on permutation tableaux of type a and bslc/wpapers/s64vortrag/kim.pdf1 / 28 on permutation tableaux...

1 / 28

On permutation tableaux of type A and B

Jang Soo Kim(Joint work with Sylvie Corteel)

University of Paris 7

64th SLC, Lyon, March 30, 2010

2 / 28

Permutation tableaux

First introduced by Postnikov in his study of totally nonnegativeGrassmanian.

2 / 28



There are many bijections between permutation tableaux andpermutations.

2 / 28




A connection with partially asymmetric exclusion process (PASEP)

2 / 28




A connection with partially asymmetric exclusion process (PASEP)

Type B Permutation tableaux defined by Lam and Williams

3 / 28

Ferrers diagram

123

45

678

910

3 / 28

Ferrers diagram

123

45

678

910

3

5

8

10

9 7 6 4 2 1

4 / 28

Permutation tableauEach column has at least one 1.

4 / 28

Permutation tableauEach column has at least one 1.There is no configuration like

1...

1 · · · 0

4 / 28


1...

1 · · · 0

0 0 1 0 0 10 0 0 1 1 10 0 0 0 10 0 1

10 NO

4 / 28


1...

1 · · · 0

0 0 1 0 0 10 0 0 1 1 10 0 0 0 10 0 1

10 NO

0 0 1 0 0 10 0 0 1 1 10 1 0 0 10 0 1

10 NO

4 / 28


1...

1 · · · 0

0 0 1 0 0 10 0 0 1 1 10 0 0 0 10 0 1

10 NO

0 0 1 0 0 10 0 0 1 1 10 1 0 0 10 0 1

10 NO

0 0 1 0 0 10 0 0 1 1 10 0 0 0 10 1 1

10 YES!

4 / 28


1...

1 · · · 0

0 0 1 0 0 10 0 0 1 1 10 0 0 0 10 0 1

10 NO

0 0 1 0 0 10 0 0 1 1 10 1 0 0 10 0 1

10 NO

0 0 1 0 0 10 0 0 1 1 10 0 0 0 10 1 1

10 YES!

A restricted 0 is1...0

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1...

1 · · · 0

0 0 1 0 0 10 0 0 1 1 10 0 0 0 10 0 1

10 NO

0 0 1 0 0 10 0 0 1 1 10 1 0 0 10 0 1

10 NO

0 0 1 0 0 10 0 0 1 1 10 0 0 0 10 1 1

10 YES!

A restricted 0 is1...0

An unrestricted row has no restricted 0.

5 / 28

The alternative representation

Topmost 1 is

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Topmost 1 is

Rightmost restricted 0 is

0 0 1 0 0 10 0 0 1 1 10 0 0 0 10 1 1

1

0

12 9 8 6 5 31

2

4

7

10

11

13

12 9 8 6 5 31

2

4

7

10

11

13

5 / 28


Topmost 1 is


0 0 1 0 0 10 0 0 1 1 10 0 0 0 10 1 1

1

0

12 9 8 6 5 31

2

4

7

10

11

13

12 9 8 6 5 31

2

4

7

10

11

13

In the alternative representation,

5 / 28


Topmost 1 is


0 0 1 0 0 10 0 0 1 1 10 0 0 0 10 1 1

1

0

12 9 8 6 5 31

2

4

7

10

11

13

12 9 8 6 5 31

2

4

7

10

11

13


Each column has exactly one

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Topmost 1 is


0 0 1 0 0 10 0 0 1 1 10 0 0 0 10 1 1

1

0

12 9 8 6 5 31

2

4

7

10

11

13

12 9 8 6 5 31

2

4

7

10

11

13



No arrow points to another.

5 / 28


Topmost 1 is


0 0 1 0 0 10 0 0 1 1 10 0 0 0 10 1 1

1

0

12 9 8 6 5 31

2

4

7

10

11

13

12 9 8 6 5 31

2

4

7

10

11

13




Unrestricted row ⇔ row without

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Topmost 1 is


0 0 1 0 0 10 0 0 1 1 10 0 0 0 10 1 1

1

0

12 9 8 6 5 31

2

4

7

10

11

13

12 9 8 6 5 31

2

4

7

10

11

13




Unrestricted row ⇔ row without

First introduced by Viennot (alternative tableau) and studied more byNadeau

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The bijection Φ of Corteel and Nadeau

12 9 8 6 5 312

4

7

101113

1, 10, 11, 13

6 / 28


12 9 8 6 5 31

2

4

7

10

11

13

1, 12, 10, 11, 13

6 / 28


12 9 8 6 5 31

2

4

7

10

11

13

9, 1, 12, 10, 11, 13

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12 9 8 6 5 31

2

4

7

10

11

13

9, 2, 7, 8, 1, 12, 10, 11, 13

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12 9 8 6 5 31

2

4

7

10

11

13

9, 4, 6, 2, 7, 8, 1, 12, 10, 11, 13

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12 9 8 6 5 31

2

4

7

10

11

13

9, 4, 6, 5, 2, 7, 8, 1, 12, 10, 11, 13

6 / 28


12 9 8 6 5 31

2

4

7

10

11

13

9, 4, 6, 5, 2, 7, 8, 3, 1, 12, 10, 11, 13

7 / 28

Statistics

Decompose π as σ1τ .

π =

σz }| {

4, 6, 5, 2, 8, 3, 1,

τz }| {

9, 7, 12, 10, 11, 1312 9 8 6 5 3

12

4

7101113

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Statistics


π =

σz }| {

4, 6, 5, 2, 8, 3, 1,

τz }| {

9, 7, 12, 10, 11, 13

The RL-minima (right-to-leftmimina) of π :

4, 6, 5, 2, 8, 3, 1, 9, 7, 12, 10, 11, 13

12 9 8 6 5 312

4

7101113

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Statistics


π =

σz }| {

4, 6, 5, 2, 8, 3, 1,

τz }| {

9, 7, 12, 10, 11, 13


4, 6, 5, 2, 8, 3, 1, 9, 7, 12, 10, 11, 13

The RL-maxima (right-to-leftmaxima) of σ :

σz }| {

4, 6, 5, 2, 8, 3, 1, 9, 7, 12, 10, 11, 13

12 9 8 6 5 312

4

7101113

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Statistics


π =

σz }| {

4, 6, 5, 2, 8, 3, 1,

τz }| {

9, 7, 12, 10, 11, 13


4, 6, 5, 2, 8, 3, 1, 9, 7, 12, 10, 11, 13


σz }| {

4, 6, 5, 2, 8, 3, 1, 9, 7, 12, 10, 11, 13

12 9 8 6 5 312

4

7101113

Proposition (Corteel & Nadeau, Nadeau)Let π = σ1τ and Φ(π) = T . Then

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Statistics


π =

σz }| {

4, 6, 5, 2, 8, 3, 1,

τz }| {

9, 7, 12, 10, 11, 13


4, 6, 5, 2, 8, 3, 1, 9, 7, 12, 10, 11, 13


σz }| {

4, 6, 5, 2, 8, 3, 1, 9, 7, 12, 10, 11, 13

12 9 8 6 5 312

4

7101113


the unrestricted rows of T ⇔ the RL-minima of π

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Statistics


π =

σz }| {

4, 6, 5, 2, 8, 3, 1,

τz }| {

9, 7, 12, 10, 11, 13


4, 6, 5, 2, 8, 3, 1, 9, 7, 12, 10, 11, 13


σz }| {

4, 6, 5, 2, 8, 3, 1, 9, 7, 12, 10, 11, 13

12 9 8 6 5 312

4

7101113


the unrestricted rows of T ⇔ the RL-minima of π

the columns with 1 in the first row of T ⇔ the RL-maxima of σ

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Nadeau’s bijective proof of a theorem of Corteel and Nadeau

Theorem

X

T∈PT (n)

xurr(T)−1ytopone(T) = (x + y)n−1 = (x + y)(x + y + 1) · · · (x + y + n − 2).

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Theorem

X

T∈PT (n)


c(n, k) : the number π ∈ Sn with k cycles

(x + y)n−1 =X

i,j

c(n − 1, i + j)

i + ji

!

xiyj

#{T ∈ PT (n) : urr(T) − 1 = i, topone(T) = j} = c(n − 1, i + j)

i + ji

!

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Theorem

X

T∈PT (n)



(x + y)n−1 =X

i,j

c(n − 1, i + j)

i + ji

!

xiyj


i + ji

!

If T ↔ π = σ1τ ,

urr(T) − 1 = RLmin(τ ) = i

topone(T) = RLmax(σ) = j

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Theorem

X

T∈PT (n)



(x + y)n−1 =X

i,j

c(n − 1, i + j)

i + ji

!

xiyj


i + ji

!

If T ↔ π = σ1τ ,



τ is a set of i cycles and σ is a set of j cycles.

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Theorem

X

T∈PT (n)



(x + y)n−1 =X

i,j

c(n − 1, i + j)

i + ji

!

xiyj


i + ji

!

If T ↔ π = σ1τ ,



τ is a set of i cycles and σ is a set of j cycles.

τ ∪ σ is a permutation of {2, 3, . . . , n} with i + j cycles.

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Another bijective proof

TheoremWe have

X

T∈PT (n)

xurr(T)−1ytopone(T) = (x + y)n−1.

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TheoremWe have

X

T∈PT (n)


Let x, y be any positive integers and let N = n + x + y − 2.

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TheoremWe have

X

T∈PT (n)


Let x, y be any positive integers and let N = n + x + y − 2.

Given T ∈ PT (n), we construct T ′ ∈ PT (N) as follows.

{x − 1

} y − 1

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{x − 1

} y − 1

T ′ satisfies the following.

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{x − 1

} y − 1

T ′ satisfies the following.1 The first y steps are south and the first y rows are unrestricted.

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{x − 1

} y − 1

T ′ satisfies the following.1 The first y steps are south and the first y rows are unrestricted.2 The last x − 1 steps are west and the last x − 1 columns have ↑’s in the first

row.

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{x − 1

} y − 1


row.

π′ = Φ(T ′) satisfies the following. (π′ = σ1τ )

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{x − 1

} y − 1


row.

π′ = Φ(T ′) satisfies the following. (π′ = σ1τ )1 1, 2, . . . , y are RL-minima of π′

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{x − 1

} y − 1


row.


2 N, N − 1, . . . , N − x + 2 are RL-maxima of σ

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{x − 1

} y − 1


row.


2 N, N − 1, . . . , N − x + 2 are RL-maxima of σ

N, N − 1, . . . , N − x + 2, 1, 2, . . . , y are arranged in this order in π′.

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Unrestricted Columns

An unrestricted column of T ∈ PT (n) is a column without 0 as follows.

1 · · · 0

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1 · · · 0

urc(T) : the number of unrestricted columns of T .

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1 · · · 0


Let

Pt(x) =X

n≥0

0

@

X

T∈PT (n)

turc(T)

1

A xn.

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1 · · · 0


Let

Pt(x) =X

n≥0

0

@

X

T∈PT (n)

turc(T)

1

A xn.

TheoremWe have

Pt(x) =1 + Et(x)

1 + (t − 1)xEt(x),

whereEt(x) =

X

n≥1

n(t)n−1xn.

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The case t = 2 : connected permutations

Corollary

X

n≥0

X

T∈PT (n)

2urc(T)xn =1x

1 −1

P

n≥0 n!xn

!

.

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Corollary

X

n≥0

X

T∈PT (n)

2urc(T)xn =1x

1 −1

P

n≥0 n!xn

!

.

π = π1 · · ·πn ∈ Sn is a connected permutation if there is no k < nsatisfying π1 · · ·πk ∈ Sk.

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Corollary

X

n≥0

X

T∈PT (n)

2urc(T)xn =1x

1 −1

P

n≥0 n!xn

!

.


CP(n) : the set of connected permutations in Sn

X

n≥0

#CP(n)xn = 1 −1

P

n≥0 n!xn

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Corollary

X

n≥0

X

T∈PT (n)

2urc(T)xn =1x

1 −1

P

n≥0 n!xn

!

.



X

n≥0

#CP(n)xn = 1 −1

P

n≥0 n!xn

Corollary

X

T∈PT (n)

2urc(T) = #CP(n + 1).

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Corollary

X

n≥0

X

T∈PT (n)

2urc(T)xn =1x

1 −1

P

n≥0 n!xn

!

.



X

n≥0

#CP(n)xn = 1 −1

P

n≥0 n!xn

Corollary

X

T∈PT (n)

2urc(T) = #CP(n + 1).

Combinatorial proof?

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Shift-connected permutations

A shift-connected permutation is π = π1 · · ·πn ∈ Sn with πj = 1 forsome j ∈ [n] such that

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there is no integer i < j with

πiπi+1 · · ·πj ∈ Sj−i+1

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πiπi+1 · · ·πj ∈ Sj−i+1

SCP(n) : the set of shift-connected permutations in Sn

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πiπi+1 · · ·πj ∈ Sj−i+1

SCP(n) : the set of shift-connected permutations in Sn

Proposition

#CP(n) = #SCP(n)

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A bijection between Sn \ CP(n) and Sn \ SCP(n)Given π = π1 · · ·πn ∈ Sn \ CP(n), define π

′ ∈ Sn \ SCP(n) as follows.

14 / 28


′ ∈ Sn \ SCP(n) as follows.Find the smallest integer k < n such that

σ = π1 · · ·πk ∈ Sk

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σ = π1 · · ·πk ∈ Sk

Decompose π asπ = στ (k + 1)ρ

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σ = π1 · · ·πk ∈ Sk


Define

σ+ = σ

+1 · · ·σ+

k , σ+i = σi + 1

π′ = τσ

+1ρ

14 / 28



σ = π1 · · ·πk ∈ Sk


Define

σ+ = σ

+1 · · ·σ+

k , σ+i = σi + 1

π′ = τσ

+1ρ

Example

14 / 28



σ = π1 · · ·πk ∈ Sk


Define

σ+ = σ

+1 · · ·σ+

k , σ+i = σi + 1

π′ = τσ

+1ρ

Example

Let π ∈ Sn \ CP(n) be

π =

σz }| {

4, 2, 5, 1, 3,

τz}|{

7 ,

k + 1z}|{

6 ,

ρz}|{

9, 8

14 / 28



σ = π1 · · ·πk ∈ Sk


Define

σ+ = σ

+1 · · ·σ+

k , σ+i = σi + 1

π′ = τσ

+1ρ

Example

Let π ∈ Sn \ CP(n) be

π =

σz }| {

4, 2, 5, 1, 3,

τz}|{

7 ,

k + 1z}|{

6 ,

ρz}|{

9, 8

Then π′ ∈ Sn \ SCP(n) is

π′ =

τz}|{

7 ,

σ+

z }| {

5, 3, 6, 2, 4,

1z}|{

1 ,

ρz}|{

9, 8

15 / 28

A combinatorial proof of∑

T∈PT (n)

2urc(T) = #CP(n + 1)

PropositionP

T∈PT (n) 2urc(T) is the number of T ∈ PT (n + 1) without a column containing1 only in the first row.

15 / 28


T∈PT (n)

2urc(T) = #CP(n + 1)

PropositionP


Proposition

15 / 28


T∈PT (n)

2urc(T) = #CP(n + 1)

PropositionP


Proposition

For π = Φ(T),

15 / 28


T∈PT (n)

2urc(T) = #CP(n + 1)

PropositionP


Proposition

For π = Φ(T),

T has a column which has a 1 only in the first row if and only if

15 / 28


T∈PT (n)

2urc(T) = #CP(n + 1)

PropositionP


Proposition

For π = Φ(T),


π 6∈ SCP(n)

15 / 28


T∈PT (n)

2urc(T) = #CP(n + 1)

PropositionP


Proposition

For π = Φ(T),


π 6∈ SCP(n)

Proposition

X

T∈PT (n)

2urc(T) = #SCP(n + 1) = #CP(n + 1)

16 / 28

The case t = −1 : sign-imbalance

Corollary

P−1(x) =X

n≥0

X

T∈PT (n)

(−1)urc(T)xn =1 − x

1 − 2x + 2x2

=12·

„

11 − (1 + i)x

+1

1 − (1 − i)x

«

16 / 28


Corollary

P−1(x) =X

n≥0

X

T∈PT (n)

(−1)urc(T)xn =1 − x

1 − 2x + 2x2

=12·

„

11 − (1 + i)x

+1

1 − (1 − i)x

«

DefinitionFor T ∈ PT (n), define the sign of T by

sgn(T) = (−1)urc(T).

16 / 28


Corollary

P−1(x) =X

n≥0

X

T∈PT (n)

(−1)urc(T)xn =1 − x

1 − 2x + 2x2

=12·

„

11 − (1 + i)x

+1

1 − (1 − i)x

«

DefinitionFor T ∈ PT (n), define the sign of T by

sgn(T) = (−1)urc(T).

Corollary

X

T∈PT (n)

sgn(T) =(1 + i)n + (1 − i)n

2=

8

<

:

(−1)k · 22k, if n = 4k or n = 4k + 1,

0, if n = 4k + 2,(−1)k+1 · 22k+1

, if n = 4k + 3.

17 / 28

Relation between the sign-imbalance of SYT

The sign of a standard Young tableau is defined as follows.

sgn

1 2 5

3 4

!

= sgn(12534) = 1

17 / 28



sgn

1 2 5

3 4

!

= sgn(12534) = 1

Stanley conjecturedX

T∈SYT (n)

sgn(T) = 2⌊n2⌋

17 / 28



sgn

1 2 5

3 4

!

= sgn(12534) = 1


T∈SYT (n)

sgn(T) = 2⌊n2⌋

Proved by Lam and Sjöstrand independently

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sgn

1 2 5

3 4

!

= sgn(12534) = 1


T∈SYT (n)

sgn(T) = 2⌊n2⌋


Generalized to skew SYTs by Kim

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sgn

1 2 5

3 4

!

= sgn(12534) = 1


T∈SYT (n)

sgn(T) = 2⌊n2⌋


Generalized to skew SYTs by Kim

If n 6≡ 2 mod 4,˛

˛

˛

˛

˛

˛

X

T∈PT (n)

sgn(T)

˛

˛

˛

˛

˛

˛

=

˛

˛

˛

˛

˛

˛

X

T∈SYT (n)

sgn(T)

˛

˛

˛

˛

˛

˛

= 2⌊n2⌋.

18 / 28

Shifted Ferrers diagram

3

5

8

10

9 7 6 4 2 1

−9

−7

−6

−4

−2

−1

3

5

8

10

9 7 6 4 2 1

The yellow cells are the diagonal cells.

18 / 28

Shifted Ferrers diagram

3

5

8

10

9 7 6 4 2 1

−9

−7

−6

−4

−2

−1

3

5

8

10

9 7 6 4 2 1

The yellow cells are the diagonal cells.

The row containing the diagonal cell in Column d is labeled with −d.

19 / 28

Type B permutation tableauxEach column has at least one 1.

19 / 28

Type B permutation tableauxEach column has at least one 1.There is no configuration like

1...

1 · · · 0

or 1 · · · 0

19 / 28


1...

1 · · · 0

or 1 · · · 0

0

0 0

1 0 10 0 0 0

0 0 0 1 10 0 0 0 0 0

0 0 0 0 0 1

1 0 1 10 0 0 1

0 0 1

NO

19 / 28


1...

1 · · · 0

or 1 · · · 0

0

0 0

1 0 10 0 0 0

0 0 0 1 10 0 0 0 0 0

0 0 0 0 0 1

1 0 1 10 0 0 1

0 0 1

NO

0

0 0

1 0 0

0 0 0 0

0 0 0 1 10 0 0 0 0 0

0 0 0 0 0 1

1 0 1 10 0 0 1

0 1 1

NO

19 / 28


1...

1 · · · 0

or 1 · · · 0

0

0 0

1 0 10 0 0 0

0 0 0 1 10 0 0 0 0 0

0 0 0 0 0 1

1 0 1 10 0 0 1

0 0 1

NO

0

0 0

1 0 0

0 0 0 0

0 0 0 1 10 0 0 0 0 0

0 0 0 0 0 1

1 0 1 10 0 0 1

0 1 1

NO

0

0 0

1 0 10 0 0 0

0 0 0 1 10 0 0 0 0 0

0 0 0 0 0 1

1 0 1 10 0 0 1

0 1 1

YES!

20 / 28

The alternative representationTopmost 1 is

20 / 28



20 / 28



Cut off the diagonal cells.

20 / 28




0

0 0

1 0 10 0 0 0

0 0 0 0 10 0 0 0 0 0

0 0 0 0 0 1

1 0 1 10 0 0 10 1 1

−10

−9

−8

−6

−3

−2

1

4

5

7

11

10 9 8 6 3 2−10

−9

−8

−6

−3

−2

1

4

5

7

11

10 9 8 6 3 2


20 / 28




0

0 0

1 0 10 0 0 0

0 0 0 0 10 0 0 0 0 0

0 0 0 0 0 1

1 0 1 10 0 0 10 1 1

−10

−9

−8

−6

−3

−2

1

4

5

7

11

10 9 8 6 3 2−10

−9

−8

−6

−3

−2

1

4

5

7

11

10 9 8 6 3 2


The diagonal line acts like a mirror!

21 / 28

A theorem of Lam and Williams

Theorem (Lam and Williams)

X

T∈PT B(n)

xurr(T)−1zdiag(T) = (1 + z)n(x + 1)n−1

21 / 28



X

T∈PT B(n)


diag(T) : the number of diagonal cells with 1

21 / 28



X

T∈PT B(n)



urr(T) : the number of unrestricted rows, i.e. without

1...0

and 0

21 / 28



X

T∈PT B(n)



urr(T) : the number of unrestricted rows, i.e. without

1...0

and 0

TheoremX

T∈PT B(n)

xurr(T)−1ytop0,1(T)zdiag(T) = (1 + z)n(x + y)n−1

22 / 28

Generalization of a theorem of Lam and WilliamsFor T ∈ PT B(n) with the topmost nonzero row labeled m,

top0,1(T) = (# 1s in Row m except in the diagonal)

+ (# rightmost restricted 0s in Column −m)

= # arrows in Row m and Column −m

0

0 0

1 0 10 0 0 0

0 0 0 0 10 0 0 0 0 0

0 0 0 0 0 1

1 0 1 10 0 0 1

0 1 1

−10

−9

−8

−6

−3

−2

1

4

5

7

11

10 9 8 6 3 2−10

−9

−8

−6

−3

−2

1

4

5

7

11

10 9 8 6 3 2

23 / 28

A type B extension of Corteel and Nadeau’s bijection

−10

−9

–8−6

−3

−2

1

45

7

11

10 9 8 6 3 2

–8,4,11

23 / 28


−10

−9

−8

−6

−3

−2

1

4

5

7

11

10 9 8 6 3 2

7, 10,−8, 4, 11

23 / 28


−10

−9

−8

−6

−3

−2

1

4

5

7

11

10 9 8 6 3 2

9, 7, 10,−8, 4, 11

23 / 28


−10

−9

−8

−6

−3

−2

1

4

5

7

11

10 9 8 6 3 2

9, 7, 10, –3, 5 − 8, 4, 11

23 / 28


−10

−9

−8

−6

−3

−2

1

4

5

7

11

10 9 8 6 3 2

9, 7, 10,−3, 5 − 8, 6, 4, 11

23 / 28


−10

−9

−8

−6

−3

−2

1

4

5

7

11

10 9 8 6 3 2

9, 7, 10, 1,−3, 5 − 8, 6, 4, 11

23 / 28


−10

−9

−8

−6

−3

−2

1

4

5

7

11

10 9 8 6 3 2

9, 7, 10, 2, 1,−3, 5 − 8, 6, 4, 11

24 / 28

Some properties of the type B bijection

Proposition

Then,

−10

−9

−8

−6

–3−2

1

457

11

10 9 8 6 3 2

π =

σz }| {

9, 7, 10,

τz }| {

6, 2, 1, –3, 5,

mz}|{

−8 ,

ρz }| {

4, 11

24 / 28


Proposition

π = ΦB(T)

Then,

−10

−9

−8

−6

–3−2

1

457

11

10 9 8 6 3 2

π =

σz }| {

9, 7, 10,

τz }| {

6, 2, 1, –3, 5,

mz}|{

−8 ,

ρz }| {

4, 11

24 / 28


Proposition

π = ΦB(T)

Row m is the topmost nonzerorow of T

Then,

−10

−9

−8

−6

–3−2

1

457

11

10 9 8 6 3 2

π =

σz }| {

9, 7, 10,

τz }| {

6, 2, 1, –3, 5,

mz}|{

−8 ,

ρz }| {

4, 11

24 / 28


Proposition

π = ΦB(T)

Row m is the topmost nonzerorow of TDecompose π = στmρ

Then,

−10

−9

−8

−6

–3−2

1

457

11

10 9 8 6 3 2

π =

σz }| {

9, 7, 10,

τz }| {

6, 2, 1, –3, 5,

mz}|{

−8 ,

ρz }| {

4, 11

24 / 28


Proposition

π = ΦB(T)


min(π) = m

Then,

−10

−9

−8

−6

–3−2

1

457

11

10 9 8 6 3 2

π =

σz }| {

9, 7, 10,

τz }| {

6, 2, 1, –3, 5,

mz}|{

−8 ,

ρz }| {

4, 11

24 / 28


Proposition

π = ΦB(T)


min(π) = mthe last element of σ is > |m|

Then,

−10

−9

−8

−6

–3−2

1

457

11

10 9 8 6 3 2

π =

σz }| {

9, 7, 10,

τz }| {

6, 2, 1, –3, 5,

mz}|{

−8 ,

ρz }| {

4, 11

24 / 28


Proposition

π = ΦB(T)


min(π) = mthe last element of σ is > |m|each element of τ is < |m|

Then,

−10

−9

−8

−6

–3−2

1

457

11

10 9 8 6 3 2

π =

σz }| {

9, 7, 10,

τz }| {

6, 2, 1, –3, 5,

mz}|{

−8 ,

ρz }| {

4, 11

24 / 28


Proposition

π = ΦB(T)



Then,1 Columns without

↔ negative integers in π

−10

−9

−8

−6

–3−2

1

457

11

10 9 8 6 3 2

π =

σz }| {

9, 7, 10,

τz }| {

6, 2, 1, –3, 5,

mz}|{

−8 ,

ρz }| {

4, 11

24 / 28


Proposition

π = ΦB(T)





2 Unrestricted rows of T↔ RL-minima of π

−10

−9

−8

−6

–3−2

1

457

11

10 9 8 6 3 2

π =

σz }| {

9, 7, 10,

τz }| {

6, 2, 1, –3, 5,

mz}|{

−8 ,

ρz }| {

4, 11

24 / 28


Proposition

π = ΦB(T)






3 Columns with in Row m↔ RL-maxima of σ

−10

−9

−8

−6

–3−2

1

457

11

10 9 8 6 3 2

π =

σz }| {

9, 7, 10,

τz }| {

6, 2, 1, –3, 5,

mz}|{

−8 ,

ρz }| {

4, 11

24 / 28


Proposition

π = ΦB(T)






3 Columns with in Row m↔ RL-maxima of σ

4 Rows with in Column |m|↔ RL-minima of τ

−10

−9

−8

−6

–3−2

1

457

11

10 9 8 6 3 2

π =

σz }| {

9, 7, 10,

τz }| {

6, 2, 1, –3, 5,

mz}|{

−8 ,

ρz }| {

4, 11

25 / 28

GeneralizationTheorem

X

T∈PT B(n)


25 / 28

GeneralizationTheorem

X

T∈PT B(n)


m

|m| {

y − 1

{

x − 1

26 / 28

Zigzag maps

0 0 1 0 0 1

0 0 0 1 1 1

0 0 0 0 10 1 1

10

12 9 8 6 5 31

2

4

7

10

11

13

26 / 28

Zigzag maps

0 0 1 0 0 1

0 0 0 1 1 1

0 0 0 0 10 1 1

10

12 9 8 6 5 31

2

4

7

10

11

13

12 9 8 6 5 31

2

4

7

10

11

13

26 / 28

Zigzag maps

0 0 1 0 0 1

0 0 0 1 1 1

0 0 0 0 10 1 1

10

12 9 8 6 5 31

2

4

7

10

11

13

12 9 8 6 5 31

2

4

7

10

11

13

12 9 8 6 5 31

2

4

7

10

11

13

26 / 28

Zigzag maps

0 0 1 0 0 1

0 0 0 1 1 1

0 0 0 0 10 1 1

10

12 9 8 6 5 31

2

4

7

10

11

13

12 9 8 6 5 31

2

4

7

10

11

13

12 9 8 6 5 31

2

4

7

10

11

13

TheoremThe zigzag map on the alternative representation is the same as ϕ ◦ Φ.

26 / 28

Zigzag maps

0 0 1 0 0 1

0 0 0 1 1 1

0 0 0 0 10 1 1

10

12 9 8 6 5 31

2

4

7

10

11

13

12 9 8 6 5 31

2

4

7

10

11

13

12 9 8 6 5 31

2

4

7

10

11

13


Example

26 / 28

Zigzag maps

0 0 1 0 0 1

0 0 0 1 1 1

0 0 0 0 10 1 1

10

12 9 8 6 5 31

2

4

7

10

11

13

12 9 8 6 5 31

2

4

7

10

11

13

12 9 8 6 5 31

2

4

7

10

11

13


Example

Φ(T) = 4, 6, 5, 2, 8, 3, 1, 9, 7, 11, 12, 10

26 / 28

Zigzag maps

0 0 1 0 0 1

0 0 0 1 1 1

0 0 0 0 10 1 1

10

12 9 8 6 5 31

2

4

7

10

11

13

12 9 8 6 5 31

2

4

7

10

11

13

12 9 8 6 5 31

2

4

7

10

11

13


Example

Φ(T) = 4, 6, 5, 2, 8, 3, 1, 9, 7, 11, 12, 10

ϕ ◦ Φ(T) = (4, 6, 5, 2, 8, 3, 1)(9, 7)(11, 12, 10)

27 / 28

The zigzag map on the type B alternative representation

−10

−9

−8

−6

−3

−2

1

4

5

7

11

10 9 8 6 3 2

TheoremThe zigzag map on the type B alternative representation is ϕ ◦ ΦB.

27 / 28


−10

−9

−8

−6

−3

−2

1

4

5

7

11

10 9 8 6 3 2


Example

27 / 28


−10

−9

−8

−6

−3

−2

1

4

5

7

11

10 9 8 6 3 2


Example

ΦB(T) = 9, 7, 10, 6, 2, 1,−3, 5, –8, 4, 11

27 / 28


−10

−9

−8

−6

−3

−2

1

4

5

7

11

10 9 8 6 3 2


Example

ΦB(T) = 9, 7, 10, 6, 2, 1,−3, 5, –8, 4, 11

ϕ ◦ ΦB(T) = (9, 7, 10, 6, 2, 1,−3, 5,−8)(4)(11)

28 / 28

Further study

Find a combinatorial proof of the following:

X

T∈PT (n)

sgn(T) =(1 + i)n + (1 − i)n

2=

8

<

:

(−1)k · 22k, if n = 4k or n = 4k + 1,

0, if n = 4k + 2,(−1)k+1 · 22k+1

, if n = 4k + 3.

28 / 28

Further study


X

T∈PT (n)

sgn(T) =(1 + i)n + (1 − i)n

2=

8

<

:

(−1)k · 22k, if n = 4k or n = 4k + 1,

0, if n = 4k + 2,(−1)k+1 · 22k+1

, if n = 4k + 3.

If n 6≡ 2 mod 4, then˛

˛

˛

˛

˛

˛

X

T∈PT (n)

sgn(T)

˛

˛

˛

˛

˛

˛

=

˛

˛

˛

˛

˛

˛

X

T∈SYT (n)

sgn(T)

˛

˛

˛

˛

˛

˛

= 2⌊n2⌋.

28 / 28

Further study


X

T∈PT (n)

sgn(T) =(1 + i)n + (1 − i)n

2=

8

<

:

(−1)k · 22k, if n = 4k or n = 4k + 1,

0, if n = 4k + 2,(−1)k+1 · 22k+1

, if n = 4k + 3.


˛

˛

˛

˛

˛

X

T∈PT (n)

sgn(T)

˛

˛

˛

˛

˛

˛

=

˛

˛

˛

˛

˛

˛

X

T∈SYT (n)

sgn(T)

˛

˛

˛

˛

˛

˛

= 2⌊n2⌋.

Find a type B analog of the above formula.

28 / 28

Further study


X

T∈PT (n)

sgn(T) =(1 + i)n + (1 − i)n

2=

8

<

:

(−1)k · 22k, if n = 4k or n = 4k + 1,

0, if n = 4k + 2,(−1)k+1 · 22k+1

, if n = 4k + 3.


˛

˛

˛

˛

˛

X

T∈PT (n)

sgn(T)

˛

˛

˛

˛

˛

˛

=

˛

˛

˛

˛

˛

˛

X

T∈SYT (n)

sgn(T)

˛

˛

˛

˛

˛

˛

= 2⌊n2⌋.

Find a type B analog of the above formula.

Thank you for your attention!

on permutation tableaux of type a and bslc/wpapers/s64vortrag/kim.pdf1 / 28 on permutation tableaux...

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