Available at: http://www.ictp.trieste.it/~pub-off IC/2000/43

United Nations Educational Scientific and Cultural Organizationand

International Atomic Energy AgencyTHE ABDUS SALAM INTERNATIONAL CENTRE FOR THEORETICAL PHYSICS

BRAID RELATIONS IN THE YOKONUMA-HECKE ALGEBRA

J. Juyumaya1

Universidad de Valparaiso, Gran Bretana 1041, Valparaiso, Chileand

The Abdus Salam International Centre for Theoretical Physics, Trieste, Italy

and

S. Senthamarai Kannan2

SPIC Mathematical Institute, 92, GN.Road, Chennai-17, Indiaand

The Abdus Salam International Centre for Theoretical Physics, Trieste, Italy.

Abstract

In this note, we prove a theorem on another presentation for the algebra of the endomorphismsof the permutation representation (Yokonuma-Hecke algebra) of a simple Chevalley group withrespect to a maximal unipotent subgroup. This presentation is done using certain non-standardgenerators.

MIRAMARE - TRIESTE

October 2000

Regular Associate of the Abdus Salam ICTP.E-mail: juyumaya@uv.cl; juyumaya@ictp.trieste.it

2E-mail: kannan@smi.ernet.in; kannan@ictp.trieste.it

1. INTRODUCTION

Let G be a simple Chevalley group defined over Fq. In this manuscript, we prove a theoremon a new presentation for the algebra of endomorphisms Yn(q) associated to the induced repre-sentation of the trivial representaion of U up to G, where U is a maximal unipotent subgroupof G. In [?], this theorem was proved for the case when the Cartan matrix of G is symmetric,that is when G is of type Al, Dl, E6, E7 or E8. In this manuscript, we prove the theorem forthe other simple Chevalley groups also. More precisely, we prove the nonstandard presentationtheorem for the simple Chevalley groups of type Bl, Cl, F 4 and G2.

In [?], T. Yokonuma has given a description (presentation) of this algebra Yn(q) in terms ofthe standard generators, that is, in terms of generators given by the double cosets (see 11.30[?]).So, we call the algebra Yn(q), the Yokonuma-Hecke algebra. The presentation of Yokonuma isanalogous to the classical presentation of the Iwahori-Hecke algebra (see [?]).

In Theorem 2.18[?], the first author of this article has proved that this algebra Yn(q) hasa presentation with non standard generators for the simple Chevalley groups G whose Cartanmatrix is symmetric. This presentation uses non-standard generators defined by a pre-fixednon-trivial additive character of Fq, and a certain non-trivial linear combination involving thestandard basis of Yn(q) (see Definition ??). Originally, these generators were defined in ageometrical way for the group GLn(Fq), that is, like Fourier Transforms on the space of functionsof flags vectors on Fnq. As an application of our main theorem, we recall that abstracting thepresentation in the case when G is of type Al, it is posible to define a certain finite dimensionalalgebra, involving braids and ties, which give new matrix representation for the Artin group oftype A, see [?] . It is a natural question to study the representation for the Artin groups oftypes Bl, Cl, F4 and G2 that arising from our theorem.

The aim of this note is to prove that the above mentioned non standard generators give apresentation for the algebra Yn(q) for the simple Chevalley groups of type Bl, Cl, F 4 and G2.

For more precise statement, see Theorem 2.The layout of this manuscript is as follows:Section 2 consists of preliminaries and statement of the main Theorem (for a more precise

statement, see Theorem 2.) Section 3 consists of the proof for the case when G is of type Bl, Clor F4. Section 4 consists of the proof for the case when G is of type G2.

2. PRELIMINARES AND STATEMENT OF THE MAIN RESULT

2.1. Let k denote a finite field with q elements. Let G be a simple simply connected Chevalleygroup defined over k. Let T be a "maximally split" torus of G. Let B be a Borel subgroup ofG containing T. Let U be the unipotent radical of B. We will denote the rank of G by l.

We denote the set of all roots with respect to T by Φ.Let A be the set of all simple roots with respect to T and B. Let N be the normaliser of T

in G and let W = N/T be the Weyl group of G with S = {sα : α G A} being the set of simplereflections. The pair (W, S) is a Coxeter system and we have the presentation:

W = (sa : (sαsβ)mαβ = l,a,l3eA),

where mαβ denote the order of sαsβLet π be the canonical homomorphism from N onto W. Using π, we have an action of the

Weyl group W on T: (w,t) 1—> w(t) := uitui~l, where weiVis such that π(ω) = w.We recall that for any root α G Φ, there is an ωα e N such that π(ωα) = sα and there is a

homomorphism φα : SL2 —> G such that

) h() f( r

2.2. Let Yn(q) be the algebra of endomorphism of the induced (permutation) representationIndUG1, over the field of complex numbers. We call the algebra Yn(q) as the Yokonuma-Heckealgebra.

From the Bruhat decomposition, G = YineN UnU, we have that the standard basis of theYokonuma-Hecke algebra is parametrised by N. Let {Rn | n G N} be the standard basis.

If n = ωα, we denote Rn by Rα.If n = t G T, we call the elements Rt in Yn(q) operators of homothety corresponding to t. In

the case t = hα(r), we denote Rt by Hα(r). With these notations, we define Eα by

Eα :=a

It is clear that the Eα's commute among themselves, and a direct computation shows that

(1) Eα2 = (q-l)Ea.

Now, we recall a Theorem due to T. Yokonuma.

Theorem 1. (See [?]) The Yokonuma-Hecke algebra Yn(q) is generated, as an algebra, byRα(α G Φ), and the homotheties Rt(t G T). Moreover, these generators with the relationsbelow define a presentation for Yn(q).

( ??.1) Rα2 = qHa(—l) + RαEα (quadratic relation)

( ??.2) RαRβRαRβ • • • = RβRαRβR α • • (braid relation)

( ??.3) RtRα = RaRt', where t' = uatu~l (teT)

( ??.4) RuRv = Ruv (u,veT).

2.3. In the following, we fix a non-trivial additive character ψ of (k,+). For any α in Φ, wedefine Ψ α as the following linear combination of elements in Yn(q),

ψ(r)Hα(r).

From a direct computation, we have that Ψα commutes with Eα, and

(2) ΨαEα = -Ea.

Definition 1. Let α G Ψ. We define the element Lα, as

Our main goal is to prove the following Theorem.

Theorem 2. The Yokonuma-Hecke algebra Yn(q) is generated (as an algebra), by Lα (α G Φ),and the homotheties Rt (teT). Moreover, these generators with the relations below define apresentation for Yn(q).

( ??.1) L2

α = 1 — q~l (Eα — LαEα) (quadratic relation)

( ??.2) LαLβLαLβ • • • = LβLαLβL α • • (braid relation)

( ??.3) RtLα = LaRv, where t' = uoatuj-1 (teT)

( ??.4) RuRv = Ruv (u,veT).

To prove this Theorem, we introduce some notations and one useful Proposition. We denoteby Eαw the effect of w on E α arising from the action of the Weyl group W on T. That is,

where γ is the root defined by w(α) = γ.In the similar way, we denote by Ψwα the effect of w on Ψα.

4

Proposition 3. Let s be the reflection corresponding to α, and let β e Φ. We have

Eαs = Eα

= -Ea = EαΨα

(-l) - Eα.

Proof. The proof of the assertions in ??.1 is an inmediate consequence of Yokonuma's Theorem,part 1.3 and the proofs of ??.2, ??.3 and ??.4 are straightforward computations. D

2.4. We are now going to sketch the proof of Theorem ?? for the simple Chevalley groups oftype Bl, Cl, F4 and G2. The only statement of Theorem ?? that involves the Dynkin diagram ofthe group is the statement about the braid relation, that is ??.2. Since Theorem ?? was provedfor the cases of type Al, Dl, E6, E7 and E8 in [?], to prove the Theorem, we need to proveonly ??.2 for the cases when G is of type Bl, Cl, F4 and G2. In Section 2, we prove ??.2 forthe case when G is of type Bl, Cl and F4. In Section 3, we prove ??.2 for the case when G is oftype G2. The method of proof involves the one parameter subgroups Hα(t), t G kx, α G Φ, andsome automorphisms of the two dimensional torus kx x k x

3. CASES Bh Q AND F 4

3.1. Let A = {α 1 • •, ai-i,ai} denote the set of all simple roots of type Bl. So, the Dynkindiagram is as follows:

ai a2 a j3

where α = ai-i and β = αl. Let s (respectively s') be the reflection corresponding to the roota (respectively β).

Notice that the simple roots α1, . . . , Cty_i of Bl turn to the set of simple roots of A\_i and sofrom Theorem 2.12[?], we deduce:

LαiLαj = LαjLαi if \i-j\>l

Lαi Lαj Lαi = Lαj Lαi Lαj if |i — j | = 1.

Therefore, to prove Theorem ??, we need to prove only the relation LαLβLαLβ = LβLαLβLα.In the proof of this braid relation, we will use the following lemma. The same proof holds forthe cases: Q and F 4 . ( The only difference is α = αl, β = cty_i in the case of Cl and α = α2,[3 = α3 in the case of F 4 ) .

Lemma 4. We have

= Eβ

sEα =

Proof. We now prove ??.1. We have s'(a) = α + 2β and so we have E^ = Eα+2β. Hence, wehave

E«EP = E Hα+2β(t) E Hβ(r)tekx rekx

J2 Ha{t)Hp{t2-r)(t,r)ekxxkx

J2 Hα(t)Hβ(r) = EαEβ,(t,r)ekxxkx

since the map (t, r) i—> (t, t2 -r) is an automorphism of kx x kx. This proves that Es

a Eβ = EαEβ.The equality EβsEα = EαEβ follows from the fact that s(β) = α+β and the map (t, r) i—> (tr, r)

is an automorphism of kx x kx.We now prove ??.2. We have s(s'(a)) = s(α + 2β) = - a + 2(β + α) = α + 2β = s'(a). This

proves that (E^)s = E*.Proof of (E^)3' = Eβs follows from the fact:

s'{s{f3)) = s'(a + f3) = {a + 2f3)-f3 = a + f3 = s{f3).

Proof of ??.3 follows from the facts that s(β) = α + β, s'(a) = α + 2β and the map (t, r) i—>(tr,tr2) is an automorphism of kx x kx.

Proof of ??A follows from the facts that

s'(s(s'(a))) = s'(s(a + 2β)) = s'{a + 2β) = (α + 2β) - 2β = α

ands(s'(s(/3))) = s{s'{a + β)) = s(α + β) = -a + α + β = β.

Proof of ?? .5 is similar to the proof of ??.2. We note here that Ψ does not play an importantrole in this situation.

We now prove ??.6. We have s'(a) = α + 2β and hence, we have

Since s(β) = α + β, we have (i?^)(-l) s = i ? « ( - l ) ^ ( - l ) . D

We now prove the following Lemma which will complete the proof of Theorem ?? for thecases when G is of type Bl, Cl and F 4 .

Lemma 5. LαLβLαLβ = LβLαLβLα.

Proof. First, we compute the products:pαβ := q2LαLβ, and pβα := q2LβLα.From the definition of Lα and Lβ, we have

a b c

Notice that b (respectively d) is obtained from EαRβΨβ (respectively RαΨ αRβΨ β) using propo-sition 3.

Now, we compute p2

αβ:

p2αβ = a2 + b2 + c2 + d2 + ab + ac + ad + ba + bc + bd + ca + cb + cd + da + db + dc.

In the same way, we obtain an analogous expression for pβα2, but in the symbols a', b', d andd!.

The proof of this Lemma is as follows. In the expression of p2

αβ and pβα2, we first bring themonomials 1, Raj Rsi RaRfli RpRa^ RaRpRa^ RflRaRfli &nd RaRsRaRs = RsRaRsRa to the

left. After this procedure, we will check that the coefficients of these monomials in p2αβ with the

corresponding coefficients in p2

βα are the same. Then, the lemma follows.Let X0, Xα, Xβ, Xαβ, Xβα, Xαβα, Xβαβ and Xαβαβ be the coefficient of 1, Rα, Rβ, RαRβ,

RβRα, RαRβRα, RβRαRα and RαRβRαRβ, respectively in pα

β2. Let Y0, Yα, Yβ, Yαβ, Yβα,Yαβα, Yβαβ and Yαβαβ be the coefficient of 1, Rα, Rβ, RαRβ, RβRα, RαRβRα, RβRαRα andRαRβRαRβ respectively in pβα2.

We need to prove that X γ = Yγ for all γ (words in α and β) as above. To do this, we willcompute Xγ and Yγ using essentially the Lemma ??. Now, as the computations are all verysimilar, we are going to compute only X0, Xαβα, Yαβα, Xα,Yα, Xαβ, Yαβ, Xαβαβ and Yαβαβ.

Computation of X0 and Y0. It is easy to see that the terms contributing to the constantcoefficient in the expression of p2

αβ (resp. in pia ) are only a2, b2 and c2 (resp. (a') 2, (b')2 and

(c')2).We have a2 = (EαEβ)2 = (EβEα)2 = (a')2.We now compute b2.We have

From the observation ??.1 of Theorem ??, we have Rβ2 = qHp(-l) + RβEβ.

Thus, the constant coefficient yielded by b2 is qHp{-l){Es^p)s>E^p.By a similar computation, it is easy to see that (c')2 yields the same constant coefficient as

that yielded by b2.A similar proof shows that the constant coeffiicient yielded by c2 and that yielded by (b1)2

are the same and both are equal to qHa(—l)(E%^>a)sEl^>a.

Thus, we have X0 = Y0.Computation of Xαβα and Yαβα. It is clear that the terms having RαRβRα in pαβ (re-

spectively pβα) is only dc (respectively b'd'). We have dc = b'd'. Namely,

b'd' =

(from??.4)

= dc.

he terms having R in pαβ2arComputation of Xα and Yα. The terms having Rα in pα

β2 are: c2, ac, ca, and db. We now

compute c2. We have

c2 = (RαΨ

= R2αΨs

αEβsΨαEβ (from proposition ??)

= (qHa(-l) + RαEα)ΨαsEβsΨαEβ (from ??.1).

Hence, c2 yields the coefficient EαΨsαEβsΨαEβ. Now, using proposition ?? and lemma ??, weget

EαΨαsEβsΨαEβ = -EaE^VaEp (from??.3)

= -EaEpmaEp (fromi??.l)

= -Ep(Eama)Ep

= EβEαEβ (from ??.3)

= {q-l)EaEp (from 1).

On the other hand, using Proposition ??, we get

ac = (EαEβ)(RαΨαEβ) = RαEαsEβsΨ αE β

andCM = RαΨαEβEαEβ.

Using Lemma ??, we deduce:

EαsEβsΨαEβ = ΨαEβEαEβ = -{q - 1)EαEβ.

Let us compute db,

db = {RaR^p){RpE^p)

= RaR}(ysy^E^f3 (from proposition ??).

Therefore, from ??.1, we have

db = R α ( q H

Thus, db yields the coefficient qHp{-l){^iY^E^p = qHp{-l)^a^E^p. Now,

*$EZ*p = qHp(-l)*aEZ*fop

= qHp(-l)VaEi(qHp(-l)-Ep) (from 11 A)

EαEβ (from ??.1)

Eβ (from ??.3).

Thus, db yields

q2^aEi + qEαEβ.

Therefore, we have

Xα = q2^aEi + EαEβ.It is easy to see that the terms having Rα in pβα are precisely a'b', b'a', cfd', and {b')2.Let us compute (b')2,

(b')2 = (RαEβsΨα)(RαEβsΨα)

= (qHa(-l) + RaEa)EpE%(qHa(-l) - Eα).

Hence (b')2 yield the coefficient EaE^(qHa(-l) - Eα) = qEαEβEβs - Eα2EβEβs. Then (b')2

yield precisely(q - 1)EαEβ.

Now, we have a'b' = (EβEα)(RαEβsΨα) = RαEβ

sEαsEβsΨα. Therefore, using Lemma ??, wededuce that a'b' yield the coefficient

EβsEα

sEβ

sΨα = -{q - 1)EαEβ.

It is easy to see that b'a' also yield the same coefficient of a'b'.Let us now compute c'd',

c'd' = (RβΨβE

8

That is,

c'd' = qR

Thus, c'd' yields the coefficient

Now, from the observation ??.4 of Proposition ??, and the observation ??.2 of Lemma ??, wehave

Thus, we have proved Yα = q2E^a + EαEβ = Xα.Computation of Xαβ and Yαβ. First, notice that there is only one term having RαRβ in

pβα2, which is b'c'. Now,

b'c' = (RαEβsΨα

Eα (from ??.2)

(from ??.1)

(from ??.3).

Therefore, we have Yαβ = —^*s

aEaEp.On the other side, the only terms having the monomial RαRβ in p2

αβ are: ad, cb, cd, da and

db. We have:

(from ??.3)

da =

cb =

= RαRβEαEβ (from ??.1).

9

From the observation ??.1 of Theorem ??, it is easy to see that db yields the coefficient

Again from the observation ??.1 of Theorem ??, we deduce that cd yield the coefficient

Thus, we have Xαβ = Yαβ = -^EaEp = EαEβ.

Computation of Xαβαβ and Yαβαβ. It is easy to see that d2 is the only term that yields

Xαβαβ and this coefficient is *« */?((*« ^pYY)

Also, it is easy to see that Yαβαβ is equal to ^^^^a^i. By Lemma ??, it is clear that

((^Sa)sY = *« and (^Y = *8p. Hence, we have Xafiafi = Yafiafi. U

4. CASE G2

Let Π = {α, β} be a system of positive simple root of Φ. Let us put

a (3G2: o < o

the Dynkin diagram. Let W denote the Weyl group of G2. Let s (respectively s') denote thereflection corresponding to the root α (respectively β). We have, (a,/3) = —1 and (f3,a) = —3.Hence, we have

(3) s(β)=3α + β, s'(a)=a + β.

In the proof of the braid relation of type G2 (Lemma ??), we will use the following Lemma.

Lemma 6. We have,

11 A) Hα(

= EαEβ,E

(??.7)(

(??.9)Ψ

10

Proof. We have s(s'(a)) = s(α + β) = -a + (3α + β) = 2α + β. Therefore, we have s'{s{s'{a))) =s'(2a + β) = 2(α + β) — β = 2α + β = s(s'(aj). Using a similar argument, it is easy to see thats(s'(s(p))) = s'(s(J3)). These observations prove ??.1 and ??.2.

Now, we will prove ??.3. We have s'(o;) = α + β, and so we have Es

a = J2rekx Hα(r)Hβ(r).Then

t£kx r£k

r,t£k

In the above assertion, we use the fact that (t, r) i—> (rt, r) is an automorphism of kx x kProof of the other assertion of ??.3 is similar to this proof.We now prove 11 A.We have

Ha(-t3)Hp(t2)

x

= E3α+2β

We note that here, we use the fact that t1—^ —t is a bijection of kx onto itself.Once again using this fact, we have HβEβ = Eβ. Therefore, we have

This proves 11 A.We now prove ??.5. We have

Hα(ts3 )Hβ(ts )(t,s)£kxxkx

Here, we use the fact that (t,s) i—> (ts3,ts2) is an automorphism of the group kx x kx.Similarly, the other assertion E3α+βE3α+2β = EβE3α+2β of ??.5 follows from the fact that

(t, s) i-> (ts, s~ 1) is an automorphism of kx x kx.We now prove the assertion ??.6.We first compute (* a**) s '(* / 3*^) JE;3a+2/3-

We have * « * * = qHa(-l) - Eα and s'(a) = α + β. Therefore, we have ( Ψ α Ψ α

qHa+p(—l) — Eα+β.

We also have *^*^' = qH^-l) - Eβ.Thus, we get

«)*' =

Then using ??.4, we have

We now prove that q2E3α+2β — qEβE3α+2β — EαEβ is s- invariant.To prove this, we prove each of the summand is s- invariant.

11

First, we have s(3α + 2β) = - 3 a + 2(3α + β) = 3α + 2β and so we have E3sα+2β = E3α+2β.Secondly, we have

β. (from ??.5)

Thirdly, we have

(t,s)£kxxkx

= EαEβ.

Here, we use the fact that (t, s) i-> (t~ls3, s) is an automorphism of the group kx x kx.Thus, we have proved ??.6.We now prove ??.7.First, we prove (EαEβ)

w = EαEβ for any w eW.Since the Weyl group of G2 is generated by s and s', it is sufficient to prove that

(EαEβ)s = EαEβ = (EαEβ

For a a proof of the first equality, we have

)s

(t,s)£kxxkx

Hα(t)Hα(s)

(t,s)ekxxkx

We note that in this proof, we use the fact that the map (t, s) 1—> (t~ls3, s) is an automorphismof kx xkx.

The proof of the second equality follows from the fact that the map (t,s) i-> (^is" 1 ) is anautomorphism of kx x kx.

We now prove that ^EaEp = -EaEp for any w G W..Since (EαEβ)w = EαEβ, we have

= (-EaEp)w (from (6.5) )

We now prove ??.8.We have

— Ha(—l)Ea+/3

t£k

t£kx

^ Ha(t)Hp(-t)t£kx

t£k

12

We now prove ??.9.We have s'(a) = α + β, and so we have

(t,s)ekxxk

Here, we use the fact that the map (t,s) i—> (t,ts) is an automorphism of kx x kx.On the other hand, we have

Hence, we have

V^Ep + VeV'pEf} = (*« - Va)E(} = 0.

Thus, we have proved ??.9. D

Lemma 7. We have LαLβLαLβLαLβ = LβLαLβLαLβLα.

Proof. Set pαβ = q3LαLβLα, and pβα = q3LβLαLβ. Then, one can re-write the braid relationof the lemma as

(??•!) Paf3Pf3a=Pf3aPaf3-

According to Proposition ??, Lemma ?? and ??, we obtain

Pa/3 = (g ~ l)EaE,3 + qHa(-l)yay8

aE8p- RaEaEfj + RpEaEJmp

and

We are now going to compare the coefficients of the monomials on Rα and Rβ obtained inboth sides of equation ??.1. For any word γ in α and β, let X γ (resp. Yγ) be the coefficient ofRγ in the expression of L.H.S (resp. R.H.S) of ??.1.

To prove the Lemma, it is sufficient to prove that X γ = Yγ for all words γ in α and β.Computation of Xαβαβαβ and Yαβαβαβ. On the left in the product of ??.1 the monomial

RαRβRαRβRαRβ appears only in the multiplication gg', and then the coefficient Xαβαβαβ ofthis monomnial is ' / '

We have

(from??.2)

^ ' ^ ) ^ a ) . (from ??.2)

This is the coefficient Yαβαβαβ of RαRβRαRβRαRβ on the right of ??.1. Notice that we have

13

Computation of Xαβαβα and Yαβαβα. It is to check that the monomial Roccurs only in the product ge' on the left of ??.1, and only in the product f'g on the rightof ??.1. Now, the coefficient Xαβαβα is

Xafiafia

(notice that from ??.1, (((Es^s')s)s')s = Eβ).Computation of Xαβαβ and Yαβαβ. On the right of the equation ??.1 the monomial

RαRβRαRβ appears only in the product fe. We now compute this coefficient.

We have s'(s(s'(s(P)))) = s(β) and so (((Esp)s')s)' = Eβs. Using this observation, we have:

(from above)

)8Y (since Eα

s = Eα)

= (-l)4EaEf3 (from ??.9)

Hence, we have

On the other side, the terms on the left of the equation ??.1. that contain the monomialRαRβRαRβ, are the products: eg', ef, gc', eg', and gf.

We now prove that eg' yields the coefficient EαEβ.We have

eg' = -RaRpRaRp(((EaEpy'yy'm*ay'*p

= -RaRpRM(-l)3EaEp). (from??.7)

Therefore, the coefficient of RαRβRαRβ in the expression of eg' is

( ??.3) -EaEp.

We now prove that ef yields the coefficient EαEβ. We have

ef = R α R β

1)4EαEβ. (from??.7)

Therefore, the coefficient yielded by ef is

??.4) EαEβ.

By using ??.7, it is easy to see that gd yields the coefficient

??.5) -(-lfEaEp = EαEβ.

Now, we compute the coefficient yielded by eg'.We have

eg' = { R a R f i m i m f i E '

14

Now, notice thatget

But, we are interested in computing only the coefficient of Rcomputations, we first compute the E's (without (Ψ's) in coefficient ofeg'

since E% = Eβ. We now compute the coefficient together with Ψ's.

β. Then using ??.1, we

β. From the aboveRβ in the product

Therefore, eg' yields the coefficient

-EaEp.

By a similar computation, we can see that the coefficient of RαRβRαRβ in the expression ofgf is equal to

Summing up these five coefficients (from ??.3 to ??.7), we have

From the observations ??.2 and ??.8, we have Xαβαβ = Yαβαβ.Computation of Xαβα and Yαβα. In the left hand side of ??.1, the products that contain

the monomials RαRβRα are: ce', ed', ee', ga', gb' and gd!.Since EαEβ is occurring in c, by using ??.7, it is easy to see that the coefficient of RαRβRα

in the expression of the product ce' is

( ??.9) -{-l)2EaE} = -(q - 1)EαEβ.

Since EαEβ is also occurring in a', one can check that the coefficient of RαRβRα in theexpression of ga' to be equal to

( ??.10) (-f)3((? - 1)EαEβ = -{q - 1)EαEβ.

We now compute ed'. We have

ed' = R α

= RaRpRa{-lfEaE} (from??.7)

Therefore, ed' yields the coefficient:

??.11) -{q-l)EaEp.

We now compute the coefficient of RαRβRα in the expression of ee'.

We have

ee' =

We also have

RαRβ2Rα = qRaHp(-l)RaEaEp + RαRβRα(Eβ)sEαEβ. (from ??.1)

We are interested in computing only the coefficient of RαRβRα in the expression of ee'. Wefirst compute only the product of E's (without Ψ's).

15

This coefficient is equal to

since E% = Eβ.

We now compute the coefficient of RαRβRα (together with the product of Ψ's) in the expres-sion of ee'.

The coeficient of RαRβRα in the expression of ee' is

^ ) s E p = (-l)4EaEJ ( from??.7)

= (q - 1)EαEβ. > (from 1)

Hence ee' yields the coefficient

( ??.12) (q - 1)EαEβ.

By a similar computation, one can check that the coeficient of RαRβRα in the expression ofgd' is

( ??.13) (q - 1)EαEβ.

We now compute gb'.It is easy to see that gb' yields the coefficient

Summing up all these coefficients (using the observations from ??.9 to ??.14): it is easy tosee that the sum of the coefficients coming from ce', ed' and ga' is — 3(q — 1)EαEβ and that thesum of the coefficients coming from ee' and gd' is 2(q — 1)EαEβ.

Therefore, we have

- 1)EαEβ) + 2(q -

Thus, we have

(??.15) Xαβα = -{q - 1)EαEβ +

Now, we will compute Yαβα. The products on the right that contain the monomials RαRβRα

are: a'g, b'g, d'f, d'g, f'c, and / ' / .Computations are similar to the computations of Xαβα.Since a' contains EαEβ as a factor, by using ??.7, it is easy to see that the coefficient of

RαRβRα in the expression of a'g is

( ??.16) (-I)3((? - 1)EαEβ = -{q - 1)EαEβ.

By the same argument, it is easy to see that the coefficient of RαRβRα in the expression off'c is

( ??.17) "(-1)2(<? - 1)EαEβ = -{q - 1)EαEβ.

(Here, we use the fact that EαEβ is a factor of c).We now compute d'f.We have

d'f = R α R

Therefore, by using ??.7, it is easy to see that the coefficient of RαRβRα in the expression ofd'f is

( ??.18) (-I)3((? - 1)EαEβ = -{q - 1)EαEβ.

By a similar computation in ee', using ??.1, and ??.7, one can check that the coefficient ofin the expression of / ' / is equal to

(-I) 4 ((? - 1)EαEβ = {q- 1)EαEβ.

16

By a similar computation in gd', using ??.1 and ??.7, one can check that the coefficient ofin the expression of d'g is equal to

We now compute the coefficient of RαRβRα in the expression of b'g.Using Theorem ??, it is easy to see that the coefficient of RαRβRα in the expression of the

product b'g is:

Then, from ??.8 we obtain that the coefficient of RαRβRα in the expression of the product b'gis

Summing up these coefficients (using observations ??.16 to ??.21), we have

??.22) Yapa = -{q - l)EaEp + ^ ( - ^ ( ( ( ( ( ^ ^ y y ( ( ^ Y ^ W

To prove Xαβα = Yαβα, from the observations ??.15 and ??.22, it is sufficient to prove that

We now prove this assertion.By a similar proof of ??.1, it is easy to see that

s(s'(s(s'(m) = "(3a + 2β) = s'(s(s'

Hence, we have

- Ep)8Y (from Theorem ??)

= q{Ha{-l)f{Hp{-l)f-{{Epyy' (since s's(/5) = 3a+ 2/5)

Using this and fact that (Ha(-l))2 = (H^-l))2 = 1, we have

(from??.8))

- (EaEp)8' (from??.7)

Eβ (since Esp =Eβ)

= VpVpEZ (from Theorem ??).

Thus, we have proved XΑΒΑ = Yαβα.Computation of Xα and Yα. We first compute Xα. On the left of ??.1, the terms containing

the monomials Rα are: (a + 6)d', c(a' + 6'), cd', de', ecf, ee', and f̂/'.Now, we deduce that de' yields q(q — 1)EαEβ. In fact,

de' = {RpEaEimp){RpRa^aEp)

= R}Ra{Eiy{{Ejs'2)y{^y'^aEp)

= RlRa{Ei)8Ea{%Y^8^!aEp (since (s')2 = land££ = Ea).

Now, using ??.1, it is easy to see that de' yields the coefficient qHp{-l){Ei)sEa{^s

p)s' ^s

p^ αEβ.

We also have {Ea )sEαEβ = (q — 1)EαEβ. Therefore, using Lemma ?? we deduce that de' yieldsthe coefficent

( ??.23) (~lfq(q ~ 1)EαEβ = -q(q - 1)EαEβ.

In the same way, one can check that ee' yields the same coefficient

( ??.24) (-f)4<?(<? - 1)EαEβ = q(q - 1)EαEβ.

Since EαEβ is a factor of c', using ??.7 and the fact that Rβ2 = qHp(—l) + RβEβ, it is easy

to see that the coefficient of Rα in the expression of ed is

( ??.25) (-1)3(<?(<? - 1))EαEβ = -q{q - 1)EαEβ.

We now compute the coefficient of Rα in the expression of cd'. We have

cd' = (RαEαEβ)(RαEβEβsΨα)

= RaEaEpEpEpi&a

= (qHa(-l) + RαEα)EαEβsEβEβsΨα.

Therefore, the coefficient of Rα in the expression of cd' is EαEαEβsEβEβsΨ α, which turns out tobe equal to

( ??.26). (-l)2(q - 1)3EαEβ = (q- 1)3EαEβ.

(by using ??.7).We now compute the coefficient of Rα in the expression of c(a' + b').Since EαEβ is a factor of c, using ??.7, it is easy to see that the coefficient of Rα in the

expression of ca' yields —{q — 1)3Eα. Now, again using ??.7, it is easy to see that the coefficientof Rα in the expression of cb' is (—1)3q(q — 1)EαEβ = —q(q — 1)EαEβ.

Therefore c(a' + b') yields the coefficient

( ??.27) -((q - If + q(q - 1))EαEβ = -(q - 1)(q2 - q + 1)EαEβ.

We now compute the coefficient of Rα in the expression of (a + b)d'.Since EαEβ is a factor of a, using ??.7, it is easy to see that —{q — 1)3EαEβ is the coefficient

of Rα in the expression of ad'.On the other hand, we have

bd' = (qHa(-iyVayaEsp)(RaEpE^Va)

= qRaHa(-iyVayaEpEpE^Va (since s2 = 1 and Ha(-l)s = Ha(-1))

= q(q - l)RaHa{-l)mamaEpEs^a.

Thus, (a + b)d' yields

( ??.28) -{q - l)3EaEp + q{q - l)Ha{-l)^a^8

aEpE^a.

We now compute the coefficient of Rα in the expression of gf. We have

g f = { R i '

Using R2α = qHa{-\) + RαEα, we get

f = qRaRlHa{-iy\^py'^ay\Es^ay'^p + R

Now, using R2β = qHp{—l) + RβEβ, and using the fact that

-l)y' = Ha(-l)Hp(-l)2 =

one can see that the coefficient of Rα in the expression of gf is

18

Therefore, using the observations from ??.23 to ??.29, we have( ??.30)

Xα = -{q - 1)(q2 + 1)EαEβ + q(q -

We now compute Yα. On the right of ??.1, the terms having the monomials Rα are: (a' + b')c,d'(a + b), d'c,Jf,f'd,f'f, and e'g.

Since EαEβ is a factor of c, using ??.7 and the facts that Eα2 = (q — 1)Eα, Eβ2 = (q — 1)Eβ, it

is easy to see that the coefficient of Rα in the expression of a'c is —{q — 1)3EαEβ. Again, since

EαEβ is a factor of c, using ??.7 and the facts that Eα2 = (q — 1)Eα, Hp{—l)Ep = Eβ, it is easy

to see that the coefficient of Rα in the expression of b'c is

(-lfq(q - 1)EαEβ = -(q(q - 1))EαEβ.

Thus, the coefficient of Rα in the expression of (a' + b')c is

( ??.31) -(((? - 1)3 + q(q - 1))EαEβ = -{q - 1)(q2 - q + 1)EαEβ.

We now compute the coefficient of Rα in the expression of d'(a + b). Since EαEβ is a factor

of a using ??.7 and the fact that Eβ2 = (q — 1)Eβ, it is easy to check that the coefficient of Rα

in the expression of d'a is —{q — 1)3EαEβ. Also, it is clear that the coefficient of Rα in the

expression of d'b is

α s E β s ) = q(q ~

Thus, the coefficient of Rα in the expression of d'{a + b) is

( ??.32) -{q - 1)3EαEβ + q(q - l ) ^ ( - l ) ( * a ) 2 ^ ^ ( ^ ) 2 .

We now compute the coefficient of Rα in the expression of d'c. We have

d'c =

Using Rα2 = qHa(—l) + RαEα and ??.7, it is easy to see that the coefficient of Rα in theexpression of d'c is

Let us compute the coefficient the yields f'd. We have

f'd = ( R α

R

Therefore from ??.1, we get

f'd = qR

We have Eαs = Eα and so we have

= Hp{-l){EaEp)8

(from??.7)

Therefore, EαEβ is a factor of the coefficient of Rα in the expression of f'd and hence using ??.7

and the fact that Eβ2 = (q — 1)Eβ, it is easy to see that the coefficient of Rα in the expression

of f'd is

19

Since EαEβ is a factor of d, using ??.7, 1.1 and the fact that Eα2 = (q — 1)Eα, it is easy to seethat the coefficient of Rα in the expression of d f is

( ??.35) (-l)3q(q ~ 1)EαEβ = -(q(q - 1))EαEβ.

We now compute the coefficient of Rα in the expression of the product / ' / . We have

f'f = ( R ' ^

Thus, from ??.1 we deduce that / ' / yields the coefficient

Using ??.7, we deduce that / ' / yields

??.36) {~l)\{q ~ 1)EαEβ = q(q - 1)EαEβ.

We now compute the coefficient of Rα in the expression of e'g. We have

e'g =

Using twice the relation ??.1 we deduce that e'g yields the coefficient

q2

which can be written by

(Here, we use (H^-l)(Ha(-l))s'y = (H^-l))2(Ha(-l))s = Ha{-l)s = Ha{-1)).Therefore, using the observations from ??.31 to ??.37, we conclude that

Yα = -qiq-lf+qiq

= Xα (from ??.30).

Computation of Xαβ and Yαβ. The products on the left of equation ??.1 involving themonomial RαRβ are: (a + b)f, cd, cf, e{a! + b'), ed, eg', dg', gd', and gf.

We now compute the coefficient of RαRβ in the expression of af.Since EαEβ is a factor of a, using ??.7, and the fact that Eβ2 = (q — 1)Eβ, it is easy to see

that the coefficient of RαRβ in the expression of af is

( ??.38) (-l)2(q - 1)2EαEβ = (q- 1)2EαEβ.

We now compute bf. We have

bf = qHa(-l)yays

aE^RaRp(E^ay'^p

-iyy\*aKy\E^y'^p (from ??.3).

Therefore, using the fact that (Ha(-l)s)s>Es^ = -Ha(-l)Ep and the fact that

it is easy to see that the coefficient of RαRβ in the expression of bf is

We now compute cd.

Since EαEβ is a factor of c, using ??.7 and the facts that Eα2 = (q — 1)Eα and Eβ2 = (q

it is easy to see that the coefficient of RαRβ in the expression of cd is

We now compute cf.

20

We have

cf = -RaE

= -RlRpEaEp{Es^a)s' (from??.3)

= (qHa(-l) + RaEa)RpEaEp(E^a)s'*p. (from ??.1)

But, we are interested only in the coefficient of RαRβ, which is equal to

-EiEaEp{E8^!aY^p = (-l)3EaEp(EaE8

pY (from 77.7)

= {{-l)\ElE^y' (from??.7)

= -{{q-l){EaEpE8

pY (since Eα2 = (q-l)Ea

aE}yy' (from??.7)

= -{q-l)2EaEp (from??.7).

Therefore, the coefficient of RαRβ in the expression of cf is

We now compute ec'.We have

ec' = -

^-l) + RpEp)^iPsipEaYEαEβ (from ?? .1)

But, we are interested only in the coefficient of RαRβ, which is equal to

i l = {-lfEaE}Ei (from??.7)

= -{q-l)EaEpEi (since Eβ2 = (q -

^ ' (from??.7)

= -{q - 1)2EαEβ > (from ??.7).

Therefore, the coefficient of RαRβ in the expression of ec' is

We now compute eg'.We have

= R2

aRp{(qHp{-l) + RpEp)aY{{{^pEaYyY^i^p (from ??.l and ??.3).

Using Rα2 = qHa(—l) + RαEα, it is easy to see that the coefficient of RαRβ in the expressionof eg' is

((£" ) s ) sHere, we first consider the term Es

a ((-£"« ) s ) s • This can be written as

21

Now, from the above two observations, using ??.7, it is easy to see that the coefficient ofRαRβ in the expression of eg' is

( ??.43) {-lfqEaEp = -qEaEp.

We now compute ea'.We have

ea' = RaRp((q-l)^VpE2

aEp

= RaRp(q-l)(-l)2EaEp (from??.7)

= (q-l)2EaEp (since Eα2 = (q-l)Ea)

Therefore, the coefficient of RαRβ in the expression of ea' is

We now compute the coefficient of RαRβ in the expression of eb'. Using the fact that EαEαEαEβ, we have

eb' = RaRfiqHfi{-l)m2

i{mamfi)s'EaEfi

= RaRpqHp(-l)(-l)4EaEp (from ??.7)

Therefore, the coefficient of RαRβ in the expression of eb' is

??.45) qEαEβ.

We now compute dg'.We have

dg' = RpEaEs^VpRpR

= RaRp{(qHp{-l) + RpEp)aY{{{EaEp^pYyY{^aY^p (from ??.f, ??.3)

= RaRp((qHp(-l) + RpEpyy\-l)4EaEp (from ??.7)

Therefore, the coefficient of RαRβ in the expression of dg' is

??.46) qEαEβ.

Here, we use the fact that Hp{—l)Ep = Eβ.We now compute RαRβ in the expression of gd!.We have

gd' = RaRpRa{^py^aRaEpEs^a

Using the quadratic relation Rα2 = qHa(—l) + RαEα and the fact that s2 = 1, it is easy tosee that the coefficient of RαRβ in the expression of gd! is

We now compute the coefficient of RαRβ in the expression of gf.We have

Using R2

α = qHa(-l) + RαEα and Rβ2 = qHp{-l) + RβEβ and the fact that Ha{-iy'mpEp

Ha(—l)E/3, it is easy to see that the coefficient of RαRβ in the expression of gf is

22

Using the observations ??.39, ??.48, and the observation ??.9 (of Lemma ??), it is easy tosee that the sum of coefficients yielded by bf and gf is equal to

Therefore, summing up all the other coefficients (using the observations from ??.38 to ??.48),we have

( ??.49) Xap = (q2-q + l)EaEp + qHa(-l)*a*8

a*Z*pEpE8

p.

We now compute the coefficient of RαRβ in the expression of pβαpαβ.In the product pβαpαβ, the terms involving RαRβ are a'e, b'e, d'd, d'e, fa, fb and f'd.We now compute a'e.Since EαEβ is a factor of a', using ??.7, and the fact that Eα2 = (q — 1)Eα, it is easy to see

that the coefficient of RαRβ in the expression of a'e is

{-\)2{q-l)2EaEp.

We now compute b'e.We have

b'e = q p { ) ^ $ i a ^ i p a

; \ i ' ^ p E a (from ??.3)

^ i m f i (from ?? . l and ??.3)

= RaRp(-l)4q((EaEpHp(-l)yy' (from ??.7)

= RaRpq((EaEp)sy' (since Hp(-l)Ep = Eβ)

= RαRβqEαEβ (from ??.7))

Here, we use the fact that

Ea{(Eiyy' = Ea{(Eay'y (from??.1)

= {EaEs;)s (since Eαs = Eα)

= EαEβ (from ??.3 and ??.7).

Therefore, the coefficient of RαRβ in the expression of b'e is

We now compute d'd.We have

d'd = R

(from??.3)

- 1)2EαEβ (from ??.7 and Eβ3 = (q - 1)2EαEβ)

Therefore, the coefficient of RαRβ in the expression of d'd is

??.52) (q - 1)2EαEβ.

We now compute the coefficient of RαRβ in the expression of d'e.We have

d'e = RaE^Es^aRaR^im pEa

= R2

aRp((EpEs^ayy'^pEa (from 11.3)

= (qHa(-l) + RaEa)R^((E^EpVayy'^^Ea (from 11.1)

23

But, we are interested in computing only the coefficient of RαRβ, which is equal to

{EiEa){(EpE^a)s)s'mi^p = (-f)3((? - l)2EaEp (from 11.3 and 11.7).

Therefore, the coefficient of RαRβ in the expression of d'e is

??.53) -{q - 1)2EαEβ.

We now compute fa.We have

fa = RaRp(Es^a)s'mp(q-l)(EaEp)

= RaRp(-l)2(q-l)2((EaE})sy' (from ??.7)

Therefore, the coefficient of RαRβ in the expression of fa is

We now compute f'b.A straightforward computation shows that the coefficient of RαRβ in the expression of f'b is

We now compute the coefficient of RαRβ in the expression of f'd.We have

f'd = RaRp{E^ay'm^{EaEi)m^

= RaR}{{Elimay'mpy\EaEi)mp (from??.3)

= Ra(qHa(-l) + RpEp)((Es^ay'yp)s' (EαEα

But, we are interested in computing only the coefficient of RαRβ, which is equal to

Ep{{Es^ay'mpy\EaEi)mp = {-lf{{EaE})s)Ei (from 11.7 and (s')2 = 1)

= -{q-l){EaEp)sEi (since Eβ2 = {q-l)Ep)

= -(q-l)(EaEpEa)s' (from ??.7)

= -(q-l)2(EaEp)s' (since Eα2 = (q-l)Ea)

= -{q-l)2EaEp (from??.7).

Therefore, the coefficient of RαRβ in the expression of f'd is

( ??.56) -(q - 1)2EαEβ.

Summing up these coefficients (using the observations from ??.50 to ??.56), we have

( 11.57) Yafi = {q2-q + l)EaEp + qHa(-l)*aK*«*pE8p(E8p)s'.

We have Eαs = E3α+β and {E^)3' = E3α+2β.Using a similar proof ??.5, it is easy to see that

( ??.58) El3(E/3)sEl3Esa+l3 = Esa+l3Esa+2/3 = ESp{{Ep)s)s .

Using the observations, ??.49, ??.57 and ??.58, it is easy to see that Xαβ = Yαβ.We now prove that

X0 = Y0.

The terms yielding the constant coefficients in the expression of pαβpβα are aa!, ab', ba!, bb',cd', dd, ee' and // ' .

The terms yielding the constant coefficients in the expression of pβαpαβ are a1 a, a'b, b'a, b'b,dd, d'c, e'e and / ' / .

24

It is easy to see that these terms yield the following coefficients.

aa' = a'a = (q - 1)4EαEβ

ab' = b'a = q(q - 1)EαEβ

bd = a'b = q(q - 1)EαEβ

bb' = b'b = q2

cd' = d'c = q(q - 1)2EαEβ

dd = c'd = q(q - 1)2EαEβ

ff = e'e = q 2 H a { - i $ g

Hence, we have

X0 = Y0.

Since the computations of Xβ, Xβα, Xβαβ, Xβαβα, Xβαβαβ are similar to the computationsof Xα, Xαβ, Xαβα, Xαβαβ, Xαβαβα respectively and the same is true for the Y's, we will onlyquote the coefficients.

We first quote the terms involving Rβ in the expression of pαβpβα and the terms involvingRβ in the expression of pβαpαβ.

The terms involving Rβ in the expression of pαβpβα are ac', 6c', cf, da', db', dc', eg' fd' and

On the other hand, the terms involving Rβ in the expression of pβαpαβ are a'd, b'd, c'a, db,c'd, d'e, e'c, e'e and g' f.

The coefficients yielded by these are as follows:

ac' = c'a = -(q - 1)2EαEβ

be' = c'b = -(q(q-l))EaEg

cf = ec' = -{q- 1)EαEβ

da' = a'd = -(q - 1)2EαEβ

db' = b'd=-{q{q-l))EaEg

dc' = c'd = {q- 1)2EαEβ

eg' = g'f = q2Hg{-l)^a^ay'^g^yy'^g{Esy

fd' = d'e = -(q-l)EaEg

ff = e'e = q{q-l)EaEg.

We note that here, we write the only coefficients (not with the monomials).Therefore, we have

Xβ = Yβ.

We now do the same for RβRα.The terms involving RβRα in the expression of pαβpβα are (a + b)e', dd', de', f{a' + b') and

fd'.On the other hand, the terms involving RβRα in the expression of pβαpαβ are a'f, b'f, dc,

c'f, d'g, e'a, e'b, e'c, e'g, g'd and g'f.

25

The coefficient yielded by these are as follows:

ae' = e'a = (q - 1)2EαEβ

be' = e'b = q(EgEsg-(q-l)EaEg)

dd' = e'c = (q- 1)2EαEβ

de! = c'f = -{q{q-l))EaEg

fa' = a' f = (q — 1)2EαEβ

fb' = b'f = qEαEβ

fd' = e'c = -{q-l)2EaEg

g d = —(<jf ) = EαEβ.

(We note that we write only the coefficients)Hence, we have

We now do the same for Rβαβ.The terms involving RβRαRβ in the expression of ag', bg', df, dg', fc' and // ' .On the other hand, terms involving RβRαRβ in the expression of pβαpαβ are de, e'd, e'e, g'a,

g'b, g'd.The coefficients yielded by these are as follows:

ag' = g'a = -{q-l)EaEg

bg' = g'b = qHa{-i){msgmay'mgmams

aEsg

dg' = g'd = (q-l)EaEg

df = e'd = -{q-l)EaEg

fc' = c'e = -{q-l)EaEg

ff = e'e = {q-l)EaEg.

Hence, we have

Xβαβ = Yβαβ.

We now do the same for RβRαRβRα.

The terms involving RβRαRβRα in the expression ofpαβpβα is fe! only.On the otherhand, the terms involving RβRαRβRα are c'g, e'f, e'g, g'c and g'f.The coefficients yielded by these are as follows:

fe' = e'f = EαEβ,

c'g = -e'g = EαEβ,

g'c = -g'f = EαEβ.

Hence, we have

Xga8a = ^

We now do the same for RβRαRβRαRβ.The only term involving RβRαRβRαRβ in the expression of pαβ is fg'.On the otherhand, the only term involving RβRαRβRαRβ is g'e.The coefficients yielded by these are

fg' = g'e = *g*g(^gyy'^Sa)s*SaEa.

•

26

AcknowledgementsThis work was done within the framework of the Associateship Scheme of the Abdus Salam

International Centre for Theoretical Physics, Trieste, Italy. J. J. was supported in part by DIPUV01-99 and Fondecyt 1000013 and S.S.K. was supported by an ICTP fellowship. The authorsthank the Abdus Salam ICTP for hospitality during their stay.

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