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### Article

Kyungpook Mathematical Journal 2022; 62(1): 1-28

Published online March 31, 2022

Copyright © Kyungpook Mathematical Journal.

### On the Decomposition of Cyclic G-Brauer's Centralizer Algebras

Annamalai Vidhya and Annamalai Tamilselvi∗

Ramanujan Institute for Advanced Study in Mathematics, University of Madras, Chennai - 600 005, India
e-mail : vidhyamath@gmail.com and tamilselvi.riasm@gmail.com

Received: January 20, 2020; Accepted: February 8, 2021

In this paper, we define the G-Brauer algebras $DfGx$, where G is a cyclic group, called cyclic G-Brauer algebras, as the linear span of r-signed 1-factors and the generalized m,k signed partial 1-factors is to analyse the multiplication of basis elements in the quotient $If→G(x,2k)$. Also, we define certain symmetric matrices $T→m,k[λ](x)$ whose entries are indexed by generalized m,k signed partial 1-factor. We analyse the irreducible representations of $DfGx$ by determining the quotient $If→G(x,2k)$ of $DfGx$ by its radical. We also find the eigenvalues and eigenspaces of $T→m,k[λ](x)$ for some values of m and k using the representation theory of the generalised symmetric group. The matrices $Tm,k[λ](x)$ whose entries are indexed by generalised m,k signed partial 1-factors, which helps in determining the non semisimplicity of these cyclic G-Brauer algebras $DfGx$, where $G=ℤr$.

Keywords: G-Brauer algebras, centraliser algebras, eigenvalues

The invariant theory of classical groups, algebraic Lie theory, algebraic number theory, knot theory, integrable models and statistical mechanics, quantum computing are the few areas of diagram algebras arising in different areas of mathematics and physics. In order to characterise invariants of classical groups acting on tensor powers of the vector representations, Brauer [2] introduced a new class of algebras called Brauer algebras. The Brauer algebras used graphs to represent its basis. Hence it can be considered as a class of diagram algebras, that are finite dimensional algebras whose basis consists of diagrams. These basis have interesting combinatorial properties to be studied in their own right.

Parvathi and Kamaraj [10] introduced a new class of algebras called signed Brauer algebras $Sf(x)$ which are a generalization of Brauer algebras. Parvathi and Selvaraj [12] studied signed Brauer algebras as a class of centraliser algebras, which are the direct product of orthogonal groups over the field of real numbers $ℝ$. Parvathi and Savithri [11] introduced a new class of algebras called G-Brauer algebras $DfGx$, where G is abelian, which are a generalization of signed Brauer algebras $Sf(x)$ introduced by [10] and Brauer algebras.

Brown [3, 4], Hanlon and Wales [6, 7] and Wenzl [14] studied the Brauer algebras by using diagrams to represent its basis and Young diagrams to represent its irreducible representations. Brown has not discussed the structure of Brauer algebras when the radical is non zero. This study was carried out by Hanlon and Wales in [6]. They determined the structure of the radical of a non-semisimple Brauer algebras by introducing the notion of 1-factor, m,k-partial 1-factor and the combinatorially defined matrix $Tm,kλ(x)$. In [7], they used these matrices to find the eigen values and eigen vectors corresponding to Brauer's centraliser algebras.

However for signed Brauer algebras, the eigen values corresponding to a non-semisimple signed Brauer algebras have not been dealt completely. This motivated us to study the eigen values for the signed Brauer algebras [13] and G-Brauer algebras where $G=ℤr$.

In this paper, we analyse the irreducible representations of $DfGx$ by determining the quotient $I→fG(x,2k)$ of $DfGx$ by its radical. We also find the eigenvalues and eigenspaces of certain symmetric matrices $T→m,k[λ](x)$ for some values of m and k using the representation theory of the generalised symmetric group. The matrices $Tm,k[λ](x)$ whose entries are indexed by generalised m,k signed partial 1-factors, which helps in determining the non semisimplicity of these cyclic G-Brauer algebras $DfGx$, where $G=ℤr$. In this paper G refers to $ℤr, r>0$.

We begin by recalling some known results in the representation theory of the generalised symmetric group [1, 5, 8].

Definition 2.1. For each standard multi-tableaux [s] and [t] of shape [λ],

$m[s][t]=π∑ σ∈R [t] σ, π∈D([t])$

where $π[s]=[t]$, $D([t])$ is the left coset representative of $R[t]$, row stabiliser of [t].

Definition 2.2. For each multi-partition $[λ]$, there exists two sided ideals $K[λ]$ and $K[λ]¯$ of $S^^m$, group algebras of the generalised symmetric group.

• 1. $K[λ]=$ Linear span ${m[s][t]/[s]$ and [t] are standard multi-tableaux of shape $[μ]$ for $[μ]⊵[λ]}$.

• 2. $K[λ]¯=$ Linear span ${m[s][t]/[s]$ and [t] are standard multi-tableaux of shape $[μ]$ for $[μ]⊳[λ]}$.

• 3. $S^^m[λ]=$ Linear span ${m[s][t]/[s]$ and [t] are standard multi-tableaux of shape $[λ]}$.

Remark 2.3. • For each multi-partition $[λ]$ of m, let $S→[λ]$ denote the Specht module corresponding to $[λ]$ and let $d[λ]$ denote the dimension of $S→[λ]$.

• The ideal $S^^m[λ]$ considered as a vector space of linear transformations of $S→[λ]$ is the full matrix algebras $End(S→[λ])$.

Definition 2.4. There exists left ideals $I→1,I→2,…,I→d[λ]$, right ideals $J→1,J→2$, $…,J→d[λ]$ and the unique minimal two sided ideal $S^^m[λ]$ of $S^^m$, group algebras of the generalised symmetric group. $S^^m[λ]$ can be written either as direct sum of simple left ideals

$S^^m[λ]=I→1⊕⋯⊕I→d[λ],$

or as a direct sum of simple right ideals

$S^^m[λ]=J→1⊕⋯⊕J→d[λ]$

where each $I→i$ is a left ideal of $S^^m$ for which multiplication on the left gives a representation isomorphic to $S→[λ]$ and each $J→i$ is a right ideal of $S^^m$ for which right multiplication is isomorphic to $S→[λ]$.

There exists a basis $A1,…,Ad[λ]$ for $S→[λ]$ with respect to which the matrices $Ψ[λ](σ)$ for $σ∈ S^^m$ acting on $S→[λ]$ are orthogonal. i.e. $Ψ[λ](σ−1)=Ψ[λ](σ)t$. For any elements $xi,yj∈ S^^m[λ],1≤i,j≤d[λ]$, choose $xi,yj$ so that

Definition 3.1. A r-signed 1-factor on 2f vertices is a signed diagram with f vertices arranged in two rows and f, r-signed edges such that each vertex is incident to exactly one r-signed edge, labeled by the primitive $rth$-root of unity $ξ1,ξ2,…,ξr$. A r-signed edge is said to be an i-edge if it is labeled by $ξi$. The set of all r-signed 1-factor on 2f vertices is denoted by $PfG$, where G is assumed to be a cyclic group.

A r-signed 1-factor $δ∈PfG$ will be represented as a diagram having two rows of f vertices each, the f vertices in the top row are labeled by $1,2,…,f$ from left to right and the f vertices in the bottom row are labeled by $f+1,f+2,…,2f$ from left to right. There are $rf(2f−1)!!=rf⋅1⋅3⋯(2f−1)$ ways of joining these 2f vertices which is incident to exactly to one r-signed edge.

Example 3.2. The 27 r-signed 1-factors in $P2ℤ3$ is as follows.

An r-signed edge of $δ∈PfG$ is called r-signed horizontal edge if it joins two vertices in the same row of $δ∈PfG$.

An r-signed edge of $δ∈PfG$ is called r-signed vertical edge if it joins two vertices in different rows of $δ∈PfG$.

Definition 3.3. Let $V→f$ be the vector space over a field K with $PfG$ as its basis.

Definition 3.4. Let $V→f(2k)=$ Linear span${δ∈PfG/$the number of r-signed horizontal edges in $δ≥2k}$, $k=0,1,…,f2$ which is a subspace of $V→f$. $V→f(2k)$ can be written as direct sum of vector subspaces $V→f*(2m),k≤m≤f2$ spanned by all r-signed 1-factors having exactly 2m r-signed horizontal edges.

$Vf→(2k)=Vf→*(2k)⊕Vf→*(2k+2)⊕…⊕Vf→*2f2.$

In the following, we make $V→f$ as an algebra over the field K(x), where K is any field and x is an indeterminate, by defining composition of two elements $δ1,δ2∈PfG$.

For $δ1,δ2∈PfG$, the graph $U→δ2δ1$ with 3f vertices arranged in three rows with the first row, the top row of $δ1$, the second row is obtained by identifying the vertices in the bottom row of $δ1$ with the vertices in the top row of $δ2$ and the third row, the bottom row of $δ2$. The graph $U→δ2δ1$ consists of exactly f, r-signed paths $P1,P2,…,Pf$, some number $mi(δ1,δ2))$ of i-cycles for $i=1,2,…,r$ such that

• 1. The r-signed path $Pi$ contains one or more r-signed edges. The initial and endpoint of $Pi$ does not meet each other.

• 2. Each i-cycle is of even length consisting some number kj of ij-edges, $1≤ij≤r$ with $∑kjij≡i( mod r)$, entirely of vertices lying in the middle row.

Example 3.5. For $δ1,δ2∈P5ℤ3$, the diagram in $U→δ2δ1$ is

Definition 3.6. Let $δ1$ and $δ2$ be r-signed 1-factors in $PfG$. Define the composition of r-signed diagrams $δ1∘δ2$ to be the r-signed 1-factor in the following way

• 1. The top (respectively bottom) row of $δ1∘δ2$ have the same r-signed horizontal edges in the top (respectively bottom) row of $δ1$ (respectively $δ2)$.

• 2. The vertices u and v are adjacent if and only if there is a r-signed path $Pi$ in $U→δ2δ1$ joining u to v and an edge joining u to v is an i-edge if the path contains some number kj of ij-edges, $1≤ij≤r$ with $∑kjij≡i( mod r)$.

Definition 3.7. The cyclic G-Brauer algebra $DfGx$ is an associative algebras over the field K[x] with basis $PfG$ and the multiplication * of r-signed 1-factors given by

This algebras $DfGx$ is called the G-Brauer algebras defined in [11], when $G=ℤr$.

Example 3.8. For $δ1,δ2∈P5ℤ3$, the diagram in $δ1*δ2$ is

### 4. The Structure of Ideals $IfG(x,2k)$

Let $IfG(x,2k)=$Linear span${δ∈PfG/$number of r-signed horizontal edges in $δ≥2k}$. Clearly by the multiplication defined above $IfG(x,2k)$ is an ideal of $DfGx$.

Let $I→fG(x,2k)=$Linear span${δ∈PfG/$number of r-signed horizontal edges in δ is equal to $2k}$. $I→fG(x,2k)$ denotes the quotient $IfG(x,2k)/IfG(x,2k+2)$.

To describe the structure of the quotients $I→fG(x,2k)$ in terms of the eigenvalues and eigenspaces of certain matrices.

Definition 4.1. A generalised m,k signed partial 1-factor on f=m+2k vertices is a r-signed diagram whose vertices are arranged in a single row with k, r-signed horizontal edges and m free vertices.

Let $Pm,kG$ denotes the set of all generalised m,k signed partial 1-factors and let $Vm,kG$ be the real vector space with basis $Pm,kG$.

The generalised symmetric group on m symbols ${1,2,…,m}$ is denoted by $SmG$.

Let us now define $π∈SmG$ by $π(i)=(τ(i),σ(i))$ where $σ∈Sm$ and the function $τ:m_→r_$, where $m_$ denotes the set ${1,2,…,m}$ and $r_$ denotes the set ${ξ,ξ2,…,ξr}$, $ξi$'s are primitive rth root of unity.

Definition 4.2. Let f1 (respectively f2) be generalised m,k signed partial 1-factors with the free vertices of f1 (respectively f2) is labeled by $α1<α2<⋯<αm$ (respectively $β1<β2<⋯<βm$).

The union of f1 and f2 is a r-signed graph obtained by identifying i-th vertex of f1 with the i-th vertex of f2 consists some number $mi(f1,f2)$ of disjoint i-cycles together with m disjoint r-signed paths $P1,…,Pm$ whose endpoints are in the set ${α1,α2,…,αm,β1,β2,…,βm}$.

Definition 4.3. Let $f1,f2∈Pm,kG$. Define an inner product $$ on $Vm,kG$ as follows.

• 1. If any r-signed path $Pi$ joins a αj to a αi (or equivalently a βj to a βi) then $=0$.

• 2. If each r-signed path $Pi$ joins βi to $ασi$ and some number kj of ij edges $1≤ij≤r$ with , then

Note. $=$, where ✠ is the anti-isomorphism defined on the algebras $KSmG$ by $σ→σ−1,σ∈SmG$.

Proposition 4.4. Let f=m+2k. Then the quotient $I→fG(x,2k)$ is isomorphic as algebras to $(Vm,kG⊗Vm,kG⊗KSmG,⋅)$, where

$(a⊗b⊗π1)⋅(c⊗d⊗π2)=a⊗d⊗(π1π2),a,b,c,d∈Pm,kG,π1,π2∈SmG.$

Proof. The proof follows as in [6]. Instead of m,k partial 1-factors and symmetric group, the generalised m,k signed partial 1-factors and the generalised symmetric group are used to prove the theorem, we give it here for the sake of completion.

As a vector space $I→fG(x,2k)$ has basis the set of all r-signed 1-factors with exactly 2k r-signed horizontal edges.

Consider the linear map $ϕ:Vm,kG⊗Vm,kG⊗KSmG→I→fG(x,2k)$.

Given $f1,f2∈Pm,kG$, we define $ϕ(f1⊗f2⊗σ)$ to be the r-signed 1-factor on 2f vertices in the following way.

Let f1 be the generalised m,k signed partial 1-factor with free vertices $α1<α2<⋯<αm$ and let f2 be the generalised m,k signed partial 1-factor with free vertices $β1<β2<⋯<βm$ and given $σ∈SmG$ such that

• 1. A r-signed horizontal edge joining i to j in the top row if and only if i and j are joined by an r-signed horizontal edge in f1.

• 2. A r-signed horizontal edge joining f+i to f+j in the bottom row if and only if i and j are joined by an r-signed horizontal edge in f2.

• 3. A r-signed vertical edge joining αi to $βσ(i)$ for $i=1,2,⋯m$.

The linear map ϕ defined in this way is clearly 1-1 and onto. Hence it is a vector space isomorphism of $Vm,kG⊗Vm,kG⊗KSmG$ onto $I→fG(x,2k)$.

It remains to show that ϕ is multiplicative.

Let us now assume that $d1=a⊗b⊗π1$ and $d2=c⊗d⊗π2$ where a,b,c,d be the generalised m,k signed partial 1-factors with free vertices $α1<α2<⋯<αm$, $β1<β2<⋯<βm$ , $γ1<γ2<⋯<γm$, $ψ1<ψ2<⋯<ψm$ respectively and $π1,π2∈SmG$.

Consider the product in $I→fG(x,2k)$

$ϕ(d1)∘ϕ(d2)=x∑ i=1rimi(d1,d2)d3, d3∈DfGx.$

Case 1. Suppose there is a r-signed path joining αi to αj or ψi to ψj in $U→ϕ(d2)ϕ(d1)$ then $ϕ(a⊗b⊗π1)∘ϕ(c⊗d⊗π2)=0=ϕ((a⊗b⊗π1)⋅(c⊗d⊗π2))$. Therefore

$ϕ(d1⋅d2)=ϕ(d1)∘ϕ(d2).$

Case 2. Suppose there is a r-signed path joining αi to $ψσ(i)$ in $U→ϕ(d2)ϕ(d1)$ for $i=1,2,…,m$, then $ϕ(d1)∘ϕ(d2)=x∑ i=1rimi(δ1,δ2)d3=ϕ(d1⋅d2)$.

Hence ϕ is an algebra isomorphism.

### 5. The Structure of Ideals $I→fG(x,2k)$

To describe the structure of the ring $I→fG(x,2k)$ in terms of the eigenvalues of certain matrices.

Definition 5.1. Let $T→m,k[λ](x)$ be the $(pd[λ])$-by-$(pd[λ])$ matrix which is $d[λ]$-by-$d[λ]$ blocks of p-by-p matrices where p is the number of generalised m,k signed partial 1-factors. The matrices in the each block are indexed by pairs of generalised m, k signed partial 1-factors being $ψ[λ]()$, for the corresponding r-signed 1-factor.

Let $N→[λ]$ denotes the null space of $T→m,k[λ](x)$, the matrix corresponding to generalised m,k signed partial 1-factors and the multi partition [λ] and $R→[λ]$ denotes the range of $T→m,k[λ](x)$, the matrix corresponding to generalised m,k signed partial 1-factors and the multi partition [λ].

Note. If $=x∑ i=1rimi)(b,c)σ$ then $=x∑ i=1rimi(c,b)σ−1$. So the matrix $T→m,k[λ](x)$ is symmetric.

Choose a basis $u(1),…,u(n)$ for $N→[λ]$ and an orthonormal basis of eigenvectors $v(1),…,v(r)$ for the nonzero eigenvalues $μ(1),…,μ(r)$.

Definition 5.2. For given any left ideal $I→t$ and a generalised m, k signed partial 1-factor d, define

Lemma 5.3. $V→L(I→t,d)$ is a left ideal of $I→fG(x,2k)$.

Proof. The proof follows as in [6]. Instead of m,k partial 1-factors and symmetric group, the generalised m,k signed partial 1-factors and the generalised symmetric group are used to prove the theorem, we give it here for the sake of completion.

For $c⊗d⊗x∈V→L(I→t,d)$, choose $δ∈I→fG(x,2k)$ such that $δ*(c⊗d⊗x)$ not equal to zero, that is $δ*(c⊗d⊗x)$ and $c⊗d⊗x$ have same number of r-signed horizontal edges.

Since $x∈I→t$ and $I→t$ is a left ideal of $S^^m$, $y=ππ1x∈I→t$. Hence

$(a⊗b⊗π)*(c⊗d⊗x)=x∑ i=1rimi(b,c)(a⊗d⊗y)∈V→L(I→t,d).$

Therefore $V→L(I→t,d)$ is a left ideal of $I→fG(x,2k)$.

Define $W→L(I→t,d)⊂V→L(I→t,d)$ to be the linear span of all

$∑(u)(c,i)c⊗d⊗A→i,t,$

where u is in $N→[λ]$ and $A→i,t$ is the basis element of $I→t$ corresponding to the basis element $A→i$ in $S→[λ]$. i.e. the linear span of the set of all elements mapped to zero in $I→fG(x,2k)$.

Proposition 5.4. Suppose $v=∑(v)(c,i)c⊗d⊗A→i,t∈V→L(I→t,d)$. Let a, b be generalised m,k signed partial 1-factors. For any $σ∈SmG$, $(a⊗b⊗σ)v=a⊗d⊗σ∑γjA→j,t$, where γj is the (b, j) entry of $T→m,k[λ](x)(v)$.

Proof. The proof follows as in [6]. Instead of m,k partial 1-factors and symmetric group, the generalised m,k signed partial 1-factors and the generalised symmetric group are used to prove the theorem, we give it here for the sake of completion.

$(a⊗b⊗σ)*∑v(c,i)c⊗d⊗A→i,t=a⊗d⊗σ∑v(c,i)A→i,t =a⊗d⊗σ∑(v)c,ix∑ i=1rimi(b,c)σ1Ai,t =a⊗d⊗∑γjA→j,t$

where $σ1Ai,t=Aj,t$ and γj is the coefficient of $A→j,t$ in $∑v(c,i)A→j,t$.

By definition of $T→m,k[λ](x)$, the coefficient of $A→j,t$ in $A→i,t$ is the (b, j), (c, i) entry of $T→m,k[λ](x)$. Thus γj is the (b,j) entry of $T→m,k[λ](x)(v)$.

Proposition 5.5.

• 1. $If→G(x,2k)W→L(I→t,d)=0$.

• 2. $I→(v)= V → L( I → t,d)$ for any v in $V→L(I→t,d)$ not in $W→L(I→t,d)$.

• 3. $V→L(I→t,d)/W→L(I→t,d)$ is irreducible as a left $I→fG(x,2k)$ module.

Proof. The proof follows as in [6]. Instead of m,k partial 1-factors and symmetric group, the generalised m,k signed partial 1-factors and the generalised symmetric group are used to prove the theorem. We give it here for the sake of completion.

Suppose w is a generating element of $W→L(I→t,d)$ and γj is the (b,j) entry of $T→m,k[λ](x)(v)$. By the definition of $W→L(I→t,d)$, γj are all 0, for any $(a⊗b⊗σ)$. Hence $I→fG(x,2k)W→L(I→t,d)=0$.

Suppose v is in $V→L(I→t,d)$ but not in $W→L(I→t,d)$. Choose γj, the (b,j) entry of $T→m,k[λ](x)(v)$ is not 0. Then $a⊗b⊗σ(v)$ is not 0 Note that a and σ were arbitrary. Since $I→t$ is an irreducible $S^^m$ module, the images under $σ∈ S^^m$ of any nonzero vector in $I→t$ generate all of $I→t$. Hence vectors of the form $(a⊗b⊗σ)*∑v(c,j)c⊗d⊗A→j,t$ generate all of $V→L(I→t,d)$. Hence $I→(v)= V → L( I → t,d)$.

The last part of the proof from the above two.

Let $W→L[λ]=⊕W→L(I→t,d)$ and . By Proposition 5.4, $W→L[λ]$ is a nilpotent left ideal of $I→fG(x,2k)$.

Recall that $S^^m[λ]$ can also be written as a direct sum of right ideals $J→1,…,J→d[λ]$.

For given any right ideal $J→t$ and a generalised m,k signed partial 1-factor a, define

Lemma 5.6. $V→R(J→t,c)$ is a right ideal of $I→fG(x,2k)$.

Proof. For $c⊗d⊗x∈V→R(J→t,c)$, choose $δ∈I→fG(x,2k)$ such that $(c⊗d⊗x)*δ$ not equal to zero, that is $(c⊗d⊗x)*δ$ have same number of signed horizontal edges as in $c⊗d⊗x$.

Since $x∈J→t$ and $J→t$ is a right ideal of $S^^m$, $y=xπ1π∈J→t$,

$(c⊗d⊗x)*(a⊗b⊗π)=x∑ i=1rimi(d,a)(c⊗b⊗y)∈V→R(J→t,c).$

Therefore $V→R(J→t,c)$ is a right ideal of $I→fG(x,2k)$.

Define $W→R(J→t,a)⊂V→R(J→t,a)$ to be the linear span of all

$∑(u)(c,i)a⊗b⊗A→j,t$

where $utT→m,k[λ](x)=0$ and $A→j,t$ is as before.

The same proofs used in Propositions 5.4 and 5.5 shows that

• 1. $W→R(J→t,a)∘(c⊗d⊗σ)=0$,

• 2. $V→R(J→t,a)/W→R(J→t,a)$ is an irreducible right $I→fG(x,2k)$ module.

Define $W→R[λ]=⊕W→R(J→t,a)$ and define $W→[λ]$ to be the nilpotent 2-sided ideal $W→[λ]=W→L[λ]+W→R[λ]$.

Definition 5.7. Define $D→[λ]$ to be the 2-sided ideal of $I→fG(x,2k)$ given by the linear span of all vectors $a⊗b⊗x$, where a and b are arbitrary and $x∈ S^^m[λ]$.

Note that $I→fG(x,2k)$ is the direct sum of the $D→[λ]$.

Proposition 5.8. $D→[λ]/W→[λ]$ is canonically isomorphic to the full matrix ring $End(R→[λ])$. Recall that $R→[λ]$ is the range of $T→m,k[λ](x)$.

Proof. Instead of m,k partial 1-factors and symmetric group, the generalised m,k signed partial 1-factors and the generalised symmetric group are used to prove the theorem. We give it here for the sake of completion.

Given eigen vectors v(r) and v(s) define

$Z(v(r),v(s))=μ(r) μ(s) −1∑v(r)a,iv(s) b,ja⊗b⊗xiyj.$

Taking the product of $Z(v(r),v(s))$ and $Z(v(t),v(u))$ we obtain

$Z(v(r),v(s))Z(v(t),v(u)) = μ (r) μ (s) μ (t) μ (u) −1 ∑ v(r) a,i v (u) d,la⊗d⊗ v (s) b,j v (t) c,k{xiyjxkyl} = μ (r) μ (s) μ (t) μ (u) −1∑ v(r) a,i v (u) d,l a⊗d⊗xiyj∑ v(s) b,j ∑ v(t)c,k xk yl.$

Now $∑v(t)c,kxk=∑γrxr$ where γr is the (b,r) entry of $T→m,k[λ](x)v(t)$. Since v(t) is an eigenvector with eigenvalue $μ(t)$, $γr=μ(t)v(t)b,r$. So

$xiyj∑ v(s)b,j∑ v(t)c,k xkyl=xiyj∑ v(s)b,j∑ γr xryl=μ(t)∑v(s) b,j v (t) b,rxiyjxryl=μ(t)xiyl∑ v(s)b,jv(t) b,j by equation1.1=μ(t)xiylδs,t by the orthonormality of the v(i).$

Therefore,

$Z(v(r),v(s))Z(v(t),v(u))= μ (r) μ (s) μ (t) μ (u) −1∑ v(r) a,i v (u) d,la⊗d⊗μ(t)xiylδs,t =δs,tZ(v(r),v(u)).$

Hence the subspace of $D→[λ]$ spanned by the $Z(v(r),v(s))$ is isomorphic to $End(R→[λ])$.

The ideal $D→[λ]=Vm,kG⊗Vm,kG⊗ S^ ^mλ$ is isomorphic as a vector space to $(Vm,kG⊗S→[λ])⊗(Vm,kG⊗S→[λ])$ via the linear map f sending $(c⊗A→i)⊗(d⊗A→j)$ to $c⊗d⊗xiyj$.

Writing $Vm,kG⊗S→[λ]$ as $N→[λ]⊕R→[λ]$ we have, from Propositions 5.4, 5.5 and 5.8, that

A. $f(N→[λ]⊗(Vm,kG⊗S→[λ])+(Vm,kG⊗S→[λ])⊗N→[λ])$ is contained in the radical of $DfGx(2k)$

B. $f(R→[λ]⊗R→[λ])$ is a full matrix ring.

The next theorem follows immediately from A and B.

Theorem 5.9. With notation as above:

• 1. Let $W→[λ]=fN→[λ]⊗(Vm,kG⊗S→[λ])+(Vm,kG⊗S→[λ])⊗N→[λ]$. Then $W→[λ]$ is the intersection of the radical of $I→fG(x,2k)$ with $D→[λ]$.

• 2. $D→[λ]/W→[λ]$ is a full matrix ring which is canonically isomorphic to $End(R→[λ])$.

### 6. Eigen Values for $T→m,k[λ](x)$ when m=0 and f Is Even

In this section, we determine the eigen values of in terms of representation of $SfG$, the generalised symmetric group on f points. Here we deal with the case s=f/2, the number of r-signed horizontal edges, when f is even.

Let Fs be the set of all r-signed 1-factors on f points arranged in a single row with exactly $s=f2$ horizontal edges. Let $T→s(x)δiδj$ be the $Fs×Fs$ matrix whose $(δi,δj)$ entry is $x∑ k=1rkmk(δi,δj)$ where mk is the number of k-cycles in $δi∪δj$.

Example 6.1. When r=3, f=2 and s=1.

The eigen values are $x(x−1)x2+x+1$, $−x(x−1)x2+x+1$ and $x3+x2+x$.

A generalised permutation $σ∈SfG$ induces a signed permutation of Fs by permuting the i-edges of r-signed 1-factors. If p and q are joined in δ, then σ(p) and σ(q) are joined in σ(δ), $σ∈SfG$.

Suppose $δ1,δ2∈Fs$ and if C1 is a connected component of $δ1∪δ2$, then $σ(C1)$ is a connected component of $σ(δ1)∪σ(δ2)$. In particular the number and size of $δ1∪δ2$ and $σ(δ1)∪σ(δ2)$ are the same.

Let Vs be the vector space with basis Fs. For $σ∈SfG$, let $Pσ$ be the generalised permutation matrix corresponding to the generalised permutation of Fs induced by $σ∈SfG$. In particular if $σ(δi)=δj$ and $σ(i)=(ξk,j)$, then $Pσ$ has a $ξk$ in the $Pδiδj$ and $0′$s elsewhere in the δi row and δj column. Hence $Pσ$ and $T→s(x)$ commutes to give

$PσT→s(x)=T→s(x)Pσ.$

The generalised permutation module has a decomposition as an $SfG$ module into irreducible subspaces corresponding to irreducible representations of $SfG$. The irreducible representations of $SfG$ are indexed by multi partitions [λ] of f. The irreducibles which occur as constituents of the generalised permutation module are indexed by even multi partitions [λ] of f. Furthermore the multiplicity of each representation is 1. This means that

$V=V1+V2+…Vn,$

where $V1,V2,…,Vn$ are invariant subspaces of $Pσ,∀σ∈SfG$ and n is the number of even multi partitions of f. As the irreducibles are distinct, $T→s(x)Vi⊆Vi$. As each Vi is irreducible, $T→s(x)$ restricted to Vi is a scalar, which is denoted by hi(x)I. In order to find the eigen values for $T→s(x)$, it is necessary to determine the scalars for $T→s(x)$ restricted to Vi. The multiplicity will be $dimVi$.

We determine these scalars hi(x) in terms of the multi partition associated with the representation and the location of certain integers on a grid. Let Δ be the grid and place the integer r(2j-i-1) in the position row and column. It is convenient to place the diagram of the even multi partition [λ] on the grid Δ.

Let [λ] be a even multi partition of f with every partitions of [λ] into even parts. Let $[λ]=(λ(1),λ(2),…,λ(r))$ where $λ(j)$ is a partition of mj with length lj where each $λi(j),i=1,2,…,lj,j=1,2,…,r$ is even, such that $∑im i=f$. Let $d=[d(1),d(2),…,d(r)]$. The diagram $d(i)$ corresponding to even partition $λ(i)$ on Δ is

There are exactly s=f/2 number of integers in Δ contained inside the boundary of d, the diagram of the even multi partition [λ].

Theorem 6.2. Let $[λ]=(λ(1),λ(2),…,λ(r))$ be an even multi partition of f. Let V[λ] be the subspace of Vi associated to the multi partition [λ] and $hs(x)=h[λ](x)$. Then

$h[λ](x)=∏ i=1r∏ d∈d (i)∏ k,lsi (x r+ξi(r−1)xr−1+…+ξix+akl(i)),$

where $|.|$ denotes modulus, $aij(i)$ are in the diagram d(i) of shape and

$h[λ](x)$ is a polynomial of degree r for the multipartition $[λ]=(λ(1),λ(2),…,λ(r))$.

Proof. The proof follows as in the approach of [6]. Instead of 1-factors, even partition standard tableaux and symmetric group, the r-signed 1-factors, multi partitions multi standard tableaux and generalised symmetric group are used to prove the theorem here. This is proved by induction on s=f/2. We give the proof here for the sake of completion.

Let

$d[λ](x)=∏ i=1r∏ d∈ d (i)∏ k,lsi (x r+ξi(r−1)xr−1+…+ξix+akl(i)),$

where $akl(i)$ are in the diagram d(i) of shape $λ(i)$. We must show that $h[λ](x)=d[λ](x)$. If s=1, then there are r possible even multi partitions of f, in the r tuple $(2,∅,∅,…,∅),(∅,2,∅,…,∅),…,(∅,∅,…,∅,2)$. $T→s(x)$ restricted to Vi is $si(xr+ξi(r−1)xr−1+…+ξix)(1)$. Since the dimension of Vi's are one, the multiplicity of each hs(x) will be one. Therefore $h[λ](x)=d[λ](x)$ for s=1. We suppose that the theorem is true for all even multi partitions of f of size smaller than 2s.

Let [λ] be an even multi partition of f=2s. Let [d] be the diagram of shape [λ] and t be the standard Young multi-tableau with $1,2,…,f$ are placed consecutively in each row in the diagram of [λ]. Let

$et=∑ε(σ)στ,∀σ∈Ct,τ∈Rt.$

where $C[t]$ is the column stabiliser of t and Rt is the row stabiliser of t, these are the two subgroups of $SfG$. For any $v∈V,|etv|∈V[λ]$. Furthermore et affords the representation corresponding to [λ].

Let $δ0$ be the r-signed 1-factor on ${1,2,…,f}$ whose lines joins 2i-1 to 2i, for $i=1,2,…,s$. We will show that $|etδ0|$ has a nonzero $δ0$ coefficient u and $|Tr(x)etδ0|$ has a nonzero $δ0$ coefficient, say $p[λ](x)u$. As $Tr(x)$ acts as a scalar on $V[λ]$ and $|etδ0|$ is in $V[λ]$, $p[λ](x)=h[λ](x)$.

If δi is a r-signed 1-factor in Fs,

$T→s(x)δi=∑jx∑ k=1rkmk(δi,δ j)δj$

where mk is the number of k cycles formed in $δi∪δj$ for which δj is in Fs. Therefore,

$|T→s(x)στδ0|=x∑ k=1rkmk (στδ0 ,δ0 ).$

Let $up[λ](x)$ be the $δ0$ component of $|T→s(x)e[t]δ0|$. Then

Some of the terms in $up[λ](x)$ gives the same expression. In particular, let $Rt0$ be the subgroup of Rt which fixes δ0 and let r0 be its order. That is, if $τ1∈Rt0$, $τ1δ0=δ0$. Now for τ in Rt, $mk(σττ1δ0,δ0)=mk(στδ0,δ0),∀k=1,2,…,r$. Let $Rt/Rt0$ be the set of right coset representatives for $Rt0$ in Rt. Therefore, equation 1.1 gives

Let $Ct0$ be the subgroup of Ct which fixes δ0 and c0 be its order. Furthermore, all $σ∈Ct0$ are even as the generalised signed permutation in each odd column is identical to the generalised signed permutation in the column to its immediate right as the line joining 2i-1 to 2i is preserved. Let $Ct/Ct0$ be the set of left coset representatives for $Ct0$ in Ct. Therefore, equation 1.2 gives

The argument above shows that $T→s(x)$ restricted to $V[λ]$ is a scalar $p[λ](x)$. We need only to show that $p[λ](x)=d[λ](x)$. We may also assume that for the diagrams $[λ*]$ of smaller size, we have $d[λ*](x)=p[λ*](x)=h[λ*](x)$. Using equation 1.3, we get

It is clear from the definition of $Ct0$ that $Ct0=Ct0(1)×Ct0(2)×…×Ct0(r)$ consists of all generalised signed permutations in Ct, where $Ct0(j),j=1,2,…,r−1$ permutes the 2i-1 column in the same way as 2i column in the residue with sign changes $ξk,k=1,2,…,r$ for $i=1,2,…,λlj(j)/2, j=1,2,…,r−1$ and $Ct0(r)$ permutes the 2i-1 column in the same way as 2i column in the residue without sign changes for $i=1,2,…,λlr(r)/2$.

Coset representatives may be chosen which fix the odd numbered column pointwise in all the residues and permutes the even numbered column with sign change $ξk,k=1,2,…,r$ in $1,2,…,r−1$ residue and with sign change ξr in residue. The coset representatives in $Ct/Ct0$ are precisely the r-signed permutations acting on even numbered columns. This is a full set as any element of Ct, which is a product of generalised signed permutations in $Ct0$ followed by generalised signed permutation in $Ct(1)×Ct(2)×…×Ct(r)$ moving only elements in even numbered columns.

The choices for coset representatives of $Rt/Rt0$ are not as natural. The group Rt is a direct product of groups $Ri(j),∀ j=1,2,…,r$, where $Ri(j)$ permutes only elements in the i-th row of the j th residue $j=1,2,…,r−1$, $Ri(r)$ permutes only elements in the i-th row of the r th residue with sign changes $ξk, k=1,2,…,r$ and fixes all the other elements. Also $Rt0$ is a direct product of groups $R0i(j),∀ j=1,2,…,r$ where .

Coset representatives may be chosen as products $∏ j=1 lrr1r2…rlj$ where ri's are coset representatives for $Ri(j)/R0i(j),j=1,2,…,r$

There arises the following two cases :

• 1. The $sth$ horizontal edge lies in the -residue, $j=1,2,…,r−1$

• 2. The $sth$ horizontal edge lies in the residue.

Case 1. If $sth$ horizontal edge lies in the -residue, $j=1,2,…,r−1$ In order to prove the theorem for a fixed j we concentrate on $(lj,λlj(j)−1)$ and the $(lj,λlj(j))$ position. For convenience we call it the position as . In order to evaluate , it is convenient to place the lines from δ0 in the diagram t.

Pictured this way $τδ0$ is a diagram with lines all in the same row. Coset representatives for $Ri(j)/R0i(j)$ may be picked anyway for $i=1,2,…,lj−1$. We choose coset representatives for $Rlj(j)/Rlj0(j)$ by first restricting a group $Rlj(j)$ to a group $Rlj*(j)$, the subgroup of $Rlj(j)$ fixing . Let $Rlj0*(j)$ be the subgroup of $Rlj*$ fixing the r-signed 1-factor δ0. Let Y be the set of representatives for $Rlj*(j)/Rlj0*(j)$. Let τi be the r-signed transpositions in $Rlj(j)$ interchanging 2s-1 and and for i=0, τ0 is the identity. The elements τiY are a full set of representatives of $Rlj(j)/Rlj0(j)$.

We also wish to choose the coset representatives appropriately for the subgroup of Ct moving elements in the $λlj(j)$ column only. Denote this subgroup by $Clj(j)$. Let $Clj*(j)$ be the subgroup of $Clj(j)$ fixing the entry b. Let Z be the set consisting of r-signed transpositions $σiξk,∀ k=1,2,…,r$ where $σiξk$ interchanges b with the entry above it in the row with sign change ξk in and for i=0, $σ0ξk$ is the identity with sign change $ξk,k=1,2,…,r$ in b. Coset representatives for $Clj(j)/Clj0(j)$ may be taken to be $σ′Clj−1(j),σ′∈Z$.

Now let $Ct*$ be the generalised signed permutations in Ct fixing a and b and $Rt*$ be the generalised signed permutations in Rt fixing . Let be the corresponding stabilisers of δ0 fixing a and b. Choose coset representatives L and M for $Ct*/Ct0*$ and $Rt*/Rt0*$. Now choose coset representatives for $Ct/Ct0$ as $σ′σ,σ′∈Z,σ∈L$.

Coset representatives for $Rt/Rt0$ can be chosen as $τiτ,τ∈M$. The coset representatives appearing in equation 1.4 are

$mk(στδ0,δ0)=mk(σ′στiτδ0,δ0),σ′∈Z,σ∈L,τ∈M,k=1,2,…,r.$

Hence the coset representatives appearing in equation 1.4 becomes

Now we concentrate on the inner sum

Therefore we can write Q as $Q=∑ l=1rQmnξl$, where

To evaluate the inner sum we deal with the following four cases.

• 1. m and n both zero

• 2. m=0 and n≠ 0

• 3. m≠ 0 and n=0

• 4. m and n both non-zero

Subcase 1. For m and n both zero, equation 1.5 becomes

In this case both σ and τ fixes a and b. Let σ*, τ* be the corresponding restrictions of σ and τ to $S→f−2$. Let $δ0*$ be the restriction of δ0 with ${a,b}$ omitted and let $mk′,k=1,2,…,r$ be the corresponding inner product on r-signed 1-factors of size f-2. The connected component of $σ0ξlστδ0∪δ0$ is precisely the orbits of $σ*τ*δ0*∪δ0*$ with ${a,b}$ adjoined labeled by $ξl$. Therefore $ml(σ0ξlστδ0,δ0)=ml′(σ*τ*δ0*,δ0*)+1$ and $mk(σ0ξlστδ0,δ0)=mk′(σ*τ*δ0*,δ0*)$ for k ≠ l and $k=1,2,…,r$. Hence by equation 1.6, we get

Since $ε(σ)=ε(σ*)$ and from the above equation $Q00ξl=ξljxlp[λ*](x)$, where $[λ*]$ is [λ] with $λlj(j)$ is replaced by $λlj(j)−2$. We know by induction that $p[λ*](x)=d[λ*](x)$. Therefore,

$Q00ξl=ξljxld[λ*](x).$

Subcase 2. For m=0 and n ≠ 0, equation 1.5 becomes

Suppose n is fixed between 1 and $λlj(j)−2$. For $σ∈L,τ∈M$, $σ*,τ*,δ0*,mk′,k=1,2,…,r$ be the corresponding restrictions to the diagrams for $[λ*]$ of size f-2.

We will show that .

Let c be the entry in $(lj,n)$. Suppose $cξp$ is joined in $στδ0$ to $dξq$. Note that $στnτδ0$ is the same as $σ*τ*δ0*$ except ${a,b}$ has been added and the lines from $(cξi)$ to $(dξi)$ is replaced by two lines one from $(cξp)$ to b and another from $(dξq)$ to a. It is now clear that all orbits of $σ*τ*δ0*∪δ0*$ not containing $(cξp)$ and $(dξq)$ is the orbit of $σ0ξlστnτδ0∪δ0$.

The orbit containing $(cξp)$ and $(dξq)$ together with ${a,b}$ is the orbit of $σ0ξlστnτδ0∪δ0$. This shows that $mk(σ0ξlστnτδ0,δ0)=mk′(σ*τ*δ0*,δ0*)$, for all $k=1,2,…,r$.

Therefore, by equation 1.8, we get

Since $ε(σ)=ε(σ*)$ and from the above equation $Q0nξl=p[λ*](x)$, where *] is [λ] with $λlj(j)$ is replaced by $λlj(j)−2$. We know by induction that $p[λ*](x)=d[λ*](x)$. Therefore,

$Q0nξl=d[λ*](x).$

Subcase 3. For m ≠ 0 and n=0, equation 1.5 becomes

We will show that $mk(σmξlστδ0,δ0)=mk′(σ*τ*δ0*,δ0*)$, for all $k=1,2,…,r$. For $m=1,2,…,lj−1$, we show that each is $Qm0ξl=−d[λ*](x)$.

We need to consider the orbits of $σmξlστδ0∪δ0$ where $σmξl$ is the r-signed transposition interchanging b with entry above it in the $mth$ row and $λlj(j)$ column, labeled by ξl. Again we have to consider the restricted term $σ*τ*δ0*$. Let $cξp$ be the entry in $(m,λlj(j))$ position in $σ*τ*δ0*$ joined to entry $dξq$ in $σ*τ*δ0*$. The lines in $σmξlστδ0$ are precisely the lines in $σ*τ*δ0*$ except line from $cξp$ to $dξq$ is replaced by a line from $dξq$ to b and one from a to $cξp$.

Again the orbits of $σ*τ*δ0*∪δ0*$ are those of $σmξlστδ0∪δ0$ except for this one orbit through $cξp$ and $dξq$. This shows that .

Note $ε(σmξlσ)=−ε(σ*)=−ε(σ)$.

Therefore, by equation 1.10, we get

Hence by the above equation, we get $Qm0ξl=−p[λ*](x)$, where $[λ*]$ is [λ] with $λlj(j)$ is replaced by $λlj(j)−2$. We know by induction that $p[λ*](x)=d[λ*](x)$. Therefore,

$Qm0ξl=−d[λ*](x).$

Subcase 4. In this case both $m≠0$ and n ≠ 0, we wish to show that $Qmnξl=0,∀l$.

The term in $Qmnξl$, for a fixed $m,n,σ,τ$ with m and n not equal to zero is

$ε(σmξlσ)x∑ k=1rkmk(σmξl στnτδ0,δ0).$

We will show that how to combine the terms for a fixed m into disjoint subsets of size two. The sum over each of these subsets will be zero and so the sum over all these terms in $Qmnξl$ will be zero.

In order to choose the subsets, we suppose $m,σ,τ,n$ are chosen with both m and n both being not zero. Let $(a′)ξp$ be the endpoint for the line joined in $σmξlστnτδ0$ to a and $(b′)ξq$ be the endpoint for the line joined in $σmξlστnτδ0$ to b. The point $(a′)ξp$ must be left of a, since n≠ 0.

Suppose $(a′)ξp$ and $(b′)ξq$ are in the even numbered column.

Let $σ′=((b′)ξq,(a′)ξp)σ$.

As $ε((b′)ξq,(a′)ξp)=ξk$, $ε(σmξlσ′)=ε(σmξl((b′)ξq,(a′)ξp)σ)=ξkε(σmξlσ)$. The orbits of $σmξl(((b′)ξq),(a′)ξp)στnτδ0∪δ0$ is the same as the orbits of $σmξlστnτδ0∪δ0$ except $(a′)ξp$ is the endpoint of b and $(b′)ξq$ is the endpoint of a. Also the lengths of the orbits and the number of orbits of both were same but their signs were opposite. Therefore they cancel each other, such terms in $Qmnξl$ cancels and the sum of those terms will be zero.

Suppose $(a′)ξp$ and $(b′)ξq$ are in the odd numbered column. Instead of σ we start with $σ′=((b′)ξq,(a′)ξp)$, the same result holds using the r-signed transpositions $((b*)ξq,(a*)ξp)$ where $(a*)ξp$ is to the immediate right of $(a′)ξp$ in the even numbered column and $(b*)ξq$ is to the immediate right of $(b′)ξq$ in the even numbered column.

Suppose $(a′)ξp$ and $(b′)ξq$ are in different columns. Let $cξk$ be the position $(j,λlj(j))$. Note that $cξk$ and $(b′)ξq$ are joined in $στnτδ0$. This means $(στnτ)−1(cξk)$ and $(στnτ)−1((b′)ξq)$ are in the same row. Let $d′$ be the entry such that $στnτ(d′)$ is in the same row as c and $(b′)ξq$ and in the same column as $(a′)ξp$. As $(a′)ξp$ and $(b′)ξq$ are in different columns, $(b′)ξq$ is not $d′$. Denote the point joined to $d′$ in $στnτδ0$ by $(c′′)$. Note that $(στnτ)−1(c′′)$ is in the same row as $(στnτ)−1((b′)ξq)$,$(στnτ)−1(c)$ and $(στnτ)−1(d′)$. Let $τ′$ be the coset representative in M for which $τkτ′δ0=((στnτ)−1((b′)ξq),(στnτ)−1(c′′))τnτδ0$ for which $τkτδ0$ is the same as $τnτδ0$ except that $(στnτ)−1(c)$ is joined to $(στnτ)−1(c′′)$ and $(στnτ)−1(d′)$ is joined to $(στnτ)−1(b′)ξq)$.

Assume now that $(a′)ξp$ and $(b′)ξq$ are in different even numbered column. We examine the terms in the sum for the r-signed transposition τn interchanges 2r-1 and $f−λlj(j)+n$ and for the r-signed transposition τk interchanges 2r-1 and $f−λlj(j)+k$ corresponding to $σ′(d′,(a′)ξp)στkτ$ and to $σ′στnτ$. Note that $ε(σmξl(d′,(a′)ξp)σ)=ξpjε(σmσ)$. The orbits of $σm(d′,(a′)ξp)στkτδ0∪δ0$ and the orbits of $σmξlστnτδ0∪δ0$ are the same except the ones through ${a,b},{(a′)ξp,(b′)ξq}$ and ${d′,c′′}$. $(a′)ξp$ is joined in $σmξl(d′,(a′)ξp)στkτδ0$ to $c′$ and $(a′)ξp$ is joined in $σmξlστnτδ0$ to a and b is joined to $(b′)ξq$. The orbits of those terms were same and their signs were different, therefore they cancels each other. If $(a′)ξp$ and $(b′)ξq$ are in odd numbered column use the above result and this proves sum of those terms will be zero. Hence,

$Qmnξl=0.$

Thus, by equations 1.7, 1.9, 1.11 and 1.12, we get

$Q=∑ l=1rQ00ξl+∑ l=1r∑ j=1 λ lj (j) −2Q0jξl+∑ l=1r∑ i=1 lj−1Qi0ξl+∑ l=1r∑ i,jQijξl =(xr+ξ(r−1)jxr−1+…+ξjx+2(λlj(j)−2)−2(lj−1))d[λ*](x)$

From the definition of $d[λ](x)$ and $d[λ*](x)$,

$d[λ](x)=(xr+ξj(r−1)xr−1+…+ξjx+2(λlj(j)−2)−2(lj−1))d[λ*](x) =h[λ](x).$ $h[λ](x)=d[λ](x)=∏ i=1j∏ d∈ d (i)∏ k,lsi (xr+ξi(r−1)xr−1+…+ξix+akl(i)).$

Case 2. If $sth$ horizontal edge lies in the $rth$ residue, we may concentrate on the $(lr,λlr(r)−1)$ and the $(lr,λlr(r))$ positions. For convenience, we call it the position as a and b. In order to evaluate , place the lines from δ0 in the diagram t. Coset representatives for $Ri(r)/R0i(r)$ may be chosen anyway for $i=1,2,…,lr−1$. We choose coset representatives for $Rlr(r)/R0m(r)$ by restricting to a group $Rlr*(r)$, the subgroup of $Rlr(r)$ fixing a and b. Let $Rlr0*(r)$ be the subgroup of $Rlr*(r)$ fixing the r-signed 1-factor δ0. Choose Y a set of representatives for $Rlr*(r)/Rlr0*(r)$. Let Y' be the set consisting of r-signed transpositions of $Rlr(r)$ where $τjξk$ interchanges 2r-1 and $2r−λlr(r)+j$ with the sign change ξk in 2r-1 for $j=1,2,…,λlr(r)−2$. For $i=0,τ0ξk$ is the identity with the sign change ξk in 2r-1. The elements $τ′Y,τ′∈Y′$ is the set of representatives for $Rlr(r)/Rlr0(r)$.

We also wish to choose the coset representatives for the subgroup of Ct moving $λlr(r)$ columns only. Denote this subgroup by $Clr(r)$. Let $Clr*(r)$ be the subgroup of $Clr(r)$ fixing b. Let σi be the r-signed transpositions interchanging b with the entry above it in the $ith$ row for $i=0,1,…,lr−1$. Coset representatives for $Clr(r)/Clr0(r)$ may be chosen as $σiClj−1(r)$ where $Clj−1(r)$ interchanges $λlj−1(r)$ column.

Now let $Ct*$ and $Rt*$ be the subgroups of Ct and Rt fixing a and b respectively. Let $Ct0*$ and $Rt0*$ be the stabilisers of δ0 fixing a and b. Choose coset representatives L and M for $Ct*/Ct0*$ and $Rt*/Rt0*$. Now choose coset representatives for $Ct/Ct0$ as $σiσ,σ∈L$. Coset representatives for $Rt/Rt0$ can be chosen as $τ′τ,τ′∈Y′,τ∈M$. The coset representatives appearing in equation 1.4 are

$mk(στδ0,δ0)=mk(σiστ′τδ0,δ0),$

where

Hence the coset representatives appearing in equation 1.4 becomes

Now we concentrate on the inner sum

We can also write Q as $Q=∑ k=1rQmnξl$, where

To evaluate the inner sum we deal with the following four cases.

• 1. m and n both zero

• 2. m=0 and n≠ 0

• 3. m ≠ 0 and n=0

• 4. m and n both non-zero

Subcase 1. For m=0 and n=0, equation 1.13 becomes

In this case both σ and τ fixes a and b. Let $σ*,τ*$ be the corresponding restrictions of σ and τ to $S→f−2.$ Let $δ0*$ be the restriction of δ0 with ${a,b}$ omitted and let be the corresponding inner products on r-signed 1-factors of size f-2. The connected component of $στ0ξlτδ0∪δ0$ is precisely the orbits of $σ*τ*δ0*∪δ0*$ with ${a,b}$ adjoined with sign change ξl. Therefore $ml(στ0ξlτδ0,δ0)=ml′(σ*τ*δ0*∪δ0*)+1$ and $mk(στ0ξlτδ0,δ0)=mk′(σ*τ*δ0*,δ0*)$ for $k≠l$ and $k=1,2,…,r$. Hence by equation 1.14, we get

Since $ε(σ)=ε(σ*)$, $Q00ξl=xlp[λ*](x)$, where *] is [λ] with $λlr(r)$ replaced by $λlr(r)−2$. We know by induction $p[λ*](x)=d[λ*](x)$ and so we get

$Q00ξl=xld[λ*](x).$

Subcase 2. For m=0 and n ≠ 0, equation 1.13 becomes

Suppose n is fixed between 1 and $λlr(r)−2$. For $σ∈L,τ∈M$, $σ*,τ*,δ0*,mk′,k=1,2,…,r$ be the corresponding restrictions to the diagrams for *] of size f-2.

We will show that $mk(στnτδ0,δ0)=mk′(σ*τ*δ0*,δ0*)$, for all $k=1,2,…,r$.

Let c be the entry in $(lr,n)$ and $τnξl$ be the r-signed transpositions in $Y′$ which interchanges 2r-1 and $f−λlr(r)+j$. Suppose $cξp$ is joined in $σ*τ*δ0$ to $dξq$. Note that $στnτδ0$ are the same as $σ*τ*δ0*$ except ${a,b}$ has been added and the lines from $cξp$ to $(dξq)$ is replaced by two lines, one from $(c)ξp$ to b and another from $(d)ξq$ to a.

It is now clear that all orbits of $σ*τ*δ0*∪δ0*$ not containing $(c)ξp$ and $(d)ξq$ is the orbit of $στnτδ0∪δ0$. The orbit containing $(c)ξp$ and $(d)ξq$ together with ${a,b}$ is an orbit of $στnτδ0∪δ0$. This shows that .

Therefore, by equation 1.16, we get

Since $ε(σ)=ε(σ*)$ and from the above equation $Q0nξl=p[λ*](x)$, where *] is [λ] with $λlj(j)$ is replaced by $λlj(j)−2$. We know by induction that $p[λ*](x)=d[λ*](x)$. Therefore,

$Q0nξl=d[λ*](x).$

Subcase 3. For m ≠ 0 and n=0, equation 1.13 becomes

We will show that . For $m=1,2,…,lr−1$ and show that each is $−d[λ*](x)$.

We need to consider the orbits of $σmστξlτδ0∪δ0$ where σi is the r-signed transposition interchanges b with entry above it in the $mth$ row and $λlr(r)$ column. Again we have to consider the restricted term $σ*τ*δ0*$. Let c be the entry in $(i,λlr(r))$ position, $cξp$ is joined in $σ*τ*δ0*$ to entry $dξq$ in $σ*τ*δ0*$. The lines in $σmστ0ξlτδ0$ are precisely the lines in $σ*τ*δ0*$ except line from $cξp$ to $dξq$ is replaced by a line from $dξq$ to b and one from a to $cξp$.

Again the orbits of $σ*τ*δ0*∪δ0*$ are those of $σmστ0ξlτδ0∪δ0$ except for this one orbit through $cξp$ and $dξq$. This shows that .

Note $ε(σmσ)=−ε(σ*)=−ε(σ)$.

Therefore, by equation 1.18, we get

Hence by the above equation, we get $Qm0ξl=−p[λ*](x)$, where $[λ*]$ is $[λ]$ with $λlj(j)$ is replaced by $λlj(j)−2$. We know by induction that $p[λ*](x)=d[λ*](x)$. Therefore,

$Qm0ξl=−d[λ*](x).$

Subcase 4. In this case both m ≠ 0 and n ≠ 0, we wish to show that $Qmnξl=0,∀l$.

The term in $Qmnξl$, for a fixed $m,n,σ,τ$ with m and n not equal to zero is $ε(σmξlσ)x∑ k=1rkmk(σmξl στnτδ0,δ0).$ We will show that how to combine the terms for a fixed m into disjoint subsets of size two. The sum over each of these subsets will be zero and so the sum over all these terms in $Qmnξl$ will be zero.

In order to choose the subsets, we suppose $m,σ,τ,n$ are chosen with both m and n both being not zero. Let $(a′)ξp$ be the endpoint for the line joined in $σmξlστnτδ0$ to a and $(b′)ξq$ be the endpoint for the line joined in $σmξlστnτδ0$ to b.

The point $(a′)ξp$ must be left of a, since $n≠0$.

Suppose $(a′)ξp$ and $(b′)ξq$ are in the even numbered column.

Let $σ′=((b′)ξq,(a′)ξp)σ$.

As $ε((b′)ξq,(a′)ξp)=ξk$, $ε(σmξlσ′)=ε(σmξl((b′)ξq,(a′)ξp)σ)=ξkε(σmξlσ)$. The orbits of $σmξl(((b′)ξq),(a′)ξp)στnτδ0∪δ0$ is the same as the orbits of $σmξlστnτδ0∪δ0$ except $(a′)ξp$ is the endpoint of b and $(b′)ξq$ is the endpoint of a. Also the lengths of the orbits and the number of orbits of both were same but their signs were opposite. Therefore they cancel each other, such terms in $Qmnξl$ cancels and the sum of those terms will be zero.

Suppose $(a′)ξp$ and $(b′)ξq$ are in the odd numbered column. Instead of σ we start with $σ′=((b′)ξq,(a′)ξp)$, the same result holds using the r-signed transpositions $((b*)ξq,(a*)ξp)$ where $(a*)ξp$ is to the immediate right of $(a′)ξp$ in the even numbered column and $(b*)ξq$ is to the immediate right of $(b′)ξq$ in the even numbered column.

Suppose $(a′)ξp$ and $(b′)ξq$ are in different columns. Let $cξk$ be the position $(j,λlj(j))$. Note that $cξk$ and $(b′)ξq$ are joined in $στnτδ0$. This means $(στnτ)−1(cξk)$ and $(στnτ)−1((b′)ξq)$ are in the same row. Let $d′$ be the entry such that $στnτ(d′)$ is in the same row as c and $(b′)ξq$ and in the same column as $(a′)ξp$. As $(a′)ξp$ and $(b′)ξq$ are in different columns, $(b′)ξq$ is not $d′$. Denote the point joined to $d′$ in $στnτδ0$ by $(c′′)$. Note that $(στnτ)−1(c′′)$ is in the same row as $(στnτ)−1((b′)ξq)$,$(στnτ)−1(c)$ and $(στnτ)−1(d′)$. Let $τ′$ be the coset representative in M for which $τkτ′δ0=((στnτ)−1((b′)ξq),(στnτ)−1(c′′))τnτδ0$ for which $τkτδ0$ is the same as $τnτδ0$ except that $(στnτ)−1(c)$ is joined to $(στnτ)−1(c′′)$ and $(στnτ)−1(d′)$ is joined to $(στnτ)−1(b′)ξq)$.

Assume now that $(a′)ξp$ and $(b′)ξq$ are in different even numbered column. We examine the terms in the sum for the r-signed transposition τn interchanges 2r-1 and $f−λlj(j)+n$ and for the r-signed transposition τk interchanges 2r-1 and $f−λlj(j)+k$ corresponding to $σ′(d′,(a′)ξp)στkτ$ and to $σ′στnτ$. Note that $ε(σmξl(d′,(a′)ξp)σ)=ξpjε(σmσ)$. The orbits of $σm(d′,(a′)ξp)στkτδ0∪δ0$ and the orbits of $σmξlστnτδ0∪δ0$ are the same except the ones through ${a,b},{(a′)ξp,(b′)ξq}$ and ${d′,c′′}$. $(a′)ξp$ is joined in $σmξl(d′,(a′)ξp)στkτδ0$ to $c′$ and

$(a′)ξp$ is joined in $σmξlστnτδ0$ to a and b is joined to $(b′)ξq$. The orbits of those terms were same and their signs were different, therefore they cancels each other. If $(a′)ξp$ and $(b′)ξq$ are in odd numbered column use the above result and this proves sum of those terms will be zero. Hence,

$Qmnξl=0.$

Thus, by equations 1.15, 1.17, 1.19 and 1.20, we get

$Q=∑l=1r Q 00 ξl+∑l=1r ∑j=1λ lr −2 Q 0j ξl+∑l=1r ∑i=1 lr−1 Q i0 ξl+∑i,j Q ij ξl=(x r+x r−1+…+x+2(λ l r (r)−2)−2(l r−1))d[λ*](x)$

From the definition of $d[λ](x)$ and $d[λ*](x)$,

$d[λ](x)=(xr+xr−1+…+x+2(λlr(r)−2)−2(lr−1))d[λ*](x) =h[λ](x).$ $h[λ](x)=d[λ](x)=∏ i=1r∏ d∈ d (i)∏ k,l(x r +ξi(r−1) x+…+ξi x+akl (i)).$

Therefore,

$h[λ](x)=∏ i=1r∏ d∈d (i)∏ k,l(x r +ξ i(r−1) x r−1 +…+ξi x+a kl (i)).$

Hence the proof.

Corollary 6.3. [6] Let $[λ]=(λ1,λ2,…,λm$ be a partition of f with all λj even. Let $V[λ]$ be the subspace of Vi associated to the partition λ and $hλ(x)=hi(x)$. Then

$hλ(x)=∏(x+aij),$

where aij are in the diagram d of shape λ.

Proof. The proof follows from the above theorem for r=1.

Corollary 6.4. Let $[λ]=(λ1(1),λ2(1),…,λl(1),λ1(2),λ2(2),…,λm(2))$ be an even bi-partition of f. Let $V[λ]$ be the subspace of Vi associated to the bi-partition [λ] and $hr(x)=h[λ](x)$. Then

$h[λ](x)=∏(x2−x+aij(1))∏(x2+x+aij(2)),$

where $aij(1)$ are in the diagram $d(1)$ of shape $λ(1)$ and $aij(2)$ are in the diagram $d(2)$ of shape $λ(2)$.

Proof. The proof follows from the above theorem for r=2.

For r=3 and f=2, there are three even multipartitions of 2. They are $λ1=$ , $λ2=$ and $λ3=$ .

$hλ1=|x3+ξ2x2+ξx|=x3+ξ2x2+ξxx3 +ξ2 x2 +ξx¯ =x(x−1)x2+x+1hλ2=|x3+ξx2+ξ2x|=x3+ξx2+ξ2xx3 +ξx2 +ξ2 x¯ =−x(x−1)x2+x+1hλ3=|x3+x2+x|=x3+x2+xx3 +x2 +x¯ =x3+x2+x$

which is same as in example 6.1.

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