English

Decompositions of the Boolean lattice into rank-symmetric chains

Combinatorics 2015-09-25 v1

Abstract

The Boolean lattice 2[n]2^{[n]} is the power set of [n][n] ordered by inclusion. A chain c0...ckc_{0}\subset...\subset c_{k} in 2[n]2^{[n]} is rank-symmetric, if ci+cki=n|c_{i}|+|c_{k-i}|=n for i=0,...,ki=0,...,k; and it is symmetric, if ci=(nk)/2+i|c_{i}|=(n-k)/2+i. We show that there exist a bijection p:[n](n/2)[n](n/2)p: [n]^{(\geq n/2)}\rightarrow [n]^{(\leq n/2)} and a partial ordering << on [n](n/2)[n]^{(\geq n/2)} satisfying the following properties: (i) \subset is an extension of << on [n](n/2)[n]^{(\geq n/2)}; (ii) if C[n](n/2)C\subset [n]^{(\geq n/2)} is a chain with respect to <<, then p(C)Cp(C)\cup C is a rank-symmetric chain in 2[n]2^{[n]}, where p(C)={p(x):xC}p(C)=\{p(x): x\in C\}; (iii) the poset ([n](n/2),<)([n]^{(\geq n/2)},<) has the so called normalized matching property. We show two applications of this result. A conjecture of F\"{u}redi asks if 2[n]2^{[n]} can be partitioned into (nn/2)\binom{n}{\lfloor n/2\rfloor} chains such that the size of any two chains differ by at most 1. We prove an asymptotic version of this conjecture with the additional condition that every chain in the partition is rank-symmetric: 2[n]2^{[n]} can be partitioned into (nn/2)\binom{n}{\lfloor n/2\rfloor} rank-symmetric chains, each of size Θ(n)\Theta(\sqrt{n}).

Keywords

Cite

@article{arxiv.1509.07346,
  title  = {Decompositions of the Boolean lattice into rank-symmetric chains},
  author = {Istvan Tomon},
  journal= {arXiv preprint arXiv:1509.07346},
  year   = {2015}
}