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This is my first question on mathoverflow! It relates to a project I'm undertaking with a student.

Work by Tamura (extending results by Luo and Stong) shows that for any closed 3-manifold $M$ and any rational number $4.5 < r < 6$ there is a triangulation $T$ of $M$ for which that the average edge-degree $\mu(T)$ is $r$. Here the degree of an edge $e$ is the number of 3-simplices having $e$ as a face.

Now, fix a closed 3-manifold $M$ and consider the (necessarily finite) set of all triangulation of $M$ containing at most $K$ 3-simplices. Call this set $\mathcal{T}_K(M)$.

QUESTION: Does anyone know of results concerning the "distribution" of the average edge-degree $\mu(T)$ for $T\in \mathcal{T}_K(M)$? That is, what is known about the fraction of triangulations from $\mathcal{T}_K(M)$ for which the edge-degree lies in a given small interval $[r - \epsilon, r+\epsilon] \subset (4.5,6)$? Results concerning any closed 3-manifold $M$ would be fine.

I suspect this is extremely difficult to answer in a precise way. However, I'd be very interested in knowing any asymptotic ($K\rightarrow \infty$) or approximate results as well. Thanks for your help!

NOTE: For the project we plan to use the Metropolis algorithm to sample from $\mathcal{T}_K(M)$ where $M$ is the 3-torus. Using these samples we hope to empirically estimate the distribution in question.

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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. –  Henry Segerman Jul 22 '12 at 2:56
    
@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 $M$! –  Aaron Trout Jul 22 '12 at 3:32
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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. –  Bruno Martelli Jul 22 '12 at 10:11
    
@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. –  Aaron Trout Jul 22 '12 at 18:21
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1 Answer

up vote 6 down vote accepted

Below is the data from Ben Burton's census of triangulations of $S^3$ with up to 9 tetrahedra. An Euler characteristic argument shows that the average edge degree is equal to $6\left(\frac{T}{V+T}\right)$, where $T$ is the number of tetrahedra and $V$ is the number of vertices, so the numbers below are all of that form. The data are given in (average edge degree, frequency) pairs. As Bruno guessed, it looks like the distribution weights heavily towards 6.

1 tetrahedron:

  • (2.0000000000000000, 1)
  • (3.0000000000000000, 1)

2 tetrahedra:

  • (2.0000000000000000, 1)
  • (2.3999999999999999, 1)
  • (3.0000000000000000, 1)
  • (4.0000000000000000, 3)

3 tetrahedra:

  • (2.5714285714285716, 1)
  • (3.0000000000000000, 2)
  • (3.6000000000000001, 9)
  • (4.5000000000000000, 20)

4 tetrahedra:

  • (2.6666666666666665, 2)
  • (3.0000000000000000, 4)
  • (3.4285714285714284, 16)
  • (4.0000000000000000, 48)
  • (4.7999999999999998, 128)

5 tetrahedra:

  • (2.7272727272727271, 1)
  • (3.0000000000000000, 4)
  • (3.3333333333333335, 23)
  • (3.7500000000000000, 110)
  • (4.2857142857142856, 468)
  • (5.0000000000000000, 1297)

6 tetrahedra:

  • (2.7692307692307692, 1)
  • (3.0000000000000000, 5)
  • (3.2727272727272729, 36)
  • (3.6000000000000001, 199)
  • (4.0000000000000000, 1103)
  • (4.5000000000000000, 4931)
  • (5.1428571428571432, 13660)

7 tetrahedra:

  • (2.7999999999999998, 1)
  • (3.0000000000000000, 3)
  • (3.2307692307692308, 39)
  • (3.5000000000000000, 301)
  • (3.8181818181818183, 2186)
  • (4.2000000000000002, 13380)
  • (4.6666666666666667, 62657)
  • (5.2500000000000000, 169077)

8 tetrahedra:

  • (2.8235294117647061, 1)
  • (3.0000000000000000, 3)
  • (3.2000000000000002, 51)
  • (3.4285714285714284, 446)
  • (3.6923076923076925, 3870)
  • (4.0000000000000000, 28826)
  • (4.3636363636363633, 180128)
  • (4.7999999999999998, 829753)
  • (5.3333333333333333, 2142197)

9 tetrahedra:

  • (2.8421052631578947, 1)
  • (3.0000000000000000, 3)
  • (3.1764705882352939, 50)
  • (3.3750000000000000, 567)
  • (3.6000000000000001, 6046)
  • (3.8571428571428572, 54876)
  • (4.1538461538461542, 422860)
  • (4.5000000000000000, 2612407)
  • (4.9090909090909092, 11673471)
  • (5.4000000000000004, 28691150)
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If one was interested, Ben also has a census of all 3-manifold triangulations (not just 3-spheres) up to 11 tetrahedra. It's unfortunately too enormous for a server to distribute publically but I believe it's available. This is different from the 11 tetrahedron census in Regina which only contains one vertex triangulations. –  Ryan Budney Jul 23 '12 at 0:56
    
@Henry Segerman: Thanks! This helps give an idea for what to expect from larger triangulations. –  Aaron Trout Jul 23 '12 at 2:23
    
@Ryan Budney: Thanks! I am certainly interested. I'd heard about Regina but somehow hadn't realized people used it to create censuses. I believe I found what you're referring to at regina.sourceforge.net/data.html. –  Aaron Trout Jul 23 '12 at 2:31
    
Yes , those are the minimal triangulations. So for example none of those triangulations support a Pachner a 4->1 move. –  Ryan Budney Jul 23 '12 at 3:33
    
@Ryan Budney: Thanks for pointing that out! I hadn't noticed the "minimal" adjective there and spoke too quickly. Perhaps I will contact Ben Burton about obtaining a copy of the complete census. –  Aaron Trout Jul 23 '12 at 3:45
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