Exit paths, part 2

In this post we continue on a previous topic ("Exit paths, part 1," 2017-08-31) and try to define a constructible sheaf via universality. Let $X$ be an $A$-stratified space, that is, a topological space $X$ and a poset $(A,\leqslant)$ with a continuous map $f:X\to A$, where $A$ is given the upset topology relative to its ordering $\leqslant$. Recall the full subcategory $\Sing^A(X)\subseteq \Sing(X)$ of exit paths on an $A$-stratified space $X$.

Proposition: If $X\to A$ is conically stratified, $\Sing^A(X)$ is an $\infty$-category.

Briefly, a stratification $f:X\to A$ is conical if for every stratum there exists a particular embedding from a stratified cone into $X$ (see Lurie for "conical stratification" and Ayala, Francis, Tanaka for "conically smooth stratified space," which seem to be the same). We will leave confirming the described stratification as conical to a later post.

This proposition, given as part of Theorem A.6.4 in Lurie, has a very long proof, so is not repeated here. Lurie actually proves that the natural functor $\Sing^A(X)\to N(A)$ described below is a (inner) fibration, which implies the unique lifting property of $\Sing^A(X)$ via the unique lifting property of $N(A)$ (and we already know nerves are $\infty$-categories).

Example: The nerve of a poset is an $\infty$-category. Being a nerve, it is already immediate, but it is worthwhile to consider the actual construction. For example, if $A = \{a\leqslant b\leqslant c \leqslant d\}$ is the poset with the ordering $\leqslant$, then the pieces $N(A)_i$ are as below.

It is immediate that every 3-horn can only be filled in one unique way (as there is only one element of $N(A)_3$), as well as that every 2-horn can be filled in one unique way (as every sequence of two composable morphisms appears as a horn of exactly one element of $N(A)_2$).

In Appendix A.9 of Higher Algebra, Lurie says that there is an equivalence of categories $$(A\text{-constructible sheaves on }X)   \cong   \left[(A\text{-exit paths on }X),\mathcal S\right],$$ given that $X$ is conically stratified, and for $\mathcal S$ the $\infty$-category of spaces (equivalently $N(Kan)$, the nerve of all the simplicial sets that are Kan complexes). So, instead of trying to define  a particular constructible sheaf on $X = \Ran^{\leqslant n}(M)\times \R_{\geqslant 0}$, (as in previous posts "Stratifying correctly," 2017-09-17 and "A constructible sheaf over the Ran space," 2017-06-24) we will try to make a functor that takes an exit path of $X$ and gives back a space.

Fix $n\in \Z_{>0}$ and set $X = \Ran^{\leqslant n}\times \R_{\geqslant 0}$. Let $SC$ be the category of simplicial complexes and simplicial maps, with $SC_n$ the full subcategory of simplicial complexes with at most $n$ vertices. There is a map
$$\begin{array}{r c l} g\ :\ X & \to & SC_n \\ (P,t) & \mapsto & VR(P,t), \end{array}$$
allowing us to say
$$X = \bigcup_{S\in SC_n}g^{-1}(S).$$
Here we consider that two elements $P_i,P_j\in P$ give an edge of $VR(P,t)$ whenever $t>d(P_i,P_j)$ (this is chosen instead of $t\geqslant d(P_i,P_j)$ so that the boundaries of the strata ``facing downward," with respect to the poset ordering, are open). Now we define a stratifying poset $A$ for $X$.

Definition: Let $A = \{a_S\ :\ S\in SC_n\}$ and define a relation $\leqslant$ on $A$ by
$$\left(a_S\leqslant a_T\right)\ \ \Longleftarrow\ \ \left( \begin{array}{c} \exists\ \sigma\in \Sing(X)_1\ \text{such that}\\ g(\sigma(0))=S,\ g(\sigma(t>0))=T. \end{array} \right)$$
Let $(A,\leqslant)$ be the poset generated by relations of the type given above.

We claim that $f:X\to A$ given by $f(P,t)=a_{g(P,t)}$ is a stratifying map, that is, continuous in the upset topology on $A$. To see this, take the open set $U_S = \{a_T\in A\ :\ a_S\leqslant a_T\}$ in the basis of the upset topology of $A$, for any $S\in SC_n$, and consider $x\in f^{-1}(U_S)$. If for all $\epsilon>0$ we have $B_X(x,\epsilon)\cap f^{-1}(U_S)^C\neq \emptyset$, then there exists $T_\epsilon\in SC_n$ with $B_X(x,\epsilon)\cap f^{-1}(a_{T_\epsilon})\neq\emptyset$, for $S\not\leqslant T_\epsilon$ (as $T_\epsilon\not\in U_S$). This means there exists $\sigma\in \Sing(X)_1$ with $\sigma(0)=x$ and $\sigma(t>0)\in f^{-1}(a_{T_\epsilon})$, which in turn implies $S\leqslant T_\epsilon$, a contradiction. Hence $f$ is continuous, so $f:X\to A$ is a stratification.

As all morphisms in $\Sing(X)$ are compsitions of the face maps $s_i$ and degenracy maps $d_i$, so are all morphisms in $\Sing^A(X)$. There is a natural functor $F:\Sing^A(X)\to N(A)$ defined in the following way:
$$\begin{array}{r r c l} %% %% L1 %% \text{objects} & \left( \begin{array}{c} \sigma:|\Delta^k|\to X \\ a_0\leqslant \cdots \leqslant a_k\subseteq A \\ f(\sigma(t_0,\dots,t_i\neq 0,0,\dots,0)) = a_i \end{array} \right) & \mapsto & \left( a_0\to\cdots\to a_k\in N(A)_k\right) \\[20pt] %% %% L2 %% \text{face maps} & \left( \begin{array}{c} \left( \begin{array}{c} \sigma:|\Delta^k|\to X \\ a_0\leqslant \cdots \leqslant a_k\subseteq A \end{array} \right)\\[10pt] \downarrow \\[10pt] \left( \begin{array}{c} \tau:|\Delta^{k+1}|\to X \\ a_0\leqslant \cdots \leqslant a_i\leqslant a_i\leqslant \cdots a_k\subseteq A \end{array} \right) \end{array} \right) & \mapsto & \left(\begin{array}{c} \left(a_0\to\cdots \to a_k\right)\\[10pt] \downarrow\\[10pt] \left(a_0\to\cdots \to a_i\xrightarrow{\text{id}}a_i\to\cdots \to a_k\right) \end{array}\right)\\[40pt] %% %% L3 %% \text{degeneracy maps} & \left( \begin{array}{c} \left( \begin{array}{c} \sigma:|\Delta^k|\to X \\ a_0\leqslant \cdots \leqslant a_k\subseteq A \end{array} \right)\\[10pt] \downarrow \\[10pt] \left( \begin{array}{c} \tau:|\Delta^{k-1}|\to X \\ a_0\leqslant \cdots \leqslant a_{i-1}\leqslant a_{i+1}\leqslant \cdots a_k\subseteq A \end{array} \right) \end{array} \right) & \mapsto & \left(\begin{array}{c} \left(a_0\to\cdots \to a_k\right)\\[10pt] \downarrow\\[10pt] \left(a_0\to\cdots \to a_{i-1}\xrightarrow{\circ}a_{i+1}\to\cdots \to a_k\right) \end{array}\right) \end{array}$$
As all maps in $\Sing^A(X)$ are generated by compositions of face and degeneracy maps, this completely defines $F$. Naturality of $F$ follows precisely because of this.

A poset (which can be viewed as a directed simple graph) may be naturally viewed as a 1-dimensional simplicial set, moreover an $\infty$-category (by virtue of being a \emph{simple} graph, with no multi-edges or loops). Hence there is a natural map, the inclusion, that takes $N(A)$ into $N(\mathcal Kan) = \mathcal S$.  Finally, Construction A.9.2 of Lurie describes a map that takes a functor from $A$-exit paths into spaces and gives back an $A$-constructible sheaf over $X$, which Theorem A.9.3 shows to be an equivalence, given the following conditions:

• $X$ is paracompact,
• $X$ is locally of singular shape,
• the $A$-stratification of $X$ is conical, and
• $A$ satisfies the ascending chain condition.

The first condition is satisfied as both $\Ran^{\leqslant n}(M)$ and $\R_{\geqslant 0}$ are locally compact and second countable. The last condition is satisfied because $A$ is a finite poset. We already mentioned that the conical property will be checked later, as will the singular shape property. Unfortunately, Lurie gives a definition of singular shape only for $\infty$-topoi, so some work must be done to translate this into our simpler setting. However, in the introduction to Appendix A, Lurie says that if $X$ is "sufficiently nice" and we assume some "mild assumptions" about $A$, then the described categorical equivalence follows, so it seems there is hope that everything will work out well in the end.

References: Stacks Project, Lurie (Higher algebra, Appendix A), Ayala, Francis and Tanaka (Local structures on stratified spaces, Sections 2 and 3)

Ordering simplicial complexes

In the context of trying to make a constructible sheaf over the Ran space, we have made several attempts to stratify $X=\Ran^{\leqslant n}(M)\times \R_{\geqslant0}$ correctly, the hope being for each stratum to have a unique simplicial complex (the Vietoris-Rips complex of the elements of $X$). In this post we make some observations and examine what it means to move around in $X$.

We use the convention that a Vietoris--Rips complex $VR(P,t)$ of an element $(P,t)\in X$ contains an edge $(P_i,P_j)$ iff $d(P_i,P_j)>t$ (as opposed to $d(P_i,P_j)\geqslant t$).

Observation 1: The VR complex $VR(P,t)$ is completely described by its 1-skeleton $sk_1(VR(P,t))$, as having a complete subgraph $K_\ell\subseteq sk_1(VR(P,t))$ is equivalent to $VR(P,t)$ having an $(\ell-1)$-cell spanning that subgraph. The 1-skeleton is a simple graph $G=(V,E)$ on $k$ vertices, so if we can order simple graphs with $\leqslant n$ vertices, we can order VR complexes of $\leqslant n$ vertices.

Let $\Gamma_k$ be the collection of simple gaphs on $k$ vertices. From now on we talk about an element $(P=\{P_1,\dots,P_k\},t)\in X$, a $k$-vertex VR complex $S=VR(P,t)$, and its 1-skeleton $G=sk_1(VR(P,t))\in \Gamma_k$ interchangeably. Consider the following informal defintion of how the stratification of $X$ should work.

Definition: A VR complex $S$ is ordered lower than another VR complex $T$ if there is a path from the stratum of type $S$ to the stratum of type $T$ that does not pass through strata of type $R$ with $|V(R)|<|V(S)|$ or $|E(R)|<|E(S)|$. If $S$ is ordered lower than $T$ and we can move from the stratum of type $S$ to the stratum of type $T$ without passing through another stratum, then we say that $S$ is directly below $T$.

To gain intuition of what this ordering means, consider the ordering on the posets $B_k'$, as defined in a previous post ("Stratifying correctly," 2017-09-17) and the 1-skeleta of the VR-complexes mapped to their elements. A complete description for $k=1,2,3,4$ and partially for $k=5$ is given below, with arrows $S\to T$ indicating the minimal number of directly below relationships. That is, if $S\to R$ but also $S\to T$ and $T\to R$, then $S\to R$ is not drawn.

The orderings on each $B_k'$ are clear and can be found in an algorithmic manner. However, it is more difficult to see which $S$ at level $k$ are directly below which $T$ at level $k+1$. The green arrows follow no clear pattern.

Observation 2: If $G\in \Gamma_k$ has an isolated vertex and $t>0$, then it can be directly below $H\in \Gamma_{k+1}$ only if $|E(H)|=|E(G)|+1$. In general, if the smallest degree of a vertex of $G\in \Gamma_k$ is $d$ and $t>0$, then $G$ can be directly below $H\in \Gamma_{k+1}$ only if $|E(H)|=|E(G)|+d+1$.

Recall the posets $B_k'$ are made by quotienting the nodes of the hypercube $B_k=\{0,1\}^{k(k-1)/2}$ by the action of $S_k$, where an element of $B_k$ is viewed as a graph $G\in \Gamma_k$ having an edge $(i,j)$ if the coordinate corresponding to the edge $(i,j)$ is 1 (there are $k(k-1)/2$ pairs $(i,j)$ of a $k$-element set).

Observation 3: It is not clear that $G$ not being ordered lower than $H$ in the hypercube context (order increases when increasing in any coordinate) implies that the VR complex of $G$ is not ordered lower than the VR complex of $H$ in $X$. No counterexample exists in the example given above, but this does not seem to exclude the possibility.

If any conclusion can be made from this, it is that this may not be the best approach to take when stratifying $X$.

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 $h_k$.