SLIDE 1 A Factorization Theorem for m-rook Placements
Nicholas Loehr
Department of Mathematics, Virginia Tech Blacksburg, VA 24061-0123 and Jeffrey Remmel Department of Mathematics, UCSD La Jolla, CA, 92093-0112 and Kenneth Barrese and Bruce Sagan Department of Mathematics, Michigan State University East Lansing, MI 48824-1027 www.math.msu.edu/˜sagan
SLIDE 2
The Factorization Theorem m-rook Placements Arbitrary Ferrers Boards Other Work
SLIDE 3
Outline
The Factorization Theorem m-rook Placements Arbitrary Ferrers Boards Other Work
SLIDE 4
Consider the board gotten by tiling the first quadrant with squares Q = . . . . . .
SLIDE 5
Consider the board gotten by tiling the first quadrant with squares Q = . . . . . . Let N be the non-negative integers. A partition is a weakly increasing sequence of elements of N, denoted B = (b1, . . . , bn).
SLIDE 6
Consider the board gotten by tiling the first quadrant with squares Q = . . . . . . Let N be the non-negative integers. A partition is a weakly increasing sequence of elements of N, denoted B = (b1, . . . , bn). We also let B stand for the Ferrers board obtained by taking the lowest bj squares of Q in column j for all j.
SLIDE 7
Consider the board gotten by tiling the first quadrant with squares Q = . . . . . . Let N be the non-negative integers. A partition is a weakly increasing sequence of elements of N, denoted B = (b1, . . . , bn). We also let B stand for the Ferrers board obtained by taking the lowest bj squares of Q in column j for all j. Ex. B = (1, 3, 3) =
SLIDE 8
For any board B, a rook placement is a subset of B having no two squares in the same row or column.
SLIDE 9
For any board B, a rook placement is a subset of B having no two squares in the same row or column. The kth rook number of B is rk(B) = number of placements of k rooks on B.
SLIDE 10
For any board B, a rook placement is a subset of B having no two squares in the same row or column. The kth rook number of B is rk(B) = number of placements of k rooks on B. Ex. B = (1, 3) = r0(B) = 1, r1(B) = 4, r2(B) = 2.
SLIDE 11
For any board B, a rook placement is a subset of B having no two squares in the same row or column. The kth rook number of B is rk(B) = number of placements of k rooks on B. Let x is a variable and n ∈ N. The corresponding falling factorial is x ↓n= x(x − 1) · · · (x − n + 1). Ex. B = (1, 3) = r0(B) = 1, r1(B) = 4, r2(B) = 2.
SLIDE 12 For any board B, a rook placement is a subset of B having no two squares in the same row or column. The kth rook number of B is rk(B) = number of placements of k rooks on B. Let x is a variable and n ∈ N. The corresponding falling factorial is x ↓n= x(x − 1) · · · (x − n + 1).
Theorem (Factorization Theorem: Goldman-Joichi-White)
For any Ferrers board B = (b1, . . . , bn) we have
n
rk(B)x ↓n−k=
n
(x + bj − j + 1). Ex. B = (1, 3) = r0(B) = 1, r1(B) = 4, r2(B) = 2.
SLIDE 13 For any board B, a rook placement is a subset of B having no two squares in the same row or column. The kth rook number of B is rk(B) = number of placements of k rooks on B. Let x is a variable and n ∈ N. The corresponding falling factorial is x ↓n= x(x − 1) · · · (x − n + 1).
Theorem (Factorization Theorem: Goldman-Joichi-White)
For any Ferrers board B = (b1, . . . , bn) we have
n
rk(B)x ↓n−k=
n
(x + bj − j + 1). Ex. B = (1, 3) = r0(B) = 1, r1(B) = 4, r2(B) = 2.
2
rk(B)x ↓n−k= 1 · x ↓2 +4 · x ↓1 +2 · x ↓0
SLIDE 14 For any board B, a rook placement is a subset of B having no two squares in the same row or column. The kth rook number of B is rk(B) = number of placements of k rooks on B. Let x is a variable and n ∈ N. The corresponding falling factorial is x ↓n= x(x − 1) · · · (x − n + 1).
Theorem (Factorization Theorem: Goldman-Joichi-White)
For any Ferrers board B = (b1, . . . , bn) we have
n
rk(B)x ↓n−k=
n
(x + bj − j + 1). Ex. B = (1, 3) = r0(B) = 1, r1(B) = 4, r2(B) = 2.
2
rk(B)x ↓n−k= 1 · x ↓2 +4 · x ↓1 +2 · x ↓0 = x2 + 3x + 2
SLIDE 15 For any board B, a rook placement is a subset of B having no two squares in the same row or column. The kth rook number of B is rk(B) = number of placements of k rooks on B. Let x is a variable and n ∈ N. The corresponding falling factorial is x ↓n= x(x − 1) · · · (x − n + 1).
Theorem (Factorization Theorem: Goldman-Joichi-White)
For any Ferrers board B = (b1, . . . , bn) we have
n
rk(B)x ↓n−k=
n
(x + bj − j + 1). Ex. B = (1, 3) = r0(B) = 1, r1(B) = 4, r2(B) = 2.
2
rk(B)x ↓n−k= 1 · x ↓2 +4 · x ↓1 +2 · x ↓0 = x2 + 3x + 2 = (x + 1)(x + 2)
SLIDE 16 For any board B, a rook placement is a subset of B having no two squares in the same row or column. The kth rook number of B is rk(B) = number of placements of k rooks on B. Let x is a variable and n ∈ N. The corresponding falling factorial is x ↓n= x(x − 1) · · · (x − n + 1).
Theorem (Factorization Theorem: Goldman-Joichi-White)
For any Ferrers board B = (b1, . . . , bn) we have
n
rk(B)x ↓n−k=
n
(x + bj − j + 1). Ex. B = (1, 3) = r0(B) = 1, r1(B) = 4, r2(B) = 2.
2
rk(B)x ↓n−k= 1 · x ↓2 +4 · x ↓1 +2 · x ↓0 = x2 + 3x + 2 = (x + 1)(x + 2) = (x + b1)(x + b2 − 1).
SLIDE 17
Outline
The Factorization Theorem m-rook Placements Arbitrary Ferrers Boards Other Work
SLIDE 18
Fix a positive integer m.
SLIDE 19
Fix a positive integer m. We partition Q into levels where the jth level consists of rows jm + 1, jm + 2, . . . , (j + 1)m.
SLIDE 20 Fix a positive integer m. We partition Q into levels where the jth level consists of rows jm + 1, jm + 2, . . . , (j + 1)m.
level 0
. . . . .
SLIDE 21 Fix a positive integer m. We partition Q into levels where the jth level consists of rows jm + 1, jm + 2, . . . , (j + 1)m.
level 0
. . . . . An m-rook placement on B ⊆ Q is a subset of B having no two squares in the same level or column.
SLIDE 22 Fix a positive integer m. We partition Q into levels where the jth level consists of rows jm + 1, jm + 2, . . . , (j + 1)m.
level 0
. . . . . An m-rook placement on B ⊆ Q is a subset of B having no two squares in the same level or column. The kth m-rook number of B: rk,m(B) = number of m-rook placements on B with k rooks.
SLIDE 23 Fix a positive integer m. We partition Q into levels where the jth level consists of rows jm + 1, jm + 2, . . . , (j + 1)m.
level 0
. . . . . An m-rook placement on B ⊆ Q is a subset of B having no two squares in the same level or column. The kth m-rook number of B: rk,m(B) = number of m-rook placements on B with k rooks.
B = (1, 2, 3) =
SLIDE 24 Fix a positive integer m. We partition Q into levels where the jth level consists of rows jm + 1, jm + 2, . . . , (j + 1)m.
level 0
. . . . . An m-rook placement on B ⊆ Q is a subset of B having no two squares in the same level or column. The kth m-rook number of B: rk,m(B) = number of m-rook placements on B with k rooks.
B = (1, 2, 3) = ∴ r2,2 = 3 : R R R R R R
SLIDE 25
The m-rook placements are realated to Cm ≀ Sk where Cm is the m cyclic group and Sk is the kth symmetric group,
SLIDE 26 The m-rook placements are realated to Cm ≀ Sk where Cm is the m cyclic group and Sk is the kth symmetric group, e.g., rk,m(
k
SLIDE 27 The m-rook placements are realated to Cm ≀ Sk where Cm is the m cyclic group and Sk is the kth symmetric group, e.g., rk,m(
k
- mk, . . . , mk) = (mk)(mk − m) · · · (m)
SLIDE 28 The m-rook placements are realated to Cm ≀ Sk where Cm is the m cyclic group and Sk is the kth symmetric group, e.g., rk,m(
k
- mk, . . . , mk) = (mk)(mk − m) · · · (m) = mkk!
SLIDE 29 The m-rook placements are realated to Cm ≀ Sk where Cm is the m cyclic group and Sk is the kth symmetric group, e.g., rk,m(
k
- mk, . . . , mk) = (mk)(mk − m) · · · (m) = mkk! = #Cm ≀ Sk.
SLIDE 30 The m-rook placements are realated to Cm ≀ Sk where Cm is the m cyclic group and Sk is the kth symmetric group, e.g., rk,m(
k
- mk, . . . , mk) = (mk)(mk − m) · · · (m) = mkk! = #Cm ≀ Sk.
Define the m-falling factorials by x ↓n,m= x(x − m)(x − 2m) · · · (x − (n − 1)m).
SLIDE 31 The m-rook placements are realated to Cm ≀ Sk where Cm is the m cyclic group and Sk is the kth symmetric group, e.g., rk,m(
k
- mk, . . . , mk) = (mk)(mk − m) · · · (m) = mkk! = #Cm ≀ Sk.
Define the m-falling factorials by x ↓n,m= x(x − m)(x − 2m) · · · (x − (n − 1)m). Given an integer m, define the mod m floor function by ⌊n⌋m = largest multiple of m which is less than or equal to n.
SLIDE 32 The m-rook placements are realated to Cm ≀ Sk where Cm is the m cyclic group and Sk is the kth symmetric group, e.g., rk,m(
k
- mk, . . . , mk) = (mk)(mk − m) · · · (m) = mkk! = #Cm ≀ Sk.
Define the m-falling factorials by x ↓n,m= x(x − m)(x − 2m) · · · (x − (n − 1)m). Given an integer m, define the mod m floor function by ⌊n⌋m = largest multiple of m which is less than or equal to n. Ex. ⌊17⌋3 = 15 since 15 ≤ 17 < 18.
SLIDE 33 The m-rook placements are realated to Cm ≀ Sk where Cm is the m cyclic group and Sk is the kth symmetric group, e.g., rk,m(
k
- mk, . . . , mk) = (mk)(mk − m) · · · (m) = mkk! = #Cm ≀ Sk.
Define the m-falling factorials by x ↓n,m= x(x − m)(x − 2m) · · · (x − (n − 1)m). Given an integer m, define the mod m floor function by ⌊n⌋m = largest multiple of m which is less than or equal to n. Ex. ⌊17⌋3 = 15 since 15 ≤ 17 < 18.
Theorem (Briggs-Remmel)
Let B = (b1, . . . , bn) be a Ferrers board with ⌊b1⌋m < ⌊b2⌋m < · · · < ⌊bn⌋m. Then
n
rk,m(B)x ↓n−k,m=
n
(x + bj − (j − 1)m).
SLIDE 34
Outline
The Factorization Theorem m-rook Placements Arbitrary Ferrers Boards Other Work
SLIDE 35
Fix m > 0.
SLIDE 36
Fix m > 0. Define the ith zone, zi = zi(B), of a Ferrers board B = (b1, . . . , bn) to be the subsequence of all bj with ⌊bj⌋m = im.
SLIDE 37 Fix m > 0. Define the ith zone, zi = zi(B), of a Ferrers board B = (b1, . . . , bn) to be the subsequence of all bj with ⌊bj⌋m = im.
- Ex. Suppose m = 3 and B = (1, 1, 2, 3, 5, 7)
SLIDE 38 Fix m > 0. Define the ith zone, zi = zi(B), of a Ferrers board B = (b1, . . . , bn) to be the subsequence of all bj with ⌊bj⌋m = im.
- Ex. Suppose m = 3 and B = (1, 1, 2, 3, 5, 7)
∴ z0 = (1, 1, 2), z1 = (3, 5), z2 = (7).
SLIDE 39 Fix m > 0. Define the ith zone, zi = zi(B), of a Ferrers board B = (b1, . . . , bn) to be the subsequence of all bj with ⌊bj⌋m = im. Given a zone z = (bi, . . . , bj) define its remainder to be r(z) =
j
(bt − ⌊bt⌋m).
- Ex. Suppose m = 3 and B = (1, 1, 2, 3, 5, 7)
∴ z0 = (1, 1, 2), z1 = (3, 5), z2 = (7).
SLIDE 40 Fix m > 0. Define the ith zone, zi = zi(B), of a Ferrers board B = (b1, . . . , bn) to be the subsequence of all bj with ⌊bj⌋m = im. Given a zone z = (bi, . . . , bj) define its remainder to be r(z) =
j
(bt − ⌊bt⌋m).
- Ex. Suppose m = 3 and B = (1, 1, 2, 3, 5, 7)
∴ z0 = (1, 1, 2), z1 = (3, 5), z2 = (7). r(z0) = 1 + 1 + 2 = 4, r(z1) = 0 + 2 = 2, r(z2) = 1.
SLIDE 41 Fix m > 0. Define the ith zone, zi = zi(B), of a Ferrers board B = (b1, . . . , bn) to be the subsequence of all bj with ⌊bj⌋m = im. Given a zone z = (bi, . . . , bj) define its remainder to be r(z) =
j
(bt − ⌊bt⌋m).
- Ex. Suppose m = 3 and B = (1, 1, 2, 3, 5, 7)
∴ z0 = (1, 1, 2), z1 = (3, 5), z2 = (7). r(z0) = 1 + 1 + 2 = 4, r(z1) = 0 + 2 = 2, r(z2) = 1.
Theorem (Barrese-Loehr-Remmel-S)
Let B = (b1, . . . , bn) be any Ferrers board. Then
n
rk,m(B)x ↓n−k,m=
n
(x + ⌊bj⌋m − (j − 1)m + ǫj) where ǫj = r(zi) if bj is the last column in zone zi, else.
SLIDE 42 n
rk,m(B)x ↓n−k,m=
n
(x + ⌊bj⌋m − (j − 1)m + ǫj) where ǫj = r(zi) if bj is the last column in zone zi, else.
SLIDE 43 n
rk,m(B)x ↓n−k,m=
n
(x + ⌊bj⌋m − (j − 1)m + ǫj) where ǫj = r(zi) if bj is the last column in zone zi, else. For the proof, it suffices to show the identity when x is a multiple
SLIDE 44 n
rk,m(B)x ↓n−k,m=
n
(x + ⌊bj⌋m − (j − 1)m + ǫj) where ǫj = r(zi) if bj is the last column in zone zi, else. For the proof, it suffices to show the identity when x is a multiple
- f m. One adds an x-by-n rectangle of boxes below B and then
counts the number of m-roook placements with n rooks on the extended board in two different ways.
SLIDE 45 n
rk,m(B)x ↓n−k,m=
n
(x + ⌊bj⌋m − (j − 1)m + ǫj) where ǫj = r(zi) if bj is the last column in zone zi, else. For the proof, it suffices to show the identity when x is a multiple
- f m. One adds an x-by-n rectangle of boxes below B and then
counts the number of m-roook placements with n rooks on the extended board in two different ways.
- Ex. Suppose m = 3 and B = (1, 1, 2, 3, 5, 7)
∴ z0 = (1, 1, 2), z1 = (3, 5), z2 = (7). r(z0) = 1 + 1 + 2 = 4, r(z1) = 0 + 2 = 2, r(z2) = 1.
SLIDE 46 n
rk,m(B)x ↓n−k,m=
n
(x + ⌊bj⌋m − (j − 1)m + ǫj) where ǫj = r(zi) if bj is the last column in zone zi, else. For the proof, it suffices to show the identity when x is a multiple
- f m. One adds an x-by-n rectangle of boxes below B and then
counts the number of m-roook placements with n rooks on the extended board in two different ways.
- Ex. Suppose m = 3 and B = (1, 1, 2, 3, 5, 7)
∴ z0 = (1, 1, 2), z1 = (3, 5), z2 = (7). r(z0) = 1 + 1 + 2 = 4, r(z1) = 0 + 2 = 2, r(z2) = 1.
n
rk,m(B)x ↓n−k,m= (x + 0 − 0 + 0)(x + 0 − 3 + 0)(x + 0 − 6 + 4) ·(x + 3 − 9 + 0)(x + 3 − 12 + 2)(x + 6 − 15 + 1).
SLIDE 47 n
rk,m(B)x ↓n−k,m=
n
(x + ⌊bj⌋m − (j − 1)m + ǫj) where ǫj = r(zi) if bj is the last column in zone zi, else. For the proof, it suffices to show the identity when x is a multiple
- f m. One adds an x-by-n rectangle of boxes below B and then
counts the number of m-roook placements with n rooks on the extended board in two different ways.
- Ex. Suppose m = 3 and B = (1, 1, 2, 3, 5, 7)
∴ z0 = (1, 1, 2), z1 = (3, 5), z2 = (7). r(z0) = 1 + 1 + 2 = 4, r(z1) = 0 + 2 = 2, r(z2) = 1.
n
rk,m(B)x ↓n−k,m= (x + 0 − 0 + 0)(x + 0 − 3 + 0)(x + 0 − 6 + 4) ·(x + 3 − 9 + 0)(x + 3 − 12 + 2)(x + 6 − 15 + 1). Note: Our theorem implies both Goldman-Joichi-White and Briggs-Remmel.
SLIDE 48
Outline
The Factorization Theorem m-rook Placements Arbitrary Ferrers Boards Other Work
SLIDE 50
- 1. p, q-analogues. There is a p, q-analogue of our theorem where p
and q keep track of a pair of statistics (related to the inv statistic
- n permutations) on m-rook placements.
SLIDE 51
- 1. p, q-analogues. There is a p, q-analogue of our theorem where p
and q keep track of a pair of statistics (related to the inv statistic
- n permutations) on m-rook placements.
- 2. Weighted file placements.
SLIDE 52
- 1. p, q-analogues. There is a p, q-analogue of our theorem where p
and q keep track of a pair of statistics (related to the inv statistic
- n permutations) on m-rook placements.
- 2. Weighted file placements. In the Briggs-Remmel Theorem
n
rk,m(B)x ↓n−k,m=
n
(x + bj − (j − 1)m) we fixed the sum side and modified the product to work for all Ferrers boards.
SLIDE 53
- 1. p, q-analogues. There is a p, q-analogue of our theorem where p
and q keep track of a pair of statistics (related to the inv statistic
- n permutations) on m-rook placements.
- 2. Weighted file placements. In the Briggs-Remmel Theorem
n
rk,m(B)x ↓n−k,m=
n
(x + bj − (j − 1)m) we fixed the sum side and modified the product to work for all Ferrers boards. Alternatively, for an arbitrary board, one can fix the product side and expand it in terms of the falling factorial basis:
n
(x + bj − (j − 1)m) =
n
fk,m(B)x ↓n−k,m for certain constants fk,m(B) and ask what they count.
SLIDE 54
- 1. p, q-analogues. There is a p, q-analogue of our theorem where p
and q keep track of a pair of statistics (related to the inv statistic
- n permutations) on m-rook placements.
- 2. Weighted file placements. In the Briggs-Remmel Theorem
n
rk,m(B)x ↓n−k,m=
n
(x + bj − (j − 1)m) we fixed the sum side and modified the product to work for all Ferrers boards. Alternatively, for an arbitrary board, one can fix the product side and expand it in terms of the falling factorial basis:
n
(x + bj − (j − 1)m) =
n
fk,m(B)x ↓n−k,m for certain constants fk,m(B) and ask what they count. The fk,m(B) are weight generating functions for certain file placements (where two rooks can occupy the same row but not the same column) as shown by Haglund.
SLIDE 56
- 3. Rook equivalence. Call boards B, B′ rook equivalent if
rk(B) = rk(B′) for all k.
SLIDE 57
- 3. Rook equivalence. Call boards B, B′ rook equivalent if
rk(B) = rk(B′) for all k. Recall for Ferrers boards
n
rk(B)x ↓n−k=
n
(x + bj − j + 1).
SLIDE 58
- 3. Rook equivalence. Call boards B, B′ rook equivalent if
rk(B) = rk(B′) for all k. Recall for Ferrers boards
n
rk(B)x ↓n−k=
n
(x + bj − j + 1). So Ferrers boards B and B′ are equivalent iff {{b1 − 1, b2 − 2, . . . , bn − n}} = {{b′
1 − 1, b′ 2 − 2, . . . , b′ n − n}}
as multisets.
SLIDE 59
- 3. Rook equivalence. Call boards B, B′ rook equivalent if
rk(B) = rk(B′) for all k. Recall for Ferrers boards
n
rk(B)x ↓n−k=
n
(x + bj − j + 1). So Ferrers boards B and B′ are equivalent iff {{b1 − 1, b2 − 2, . . . , bn − n}} = {{b′
1 − 1, b′ 2 − 2, . . . , b′ n − n}}
as multisets. One can ask for canonical equivalence class representatives.
SLIDE 60
- 3. Rook equivalence. Call boards B, B′ rook equivalent if
rk(B) = rk(B′) for all k. Recall for Ferrers boards
n
rk(B)x ↓n−k=
n
(x + bj − j + 1). So Ferrers boards B and B′ are equivalent iff {{b1 − 1, b2 − 2, . . . , bn − n}} = {{b′
1 − 1, b′ 2 − 2, . . . , b′ n − n}}
as multisets. One can ask for canonical equivalence class representatives.
Theorem (Foata-Sch¨ utzenberger)
Every Ferrers board is rook equivalent to a unique strictly increasing board.
SLIDE 61
- 3. Rook equivalence. Call boards B, B′ rook equivalent if
rk(B) = rk(B′) for all k. Recall for Ferrers boards
n
rk(B)x ↓n−k=
n
(x + bj − j + 1). So Ferrers boards B and B′ are equivalent iff {{b1 − 1, b2 − 2, . . . , bn − n}} = {{b′
1 − 1, b′ 2 − 2, . . . , b′ n − n}}
as multisets. One can ask for canonical equivalence class representatives.
Theorem (Foata-Sch¨ utzenberger)
Every Ferrers board is rook equivalent to a unique strictly increasing board. One can also ask for the number of elements in an equivalence class and it turns out to be a product of binomial coefficents.
SLIDE 62
- 3. Rook equivalence. Call boards B, B′ rook equivalent if
rk(B) = rk(B′) for all k. Recall for Ferrers boards
n
rk(B)x ↓n−k=
n
(x + bj − j + 1). So Ferrers boards B and B′ are equivalent iff {{b1 − 1, b2 − 2, . . . , bn − n}} = {{b′
1 − 1, b′ 2 − 2, . . . , b′ n − n}}
as multisets. One can ask for canonical equivalence class representatives.
Theorem (Foata-Sch¨ utzenberger)
Every Ferrers board is rook equivalent to a unique strictly increasing board. One can also ask for the number of elements in an equivalence class and it turns out to be a product of binomial coefficents. We have analogous results in our setting.
SLIDE 63
THANKS FOR LISTENING!
