On the number and size of holes in the growing ball of first-passage percolation

TL;DR

This paper investigates the structure of holes in the growing ball of first-passage percolation on integer lattices, establishing lower bounds on the number and size of holes, and conditional upper bounds under a curvature assumption.

Contribution

It proves that non-deterministic weights lead to many holes with large volume in the growing ball, and provides bounds on their number and size, advancing understanding of the geometric complexity.

Findings

Number of holes grows at least as $t^{d-1}$ for large $t$

Maximum hole volume is at least logarithmic in $t$

Conditional bounds on the number of holes and their size depend on curvature assumptions

Abstract

First-passage percolation is a random growth model defined on $\mathbb{Z}^d$ using i.i.d. nonnegative weights $(\tau_e)$ on the edges. Letting $T(x,y)$ be the distance between vertices $x$ and $y$ induced by the weights, we study the random ball of radius $t$ centered at the origin, $B(t) = \{x \in \mathbb{Z}^d : T(0,x) \leq t\}$. It is known that for all such $\tau_e$, the number of vertices (volume) of $B(t)$ is at least order $t^d$, and under mild conditions on $\tau_e$, this volume grows like a deterministic constant times $t^d$. Defining a hole in $B(t)$ to be a bounded component of the complement $B(t)^c$, we prove that if $\tau_e$ is not deterministic, then a.s., for all large $t$, $B(t)$ has at least $ct^{d-1}$ many holes, and the maximal volume of any hole is at least $c\log t$. Conditionally on the (unproved) uniform curvature assumption, we prove that a.s., for all large $t$, the number of holes is at most $(\log t)^C t^{d-1}$, and for $d=2$, no hole in $B(t)$ has volume larger than $(\log t)^C$. Without curvature, we show that no hole has volume larger than $Ct \log t$.