Rook placements on South-West permutationdiagrams

Herman Goulet-Ouellet˚

September 15, 2014

Abstract

We study rook placements over South-West diagrams of permutationsof given length. Our main motivation comes from an article from Foataand Schützenberger, [3], and a serie of papers published in the 70s byGoldman, Joichi and White about Rook Theory, [5, 7]. We talk aboutrook-equivalence in Sn and give some results about equivalence classes.We also propose a labelling of Ferrers board with many interesting prop-erties. Finally, we examine some q-analogues for this problem.

Résumé

On étudie les placements de tours sur les diagrammes Sud-Ouest depermutations. Notre motivation principale provient d’un article de Foataet Schützenberger, [3], ainsi que d’une série d’articles publiées dans lesannées 70 par Goldman, Joichi et White, au sujet de la théorie des place-ments de tours, [5, 7]. On aborde la relation d’équivalence induite par lesplacements de tours dans Sn, et on donne quelques résultats au sujet desclasses équivalences. On propose ensuite une fonction d’étiquettage sur lesdiagrammes de Ferrers, dont on exhibe quelques propriétés intéressantes.Finalement, on examine brièvement la question des q-analogues pour lesplacements de tours.

1 IntroductionThe main problem is to tell when two given chess boards are rook-equivalent,that is when the number of ways of placing k nontaking rooks are the sameon both boards for all k. Foata and Schützenberger in [3], and later Goldman,Joichi and White in [5], made a lot of progress by giving a complete answer forthe particular case of Ferrers boards. Goldman, Joichi and White also unveiledin [7] a link between rook-equivalence and graph coloration.

˚This work is supported by a grant from NSERC for an undergraduate summer intern-ship. The internship was supervised by the Laboratoire de Combinatoire et d‘InformatiqueMathématique (LaCIM), UQAM.

1

Our goal is to extend the complete classification of rook-equivalence overFerrers boards to a larger class that contains Ferrers board, called permutationdiagrams. These diagrams are related, among others, to Schubert calculus (see[9]). They have been studied in many papers, notably in [1] by Morales andLewis, and in [8] by the same authors along with Klein.

Section 3 is mainly devoted to an "extension operation", in which we adaptmost of the existing results concerning Ferrers boards to the new framework ofpermutation diagrams. We try to generalize the statements whenever possible.We check how the Ferrers part interact with other permutation diagrams. Withexisting tools, we also give a very precise condition for the membership of apermutation to what we call dominant classes: a permutation is rook-equivalentto a dominant permutation (that is, permutation whose diagram is a Ferrersboard) if and only if its rook polynomial has only integer roots.

Section 4 adresses the study of rook-equivalence classes in Sn. We startby showing that there are 2n´1 classes in Sn containing a dominant permu-tation. We then study a type of classes we call "mixed" (composed of bothcovexillary and non-covexillary permutations), about which we give a conjec-ture (Conjecture 2). As an evidence of this conjecture, we give a method thatallows to construct an infinite collection of permutations for which the conjec-ture is proved. More precisely, we show that the following permutation is anon-covexillary member of a mixed class:

ω “ pρ2m`2 ‘ ρ1q a pρm ‘ ρ1q.

The 5th section is devoted to a labelling of Ferrers boards with many interest-ing properties, mainly related to the work of Foata, Schützenberger, Goldman,Joichi and White. The labelling was inspired by a paper of Ding, [2]. Ourlabelling is a function from B to N defined by

ΦBpi, jq “ n` i´ j.

Finally we address in Section 6 the question of q-analogues for rook place-ments, mainly studied by Garsia, Remmel, Haglund, and many others. Using aq-analogue defined by Garsia and Remmel, we give a conjecture concerning thenew equivalence relation induced by q rook numbers.

AcknowledgementsFirstly, I want to thank Alejandro Morales for his great work as a supervisor, forhis precious help in the making of this document and for proposing the projectof studying rook equivalence of permutation diagrams. I wish to thank FrançoisBergeron for sponsoring and supervising this internship. I thank also NadiaLafrenière who provided many helpful suggestions on both the form and thecontent of this document, and Nancy Wallace for discussing many proofs andpropositions.

Many thanks to Bruce Sagan for taking the time to give us some precioushints: the labelling presented in Section 5 mainly comes from his suggestions. I

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also thank Joel Brewster Lewis: Proposition 1 has been inspired by one of hissuggestions.

2 Preliminaries

2.1 Boards and South-West diagramsWe use the following convention through this document: the integer intervalti P N : a ď i ď bu will be noted ra, bs. If a “ 1, we simply note rbs in place ofr1, bs.

We call a board a subset B of rns ˆ rns. A board B Ď rns ˆ rns is calledproper if it satisfies:

1. for all pi, jq P B, i ă j;

2. B defines a transitive relation on rns, that is pi, jq P B and pj, kq P Bimplies pi, kq P B.

Defining Tn “ tpi, jq P rns ˆ rns : i ă ju, it follows that any transitive relationB on rns is proper as soon as B Ď Tn.

Moreover, if a proper board B has the property that whenever t P B, alltiles North or East of t are also in B, then B will be called a Ferrers board. Thetuple λ “ pλ1, . . . , λnq, where λi is the number of boxes in the ith row of B, iscalled the shape of B. By definition, λ is always decreasing, and if λ is strictlydecreasing except for its zero entries, we say that B is a decreasing Ferrersboard. Remark that because λ is a partition, this definition indeed correspondsto the classical definition of Ferrers diagrams, up to the fact that our Ferrersboards will be right-aligned, decreasing, and strictly upper triangular. Theseconventions will be useful when dealing with permutation diagrams.

Given a permutation σ of length n (written in one-line notation), we definethe following n-board to be the South-West diagram of σ:

Dσ “ tpi, σpjqq : i ă j, σpiq ă σpjqu.

Such a board can be obtained by placing n points pi, σpiqq on rnsˆrns (usingmatrix notation) and crossing all tiles South and West of those points, hencethe name South-West diagram. Figure 1 gives an example.

Some may see the resemblance between South-West diagrams and Rothediagrams (which are in fact South-East diagrams). As we will exclusively beconcerned with the former, we will allow ourselves to recycle some of the termsusually defined for Rothe diagrams. For instance, a permutation σ will bereferred to as dominant if Dσ has the shape of a partition, that is if Dσ is infact a Ferrers board, as defined above. The reason why Dσ should necessarilybe strictly upper triangular will be given by Lemma 1.

Example 1. The permutation p 2 3 4 5 1 q is dominant, and the shape of its di-agram is p3, 2, 1, 0, 0q. On the other hand, the permutation p 5 1 2 3 4 q is notdominant, even though the shape of its diagram appears to a partition: in fact,its shape is p0, 3, 2, 1, 0q which is not a partition. See Figure 1 for illustrations.

3

Figure 1: The South-West diagrams of p 2 3 4 5 1 q and p 5 1 2 3 4 q. The leftmostdiagram is dominant, whereas the other is not.

We will say that a permutation σ is covexillary if the rows of Dσ (and,consequently its columns) can be totally ordered by inclusion, that is if forevery i, j we either have pi, kq P Dσ whenever pj, kq P Dσ or the converse. Itis well known that covexillary permutations are exactly those for which thereexists no subsequence order-isomorphic to p 3 4 1 2 q. We will say that such apermutation avoids the pattern p 3 4 1 2 q. Note that every dominant permutationis covexillary, because its rows are naturally ordered by inclusion from bottomto top.

Example 2. The permutation p 5 8 1 6 3 7 2 4 q is non-covexillary because, amongstothers, the subsequence p5, 8, 1, 3q is order-isomorphic to p 3 4 1 2 q. See Figure 2for an illustration.

The code of σ, noted cσ, is a tuple of length n in which the ith element isthe number of points pi, kq in Dσ. It can be shown (see [9], chapter 2) that apermutation σ is dominant if and only if cσ is a partition (that is, decreasing).For instance, the permutation σ “ p 2 3 4 5 1 q is dominant and its code is cσ “p3, 2, 1, 0, 0q. If cσ is strictly decreasing (with the exception of zero entries), thenσ will be referred to as strictly dominant.

A more thourough treatement of these notions (though defined for Rothe dia-grams instead) may be found in a book written by Manivel, Fonctions Symétriques,polynômes de Schubert et lieux de dégénérescence [9].

2.2 Rook placementsRook Theory mainly consists in counting the number of ways of placing non-taking rooks on arbitrary boards. Define for a board B the set of all k rooksplacements on B:

RkpBq “ tR Ď B : #R “ k, no two pi, jq P R sharing a coordinateu.

We will call an element R of RkpBq a k rooks placement over B, whereasan element pi, jq of R will be called a rook. We conveniently represent a rooksplacement R P RkpBq by placing k nontaking rooks on B.

Similarly, we define the set of all k-hit rooks placements over a board B:

HkpBq “ tτ P Sn : #ti : pi, τiq P Bu “ ku.

4

R

R

R

R R

R

R

R

R

R

R

R

Figure 2: A 4-rook placement and 4-hit rook placement on the South-Westdiagram of p 5 8 1 6 3 7 2 4 q.

HkpBq is the set of all n-rook placements over rnsˆ rns (elements of Sn) forwhich k rooks lie on B. An element H P HkpBq is called a k hit placements.We give an example of a k rooks placement and k hit placement in Figure 2.

We will write rkpBq :“ #RkpBq and hkpBq :“ #HkpBq. Furthermore, ifB “ Dσ for a certain σ, we will write Rkpσq in place of RkpDσq, and so on,without ambiguity.

Finally, we will be interested in the relation of rook equivalence between twon-board B,C, which occurs if rkpBq “ rkpCq for all k ě 0. We will simply writeB „ C and σ „ τ for two permutations σ, τ P Sn if Dσ „ Dτ .

2.3 Multiset and tuplesWe give some basic definitions about tuples and multisets, without too muchformalism.

A tuple T “ pt1, . . . , tkq is an ordered collection of elements (not necessarilydistincts). If U “ pu1, . . . , ulq is another tuple, the concatenation of T and U isthe tuple

TU “ pt1, . . . , tk, u1, . . . , ulq

We write the concatenation of a finite family of tuples T1, . . . , Tn

nź

i“1

Ti “ T1 . . . Tn.

Tn represents the concatenationś

iPrns T . If n “ 0, then T 0 “ ε, where ε isthe empty tuple.

Moreover, the indicator of T , noted 1T is a function from T to t0, 1u definedby

1T ptq “

#

1 if t P T0 otherwise

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For the notion of multiset, we mainly use the definition of Stanley in [11].Let A be a set. A multiset M on A is a function ν : AÑ N satisfying

ÿ

aPA

νpaq ă 8

The sumř

aPA νpaq is called the cardinality of M , noted #M , and M itselfwill be noted

M “ tpa, νpaqq : a P Au

3 Permutation diagrams

3.1 Dominant permutations and Ferrers boardsThis subsection is mainly technical. It aims to clarify the definition of dominantpermutation and to show the relevance of our unusual definition of Ferrers board.We start by a very simple lemma coming from the book of Laurent Manivel,Fonctions symétriques, polynômes de Schubert et lieux de dégénérescence [9]:

Lemma 1. Let σ P Sn and cσ “ pc1, . . . , cnq. Then for all ci P cσ,

ci ď n´ i

Proof. Recall that

Dσ “ tpi, σpjqq : i ă j, σpiq ă σpjqu.

For a fixed i, the set tj : i ă j ď nu has n´ i elements, and thus there canbe no more than n´ i elements in Dσ with the first coordinate being i.

This lemma explains why we defined our Ferrers board to be strictly uppertriangular: the diagram of a partition that is not strictly upper triangular cannot be the diagram of a permutation.

We now give an explicit formula to compute the dominant permutation cor-responding to a given Ferrers board:

Proposition 1. Let B Ď Tn be a Ferrers board of shape λ “ pλ1, . . . , λnq.Define recursively the sequence of sets tAiu for i P rns:

Ai “ tj P rn´ λis : @k P ri´ 1s, j ‰ maxpAkqu.

Define the map σ by σpiq “ maxpAiq, i P rns. Then σ P Sn and Dσ “ B.

Proof. The proof is in two parts: we first show that σ P Sn, then we show thatDσ “ B.

Because B Ď Tn, we have 0 ď λi ď n´ i, and thus

0 ď λi ď n´ i ðñ i´ n ď ´λi ď 0 ðñ i ď n´ λi ď n. (1)

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As j P Ai implies j P rn´ λis, the range of σ is indeed in rns. Furthermore,the set tmaxpAjq : j P ri´1su has i´1 elements, whereas, by equation 1, rn´λishas at least i elements. Therefore Ai ‰ ∅. Thus the maximum of Ai exists forall i P rns and σ is well defined as a map from rns onto itself.

Take l,m P rns and asssume l ă m. Because σpmq “ maxpAmq by definition,we have σpmq P Am. By construction of Am, σpkq ‰ σpmq whenever k ă m.In particular, we must have σplq ‰ σpmq as we assumed l ă m. It shows thatσ : rns ãÑ rns, that is σ is an injection from rns to rns. Because its range isequal its domain, σ is also a bijection and is well defined as a permutation ofSn, as desired.

To show that Dσ “ B, we start by showing that for a given i P rns, pi, σpjqq PDσ if and only if n´ λi ă σpjq ď n. Recall the definition of Dσ:

Dσ “ tpi, σpjqq : i ă j, σpiq ă σpjqu.

Start with the "if" part, and suppose pi, σpjqq P Dσ. Then j ą i by definitionof Dσ. We already know that σpjq ď n, and thus we just need to show thatσpjq ą n ´ λi. Suppose that σpiq ă σpjq ď n ´ λi. Then σpjq P rn ´ λis, andas j ą i, σpjq ‰ σpkq for all k P ri ´ 1s. Therefore we have σpjq P Ai, but wealso have σpjq ą σpiq “ maxpAiq, which is a contradiction. So we indeed haven´ λi ă σpjq ď n whenever pi, σpjqq P Dσ.

For the "only if" part, suppose σpjq ą n ´ λi. Because σpjq P Aj , we haveσpjq ď n´ λj . Therefore,

n´ λi ă σpjq ď n´ λj ñ λj ă λi,

and because λ is a partition, it implies that j ą i. Thus, whenever n ´ λi ăσpjq ď n, we have pi, σpjqq P Dσ.

That being said, we can compute the ith element of cσ:

ci “ #tj : pi, σpjqq P Dσu “ #tn´ λi ă j ď nu “ λi,

as desired. Suppose pi, jq P Dσ, we can show that for k ă i and l ą j, bothpi, lq and pk, jq are in Dσ. As pi, jq P Dσ, n ´ λi ă j ď n, and thus j ă l ď nimplies pi, lq P Dσ. On the other hand, as k ă i, n´ λk ă n´ λi ă j and thusn´ λk ă j ď n, and we have pk, jq P Dσ as desired.

We find that each Ferrers board is the diagram of dominant permutation,and that each dominant permutation has for diagram a Ferrers board. It is nothard to see also that our Ferrers board are in fact Dyck path in rns ˆ rns, andit is well-known that these are counted by Catalan numbers. Thus, we have thefollowing corollary:

Corollary 1. The number of dominant permutations in Sn is equal to the nthCatalan number,

Cn “1

n` 1

ˆ

2n

n

˙

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3.2 Skew and direct sum of permutationsWe define the skew and direct sum of two permutations as follow:

Definition 1. Let σ P Sn, τ P Sm. Then the skew sum τ a σ is

pτ a σqpiq “

#

τpiq ` n if 1 ď i ď m,

σpi´mq if m` 1 ď i ď n`m.

The direct sum τ ‘ σ is

pτ ‘ σqpiq “

#

τpiq if 1 ď i ď m,

σpi´mq `m if m` 1 ď i ď m` n.

We give two properties that show how rook-equivalence behaves with skewsum:

Proposition 2. Let σ P Sn, τ P Sm and let ρk “ p k k´1 ... 1 q be the kth orderreversing permutation. Then,

1. rkpσ a τq “ř

i`j“k ripσqrjpτq;

2. rkpσ a ρkq “ rkpσq “ rkpρk a σq.

The convolution property comes from the fact that the diagram of σ a τ isa disjoint union of Dσ and Dτ . Therefore, placing j rooks on Dσ and then irooks on Dτ yields an i ` j rooks placement on Dσaτ . The second propertyis an immediate consequence of the South-West construction of permutationdiagrams.

3.3 Factoring the rook polynomialThe following polynomial, as defined by Goldman, Joichi and White in [5], willbe useful for studying rook-equivalence:

Definition 2. Let σ P Sn. The rook polynomial of σ is given by

χσpxq “ÿ

kě0

rkpσqpxqn´k.

where pxqj “ xpx´ 1q . . . px´ j ` 1q refers to the j falling factorial of x.

The next theorem originates from [5], but was expressed in a different nota-tion. We rephrase it in terms of permutation diagrams:

Theorem 1 (Goldman-Joichi-White). For every dominant permutation σ P Sn

of code cσ “ pc1, . . . , cnq,

χσpxq “nź

i“1

px` ci ´ n` iq.

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We can adapt this result for all covexillary permutations. First we need adefinition and a lemma:

Definition 3. Let σ P Sn, we say that the following set is a row of Dσ:

li “ tk : pi, kq P Dσu

We say that row i is smaller than row j (written li ď lj) either if li “ lj as aset and i ě j, or if li Ă lj. We denote by si the number of rows strictly smallerthan row i in Dσ.

Remark 1. For a covexillary permutation σ, the partial order defined above is astrict total order, since by definition the rows of Dσ can be ordered by inclusion.Moreover, for σ dominant, we have si “ n´ i because the rows are ordered frombottom to top.

We now give a more general form of Theorem 1:

Proposition 3. For any covexillary permutation σ P Sn, let cσ “ pc1, . . . , cnq,then

χσpxq “nź

i“1

px` ci ´ siq.

Proof. We use the same idea Goldman, Joichi and White used to prove Theorem1 in [5].

Fix x P N and letX “ rns ˆ rn` 1, n` xs

For B “ Dσ \X, we show that:nź

i“1

px` ci ´ siq “ rnpBq “ÿ

kě0

rkpσqpxqn´k

The rows of B can be totally ordered by inclusion because σ is covexillaryand the rows of X are all equal as sets. Thus, if li is the ith row of B, we canwrite:

li1 ă ¨ ¨ ¨ ă lin ,

for some i1, . . . , in P rns. Because the choice of a rooks placement R P RnpBqis equivalent to a choice of one element in each li, no two equal, we can countrnpBq by counting the choices row by row. We know that #li “ x`ci. Startingwith li1 , we have x ` ci1 choices, si1 being zero, as desired. Continuing up torow lik , we have x`cik´sik choices, as we cannot choose one of the sik elementsalready chosen. Thus,

rnpBq “nź

k“1

px` cik ´ sikq “nź

i“1

px` ci ´ siq

For the second part, let

RknpBq “ tR P RnpBq : pRXDσq P Rkpσqu

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Because for each R P RnpBq, the intersection pRXDσq is a k rooks placementover Dσ for some k, we have that

RnpBq “ğ

kě0

RknpBq ñ rnpBq “ÿ

kě0

#RknpBq

We can choose R P RknpBq by first placing k rooks in rkpσq ways, and thenplacing n´ k rooks in the n´ k remaining rows of X. This last operation canbe done in pxqn´k ways, as we have x free boxes on the first row, x´ 1 on thesecond, and so on. Thus,

RnpBq “ÿ

kě0

#RknpBq “ÿ

kě0

rkpσqpxqn´k,

completing the proof.

There is a straightforward corollary of Theorem 1: if for a given permuta-tion σ there exists a dominant permutation τ such that σ „ τ , then all theroots of χσ are in N. We want to show that the converse is also true. Withthat, we completely characterize any permutation rook-equivalent to a domi-nant permutation. First we need a theorem from Goldman, Joichi and White(see [7]):

Theorem 2 (Goldman-Joichi-White). Let B Ď rns ˆ rns be a proper board.Define the graph

ΓpBq “ pV, Eq “ prns, tpi, jq R B : i ă juq (2)

If CΓpBq is the chromatic polynomial of ΓpBq, then CΓpBq “ χB.

The next proposition states that for permutation diagrams, we can take theinversion graph instead of the graph Γ of Theorem 2. Moreover, we also showthat in such case, the diagram need not be proper. The proposition was statedin [1], for which we should credit Axel Hultman who wrote the appendix wherethe theorem appears.

Proposition 4 (Hultman). Let σ P Sn be covexillary and Gσ be the inversiongraph of σ, i.e.

Gσ “ pV, Eq “ prns, tpi, jq : i ă j, σpjq ă σpiquq,

Then,CGσ pxq “ χσpxq,

where CGσ is the chromatic polynomial of Gσ.

Sketch of proof.

Step 1: show that the relation stands for σ whenever it stands for σ a ρn.

Step 2: Dτ is a proper board, and thus χτ pxq “ CΓpτqpxq by Theorem 2.

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Step 3: let B “ tpi, jq : pi, σpjqq P Dσu, then B „ Dσ.

Step 4: ΓpBq “ Gτ pxq.

We can now show that permutations whose rook polynomial has only integerroots are rook-equivalent to a dominant permutation. This is the main resultof the section:

Proposition 5. Let σ P Sn. Then there exists a dominant permutation τ suchthat τ „ σ if and only if χσ factors completely in N.

Proof. The "only if" direction is an immediate consequence of Theorem 1.For the "if" direction, take a permutation σ for which the rook polynomial

χσ has only integer roots. Since the rook polynomial is also the chromaticpolynomial of the graph ΓpDσq, we know that j P N is a root of χσ if and onlyif 0 ď j ď c, where c is the chromatic number of Γ. Therefore, the roots of χσform a multiset M of n elements on r0, cs, where νpkq ą 0 for all k P r0, cs.

Write M in decreasing order into a tuple pt1, . . . , tnq. We then define thetuple λ “ pλ1, . . . , λnq where λi “ n ´ i ´ ti. We wish to show that λ is theshape of a Ferrers board in rns ˆ rns. We can check that λ is a decreasing tuplesince we either have ti`1 “ ti or ti`1 “ ti ´ 1, and in both cases λi`1 ď λi.Additionnaly, we have λi ď n ´ i since ti ě 0. Because χσ is a chromaticpolynomial, χσp0q “ 0 and therefore tn “ 0. Because of that λn “ n ´ n “ 0.Because λ is decreasing, we have 0 ď λ ď n´ i and λ is the shape of a Ferrersboard.

Using Proposition 1, we are able to find τ P Sn such that Dτ is Ferrersboard with shape λ. Using Theorem 1, the roots of χτ are exactly the integers´pλi ´ n` iq. Furthermore, for all i,

νp´pλi ´ n` iqq “ νp´pn´ i´ ti ´ n` iqq “ νptiq,

which means that the roots of χσ are exactly the roots of χτ , and thereforeχσ “ χτ as desired.

Foata and Schützenberger have shown in [3] that any Ferrers board is rook-equivalent to a unique decreasing Ferrers board. The next proposition wasstated as a corollary in [3]:

Proposition 6 (Foata-Schützenberger). Two decreasing Ferrers boards of re-spective shape λ and µ are rook-equivalent if and only if λ “ µ.

Moreover, because the rook polynomial of a covexillary has only integerroots, we know from Proposition 5 that every covexillary permutation is rook-equivalent to a unique strictly dominant permutation. One can now ask how tofind, for any covexillary permutation, the strictly dominant permutation rook-equivalent. Using Proposition 1, we only need to show how to find the shapeof the corresponding Ferrers board. The construction we give will make use ofthis lemma:

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Lemma 2. Let σ P Sn be a covexillary permutation and let cσ “ pc1, . . . , cnq.Then

si ´ n ă ci ď si p1 ď i ď nq.

Proof. For the first part, observe that si ă n, and thus si ´ n ă 0 ď ci bydefinition.

For the second part, we suppose ci ą si. Let A “ tj : lj ą li or lj “ liu.Since the order defined of the rows of Dσ is linear, #A “ n ´ si. Lettingk “ n´ si, we realize that, since

j P Añ li Ď lj ,

there are at least k rows lj for which cj ą si “ n´ k. Then by the pigeonholeprinciple, there must exists an index m such that m ě k and cm ą n´ k. Butby a preceeding lemma,

cm ď n´m ď n´ k, (3)

yielding a contradiction. Therefore, si ´ n ă ci ď si as desired.

We now give an explicit way of constructing an increasing Ferrers boardequivalent to the diagram of a covexillary permutation:

Proposition 7. Let σ P Sn be a covexillary permutation with cσ “ pc1, . . . , cnq.Let A “ tci ´ siu and define on A the multiset M “ tpa, νpaqq : a P Au, where

νpaq “ #ti P rns : ci ´ si “ au

Then, define the tuples

T “n´1ź

j“0

p´jqνp´jq´1M p´jq, U “nź

k“1

pk ´ nq1M pk´nq (4)

Let TU “ pv1, . . . , vnq, the following tuple is the shape of a strictly Ferrersboard B:

λ “ pv1 ` n´ 1, . . . , vn ` n´ nq,

with B „ Dσ.

Proof. Because of Lemma 2, we know that ´n ă ci ´ si ď 0. Therefore, usingProposition 3:

χσpxq “nź

s“1

px` s´ nqνps´nq

We wish to show that TU is an ordering of M in the following sense: ifT “ pt1, . . . , tlq and U “ pu1, . . . , umq, then

@a P A,#tj : tj “ au `#tj : uj “ au “ νpaq

Take a P A. Then by Lemma 2, ´n ă a ď 0 and 0 ď ´a ď n´ 1. Thereforewe can write

#tj : tj “ au “ νpaq ´ 1M paq “ νpaq ´ 1 (5)

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For the same reason, 1 ď a ` n ď n and #tj : uj “ au “ 1M paq “ 1.Therefore, we have

#tj : tj “ au `#tj : uj “ au “ νpaq ´ 1` 1 “ νpaq

for all a P A. Moreover, for any s R A, both νpsq and 1M psq are zero, and s isnot in T or U by definition.

That being said, we want to show that λ is a decreasing partition. Remarkthat, by definition, we can write λ in two parts:

λ “ pt1 ` n´ 1, . . . , tl ` n´ lqpu1 ` n´ pl ` 1q, . . . , um ` n´ pl `mqq

By definition of T , we have that ti ě ti`1. Therefore, ti ` n ´ i ą ti`1 `

n ´ pi ` 1q and λ is strictly decreasing on its first part. Additionally we knowby Proposition 3 that νp´kq ą 0 whenever χσpkq “ 0, and since by Proposition4 χσ is a chromatic polynomial, it implies that νp´jq ą 0 for all j P rks. Thuswe have U “ p´m` 1,´m` 2, . . . , 0q. Because m` l “ n:

ui ` n´ pl ` iq “ ´m` i` n´ l ´ i “ 0,

for all i. As U is increasing and a P U for all a P A, we find that ´m ` 1 “minpAq. So we have

tl ` n´ l ě ´m` 1` n´ l ě ´m` n´ l ´ 1 “ 0,

and because the first part of λ is strictly decreasing, the positivity of λ follows.Besides, since vi ď 0, we have λi “ vi`n´ i ď n´ i. So we know for a fact thatλ is the shape of a decreasing Ferrers board B Ď rnsˆrns. Furthermore, becausethe tuple pλ1 ´ n` 1, . . . , λn ´ n` nq is the tuple TU , which is a reordering ofM , it follows from Theorem 1 that B „ Dσ, completing the proof.

Example 3. Take the covexillary permutation σ “ p 7 1 3 6 2 4 5 q. The code ofσ is cσ “ p0, 5, 3, 0, 2, 1, 0q and the values si for 1 ď i ď 7 are p2, 6, 5, 1, 4, 3, 0q.The multiset on tci ´ siu is tp0, 1q, p´1, 2q, p´2, 4qu, and we have the two fol-lowing tuples:

T “ p´1,´2,´2,´2q, U “ p´2,´1, 0q

The shape of the corresponding decreasing Ferrers board is the tuple

λ “ p5, 3, 2, 1, 0, 0, 0q

Now using the recursive formula of Proposition 1, we compute the successivevalues of the dominant permutation τ :

τp1q “ maxtj P r7´ 5su “ 2

τp2q “ maxtj P r7´ 3s, j ‰ 2u “ 4

τp3q “ maxtj P r7´ 2s, j ‰ 2, 4u “ 5

τp4q “ maxtj P r7´ 1s, j ‰ 2, 4, 5u “ 6

τp5q “ maxtj P r7s, j ‰ 2, 4, 5, 6u “ 7

τp6q “ maxtj P r7s, j ‰ 2, 4, 5, 6, 7u “ 3

τp7q “ maxtj P r7s, j ‰ 2, 3, 4, 5, 6, 7u “ 1

13

0 - 2 = -25 - 6 = -13 - 5 = -20 - 1 = -12 - 4 = -21 - 3 = -20 - 0 = 0

Figure 3: To the right is the diagram of σ “ p 7 1 3 6 2 4 5 q along with the valuesci ´ si. To the left is the stricly dominant permutation equivalent to σ.

The permutation τ “ p 2 4 5 6 6 3 1 q is strictly dominant and is rook-equivalentto σ. Both permutations can be seen in Figure 3

4 Rook-equivalence in Sn

Let Rn be the quotient set of Sn by rook-equivalence, that is the set of allrook-equivalence classes in Sn. We will denote by rσs the class of σ in Rn. Thepurpose of this section is to give some results and conjectures about this set.

Using results of the preceeding section, we are able to count the numberof dominant classes, that is classes containing dominant permutations. Moreprecisely:

Proposition 8. Let Fn “ trσs P Rn : σ is dominantu. Then #Fn “ 2n´1.

Proof. We know from Proposition 6 that we can index dominant classes usingthe strictly dominant permutation they contain: if σ and τ are both strictlydominant, then rσs ‰ rτ s whenever σ ‰ τ . Moreover, because of Proposition7,we know that every dominant class contain such an element. Therefore, itsuffices to show that there exist 2n´1 strictly dominant permutations in Sn.

By Proposition 1 and Lemma 1, we only need to prove that the number ofdecreasing Ferrers boards B Ď Tn is 2n´1. We know that a decreasing Ferrersboards always have k non-empty rows with 0 ď k ď n ´ 1, and at most n ´ 1columns. Therefore, there are

`

n´1k

˘

Ferrers boards B Ď Tn with k non-emptylines, as we have to choose the k columns where B will be decreasing. It impliesthat

#Fn “n´1ÿ

k“0

ˆ

n´ 1

k

˙

“ 2n´1

So far, we don’t have any expression for #Rn, but we observe that #Fnseems to grow much slower than #Rn. For instance, we know that for n “ 5,we have 16 dominant classes, over a total of 22 classes, wheras for n “ 9 thereare only 256 dominant classes over a total of 6109. This might be a consequence

14

Figure 4: A non-covexillary member of the first mixed class and its strictlydominant equivalent.

of the Stanley-Wilf theorem (see [10]), which in short says that there exists avalue c such that the number of covexillary permutations in Sn is less than orequal to cn. However, this is not an immediate corollary, as we do not knowhow many permutations can be in the same equivalence class.

Based on this evidence, we give the following conjecture:

Conjecture 1. Let Rn be the set Sn modulo rook-equivalence, and Fn the setof all classes in Rn containing a dominant permutation. Then

limnÑ8

#Fn#Rn

“ 0.

Proposition 3 implies that the rook polynomial of a covexillary permutationhas only integer roots. We found that the converse is not true: there are non-covexillary permutations whose rook polynomial factors in N. For instance, thenon-covexillary permutation σ “ p 6 5 4 3 7 1 2 q has the following rook polyno-mial:

σpxq “ xpx´ 1qpx´ 2qpx´ 3q3px´ 4q.

By Proposition 5, it follows that there are non-covexillary permutations rook-equivalent to covexillary permutations. An equivalence class containing bothcovexillary and non-covexillary members is a mixed class. We verified thatthere are no mixed classes in Rn up to n “ 6.

Example 4. We illustrate in Figure 4 one of the four non-covexillary membersof the only mixed class of R7 along with its strictly dominant equivalent. Ifρn is the nth order reversing permutation, then observe that the non-covexillarymember showed in Figure 4 can be written as a skew sum of τ and σ where τand σ are themselves direct sums of the form ρk ‘ ρ1.

We now give a method to construct an infinite collection of mixed classes.First, we need the following lemma:

Lemma 3. Let τ P Sm`1 “ ρm‘ ρ1 (that is, Dτ is a mˆ 1 vertical rectangle).Then for any permutation σ P Sn,

χσaτ pxq “ pxqm`1χσpx´m´ 1q `mpxqmχσpx´mq.

15

Proof. We first observe that

pxqkpx´ kql “k´1ź

i“0

px´ iql´1ź

j“0

px´ k ´ jq “k`l´1ź

i“0

px´ iq “ pxqk`l. (6)

Also recall the convolution formula of Proposition 2:

rkpτ a σq “ÿ

i`j“k

ripτqrjpσq.

We also know that r0pτq “ 1, r1pτq “ m and rkpτq “ 0 if k ą 1. Therefore,we have

rkpτ a σq “ rkpσq `m ¨ rk´1pσq,

whenever k ‰ 0.Suppose σ P Sn. Using the last relation in the rook polynomial, we find

χτaσpxq “ÿ

kě0

rkpτ a σq ¨ pxqn`m`1´k

“pxqn`m`1 `ÿ

kě1

`

rkpσq `m ¨ rk´1pσq˘

pxqn`m`1´k

“pxqn`m`1 `ÿ

kě1

rkpσq ¨ pxqn`m`1´k

`mpxqn`m `mÿ

kě1

rkpσq ¨ pxqn`m´k.

Using Equation 6, we are able to rearrange the last expression to obtain:

χτaσpxq “pxqm`1

˜

px´m´ 1qn `ÿ

kě1

rkpσqpx´m´ 1qn´k

¸

`mpxqm

˜

px´mqn `ÿ

kě1

rkpσq ¨ px´mqn´k

¸

“pxqm`1χσpx´m´ 1q `mpxqmχσpx´mq

as desired.

We now give our construction:

Proposition 9. Let τ P Sm`1 be the direct sum ρm ‘ ρ1, and σ P Sn`1 thedirect sum ρn ‘ ρ1. Let ω “ σ a τ . If n “ 2m ` 2, then rωs P Rn`m`2 is amixed class, ω itself being one of its non-covexillary elements.

Proof. We know that χσpxq “ pxqn`1 ` npxqn. Using this with Lemma 3, wehave

χωpxq “ pxqm`1

`

px´m´1qn`1`npx´m´1qn˘

`mpxqm`

px´mqn`1`npx´mqn˘

.

16

Using again Equation (6) yields

χωpxq “ pxqn`m`2 ` pn`mqpxqn`m`1 `mnpxqn`m.

Factoring pxqn`m, we find

χωpxq “ pxqm`n`

px´m´ nqpx´m´ n´ 1q ` pn`mqpx´m´ nq `mn˘

.

Assuming n “ 2m` 2, we can simplify to

χωpxq “ px´m´ 2qpx´ 2m´ 1qpxq3m`2

and therefore χω factors in N. Furthermore, we can prove that ω is non-covexillary. Using the definitions of skew and direct sum, we compute:

ωp1q “ pρ2m`2 ‘ ρ1qp1q `m` 1 “ ρ2m`2p1q `m` 1 “ 3m` 3

ωp2m` 3q “ pρ2m`2 ‘ ρ1qp2m` 3q `m` 1 “ ρ1p1q ` 3m` 3 “ 3m` 4

ωp2m` 4q “ pρm ‘ ρ1qp1q “ ρmp1q “ m

ωp3m` 4q “ pρm ‘ ρ1qpm` 1q “ ρ1p1q `m “ m` 1,

Thus, the sequence pωp1q, ωp2m`3q, ωp2m`4q, ωp3m`4qq is order-isomorphicto p 3 4 1 2 q and ω is non-covexillary as desired.

The proof of Proposition 9 is constructive, so we can give an example of sucha mixed class:

Example 5. Fix m “ 3 and take

ω “ pρ8 ‘ ρ1q a pρ3 ‘ ρ1q “ p 12 11 10 9 8 7 6 5 13 3 2 1 4 q .

Then, using Equation (4), we can factor χω immediately:

χω “ px´ 5qpx´ 7qpxq11.

According to Proposition 7, the corresponding strictly dominant permutationis the one whose diagram has shape λ “ p7, 4, 0 . . . , 0q in S13. We illustrate bothpermutations at Figure 5.

One can easily see that for any m, the strictly dominant permutation σ P rωswill satisfy

cσ “ p2m` 1,m` 1, 0, . . . , 0q.

We now give a conjecture concerning mixed classes:

Conjecture 2. Let rωs P Rn be a mixed class. Then for any permutationsσ P rωs, σ contains the pattern p 3 4 1 2 q 4k times, k P N.

This conjecture has been verified in Rn up to n “ 9. Furthermore, wecan prove that the non-covexillary permutation constructed using Proposition9 contains the pattern p 3 4 1 2 q 4k times. Indeed, it suffices to show that thenumber of unordered tilesmp2m`2q ” 0 mod 4, which is fairly straightforward.

We have already checked that the converse of the conjecture is false. Thatis, containing the pattern p 3 4 1 2 q 4k times, k P N, might be a necessary butinsufficient condition for a non-covexillary permutation to be in a mixed class.We wonder what a necessary and sufficient condition would be.

17

Figure 5: A non-covexillary member of a mixed class constructed using Propo-sition 9, along with its strictly dominant equivalent.

5 A labelling of Ferrers boardThe purpose of this section is to give a labelling of Ferrers board (or, equiv-alently, of dominant permutations) that regroups many interesting properties.For instance, this labelling gives a factorization of the rook polynomial, whilecaracterising the equivalence class of the diagram. It also tells us the greatestnumber of rooks we can place on this diagram, while identifying a "canonical"k rooks placement. We think this labelling could be used to build an explicitbijection between rooks placements on equivalent Ferrers boards.

Definition 4. Let σ P Sn be a dominant permutation. Let D1σ “ Dσ Y tpi, n`1q : i P rnsu be the right extension of Dσ. That is, D1σ is the board obtained byadding a n column at the right of Dσ. We define the map Φσ : D1σ Ñ N by

Φσ : pi, jq ÞÑ n` i´ j.

Φσ is called the labelling of Dσ, whereas Φσptq is called the label of t.

Example 6. One can compute the labelling of a Ferrers board by filling all thetiles in the kth principal diagonal (starting in the upper right corner) with k’s.We illustrate the labelled diagram of the dominant permutation σ “ p 3 5 4 2 7 6 8 1 q

in Figure 6.

This particular labelling has many interesting properties. For instance, wecan rewrite Theorem 1 using the labelling function:

Proposition 10. If σ P Sn is dominant, let

Mσ “ tpi, jq : pi, kq P Dσ ñ Φσpi, kq ď Φσpi, jqu.

Then:χσpxq “

ź

tPMσ

px` Φσptq ´ n` 1q.

18

12233

344

44

5

55

56

6

01234567

Figure 6: The labelled diagram of σ “ p 3 5 4 2 7 6 8 1 q. The red labelled cellsare those for which the labelling function is maximal on the corresponding row.

Proof. Because σ is dominant, every tile t P M is always the leftmost tile ofD1σ. If t is the leftmost tile of the ith row, it follows that t “ pi, n´ ci` 1q, andtherefore

Φσptq ´ n` 1 “ n` i´ n` ci ´ 1´ n` 1 “ ci ´ n` i.

Thus, by Theorem 1, the proposition stands.

Example 7. Coming back to Example 6, we can use the labelled diagram tocompute the rook polynomial. In Figure 6, the labels of the cells t P Mσ areshown in red. Using the last proposition, we compute row by row:

χσpxq “ px´2qpx´3qpx´2qpx´1qpx´2qpx´1qpx´1qx “ xpx´1q3px´2q3px´3q.

In [2], Ding remarked that, for any Ferrers board B, there exists a "canoni-cal" k-rook placement, in the sense that this rook placement is always in RkpBqwhenever RkpBq is non-empty. This canonical rook placement can be charac-terized by our labelling function:

Proposition 11. For every dominant permutation σ, Rkpσq ‰ ∅ implies thatR P Rkpσq where

R “ tt P Dσ : Φσptq “ ku.

Thus we can place k non-attacking rooks on Dσ if and only if k tiles of Dσ

are labelled k.

Proof. We first remark that the set tt P Dσ : Φσptq “ ku is a k-rook placementon Dσ for all k (with the convention that R0pσq “ t∅u). Indeed, as we fixalternatively i or j, there exists only one solution to the equation Φσpi, jq “ k.Therefore we have at most one tile labelled k for a given row or column.

The "if" direction is then immediate, for if k tiles in Dσ are labelled k, therook placement tt P Dσ : Φσptq “ ku is a k-rook placement.

For the "only if" direction, we need only to show that Rkpσq ‰ ∅ impliesthat at least k tiles in Dσ are labelled k. Begin by taking a k-rook placement

19

R over Dσ. For 1 ď i ď k, let ai be the number of rooks in R that are aboveor on the ith row, and bi be the number of rooks in R below the ith row. Itfollows that ai ` bi “ k. Suppose that the number ci of tiles in row i satisfiesci “ bi. Then we must have ai ă i for there can be no rooks in the ith row.Therefore bi ą k ´ i and we have pi, n ´ k ` iq P Dσ. On the other hand,suppose that ci ą bi. Then, because ai ď i, we have ci ą bi ě k ´ i and, again,pi, n´ k ` iq P Dσ. Labelling this particular tile yields:

Φσpi, n´ k ` iq “ n` i´ n` k ´ i “ k,

as desired.

Example 8. Taking again σ “ p 3 5 4 2 7 6 8 1 q, we see directly from Figure 6that it is impossible to place more than 4 rooks over Dσ, because there are lessthan 5 tiles labelled 5 in Dσ.

Also remark that we can never have more than k tiles labelled k in Dσ, forthe kth principal diagonal of the grid (starting from the upper right corner) hask elements.

It follows directly from the two last propositions that rook-equivalence fordominant permutations may be reduced to isomorphism of the labelling func-tions. To show this, we first need a lemma:

Lemma 4. Let σ and τ be two dominant permutations such that σ „ τ . Thenthe maximum of Φσ restricted to Dσ is equal to the maximum of Φτ restrictedto Dτ .

Proof. Call m the maximum of Φσ restricted to Dσ and l the maximum of Φτrestricted to Dτ . With no loss of generality, we can assume m ą l. Becauseσ „ τ , the pm` 1qth row of Dτ must be empty, or else px`m´ n` 1q wouldnot be a factor of pτ whereas it will be one of pσ, and that would contradict thefact that σ „ τ .

But it is also clear that px`m´ n` 1q2 is not a factor of pτ because m` 1is the only solution to Φτ px, n ` 1q “ m. It follows that the pm ` 1qth row ofpσ must not be empty, and therefore we must have pm` 1, nq P Dτ . Labellingthis tile yields

Φσpm` 1, nq “ n´ n`m` 1 “ m` 1,

which contradict the fact that m is the maximum of Φσ over Dσ. Thereforem “ l as desired.

We can now prove the following proposition:

Proposition 12. Two dominant permutations σ, τ P Sn are equivalent if andonly if Φσ – Φτ , in the sense that there exists a bijection ψ : Dσ Ñ Dτ suchthat

Φσ “ Φτ ˝ ψ (7)

20

Proof. We proceed by induction over the number of tiles s of the boards. Ourbase case, s “ 2, is easy because the only dominant permutations in Sn with2 tiles are those with shape p2, 0, . . . , 0q and p1, 1, 0, . . . , 0q. They are rook-equivalent, and we indeed have a bijection ψ defined by

ψp1, nq “ p1, nq, ψp1, n´ 1q “ p2, nq,

that satisfies Equation (7) for those two permutations. Therefore the propositionstands for every dominant permutation whose diagram has two tiles.

Take σ and τ in Sn with s tiles (assuming s ě 2). By induction we assumeany two dominant permutations σ1, τ 1 with less than s tiles are rook equivalentif and only if we can find a bijection from Dσ1 to Dτ 1 satisfying Equation (7).

For the "if" direction, suppose there exists a bijective map ψ : Dσ Ñ Dτ

satisfying Equation (7). Then the maximum of both labelling functions Φσ,Φτover Dσ and Dτ is the same. Let m be that value and take any tile t “ pi, jqin Dσ labelled m. By definition of ψ, ψptq is also labelled m. Because m ismaximal, both t and ψptq must be the leftmost tile of some row in Dσ and Dτ ,and therefore we can use Proposition 10 to say that

χσpxq “ px`m´ n` 1qP pxq, χτ pxq “ px`m´ n` 1qQpxq, (8)

where P and Q are two polynomials.Remark that because m is maximal, there can be no cell in Dσ South of t,

for Φσptq “ m would then not be maximal. The same remark also applies toψptq in Dτ . It follows that both Dσ ´ ttu and Dτ ´ tψptqu are Ferrers board.Because they are respective subsets of Dσ and Dτ , Proposition 1 tells us we canfind two dominant permutations τ 1, σ1 P Sn satisfying

Dσ1 “ Dσ ´ ttu, Dτ 1 “ Dτ ´ tψptqu (9)

Using Proposition 10 again, the rook polynomial of σ1 and τ 1 can easily beexpressed in terms of the polynomials P and Q of Equation (8), for we knowthat Dσ1 and Dτ 1 differ respectively from Dσ and Dτ only by the row on whicht, respectively ψptq, lies. Moreover, for this particular row, the leftmost cell isnow labelled m´ 1, and thus

χσ1pxq “ px`m´ nqP pxq, χτ 1pxq “ px`m´ nqQpxq.

By construction, the restriction of ψ to Dσ ´ ttu still satisfies Equation (7),and thus, by induction, σ1 „ τ 1. It follows that P “ Q, and therefore, becauseof Equation (8), we also have σ „ τ .

For the "only if" direction, we now suppose that σ and τ are rook-equivalent.By Lemma 4, callm the maximum of both Φσ restricted toDσ and Φτ restrictedto Dτ . Take two tiles t, u respectively in Dσ and Dτ such that Φσptq “ Φτ puq “m. Using Proposition 1, we can find two dominant permutations σ1τ 1 P Sn suchthat

Dσ1 “ Dσ ´ ttu, Dτ 1 “ Dτ ´ tuu

21

12233

344

44

55

55

66

01234567

Figure 7: The labelled diagram of τ “ p 2 3 5 6 8 7 4 1 q.

As we already said, the set Mσ1 equals Mσ except for the row where weremove the tile t. As the same remark applies for τ 1 and τ , we can give boththe polynomials of τ 1 and σ1 using Proposition 10:

χσ1pxq “

ˆ

x`m´ n

x`m´ n` 1

˙

χσpxq, χτ 1pxq “

ˆ

x`m´ n

x`m´ n` 1

˙

χτ pxq.

As we assumed that σ „ τ , we have χσ “ χτ and therefore σ1 „ τ 1. Thus,by induction, we can find a bijective map ψ1 : Dσ1 Ñ Dτ 1 satisfying Equation(7). Then it suffices to extend such a map to all Dσ by defining

ψpi, jq “

#

u if pi, jq “ t

ψ1pi, jq otherwise,

The map ψ is indeed bijective since ψ1 is and u R Dτ 1 . Moreover, becauseΦσptq “ Φτ puq, ψ also satisfies Equation (7). That is, we found the desiredmap, completing the proof.

Example 9. One may compare the labelled diagram of τ “ p 2 3 5 6 8 7 4 1 q, inFigure 9, with the labelled diagram of Figure 6. Since they have the same numberof cells labelled k for all k, we can find a bijective map from one diagram to theother that preserves the labelling. Thus, by the preceeding proposition, τ and thepermutation of Example 6 are rook-equivalent.

Corollary 2. Let σ, τ P Sn such that σ „ τ . Let m be the maximum of bothΦσ and Φτ , and let t P Dσ and u P Dτ such that Φσptq “ Φτ puq “ m. Then forall k, the number of k rooks placements with a rook on t is equal to the numberof k rooks placements with a rook on u.

Proof. Remark that

rkpσq “ #tR P Rkpσq : t P Ru `#tR P Rkpσq : t R Ru

As the same remark apply for τ and because rkpσq “ rkpτq, it suffices toshow that

#tR P Rkpσq : t R Ru “ #tR P Rkpτq : u R Ru (10)

22

We can count these by counting the number of k ´ 1 rooks placements onthe boards obtained by deleting respectively the cells t and u. Let σ1 and τ 1 bethe dominant permutations realizing those boards. By Proposition 12, we knowthere exists a bijection between Dσ and Dτ that preserves the labelling, and wecan restrict it to Dσ1 and Dτ 1 . Again, by Proposition 12, it follows that σ1 „ τ 1,and therefore rk´1pσ

1q “ rk´1pτ1q as desired.

6 On q-analogues of rook placementsIn [6], Haglund established a connection between Rook Theory and n ˆ n ma-trices in Fq with certain entries restricted to zero.The case where the restrictedentries have the shape of a permutation diagram have been studied in [8] and [1].The connection between Rook Theory and restricted matrices over Fq was madethrough a q-analogue of the numbers rkpBq defined by Garsia and Remmel in[4]. We start by giving a formal definition of Garsia and Remmel’s q-analoguefor rook numbers:

Definition 5. Let B be a board and take a rooks placement R over B. Let

NR “ tpi, jq : pi, kq R R if k ď j, pl, jq R R if l ě iu.

The Garsia-Remmel number of R as a rook placement over B is

GRpR,Bq “ #pNR XBq.

The q-rook numbers of B are the polynomials

rkpB, qq “ÿ

RPRkpBq

qGRpR,Bq.

If B “ Dσ for a certain permutation σ, we will then write rkpσ, qq in placeof rkpB, qq without ambiguity.

Thus, we define a q-analogue of the rook polynomial:

Definition 6. Let B Ď rns ˆ rns, we define the q-rook polynomial of B:

χBpx, qq “ÿ

kě0

rkpσ, qqrxsqrx´ 1sq . . . rx´ pn´ kq ` 1sq,

where rxsq “ p1´ qxq{p1´ qq. If B “ Dσ for σ P Sn, we note χσpx, qq.

This q-analogue of Garsia and Remmel is a good q-analogue in many ways.Firstly, it is clear that rkpB, 1q “ rkpBq. Furthermore, Garsia and Remmelproved, also in [4], the following formula:

Theorem 3 (Garsia-Remmel). For any Ferrers board B of shape λ “ pλ1, . . . , λnq,

χBpx, qq “nź

i“1

rx` λi ´ n` isq.

23

We now consider q-rook equivalence classes in Sn. More precisely:

Definition 7. We shall say that σ „q τ if rkpσ, qq “ rkpτ, qq for all k. Equiva-lently, σ „q τ if and only if χσpx, qq “ χτ px, qq.

In [8], Brewster, Klein and Morales essentially remarked that for covexil-lary permutations, q-rook-equivalence is the same as ordinary rook-equivalence.Therefore, it is clear that there are also 2n´1 dominant classes in Sn under q-rook-equivalence. Interestingly, q-rook-equivalence seems to be an helpful toolto deal with mixed classes. Based on our observations, q-rook-equivalence isstrong enough to split mixed classes, and precise enough so it regroups the per-mutations according to the number of times they contain the pattern p 3 4 1 2 q.The following conjecture is verified in Sn up to n “ 9:

Conjecture 3. Let σ, τ P Sn with σ covexillary and τ non-covexillary. Thenσ q τ .

Whenever σ τ , the conjecture is true, for σ τ implies σ q τ as well.Thus, the difficult part of this conjecture is to explain how it tears apart themixed classes.

Proposition 13. Conjecture 3 stands if τ “ pρ2m`2 ‘ ρ1q a pρm ‘ ρ1q.

Proof. Assume τ “ pρ2m`2‘ρ1qapρm‘ρ1q and take σ P S3m`4 covexillary suchthat σ „ τ . Since q-rook-equivalence is the same as ordinary rook-equivalence,we can consider without losing generality that σ is dominant. We will show thatr1pσ, qq ‰ r1pτ, qq, for in particular the coefficient of qc´1, where c “ #Dσ “

#Dτ , is 1 in r1pσ, qq whereas it is 2 for r1pτ, qq.Since σ is dominant, it is the shape of a Ferrers board, and therefore placing

a rook on the upper right corner of Dτ is the only 1 rook placement for whichthe Garsia-Remmel number is c ´ 1. Thus the coefficient of qc´1 in r1pσq is 1as desired.

By definition, τ is a disjoint union of two rectangles. Thus, there is two 1 rookplacement for which the Garsia-Remmel number is c´1: we can either place onerook on the upper right corner of the first or of the second rectangle. Thereforethe coefficient of qc´1 is 2 in r1pτ, qq, and r1pσ, qq ‰ r1pτ, qq, completing theproof.

We now give an example to show how mixed classes behave under q-rook-equivalence:

Example 10. Let σ “ p 4 5 6 7 9 8 3 2 1 q. The class rσs P R9 is a mixed and σis its strictly dominant element. We spare the reader the inconvenience of anexhaustive list of the 438 elements of rσs. Note that

χσpxq “ pxq9 ` 14pxq8 ` 55pxq7 ` 65pxq6 ` 16pxq5.

Of the 438 elements of rσs, 416 are covexillaries, 20 contain the patternp 3 4 1 2 q 4 times, and only 2 contain it 8 times. To exemplify, we give τ “

24

r0pσ, qq q14

r1pσ, qq q13 ` 2q12 ` 3q11 ` 4q10 ` 4q9

r2pσ, qq q11 ` 3q10 ` 7q9 ` 12q8 ` 14q7 ` 12q6 ` 6q5

r3pσ, qq q8 ` 4q7 ` 10q6 ` 16q5 ` 18q4 ` 12q3 ` 4q2

r4pσ, qq q4 ` 4q3 ` 6q2 ` 4q ` 1

r0pτ, qq q14

r1pτ, qq q13 ` 2q12 ` 5q11 ` 2q10 ` 2q9 ` 2q8

r2pτ, qq q11 ` 5q10 ` 10q9 ` 10q8 ` 11q7 ` 10q6 ` 5q5 ` 2q4 ` q3

r3pτ, qq 4q8 ` 8q7 ` 11q6 ` 14q5 ` 13q4 ` 8q3 ` 5q2 ` 2q

r4pτ, qq q6 ` 2q5 ` 3q4 ` 4q3 ` 3q2 ` 2q ` 1

r0pω, qq q14

r1pω, qq q13 ` 2q12 ` 4q11 ` 3q10 ` 3q9 ` q8

r2pω, qq q11 ` 4q10 ` 8q9 ` 12q8 ` 13q7 ` 9q6 ` 6q5 ` 2q4

r3pω, qq 2q8 ` 6q7 ` 12q6 ` 15q5 ` 14q4 ` 10q3 ` 5q2 ` q

r4pω, qq q5 ` 3q4 ` 4q3 ` 4q2 ` 3q ` 1

Table 1: The respective non-zero q-rook numbers of σ “ p 4 5 6 7 9 8 3 2 1 q, τ “p 4 5 8 9 6 3 2 1 7 q and ω “ p 3 8 7 6 5 1 9 2 4 q.

p 4 5 8 9 6 3 2 1 7 q containing the pattern p 3 4 1 2 q 4 times, and ω “ p 3 8 7 6 5 1 9 2 4 q

containing it 8 times. We illustrate σ, τ and ω in Figure 8.All the covexillaries elements have the same q-rook polynomial as σ, all the

elements containing p 3 4 1 2 q 4 times have the same rook polynomial as τ and allthe elements containing it 8 times have the same rook polynomial as ω. Insteadof giving the polynomials (which are fairly long), we list the corresponding q-rooknumbers in Table 1.

References[1] J. Brewster Lewis and A. H. Morales. Combinatorics of diagrams of per-

mutations. ArXiv e-prints, May 2014.

[2] Kequan Ding. Invisible permutations and rook placements on a ferrersboard. Discrete Math., 139(1-3):105–127, May 1995.

[3] D. Foata and M. P. Schützenberger. On the rook polynomials of Ferrersrelations. In Combinatorial theory and its applications, II (Proc. Colloq.,Balatonfüred, 1969), pages 413–436. North-Holland, Amsterdam, 1970.

[4] A.M Garsia and J.B Remmel. Q-counting rook configurations and a formulaof frobenius. Journal of Combinatorial Theory, Series A, 41(2):246 – 275,1986.

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Figure 8: From left to right: the diagram of p 4 5 6 7 9 8 3 2 1 q, p 4 5 8 9 6 3 2 1 7 q

and p 3 8 7 6 5 1 9 2 4 q. The three permutations are rook-equivalent.

[5] Jay R. Goldman, J. T. Joichi, and Dennis E. White. Rook theory. i.: Rookequivalence of ferrers boards. Proceedings of the American MathematicalSociety, 52(1):pp. 485–492, 1975.

[6] James Haglund. q-rook polynomials and matrices over finite fields. Ad-vances in Applied Mathematics, 20(4):450 – 487, 1998.

[7] Dennis E. White Jay R. Goldman, J. T. Joichi. Rook polynomials and thechromatic structure of graphs. Journal of combinatorial theory, 1975.

[8] AaronJ. Klein, JoelBrewster Lewis, and AlejandroH. Morales. Countingmatrices over finite fields with support on skew young diagrams and com-plements of rothe diagrams. Journal of Algebraic Combinatorics, 39(2):429–456, 2014.

[9] Laurent Manivel. Fonctions Symétriques, polynômes de Schubert et lieuxde dégénérescence. Société Mathématique de France, 1986.

[10] Adam Marcus and Gábor Tardos. Excluded permutation matrices andthe stanley–wilf conjecture. Journal of Combinatorial Theory, Series A,107(1):153 – 160, 2004.

[11] Richard P. Stanley. Enumerative Combinatorics. Wadsworth &Brooks/Cole Advanced Books & Software, 1986.

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