A constructible sheaf over the Ran space

Let $M$ be a manifold. The goal of this post is to show that the sheaf $\mathcal F_{(P,t)}=\text{Rips}(P,t)$ valued in simplicial complexes over $\Ran^{\geqslant}(M)\times \R_{\geqslant 0}$ is constructible, a goal not quite achieved (see "A naive constructible sheaf," 2017-12-19 for a solution to the problems encountered here). This space will be described using filtered diagram of open sets, with the sheaf on consecutive differences of the diagram giving simplicial complexes of the same homotopy type.

Definition: Let $P=\{P_1,\dots,P_n\}\in \Ran(M)$. For every collection of open neighborhoods $\{U_i\owns P_i\}_{i=1}^n$ of the $P_i$ in $M$, there is an open neighborhood of $P$ in $\Ran(M)$ given by
$$\Ran(\{U_i\}_{i=1}^n) = \left\{Q\in \Ran(M)\ :\ Q\subset \bigcup_{i=1}^n U_i,\ Q\cap U_i\neq \emptyset\right\}.$$ Moreover, these are a basis for any open neighborhood of $P$ in $\Ran(M)$.

Sets

We begin with a few facts about sets. Let $X$ be a topological space.

Lemma 1: Let $A,B\subset X$. Then:

• (a) If $A\subset B$ is open and $B\subset X$ is open, then $A\subset X$ is open.
• (b) If $A\subset B$ is closed and $B\subset X$ is open, then $A\subset X$ is locally closed.
• (c) If $A\subset B$ is open and $B\subset X$ is locally closed, then $A\subset X$ is locally closed.
• (d) If $A\subset B$ is locally closed and $B\subset X$ is locally closed, then $A\subset X$ is locally closed.

Proof: For part (a), first recall that open sets in $B$ are given by intersections of $B$ with open sets of $A$. Hence there is some $W\subset X$ open such that $A = B\cap W$. Since both $B$ and $W$ are open in $X$, the set $A$ is open in $X$.
     For part (b), since $A\subset B$ is closed, there is some $Z\subset X$ closed such that $A=B\cap Z$. Since $B$ is open in $X$, $A$ is locally closed in $X$.

     For parts (c) and (d), let $B = W_1\cap W_2$, for $W_1\subset X$ open and $W_2\subset X$ closed. For part (c), again there is some $W\subset X$ open such that $A = B\cap W$. Then $A = (W_1\cap W_2)\cap W = (W\cap W_1)\cap W_2$, and since $W\cap W_1$ is open in $X$, the set $A$ is locally closed in $X$.

     For part (d), let $A = Z_1\cap Z_2$, where $Z_1\subset B$ is open and $Z_2\subset B$ is closed. Then there exists $Y_1\subset X$ open such that $Z_1 = B\cap Y_1$ and $Y_2\subset X$ closed such that $Z_2 = B\cap Y_2$. So $A = Z_1\cap Z_2 = (B\cap Y_1) \cap (B\cap Y_2) = (B\cap Y_1)\cap Y_2$, where $(B\cap Y_1)\subset X$ is open and $Y_2\subset X$ is closed. Hence $A\subset X$ is locally closed. $\square$

Lemma 2: Let $U\subset X$ be open and $f:X\to \R$ continuous. Then $\bigcup_{x\in U}\{x\}\times (f(x),\infty)$ is open in $X\times \R$.

Proof: Consider the function
$$\begin{array}{r c l} g\ :\ X\times \R & \to & X\times \R, \\ (x,t) & \mapsto & (x,t-f(x)). \end{array}$$ Since $f$ is continuous and subtraction is continuous, $g$ is continuous (in the product topology). Since $U\times (0,\infty)$ is open in $X\times \R$, the set $g^{-1}(U\times (0,\infty))$ is open in $X\times \R$. This is exactly the desired set. $\square$

Filtered diagrams

Definition: A filtered diagram is a directed graph such that

• for every pair of nodes $u,v$ there is a node $w$ such that there exist paths $u\to w$ and $v\to w$, and
• for every multi-edge $u\stackrel{1,2}\to v$, there is a node $w$ such that $u\stackrel 1\to v\to w$ is the same as $u\stackrel2\to v\to w$.

For our purposes, the nodes of a filtered diagram will be subsets of $\Ran^n(M)\times \R_{\geqslant 0}$ and a directed edge will be open inclusion of one set into another set (that is, the first is open inside the second). Although we require below that loops $u\to u$ be removed, we consider the first condition above to be satisfied if there exists a path $u\to v$ or a path $v\to u$.

Remark: In the context given,

• edge loops $U\to U$ and path loops $U\to\cdots\to U$ may be replaced by a single node $U$ ($U\subseteq U$ is the identity),
• multi-edges $U\stackrel{1,2}\to V$ may be replaced by a single edge $U\to V$ (inclusions are unique), and
• multi-edges $U\to V\to U$ may be replaced by a single node $U$ (if $U\subseteq V$ and $V\subseteq U$, then $U=V$).

A diagram with all possible replacements of the types above is called a reduced diagram.

Lemma 3: In the context above, a reduced filtered diagram $D$ of open sets of any topological space $X$ gives an increasing sequence of open subsets of $X$, with the same number of nodes.

Proof: Order the nodes of $D$ so that if $U\to V$ is a path, then $U$ has a lower index than $V$ (this is always possible in a reduced diagram). Let $U_1,U_2,\dots,U_N$ be the order of nodes of $D$ (we assume we have finitely many nodes). For every pair of indices $i,j$, set
$$\delta_{ij} = \begin{cases} \emptyset & \text{ if }U_i\to U_j\text{ is a path in }D, \\ U_i & \text{ if }U_i\to U_j\text{ is not a path in }D. \end{cases}$$ Then the following sequence is an increasing sequence of nested open subsets of $X$:
$$U_1 \to \delta_{12} \cup U_2 \to \delta_{13}\cup \delta_{23}\cup U_3 \to \cdots \to \underbrace{\left(\bigcup_{i=1}^{j-1}\delta_{ij} \right)\cup U_j}_{V_j} \to \cdots \to U_N.$$ Indeed, if $U_i \to U_j$ is a path in $D$, then $U_i$ is open in $V_j$, as $U_i\subset V_j$. If $U_i\to U_j$ is not a path in $D$, then $U_i$ is still open in $V_j$, as $U_i\subset V_j$. As unions of opens are open, and by Lemma 1(a), $V_{j-1}$ is open in $V_j$ for all $1<j<N$. $\square$

Remark 4: Note that every consecutive difference $V_j\setminus V_{j-1}$ is a (not necessarily proper) subset of $U_j$.

Definition 5: For $k\in \Z_{>0}$, define a filtered diagram $D_k$ over $\Ran^k(M)\times \R_{\geqslant 0}$ by assigning a subset to every corner of the unit $N$-hypercube in the following way: for the ordered set $S=\{(i,j) \ :\ 1\geqslant i<j\geqslant k\}$ (with $|S|=N=k(k-1)/2$), write $P=\{P_1,\dots,P_k\}\in \Ran^k(M)$, and assign
$$(\delta_1,\dots,\delta_k) \mapsto \left\{ (P,t)\in \Ran^k(M)\times \R_{\geqslant 0}\ :\ t>d(P_{(S_\ell)_1},P_{(S_\ell)_2}) \text{ whenever }\delta_\ell=0,\ \forall\ 1\leqslant \ell\leqslant k \right\},$$ where $\delta_\ell\in \{0,1\}$ for all $\ell$, and $d(x,y)$ is the distance on the manifold $M$ between $x,y\in M$. The edges are directed from smaller to larger sets.

Remark 6: This diagram has $2^{k(k-1)/2}$ nodes, as $k(k-1)/2$ is the number of pairwise distances to consider. Moreover, the difference between the head and tail of every directed edge is elements $(P,t)$ for which $\text{Rips}(P,t)$ is constant. 

Example: For example, if $k=3$, then $2^{3\cdot2/2}=8$, and $D_3$ is the diagram below. For ease of notation, we write $\{t>\cdots\}$ to mean $\{(P,t)\ :\ P=\{P_1,P_2,P_3\}\in \Ran^3(M),\ t>\cdots\}$.

The diagram of corresponding Vietoris-Rips complexes introduced at each node is below. Note that each node contains elements $(P,t)$ whose Vietoris-Rips complex may be of type encountered in any paths leading to the node.

Lemma 7: In the filtered diagram $D_k$, every node is open inside every node following it.

Proof: The left-most node of $D_k$ may be expressed as
$$\{(P,t)\ :\ P=\{P_1,\dots,P_k\}\in \Ran^k(M), t>d(P_i,P_j)\ \forall\ P_i,P_j\in P\} = \bigcup_{P\in \Ran^k(M)} \{P\}\times \left(\max_{P_i,P_j\in P}\{d(P_i,P_j)\},\infty\right).$$ Applying a slight variant of Lemma 2 (replacing $\R$ by an open ray that is bounded below), with the $\max$ function continuous, we get that the left-most node is open in the nodes one directed edge away from it. Repeating this argument, we get that every node is open inside every node following it. $\square$

The constructible sheaf

Recall that a constructible sheaf can be given in terms of a nested cover of opens or a cover of locally closed sets (see post "Constructible sheaves," 2017-06-13). The approach we take is more the latter, and illustrates the relation between the two. Let $n\in \Z_{>0}$ be fixed.

Definition: Define a sheaf $\mathcal F$ over $X=\Ran^{\leqslant n}(M)\times \R_{\geqslant 0}$ valued in simplicial complexes, where the stalk $\mathcal F_{(P,t)}$ is the Vietoris--Rips complex of radius $t$ on the set $P$. For any subset $U\subset X$ such that $\mathcal F_{(P,t)}$ is constant for all $(P,t)\in U$, let $\mathcal F(U)=\mathcal F_{(P,t)}$.

Note that we have not described what $\mathcal F(U)$ is when $U$ contains stalks with different homotopy types. Omitting this (admittedly large) detail, we have the following:

Theorem: The sheaf $\mathcal F$ is constructible.

Proof: First, by Remark 5.5.1.10 in Lurie, we have that $\Ran^{n}(M)$ is open in $\Ran^{\leqslant n}(M)$. Hence $\Ran^{\leqslant n-1}(M)$ is closed in $\Ran^{\leqslant n}(M)$. Similarly, $\Ran^{\leqslant n-2}(M)$ is closed in $\Ran^{\leqslant n-1}(M)$, and so closed in $\Ran^{\leqslant n}(M)$, meaning that $\Ran^{\leqslant k}(M)$ is closed in $\Ran^{\leqslant n}(M)$ for all $1\leqslant k\leqslant n$. This implies that $\Ran^{\geqslant k}(M)$ is open in $\Ran^{\leqslant n}(M)$ for all $1\leqslant k\leqslant n$, meaning that $\Ran^k(M)$ is locally closed in $\Ran^{\leqslant n}(M)$, for all $1\leqslant k\leqslant n$.
     Next, for every $1\leqslant k\leqslant n$, let $V_{k,1}\to \cdots \to V_{k,N_k}$ be a sequence of nested opens covering $\Ran^k(M)\times \R_{\geqslant 0}$, as given in Definition 5 and flattened by Lemma 3. The sets are open by Lemma 7. This gives a cover $\mathcal V_k = \{V_{k,1},V_{k,2},\setminus V_{k,1},\dots,V_{k,N_k}\setminus V_{k,N_k-1}\}$ of $\Ran^k(M)\times \R_{\geqslant 0}=V_{k,N_k}$ by consecutive differences, with $V_{k,1}$ open in $V_{k,N_k}$ and all other elements of $\mathcal V_k$ locally closed in $V_{k,N_k}$, by Lemma 1(b). By Lemma 1 parts (c) and (d), every element of $\mathcal V_k$ is locally closed in $\Ran^{\leqslant n}(M)\times \R_{\geqslant 0}$, and so $\mathcal V = \bigcup_{k=1}^n \mathcal V_k$ covers $\Ran^{\leqslant n}(M)\times \R_{\geqslant 0}$ by locally closed subsets.
     Finally, by Remarks 4 and 6, over every $V\in\mathcal V$ the function $\text{Rips}(P,t)$ is constant. Hence $\mathcal F|_V$ is a locally constant sheaf, for every $V\in \mathcal V$. As the $V$ are locally closed and cover $X$, $\mathcal F$ is constructible. $\square$

References: Lurie (Higher algebra, Section 5.5.1)