Journal of Combinatorial Theory, Series A 124 (2014) 251–262

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Journal of Combinatorial Theory,Series A

www.elsevier.com/locate/jcta

Annular noncrossing permutations and minimaltransitive factorizations

Jang Soo Kim a, Seunghyun Seo b, Heesung Shin c

a Department of Mathematics, Sungkyunkwan University, Suwon, 440-746,South Koreab Department of Mathematics Education, Kangwon National University,Hyoja-dong, Chuncheon, 200-701, South Koreac Department of Mathematics, Inha University, 100 Inharo, Nam-gu,Incheon, 402-751, South Korea

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

Article history:Received 19 March 2012Available online 4 March 2014

Keywords:Minimal transitive factorizationAnnular noncrossing permutationCombinatorial proof

We give a combinatorial proof of Goulden and Jackson’sformula for the number of minimal transitive factorizationsof a permutation when the permutation has two cycles. Weuse the recent result of Goulden, Nica, and Oancea on thenumber of maximal chains of annular noncrossing partitionsof type B.

© 2014 Elsevier Inc. All rights reserved.

1. Introduction

Given an integer partition λ = (λ1, λ2, . . . , λ�) of n, denote by αλ the permutation

(1 . . . λ1)(λ1 + 1 . . . λ1 + λ2) . . . (n− λ� + 1 . . . n)

of the set {1, 2, . . . , n} expressed in cycle notation. Throughout this paper we multiplypermutations from left to right, i.e., (πσ)(i) = σ(π(i)). Let Fλ be the set of all (n + �−2)-tuples (η1, . . . , ηn+�−2) of transpositions such that

E-mail addresses: jangsookim@skku.edu (J.S. Kim), shyunseo@kangwon.ac.kr (S. Seo),shin@inha.ac.kr (H. Shin).

http://dx.doi.org/10.1016/j.jcta.2014.02.0050097-3165/© 2014 Elsevier Inc. All rights reserved.

252 J.S. Kim et al. / Journal of Combinatorial Theory, Series A 124 (2014) 251–262

(1) η1 · · · ηn+�−2 = αλ, and(2) the group 〈η1, . . . , ηn+�−2〉 generated by η1, . . . , ηn+�−2 acts transitively on

{1, 2, . . . , n}.

Such tuples are called minimal transitive factorizations of the permutation αλ of type λ,which are related to the branched covers of the sphere studied by Hurwitz [8].

The following formula was first published by Hurwitz [8] in 1891 with a sketch of proofwhich was completed by Strehl [15]:

|Fλ| = (n + �− 2)!n�−3�∏

i=1

λλii

(λi − 1)! . (1)

Goulden and Jackson [4] also proved (1) using algebraic methods. Bousquet-Mélou andSchaeffer [3] proved a more general formula than (1) and obtained (1) using the principleof inclusion and exclusion.

If λ = (n), the formula (1) yields

|F(n)| = nn−2, (2)

and there are several combinatorial proofs of (2) [2,6,12].If λ = (p, q), the formula (1) yields

|F(p,q)| = pq

p + q

(p + q

q

)ppqq. (3)

There are several papers about bijective proofs of (3): Kim and Seo [10] for the case(p, q) = (1, n− 1), and Rattan [14] for the cases (p, q) = (2, n− 2) and (p, q) = (3, n− 3).For the general case λ = (p, q), Irving [9] found a bijective proof using planar maps andtrees. We also note that Arnol’d [1] gave an analytic proof of (3).

When there is only one cycle, i.e., λ = (n), there is a bijection between minimaltransitive factorizations and maximal chains in the poset of noncrossing partitions. Itis natural to ask a similar combinatorial explanation for λ = (p, q). Goulden, Nica, andOancea [5] showed that the number of maximal chains in the poset NCB(p, q) of annularnoncrossing partitions of type B is

(p + q

q

)ppqq +

∑c�1

2c(p + q

p− c

)pp−cqq+c. (4)

Interestingly it turns out that half of the sum in (4) is equal to the number in (3):

∑c

(p + q

p− c

)pp−cqq+c = pq

p + q

(p + q

q

)ppqq. (5)

c�1

J.S. Kim et al. / Journal of Combinatorial Theory, Series A 124 (2014) 251–262 253

In this paper we give another combinatorial proof of (3) by finding a two-to-onemap from the set of minimal transitive factorizations of type (p, q) to the set of cer-tain maximal chains in the poset NCB(p, q). The rest of this paper is organized asfollows. In Section 2 we recall the poset SB

nc(p, q) of annular noncrossing permutationsof type B which is isomorphic to the poset NCB(p, q) of annular noncrossing partitionsof type B, and show that the number of connected maximal chains in SB

nc(p, q) is equalto 2pq

p+q

(p+qq

)ppqq. In Section 3 we prove that there is a two-to-one map from the set of

connected maximal chains in SBnc(p, q) to F(p,q).

2. Connected maximal chains

A signed permutation is a permutation σ on {±1, . . . ,±n} satisfying σ(−i) = −σ(i)for all i ∈ {1, . . . , n}. We denote by Bn the set of signed permutations on {±1, . . . ,±n}.

We will use the two notations

[a1 a2 . . . ak] = (a1 a2 . . . ak −a1 −a2 . . .−ak),

((a1 a2 . . . ak)) = (a1 a2 . . . ak)(−a1 −a2 . . .−ak),

and call [a1 a2 . . . ak] a zero cycle and ((a1 a2 . . . ak)) a paired nonzero cycle. We also callthe cycles εi := [i] = (i −i) and ((i j)) type B transpositions, or simply transpositions ifthere is no possibility of confusion.

For π ∈ Bn, the absolute length �B(π) is defined to be the smallest integer k such thatπ can be written as a product of k type B transpositions. The absolute order on Bn isdefined by

π � σ ⇔ �B(σ) = �B(π) + �B(π−1σ

).

From now, we fix positive integers p and q. The poset SBnc(p, q) of annular noncrossing

permutations of type B is defined by

SBnc(p, q) := [ε, γp,q] = {σ ∈ Bp+q: ε � σ � γp,q} ⊆ Bp+q,

where ε is the identity in Bp+q and γp,q = [1 . . . p][p+ 1 . . . p+ q]. Fig. 1 shows the Hassediagram for SB

nc(2, 1). Nica and Oancea [13, (2.2)] showed that SBnc(p, q) is a graded poset

with rank function

rank(σ) = �B(σ) = (p + q) − (# of paired nonzero cycles of σ). (6)

Nica and Oancea [13] showed that σ ∈ SBnc(p, q) if and only if σ can be drawn

without crossing arrows inside an annulus in which the outer circle has integers1, 2, . . . , p,−1,−2, . . . ,−p in clockwise order and the inner circle has integers p + 1,p + 2, . . . , p + q,−p − 1,−p − 2, . . . ,−p − q in counterclockwise order, see Fig. 2. They

254 J.S. Kim et al. / Journal of Combinatorial Theory, Series A 124 (2014) 251–262

Fig. 1. The Hasse diagram for SBnc(2, 1).

Fig. 2. π = ((1 5 6))((2 3 4)) and σ = [1 5 −7 2]((3 4)) in SBnc(4, 3).

also showed that SBnc(p, q) is isomorphic to the poset NCB(p, q) of annular noncrossing

partitions of type B. Here annular noncrossing partitions are those obtained from an-nular noncrossing permutations by changing all zero cycles into a single block and eachnonzero cycle into a block.

A paired nonzero cycle ((a1 a2 . . . ak)) is called connected if the set {a1, . . . , ak} inter-sects with both {±1, . . . ,±p} and {±(p+ 1), . . . ,±(p+ q)}, and disconnected otherwise.A zero cycle is always considered to be disconnected. For σ ∈ SB

nc(p, q), the connectivityof σ is the number of connected paired nonzero cycles of σ.

We say that a maximal chain C = {ε = π0 < π1 < · · · < πp+q = γp,q} of SBnc(p, q)

is disconnected if the connectivity of each πi is zero. Otherwise, C is called connected.Denote by CM(SB

nc(p, q)) the set of connected maximal chains of SBnc(p, q).

We need the following theorem.

Theorem 1. (See [5, Theorem 5.3].) The number of disconnected maximal chains ofSB

nc(p, q) is equal to

(p + q

)ppqq (7)

q

J.S. Kim et al. / Journal of Combinatorial Theory, Series A 124 (2014) 251–262 255

and the number of connected maximal chains of SBnc(p, q) is equal to

∑c�1

2c(p + q

p− c

)pp−cqq+c. (8)

Remark 1. Theorem 5.3 in [5] only states that the number of maximal chains ofNCB(p, q) ∼= SB

nc(p, q) equals (7) plus (8). However, their proof also shows that thenumber of disconnected maximal chains of NCB(p, q) is equal to (7).

We now prove the following identity that appears in the introduction. The proof isdue to Krattenthaler [11].

Lemma 2. We have∑c�1

c

(p + q

p− c

)pp−cqq+c = pq

p + q

(p + q

q

)ppqq. (9)

Proof. Since c = p · q+cp+q − q · p−c

p+q , we have

p∑c=0

c

(p + q

p− c

)pp−cqq+c =

p∑c=0

(p · q + c

p + q− q · p− c

p + q

)(p + q

p− c

)pp−cqq+c

=p∑

c=0

((p + q − 1p− c

)pp−c+1qq+c −

(p + q − 1p− c− 1

)pp−cqq+c+1

)

=(p + q − 1

p

)pp+1qq = pq

p + q

(p + q

p

)ppqq. �

By Theorem 1 and Lemma 2, we get the following.

Corollary 3. The number of connected maximal chains of SBnc(p, q) is equal to

2pqp + q

(p + q

q

)ppqq. (10)

For example, there are 16 = 2·2·12+1

(31)2211 connected maximal chains of SB

nc(2, 1), seeFig. 3. By Corollary 3, in order to prove (3) combinatorially it is sufficient to find a 2–1map from CM(SB

nc(p, q)) to F(p,q). We will find such a map in the next section.There is a natural correspondence between maximal chains and factorizations. For

a maximal chain C = {ε = π0 < π1 < · · · < πp+q = γp,q} of SBnc(p, q), we define

ϕ(C) = (τ1, τ2, . . . , τp+q), where τi = π−1i−1πi. Note that each τi is a type B transposition

and πi = τ1τ2 · · · τi for all i = 1, 2, . . . , p+ q. The map ϕ will be used in the next section.

Remark 2. One can check that the factorizations ϕ(C) coming from connected maximalchains C in SB

nc(p, q) are precisely the minimal factorizations of γp,q in the Weyl group Dn.

256 J.S. Kim et al. / Journal of Combinatorial Theory, Series A 124 (2014) 251–262

Fig. 3. Connected maximal chains in SBnc(2, 1).

Thus Corollary 3 can be restated as follows: the number of minimal factorizations of γp,qin Dn is equal to 2pq

p+q

(p+qq

)ppqq. Goupil [7, Theorem 3.1] also proved this result by finding

a recurrence relation.

Remark 3. Since the proof of Lemma 2 is a simple manipulation, it is possible to constructa combinatorial proof for the identity in Lemma 2. Together with the result in Section 3we get a combinatorial proof of (10).

3. A 2–1 map from CM(SBnc(p, q)) to F(p,q)

Recall that a minimal transitive factorization of αp,q = (1 . . . p)(p + 1 . . . p + q) is asequence (η1, . . . , ηp+q) of transpositions in Sp+q such that

(1) η1 · · · ηp+q = αp,q, and(2) 〈η1, . . . , ηp+q〉 acts transitively on Sp+q,

and F(p,q) is the set of minimal transitive factorizations of αp,q.In this section we will prove the following theorem.

Theorem 4. There is a 2–1 map from the set of connected maximal chains in SBnc(p, q)

to the set F(p,q) of minimal transitive factorizations of αp,q.

In order to prove Theorem 4 we need some definitions.

Definition 5 (Two maps (·)+ and | · |). We introduce the following two maps.

(1) The map (·)+ : Bn → Bn is defined by

σ+(i) ={ |σ(i)| if i > 0,−|σ(i)| if i < 0.

(2) The map | · | : Bn → Sn is defined by |σ|(i) = |σ(i)| for all i ∈ {1, . . . , n}.

J.S. Kim et al. / Journal of Combinatorial Theory, Series A 124 (2014) 251–262 257

It is easy to see that two maps in Definition 5 are group homomorphisms.

Definition 6. A (p + q)-tuple (τ1, . . . , τp+q) of transpositions in Bp+q is called a minimaltransitive factorization of type B of γp,q = [1 . . . p][p + 1 . . . p + q] if it satisfies

(1) τ1 . . . τp+q = γp,q,(2) 〈|τ1|, . . . , |τp+q|〉 acts transitively on Sp+q.

Denote by F (B)(p,q) the set of minimal transitive factorizations of type B of γp,q.

Definition 7. A (p + q)-tuple (σ1, . . . , σp+q) of transpositions in Bp+q is called a positiveminimal transitive factorization of type B of βp,q = ((1 . . . p))((p + 1 . . . p + q)) if itsatisfies

(1) σ1 . . . σp+q = βp,q,(2) 〈|σ1|, . . . , |σp+q|〉 acts transitively on Sp+q,(3) σi = σ+

i for all i = 1, . . . , p + q.

Denote by F+(p,q) the set of positive minimal transitive factorizations of type B of βp,q.

For the rest of this section we will prove the following:

(1) The map ϕ : CM(SBnc(p, q)) → F (B)

(p,q) is a bijection (Lemma 8).(2) There is a 2–1 map (·)+ : F (B)

(p,q) → F+(p,q) (Proposition 10).

(3) There is a bijection | · | : F+(p,q) → F(p,q) (Lemma 9).

By the above three statements the composition |ϕ+| := | · | ◦ (·)+ ◦ ϕ is a 2–1 map fromCM(SB

nc(p, q)) to F(p,q), which completes the proof of Theorem 4. Since the proofs of thefirst and the third statements are simpler, we will present these first.

Lemma 8. The map ϕ : CM(SBnc(p, q)) → F (B)

(p,q) is a bijection.

Proof. Given a connected maximal chain C = {ε = π0 < π1 < · · · < πp+q = γp,q}in SB

nc(p, q), the elements in the sequence ϕ(C) = (τ1, . . . , τp+q) are transpositions withτ1 · · · τp+q = γp+q. Since C is connected, 〈|τ1|, . . . , |τp+q|〉 acts transitively on Sp+q. Thusϕ(C) ∈ F (B)

(p,q). Conversely, if τ = (τ1, . . . , τp+q) ∈ F (B)(p,q), then ϕ−1(τ) = {ε = π0 < π1 <

· · · < πp+q = γp,q}, where πi = τ1 · · · τi, is a connected maximal chain in SBnc(p, q). �

Lemma 9. There is a bijection | · | : F+(p,q) → F(p,q).

Proof. Let (σ1, . . . , σp+q) ∈ F+(p,q). Since σ+

i = σi, each σi can be written as σi =((j k)) for some positive integers j and k rather than [i]. In this case we let ηi = |σi| =

258 J.S. Kim et al. / Journal of Combinatorial Theory, Series A 124 (2014) 251–262

(j k) ∈ Sp+q. Then the map | · | : F+(p,q) → F(p,q) sending (σ1, . . . , σp+q) to (η1, . . . , ηp+q)

is a bijection. �For (τ1, τ2, . . . , τp+q) ∈ F (B)

(p,q), we define (τ1, τ2, . . . , τp+q)+ = (τ+1 , τ+

2 , . . . , τ+p+q). For

the remainder of this section we will prove the following proposition.

Proposition 10. The map (·)+ : F (B)(p,q) → F+

(p,q) is a 2–1 map.

Recall εi = [i] = (i −i). We write ((i j)) := ((i −j)). It is easy to see that fori, j ∈ {±1, . . . ,±(p + q)}, we have

[i j] = ((i j))εi = εj((i j)) = εi((i j)) = ((i j))εj . (11)

We now consider graphs on {1, 2, . . . , p+ q} with multiple edges or loops allowed. Fora transposition κ = ((i j)) we define e(κ) to be an edge connecting |i| and |j|. We alsodefine e(εi) to be a loop at i.

Lemma 11. Suppose τ = (τ1, τ2, . . . , τp+q) ∈ F (B)(p,q). Let τk = ((u v)) be the last connected

transposition in τ , i.e., k is the largest integer for which τk is connected. Then we havethe following.

(1) The vertices |u| and |v| are connected in the graph with edges e(τ1), e(τ2), . . . , e(τk−1).(2) The graph with edges e(τ1), . . . , e(τk−1), e(τk+1), . . . , e(τp+q) is a tree.(3) Every transposition τi is of the form ((a b)) rather than [a] = (a,−a).(4) τ1 · · · τk · · · τp+q = ((a1 . . . ap))((ap+1 . . . ap+q)), where ai is either i or −i for each

i = 1, 2, . . . , p + q.

Proof. Proof of Item (1): Suppose that |u| and |v| are not connected in the graph withedges e(τ1), e(τ2), . . . , e(τk−1). Then τ+

1 . . . τ+k has a connected cycle which contains both

|u| and |v|. Since all of τk+1, . . . , τp+q are disconnected, τ+1 . . . τ+

p+q = βp,q has a connectedcycle, which is a contradiction.

Proof of Items (2) and (3): Since 〈|τ1|, . . . , |τp+q|〉 acts transitively on Sp+q, the graphon {1, 2, . . . , p + q} with edges e(τ1), . . . , e(τp+q) is connected. By Item (1), the graphis still connected if we remove the edge e(τk). Since the resulting graph is a connectedgraph with p+ q vertices and p+ q− 1 edges and the resulting graph is a tree. This alsoimplies Item (3) because none of the edges is a loop.

Proof of Item (4): Since τ+1 . . . τ+

p+q = βp,q, we have that τ1 · · · τk · · · τp+q is equalto either ((a1 . . . ap))((ap+1 . . . ap+q)) or [a1 . . . ap][ap+1 . . . ap+q]. Thus it is sufficient toshow that τ1 · · · τk · · · τp+q has no zero cycles. Notice that τ1 · · · τk · · · τp+q has zero cycles.Suppose that τ1 . . . τi−1 has no zero cycles and τi = ((a b)). Then we have three cases.Case 1: a, b,−a,−b are contained in four distinct cycles of τ1 . . . τi−1. Case 2: a and b arecontained in the same cycle of τ1 . . . τi−1. Case 3: a and −b are contained in the same

J.S. Kim et al. / Journal of Combinatorial Theory, Series A 124 (2014) 251–262 259

cycle of τ1 . . . τi−1. Among these three cases, τ1 . . . τi has zero cycles in Case 3 only. ByItems (1) and (2), Case 2 or Case 3 can only happen when i = k. Since τ1 · · · τk · · · τp+q

has zero cycles we must have Case 3 when i = k. Then τ1 . . . τk has zero cycles butτ1 . . . τk−1τk does not. Repeating the same argument, we get that τ1 · · · τk · · · τj has nozero cycles for j = k + 1, k + 2, . . . , p + q, which finishes the proof of Item (4). �

Let (τ1, τ2, . . . , τp+q) ∈ F (B)(p,q). By Items (2) and (3) in Lemma 11 and the fact

τ+1 . . . τ+

p+q = γ+p,q = βp,q, we have (τ1, τ2, . . . , τp+q)+ ∈ F+

(p,q). Thus (·)+ is a well-definedmap from F (B)

(p,q) to F+(p,q). We show that this map is surjective.

Lemma 12. The map (·)+ : F (B)(p,q) → F+

(p,q) is surjective.

Proof. Suppose (σ1, σ2, . . . , σp+q) ∈ F+(p,q). Since σ1σ2 . . . σp+q = βp,q and γp,q =

βp,qε1εp+1, we have

γp,q = σ1σ2 . . . σp+qε1εp+1. (12)

By (11), for given u and v, if � is the largest integer with σ� = ((u v)), we have

σ1σ2 . . . σp+qεu = σ1 · · ·σ�−1σ�σ̃�+1 . . . σ̃p+qεv, (13)

where

σ̃i ={σi if σi has either u or v,

σi otherwise.

Since 〈|σ1|, |σ2|, . . . , |σp+q|〉 acts transitively on Sp+q, we can find integers 1 =a0, a1, . . . , ak = p + 1 such that ((ai−1, ai)) ∈ {σ1, σ2, . . . , σp+q} for all 1 � i � k.Using this fact and the relation in (13), we can rewrite (12) as

γp,q = (τ1τ2 . . . τp+qεp+1)εp+1 = τ1τ2 . . . τp+q,

where τi = σi or σi for all i = 1, 2, . . . , p + q. Hence, (τ1, τ2, . . . , τp+q)+ =(σ1, σ2, . . . , σp+q) and (·)+ is surjective. �

To complete the proof of Proposition 10 it remains to show that (·)+ is two-to-one.This follows from the next lemma.

Lemma 13. Let τ = (τ1, τ2, . . . , τp+q) and ρ = (ρ1, ρ2, . . . , ρp+q) be distinct elements inF (B)

(p,q) satisfying τ+ = ρ+. Then, for each i = 1, 2, . . . , p + q, we have ρi = τi if τi isconnected, and ρi = τi otherwise.

260 J.S. Kim et al. / Journal of Combinatorial Theory, Series A 124 (2014) 251–262

Proof. Let τk be the last connected transposition in τ . Since τ+ = ρ+, we have ρi = τior ρi = τi for all i ∈ {1, 2, . . . , p+ q} and ρk is also the last connected transposition in ρ.

For 1 � i � p+ q and η = (η1, η2, . . . , ηp+q) ∈ F (B)(p,q), define Ti(η) = (η′1, η′2, . . . , η′p+q),

where η′j = εiηjεi. Equivalently, η′j = ηj if ηj contains i, and η′j = ηj otherwise. Clearly,Ti(η) is a factorization of εiγp+qεi.

We claim that there exists a subset S = {s1 < s2 < · · · < sr} ⊆ {2, 3, . . . , p + q}such that η = (η1, η2, . . . , ηp+q) := (Ts1 ◦ Ts2 ◦ · · · ◦ Tsr)(τ) satisfies ηi = ρi for eachi ∈ {1, 2, . . . , p + q} \ {k}. We construct such a subset S as follows. Consider the graphG with vertex set {1, 2, . . . , p+q} and edge set {e(τ1), . . . , e(τk−1), e(τk+1), . . . , e(τp+q)}.By Lemma 11, G is a tree. For given j ∈ {2, 3, . . . , p + q}, we determine whether S

contains j in the following way. Let e(τi1), e(τi2), . . . , e(τit) be the edges of the uniquepath between 1 and j in G. Then j ∈ S if and only if there are odd number of integers� ∈ {1, 2, . . . , t} satisfying τi� = ρi� . We now show that the set S constructed in this waysatisfies the condition of the claim. Let i ∈ {1, 2, . . . , p + q} \ {k}. Suppose that e(τ) isthe edge connecting a and b. Then τi = ((c d)), where c ∈ {a,−a} and d ∈ {b,−b}. Bythe construction of S, if τi = ρi, then S contains both a and b, or S contains neithera nor b. If τi = ρi, then S contains only one of a and b. Thus in either case we haveηi = ρi. This implies the claim.

Now we can assume that η = (η1, η2, . . . , ηp+q) := (Ts1 ◦ Ts2 ◦ · · · ◦ Tsr)(τ) satisfiesηi = ρi for each i ∈ {1, 2, . . . , p+ q}\{k}, where {s1 < s2 < · · · < sr} ⊆ {2, 3, . . . , p+ q}.Since η+ = ρ+, we have ηk = ρk or ηk = ρk. If ηk = ρk, then by Lemma 11, η isa factorization of ((a1 . . . ap))((ap+1 . . . ap+q)), where ai is either i or −i for each i =1, 2, . . . , p + q. On the other hand η is also a factorization of εs1 . . . εsrγp,qεsr . . . εs1 ,which is a contradiction. Thus we must have ηk = ρk. Then we have η = ρ, whichimplies εs1 . . . εsrγp,qεsr . . . εs1 = γp,q. This can only happen when {s1, . . . , sr} = ∅ or{s1, . . . , sr} = {p+1, p+2, . . . , p+q}. Since ρ �= τ we get ρ = (Tp+1◦Tp+2◦· · ·◦Tp+q)(τ).Thus for each i = 1, 2, . . . , p+ q, we have ρi = εp+1 . . . εp+qτiεp+q . . . εp+1. Using the factthat εj((a b))εj = ((a b)) if j /∈ {|a|, |b|} and εj((a b))εj = ((a b)) if j ∈ {|a|, |b|}, weobtain ρi = τi if τi is connected, and ρi = τi otherwise. �Example. Let σ = ( ((3 4)), ((4 5)), ((2 3)), ((2 5)), ((1 2)) ) ∈ F+

(3,2). Since γ3,2 =β3,2 ε1 ε4, we can obtain a factorization of γ3,2 from σ as follows:

γ3,2 = [1 2 3][4 5] = ((3 4)) ((4 5)) ((2 3)) ((2 5)) ((1 2)) ε1 ε4

= ((3 4)) ((4 5)) ((2 3)) ((2 5)) ((1 2)) ε2 ε4

= ((3 4)) ((4 5)) ((2 3)) ((2 5)) ((1 2)) ε3 ε4

= ((3 4)) ((4 5)) ((2 3)) ((2 5)) ((1 2)) ε4 ε4

= ((3 4)) ((4 5)) ((2 3)) ((2 5)) ((1 2)).

J.S. Kim et al. / Journal of Combinatorial Theory, Series A 124 (2014) 251–262 261

Thus τ = ( ((3 4)), ((4 5)), ((2 3)), ((2 5)), ((1 2)) ) ∈ F (B)(3,2) satisfies τ+ = σ. The factor-

ization τ ′ = ( ((3 4)), ((4 5)), ((2 3)), ((2 5)), ((1 2)) ) obtained by toggling the connectedtranspositions of τ also satisfies (τ ′)+ = σ.

4. Final remarks

In this paper we have found a combinatorial proof of (1) for a permutation havingtwo cycles by considering the poset of noncrossing partitions of type B. It is natural toask if our approach can be extended to the case of a permutation having more than twocycles. However, in this case it seems difficult to find a natural poset corresponding tominimal transitive factorizations.

An anonymous referee has informed us that Duchi, Poulalhon, and Schaeffer recentlyfound a bijective proof of (1).

Acknowledgments

The authors would like to thank Christian Krattenthaler for the proof of Lemma 2and the anonymous referees for their valuable comments and suggestions.

For the first author, this research was supported by Basic Science Research Programthrough the National Research Foundation of Korea (NRF) funded by the Ministry of Ed-ucation (NRF-2013R1A1A2061006). For the second author, this research was supportedby Basic Science Research Program through the National Research Foundation of Korea(NRF) funded by the Ministry of Education (2011-0008683). For the third author, this re-search was supported by Basic Science Research Program through the National ResearchFoundation of Korea (NRF) funded by the Ministry of Science, ICT & Future Planning(NRF-2012R1A1A1014154) and INHA UNIVERSITY Research Grant (INHA-44756).

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