Recall that a (combinatorial) simplicial complex $X$ is said to be $n$-dimensional if it contains at least one face of dimension $n$ and no faces of dimension $n+1$. Further, an $n$-dimensional finite simplicial complex $X$ is said to be pure if every face $\delta$ of dimension $k\lt n$ is a face of some $n$-dimensional face of $X$.

For simplicial sets, we have a different definition for dimension given as:

A simplicial set $S$ is said to be $n$-skeletal if the inclusion $\operatorname{Sk}^n S \subseteq S$ of the $n$-skeleton of $S$ is an isomorphism. We say that $S$ is $m$-dimensional if $m$ is the smallest number for which $S$ is $m$-skeletal.

Given a simplicial complex $X$ as a simplicial set, we see that the associated simplicial set $\Delta[X]$ is $n$-dimensional if and only if $X$ is $n$-dimensional.

Is there a useful notion extending the definition of purity to those simplicial sets with only finitely many nondegenerate simplices?

The property that I'm hoping for in such an extension is sort of a generalization of the "prism decomposition" for products of simplices. That is, for the product of two simplicial sets $S$ and $T$, pure of dimension $s$ and $t$ respectively, I would like the product $S\times T$ to be pure of dimension $s+t$. It is certainly the case that the product has dimension $s+t$ (this can be seen working simplex by simplex), but without a proper definition of purity, we can't give a real generalization.

  • $\begingroup$ Perhaps something like: If $S$ is an $n$-dimensional simplicial set, it is pure if every nondegenerate $k$-simplex $\delta \in S_k$ for $k<n$ is a face of some nondegenerate $n$-simplex $\gamma\in S_n$? $\endgroup$ Dec 31, 2011 at 13:25

1 Answer 1


I [EDIT: almost] would follow the suggestion you made in your comment.

[EDIT: below the line was my original answer, which is wrong, as pointed out by Karol in the comments. Here is an answer which is less functorial, but I hope more correct.]

  • It would be nice to define pure dimension internal to categories other than $Set$. Unfortunately the definition so far need the internal logic to satisfy excluded middle, and I'm not sure how to get around this. My intuition is that a finitely extensive regular category should be the minimum requirement.

Assume $S$ is $n$-dimensional (i.e. $n$-skeletal). Let $ND(S_n) \subset S_n$ be the non-degenerate $n$-simplices (in Harry's original example this is a finite set), that is, the compliment of the joint image of $$ s_{i_1}\ldots s_{i_{n-k}} : \coprod_{k = 0}^{n-1} S_k \to S_n $$ (we can do this step for simplicial objects in a Boolean topos, or more generally in a regular category with complements of all subobjects and finite coproducts). This defines a presheaf on $\tilde\Delta_n$, the subcategory of $\Delta$ consisting of objects $\le n$ and only the coface maps. The $n$-simplices are $ND(S_n)$ and all lower dimensional simplices are the same as for $S$. Unfortunately this construction is not functorial, because a map of simplicial sets may send non-degenerate simplices to degenerate ones.

Definition: A presheaf $S'$ on $\tilde\Delta_n$ is of pure dimension if the collection of maps

$$ d_{i_1}\ldots d_{i_{n-k}} : S'_n \to \coprod_{k = 0}^{n-1} S'_k $$

is jointly regular epimorphic. This step works in any regular category with finite coproducts. This differs from Harry's suggestion in that we ask that all $k$-simplices are a face of a non-degenerate $n$-simplex.

I'll have a think about the proof that pure dimension, defined this way, is additive. It is here that I think the extensivity should be used.

I believe there is a functor $$ R:Set^{\Delta^{op}} \to Set^{\tilde\Delta^{op}} $$ (where $\tilde\Delta$ is the subcategory of $\Delta$ with the same objects but only coface maps) that takes a simplicial set and removes all degenerate simplices. If I am not mistaken, this should be adjoint to one of the adjoints to the restriction along the inclusion $\tilde\Delta \to \Delta$, the one which adds the smallest number of degenerate simplices. (Check this, I am in a rush to leave on an interstate drive)

There are analogues $R_n$ for the finite subcategories of $\Delta$ consisting of all objects $\le n$.

Assume $S$ is $n$-dimensional. Define it to be of pure dimension if $S' := R_n sk_n S$ satisfies:

  • for all $m \le n$, the face maps $d_i : S'_m \to S'_{m+1}$ are jointly surjective.

I think the functors $R,R_n$ preserve products, as do the skeleton functors, so showing that dimension is additive should be ok.

  • $\begingroup$ There is no such functor as your $R$. You cannot just "remove all degenerate simplices" since some degenerate simplices may be faces of non-degenerate ones. You can consider all simplices which are faces of non-degenerate simplices, they form so called "core" of a simplicial set. However, the core is not functorial. Of course both adjoints you mention exist, but they go the opposite direction than your $R$. $\endgroup$ Jan 2, 2012 at 19:34
  • $\begingroup$ Hmm, you're right. :/ $\endgroup$
    – David Roberts
    Jan 2, 2012 at 23:10
  • $\begingroup$ What is the reason for the different definition? It seems like your condition might even be equivalent (didn't check). $\endgroup$ Jan 3, 2012 at 0:22
  • $\begingroup$ The original definition simply doesn't work. There is no such object as $S'$. $\endgroup$
    – David Roberts
    Jan 3, 2012 at 2:44

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