DEDICATED TO DOMINIQUE FOATA. Abstract. In classical rook theory there is a fundamental relationship between the

Transcription

1 ROOK THEORY FOR PERFECT MATCHINGS J Haglund and J B Remmel April 4, 00 DEDICATED TO DOMINIQUE FOATA Abstract In classical roo theory there is a fundamental relationship between the roo numbers and the hit numbers relative to any board In that theory the -th hit number of a board B can be interpreted as the number of permutations whose intersection with B is of size In the case of Ferrers boards there are q-analogues of the hit numbers and the roo numbers developed by Garsia and Remmel [GaRe], Dworin [D], [D] and Haglund [H] In this paper we develop a roo theory appropriate for shifted partitions, where hit numbers can be interpreted as the number of perfect matchings in the complete graph whose intersection with the board is of size We show there is also analogous q-theory for the roo and hit numbers for these shifted Ferrers boards Introduction Perfect Matchings and Roo Boards In classical roo theory there is a fundamental relationship between the roo numbers and the hit numbers relative toany board A board B is a subset of the n n board A n pictured in Fig n n- = A n n- n Figure 99 Mathematics Subect Classication 05A05, 05A0 Key words and phrases Roo theory, perfect matching, q-analogue The wor of the rst author was supported in part by NSF grant DMS-9674 Typeset by AMS-TE

2 J HAGLUND AND J B REMMEL Given a board B A n,we let R (B) denote the set of all element subsets p of B such that no two elements of p lie in the same row or column Such a set p is called a roo placement of nonattacing roos on B and r (B) =R (B) is called the -th roo number of B For eample, if B A 4 is the board consisting of all shaded squares in Fig, then r 0 (B) =,r (B) =6,r (B) = 0, r (B) = 4, and r 4 (B) =0 4 B 4 Figure Given a permutation in the symmetric group S n,we identify with the roo placement p = f(i; ) : (i) =g We then dene H ;n (B) to be the set of all S n such that p \B = and we call h ;n (B) =H ;n (B) the -th hit number of B relative toa n One can easily prove the following formula which relates the roo numbers r (B) to the hit numbers h ;n (B) for any board B A n n =0 h ;n (B)(z +) = n =0 r (B)(n, )!z : () That is, it is easy to see that the left-hand side of () equals the sum (T;p ) T p \B z T : () However, the right-hand side of () also counts () since we can rst pic T R (B) and then etend it to a placement p for some S n in (n, )! ways Replacing z by z, in () gives the following classical formula of Riordan and Kaplansy [KaRi] n =0 h ;n (B)z = n =0 r (B)(n, )!(z, ) : () Garsia and Remmel [GaRe] gave aq-analogue of the roo numbers and hit numbers for a certain collection of boards B A n called Ferrers boards Let A(a ;a ;::: ;a n ) denote the board B contained in A n consisting of all squares f(i; ) : a i gfor eample, A(; ; ; ) is pictured in Fig

3 4 = A(,,, ) 4 Figure Thus A(a ;a ;::: ;a n ) denotes the board whose column heights reading from left to right are a ;a ;::: ;a n We shall call a board A(a ;a ;::: ;a n ) A n a syline board A(a ;a ;::: ;a n ) is called a Ferrers board if a a a n Let F = A(a ;a ;::: ;a n ) be some ed Ferrers board contained in A n Given a placement p R (B), let each roor cancel all squares to its right and all squares below r We let u F (p) denote the number of squares of F which are uncancelled, ie the number of squares which are neither in p nor cancelled by a roo in p For eample, if F = A(; ; ; ; 4; 4) and p is the placement ofr (F ) consisting of the squares containing an in Fig 4, then we put dots in the squares which are cancelled by arooinp Then u F (p) =5isthenumber of uncancelled squares, ie the squares which contain neither a dot nor an u F (p) = 5 Figure 4 Garsia and Remmel [GaRe] then dened a q-analogue of r (F )by setting r (F; q) = pr (F ) q u F (p) : (4) Garsia and Remmel proved [GaRe] that if F = A(a ;::: ;a n ) where 0 a a n n, then ny [ + a i + i, ] = n =0 r n, (F; q)[] # (5)

4 4 J HAGLUND AND J B REMMEL where [n] =+q + :::+ q n, =,qn,q and [n] # =[n][n, ] [n, + ] We also dene [n]! = [n][n, ] [][] and n = [n]! []![n, ]! : Garsia and Remmel also dened a q-analogue of the hit numbers, h ;n (F; q), for Ferrers boards by the formula n =0 h ;n (F; q)z = n =0 r (F; q)[n, ]! ny i=n,+ (z, q i ): Garsia and Remmel proved that h ;n (F; q) is a polynomial in q with nonnegative coecients In fact, they proved that there is a statistic s F (p) such that h ;n (F; q) = ph ;n (F ) q s F (p) : (6) However, Garsia and Remmel did not provide a direct description of s F (p) but only dened s F (p) indirectly via a recursive denition Later Dworin [D], [D] and Haglund [H] independently gave direct descriptions of statistics s F;d (p) and s F;h (p) on p H ;n (F ) such that h ;n (F; q) = ph ;n (F ) q s F;d(p) = ph ;n (F ) q s F;h(p) : The Dworin statistic s F;d (p) and the Haglund statistic s F;h (p) have very similar descriptions Given a placement p H ;n (F ), rst let each roo cancel all squares to its right Then for each roor =(i; ) which is not in F, r cancels all squares below r which are not in F Finally for each roor =(i; ) inf, in the Dworin statistic the roo cancels all squares below r, plus all squares o the board in its column, and s F;d (p) is the number of uncancelled squares In the Haglund statistic, each roorin F cancels all squares in F which lie above r, plus all squares o the board in its column, and s F;h (p) is the number of uncancelled squares For eample, in Fig 5, we picture the two types of cancellations for a placement p H ;6 (F ) where F = A(; 4; 4; 4; 4; 4) Once again, we indicate the squares of the placement by placing an in those squares and we indicate the cancelled squares by placing a dot in the cancelled squares We should note that the methods of proof employed by Dworin [D] and Haglund [H] are very dierent and up until now there was no nown weight preserving biection which shows that both statistics give rise to the same q-analogue of the hit numbers for Ferrers boards (As part of our research for this article we discovered such a biection, which we describe in section 5) Indeed, it is easy to see that the denitions of s F;d (p) and s F;h (p) mae sense for any syline board F = A(a ;::: ;a n ) However, Dworin proved combinatorially that for any syline board F = A(a ;::: ;a n ) and any permutation S n, ph (F ) q s F;d(p) = ph ((F )) q s (F );d (p) (7)

5 5 Dworin Cancellation Haglund Cancellation s F,d (p) = 9 s F,h (p) = 0 Figure 5 where (F )=A(a () ;::: ;a (n) ) Haglund showed that (7) does not always hold if s F;d (p) and s (F );d (p) are replaced by s F;h (p) and s (F );h (p) respectively The main purpose of this paper is to prove analogues of the results described above where we replace permutations by perfect matchings Our wor was initially inspired by unpublished wor of Reiner and White [ReWh], who suggested that one consider the board B n pictured in Fig 6 n- n- n- n = B n Figure 6 Note that for the board A n, a roo placement p is ust a partial permutation, ie a set of squares of A n that can be etended to a permutation p for some S n For the board B n,we replace permutations by perfect matchings of the complete graph K n on vertices ; ;::: ;n That is, for each perfect matching m of K n consisting of n pairwise verte disoint edges in K n,we let p m = f(i; ) : i<and fi; g mg where (i; ) denotes the square in row i and column of B n

6 6 J HAGLUND AND J B REMMEL according to the labeling of rows and columns pictured in Fig 6 For eample, p m is pictured in Fig 7 where m = ff; 4g; f; 7g; f; 5g; f6; 8gg is a perfect matching of K Figure 7 For the board B n,wethus dene a roo placement to be a subset of some p m for a perfect matching m of K n Given a board B B n,we let M (B) denote the set of element roo placements of B and let m (B) =M (B) Similarly, we let F ;n (B) =fp m : p m \ B = and m is a perfect matching of K n g and let f ;n (B) =F ;n (B) We call m (B) the -th roo number of B and f ;n (B) the -th hit number of B One can prove that n =0 f ;n (B)z = n =0 in much the same way that one proved () Here we let n!! = ny (i, ) and [n]!! = m (B)(n, )!!(z, ) (8) ny [i, ]: The analogue of a syline board in this setting is a board B(a ;a ;::: ;a n )= f(i; i + ) : a i g Thus B(a ;a ;::: ;a ) is the board whose row lengths are a ;a ;::: ;a respectively Wesay that B(a ;::: ;a ) is a shifted Ferrers board if n, a a a 0, and the non-zero entries of a ;::: ;a are strictly decreasing For eample, B(5; ; ; ; 0; 0; 0) B 8 is pictured in Fig 8 We note that if we identify a board B B n with the graph G B =(V;E B ) where V = f;::: ;ng and E B = ffi; g :(i; ) Bg, then the graph of a shifted Ferrers board is called a threshold graph in the graph theory literature Our investigation of roo numbers and hit numbers was, in part, motivated by trying to nd a q-analogue of the following formula of Reiner and White which

7 7 = B(5,,,, 0, 0, 0) Figure 8 holds for any shifted Ferrers board F = B(a ;::: ;a ) B n Y ( + a n,i, i +)= =0 m (F )() ##, : (9) Here () ## = (, )(, 4) (, + ) We can dene q-roo numbers for which aq-analogue of Reiner and White's formula (9) holds as follows We say that roo (i; ) with i<in a roo placement roo-cancels all cells (i; s) inb n with i<s<and all cells (t; ) and (t; i) with t<i For eample the cells roo-cancelled by (4; 7) in B 8 are pictured in Fig Figure 9 Given a shifted Ferrers board F = B(a ;::: ;a ) B n and a placement p M (F ), we let u F (p) denote the number of cells of F which are neither in p nor roo-cancelled by arooinp Then we dene m (F; q) = pm (B) q u F (p) : (0)

8 8 J HAGLUND AND J B REMMEL This given, we shall prove the following q-analogue of (9) Y [ + a n,i, i +]= =0 m (F; q)[] ##, () where [] ## =[][, ] [, + ] We dene the q-analogue of the hit numbers for F by dening f ;n (F; q) via the formula n =0 f ;n (F; q)z = n =0 m (F; q)[n, ]!! ny i=n,+ (z, q i, ): () We shall show that one can dene a Dworin type statistic t F (p) for p F (F ) such that f ;n (F; q) = q t F (p) () pf (F ) so that () ensures that the f ;n (F; q) are polynomials in q with nonnegative coecients We note that the q-roo numbers m (F; q) appear as a special case of a more general roo placement model due to Remmel and Wachs [ReWa] Our results suggest that there is a natural etension of q-hit numbers that can be dened in their model However, there is no obvious way to dene the analogue of our perfect matchings in the Remmel-Wachs model much less how one could nd a statistic The outline of this paper is as follows In section we develop basic results for the q = case of roo numbers and hit numbers for shifted Ferrers boards In section we dene natural q-analogues of the roo and hit numbers for shifted Ferrers boards and prove some basic identities that these numbers satisfy Our basic denition of the q-hit numbers for shifted Ferrers boards is algebraic However we also dene a combinatorial interpretation of these numbers In section we prove the combinatorial interpretation and the algebraic denition of the q-hit numbers for shifted Ferrers boards are the same Section 4 contains a number of algebraic identities satised by the q-roo and q-hit numbers for shifted Ferrers boards, which are used in the proofs of our theorems In section 5 we introduce new families of statistics for the q-hit numbers in both the classical Ferrers board and shifted Ferrers board case so that q-counting permutations/perfect matchings with respect to these statistics generate the corresponding q-hit numbers In the classical case this will give a direct proof that the statistics introduced by Dworin [D], [D] and Haglund [H] give rise to the same q-hit numbers Basic results for roo numbers and hit numbers for boards in B n In this section, we shall prove anumber of basic results for the hit numbers and roo numbers for boards contained in B n Let PM(B n )=fp m : m is a perfect matching of K n g It is easy to see that PM(B n ) = n!! That is, there are n, choices for an edge that contains verte, ie f;ig; i = ;::: ;n If we pic an edge f;g, then the number of ways to complete f;g to a perfect matching of K n is clearly ust the number of perfect matchings on the complete graph on vertices f;::: ;ng,f;g Thus PM(B n ) =(n, )PM(B n, ) and hence

9 PM(B n ) = (n, ) = n!! by induction More generally, it follows that if we are given pairwise verte disoint edges fi ; g;::: ;fi ; g in K n, then the number of ways to etend fi ; g;::: ;fi ; g to a perfect matching of K n is equal to PM(B n, ) =(n, )!! Now recall that given a board B B n, F ;n (B) =fp m PM(B n ): p m \ B = g and the -th hit number of B is f ;n (B) =F ;n (B) A set p B is a roo placement of B if p B \ p m for some p m PM(B n ) We let M (B) denote the set of all -element roo placements of B and we dene m (B) =M (B) to be the -th roo number of B Our rst result is the analogue of () for B n Theorem Let B be aboard inb n Then n =0 f ;n (B)(z +) = n =0 Proof It is easy to see that the left-hand side of (4) is ust (T;pm) T pm\b pmpm(b n ) m (B)(n, )!!z : (4) z T : (5) On the other hand, for each roo placement T B, there are (n, )!! ways to etend T to a perfect matching p m PM(B n )ift = so that the right-hand side of (4) is also equal to (5) Note if we replace z by z, in (4), we get the following analogue of the Riordan-Kaplansy formula () for any B B n n =0 f ;n (B)z = n =0 m (B)(n, )!!(z, ) : (6) Net we prove anumber of simple recursions for the roo numbers and hit numbers of B n -boards To this end, given a board B B n and a pair (i; ) B with i<,we dene two boards, B=(i; ) and B=(i; ) B=(i; ) is ust the board which is the result of removing the square (i; ) from B B=(i; ) is the board contained in B n, which is obtained as follows First let C n denote the set of (i;) all squares of B n which have either i or as a coordinate It is easy to see that B n, C n will be a copy ofb (i;) n, ecept that it will involve the coordinates f;::: ; ng, fi; g instead of f;::: ; n, g Thus we can isomorphically map the resulting board onto B n, by sending a coordinate to ' i; () where ' i; () = 8 >< >:,, if <i if i<< if <: Then B=(i; ) =f(' i; (s);' i; (t)) : (s; t) B, C n g: (i;) This process is pictured in Fig 0 for the board B = B(6; 4; ; ; 0; 0; 0) and (i; ) =(; 5) In Fig 0, we construct B=(; 5) and B=(; 5) and we indicate the cells in B 8 which have a coordinate equal to or 5 by placing dots in those squares This given, we have the following 9

10 0 J HAGLUND AND J B REMMEL B/(,5) 5 B/(,5) Figure 0 Theorem For any board B B n and (i; ) B, (i) m (B) =m (B=(i; )) + m, (B=(i; )): (7) (ii) f ;n (B) =f ;n (B=(i; )) + f,;n, (B=(i; )), f ;n, (B=(i; )): (8) Proof For recursion (i), we simply classify the -element roo placements p according to whether (i; ) p That is, let M (i;) (B) =fp M (B) :(i; ) pg Then it is easy to see that M (B=(i; )) = M (B),M (i;) (B) Moreover ' i; induces a : correspondence between M (i;) (B) and M, (B=(i; )) That is, if p M (i;) (B), then we let ' i; (p) =f(' i; (s);' i; (t)) : (s; t) p,f(i; )gg Recursion (i) easily follows To prove recursion (ii), we again partition the p m F ;n (B) into two sets according to whether (i; ) p m Let F (i;) ;n (B) =fp m F ;n (B) :(i; ) p m g Again ' i; induces a : correspondence between F (i;) ;n (B) and F,;n,(B=(i; )) where if p m F (i;) ;n (B), then ' i;(p m )=f(' i; (s);' i; (t)) : (s; t) p m,f(i; )gg Net consider F ;n (B=(i; )) Note that F ;n (B=(i; )),F (i;) ;n (B=(i; )) = F ;n(b), F (i;) ;n (B) That is, if (i; ) = p m, then p m \ B = p m \ B=(i; ) By the same

11 argument asabove ' i; induces a : correspondence between F (i;) ;n (B=(i; )) and F ;n, ((B=(i; ))=(i; )) However, it is easy to see that (B=(i; ))=(i; ) =B=(i; ) Thus F ;n (B=(i; )) = F ;n (B), F (i;) ;n (B) + F ;n,(b=(i; )) or equivalently F ;n (B), F (i;) ;n (B) = f ;n(b=(i; )), f ;n, (B=(i; )) Since F (i;) ;n (B) = F,;n, (B=(i; )) = f,;n, (B=(i; )), recursion (ii) follows There is one other fundamental recursion for the hit numbers which we shall state since the q-analogue of this recursion will play a crucial role for our combinatorial interpretation of the q-hit numbers Theorem Suppose that B isaboard contained inb n such that B \f(i; n) : i n, g = ; (Thus B contains no elements in the last column of B n ) Then f ;n (B) = f ;n, (B=(i; n)): (9) Proof: Note that every p m PM(B n )must contain a square in the last column of B n since every perfect matching m of K n must contain one S edge of the form fi; ng with i n, Thus F ;n (B) can be partitioned into F (i;n) ;n (B) since B contains no elements in the last column of B n But ' i;n induces a : correspondence between F (i;n) ;n (B) and F ;n,(b=(i; n)) for i =;::: ;n, Hence (9) immediately follows We end P this section with a proof of the factorization formula (9) for the roo polynomial m =0 (B)() ##, for shifted Ferrers boards Reiner and White's n original proof of (9) was recursive We will give a biective proof of (9) for a slightly larger family of boards which we call nearly Ferrers boards That is, we say a board B B n is nearly Ferrers if for all (i; ) B, the squares f(s; ) :s<ig[f(t; i) :t<ig are also in B It is easy to see that every shifted Ferrers board F B is nearly Ferrers Moreover, you can construct a nearly Ferrers board by starting with a full triangle of squares i = f(s; t) :s<tig and then adding any columns to the right of i of height i See Fig for such an eample Figure

12 J HAGLUND AND J B REMMEL Theorem 4 Let B be anearly Ferrers board B n and let a i denote the number of squares in B that lie in row i for i =;::: ;n, Then Y ( + a n,i, i +)= n =0 m (B)() ##, : (0) Proof We let B n; denote the board B n with columns of height n, added to the right ofb n ; see Fig n+ n+ n- n- n- n Figure : The board B n; Wewant to consider the set of all placements of nonattacing roos in B n; but we have to dene the set of squares that a roo in a square (i; ) attacs Now if (i; ) B n, then a roo r in (i; ) attacs all cells in row i and column plus all cells in A n = f(s; t) B (i;) n : fs; tg\fi; g =g However, if (i; ) B n;,b n, then the cells that a roo in (i; ) attacs in a roo placement p depends on the other roos in p \ (B n;, B n ) That is, if (i; ) is the position of the lowest roo r in p \ (B n;, B n ), then r attacs all cells in row i and column other than (i; ) plus all cells in the column, ifn +<If =n + then r attacs all cells in row i and column plus all cells in column n + In general, if (i; ) is the position of the -th lowest roo r in p \ (B n;, B n ), then r attacs all cells in row i and column other than (i; ) plus all cells in the rst column in the following list of columns,;,;:::;n; n+; n+,;::: ;+ that contain a square which is not attaced by anyofthe, lower roos in B n;, B n Note that this means that each roor in p \ (B n;, B n ) will attac all cells in two columns of B n;, B n That is, if r is in cell (i; ), r attacs all cells in column It then loos for the rst column s>n to the left of column which has a cell that is not attaced by alower roo in p\(b n;,b n ) If there is no such column s, then r starts at column n+ and loos for the right most column s which has a square which is not attaced by anylower roo in p \ (B n;, B n ) Note that we are guaranteed that such a column s eists if 4n, Then r attacs all cells in

13 column s as well Our denition of nearly Ferrers board also ensures that each roo r p that lies in B also attacs the squares in two columns of B which lie above r, namely, the squares in column i and For eample, consider the placement p pictured in Fig consisting of roos, r in (7; 0), r in (5; ), and r in (; 7) We have indicated all cells attaced by r i by placing an i in the cell ,,, r,,,, r r,,, Figure Now let B be a board contained in B n and assume that 4n, Let N n; (B) denote the set of all placements p of n, roos in B n; such that no cell which contains a roo in p is attaced by another roo in p and any roor in B n \ p is an element ofb We claim that (0) arises from two dierent ways of counting N n; (B) That is, the number of ways to place a roo r in row is ust + a Then r attacs two cells in row n, ofb [ (B n;, B n ) so that there will be + a n,, ways to place a roo r n, in row n, Net in row n,, r and r n, together will attac four cells so that there will be + a n,, 4ways to place a roo r n, in row n, Continuing on in this way, it is easy to see that N n; (B) = Y ( + a n,i, i +): On the other hand, suppose that we a placement p of nonattacing roos on B Thus p M (B) We claim that the number of ways to etend p to a placement q N n; (B) such that q \ B n = p is (, ) (, (n,, ) + ) That is, there are n,, rows in B n;, B n,say d < <d, n, which have no cells which are attaced by roos in p Now for the lowest such row d,,wehave choices of where to place a roo in B n;, B n that lies in

14 4 J HAGLUND AND J B REMMEL d, The roo in row d, will attac eactly two cells in B n;, B n that lie in row d,, so that there will be, choices of where to place a roo in row d,, The roos in rows d, and d n,, will attac a total of four roos in (B n;, B n ) that lie in row d n,, so that there will be a total of, 4ways to place a roo in row d n,, Continuing on in this way, it is easy to see that there are a total of () ##, ways to etend p to a roo placement q N n; (B) such that q \ B n = p Thus N n; (B) = n =0 m (B)() ##, : Now suppose that we set =n, in (0) Then (n, ) ##, = 0 for =0;::: ;n, Thus the only term that survives on the right-hand side of (0) is m n (B)(n, ) ## n, Note (n, ) ## n, = n, (n, )! Thus the following is an immediate corollary of Theorem 4 Corollary Let B be anearly Ferrers board B n and for i =;::: ;n,, let a i be the number of squares in row i that are inb Then the number of perfect matchings of the graph G B =(f;::: ;ng; ffi; g :(i; ) Bg) is Y (a n,i, (n, i))= n, (n, )!: q-roo numbers and q-hit numbers for boards in B n In this section we shall dene q-roo numbers and q-hit numbers for boards in B n and prove some of their basic properties Let B be any board contained in B n For any roorin a square (i; ), we say that r roo-cancels squares f(r; i):r < ig[f(i; s) :i + s < g[f(t; ) :t < ig For eample, the squares that are roo-cancelled byarooin(4; 7) in B 8 are pictured in Fig 9 with a dot in them Thus the squares roo-cancelled by a roo r in cell (i; ) is ust the squares (a; b) which are attaced by r such that a + b<i+ Net for any roo placement p M (B) for some, we let u B (p) denote the number of squares in B, p that are not roo-cancelled by anyrooinp We then dene m (B;q) for >0by m (B;q) = q ub(p) : () pm (B) We dene m 0 (B;q) =q B We call m (B;q) the -th q-roo number of B We shall dene the -th hit number of B, f ;n (B;q), for any board B B n by the formula n =0 f ;n (B;q)z = n =0 m (B;q)[n, ]!! ny i=n,+ (z, q i, ): () Note for = 0, the product Q n i=n,+ (z, qi, ) is equal to by denition We shall call f ;n (B;q) the -th hit number of B relative to B n For eample consider

15 the shifted Ferrers board B(; ; 0) B 4 Then m 0 (B;q) =q since B has squares m (B;q) =+q + q since there are three roo placements in M (B) pictured in Fig 4, and m (B;q) = 0 since M (B) =; Thus =0 f ;4 (B(; ; 0);q)z = =0 m (B(; ; 0);q)[, ]!! = q [][] + ( + q + q )[](z, q ) Y i=,+ (z, q i, ) =(+q + q )z: Thus f 0;4 (B(; ; 0);q)=f ;4 (B(; ; 0);q)=0andf ;4 (B(; ; 0);q)=+q + q 5 u B(,,0) (p )= u B(,,0) (p )= u B(,,0) (p )=0 Figure 4 We should note that in general the f ;n (B;q) are not polynomials in q with nonnegative coecients That is, consider the board B pictured in Fig 5, which gives the roo placements of M (B), the roo placement inm (B), and the corresponding values of u B (p) u B (p )= u B (p )= u B (p )= u B (p 4 )=0 Figure 5 Thus m 0 (B;q) =q, m (B;q) =q + q, and m (B;q) = Hence f ;4 (B;q)z = m (B;q)[, ]!! =0 =0 i=,+ Y (z, q i, ) = q [] + (q + q )(z, q )+(z, q )(z, q) = q +(q + q, q )z + z

16 6 J HAGLUND AND J B REMMEL so that f 0;4 (B;q) =q, f ;4 (B;q) =q + q, q, and f ;4 (B;q) = As mentioned in the introduction, the main result of this paper is to show that if F is a shifted Ferrers board of B n, then the q-hit numbers f ;n (F; q) are polynomials in q with nonnegative coecients Indeed, we can dene an analogue t F (p) of the Dworin statistic s F;d (p) for boards contained in A n such that f ;n (F; q) = pf ;n (F ) q t F (p) : () Let B be any board contained in B n and suppose that we are given a placement p F ;n (B) If roo r is on cell (i; ) p \ B, then r p m -cancels all squares f(r;i): r<ig[f(i; s) : i + s<g [f(t; ) :t<ig[f(u; ) : u>and (u; ) = Bg: That is, if r is on B, then it p m -cancels all squares s to the left of r that it roocancels, and also all squares above r as in the denition of u F, plus all squares in its column which are below r and not in B However, if a roo r is on (i; ) and (i; ) = B, then r p m -cancels all squares in f(r;i): r<ig[f(i; s) : i + s<g[f(t; ) :t<iand (t; ) = Bg: That is, if r is o the board, r p m -cancels the same squares to the left of r that it roo-cancels plus squares in its column which lie above r and are o the board We then let t B (p) be the number of squares in B n, p which are not p m -cancelled For eample, for the placement p F ;0 (B(9; 7; 5; 4; ; 0; 0; 0; 0)) pictured in Fig 6, wehave put dots in all the p m -cancelled squares There are a total of uncancelled squares so that t F (p) = = p t f (p) = Figure 6

17 The main goal of this paper is to prove that if B is a shifted Ferrers board contained in B n and f ;n (B;q) is dened via (), then () holds For any board B B n, let ~f ;n (B;q) = q tb(p) : (4) pf ;n (B) There are two simple recursions that are satised by the ~ f ;n (B;q) which are q-analogues of (8) and (9) Theorem 5 Suppose that B is a board contained inb n such that B \f(; n) : n, g = f(; n) : ig; where i Thus in the last column of B n, B contains eactly the squares (; n); (; n);::: ;(i; n) Let =(i; n) Then ~f ;n (B;q) =q ~ f ;n (B=; q)+ ~ f,;n, (B=; q), q ~ f;n, (B=; q): (5) Proof Just as in the proof of Theorem :ii, we partition F ;n (B) into two sets, F (i;n) ;n (B) and F ;n(b), F (i;n) (i;n) ;n (B) Now ifp F ;n (B), then the roo r on (i; n) p m -cancels all cells (; n) such that 6= i since (i; n) is the lowest cell of B in column n It follows that ' i;n induces a weight preserving biection between (B) and F,;n,(B=) so that F (i;n) ;n t q B (p) = pf ;n (i;n) (B) pf,;n, (B=) 7 q t B=(p) = ~ f,;n, (B=; q): (6) Again it is the case that F ;n (B=(i; n)),f (i;n) ;n (B=(i; n)) = F ;n(b),f (i;n) ;n (B) since if (i; n) = p m for some p m PM(B n ), then p m \ B = p m \ B=(i; n) However, there is a dierence between t B=(i;n) (p m ) and t B (p m ) for such p m That is, p m contains one roo r in the last column of B n Sayris on square (; n) Now if <i, then r is on both B and B=(i; n) However relative tob, r p m -cancels all cells (s; n) with s<or s>i Relative tob=(i; n), r p m -cancels all cells (s; n) with s<or s i That is, (i; n) is not p m -cancelled relative tob but it is p m -cancelled relative tob=(i; n) Similarly, if>iso that (; n) = B and (; n) = B=(i; n), (i; n) is not p m -cancelled relative tob but it is cancelled relative tob=(i; n) If r 0 is any rooinp,frg then it is easy to see that r 0 p m -cancels the same squares relative tob that it p m -cancels relative tob=(i; n) It follows that for all p F ;n (B), F (i;n) ;n (B) =F ;n(b=), F (i;n) ;n (B=); t B (p) =+t B= (p): (7) Net suppose that p F (i;n) ;n (B=) Then the roo r on (i; n) inp does not p m -cancel any squares in the last column relative tob=(i; n) so there are n, uncancelled squares in the last column of B n This given, it is easy to see that ' i;n

18 8 J HAGLUND AND J B REMMEL induces a : correspondence between F (i;n) ;n (B=) and F ;n,((b=))=) = F ;n, (B=) which shows that q pf ;n (i;n) (B=) Thus by (7) and (8), Hence q ~ f ;n (B=; q) =q q pf ;n (B),F ;n (i;n) (B) q pf ;n (i;n) (B=) t B= (p) = q n, pf ;n, (B=) q t B=(p) = q n, ~ f;n, (B=; q): (8) t B= (p) + q = q ~ f;n, (B=; q)+ Clearly (5) follows immediately from (6) and (0) We have the following analogue of Theorem t q B= (p) pf ;n (B=),F ;n (i;n) (B=) q pf ;n (B),F ;n (i;n) (B) t B (p) : (9) t B (p) = q ~ f ;n (B=; q), q ~ f;n, (B=; q): (0) Theorem 6 Suppose that B is any board contained inb n such that B has no cells in the last column, then for any ~f ;n (B) = q n,i, ~ f,;n, (B=(i; n)): () Proof As in the proof of Theorem, we partition F ;n (B) into S F (i;n) ;n (B) For a placement p F (i;n) ;n (B), the roo r on (i; n) inpp m-cancels all squares (; n) with <isince there are no cells in B in the last column Thus there are n,, i uncancelled squares in the last column of B n relative top It is then easy to see that the : correspondence that ' i;n induces between F (i;n) (B) and F ;n, (B=(i; n)) proves that for i =;::: ;n,, q pf ;n (i;n) (B) t B (p) = q n,i, pf ;n, (B(i;n)) q t B=(i;n) (p) ;n = q n,i, ~ f;n, (B(i; n)): () Thus () holds It is easy to chec that for all boards B B and for all f0; g, f ; (B;q) = ~f ; (B;q) Thus to prove that f ;n (B;q) = f ~ ;n (B;q) for all nearly Ferrers boards

19 B B n and all f0;::: ;ng, we only need show that the analogues of Theorems 5 and 6 hold for all shifted Ferrers boards B when f ~ ;n (B;q) is replaced by f ;n (B;q) First we shall show that the analogue of Theorem 5 follows from the following simple recursion for the m (B;q)'s We shall say that a square (i; ) of a board B B n is a corner square ofb if B \A i; = ; where A i; = f(s; t) B n : fs; tg \fi; g = and s + t > i + g Note that A i; represents the squares S B n such that a roo on S could roo-cancel the cell (i; ) relative totheu B (p) statistic If B is a shifted Ferrers board, it is easy to see that (i; ) is a corner square if and only if there are no squares in B to the south-east of (i; ) inb n Theorem 7 Let B be aboard contained inb n and =(i; ) be acorner square of B Then for any, m (B;q) =qm (B=; q)+m, (B=; q): () Proof Set M (i;) (B) =fp M (B) :(i; ) pg First we partition the roo placements of M (B) into two sets, namely M (i;) (B) and M (B), M (i;) (B) Now ifp M (i;) (B), then the roo r on (i; ) roo-cancels all squares in B in Ci; n \ B since (i; ) is a corner square Thus ' i; induces a : weight preserving correspondence between M (i;) (B) and M, (B=) Hence it follows that pm (i;) (B) q u B(p) = m, (B=; q): (4) If p M (B), M (i;) (B), then cell (i; ) is not roo-cancelled by anyrooinp since (i; ) is a corner cell Thus u B (p) =+u B= (p) Hence pm (B),M (i;) (B) q u B (p) = pm (B=) 9 q +u B=(p) = qm (B=; q): (5) Corollary Let B be aboard contained inb n and let =(i; ) be acorner square ofb Then for any, f ;n (B;q) =qf ;n (B=; q)+f,;n, (B=; q), q f ;n, (B=; q): (6) Proof By (), = q n =0 + f ;n (B;q)z = n =0 n =0 = n n =0 =0 m (B;q)[n, ]!! qm (B=; q)[n, ]!! ny ny i=n,+ i=n,+ m, (B=; q)[n,, (, )]!!(z, q ) (z, q i, ) (z, q i, ) n, Y i=n,+,(,) n, f ;n (B=; q)z +(z,q ) m (B=; q)[n,,]!! = q n =0 =0 n, Y i=n,+, (z, q i, ) (z,q i, ) n, f ;n (B=; q)z +(z, q ) f ;n, (B=; q)z : (7) =0

20 0 J HAGLUND AND J B REMMEL Taing the coecient ofz on both sides of (7) yields (6) Note that the recursion (6) which holds for the f (B;q)'s represents a more general recursion than the recursion (5) which holds for the f ~ ;n (B;q)'s We could prove that f ;n (B;q) = f ~ ;n (B;q) for all shifted Ferrers boards B if we could give a direct combinatorial proof of the analogue of (6) for the f ~ ;n (B;q)'s However we have not been able to nd such a direct combinatorial proof The method of proof of Theorem 5 does not etend for arbitrary corner squares even for shifted Ferrers boards For eample consider the board B = B(; ; 0) B 4 In our proof of Theorem 5, we showed that if B was a shifted Ferrers board and is the corner square in the last column of B, then q tb(p) = f ~,;n, (B=; q) pf (B) and pf (B),F (B) q tb(p) = q ~ f ;n (B=; q), q ~f;n, (B=; q): Now suppose =(; ) One can see from Fig 7 that pf (B) q tb (p) = q and pf (B),F (B) q tb(p) =+q: p = p = p = t B (p ) = t B (p ) = t B (p ) = 0 Figure 7 However one can easily calculate ~ f 0; (B=; q) = ~ f 0; (;;q)=, ~ f; (B=; q) =0, and ~ f ;4 (B=; q) =+q Thus pf (B) q tb (p) 6= ~ f 0; (B=; q) and pf (B),F (B) q tb(p) 6= q ~ f ;4 (B=; q), q ~ f; (B=; q): Our inability to give a direct proof of the analogue of recursion (6) for the ~ f ;n (B;q)'s forced us to tae a dierent path of proof to establish the equality ofthef ;n (B;q)

21 and ~ f ;n (B;q) for shifted Ferrers boards Namely, we show that the f ;n (B;q)'s satisfy the analogue of the recursion () which holds for the ~ f ;n (B;q)'s Unfortunately it is not at all straightforward to show that the f ;n (B;q)'s satisfy the analogue of () Indeed most of section will be devoted to proving such a recursion In preparation for this proof, we shall end this section by proving a number of identities for the m (B;q)'s which will be used in section We start with a q-analogue of Theorem 4 Theorem 8 Let B be anearly Ferrers board contained inb n such that B has a i cells in row i Then Y [ + a n,i, i +]= n =0 m (B;q)[] ##, : (8) Proof As in the proof of Theorem 4, we shall consider roo placements in N n; (B) Now suppose that r is a roo in p where p N n; (B) Then if r is on (i; ) B, then we say r N -cancels all cells in f(r;): r<ig[f(i; s) : i + s<g[f(t; i) : t<ig [f(i; u) : u>and (u; i) = Bg: Note that the rst three sets in this union are the same cells that r roo-cancels relative to the u B (p \ B) statistic and the last set in the union is all the cells to the right ofr which are not in B Ifr is on (i; ) B n;,b n, let A n denote the set (i;) of cells attaced by r as dened in Theorem 4 Then r N -cancels all cells in A n (i;) that lie in a row s with s<iplus all cells in row i that are either in B n, B or to the right of(i; ) We let u N (p) denote the number of squares in B n;, p which are not N -cancelled by anyrooinp We claim that (8) results by computing the sum q u N (p) (9) pn n; (B) in two dierent ways Consider the ways to place a roo r in row n, If we place the roo in the rightmost position in B, then r will N -cancel all cells in row n, If we place r in the net to rightmost position in B, then r will cancel all but one cell in row n, As we continue to move r to the left in B in row n,, we increase the number of uncancelled cells in row n, by one until we reach the leftmost cell in row n, ofb where we would have a, uncancelled cells Net consider the placement ofr n, in cell (n, ; n + ) In that case, we would have a total of a uncancelled cells in row n,, namely the cells that lie in row n, and in B Then as we move r successively to the right, we would increase the number of uncancelled cells by one until we reach the rightmost position, namely (n, ; n + ), where we would have a total of + a, uncancelled cells Thus the factor of (8) contributed by the possible placements of r in row n, is+q + :::+ q +a, =[ + a ] Note that if r is placed in a cell in B, our denition of nearly Ferrers board ensures that it will N -cancel eactly two cells in B in every row i with i < If

22 J HAGLUND AND J B REMMEL r is placed in B n;,b n, then it will N -cancel eactly two cells in B n;,b n in every row i with i<n, Thus when we consider the placement of a roo r n, in row n,, we can use the same argument to prove that the factor of (9) contributed by the possible placement ofr n, is [ + a n,, ] Once again if r n, is placed in a cell in B, it will N -cancel an additional two cells in B in each row i with i<n, and if r n, is placed in a cell in B n;,b n, it will N -cancel an additional two cells in B n;, B n in each row i with i<n, Hence the factor of (9) contributed by the possible placement of a roo r n, in row n, is [ + a n,, 4] Continuing on in this way, it is easy to see that pn n; (B) q u N (p) = Y [ + a n,i, i +]: (40) Net suppose that we a placement p M (B) and we consider the sum p 0 N n; (B) p 0 \B=p q u N (p) : It is easy to chec that our denitions ensure that for any p 0 N n; (B) such that p 0 \ B = p, the number of squares of B n, p that are not N -cancelled by some roo in p 0 is ust u B (p) Moreover, by the same type of argument that we used above, the factor of (4) that arises from the possible placements of n,, roos in B n;, B n is ust [][, ] [, (n,, )+]=[] ##, Thus it follows that pn n; (B) q u N (p) = = n =0 pm (B) n =0 q u B(p) [] ##, m (B;q)[] ##, : We end this section by proving three recursions for the m (B;q), where B is a shifted Ferrers board or nearly Ferrers board which has no cells in the last column of B n Theorem 9 Suppose that B isaboard contained inb n which has no cells in the last column of B n Let =(i; r) be the cell which is at the bottom of the rightmost column of B Then (a) if B is a nearly Ferrers board, m (B=(i; n);q)=q i, m (B=; q)+ (b) if B is a shifted Ferrers board, i, = q i,, m, ((B=)=(; n, );q); (4) m (B=(r; n);q)=[r, ]m, (B=; q)+q r,, m (B=; q); (4)

23 (c) if B is a shifted Ferrers board, = q n,, m (B=(; n);q) =[n,, ]m (B;q), (q, q n,, )m + (B;q): (4) Proof Before proceeding with the proof of these three recursions, it will be useful to see the relations between three boards mentioned in recursion (a) and (b) It is easy to see from Fig 8 that B=(i; n) is ust the board B= with an etra column of height i, added in column r, The board B=(r; n) is ust the board B with the last column removed For recursion (a), we simply classify the roo placements p of m (B=(i; n)) according to whether or not p has a roo in the last column of B=(i; n) That is, if p M (;r,) (B=(i; n)) where i,, then the roo on square (; r, ) in p will cancel all but i,, squares in the last column It follows that q u B=(i;n) (p) = q i,, q u (B=(i;n))=(;r,) (' ;r,(p)) : Clearly (B=(i; n))=(; r, ) is the same board as (B=(i; r))=(; n, ) Thus i, = pm (;r,) (B=(i;n)) = i, q i,, = q u B=(i;n) (p) p 0 M, ((B=)=(;n,)) = i, = q u (B=)=(;n,) (p0 ) q i,, m, ((B=)=(; n, );q): (44) On the other hand given a p M (B=(i; n)) having no roo in column r,, all the squares in column r, will not be roo-cancelled so that u B=(i;n) (p) =q i, u B= (p): Thus pm (B=(i;n)), S i, = M (;r,) (B=(i;n)) q u B=(i;n) (p) = q i, m (B=; q): (45) Combining (44) and (45) yields (4) as desired For recursion (b), note that the shifted Ferrers board B= is the board which results by removing the last column and the rst row from B Thus B= is the result of removing the rst row from B=(r; n) It follows that recursion (b) can be rephrased as follows Suppose that D is a shifted Ferrers board with r, columns

24 4 J HAGLUND AND J B REMMEL B = ẋ B/ α = 7 B = B = B/ (i, n) = B/ (r, n) = Figure 8 and C is the shifted Ferrers board that results from removing the rst row ofd Then m (D; q) =[r, ]m, (C; q)+q r,, m (C; q): (46) Once we have rephrased recursion (b) in this way, it is simple to prove Namely we simply partition the elements p of M (D) depending on whether or not p has a roo in the rst rowofd That is, let M (D) =fp M (D) : p has a roo in the rst rowg Now ifp M (D), M (D), then p has all roos below the rst row Since each of these roos roo-cancel two squares in row, there will be r,, uncancelled squares in the rst row Of course, the board C is ust the rows of D below row

25 5 so that pm (D),M (D) q ud(p) = q r,, pm (C) q u C (p) = q r,, m (C; q): (47) Net suppose that p 0 M, (C) We can thin of p 0 as a roo placement ind with no roos in the rst row There will be r,, (, ) = r, uncancelled squares in the rst row ofd Thus we can etend p 0 to a placement p M (D) in r, ways by placing a roo r in one of these r, uncancelled squares in the rst row ofd If we placed r in the i-th uncancelled square in row starting from the right, r will roo-cancel all squares to its left and leave i, uncancelled squares in row It follows that pm (D) q ud(p) =(+q + :::+ q r,, ) p 0 M, (C) q u C(p 0 ) =[r, ]m, (C; q): (48) Hence (46) holds We do not have a simple combinatorial proof of recursion (c) Instead we shall prove recursion (c) by induction, rst on n and then on the number of squares in B It is easy to verify that recursion (c) holds for all boards B B Thus assume that (c) holds for all boards B 0 B n, NowifB is the empty board contained in B n, then it is easy to see that both sides of (4) are zero if If =0, B=(; n) is the empty board for all so that m 0 (B=(; n);q)=m 0 (B;q) = and m (B;q) = 0 Thus in that case (4) becomes = q, =[n, ]: Thus (4) holds for the empty board for all n Finally by induction, assume that (4) holds for all shifted Ferrers boards with less than t squares and that B B n is a shifted Ferrers board with t squares which has no squares in the last column of B n Let =(i; r) denote the corner square in the rightmost column of B Applying recursion () and then recursion

26 6 J HAGLUND AND J B REMMEL (c) to B= and B= by induction, we nd that [n,, ]m (B;q), (q, q n,, )m + (B;q) = q([n,, ]m (B=; q), (q, q n,, )m + (B=; q)) +[n,, ]m, (B=; q), (q, q n,, )m (B=; q) = q = q, m ((B=)=(; n);q) +[(n,),,(,)]m, (B=; q),(q (n,),,q (n,),,(,) )m (B=; q) + n, = +(q (n,),, q (n,),,(,), q + q n,, )m (B=; q) = q = q, m ((B=)=(; n);q) q n,, m, ((B=)=(; n, );q), (q, q n, )m (B=; q): (49) We would also lie to apply recursion () to the left-hand side of (4) but this requires some care That is, if < i, then the image of =(i; r) under ' ;n is =(i, ;r, ) which will still be the rightmost corner square of B=(; n) Similarly if i<<r, then the image of under ' ;n is =(i; r, ) will also be the rightmost corner square of B=(; n) If >r, then (; n) only attacs empty squares so that is the rightmost corner cell of B=(; n) See Fig 9 It is easy to see that if <i, then B=(; n)= =(B=)=(; n) and B=(; n)= = (B=)=(; n, ) If i<<r, then B=(; n)= =(B=)=(; n) and B=(; n)= = (B=)=(, ; n, ) Finally if > r, then B=(; n)= =(B=)=(; n) and B=(; n)= =(B=)=(, ; n, ) This given, we can apply recursion (a) to obtain the following + = + q, m (B=(; n);q) + =r+ r, i, = =i+ = q,i m (B=(i; n);q)+q,r m (B=(r; n);q) q, (qm ((B=)=(; n);q)+m, ((B=)=(; n, );q)) q, (qm ((B=)=(; n);q)+m, ((B=)=(, ; n, );q)) q, (qm ((B=)=(; n);q)+m, ((B=)=(, ; n, );q)): (50) Comparing the right-hand sides of (49) and (50), we can prove (4) if we can show

27 B = α α B = B = β B/ (, n) with < i α B/ (, n) with > r γ B/ (, n) with i < < r 7 Figure 9

28 8 J HAGLUND AND J B REMMEL that q,i m (B=(i; n);q)+q,r m (B=(r; n);q) + n, = i, = r, =i n, =r, q, m, ((B=)=(; n, );q) q n,, m, ((B=)=(; n, );q) q n,, m, ((B=)=(; n, );q) = q n,i m ((B=)(i; n);q)+q n,r m ((B=)(r; n);q) q n,, m ((B=)=(; n, );q), (q, q n, )m (B=; q): (5) It is easy to see that (B=)=(i; n) = B=(i; n) and (B=)=(r; n) = B=(r; n) since both (i; n) and (; n) attac Thus (5) is equivalent to i, = (q,, q n,, )m, ((B=)=(; n, );q) + r, =i (q n,,, q n,, )m, ((B=)=(; n, );q) =(q n,i, q n,i, )m (B=(i; n);q)+(q n,r, q n,r, )m (B=(r; n);q) Dividing both sides by q, gives (q +) i, = q n,, m, ((B=)=(; n, );q) + r, =i q n,, m, ((B=)=(; n, );q), (q, q n, )m (B=; q): (5) = q n,i, m (B=(i; n);q)+q n,r, m (B=(r; n);q), ( + q)q n, m (B=; q): (5) But we can now apply recursions (a) and (b) to the rst two terms on the right-hand side of (5) to show that the right-hand side of (5) is q n, m (B=; q)+ i, = q n,, m ((B=)=(; n, );q) + q n,r, [r, ]m, (B=; q)+q n,, m (B=; q), ( + q)q n, m (B=; q): (54)

29 Now replacing the right-hand side of (5) by (54) and collecting terms we get that (4) is equivalent to proving r, = q n,, m, ((B=)=(; n, );q) = q n,r, [r, ]m, (B=; q), (q n,, q n,, )m (B=; q): (55) Note however that by induction n, = q n,, m, ((B=)=(; n, );q) =[n,, (, )]m, (B=; q), (q n,, q n,5,(,) )m (B=; q) =[n,, ]m, (B=; q), (q n,, q n,, )m (B=; q): (56) Moreover since B had only r, columns then B= has at most r, columns Thus (B=)=(; n, )) = B= for r, since (; n, ) will only attac cells in empty columns Hence n, =r, q n,, m, ((B=)=(; n, );q) = n, =r, q n,, m, (B=; q) =[n,, r]m, (B=; q): Thus subtracting [n,, r]m, (B=; q) from both sides of (56) yields (55) as desired Main Theorem In this section we prove our main result, namely that f (B;q) = f ~ (B;q) for all shifted Ferrers boards B We start by proving two identities which hold for any board n Theorem 0 If B isaboard, B B n, 0 n, and is the q-binomial coecient base q, then f ;n (B;q) = m (B;q)[n, ]!!(,), Proof Recall the q-binomial theorem [A]: Y m, i=0 ( + q i )= m =0 q ( ) m : q q q (,)(n,,) : Theorem 0 follows by applying this to the product on the right-hand side of () 9

30 0 J HAGLUND AND J B REMMEL Theorem If B is a board, B B n, then for 0 n, f ;n (B;q) z, q (,)(,) q p = m p (B;q)[n, p]!! q (p,)(n,,p) (z; q ) p, (,) p, ; (57) p q where (z; q) =(, z)(, zq) (, zq, ) Proof Using Theorem 0, the left-hand side of (57) equals z, q (,)(,) q p = m p (B;q)[n, p]!!(,) p, p p = m p (B;q)[n, p]!!(,) p, p p q m p (B;q)[n, p]!!(,) p, p q q (p,)(n,p,) p q (,)(,)+(p,)(n,p,) (,), z, q q p, u=0 = m p (B;q)[n, p]!!(,) p, p p p, q u(,)+(p,,u)(n,p,,u) (,) u z u u q p, q q (p,)(n,p,) = m p (B;q)[n, p]!!(,) p, p p using the q-binomial theorem u=0 p, q u,u (,) u z u u q q q (p,)(n,p,) (z; q ) p, Theorem If B is a shifted Ferrers board, B B n, then f ;n (B;q) = q n,i, f ;n, (B=(i; n);q): Proof We start by setting z = q, in eq (57) to get f ;n (B;q) q q (,)(n,,) = m (B;q)[n, ]!!, m + (B;q)[n,, ]!![ +] q q n,, (, q, ) = m (B;q)[n, ]!!, m + (B;q)[n,, ]!![ +] q q n,, (q, ): (58)

31 On the other hand if B has less than n, columns in B n then ( = q n,i, f ;n, (B=(i; n);q))z, Setting z = q, in (59) we get ( q n,i, f ;n, (B=(i; n);q) q q (,)(,) q n,i, f ;n, (B=(i; n);q)) q (,)(n,,) q = q n,i, f ;n, (B=(i; n);q) = q z, q (,)(,) : (59) q q (,)(n,,) q n,i, m (B=(i; n);q)[n,, ]!! (60) where the last equality follows by using the special case of (57) with z = for the boards B=(i; n) Comparing (59) and (60), we get that if B has less than n, columns, f ;n (B;q) if we can show that q (,)(n,,) q = ( q n,i, m (B=(i; n);q) q n,i, f ;n, (B=(i; n);q)) q (,)(n,,) (6) q =[n,, ]m (B;q), q+, q, qn,, (q, )m + (B;q) =[n,, ]m (B;q), (q, q n,, )m + (B;q): Since this is equation (4), (6) holds for all assuming B is contained in the rst n, columns of B n If = n this reduces to f n;n (B;q) = q n,i, f n;n, (B=(i; n);q)): (6)

32 J HAGLUND AND J B REMMEL If = n, (6) reduces to f n;n (B;q) = n n, q q n,,(n,) + f n,;n (B;q) q n,i, f n;n, (B=(i; n);q) + n n, q q n,,(n,) q n,i, f n,;n, (B=(i; n);q): (6) Thus we can use (6) to cancel the rst terms on both sides of (6) to get f n,;n (B;q) = Continuing in this manner we get for all f ;n (B;q) = q n,i, f n,;n, (B=(i; n);q): q n,i, f ;n, (B=(i; n);q) Corollary If B is a shifted Ferrers board, B B n, and 0 n, then () holds, ie f ;n (B;q) = ~ f ;n (B;q): Proof If B has no cells in the last column of B n, then f ;n (B;q) and f ~ ;n (B;q) satisfy the same recursion by Theorem's 6 and 0 If B has at least one cell in the last column of B n, then they both satisfy the same recursion by Theorem 5 and Corollary If B is the empty board, then f ~ ;n (B;q) =( = 0)[n]!! For this board, m (B;q) =( = 0) and so by (), f ;n (B;q) =( = 0)[n]!! Since the f ;n and the f ~ ;n satisfy the same recursion with the same initial conditions, they are equal for all B 4 Algebraic Identities In this section we proveanumber of algebraic identities for the m and the f In many cases these are analogues for nearly Ferrers boards of nown identities for q- roo and q-hit numbers We use the notation (a; q) =(,a)(,aq) (,aq, ) Theorem If B B n is a nearly Ferrers board with b i squares in row i for i =;::: ;n,, and 0 n,, then [][4] []m, (B;q) = =0 Y q (, ) (,) (,) q [ + b i, i +]:

33 Proof By Theorem 8, the right-hand side above equals 0 = s0 q q (, ) (,) (,) s0[][, ] [, s +]m,s (B;q) m,s (B;q) s [][, ] [, s +],s = m,s (B;q) [(u + s)] [u +] s0 = s0 m,s (B;q),s u=0 u=0 s = s0 m,s (B;q)[][4] [s] = s0 m,s (B;q)[][4] [s] by the q-binomial theorem q (,) (,s) [][4] [s] u + s q (, ) (,) (,) q q (,s,u ) (,) (,s,u) q [, s][, s, ] [, s, u +] q (,s,u ) (,) u [][4] [u] s s q (,) (,s),s u=0 q (,) (,s) (,) (,s) = [][4] []m, (B;q), s q (,s,u ) (,) u u q,s u=0, s q (u ) (,) u u q Theorem 4 If B B n is a nearly Ferrers board with b i squares in row i for i =;::: ;n,, then [n]!! 0 f ;n (B;q)[], [, n + +] = Proof By Theorem 0, f ;n (B;q) = so the left-hand side of (64) equals [n]!! 0 m (B;q)[n, ]!!(,), [], [, n + +] Y m (B;q)[n, ]!!(,), [ + b i, i +]: (64) q q (,)(n,,) ; q q (,)(n,,)

34 4 J HAGLUND AND J B REMMEL = 0 [n, ]!! m (B;q) [n]!! (,), q (,)(n,,) [], [, n + +] q =0 = 0 [n, ]!! m (B;q) [n]!! =0 (q, ; q ) (q ; q ) (,q ) q,( ) (,), q (,)(n,,) [][, ] [, (n, )] (q,n+ ; q ) (q,4n+4 ; q ) =[][, ] [, 4n +4] 0 m (B;q) [n, ]!! (,) q (n,) [n]!! q, ; q,n+ ; q +,n ; q q,4n+4 where in the last equalitywehave used the fact that,(,)+(n,),n+, + = ( +, n)+(n, ) Using the q-vandermonde convolution for the sum of a terminating [GR, p6], the equation above equals [][, ] [, 4n +4] 0 m (B;q) [n, ]!! (,) q (n,) (q,n ; q ) [n]!! (q,4n+4 ; q ) = 0 m (B;q)[][, ] [, (n,, )+] C; where C = [n, ]!! (,) q(n,) [n]!! (, q) (, q,n )(, q,n ) (, q,,n ) = [n, ]!! [n]!! q (n,) (, q) q,n+,n+:::+,,n (q, )(q n,, ) (q n,+, )(,) = [n, ]!! q (n,)+,n [n, ][n, ] [n, +]=: [n]!! Theorem 4 now follows from Theorem 8

35 5 Corollary 4 If B isanearly Ferrers board, B B n, then n =0 f ;n (B;q) =[n]!!: Proof Letting!in the left-hand side of Theorem 4 we get [n]!! n =0 f ;n (B;q) (, q) = (, q) : Remar: Corollaries and 4 together show that for any ed shifted Ferrers board B B n, the statistic t f (B) has what could be called the \Mahonian" property for perfect matchings, ie its distribution is [n]!! Theorem 5 If B B n is a nearly Ferrers board with b i squares in row i for i =;::: ;n,, then f ;n (B;q) = n, s=0 n += q (n,,s ) (,) n,,s n,, s q Y [s]!! [][4] [n, +s] Proof Using Theorem 8, the right-hand side above equals n, s=0 n += q (n,,s ) (,) n,,s n,, s q [s]!! [][4] [n, +s] 0 [n, +s + b i, i +]: m (B;q)[n, +s], : The coecient ofm (B;q) above is clearly zero unless n, +s (n,, ), or n, s, or n, s and since s n, we have n, (n, ) = Thus the right-hand side of Theorem 5 equals m (B;q), u=0 (s=n,+u) n += q (n,,(n,+u) ) (,) n,,(n,+u) [n, + u]!!,, u q [][4] [u] = m (B;q)[n, ]!! n +=,, q u=0 q (u ) (q,(,) ; q ) u (q n+,++ ; q ) u q (,+u ) (,) (,+u) (qn,+ ; q ) u (q ; q ) u (,q (,) ) u :