Barycentric Subdivision

We start off with some definitions.

Simplicial Complexes

Definition: Recall that a simplicial complex $K$ is a set of simplices such that:

every face of a simplex of $K$ is also in $K$ ,
and the non-empty intersection of any two simplices $\sigma_1,\, \sigma_2 \in K$ is a face of both $\sigma_1$ and $\sigma_2$ .

As basic examples, the following are simplicial complexes. In fact, they are all simplexes. A simplicial complex would be the union of some of these simplexes, possibly attached. $\text{[math]}$ These simplices are the $0$ -simplex, the $1$ -simplex, the $2$ -simplex and the $3$ -simplex.

Subdivision

We now want to take a simplicial complex and cut it up into smaller pieces. This is what is known as a subdivision.

Definition: Given a simplicial complex $K$ , a subdivision of $K$ is a simplicial complex $K'$ such that:

each simplex of $K'$ is contained in a simplex of $K$
and each simplex of $K$ is a finite union of simplicies in $K'$ .

We're going to look at a particular type of subdivision, called the barycentric subdivision. To define this I need to first talk about the barycenter.

Definition: Given a set of (indpendent) points $\left\{v_0, v_1, \ldots, v_n\right\}$ , we can let our simplex be the set of points $x$ such that $\text{[math]}$ So each point $x$ in the simplex can be associated to the vector $(\lambda_i)$ . These $(\lambda_i)$ are known as the barycentric coordinates of $x$ .

This allows us to define a subdivision based off barycenters. Let $\sigma$ be a simplex with verticese $v_0, v_1, \ldots, v_n$ . The barycenter of $\sigma$ is the point: $\text{[math]}$ In fact, it is exactly the point in the interior of $\sigma$ whose barycenter coordinate with respect to the vertices of $\sigma$ are equal.

Let's look at an example with our $2$ -simplex from above (the triangle). $\text{[math]}$

Permutations

It turns out you can associate a permutation naturally to faces of the barycentric subdivision of a simplex! To do this, you can give an ordering on the barycentric coordinates so that $\sigma_i$ is the position of the $i$ th coordinate and by reading equal coordinates from left to right. The best way to do this is through a few examples.

If we let $(0,1,0)$ be our barycentric coordinate, then we associate it to the permutation $132$ since the first and third entries are equal (both $0$ ) and the second entry is greater than $0$ .