Recursions and divisibility properties for combinatorial Macdonald polynomials

. For each integer partition µ , let (cid:101) F µ ( q, t ) be the coefﬁcient of x 1 · · · x n in the modiﬁed Macdonald polynomial (cid:101) H µ . The polynomial (cid:101) F µ ( q, t ) can be regarded as the Hilbert series of a certain doubly-graded S n -module M µ , or as a q, t -analogue of n ! based on permutation statistics inv µ and maj µ that generalize the classical inversion and major index statistics. This paper uses the combinatorial deﬁnition of (cid:101) F µ to prove some recursions characterizing these polynomials, and other related ones, when µ is a two-column shape. Our result provides a complement to recent work of Garsia and Haglund, who proved a different recursion for two-column shapes by representation-theoretical methods. For all µ , we show that (cid:101) F µ ( q, t ) is divisible by certain q -factorials and t -factorials depending on µ . We use our recursion and related tools to explain some of these factors bijectively. Finally, we present fermionic formulas that express (cid:101) F (2 n ) ( q, t ) as a sum of q, t -analogues of n !2 n indexed by perfect matchings.


Introduction
The Macdonald polynomials P µ [Mac88] play a prominent role in algebraic combinatorics and the theory of symmetric functions.The polynomials P µ , as µ ranges over all integer partitions of n, form a basis for the vector space of symmetric polynomials of degree n with coefficients in the field Q(q, t).By specializing the parameters q and t in various ways, one can obtain many classical symmetric function bases from the Macdonald basis.For instance, P µ | t=1 is the monomial symmetric polynomial m µ ; P µ | q=1 is the elementary symmetric polynomial e µ ; P µ | q=t is the Schur symmetric polynomial s µ ; and P µ | q=0 is a Hall-Littlewood polynomial [Mac95].
Garsia, Haglund, and Haiman (among others) studied modified versions of Macdonald polynomials, denoted H µ , that have remarkable connections to representation theory and combinatorics [GH93,Hag04,HHL05a].To compute the modified polynomials H µ from the original polynomials P µ , one first multiplies each P µ by a certain scalar in Q(q, t) to obtain the integral forms J µ , then applies a suitable plethystic transform to obtain H µ , and finally sets H µ = t n(µ) H µ | t→1/t .See [GH98,HHL05b] for more details.One can also give a direct characterization of H µ in terms of triangularity and normalization axioms [HHL05a,HHL05b], or one can define H µ to be the eigenfunctions of the linear operator ∆ 1 [GH96].
In [GH93], Garsia and Haiman defined doubly-graded S n -modules M µ and conjectured that the Frobenius series for these modules were none other than the modified Macdonald polynomials H µ .This conjecture was ultimately proved by Haiman using deep results from commutative algebra and algebraic geometry [Hai01,Hai02].Let F µ ∈ Q(q, t) be the coefficient of x 1 x 2 • • • x n in H µ ; equivalently, F µ = H µ , h (1 n ) .It follows from Haiman's result that F µ is the Hilbert series of the doubly-graded module M µ ; i.e., F µ (q, t) = i≥0 j≥0 dim(M (i,j) µ )q i t j , where M (i,j) µ denotes the bihomogeneous component of M µ of bidegree (i, j).In a groundbreaking paper [Hag04], Haglund conjectured a combinatorial formula for F µ of the form where inv µ , maj µ : S n → N are suitable permutation statistics parametrized by µ.This formula leads directly to a combinatorial prescription for the expansion of H µ into fundamental quasisymmetric polynomials.These formulas, which are described in more detail in §2 below, were proved by Haglund, Haiman, and Loehr [HHL05a,HHL05b].Haglund's formula opened the way to new explorations of the combinatorial properties of Macdonald polynomials and the associated Hilbert series F µ .A problem of ongoing interest involves the search for recursions characterizing the polynomials F µ , as µ ranges over all integer partitions.One such recursion follows from a version of the Pieri rules for Macdonald polynomials [Mac95].This recursion has the form where we sum over all partitions ν that can be obtained from µ by removing a single corner box, and c µ,ν (q, t) ∈ Q(q, t) is a complicated quotient of two polynomials built from the arm and leg numbers of the cells in µ and ν [GH98].While this recursion is very striking, the presence of rational coefficients makes it difficult to use in combinatorial settings.For instance, there is presently no known combinatorial proof of this recursion based on (1), nor can one easily deduce the polynomiality of F µ from (2).In a very recent paper [GHar], Garsia and Haglund found a new recursion valid for polynomials F µ indexed by shapes µ with two columns.They proved this recursion using a multitude of clever representation-theoretical manipulations.In related work, M. Yoo used (1) to derive a combinatorial formula for polynomials F µ indexed by partitions µ of hook shape [Yoo09].Some results for two-row shapes and rectangles appear in J. Bandlow's doctoral dissertation [Ban07].
The Garsia-Haglund recursion builds up a given shape µ from smaller shapes ν obtained by removing one of the inner corners of µ.In this paper, we shall present different recursions characterizing the polynomials F µ (where µ has two columns) that are based on removing the bottom row of µ.Our recursions are stated in Theorem 6.10 and Theorem 6.16 below.We will give a fully combinatorial proof of our recursions using bijections and combinatorial operations on permutations weighted by the statistics inv µ and maj µ .For our approach, it is more convenient to use F µ (q, t) = t n(µ) F µ (q, 1/t); this entails replacing maj µ by a related statistic comaj µ , which is defined in §2. For One consequence of our recursion is a combinatorial explanation of the fact that F (2 m ) (q, t) is divisible by [m]! t .We will prove the following more general divisibility property in §4.
As we will see, (a) follows bijectively from a certain combinatorial move on permutations.Then (b) follows algebraically from (a) using the known symmetry property H µ (X; q, t) = H µ (X; t, q) of Macdonald polynomials [Mac95].One of our motivations for developing the recursions below is to obtain a greater understanding of why (b) holds at the combinatorial level.Proving (b) bijectively for general µ appears to be a difficult problem.We hope that the special cases considered here will yield some clues to the even more challenging problem of giving a bijective proof of the general symmetry property 2 Combinatorial Formula for F µ This section reviews Haglund's combinatorial formula for F µ (q, t) and defines a related polynomial F µ (q, t).Definition 2.1 A partition of a positive integer n is a sequence µ = (µ 1 , µ 2 , . . ., µ k ) with for some k.Let Par(n) be the set of all partitions of n.Definition 2.2 A standard filling T of shape µ ∈ Par(n) is a placement of the integers 1, 2, . . ., n in the Ferrers diagram of µ with each integer used exactly once.A filling can be identified with a permutation w ∈ S n by reading the rows of the diagram from left to right, beginning with the shortest row.The column words of a filling are obtained by reading the columns from top to bottom.
In order to define the statistics inv µ (T ) and comaj µ (T ), we first recall some classical permutation statistics defined for w ∈ S n .
Definition 2.4 Fix w = w 1 w 2 • • • w n ∈ S n .An inversion of w is a pair i < j such that w i > w j .Note that i and j do not need to be consecutive.Let inv(w) be the number of inversions of w.A descent of w is an index i < n with w i > w i+1 .The descent set of w is Des(w) = {i : w i > w i+1 }.The major index of w is maj(w) = i∈Des(w) i.Similarly, define the ascent set Asc(w) = {i : w i < w i+1 } and the co-major index comaj(w) = i∈Asc(w) i. Definition 2.10 For all partitions µ, let F µ be the set of all standard fillings of the shape µ.
Haglund's combinatorial formula [Hag04] for F µ (q, t) is Note that comaj µ (T ) can also be written as the sum of the co-major indices of the column words of T .
In the following, we will use the specialization

The Inversion Flip Operation
This section describes a combinatorial operation on fillings called the inversion flip that will allow us to partially sort the bottom row of a filling T ∈ F µ .A similar operation on fillings that may have repeated symbols was studied by Bandlow [Ban07, Section 5.5].
Definition 3.1 Suppose µ ∈ Par(n) and i ∈ N + satisfy µ i = µ i+1 .Define the inversion flip move s i : F µ → F µ as follows: • Given T ∈ F µ , let a (resp.b) be the entry of T at the bottom of column i (resp.i + 1).
• Switch entries a and b in the bottom row as shown here: • If a, c, d and b, c, d are either both inversion triples or both not inversion triples, the move is complete.Otherwise, apply s i recursively to the filling T of (µ 2 , µ 3 , . ..) obtained by ignoring the bottom row of T .
Proof: Both properties follow directly from the definition of s i . 2 Remark 3.3 The s i 's do not satisfy the braid relations in general; i.e., when µ i = µ i+1 = µ i+2 , we may have This can be seen in Figure 1.
Proposition 3.4 Given µ ∈ Par(n) and i ∈ N + with µ i = µ i+1 , let T ∈ F µ have entries a and b in the bottom row of columns i and i + 1, respectively.Then: Proof: (cf.[Ban07, Section 5.5]) Since every column contributes independently to comaj µ and maj µ , to prove (a) and (b) it is sufficient to consider a filling T of shape µ = (2 n ) and i = 1.By 3.2(a) we may also assume that the bottom row of T is increasing.We must show that comaj µ (s 1 (T )) = comaj µ (T ), which automatically yields maj µ (s 1 (T )) = maj µ (T ).This result is true when n = 1.
It is also possible that In this case, if a, c, d formed an inversion triple in T , then b, c, d forms an inversion triple in then the location and number of the ascents are preserved, so comaj µ (s 1 (T )) = comaj µ (T ).If a < d < c < b, the number of ascents is preserved, but the column in which the ascent is located changes.However, since the columns are of equal height, , the location and number of ascents are preserved.If a < c < d < b, the number of ascents are preserved, but the column in which the ascent is located changes.As before, since the columns are of equal height, comaj µ (s 1 (T )) = comaj µ (T ).By induction, this result holds for any number of rows.
To show that inv µ (s i (T )) = inv µ (T ) + 1, we must show that the inversion flip does not affect the total number of inversion triples, excluding the triple a, b, ∞.By the definition of the inversion flip, it is sufficient to consider triples positioned as shown: since all other triples in T will be preserved.Once again, there are two possibilities.First, if the inversion triples themselves are preserved.On the other hand, if to show that the total number of inversion triples is preserved requires a tedious case analysis.We present several cases here and leave the remainder to the reader.First, suppose z < a < b < c < d.Then none of z < a < c, z < b < c, z < a < d, and z < b < d are inversion triples.Next, if a < z < b < c < d, then a < z < c is an inversion triple in T , and z < b < c is not an inversion triple in s i (T ).On the other hand, z < b < d is not an inversion triple in T , but a < z < d is an inversion triple in s i (T ).Thus, the total number of inversion triples is preserved.The remaining cases are similar.2 4 Divisibility Properties for F µ We now prove Theorem 1.
First we prove (a) bijectively by using the inversion flip.We begin by considering a filling T of shape µ = (m n ).For 1 ≤ i ≤ m, let k i be the ith smallest entry in the bottom row of T , and let c 1 , c 2 , . . ., c m be the cells in the bottom row of µ.
Continue likewise to move each k i into cell c i in decreasing order.The sequence of inversion flips used to sort the k i 's gives rise to a permutation w ∈ S m , by replacing each s j in the sequence by the basic transposition (j, j + 1).Let T be the filling that results from sorting T in this way.By 3.4(c), inv µ (T ) = inv µ (T ) + inv(w) and comaj µ (T ) = comaj µ (T ).Hence, T ∈Fµ with increasing row 1 q invµ(T ) t comaj µ (T ) = [m]!q T ∈Fµ with increasing row 1 q invµ(T ) t comaj µ (T ) .This result can be extended to µ ∈ Par(n) with µ = (1 m1 2 m2 • • • p mp ) by applying the above argument to each set of m i columns of equal height.Thus F µ (q, t) is divisible by It follows that F µ (q, t) = t n(µ) F µ (q, 1/t) is divisible by the same factors.We remark that the divisibility property 1.1(a) for rectangles is also a consequence of [Ban07], Corollary 3 in Section 5.5.
Part (b) of 1.1 follows algebraically from (a) and the fact that F µ (q, t) = F µ (t, q).
Considering the first two columns, the transposition associated with s 1 is (1, 2), so the permutation generated by the inversion flip is 12 ∈ S 2 .Considering the last three columns, the transpositions associated with s 3 and s 4 are (3, 4) and (4, 5), respectively.This gives us the permutation 534 of {3, 4, 5}.

The Cyclic Shift Operation
We wish to obtain a combinatorial explanation for part (b) of Theorem 1.1, at least in the case where µ is a two-column rectangle.For this, we must first define another bijection on standard fillings.(c) inv µ (cyc(T )) = inv µ (T ).
Proof: This can be seen by considering what happens to the cell c containing n in a filling.
The ascent between n and the cell above (if any) is shifted downward by a unit, while the descent between n and the cell below are shifted up one unit.All other ascents and descents are unaffected.All triples in T which do not involve n are preserved.Triples in T which include n have the form a < b < n.The corresponding triple in T will be 1 < a + 1 < b + 1.Thus, the status of the triple will be preserved.The assumption that n is not in the bottom row is needed to ensure that no triple involving n also involves an ∞ below the bottom row.Then Proof: First notice that when 2n is in the bottom row of T , because the ascent which occurs in T between 2n and the entry above it is lost.It follows from 3.4 that

Recursions for Two-Column Shapes
In this section, we consider what happens when we fix the bottom row of the fillings of (2 n ) to be 1 a .This will lead to a recursion characterizing the polynomials F (2 n ) (q, t).

Augmentation
We consider next what A a does to inv µ and comaj µ and how it interacts with cyc and s 1 .Here and below, for any statement P , we write χ(P ) = 1 if P is true and χ(P ) = 0 if P is false.Proposition 6.4For all n ≥ 2, all a with 2 ≤ a ≤ 2n, and all T ∈ F (2 n−1 ) with bottom row x z , Proof: Let T ∈ F (2 n−1 ) have bottom row x z , and let 2 ≤ a ≤ 2n.The bottom two rows of A a (T ) look like: x z 1 a since the two entries in the bottom row of A a (T ) do not form an inversion triple, and 1, x , z form an inversion triple in A a (T ) iff ∞, x, z form an inversion triple in T .The ascents in the top n − 1 rows of A a (T ) are the same as the ascents in T .There cannot be an ascent between the two lowest cells in column 1 of A a (T ) since the lowest cell in this column contains 1.If a is greater than the entry directly above it in A a (T ), that entry is z + 1, giving a new ascent, so If a is less than the entry directly above it in A a (T ), that entry must be z + 2, and thus comaj (2 n ) (A a (T )) = comaj (2 n−1 ) (T ). 2 Definition 6.5 For all n ≥ 2 and all a with 2 ≤ a ≤ 2n, define cyc : Then, by 5.3 and 6.4, By 5.3 and 6.4, inv (2 n ) (cyc j (A a (T ))) = inv (2 n−1 ) (T ).

Recursions for Two-Column Rectangles
We can compute R (2 n ),a for 2 ≤ a ≤ n + 1 by the recursion below.A nicer version of this recursion appears in 6.10 below.
Proof: To prove the theorem bijectively, we will decompose and Using 3.2, 5.2, and 5.5, one sees that each T ∈ F (2 n ),a can be written as cyc j (s 1 (A a (U ))) for exactly one choice of j ∈ {0, 1, . . ., 2n − 3}, ∈ {0, 1}, b ∈ {2, . . ., n}, and U ∈ F (2 n−1 ),b .To find j, , b, and U , delete the bottom row of T , renumber the remaining entries to be {1, 2, . . ., 2n − 2}, cyclically shift down until 1 appears in the new bottom row, and then apply s 1 if needed to get 1 in the lower-left corner.For example, if n = 5 and a = 4, the filling for all i, j in the indicated ranges.Before continuing the proof, we give an example that illustrates the main ideas in the calculations that follow.Notice that q inv (2 3 ) (U ) t comaj (2 3 ) (U ) = q 2 t 1 .In Figure 2, the fillings in bold are elements of X 3 derived from U , while the rest are elements of Y 3 derived from U .The generating functions for these elements in X 3 and Y 3 are [3] t (1 + t)q 2 t 1 and [3] t (qt 2 + qt 3 )q 2 t 1 , respectively.
Returning to the proof, we first consider the case where 2 ≤ b < n and a ≤ b+1.Let U ∈ F (2 n−1 ),b .By 6.6, comaj  q 2 t q 2 t 2 q 2 t 3 q 2 t 4 q 3 t 5 2 2 2 7 7 7 6 6 6 5 5 5 3 3 3 8 8 8 1 1 1 4 4 4 q 3 t 4 q 3 t 5 q 3 t 3 q 3 t 4 q 2 t 2 q 2 t 3 We now compute 2n−3 i=0 t comaj (2 n ) (cyc i (Aa(U ))) by splitting the sum into the ranges 0 For the remaining values of i, we obtain Thus, the generating function for . Similarly, we can use 6.6 and 5.6 to obtain: The bottom row of cyc m+2n−x (T ) is x z where x = m + 2n and z = m + 2n − x + z.Then cyc m+2n−x+1 (T ) has bottom row 1 b with b = m + 2n − x + z + 1.The ascent involving x and the entry directly above it in cyc m+2n−x (T ) is lost when cyc is applied, so Since m + 2n is in the bottom row of cyc m+2n−x (T ), We can calculate R (1 m 2 n ),a for 2 ≤ a ≤ m + 2n by the recursion below.
Theorem 6.16For all m ≥ 1, n ≥ 2, and 2 ≤ a ≤ m + 2n, Then there exists a unique U ∈ F (1 m 2 n−1 ) such that T = A a (cyc i (U )) and the bottom row of U is either 1 b or b 1 for some b with 2 ≤ b ≤ m + 2n − 2 and some i with If the bottom row of U is 1 b , then by 6.14 and 5.3 If the bottom row of U is b 1 , then by 6.14 and 5.3, Let U = cyc m+2n−1−b (U ), then, by 6.15, Thus the second term in the recursion accounts for fillings T arising from those U having a decreasing bottom row. 2 Remark 6.17 We saw earlier that The recursion above easily yields that every R (1 m 2 n ),a (q, t) is divisible by [m]! t .Computations suggest that these polynomials are also divisible by [n − 1]! t , but this fact does not seem to be evident from the form of our recursion.In particular (in contrast to the situation for two equal columns), we have not given a combinatorial explanation of the divisibility of F (1 m 2 n ) (q, t) by [n]! t .

Fermionic Formula
In the case of two-column rectangles, we can use our earlier recursion to deduce some fermionic formulas for F (2 n ) (q, t) and R (2 n+1 ,a) (q, t).For each n ≥ 1, define Iteration of this recursion immediately yields the following fermionic formula for R (2 n+1 ),a .

Now let PM(K 2n
) denote the set of perfect matchings of the complete graph K 2n on 2n vertices.By definition, an element M ∈ PM(K 2n ) is a set partition of {1, 2, . . ., 2n} into n blocks of size 2.There is a bijection f : P n → PM(K 2n ) that can be defined recursively as follows.For n = 1, f ((2)) = {{1, 2}}.Given n > 1 and (a 1 , . . ., a n ) ∈ P n , recursively compute M = f (a 2 , . . ., a n ) ∈ PM(K 2n−2 ).Relabel the perfect matching M to use the vertex set {1, 2, . . ., 2n} ∼ {1, a 1 }, and add the edge {1, a 1 } to obtain the perfect matching M = f (a 1 , . . ., a n ).We can use this bijection to change the indexing set for the summation in 7.1 from P n to PM(K 2n ).We thereby see that R (2 n+1 ),a can be written in the form where [2 n ] M,a denotes a q, t-analogue of 2 n that depends on the matching M and the integer a.In particular, we have such an expression for F (2 n ) = R (2 n+1 ),2 .This yields a q, t-analogue of the enumeration formula | PM(K 2n )| = (2n)!n!2 n .It is also natural to incorporate a into the matching M by summing over perfect matchings on 2n + 2 vertices that contain the edge {1, a}.This viewpoint produces a formula of the form We calculate (a 0 , a 1 , a 2 , a 3 ) = f −1 (M ) = (6, 3, 4, 2).Referring to 7.1, we see that [2 3 ] M = (t 2 + qt 3 )(1 + q)(t + qt).

Future Work
The computations in [Nie10] suggest that the recursions considered here may extend to polynomials F µ indexed by partitions µ with more than two columns.In particular, the case of a three-column rectangle appears to be tractable, although the required combinatorial manipulations become substantially more involved compared to two-column rectangles.These extensions of the current work, which are treated in the latter part of [Nie10], will be the subject of future papers.

Definition 6. 2
For n ≥ 1, T ∈ F (2 n ) and 2 ≤ a ≤ 2n + 2, we define the a-augmentation of T to be the filling A a (T ) ∈ F (2 n+1 ),a which is obtained by first relabeling the entries of T as follows: c ∈ T is replaced by c + 1 if c < a − 1 and by c + 2 if c ≥ a − 1; and then placing the relabeled filling over the new bottom row 1 a .Example 6.3For a = 3 and T ∈ F (2 3 ) given by
It follows that F (2 n ),a is the disjoint union of the sets X b and Y b just defined.By 5.6 and 6.6, inv (2 n )