Maximum Persistent Betti Numbers of Čech Complexes

TL;DR

This work addresses the problem of bounding the number of $p$-dimensional holes in a Čech complex that persist over a fixed interval from radius $1$ to $1+\varepsilon$ for $p<d$ and $n$ points in ${\mathbb R}^d$. The authors present a self-contained geometric construction, the snap complex, built from a uniform space partition, and show that the persistent Betti number $\beta\beta_p({\check{C}}_{1},{\check{C}}_{1+\varepsilon})$ is bounded above by $\dim H_p(Q_1)$, which they prove grows linearly with $n$ via a packing argument. The bound extends to Alpha and Vietoris–Rips complexes and aligns with sparse-filtration results, while offering a direct, elementary proof that avoids interleaving machinery. Overall, the paper establishes a linear-in-$n$ bound on the number of long-persisting $p$-dimensional holes, clarifying the topological complexity of fixed-interval persistence in Euclidean filtrations and informing practical topological data analysis and stochastic topology.

Abstract

This note proves that only a linear number of holes in a Čech complex of $n$ points in $\mathbb{R}^d$ can persist over an interval of constant length. Specifically, for any fixed dimension $p < d$ and fixed $\varepsilon > 0$, the number of $p$-dimensional holes in the Čech complex at radius $1$ that persist to radius $1 + \varepsilon$ is bounded above by a constant times $n$, where $n$ is the number of points. The proof uses a packing argument supported by relating the Čech complexes with corresponding snap complexes over the cells in a partition of space. The argument is self-contained and elementary, relying on geometric and combinatorial constructions rather than on the existing theory of sparse approximations or interleavings. The bound also applies to Alpha complexes and Vietoris-Rips complexes. While our result can be inferred from prior work on sparse filtrations, to our knowledge, no explicit statement or direct proof of this bound appears in the literature.

Figures (2)

• Figure 1: Left: a portion of a $2$-cycle in which $x$ and $y$ belong to two triangles that share an edge different from the edge connecting $x$ to $y$. For better visibility, we shade the triangles in the stars of $x$ and $y$dark, light, and in between depending on whether they share such an edge, they belong to both stars, and neither, respectively. Middle: the contraction of the edge connecting $x$ to $y$ produces two bi-gons and two triangles that share the same three vertices. Right: these bi-gons and duplicate triangles are removed.
• Figure 2: Left: the Čech complex for six points inside four cells in the partition of the plane. Assuming the two triangles are isosceles, right-angled, and have smallest enclosing circles of radius $1+\varepsilon{\hbox{$\varepsilon$}}$, the narrow rectangle between the two triangles has a smallest enclosing circle with radius strictly larger than $1+\varepsilon{\hbox{$\varepsilon$}}$. Hence, the boundary of the convex hexagon that passes through the six points is a non-trivial $1$-cycle in ${\check{C}}_{1+\varepsilon{\hbox{$\varepsilon$}}}{\hbox{${\check{C}}_{1+\varepsilon{\hbox{$\varepsilon$}}}$}}$, and for $\varepsilon{\hbox{$\varepsilon$}} \leq \sqrt{2}-1$, it already exists in ${\check{C}}_{1}{\hbox{${\check{C}}_{1}$}}$. Right: the image of the hexagon is a quadrangle in the snap complex. Its boundary is a trivial $1$-cycle in $Q_{1+\varepsilon{\hbox{$\varepsilon$}}}$ because the rectangle collapses to a single edge shared by the images of the two triangles.

Theorems & Definitions (13)

• Lemma 1
• proof
• Definition 1
• Lemma 2
• Lemma 3
• proof
• Definition 2
• Lemma 4
• proof
• Corollary 5