Let M be a manifold. The goal of this post is to show that the sheaf F(P,t)=Rips(P,t) valued in simplicial complexes over Ran⩾(M)×R⩾0 is constructible, a goal not quite achieved (see "A naive constructible sheaf," 2017-12-19 for a solution to the problems encountered here). This space will be described using filtered diagram of open sets, with the sheaf on consecutive differences of the diagram giving simplicial complexes of the same homotopy type.
Definition: Let P={P1,…,Pn}∈Ran(M). For every collection of open neighborhoods {Ui∋Pi}ni=1 of the Pi in M, there is an open neighborhood of P in Ran(M) given by
Ran({Ui}ni=1)={Q∈Ran(M) : Q⊂n⋃i=1Ui, Q∩Ui≠∅}.Moreover, these are a basis for any open neighborhood of P in Ran(M).
Proof: The left-most node of Dk may be expressed as
{(P,t) : P={P1,…,Pk}∈Rank(M),t>d(Pi,Pj) ∀ Pi,Pj∈P}=⋃P∈Rank(M){P}×(maxPi,Pj∈P{d(Pi,Pj)},∞). Applying a slight variant of Lemma 2 (replacing R by an open ray that is bounded below), with the max function continuous, we get that the left-most node is open in the nodes one directed edge away from it. Repeating this argument, we get that every node is open inside every node following it. ◻
Definition: Let P={P1,…,Pn}∈Ran(M). For every collection of open neighborhoods {Ui∋Pi}ni=1 of the Pi in M, there is an open neighborhood of P in Ran(M) given by
Ran({Ui}ni=1)={Q∈Ran(M) : Q⊂n⋃i=1Ui, Q∩Ui≠∅}.Moreover, these are a basis for any open neighborhood of P in Ran(M).
Sets
We begin with a few facts about sets. Let X be a topological space.
Lemma 1: Let A,B⊂X. Then:
Lemma 1: Let A,B⊂X. Then:
- (a) If A⊂B is open and B⊂X is open, then A⊂X is open.
- (b) If A⊂B is closed and B⊂X is open, then A⊂X is locally closed.
- (c) If A⊂B is open and B⊂X is locally closed, then A⊂X is locally closed.
- (d) If A⊂B is locally closed and B⊂X is locally closed, then A⊂X is locally closed.
Proof: For part (a), first recall that open sets in B are given by intersections of B with open sets of A. Hence there is some W⊂X open such that A=B∩W. Since both B and W are open in X, the set A is open in X.
For part (b), since A⊂B is closed, there is some Z⊂X closed such that A=B∩Z. Since B is open in X, A is locally closed in X.
For part (b), since A⊂B is closed, there is some Z⊂X closed such that A=B∩Z. Since B is open in X, A is locally closed in X.
For parts (c) and (d), let B=W1∩W2, for W1⊂X open and W2⊂X closed. For part (c), again there is some W⊂X open such that A=B∩W. Then A=(W1∩W2)∩W=(W∩W1)∩W2, and since W∩W1 is open in X, the set A is locally closed in X.
For part (d), let A=Z1∩Z2, where Z1⊂B is open and Z2⊂B is closed. Then there exists Y1⊂X open such that Z1=B∩Y1 and Y2⊂X closed such that Z2=B∩Y2. So A=Z1∩Z2=(B∩Y1)∩(B∩Y2)=(B∩Y1)∩Y2, where (B∩Y1)⊂X is open and Y2⊂X is closed. Hence A⊂X is locally closed. ◻
Lemma 2: Let U⊂X be open and f:X→R continuous. Then ⋃x∈U{x}×(f(x),∞) is open in X×R.
Proof: Consider the function
g : X×R→X×R,(x,t)↦(x,t−f(x)).Since f is continuous and subtraction is continuous, g is continuous (in the product topology). Since U×(0,∞) is open in X×R, the set g−1(U×(0,∞)) is open in X×R. This is exactly the desired set. ◻
Lemma 2: Let U⊂X be open and f:X→R continuous. Then ⋃x∈U{x}×(f(x),∞) is open in X×R.
Proof: Consider the function
g : X×R→X×R,(x,t)↦(x,t−f(x)).Since f is continuous and subtraction is continuous, g is continuous (in the product topology). Since U×(0,∞) is open in X×R, the set g−1(U×(0,∞)) is open in X×R. This is exactly the desired set. ◻
Filtered diagrams
Definition: A filtered diagram is a directed graph such that
- for every pair of nodes u,v there is a node w such that there exist paths u→w and v→w, and
- for every multi-edge u1,2→v, there is a node w such that u1→v→w is the same as u2→v→w.
For our purposes, the nodes of a filtered diagram will be subsets of Rann(M)×R⩾0 and a directed edge will be open inclusion of one set into another set (that is, the first is open inside the second). Although we require below that loops u→u be removed, we consider the first condition above to be satisfied if there exists a path u→v or a path v→u.
Remark: In the context given,
Remark: In the context given,
- edge loops U→U and path loops U→⋯→U may be replaced by a single node U (U⊆U is the identity),
- multi-edges U1,2→V may be replaced by a single edge U→V (inclusions are unique), and
- multi-edges U→V→U may be replaced by a single node U (if U⊆V and V⊆U, then U=V).
A diagram with all possible replacements of the types above is called a reduced diagram.
Lemma 3: In the context above, a reduced filtered diagram D of open sets of any topological space X gives an increasing sequence of open subsets of X, with the same number of nodes.
Proof: Order the nodes of D so that if U→V is a path, then U has a lower index than V (this is always possible in a reduced diagram). Let U1,U2,…,UN be the order of nodes of D (we assume we have finitely many nodes). For every pair of indices i,j, set
δij={∅ if Ui→Uj is a path in D,Ui if Ui→Uj is not a path in D.Then the following sequence is an increasing sequence of nested open subsets of X:
U1→δ12∪U2→δ13∪δ23∪U3→⋯→(j−1⋃i=1δij)∪Uj⏟Vj→⋯→UN.Indeed, if Ui→Uj is a path in D, then Ui is open in Vj, as Ui⊂Vj. If Ui→Uj is not a path in D, then Ui is still open in Vj, as Ui⊂Vj. As unions of opens are open, and by Lemma 1(a), Vj−1 is open in Vj for all 1<j<N. ◻
Remark 4: Note that every consecutive difference Vj∖Vj−1 is a (not necessarily proper) subset of Uj.
Definition 5: For k∈Z>0, define a filtered diagram Dk over Rank(M)×R⩾0 by assigning a subset to every corner of the unit N-hypercube in the following way: for the ordered set S={(i,j) : 1⩾i<j⩾k} (with |S|=N=k(k−1)/2), write P={P1,…,Pk}∈Rank(M), and assign
(δ1,…,δk)↦{(P,t)∈Rank(M)×R⩾0 : t>d(P(Sℓ)1,P(Sℓ)2) whenever δℓ=0, ∀ 1⩽ℓ⩽k},where δℓ∈{0,1} for all ℓ, and d(x,y) is the distance on the manifold M between x,y∈M. The edges are directed from smaller to larger sets.
Remark 6: This diagram has 2k(k−1)/2 nodes, as k(k−1)/2 is the number of pairwise distances to consider. Moreover, the difference between the head and tail of every directed edge is elements (P,t) for which Rips(P,t) is constant.
Example: For example, if k=3, then 23⋅2/2=8, and D3 is the diagram below. For ease of notation, we write {t>⋯} to mean {(P,t) : P={P1,P2,P3}∈Ran3(M), t>⋯}.
δij={∅ if Ui→Uj is a path in D,Ui if Ui→Uj is not a path in D.Then the following sequence is an increasing sequence of nested open subsets of X:
U1→δ12∪U2→δ13∪δ23∪U3→⋯→(j−1⋃i=1δij)∪Uj⏟Vj→⋯→UN.Indeed, if Ui→Uj is a path in D, then Ui is open in Vj, as Ui⊂Vj. If Ui→Uj is not a path in D, then Ui is still open in Vj, as Ui⊂Vj. As unions of opens are open, and by Lemma 1(a), Vj−1 is open in Vj for all 1<j<N. ◻
Remark 4: Note that every consecutive difference Vj∖Vj−1 is a (not necessarily proper) subset of Uj.
Definition 5: For k∈Z>0, define a filtered diagram Dk over Rank(M)×R⩾0 by assigning a subset to every corner of the unit N-hypercube in the following way: for the ordered set S={(i,j) : 1⩾i<j⩾k} (with |S|=N=k(k−1)/2), write P={P1,…,Pk}∈Rank(M), and assign
(δ1,…,δk)↦{(P,t)∈Rank(M)×R⩾0 : t>d(P(Sℓ)1,P(Sℓ)2) whenever δℓ=0, ∀ 1⩽ℓ⩽k},where δℓ∈{0,1} for all ℓ, and d(x,y) is the distance on the manifold M between x,y∈M. The edges are directed from smaller to larger sets.
Remark 6: This diagram has 2k(k−1)/2 nodes, as k(k−1)/2 is the number of pairwise distances to consider. Moreover, the difference between the head and tail of every directed edge is elements (P,t) for which Rips(P,t) is constant.
Example: For example, if k=3, then 23⋅2/2=8, and D3 is the diagram below. For ease of notation, we write {t>⋯} to mean {(P,t) : P={P1,P2,P3}∈Ran3(M), t>⋯}.
The diagram of corresponding Vietoris-Rips complexes introduced at each node is below. Note that each node contains elements (P,t) whose Vietoris-Rips complex may be of type encountered in any paths leading to the node.
Lemma 7: In the filtered diagram Dk, every node is open inside every node following it.
Proof: The left-most node of Dk may be expressed as
{(P,t) : P={P1,…,Pk}∈Rank(M),t>d(Pi,Pj) ∀ Pi,Pj∈P}=⋃P∈Rank(M){P}×(maxPi,Pj∈P{d(Pi,Pj)},∞). Applying a slight variant of Lemma 2 (replacing R by an open ray that is bounded below), with the max function continuous, we get that the left-most node is open in the nodes one directed edge away from it. Repeating this argument, we get that every node is open inside every node following it. ◻
The constructible sheaf
Recall that a constructible sheaf can be given in terms of a nested cover of opens or a cover of locally closed sets (see post "Constructible sheaves," 2017-06-13). The approach we take is more the latter, and illustrates the relation between the two. Let n∈Z>0 be fixed.
Definition: Define a sheaf F over X=Ran⩽n(M)×R⩾0 valued in simplicial complexes, where the stalk F(P,t) is the Vietoris--Rips complex of radius t on the set P. For any subset U⊂X such that F(P,t) is constant for all (P,t)∈U, let F(U)=F(P,t).
Note that we have not described what F(U) is when U contains stalks with different homotopy types. Omitting this (admittedly large) detail, we have the following:
Theorem: The sheaf F is constructible.
Proof: First, by Remark 5.5.1.10 in Lurie, we have that Rann(M) is open in Ran⩽n(M). Hence Ran⩽n−1(M) is closed in Ran⩽n(M). Similarly, Ran⩽n−2(M) is closed in Ran⩽n−1(M), and so closed in Ran⩽n(M), meaning that Ran⩽k(M) is closed in Ran⩽n(M) for all 1⩽k⩽n. This implies that Ran⩾k(M) is open in Ran⩽n(M) for all 1⩽k⩽n, meaning that Rank(M) is locally closed in Ran⩽n(M), for all 1⩽k⩽n.
Next, for every 1⩽k⩽n, let Vk,1→⋯→Vk,Nk be a sequence of nested opens covering Rank(M)×R⩾0, as given in Definition 5 and flattened by Lemma 3. The sets are open by Lemma 7. This gives a cover Vk={Vk,1,Vk,2,∖Vk,1,…,Vk,Nk∖Vk,Nk−1} of Rank(M)×R⩾0=Vk,Nk by consecutive differences, with Vk,1 open in Vk,Nk and all other elements of Vk locally closed in Vk,Nk, by Lemma 1(b). By Lemma 1 parts (c) and (d), every element of Vk is locally closed in Ran⩽n(M)×R⩾0, and so V=⋃nk=1Vk covers Ran⩽n(M)×R⩾0 by locally closed subsets.
Finally, by Remarks 4 and 6, over every V∈V the function Rips(P,t) is constant. Hence F|V is a locally constant sheaf, for every V∈V. As the V are locally closed and cover X, F is constructible. ◻
References: Lurie (Higher algebra, Section 5.5.1)
Definition: Define a sheaf F over X=Ran⩽n(M)×R⩾0 valued in simplicial complexes, where the stalk F(P,t) is the Vietoris--Rips complex of radius t on the set P. For any subset U⊂X such that F(P,t) is constant for all (P,t)∈U, let F(U)=F(P,t).
Note that we have not described what F(U) is when U contains stalks with different homotopy types. Omitting this (admittedly large) detail, we have the following:
Theorem: The sheaf F is constructible.
Proof: First, by Remark 5.5.1.10 in Lurie, we have that Rann(M) is open in Ran⩽n(M). Hence Ran⩽n−1(M) is closed in Ran⩽n(M). Similarly, Ran⩽n−2(M) is closed in Ran⩽n−1(M), and so closed in Ran⩽n(M), meaning that Ran⩽k(M) is closed in Ran⩽n(M) for all 1⩽k⩽n. This implies that Ran⩾k(M) is open in Ran⩽n(M) for all 1⩽k⩽n, meaning that Rank(M) is locally closed in Ran⩽n(M), for all 1⩽k⩽n.
Next, for every 1⩽k⩽n, let Vk,1→⋯→Vk,Nk be a sequence of nested opens covering Rank(M)×R⩾0, as given in Definition 5 and flattened by Lemma 3. The sets are open by Lemma 7. This gives a cover Vk={Vk,1,Vk,2,∖Vk,1,…,Vk,Nk∖Vk,Nk−1} of Rank(M)×R⩾0=Vk,Nk by consecutive differences, with Vk,1 open in Vk,Nk and all other elements of Vk locally closed in Vk,Nk, by Lemma 1(b). By Lemma 1 parts (c) and (d), every element of Vk is locally closed in Ran⩽n(M)×R⩾0, and so V=⋃nk=1Vk covers Ran⩽n(M)×R⩾0 by locally closed subsets.
Finally, by Remarks 4 and 6, over every V∈V the function Rips(P,t) is constant. Hence F|V is a locally constant sheaf, for every V∈V. As the V are locally closed and cover X, F is constructible. ◻
References: Lurie (Higher algebra, Section 5.5.1)