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 $\sigma_i\cap\sigma_j\in\Delta$.

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    $\begingroup$ 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? $\endgroup$ Sep 7, 2021 at 22:15
  • $\begingroup$ @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. $\endgroup$ Sep 7, 2021 at 23:09
  • $\begingroup$ 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. $\endgroup$ Sep 7, 2021 at 23:56
  • $\begingroup$ @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 $\sigma_i\cap\sigma_j$? $\endgroup$ Sep 8, 2021 at 3:26
  • $\begingroup$ 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. $\endgroup$ Sep 8, 2021 at 3:49

1 Answer 1


Lemma. The intersection of two maximal simplices is a simplex, and it is the convex hull of its vertices.

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.

  • $\begingroup$ What is meant here by "maximal simplex?" This is unclear from the question, and isn't even really clear from the discussion under it. $\endgroup$ Sep 8, 2021 at 17:04
  • $\begingroup$ The simplices $\sigma_i$ are the maximal ones. $\endgroup$ Sep 8, 2021 at 18:00
  • $\begingroup$ Noting $F_i\cap F_j =F'_i\cap F'_j$ is really brilliant! Thanks. $\endgroup$ Sep 9, 2021 at 4:44

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