The adjoint representation of quantum algebra U q

Transcription

1 Journal of Nonlinear Mathematical Physics Volume * Number * 20** 1 14 Article The adjoint representation of quantum algebra U q sl2 Č Burdík a O Navrátil b and S Pošta c a Department of Mathematics Czech Technical University Faculty of Nuclear Sciences and Physical Engineering Czech Republic a burdik@kmlinuxfjficvutcz c severin@km1fjficvutcz b Department of Mathematics Czech Technical University Faculty of Transportation Sciences Czech Republic navratil@fdcvutcz Received Month * 200*; Accepted in Revised Form Month * 200* Abstract Starting from any representation of the Lie algebra g on the finite dimensional vector space V we can construct the representation on the space AutV These representations are of the type of ad That is one of the reasons why it is important to study the adjoint representation of the Lie algebra g on the universal enveloping algebra Ug A similar situation is for the quantum groups U q g In this paper we study the adjoint representation for the simplest quantum algebra U q sl2 in the case that q is not a root of unity 1 Introduction It is a well-known fact that the adjoint representation of the algebra sl2 on its enveloping algebra U sl2 is fully reducible and can be decomposed into a direct sum of finite dimensional representations 1 Specifically let the generators e f and h fulfil the commutation relations h e = 2e h f = 2f e f = h Then the center U sl2 is generated by an element C = ef + fe h2 and for any n k N 0 N 0 = the vector spaces V nk = ad Usl2 e n C k Copyright c 2008 by Č Burdík O Navrátil and S Pošta

2 2 Č Burdík O Navrátil and S Pošta are invariant with respect to the adjoint representation and 2n + 1 dimensional Moreover we can write U sl2 = nk=0 V nk 11 A similar situation is for complex semisimple Lie algebras see eg On the other hand in the case of the quantum group U q sl2 there exists an element a for which the space ad Uqsl2a is infinite dimensional The aim of this paper is to find the decomposition of the quantum group U q sl2 where q is not a root of unity which is similar to the decompositio1 2 The quantum group U q sl2 Let as have the quantum group U q sl2 5 which is generated by the elements E F K and K 1 fulfilling the relations KE = q 2 EK KF = q 2 F K EF F E = K K 1 q q 1 KK 1 = K 1 K = 1 21 The coproduct the antipod S and the counit ɛ are defined by E = E K + 1 E SE = EK 1 ɛe = 0 F = F 1 + K 1 F SF = KF ɛf = 0 K = K K SK = K 1 ɛk = 1 K 1 = K 1 K 1 S K 1 = K ɛ K 1 = 1 It is a well-known fact that the basis in U q sl2 consists of the elements E F n 2 K n 3 where n 2 N 0 and n 3 Z The adjoint action of the Drinfeld Jimbo algebras is given by the formula see eg 6 ad a b = a 1 bs a 2 In particular in the case of the algebra U q sl2 we have ad E a = EaK 1 aek 1 ad F a = F a K 1 akf ad K a = KaK 1 ad K 1a = K 1 ak 22 This shows that ad E E = ad F KF = 0 holds We denote = E Y = KF Z = ad E Y = q 2 EF F E = K K 1 q q 1 q 1 q q 1 EF 23 and take the element C = EF + q 1 K + qk 1 q q

3 The adjoint representation of quantum algebra U q sl2 3 which generates the center of U q sl2 From equations 23 and 24 we obtain K = q q 1 q + q 1 qz + q q 1 C 25 Notice that the element K 1 can not be to expressed by these equations 3 The new basis in U q sl2 The next step is to express U q sl2 using the elements Y Z C and W = K 1 From relations and the commutation relations 21 we obtain C = C CY = Y C CZ = ZC CW = W C W = q 2 W W Y = q 2 Y W W Z = ZW Z = q 2 Z + q q 1 2 C Y Z = q 2 ZY + q q 1 2 Y C Y Y + q q 1 q + q 1 qz + q q 1 C Z = 0 qz + q q 1 C W q + q 1 q q 1 = 0 qz + q q 1 C q 1 Z q q 1 C + q q + q 1 2 q + q 1 2 Y + q q 1 = 0 31 The original elements E F K and K 1 in U q sl2 can be written in the following way: E = F = W Y K = q q 1 q + q 1 qz + q q 1 C K 1 = W If we denote by T a free algebra generated by Y Z C and W and take the two-sided ideal J in T generated by relations 31 the algebra U q sl2 is equal to T /J It is easy to show the following analogy of the Poincaré Birkhoff Witt theorem: Lemma The elements Z n 3 where n k 0 n 2 = 0 and n 3 n 5 = 0 form the basis in U q sl2 4 The adjoint representation in the new basis The adjoint representation 22 can be rewritten in the form ad E a = aw aw ad F a = W Y a W ay ad K a = q q 1 q + q 1 qz + q q 1 C aw ad K 1a = q q 1 q + q 1 W a qz + q q 1 C 41

4 4 Č Burdík O Navrátil and S Pošta 41 The finite dimensional representations From relations 31 and 41 it is possible by direct calculations to show that the vector space V N with the basis Z n 3 where n 2 = 0 and + n 2 + n 3 + n 4 N is invariant with respect to the adjoint representation for any N The dimension of this space is dim V N = 1 6 N + 1N + 22N + 3 In the space V N there are the highest weight elements + n 4 N It is well known see eg 5 p 61 that the invariant subspace V n1 n 4 = ad Uq sl2 C n 4 has the dimension and is irreducible Since = 1 6 N + 1N + 22N + 3 = dim V N +n 4 N the subspace V fin with the basis Z n 3 n 2 = 0 can be decomposed in to a direct sum of invariant subspaces V fin = n 4 V n1 n 4 The eigenvalue of ad C on the space V n1 n 4 is determined by the relation ad C = q2+1 + q 2 1 q q The infinite dimensional representations By direct calculations we obtain the action of the adjoint representation on the elements of the basis and for n 5 1 of the following form in the formulas where 1 resp 1 occurs we suppose that > 0 resp n 2 > 0: ad E W n 5 = 1 q 2n 5 n1+1 W n5+1 ad F W n 5 = q 2+n 5 +1 n 5 q q 1 n1 1 W n q 2+2n 5 1 C n4+1 qn n 5 q q 1 n1 1 W n5 1 ad K W n 5 = q 2 ad K 1 W n 5 = q 2 ad E Y n 2 W n 5 = q n 5 1 n 5 q q 1 Y n2 1 W n q 2n 2 q 2n 5 Y n2 1 C n q 2n 2 n n 2 n 5 q q 1 Y n2 1 W n5 1 ad F Y n 2 W n 5 = q 2n q 2n 5 Y n2+1 W n5+1

5 The adjoint representation of quantum algebra U q sl2 5 ad K Y n 2 = q 2n 2 ad K 1 Y n 2 = q 2n 2 By using these relations we obtain ad C = q2 2n q 2+2n 5 1 q q q 2 n 5 n q n 5 n1 n 5 q q ad C Y n 2 = q2n 2 2n q 2n 2+2n 5 1 q q q 2n 2 n 5 n q n 2 n 5 n2 n 5 q q 1 2 Y n The eigenvalues and the eigenfunctions of ad C Since V fin is invariant subspace we can define factor representation of ad on the space W = U q sl2 /Vfin We denote this representation by the same symbol ad no confusion arises from this abuse of notation In the space W we introduce the notation T n = n 1! W n for n 1 and the basis consisting of elements T n5 T n5 = n 5 1! = n 5 1! q 2n 2n 5 In this basis the representation has the form ad E T n5 = q n 5 q q 1 +1 T n5 +1 ad F T n5 C n 4 = q 2+n 5 +1 q q 1 n1 1 T n q +n 5 n1 n 5 q q 1 n1 1 T n5 C n4+1 qn n 5 n5 1 q q 1 n1 1 T n5 1 ad K T n5 C n 4 = q 2 T n5 ad K 1 T n5 = q 2 T n5 ad E Tn5 C n 4 = q 2n 2+n 5 +1 q q 1 T n5 +1Y n2 1 q n 2+n 5 n2 n 5 q q 1 T n5 Y n2 1 C n qn 5 1 2n 2 n 5 n5 1 q q 1 T n5 1Y n2 1 ad F Tn5 C n 4 = q n 5 q q 1 T n5 +1Y n2+1

6 6 Č Burdík O Navrátil and S Pošta ad K Tn5 = q 2n 2 T n5 ad K 1 Tn5 = q 2n 2 T n5 ad C T n5 = q2 2n q 2+2n 5 1 q q 1 2 T n5 + q 2 T n q n 5 q q 1 2 T n ad C Tn5 = q2n 2 2n q 2n 2+2n 5 1 q q 1 2 T n5 + q 2n 2 T n q n 2 n 2 n 5 q q 1 2 Tn To find the eigenfunctions of ad C we add certain vectors of the form N 1 =0 n 5 =1 n 4 =0 a n1 n 5 n 4 T n5 and N 2 n 2 =0 n 5 =1 n 4 =0 b n2 n 5 n 4 T n5 to the space W and denote this new space by W In the space W we find the elements N 1 0 N 4 S = 0 N 2 N 4 S = k k=0 r=0 k=0 r=0 N 1 +b N 1N 4 S a N 1N 4 S kr T N1 +S+2kC N 4+2r + kr T N1 +S+2k+1C N 4+2r+1 k c N 2N 4 S kr T N2 +S+2kC N4+2r + +d N 2N 4 S kr T N2 +S+2k+1C N 4+2r+1 Y N 2 where N 1 + S 1 N 2 + S 1 and a N 1N 4 S 00 = 1 for which the relations ad C N 1 0 N 4 S = q2s 1 + q 2S+1 q q 1 2 N 1 0 N 4 S ad C 0 N 2 N 4 S = q2s 1 + q 2S+1 q q N 2 N 4 S are valid For the constants a N 1N 4 S kr and b N 1N 4 S kr we obtain the system of equations a N 1N 4 S k+10 2S + 2k q 2N 1 a N 1N 4 S k0 = 0 a N 1N 4 S 2S + 2k + 1 = q N 1 b N 1N 4 S S + 2k + 1 q q 1 2 k+1k+1 a N 1N 4 S k+1r b N 1N 4 S kk b N 1N 4 S k+1r kk 2S + 2k q 2N 1 a N 1N 4 S kr = = q N 1 b N 1N 4 S kr 1 S + 2k + 1 q q 1 2 for 1 r k 2k + 1 2S + 2k = q N 1 a N 1N 4 S S + 2k q q 1 2 kk 2k + 3 2S + + q 2N 1 b N 1N 4 S = q N 1 a N 1N 4 S k+1r kr = S + q q 1 2 for 0 r k

7 The adjoint representation of quantum algebra U q sl2 7 The constants c N 2N 4 S kr and d N 2N 4 S kr fulfil a similar system of equations The system of equations for a kr and b kr for simplicity we omit the upper indices can be solved in the following way: Let a 00 = 1 If we know any a st for 0 t s k and b st for 0 t s k 1 we find b kr from the equations 2k + 1 2S + 2k bkk = q N 1 q q 1 2 S + 2k akk 2k + 1 2S + 2k bkr = q N 1 q q 1 2 S + 2k akr q 2N 1 b k 1r for 0 r k 1 42 If we know any a st and b st for 0 t s k we find a k+1r from the equations 2S + 2k + 1 ak+10 = q 2N 1 a k0 2S + 2k + 1 ak+1r = q N 1 q q 1 2 S + 2k + 1 bkr 1 q 2N 1 a kr for 1 r k 43 2S + 2k + 1 ak+1k+1 = q N 1 q q 1 2 S + 2k + 1 bkk As far as S > 0 S +k 0 for any k and from equations 42 and 43 we can evaluate a kr and b kr with the help of a 00 = 1 For S = L 0 we obtain zero in 42 for k = L and we are not able to find b Lr For L = 0 and k = 0 equation 42 is trivially valid and gives no other condition for b 00 For L 1 the systems 42 and 43 look like 2k + 1 2L 2k bkk = q N 1 q q 1 2 L 2k akk 2k + 1 2L 2k bkr = q N 1 q q 1 2 L 2k akr + q 2N 1 b k 1r 2L 2k 1 ak+10 = q 2N 1 a k0 for 0 r k L 2k 1 ak+1r = q N 1 q q 1 2 L 2k 1 bkr 1 + q 2N 1 a kr for 1 r k 2L 2k 1 ak+1k+1 = q N 1 q q 1 2 L 2k 1 bkk For 0 k L 1 and k = L the system 42 is reduced to a LL = 0 b L 1r = q N 1 q q 1 2 L alr for 0 r L 1 45 In the appendix we show that the system 44 has a solution for which a 00 = 1 and the relations 45 are a consequence of it We make the following choice The constants b Lr 0 r L which are not determined by the system are set to zero The constants a kr and b kr for k L are uniquely determined by equations 43 for k L and by equations 42 for k L + 1 Just this element is denoted by N 1 0 N 4 S for S 0 Analogously we define the elements 0 N 2 N 4 S

8 8 Č Burdík O Navrátil and S Pošta 422 The decomposition of the adjoint representation in Ŵ Next we introduce the space Ŵ W which is generated by the elements N 1 0 N 4 S and 0 N 2 N 4 S Now we find the action of the adjoint representation on these vectors We obtain ad E N 1 0 N 4 S = q N 1 S q q 1 N1+1 k q 2k a N 1N 4 S kr T N1 +S+2k+1C N4+2r + If we denote k=0 r=0 a kr = q 2k a N 1N 4 S kr b kr = q 2k 1 b N 1N 4 S kr +q 2k 1 b N 1N 4 S kr T N5 +S+2k+2C N 4+2r+1 we can easily show that if a N 1N 4 S kr and b N 1N 4 S kr fulfill equations 42 and 43 for N 1 and S then a kr b kr fulfil these systems for N and S In addition for S = L 0 we have b Lr = 0 Since a 00 = a N 1N 4 S 00 = 1 ad E N 1 0 N 4 S = q q 1 q N 1 S N N 4 S 46 Similarly we show ad F 0 N 2 N 4 S = q q 1 q N 2 S 0 N N 4 S 47 If we apply ad F on N 1 0 N 4 S where N 1 1 similar but the more cumbersome calculations gives ad F N 1 0 N 4 S = qn 1+S 1 N 1 S S + N 1 1 q q 1 N N 4 S 48 and for action ad E on 0 N 2 N 4 S we can show that for N 2 1 ad E 0 N 2 N 4 S = qn 2+S 1 N 2 S S + N 2 1 Moreover we have ad K N 1 0 N 4 S = q 2N 1 N 1 0 N 4 S ad K 0 N 2 N 4 S = q 2N 2 0 N 2 N 4 S q q 1 0 N N 4 S 49 The representation preserves the value of N 4 and S but changes N 1 and N 2 Nevertheless we see that the lowest and highest weight vectors are ad F N 1 0 N 4 S = 0 when S = N 1 1 or S = 1 N 1 0 ad E 0 N 2 N 4 S = 0 when S = N 2 1 or S = 1 N 2 0

9 The adjoint representation of quantum algebra U q sl2 9 From the computations above it follows that for N 4 0 the subspaces Ŵ defined as Ŵ S0N4 S = ad Uq sl2 S 0 N 4 S = { N 1 0 N 4 S ; N 1 S 1 } for S 1 Ŵ 0SN4 S = ad Uqsl2 0 S N 4 S = { 0 N 2 N 4 S ; N 2 S 1 } for S 1 Ŵ 1 S0N4 S = ad Uq sl2 1 S 0 N 4 S = { N 1 0 N 4 S ; N 1 1 S } for S 0 Ŵ 01 SN4 S = ad Uqsl2 0 1 S N 4 S = { 0 N 2 N 4 S ; N 2 1 S } for S are invariant Except the elements from the subspaces 410 in the space Ŵ there are the elements of the form N 1 0 N 4 S with 0 N 1 < S and 0 N 2 N 4 S where 0 N 2 < S For the vectors 0 0 N 4 S where N 4 0 and S 1 we have from relations 46 and 47 ad E k 0 0 N 4 S = q q 1 k q S kk 1/2 k 0 N 4 S ad F k 0 0 N 4 S = q 1 q k q S kk 1/2 0 k N 4 S For N 4 0 and S 1 we consider the invariant subspace Ŵ 00N4 S = ad Uq sl2 0 0 N 4 S This subspace contains all vectors N 1 0 N 4 S and 0 N 2 N 4 S and is reducible It is clear that its irreducible components are ŴS0N 4 S and Ŵ0SN 4 S If we bring together the previous results we obtain Proposition The space Ŵ can be expressed as a direct sum of the invariant subspaces in the form Ŵ = N 4 =0 N 1 =1 ŴN1 0N 4 1 N 1 Ŵ0N 1 N 4 1 N 1 S=1 Ŵ 00N4 S The subspaces Ŵ00N 4 S are reducible their irreducible components are ŴS0N 4 S and Ŵ 0SN4 S and the dimension of the factor space Ŵ00N 4 S/ Ŵ S0N4 S + Ŵ0SN 4 S is 2S The infinite dimensional representation for the adjoint representation of U q sl2 In section 421 we introduced the factor space W its modification Ŵ and factor representation on the space Ŵ In this section we return to the adjoint representation on the U q sl2 This representation acts on the vectors T n5 and T n5 for n 5 > 1

10 10 Č Burdík O Navrátil and S Pošta as was described in 421 but for n 5 = 1 we obtain ad F T 1 = q 2+2 ad E T1 = q 2n 2+2 q q 1 1 T 2 + +q +1 1 q q 1 1 T 1 +1 q q 1 T 2 1 q n 2+1 n 2 1 q q 1 T n1 1 q q 1 1 2n2 1 q q 1 1 To get the eigenvectors and eigenvalues of the operator of ad C back in the game we modify U q sl2 as we did in the case of the space W This space will be denoted by Û q sl2 It is easy to show that for N 1 + S > 1 the representation is defined by relations and 410 but for N 1 + S = 1 we have relations and 2N1 1 ad F N 1 0 N 4 1 N 1 = q q 1 N1 1 C N 4 2N2 1 ad E 0 N 2 N 4 1 N 2 = Consequently the spaces q q 1 Y N 2 1 C N 4 V N1 0N 4 1 N 1 = ad Uq sl2 N 1 0 N 4 1 N 1 V 0N2 N 4 1 N 2 = ad Uqsl2 0 N 2 N 4 1 N 2 are no longer irreducible but contain the finite dimensional invariant subspaces 411 ad Uq sl2 N 1 1 C N 4 and ad Uq sl2 Y N 2 1 C N 4 If we now define for any N 1 1 S 1 and N 4 0 the invariant subspaces of Ûq sl2 V N1 1N 4 = ad Uqsl2 N 1 1 C N 4 V N1 0N 4 1 N 1 = ad Uq sl2 N 1 0 N 4 1 N 1 V 0N1 N 4 1 N 1 = ad Uq sl2 0 N 1 N 4 1 N 1 V S0N4 S = ad Uqsl2 S 0 N 4 S V 0SN4 S = ad Uq sl2 0 S N 4 S V 00N4 S = ad Uq sl2 0 0 N 4 S we can formulate the following theorem Theorem The space Ûq sl2 can be decomposed to Û q sl2 = V N1 0N 4 1 N 1 + V 0N1 N 4 1 N 1 N 4 =0 N 1 =1 S=1 V 00N4 S Moreover V N1 1N 4 = V N1 0N 4 1 N 1 V 0N1 N 4 1 N 1 is the irreducible subspace of dimension 2N 1 1 and the subspace V 00N4 S has the invariant infinite subspaces V S0N4 S and V 0SN4 S and the factor space V 00N4 S/ V S0N4 S+V 0SN4 S has the dimension 2S 1

11 The adjoint representation of quantum algebra U q sl Conclusion In the presented paper we have studied the decomposition of the adjoint representation of the quantum group U q sl2 By construction we modified this quantum group Uq sl2 by adding some formal series Let us mention that this deformation can be obtained from U h sl2 that contains the formal series The quantum group Uq sl2 is a subgroup in U h sl2 where we restrict ourselves to the special formal series especially to the formal series for K K 1 and E F = K K 1 q q 1 We showed that the decomposition of the quantum algebra Ûq sl2 is more complicated than the non-deformed case The modified space Ûq sl2 contains finite and infinite dimensional invariant subspaces and in addition this space is not fully reducible ie we cannot write it as a direct sum of invariant irreducible subspaces Of much interest is a duality between the subspaces Ûq sl2 with S > 0 and S 0 It would be useful to study whether this duality is a property of U q sl2 only or it is a general property for all quantum groups U q g Appendix In this appendix we show that the system 44 has a solution for which a 00 = 1 and that the solution fulfils relation 45 First we define the new variables α kr and β kr by the relations a kr = q 2kN 1 q q 1 2r αkr b kr = q 2k+1N 1 q q 1 2r+2 βkr With respect to these new variables the system 44 has the form 2k + 1 2L 2k βkk = L 2k α kk 2k + 1 2L 2k βkr = L 2k α kr + β k 1r 0 r k 1 2L 2k 1 αk+10 = α k0 2L 2k 1 αk+1r = L 2k 1 β kr 1 + α kr 1 r k 51 2L 2k 1 αk+1k+1 = L 2k 1 β kk which is valid for 0 k L 1 and the conditions 45 are of the form α LL = 0 β L 1r = L α Lr for 0 r L 1 52 For k L 1 we obtain from the first and last equations of 51 L 2k β kk = α kk 2k + 1 2L 2k L 2k 1 L 2k α k+1k+1 = α kk 2k + 1 2L 2k 1 2L 2k

12 12 Č Burdík O Navrátil and S Pošta which results in α kk = 1 2k! β kk = 2L 2k! 2L! L! L 2k! 1 2L 2k 1! L! 2k + 1! 2L! L 2k 1! 53 In particular from 53 we obtain α kk = 0 for k > 1 2 L and β kk = 0 for k > 1 2 L 1 and thus a LL = 0 Next for k r we put 2r 2L 2k 1 2r + 1 2L 2k 2 α kr = R kr β kr = S kr 2k 2L 2r 1 2k + 1 2L 2r 2 and we define 0 = 1 = 1 we obtain from 51 the system R k+10 R k0 = 0 R k+1r R kr = S k+1r S kr = 2k 2k + 1 2r 1 2L 2k 2 2r 2L 2k 1 2L 2r 1 2L 2r L 2k 1 Skr 1 r 1 2k + 1 2r 2L 2k 3 2r + 1 2L 2k 2 2L 2r 2 L 2k 2 Rk+1r r 0 2L 2r 1 This systems results in R k0 = 1 R kr = S kr = 2r 1 2r 2L 2r 1 k 1 2L 2r 2s 2s + 1 s=r 1 2L 2s 2 L 2s 1 S sr 1 r 1 2L 2s 1 2r 2L 2r 2 2r + 1 2L 2r 1 54 k s=r and the following relations: 2s 1 2L 2s 1 L 2s R sr r 0 2s 2L 2s R k0 r = 0 and S k r 1 = 0 for all k k 0 R kr = 0 for all k > k 0 S k0 r = 0 and R k r = 0 for all k > k 0 S kr = 0 for all k > k 0 To prove 52 it suffices to show the relation L L S L 10 = R L0 = 55 2L 2L

13 The adjoint representation of quantum algebra U q sl2 13 and that for any r 1 r 1 2 L we have R L r+1r = S L rr = 0 56 First we prove 55 In accordance with 54 we have L S L 10 = + 2L L 1 2L 2 2L 1 s=1 2s 1 2L 2s 1 L 2s 2s 2L 2s If we in substitute s L s in the sum we will discover that equation 55 holds The conditions R L r+1r = S L 1r = 0 for 1 r 1 2 L are equivalent to the equations L r s=r 1 L r s=r 2s 2L 2s 2 L 2s 1 S sr 1 = 0 2s + 1 2L 2s 1 2s 1 2s 2L 2s 1 2L 2s L 2s R sr = 0 57 To prove these equations we use 54 After an appropriate arrangement of the terms in the sums we can deduce that 57 is valid if the following statement holds: Let 1 r 1 2 L Then for L = 2J we have S J+sr 1 S J s 1r 1 = 0 for 0 s J r R J+sr R J sr = 0 for 1 s J r 58 and for L = 2J + 1 is S J+sr 1 S J sr 1 = 0 for 1 s J r + 1 R J+s+1r R J sr = 0 for 0 s J r 59 We will study the first case 58 Let 1 r J The relation R J+sr R J sr = 0 for 1 s J r with the use of 54 is equivalent to the equation J+s 1 t=j s 2t 4J 2t 2 2J 2t 1 S tr 1 S 2J t 1r 1 = 0 2t + 1 4J 2t 1 which is the consequence of the relation S J+sr 1 S J s 1r 1 = 0 for 0 s J r On the other hand for 2 r the equality S J+sr 1 S J s 1r 1 = 0 for 0 s J r is equivalent to the equation J+s t=j s 2t 1 4J 2t 1 2J 2t R tr 1 R 2J tr 1 = 0 2t 4J 2t and this is the consequence of the relation R J+sr 1 R J sr 1 = 0 for 1 s J r + 1 It follows from this consideration that for L = 2J all relations 58 are result of the relations S J+s0 S J s 10 = 0 for 0 s J 1 Similarly we can see that for L = 2J+1 the validity 58 follows from S J+s0 S J s0 = 0 for 1 s J

14 14 Č Burdík O Navrátil and S Pošta For L = 2J with the use of 54 the condition S J+s0 S J s 10 = 0 is equivalent to the equation J+s r=j s 2r 1 4J 2r 1 2J 2r = 0 2r 4J 2r which we can easily prove by substitution r 2J r and for L = 2J + 1 the condition S J+s0 S J s0 = 0 is equivalent to the equation J+s r=j s+1 2r 1 4J 2r + 1 2J 2r + 1 = 0 2r 4J 2r + 2 which is easily seen to hold after substitution r 2J r + 1 Acknowledgments The authors are grateful to M Havlíček and AU Klimyk for valuable and useful discussions This work was partially supported by GAČR 201/05/0857 and by research plan MSM References 1 Dixmier J Algebres Enveloppantes Gauthier-Villars Editeur Paris Kostant B Lie group representations on polynomial rings Am J Math Phys Kirillov A Elementy Teorii Predstavlenij Nauka Moscow 1972 English translation: Elements of the theory of representations Springer-Verlag Flath D Decomposition of the enveloping algebra of sl 3 J Math Phys Klimyk A and Schmüdgen K Quantum Groups and Their Representations Springer-Verlag Berli997 6 Joseph A and Letzler G Local finiteness of the adjoint action for quantized enveloping algebras J algebra