# Intersection of two simplices is a convex hull

My definition of $$d$$-dimensional simplex (also called $$d$$-simplex) is a convex hull of $$d+1$$ many points in $$\Bbb R^d$$ (or larger space) which are affinely independent. Also, a face of a simplex is a convex hull of arbitrary subset of vertices of a simplex $$\sigma$$.

Currently, I'm proving the triangulation of prism is actually a triangulation. More precisely, if $$\sigma$$ is a $$d$$-simplex, then we define a prism $$P = \sigma\times[0,1]$$. If we let $$v_0,...,v_d$$ be the vertices of $$\sigma$$ and $$v'_i$$ be a vertice above $$v_i$$, then define a simplex $$\sigma_i$$ by the convex hull of $$\{v_0,v_1,...,v_i,v_i',...,v_d'\}$$.

In this situation, I want to show that if $$F_1,F_2$$ are faces of some simplices $$\sigma_i,\sigma_j$$ respectively, then their intersection is a convex hull of vertices both contained in $$F_1$$ and $$F_2$$. For example, if $$F_1$$ is a convex hull of $$v_0,w_1,w_2,w_3$$ and $$F_2$$ is a convex hull of $$w_2,w_3,w_4$$, then $$F_1\cap F_2$$ is a convex hull of $$w_2,w_3$$.

One known is that if the set of points we consider (in my example, $$v_0,w_1,w_2,w_3,w_4$$) are affinely independent, then the statement is true. But in my case, the set of points may not be affinely independent. Could you help?

Edit: I'll further assume I know 1. the interiors of the simplices $$\sigma_0,...,\sigma_d$$ are disjoint. 2. $$\bigcup_{i=0}^d\sigma_i = P$$.

Edit: A nonempty collection $$\Delta$$ of simplies is called simplicial complex (1) if $$\sigma\in\Delta$$ then all the faces of $$\sigma$$ is also contained in $$\Delta$$, (2) If $$\sigma_i,\sigma_j\in\Delta$$ then $$\sigma_i\cap\sigma_j\in\Delta$$.

• Why don't you know that the vertices are affinely independent? If the original simplex has affinely independent vertices in $\mathbb{R}^d$, then the vertices of each simplex $\sigma_j$ in the prism in $\mathbb{R}^{d+1}$ will also be independent, won't they? Sep 7, 2021 at 22:15
• @JohnPalmieri True, but if I take some vertices in $\sigma_i$ and some vertices in $\sigma_j$ ($i\neq j$), then the total vertex set I'm considering may not be affinely independent. For example, for $d =1$ case, $P$ is just a square with one diagonal. If I take two vertical $1$-simplex, then there are total $4$ vertices so they are affinely dependent. Sep 7, 2021 at 23:09
• The intersection $F_1 \cap F_2$ will be contained in the intersection $\sigma_i \cap \sigma_j$, so you should only need to consider intersections of the maximal simplices, and you can work those out explicitly. Sep 7, 2021 at 23:56
• @JohnPalmieri I misunderstood something. What I need to show is the convex hull of vertex set that is both contained in $F_1$ and $F_2$ is same as $F_1\cap F_2$. It's not clear to me that it suffices to consider intersections of the maximal simplices. Once I prove $\sigma_i\cap\sigma_j$ is a face of both simplices, why can I say that $F_1\cap F_2$ is a face of $\sigma_i\cap\sigma_j$? Sep 8, 2021 at 3:26
• According to your original post, it suffices to show that the involved vertices are affinely independent. I believe that this holds when $F_1$ and $F_2$ are maximal simplices, and therefore when $F_1$ and $F_2$ are faces of maximal simplices. Sep 8, 2021 at 3:49

I'm not going to prove this, and it could be that a careful proof would also solve the general problem you're asking. If I wanted to prove it, I might resort to coordinates: assume that the original simplex $$\sigma$$ is the standard $$d$$-simplex with its standard barycentric coordinates. Then each point in $$\sigma \times I$$ has coordinates, and the job would be to identify the points in $$\sigma_{i}$$, $$\sigma_{j}$$, and their intersection. Maybe we wouldn't need to resort to coordinates and could determine the intersection of $$\sigma_i$$ and $$\sigma_j$$ from their vertices.
Anyway, assume that the lemma is true and suppose that $$F_{i}$$ and $$F_{j}$$ are faces of $$\sigma_{i}$$ and $$\sigma_{j}$$, respectively, and let $$F = \sigma_{i} \cap \sigma_{j}$$. Then $$F_{i} \cap F_{j} \subseteq F$$, and rather than considering $$F_{i} \cap F_{j}$$, we can instead let $$F_{i}' = F_{i} \cap F$$ and $$F_{j}' = F_{j} \cap F$$, and consider $$F_{i} \cap F_{j} = F_{i}' \cap F_{j}'$$. Now $$F_{i}'$$ and $$F_{j}'$$ are both contained in $$F$$, and so all of the involved vertices are in general position, so $$F_{i} \cap F_{j} = F_{i}' \cap F_{j}'$$ is the convex hull of the vertices in the intersection.
• The simplices $\sigma_i$ are the maximal ones. Sep 8, 2021 at 18:00
• Noting $F_i\cap F_j =F'_i\cap F'_j$ is really brilliant! Thanks. Sep 9, 2021 at 4:44