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. 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.

\[

\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 $\Ran(M)$.

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 $X$.

\[

\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 $1<j<N$. $\square$

\[

(\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.

\[

\{(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. $\square$

**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**

Definition:Definition:

\[

\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. $\square$

\[

\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. $\square$

**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.

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. $\square$

**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 $\mathcal F(U)=\mathcal F_{(P,t)}$.**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)