Ordering simplicial complexes with unlabeled vertices

The goal of this post is to describe a partial order on the collection of simplical complexes with $\leqslant n$ unlabeled vertices that is nice in the context of the space $X=\Ran^{\leqslant n}(M)\times \R_{>0}$.

First note that there is a natural order on (abstract) simplicial complexes, given by set inclusion. Interpreting elements of $X$ as simplicial complexes induces a more restrictive order, as new vertices must "split off" from existing ones rather than just be introduced anywhere. Also note that the category usually denoted by $SC$ of simplicial complexes and simplicial maps contains objects with unordered vertices. Here we assume an order on them and consider the action of the symmetric groups to remove the order.

Definition: Let $SC_k$, for some positive integer $k$, be the collection of simplicial complexes with $k$ uniquely labeled vertices. This collection is a poset, with $S\leqslant T$ iff $\sigma\in T$ for every $\sigma\in S$.

The symmetric group on $k$ elements acts on $SC_k$ by permuting the vertices, and taking the image under this action we get $SC_k/S_k$, the collection of simplicial complexes with $k$ unlabeled vertices. This set also has a partial order, with $S\leqslant T$ in $SC_k/S_k$ iff $S'\leqslant T'$ in $SC_k$, for some $S'\in q_k^{-1}(S)$ and $T'\in q_k^{-1}(T)$, where $q_k:SC_k \twoheadrightarrow SC_k/S_k$ is the quotient map.

Definition: For all $i=1,\dots,k$, let $s_{k,i}$ be the $i$th splitting map, which splits the $i$th vertex in two. That is, if the vertices of $S\in SC_k$ are labeled $v_1,\dots,v_k$, then $s_{k,i}$ is defined by
\[\begin{array}{r c l}
s_{k,i}\ :\ SC_k & \to & SC_{k+1}, \\
S & \mapsto & \left\langle S'\cup \{v_i,v_{i+1}\} \cup \displaystyle \bigcup_{\{v_i,w\}\in S'} \{v_{i+1},w\} \right\rangle ,
\end{array}\]where $S'$ is $S$ with $v_j$ relabeled as $v_{j+1}$ for all $j>i$, and $\langle T\rangle$ is the simplicial complex generated by $T$.

By "generated by $T$" we mean generated in the Vietoris-Rips sense, that is, if $\{v_a,v_b\}\in T$ for all $a,b$ in some indexing set $I$, then $\{v_c\ :\ c\in I\}\in \langle T\rangle$. The $i$th splitting map is essentially the $i$th face map used for simplicial sets.

Let $A = \bigcup_{k=1}^n SC_k/S_k$. The splitting maps induce a partial order on $A$, with $S\leqslant T$, for $S\in SC_k/S_k$ and $T\in SC_{k+1}/S_{k+1}$, iff $s_{k,i}(S')\leqslant T'$ in $SC_k$, for some $S'\in q_k^{-1}(S)$, $T'\in q_{k+1}^{-1}(T)$, and $i\in \{1,\dots,k\}$. This generalizes via composition of the splitting maps to any pair $S,T\in A$, and is visually decribed by the diagram below.

Now, let $M$ be a smooth, compact, connected manifold embedded in $\R^N$, and $X=\Ran^{\leqslant n}(M)\times \R_{>0}$. Let $f:X\to A$ be given by $(P,t)\mapsto VR(P,t)$, the Vietoris-Rips complex around the points of $P$ with radius $t$.

Proposition: The map $f:X\to A$ is continuous.

Proof: Let $S\in A$ and $U_S \subseteq A$ be the open set based at $S$. Take any $(P,t)\in f^{-1}(U_S)\subseteq X$, for which we will show that there is an open ball $B\owns (P,t)$ completely within $f^{-1}(U_S)$.

Case 1: $t\neq d(P_i,P_j)$ for all pairs $P_i,P_j\in P$. Then set
\[\epsilon = \min\left\{t, \min_{i<j} |t-d(P_i,P_j)|, \min_{i<j} d(P_i,P_j) \right\}.\]Set $B = B^{\Ran^{\leqslant n}(M)}_{\epsilon/4}(P) \times B^{\R_{>0}}_{\epsilon/4}(t)$, which is an open neighborhood of $(P,t)$ in $X$. It is immediate that $f(P',t')$, for any other $(P',t')\in B$, has all the simplices of $f(P,t)$, as $\epsilon \leqslant |t-d(P_i,P_j)|$ for all $i<j$. If $P_i$ has split in two in $P'$, then for every simplex containing $P_i$ in $f(P,t)$ there are two simplices in $f(P't')$, with either of the points into which $P_i$ split. That is, there may be new simplices in $f(P',t')$, but $f(P',t')$ will be in the image of the splitting maps. Equivalently, $f(P,t)\leqslant f(P',t')$ in $A$, so $B\subseteq f^{-1}(U_S)$.

Case 2: $t= d(P_i,P_j)$ for some pairs $P_i,P_j\in P$. Then set
\[\epsilon = \min\left\{t, \min_{i<j \atop t\neq d(P_i,P_j)} |t-d(P_i,P_j)|,\ \min_{i<j} d(P_i,P_j) \right\},\]and define $B$ as above. We are using the definition of Vietoris-Rips complex for which we add an edge between $P_i$ and $P_j$ whenever $t>d(P_i,P_j)$. Now take any $(P',t')\in B$ such that its image and the image of $(P,t)$ under $f$ are both in $SC_k/S_k$. Then any points $P_i,P_j \in P$ with $d(P_i,P_j)=t$ that have moved around to get to $P'$, an edge will possibly be added, but never removed, in the image of $f$ (when comparing with the image of $(P,t)$). This means that we have $f(P,t)\leqslant f(P',t')$ in $SC_k/S_k$, so certainly $f(P,t)\leqslant f(P',t')$ in $A$. The same argument as in the first case holds if points of $P$ split. Hence $B\subseteq f^{-1}(U_S)$ in this case as well.  $\square$

This proposition shows in particular that $X$ is poset-stratified by $A$

Towards a sheaf of simplicial complexes

The goal of this post is to describe a new stratification of $\Ran^n(M)\times \R_{\geqslant 0}$ that builds on the ideas from a previous post (see "The point-counting stratification of the Ran space is conical (really though) ," 2017-11-15) and some newer ones.

Let $SC_n$ be the set of simplicial complexes on $n$ ordered vertices. There is a natural partial order on $SC_n$ given by inclusion of sets, viewing every simplex as a subset of the power set $\mathbf P(\{1,\dots,n\})$. The symmetric group $S_n$ has a natural action on $SC_n$ and $SC_n/S_n$ has an induced partial order as well. Hence we have a map
\[\begin{array}{r c l}
f\ :\ \Ran^n(M)\times \R_{\geqslant 0} & \to & SC_n/S_n, \\
(P,t) & \mapsto & VR(P,t),
\end{array}\]
where $VR(P,t)$ is the Vietoris-Rips complex on $P$ with radius $t$. We include a $k$-cell in $VR(P,t)$ at the vertices $\{P_0,\dots,P_k\}\subset P$ if $d(P_i,P_j)<t$ for all $0\leqslant i<j\leqslant k$. Because we have strict inequality, the map is continuous in the upwards-directed, or Alexandrov topology on $SC_n/S_n$. Indeed, taking the preimage of an open set $U_S$ in $SC_n/S_n$ based at some simplicial complex $S$ (such $U_S$ form the basis of topology on $SC_n/S_n$), there is an open ball of radius $\min_{i<j} d(P_i,P_j)/2$ in the $\Ran^n(M)$ component and $\min_{(P_i,P_j)\subset f(P,t)} |t-d(P_i,P_j)|$ in the $\R_{\geqslant 0}$ component around any $(P,t)\in f^{-1}(U_S)$.

Remark: The above shows that $\Ran^n(M)\times \R_{\geqslant 0}$ is poset-stratified by $SC_n/S_n$, in the sense of Definition A.5.1 of Lurie. However, the strata are all of the same dimension, so there is no chance of this being a conical stratification, in the sense of Definition A.5.5 of Lurie. We hope to fix that with a different stratification.

Definition: Construct a poset $(A,\leqslant_A)$ in the following way:

• $SC_n/S_n \subset A$, with $S\leqslant_A T$ whenever $S\leqslant_{SC_n/S_n} T$ ,
• for every $S\neq T\in SC_n/S_n$, let $a_{ST}\in A$ with $a_{ST}\leqslant_A S$ and $a_{ST}\leqslant_A T$,
• for every $\{S_1,\dots,S_{k>2}\}\subset SC_n/S_n$, let $a_{S_1\cdots S_k}\in A$ with $a_{S_1\cdots S_k} \leqslant_A a_{S_1\cdots \widehat{S_i}\cdots S_k}$ for all $1\leqslant i\leqslant k$.

Define a map into $(A,\leqslant_A)$ in the following way:
\[\begin{array}{r c l}
g\ :\ \Ran^n(M)\times \R_{\geqslant 0} & \to & A, \\
(P,t) & \mapsto & \begin{cases}
S, & \text{ if $(P,t)\in \text{int}(f^{-1}(S))$ for some }S\in SC_n/S_n, \\
a_{S_1\cdots S_k}, & \text{ if }(P,t)\in \text{cl}(f^{-1}(T))\ \iff\ T\in \{S_1,\dots,S_k\}.
\end{cases}
\end{array}\]

We now claim that $g$ is a stratifying map.

Proposition: The map $g$ is continuous.

Proof: Since $\text{int}(f^{-1}(S))\cap \text{int}(f^{-1}(T)) = \emptyset$ for all $S\neq T\in SC_n/S_n$, the open sets $U_S\subseteq A$ based at $S$ all have open preimage $g^{-1}(U_S) \subseteq X$. Now take $(P,t)\in g^{-1}(U_{a_{S_1\cdots S_k}})$, for $k\geqslant 2$. If every open ball around $(P,t)\in X$ intersects $X_{a_{\mathbf T}}$, for some $\mathbf T \subseteq SC_n/S_n$, then $(P,t)$ must be in the closure of $f^{-1}(T)$, for every $T\in \mathbf T$. Hence the only possible such $\mathbf T$ are $\mathbf T\subseteq \{S_1,\dots,S_k\}$, so $g^{-1}(U_{a_{S_1\cdots S_k}})$ is open in $X$. $\square$

The next step would be to show that this stratification is conical, though it is not clear yet if it is.

References: Lurie (Higher Algebra, Appendix A)

Stratifying correctly

In a previous blog post ("A constructible sheaf over the Ran space," 2017-06-24) it was claimed that there was a particular constructible sheaf over $\Ran^{\leqslant n}(M)\times \R_{\geqslant 0}$. However, the proof actually uses finite ordered subsets of $M$ to make the stratification, rather than finite unordered subsets. This means that the sheaf is actually over $M^{\times n}\times \R_{\geqslant 0}$, and in this post we try to fix that problem.

Let $\Delta_n$ be the "fat diagonal" of $M^{\times n}$, that is, the collection of $P\in M^{\times n}$ for which at least two coordinates are the same. For every $k>0$, there is an $S_k$ action on $M^{\times k}\setminus \Delta_k$, quotienting by which we get a map
\[M^{\times k}\setminus \Delta_k \xrightarrow{\ q_k\ }\Ran^k(M)\]
to the Ran space of degree $k$. The stratification of $M^{\times k}\times \R_{\geqslant 0}$ given in the previous post will be pushed forward to a stratification of $\Ran^k(M)\times \R_{\geqslant 0}$, for all $0<k\leqslant n$. A large part of the work already has been done, it remains to put everything in the right order and check openness. The process is given as follows:

1. Stratify $\Ran^{\leqslant n}(M)\times \R_{\geqslant 0}$ into $n$ pieces, each being $\Ran^k(M)\times \R_{\geqslant 0}$.
2. Stratify $(M^{\times k}\setminus \Delta_k)\times \R_{\geqslant 0}$ as in the previous post.
3. Quotient by $S_k$-action to get stratification of $\Ran^k(M)\times \R_{\geqslant 0}$.

Step 1

As stated in the proof of the Theorem, $\Ran^{\geqslant k}(M)\times \R_{\geqslant 0}$ is open inside $\Ran^{\leqslant n}(M)\times \R_{\geqslant 0}$, allowing us to make a stratification $f:\Ran^{\leqslant n}(M)\times \R_{\geqslant 0}\to A$, where $A$ is the poset

 

where the tail of an arrow is ordered lower than the head. The map $f$ sends $\Ran^k(M)\times \R_{\geqslant 0}$ to $a_k$, which is a continuous map in the upset topology on $A$.

Step 2

As stated in Definition 5, we have a stratification $g_k:(M^{\times k}\setminus \Delta_k) \times \R_{\geqslant 0}\to B_k$, where $B_k$ may be viewed as a directed graph $B_k=(V_k,E_k)$. The vertex set is $V_k = \{0,1\}^{k(k+1)/2}$, whose elements are strings of 1 and 0, and the edge set $E_k$ contains $v\to v'$ iff $d_H(v,v')=1$ and $d_H(v,0)<d_H(v',0)$, for $d_H$ the Hamming distance. Let $U_v\subset B_k$ denote the upset based at $v$, that is, all elements $v'\in B_k$ with $v\leqslant v'$.

Order all distinct pairs $(i,j)\in \{1,\dots,k\}^2$, of which there are $k(k+1)/2$. Under the stratifying map $g_k$, each upset $U_v$ based at the vertex $v\in\{0,1\}^{k(k+1)/2}$ receives elements $(P,t)\in (M^{\times k}\setminus \Delta_k)\times \R_{\geqslant 0}$ satisfying $t>d(P_i,P_j)$ whenever the position representing $(i,j)$ in $v$ is 1. For example, when $k=3$,

To check that $g_k$ is continuous in the upset topology, we restate Lemma 2 in a clearer way.

Lemma 1: Let $U\subset X$ be open and $\varphi:X\to A\subset \R_{\geqslant 0}$ continuous, with $|A|<\infty$. Then
\[\bigcup_{x\in U}\{x\}\times (\varphi(x),\infty)\ \subseteq\ X\times (z',\infty)\]
is open, for any $z'\leqslant z:= \min_{x\in U}\{\varphi(x)\}$.

Proof: Consider the function
\[\begin{array}{r c l}
\psi\ :\ X\times (z',\infty) & \to & X\times (-\infty,z), \\
(x,t) & \mapsto & (x,\varphi(x)-t).
\end{array}\]
Since $\varphi$ is continuous and subtraction is continuous, $\psi$ is continuous (in the product topology). Since $U\times (-\infty,0)$ is open in $X\times (-\infty,z)$, the set $\psi^{-1}(U\times (-\infty,0))$ is open in $X\times (z',\infty)$. For any $x\in U$ and $t=\varphi(x)$, we have $\varphi(x)-t =0$. For any $x\in U$ and $t\to \infty$, we have $\varphi(x)-t\to -\infty$. It is immediate that all other $t\in (\varphi(x),\infty)$ give $\varphi(x)-t\in (-\infty,0)$. Hence $\psi^{-1}(U\times (-\infty,0))$ is the collection of points $(x,t)$ with $t\in (\varphi(x),\infty)$, which is then open in $X\times [0,z')$. $\square$

Applying Lemma 1 to $U=X=M^{\times k}\setminus \Delta_k$ and $\varphi(P) = \max_{i\neq j}\{d(P_i,P_j)\}$, which is continuous, gives that $g_k^{-1}(U_{11\cdots1})\subseteq M^{\times k}\setminus \Delta_k$ is open. This also works to show that $g_k^{-1}(U_v)\subseteq g_k^{-1}(U_{v'})$ is open, for any $v'\leqslant v$, by limiting the pairs of indices iterated over by the function $\varphi$. Hence $g_k$ is continuous.

Step 3

The symmetric group $S_k$ acts on $(M^{\times k}\setminus \Delta_k)\times \R_{\geqslant 0}$ by permuting the order of elements in the first factor. That is, for $\sigma\in S_k$, we have
\[\sigma (P=\{P_1,\dots,P_k\},t) = (\{P_{\sigma(1)},\dots,P_{\sigma(k)}\},t).\]
Note that $((M^{\times k}\setminus \Delta_k)\times \R_{\geqslant 0})/S_k = \Ran^k(M)\times \R_{\geqslant 0}$.

Remark: Graph isomorphism for two graphs with $k$ vertices may also be viewed as the equivalence relation induced by $S_k$ acting on $\Gamma_k = \{$simple vertex-labeled graphs with $k$ vertices$\}$. First, let $G_v$ be the (unique) graph first introduced at element $v\in B_k$ by $g_k$. That is, we have $G_v =VR(P,t)_1$ (the ordered 1-skeleton of the Vietoris--Rips complex on the set $P$ with radius $t$) whenever $g_k((P,t))\in U_v$ and $g_k((P,t))\not\in U_{v'}$ for any $v'\leqslant v$, $v'\neq v$. Then the elements of $B_k$ are in bijection with the elements of $\Gamma_k$ (given by $v\leftrightarrow G_v$), so we have $B_k/S_k = B_k'$. Recall that $v\leqslant v'$ in $B_k$ iff adding an edge to $G_v$ gives $G_{v'}$. In $B_{k'}$, this becomes a partial order on equivalence classes $[w] = \{v\in B_k\ :\ \sigma G_v=G_w\ $for some $\sigma\in S_k\}$. We write $[w]\leqslant [w']$ iff there is a collection of pairs $\{(v_1,v_1'),\dots,(v_\ell,v_\ell')\}$ such that $v_i\leqslant v_i'$ for all $i$, and $\{v_1,\dots,v_\ell\} = [w]$ and $\{v_1',\dots,v_\ell'\} = [w']$ (there may be repetition among the $v_i$ or $v_i'$).

By the universal property of the quotient, there is a unique map $h_k:\Ran^k(M)\times \R_{\geqslant 0}\to B_k'$ that makes the following diagram commute.

This will be our stratifying map. To check that $h_k$ is continuous take $U\subseteq B_k'$ open. As $\pi$ is the projection under a group action, it is an open map, so $\pi^{-1}(U)\subseteq B_k$ is open. Since $g_k$ is continuous in the upset topology, $g_k^{-1}(\pi^{-1}(U))$ is open. Again, $S_k\curvearrowright$ is the projection under a group action, so $(S_k\curvearrowright)(g_k^{-1}(\pi^{-1}(U)))$ is open, giving continuity of $h_k$.