An equivalence of multistatistics on permutations

We prove a conjecture of J.-C. Novelli, J.-Y. Thibon, and L. K. Williams (2010) about an equivalence of two triples of statistics on permutations. To prove this conjecture, we construct a bijection through different combinatorial objects, starting with a Catalan based object related to the PASEP. As a byproduct of this research, we also provide a new co-sylvester class-preserving bijection on permutations.


Introduction
The algebra of noncommutative symmetric functions Sym [7] has been studied in algebraic combinatorics during the past twenty years.It is a graded algebra Sym = n≥0 Sym n , where Sym n is of dimension 2 n−1 for n ≥ 1, so that its bases are indexed by integer compositions, that is, finite sequences of positive integers of sum n.There are many purely combinatorial problems related to this algebra such as, for example, the explicit description of the relations between different bases through their transition matrices.In this paper, we shall be interested in the basis introduced by Tevlin in [10], the so-called monomial basis of Sym, for which the transition matrices M (n) with the ribbon basis have been described in [6].
In this last paper, the authors prove that the coefficient M (n) I,J , indexed by two integer compositions I and J, is equal to the number of permutations satisfying GC(σ) = I and Rec(σ) = J, where GC and Rec are two statistics that will be recalled later.Some properties of this basis correspond to properties of the PASEP (Partially Asymmetric Simple Exclusion Process), a physical model in which particles hop back and forth (and in and out) of a one-dimensional lattice.More precisely, in the study of the basis of Tevlin, the sum of the coefficients of the rows of the transition matrices M (n) corresponds to the unnormalized steady-state probabilities of one state of the PASEP with n−1 sites.
In the combinatorial version of the PASEP, the steady-state probabilities are computed through the enumeration of the so-called permutation tableaux, an object introduced in [3].These objects are certain fillings of Ferrers diagrams.There is a simple bijection between a state of the PASEP with N sites and a Ferrers diagram of semiperimeter N such that the steady-state probability of this state of the PASEP corresponds to the sum of the fillings of the Ferrers diagram as permutation tableaux.Moreover, in [9] the authors present a bijection between permutation tableaux and permutations such that the permutation tableaux of a given shape are sent to the permutations of a given GC.
In the PASEP context, there exists a natural q-statistic on permutation tableaux that becomes the number of 31−2 patterns on permutations (denoted by tot) thanks to the bijection of [9].It is then natural to define a q-analog of the basis of Tevlin as the functions whose transition matrices M (n) (q) with the ribbon basis are such that M (n) I,J (q) is the sum of the q tot(σ) for all σ satisfying GC(σ) = I and Rec(σ) = J.
In [8], the authors studied those matrices in an algebraic way.However, the statistics GC and tot on permutations were not appropriate for an algebraic study.So the authors of [8] built suitable matrices M (n) (q) for their algebraic purpose through other statistics on permutations also recalled later (LC, Rec, and α).Then, the coefficient M (n) I,J (q) is the sum of the q α(σ) for all σ satisfying LC(σ) = I and Rec(σ) = J.They conjectured that their matrices are the same as the M (n) (q), or equivalently that the triples of statistics (GC, Rec, tot) and (LC, Rec, α) are equidistributed.
The aim of this paper is to prove this conjecture bijectively.As the triple of statistics (LC, Rec, α) has a natural description on subexcedent functions, in order to gain in readability, we build a bijection from permutations to subexcedent functions sending the triple of statistics (GC, Rec, tot) to (LC, DC, α).The global bijection is described as a sequence of bijections through different combinatorial objects: (1) P ←→ SF, where P, LH, LLH, DWSF, and SF are respectively Permutations, Laguerre Histories, Large Laguerre Histories, Decreasing Weighted Subexcedent Functions, and Subexcedent Functions which are all recalled or defined below.Laguerre histories and large Laguerre histories are both weighted Motzkin paths with similar conditions on the weights.In order to avoid confusion and to simplify the description of the map ψ we shall use another common representation on those objects in terms of Weighted Dyck Paths (WDP).Diagram (1) becomes: (2) P Under each bijection in the diagram we indicate in which subsection of the paper it is presented.
In the last part of the paper we use the connection between large Laguerre histories and permutations through a variation of ψ F V that we call ψ 0 F V to define another triple of statistics close to (GC, Rec, tot) also giving a combinatorial interpretation for the coefficients of M (n) (q).By combining ψ 0 F V with ψ F V and ψ through the following diagram (3) P we define a new bijection on permutations preserving the co-sylvester classes, classes inherited from a monoid structure related to the combinatorics of binary search trees [5].
In Section 2 we give the background and notations needed for the rest of the paper.In Section 3 we describe the bijections ψ F V and ψ which are natural bijections leading to objects where the statistics GC and tot are naturally defined.The object obtained is a pair consisting of a Catalan object and a weight.In Section 4 we begin by building a Catalan object associated with a weight from subexcedent functions through φ 1 and then describe φ 2 , a Catalan bijection between decreasing subexcedent functions and Dyck paths.In Section 5 we prove the conjectures of [8] and give some properties associated with the global bijection.Finally, in Section 6 we introduce a new co-sylvester classpreserving bijection and a new triple of statistics.

Permutations, compositions, and subexcedent functions.
Let us first fix our notations concerning permutations.We represent a permutation σ as a word σ 1 σ 2 • • • σ n such that σ i = σ(i).For all the forthcoming examples we fix τ = 528713649.
A recoil of a permutation σ is a value i such that i + 1 is on the left of i.The recoil set of σ is the set of the values of recoils.For example, the recoil set of τ is {1, 4, 6, 7}.A 31−2 pattern of σ is a pair (i, j) such that j > i + 1 and σ i+1 < σ j < σ i .We denote by tot(σ) the number of 31−2 patterns of σ.For example, tot(τ ) = 5.We shall often need the number of times that a value appears as a 2 in a 31 −2 pattern.We define tot k (σ) as the number of times k appears as a 2 in a 31−2 pattern in σ.
A composition of an integer n is a sequence I = (i 1 , . . ., i r ) of positive integers of sum n.The integer r is called the length of the composition.The descent set of I is Des The major index maj(I) of a composition is the sum of the values in the descent set of I.For example, maj(1, 3, 2, 1, 2) = 18.
The recoil composition Rec(σ) of a permutation σ ∈ S n is the composition of n whose descent set is the recoil set of σ.For example, The Genocchi descent set [6] of a permutation σ ∈ S n is the set of values immediately followed by a smaller value (it is sometimes called the descent tops of σ [9]).The Genocchi composition of descents (or G-composition, for short) GC(σ) of a permutation is the integer composition I of n whose descent set is {d−1 | d ∈ GDes(σ)}.For example, we have GDes(τ ) = {5, 6, 7, 8} and GC(τ ) = (4, 1, 1, 1, 2).
About the statistic LC, the definition on permutations given in [8] starts by taking the Lehmer code of the inverse permutation, a bijection between permutations and subexcedent functions.We shall directly define it on subexcedent functions as the other two statistics of the triple (LC, Rec, α) are easily defined on this object.
A subexcedent function of size n is a word u of size n on the alphabet of nonnegative integers such that for each i ∈ {1, . . ., n} we have u i ≤ n − i.We denote by SF n the set of subexcedent functions of size n.They are enumerated by n! as they are in bijection with permutations through the Lehmer code.The statistic LC is defined on a subexcedent function u as follows.
• Set S = ∅ and read u from right to left.At each step, if the entry k is strictly greater than the size of S, add the (k −|S|)-th element of the sequence [1, n] not considering the elements of S.
• The set S is the descent set of a composition C, and LC(σ) is the mirror image C of C.
For example, with u = 315503200, S is ∅ at first, then the set {2} at the third step as the third letter from the right is a 2 and S was empty, then {2, 3} (fourth step), then {2, 3, 5} (sixth step), then {2, 3, 4, 5} (seventh step).Hence C is (2, 1, 1, 1, 4), so that LC(u) = (4, 1, 1, 1, 2).Define the number of inversions of a subexcedent function as the sum of its values.We also define the descent set of a subexcedent function as (4) Des and the descent composition of u (denoted by DC(u)) as the composition of size n whose descent set is Des(u).One easily checks that DC(315503200) = (1, 3, 2, 1, 2).This definition of the descent composition of a subexcedent function corresponds to the recoil composition of the permutation before taking the Lehmer code of its inverse.
2.2.Different weighted paths.Recall that a Dyck path of size n is a path with n increasing steps and n decreasing steps that never goes below the horizontal axis.A Motzkin path of size n is a path with n steps among increasing steps, decreasing steps, and horizontal steps that never goes below the horizontal axis.For any path P we denote by P i the i-th step of this path.We call the height of a step the distance between the beginning of this step and the horizontal axis.A weight associated with a path is a nonnegative integer vector with the same size as the size of the path.
Definition 2.1.A Laguerre history L of size n is a weighted Motzkin path of size n with two kinds of horizontal steps (−→ and ).The weight w is a sequence of integers of size n satisfying for all i ≤ n, A large Laguerre history of size n − 1 is a weighted Motzkin path of size n − 1 with two kinds of horizontal steps where for any i ≤ n − 1, the weight w satisfies 0 ≤ w i ≤ h i .
Figure 1 represents a Laguerre history of size 9 and a large Laguerre history of size 8.For a better readability we only wrote the strictly positive w i .Both objects are in bijection and enumerated by n!.A common way to build a bijection between these is to represent each by a weighted Dyck path.Definition 2.2.A weighted Dyck path of size n is a Dyck path associated with a weight both of size n.For all i, the weight satisfies w i ≤ (h i − 1)/2 where h i is the height of the 2i-th step of the Dyck path.
An example of a weighted Dyck path of size 9 is given in Figure 2.For a better readability we grouped the steps two by two between vertical dashed lines and only wrote the strictly positive w i above steps D 2i−1 and D 2i .In this representation, h i is exactly the height of the meeting point of the 2i − 1-st and the 2i-th steps of the Dyck path.The map sending a Laguerre history to a weighted Dyck path is described by the following algorithm.Then the weighted Dyck path D associated with L satisfies The weight remains the same.
To build a weighted Dyck path from a large Laguerre history L, start by applying Algorithm 2.3 to L then add an extra / step at the beginning and \ step at the end.
Applying the corresponding algorithm to both paths of Figure 1 gives the weighted Dyck path of Figure 2.
The fact that both objects are enumerated by n! comes from the Françon-Viennot bijection.
2.3.The Françon-Viennot bijection.The Françon-Viennot bijection, first described in [4], is a bijection between permutations and Laguerre histories.We shall describe a bijection between permutations and weighted Dyck paths (defined below) that can be obtained by applying Algorithm 2.3 to the result of the version of Corteel of the Françon-Viennot bijection described in [2].
This bijection corresponds to the first part of Diagram (2).
In order to compute each step of the Dyck path, the Françon-Viennot bijection compares the values of σ with both of their neighbors.In order to do so with both the first and the last values of σ we use the convention σ 0 = 0 and The weight is built as follows: w k is equal to tot k (σ), the number of 31−2 patterns where k stands for the 2.
Let us also describe the reverse algorithm building a permutation from a weighted Dyck path (D, w).
Algorithm 2.5.The permutation σ is built iteratively by For the weighted Dyck path of Figure 2, the previous algorithm gives:

Permutations to weighted Dyck paths
Let us describe what happens to the triple of statistics on permutations through the Françon-Viennot bijection.• The total weight of (D, w) (denoted by tw(D, w)) is the sum of the values of w.
and the descent composition DC(D, w) is the composition of n whose descent set is Des(D, w).
On the weighted Dyck path (D, w) of Figure 2 we have tw(D, w) = 5.The positions i such that D 2i−1 = \ are GDes(D, w) = {5, 6, 7, 8} so that GC(D, w) = (4, 1, 1, 1, 2) and one can check that DC(D, w) = (1, 3, 2, 1, 2).Proposition 3.2.Let σ be a permutation of size n and (D, w) = ψ F V (σ).We have the following properties: • tw(D, w) = tot(σ); Proof.The first two assertions come directly from our description of ψ F V .To prove the last one, it is easier to work with ψ −1 F V that builds a permutation σ from a weighted Dyck path (D, w), as described in Algorithm 2.5.Let k ∈ {1, . . ., n − 1}.If w k < w k+1 , then k is on the left of k + 1 in σ so it is not a recoil for σ.Using the same idea, if w k < w k+1 then k is a recoil for σ.Now, when w k = w k+1 , we have a recoil in k if and only if there is a new • on the left of k when we place it in σ which happens if and only if D 2k = /.
We now have a representation of the triple of statistics on the weighted Dyck paths but the definition of GC on those objects is not very natural.Indeed, the fact that i belongs to GDes(D, w) or not depends on the (i + 1)-st group of steps.To simplify the second part of the construction, we transform the Dyck paths to obtain a more suitable statistic.We extend this definition to weighted Dyck paths by carrying the weight.Note that it is an involution.An example is given in Figure 3 where we apply ψ to the weighted Dyck path of Figure 2.

An involution on
Definition 3.3 is a visual way of defining ψ, a more formal way would be by using the following lemma.Lemma 3.4.Let D be a Dyck path, we have: Proof.If D 2i = D 2i+1 then ψ does not change those step so the result is clear.Otherwise D 2i = D 2i+1 and then ψ sends one to the other which ends the proof.This involution corresponds to the second part of Diagram (2).
This bijection gives a new bijection between Laguerre histories and large Laguerre histories sending the equivalent of the GC statistic to a very similar one.Unfortunately this bijection is not intuitive on Laguerre objects.
Let us introduce the new statistics on weighted Dyck paths corresponding to DC and GC after applying ψ.Definition 3.5.The statistics GC 0 and DC 0 are defined as follows.
• The Genocchi descent set of type 0 of a weighted Dyck path of size n is and the Genocchi composition of descent of type 0 as the composition of n whose descent set is GDes 0 (D, w).• The descent set of type 0 of (D, w) is and the descent composition of type 0 (denoted by DC 0 (D, w)) is the composition of n whose descent set is Des 0 (D, w).Proposition 3.6.Let (D, w) be a weighted Dyck path, then • tw(ψ(D, w)) = tw(D, w); Proof.As ψ does not change the weight of the Dyck path, the total weight is carried.
For the two other points, the statistics are really close but just differ on the index of the considered steps, so Lemma 3.4 ends the proof.

Subexcedent functions to weighted Dyck paths
The aim of this section is to build a bijection from subexcedent functions to weighted Dyck paths sending the triple (LC, DC, inv − maj(LC)) to (GC 0 , DC 0 , tw) where the statistic inv − maj(LC) corresponds to the statistic α in the introduction.To do this, we build a bijection from subexcedent functions to an intermediate object: decreasing weighted subexcedent functions which are represented by a Catalan object and a weight, and then build a Catalan bijection between decreasing subexcedent functions and Dyck paths.• A subexcedent function u of size n is decreasing if the word obtained by removing all its zeroes is a strictly decreasing word.• A weight associated with a decreasing subexcedent function is a word w of size n such that for all k ∈ {1, . . ., n}, w k is smaller than or equal to the number of i < k such that 0 < u i < n − k ( i.e., the number of positive values on the left of k that could be at position k).
For example, the subexcedent function u = 540300200 is decreasing.About the associated weight, we can define a maximum weight as the weight for which each value is maximum.The maximum weight of u is 012221000 so the weight 002201000 is correct since it is smaller than the maximum weight component-wise, whereas 000002000 is not.
We say that a value of a subexcedent function is weighted if the corresponding weight is nonzero.
The decreasing subexcedent functions are indeed Catalan objects since one can build a bijection with nondecreasing parking functions, the nondecreasing words whose i-th value is smaller than or equal to i and whose first value is equal to one.Algorithm 4.2.Let u be a decreasing subexcedent function.
• Let v be the mirror-image of u, • replace each 0 by the first nonzero value to its left (the first zeros do not change), • add one to each value.
This map corresponds to the last part of Diagram (2).
The map φ 1 is a well-defined function from subexcedent functions to decreasing subexcedent functions.
Proof.It is clear that φ 1 is well-defined for every subexcedent function u and that the algorithm stops (for each pivot we have at most n − 1 − pivot swaps).
The result is decreasing in the sense of Definition 4.1 because the algorithm sorts the subexcedent function.
To prove that the result is a decreasing weighted subexcedent function, we need to prove that the weight satisfies the constraints of Definition 4.1.The value of the weight at position j was increased at most once per pivot that ended on its left, so it is smaller than the number of non-zero values on the left of j at the end of the algorithm.Moreover, it was increased only if the pivot was at position j, which is possible only if the pivot is smaller than or equal to n − j.So w satisfies the constraints of the definition.The position i of the pivot is the rightmost position such that there exists a strictly smaller weighted value to its right, i.e., the rightmost position i such that there exists k > i with w k > 0 and u i > u k .
The pivot swapped with the nearest weighted value to its right which is always strictly smaller than the pivot, i.e., the position i is the smallest i > k such that w i > 0.
Proof.We start by proving the second part of the lemma.We prove by induction on each different pivot that, with the notations of Algorithm 4.3, for all i < j < k, we have w j = 0.For the first pivot there is no problem as there are no weighted values to its left.
Assume the property is true when we change the value of the pivot.The weighted values are necessarily greater than or equal to the values on their left (up to the position where the previous pivot ended) so the new pivot must swap with all these weighted values.This ends the proof of this part of the lemma.
We also prove the other part of the lemma by induction.After the first exchange, the property is necessarily satisfied because of Step 2 of the algorithm.Assume that the property is true just after the exchange and for all the previous exchanges.There are two possibilities.
• If the value the pivot has just changed, there is no weighted values to its right since it is the greatest value after the previous pivot and it begins at the starting position of the previous pivot or to its right.As there are no weighted values to its right, after its exchange the pivot is the value greater than the rightmost weighted value on its left.• If the pivot stays the same, the property also holds since after the exchange, with the notations of Algorithm 4.3, u k is smaller than or equal to each weighted value to its right since the pivot swapped with all these weighted values.Moreover, each value between u i and u k is smaller than or equal to u k .Those cases prove that the property still holds after each exchange.
The previous lemma shows that the map φ 1 is injective by proving that we can find the previous pivot and the value with which it is swapped at each step.We shall admit for now that decreasing weighted subexcedent functions are enumerated by n! (which is proved in the next section) which proves the following proposition: Proposition 4.6.The map φ 1 is a bijection.
Note that the inverse map comes straightforwardly from Lemma 4.5.

The statistics through φ 1 .
Let us first define the new statistics on the decreasing weighted subexcedent functions.Definition 4.7.Let (u, w) be a decreasing weighted subexcedent function of size n.
• The number of inversions inv(u, w) is the sum of the values of u and w. • The descent set of (u, w) is and its descent composition DC(u, w) is the composition of n whose descent set is Des(u, w).• The statistic LC on (u, w) is the composition LC(u).Remark 4.8.We shall make some remarks on the previous definitions.
(1) The previous definitions are still correct if the weighted subexcedent function is not decreasing.Moreover, if we associate a null weight with a subexcedent function, those definitions give the same statistics as the usual ones on subexcedent functions.
(2) Note that on a decreasing subexcedent function, the mirror composition of the statistic LC exactly corresponds to the composition whose descent set is the set of values of u.By defining the total weight statistic on the decreasing weighted subexcedent function as the sum of the values of its weight, we have directly tw(u, w) = inv(u, w) − maj(LC(u, w)).
Proposition 4.9.Let u be a subexcedent function, then: To prove the last point of this proposition we shall consider the exchanges of Algorithm 4.3 as a succession of elementary exchanges described by the following algorithm where i and k come from the notations of Algorithm 4.3.Algorithm 4.10.Set j 1 equal to i and j 2 equal to k.
For example, if at some point of Algorithm 4.3, we have u = . . .4106 . . .and we have to exchange 6 with 4, the exchanges of Algorithm 4.10 would be: • The inv statistic is not modified since each decrementation of a value of u is balanced by a incrementation of a value of w.
In order to prove that the descent set of a weighted subexcedent function does not change when we are doing an exchange of Algorithm 4.3, we have to study different cases depending on whether there is a descent at positions i − 1, i, k − 1, or k or not.Let (u (1) , w (1) ) be the weighted subexcedent function before the exchange and (u (2) , w (2) ) the one after it.
• We start by considering what happens at position k.
-As u k is the rightmost occurrence of the pivot, the only way to not have a descent at k is that w k+1 .But in this case Lemma 4.5 tells us that the pivot was at position k + 1 at the previous step, so u k+1 + 1 and then after the exchange u k+1 so k is not a descent.
k so k − 1 is not a descent for (u (1) , w (1) ) and it is necessarily also the case for (u 2 , w (2) ).
• If i = k − 1, it is possible to have a descent or not in i.
-As u (1) i < u k , the only way to have a descent in i is to have w k and then w k and then w k so i is not a descent either for (u (2) , w (2) ).
• For the descent in i when i < k − 1, we have w i+1 so i is always a descent of (u (1) , w (1) ) in this case.Moreover, we still have w i+1 so it is also the case for (u (2) , w (2) ).
• For the position i − 1, as w is not strictly greater than the pivot, otherwise the same argument gives the same result) (u (1) , w (1) ) has a descent in i − 1 if and only if it is the case for (u (2) , w (2) ).
We have shown that in any possible situation, the descent set is constant after each exchange of Algorithm 4.3.
To prove that the LC statistic does not change at each step we prove that it is also true after each elementary exchange of Algorithm 4.10.As in the first step we have u j 1 > u j 1 +1 and in the second step we have u j 2 −1 ≤ u j 2 we only have to prove that two subexcedent functions u (1)  and u (2) which differ only at positions j and j + 1 such that, if u (1) = . . .ab . . .with a > b, u (2) = . . .b+1 a . .., then LC(u (1) ) = LC(u (2) ).There are three different cases.Let S be the set obtained during the computation of the two LC statistics before considering the value at position j + 1.
• In u (1) , if b ≤ |S| and a ≤ |S|, then after considering those two values in u (1) , the set used to compute LC(u (1) ) is still S.In u (2) we also have a ≤ |S| and b + 1 ≤ a ≤ |S| so the set has not changed either.• In u (1) , if b ≤ |S| and a > |S|, let α be the (a − |S|)-th letter of [1, n], then after considering those two values, the set is equal to S ∪ {α}.In u (2) , after considering a, the set is equal to S ∪ {α} and then b + 1 ≤ |S| + 1 so the set does not change.• In u (1)   Lemma 4.12.In the path D built in Algorithm 4.11, let h i be the height be the height of the 2i-th step of D. Then, for all i, we have (h i − 1)/2 equal to the number of values smaller than or equal to n − i in u to the left of i.
Proof.Let us call W the maximal weight associated with u.Our aim is to prove that for all i we have (h i − 1)/2 = W i .We prove this lemma by induction.For i = 1, we have (h 1 − 1)/2 = 0 = W 0 and the property holds.Assume the property for a given i.We have W i+1 = W i + 1 if and only if n − i is not a value in u and u i = 0 so that all the values on the left of u i that count for W i also count for u i+1 and it is also the case for u i .In this case we have D 2i = D 2i+1 and so h i+1 = h i + 2 so the property holds.With the same idea, we have W i+1 = W i − 1 if and only if n − i is a value in u and u i = 0, in this case we have h i+1 = h i − 2. Finally, in the other two cases we have Proposition 4.13.The map φ 2 is well-defined from decreasing weighted subexcedent functions to weighted Dyck paths and is a bijection.
Proof.The image of a decreasing weighted subexcedent function is a path.Lemma 4.12 proves that the height of the path is always nonnegative and that h n = 1.As D 2n = \, the path is a Dyck path.Lemma 4.12 also gives us that the constraints on the weight of a decreasing weighted subexcedent function correspond to the constraints of its image.
As the constraints for the weights correspond exactly from one object to the other, we only need to prove that φ 2 is a bijection from decreasing subexcedent functions to Dyck paths to prove that φ 2 is a bijection on the weighted objects.
As decreasing subexcedent functions and Dyck paths are both enumerated by the Catalan numbers, we only need to prove that this map is injective.The only way to have the same image from two decreasing subexcedent functions is that the nonzero values and their positions are fixed, but as the subexcedent functions are decreasing, there is no choice in the order of those values.
Note that by proving that φ 2 is a bijection, we proved that decreasing weighted subexcedent functions are enumerated by n!, so this finishes the proof of Proposition 4.6.

The statistics through φ 2 .
Proposition 4.14.Let (u, w) be a decreasing weighted subexcedent function.We have Proof.As the weight is carried without being changed, we have tw(u, w) = tw(φ 2 (u, w)).For the descent set, as the weight does not change and the positions of the zeroes in u correspond to the \ steps at odd positions in D, we also have DC(u, w) = DC(φ 2 (u, w)).Moreover, LC(u, w) is the mirror composition of the composition whose descent set is the nonzero values in u and GC 0 (D, w) is related to the \ steps at even positions in D, so that we also have LC(u, w) = GC 0 (φ 2 (u, w)).

Global result
5.1.Proof of previous conjectures.We now have all the tools we need to prove the conjectures of [8].To do so, we prove Conjecture 6.2 which is the equidistribution of both triples of statistics.
Theorem 5.1.Let I and J be two compositions of n.We have Proof.As a composition of bijections, the map Ψ is a bijection between subexcedent function and permutations.Let u be a subexcedent function.Applying Propositions 4.9, 4.14, 3.6, and 3.2 give ( 12) which proves the theorem.
Remark 5.2.The other conjectures of [8] are directly obtained from this one.
Proving those conjectures proves that it is possible to interpret the coefficients of the transitions matrices of [8] in terms of the triple of statistics (GC, Rec, tot).It also gives another q-refinement of the steadystate probabilities of the PASEP with statistics that arise from the combinatorics of the PASEP.

5.2.
Other properties of this bijection.In addition to the three statistics we have studied, the bijection we defined from subexcedent functions to permutations also carries another statistic.The same statistic on subexcedent functions is defined as the positions containing zeroes.
Note that the definition of the left-to-right maxima of a subexcedent function corresponds to its definition on permutations after taking the Lehmer code of the inverse of the permutation.Proposition 5.4.Let u be a subexcedent function.Its left-to-right maxima corresponds to the left-to-right maxima of Φ(u).
To prove this proposition, we need to see what happens to this statistic through the different bijections.First we need the following lemma that gives an equivalent definition of the left-to-right maxima in terms of descents and 31−2 patterns.Lemma 5.5.Let σ be a permutation.A value k = σ j is a left-to-right maximum of σ if and only if σ j−1 < σ j and tot k (σ) = 0.
Proof of Lemma 5.5.If k is a left-to-right maximum of σ, we necessarily have σ j−1 < σ j and there are no values greater than k to its left.So tot k (σ) = 0. Conversely, assume there exists i < j such that σ i > σ j and let i be the rightmost position such that σ i > σ j with i < j.Then σ i > σ i+1 < σ j and this is a 31−2 pattern where k stands for the 2.
With this lemma we are now able to follow what happens to the left-to-right maxima through the various bijections.
Proof of Proposition 5.4.Let σ be a permutation and k a value of σ.Lemma 5.5 shows that if we apply the Françon-Viennot bijection, the value k is a right to left maximum if and only if D 2k = \ and w k = 0 where (D, w) = ψ F V (σ).Thanks to Lemma 3.4 the value k is a right to left maximum if and only if D 0 2k+1 = \ and w 0 k = 0 where (D 0 , w 0 ) = ψ • ψ F V (σ).Moreover, let (u, w) be the nondecreasing weighted subexcedent function obtained after applying φ −1 2 to (D 0 , w 0 ).Then k is a right to left maximum of σ if and only if u k = 0 and w k = 0.
Finally, we need to prove that a value at position k in a subexcedent function v is equal to zero if and only if the value at the same position after applying the sorting function φ 2 is also zero with a null weight.In Algorithm 4.3 one can see that no pivot shall exchange with a value equal to zero so the result at position k shall also be zero with a null weight.Conversely, if the value at position k in φ 1 (u) has a weight equal to zero, it means that the value never changed during the algorithm so the value is also equal to zero in u.

Another convention
6.1.Another combinatorial interpretation.In this part of the paper we change the convention on permutations such that σ 0 = σ n+1 = 0.In the definition of the Genocchi descent set, we compare each value to its successor so we need to define a new statistic associated with this new convention.Definition 6.1.Let σ ∈ S n .We define the Genocchi descent set of type 0 as Note that σ n and n always belong to GDes 0 (σ).We also define the Genocchi composition of descents of type 0 (denoted by GC 0 ) as the composition of n whose descent set is GDes 0 (σ)\{n}.
The Françon-Viennot bijection still exists with this convention and builds a large Laguerre history.As explained in Section 2.2, to obtain a weighted Dyck path from a large Laguerre history one needs to send each step of the large Laguerre history to two steps of the Dyck path and then add an increasing step at the beginning of the path and a decreasing one at the end.In order to directly build the weighted Dyck path from the permutation, we apply Algorithm 2.4 on the n − 1 first values (values from 1 to n−1) of the permutation and then add an up step at the beginning of the path and a down step at the end.We also add a 0 at the end of the weight.For example, with σ = 528971364 we obtain the weighted Dyck path of Figure 5.
Note that we do not need to consider n in the algorithm as it is always greater than both its neighbors with this convention.With this convention we have the following proposition where we denote by ψ 0 F V the map defined by the algorithm above.The inverse map is built as follows.Apply ψ −1 F V to the path obtained after removing the first and last steps.At the end, instead of removing the last • in the obtained word, replace it by n.Proposition 6.2.Let σ ∈ S n , we have: ).The proof of this statement is the same as the one of Proposition 3.2, the only difference being that we have to add one to the index of the steps in the path as we add an extra step at the beginning when applying the Françon-Viennot bijection with this convention.
From this proposition we derive the following proposition.
Corollary 6.3.Let I and J be two compositions of n, we have: The proof of this statement is based on the same approach as the proof of Theorem 5.1, using the bijection Ψ = (ψ 0 F V ) −1 • ψ • ψ F V .This result gives another combinatorial interpretation of the coefficients of the transition matrices described previously.It also implies another way of refining the PASEP in terms of statistics close to the usual one used to describe its combinatorial behavior.

6.2.
A co-sylvester class preserving bijection.By construction, ψ preserves the recoil classes (that is the permutations having the same recoils composition).It happens that ψ preserves smaller sets known as co-sylvester classes of permutations, which have been described in [5] with sylvester classes.A co-sylvester class is represented by a binary search tree.Definition 6.4.A binary search tree is a labeled binary tree such that for each node, all the values in the left (resp.right) subtree are smaller (resp.greater) than the value at the node.
We associate a binary search tree with a permutation through the following algorithm.An example of this algorithm is given in Figure 6.In σ = 528971364 the first value is 5.The left subtree is built by finding the first value of [4] and the right subtree is built by finding the first value of {6, 7, 8, 9} in σ.We say that two permutations σ and τ belong to the same cosylvester class if their associated binary search trees are equal.
There exists another way to build the tree by reading the permutation from left to right and adding each value to its only possible position in the binary search tree but we shall only need Algorithm 6.5.
The usual way to prove that two binary search trees are equal is to do it inductively by considering the roots and then considering both subtrees.We describe here another way to prove it without using induction.Lemma 6.6.Let A and A ′ be two binary search tree.We have A = A ′ if and only if they have the same root in their restriction to any interval.
The restriction of a binary search tree to an interval always gives a new binary search tree so the root of this tree is well-defined.
Proof of Lemma 6.6.If A = A ′ , the property is clearly true.Let A and A ′ be two binary search tree such that for any interval I they give the same root when restricted to I. In particular A and A ′ have the same root (if we take I to be the interval of all values in them).The proof of the lemma comes by induction by applying the property on the intervals of values smaller than the root and greater than the root which gives the two subtrees of the root.Theorem 6.7.Let σ be a permutation.Then σ and Ψ(σ) belong to the same co-sylvester class.
In order to prove this theorem, we need to define a co-sylvester class on weighted Dyck paths.To do so, let us describe the following recursive algorithm building a binary search tree from a weighted Dyck path.Algorithm 6.8.Start with I = {1, 2, • • • , n}.Among the i ∈ I such that w i is minimal, the root of the current tree is equal to the smallest i such that D 2i = \.If there are no such positions, let i be the maximal value of I among the ones of minimal weight.Apply this algorithm to the interval of values smaller than i in I to obtain the left subtree and to the values greater than i for the right subtree.
An example of a Dyck path and its corresponding tree is given in Figure 7.At the first step, the i such that the weight is minimal are {1, 2, 5, 7, 8, 9}.Among these, 5 is the minimal one such that D 2i = \.To obtain the left subtree, we apply the algorithm to I = {1, 2, 3, 4} on the path.The minimal weights are for i in {1, 2} but there is no i in this set such that D 2i = \ so i = 2. Apply this algorithm on {1} to obtain the left subtree and {3, 4} for the right subtree.Remark 6.9.Note that we could have applied this algorithm on any weighted path with 2n steps.It does not need to stay above the horizontal axis nor to end at position (2n, 0).
The idea of the algorithm came after discussing with Aval, Boussicault, and Zubieta [1].Lemma 6.10.Let σ be a permutation and let A be the the binary search tree associated with the co-sylvester class of σ.Applying Algorithm 6.8 to ψ F V (σ) also gives A.
Moreover, let (D, w) = ψ 0 F V (σ).Applying Algorithm 6.8 to the weighted path obtained from (D, w) by removing the first step of D and adding a decreasing step at its end also gives the binary search tree A.
In order to prove this lemma we need the following lemma.Lemma 6.11.Let σ be a permutation and i < j < k be such that σ j < σ k < σ i ( i.e., they form a 3−1−2 pattern), then σ k stands for the 2 in a 31−2 pattern where the 1 is to the right of σ i .
Proof of Lemma 6.11.Among the values greater than σ k on the left of σ j , let σ r be the rightmost one.Necessarily, σ r > σ k > σ r+1 and r ≥ i so it defines a 31−2 pattern of σ to the right of σ i where σ k is the 2.
Proof of Lemma 6.10.To prove both assertions of this lemma, we use Lemma 6.6.Let A be the binary search tree of σ and A ′ be the binary search tree obtained after applying Algorithm 6.8 to (D, w) = ψ F V (σ).Let I be an interval of values of σ.
In A, the root of the tree obtained from the values in I corresponds to the first value of I appearing in σ.We now have to prove that this value corresponds to the position returned by Algorithm 6.8 for the same interval on (D, w).In the rest of the proof, consider σ = where σ i is the first value of I in σ and σ j is another value of I.
In order to prove that we have to search a step of minimal weight, we need to prove that for any 31 − 2 pattern where σ i is the 2, we have a 31 − 2 pattern where σ j is also the 2. Indeed, for any 31 − 2 pattern where σ i is the 2, the value representing the 3 (resp.the 1) is necessarily greater (resp.smaller) than all values in I so σ j is also between them and so is involved as a 2 in a 31−2 pattern.
If σ i > σ i−1 (so D 2i = \) and if σ j−1 < σ j with σ j < σ i then applying Lemma 6.11 proves that there is a 31−2 pattern to the right of σ i where σ j stands for the 2 and so the weight associated with σ j in (D, w) is strictly greater than the one associated with σ i .This proves that if D 2i = \, it is the leftmost decreasing step in even position of minimal weight in the interval I.
If σ i < σ i−1 (so D 2i = /) and if σ j > σ i then σ i−1 σ i • • • σ j is a 31−2 pattern because σ i−1 is greater than any value in I.This proves that D 2i is the rightmost even step among the ones of minimal weight.Moreover, if σ i > σ j and σ j−1 < σ j , we can again use Lemma 6.11 to prove that there is a 31 −2 pattern where σ j is a 2 and where the 1 is on the right of σ i , so w i < w j and there are no decreasing steps of minimal weight in position 2k for k in I.
We now use Lemma 6.10 to prove that A = A ′ .The proof of the second part of the lemma is the same after noticing that ψ 0 F V corresponds to ψ F V where we remove the last step of the path and add an increasing step at the beginning.
The proof of Theorem 6.7 follows directly.
Proof of Theorem 6.7.As it can be seen in Equation (7) in the proof of Lemma 3.6, we have ψ(D, w) 2i+1 = D 2i so both ways of building a binary search tree give the same tree if applied before ψ for the first one and after it for the second one.The statement comes from applying Lemma 6.10.

Figure 1 .
Figure 1.A Laguerre history of size 9 and a large Laguerre history of size 8.

Algorithm 2 . 3 .
Let L be a Laguerre history of size n and i ∈ [n].
Dyck paths.Definition 3.3.Let D be a Dyck path of size n.The Dyck path ψ(D) is obtained by sending every pair of steps /\ or \/ at positions 2i, 2i+1 to one another.

4. 1 .
Subexcedent functions to decreasing weighted subexcedent functions.Definition 4.1.Let us define a decreasing subexcedent function of size n and a weight for it.

Algorithm 4 . 3 .
Let u be a subexcedent function of size n.Set w the weight, to 0 n .•Step 1: define the pivot as the greatest value in u such that one of its occurrences has smaller or equal nonzero values to its left.If the pivot is not defined, the algorithm stops.Otherwise, let k be the position of the rightmost occurrence of the pivot in u. • Step 2: among the values smaller than or equal to the pivot on its left, let i be the position of the rightmost occurrence of the largest one.Modify the subexcedent function by decrementing u i by 1 and then swapping u i with u k .Modify the weight by incrementing w k .Go back to Step 1.

Lemma 4 . 5 .
With notations ofAlgorithm 4.3, consider  Step 2. After the exchange, we can recover both positions i and k.
, if b > |S| then the set is changed to S ∪ {β} where β is the (b − |S|)-th letter of[1, n]\S.Then as a > b, we have a > |S| + 1 and the set changes again to S ∪ {β, α} where α is the (a − |S| − 1)-th letter of[1, n]\(S ∪ {β}).For u(2) , we still have a > |S| so we add to S the (a − |S|)-th value of[1, n]\S which is also α because we necessarily have β < α.Then when we read b + 1, we have b + 1 > |S| + 1 so we add to the set the (b − |S|)-th value of [1, n]\(S ∪ {α}) which is again β, so the final set is the same in both cases.This proves the last point of the proposition.4.2.Decreasing weighted subexcedent functions to weighted Dyck paths.Let us now describe the final bijection between decreasing weighted subexcedent functions and weighted Dyck paths that corresponds to the third part of Diagram (2describe it, we build a Dyck path from a decreasing subexcedent function and carry the weight without modifying it.Then we show that it sends the triple of statistics of the decreasing weighted subexcedent functions to the triple of statistics of the weighted Dyck paths.4.2.1.Description of the bijection φ 2 between decreasing weighted subexcedent functions and weighted Dyck paths.The map φ 2 on a decreasing weighted subexcedent function of size n is built by carrying the weight and building the Dyck path D from the decreasing subexcedent function u as follows.

Definition 5 . 3 .
Define the left-to-right maxima of a permutation as the values with only smaller values to their left.

Algorithm 6 . 5 .
At the beginning fix I = {1, 2, • • • , n}.The root of the current tree is the value k of I such that k is the first value of σ |I , the restriction of σ to the values in I. Apply this algorithm to the interval of values smaller than k in I to obtain the left subtree and to the values greater than k for the right subtree.

Figure 7 .
Figure 7. Binary search tree associated with a weighted Dyck path.