Packing the Boolean lattice with copies of a poset

TL;DR

This paper proves that for large n, the Boolean lattice can be nearly perfectly packed with copies of a given poset P, covering almost all elements, confirming a conjecture by Lonc from 1991.

Contribution

It establishes the existence of near-complete packings of the Boolean lattice with copies of any poset P for large n, confirming a longstanding conjecture.

Findings

Almost all elements of the Boolean lattice can be covered by copies of P.

If |P| divides 2^n - 2, the lattice minus the minimum and maximum can be partitioned into copies of P.

The result confirms Lonc's conjecture from 1991.

Abstract

Let $P$ be a partially ordered set. We prove that if $n$ is sufficiently large, then there exists a packing $\mathcal{P}$ of copies of $P$ in the Boolean lattice $(2^{[n]},\subset)$ that covers almost every element of $2^{[n]}$: $\mathcal{P}$ might not cover the minimum and maximum of $2^{[n]}$, and at most $|P|-1$ additional points due to divisibility. In particular, if $|P|$ divides $2^{n}-2$, then the truncated Boolean lattice $2^{[n]}-\{\emptyset,[n]\}$ can be partitioned into copies of $P$. This confirms a conjecture of Lonc from 1991.