Rook Theory and Simplicial Complexes

Rook theory and simplicial complexes

Ira M. Gessel

Department of Mathematics Brandeis University

A Conference to Celebrate The Mathematics of Michelle Wachs University of Miami, Coral Gables, Florida January 6, 2015 In Celebration of the Mathematics of

Michelle Wachs We use Cartesian numbering.

Rook numbers

We start with [n] × [n], where [n] = {1, 2,..., n}: Rook numbers

We start with [n] × [n], where [n] = {1, 2,..., n}:

(4,2)

We use Cartesian numbering. A board is a subset of these n2 squares: In our example, r0 = 1, r1 = 5, r2 = 6, r3 = 1, r4 = r5 = 0.

The rook number rk is the number of ways to put k non-attacking rooks on the board, that is, the number of ways to choose k squares from the board with no two in the same row or column. The rook number rk is the number of ways to put k non-attacking rooks on the board, that is, the number of ways to choose k squares from the board with no two in the same row or column.

In our example, r0 = 1, r1 = 5, r2 = 6, r3 = 1, r4 = r5 = 0. Hit numbers

We can identify a permutation π of [n] = {1, 2,..., n} with the set of ordered pairs { (i, π(i)) : i ∈ [n] } ⊆ [n] × [n], and we can represent such a set of ordered pairs as a set of n squares from [n] × [n], no two in the same row or column.

1 2 3 4 5

1 2 3 4 5 This is the permutation 4 5 1 3 2 . (The rows are i and the columns are π(i).) The squares of a permutation that are on the board are called hits of the permutation. So this permutation has just one hit:

1 2 3 4 5

The hit number hk is the number of permutations of [n] with k hits. Examples

For the board

hk is the number of permutations with k ﬁxed points, and in particular, h0 is the number of derangements. For the upper triangular board

hk is the number of permutations with k excedances, an Eulerian number. Multiplying by tj and summing on j gives

X i X j hi (1 + t) = t rj (n − j)!. i j

So replacing t with t − 1 shows that the hit numbers are determined by the rook numbers.

The fundamental identity

X i h = r (n − j)!. i j j i Proof: Count pairs (π, H) where H is a j-subset of the set of hits of π. Picking π ﬁrst gives the left side. Picking H ﬁrst gives the right side, since a choice of j nonattacking rooks can be extended to a permutation of [n] in (n − j)! ways. Multiplying by tj and summing on j gives

X i X j hi (1 + t) = t rj (n − j)!. i j

So replacing t with t − 1 shows that the hit numbers are determined by the rook numbers.

The fundamental identity

X i X j hi (1 + t) = t rj (n − j)!. i j The fundamental identity

X i X j hi (1 + t) = t rj (n − j)!. i j

So replacing t with t − 1 shows that the hit numbers are determined by the rook numbers. Matching numbers

Let G be a graph with vertex set [2n].

Let mk be the number of k-matchings in G, that is, the number of sets of k vertex-disjoint edges in G (analogous to rook numbers). For any complete matching M of [2n], a hit of M is an edge of M that is in G. The hit number hi of G is the number of complete matchings of [2n] with i hits. For example, this complete matching has two hits:

1 2

6 3

5 4 The number of complete matchings of the complete graph K2l , i.e., the number of partitions of [2l] into blocks of size 2, is (2l − 1)!! = 1 · 3 ··· (2l − 1) = (2l)!/2l l!. Then by the same reasoning as for ordinary rook numbers we have X i h = m (2n − 2j − 1)!!. i j j i A sufﬁcient condition for ∆ to be a rook complex is that every face of size k is covered by the same number of faces of size k + 1.

Rook complexes

Let us deﬁne a rook complex to be a simplicial complex ∆ on a set A (every subset of an element of ∆ is an element of ∆) with the property that every facet (maximal face) has size n, and there are integers c0, c1,..., cn such that if U ∈ ∆ with |U| = j then the number of facets containing U is cn−j . We will call c0,..., cn the factorial sequence for ∆. (The faces of ∆ correspond to placements of nonattacking rooks.) A sufﬁcient condition for ∆ to be a rook complex is that every face of size k is covered by the same number of faces of size k + 1. For matching numbers, I A is the set of 2-subsets of [2n] I ∆ is the set of disjoint subsets of A (matchings) I cj = (2j − 1)!! Note that in both cases, we have a sequence of rook complexes indexed by a nonnegative integer n, and the factorial sequence is independent of n. In our next example, the factorial sequence depends on n.

So for rook numbers, I A is [n] × [n] I ∆ is the set of subsets of A in which two different ordered pairs must differ in both coordinates (i.e., sets of nonattacking rooks) I cj = j! Note that in both cases, we have a sequence of rook complexes indexed by a nonnegative integer n, and the factorial sequence is independent of n. In our next example, the factorial sequence depends on n.

So for rook numbers, I A is [n] × [n] I ∆ is the set of subsets of A in which two different ordered pairs must differ in both coordinates (i.e., sets of nonattacking rooks) I cj = j! For matching numbers, I A is the set of 2-subsets of [2n] I ∆ is the set of disjoint subsets of A (matchings) I cj = (2j − 1)!! Note that in both cases, we have a sequence of rook complexes indexed by a nonnegative integer n, and the factorial sequence is independent of n. So for rook numbers, I A is [n] × [n] I ∆ is the set of subsets of A in which two different ordered pairs must differ in both coordinates (i.e., sets of nonattacking rooks) I cj = j! For matching numbers, I A is the set of 2-subsets of [2n] I ∆ is the set of disjoint subsets of A (matchings) I cj = (2j − 1)!! Note that in both cases, we have a sequence of rook complexes indexed by a nonnegative integer n, and the factorial sequence is independent of n. In our next example, the factorial sequence depends on n. In other words, given any rooted forest on [n + 1] with j edges (and therefore n + 1 − j trees), the number of rooted trees containing it is (n + 1)n−j . This can be proved by Jim Pitman’s method for counting trees.

Forest complexes

Let A = [n + 1] × [n + 1] which we think of as directed edges, and let ∆ be the set of subsets of A that are forests of rooted trees (with all edges directed towards the roots). Then ∆ is a j rook complex with cj = (n + 1) . In other words, given any rooted forest on [n + 1] with j edges (and therefore n + 1 − j trees), the number of rooted trees containing it is (n + 1)n−j . Forest complexes

The fundamental identity for rook complexes

Let ∆ be rook complex on the set |A| with factorial sequence c0,..., cn. For any B ⊆ A we deﬁne the rook numbers ri to be the number of elements of ∆ of size i contained in B, and we deﬁne the hit number hi of B to be the number of facets of ∆ that intersect B in i points. Then by exactly the same reasoning as in the previous two cases, we have X i h = r c . i j j n−j i For example, in the forest complex with n = 2 (forests on 3 points), let B be the directed graph (in this case a rooted tree)

Then r0 = 1, r1 = 2, and r2 = 1 and also h0 = 4, h1 = 4, and h2 = 1. Goldman, Joichi, and White (1975) deﬁned the factorial rook polynomial of B to be X X FB(x) = rj x(x − 1) ··· (x − (n − j) + 1) = rj x↓n−j . j j

Factorial rook polynomials and reciprocity

Let’s return to ordinary rook numbers. 1. Factorization theorems. Goldman, Joichi, and White proved that for Ferrers boards the factorial rook polynomial factors into linear factors. 2. Reciprocity theorems. There is a simple relation between the factorial rook polynomial of B and of its complement.

Factorial rook polynomials and reciprocity

Let’s return to ordinary rook numbers. Goldman, Joichi, and White (1975) deﬁned the factorial rook polynomial of B to be X X FB(x) = rj x(x − 1) ··· (x − (n − j) + 1) = rj x↓n−j . j j

Why is it useful? 2. Reciprocity theorems. There is a simple relation between the factorial rook polynomial of B and of its complement.

Factorial rook polynomials and reciprocity

Why is it useful? 1. Factorization theorems. Goldman, Joichi, and White proved that for Ferrers boards the factorial rook polynomial factors into linear factors. Factorial rook polynomials and reciprocity

Why is it useful? 1. Factorization theorems. Goldman, Joichi, and White proved that for Ferrers boards the factorial rook polynomial factors into linear factors. 2. Reciprocity theorems. There is a simple relation between the factorial rook polynomial of B and of its complement. Given hit numbers h0,..., hn for a board B in [n] × [n], the hit ¯ numbers hi for the complementary board B in [n] × [n] are ¯ given by hi = hn−i . Since the rook numbers and hit numbers determine each other, there must be a relation between the rook numbers of a board and its complement. The factorial rook polynomial enables us to express this relation in an elegant way. P i In the fundamental identity i hi j = rj (n − j)!, we multiply x both sides by n−j and sum on j. Applying Vandermonde’s theorem gives

X X x + i F (x) = r x↓ = h . B j n−j i n j i

So the coefﬁcients of FB(x) in the basis {x↓j } for polynomials are the rook numbers for B, and the coefﬁcients of FB(x) in the x+i basis { n }0≤i≤n for polynomials of degree at most n are the hit numbers for B. Chow’s reciprocity theorem was actually for the more general “cover polynomial”.

So the factorial rook polynomial for the complementary board B is X x + i F (x) = h B n−i n i X x + n − i = h i n i X −x − 1 + i = (−1)n h i n i n = (−1) FB(−x − 1).

This is the reciprocity theorem for factorial rook polynomials, due to Timothy Chow (proved by a different method). So the factorial rook polynomial for the complementary board B is X x + i F (x) = h B n−i n i X x + n − i = h i n i X −x − 1 + i = (−1)n h i n i n = (−1) FB(−x − 1).

This is the reciprocity theorem for factorial rook polynomials, due to Timothy Chow (proved by a different method). Chow’s reciprocity theorem was actually for the more general “cover polynomial”. An example of the factorial rook polynomial

Let’s consider the upper triangular board

Here hk is the number of permutations with k excedances (the Eulerian number An,k+1), where an excedance of a permutation π is an i for which π(i) > i. The rook number rk is the Stirling number of the second kind S(n, n − k). The factorial rook polynomial is

X X n S(n, n − j)x↓n−j = S(n, j)x↓j = x j j so we have the formula X x + i xn = A . n,i+1 n i The reciprocity theorem gives

X x + i F (x) = (−1)n(−x − 1)n = (x + 1)n = A . B n,i n i Generalized factorial rook polynomials

We can construct a similar “factorial rook polynomial” for any rook complex. Given the fundamental identity

X i h = r c , i j j n−j i

we multiply both sides by some polynomial un−j (x) (to be determined) and sum on j. We get

X X X i r c u (x) = h u (x). j n−j n−j i j n−j j i j Generalized factorial rook polynomials

We can construct a similar “factorial rook polynomial” for any rook complex. Given the fundamental identity

X i h = r c , i j j n−j i

we multiply both sides by some polynomial un−j (x) (to be determined) and sum on j. We get

X X X i r c u (x) = h u (x). j n−j n−j i j n−j j i j | {z } Pi (x) Generalized factorial rook polynomials

We can construct a similar “factorial rook polynomial” for any rook complex. Given the fundamental identity

X i h = r c , i j j n−j i

we multiply both sides by some polynomial un−j (x) (to be determined) and sum on j. We get

X X X i r c u (x) = h u (x). j n−j n−j i j n−j j i j | {z } Pi (x)

To get a nice reciprocity theorem, we want Pi (x) to have the n property that Pn−i (x) = (−1) Pi (−x − γ) for some γ. x x+i We could just take un−j (x) to be n−j , so that Pi (x) = n , as we did for the factorial rook polynomial, and we would get a reciprocity theorem. But it doesn’t seem that this gives the nicest results (e.g., factorization theorems). then we should take

j−1 1 Y uj (x) = (x − βi) cj i=0 which gives

Qn−i−1(x − jβ) Qi−1(x + α + jβ) P (x) = j=0 j=0 . i Qn−1 j=0 (α + jβ)

and n Pn−i (x) = (−1) Pi (−x − α).

It turns out that if the factorial sequence is of the form

j−1 Y cj = (α + βi) i=0 which gives

Qn−i−1(x − jβ) Qi−1(x + α + jβ) P (x) = j=0 j=0 . i Qn−1 j=0 (α + jβ)

and n Pn−i (x) = (−1) Pi (−x − α).

It turns out that if the factorial sequence is of the form

j−1 Y cj = (α + βi) i=0 then we should take

j−1 1 Y uj (x) = (x − βi) cj i=0 It turns out that if the factorial sequence is of the form

j−1 Y cj = (α + βi) i=0 then we should take

j−1 1 Y uj (x) = (x − βi) cj i=0 which gives

Qn−i−1(x − jβ) Qi−1(x + α + jβ) P (x) = j=0 j=0 . i Qn−1 j=0 (α + jβ) and n Pn−i (x) = (−1) Pi (−x − α). Note that FB(x) has integer coefﬁcients as a polynomial in x, but Pi (x) does not.

Qj−1 To recap, we have cj = i=0(α + βi) and

n−j−1 X Y X FB(x) := rj (x − βi) = hi Pi (x), j i=0 i where Qn−i−1(x − jβ) Qi−1(x + α + jβ) P (x) = j=0 j=0 i Qn−1 j=0 (α + jβ) and we have the reciprocity theorem

n FB(x) = (−1) FB(−x − α). Qj−1 To recap, we have cj = i=0(α + βi) and

n−j−1 X Y X FB(x) := rj (x − βi) = hi Pi (x), j i=0 i where Qn−i−1(x − jβ) Qi−1(x + α + jβ) P (x) = j=0 j=0 i Qn−1 j=0 (α + jβ) and we have the reciprocity theorem

n FB(x) = (−1) FB(−x − α).

Note that FB(x) has integer coefﬁcients as a polynomial in x, but Pi (x) does not. The special case α = β = 1 corresponds to the ordinary factorial rook polynomial. The case α = 1, β = 2 corresponds to a “factorial matching polynomial” introduced (with a different normalization) by Reiner and White, and further studied by Haglund and Remmel. A canonical example: Colored permutations

We can think of a nonattacking rook placement in [n] × [n] as the set of edges of a digraph on [n] in which every vertex has indegree at most one and outdegree at most one. So the facets (permutations) are digraphs in which every vertex has indegree one and outdegree one:

3 Now we generalize this by coloring each edge with an integer modulo β, but we require that the sum of the colors around any cycle must be congruent to one of 0, 1, . . . , α − 1 modulo β. (So 1 ≤ α ≤ β.) This gives a rook complex with

ck = α(α + β)(α + 2β) ··· (α + (k − 1)β)

The basic idea behind this formula is that when we add an edge that does not close a cycle, there are β ways to color it, but an edge that closes a cycle can be colored in α ways. (The case α = β was considered by Briggs and Remmel.) This ﬁts into our general formula with α = n + 1 and β = 0. Qn−j−1 n−j In this case, the product i=0 (x − βi) is just x so our generalized factorial rook polynomial, reduces to an “ordinary” P n−j rook polynomial j rj x .

Forests

Recall that the forest complex consists of subsets of [n + 1] × [n + 1] (directed edges) that are forests of rooted trees (with all edges directed towards the roots). This is a rook j complex with cj = (n + 1) . (Note that cj depends on n, unlike our other examples.) Qn−j−1 n−j In this case, the product i=0 (x − βi) is just x so our generalized factorial rook polynomial, reduces to an “ordinary” P n−j rook polynomial j rj x .

Forests

Recall that the forest complex consists of subsets of [n + 1] × [n + 1] (directed edges) that are forests of rooted trees (with all edges directed towards the roots). This is a rook j complex with cj = (n + 1) . (Note that cj depends on n, unlike our other examples.) This ﬁts into our general formula with α = n + 1 and β = 0. Qn−j−1 n−j In this case, the product i=0 (x − βi) is just x so our generalized factorial rook polynomial, reduces to an “ordinary” P n−j rook polynomial j rj x . We have

X X xn−i (x + n + 1)i F (x) = r xn−j = h B j i (n + 1)n j i and the reciprocity theorem

n FB(x) = (−1) FB(−x − n − 1), due to Bedrosian and Kelmans (further generalized by Pak and Postnikov). In the example we looked at before, with n = 2 (forests on 3 points), where B is the directed graph

and r0 = 1, r1 = 2, and r2 = 1; and h0 = 4, h1 = 4, and h2 = 1 we have

2 2 FB(x) = x + 2x + 1 = (x + 1) x2 x(x + 3) (x + 3)2 = 4 + 4 + 1 9 9 9 and 2 2 FB(x) = FB(−x − 3) = (x + 2) = x + 4x + 4. 2 FB(x) = x + 4x + 4. is the factorial rook polynomial for the complementary digraph: We deﬁne the rook “numbers” and hit “numbers” as before, and we still have the fundamental identity

X i h = r c , i j j n−j i so we can deﬁne the factorial rook polynomials as before (but α can be a weight).

Weighted rook complexes

A weighted rook complex is a rook complex ∆ in which every face is assigned a weight, and there are quantities c0, c1,..., cn such that if U ∈ ∆ with |U| = j then the sum of the weights of the facets containing U is cn−j . Weighted rook complexes

X i h = r c , i j j n−j i so we can deﬁne the factorial rook polynomials as before (but α can be a weight). One of the most interesting examples (and the inspiration for this theory) consists of ordinary rook placements (viewed as digraphs in which every vertex has indegree and outdegree at most 1) in which the weight of a digraph with i cycles is αi . Here j cj = α(α + 1) ··· (α + j − 1) = α↑ . and X FB(x) = rk x↓n−k . k This polynomial was introduced by Chung and Graham in 1995 under the name cover polynomial. We have the generating function

∞ P i X m + α − 1 hn−i t F (m)tm = i m B (1 − t)n+α m=0 and the reciprocity theorem

The hit numbers (which are now polynomials in α) are related to the cover polynomial by

i X (x + α)↑ x↓ − F (x) = h n i . B i α↑n i The hit numbers (which are now polynomials in α) are related to the cover polynomial by

i X (x + α)↑ x↓ − F (x) = h n i . B i α↑n i We have the generating function

∞ P i X m + α − 1 hn−i t F (m)tm = i m B (1 − t)n+α m=0 and the reciprocity theorem

n FB(x) = (−1) FB(−x − α). For example, in the upper triangular board, there are no cycles n within the board, so the factorial rook polynomial is FB(m) = x as before, but now the hit polynomial counts permutations by excedances and cycles. For ordinary rook polynomials, permuting the rows or columns doesn’t change the rook numbers or hit numbers. But they do change when we keep track of cycles. There is a beautiful result of Morris Dworkin giving a sufﬁcient condition for the cover polynomial of a permuted Ferrers board to factor.

The Goldman-Joichi-White result on factorization of F(x) for Ferrers boards extends directly. There is a beautiful result of Morris Dworkin giving a sufﬁcient condition for the cover polynomial of a permuted Ferrers board to factor.

The Goldman-Joichi-White result on factorization of F(x) for Ferrers boards extends directly. For ordinary rook polynomials, permuting the rows or columns doesn’t change the rook numbers or hit numbers. But they do change when we keep track of cycles. The Goldman-Joichi-White result on factorization of F(x) for Ferrers boards extends directly. For ordinary rook polynomials, permuting the rows or columns doesn’t change the rook numbers or hit numbers. But they do change when we keep track of cycles. There is a beautiful result of Morris Dworkin giving a sufﬁcient condition for the cover polynomial of a permuted Ferrers board to factor. For a permutation σ, let σ(Tn) be Tn with its rows permuted by σ, so there are σ(i) squares in row i.

Let Tn be the staircase board {{i, j} : 1 ≤ i ≤ j ≤ n }.

n Its cover polynomial is FTn (x, α) = (x + α) . Let Tn be the staircase board {{i, j} : 1 ≤ i ≤ j ≤ n }.

n Its cover polynomial is FTn (x, α) = (x + α) .

For a permutation σ, let σ(Tn) be Tn with its rows permuted by σ, so there are σ(i) squares in row i. A special case of Dworkin’s theorem: If σ is a noncrossing c n−c permutation with c cycles, then Fσ(Tn) = (x + α) (x + 1) .

As a consequence, the generating polynomial An,c(t, α) for permutations π of [n] according to the cycles of π and excedances of σ ◦ π is given by ∞ An,c(t, α) X m + α − 1 = (m + α)c(m + 1)n−ctm. (1 − t)n+α m m=0 A noncrossing permutation with one cycle looks like this:

1 2 3 4 5

In generally, a noncrossing permutation is made from a noncrossing partition by making each block into a cycle of this type:

What is a noncrossing permutation? In generally, a noncrossing permutation is made from a noncrossing partition by making each block into a cycle of this type:

What is a noncrossing permutation? A noncrossing permutation with one cycle looks like this:

1 2 3 4 5 What is a noncrossing permutation? A noncrossing permutation with one cycle looks like this:

1 2 3 4 5

In generally, a noncrossing permutation is made from a noncrossing partition by making each block into a cycle of this type: So the number of noncrossing permutations of [n] is the 1 2n Catalan number Cn = n+1 n and the number of noncrossing permutations of [n] with c cycles is the Narayana number 1 n n n c c−1 . In particular, we have

n−i i X n−j X x (x + u1 + ··· + un+1) FB(x) = rj x = hi n (u1 + ··· + un+1) j i

and the reciprocity theorem is

n FB(x) = (−1) FB(−x − u1 − · · · − un+1), which is Pak and Postnikov’s generalization of Bedrosian and Kelmans’s reciprocity theorem.

Weighted Forests

We can make the forest complex into a weighted rook complex by deﬁning the weight of a rooted forest on [n + 1] to be j1 jn+1 u1 ··· un+1, where ji is the indegree of vertex i. Here j cj = (u1 + ··· + un+1) , and all of our formulas apply with α = u1 + ··· + un+1 and β = 0. Weighted Forests

n−i i X n−j X x (x + u1 + ··· + un+1) FB(x) = rj x = hi n (u1 + ··· + un+1) j i

and the reciprocity theorem is

n FB(x) = (−1) FB(−x − u1 − · · · − un+1), which is Pak and Postnikov’s generalization of Bedrosian and Kelmans’s reciprocity theorem.