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

Published online March 31, 2022 https://doi.org/10.5666/KMJ.2022.62.1.1

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 IfG(x,2k). Also, we define certain symmetric matrices Tm,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 IfG(x,2k) of DfGx by its radical. We also find the eigenvalues and eigenspaces of Tm,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=r.

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 . 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=r.

In this paper, we analyse the irreducible representations of DfGx by determining the quotient IfG(x,2k) of DfGx by its radical. We also find the eigenvalues and eigenspaces of certain symmetric matrices Tm,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=r. In this paper G refers to r,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 σR [t] σ,πD([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 I1,I2,,Id[λ], right ideals J1,J2, ,Jd[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]=I1Id[λ],

or as a direct sum of simple right ideals

S^^m[λ]=J1Jd[λ]

where each Ii is a left ideal of S^^m for which multiplication on the left gives a representation isomorphic to S[λ] and each Ji 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](σ1)=Ψ[λ](σ)t. For any elements xi,yj S^^m[λ],1i,jd[λ], choose xi,yj so that

xiyjxryl=xiyl, ifj=r;0,otherwise.

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 δPfG 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(2f1)!!=rf13(2f1) 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 P23 is as follows.

An r-signed edge of δPfG is called r-signed horizontal edge if it joins two vertices in the same row of δPfG.

An r-signed edge of δPfG is called r-signed vertical edge if it joins two vertices in different rows of δPfG.

Definition 3.3. Let Vf be the vector space over a field K with PfG as its basis.

Definition 3.4. Let Vf(2k)= Linear span{δPfG/the number of r-signed horizontal edges in δ2k}, k=0,1,,f2 which is a subspace of Vf. Vf(2k) can be written as direct sum of vector subspaces Vf*(2m),kmf2 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 Vf 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,δ2PfG.

For δ1,δ2PfG, the graph Uδ2δ1 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 Uδ2δ1 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, 1ijr with kjiji(modr), entirely of vertices lying in the middle row.

Example 3.5. For δ1,δ2P53, the diagram in Uδ2δ1 is

Definition 3.6. Let δ1 and δ2 be r-signed 1-factors in PfG. Define the composition of r-signed diagrams δ1δ2 to be the r-signed 1-factor in the following way

  • 1. The top (respectively bottom) row of δ1δ2 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 Uδ2δ1 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, 1ijr with kjiji(modr).

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

δ1*δ2=x i=1rimi(δ1,δ2)(δ1δ2), where G=r

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

Example 3.8. For δ1,δ2P53, the diagram in δ1*δ2 is

Let IfG(x,2k)=Linear span{δPfG/number of r-signed horizontal edges in δ2k}. Clearly by the multiplication defined above IfG(x,2k) is an ideal of DfGx.

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

To describe the structure of the quotients IfG(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 πSmG by π(i)=(τ(i),σ(i)) where σSm 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,f2Pm,kG. Define an inner product <f1,f2> 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 <f1,f2>=0.

  • 2. If each r-signed path Pi joins βi to ασi and some number kj of ij edges 1ijr with ij=1rkjijτ(i)(mod r), then

<f1,f2>=x i=1rimi(f1,f2)π, where π=(τ,σ)SmG,σSm.

Note. <f1,f2>=<f2,f1>, where ✠ is the anti-isomorphism defined on the algebras KSmG by σσ1,σSmG.

Proposition 4.4. Let f=m+2k. Then the quotient IfG(x,2k) is isomorphic as algebras to (Vm,kGVm,kGKSmG,), where

(abπ1)(cdπ2)=ad(π1<b,c>π2),a,b,c,dPm,kG,π1,π2SmG.

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 IfG(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,kGVm,kGKSmGIfG(x,2k).

Given f1,f2Pm,kG, we define ϕ(f1f2σ) 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 σSmG 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,kGVm,kGKSmG onto IfG(x,2k).

It remains to show that ϕ is multiplicative.

 i.e ϕ(d1d2)=ϕ(d1)ϕ(d2),d1,d2Vm,kGVm,kGKSmG.

Let us now assume that d1=abπ1 and d2=cdπ2 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,π2SmG.

Consider the product in IfG(x,2k)

ϕ(d1)ϕ(d2)=x i=1rimi(d1,d2)d3,d3DfGx.

Case 1. Suppose there is a r-signed path joining αi to αj or ψi to ψj in Uϕ(d2)ϕ(d1) then ϕ(abπ1)ϕ(cdπ2)=0=ϕ((abπ1)(cdπ2)). Therefore

ϕ(d1d2)=ϕ(d1)ϕ(d2).

Case 2. Suppose there is a r-signed path joining αi to ψσ(i) in Uϕ(d2)ϕ(d1) for i=1,2,,m, then ϕ(d1)ϕ(d2)=x i=1rimi(δ1,δ2)d3=ϕ(d1d2).

Hence ϕ is an algebra isomorphism.

To describe the structure of the ring IfG(x,2k) in terms of the eigenvalues of certain matrices.

Definition 5.1. Let Tm,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](<b,c>), for the corresponding r-signed 1-factor.

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

Note. If <b,c>=x i=1rimi)(b,c)σ then <c,b>=x i=1rimi(c,b)σ1. So the matrix Tm,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 It and a generalised m, k signed partial 1-factor d, define

VL(It,d)= Linear Span {cdx/cPm,kG,xIt}.

Lemma 5.3. VL(It,d) is a left ideal of IfG(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 cdxVL(It,d), choose δIfG(x,2k) such that δ*(cdx) not equal to zero, that is δ*(cdx) and cdx have same number of r-signed horizontal edges.

(abπ)*(cdx)=(adπ<b,c>x),πSmG,xIt          =(adπx i=1rimi(b,c)π1x),               where π,π1SmG,xIt          =x i=1rimi(b,c)(adππ1x),              (ππ1SmG by the definition of SmG)

Since xIt and It is a left ideal of S^^m, y=ππ1xIt. Hence

(abπ)*(cdx)=x i=1rimi(b,c)(ady)VL(It,d).

Therefore VL(It,d) is a left ideal of IfG(x,2k).

Define WL(It,d)VL(It,d) to be the linear span of all

(u)(c,i)cdAi,t,

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

Proposition 5.4. Suppose v=(v)(c,i)cdAi,tVL(It,d). Let a, b be generalised m,k signed partial 1-factors. For any σSmG, (abσ)v=adσγjAj,t, where γj is the (b, j) entry of Tm,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.

(abσ)*v(c,i)cdAi,t=adσv(c,i)<b,c>Ai,t              =adσ(v)c,ix i=1rimi(b,c)σ1Ai,t              =adγjAj,t

where σ1Ai,t=Aj,t and γj is the coefficient of Aj,t in v(c,i)<b,c>Aj,t.

By definition of Tm,k[λ](x), the coefficient of Aj,t in <b,c>Ai,t is the (b, j), (c, i) entry of Tm,k[λ](x). Thus γj is the (b,j) entry of Tm,k[λ](x)(v).

Proposition 5.5.

  • 1. IfG(x,2k)WL(It,d)=0.

  • 2. I(v)= V L( I t,d) for any v in VL(It,d) not in WL(It,d).

  • 3. VL(It,d)/WL(It,d) is irreducible as a left IfG(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 WL(It,d) and γj is the (b,j) entry of Tm,k[λ](x)(v). By the definition of WL(It,d), γj are all 0, for any (abσ). Hence IfG(x,2k)WL(It,d)=0.

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

The last part of the proof from the above two.

Let WL[λ]=WL(It,d) and . By Proposition 5.4, WL[λ] is a nilpotent left ideal of IfG(x,2k).

Recall that S^^m[λ] can also be written as a direct sum of right ideals J1,,Jd[λ].

For given any right ideal Jt and a generalised m,k signed partial 1-factor a, define

VR(Jt,a)= Linear Span {abx/bPm,kG,xJt}.

Lemma 5.6. VR(Jt,c) is a right ideal of IfG(x,2k).

Proof. For cdxVR(Jt,c), choose δIfG(x,2k) such that (cdx)*δ not equal to zero, that is (cdx)*δ have same number of signed horizontal edges as in cdx.

(cdx)*(abπ)=(cbx<d,a>π),πSmG,xJt          =(cbxx i=1rimi(d,a)π1π),               where π,π1SmG,xJt          =x i=1rimi(d,a)(cbxπ1π),              (π1πSmG by the definition of SmG)

Since xJt and Jt is a right ideal of S^^m, y=xπ1πJt,

(cdx)*(abπ)=x i=1rimi(d,a)(cby)VR(Jt,c).

Therefore VR(Jt,c) is a right ideal of IfG(x,2k).

Define WR(Jt,a)VR(Jt,a) to be the linear span of all

(u)(c,i)abAj,t

where utTm,k[λ](x)=0 and Aj,t is as before.

The same proofs used in Propositions 5.4 and 5.5 shows that

  • 1. WR(Jt,a)(cdσ)=0,

  • 2. VR(Jt,a)/WR(Jt,a) is an irreducible right IfG(x,2k) module.

Define WR[λ]=WR(Jt,a) and define W[λ] to be the nilpotent 2-sided ideal W[λ]=WL[λ]+WR[λ].

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

Note that IfG(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 Tm,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) 1v(r)a,iv(s) b,jabxiyj.

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,lad v (s) b,j v (t) c,k{xiyj<b,c>xkyl}  = μ (r) μ (s) μ (t) μ (u) 1 v(r) a,i v (u) d,l  adxiyj v(s) b,j v(t)c,k <b,c>xk yl.

Now v(t)c,k<b,c>xk=γrxr where γr is the (b,r) entry of Tm,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 <b,c>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,jbyequation1.1=μ(t)xiylδs,tbytheorthonormalityofthev(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,ladμ(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,kGVm,kG S^ ^mλ is isomorphic as a vector space to (Vm,kGS[λ])(Vm,kGS[λ]) via the linear map f sending (cAi)(dAj) to cdxiyj.

Writing Vm,kGS[λ] as N[λ]R[λ] we have, from Propositions 5.4, 5.5 and 5.8, that

A. f(N[λ](Vm,kGS[λ])+(Vm,kGS[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,kGS[λ])+(Vm,kGS[λ])N[λ]. Then W[λ] is the intersection of the radical of IfG(x,2k) with D[λ].

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

In this section, we determine the eigen values of Ts(x),s=f2 and f is even  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 Ts(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 δiδj.

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

T1(x) =x3x2xx2xx3xx3x2

The eigen values are x(x1)x2+x+1, x(x1)x2+x+1 and x3+x2+x.

A generalised permutation σSfG 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), σSfG.

Suppose δ1,δ2Fs and if C1 is a connected component of δ1δ2, then σ(C1) is a connected component of σ(δ1)σ(δ2). In particular the number and size of δ1δ2 and σ(δ1)σ(δ2) are the same.

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

PσTs(x)=Ts(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,σSfG and n is the number of even multi partitions of f. As the irreducibles are distinct, Ts(x)ViVi. As each Vi is irreducible, Ts(x) restricted to Vi is a scalar, which is denoted by hi(x)I. In order to find the eigen values for Ts(x), it is necessary to determine the scalars for Ts(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 i th  row and 2j th  column. It is convenient to place the diagram of the even multi partition [λ] on the grid Δ.

Residue iColumn No.:12345678902rξi4rξi6rξirξirξi3rξi5rξiΔ=2rξi02rξi4rξi3rξirξirξi3rξi4rξi2rξi02rξi,

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 dd (i) k,lsi (x r+ξi(r1)xr1++ξix+akl(i)),

where |.| denotes modulus, aij(i) are in the diagram d(i) of shape λ(i) for i=1,2,,r and

si =1,if1i<r21,ifr2ir.  

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(r1)xr1++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). Ts(x) restricted to Vi is si(xr+ξi(r1)xr1++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,σCt,τRt.

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 vV,|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,

Ts(x)δi=jx k=1rkmk(δi,δ j)δj

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

|Ts(x)στδ0|=x k=1rkmk (στδ0 ,δ0 ).

Let up[λ](x) be the δ0 component of |Ts(x)e[t]δ0|. Then

up[λ](x) =ε(σ)x k=1rkmk(στδ0,δ0).

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 τ1Rt0, 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

up[λ](x) =si ε(σ)x k=1rkmk (στδ0 ,δ0 )r0,σCt,τRt/Rt0.

Let Ct0 be the subgroup of Ct which fixes δ0 and c0 be its order. Furthermore, all σCt0 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

up[λ](x) =ε(σ)x k=1rkmk (στδ0 ,δ0 )r0c0,σCt/Ct0,τRt/Rt0.

The argument above shows that Ts(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

p[λ](x) =ε(σ)x k=1rkmk(στδ0,δ0),σCt/Ct0,τRt/Rt0.

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,,r1 permutes the 2i-1 column in the same way as 2i column in the j th  residue with sign changes ξk,k=1,2,,r for i=1,2,,λlj(j)/2,j=1,2,,r1 and Ct0(r) permutes the 2i-1 column in the same way as 2i column in the r th  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,,r1 residue and with sign change ξr in r th  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,,r1, 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 R0i(j)=Ri(j)Rt0, for j=1,2,,r.

Coset representatives may be chosen as products j=1 lrr1r2rlj 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 j th -residue, j=1,2,,r1

  • 2. The sth horizontal edge lies in the r th  residue.

Case 1. If sth horizontal edge lies in the j th -residue, j=1,2,,r1 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 a and b. In order to evaluate mk(στδ0,δ0),k=1 to r, 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,,lj1. 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 a and b. 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 2sλlj(j)+i for i=1,2,,λlj(j)2 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 i th  row with sign change ξk in b, for i=1,2,,lj1 and k=1,2,,r 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 σClj1(j),σZ.

Now let Ct* be the generalised signed permutations in Ct fixing a and b and Rt* be the generalised signed permutations in Rt fixing a and b. Let Ct0* and Rt0* 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 σσ,σZ,σL.

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

mk(στδ0,δ0)=mk(σστiτδ0,δ0),σZ,σL,τM,k=1,2,,r.

Hence the coset representatives appearing in equation 1.4 becomes

p[λ](x) = n,σZε( σ σ)x k=1rkmk( σ στnτδ0,δ0) , for σL,τM. 

Now we concentrate on the inner sum

Q=ε(σσ)x k=1rkmk(σστnτδ0,δ0), for σL,τM.

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

Qmnξl=ε(σmξlσ)x k=1rkmk(σmξk στnτδ0,δ0), for σL,τM,

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

Q00ξl=ε(σ0ξlσ)x k=1rkmk(σ0ξl στδ0,δ0), for σL,τM.

In this case both σ and τ fixes a and b. Let σ*, τ* be the corresponding restrictions of σ and τ to Sf2. 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 σ0ξlστδ0δ0 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

Q00ξl=ε(σ*)ξljxlx k=1rmk(σ*τ*δ0*,δ0*), for σ*L and τ*M.

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

Q0nξl=ε(σ0ξlσ)x k=1rkmk(σ0ξl στnτδ0,δ0), for σL,τM.

Suppose n is fixed between 1 and λlj(j)2. For σL,τM, 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 mk(σ0ξlστnτδ0,δ0)=mk(σ*τ*δ0*,δ0*), for all k=1,2,,r.

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 σ0ξlστnτδ0δ0.

The orbit containing (cξp) and (dξq) together with {a,b} is the orbit of σ0ξlστnτδ0δ0. This shows that mk(σ0ξlστnτδ0,δ0)=mk(σ*τ*δ0*,δ0*), for all k=1,2,,r.

Therefore, by equation 1.8, we get

Q0nξl=ε(σ0ξlσ)x k=1rkmk(σ*τ*δ0* ,δ0* ), for σ*L and τ*M.

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

Qm0ξl=ε(σmξlσ)x k=1rkmk(σmξl στδ0,δ0), for σL,τM.

We will show that mk(σmξlστδ0,δ0)=mk(σ*τ*δ0*,δ0*), for all k=1,2,,r. For m=1,2,,lj1, we show that each is Qm0ξl=d[λ*](x).

We need to consider the orbits of σmξlστδ0δ0 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 σmξlστδ0δ0 except for this one orbit through cξp and dξq. This shows that mk(σmξlστδ0,δ0)=mk(σ*τ*δ0*,δ0*), for all k=1,2,,r.

Note ε(σmξlσ)=ε(σ*)=ε(σ).

Therefore, by equation 1.10, we get

Qm0ξl=ε(σ)x k=1rkmk(σ*τ*δ0*,δ0*), for σ*L and τ*M.

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 m0 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)στnτδ0δ0 is the same as the orbits of σmξlστnτδ0δ0 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 τkτδ0=((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 fλlj(j)+n and for the r-signed transposition τk interchanges 2r-1 and fλlj(j)+k corresponding to σ(d,(a)ξp)στkτ and to σστnτ. Note that ε(σmξl(d,(a)ξp)σ)=ξpjε(σmσ). The orbits of σm(d,(a)ξp)στkτδ0δ0 and the orbits of σmξlστnτδ0δ0 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 lj1Qi0ξl+ l=1r i,jQijξl=(xr+ξ(r1)jxr1++ξjx+2(λlj(j)2)2(lj1))d[λ*](x)

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

d[λ](x)=(xr+ξj(r1)xr1++ξjx+2(λlj(j)2)2(lj1))d[λ*](x)  =h[λ](x). h[λ](x)=d[λ](x)= i=1j d d (i) k,lsi (xr+ξi(r1)xr1++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 mk(στδ0,δ0), for all k=1,2,,r, place the lines from δ0 in the diagram t. Coset representatives for Ri(r)/R0i(r) may be chosen anyway for i=1,2,,lr1. 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 τjξk,for all k=1,2,,r of Rlr(r) where τjξk interchanges 2r-1 and 2rλlr(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 τY,τY 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,,lr1. Coset representatives for Clr(r)/Clr0(r) may be chosen as σiClj1(r) where Clj1(r) interchanges λlj1(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,σL. Coset representatives for Rt/Rt0 can be chosen as ττ,τY,τM. The coset representatives appearing in equation 1.4 are

mk(στδ0,δ0)=mk(σiσττδ0,δ0),

where σL,τY,τM and for all k=1,2,,r.

Hence the coset representatives appearing in equation 1.4 becomes

p[λ](x)=ε(σiσ)x k=1rkmk (σi σττδ0 ,δ0 ),  for  σL,τM.

Now we concentrate on the inner sum

Q=ε(σiσ)x k=1rkmk(σiσττδ0,δ0),  for  σL,τM.

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

Qmnξl=ε(σ)x k=1rkmk(σmστnξl τδ0,δ0),  for σL,τM.

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

Q00ξl=ε(σ)x k=1rkmk(στ0ξl τδ0,δ0), for σL,τM.

In this case both σ and τ fixes a and b. Let σ*,τ* be the corresponding restrictions of σ and τ to Sf2. Let δ0* be the restriction of δ0 with {a,b} omitted and let mk, for all k=1,2,,r be the corresponding inner products on r-signed 1-factors of size f-2. The connected component of στ0ξlτδ0δ0 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 kl and k=1,2,,r. Hence by equation 1.14, we get

Q00ξl=ε(σ)xlx k=1rkmk(σ*τ*δ0*,δ0*), for σ*L and τ*M.

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

Q0nξl=ε(σ)x k=1rkmk(στnτδ0,δ0), for σL,τM.

Suppose n is fixed between 1 and λlr(r)2. For σL,τM, 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 fλlr(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 στnτδ0δ0. The orbit containing (c)ξp and (d)ξq together with {a,b} is an orbit of στnτδ0δ0. This shows that mk(στnτδ0,δ0)=mk(σ*τ*δ0*,δ0*), for all k=1,2,,r.

Therefore, by equation 1.16, we get

Q0nξl=ε(σ*)x k=1rkmk(σ*τ*δ0*,δ0*), for σ*L and τ*M.

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

Qi0ξl=ε(σ)x k=1rkmk(σmστ0ξl τδ0,δ0), for σL,τM.

We will show that mk(σmστ0ξlτδ0,δ0)=mk(σ*τ*δ0*,δ0*), for all k=1,2,,r. For m=1,2,,lr1 and show that each is d[λ*](x).

We need to consider the orbits of σmστξlτδ0δ0 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 σmστ0ξlτδ0δ0 except for this one orbit through cξp and dξq. This shows that mk(σmστ0ξlτδ0,δ0)=mk(σ*τ*δ0*,δ0*), for all k=1,2,,r.

Note ε(σmσ)=ε(σ*)=ε(σ).

Therefore, by equation 1.18, we get

Qm0ξl=ε(σ)x k=1rkmk(σ*τ*δ0*,δ0*), for σ*L and τ*M.

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 n0.

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)στnτδ0δ0 is the same as the orbits of σmξlστnτδ0δ0 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 τkτδ0=((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 fλlj(j)+n and for the r-signed transposition τk interchanges 2r-1 and fλlj(j)+k corresponding to σ(d,(a)ξp)στkτ and to σστnτ. Note that ε(σmξl(d,(a)ξp)σ)=ξpjε(σmσ). The orbits of σm(d,(a)ξp)στkτδ0δ0 and the orbits of σmξlστnτδ0δ0 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 lr1 Q i0 ξl+i,j Q ij ξl=(x r+x r1++x+2(λ l r (r)2)2(l r1))d[λ*](x)

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

d[λ](x)=(xr+xr1++x+2(λlr(r)2)2(lr1))d[λ*](x)  =h[λ](x). h[λ](x)=d[λ](x)= i=1r d d (i) k,l(x r +ξi(r1) x++ξi x+akl (i)).

Therefore,

h[λ](x)= i=1r dd (i) k,l(x r +ξ i(r1) x r1 ++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)=(x2x+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(x1)x2+x+1hλ2=|x3+ξx2+ξ2x|=x3+ξx2+ξ2xx3 +ξx2 +ξ2 x¯=x(x1)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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