The generalized Delta conjecture at t=0
aa r X i v : . [ m a t h . C O ] J a n THE GENERALIZED DELTA CONJECTURE AT t = 0 MICHELE D’ADDERIO, ALESSANDRO IRACI, AND ANNA VANDEN WYNGAERD
Abstract.
We prove the cases q = 0 and t = 0 of the generalized Delta conjecture of Haglund,Remmel and Wilson [10] involving the symmetric function ∆ h m ∆ ′ e n − k − e n . Our theorem generalizesrecent results by Garsia, Haglund, Remmel and Yoo [6]. This proves also the case q = 0 of our recent generalized Delta square conjecture [3]. Contents
1. Introduction 12. The generalized Delta conjecture 23. Symmetric functions 43.1. Notation 43.2. Some useful identities 63.3. A crucial theorem at t = 0 Introduction
In [10] Haglund Remmel and Wilson stated the so called
Delta conjecture , which can be written as ∆ ′ e n − k − e n = X P ∈ PLD (0 ,n ) q dinv ( P ) t area ( P ) x P , where on the left hand side we have one of the Delta operators introduced in [1] applied to the symmetricfunction e n , and on the right hand side we have a combinatorial formula given in terms of labelledDyck paths (see Sections 2 and 3 for precise definitions). This formula generalizes the so called Shuffleconjecture in [9] (which is the case k = 0 ), recently proved in [2] by Carlsson and Mellit.The Delta conjecture already attracted quite a bit of interest, and several special cases have beenproved (e.g. see [4] and references therein). In particular, the special cases q = 0 and t = 0 have beenrecently proved in [6] by Garsia, Haglund, Remmel and Yoo. To this day, the full conjecture remainswidely open.In the same [10], the authors formulated a more general conjecture, that we call generalized Deltaconjecture , and that can be stated as ∆ h m ∆ ′ e n − k − e n = X P ∈ PLD ( m,n ) q dinv P t area ( P ) x P , where now on the left hand side we act with another Delta operator, while on the right hand side wesum over partially labelled Dyck paths (again, see Sections 2 and 3 for precise definitions). The Deltaconjecture is simply the case m = 0 of this one.In [4] we proved the so called Schröder case , i.e. the case h· , e n − d h d i , of the generalized Delta conjec-ture. he main result of this paper is to prove the special cases q = 0 and t = 0 of the generalized Deltaconjecture: Theorem 1.1.
For m, n, k ∈ N , m ≥ and n > k ≥ , we have both ∆ h m ∆ ′ e n − k − e n (cid:12)(cid:12)(cid:12) t =0 = X P ∈ PLD ( m,n ) area ( P )=0 q dinv ( P ) x P and ∆ h m ∆ ′ e n − k − e n (cid:12)(cid:12)(cid:12) q =0 = X P ∈ PLD ( m,n ) dinv ( P )=0 t area ( P ) x P . Notice that this result generalizes the main result in [6].
Remark . We would like to emphasize that our proof is independent of the results in [6], henceproviding a further new proof of the Delta conjecture at q = 0 or t = 0 , after the alternative proofs in[11] and in [3].It should also be noticed that our proof has the peculiar property that it does not specialize to thecase m = 0 : for our argument to go through, the full generalized Delta conjecture at t = 0 or q = 0 isneeded.Finally, in [3] we proposed what we called a generalized Delta square conjecture , which extends the square conjecture of Loehr and Warrington [14], recently proved by Sergel [16], and it reduces to thegeneralized Delta conjecture when q = 0 : hence we proved also the case q = 0 of this newer conjecture.The paper is organized in the following way. In Section 2 we recall the combinatorial definitionsneeded for stating the generalized Delta conjecture. In Section 3 we introduce some notation and weprove the identities of symmetric function theory needed in the following sections. In Section 4 we recallsome results from Wilson [18] about ordered set partitions, and determine some properties that we willneed in Section 5 to show the equivalence between the cases q = 0 and t = 0 of the generalized Deltaconjecture, and to reduce their proof to an identity involving ordered set partitions. In Section 6 weestablish the main recursive combinatorial steps, that we will use in Section 7 to complete the proof ofour main results. We conclude the article with some open problems2. The generalized Delta conjecture
We refer to Section 3 for notations and definitions concerning symmetric functions.
In [10], the authors conjectured a combinatorial interpretation for the symmetric function ∆ h m ∆ ′ e n − k − e n in terms of partially labelled decorated Dyck paths, known as the generalized Delta conjecture becauseit reduces to the Delta conjecture when m = 0 . We give the necessary definitions. Definition 2.1. A Dyck path of size n is a lattice path going from (0 , to ( n, n ) , using only north andeast unit steps and staying weakly above the line x = y (also called the main diagonal ). The set of Dyckpaths of size n will be denoted by D ( n ) . A partially labelled Dyck path is a Dyck path whose verticalsteps are labelled with (not necessarily distinct) non-negative integers such that the labels appearing ineach column are strictly increasing from bottom to top, and does not appear in the first column. Theset of partially labelled Dyck paths with m zero labels and n nonzero labels is denoted by PLD ( m, n ) .Partially labelled Dyck paths differ from labelled Dyck paths only in that is allowed as a label inthe former and not in the latter. Definition 2.2.
We define for each D ∈ PLD ( m, n ) a monomial in the variables x , x , . . . : we set x P := n + m Y i =1 x l i ( P ) where l i ( D ) is the label of the i -th vertical step of D (the first being at the bottom), where we conven-tionally set x = 1 . The fact that x does not appear in the monomial explains the word partially . Definition 2.3.
Let D be a (partially labelled) Dyck path of size n + m . We define its area word to bethe string of integers a ( D ) = a ( D ) · · · a n + m ( D ) where a i ( D ) is the number of whole squares in the i -throw (counting from the bottom) between the path and the main diagonal. efinition 2.4. The rises of a Dyck path D are the indices Rise ( D ) := { ≤ i ≤ n + m | a i ( D ) > a i − ( D ) } , or the vertical steps that are directly preceded by another vertical step. Taking a subset DRise ( D ) ⊆ Rise ( D ) and decorating the corresponding vertical steps with a ∗ , we obtain a decorated Dyck path , andwe will refer to these vertical steps as decorated rises . Definition 2.5.
Given a partially labelled Dyck path, we call zero valleys its vertical steps with label (which are necessarily preceded by an horizontal step, that is why we call them valleys).The set of partially labelled decorated Dyck paths with m zero labels, n nonzero labels and k decoratedrises is denoted by PLD ( m, n ) ∗ k . See Figure 1 for an example.
13 046 026 ∗ ∗
Figure 1.
Example of an element in
PLD (2 , ∗ .We define two statistics on this set. Definition 2.6.
We define the area of a (partially labelled) decorated Dyck path D as area ( D ) := X i DRise ( D ) a i ( D ) . For a more visual definition, the area is the number of whole squares that lie between the path and themain diagonal, except for the ones in the rows containing a decorated rise. For example, the decoratedDyck path in Figure 1 has area .Notice that the area does not depend on the labels. Definition 2.7.
Let D ∈ PLD ( m, n ) . For ≤ i < j ≤ n + m , we say that the pair ( i, j ) is an inversion if • either a i ( D ) = a j ( D ) and l i ( D ) < l j ( D ) ( primary inversion ), • or a i ( D ) = a j ( D ) + 1 and l i ( D ) > l j ( D ) ( secondary inversion ),where l i ( D ) denotes the label of the vertical step in the i -th row.Then we define dinv ( D ) := { ≤ i < j ≤ n + m | ( i, j ) is an inversion } . For example, the decorated Dyck path in Figure 1 has primary inversion (the pair (2 , ) and secondary inversions (the pairs (2 , and (5 , ), so its dinv is .Notice that the decorations on the rises do not affect the dinv. Definition 2.8.
We define a formal series in the variables x = ( x , x , . . . ) and coefficients in N [ q, t ] PLD x,q,t ( m, n ) ∗ k := X D ∈ PLD ( m,n ) ∗ k q dinv ( D ) t area ( D ) x D . The following conjecture is stated in [10].
Conjecture 2.9 (Generalized Delta) . For m, n, k ∈ N , m ≥ and n > k ≥ , ∆ h m ∆ ′ e n − k − e n = PLD x,q,t ( m, n ) ∗ k . . Symmetric functions
For all the undefined notations and the unproven identities, we refer to [5, Section 1], where definitions,proofs and/or references can be found. In the next subsection we will limit ourselves to introduce somenotation, while in the following one we will recall some identities that are going to be useful in thesequel. In the third subsection we prove a crucial theorem, that we are going to use in the fourth andfinal subsection, where we will prove the main results on symmetric functions of this work.For more references on symmetric functions cf. also [15], [17] and [12].3.1.
Notation.
We denote by
Λ = L n ≥ Λ ( n ) the graded algebras of symmetric functions with coeffi-cients in Q ( q, t ) , and by h , i the Hall scalar product on Λ , which can be defined by saying that the Schurfunctions form an orthonormal basis.The standard bases of the symmetric functions that will appear in our calculations are the monomial { m λ } λ , complete { h λ } λ , elementary { e λ } λ , power { p λ } λ and Schur { s λ } λ bases. We will use implicitly the usual convention that e = h = 1 and e k = h k = 0 for k < . For a partition µ ⊢ n , we denote by e H µ := e H µ [ X ] = e H µ [ X ; q, t ] = X λ ⊢ n e K λµ ( q, t ) s λ (1)the (modified) Macdonald polynomials , where e K λµ := e K λµ ( q, t ) = K λµ ( q, /t ) t n ( µ ) with n ( µ ) = X i ≥ µ i ( i − (2)are the (modified) Kostka coefficients (see [12, Chapter 2] for more details).The set { e H µ [ X ; q, t ] } µ is a basis of the ring of symmetric functions Λ . This is a modification of thebasis introduced by Macdonald [15].If we identify the partition µ with its Ferrers diagram, i.e. with the collection of cells { ( i, j ) | ≤ i ≤ µ j , ≤ j ≤ ℓ ( µ ) } , then for each cell c ∈ µ we refer to the arm , leg , co-arm and co-leg (denotedrespectively as a µ ( c ) , l µ ( c ) , a µ ( c ) ′ , l µ ( c ) ′ ) as the number of cells in µ that are strictly to the right, above,to the left and below c in µ , respectively.We set M := (1 − q )(1 − t ) and we define for every partition µB µ := B µ ( q, t ) = X c ∈ µ q a ′ µ ( c ) t l ′ µ ( c ) (3) T µ := T µ ( q, t ) = Y c ∈ µ q a ′ µ ( c ) t l ′ µ ( c ) (4) Π µ := Π µ ( q, t ) = Y c ∈ µ/ (1) (1 − q a ′ µ ( c ) t l ′ µ ( c ) ) (5) w µ := w µ ( q, t ) = Y c ∈ µ ( q a µ ( c ) − t l µ ( c )+1 )( t l µ ( c ) − q a µ ( c )+1 ) . (6)We will make extensive use of the plethystic notation (cf. [12, Chapter 1]).We have for example the addition formulas e n [ X + Y ] = n X i =0 e n − i [ X ] e i [ Y ] and h n [ X + Y ] = n X i =0 h n − i [ X ] h i [ Y ] . (7)We will also use the symbol ǫ for(8) f [ ǫX ] = ( − k f [ X ] for f [ X ] ∈ Λ ( k ) , so that, in general, f [ − ǫX ] = ωf [ X ] (9)for any symmetric function f , where ω is the fundamental algebraic involution which sends e k to h k , s λ to s λ ′ and p k to ( − k − p k .Recall the Cauchy identities h n [ XY ] = X λ ⊢ n s λ [ X ] s λ [ Y ] and h n [ XY ] = X λ ⊢ n h λ [ X ] m λ [ Y ] . (10) e define the nabla operator on Λ by ∇ e H µ := T µ e H µ for all µ, (11)and we define the delta operators ∆ f and ∆ ′ f on Λ by ∆ f e H µ := f [ B µ ( q, t )] e H µ and ∆ ′ f e H µ := f [ B µ ( q, t ) − e H µ , for all µ. (12)Observe that on the vector space of symmetric functions homogeneous of degree n , denoted by Λ ( n ) , theoperator ∇ equals ∆ e n . Moreover, for every ≤ k ≤ n , ∆ e k = ∆ ′ e k + ∆ ′ e k − on Λ ( n ) , (13)and for any k > n , ∆ e k = ∆ ′ e k − = 0 on Λ ( n ) , so that ∆ e n = ∆ ′ e n − on Λ ( n ) .For a given k ≥ , we define the Pieri coefficients c ( k ) µν and d ( k ) µν by setting h ⊥ k e H µ [ X ] = X ν ⊂ k µ c ( k ) µν e H ν [ X ] , (14) e k (cid:20) XM (cid:21) e H ν [ X ] = X µ ⊃ k ν d ( k ) µν e H µ [ X ] , (15)where ν ⊂ k µ means that ν is contained in µ (as Ferrers diagrams) and µ/ν has k lattice cells, and thesymbol µ ⊃ k ν is analogously defined. The following identity is well-known: c ( k ) µν = w µ w ν d ( k ) µν . (16)The following summation formula is also well-known (e.g. cf. [5, Equation 1.35]):(17) X ν ⊂ µ c (1) µν = B µ , while the following one is proved right after Equation (5.4) in [5]: for α ⊢ n ,(18) X ν ⊂ ℓ α c ( ℓ ) αν T ν = e n − ℓ [ B α ] . Recall also the standard notation for q -analogues: for n, k ∈ N , we set [0] q := 0 , and [ n ] q := 1 − q n − q = 1 + q + q + · · · + q n − for n ≥ , (19) [0] q ! := 1 and [ n ] q ! := [ n ] q [ n − q · · · [2] q [1] q for n ≥ , (20)and (cid:20) nk (cid:21) q := [ n ] q ![ k ] q ![ n − k ] q ! for n ≥ k ≥ , while (cid:20) nk (cid:21) q := 0 for n < k. (21)Recall also (cf. [17, Theorem 7.21.2]) that h k [[ n ] q ] = (cid:20) n + k − k (cid:21) q for n ≥ and k ≥ , (22)and e k [[ n ] q ] = q ( k ) (cid:20) nk (cid:21) q for all n, k ≥ . (23) .2. Some useful identities.
We will use the following form of
Macdonald-Koornwinder reciprocity :for all nonempty partitions α and β e H α [ M B β ]Π α = e H β [ M B α ]Π β . (24)The following identity is also known as Cauchy identity : e n (cid:20) XYM (cid:21) = X µ ⊢ n e H µ [ X ] e H µ [ Y ] w µ for all n. (25)We need the following well-known proposition. Proposition 3.1.
For n ∈ N we have e n [ X ] = e n (cid:20) XMM (cid:21) = X µ ⊢ n M B µ Π µ e H µ [ X ] w µ . (26) Moreover, for all k ∈ N with ≤ k ≤ n , we have h k (cid:20) XM (cid:21) e n − k (cid:20) XM (cid:21) = X µ ⊢ n e k [ B µ ] e H µ [ X ] w µ , (27) and ω ( p n [ X ]) = [ n ] q [ n ] t X µ ⊢ n M Π µ e H µ [ X ] w µ . (28)We will make use of [8, Theorem 2.6], i.e. for any A, F ∈ Λ homogeneous(29) X µ ⊢ n Π µ F [ M B µ ] d Aµν = Π ν (cid:0) ∆ A [ MX ] F [ X ] (cid:1) [ M B ν ] , where d Aµν is the generalized Pieri coefficient defined by(30) X µ ⊃ ν d Aµν e H µ = A e H ν . We will use the following theorem from [5].
Theorem 3.2 ([5, Theorem 3.1]) . For b, k ≥ and m ≥ , we have X γ ⊢ b e H γ [ X ] w γ h k [(1 − t ) B γ ] e m [ B γ ] = (31) = m X j =0 t m − j k X s =0 q ( s ) (cid:20) s + js (cid:21) q (cid:20) k + j − s + j − (cid:21) q h s + j (cid:20) X − q (cid:21) h m − j (cid:20) XM (cid:21) e b − s − m (cid:20) XM (cid:21) . A crucial theorem at t = 0 . We introduce the following notation: for i ≥ , j ≥ and m ≥ (32) A ( i, j, m ) := X µ ⊢ i + j e m [ B µ ] e i [ B µ ] e H µ [ X ] w µ . It is easy to see that A ( i, j, m ) is symmetric in q and t .The goal of this subsection is to prove the following theorem. Theorem 3.3.
For i ≥ , j ≥ and m ≥ , we have A ( i, j, m ) | q =0 = i X s =0 t ( i − s ) (cid:20) mi − s (cid:21) t (cid:20) s + ms (cid:21) t h s + m (cid:20) X − t (cid:21) e i + j − s − m (cid:20) XM (cid:21)(cid:12)(cid:12)(cid:12)(cid:12) q =0 . (33) Moreover, we get an equivalent identity if we exchange q and t everywhere in this formula.Proof. The last statement follows easily from the symmetry of A ( i, j, m ) in q and t .So we will prove the formula (33) with q and t interchanged.Recall from (27) that(34) h i (cid:20) XM (cid:21) e j (cid:20) XM (cid:21) = X µ ⊢ i + j e i [ B µ ] e H µ [ X ] w µ . e want a formula for(35) ∆ e m (cid:18) h i (cid:20) XM (cid:21) e j (cid:20) XM (cid:21)(cid:19)(cid:12)(cid:12)(cid:12)(cid:12) t =0 = X µ ⊢ i + j e m [ B µ ] e i [ B µ ] e H µ [ X ] w µ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) t =0 . We start by observing that(36) h i (cid:20) XM (cid:21) e j (cid:20) XM (cid:21)(cid:12)(cid:12)(cid:12)(cid:12) t =0 = h i (cid:20) X − q (cid:21) e j (cid:20) XM (cid:21)(cid:12)(cid:12)(cid:12)(cid:12) t =0 The following proposition is proved in [7].
Proposition 3.4 ([7, Proposition 2.6]) . For i ≥ and j ≥ we have (37) h i (cid:20) X − q (cid:21) e j (cid:20) XM (cid:21) = X µ ⊢ i + j e H µ [ X ] w µ i X r =1 (cid:20) i − r − (cid:21) q q ( r ) + r − ir ( − i − r h r [(1 − t ) B µ ] . We get ∆ e m (cid:18) h i (cid:20) XM (cid:21) e j (cid:20) XM (cid:21)(cid:19)(cid:12)(cid:12)(cid:12)(cid:12) t =0 = ∆ e m (cid:18) h i (cid:20) X − q (cid:21) e j (cid:20) XM (cid:21)(cid:19)(cid:12)(cid:12)(cid:12)(cid:12) t =0 (using (37)) = i X r =1 (cid:20) i − r − (cid:21) q q ( r ) + r − ir ( − i − r X µ ⊢ i + j e H µ [ X ] w µ h r [(1 − t ) B µ ] e m [ B µ ] (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) t =0 (using (31)) = i X r =1 (cid:20) i − r − (cid:21) q q ( r ) + r − ir ( − i − r ×× r X s =0 q ( s ) (cid:20) s + ms (cid:21) q (cid:20) r + m − s + m − (cid:21) q h s + m (cid:20) X − q (cid:21) e i + j − s − m (cid:20) XM (cid:21)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) t =0 = i X s =0 q ( s ) (cid:20) s + ms (cid:21) q h s + m (cid:20) X − q (cid:21) e i + j − s − m (cid:20) XM (cid:21)(cid:12)(cid:12)(cid:12)(cid:12) t =0 ×× i X r =max(1 ,s ) (cid:20) i − r − (cid:21) q q ( r ) + r − ir ( − i − r (cid:20) r + m − s + m − (cid:21) q = i X s =0 q ( i − s ) (cid:20) mi − s (cid:21) q (cid:20) s + ms (cid:21) q h s + m (cid:20) X − q (cid:21) e i + j − s − m (cid:20) XM (cid:21)(cid:12)(cid:12)(cid:12)(cid:12) t =0 , where in the last equality we used the following lemma: Lemma 3.5.
For s ≥ , m ≥ and i ≥ , we have (38) i X r =1 q ( s ) (cid:20) i − r − (cid:21) q q ( r ) + r − ir ( − i − r (cid:20) r + m − s + m − (cid:21) q = q ( i − s ) (cid:20) mi − s (cid:21) q . Proof.
This is none other than [5, Lemma 3.6] with a replaced by − s and s replaced by m + s . Noticethat, even if it is stated for a ≥ , the proof in [5] actually works for a ≥ − i . (cid:3) This completes the proof of (33) with q and t interchanged. (cid:3) Red and blue formulae.
The following notation will be useful: for n ≥ , m ≥ and ≤ k ≤ n ,we set(39) t C ( m ) n,k := ∆ h m ∆ ′ e k − e n (cid:12)(cid:12)(cid:12) q =0 and q C ( m ) n,k := ∆ h m ∆ ′ e k − e n (cid:12)(cid:12)(cid:12) t =0 . Remark . Since ∆ h m ∆ ′ e k − e n is symmetric in q and t , we have(40) t C ( m ) n,k = q C ( m ) n,k (cid:12)(cid:12)(cid:12) q = t . The goal of this subsection is to prove the following theorem. heorem 3.7. For ≤ j < n , ≤ k ≤ n and m ≥ , we have the blue formula h ⊥ j q C ( m ) n,k = min( j,k − X s =0 q ( j − s ) (cid:20) mj − s (cid:21) q (cid:20) s + ms (cid:21) q q s · q C ( s + m ) n − j,k − s (41) + min( j,k − X s =0 q ( j − s ) (cid:20) mj − s (cid:21) q (cid:20) s + m − s − (cid:21) q s + m − X r =0 q r ( q C ( r ) n − j,k − s + q C ( r ) n − j,k − s +1 )+ χ ( j ≥ k ) q ( j − k ) (cid:20) mj − k (cid:21) q (cid:20) k + m − k − (cid:21) q k + m − X r =0 q r · q C ( r ) n − j, , and the red formula h ⊥ j q C ( m ) n,k = min( j,k − X s =0 q ( j − s ) (cid:20) mj − s (cid:21) q (cid:20) s + ms (cid:21) q · q C ( s + m ) n − j,k − s (42) + min( j,k − X s =0 q ( j − s ) (cid:20) mj − s (cid:21) q (cid:20) s + m − s − (cid:21) q q k − s s + m − X r =0 q r ( q C ( r ) n − j,k − s + q C ( r ) n − j,k − s +1 )+ χ ( j ≥ k ) q ( j − k ) (cid:20) mj − k (cid:21) q (cid:20) k + m − k − (cid:21) q k + m − X r =0 q r · q C ( r ) n − j, . (43) Moreover, replacing q by t everywhere these formulae still hold.Proof. The last statement follows immediately from Remark 3.6.We start with a remark.
Remark . Observe that(44) t C ( m ) n,k = ∆ h m ∆ ′ e k − e n (cid:12)(cid:12)(cid:12) q =0 = [ k ] t [ n ] t ∆ h m ∆ e k ω ( p n ) | q =0 . The argument to prove this is the same as the one appearing in the proof of Theorem 5.1 in [3] in thecase m = 0 .For every j such that ≤ j < n , using (28), we have h ⊥ j ∆ h m ∆ e k ω ( p n ) = [ n ] q [ n ] t X µ ⊢ n M Π µ w µ h m [ B µ ] e k [ B µ ] h ⊥ j e H µ [ X ] (using (14)) = [ n ] q [ n ] t X µ ⊢ n M Π µ w µ h m [ B µ ] e k [ B µ ] X ν ⊂ j µ c ( j ) µν e H ν [ X ] (using (16)) = [ n ] q [ n ] t X ν ⊢ n − j M e H ν [ X ] w ν X µ ⊃ j ν Π µ h m [ B µ ] e k [ B µ ] d ( j ) µν = [ n ] q [ n ] t X ν ⊢ n − j M e H ν [ X ] w ν Π ν (∆ e j h m [ X/M ] e k [ X/M ]) (cid:12)(cid:12) X = MB ν (using (27)) = [ n ] q [ n ] t X ν ⊢ n − j M e H ν [ X ] w ν Π ν X γ ⊢ k + m e j [ B γ ] e m [ B γ ] e H γ [ M B ν ] w γ , where in the fourth equality we used (29) with A [ X ] = e j [ X/M ] and F [ X ] = h m [ X/M ] e k [ X/M ] .Specializing at q = 0 , we get h ⊥ j [ k ] t [ n ] t ∆ h m ∆ e k ω ( p n ) | q =0 == [ k ] t X ν ⊢ n − j M e H ν [ X ] w ν Π ν X γ ⊢ k + m e j [ B γ ] e m [ B γ ] e H γ [ M B ν ] w γ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) q =0 (using (33)) = j X s =0 [ k ] t t ( j − s ) (cid:20) mj − s (cid:21) t (cid:20) s + ms (cid:21) t X ν ⊢ n − j M e H ν [ X ] w ν Π ν h s + m [ B ν ] e k − s [ B ν ] | q =0 . (45) n the outer sum of (45), if s = k , we get [ k ] t t ( j − k ) (cid:20) mj − k (cid:21) t (cid:20) k + mk (cid:21) t X ν ⊢ n − j M e H ν [ X ] w ν Π ν h k + m [ B ν ] | q =0 == t ( j − k ) (cid:20) mj − k (cid:21) t (cid:20) k + m − k − (cid:21) t X ν ⊢ n − j M e H ν [ X ] w ν Π ν [ k + m ] t h k + m [ B ν ] | q =0 = t ( j − k ) (cid:20) mj − k (cid:21) t (cid:20) k + m − k − (cid:21) t X ν ⊢ n − j M e H ν [ X ] w ν Π ν k + m − X r =0 t r B ν h r [ B ν ] | q =0 (using (26)) = t ( j − k ) (cid:20) mj − k (cid:21) t (cid:20) k + m − k − (cid:21) t k + m − X r =0 t r ∆ h r e n − j | q =0 = t ( j − k ) (cid:20) mj − k (cid:21) t (cid:20) k + m − k − (cid:21) t k + m − X r =0 t r C ( r ) n − j, , where in the second equality we used the following lemma: Lemma 3.9.
For a ≥ and any nonempty partition ν , we have (46) [ a ] t h a [ B ν ] | q =0 = a − X r =0 t r B ν h r [ B ν ] | q =0 . Proof of the Lemma.
Using (22), we have [ a ] t h a [ B ν ] | q =0 = [ a ] t (cid:20) ℓ ( ν ) + a − a (cid:21) t (47) = [ ℓ ( ν ) + a − t (cid:20) ℓ ( ν ) + ( a − − a − (cid:21) t = t a − [ ℓ ( ν )] t (cid:20) ℓ ( ν ) + ( a − − a − (cid:21) t + [ a − t (cid:20) ℓ ( ν ) + ( a − − a − (cid:21) t (by induction) = a − X r =0 t r [ ℓ ( ν )] t (cid:20) ℓ ( ν ) + r − r (cid:21) t (using (22)) = a − X r =0 t r B ν h r [ B ν ] | q =0 . (cid:3) On the other hand, again in the outer sum of (45), if s < k , then the corresponding internal sum gives X ν ⊢ n − j M e H ν [ X ] w ν Π ν h s + m [ B ν ] e k − s [ B ν ] | q =0 = 1[ k − s ] t ∆ h s + m ∆ e k − s ω ( p n − j ) (cid:12)(cid:12) q =0 , so that h ⊥ j t C ( m ) n,k = min( j,k − X s =0 t ( j − s ) (cid:20) mj − s (cid:21) t (cid:20) s + ms (cid:21) t [ k ] t [ k − s ] t t C ( s + m ) n − j,k − s + χ ( j ≥ k ) t ( j − k ) (cid:20) mj − k (cid:21) t (cid:20) k + m − k − (cid:21) t k + m − X r =0 t r · t C ( r ) n − j, . he first sum can be developed in two ways: either using [ k ] t = t s [ k − s ] t + [ s ] t or using [ k ] t = t k − s [ s ] t +[ k − s ] t . Using the first formula for [ k ] t we get h ⊥ j t C ( m ) n,k = min( j,k − X s =0 t ( j − s ) (cid:20) mj − s (cid:21) t (cid:20) s + ms (cid:21) t t s · t C ( s + m ) n − j,k − s + min( j,k − X s =0 t ( j − s ) (cid:20) mj − s (cid:21) t (cid:20) s + ms (cid:21) t [ s ] t [ k − s ] t t C ( s + m ) n − j,k − s + χ ( j ≥ k ) t ( j − k ) (cid:20) mj − k (cid:21) t (cid:20) k + m − k − (cid:21) t k + m − X r =0 t r · t C ( r ) n − j, . Now (cid:20) s + ms (cid:21) t [ s ] t [ k − s ] t t C ( s + m ) n − j,k − s = (cid:20) s + m − s − (cid:21) t X ν ⊢ n − j M e H ν [ X ] w ν Π ν [ s + m ] t h s + m [ B ν ] e k − s [ B ν ] | q =0 (using (46)) = (cid:20) s + m − s − (cid:21) t X ν ⊢ n − j M e H ν [ X ] w ν Π ν s + m − X r =0 t r B ν h r [ B ν ] e k − s [ B ν ] (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) q =0 (using (26)) = (cid:20) s + m − s − (cid:21) t s + m − X r =0 t r ∆ h r ∆ e k − s e n − j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) q =0 (using (13)) = (cid:20) s + m − s − (cid:21) t s + m − X r =0 t r (∆ h r ∆ ′ e k − s − e n − j + ∆ h r ∆ ′ e k − s e n − j ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) q =0 = (cid:20) s + m − s − (cid:21) t s + m − X r =0 t r ( t C ( r ) n − j,k − s + t C ( r ) n − j,k − s +1 ) , which gives precisely the blue formula (41) with q replaced by t , i.e. h ⊥ j t C ( m ) n,k = min( j,k − X s =0 t ( j − s ) (cid:20) mj − s (cid:21) t (cid:20) s + ms (cid:21) t t s · t C ( s + m ) n − j,k − s (48) + min( j,k − X s =0 t ( j − s ) (cid:20) mj − s (cid:21) t (cid:20) s + m − s − (cid:21) t s + m − X r =0 t r ( t C ( r ) n − j,k − s + t C ( r ) n − j,k − s +1 )+ χ ( j ≥ k ) t ( j − k ) (cid:20) mj − k (cid:21) t (cid:20) k + m − k − (cid:21) t k + m − X r =0 t r · t C ( r ) n − j, . The proof of the red formula (42) with q replaced by t is analogous, except that it uses the identity [ k ] t = t k − s [ s ] t + [ k − s ] t instead.Replacing t by q everywhere, and using Remark 3.6, gives the blue formula (41) and the red formula(42), completing the proof of our theorem. (cid:3) A useful lemma.
The following lemma will be useful.
Lemma 3.10.
For ≤ k ≤ n and m ≥ we have (49) h ⊥ n q C ( m ) n,k = q ( n − k ) (cid:20) mn − k (cid:21) q (cid:20) m + k − k − (cid:21) q . Proof.
In [4, Theorem 4.7] we proved the case h· , e n − d h d i of the generalized Delta conjecture. So, inparticular, we have the case h ∆ h m ∆ ′ e k − e n , h n i = h ⊥ n ∆ h m ∆ ′ e k − e n . Specializing to t = 0 we get that h ⊥ n q C ( m ) n,k is the sum of q dinv ( D ) over all D ∈ PLD ( m, n ) ∗ n − k of area whose dinv reading word is · · · n . Evaluating at t = 0 the formula for h ∆ h m ∆ ′ e k − e n , h n i given in [4, Theorem 4.7] yields thedesired polynomial. (cid:3) . Ordered set partitions
The following definitions are analogous to the definitions in [18], except that is added to the list ofelements. Definition 4.1.
Let m ∈ N , and let α = ( α , α , . . . , α l ) ∈ N l be a weak composition of n ∈ N . An ordered multiset partition of type ( m, α ) is a ordered partition of the multiset M := { m }∪{ i α i | ≤ i ≤ l } into sets called blocks . So even though an element might appear multiple times, it appears at most oncein each block. The set of such ordered multiset partitions of type ( m, α ) with m + k blocks is denotedby OP ( m, α ) k , and we set OP ( m, n ) k := ∪ α OP ( m, α ) k where α runs over all weak compositions of n .We will represent elements of OP ( m, n ) k in the following way. Example 4.2.
Let m = 2 , n = 6 and k = 4 . To represent an element of OP (2 , , we separate theblocks with a | , for example | | | , where by convention we choose to always write elements in the same block in strictly decreasing order.The type of this ordered multiset partition is (3 , , .Another useful way to represent an element of OP (2 , is to indicate that elements belong to thesame block by adding a ∗ between them: e.g. the previous example gets written as ∗ ∗ ∗ ∗ . Definition 4.3.
Given π ∈ OP ( m, n ) k , we define inv ( π ) as the number of pairs ( i, j ) such that(i) i appears in π to the left of j ;(ii) i and j are in distinct blocks;(iii) j < i ;(iv) j is minimal in its block. Example 4.4.
The inv of the ordered multiset partition | | | is 4. Indeed, reading theelements from left to right, the second creates inv with the first two ’s and the last creates inv withthe and the first . Definition 4.5.
Given π ∈ OP ( m, n ) k , a primary diagonal inversion of π is a pair ( i, j ) such that(i) i appears in π to the left of j ;(ii) i and j are in distinct blocks;(iii) j < i ;(iv) if i is the h -th smallest element in its block, then so is j .A secondary diagonal inversion is a pair ( i, j ) such that(i) i appears in π to the left of j ;(ii) i and j are in distinct blocks;(iii) i < j ;(iv) if i is the h -th smallest element in its block, then j is the ( h + 1) -th smallest element in its block.We define dinv ( π ) to be the number of diagonal inversions of π (primary or secondary). Example 4.6.
The dinv of the ordered multiset partition | | | is 7. Indeed, reading theelements from left to right, the second creates primary dinv with the second , the two createsecondary dinv with the minimal elements of the blocks to their left ( ), and the creates secondarydinv with the first . Definition 4.7.
Let σ = σ · · · σ n be a word of integers. A descent of σ is an i ∈ { , , . . . , n − } suchthat σ i > σ i +1 . The set of descents of σ is denoted by Des ( σ ) . Definition 4.8.
Given π ∈ OP ( m, n ) k , consider the word σ = σ σ · · · σ m + n obtained from π by writingthe blocks one after the other, from left to right, with the elements of each block in decreasing order. Wedefine another word w , recursively. Set w = 0 and w i = w i − + χ ( σ i is minimal in its block ) , where χ is the function defined as χ ( P ) = 1 if the statement P is true, and χ ( P ) = 0 otherwise. We define the major index of π as maj ( π ) := X i ∈ Des ( σ ) w i . Example 4.9.
The maj of the ordered multiset partition | | | is 8. Indeed, in this case σ = 1 0 1 3 2 0 2 1 , Des ( σ ) = { , , , } , and the corresponding w word is . efinition 4.10. Let π ∈ OP ( m, n ) k and ( m, α ) its type, where α = ( α , ..., α l ) . We define the monomial x π := x α · · · x α l l . Definition 4.11.
We define two subsets of OP ( m, n ) k , denoted OP L ( m, n ) k and OP R ( m, n ) k that con-sist of the ordered multiset partitions that do not contain a zero in the leftmost and rightmost block,respectively. Proposition 4.12.
There exists a bijection ξ : S ⊆ PLD ( m, n ) ∗ n − k → OP R ( m, n ) k where S is the subset of all partially labelled decorated Dyck paths of area that touch the diagonal m + k + 1 times, and it preserves the dinv statistic.Proof. A partially labelled decorated Dyck path has area if and only if all its rises are decorated andthe only vertical steps that are not rises start from the diagonal. This implies that elements of area in PLD ( m, n ) ∗ n − k touch the diagonal m + k + 1 times and a stretch of vertical steps is always followed bythe same number of horizontal steps. Thus, taking the “blocks” of labels that label the different verticalstretches, from right to left, we obtain an ordered multiset partition that completely determines the path.Notice that the rightmost block does not contain a since the first label of a partially labelled Dyckpath is never . For example the path in Figure 2 corresponds under ξ to the ordered multiset partition | | | in OP R (3 , .
23 014 05 0 ∗ ∗∗ ∗
Figure 2.
A partially labelled decorated Dyck path of area .It is immediate to check that diagonal inversions are mapped to diagonal inversions, hence the mappreserves the dinv statistic. (cid:3) Proposition 4.13.
There exists a bijection η : T ⊆ PLD ( m, n ) ∗ n − k → OP R ( m, n ) k where T is the set of all partially labelled decorated Dyck paths of dinv , and it maps area to maj.Proof. First we write the labels in the order in which they appear going top to bottom in the Dyck path,and then we declare that two consecutive letters belong to the same block if and only if they form adecorated rise in the path. In this way we get an ordered set partition with m + k blocks by construction.For example the path in Figure 3 corresponds under η to the ordered multiset partition | | | in OP R (2 , .Now we compute the area in the following way: for each (not necessarily decorated) rise (whichcorresponds to a descent in the word w of the image), we count the number of steps that are notdecorated rises and that lie strictly above it (which corresponds to the number of blocks to the left ofthe corresponding letter in the image). If we sum all these values, then we get the area, since each stepthat is not a decorated rise is counted a number of times equal to its height (i.e. the vertical steps belowit) minus the number of ascents (i.e. the horizontal steps below it) in the reverse pmaj reading word (i.e.the sequence of the labels going top to bottom), which is exactly the number of squares in its row. Butthe sum is exactly the one given in the definition of maj, so the two statistics match. (cid:3)
45 06 013 ∗ ∗ ∗∗
Figure 3.
A partially labelled decorated Dyck path with dinv .4.1. Standardization.Definition 4.14.
Given a partially labelled Dyck path D , we define its dinv reading word to be theword obtained by reading its positive labels along the diagonals, from bottom to top, from left to right.While we define its pmaj reading word to be the word obtained by reading its positive labels along therows, from bottom to top. Example 4.15.
For example, consider the partially labelled Dyck path of Figure 2: its dinv readingword is , while its pmaj reading word is . Definition 4.16.
We define the set
SOP ( m, n ) k of standardized ordered multiset partitions to be the set OP ( m, (1 , , . . . , k ⊂ OP ( m, n ) k of all ordered multiset partitions of type ( m, (1 , , . . . , with m + k blocks. We define SOP L ( m, n ) k and SOP R ( m, n ) k as the subsets of SOP ( m, n ) k of the partitions that donot contain a in their leftmost or rightmost block, respectively.Given an ordered multiset partition π , we define its dinv (resp. maj) reading word as the dinv (resp.pmaj) reading word of the corresponding partially labelled Dyck path ξ − ( π ) (possibly allowing a blankin the bottom left corner), where ξ is the obvious extension of the map in Proposition 4.12. We defineits inv reading word as its dinv reading word. Definition 4.17.
Given an ordered multiset partition π ∈ OP ( m, n ) we can define the standardization with respect to a statistic (it being inv, dinv, or maj) as the unique standard multiset partition obtainedfrom π by replacing its ’s by , , . . . α , its ’s by α + 1 , . . . , α + α and so on, in such a way that thereverse reading word (with respect to the relevant statistic) of the standardization is an α -shuffle (i.e. itis in , . . . , α (cid:1) α + 1 , . . . , α + α (cid:1) . . . ). Example 4.18.
Consider the ordered multiset partition | | | ∈ OP R (2 , (3 , , , . Itsstandardization with respect to both dinv and inv is | | | , whose reverse dinv (or inv) readingword is ∈ , , (cid:1) , (cid:1) , while its standardization with respect to maj is | | | ,whose reverse pmaj reading word is ∈ , , (cid:1) , (cid:1) .The verification of the following proposition is straightforward, hence left to the reader. Proposition 4.19.
The standardization with respect to one of the statistics dinv, inv or maj preservesthat statistic.
Using the well-known theory of shuffles (see e.g. [12, Chapter 6]), and the previous proposition, it isclear that X π ∈ OP L ( m,n ) k q dinv ( π ) x π = X π ∈ SOP L ( m,n ) k q dinv ( π ) Q ides ( σ ( π )) , where σ ( π ) denotes the dinv reading word of π and ides ( σ ) is the descent set of σ − . imilarly X π ∈ OP L ( m,n ) k q inv ( π ) x π = X π ∈ SOP L ( m,n ) k q inv ( π ) Q ides ( σ ( π )) , X π ∈ OP R ( m,n ) k q inv ( π ) x π = X π ∈ SOP R ( m,n ) k q inv ( π ) Q ides ( σ ( π )) , X π ∈ OP R ( m,n ) k q maj ( π ) x π = X π ∈ SOP R ( m,n ) k q maj ( π ) Q ides ( e σ ( π )) , where e σ ( π ) denotes the maj reading word of π .4.2. Insertion maps.
In [18], Wilson proved that, given a composition α , the three statistics definedabove are equidistributed over the set of ordered multiset partitions of type α . He did so by constructingbijections using Carlitz’s insertion method. We will describe these maps, but will not argue why theyare bijective: the interested reader can find the argument in [18]. Theorem 4.20.
There exist bijections φ : OP ( m, n ) k → OP ( m, n ) k θ : OP ( m, n ) k → OP ( m, n ) k such that for all π ∈ OP ( m, n ) k inv ( φ ( π )) = dinv ( π ) inv ( θ ( π )) = maj ( π ) . Both these maps are constructed following the same method. Start with a path in π ∈ OP ( m, n ) k oftype α = ( α , α , . . . ) .Remove all the nonzero letters of π one after the other, always removing a maximal letter (in specificorder) and recording(1) the amount the statistic (dinv for φ and maj for θ ) goes down by (this may be zero);(2) if the number of blocks goes down or not.Repeating this process of deleting a maximal letter and recording this data, we obtain, after n deletions,the ordered multiset partition | | · · · | .Next, we build the image of π , starting from | | · · · | and inserting α ’s then α ’s and so on.We make sure the i -th insertion happens in a place such that inv goes up by the same amount as it wentdown after the i -th deletion. Furthermore we make sure the i -th insertion changes the number of blocksif and only if the i -th deletion did.If α l is the last nonzero component of α , let α − be the weak composition obtained from α by putting α l equal to zero.We describe insertion methods for each statistic. That is, given a multiset D of size α l of pairs ofdata ( c, b ) where c is a nonnegative integer representing the change in the statistic and b a bit indicatingwether the number of blocks changes ( b = 1 ) or not ( b = 0 ); we detail how to insert α l l ’s into a π ∈ OP ( m, n − α l ) k of type α − such that each insertion is compatible with the ( c, b ) data. • Insertion for inv. (1) Label the blocks from to m + k − , from right to left.(2) Define gaps to be the positions before the first block, between two existing blocks and afterthe last block. Label the gaps from to m + k , right to left.We pick the elements of D one by one from biggest c to smallest and picking ( c, before ( c, .For each of the ( c, , we insert l into the block labelled c , thus not changing the number ofblocks and augmenting the inv by c . For each of the ( c, ’s we create a new block containingonly an l in the position of the gap labelled c , thus augmenting the inv by c . • Insertion for dinv. (1) Let h be the size of the biggest block. Label the blocks from to m + k − starting withthe blocks of size h , from left to right, then the blocks of size h − , again from left to right,etc.(2) Label the gaps in the same way as for the inv insertion.Insert the α l l ’s in the same way as for the inv, with respect to this new labelling. Insertion for maj.
This is where the representation of ordered multiset partitions is more natural.Let σ be the word of nonegative integers obtained from π by disregarding the ∗ ’s. We say thatthe position between two letters of an ordered multiset partition is a descent if the index of theletter directly preceding it is a descent of σ . We always consider the last gap (i.e. the positionafter the last block), to be a descent as well.Label the m + k + 1 gaps from to m + k starting with the descents, from right to left, followedby the gaps that are not descents, from left to right.Define D ′ from D by replacing all the ( c, ’s by ( c + 1 , ’s and pick its elements one by onestarting with the elements having a maximal first component and picking ( x, before ( x, .(1) For each ( c + 1 , , insert it into the gap numbered c + 1 , move all the ∗ ’s to the right of theinserted l one descent to the left and add a ∗ in the rightmost descent between two letters.This creates c maj and no new block.(2) For each ( c, , insert it in the gap numbered i and move all the ∗ ’s to the right of theinserted n one descent to the left. This creates c maj and a new block.One can easily deduce from these insertion methods, in which order to delete α l l ’s in a partition π ∈ OP ( m, n ) k of type α and which data to record. Example 4.21.
We go through an example for φ . Start from an element in OP (3 , of type (2 , , : | | | | | . First, we delete the nonzero letters starting with the whose deletion has the smallest influence on thedinv. Deleted letter number of blocks changes dinv goes down by new partition3 No 1 | | | | | | | | | | | | | | | | | | | | | | | Then we do the same thing but backwards and looking at the inv instead of the dinv. We start from | | . Inserted letter number of blocks changes inv goes up by new partition1 No 2 | | | | | | | | | | | | | | | | | | | | | | | | | | So we get that φ (0 | | | | |
32) = 310 | | | | | . Indeed we check that dinv (0 | | | | |
32) = inv (310 | | | | |
2) = 14 . We look now at some restrictions of the maps φ and θ . Lemma 4.22.
We have(i) φ ( OP R ( m, n ) k ) = OP R ( m, n ) k ;(ii) φ ( OP L ( m, n ) k ) = OP L ( m, n ) k ;(iii) θ ( OP R ( m, n ) k ) = OP R ( m, n ) k .Proof. As φ and θ are bijections, it is enough to show the inclusions ⊆ in order to prove the equalities.We prove (i) and (iii) simultaneously.Take an element in OP R ( m, n ) k , i.e. an ordered multiset partition whose rightmost block does notcontain a zero. So all the letters in the rightmost block will be removed during the deletion phase of thealgorithm. When the last letter of the rightmost block gets deleted, the number of blocks goes down and he dinv and the maj goes down by units. Indeed, the rightmost gap is labelled in both the insertionmethods of dinv and maj.Starting from | | · · · | and inserting the nonzero letters, a necessary and sufficient conditionto obtain a path whose rightmost block does not contain a zero, is that at some point we must have ablock-creating insertion in the rightmost gap. Since in the inv insertion, the rightmost gap is labelled ,this means that we must have recorded a block-creating deletion with change of the statistic. By theargument above, this is always the case when starting from an element in OP R ( m, n ) k .The argument for (ii) is very similar. Starting from an element in OP L ( m, n ) k , we will delete all theletters in the leftmost block. Let h + 1 be the number of blocks right before we delete the last letter, callit i , in this block. By deleting i , the number of blocks goes down by and the dinv goes down by thelargest value possible, h , or the number of blocks in the resulting partition. We use the fact that, in ouralgorithm, the number of blocks right before deleting i is always equal to the number of blocks directlyafter inserting i . This implies that when we insert i , the fact that this must create a block and h inv,forces us to create a new block containing only i in the gap to the left of all the h existing blocks. Thus,the obtained partition will not contain a zero in its leftmost block. (cid:3) Reductions
Combining Theorem 4.20 with Lemma 4.22, we get immediately the following corollary.
Corollary 5.1.
We have (50) X π ∈ OP R ( m,n ) k q inv ( π ) x π = X π ∈ OP R ( m,n ) k q dinv ( π ) x π = X π ∈ OP R ( m,n ) k q maj ( π ) x π . Combining the previous corollary with Proposition 4.12, Proposition 4.13 and Remark 3.6, we getimmediately the following equivalence.
Theorem 5.2.
For m, n, k ∈ N , m ≥ and n > k ≥ , (51) ∆ h m ∆ ′ e n − k − e n (cid:12)(cid:12)(cid:12) t =0 = PLD x,q, ( m, n ) ∗ k if and only if ∆ h m ∆ ′ e n − k − e n (cid:12)(cid:12)(cid:12) q =0 = PLD x, ,t ( m, n ) ∗ k . So, in order to prove both cases q = 0 and t = 0 of the generalized Delta conjecture, it will be enoughto prove (51): this is what we will do in the rest of this article.6. Combinatorial recursions
Let
OPi
Lx,q ( m, n ) k := X π ∈ OP L ( m,n ) k q inv ( π ) x π = X π ∈ SOP L ( m,n ) k q inv ( π ) Q ides ( σ ( π )) be the q -enumerator of OP L ( m, n ) k for the inv statistic. Remark . Extending in the obvious way the map ξ − : OP R ( m, n ) k → PLD ( m, n ) ∗ n − k from Proposi-tion 4.12 to a map e ξ − from the all set OP ( m, n ) k , it is easy to see that OPi
Lx,q ( m, n ) k = P P ∈ S q dinv ( P ) x P ,where S is the set of Dyck paths of size m + n of area , labelled with nonnegative integers so that thelabels are strictly increasing along the columns, with the restriction that the rightmost column does notcontain a (for the partially labelled Dyck paths the restriction was on the leftmost column), with m labels equal to and m + k columns; for these objects the corresponding monomial and the dinv aredefined analogously as for the partially labelled Dyck paths.By a standard argument using LLT polynomials (see for example the argument sketched in [3, Re-mark 3.13]), it is easy to see that our sum over S is indeed a symmetric function, therefore OPi
Lx,q ( m, n ) k is a symmetric function as well. emma 6.2. For ≤ j < n , ≤ k ≤ n and m ≥ h ⊥ j OPi
Lx,q ( m, n ) k = k X s =0 q ( j − s ) (cid:20) mj − s (cid:21) q (cid:20) m + ss (cid:21) q OPi
Lx,q ( m + s, n − j ) k − s + k X s =0 q ( j − s ) (cid:20) mj − s (cid:21) q q m (cid:20) m + s − s − (cid:21) q × m + s X r =1 (cid:16) OPi
Lx,q ( m + s − r, n − j ) k − s + OPi
Lx,q ( m + s − r, n − j ) k − s +1 (cid:17) and OPi
Lx,q ( m, k = δ k, δ m, . Proof.
We will show this combinatorial recursion for the (finite) set
SOP L ( m, n ) k .Let us start with the second identity. The set SOP L ( m, k consists of the standard ordered setpartitions with m ’s, m + k blocks, and no non-zero elements, with no in the leftmost block. If m > then there must be a in the leftmost block, so the set is empty; if k > then there must be a blockcontaining a non-zero element, but there are none. It follows that the set is non-empty (and consists ofthe empty partition alone) if and only if m = k = 0 . Recalling that inv ( ∅ ) = 0 , we get the desired initialconditions.When applying h ⊥ j we should only consider set partitions π such that , , . . . , j appear in σ ( π ) indecreasing order, which means that no pair of these labels form an inversion in π (as h ⊥ j acts as on thecomplement). Let us call small cars these j labels.Let s be the number of small cars that are the minimum of their block (equivalently, whose blockdoes not contain a ). The recursive step consists of removing the small cars if their block contains a ,and turning them into ’s otherwise; next we standardize the resulting partition. If in this way we get apartition that contains a in its leftmost block, then we delete ’s from the leftmost blocks until it doesnot anymore. See Example 6.3.First we remove the j − s small cars from blocks that also contain ’s. The contribution of each of theselabels is equal to the number of ’s in blocks to its right, and since their block also contains a , thesecontributions must be different numbers ranging from to m − , which are q -counted by q ( j − s ) (cid:2) mj − s (cid:3) q .Now we have to distinguish two cases. If the leftmost block does not contain a small car, then thecontributions of the small cars that are the minimum of their block depend only on their interlacing withthe ’s, and these are taken into account by the factor (cid:2) m + ss (cid:3) q . If the leftmost block does contain a smallcar, then its contribution must be exactly q m ; the contribution of the remaining s − small cars is takeninto account by the factor (cid:2) m + s − s − (cid:3) q .After the standardization, in the former case we are left with a partition in OP L ( m + s, n − j ) k − s andwe are done. In the latter case, say that the resulting partition starts with r consecutive singletonsfollowed by a block whose minimum is non-zero, or r − singletons followed by a non-singleton blockwhose minimum is (in both cases r ≥ , because it had a small car in its leftmost block). Then weremove these ’s, and we are left with a partition in OP L ( m + s − r, n − j ) k − s or OP L ( m + s − r, n − j ) k − s +1 depending whether the rightmost of these ’s is a singleton or not (if it is, the partition loses r blocks;otherwise, it loses r − ).This completes the proof. (cid:3) Example 6.3.
Let π = 2 | | | | | | and j = 4 . First we delete all the small cars that liein the same block of some , getting | | | | | | ( inv decreases by ). Then we turn theother small cars to ’s, getting | | | | | | ( inv decreases by ). Next we delete the starting ’s, getting | | | | ( inv doesn’t change). Finally we standardize, getting | | | | ( inv doesn’t change). This completes the recursive step.Let OPd
Rx,q ( m, n ) k := X π ∈ OP R ( m,n ) k q dinv ( π ) x π = X π ∈ SOP R ( m,n ) k q dinv ( π ) Q ides ( σ ( π )) be the q -enumerator for the dinv statistic. Remark . Using Proposition 4.12, we know that
OPd
Rx,q ( m, n ) k = PLD x,q,t ( m, n ) ∗ n − k (cid:12)(cid:12) t =0 . s it is known that PLD x,q,t ( m, n ) ∗ n − k is a symmetric function (see for example a similar argumentsketched in [3, Remark 3.13]), this shows that OPd
Rx,q ( m, n ) k is a symmetric function as well.The following lemma shows that OPd
Rx,q ( m, n ) k satisfies the red formula (42) as q C ( m ) n,k does in Theo-rem 3.7. Lemma 6.5.
For ≤ j < n , ≤ k ≤ n and m ≥ h ⊥ j OPd
Rx,q ( m, n ) k = k X s =0 q ( j − s ) (cid:20) mj − s (cid:21) q (cid:20) m + ss (cid:21) q q s OPd
Rx,q ( m + s, n − j ) k − s + k X s =0 q ( j − s ) (cid:20) mj − s (cid:21) q (cid:20) m + s − s − (cid:21) q × m + s X r =1 (cid:16) OPi
Lx,q ( m + s − r, n − j ) k − s + OPi
Lx,q ( m + s − r, n − j ) k − s +1 (cid:17) and OPd
Rx,q ( m, k = δ k, δ m, . Proof.
We will show the combinatorial recursion for the (finite) set
SOP R ( m, n ) k . Let us start with thesecond identity. The set SOP R ( m, k consists of the ordered set partitions with m ’s, m + k blocks, andno non-zero elements, with no in the rightmost block. As before, the set is non-empty (and consistsof the empty partition alone) if and only if m = k = 0 . Recalling that dinv ( ∅ ) = 0 , we get the desiredinitial conditions.When applying h ⊥ j we should only consider set partitions π such that , , . . . , j appear in σ ( π ) indecreasing order, which means that no pair of these labels form a diagonal inversion in π (as h ⊥ j acts as on the complement). Let us call small cars these j labels.Let once again s be the number of small cars that are the minimum of their block. The recursive stepconsists of removing the small cars if their block contains a , turning them into ’s otherwise, and thenstandardizing the resulting partition. If in this way we get a partition that contains a in its rightmostblock, then we will conclude using a different argument.First we remove the j − s small cars from blocks that also contain ’s. The contribution of each ofthese small cars is equal to the number of ’s in blocks to its left, and since their block also containsa , these contributions must be different numbers ranging from to m − , which are q -counted by q ( j − s ) (cid:2) mj − s (cid:3) q .Now we have to distinguish two cases. If the rightmost block does not contain a small car, then thecontributions of the small cars that are the minimum of their block depend only on their interlacing withthe ’s, and these are taken into account by the factor (cid:2) m + ss (cid:3) q . If the rightmost block does contain asmall car, then its contribution must be exactly q m ; the contribution of the remaining s − small carsis taken into account by the factor (cid:2) m + s − s − (cid:3) q .After standardizing, in the former case we’re left with a partition in OP R ( m + s, n − j ) k − s and weare done. In the latter case, we remove the resulting from the rightmost block, and then we move thewhole block to the leftmost position, getting a partition in OP L ( m + s − , n − j ) k − s if that block wasnot a singleton, and a partition in OP ( m + s − , n − j ) k − s if it were. In both cases, the dinv decreasesby exactly k − s (which is the number of blocks in the resulting partition whose minimum is not a ): ifthe rightmost block was not a singleton, then the primary (resp. secondary) dinv involving the non-zeroelements of that block becomes secondary (resp. primary) dinv when we move the block to the left, andthe only lost contribution is the one coming from the deleted . Now, by Theorem 4.20 and Lemma 4.22we know that in OP L ( m + s, n − j ) k − s and in OP ( m + s − , n − j ) k − s the statistics dinv and inv havethe same distribution, so we can use the q -enumerator with respect to inv instead.We also have that SOP ( m + s − , n − j ) k − s = SOP L ( m + s − , n − j ) k − s ⊔ m + s G r =2 (cid:16) SOP L ( m + s − r, n − j ) k − s ⊔ SOP L ( m + s − r, n − j ) k − s +1 (cid:17) up to a natural statistic-preserving bijection) by deleting all the consecutive singletons on the left,and possibly the from the leftmost non- -singleton block, if there is any. Notice that this operationdoes not change the inv .It follows that the remaining factor is m + s X r =1 (cid:16) OPi
Lx,q ( m + s − r, n − j ) k − s + OPi
Lx,q ( m + s − r, n − j ) k − s +1 (cid:17) as claimed.This completes the proof. (cid:3) Example 6.6.
Let π = 2 | | | | | | and j = 4 . First we delete all the small cars that lie inthe same block of some , getting | | | | | | ( dinv decreases by ). Then we turn the othersmall cars to ’s, getting | | | | | | ( dinv decreases by ). Next we delete the rightmost ’s, which is a singleton, getting | | | | | ( dinv decreases by ). Now we standardize andthen apply the bijection mapping dinv to inv , and in the image we delete the starting (leftmost) ’s, aswe did in the previous recursion. This completes the recursive step.If the rightmost block was not a singleton small car, say if π = 92 | | | | | | , thenwhile deleting it we also move the rest of the block to the leftmost position, and at that step we get | | | | | | ( dinv decreases by the same amount as before), and now we know that theresulting partition does not have a in its leftmost block, so we are already done.7. Main results
In this section we prove the main results of this work.
Theorem 7.1.
OPi
Lx,q ( m, n ) k = q m · q C ( m ) n,k .Proof. We proceed by induction on n . The base case n = 0 (or n = 1 ) is straightforward to check.Let n ≥ . We are going to use that if two symmetric functions f, g ∈ Λ ( n ) with n > are such that h ⊥ j f = h ⊥ j g for all ≤ j ≤ n , then it is not hard to see that we must have f = g (cf. [13, Lemma 3.6]).For j = n , the expression for h ⊥ n q C ( m ) n,k from Lemma 3.10 coincides with the polynomial we get for OPi
Lx,q ( m, n ) k from the recursion.For ≤ j < n , comparing Lemma 6.2 and the blue formula (41) multiplied by q m , and using theinduction on n , we conclude that h ⊥ j OPi
Lx,q ( m, n ) k = h ⊥ j q m · q C ( m ) n,k .Therefore we proved that h ⊥ j OPi
Lx,q ( m, n ) k = h ⊥ j q m · q C ( m ) n,k for ≤ j ≤ n , which implies OPi
Lx,q ( m, n ) k = q m · q C ( m ) n,k , completing the proof of the theorem. (cid:3) We are finally in a position to prove (51).
Theorem 7.2.
OPd
Rx,q ( m, n ) k = q C ( m ) n,k .Proof. The proof is similar to the proof of Theorem 7.1.We proceed by induction on n . The base case n = 0 (or n = 1 ) is straightforward to check.Let n ≥ . We are going to use that if two symmetric functions f, g ∈ Λ ( n ) with n > are such that h ⊥ j f = h ⊥ j g for all ≤ j ≤ n , then it is not hard to see that we must have f = g (cf. [13, Lemma 3.6]).The case j = n is dealt with in the proof of Lemma 3.10.For ≤ j < n , consider the formula in Lemma 6.5.Since we know that OPi
Lx,q ( m + s − r, n − j ) k − s = q m + s − r · q C ( m + s − r ) n − j,k − s (same for the term with theextra +1 ) from Theorem 7.1, and that q C ( m + s − r ) n − j,k − s = OPd
Rx,q ( m + s − r, n − j ) k − s (same for the term withthe extra +1 ) by induction on the degree, we can replace it with m + s X r =1 q m + s − r (cid:16) OPd
Rx,q ( m + s − r, n − j ) k − s + OPd
Rx,q ( m + s − r, n − j ) k − s +1 (cid:17) . Comparing the resulting formula with the red formula (42), and using the induction on n , we concludethat h ⊥ j OPd
Rx,q ( m, n ) k = h ⊥ j q C ( m ) n,k .Therefore we proved that h ⊥ j OPd
Rx,q ( m, n ) k = h ⊥ j q C ( m ) n,k for ≤ j ≤ n , which implies OPd
Rx,q ( m, n ) k = q C ( m ) n,k , completing the proof of the theorem. (cid:3) he following theorem is the main result of the present article. Theorem 7.3.
For m, n, k ∈ N , m ≥ and n > k ≥ , we have both ∆ h m ∆ ′ e n − k − e n (cid:12)(cid:12)(cid:12) t =0 = PLD x,q, ( m, n ) ∗ k and ∆ h m ∆ ′ e n − k − e n (cid:12)(cid:12)(cid:12) q =0 = PLD x, ,t ( m, n ) ∗ k . Combining this theorem with the fact that the generalized Delta conjecture at q = 0 is equivalent tothe generalized Delta square conjecture of [3, Conjecture 3.12] at q = 0 (see the proof of [3, Theorem 5.1]for an argument), we get the following corollary, which generalizes [3, Theorem 5.1]. Theorem 7.4.
The generalized Delta square conjecture [3, Conjecture 3.12] at q = 0 holds true. Open problems
Combining Theorem 7.1 and Theorem 7.2 with Theorem 4.20 and Lemma 4.22, we proved that q m · OPd
Rx,q ( m, n ) k = OPd
Lx,q ( m, n ) k . It would be interesting to find a direct (bijective?) combinatorial proof of this identity. Notice thatsuch a proof would reduce Theorem 7.2 to Theorem 7.1.Observe also that the generalized Delta square conjecture [3, Conjecture 3.12] is not symmetric in q and t , so our results leave open the case t = 0 . References [1] F. Bergeron, A. M. Garsia, M. Haiman, and G. Tesler,
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