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)

Towards a sheaf of simplicial complexes

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

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

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

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

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

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

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

Proposition: The map $g$ is continuous.

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

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

References: Lurie (Higher Algebra, Appendix A)

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).

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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")

Splitting points in two

The goal of this post is to expand upon some final ideas in a previous post ("Atempts at proving conical stratification," 2017-10-27). Let $M$ be a compact smooth $m$-manifold embedded in $\R^N$, and fix $n\in \Z_{>0}$. Let $X = \Ran^{\leqslant n}(M)$ and $f:X\to A=\{1,\dots,n\}$ the usual point-counting stratification. Let $$\begin{align*} B^X_\epsilon(P) & = \left\{Q\in X\ :\ 2d_M(P,Q) =\sup_{p\in P}\inf_{q\in Q} d_M(p,q) + \sup_{q\in Q}\inf_{p\in P}d_M(p,q) < 2\epsilon\right\}, \\ B^M_\epsilon(p) & = \left\{q\in M\ :\ d_M(p,q) <\epsilon\right\},\\ B^{\R^m}_\epsilon(0) & = \left\{x\in \R^m\ :\ d(0,x)<\epsilon\right\} \end{align*}$$ be open balls in their respective spaces. We use $d_M$ for distance on $M$ and $d$ for distance in $\R^N$. Since $M$ is an $m$-manifold, we will work in charts in $\R^m$ when necessary.

Proposition: The stratification $f:X\to A$ is conical in the top two strata $\Ran^n(M)$ and $\Ran^{n-1}(M)$.

Proof: Let $P = \{P_1,\dots,P_n\}\in \Ran^n(M)$ and $2\epsilon = \min_{1\leqslant i <j\leqslant n}d(P_i,P_j)$. Let $Y = \emptyset$ which has a natural $(A_{>n} = \emptyset)$-stratification with $C(Y) = \{*\}$ having a natural $(A_{\geqslant n} = \{n\})$-stratification. Let $Z = B^X_\epsilon(P) = \prod_{i=1}^nB^M_\epsilon(P_i)$, for which the identity map $Z\times \{*\} \cong Z\hookrightarrow X$ is an open embedding. Hence $X$ is stratified at every $P\in \Ran^n(M)$.

Let $P = \{P_1,\dots,P_{n-1}\}\in \Ran^{n-1}(M)$ and $2\epsilon = \min_{1\leqslant i <j\leqslant n-1}d(P_i,P_j)$. Let $$Y = \coprod_{i=1}^{n-1} \mathbf P\partial B^{\R^m}_{\epsilon/2}(0), \hspace{1cm} Z = B^{\R^m}_{\epsilon/2}(0),$$ where $\mathbf P\partial B$ is the projectivization of the sphere, so may be viewed as a collection of unique pairs $\{\vec v, -\vec v\}$. Then the cone $C(Y)$ may be viewed as a collection of pairs $\{\vec v,t>0\}$ along with the singleton $\{0\}$, with the usual cone topology. Define a map $$\begin{array}{r c l} \varphi\ :\ Z\times C(Y) & \to & X, \\ (x,\vec v,t) & \mapsto & \{x+t\vec v,x-t\vec v\} ,\\ (x,0) & \mapsto & \{x\}. \end{array}$$ Note that $B^X_{\epsilon/2}(P) \subseteq \im(\varphi)\subseteq B^X_\epsilon(P)$. This map is injective as every pair of points on $M$ within an $\epsilon/2$-radius of $P_i$ is uniquely defined by their midpoint (the element of $Z$), a direction from that midpoint (the element of $Y$) and a distance from that midpoint (the cone component $t\in [0,1)$). By construction $\varphi$ is continuous and an embedding. The map takes open sets to open sets, so we have an open embedding into $X$. Hence $X$ is conically stratified at every $P\in \Ran^{n-1}(M)$. $\square$

The problem with generalizing this to $P\in \Ran^k(M)$ for all other $k$ is that an $(n-k+1)$-tuple of points has no unique midpoint. It does have a unique centroid, but it is not clear what the $[0,1)$ component of the cone should then be.

Proposition: The space $X$ is of locally singular shape.

Proof: First note that every $P\in X$ has an open neighborhood that is homemorphic to an open ball of dimension $mn$ (see Equation (1) of previous post "Attempts at proving conical stratification," 2017-10-27). Hence we may cover $X$ by contractible sets. By Remark A.4.16 of Lurie, $X$ will be of locally singular shape if every element of the cover is of singular shape. Since all elements of the cover are contractible, by Remark A.4.11 of Lurie we only need to check if the topological space $*$ is of singular shape. Finally, Example A.4.12 of Lurie gives that $*$ has singular shape. $\square$

References: Lurie (Higher algebra, Appendix A)