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Every surface can be triangulated in such a way that at most 7 trianlges meet at one vertex. Every surface can be decomposed in squares such that at every vertex at most 5 suqares meet. For surfaces of genus more than 1 this is the low bound.

What happen in higher dimensions, for example for 3 and 4-manifolds, ect...? It should be easy to show that for every dimension $n$ there are numbers $S(n)$ and and $C(n)$ such that every manifold $M^n$ admits a simplicial decomposition with at most $S(n)$ simplexes at every vertex and a cubical decomposition with at most $C(n)$ cubes at every vertex. The refference of Gil below confirms this for $n=3$.

Here are three questions (I suspect they are hard).

1) Can it be proven that $C(n)>2^n$?

2) Can it be proven that $S(n)>\frac{Vol(S^n)}{Vol(\Delta^n)}$, where $\Delta^n$ is the spherical tetrahedron with edge of length $\frac{\pi}{3}$ in the unit sphere $S^n$.

3) Is there any reasonable estimation for $C(n)$ and $S(n)$ from above?

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  • $\begingroup$ Isn't the hyperbolic plane a surface? Tilings of the hyperbolic plane can have as many triangles at a vertex as one wishes. $\endgroup$
    – Joseph Malkevitch
    Jan 2, 2010 at 20:07
  • $\begingroup$ Sure Joseph, but 7 is the minimum and this is what I am looking for. $\endgroup$
    – Dmitri Panov
    Jan 3, 2010 at 0:42

2 Answers 2

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The paper of Cooper and Thurston: Triangulating 3-manifolds using five vertex link types. Topology 27 (1988), no. 1, 23--25, is relevant. From the review: "It is known that, for any dimension n, there is a finite set of link types such that every n-manifold has a triangulation in which the link of each vertex is in this set." This does not answer the specific question, and also does not deal the case of cubed-manifolds but it can be a good place to start.

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  • $\begingroup$ Gil, thanks for your answer I will have a look, curious to know about 5 types! By the way, the finiteness of types for cubes can be deduced quickly from finitenss for simplexes. The point is that a simplex of dimension $n$ can be decomposed in $n!$ cubes in a canoncial way. So every simplicial decomposition produces a "cubisation", somewhat similar to the baricentric decomposition $\endgroup$
    – Dmitri Panov
    Jan 2, 2010 at 19:49
  • $\begingroup$ Dear Dimitri, unfortunately I cannot answer your precise question and also I dont have a clear intuition what the answer should be. Regarding cubical structure and your conjecture I wonder if already the n dimensional torus can have a cubical structure where the degree of every vertex is less than 2^d. $\endgroup$
    – Gil Kalai
    Jan 2, 2010 at 20:07
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For n = 3, Cooper and Thurston show (see Gil's answer for citation) that any 3-manifold can be paved with cubes using only 3 vertex link types. These three types have 6, 8 and 10 cubes at each vertex. Cooper and Thurston then subdivide the cubes into simplices to show that any 3-manifold can be triangulated with 5 vertex link types. A slightly more clever subdivision strategy shows that in fact one only needs 3 vertex link types in a triangulation. The largest of these types has 30 simplices around a vertex.

C&T use a result of Montesinos which says that any 3-manifold branch covers the 3-sphere with branch locus the borromean rings. This implies that any 3-manifold has a Euclidean orbifold-ish structure where the non-smooth points have neighborhoods isomorphic to an interval cross a 2-dimensional cone with cone angle 3/2 pi or 5/2 pi. If an analogous result were true in higher dimensions then one could show that any n-manifold could be paved with hypercubes using only 3 vertex link types. The largest vertex link would use 5/4 * 2^n cubes.

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  • $\begingroup$ Kevin thanks for this explantion! Just a nitpick, a cone points with angles 5/2 pi and 3/2 pi should not be called orbifold points, orbifold points are points with angle 2pi/n. Also 5/4 2^n looks quite optimistic for me... But it is surelly worth to think about generalisation of their construction for high dim. $\endgroup$
    – Dmitri Panov
    Jan 3, 2010 at 13:59
  • $\begingroup$ Yes, bad choice of terminology on my part. I'll edit my answer to fix it. $\endgroup$
    – Kevin Walker
    Jan 3, 2010 at 15:12

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