Barycentric subdivision of a simplex is a geometric simplicial complex

Question

My question is part of a question which has been asked a few years ago here: Barycentric subdivisions of simplices yield a simplicial complex, but has not been answered to my satisfaction:

The barycenter $b_\sigma$ of an $n$-simplex $\sigma$ with vertices $v_0, \ldots, v_n$ is the point $$\frac{1}{n+1}(v_0 + \cdots + v_n).$$ Show that:

(a) For any strictly increasing sequence $\sigma_0 \subset \sigma_1 \subset \ldots \subset \sigma_k$ of faces of $\sigma$, the barycenters $b_{\sigma_0}, b_{\sigma_1}, \ldots, b_{\sigma_k}$ form the vertices of a $k$-simplex.

(b) These simplices form a simplicial complex.

I already managed to proved (a) in all details, but I did not succeed with (b).

We have to show that if $\sigma_1, \sigma_2$ lie in the barycentric subdivision, then $\sigma_1 \cap \sigma_2$ is a face of both $\sigma_1$ and $\sigma_2$. The answer given to the original question is intuitively right: The intersection of the convex hulls of two sets of barycenters is exactly the convex hull of their intersection. But this is not true in general for all sets! However, it seems to be true in this special case.

Does anybody know a technically clean proof for this fact or a different approach which leads to a proof of (b)?

Answer 1

In my own search for a rigourous proof I found that it is much easier to use an equivalent condition for what constitutes a simplicial complex; namely

A set of simplicies $\mathcal{S}$ in some euclidean space is a simplicial complex if and only if

1. The faces of simplicies in $\mathcal{S}$ are also in $\mathcal{S}$, and
2. The interiors of distinct elements of $\mathcal{S}$ are disjoint.

The fact that the face of an element of the barycentric subdivision is also an element of the barycentric subdivision is practically trivial so I will roughly lay out a proof of condition 2.

So, let $\text{Bary}(\mathcal{S})$ be the barycentric subdivision and fix $\upsilon,\rho \in \text{Bary}(\mathcal{S})$ such that $\text{int}(\upsilon) \cap \text{int}(\rho) \neq \emptyset$. We wish to show that $\upsilon = \rho$ which proves condition 2 and that $\text{Bary}(\mathcal{S})$ is a simplicial complex.

Take $\sigma_0 \subset \dots \subset \sigma_m \in \mathcal{S}$ and $\tau_0 \subset \dots \subset \tau_n \in \mathcal{S}$ such that $\upsilon = \text{Convex}(\{b_{\sigma_0},\dots,b_{\sigma_m}\})$ and $\rho = \text{Convex}(\{b_{\tau_0},\dots,b_{\tau_n}\})$.

Proceeding by induction assume that for some $k \in \{0,\dots,\min\{m,n\}-1\}$ we have $\text{int}(\text{Convex}(\{b_{\sigma_0},\dots,b_{\sigma_{m-k}}\})) \cap \text{int}(\text{Convex}(\{b_{\tau_0},\dots,b_{\tau_{n-k}}\})) \neq \emptyset$ and $\sigma_{m-i} = \tau_{n-i}$ for all $i \in \{0,\dots,k-1\}$.

The case $k=0$ is trivial. It is not too difficult to see, by explicitly writing out barycentric coordinates and rearranging, that $\text{int}(\text{Convex}(\{b_{\sigma_0},\dots,b_{\sigma_{m-k}}\})) \subseteq \text{int}(\sigma_{m-k})$ and $\text{int}(\text{Convex}(\{b_{\tau_0},\dots,b_{\tau_{n-k}}\})) \subseteq \text{int}(\tau_{n-k})$. So, by assumption and since $\mathcal{S}$ is a simplicial complex, we have $\sigma_{m-k} = \tau_{n-k}$.

Next, fix $p \in \text{int}(\text{Convex}(\{b_{\sigma_0},\dots,b_{\sigma_{m-k}}\})) \cap \text{int}(\text{Convex}(\{b_{\tau_0},\dots,b_{\tau_{n-k}}\}))$. If we take the ray from $b_{\sigma_{m-k}} = b_{\tau_{n-k}}$ through $p$ then on the one hand we find it intersects $\text{int}(\text{Convex}(\{b_{\sigma_0},\dots,b_{\sigma_{m-k-1}}\}))$ and on the other it intersects $\text{int}(\text{Convex}(\{b_{\tau_0},\dots,b_{\tau_{n-k-1}}\}))$. Again, one can write out $p$ in barycentric coordinates to explicitly find the intersection points.

However, we have $\text{int}(\text{Convex}(\{b_{\sigma_0},\dots,b_{\sigma_{m-k-1}}\})) \subseteq \sigma_{m-k-1}$ which is a proper face of $\sigma_{m-k}$ and likewise $\text{int}(\text{Convex}(\{b_{\tau_0},\dots,b_{\tau_{n-k-1}}\})) \subseteq \tau_{n-k-1}$ which is a proper face of $\tau_{n-k}$. But, $\sigma_{m-k} = \tau_{n-k}$ and, since this is a convex set, the ray from the interior point $b_{\sigma_{m-k}} = b_{\tau_{n-k}}$ through $p$ must intersect the boundary in at most one point.

Hence, the intersections in $\text{int}(\text{Convex}(\{b_{\sigma_0},\dots,b_{\sigma_{m-(k+1)}}\}))$ and $\text{int}(\text{Convex}(\{b_{\tau_0},\dots,b_{\tau_{n-(k+1)}}\}))$ are in fact the same so that $\text{int}(\text{Convex}(\{b_{\sigma_0},\dots,b_{\sigma_{m-(k+1)}}\})) \cap \text{int}(\text{Convex}(\{b_{\tau_0},\dots,b_{\tau_{n-(k+1)}}\})) \neq \emptyset$.

As a result we have by induction that $\sigma_{m-i} = \tau_{n-i}$ for all $i \in \{0,\dots,\min\{m,n\}\}$. If $m < n$ then we would have $\{b_{\sigma_0},\dots,b_{\sigma_m}\} \subset \{b_{\tau_0},\dots,b_{\tau_n}\}$ so that $\upsilon$ is a proper face of $\rho$ which contradicts the fact that $\text{int}(\upsilon) \cap \text{int}(\rho) \neq \emptyset$. Likewise, we cannot have $n < m$.

As such, $m = n$ so that $\sigma_i = \tau_i$ and $b_{\sigma_i} = b_{\tau_i}$ for all $i \in \{0,\dots,n\}$. Thus, $\upsilon = \rho$ and therefore $\text{Bary}(\mathcal{S})$ is a simplicial complex.