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°

General Form of Non-Symmetric Spin Models

TAKUYA IKUTA [email protected]

Kobe Gakuin Women’s College, Hayashiyama, Nagata-ku, Kobe, 653-0861 Japan

KAZUMASA NOMURA [email protected]

College of Liberal Arts and Sciences, Tokyo Medical and Dental University, Kohnodai, Ichikawa, 272-0827 Japan Received August 20, 1998; Revised October 14, 1998

Abstract. A spin model (for link invariants) is a square matrix W with non-zero complex entries which satisfies certain axioms. Recently (Jaeger and Nomura, J. Alg. Combin. 10 (1999), 241–278) it was shown thattWW1is a permutation matrix (the order of this permutation matrix is called the “index” of W ), and a general form was given for spin models of index 2. In the present paper, we generalize this general form to an arbitrary index m. In particular, we give a simple form of W when m is a prime number.

Keywords: spin model, association scheme, Bose-Mesner algebra

1. Introduction

Spin models were introduced by Vaughan Jones [7] to construct invariants of knots and links. A spin model is essentially a square matrix W with nonzero entries which satis- fies two conditions (type II and type III conditions). In his definition of a spin model, Jones considered only symmetric matrices. It was generalized to non-symmetric case by Kawagoe-Munemasa-Watatani [8].

Recently, Fran¸cois Jaeger and the second author [6] introduced the notion of “index” of a spin model. For every spin model W , the transposetW is obtained from W by a permutation of rows. Letσ denote the corresponding permutation of X = {1, . . . ,n}(n is the size of W ). Then the index m is the order ofσ. In [6], it was shown that X is partitioned into m subsets X0, X1, . . . ,Xm1 such that W(x,y) = ηijW(y,x) holds for all xXi, yXj. Moreover, the case of m=2 was deeply investigated, and a general form of spin models of index 2 was given.

In the present paper, we investigate the structure of spin models of an arbitrary index m.

In Section 4, we show that W is decomposed into blocks Wi j, and Wi jsplits into Kronecker product of two matrices Si jand Ti j(Proposition 4.3). In Section 5, we give conditions on Ti j (Propositions 5.1 and 5.5). In Section 6, we apply this general form to some special cases (Propositions 6.1 and 6.2). In particular, we give a simple form of W when the index m is a prime number (Corollary 6.3).

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2. Preliminaries

In this section, we give some basic materials concerning spin models and association schemes. For more details the reader can refer to [4–7].

Let X be a finite non-empty set with n elements. We denote by MatX(C)the set of square matrices with complex entries whose rows and columns are indexed by X . For WMatX(C)and x, yX , the(x,y)-entry of W is denoted by W(x,y).

A type II matrix on X is a matrix WMatX(C)with nonzero entries which satisfies the type II condition:

X

xX

W(a,x)

W(b,x)=a,b (for all a, bX). (1)

Let WMatX(C)be defined by W(x,y) =W(y,x)1. Then type II condition is written as W W =n I (I denotes the identity matrix). Hence, if W is a type II matrix, then W is non-singular with W1 =n1W. It is clear that W1 andtW are also type II matrices.

A type II matrix W is called a spin model on X if W satisfies type III condition:

X

xX

W(a,x)W(b,x)

W(c,x) =D W(a,b)

W(a,c)W(c,b) (for all a, b, cX) (2) for some nonzero complex number D. The number D is called the loop variable of W . Set- ting b=c in (2),P

xXW(a,x)=DW(b,b)1holds, so that the diagonal entries W(b,b) is a constant, which is called the modulus of W .

For a spin model W with loop variable D, any nonzero scalar multipleλW is a spin model with loop variableλ2D. Usually W is normalized so that D2=n, but we allow any nonzero value of D in this paper to simplify our arguments.

Observe that, for any spin models Wion Xiwith loop variable Di(i =1,2), their tensor (Kronecker) product W1W2is a spin model with loop variable D=D1D2. Conversely, it is not difficult to show that, if W1W2and W1are spin models, then W2must be a spin model.

A(class d)association scheme on X is a partition of X×X with nonempty relations R0,R1, . . . ,Rd, where R0= {(x,x)|xX}which satisfy the following conditions:

(i) For every i in{0,1, . . . ,d}, there exists i0in{0,1, . . . ,d}such that Ri0 = {(y,x)| (x,y)Ri}.

(ii) There exist integers pki j(i , j , k∈ {0,1, . . . ,d}) such that for every(x,y)Rk, there are precisely pi jk elements z such that(x,z)Riand(z,y)Rj.

(iii) pki j= pkj ifor every i , j in{0,1, . . . ,d}.

Let Aidenote the adjacency matrix of the relation Ri, so AiMatX(C)is a{0,1}-matrix whose(x,y)-entry is equal to 1 if and only if(x,y)Ri. Clearly A0=I , AiAj=δi,jAi

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(entry-wise product),Pd

i=0Ai = J (all 1’s matrix), and AiAj =Pd

k=0pki jAkhold. The linear spanA of{A0,A1, . . . ,Ad}becomes a subalgebra of MatX(C), called the Bose- Mesner algebra of the association scheme. Observe that A is closed under entry-wise product,Ais closed under transposition A7→ tA, andAcontains I , J .

3. Associated permutation

Let W be a spin model on X . Then there exists an association scheme R0, . . . ,Rd on X such that the corresponding Bose-Mesner algebraAcontains W ([5] Theorem 11). In [6], it was shown thattWW1 =As(the adjacency matrix of Rs) for some s∈ {0,1, . . . ,d}, and moreover As is a permutation matrix ([6] Proposition 2). Letσ denote the corresponding permutation on X , so that As(x,y) =1 if y =σ(x)and As(x,y) =0 otherwise. The order m ofσ is called the index of W .

Observe that m=1 if and only if W is symmetric. Also observe that, for two spin models Wi of index mi (i =1,2), the index of W1W2is equal to the least common multiple of m1and m2. In particular, tensor product of a spin model of index m with any symmetric spin model has index m.

Lemma 3.1

(i) W(x, σ(x))=W(y, σ (y)) (x,yX). (ii) W(y,x)=W(σ(x),y) (x,yX). (iii) Every orbit ofσ has length m.

Proof:

(i) Observe that, since WA, W is written as a linear combination W =Pd

i=0tiAi, so W(x,y)=ti for(x,y)Ri. Since(x, σ (x))Rs (for every xX ), it holds that W(x, σ (x))=ts =W(y, σ (y)).

(ii) W(y,x)=tW(x,y)=(AsW)(x,y)=W(σ(x),y).

(iii) Pick any i (0<i<m). Since Aisis a linear combination of A0, . . . ,Adand since Ais is a permutation matrix, we get Ais = Aj for some j 6=0. Observe that the diagonal entries of Aj are all zero since j 6= 0. This means thatσi (which corresponds the permutation matrix Aj) has no fixed point on X . We have shown thatσifixes no point (0<i <m). Thus every orbit ofσmust have length m. 2 Lemma 3.2 There is a partition X=X0∪ · · · ∪Xm1such that(for all i , j ∈ {0, . . . , m−1})

W(x,y)=ηijW(y,x) (for all xXi,yXj), (3) whereηdenotes a primitive m-root of unity. Moreover,for every i, σ (Xi)=Xj holds for some j .

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Proof: The existence of such a partition follows from [6] Proposition 3. As in the proof of Lemma 3.1(i), we have(x, σ (x))Rsand W(x, σ (x))=ts for all xX . Then there exists s0such that(σ (x),x)Rs0, so that W(σ (x),x)=ts0. Now pick any xXi. Then σ(x)Xj for some j . On the other hand, W(x, σ(x))=ηijW(σ(x),x). These imply ηij =tsts01. This means that j is independent of the choice of xXi, so thatσ(Xi)=Xj. 2 We fix a primitive m-root of unityη, and let X0, . . . ,Xm1be the partition of X given in Lemma 3.2. We identify the index set{0,1, . . . ,m−1}with Zm=Z/mZ. By Lemma 3.2, there is a permutationπ on Zmsuch thatσ(Xi)= Xπ(i)(iZm). Let t denote the order ofπ, and set k=m/t.

Lemma 3.3 π(i)i =π(j)j for all i, jZm.

Proof: Pick any xXi, yXj. We haveσ(x)Xπ(i),σ(y)Xπ(j). By Lemma 3.2, W(x, σ(x)) = ηi−π(i)W(σ (x),x)and W(y, σ(y)) = ηi−π(j)W(σ(y),y). On the other hand, W(x, σ(x)) = W(y, σ (y))by Lemma 3.1(i), and also W(σ (x),x) = W(x,x) = (the modulus of W) =W(y,y)=W(σ (y),y)by Lemma 3.1(ii). These implyηi−π(i) =

ηj−π(j). 2

Lemma 3.4 There exists an automorphismϕof the additive group Zmsuch thatπ(ϕ(i))= ϕ(i +k)for all iZm. Moreover,W(x,y) =ϕ(1))ijW(y,x)for every xXϕ(i), yXϕ(j).

Proof: Set k0=π(0). Then π(i) = i +k0 (iZm) by Lemma 3.3. Thus k0Zm = {0,k0,2k0, . . . , (t −1)k0}is an orbit ofπ. Note that every orbit ofπ has length t, and hence the number of orbits ofπis equal to k =m/t (in particular, k must be an integer).

Clearly k0Zmis the unique subgroup of Zmof order t , so k0Zm =kZm. Hence there is an automorphismϕof the additive group kZmsuch thatϕ(k)=k0.

We claim thatϕ can be extended to an automorphism of Zm. In fact, for any cyclic group G and for any subgroup H of G, any automorphism of H can be extended to an automorphism of G. This fact can be easily shown when G is a cyclic p-group. For general case, decompose G into the Sylow subgroups.

Now we have an automorphismϕof Zmsuch thatϕ(k)=k0. Sinceπ(i)=i+k0for all iZm, we getπ(ϕ(i))=ϕ(i)+k0=ϕ(i)+ϕ(k)=ϕ(i+k).

Let xXϕ(i), yXϕ(j). Then, by Lemma 3.2, W(x,y)=ηϕ(i)−ϕ(j)W(y,x)holds for all xXϕ(i), yXϕ(j). Hereϕ(i)ϕ(j) = ϕ(i ·1)ϕ(j ·1) = iϕ(1)jϕ(1) = ϕ(1)(ij). Hence W(x,y)=ϕ(1))ijW(y,x). 2 Thus, by reordering the indices{0,1, . . . ,m−1}byϕ, and by replacingηwithηϕ(1), we may assume that

π(i)=i+k (iZm). (4)

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4. General form of W

We use the notation of the previous section. We also use the notation:

γk(`,i)=η−`i−(k/2)`(`−1). (5)

Proposition 4.1 Let i,jZmand xXi,yXj. Then for`, `0Z,

W(σ`(x), σ`0(y))=γk(``0,ij)W(x,y). (6) Proof: Assume` ≥ 0 and`0 ≥ 0. First we consider the case of`0 = 0. We proceed by induction on `. Obviously (6) holds for` = 0. By Lemma 3.1(ii) and Lemma 3.2, W(y,x)=W(σ(x),y)and W(y,x)=ηjiW(x,y). Hence W(σ(x),y)=ηjiW(x,y), so (6) holds for`=1. Now assume` >1. Notingσ(x)Xπ(i)=Xi+kand using induction,

W(σ`(x),y)=W(σ`−1(σ(x)),y)

=γk(`−1, (i+k)j)W(σ(x),y)

=γk(`−1, (i+k)j)ηjiW(x,y)

=γk(`,ij)W(x,y).

Hence (6) holds for `0 = 0. Now suppose `0 > 0. Notingσ`0(y)Xj+`0k and using Lemma 3.2,

W(σ`(x), σ`0(y))=γk(`,i(j+`0k))W(x, σ`0(y))

=γk(`,i(j+`0k))ηi−(j+`0k)W(σ`0(y),x)

=γk(`,i(j+`0k))ηi−(j+`0k)γk(`0,ji)W(y,x)

=γk(`,i(j+`0k))ηi−(j+`0k)γk(`0,ji)ηjiW(x,y)

=γk(``0,ij)W(x,y).

Thus (6) holds for non-negative integers`,`0. Sinceσ−`(x)Xi−`k,

W(x,y)=W(σ`−`(x)),y)=γk(`, (i`k)j)W(σ−`(x),y).

Hence

W(σ−`(x),y)=γk(`,i`kj)1W(x,y)

=η`(i−`kj)+(k/2)`(`−1)W(x,y)

=η`(ij)+(k/2)`(`+1)W(x,y)

=γk(−`,ij)W(x,y).

Sinceσ−`0(y)Xj−`0k,

W(x,y)=W(x, σ`0−`0(y)))=γk(−`0,i(j`0k))W(x, σ−`0(y)).

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Hence

W(x, σ−`0(y))=γk(−`0,ij+`0k)1W(x,y)

=γk(`0,ij)W(x,y).

Sinceσ`0(y)Xj+`0k,

W(σ−`(x), σ`0(y))=γk(−`,i(j+`0k))W(x, σ`0(y))

=γk(−`,ij`0k)γk(−`0,ij)W(x,y)

=γk(−``0,ij)W(x,y).

Similarly, we can show that

W(σ`(x), σ−`0(y))=γk(`+`0,ij)W(x,y), and

W(σ−`(x), σ−`0(y))=γk(−`+`0,ij)W(x,y).

This completes the proof of (6). 2

Lemma 4.2 If m is even,then k is even.

Proof: We apply Proposition 4.1 for` = m, `0 = 0 and i = j . Then (6) implies γk(m,0) = 1, and this becomesm/2)k(m1) = 1. Observe thatηm/2 = −1, sinceη is a primitive m-root of unity and m is even. Hence(−1)k(m1) = 1, so that k must be

even. 2

For iZm, set 1i =

t1

[

h=0

Xi+hk.

Observe that|1i| =t(n/m)=tn/(kt)=n/k, and that

X =

k1

[

i=0

1i,

Sinceσ(1i)=1i,1iis partitioned intoσ-orbits Yαi: 1i =

[r α=1

Yαi (i =0, . . . ,k−1),

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where r = |1i|/m = n/(mk). Observe that|Yαi| = m and |YαiXi| = k. We choose representative elements

yiαYαiXi (i =0, . . . ,k−1, α=1, . . . ,r).

Then X

σ`¡

yαi¢ ¯¯i =0, . . . ,k−1, α=1, . . . ,r, `=0, . . . ,m−1ª

, (7)

and W¡

σ`¡ yαi¢

, σ`0¡ yβj¢¢

=γk(``0,ij)W¡ yαi,yβj¢

(8) for`,`0Zm, i , j =0, . . . ,k−1 andα, β =1, . . . ,r .

We define square matrices Ti jof size r and Si jof size m(i,j =0, . . . ,k−1)by Tij(α, β)=W¡

yαi,yβj¢

(α, β=1, . . . ,r),

Si j(`, `0)=γk(``0,ij) (`, `0=0, . . . ,m−1).

For subsets A, B of X , let W|A×Bdenote the restriction (submatrix) of W on A×B. For two matrices S, T , we denote the Kronecker product by ST .

Proposition 4.3 For i, j =0, . . . ,k−1, W|Yi

α×Yβj =Ti j(α, β)Si j (α, β=1, . . . ,r), and

W|1i×1j =Si jTi j. (9)

Proof: Clear. 2

Thus W decomposes into blocks Wij=W|1i×1j (i,j =0, . . . ,k−1), and each block has the form Wi j=Si jTi j(i,j=0, . . . ,k−1).

5. Type II and Type III conditions

Let m, k, t, r be positive integers with m=kt.

Let Ti j (i,j =0, . . . ,k−1)be any matrices of size r with nonzero entries, and let Si j

(i,j =0, . . . ,k−1)be the matrix of size m defined by Si j(`, `0)=γk(``0,ij) (`, `0=0, . . . ,m−1),

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whereγkis defined by (5) for a primitive m-root of unityη. Now set Wi j=Si jTi j (i,j =0, . . . ,k−1),

and let W be the matrix of size n =kmr whose(i,j)block is Wij(i,j =0, . . . ,k−1). We index the rows and the columns of W by the set:

X = {[i, `, α]|0≤ik−1,0≤`m−1,1≤αr}, so that

W([i, `, α],[ j, `0, β])=Si j(`, `0)Ti j(α, β). (10) Proposition 5.1 W is a type II matrix if and only if Ti j is a type II matrix for all i,

j ∈ {0, . . . ,k−1}.

Proof: The type II condition (1) for a=[i1, `1, α1], b=[i2, `2, α2] becomes

k1

X

i=0 m1

X

`=0

Xr α=1

W([i1, `1, α1],[i, `, α])

W([i2, `2, α2],[i, `, α]) =i1,i2δ`1,`2δα12. (11) Using (10), we rewrite the left-hand-side as follows:

l.h.s.=

k1

X

i=0 m1

X

`=0

Xr α=1

γk(`1`,i1i)Ti1i1, α) γk(`2`,i2i)Ti2i2, α)

=η−`1ii+`2i2−(k/2)(`1−`2)(`1+`21)

k1

X

i=0

η(`1−`2)i Xr α=1

Tiii1, α) Ti2i2, α)

mX1

`=0

η(i1i2+k(`1−`2))`.

Observe that, sinceηis a primitive m-root of unity,

m1

X

`=0

η((i1i2)+k(`1−`2))`=

½m if(i1i2)+k(`1`2)≡0 (mod m), 0 otherwise.

Observe that(i1i2)+k(`1`2)≡ 0 (mod m)if and only if i1 = i2 and`1`2

(mod t), since 0≤i1,i2k1 and m=kt .

Now suppose that Ti jare type II (i , j =0, . . . ,k−1). We must show that the l.h.s. of (11) becomes zero for [i1, `1, α1]6=[i2, `2, α2]. By the above observation, we may assume that i1 =i2and`1`2 (mod t). We set`1`2 =t s. Ifα1 6=α2, then l.h.s. of (11) vanishes by type II condition for Ti1i. Hence we may assumeα1 = α2. Thus we have i1=i2,α1 =α2,`1`2≡0 (mod t)and`16=`2. Hence

l.h.s.=mrη−`1i1+`2i2−(k/2)(`1−`2)(`1+`21)

k1

X

i=0

ηtsi.

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Observe thatηtis a primitive k-root of unity. So,Pk1

i=0t)si =0, since s6≡0 (mod k). We have shown that W is type II.

Next suppose that W is type II. Pick any distinctα1,α2∈ {1, . . . ,r}. From (11) at i1=i2 and`1`2 (mod t), we obtain

k1

X

i=0

η(`1−`2)i Xr α=1

Tiii1, α) Ti2i2, α) =0. Setting

Ki = Xr α=1

Ti1i1, α) Ti1i2, α)

and considering the case`2=0, the above equation implies

k1

X

i=0

¡η`1¢i

Ki =0 (`1=0,t,2t, . . . , (k−1)t),

or equivalently

k1

X

i=0

t)eiKi =0 (e=0,1, . . . ,k−1).

Observe thatt)e(e=0,1, . . . ,k−1) are distinct, sinceηtis a primitive k-root of unity.

Hence Ki =0 (i =0,1, . . . ,k−1) by Vandermonde determinant. Thus Xr

α=1

Ti1i1, α)

Ti1i2, α) =0 (i =0, . . . ,k−1),

so that Ti1i is type II. 2

Lemma 5.2 Assume k is even when m is even. Then the matrix W satisfies the type III condition(2)if and only if the following equation holds for all i1,i2,i3 ∈ {0, . . . ,k−1} and for allα1, α2, α3∈ {1, . . . ,r}:

k1

X

i=0

Ãm1 X

`=0

ηk`γk(`,ii1i2+i3)

r X

α=1

Ti1,i1, α)Ti2,i2, α) Ti3,i3, α)

!

=D Ti1,i21, α2) Ti1,i31, α3)Ti3,i23, α2).

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Proof: The type III condition (2) for a = [i1, `1, α1], b =[i2, `2, α2], c =[i3, `3, α3] becomes

k1

X

i=0 m1

X

`=0

Xr α=1

γk(`1`,i1i)γk(`2`,i2i)

γk(`3`,i3i) ·Ti1,i1, α)Ti2,i2, α) Ti3,i3, α)

=D γk(`1`2,i1i2)

γk(`1`3,i1i3k(`3`2,i3i2)· Ti1,i21, α2) Ti1,i31, α3)Ti3,i23, α2). By a direct (but somewhat long) computation, we obtain

γk(`1`,i1i)γk(`2`,i2i)

γk(`3`,i3i) ·γk(`1`3,i1i3k(`3`2,i3i2) γk(`1`2,i1i2)

=ηk(`− ˆ`)γk(`− ˆ`,i− ˆi),

where`ˆ=`1+`2`3,iˆ=i1+i2i3. So the type III condition becomes

k1

X

i=0

Ãm1 X

`=0

ηk(`− ˆ`)γk(`− ˆ`,i− ˆi)

r X

α=1

Ti1,i1, α)Ti2,i2, α) Ti3,i3, α)

!

=D Ti1,i21, α2) Ti1,i31, α3)Ti3,i23, α2). To complete our proof, we must show that

m1

X

`=0

ηk(`− ˆ`)γk(`− ˆ`,i− ˆi)=

m1

X

`=0

ηk`γk(`,i− ˆi).

To show this, it is enough to show thatγk(`+m,j)=γk(`,j)holds for all j ,`. γk(`+m,j)=η−(`+m)j−(k/2)(`+m)(`+m1)

=γk(`,j)ηm jkm`η−(k/2)m(m1)

=γk(`,j)ηkm(m1)/2.

When m is odd,(m−1)/2 is an integer. When m is even, k is even by our assumption, and

so k/2 is an integer. Thusηkm(m1)/2=1. 2

Lemma 5.3 For all u,s(0≤ut−1,0≤sk−1), γk(u+st,j)=((−1)t1ηt j)sγk(u,j).

Proof: We computeγk(u+st,j)as follows.

γk(u+st,j)=η−(u+st)j−(k/2)(u+st)(u+st1)

=ηu j−(k/2)u(u1)ηst jη−(kt)suη−(k/2)st(st1)

=γk(u,j)ηst jη−(k/2)st(st1).

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So it is enough to show that

η−(k/2)st(st1)=(−1)(t1)s. (12)

If t is even, then m is even and soη(m/2)= −1. Hence, noting st−1 is odd, η−(k/2)st(st1)

η−(m/2)¢(st1)s

(−1)(st1)¢s

=(−1)s, so (12) holds. Next assume t is odd. If s is even, then

η−(k/2)st(st1)=η−(kt)(s/2)(st1)=ηm(s/2)(st1)=1. If s is odd, then st−1 is even. Hence

η−(k/2)st(st1)=η−(kt)s(st1)/2 =m)s(st1)/2=1.

Therefore (12) holds in each case. 2

Lemma 5.4 (i) If t is odd,then

m1

X

`=0

ηk`γk(`,j)=



 k

t1

X

u=0

ηu jku(u+1)/2 if j≡0 (mod k),

0 otherwise.

(ii) If t and k are even,then

m1

X

`=0

ηk`γk(`,j)=



 k

t1

X

u=0

ηu jku(u+1)/2 if jk

2 (mod k),

0 otherwise.

Proof: Using Lemma 5.3, we proceed as follows.

m1

X

`=0

ηk`γk(`,j)=

t1

X

u=0 k1

X

s=0

ηk(u+st)γk(u+st,j)

=

t1

X

u=0 k1

X

s=0

η−(ku+ms)((−1)t1ηt j)sγk(u,j)

= Ãk1

X

s=0

((−1)t1ηt j)s

t1 X

u=0

ηkuγk(u,j)

! .

If t is odd, then the first factor becomes

k1

X

s=0

t j)s=

k1

X

s=0

t)j s =

(k if j ≡0 (mod k), 0 otherwise.

(12)

Suppose t and k are even. In this case, m is also even, so thatη(kt/2)=η(m/2)= −1. Hence the first factor becomes

k1

X

s=0

((−1t j)s =

k1

X

s=0

¡η(kt/2)ηt j¢s

=

k1

X

s=0

t)((k/2)−j)s

=



k ifk

2 − j≡0 (mod k), 0 otherwise.

Now the result follows by

ηkuγk(u,j)=ηu jku(u+1)/2. 2

Proposition 5.5 Assume k is even when m is even. Then the matrix W satisfies the type III condition(2)if and only if the following equation holds for all i1,i2,i3∈ {0, . . . ,k−1} and for allα1, α2, α3∈ {1, . . . ,r}:

Ãt

1

X

u=0

ηu(i−ˆi)−ku(u+1)/2

r X

α=1

Ti1,i1, α)Ti2,i2, α) Ti3,i3, α)

!

=(D/k) Ti1,i21, α2) Ti1,i31, α3)Ti3,i23, α2),

whereˆi=i1+i2i3,and i denotes the integer in{0, . . . ,k−1}such that

i



ˆi (mod k) if t is odd, ˆi+k

2 (mod k) if t is even.

Proof: This is a direct consequence of Lemmas 5.2 and 5.4. 2 6. Some special cases

We use the notation in Section 4.

Proposition 6.1 Suppose k =1. Then m is odd,and W =ST,

where S is a spin model of size m and index m which is given by S(`, `0)=η−(1/2)(`−`0)(`−`01) (`, `0=0,1, . . . ,m−1), and T is a symmetric spin model of size n/m.

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