Long Symmetric Chains in the Boolean Lattice

Let [n] = {1, 2, . . . , n} be a set with n elements, and let 2[n] denote the poset of all subsets of [n] ordered by inclusion. In other words, 2[n] is the Boolean lattice of order n or the n-dimensional hypercube. It is easy (for example using a symmetric chain decomposition [1, Theorem 3.1.1]) to find n disjoint skipless (saturated) symmetric chains of length n − 2, that is n disjoint chains that have exactly one element from each rank starting with an element of rank 1 and ending with an element of rank n − 1. What if you start by finding k such symmetric chains without any prior plan, and only making sure that each new symmetric chain is disjoint from the ones before? Will you then be able to find an additional n−k disjoint symmetric chains or might you get stuck at some point? In this paper we show that as long as k ≤ n − 3, you will be able to complete the project. Moreover, it is possible that given n− 1 disjoint skipless chains of length n− 2 in 2[n], there is no other such chains disjoint from the given collection, and given any n − 2 such chains it is always possible to find at least one more (but maybe not two more). Recall that a cutset in a poset is a subset of the poset that meets every maximal chain [4, 7]. In the above situation the non-existence of any more disjoint skipless symmetric chains of length n − 2 is equivalent to saying that the subsets in the given symmetric chains form a cutset for 2[n]. In other words our results imply that there is a cutset of 2[n] consisting of n − 1 disjoint skipless symmetric chains of length n − 2 while there is no