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Sunday, December 3, 2017

Ordering simplicial complexes with unlabeled vertices

The goal of this post is to describe a partial order on the collection of simplical complexes with n unlabeled vertices that is nice in the context of the space X=Rann(M)×R>0.

First note that there is a natural order on (abstract) simplicial complexes, given by set inclusion. Interpreting elements of X as simplicial complexes induces a more restrictive order, as new vertices must "split off" from existing ones rather than just be introduced anywhere. Also note that the category usually denoted by SC of simplicial complexes and simplicial maps contains objects with unordered vertices. Here we assume an order on them and consider the action of the symmetric groups to remove the order.

Definition: Let SCk, for some positive integer k, be the collection of simplicial complexes with k uniquely labeled vertices. This collection is a poset, with ST iff σT for every σS.

The symmetric group on k elements acts on SCk by permuting the vertices, and taking the image under this action we get SCk/Sk, the collection of simplicial complexes with k unlabeled vertices. This set also has a partial order, with ST in SCk/Sk iff ST in SCk, for some Sq1k(S) and Tq1k(T), where qk:SCkSCk/Sk is the quotient map.

Definition: For all i=1,,k, let sk,i be the ith splitting map, which splits the ith vertex in two. That is, if the vertices of SSCk are labeled v1,,vk, then sk,i is defined by
sk,i : SCkSCk+1,SS{vi,vi+1}{vi,w}S{vi+1,w},where S is S with vj relabeled as vj+1 for all j>i, and T is the simplicial complex generated by T.

By "generated by T" we mean generated in the Vietoris-Rips sense, that is, if {va,vb}T for all a,b in some indexing set I, then {vc : cI}T. The ith splitting map is essentially the ith face map used for simplicial sets.

Let A=nk=1SCk/Sk. The splitting maps induce a partial order on A, with ST, for SSCk/Sk and TSCk+1/Sk+1, iff sk,i(S)T in SCk, for some Sq1k(S), Tq1k+1(T), and i{1,,k}. This generalizes via composition of the splitting maps to any pair S,TA, and is visually decribed by the diagram below.

Now, let M be a smooth, compact, connected manifold embedded in RN, and X=Rann(M)×R>0. Let f:XA be given by (P,t)VR(P,t), the Vietoris-Rips complex around the points of P with radius t.

Proposition: The map f:XA is continuous.

Proof: Let SA and USA be the open set based at S. Take any (P,t)f1(US)X, for which we will show that there is an open ball B(P,t) completely within f1(US).

Case 1: td(Pi,Pj) for all pairs Pi,PjP. Then set
ϵ=min{t,mini<j|td(Pi,Pj)|,mini<jd(Pi,Pj)}.Set B=BRann(M)ϵ/4(P)×BR>0ϵ/4(t), which is an open neighborhood of (P,t) in X. It is immediate that f(P,t), for any other (P,t)B, has all the simplices of f(P,t), as ϵ|td(Pi,Pj)| for all i<j. If Pi has split in two in P, then for every simplex containing Pi in f(P,t) there are two simplices in f(Pt), with either of the points into which Pi split. That is, there may be new simplices in f(P,t), but f(P,t) will be in the image of the splitting maps. Equivalently, f(P,t)f(P,t) in A, so Bf1(US).

Case 2: t=d(Pi,Pj) for some pairs Pi,PjP. Then set
ϵ=min{t,mini<jtd(Pi,Pj)|td(Pi,Pj)|, mini<jd(Pi,Pj)},and define B as above. We are using the definition of Vietoris-Rips complex for which we add an edge between Pi and Pj whenever t>d(Pi,Pj). Now take any (P,t)B such that its image and the image of (P,t) under f are both in SCk/Sk. Then any points Pi,PjP with d(Pi,Pj)=t that have moved around to get to P, an edge will possibly be added, but never removed, in the image of f (when comparing with the image of (P,t)). This means that we have f(P,t)f(P,t) in SCk/Sk, so certainly f(P,t)f(P,t) in A. The same argument as in the first case holds if points of P split. Hence Bf1(US) in this case as well. 

This proposition shows in particular that X is poset-stratified by A

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