schützenberger's jeu de taquin · schützenberger’sjeudetaquin valentinféray institut für...

Schützenberger’s jeu de taquin

Valentin Féray

Institut für Mathematik, Universität Zürich

Conference dedicated tothe scientific legacy of Marcel-Paul Schützenberger

Bordeaux, March 22nd, 2016

V. Féray (Zurich) (I-Math, UZH)Jeu de taquin MPS 2016 1 / 14

Outline

1 Jeu de taquin

2 Knuth equivalence and Lascoux-Schützenberger’s plactic monoid

3 Motivation: product of Schur functions

4 Jeu de taquin in recent research work

Jeu de taquin Operation on tableaux

(Straight) Young diagram: set of boxespacked in the north-west corner.

(Straight) Young tableau: filling of aYoung diagram

weakly increasing along rows;strictly increasing along columns.

1 1 2 42 3 3 54 4 56

Skew diagram: difference setbetween two Young diagrams

Skew tableau: filling of a skewdiagram with the same condi-tions

2 43 3 54 5

(Straight) Young diagram: set of boxespacked in the north-west corner.

(Straight) Young tableau: filling of aYoung diagram

weakly increasing along rows;strictly increasing along columns.

1 1 2 42 3 3 54 4 56

Skew diagram: difference setbetween two Young diagrams

Skew tableau: filling of a skewdiagram with the same condi-tions

2 43 3 54 5

Jeu de taquin

Start with a skew tableau and color in red one of its inner corner.

2 43 3 54 5

7→2 43 3 54 5

7→2 3 43 54 5

7→2 3 43 5 54

Move the red box according to the following rule:

xy 7→

if x < y ;

y x if y ≤ x .

until the red box is outside the diagram; then erase the red box.

→ The rule ensures that we keep a tableau.

Jeu de taquin

2 43 3 54 5

7→2 43 3 54 5

7→2 3 43 54 5

7→2 3 43 5 54

xy 7→

if x < y ;

y x if y ≤ x .

Jeu de taquin

2 43 3 54 5

7→2 43 3 54 5

7→2 3 43 54 5

7→2 3 43 5 54

xy 7→

if x < y ;

y x if y ≤ x .

Jeu de taquin

2 43 3 54 5

7→2 43 3 54 5

7→2 3 43 54 5

7→2 3 43 5 54

xy 7→

if x < y ;

y x if y ≤ x .

Rectification of a skew-tableau

Start with a skew tableau T . Successively choose inner corners and applyjeu de taquin until you get a tableau.

2 43 3 54 5

7→2 3 43 5 54

7→2 3 43 5 5

2 4 7→2 3 4

2 3 5 54 7→

2 2 3 43 5 54

TheoremThe final straight tableau does not depend on the order in which we chooseinner corners.

This final tableau is called the rectification of T and denoted Rect(T ).

2 43 3 54 5

7→2 3 43 5 54

7→2 3 43 5 5

2 4 7→2 3 4

2 3 5 54 7→

2 2 3 43 5 54

2 43 3 54 5

7→2 3 43 5 54

7→2 3 43 5 5

7→2 3 4

2 3 5 54 7→

2 2 3 43 5 54

2 43 3 54 5

7→2 3 43 5 54

7→2 3 43 5 5

2 4 7→2 3 4

2 3 5 54 7→

2 2 3 43 5 54

2 43 3 54 5

7→2 3 43 5 54

7→2 3 43 5 5

2 4 7→2 3 4

2 3 5 54 7→

2 2 3 43 5 54

Glimpse in the proof Knuth equivalence and the plactic monoid

Reading word of a skew tableau (from bottom to top, from left to right):

2 43 3 54 5

7→ w(T ) = 24533524.

DefinitionTwo words are Knuth equivalent if one can be obtained from the other byelementary Knuth transforms

y z x ≡ y x z (if x < y ≤ z), x z y ≡ z x y (if x ≤ y < z).

xy z 7→ x z

x z 7→ x yz

yzx ≡ yxz xzy ≡ zxy

2 43 3 54 5

7→ w(T ) = 24533524.

xy z 7→ x z

x z 7→ x yz

yzx ≡ yxz xzy ≡ zxy

2 43 3 54 5

7→ w(T ) = 24533524.

LemmaIf one obtains T ′ from T by jeu de taquin, then w(T ) and w(T ′) areKnuth equivalent.

2 43 3 54 5

7→ w(T ) = 24533524.

The monoid of words up to Knuth equivalence is the plactic monoid,studied by Lascoux and Schützenberger.

Glimpse in the proof Back to rectification

Uniqueness of the rectification

TheoremEach Knuth equivalence class contains the reading word of a uniquestraight tableau.

Corollary

Rect(T ) is the unique straight tableau whose reading word is in the sameclass as w(T ) (and thus does not depend on the successive choices ofcorners).

Motivation: product of Schur functions Definitions

If T is a tableau, write xT = x#1 in T1 · x#2 in T

2 · · ·

1 1 2 32 3 3 54 5 56

, xT = x21 x2

2 x33 x4 x

35 x6.

Then if λ is a diagram, the associated Schur function is

sλ :=∑T

where the sum is taken over all tableaux of shape λ.

Lemmasλ is symmetric and the family (sλ)λ straight is a basis of the symmetricfunction ring.

Problem

Describe Littlewood-Richardson coefficients cλµ,ν defined by

sµsν =∑

λ cλµ,νsλ.

2 · · ·

1 1 2 32 3 3 54 5 56

, xT = x21 x2

2 x33 x4 x

35 x6.

sλ :=∑T

Problem

sµsν =∑

λ cλµ,νsλ.

2 · · ·

1 1 2 32 3 3 54 5 56

, xT = x21 x2

2 x33 x4 x

35 x6.

sλ :=∑T

Problem

sµsν =∑

λ cλµ,νsλ.

Motivation: product of Schur functions Product of tableaux

Tableaux

∗ product

T 7→xT //

T 7→w(T ) &&

PolynomialsUsual multiplication

Plactic monoidConcatenation

Tableaux∗ product

T 7→xT //

T 7→w(T ) &&

Definition

T ∗U :=Rect(�UT

). Example: 1 2 2

2 4 ∗ 1 32 = Rect

1 2 22 4

Tableaux∗ product

T 7→xT //

T 7→w(T ) &&

Proposition

Denote Sµ =∑

T ofshape µ

T . In the algebra of tableaux

Sµ ∗ Sν =∑λ

cλµ,ν Sλ.

→ enables to describe combinatorially products of Schur functions.

Remarkable fact: the algebra of the plactic monoid contains the symmetricfunction ring as a subalgebra.

Tableaux∗ product

T 7→xT //

T 7→w(T ) &&

Proposition

Denote Sµ =∑

T ofshape µ

T . In the algebra of tableaux

Sµ ∗ Sν =∑λ

cλµ,ν Sλ.

→ enables to describe combinatorially products of Schur functions.

Remarkable fact: the algebra of the plactic monoid contains the symmetricfunction ring as a subalgebra.

Jeu de taquin in recent research work

1 in algebraic combinatorics (Pechenik, Yong 2015);

2 in bijective combinatorics (Fang 2015);

3 in probabilistic combinatorics (Romik, Śniady 2015).

Jeu de taquin in recent research work Algebraic combinatorics

Pechenik-Yong’s combinatorial rule

Littlewood-Richardson coefficients also appear in the “cohomologyring” of the Grassmannian.

Pechenik and Yong (2015) consider an analogue ofLittlewood-Richardson coefficients in the “equivariant K-theory” of theGrassmannian.

They find a combinatorial rule using enriched tableaux and adaptingjeu de taquin to these tableaux.

Jeu de taquin in recent research work Algebraic combinatorics

Pechenik-Yong’s tableaux and jeu de taquin

12 21 22

21 3221 32 42

T (11) =

•11 12•11 11

•11 11

11 1211 •12

11 •12

11 •1211 •12

11 •12

11 1211 •21

11 •21

11 •2111 •21

11 •21

11 1211

H5.3/B2/T3

t1t2· t3t4 ·

(− t5

t1t2· t3t4 ·

∅/B1/∅H3/∅/∅H3/∅/∅

H1/∅/∅H3/∅/∅H3/∅/∅

1− t6t7

Delete •’s

→ jeu de taquin is a robust technique!

Jeu de taquin in recent research work Bijective combinatorics

Bijective proof of character identities (Fang)

Denote f λ the number of standard tableaux of shape λ.

Theorem (Frobenius, Murnaghan-Nakayama)

2 f λ

∑(i ,j)∈λ

(j − i)

)(f λ/(2) − f λ/(1,1)

Both sides compute symmetric group characters on transpositions. . .

Fang (2015) gave a bijective proof using jeu de taquin

1 2 5 9 113 7 104 8 136 12

c©Fang

Jeu de taquin in recent research work Bijective combinatorics

Bijective proof of character identities (Fang)

Denote f λ the number of standard tableaux of shape λ.

Theorem (Frobenius, Murnaghan-Nakayama)

2 f λ

∑(i ,j)∈λ

(j − i)

)(f λ/(2) − f λ/(1,1)

Both sides compute symmetric group characters on transpositions. . .

Fang (2015) gave a bijective proof using jeu de taquin

1 2 5 9 113 7 104 8 136 12

c©Fang

Jeu de taquin in recent research work Probabilistic combinatorics

Jeu de taquin on random infinite tableaux (Romik, Śniady)

A random infinite tableau.

Romik and Śniady (2015) show that the jeu de taquin paths arealmost surely asymptotically straight lines,with a random direction whose distribution is explicit.

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