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Friday, August 11, 2017

The Ran space and singularity sets

Fix a manifold M along with an embedding of M into RN and set X=Ran(M)×R0. The goal of this post is to show that every (P,t)X has an open neighborhood that contains no points of the type (Q,d(Qi,Qj)), for some ij. The collection of all such elements of X is called the singularity set of X, as the Vietoris-Rips complex at Q with such a radius changes at such elements.

Following Lurie, given a collection of open sets {Ui}ki=1 in M, set
Ran({Ui}ki=1)={PRan(M) : Pki=1Ui, PUi  i}.
The topology on Ran(M) is the smallest topology for which every Ran({Ui}ki=1) is open, for any {Ui}ki=1, for any k. The topology on the product X is the product topology.

Remark: Note that the Ran space Ran(M) by itself can be split up into the pieces Rank(M), with "singularities" viewed as when a point splits into two (or more) points, or two (or more) combine into one. Then every element of Ran(M) is on the edge of the singularity set, as any neighborhood of a single point on the manifold contains two points on the manifold.

Fix (P,t)X not in the singularity set of X, with P=(P1,,Pk), for 1kn. Set
μ=min{t,min1i<jk{|td(Pi,Pj)|}},
with distance d being Euclidean distance in RN. The quantity μ should be thought of as the upper bound on how "far" we may move from (P,t) without hitting the singularity set.

Proposition: Let (P,t) be as above and t,α,β>0 such that α+β=μ. Then
U=Ran({B(Pi,α/2)}ki=1)×(tβ,t+β)
is an open neighborhood of (P,t) in X and does not contain any points of the singularity set of X.

If t=0, then having [0,β) as the second component of U, with α+β=mini,jd(Pi,Pj) works as the open neighborhood of (P,t). The balls B(x,r) are N-dimensional in RN. The proof is mostly applications of the triangle inequality.

Proof: By construction we have that U is open in X and that it contains (P,t). For (Q,s)U any other element, we have three cases. We will show that the distance between any two Qa,QbQ is never s. Fix distinct indices ,m{1,,k}.
  1.  Case 1: Qa,QbB(P,α/2). The situation looks as in the diagram below.
    Observe that d(Qa,Qb)d((Qa,P)+d(Qb,P)<α=μβtβ. Hence d(Qa,Qb)<s.
  2. Case 2: QaB(P,α/2),QbB(Pm,α/2),d(P,Pm)>t. The situation looks as in the diagram below.
    Observe that d(P,Pm)d(P,Qb)+d(Pm,Qb)d(P,Qa)+d(Qa,Qb)+d(Pm,Qb)<α+d(Qa,Qb). Since d(P,Pm)>t, the definition of μ gives us that μd(P,Pm)t, so combining this with the previous inequality, we get d(Qa,Qb)>d(P,Pm)αμ+t(μβ)=t+β. Hence d(Qa,Qb)>s.
  3. Case 3: QaB(P,α/2),QbB(Pm,α/2),d(P,Pm)<t. The situation looks as in the diagram below.
    Observe that d(Qa,Qb)d(Pm,Qb)+d(Pm,Qa)d(P,Qa)+d(P,Pm)+d(Pm,Qa)<α+d(P,Pm). Since d(P,Pm)<t, the definition of μ gives us that μtd(P,Pm), so combining this with the previous inequality, we get d(Qa,Qb)<μβ+tμ=tβ. Hence d(Qa,Qb)<s.
As an extension, it would be nice to show that the Vietoris--Rips complex of every element in U is homotopy equivalent. This seems to be intuitively true, but a similar case analysis as above seems daunting.

References: Lurie (Higher Algebra, Section 5.5.1)

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