TITLE: Simplicial and cubical decompositions of low valence
QUESTION [7 upvotes]: 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?

REPLY [6 votes]: 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.