at 18:21 | comment | added | Aaron Trout | @Bruno Martelli: Thanks for the comment! I'd heard of the super/non-super exponential growth question while reading some papers on simplicial quantum gravity. (My question is related to the distribution of Einstien-Hilbert-Regge actions for triangulations with unit length edges.) I had hoped that someone knew of results concerning the proportion of triangulations for which $μ(T)$ lies in some fixed interval, say $[5.1−\epsilon,5.1+\epsilon]$, without necessarily understanding the total number of triangulations. Since even the total number is so elusive, perhaps this is just too difficult. | |
| Jul 22, 2012 at 10:11 | comment | added | Bruno Martelli | It is typically impossible to have any asymptotic on triangulations because we don't even know if the number of triangulations of a 3-manifold is exponential or super-esponential. The average degree should be linked to the ratio between the number of vertices and the number of simplexes in the triangulations. My guess is that a random triangulation has much fewer vertices than simplexes and hence the average edge-degree tends to 6, but I have no arguments. | |
| Jul 22, 2012 at 3:32 | comment | added | Aaron Trout | @Henry Segerman: Thanks Henry! I also know of a census by Lutz (arxiv.org/abs/math/0604018) of all combinatorial 3-manifolds with at most 10 vertices, but I hadn't heard of Burton's work. I'm more focused on the large $K$ cases but it would be wonderful to know exact answers for small K and at least one $M$! | |
| Jul 22, 2012 at 2:56 | comment | added | Henry Segerman | Ben Burton has a census of all triangulations of $S^3$ up to something like 11 tetrahedra. So it would be possible to calculate directly for these cases. I'll see what I can do. | |
| Jul 22, 2012 at 1:43 | history | edited | Aaron Trout | CC BY-SA 3.0 |
edited title
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| Jul 22, 2012 at 1:33 | history | edited | Aaron Trout | CC BY-SA 3.0 |
edited body; edited title
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| Jul 22, 2012 at 1:24 | history | edited | Aaron Trout | CC BY-SA 3.0 |
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| Jul 21, 2012 at 19:06 | history | edited | Aaron Trout | CC BY-SA 3.0 |
added 235 characters in body; edited title; edited title; edited title
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| Jul 21, 2012 at 18:49 | history | asked | Aaron Trout | CC BY-SA 3.0 |