Artin gluing a sheaf 4: a single sheaf in two ways

The goal of this post is to give an alternative perspective on making a sheaf over $X = \Ran^{\leqslant n}(M)\times \R_{\geqslant 0}$, alternative to that of a previous post ("Artin gluing a sheaf 3: the Ran space," 2018-02-05). We will have one unique sheaf on all of $X$, valued either in simplicial complexes or simplicial sets.

Remark: Here we straddle the geometric category $SC$ of simplicial complexes and the algebraic category $\sSet$ of simplicial sets. There is a functor $[\ \cdot\ ]:SC\to \sSet$ for which every $n$-simplex in $S$ gets $(n+1)!$ elements in $[S]$, representing all the ways of ordering the vertices of $S$ (which we would like to view as unordered, to begin with).

Recall from previous posts:

• maps $f:X\to SC$ and $g = [f]:X\to \sSet$,
• the $SC_k$-stratification of $\Ran^k(M)\times \R_{\geqslant 0}$,
• the point-counting stratification of $\Ran^{\leqslant n}(M)$,
• the combined (via the product order) $SC_{\leqslant n}$-stratification of $\Ran^{\leqslant n}(M)\times \R_{\geqslant 0}$,
• an induced (by the $SC_k$-stratification) cover by nested open sets $B_{k,1},\dots,B_{k,N_k}$ of $\Ran^k(M)\times \R_{\geqslant 0}$,
• a corresponding induced total order $S_{k,1},\dots,S_{k,N_k}$ on $f(\Ran^k(M)\times \R_{\geqslant0})$.

The product order also induces a cover by nested opens of all of $X$ and a total order on $f(X)$ and $g(X)$. We call a path $\gamma:I\to X$ a descending path if $t_1<t_2\in I$ implies $h(\gamma(t_1))\geqslant h(\gamma(t_2))$ in any stratified space $h:X\to A$. Below, $h$ is either $f$ or $g$.

Lemma: A descending path $\gamma:I\to X$ induces a unique morphism $h(\gamma(0))\to h(\gamma(1))$.

Proof: Write $\gamma(0) = \{P_1,\dots,P_n\}$ and $\gamma(1) = \{Q_1,\dots,Q_m\}$, with $m\leqslant n$. Since the path is descending, points can only collide, not split. Hence $\gamma$ induces $n$ paths $\gamma_i:I\to M$ for $i=1,\dots,n$, with $\gamma_i$ the path based at $P_i$. This induces a map $h(\gamma(0))_0\to h(\gamma(1))_0$ on 0-cells (vertices or 0-objects), which completely defines a map $h(\gamma(0))\to h(\gamma(1))$ in the desired category. $\square$

Our sheaves will be defined using colimits. Fortunately, both $SC$ and $\sSet$ have (small) colimits. Finally, we also need an auxiliary function $\sigma:\Op(X)\to SC$ that finds the correct simplicial complex. Define it by \[ \sigma(U)  = \begin{cases}
S_{k,\ell} & \text{ if } U\neq\emptyset, \text{ for } k = \max\{1\leqslant k'\leqslant n\ :\ U\cap \Ran^k(M)\times \R_{\geqslant 0}\neq \emptyset\}, \\ & \hspace{2.23cm} \ell = \max\{1\leqslant \ell'\leqslant N_k\ :\ U \cap B_{k,\ell'}\neq\emptyset\},\\
* & \text{ if }U= \emptyset.
\end{cases} \]

Proposition 1: Let $\mathcal F$ be the function $\Op(X)^{op}\to SC$ on objects given by \[ \mathcal F(U) = \colim\left(\sigma(U)\rightrightarrows S\ :\ \text{every }\sigma(U)\to S \text{ is induced by a descending }\gamma:I\to U\right). \] This is a functor and satisfies the sheaf gluing conditions.

Proof: We have a well-defined function, so we have to describe the restriction maps and show gluing works. Since $V\subseteq U\subseteq X$, every $S$ in the directed system defining $\mathcal F(V)$ is contained in the directed system defining $\mathcal F(U)$. As there are maps $\sigma(V)\to \mathcal F(V)$ and $S\to \mathcal F(V)$, for every $S$ in the directed system of $V$, precomposing with any descending path we get maps $\sigma (U)\to \mathcal F(V)$ and $S\to \mathcal F(V)$, for every $S$ in the directed system of $U$. Then universality of the colimit gives us a unique map $\mathcal F(U)\to \mathcal F(V)$. Note that if there are no paths (decending or otherwise) from $U$ to $V$, then the colimit over an empty diagram still exists, it is just the initial object $\emptyset$ of $SC$.

To check the gluing condition, first note that every open $U\subseteq X$ must nontrivially intersect $\Ran^n(M)\times \R_{\geqslant 0}$, the top stratum (in the point-counting stratification). So for $W = U\cap V$, if we have $\alpha\in \mathcal F(U)$ and $\beta \in \mathcal F(V)$ such that $\alpha|_W = \beta|_W$ is a $k$-simplex, then $\alpha$ and $\beta$ must have been $k$-simplices as well. This is because a simplicial takes a simplex to a simplex, and we cannot collide points while remaining in the top stratum. Hence the pullback of $S\owns \alpha$ and $T\owns \beta$ via some induced maps (by descending paths) from $U$ to $W$ and $V$ to $W$, respectively, will restrict to the identity on the chosen $k$-simplex. Hence the gluing condition holds, and $\mathcal F$ is a sheaf. $\square$

Functoriality of $[\ \cdot\ ]$ allows us to extend the proof to build a sheaf valued in simplicial sets.

Proposition 2: Let $\mathcal G$ be the function $\Op(X)^{op}\to \sSet$ on objects given by \[ \mathcal G(U) = \colim\left([\sigma(U)]\rightrightarrows S\ :\ \text{every }[\sigma(U)]\to S \text{ is induced by a descending }\gamma:I\to U\right). \] This is a functor and satisfies the sheaf gluing conditions.

Remark: The sheaf $\mathcal G$ is non-trivial on more sets. For example, any path contained within one stratum of $X$ induces the identity map on simplicial sets (though not on simplicial complexes). Hence $\mathcal G$ is non-trivial on every open set contained within a single stratum.

References: nLab (article "Simplicial complexes"), n-category Cafe (post "Simplicial Sets vs. Simplicial Complexes," 2017-08-19)

Artin gluing a sheaf 3: the Ran space

The goal of this post is to extend earlier ideas, of a sheaf defined on $\Conf_n(M)\times \R_{\geqslant 0}$, to a family of sheaves defined on $\bigcup_{k=1}^n \Conf_n(M)\times \R_{\geqslant 0} = \Ran^{\leqslant n}(M)\times \R_{\geqslant 0}$.

Recall our main map $f:\Conf_n(M)\times \R_{\geqslant 0} \tov{VR(-)} SC_n \tov{\Hom(\Delta^\bullet,-)} \sSet$. Following Definition 1 and Proposition 2 in a previous post ("Artin gluing a sheaf 2: simplicial sets and configuration spaces," 2018-01-31), define a sheaf $\mathcal F_k$ on $X_k$ by \[ \mathcal F_k(U) = \begin{cases}
S_{k,\max\{1\leqslant \ell\leqslant N_k\ :\ U\cap B_{\ell}\neq \emptyset\}} & \text{ if $U$ is good,}\\
S_\emptyset & \text{ else if }U\neq\emptyset,
\end{cases} \hspace{2cm} (1) \] for all $k=1,\dots,n$. We have assumed a total order on all simplicial complexes on $k$ vertices, induced by a cover $U_k,\dots,U_{k,N_k}$ of nested opens of $X_k$. This induces a total order $S_{k,1},\dots,S_{k,N_k}$ on the image of $\Ran^k(M)\times \R_{\geqslant 0}$ in $\sSet$, and by the product order, a total order on all of $\sSet' := f(\Ran^{\leqslant n}(M)\times \R_{\geqslant 0})$.

A small example

Let $n=3$, so $X = \Ran^{\leqslant 3}(M)\times \R_{\geqslant 0}$. We already have $\mathcal F_1,\mathcal F_2,\mathcal F_3$ on $X_1,X_2,X_3$, respectively, and we will extend them from the top down to sheaves over all of $X$, as in the diagram below.

The map $i$ will be the inclusion of an open set into a larger one, and $j$ the inclusion of a closed set into a larger one. Recall that the pullback of two sheaves is defined equivalently by a map of sheaves on the boundary of the open nd closed sets. With that in mind, for $U\subseteq X_2\cup X_3$ good, the pullback square

defines $\mathcal F_{d_0}$, where the $d_0$ indicates the face map that skips the $0$th spot. The sheaf $\mathcal F_{d_1}$ is defined similarly, but by the face map $d_1$, and $\mathcal F_{d_2}$ by the face map $d_2$. For each of these three sheaves on $X_3\cup X_2$, we have two other sheaves, based on where the single point maps to. However, we note that for $U\subseteq X$ good and $U\cap X_1\neq\emptyset$, \[ \left((i_*\mathcal F_{d_0}\times j_*\mathcal F_1)(U) \text{ defined by } d_0\right)
\ \ =\ \
\left((i_*\mathcal F_{d_1}\times j_*\mathcal F_1)(U) \text{ defined by } d_0\right), \] where $\times$ denotes the pullback over the appropriate sheaf, and similarly for the other sheaves on good sets intersecting $X_1$. We now have 6 unique shaves on all of $X$.

Generalizing

Now let $n$ be any positive integer, and $X = \Ran^{\leqslant n}(M)\times \R_{\geqslant 0}$. We reverse the indexation of the $\mathcal F_k$ and $X_k$ above to make notation less cumbersome (so now $\mathcal F_k$ is $\mathcal F_{n-k+1}$ from (1), over $X_k = \Ran^{n-k+1}(M)\times \R_{\geqslant 0}$). Define pullback sheaves $\mathcal F_{d_{\ell_1}}$ for $\ell_1=0,\dots,n$ on $X_2\cup X_2$ by the diagram

At the $k$th step, for $1<k<n$, we have sheaves $\mathcal F_{d_{\ell_1}\cdots d_{\ell_{k-1}}}$ over $\bigcup_{m=1}^k X_m$, defined by sequences of face maps $d_{\ell_{k-1}}$ when going from $X_k$ to $X_{k-1}$ and so on, where $\ell_m\in \{0,\dots,n-m+1\}$. Define pullback sheaves $\mathcal F_{d_{\ell_1}\cdots d_{\ell_{k-1}}d_{\ell_k}}$, for $\ell_k = 0,\dots,n-k+1$ on $\bigcup_{m=1}^{k+1} X_k$ by the diagram

At the end of this inductive process, we have $n!$ distinct sheaves $\mathcal F_{d_{\ell_1}\cdots d_{\ell_{n-1}}}$ on all of $X$. Note there is a sheaf map $\mathcal F_{d_{\ell_1}\cdots d_{\ell_i}\cdots d_{\ell_{n-1}}} \to \mathcal F_{d_{\ell_1}\cdots d_{\ell_i'}\cdots d_{\ell_{n-1}}}$, given on $U$ good by \[ \mathcal F_{d_{\ell_1}\cdots d_{\ell_i}\cdots d_{\ell_{n-1}}}(U) = S \mapsto \begin{cases}
S & \text{ if } |S_0| \leqslant n-i, \\
(\ell_i\ \ell_i')(S) & \text{ else,}
\end{cases} \] where $(\ell_i\ \ell_i')\in \mathfrak S_n$ (the symmetric group on the numbers $0,\dots,n-1$) is the transposition swaps the $\ell_i$ and $\ell_i'$ indices of $S_0$, the 0-cells of $S$, inducing a map of simplicial sets. If the two sheaves differ in only two indices $\ell_i\neq \ell_i'$ and $\ell_j\neq \ell_j'$, with $i<j$, then we get $S\mapsto (\ell_j\ \ell_j')_{d_{\ell_{i-1}} \cdots d_{\ell_j}}(\ell_i\ \ell_i')(S)$. Here $(\ell_j\ \ell_j')_{d_{\ell_{i-1}} \cdots d_{\ell_j}}$ is the element of $\mathfrak S_{n-i}$ found by taking $(\ell_j\ \ell_j')$ from $\mathfrak S_{n-j}$ to $\mathfrak S_{n-i}$ by the sequence of group inclusion maps induced by the face maps $d_{\ell_j},\dots,d_{\ell_{i-1}}$.

Remark: This construction is not the most satisfying, for several reasons:

• we do not have a single sheaf, rather a family of sheaves, and
• the use of "good" sets leaves something to be desired, as we should be able to consider larger sets.

Both will hopefully be remedied in a later post.

A naive constructible sheaf

In this post we describe a constructible sheaf over $X=\Ran^{\leqslant n}(M)\times \R_{>0}$ valued in simplicial complexes, for a compact, smooth, connected manifold $M$. We note however that it does not capture all the information about the underlying space. Thanks to Joe Berner for helpful ideas.

Recall the category $SC$ of simplicial complexes and simplicial maps, as well as the full subcategories $SC_n$ of simplicial complexes with $n$ vertices (the vertices are unordered). Let $A = \bigcup_{k=1}^n SC_n$ with the ordering $\leqslant_A$ as in a previous post ("Ordering simplicial complexes with unlabeled vertices," 2017-12-03), and $f:X\to A$ the stratifying map. Let $\{A_k\}_{k=1}^N$ be a cover of $X$ by nested open sets of the type $f^{-1}(U_S) = f^{-1}(\{T\in A\ :\ S\leqslant_A T\})$, whose existence is guaranteed as $A$ is finite. Note that $f(A_1)$ is a singleton containg the complete simplex on $n$ vertices.

Remark: For every simplicial complex $S\in A$, there is a locally constant sheaf over $f^{-1}(S)\subseteq X$. Given the cover $\{A_k\}$ of $X$, denote this sheaf by $\mathcal F_k \in \Shv(A_k\setminus A_{k-1})$ and its value by $S_k\in SC$.

Let $i^1:A_1\hookrightarrow A_2$ and $j^2:A_2\setminus A_1 \hookrightarrow A_2$ be the natural inclusion maps . Note that $A_1$ is open and $A_2\setminus A_1$ is closed in $A_2$. The maps $i^1,j^2$ induce direct image functors on the sheaf categories\[i^1_*:\Shv(A_1) \to \Shv(A_2),
\hspace{1cm}
j^2_*:\Shv(A_2\setminus A_1) \to \Shv(A_2).\]The induced sheaves in $\Shv(A_2)$ are extended by 0 on the complement of the domain from where they come. Note that since $A_2\setminus A_1\subseteq A_2$ is closed, $j^2_*$ is the same as $j^2_!$, the direct image with compact support. We then have the direct sum sheaf $i^1_*\mathcal F_1 \oplus j_*^2\mathcal F_2 \in \Shv(A_2)$, which we interpret as the disjoint union in $SC$. Then\[\left(i_*^1\mathcal F_1 \oplus j_2^*\mathcal F_2\right)(U) = \begin{cases}
S_1 & \text{ if }U\subseteq A_1, \\
S_2 & \text{ if }U\subseteq A_2\setminus A_1, \\
S_1\sqcup S_2 & \text{ else,}
\end{cases}
\hspace{1cm}
\left(i_*^1\mathcal F_1 \oplus j_2^*\mathcal F_2\right)_{(P,t)} = \begin{cases}
S_1 & \text{ if } (P,t)\in A_1, \\
S_2 & \text{ if }(P,t)\in \text{int}(A_2\setminus A_1), \\
S_1\sqcup S_2 & \text{ else,}
\end{cases}\]for $U\subseteq A_2$ open and $(P,t)\in A_2$. Generalizing this process, we get a sheaf on $X$. The diagram

may be helpful to keep in mind. We use the fact that direct sums commute with colimits (used in the definition of the direct image sheaf) to simplify notation. We then get sheaves\[\begin{array}{r c l}
\mathcal F^1 & \in & \Shv(A_1), \\
i_*^1\mathcal F^1 \oplus j_*^2 \mathcal F^2 & \in & \Shv(A_2), \\
i_*^2i_*^1\mathcal F^1 \oplus i_*^2j_*^2 \mathcal F^2 \oplus j_*^3 \mathcal F^3 & \in & \Shv(A_3), \\
i_*^3i_*^2i_*^1\mathcal F^1 \oplus i_*^3i_*^2j_*^2 \mathcal F^2 \oplus i_*^3j_*^3 \mathcal F^3 \oplus j_*^4 \mathcal F^4 & \in & \Shv(A_4),
\end{array}\]and finally\[i_*^{N-1\cdots 1}\mathcal F^1 \oplus \left(\bigoplus_{k=2}^{N-1} i_*^{N-1\cdots k}j_*^k \mathcal F^k \right) \oplus j_*^N \mathcal F^N \in \Shv(A_N=X),\]where $i_*^{N-1\cdots k}$ is the composition $i_*^{N-1} \circ i_*^{N-2} \circ \cdots \circ i_*^k$ of direct image functors. Call this last sheaf simply $\mathcal F \in \Shv(X)$. Each $i_*^k$ extends the sheaf by 0 on an ever larger domain, so every summand in $\mathcal F$ is non-zero on exactly one stratum as defined by $f:X\to A$. We now have a functor $\mathcal F:Op(X) \to SC$ defined by\[\mathcal F(U) = \bigsqcup_{k=1}^N S_k \delta_{U,A_K\setminus A_{k-1}},
\hspace{1cm}
\mathcal F_{(P,t)} = \bigsqcup_{k=1}^N S_k \delta_{(P,t),\text{cl}(,A_K\setminus A_{k-1})},\]where $\delta_{U,V}$ is the Kronecker delta that evaluates to the identity if $U\cap V \neq \emptyset$ and zero otherwise.

Remark: The sheaf $\mathcal F$ is $A$-constructible, as $\mathcal F|_{f^{-1}(S)}$ is a constant sheaf evaluating to the simplicial complex $S\in A$. However, if we want the cohomology groups to capture how the simplicial complexes change between strata, then we must use a different approach - all groups die when leaving a stratum because of the extension by zero construction.

References: nLab (article "Simplicial complexes")

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 $A$

Perspectives on the Ran space

This post combines the finite subset approach with the mapping space approach of the Ran space, in the context of stratifications. The goal is to understand the colimit construction of the Ran space, as that leads to more powerful results.

Topology

Let $X,Y$ be topological spaces.

Definition: The mapping space of $X$ with respect to $Y$ is the topological space $X^Y = \{f:Y\to X$ continuous$\}$. The topology on $X^Y$ is the compact-open topology which has as basis finite intersections of sets
\[\{f\in X^Y\ :\ f(K)\subseteq U\}, \hspace{3cm} (1)\]
for all $K\subseteq Y$ compact and all $U\subseteq X$ open.

Now fix a positive integer $n$.

Definition: The Ran space of $X$ is the space $\Ran^{\leqslant n}(X) = \{P\subseteq X\ :\ 0<|P|\leqslant n\}$. The topology on $\Ran^{\leqslant n}(X)$ is the coarsest which contains
\[\left\{P\in \Ran^{\leqslant n}(X)\ :\ P\subseteq \bigcup_{i=1}^k U_i,\ P\cap U_i\neq \emptyset\ \forall\ i \right\} \hspace{3cm} (2)\]
as open sets, for all nonempty finite collection of parwise disjoint open sets $\{U_i\}_{i=1}^k$ in $X$.

From now on, we let $I$ be a set of size $n$ and $M$ be a compact, smooth, connected $m$-manifold. There is a natural map
\[\begin{array}{r c l}
\varphi\ :\ M^I & \to & \Ran^{\leqslant n}(M), \\
(f:I\to M) & \mapsto & f(I).
\end{array}\]
This map is surjective, and for $n>1$, is not injective.

Proposition 1: The map $\varphi$ is continuous and an open map.

Proof: For continuity, take an open set $U\subseteq \Ran^{\leqslant n}(M)$ as in (2) and consider $\varphi^{-1}(U)$. We use the fact that $\{*\}\subset I$ is a compact (in fact open and closed) subset of $I$ and that all the $U_i$ are open, as is their union. Observe that
\begin{align*}\varphi^{-1}(U) & = \left\{f\in M^I \ :\ f(I)\subset \bigcup_{i=1}^k U_i,\ f(I)\cap U_i\neq \emptyset\ \forall\ i \right\} \\
& = \left\{f\in M^I\ :\ f(I)\subset \bigcup_{i=1}^k U_i\right\} \cap \bigcap_{i=1}^k \left\{f\in M^I\ :\ f(*\in I) \in U_i\right\},\end{align*}
which is a finite intersection of sets of the type (1), and so $\varphi^{-1}(U)$ is open in $M^I$.

For openness, take an open set $V$ as in (1), so $V = \bigcap_{i=1}^k \{f\in M^I\ :\ f(K)\subseteq U_i\}$ for different subsets $K\subseteq I$. By Lemma 1, we may assume that the $U_i$ are pairwise disjoint. For each $U_i$, let $\{U_{i,j}\}_{j=1}^\infty$ be a sequence of increasing open sets in $U_i$ such that $U_{i,j}\subseteq U_{i,j+1}$ and $U_{i,j}\xrightarrow{\ j\to\infty\ } U_i$. Then
\[\varphi (V) = \underbrace{\left\{P\in M\ :\ P\subset \bigcup_{i=1}^k U_i,\ P\cap U_i\neq \emptyset\ \forall\ i\right\}}_{f\in M^I\text{\ with image completely in the\ }U_i} \cup \underbrace{\bigcap_{i=1}^k \bigcup_{j=1}^\infty \left\{ P\in M\ :\ P\subset U_{i,j}\cup \left(\overline{U_{i,j}}\right)^c,\ P\cap U_{i,j}\neq \emptyset,\ P \subset \left(\overline{U_{i,j}}\right)^c \neq \emptyset\right\}}_{f\in M^I\text{\ with image partially in the\ }U_i}.\]
Note that $U_{i,j}$ and $\left(\overline{U_{i,j}}\right)^c$, the complement of the closure of $U_{i,j}$ are both open and disjoint in $M$. Since infinite unions and finite intersections of elements in the topology are also open, we have that $\varphi(V)$ is open in $\Ran^{\leqslant n}(M)$. $\square$

The above proposition says that we may talk equivalently about the compact-open topology on $M^I$ and the Ran space topology on $\Ran^{\leqslant n}(M)$. Viewing the Ran space as a function space allows for more general terminology to be applied.

Lemma 1: Let $U_i\subseteq M$ be open, for $i=1,\dots,k$. Then $\bigcap_{i=1}^k\{f\in M^I\ :\ f(K)\subseteq U_i\}$ may be written as a union of intersections $\bigcap_{j=1}^\ell \{f\in M^I\ :\ f(K)\subseteq V_j\}$ with the $V_j$ open, pairwise disjoint, and $\ell\leqslant k$.

Proof: It suffices to prove this in the case $k=2$. Let $U,V\subseteq M$ open and suppose than $U\cap V\neq \emptyset$. Note that $U\setminus V$ and $V\setminus U$ are separated (that is, $(U\setminus V) \cap \overline{V\setminus U} = \emptyset$ and $(V\setminus U)\cap \overline{U\setminus V} = \emptyset$), and since $\R^N$ is a completely normal space (equivalently, satisfies the $T5$ axiom), there exist disjoint open sets $A,B$ with $U\setminus V\subseteq A$ and $U\setminus V\subseteq B$. So for $A' = A\cap (U\cup V)$ and $B' = B\cap (U\cup V)$, we have
\begin{align*} \{f\in M^I\ &:\ f(K)\subseteq U\} \cap \{f\in M^I\ :\ f(K)\subseteq V\} \\
& = \left(\{f\in M^I\ :\ f(K)\subseteq U\setminus V\} \cap \{f\in M^I\ :\ f(K)\subseteq V\setminus U\}\right) \cup \{f\in M^I\ :\ f(K)\subseteq U\cap V \} \\
& = \left(\{f\in M^I\ :\ f(K)\subseteq A'\} \cap \{f\in M^I\ :\ f(K)\subseteq B'\}\right) \cup \{f\in M^I\ :\ f(K)\subseteq U\cap V \}, \end{align*}
for $A',B',U\cap V$ open, and $A'\cap B' = \emptyset$.  $\square$

Note that in the last calculation of the proof, the intersection of sets in the second line is smaller than the intersection of sets in the last line (as $U\setminus V \subsetneq A$ and $V\setminus U\subsetneq B$). However, all the extra ones in the third line appear in the set $\{f\in M^I\ :\ f(K)\subseteq U\cap V\}$.

Stratifications

Now we compare stratifications on $M^I$ and $\Ran^{\leqslant n}(M)$. As before, $I$ is a set of size $n$.

Corollary: An image-constant $A$-stratification on $M^I$ is equivalent to an $A$-stratification on $\Ran^{\leqslant n}(M)$.

This follows from Proposition 1. By image-constant we mean if $\alpha,\beta\in M^I$ have the same image (that is, $\alpha(I)=\beta(I)$), then $\alpha,\beta$ are sent to the same element of $A$.

Proof: If we start with a continuous map $f:M^I\to A$, setting $g(P) = f(I\to M)$ whenever $(I\to M) \in \varphi^{-1}(P)$ is continuous, as $\varphi(f^{-1}(U))$ is open, by continuity of $f$ and openness of $\varphi$. The assignment $g(P) = f(I\to M)$ whenever $(I\to M) \in \varphi^{-1}(P)$ is well defined, as the stratification is image-constant, so any continuous map from $M^I$ must send every element of $\varphi^{-1}(P)$ to the same place.

Conversely, if we start with a continuous map $g:\Ran^{\leqslant n}(M)\to A$, setting $f(I\to M) = g(\varphi(I\to M))$ is continuous, as $\varphi^{-1}(g^{-1}(U))$ is open, by continuity of $g$ and continuity of $\varphi$. This map is image-constant, as $\varphi(\alpha:I\to M) = \alpha(I)$. $\square$

Next we consider a particular stratification of $M^I$, adapted from Example 3.5.17 of Ayala-Francis-Tanaka, simplified with $P=\{*\}$. That is, the example begins with a stratified space $M\to P$ and proceeds to construct another stratification $M^I\to P'$, but we only consider the trivial stratification $M\to \{*\}$.

Definition: Given $M$ and $I$, let the poset $\mathcal P(I)$ of coincidences on $I$ be the set of equivalence relations on $I$, ordered by reverse set inclusion. Let $f_I:M^I\to \mathcal P(I)$ be the natural stratification that takes a map $\alpha: I\to M$ to the equivalence relation on $I$ describing which elements of $I$ coincide in the image of $\alpha$.

Example: An element of $\mathcal P(I)$ is a subset of $I\times I$ always containing $(a,a)$ for every $a\in I$ (reflexivity), and satisfying the symmetry and transitivity conditions. For example, if $|I|=3$ or 4, then $\mathcal P(I)$ is ordered as in the diagrams below, with order increasing from left to right. We simplify things by writing $[x_1,\dots,x_k]$ for the collection $(x_i,x_j)$ of all $i\neq j$ (the equivalence class).

To check that the map $f_I:M^I\to \mathcal P(I)$ is continuous, we first note that an element $U_{[x_1],\dots,[x_k]}$ in the basis of the upwards-directed topology on $\mathcal P(I)$ contains images of $\alpha\in M^I$ whose images have at most the elements of each equivalence class $[x_i]$ coinciding. Hence
\[ f_I^{-1}(U_{[x_1],\dots,[x_k]}) = \bigcup_{U_1,\dots,U_k\subseteq M \atop \text{open, disjoint}}\ \bigcap_{i=1}^k \left\{\alpha\in M^I\ :\ \alpha(K = \{x\in [x_i]\}) \subseteq U_i\right\},\]
which is an open set in the compact-open topolgy on $M^I$.

The Ran space as a colimit

Beilinson-Drinfeld (Section 3.4) and Ayala-Francis-Tanaka (Section 3.7) describe the Ran space as a colimit, the former of a functor into topological spaces, the latter of a functor into stratified spaces. See Mac Lane for a full treatment of colimits. Both BD and AFT use the category $\Fin^{surj,\leqslant n}$ of finite sets and surjections, that is,
\begin{align*}\Obj(\Fin^{surj,\leqslant n}) & = \{I\in \Obj(\Set)\ :\ 0<|I|\leqslant n\}, \\
\Hom_{\Fin^{surj,\leqslant n}}(I,J) & = \begin{cases}
\emptyset, & \text{\ if\ } |I|<|J|, \\
\left\{\text{surjections\ }I\to J\right\}, & \text{\ if\ } |I|\geqslant |J|.
\end{cases}\end{align*}
AFT uses more involved terminology, with "conically smooth" stratified spaces instead of just poset-stratified. They use a category $\Strat$, which for our purposes we may define as
\begin{align*}
\Obj(\Strat) & = \{\text{poset-stratified topological spaces }X\xrightarrow{ f } A\}, \\
\Hom_{\Strat}(X\xrightarrow{ f } A,Y\xrightarrow{ g } B) & = \{(\mu\in \Hom_{\Top}(X,Y), \nu \in \Hom_{\Set}(A,B)\ :\ g\circ \mu = \nu\circ f\}.
\end{align*}

Remark:  There is a natural functor $\mathcal F_M:(\Fin^{surj,\leqslant n})^{op} \to \Top$, given by $I\mapsto M^I$. A surjection $s:I\to J$ induces a map $M^J\to M^I$, with $(f:J\to M)\mapsto (f\circ s :I\to M)$. BD use this to declare that $\Ran^{\leqslant n}(M) = \colim(\mathcal F_M)$.

Remark: There is also a natural functor $\mathcal G_M:(\Fin^{surj,\leqslant n})^{op} \to \Strat$, given by $I\mapsto (M^I\to \mathcal P(I))$. AFT use this to declare that $(\Ran^{\leqslant n}(M)\to \{1,\dots,n\}) = \colim(\mathcal G_M)$.

The construction of AFT is even more general, as they consider the Ran space of an already stratified space. Here we use their result for $M\to \{*\}$ trivially stratified.

References: Ayala, Francis, and Tanaka (Local structures on stratified spaces, Sections 3.5 and 3.7), Beilinson and Drinfeld (Chiral algebras, Section 3.4), Mac Lane (Categories for the working mathematician, Chapter III.3)

The point-counting stratification of the Ran space is conical (really though)

This post is meant to correct the issues of a previous post ("The point-counting stratification of the Ran space is conical," 2017-11-06). The setting is the same as before.

A key object here will be the Ran space of a disconnected space. If $M$ has $k$ connected components, then $\Ran^{\leq n}(M)$, for $k\leq n$, has $2^k$ connected components (as we must choose whether or not to have at least one point in each component). When $M$ is disconnected with $k\leq n$ components, let $\Ran^{\leq n}(M)'$ be the largest connected component of $\Ran^{\leq n}(M)$, that is, the one with at least one point in every component of $M$.

Proposition 1: The point-counting stratification $f:X\to A$ is conical.

Proof: Fix $P=\{P_1,\dots,P_k\}\in \Ran^k(M)\subseteq \Ran^{\leqslant n}(M)$ and set $2\epsilon = \min_{i<j}d(P_i,P_j)$. Set
\[
Z := \prod_{i=1}^k B^{\R^m}_{\epsilon/3}(0),
\hspace{.5cm}
Y := \Bigg\{(Q^1,\dots,Q^k)\in \underbrace{\Ran^{\leq n}\left(\coprod_{i=1}^k B^{\R^m}_{2\epsilon/3}(0)\right)'}_{W}\ :\ \overbrace{\sum_{i=1}^k \mathbf d(\{0\},Q^i)=\frac{2\epsilon}3}^{\text{the cone condition}}\ ,\ \hspace{-5pt}\underbrace{\sum_{j=1}^{|Q^i|}Q_j^i =0}_{\text{the centroid condition}}\hspace{-5pt}\ \forall\ i\Bigg\}.
\]
The cone condition ensures the right topology at the cone point of $C(Y)$. The centroid condition ensures injectivity of $\varphi_P$ when multiplying by $Z$. Reasoning for the constants $\epsilon/3$ and $2\epsilon/3$ is given in Proposition 2. The balls $B_{\epsilon/3}$ in $Z$ are open and the balls $B_{2\epsilon/3}$ in $Y$ are closed. The collection $Q^i$ is all the points in the $i$th ball $B_{2\epsilon/3}$. The product $Z\times C(Y)$ is given the topology for which a set $U$ is open if $\varphi_P(U)$ is open in the subspace topology on $\im(\varphi_P)\subseteq X$.

If $k=n$, then $Y=\emptyset$ because of the cone and centroid conditions. Then $C(Y)=*$ and $Z\times * \cong Z$ is by construction an open set in $X$, so $X$ is conically stratified at $P$. If $k<n$, then $Y$ has a natural $A_{>k}$ stratification\[(Q^1,\dots,Q^k) \mapsto \sum_{i=1}^k |Q^i| \in \{k+1,\dots,n\},\]
with the image landing in $A_{>k}=\{k+1,\dots,n\}$ because there are at least $k$ points in every element of $W$, and to satisfy the cone condition at least one $Q^i$ must have at least two points. We now claim that
\[\begin{array}{r c l}
\varphi_P\ :\ Z \times C(Y) & \to & X, \\
(R,(Q^1,\dots,Q^k),t\neq 0) & \mapsto & \{P_1+R_1+tQ^1,\dots,P_k+R_k+tQ^k\},\\
(R,*,0) & \mapsto & \{P_1+R_1,\dots,P_k+R_k\}
\end{array}\]
is an open embedding. Here $t\in [0,1)$ is the cone component and $P_i+R_i+Q^i$ is the collection of all $P_i+R_i+Q^i_j,$ for $j=1,\dots,|Q^i|$. Injectivity follows from the cone condition and centroid condition on $Y$: it is clear that for fixed $R\in Z$ the map $\varphi_P$ is injective, as we are taking points at different distances from $P_i+R_i$ (the cone condition). Because of the centroid condition, moving around to different $R$ means the centroid will also move, so we will not get a collection of points we previously had.

Further, $Z\times C(Y)$ gets mapped homeomorphically into $X$ because of the topology forced upon it - continuity and openness of the map follow immediately. Finally, since $\im(\varphi_P)$ is open in $X$, $\varphi_P$ is an open embedding, so $X$ is conically stratified at $P$. $\square$

The following proposition gives a better idea of where $\im(\varphi_P)$ actually lands in $X$.

Proposition 2: For $\varphi_P$ as above, we have $B^X_{2\epsilon/3k}(P)\subseteq \im(\varphi_P)\subseteq B^X_{\epsilon}(P)$.

Proof: For the first inclusion, take $S\in B^X_{2\epsilon/3k}(P)$ and for every $1\leq i\leq k$ set
\begin{align*}S^i & := \{s\in S\ :\ d(s,P_i)<d(s,P_j)\ \forall\ j\neq i\}, & (\text{points closest to $P_i$}) \\
T_i & := \frac1{|S^i|}\sum_{j=1}^{|S^i|} S^i_j, & (\text{centroid of the $S^i_j$}) \\
c_i & := \mathbf d(\{T_i\},S^i). & (\text{distance from the $S^i_j$ to their centroid})\end{align*}
If $|S|=k$, then $c_i=0$ for all $i$, and $t=0$, so we are at the cone point, and $S = \varphi_P(T-P,*,0)$. If $|S|>k$, then $0<\sum_i c_i <2\epsilon/3$, as $c_i < 2\epsilon/{3k}$ for all $i$, so there is some $t'\in (0,1)$ such that $\sum_i c_i = t'2\epsilon/3$. Then
\begin{align*}\varphi_P\left(T-P, \left( {\textstyle \frac1{t'}} (S^1-T_1),\dots, {\textstyle \frac1{t'}}(S^k-T_k) \right), t'\right)
& = \left\{ P_1+(T-P)_1 + t'{\textstyle \frac1{t'}}(S^1-T_1), \dots, P_k+(T-P)_k + t'{\textstyle \frac1{t'}}(S^k-T_k)\right\} \\
& = \left\{ P_1+T_1-P_1 + S^1-T_1, \dots, P_k+T_k-P_k + S^k-T_k\right\} \\
& = \left\{S^1, \dots, S^k\right\} \\
& = S.\end{align*}
Note that since $T_i$ is the centroid of the $S^i_j$, and $S^i \subset B^M_{2\epsilon/3k}(P_i)$, and the centroid of a collection of points is in their convex hull, we also have $T_i\in B^M_{2\epsilon/3k}(P_i)$. Since $\frac{2\epsilon}{3k}<\frac{\epsilon}3$ when $k>2$, we have that $P_i-T_i\in B^{\R^m}_{\epsilon/3}(0)$. If $k\leq 2$, then use $k+2$ instead of $k$. Finally, since $\sum_i c_i/t' = 2\epsilon/3$ and $c_i,t\geq 0$, we have that $c_i/t'\leq 2\epsilon/3$, meaning that $\frac1{t'}(S^i-T_i)\subset B^{\R^m}_{2\epsilon/3}(0)$. Hence the argument of $\varphi_P$ given above is in the domain of $\varphi_P$.

For the second inclusion, first fix $i$. For an element in the image of $\varphi_P$, note that $d(P_i,P_i+R_i)\leqslant \epsilon/3$ and $\mathbf d(\{R_i\},tQ^i)<\mathbf d(\{R_i\},Q^i)\leq 2\epsilon/3$. Since $d(P_i,R_i) = \mathbf d(\{P_i\},\{R_i\})$, we have that
\begin{align*}\mathbf d(\{P_i\},P_i+R_i+tQ^i) & \leqslant \mathbf d(\{P_i\},\{P_i+R_i\})+\mathbf d(\{P_i+R_i\},P_i+R_i+tQ^i) & (\text{triangle inequality}) \\
& = \mathbf d(\{0\},\{R_i\}) + \mathbf d(\{0\},tQ^i) & (\text{linearity of $\mathbf d$}) \\
& = \mathbf d(\{0\},\{R_i\}) + t\mathbf d(\{0\},Q^i) & (\text{linearity of $\mathbf d$}) \\
& < \frac{\epsilon}3 +\frac{2\epsilon}3 & (\text{assumption}) \\
& = \epsilon.\end{align*} $\square$

The following diagram describes the last calculation in the proof.

The point-counting stratification of the Ran space is conical

Note: There are problems with the proof here, in particular with making the map $\varphi$ an embedding. The mistakes are corrected in a later post ("The point-counting stratification of the Ran space is conical (really though)," 2017-11-15).

---

This post completes the effort of several previous posts to show that $f:\Ran^{\leqslant n}(M)\to A=\{1,\dots,n\}$ is a conically stratified space, where $f$ is the point-counting map, for $M$ a compact smooth $m$-manifold embedded in $\R^N$.

Remark: Since $M$ is a manifold, we will work on $M$ or through charts in $\R^m$, as necessary, without explicitly mentioning the charts or domains. Balls $B^M_\lambda, B^{\R^m}_\lambda$ of radius $\lambda$ will be closed and $\mathcal B^{\R^m}_\lambda,\mathcal B^X_\lambda$ will be open. We write $d$ for distance between points of $M$ (or $\R^m$) and $\mathbf d$ for distance between finite subsets of $\R^m$. This is essentially the definition given by Remark 5.5.1.5 of Lurie: \[ \mathbf d(P,Q) = \frac 12 \left(\sup_{p\in P}\inf_{q\in Q} d(p,q) + \sup_{q\in Q}\inf_{p\in P}d(p,q)\right). \] We add the $\frac 12$ so that $\mathbf d(\{p\},\{q\}) = d(p,q)$. Note also $\sup,\inf$ may be replaced by $\max,\min$ in the finite case.

Remark: In our context, given $P\in X$, $\mathbf d$ may be thought of as how far away have new points split off from the $P_i$. That is, if $Q\in X$ is close to $P$ representing the $P_i$ splitting up, then $\mathbf d(P,Q)$ is (half) the sum of the distance to the farthest point splitting off from the $P_i$ and to the farthest point among every $P_i$'s closest point. The diagram below gives the idea.

Then the distance between $P$ and $Q$ is given by
\begin{align*} \mathbf d(P,Q) & = \frac12 \left(\sup_{P_i}\left\{\inf_{Q_j}\left\{d(P_i,Q_j)\right\}\right\} + \sup_{Q_j}\left\{\inf_{P_i}\left\{d(P_i,Q_j)\right\}\right\}\right) \\
& = \frac12\left(\sup \left\{\inf\left\{a,b,c\right\}, \inf\left\{d,e,f,g\right\}\right\} + \sup \left\{ a,b,c,d,e,f,g \right\}\right) \\
& = \frac12\left( \sup\left\{ a,g \right\} + c \right) \\
& = \frac12(a+c). \end{align*}

Now we move on to the main result.

Proposition: The point-counting stratification $f:X\to A$ is conical.

Proof: Fix $P=\{P_1,\dots,P_k\}\in \Ran^k(M)\subseteq \Ran^{\leqslant n}(M)$ and set $2\epsilon = \min_{i<j}d(P_i,P_j)$. Set \[ Z = \prod_{i=1}^k \mathcal B^{\R^m}_\epsilon(0),
\hspace{2cm}
Y = \coprod_{\sum \ell_i=n \atop \sum t_i = \epsilon} \prod_{i=1}^k\ \left\{Q\in \Ran^{\ell_i}(B^{\R^m}_{t_i}(0))\ :\ \textbf d(0,Q) = t_i,\ \textstyle \sum Q_j = 0 \right\}, \] both of which are topological spaces. The first condition on elements of $Y$ is the cone condition, which ensures the right topology at the cone point in $C(Y)$. The second condition on $Y$ is the centroid condition, which ensures that the point to which 0 maps to (under $\varphi$) is the centroid of points splitting off it, so that we don't overcount when multiplying by $Z$. For $C(Y) = (Y\times [0,1))/(Y\times \{0\})$ the cone of $Y$, define a map \[ \begin{array}{r c l}
\varphi\ :\ C(Y)\times Z & \to & X, \\
\left(\Ran^{\ell_i}(B_{t_i}^{\R^m}(0)),t,R\right) & \mapsto & \Ran^{\ell_i}(B_{tt_i}^M(R_i)),
\end{array} \] where $t\in [0,1)$ is the cone component and $R=\{R_1,\dots,R_k\}\in Z$ is an element of $\Ran^k(M)$ near $P$. It is sufficient to describe where the $\Ran^{\ell_i}$ map to, as all the $Q$ in a fixed $\Ran^{\ell_i}$ map in the same way into $X$.

The map $\varphi$ is continuous by construction, injective by the centroid condition, and a homeomorphism onto its image by the cone condition. Hence $\varphi$ is an embedding, and since the image is open, it is an open embedding. Note that we are taking "open embedding" to mean an embedding whose image is open. Hence every $P\in X$ satisfies Definition A.5.5 of Lurie, so $f:X\to A$ is conically stratified.  $\square$

Remark: Observe that $\mathcal B^X_{\epsilon/k}(P)\subseteq \im(\varphi) \subseteq \mathcal B^X_\epsilon(P)$, both inclusions coming from the $\sum t_i=\epsilon$ condition.

Combined with the proposition of a previous post ("Splitting points in two," 2017-11-02) and Theorem A.9.3 of Lurie, it follows that $A$-constructible sheaves on $X$ are equivalent to functors of $A$-exit paths on $X$ to the category $\mathcal S$ of spaces. A previously given construction (in "Exit paths, part 2," 2017-09-28) gives such a functor, indicating that there exists an $A$-constructible sheaf on $X$.

Next steps may involve applying this approach to the space $\Ran^{\leqslant n}(M)\times \R_{\geqslant 0}$, which was the motivator for all this, or continuing with Lurie's work to see how far this can be taken.

References: Lurie (Higher Algebra, Appendix A), nLab (article "Embedding of topological spaces")

Attempts at proving conical stratification

This post chronicles several attempts and failures to show that $X=\Ran^{\leqslant n}(M)$ is conically stratified. Here $M$ will be a smooth, compact manifold of dimension $m$, embedded in $\R^N$ for $N\gg 0$. Recall that a stratified space $f:X\to A$ is conically stratitifed at $x\in X$ if there exist:

1. a stratified space $g:Y\to A_{>f(x)}$,
2. a topological space $Z$, and
3. an open embedding $Z\times C(Y)\hookrightarrow X$ of stratified spaces whose image contains $x$.

The cone $C(Y)$ has a natural stratification $g':C(Y)\to A_{\geqslant f(x)}$, as does the product $Z\times C(Y)$. The space $X$ itself is conically stratified if it is conically stratfied at every $x\in X$.

Let $P=\{P_1,\dots,P_k\}\in \Ran^k(M)\subseteq \Ran^{\leqslant n}(M)=X$, and $2\epsilon = \min_{1\leqslant i<j\leqslant k}\{d(P_i,P_j)\}$.

Observations

Observation 1: When $M=I = (0,1)$, the interval, we can visualize what $\Ran^{\leqslant 3}(M)$ looks like via the construction $\Ran^{\leqslant 3}(M) = (M^3\setminus \Delta_3)/ S_3$, to gain some intuition about what the Ran space looks like in general. 

A drawback is that $\dim(M)=1$, which masks the problems in higher dimensions.

Observation 2: An open neighborhood of $P\in X$ looks like
\[\coprod_{\sum \ell_i = n \atop \ell_i\in \Z_{>0}}\ \prod_{i=1}^k \Ran^{\leqslant \ell_i}(B^M_\epsilon(P_i)) = B^X_{\epsilon/2}(P) \times \coprod_{\sum \ell_i = n \atop \ell_i\in \Z_{>0}}\ \prod_{i=1}^k \Ran^{\leqslant \ell_i}(B^M_{\epsilon/2}(P_i)),\hspace{2cm} (1) \]
for $B^M_\epsilon(x) = \{y\in M\ :\ d_M(x,y)<\epsilon\}$ the open ball of radius $\epsilon$ around $x\in M$, and similarly for $P\in X$. Most attempts to prove conical stratification are based around expressing these as $Z\times C(Y)$, usually for $Z=B_{\epsilon/2}^X(P)$.

Observation 3: When $k<n$, the "steepest" direction from $P_i$ into the highest stratum of $X$ is given by $P_i$ splitting into $n-k+1$ points uniformly distributed on $\partial B^M_t(P_i)$. Hence the $[0,1)$ part of the cone (recall $C(Y)=Y\times [0,1)/\sim$) should be along $t\in [0,1)$.

Attempts

Attempt 1: Use more resrictive (but better described) AFT definition.

Ayala-Francis-Tanaka describe $C^0$ stratified spaces, a special type of stratified space. Any space that has a cover by topological manifolds is a $C^0$ stratified space, however it seems that $X$ cannot be covered by topological manifolds. Even further, each element in the cover must have the trivial stratification, and since we must have overlaps, $f:X\to A$ will have $A=\{*\}$, which is not what we want.

Attempt 2: Stratify $\Ran^{\leqslant n}(M)\times \R_{\geqslant 0}$ instead.

This is more difficult, but was the original impetus, with strata defined by collecting the Vietoris-Rips complexes $VR(P,t)$ of the same type. The problem is that this space has strata next to each other of the same dimension, which does not conform to a standard definition of stratification, and so doesn't admit a conical stratification. Dimension counting and requiring an open embedding $Z\times C(Y)\hookrightarrow X$ shows this is impossible at the boundary point between two such strata.

Weinberger gives some standard stratifed space types, among them a manifold stratified space, a manifold stratified space with boundary, and a PL stratified space, but $X\times \R_{\geqslant 0}$ is none of these.

Attempt 3: Naively describe the neighborhood of $P$ as a cone. 

This is the most direct attempt to write (1) as $Z\times C(Y)$. If we say
\[ C(Y) = \underbrace{\coprod_{\sum \ell_i = n \atop \ell_i\in \Z_{>0}}\ \prod_{i=1}^k \Ran^{\leqslant \ell_i}(\partial B^M_{t}(P_i))}_{Y} \times [0,\epsilon/2) \Bigg/\sim, \]
then we miss points splitting off at different "speeds". That is, in this presentation $P_i$ can only split into points that are all the same distance away from it. Between such a collection of points and $P_i$ are points that are some closer, some the same distance away, and those are not accounted for.

Moreover, using $Z=B^X_{\epsilon/2}(P)$, leads to overcounting, and the map into $X$ would not be injective.

Attempt 4: Iterate over different number of points at common radius.

This came out of an attempt to fix the previous attempt. As in a previous post ("The Ran space is locally conical," 2017-10-22), let $E_\ell$ be the collection of distinct partitions of $\ell$ elements, and for $e\in E_\ell$, let $T(e)$ be the collection of distinct total orderings of $e$. A candidate for $Z\times C(Y)$ would then be

with $t_{i,0} = \epsilon$ and $t_{i,j>0}$ the chosen element of $(0,t_{i,j-1})$. The open embedding $Z\times C(Y) \to X$ would be the inclusion on the $C(Y)$ component, and would scale every factor in the $Z$ component to a neighborhood of $P_i$ of radius $t_{i,|\tau_i|}$. However, this embedding is not continuous, because a point in $\Ran^k(M)$ is next to a point in $\Ran^{n}(M)$, where $P_i$ has split off into $n-k$ points, but the radius of $B^M_\epsilon(P_i)$ in $\Ran^k(M)$ is $\epsilon$, while in $\Ran^n(M)$ it is the shortest distance from one of the new points to $P_i$.

Attempt 5: Iterate over common radii, but only "antipodal" points.
This was an attempt to fix the previous attempt and combine it with the naive description. In fact, this approach works when $k=1$ and $n=2$. Then $P = \{P_1\}$, and
\[B^M_\epsilon(P_1) \times \left.\left(\mathbf P\partial B^M_t(P_1) \times [0,1)\right)\right/\sim\]
maps into $B^X_\epsilon(P_1)$ by first scaling $[0,1)$ down to $[0,\epsilon-d_M(P,P_1))$, where $P\in B^M_\epsilon(P_1)$ is the chosen point. The object $\mathbf P\partial B^M_t(P_1)$ is the projectivization of the boundary of the open $\dim(M)$-ball of radius $t$ around $P_1$ on $M$. That is, every element in it is a pair of antipodal points on the boundary of this ball that are exactly $t\in [0,\epsilon-d_M(P,P_1))$ away from $P_1$.

This works because every pair of points in a contractible neighborhood of $P_1$ is described uniquely by a pair $(P,v)$, for $P$ the midpoint of the two points and $v$ the $\dim(M)$-vector giving the direction of the points from $P$ (this may rely on working in charts, which is fine, as $M$ is a manifold). However, trying to generalize to more than two points fails because $\ell>2$ points in general are not equally distributed on a sphere. If instead of using the "antipodal" property we take a point from which all $\ell$ points are equidistant, this point may not be in the $\epsilon$-neighborhood of $P_1$.

Possible solutions

Solution 1: Instead of a smooth manifold, let $M$ be a simplicial complex. Then $\Ran^{\leqslant n}(M)$ should also be a simplicial complex. Then it may be possible to apply a general theorem to find appropriate cones.

Solution 2: Extend the only partially successful attempt, Attempt 5. Extend by describing a point splitting off into $\ell$ pieces as a sequence of points splitting into 2 pieces. Or, extend by using the centroid of $\ell$ points instead of the midpoint.

Solution 3: Weaken definition of "conically stratifed" to exclude either open embedding condition or $A_{>f(x)}$ stratification of $Y$, though this would involve following out Lurie's proof to see what can not be concluded.

References: Lurie (Higher algebra, Appendix A), Ayala, Francis and Tanaka (Local structures on stratified spaces, Sections 2 and 3), Weinberger (The classification of topologically stratified spaces)

The Ran space is locally conical

In this post we show that every point in the Ran space $\Ran^{\leqslant n}(M)$, for $M$ a compact, smooth embedded manifold, is the base of a cone in $\Ran^{\leqslant n}(M)$. Let $\dim(M) = m$ and let $P=\{P_1,\dots,P_k\}\in \Ran^k(M)\subseteq \Ran^{\leqslant n}(M)$. We write $d(x,y)$ for distance in Euclidean space $\R^N$ where $M$ is embedded, and $d_M(x,y)$ for distance on the embedded manifold $M$ (note $d\leqslant d_M$). Define the following objects:
\begin{align*}
N_\epsilon(x) & = \{z\in M\ :\ d_M(x,z)<\epsilon\}, \\
E_n & = \{\text{distinct partitions of an unlabeled set of $n$ elements}\}, \\
T(e) & = \{\text{distinct total orderings of }e\in E_n\}.
\end{align*}
We write $\tau=(\tau_1<\cdots<\tau_{|\tau|})$ for an element $\tau\in T(e)$.

Example: Let $n=4$, so then
\[
E_4 = \Big\{\{\{*\},\{*\},\{*\},\{*\}\},\hspace{10pt}
\{\{*,*\},\{*\},\{*\}\},\hspace{10pt}
\{\{*,*\},\{*,*\}\},\hspace{10pt}
\{\{*,*,*\},\{*\}\},\hspace{10pt}
\{\{*,*,*,*\}\}\Big\}.
\]
By stacking the $*$ on top of one another to indicate containment in a single set, and for order increasing from left to right, we have the following distinct total orderings for every element of $E_4$.

Set $\epsilon = \min_{1\leqslant i<j\leqslant k}\{d(P_i,P_j)\}$, $t_0\in(0,\epsilon/2)$, and $t_{j>0}\in (0,t_{j-1})$. By construction, the object
\begin{align*}
C_P & = \{P\} \cup \coprod_{\genfrac{}{}{0pt}{}{\sum \ell_i=n-k}{\ell_i \in \Z_{>0}}}\ \prod_{i=1}^k\  \coprod_{\genfrac{}{}{0pt}{}{\tau\in T(e)}{e\in E_{\ell_i}}}\  \prod_{j=1}^{|\tau|} \Ran^{|\tau_j|}\left(\partial N_{t_j}(P_i)\right) \times (0,t_{j-1}) \\
& = \left.\coprod_{\genfrac{}{}{0pt}{}{\sum \ell_i=n-k}{\ell_i \in \Z_{>0}}}\ \prod_{i=1}^k\ \coprod_{\genfrac{}{}{0pt}{}{\tau\in T(e)}{e\in E_{\ell_i}}}\left( \Ran^{|\tau_1|} (\partial N_{t_0}(P_i))\times \prod_{j=2}^{|\tau|} \Ran^{|\tau_j|}\left(\partial N_{t_j}(P_i)\right) \times (0,t_{j-1})\right) \times [0,\epsilon/2)\right/\sim
\end{align*}
is an open cone based at $P$ sitting inside $\Ran^{\leqslant n}(M)$. Here $\sim$ is the equivalence relation of all elements with $t_0=0$, with $[0,\epsilon/2)\owns t_0$ representing the unit interval in the usual cone construction. Moreover, given the point-counting stratification $f:\Ran^{\leqslant n}(M)\to A$, there is a natural stratification $g:C_p\to A_{\geqslant f(P)}$, with $P\in C_P$ the only element mapping to $f(P)$ under $g$.

The next step is to show that $P$ has an open neighborhood in $\Ran^{\leqslant n}(M)$ that is the image of an open embedding $Z\times C_P$, for some topological space $Z$. The obvious choice $Z = \prod_{i=1}^k N_{\epsilon/2}(P_i)$ does not work, because we double count points in higher strata, so we do not have an embedding.

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 $X$.

The Ran space and singularity sets

Fix a manifold $M$ along with an embedding of $M$ into $\R^N$ and set $X=\Ran(M)\times \R_{\geqslant 0}$. The goal of this post is to show that every $(P,t)\in X$ has an open neighborhood that contains no points of the type $(Q,d(Q_i,Q_j))$, for some $i\neq j$. The collection of all such elements of $X$ is called the singularity set of $X$, as the Vietoris-Rips complex at $Q$ with such a radius changes at such elements.

Following Lurie, given a collection of open sets $\{U_i\}_{i=1}^k$ in $M$, set
\[\Ran(\{U_i\}_{i=1}^k) = \left\{P\in \Ran(M)\ :\ P\subset \bigcup_{i=1}^k U_i,\ P\cap U_i\neq \emptyset\ \forall\ i\right\}.\]
The topology on $\Ran(M)$ is the smallest topology for which every $\Ran(\{U_i\}_{i=1}^k)$ is open, for any $\{U_i\}_{i=1}^k$, for any $k$. The topology on the product $X$ is the product topology.

Remark: Note that the Ran space $\Ran(M)$ by itself can be split up into the pieces $\Ran^k(M)$, with "singularities" viewed as when a point splits into two (or more) points, or two (or more) combine into one. Then every element of $\Ran(M)$ is on the edge of the singularity set, as any neighborhood of a single point on the manifold contains two points on the manifold.

Fix $(P,t)\in X$ not in the singularity set of $X$, with $P=(P_1,\dots,P_k)$, for $1\leqslant k\leqslant n$. Set
\[\mu = \min\left\{t,\min_{1\leqslant i<j\leqslant k}\left\{|t-d(P_i,P_j)|\right\}\right\},\]
with distance $d$ being Euclidean distance in $\R^N$. The quantity $\mu$ should be thought of as the upper bound on how "far" we may move from $(P,t)$ without hitting the singularity set.

Proposition: Let $(P,t)$ be as above and $t,\alpha,\beta>0$ such that $\alpha+\beta=\mu$. Then
\[U=\Ran\left(\{B(P_i,\alpha/2)\}_{i=1}^k\right) \times \left(t-\beta,t+\beta\right)\]
is an open neighborhood of $(P,t)$ in $X$ and does not contain any points of the singularity set of $X$.

If $t=0$, then having $[0,\beta)$ as the second component of $U$, with $\alpha+\beta=\min_{i,j}d(P_i,P_j)$ works as the open neighborhood of $(P,t)$. The balls $B(x,r)$ are $N$-dimensional in $\R^N$. The proof is mostly applications of the triangle inequality.

Proof: By construction we have that $U$ is open in $X$ and that it contains $(P,t)$. For $(Q,s)\in U$ any other element, we have three cases. We will show that the distance between any two $Q_a,Q_b\in Q$ is never $s$. Fix distinct indices $\ell,m\in \{1,\dots,k\}$.

1.  Case 1: $Q_a,Q_b\in B(P_\ell,\alpha/2)$. The situation looks as in the diagram below.
   
   Observe that $d(Q_a,Q_b)\leqslant d((Q_a,P_\ell)+d(Q_b,P_\ell) <\alpha = \mu-\beta \leqslant t-\beta$. Hence $d(Q_a,Q_b)<s$.
2. Case 2: $Q_a\in B(P_\ell,\alpha/2), Q_b\in B(P_m,\alpha/2), d(P_\ell,P_m)>t$. The situation looks as in the diagram below.
   
   Observe that $d(P_\ell,P_m)\leqslant d(P_\ell,Q_b)+d(P_m,Q_b)\leqslant d(P_\ell,Q_a)+d(Q_a,Q_b)+d(P_m,Q_b) < \alpha+d(Q_a,Q_b)$. Since $d(P_\ell,P_m)>t$, the definition of $\mu$ gives us that $\mu \leqslant d(P_\ell,P_m)-t$, so combining this with the previous inequality, we get $d(Q_a,Q_b) > d(P_\ell,P_m)-\alpha\geqslant \mu+t-(\mu-\beta)=t+\beta$. Hence $d(Q_a,Q_b)>s$.
3. Case 3: $Q_a\in B(P_\ell,\alpha/2), Q_b\in B(P_m,\alpha/2), d(P_\ell,P_m)<t$. The situation looks as in the diagram below.
   
   Observe that $d(Q_a,Q_b)\leqslant d(P_m,Q_b) + d(P_m,Q_a) \leqslant d(P_\ell,Q_a)+d(P_\ell,P_m)+d(P_m,Q_a)<\alpha+d(P_\ell,P_m)$. Since $d(P_\ell,P_m)<t$, the definition of $\mu$ gives us that $\mu\leqslant t-d(P_\ell,P_m)$, so combining this with the previous inequality, we get $d(Q_a,Q_b)<\mu-\beta+t-\mu = t-\beta$. Hence $d(Q_a,Q_b)<s$. $\square$

As an extension, it would be nice to show that the Vietoris--Rips complex of every element in $U$ is homotopy equivalent. This seems to be intuitively true, but a similar case analysis as above seems daunting.

References: Lurie (Higher Algebra, Section 5.5.1)

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)

Constructible sheaves

Let $X$ be a topological space with an open cover $\mathcal U = \{U_i\}$, and category $Op(X)$ of open sets of $X$. The goal is to define constructible sheaves and consider some applications. Thanks to Joe Berner for helpful pointers in this area.

Definition: Constructible subsets of $X$ are the smallest family $F$ of subsets of $X$ such that

• $Op(X)\subset F$,
• $F$ is closed under finite intersections, and
• $F$ is closed under complements.

This idea can be applied to sheaves. Recall that a locally closed subset of $X$ is the intersection of an open set and a closed set.

Definition: A sheaf $\mathcal F$ over $X$ is constructible if there exists, equivalently,

• a filtration $\emptyset=U_0\subset \cdots \subset U_n=X$ of $X$ by opens such that $\mathcal F|_{U_{i+1}\setminus U_i}$ is constant for all $i$, or
• a cover $\{V_i\}$ of locally closed subsets of $X$ such that $\mathcal F|_{V_i}$ is constant for all $i$.

Since the category of abelian sheaves over a topological space has enough injectives, we may consider an injective resolution of a sheaf $\mathcal F$ rather than the sheaf itself. The resolution may be considered as living inside the derived category of sheaves on $X$.

Definition: Let $A$ be an abelian category.

• $C(A)$ is the category of cochain complexes of $A$, 
• $K(A) = C(A)$ modulo cochain homotopy, and
• $D(A) = K(A)$ modulo $F\in K(A)$ such that $H^n(F)=0$ for all $n$, called the derived category of $A$.

Next we consider an example. Recall the Ran space $\Ran(M) = \{X\subset M\ :\ 0<|X|<\infty\}$ of non-empty finite subsets of a manifold $M$ and the Čech complex of radius $t>0$ of $P\in \Ran(M)$, a simplicial complex with $n$-cells for every $P'\subset P$ of size $n+1$ such that $d(P'_1,P'_2)<t$ for all $P'_1,P'_2\in P'$.

Example: Consider the subset $\Ran^{\leqslant 2}(M) = \{X\subset M\ :\ 1\leqslant |X|\leqslant 2\}$ of the Ran space. Decompose $X=\Ran^{\leqslant 2}(M)\times \R_+$ into disjoint sets $U_\alpha\cup U_\beta$, where
\[
U_\alpha = \underbrace{\left(\Ran^1(M)\times \R_+\right)}_{U_{\alpha,1}} \cup \underbrace{\bigcup_{P\in \Ran^2(M)}\{P\}\times (d_M(P_1,P_2),\infty)}_{U_{\alpha,2}},
\hspace{1cm}
U_\beta = \bigcup_{P\in \Ran^2(M)} \{P\} \times (0,d_M(P_1,P_2)],
\]
with $d_M$ the distance on the manifold $M$. The idea is that for every $(P,t)\in U_\alpha$, the Čech complex of radius $t$ on $P$ has the homotopy type of a point, whereas on $U_\beta$ has the homotopy type of two points. With this in mind, define a constructible sheaf $F\in\text{Shv}(\Ran^{\leqslant 2}(M)\times \R_+)$ valued in simplicial complexes, with $F|_{U_\alpha}$ and $F|_{U_\beta}$ constant sheaves. Set
\[
F_{(P,t)\in U_\alpha} = F(U_\alpha) = \left(0\to \{*\} \to 0\right),
\hspace{1cm}
F_{(P,t)\in U_\beta} = F(U_\beta) = \left(0\to \{*,*\}\to 0\right).
\]
Note that the chain complex $F(U_\alpha)$ is chain homotopic to $0\to \{-\}\to \{*,*\}\to 0$, where $-$ is a single 1-cell with endpoints $*,*$. To show that this is a constructible sheaf, we need to filter $\Ran^{\leqslant 2}(M)\times \R_+$ into an increasing sequence of opens. For this we use a distance on $\Ran^{\leqslant 2}(M)\times \R_+$, given by $d((P,t),(P',t'))=d_{\Ran(M)}(P,P')+d_\R(t,t'),$ where $d_\R(t,t')=|t-t'|$ and
\[
d_{\Ran(M)}(P,P')=\max_{p\in P}\left\{\min_{p'\in P'}\left\{d_M(p,p')\right\}\right\} + \max_{p'\in P'}\left\{\min_{p\in P}\left\{d_M(p,p')\right\}\right\}.
\]
Note that $U_\alpha$ is open. Indeed, for $(P,t)\in U_{\alpha,1}$, every other $P'\in \Ran^1(M)$ close to $P$ is also in $U_{\alpha,1}$, and if $P'\in \Ran^2(M)$ is close to $P$, then the non-zero component $t\in\R_+$ still guarantees the same homotopy type. The set $U_{\alpha,2}$ is open as well, so $U_\alpha$ is open. The whole space is open, so a filtration $\emptyset\subset U_\alpha\subset X$ works for us.

References: Hartshorne (Algebraic geometry, Section II.3), Hartshorne (Residues and Duality, Chapter IV.1), Kashiwara and Schapira (Sheaves on manifolds, Chapters 2 and 8), Lurie (Higher algebra, Section 5.5.1)