Equality of multiplicity free skew characters

DOI 10.1007/s10801-008-0158-8

Equality of multiplicity free skew characters

Christian Gutschwager

Received: 12 June 2008 / Accepted: 9 October 2008 / Published online: 3 November 2008

© Springer Science+Business Media, LLC 2008

Abstract In this paper we show that two skew diagramsλ/μandα/βcan represent the same multiplicity free skew character[λ/μ] = [α/β]only in the the trivial cases whenλ/μandα/βare the same up to translation or rotation or ifλ=αis a staircase partitionλ=(l, l−1, . . . ,2,1)andλ/μandα/βare conjugate of each other.

Keywords Skew characters·Symmetric group·Skew Schur functions·Schubert calculus

1 Introduction

The question under which circumstances two different skew diagramsλ/μandα/β give rise to the same skew character[λ/μ] = [α/β]has lately received much attention and is by [1, Theorem 4.2] equivalent to the question under which circumstances two products of Schubert classesσ_α1·σ_α2 andσ_β1·σ_β2 are equal.

Trivial cases for equality of skew characters[λ/μ] = [α/β]are given if the skew diagramsλ/μandα/β are the same up to translation or rotation.

In [5] and later in [4] a method for constructing skew diagrams with nontrivial equality of their corresponding skew characters was presented. The fact that for a staircase partitionλ=(l, l−1, . . . ,2,1)the skew diagram λ/μand its conjugate (λ/μ)give rise to the same skew character was proved in [5, Theorem 7.32].

On the other hand new necessary conditions for two skew diagramsλ/μandα/β to give rise to the same skew character have been given recently in [3] and [2].

In [1] we classified the skew diagrams λ/μ whose corresponding skew char- acter [λ/μ] is multiplicity free which means that in the decomposition [λ/μ] = c(λ;ν, μ)[ν]all coefficientsc(λ;ν, μ)are either 0 or 1.

C. Gutschwager (

Institut für Algebra, Zahlentheorie und Diskrete Mathematik, Leibniz Universität Hannover, Welfengarten 1, 30167 Hannover, Germany

e-mail:[email protected]

Using algebraic arguments Reiner et al. showed in [5] that equality of [λ/μ] = [α/β]whenλ/μdecays into partitions is only possible in the trivial cases.

We will examine in this paper the case whenλ/μandα/β are connected skew diagrams and[λ/μ] = [α/β]is multiplicity free. We will see that the only nontrivial case for[λ/μ] = [α/β]is whenλ=αis a staircase partition and additionallyλ/μ= (α/β).

2 Notation and Littlewood-Richardson-symmetries

We mostly follow the standard notation in [6] or [7]. A partitionλ=(λ1, λ2, . . . , λl) is a weakly decreasing sequence of non-negative integers where only finitely many of theλ_i are positive. We regard two partitions as the same if they differ only by the number of trailing zeros and call the positive λ_i the parts ofλ. The length is the number of positive parts and we writel(λ)=l for the length and |λ| =

iλi

for the sum of the parts. With a partitionλwe associate a diagram, which we also denote byλ, containingλ_ileft-justified boxes in thei-th row and we use matrix-style coordinates to refer to the boxes.

Definition 2.1 We will call a partitionλwithndifferent parts adp=npartition or writedp(λ)=n.

On the set of partitions we define the lexicographic order in the following way.

For two partitionsλ, μ, we sayλ < μif there is aniwithλ_i< μ_i and forj < iwe haveλj=μj.

The conjugateλofλis the diagram which hasλ_i boxes in thei-th column.

Forμ⊆λwe define the skew diagramλ/μas the difference of the diagramsλand μdefined as the difference of the set of the boxes. Rotation ofλ/μby 180^◦yields a skew diagram(λ/μ)^◦which is well defined up to translation. A skew tableauT is a skew diagram in which positive integers are written into the boxes. We refer with T (i, j )to the entry in box(i, j ). A semistandard tableau of shapeλ/μis a filling of λ/μwith positive integers such that the following expressions hold for all(i, j )for which they are defined:T (i, j ) < T (i+1, j )andT (i, j )≤T (i, j+1). The content of a semistandard tableauT isν=(ν₁, . . .)if the number of occurrences of the entry iinT isν_i. The reverse row word of a tableauT is the sequence obtained by reading the entries ofT from right to left and top to bottom starting at the first row. Such a sequence is said to be a lattice word if for alli, n≥1 the number of occurrences of iamong the firstnterms is at least the number of occurrences ofi+1 among these terms. The Littlewood-Richardson (LR) coefficientc(λ;μ, ν)equals the number of semistandard tableaux of shapeλ/μwith contentνsuch that the reverse row word is a lattice word. We will call those tableaux LR tableaux. The LR coefficients play an important role in different contexts (see [6] or [7]).

The irreducible characters[λ]of the symmetric groupSnare naturally labeled by partitionsλn. The skew character[λ/μ]of a skew diagramλ/μis defined by the LR coefficients:

[λ/μ] =

ν

c(λ;μ, ν)[ν].

There are many known symmetries of the LR coefficients.

We have thatc(λ;μ, ν)=c(λ;ν, μ). Translation symmetry gives[λ/μ] = [α/β] if the skew diagrams ofλ/μandα/β are the same up to translation while rotation symmetry gives[(λ/μ)^◦] = [λ/μ]. Another well known symmetry is the conjugation symmetryc(λ;μ, ν)=c(λ;μ, ν).

We say that a skew diagramDdecays into the disconnected skew diagramsAand Bif no box ofA(viewed as boxes inD) is in the same row or column as a box ofB.

We writeD=A⊗BifDdecays intoAandB. A skew diagram is connected if it does not decay.

A skew character whose skew diagramDdecays into disconnected (skew) dia- gramsA,Bis equivalent to the product of the characters of the disconnected dia- grams induced to a larger symmetric group. We have

[D] =([A] × [B])↑^S_Sⁿ_n⁺_×^m_Sm=: [A] ⊗ [B]

with|A| =n,|B| =m. IfD=λ/μandA,Bare proper partitionsα, βwe have:

[λ/μ] =

ν

c(λ;μ, ν)[ν] =

ν

c(ν;α, β)[ν] = [α] ⊗ [β].

The product of two Schubert classes σα, σβ in the cohomology ring H^∗(Gr(l,Cⁿ),Z)of the GrassmannianGr(l,Cⁿ)ofl-dimensional subspaces ofCⁿ is given by:

σ_α·σ_β=

ν⊆((n−l)^l)

c(ν;α, β)σ_ν.

In [1, Section 4] we established a close connection between the Schubert product and skew characters. To use this relation later on we define the Schubert product for characters in the obvious way as a restriction of the ordinary product:

[α]_(kl)[β] :=

ν⊆(k^l)

c(ν;α, β)[ν].

Since translation ofλ/μdoes not change the corresponding skew character, we may assume thatμi< λi,μi ≤λi+1for each 1≤i≤l(λ), which means thatλ/μ doesn’t have empty rows or columns. We call such a skew diagram a basic skew diagram.

For a basic skew diagramλ/μ we define two lattice paths from the lower left corner to the upper right corner. The outer lattice path starts to the right, follows the shape ofλand ends upwards in the corner, while the inner lattice path starts upwards, follows the shape ofμand ends with a segment to the right. Withs_inwe refer to the length of the shortest straight segment of the inner lattice path whiles_outis the length of the shortest straight segment of the outer lattice path.

In [1, Theorem 3.8] we classified the multiplicity free skew characters.

Theorem 2.2 ([1, Theorem 3.8]) Letλ/μbe a basic skew diagram which is neither a partition nor a rotated partition. Then[λ/μ]is multiplicity free if and only if up

to rotation ofλ/μ μis a rectangle and additionally one of the following conditions holds:

1. sin=1

2. sin=2 andλis adp=3 partition 3. λis adp=3 partition andsout=1 4. λis adp=2 partition.

3 Equality of skew characters

3.1 Notation and preliminary

In this section we prove that two connected skew diagrams λ/μ,α/β which give rise to the same multiplicity free skew character have (up to translation or rotation) to be the same or bothλandαare the same staircase partitionλ=α=(n, n−1, n−2, . . . ,2,1)and additionallyμis the conjugate ofβ:μ=β.

Reiner et al. proved in [5, Section 6] the following, using the Jacobi-Trudi deter- minant but no LR combinatorics:

Lemma 3.1 ([5, Section 6]) LetA¹ andA² be skew diagrams andA¹ decay into partitions. Let[A¹] = [A²].

Then the equality is trivial i.e. up to translation or rotationA¹ andA² are the same.

Here translation or rotation does not require the entire skew diagram to be trans- lated or rotated but also includes the case when the partitions into whichA¹decays are translated or rotated independent of each other.

We will use this lemma always to argue thatA¹=γ⊗δwith partitionsγ , δand [A¹] = [A²]requiresA²=γ⊗δ.

To exclude the trivial cases for[λ/μ] = [α/β]we will in the following always assume thatμis a rectangle and bothλ/μandα/βare basic skew diagrams.

Furthermore we will use the following notation:

λ=(λ^l₁¹, λ^l₂², . . .)withl=l(λ)=

il_i andλ_i> λ_i+1,μ=(μ^m₁), α=(α^a₁¹, α^a₂², . . .)witha=l(α)=

ia_iandα_i> α_i+1,β=(β₁^b).

Before we begin with the proofs we state some additional facts.

Recall that the parts of a skew diagram A=γ /δ are the numbers γ_i −δ_i (1≤i ≤l(γ )) and so are the number of boxes in the rows of A. Furthermore the heights of a skew diagram A are the number of boxes in the columns of A and so are the parts of the conjugated skew diagram. For example the skew dia- gramA=(7²,5,3,2)/(4,2²,1)= has the parts 3,5,3,2,2 and heights 1,2,3,2,3,2,2.

It is known, that skew diagrams which give rise to the same skew character have to have the same parts and heights in the same quantity. This follows from the fact, that the LR tableau of the skew diagramAobtained by filling every column with the

entries 1 to the height of the column (e.g. ) has as contentνthe lexicographic biggest partition whose corresponding character[ν]appears in the decomposition of [A]. Clearly this is the partition obtained by reordering the heights ofAto form a partition. So skew diagrams which give rise to the same skew character have to have the same heights in the same quantity and by conjugation the same holds true for the parts.

In the following we will assume that[λ/μ] = [α/β]. Since we need the same number of rows and columns inλ/μ andα/β we need λ₁=α₁ and a=l if we assume[λ/μ] = [α/β].

If we remove from an LR tableau of shapeλ/μcontainingλ₁ entries 1 all the boxes with entry 1 and replace every entryi >1 byi−1 we obtain an LR tableau of a shape which is obtained fromλ/μby removing the top box from every column.

This gives us a 1−1 relation between the characters[ν] ∈ [λ/μ]with maximal first partν1=λ1and arbitrary characters[ξ] ∈ [ˆλ/μ]withλ/μˆ the skew diagram obtained by removing the top boxes in every column ofλ/μsoλˆ=(λ^l₁¹⁻¹, λ^l₂², . . .)and the 1−1 relation is given byξ_i=ν_i+1. So we have the following lemma:

Lemma 3.2 Let[λ/μ] = [α/β].

Letλ/μ (resp.ˆ α/β )ˆ be the skew diagram obtained fromλ/μ (resp.α/β )by re- moving the topiboxes from every column ofλ/μ (resp.α/β ). Letλ/μ (resp.˜ α/β )˜ be the skew diagram obtained fromλ/μ (resp.α/β )by removing in every row ofλ/μ (resp.α/β )the leftiboxes.

Then[ˆλ/μ] = [ ˆα/β]and[˜λ/μ] = [ ˜α/β].

Proof [ˆλ/μ] = [ ˆα/β]follows from the above 1−1 relation.[˜λ/μ] = [ ˜α/β]follows

then by conjugation symmetry.

In most cases we remove the boxes untilλ/μˆ decays into two disconnected skew diagrams so that we can use Lemma3.1.

If we say in the following that we removeljresp.λi boxes from the top resp. left ofλ/μwe always mean that we remove in every column lj resp. in every row λi

boxes.

In [1] we proved the following:

Theorem 3.3 ([1, Theorem 4.2]) Letμ, λbe partitions withμ⊆λ⊆(k^l)with some fixed integersk, l. Setλ⁻¹=(k^l)/λ.

Then: The coefficient of[α]in[λ/μ]equals the coefficient of[α⁻¹] = [(k^l)/α]in [μ]_(kl)[λ⁻¹].

Fork≥μ1+ν1, l≥l(μ)+l(ν)the Schubert product[μ]_(kl)[ν]is the ordinary product[μ] ⊗ [ν].

Lemma 3.4 Let[A¹] = [A²]withA¹being an arbitrary skew diagramA¹=γ /δ having a partγ₁and a heightl(γ ) (see Figure1).

ThenA¹=A²orA¹=(A²)^◦.

Fig. 1 Lemma3.4:A¹

Proof SinceA¹andA²have the same heights and parts we have also the partγ₁and the heightl(γ )inA². The lemma follows now from Lemma3.1and Theorem3.3.

3.2 Proofs

The proof is arranged as follows. We first prove that [λ/μ] = [α/β] and dp(λ)=2 requiresλ/μ=α/βorλ/μ=(α/β)^◦. Next we prove that[λ/μ] = [α/β] anddp(λ)=3 requiresdp(α)=3. Then we prove thatλ/μ=α/βis required if both λandαaredp=3 partitions and[λ/μ] = [α/β].

Then we examine the cases when bothλandαhave more than 3 different parts and[λ/μ] = [α/β]is multiplicity free.

Theorem 3.5 Let[λ/μ] = [α/β]andλbe adp=2 partition.

Thenαis also adp=2 partition andα/β=λ/μorα/β=(λ/μ)^◦.

Proof We will assume thatαhasn≥2 different parts,dp(α)=n≥2, and show first thatn=2.

Case 1:λ₂> μ₁, l₁> m. This case is covered by Lemma3.4.

So we have either Case 2:l₁≤mor Case 3:λ₂≤μ₁. We cannot have both at the same time withoutλ/μdecaying into two disconnected rectangles. It is sufficient to examine only the Case 2 since Case 3 follows then by conjugation symmetry.

Case 2:l1≤m, λ2> μ1. By comparing the heights of lengthland the partsλ1in λ/μandα/βwe geta1≤b, α_n> β1.

Sinceλ2> μ1we haveλ2−μ1times the heightlinλ/μand so needλ2−μ1= α_n−β1.

If we remove inλ/μthe leftλ2−μ1boxes the remaining skew diagram is:

λ/μ˜ =((λ₁−μ₁−(λ₂−μ₁)

=λ1−λ2

)^l1)⊗((λ₂−λ₂+μ₁

=μ1

)^l−m).

If we remove inα/β the leftλ2−μ1=α_n−β1< α_n boxes the remaining skew diagram has to decay by Lemmas3.1and3.2but decays only forb≥l−a_n and is then (see Figure2):

˜

α/β=((λ1−α_n)^a1, (α2−α_n)^a2, . . . (α_n−1−α_n)^an−1)⊗((β1)^l−b).

By Lemmas3.1and3.2λ/μ˜ andα/β˜ have to be related by translation or rotation if[˜λ/μ] = [ ˜α/β]and so we have[λ/μ] = [α/β]forn≥3.

Fig. 2 Case 2:λ/μ˜ andα/β˜

Forn=2 we have:

˜

α/β=((λ1−α2)^a1)⊗(β₁^l−b).

By Lemmas3.1and3.2we have either

Case 2.1: (λ₁−λ₂)^l1=(λ₁−α₂)^a1, μ^l₁^−m=β₁^l−b or

Case 2.2: (λ1−λ2)^l1 =β₁^l−b, μ^l−m₁ =(λ1−α2)^a1. In Case 2.1 we immediately getλ/μ=α/β.

In Case 2.2 we have

λ1−λ2=β1, λ1−α2=μ1, b=l−l1, l−a1=m

which meansλ/μ=(α/β)^◦.

Theorem 3.6 Letλbe adp=3 partition and[λ/μ] = [α/β].

Thendp(α)=3.

Proof By Theorem3.5αcannot have only 2 different parts. So we will assume, that dp(α)=n≥4 and show that we get contradictions.

Case 1:l1≤m, λ3≤μ1. Since we do not have partsλ1or heightsl inλ/μwe also havea1≤bandαn≤β1.

Forλ/μto be connected we needλ2> μ1andl1+l2> m.

The cases Case 1.1:λ₃≤λ₂−μ₁and Case 1.2:l₁≤l₁+l₂−mare related by conjugation symmetry so it is sufficient to consider only Case 1.1.

Case 1.1:λ3≤λ2−μ1. If we remove inλ/μ λ3boxes from the left and in total l·λ3boxes the remaining skew diagram is((λ1−λ3)^l1, (λ2−λ3)^l2)/μand is not a partition and decays (in the caseλ₂−μ₁=λ₃) only afterλ₃boxes from the left are removed and is connected if onlyλ₃−1 boxes are removed from the left.

If we remove inα/β λ3boxes from the left the remaining skew diagramα/β˜ has to be by Lemma3.2and Theorem3.5adp=2 partition with a rectangle removed.

This cannot be obtained. The best we can obtain is the following. If we demand that

˜

α/βis adp=2 partition with or without another partition removed and demand that removingλ₃boxes from the left removes in totall·λ₃boxes we needn=4, α4=λ₃ as well as eitherb=a1, α1−β1−λ3=0 orb=a1+a2, α2−β1−λ3=0 (see Figure3. The caseb=a1+a2+a3, α3−β1−λ3=0 can be excluded, because in this case eitherα/β decays or has a heightl.). In both cases removingλ₃−1 boxes from the left ofα/β gives a decaying skew diagram.

Fig. 3 Case 1.1:λ/μ˜ andα/β˜ ⁱ

Fig. 4 Case 1.3:λ/μ˜ andλ/μˆ

Case 1.3:λ₃> λ₂−μ₁, l₁> l₁+l₂−m.

If we remove inλ/μ λ₂−μ₁boxes from the left and in totall·(λ₂−μ₁)boxes the remaining skew diagramλ/μ˜ decays into a rectangle and adp=2 partition. If we remove onlyλ₂−μ₁−1 boxes from the left the remaining skew diagram does not decay. We have (see Figure4):

λ/μ˜ =((λ1−λ2)^l1)⊗(μ^l₁^1+l2−m, (λ3−λ2+μ1)^l3). (1) If we remove inλ/μ l1+l2−mboxes from the top which are in totalλ1·(l1+ l2−m)boxes the remaining skew diagramλ/μˆ decays also into a rectangle and a dp=2 partition. Again removing onlyl1+l2−m−1 boxes from the top gives a connected skew diagram. We have (see Figure4):

λ/μˆ =((λ1−μ1)^m−l2, (λ2−μ1)^l2)⊗(λ^l₃³). (2) If we remove inα/β λ2−μ1boxes from the left the remaining skew diagram has by Lemma3.2also to decay into a rectangle and adp=2 partition. To obtain this and remove in totall·(λ2−μ1)boxes and also be in the situation that after removing λ2−μ1−1 boxes from the left the remaining skew diagramα/β˜ does not decay we needn=4, α4=λ₂−μ₁as well as eithera₁≤b < a₁+a₂, α₂−β₁=λ₂−μ₁or a₁+a₂≤b < a₁+a₂+a₃, α₃−β₁=λ₂−μ₁(see Figure5).

With the same reasoning the skew diagramα/βˆ obtained by removingl₁+l₂−m boxes from the top ofα/βmust decay into a rectangle and a skew diagram and we needn=4, a1=l₁+l₂−mas well as eitherα₃≤β₁< α₂, a₁+a₂−b=l₁+l₂−m orα₄≤β₁< α₃, a₁+a₂+a₃−b=l₁+l₂−m(see Figure6).

Fig. 5 Case 1.3: The two cases ofα/β˜ ⁱ

Fig. 6 Case 1.3: The two cases ofα/βˆ ⁱ

By Lemmas3.1and3.2the skew diagramsλ/μˆ resp.λ/μ˜ andα/βˆ resp.α/β˜ have to decay into the same partitions.

In the Case 1.3.1:a₁≤b < a₁+a₂, α₂−β₁=λ₂−μ₁we have:

˜

α/β¹=((λ1−α2)^a1)⊗((α2−λ2+μ1

=β1

)^a1+a2−b, (α3−α2+β1)^a3).

Comparing with (1) gives:

α2=λ2, β1=μ1, α3=λ3. (3)

In the Case 1.3.2:a₁+a₂≤b < a₁+a₂+a₃, α₃−β₁=λ₂−μ₁we have:

˜

α/β²=((λ₁−α₃)^a1, (α₂−λ₂−β₁+μ₁)^a2)⊗(β₁^a1+a2+a3−b).

Comparing with (1) gives:

α₃=λ₁−μ₁, α₃−β₁=λ₂−μ₁, α₂=λ₃+β₁. (4) In the Case 1.3.x.1:α₃≤β₁< α₂, a₁+a₂−b=l₁+l₂−mwe have:

ˆ

α/β¹=((α2−β1)^a2)⊗(α₃^a3, α^a₄⁴).

Comparing with (2) gives:

α₃≤β₁, α₂=λ₃+β₁, α₄=λ₂−μ₁. (5)

In the Case 1.3.x.2:α₄≤β₁< α₃, a₁+a₂+a₃−b=l₁+l₂−mwe have:

ˆ

α/β²=((α₂−β₁)^a2, (α₃−β₁)^a3)⊗(α₄^a4).

Comparing with (2) gives:

α2=λ1−μ1+β1, α3=λ2+β1−μ1, α4=λ3. (6) From the equations of Case 1.3.1.1 ((3) and (5)) followsα3+β1=λ3+β1= α₂=λ₂=α₄+μ₁=α₄+β₁, butα₃=α₄.

From the equations of Case 1.3.1.2 ((3) and (6)) followsα3=λ3=α4, butα3= α4.

From the equations of Case 1.3.2.1 ((4) and (5)) follows 0≥α3−β1=λ2−μ1, but since we are in the casel₁≤mandλ₃≤μ₁,λ/μdecays forλ₂−μ₁≤0.

From the equations of Case 1.3.2.2 ((4) and (6)) followsα₃=λ₁−μ₁=α₂−β₁= λ3=α4, butα4=α3.

These contradictions finish Case 1.

Case 2:l₁> m, λ₃≤μ₁and Case 3:l₁≤m, λ₃> μ₁are related by conjugation symmetry so it is sufficient to consider only Case 2.

Case 2:l1> m, λ3≤μ1. Since we have the partλ1l1−mtimes inλ/μwe need l1−m=a1−b. Removingl1−mboxes from the top ofλ/μgives a connected skew diagramλ/μˆ forλ₂> μ₁with

λ/μˆ =(λ^m₁, λ^l₂², λ^l₃³)/μ

or a decaying skew diagramλ/μˆ forλ2≤μ1with

λ/μˆ =((λ1−μ1)^m)⊗(λ^l₂², λ^l₃³).

Forλ2> μ1removingl1−m=a1−b boxes from the top ofα/β must yield a connected skew diagramα/βˆ and we get:

ˆ

α/β=(α₁^b, α^a₂², α^a₃³, α^a₄⁴, . . .)/β.

Since we are now in Case 1 withl₁≤m, λ₃≤μ₁we get from the above[λ/μ] = [α/β].

Forλ2≤μ1removingl1−m=a1−bboxes from the top ofα/β must yield a decaying skew diagramα/βˆ and we get:

λ/μˆ =((α₁−β₁)^b)⊗(α₂^a2, α₃^a3, α₄^a4, . . .).

Lemma3.1gives[λ/μ] = [α/β].

Case 4:l1> m, λ3> μ1is covered by Lemma3.4.

Theorem 3.7 Letλ, αbedp=3 partitions and[λ/μ] = [α/β].

Thenλ/μ=α/β.

Proof Case 1:l₁≤m, λ₃≤μ₁. Since we do not have heightsland partsλ₁inλ/μ we also needa₁≤b, α₃≤β₁.

Forλ/μto be connected we needλ₂> μ₁, l₁+l₂> m.

Case 1.1:l₁> l₁+l₂−mand Case 1.2:λ₃> λ₂−μ₁are related by conjugation symmetry so it is sufficient to consider only Case 1.1.

Case 1.1:l₁> l₁+l₂−m.

Removingl₁+l₂−mboxes from the top ofλ/μgives:

λ/μˆ =(λ^l₃³)⊗((λ1−μ1)^m−l2, (λ2−μ1)^l2).

If we remove onlyl1+l2−m−1 boxes from the top ofλ/μthe remaining skew diagram does not decay, so we needl1+l2−m=a1+a2−b, b > a2and removing a1+a2−bboxes from the top ofα/β gives then:

ˆ

α/β=(α^a₃³)⊗((λ₁−β₁)^b−a2, (α₂−β₁)^a2).

This givesλ/μ=α/β.

Case 1.3:l1≤l1+l2−m⇔m≤l2, λ3≤λ2−μ1.

Removingl1 boxes from the top ofλ/μremoves in totall1·λ1boxes and gives a skew diagram which either decays into two disconnected rectangles (forl1=l1+ l2−m) or is(λ^l₂², λ^l₃³)/μ(forl1< l1+l2−m). We needa1≤l1because removing l1boxes from the top ofα/βmust either yield a decaying skew diagram or adp=2 partition with a rectangle removed. But since removing the topl1boxes ofα/β has to remove in totall1·λ1boxes we need alsoa1≥l1and soa1=l1.

Removingl1+l2−m−1 boxes from the top of λ/μyields a connected skew diagram but removing the topl1+l2−mboxes yields:

λ/μˆ =(λ^l₃³)⊗((λ2−μ1)^m).

Sinceα/β must also decay after removing l₁+l₂−mboxes from the top but not after removing less boxes we geta₁+a₂−b=l₁+l₂−mand soa₂−b=l₂−m.

In an analogous way by removingλ₃resp.λ₂−μ₁ boxes from the left we get α₃=λ₃andα₂−β₁=λ₂−μ₁.

We will first examine the Case 1.3.1:μ₁=β₁and show that this gives[λ/μ] = [α/β]. This covers by conjugation symmetry also the Case 1.3.2:m=b.

For the following construction we only need:

α2−β1=λ2−μ1, a1=l1, m≤l2, b≤a2. (7) Without loss of generality we may also assume thatμ₁< β₁and setβ₁=μ₁+n which givesα₂=λ₂+n.

Case 1.3.1.1:λ₁−(λ₂+n)+1≤μ₁.

We can in λ write entries 1 to l₁ into the columnsλ₂+n toλ₁ which are in totalλ₁−(λ₂+n)+1 columns (see Figure7). If we place the remaining entries so that they obey the LR rule we get an LR filling ofλwith contentμwhere the box (l₁, λ₂+n)is filled. So there is a character[ν]in[λ/μ]withν not containing the box(l₁, λ₂+n).

Fig. 7 LR filling ofλin the Case 1.3.1.1

Fig. 8 LR filling ofαin the Case 1.3.1.2

Inαthere area₂≥bboxes below the box(l₁, λ₂+n)=(a₁, α₂)so in every LR filling ofαwith contentβthe box(l₁, λ₂+n)remains empty. So for every character [ν]in[α/β],νcontains a box in position(l₁, λ₂+n). This gives[λ/μ] = [α/β].

Case 1.3.1.2:λ₁−(λ₂+n)≥μ₁⇔λ₁−λ₂≥μ₁+n=β₁.

Let us now construct an LR filling of α with content β which leaves the box (l₁+1, λ2+1)empty and so gives[λ/μ] = [α/β].

Place the entriesbinto the the rowslandl₁+a₂with at least onebin rowl(see Figure8). Now place for 1< i < btheiabovei+1 if possible. Forb−i=a₃we cannot place the entryiabove the entryi+1 unless there areα2−α3entriesbin row l1+a2. If there are less thanα2−α3entriesbin rowl1+a2we place the entryiin rowl1+a2directly left to theb. If we now placeβ1−1 entries 1 into the rightβ1−1 columns we only get to columnλ1−(β1−1)+1=λ1−β1+2≥λ2+2. If we now place the remaining 1 atop of one of the entries 2 which are in rowl1+a2+a3−b+2 (in one of the columns having in rowa1+a2an entryb−a3instead of an entryb) then the 1 is placed in rowl1+a2+a3−b+1≥l1+2 where we useda2≥band a3≥1. Also the highest position of a 2 in this filling is in rowl1+a2−b+2 and so below rowl1+2. So the box(l1+1, λ2+1)is empty in this LR filling. So we have a character[ν]in[α/β]withν having a box in position(l1+1, λ2+1). But since (l1+1, λ2+1)is not inλthere is no character[ν]in[λ/μ]withνcontaining a box in position(l1+1, λ2+1)and so[λ/μ] = [α/β].

Case 1.3.3:μ₁=β₁, m=b.

Since we havea₂−b=l₂−mandm=b we have alsoa₂=l₂and from this followsl₃=l−l₁−l₂=l−a₁−a₂=a₃.

Also fromα₂−β₁=λ₂−μ₁andβ₁=μ₁followsα₂=λ₂.

Together withl₁=a₁andλ₃=α₃which we proved above, this gives the desired λ/μ=α/βand so finishes the casel₁≤m, λ₃≤μ₁.

Case 2:l₁> m, λ₃≤μ₁and Case 3:l₁≤m, λ₃> μ₁are related by conjugation symmetry so it is sufficient to consider only Case 2.

Case 2:l1> m, λ3≤μ1.

Since we havel1−mtimes the partλ1inλ/μwe needl1−m=a1−b.

The skew diagramλ/μˆ obtained after removingl1−mboxes from the top ofλ/μ decays forλ₂≤μ₁but is connected forλ₂> μ₁. So we haveλ₂≤μ₁if and only if α₂≤β₁.

Case 2.1:λ₂≤μ₁.

After removingl1−m=a1−bboxes from the top ofλ/μandα/β we have the skew diagrams

λ/μˆ =((λ₁−μ₁)^m)⊗(λ^l₂², λ^l₃³) and

ˆ

α/β=((λ1−β1)^b)⊗(α₂^a2, α^a₃³) which gives by Lemma3.1λ/μ=α/β.

Case 2.2:λ2> μ1.

Removingl1−m=a1−bboxes from the top ofλ/μandα/βgives λ/μˆ =(λ^m₁, λ^l₂², λ^l₃³)/(μ^m₁)

and

ˆ

α/β=(λ^b₁, α₂^a2, α^a₃³)/(β₁^b).

Using the result of Case 1:l1≤m, λ3≤μ1givesα/β=λ/μand finishes Case 2.

Case 4:l1> m, λ3> μ1is covered by Lemma3.4.

We will now compare the multiplicity free skew characters[λ/μ]and[α/β]when bothλandαhave more than 4 different parts and assume for the following lemmas thatdp(λ), dp(α)≥4.

There are 4 cases when[λ/μ]is multiplicity free andλhas more than 4 different parts and we will compare them against each other (see Figure9).

The first lemma covers by conjugation symmetry also the case whenμ₁=β₁=1.

Lemma 3.8 Letλ/μandα/βbe skew diagrams withm=b=1 and[λ/μ] = [α/β].

Thenλ/μ=α/β.

Proof We have inλ/μ l1−1 times the partλ1. Since we need inα/βthe partλ1also a₁−1 times we geta₁=l₁.

If we removel₁boxes from the top ofλ/μwe either get a connected skew diagram for λ₂> μ₁, l₂>1 orλ₃> μ₁, a disconnected skew diagram for λ₂> μ₁≥λ₃, l2=1 or a partition forλ2≤μ1(see Figure10). Obviously the same must apply for

ˆ

α/βif[λ/μ] = [α/β].

Suppose we get a connected skew diagramλ/μˆ withλˆhaving less than 4 different parts. Then we can use Theorems3.6and3.7to getλ/μ=α/β.

Fig. 9 The four multiplicity free cases withλadp=n >3 partition

Fig. 10 Lemma3.8: The three cases forλ/μˆ

Ifλ/μˆ is connected butλˆ has 4 or more different parts we can iterate this process until we reach the case where λˆ has less than 4 different parts orλ/μˆ is either a disconnected skew diagram or a partition.

Supposeλ/μˆ is a partition. If we remove onlyl₁−1 boxes from the top ofλ/μ andα/βto getλ/μˆ andα/βˆ we get:

λ/μˆ =(λ₁−μ₁)⊗(λ^l₂², λ^l₃³, λ^l₄⁴, . . .) and

ˆ

α/β=(λ₁−β₁)⊗(α^a₂², α₃^a3, α₄^a4, . . .).

This givesλ/μ=α/β.

So now supposeλ/μˆ is a disconnected skew diagram. Then we have:

λ/μˆ =(λ2−μ1)⊗(λ^l₃³, λ^l₄⁴, . . .) and

ˆ

α/β=(α2−β1)⊗(α^a₃³, α₄^a4, . . .).

This givesλ^l_iⁱ=α_i^aifori≥3 andλ₂−μ₁=α₂−β₁.

Forμ₁=β₁we getλ₂=α₂and since we have in this case alsol₂=a₂=1 we getλ/μ=α/β.

Forμ₁=β₁ we can use the construction of LR fillings following equation (7)

(page225) and get[λ/μ] = [α/β].

Lemma 3.9 Letλ/μandα/βbe skew diagrams withμ=(1)=β,m=1=β₁and [λ/μ] = [α/β].

Thenλ=(l, l−1, l−2, . . . ,2,1)is a staircase partition andλ/μis the conjugate ofα/β, α/β=λ/μ.

Proof Supposel₁>1. Then we have the partλ₁ l₁−1 times inλ/μand so need a₁=b+l₁−1. The smallest height inλ/μis eitherl₁(in the caseλ₂≥μ₁) orl₁−1 (forλ₂< μ₁). The smallest height inα/βis eithera₁=b+l₁−1> l₁orl−b. For λ₂≥μ₁this givesl−b=l₁and sol=b+l₁=a₁+1 which means thatαhas only 2 different parts. Forλ₂< μ₁this givesl−b=l₁−1 and sol=l₁+b−1=a₁ which means thatαis a rectangle.

So we havel₁=1 and sinceλ/μdoes not decay we needλ₂> μ₁.

If we remove the top box of λ/μand α/β we get a connected skew diagram for eitherl₂>1 or forλ₃> μ₁. Suppose we are in this case then if the new skew diagrams have less than 4 different parts we can use Theorem3.7which then gives that[λ/μ] = [α/β]. If the new skew diagram has 4 or more different parts we get λ₂=α₂, becauseλ₂ is the number of columns in the new skew diagram, and so a₁=1. Since we have the part λ₁−1 in α/β we needλ₂=λ₁−1, because we need a the partλ₁−1 also inλ/μ. We can repeat the above argument until the skew diagram we obtain after removing the top box decays.

So we now assume that the skew diagramsλ/μˆ andα/βˆ obtained after removing the top box ofλ/μandα/β decay and so we needl2=1, λ3≤μ1. Ifλ/μˆ decays we have:

λ/μˆ =(λ₂−μ₁)⊗(λ^l₃³, λ^l₄⁴, . . .).

Sinceα/βˆ must also decay we needα_n=1 anda=b+a_n+1 ifα/βhasndifferent parts. We then have:

ˆ

α/β=(1^an)⊗((α 1−1)^a1−1

fora₁>1

, (α2−1)^a2, (α3−1)^a3, . . .).

Since we have height 1 inλ/μ and the smallest height in α/β is either a₁ or l−b=a_n+1>1 we needa₁=1.

Comparingλ/μˆ withα/βˆ gives λ₂−μ₁=1, an=1, λ^l_iⁱ=(α_i−1−1)^ai−1 for i=3, . . . , n. Since we have again the partλ₁−1 inα/βwe needλ₂=λ₁−1 and so μ₁=λ₁−2.

Suppose we havel_i =a_i =1 and α_i =λ_i for 1≤i < p and fixedp≥2. This holds true forp=2. Sincel_i+1=a_i fori≥2 andl₂=1 we have alsol_p=1.

Suppose we haveλ_p> λ_p+1+1 whereλ_p+1=0 is allowed.

We now construct an LR filling of λ with content μ. We place in λ entries 1 into every column of λ but not into the columns λ_p+1+1 and λ_p+1+2 and so

Fig. 11 Lemma3.9: LR fillings ofλandα

we get an LR filling which leaves the box(p, λ_p+1+2)empty (see Figure 11).

This means that there is a character[ν] ∈ [λ/μ]withν containing a box in position (p, λ_p+1+2). Sinceα_p=λ_p+1+1 the box(p, λ_p+1+2)is not inαand so there is no character[ν] ∈ [α/β]withν containing a box in position(p, λ_p+1+2). This gives[λ/μ] = [α/β]. So we needλp=λp+1+1=αp.

So now suppose we havea_p>1.

Placing in α the entries into the rows 1 top−1 andp+2 to l gives an LR filling which leaves the box(p+1, αp)empty (see Figure 11) and so we have a character[ν] ∈ [α/β]withνcontaining a box in position(p+1, αp). Sincel_p=1 the p+1th row inλhas onlyλp+1< λp=αpboxes which and so there is no character [ν] ∈ [λ/μ]withν containing a box in position(p+1, αp)and so[λ/μ] = [α/β] fora_p>1 and so we needa_p=1.

It now follows by induction thatα=λhas to be a staircase partitionλ=(l, l− 1, . . . ,2,1)which then also givesb=μ₁and soλ/μ=(α/β). The following lemma covers by conjugation also the case whenμ1=1, b=a− 1=l−1.

Lemma 3.10 Letλ/μandα/βbe skew diagrams withm=1, β1=α₁−1=λ₁−1, b >1.

Then[λ/μ] = [α/β].

Proof We havea₁> bsince otherwiseα/βwould decay.

Removinga1−bboxes from the top ofα/βgivesα/βˆ which decays into ˆ

α/β=(1^b)⊗(α₂^a2, α^a₃³, α^a₄⁴, . . .).

Sinceμ=(μ1)we have that ifλ/μdecays aftera1−bboxes are removed from the top and in total(a₁−b)·λ₁boxes it decays into

λ/μˆ =(λ₁−μ₁)⊗(λ^l₂², λ^l₃³, λ^l₄⁴, . . .).

Sinceb >1 we get[λ/μ] = [α/β].

The following lemma covers by conjugation also the case when μ₁=1, β1= α1−1=λ1−1.

Lemma 3.11 Letλ/μand α/β be skew diagrams withμ1>1=m, b=l−1= a−1.

Then[λ/μ] = [α/β].