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$\newcommand{\sset}{\mathsf{sSet}}\newcommand{\poset}{\mathsf{Poset}}\newcommand{\p}{\mathscr{P}^{\mathsf{nd}}}\newcommand{\N}{\mathcal{N}}\newcommand{\sd}{\operatorname{sd}}$Following this paper, I encountered two different subdivision functors on $\sset$, and while there is a clear mathematical difference in their definitions it is rather unclear to me how these functors are "actually" different. I use different notation than what appears in the paper.


Definitions:

Let $\N:\poset\to\sset$ denote the (restriction of) usual nerve functor, where posets are viewed as categories in the standard way, given by $\N(K)_n=\poset([n],K)$ where $[n]=[0<1<2<\cdots<n]$ for each $n\in\Bbb N_0$. We consider the functor $\p:\sset\to\poset$ which, on objects, maps the simplicial set $X$ to the set of its nondegenerate simplices (in any and all degrees) ordered by the face relation $\sigma\preceq\tau$ iff. there exists an injection $f$ in $\Delta$ such that $X_f(\tau)=\sigma$. On arrows, this functor can be defined and proven to be a functor using repeated applications of the Eilenberg-Zilber lemma.

I consider the first "naive" subdivision functor $S:\sset\to\sset$ which is defined to be the composite $\N\p$. Then there is the second subdivision functor, which is the one that is actually considered the simplicial subdivision, $\sd:\sset\to\sset$ which could be defined as the left Kan extension of $S\circ\Delta^\bullet$ along $\Delta^\bullet$ or as the unique cocontinuous functor agreeing with $S$ on the $\Delta^n$ or as the coend $X\mapsto\int^{n\in\Delta}X_n\otimes S\Delta^n$ or as $\sd X=\varinjlim_{(n,\sigma)\in\mathrm{el}(X)}S\Delta^n$. These all agree up to isomorphism. This is left adjoint with right adjoint $\Gamma(X)_n:=\sset(S\Delta^n,X)$

By easy abstract nonsense, $\sd\Delta^n\cong S\Delta^n$ naturally in $n$. In fact there is a canonical comparison natural transformation $\pi:\sd\implies S$ which can be shown to always be surjective; $\pi_X$ is an isomorphism if $X=\N(Q)$ for some poset $Q$. It can be shown (a bit more care than is usually given is required) that $|S\Delta^n|\cong|\Delta^n|$ naturally in $n$ and it follows that $|\sd(X)|\cong|X|$ naturally in the simplicial set $X$.


The actual question: $\pi$ surjects, but what is stopping it from being injective in general? How does $\sd$ differ from $S$ in "what it is actually doing"? The functor $S$ handles data about chains of subfaces of the n.d. simplices and the functor $\sd$ handles data about chains of face maps glued together when the images of two simplices under these maps agree; loosely, $\sd$ is also bundling equivalence classes of chains of subfaces of the n.d. simplices and both are fairly reasonable (to my mind) attempts at a notion of subdivision; however, only one has succeeded and earned its title. Jardine refers to $S$ as a "classifying space" functor instead.

Usually when trying to understanding something simplicial, it helps to take geometric realisations and view what is happening, but since $|\sd|\cong|\cdot|$ and in all the simple examples I've tried, $|S|\cong|\cdot|$ as well, I can't really see what simplices $\sd$ has that $S$ does not. The question could be restated as: what obstructs $\pi$ from being an isomorphism?

I would appreciate any insights or good examples as to how $\sd$ differs from $S$. For example, that might mean showing $S$ fails to be cocontinuous (for all I know, $S$ is cocontinuous) with some explanation as to why. Of course, having a cocontinuous functor is very useful, but I wonder if there are intuitions/motivations for $\sd$ that are less centred on pure convenience.

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  • $\begingroup$ It is not true that |sd(X)| and |X| are naturally homeomorphic for simplicial sets X. No choices of homeomorphisms for X = \Delta^2 and X = \Delta^1 are simultaneously compatible with both degeneracy maps \Delta^2 \to \Delta^1. A weaker, quasi-naturality statement appears as Theorem 2.3.2 in mn.uio.no/math/personer/vit/rognes/papers/aoms186-nocrop.pdf . $\endgroup$ Commented Oct 17, 2023 at 14:26
  • $\begingroup$ @JohnRognes Thank you for flagging that the canonical homeomorphism(s) isn't/aren't actually natural and in fact cannot be. That's what I get for checking naturality on face maps and not on degeneracy maps. I'll have a think about why that is. I'll also have a deeper think about the two answers given; thanks for sharing the paper you cowrote. $\endgroup$
    – FShrike
    Commented Oct 18, 2023 at 0:22
  • $\begingroup$ I posted the counterexample for naturality here: mathoverflow.net/a/139397/3969 $\endgroup$ Commented Oct 18, 2023 at 7:53
  • $\begingroup$ @JohnRognes A lack of naturality would ruin the obvious argument for $|\mathrm{sd}(X)|\cong|X|$ (just unnaturally in $X$, on a case by case basis). Is this even true? I expect it should be true for the so-called simplicial complexes $X$, but am hesitant $\endgroup$
    – FShrike
    Commented Aug 7 at 12:48
  • $\begingroup$ *to clarify: I know it is true for simplicial complexes, but don't know if it is true in total generality for all simplicial sets $\endgroup$
    – FShrike
    Commented Aug 7 at 15:11

2 Answers 2

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Lemma 2.2.11 in mn.uio.no/math/personer/vit/rognes/papers/aoms186-nocrop.pdf shows that the surjection you denote $sd(X) \to S(X)$ is an isomorphism if and only if $X$ is a non-singular simplicial set, i.e., if and only if each non-degenerate simplex in $X$ is embedded. It may be interesting to determine for which $X$ the initial non-singular simplicial set under $sd(X)$ is precisely $S(X)$. In the book above, we called this the desingularization of $sd(X)$.

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Let us compare what both functors do on $X=S^2=\Delta^2/\partial \Delta^2$.

There are only two nondegenerate simplices, so the first functor sends it to an interval. The second functor sends it to the barycentric subdivision of $\Delta^2$ modulo its boundary, i.e. a 2-sphere.

The map between the two sends the boundary to one endpoint of the interval and the new inner point in the barycentric subdivision of $\Delta^2$ to the other point. It looks a bit like the height on the usual visualization of the 2-sphere.

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    $\begingroup$ Excellent example; worth also noting that $S^1 = \Delta^1/\delta \Delta^1$ is already fairly illustrative, and is easier to draw. $\endgroup$ Commented Oct 17, 2023 at 19:05
  • $\begingroup$ Thanks for this example. I'm having a bit of trouble verifying the action of $\pi$ here; it seems like I would need to resort to an explicit calculation of the simplices of the colimit defining $\operatorname{sd}(X)$ which I'm too tired to accurately do right now. Is there an easier way without getting one's hands dirty? $\endgroup$
    – FShrike
    Commented Oct 18, 2023 at 18:36
  • $\begingroup$ Peters example is here slightly simpler. We have $S^1 = \Delta^1 \cup_{\partial \Delta^1} pt$. Since sd is a left adjoint, it is compatible with pushouts, so $sd(S^1) = sd(\Delta^1) \cup_{sd(\partial \Delta^1)} sd(pt)$, e.g. we take the qoutient of $sd(\delta^1)$ by the subdivision of the boundary. To see what $\pi$ does, we can use that the target is the nerve of a poset. Thus we only have to figure out what happens on zero-simplices and then a simplex $0,1,,..,n$ will be send to the chain $\pi(0),..,\pi(n)$, $\endgroup$ Commented Oct 18, 2023 at 19:03

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