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The Dominance Order for Permutations Niccoló Castronuovo 1 Introduction

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The Dominance Order for Permutations Niccoló Castronuovo 1 Introduction PU. M. A. Vol. 25 (2015), No.1, pp. 45 62 The dominance order for permutations Niccoló Castronuovo Dipartimento di Scienze Fisiche, Informatiche, Matematiche Università di Modena e Reggio Emilia Italy email: [email protected] (Received: September 14, 2014, and in revised form March 7, 2015) Abstract. We dene an order relation over Sn considering the Robinson-Schensted bijection and the dominance order over Young tableaux. This order relation makes Sn(k k − 1:::3 2 1) -the set of permutations of length n that avoid the pattern k k − 1:::3 2 1, k ≤ n- a principal lter in Sn. We study in detail these order relations on Sn(321) and Sn(4321), nding order-isomorphisms between these sets and sets of lattice paths. Mathematics Subject Classication(2010). 05A05, 05A19. Keywords: Robinson-Schensted bijection, Dyck path, 321-avoiding permutation, Young tableau, dominance order. 1 Introduction The study of particular order relations over sets of combinatorial objects can be a powerful tool to understand the structure of such sets. The present paper focuses on the set Sn of permutations of length n and his subset Sn(k k − 1:::3 2 1), namely, the set of permutations that avoid a descending pattern of length k (k ≤ n). The order relation that we study over this set is obtained by considering the dominance order E over T ab(n), the set of standard Young tableaux with n boxes, and the product order E × E over the set SY T (n) of pairs of standard Young tableaux of the same shape. This last relation induces, via the Robinson-Schensted bijection, an order relation over , called again dominance order, and hence, E Sn by restriction, over Sn(k k − 1:::3 2 1). The choice of this order E has two reasons: the rst one is the importance of the dominance order over the set of standard Young tableaux in representation theory (see e.g. [5]), the second one is that, as we will see, the posets have some interesting properties when or . (Sn(k k − 1:::3 2 1); E) k = 3 k = 4 In Section 2 we study the order over . The structure of the partially ordered set E Sn(k k − 1:::3 2 1) is quite tangled (for , it is neither a lattice, nor a graded poset), however, (Sn(k k − 1:::3 2 1); E) k ≥ 4 it has some remarkable properties, for example the fact that Sn(k k − 1:::3 2 1) is a principal lter in . (Sn; E) In Section 3 we restrict our attention to the case of k = 3. In the literature, several bijections between the set Dn of Dyck paths of semilength n and the subset of Sn of permutations avoiding a pattern of length 3 appear. Most of them are based on the bijections given in [11], [12] and [13]. In this paper, we consider the bijection φ given in [8] between Sn(321) and Dn. This bijection is an order isomorphism between and the distributive lattice (endowed with the following order: (Sn(321); E) Dn 45 DOI: 10.1515/puma-2015-0004 46 N. CASTRONUOVO a path is smaller than a path if and only lies above ). In particular, turns out to f g f g (Sn(321); E) be a distributive lattice. We show that the dominance order over Sn(321) can be described in terms of Knuth equivalences and use the map φ to nd some interesting partitions of the sets Sn(321) and Dn. In Section 4 we dene a bijection between Sn(4321) and a suitable subset of the set of Motzkin paths of length 2n. This bijection, restricted to the subset Sn(321) of Sn(4321) turns out to be nothing but the map φ dened in section 3. 2 An order over the set of permutations avoiding k k-1...321 We begin by recalling some basic notion about standard Young tableaux. A partition of the positive integer n is a non-increasing sequence of non-negative integers (λ1; λ2; λ3; :::) such that Pk ( is the greatest integer such that ). A partition of can be identied i=1 λi = n k λk > 0 n with a Ferrers diagram with n boxes, namely, a left-justied array of n empty boxes such that each row contains at most as many boxes as the preceding one. The Ferrers diagram corresponding to the partition (λ1; :::; λk) has k rows of length λ1; λ2; :::; λk, respectively. A standard Young Tableau P is a completion of a Ferrers diagram with the integers f1; 2; :::; ng such that it is strictly increasing along rows and columns; the corresponding partition is called the shape of P . For a detailed introduction to the combinatorics of Young tableaux and their applications in representation theory see e.g. [10]. Given a tableau P , for every 1 ≤ j ≤ n let Pj be the tableau induced by the elements f1; 2; :::; jg of P and denote by shPj the shape of the tableau Pj. We can identify the tableau P with the sequence of shapes (shP1; shP2; :::; shPn), with shPi ⊆ shPi+1, and, conversely, every such sequence gives rise to a standard tableau, since we have an order for the insertion of elements. Example 2.1 The standard Young tableau 1 2 3 7 4 5 6 can be identied with the sequence 0 1 B C B C B C B ; ; ; ; ; ; C B C @ A We recall that the dominance order between partitions of n is given by λ E τ if and only if λ1 + ::: + λj ≤ τ1 + ::: + τj for all j (for a detailed study of the lattice of partitions, see [6]). This order relation can be extended to an order over the set T ab(n) of standard Young tableaux with n boxes, called dominance order over standard tableaux, by setting P E Q (1) THE DOMINANCE ORDER FOR PERMUTATIONS 47 if and only if for all . This order is used in the representation theory of the shPj E shQj j ≤ n symmetric group and of the general linear group, see e.g. [5]. Example 2.2 The tableau 0 1 B C B C 1 2 B C = B ; ; ; ; C 3 5 B C @ A 4 is greater than the tableau 0 1 B C B C 1 3 B C = B ; ; ; ; C 2 5 B C @ A 4 Note that, in general, the dominance order dened in (1) does not yield a lattice structure over the set of standard Young tableaux. Consider for example the tableaux P = 1 2 3 7 and P 0 = 1 2 3 4 5 4 5 6 6 7 Then two incomparable lower bounds of P and P 0 are 1 2 3 7 and 1 2 3 . Since P covers 4 5 4 5 6 6 7 both these lower bounds, the inmum of P and P 0 does not exist. It will be useful to describe explicitly the covering relation associated with the dominance order. Proposition 2.3 If P , P 0 are two standard tableaux with the same number of boxes, P 0 covers P in the dominance order (in symbols, P 0 P ) if and only if P is obtained from P 0 by one of the following operations: (S) if shP = shP 0, choose one box lled with the integer i whose position is right and above the box lled with i + 1 and swap these two boxes. (M) if shP 6= shP 0, move one corner-box (i; j) lled with the integer k to the rst row below the i-th whose length is smaller than j, with the further condition that between the starting and the ending position of k there are no corner-boxes lled with an integer greater than k. 48 N. CASTRONUOVO Proof. First of all, note that the covering relation in the lattice of integer partitions with the dominance order is the following: the partition σ covers the partition τ if and only if the Ferrers diagram of τ is obtained by the Ferrers diagram of σ by moving one corner-box in position (i; j) to the rst row below the i-th whose length is smaller than j (see [6]). Now we proceed by induction on the number of boxes n. If n = 1; 2; 3 the proposition is trivial. Suppose it true for tableaux with at most n − 1 boxes and suppose that P and P 0 are two standard tableaux with boxes such that 0. Consider and 0 . Since 0 covers , 0 covers n P ≺ P Pn−1 Pn−1 P P Pn−1 Pn−1 and then, by the induction hypothesis, one of the following possibilities is true: is obtained from 0 by applying operation (S). • Pn−1 Pn−1 0 0 In this case, the shapes of P and P must be equal and P and P are obtained from Pn−1 and 0 , respectively, by adding one box (lled with ) in the same position. So we get from 0 Pn−1 n P P by applying operation (S). is obtained from 0 by applying the operation (M) to the box lled with the integer , • Pn−1 Pn−1 j j ≤ n − 1. In this case, and 0 are obtained from and 0 , respectively, by adding the box lled P P Pn−1 Pn−1 with n. Since P 0 covers P , the new box must be in a position not between the starting and the ending position of the box lled with j. So we get P from P 0 by applying the operation (M).
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