Combinatorics of the group of parity alternating permutations

  • Published on
    26-Jun-2016

  • View
    214

  • Download
    2

Embed Size (px)

Transcript

  • Advances in Applied Mathematics 44 (2010) 225230

    Contents lists available at ScienceDirect

    Cp

    Sh

    De

    a

    ArReAcAv

    M0520

    KePeEuSig

    1.

    trpeThpea

    aiexfu

    01doAdvances in Applied Mathematics

    www.elsevier.com/locate/yaama

    ombinatorics of the group of parity alternatingermutations

    inji Tanimoto

    partment of Mathematics, Kochi Joshi University, Kochi 780-8515, Japan

    r t i c l e i n f o a b s t r a c t

    ticle history:ceived 31 March 2008cepted 8 July 2009ailable online 11 September 2009

    SC:A05B30

    ywords:rmutationslerian numbersned Eulerian numbers

    A permutation is called parity alternating if its entries assume evenand odd integers alternately. Parity alternating permutations forma subgroup of the symmetric group. This paper deals with thenumber of permutations classied by the ascent number. Thesenumbers have a close relationship to signed Eulerian numbers.The present approach is based on a study of permutations thatare not parity alternating. It is proved that the number of evenpermutations is equal to that of odd ones in it. Hence signedEulerian numbers are described by parity alternating permutations.

    2009 Elsevier Inc. All rights reserved.

    Introduction

    A permutation of [n] = {1,2, . . . ,n} will be called a parity alternating permutation (PAP) if its en-ies assume even and odd integers alternately, such as 72581634. It is readily checked that suchrmutations form a subgroup of the symmetric group of degree n, and it will be denoted by Pn .e objective of this paper is to study combinatorial properties of those permutations with respect tormutation statistics. The approach, which complements that in [5], is self-contained and based onstudy of the set of permutations that are not parity alternating.Let = a1a2 an be a permutation of [n]. An ascent (or descent) of is an adjacent pair such that

    < ai+1 (or ai > ai+1) for some i (1 i n1). Let E(n,k) be the set of all permutations of [n] withactly k ascents, where 0 k n 1. Its cardinality, |E(n,k)|, is the classical Eulerian number. Let usrther put

    P(n,k) = Pn E(n,k),

    E-mail address: tanimoto@cc.kochi-wu.ac.jp.

    96-8858/$ see front matter 2009 Elsevier Inc. All rights reserved.

    i:10.1016/j.aam.2009.07.002

  • 226 S. Tanimoto / Advances in Applied Mathematics 44 (2010) 225230

    th

    thPA

    usev

    on

    wre

    thPA

    (schon

    (a

    (b(

    era1anid

    Wis

    W

    Abi(1

    Thtipre set of PAPs with k ascents, and

    E(n,k) = E(n,k) \ P(n,k),

    e set of all permutations that are not PAPs in E(n,k). Remark that, when n is odd, a1 and an in aP = a1a2 an are odd entries.An inversion of a permutation = a1a2 an is a pair (i, j) such that 1 i < j n and ai > a j . Letdenote by inv() the number of inversions in . A permutation is called even or odd if it has anen or odd number of inversions, respectively.The signed Eulerian number Dn,k is the difference of the cardinalities of even permutations and oddes in E(n,k). The recurrence relation for signed Eulerian numbers is

    Dn,k ={

    (n k)Dn1,k1 + (k + 1)Dn1,k if n is odd,Dn1,k1 Dn1,k if n is even, (1)

    ith initial condition D1,0 = 1, as shown in [1,2,6]. It was conjectured in [2], and proved in [1] and [6],spectively, by a generating function argument and a combinatorial involution.The purpose of this paper is to show that in E(n,k) the number of even permutations is equal to

    at of odd ones. Therefore, in order to investigate signed Eulerian numbers, one has only to considerPs or the subset P(n,k).For our study we use an operator introduced in [3]. It is useful for a study of permutation statistics

    ee [35]). The operator is dened by adding one to all entries of a permutation a1a2 an , but byanging n + 1 into one. However, when n appears at either end of a permutation, it is removed ande is put at the other end. Formally we dene as follows.

    ) (a1a2 an) = b1b2 bn , where bi = ai + 1 for 1 i n and n + 1 is replaced by one at theposition.

    ) (a1 an1n) = 1b1b2 bn1, where bi = ai + 1 for 1 i n 1.c) (na1 an1) = b1b2 bn11, where bi = ai + 1 for 1 i n 1.

    The operator preserves the ascent and descent numbers. In what follows we will use only prop-ties (a) and (b), since we exclusively deal with permutations of the form a1a2 an1n, wherea2 an1 are permutations of [n 1]. We call those permutations canonical. For a permutation d a positive integer , we denote successive applications of by = ( 1), 0 being theentity operator on permutations.It is easy to see that in the case of even n the operator is a map on E(n,k) and on P(n,k).hen n is odd, however, this is not the case. Even if = 1476523 is a PAP, for example, = 2517634not a PAP.Next we observe how the number of inversions of a permutation behaves when is applied.hen n appears at either end of a permutation as in (b) or (c), it is evident that

    inv( ) = inv().

    s for the case (a), i.e., ai = n for some i (2 i n 1), we get (a1a2 an) = b1b2 bn and= 1. In this case, n i inversions (i, i + 1), . . . , (i,n) of vanish and, in turn, i 1 inversions, i), . . . , (i 1, i) of occur. Hence the difference between the numbers of inversions is given by

    inv( ) inv() = (i 1) (n i) = 2i (n + 1). (2)

    is means that, when n is even, each application of the operator changes the parity of permuta-ons as long as n remains in the interior of permutations. If n is odd, however, the operator also

    eserves the parity of all permutations of [n].

  • S. Tanimoto / Advances in Applied Mathematics 44 (2010) 225230 227

    2.

    on

    wta

    inexev

    apaith

    Lesuhan

    Previnriwenas

    thpaanofw

    Ifodtio

    wesse

    LenPermutations that are not PAPs

    In order to study canonical permutations by means of , we introduce another operator. Based , let us dene an operator on canonical permutations of E(n,k) by

    (a1a2 an1n) = nan1(a1a2 an1n) = b1b2 bn1n,

    here b1 = n an1 and bi = ai1 + (n an1) (mod n) for i (2 i n 1). For a canonical permu-tion and a positive integer , is similarly dened.It is easy to see that in the case of even n the operator is a map on canonical permutations

    E(n,k) and in P(n,k). On the other hand, when n is odd, this is not the case. = 1436527, forample, is a canonical PAP, but = 5621437 is not a PAP. In this section we deal with the case ofen n.A PAP sequence aiai+1 a j is a subsequence of a permutation in which even and odd integerspear alternately. A PAP sequence aiai+1 a j is called maximal, if both ai1ai a j (i 2) and a ja j+1 ( j n 1) are no longer PAP sequences. The length of a subsequence of a permutation ise number of its entries.

    mma 1. Let n be an even positive integer and let = a1 aiai+1 an1n be a canonical permutation. As-me that is not a PAP and C = ai+1 an1n is a maximal PAP sequence. Then all of {, , . . . , ni1}ve the same parity and the rst entries of the permutations in {, 2, . . . , ni1} are all odd. Buti has a different parity from that of {, , . . . , ni1} and the rst entry of ni is even.

    oof. First we examine the parity of . Remark that inv( ) = inv(), because is canonical. How-er, each additional application of changes the parity of permutations as long as n lies in theterior of permutations, as shown in Section 1. Therefore, after the entry an1 of becomes n at theght end of by the application of nan1 , the parity of has changed n an1 1 times. Hence,hen an1 is even or the length of C is one, the parity of is different from that of and the rsttry of is an even n an1. On the other hand, when an1 is odd, the parity of is the samethat of and the rst entry of is odd.In the latter case, let = b1b2 bn1n, where bn1 = an2 + (n an1) (mod n). If an2 is even,

    en bn1 is odd. So the above argument can be applied to , implying that {, 2} have the samerity. Hence all of {, , 2} have the same parity. Moreover, the rst entry of 2 is odd. But if2 is odd or the length of C is two, then bn1 is even and 2 has a different parity from that{, }. Moreover, the rst entry of 2 is even. For a maximal PAP sequence C = ai+1 an1n

    ith length n i, the lemma follows by employing this argument n i times. By Lemma 1, when is not a PAP, it eventually changes the parity by repeated applications of .

    is a PAP, however, the parity of {, , . . . , n1} remains the same and their rst entries are alld. Remark that n = holds, since each entry of returns to the original position after n applica-ns of and both n and are canonical, i.e., their last entry is n.When is canonical but not a PAP, it can be written as

    = a1 a ja j+1 aiai+1 an1n = A B C, (3)

    here A = a1 a j and C = ai+1 an1n are maximal PAP sequences and B = a j+1 ai is not nec-sarily a PAP sequence and may be empty. A (or C ) will be called the rst (or last) maximal PAPquence of . A permutation 45316278, for example, is expressed by 45 316 278.

    mma 2. In addition to the assumptions of Lemma 1, suppose that the rst entry a1 of is even and leti = c1 cnicni+1 cn1n. Then c1 cni is a maximal PAP sequence.

  • 228 S. Tanimoto / Advances in Applied Mathematics 44 (2010) 225230

    Prisa1alwfois

    mseenth

    3.

    odbepritca

    PAA

    Th

    Pr

    theacapeby

    mnoin

    riisithhaoof. Let us put = a1 aiai+1 an1n = A B C and = b1b2 bn1n, where C = ai+1 an1nthe last maximal PAP sequence of and A is the rst one. Then both b1 = n an1 and b2 =+ (n an1) (mod n) have the same parity, because a1 is even. By each application of to ,l entries {n,an1, . . . ,ai+1} in C move to the left end of permutations one by one in this order. Ife put ni = c1 cnicni+1 cn1n, then c1 cni is a PAP sequence and it is maximal. Thisllows from the facts that ai+1 an1n is a PAP sequence of and that the parity of cni and cni+1the same, since b1 and b2 of have the same parity. Lemma 2, together with Lemma 1, states that the last maximal PAP sequence of = A B C

    oves into the rst maximal PAP sequence of the permutation ni with opposite parity and bothquences have the same length. Moreover, the rst entry of ni is even. On the other hand, alltries of A and B move to the right by n i positions by means of an application of ni , althougheir values have changed.

    Combinatorics of E(n,k)

    The main purpose of this paper is to show that in E(n,k) the numbers of even permutations andd ones are equal. The proof is divided into two parts by the parity of n. The case of odd n (Part Ilow) was proved in a part of Lemma 4.2 of [5], but it contained a certain obscure point. So weovide another, more plausible, proof based upon only Lemmas 1 and 2. Hence, different from [5],is self-contained and does not utilize the recurrence relation (1) for signed Eulerian numbers. These of even n can easily be deduced from the odd case.For a permutation = A B C expressed by (3), we denote by A the length of the rst maximalP sequence A, and similarly for the lengths of B and the last maximal PAP sequence C . Note that > 0 and C > 0 for E(n,k).

    eorem 3. In E(n,k) the total number of even permutations is equal to that of odd ones.

    oof. Part I (odd case).Let n be an odd integer. In order to consider the set of permutations in E(n,k), let us introduce

    e set of all canonical permutations in E(n+1,k+1), which is denoted by (n+1,k+1), and withch a1a2 an in E(n,k) we associate a1a2 an(n + 1) in (n + 1,k + 1). Since n + 1 is even, wen apply Lemma 1 to canonical permutations of (n + 1,k + 1). Then we are able to nd canonicalrmutations with even entry a1, since permutations in (n + 1,k + 1) are not PAPs. We denote (n + 1,k + 1) the set of such particular canonical ones.A permutation = a1a2 an(n + 1) in (n + 1,k + 1) can be written as = A B C by using

    aximal PAP sequences A and C , as in (3). So we have n+1 = A+ B+ C. The sequence B needst be a PAP sequence and may be empty. As proved in Lemma 1, does not turn into a permutation(n + 1,k + 1) with opposite parity until the operator is applied C times to .Suppose that 1, 2, . . . , m are all of even canonical permutations in (n + 1,k + 1), i.e., theirst entries are even. For each i (1 i m) let us dene i as the smallest positive integer such that= i i becomes an odd permutation in (n + 1,k + 1). The correspondence between i and ia bijection and thus we get all odd ones 1, 2, . . . , m by this procedure. For i (1 i m) let= Ai Bi Ci and i = Ai B i C i be expressions by maximal PAP sequences. Then it follows thate PAP sequence Ci turns into the PAP sequence Ai of i and i = Ci = Ai from Lemma 2. So weve

    mAi=

    mCi. (4)i=1 i=1

  • S. Tanimoto / Advances in Applied Mathematics 44 (2010) 225230 229

    w

    Fr

    Fonual

    sePAth

    Coeqod

    a1

    ((i(ii

    Oty

    ((i(ii

    th

    eaTh

    anofO(aarSince each odd i is changed into a certain even permutation in (n+ 1,k + 1) in a similar way,e obtain an analogous equality

    mi=1

    Ai =mi=1

    C i. (5)

    om the above equalities we also get

    mi=1

    Bi =mi=1

    B i.

    r each i (1 i m) the permutations i remain even, whenever < Ci. Noting 0i = i , thember of all even permutations in (n + 1,k + 1) is given by the sum (4). Similarly, the number ofl odd permutations in (n + 1,k + 1) is given by the sum (5).Now we know that the last maximal PAP sequence Ci of i changes into the rst maximal PAP

    quence Ai of i and the last maximal PAP sequence Ci of i , in turn, changes into the rst maximal

    P sequence A j of some even j . So both sums of the numbers of the entries that are involved withis exchange must be equal:

    mi=1

    Ci =mi=1

    C i.

    nsequently, it follows from (4) and (5) that the number of even permutations in (n + 1,k + 1) isual to that of odd ones. Namely, in E(n,k), the number of even permutations is equal to that ofd ones, which completes the proof of Part I.Part II (even case).The case of even n can easily be derived from the former part. We divide all even permutationsa2 an of E(n,k) into the three types by the position of n:

    i) ai = n for some i (2 i n 1);i) an = n;i) a1 = n.

    n the other hand, we divide all odd permutations a1a2 an of E(n,k) into the following threepes by the position of one:

    i) ai = 1 for some i (2 i n 1);i) a1 = 1;i) an = 1.

    Let = a1a2 an be an even permutation of type (i). Then, using property (a) of , we seeat is an odd one of type (i), because the difference of the numbers of inversions betweenand is |n + 1 2i| by (2) and it is odd by assumption. Since is a bijection, we see that toch permutation of type (i) in E(n,k) corresponds only one permutation of type (i) in E(n,k).erefore, both types have the same cardinality.Let = a1a2 an1n be of type (ii), where a1a2 an1 is a permutation of [n 1]. Since iseven permutation, the permutation a1a2 an1 E(n 1,k 1) is also even. Hence the numberelements in E(n,k) of type (ii) is the cardinality of all even permutations of E(n 1,k 1).

    n the other hand, the set E(n,k) of type (ii) consists of all odd = 1a2a3 an and hence2 1)(a3 1) (an 1) are all odd in E(n 1,k 1). Using Part I, we see that both cardinalities

    e the same.

  • 230 S. Tanimoto / Advances in Applied Mathematics 44 (2010) 225230

    nthofwca

    re

    im

    by

    4.

    sian

    Th

    ho

    Prsi

    (nti

    Thdi

    Prth

    Re

    [1[2[3[4[5[6Similar arguments can be applied to types (iii) and (iii). Let = na1a2 an1 be of type...

Recommended

View more >