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

Hans
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one potato two potato
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    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? – John Palmieri Sep 07 '21 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. – one potato two potato Sep 07 '21 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. – John Palmieri Sep 07 '21 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 $\sigma_i\cap\sigma_j$? – one potato two potato Sep 08 '21 at 03:26
  • 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. – John Palmieri Sep 08 '21 at 03:49
  • @JohnPalmieri It's almost straightforward that the involved vertices are affinely independent since it's a subset of some simplex $\sigma_i$. What I can't show is that the intersection of two faces $F_1\subset\sigma_i,F_2\subset\sigma_j$ is again a face of $\sigma_i$ and $\sigma_j$. By definition, equivalent to showing that $F_1\cap F_2$ is a convex hull of vertices both contained in $F_1$ and $F_2$. What I'm showing is $\sigma_i$'s and its faces form a simplical complex. – one potato two potato Sep 08 '21 at 03:56
  • So should you delete, or at least edit, your second-to-last paragraph? – John Palmieri Sep 08 '21 at 04:48
  • @JohnPalmieri I still can't understand your confusion/argument about the question/post. Could you clarify what part of my question/post is confusing? The collection of all $\sigma_i$ and its faces form a simplicial complex. That is my question. – one potato two potato Sep 08 '21 at 05:00
  • You say in your post that the statement is true if the set of points is affinely independent, and I have tried to argue why the points should indeed be independent. You in fact seemed to agree that the points are affinely independent a few comments ago. That’s my confusion: why isn’t the matter now settled? – John Palmieri Sep 08 '21 at 05:23
  • @JohnPalmieri Let we write that part formally : Let $V$ be a set of points that is affinely independent. If $F\subset V$ and $E\subset V$, then $\operatorname{conv}(F)\cap\operatorname{conv}(E) =\operatorname{conv}(F\cap E)$ where 'conv' is a convex hull of $(-)$. In our case, $E\cup F$ may not be affinely independent as I mentioned in the first comment. – one potato two potato Sep 08 '21 at 05:30

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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.

John Palmieri
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