Triangulations of the sphere

This article describe various general methods for triangulating the sphere i.e. giving the sphere the structure of a polyhedron or a geometric realization of a simplicial complex.

Description
This triangulation sends the sphere to the generalized version of the octahedron. One way of viewing this is as the map sending the sphere to the $$L^1$$-sphere.

The structure of the polyhedron (as a geometric realization) is given as follows:


 * There are $$2^{n+1}$$ simplices of maximal dimension, each of dimension $$n$$
 * For a general simplex of maximal dimension, pick a sign sequence (a sequence of length $$n$$ made up of $$+$$s and $$-$$s. The corresponding simplex is the convex hull of the points $$(0,0,\ldots,\pm 1,0,0,\ldots,0)$$ where the sign is chosen as per the sign sequence.
 * Every simplex is contained in a simplex of maximal dimension (i.e. the simplicial complex is full)

Combinatorial information about this triangulation:


 * Number of vertices: $$2n$$
 * Number of maximal simplices: $$2^n$$

Advantage
The key advantages of the octahedral triangulation for computational purposes: All the coordinates are $$0,\pm 1$$ and the sphere is embedded in just one higher dimension.

Disadvantage
The triangulation is wasteful in terms of the number of maximal simplices.

The simplex boundary triangulation
This triangulation uses the fact that $$S^n$$ is homeomorphic tothe boundary of the $$(n+1)$$-simplex. The standard embedding of the $$(n+1)$$-simplex in Euclidean space is in $$\R^{n+2}$$, as the convex hull of the points $$(0,0,\ldots,1,0,0,\ldots,0)$$. Thus, its boundary can be described as follows:


 * There are exactly $$(n+2)$$ maximal simplices, each of dimension $$n$$
 * The maximal simplices are obtained as convex hulls of points $$(0,0,\ldots,1,0,0,\ldots,0)$$, each convex hull leaving out exactly one point.

Advantage
There are fewer simplices (in fact, the fewest possible) but for computational ease, we need to embed in Euclidean space that is one dimension more than necessary (we can embed with this triangulation in $$\R^{n+1}$$, but the coordinates get substantially messier.