# Simplicial and cubical decompositions of low valence

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?

• Isn't the hyperbolic plane a surface? Tilings of the hyperbolic plane can have as many triangles at a vertex as one wishes. – Joseph Malkevitch Jan 2 '10 at 20:07
• Sure Joseph, but 7 is the minimum and this is what I am looking for. – Dmitri Panov Jan 3 '10 at 0:42

• 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 – Dmitri Panov Jan 2 '10 at 19:49