The uniqueness of plethystic factorisation
Chris Bowman Rowena Paget


The uniqueness of  plethystic  factorisation]The uniqueness of   plethystic  factorisation 


 
[
    Rowena Paget
    December 30, 2023
=====================



§ INTRODUCTION

Let s_λ and s_μ denote the Schur functions labelled by the partitions  and μ.  
There are three ways of “multiplying" this pair of functions together in order to obtain a new symmetric function; these are the   Littlewood–Richardson, Kronecker, and plethysm products. 
 The primary purpose of this  paper is to address the most fundamental question one can ask of  such a  product: “does  
it  factorise uniquely?".    


For the Littlewood–Richardson product, this question  was answered by  Rajan  <cit.>.  
We solve this question  for   the most difficult  and mysterious 
    of    these products, the plethysm product (which we denote ∘) as follows.
 
 Let μ,ν,  π,ρ be arbitrary partitions.  
   If 
 s_ν∘ s_μ = s_ρ∘ s_π
 then either ν=ρ and μ=π;  
  or we are  in one of five exceptional cases, 

    [                                                   s_(2,1^2)∘ s_(1)= s_(1^2)∘ s_(1^2),            s_(3,1)∘ s_(1)= s_(1^2)∘ s_(2),;                                                s_(2,1^2)∘ s_(2)  = 
    s_(1^2)∘ s_(3,1), s_(2,1^2)∘ s_(1^2)  = 
    s_(1^2)∘ s_(2,1^2),;                                                                s_ν∘ s_(1)=s_(1)∘ s_ν. ]

    

 
 In general, the decomposition of a plethysm product will have very, very many constituents. 
We  ask: “when is the plethysm product of two Schur functions indecomposable?". 
We prove that in fact such a product is always decomposable, and even inhomogeneous, 
except for some obvious exceptions.  The analogous result  for the Kronecker product was obtained by   Bessenrodt and Kleshchev  <cit.>. 
   


Let μ,ν be  partitions.  
The product 
  s_ν∘ s_μ 
 is decomposable and inhomogeneous except in the following exceptional cases:
 
    s_(1^2)∘ s_(1^2)= s_(2,1^2),
      
     s_(1^2)∘ s_(2)= s_(3,1),
        
      s_ν∘ s_(1)=s_ν=
       s_(1)∘    s_ν  .


 

 
 Understanding and decomposing the  Kronecker and   plethystic products of pairs of Schur functions was identified by Richard Stanley as two of the most important open problems in   algebraic combinatorics <cit.>. 
Almost nothing is known about general constituents of plethysm products;  however the maximal
 terms in the dominance ordering are now well-understood   <cit.>.
 Our proof of    Theorems A and B proceeds  by careful analysis  of these maximal terms.   
   


 
Outside of combinatorics,  
 plethysm products arise naturally  in the representation theory of symmetric and general linear groups.  
In quantum information  theory, the positivity of  
constituents  in a  plethysm product of two Schur functions      is equivalent to the existence of  
quantum states with certain spectra, margins, and occupation numbers <cit.>.  
Decomposing     Kronecker  and  plethystic  products  of Schur functions is  the  central plank of Geometric Complexity Theory, an approach that seeks to settle the P versus NP problem <cit.>;  this approach was recently shown to require not only knowledge of the positivity but also precise information on the actual multiplicities of the  constituents  of the products s_ν∘ s_μ <cit.>.  
  

 
 

§ PARTITIONS, SYMMETRIC FUNCTIONS
  AND MAXIMAL TERMS IN PLETHYSM
 
 
  We define a composition λ n to be a   finite  sequence  of non-negative integers (λ_1,λ_2, …) whose sum, |λ| = λ_1+λ_2 + …, equals n.
  If the sequence (λ_1,λ_2, …) is  weakly decreasing,  we say that  is a partition and write ⊢ n.       Given  λ a partition of n, the Young diagram   is defined to be the configuration of nodes

    []={(r,c) |  1≤  c≤λ_r}.

We say that a partition is linear if it consists only of one row, or one column.  
 The conjugate partition, ^T, is the partition obtained by interchanging the rows and columns of .  The number of non-zero parts of a partition, ,  is called its length, ℓ(λ); the size of the largest part is called the width, w(); the sum of all the parts of λ is called its  size.

 Given two partitions  and μ, we let  + μ  and  ⊔μ denote the partitions obtained by adding the partition horizontally and vertically respectively.    In more detail
 
    +μ=(_1+μ_1,_2+μ_2,_3+μ_3,… )

 and ⊔μ is the partition whose multiset of parts is the disjoint union of the multisets of parts of  and μ. We have that  
  
    ⊔μ= (^T+μ^T)^T.

Finally we remark that, in this paper, the partition ⊔μ is usually equal to 
    (_1,_2,…,_ℓ(),μ_1,μ_2,…,μ_ℓ(μ)).
 In other words, we often do not need to reorder the multisets of parts — this is simply because _ℓ()≥μ_1 in most cases.  

 We   now recall the dominance ordering on partitions. Let ,μ be partitions. 
 We  write  
 ⊵μ  if 
 
    ∑_1≤ i ≤ k_i ≥∑_1≤ i ≤ kμ_i   for all  k≥ 1.

 If ⊵μ and ≠μ we write μ.  The dominance ordering  is a partial ordering on the set of  partitions of a given size.  This partial order can be refined into a total ordering as follows: 
  we write 
≻μ  if 
 
    _k >μ_k for some k≥ 1 and  _i = μ_i  for all  1≤ i ≤ k-1.

  We refer to   ≻  as the lexicographic ordering. 
    We now define the  transpose-lexicographic ordering  as follows:
    
    ≻_Tμ if and only if ^T≻μ^T .

  We emphasise that this total ordering is not simply the opposite ordering to the lexicographic ordering;  minimality with respect to ≻ is not equivalent to maximality with respect to ≻_T.  
  
   Let   λ be  a partition  of n. 
 A Young tableau of shape λ may be defined as  a map
 : [] →.  Recall that the tableau  is semistandard if  (r,c-1)≤(r,c)   and
(r-1,c)< (r,c)  for all (r,c)∈ [].  
 We let _k = |{ (r,c)∈ [] |(r,c)=k}| for k∈.  
 We refer to the composition α=(_1,_2,_3,…) as the weight of the tableau .  
 We denote the set of all  tableaux of shape  by _(),  and the subset of those having weight α by _(,α).    
The Schur function s_, for  a partition of n, may be defined as follows:
   
    s_ = ∑_c
       α n
         
    | _(,α)
    | x^α       where       x^α= x_1^α_1 x_2^α_2 x_3^α_3… .


The  plethysm product of two symmetric functions 
 is defined in <cit.> or   <cit.>. 
 The plethysm product of two Schur functions  is again a symmetric function  and so can be rewritten as a linear combination of Schur functions:

    s_ν∘ s_μ 
     =∑_α p(ν,μ,α) s_α

  such that p(ν,μ,α) ≥ 0.   
We say that the product is homogeneous if there is precisely one partition, α, such that  p(ν,μ,α) > 0; we say that the product is 
 indecomposable if, in addition,   p(ν,μ,α) =1.  
We now recall the role  conjugation – often called the ω involution – plays   in plethysm (see, for example, <cit.>). For μ⊢ m, ν⊢ n, and α⊢ mn  we have that 

    p( ν,μ,α)=
    
    p( ν,μ^T,α^T)				   if m is even
    
    p( ν^T,μ^T,α^T)				   if m is odd.

Throughout this paper we shall let μ,ν,  π,ρ be partitions of m, n, p and q respectively.  In order to keep track of the effect of this conjugation 
when comparing products 
s_ν∘ s_μ and s_ρ∘ s_π,  
 we set 

    ν^M=ν   if m is even
    ν^T    if m is odd    ρ^P=ρ   if p is even
    ρ^T    if p is odd

 Given   a total ordering, >, on partitions we let   
    max_>  (s_ν∘ s_μ)

 denote the  unique  partition, λ,  such that p(ν,μ,λ)≠ 0 and  p(ν,μ,α)=0 for all α >  λ.  We shall use this with both the lexicographic ≻ and transpose-lexicographic ≻_T orderings .   By <ref> we have that

    max_≻_T  (s_ν∘ s_μ)  
     =  ( max_≻ (s_ν^M∘ s_μ^T))^T   .

 
 









 The following theorems will be incredibly important in our arguments.



Let μ,  ν be partitions of m and n respectively.  
The unique maximal terms of 
s_ν∘ s_μ
 in the lexicographic and transpose lexicographic ordering are as follows :
  
    max_≻ (s_ν∘ s_μ)  =
    (nμ_1,nμ_2,…, nμ_ℓ(μ) -1,nμ_ℓ(μ)-n+ν_1, ν_2, … ,ν_ℓ(ν)) ,

  
    max_≻_T  (s_ν∘ s_μ)  =
     (nμ_1^T,nμ_2^T,…, nμ^T_μ_1 -1,nμ^T_μ_1-n+ν_1^M, ν_2^M, … ,ν_ℓ(ν^M)^M) )^T.

 Moreover, we have that  
    p(ν,μ, max_≻ (s_ν∘ s_μ) )=1=
     p(ν,μ,  max_≻_T  (s_ν∘ s_μ)   ).
  


  
When μ=(m), Theorem <ref> shows that
  
    max_≻ (s_ν∘ s_(m)) = (nm-n) + ν,    max_≻_T  (s_ν∘ s_(m))  =
     ((n^m-1) ⊔ν^M)^T.




Sometimes we shall use the dominance ordering   to compare the summands of s_ν∘ s_μ, and then there will, in general, be many  (incomparable) maximal partitions. To understand these summands, we require some further definitions.    We place a lexicographic  ordering, ≺, on the set of semistandard Young tableaux as follows.
  Let ≠ be semistandard μ-tableaux,   and consider the leftmost column in which  and  differ.  We write 
 ≺ if  the greatest entry not appearing in both columns lies in .  Following <cit.>, we define a plethystic  tableau of shape μ^ν  and weight α to be  a map
 
    : [ν] →_(μ)
 such that the total  number of occurrences of
k in the tableau entries of  is α_k for each k.
 We say that such a tableau is semistandard if  (r,c-1) ≼(r,c)   and
(r-1,c)  ≺(r,c)  for all (r,c)∈ [ν].  
   We denote the set of all plethystic  tableaux of shape μ^ν and weight α by by (μ^ν,
  α).    
 

 


 
The maximal partitions α in the dominance order such that s_α is a constituent of s_ν∘ s_μ are precisely the maximal weights of the plethystic semistandard tableaux of shape μ^ν. Moreover if α is such a maximal partition then p( ν,μ,α) is equal to |(μ^ν,
  α)|.    

Finally, we recall the one known case in which every term in a plethystic product is both maximal and minimal in the dominance ordering.
Given α a partition of n with distinct parts, we let  
2[α] denote the unique partition of 2n  whose leading diagonal hook-lengths  are 2α_1, …, 2α_ℓ(α)  and whose ith row has length α_i+i for  1≤ i ≤ℓ(α).
(An example follows.) We have the decomposition

    s_(1^n)∘ s_(2)=∑_α  s_2[α],

where the sum is over all partitions α  of n into distinct parts.    
 This decomposition is given in <cit.> and  <cit.>.  We observe that for n>2 this product is never homogeneous (for example α=(n) and α=(n-1,1) both label summands).  


For n=5 the decomposition obtained is

    s_(1^5)∘ s_(2)=
    s_2[(3,2)]+ s_2[(4,1)]+s_2[(5)] 
    = s_(4^2,2)+s_(5,3,1^2)+s_(6,1^4).

We picture these partitions (and the manner in which they are formed) in <ref> below.  We remark that 

    s_(1^5)∘ s_(1^2)=
      s_(4^2,2)^T+s_(5,3,1^2)^T+s_(6,1^4)^T
    =s_(3^2,2^2) +s_(4,2^2,1^2) +s_(5,1^5)

by <ref> simply because m=2 is even.  
 



 

 


§ DECOMPOSABILITY AND HOMOGENEITY OF PLETHYSM 


In this section, we prove Theorem B of the introduction: namely we classify all decomposable/homogeneous   plethystic products of Schur functions.  
This also serves to remove the homogeneous products from consideration in the proof of Theorem A.  



Let μ,ν be  partitions of m  and n, respectively.  
The product 
  s_ν∘ s_μ 
 is decomposable and inhomogeneous except in the following cases:
 
    s_(1^2)∘ s_(1^2)= s_(2,1^2),
      
     s_(1^2)∘ s_(2)= s_(3,1),
        
      s_ν∘ s_(1)=s_ν,
        
     s_(1)∘    s_μ  =s_μ.





That the listed products are homogeneous is obvious.  We assume that m, n ≠ 1 and

    max_≻(s_ν∘ s_μ ) = max_≻_T(s_ν∘ s_μ ).

We shall show that this implies   
 that ν =(1^2) and μ⊢ 2.
  We first assume that μ is non-linear, that is μ is neither (m) nor (1^m). 
We set k=ℓ(μ ).  
 We draw a horizontal line across the Young diagrams of max_≻(s_ν∘ s_μ )
  and ( max_≻(s_ν^M∘ s_μ^T ))^T  so that the partitions below each of these lines  each have strictly fewer than n nodes in total and are maximal with respect to this property.  
For  max_≻(s_ν∘ s_μ ), this line is drawn between the
 kth   and (k+1)th rows (even though the (k+1)th row might be zero).   
  For ( max_≻(s_ν^M∘ s_μ^T ))^T, this line is drawn at some point after the (n(k-1)+1)^th row.  
Since k  < n(k-1)+1 for n>1, we see that max_≻(s_ν∘ s_μ )
≠  ( max_≻(s_ν^M∘ s_μ^T ))^T as required.  

   It remains to consider the case that μ is linear and we assume (by conjugating if necessary) that
  μ=(m). Then, as we saw in Example <ref>,
 
    max_≻(s_ν∘ s_(m))=(mn-n)+ ν,    
    ( max_≻(s_ν^M∘ s_(1^m) ))^T = ( (m-1)^n)+(ν^M)^T.

Therefore row n of max_≻(s_ν∘ s_(m))  has length ν_n which is at most 1, and  row n of 
( max_≻(s_ν^M∘ s_(1^m) ))^T has length at least m-1. 
Since we are considering only m ≥ 2, we conclude that m=2 and ν_n=1, that is ν=(1^n).
From the closed formula for the decomposition of s_(1^n)∘ s_(2)  in <ref>, and the resulting decomposition  of its plethystic conjugate s_(1^n)∘ s_(1^2), we observe that the product is  homogeneous if and only if n=1,2. 
 

 


If  s_ν∘ s_(1)=s_ρ∘ s_π or s_(1)∘ s_μ=s_ρ∘ s_π then 
 either: π=(1^2) and ρ is a partition of 2; or  at least one of ρ or π has size 1.  


Therefore in the remainder of the paper, we can and will assume that none of the indexing partitions in our plethystic products are equal to (1)⊢ 1.  

 



§ UNIQUE FACTORISATION OF   PLETHYSM  
 


A quick scan of the diagrams in <ref> tells us that the maximal terms in the  product under the lexicographic and transpose-lexicographic orderings encode a great deal of information concerning the multiplicands of the product. 
 We might even think that these maximal terms are enough to uniquely determine the multiplicands.  
 In fact, this is not the case (as the following example shows).  

 
 Consider the plethysm products
 
    s_(3^3,2,1)∘ s_(1^2)     and  
    s_(2,1)∘ s_(4,1^4).

Both have the same maximal terms in the lexicographic  and transpose-lexicographic orderings, namely those labelled by (12,3^3,2,1) and (15,3^2,2,1)^T. 
<ref> depict  how these two partitions can be seen to be maximal in the lexicographic and  transpose-lexicographic orderings using <ref>.  




This puts a scupper on our plans to determine uniqueness solely using   maximal terms in the lexicographic and transpose-lexicographic orderings.  
 Now, we notice that the  plethysm products
 s_(3^3,2,1)∘ s_(1^2)
and s_(2,1)∘ s_(4,1^4)  can still  be distinguished by looking at the maximal terms for both products in  the   dominance ordering.  
For example, 
(11, 4, 4, 3, 2) labels a maximal term that appears in  s_(3^3,2,1)∘ s_(1^2) but it is not a maximal term in  and  s_(2,1)∘ s_(4,1^4). Similarly, (11, 4, 3, 3, 3)   labels a maximal term in s_(2,1)∘ s_(4,1^4) but not in s_(3^3,2,1)∘ s_(1^2).

 
 
Our method of proof will proceed to distinguish plethysm products by first using maximal terms in the lexicographic ordering and  only when necessary considering  the broader family of terms which are maximal in the dominance ordering. We first consider the case where μ consists of a single row. 
 
  
 Let μ,ν,  π,ρ be  partitions of m,n,p,q>1 respectively. 
 We suppose that   μ=(m).  
   If 
 
    s_ν∘ s_μ = s_ρ∘ s_π

 then either ν=ρ and μ=π or we are in the exceptional case 
 
    s_(2,1^2)∘ s_(2)  = 
    s_(1^2)∘ s_(3,1).
     

 
 
From the set-up, we know mn=pq. We set ℓ(π)=c+1 for some c≥ 0.  
 By assumption, we have that 
 
    max_≻(s_ν∘ s_(m))
        =
       max_≻(s_ρ∘ s_π)
     max_≻(s_ν^M∘ s_(1^m))
        =
       max_≻(s_ρ^P∘ s_π^T).

 As a warm-up, we first consider the case where π is linear.  
 If μ=(m) and π=(p)  then (see Example <ref>) <ref> says that
(n^m-1) ⊔ (ν^M)^T=(q^p-1) ⊔ (ρ^P)^T. 
By comparing   widths  we deduce that      q=n. This   implies m=p and then ν=ρ.
  Now,  suppose that  μ=(m) and π=(1^p).   
Then max_≻(s_ν∘ s_(m))=(nm-n)+ν which, as m ≥ 2 and ν has size n, has final column of  length 1. For   <ref> to hold,  the same to be true of  max_≻(s_ρ∘ s_(1^p))=(q^p-1) ⊔ρ; this implies p=2. Similarly, comparing the final columns of max_≻(s_ν^M∘ s_(1^m)) = (n^m-1) ⊔ν^M and max_≻(s_ρ^P∘ s_(p)) = (qp-q)+ρ^P also shows that m=2. Hence n=q and we obtain a contradiction from comparing the widths of 
(n) ⊔ν^M and (q)+ρ^M.

We now assume that π is non-linear so   π_1>1 and c>0.  By <ref>, 

    (n^m-1) ⊔ν^M 
    = (qπ_1^T,qπ_2^T,…, qπ_π_1-1^T,qπ_π_1^T-q+ρ_1^M,ρ_2^M,…  ).

 Since m≥ 2   and π_1>1, it follows that n=qπ_1^T=q(c+1)   and, as mn=pq, p=(1+c)m.  

   If ν^M=(n) then the left hand side of <ref> is (n^m).   
  Since  q<n, comparing the width in  <ref> shows that ρ^P=(q) and  that  π = (m^c+1).  
  This implies that max_≻(s_ν∘ s_(m)) is a hook partition whereas
   max_≻   ( s_ρ∘ s_ (m^c+1 )	) has second row of width at least q(m-1)>1, a contradiction.  
   
 Therefore we can assume that  ν^M≠ (n).  
Then  <ref> implies  that the first m-1   rows of π^T are all equal to n/q=c+1 and  therefore  
 π=((m-1)^c+1) + π' for some π'⊢ c+1.  
 In particular, π_1-π_2 ≤ c+1.  
We now consider <ref>: the difference between the first and second rows of   
max_≻(s_ν∘ s_μ) is 

    ((m-1)n+ν_1)-ν_2

whereas the difference between the first and second rows of max_≻(s_ρ∘ s_π) is 
 less than or equal to q× (π_1-π_2  +1)	 = n+q.     
  Therefore the necessary inequality 

    (m-1)n+ν_1 -ν_2 ≤ n+q

implies that m=2 (since q<n).  For the remainder of the proof μ=(2) and π= (1^c+1) + π'
⊢ 2(c+1) and therefore ρ^P=ρ and ν^M=ν.     

We first consider the case c>1. Here we have that ℓ(π)=c+1>2 and so 
the difference between the first and second rows of max_≻(s_ρ∘ s_π) is 
 q× (π_1-π_2)=q(π'_1-π'_2)≤ q(1+c)=n. On the other hand, for   max_≻(s_ν∘ s_(m))=(n)+ν the difference is at least n. For equality, we require π'=(c+1), that is π=(c+1, 1^c). 
Then <ref> becomes (n)+ν=(q(c+1)+q, q^c-1) ⊔ρ and we find ν=(q^c) ⊔ρ.
We now employ the dominance ordering to examine the case 
 
    π =(c+2,1^c)     ν=(q^c) ⊔ρ.
 


A necessary condition for   ( (c+2,1^c)^ρ, α))≠∅  
  is that 
α_1+α_2≤ q(c+3).  To see this, simply note that if ∈( (c+2,1^c)^ρ, α)), then 

    : [ρ] →_((c+2,1^c))

and the maximum number of entries equal to 1 or 2 in a semistandard Young tableaux of shape (c+2,1^c) is equal to 
(c+2)+1=c+3 (the sum of the lengths of the first and second rows of (c+2,1^c)).  
Thus p(ρ, (c+2,1^c),α) =0 for any α such that α_1+α_2 >q(c+3) by <ref>.  
We shall now construct  a plethystic tableau ∈((2)^(q^c) ⊔ρ,β) with β_1+β_2>q(c+3). This tableau will either be of maximal possible weight or there  exists another plethystic tableau of the same shape but of weight β' β; in either case,  for a partition  for γ∈{β,β'}, 0 ≠ p((q^c) ⊔ρ,(2),γ ) whereas  p(ρ,(c+2,1^c),γ ) =0  (by <ref>), 
 providing us with the necessary contradiction.
  Let ∈((2)^ (q^c) ⊔ρ, β) be the plethystic tableau such that 
 
    (a,b)=
    (2;2)    if (a,b) is the least dominant (lowest) removable node of (q^c) ⊔ρ
    (1;a)    otherwise.
 
This tableau has weight β with β_1=q(c+2)-1 and β_2=q+2 and so β_1+β_2=   
q(c+3)+1 as required. 
 

Finally, we consider the case c=1. Here 
μ=(2) and π⊢ 2(c+1)=4 is either (3,1) or (2,2).
 In the (2^2) case, comparing the widths of the partition on the left and right   of <ref> we see that ν_1=0, a contradiction.  
 In the (3,1) case, comparison of maximal terms again reveals that ν=(q)  ⊔ρ. Now

    s_ρ∘ s_(3,1)=
    s_ρ∘ (s_(1^2)∘ s_(2))=
    (s_ρ∘  s_(1^2))∘ s_(2).
 
We observe that  max_≻(s_ρ∘ s_(1^2))=(q) ⊔ρ, but s_ρ∘ s_(1^2) is decomposable unless ρ=(1^2) by <ref>. For ρ≠ (1^2), we deduce that 
s_(q) ⊔ρ∘ s_(2) is properly contained in s_ρ∘ s_(3,1). Thus we have q=2, ρ=(1^2) and ν=(2,1^2), as required. 
     

 
We may conjugate (applying <ref>) to complete the  case where μ is linear.

 
 Let μ,ν,  π,ρ be  partitions of m,n,p,q>1 respectively. 
 We suppose that   μ=(1^m).  
   If 
 
    s_ν∘ s_μ = s_ρ∘ s_π

 then either ν=ρ and μ=π or we are in the exceptional case 
 
    s_(2,1^2)∘ s_(1^2)  = 
    s_(1^2)∘ s_(2,1^2).
     
 


Let μ,ν,  π,ρ be arbitrary partitions of m,n,p,q>1 respectively.   
 We now consider what the condition

    max_≻(s_ν∘ s_μ)
     =
       max_≻(s_ρ∘ s_π)

  tells us about this quadruple of partitions.  
  We first suppose that ℓ(μ)=ℓ(π)=k.  This implies that ℓ(ν)=ℓ(ρ)=ℓ, say. Furthermore,
  
    ( nμ_1,nμ_2,…, nμ_k-1, nμ_k-n+ν_1, ν_2,  …, ν_ℓ)
    
    =
        ( qπ_1,qπ_2,…, qπ_k-1, qπ_k-q+ρ_1, ρ_2,  …, ρ_ℓ).

We set d=(n,q),  
e=(m,p)  and set n=n'd, q=q'd, m=m'e, p=p'e.   
Since 
mn=pq, we note that m'n'ed=p'q'ed and so m'n'=p'q'.  
Since m' and p' are coprime, as are n' and q',  it follows that m'=q' and p'=n'.  Thus 

    m=q'e   
    n=n'd    
    q=q'd 
       
    p=n'e.

From <ref>, we observe that nμ_i=π_i q implies n'μ_i = q' π_i, and so we can set 
α_i:=μ_iq'
 =π_in'∈ for all 1≤ i ≤ k-1.  
Now, μ⊢ m=q'e and so the final row length satisfies

    μ_k=q'e - ∑_i=1^k-1 q' α_i = q'
    (e- ∑_i=1^k-1α_i ) _α_k.

We have a partition 
(α_1,…,α_k)⊢ e with q'α=μ, and, in a similar fashion,  we deduce that  n'α=π.   
Without loss of generality, we now assume that n≥ q.  We plug in our equalities 
π=n'α and μ=q'α back into <ref> and to show  that

    ρ_i=ν_i for i ≥ 2 and ν_1 = (n-q)+ρ_1.

We immediately obtain the following corollary.  


 Let μ,ν,  π,ρ be  partitions of m,n,p,q>1, respectively. 
 We suppose that   ℓ(π)=ℓ(μ).  If 
 
    s_ν∘ s_μ = s_ρ∘ s_π

 then  ν=ρ and μ=π.  


By the discussion above, we know that we are dealing with a quadruple

    μ=q'α,
         ν=ρ+(n-q),
         π=n'α,	
         ρ	.

Comparing the width of the  partitions on the left  and right  of 

    max_≻(s_ν^M∘ s_μ^T)=
     max_≻(s_ρ^P∘ s_π^T)

we deduce that  ℓ(μ)n=ℓ(π)q.  Thus n=q, ν=ρ, q'=n' and thus 
μ=π, as required.  


We now consider the case 
 where the  lengths of the partitions μ and π (and hence ν and ρ) differ.
 We suppose (without loss of generality) that ℓ(μ)<ℓ(π).  
 We set ℓ(μ)=k and ℓ(π)=k+c for some c≥ 1.  Thus ℓ(ρ)+c=ℓ(ν)=ℓ, say.  
Observe that max_≻(s_ν∘ s_μ)
 =
   max_≻(s_ρ∘ s_π) if and only if  the partitions 

    0.9[         (nμ_1             …        nμ_k-1    nμ_k-n+ν_1           ν_2             …           ν_c         ν_c+1         ν_c+2             …          ν_ℓ);         (qπ_1             …        qπ_k-1          qπ_k        qπ_k+1             …      qπ_k+c-1 q π_k+c-q+ρ_1           ρ_2             …       ρ_ℓ-c). ]

 coincide.  
We deduce that 
 
    μ=q'(α_1,…,α_k),	        π = n' (α_1,…, α_k-1) ⊔(π_k,…, π_k+c)
 
for α⊢ e,   (π_k,…, π_k+c)⊢ n'α_k  and 

    ν=(qπ_k - n(q'α_k-1)) ⊔ q(π_k+1,…, π_k+c-1)⊔ ( q(π_k+c-1)+ρ_1)⊔ (ρ_2, ρ_3, …ρ_ℓ-c)

and, in order for ν to be a partition, we need
    
    qπ_k - n(q'α_k-1)  ≥ q π_k+1

which, rearranging, gives 
   
    q(π_k - π_k+1)   ≥ n(μ_k-1).

We are now ready to complete our proof of Theorem A.



 
  
 Let μ,ν,  π,ρ be  partitions of m,n,p,q>1, respectively. 
 We suppose that both μ and π are non-linear  and ℓ(π)>ℓ(μ).     If 
 
    s_ν∘ s_μ = s_ρ∘ s_π

 then   ν=ρ and μ=π.
 
 
 
 
 
We set ℓ(μ)=k≥ 2 and ℓ(π)=k+c for c≥ 1.  
We first see what can be deduced from max_≻(s_ν∘ s_μ)
 =
   max_≻(s_ρ∘ s_π).  
From <ref> we have that

    μ=q'(α_1,…,α_k)	        π = n' (α_1,…, α_k-1) ⊔(π_k,…, π_k+c)
 
for α⊢ e and   (π_k,…, π_k+c)⊢ n'α_k,
and, from <ref>,  we  
deduce that |ρ|<|ν| and so q=q'd<n'd=n which implies q'<n'.  
From <ref> this implies that μ_1=q'α_1<n'α_1=π_1 in other words ℓ(μ^T) < ℓ(π^T).  


 We now see what can be deduced from max_≻(s_ν^M∘ s_μ^T)
 =
   max_≻(s_ρ^P∘ s_π^T).  
We have already concluded that ℓ(μ^T)< ℓ(π^T).  Therefore applying <ref> (but with the partitions 
μ^T, ν^M, π^T and ρ^P) we deduce that 

    μ^T=q'(β_1,…, β_μ_1)
        π^T= n'	(β_1,…, β_μ_1-1) ⊔ (π^T_μ_1,…, π^T_π_1)

for some β⊢ e and 
(π^T_μ_1,…, π^T_π_1)⊢ n'β_μ_1. 
 
From <ref> we deduce that μ can be built from boxes of size q'× q'.  In other words,

    μ=q'(γ_1,γ_1,…, γ_1_q',γ_2,γ_2,…, γ_2_q', …).

for some γ⊢ m /q'^2.  Since γ might have repeated parts, we write γ in the form

    γ=(a_1^b_1,a_2^b_2,…, a_x^b_x)

where a_1>a_2>… >a_x, so

    γ^T=((b_1+…+b_x )^a_x,(b_1+…+b_x-1)^a_x-1-a_x,…, b_1^a_1-a_2 ).

Now,  <ref> reveals  that

    π = (n'a_1,… ,n'a_1_b_1q',n'a_2,… ,n'a_2_b_2q',…, 
    n'a_x,… ,n'a_x_b_xq'-1,
    π_k,
    … , π_k+c)
 where (π_k,
  … , π_k+c)⊢ n'a_x and, from <ref>,

    π^T =    ( (n'(b_1+…+b_x ))^ q' a_x,(n'(b_1+…+b_x-1 ))^  q'( a_x-a_x-1), … 
     (n'b_1)^q'(a_1-a_2)-1) ⊔
       
    ( π^T_μ_1,  … ,  π^T_π_1)

where 
(π^T_μ_1, … ,  π^T_π_1)⊢ n'b_1.  
 By looking at the first row of π^T we deduce that provided x 1 the last part of π is q'a_x and that it appears with multiplicity n'b_x.  This implies that 

    (π_k, …, π_k+c)
    =
    (…, q'a_x,… q'a_x_n'b_x)⊢ n'a_x.

But the sum over these final n'b_x rows is 
q'a_a × n' b_x 
which implies q'=1 and b_x=1 and that 

    (π_k, …, π_k+c)
    =
    (  a_x, … , a_x_n' )⊢ n'a_x.

Now we input this into <ref> to deduce that 


  


    ℓ(π)
    
    =b_1+… +b_x -1+n'.
 

  On the other hand by <ref> we know that

    ℓ(π)=π^T_1= 
     n' (b_1+…+b_x )

Therefore 

    n'(b_1+…+b_x -1)
    =
    b_1+b_2+… +b_x-1

and thus n'=1 or b_1+b_2+… +b_x =1.  If n'=1 then n=q, contrary to our earlier observation that q<n.

If  b_1+b_2+… +b_x =1, then ℓ(γ)=ℓ(α)=ℓ(μ)=1, contrary to our assumption that μ is non-linear.




Finally, it remains to consider the x=1 case.  This is the case in which γ=(a^b) is a rectangle.  
Here we have that 
μ=q'(a^bq'), μ^T=q'(b^aq')
and 

    π = ((n'a)^q'b-1)⊔( π_k,… ,π_k+c)      for ( π_k,… ,π_k+c)⊢ n'a
    π  =  ((q'a-1)^n'b)+( π_μ_1^T,… ,π_π_1^T)^T      for ( π_μ_1^T,… ,π_π_1^T)⊢ n'b.

Now, recall that q'<n'; 







and so 

    q'b-1<q'b <n'b

and so the  rectangle  in  <ref> is at least 2 rows shorter than that in <ref>.


This implies that q'=1 and a or b=1 and so μ is linear, a contradiction.  
 
 
 
 
We have now classified all possible equalities between  products s_ν∘ s_μ = s_ρ∘ s_π where neither, one, or both of π and μ are linear partitions.  
This completes the proof of Theorem A.  


We would like to thank Cedric Lecouvey for
 bringing to our attention the question of unique factorisability of products of Schur  functions and in particular for introducing us to  Rajan's  result for Littlewood–Richardson products.  
 

   
   



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