Abstract 1 Introduction 2 Background 3 Setup 4 Lifting 5 Apex Representatives 6 Four Intervals 7 Zigzag Representatives References

Apex Representatives

Tamal K. Dey ORCID Purdue University, West Lafayette, IN, USA Tao Hou ORCID University of Oregon, Eugene, OR, USA Dmitriy Morozov ORCID Lawrence Berkeley National Laboratory, Berkeley, CA, USA
Abstract

Given a zigzag filtration, we want to find its barcode representatives, i.e., a compatible choice of bases for the homology groups that diagonalize the linear maps in the zigzag. To achieve this, we convert the input zigzag to a levelset zigzag of a real-valued function. This function generates a Mayer–Vietoris pyramid of spaces, which generates an infinite strip of homology groups. We call the origins of indecomposable (diamond) summands of this strip their apexes and give an algorithm to find representative cycles in these apexes from ordinary persistence computation. The resulting representatives map back to the levelset zigzag and thus yield barcode representatives for the input zigzag. Our algorithm for lifting a p-dimensional cycle from ordinary persistence to an apex representative takes O(pmlogm) time. From this we can recover zigzag representatives in time O(logm+C), where C is the size of the output.

Keywords and phrases:
zigzag persistent homology, Mayer–Vietoris pyramid, cycle representatives
Funding:
Tamal K. Dey: partially supported by NSF grants CCF-2437030 and DMS-2301360.
Tao Hou: supported by NSF fund CCF 2439255.
Dmitriy Morozov: supported by the Director, Office of Science, Office of Advanced Scientific Computing Research, of the U.S. Department of Energy under Contract No. DE-AC02-05CH11231.
Copyright and License:
[Uncaptioned image] © Tamal K. Dey, Tao Hou, and Dmitriy Morozov; licensed under Creative Commons License CC-BY 4.0
2012 ACM Subject Classification:
Theory of computation Computational geometry
; Mathematics of computing Algebraic topology
Related Version:
Full Version: https://arxiv.org/abs/2502.17704 [8]
Editors:
Oswin Aichholzer and Haitao Wang
Figure 1: A zigzag on the top. A total complex K on the left. The support of each simplex in a zigzag, T(σ), shown as an interval. The corresponding prism K¯ on the bottom. A 1-cycle in K[11,17] and its lift in K¯[11,17] are highlighted in red. The lifted cycle is an apex representative; its slices (pairs of vertices in red) are representatives of a bar in the original zigzag.

1 Introduction

In topological data analysis, one often wants to not only describe the distribution of topological features in data, but also to identify individual features directly in the input. For persistent homology, a natural starting point are the cycle representatives of homology classes. Specifically, one wants to find barcode representatives, a consistent choice of bases for the input sequence of homology groups. For ordinary persistence, this comes directly from the computation: a representative at the start of the bar includes into every consecutive space and therefore remains a valid choice of the basis element.

For zigzag persistence, the situation is more complicated. Because simplices can enter and leave the complex, a representative cycle at one step in the zigzag filtration may not exist at a later step. So even though zigzag persistence algorithms keep track of representative cycles, the specific choices that they make do not always produce compatible barcode representatives. To cope with this problem, Dey et al. [9] introduced an algorithm that keeps track of additional information that allows it to report consistent barcode representatives. For a zigzag filtration with m insertions and deletions, the algorithm requires O(m3) preprocessing and O(m2) time to report one or all representatives for any single bar, and O(m3) time to report all representatives.

On the surface, this is as well as one can hope to do: there are O(m) bars, over O(m) spaces, with each representative consisting of O(m) simplices. But we can do better. Every zigzag filtration is isomorphic to a levelset zigzag filtration of some function. The function generates the Mayer–Vietoris pyramid [5], which decomposes into diamond summands [2]. Each diamond has a single space as its origin, its apex: this space maps into every other space in the diamond. By finding the diamond’s representative in the apex – an apex representative – we can quickly recover its representative in every other space, including those that fall in the levelset zigzag, and therefore in the original zigzag; see Figure 1. All the information can be recovered from an ordinary persistence computation. It takes matrix multiplication time, O(mω), to compute persistence [16], and an extra O(pmlogm) time to recover a p-dimensional apex representative of size O(pm). After the preprocessing, zigzag representatives can be recovered in O(logm+C) time, where C is the size of the output.

2 Background

We assume familiarity with simplicial and cell complexes, homology, persistent homology, including zigzag persistence. We only recap select topics to establish notation. Any gaps can be filled with the standard literature [14, 12, 13].

Throughout the paper, we take care to work with arbitrary field coefficients, but abuse notation as follows. Cell σz means the coefficient of σ in z is non-zero, σ,z0. If α is a chain in some cell complex K, then αL means that α is supported on the subcomplex LK, i.e., σ(KL),σ,α=0. Similarly, zL refers to the restriction of chain z to L, i.e., zL=σLσ,zσ.

Persistence.

Given a filtered cell complex K, let D be a matrix that represents its boundary operator, with rows and columns ordered according to the filtration. To find ordinary persistence pairing, one can compute a decomposition, R=DV, where matrix R is reduced, meaning its pivots – lowest non-zero elements in its columns – appear in unique rows, and matrix V is full-rank upper-triangular. We index the columns and rows of the matrices by the (totally ordered) cells of the input complex. The persistence pairing is given by the pivots of matrix R: a p-dimensional homology class created by the addition of simplex σ is destroyed by the addition of simplex τ iff lowR[τ]=σ, where lowR[τ] returns the pivot in the column R[τ]. By definition, the corresponding column V[τ] stores a (p+1)-dimensional chain whose boundary is the cycle R[τ].

The decomposition R=DV is not unique: multiple matrices R and V satisfy it. However, the original persistence algorithm [10] performs operations in a lazy fashion, subtracting columns from left to right only if their pivots collide. We call this a lazy reduction. It has the following special properties that simplify our algorithms.

Lemma 1 (Lemma 1 in [17]).

Assume decomposition R=DV is obtained via the lazy reduction, and σi=lowR[τi] and σj=lowR[τj] are such that τi<τj. If σi<σj, then entry V[τi,τj]=0.

Proof.

Induction on V; see [17].

 Remark 2.

The authors of [17] simplify the statement of the preceding lemma by abusing the notation. They assume that if R[τ]=0, then lowR[τ]=σ¯, a special imaginary cell that precedes every cell in the complex. In particular, it follows that in a lazy reduction, entry V[τi,τj]=0 if R[τi]=0.

The contrapositive of the lemma is the following corollary.

Corollary 3.

Assume decomposition R=DV is obtained via the lazy reduction, and σi=lowR[τi] and σj=lowR[τj] are such that τi<τj. If V[τi,τj]0, then σj<σi<τi<τj.

Zigzag persistence.

We start with a zigzag of simplicial complexes,

K0K1K2K3K4Kn, (1)

where every consecutive pair of complexes vary by at most one simplex, i.e., Ki can be equal to Ki+1 in the sequence. It is convenient to assume that every simplex σ appears and disappears exactly once in this sequence, and therefore is present during some interval T(σ)=[i,j], i.e., σKk iff kT(σ). This assumption can be made without loss of generality by assuming that the union K of all simplices across all times, K=iKi forms a Δ-complex. The full definition of this object is too verbose – we refer the reader to [7], which adapts the classical concept of Δ-complexes in [14, p. 103] to our case of zigzag filtration – but informally it relaxes the requirement that in a simplicial complex two simplices intersect in a common face. All the familiar notions of simplicial homology work the same, but now instead of having a simplex supported over multiple intervals in the zigzag, we can have multiple simplices, with the same boundary, each supported on a single interval. We use m=|K| to denote the size of the input complex.

 Remark.

To simplify the notation we let min(σ)=minT(σ) and max(σ)=maxT(σ).

 Remark 4.

We assume that max(σ)>min(σ), i.e., T(σ) is not a single point. We can assume this without loss of generality because we can always pad the zigzag with extra copies of a space.

Applying homology, we get a zigzag of homology groups, where we suppress dimension for simplicity,

𝖧(K0)𝖧(K1)𝖧(K2)𝖧(K3)𝖧(K4)𝖧(Kn). (2)

Just like with ordinary persistence, this zigzag decomposes into k bars or persistence intervals [3]. Let {αi1,,αik} be a choice of elements in 𝖧(Ki) such that the non-zero elements form a basis for 𝖧(Ki). We say that such bases are compatible across the zigzag if the maps in Equation 2 diagonalize, i.e., α2i1jα2ijα2i+1j. For every j, αj are non-zero over a single persistence interval [bj,dj]. Cycles zij that give a set of compatible bases αij=[zij] are called zigzag representatives, with zj being the representatives of the j-th bar in the zigzag barcode.

Real-valued function.

A convenient setting for persistence is that of a Morse-like [4] real-valued function f:𝕏. We denote with 𝕏ab=f1[a,b] the preimage of an interval and allow the endpoints to be infinite, a,b=±, in which case the interval is understood to be open at the infinite ends. We denote the following pairs of spaces:

𝕏[b,d]=(𝕏bd,), 𝕏(b,d]=(𝕏d,𝕏b),
𝕏[b,d)=(𝕏b,𝕏d), 𝕏(b,d)=(𝕏,𝕏b𝕏d).

Let a1<<an be the critical values of the function f. Let si be regular values interleaved with the critical values: s0<a1<s1<a2<<sn1<an<sn. The following constructions play an important role in the theory of persistent homology:

  • Extended persistence (EP):

    0 𝖧(𝕏[,s0])𝖧(𝕏[,sn])=𝖧(𝕏)
    𝖧(𝕏[,sn))𝖧(𝕏[,s1))𝖧(𝕏[,s0))=0.
  • Levelset zigzag (LZZ):

    0𝖧(𝕏[s0,s0])𝖧(𝕏[s0,s1])𝖧(𝕏[s1,s1])𝖧(𝕏[sn1,sn])𝖧(𝕏[sn,sn])0.

Carlsson et al. [5] showed that these two sequences contain the same information by arranging the four types of spaces into a Mayer–Vietoris Pyramid, see Figure 2(left). Once homology is applied, the pyramid unrolls into an infinite Mayer–Vietoris strip of homology groups, where every diamond belongs to the Mayer–Vietoris long exact sequence, making it exact in the terminology of [3], see Figure 2(right). This allows one to translate the decomposition between any two paths that differ by a single diamond (and therefore between any pair of paths via composition).

Figure 2: Left: Mayer–Vietoris Pyramid arranges spaces 𝕏[b,d],𝕏(b,d],𝕏[b,d),𝕏(b,d) in a way where every diamond in the pyramid belongs to the Mayer–Vietoris long exact sequence. Right: Four types of flush diamond indecomposables in the infinite strip of homology groups; the apex of each diamond is marked.

Bendich et al. [2] further showed that the translation rules in [5] hold at the level of the basis elements of the individual paths, which means the entire pyramid decomposes as a direct sum of (indecomposable) pointwise 1-dimensional diamond summands, flush with its boundaries, shown in blue in Figure 2(right). The decomposition of the levelset zigzag and extended persistence are simply slices through these flush diamond summands. Bauer et al. [1] study the Mayer–Vietoris Pyramid in the cohomological setting.

We call the bottom-most space in the support of a diamond its apex. The homology class assigned to the diamond in that space, an apex class, and any cycle that belongs to such a class, an apex representative. We note that just like the choice of the basis for ordinary persistence is not unique, neither is the choice of the apex classes. An apex class is characterized by lying outside the image of the maps into the apex; specifically, for the four types of apexes their classes satisfy:

[z]𝖧(𝕏[b,d]) s.t. [z]im(𝖧(𝕏[b+ε,d])𝖧(𝕏[b,d])) and [z]im(𝖧(𝕏[b,dε])𝖧(𝕏[b,d]))
[z]𝖧(𝕏(b,d]) s.t. [z]im(𝖧(𝕏(bε,d])𝖧(𝕏(b,d])) and [z]im(𝖧(𝕏(b,dε])𝖧(𝕏(b,d]))
[z]𝖧(𝕏[b,d)) s.t. [z]im(𝖧(𝕏[b+ε,d))𝖧(𝕏[b,d))) and [z]im(𝖧(𝕏[b,d+ε))𝖧(𝕏[b,d)))
[z]𝖧(𝕏(b,d)) s.t. [z]im(𝖧(𝕏(b+ε,d))𝖧(𝕏(b,d))) and [z]im(𝖧(𝕏(b,dε))𝖧(𝕏(b,d)))

Because there is a map from the apex to every space in the diamond, we can recover a representative in any space by simply mapping an apex representative forward. All the maps in the pyramid are inclusions, except for the boundary homomorphism connecting two consecutive dimensions of homology in the strip. We say more about this translation in Section 7, but the overarching point is that given an apex representative, it is straightforward to recover every other representative, and levelset zigzag representatives in particular.

3 Setup

Prism.

Given a zigzag in Equation 1, we define a larger cell complex K¯, called a prism:

K¯={σ×[i,i+1][i,i+1]T(σ),i}{σ×iiT(σ),i}, (3)

together with a function f¯:|K¯| on its underlying space, defined by the projection onto the second component, f¯(x×t)=t. See Figure 3. We observe that the prism consists of two types of cells: horizontal cells σ×[i,i+1] and vertical cells σ×i.

Figure 3: A Δ-complex K, with the times T(σ) of its simplices illustrated, together with the corresponding prism K¯ and its projection f¯ onto the second coordinate.

We define an integer interlevel set, for i,j, K¯ij={σ×TK¯T[i,j]}. We note that its underlying space |K¯ij|=f¯1[i,j]. Then for any pair of integers b,d, we get four types of pairs of subspaces,

K¯[b,d]=K¯bd, K¯(b,d]=(K¯1d,K¯1b), K¯[b,d)=(K¯bn+1,K¯dn+1), K¯(b,d)=(K¯,K¯1bK¯dn+1).

Applying homology to each pair, we get relative homology groups that appear in the four quadrants of the Mayer–Vietoris pyramid: 𝖧(K¯[b,d]),𝖧(K¯(b,d])),𝖧(K¯[b,d)),𝖧(K¯(b,d)).

Because even-indexed spaces in Equation 1 include into odd-indexed spaces, K2i1K2iK2i+1, we have an isomorophism of homology groups 𝖧(K¯[2i,2i+1])𝖧(K¯[2i+1,2i+1])𝖧(K¯[2i+1,2i+2]), induced by deformation retractions. It follows that the levelset zigzag [5] of function f¯,

𝖧(K¯[0,0])𝖧(K0)𝖧(K¯[0,1])𝖧(K¯[1,1])𝖧(K¯[1,2])𝖧(K1)𝖧(K¯[2,2])𝖧(K2)𝖧(K¯[n,n])𝖧(Kn)

is isomorphic to our starting zigzag in Equation 2. Explicitly, the isomorphism is given by the following bijection between the four types of intervals:

(𝖧(K2i),𝖧(K2j)) (𝖧(K¯[2i,2i]),𝖧(K¯[2j,2j])) (open-open)
(𝖧(K2i),𝖧(K2j+1)) (𝖧(K¯[2i,2i]),𝖧(K¯[2j+1,2j+2])) (open-closed)
(𝖧(K2i+1),𝖧(K2j)) (𝖧(K¯[2i,2i+1]),𝖧(K¯[2j,2j])) (closed-open)
(𝖧(K2i+1),𝖧(K2j+1)) (𝖧(K¯[2i,2i+1]),𝖧(K¯[2j+1,2j+2])) (closed-closed)

Intervals.

Because we perform computation on the input complex K, rather than the prism, we need a notation for different intervals and pairs of intervals. The connection between these and the corresponding intervals in K¯ will be made clear in Section 6. We denote the sub- and super-level sets:

K1i={σKmin(σ)i}, Kjn+1={σKmax(σ)j},

and define four pairs of spaces:

K[b,d]=Kbn+1K1d, K(b,d]=(K1d,K1b),
K[b,d)=(Kbn+1,Kdn+1), K(b,d)=(K,K1bKdn+1).

Because every step of the input zigzag is a simplicial complex, it follows that for any σ,τK,

στmin(σ)<min(τ)<max(τ)<max(σ). (4)

The following relationships between sub- and super-levelsets follow immediately:

αK1x αK1x (5)
αKKxn+1 αK1x (6)

Cone.

To compute the persistence decomposition of the zigzag, we follow the algorithm of Dey and Hou [7], who use a construction similar to extended persistence [6] on the cone K^=ωK over the input complex K, where indicates a join with the cone vertex ω. Every simplex σ in the base space K gets value min(σ) from the input zigzag. Every simplex σ^ in the cone K^K gets value max(σ) from the input zigzag. We then define a filtration, where the base space simplices come first, ordered by increasing value, and the cone simplices come second, ordered by decreasing value. Specifically,

0𝖧(K11)𝖧(K1n)=𝖧(K)𝖧(K,Knn+1)𝖧(K,K1n+1)𝖧(K,K)=0,

which in reduced homology is isomorphic to

0𝖧(K11)𝖧(K)𝖧(KωKnn+1)𝖧(KωK1n+1)𝖧(K^)=0. (7)

The latter is an ordinary filtration and we compute its persistence via the R=DV decomposition of the boundary map of the cone K^. Crucially, the resulting bars are in one-to-one correspondence with the bars in the decomposition of the input zigzag in Equation 2.

4 Lifting

Because prism K¯ has so many cells, it is expensive to process directly. We want to perform computation on the cone K^ instead. To recover the apex representatives for the prism, we need to lift cone cycles in K^ to prism cycles in K¯. In Section 6, we explain what exact cycles to lift to get different apex representatives. Meanwhile, we first describe an algorithm that given a (relative) cycle z in KI produces a cycle z¯ in K¯I, where I is one of the four types of intervals, (b,d], [b,d), (b,d), [b,d]. Algorithm 1 in this section is a reinterpretation of an inefficient, but easier-to-understand Lift-Cycle-easy algorithm in the full version of this paper [8]. We encourage the reader to go through through that algorithm first.

 Remark.

A p-chain z¯ in K¯ consists of two types of p-cells: τ×t, where τ is a p-simplex, and σ×[t1,t2], where σ is a (p1)-simplex.

 Remark.

We use an abridged notation, cσ×[i,j], to represent the chain k[i,j1]cσ×[k,k+1]. This choice is guided by computational efficiency: the former requires constant space vs. the O(ji) space required for the latter.

Algorithm 1 takes a p-cycle z, a direction expressed as a pair of values (s,f), and an initial (p1)-cycle winit. We call the pair (s,f) a direction because the order of the values specifies whether we process the cycle in increasing (s<f) or decreasing (f<s) order. The algorithm stretches the cycle from K¯[s,s] to K¯[f,f], covering K¯I; see Figure 1.

Algorithm 1 Lift-Cycle(z, (s,f), winit.)

Correctness.

The correctness of Algorithm 1 follows from the following claim about the boundary structure, which will be important in Section 6. We let zxy be the restriction of cycle z to the simplices τ whose times tτ lie in the interval [x,y], i.e., zxy=τz,tτ[x,y]τ,zτ.

Claim 5.

Given input cycle zK, the boundary of the lifted cycle z¯ satisfies z¯=winit×s+wfinal×f, where wfinal=winit+zbd.

Proof.

Simplex σ generates a cell σ×[s,l]z¯ iff σ,winit0 (Line 5). This cell contributes σ×s to the boundary of z¯. Simplex σ generates a cell σ×[l,f]z¯ iff σ,winit+τδσzbdσ,τ0. By definition this is the case only for simplices in wfinal=winit+zbd. It follows that if winitK[s,s], then z¯ is a cycle in K¯I. What is not immediate is why the chains σ×[l,tτ] and σ×[l,f] in Lines 9 and 13 are in the prism K¯.

Claim 6.

Chains σ×[l,tτ] and σ×[l,f] added in Lines 9 and 13 of Algorithm 1 are present in K¯.

Proof.

Without loss of generality, assume s<f. There are three types of chains:

  1. 1.

    σ×[s,tτ1]. σ has a non-zero coefficient on this interval iff σ,winit0 (Line 5), in which case σK[s,s]. Existence of coface τ1 implies min(σ)stτ1<max(σ).

  2. 2.

    σ×[tτi,tτi+1]. From Equation 4, min(σ)<tτitτi+1<max(σ).

  3. 3.

    σ×[tτi,f]. Suppose that this chain is not in K¯. This means x=max(σ)[tτi,f). Equation 4 implies that σ has no cofaces in [x,f]. Therefore, σwinit+zsx and σzxf. Therefore, σz. Since we assumed z is a relative cycle in KI, it cannot have any boundary inside the interval itself – a contradiction.

 Remark.

It may seem strange that we state nothing about the relationship between z and z¯. This is because we exploit additional properties of the algorithm in Section 6.

Running time.

Let m be the size of the input p-cycle z. Algorithm 1 requires O(pmlogm) time. It goes through all simplices σ that are faces of some simplex τz; there are at most (p+1)m such (p1)-simplices. For each one, the algorithm sorts its cofaces by their times tτ. Although there is no bound on the number of cofaces of one simplex, the total number of the cofaces to sort is again at most (p+1)m. Individual operations in the update take constant time, so the overall running time of Algorithm 1 is O(pmlogm). The size of the lifted cycle is O(pm).

5 Apex Representatives

We give a characterization of apex representatives in the prism.

Lemma 7 (Closed endpoint).
  1. 1.

    If a relative cycle z¯, with [z¯]𝖧(K¯(b,d]), contains τ×d with minτ=d, then

    [z¯]im(𝖧(K¯(b,d1])𝖧(K¯(b,d])).
  2. 2.

    If a relative cycle z¯, with [z¯]𝖧(K¯[b,d)), contains τ×b with maxτ=b, then

    [z¯]im(𝖧(K¯[b+1,d))𝖧(K¯[b,d))).
  3. 3.

    If an absolute cycle z¯, with [z¯]𝖧(K¯[b,d]), contains σ×b and τ×d with maxσ=b and minτ=d, then

    [z¯]im(𝖧(K¯[b+1,d])𝖧(K¯[b,d]))and[z¯]im(𝖧(K¯[b,d1])𝖧(K¯[b,d])).

Proof.

All the statements follow the same proof, except for the endpoints of the intervals, so we only show the first. Suppose there is some class [α]𝖧(K¯(b,d1]) that maps to [z¯]𝖧(K¯(b,d]). Then z¯=α+β for some chain β. Since τz¯ and τα (since minτ=d>d1), τ must be in β. But from Equation 4, τ has no cofaces in K1d – a contradiction.

Lemma 8 (Open endpoint).
  1. 1.

    If the boundary z¯ of a relative cycle z¯, with [z¯]𝖧(K¯(b,d]), contains σ×b with minσ=b, then [z¯]im(𝖧(K¯(b1,d])𝖧(K¯(b,d])).

  2. 2.

    If the boundary z¯ of a relative cycle z¯, with [z¯]𝖧(K¯[b,d)) contains σ×d with maxσ=d, then [z¯]im(𝖧(K¯[b,d+1))𝖧(K¯[b,d))).

  3. 3.

    If the boundary z¯ of a relative cycle z¯, with [z¯]𝖧(K¯(b,d)), contains σ×b with minσ=b and τ×d with maxτ=d, then

    z¯im(𝖧(K¯(b1,d))𝖧(K¯(b,d)))andz¯im(𝖧(K¯(b,d+1))𝖧(K¯(b,d))).

Proof.

All the statements follow the same proof, except for the endpoints of the intervals, so we only show the first. From the long exact sequence of the triple [14, p. 118], K1b1K1bK1d, we know that the image in the claim is equal to the kernel of the map induced by the boundary,

im(𝖧(K¯(b1,d])𝖧(K¯(b,d]))=ker(𝖧(K¯(b,d])𝖧1(K¯(b1,b])).

For [z¯] to be in the kernel, z¯ needs to be a relative boundary in K¯(b1,b], i.e., z¯=β+γ for some βK1b and γK1b1. Since σz¯ and σγ, σ must be in β. But from Equation 4, σ has no cofaces in K1b – a contradiction.

Putting Lemmas 7 and 8 together, we get the following theorem.

Theorem 9.
  1. 1.

    If a relative cycle z¯, with [z¯]𝖧(K¯(b,d]), contains τ×d with minτ=d, and its boundary z¯ contains σ×b with minσ=b, then z¯ is an apex representative.

  2. 2.

    If a relative cycle z¯, with [z¯]𝖧(K¯[b,d)), contains τ×b with maxτ=b, and its boundary z¯ contains σ×d with maxσ=d, then z¯ is an apex representative.

  3. 3.

    If an absolute cycle z¯, with [z¯]𝖧(K¯[b,d]), contains σ×b and τ×d with maxσ=b and minτ=d, then z¯ is an apex representative.

  4. 4.

    If the boundary z¯ of a relative cycle z¯, with [z¯]𝖧(K¯(b,d)), contains σ×b with minσ=b and τ×d with maxτ=d, then z¯ is an apex representative.

6 Four Intervals

We assume R=DV is the decomposition of the boundary matrix D of the cone K^ filtration in Equation 7. We use σ,τ to refer to simplices in the base space K, and σ^,τ^ for the simplices in the cone K^K.

Summary.

The main content of this section is summarized in Table 1: lifting the stated cycles z in K, derived from the lazy reduction, with the given arguments to the algorithm Lift-Cycle, produces apex representatives in K¯. The results of the section are more general, but also more verbose; they derive the expression for the cycles for any reduction.

Table 1: Summary of the chains and arguments used in different cases of the Lift-Cycle algorithm, derived from the lazy reduction.
EP type LZZ type Apex z (s,f) winit
Ordinary closed-open K(b,d] V[τ] (d,b) 0
Relative open-closed K[b,d) R[τ^]K (b,d) 0
Extended (d,b) open-open K[b,d] R[τ^] (d,b) 0
Extended (b,d) closed-closed K(b,d) V[τ^]K (b,d) R[τ^]

6.1 Ordinary (closed-open)

Throughout this subsection, we assume the following setting:

a pair (σ,τ) in the cone filtration, born at b and dying at d, i.e., min(σ)=b<d=min(τ).
Claim 10.

Let z=V[τ]. Then [z]𝖧(K(b,d]). Furthermore, τz and σz=R[τ].

Proof.

Since V is upper-triangular with all diagonal entries non-zero, τ is the latest simplex in V[τ], i.e., max{min(τ)τV[τ]}=min(τ)=d. Therefore, V[τ]K1d. By definition, the boundary V[τ]=R[τ]. σ is its latest simplex, i.e., max{min(σ)σR[τ]}=min(σ)=b. Therefore, V[τ]K1b. It follows that if z=V[τ], then [z]𝖧(K1d,K1b)=𝖧(K(b,d]).

Claim 11.

Relative cycle z¯ produced by the call Lift-Cycle(V[τ],(d,b),0) is an apex representative in 𝖧(K¯(b,d]).

Proof.

Let z=V[τ].

  1. 1.

    σ×bz¯, min(σ)=b.

    Because σz (Claim 10) and min(σ)=b, its coboundary δσKbn+1. Meanwhile, zK1d. Therefore, all cofaces of σ in z intersect [b,d]. Therefore, σ(z[b,d]). It follows from Claim 5 that

    σ×bz¯=wfinal×b=(z[b,d])×b.
  2. 2.

    τ×dz¯, min(τ)=d.

    This follows immediately since τz (Claim 10) and T(τ)[b,d]={d}.

It follows from Theorem 9 that z¯ is an apex representative in 𝖧(K¯(b,d]).

6.2 Relative (open-closed)

Throughout this subsection, we assume the following setting:

a pair (σ^,τ^) in the cone filtration, born at d and dying at b, i.e., max(σ)=d>b=max(τ).
Claim 12.

Let chain z consist of the boundary of the cone simplices in V[τ^] restricted to the base space,

z=(V[τ^](K^K))K.

Then [z]𝖧(K[b,d)). Furthermore, τz and σz.

Proof.

Let γ=V[τ^](K^K), i.e., z=(γ)K.

  1. 1.

    zKbn+1, τz.

    Because γ consists of only cone simplices, every simplex τz is in the boundary of some cone simplex τ^. Because τ^ is the latest simplex in V[τ^], we have that τ^ is added before τ^. Therefore, max(τ)max(τ)=b. Because τ^ is the only simplex in γ with τ in its boundary, τγK=z.

  2. 2.

    zKdn+1.

    Let

    α=z=(γ(K^K))

    be the shared boundary between the base space and the cone parts of γ. We note that because cone simplices can only be in the boundary of cone simplices, (V[τ^])(K^K)=γ(K^K). Therefore, all the cone simplices σ^γ(K^K)) are such that max(σ)max(σ)=d. Therefore, any simplex in the shared boundary α is in Kdn+1.

    Putting the first two sub-claims together, [z]𝖧(Kbn+1,Kdn+1)=𝖧(K[b,d)).

  3. 3.

    σz.

    (σ^,τ^) is a persistence pair, therefore, σ^V[τ^]. Because cone simplices can only be in the boundaries of cone simplices, σ^γ. Because σ^ is the only simplex in (γ)(K^K) that has σ in its boundary, it follows that

    σz=((γ)K)=((γ)(K^K)).

 Remark 13 (Lazy reduction).

If R=DV is obtained by lazy reduction, it follows from Corollary 3 that V[τ^] consists only of cone simplices. In other words, the chain in the prior claim simplifies to

z=(V[τ^])K=R[τ^]K.
Claim 14.

Relative cycle z¯ produced by the call Lift-Cycle(R[τ^]K,(b,d),0) is an apex representative in 𝖧(K¯[b,d)).

Proof.

Let z=R[τ^]K.

  1. 1.

    σ×dz¯, max(σ)=d.

    Because σz (Claim 12) and max(σ)=d, its coboundary δσK1d. Meanwhile, zKbn+1. Therefore, all cofaces of σ in z intersect [b,d]. Therefore, σ(z[b,d]). It follows from Claim 5 that

    σ×dz¯=wfinal×d=(z[b,d])×d.
  2. 2.

    τ×bz¯, max(τ)=b.

    This follows immediately since τz (Claim 12) and T(τ)[b,d]={b}.

It follows from Theorem 9 that z¯ is an apex representative in 𝖧(K¯[b,d)).

6.3 Extended (open-open)

Throughout this subsection, we assume the following setting:

a pair (σ,τ^) in the cone filtration born at d, dying at b, i.e., min(σ)=d>b=max(τ).
Claim 15.

Let chain c consist of those simplices in V[τ^] that are either in the cone, or in Kbn+1, i.e.,

c=V[τ^](K^K)+V[τ^]Kbn+1.

Let z=c; then [z]𝖧(K[b,d]). Furthermore, σ,τz.

Proof.

We split the chain V[τ^] into three parts:

α=V[τ^](KKbn+1), β=V[τ^]Kbn+1, γ=V[τ^](K^K),

and re-write its boundary accordingly: R[τ^]=V[τ^]=α+β+γ. c=(β+γ)=V[τ^]α.

  1. 1.

    z=cK1d, σz.

    V[τ^]K1d because σ is the latest simplex in R[τ^]. Putting together Equations 6 and 5, αKKbn+1 implies αK1bK1d. Therefore, z=V[τ^]αK1d.

    Because σV[τ^], but σα, σz.

  2. 2.

    zKbn+1, τz.

    Since chain βKbn+1, its boundary βKbn+1. Since boundaries V[τ^],α,βK, so is γ. Every simplex in γ has the form ωτ, where τKbn+1. Therefore, γKbn+1. It follows that z=β+γKbn+1.

    Because max(τ)=b, it has no cofaces in Kbn+1. Therefore, τβ. Because τ^ is the only simplex in γ that has τ in its boundary, τγ. Therefore, τz=β+γ.

Therefore, [z]𝖧(Kbn+1K1d)=𝖧(K[b,d]).

 Remark 16 (Lazy reduction).

Furthermore, if R=DV comes from the lazy reduction, then the next claim proves that in the previous claim, chain c=V[τ^]. In other words, we can use z=R[τ^]=V[τ^] directly.

Claim 17 (Lazy reduction).

If R=DV is obtained from the lazy reduction, then all the base space simplices in V[τ^] are in Kbn+1, i.e., V[τ^]KKbn+1.

Proof.

Let τ be a simplex in V[τ^]K. In this case, for the lazy reduction, because V[τ,τ^]0, from Corollary 3, taking σj=σ and τi=τ, we get

max(τ)>min(τ)>min(σ)=d>b.

Claim 18.

Absolute cycle z¯ produced by the call Lift-Cycle(R[τ^],(d,b),0) is an apex representative in 𝖧(K¯[b,d]).

Proof.

Let z=R[τ^]. Because σ,τz (Claim 15) and min(σ)=d and max(τ)=b, both simplices σ×d and τ×b are in the lifted cycle z¯. It follows from Theorem 9 that z¯ is an apex representative in 𝖧(K¯[b,d]).

6.4 Extended (closed-closed)

Throughout this subsection, we assume the following setting:

a pair (σ,τ^) in the cone filtration, born at b, dying at d, i.e., min(σ)=b<d=max(τ).
Claim 19.

Let z=V[τ^]Kbn+1. Then [z]𝖧(K(b,d)). Furthermore,

σwinit=(V[τ^](K^K)+V[τ^]Kbn+1)K[b,b]

and τ(winit+z)=(V[τ^](K^K)).

Proof.

We split the chain V[τ^] into three parts:

α=V[τ^](KKbn+1), z=β=V[τ^]Kbn+1, γ=V[τ^](K^K),

and re-write its boundary accordingly: R[τ^]=V[τ^]=α+β+γ.

  1. 1.

    zK1bKdn+1.

    Putting together Equations 6 and 5, αKKbn+1 implies αK1b. Because σ is the latest simplex in R[τ^] and min(σ)=b, R[τ^]K1b. It follows that (β+γ)=R[τ^]αK1b. Since γKdn+1, we have z=(β+γ)γK1bKdn+1. Therefore, [z]𝖧(K(b,d)).

  2. 2.

    σwinit=(β+γ)K[b,b].

    We already saw that (β+γ)K1b. To show that it is also in Kbn+1, we note that since βKbn+1, its boundary βKbn+1. γKdn+1Kbn+1. Therefore, (β+γ)Kbn+1.

    Because min(σ)=b, Equation 4 implies that σ cannot be in the boundary of any simplex τα since max(τ)<b. Since σR[τ^] and winit=(β+γ)=R[τ^]α, it follows that σwinit.

  3. 3.

    τγ.

    Because τ^ is the only simplex in γ that has τ in its boundary, τγ.

Claim 20 (Lazy reduction).

If R=DV is obtained from the lazy reduction, then z=V[τ^]K is in K(b,d), σwinit=R[τ^]=V[τ^]K[b,b], and τ(winitz).

Proof.

We only need to show that, in the case of the lazy reduction, α=V[τ^](KKbn+1)=0, i.e., V[τ^]K=V[τ^]Kbn+1. Then the claim follows from the previous Claim 19.

The proof is analogous to the proof of Claim 17. Let τ be a simplex in V[τ^]K. In this case, for the lazy reduction, because V[τ,τ^]0, from Corollary 3, taking σj=σ and τi=τ, we get max(τ)>min(τ)>min(σ)=b.

 Remark 21.

V[τ^]K can be empty. For example, if K is a single vertex that appears at b and then disappears at d. This is the reason why we need to bootstrap Algorithm 1 with an initial cycle winit.

Claim 22.

Relative cycle z¯ produced by the call Lift-Cycle(V[τ^]K,(b,d),R[τ^]) is an apex representative in 𝖧(K¯(b,d)).

Proof.

  1. 1.

    σ×bz¯, min(σ)=b.

    Simplex σ, with min(σ)=b, is in the initial cycle winit (Claim 19). Therefore, σ×b is in the boundry z¯ of the lifted cycle.

  2. 2.

    τ×dz¯, max(τ)=d.

    Simplex τ, with max(τ)=d, is in winit+z (Claim 19). Its coboundary δτK1d. Meanwhile, zKbn+1 (Claim 19). Therefore, all cofaces of τ in z intersect [b,d]. Therefore, τwinit+(z[b,d])=wfinal. It follows from Claim 5 that

    τ×dwfinal×dz¯.

It follows from Theorem 9 that z¯ is an apex representative in 𝖧(K¯(b,d)).

7 Zigzag Representatives

To solve our original problem – to find zigzag representatives – all that remains is to map an apex representative into an appropriate space in the zigzag. Because we assumed in Remark 4 that no simplex is supported on a single space, the death and the birth for any given interval are distinct. Suppose we want to map an apex representative z¯, with [z¯]𝖧(K¯I), into a zigzag representative in 𝖧(K¯[i,i])𝖧(Ki).

If i is odd, then the p-cycle z¯ may contain a vertical cell τ×i, which does not map into (p1)-cells in the levelset. To sidestep this complication, we perturb i. Let x=i±εI be a real value near i that lies inside our birth-death interval. We assume x=i+ε; the other case is symmetric. We implicitly subdivide K¯ at x and extend our previous notation to allow for the interlevel sets ending at x, e.g., K¯ix. We note that K¯ix deformation retracts onto K¯ii, with the homotopy following the second (time) coordinate, and K¯xx includes into K¯ii once the cell times are shifted from x to i. We want to compose three maps:

𝖧p(K¯I)𝖧p(K¯(i,x))𝖧p1(K¯[x,x])𝖧p1(Ki).

Because K¯I is an apex of a diamond, and K¯(i,x) is a space in the diamond, the first map is an inclusion (of pairs), so z¯ remains a relative cycle in K¯(i,x). Let

z¯=ατ(τ×tτ)+ασ(σ×[tσ1,tσ2]),

then the second (boundary) map takes z¯ to

z¯[x,x]=(σ×[tσ1,tσ2])z¯x[tσ1,tσ2]ασ(σ×x),

which maps into Ki by dropping the second coordinate,

z(i)=(σ×x)z¯[x,x]ασσ.

By storing the apex cycle z¯ in an interval tree [11, 15], we can retrieve any zigzag representative of size C in time O(logm+C).

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