cvut.czContents 1. Introduction. 2 2. Basic preliminary. 4 3. PBW theorem for universal enveloping...

�ESKÉ VYSOKÉ U�ENÍ TECHNICKÉFakulta jaderná a fyzikáln¥ inºenýrská

katedra matematiky

DIPLOMOVÁPRÁCE

Studiumvlastností Fairliehoalgebry

Vypracoval: Severin Po²taVedoucí práce: Ing. Prof. Miloslav Havlí£ek

�kolní rok: 1997/98

CzechTechnical UniversityFaculty of nuclear sciences and physical engineering

Department of mathematics

MasterThesis

Matrix rereprestationsofFairlie algebra

Author: Severin Po²taCoordinator: Prof. Miloslav Havlí£ekSchool year: 1997/98

Prohla²uji, ºe jsem diplomovou práci vypracoval samostatn¥a uvedl v²echnu pouºitou literaturu.

V praze 24. 4. 1998

I state that this thesis was done exclusivelyby myself and the bibliography is complete.

Prague, April 24, 1998

Contents1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22. Basic preliminary . . . . . . . . . . . . . . . . . . . . . . . . . . 43. PBW theorem for universal enveloping algebras . . . . . . . . . . . . . 64. Fairlie's quantum deformation of so(3) . . . . . . . . . . . . . . . . . 95. Matrix representations not at root of unity . . . . . . . . . . . . . . . . 146. Fairlie algebra as a special case of quadratic algebra . . . . . . . . . . . . 337. Hopf algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . 348. Extension of sl(2) and its representations . . . . . . . . . . . . . . . . 359. Algebra homomorphism from deformation of so(3) into deformation of sl(2) . . 3610. Finite dimensional representations obtained using algebra homomorphism . . 3711. Tensor product of representations . . . . . . . . . . . . . . . . . . . 4112. Infinite dimensional representations obtained using algebra homomorphism . 4213. Other infinite dimensional representations . . . . . . . . . . . . . . . 4414. Finite dimensional representations at root of unity . . . . . . . . . . . . 4615. Representations at root of unity obtained using homomorphism . . . . . . 4816. Other representations at root of unity . . . . . . . . . . . . . . . . . 54

1. IntroductionThe main aim of this work is to explicitly express all irreducible finite matrixrepresentations of the associative algebra Uq(so3) what is called the Fairlie'squantum deformation of so3, and point out some other properties of this algebra(Poincare-Birkhoff-Witt property, homomorphism into Uq(sl2) and its connectionto the representations, tensor product of representations, infinite dimensionalrepresentations, center, ...). It is also pointed out the big differences between the caseswhen parameter q is/is not root of unity.Uq(so3) is a infinitedimensional quotient algebra, it is obtained by q-deformation of thestandard commutation relations

[I1;I2]=I3; [I2;I3]=I1; [I3;I1]=I2 (1.1)

of the Lie algebra so3. So, Uq(so3) is defined as the complex associative algebra withunit element generated by the elements I1, I2, I3 satisfying the defining relations

[I1;I2]q:=q1=2I1I2�q�1=2I2I1=I3; (1.2)

[I2;I3]q:=q1=2I2I3�q�1=2I3I2=I1; (1.3)

[I3;I1]q:=q1=2I3I1�q�1=2I1I3=I2: (1.4)

Uq(so3) can also be given (more precisely is isomorphic to the algebra generated) bythese three relations:

J1=J3J2�qJ2J3;J2=J3J1�q�1J1J3; (1.5)J3=J2J1�qJ1J2:

Together with results of [2] the complete list of all irreducible finite matrix*-representations for all possible q is reached. However, we do not restrict to*-representations, the case q 6= p

p1 is solved completely. We also provide the proof of

Poincare-Birkhoff-Witt theorem for this algebra & partial discussion about PBWfeature of quadratic algebras generated by 3 relations without quadratic terms whichleads into this type of deformation.

Quantum deformations were first discovered by theoretical physicists to occur assymmetries of integrable 1+1-dimensional systems, particularly through the quantuminverse scattering mechanism. The strong relations of quantum deformations to theYang-Baxter equation and transfer matrices in statistical mechanical models was alsocrucial in their development (see for example R. J. Baxter: Exactly solved modelsin statistical mechanics, 1st ed., Academic Press, London 1982). Between thesedeformations some of them are especially important, namely Drinfeld and Jimbo's1-parameter deformations of universal enveloping algebras of semi-simple Lie algebraswhich are Hopf algebras; Faddeev, Reshetikhin and Takhajan developed the theory ofmatrix quantum groups; Woronowicz independently initiated the study of quantumgroups from a C�-algebra point of view and so on.However, quantum deformations are not always seen from the right point of view (seefor example [10], [11]). It should be realized that using these deformations is not onlyintroducing some new free parameters which can be derived from experiments and thenproperly adjusted -- the main reason are the new kinds representations using them onecan derive usually generally new results which are not ``continuously'' related to theclassical ones.More recently it has been realized that quantum deformations occur as symmetries of alarge number of systems in mathematical physics (see for example A. Tsuchiya, E.

2

Eguchi, and M. Jimbo, Infinite analysis: Proccedings the R. I. M. S. research project,June-August 1991, Advanced series in mathematical physics, vol. 16, Kyoto, R. I.M. S., World Scientific, Singapore, 1992, also published in Int. J. Mod. Physics A71992, proceedings supplement 1 A, B <or> T. I. Curtright, D. B. Fairlie, and C. K.Zachos (eds.), Proceedings of the ArgonneWorkshop: Quantum groups, ANL,Illinois, World Scientific, Singapore, 1991). Notably quantum enveloping deformedalgebras occur as symmetries in quantum spin chains and solvable lattice models (seefor example M. Jimbo and T. Miwa, Algebraic analysis of solvable lattice models,preprint RIMS-981, R. I. M. S. Kyoto May 1994, appeared also in a book in the CBMSRegional Conference Series in Mathematics), in two dimensional conformal field theory(see for instance J. Fuchs, A�ne Lie algebras and quantum groups, CambridgeUniversity Press, Cambridge 1992), and massive integrable systems (see F. A. Smirnov,Form factors in completely integrable models of quantum �eld theory, World

Scienti�c, Singapore, 1992). Quantum deformations also play an important role in thetheory of link invariants and knots (see C. N. Yang and M. L. Ge (eds.), Braid groups,knot theory and statistical mechanics, Advanced series in math physics, vol. 9, WorldScientific, Singapore 1989).One can say that by now the use of quantum deformations in theoretical andmathematical physics has become so wide-spread that almost every day new preprintsappear describing their structure and applications.

Let us briefly compare results of this paper with results reached by other authors. Thefirst mention about representations of Uq(so3) appears in [4], one concrete irreduciblematrix representation for each dimension is constructed according to analogy withnon-deformed case. One concrete kind of irreducible representation appears in [6], in[5] there is constructed so called ``Fock'' representation (it is representation of moregeneral case quadratic algebra and one represetation constructed in this paper is aspecial case of representation constructed there). Classification of *-representations iscompletely done in [2] (however *-representation is a special case of representationsconsidered here), some *-representations are obtained another way in [15]. Irreduciblerepresentations which can be obtained using algebra homomorphism into Uq(sl2) (seechapter 9) are described in [7].The results in chapters 2-7 are quite original, sum up and finish partial results ofprevious papers and are ready for publication, the results in chapters 8-16 have beenreached in close cooperation with M. Havlí£ek and A. U. Klimyk and submitted intoJournal Physics A.

Let us do the brief look on the contents of each chapter. In chapter 1 we start withbasic preliminary, set up notation and give basic definitions which are valid to the restof the work.The chapter 2 gives a standard approach to the Poincare-Birkhoff-Witt theorem foruniversal envelopping algebras of simple Lie algebras.This approach is needed in chapter 4 where it is applied on quantum generalization ofLie algebra - quotient algebra of tensor algebra generated by two-sided ideal whichconsists of the relations used to define Uq(so3). It is pointed out that from thisapproach is clear that this and only this type of deformation in the wide class of evenmulti-parameter deformation leads to the algebra which is PBW-type i. e. which hasthe basis generated by all well ordered monomials. This fact can be used to give reasonswhy this type of deformation of the algebra so3 can be considered as more importantthan the others because the PBW feature is also important in physical aplications.

3

In the chapter 3 there is a detailed look on the definition of Uq(so3).The chapter 5 explains the procedure how to reach all irreducible matrixrepresentations of this algebra for the case when q not root of unity. It is also clarifiedwhy this proccess is not successful for the case when q is root of unity. The rest of thechapter contains important theorems about using the derived representations for thecase when q is root of unity. Everything in the chapter is done constructive way (i. e.reducibility of representations is shown as decomposing the reducible representationinto irreducible ones or finding concrete invariant subspace).The chapters 6 and 7 connect the work to the related topics - Hopf and quadraticalgebras. However it is not clear yet if this algebra has or has not Hopf structure. Hopfalgebra structure is not known on Uq(so3), however, it can be embedded into the Hopfalgebra Uq(sl3) as a Hopf coideal9. This embedding is very important for the possibleapplication in spectroscopy.The chapter 8 gives alternative way how to obtain all representations of the Uq(so3)when q is root of unity. It contanis some information about the center of the algebra,about the inner automorphisms and also gives list for some low dimensions but thecomplete list for any n is not reached yet.In the chapter 9 an algebra homomorphism from the algebra Uq(so3) to the extension

Uq(sl2) of the Hopf algebra Uq(sl2) is constructed. Not all irreducible representations of

Uq(sl2) can be extended to representations of Uq(sl2). Composing the homomorphism

with irreducible representations of Uq(sl2) we obtain representations of Uq(so3). Not allof these representations of Uq(so3) are irreducible. Reducible representations of Uq(so3)are decomposed in the next chapters into irreducible components. In this way weobtain all irreducible representations of Uq(so3) when q is not a root of unity. A part ofthese representations turn into irreducible representations of the Lie algebra so3 whenq!1.One of exciting aspects of this quantum deformation is the presence of new types ofrepresentations that have no classical analogues. It seems that for the case when q isroot of unity the number of representations with no classical analogue is increasingrapidly.Using the homomorphism it is shown in the next chapter how to construct tensorproducts of finite dimensional representations of Uq(so3). Irreducible representations ofUq(so3) when q is a root of unity are constructed. Part of them are obtained from

irreducible representations of Uq(sl2) by means of the homomorphism .

2. Basic preliminaryWe start by recalling basic definitions of some elementary algebraic structures usedin the following parts: associative algebras, free algebras, lie algebras, universalenveloping algebras to set up my notation.Let N,Z,R and C denote the sets of all integers, whole numbers, real numbers andcomplex numbers, resp. Let bn denote the set f1;2;:::;ng.2.1. De�nition. Ring. A Ring (R;+;�) is a non-empty set R with 2 operations: anaddition + and a multiplication operation �, such that1. (R;+) is a commutative group,2. the multiplication is associative,3. the multiplication and addition are distributive, i. e.

8a;b;c2R:(a+b)�c=a�c+b�c^a�(b+c)=a�b+a�c:4

The identity element of (R;+) is usually denoted as 0. If a ring R has a unity element 1(such that 1�x=x=x�1) then R is called a ring with unity (R;+;�;1). A ring is said to becommutative if ab=ba holds for each a;b2R.2.2. De�nition. Module. Let R be a ring. A left-module (M;R;+;�) over the ring R (orR-module) is an abelian group (M;+) and an operation �:R�M!M) such that

(a+b)�u=a�u+b�u^a�(u+v)=a�u+a�v 8a;b2R;u;v2M:

2.3. De�nition. Algebra. Let R be a commutative ring. An algebra (A;R;m) overR (orR-algebra) is an R-module A with a bilinear product map A�A!A. An algebra is saidto be associative if

m(x;m(y;z))=m(m(x;y);z) 8x;y;z2A:If A has an element 1 suchthatm(x;1)=x=m(1;x) for all x, then the algebra is calledan R-algebra with unity.

2.4. De�nition. Free algebra. Let n2N, X=x1;:::;xn be set of n distinct letters.Consider the set of finite (noncommutative) monomials (or, more precisely, finitesequences) in the elements of X[1 (strings of letters in the alphabet X). The setof (finite) linear combinations of these monomials with coefficients in a ring R(noncommutative polynomials) R[[x1;:::;xn]] can be given the structure of anassociative algebra (R[[X]];R;�) by defining the multiplication of any pair of monomialsxi1 ;:::;xir and xj1 ;:::;xjs in an obvious way to be

(xi1 :::xir)�(xj1 :::xjs)=xi1 :::xir �xj1 :::xjs;xi1 :::xir �1=xi1:::xir ;1�xi1 :::xir=xi1 :::xir :

This R-algebra is called free R-algebra on the generators in the setX.

2.5. De�nition. Ideal. Let A be an algebra. A subset B�A is called a subalgebra of Aif B also forms an algebra. A subalgebra I of A is called a left (resp. right) ideal, if8x2I;A�x�I (resp. x�A�I). A subalgebra which is both a left and a right ideal, iscalled a two sided ideal. If I 6=A then I is called proper ideal of A. If I 6=fg then I isnontrivial ideal of A.

2.6. Lemmma. Quotient algebra. Let A be an R-algebra with a two-sided ideal I. Thequotient set A=I which contains of all classes denoted by equivalence x�y,x�y2I isan algebra called the quotient algebra of A by I.

2.7. De�nition. Centre. Let A be an R-algebra. The centre Z(A) of A is defined to bethe set fx2Ajx�y=y�x 8y2Ag.2.8. De�nition. Tensor algebra. Let V be a vector space on field0 C . For each integer

n let us define Tn(V )=nNi=1

V . Define T 0(V )=C . Because the tensor product is

associative, there is a bilinear map T r(V )�T s(V )!T r+s(V ) using this we can obtainthe space

T (V )=+1Ln=0

Tn(V )=C�(V )�(V V )�::::

This space forms structure of C -algebra called tensor algebra T (V ) of V . The productin T (V ) is again denoted by .and and

2.9. De�nition. Lie algebra. A Lie algebra (g;[:;:]) is a C -algera with a C -bilinearproduct map g�g!g satisfying antisymmetry and Jacobi identity

5

[x;x]=0 8x2g;[x;[y;z]]+[y;[z;x]]+[z;[x;y]]=0 8x;y;z2g:

2.10. De�nition. Universal enveloping algebra. Let (g;[:;:]) be a Lie algebra. Let T (g)be the tensor algebra of the C -vector space g. Let us consider then the following ideal inT (g):

I=linfa(xy�yx�[x;y])bjx;y2g;a;b2T (g)g:Let's consider the quotient set U(g)=T (g)=I, denote classes of equivalence of x2T (g)by i(x). Then U(g) with multiplication

i(x)i(y)= i(xy)8x;y2T (g):is called the universal enveloping algebra of g.2.11. Lemmma. U(g) is an associative infinite dimensional C -algebra with unity.Proof. Let x;x0;y;y02U(g), let i(x)= i(x0), i(y)= i(y0). Then9k1;k22I:x=x0+k1^y=y0+k2. According to 2.10 we have

i(x0)�i(y0)= i(x0 �y0)= i((x+k1)�(y+k2))= i(x�y+x�k2+k1�y+k1�k2)== i(x�y+something2K)= i(x�y):

Multiplication in U(g) is associative because multiplication in T (g) is associative.

3. PBW theorem for universal enveloping algebras3.1. Lemmma. Let (A;�) be an associative C -algebra. Defining the commutator[:;:]:A�A!A

[x;y]=x�y�y�x;we obtain Lie algebra L(A)=(A;[:;:]).This result can be reversed in some sense:3.2. Theorem. PBW theorem about Lie algebras. For each Lie algebra g there existsan associative algebra A (over the same field) such that g is issomorphic to somesub-algebra of L(A).We need following definitons & lemma's for the proof.3.3. Lemmma. Defining a Lie bracket in natural way

[i(x);i(y)]= i(x)i(y)�i(y)i(x);U(g) can be considered as a Lie algebra.3.4. De�nition. Let g be a finite dimensional Lie algebra with basis (e1;:::;en).Any x2T (g) we call monomial, 9k2N 9i1;:::;ik2bn:x=ei1ei2:::eik .A monomial is ordered, when i15 :::5 ik holds.

3.5. De�nition. Let (i1;:::;ik)2Nk . Every pair (a;b)2bk2 where ia>ib is called inversein permutation (i1;:::;ik).3.6. Lemmma. Each class in U(g) contains at least 1 element which is linearcombination of ordered monomials.Proof. It is sufficient to show that each monomial ei1:::eik is able to be written asordered monomial + something from ideal.We will prove lemma by induction according to monomial length and (then) accordingto number of inverses in permutation (i1;:::;ik):Lemma is certainly valid for ordered monomials. Now let's take any monomial

ei1:::eijeij+1:::eik;where ij>ij+1. Then

6

ei1:::eijeij+1:::eik == ei1:::(eijeij+1�[eij ;eij+1]�eij+1eij):::eik++ei1:::[eij ;eij+1]:::eik+ei1:::eij+1eij:::eik ;

the first term is from ideal, we can use induction hypothesis acting on the second (itis shorter) and the third (it contains certainly less inverses). Lemma is fulfilled byinduction.3.7. Lemmma. Any set of classes in U(g) generated by different ordered monomials islinear independent.Proof. We will prove it by contradiction. Let

�1i(ei1;1:::ei1;k1 )+�2i(ei2;1:::ei2;k2 )+:::+�mi(eim;1:::eim;km)= i(0-):

We want to show

�1= :::=�m=0:

Let B denote span of all ordered monomials. Clearly it is a subspace of T (g).We will construct linear mapping �:T (g)!B for which following is true:

i) 8m2N 8i15 :::5 im2bn:�(ei1:::eim)=ei1:::eimii) 8m2N 8i1;:::;im2bn;ij= ij+1:�(ei1:::eijeij+1:::eim)==�(ei1:::eij+1eij:::eim)+�(ei1:::[eij ;eij+1]:::eim):

If � exists, it is obviously constant on all classes, it means

8x2 i(y)2T (g)=I :�(x)=�(y):Consequently it induces mapping ~�:T (g)=I!B defined by

8x2T (g):~�(i(x))=�(x):~� is linear, in addition

8m2N 8i15 :::5 im2bn:~�(i(ei1:::eim))=�(ei1:::eim)=ei1:::eim;applying ~� on original linear combination yields

0-=~�(i(0-))= ~�(�1i(ei1;1:::ei1;k1 )+�2i(ei2;1:::ei2;k2)++:::+�mi(eim;1:::eim;km

))=�1ei1;1:::ei1;k1+�2ei2;1:::ei2;k2++:::+�meim;1:::eim;km

which is contradiction (any set of different ordered monomials is linear independent).NowWe will show that mapping � fulfilling i) and ii) can be constructed.� is well defined on ordered monomials by i). Let's take any monomial

ei1:::eijeij+1:::eik;where ij>ij+1.Assume � is well defined on a subspace of T (g) consisting of span of all monomialsshorter than this one and of all monomials which have equal length but less inverses.Now define

�(ei1:::eijeij+1:::eik)==�(ei1:::eij+1eij:::eik)+�(ei1:::[eij ;eij+1]:::eik)

(we know how to apply � on last two monomials: the first has less inverses and thesecond is shorter).We have to make sure that the definition is correct. Original monomial could containanother inverse il>il+1. We must consider 2 cases:1) j+1<l (or l+1<j) (inverses are farther),2) j+1= l (or l+1=j) (one element is common to both inverses).(Let's consider first alternatives in both cases, proof of the other is similar.) In bothcases 1), 2) we need to show

7

�(ei1:::eij+1eij:::eik)+�(ei1:::[eij;eij+1]:::eik)==�(ei1:::eil+1eil:::eik)+�(ei1:::[eil;eil+1]:::eik):

Case 1). We know

�(ei1:::eij+1eij:::eileil+1:::eik)++�(ei1:::[eij ;eij+1]:::eileil+1:::eik)==�(ei1:::eij+1eij:::eil+1eil:::eik)++�(ei1:::eij+1eij:::[eil;eil+1]:::eik)++�(ei1:::[eij ;eij+1]:::eil+1eil:::eik)++�(ei1:::[eij ;eij+1]:::[eil;eil+1]:::eik);

and likewise for the second expression.Case 2). We need to show

�(ei1:::eijeil+1eil:::eik)+�(ei1:::eij[eil;eil+1]:::eik)==�(ei1:::eileijeil+1:::eik)+�(ei1:::[eij ;eil]eil+1:::eik):

Left side can be written as

�(ei1:::eijeil+1eil:::eik)+�(ei1:::eij[eil;eil+1]:::eik)==�(ei1:::eil+1eijeil:::eik)+�(ei1:::[eij ;eil+1]eil:::eik)++�(ei1:::eij[eil ;eil+1]:::eik)=�(ei1:::eil+1eileij:::eik)++�(ei1:::eil+1[eij ;eil]:::eik )+�(ei1:::[eij ;eil+1]eil:::eik)++�(ei1:::eij[eil;eil+1]:::eik );

right side as

�(ei1:::eileijeil+1:::eik)+�(ei1:::[eij;eil]eil+1:::eik)==�(ei1:::eileil+1eij:::eik)+�(ei1:::eil[eij ;eil+1]:::eik)++�(ei1:::[eij ;eil]eil+1:::eik)=�(ei1:::eil+1eileij:::eik)++�(ei1:::[eil ;eil+1]eij:::eik )+�(ei1:::eil[eij ;eil+1]:::eik )++�(ei1:::[eij ;eil]eil+1:::eik):

The first terms are identical. In next 3 terms brackets can be combined from baseelements:

[eij ;eil]=

nXq=1

�qeq;

[eij ;eil+1]=nX

q=1

�qeq;

[eil;eil+1]=nX

q=1

qeq:

If we put these sums into left and right sides, for each q2bn we can from inductionhypothesis reduce inverses on both sides. For example, if q= il+1 then we have on theleft side

�(:::eil+1eq:::)=�(:::eqeil+1:::)+�(:::[eil+1 ;eq]:::);the first term is equal with that one on the right side, the second term remains.If we now put all terms on the left side, we have

�(:::[eil+1 ;[eij;eil]]:::)+�(:::[[eij ;eil+1];eil]:::)+�(:::[eij ;[eil;eil+1]]:::);but it is zero because of Jacobi identity.3.8. Lemmma. U(g) and the subset B of T (g) consisting of all ordered monomials areisomorphic.

8

Proof. It can be easily seen that ~� defined in previous proof is bijection: It is onto --

for each linear combination of ordered monomials x=kX

j=1

�jxj there exists a class

~x=

kXj=1

�ji(xj ) such that ~�(~x)=x. It is injection -- for any class i(x) we know that x can

be considered as linear combination of ordered monomials x=kX

j=1

�jxj but i(x)=0-

implies �j=0 for each j.

Now we can finish the proof of PBW theorem.

Proof. If we take a subset B of T (g) consisting of all ordered monomials, we can easydefine associative multiplication on B as in previous lemma. Now let's take all elementsof B corresponding to base elements of g � there exists a natural isomorphism betweenthis subspace and g.

4. Fairlie's quantumdeformation of so(3)

It is well-known that the Lie algebras sl2 and so3 of the Lie groups SL(2;C ) and SO(3),respectively, are isomorphic. But these algebras differ from each other if we considertheir embedding to the wider Lie algebra sl3. There is no automorphism of sl3 whichtransfers the embedding sl2� sl3 to the embedding so3�sl3. Note that the embeddingso3�sl3 is of great importance for nuclear physics: it is used in spectroscopy.

The definition of the q-analogue of the universal enveloping algebra U(sl2) iswell-known. It is the quantum algebra Uq(sl2) which is a Hopf algebra. If we wish tohave a q-analogue of the universal enveloping algebra so3 such that at q!1 we obtainthe classical embedding so3�sl3, then the algebra sl2 is not appropriate for this role.By other words, an algebra Uq(so3) must differ from Uq(sl2). This algebra Uq(so3) iswell. It is the associative algebra generated by three elements I1, I2 and I3 satisfyingthe relations

q1=2I1I2�q�1=2I2I1=I3; (4.1)q1=2I2I3�q�1=2I3I2=I1; (4.2)q1=2I3I1�q�1=2I1I3=I2: (4.3)

Such (and more general) deformation of the commutator [Ii;Ij]=IiIj�IjIi was definedat 1967 by R. Santilli (see also Refs. [13]) under studying a generalization of the Lietheory. Afterwards (in 1990), the algebra Uq(so3) with commutation relations(4.1)--(4.3) was determined by D. Fairlie (see [4]). An algebra which can be reduced toUq(so3) was defined in 1986 by M. Odesski (see [14]).The algebra Uq(so3) can be also defined using slightly different relations (1.5):

4.1. De�nition. Let's take three dimensional vector space g=linfI1;I2;I3g. Let q 6=0.Then take following subspace of tensor algebra T (g):

I=lin(fx(I1�I3I2+qI2I3)yjx;y2T (g)g[[fx(I2�I3I1+q�1I1I3)yjx;y 2T (g)g[[fx(I3�I2I1+qI1I2)yjx;y2T (g)g):

I is clearly ideal in T (g). A quotient algebra T (g)=I is what we call quantumdeformation of so(3) (the algebra Uq(so3)).

9

Fairlie gave finite dimensional irreducible representations of the algebra Uq(so3) whichat q!1 give the well-known finite dimensional irreducible representations of the Liealgebra so3. These representations are given by integral or half-integral non-negativenumbers. Odesski also gave some classes of irreducible representations.It was shown (see [2]) that the algebra Uq(so3) has irreducible finite dimensionalrepresentations which have no classical analogue (that is, which do not admit the limitq!1). It was not clear why such strange representations of the algebra Uq(so3) appear.

Construction of a homomorphism from Uq(so3) to the algebra Uq(sl2) gives clear answerto this question.We construct a homomorphism from Uq(so3) to the algebra Uq(sl2) which isan extension of the well-known quantum algebra Uq(sl2) (note that there is nohomomorphism from Uq(so3) to Uq(sl2)). Irreducible finite dimensional representationsof Uq(sl2) (but not all) can be extended to finite dimensional representations of

the algebra Uq(sl2). Composing a homomorphism Uq(so3)! Uq(sl2) with these

representations of Uq(sl2), we obtain representations of the algebra Uq(so3). But some

of irreducible representations of Uq(sl2) lead to reducible representations of thealgebra Uq(so3). Decomposing these reducible representations of Uq(so3) we obtainirreducible representations of this algebra which have no analogue for the Lie algebraso3. If q is not a root of unity, then in this way we obtain all finite dimensionalirreducible representations of Uq(so3). But there are infinite dimensional irreduciblerepresentations of Uq(so3) which cannot be obtained in this way.

Existence of the homomorphismUq(so3)! Uq(sl2) allows us to define tensor productsof representations of the algebra Uq(so3) which is not a Hopf algebra.

Using the homomorphismUq(so3)! Uq(sl2) and irreducible representations of

Uq(sl2) we obtain representations of Uq(so3) when q is a root of unity. Takingirreducible representations of Uq(so3) obtained in this way and decomposing reduciblerepresentations, we obtain several series of irreducible representations of Uq(so3). Inaddition, we construct irreducible representations of Uq(so3) which cannot be derived

from Uq(sl2).When q is not a root of unity, then each irreducible (finite or infinite dimensional)representation of Uq(so3) is equivalent to one of the representations constructed in thenext chapters. For the irreducible representations of Uq(so3) when q is a root of unitywe have no proof of similar assertion. The reason of this is that in this case there aremany classes of irreducible representations and a proof of completeness of irreduciblerepresentations becomes very tedious.Let us remark that in Ref. [14] there were constructed irreducible finite dimensionalrepresentations of Uq(so3) when q is not a root of unity and a part of irreducible infinitedimensional representations. In Refs. [2] and [15], there were constructed irreduciblerepresentations of Uq(so3) which satisfy the conditions of �-representations (that is,such that T (I�j )=�T (Ij ), j=1;2). These �-representations are a part of irreduciblerepresentations of Uq(so3) constructed in this paper. Also remark irreduciblerepresentations of Uq(so3) for q a root of unity in Ref. [7], where a part of irreduciblerepresentations for this case were constructed.Note that in all Refs. above there are no relations of representations of Uq(so3) to

representations of Uq(sl2). This relation makes representations of Uq(so3) clear andunderstandable.

10

Now we can ask the first important question: if PBW holds for this type of algebra, itmeans if any set of classes in T (g)=I generated by different ordered monomials is linearindependent. The answer is yes, we can say even more:

Let �1;�2;�3;�1;�2;�3; 1; 2; 3 are non-zero complex numbers. Let us considerfollowing set:

I1=lin(fx(�1I1��1I3I2+ 1I2I3)yjx;y2T (g)g[[fx(�2I2��2I3I1+ 2I1I3)yjx;y2T (g)g[[fx(�3I3��3I2I1+ 3I1I2)yjx;y2T (g)g):

Without loss of generality we can assume �1=�2=�3=1 (otherwise we could divideall three expressions by �1, �2, �3 resp. and the span wouldn't change).Another simplification can be achieved considering new basis (J1;J2;J3):

J1=1p�2�3

I1;

J2=1p�1�3

I2;

J3=1p�1�2

I3:

Then the set I1 is given by formula

I1=lin(fx(J1�J3J2+�23J2J3)yjx;y2T (g)g[[fx(J2�J3J1+�13J1J3)yjx;y 2T (g)g[[fx(J3�J2J1+�12J1J2)yjx;y2T (g)g);

where �13= 2;�12= 3;�23=�1.4.2. Lemmma. Every class T (g)=I1 contains at least one element which is linearcombination of ordered monomials.Proof. Very simliar to the proof of 3.6It is sufficient to show that each monomial Ji1:::Jik is able to be written as orderedmonomial + something from ideal.We will prove lemma by induction according to monomial length and (then) accordingto number of inverses in permutation (i1;:::;ik):Lemma is certainly valid for ordered monomials. Now let's take any monomial

Ji1:::JijJij+1:::Jik;where ij>ij+1. Then

Ji1:::JijJij+1:::Jik ==Ji1:::(JijJij+1�[Jij;Jij+1]��ij+1ijJij+1Jij):::Jik++Ji1:::[Jij ;Jij+1]:::Jik+�ij+1ijJi1:::Jij+1Jij:::Jik;

the first term is from ideal, we can use induction hypothesis acting on the second (itis shorter) and the third (it contains certainly less inverses). Lemma is fulfilled byinduction.4.3. Theorem. Any set of classes in T (g)=I1 generated by different ordered monomials

is linear independent,9q2C ;q 6=0 such that q= 1= 3=1

2.

Proof. ): Because the tensor product in T (g)=I1 is associative, following must hold:

i(J3)�(i(J2)�i(J1))=(i(J3)�i(J2))�i(J1):Reordering the left side we have

11

i(J3)�(i(J2)�i(J1))= i(J3)�i(�12J1J2+J3)==�12i((J3J1)J2)+i(J3J3)=�12(i(�13J1J3+J2)�i(J2))+i(J3J3)==�12�13i(J1(J3J2))+�12i(J2J2)+i(J3J3)==�12�13(i(J1)�i(�23J2J3+J1))+�12i(J2J2)+i(J3J3)==�12�13�23i(J1J2J3)+�12�13i(J1J1)+�12i(J2J2)+i(J3J3):

On the right side we have

(i(J3)�i(J2))�i(J1)= i(�23J2J3+J1)�i(J1)==�23i(J2(J3J1))+i(J1J1)=�23(i(J2)�i(�13J1J3+J2))+i(J1J1)==�23�13i((J2J1)J3)+�23i(J2J2)+i(J1J1)==�23�13(i(�12J1J2+J3)�i(J3))+�23i(J2J2)+i(J1J1)==�23�13�12i(J1J2J3)+�23�13i(J3J3)+�23i(J2J2)+i(J1J1):

According to assumption the classes generated by different ordered monomials arelinear independent so that the corresponding coefficients on both sides must be equal.

�12�13�23 = �23�13�12�12�13 = 1

�12 = �231 = �23�13:

This system of equations (all coef. are non-zero) is fulfilled if and only if there is aq2C;q 6=0 such that

�23 = q�13 = q�1

�12 = q:

(: Similar to the proof of the 3.7 We will prove it by contradiction. Let

�1i(Ji1;1:::Ji1;k1)+�2i(Ji2;1:::Ji2;k2)+:::+�mi(Jim;1:::Jim;km)= i(0-):

We want to show �1= :::=�m=0:Let B denote span of all ordered monomials. Clearly it is a subspace of T (g).We will construct linear mapping �:T (g)!B for which following is true:

i) 8m2N 8i15 :::5 im2bn:�(Ji1:::Jim)=Ji1:::Jimii) 8m2N 8i1;:::;im2bn;ij= ij+1:�(Ji1:::JijJij+1:::Jim)==�ij+1ij�(Ji1:::Jij+1Jij:::Jim)+�(Ji1:::[Jij ;Jij+1]:::Jim):

If � exists, it is obviously constant on all classes, it means

8x2 i(y)2T (g)=I1 :�(x)=�(y):Consequently it induces mapping ~�:T (g)=I1!B defined by

8x2T (g):~�(i(x))=�(x):~� is linear, in addition

8m2N 8i15 :::5 im2bn:~�(i(Ji1:::Jim))=�(Ji1:::Jim)=Ji1:::Jim;applying ~� on original linear combination yields

0-=~�(i(0-))= ~�(�1i(Ji1;1:::Ji1;k1)+�2i(Ji2;1:::Ji2;k2)+:::++�mi(Jim;1:::Jim;km

))=�1Ji1;1:::Ji1;k1+�2Ji2;1:::Ji2;k2+:::++�mJim;1:::Jim;km

which is contradiction (any set of different ordered monomials is linear independent).Now we will show that mapping � fulfilling i) and ii) can be constructed.� is well defined on ordered monomials by i). Let's take any monomial

Ji1:::JijJij+1:::Jik;12

where ij>ij+1:Assume � is well defined on a subspace of T (g) consisting of span of all monomialsshorter than this one and of all monomials which have equal length but less inverses.Now define

�(Ji1:::JijJij+1:::Jik)=�ij+1ij�(Ji1:::Jij+1Jij:::Jik)++�(Ji1:::[Jij ;Jij+1]:::Jik);

(we know how to apply � on last two monomials: the first has less inverses and thesecond is shorter).We have to make sure that the definition is correct. Original monomial could containanother inverse il>il+1. We must consider 2 cases:1) j+1<l (or l+1<j) (inverses are farther),2) j+1= l (or l+1=j) (one element is common to both inverses).(Let's consider first alternatives in both cases, proof of the other is similar.) In bothcases 1), 2) we need to show

�ij+1ij�(Ji1:::Jij+1Jij:::Jik)+�(Ji1:::[Jij;Jij+1]:::Jik)==�il+1il�(Ji1:::Jil+1Jil:::Jik)+�(Ji1:::[Jil;Jil+1]:::Jik):

Case 1). We know

�ij+1ij�(Ji1:::Jij+1Jij:::JilJil+1:::Jik)++�(Ji1:::[Jij;Jij+1]:::JilJil+1:::Jik)==�ij+1ij�il+1il�(Ji1:::Jij+1Jij:::Jil+1Jil:::Jik)++�ij+1ij�(Ji1:::Jij+1Jij:::[Jil;Jil+1]:::Jik)++�il+1il�(Ji1:::[Jij;Jij+1]:::Jil+1Jil:::Jik)++�(Ji1:::[Jij ;Jij+1]:::[Jil;Jil+1]:::Jik);

and likewise for the second expression:

�il+1il�(Ji1:::JijJij+1:::Jil+1Jil:::Jik)++�(Ji1:::JijJij+1:::[Jil;Jil+1]:::Jik)==�il+1il�ij+1ij�(Ji1:::Jij+1Jij:::Jil+1Jil:::Jik)++�il+1il�(Ji1:::[Jij;Jij+1]:::Jil+1Jil:::Jik)++�ij+1ij�(Ji1:::Jij+1Jij:::[Jil;Jil+1]:::Jik)++�(Ji1:::[Jij ;Jij+1]:::[Jil;Jil+1]:::Jik):

Case 2). Corollary ij=3;ij+1= il=2;il+1=1. We need to show

�12�(Ji1:::J3J1J2:::Jik)+�(Ji1:::J3[J2;J1]:::Jik)==�23�(Ji1:::J2J3J1:::Jik)+�(Ji1:::[J3;J2]J1:::Jik):

At this moment we're applying assumption �12=q;�13=q�1;�23=q on the left side...

�12�(Ji1:::J3J1J2:::Jik)+�(Ji1:::J3[J2;J1]:::Jik)==�12�13�(Ji1:::J1J3J2:::Jik)+�12�(Ji1:::[J3;J1]J2:::Jik)++�(Ji1:::J3[J2;J1]:::Jik)=�12�13�23�(Ji1:::J1J2J3:::Jik)++�12�13�(Ji1:::J1[J3;J2]:::Jik)+�12�(Ji1:::[J3;J1]J2:::Jik)++�(Ji1:::J3[J2;J1]:::Jik)=q�(Ji1:::J1J2J3:::Jik)++�(Ji1:::J1J1:::Jik)+q�(Ji1:::J2J2:::Jik)++�(Ji1:::J3J3:::Jik);

and similarly on the right side...

�23�(Ji1:::J2J3J1:::Jik)+�(Ji1:::[J3;J2]J1:::Jik)==�23�13�(Ji1:::J2J1J3:::Jik)+�23�(Ji1:::J2[J3;J1]:::Jik)++�(Ji1:::[J3;J2]J1:::Jik)=�23�13�12�(Ji1:::J1J2J3:::Jik)++�23�13�(Ji1:::[J2;J1]J3:::Jik)+�23�(Ji1:::J2[J3;J1]:::Jik)+

13

+�(Ji1:::[J3;J2]J1:::Jik)=q�(Ji1:::J1J2J3:::Jik)++�(Ji1:::J3J3:::Jik)+q�(Ji1:::J2J2:::Jik)++�(Ji1:::J1J1:::Jik):

Both sides are identical, theorem holds.

4.4. Theorem. If 9q2C;q 6=0 such that �23=q^�13=q�1^�12=q then U(g) and thesubspace B containing all ordered monomials from T (g) are isomorphic.

Proof. See 3.8

The finishing of the proof of PBW theorem for deformation case is identical tonon-deformed one.

5. Matrix representations not at root of unity

5.1. De�nition. Representation. Let V be a C -vector space and let A beC -algebra. Denote set of all endomorphisms on V by End(V ). If ':A!End(V ) is ahomomorphism, that means

'(x�y)='(x)'(y)8x;y2A;then it is called a (linear) representation of A on V . The vector space V is also calledA-module.

5.2. De�nition. Submodule. Let V be an A-module. If V contains a proper linearsubspaceW such thatW is closed under the action of A, that is '(A)W �W , thenW issaid to carry a subrepresentation of A andW is a A-submodule of V .

5.3. De�nition. Irreducibility. If an A-module has no proper, non-trivial submodules,it is called irreducible, otherwise it is called reducible.

The main result of this section is covered in the following theorem:

5.4. Theorem. Let q2C , qk 6=1 8k2N. Let us consider following matrices of dimensionn:

J3=

0BBBB@�0

. . .�k

. . .�n�1

1CCCCA; J1=0BBBBB@

0 �0

0. . . �k

k. . .

. . . �n�2 n�2 K

1CCCCCA;

J2=J3J1�q�1J1J3;

where �k, �k, k andK are in this table:

14

�k �k k K

1)��n�1

2 +k�q

�q2(q2(k+1�n)�1)(q2k+2�1)(q2k�n+3+1)(q2�1)2

1q2k+1�n+1 0

2) +i 1+q2n�1�2k

qn�32�k(q2�1)

q1�2n(q2(2n�1�k)�1)(q2(k+1)�1)(q2n�3�2k�1)(q2�1)2

q2n�1�2k

q2n�1�2k�1 +iq32�n(q2n�1)(q�1)(q2�1)

3) +i 1+q2n�1�2k

qn�32�k(q2�1)

q1�2n(q2(2n�1�k)�1)(q2(k+1)�1)(q2n�3�2k�1)(q2�1)2

q2n�1�2k

q2n�1�2k�1 �iq32�n(q2n�1)(q�1)(q2�1)

4) �i 1+q2n�1�2k

qn�32�k(q2�1)

q1�2n(q2(2n�1�k)�1)(q2(k+1)�1)(q2n�3�2k�1)(q2�1)2

q2n�1�2k

q2n�1�2k�1 +iq32�n(q2n�1)(q�1)(q2�1)

5) �i 1+q2n�1�2k

qn�32�k(q2�1)

q1�2n(q2(2n�1�k)�1)(q2(k+1)�1)(q2n�3�2k�1)(q2�1)2

q2n�1�2k

q2n�1�2k�1 �iq32�n(q2n�1)(q�1)(q2�1)

Then following theorem holds: Every finite dimensional irreducible matrixrepresentation is equivalent to one of listed above.

To proof this theorem we need following (helping) theorems & lemmas.

5.5. De�nition.. Let q2C�f0;�1;1g, �2C . We put [�]q=q��q��q�q�1 .

5.6. Lemmma. Let n2N, q2C �f0;�1;1g, V be finite dimensional complex vectorspace, let the operators J1;J2;J3:V !V fulfil commutation relations

J2J1�q J1J2 = J3;J3J1�q�1J1J3 = J2; (5.1)J3J2�q J2J3 = J1;

let x2V and �2C such that J3x=[�]qx.

Then

J3(J1�q��J2)x=[��1]q(J1�q��J2)x:Proof.

J3J1x�q��J3J2x=(J2+q�1J1[�]q)x�q��(J1+qJ2[�]q)x==(1�q��+1[�]q)J2+(�q��+q�1[�]q)J1x=([��1]q�(�q��))J2+([��1]q)J1x:

5.7. Note. In [9] the commutation relations have slightly different form:

~qXY �1

~qY X = Z

~qY Z�1

~qZY =X (5.2)

~qZX�1

~qXZ = Y:

It can be easily seen that transformation between our relations and those presented in[9] is

J3= iX;J2=~qY;J1= iZ;q=~q�2:

5.8. Lemmma. Let �2C , q2C�f0;1;�1g. Pak 9�2C :[�]q=�.Proof. We need to solve equation

15

q��q��q�q�1 = �

q2��1��q�(q�q�1) = 0

q� =�(q�q�1)�

p�2(q�q�1)2+4

2| {z }6=0

� =

ln

�(q�q�1)�

p�2(q�q�1)2+4

2

!lnq

Any of signs can be choosen. General power q� as function of � maps onto C�f0g forany q2C�f0;1;�1g so that � always exists.5.9. Lemmma. 8�;�;k;l2C :1) [�]q=[�]q , q�=q�_q�=�q��,2) [�]q=[�+k]q , qk=1_q2�+k=�1,3) [�+k]q=[�+l]q , ql�k=1_q2�+k+l=�1.Proof. 1)

[�]q = [�]q,q��q�� = q��q��,q��q� = q��q��,

q�(1�q��) = �q��(1�q�� ),q�=q�_q�=�q��

2) according to 1)

[�]q=[�+k]q,q�=q�+k_q�=q��k,qk=1_q2�+k=�1:3) directly from 2).5.10. Lemmma. 8q2C�f0;1;�1g 8c2C :1) [c]q=q

c�1+1

q[c�1]q,

2) [c]q=q�(c�1)+q[c�1]q,

3) (q[c]q�[c+1]q)(1

q[c]q�[c+1]q)=1,

Proof. By inspection.5.11. Lemmma. Let q2C �f0;1;�1g, qk 6=1 8k2N. Let �2C , k1;k22Z, k1 6=k2,[�+k1]q=[�+k2]q. Then

8k3;k42Z:[�+k3]q=[�+k4]q,k3+k4=k1+k2:

Proof.

q2(�+k3)+k4�k3=�1,q2�+k1+k2+k3+k4�(k1+k2)=�1,qk3+k4�(k1+k2)=1,,k3+k4�(k1+k2)=0:

5.12. Lemmma. Let q2C�f0;1;�1g, qk 6=1 8k2N. Let �2C . Then 9�2C :[�]q=[�]qand the numbers [�+k]q are for all k2N0 mutually different. (Similar lemma is valid

for the numbers [��k]q, k2N0.)Proof. If the condition is not met for �, then there exist k1;k22N0 such that[�+k1]q=[�+k2]q. According to 5.11 it is sufficient to put �=�+k1+k2.

16

5.13. Theorem. Let n2N, q2C�f0g;qk 6=1 8k2N, �2C ;�2C .Let �

q4�=q�2n+2^�=0�_��2=

�[n+1]q[n�1]q�1q

^[�+n]q=[�+n�1]q�:

Then the matrices (we write non-zero elements only)

J3=

0BBBBB@[�]q

. . .[�+k]q

. . .[�+n�1]q

1CCCCCA;

J1=

0BBBBBBBBBBBBBBB@

0�q� [2�]q[1]qq�+1+q�(�+1)

1q2�+1 0

. . .

0�q�+k[2�+k]q[k+1]qq�+k+1+q�(�+k+1)

1q2(�+k)+1

0. . .

0�q�+n�2[2�+n�2]q[n�1]q

q�+n�1+q�(�+n�1)

1q2(�+n�2)+1

�q2(�+n�1)+1

1CCCCCCCCCCCCCCCA;

J2=

0BBBBBBBBBBBBBBB@

0q�1[2�]q[1]qq�+1+q�(�+1)

1q�+q�� 0

. . .

0q�1[2�+k]q[k+1]qq�+k+1+q�(�+k+1)

1q�+k+q�(�+k) 0

. . .

0q�1[2�+n�2]q [n�1]qq�+n�1+q�(�+n�1)

1q�+n�2+q�(�+n�2)

�q�+n�1+q�(�+n�1)

1CCCCCCCCCCCCCCCA;

fulfil comutation relations (5.1)

Proof. Let's verify the relation J3J1�q�1J1J3=J2. It is sufficient to compute elementsabove and below diagonal and in the last column. The others are clearly equal to zero.� Elements above the diagonal, row i+1 (i=0;:::;n�2), column i:

[�+i+1]q1

q2(�+i)+1�q�1 1

q2(�+i)+1[�+i]q=

q�+i

q2(�+i)+1=

1

q�+i+q�(�+i): OK

� Elements above the diagonal, row i, column i+1, i=0;:::;n�3:

[�+i]q�q�+i[2�+i]q[i+1]qq�+i+1+q�(�+i+1)

�q�1�q�+i[2�+i]q [i+1]q

q�+i+1+q�(�+i+1)[�+i+1]q=

=�q�+i[2�+i]q[i+1]qq�+i+1+q�(�+i+1)

([�+i]q�q�1[�+i+1]q)=

17

=�q�+i[2�+i]q[i+1]qq�+i+1+q�(�+i+1)

(�q�1q�(�+i))= q�1[2�+i]q[i+1]qq�+i+1+q�(�+i+1)

: OK

� Elements in the last column n�1, row i, i=0;:::;n�1:If i 6=n�1, element is clearly zero, if i=n�1, we have

[�+n�1]q�

q2(�+n�1)+1�q�1 �

q2(�+n�1)+1[�+n�1]q=

=�

q2(�+n�1)+1([�+n�1]q�q�1[�+n�1]q)=

=�

q2(�+n�1)+1([�+n�1]q�[�+n]q+q�+n�1):

It is fulfilled if the following is true:

�=0_[�+n�1]q=[�+n]q;

but this yields from assumption. OK.

Now let's take the relation J3J2�qJ2J3=J1. It suffices to verify elements above andbelow the diagonal and in the last column. The others are clearly equal to zero.

� Elements below the diagonal, row i+1 (i=0;:::;n�2), column i:

[�+i+1]q1

q�+i+q�(�+i)�q 1

q�+i+q�(�+i)[�+i]q=

q�(�+i)

q�+i+q�(�+i)=

=1

q2(�+i)+1: OK

� Elements above the diagonal except the last one, row i, column i+1, i=0;:::;n�3:

[�+i]qq�1[2�+i]q[i+1]qq�+i+1+q�(�+i+1)

�q�q�1[2�+i]q[i+1]q

q�+i+1+q�(�+i+1)[�+i+1]q=

=q�1[2�+i]q[i+1]qq�+i+1+q�(�+i+1)

([�+i]q�q[�+i+1]q)=q�1[2�+i]q[i+1]qq�+i+1+q�(�+i+1)

(�qq�+i)=

=�q�+i[2�+i]q[i+1]qq�+i+1+q�(�+i+1)

: OK

� Elements in the last column, column n�1, row i, i=0;:::;n�1:If i 6=n�1 the element is trivially zero, if i=n�1 then we have

[�+n�1]q�

q�+n�1+q�(�+n�1)�q �

q�+n�1+q�(�+n�1)[�+n�1]q=

=�

q�+n�1+q�(�+n�1)([�+n�1]q�q[�+n�1]q)=

=�

q�+n�1+q�(�+n�1)([�+n�1]q�[�+n]q+q�+n�1): OK

This is true if following condition holds:

�=0_[�+n�1]q=[�+n]q;

but this follows from the assumption (see previous relation ").The last relation J2J1�qJ1J2=J3. If we multiply two matrices with non-zero elementsabove and under the diagonal and in the right lower corner non-zero elements canappear theoretically at these places:

18

0BBBBBBBBBBBB@

? 0 ? 0 0 0 0 00 ? 0 ? 0 0 0 0? 0 ? 0 ? 0 0 0

0 ? 0 ? 0. . . 0 0 0

0 0 ? 0 ? ? 0 0

0 0 0 ? 0. . . 0 ? 0

0 0 0 0 ? ? 0 ?

0 0 0 0 0. . . 0 ? ?

0 0 0 0 0 ? ? ?

1CCCCCCCCCCCCA:

� Elements under the diagonal, row i, column i�2, where i=2;3;:::;n�1. Elementcontaining � is out of interest here.

1

q�+i�1+q�(�+i�1)� 1

q2(�+i�2)+1+q�1[2�+i]q[i+1]qq�+i+1+q�(�+i+1)

�0�

�q

1

q2(�+i�1)+1� 1

q�+i�2+q�(�+i�2)+�q�+i[2�+i]q[i+1]qq�+i+1+q�(�+i+1)

�0!=

=q�+i�1�qq�+i�2

(q2(�+i�1)+1)(q2(�+i�2)+1)=0: OK

� Elements above the diagonal, row i, column i+2, i=0;:::;n�3:If i 6=n�4 then we have

1

q�+i�1+q�(�+i�1)�0+ q�1[2�+i]q[i+1]q

q�+i+1+q�(�+i+1)��q

�+i+1[2�+i+1]q[i+2]qq�+i+2+q�(�+i+2)

+

+�

q�+n�1+q�(�+n�1)�0�q

�1

q2(�+i�1)+1�0+

+�q�+i[2�+i]q[i+1]qq�+i+1+q�(�+i+1)

�q�1[2�+i+1]q[i+2]qq�+i+2+q�(�+i+2)

+�

q2(�+n�1)+1�0�=

=q�1[2�+i]q[i+1]q[2�+i+1]q[i+2]q(�q�+i+1�q(�q�+i))

(q�+i+1+q�(�+i+1))(q�+i+2+q�(�+i+2))=0: OK

If i=n�4 we will get the same plus the term containing �:

:::+�

q�+n�1+q�(�+n�1)� 1

q2(�+n�2)+1�q�:::+

�

q2(�+n�1)+1�

� 1

q�+n�2+q�(�+n�2)

�=0+

�(q�+n�1�qq�+n�2)(q2(�+n�1)+1)(q2(�+n�2)+1)

=0: OK

� Diagonal elements. For i=0 we have

q�1[2�]q[1]qq�+1+q�(�+1)

� 1

q2�+1�q �q

� [2�]q[1]qq�+1+q�(�+1)

� 1

q�+q��=

=[2�]q[1]q(q

�1�q(�q�)q�)(q�+1+q�(�+1))(q2�+1)

=q� [2�]q[1]q(q

2�+2+1)

(q2�+2+1)(q2�+1)=q� [2�]q [1]q(q2�+1)

=

=q��(q4��1)

(q2�+1)(q�q�1) =[�]q: OK

For i=1;2;:::;n�3we have1

q�+i�1+q�(�+i�1)��q

�+i�1[2�+i�1]q[i]qq�+i+q�(�+i)

+q�1[2�+i]q[i+1]qq�+i+1+q�(�+i+1)

� 1

q2(�+i)+1�

19

�q� 1

q2(�+i�1)+1�q�1[2�+i�1]q[i]qq�+i+q�(�+i)

+�q�+i[2�+i]q[i+1]qq�+i+1+q�(�+i+1)

� 1

q�+i+q�(�+i)+

+�

q2(�+n�1)+1�0�=(q�+iq�+i�1(�q�+i�1)�qq�1q�+i)[2�+i�1]q[i]q

(q2(�+i�1)+1)(q2(�+i)+1)+

+(q�+i+1q�1�qq�+i+1(�q�+i)q�+i)[2�+i]q[i+1]q

(q2(�+i�1)+1)(q2(�+i)+1)=

=q�+i+2

(q2(�+i)+1)(q2�1)2�(q2�+i�1�q�(2�+i�1))(qi�q�i)(�q2(�+i�1)�1)

q2(�+i�1)+1

�+

+q�+i+2

(q2(�+i)+1)(q2�1)2�(q2�+i�q�(2�+i))(qi+1�q�(i+1))(1+q2(�+i+1))

q2(�+i+1)+1

�=

=q�+i+2

(q2(�+i)+1)(q2�1)2 �

��(q2�+i�1�q�(2�+i�1))(qi�q�i)(�1)(q2�+i�q�(2�+i))(qi+1�q�(i+1))

�=

=�q3�+3i+1+q��+3+i+q3�+i+1�q��i+3

(q2(�+i)+1)(q2�1)2 +

+q3�+3i+3�q3�+i+1�q��+i+3+q�i��+1

(q2(�+i)+1)(q2�1)2 =

=q3�+3i+1(q2�1)+(q2�1)(�q��i+1)

(q2(�+i)+1)(q2�1)2 =q(q2�1)q�(�+i)(q4(�+i)�1)

(q2(�+i)+1)(q2�1)2 =

=qq�(�+i)(q2(�+i)�1)

q2�1 =[�+i]q: OK

For i=n�2 we can use calculation " for i=1;:::;n�3 and the fact that

�

q�+n�1+q�(�+n�1)� 1

q2(�+n�2)+1�q �

q2(�+n�1)+1� 1

q�+n�2+q�(�+n�2)=

=�(q�+n�1�qq�+n�2)

(q2(�+n�1)+1)(q2(�+n�2)+1)=0: OK

Diagonal element when i=n�1:1

q�+n�2+q�(�+n�2)��q

�+n�2[2�+n�2]q[n�1]qq�+n�1+q�(�+n�1)

+�

q�+n�1+q�(�+n�1)�

� �

q2(�+n�1)+1�q� 1

q2(�+n�2)+1�q�1[2�+n�2]q[n�1]qq�+n�1+q�(�+n�1)

+�

q2(�+n�1)+1�

� �

q�+n�1+q�(�+n�1)

�=

1

q�+n�1+q�(�+n�1)��(�1)[2�+n�2]q[n�1]q

1

+

+�2(1�q)

q2(�+n�1)+1

�;

now we must consider two cases: if � 6=0, it implies �=�[n�1]q[n+1]q�1

qand

furthermore we know [�+n�1]q=[�+n]q, q2�+2n�1+1=0, q�2��n+2=

=�qn+1, [2�+n�2]q=[n+1]q, hence

20

:::=1

q�+n�1+q�(�+n�1)��[n+1]q[n�1]q+

(�[n�1]q[n+1]q�1)(1�q)q(�q�1+1)

�| {z }

=1

=

=1

q�+n�1+q�(�+n�1)=

q�+n�1

q2(�+n�1)+1=q�+n�1

�q�1+1=

q�+n�1(1+q�1

(1�q�1)(1+q�1) =

=q�+n�1+q�+n�2

(1�q�2) =q�+n+q�+n�1

(q�q�1) =q�+n�q�(�+n)

(q�q�1) =[�+n]q=

=[�+n�1]q: OKIf �=0, then from the assumption we know q4�=q�2n+2, hence

:::=(�1)[2�+n�2]q[n�1]qq�+n�1+q�(�+n�1)

=�(q2�+n�2�q�(2�+n�2))(qn�1�q�(n�1))

(q�+n�1+q�(�+n�1))(q�q�1)2 =

=q�2�+1+q2��1�q2�+2n�3�q�2��2n+3

(q�+n�1+q�(�+n�1))(q�q�1)2 =

=q�2�+1+q2��1+

inserted 0z }| {(�q2�+2n�1�q�2��2n+1+q2�+2n�1+q�2��2n+1)

(q�+n�1+q�(�+n�1))(q�q�1)2 �

� q2�+2n�3+q�2��2n+3

(q�+n�1+q�(�+n�1))(q�q�1)2 =

=q�2�(q+q4��1�q4�+2n�1�q�2n+1)+(q2�+2n�1+q�2��2n+1

(q�+n�1+q�(�+n�1))(q�q�1)2 �

� q2�+2n�3+q�2��2n+3)

(q�+n�1+q�(�+n�1))(q�q�1)2 =

=q�2�(q4�(q�1�q2n�1)�q�2n+2(q�1�q2n�1))

(q�+n�1+q�(�+n�1))(q�q�1)2 +

+(q2�+2n�2�q�2��2n+2)(q�q�1)(q�+n�1+q�(�+n�1))(q�q�1)2 =

q�2�=0 from assump.z }| {(q4��q�2n+2)(q�1�q2n�1)

(q�+n�1+q�(�+n�1))(q�q�1)2 +

+(q�+n�1�q�(�+n�1))(q�+n�1+q�(�+n�1))(q�q�1)

(q�+n�1+q�(�+n�1))(q�q�1)2 =[�+n�1]q: OK

� Column n�2, row n�1:1

q�+n�2+q�(�+n�2)�0+ �

q�+n�1+q�(�+n�1)� 1

q2(�+n�2)+1�

�q� 1

q2(�+n�2)+1�0+ �

q2(�+n�1)+1� 1

q�+n�2+q�(�+n�2)

�=

=�(q�+n�1�qq�+n�2)

(q2(�+n�1)+1)(q2(�+n�2)+1)=0: OK

� Column n�1, row n�2:q�1[2�+n�2]q[n�1]qq�+n�1+q�(�+n�1)

� �

q2(�+n�1)+1�

�q�q�+n�2[2�+n�2]q[n�1]qq�+n�1+q�(�+n�1)

� �

q�+n�1+q�(�+n�1)=

21

=�[2�+n�2]q[n�1]q� q�1+qq�+n�2q�+n�1

(q�+n�1+q�(�+n�1))(q2(�+n�1)+1)

�=

=�[2�+n�2]q[n�1]qq�1� 1+q2�+2n�1

(q�+n�1+q�(�+n�1))(q2(�+n�1)+1)

�=0;

because either �=0 or 1+q2�+2n�1=0, [�+n]q=[�+n�1]q.The proof is finished.5.14. Theorem. Let n2N, q2C�f0g, qk 6=1 8k2N. Let ' n-dimensionalrepresentation U(g) on a complex vector space V (dimV =n). Denote the operatorsresp. their matrices '(J1);'(J2);'(J3) simply as J1;J2;J3.Then there exist a basisX=(x0;:::;xn�1) of the space V such that matrices of operatorsJ1;J2;J3 have the form from 5.4Proof. We are in complex case that implies there exists at least one eigenvalue of J3,denote it �. From 5.8 and 5.12 implies that exists �2C such that �=[�]q and the

numbers [��k]q, k2N0 are mutually different.

Denote y0 eigenvector corresponding to �. Let yi+1=(J1�q�(��i)J2)yi 8i2N0. From5.6 follows that 9i02N0:yi=0- for i>i0 and y0;:::;yi0 are eigenvectors of J3 belonging todifferent eigenvalues which implies they are linear independent.Now let �=��i0, x0=yi0 , xi+1=(J1+q�+iJ2)xi 8i2N0. From the formula for yi0follows that (J1�q��J2)x0=0-.Let us search in the sequence x0;x1;::: the first vector xm+1 (m2N0) such that it is a

linear combination of previous ones i. e. xm+1=mXi=0

�ixi.

We need to show thatm=n�1. The casem>n�1 is impossible (it implies in Vthere are more than n linear independent vectors); the casem<n�1 is impossibletoo because we will now show that the vectors x0;:::;xm would compose an invariantsubspace of all three operators J1;J2;J3 (contradiction with irreducibility ofrepresentation).

Let C3=(q�q�1)J1J2J3+(J21�q�1J22+J23 ). Operator C3 commutes with the operatorsJ1;J2;J3. According to Schur lemma 9�2C 8x:C3x=�x.The following is true

(J1�q�(�+k+1)J2)xk+1=�(q�(�+k+1)[�+k]q+[�+k]2q��)xk8k2N0 (5.3)

because

0-=C3xk��xk=(q�q�1)J1J2[�+k]qxk+(J21�q�1J22 )xk+[�+k]2qxk��xkand

(J1�q�(�+k+1)J2)xk+1=(J1�q�(�+k+1)J2)(J1+q�+kJ2)xk==(J21�q�1J22 )xk�q�(�+k+1)(J3+qJ1J2)xk+q�+kJ1J2xk==(J21�q�1J22 )xk�q�(�+k+1)[�+k]qxk+(�q�(�+k)+q�+k)J1J2xk==(J21�q�1J22 )xk�q�(�+k+1)[�+k]qxk+(q�q�1)[�+k]qJ1J2xk;

it yields

(q�q�1)J1J2[�+k]qxk+(J21�q�1J22 )xk==(J1�q�(�+k+1)J2)xk+1+q�(�+k+1)[�+k]qxk

which implies

0-=(J1�q�(�+k+1)J2)xk+1+q�(�+k+1)[�+k]qxk+[�+k]2qxk��xk;22

which is (5.3).

The next equation we need is

0-=(q��1[�]q�[�]2q+�)x0: (5.4)

The proof is similar..

0-=(J1+q��1J2)0-=(J1+q��1J2)(J1�q��J2)x0;0-=C3x0��x0=(q�q�1)J1J2[�]qx0+(J21�q�1J22 )x0+[�]2qx0��x0

and

(J1+q��1J2)(J1�q��J2)x0==(J21�q�1J22 )x0+q��1(J3+qJ1J2)x0�q��J1J2x0==(J21�q�1J22 )x0+q��1[�]qx0+(q��q��)J1J2x0==(J21�q�1J22 )x0+q��1[�]qx0+(q�q�1)[�]qJ1J2x0;

it yields

(q�q�1)[�]qJ1J2x0+(J21�q�1J22 )x0==(J1+q��1J2)(J1�q��J2)x0�q��1[�]qx0;

hence

0-=(J1+q��1J2)(J1�q��J2)x0�q��1[�]qx0+[�]2qx0��x0:From definition of xi we have

(J1+q�+kJ2)xk=xk+1 8k2N0: (5.5)

Put x�1=0-. Then considering (5.3) and (5.4) we have 8k2N0 two equations

(q�+k+q�(�+k))J2xk=xk+1+(q�(�+k)[�+k�1]q+[�+k�1]2q��)xk�1; (5.6)

(q�(�+k)+q(�+k))J1xk==q�(�+k)xk+1�q�+k�(q�(�+k)[�+k�1]q+[�+k�1]2q��)xk�1: (5.7)

For k2f0;1;2;:::;m�1gwe can both equations (5.6) and (5.7) divide by constantq�+k+q�(�+k), it is not equal to zero (if it is the right side would be equal to 0- but thereis a nontrivial linear combination (coefficient by xk+1 is 1), so that it would implyxk+1=0- but it is not true because xk+1 is eigenvector of J3). (The vectors x0;:::;xm�1map to linfx0;:::;xmg for all three operators.)For k=m it is clear that we can divide both sides only when at least one �i is non-zero.For all �i=0 we must show another way that the coefficient we're dividing by is notequal to zero (on the right side there is not nontrivial combination more).

Let us consider all �i=0. It implies xm=0-.

For k=m we have xm+1=0-. We need to show q�+m+q�(�+m) 6=0.

From (5.3) for k=m we have

(J1�q�(�+m+1)J2)xm+1=0-=�(q�(�+m+1)[�+m]q+[�+m]2q��)xm;therefore

�=q�(�+m+1)[�+m]q+[�+m]2q :

From (5.4) we have

�=�q��1[�]q+[�]2q:

Apllying definition of [�]q we get

23

q�(�+m+1)q�+m�q�(�+m)

q�q�1 +q2(�+m)�2+q�2(�+m)

(q�q�1)2 =

=q2��2+q�2�(q�q�1)2 �q��1 q

��q��q�q�1 ;

hence

q�2m(1�q2+2m)=q4�(1�q2m+2);

i. e.

q2+2m=1_q2(2�+m)=1:

From the assumption we know qk 6=0 8k2N, it yields q2+2m=1 is not true. So let'sconsider the second case q2(2�+m)=1.Now if q�+m+q�(�+m)=0 we can square it)q2(2�+m)+2m=1, and applying previousfact q2(2�+m)=1 we have q2m=1 which is contradiction.Thus, q�+m+q�(�+m) 6=0 is always true, we can divide (5.6) and (5.7) by this constantand we see all vectors map back to the span linfx0;:::;xmg.Therefore linfx0;:::;xmg is invariant subspace of operators J1;J2;J3.From the equations (5.6) and (5.7) we can get for k=0;:::;m=n�1matrices ofoperators J1;J2. Matrix of J3 is trivial; if we show �0= :::=�n�2=0 the proof will befinished.The matrices must hold commutation relations. From the relation J3J1�q�1J1J3=J2we will get comparing the first column 8k2bn�1 the equations

�kq2(�+n�1)+1

[�+k]q�q�1�k

q2(�+n�1)+1[�+n�1]q=

�kq�+n�1

q2(�+n�1)+1;

they are fulfilled if and only if �k=0 or

[�+k]q�q�1[�+n�1]q=q�+n�1i. e.

[�+k]q=[�+n]q:

Thus for �0;:::;�n�1 the following condition holds:

8k2bn�1:�k=0_[�+k]q=[�+n]q:

If all �k=0 the proof is finished. Let us assume that for any k2bn�1 is �k 6=0. Then wehave [�+n]q=[�+k]q.

It implies �i=0 for the others i2bn�1, i 6=k. (If for any i02bn�1, i0 6=k was �i0 6=0 then[�+n]q=[�+i0]q, 5.11 would yield i0+n=k+n) i0=k which is contradiction.)Thus only one from �i is non-zero.From the first relation J2J1�qJ1J2=J3 we will get comparing elements in the rowsi=1;2;:::;n�4 resp. i=n�2

1

q�+i�1+q�(�+i�1)� �i�1q2(�+n�1)+1

�q 1

q2(�+i�1)+1� �i�1q�+n�1+q�(�+n�1)

=

=�i�1� q�+i�1�qq�+n�1(q2(�+n�1)+1)(q2(�+i�1)+1)

�should be

= 0;

it can't be true for �i�1 6=0 because q�+i�1=q�+n,qi�1=qn, i�1=n, which is nottrue. Thus �j=0 for j=0;:::;n�5, j=n�3.We must examine �j for j=n�4 a j=n�2:All �n�2k, k=1;2;::: are equal to zero, if this was not true the following would hold

24

[�+n�2k]q=[�+n]q,q2(�+n�k)=�1,q(�+n�k)+q�(�+n�k)=0;

but this is a number we were dividing by in the first part of the proof -- it is non-zero.

Thus the only non-zero is �n�1.Let's compare right lower element (n�1;n�1) in the first relation J2J1�qJ1J2=J3. Itwill give us the condition on �n�1.

1

q�+n�2+q�(�+n�2)��q

�+n�2[2�+n�2]q[n�1]qq�+n�1+q�(�+n�1)

+

+�n�1

q�+n�1+q�(�+n�1)� �n�1q2(�+n�1)+1

�

�q� 1

q2(�+n�2)+1� q�1[2�+n�2]q[n�1]qq�+n�1+q�(�+n�1)

+

+�n�1

q2(�+n�1)+1krat

�n�1q�+n�1+q�(�+n�1)

�?=[�+n�1]q:

Left side is equal to

=�

[n+1]qz }| {[2�+n�2]q[n�1]q

(q�+n�1+q�(�+n�1))+

�2n�1(1�q)(q�+n�1+q�(�+n�1))(q2(�+n�1)+1)

=

=�(

�q�n�1z }| {q2�+n�2�q�2��n+2)(qn�1�q�n+1)

(q�+n�1+q�(�+n�1))(q�q�1)2 +

+�2n�1(1�q)

(q�+n�1+q�(�+n�1))(q2(�+n�1)| {z }�q�1

+1)=

=q�+n�1

(q2(�+n�1)| {z }�q�1

+1)

��(�q�n�1�(�qn+1))(qn�1�q�n+1)

(q�q�1)2 +�2n�1(1�q)(�q�1+1)

�=

=q�+n�1

1�q�1��[n+1]q[n�1]q��n�12q

�;

Therefore following must be true

q�+n�1

1�q�1��[n+1]q[n�1]q��n�12q

�= [�+n�1]q=[�+n]q=

q�+n

q�+n�1z }| {�q�(�+n)q�q�1

i. e.��[n+1]q[n�1]q��n�12q

�=q�1q

q+1

q�q�1 =1

�[n+1]q [n�1]q = 1+�2n�1q

) �[n+1]q[n�1]q�1q

= �2n�1:

The proof is finished.

5.15. Lemmma. Let q2C�f0;1;�1g, n2N, c2C , c4=1.

Let c2q�n+1+2k+1 6=0 for k2f0;1;:::;n�1g.Let q�=cq

�n+12 . Then the matrices from theorem 5.13 have following form:

25

J3=

0BBBBBB@

k=0. . .

c2q�n+1+2k�1cq

�n+12 +k�1(q2�1)

. . .k=n�1

1CCCCCCA;

J1=

0BBBBB@0 k=0

k=0. . . �q2(q2(k+1�n)�1)(q2k+2�1)

(c2q2k�n+3+1)(q2�1)21

c2q2k+1�n+1. . . k=n�2

k=n�2 0

1CCCCCA;

J2=J3J1�q�1J1J3=

=

0BBBBBB@

0 k=0

k=0. . . q

n+12 �k(q2(k+1�n)�1)(q2k+2�1)c(c2q2k�n+3+1)(q2�1)2

cq�n+1

2+k

c2q2k+1�n+1. . . k=n�2

k=n�2 0

1CCCCCCA:

Proof. The conditions are because of there must be non-zero elements in denominators.5.16. Lemmma. Let q2C�f0;1;�1g, n2N, c2C , c4=1.Let c2=�1, n odd. Then the matrices written above in 5.15 don't have sense.Proof. c2q2k+1�n 6=�18k2f0;:::;n�1g,q1�n;q3�n;:::;qn�3;qn�1 6= �1

c2 =1 now, butthis is always not tru for n odd (it runs accross q0=1).5.17. Lemmma. Let q2C�f0;1;�1g, n2N, c2C , c4=1.Let n odd, c2=1 i. e. c=�1.Let q2;q4;:::;qn�1 6=�1.Then the matrices written in 5.15 have sense and form

J3=c

0BBBB@��n�1

2

�q. . .

. . . �n�12

�q

1CCCCA;

J1=

0BBBBB@0 k=0

k=0. . . �q2(q2(k+1�n)�1)(q2k+2�1)

(q2k�n+3+1)(q2�1)21

q2k+1�n+1. . . k=n�2

k=n�2 0

1CCCCCA:Proof. c2q2k+1�n= q2k+1�n 6=�1, q1�n;q3�n;:::;qn�3;qn�1 6==�1,1;q2;:::;qn�3;qn�1 6=�1 OK.5.18. Lemmma. Let q2C�f0;1;�1g, n2N, c2C , c4=1.Let c2q�n+2;c2q�n+3;:::;c2qn�2 6=�1^q1;q2;:::;qn�1 6=1.Then J3 written in 5.15 has sense and mutually different eigenvalues on the diagonal.Proof.

26

[�+k]q=[�+l]q 8k>l;k;l2f0;:::;n�1g,q2�+k+l=�1_qk�l==1 8k>l;k;l2f0;:::;n�1g,c2q�n+1+k+l=�1_qk�l==1 8k>l;k;l2f0;:::;n�1g, theconditioninassignment:

5.19. Lemmma. Let q2C�f0;1;�1g, n2N, c2C , c4=1.Let c2=�1;n=2. Then J3 written in 5.15 has sense and does NOT have mutuallydifferent eigenvalues on the diagonal.Proof. From previous lemma we know the condition c2q�n+1+k+l=�1 so that wechoose k= n

2�1, l= n2 for n even and k=

�n2

��1, l=�n2 �+1 for n odd.

Then we have c2q�n+1+k+l=c2�1=�1) [�+k]q=[�+l]q.

5.20. Lemmma. Let q2C�f0;1;�1g, n2N, c2C , c4=1. Let c2=1.Let q1;q2;:::;qn�2 6=�1^q1;q2;:::;qn�1 6=1.Then J3 written 5.15 has sense and mutually different eigenvalues on the diagonal.Proof. From 5.18 for c2=1.

5.21. Lemmma. Let q2C�f0;1;�1g, n2N, c2C , c4=1.Let c2=1.Let q2;q4;:::;qn�1 6=�1 for n odd resp. let q1;q3;:::;qn�1 6=�1 for n even.Then the matrices J1, J2, J3 written in 5.15 have sense.Proof. Clearly from 5.15

5.22. Lemmma. Let q2C�f0;1;�1g, n2N, c2C , c4=1.Let c2=1.Let q1;q2;:::;qn�1 6=�1^q1;q2;:::;qn�1 6=1.Then J1, J2, J3 written in 5.15 have sense and J3 has mutually different eigenvalues onthe diagonal.

Proof. From 5.18 for c2=1.5.23. Theorem. Let q2C�f0;1;�1g, n2N, c2C , c4=1.Let c2=1.Let q1;q2;:::;qn�1 6=�1^q1;q2;:::;qn�1 6=1.

Then J1, J2, J3 written in 5.15 have sense and it is an irreducible representation ofFairlie algebra.Proof. From 5.22 we have: J1, J2, J3 written in 5.15 have sense and J3 has mutuallydifferent eigenvalues on the diagonal.We will prove equivalent proposition that every non-zero vector is cyclic:

8x 6=0-8y9 polynomial P in variables J1;J2;J3 such that Px=y:

Let x 6=0-. Let x=n�1Xj=0

�jej .

Without loss of generality we can choose y=ei0 for any appropriate i02bn�1.Let j02bn�1 such that �j0 6=0 (because of x 6=0- it always exists).

Then obviously

1

�j0

n�1Yk=0k 6=j0

J3�c��n�1

2+k

�q

c

��n�1

2+j0

�q

�c��n�1

2+k

�q

x=1

�j0�j0ej0 =ej0 :

If i0=j0 we have finished, otherwise it is sufficient to show that

27

8j2f0;:::;n�2g9�;�2C ;� 6=0: (J1+�J2)ej=�ej+1^8j2f1;:::;n�1g9�;�2C ;� 6=0: (J1+�J2)ej=�ej�1:

(After showing this we can easily go from ej0 to ei0.)For any �2C is

(J1+�J2)ej=

0BBBBBBBBBBBBBB@

0...0

(�q2+�cqn+12 �j+1)(q2(j�n)�1)(q2j�1)

(q2j�n+1+1)(q2�1)20

1+�cq�n+1

2 +j

q2j+1�n+10...0

1CCCCCCCCCCCCCCA;

so that if we choose � properly we can achieve zero at least on one of these two places.Will the other stay non-zero?First of all it is clear that (q2(j�n)�1)(q2j�1) 6=0 for any j=1;:::;n�1. Theseconditions mean

q2;q4;:::;q2n�2 6=1^q2(1�n);q2(2�n);:::;q2(�1) 6=1;

the second line follows from the first one and the first one from q1 6=�1,..., qn�1 6=�1squared.For the index-growing j=1;2;:::;n�2 (for j=0 we can choose �=0) we want

q2=�cqn+12 �j+1 i. e. �=

1

cq2�

n+12 +j�1=cq1+j�

n+12 (c=

1

c).

Now we can make a test if the second one is 6=0:

1+�cq�n+1

2 +j=1+c2q1+j�n+12 +�n+1

2 +j=1+q1+2j�n 6=0?

I. e. q3�n;q5�n;:::;qn�3 6=�1?It is true because it is condition in the assumption.For decreasing index j=1;2;:::;n�2 (for j=n�1 we can choose �=0) we choose � suchthat

1+�cq�n+1

2 +j=0 i. e. �=�cq n�12 �j :

Then �q2+�cq n+12 �j+1=�q2�c2q n+12 �j+1+n�1

2 �j=�q2�qn+1�2j==�q2(1+qn�1�2j) 6=0 similarly as in previous case -- from the assumption follows thatthis is true.The result is: it is really irreducible representation.5.24. Theorem. Let q2C�f0;1;�1g, n2N, c2C , c4=1.Let c2=1 tj. c=�1. Let q1;q2;:::;qn�1 6=�1^q1;q2;:::;qn�1 6=1.Then J3, J1, J2 written in 5.15 resp. in 5.17 have sense and they are two equivalentireducible representations of Fairlie algebra.Proof. Let

R=PD=

0B@ 1. . .

11

1CA0B@�0

�1. . .

�n�1

1CA=

0B@ �n�1. . .

�1�0

1CA;28

we can easily compute

R�1=

0BB@1�0

. . .

1�n�2

1�n�1

1CCA:It is immediatelly clear that

RJ(+1)3 R�1=PDJ (+1)

3 D�1P�1=PJ (+1)3 DD�1P�1=PJ (+1)

3 PT =J(�1)3 :

Let's take

X=

0BBBBB@0 a0

b0 0. . .

. . . . . . . . .. . . 0 an�2

bn�2 0

1CCCCCA:Then

RXR�1=

0BBBBBBBB@

0 bn�2�n�1

�n�2

an�2�n�2

�n�10

. . .

. . . . . . . . .. . . 0 b1

�2�1

a1�1�2

0 b0�1�0

a0�0�1

0

1CCCCCCCCA:

) we have equations for �k, k=0;:::;n�2:If we compare elements above diagonal we will get condition (k=0;:::;n�2)

bk�k+1

�k=an�2�k (5.8)

i. e.

1

q2k+1�n+1

�k+1

�k=�q2(q2(n�2�k+1�n)�1)(q2(n�2�k)+2�1)

(q2(n�2�k)+3�n+1)(q2�1)2 ;

comparing elements under the diagonal gives us the second condition (k=0;:::;n�2)ak

�k�k+1

=bn�2�k (5.9)

i. e.

�q2(q2(k+1�n)�1)(q2k+2�1)(q2k+3�n+1)(q2�1)2

�k�k+1

=1

q2(n�2�k)+1�n+1:

From (5.8) we have

�k+1

�k=(q2k+1�n+1)

�q2(q�2�2k�1)(q2n�2�2k�1)(qn�1�2k+1)(q2�1)2 ;

from (5.9) we have

�k�k+1

=1

qn�3�2k+1

(q2k+3�n+1)(q2�1)2�q2(q2(k+1�n)�1)(q2k+2�1) :

It is consistent i. e. it is true that

1=�k+1

�k

�k�k+1

=q2n�2k�2�2k�2�n+1+2k�n+3+2k=1:

29

It is sufficient to solve (5.8), (5.9) is automatically fulfilled. (5.8) is set of equationsfor k=0;:::;n�2 that means it is undetermined, we put �0=any number 6=0,

�k+1=�kan�2�k

bk(bk 6=0 OK).

5.25. Lemmma. Let q2C�f0;1;�1g, n2N.Let us consider the matrices written in 5.13 for the case when following condition isfulfilled:

[�+n�1]q=[�+n]q^�2=�[n�1]q[n+1]q+1

q

Then the matrices have sense, q2n�1�2k 6=1 for k=0;1;:::;n�1 i. e.q3 6=1;q5 6=1;:::;q2n�1 6=1 and they look like this:

J3=�i

0BBBBBBBBB@

1+q2n�1

qn�32 (q2�1)

. . .1+q2n�1�2k

qn�32�k(q2�1)

. . .1+q

q�12 (q2�1)

= q12

q�1

1CCCCCCCCCA;

J1=

0BBBBB@0 �0

0. . . �k

k. . .

. . . �n�2 n�2 K

1CCCCCA;where

�k=q1�2n(q2(2n�1�k)�1)(q2(k+1)�1)

(q2�1)2(q2n�3�2k�1) ; k=q2n�1�2k

q2n�1�2k�1 ;K=�iq32�n(q2n�1)(q�1)(q2�1) ;

J2=

0BBBBB@0 �00 00

. . . �0k 0k

. . .. . . �0n�2 0n�2 K 0

1CCCCCA;where

�0k=q�

12�n�k(q2(2n�1�k)�1)(q2(k+1)�1)

(q2�1)2(q2n�3�2k�1) ; 0k=qn�

12�k

q2n�1�2k�1 ;K0=�iq1�n(q2n�1)(q�1)(q2�1) :

Proof. The conditions in 5.13 say that

[�+n�1]q=[�+n]q,q=1_q2�+2n�1=�1,q2�+2n�1=�1,q�=

=�iq�n+12 ;

�2=� [n�1]q [n+1]q+1

q ,�2=�q1�2n(q4n�2q2n+1)(q2�1)2 =�q1�2n(q2n�1)2

(q2�1)2 ,�=

=�iq12�n(q2n�1)

q2�1 :

5.26. Lemmma. The same assumptions as in 5.25 Then J3 has mutually differenteigenvalues on the diagonal, q2 6=1;q4 6=1;:::;q2n�4 6=1.

30

Proof. We must test condition 1+q2n�1�2k=1+q2n�1�2l for l;k=0;:::;n�1.WLOG take 05 l<k5n�1. Then

q�2k=q�2l,q�2k+2l=1,q2(l�k)=1,q2(k�l)=1 (5.10)

and because of 05 l<k5n�1 we know that

k�l2f1;2;:::;n�2g;and sufficient for fulfilling (5.10) is to ensure

q2(1) 6=1;q2(2) 6=1;:::;q2(n�2) 6=1,q2 6=1;:::;q2n�4 6=1:

5.27. Lemmma. The same assumptions as in 5.25 If the matrices exists (conditions in5.25 are met) and

q2=1_q4=1_:::_q2(2n�1)=1

then the representation is reducible.Proof. There is obviously zero above the diagonal on certain (but same) place inmatrices J1;J2;J3. Thus linfei0;:::;en�1g is an invariant subspace for some i02bn�1.5.28. Lemmma. The same assumptions as in 5.25 If the matrices exists (conditions in5.25 are met) and

q2 6=1^q4 6=1^:::^q2(2n�1) 6=1

then the representation is irreducible.Proof. We will prove that every x 6=0- is cyclic vector. Lemma 5.26 says that J3 hasmutually different eigenvalues on the diagonal. Thus there exists some polynomial P invariable J3 and i02bn�1 such that Px=ei0 .J1 has everywhere above and below the diagonal non-zero elements, thus for anyi02f0;:::;n�2g there is a polynomial Q in variables J3 and J1 and some � 6=0 such that

Qei0 =�ei0+1:

For any i02f1;:::;n�1g the similar is true: there is some Q and � such that

Qei0 =�ei0�1:Hence now can be seen that for any given x 6=0- and for given y there exists polynomialS in variables J1;J3 such that Sx=y.5.29. Lemmma. The same assumptions as in 5.25 If the matrices exists (conditions in5.25 are met) and the condition in 5.28 is met (representations are irreducible) then allfour representations (for all combinations of �) are mutually nonequivalent.Proof. One can easily see that either eigenvalues of J3 are different or trace of J1 isdifferent.5.30. Lemmma. Let's consider the same assumptions as in 5.15 Let c2=�1 i. e. c=�i.Then

J1=

0BBBB@0 �0

0. . . �k k

. . . �n�2 n�2 0

1CCCCA;where

�k=�q2(q2(k+1�n)�1)(q2k+2�1)

(1�q2k�n+3)(q2�1)2 ; k=1

1�q2k+1�n ;

and 8k2f0;:::;n�2g:�n�2�k k

=�k

n�2�k.

31

Proof.

�n�2�k k

=�q2(q2(�1�k)�1)(q2n�2�2k�1)(1�q2k+1�n)

(1�qn�1�2k)(q2�1)2 =

=q�2k�n�1(q2(k+1)�1)(q2(k+1)�q2n)

(q2�1)2 =

=�q2(q2(k+1�n)�1)(q2k+2�1)(1�qn�3�2k)

(1�q2k�n+3)(q2�1)2 =�k

n�2�k:

5.31. Lemmma. Let's consider the same assumptions as in 5.15 Let c2=�1 i. e. c=�i.Then representation is completely reducible.

Proof. We know n is even (otherwise matrices don't have sense). Let r0;r1;:::;rn2�1 benon-zero complex numbers. Let

X=

0BBBBBB@

r0 �r0. . . . . .

rn2�1 �rn2�11 1

. . . . . .1 1

1CCCCCCA:

Then

X�1=1

2

0BBBBBBBBB@

1r0

1. . . . . .

1rn2�1

1

� 1rn2�1

1

. . . . . .� 1

r01

1CCCCCCCCCA;

X�1J1=1

2

0BBBBBBBBBBBBBBBBBB@

�0r0

n�2 0r1

. . . . ..

�n�2. . .

�n2�2

rn2�2

n2

. . .

n2�2

rn2�1

n2�1

�n2�1

rn2�1

�n2

� n2�2

rn2�1

n2�1 ��n

2�1

rn2�1

�n2

. . . ��n2�2

rn2�2

n2

. . .

� 0r1

. . . . . . �n�2��0

r0 n�2

1CCCCCCCCCCCCCCCCCCA

;

finally

32

X�1J1X=1

2

0BBBBBBBBBBB@

a10 a20

a30. . . . . . a40

. . . a1m a2m . . .

a3m A B a4ma4m �B �A a3m

. . . a2m a1m. . .

a40 . .. . . . a30

a20 a10

1CCCCCCCCCCCA;

wherem=n

2�2 and

a1k=�krk+1+ n�2�krk

rk; a2k=

��krk+1+ n�2�krkrk

;

a3k=�n�2�krk+1+ krk

rk+1; a4k=

�n�2�krk+1� krkrk+1

:

A=�n2�1+ n

2�1r

2n2�1

rn2�1

;

B=�n2�1� n2�1r

2n2�1

rn2�1

:

If we now choose rk to satisfy conditions (see 5.30)

r2n2�1=

�n2�1

n2�1

;

rkrk+1

= �n�2�k

k8k2f0;1;:::;n2�2g;

we see that X�1J1X has block-diagonal form. Because J3 is diagonal and J2 isuniquely determined by J1 and J3, proof is finished.

In the next two paragraphs we pose two open questions which are together withdetailed examination of representations for the case qk=1 of my interest in the future.

6. Fairlie algebra as a special case of quadratic algebra6.1. De�nition. Homogeneous quadratic algebra. Let V be a vector space and T (V )its tensor algebra. Fix a subspace R�T 2(V )=V V and consider the two-sided ideal Igenerated by R:

I=linfxryjr2R;x;y2T (V )g:The quotient algebra U =T (V )=I is called a (homogeneous) quadratic algebra.6.2. De�nition. Nonhomogeneous quadratic algebra. Similarly define

nonhomogeneous quadratic algebra. Consider the natural filtration F i(V )=iL

n=0Tn(V ).

Then fix a subspace P �F 2(V )=C�(V )�(V V ). Let I be a two-sided ideal generatedby P :

I=linfxpyjp2P;x;y2T (V )g:Then the quotient algebra U =T (V )=I is called a nonhomogeneous quadratic algebra.6.3. Lemmma.airlies deformation of SU(2) is nonhomogeneous quadratic algebra.Proof. The subspace

P =linfI1�I3I2+qI2I3;I2�I3I1+q�1I1I3;I3�I2I1+qI1I2g:Open question. In [8] there is introduced special class of quadratic algebras (so calledKozsul type) for which can be formulated a condition which completely determines if

33

some algebra is of PBW type or not. The open question is if Fairlie algebra is of Kozsultype.

7. Hopf algebras

7.1. De�nition. Coalgebra. Let R be a ring. A coalgebra (C;+;R;4) over R(R-coalgebra) is a R-module C with a linear map4: C!CC called the coproduct(or comultiplication map). If

((id�4)�4)(x)=((4�id)�4)(x)8x2Cthen the coproduct4 and the coalgebra is called coassociative. If there exists a map ":C!R such that

(("�id)�4)(x)=((id�")�4)(x)8x2Cthen the map " is called a counit map and the coalgebra is called a coalgebra withcounit. A coalgebra C is called cocommutative if

(��4)(x)=4(x)8x2C;where �: CC!CC is a twist map, i. e.

�(xy)=yx8x;y2C:7.2. De�nition. Bialgebra. Let B be an R-algebra. B has the structure of a bialgebra(B;+;m;4) if (B;+;m) is an R-algebra and4 is a coproduct map that is an algebrahomomorphism.

7.3. De�nition. Antipode. Let (B;+;R;m;4;�;") be a bialgebra with unity � andcounit " homomorphisms. An algebra antiautomorphism S: B!B which satisfies

(S�m)(xy)=m(S(y)S(x))8x;y2B(m�(S�id)�4)(a)=(��")(a)=(m�(id�S)�4)(a)8a2B

is called antipode map.

7.4. De�nition. Hopf algebra. A Hopf algebra (H;+;R;m;4;�;";S) is a bialgebra witha unity �: R!H, counit ": H!R and an antipode antiautomorphism S.

7.5. Lemmma. Let g be a C -Lie algebra and U(g) its universal enveloping algebra.Then U(g) has the structure of a Hopf algebra by extending the maps

4(x)=x1+1x;4(��1)=��11;�(�)=�1;"(x)=0; 8x2g;�2CS(x)=�x;"(��1)=�;S(1)=1

homomorphically to all elements of U(g).

Proof. Clearly all the Hopf algebra maps are algebra homomorphisms. For thecoproduct following holds:

4(x�y�y�x)=4(x)�4(y)�4(y)�4(x)==(x�y)1+xy+yx+1(x�y)�(y�x)1�yx�xy�1(y�y)==(x�y�y�x)1+1(x�y�y�x)=4([x;y]):

Open Question. Can be Fairlie algebra given the structure of a Hopf algebra by suitablecoproduct map?

34

8. Extension of sl(2) and its representationsIt follows from the relations (4.1)--(4.3) and we have shown that for the algebra Uq(so3)the Poincar�e-Birkhoff-Witt theorem is true and this theorem can be formulated as:The elements Ik1 I

m2 I

n3 , k;m;n=0;1;2;��, form a basis of the linear space Uq(so3).

Indeed, by using the relations (4.1)--(4.3) any product Ij1Ij2 ��Ijs, j1;j2;��;js=1;2;3,can be reduced to a sum of the elements Ik1 I

m2 I

n3 with complex coefficients.

Note that by (4.3) the element I3 is not independent: it is determined by the elementsI1 and I2. Thus, the algebra Uq(so3) is generated by I1 and I2, but now instead ofquadratic relations (4.1)--(4.3) we must take the relations

I1I22�(q+q�1)I2I1I2+I22I1=�I1; (8.1)

I2I21�(q+q�1)I1I2I1+I21I2=�I2; (8.2)

which are obtained if we substitute the expression (4.3) for I3 into (4.1) and (4.2). Theequation I3=q1=2I1I2�q�1=2I2I1 and the relations (8.1) and (8.2) restore the relations(4.1)--(4.3).Remark that the definition of Uq(so3) by means of relations (8.1) and (8.2) was used forthe embedding of Uq(so3) to Uq(sl3). The relations (8.1) and (8.2) differ from Serre'srelations in the definition of quantum algebras by V. Drinfeld and M. Jimbo byappearance of non-vanishing right hand sides.The algebra Uq(so3) is closely related to (but not coincides with) the quantum algebraUq(sl2). The last algebra is generated by the elements qH , q�H , E, F satisfying therelations

qHq�H=q�HqH=1; qHEq�H=qE; qHFq�H=q�1F; (8.3)

[E;F ]:=EF�FE= q2H�q�2H

q�q�1 :

Note that Uq(sl2) is the associative algebra equipped with a Hopf algebra structure (acomultiplication, a counit and an antipode). In particular, the comultiplication � isdetermined by the formulas

�(q�H)=q�Hq�H; �(E)=EqH+q�HE;�(F )=FqH+q�HF:

In order to relate the algebras Uq(so3) and Uq(sl2) we need to extend Uq(sl2) by the

elements (qkqH+q�kq�H)�1 in the sense of Ref. 10. We denote by Uq(sl2) theassociative algebra with unit element generated by the elements

qH ; q�H ; E; F; (qkqH+q�kq�H)�1; k2Z;satisfying the defining relations of the algebra Uq(sl2) and the following naturalrelations:

(qkqH+q�kq�H)�1(qkqH+q�kq�H)=(qkqH+q�kq�H)(qkqH+q�kq�H)�1==1; (8.4)q�H(qkqH+q�kq�H)�1=(qkqH+q�kq�H)�1q�H; (8.5)(qkqH+q�kq�H)�1E=E(qk+1qH+q�k�1q�H)�1; (8.6)(qkqH+q�kq�H)�1F =F (qk�1qH+q�k+1q�H)�1: (8.7)

Note that the algebra Uq(sl2) has finite dimensional irreducible representations

Tl�T (1)l , T

(�1)l , T

(i)l , T

(�i)l , l=0;12;1;

32;��, acting on the vector spacesHl with base

vectors xm,m=�l;�l+1;��;l.These representations are given by the formulas

T(1)l (qH)xm=qmxm; T

(1)l (E)xm=[l�m]xm+1; (8.8)

35

T(1)l (F )xm=[l+m]xm�1; (8.9)

where a number in square brackets means a q-number, defined by the formula

[a]= qa�q�aq�q�1 ;

and by the formulas

T(�1)l (qH)xm=�qmxm; T

(�1)l (E)=T

(1)l (E); T

(�1)l (F )=T

(1)l (F ); (8.10)

T(i)l (qH)xm=iqmxm; T

(i)l (E)=T

(1)l (E); T

(i)l (F )=�T (1)

l (F ); (8.11)

T(�i)l (qH)xm=�iqmxm; T

(�i)l (E)=T

(1)l (E); T

(�i)l (F )=�T (1)

l (F ): (8.12)

The representations T(1)l , T

(�1)l , T

(i)l , T

(�i)l , l=0;12;1;

32;��; are pairwise non-equivalent,

and any finite dimensional irreducible representation of Uq(sl2) is equivalent to one ofthese representations (see, for example, Ref. [17], Chapter 3).

Now we wish to extend these representations of Uq(sl2) to the representations of Uq(sl2)by using the relation

T ((qkqH+q�kq�H)�1):=(qkT (qH)+q�kT (q�H))�1: (8.13)

Clearly, only those irreducible representations T of Uq(sl2) can be extended to

Uq(sl2) for which the operators qkT (qH)+q�kT (q�H) are invertible. From formulas(8.8)--(8.9) it is clear that these operators are always invertible for the representations

T(1)l , T

(�1)l , l=0;12;1;

32;��; and for the representations T

(i)l , T

(�i)l , l= 1

2 ;32;

52;��. For the

representations T(i)l , T

(�i)l , l=0;1;2;��, some of these operators are not invertible

since they have zero eigenvalue. Denoting the extended representations by the samesymbols, we can formulate the following statement:8.1. Theorem. The algebra Uq(sl2) has the irreducible finite dimensional

representations T(1)l , T

(�1)l , l=0;12 ;1;

32;��; and T

(i)l , T

(�i)l , l= 1

2 ;32;

52;��. Any

irreducible finite dimensional representation of Uq(sl2) is equivalent to one of theserepresentations.

9. Algebrahomomorphism fromdeformation of so(3) intodeformation of sl(2)

The aim of this chapter is to give (in an explicit form) the homomorphism of

the algebra Uq(so3) to Uq(sl2). This homomorphism is described by the followingproposition:9.1. Theorem. There exists a unique algebra homomorphism : Uq(so3)! Uq(sl2) suchthat

(I1)=i

q�q�1 (qH�q�H); (9.1)

(I2)=(E�F )(qH+q�H)�1; (9.2) (I3)=(iqH�1=2E+iq�H�1=2F )(qH+q�H)�1; (9.3)

where qH+a:=qHqa for a2C.Proof. In order to prove this proposition we have to show that

q1=2 (I1) (I2)�q�1=2 (I2) (I1)= (I3);q1=2 (I2) (I3)�q�1=2 (I3) (I2)= (I1); (9.4)q1=2 (I3) (I1)�q�1=2 (I1) (I3)= (I2):

36

Let us prove the first relation. (Other relations are proved similarly.) Substitutingthe expressions (9.1)--(9.3) for (Ii), i=1;2;3, into the first relation we have (aftermultiplying both sides of equality by (qH+q�H) on the right) the relation

q(E�F )EqH(qqH+q�1q�H)�1+q(E�F )Fq�H(q�1qH+qq�H)�1��qE2qH(qqH+q�1q�H)�1�q�1FEq�H(qqH+q�1q�H)�1+

+q�1EFqH(q�1qH+qq�H)�1+qF 2q�H(q�1qH+qq�H)�1=iq2H�q�2H

q�q�1 :

The formula (9.4) is true if and only if this relation is correct. We multiply both itssides by (qqH+q�1q�H)(q�1qH+qq�H) on the right and obtain the relation in thealgebra Uq(sl2) (that is, without the expressions (qkqH+q�kq�H)�1). This relation iseasily verified by using the defining relations of the algebra Uq(sl2). Proposition isproved.

10. Finite dimensional representations obtained using algebrahomomorphismWe assume in this and following three chapters that q is not a root of unity.If T is a representation of the algebra Uq(sl2) on a linear space V , then the mappingR: Uq(so3)!V defined as the composition R=T � , where is the homomorphismfrom theorem 9.1, is a representation of Uq(so3).Let us consider the representations

R(1)l =T

(1)l � ; R

(�1)l =T

(�1)l � ; R

(i)l =T

(i)l � ; R

(�i)l =T

(�i)l �

of Uq(so3), where T(1)l , T

(�1)l , T

(i)l , T

(�i)l are the irreducible representations of Uq(sl2)

from theorem 8.1.Using formulas for the representations T

(�1)l of Uq(sl2) and the expressions (9.1)--(9.3)

for (Ij), j=1;2;3, we find that

R(1)l (I1)xm=i[m]xm;

R(1)l (I2)xm= 1

qm+q�m ([l�m]xm+1�[l+m]xm�1);

R(1)l (I3)xm= iq1=2

qm+q�m (qm[l�m]xm+1+q�m[l+m]xm�1)

for the representation R(1)l and

R(�1)l (I1)xm=�i[m]xm; R

(�1)l (I2)=�R(1)

l (I2); R(�1)l (I3)=R

(1)l (I3):

Denoting the vectors xm by x�m for the representations R(�1)l we easily find that

the matrices of the representation R(�1)l in the basis containing vectors x�m,

m=�l;�l+1;��;l, coincide with the corresponding matrices of the representation R(1)l .

Thus, the non-equivalent representations T(1)l and T

(�1)l of the algebra Uq(sl2) lead to

equivalent representations of Uq(so3).

For the representations R(i)l and R

(�i)l we have

R(i)l (I1)xm=�qm+q�m

q�q�1 xm;

R(i)l (I2)xm=i [l�m]

qm�q�mxm+1+i [l+m]qm�q�mxm�1;

R(i)l (I3)xm=� iqm+1=2[l�m]

qm�q�m xm+1� iq�m+1=2[l+m]qm�q�m xm�1

and

37

R(�i)l (I1)xm= qm+q�m

q�q�1 xm;

R(�i)l (I2)xm=�i [l�m]

qm�q�mxm+1�i [l+m]qm�q�mxm�1;

R(�i)l (I3)xm=� iqm+1=2[l�m]

qm�q�m xm+1� iq�m+1=2[l+m]qm�q�m xm�1:

10.1. Theorem. The representations R(1)l of Uq(so3) are irreducible. The

representations R(i)l and R

(�i)l are reducible.

Proof. To prove the first part of the proposition we first note that since q is not a root

of unity, the eigenvalues i[m],m=�l;�l+1;��;l, of the operator R(1)l (I1) are pairwise

different.Let V be an invariant subspace of the space Hl of the representation R

(1)l , and let

v�Xmi

�ixmi 2V , where xmi are eigenvectors of R(1)l (I1). Then xmi 2V . We prove this

for the case when v=�1xm1+�2xm2 . (The case of more number of summands is proved

similarly.) We have R(1)l (I1)v=i�1[m1]xm1+i�2[m2]xm2 . Since

v=�1xm1+�2xm2 2V; v0� i�1[m1]xm1+i�2[m2]xm2 2Vone derives that

i[m1]v�v0=i�2([m1]�[m2])xm2 2V:Since [m1] 6=[m2], then xm2 2V and hence xm1 2V .In order to prove that V =Hl we obtain from the above formulas for R(1)

l (I2)xm and

R(1)l (I3)xm that

(R(1)l (I3)�iqm+1=2R

(1)l (I2))xm=iq1=2xm�1;

(R(1)l (I3)+iq�m+1=2R

(1)l (I2))xm=iq1=2xm+1:

Since V contains at least one basis vector xm, it follows from these relations that Vcontains the vectors xm�1 , xm�2 ,��,x�l and the vectors xm+1 ,xm+2 ,�� ,xl. Thismeans that V =Hl and the representation R

(1)l is irreducible.

Let us show that the representations R(i)l are reducible. The eigenvalues of the operator

R(i)l (I1) are

�qm+q�m

q�q�1 ; m=�l;�l+1;��;l;that is, every spectral point has multiplicity 2. Namely, the pairs of vectors xm andx�m are of the same eigenvalue. Let V1 be the subspace of the representation spaceHl

spanned by the vectors

x12+ix�1

2; x3

2�ix�3

2; x5

2+ix�5

2; x7

2�ix�7

2; �� ; (10.1)

and let V2 be the subspace spanned by the vectors

x12�ix�1

2; x3

2+ix�3

2; x5

2�ix�5

2; x7

2+ix�7

2; �� : (10.2)

We denote the vectors (10.1) by

x012; x03

2; x05

2; x07

2; �� (10.3)

and the vectors (10.2) by

x0012; x003

2; x005

2; x007

2; ��: (10.4)

38

Then

R(i)l (I1)x0m=�qm+q�m

q�q�1 x0m; R(i)l (I1)x00m=�qm+q�m

q�q�1 x00m:

We also have

R(i)l (I2)x

012=i

[l�12 ]

q1=2�q�1=2x32+i

[l+12 ]

q1=2�q�1=2x�12+

[l+12 ]

q1=2�q�1=2x12+

[l�12 ]

q1=2�q�1=2x�32=

=[l+1

2 ]

q1=2�q�1=2x012+i

[l�12 ]

q1=2�q�1=2x032:

We derive similarly that

R(i)l (I2)x001

2=� [l+1

2 ]

q1=2�q�1=2x0012+i

[l�12 ]

q1=2�q�1=2x0032

and that

R(i)l (I2)x

0m=i [l�m]

qm�q�mx0m+1+i [l+m]

qm�q�mx0m�1; m> 1

2 ;

R(i)l (I2)x00m=i [l�m]

qm�q�mx00m+1+i [l+m]

qm�q�mx00m�1; m> 1

2 :

Thus, the subspaces V1 and V2 are invariant with respect to the operatorsR(i)l (I1) and

R(i)l (I2). This means that they are invariant with respect to the representation R

(i)l .

It is proved similarly that the subspace V1 of the space Hl of the representation R(�i)l

spanned by the vectors (10.1) and the subspace V2 of Hl spanned by the vectors (10.2)

are invariant with respect to the operators R(�i)l (I1) and R

(�i)l (I2). That is, the

representation R(�i)l is also reducible. Proposition is proved.

Let R(i;+)n and R

(i;�)n , n= l+1

2=dimV1=dimV2, be the representations of Uq(so3) which

are restrictions of R(i)l to the subspaces V1 and V2, respectively. Denoting the vectors

(10.3) of the subspace V1 by

x1; x2; x3; x4; ��; xn�xl+12; (10.5)

respectively, we have

R(i;+)n (I1)xk=�qk�1=2+q�k+1=2

q�q�1 xk;

R(i;+)n (I2)x1=

[n]q1=2�q�1=2x1+i [n�1]

q1=2�q�1=2x2;

R(i;+)n (I2)xk=i [n�k]

qk�1=2�q�k+1=2xk+1+i [n+k�1]qk�1=2�q�k+1=2xk�1; k 6=1:

For the operator R(i;+)n (I3) we have

R(i;+)n (I3)x1=� [n]

q1=2�q�1=2x1�i q[n�1]q1=2�q�1=2x2;

R(i;+)n (I3)xk=�i qk[n�k]

qk�1=2�q�k+1=2xk+1�i q�k+1[n+k�1]qk�1=2�q�k+1=2xk�1; k 6=1:

Denoting the vectors (10.4) of the subspace V2 by the symbols (10.5), respectively, weobtain

R(i;�)n (I1)xk=�qk�1=2+q�k+1=2

q�q�1 xk;

R(i;�)n (I2)x1=� [n]

q1=2�q�1=2x1+i [n�1]q1=2�q�1=2x2;

R(i;�)n (I2)xk=R

(i;+)n (I2)xk; k 6=1:

For the operator R(i;�)l (I3) we find that

R(i;�)n (I3)x1=

[n]q1=2�q�1=2x1�i q[n�1]

q1=2�q�1=2x2;

39

R(i;�)n (I3)xk=R

(i;+)n (I3)xk; k 6=1:

Let now R(�i;+)n and R

(�i;�)n , n= l+ 1

2 , be the representations of Uq(so3) which are

restrictions of the representation R(�i)l to the subspaces V1 and V2, respectively.

Introducing the vectors similar to the vectors (10.5), for the representation R(�i;+)n we

have

R(�i;+)n (I1)xk=

qk�1=2+q�k+1=2

q�q�1 xk; R(�i;+)n (I2)=�R(i;+)

n (I2);

R(�i;+)n (I3)x1=

[n]q1=2�q�1=2x1+i q[n�1]

q1=2�q�1=2x2;

R(�i;+)n (I3)xk=i qk[n�k]

qk�1=2�q�k+1=2xk+1+i q�k+1[n+k�1]qk�1=2�q�k+1=2xk�1; k 6=1:

For the representation R(�i;�)l we obtain

R(�i;�)n (I1)xk=

qk�1=2+q�k+1=2

q�q�1 xk; R(�i;�)n (I2)=�R(i;�)

n (I2);

R(�i;�)n (I3)x1=� [n]

q1=2�q�1=2x1+i q[n�1]q1=2�q�1=2x2;

R(�i;�)n (I3)xk=R

(�i;+)n (I3)xk:

Thus, we constructed the representations R(i;+)n , R

(i;�)n , R

(�i;+)n and R

(�i;�)n of the

algebra Uq(so3). The following theorem characterizes them.

10.2. Theorem. The representations R(i;+)n , R

(i;�)n , R

(�i;+)n and R

(�i;�)n are irreducible

and pairwise nonequivalent. For any l the representation R(1)l is not equivalent to any

of these representations.

Proof. The irreducibility is proved exactly in the same way as in 10.1. Equivalencerelations may exist only for irreducible representations of the same dimension.

That is, we have to show that under fixed n no pair of the representations R(i;+)n ,

R(i;�)n , R

(�i;+)n and R

(�i;�)n is equivalent. It follows from the above formulas that

the operators R(i;+)n (I1) and R

(i;�)n (I1), as well as the operators R

(�i;+)n (I1) and

R(�i;�)n (I1), have the same set of eigenvalues. Moreover, the spectrum of the first pair

of operators differs from that of the second pair. Hence, no of representations R(i;+)n and

R(i;�)n is equivalent to R(�i;+)

n or R(�i;�)n . The representations R(i;+)

n and R(i;�)n are

not equivalent since the operators R(i;+)n (I2) and R

(i;�)n (I2) have different traces (for

equivalent representations these operators must have the same trace). For the same

reason, the representations R(�i;+)n and R

(�i;�)n are not equivalent. The last assertion of

the theorem follows from the fact that the spectrum of the operator R(1)l (I1) differs

from the spectra of the operators R(i;+)n (I1), R

(i;�)n (I1), R

(�i;+)n (I1) and R

(�i;�)n (I1).

Theorem is proved.

Clearly, the reducible representations R(i)n and R

(�i)n decomposes into irreducible

components as

R(i)n =R

(i;+)n �R(i;�)

n ; R(�i)n =R

(�i;+)n �R(�i;�)

n : (10.6)

10.3. Theorem. Every irreducible finite dimensional representation of Uq(so3) is

equivalent to one of the representations R(1)l , R

(i;+)n , R

(i;�)n , R

(�i;+)n , R

(�i;�)n . That is,

these representations exhaust, up to equivalence, all irreducible finite dimensional

40

representations of Uq(so3).

Proof. It can be easily seen that representations R(1)l , R(i;+)

n , R(i;�)n , R(�i;+)

n , R(�i;�)n

are equivalent to the representations described in the theorem 5.4. Thus the prooffollows from this theorem.

11. Tensor product of representationsAs mentioned above, no Hopf algebra structure is known for the algebra Uq(so3).Therefore, we cannot construct tensor product of finite dimensional representations ofUq(so3) by using a comultiplication as we do in the case of the quantum algebra Uq(sl2).However, we may construct some tensor product representations by using the algebrahomomorphism of 9.1.First we determine which tensor products of irreducible representations of Uq(sl2)

can be extended to representations of the algebra Uq(sl2). Verifying for which tensorproducts T =T 0T 00 of irreducible representations of Uq(sl2) the operators

qkT (qH)+q�kT (q�H); k2Z;are invertible, we conclude that only the tensor products

T(�1)l T (�1)

l0 ; T(�1)l T (�1)

l0 ; l;l0=0;12;1;3

2;��;

T(�1)l T (�i)

l0 ; T(�1)l T (�i)

l0 ; l=0;1;2;��; l0= 12 ;

32;

52 ;��;

T(�i)l T (�1)

l0 ; T(�i)l T (�1)

l0 ; l= 12 ;

32;

52;��; l0=0;1;2;��;

T(�i)l T (�i)

l0 ; T(�i)l T (�i)

l0 ; l;l0= 12;32;52;��;

can be extended to the algebra Uq(sl2). Taking into account the decompositions oftensor products of irreducible representations of Uq(sl2) (see, for example, the end ofSubsection 3.2.1 and Proposition 3.22 in [17]) we find that

T(!)l T (!0)

l0 'T (!!0)l+l0 �T (!!0)

l+l0�1��T (!!0)jl+l0j ; (11.1)

T(!)l T (�i)

l0 'T (�!i)l+l0 �T (�!i)

l+l0�1��T (�!i)jl+l0j ; (11.2)

T(�i)l T (!)

l0 'T (�!i)l+l0 �T (�!i)

l+l0�1��T (�!i)jl+l0j ; (11.3)

T(!i)l T (!0i)

l0 'T (�!!0)l+l0 �T (�!!0)

l+l0�1 ��T (�!!0)jl+l0j ; (11.4)

where !;!0=�1.Now we define tensor products of representations of Uq(so3) corresponding to the above

tensor product representations of Uq(sl2) as

RR0=(TT 0)� ;where R=T � and R0=T 0� . Taking into account the definitions of tensor productsof representations of Uq(sl2) by means of the comultiplication and the definition of themapping we have

(RR0)(I1)=(TT 0)� (I1)= iq�q�1

�T (qH)T 0(qH)�T (q�H )T 0(q�H)

�:

Similarly,

(RR0)(I2)=(T (E)T 0(qH)+T (q�H)T 0(E)�T (F )T 0(qH)��(T (q�H )T 0(F ))�(T (qH )T 0(qH)+T (q�H )T 0(q�H))�1:

Composing both sides of the relations (11.1)--(11.4) with the mapping of 9.1, we findthe decomposition into representations of Uq(so3) for the tensor products

41

R(1)l R(1)

l0 ; R(1)l R(�i)

l0 ; R(�i)l0 R(1)

l ; R(�i)l R(�i)

l0 ; R(�i)l R(�i)

l0 ;

where the second and the third tensor products are defined only for l=0;1;2;��. (Notethat the representations R

(�i)l are defined only for l= 1

2 ;32;

52;��.) We have

R(1)l R(1)

l0 'R(1)l+l0�R(1)

l+l0�1��R(1)jl�l0j;

R(1)l R(�i)

l0 'R(�i)l+l0�R(�i)

l+l0�1��R(�i)jl�l0j;

R(�i)l R(1)

l0 'R(�i)l+l0�R(�i)

l+l0�1��R(�i)jl�l0j;

R(!i)l R(!0i)

l0 'R(1)l+l0�R(1)

l+l0�1��R(1)jl�l0j:

In these formulas the representations R(�i)l are reducible. Unfortunately, our definition

of tensor products of representations of Uq(so3) does not allow to determine the tensor

products containing the irreducible representations R(�i;�)n and R

(�i;�)n .

12. Infinite dimensional representations obtained usingalgebra homomorphism

By using the homomorphism : Uq(so3)! Uq(sl2) from 9.1 and infinite dimensional

irreducible representations of the algebra Uq(sl2) we can construct infinite dimensionalirreducible representations of the algebra Uq(so3).Let us first describe irreducible infinite dimensional representations of the algebraUq(sl2). Note that by an infinite dimensional representation T of Uq(sl2) we mean ahomomorphism of Uq(sl2) into the algebra of linear operators (bounded or unbounded)on a Hilbert space, defined on an everywhere dense invariant subspace D, such that theoperator T (qH) can be diagonalized, has a discrete spectrum and its eigenvectorsbelong to D. Infinite dimensional representations T of Uq(so3) are described in thesame way replacing the operator T (qH) by T (I1).Two representations T and T 0 of Uq(sl2) on spacesH and H 0, respectively, are called(algebraically) equivalent if there exist everywhere dence invariant subspaces V �Hand V 0�H 0 and a one-to-one linear operator A: V !V 0 such that AT (a)v=T 0(a)Av forall a2Uq(sl2) and v2V . Equivalence of infinite dimensional representations of Uq(so3)is defined in the same way.Let � be a fixed complex number such that 0�Re�<1, and let H� be a complex Hilbertspace with the orthonormal basis

xm; m=n+�; n=0;�1;�2;��: (12.1)

For every complex number a we construct the representation Ta� on the Hilbert spaceH� defined by

Ta�(qH)xm=qmxm; Ta�(E)xm=[a�m]xm+1; Ta�(F )xm=[a+m]xm�1;where [a�m] is the q-number (see, for example, Ref. [18]). The equivalence relations inthe set of the representations Ta� can be extracted from Ref. [18].Note that the representation Ta� is irreducible if and only if a 6=��(modZ).All the representations Ta� can be extended to representations of the algebraUq(sl2) except for the case when �=�i�=2� , where q=e� . (We suppose below that� 6=�i�=2� .) We denote these extended representations by the same symbols Ta�.The formula Ra�=Ta�� associates with every irreducible representation Ta�,� 6=�i�=2� , of Uq(sl2) a representation of the algebra Uq(so3).

42

Let � 6=�i�=2� and � 6=�i�=2�+ 12 . Then for the representations Ra� of Uq(so3) we

have

Ra�(I1)xm=i[m]xm; (12.2)Ra�(I2)xm= 1

qm+q�mf[a�m]xm+1�[a+m]xm�1g; (12.3)

Ra�(I3)xm= iq1=2

qm+q�mfqm[a�m]xm+1+q�m[a+m]xm�1g: (12.4)

If �=i�=2�+ 12, then denoting the basis elements xm,m=n+�, n2Z, by xn+1

2, n2Z,

respectively, we obtain

Ra�(I1)xk=�qk+q�k

q�q�1 xk;

Ra�(I2)xk=i [a0�k]qk�q�kxk+1+i [a0+k]

qk�q�kxk�1;

Ra�(I3)xk=� iqk+1=2[a0�k]qk�q�k xk+1� iq�k+1=2[a0+k]

qk�q�k xk�1;

where a0=a+i�=2� and k=n+12 . If �=�i�=2�+1

2 , then using the same notations forbasis elements we obtain

R0a�(I1)xk=qk+q�k

q�q�1 xk;

R0a�(I2)xk=�i [a0�k]qk�q�kxk+1�i [a0+k]

qk�q�kxk�1;

R0a�(I3)xk=� iqk+1=2[a0�k]qk�q�k xk+1� iq�k+1=2[a0+k]

qk�q�k xk�1

(to distinquish these representations from the previous ones we supplied Ra� by prime).12.1. Theorem. The representations Ra� of Uq(so3) are irreducible for irreducible

representations Ta�, � 6=�i�=2�+12 , of Uq(sl2). The representations Ra�, �=i�=2�+1

2 ,

and R0a�, �=�i�=2�+12, are reducible.

Proof. is given in the same way as in the case of 10.1.As in the case of finite dimensional representations in previous chapter, decomposingthe representations Ra�, �=�i�=2�+ 1

2 , we obtain irreducible infinite dimensional

representations of Uq(so3) which will be denoted by R(i;�)a0 and R

(�i;�)a0 , a0=a+i�=2� .

In the basis

xn; n=1;2;3;��;they are given by the formulas

R(i;�)a0 (I1)xk=�qk�1=2+q�k+1=2

q�q�1 xk;

R(i;�)a0 (I2)x1=� [a0]

q1=2�q�1=2x1+i [a0�1]q1=2�q�1=2x2;

R(i;�)a0 (I2)xk=i [a0�k]

qk�1=2�q�k+1=2xk+1+i [a0+k�1]qk�1=2�q�k+1=2xk�1; k 6=1:

R(i;�)a0 (I3)x1=� [a0]

q1=2�q�1=2x1�i q[a0�1]q1=2�q�1=2x2;

R(i;�)a0 (I3)xk=�i qk[a0�k]

qk�1=2�q�k+1=2xk+1�i q�k+1[a0+k�1]

qk�1=2�q�k+1=2xk�1; k 6=1:

and by the formulas

R(�i;�)a0 (I1)xk=

qk�1=2+q�k+1=2

q�q�1 xk; R(�i;�)a0 (I2)=�R(i;�)

a0 (I2);

R(�i;�)a0 (I3)x1=� [a0]

q1=2�q�1=2x1+i q[a0�1]q1=2�q�1=2x2;

R(�i;�)a0 (I3)xk=i qk[a0�k]

qk�1=2�q�k+1=2xk+1+i q�k+1[a0+k�1]

qk�1=2�q�k+1=2xk�1; k 6=1:

12.2. Theorem. The representations R(i;�)a0 R

(�i;�)a0 are irreducible and pairwise

43

nonequivalent. For any a the irreducible representation Ra� is not equivalent to someof these representations.Proof. is given in the same way as in the finite dimensional case (see the proof of 10.2).

The algebra Uq(sl2) has also irreducible infinite dimensional representations withhighest weights or with lowest weights. They are classified in Ref. [18]. All of these

representations T can be extended to the algebra Uq(sl2). Using the compositionR=T � we obtain the corresponding representations R of Uq(so3). As above, it can be

easily proved that to nonequivalent representations T of Uq(sl2) with highest or lowestweight there correspond nonequivalent irreducible representations of Uq(so3). We givea list of these representations.Let l= 1

2 ;1;32;2;��. We denote byR+

l the representation of Uq(so3) acting on theHilbert space Hl with the orthonormal base vectors xm,m= l;l+1;l+2;��, and givenby formulas (12.2)--(12.4) with a=�l. By R�l we denote the representation of

Uq(so3) acting on the Hilbert space Hl with the orthonormal base vectors xm,m=�l;�l�1;�l�2;��, and given by formulas (12.2)--(12.4) with a= l.Now let a 6=0 (modZ) and a 6= 1

2 (modZ). We denote byHa the Hilbert space with theorthonormal basis xm,m=�a;�a+1;�a+2;��. On this space the representationR+a acts which is given by formulas (12.2)--(12.4). On the Hilbert space Ha with the

orthonormal basis xm,m=a;a�1;a�2;��, the representation R�a acts which is given byformulas (12.2)--(12.4).

12.3. Theorem. The above representations R�l and R�a are irreducible and pairwisenonequivalent.Proof. of this proposition is contained in Ref. [9].

13. Other infinitedimensional representationsThe algebra Uq(so3) has also irreducible infinite dimensional representations which

cannot be obtained from representations of Uq(sl2). We describe these representationsin this section.Let H be the infinite dimensional vector space with the basis xm,m=0;�1;�2, ��,and let �=q� be a nonzero complex number such that 0�Re� <1. Then a directcalculation shows that the operators Q+

� (I1) and Q+� (I2) given by the formulas

Q+� (I1)xm= �qm+��1q�m

q�q�1 xm; (13.1)

Q+� (I2)xm= 1

q�q�1xm+1+1

q�q�1 xm�1

satisfy the relations (8.1) and (8.2) and hence determine a representation of Uq(so3)which will be denoted by Q+

� . Similarly, the operators Q�� (I1) and Q�� (I2) given on the

space H by

Q�� (I1)xm=��qm+��1q�m

q�q�1 xm; Q�� (I2):=Q+� (I2)

determine a representation of Uq(so3) which is denoted by Q�� . The operators Q�� (I3)

can be calculated by means of formula (1.2).

13.1. Theorem. If � 6=1 and � 6=q1=2, then the representations Q+� and Q�� are

irreducible. The representations Q�1 and Q�pq are reducible.

Proof. The first part is proved in the same way as that of 10.1. Let us prove the secondpart. The representations Q�1 and Q�pq are the only representations in the set fQ�� g for

44

which the operator Q�� (I1) has not a simple spectrum. The operators Q�1 (I1) has thespectrum

��; q�2+q2; q�1+q; 2; q+q�1; q2+q�2; ��:Thus, only the spectral point 2 has multiplicity 1. All other points have multiplicity 2.Let V1 and V2 be the vector subspaces ofH with the bases

x0; x0m=xm�x�m; m=1;2;��;

and

x00m=xm+x�m; m=1;2;��;respectively. These basis vectors are eigenvectors of the operator Q�1 (I1):

Q�1 (I1)x0m=�qm+q�m

q�q�1 x0m; Q�1 (I1)x00m=�qm+q�m

q�q�1 x00m;

and

Q�1 (I2)x0=1

q�q�1x01; Q�1 (I2)x

001 =

1q�q�1 x

002 ;

Q�1 (I2)x0m= 1

q�q�1 x0m+1+

1q�q�1x

0m�1; m>0;

Q�1 (I2)x00m= 1

q�q�1 x00m+1+

1q�q�1x

00m�1; m>1:

Thus, the subspaces V1 and V2 are invariant with respect to the representation Q+1 (and

the representation Q�1 ). We denote the subrepresentations of Q�1 realized on V1 and V2

by Q1;�1 and Q2;�

1 , respectively.The eigenvalues of the operators Q�pq(I1) are

��; q�3=2+q3=2; q�1=2+q1=2; q1=2+q�1=2; q3=2+q�3=2; ��:Thus, every spectral point has multiplicity 2. We denote byW1 andW2 the vectorsubspaces ofH spanned by the basis vectors

x01=2=x0�x�1; x03=2=x1�x�2;��;x0m+12=xm�x�m�1;�� (13.2)

and

x001=2=x0+x�1; x003=2=x1+x�2;��;x00m+1

2=xm+x�m�1;��; (13.3)

respectively. These basis vectors are eigenvectors of the operator Q�pq(I1):

Q�pq(I1)x0m+1

2=�qm+1=2+q�m�1=2

q�q�1 x0m+1

2; (13.4)

Q�pq(I1)x00m+1

2=�qm+1=2+q�m�1=2

q�q�1 x00m+1

2(13.5)

and

Q�pq(I2)x012=� 1

q�q�1x012+ 1

q�q�1 x032; (13.6)

Q�pq(I2)x0m+1

2= 1

q�q�1x0m+3

2+ 1

q�q�1 x0m�1

2; m>0; (13.7)

Q�pq(I2)x0012= 1

q�q�1x0012+ 1

q�q�1 x0032; (13.8)

Q�pq(I2)x00m+1

2= 1

q�q�1x00m+3

2+ 1

q�q�1 x00m�1

2; m>0: (13.9)

Thus, the subspacesW1 andW2 are invariant with respect to the representations Q�pq.

We denote the subrepresentations of Q�pq realized onW1 andW2 by Q1;�pq and Q2;�p

q ,

respectively. Proposition is proved.13.2. Theorem. The representations Q1;�

1 , Q2;�1 , Q1;�p

q and Q2;�pq are irreducible and

45

pairwise nonequivalent. For any admissible value of � the representation Q+� (as well

as the representation Q�� ) is not equivalent to some of these representations.Proof. Proof is similar to that of 10.2 if to take into account spectra of the operatorsQ1;�1 (I1), Q

2;�1 (I1), Q

1;�pq (I1), Q

2;�pq (I1), Q

�� (I1) and traces of the operators Q1;�

1 (I2),

Q2;�1 (I2), Q

1;�pq (I2), Q

2;�pq (I2).

14. Finite dimensional representations at root of unityEverywhere below q is a root of unity, that is, there is a smallest positive integer p suchthat qp=1. We suppose that p 6=1;2. We introduce the number p0 setting p0=p if p isodd and p0=p=2 if p is even.As in the case of the algebra Uq(sl2) (see Ref. [17], Chapter 3), if q is a root of unity,then we claim Uq(so3) is a finite dimensional vector space over the center of Uq(so3).If q is a primitive root of unity, then this assertion is stated in Ref. [14]. If q is any rootof unity, then this assertion may be proved in the following way. If qp=1, then weclaim that the center C of Uq(so3) contains the elements

Pp=Ipj+aI

p�2j +bIp�4j +��+dIrj ; j=1;2;3;

where r=0 if p is even and r=1 if p is odd and a;b;��;d are certain fixed complexnumbers expressed in terms of q. (They are the polynomials P defined in Ref.[14] if q is a primitive root of unity. Unfortunately, we could not find the explicitexpressions for the coefficients a;b;��;d. But note that P3=I3j+Ij, P4=I4j+I2j andP5=I

5j+(1+(q+q�1)�1I3j+(q+q�1)�1Ij.) Therefore, Isj , s>n, can be reduced to the

linear combination of Iij , i<n, with coefficients from the center C. Now our assertionfollows from this and from Poincar�e--Birkhoff--Witt theorem for Uq(so3). Thus,following seems to be true:14.1. Theorem. If q is a root of unity, then any irreducible representation of Uq(so3) isfinite dimensional.Proof. (using the claim) Let T be an irreducible representation of Uq(so3). Then Tmaps central elements into scalar operators. Since according to our claim the linearspace Uq(so3) seems to be finite dimensional over the center C with the basis Ik1 I

m2 I

n3 ,

k;m;n<p, then for any a2Uq(so3) we have T (a)=X

k;m;n<p

T (Ik1 Im2 I

n3 ). Hence, if v is a

nonzero vector of the representation space V, then T (Uq(so3))v=V and V is finitedimensional. If our claim is valid, theorem is proved.Taking into account 14.1, below we consider only finite dimensional representations ofUq(so3).In order to find irreducible representations of Uq(so3) for q a root of unity, we use thesame method as before, that is, we apply the homomorphism from 9.1 and irreduciblerepresentations of the algebra Uq(sl2) for q a root of unity.

Let us find irreducible representations of Uq(sl2) for q a root of unity. The quantumalgebra Uq(sl2) for q a root of unity has the following irreducible representations (seeRef. [17], Subsection 3.3.2):

(a) The representations T(1)l , T

(�1)l , T

(i)l , T

(�i)l , 2l<p0, given by the formulas

(8.8)--(8.12).(b) The representations Tab�, a;b;�2C , � 6=0, acting on a p0-dimensional vector spaceH with the basis xj , j=0;1;2;��;p0�1, and given by the formulas

46

Tab�(qH)xi=q�i�xi; Tab�(F )xp0�1=bx0; (14.1)Tab�(F )xi=xi+1; i<p

0�1; Tab�(E)x0=axp0�1; (14.2)

Tab�(E)xi=�ab+[i]�

2q1�i��2qi�1

q�q�1

�xi�1; i>0: (14.3)

The representations Tab� with (a;b)=(0;0) and �=�qn, n=0;1;2;��;p0�2, are reducibleand must be taken out from this set.

(c) The representations T 00b�, b;�2C , � 6=0, acting on a p0-dimensional vector spaceHwith the basis xj , j=0;1;2;��;p0�1, and given by the formulas

T 00b�(qH)xi=qi��1xi; T 00b�(E)xp0�1=bx0; (14.4)

T 00b�(E)xi=xi+1; i<p0�1; T0b�(F )x0=0; (14.5)

T 00b�(F )xi=[i]�2q1�i��2qi�1

q�q�1 xi�1; i>0: (14.6)

The representations T 000� with �=�qn, n=0;1;2;��;p0�2, are reducible and must betaken out from this set.14.2. Note. In the set of representations (a)--(c) there exist equivalent representations(see, for example, Propositions 3.17 and 3.18 in Ref. [17]).

14.3. Note. In Ref. [17], Subsection 3.3.2, irreducible representations of the algebragenerated by the elements E;F;K:=q2H;K�1:=q�2H 2Uq(sl2) are given. Clearly, thisalgebra is a subalgebra in Uq(sl2). It is easy to generalize the results of Subsection 3.3.2in Ref. [17] for Uq(sl2). Let us note that the algebra Uq(sl2) has a unique automorphism' such that '(qH)=iqH , '(E)=�E and '(F )=F . (If q is not a root of unity, then

this automorphism transforms the representations T(1)l to the representations T

(i)l ,

respectively.) Therefore, the mapping ~T�a;b;�=Tab��' is also a representation ofUq(sl2). We have

~Tab�(qH)xi=iq�i�xi; ~Tab�(F )xp0�1=bx0; (14.7)~Tab�(F )xi=xi+1; i<p

0�1; ~Tab�(E)x0=axp0�1; (14.8)

~Tab�(E)xi=�ab�[i]�2q1�i��2qi�1

q�q�1

�xi�1; i>0: (14.9)

However, it is easy to see by comparing (14.1)--(14.3) with (14.7)--(14.9) that the

representation ~Tab� is equivalent to Ta;b;i�. This means that for q a root of unity we donot obtain new representations of Uq(sl2) from Tab� applying the automorphism ' as in

the case of the representations T(1)l .

We have described irreducible representations of the algebra Uq(sl2). Now we wish to

extend these representations to obtain representations of the algebra Uq(sl2) by usingthe relation

T ((qkqH+q�kq�H)�1):=(qkT (qH)+q�kT (q�H))�1:

Clearly, only those irreducible representations T of Uq(sl2) can be extended to

Uq(sl2) for which the operators qkT (qH)+q�kT (q�H) are invertible. From formulas(8.8)--(8.12) it is clear that these operators are always invertible for the irreducible


(�1)l , l=0;12;1;

32;��;p

0�12 , and for the irreducible representations

T(i)l , T

(�i)l , l= 1

2;32 ;

52;��;p

0�12 (or p0�2

2 ). (For the representations T(i)l , T

(�i)l , l=0;1;2;��,

some of these operators are not invertible since they have zero eigenvalue.) We denote

the extended representations by the same symbols T(1)l , T

(�1)l , T

(i)l , T

(�i)l , respectively.

Similarly, the representation Tab� (and the representation T 00b�) can be extended to a

47

representation of the algebra Uq(sl2) if and only if � 6=�iqk, k2Z.14.4. Theorem. The algebra Uq(sl2) for q a root of unity has the irreducible


(�1)l , l=0;12;1;

32;��;p

0�12 , the irreducible representations T

(i)l ,

T(�i)l , l= 1

2;32;52;��p0�1

2(or p0�2

2), and the irreducible representations Tab�, T 00b�,

� 6=�iqk, k2Z. Any irreducible representation of Uq(sl2) for q a root of unity isequivalent to one of these representations.

15. Representations at root of unity obtained usinghomomorphismAs in previous sections, we shall obtain representations of Uq(so3) for q a root of unity

by applying the homomorphism from 9.1. Namely, if T is a representation of Uq(sl2),then

R=T � (15.1)

is a representation of Uq(so3). As in previous sections, application of this method to the

pair of the irreducible representations T(1)l and T

(�1)l of Uq(sl2) leads to the same

representation of Uq(so3) which will be denoted by R(1)l . Applying the formula (15.1)

to the irreducible representations T(i)l and T

(�i)l of Uq(sl2) give the representations of

Uq(so3) which will be denoted by R(i)l and R

(�i)l , respectively.

vetaproposition8 The representations R(1)l of Uq(so3) are irreducible. The


(�i)l are reducible.

Proof. of this proposition is the same as that of 10.1.Repeating word-by-word the reasoning of previous sections, we decompose the


(�i)l into the direct sums of representations of Uq(so3) which

are denoted by R(�i;+)n and R

(�i;�)n :

R(i)l =R

(i;+)n �R(i;�)

n ; R(�i)l =R

(�i;+)n �R(�i;�)

n ; n= l+12 :

Moreover, the representations R(�i;+)n and R

(�i;�)n are given in the appropriate bases

x1, x2, ��, xn by the corresponding formulas of previous sections.

15.1. Theorem. The representations R(i;+)n , R

(i;�)n , R

(�i;+)n , R

(�i;�)n , n=1;2;3;��;p0

2(or

p0�12 ) are irreducible and pairwise nonequivalent. For any l, l=0; 12 ;1;

32;��;p

0�12 , the

representation R(1)l is not equivalent to some of these representations.

Proof. is the same as that of 10.2.Now we apply formula (15.1) to the representations Tab� and T 00b�. As a result, weobtain the representations

Rab�=T�a;b;�i�� ; R00b�=T00;b;�i�

given in the bases xj , j=0;1;2;��;p0�1, by the formulas

Rab�(I1)xi=�1

q�q�1 (q�i�+qi��1)xi; (15.2)

Rab�(I2)x0=i

��1 (axp0�1+x1); (15.3)

Rab�(I2)xp0�1= iq�p0+1��qp0�1��1

�bx0+

48

+�ab+[p0�1]q�p

0+2�2�qp0�2��2

q�q�1

�xp0�2

�; (15.4)

Rab�(I2)xi=i

q�i��qi��1

��ab+[i]q

�i+1�2�qi�1��2

q�q�1

�xi�1+

+xi+1

�; 0<i<p0�1: (15.5)

and by the formulas

R00b�(I1)xi=1

q�q�1 (q�i�+qi��1)xi; R00b�(I2)x0=�i

��1x1;

R00b�(I2)xp0�1=�i

q�p0+1��qp0�1��1

�bx0+[p0�1]q�p

0+2�2�qp0�2��2

q�q�1 xp0�2

�;

R00b�(I2)xi=�i

q�i��qi��1

�xi+1+[i]q

�i+1�2�qi�1��2

q�q�1 xi�1

�; 0<i<p0�1:

The operatorsRab�(I3) and R00b�(I3) can be calculated by means of the relation

R(I3)=q1=2R(I1)R(I2)�q�1=2R(I2)R(I1):Recall that the representations Rab� and R

00b� are determined for � 6=0 and � 6=�qk,

k2Z.It is seen from the above formulas that

R00b�(I1)=R0;b;��(I1); R00b�(I2)=R0;b;��(I2);that is, the representations R0;b;�� and R00b� are equivalent. For this reason, weconsider below only the representations Rab�.In order to study the representations Rab� of Uq(so3) we consider the spectrum of theoperator Rab�(I1). It coincides with the set of points

��+��1

q�q�1 ; �q�1�+q��1

q�q�1 ; �q�2�+q2��1

q�q�1 ; �� ;�q1�p0

�+qp0�1��1

q�q�1 : (15.6)

It is easy to see that there exist coinciding points in this set if and only if � is equal toone of the numbers

�q1=2; �q3=2; �q5=2; �� ;�q(p0�1)=2 (or �q(p0�2)=2):(Here we have to take �q(p0�1)=2 if p0 is even and �q(p0�2)=2 if p0 is odd.) Moreover, the

set (15.6) splits into pairs of coinciding points if and only if �=�q(p0�1)=2. In all othercases there exists at least one spectral point which coincides with no other point. Inparticular, if �=�q(p0�2)=2, then in this set there exists only one eigenvalue withmultiplicity 1. In all other cases there are more than one eigenvalues with multiplicity1.15.2. Theorem. If � 6=�q(p0�1)=2 for even p0 and � 6=�q(p0�2)=2 for odd p0, then therepresentation Rab� is irreducible.Proof. Let � 6=�q(p0�1)=2 for even p0 and � 6=�q(p0�2)=2 for odd p0. We distinguish twocases: when the spectrum of the operator Rab�(I1) is simple and when there exists atlist one spectral point of this operator having multiplicity 2. In the first case the proofis the same as the first part of the proof of 10.1. For the second case, we give a proofonly for �=q1=2. (Proofs for other values of q are similar.) Then in the set (15.6)

there are only two coinciding points ��+��1

q�q�1 and �q�1�+q��1

q�q�1 corresponding to the

eigenvectors x0 and x1. Let V be an invariant subspace of the representation space H.As in the proof of 10.1, it is shown that V is a linear span of eigenvectors of the operatorRab�(I1), that is, a certain part of the vectors xi, i 6=0;1, �0x0+�1x1, �0x0+�1x1

49

constitutes a basis of V . Let V contain some basis vector xj . Then as in the proofof 10.1, acting successively upon xj by certain linear combinations of the operatorsRab�(I2) and Rab�(I3) we generate all the vectors xi, i=0;1;��;12(p0�1). This meansthat V =H and the representation Rab� is irreducible. If V contains no vector xj ,j 6=0;1, then some linear combination �0x0+�1x1 belongs to V . Then the vectorv=Rab�(I2)(�0x0+�1x1) belongs to V . Since v contains the summand �x2 withnonzero coefficient �, then x22V . This is a contradiction. Hence, the representationRab� is irreducible. Proposition is proved.Let p0 be even. Let us study the representations Rab� for �=�q(p0�1)=2. For�=q(p

0�1)=2 we haveRab�(I1)xi=

�1q�q�1 (q�i+(p0�1)=2+qi�(p

0�1)=2)xi; (15.7)

Rab�(I2)x0=c(p0�1)=2(axp0�1+x1); (15.8)

Rab�(I2)xp0�1=�c(p0�1)=2�(ab+[p0�1]2)xp0�2+bx0

�; (15.9)

Rab�(I2)xi=c�i+(p0�1)=2�(ab+[i]2)xi�1+xi+1

�; (15.10)

where

cj=i

qj�q�j :

The operator Ra;b;(p0�1)=2(I1) has the spectrum�1

q�q�1 (q�i+(p0�1)=2+qi�(p0�1)=2); i=0;1;2;��;p0�1;

that is, if p0 is even, then all spectral points are of multiplicity 2.We assume that ab 6=�[j]2, j=0;1;��;p0�1, and go over from the basis contaningvectors xi to the basis contaning vectors x�i , where

x�i =iY

j=0

(ab+[j]2)�1=2xi; i=0;1;2;��;p0�1:

Then the formula (15.7) does not change and the formulas (15.8)--(15.10) turn into

Rab�(I2)x�0=c(p0�1)=2

�a

p0�1Yj=1

(ab+[j]2)1=2x�p0�1+(ab+1)1=2x�1

�;

Rab�(I2)x�p0�1=�c(p0�1)=2�(ab+1)1=2x�p0�2+

bp0�1Yj=1

(ab+[j]2)1=2

x�0

�;

Rab�(I2)x�i =c�i+(p0�1)=2�(ab+[i]2)1=2x�i�1+(ab+[i+1]2)1=2x�i+1

�:

We split the representation space H into the direct sum of two linear subspacesH1 andH2 spanned by the basis vectors x0j , j=0;1;2;��;12(p0�2), and x00j , j=0;1;2;��;12(p0�2),where

x0j=x�j+i(�1)�j�1+p0=2x�p0�j�1; x00j =x

�j+i(�1)�j+p0=2x�p0�j�1:

Then as in previous sections, we derive

Ra;b;(p0�1)=2(I1)x0j=�1

q�q�1 (q�j+(p0�1)=2+qj�(p

0�1)=2)x0j ;

Ra;b;(p0�1)=2(I1)x00j =�1

q�q�1 (q�j+(p0�1)=2+qj�(p0�1)=2)x00j

for the operator Ra;b;(p0�1)=2(I1) and

Ra;b;(p0�1)=2(I2)x0j=c�j+(p0�1)=2�(ab+[j+1]2)1=2x0j+1+(ab+[j]2)1=2x0j�1

�;

50

Ra;b;(p0�1)=2(I2)x00j =c�j+(p0�1)=2�(ab+[j+1]2)1=2x00j+1+(ab+[j]2)1=2x00j�1

�;

where j 6=0;p0

2 �1,Ra;b;(p0�1)=2(I2)x0p0

2 �1= 1

q1=2�q�1=2 (ab+[p0

2]2)1=2x0p0

2 �1+

+ iq1=2�q�1=2 (ab+[p

0

2 �1]2)1=2x0p02 �2

;

Ra;b;(p0�1)=2(I2)x00p02 �1

=� 1q1=2�q�1=2 (ab+[p

0

2 ]2)1=2x00p0

2 �1+

+ iq1=2�q�1=2 (ab+[p

0

2 �1]2)1=2x00p02 �2

;

Ra;b;(p0�1)=2(I2)x00=c(p0�1)=2

�a

p0�1Yj=1

(ab+[j]2)1=2x�p0�1+(ab+1)1=2x�1

��

�i(�1)(p0�2)=2c(p0�1)=2�(ab+1)1=2x�p0�2+

bp0�1Yj=1

(ab+[j]2)1=2

x�0

�:

When

a

p0�1Yj=1

(ab+[j]2)1=2= bp0�1Yj=1

(ab+[j]2)1=2

; (15.11)

then the last relation reduces to

Ra;b;(p0�1)=2(I2)x00=(�1)(p0�2)=2

q(p0�1)=2�q�(p0�1)=2 a

p0�1Yj=1

(ab+[j]2)1=2x00�

�c(p0�1)=2(ab+1)1=2x01:Similarly, if the condition (15.11) is fulfilled, then

Ra;b;(p0�1)=2(I2)x000 =(�1)p0=2

q(p0�1)=2�q�(p0�1)=2 a

p0�1Yj=1

(ab+[j]2)1=2x000+

+c(p0�1)=2(ab+1)1=2x001 :Thus, the subspacesH1 and H2 are invariant with respect to the representationRa;b;(p0�1)=2 if the condition (15.11) is fulfilled. We denote the corresponding

subrepresentations by R1;+a;b;(p0�1)=2 and R

2;+a;b;(p0�1)=2, respectively.

Similarly, if �=�q(p0�1)=2, thenRa;b;�(p0�1)=2(I1)=�Ra;b;(p0�1)=2(I1); Ra;b;�(p0�1)=2(I2)==�Ra;b;(p0�1)=2(I2)

and the subspacesH1 and H2 are invariant with respect to the representationRa;b;�(p0�1)=2 if the condition (15.11) is fulfilled. We denote the corresponding

subrepresentations by R1;�a;b;�(p0�1)=2 and R

2;�a;b;�(p0�1)=2, respectively.

15.3. Theorem. Let the condition (15.11) is satisfied. Then the representations

Ri;+a;b;(p0�1)=2 and R

i;�a;b;�(p0�1)=2, i=1;2, of the algebra Uq(so3) are irreducible

and pairwise nonequivalent. If the condition (15.11) is not satisfied, then the

51

representations Ra;b;(p0�1)=2 and Ra;b;�(p0�1)=2 are irreducible.Proof. is similar to that of the previous propositions and we omit it.

Remark that the representations Ri;+a;b;(p0�1)=2 and R

i;�a;b;�(p0�1)=2, i=1;2, have two

nonzero diagonal matrix elements x#p02 �1

(Rxp0

2 �1) and x#0 (Rx0).

Let now p0 be odd and �=q(p0�2)=2. For this value of � we have

Rab�(I1)xi=�1

q�q�1 (q�i+(p0�2)=2+qi�(p

0�2)=2)xi;

Rab�(I2)x0=c(p0�2)=2(axp0�1+x1);Rab�(I2)xp0�1=�cp0=2((ab+�[p0�1][p0])xp0�2+bx0);Rab�(I2)xi=c�i+(p0�2)=2((ab+�[i][i+1])xi�1+xi+1);

where �=1 for p0=p=2, �=�1 for p0=p and cj is such as in (15.7)--(15.10). Theoperator Ra;b;(p0�2)=2(I1) has the spectrum

�1q�q�1 (q

�i+(p0�2)=2+qi�(p0�2)=2); i=0;1;2;��;p0�1;

that is, all spectral points are of multiplicity 2 except for the point�(qp0=2+q�p0=2)=(q�q�1) which is of multiplicity 1.

We assume that ab 6=��[j][j+1], j=0;1;��;p0�1, and go over from the basis containingvectors xi to the basis containing vectors x�i , where

x�i =iY

j=0

(ab+�[j][j+1])�1=2xi; i=0;1;2;��;p0�1:

Then

Rab�(I1)x�i =�1

q�q�1 (q�i+(p0�2)=2+qi�(p0�2)=2)x�i ;

Rab�(I2)x�0=c(p0�2)=2

�a

p0�1Yj=1

(ab+�[j][j+1])1=2x�p0�1+(ab+�[2])1=2x�1

�;

Rab�(I2)x�p0�1=�cp0=2((ab+�[p0�1][p0])1=2x�p0�2+

+b

p0�1Yj=1

(ab+�[j][j+1])�1=2x�0);

Rab�(I2)x�i =c�i+(p0�2)=2((ab+�[i][i+1])1=2x�i�1+(ab+�[i+1][i+2])1=2x�i+1);

where �=q(p0�2)=2. Let H1 andH2 be two linear subspaces of the representation space

H spanned by the basis vectors

x0j=x�j+i(�1)jx�p0�j�2; j=0;1;2;��;p0�32 ;

and the basis vectors

x00j =x�j+i(�1)j+1x�p0�j�2; j=0;1;2;��;p0�32 ;

respectively. Then the operator Ra;b;(p0�2)=2(I1) acts on the basis elements x0j and x00j as

on the vectors xj and

Ra;b;(p0�1)=2(I2)x0j = c�j+(p0�2)=2��(ab+�[j+1][j+2])1=2x0j+1+(ab+�[j][j+1])1=2x0j�1

�;

Ra;b;(p0�2)=2(I2)x00j = c�j+(p0�2)=2��(ab+�[j+1][j+2])1=2x00j+1+(ab+�[j][j+1])1=2x00j�1

�;

52

where j 6=0;p0�32 ,

Ra;b;(p0�2)=2(I2)x0p0�32

= (�1)(p0�3)=2

q1=2�q�1=2 (ab+�[p0�12 ][p

0+12 ])1=2x0p0�3

2

+

+ iq1=2�q�1=2 (ab+�[

p0�12 ][p

0�32 ])1=2x0p0�5

2

;

Ra;b;(p0�2)=2(I2)x00p0�32

=� (�1)(p0�3)=2

q1=2�q�1=2 (ab+�[p0�12 ][p

0+12 ])1=2x00p0�3

2

+

+ iq1=2�q�1=2 (ab+�[

p0�12

][p0�32

])1=2x00p0�52

;

Ra;b;(p0�2)=2(I2)x00= c(p0�2)=2

�a

p0�1Yj=1

(ab+�[j][j+1])1=2x�p0�1+

+(ab+�[2])1=2x�1�i(ab+�[2])1=2x�p0�1�i(ab+�[p0�2][p0�1])1=2x�p0�3�;

Ra;b;(p0�2)=2(I2)x000 = c(p0�2)=2

�a

p0�1Yj=1

(ab+�[j][j+1])1=2x�p0�1+

+(ab+�[2])1=2x�1+i(ab+�[2])1=2x�p0�1+i(ab+�[p0�2][p0�1])1=2x�p0�3�:

If

a

p0�1Yj=1

(ab+�[j][j+1])1=2+i(ab+�[2])1=2=0; (15.12)

(ab+�[2])1=2p0�1Yj=1

(ab+�[j][j+1])1=2=ib; (15.13)

then

Ra;b;(p0�2)=2(I2)xp0�1=�bcp0=2

p0�1Yj=1

(ab+�[j][j+1])1=2

x00;

Ra;b;(p0�2)=2(I2)x00=i(ab+�[2])1=2

q(p0�2)=2�q�(p0�2)=2x01+cx

0p0�1;

Ra;b;(p0�2)=2(I2)x000 =i(ab+�[2])1=2

q(p0�2)=2�q�(p0�2)=2x001 ;

where c is a nonzero coefficient easily determined from the above formulas. Hence, thesubspacesH1+C xp0�1 and H2 of the representation space are invariant with respect tothe representation Ra;b;(p0�2)=2 (we denote these subrepresentations by R1

a;b;(p0�2)=2and R2

a;b;(p0�2)=2, respectively). Remark that

dimH1+C xp0�1= 12(p

0+1); dimH2=12(p

0�1):If

a

p0�1Yj=1

(ab+�[j][j+1])1=2�i(ab+�[2])1=2=0; (15.14)

(ab+�[2])1=2p0�1Yj=1

(ab+�[j][j+1])1=2=�ib; (15.15)

53

then

Ra;b;(p0�2)=2(I2)xp0�1=�bcp0=2

p0�1Yj=1

(ab+�[j][j+1])1=2

x000 ;

Ra;b;(p0�2)=2(I2)x00=i(ab+�[2])1=2

q(p0�2)=2�q�(p0�2)=2x01;

Ra;b;(p0�2)=2(I2)x000 =i(ab+�[2])1=2

q(p0�2)=2�q�(p0�2)=2x001+cxp0�1;

where c is a nonzero coefficient. Hence, now the subspacesH1 and H2+Cxp0�1 ofthe representation space are invariant. We denote the subrepresentations on thesesubspaces by R1

a;b;(p0�2)=2 and R2a;b;(p0�2)=2, respectively). Note that the representation

R1a;b;(p0�2)=2 is not equivalent to R

2a;b;(p0�2)=2 (and the representation R2

a;b;(p0�2)=2 is

not equivalent to R1a;b;(p0�2)=2) since the parameters a and b determining these

representations satisfy different equations.If a and b do not satisfy the relations (15.12) and (15.13) or the relations (15.14) and(15.15), then the representation Ra;b;(p0�2)=2 is irreducible.

Let now p0 be odd and �=�q(p0�2)=2. In this case, the representation Ra;b;�(p0�2)=2 isirreducible if a and b do not satisfy the relations (15.12) and (15.13) or the relations(15.14) and (15.15). If a and b satisfy the relations (15.12) and (15.13), thenRa;b;�(p0�2)=2 is a reducible representation and decomposes into the direct sumof two subrepresentations acting on the subspacesH1+Cxp0�1 andH2. Thesesubrepresentations are denoted by R1

a;b;�(p0�2)=2 and R2a;b;�(p0�2)=2, respectively, and

are determined as

Ria;b;�(p0�2)=2(I1)=�Ri

a;b;(p0�2)=2(I1); Ria;b;�(p0�2)=2(I2)=

=�Ria;b;(p0�2)=2(I2); i=1;2:

Similarly, if a and b satisfy the relations (15.14) and (15.15), then Ra;b;�(p0�2)=2 is areducible representation and decomposes into the direct sum of two subrepresentationsacting on the subspacesH1 and H2+C xp0�1. These subrepresentations are denoted by

R1a;b;�(p0�2)=2 and R

2a;b;�(p0�2)=2, respectively, and are determined as

Ria;b;�(p0�2)=2(I1)=�Ri

a;b;(p0�2)=2(I1); Ria;b;�(p0�2)=2(I2)=

=�Ria;b;(p0�2)=2(I2); i=1;2:

15.4. Theorem. Let the conditions (15.12) and (15.13) are satisfied. Then therepresentations R1

a;b;(p0�2)=2, R2a;b;(p0�2)=2, R

1a;b;�(p0�2)=2 and R

2a;b;�(p0�2)=2 are

irreducible and pairwise nonequivalent. If the conditions (15.14) and (15.15)

are satisfied, the representations R1a;b;(p0�2)=2, R

2a;b;(p0�2)=2, R

1a;b;�(p0�2)=2 and

R2a;b;�(p0�2)=2 are irreducible and pairwise nonequivalent.

Proof. is similar to that of the previous propositions and we omit it.

16. Other representations at root of unityIn the previous section we described irreducible representations of Uq(so3) obtained

from irreducible representations of the algebra Uq(sl2) for q a root of unity. However, atq a root of unity the algebra Uq(so3) has irreducible representations which cannot be

54

derived from those of Uq(sl2). They are obtained as irreducible components of therepresentations Q� from Section VII when one put q equal to a root of unity. Wedescribe these representations of Uq(so3) in this section.

Let �=q� be a nonzero complex number such that 0�Re� <1 and let H be thep0-dimensional complex vector space with basis

xm; m=0;1;2;��;p0�1:We define on this space the operatorsQ0�(I1) and Q

0�(I2) determined by the formulas

Q0�(I1)xm= �qm+��1q�m

q�q�1 xm;

Q0�(I2)x0=1

q�q�1 x1+1

q�q�1xp0�1;

Q0�(I2)xp0�1=1

q�q�1xp0�2+ 1q�q�1 x0;

Q0�(I2)xm= 1q�q�1xm�1+ 1

q�q�1xm+1; m 6=0;p0�1:A direct computation shows that these operators satisfy the relations (8.1) and (8.2)and hence determine a representation of Uq(so3) which will be denoted byQ0�.

16.1. Theorem. If � 6=1 and � 6=q1=2, then the representation Q0� is irreducible.Proof. of this proposition is the same as that of the first part of 10.1.The representations Q01 and Q

0pq are studied in the same way as the representations Q1

and Qpq in previous sections. This study leads to the irreducible representations ofUq(so3) which are described below. (Note that the description of these representationsfor p0 even and for p0 odd is deferent.)Let p0 be odd. We denote byHr and Hs, r=

12(p

0+1), s= 12(p

0�1), the complex vectorspaces with the bases

x0; x1; x2; �� ; x12 (p

0�1) and x1; x2; �� ; x12 (p

0�1);

respectively. Four representations Q�;�1 act on the spaceHr and are given by theformulas

Q+;�1 (I1)xm= qm+q�m

q�q�1 xm; m=0;1;2;��; 12(p0�1); (16.1)

Q+;�1 (I2)x1

2 (p0�1)=� 1

q�q�1x12 (p

0�1)+1

q�q�1 x12 (p

0�3); (16.2)

Q+;�1 (I2)xm= 1

q�q�1 xm+1+1

q�q�1 xm�1; m< 12(p

0�1); (16.3)

and by the formulas

Q�;�1 (I1)xm=�qm+q�m

q�q�1 xm; m=0;1;2;��;12(p0�1); (16.4)

Q�;�1 (I2):=Q+;�1 (I2): (16.5)

Note that the upper sign corresponds to the representations Q+;+1 and Q�;+1 and the

lower sign to the representations Q+;�1 and Q�;�1 .

On the space Hs, four representations Q�;�1 act by the corresponding formulas

(16.1)--(16.5), but nowm runs over the values 1;2;3;��;12(p0�1).Let nowH 0

r and H0s, r=

12(p0+1), s= 1

2(p0�1), be the complex vector spaces with the

bases

xm+12; m=0;1;2;��; 12(p0�1); and xm+1

2; m=0;1;2;��; 12(p0�3);

respectively. The four representations Q�;�pq act on the space H 0

r and are given by the

formulas

55

Q+;�pq (I1)xm+1

2= qm+1=2+q�m�1=2

q�q�1 xm+12; m=0;1;2;��;12(p0�1); (16.6)

Q+;�pq (I2)x1

2=� 1

q�q�1x12+ 1

q�q�1 x32; (16.7)

Q+;�pq (I2)xm+1

2= 1

q�q�1xm+32+ 1

q�q�1xm�12; m 6=0; (16.8)

where xm+32�0 ifm= 1

2(p0�1), and by the formulas

Q�;�pq (I1)xm+1

2=�qm+1=2+q�m�1=2

q�q�1 xm+12; m=0;1;2;��;12(p0�1); (16.9)

Q�;�pq (I2):=Q

+;�pq (I2): (16.10)

On the space H 0s, four representations �Q�;�p

q act by the corresponding formulas

(16.6)--(16.10), but nowm runs through the values 0;1;2;��;12(p0�3).Let now p0 be even. We denote byHr and Hs, r=

12(p

0+2), s= 12(p

0�2), the complexvector spaces with the bases

x0; x1; x2; �� ; x12p

0 and x1; x2; �� ; x12 (p

0�2);

respectively. The representations Q1;�1 and Q2;�

1 act onHr and Hs, respectively, whichare given by the formulas

Qi;�1 (I1)xm=�qm+q�m

q�q�1 xm; i=1;2;

Qi;�1 (I2)xm= 1

q�q�1 xm+1+1

q�q�1xm�1; i=1;2;

where xm+1 or xm�1 must be put equal to 0 if the corresponding vector does not exist.Let Hp0=2 be the complex vector space with the basis

xm+12; m=0;1;2;��; 12(p0�2):

Four representations Q�;�pq act on this space which are given by the formulas

Q+;�pq (I1)xm+1

2= qm+1=2+q�m�1=2

q�q�1 xm+12;

Q+;�pq (I2)x1

2=� 1

q�q�1x12+ 1

q�q�1x32;

Q+;�pq (I2)x1

2 (p0�2)=� 1

q�q�1x12 (p

0�2)+1

q�q�1x12 (p

0�4);

Q+;�pq (I2)xm+1

2= 1

q�q�1xm�12+ 1

q�q�1xm+32; m 6= 1

2;x1

2 (p0�2);

and by the formulas

Q�;�pq (I1)xm+1

2=�qm+1=2+q�m�1=2

q�q�1 xm+12;

Q�;�pq (I2)= Q

+;�pq (I2):

Let us mention peculiarities of the representations described above. The operatorsQ�;�1 (I2), Q

�;�pq (I2), �Q

�;�pq and Q�;�p

q (I2) have nonzero diagonal matrix elements and

nonzero traces. Moreover, the operators Q�;�pq (I2) have two such diagonal elements.

Spectra of the operators Q�;�1 (I1), Q�;�pq (I1), Q

1;�1 (I1), Q

2;�1 (I1) and Q

�;�pq (I1) are not

symmetric with respect to the zero point.16.2. Theorem. The representations Q�;�1 , Q�;�p

q , �Q�;�pq , Q1;�

1 , Q2;�1 , Q�;�p

q are

irreducible and pairwise nonequivalent. No representation Q0� is equivalent to any ofthese representations.Proof. is the same as that of 10.1.

56

Bibliography[1] M. Havlí£ek, A. U. Klimyk, S. Po²ta, Representations of the cyclically symmetric

q-deformed algebra so(3). (submitted to J. Phys. A, in preparation)[2] Bagro O. V., Krugl�k S. A., Predstavleni� algebr D. Fairlie, preprint,

Kiev, 1996 g.[3] Nathan Jacobson, Lie algebras, Interscience Publishers, A division of John Wiley

and Sons, New York� London 1963.[4] D. B. Fairlie, Quantum deformations of SU(2), J. Phys. A: Math Gen. 23 (1990)

L183-L187.[5] Fabian H. L. Essler and Vladimir Rittenberg, Representations of the quadratic

algebra and partially asymmetric diffusion with open boundaries. Preprint ofPhysikalisches Institut der Universität Bonn, Germany, 1995.

[6] A. S. �edanov, Skryta� simmetri� polinomov Aski-Vil~sona,Teoretiqeska� i matematiqeska� fizika 89, No. 2, no�br~, 1991 g.

[7] M. Havlí£ek, A. U. Klimyk, E. Pelantová, Fairlie algebra Uq(so3): tensorproducts, oscillator realizations, root of unity, Hadronic Journal 20, 1997.

[8] A. Braverman, D. Gaitsgory, Poincaré-Birkhoff-Witt theorem for quadraticalgebras of Koszul type, preprint of School of Mathematical Sciences, Tel AvivUniversity, Israel, 1994.

[9] A. M. Gavrilik, A. U. Klimyk, Representations of the q-deformed algebrasUq(so2;1) and Uq(so3;1), J. Math. Phys 35 (2), February 1994.

[10] J. V. Kadeisvili, An Introduction the The Lie-Santilli Isotopic Theory,Mathematical Methods in the Applied Sciences, Vol. 19, pp. 1349-1395, 1996.

[11] R. M. Santilli, Problematic aspects of classical and quantum deformations,preprint No 03-65/1997 of institute for Basic Research, Palm Harbor, USA.

[12] R. M. Santilli, Isotopic Grand unification with the Inclusion of Gravity,contributed paper for the VIII Marcel Grossmann Meeting on General Relativity,Jerusalem, June 23-27, 1997

[13] R. M. Santilli, Nouvo Cim. 51, 570 (1967).[14] M. Odesski, Function Analysis Appl. 20, No. 2, 78 (1986).[15] Yu. S. Samoilenko and L. B. Turovska, Semilinear relations and *-representations

of deformations of so(3), in book Quantum Groups and Quantum Spaces, BanachCenter Publications, vol. 40, Warzsaw, 1997, pp. 21-40.

[16] J. Dixmier, Algébres Enveloppantes, Gauthier-Villars, Paris, 1974.[17] A. Klimyk and K. Schmüdgen, Quantum Groups and Their Representations,

Springer, Berlin, 1997.[18] I. M. Burban and A. U. Klimyk, J. Phys. A 26, 2139 (1993).

57

cvut.czContents 1. Introduction. 2 2. Basic preliminary. 4 3. PBW theorem for universal enveloping...

Documents

Transcript of cvut.czContents 1. Introduction. 2 2. Basic preliminary. 4 3. PBW theorem for universal enveloping...