A quadruple set-valued equidistribution over permutations

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(1)D EPARTMENT OF M ATHEMATICS N ATIONAL TAIWAN N ORMAL U NIVERSITY. A thesis submitted in partial fulfilment of the requirements for the degree of. M ASTER OF S CIENCE in. M ATHEMATICS. A quadruple set-valued equidistribution over permutations. Author Hsiang-Ren S HIH. Supervisor Dr. Sen-Peng E U. August, 2019.

(2) Abstract In this paper we give a detailed constructive proof of an equidistribution between two quadruples of set-valued statistics. (sort, Cyc, Lmap, Lmal) ∼ (inv, Lmap, Rmil, Rmip) over the set of permutations, where sort, Cyc, Lmap, Lmal stand for the statistics sorting index, cycle set, left to right maximal place set, left to right maximal letter set and inv, Lmap, Rmil, Rmip stand for the statistics inversion, left to right maximal place set, right to left minimum letter set, right to left minimum place set respectively. Our main result will be proved by way of a bijection F : Sn → Sn , which is a composition of four mappings. Key words: Permutations, sorting index, Cycle, Lmap, Lmal, inversion, Rmil, Rmip.

(3) Contents 1. 2. 3. 4. Introduction 1.1 Notation of a permutation . . . . . . . . . 1.2 Transposition array . . . . . . . . . . . . . 1.3 Words . . . . . . . . . . . . . . . . . . . . . 1.4 Statistics and Set valued statistics . . . . . 1.5 Foata-Han and Chen-Gong-Guo’s results 1.6 Our main result . . . . . . . . . . . . . . . Four bijections 2.1 The bijection A 2.2 The bijection R . 2.3 The bijection C . 2.4 The bijection B .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. 1 2 2 3 4 5 6. . . . .. . . . .. . . . .. . . . .. . . . .. . . . .. . . . .. 9 9 11 12 18. Proof of the Main Result 3.1 sort(σ) = inv( F (σ)) . . . . . . . . . . . . . . . . . . 3.2 Cyc(σ) = Lmap( F (σ)) . . . . . . . . . . . . . . . . . 3.3 Lmap(σ) = Rmil( F (σ)) and Lmal(σ ) = Rmip( F (σ)) 3.4 Proof of the main result . . . . . . . . . . . . . . . .. . . . .. . . . .. . . . .. . . . .. . . . .. . . . .. 22 22 23 28 36. . . . .. . . . .. . . . .. . . . .. . . . .. . . . .. . . . .. . . . .. . . . .. Discussion and concluding remarks. Bibliography. . . . .. . . . .. . . . .. . . . .. . . . .. . . . .. . . . .. . . . .. . . . .. . . . .. 37 40.

(4) Chapter 1. Introduction The study of statistics of combinatorial structures is one of the core themes in algebraic and enumerative combinatorics. The reason is that one of the fundamental goal of combinatorics is to understand combinatorial structures, and a result of statistic not only reveals an inner relation but also helps us understand the combinatorial structures more. As the set of permutations is the most fundamental combinatorial structure, there have been numerous important work on the statistics of permutations. The study of statistics of permutations can trace back to Euler (or even further). Many mathematicians made important contributions in this topic. To name a few we have Euler [9], McMahon [19], Stanley [22], Foata [10, 11, 12], Foata and Schützenberger [14, 15, 16], Foata and Han [13], Humpherey [17], Carlitz [3] and many others. One may think that this field permuations statistics seems old and maybe there is nothing new to say. This is very misleading – in fact, the study of permutation statistics is still very active and new important results appear occasionally. For example, the generating function on (maj, exc) is not known until 2007 [21], the concepts of sorting index [20], set valued statistic [13], folding phenomenon [7, 6] are all relatively new. A recent highlight is the connection between polytopal geometry and permutation statistics [1]. Also, there are still many of unsolved problems [2, 5]. In this paper, we will give a detailed constructive proof of an equidistribution between two quadruples of set-valued statistics. (sort, Cyc, Lmap, Lmal) ∼ (inv, Lmap, Rmil, Rmip) over the set of permtuations. The definitions and notations will be given shortly. 1.

(5) 1.1. Notation of a permutation. A permutation on [n] := {1, 2, . . . , n} is a bijection on [n]. We denote by Sn the set of all permutations on [n]. The 2-line notation of a permutation σ ∈ Sn is 1 2 ··· n , σ1 σ2 · · · σn in which σ(i ) = σi . By deleting the upper row we obtain its one-line notation σ = σ1 σ2 . . . σn . A l-cycle C = (c1 , c2 , . . . , cl ) is a permutation with C (ci ) = ci+1 for i = 1, . . . , l − 1 and C (cl ) = c1 . It is well known that we can write a permutation σ ∈ Sn as σ = C1 C2 · · · Ck , a product of disjoint cycles. In this case we say C1 C2 · · · Ck the cycle notation of σ.. 1.2. Transposition array. We need one more notation of a permutation, namely the transposition array, which plays a central role in this thesis. A 2-cycle is called a transposition. It is also known that σ ∈ Sn can be written as σ = ( p1 , 1)( p2 , 2) · · · ( pn , n), a composition of n transpositions. We call p1 p2 . . . pn the transposition array of σ and denote it by TA(σ). Let us look an example. Example 1.2.1. For the permutation in its two line notation 1 2 3 4 5 6 7 8 σ= , 7 8 2 4 6 5 1 3 its one-line notation is 78246513 and cycle notation (1, 7)(2, 8, 3)(4)(5, 6). To represent σ into transposition array we do the following:. 2.

(6) . . ⇒ . ⇒ . ⇒ . ⇒ . ⇒ . ⇒ . ⇒. 1 2 3 4 5 6 7 8 7 8 2 4 6 5 1 3. . 1 2 3 4 5 6 7 8 7 3 2 4 6 5 1 8. . 1 2 3 4 5 6 7 8 1 3 2 4 6 5 7 8. . 1 2 3 4 5 6 7 8 1 3 2 4 5 6 7 8. . 1 2 3 4 5 6 7 8 1 3 2 4 5 6 7 8. . 1 2 3 4 5 6 7 8 1 3 2 4 5 6 7 8. . 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8. . 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8. . (2, 8). (1, 7)(2, 8). (5, 6)(1, 7)(2, 8). (5, 5)(5, 6)(1, 7)(2, 8). (4, 4)(5, 5)(5, 6)(1, 7)(2, 8). (2, 3)(4, 4)(5, 5)(5, 6)(1, 7)(2, 8). (2, 2)(2, 3)(4, 4)(5, 5)(5, 6)(1, 7)(2, 8). (1, 1)(2, 2)(2, 3)(4, 4)(5, 5)(5, 6)(1, 7)(2, 8). Hence TA(σ) = 12245512. Note that only the following two actions are performed on the digits of the 2−line notation when we constuct the transposition array: (i) exchange a with a smaller digit b, in which b is to the right of a. (ii) exchange a with a bigger digit b, in which b is to the left of a. 1.3. Words. We need two more sets in this thesis. Let CCn := { a1 a2 . . . an : 0 ≤ ai ≤ i − 1} and RCCn := { a1 a2 . . . an : 0 ≤ ai ≤ n − i }. 3.

(7) In other words, the first place of an element w ∈ CCn is 0, the second place can be 0, 1, while the third place can be 0, 1 or 2, etc. And the elements of RCCn is the reversal of the elements of CCn . It is clear that. |CCn | = |RCCn | = n!. Example 1.3.1. we have CC3 = {000, 001, 002, 010, 011, 012} and RCC3 = {000, 100, 200, 010, 110, 210}.. 1.4. Statistics and Set valued statistics. Let σ = σ1 σ2 . . . σn = C1 C2 · · · Ck ∈ Sn with the transposition array TA(σ) = p1 p2 . . . pn . We define its inversion stastics by inv(σ) := |{(i, j) ∈ [n] × [n] : i < j and σi > σj }|, its sorting index by n. sort(σ) :=. ∑ ( i − p i ),. i =1. its cycle (set-valued) index by Cyc(σ) := {i ∈ [n] : i is the minimal number of Cj , for some j = 1, 2, . . . , k }. We also define the following four set-valued indices: Lmap(σ) := {i ∈ [n] : σi > σj , for j = 1, 2, . . . , i − 1}, Lmal(σ) := {σi ∈ [n] : σi > σj , for j = 1, 2, . . . , i − 1}, Rmip(σ) := {i ∈ [n] : σi < σj , for j = i + 1, i + 2, . . . , n}, Rmil(σ) := {σi ∈ [n] : σi < σj, for j = i + 1, i + 2, . . . , n}. Example 1.4.1. For the permutation σ = 78246513 = (1, 7)(2, 8, 3)(4)(5, 6) with its transposition array TA(σ) = 12245512, its inversion index is inv(σ) = 20 as there are 20 pairs of (i, j) with σi > σj , namely (1, 2), (1, 3), . . . , (6, 8). The sortin index can be calculated by sort(σ) = (1 − 1) + (2 − 2) + (3 − 2) + (4 − 4) + (5 − 5) + (6 − 5) + (7 − 1) + (8 − 2). = 14. The cycle index is Cyc(σ) = {1, 2, 4, 5}, and Lmap(σ) = {1, 2},Lmal(σ) = {7, 8}, Rmip(σ) = {7, 8} and Rmil(σ) = {1, 3}. 4.

(8) 1.5. Foata-Han and Chen-Gong-Guo’s results. For σ ∈ Sn , its Lehmer code [18], defined by Lehmer, is Leh(σ) = a1 a2 . . . an , where ai = |{ j : 1 ≤ j ≤ i, σj ≤ σi }|, for 1 ≤ i ≤ n. From Lehmer code Foata and Han defined the A-code [13] of σ by A-code(σ) = Leh(σ−1 ). Foata and Han also defined the B-code of a σ ∈ Sn . Let σ = C1 C2 · · · Ck be the cycle notation, the its B-code is B-code(σ) = b1 b2 . . . bn , where bi is determined in the following way. Suppose i ∈ Cj , for some j ∈ {1, . . ., k }, Write Cj as (c ji1 , . . . , c jil , i ), then bi := max{1 ≤ m ≤ l : c jim < i }. Also if c jim < i for 1 ≤ m ≤ l, then bi := i. In other word, bi is to search, to the left of i, the first digit not greater than i. By way of the bijections φ : Sn → Sn and i : Sn → Sn , where φ := (B-code)−1 ◦ A-code and. i ( σ ) = σ −1 ,. Foata and Han [13] proved the following theorem. Theorem 1.5.1. ([13]) We have. (Cyc, Lmap) ∼ (Lmap, Rmil), that is, two pairs of triple statistics (Cyc, Lmap) and (Lmap, Rmil) are equidistributed over Sn . It is extend by Chen, Gong and Guo [4] to the following. Theorem 1.5.2. ([4]) We have. (sort, Cyc, Lmap) ∼ (inv, Lmap, Rmil), that is, two pairs of triple statistics (sort, Cyc, Lmap) and (inv, Lmap, Rmil) are equidistributed over Sn . 5.

(9) Their strategy of proof can be depicted by the following diagram: φ −1. i.  Sn  −→  Sn  −→  Sn  sort inv inv  Cyc   Rmil   Lmap  Lmap Lmap Rmil. 1.6. Our main result. The main result of this paper is to extend the above result even further into an equidistribution of a pair of quadruple set-valued statistics. Theorem 1.6.1 (Main result I). We have. (sort, Cyc, Lmap, Lmal) ∼ (inv, Lmap, Rmil, Rmip) are equidistributed over Sn Example 1.6.2. We give a full table for n = 2, 3 and 4 to show that this result is really nontrivial. σ sort(σ) Cyc(σ) Lmap(σ) Lmal(σ) inv(σ) Lmap(σ) Rmil(σ) Rmip(σ) 12 0 1, 2 1, 2 1, 2 0 1, 2 1, 2 1, 2 21. 1. 1. 1. 2. 1. 1. 1. 2. σ sort(σ) Cyc(σ) Lmap(σ) Lmal(σ) inv(σ) Lmap(σ) Rmil(σ) Rmip(σ) 123 0 1, 2, 3 1, 2, 3 1, 2, 3 0 1, 2, 3 1, 2, 3 1, 2, 3 132. 1. 1, 2. 1, 2. 1, 3. 1. 1, 2. 1, 2. 1, 3. 213. 1. 1, 3. 1, 3. 2, 3. 1. 1, 3. 1, 3. 2, 3. 231. 2. 1. 1, 2. 2, 3. 2. 1, 2. 1. 3. 312. 3. 1. 1. 3. 2. 1. 1, 2. 2, 3. 321. 2. 1, 2. 1. 3. 3. 1. 1. 3. 6.

(10) σ sort(σ) Cyc(σ) Lmap(σ) Lmal(σ) inv(σ) Lmap(σ) Rmil(σ) Rmip(σ ) 1234 0 1, 2, 3, 4 1, 2, 3, 4 1, 2, 3, 4 0 1, 2, 3, 4 1, 2, 3, 4 1, 2, 3, 4 1243. 1. 1, 2, 3. 1, 2, 3. 1, 2, 4. 1. 1, 2, 3. 1, 2, 3. 1, 2, 4. 1324. 1. 1, 2, 4. 1, 2, 4. 1, 3, 4. 1. 1, 2, 4. 1, 2, 4. 1, 3, 4. 1342. 2. 1, 2. 1, 2, 3. 1, 3, 4. 2. 1, 2, 3. 1, 2. 1, 4. 1423. 3. 1, 2. 1, 2. 1, 4. 2. 1, 2. 1, 2, 3. 1, 3, 4. 1432. 2. 1, 2, 3. 1, 2. 1, 4. 3. 1, 2. 1, 2. 1, 4. 2134. 1. 1, 3, 4. 1, 3, 4. 2, 3, 4. 1. 1, 3, 4. 1, 3, 4. 2, 3, 4. 2143. 2. 1, 3. 1, 3. 2, 4. 2. 1, 3. 1, 3. 2, 4. 2314. 2. 1, 4. 1, 2, 4. 2, 3, 4. 2. 1, 2, 4. 1, 4. 3, 4. 2341. 3. 1. 1, 2, 3. 2, 3, 4. 3. 1, 2, 3. 1. 4. 2413. 4. 1. 1, 2. 2, 4. 3. 1, 2. 1, 3. 3, 4. 2431. 3. 1, 3. 1, 2. 2, 4. 4. 1, 2. 1. 4. 3124. 3. 1, 4. 1, 4. 3, 4. 2. 1, 4. 1, 2, 4. 2, 3, 4. 3142. 4. 1. 1, 3. 3, 4. 3. 1, 3. 1, 2. 2, 4. 3214. 2. 1, 2, 4. 1, 4. 3, 4. 3. 1, 4. 1, 4. 3, 4. 3241. 3. 1, 2. 1, 3. 3, 4. 4. 1, 3. 1. 4. 3412. 4. 1, 2. 1, 2. 3, 4. 4. 1, 2. 1, 2. 3, 4. 3421. 5. 1. 1, 2. 3, 4. 5. 1, 2. 1. 4. 4123. 6. 1. 1. 4. 3. 1. 1, 2, 3. 2, 3, 4. 4132. 4. 1, 3. 1. 4. 4. 1. 1, 2. 2, 4. 4213. 5. 1, 2. 1. 4. 4. 1. 1, 3. 3, 4. 4231. 3. 1, 2, 3. 1. 4. 5. 1. 1. 4. 4312. 5. 1. 1. 4. 5. 1. 1, 2. 3, 4. 4321. 4. 1, 2. 1. 4. 6. 1. 1. 4. An exemplary example is in boldface. We can see that. (sort, Cyc, Lmap, Lmal)(3214) = (2, {1, 2, 4}, {1, 4}, {3, 4}),. 7.

(11) which corresponds to. (inv, Lmap, Rmil, Rmip)(2314) = (2, {1, 2, 4}, {1, 4}, {3, 4}). To prove our theorem it suffices to find a bijection F : Sn → Sn such that (sort, Cyc, Lmap, Lmal)(σ) = (inv, Lmap, Rmil, Rmip)( F (σ)). As the table shows, F should send 3214 to 2314, 1234 to 1234, etc. Note that F is not an involution, as 2314 will be sent to 3124 (rather than 3214). During the investigation we found ourselves very much want to know how these statistics vary and what really happens among these permutations. It turns out that we will give a very close investigation and our main result will be proved by the bijection F : Sn → Sn , which is a composition of four mappings A, R, C, and B. The rest of the thesis is organized as follows. In the next section we introduce the four bijections needed for the proof. In Section 3 we prove our main result. Discussion and concluding remarks are put in Section 4.. 8.

(12) Chapter 2. Four bijections As we said in the introduction, we will find a bijection F : Sn → Sn composed of four bijections. In fact, we will see that F = B ◦ C ◦ R ◦ A. In this chapter we will define these bijections A, R, C, B and prove that they are well-defined.. 2.1. The bijection A. Firstly we look at the mapping A. Recall that CCn := { a1 a2 . . . an : 0 ≤ ai ≤ i − 1}. Let A : Sn −→ CCn be defined as follows. For σ ∈ Sn with TA(σ) = p1 p2 . . . pn , define A(σ) := (1 − p1 )(2 − p2 ) . . . (n − pn ). Example 2.1.1. A : S3 −→ CC3 σ 123. TA(σ) 123. 132. 122. 001. 213. 113. 010. 231. 112. 011. 312. 111. 012. 321. 121. 002. 9. A(σ) 000.

(13) It is clear that from the definition of TA(σ ) we have 1 ≤ pi ≤ i, so i − 1 ≥ i − pi ≥ 0 and A(σ) ∈ CCn , hence A is well-defined. Now we prove the mapping A is an injection. Lemma 2.1.2. The mapping A is injective. Proof. Let σ, σ0 ∈ Sn and A(σ) = a1 a2 . . . an 6= a10 a20 . . . a0n = A(σ0 ). From the definition of A we know A(σ) 6= A(σ0 ) implies TA(σ) = p1 p2 . . . pn 6= p10 p20 . . . p0n = TA(σ0 ). Now we define k = max{i ∈ [n] : pi 6= pi0 }. We will prove that in the one-line notations of σ and σ0 , their positions of k are different. Hence σ 6= σ0 and the mapping A is injective. First note that in the one-line notation of τ := ( pk+1 , k + 1)( pk+2 , k + 2) · · · ( pn , n) = ( p0k+1 , k + 1)( p0k+2 , k + 2) · · · ( p0n , n), the positions of 1, 2, . . . , n are of the same. However, ω := ( pk , k)( pk+1 , k + 1)( pk+2 , k + 2) · · · ( pn , n) is to switch pk and k in τ, and ω 0 := ( p0k , k )( p0k+1 , k + 1)( p0k+2 , k + 2) · · · ( p0n , n) is to switch p0k and k in τ. Since pk 6= p0k , the positions of k in ω and ω 0 are different. Note that if we compose by ( pk−1 , k − 1), ( pk−2 , k − 2), . . . , ( p1 , 1) successively to ω, the position of k in the one-line notation will not change. Similary when we compose ( p0k−1 , k − 1), ( p0k−2 , k − 2), . . . , ( p10 , 1) successively to the left of ω 0 , the position of k in its one-line notation will not change either. Hence the positions of k in σ and σ0 are different and we are done. Also, the mapping A is a surjection. Lemma 2.1.3. The mapping A is surjective.. 10.

(14) Proof. For a1 a2 . . . an ∈ CCn , let τ = (1 − a1 )(2 − a2 ) . . . (n − an ) ≡ p1 p2 . . . pn with 1 ≤ pi ≤ i. Let σ = ( p1 , 1)( p2 , 2) · · · ( pn , n) and we have A(σ) = a1 a2 . . . an . Hence the mapping A is onto. Here we use the fact that if 1 ≤ pi ≤ i doesn’t hold, then TA(σ) 6= p1 . . . pn . Proposition 2.1.4. The mapping A is a bijection. Proof. From the Lemmas 2.1.2 and 2.1.3, the mapping A is a bijection. . 2.2. The bijection R. The bijection R is easier. Let R : CCn −→ RCCn defined by R( a1 a2 . . . an ) = an an−1 . . . a1 = b1 b2 . . . bn . That is, we also set bi = an−i+1 . Example 2.2.1. R : CC3 −→ RCC3 σ 000. R(σ) 000. 001. 100. 010. 010. 011. 110. 012. 210. 002. 200. For a1 a2 . . . an ∈ CCn , 0 ≤ ai ≤ i − 1, 0 ≤ bi = an−i+1 ≤ n − i, hence R( a1 a2 . . . an ) ∈ RCCn and R is well-defined. Proposition 2.2.2. The mapping R is a bijection. We first prove R is injective. Let a1 a2 . . . an and a10 a20 . . . a0n ∈ CCn with a1 a2 . . . an 6= a10 a20 . . . a0n . There must exist k ∈ [n] such that ak 6= a0k , hence R( a1 a2 . . . an ) = an an−1 . . . a1 6= a0n a0n−1 . . . a10 = R( a10 a20 . . . a0n ). 11.

(15) This means R is an injection. Now the surjectivity. For a1 a2 . . . an ∈ RCCn , we have 0 ≤ ai ≤ n − i. Hence an an−1 . . . a1 ∈ CCn and R( an an−1 . . . a1 ) = a1 a2 . . . an , which means R is a surjection. Therefore R is a bijection. . 2.3. The bijection C. The bijection C is the most complicated among four. Let C : RCCn −→ CCn be defined by C := C1 ◦ C2 ◦ · · · C n , a composition of n mappings, where Ci is defined by C i ( a 1 a 2 . . . a n ) = a 1 a 2 . . . a i −1 a i +1 a i +2 . . . a i + a i −1 a i + a i a i a i + a i +1 . . . a n . | {z } Note the effect of Ci on the underbraced subword. In other words, Ci moves the i-th position letter ’ai ’ to the right with the distance ai . Example 2.3.1. C : RCC3 −→ CC3 0. 00. 00. σ 000. C3 (σ ) ≡ σ0 000. C2 (σ ) ≡ σ 000. 100. 100. 100. 010. 010. 010. 001. 001. 110. 110. 101. 011. 210. 210. 201. 012. 200. 200. 200. 002. C1 (σ ) = C (σ ) 000. Lemma 2.3.2. The mapping C is well defined. Proof. For a1 a2 . . . an ∈ RCCn we have ai ≤ n − i and there are n − i digits to the right of ai . Hence the action of Ci is at least performable. Now let j ∈ [n]. By the definition of Ci we know that C j+1 ◦ · · · ◦ C n will not affect the position of a j , and not until C j act on this word does the position of a j move to the right with a j positions. Moreover, those letters moving from the right of a j to the left of a j will not change the relative order with a j when we perform C1 ◦ C2 ◦ · · · ◦ C j−1 .. 12.

(16) Since there are at least a j letters to the left of a j , hence the final position of the letter a j will be larger than the number a j . That is, in b1 b2 . . . bn := C ( a1 a2 . . . an ) we will have bi < i, for all i ∈ [n]. Therefore, b1 b2 . . . bn ∈ CCn and C is well-defined.. . To prove C is a bijection we will show that there indeed exists an inverse function D. Define D : CCn −→ RCCn by D : = D n ◦ D n −1 ◦ · · · ◦ D 1 , where Di is defined by D i ( a 1 a 2 . . . a n ) = a 1 a 2 . . . a i − a i −1 a i a i − a i a i − a i +1 . . . a i −1 a i +1 a i +2 . . . a n . | {z } In other words, Di moves the i-th letter ai to the left of distant ai . Example 2.3.3. D : CC3 −→ RCC3 0. 00. 00. σ 000. D1 (σ ) ≡ σ0 000. D2 (σ ) ≡ σ 000. 010. 010. 100. 100. 001. 001. 001. 010. 011. 011. 101. 110. 012. 012. 102. 210. 002. 002. 002. 200. D3 (σ ) = D (σ ) 000. Lemma 2.3.4. The mapping D is well defined. Proof. For a1 a2 . . . an ∈ CCn we have ai ≤ i − 1 and there are i − 1 letters to the left of ai , hence the action of Di make senses. Let j ∈ [n]. From the defition of Di we know D j−1 ◦ · · · ◦ D1 will not affect the position of a j . And not until D j applies on the word does the letter a j move to the left with distance a j . Moreover, those letters moves from the left of a j to the right of a j will not change their relative positions with a j when acted by D n ◦ D n−1 ◦ · · · ◦ D j+1 .. 13.

(17) Since there are at least a j letters to the right of a j , hence the final position of the letter a j will be less than the number n − a j . In other words, in b1 b2 . . . bn := D ( a1 a2 . . . an ) we have bi ≤ n − i, for all i ∈ [n]. Hence b1 b2 . . . bn ∈ RCCn . This proves that D is well-defined. From the definition one may wonder that there are some symmetry between C and D, and they are inverse to each other. Set τ = a1 a2 . . . an ∈ RCCn and suppose D ◦ C (τ ) = b1 b2 . . . bn . Our goal is to prove the following. Proposition 2.3.5. The mappings C and D are inverse to each other. Namely, D ◦ C = idRCCn and C ◦ D = idCCn . In other words, we have a1 . . . an = b1 . . . bn and both C and D are bijections. However this fact is never easy to prove. The bulk of this section is to prove this fact. The stategy of the proof is by Induction by combining the following two lemmas, one for the base case and the other for the inductive step. Lemma 2.3.6. We have bn = an . Proof. Suppose C ( a1 a2 . . . an ) = . . . an ai1 ai2 . . . aik . Note that the effect of C is to move letters to the right, and initially an is at the rightmost position. Hence after the action of C, the letters ai1 , ai2 , . . ., aik to the right of an must been moved before, which means a i1 , a i2 , . . . , a i k > 0 as numbers. Now we look at D (. . . an ai1 ai2 . . . aik ). Since the first n − k − 1 movements only move the letters to the left of an , hence D n−(k+1) ◦ · · · ◦ D1 (C (τ )) = . . . an ai1 ai2 . . . aik . As a1 . . . an ∈ RCCn , an = 0, therefore D n−k ◦ D n−(k+1) ◦ · · · ◦ D1 (C (τ )) = . . . an ai1 ai2 . . . aik . 14.

(18) From previous discussion we have ai1 > 0, hence we have ... (n−k)+1 n−k 1 D ◦D ◦ · · · ◦ D (C (τ )) = a n a i2 . . . a i k . a i1 What matters is that ai1 moves to the left to the an and the actual position of ai1 is not important. Repeat the process on ai2 , . . ., aik result in ... n 1 an . D ◦ · · · ◦ D (C (τ )) = ai1 , ai2 , . . ., aik Therefore in D ◦ C (τ ) we have bn = an and the lemma is proved.. . Now the second lemma for the inductive step. Lemma 2.3.7. Suppose we have bk = ak for k = m + 1, . . . , n. Then bm = am . Proof. For convenients sake, we denote some permutation in the letters { am+1 , . . . , an } by [ am+1 , . . . , an ]. The actural order of the letters is irrelavent. Since the action of C is to move letters to the right, hence C m +1 ◦ · · · ◦ C n ( τ ) = a 1 . . . a m [ a m +1 , . . . , a n ] . This means that the actions of C m+1 , . . . , C n will not affect a1 , . . . , am . Meanwhile in this stage we do not care about how do C m+1 , . . . , C n affect am+1 , . . . , an and we can regard am+1 , . . . , an as a whole. For the action of C m , we separate am+1 , . . . , an into two groups, one group contains those moved from the right of am to its left after the action of C m , the other contains those staying to the right of am , as C m ◦ · · · ◦ C n ( τ ) = a 1 . . . a m −1 ( s 1 , . . . , s a m ) a m ( s a m +1 , . . . , s n − m ) We have { am+1 , . . . , an } = {s1 , . . . , sn−m }, and we know, from the definition of C m , among am+1 , . . . , an there are am letters moved from the right of am (as a letter) to the left of it. Now we look at the action of C1 ◦ · · · ◦ C m−1 and separate a1 , . . . , am−1 into two groups, one contains those moved from the left to the right of am after the action, the other contains the rest, as ! s a m +1 , . . . , s n − m s1 , . . . , s am 1 n 0 0 C ◦ · · · ◦ C (τ ) = am . t1 , . . . , t β t1 , . . . , t α We have. 0. 0. { a 1 , . . . , a m −1 } = { t 1 , . . . , t α , t 1 , . . . , t β } , 15.

(19) 0. 0. in which t1 , . . . , t β are those moved from the left to the right of am , and t1 , . . . , tα are those stay to the left. It must be emphasized that in this step what really matters is am . For the other letters we consider the relative order with am before and after the action of C. With this there are four possibilities. In the following we will apply D on C (τ ) and use induction to explain two facts: (i) Those to the right of am will stay in the right of it after the action D ◦ C. (ii) Those to the left of am will stay in the left of it after the action D ◦ C. These two cases will be proved respectively in the following two lemmas. Lemma 2.3.8. Those to the right of am will stay in the right of it after the action D ◦ C. Proof. We first investigate the result after the action of D am +α ◦ · · · ◦ D1 for those to the left of am (namely, {s1 , . . . , s am , t1 , . . . , tα }) in C (τ ). Suppose that after the action of D am +α ◦ · · · ◦ D1 there exists si ∈ {s1 , . . . , s am } staying to the left of t j ∈ {t1 , . . . , tα }. From the effect of D we know the D n ◦ · · · ◦ D am +α+1 will only move am and those letters in C (τ ) to the right of am . This means the relative order of si and t j is not changed. In fact, it is easy to see si will be still staying to the left of t j in D (C (τ )). Now we let a letter s in D (C (τ )) with s ∈ {s1 , . . . , sn−m } and s be the left most letter in any letter in {s1 , . . . , sn−m }. We know that, in D (C (τ )), to the right of s there are not only n − m − 1 letters. { s1 , . . . , s n − m } \ { s }, but also the letter t j – since t j is to the right of si and s is the leftmost letter among {s1 , . . ., sn−m }, hence t j is to the right of s. Hence we know that there are at least n − m letters to the right of s. This means that in D (C (τ )), s can be at one of the first m positions and this is a contradiction since the letters in {s1 , . . ., sn−m } are to the right of am in τ and after the action of D ◦ C (τ ) they will be back to the initial positions. Hence, after the action of D am +α ◦ · · · ◦ D1 letters in {s1 , . . ., s am } will be to the right of any letters in {t1 , . . ., tα }, as ! s , . . . , s n − m a + 1 m 0 0 . D am +α ◦ · · · ◦ D1 (C (τ )) = [t1 , . . . , tα ][s1 , . . . , s am ] am t1 , . . . , t β. 16.

(20) We therefore apply D am +α+1 on it and obtain D am +α ◦ · · · ◦ D1 (C (τ )) = [t1 , . . . , tα ] am [s1 , . . . , s am ]. !. s a m +1 , . . . , s n − m 0 0 t1 , . . . , t β. .. Finally we consider the letters s am +1 , . . . , sn−m . Again, suppose that after the action of D there exists si ∈ {s am +1 , . . . , s an−m } to the left of am . Similarly to above, we let s ∈ {s1 , . . . , sn−m } to be the letter in D (C (τ )) and be the leftmost letter among {s1 , . . ., sn−m }. We know that in D (C (τ )) to the right of s there are not only n − m − 1 letters {s1 , . . . , sn−m } \ {s} but also am – since am is to the right of si and s is the left most letter among {s1 , . . . , sn−m } and therefore am is to the right of s. Hence there are at least n − m letters to the right of s, which means that in D (C (τ )) the letter s can only occupy one of the first m positions, a contradiction from the fact that since letters in {s1 , . . . , sn−m } are to the right of am in τ and after the action of D ◦ C they should be back to the places. Therefore we reach the conclusion that after the action of D, letters in. { s a m +1 , . . . , s a n − m } will stay to the right of am .. . Lemma 2.3.9. Those to the left of am will stay in the left of it after the action D ◦ C. 0. 0. Proof. We suppose t1 , . . . , t β are to the left of am in C (τ ). In the following 0. 0. 0. we focus on t1 . The discussion of t2 , . . . , t β is similar. Suppose in C (τ ), among elements {s am +1 , . . . , s an−m } there are j1 letters 0 to the left of t1 , say s am +1 , s am +2 , . . . , s am + j1 . From the discussion of the previous lemma we know that after the action of D the letters s am +1 , s am +2 , . . . , s am + j1 are still to the right of am . We look at 0 the instance right before D acts on t1 , which will be 0 s s1 , . . ., s am am + j1 +1 , . . . , sn−m am +α+1+ j1 1 0 0 t1 D ◦ · · · ◦ D (C (τ )) = [t1 , . . . , tα ] am sa +1 , . . ., sa + j . t ,...,t m. m. 1. 2. β. Since we suppose that in C (τ ) the letters s am +1 , s am +2 , . . . , s am + j1 are to the 0 left of t1 , hence this shows the instant that the movements of s am +1 , s am +2 , . . ., s am + j1 via the action of D are already done. 17.

(21) 0. Now we act by D to move t1 . We need to see how the effect of C on 0 0 0 t1 in C (τ ). Note that t1 is to the left of am in τ, hence t1 moves later than am , s1 , . . . , sn−m when we perform C (τ ). We infer that when C acts on τ, 0 when acting on t1 , its right-hand side element am , s1 , . . . , s am , s am +1 , . . . , s am + j1 0 will move to the left hand side of t1 . This means as a number 0. t1 ≥ 1 + am + j1 . 0. Hence when applied by D, t1 will move to the left of am , namely D am +α+1+ j1 +1 ◦ · · · ◦ D1 (C (τ )) =. . t1 , . . ., tα 0 t1. 0. . am. . s1 , . . ., s am s am +1 , . . ., s am + j1. . s am + j1 +1 , . . . , sn−m 0 0 t2 , . . . , t β. 0. Now we know t1 is moved to the left of am . For t2 , . . . the arguments are 0 0 similar and we reach to the result that in D ◦ C (τ ), letters t1 , . . . , t β will move to the left of am . Proof of Lemma 2.3.7 The proof of Lemma 2.3.7 is done by combining the above two lemmas, namely ! t1 , . . ., tα s1 , . . ., s am 0 0 D (C (τ )) = am . t1 , . . ., t β s am +1 , . . ., sn−m and we have am = bm .. . Finally we can reach our goal. Proof of Proposition 2.3.5. By apply the above lemmas and induction on k we prove D ◦ C is an identity mapping on RCCn . The fact that C ◦ D is an identity mapping on CCn is done similarly and both C and D are bijections. . 2.4. The bijection B. The last ingredient is the function B. Let B : CCn −→ Sn defined by B( a1 a2 . . . an ) = B1 ◦ B2 ◦ · · · ◦ Bn (123 . . . n), where Bi is defined by Bi (σ1 . . . σi−1 σi σi+1 . . . σn ) = σ1 . . . σi−1 σi+1 . . . σi+an−i+1 σi σi+an−i+1 +1 . . . σn .. 18. ..

(22) In other words, initially we start from 123 . . . n. The action Bn moves n to the right with a1 steps , then Bn−1 moves n − 1 to the right with a2 steps and so on. Generally Bk moves k to the right with an−k+1 steps. Example 2.4.1. B : CC3 −→ S3 0. 0. σ 000. B3 (123) ≡ τ 123. B2 ( τ ) ≡ τ 123. B1 ( τ ) = B ( σ ) 123. 010. 123. 132. 132. 001. 123. 123. 213. 011. 123. 132. 312. 012. 123. 132. 321. 002. 123. 123. 231. Lemma 2.4.2. The mapping B is well defined. Proof. For any i ∈ [n] each of Bi+1 , Bi+2 , . . . , Bn is to move letters to the right of i further to the right. That is, not until the action of Bi , the letter i will not move. Meanwhile, at the moment i is acted by Bi the letters to the right of i are letters i + 1, i + 2, . . . , n and there are n − i of them. We know Bi moves i to the right with an−i+1 step, and an−i+1 is the (n − i + 1)-th letter of some word in CCn . Hence from the definition of CCn we know 0 ≤ an−i+1 ≤ (n − i + 1) − 1 = n − i, which means that for the letter i (to its right there are n − i letters) the action of Bi on it will not move it to the right for more than n − i steps. As B is a rearrangment of [n], we have B( a1 a2 . . . an ) ∈ Sn and B is well-defined. Lemma 2.4.3. The mapping B is injective. Proof. For i ∈ [n], at the moment that i is acted by Bi , the letters to the right of i will be i + 1, i + 2, . . . , n. When i moves an−i+1 steps to the right, it crosses letters larger than i. Also the action B1 ◦ · · · ◦ Bi−1 will not affect letters greater or equal to i. In other words, after the action of Bi , the relative position between i and those greater than i will not change. Hence we know in B( a1 . . . an ) to the left of i there will be an−i+1 letters greater than i. 19.

(23) Now suppose a1 a2 . . . an , b1 b2 . . . bn ∈ CCn , and a1 . . . an 6= b1 . . . bn . If an−i+1 6= bn−i+1 for some i ∈ [n], then in B( a1 . . . an ) there will be an−i+1 letters larger than i to the left of i. However in B(b1 . . . bn ) there will be bn−i+1 letters larger than i to the left of i. As an−i+1 6= bn−i+1 we know B( a1 . . . an ) 6= B(b1 . . . bn ) and hence B is an injection.. . Lemma 2.4.4. The mapping B is surjective. Proof. For σ = σ1 . . . σn ∈ Sn , let an−i+1 is the number of letters to the left of i and larger than i in σ. Since there are n − i numbers in [n] greater than i, hence 0 ≤ an−i+1 ≤ n − i, that is 0 ≤ ai ≤ i − 1 for all i. Hence we have a1 a2 . . . an ∈ CCn . Also, from the arguments above we know that in B( a1 . . . an ) the number of letters to the left of i and greater than i will be an−i+1 for all i. Hence B ( a1 . . . a n ) = σ and B is a surjection.. . By combining the above two lemmas, we have the following. Proposition 2.4.5. The mapping B is a bijection. Now we have all the four bijections A, R, C, B at hand. In the next section we will prove that the composition bijection F := B ◦ C ◦ R ◦ A is exactly what we need for proving our equidistribution.. 20.

(24) Example 2.4.6. F : S4 −→ S4 σ 1234. TA(σ) 1234. A(σ) 0000. R ◦ A(σ) 0000. C ◦ R ◦ A(σ) 0000. F (σ) 1234. 1243. 1233. 0001. 1000. 0100. 1243. 1324. 1224. 0010. 0100. 0010. 1324. 1342. 1223. 0011. 1100. 0110. 1423. 1423. 1222. 0012. 2100. 0120. 1432. 1432. 1232. 0002. 2000. 0020. 1342. 2134. 1134. 0100. 0010. 0001. 2134. 2143. 1133. 0101. 1010. 0101. 2143. 2314. 1124. 0110. 0110. 0011. 3124. 2341. 1123. 0111. 1110. 0111. 4123. 2413. 1122. 0112. 2110. 0121. 4132. 2431. 1132. 0102. 2010. 0021. 3142. 3124. 1114. 0120. 0210. 0012. 3214. 3142. 1113. 0121. 1210. 0112. 4213. 3214. 1214. 0020. 0200. 0002. 2314. 3241. 1213. 0021. 1200. 0102. 2413. 3412. 1212. 0022. 2200. 0022. 3412. 3421. 1112. 0122. 2210. 0122. 4312. 4123. 1111. 0123. 3210. 0123. 4321. 4132. 1131. 0103. 3010. 0013. 3241. 4213. 1211. 0023. 3200. 0023. 3421. 4231. 1231. 0003. 3000. 0003. 2341. 4312. 1121. 0113. 3110. 0113. 4231. 4321. 1221. 0013. 3100. 0103. 2431. 21.

(25) Chapter 3. Proof of the Main Result Recall that our goal is to prove that. (sort, Cyc, Lmap, Lmal) ∼ (inv, Lmap, Rmil, Rmip) over the symmetric group Sn . In the previous chapter we introduce four bijections A, R, C, B. In this section we will consider the bijection F : Sn → Sn defined by F := B ◦ C ◦ R ◦ A and show that F has the desired property. Namely, we will prove that for σ ∈ Sn the following properties hold. (i) sort(σ) = inv( F (σ)), (ii) Cyc(σ) = Lmap( F (σ)), (iii) Lmap(σ) = Rmil( F (σ)), (iv) Lmal(σ) = Rmip( F (σ)). The first item will be in Section 3.1. The second item will be treated in Section 3.2, and the final two are in Section 3.3. After these our main result is proved.. 3.1. sort(σ ) = inv( F (σ)). First we prove is the following. Theorem 3.1.1. We have sort(σ) = inv( F (σ)). 22.

(26) Proof. Given σ ∈ Sn , set TA(σ) = p1 p2 . . . pn . From the definition of A we have A(σ) = (1 − p1 )(2 − p2 ) . . . (n − pn ) ≡ a1 a2 . . . an . And here sort(σ) = ∑in=1 (i − pi ) = ∑in=1 ai . We look at the action of C ◦ R on A(σ). Let C ◦ R ◦ A(σ) = b1 b2 . . . bn , since the effects of R : CCn −→ RCCn and C : RCCn −→ CCn are to change the positions of letters rather than their magnitude, hence ∑in=1 ai = ∑in=1 bi . Now we let F (σ) = B(C ◦ R ◦ A(σ)) = B(b1 b2 . . . bn ) = τ = τ1 τ2 . . . τn and consider inv(τ ) := inv( F (σ)). First we can write n. inv(τ ) = |{(i, j) : i < j, τi > τj }| =. ∑ |{i : i < j, τi > τj }|.. j =1. From the property of the mapping B we know that for any j ∈ [n], there are bn− j+1 letters to the left of j and greater than j. Hence n. ∑ |{i : i < j, τi > τj }| =. j =1. n. ∑ bn − j + 1 =. j =1. n. ∑ bj .. j =1. Again since R and C are just rearrangements of letters, hence n. ∑ bj =. j =1. n. ∑ a j = sort(σ). j =1. and inv( F (σ)) = sort(σ), as desired.. 3.2. . Cyc(σ ) = Lmap( F (σ )). Our second proposition is the following. Theorem 3.2.1. We have Cyc(σ ) = Lmap( F (σ)). For σ ∈ Sn , we let A(σ) = a1 a2 . . . an . We need the following lemmas to prove this theoerm. Lemma 3.2.2. We have Cyc(σ) = {i ∈ [n] : ai = 0} 23.

(27) Proof. First let us see how a transposition (i, j) affects the cycle notation σ = C1 C2 · · · Ck . What we need here is the case that i, j are not in the same cycle. The following fact is easy and we contain it here for completeness. Lemma 3.2.3. Suppose i, j are not in the same cycle of the σ = C1 C2 · · · Ck . Then the action of (i, j) on σ will combine two cycles which i, j respectively belongs to into one cycle. Proof. Since C1 , . . . , Ck are disjoint cycles, without loss of generality we can suppose i is in the cycle C1 = (i, x1 , . . . , xs ) and j is in C2 = ( j, y1 , . . . , yt ). A simple calculation shows that. (i, j) ◦ C1 · · · Ck = (i, x1 , . . . , xs , j, y1 , . . . , yt )C3 C4 · · · Ck . . and we are done.. We continue the proof of Lemma. Write σ = ( p1 , 1)( p2 , 2) · · · ( pn , n), 0 ≤ pi ≤ i, for all i ∈ [n] and we regard each ( pi , i ) as a transposition and discuss how this chain of composition acting on the identity (1)(2) · · · (n). For ( pn , n) ◦ (1) · · · (n) we have two cases. (i) Case 1, if pn = n then nothing happens. hence ( pn , n) ◦ (1) · · · (n) = (1) · · · (n). Since pi ≤ i < n, for i = 1, . . . , n − 1, hence ( p1 , 1), . . . , ( pn−1 , n − 1) will not merge the cicle (n) with any other circle. This means in the cycle notation of ( p1 , 1) · · · ( pn , n) the smallest number in the cycle containing n is n itself, hence n ∈ Cyc(σ). (ii) Case 2, if pn < n. Now (n) will merge with the cycle in which pn belongs to. We will explain in the following that n will continue to stay in the same cycle with pn . As pn < n hence n ∈ / Cyc(σ). Now we apply ( pn−1 , n − 1). First we note that no matter ( pn , n) is in which case above, 1, . . . , n − 1 are the smallest letter in their own cycles. Again we consider two cases: (i) if pn−1 = n − 1 then nothing happens. Since pi ≤ i < n − 1, for i = 1, . . . , n − 2, the transposition ( p1 , 1), . . . , ( pn−2 , n − 2) will not merge the circle containing n − 1 with any other cycle. Hence n − 1 ∈ Cyc(σ). (ii) pn−1 < n − 1. As 1, . . . , n − 1 are the smallest number in their own cycles, n − 1 and pn−1 are in different cycles and ( pn−1 , n − 1) will merge these cycles into one cycle in which the smallest element is pn−1 . Later we will continue to explain that n − 1 and pn−1 will remain in the same cycle till the end. Hence n − 1 ∈ / Cyc(σ). We continue the same arguments for ( pn−2 , n − 2), . . ., ( p1 , 1). Generally, when we apply ( pi , i ), we first note that no matter how does ( pi+1 , i + 24.

(28) 1) ◦ · · · ◦ ( pn , n) do, 1, . . . , i are the smallest letter in their own cycles. We look at two cases similarly: (i) if pi = i then nothing happens and the by the same argument i ∈ Cyc(σ). (ii) if pi < i, then ( pi , i ) merge the cycles containing pi and i respectively into one, in which the smallest letter is pi . These two numbers remain in the same cycle till the end and i ∈ / Cyc(σ). Hence, for i ∈ [n], the effect of the action by ( pi , i ) is either staying invariant or merging two cycles. Especially it won’t separate any two letters already in the same cycle. Therefore, given i ∈ [n], (i) if pi = i, since ( p1 , 1) ◦ · · · ◦ ( pi−1 , i − 1) will not affect the cycle containing i, hence i ∈ Cyc(σ). (ii) if pi 6= i, then i will get into the circle containing pi and will remain so till the end. As pi < i we have i ∈ / Cyc(σ). Hence Cyc(σ) = {i ∈ [n] : pi = i }, where p1 . . . pn = TA(σ). From the definition of A we may set A(σ) = a1 . . . an , then Cyc(σ ) = {i ∈ [n] : pi − i = 0}. = { i ∈ [ n ] : a i = 0} . and the lemma is proved. Example 3.2.4. A(σ) = 0101134, TA(σ) = 1133433. h3, 7i h3, 6ih3, 7i h4, 5ih3, 6ih3, 7i h3, 4ih4, 5ih3, 6ih3, 7i h3, 3ih3, 4ih4, 5ih3, 6ih3, 7i h1, 2ih3, 3ih3, 4ih4, 5ih3, 6ih3, 7i h1, 1ih1, 2ih3, 3ih3, 4ih4, 5ih3, 6ih3, 7i. = (1)(2)(37)(4)(5)(6) = (1)(2)(376)(4)(5) = (1)(2)(376)(45) = (1)(2)(37645) = (1)(2)(37645) = (12)(37645) = (12)(37645) ⇒ Cyc(σ) = {1, 3}. Now let us look at Lmap( F (σ)). Lemma 3.2.5. As F = B ◦ C ◦ R ◦ A, suppose that C ◦ R ◦ A(σ) = b1 b2 . . . bn ∈ CCn and let B(b1 . . . bn ) = τ = τ1 . . . τn . then Lmap(τ ) = {n − k + 1 ∈ [n] : ωk = 0}, where ω = D (b1 . . . bn ). 25.

(29) Proof. Our strategy is as follows. Firstly, from bi we will see which letters be in Lmal(τ ). Secondly we will trace the position of i via the function D. From property of B we know after its action there are bn−i+1 numbers greater than i and to the left of i. Hence i ∈ Lmal(τ ) if and only if bn−i+1 = 0. We need to know where does B send these i’s. We need to see the action of B = B1 ◦ · · · ◦ Bn on 123 . . . n, as well as D = D n ◦ · · · ◦ D1 on b1 . . . bn . Note first that if i ∈ [n] is at the j-th place of 123 . . . n from the left, then bn−i+1 is at the j-th place of b1 . . . bn from the right. Now let us look at the first step. Bn will move n of 123 . . . n to the right by distance b1 , and D1 will move b1 to the left by distance b1 . Since the distance moved are of the same, hence the following property hold: For any i ∈ [n], if i is at the j-th place from the left in Bn (123 . . . n), Then bn−i+1 is at the j-th placce from the right in D1 (b1 . . . bn ). We generalize the above observation. From k = n − 1 to k = 1, the map Bk will move k of Bk+1 ◦ · · · ◦ Bn (123 . . . n) to the right by distance bn−k+1 , and D n−k+1 moves bn−k+1 to the left by distance bn−k+1 . Then, for any i ∈ [n], if i is at the j-th place from the left in Bk ◦ · · · ◦ Bn (123 . . . n), then bn−i+1 is at the j-th place from the right in D n−k+1 ◦ · · · ◦ D1 (b1 . . . bn ). Therefore after the action of B on 123 . . . n and that of D on b1 . . . bn , we know that for any i ∈ [n], if it is at the j-th place from the left in B(123 . . . n), then bn−i+1 is at the j-th place from the right in D (b1 . . . bn ). Now we can complete our proof. Since i ∈ Lmal(σ) iff bn−i+1 = 0, hence Lmap( B(b1 . . . bn )) = {k ∈ [n] : τk ∈ Lmal(τ )}. = {k ∈ [n] : bn−τk +1 = 0}. We know that τk is at the k-th place from the left in B(b1 . . . bn ) = τ, hence bn−τk +1 is at the k-th place from the right in D (b1 . . . bn ), or (n − k + 1)-th place from the left. Let D (b1 . . . bn ) = ω = ω1 . . . ωn , we then have bn−τk +1 = ωn−k+1 , and. {k ∈ [n] : bn−τk +1 = 0} = {k ∈ [n] : ωn−k+1 = 0} = { n − k + 1 ∈ [ n ] : ω k = 0} That is, Lmap( B(b1 . . . bn )) = {n − k + 1 ∈ [n] : ωk = 0}, where ω = D (b1 . . . bn ). This ends the proof. 26. .

(30) Example 3.2.6. τ = 00104205 ≡ c8 c7 c6 c5 c4 c3 c2 c1. (00104205) (00104205) (00104205) (00104205) (00104205) (00104205) (00104205) (00104205). B8 (12345678) B7 (12345678) B6 (12345678) B5 (12345768) B4 (12345768) B3 (12357684) B2 (12573684) B1 (12573684). = 12345678 = 12345678 = 12345768 = 12345768 = 12357684 = 12573684 = 12573684 = 25736184. D8 D7 D6 D5 D4 D3 D2 D1. ⇒ 00104205 ⇒ 00104205 ⇒ 01004205 ⇒ 01004205 ⇒ 40100205 ⇒ 40120005 ⇒ 40120005 ⇒ 40512000. ≡ c 8 c7 c6 c5 c4 c3 c2 c1 ≡ c8 c 7 c6 c5 c4 c3 c2 c1 ≡ c8 c 6 c7 c5 c4 c3 c2 c1 ≡ c8 c6 c7 c 5 c4 c3 c2 c1 ≡ c 4 c8 c6 c7 c5 c3 c2 c1 ≡ c4 c8 c6 c 3 c7 c5 c2 c1 ≡ c4 c8 c6 c3 c7 c5 c 2 c1 ≡ c4 c8 c 1 c6 c3 c7 c5 c2. Proof of Theorem 3.2.1. For σ ∈ Sn , let A(σ) = a1 . . . an and from Lemma 3.2.2 we know Cyc(σ) = {i ∈ [n] : ai = 0}. Since R( A(σ)) = an . . . a1 and D ◦ C = idRCCn , we have D ◦ C ◦ R ◦ A(σ) = an . . . a1 . Now let D (C ◦ R ◦ A(σ)) = ω = ω1 . . . ωn and we will have. { i ∈ [ n ] : a i = 0 } = { n − i + 1 ∈ [ n ] : a n − i +1 = 0 } = { n − i + 1 ∈ [ n ] : ωi = 0} Plug C ◦ R ◦ A(σ) into b1 . . . bn into Lmap( B(b1 . . . bn )) = {n − k + 1 ∈ [n] : ωk = 0} from Lemma 3.2.5, we have. {n − i + 1 ∈ [n] : ωi = 0} = Lmap( B(b1 . . . bn )) = Lmap( B(C ◦ R ◦ A(σ))) = Lmap( F (σ)). Hence Cyc(σ) = Lmap( F (σ)). . and the proof is done.. 27.

(31) 3.3. Lmap(σ) = Rmil( F (σ)) and Lmal(σ) = Rmip( F (σ )). In this section we will prove the following. Theorem 3.3.1. We have Lmap(σ) = Rmil( F (σ)) and Lmal(σ) = Rmip( F (σ)). The proof is quite complicated and is feasible by way of TA(σ). More precisely, our strategy is the following and each step is a Lemma. (i) Lmap(σ) ⊆ Rmil( TA(σ)) and Lmal(σ) ⊆ Rmip( TA(σ)). (ii) Rmil( TA(σ)) ⊆ Lmap(σ) and Rmip( TA(σ)) ⊆ Lmal(σ). (iii) Rmil( TA(σ)) ⊆ Rmil( F (σ)) and Rmip( TA(σ)) ⊆ Rmip( F (σ)). (iv) Rmil( F (σ)) ⊆ Rmil( TA(σ)) and Rmip( F (σ)) ⊆ Rmip( TA(σ)). If so, then from first two items we have Lmap(σ) = Rmil( TA(σ)),. Lmal(σ) = Rmip( TA(σ)),. while from the last two items we have Rmil( TA(σ)) = Rmil( F (σ)),. Rmip( TA(σ)) = Rmip( F (σ)). . And the proof is completed.. For convienient’s sake in the following write TA(σ) = p1 p2 . . . pn in 2-line notation 1 2 ... n TA(σ) = . p1 p2 . . . p n Note that the 2-line notation are also applied on words in CCn and RCCn . We also define Rmip, Rmil of TA(σ) by Rmip( TA(σ)) := {i ∈ [n] : pi < p j , for j = i + 1, . . . , n}, Rmil( TA(σ)) := { pi : pi < p j , for j = i + 1, . . . , n}. Lemma 3.3.2. We have Lmap(σ) ⊆ Rmil( TA(σ)),. Lmal(σ) ⊆ Rmip( TA(σ)).. 28.

(32) Proof. For σ = σ1 . . . σn ∈ Sn we let TA(σ) = p1 . . . pn . Suppose i ∈ Lmap(σ) for some i ∈ [n] and meanwhile we have σi ∈ Lmal(σ), that is 1 ... i−1 i i+1 ... n σ= σi ... (< σi ) For σi 6= n, the calculation of TA(σ) is by way of 2-line notation of σ: starting from n, consider the position of n is pn and exchange n with the letter in the n-th place. We can successively perform similar operation with respect to n − 1, n − 2, . . . 1. Recall that when doing the exchange we are to exchange some letter a with some smaller letter to the right, or exchange a with some larger letter to the left. That is, when we compute pn , since σ1 , . . . , σi−1 are smaller than σi , the position of n is greater than i and therefore pn > i. Also n will be exchanged with some smaller letter b to its right. The position of b will still be larger than i and the exchange will not change the facts that (i) σi is at the i-th position and (ii) letters to the left of σ(i ) are all smaller than σ(i ). Recurrently we have that pn , pn−1 , . . . , pσi +1 are all greater than i; and when computing pσi the letter σi remains at the i-th position. Therefore pσi = i and ... σi σi + 1 . . . n TA(σ) = ... i (> i ) Therefore i ∈ Rmil( TA(σ)) and σi ∈ Rmip( TA(σ)), that is Lmap(σ) ⊆ Rmil( TA(σ)),. Lmal(σ) ⊆ Rmip( TA(σ)). For σi = n, we have pn = i, that is ... TA(σ) = .... σi i. . Therefore i ∈ Rmil( TA(σ)) and σi ∈ Rmip( TA(σ)), that is Lmap(σ) ⊆ Rmil( TA(σ)),. Lmal(σ) ⊆ Rmip( TA(σ)). . and the proof is completed. Now we look at the other direction of inclusions. Lemma 3.3.3. We have Rmil( TA(σ)) ⊆ Lmap(σ),. Rmip( TA(σ)) ⊆ Lmal(σ). 29.

(33) Proof. For TA(σ) = p1 . . . pn , suppose i ∈ Rmip(σ) for some i ∈ [n] and pi ∈ Rmil(σ). That is, ... i i+1 ... n TA(σ) = ... pi (> pi ) We reverse the steps of computing TA(σ) and compute σ from TA(σ). Starting from 123 . . . n. As 1 is to move to p1 -th position and since 1 ≤ pi ≤ i, hence the letter 1 (which is at the first place) didn’t move rightward and the letters to the right of 1 remain in their places. Similarly, letters i, . . . , n remain intact until move i − 1 to pi−1 -th position. The next step will move i leftword to pi -th position and here those letters greater than i will lie to the right of i and those to the left of i are smaller than i. Now we move i + 1 to the pi+1 -th place. Since pi+1 > pi , we know that i + 1 exchanges with some letter to the right of i, hence those to the left of i are still smaller than i. Similarly when we move i + 2, . . . , n, those to the left of i keep smaller than i. Therefore in the end we have   ... pi pi + 1 . . . n  ... σ= (< i ) i i+1...n and i ∈ Lmal(σ), pi ∈ Lmap(σ).. . Now we prove the third lemma. Lemma 3.3.4. We have Rmil( TA(σ)) ⊆ Rmil( F (σ)),. Rmip( TA(σ)) ⊆ Rmip( F (σ)). Proof. Since 1 ≤ p j ≤ j, we suppose i + x ∈ Rmip( TA(σ)) and pi+ x = i ∈ Rmil( TA(σ)) for some x ≥ 0. That is ... i+x i+x+1 ... n TA(σ) = . ... i ( >i ) We let A(σ) = a1 . . . an . Since ai = i − pi for i ∈ [n] and p j > i ⇒ a j = j − p j < j − i for j = i + x + 1, . . . , n, we have A(σ) =. ... i+x i+x+1 i+x+2 ... n ... x < x+1 < x+2 ... < n−i 30. ..

(34) As R ◦ A(σ) = an . . . a1 , we have 1 . . . n − ( i + x + 1) + 1 n − ( i + x ) + 1 . . . R ◦ A(σ) = . < n−i ... < x+1 x ... Note that C n−(i+ x)+2 ◦ · · · ◦ C n is to move rightward those letters to the right of x (which is at the (n − (i + x ) + 1)-th position) and it will not affect letters at the 1, . . . , n − (i + x ) + 1 positions. Hence we have C n−(i+ x)+2 ◦ · · · ◦ C n ( R ◦ A(σ)) = 1 . . . n − ( i + x + 1) + 1 n − ( i + x ) + 1 . . . . < n−i ... < x+1 x ... Similarly note that C n−(i+ x)+1 moves x (at the (n − (i + x ) + 1)-th position) to the right by x and leaves it at the n − (i + x ) + 1 + x = (n − i + 1)-th position. That is C n−(i+ x)+1 ◦ · · · ◦ C n ( R ◦ A(σ)) = 1 . . . n − ( i + x + 1) + 1 . . . n − i + 1 . . . . < n−i ... < x+1 ... x ... Again, C n−(i+ x) moves rightward the letter at the (n − (i + x ))-th position. But as the distance it moves is smaller than x + 1 it will not affect x and x remains intact. Again C n−(i+ x)−1 moves rightward the letter at the (n − (i + x ) − 1)-th position by a distance less than x + 2. The x is still at the (n − i + 1)-th position and C n−(i+ x)−1 will not allow any letter to move to the right of x. Successively by the same argument we obtain that, those letters to the left of x (which is at the n − (i + x ) + 1-th position in R ◦ A(σ)) will not move to the right of x (which is moved by C to the (n − i + 1)-th position). Hence, in C ◦ R ◦ A(σ) the (n − i + 1)-th letter is exactly x and we have ... n−i+1 ... C ( R ◦ A(σ)) = ... x ... Moreover, we will see those letters to the right of x (which itself is at the (n − i + 1)-th position of C ◦ R ◦ A(σ)). We use function D. First, D n−i ◦ · · · ◦ D1 is to move leftward those letters to the left of x (at the (n − i + 1)-th position) and it will not affect x and those letters to its right. That is, 31.

(35) D. n −i. . 1. ◦ · · · ◦ D (C ◦ R ◦ A(σ)) =. ... n−i+1 ... ... x .... .. Now, D n−i+1 moves the (n − i + 1)-position letter x to the left with distance x, and x is now at the position n − (i + x ) + 1. This is reasonable since we suppose i + x ∈ Rmip( TA(σ)) and therfore i + x ∈ [n] and n − (i + x ) + 1 > 0. Currently we have D n−i+1 ◦ · · · ◦ D1 (C ◦ R ◦ A(σ)) = . . . n − (i + x ) + 1 . . . n − i + 2 . . . n . ... x ... ( ? ) Hence this x is back to its position in R ◦ A(σ). Now we claim that the (n − i + 1 + j)-th position letter in C ◦ R ◦ A(σ) is smaller than x + j. Suppose not, then this letter will move from the right of x to the left of x after the action of D on C ◦ R ◦ A(σ). This will make x not be at the (n − (i + x ) + 1)-th position of D (C ◦ R ◦ A(σ)). But by the previous lemma we have D ◦ C = idRCCn and D (C ◦ R ◦ A(σ)) = R ◦ A(σ), which means x is at the (n − (i + x ) + 1)-th position, a contradiction. Hence we know that in C ◦ R ◦ A(σ), the letter at the (n − i + 1 + j)-th position will be less than x + j . That is, ... n−i+1 n−i+2 n−i+3 ... n C ( R ◦ A(σ)) = . ... x < x+1 < x+2 ... < x+i−1 Now we investigate the effect of B on 123 . . . n. Bn−(n−i)+1 ◦ · · · ◦ Bn moves i + 1, . . . , n to the right and does not affect the positions of i and those letters to the left of i, namely 1 ... i i+1 ... n i +1 n B ◦ · · · ◦ B (123 . . . n) = . 1 ... i ... Note that Bi is to move i to the right with distance x to the (i + x )-th position, that is 1 ... i−1 ... i+x ... Bi ◦ · · · ◦ Bn (123 . . . n) = . 1 ... i−1 ... i ... Similarly Bi−1 moves i − 1 to the right by distance < x + 1, hence i − 1 will not move to the right of i, which means that i remains at the (i + x )position. As Bi−2 moves i − 2 to the right by distance < x + 2 and i stays 32.

(36) at the (i + x )-th position, the letter i − 2 will not move to the right of i and i remains at the (i + x )-th position. Recurrently by the same arguments the action of B will leave the letter i at the (i + x )-th position and leave those letters smaller than i staying to the left of i. That is   ... i+x ... , ... F (σ ) = B(C ◦ R ◦ A(σ)) =  i ... 1...i−1 which means i ∈ Rmil( F (σ)) and i + x ∈ Rmip( F (σ)). Hence Rmil( TA(σ)) ⊆ Rmil( F (σ)) and Rmip( TA(σ)) ⊆ Rmip( F (σ)).. The last lemma we need is the following. Lemma 3.3.5. We have Rmil( F (σ)) ⊆ Rmil( TA(σ)),. Rmip( F (σ)) ⊆ Rmip( TA(σ)).. Proof. Let i ∈ Rmil( F (σ)). It is easy to see that the position of i is greater or equal to i (If not, then there exists a letter in 1, . . . , i − 1 which is to the right of i and then we have i ∈ / Rmil( F (σ)), which is absurd.) We suppose the position of i is i + x for some x ≥ 0 and i + x ∈ Rmip( F (σ)). That means   ... i+x ... . ... F (σ) = B(C ◦ R ◦ A(σ)) =  i ... 1...i−1 We investigate C ◦ R ◦ A(σ). From above we know that after the action of B, i moves to the (i + x )-th position and those letters smaller than i are to the left of i. Hence i does not move after moved Bi . We infer that Bi moves i by a distance x, so the (n − i + 1)-th position in C ◦ R ◦ A(σ) is x. That means ... n−i+1 ... C ◦ R ◦ A(σ) = . ... x .... 33.

(37) Let us look the letters in C ◦ R ◦ A(σ) and to the right of x (which is now at the (n − i + 1)-position). If the letter at the (n − i + 1 + j)-th position is greater or equal to x + j, then under the action of B, the letter i − j will move to the right of i and then the position of i in F (σ) will not be the (i + x )-th, a contradiction. Hence the (n − i + 1 + j)-th letter in C ◦ R ◦ A(σ) will be less than x + j and we have ... n−i+1 n−i+2 n−i+3 ... n C ◦ R ◦ A(σ) = . ... x < x+1 < x+2 ... < x+i−1 We use function D again and apply D on C ◦ R ◦ A(σ). D n−i ◦ · · · ◦ D1 is to move leftward those letters to the left of x (which is at the (n − i + 1)-th position) and keeps x and the letters to the right of x intact. Hence D n−i ◦ · · · ◦ D1 (C ◦ R ◦ A(σ)) = ... n−i+1 n−i+2 n−i+3 ... n . ... x < x+1 < x+2 ... < x+i−1 Again D n−i+1 moves x (at the (n − i + 1)-th position) to the left by x and results in D n−i+1 ◦ · · · ◦ D1 (C ◦ R ◦ A(σ)) = . . . n − (i + x ) + 1 . . . n − i + 2 n − i + 3 . . . n . ... x ... < x+1 < x+2 ... < x+i−1 As D n−i+2 moves the letter at the (n − i + 2)-th position to the left by a distance less than x + 1 and so x (at the (n − (i + x ) + 1)-th position) will not change. Again D n−i+3 moves the letter at the (n − i + 3)-th position to the left by a distance less than x + 2 and will keep x intact. By the same arguments we know that the letters to the right of x (which is originally at the (n − i + 1)-th position of C ◦ R ◦ A(σ)) will stay to the right of x (which is now at the (n − (i + x ) + 1)-th position after the action of D), as . . . n − (i + x ) + 1 . . . D (C ◦ R ◦ A(σ)) = . ... x ... Now we apply C on D (C ◦ R ◦ A(σ)) to observe those letters to the left of x (which is at the (n − (i + x ) + 1)-th position of D (C ◦ R ◦ A(σ))). Firstly, the effect of C n−(i+ x)+2 ◦ · · · ◦ C n is to move rightward those letters to the right of x, hence it does not affect x and the letters to the left of x. We have now . . . n − (i + x ) + 1 . . . n−(i + x )+2 n C ◦ · · · ◦ C ( D ◦ C ◦ R ◦ A(σ)) = . ... x ... 34.

(38) Next, C n−(i+ x)+1 move x (at the (n − (i + x ) + 1)-th position) to the right by x and results in the (n − i + 1)-th position. Now we have C n−(i+ x)+1 ◦ · · · ◦ C n ( D ◦ C ◦ R ◦ A(σ)) = 1 . . . n − (i + x ) . . . n − i + 1 . . . ( ? ) ... x ... Note that x comes back to the position it was originally in C ◦ R ◦ A(σ). We claim below that the letter at the (n − (i + x ) + 1 − j)-th position of D ◦ C ◦ R ◦ A(σ) will be less than x + j. Suppose not, then this number will move from the left of x to the right of x after the action of C on D ◦ C ◦ R ◦ A(σ). This will make x not be at the (n − i + 1)-th position of C ( D ◦ C ◦ R ◦ A(σ)). However from the previous lemma we know C ◦ D = idCCn and C ( D ◦ C ◦ R ◦ A(σ)) = C ◦ R ◦ A(σ), so x will be at the (n − i + 1)-th position of C ( D ◦ C ◦ R ◦ A(σ)), a contradiction. Therefore we obtain that the letter at the (n − (i + x ) + 1 − j)-th position of D ◦ C ◦ R ◦ A(σ) will be less than x + j. That is, D ◦ C ◦ R ◦ A(σ) = 1 . . . n − ( i + x + 2) + 1 n − ( i + x + 1) + 1 n − ( i + x ) + 1 . . . . < n−i ... < x+2 < x+1 x ... From lemma we know D ◦ C = idRCCn and hence D ◦ C ◦ R ◦ A(σ) = R ◦ A(σ) and we have R ◦ A(σ) = 1 . . . n − ( i + x + 2) + 1 n − ( i + x + 1) + 1 n − ( i + x ) + 1 . . . . < n−i ... < x+2 < x+1 x ... Since the inverse of R is R itself, we consider R ◦ R ◦ A and obtain ... i+x i+x+1 i+x+2 ... n A(σ) = . ... x < x+1 < x+2 ... < n−i Let A(σ) = a1 . . . an , TA(σ) = p1 . . . pn . By defitnion pi+ x = (i + x ) − ai+ x = (i + x ) − x = i and ai+ x+ j < x + j ⇒ pi+ x+ j = (i + x + j) − ai+ x+ j > (i + x + j) − ( x + j) = i, which means ... i+x i+x+1 ... n TA(σ) = . ... i ( >i ). 35.

(39) That is, i ∈ Rmil( TA(σ)), and i + x ∈ Rmip( TA(σ)). Hence Rmil( TA(σ)) ⊇ Rmil( F (σ)), Rmip( TA(σ)) ⊇ Rmip( F (σ)). . and we are done.. 3.4. Proof of the main result. Now all the ingredients for the proof of the main theorem are ready. Proof of Theorem 1.6.1 We define F : Sn → Sn by F := B ◦ C ◦ R ◦ A, and for σ ∈ Sn we have sort(σ ) = inv( F (σ)) (Theorem 3.1.1), Cyc(σ) = Lmap( F (σ)) (Theorem 3.2.1) Lmap(σ) = Rmil( F (σ)) and Lmal(σ) = Rmip( F (σ)) (Theorem 3.3.1). Hence. (sort, Cyc, Lmap, Lmal) ∼ (inv, Lmap, Rmil, Rmip) . and the proof is completed.. 36.

(40) Chapter 4. Discussion and concluding remarks It is somewhat unexpected and disappointing to us that after the completion of this work we find our result can be recovered also by combining that of Chen, Gong and Guo [4] and Eu, Lo and Wong [8] together with the inverse mapping between Sn . We explain it here. First of all, from the work of Eu, Lo and Wong [8] we know. (sort, Cyc, Lmap, Lmal)σ = (inv, Rmil, Lmap, Lmal)φ−1 σ for all σ ∈ Sn . For σ ∈ Sn , if i ∈ Lmal(σ ) then σj < σi for j < i. We look at σ−1 . As σ−1 (σi ) = i, which means in σ−1 the σi -th position is i. Moreover in σ−1 the position of any letter j smaller than i in σ−1 will be less than σi . In other words, in σ−1 the letters smaller than i will be to the left of i, hence i ∈ Rmip(σ−1 ). On the other hand, suppose i ∈ Rmip(σ−1 ), means that for any j with − 1 σ i > σ−1 j we have i > j. Look at (σ−1 )−1 = σ and we know that in σ the position of i is σ−1 i and for any j to the left of i in σ there is σ−1 j < σ−1 i and hence i > j. In other words, in σ the letters to the left of i is smaller than i and i ∈ Lmal(σ). Combine the above two paragraphs we get Lmal(σ) = Rmip(σ−1 ). Together with Theorem 1.5.2 we obtain. 37.

(41)    . Sn sort Cyc Lmap Lmal. φ −1.  −→  Sn inv   Rmil     Lmap Lmal. i.  −→ . .   .  , . Sn inv  Lmap   Rmil Rmip. which is exactly what we’ve done in this paper. However, both proofs of Chen, Gong and Guo [4] and Eu, Lo and Wong [8] are rather involved and so far it is still not clear to us that if our bijection is equivalent to the composition above. Computer experiment shows until n = 5 both bijections are equivalent. We hope to clarify it in the near future. Problem 1. Is our bijection F equivalent to φ−1 ◦ i? A natural question arises once we have a bijection on Sn . If we apply F continuously, it is clear that each σ belongs to an orbit σ → F (σ) → F ◦ F (σ) → · · · → F ◦ F ◦ · · · ◦ F (σ) → σ. Elements in each orbit are equivalent, and the set Sn can be partitioned into equivalent classes (namely, orbits) under this equivalent relation. Hence it is natural to ask what the orbit structure is. A computer program gives the following data. n 3 4. 5. i 1 3 1 3 9 1 3 9 12 19 34. numbers of orbits with length i in Sn 3 1 6 3 1 10 9 2 1 1 1. n. i. 6. 1 3 4 9 12 19 34 136 142 145. numbers of orbits with length i in Sn 20 25 1 6 2 2 2 1 1 1. Problem 2. What is the orbit structure Sn under this equivalent relation? 38.

(42) n. i. 7. 1 3 4 7 9 12 13 14 19 24. numbers of orbits with length i in Sn 35 68 3 1 18 6 1 2 6 1. n. i. 7. 33 34 61 125 136 142 145 206 2894. numbers of orbits with length i in Sn 1 6 1 1 2 2 2 1 1. Conjecture 1. There are (b nn c) fixed points of F on Sn . 2. Conjecture 2. The fixed points of F are the permutations which are both involution and become a SYT which has less than two rows by RSK algorithm. In enumerative combinatorics an important goal is the generating function. For our case it seems not trivial to write down the generating function, as three of the statistics are set-valued and should be encoded by nomomials. Problem 3. What can we say about the generating function F ( x, y, z, w) =. ∑. xsort(σ) yCyc(σ) zLmap(σ) wLmal(σ) .. σ ∈Sn. The final problem arises from many instances in the study of statistic. In the study of permutation statistics sometimes an equidistribution identity over Sn holds also for some proper subset A ⊂ Sn . Problem 4. Can our equidistribution result also hold for some interesting proper subset A ⊂ Sn ? We leave these problems for future study and also to interested readers.. 39.

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