Given two abstract simplicial complexes $\mathcal{K}$ and $\mathcal{L}$, what is the definition of their product $\mathcal{K} \times \mathcal{L}$, as another abstract simplicial complex? Basically I'm looking for the definition of "product" such that $|\mathcal{K} \times \mathcal{L}|$ is homeomorphic to $|\mathcal{K}| \times |\mathcal{L}|$, if that makes sense ($|\cdot|$ denotes geometric realization; not sure if this is standard usage). I understand how to obtain their product when considering their geometric realizations, but is there a nice combinatorial definition of the product of two abstract simplicial complexes?

For example, it doesn't make sense to simply take their Cartesian product $\mathcal{K} \times \mathcal{L}$ as sets. So then what do we do with $\mathcal{K}$ and $\mathcal{L}$?


This is more subtle than it might first appear. First of all, how do we triangulate $\Delta^n \times \Delta^m$? One answer is to use barycentric subdivision: see Q5 here. However, there is also a triangulation whose vertices are pairs of vertices in the two factors. Either way, once have decided on a coherent procedure for triangulating $\Delta^n \times \Delta^m$, it is a relatively straightforward matter to extend this to a procedure for triangulating products of abstract simplicial complexes in general.

Let me describe the second option in more detail. Let $K$ and $L$ be abstract simplicial complexes and choose a linear ordering of the vertices. We define $K \otimes L$ to be the following abstract simplicial complex:

  • Its vertices are pairs $(x, y)$, where $x$ is a vertex of $X$ and $y$ is a vertex of $Y$.
  • An $n$-simplex in $K \otimes L$ is a set $\{ (x_0, y_0), \ldots, (x_n, y_n) \}$ such that $x_0 \le \cdots \le x_n$, $y_0 \le \cdots \le y_n$, $\{ x_0, \ldots, x_n \}$ is a simplex in $K$ and $\{ y_0, \ldots, y_n \}$ is a simplex in $L$. (We allow the possibility that the corresponding simplices in $K$ and $L$ are of dimension $< n$, but the sum of their dimensions is always $\ge n$.)

For example, $\Delta^1 \otimes \Delta^1$ corresponds to $⧄$. (If we had neglected the ordering, we would instead get $\Delta^3$!) There are evident simplicial maps $K \otimes L \to K$ and $K \otimes L \to L$, and their geometric realisations induce a homeomorphism $| K \otimes L | \to | K | \times | L |$, as required. (As before, first verify the claim for the standard simplices.)

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    $\begingroup$ This is not the same as the categorical product of abstract simplicial complexes, right? I think taking the categorical product of $\Delta^1$ with itself does indeed turn out to be $\Delta^3$. $\endgroup$ – Vikram Saraph Mar 30 '14 at 2:04
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    $\begingroup$ That's right. This is actually the product in the category of simplicial sets. $\endgroup$ – Zhen Lin Mar 30 '14 at 8:00
  • $\begingroup$ Amid all the accidental deleting of comments, I'll just post this here: "Does $|\mathcal{K} \otimes \mathcal{L}|$ embed into $|\mathcal{K} \times \mathcal{L}|$?" Answer: yes it does, since this is the case for the standard simplexes! $\endgroup$ – Vikram Saraph Mar 21 '15 at 19:11

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