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