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This question concerns with the triangulation of a prism as follows. Let $\sigma$ be a $d$-simplex. 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 $\{v_0,v_1,...,v_i,v_i',...,v_d'\}$.

How do we show $\displaystyle\bigcup_{i=0}^d\sigma_i=P$? Specifically, $\forall v\in P$,$v$ is a convex combination of all the $2(d+1)$ vertices. How do we show $v$ has to be a convex combination of the convex combination of the vertices of $\sigma_i$ for some $i$.

Hans
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1 Answers1

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Depending on your background the following may not be helpful, but it is the slickest proof for your claim I know.

Consider the totally ordered set $[d] = \{0 < \dots < d\}$, which represents the $d$-simplex $\Delta^d$. Then consider the partially ordered set $[d]\times[1]$ with order relation $(i,k) \leq (j,l)$ iff $i\leq j$ and $k\leq l$. This represents the prism $\Delta^d\times\Delta^1$. It has totally ordered subsets of the form $Z_p=\{(0,0) < \dots < (p,0) < (p,1) < \dots < (d,1)\}$ for $p=0 \dots d$, which represent your simplices $\sigma_p$. Now note that $$[d]\times[1] = \bigcup\limits_{p=0}^d Z_p$$ as partially ordered sets (every vertex is contained in the union and so is every $\leq$ relation). The composite functor $\lvert N(-)\rvert:\mathsf{Poset}\longrightarrow\mathsf{sSet}\longrightarrow\mathsf{Top}$ preserves the unions in question* as well as finite products* and since $\lvert N([d])\rvert=\Delta^d$ this gives the desired claim $$\Delta^d\times\Delta^1 = \lvert N([d])\rvert\times\lvert N([1])\rvert \cong \lvert N([d])\times N([1]) \rvert = \left\lvert N\left(\bigcup\limits_{p=0}^d Z_i\right)\right\rvert = \bigcup\limits_{p=0}^d\sigma_p$$

*To be completely honest: This claim is highly non-obvious. The functor consists of a left adjoint (geometric realization) after a right adjoint (nerve), so a priori we cannot expect it to preserve unions or products.

Edit The claim about the unions is even worse than I thought and my previous answer included an error. The claim is true though. According to a friend, one way to see this is by saying that the nerve preserves pushouts along Dwyer-maps. Another way is to say that the simplices are dense in $\mathsf{Poset}$ and in $\mathsf{sSet}$ and in this situation the nerve preserves the colimits over the canonical cocones. By cofinality and the fact that the nerve preserves monos, these can be regarded as colimits over the non-degenerate subsimplices, which is the union in question.

Jonas Linssen
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  • Thank you for your answer. Let me study it along with the new concepts. But is there an elementary proof? The result seems to be elementary enough for an elementary proof. – Hans Jul 29 '23 at 00:27
  • Your answer might go better on the older duplicate of this question. I'm considering writing my own simplicial answer (the existing answer there took my main idea, but I have another one!). Then we can, amusingly, have three high-level answers to what should be an elementary geometry problem – FShrike Sep 05 '23 at 10:36
  • At any rate, I'd be interested in an elaboration on how you expect the nerve, or the realisation-nerve composite, to preserve unions here. "in this situation" is tantalising – FShrike Sep 05 '23 at 10:37