Simplexes, Barycentric Subdivisionn

Centers

As mentioned in the previous section, the center of an n-simplex has barycentric coordinates equal to 1/(n+1). Pull back to barycentric space and find the center of the preimage, having coordinates 1/(n+1) across the board. This is the midpoint of a segment (1 dimension), the center of an equilateral triangle (2 dimensions), the center of a tetrahedron (3 dimensions), and so on.

Although it is not immediately obvious, a line drawn from a vertex to the center of the opposite face contains the center of the simplex. To see this, pull back to the preimage, where the simplex is symmetric. In this case the relationship is clear. Apply the linear map in the forward direction, which carries lines to lines, and the center is indeed coliniear with the vertex and the center of the opposite face.

Barycentric Subdivision

Creating a barycentric subdivision is a recursive process. Let's start at the bottom. The midpoint of a segment cuts the segment in half, implementing a barycentric subdivision in 1 dimension.

Given a triangle, subdivide its sides, i.e. establish the midpoints, and draw lines from the center to the midpoints, and to the corners. This cuts the triangle into 6 smaller triangles. Note that each of the smaller triangles is itself a simplex.

Given a pyramid in 3 space, subdivide each triangular face as above, then draw walls from the center to the lines that subdivide each triangle. This partitions the simplex into 24 smaller simplexes.

In general, an n-simplex is cut into (n+1)! smaller simplexes.

Shrinking Diameters

The diameter of a set in a metric space is the greatest distance between two points in the set. Given x and y in a simplex, pull them apart until they run into walls. Then slide them along the walls until they hit edges. Then slide them along edges until they reach vertices. Therefore the diameter of a simplex is the length of the longest edge, i.e. the greatest distance between two vertices.

Subdivision cuts a 1-simplex in half, and the two smaller simplexes have half the diameter.

Moving to a triangle, let the longest side run along the x axis, from the origin to x = d (where d is the diameter). The center, call it p, is the average of the three vertices. Lines emanate from p to the corners and midpoints, creating 6 smaller triangles. Suppose the longest edge of any of these triangles is contained in an edge of the larger triangle. Then the diameter is cut in half, as described above. Surely we can do better than this.

Let a segment start at p and head towards one of the sides. One can always slide the other end of the segment along the side, until it reaches a corner. Therefore the longest edge in any of the subsimplexes runs from p to a corner.

Let the longest edge run from p to the origin. This edge gets longer if we stretch the other edge, incident to the origin, until it too has a length of d. The distance to p is longer still if we shrink the face opposite the origin, until it is very small. Now the average of the vertices is d+d+0 over 3. The diameter is multiplied by 2/3.

Now suppose the longest edge of the original simplex, having length d, is contained in the face opposite the origin. Stretch the simplex along the line from the origin, through p, to the center of the opposite face. This increases the distance from 0 to p. Keep stretching until one of the edges incident to the origin has length d. This reproduces the earlier case. Therefore the distance is still bounded below d times 2/3.

One can generalize the above, so that the diameter of an n-simplex, when subdivided, is multiplied by n/(n+1). Repeat this process, and the simplexes shrink to 0 geometrically.