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)

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.

Exit paths, part 2

In this post we continue on a previous topic ("Exit paths, part 1," 2017-08-31) and try to define a constructible sheaf via universality. Let $X$ be an $A$-stratified space, that is, a topological space $X$ and a poset $(A,\leqslant)$ with a continuous map $f:X\to A$, where $A$ is given the upset topology relative to its ordering $\leqslant$. Recall the full subcategory $\Sing^A(X)\subseteq \Sing(X)$ of exit paths on an $A$-stratified space $X$.

Proposition: If $X\to A$ is conically stratified, $\Sing^A(X)$ is an $\infty$-category.

Briefly, a stratification $f:X\to A$ is conical if for every stratum there exists a particular embedding from a stratified cone into $X$ (see Lurie for "conical stratification" and Ayala, Francis, Tanaka for "conically smooth stratified space," which seem to be the same). We will leave confirming the described stratification as conical to a later post.

This proposition, given as part of Theorem A.6.4 in Lurie, has a very long proof, so is not repeated here. Lurie actually proves that the natural functor $\Sing^A(X)\to N(A)$ described below is a (inner) fibration, which implies the unique lifting property of $\Sing^A(X)$ via the unique lifting property of $N(A)$ (and we already know nerves are $\infty$-categories).

Example: The nerve of a poset is an $\infty$-category. Being a nerve, it is already immediate, but it is worthwhile to consider the actual construction. For example, if $A = \{a\leqslant b\leqslant c \leqslant d\}$ is the poset with the ordering $\leqslant$, then the pieces $N(A)_i$ are as below.

It is immediate that every 3-horn can only be filled in one unique way (as there is only one element of $N(A)_3$), as well as that every 2-horn can be filled in one unique way (as every sequence of two composable morphisms appears as a horn of exactly one element of $N(A)_2$).

In Appendix A.9 of Higher Algebra, Lurie says that there is an equivalence of categories \[(A\text{-constructible sheaves on }X)   \cong   \left[(A\text{-exit paths on }X),\mathcal S\right],\] given that $X$ is conically stratified, and for $\mathcal S$ the $\infty$-category of spaces (equivalently $N(Kan)$, the nerve of all the simplicial sets that are Kan complexes). So, instead of trying to define  a particular constructible sheaf on $X = \Ran^{\leqslant n}(M)\times \R_{\geqslant 0}$, (as in previous posts "Stratifying correctly," 2017-09-17 and "A constructible sheaf over the Ran space," 2017-06-24) we will try to make a functor that takes an exit path of $X$ and gives back a space.

Fix $n\in \Z_{>0}$ and set $X = \Ran^{\leqslant n}\times \R_{\geqslant 0}$. Let $SC$ be the category of simplicial complexes and simplicial maps, with $SC_n$ the full subcategory of simplicial complexes with at most $n$ vertices. There is a map
\[\begin{array}{r c l}
g\ :\ X & \to & SC_n \\
(P,t) & \mapsto & VR(P,t),
\end{array}\]
allowing us to say
\[X = \bigcup_{S\in SC_n}g^{-1}(S).\]
Here we consider that two elements $P_i,P_j\in P$ give an edge of $VR(P,t)$ whenever $t>d(P_i,P_j)$ (this is chosen instead of $t\geqslant d(P_i,P_j)$ so that the boundaries of the strata ``facing downward," with respect to the poset ordering, are open). Now we define a stratifying poset $A$ for $X$.

Definition: Let $A = \{a_S\ :\ S\in SC_n\}$ and define a relation $\leqslant$ on $A$ by
\[ \left(a_S\leqslant a_T\right)\ \ \Longleftarrow\ \ \left(
\begin{array}{c}
\exists\ \sigma\in \Sing(X)_1\ \text{such that}\\
g(\sigma(0))=S,\ g(\sigma(t>0))=T.
\end{array}
\right)\]
Let $(A,\leqslant)$ be the poset generated by relations of the type given above.

We claim that $f:X\to A$ given by $f(P,t)=a_{g(P,t)}$ is a stratifying map, that is, continuous in the upset topology on $A$. To see this, take the open set $U_S = \{a_T\in A\ :\ a_S\leqslant a_T\}$ in the basis of the upset topology of $A$, for any $S\in SC_n$, and consider $x\in f^{-1}(U_S)$. If for all $\epsilon>0$ we have $B_X(x,\epsilon)\cap f^{-1}(U_S)^C\neq \emptyset$, then there exists $T_\epsilon\in SC_n$ with $B_X(x,\epsilon)\cap f^{-1}(a_{T_\epsilon})\neq\emptyset$, for $S\not\leqslant T_\epsilon$ (as $T_\epsilon\not\in U_S$). This means there exists $\sigma\in \Sing(X)_1$ with $\sigma(0)=x$ and $\sigma(t>0)\in f^{-1}(a_{T_\epsilon})$, which in turn implies $S\leqslant T_\epsilon$, a contradiction. Hence $f$ is continuous, so $f:X\to A$ is a stratification.

As all morphisms in $\Sing(X)$ are compsitions of the face maps $s_i$ and degenracy maps $d_i$, so are all morphisms in $\Sing^A(X)$. There is a natural functor $F:\Sing^A(X)\to N(A)$ defined in the following way:
\[\begin{array}{r r c l}
%%
%% L1
%%
\text{objects} & \left(
\begin{array}{c}
\sigma:|\Delta^k|\to X \\
a_0\leqslant \cdots \leqslant a_k\subseteq A \\
f(\sigma(t_0,\dots,t_i\neq 0,0,\dots,0)) = a_i
\end{array}
\right) & \mapsto & \left( a_0\to\cdots\to a_k\in N(A)_k\right) \\[20pt]
%%
%% L2
%%
\text{face maps} & \left(
\begin{array}{c}
\left(
\begin{array}{c}
\sigma:|\Delta^k|\to X \\ a_0\leqslant \cdots \leqslant a_k\subseteq A
\end{array}
\right)\\[10pt]
\downarrow \\[10pt]
\left(
\begin{array}{c}
\tau:|\Delta^{k+1}|\to X \\ a_0\leqslant \cdots \leqslant a_i\leqslant a_i\leqslant \cdots a_k\subseteq A
\end{array}
\right)
\end{array}
\right) & \mapsto &
\left(\begin{array}{c}
\left(a_0\to\cdots \to a_k\right)\\[10pt]
\downarrow\\[10pt]
\left(a_0\to\cdots \to a_i\xrightarrow{\text{id}}a_i\to\cdots \to a_k\right)
\end{array}\right)\\[40pt]
%%
%% L3
%%
\text{degeneracy maps} & \left(
\begin{array}{c}
\left(
\begin{array}{c}
\sigma:|\Delta^k|\to X \\ a_0\leqslant \cdots \leqslant a_k\subseteq A
\end{array}
\right)\\[10pt]
\downarrow \\[10pt]
\left(
\begin{array}{c}
\tau:|\Delta^{k-1}|\to X \\ a_0\leqslant \cdots \leqslant a_{i-1}\leqslant a_{i+1}\leqslant \cdots a_k\subseteq A
\end{array}
\right)
\end{array}
\right) & \mapsto &
\left(\begin{array}{c}
\left(a_0\to\cdots \to a_k\right)\\[10pt]
\downarrow\\[10pt]
\left(a_0\to\cdots \to a_{i-1}\xrightarrow{\circ}a_{i+1}\to\cdots \to a_k\right)
\end{array}\right)
\end{array}\]
As all maps in $\Sing^A(X)$ are generated by compositions of face and degeneracy maps, this completely defines $F$. Naturality of $F$ follows precisely because of this.

A poset (which can be viewed as a directed simple graph) may be naturally viewed as a 1-dimensional simplicial set, moreover an $\infty$-category (by virtue of being a \emph{simple} graph, with no multi-edges or loops). Hence there is a natural map, the inclusion, that takes $N(A)$ into $N(\mathcal Kan) = \mathcal S$.  Finally, Construction A.9.2 of Lurie describes a map that takes a functor from $A$-exit paths into spaces and gives back an $A$-constructible sheaf over $X$, which Theorem A.9.3 shows to be an equivalence, given the following conditions:

• $X$ is paracompact,
• $X$ is locally of singular shape,
• the $A$-stratification of $X$ is conical, and
• $A$ satisfies the ascending chain condition.

The first condition is satisfied as both $\Ran^{\leqslant n}(M)$ and $\R_{\geqslant 0}$ are locally compact and second countable. The last condition is satisfied because $A$ is a finite poset. We already mentioned that the conical property will be checked later, as will the singular shape property. Unfortunately, Lurie gives a definition of singular shape only for $\infty$-topoi, so some work must be done to translate this into our simpler setting. However, in the introduction to Appendix A, Lurie says that if $X$ is "sufficiently nice" and we assume some "mild assumptions" about $A$, then the described categorical equivalence follows, so it seems there is hope that everything will work out well in the end.

References: Stacks Project, Lurie (Higher algebra, Appendix A), Ayala, Francis and Tanaka (Local structures on stratified spaces, Sections 2 and 3)

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 $X$.

Stratifying correctly

In a previous blog post ("A constructible sheaf over the Ran space," 2017-06-24) it was claimed that there was a particular constructible sheaf over $\Ran^{\leqslant n}(M)\times \R_{\geqslant 0}$. However, the proof actually uses finite ordered subsets of $M$ to make the stratification, rather than finite unordered subsets. This means that the sheaf is actually over $M^{\times n}\times \R_{\geqslant 0}$, and in this post we try to fix that problem.

Let $\Delta_n$ be the "fat diagonal" of $M^{\times n}$, that is, the collection of $P\in M^{\times n}$ for which at least two coordinates are the same. For every $k>0$, there is an $S_k$ action on $M^{\times k}\setminus \Delta_k$, quotienting by which we get a map
\[M^{\times k}\setminus \Delta_k \xrightarrow{\ q_k\ }\Ran^k(M)\]
to the Ran space of degree $k$. The stratification of $M^{\times k}\times \R_{\geqslant 0}$ given in the previous post will be pushed forward to a stratification of $\Ran^k(M)\times \R_{\geqslant 0}$, for all $0<k\leqslant n$. A large part of the work already has been done, it remains to put everything in the right order and check openness. The process is given as follows:

1. Stratify $\Ran^{\leqslant n}(M)\times \R_{\geqslant 0}$ into $n$ pieces, each being $\Ran^k(M)\times \R_{\geqslant 0}$.
2. Stratify $(M^{\times k}\setminus \Delta_k)\times \R_{\geqslant 0}$ as in the previous post.
3. Quotient by $S_k$-action to get stratification of $\Ran^k(M)\times \R_{\geqslant 0}$.

Step 1

As stated in the proof of the Theorem, $\Ran^{\geqslant k}(M)\times \R_{\geqslant 0}$ is open inside $\Ran^{\leqslant n}(M)\times \R_{\geqslant 0}$, allowing us to make a stratification $f:\Ran^{\leqslant n}(M)\times \R_{\geqslant 0}\to A$, where $A$ is the poset

 

where the tail of an arrow is ordered lower than the head. The map $f$ sends $\Ran^k(M)\times \R_{\geqslant 0}$ to $a_k$, which is a continuous map in the upset topology on $A$.

Step 2

As stated in Definition 5, we have a stratification $g_k:(M^{\times k}\setminus \Delta_k) \times \R_{\geqslant 0}\to B_k$, where $B_k$ may be viewed as a directed graph $B_k=(V_k,E_k)$. The vertex set is $V_k = \{0,1\}^{k(k+1)/2}$, whose elements are strings of 1 and 0, and the edge set $E_k$ contains $v\to v'$ iff $d_H(v,v')=1$ and $d_H(v,0)<d_H(v',0)$, for $d_H$ the Hamming distance. Let $U_v\subset B_k$ denote the upset based at $v$, that is, all elements $v'\in B_k$ with $v\leqslant v'$.

Order all distinct pairs $(i,j)\in \{1,\dots,k\}^2$, of which there are $k(k+1)/2$. Under the stratifying map $g_k$, each upset $U_v$ based at the vertex $v\in\{0,1\}^{k(k+1)/2}$ receives elements $(P,t)\in (M^{\times k}\setminus \Delta_k)\times \R_{\geqslant 0}$ satisfying $t>d(P_i,P_j)$ whenever the position representing $(i,j)$ in $v$ is 1. For example, when $k=3$,

To check that $g_k$ is continuous in the upset topology, we restate Lemma 2 in a clearer way.

Lemma 1: Let $U\subset X$ be open and $\varphi:X\to A\subset \R_{\geqslant 0}$ continuous, with $|A|<\infty$. Then
\[\bigcup_{x\in U}\{x\}\times (\varphi(x),\infty)\ \subseteq\ X\times (z',\infty)\]
is open, for any $z'\leqslant z:= \min_{x\in U}\{\varphi(x)\}$.

Proof: Consider the function
\[\begin{array}{r c l}
\psi\ :\ X\times (z',\infty) & \to & X\times (-\infty,z), \\
(x,t) & \mapsto & (x,\varphi(x)-t).
\end{array}\]
Since $\varphi$ is continuous and subtraction is continuous, $\psi$ is continuous (in the product topology). Since $U\times (-\infty,0)$ is open in $X\times (-\infty,z)$, the set $\psi^{-1}(U\times (-\infty,0))$ is open in $X\times (z',\infty)$. For any $x\in U$ and $t=\varphi(x)$, we have $\varphi(x)-t =0$. For any $x\in U$ and $t\to \infty$, we have $\varphi(x)-t\to -\infty$. It is immediate that all other $t\in (\varphi(x),\infty)$ give $\varphi(x)-t\in (-\infty,0)$. Hence $\psi^{-1}(U\times (-\infty,0))$ is the collection of points $(x,t)$ with $t\in (\varphi(x),\infty)$, which is then open in $X\times [0,z')$. $\square$

Applying Lemma 1 to $U=X=M^{\times k}\setminus \Delta_k$ and $\varphi(P) = \max_{i\neq j}\{d(P_i,P_j)\}$, which is continuous, gives that $g_k^{-1}(U_{11\cdots1})\subseteq M^{\times k}\setminus \Delta_k$ is open. This also works to show that $g_k^{-1}(U_v)\subseteq g_k^{-1}(U_{v'})$ is open, for any $v'\leqslant v$, by limiting the pairs of indices iterated over by the function $\varphi$. Hence $g_k$ is continuous.

Step 3

The symmetric group $S_k$ acts on $(M^{\times k}\setminus \Delta_k)\times \R_{\geqslant 0}$ by permuting the order of elements in the first factor. That is, for $\sigma\in S_k$, we have
\[\sigma (P=\{P_1,\dots,P_k\},t) = (\{P_{\sigma(1)},\dots,P_{\sigma(k)}\},t).\]
Note that $((M^{\times k}\setminus \Delta_k)\times \R_{\geqslant 0})/S_k = \Ran^k(M)\times \R_{\geqslant 0}$.

Remark: Graph isomorphism for two graphs with $k$ vertices may also be viewed as the equivalence relation induced by $S_k$ acting on $\Gamma_k = \{$simple vertex-labeled graphs with $k$ vertices$\}$. First, let $G_v$ be the (unique) graph first introduced at element $v\in B_k$ by $g_k$. That is, we have $G_v =VR(P,t)_1$ (the ordered 1-skeleton of the Vietoris--Rips complex on the set $P$ with radius $t$) whenever $g_k((P,t))\in U_v$ and $g_k((P,t))\not\in U_{v'}$ for any $v'\leqslant v$, $v'\neq v$. Then the elements of $B_k$ are in bijection with the elements of $\Gamma_k$ (given by $v\leftrightarrow G_v$), so we have $B_k/S_k = B_k'$. Recall that $v\leqslant v'$ in $B_k$ iff adding an edge to $G_v$ gives $G_{v'}$. In $B_{k'}$, this becomes a partial order on equivalence classes $[w] = \{v\in B_k\ :\ \sigma G_v=G_w\ $for some $\sigma\in S_k\}$. We write $[w]\leqslant [w']$ iff there is a collection of pairs $\{(v_1,v_1'),\dots,(v_\ell,v_\ell')\}$ such that $v_i\leqslant v_i'$ for all $i$, and $\{v_1,\dots,v_\ell\} = [w]$ and $\{v_1',\dots,v_\ell'\} = [w']$ (there may be repetition among the $v_i$ or $v_i'$).

By the universal property of the quotient, there is a unique map $h_k:\Ran^k(M)\times \R_{\geqslant 0}\to B_k'$ that makes the following diagram commute.

This will be our stratifying map. To check that $h_k$ is continuous take $U\subseteq B_k'$ open. As $\pi$ is the projection under a group action, it is an open map, so $\pi^{-1}(U)\subseteq B_k$ is open. Since $g_k$ is continuous in the upset topology, $g_k^{-1}(\pi^{-1}(U))$ is open. Again, $S_k\curvearrowright$ is the projection under a group action, so $(S_k\curvearrowright)(g_k^{-1}(\pi^{-1}(U)))$ is open, giving continuity of $h_k$.