September,2011 O.Azenhas,A.Conﬂitti,R.Mamede RibbonSchurfunctionswithfullsupportandSchurpositivity

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Ribbon Schur functions with full support

Ribbon Schur functions with full support and Schur positivity

O. Azenhas, A. Conflitti, R. Mamede

CMUC, Universidade de Coimbra 67th S´eminaire Lotharingien de Combinatoire

Joint session with

XVII Incontro Italiano di Combinatoria Algebrica Bertinoro

September, 2011

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Outline

1 Maximal support and Schur positivity

2 Classification of ribbon Schur functions with interval support

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Ribbon Schur functions with full support Maximal support and Schur positivity

Schur functions and support

The Schur functions are considered to be the most important basis for the ring of symmetric functions.

Given partitionsµ⊆λ,A:=λ/µ s_A=X

ν

c_A^νs_ν,

where c_A^ν is the number of SSYT of shapeAand contentν, satisfying the Littlewood-Richardson rule.

A= 3321/211 =

s_A=s₂₁₁₁+2s₂₂₁+s₃₁₁+s₃₂

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Schur functions and support

The Schur functions are considered to be the most important basis for the ring of symmetric functions.

Given partitionsµ⊆λ,A:=λ/µ s_A=X

ν

c_A^νs_ν,

where c_A^ν is the number of SSYT of shapeAand contentν, satisfying the Littlewood-Richardson rule.

A= 3321/211 =

s_A=s₂₁₁₁+2s₂₂₁+s₃₁₁+s₃₂

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Schur functions and support

A= 3321/211 =

sA=s2111+2s221+s311+s32

r(A) is the partition consisting of the row lengths of A, and c(A) is defined similarly. The support of A, considered as a subposet of the dominance lattice, has a top elementr(A)⁰ and a bottom element c(A),

sA= X

c(A)ν⁰r(A)⁰

c_A^νsν.

supp(A) ={ν⁰ :c_A^ν >0} ⊆[c(A),r(A)⁰] =

[221; 41]

1 1 2 2 1

1 1 2 3

4 .

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Schur functions and support

A= 3321/211 =

sA=s2111+2s221+s311+s32

sA= X

c(A)ν⁰r(A)⁰

c_A^νsν.

supp(A) ={ν⁰ :c_A^ν>0}

⊆[c(A),r(A)⁰] =

[221; 41]

1 1 2 2 1

1 1 2 3

4 .

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Schur functions and support

A= 3321/211 =

sA=s2111+2s221+s311+s32

sA= X

c(A)ν⁰r(A)⁰

c_A^νsν.

supp(A) ={ν⁰ :c_A^ν>0} ⊆[c(A),r(A)⁰] =

[221; 41]

1 1 2 2 1

1 1 2 3

4 .

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Schur positivity

Given skew shapes A and B, when is s_A−s_B Schur positive?

sA−sB is Schur positive only ifsupp(B)⊆supp(A).

A= 3321/211 = ,B= 3311/21 =

sA =s32+s211+2s221+s311, sB =s32+s211+1s221+s311

suppA=suppB,

sA−sB =s221is Schur positive butsB−sA=−1s221is not.

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Schur positivity

A= 3321/211 = ,B= 3311/21 =

sA =s32+s211+2s221+s311, sB =s32+s211+1s221+s311

suppA=suppB,

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Schur positivity

A= 3321/211 = , B= 3311/21 =

sA =s32+s211+2s221+s311, sB =s32+s211+1s221+s311

suppA=suppB,

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Skew shape equivalences

Skew shapes yielding the same Schur function AanB are said to be Schur equivalentifsA=sB

[A] ={B :s_A=s_B}

A= andB = are not Schur equivalent but

Aand its antipodal rotationA^π= are.

Skew shapes yielding the same support

AanB are said to be support equivalentif suppA=suppB bAc={B:suppB =suppA}

A,B andA^π are support equivalent.

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Skew shape equivalences

Skew shapes yielding the same Schur function AanB are said to be Schur equivalentifsA=sB

[A] ={B :s_A=s_B}

A= andB = are not Schur equivalent but

Aand its antipodal rotationA^π= are.

Skew shapes yielding the same support

AanB are said to be support equivalentif suppA=suppB bAc={B:suppB =suppA}

A,B andA^π are support equivalent.

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Partial orders on skew shape classes

PN is the poset of all Schur equivalence classes [A] such thatAhas N boxes.

[A]≥s [B] if sA−sB is Schur positive

Supp_N is the poset of all support equivalence classesbAcsuch that AhasN boxes.

bAc ≥supp bBcif the support ofB is contained in that of A

A= andB =

[B]<s [A] in P5

bBc=bAc in Supp5

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Maximal supports among connected skew shapes

InMaximal supports and Schur-positivity among connected skew shapes arXiv:1107.4373 P. R. W. MacNamara, S. van Willigenburg classify the maximal connected skew shapes ofSupp_N.

Theorem

An elementbRcinSupp_N is a maximal connected element iff R is a ribbon in which the lengths of any two empty rows differ by at most one and the lengths of any two nonempty columns differ by at most one.

ThesuppR is the full interval.

R= (2,3,3) = bRc=bR^πc

R⁰ = bRc⁰ =bR⁰c

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Maximal supports among connected skew shapes

InMaximal supports and Schur-positivity among connected skew shapes arXiv:1107.4373 P. R. W. MacNamara, S. van Willigenburg classify the maximal connected skew shapes ofSupp_N.

Theorem

An elementbRcinSupp_N is a maximal connected element iff R is a ribbon in which the lengths of any two empty rows differ by at most one and the lengths of any two nonempty columns differ by at most one.

ThesuppR is the full interval.

R= (2,3,3) = bRc=bR^πc

R⁰ = bRc⁰ =bR⁰c

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Ribbon shapes with full support

PROBLEM: What are the ribbon shapesR= (r₁, . . . ,r_s) whose support consists of the whole interval in the dominance lattice?

What are the ribbon shapes R= (r1, . . . ,rs) with suppR= [(rk₁, . . . ,rk_s); (P

j≥1rj−s+ 1,s−1)]? s−1 the number of rows with length two in R

R= (2,3,3) =

X X

suppR= [3²2; 6 2]

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Ribbon shapes with full support

PROBLEM: What are the ribbon shapesR= (r₁, . . . ,r_s) whose support consists of the whole interval in the dominance lattice?

What are the ribbon shapes R= (r1, . . . ,rs) with suppR= [(rk₁, . . . ,rk_s); (P

j≥1rj−s+ 1,s−1)]?

s−1 the number of rows with length two in R

R= (2,3,3) =

X X

suppR= [3²2; 6 2]

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R= (32522271)

(7 5 3 2 2 2 2 1)ξ= (8 8 8)(24−7,7).

8 7 6 54 3 2

1

8 7 6

7 6 54 1

8

54 3 2

3 21

8 7 6

1

8 7 654321

54 3 2

There is vertical space to put the last string of length 8. P2

i=1ξi−P2

i=1ri= (8 + 8)−(7 + 5) = 4>p= 3, ξ3= 8<P

i≥3ri−p= 3 + 2 + 2 + 2 + 2 + 1−3, (8 8 8)∈supp(R) ξ3= 8 = (P

i≥3ri−p)−1

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R= (32522271)

(7 5 3 2 2 2 2 1)ξ= (8 8 8)(24−7,7).

8 7 6 54 3 2

1

8 7 6

7 6 54 1

8

54 3 2

3 21

8 7 6

1

8 7 654321

54 3 2

i=1ξi−P2

i=1ri= (8 + 8)−(7 + 5) = 4>p= 3, ξ3= 8<P

i≥3ri−p= 3 + 2 + 2 + 2 + 2 + 1−3, (8 8 8)∈supp(R) ξ3= 8 = (P

i≥3ri−p)−1

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R= (32522271)

(7 5 3 2 2 2 2 1)ξ= (8 8 8)(24−7,7).

8 7 6 54 3 2

1

8 7 6

7 6 54 1

8

54 3 2

3 21

8 7 6

1

8 7 654321

54 3 2

i=1ξi−P2

i=1ri= (8 + 8)−(7 + 5) = 4>p= 3, ξ3= 8<P

i≥3ri−p= 3 + 2 + 2 + 2 + 2 + 1−3, (8 8 8)∈supp(R) ξ3= 8 = (P

i≥3ri−p)−1

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R= (32522271)

(7 5 3 2 2 2 2 1)ξ= (8 8 8)(24−7,7).

8 7 6 54 3 2

1

8 7 6

7 6 54 1

8

54 3 2

3 21

8 7 6

1

8 7 654321

54 3 2

i=1ξi−P2

i=1ri= (8 + 8)−(7 + 5) = 4>p= 3, ξ3= 8<P

i≥3ri−p= 3 + 2 + 2 + 2 + 2 + 1−3, (8 8 8)∈supp(R) ξ3= 8 = (P

i≥3ri−p)−1

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R= (32522271)

(7 5 3 2 2 2 2 1)ξ= (8 8 8)(24−7,7).

8 7 6 54 3 2

1

8 7 6

7 6 54 1

8

54 3 2

3 21

8 7 6

1

8 7 654321

54 3 2

i=1ξi−P2

i=1ri= (8 + 8)−(7 + 5) = 4>p= 3, ξ3= 8<P

i≥3ri−p= 3 + 2 + 2 + 2 + 2 + 1−3, (8 8 8)∈supp(R) ξ3= 8 = (P

i≥3ri−p)−1

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R= (32522271)

(7 5 3 2 2 2 2 1)ξ= (8 8 8)(24−7,7).

8 7 6 54 3 2

1

8 7 6

7 6 54 1

8

54 3 2

3 21

8 7 6

1

8 7 654321

54 3 2

There is vertical space to put the last string of length 8.

P2

i=1ξi−P2

i=1ri= (8 + 8)−(7 + 5) = 4>p= 3, ξ3= 8<P

i≥3ri−p= 3 + 2 + 2 + 2 + 2 + 1−3, (8 8 8)∈supp(R) ξ3= 8 = (P

i≥3ri−p)−1

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R

= (662322) (6

²

3 2

³

) (7 7 7) (21, 21

−

5)

1 2 3 4 5 6 1 2 3 4 5 7 6 7

1 2 3 4 5 6 1 2 3 4 5 6 7 7

There are enough boxes to put the last string of length 7 but not enough vertical space: a row of length two remains.

ξ1−r1+ξ2−r2= 7−6+7−6≤3−1 p= 3 ξ3= 7≥2+3+2+2−2 ξ= (777)∈/ suppR

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R

= (662322) (6

²

3 2

³

) (7 7 7) (21, 21

−

5) 1 2

3 4 5 6 1 2 3 4 5 7 6 7

1 2 3 4 5 6 1 2 3 4 5 6 7 7

ξ1−r1+ξ2−r2= 7−6+7−6≤3−1 p= 3 ξ3= 7≥2+3+2+2−2 ξ= (777)∈/ suppR

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R

= (662322) (6

²

3 2

³

) (7 7 7) (21, 21

−

5) 1 2

3 4 5 6 1 2 3 4 5 7 6 7

1 2 3 4 5 6 1 2 3 4 5 6 7 7

ξ1−r1+ξ2−r2= 7−6+7−6≤3−1 p= 3 ξ3= 7≥2+3+2+2−2 ξ= (777)∈/ suppR

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R

= (662322) (6

²

3 2

³

) (7 7 7) (21, 21

−

5) 1 2

3 4 5 6 1 2 3 4 5 7 6 7

1 2 3 4 5 6 1 2 3 4 5 6 7 7

ξ1−r1+ξ2−r2= 7−6+7−6≤3−1 p= 3 ξ3= 7≥2+3+2+2−2 ξ= (777)∈/ suppR

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R

= (662322) (6

²

3 2

³

) (7 7 6 1) (7 7 7) (21, 21

−

5)

1 2 3 4 5 6 1 2 3 4 5 7

6 7

1 2 3 4 5 6 1 2 3 4 5 7 6 7

21 43 65

There is not enough vertical space to put a third string of length 7 but there is enough vertical space to put two more strings: one of length 6 and another of length 1.

ξ1−r1+ξ2−r2= 7−6+7−6 = 2 p= 3 ξ3= 6 = 2+3+2+2−3 (777)∈/ suppR (7761)∈suppR

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R

= (662322) (6

²

3 2

³

) (7 7 6 1) (7 7 7) (21, 21

−

5) 1 2

3 4 5 6 1 2 3 4 5 7

6 7

1 2 3 4 5 6 1 2 3 4 5 7 6 7

21 43 65

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R

= (662322) (6

²

3 2

³

) (7 7 6 1) (7 7 7) (21, 21

−

5) 1 2

3 4 5 6 1 2 3 4 5 7

6 7

1 2 3 4 5 6 1 2 3 4 5 7 6 7

21 43 65

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R

= (662322) (6

²

3 2

³

) (7 7 6 1) (7 7 7) (21, 21

−

5) 1 2

3 4 5 6 1 2 3 4 5 7

6 7

1 2 3 4 5 6 1 2 3 4 5 7 6 7

21 43 65

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R

= (662322) (6

²

3 2

³

) (7 7 6 1) (7 7 7) (21, 21

−

5) 1 2

3 4 5 6 1 2 3 4 5 7

6 7

1 2 3 4 5 6 1 2 3 4 5 7 6 7

21 43 65

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R= (662322) (6²3 2³)(7 7 6 1)(7 7 7)(8 7 6) (21,21−5)

1 2 3 45 6 1 2 34 7 8 6 7 5

1 2 3 45 6 1 2 34 7 8

76 5

1 2 34 5 6 1 23 4 7 8

7 6 5

65 4 321

ξ1−r1+ξ2−r2= 8−6+7−6 = 3 =p= 3 ξ3= 6 = 2+3+2+2−3 (777)∈/ suppR, (7761)∈suppR, (8 7 6)∈suppR

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R= (662322) (6²3 2³)(7 7 6 1)(7 7 7)(8 7 6) (21,21−5)

1 2 3 45 6 1 2 34 7 8 6 7 5

1 2 3 45 6 1 2 34 7 8

76 5

1 2 34 5 6 1 23 4 7 8

7 6 5

65 4 321

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R= (662322) (6²3 2³)(7 7 6 1)(7 7 7)(8 7 6) (21,21−5)

1 2 3 45 6 1 2 34 7 8 6 7 5

1 2 3 45 6 1 2 34 7 8

76 5

1 2 34 5 6 1 23 4 7 8

7 6 5

65 4 321

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R= (662322) (6²3 2³)(7 7 6 1)(7 7 7)(8 7 6) (21,21−5)

1 2 3 45 6 1 2 34 7 8 6 7 5

1 2 3 45 6 1 2 34 7 8

76 5

1 2 34 5 6 1 23 4 7 8

7 6 5

65 4 321

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R= (662322) (6²3 2³)(7 7 6 1)(7 7 7)(8 7 6) (21,21−5)

1 2 3 45 6 1 2 34 7 8 6 7 5

1 2 3 45 6 1 2 34 7 8

76 5

1 2 34 5 6 1 23 4 7 8

7 6 5

65 4 321

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R= (662322) (6²3 2³)(7 7 6 1)(7 7 7)(8 7 6) (21,21−5)

1 2 3 45 6 1 2 34 7 8 6 7 5

1 2 3 45 6 1 2 34 7 8

76 5

1 2 34 5 6 1 23 4 7 8

7 6 5

65 4 321

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Ribbon shape LR fillings

Lemma

Letξ= (ξ1, . . . , ξt) be a partition in the Schur interval [(rk₁, . . . ,rk_s); (P

j≥1rj−s+ 1,s−1)] but not in the support ofR.

Then there exists an 1≤i ≤t−1 such that ifp≥1 is the number of rows with length two among the columns indexed byS ={ki+1, . . . ,ks}, one has

ξi+1≥X

q∈S

rq−p+ 1



⇒

i

X

j=1

(ξj−rk_j)≤p−1



. (1) This implies that the numberpof rows of length two, among the

adjacent columns indexed byS in R, can not be shortened by what remainsPi

j=1(ξj−rk_j).

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Classification of ribbon Schur functions with interval support

Theorem

LetR= (r1, . . . ,rs),s≥2, be a ribbon. Then suppR$[(rk₁, . . . ,rk_s); (P

j≥1

rj−s+ 1,s−1)] if and only if for some 1≤i≤s−2 withp>0 rows of length two among the columns indexed by{ki+1, . . . ,ks}, there existg1, . . . ,gi ≥0 withPi

j=1gj =p−1, such that

r_k₁+g₁≥

s

X

j=i+1

r_k_j −p+ 1 ...

r_k_i−1+g_i−1≥

s

X

j=i+1

r_k_j −p+ 1

r_k_i+g_i≥

s

X

j=i+1

r_k_j −p+ 1

Moreover (rk1+g1, . . . ,rk_i +gi,Ps

j=i+1rk_j −p+ 1)≥ ∈/ suppR.

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R

= (662322) (6

²

3 2

³

) (7 7 7) (21, 21

−

5)

1 2 3 4 5 6 1 2 3 4 5 7 6 7

1 2 3 4 5 6 1 2 3 4 5 6 7 7

ξ

= (777)

∈/suppR

ξ₁−r₁

+ξ

₂−r₂

= 7

−6+7−6≤

3−1

ξ₃

= 7

≥

2+3+2+2−2

r1

+ 1 = 6 + 1

≥

2 + 3 + 2 + 2

−

2

r2

+ 1 = 6 + 1

≥

2 + 3 + 2 + 2

−

2 (876)

∈suppR

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R

= (662322) (6

²

3 2

³

) (7 7 7) (21, 21

−

5) 1 2

3 4 5 6 1 2 3 4 5 7 6 7

1 2 3 4 5 6 1 2 3 4 5 6 7 7

ξ

= (777)

∈/suppR

ξ₁−r₁

+ξ

₂−r₂

= 7

−6+7−6≤

3−1

ξ₃

= 7

≥

2+3+2+2−2

r1

+ 1 = 6 + 1

≥

2 + 3 + 2 + 2

−

2

r2

+ 1 = 6 + 1

≥

2 + 3 + 2 + 2

−

2 (876)

∈suppR

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R

= (662322) (6

²

3 2

³

) (7 7 7) (21, 21

−

5) 1 2

3 4 5 6 1 2 3 4 5 7 6 7

1 2 3 4 5 6 1 2 3 4 5 6 7 7

ξ

= (777)

∈/suppR

ξ₁−r₁

+ξ

₂−r₂

= 7

−6+7−6≤

3−1

ξ₃

= 7

≥

2+3+2+2−2

r1

+ 1 = 6 + 1

≥

2 + 3 + 2 + 2

−

2

r2

+ 1 = 6 + 1

≥

2 + 3 + 2 + 2

−

2 (876)

∈suppR

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R

= (662322) (6

²

3 2

³

) (7 7 7) (21, 21

−

5) 1 2

3 4 5 6 1 2 3 4 5 7 6 7

1 2 3 4 5 6 1 2 3 4 5 6 7 7

ξ

= (777)

∈/suppR

ξ₁−r₁

+ξ

₂−r₂

= 7

−6+7−6≤

3−1

ξ₃

= 7

≥

2+3+2+2−2

r1

+ 1 = 6 + 1

≥

2 + 3 + 2 + 2

−

2

r2

+ 1 = 6 + 1

≥

2 + 3 + 2 + 2

−

2 (876)

∈suppR

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Examples

Ribbons whose column and row lengths differ in one unity have full support

[(t

^m,

(t

−

1)

ⁿ

); (mt +

n(t−

1)

−m−n

+ 1,

m

+

n−

1)].

The support of a ribbon

R

= (r

₁,r₂,r₃

) has full interval except when

r

= (r

1,r2,r3

) or

r

= (r

2,r3,r1

) with

r1 ≥r2

+

r3

.

R

= (6222276), (7662222).

6 + 2

≥

2 + 2 + 2 + 2

−

2 7, 6

≥

2 + 2 + 2 + 2

−

2

.

Then

ξ

= (6 + 2, 7, 6, 2 + 2 + 2 + 2

−

2)

∈/ supp(R).

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1 O. Azenhas, The admissible interval for the invariant factors of a product of matrices, Linear and Multilinear Algebra 46 (1999), 51-99.

2 P. R. W. McNamara, Necessary conditions for Schur-positivity, J Algebr Comb 28 (2008), 495-507.

3 P. R. W. MacNamara, S. van Willigenburg, Maximal supports and Schur-positivity among connected skew shapes, arXiv:1107.4373 [math.CO, RT]

4 P. R. W. MacNamara, S. van Willigenburg,Maximal supports and Schur-positivity among connected skew shapesarXiv:1107.4373

5 S. van Willigenburg, Equality of Schur and skew Schur functions, Ann. Comb. 9 (2005), 355–362 .

6 I. Zaballa. Increasing and Decreasing Littlewood-Richardson Sequences and Duality, preprint, University of Basque Country, 1996.