一些排列類上的折疊現象

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國立高雄大學應用數學系

博士論文

Folding Phenomena on some classes of permutations

一些排列類上的折疊現象

研究生：丁建太撰

指導教授：游森棚、陳晴玉

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致謝

感謝天恩師德！在此論文完成之際，首先誠摯的感謝我的指導教授陳晴玉教授與游森棚教授(教官)，在兩位老師嚴謹的治學和耐心的指導下，使我能順利完成博士班的學業，並獲得博士學位。在博士班進修期間，受到很多貴人的幫忙，但最要感謝的是我的指導教授游森棚教官，教官不辭勞苦的指導課業，循循善誘的引導我做研究，尤其在教學技巧與方法上，更是我學習的典範，這對於日後在工作上助益甚大。另外在課餘，不管在工作與生活上遭遇到挫折和困難時，教官總是給予我很多的鼓勵、協助與建議，今若有小小的成就，皆要感恩於教官。教官，謝謝您。在這也要感謝傅東山教授與潘業忠教授，給我很多課業與研究上的指導與協助，亦要謝謝我的口試委員張惠蘭教授與嚴志弘教授，對我的論文提出的指導與審閱，還要感謝多年同窗同學的好友立志，一起上課、研究與討論，口試當天特地請假過來幫忙與打氣，謝謝學姐顏珮嵐博士、學長羅元勳教授、一同修課與討論的澤初與維迦、系辦助理千惠姐與雅鳳姐、傅明宇教授與蔡震寰教授，謝謝您們。我還要感謝高大應數系，提供頂尖的師資與良好的學習研究環境，讓我可以順利完成博士學業。最後，要感謝我親愛的家人，爸爸、媽媽、老婆嫦慧、三個可愛的小寶貝柏廷、庭曦與柏爵，總在背後默默地鼓勵與支持，讓我有前進的動力來完成學業。再次對所有關心、指導與幫助我的家人、師長、同學與朋友們，表達最誠摯的感謝與敬意，一切的一切均銘記於心。丁建太謹誌於國立高雄大學應用數學系中華民國一Ｏ六年七月

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一些排列類上的折疊現象

指導教授：游森棚教授國立臺灣師範大學數學系指導教授：陳晴玉教授國立高雄大學應用數學系學生：丁建太國立高雄大學應用數學系摘要令 𝑋_𝑛 是一個長度為 𝑛 的某特殊排列族所成的集合。其上有第一個統計量 stat_1 與第二個統計量 stat_2。若下式成立 ∑ (−1)𝑠𝑡𝑎𝑡1(𝜋)𝑞𝑠𝑡𝑎𝑡2(𝜋) = 𝑓(𝑞) ∑ 𝑞2∙𝑠𝑡𝑎𝑡2(𝜋) 𝜋∈𝑋𝑛 𝜋∈𝑋2𝑛 𝑜𝑟 𝑋2𝑛+1 其中 𝑓(𝑞) 是一個有理函數，我們就說這個 𝑋_𝑛 集合在此兩個統計量之下有折疊現象。這篇論文的主要結果是證明三個排列族有折疊現象，包括交錯排列、Baxeter 排列和 involutions 排列。關鍵字：折疊現象、交錯排列、Baxeter 排列、involutions 排列

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Folding Phenomena on some classes of permutations

Advisors: Professor Sen-Peng Eu Department of Mathematics National Taiwan Normal University

Professor Ching-Yu Chen Department of Applied Mathematics

National University of Kaohsiung

Student: Chien-Tai Ting Department of Applied Mathematics

National University of Kaohsiung

ABSTRACT

Let 𝑋_𝑛 be a family of objects of size 𝑛 with statistics 𝑠𝑡𝑎𝑡₁ and 𝑠𝑡𝑎𝑡₂. We say (𝑋_𝑛, 𝑠𝑡𝑎𝑡₁, 𝑠𝑡𝑎𝑡₂) exhibits a folding phenomenon if

∑ (−1)𝑠𝑡𝑎𝑡1(𝜋)𝑞𝑠𝑡𝑎𝑡2(𝜋) = 𝑓(𝑞) ∑ 𝑞2∙𝑠𝑡𝑎𝑡2(𝜋)

𝜋∈𝑋𝑛

𝜋∈𝑋2𝑛 𝑜𝑟 𝑋2𝑛+1

for some rational function 𝑓(𝑞).

Among other miscellaneous results, the main parts of this thesis are three examples of this folding phenomena, namely, the alternating permutations, the Baxeter permutations, and the involutions.

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Folding Phenomena on some classes of permutations

Chien-Tai Ting

A DISSERTATION

in

Mathematics

Presented to the Faculties of the National University of Kaohsiung in

Partial Fulfillment of the Requirements for the Degree of Doctor of

Philosophy

June, 2017

Supervisor of Dissertation

Sen-Peng Eu, Colonel Professor in Mathematics

Ching-Yu Chen, Professor in Mathematics

Graduate Group Chairperson

Tung-Shan Fu, Professor in Mathematics

Dissertation Committee:

Sen-Peng Eu, Colonel Professor in Mathematics

Ching-Yu Chen, Professor in Mathematics

Tung-Shan Fu, Professor in Mathematics

Hui-Lan Chang, Associate Professor in Mathematics

Yeh-Jong Pan, Associate Professor in Mathematics

Chi-Hung Yen, Associate Professor in Mathematics

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ABSTRACT

Folding Phenomena on some classes of permutations

Let Xn be a family of permutations of size n with statistics stat1 and stat2. We

say (Xn, stat1, stat2) exhibits a folding phenomenon if

X

π∈X2n or X2n+1

(−1)stat1(π)_qstat2(π)_{= f (q)} X π∈Xn

q2·stat2(π)_,

for some rational function f (q).

Among other miscellaneous ones, the main results of this thesis are three in-stances of this folding phenomenon on three families of pattern avoiding permutations, namely, the alternating permutations, the Baxeter permutations, and the involutions.

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Chapter 1 Introduction

Let Sn be the set of permutations of {1, 2, . . . , n}. a permutation σ = σ1σ2. . . σn ∈

S_n _{is 123-avoiding if there exists no 1 ≤ i < j < k ≤ n such that σ}_i _{< σ}_j _{< σ}_k_. Simion and Schmidt [29] proved that the number of 123-avoiding permutations of length n is the Catalan number Cn = _n+11 2n_n

.

The sign of σ, denoted by sign(σ), is defined by sign(σ) = (−1)inv(σ)_{, where inv(σ) =}

|{(σi, σj) : i < j and σi > σj}| is the number of inversions of σ. For a subset P ⊆ Sn,

the sign-imbalance I(P) of P is defined by I(P) =X

σ∈P

sign(σ).

The first thread of this thesis is the following curious identity on 123-avoiding permutations in Sn proved also in [29]:

I(Sn(123)) =

( Cn−1

2 if n odd,

0 if n even.

Note that everything is cancelled out when n is even. However, when n is odd, the cardinality after the cancellation also (surprisingly) turns out to be a Catalan number, roughly indexed by ‘one half’ of n. Similar observation holds also for 321-avoiding permutations Sn(321).

The second thread is a refinement of the above observation over Sn(321). In [2]

Adin and Roichman derived such a refinement with respect to the statistic ldes(σ) of the permutations σ ∈ Sn(321), where ldes(σ) = max{i : σi > σi+1, 1 ≤ i ≤ n − 1} is

the last descent of σ. They obtained X σ∈S2n+1(321) sign_{(σ) · q}ldes(σ) = X σ∈Sn(321) q2·ldes(σ) _{(n ≥ 0),} X σ∈S2n(321) sign_{(σ) · q}ldes(σ) _{= (1 − q)} X σ∈Sn(321) q2·ldes(σ) _{(n ≥ 1).}

By letting q = 1 these two identities reduce to Simon and Schmidt’s result. One notes that in both identities the indices of the left hand side are roughly twice of that of the right hand side. Also on the right hand side the statistic ldes are ‘doubled’.

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Our experiments reveal that the following phenomenon (up to small variations) occurs in several places:

The signed enumerator of objects of size 2n is essentially equal to the ordinary enumerator of objects of size n.

This motivates us to consider the following general setting. Let Xn be a family

of objects (permutations) of size n with statistics stat1 and stat2. We may say that

(Xn, stat1, stat2) exhibits a folding phenomenon if

X π∈X2n or X2n+1 (−1)stat1(π) qstat2(π) = f (q) X π∈Xn q2·stat2(π) , for some rational function f (q).

We emphasize that the folding phenomenon rarely occurs and the proof depends very much on the nature of the permutation family at hand. It makes the fold-ing phenomena elusive and interestfold-ing. Among other missalaneous results, the main parts of this thesis is to investigate three families of permutations and prove a fold-ing phenoemnon for each. Namely, we will investigate the followfold-ing three familes of permutations: the alternating permutations, the Baxeter permutations, and the involutions. We will see that the proofs differ greatly.

The rest of the paper contains four published papers. Chapter 2 and 3 are about alternating permutations. Chapter 4 considers Baxter permutations and Chapter 5 is on involutions. Each chapter is self-contained and here we will not repeat the intro-duction as readers will have no difficulty following the introintro-ductions of each chapter. We present the papers as they are published and list the names of the journals here.

(i) Chapter 2: Sign-imbalance of alternating permutations avoiding a pattern of length three. Ars Combin. 109 (2013), 371-382. (joint with Yeh-Jong Pan) (ii) Chapter 3: Sign-balance identities of Adin-Roichman type on 321-avoiding

alternating permutations. Discrete Math. 312 (2012), no. 15, 2228-2237. (joint with Sen-Peng Eu, Tung-Shan Fu, and Yeh-Jong Pan)

(iii) Chapter 4: Baxter permutations, maj-balances, and positive braids. Electron.

J. Combin. 19 (2012), no. 3, Paper 26, 17 pp. (joint with Sen-Peng Eu,

Tung-Shan Fu, and Yeh-Jong Pan)

(iv) Chapter 5: Two refined major-balance identities on 321-avoiding involutions.

European J. Combin. 49 (2015), 250-264. (joint with Sen-Peng Eu, Tung-Shan

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Chapter 2 Alternating permutations I

2.1 Introduction

Let Sn be the symmetric group of all permutations on [n] = {1, . . . , n}. A

permuta-tion σ = σ1· · · σn ∈ Snis alternating if σ1 > σ2 < σ3 > σ4 < · · · . These permutations

(starting with a descent) are also known as “down-up” permutations, while those

“up-down” permutations (starting with an ascent) are called reverse alternating. Let Altn

denote the set of alternating permutations in Sn.

For a permutation σ, the sign of σ, denoted by sign(σ), is defined by sign(σ) = (−1)inv(σ)_{, where inv(σ) = |{(σ}

i, σj) : i < j and σi > σj}| is the number of inversions

of σ.

For a subset P ⊆ Sn, the sign-imbalance I(P) of P is defined by

I(P) =X

σ∈P

sign(σ).

It is obvious that I(Sn) = 0 for n ≥ 2. Also it is not hard to see that I(Altn) = ±1 or

0 (e.g., by the following sign-reversing involution on Altn: for a permutation σ ∈ Altn,

find the least integer i such that 2i − 1 and 2i are not adjacent in σ, then interchange 2i − 1 and 2i). In this paper we shall consider two classes of Altn, the alternating

permutations avoiding each pattern of length three and the alternating permutations of genus 0 (see section 2.1.2. for definition). Meanwhile, we shall give a simpler proof of Dulucq-Simion’s result on the number of alternating permutations of genus 0.

2.1.1 Alternating permutations avoiding a pattern of length

three

For a permutation ω = ω1· · · ωt ∈ St (t ≤ n), we say that σ contains an ω-pattern

if there are indices i1 < i2 < · · · < it such that σij < σik if and only if ωj < ωk.

Otherwise, σ is ω-avoiding. For a subset P ⊆ Sn, let P(ω) denote the set of

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Simion and Schmidt showed in their seminal paper [29] that |Sn(ω)| = Cn := 1 n+1 2n n

, the Catalan number, for each ω ∈ S3. Later, Mansour [22] proved that the

cardinality of Altn(ω) also coincides with the Catalan number (with indices shifted)

for each ω ∈ S3. As for sign-imbalance, Simion-Schmidt [29] also determined the

sign-imbalance of the set of 123-avoiding permutations in Sn, namely,

I(Sn(123)) =

( Cn−1

2 if n odd,

0 if n even.

The first main result of this paper is to completely determine the sign-imbalances of Altn(ω) for each ω ∈ S3. By establishing a bijection between Altn(132) and

bi-nary trees, we first determine I(Altn(132)) in Theorems 2.2.1 by a parity-reversing

involution on binary trees. _{The sign-imbalances I(Alt}n(213)), I(Altn(231)), and

I(Altn(312)) can be derived from I(Altn(132)) and we shall show them in

Theo-rem 2.3.1-2.3.3. Moreover, I(Altn(123)) can be derived from Theorem 2.3.4, and a

further refinement of I(Altn(321)) will be discussed in S.-P. Eu, T.-S. Fu, Y.-J. Pan

and C.-T. Ting [16].

We make a brief list of I(Altn(ω)) for each ω ∈ S3 in the following table. It

sug-gests that the result involves only Catalan numbers Cn and Fine numbers Fn. The

Catalan numbers Cn have the generating function C(x) :=

P

n≥0Cnxn = 1− √

1−4x 2x ,

and the Fine numbers {Fn}n≥0 = {1, 0, 1, 2, 6, 18, · · ·} can be defined by the

gener-ating function F (x) =P_n_≥0Fnxn = _x1 1− √

1−4x

3−√1−4x. Both are ubiquitous in enumerative

combinatorics. We refer readers to Deutsch and Shapiro’s survey [10] and Stanley [32] for more information.

I(Altn(σ3))\n 0 1 2 3 4 5 6 7 8 9 10 11 12 13

I(Altn(123)) 1 1 -1 0 1 -1 0 0 -2 2 0 0 5 -5 Catalan

I(Altn(132)) 1 1 -1 0 0 1 -1 2 -2 6 -6 18 -18 57 Fine

I(Altn(213)) 1 1 -1 1 0 2 -1 5 -2 14 -6 42 -18 132 Catalan, Fine

I(Altn(231)) 1 1 -1 0 2 1 -5 -2 14 6 -42 -18 132 57 Catalan, Fine

I(Altn(312)) 1 1 -1 -1 2 2 -5 -5 14 14 -42 -42 132 132 Catalan

I(Altn(321)) 1 1 -1 0 0 -1 1 0 0 2 -2 0 0 -5 Catalan

For example, from the table we have

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2.1.2 Alternating permutations of genus zero

A hypermap of size n is an ordered pair of permutations of Sn such that the group

they generate is transitive on [n]. Let z(σ) be the number of cycles of a permutation σ. Given a hypermap (σ, α) ∈ Sn×Sn, the genus g of (σ, α) is defined by the relation

z(σ) + z(α) + z(α−1_{σ) = n + 1 − 2g.}

Based on this notion, Dulucq-Simion [11] considered the genus g(α) of a permutation α ∈ Sn with σ restricted to the n-cycle 2 . . . n 1 (in cycle notation σ = (1 2 . . . n)),

namely,

z(α) + z(α−1_{· (1 2 . . . n)) = n + 1 − 2g(α).}

For example, if α = 2 3 1 5 4 (in cycle notation α = (1 2 3)(4 5)), then α−1_{σ =}

(1 4)(2)(3)(5), and hence g(α) = 0. They proved that the number of alternating (resp. reverse alternating) permutations of genus zero coincides with the small (resp. large) Schr¨oder numbers. The large Schr¨oder numbers {Rn}n≥0 = {1, 2, 6, 22, 90, . . . }

have the ordinary generating function 1−x−√_2x1−6x+x2, and the small Schr¨oder numbers {Sn}n≥0= {1, 1, 3, 11, 45, . . . } are defined by S0 = 1 and Sn= Rn/2 for n ≥ 1.

Let D(0)n and U(0)n denote the sets of alternating and reverse alternating

permu-tations of genus zero in Sn, respectively. By encoding a permutation of genus zero

by a word in an alphabet of four letters, Dulucq-Simion [11] obtained a system of grammars for the language formed by the words corresponding to the permutations in D(0)n and U

(0)

n for n ≥ 0, from which they derived the cardinalities of D (0)

n and U (0) n .

Meanwhile, we shall present a simpler proof of this result by means of generating functions (see Theorem 2.4.4). The second main result of this paper is to determine the sign-imbalance of D(0)_n and U(0)_n , namely,

I(D(0)2n) = I(D (0)

2n+1) = (−1)n and I(U(0)n ) = 0.

2.1.3 Organization of the paper.

The rest of this paper is organized as follows. Section 2.2 is devoted to I(Altn(132)).

Section 2.3 deals with I(Altn(ω)) for other patterns ω ∈ S3 except for 321. The

sign-imbalance of alternating permutations of genus zero will be discussed in section 2.4.

Though the result may be solved algebraically, our exposition here is deliberately bijective. The basic strategy is to translate the problem to trees and apply a sign-reversing involution.

2.2 Sign-imbalance of Alt

_n

(132)

In this section, we shall determine the sign-imbalance of the set Altn(132) and obtain

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Theorem _{2.2.1. For n ≥ 1, the following identities hold.} (i) I(Alt2n−1(132)) = Fn−1,

(ii) I(Alt2n(132)) = −I(Alt2n−1(132)).

To prove Theorem 2.2.1 we first translate the problem to binary trees. Let Bn

be the set of binary trees with n vertices. It is known that |Bn| = Cn. For a binary

tree T ∈ Bn, the leftmost path of T is defined by the maximal sequence v1, . . . , vk of

vertices of T such that v1 is the root and vi is the left child of vi−1, for 2 ≤ i ≤ k.

Let lmpb(T ) denote the number of vertices in the leftmost path of T . Note that lmpb(T ) = 1 if the root has no left child.

Observe that |Alt2n−1(132)| = Cn (see [22, Theorem 2.2]). Our approach is to

establish a bijection Γ between Alt_2n−1_{(132) and B}n such that for a σ ∈ Alt2n−1(132),

the sign of σ is opposite to the parity of lmpb(Γ(σ)). Then we can determine the sign-imbalance of Alt_2n−1_{(132) by an involution Ψ on B}n that reverses the parity of

lmpb(T ) if T is a binary tree except for the fixed points of Ψ.

Let Γ be a mapping from Alt_2n−1_{(132) to B}n defined inductively as follows. Given

a σ = σ1· · · σ2n−1 ∈ Alt2n−1(132), we factorize σ as σ = σ1p1· · · pn−1, where the

subword pi = σ2iσ2i+1 consists of an adjacent pair of elements, for 1 ≤ i ≤ n − 1.

For convenience, denote pi = (σ2i, σ2i+1) and let p0 = (0, σ1). We shall associate

σ with a binary tree Γ(σ) ∈ Bn with vertices labeled by p0, p1, . . . , pn−1 and write

Γ(σ) = Γ(p0p1· · · pn−1). Suppose that pj (0 ≤ j ≤ n − 1) is the subword which

con-tains the greatest element of σ. Then take pj as the root of Γ(σ) and put Γ(p0· · · pj−1)

and Γ(pj+1· · · pn−1) as the left subtree and the right subtree, respectively (see

Exam-ple 2.2.2).

Example _{2.2.2. Take a permutation σ = 9 8 10 7 11 4 5 3 6 1 2 ∈ Alt}11(132), and

fac-torize σ as σ = p0· · · p5 = (0, 9)(8, 10)(7, 11)(4, 5)(3, 6)(1, 2). The corresponding tree

Γ(σ), along with the vertex-labeling, is shown in Figure 2.1.

(1,2) (7,11) (8,10) (3,6)

(0,9) (4,5)

Figure 2.1: The binary tree corresponding to σ = 9 8 10 7 11 4 5 3 6 1 2.

Proposition 2.2.3. Γ is a bijection between Alt_2n−1_{(132) and B}nsuch that a

permuta-tion σ = σ1· · · σ2n−1 ∈ Alt2n−1(132) is carried to a tree Γ(σ) ∈ Bn with lmpb(Γ(σ)) =

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Proof. To show that Γ is a bijection, it remains to find Γ−1_{. Given a binary tree}

T ∈ Bn, we shall recover the word Γ−1(T ) by defining inductively a vertex-labeling

of T starting with the set S = {1, . . . , 2n − 1}. Let T1 and T2 be the right and left

subtrees of the root of T , respectively. If T1 contains k vertices (possibly empty),

then we label T1 with S1 which consists of the least 2k elements in S, label the root

of T by the pair (x, y), where x is the least element and y is the greatest element in S − S1, and label T2 with S2 = S − S1− {x, y}. Note that |S2| is odd, and inductively

the leftmost vertex of T will be labeled by a single element (and then we attach 0 to the left).

Observe that whenever lmpb(T ) = t the elements 2n − 1, 2n − 2, . . . , . . . , 2n − t of σ appear in the leftmost path of T accordingly, and the last vertex of this path is labeled by (0, σ1), where σ1 = 2n − t. The second assertion follows.

With Proposition 2.2.3, the following lemma leads to the fact that the sign of σ is opposite to the parity of lmpb(Γ(σ)).

Lemma 2.2.4. For every σ = σ1· · · σ2n−1 ∈ Alt2n−1(132), the statistic inv(σ) has the

opposite parity of σ1.

Proof. For 1 ≤ i ≤ n − 1, let pi = (σ2i, σ2i+1) be the ith ascent of σ. For each pi and

2i + 1 < j ≤ 2n − 1, we observe that σ2i+1 > σj if and only if σ2i > σj since σ is

132-avoiding. It follows that each pi contributes even number of inversions to inv(σ),

and the parity of inv(σ) depends on the number of inversions due to σ1, which is equal

to σ1− 1.

It is known that the binary trees T ∈ Bn can be transformed into plane trees G

with n edges such that the degree of root of G equals lmpb(T ) (similar to leftmost child next right sibling, but here we use rightmost child next left sibling, see Figure 2.2). Moreover, among many other objects, the nth Fine number Fn counts the number

of plane trees with n edges where the root is of even degree (e.g., see [10]). Thus Fn

counts the number of binary trees T ∈ Bn with even lmpb(T ).

a

b

h

d

g

c

e i

f

a

b

d

c

h g f e

i

Figure 2.2: A binary-tree representation for plane trees.

For a vertex x of a rooted tree T , let τ (x) denote the subtree of T rooted at x. Before proceeding the proof of Theorem 2.2.1, we shall define an involution Ψ on Bn

such that it serves the needs of parity-reversing. Given a T ∈ Bn, we construct Ψ(T )

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(i) lmpb(T ) is even. Let u and v be the left and right children of the root of T , respectively. Let x and y be the left and right children of u, respectively. The tree Ψ(T ) is constructed from T as follows. Separate the subtrees τ (v), τ (x) and τ (y) from T , and then change u to be the right child of the root. Attach τ (y) and τ (v) to u as the left and right subtrees of u, and attach τ (x) to the root as the left subtree of the root (see Figure 2.3).

(ii) lmpb(T ) is odd. If the root has no right child, then let Ψ(T ) = T (i.e., T is a fixed point), otherwise Ψ(T ) is constructed by reversing the operation using in (i).

u

Ψ

u

x

y

v

y

x

Figure 2.3: An example for the the map Ψ.

Let Fn denote the set of fixed points of the requested involution Ψ, namely,

Fn= {T ∈ Bn : lmpb(T ) is odd, and the root has no right child}.

Lemma _{2.2.5. For n ≥ 1, we have}

|Fn| = Fn−1.

Proof. Given a T ∈ Fn, it is easy to see that T is in one-to-one correspondence to the

tree T′ _{∈ B}

n−1 with even lmpb(T′), where T′ is obtained from T with the root vertex

removed. The assertion follows.

We remark that Theorem 2.2.1.(i) is essentially proved by the following bijective result.

Proposition _{2.2.6. Ψ is an involution on the set B}n such that

(i) Fn is the set of fixed points, and

(ii) lmpb(Ψ(T )) has the opposite parity of lmpb(T ) if T ∈ Bn− Fn.

Proof. It is a routine to check that Ψ is an involution on Bn. The assertion (i) is

trivial by the definition of Ψ. For a T ∈ Bn− Fn, we have lmpb(Ψ(T )) = lmpb(T ) − 1

if lmpb(T ) is even, and lmpb(Ψ(T )) = lmpb(T ) + 1 if lmpb(T ) is odd. The proof is completed.

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Now we are able to evaluate the sign-imbalance of the set Altn(132).

Proof of Theorem 2.2.1. (i) For the case of odd length, we have

I(Alt2n−1(132)) = X σ∈Alt2n−1(132) (−1)inv(σ) = − X T∈Bn

(−1)lmpb(T ) (by Proposition 2.2.3 and Lemma 2.2.4)

= − X

T∈Fn

(−1)lmpb(T ) (by Proposition 2.2.6) = Fn−1.

(ii) For the case of even length, given a σ = σ1· · · σ2n ∈ Alt2n(132), observe

that σ2n = 1, otherwise there will be a 132-pattern (1, σ2n−1, σ2n) in σ. There is an

immediate bijection between Alt2n(132) and Alt2n−1(132) for which σ corresponds to

the permutation ω = ω1· · · ω2n−1 ∈ Alt2n−1(132), where ωi = σi−1 for 1 ≤ i ≤ 2n−1.

Note that σ has the opposite parity of ω since inv(σ) = inv(ω) + 2n − 1. Hence I(Alt2n(132)) = −I(Alt2n−1(132)). The proof is completed. ✷

2.3 Sign-imbalance of Alt

_n

(ω) for the other

pat-terns

ω of length three

In this section, we derive the sign-imbalance of the set Altn(ω), for ω ∈ {213, 231, 312, 123}.

Theorem _{2.3.1. For n ≥ 1, the following identities hold.} (i) I(Alt2n(213)) = −Fn−1,

(ii) I(Alt2n−1(213)) = Cn.

Proof. (i) Under the operations of complement and reverse, there is an immediate

bijection between Alt2n(213) and Alt2n(132). Namely, to each ω = ω1· · · ω2n ∈

Alt_2n(213) there corresponds a σ = σ1· · · σ2n∈ Alt2n(132), where σi = 2n+1−ω2n+1−i

(1 ≤ i ≤ 2n). Moreover, inv(σ) = inv(ω) since the pair (σi, σj) is an inversion of σ

if and only if the pair (ω_2n+1−j, ω_2n+1−i_{) is an inversion of ω. Hence I(Alt}2n(213)) =

I(Alt2n(132)) = −Fn−1.

(ii) For every σ = σ1· · · σ2n−1 ∈ Alt2n−1(213), we factorize σ as σ = σ1p1· · · pn−1,

where the subword pi = (σ2i, σ2i+1) is the ith ascent. For each pi and 1 ≤ j ≤ 2i − 1,

observe that σj > σ2i if and only if σj > σ2i+1 since σ is 213-avoiding. It follows that

each pi contributes even number of inversions to inv(σ). Hence inv(σ) is even, and

I(Alt2n−1(213)) = |Alt2n−1(213))| = Cn.

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(i) I(Alt2n−1(231)) = (−1)n−1Fn−1,

(ii) I(Alt2n(231)) = (−1)nCn.

Proof. (i) Under the operations of reverse, there is an immediate bijection between

Alt_2n−1(231) and Alt_2n−1(132). Namely, to each ω = ω1· · · ω2n−1 ∈ Alt2n−1(231) there

corresponds a σ = σ1· · · σ2n−1 ∈ Alt2n−1(132), where σi = ω2n−i (1 ≤ i ≤ 2n − 1).

Moreover, inv(σ) = 2n−1₂ _{−inv(ω) since for i < j, σ}i > σj if and only if ω_2n−j < ω_2n−i.

It follows that inv(σ) has the same parity of inv(ω) if n is odd, and has the opposite parity of inv(ω) if n is even. Hence I(Alt2n−1(231)) = (−1)n−1I(Alt2n−1(132)) =

(−1)n−1_F n−1.

(ii) For every σ = σ1· · · σ2n ∈ Alt2n(231), we factorize σ as σ = q1· · · qn, where the

subword qi = (σ_2i−1, σ2i) is the ith descent. For each qiand j ≤ 2i−2, we observe that

σj > σ2i−1if and only if σj > σ2isince σ is 231-avoiding. It follows that each qi, along

with the inversion of qi itself, contributes an odd number of inversions to inv(σ).

Hence inv(σ) has the same parity of n, and I(Alt2n(231)) = (−1)n|Alt2n(231))| =

(−1)n_C n.

Theorem _{2.3.3. For n ≥ 1, the following identities hold.} (i) I(Alt2n−1(312)) = (−1)n−1Cn−1,

(ii) I(Alt2n(312)) = (−1)nCn.

the two sets Alt_2n−1(312) and Alt_2n−1(213). Namely, to each ω = ω1· · · ω2n−1 ∈

Alt_2n−1(312) there corresponds a σ = σ1· · · σ2n−1 ∈ Alt2n−1(213), where σi = ω2n−i

(1 ≤ i ≤ 2n − 1). Moreover, inv(σ) = 2n−12

− inv(ω) since for i < j, σi > σj if

and only if ω_2n−j < ω_2n−i. It follows that inv(σ) has the same parity of inv(ω) if n is odd, and has the opposite parity of inv(ω) if n is even. Hence I(Alt2n−1(312)) =

(−1)n−1_I(Alt

2n−1(213)) = (−1)n−1Cn−1.

(ii) For every σ = σ1· · · σ2n+1 ∈ Alt2n+1(312), we observe that the greatest element

of σ is σ2n+1 = 2n + 1 (otherwise there will be a 312-pattern (2n + 1, σ2n, σ2n+1) in

σ). Then removing the element σ2n+1 from σ results in a member σ′ ∈ Alt2n(312)

with inv(σ′_{) = inv(σ). This is a parity-preserving bijection between Alt}

2n+1(312) and

Alt2n(312). Hence I(Alt2n(312)) = I(Alt2n+1(312)). The proof is completed.

To determine I(Alt2n−1(123)) we first quote the following theorem for the case

I(Alt2n−1(321)).

Theorem _{2.3.4 ([16], Corollary 3.4.). For n ≥ 1, the following identities hold.}

(i) I(Alt2n(321)) =    (−1)n−12 C_n−1 2 n odd 0 n even,

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(ii) I(Alt2n−1(321)) = −I(Alt2n(321)).

✷

Theorem 2.3.5. The following identities hold.

(i) For n ≥ 1, I(Alt2n−1(123)) =

   (−1)n−12 C_n−1 2 n odd 0 n even.

(ii) For n ≥ 2, I(Alt2n(123)) = −I(Alt2n+1(123)).

Alt_2n−1(123) and Alt_2n−1(321). Namely, to each ω = ω1· · · ω2n−1 ∈ Alt2n−1(123) there

corresponds a σ = σ1· · · σ2n−1 ∈ Alt2n−1(321), where σi = ω2n−i (1 ≤ i ≤ 2n − 1).

Moreover, inv(σ) = 2n−1₂ _{− inv(ω) since for i < j, σ}i > σj if and only if ω2n−j <

ω_2n−i. It follows that inv(σ) has the same parity of inv(ω) if n is odd, and has the opposite parity of inv(ω) if n is even. By Theorem 2.3.4, we have I(Alt2n−1(123)) =

I(Alt2n−1(321)) = (−1) n−1

2 C_n−1 2 .

(ii) For n ≥ 2 and for every σ = σ1· · · σ2n+1 ∈ Alt2n+1(123), we observe that

σ2n = 1, and that σ2i−1 > σ2i+1, for 2 ≤ i ≤ n (otherwise there will be 123-patterns

in σ). Then removing the element σ2nfrom σ and subtracting 1 from the other entries

leads to a member σ′ _{∈ Alt}

2n(123) with inv(σ′) = inv(σ) − (2n − 1). This is a

parity-reversing bijection between Alt2n+1(123) and Alt2n(123). Hence I(Alt2n(123)) =

−I(Alt2n+1(123)). The proof is completed.

2.4 Sign imbalance of alternating permutations of

genus zero

Recall that D(0)_n and U(0)_n denote the set of alternating and reverse alternating per-mutations of genus zero in Sn, respectively. In this section, we enumerate D(0)n and

U(0)_n by the method of generating functions for completeness and determine the sign-imbalance of the two sets D(0)n and U(0)n .

We say that an m-cycle of a permutations α ∈ Sn is increasing if its elements are

expressible as i < α(i) < α2_{(i) < · · · < α}m−1_{(i). It is known that a permutation of}

genus zero can be completely characterized as follows, see [?, ?].

Lemma _{2.4.1. Let α ∈ S}n. Then g(α) = 0 if and only if the cycle decomposition of

α gives a noncrossing partition of [n], and each cycle of α is increasing. The following observation is an immediate consequence of Lemma 2.4.1. Proposition _{2.4.2. For n ≥ 1, the following identities hold.}

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(i) |D(0)2n| = |D (0) 2n+1|, (ii) |U(0)2n| = |U (0) 2n+2|.

Proof. (i) For every α = α1· · · α2n+1 ∈ D(0)2n+1, we observe that if α2n+1 6= 2n + 1

then by Lemma 2.4.1 and the fact α2n < α2n+1 there will be a crossing in the cycle

decomposition of α. Hence α2n+1 = 2n + 1 and the subword α1· · · α2n ∈ D(0)2n. This

establishes a bijection between the two sets D(0)_2n and D(0)_2n+1.

(ii) For every α = α1· · · α2n+2 ∈ U(0)2n+2, by the same argument as in (i) we observe

that α2n+2 = 2n+2 and the subword α1· · · α2n+1 ∈ U(0)2n+1. This establishes a bijection

between the two sets U(0)_2n+1 and U(0)_2n+2. We assume |D(0)0 | = |U

(0)

1 | = 1, and define the generating functions for |D (0) 2n| and |U(0)2n+1| S = S(x) =X i≥0 |D(0)2i |xi, R = R(x) = X i≥0 |U(0)2i+1|xi. (2.4.1)

Proposition 2.4.3. The following relations hold. (i) R − 1 = 2(S − 1),

(ii) R − 1 = 2xRS.

Proof. To prove (i), is suffices to show that |U(0)2n+1| = 2|D (0)

2n|, for n ≥ 1. For every

permutation α = α1· · · α2n+1 ∈ U(0)2n+1, by Lemma 2.4.1 we observe that either α1 = 1

or α1 = 2. Moreover, the elements 1 and 2 are not adjacent in α. Let U(0)2n+1 be

partitioned into two sets A1 and A2, where α ∈ A1 if α1 = 1 and α ∈ A2 if α1 = 2.

There is an immediate bijection between A1 and A2 by interchanging the elements

1 and 2 of the permutation α. For every β = β1· · · β2n ∈ D(0)2n, we associate β

with a permutation α = α1· · · α2n+1 ∈ S2n+1, where α1 = 1 and αi = βi−1 + 1 for

2 ≤ i ≤ 2n + 1. We observe that α ∈ U(0)2n+1 and this establishes a bijection between

D(0)_2n and A1. The assertion (i) follows.

(ii) For every α = α1· · · α2n+1 ∈ A1, we have α1 = 1 and α2k+1 = 2 for some

1 ≤ k ≤ n. We factorize α as α = α1µα2k+1ν, where µ = α2· · · α2k and ν =

α2k+2· · · α2n+1. By Lemma 2.4.1, the entries in ν are greater than the entries in µ.

We observe that the word ν is a down-up permutation of length 2n − 2k, and that upon normalized ν is in one-to-one correspondence to the member β = β1· · · β2n−2k ∈

D(0)_2n−2k, where βi = α2k+1+i− 2k − 1, for 1 ≤ i ≤ 2(n − k). Moreover, the word µ is a

down-up permutation of length length 2k−1, and µ is in one-to-one correspondence to the member γ = γ1· · · γ2k−1 ∈ U(0)2k−1, where γj = 2k + 2 − α_2k+1−j, for 1 ≤ j ≤ 2k − 1.

The above argument works well for the case α ∈ A2. This proves the assertion (ii).

Now, we prove the following theorem.

Theorem _{2.4.4 (Dulucq-Simion). For n ≥ 1, the following identities hold.} (i) |D(0)2n| = |D

(0)

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(ii) |U(0)2n+1| = |U (0)

2n+2| = Rn.

Proof. By Proposition 2.4.3, we derive that the generating function R satisfies the

equation R = 1 + xR + x2_R2_{. Solving this equation leads to}

R = 1 − x − √

1 − 6x + x2

2x ,

which coincides with the generating function for large Schr¨oder numbers. Hence we have |U(0)2n+1| = Rn. Moreover, it follows from the relation R = 2S − 1 that

|D(0)2n| = Sn.

Theorem _{2.4.5. For n ≥ 1, we have} (i) I(D(0)2n) = I(D

(0)

2n+1) = (−1)n,

(ii) I(U(0)n ) = 0.

Proof. Let γ ∈ D(0)2n be the permutation with cycle decomposition γ = (1 2)(3 4) · · · (2n−

1 2n). Note that sign(γ) = (−1)n_{. We come up with a sign-reversing involution on}

the set D(0)_2n _{− {γ}.}

For a α = α1· · · α2n ∈ D(0)2n − {γ}, find the least integer j such that (α2j−1 α2j) 6=

(2j 2j−1), say α2k = 2j−1 for some k > j. Then either α2k−1= 2k−1 or α2k−1 = 2k.

Moreover, the elements 2k −1 and 2k are not adjacent in α. The requested involution is obtained by interchanging the elements 2k − 1 and 2k in α.

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Chapter 3 Alternating permutations II

3.1 Introduction

Let Sn be the set of all permutations of {1, . . . , n}. For a σ = σ1· · · σn ∈ Sn, the

sign of σ, denoted by sign(σ), is defined to be sign(σ) = (−1)inv(σ)_{, where inv(σ) =}

Card{(σi, σj) : i < j and σi > σj} is the inversion number of σ. Note that sign(σ) = 1

(respectively, −1) if σ is even (respectively, odd). For a subset P ⊆ Sn, the

sign-balance I(P) of P is defined by

I(P) =X

σ∈P

sign(σ),

i.e., the difference between the number of even and odd permutations in P.

The problem of sign-balance was first studied by Simion and Schmidt on sets of patten-avoiding permutations in Sn. Let Sn(321) denote the set of 321-avoiding

permutations in Sn. (A permutation is called 321-avoiding if it contains no decreasing

subsequence of length three.) Simion and Schmidt [29] determined the sign-balance of Sn(321) I(Sn(321)) =    Cn−1 2 if n odd 0 if n even, (3.1.1) where Cn= _n+11 2n_n

is the nth Catalan number. Making use of a multivariate gener-ating function, Adin and Roichman [2] proved a refinement of Eq. (4.1.1) with respect to the statistic ldes(σ) of the permutations σ ∈ Sn(321), where ldes(σ) = max{i :

σi > σi+1, 1 ≤ i ≤ n − 1} is the last descent of σ.

Theorem 3.1.1 (Adin-Roichman). The following identities hold. X σ∈S2n+1(321) sign_{(σ) · q}ldes(σ) = X σ∈Sn(321) q2·ldes(σ) _{(n ≥ 0),} X σ∈S2n(321) sign_{(σ) · q}ldes(σ) _{= (1 − q)} X σ∈Sn(321) q2·ldes(σ) _{(n ≥ 1).}

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Shortly after this result had appeared on ArXiv, Reifegerste [25] gave a bijective proof of Theorem 4.1.1 in terms of standard Young tableaux with at most two rows, and proved an analogous refinement respecting the length of the longest increasing subsequence of σ ∈ Sn(321). Mansour [23] found variation of the identities for

132-avoiding permutations.

In this paper, we investigate refined sign-balance properties on 321-avoiding alter-nating permutations. A permutation σ = σ1· · · σn ∈ Sn is alternating (or down-up)

if σ1 > σ2 < σ3 > σ4 < · · · . Let Altn(321) denote the set of 321-avoiding alternating

permutations in Sn, and let lead(σ) = σ1, the first entry of σ. One of our main results

is the following.

Theorem _{3.1.2. For all n ≥ 1, we have}

(i) X σ∈Alt4n+2(321) sign_{(σ) · q}lead(σ) _{= (−1)}n+1 X σ∈Alt2n(321) q2·lead(σ) (ii) X σ∈Alt4n+1(321) sign_{(σ) · q}lead(σ) _{= (−1)}n X σ∈Alt2n(321) q2·lead(σ) (iii) X σ∈Alt4n(321) sign_{(σ) · q}lead(σ) _{= (−1)}n+1_{(1 − q)} X σ∈Alt2n(321) q2(lead(σ)−1) (iv) X σ∈Alt4n−1(321) sign_{(σ) · q}lead(σ) _{= (−1)}n_{(1 − q)} X σ∈Alt2n(321) q2(lead(σ)−1).

We also prove an analogous result respecting the statistic end(σ), the last entry, of σ.

Theorem _{3.1.3. For all n ≥ 1, we have}

(i) X σ∈Alt4n+2(321) sign_{(σ) · q}end(σ) _{= (−1)}n+1 X σ∈Alt2n(321) q2·end(σ)+1 (ii) X σ∈Alt4n+1(321) sign_{(σ) · q}end(σ) _{= (−1)}n X σ∈Alt2n(321) q2·end(σ)+1 (iii) X σ∈Alt4n(321) sign_{(σ) · q}end(σ) _{= (−1)}n_{(1 − q)} X σ∈Alt2n(321) q2·end(σ) (iv) X σ∈Alt4n−1(321) sign_{(σ) · q}end(σ) _{= (−1)}n+1_{(1 − q)} X σ∈Alt2n(321) q2·end(σ).

To prove the above two theorems, we establish a sign-reversing involution on Altn(321), which is defined in terms of plane trees. In fact, the notion of sign-balance

has been extended to plane trees. Let Tn be the set of plane trees with n edges.

Regarding the parity of the number of leaves, Eu, Liu and Yeh [15] determined the sign-balance of Tn I(Tn) =    (−1)n+12 C_n−1 2 if n odd 0 if n even. (3.1.2)

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Chen, Shapiro and Yang [6] gave an elegant combinatorial proof of this result by establishing a sign-reversing involution on Tn. Their involution turns out to reverse

the parity of another statistic hsum(T ) of a plane tree T , the total height of all vertices of T . In this paper, we extend their method to develop a new involution on Tn that reverses the parity of hsum(T ) while preserving the length of the leftmost

(respectively, rightmost) path of T , which proves refined sign-balance properties on Tn (Theorems 3.4.2 and 3.4.3) and paves the way to a proof of Theorems 3.1.2 and

3.1.3.

This paper is organized as follows. In section 2, we establish a connection between 321-avoiding alternating permutations of even length and plane trees. In section 3, we review the involution given by Chen, Shapiro and Yang. In section 4, we develop an involution that proves refined sign-balance results on plane trees. In section 5, we prove Theorems 3.1.2 and 3.1.3.

3.2 A bijection between 321-avoiding alternating

permutations and plane trees

In this section, we establish a simple bijection between Alt2n(321) and Tn, with Dyck

paths as an intermediate stage, such that a permutation σ ∈ Alt2n(321) is carried to

a tree T ∈ Tn with hsum(T ) = inv(σ).

3.2.1 The first stage of the bijection between Alt

_2n

(321) and

T

n

The first stage is a special case of the Fran¸con-Viennot [19] bijection between per-mutations and lattice paths such that the inversion number of the permutation is translated to the area under the path.

A Dyck path of length n is a lattice path from (0, 0) to (n, n), using north step (0, 1) and east step (1, 0), that never goes below the line y = x. Let N and E denote a north step and an east step, respectively. Let Qn = ENEN · · · EN be the specific

lattice path from (0, 0) to (n, n) formed by the steps E and N alternatingly. Note that Qn goes between the line y = x and the line y = x − 1.

Let Cn be the set of Dyck paths of length n. It is known that |Cn| = Cn. For

every path P ∈ Cn, we define the statistic area(P ) of P to be the number of unit

squares in the polyomino (P, Qn) enclosed by P and Qn. For example, Figure 3.1(a)

shows a polyomino (P, Q4) with the lower path Q4 = ENENENEN, the upper path

P = NNENNEEE, and area(P ) = 8.

Proposition 3.2.1. There is a bijection Ω between the two sets Alt2n(321) and Cn

such that a permutation σ ∈ Alt2n(321) is carried to a path Ω(σ) ∈ Cn with inv(σ) =

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Proof. Given a σ = σ1· · · σ2n ∈ Alt2n(321), we label the ith step of Qn by σi, for

1 ≤ i ≤ 2n. The corresponding path Ω(σ) is constructed from Qn by rearranging

the steps of Qn according to the numerical order of their labels. Since σ is

321-avoiding and alternating, the odd entries σ1 < σ3 < · · · < σ2n−1 (respectively, even

entries σ2 < σ4 < · · · < σ2n) are increasing, which are on the n east (respectively,

north) steps of the path Ω(σ), accordingly. Then Ω(σ) is a path from (0, 0) to (n, n). Moreover, since σ_2i−1 > σ2i, there are always at least i north steps preceding the ith

east step in Ω(σ). Hence Ω(σ) ∈ Cn.

To find Ω−1_{, given a P ∈ C}

n, we label the steps of P from 1 to 2n. With the

polyomino (P, Qn), we assign the ith step of Qn the label zi of the opposite step

across the polyomino, for 1 ≤ i ≤ 2n. Then we recover the word Ω−1_{(P ) = z}

1· · · z2n ∈

Alt_2n(321).

Note that every inversion (σi, σj) of σ is in one-to-one correspondence to the unit

square at the intersection of the column with label σi and the row with label σj of

the polyomino (Ω(σ), Qn). Hence inv(σ) = area(Ω(σ)).

Example _{3.2.2. Take a σ = 3 1 6 2 7 4 8 5 ∈ Alt}8(321). The paths Ω(σ) and Q4, along

with their step-labeling, are shown in Figure 3.1(a). Note that area(Ω(σ)) = inv(σ) = 8. (a) (b) 3 3 2 1 7 8 5 4 4 5 6 7 8 2 6 1

Figure 3.1: The polyomino (Ω(σ), Q4) and plane tree Π ◦ Ω(σ) corresponding to the

per-mutation σ = 3 1 6 2 7 4 8 5 ∈ Alt8(321).

3.2.2 The second stage of the bijection between Alt

_2n

(321)

and

_T

_n

The second stage is an adaptation of the standard plane-tree representation of per-mutations.

For a tree T ∈ Tnand a vertex x ∈ T , the height of x, denoted by height(x), is the

distance from the root to x. We define the statistic hsum(T ) of T , the total height of all vertices of T , as

hsum(T ) =X

x∈T

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Proposition _{3.2.3. There is a bijection Π between the two sets C}n and Tn such that

a path P ∈ Cn is carried to a tree Π(P ) ∈ Tn with area(P ) = hsum(Π(P )).

Proof. A well-known bijection Π : Cn → Tn is constructed as follows. For every path

P ∈ Cn, to each north step there corresponds an edge traversed (in preorder) on the

way down, and to each east step there corresponds an edge traversed on the way up. Let x1, . . . , xn be the non-root vertices of Π(P ) in preorder. We observe that

for each i (1 ≤ i ≤ n), the height of xi in Π(P ) equals the number of unit squares

in the ith row (from bottom to top) of the polyomino (P, Qn). Hence area(P ) =

hsum(Π(P )).

Example _{3.2.4. Take a P = NNENNEEE ∈ C}4, i.e., the upper path of the polyomino

shown in Figure 3.1(a). The corresponding tree Π(P ) is shown in Figure 3.1(b), where area(P ) = hsum(Π(P )) = 8.

With the composition Π ◦ Ω : Alt2n(321) → Tn, each permutation σ ∈ Alt2n(321)

can be represented by a plane tree T = Π ◦ Ω(σ) with inv(σ) = hsum(T ).

3.3 A sign-reversing involution on plane trees

In this section, we review Chen-Shapiro-Yang’s involution φ on Tn and show that the

involution reverses the parity of the statistic hsum(T ) of T . First, we describe the fixed points of the involution φ in case n is odd.

Legal trees. _{We specify a subset of T}_2n+1_{, constructed from T}n as follows. For every

T ∈ Tn, we associate T with a tree λ(T ) ∈ T2n+1 by attaching to each vertex of T a

leaf as the first child of the vertex. We observe that the resulting trees λ(T ) can be characterized by the following condition

(A) a vertex is a leaf if and only if it is the first child of an internal vertex. Let Gn⊆ Tn be the set of trees that satisfy condition (A). Notice that

|G2n| = 0 and |G2n+1| = |Tn| = Cn, (3.3.1)

and that the map λ is a bijection between Tn and G2n+1.

Lemma _{3.3.1. For every tree T ∈ G}_2n+1, the statistic hsum(T ) has the opposite parity of n.

Proof. Note that T contains n + 1 leaves and n + 1 internal vertices. Let v1, . . . , vn+1

be the leaves of T . For 1 ≤ i ≤ n + 1, let ui be the parent of vi. Then u1, . . . , un+1

are distinct, and height(vi) = height(ui) + 1. We have

hsum(T ) = n+1 X i=1 height(ui) + n+1 X i=1 height(vi) = 2 n+1 X i=1 height(ui) ! + n + 1, as required.

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For any non-root vertex x of a plane tree T , we say that x is legal if either x is a leaf and it is also the first child of its parent, or x is an internal vertex but it is not the first child of its parent. Otherwise, x is illegal. The tree T is said to be legal if T contains at least one edge and every non-root vertex of T is legal. In other word, Gn

consists of all the legal trees in Tn for all n ≥ 1. By Eq. (3.3.1), a legal tree has odd

number of edges necessarily. Let τ (x) denote the subtree of T rooted at x, consisting of x and the descendants of x.

The involution φ. Making use of the construction given in [6, Theorem 2.1], the map φ acting on plane trees has the following consequence.

Theorem _{3.3.2. For all n ≥ 1, there is an involution φ on the set T}n − Gn that

reverses the parity of the statistic hsum(T ) of plane trees T .

Proof. Given a T ∈ Tn−Gn, find the last illegal vertex, say v, in right-to-left preorder.

Let u be the parent of v. There are two possibilities for the vertex v. (1) v is a leaf, and it has siblings, say w1, . . . , wk, to the left of v. (2) v is an internal vertex, and it

is the first child of u.

For case (1), we separate the subtrees τ (w1), . . . , τ (wk) from T , and then construct

the tree φ(T ) by attaching the subtrees τ (w1), . . . , τ (wk) to v in the same order. Note

that the vertex v remains to be the last illegal vertex of φ(T ) in right-to-left preorder. This operation makes the height of every vertex x ∈ τ(w1) ∪ · · · ∪ τ(wk) increase by

1. Hence hsum(φ(T )) = hsum(T ) +Pk_i=1_{|{x : x ∈ τ(w}i)}|.

For case (2), the tree φ(T ) is constructed by reversing the operation in case (1). In this case, we have hsum(φ(T )) = hsum(T ) −Pki=1|{x : x ∈ τ(wi)}|.

We observe that the first subtree τ (w1) = w1 is a single vertex, and the other

sub-trees τ (w2), . . . , τ (wk) are legal trees, each of which contains even number of vertices.

Hence the term Pk_i=1_{|{x : x ∈ τ(w}i)}| is odd, and the statistic hsum(φ(T )) has the

opposite parity of hsum(T ). The assertion follows.

Example 3.3.3. Given the tree T shown in Figure 3.2(a), the white vertices are the legal vertices while the black vertices are the illegal vertices, where the vertex v is the last illegal vertex in right-to-left preorder. The corresponding tree φ(T ) is shown in Figure 3.2(b).

The involution φ leads immediately to the sign-balance of Altn(321).

Corollary _{3.3.4. For n ≥ 1, the following identities hold.}

(i) I(Alt2n(321)) =    (−1)n−12 C_n−1 2 n odd 0 n even,

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(a) (b)

v

u

v

φ

Figure 3.2: An example for the involution φ on illegal plane trees. Proof. (i) For the case of even length, we have

I(Alt2n(321)) = X σ∈Alt2n(321) (−1)inv(σ) = X T∈Tn

(−1)hsum(T ) (by Propositions 3.2.1 and 3.2.3)

= X

T∈Gn

(−1)hsum(T ) (by Theorem 3.3.2) =    (−1)n+12 C_n−1 2 if n odd

0 if n even. (by Eq. (3.3.1) and Lemma 3.3.1). (ii) For every σ = σ1· · · σ2n ∈ Alt2n(321), we observe that the greatest entry is

σ_2n−1 = 2n (otherwise there will be a 321-pattern (2n, σ_2n−1, σ2n) in σ), and that

σ_2n−2 < σ2n (otherwise there will be a 321-pattern (σ2n−3, σ2n−2, σ2n) in σ). Then

removing the entry σ_2n−1 from σ results in a member σ′ _{∈ Alt}

2n−1(321) with inv(σ′) =

inv_{(σ) − 1. This is a sign-reversing bijection between Alt}_2n−1(321) and Alt2n(321).

Hence I(Alt2n−1(321)) = −I(Alt2n(321)). The proof is completed.

Remarks. Shattuck [28] studied a q-polynomial of generating function for the inver-sion number of Catalan words, which can be translated from Dyck paths with north step and east step replaced by 1 and 0, respectively. Corollary 3.3.4(i) is essentially equivalent to Shattuck’s result [28, Theorem 4.1].

3.4 Refined sign-balance results on plane trees

In this section, we establish a new involution Φ on Tnthat proves refined sign-balance

results on plane trees and paves the way to a proof of Theorems 3.1.2 and 3.1.3. For a T ∈ Tn, by the leftmost (respectively, rightmost) path of T we mean the

path that connects the root and the first (respectively, last) leaf in preorder. Let lmp(T ) (respectively, rmp(T )) be the number of vertices on the leftmost (respectively, rightmost) path of T . Note that lmp(T ) ≥ 2 and rmp(T ) ≥ 2 if T contains at least

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one edge. By the bijection Π ◦ Ω constructed from Propositions 3.2.1 and 3.2.3, we have the following observation.

Lemma _{3.4.1. For every σ ∈ Alt}_2n_{(321) and the corresponding tree T = Π ◦ Ω(σ) ∈} Tn, the following relations hold.

(i) lead(σ) = lmp(T ),

(ii) end(σ) = 2n − rmp(T ) + 1.

With Lemma 3.4.1, the identities (i) and (iii) of Theorem 3.1.2 can be deduced from the following identities for plane trees.

Theorem _{3.4.2. For all n ≥ 1, the following identities hold.}

(i) X T∈T2n+1 (−1)hsum(T )qlmp(T ) _{= (−1)}n+1 X T∈Tn q2·lmp(T ). (ii) X T∈T2n (−1)hsum(T )_qlmp(T )_{= (−1)}n+1_{(1 − q)} X T∈Tn q2(lmp(T )−1)_.

The identities (i) and (iii) of Theorem 3.1.3 can be deduced from the following identities.

Theorem _{3.4.3. For all n ≥ 1, the following identities hold.}

(i) X T∈T2n+1 (−1)hsum(T )q−rmp(T )_{= (−1)}n+1 X T∈Tn q−2·rmp(T ). (ii) X T∈T2n (−1)hsum(T )q−rmp(T ) _{= (−1)}n_{(1 − q)} X T∈Tn q−2·rmp(T )+1.

The above two theorems can be proved by the same method. In the following, we focus on the proof of Theorem 3.4.2. We shall establish an involution Φ on plane trees T that reverses the parity of hsum(T ) while preserving the statistic lmp(T ), making use of the involution φ in Theorem 3.3.2 as building blocks.

For 2 ≤ k ≤ n + 1, we define

Tn,k = {T ∈ Tn : lmp(T ) = k}.

For a T ∈ Tn,k, let u1, . . . , uk be the vertices on the leftmost path of T from root

to leaf. The other n − k + 1 edges of T are decomposed into k subtrees R1, . . . , Rk

(possibly with no edges), where Ri is rooted at ui (1 ≤ i ≤ k), with the convention

that the last subtree Rk= ukis always a single vertex. The tree T is fully described by

subtree-sequence R1, . . . , Rk under the leftmost path and we write T = (R1, . . . , Rk).

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3.4.1 The construction of the involution

_{Φ on T}

_2n+1

Recall that for all n ≥ 0 the map λ : Tn → G2n+1 carries a tree T ∈ Tn to a legal tree

λ(T ) ∈ G2n+1 by attaching to each vertex of T a leaf as the first child of the vertex.

Note that for n = 0, the unique tree in T0 (respectively, G1) consists of a single vertex

(respectively, edge).

Generalized legal trees. _{Based on the map λ, we specify a subset of T}2n+1,2k,

constructed from Tn,k, which serves as the set of fixed points of the map Φ.

Given a T = (R1, . . . , Rk) ∈ Tn,k, we associate T with a tree Λ(T ) ∈ T2n+1,2k as

follows. Let v1, . . . , v2k be the vertices on the leftmost path of Λ(T ) from root to leaf,

and let the subtrees under the leftmost path of Λ(T ) be indexed by S1, . . . , S2k, where

Sj is rooted at vj. For 1 ≤ i ≤ k, each pair (S2i−1, S2i) is defined from the subtree

Ri of T by the following procedure. If Ri is a single vertex then both of S2i−1 and

S2i are a single vertex. Otherwise, Ri contains at least one edge. Let wi be the last

child of the root ui of Ri. Removing the edge uiwi from Ri results in two trees Tui

and Twi, where Twi = τ (wi) is the subtree of Ri rooted at wi, and Tui = Ri− τ(wi)

consists of the remainder of Ri, rooted at ui. Then we find the corresponding legal

trees of Tui and Twi under the map λ, and assign S2i−1 = λ(Tui) and S2i = λ(Twi).

The tree Λ(T ) is constructed. See Figure 3.3 for an illustration.

v1 v3 v4 v5 v6 v7 v8 u1 w1 u2 u3 u4 w3 Tw1 Tu1 λ Tu3 Tw3 Λ S2 S1 S6 S5 v2

Figure 3.3: An example for the construction of the map Λ.

For 2 ≤ k ≤ n + 1, we define the set G2n+1,2k ⊆ T2n+1,2k by

G2n+1,2k = {Λ(T ) : T ∈ Tn,k}.

It is straightforward to prove that the map Λ−1 _{can be obtained by a reverse}

proce-dure, so Λ is a bijection between Tn,kand G2n+1,2k. Hence |G2n+1,2k| = |Tn,k|. Moreover,

for the trees in G2n+1,2k, their subtree-sequence S1, . . . , S2k under the leftmost path

can be characterized by the following conditions.

(B1) For 1 ≤ i ≤ k − 1 and for each pair (S2i−1, S2i), either both of S2i−1 and S2i are

legal trees, or both of S_2i−1 and S2i are a single vertex.

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For convenience, a vertex u of a plane tree T is colored black if u is of odd height and colored white otherwise. It is clear that the parity of hsum(T ) is the same as the parity of the number of black vertices of T .

Lemma _{3.4.4. For every T ∈ G}2n+1,2k, the statistic hsum(T ) has the opposite parity

of n, independent of k.

Proof. Let T′ _{= Λ}−1_{(T ) ∈ T}

n,k. Note that T′ contains n + 1 vertices. By the

construction of the map Λ, we observe that every vertex of T′ _{is associated with a}

black vertex and a white vertex in T in the following manner. For the vertices ui on

the leftmost paths of T′_{, we associate u}

i with the two vertices v_2i−1, v2i∈ T . For each

subtree Ri ⊆ T′ with a non-empty edge set, we associate the last child wi of the root

ui of Ri with the first leaf of S_2i−1 and the first leaf of S2i. For the other vertices

x ∈ Ri− {ui, wi}, x is attached a leaf as its first child in S_2i−1∪ S2i, so we associate x

with itself and its first child in S_2i−1_{∪ S}2i. This exhausts the vertex set of T . Hence

T contains n + 1 black vertices, and the assertion follows.

The involution _{Φ on T}2n+1. Given a T ∈ T2n+1 with lmp(T ) = m, let R1, . . . , Rm

be the subtrees under the leftmost path of T . We define the corresponding tree Φ(T ) ∈ T2n+1,m according to the following two cases.

Case 1. There exist some illegal trees in {R1, . . . , Rm} with a non-empty edge

set. Then find the last illegal tree, say Rj, with at least one edge. The tree Φ(T )

is obtained from T by replacing Rj by its counterpart φ(Rj) under the involution

φ. Note that φ(Rj) is the last illegal tree with a non-empty edge set in the

subtree-sequence of Φ(T ), and that hsum(Φ(T )) has the opposite parity of hsum(T ).

Case 2. Every tree in {R1, . . . , Rm} is either a single vertex or a legal tree. We

partition the sequence R1, . . . , Rm into adjacent pairs (R1, R2), (R3, R4), · · · . There

are two subcases.

(2a) There exist some pairs (R_2i−1, R2i) with i ≤ ⌊m₂−1⌋ such that one of the two

trees is a single vertex and the other is a legal tree. Then we find the last pair (R_2i−1, R2i) (i.e., with the largest number i ≤ ⌊m−1₂ ⌋) with such a property.

The tree Φ(T ) is obtained from T by interchanging R_2i−1 and R2i. We observe

that there are odd number of vertices in R_2i−1_{∪ R}2i and that this operation

interchanges black and white colors of these vertices. Hence hsum(Φ(T )) has the opposite parity of hsum(T ). Moreover, (R2i, R2i−1) is also the last pair with

such a property in Φ(T ). Note that the restriction i ≤ ⌊m−12 ⌋ excludes the

possibility of interchanging the last two trees Rm−1 and Rm when m is even, so

the length of leftmost path is preserved. (2b) For 1 ≤ i ≤ ⌊m−1

2 ⌋ and for each pair (R2i−1, R2i), either both of R2i−1 and R2i

are legal trees, or both of R_2i−1 and R2i are a single vertex. Then let Φ(T ) = T ,

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Example 3.4.5. Given the tree T = (R1, . . . , R6) ∈ T13,6 shown on the left of Figure

3.4, note that R3 is the only illegal tree with a non-empty edge set in {R1, . . . , R6}.

By Case 1 of the map Φ, the tree Φ(T ) is obtained from T by replacing R3 by φ(R3),

shown on the right of Figure 3.4.

Φ

φ(R3)

R3

Figure 3.4: An example for Case 1 of the map Φ.

Example 3.4.6. Given the tree T = (R1, . . . , R6) ∈ T13,6 shown on the left of Figure

3.5, each tree in {R1, . . . , R6} is either a single vertex or a legal tree. However,

(R3, R4) is the only pair in {(R1, R2), (R3, R4)} such that one is a single vertex and

the other is a legal tree. By Case (2a) of the map Φ, the tree Φ(T ) is obtained from T by interchanging R3 and R4, shown on the right of Figure 3.5.

R3

R4

Φ

Figure 3.5: An example for Case (2a) of the map Φ.

The involution Φ on T2n+1 has the following refinement.

Proposition 3.4.7. The following facts hold.

(i) For 2 ≤ k ≤ n + 1, the map Φ induces an involution on the set T2n+1,2k such

that G2n+1,2k is the set of fixed points.

(ii) The map Φ induces an involution on the set T2n+1,2 with no fixed points.

(iii) For 2 ≤ k ≤ n + 1, the map Φ induces an involution on the set T2n+1,2k−1 with

no fixed points.

Proof. (i) Given a fixed point T = (R1, . . . , R2k) ∈ T2n+1,2k under the map Φ, we shall

verify that the subtrees R1, . . . , R2k satisfy conditions (B1) and (B2) for the trees in

G2n+1,2k. It follows immediately from Case (2b) of the map Φ that condition (B1) is

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Suppose on the contrary that R_2k−1contains at least one edge. By Case 1 of the map Φ, R_2k−1 is legal, and then it contains odd number of edges. Since there are totally 2n − 2k + 2 edges in the trees R1∪ · · · ∪ R2k−1, there are odd number of members in

{R1, . . . , R2k−2} being legal trees, which implies that there exists a pair (R2i−1, R2i)

such that one of which is a single vertex and the other is a legal tree. By Case (2a) of the map Φ, T is not a fixed point, a contradiction. Hence R_2k−1 contains no edges. The assertion (i) follows.

(ii) For every T = (R1, R2) ∈ T2n+1,2, note that R2 is a single vertex and that R1

is a tree with 2n edges, which is an illegal tree. By Case 1 of the map Φ, T is not a fixed point.

(iii) Given a T = (R1, . . . , R2k−1) ∈ T2n+1,2k−1, suppose that T is a fixed point

under the map Φ. By Case 1 of the map Φ, each tree Ri is either a single vertex or a

legal tree. Since there are totally 2n − 2k + 3 edges in the trees R1∪ · · · ∪ R2k−2, there

are odd number of members in {R1, . . . , R2k−2} being legal trees, which implies that

there exists a pair (R_2i−1, R2i) such that one of which is a single vertex and the other

is a legal tree. By Case (2a) of the map Φ, T is not a fixed point, a contradiction. The assertion (iii) follows.

Example _{3.4.8. The map Φ acting on the set T}₅ is illustrated in Figure 3.6, where T5 is partitioned into T5,i, for 2 ≤ i ≤ 6, with respect to lmp(T ) = i. The trees shown

in the upper (respectively, lower) cell of T5,i have even (respectively, odd) hsum(T ),

where the trees that are counterparts of each other under the map Φ are arranged accordingly. Note that the map Φ restricted to T5,2, T5,3 and T5,5 has no fixed points,

and that G5,4 and G5,6 are the sets of fixed points of the map Φ restricted to T5,4 and

T5,6, respectively.

Now we are able to prove the first identity of Theorem 3.4.2.

Proof of Theorem 3.4.2(i). By Lemma 3.4.4 and Proposition 3.4.7, we have

X T∈T2n+1 (−1)hsum(T )qlmp(T ) = 2n+2_X j=2   X T∈T2n+1,j (−1)hsum(T )qlmp(T )   = n+1 X k=2   X T∈G2n+1,2k (−1)n+1q2k   = (−1)n+1 n+1 X k=2 |Tn,k|q2k = (−1)n+1 n+1 X k=2   X T∈Tn,k q2·lmp(T )   = (−1)n+1 X T∈Tn q2·lmp(T ), as required.

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Λ G5,4 G5,6 T5,4 T5,5 T5,3 Φ T5,2 T5,6 − + T2 T2,3 T2,2 Λ

Figure 3.6: A picture of the map Φ on T5.

3.4.2 The involution

_{Φ on T}

_2n

The construction of the involution Φ on the set T2nis exactly the same as construction

on the set T2n+1. For the fixed points of the map Φ, we specify two families of subsets

{H2n,2k−1: 2 ≤ k ≤ n + 1} and {H2n,2k−2: 2 ≤ k ≤ n + 1} of T2n.

For the first family, the set H2n,2k−1 is constructed from G2n+1,2k as follows. Given

a T = (R1, . . . , R2k) ∈ G2n+1,2k, recall that last two subtrees R2k−1and R2kare a single

vertex. We associate T with a tree Γ(T ) ∈ T2n,2k−1in the form Γ(T ) = (R1, . . . , R2k−1)

by removing R2k from T , in other word, deleting the first leaf in preorder.

H2n,2k−1 = {Γ(T ) : T ∈ G2n+1,2k}.

Note that Γ is a bijection between G2n+1,2k and H2n,2k−1 since the trees T ∈ G2n+1,2k

can be recovered from Γ(T ) ∈ H2n,2k−1 by attaching a vertex to the first leaf of Γ(T ).

Hence |H2n,2k−1| = |G2n+1,2k| = |Tn,k|. Since R2k is a black vertex in T , the statistics

hsum(T ) and hsum(Γ(T )) have opposite parities. By Lemma 3.4.4, we have following fact.

Lemma _{3.4.9. For every T ∈ H}_2n,2k−1, the statistic hsum(T ) has the same parity of n, independent of k.

By the same argument as in the proof of Proposition 3.4.7, the map Φ on the set T2n,2k−1 has the following immediate consequence.

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Proposition _{3.4.10. For 2 ≤ k ≤ n + 1, H}_2n,2k−1 is the set of fixed points of the involution Φ on the set T2n,2k−1.

For the second family, the set H2n,2k−2 is constructed from Tn,k by a modified

construction of the map Λ for the set G2n+1,2k. Given a T = (R1, . . . , Rk) ∈ Tn,k,

we shall associate T with a tree Λ′_{(T ) ∈ T}

2n,2k−2. Let S1, . . . , S2k−2 be the subtrees

under the leftmost path of Λ′_{(T ). For 1 ≤ i ≤ k − 2, each pair (S}

2i−1, S2i) is defined

by the subtree Ri of T in exactly the same way as in the map Λ. That is, if Ri is

a single vertex then both of S_2i−1 and S2i are a single vertex, otherwise we assign

S_2i−1 = λ(Tui) and S2i = λ(Twi), where wi is the last child of the root ui of Ri, and

Tui and Twi are the two trees when the edge uiwi is removed from Ri. For the last

pair (S_2k−3, S_2k−2), we assign S_2k−3 = λ(Rk−1) and assign S2k−2 a single vertex. See

Figure 3.7 for an illustration.

u1 w1 u2 u3 u4 w3 v1 v2 v3 v4 v5 v6 Λ′ S2 S1 S5 Tw1 Tu1 λ R3

Figure 3.7: An example for the construction of the map Λ′_.

H2n,2k−2= {Λ′(T ) : T ∈ Tn,k}.

Note that Λ′ _{is a bijection between T}

n,k and H2n,2k−2. Hence |H2n,2k−2| = |Tn,k|.

Moreover, for the trees in H2n,2k−2, their subtree-sequence S1, . . . , S2k−2 under the

leftmost path can be characterized by the following conditions.

(C1) For 1 ≤ i ≤ k − 2 and for each pair (S2i−1, S2i), either both of S2i−1 and S2i are

legal trees, or both of S_2i−1 and S2i are a single vertex.

(C2) S_2k−3 is a legal tree but S_2k−2 is a single vertex.

Lemma _{3.4.11. For every T ∈ H}_2n,2k−2, the statistic hsum(T ) has the opposite parity of n, independent of k.

Proof. Let T′ _{= Λ}′−1_{(T ) ∈ T}

n,k with subtree-sequence R1, . . . , Rk. We associate the

vertex in Rk of T′ with the single vertex in S_2k−2 of T , which is black. We associate

each vertex v of Rk−1 ⊆ T′ with itself and its first child in S2k−3⊆ T . For the other

vertices x of T′_{, we associate x with a black vertex and a white vertex in T in the}

same manner as in the proof of Lemma 3.4.4. Hence T contains n + 1 black vertices, and the assertion follows.

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Proposition _{3.4.12. For 2 ≤ k ≤ n + 1, H}_2n,2k−2 is the set of fixed points of the involution Φ on the set T2n,2k−2.

Proof. Take m = 2k−2 in the construction of the map Φ. For a T = (R1, . . . , R2k−2) ∈

T2n,2k−2, the restriction i ≤ ⌊m−12 ⌋ = k−2 excludes the possibility of interchanging the

last two trees R_2k−3 and R_2k−2, so the length of leftmost path of T is preserved under the map Φ. We shall verify that if T is a fixed point then the subtrees R1, . . . , R2k−2

satisfy conditions (C1) and (C2) for the trees in H2n,2k−2.

By Case (2b) of the map Φ, they satisfy condition (C1), which implies that there are even number of members in {R1, . . . , R2k−4} are legal trees. Since there are totally

2n − 2k + 3 edges in R1∪ · · · ∪ R2k−3 (while R2k−2 is a single vertex), the tree R2k−3

contains odd number of edges. By Case 1, R_2k−3is a legal tree. This verifies condition (C2). The proof is completed.

Example _{3.4.13. The map Φ acting on the set T}4 is illustrated in Figure 3.8, where

H4,i is the set of fixed points of the map Φ restricted to the set T4,i, for 2 ≤ i ≤ 5.

Note that H4,3 and H4,5 are in one-to-one correspondence to G5,4 and G5,6 under the

map Γ, respectively, and that H4,2 and H4,4 are in one-to-one correspondence to T2,2

and T2,3 under the map Λ′, respectively.

Γ − + Λ T4,5 T4,4 T4,3 T4,2 Φ Λ′ Λ′ T2,2 T2,3 H_4,4 H4,3 H4,5 H4,2 G5,4 G_5,6 Γ T2 T2,3 T2,2

Figure 3.8: _{A picture of the map Φ on T}4.

Now we are able to prove the second identity of Theorem 3.4.2.

一些排列類上的折疊現象

國立高雄大學應用數學系

博士論文

Folding Phenomena on some classes of permutations

一些排列類上的折疊現象

研究生：丁建太 撰

指導教授：游森棚、陳晴玉

致 謝

一些排列類上的折疊現象

Folding Phenomena on some classes of permutations

Folding Phenomena on some classes of permutations

Chien-Tai Ting

A DISSERTATION

in

Mathematics

Presented to the Faculties of the National University of Kaohsiung in

Partial Fulfillment of the Requirements for the Degree of Doctor of

Philosophy

June, 2017

Supervisor of Dissertation

Sen-Peng Eu, Colonel Professor in Mathematics

Ching-Yu Chen, Professor in Mathematics

Graduate Group Chairperson

Tung-Shan Fu, Professor in Mathematics

Dissertation Committee:

Sen-Peng Eu, Colonel Professor in Mathematics

Ching-Yu Chen, Professor in Mathematics

Tung-Shan Fu, Professor in Mathematics

Hui-Lan Chang, Associate Professor in Mathematics

Yeh-Jong Pan, Associate Professor in Mathematics

Chi-Hung Yen, Associate Professor in Mathematics

Folding Phenomena on some classes of permutations

Contents

Chapter 1

Introduction

Chapter 2

Alternating permutations I

2.1

Introduction

2.1.1

Alternating permutations avoiding a pattern of length

three

2.1.2

Alternating permutations of genus zero

2.1.3

Organization of the paper.

2.2

Sign-imbalance of Alt

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a

b

h

d

g

c

e i

f

a

b

d

c

h g f e

i

u

Ψ

u

x

y

v

v

y

x

2.3

Sign-imbalance of Alt

(ω) for the other

pat-terns

ω of length three

2.4

Sign imbalance of alternating permutations of

genus zero

研究生：丁建太撰

致謝

_T

_{Φ on T}

_{Φ on T}