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Chain Lengths in the Dominance Lattice June 8, 2013 1 Introduction Chain Lengths in the Dominance Lattice June 8, 2013 Edward Early St. Edward’s University Austin, TX [email protected] Abstract We construct the largest union of two or three chains in the lattice of partitions of n under the dominance order. This construction provides a framework for finding the largest union of four or more chains. We also see how these chain lengths relate to a theorem of Gansner and Saks and the problem of finding the largest antichain. 1 Introduction Let Pn denote the poset of partitions of the positive integer n, ordered by dominance (or majorization). That is, µ ν if µ1 + µ2 + + µk ν1 + ν2 + + νk for all k, where parts are indexed in nonincreasing≤ order and trailing··· zeros≤ are added··· as needed. This poset is a lattice and is self-dual under conjugation of partitions, where conjugation is accomplished by transposition of the Young diagram. The poset Pn is not graded (ranked) for n 7, since there exist maximal chains from (n) to (1n) of all lengths from 2n 3 to cn3/2 [2,≥ 9]. − Given any poset P , there exists a partition λ(P ) such that the sum of the first k parts of λ is the maximum number of elements in a union of k chains in P . In fact, the conjugate of λ has the same property for antichains [1, 5, 7, 8]. Let λk(P ) denote the kth part of this partition. m+1 The length h(Pn) of the longest chain in Pn has been known for some time. If n = 2 + r, 3 3 m m m m 0 r m, then h(Pn) = 3− + rm [9]. In other words, λ1(Pn) = 3− + rm +1. Our main≤ results≤ are the following theorems. Theorem 1. For n> 16, λ (P )= λ (P ) 6. 2 n 1 n − Theorem 2. For n> 22, λ (P )= λ (P ) 6. 3 n 2 n − We can define a rank function r on a graded poset such that if x y, then r(y) r(x) is the length of a maximal chain from x to y. The Hasse diagram of≤ a ranked poset− can have all of the elements of the same rank on a horizontal line, giving every element a well-defined level. An unranked poset can also have this property, as illustrated in Figure 1. In such a poset, we can draw a Hasse diagram where every element (except those on the top and bottom levels) covers something on the level below and is covered by something on the 1 Figure 1: An unranked but leveled poset. level above, thus giving each element a well-defined level on which to be drawn. Call such a poset leveled. Every element of a leveled poset P is in one of the longest chains of P . Consider the subposet Qn of Pn consisting of the partitions that appear in chains of length h(Pn). Conjugation of partitions shows that Qn is self-dual, since conjugation takes a de- creasing chain to an increasing chain of the same length. Whether Qn is a graded lattice remains an open question, but for our purposes it will suffice to use a weaker statement: for µ Qn, define r(µ) to be the length of the longest chain from (n) down to µ; for µ (1∈n), (n) , µ is covered by an element ν such that r(ν) = r(µ) 1 and covers an ele- 6∈ { } − ment ρ such that r(ρ)= r(µ) + 1. In other words, Qn is leveled. Note that the top element is level 0, and the levels increase as we move down. Given any poset, the subposet of elements that appear in chains of maximum length will be leveled, and this is a maximal leveled subposet. It is maximal in the sense that it has the largest possible number of levels and the addition of any other elements would make it no longer leveled. This observation turns out to be useful for computing Greene-Kleitman partitions of certain posets [3, 4]. By temporarily discarding all elements that are not on longest chains, we can focus on the elements that really matter for our computations. We will know that the chains we find really are the longest possible, despite the lack of a true rank function, by showing that there is an antichain decomposition that puts the discarded elements onto levels already used in the chains. A nonconstructive proof of Theorem 1 was given in [3] using an elegant lemma. Lemma 1. If a leveled poset has at least two elements on every level, then it has two disjoint chains of maximum length. The analogous lemma for three chains is less elegant to state, does not apply to all leveled posets, is tedious to prove, and does not generalize in any usable way to four or more chains. Moreover, the version of Theorem 2 stated in [3] is only for n> 135, so the proof given here makes a significant improvement on the lower bound, to the point that we can now compute λ3(Pn) for all n. The proofs of Theorems 1 and 2 we will give are constructive. While some casework is required to obtain the smallest n, we will see that an underlying pattern can be exhibited for large values of n without the cases, allowing some progress toward generalizing to unions of more than three chains. The covering pairs in Pn come in two types. We say µ covers ν by an H-step if there exists i such that ν = µ 1, ν = µ + 1, and ν = µ for k / i, i +1 . In terms of Young i i − i+1 i+1 k k ∈ { } diagrams, this corresponds to moving a box one row down (and some distance to the left). 2 The other type is a V-step, which is an H-step on the conjugate, and corresponds to moving a box one column to the left (and some distance down). Chains from (n) to (1n) consisting of H-steps followed by V-steps are maximal [9]. Not all maximal chains are of this form, however, so when we intersperse H and V steps it will be necessary to keep track of levels carefully. 2 Proof of Theorem 1 We will prove Theorem 1 by showing that there exist disjoint chains in Qn of lengths h(Pn) and h(P ) 6. Since Q is a subposet of P , these are also chains in P . For n 5, the top n − n n n ≥ three levels of Pn (and Qn) consist only of (n), (n 1, 1), and (n 2, 2); the bottom three n n 2 n 4 − − levels will be (1 ), (2, 1 − ), and (2, 2, 1 − ). Since these six elements of Pn are in maximal antichains of size 1 (that is, each of those six partitions is comparable to any other partition in Pn, so they cannot appear in antichains of 2 elements), the two chains will have the largest possible union, thus giving λ2(Pn). n 6 To that end, we seek disjoint chains in Qn from (n 2, 1, 1) and (n 3, 3) to (2, 2, 2, 1 − ) n 3 − − and (3, 1 − ). Let Qn∗ denote Qn without the top three and bottom three elements, so our goal is to construct disjoint maximal chains in Qn∗ . Since Qn∗ is self-dual, we will construct chains to the halfway point that end with a self-conjugate partition or a partition that covers its conjugate when the number of elements in each chain is odd or even, respectively. The rest of the chain is then the conjugate of the chain already constructed. m3 m m3 m Since λ1(Pn)= 3− + rm + 1, and 3− is even, and λ2(Pn) has the same parity as λ1(Pn) (since they differ by 6 in the cases we are considering), the number of levels in Qn∗ will be even only when m and r are both odd. If m or r is even, then any self-conjugate partitions in Qn∗ will be on its middle level (note that there can be non-self-conjugate partitions on the middle level whose conjugates are also on that level). As a first approximation of the desired chains, take the following construction. Chain L starts at (n 2, 1, 1). At every step, we take the bottommost (in the Young diagram) possible H-step.− That is, keep the large parts intact as long as possible, so the next few partitions are (n 3, 2, 1), (n 4, 3, 1), (n 4, 2, 2), and (n 4, 2, 1, 1). − − − − Chain R starts at (n 3, 3). At every step, we take the topmost possible H-step. That is, keep the number of− parts as small as possible, so the next few partitions are (n 4, 4), − (n 5, 5), (n 6, 6), and (n 7, 7). − − − Both chains will reach (m, m 1,...,r +1, r, r, r 1,..., 2, 1), which is at least the halfway − − point, so only minor modifications will be needed to turn these into the desired chains. The following lemmas will help us ensure that our chains are disjoint. Lemma 2. If µ = (µ1,µ2,...,µk) R, then µi µi+1 2 for 1 i k 2. In other words, only the last difference can be∈ greater than 2.− Moreove≤ r, excluding≤ ≤ the last− difference, 3 µ cannot have more than one difference equal to 2. Proof. By construction, we are always using the topmost possible H-step.
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