In this post we show that every point in the Ran space Ran⩽n(M), for M a compact, smooth embedded manifold, is the base of a cone in Ran⩽n(M). Let dim(M)=m and let P={P1,…,Pk}∈Rank(M)⊆Ran⩽n(M). We write d(x,y) for distance in Euclidean space RN where M is embedded, and dM(x,y) for distance on the embedded manifold M (note d⩽dM). Define the following objects:
Nϵ(x)={z∈M : dM(x,z)<ϵ},En={distinct partitions of an unlabeled set of n elements},T(e)={distinct total orderings of e∈En}.
We write τ=(τ1<⋯<τ|τ|) for an element τ∈T(e).
Example: Let n=4, so then
E4={{{∗},{∗},{∗},{∗}},{{∗,∗},{∗},{∗}},{{∗,∗},{∗,∗}},{{∗,∗,∗},{∗}},{{∗,∗,∗,∗}}}.
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 E4.
Set ϵ=min1⩽i<j⩽k{d(Pi,Pj)}, t0∈(0,ϵ/2), and tj>0∈(0,tj−1). By construction, the object
CP={P}∪∐∑ℓi=n−kℓi∈Z>0 k∏i=1 ∐τ∈T(e)e∈Eℓi |τ|∏j=1Ran|τj|(∂Ntj(Pi))×(0,tj−1)=∐∑ℓi=n−kℓi∈Z>0 k∏i=1 ∐τ∈T(e)e∈Eℓi(Ran|τ1|(∂Nt0(Pi))×|τ|∏j=2Ran|τj|(∂Ntj(Pi))×(0,tj−1))×[0,ϵ/2)/∼
is an open cone based at P sitting inside Ran⩽n(M). Here ∼ is the equivalence relation of all elements with t0=0, with [0,ϵ/2)∋t0 representing the unit interval in the usual cone construction. Moreover, given the point-counting stratification f:Ran⩽n(M)→A, there is a natural stratification g:Cp→A⩾f(P), with P∈CP the only element mapping to f(P) under g.
The next step is to show that P has an open neighborhood in Ran⩽n(M) that is the image of an open embedding Z×CP, for some topological space Z. The obvious choice Z=∏ki=1Nϵ/2(Pi) does not work, because we double count points in higher strata, so we do not have an embedding.
Nϵ(x)={z∈M : dM(x,z)<ϵ},En={distinct partitions of an unlabeled set of n elements},T(e)={distinct total orderings of e∈En}.
We write τ=(τ1<⋯<τ|τ|) for an element τ∈T(e).
Example: Let n=4, so then
E4={{{∗},{∗},{∗},{∗}},{{∗,∗},{∗},{∗}},{{∗,∗},{∗,∗}},{{∗,∗,∗},{∗}},{{∗,∗,∗,∗}}}.
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 E4.
Set ϵ=min1⩽i<j⩽k{d(Pi,Pj)}, t0∈(0,ϵ/2), and tj>0∈(0,tj−1). By construction, the object
CP={P}∪∐∑ℓi=n−kℓi∈Z>0 k∏i=1 ∐τ∈T(e)e∈Eℓi |τ|∏j=1Ran|τj|(∂Ntj(Pi))×(0,tj−1)=∐∑ℓi=n−kℓi∈Z>0 k∏i=1 ∐τ∈T(e)e∈Eℓi(Ran|τ1|(∂Nt0(Pi))×|τ|∏j=2Ran|τj|(∂Ntj(Pi))×(0,tj−1))×[0,ϵ/2)/∼
is an open cone based at P sitting inside Ran⩽n(M). Here ∼ is the equivalence relation of all elements with t0=0, with [0,ϵ/2)∋t0 representing the unit interval in the usual cone construction. Moreover, given the point-counting stratification f:Ran⩽n(M)→A, there is a natural stratification g:Cp→A⩾f(P), with P∈CP the only element mapping to f(P) under g.
The next step is to show that P has an open neighborhood in Ran⩽n(M) that is the image of an open embedding Z×CP, for some topological space Z. The obvious choice Z=∏ki=1Nϵ/2(Pi) does not work, because we double count points in higher strata, so we do not have an embedding.
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