Bijections on m-level Rook Placements
Kenneth Barrese, Nicholas Loehr, Jeffrey Remmel, Bruce E. Sagan
aa r X i v : . [ m a t h . C O ] A ug Bijections on m -level Rook Placements Kenneth Barrese
Department of Mathematics, Michigan State University,East Lansing, MI 48824-1027 [email protected]
Nicholas Loehr ∗ Department of Mathematics, Virginia TechBlacksburg, VA 24061-0123 [email protected] andDepartment of Mathematics, United States Naval AcademyAnnapolis, MD 21402-5002 [email protected]
Jeffrey Remmel
Department of Mathematics, UCSDLa Jolla, CA, 92093-0112 [email protected]
Bruce E. Sagan
Department of Mathematics, Michigan State University,East Lansing, MI 48824-1027 [email protected]
October 2, 2018
Abstract
Suppose the rows of a board are partitioned into sets of m rows called levels. An m -level rookplacement is a subset of the board where no two squares are in the same column or the same level. Weconstruct explicit bijections to prove three theorems about such placements. We start with two bijectionsbetween Ferrers boards having the same number of m -level rook placements. The first generalizes a mapby Foata and Sch¨utzenberger and our proof applies to any Ferrers board. This bijection also preserves the m -inversion number statistic of an m -level rook placement, defined by Briggs and Remmel. The secondgeneralizes work of Loehr and Remmel. This construction only works for a special class of Ferrers boards,but it yields a formula for calculating the rook numbers of these boards in terms of elementary symmetricfunctions. Finally we generalize another result of Loehr and Remmel giving a bijection between boardswith the same hit numbers. The second and third bijections involve the Involution Principle of Garsiaand Milne. Rook theory is the study of the numbers r k ( B ), which count the number of ways to place k non-attackingrooks on a board B . It originated with Kaplansky and Riordan [KR46] who studied the connections betweenrook placements and elements of the symmetric group S n . We will focus on a particular type of board: aFerrers board is a board where the columns are bottom justified and their heights form a weakly increasingsequence. Foata and Sch¨utzenberger [FS70] characterized the equivalence classes of Ferrers boards by aunique increasing representative. They did so by constructing explicit bijections between rook placementson boards in a class and rook placements on the unique representative board. The rook polynomial of a boardis the generating function for the numbers r k ( B ) in the falling factorial basis for the ring of polynomials. Thetheorem of Foata and Sch¨utzenberger was later proved as an elegant corollary to the Factorization Theoremof Goldman, Joichi, and White [GJW75], which gave a complete factorization of the rook polynomial of aFerrers board over the integers. Loehr and Remmel [LR09] constructed a bijection between rook placementson rook equivalent Ferrers boards using the Garsia-Milne Involution Principle [GM81], which also impliedthe Factorization Theorem. Later in the paper, they presented a similar bijection for the sets counted by the ∗ This work was partially supported by a grant from the Simons Foundation ( = (1 , , ,
4) =
R R R
Figure 1: A Ferrers board B and a placement of three rooks on B .hit numbers of rook equivalent Ferrers boards. Briggs and Remmel [BR06] generalized the notion of rookplacements to m -level rook placements. These correspond to elements of C m ≀ S n , the wreath product ofthe cyclic group of order m with the symmetric group on n elements, in the same way that traditional rookplacements correspond to elements of S n . Using these placements and the concept of flag descents developedby Adin, Brenti, and Roichman [ABR01], Briggs and Remmel were able to generalize a formula of Frobeniusto C m ≀ S n .The purpose of this paper is to generalize the bijection of Foata and Sch¨utzenberger and those of Loehrand Remmel to m -level rook placements. The remainder of this section gives the background terminologynecessary to begin this task. In Section 2 we generalize the bijection used by Foata and Sch¨utzenberger.Although this bijection is the composition of many intermediary bijections, and is therefore not direct, itdoes provide an explicit bijection between m -level rook placements on arbitrary m -level rook equivalentFerrers boards. We will need this bijection again in Section 5. In Section 3 we show that the bijectionprovided in Section 2 preserves the m -inversion number of an m -level rook placement, as defined by Briggsand Remmel. In Section 4 we generalize a construction of Loehr and Remmel. In this case the bijectioncan only be specified for singleton boards, a subset of all Ferrers boards. However, the construction leads toan explicit calculation of the m -level rook numbers for such boards using elementary symmetric functionsand Stirling numbers of the second kind. Furthermore, this bijection also preserves the m -inversion numberof m -level rook placements. In Section 5, we generalize a second bijection of Loehr and Remmel, and indoing so prove that any two m -level rook equivalent Ferrers boards have the same hit numbers. The last twobijections involve the Garsia-Milne Involution Principle [GM81]. Finally, in Section 6 we present an openproblem about counting the number of Ferrers boards in m -level rook equivalence classes.A board is any finite subset of Z + × Z + where Z + is the positive integers. Given an integer partition0 ≤ b ≤ b ≤ · · · ≤ b n , the corresponding Ferrers board is B = { ( i, j ) ∈ Z + × Z + | ≤ i ≤ n and j ≤ b i } . Usually B is denoted by B = ( b , b , . . . , b n ). Graphically, one represents a Ferrers board as an array ofsquare cells, where the i th column contains b i cells. See the diagram on the left in Figure 1 for the board(1 , , , i, j ) to denote the cell in the i th column and j th row of B . Note thatthis is neither the English nor the French style of writing Ferrers diagrams, but is the standard conventionin modern rook theory literature. It is useful because we usually consider placing rooks on the board fromleft to right, and enumerating the number of such placements is facilitated by our convention.For any non-negative integer k , a placement of k rooks on B is a subset of the cells of B of cardinality k which contains no more than one cell from any row or column of B . Graphically, this corresponds to placingrooks in the cells of B so no two rooks attack each other. See the diagram on the right in Figure 1 for aplacement of three rooks on (1 , , , m is a fixed positive integer. We define ⌈ j ⌉ m to be the least multiple of m greater than or equal to j and call it the m -ceiling of j . Similarly, let ⌊ j ⌋ m be the greatest multiple of m less than or equal to j , and call this the m -floor of j . Given a positive integer p , let L p ⊂ Z + × Z + bedefined by: L p = { ( i, j ) ∈ Z + × Z + | ⌈ j ⌉ m = pm } . Then the p th level of B is B ∩ L p . Thus, the first level of B consists of the first m rows, the second levelconsists of the next m rows, and so forth. Note that for ( i, j ) ∈ B , ⌈ j ⌉ m = pm if and only if ( i, j ) is in the p th level of B . 2 = RR B S = R R l ( B S ) = R R
Figure 2: On the left, a placement of two 2-level rooks on B . In the middle, the corresponding placementfrom Lemma 3 of two 2-level rooks on singleton board B S . On the right, the placement on l ( B S ) fromLemma 5.For any non-negative integer k , an m -level rook placement of k rooks on B is a subset of cardinality k ofthe cells of B which contains no more than one cell from any given level or column of B . See Figure 2 forthree 2-level rook placements where thickened lines demarcate where levels begin and end; the numbering ofthe boards can be ignored for now. An m -level rook is a rook placed so that it is the only rook in its leveland column. The k th m -level rook number of B is r k,m ( B ) = the number of m -level rook placements of k rooks on B .Two boards are m -level rook equivalent if their m -level rook numbers are equal for all k . Note that m -levelrook placements are always rook placements. Furthermore, when m = 1 rook placements and m -level rookplacements are equivalent.The i th column of B terminates in level p if p is the largest integer such that the i th column has non-empty intersection with L p . A singleton board is any Ferrers board such that, for each positive integer p ,the set of all columns b i terminating in level p contains at most one i such that b i m . The Ferrersboard on the left in Figure 2 is not a singleton board, as two different columns terminate in the second levelwithout having 2 cells in that level, while the Ferrers boards in the middle and on the right are singletonboards. In order to produce bijections between m -level rook placements on Ferrers boards, it is convenient to restrictour attention to singleton boards. In order to do this we prove the following two lemmas. First we showthat for every Ferrers board there is a unique singleton board which has the same number of cells at eachlevel. Then we prove that there is a bijection between the rook placements on a Ferrers board and those onthe singleton board guaranteed in the first lemma. These lemmas together imply that every Ferrers board is m -level rook equivalent to a singleton board and that there is an explicit bijection between the correspondingrook placements. Lemma 1.
Given a Ferrers board B , there exists a unique singleton board B S which has the same numberof cells at each level as B .Proof. Let B have l p cells in the p th level. In order for B S to be a singleton board with l p cells in the p thlevel, the cells of the p th level must be arranged uniquely as follows. If l p = cm + r with 0 ≤ r < m , thenlevel p of B S must have one column with r cells followed on the right by c columns with a full m cells in3he level. This is because a singleton board may have at most one column which intersects a given levelnon-trivially in fewer than m cells. Thus B S must be unique if it exists.In order to show that B S exists, we shall construct it. Arrange each level as specified above and line upthe furthest right column in each level to create the furthest right column of B S . This yields a Ferrers boardbecause every column which has any cells in the p th level of B must have a full m cells in the ( p − B . Thus the total number of columns in the p th level of B S will be less than or equal to the number ofcolumns in the ( p − B S containing m cells at that level. Hence a singleton board B S exists andis unique.Ignoring the rook placement, Figure 2 shows a board B and its corresponding board B S . Since we knowthat an arbitrary Ferrers board B has the same number of cells at each level as a unique singleton board B S , we wish to provide an explicit bijection between rook placements on the two boards. In order to do sowe require the following numbering on a Ferrers board. Definition 2. A level numbering of board B assigns a number to each cell of B in the following way.Proceeding level by level in B , number the cells in the level by numbering each column from bottom to top,starting with the rightmost column and working left. In each level begin the numbering with 1. Figure 2 presents two examples of this numbering, on the left and middle boards, and also illustrates thebijection of the next lemma.
Lemma 3.
Given a Ferrers board B , there is an explicit bijection between m -level rook placements of k rookson B and m -level rook placements of k rooks on B S , where B S is as constructed in Lemma 1.Proof. Give both B and B S a level numbering as shown in Figure 2. Since both boards have the samenumber of cells in each level, corresponding levels will each be numbered with the same set of numbers.Given any m -level rook placement on B , place rooks on B S initially so that each rook occupies the samenumbered cell in the same level as it does in B . This may not provide an m -level rook placement on B S since two rooks could end up in the same column, so we will modify it as follows.Notice that if a rook in column i and level p of B is not in the same cell in B S , then column i mustbe to the left of a column of B that intersects L p in less than m cells. Furthermore, if the rook ends upin column i ′ in B S , then all columns in the interval [ i, i ′ ] have a full m cells in levels below p in B . Thus,if any of the rooks that move create a column with two or more rooks, there will be exactly two rooks inthe column and the upper rook will have moved while the lower rook remained stationary. To rectify thesituation, whenever a rook is moved from column i in B to column i ′ in B S , move all other rooks in columnsin the interval ( i, i ′ ] one column to the left, preserving their row. This is possible since, in both B and B S ,these columns must contain m cells in all levels lower than the upper rook in order for the upper rook tohave been in that column in B . Rearranging the rooks at each level in this fashion provides a function from m -level rook placements on B to m -level rook placements on B S . Figure 2 illustrates this map on a rookplacement, including moving a lower rook one column to the left.To see that this is a bijection, use the level numbering to produce a set of rooks on B from those on B S .All the rooks will return to their initial positions once the appropriate right shift is applied. Similarly onecan show that applying the map first to B S and then to B is the identity. Thus we have a bijection between m -level rook placements on B and m -level rook placements on B S .Lemma 1 and Lemma 3 guarantee that every Ferrers board is m -level rook equivalent to a singletonboard. Additionally, there is an explicit bijection between m -level rook placements on the two boards. Thispermits us to restrict our attention to singleton boards henceforth. l -operator Transposition of boards plays a central role in the Foata-Sch¨utzenberger construction of bijections betweenrook-equivalent Ferrers boards when m = 1. We will need a generalization of this operation for arbitrary m and this is given in the next definition. 4igure 3: The dashed line goes through the cells counted by the 2-arm length of the fourth column and firstlevel, and the shaded cells are counted by the corresponding 2-leg length. Definition 4.
Given a Ferrers board B , the l -operator applied to B is defined as follows. If t is the largestindex of a non-empty level of B and the number of cells in the p th level of B is l p , then l ( B ) = ( l t , l t − , . . . , l ) . Figure 2 contains an example board B S as well as l ( B S ). The fact that l ( B ) is a Ferrers board comesfrom the proof of Lemma 1. In particular, if B is a Ferrers board then its p th level must fit above its ( p − ⌊ l p ⌋ m ≤ ⌊ l p − ⌋ m , with strict inequality if l p m . It follows that l ( B ) is a weakly increasing sequence and so l ( B ) is aFerrers board and, because of the strict inequality for non-multiples of m , a singleton board.To see that the l -operator is a generalization of transposition, note that if m = 1 then the levels of B areindividual rows and these become the columns of l ( B ). Furthermore, when restricted to the set of singletonboards the l -operator is an involution. This is shown in Proposition 7.4 of [BLRS13]. Thus, the l -operatoris a surjection from the set of Ferrers boards onto the set of singleton boards with B = l ( l ( B )) when B issingleton. We now provide a bijection between m -level rook placements on B and m -level rook placementson l ( B ) to generalize the well-known bijection for transposition. Lemma 5.
Given a singleton board B and a non-negative integer k , there is an explicit bijection between m -level rook placements of k rooks on B and m -level rook placements of k rooks on l ( B ) .Proof. Give B a level numbering, then number the columns of l ( B ) from bottom to top beginning with thenumber 1 in each column. Note that in this case the numbering of a level of B will consist of the same setof numbers as the numbering of the corresponding column of l ( B ). Assume that B has t non-empty levels.For a given m -level rook placement of k rooks on B , place rooks on l ( B ) in the following way. If a rook wasin the cell numbered n of level p in B , then place a rook in the cell numbered n in column t − p + 1 in l ( B ).See Figure 2 for an example of this map for a 2-level placement.We must show that this gives a valid m -level rook placement on l ( B ). If two rooks end up in the samecolumn of l ( B ) they must have originated in the same level of B , contradicting having an m -level rookplacement on B . Similarly, if two rooks end up in the same level of l ( B ), then they must have originated inthe same column of B , since B is a singleton board.The inverse of this map acts as follows. If a rook is in the cell numbered a of column t − p + 1 in l ( B )then it is placed in the cell numbered a in level p of B . The proof that this gives a rook placement is similarto the one in the previous paragraph and so is omitted.Note that Lemma 3 and Lemma 5 combine to provide an explicit bijection between m -level rook place-ments on any Ferrers board B and on its m -transpose, l ( B ) = l ( B S ). l -operator For any set S , let S be the cardinality of S . Given a column i and a level p define the m -arm length ofcolumn i , level p by arml m ( i, p ) = { ( i, j ′ ) ∈ B | ( i, j ′ ) is strictly above level p } . = l , ( B ) =Figure 4: On the left, B , is shaded within B = (1 , , , l , is not permissible for B since ⌊ arml m (4 , ⌋ m < legl m (3 , l , ( B ) will not be a singleton board. On the right the shadedcells in l ( B , ) illustrate this; in this case l , ( B ) is not even a Ferrers board.In Figure 3 the cells counted by the 2-arm length of column 4, level 1 have a dashed line through them.(Reflecting our boards to put them in English notation will result in the arm being the usual set of squareswhen m = 1.) We let arml m ( i, p ) = ∞ if the number of columns in B is less than i , for reasons detailed inLemma 7.Similarly, define the m -leg length of column i , level p to belegl m ( i, p ) = { ( i ′ , j ′ ) ∈ B | ( i ′ , j ′ ) is in level p and i ′ < i } . The cells counted by the 2-leg length of column 4, level 1 are shaded in Figure 3. As before, this is equivalentto the usual notion of leg length in the m = 1 case. We also let legl m ( i,
0) = ∞ by convention.Since the l operation generalizes the transposition of a Ferrers board, one would expect that some sortof local l operation would be the appropriate generalization of the local transposition introduced by Foataand Sch¨utzenberger. This is indeed the case, and we define the local l operation as follows.Given a Ferrers board B with non-empty intersection of the i th column and p th level, let B i,p denotethe subboard of B consisting of all cells in or above the p th level and in or to the left of the i th column: seeFigures 4 and 5 for examples. Note that if B is a singleton board, then B i,p is also, because the set of rowsin level p ′ of B i,p will be the same as the set of rows in level p + p ′ − B . If B is a Ferrers board then the local l operation at ( i, p ) is the result of applying the l operator to the subboard B i,p and leaving the rest of B fixed. We will denote the resulting board by l i,p ( B ).As defined above l i,p ( B ) may not be a Ferrers board, let alone a singleton board. We now develop a pairof conditions to determine if l i,p ( B ) will be a singleton board. Definition 6.
The operation l i,p is permissible for a singleton board B if arml m ( i, p ) ≤ ⌊ legl m ( i, p − ⌋ m and legl m ( i, p ) ≤ ⌊ arml m ( i + 1 , p ) ⌋ m . See Figure 4 for an example of a local l -operation not permissible for the given board, and Figure 5 fora local l -operation which is permissible. Lemma 7.
Let a singleton Ferrers board B have a non-empty intersection of the i th column and p th level.Then l i,p is permissible for B if and only if l i,p ( B ) is a singleton Ferrers board.Proof. If column i , level p in B contains fewer than m cells, then l i,p ( B ) = B since B is singleton, and thereis nothing to prove. Henceforth, assume that column i , level p in B contains m cells. We know that B , B i,p ,and l ( B i,p ) are all singleton Ferrers boards. It follows that l i,p ( B ) will be a singleton Ferrers board if andonly if these three conditions hold for the board l i,p ( B ).(a) The lowest row of level p is weakly shorter than the highest row of level p − i is weakly shorter than column i + 1; and(c) if columns i and i + 1 terminate at the same level, then the height of column i + 1 is a multiple of m .6ondition (c) is needed to ensure l i,p ( B ) will be singleton.To determine when these conditions hold, first note that applying l i,p to B exchanges arml m ( i, p ) andlegl m ( i, p ). Because B is singleton, the top row of level p − B (and in l i,p ( B )) extends left of column i by ⌊ legl m ( i, p − ⌋ m /m cells. On the other hand, the new bottom row of level p in l i,p ( B ) extends left ofcolumn i by ⌈ arml m ( i, p ) ⌉ m /m cells. Thus, condition (a) holds if and only if ⌈ arml m ( i, p ) ⌉ m ≤ ⌊ legl m ( i, p − ⌋ m . Since both sides are multiples of m , this inequality is equivalent to arml m ( i, p ) ≤ ⌊ legl m ( i, p − ⌋ m , whichis the first condition in the definition of permissibility.Now consider the heights of columns i and i + 1 in l i,p ( B ). Both column i and column i + 1 have a full m cells in level p . So, in both B and l i,p ( B ), column i + 1 extends above level p by arml m ( i + 1 , p ) cells. Onthe other hand, the new column i in l i,p ( B ) extends above level p by legl m ( i, p ) cells. So condition (b) willhold if and only if legl m ( i, p ) ≤ arml m ( i + 1 , p ) . To deal with condition (c), consider two cases. First suppose that arml m ( i + 1 , p ) is a multiple of m . Thencondition (c) must hold, and here condition (b) will hold if and only if legl m ( i, p ) ≤ ⌊ arml m ( i + 1 , p ) ⌋ m . Nowsuppose that arml m ( i + 1 , p ) is not a multiple of m . Given that condition (b) holds, the new board l i,p ( B )will be singleton if and only if the strengthened inequality legl m ( i, p ) ≤ ⌊ arml m ( i + 1 , p ) ⌋ m is true. Thus,this last inequality is equivalent to the truth of (b) and (c) in all cases. l -operation on an m -level rook placement Since there is a bijection between rook placements on B and l ( B ) when B is singleton, it stands to reasonthat it would generalize to a bijection between rook placements on B and l i,p ( B ). The following lemmamakes this precise. Lemma 8.
For a singleton board B , suppose l i,p is permissible for B . Then there is an explicit bijectionbetween m -level rook placements of k rooks on B and m -level rook placements of k rooks on l i,p ( B ) .Proof. Use the bijection induced by the l operation in Lemma 5 on the subboard transposed by l i,p , notmoving the rooks on the part of board B which is fixed. However, this may cause a rook in the transposedsubboard to occupy the same column or level of l i,p ( B ) as one of the rooks which was fixed. We deal withthis possibility next.In order for two rooks to end up in the same column, there must be rooks placed on B beneath B i,p ,so we can assume p > B which do not containrooks in B i,p , and the set of columns of l i,p ( B ) which do not contain rooks in l ( B i,p ). By our assumption on p , these two sets have the same cardinality and so we can put a canonical bijection on them by pairing theleftmost columns in each set and moving to the right. If there is a rook lower than level p in one of thesecolumns of B , use this bijection on the columns to move it to the cell in the same row of the correspondingcolumn of l i,p ( B ). After doing so, there must be at most one rook in each column of l i,p ( B ). For example,in Figure 5 the rook in (3 ,
2) is in the second column from the left of B which does not contain a rook in B , . Thus it moves to column 2, which is the second column from the left of l , ( B ) that does not containa rook in l ( B , ).If two rooks end up in the same level we treat them similarly where we can assume, without loss ofgenerality, that the i -th column is not the rightmost column of B . There is a canonical bijection betweenthe levels of B which do not contain rooks in B i,p and those of l i,p ( B ) that do not contain rooks in l ( B i,p ).Adjust the levels of all rooks to the right of column i using this bijection, fixing the column of the rook thatmoves. Furthermore, fix the height of the rook that moves within the level, that is, if the rook was in cell( x, y ), move the rook to cell ( x, y ′ ) in the appropriate level with y ≡ y ′ (mod m ). Note that since B and l i,p ( B ) are singleton boards, columns to the right of column i will contain a full m cells at any level whichcontained a rook in the subboard B i,p or l ( B i,p ).To see that this is a bijection, we construct its inverse. Recall that the l operator is an involution onsingleton boards. Thus, since B i,p is a singleton subboard, l i,p ( l i,p ( B )) = B . Similarly, applying the bijectionfrom Lemma 5 and then its inverse returns the original placement of rooks on B i,p . All that remains to7 = RRR l , ( B ) = R RR
Figure 5: On the left, B , is shaded. Here l , is permissible for B and l , ( B ) is shown on the right.check is that any rooks moved outside of B i,p return to their original cells. Since the rooks return to theiroriginal placement on B i,p , the set of columns that gain a rook in l ( B i,p ) after the first application of l willbe the same set as those that lose a rook in B i,p after the second application of l . Thus the bijection onthe columns induced by the first application of l will be the inverse of the bijection induced by the secondapplication, and any rook required to move in l i,p ( B ) will move back in l i,p ( l i,p ( B )). A similar argumentholds for levels, noting that l i,p ( B ) being singleton ensures that any level which gains a rook in l ( B i,p ) afterapplying l contains a full m cells in every column to the right of column i . Thus this yields a bijectionbetween rook placements on B and l i,p ( B ). Figure 5 illustrates this bijection. m -increasing boards Foata and Sch¨utzenberger proved there is a unique Ferrers board in every rook equivalence class whosecolumn lengths are strictly increasing and used this board as a target for their bijections. To accomplish thesame thing, we need the following definition and theorem.
Definition 9.
A Ferrers board B = ( b , b , . . . , b n ) is called m -increasing if b i +1 ≥ b i + m for all ≤ i ≤ n − . Notice that when m = 1 increasing and m -increasing are equivalent. Theorem 10 (Theorem 4.5 [BLRS13]) . Every Ferrers board is m -level rook equivalent to a unique m -increasing board. We are now almost ready to prove the main result of this section, Theorem 12 below. However, to doso we must put an order on Ferrers boards. Once we have established this order, we will be able to givean explicit bijection between m -level rook placements on an arbitrary Ferrers board B and on an m -levelrook equivalent Ferrers board which is greater than B in this order, if such a board exists. Additionally, theset of all Ferrers boards equivalent to B will have a unique maximum element under this order, namely the m -increasing board guaranteed by the previous theorem.To define this order, if B = ( b , . . . , b n ) then consider the reversal of B , B r = ( b n , . . . , b ). Now let B < B ′ if B r is lexicographically smaller than ( B ′ ) r . It is important to note that when applying Lemma 3we will always have B S ≥ B (2.1)since in B S all the cells in each level are as far to the right as possible. Lemma 11.
Given a singleton board B containing a column i and a level p with the property that arml m ( i, p ) < legl m ( i, p ) , (2.2) there is a singleton board B ′ = l i ′ ,p ( B ) with i ′ ≥ i and B ′ > B .Furthermore, if B is not m -increasing then a column i and level p satisfying equation (2.2) must exist. RR (a) R RR (b)
R R R (c)
R R R (d)Figure 6: (a) A 2-level rook placement on a Ferrers board. (b) The placement on the singleton boardobtained after applying Lemma 3. (c) The placement obtained after applying Lemma 8 using l , . (d) Theplacement obtained on a 2-increasing board after applying Lemma 8 again using l , . Proof.
To prove the first statement, let i ′ ≥ i be the maximum index such that arml m ( i ′ , p ) < legl m ( i ′ , p ).Note that by our convention on arml m , we must have that i ′ is at most the number of columns of B . Weclaim that it suffices to show that l i ′ ,p is permissible for B . This is because if l i ′ ,p is permissible for B ,then the resulting board B ′ must satisfy B ′ > B . Indeed, l i ′ ,p ( B ) increases the length of column i ′ bylegl m ( i ′ , p ) − arml m ( i ′ , p ), which must be greater than 0, and column i ′ is the rightmost column of B affectedby l i ′ ,p . Thus B ′ > B .If l i ′ ,p is not permissible for B , then we claim that we have arml m ( i ′ + 1 , p ) < legl m ( i ′ + 1 , p ) which willcontradict the maximality of i ′ and complete this part of the proof. Note thatarml m ( i ′ , p ) < legl m ( i ′ , p ) ≤ ⌊ legl m ( i ′ , p − ⌋ m . So l i ′ ,p not being permissible for B implies that ⌊ arml m ( i ′ + 1 , p ) ⌋ m < legl m ( i ′ , p ) = legl m ( i ′ + 1 , p ) − m since B is singleton and, because legl m ( i ′ , p ) is positive, i ′ cannot be the leftmost column terminating in level p .This implies the desired contradiction that arml m ( i ′ + 1 , p ) < legl m ( i ′ + 1 , p ).To prove the second statement of the theorem, note that if B is not m -increasing there are two possiblecases: either there are two adjacent columns i − , i of B which terminate at the same level, or column i − p and B has exactly r cells in the p th level of column i − r cells in the( p + 1)st level of column i where r > r > i − i both terminate at level p . Then arml m ( i, p ) = 0, by the assumption thatcolumn i terminates at level p , but legl m ( i, p ) ≥ i − p th level. Thusarml m ( i, p ) < legl m ( i, p ) as desired.Case 2: By assumption arml m ( i, p ) = r < r ≤ legl m ( i, p ) which completes the proof.We are now in a position to prove our main theorem of this section. Theorem 12.
Given any two m -level rook equivalent Ferrers boards, there is an explicit bijection between m -level rook placements of k rooks on them.Proof. Given any Ferrers board B , let B m be the unique m -increasing board in the m -level rook equivalenceclass of B guaranteed by Theorem 10. It suffices to show that there is an explicit bijection between the m -level rook placements of k rooks on B and those on B m . This is trivial if B = B m so assume B = B m .By Lemma 3, we have an explicit bijection between the placements on B and those on B S where B S ≥ B byequation (2.1). If B S = B m then we are done. Otherwise, apply the local l operator defined in Lemma 11which will give B ′ = l i,p ( B S ) with B ′ > B S and, by Lemma 8, another explicit bijection between rookplacements. We now repeat this process if necessary. Since there are only finitely many boards in an m -levelrook equivalence class and the lexicographic order increases at each stage, we must eventually terminate.And, by Lemma 11 again, termination must occur at B m . Composing all the bijections finishes the proof.See Figure 6 for a short example of this process. 9 ∗ R ∗ ∗∗ ∗ ∗∗∗∗ ∗ ∗ ∗∗∗∗ ∗ ∗∗∗ R Figure 7: A placement, π , with inv ( π ) = 19. q -Analogues Briggs and Remmel [BR06] defined p, q -analogues of the m -level rook numbers, denoted r k,m ( B ; p, q ), byassigning a monomial in p and q to each m -level rook placement of k rooks on B . Briggs and Remmel proveda factorization formula involving r k,m ( B ; p, q ) for singleton boards, which was generalized to all Ferrersboards by the present authors [BLRS13, Thm. 3.3]. In this section, we show that the bijections given earlierin this paper preserve the q -power assigned to a rook placement. This leads to bijective proofs that two m -level rook equivalent boards have the same rook polynomials r k,m ( B ; 1 , q ) for all k . Our bijections do notpreserve the p -power, however, and we leave it as an open problem to give a bijective treatment of the full p, q -analogue of m -level rook numbers. q -weight To begin, we recall that the q -weight assigned to an m -level rook placement π on a board B is the m -inversion number of π . The m -inversion number , denoted inv m ( π ), counts cells c in B satisfying the followingconditions:1. The cell c does not contain a rook.2. There is no rook above c in the same column.3. There is no rook to the left of c in the same level.For example, the 3-level rook placement shown in Figure 7 has an m -inversion number of 19; the cellscontributing to the m -inversion number are marked by stars.To motivate why this statistic is called the m -inversion number, consider the case where m = 1. Thusthe 1-inversion number counts the number of cells which do not contain a rook and are neither below nor tothe right of a rook. If B is an n by n board and σ is an element of S n , the symmetric group on the elements { , , . . . , n } , then we can associate with σ a placement of n rooks on B , π , by the convention that thereis a rook in column i and row n + 1 − p if and only if σ i = p . In this case inv ( π ) = inv( σ ) where inv( σ )is the standard inversion number of a permutation, counting the number of pairs of indices ( a, b ) with theproperty that a < b but σ ( a ) > σ ( b ).Define r k,m ( B ; q ) by r k,m ( B ; q ) = X π q inv m ( π ) (3.1)where π ranges over all m -level rook placements of k rooks on B . Define boards B and B ′ to be m -level q -rook equivalent if r k,m ( B ; q ) = r k,m ( B ′ ; q ) for all nonnegative integers k . We will give bijective proofs ofthe m -level q -rook equivalence of various boards, by showing that the bijections given earlier preserve the q -power. 10 R R R ∗∗∗ ∗ ∗ ∗
R R R R ∗∗∗ ∗ ∗ ∗
Figure 8: On the left, a placement π with inv π = 6 on a Ferrers board. On the right, the correspondingplacement on B S .We will need two formulas for the m -inversion number of an m -level rook placement π , one which addsthe contributions of each level, and another which sums over the contributions of each column. The firstformula uses the numbering of cells in each level from Definition 2. For each level p such that π has a rook R in level p , let h p ( π ) count the cells in level p with a higher number than the cell containing R . Also letNW p ( π ) be the number of rooks in π northwest of R , that is, rooks in a higher level and earlier columnthan R . For each level p containing no rook, let h p ( π ) be the total number of cells in this level, and letNW p ( π ) be the number of rooks in higher levels than p . Note that the definition for a level containing norooks can be considered as a limiting case of the one for a level containing a rook by letting the rook moveto the right until it exits the board. So in our proofs we will only consider the first case as the second onewill automatically follow using this procedure. Define, using “h” for “horizontal,”hinv p ( π ) = h p ( π ) − m · NW p ( π ) . (3.2)It is routine to check that for any board B , inv m ( π ) = P p ≥ hinv p ( π ). In particular, if a column has fewerthan m cells in level p , there can be no rook weakly west of this column in a higher level. Thus each rookcounted by NW p ( π ) removes a full m cells from the cells that would have contributed to inv m ( π ) in level p .In the example in Figure 7, hinv ( π ) = 6, hinv ( π ) = 7, hinv ( π ) = 6, and hinv ( π ) = 0.The second formula for inv m ( π ) classifies cells based on their columns. For each column i such that π has a rook R in column i , let h ′ i ( π ) count the cells in column i above R , and let NW ′ i ( π ) be the number ofrooks in π northwest of R . For each column i containing no rook, let h ′ i ( π ) be the total number of cells inthis column, and let NW ′ i ( π ) be the number of rooks in earlier columns than i . Again, the second case is alimiting instance of the first where now the rook moves down until it is off the board. Define, using “v” for“vertical,” vinv i ( π ) = h ′ i ( π ) − m · NW ′ i ( π ) . (3.3)One may check that for any rook placement π on a singleton board B , inv m ( π ) = P i ≥ vinv i ( π ). Thesingleton condition ensures that any rook counted by NW ′ i ( π ) must remove a full m cells from the cells thatwould have contributed to inv m ( π ) in column i . In the example from Figure 7, it happens that inv m ( π ) = 19is not the sum of the entries in (vinv ( π ) , . . . , vinv ( π )) = (2 , , , , , , ,
1) because B is not a singletonboard and the rook to the northwest of the rook in the rightmost column only cancels one cell in the rightmostcolumn, rather than a full 3. B to placements on B S We now prove that the bijection in Lemma 3, mapping m -level rook placements on an arbitrary board B to m -level rook placements on the singleton board B S , preserves the m -inversion number. See Figure 8 for an11 R R R ∗ ∗∗∗ ∗ ∗∗
R R R R ∗∗ ∗∗∗∗ ∗
Figure 9: On the left, a 2-level placement with a 2-inversion number of 7 on a singleton board. On the right,the corresponding placement after applying the l -operator to the board on the left.example in the case m = 3. Lemma 13. If π is a rook placement on a Ferrers board B that maps to the rook placement π S on B S whenwe apply the bijection in Lemma 3, then inv m ( π ) = inv m ( π S ) .Proof. We use (3.2) to show that hinv p ( π S ) = hinv p ( π ) for each level p . Let π ′ be the placement createdfrom π in the first stage of the map, in which all rooks remain in their original numbered cell in their level.By definition of the level numbering, h p ( π ′ ) = h p ( π ) for all p . Consider a rook that moves from column i to column i ′ > i in the first stage, and a level p below that rook that has a rook in the interval ( i, i ′ ]. Insuch a level, NW p ( π ′ ) = NW p ( π ) −
1, so hinv m ( π ′ ) = hinv m ( π ) + m . The second stage corrects for thisincrease by moving the rook in level p one column to the left, which decreases h p by m . The net effect isthat hinv p ( π S ) = hinv p ( π ) and hence inv m ( π S ) = inv m ( π ), as needed. l -operator Let B be a singleton board. We now show that the bijection from Lemma 5, which maps an m -level rookplacement π on B to an m -level rook placement l ( π ) on l ( B ), preserves the m -inversion number. Lemma 14. If π is an m -level rook placement on a singleton board B , then inv m ( π ) = inv m ( l ( π )) .Proof. The level numbering of B and the column numbering of l ( B ) induce a bijection between the squaresof B and the squares of l ( B ). It is easy to see from the definitions that a square of B contributes to inv m ( π )if and only if the corresponding square of l ( B ) contributes to inv m ( l ( π )). So the lemma is proved.The example in Figure 9 illustrates the ideas in this proof in a case where m = 2; note that the starredcells in each level of the original placement become starred cells in each column of the new placement. l -operator Next we show that the bijection in Lemma 8 preserves the m -inversion number. Lemma 15.
Given an m -level rook placement π on a singleton board B , a column i , and a level p suchthat l i,p is permissible for B , let π ′ denote the corresponding placement on l i,p ( B ) as described in Lemma 8.Then inv m ( π ) = inv m ( π ′ ) .Proof. We have already shown in § m -inversion number coming from the cellsin B i,p and l ( B i,p ) is the same. We show that the adjustments made below B i,p do not affect the m -inversionnumber, as follows. By the construction of the bijection in Lemma 8, every rook beneath B i,p is adjusted12 R R R R ∗∗∗∗∗ ∗ ∗∗ ∗∗∗ ∗
R R R R R ∗∗ ∗∗ ∗∗∗∗ ∗∗∗ ∗
Figure 10: On the left, a placement with a 2-inversion number of 12 on a singleton board. On the right, thecorresponding placement after applying the local l -operator l , to the board on the left.horizontally until there are as many columns to the left of it in l i,p ( B ) that do not contain rooks in l ( B i,p )as there were before applying the l -operator. Since the relative order of rooks in levels beneath level p ispreserved, each level beneath level p will have the same number of cells counted by the m -inversion numberbefore and after the adjustment. An analogous argument shows that the level adjustments to the right of B i,p do not change the m -inversion number, since after the adjustment each rook to the right of B i,p willhave the same number of levels which get counted for the m -inversion number above it in l i,p ( B ) as it doesin B .See Figure 10 for an example with m = 2. Note that the rook in (4 ,
2) on the left moves to (3 ,
2) keepingone column with no higher rook to the left of it, so the first level contributes 2 to the m -inversion number inboth placements. Also consider the rook in (5 ,
5) on the left which moves to (5 ,
3) maintaining its positionin the bottom row of its level. Even though there are more rooks to the northwest of it in the right-handdiagram, the number of levels above it that do not contain rooks to the left of it is unchanged, so the fifthcolumn contributes 3 to the m -inversion number. Theorem 16. If B and B ′ are m -level rook equivalent Ferrers boards, then r k,m ( B ; q ) = r k,m ( B ′ ; q ) .Proof. We can compose all the bijections, as in the proof of Theorem 12, to obtain an m -inversion-preservingbijection between m -level rook placements of k rooks on B and B ′ . By the definition of r k,m ( B ; q ) in § r k,m ( B ; q ) = r k,m ( B ′ ; q ). m -level rook placements Our next two main results will require the Garsia-Milne Involution Principle. First, we will use the InvolutionPrinciple to construct another explicit bijection between two arbitrary m -level rook placements of k rookson m -level rook equivalent singleton boards. Theorem 17 (Garsia-Milne Involution Principle [GM81]) . Consider a triple ( S, T, I ) where S is a signedset, I is a sign-reversing involution on S , and the set T of fixed points of I is required to be a subset of thepositive part S + of S . Let ( S ′ , T ′ , I ′ ) be defined similarly. Then, given an explicit sign-preserving bijection f from S to S ′ , one can construct an explicit bijection between T and T ′ . The way that Garsia and Milne define the explicit bijection is as follows. Start with an element t ∈ T ⊆ S + . If f ( t ) T ′ , then apply ( f ◦ I ◦ f − ◦ I ′ ) to f ( t ). This takes f ( t ) ∈ S ′ + to S ′− , then to S − , then to S + ,and finally back to S ′ + . Iterating this procedure must ultimately yield an element of T ′ which is consideredthe image of t under the desired bijection. 13 R WC = ∈ S RW WI ( C ) = RR Wf ( C ) = RW Wf ( I ( C )) =Figure 11: On the top left, an element in S with sign −
1. On the top right, the image under I which hassign +1. Beneath each board is its image under f . We will use the Involution Principle to construct a bijection between m -level rook placements on two m -levelrook equivalent singleton boards. We must first construct a signed set and a sign-reversing involution sothat the m -level rook placements are the fixed points under the involution. We do this as follows.Given two Ferrers boards, B and B ′ , we shall say B fits inside B ′ if juxtaposing the two boards withtheir lower right cells in the same position makes the cells of B a subset of the cells of B ′ . Figure 11 showsthat the thick bordered B = (2 ,
3) fits inside B ′ = (0 , , , n,m denote the triangular Ferrers board (0 , m, m, . . . , ( n − m ). Given a singletonboard B , fix N large enough that B fits inside ∆ N,m . If B has fewer than N columns, expand B on theleft with columns of height zero so B = ( b , b , . . . , b N − ) has the same number of columns as ∆ N,m . Fixa non-negative integer k with k < N and let the integer i vary over 0 ≤ i ≤ k . Then S will consist of allconfigurations C constructed as follows. Take ∆ N,m with B fitting inside and place white rooks W in i cellsof ∆ N,m that are outside of B so that no two white rooks are in the same column. Next, place k − i blackrooks R forming an m -level rook placement on the subboard ∆ N − i,m which is located in the columns of∆ N,m which do not contain a white rook. We will call this the inset ∆ N − i,m board . Note that the columnsof the inset ∆ N − i,m may not be contiguous.See the top left board of Figure 11 for an example of such an object C where m = 2. The singletonboard B = (0 , , ,
3) fits inside ∆ , . Here k = 3 < i = 1 white rook on the board ∆ , \ B and k − i = 2 black rooks on the board ∆ , which is represented by the grey shaded cells inside ∆ , . Therooks on ∆ , form a 2-level rook placement, but there is both a black rook and a white rook in the secondlevel of ∆ , .Note that each column of ∆ N,m may contain at most one white rook or black rook. On the other hand,a level of ∆
N,m will contain at most one black rook, but may contain any number of white rooks. Further,define the sign of such a placement to be ( − i . The sign of the placement on the top left in Figure 11 is − I on an element C ∈ S , if all rooks of C are in B , and therefore black, then C is a fixed point.Otherwise, examine the columns of C from left to right until coming to a column with a rook outside of B .If that column contains a black rook, change the rook to a white rook, increase i by one, and move everyblack rook above and to the right of the cell containing the new white rook down m cells. If that column14ontains a white rook, change it to a black rook, decrease i by one, and move every black rook to the rightand at the same level or higher as the new rook up m cells. The placement on the top right in Figure 11illustrates what happens to the board on the left under I . Similarly, I takes the placement on the right tothe placement on the left.We must show that I ( C ) will be an element of S . Clearly each column has at most one rook. We claimthat each level will still contain at most one black rook. First, suppose that a black rook is added. In thiscase all black rooks at its level or above to the right of the new rook move up one level. Furthermore, therecan be no black rooks at the same level or higher to the left of the new black rook. This is because thenew black rook was a white rook which, by definition, was above board B . Since B is a singleton board, nocolumns of B to the left of the white rook in question will terminate in the level of the white rook. Thus ifthere were a black rook at the same level or higher to the left, it too would be outside of board B , whichcontradicts the white rook being the leftmost rook outside of board B . Thus the black rooks still form an m -level placement when a black rook is added. The proof that this also holds when a black rook becomeswhite is similar.We must also check that the black rooks continue to fit on the new insert board. When a black rook isadded, the black rooks must be placed on a board ∆ N − i +1 ,m where the column in which the new black rookis placed is added to the columns in the initial inset ∆ N − i,m . Since there are no white rooks to the left of thenew black rook, there will be no omitted columns to the left of the column containing the new black rook,thus all cells of that column will be in the inset ∆ N − i +1 ,m and the new rook must be inside ∆ N − i +1 ,m . Thismeans that all the columns to the right of the new black rook that do not contain a white rook will contain m more squares in the inset ∆ N − i +1 ,m than they did in the inset ∆ N − i,m . Thus moving black rooks to theright of the new black rook up m cells will keep them within the new ∆ N − i +1 ,m . Similarly, changing a blackrook to a white rook will decrease the number of cells in the columns of ∆ N − i − ,m to the right of the newwhite rook by m , but all black rooks to the right of the new white rook and at a higher level than it aremoved down m cells, so they will be in ∆ N − i − ,m because they were in ∆ N − i,m originally. Finally, if thereare any black rooks below the level of the new white rook but to its right, they will remain in ∆ N − i − ,m because the first column in ∆ N − i − ,m to the right of the new white rook must go up to at least the level ofthe new white rook since previously it was a black rook contained in ∆ N − i,m .By construction, I is an involution. The fixed set of I will be denoted T . It is the set of all configurationswhich only have rooks on the subboard B and, by definition, these rooks must be black. As such, T is equalto the set of m -level rook placements of k rooks on B . Furthermore, if a board is not in T , then I eitherincreases or decreases the number of white rooks on the board by one. Either way I will change the sign ofthe board. And if a board is in T , then it has positive sign.Given a singleton board B ′ , define N ′ , S ′ , T ′ , and I ′ similarly for B ′ contained in ∆ N ′ ,m . Without aloss of generality, assume N = N ′ . Let B ′ = ( b ′ , b ′ , . . . , b ′ N − ). If B and B ′ are m -level rook equivalentsingleton boards we can use I and I ′ to construct an explicit bijection between m -level rook placements of k rooks on B and m -level rook placements of k rooks on B ′ . We do this by constructing a sign-preservingbijection between S and S ′ . We will need the following characterization of when two singleton boards are m -level rook equivalent.The root vector of B is ζ m = ( − b , m − b , . . . , ( N − m − b N − ) . The following result of Briggs and Remmel determines when two singleton boards are m -level rook equivalentsimply by considering their root vectors. Theorem 18 (Briggs-Remmel [BR06]) . If B = ( b , . . . , b N ) is a singleton board then N X k =0 r k,m ( B ) x ↓ N − k,m = N Y i =1 ( x + b i − ( i − m ) where x ↓ k,m = x ( x − m )( x − m ) . . . ( x − ( k − m ) . Note that the indexing in the theorem begins at 1, rather than 0, simply to be consistent with the originalstatement of the theorem. 15
R R { , } , { , , } , { } .Since the root vector contains exactly the roots of the rook polynomial, we see that two singleton boardsare m -level rook equivalent if and only if they have the same root vector, up to rearrangement, for asufficiently large N . We are now ready to apply the Garsia-Milne Involution Principle. Theorem 19.
Let B and B ′ be m -level rook equivalent singleton boards. Then there exists an explicitGarsia-Milne bijection between m -level rook placements of k rooks on B and m -level rook placements of k rooks on B ′ .Proof. By Theorem 17 and what we have already established, it suffices to find a sign-preserving bijection f : S → S ′ . We construct f as follows.For clarity of notation, let B be placed in ∆ N,m and B ′ be placed in a copy ∆ ′ N,m of ∆
N,m . Notice thatthe k th element of the root vector of B , km − b k , is the number of cells in the k th column of ∆ N,m which lieoutside of board B . Since B and B ′ are m -level rook equivalent, the root vector for B ′ is a rearrangementof the root vector for B . Therefore there is a length-preserving bijection between the columns of the setdifference ∆ N,m \ B and the columns of ∆ ′ N,m \ B ′ which takes the leftmost column of a given length in∆ N,m \ B to the leftmost column with that length in ∆ ′ N,m \ B ′ and so forth. This bijection induces abijection on the placement of the white rooks. If a white rook appears in the j th cell above B , place a whiterook in the j th cell above B ′ in the associated column.Once all the white rooks are placed, create a copy of ∆ ′ N − i,m inside of ∆ ′ N,m using the columns which donot contain a white rook. Place the black rooks on the board in relation to the ∆ ′ N − i,m subboard exactly asthey are placed on the original board in relation to the original ∆ N − i,m subboard. Each placement on thebottom of Figure 11 is the image under f of the corresponding placement on the top where B = (0 , , , B ′ = (0 , , , B . In the board on the bottom left the white rook is still at the topof the second column from the left which has two cells above B ′ .Under this map the white rooks must be placed inside ∆ ′ N,m but outside B ′ , and the black rooks areplaced inside ∆ ′ N − i,m , so f maps S to S ′ . Further this map preserves the number of white rooks placed onthe board, so it is sign preserving. Therefore we may conclude from the Involution Principle that there is anexplicit bijection between m -level rook placements of k rooks on B and m -level rook placements of k rookson B ′ .To obtain a consequence of this construction, we will need some background on symmetric functionsand Stirling numbers. For d ≤ n both non-negative integers, let e d ( x , x , . . . , x n ) denote the elementarysymmetric function of degree d in n variables , that is, e d ( x , x , . . . , x n ) = X ≤ i
R W WR R ∗∗ ∗∗ ∗ ∗∗∗ ∗∗∗ ∗∗∗∗
Figure 13: On the left, an element of S with augmented 2-inversion number 15 and sign −
1. On the right isthe image of the left placement under I , which still has augmented 2-inversion number 15, but has sign +1. i column j . See Figure 12 for the rook placement corresponding to { , } , { , , } , { } . Thus the numberof m -level rook placements of n − d rooks on ∆ n,m is m n − d S ( n, d ). The extra m n − d counts the number ofways of choosing a placement for each of the n − d rooks in the m cells of a level.It is interesting to note that the construction of I yields the following theorem giving an explicit calculationfor the m -level rook numbers of a singleton Ferrers board B . Theorem 20.
For any singleton board B = ( b , b , . . . , b N − ) fitting inside ∆ N,m , r k,m ( B ) = k X i =0 ( − i m k − i S ( N − i, N − k ) e i ( − b , m − b , . . . , ( N − m − b N − ) . Proof.
Since the fixed points of the involution I are counted by r k,m ( B ), it suffices to show that the sumcounts all elements of the set S by sign. First note that the number of ways of putting i white rooks in i different columns of ∆ N,m outside of B is e i ( − b , m − b , . . . , ( N − m − b N − ). Furthermore the numberof m -level rook placements of k − i rooks on ∆ N − i,m is m k − i S ( N − i, N − k ). Putting these two countstogether with the appropriate sign gives the sum as desired.Note that this theorem implies the previously noted result that if two boards have the same root vectorthen they are m -level rook equivalent. inv m We finish this section by showing that the bijections of the previous subsection preserve the m -inversionnumbers of m -level rook placements. Given a singleton board B , consider a configuration C in the set S consisting of i white rooks in different columns of ∆ N,m \ B , together with an m -level placement π of k − i black rooks on the inset board ∆ N − i,m . Let the augmented m -inversion number of C , denoted ainv m ( C ), bethe m -inversion number of π , as in Section 2, calculated relative to the inset board ∆ N − i,m , plus the numberof cells which lie in a column above a white rook. For example, the 2-level configurations in Figure 13 bothhave augmented m -inversion number 15. Lemma 21. If C ∈ S , then ainv m ( C ) = ainv m ( I ( C )) where I is the map from Subsection 4.1.Proof. It suffices to consider the case where I changes the leftmost rook outside B from white to black. Inthis case, let W denote the leftmost white rook in C . All squares in the column above W contributed toainv m ( C ) since they were above a white rook. These squares must still contribute to the augmented m -inversion number of I ( C ) because, as proved earlier, there can be no black rook northwest of these squares.17 RR Figure 14: The placement on Sq , corresponding to ( α , α , α ; (1 , , R in C . If R is to the left of W , the cells in this columncontributing to the augmented m -inversion number are the same in C and I ( C ). If R is to the right of W in a lower level, the new inset board ∆ N − i +1 ,m will have m more cells above R in its column. But, m ofthose cells are located in the same level as the new black rook where W was, so that the contribution of thiscolumn to ainv m ( C ) is the same as to ainv m ( I ( C )). If R is to the right of W at the same or higher level, R will move up m cells, but the new inset board ∆ N − i +1 ,m will also have m new cells in this column. Asimilar analysis shows that a column of the inset board containing no rook makes the same contribution tothe augmented m -inversion number in C and I ( C ). Finally, any column which does not intersect the insetboard ∆ N − i +1 ,m contributes the same amount to ainv m ( C ) and ainv m ( I ( C )) because none of the otherwhite rooks have changed location.Next we show that the sign-preserving bijection f : S → S ′ from the proof of Theorem 19 preserves theaugmented m -inversion number. Lemma 22. If C ∈ S and f ( C ) ∈ S ′ , then ainv m ( C ) = ainv m ( f ( C )) where f is the map from Subsection 4.1.Proof. Since f sends the placement of black rooks on the inset board ∆ N − i,m to the identical placement ofblack rooks on the inset board ∆ ′ N − i,m , the contribution to the augmented m -inversion number from theblack rooks is the same in C and f ( C ). By the way f moves the white rooks, the total number of cells abovethe white rooks in C and f ( C ) also agrees. Thus ainv m ( C ) = ainv m ( f ( C )), as desired. Theorem 23.
The explicit bijection produced by Theorem 19 preserves the m -inversion number of the board.Proof. From the previous two lemmas, we see that the Garsia-Milne Involution Principle provides a bijection g between the fixed point sets T and T ′ , which preserves the augmented m -inversion number. Recall thata configuration C ∈ T or C ′ = g ( C ) ∈ T ′ has no white rooks, and all black rooks are on the board B , notmerely on the larger board ∆ N,m which is the inset board when there are no white rooks. Let π and π ′ be theplacments on B and B ′ , respectively. Note that ainv m ( C ) is computed relative to the board ∆ N,m , whereasinv m ( π ) is computed relative to the smaller board B . But since B is a singleton Ferrers board located inthe southeast corner of ∆ N,m , every square in ∆
N,m \ B will contribute to ainv m ( C ). Since B = B ′ , weconclude that inv m ( π ) = ainv m ( C ) − N,m \ B ) = ainv m ( C ′ ) − N,m \ B ′ ) = inv m ( π ′ )for all C ∈ T . This shows that the bijection g preserves the m -inversion number of m -level rook placementscomputed relative to the boards B and B ′ . We will now use the Involution Principle to prove that two boards that are m -level rook equivalent have thesame hit numbers. We begin with some definitions. As usual, let m be a fixed positive integer.Let B be a Ferrers board and let the integer N be sufficiently large so that B fits inside a rectangularboard Sq N,m with N columns and mN rows. If α is a generator of the cyclic group C m , then C m ≀ S N = { ( α s , α s , . . . , α s N ; σ ) | ≤ s i ≤ m for each i and σ ∈ S N } . WWs = RRWI ( s ) =Figure 15: On the left, an element in S with sign +1. On the right, the image under I which has sign − ω ∈ C m ≀ S N a placement on Sq N,m by placing a rook in level N + 1 − p and column i if σ ( i ) = p . Furthermore, the rook in column i will be j cells from the bottom of the level if s i = j . SeeFigure 14 for an example with m = 2 and N = 3, where the placement corresponds to ( α , α , α ; (1 , , σ is in one line notation. Let R ( ω ) denote the rook placement corresponding to ω . Define the k th hitset of B to be H ( m ) k,N ( B ) = { R ( ω ) | ω ∈ C m ≀ S N and R ( ω ) ∩ B ) = k } . (5.1)Also define the k th hit number of B to be h ( m ) k,N = H ( m ) k,N . (5.2)In order to show that two m -level rook equivalent Ferrers boards have the same hit numbers, we useGarsia and Milne’s result again. To do so, we must construct a signed set and a sign-reversing involutionwhich has a set counted by h ( m ) k,N as its fixed set. We do this as follows.Let N be large enough that B fits inside Sq N,m and fix a non-negative integer k with k ≤ N . Then theset S will consist of all configurations C constructed as follows. Let i vary over all non-negative integerssuch that k + i ≤ N . Place k + i non-attacking black, m -level rooks R on the board B if possible. If thisis not possible then there are no elements of S corresponding to this choice of k and i . Furthermore, circle i of the rooks in the placement. Finally, consider the N − k − i columns and N − k − i levels which do notcontain a black rook as a subboard of shape Sq N − k − i,m . As in the previous section, we will call this the inset Sq N − k − i,m board . Place N − k − i non-attacking white m -level rooks, denoted by W , on the inset Sq N − k − i,m .Notice that, ignoring the color of the rooks, this is an m -level rook placement of N rooks on Sq N,m . Thusit corresponds to some element of C m ≀ S N . Let the sign of a configuration be ( − i . See Figure 15 for twoexamples of such configurations. Here m = 2 and B = (1 , ,
4) is placed fitting in Sq , . On the left, thereare no circled black rooks so i = 0 and the white rooks are placed on the shaded inset Sq , . On the rightthere is one circled black rook so i = 1 and the white rooks are on a shaded inset Sq , .In order to produce a sign-reversing involution I on such configurations C , we do the following. If B contains neither a white rook nor a circled black rook, then C is fixed by I . Otherwise, examine the columnsof B from left to right until the first white rook or circled black rook is found. If the first rook found iswhite, exchange it for a circled black rook and increase i by 1. If the first rook found is a circled black rook,exchange it for a white rook and decrease i by 1. See Figure 15 for two examples of such configurations with k = 1. It is easy to see that I is an involution and reverses signs in its 2-cycles. Also note that fixed pointshave no circled black rooks, so i = 0 and the sign of the configuration is +1. Furthermore, for a fixed pointthere are no white rooks placed on B , so the m -level placement intersects B in exactly k black rooks. Thusthe fixed points are exactly the elements of H ( m ) k,N ( B ) if one just ignores the colors of the rooks.The reader will find an example illustrating the next proof in Figure 16. This example uses the boardsfrom the example of Theorem 12 found in Figure 6. Theorem 24.
Let B and B ′ be two m -level rook equivalent Ferrers boards and N be large enough that B and B ′ both fit inside Sq N,m . Then for any non-negative integer k ≤ N , there is an explicit bijection between H ( m ) k,N ( B ) and H ( m ) k,N ( B ′ ) .Proof. As in the proof of Theorem 19, we use the Garsia-Milne Involution Principle. Construct S for B placed inside Sq N,m and S ′ for B ′ placed inside Sq ′ N,m . From what we have already done, all that remainsis to construct the sign-preserving bijection f : S → S ′ .19 RRW W R R RW W
Figure 16: On the left, a 2-level placement of white rooks, black rooks, and circled black rooks on (1 , , , , , . On the right, the corresponding placement on (3 , ,
8) under the construction in Theorem 24.Consider an element C ∈ S . The black rooks, circled and uncircled, form an m -level rook placement of k + i rooks on B . Map this to an m -level rook placement of k + i rooks on B ′ using the explicit bijectionguaranteed by Theorem 12. Furthermore, add circles to the rooks on B ′ in such a way so that if the r throok from the right on board B is circled, the r th rook from the right on board B ′ is circled. Finally, placethe white rooks on Sq ′ N,m by considering the inset Sq ′ N − k − i,m of columns and levels containing no blackrooks. Place the white rooks on this inset board in the exact same arrangement as they are in on the insetSq N − k − i,m of Sq N,m . This is easily seen to be a bijection and so the proof is complete.The next corollary follows immediately from the previous theorem.
Corollary 25.
Let B and B ′ be two m -level rook equivalent Ferrers boards and N be large enough such that B and B ′ both fit inside Sq N,m . Then for any non-negative integer k ≤ N , h ( m ) k,N ( B ) = h ( m ) k,N ( B ′ ) . The l -operator leads to a second formulation of the factorization theorem for the m -level rook polynomial ofa Ferrers board, originally found in [BLRS13]. This theorem generalized Theorem 18 from singleton boardsto all Ferrers boards. Theorem 26.
Let B = ( b , . . . , b n ) be a Ferrers board with t non-empty levels, and let l ( B ) = ( l t , l t − , . . . , l ) be the singleton board where l p is the number of cells in level p of B , as in § N greaterthan or equal to both n and t N X k =0 r k,m ( B ) x ↓ N − k,m = N Y i =1 ( x + l N − i +1 − ( i − m ) ,where l N − i +1 = 0 if N − i + 1 > t .Proof. Since r k,m ( B ) = r k,m ( l ( B )) by Lemma 5, the choice of N ensures that N X k =0 r k,m ( B ) x ↓ N − k,m = N X k =0 r k,m ( l ( B )) x ↓ N − k,m .Since the right hand side of the equation is the m -level rook polynomial of the singleton board l ( B ), Theo-rem 18 implies N X k =0 r k,m ( l ( B )) x ↓ N − k,m = N Y i =1 ( x + l N − i +1 − ( i − m ).20 B = Rl ( B ) = *Figure 17: On the left, a 2-level placement π of one rook on Ferrers board B = (1 ,
1) with wt ( π ) = −
2. Onthe right, l ( B ) with wt ( l ( π )) = −
1. The single cell counted by β ( l ( π )) is denoted with an asterisk.Combining these equations yields the desired theorem. The characterization of the rook equivalence class of a singleton board in terms of its root vector in The-orem 18 provides a way to count the number of singleton boards in a given m -level rook equivalence classas was done in [BLRS13]. However, it is an open problem to count the total number of Ferrers boards in agiven m -level rook equivalence class. If C is a singleton board, then perhaps counting the number of Ferrersboards B with l ( B ) = C would be a good start to this problem, but this too remains open.Another open question concerns a p -analogue of the m -level rook numbers. Here, p refers to a variableand not a level. When the q -analogue was introduced above, it was mentioned that Briggs and Remmelassigned to each m -level rook placement π a monomial in p and q , where the power of q turned out to beinv m ( π ). The interpretation of the power of p turns out to be less intuitive. Given a placement π of k rooksin columns c , c , . . . , c k , let β ( π ) denote the number of cells c satisfying the following conditions:1. The cell c is below a cell containing a rook.2. There is no rook to the left of c in the same level.Then the power of p associated with placement π , called the p -weight of π , iswt m ( π ) = β ( π ) − m ( c + c + · · · + c k ) . For example, the 2-level rook placement on the right in Figure 17 has β ( π ) = 1 and c = 1 so thatwt ( π ) = 1 − − p -weight of aplacement. In fact, the multiset of p -weights associated with two m -level rook equivalent Ferrers boardsmay not even be equal. Consider, for example, m = 2 and the boards B = (1 ,
1) and l ( B ) = (2), shown inFigure 17. Clearly the two boards are m -level rook equivalent, because the board on the right is obtainedby applying the l -operator to the board on the left. On either board, there is one way to place no rooks andtwo ways to place one rook. Doing this on B yields p -weights of 0 , − , − l ( B ) one obtains 0 , − , − p -analogue of m -level rook placements whichis preserved by the bijections given in this paper, or by similar bijections? If such a p -analogue exists, doesit have a more “natural” motivation? Also, is there a factorization of the p, q -analogue of the m -level rookpolynomial using the new p -analogue, similar to that given in [BLRS13]?Finally there are some open problems related to hit numbers. In [BR06], Briggs and Remmel defined the p, q -hit numbers h ( m ) n,k ( B, p, q ) for any singleton board B that fits inside the rectangular board Sq n,m by n X k =0 h ( m ) k,n ( B, p, q ) x k = n X k =0 r k,m ( B, p, q )[ m ( n − k )] ↓ n − k,m p m (( k +12 ) + k ( m − k ) ) n Y ℓ = n − k +1 ( x − q mℓ p m ( n − ℓ ) ) , (6.1)where r k,m ( B, p, q ) is the p, q -rook number defined in [BR06], [ n ] p,q = p n − q n p − q = p n − + p n − q + · · · + pq n − + q n − for any positive integer n , and [ mk ] ↓ k,m = [ mk ] p,q [ m ( k − p,q · · · [ m ] p,q . They showed that for allsingleton boards B that fit inside the rectangular board Sq n,m , h ( m ) n,k ( B, p, q ) is always polynomial in p and q with non-negative integer coefficients. In [Bri03], Briggs gave a combinatorial interpretation of the h ( m ) k,n ( B, , q ) for any singleton Ferrers board B that fits inside the rectangular board Sq n,m as follows: h ( m ) k,n ( B, , q ) = X R ( ω ) ∈ H ( m ) k,n q ξ mB ( R ( ω )) (6.2)21 R R R ******** ********** ******* **********Figure 18: An example on B = (2 , , ,
10) with m = 3 and n = 4. The cancelled cells which do not containrooks are marked with asterisks. Note there are 9 empty cells which are uncancelled, so ξ B ( R ( ω )) = 9.where ξ mB ( R ( ω )) can be calculated for any R ( ω ) as follows,1. each rook R that does not lie in B cancels all the cells in its column that lie weakly below R andoutside of B plus all the cells in its level which lie strictly to the right of R , and2. each rook R that lies in B cancels all the cells in its column that either lie weakly below R or outsideof B , and all the cells in its level which lie strictly to the right of R , and3. ξ mB ( R ( ω )) is the number of uncanceled cells in Sq n,m .For example, Figure 18 shows a case where m = 3 and n = 4. The placement is an element R ( ω ) ∈ H (3)2 , .We have put asterisks in all the cells which are canceled, which do not already contain rooks, so that ξ B ( R ( ω )) = 9. In the special case m = 1, this statistic corresponds to the statistic for hit numbers onFerrers boards due to Dworkin [Dwo96].Our bijection θ of Theorem 24 between between H ( m ) k,N ( B ) and H ( m ) k,N ( B ′ ) does not send the statistic ξ mB tothe statistic ξ mB ′ . That is, it is not alway the case that if R ( ω ) ∈ H ( m ) k,N ( B ), then ξ mB ( R ( ω )) = ξ mB ′ ( θ ( R ( ω ))).Thus we ask whether it is poosible to define a natural bijection Γ between H ( m ) k,N ( B ) and H ( m ) k,N ( B ′ ) suchthat ξ mB ( R ( ω )) = ξ mB ′ (Γ( R ( ω )))? Also, if we use (6.1) to define h ( m ) k,n ( B, p, q ) for non-singleton Ferrersboards contained in Sq n,m , can we classify the collection of such boards such that h ( m ) k,n ( B, p, q ) is always apolynomial in p and q with non-negative integer coefficients, or when h ( m ) k,n ( B, , q ) is always polynomial in q with non-negative integer coefficients? References [ABR01] Ron M. Adin, Francesco Brenti, and Yuval Roichman. Descent numbers and major indices forthe hyperoctahedral group.
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