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Sunday, November 26, 2017

Towards a sheaf of simplicial complexes

The goal of this post is to describe a new stratification of Rann(M)×R0 that builds on the ideas from a previous post (see "The point-counting stratification of the Ran space is conical (really though) ," 2017-11-15) and some newer ones.

Let SCn be the set of simplicial complexes on n ordered vertices. There is a natural partial order on SCn given by inclusion of sets, viewing every simplex as a subset of the power set P({1,,n}). The symmetric group Sn has a natural action on SCn and SCn/Sn has an induced partial order as well. Hence we have a map
f : Rann(M)×R0SCn/Sn,(P,t)VR(P,t),
where VR(P,t) is the Vietoris-Rips complex on P with radius t. We include a k-cell in VR(P,t) at the vertices {P0,,Pk}P if d(Pi,Pj)<t for all 0i<jk. Because we have strict inequality, the map is continuous in the upwards-directed, or Alexandrov topology on SCn/Sn. Indeed, taking the preimage of an open set US in SCn/Sn based at some simplicial complex S (such US form the basis of topology on SCn/Sn), there is an open ball of radius mini<jd(Pi,Pj)/2 in the Rann(M) component and min(Pi,Pj)f(P,t)|td(Pi,Pj)| in the R0 component around any (P,t)f1(US).

Remark: The above shows that Rann(M)×R0 is poset-stratified by SCn/Sn, in the sense of Definition A.5.1 of Lurie. However, the strata are all of the same dimension, so there is no chance of this being a conical stratification, in the sense of Definition A.5.5 of Lurie. We hope to fix that with a different stratification.

Definition: Construct a poset (A,A) in the following way:
  • SCn/SnA, with SAT whenever SSCn/SnT ,
  • for every STSCn/Sn, let aSTA with aSTAS and aSTAT,
  • for every {S1,,Sk>2}SCn/Sn, let aS1SkA with aS1SkAaS1^SiSk for all 1ik.
Define a map into (A,A) in the following way:
g : Rann(M)×R0A,(P,t){S, if (P,t)int(f1(S)) for some SSCn/Sn,aS1Sk, if (P,t)cl(f1(T))  T{S1,,Sk}.

We now claim that g is a stratifying map.

Proposition: The map g is continuous.

Proof: Since int(f1(S))int(f1(T))= for all STSCn/Sn, the open sets USA based at S all have open preimage g1(US)X. Now take (P,t)g1(UaS1Sk), for k2. If every open ball around (P,t)X intersects XaT, for some TSCn/Sn, then (P,t) must be in the closure of f1(T), for every TT. Hence the only possible such T are T{S1,,Sk}, so g1(UaS1Sk) is open in X.

The next step would be to show that this stratification is conical, though it is not clear yet if it is.

References: Lurie (Higher Algebra, Appendix A)

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