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.