number of subdivisions needed to compare simplicial sets to simplicial complexes? - MathOverflow

number of subdivisions needed to compare simplicial sets to simplicial complexes?

2010-11-21T15:52:42Z

This is a question in elementary geometric topology, of which I know little. It has to do with the result that geometric realizations of simplicial sets and geometric realizations of abstract simplicial complexes coincide up to homeomorphism, and with how many subdivisions are needed to effect the argument.

Here is a sketch that the realization of a simplicial set is the realization of a simplicial complex. There are functors

$$Face: [\Delta^{op}, Set] \to Pos$$

$$Nerve: Pos \to [\Delta^{op}, Set]$$

where $Face(X)$ is the poset whose elements are simplices of $X$, ordered by $x \leq y$ if $x$ is a nondegenerate face of $y$. The composite $Nerve \circ Face$ is the barycentric subdivision of $X$. This subdivision is a regular simplicial set, whose realization is a regular CW complex. Associated to a regular CW complex is a simplicial complex, also called its subdivision, whose vertices are open cells in the CW decomposition, and where simplices are finite chains $e_1 < \ldots < e_n$ where each $e_i$ is a proper face of $e_{i+1}$ ($e_i$ is contained in the closure of $e_{i+1}$). There is a theorem that the realization of this simplicial complex is homeomorphic to the space obtained by gluing together the regular CW complex.

This line of argument almost feels like overkill to me: we subdivide once, realize, and then subdivide again to get to the simplicial complex. I'd like to simplify this if possible. Define a functor

$$Flag: Pos \to SimpComplex$$

which takes a poset $P$ to the simplicial complex whose vertices are elements of $P$, and whose simplices are finite "flags" $x_1 < \ldots < x_n$. (In passing, I'll note that there is a fourth functor

$$U: SimpComplex \to Pos$$

which takes a simplicial complex to the poset of simplices ordered by inclusion; the composite $Flag \circ U$ is the barycentric subdivision functor on simplicial complexes.) My question is:

Let $X$ be any simplicial set. Is the realization of the simplicial complex $Flag \circ Face(X)$ homeomorphic to the realization of $X$?

I am not hugely confident that the answer is "yes", and even suspect there are standard counterexamples, but it seems to be true for simple examples.

Edit: I have a feeling that I botched the description of the subdivision of simplicial sets, but I'll let the experts correct me if needed.

Answer by Buschi Sergio for number of subdivisions needed to compare simplicial sets to simplicial complexes?

2010-11-21T18:35:28Z

I think yes, this is my attempt:

as you said $Nerve\circ Face(X)$ is the barycentric subdivision of $X$, then from [LW] Prop. 7.5 p. 105 the goemetric realizations are naturally homeomorphic: $|X| \cong |Nerve\circ Face(X)|$. Then it's enought to show that for a poset $X$ the goemetrical realization of the simplicial set $Nerve(X)$ and the simplical complex $Flag(X)$ coincide.

But $Flag(X)$ is maked by all nondegenerate simplexes of $Nerve(X)$ and the assert follow from the properties of the goemetrical realization (that is essentially based on the non degenerate simplexes only) [LW] T.4.6 p. 91. In other way, because $X$ is a order the nodegenerate simplexes of $Nerve(X)$ form a final subcategory of $\Delta\downarrow Nerve(X) $ then the goemetric realization of $Nerve(X)$ is the colimits of:

$\Delta\downarrow Nerve(X)\to \Delta \to Top $ restricted to this final category

and this is $|Flag(X)|$ .

[LW]. Lundell, Weingram: The topology of CW complexes

Answer by Giacomo d'Antonio for number of subdivisions needed to compare simplicial sets to simplicial complexes?

2011-03-14T12:03:39Z

I will try to write down an answer, even though I am also new to the realm of simplicial sets. So, feel free to correct me if write something wrong.

In my last work I needed to consider the barycentric subdivision of a simplicial set too, and I came up with the following solution.

The composite $\mbox{Nerve}\circ\mbox{Face}$ is the barycentric subdivision of $X$. This subdivision is a regular simplicial set, whose realization is a regular CW complex.

I think this argument is faulty. Consider the $\Delta-Set$ composed by one $1$-simplex and one $0$-simplex (topologically an $S^1$) and add degenerate simplex in order to get a simplicial set. The composite $\mbox{Nerve}\circ\mbox{Face}$ gives a segment, clearly not homeomorphic to $S^1$.

The mistake was considering face posets. Instead one should consider a small category (the face category), which reduces to the face poset in case of regular simplical sets. One defines it on the $\Delta$-set (alternatively, on the non-degenerate simplexes), the objects are the simplexes of the $\Delta$-set and the morphisms are the face maps.

With this definition of the functor $\mbox{Face}: Simp \to Cat$, the composite $\mbox{Nerve}\circ\mbox{Face}$ gives a regular simplicial set. Applying once again $\mbox{Face}$ we get a poset, and then the nerve of a poset is the same as its order complex (i.e. your $\mbox{Flag}(X)$).

Answer by John Rognes for number of subdivisions needed to compare simplicial sets to simplicial complexes?

2012-12-20T14:09:17Z

Let $X$ be a simplicial set. The partially ordered set $(X^{nd}, \leq)$ of non-degenerate simplices of $X$, with $x\le y$ if $x$ is a face of $y$, was considered by Barratt in a 1956 Princeton preprint. I write $B(X) = N(X^{nd}, \le)$ for its nerve, the Barratt nerve.

It is not quite clear to me if, in your definition of $Face \colon sSet \to Pos$, you only allow the non-degenerate simplices as elements. Let me assume that you make that restriction, so that $Nerve \circ Face(X) = B(X)$.

If $X$ is the simplicial set associated to an (ordered) simplicial complex, then $B(X)$ is indeed the barycentric subdivision of that simplicial complex. In these cases, there is a homeomorphism $|B(X)| \cong |X|$. However, if $X = \Delta^n/\partial\Delta^n$ for $n\ge1$, then $X^{nd}$ is isomorphic to $[1] = (0 < 1)$ and its nerve is contractible. So in these cases $|B(X)|$ is not homeomorphic (nor homotopy equivalent) to $|X|$.

From the definition of Kan's normal subdivision $Sd(X)$ as a coend, there is a canonical natural map $b_X \colon Sd(X) \to B(X)$. It is an isomorphism if and only if $X$ is non-singular, meaning that the representing map $\bar x \colon \Delta^n \to X$ of each non-degenerate simplex of $X$ is a cofibration. Here $n$ is the dimension of $x$.

There exists a homeomorphism $h_X \colon |Sd(X)| \cong |X|$, due to Fritsch and Puppe (1967). However, it is not natural for most maps $X \to Y$ of simplicial sets. For instance, there is no way to fix $h_{\Delta^1}$ and $h_{\Delta^2}$ so that the homeomorphisms are compatible with both of the degeneracy maps $\sigma^j \colon \Delta^2 \to \Delta^1$.

The normal subdivision $Sd(X)$ of any simplicial set is a regular simplicial set, in the sense that each representing map $\bar x \colon \Delta^n \to X$ only makes identifications along the last ($n$th) face. The geometric realization of a regular simplicial set is a regular CW complex, i.e., if the closure of each $n$-cell is an $n$-ball and the boundary of the open $n$-cell in its closure is an $(n-1)$-sphere. Any regular CW complex admits a triangulation. See e.g. Fritsch and Piccinini (1990), sections 3.4 and 4.6 for proofs.

Combining the two previous paragraphs, the realization of any simplicial set can be triangulated by a simplicial complex, but not in a natural way. Not every CW complex can be triangulated (Metzler, 1967), but the realizations of (regular) simplicial sets can.

In work of Waldhausen, Jahren and myself (Spaces of PL Manifolds and Categories of Simple Maps, Annals of Maths. Studies, to appear in 2013) we consider finite simplicial sets $X$ and prove in Proposition 2.5.8 that there are natural maps $$ Sd(Sd(X)) \to B(Sd(X)) \to Sd(X) \to X . $$ Each of these are simple, in the sense that their geometric realizations have contractible point inverses. In particular, they are simple-homotopy equivalences. The term $B(Sd(X))$ is the nerve of a partially ordered set, hence an ordered simplicial complex. So there is, indeed, a natural simple map $$ i_X \colon B(Sd(X)) \to X $$ from an ordered simplicial complex to $X$.

You mentioned geometric topology at the outset. If $X$ is a combinatorial manifold, so that $M = |B(Sd(X))|$ and $N = |X|$ are PL manifolds, then Cohen (1970) proved that any simple PL map $M \to N$ can be uniformly approximated by a PL homeomorphism (assuming the manifolds are of dimension at least 5). Hence the natural map $|i_X| \colon |B(Sd(X))| \to |X|$ is a map from an ordered simplicial complex that is arbitrarily close to a PL homeomorphism, but I have no reason to expect that this approximation can be made natural.