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...? What is the minimal number of simplexes of cubes that we It should alow be easy to meet 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 one most $S(n)$ simplexes at every vertex ? It is reasonable to guess that the number of cubes should be and a cubical decomposition with at least most $2^n$, where n is the dimensionC(n)$ cubes at every vertex. But is The refference of Gil below confirms this known for $n>2$? n=3$.
Here are three questions (Notice I suspect they are hard).
1) Can it be proven that only faces of the same dimension should $C(n)>2^n$?
2) Can it be glued, i.e., we can not glue a vertex to an interior point proven that $S(n)>\frac{Vol(S^n)}{Vol(\Delta^n)}$, where $\Delta^n$ is the spherical tetrahedron with edge of an edge).length $\frac{\pi}{3}$ in the unit sphere $S^n$.
3) Is there any reasonable estimation for $C(n)$ and $S(n)$ from above?
