DDINV, AREA, AND BOUNCE FOR (cid:126)k -DYCK PATHS
GUOCE XIN , ∗ AND YINGRUI ZHANG Abstract.
The well-known q, t -Catalan sequence has two combinatorial interpreta-tions as weighted sums of ordinary Dyck paths: one is Haglund’s area-bounce formula,and the other is Haiman’s dinv-area formula. The zeta map was constructed to connectthese two formulas: it is a bijection from ordinary Dyck paths to themselves, and ittakes dinv to area, and area to bounce. Such a result was extended for k -Dyck paths byLoehr. The zeta map was extended by Armstrong-Loehr-Warrington for a very generalclass of paths.In this paper, We extend the dinv-area-bounce result for (cid:126)k -Dyck paths by: i) givinga geometric construction for the bounce statistic of a (cid:126)k -Dyck path, which includesthe k -Dyck paths and ordinary Dyck paths as special cases; ii) giving a geometricinterpretation of the dinv statistic of a (cid:126)k -Dyck path. Our bounce construction is inspiredby Loehr’s construction and Xin-Zhang’s linear algorithm for inverting the sweep mapon (cid:126)k -Dyck paths. Our dinv interpretation is inspired by Garsia-Xin’s visual proof ofdinv-to-area result on rational Dyck paths. Mathematic subject classification : Primary 05A19; Secondary 05E99.
Keywords : q, t -Catalan numbers; sweep map; (cid:126)k -Dyck paths. Introduction
In their study of the space DH n of diagonal harmonics [4], Garsia and Haimanintroduced a q, t -analogue of the Catalan numbers, which they called the q, t -Catalansequence. There are several equivalent characterizations of the (original) q, t -Catalansequence, which includes two combinatorial formulas as weighted sums over the set D n of Dyck paths of length n : One is Haiman’s dinv-area q, t -Catalan sequence HC n ( q, t ) = (cid:88) D ∈D n q dinv ( D ) t area ( D ) ( n = 1 , , , ... );the other is Haglund’s area-bounce q, t -Catalan sequence C n ( q, t ) = (cid:88) D ∈D n q area ( D ) t bounce ( D ) ( n = 1 , , , ... ) . They are connected by a bijection, called the zeta map ζ , from D n to itself [1, 8, Section 5].More precisely, we have the bi–statistic equality:( area ( ζ ( D )) , bounce ( ζ ( D )) = ( dinv ( D ) , area ( D )) . There are many interesting results and generalizations related to the q, t -Catalan num-bers C n ( q, t ). The q, t symmetry C n ( q, t ) = C n ( t, q ) is proved as a consequence of thewell-known Shuffle theorem of Carlsson and Mellit [3], and its generalization for rational q, t Catalan numbers is proved as a consequence of the rational Shuffle theorem of Mellit[13]. But combinatorially proving these q, t symmetry properties has been intractable.
Date : November 9, 2020. a r X i v : . [ m a t h . C O ] N ov GUOCE XIN , ∗ AND YINGRUI ZHANG D. Armstrong, N. Loehr, and G. Warrington [2, Section 3.4] introduced the sweepmap for a very general class of paths, including the zeta map for Dyck paths and rationalDyck paths as special cases. They proposed a modern view by using only one statistic area and an appropriate sweep map Φ. Then related polynomials can be constructedsimilarly by defining dinv ( D ) = area (Φ( D )), and bounce ( D ) = area (Φ − ( D )). Severalclasses of polynomials constructed this way are conjectured to be jointly symmetric. See[2, Section 6].To attack the joint symmetry problem, we need a better understanding of the dinv or bounce statistic. The modern view hardly helps because the construction of the sweepmap is deceptively simple. For instance, the invertibility of the sweep map for rationalDyck paths was open for over ten years, until recently proved by Thomas-Williams in[14] for a very general modular sweep map; the bounce statistic remains mysteries forrational Dyck paths. See [6] for further references.Our main objective in this paper is two-folded. One is to give a geometric con-struction for the bounce statistic of a (cid:126)k -Dyck path; the other is to give a geometricinterpretation of the dinv statistic of a (cid:126)k -Dyck path. With a properly modified defini-tion of the area statistic, we extend the dinv sweeps to area , and area to bounce result.The bounce result is inspired by a recent work of [15], where a linear algorithm for Φ − was developed for (cid:126)k -Dyck paths. Note that Thomas-Williams’ general algorithm for Φ − is quadratic and is hard to be carried out by hand [14]. The dinv result is inspired by arecent work of [5], where a visual interpretation of dinv was introduced.1.1. The sweep map and (cid:126)k -Dyck paths.
To introduce the sweep map clearly, we usethe following three models and some notions in [15]. For a vector (cid:126)k = ( k , k , . . . , k n ) of n positive integers, denote by | (cid:126)k | = (cid:80) ni =1 k i . Denote by D (cid:126)k the set of all (cid:126)k -Dyck paths.We will see that (cid:126)k -Dyck paths reduce to ordinary Dyck paths when k i = 1 for all i , andreduce to k -Dyck paths when k i = k for all i .Model 1: Classical path model. (cid:126)k -Dyck paths are two dimensional lattice pathsfrom (0 ,
0) to ( | (cid:126)k | , | (cid:126)k | ) that never go below the main diagonal y = x , with north steps oflengths k i , 1 ≤ i ≤ n from bottom to top, and east unit steps. Each vertex is assigneda rank as follows. We start by assigning 0 to (0 , k i as wego north with a length k i step, and subtract a 1 as we go east. Figure 1 illustrates anexample of a (cid:126)k -Dyck path with (cid:126)k = (3 , , , , D ∈ D (cid:126)k , the SW-word SW ( D ) = σ σ · · · σ | (cid:126)k | + n is anatural encoding of D , where σ i is either an S k j or a W depending on whether the i -thvertex of D is the j -th South end (of the j -th North step) or a West end (of an Eaststep). The rank is then associated to each letter of SW ( D ) by assigning r = 0 to thefirst letter σ = S k and for 1 ≤ i ≤ | (cid:126)k | + n −
1, recursively assigning r i +1 to be either r i + k j if the i -th letter σ i = S k j , or r i − σ i = W . We can then form thetwo line array (cid:16) SW ( D ) r ( D ) (cid:17) . For instance for the path D in Figure 1 this gives (cid:32) SW ( D ) r ( D ) (cid:33) = (cid:32) S W S W W W S W W S S W W W W (cid:33) . INV, AREA, AND BOUNCE FOR (cid:126)k -DYCK PATHS 3 D = Figure 1.
An example of a (cid:126)k -Dyck path in model 1.Model 3: Visual path model. (cid:126)k -Dyck paths are two dimensional lattice paths from(0 ,
0) to ( | (cid:126)k | + n,
0) that never go below the horizontal axis with up steps (red arrows)(1 , k i ), 1 ≤ i ≤ n from left to right, and down steps (blue arrows) (1 , − y -coordinates). The sweep map image D of D is obtainedby reading its steps by their starting levels from bottom to top, and from right to leftwhen at the same level. This corresponds to sweeping the starting points of the stepsfrom bottom to top using a line of slope (cid:15) for sufficiently small (cid:15) >
0. The visualizationof the ranks in this model allows us to have better understanding of many results. Forinstance for the path D in Figure 1 this gives D = Figure 2.
An example of a (cid:126)k -Dyck path in model 3. The sweep dinv is13 and the red dinv is 3. Thus we have dinv ( D ) = 16 . The sweep map of a (cid:126)k -Dyck path is usually a (cid:126)k (cid:48) -Dyck path where (cid:126)k (cid:48) is obtainedfrom (cid:126)k by permuting its entries. Denote by K the set of all such (cid:126)k (cid:48) and by D K the unionof D (cid:126)k (cid:48) for all such (cid:126)k (cid:48) . The sweep map is a bijection from D K to itself.A Dyck path D ∈ D (cid:126)k may be encoded as D = ( a , a , . . . , a | (cid:126)k | + n ) with each entryeither k i or −
1. The SW -word of the D is SW ( D ) = σ σ · · · σ | (cid:126)k | + n where σ j = S k i if a j = k i and σ j = W if a j = −
1. The rank sequence r ( D ) = (0 = r , r , . . . , r | (cid:126)k | + n )of D is defined as the partial sums r i = a + a + · · · + a i − ≥
0, called startingrank (or level) of the i -th step. Geometrically, r i is just the level or y -coordinate ofthe starting point of the i -th step. We also need to consider the end rank sequence˙ r ( D ) = ( ˙ r , ˙ r , . . . , ˙ r | (cid:126)k | + n ) = ( r , r , . . . , r | (cid:126)k | + n , GUOCE XIN , ∗ AND YINGRUI ZHANG usually write S k i as S , and denote by r ( S ) and ˙ r ( S ) its starting rank and end rank,respectively. The length of S is written as (cid:96) ( S ) = ˙ r ( S ) − r ( S ).We will frequently use two orders on the arrows A and B of a Dyck path D : i) A < B (under the natural order) means that A is to the left of B in D ; ii) A < s B (under the sweep order) means that r ( A ) < r ( B ) or r ( A ) = r ( B ) and B < A .The paper is organized as follows. In this introduction, we have introduced the basicconcepts. In Section 2 we define the three statistics area , dinv and bounce for (cid:126)k -Dyckpaths and state our main result in Theorem 1. The proof of the theorem is given inthe next two sections: Section 3 proves that dinv sweeps to area , and Section 4 provesthat area sweeps to bounce . Finally Section 5 gives a summary and a conjecture on q, t symmetry. 2. Area, Dinv, and Bounce for (cid:126)k -Dyck paths
Throughout this section, (cid:126)k = ( k , k , . . . , k n ) is a fix vector of n positive integers,unless specified otherwise. We define the three statistics for (cid:126)k -Dyck paths. The area and bounce are defined using model 1, and the area and dinv are defined using model3. The two area definition are easily seen to be equivalent. We also consider the q, t symmetry property.2.1. The Area statistic for (cid:126)k -Dyck paths.
Traditionally, the area of a rational Dyckpath is defined to be the number of complete lattice cells between the path and the maindiagonal.We define the area statistic of a (cid:126)k -Dyck path D to be equal to the sum of thestarting ranks of all north steps of D . In model 1, this is the number of complete latticecells between the path and the main diagonal, and in rows containing a south end ofa north step; In model 3, this is the number of complete lattice cells between the redarrows and the horizontal axis. For example, in Figures 1 and 2, we have area ( D ) = 7.Note that some of the complete lattice cells with crosses are not counted, because theirrows do not contain a south end of a north step.This definition is closely related to the dinv statistic in the next subsection. Itagrees with the area for ordinary Dyck paths.2.2. The dinv statistic for (cid:126)k -Dyck paths.
Our dinv sweeps to area result is inspiredby [5, Proposition 4] for ( m, n )-Dyck paths. In that paper, the authors gave a geometricdescription of the dinv statistic and a representation of the area by ranks. Our area definition mimics that area formula. We follow some notations there.By abuse of notation, we will use W i (resp. S j ) for the i -th blue (resp. j -th red)arrow for a (cid:126)k -Dyck path D . Then we have area ( D ) = (cid:88) S j r ( S j ) , where the sum ranges over all red arrows S j of D . Compare this formula with [5, Theorem2] for ( m, n )-Dyck paths. INV, AREA, AND BOUNCE FOR (cid:126)k -DYCK PATHS 5
The dinv statistic of a Dyck path D ∈ D (cid:126)k needs a correction term which we callthe red dinv . More precisely, the dinv consists of two parts that can be describedgeometrically as follows.(1) Sweep dinv : Each pair ( W i , S j ) with W i < S j contributes 1 if W i sweeps S j ,denoted W i → S j , which means W i intersects S j when we move it along a line ofslope (cid:15) (with 0 < (cid:15) <<
1) to the right past S j ;(2) Red dinv : Each pair ( S i , S j ) of red arrows with S i < S j contributes ˙ r ( S j ) − ˙ r ( S i )if r ( S i ) ≥ r ( S j ) and ˙ r ( S j ) > ˙ r ( S i ), and contributes ˙ r ( S i ) − ˙ r ( S j ) if r ( S i ) < r ( S j )and ˙ r ( S j ) < ˙ r ( S i ). In other words, each pair ( S i , S j ) of red arrows contributes | ˙ r ( S j ) − ˙ r ( S i ) | if one of the two arrows can be contained in the other by movingthem along a line of slope (cid:15) .In formula we have dinv ( D ) = (cid:88) W i ˙ r ( S i ))( ˙ r ( S j ) − ˙ r ( S i ))+ (cid:88) S i
The bounce statistic was defined byHaglund for ordinary Dyck paths and extended by Loehr for k -Dyck paths. We willextend Leohr’s bounce path to that of (cid:126)k -Dyck paths with the help of an intermediaterank tableau R b ( D ), which will be proved to be the rank tableau R ( D ) in [15]. Thebounce paths for rational Dyck paths are still unknown.The bounce path is a sequence of alternating vertical moves and horizontal moves constructed with the help of an intermediate rank tableau R b ( D ) consisting of n columnswith k i + 1 cells in the i -th column. The entries in each column will be of the form a, a + 1 , a + 2 , . . . from top to bottom, so to construct R b ( D ) it suffices to determine thetop row entries.We begin at (0 ,
0) with a vertical move, and eventually end at ( | (cid:126)k | , | (cid:126)k | ) after ahorizontal move. Let v , v , · · · denote the number of passing north steps of the successivevertical moves and let h , h , ... denote the number of passing east steps of the successivehorizontal moves. These numbers are calculated in the following algorithm. Bouncing Algorithm
Input: A (cid:126)k -Dyck path D ∈ D (cid:126)k in model 1.Output: The bounce path of D , bounce ( D ), and the rank tableau R b ( D ).(1) To find v , move due north from P = (0 ,
0) until you reach the west end Q ofan east step of the Dyck path D ; the number of north steps traveled is v . Write v zeroes in turn in the first row in R , add one in the lower cells from top tobottom in each column to obtain R . Let h be the number of 1’s in R andmove due east h units to a position P .(2) Suppose in general we reached a position P i and need to find v i . Then we movenorth from P i until we reach the west end Q i of an east step of the Dyck path.Define v i to be the number of north steps traveled. Write v i (possibly equal to GUOCE XIN , ∗ AND YINGRUI ZHANG i ’s in turn in the first row in R i , add one in the lower cell from top to bottomin each column to obtain R i +1 . Let h i be the number of i + 1 in R i +1 and moveeast h i units to a position P i +1 .(3) Proceed as above until we eventually end at P f = ( | (cid:126)k | , | (cid:126)k | ). The final tableau R f is our rank tableau R b ( D ), and the bounce statistic is defined to be bounce ( D ) = (cid:88) i ≥ i × v i ( D )a weighted sum of the lengths of the vertical moves in the bounce path derivedfrom D . Equivalently, bounce ( D ) is also the sum of the entries in the first rowof R b ( D ).We illustrate the bounce statistic by the following Figure 3, where the Dyck path D is the sweep map image of the path D in Figure 1. To obtain the bounce path( bounce ( D )) and the rank tableau R b ( D ), We first find v = 2. Then we construct thetableau R with two columns. Thus h = 2 and we reach the position P = (2 , Q = P , which means v = 0,and hence R = R . It follows that h = 2 and we reach the position P = (4 , R , R , and the bouncepath ( bounce ( D )). The rank tableau R b ( D ) = R . Figure 3.
An example of the bounce path and rank tableau for the Dyckpath D = Φ( D ), where D is depicted in Figure 1.Now we need to show that the bounce path is always well-defined.Note that, for a Dyck path D ∈ D (cid:126)k , the bounce path does not necessarily returnto the diagonal x = y after each horizontal move. Consequently, it may occur that P i isthe starting point of an east step of D , so v i = 0. We claim that h i > i < f . Then we move forward to P f without a stop. Assume to the contrary that h i = 0.Then there are no i + 1 in R i , and hence no larger ranks also. This means that we havemoved (cid:80) v + v + ··· + v i − j =1 k j east steps in total. Since the lengths of the north steps is v (cid:88) j =1 k j + v + v (cid:88) j = v +1 k j + · · · + v + ··· + v i − (cid:88) j = v + ··· + v i − +1 k j = v + ··· + v i − (cid:88) j =1 k j ,P i is on the diagonal line. But then the east step starting at P i will go below the diagonalline. This contradicts the fact that D is a Dyck path. INV, AREA, AND BOUNCE FOR (cid:126)k -DYCK PATHS 7
Our bounce path reduces to Loehr’s bounce path for k -Dyck paths. Theorem 1.
The sweep map takes dinv to area and area to bounce for (cid:126)k -Dyckpaths. That is, for any Dyck path D ∈ D K with sweep map image D = Φ( D ) , wehave dinv ( D ) = area ( D ) and area ( D ) = bounce ( D ) . About the q, t -symmetry.
A vector (cid:126)k = ( k , . . . , k n ) of positive integers is alsocalled an ordered partition. Arranging its entries decreasingly gives a partition, calledthe partition λ ( (cid:126)k ) of (cid:126)k . We can define q, t -Catalan numbers of type λ similarly as follows: C λ ( q, t ) := (cid:88) λ ( (cid:126)k )= λ (cid:88) D ∈D (cid:126)k q dinv ( D ) t area ( D ) = (cid:88) λ ( (cid:126)k )= λ (cid:88) D ∈D (cid:126)k q area ( D ) t bounce ( D ) , where the sum ranges over all (cid:126)k -Dyck paths satisfying λ ( (cid:126)k ) = λ . Clearly, C k n ( q, t ) agreeswith Loehr’s higher q, t -Catalan polynomials, where k n denotes the partition consistingof n equal parts k .We investigate the q, t symmetry of C λ ( q, t ), and report as follows.We do have the q, t symmetry for partitions λ of length n = 2. We can prove thisproperty easily as follows. For (cid:126)k = ( k , k ), Dyck paths D are uniquely determined bythe two ranks ( r = 0 , r ) of the two red arrows. Let us call them the red ranks. Thepath D starts with a red arrow S k followed by k − r blue arrows W , then a red arrow S k followed by k + r blue arrows W .It is easily checked that Φ − ( D ) has red ranks ( r = 0 , k − r ) for 0 ≤ r ≤ k , butwhen r = k , Φ − ( D ) starts with S k instead of S k . It follows that the contributionof D in C λ ( q, t ) is q r t k − r by using the bounce formula. Thus we can define the map π : D (cid:55)→ π ( D ) ∈ D (cid:126)k , where π ( D ) is determined by its two red ranks (0 , k − r ). Themap π shows the q, t symmetry of C λ ( q, t ).For partitions λ of length n = 3, computer experiment suggests that C λ ( q, t ) issymmetric in q, t . We obtain explicit bounce formula as follows. The dinv formula doesnot seem nice.For (cid:126)k = ( k , k , k ), Dyck paths D ∈ D (cid:126)k are uniquely determined by their red ranks( r = 0 , r , r ). We have bounce ( D ) = (cid:40) k − r ) + r + k − r − min( r , k ) , if r + k − r ≥ r , k );2( k − r ) + (cid:100) r + k − r (cid:101) , otherwise.We have verified the q, t symmetry for almost all cases for which C λ (1 , < × .A combinatorial proof seems out of reach at this moment. We believe that it isvery hopeful to prove the q, t symmetry property in this case by MacMahon’s partitionanalysis technique.For partition λ of length n ≥
4, the q, t symmetry no longer holds. The smallestcase that violates the q, t symmetry is when λ = (3 , , , C λ ( q, t ) − C λ ( t, q ) = q t − q t − q t − q t + 2 q t + q t + 2 q t + q t − q t − q t − q t + q t . GUOCE XIN , ∗ AND YINGRUI ZHANG Another example is when λ = (3 , , , C λ ( q, t ) − C λ ( t, q ) = q t − q t − q t − q t + q t − q t + q t + q t + q t − q t − q t + q t + q t − q t + q t + q t + q t − q t − q t − q t − q t + q t . For most partitions λ of length n ≥ C λ ( q, t ) are not q, t symmetric, but we do have aconjecture stated in Section 5.3. proof that dinv sweeps to area Proposition 2 (in [5]) . The starting rank of any arrow A of D may be simply obtainedby drawing a line of slope < (cid:15) (cid:28) at the starting point of its preimage ¯ A , then countingthe lengths of the red arrows starting below the line and minus the number of the bluearrows that start below the line. In formula, we have r ( A ) = (cid:88) ¯ S< s ¯ A (cid:96) ( ¯ S ) − { ¯ W : ¯ W < s ¯ A } . Proposition 3 (Zero-row-count property in [5]) . In model , each lattice cell may containa segment of a red arrow or a segment of a blue arrow or no segment at all. The redsegment count of row j will be denoted c r ( j ) and the blue segment count is denoted c b ( j ) .We will denote by c ( j ) = c r ( j ) − c b ( j ) and refer to it the j -th row count. Observe thatin every row of a path diagram , the red segments and blue segments have to alternate.Every row must start with a red segment and end with a blue segment, and hence c ( j ) = 0 holds for all j .Proof of Theorem 1 part 1. We follow the idea in [5]. We will prove dinv ( D ) = area ( D )by induction on area ( D ).The base case is when area ( D ) = 0. Such a (cid:126)k -Dyck path D is uniquely given by: ared arrow S k followed by k blue arrows W , then a red arrow S k followed by k bluearrows W , and so on. The sweep map image D of D is clearly given by: red arrows S k n , S k n − , ..., S k followed by | (cid:126)k | blue arrows W .To show dinv ( D ) = area ( D ) in this case, we compute as follows. The area of D issimply given by area ( D ) = ( n − k n + ( n − k n − + · · · + k . The dinv formula in this case simplifies as follows. dinv ( D ) = (cid:88) j ≥ (cid:88) W i k i )( k j − k i ) . For each S j = S k j , we group the arrow S k t together with the followed k t blue arrows W ,and compute their contribution in the above dinv formula. For each t < j , we have twocases:Case 1 when k t ≥ k j : only the final k j blue arrows sweep S k j , contributing k j sweepdinvs; the red dinv is clearly 0. Thus the total contribution in this case is k j .Case 2 when k t < k j : the k t blue arrows sweep S k j , contributing k t sweep dinvs; thered dinv is clearly k j − k t . Thus the total contribution in this case is still k j . INV, AREA, AND BOUNCE FOR (cid:126)k -DYCK PATHS 9
It follows that dinv ( D ) = ( n − k n + ( n − k n − + · · · + k = area ( D ) . Now assume area ( D ) >
0. Then we choose the rightmost red arrow S with thelargest rank among the red arrows. Then S must be followed by a blue arrow, denoted W .By switching the two arrows S and W in D (denoting them by W (cid:48) S (cid:48) ), we subtract one cellfrom D and obtain another (cid:126)k -Dyck path D (cid:48) . Clearly area ( D (cid:48) ) is one less than area ( D ).Let D (cid:48) be the sweep map image of D (cid:48) . Then by the induction hypothesis, dinv ( D (cid:48) ) = area ( D (cid:48) ). We will show in Lemma 4 that dinv ( D ) − dinv ( D (cid:48) ) = area ( D ) − area ( D (cid:48) ).The theorem is then proved. Lemma 4.
We subtract one cell by switching the two arrows S and W in D and obtainanother (cid:126)k -Dyck path D (cid:48) . Let D (cid:48) be the sweep map image of D (cid:48) . We have the equation dinv ( D ) − dinv ( D (cid:48) ) = area ( D ) − area ( D (cid:48) ) . To prove Lemma 4, it suffices to prove the following Propositions 5 and 6. The formercomputes the difference area ( D ) − area ( D (cid:48) ) and the latter computes the difference dinv ( D ) − dinv ( D (cid:48) ).Let D and D (cid:48) be as above. In what follows in this section, we shall also suppose r ( S ) = i and (cid:96) ( S ) = k and use S , W , S (cid:48) , W (cid:48) described below unless specified otherwise.We have depicted in the left picture of Figure 4 the cell that we have subtracted fromthe preimage D to obtain D (cid:48) . Replacing the SW in D by the dashed arrows W (cid:48) S (cid:48) gives D (cid:48) . Clearly (cid:96) ( S ) = (cid:96) ( S (cid:48) ) = k . Let D and D (cid:48) be the sweep map images of D and D (cid:48) . Figure 4.
Contribution for the area difference and dinv difference.It is convenient to use the following notations. For any set U ⊆ N and a ∈ N , wedenote by S L U = { S | S < S , r ( S ) ∈ U } = { S | S < S (cid:48) , r ( S ) ∈ U } , where the first set is in D and the second set is in D (cid:48) . The equality is clear and we will not distinguish whetherthe set is for D or D (cid:48) .We list some similar notations as follows. S L U = { S | S < S , r ( S ) ∈ U } , S L ,a U = { S | S ∈ S L U , ˙ r ( S ) ≥ a } ,S R U = { S | S < S, r ( S ) ∈ U } , S R ,a U = { S | S ∈ S R U , ˙ r ( S ) ≥ a } , , ∗ AND YINGRUI ZHANG W L U = { W | W < S , r ( W ) ∈ U } , W R U = { W | S < W, r ( W ) ∈ U } . The notations S L U , S L ,a U , S R U and S R ,a U are same with S < S (cid:48) or S (cid:48) < S in D (cid:48) .Then the following two properties about set S L U , S L ,a U .Let U = U (cid:83) U , U (cid:84) U = ∅ , U , U ⊆ N .1 . S L U = S L U (cid:83) S L U . . | S L U /S L ,a U | = | S L U | − | S L ,a U | . Now our task is to determine the difference area ( D ) − area ( D (cid:48) ) = (cid:88) S ∈ D r ( S ) − (cid:88) S ∈ D (cid:48) r ( S )using D and D (cid:48) by means of Proposition 2. Proposition 5.
Let D (cid:48) be obtained from D by removing one area cell as above, and let D and D (cid:48) be their sweep map images. Then area ( D ) − area ( D (cid:48) ) = (cid:88) S ∈ S L{ i − } (cid:96) ( S ) − k × | S L{ i − } | + | S L{ i } | − c b ( B ) − c b ( B ) . Proof.
For each red arrow S (cid:54) = S in D , it is also in D (cid:48) . We need to compute the differenceof its corresponding ranks in D and D (cid:48) . Clearly, this difference is given by kχ ( S < s S ) − kχ ( S (cid:48) < s S ) − χ ( W < s S ) + χ ( W (cid:48) < s S ) , since all the other terms cancel.This can be simplified as − kχ ( S (cid:48) < s S < s S ) + χ ( W (cid:48) < s S < s W ) , since S (cid:48) < s S and W (cid:48) < s W . Now χ ( S (cid:48) < s S < s S ) = 1 only when S ∈ S L{ i − } and χ ( W (cid:48) < s S < W ) = 1 only when S ∈ S L{ i } .By summing over all such S , the difference becomes − k × | S L{ i − } | + | S L{ i } | . (1) Finally the difference of the rank for S in D and the rank for S (cid:48) in D (cid:48) is given by (cid:88) S (cid:54) = S (cid:96) ( S )( χ ( S < s S ) − χ ( S < s S (cid:48) )) − (cid:88) W (cid:54) = W ( χ ( W < s S ) − χ ( W < s S (cid:48) ))= (cid:88) S (cid:54) = S (cid:96) ( S ) χ ( S (cid:48) < s S < s S ) − (cid:88) W (cid:54) = W χ ( S (cid:48) < s W < s S ) , since S (cid:48) < s S and W (cid:48) < s W . Now χ ( S (cid:48) < s S < s S ) = 1 only when S ∈ S L{ i − } and χ ( W (cid:48) < s W < W ) = 1 only when W ∈ W L{ i − } (cid:83) W R{ i } .By summing over all such S and W , the difference becomes r ( S ) − r ( S (cid:48) ) = (cid:88) S ∈ S L{ i − } (cid:96) ( S ) − | W L{ i − } | − | W R{ i } | = (cid:88) S ∈ S L{ i − } (cid:96) ( S ) − c b ( B ) − c b ( B ) . (2) INV, AREA, AND BOUNCE FOR (cid:126)k -DYCK PATHS 11 where c b ( B ) , c b ( B ) denote blue segment counts in the corresponding regions in Figure4 (left picture). By Proposition 3, we have W L{ i − } = c b ( B ) and W R{ i } = c b ( B ).The proposition then follows by adding the two formulas (1) and (2). Proposition 6.
Let D (cid:48) be obtained from D by removing an area cell. Then dinv ( D ) − dinv ( D (cid:48) ) = (cid:88) S ∈ S L{ i − } (cid:96) ( S ) − k × | S L{ i − } | + | S L{ i } | − c b ( B ) − c b ( B ) Proof.
We give the dinv recursion dinv ( D ) − dinv ( D (cid:48) ) that can be stated two parts asfollows:Part 1: The difference for sweep dinv coming from ( W i → S j ) . Since D (cid:48) is obtained from D by replacing the solid arrows S , W by dashed arrows S (cid:48) , W (cid:48) , we can divide the contribution of a pair ( W i → S j ) to the difference into fourcases.(1) Both W i and S j are not in the displayed arrows. The contribution in this caseis always 0.(2) Both W i and S j are in the displayed arrows. This can only happen when ( W , S )in D (no dinv ) becomes ( W (cid:48) , S (cid:48) ) in D (cid:48) (1 dinv ). Therefore the contribution to thedifference in this case is − W i is one of the displayed arrows. This means ( W , S j ) in D becomes( W (cid:48) , S j ) in D (cid:48) . Observe that { ( W → S j ) } = c r ( T ) and { ( W (cid:48) → S j ) } = c r ( B ).Therefore the contribution to the difference in this case is c r ( T ) − c r ( B ).(4) Only S j is in the displayed arrows. This means ( W i , S ) in D becomes ( W i , S (cid:48) ) in D (cid:48) . Their contribution to the difference is 1 if W i has a blue segment in T , − W i hasa blue segment in B and 0 if W i does not have a segment in neither T or B . Thereforethe contribution to the difference in this case is c b ( T ) − c b ( B ).where c b ( B ) , c b ( T ) denote blue segment counts and c r ( B ) , c r ( T ) denote red seg-ment counts in the corresponding regions in Figure 4 (right picture).So the total contribution to the difference in this part is − c r ( T ) − c r ( B ) + c b ( T ) − c b ( B ) . (3) Part 2: For red dinv , we need to consider three cases.Case 1: S , S (cid:48) are not involved. Then ( S i , S j ) in D becomes ( S i , S j ) in D (cid:48) and the dinv difference is 0.Case 2: ( S i , S ) in D becomes ( S i , S (cid:48) ) in D (cid:48) . The red dinv for this type in D is givenby (cid:88) S< S χ ( r ( S ) ≥ r ( S ) & ˙ r ( S ) > ˙ r ( S ))( ˙ r ( S ) − ˙ r ( S ))+ (cid:88) S< S χ ( r ( S ) < r ( S ) & ˙ r ( S ) > ˙ r ( S ))( ˙ r ( S ) − ˙ r ( S )) . , ∗ AND YINGRUI ZHANG Recall that by our choice of S , r ( S ) > r ( S ) = i is impossible. Thus the sum becomes (cid:88) S ∈ S L{ i } /S L ,i + k { i } ( ˙ r ( S ) − ˙ r ( S )) + (cid:88) S ∈ S L ,i + k +1 { , ,..,i − } ( ˙ r ( S ) − ˙ r ( S )) . = (cid:88) S ∈ S L{ i } /S L ,i + k { i } ( i + k − ˙ r ( S )) + (cid:88) S ∈ S L ,i + k { , ,..,i − } ( ˙ r ( S ) − k − i ) , where we have add 0 = ( ˙ r ( S ) − k − i ) for those S with ˙ r ( S ) = k + i .The red dinv for this type in D (cid:48) is similar: (cid:88) S ∈ S L{ i,i − } /S L ,i + k − { i,i − } ( k + i − − ˙ r ( S )) + (cid:88) S ∈ S L ,i + k { , ,..,i − } ( ˙ r ( S ) − k − i + 1)= (cid:88) S ∈ S L{ i,i − } /S L ,i + k { i,i − } ( k + i − − ˙ r ( S )) + (cid:88) S ∈ S L ,i + k { , ,..,i − } ( ˙ r ( S ) − k − i + 1) , where we have add 0 = k + i − − ˙ r ( S ) for those S with ˙ r ( S ) = k + i − − (cid:88) S ∈ S L{ i − } /S L ,i + k { i − } ( k − (cid:96) ( S )) + | S L{ i } /S L ,i + k { i } | − | S L ,i + k { , ,..,i − } | + (cid:88) S ∈ S L ,i + k { i − } ( ˙ r ( S ) − k − i )= (cid:88) S ∈ S L{ i − } ( (cid:96) ( S ) − k ) − (cid:88) S ∈ S L ,i + k { i − } ( (cid:96) ( S ) − k ) + | S L{ i } /S L ,i + k { i } | − | S L ,i + k { , ,..,i − } | + (cid:88) S ∈ S L ,i + k { i − } ( (cid:96) ( S ) − k − (cid:88) S ∈ S L{ i − } ( (cid:96) ( S ) − k ) + | S L{ i } /S L ,i + k { i } | − | S L ,i + k { , ,..,i − } | − | S L ,i + k { i − } | Case 3: ( S , S j ) becomes ( S (cid:48) , S j ). The red dinv in D is (cid:88) S ˙ r ( S ))( ˙ r ( S ) − ˙ r ( S ))since r ( S ) < r ( S ) is impossible by our choice of S . Thus the sum becomes (cid:88) S ∈ S R ,i + k +1 { , ,..,i − } ( ˙ r ( S ) − ( i + k )) = (cid:88) S ∈ S R ,i + k { , ,..,i − } ( ˙ r ( S ) − ( i + k )) . The red dinv in D (cid:48) is similar: (cid:88) S (cid:48) ˙ r ( S (cid:48) ))( ˙ r ( S ) − ˙ r ( S (cid:48) )) = (cid:88) S ∈ S R ,i + k { , ,..,i − } ( ˙ r ( S ) − ( i + k − . since r ( S (cid:48) ) < r ( S ) ⇒ r ( S ) ≤ r ( S ) is impossible by our choice of S .Their difference is −| S R ,i + k { , ,..,i − } | . INV, AREA, AND BOUNCE FOR (cid:126)k -DYCK PATHS 13
So the contribution to the difference in this part is (cid:88) S ∈ S L{ i − } ( (cid:96) ( S ) − k ) + | S L{ i } /S L ,i + k { i } | − | S L ,i + k { , ,..,i − } | − | S L ,i + k { i − } | − | S R ,i + k { , ,..,i − } | = (cid:88) S ∈ S L{ i − } ( (cid:96) ( S ) − k ) + | S L{ i } | − | S L ,i + k { i } | − | S L ,i + k { , ,..,i − } | − | S L ,i + k { i − } | − | S R ,i + k { , ,..,i − } | = (cid:88) S ∈ S L{ i − } ( (cid:96) ( S ) − k ) + | S L{ i } | − | S L ,i + k { , ,..,i } | − | S R ,i + k { , ,..,i − } | = (cid:88) S ∈ S L{ i − } (cid:96) ( S ) − k × | S L{ i − } | + | S L{ i } | − c r ( T ) − c r ( T ) . (4)where c r ( T ) , c r ( T ) denote red segment counts in the corresponding regions in Figure 4(left picture). Recall that by our choice of S , we have c r ( T ) = | S L ,i + k { , ,..,i } | and c r ( T ) = | S R ,i + k { , ,..,i − } | .The formula (3) is − c r ( T ) − c r ( B ) + c b ( T ) − c b ( B ) . The proposition then follows by adding the two formulas (3) and (4), and using thefact c b ( T ) = c r ( T ) and c r ( B ) + 1 = c b ( B ), which are consequences of Proposition 3.4. Proof the area sweeps to bounce
Our proof relies on the inverting sweep map in [15]. We will quote some results forthe readers’ convenience.
Algorithm 7 (Filling Algorithm [15]) . Input: The SW-sequence SW ( D ) of a (cid:126)k -Dyck path D ∈ D (cid:126)k .Output: A tableau T = T ( D ) ∈ T (cid:126)k . (1) Start by placing a in the top row and the first column. (2) If the second letter in SW ( D ) is an S ∗ we put a on the top of the second column. (3) If the second letter in SW ( D ) is a W we place below the . (4) At any stage the entry at the bottom of the i -th column but not in row k i + 1 willbe called active . (5) Having placed , , · · · i − , we place i immediately below the smallest active entryif the i th letter in SW ( D ) is a W , otherwise we place i at the top of the first emptycolumn. (6) We carry this out recursively until , , . . . , n + | (cid:126)k | have all been placed. Algorithm 8 (Ranking Algorithm [15]) . Input: A tableau T = T ( D ) ∈ T (cid:126)k .Output: A rank tableau R ( D ) of the same shape with T . (1) Successively assign , , , ..., k to the first column indices of T from top to bot-tom; (2) For i from to n , if the top index of the i -th column is A + 1 , and the rank ofindex A is a , then assign the index A + 1 rank a . Moreover, the ranks in the i -thcolumn are successively a, a + 1 , . . . , a + k i from top to bottom. , ∗ AND YINGRUI ZHANG For instance if D is the path in Figure 3, with SW-sequence SW ( D ) = S S W W W S W S W S W W W W W , then we obtain the tableau T ( D ) and R ( D ) in Figure 5. T ( D ) = R ( D ) = Figure 5.
The filling tableau T ( D ) and the rank tableau R ( D ) of thepath in Figure 3.The following result is a summary of Lemmas 3.1, 3.2 and Theorem 2.14 in [15]. Theorem 9.
For a Dyck path D ∈ D (cid:126)k , Let D be the preimage of D on the sweep map.We obtain a Filling tableau T ( D ) and a Ranking tableau R ( D ) by Filling algorithm andRanking algorithm. The Ranking algorithm assigns every index a rank in T ( D ) and theranks are weakly increasing according to their indices. If indices , , . . . , n + | (cid:126)k | areassigned ranks r , r , . . . , r n + | (cid:126)k | , then the rank sequence of D is exactly ( r , r , . . . , r n + | (cid:126)k | ) . Now we are ready to prove that the area sweeps to bounce . Proof of Theorem 1 part 2.
Note that area ( D ) = bounce ( D ), and the ranks of the southends of D are just the first row entries of R ( D ). It suffices to show that the tableau R b ( D ) is the same as the ranking tableau R ( D ).Since both tableaux have columns of the form a, a + 1 , a + 2 , . . . from top to bottom,and have the first row weakly increasing, it suffices to show the following claim.Claim: R ( D ) has v i i ’s for i = 0 , , . . . , in its first row.We prove the claim by induction on i .The base case is when i = 0. By definition, D starts with v north steps followedby an east step. Now the filling algorithm will produce 1 , , . . . , v in the first row, with v + 1 under 1. It follows that R ( D ) has v v
0’ sincethe rank of v + 1 is already 1. The claim then holds true in this case.Assume the claim holds true for i and we need to show the case i + 1.Consider the bouncing path part P i goes north to Q i (in D ), goes east to P i +1 , andgoes north to Q i +1 (in D ). We have the following facts:i) From Q i to Q i +1 in D , there are h i east steps and v i +1 north steps, with indices (cid:80) i − j =0 ( v j + h j ) + v i + s for s = 1 , , . . . , v i +1 + h i .ii) Suppose now we have filled the indices up to Q i in the filling tableaux. Bythe induction hypothesis, R i agrees with R ( D ), so that these indices corresponds tothe ranks no more than i below the first row of R i , with the number of j + 1’s beingequal to h j for j ≤ i −
1. By the filling algorithm, the bottom indices are active onlywhen its rank is i and the cell under it has rank i + 1 in R i . This implies, the index1 + (cid:80) i − j =0 ( v j + h j ) + v i , corresponding to the west end Q i , must be filled under one of INV, AREA, AND BOUNCE FOR (cid:126)k -DYCK PATHS 15 the active i ’s. Consequently, the indices of the h i east steps in the path from Q i to Q i +1 must be filled in the h i cells of rank i + 1 in R i , and the indices of the v i +1 north steps inthe path from Q i to Q i +1 must also have rank i + 1 by the ranking algorithm. To see thatthere is no more rank i + 1 for north steps, we observe that the next index correspondsto the west end Q i +1 . By the filling algorithm, this index cannot be put below an indexof rank less than i + 1, and hence must have rank i + 2.5. Summary
We have defined the dinv , area , and bounce statistics for (cid:126)k -Dyck paths, and provedthat the sweep map takes dinv to area , and area to bounce . Such a result was firstknown by Haiman and Haglund for classical (or ordinary) Dyck paths; The result wasextended for k -Dyck paths by Loehr. Our result includes the two mentioned cases asspecial cases.The dinv sweeps to area result was also known for ( m, n ) rational Dyck paths by[11, 7, 12, 5]. Our work for (cid:126)k -Dyck paths are inspired by Garsia-Xin’s visual proof in [5].We should mention that such a result also has a parking function version for ordinaryDyck paths. See, e.g., [9]. Finding a proper extension of (cid:126)k -Dyck paths to (cid:126)k -parkingfunctions is one of our future projects.The bounce statistic was only known for classical Dyck paths, k -Dyck paths, andremains unknown for rational Dyck paths.We also investigated the q, t -symmetry of C λ ( q, t ). The symmetry is easily provedwhen the length of λ is (cid:96) ( λ ) = 2, and hopefully will be proved for the case (cid:96) ( λ ) = 3 inan upcoming paper. The symmetry no longer holds in general for n ≥
4. But computerexperiments suggest the following conjecture.
Conjecture 10.
Let λ = (( a + 1) s , a n − s ) be consisting of s copy of a + 1 ’s and n − s copy of a ’s. Then C λ ( q, t ) is q, t -symmetric. The s = 0 and s = n cases reduce to k -Dyck paths, and the conjecture holds truein these cases. References [1] G. Andrews, C. Krattenthaler, L. Orsina, and P. Papi.“ad-nilpotent b-ideals in sl ( n ) having afixed class of nilpotence: combinatorics and enumeration”. Trans. Amer. Math. Soc. 354(2002), pp.3835–3853.[2] D. Armstrong, N. A. Loehr, and G. S. Warrington, Sweep maps: A continuous family of sortingalgorithms, Adv. Math. 284 (2015), 159–185.[3] E. Carlsson and A. Mellit, A proof of the shuffle conjecture, Journal of the American MathematicalSociety Volume 31, Number 3, July 2018, Pages 661–697.[4] A. Garsia and M. Haiman, A remarkable q, t -Catalan sequence and q -Lagrange inversion, J. Alge-braic Combinatorics 5 (1996), 191–244.[5] A. Garsia and G. Xin, Dinv and Area, Electron. J. Combin., 24 (1) (2017), P1.64.[6] A. Garsia and Guoce Xin, Inverting the rational sweep map, J. Combin., 9 (2018), 659–679.[7] E. Gorsky and M. Mazin, Compactified Jacobians and q, t -Catalan Numbers I, J. Combin. TheorySer. A, 120 (2013), 49–63.[8] J. Haglund, Conjectured Statistics for the q, t -Catalan numbers, Advances in Mathematics 175(2003), 319–334.[9] J. Haglund, The q, t -Catalan numbers and the space of diagonal harmonics, with an appendix onthe combinatorics of Macdonald polynomials, AMS University Lecture Series, 2008. , ∗ AND YINGRUI ZHANG [10] Nicholas A. Loehr. Conjectured Statistics for the Higher q, t -Catalan Sequences[J]. Electronic Jour-nal of Combinatorics, 2005, 12(1):318–344.[11] Nicholas A. Loehr and Gregory S. Warrington, A continuous family of partition statistics equidis-tributed with length, J. Combinatorial Theory Ser. A, 116:379–403, 2009.[12] Mikhail Mazin, A bijective proof of Loehr-Warrington’s formulas for the statistics ctot qp and mid qp ,Annals Combin., 18:709–722, 2014.[13] A. Mellit, Toric braids and ( m, n )-parking functions, preprint (2016), arXiv:1604.07456.[14] H. Thomas and N. Williams, Sweepping up zeta, Sel. Math. New Ser., 24 (2018), 2003–2034.[15] G. Xin and Y. Zhang, On the Sweep Map for (cid:126)k -Dyck Paths, Electron. J. Combin., 26 (3) (2019),P3.63. , School of Mathematical Sciences, Capital Normal University, Beijing 100048,PR China
Email address : guoce [email protected] &2