Barycentric subdivision - do different simplices have different barycenters?

Question

I am reading the construction of a new complex $K^1$ from the complex $K$ by a process called the Barycentric subdivision. Given a simplex $A$ of $K$ generated by vertices $v_0,\cdots, v_k$, its barycenter is defined to be the point $$ \hat{A} = \frac{1}{k+1}(v_0 + \cdots + v_k). $$ To form $K^1$ we choose its vertices to be the barycenters of the simplexes of $K$. A collection $\hat{A_0}, \hat{A_1}, \cdots, \hat{A_k}$ of such barycentres form the vertices of a simplex of $K^1$ if and only if $$ A_{\sigma(0)} < A_{\sigma(1)} < \cdots < A_{\sigma(k)} $$ for some permutation $\sigma$ of the integers $0,1,\cdots, k$. My question is, isn't it that two different simplices may give the same barycenter? In this case, we might not be sure about which simplex $\hat{A_0}$ is associated about, will it affect us determine whether the chain of simplices can be increasing?

Thank you!

Answer 1

Two simplicial complexes intersect only along faces, and in particular their interiors are disjoint. Since the barycenter of a simplex lies in its interior, it cannot coincide with any other barycenters. Therefore no two simplices will share a barycenter.

(By convention, the interior of a vertex is the vertex itself.)

You cannot have (for example) two overlapping triangles arranged so that their barycenters match up: the intersection of those two triangles will not consist of faces of those triangles, so this does not follow the rules of a simplicial complex.