Thursday, November 2, 2017

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)

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