Representations and Classification of the compact quantum groups U_q(2) for complex deformation parameters
aa r X i v : . [ m a t h . QA ] F e b Representations and Classification of the compact quantumgroups U q (2) for complex deformation parameters Satyajit Guin and Bipul Saurabh
February 23, 2021
Abstract
In this article, we obtain a complete list of inequivalent irreducible representations ofthe compact quantum group U q (2) for non-zero complex deformation parameters q , whichare not roots of unity. The matrix coefficients of these representations are described interms of the little q -Jacobi polynomials. The Haar state is shown to be faithful and anorthonormal basis of L ( U q (2)) is obtained. Thus, we have an explicit description of thePeter-Weyl decomposition of U q (2). As an application, we discuss the Fourier transformand establish the Plancherel formula. We also describe the decomposition of the tensorproduct of two irreducible representations into irreducible components. Finally, we classifythe compact quantum groups U q (2). AMS Subject Classification No.: B , B , L . Keywords.
Compact quantum group, quantum U (2) group, matrix coefficients, Peter-Weyl decomposition, little q -Jacobi polynomial. In the theory of compact quantum group (CQG) developed by Woronowicz ([19],[20],[21]), thefirst nontrivial and the most accessible example is the SU q (2) for q ∈ R \ { } . It is widelystudied in the literature through different perspective. Its representation theory ([11],[12]) hasstriking similarity with that of its classical analogue SU (2). Chakraborty-Pal [2] studied itsgeometry along the line of Connes [3], and constructed a finitely summable K -homologicallynontrivial spectral triple that is equivariant under its own comultiplication action. Connesproved regularity of this spectral triple and computed its local chern character [5]. This buildsa bridge between the compact quantum groups of Woronowicz and Noncommutative geometryprogram of Connes. In a recent work [8], Woronowicz et al. defined a family of q -deformationsof SU (2) for q ∈ C \ { } . This agrees with the earlier defined SU q (2) when q is real. Note that1or q ∈ C \ R , SU q (2) is not a CQG, but a braided compact quantum group in a suitable tensorcategory. In [13], it is shown that the quantum analogue of the semidirect product constructionfor groups turns the braided quantum group SU q (2) into a genuine CQG. This CQG is thecoopposite of the compact quantum group U q (2) defined in [23]. In this article, our object ofstudy is this CQG U q (2) , q ∈ C \{ } , for generic parameter values (i,e. not a root of unity) withthe intention to prepare the ground for investigating its noncommutative geometric aspects.For real q , Koelink [10] initiated the study of U q ( n ). He begins with a deformation of thealgebra of polynomials on U ( n ). This algebra, denoted by P ol ( U q ( n )), is a Hopf ⋆ -algebra and iscompleted into a unital C ∗ -algebra C ( U q ( n )) which yields a compact quantum group in the senseof Woronowicz. The matrix coefficients and Peter-Weyl decomposition is obtained explicitly in[16]. The n = 2 case is studied further in [22], and it is shown that as a compact quantum group U q (2) is some twisted tensor product (closely related to Drinfeld’s quantum double construction)of SU q (2) and C ( T ). Note that at the level of C ∗ -algebras, C ( U q (2)) ∼ = C ( SU q (2)) ⊗ C ( T ) forreal q . U q (2) has also been studied further from representation category viewpoint (e.g. [1],[14]and references therein). Now, we turn to the complex q situation which is our object of study.The main difficulty in defining U q (2) for complex q , unlike the real case, is to establish a suitable ⋆ -structure on the Hopf algebra O ( GL q (2)). Zhang-Zhao [23] suceeded in defining the CQG U q (2) in Woronowicz’s framework following the FRT approach by taking Hayashi’s R -matrixwith n = 2. This agrees with the earlier defined U q (2) when q is real. The θ -deformation U θ (2)of the classical Lie group U (2), constructed by Connes-Dubois-Violette in [4], is a special caseof U q (2) for q = exp ( iθ ) [24]. For | q | 6 = 1 , U q (2) is a CQG of non-Kac type and for | q | = 1 itis of Kac type . While for real q there is a nondegenerate pairing between U q ( gl (2)) and theHopf algebra O ( U q (2)) associated with U q (2) (Page 316 and 440 in [9]), for non-real complexnumbers no such pairing has been established yet. Hence, it is not clear how O ( U q (2)) can beviewed as certain subset of the dual of U q ( gl (2)). This makes the study of U q (2) more involved,but at the same time interesting and worth investigating.Our goal is to understand the geometry of U q (2) along the line of Connes, and this ar-ticle is a first step towards that. For this purpose, the appropriate object to look for is anequivariant spectral triple (unbounded K -cycle). It captures not only the geometric data ofa CQG but also its comultiplication action. To construct spectral triples, one first needs toobtain explicit inequivalent irreducible corepresentations and then, the matrix coefficients needto be closely analysed. Here, the abstract realization of representation theory of U q (2) willnot be of much use since, to show that the Dirac operator has bounded commutator with thematrix coefficients, an explicit description of them is needed. That is, even if one concludesthat U q (2) has the same representation type of some classical Lie group, it does not help toproduce an equivariant Dirac operator on U q (2). Faithfulness of the Haar state h is desirable toget a faithful representation of the underlying C ∗ -algebra C ( U q (2)) on the GNS Hilbert space L ( h ). Obtaining a precise orthonormal basis of L ( h ) comes in handy in many situations,2specially in determining action of the generators on these basis elements. Bounds on the ratioof the norm of the matrix coefficients chosen appropriately put a restriction on the growth ofthe singular values of an equivariant Dirac operator, which can be crucial in obtaining optimalspectral triple and computing the spectral dimension. Thus, it is also important to obtainnorm of the matrix elements of irreducible corepresentations of U q (2). This article deals withthese problems.Brief description of our work is the following. We separate three cases namely, | q | < , | q | = 1and | q | >
1, as the representation theory of the C ∗ -algebra C ( U q (2)) is different accordingly.First we deal the | q | < C ( U q (2)), andprove faithfulness of the Haar state. Note that in [20], faithfulness of the Haar state is shownonly on a dense subalgebra and therefore, this is not automatic on the C ∗ -algebra C ( U q (2)).In Sec. 3, using actions of T we put a Z grading on the Hopf ⋆ -algebra O ( U q (2)). In Sec. 4,we describe all finite dimensional inequivalent irreducible representations of U q (2), find theirmatrix coefficients explicitly, compute their norm, and obtain the Peter-Weyl decomposition.This gives an explicit orthonormal basis of L ( U q (2)) consisting of normalized matrix elements.We have also expressed these matrix coefficients in terms of the little q -Jacobi polynomials. TheFourier transform on U q (2) is discussed, and the Plancherel formula is established. Then in Sec.5, we give an explicit decomposition of the tensor product of two irreducible representationsinto irreducible components. Although it looks similar to that in the case of SU q (2), theproof is completely different. In the case of SU q (2), one first finds formula for the tensorproduct decomposition of representations of U q ( sl ) in the Drinfeld-Jimbo side ([6],[7]), andthen through nondegenerate pairing between U q ( sl ) and the canonical Hopf algebra of SU q (2),one gets the result in the dual side. Since, no such pairing is known yet for the case of U q (2) with q ∈ C \ { } , one can not follow this line of argument. Our approach uses explicit descriptionof the matrix coefficients and L ∞ functional calculus.Next, in Sec. 6, we deal the case of | q | = 1. We define faithful C ∗ -representation of C ( U q (2))in this case, and prove faithfulness of the Haar state. Then, the Peter-Weyl decomposition andthe Plancherel formula follows from similar computations done in the case of | q | <
1, andhence we only mention the final results. This subsumes earlier investigations in ([23],[24]).Finally, in Sec. 7, we have classified the CQG U q (2) and it turns out that for nonzero complexnumbers q and q ′ which are not roots of unity, U q (2) and U q ′ (2) are isomorphic if and only if q ′ ∈ { q, q, q , q } . This also justifies that for | q | 6 = 1, it is enough to do the computations for | q | < Notations :
The following notations are used throughout the article.(i) A q denotes the C ∗ -algebra C ( U q (2)) and A q denotes the underlying Hopf ⋆ -algebra O ( U q (2)).(ii) a, b, D will always denote the generators of A q .3iii) S will always denote the coinverse of A q .(iv) t lij D k denotes the matrix coefficients of irreducible corepresentations of A q .(v) ‘ i ’ will always denote an element in N , and whenever the complex inderminate appearswe write it √− ℓ ( N ) (or ℓ ( Z )) is denoted by { e n : n ∈ N } (or n ∈ Z depending on context).(vii) N denotes the number operator e n ne n acting on ℓ ( N ) or ℓ ( Z ) depending on context.(viii) V denotes the right shift operator e n e n +1 acting on ℓ ( N ), and U denotes the unitaryshift operator e n e n +1 acting on ℓ ( Z ). Throughout the article, q denotes a non-zero complex number which is not a root of unity,and we use the notations C ∗ := C \ { } , R ∗ := R \ { } . We first recall the compact quantumgroup U q (2) as defined in [23]. The C ∗ -algebra C ( U q (2)), to be denoted by A q , is the universal C ∗ -algebra generated by a, b, D satisfying the following relations : ba = qab, a ∗ b = qba ∗ , bb ∗ = b ∗ b, aa ∗ + bb ∗ = 1 ,aD = Da, bD = q | q | − Db, DD ∗ = D ∗ D = 1 , a ∗ a + | q | b ∗ b = 1 . (2.1)The compact quantum group structure is given by the comultiplication ∆ : A q −→ A q ⊗ A q defined as follows :∆( a ) = a ⊗ a − ¯ qb ⊗ Db ∗ , ∆( b ) = a ⊗ b + b ⊗ Da ∗ , ∆( D ) = D ⊗ D. (2.2)Let O ( U q (2)) be the ⋆ -subalgebra of C ( U q (2)) generated by a, b and D . We will denote it by A q . The Hopf ⋆ -algebra structure on it is given by the following :antipode: S ( a ) = a ∗ , S ( b ) = − qbD ∗ , S ( D ) = D ∗ , S ( a ∗ ) = a, S ( b ∗ ) = − (¯ q ) − b ∗ D , counit: ǫ ( a ) = 1 , ǫ ( b ) = 0 , ǫ ( D ) = 1 . We divide the article in two cases namely, the case of | q | = 1 and | q | 6 = 1. First we restrictour attention to the case of | q | < q = 0 till Section 5. In Section 7, where we prove theclassification of U q (2), we will see that for | q | 6 = 1, it is enough to provide computations onlyfor the case of | q | <
1. The case of | q | = 1 is dealt separately in Section 6.4ix any q ∈ C ∗ with | q | < θ = π arg ( q ). The C ∗ -algebra A q = C ( U q (2)) can berealized more concretely as follows. Let H be the Hilbert space ℓ ( N ) ⊗ ℓ ( Z ) ⊗ ℓ ( Z ). Considerthe right shift operator V : e n e n +1 acting on ℓ ( N ) and the bilateral shift U : e n e n +1 acting on ℓ ( Z ). Define the following representation π of A q on H : π ( a ) = q − | q | N V ⊗ I ⊗ I, π ( b ) = q N ⊗ U ⊗ I, π ( D ) = I ⊗ e − π √− θN ⊗ U. (2.3)Let T be the Toeplitz algebra generated by the right shift operator V acting on ℓ ( N ). Denoteby t : t t the identity map on T . Then, the homomorphism t U is an isomorphismbetween C ( T ) and the C ∗ -subalgebra of B ( ℓ ( Z )) generated by the bilateral shift U acting on ℓ ( Z ). Let σ : T −→ C ( T ) be the symbol map sending V to t , and ev t : C ( T ) −→ C denotesthe evaluation map at t . Proposition 2.1.
The representation π of A q defined above is faithful.Proof : Since p − | q | N V, q N ∈ T , one can view A q ⊆ T ⊗ B ( ℓ ( Z )) ⊗ C ( T ). Consider thefollowing homomorphisms, σ ⊗ I ⊗ I : T ⊗ B ( ℓ ( Z )) ⊗ C ( T ) −→ C ( T ) ⊗ B ( ℓ ( Z )) ⊗ C ( T ) ,I ⊗ I ⊗ ev t : T ⊗ B ( ℓ ( Z )) ⊗ C ( T ) −→ C ( T ) ⊗ B ( ℓ ( Z )) . Let ϕ and ψ be the restriction of σ ⊗ I ⊗ I and I ⊗ I ⊗ ev t respectively to A q . One has ϕ ( π ( a )) = t ⊗ I ⊗ I , ϕ ( π ( b )) = 0 , ϕ ( π ( D )) = I ⊗ e − π √− θN ⊗ t ;and ψ ( π ( a )) = q − | q | N V ⊗ I , ψ ( π ( b )) = q N ⊗ U , ψ ( π ( D )) = tI ⊗ e − π √− θN . It is not difficult to see that all one dimensional irreducible representations of A q factor throughthe homomorphism ϕ ◦ π , and all infinite dimensional representations of A q described in Lemma3 . . ψ ◦ π . This proves that anyirreducible representation of A q factors through the map π . Hence, image of an element inker( π ) under any irreducible representation of A q is zero, which implies that the element is zerobecause norm of any element in a C ∗ -algebra is the supremum of the norms of the image ofthat element under all irreducible representations. This proves the claim. ✷ Remark 2.2.
Note that Lemma 3 . | q | >
1. In the case of | q | < H as ker( a ∗ ).From now on, thanks to the above proposition, we will identify a , b and D with π ( a ), π ( b )and π ( D ) respectively. Moreover, we view A q as the C ∗ -subalgebra of B (cid:0) ℓ ( N ) ⊗ ℓ ( Z ) ⊗ ℓ ( Z ) (cid:1) generated by π ( a ), π ( b ) and π ( D ). For n, l ∈ Z and m, k ∈ N , define h n, m, k, l i = a n b m ( b ∗ ) k D l if n ≥ , ( a ∗ ) − n b m ( b ∗ ) k D l if n ≤ . heorem 2.3 ([23]) . The set {h n, m, k, l i : n, l ∈ Z , m, k ∈ N } forms a linear basis of O ( U q (2)) for all q ∈ C ∗ . Theorem 2.4 ([23]) . The Haar state h : C ( U q (2)) −→ C is given by the following, h ( x ) = (1 − | q | ) ∞ X n =0 | q | n h e n, , , π ( x ) e n, , i , where { e n,r,s } denotes the standard orthonormal basis of ℓ ( N ) ⊗ ℓ ( Z ) ⊗ ℓ ( Z ) . Moreover, onehas h ( h n, m, k, l i ) = −| q | −| q | m +1) if m = k, and n = l = 0 , otherwise . In this case of | q | <
1, the Haar state is not a trace as h ( a ∗ a − aa ∗ ) = (1 − | q | ) h ( bb ∗ ) = −| q | | q | = 0 . Let ∆ T denotes the coproduct on C ( T ) and φ : A q −→ C ( T ) be the homomorphism given by φ ( a ) = 1 , φ ( b ) = 0 and φ ( D ) = t . One can check that ∆ T ◦ φ = ( φ ⊗ φ ) ◦ ∆ . Thus, T is aquantum subgroup of U q (2). Hence, T acts on A q by the formula Φ( x ) = (id ⊗ φ )∆ . In such acase, one defines the quotient space U q (2) / T as follows, C ( U q (2) / T ) = { x ∈ A q : (id ⊗ φ )∆( x ) = x ⊗ } . The conditional expectation E : A q −→ C ( U q (2) / T ) is defined as (id ⊗ ( h T ◦ φ )) ◦ ∆ , where h T denotes the Haar state on T . Lemma 2.5.
The C ∗ -algebra C ( U q (2) / T ) is the C ∗ -subalgebra of A q generated by a and b .Proof : Using the continuity of E and by the open mapping theorem, one has C ( U q (2) / T ) = E ( A q ) = E ( A q ) = E ( A q ) . (2.4)Applying the formula E = (id ⊗ ( h T ◦ φ )) ◦ ∆ , one can see that E ( a ) = a , E ( a ∗ ) = a ∗ , E ( b ) = b , E ( b ∗ ) = b ∗ and E ( D m ) = 0 for m ∈ Z \ { } . This proves that C ( U q (2) / T ) contains the ⋆ -subalgebra generated by a and b . Using the prop-erty E ( xy ) = xE ( y ) for x ∈ C ( U q (2) / T ) and y ∈ A q , we get the following, E ( h n, m, k, l i ) = h n, m, k, i E ( D l ) = h n, m, k, i if l = 0 , l = 0 . Hence, E ( A q ) is the ⋆ -algebra generated by a and b . This, along with the eqn. 2.4, proves theclaim. ✷ Denote by A q the C ∗ -subalgebra C ( U q (2) / T ). Let C ( SU | q | (2)) be the C ∗ -subalgebra of B ( ℓ ( N ) ⊗ ℓ ( Z ) ⊗ ℓ ( Z )) generated by the operators p − | q | N V ⊗ I ⊗ I and | q | N ⊗ U ⊗ I .6 emma 2.6. For | q | < , one has A q = C ( SU | q | (2)) .Proof : To see C ( SU | q | (2)) ⊆ A q , first observe that b ∗ b = | q | N ⊗ I ⊗ I . Using the spectraldecomposition of b ∗ b , one can see that p i ⊗ I ⊗ I ∈ A q ∀ i ∈ N , where p i is the rank oneprojection acting on ℓ ( N ) by p i ( e n ) = δ in e n . Hence, T n = (cid:0) P ni =0 | q | i q i p i ⊗ I ⊗ I (cid:1) b ∈ A q . Then,we have k T n − | q | N ⊗ U ⊗ I k ≤ | q | n +1 k U kk I k = | q | n +1 . This shows that lim n →∞ T n = | q | N ⊗ U ⊗ I and hence, | q | N ⊗ U ⊗ I ∈ A q . The reverse inclusionfollows from similar argument. ✷ Remark 2.7.
The above Lemma and Lemma 3 . A q = A q for q , q ∈ C ∗ with | q | , | q | <
1. It is not difficult to see that A q is same as C ( SU q (2)) mentioned in [8] andtherefore, the isomorphism of A q follows from Thm. 2 . A q and A | q | . This iscrucial in getting the faithfulness of the Haar state. Theorem 2.8.
The Haar state h on the quantum group U q (2) is faithful.Proof : To prove the claim, we will apply Lemma 2 . . h T is well-known. Moreover, the counit ǫ T , initially defined on the algebra of polynomials in t and t − , is same as evaluation at 1 and hence, it can be extended to C ( T ). It follows fromThm. 2.4 and Lemma 2.6 that the restriction h | A q of the Haar state h of A q to A q is same asthe Haar state of C ( SU | q | (2)). Hence, it follows from Thm. 1 . h | A q is a faithfulstate on A q . Combining all these facts and using Lemma 2 . ✷ O ( U q (2)) In this section we decompose A q = O ( U q (2)) in a way similar to what is done in (page 105,[9]) for O ( SL q (2)), but with respect to three different coactions. We assign three integers toeach basis element and put a Z grading on A q . Consider the Hopf ⋆ -algebra O ( T ) = C [ z, z − ]where, z ∗ = z − , ∆( z ) = z ⊗ z , S ( z ) = z − , ǫ ( z ) = 1 . Define a ⋆ -homomorphism φ T : A q −→ O ( T ) given by φ T ( a ) = z , φ T ( b ) = 0 and φ T ( D ) = 1.Using this homomorphism, one can put a natural left O ( T ) comodule structure L T : A q −→O ( T ) ⊗ A q and a right O ( T ) comodule structure R T : A q −→ A q ⊗ O ( T ) defined as follows; L T = ( φ T ⊗ id) ◦ ∆ and R T = (id ⊗ φ T ) ◦ ∆ . m, n ∈ Z , define A q [ m, n ] = { x ∈ A q : L T ( x ) = z m ⊗ x and R T ( x ) = x ⊗ z n } . Let ζ = bb ∗ . If m − n is even, define e m,n = a m + n b m − n if m + n ≥ , m ≥ n,a m + n ( b ∗ ) n − m if m + n ≥ , m ≤ n,b m − n ( a ∗ ) − m − n if m + n ≤ , m ≥ n, ( b ∗ ) n − m ( a ∗ ) − m − n if m + n ≤ , m ≤ n. (3.1) Proposition 3.1.
One has the followings.(i) A q [ m, n ] A q [ p, q ] ⊂ A q [ m + p, n + q ] .(ii) h n, m, k, l i ∈ A q [ n + m − k, n − m + k ] .(iii) A q [0 , is a commutative ⋆ -subalgebra of A q generated by D , D ∗ and bb ∗ .(iv) A q [ m, n ] is a A q [0 , -bimodule.(v) A q = ⊕ m,n ∈ Z A q [ m, n ] .(vi) If m − n is odd, then A q [ m, n ] = { } .(vii) If m − n is even, then A q [ m, n ] = e m,n A q [0 ,
0] = A q [0 , e m,n .Proof : It follows from the definition of A q [ m, n ] and the fact that {h n, m, k, l i : n, l ∈ Z , m, k ∈ N } forms a linear basis of A q . ✷ From part ( i ) of the above Lemma, we see that this decomposition of A q into A q [ m, n ]’s isa Z grading on A q . To put a Z grading, consider another ⋆ -homomorphism ψ T : A q −→ O ( T )given by ψ T ( a ) = 1 , ψ T ( b ) = 0 and ψ T ( D ) = z . Define the coaction L ψ T = ( ψ T ⊗ id) ◦ ∆ . Let A q [ m, n, r ] = { x ∈ A q [ m, n ] : L ψ T ( x ) = z r ⊗ x } . Proposition 3.2.
One has the followings.(i) A q [ m, n, r ] A q [ p, q, s ] ⊂ A q [ m + p, n + q, r + s ] .(ii) A q [ m, n, r ] ∗ = A q [ − m, − n, − r ] .(iii) h n, m, k, l i ∈ A q [ n + m − k, n − m + k, l ] .(iv) A q [0 , , is isomorphic to C [ ζ ] .(v) A q [ m, n, r ] = D r A q [ m, n,
0] = A q [ m, n, D r for r ∈ Z . vi) If m − n is even, A q [ m, n ] = ⊕ r ∈ Z A q [ m, n, r ] = ⊕ r ∈ Z D r A q [ m, n,
0] = ⊕ r ∈ Z D r e m,n C [ ζ ] .(vii) S ( A q [ m, n, r ]) = A q [ − n, − m, − r ] .Proof : Part ( i ) and ( ii ) follow from the fact that L T , R T and L ψ T are ⋆ -preserving homomor-phisms. Part ( iii ) follows immediately from the definition. Observe that D r ∈ A q [0 , , r ] for r ∈ Z . Using this and part ( iii ) of Propn. 3 .
1, we get part ( iv ). Part ( v ) follows directlyfrom part ( i ). Part ( vi ) follows from part ( iii ), part ( vii ) of Propn. 3 .
1, and the fact that {h n, m, k, l i : n, l ∈ Z , m, k ∈ N } forms a linear basis of A q . For part ( vii ), observe that S ( ζ ) = ζ , S ( D ) = D ∗ , S ( D ∗ ) = D and S ( e m,n ) = Ce − n, − m . where, C is a nonzero constant. Combining this with part ( v ), we get that S ( A q [ m, n, r ]) ⊆ A q [ − n, − m, − r ] . To get the reverse inclusion, take any y ∈ A q [ − n, − m, − r ]. From part ( ii ), we have y ∗ ∈ A q [ n, m, r ] ⇒ S ( y ∗ ) ∈ A q [ − m, − n, − r ] ⇒ S ( y ∗ ) ∗ ∈ A q [ m, n, r ] . Since S ( S ( y ∗ ) ∗ ) = y , we get the claim. ✷ Proposition 3.3. If m, n, r ∈ Z with m − n is even, A q [ m, n, r ] is a free C [ ζ ] -left (or right)module with basis e m,n D r . Moreover, if f ( ζ ) e m,n D r = e m,n D r g ( ζ ) = e m,n h ( ζ ) D r for polyno-mials f, g, h then degree of f, g and h are same.Proof : By part ( v ) of Propn. 3 .
2, one can see that e m,n D r generates A q [ m, n, r ]. Using thecommutation relations in 2.1 and Thm. 2.3, it follows that for any polynomial P , P ( ζ ) e m,n D r =0 implies P ( ζ ) = 0. This proves that e m,n D r is a C [ ζ ]-basis of A q [ m, n, r ]. Since ζ k e m,n = Ce m,n ζ k for some nonzero constant, we get the last part of the claim. ✷ Proposition 3.4. If x ∈ A q [ m, n, r ] and ( m, n, r ) = (0 , , , then h ( x ) = 0 .Proof : For z ∈ T , consider the linear functionals β z = ev z ◦ φ T and γ z = ev z ◦ ψ T on A q , where ev z is the evaluation map at z . Take any x ∈ A q [ m, n, r ]. Using the invariance property ofHaar measure, we have h ( x ) = ( β z ⊗ h ) ◦ ∆( x ) = ( ev z ⊗ h )( φ T ⊗ id) ◦ ∆( x ) = ( ev z ⊗ h ) L T ( x ) = ( ev z ⊗ h )( z m ⊗ x ) = z m h ( x ) . for all z ∈ T . Similarly, considering h ( x ) = ( h ⊗ β z ) ◦ ∆( x ) and h ( x ) = ( γ z ⊗ h ) ◦ ∆( x ), we getthat h ( x ) = z n h ( x ) and h ( x ) = z r h ( x ) . for all z ∈ T . Hence if ( m, n, r ) = (0 , , h ( x ) = 0. ✷ emark 3.5. It follows that for any x ∈ A q , h ( x ∗ ) = h ( x ) and h ( S ( x )) = h ( x ). To see this,take any x ∈ A q [ m, n, r ]. From part ( ii ) of Propn. 3 .
2, it follows that x ∗ ∈ A q [ − m, − n. − r ].Hence, if ( m, n, r ) = (0 , , h ( x ∗ ) = 0 = h ( x ). If x ∈ A q [0 , , iv ) of Propn. 3 . x = p ( ζ ) where, p is a polynomial with complex coefficients. Since ζ ∗ = ζ , we get that h ( x ∗ ) = h ( x ). The other part follows from a similar argument. In this section, we describe all the irreducible representations of U q (2) and obtain an orthonor-mal basis of L ( U q (2) , h ) in terms of the matrix coefficients of these representations. Let us fixsome notations. For l ∈ N , m, n ∈ N , α ∈ C and q ∈ C ∗ , let c = − ¯ qDb ∗ , d = Da ∗ .I l = {− l, − l + 1 , · · · , l − , l } . ( α, q ) n = n = 0 , Q n − r =0 (1 − αq r ) if n > . (cid:18) nm (cid:19) q = ( q, q ) n ( q, q ) m ( q, q ) n − m . | m + 1 | | q | = m X k =0 | q | m − k . Thus, | q | m | m + 1 | | q | = P mk =0 | q | m − k ) = −| q | m +1) −| q | . A q on the quantum plane Let O ( C q ) be the complex associative unital algebra generated by two symbols x and y satisfying xy = qyx . This is called the quantum plane. Proposition 4.1.
The association x x ⊗ Da ∗ + y ⊗ b and y x ⊗ ( − ¯ qDb ∗ ) + y ⊗ a (4.1)extends to a unique homomorphism ψ R : O ( C q ) −→ O ( C q ) ⊗ A q . Moreover, ψ R is a rightcoaction of A q on O ( C q ) .Proof : It is enough to show that image of xy and qyx under the map ψ R are same. Using the10ommutation relations, we get that ψ R ( xy ) = ( x ⊗ Da ∗ + y ⊗ b )( x ⊗ ( − ¯ qDb ∗ ) + y ⊗ a )= − ¯ qx ⊗ Da ∗ Db ∗ + y ⊗ ba + xy ⊗ Da ∗ a − ¯ qyx ⊗ bDb ∗ = − ¯ qx ⊗ D a ∗ b ∗ + qy ⊗ ab + xy ⊗ D ( aa ∗ + (1 − | q | ) b ∗ b ) − ¯ qq q | q | xy ⊗ Dbb ∗ = − ¯ q x ⊗ D b ∗ a ∗ + qy ⊗ ab + xy ⊗ Daa ∗ − | q | xy ⊗ Db ∗ b = − q ¯ qx ⊗ Db ∗ Da ∗ + qy ⊗ ab + qyx ⊗ Daa ∗ − q ¯ qxy ⊗ Db ∗ b = q ( x ⊗ ( − ¯ qDb ∗ ) + y ⊗ a )( x ⊗ Da ∗ + y ⊗ b )= ψ R ( qyx ) . To show that ψ R is a right coaction of A q , one has to check the conditions ( ψ R ⊗ id) ◦ ψ R =(id ⊗ ∆) ◦ ψ R and (id ⊗ ǫ ) ◦ ψ R = id on the generators x and y of O ( C q ), and this follows easilyfrom the definition of ψ R . ✷ Proposition 4.2.
The association x Da ∗ ⊗ x + ( − ¯ qDb ∗ ) ⊗ y and y b ⊗ x + a ⊗ y (4.2)extends to a unique homomorphism ψ L : O ( C q ) −→ A q ⊗O ( C q ) . Moreover, ψ L is a left coactionof A q on O ( C q ) .Proof : Proof is similar to that of Propn. 4.1. ✷ Let O ( C q ) l be the vector subspace of O ( C q ) spanned by degree 2 l homogeneous monomialsin x and y , i.e. O ( C q ) l = L lj = − l C y l − j x l + j . It follows from 4.1 that O ( C q ) l is invariant underthe map ψ R . Hence, the restriction of the coaction ψ R to O ( C q ) l gives a (2 l + 1)-dimensionalcorepresentation of A q which will be denoted by T l . In what follows, we will embed O ( C q ) l in A q and find a corepresentation of A q on a vector subspace of A q which is equivalent to T l . Let A q ( a, b ) be the subalgebra of A q generated by a and b . The algebra A q ( a, b ) has a canonical A q -comodule structure given by restriction of the comultiplication ∆ to A q ( a, b ). Since ba = qab ,we get a homomorphism ϑ R : O ( C q ) −→ A q ( a, b ) such that ϑ R ( x ) = b and ϑ R ( y ) = a . Proposition 4.3.
The homomorphism ϑ R : O ( C q ) −→ A q ( a, b ) is an isomorphism between theright A q -comodule algebra O ( C q ) with the right coaction ψ R and the right A q -comodule algebra A q ( a, b ) with the comultiplication ∆ .Proof : Since { x m y n : m, n ∈ N } and { b m a n : m, n ∈ N } are linear basis of O ( C q ) and A q ( a, b )respectively, the homomorphism ϑ R is an algebra isomorphism. To prove that ϑ R is an A q -comodule isomorphism, it is enough to check the condition ( ϑ R ⊗ id ) ◦ ψ R = ∆ ◦ ϑ R on thegenerators x and y of O ( C q ). Applying the formula of comultiplication of a, b and the coaction ψ R , this follows easily. ✷ ϑ L : O ( C q ) −→ A q ( a, c ) such that ϑ L ( x ) = c and ϑ L ( y ) = a , since ca = ¯ qac . This gives an isomorphism between the left A q -comodulealgebra O ( C q ) with the left coaction ψ L and the left A q -comodule algebra A q ( a, c ) with thecomultiplication ∆. For l ∈ N and i, j ∈ I l , let f lj = (cid:18) ll + j (cid:19) / | q | ϑ R ( y l − j x l + j ) = (cid:18) ll + j (cid:19) / | q | a l − j b l + j ,e li = (cid:18) ll + i (cid:19) / | q | ϑ L ( y l − i x l + i ) = (cid:18) ll + i (cid:19) / | q | a l − i c l + i ,V Rl = ⊕ lj = − l C f ( l ) j , V Ll = ⊕ li = − l C e ( l ) i ,T Rl = ∆ | V Rl , T Ll = ∆ | V Ll . Employing Propn. (4.1,4.3) and the fact that O ( C q ) l is an invariant subspace under ψ R , onegets a corepresentation T Rl : V Rl −→ V Rl ⊗ A q of A q on V Rl which is equivalent to T l . For i, j ∈ I l , let t lij denote the matrix coefficients of T Rl with respect to the basis { f lj } of V Rl . Then,we have T Rl ( f lj ) = ∆( f lj ) = l X i = − l f li ⊗ t lij . (4.3)Similarly, let w lij be the matrix coefficients of T Ll with respect to the basis { e li } of V Ll . Then,we have T Ll ( e li ) = ∆( e li ) = l X j = − l w lij ⊗ e lj . (4.4)Consider the sesquilinear forms h· , ·i R and h· , ·i L on A q defined as follows, h x, y i L = h ( x ∗ y ) , h x, y i R = h ( xy ∗ ) for x, y ∈ A q . Both the sesquilinear forms are positive definite, thanks to Thm. 2.8, and hence A q is an innerproduct space under h· , ·i R and h· , ·i L . The following proposition says that the decompositionof A q given in part (vi) of Propn. 3.2 is orthogonal. Proposition 4.4.
Let x ∈ A q [ m, n, r ] and y ∈ A q [ m ′ , n ′ , r ′ ] . If ( m, n, r ) = ( m ′ , n ′ , r ′ ) then, h x, y i L = 0 and h x, y i R = 0 . Proof : By Propn. 3.2, we have x ∗ ∈ A q [ − m, − n, − r ] and y ∗ ∈ A q [ − m ′ , − n ′ , − r ′ ]. Hence, x ∗ y ∈ A q [ m ′ − m, n ′ − n, r ′ − r ] and xy ∗ ∈ A q [ m − m ′ , n − n ′ , r − r ′ ]. Using Propn. 3.4, it nowfollows that h ( x ∗ y ) = h ( xy ∗ ) = 0. ✷ .2 The matrix coefficients and Peter-Weyl decomposition For x ∈ A q , define k x k L = h x, x i / L and k x k R = h x, x i / R . Lemma 4.5.
Let d r,s = k a r b s k L . Then, one has the followings.(i) d r,s = d r − ,s − | q | r d r − ,s +1 ∀ r ≥ (ii) d r,s = (cid:16) | q | ( r + s ) | r + s + 1 | | q | (cid:0) r + ss (cid:1) | q | (cid:17) − .Proof : For part ( i ), we have d r,s = h (( b ∗ ) s ( a ∗ ) r a r b s ) = h (( b ∗ ) s b s ( a ∗ ) r a r )= h (cid:16) ( b ∗ ) s ( a ∗ ) r − (1 − | q | b ∗ b ) a r − b s (cid:17) = h (cid:16) ( b ∗ ) s ( a ∗ ) r − ) a r − b s (cid:17) − | q | r h (cid:16) ( b ∗ ) s +1 ( a ∗ ) r − a r − b s +1 (cid:17) = d r − ,s − | q | r d r − ,s +1 . For part ( ii ), we use induction based on the formula given in part ( i ). If r = 0, we have d ,s = h (( b s ) ∗ b s ) = 1 − | q | − | q | s +1) = 11 + | q | + | q | + · · · + | q | s = 1 | q | s | s + 1 | | q | . Hence, the claim is true for all the tuples (0 , s ) where s ∈ N . Assume that the claim holds foreach tuple ( p, s ), where 0 ≤ p < r and s ∈ N . Using part ( i ), we then have d r,s = d r − ,s − | q | r d r − ,s +1 = 1 | q | r + s − | r + s | | q | (cid:0) r + s − s (cid:1) | q | − | q | r | q | r + s | r + s + 1 | | q | (cid:0) r + ss +1 (cid:1) | q | = ( | q | , | q | ) r − ( | q | , | q | ) s (1 + | q | + · · · + | q | r +2 s − )( | q | , | q | ) r + s − − | q | r ( | q | , | q | ) r − ( | q | , | q | ) s +1 (1 + | q | + · · · + | q | r +2 s )( | q | , | q | ) r + s = (1 − | q | ) Q r − i =1 (1 − | q | i ) Q si =1 (1 − | q | i )(1 − q | r +2 s ) Q r + s − i =1 (1 − | q | i ) − | q | r (1 − | q | ) Q r − i =1 (1 − | q | i ) Q s +1 i =1 (1 − | q | i )(1 − q | r +2 s +2 ) Q r + si =1 (1 − | q | i )= (1 − | q | ) Q r − i =1 (1 − | q | i ) Q si =1 (1 − | q | i ) Q r + si =1 (1 − | q | i ) (cid:16) − | q | r (1 − | q | s +2 )1 − | q | r +2 s +2 (cid:17) = (1 − | q | ) Q ri =1 (1 − | q | i ) Q si =1 (1 − | q | i )(1 − | q | r +2 s +2 ) Q r + si =1 (1 − | q | i )= 1(1 + | q | + · · · + | q | r +2 s ) (cid:0) r + ss (cid:1) | q | = 1 | q | ( r + s ) | r + s + 1 | | q | (cid:0) r + ss (cid:1) | q | . ✷ Lemma 4.6.
Let c r,s = k a r b s k R . Then, one has the followings. i) c r,s = | q | s c r − ,s − | q | s c r − ,s +1 ∀ r ≥ (ii) c r,s = | q | r (cid:16) | q | ( r + s ) | r + s + 1 | | q | (cid:0) r + ss (cid:1) | q | (cid:17) − = | q | r d r,s .Proof : For part ( i ), we have c r,s = h ( a r b s ( b ∗ ) s ( a ∗ ) r )= 1 | q | s h ( a r − b s ( b ∗ ) s aa ∗ ( a ∗ ) r − )= 1 | q | s h ( a r − b s ( b ∗ ) s ( a ∗ ) r − ) − | q | s h ( a r − b s ( b ∗ ) s bb ∗ ( a ∗ ) r − )= 1 | q | s c r − ,s − | q | s c r − ,s +1 . For part ( ii ), we use induction based on the formula given in part ( i ). If r = 0, we have c ,s = h ( b s ( b ∗ ) s ) = (cid:16) | q | s | s + 1 | | q | (cid:17) − . Observe that this is same as d ,s in the previous lemma due to normality of b . Hence, the claimis true for all the tuples (0 , s ) where s ∈ N . Assume that the claim holds for each tuple ( p, s ),where 0 ≤ p < r and s ∈ N . Then, using part ( i ), we have c r,s = 1 | q | s c r − ,s − | q | s c r − ,s +1 = | q | r − | q | s | q | ( r + s − | r + s | | q | (cid:0) r + s − s (cid:1) | q | − | q | r − | q | s | q | ( r + s ) | r + s + 1 | | q | (cid:0) r + ss +1 (cid:1) | q | = 1 − | q | | q | s (cid:16) Q r − l =1 (1 − | q | l ) | q | r − Q rl =1 (1 − | q | l + s ) ) − Q r − l =1 (1 − | q | l ) | q | r − Q rl =1 (1 − | q | l + s +1) ) (cid:17) = (1 − | q | ) | q | r − | q | s Q r − l =1 (1 − | q | l )( | q | s +1) − | q | r + s +1) ) Q r +1 l =1 (1 − | q | l + s ) )= (1 − | q | ) | q | Q rl =1 (1 − | q | l ) | q | r − Q r +1 l =1 (1 − | q | l + s ) )= (1 − | q | ) Q rl =1 (1 − | q | l ) | q | r Q r +1 l =1 (1 − | q | l + s ) )= | q | r | q | ( r + s ) | r + s + 1 | | q | (cid:0) r + ss (cid:1) | q | . Thus, c r,s = | q | r d r,s by Lemma 4.5. ✷ Proposition 4.7.
In the Hilbert space ( V Rl , h· , ·i L ) , one has the followings.(i) k f lj k = | q | l | l +1 | / | q | ; (ii) The ordered set B l := (cid:8) | q | l | l + 1 | / | q | f lj (cid:9) i ∈ I l is an orthonormal basis of V Rl ;14 iii) The matrix coefficients of T Rl with respect to B l are t lij , i, j ∈ I l . Proof : The first part follows from Lemma 4.5. For r = r ′ , using the faithful representationsdefined in 2.3, we have for n ≥ r h v n, , , a r ′ b l − r ′ ( b ∗ ) l − r ( a ∗ ) r ( v n, , ) i = h v n, , , Cv n − r + r ′ ,r − r ′ , i = 0for orthonormal basis { v n,m,k } of ℓ ( N ) ⊗ ℓ ( Z ) ⊗ ℓ ( Z ), where C is a nonzero constant. Sincethe above inner-product is automatically zero for 0 ≤ n < r , by Thm. 2.4 we get that h a r ′ b l − r ′ , a r b l − r i R = h (cid:16) a r ′ b l − r ′ ( b ∗ ) l − r ( a ∗ ) r (cid:17) = 0 . Combining this with part ( i ), we get the claim. The last part follows from eqn. 4.3. ✷ Lemma 4.8.
For m ≥ , one has the following : a m ( a ∗ ) m = m X k =0 ( − k (cid:18) mk (cid:19) | q | | q | k + k − km ( bb ∗ ) k , ( a ∗ ) m a m = m X k =0 ( − k (cid:18) mk (cid:19) | q | | q | k + k ( bb ∗ ) k . Proof : Follows from induction on m . ✷ Proposition 4.9.
For l ∈ N , let t lij , i, j ∈ I l be the matrix coefficients of the corepresentation T Rl with respect to the basis { f lj : j ∈ I l } . Then, we have(i) t lij = X m + n = l − i ≤ m ≤ l − j ≤ n ≤ l + j q n ( l − j − m ) (cid:0) ll + j (cid:1) / | q | (cid:0) ll + i (cid:1) / | q | (cid:18) l − jm (cid:19) | q | (cid:18) l + jn (cid:19) | q | a m c l − j − m b n d l + j − n . (ii) For m, n ∈ N satisfying ≤ m ≤ l − j, ≤ n ≤ l + j and m + n = l − i , let C q ( l, i, j, m, n ) = | q | ( l − j − m )( l − j − m +2 n +1)+2 n ( l + i ) (¯ q ) ( i − j )( l + i ) (cid:0) ll + j (cid:1) / | q | (cid:0) ll + i (cid:1) / | q | (cid:18) l − jm (cid:19) | q | (cid:18) l + jn (cid:19) | q | . Then, one has t lij = X m + n = l − i ≤ m ≤ l − j ≤ n ≤ l + j ( − l − j − m C q ( l, i, j, m, n ) a m ( a ∗ ) l + j − n ( b ∗ ) l − j − m b n D l + i . iii) For ζ = bb ∗ and e m,n as defined in . , one has t li,j = e − i, − j P i,j ( ζ ) D l + i for some polynomial P i,j where, deg ( P i,j ) = l − max {| i | , | j |} .Proof : For c = − ¯ qDb ∗ and d = Da ∗ , we have∆( a ) = a ⊗ a + b ⊗ c , ∆( b ) = a ⊗ b + b ⊗ d . Using Propn. (2) in (page 39, eqn. 18 in [9]), we have∆( a l − j ) = l − j X m =0 (cid:18) l − jm (cid:19) | q | a m b l − j − m ⊗ a m c l − j − m ∆( b l + j ) = l + j X n =0 (cid:18) l + jn (cid:19) | q | a n b l + j − n ⊗ b n d l + j − n since, ab = q − ba, ac = (¯ q ) − ca, bd = (¯ q ) − db . Multiplying these two expressions, we get that∆( f lj ) = ∆ (cid:16)(cid:18) ll + j (cid:19) / | q | a l − j b l + j (cid:17) = l − j X m =0 l + j X n =0 (cid:18) ll + j (cid:19) / | q | (cid:18) l − jm (cid:19) | q | (cid:18) l + jn (cid:19) | q | a m b l − j − m a n b l + j − n ⊗ a m c l − j − m b n d l + j − n = l − j X m =0 l + j X n =0 q n ( l − j − m ) (cid:18) ll + j (cid:19) / | q | (cid:18) l − jm (cid:19) | q | (cid:18) l + jn (cid:19) | q | a m + n b l − m − n ⊗ a m c l − j − m b n d l + j − n . Putting m + n = l − i , we get that∆( f lj ) = X m + n = l − i ≤ m ≤ l − j ≤ n ≤ l + j q n ( l − j − m ) (cid:18) ll + j (cid:19) / | q | (cid:18) l − jm (cid:19) | q | (cid:18) l + jn (cid:19) | q | a l − i b l + i ⊗ a m c l − j − m b n d l + j − n = X m + n = l − i ≤ m ≤ l − j ≤ n ≤ l + j q n ( l − j − m ) (cid:0) ll + j (cid:1) / | q | (cid:0) ll + i (cid:1) / | q | (cid:18) l − jm (cid:19) | q | (cid:18) l + jn (cid:19) | q | f li ⊗ a m c l − j − m b n d l + j − n . Comparing with eqn. 4.3, we get that t lij = X m + n = l − i ≤ m ≤ l − j ≤ n ≤ l + j q n ( l − j − m ) (cid:0) ll + j (cid:1) / | q | (cid:0) ll + i (cid:1) / | q | (cid:18) l − jm (cid:19) | q | (cid:18) l + jn (cid:19) | q | a m c l − j − m b n d l + j − n X m + n = l − i ≤ m ≤ l − j ≤ n ≤ l + j q n ( l − j − m ) (cid:0) ll + j (cid:1) / | q | (cid:0) ll + i (cid:1) / | q | (cid:18) l − jm (cid:19) | q | (cid:18) l + jn (cid:19) | q | a m ( − ¯ qDb ∗ ) l − j − m b n ( Da ∗ ) l + j − n = X m + n = l − i ≤ m ≤ l − j ≤ n ≤ l + j ( − l − j − m C q ( l, i, j, m, n ) a m ( a ∗ ) l + j − n ( b ∗ ) l − j − m b n D l + i . This completes part ( i ) and part ( ii ). To get the last part, one needs to consider four casesnamely, ( i ) i + j ≥ , i ≥ j ( ii ) i + j ≥ , i ≤ j ( iii ) i + j ≤ , i ≥ j and ( iv ) i + j ≤ , i ≤ j .We will prove the claim for the first case only as all the other cases are similar. t lij = X m + n = l − i ≤ m ≤ l − i ≤ n ≤ l − i ( − l − j − m C q ( l, i, j, m, n ) a m ( a ∗ ) l + j − ( l − i − m ) ( b ∗ ) l − j − m b l − i − m D l + i = X m + n = l − i ≤ m ≤ l − i ≤ n ≤ l − i ( − l − j − m C q ( l, i, j, m, n ) a m ( a ∗ ) m ( a ∗ ) i + j ( b ∗ ) i − j ( b ∗ ) l − i − m b l − i − m D l + i = X m + n = l − i ≤ m ≤ l − i ≤ n ≤ l − i ( − l − j − m ( q ) ( i + j )( i − j ) C q ( l, i, j, m, n ) a m ( a ∗ ) m ζ l − i − m e − i, − j D l + i = (cid:16) X m + n = l − i ≤ m ≤ l − i ≤ n ≤ l − i m X k =0 ( − l − j − m + k ( q ) ( i + j )( i − j ) | q | i + j )( l − i − m ) C q ( l, i, j, m, n ) (cid:18) mk (cid:19) | q | | q | k + k − km ζ l − i − m + k (cid:17) e − i, − j D l + i . Hence, t lij = R i,j ( ζ ) e − i, − j D l + i where R i,j is a polynomial of degree l − i . Applying Propn.3.3 we get the claim. ✷ Remark 4.10.
A more concrete description of t lij in terms of the little q -Jacobi polynomialsis given in the next subsection 4 . Lemma 4.11.
For l ∈ N , one has the followings.(i) f lj = t l − l,j and e li = t li, − l for i ∈ I l ; (ii) t lij = w lij for i, j ∈ I l ; (iii) t lij ∈ A q [ − i, − j, l + i ] for i, j ∈ I l . Proof : Part ( i ) and ( iii ) are immediate consequences of Propn. 4.9. To show part ( ii ), observethat ∆( t lij ) = P k ∈ I l t lik ⊗ t lkj . j = − l and using the fact that e li = t li, − l , we get that∆( e li ) = P k ∈ I l t lik ⊗ e lk . Comparing with the eqn. 4.4, we get the claim. ✷ Lemma 4.12.
Let m, n ∈ Z such that m − n is even. Then, for k , k ∈ N , (i) h e m,n ζ k , e m,n ζ k i L = | q | − m − n h e m,n ζ k , e m,n ζ k i R , where e m,n is as defined in . . (ii) For x ∈ A q [ m, n ] , one has h x, x i L = | q | − m − n h x, x i R . Proof : Let m + n ≥ m ≥ n . Using the expressions for d r,s and c r,s from the Lemmas(4.5,4.6) and the fact that ba r ( a ∗ ) r = a r ( a r ) ∗ b , we have h e m,n ζ k , e m,n ζ k i L = (cid:10) a m + n b m − n ζ k , a m + n b m − n ζ k (cid:11) L = h (cid:0) ζ k ( b ∗ ) m − n ( a ∗ ) m + n a m + n b m − n ζ k (cid:1) = h (cid:0) ( b ∗ ) m − n + k + k ( a ∗ ) m + n a m + n b m − n + k + k (cid:1) = d m + n , m − n + k + k = | q | − m − n c m + n , m − n + k + k = | q | − m − n h (cid:0) a m + n b m − n + k + k ( b ∗ ) m − n + k + k ( a ∗ ) m + n (cid:1) = | q | − m − n h e m,n ζ k , e m,n ζ k i R . In the case of m + n ≤ m ≤ n , we have h e m,n ζ k , e m,n ζ k i L = (cid:10) ( b ∗ ) n − m ( a ∗ ) − m − n ζ k , ( b ∗ ) n − m ( a ∗ ) − m − n ζ k (cid:11) L = h (cid:0) ζ k a − m − n b n − m ( b ∗ ) n − m ( a ∗ ) − m − n ζ k (cid:1) = | q | k ( − m − n ) | q | k ( − m − n ) h (cid:0) a − m − n b n − m + k + k ( b ∗ ) n − m + k + k ( a ∗ ) − m − n (cid:1) = | q | k ( − m − n ) | q | k ( − m − n ) c − m − n , n − m + k + k = | q | − m − n | q | k ( − m − n ) | q | k ( − m − n ) d − m − n , n − m + k + k = | q | − m − n | q | k ( − m − n ) | q | k ( − m − n ) h (cid:0) ( b ∗ ) n − m + k + k ( a ∗ ) − m − n a − m − n b n − m + k + k (cid:1) = | q | − m − n h (cid:0) ( b ∗ ) n − m ( a ∗ ) − m − n ζ k + k a − m − n b n − m (cid:1) = | q | − m − n h e m,n ζ k , e m,n ζ k i R . The remaining two cases follow along the line of the above calculations. This proves the firstpart of the claim. To prove the second part, take a polynomial P ( ζ ) = P nk =0 α k ζ k . Invokingthe first part of this Lemma, we have h e m,n P ( ζ ) , e m,n P ( ζ ) i L = n X k =0 n X k =0 α k α k h e m,n ζ k , e m,n ζ k i L = | q | − m − n h e m,n P ( ζ ) , e m,n P ( ζ ) i R . For r ∈ Z , from the commutation relations in 2.1, we have18 D r ) ∗ ζ k e ∗ m,n e m,n ζ k D r = ζ k e ∗ m,n e m,n ζ k . Hence, we get that h e m,n P ( ζ ) D r , e m,n P ( ζ ) D r i L = | q | − m − n h e m,n P ( ζ ) D r , P ( ζ ) e m,n D r i R . (4.5)If r = s , we have from Propn. 3.2 that e m,n P ( ζ ) D r ( e m,n P ( ζ ) D s ) ∗ ∈ A q [0 , , r − s ] and( e m,n P ( ζ ) D r ) ∗ e m,n P ( ζ ) D s ∈ A q [0 , , s − r ]. Using Propn. 3 .
4, we get that h e m,n P ( ζ ) D r , e m,n P ( ζ ) D s i L = 0 = h e m,n P ( ζ ) D r , e m,n P ( ζ ) D s i R . (4.6)Take x ∈ A q [ m, n ]. By part ( v ) of Propn. 3 .
2, we have x = P kr = − k e m,n P r ( ζ ) D r for somepolynomials P r , − k ≤ r ≤ k, k ∈ N . Using eqns. (4 . , . h x, x i L = k X r = − k h e m,n P r ( ζ ) D r , e m,n P r ( ζ ) D r i L = | q | − m − n k X r = − k h e m,n P r ( ζ ) D r , e m,n P r ( ζ ) D r i R = | q | − m − n h x, x i R . ✷ Remark 4.13.
Using the relations in 2.1, it is easy to check that the map σ : A q −→ A q defined by a → q − a, a ∗ → q a ∗ , b → b, b ∗ → b ∗ , D → D, D ∗ → D ∗ extends to an algebraautomorphism. One can now follow (Lemma 3 . h ( xy ) = h ( σ ( y ) x )for x, y ∈ A q , i,e. σ is the modular automorphism. Using this, the above lemma can be proved.However, the proof given here is more straightforward and uses only norms given in Lemmas(4.5,4.6). Theorem 4.14.
For l ∈ N , T Rl is an irreducible corepresentation of A q . Moreover, the matrix (cid:0)(cid:0) t lij (cid:1)(cid:1) of T Rl with respect to the orthonormal basis B l in Propn. . is a unitary element in M l +1 ( C ) ⊗ A q .Proof : Tha claim of unitarizability of T Rl with respect to the inner-product h· , ·i L followsalong the same line of argument in (Propn. 3 .
5, [12]). Since B l is an orthonormal basis ofthe inner-product space ( V Rl , h· , ·i L ), the matrix of T Rl with respect to B l is a unitary matrix.Irreducibility of T Rl follows along the same line of argument in the proof of Thm. 9 in ([9],page 110). ✷ Remark 4.15.
One can also use Propn. 4.9 and directly verify that S (cid:0) t lij (cid:1) = (cid:0) t lji (cid:1) ∗ , in orderto show that (cid:0) t lij (cid:1) is a unitary element in M l +1 ( C ) ⊗ A q .19or m ∈ Z , let V Rl D m = L li = − l C f ( l ) i D m and B l,m = {| q | l | l + 1 | | q | f ( l ) i D m : i ∈ I l } . Definea representation T Rl D m : V Rl D m −→ V Rl D m ⊗ A q by a linear map T Rl D m ( f li D m ) = X j ∈ I l f lj D m ⊗ t lji D m . Theorem 4.16.
For m ∈ Z and l ∈ N , T Rl D m is an irreducible corepresentation of A q andits matrix coefficients with respect to the basis B l are { t lij D m : i, j ∈ I l } . Moreover, the matrix (cid:0)(cid:0) t lij D m (cid:1)(cid:1) is a unitary element of M l +1 ( C ) ⊗ A q .Proof : For i, j ∈ I l , we have∆( t lij D m ) = ∆( t lij )∆( D m ) = ( X k ∈ I l t lik ⊗ t lkj )( D m ⊗ D m ) = X k ∈ I l t lik D m ⊗ t lkj D m . This proves that T Rl D m is a corepresentation of A q . Since T Rl D m ( f li D m ) = P j ∈ I l f lj D m ⊗ t lji D m , it follows that matrix coefficients with respect to the basis B l,m are { t lij D m : i, j ∈ I l } .Moreover, we have S ( t lij D m ) = S ( D m ) S ( t lij ) = ( D m ) ∗ ( t lji ) ∗ = ( t lji D m ) ∗ which shows that T Rl D m is a unitary corepresentation. Irreducibility follows by a similarargument used to prove Thm. 9 in [9] (see page no. 110, [9]). ✷ Theorem 4.17.
Peter-Weyl decomposition :
Given a corepresentation T of A q , let C ( T ) denotes the vector subspace of A q generated by the matrix coefficients of T . Then, we have(i) A q = L l ∈ N ,m ∈ Z C ( T Rl D m ) .(ii) h t lij D m , t l ′ i ′ j ′ D m ′ i R = | q | − j | l + 1 | − | q | δ ll ′ δ ii ′ δ jj ′ δ mm ′ (4.7) h t lij D m , t l ′ i ′ j ′ D m ′ i L = | q | i | l + 1 | − | q | δ ll ′ δ ii ′ δ jj ′ δ mm ′ (4.8)(iii) The set { T Rl D m : l ∈ N , m ∈ Z } is a complete list of irreducible mutually inequivalentcorepresentations of A q .(iv) The set (cid:8) | q | − i | l + 1 | / | q | t lij D m : l ∈ N , m ∈ Z (cid:9) is an orthonormal basis of L ( h ) .Proof : proof of part (i): Fix i, j ∈ Z . To prove the claim, it suffices to show that A q [ − i, − j ] ⊆ L l ∈ N ,m ∈ Z C ( T Rl D m ) . Since, C ( T Rl D m ) = C ( T Rl ) D m and A q [ − i, − j, r ] = A q [ − i, − j, D r for r ∈ Z , it is enoughto show that 20 q [ − i, − j, ⊆ L l ∈ N ,m ∈ Z C ( T Rl D m ) . By Proposition 3 .
2, we have A q [ − i, − j,
0] = e − i, − j C [ ζ ]. Therefore, given any n ∈ N ifwe can exhibit an element in L l ∈ N ,m ∈ Z C ( T Rl D m ) of the form e − i, − j P ( ζ ), where P is apolynomial of degree n , then we get the claim. For that, choose l n = n + max {| i | , | j |} and m n = − l n − i . By part ( ii ) of Propn. 4.9, we have t l n i,j D m n = e − i, − j P i,j ( ζ )where, P i,j is a polynomial of degree n and this completes the proof. proof of part (ii): With Lemma 4.12 at our disposal, we can follow the proof of Thm. 17 in([9], page 115). proof of part (iii):
From part ( i ) of the claim, it follows that { T Rl D m : l ∈ N , m ∈ Z } is a complete list of irreducible corepresentations of A q . Also, if two corepresentations T and T are equivalent then C ( T ) = C ( T ). By part ( ii ), we have C ( T Rl D m ) ⊥ C ( T Rl ′ D m ′ ) if( l, m ) = ( l ′ , m ′ ). This proves the assertion. proof of part (iv): Follows from part ( i ) and ( ii ). ✷ q -Jacobi polynomials and matrix coefficients Recall that the little q -Jacobi polynomials are defined as the following, P ( α,β ) n ( z ; q ) = X r ≥ ( q − n ; q ) r ( q α + β + n +1 ; q ) r ( q ; q ) r ( q α +1 ; q ) r ( qz ) r , where, ( a ; q ) m := Q m − k =0 (1 − aq k ) for any a ∈ C . Theorem 4.18.
The matrix coefficients t lij D k are expressed in terms of the little q-Jacobipolynomials in the following way :(i) for the case of i + j ≤ , i ≥ j , a − ( i + j ) c i − j (¯ q ) ( j − i )( l + j ) ( ll + j ) / | q | ( ll + i ) / | q | (cid:0) l − ji − j (cid:1) | q | P ( i − j, − i − j ) l + j ( bb ∗ ; | q | ) D l + j + k ; (ii) for the case of i + j ≤ , i ≤ j , a − ( i + j ) b j − i q ( i − j )( l + i ) ( ll + j ) / | q | ( ll + i ) / | q | (cid:0) l + jj − i (cid:1) | q | P ( j − i, − i − j ) l + i ( bb ∗ ; | q | ) D l + i + k ; (iii) for the case of i + j ≥ , i ≤ j , q ( i − j )( l + i ) ( ll + j ) / | q | ( ll + i ) / | q | (cid:0) l + jj − i (cid:1) | q | P ( j − i,i + j ) l − j ( bb ∗ ; | q | )( a ∗ ) i + j b j − i D l + i + k ;21 iv) for the case of i + j ≥ , i ≥ j , (¯ q ) ( j − i )( l + j ) ( ll + j ) / | q | ( ll + i ) / | q | (cid:0) l − ji − j (cid:1) | q | P ( i − j,i + j ) l − i ( bb ∗ ; | q | )( a ∗ ) i + j c i − j D l + j + k . Proof : The proof is purely computational and we only mention the key steps of the compu-tations for Case ( i ). All the other cases follow from similar computations. First, by inductionone verifies that ( Db ∗ ) n = (cid:16) ¯ q | q | (cid:17) n − n D n ( b ∗ ) n . Then, using this and the defining relations in2.1, the following holds, a m c l − j − m b n d l + j − n = a m c i − j + n b n d l + j − n = ( − ¯ q ) i − j + n (cid:16) ¯ q | q | (cid:17) n − n a m ( Db ∗ ) i − j D n ( b ∗ ) n b n ( a ∗ ) l + j − n D l + j − n = ( − ¯ q ) n (cid:16) ¯ q | q | (cid:17) n − n a m c i − j D n ( bb ∗ ) n ( a ∗ ) i + j + m D l + j − n = ( − ¯ q ) n (cid:16) ¯ q | q | (cid:17) n − n ( | q | ) − n ( i + j + m ) a m c i − j ( a ∗ ) i + j + m D l + j ( bb ∗ ) n = ( − ¯ q ) n (cid:16) ¯ q | q | (cid:17) n − n ( | q | ) − n ( i + j + m ) (¯ q ) − ( i − j )( i + j + m ) a − ( i + j ) c i − j a i + j + m ( a ∗ ) i + j + m D l + j ( bb ∗ ) n = ( − ¯ q ) n (cid:16) ¯ q | q | (cid:17) n − n ( | q | ) − n ( i + j + m ) (¯ q ) − ( i − j )( i + j + m ) a − ( i + j ) c i − j (cid:16) m + i + j X k =0 ( − k | q | k + k − k ( i + j + m ) (cid:18) m + i + jk (cid:19) | q | ( bb ∗ ) k + n (cid:17) D l + j . Hence, we have the following t lij = l + j X n =0 m + n = l − i ≤ m ≤ l − j q n ( i − j + n ) (cid:0) ll + j (cid:1) / | q | (cid:0) ll + i (cid:1) / | q | (cid:18) l − jl − i − n (cid:19) | q | (cid:18) l + jn (cid:19) | q | a m c l − j − m b n d l + j − n = a − ( i + j ) c i − j (cid:0) ll + j (cid:1) / | q | (cid:0) ll + i (cid:1) / | q | l + j X n =0 α n q n ( i − j + n ) (cid:18) l − jl − i − n (cid:19) | q | (cid:18) l + jn (cid:19) | q | (cid:16) l + j − n X k =0 ( − k | q | k + k − k ( l + j − n ) (cid:18) l + j − nk (cid:19) | q | ( bb ∗ ) k + n (cid:17) D l + j where, α n = ( − ¯ q ) n (cid:16) ¯ q | q | (cid:17) n − n ( | q | ) − n ( l + j − n ) (¯ q ) − ( i − j )( l + j − n ) . Now, l + j X n =0 l + j − n X k =0 ( − k α n q n ( i − j + n ) (cid:18) l − jl − i − n (cid:19) | q | (cid:18) l + jn (cid:19) | q | | q | k + k − k ( l + j − n ) (cid:18) l + j − nk (cid:19) | q | ( bb ∗ ) k + n = l + j X r =0 r X k =0 ( − k α r − k q ( r − k )( i − j + r − k ) (cid:18) l − jl − i − r + k (cid:19) | q | (cid:18) l + jr − k (cid:19) | q | | q | k + k − k ( l + j − r + k ) (cid:18) l + j − r + kk (cid:19) | q | ( bb ∗ ) r P sp =0 P s − pk =0 f ( k, p ) ξ k + p = P sr =0 (cid:16) P rk =0 f ( k, r − k ) (cid:17) ξ r . The coefficientof ( bb ∗ ) r becomes the following, r X k =0 ( − k α r − k q ( r − k )( i − j + r − k ) | q | k + k − k ( l + j − r + k ) (cid:18) l − jl − i − r + k (cid:19) | q | (cid:18) l + jr − k (cid:19) | q | (cid:18) l + j − r + kk (cid:19) | q | = r X p =0 ( − ( r − p ) α p q p ( i − j + p ) | q | ( r − p ) +( r − p ) − r − p )( l + j − p ) (cid:18) l − jl − i − p (cid:19) | q | (cid:18) l + jp (cid:19) | q | (cid:18) l + j − pr − p (cid:19) | q | = r X p =0 ( − ( r − p ) α p q p ( i − j + p ) | q | ( r − p ) +( r − p ) − r − p )( l + j − p ) (cid:18) l − jl − i − p (cid:19) | q | (cid:18) l + jl + j − r (cid:19) | q | (cid:18) rr − p (cid:19) | q | = (cid:18) l + jl + j − r (cid:19) | q | r X p =0 f α p (cid:18) l − jl − i − p (cid:19) | q | (cid:18) rr − p (cid:19) | q | | q | p ( i − j + p ) where, f α p = α p ( − r − p | q | ( r − p ) +( r − p ) − r − p )( l + j − p ) (¯ q ) − p ( i − j + p ) = ( − r − p | q | ( r − p ) +( r − p ) − r − p )( l + j − p ) (¯ q ) − p ( i − j + p ) ( − ¯ q ) p | q | − p ( l + j − p ) (cid:16) ¯ q | q | (cid:17) p − p (¯ q ) − ( i − j )( l + j − p ) = ( − r (¯ q ) ( j − i )( l + j ) | q | r + r − r ( l + j ) , i,e. independent of ‘ p ’. Thus, the coefficient of ( bb ∗ ) r finally becomes the following, (cid:18) l + jl + j − r (cid:19) | q | r X p =0 f α p (cid:18) l − jl − i − p (cid:19) | q | (cid:18) rr − p (cid:19) | q | | q | p ( i − j + p ) = (cid:18) l + jl + j − r (cid:19) | q | ( − r (¯ q ) ( j − i )( l + j ) | q | r + r − r ( l + j ) r X p =0 (cid:18) l − jl − i − p (cid:19) | q | (cid:18) rr − p (cid:19) | q | | q | p ( i − j + p ) = (cid:18) l + jl + j − r (cid:19) | q | (cid:18) l − j + rl − i (cid:19) | q | ( − r (¯ q ) ( j − i )( l + j ) | q | r + r − r ( l + j ) = (¯ q ) ( j − i )( l + j ) (cid:18) l − j + ri − j + r (cid:19) | q | (cid:18) l + jr (cid:19) | q | ( − r | q | r + r − r ( l + j ) = (¯ q ) ( j − i )( l + j ) (cid:18) l − ji − j (cid:19) | q | ( | q | l − j +1) ; | q | ) r ( | q | i − j +1) ; | q | ) r ( | q | − l + j ) ; | q | ) r ( | q | ; | q | ) r | q | r . Therefore, the matrix coefficient t lij D k is the following, t lij D k = a − ( i + j ) c i − j (¯ q ) ( j − i )( l + j ) (cid:0) ll + j (cid:1) / | q | (cid:0) ll + i (cid:1) / | q | (cid:18) l − ji − j (cid:19) | q | (cid:16) l + j X r =0 ( | q | − l + j ) ; | q | ) r ( | q | ; | q | ) r ( | q | l − j +1) ; | q | ) r ( | q | i − j +1) ; | q | ) r ( | q | bb ∗ ) r (cid:17) D l + j + k = a − ( i + j ) c i − j (¯ q ) ( j − i )( l + j ) (cid:0) ll + j (cid:1) / | q | (cid:0) ll + i (cid:1) / | q | (cid:18) l − ji − j (cid:19) | q | P ( i − j, − i − j ) l + j ( bb ∗ ; | q | ) D l + j + k . To each f ∈ A q = O ( U q (2)), consider a matrix ˆ f ( l,k ) = ( ˆ f ( l,k ) m,n ) ∈ M l +1 ( C ) defined by ˆ f ( l,k ) m,n := h ( S ( t lm,n D k ) f ) , l ∈ N , m, n ∈ I l , k ∈ Z , where h is the Haar state on U q (2). The mapping F : A q −→ c A q := M ( l,k ) ∈ N × Z M l +1 ,k ( C ) f ˆ f = ( ˆ f ( l,k ) )where each M l +1 ,k ( C ) = M l +1 ( C ), is called the Fourier transform on the quantum group U q (2). Let τ l : M l +1 ( C ) −→ C be the q -trace defined by τ l ( M ) = P li = − l | q | − i m ii for M =( m ij ). We get an inner-product on M l +1 ( C ) defined by h M, N i l := τ l ( M ∗ N ).Let L ( U q (2)) be the Hilbert space associated with A q with respect to the inner-product h x, y i L := h ( x ∗ y ), and ℓ ( \ U q (2)) be the Hilbert space completion of L ( l,k ) ∈ N × Z M l +1 ,k ( C ),where each M l +1 ,k ( C ) = M l +1 ( C ), with respect to the following inner-product h M, N i = P l ∈ N P k ∈ Z | l + 1 | | q | h M ( l,k ) , N ( l,k ) i l . Theorem 4.19.
The Fourier transform F : A q −→ c A q is a C -isomorphism. The inversetransform is given by f = X l ∈ N X k ∈ Z | l + 1 | | q | X i,j ∈ I l | q | − i ˆ f ( l,k ) j,i t lij D k , and the following Plancherel formula holds h f, g i L = X l ∈ N X k ∈ Z | l + 1 | | q | X i,j ∈ I l | q | − i ˆ f ( l,k ) j,i ˆ g ( l,k ) j,i = X l ∈ N X k ∈ Z | l + 1 | | q | h ˆ f ( l,k ) , ˆ g ( l,k ) i l . The Fourier transform implements a unitary equivalence between the Hilbert spaces L ( U q (2)) and ℓ ( \ U q (2)) .Proof : The Peter-Weyl decomposition (Thm. 4.17) shows that any f ∈ A q is expressed as afinite sum of the following form f = X l ∈ N X k ∈ Z X i,j ∈ I l c ( l ) i,j,k t ( l ) ij D k , (4.9)where the coefficients c ( l ) i,j,k are given by h f, t ( l ) ij D k i L h t ( l ) ij D k , t ( l ) ij D k i − L . Now, h t ( l ) ij D k , f i L = h (( t ( l ) ij D k ) ∗ f ) = h ( S ( t ( l ) ji D k ) f ) = ˆ f ( l,k ) j,i . (4.10)24herefore, eqn. 4.9 and part ( ii ) of Thm. 4.17 gives us the following inverse transform, f = X l ∈ N X k ∈ Z | l + 1 | | q | X i,j ∈ I l | q | − i ˆ f ( l,k ) j,i t lij D k . Now, for any g ∈ A q we get that h f, g i L = X l ∈ N X k ∈ Z | l + 1 | | q | X i,j ∈ I l | q | − i ˆ f ( l,k ) j,i h t lij D k , g i L . Since, h t lij D k , g i L = ˆ g ( l,k ) j,i by eqn. 4.10, we get the following Plancherel formula, h f, g i L = X l ∈ N X k ∈ Z | l + 1 | | q | X i,j ∈ I l | q | − i ˆ f ( l,k ) j,i ˆ g ( l,k ) j,i = X l ∈ N X k ∈ Z | l + 1 | | q | X i | q | − i X j ˆ f ( l,k ) j,i ˆ g ( l,k ) j,i = X l ∈ N X k ∈ Z | l + 1 | | q | τ l (cid:16) ( ˆ f ( l,k ) ) ∗ ˆ g ( l,k ) (cid:17) = X l ∈ N X k ∈ Z | l + 1 | | q | h ˆ f ( l,k ) , ˆ g ( l,k ) i L,l , proving that F is isometry. In order to see surjectivity of F , observe that for any φ = ( ˆ φ ( l,k ) ) ∈ c A q if we choose f = X l ∈ N X k ∈ Z X i,j ∈ I l | q | − l + j ) − | q | l +1) − | q | ˆ φ ( l,k ) i,j t lj,i D k ∈ A q , then it follows that F ( f ) ( l,k ) i,j = ˆ φ ( l,k ) i,j , and consequently F ( f ) = φ . This is because by eqn.4.10, we have F ( f ) ( l,k ) i,j = h (cid:16) S ( t li,j D k ) F ( f ) (cid:17) = h (cid:16)(cid:16) t lj,i D k (cid:17) ∗ F ( f ) (cid:17) = h t lj,i D k , F ( f ) i L = [ F ( f ) ( l,k ) i,j . This completes the proof. ✷ This section describes how the tensor product of two irreducible representations of U q (2) decom-poses into irreducible components. In the following, we denote the character of a representation T by χ ( T ). Theorem 5.1.
One has the following, l ⊗ T ≃ T l + ⊕ T l − D .
Proof : Let T l ⊗ T ≃ M n ∈ N m ∈ Z C nm T n D m . where C nm denotes the multiplicity of the corepresentation T n D m in T l ⊗ T . We have frompart ( i ) of Propn. 4.9, C nm χ ( T n D m ) = χ ( T l ⊗ T ) = χ ( T l ) χ ( T ) = (cid:16) l X i = − l t lii (cid:17) ( a + a ∗ D )= a l +1 + l X i = − l +1 (cid:16) t li − ,i − a ∗ D + t lii a (cid:17) + t lll a ∗ D. (5.1)From Lemma 4.11 and part ( i, iii ) of the Propn. 3 .
2, we have for − l + 1 ≤ i ≤ lt lll a ∗ D ∈ A q [ − l − , − l − , l + 1] t li − ,i − a ∗ D + t li,i a ∈ A q [ − i + 1 , − i + 1 , l + i ] (5.2)Hence, by Propn. 4.4 and Thm 2.8 we get that h χ ( T l ⊗ T ) , a l +1 i L = h a l +1 , a l +1 i L > . On the other hand, it follows from Thm. 4 .
17 that h χ ( T n D m ) , a l +1 i L = h a l +1 , a l +1 i L if n = l + , m = 0 ;0 otherwise ;since, a l +1 = t l + − l − , − l − by part ( i ) of Propn. 4.9. This proves that C l + , = 1. Therefore,we have χ ( T l ⊗ T ) − χ ( T l + ) = X n ∈ N , m ∈ Z ( n,m ) =( l + , χ ( C nm T n D m ) . Invoking eqns. (5 . , .
2) and Propn. 3 .
4, we have h χ ( T l ⊗ T ) − χ ( T l + ) , a l − D i L = h t l − l, − l a ∗ D + t l − l +1 , − l +1 a − t l + − l + , − l + , a l − D i L . Using Propn. 4.9, we have t l − l, − l a ∗ D + t l − l +1 , − l +1 a − t l + − l + , − l + = a l a ∗ D + a l − a ∗ Da − | q | (cid:18) l − l − (cid:19) | q | a l − b ∗ bDa − a l a ∗ D + | q | (cid:18) l l − (cid:19) | q | a l − b ∗ bD = a l − (1 − | q | b ∗ b ) D − | q | (cid:16) | q | + | q | + · · · | q | l − (cid:17) a l − b ∗ bD + | q | (cid:16) | q | + · · · | q | l − (cid:17) a l − b ∗ bD = a l − D − | q | (cid:16) | q | + | q | + · · · | q | l − (cid:17) a l − b ∗ bD + | q | (cid:16) | q | + | q | + · · · | q | l − (cid:17) a l − b ∗ bD = a l − D . h χ ( T l ⊗ T ) − χ ( T l + ) , a l − D i L = h a l − D, a l − D i L . By Thm. 4.17, we have h χ ( T n D m ) , a l − D i L = h a l − D, a l − D i L if n = l − , m = 1 ;0 otherwise . This shows that C l − , = 1. Looking at the dimension of the corepresentations T l , T l + and T l − D , we get C nm = 0 for ( n, m ) = ( l + ,
0) or ( l − , ✷ Let P = χ ( T ) = a + Da ∗ and B be the C ∗ -subalgebra of A q generated by P and D . Observethat B is a commutative C ∗ -algebra. Let α and β be the roots of the quadratic polynomial y − P y + D , given by α = P + √ P − D , β = P − √ P − D . Here, thanks to the L ∞ -functional calculus, we choose a square root of the normal operator P − D in the commutative von-Neumann algebra B ′′ ⊆ C ( U q (2)) ′′ . Therefore, α and β commute with each other. Lemma 5.2.
For l ∈ N , one has the following, χ ( T l ) = P lr =0 α l − r β r . Proof : This obviously holds for l = 0. Assume that it holds for l ∈ { , , , · · · , n − } . ByThm. 5 .
1, and the fact that χ ( T ) = P = α + β and D = αβ , we have χ ( T n ) = χ ( T n − ) χ ( T ) − χ ( T n − ) D = (cid:16) n − X r =0 α n − − r β r (cid:17) ( α + β ) − αβ (cid:16) n − X r =0 α n − − r β r (cid:17) = n − X r =0 α n − r β r + n − X r =0 α n − − r β r +1 − n − X r =0 α n − − r β r +1 = n X r =0 α n − r β r . ✷ Theorem 5.3.
The following decomposition holds, T l D m ⊗ T l D n ≃ T | l + l | D m + n ⊕ T | l + l |− D m + n +1 ⊕ · · · ⊕ T | l − l | D m + n +2 min { l ,l } . Proof : Without loss of generality, assume that l ≥ l . To prove the assertion, it is enough toshow that χ ( T l D m ⊗ T l D n ) = χ ( T l + l D m + n ) + χ ( T l + l − D m + n +1 ) + · · · + χ ( T l − l D m + n +2 l ) . χ ( T D r ) = χ ( T ) D r for any corepresentation T of A q and r ∈ Z , we need to show that χ ( T l ⊗ T l ) = χ ( T l + l ) + χ ( T l + l − D ) + · · · + χ ( T l − l D l ) . Now, l X r =0 χ ( T l + l − r D r ) = l X r =0 2 l +2 l − r X s =0 α r β r α l +2 l − r − s β s = l X r =0 2 l +2 l − r X s =0 α l +2 l − r − s β r + s + l X r = l +1 2 l +2 l − r X s =0 α l +2 l − r − s β r + s = l X r =0 2 l X s =0 α l +2 l − r − s β r + s + l − X r =0 2 l +2 l − r X s =2 l +1 α l +2 l − r − s β r + s + l X r = l +1 2 l +2 l − r X s =0 α l +2 l − r − s β r + s (5.3)Let r ′ = 2 l − r and s ′ = 2 r + s − l . By a change of variable, we get that l − X r =0 2 l +2 l − r X s =2 l +1 α l +2 l − r − s β r + s = l X r ′ = l +1 2 l X s ′ =2 l +2 l − r ′ +1 α l +2 l − r ′ − s ′ β r ′ + s ′ . Putting this in eqn. 5.3, we finally have the following l X r =0 χ ( T l + l − r D r )= l X r =0 2 l X s =0 α l +2 l − r − s β r + s + l X r = l +1 2 l X s =0 α l +2 l − r − s β r + s = l X r =0 2 l X s =0 α l +2 l − r − s β r + s = l X r =0 α l − r β r l X s =0 α l − s β s = χ ( T l ⊗ T l ) , and this completes the proof. ✷ | q | = 1 This section deals with the case of | q | = 1 and q not a root of unity. Let K be the Hilbert space ℓ ( Z ) ⊗ ℓ ( Z ) ⊗ ℓ ( Z ) and U : e n e n +1 be the bilateral shift acting on ℓ ( Z ). Define the28epresentation π of the C ∗ -algebra C ( U q (2)) on K as follows : π ( a ) = U ⊗ I ⊗ U ∗ + U ,π ( b ) = q N ⊗ U ⊗ U ∗ − U i ,π ( D ) = U ⊗ I ⊗ U. (6.1)One can view C ( U q (2)) as a C ∗ -subalgebra of B (cid:0) ℓ ( Z ) ⊗ ℓ ( Z ) (cid:1) ⊗ C ( T ) by identifying U withthe function t : t t . For t ∈ T , let ψ t : B (cid:0) ℓ ( Z ) ⊗ ℓ ( Z ) (cid:1) ⊗ C ( T ) −→ B (cid:0) ℓ ( Z ) ⊗ ℓ ( Z ) (cid:1) be thehomomorphism given by the restriction of I ⊗ I ⊗ ev t to C ( U q (2)), where ev t is the evaluationat t . Proposition 6.1.
The representation π of C ( U q (2)) defined above is faithful.Proof : It is enough to show that all irreducible representations of C ( U q (2)) factor through therepresentation π . Let ρ be an irreducible representation of C ( U q (2)). Using the descriptiongiven in (Thm. 3 .
4, [23]), it is not difficult to establish that if ρ ( b ) = 0 then ρ factors through ψ ◦ π , if ρ ( a ) = 0 then ρ factors through ψ √− ◦ π , and if ρ ( a ) , ρ ( b ) = 0 then ρ factors through ψ t ◦ π for some t = 1 , √− ✷ Theorem 6.2 ([23]) . The Haar state h : C ( U q (2)) −→ C is given as follows : h ( h n, m, k, l i ) = m +1 if m = k, and n = l = 0 , otherwise . It is known that in this case the Haar state is trace [23]. Consider the homomorphism φ : C ( U q (2)) −→ C ( T ) given by φ ( a ) = z , φ ( b ) = 0, and φ ( D ) = 1. This gives a T -coactionΦ : C ( U q (2)) −→ C ( U q (2)) ⊗ C ( T ) on C ( U q (2)) defined by Φ( x ) = (id ⊗ φ ) ◦ ∆ . The quotientspace T(cid:31) U q (2) (cid:30)T is defined as follows : C ( T(cid:31) U q (2) (cid:30)T ) = { x ∈ C ( U q (2)) : ( φ ⊗ id ⊗ φ )(∆ ⊗ id)∆( x ) = 1 ⊗ x ⊗ } . In such a case, the conditional expectation E : C ( U q (2)) −→ C ( T(cid:31) U q (2) (cid:30)T ) is defined to bethe map (( h T ◦ φ ) ⊗ id ⊗ ( h T ◦ φ ))(∆ ⊗ id) ◦ ∆ . Lemma 6.3.
The C ∗ -algebra C ( T(cid:31) U q (2) (cid:30)T ) is the C ∗ -subalgebra of C ( U q (2)) generated by D .Proof : Let O ( U q (2)) be the ⋆ -subalgebra of C ( U q (2)) generated by a, b and D . Then, we have C ( T(cid:31) U q (2) (cid:30)T ) = E ( C ( U q (2)) = E ( O ( U q (2))) = E ( O ( U q (2))) . (6.2)29ne has the following, ( h T ◦ φ )( t li,j D m ) = l = 0 , l = 0 . Hence, E ( t li,j D m ) = X r,s ∈{− l,...,l } ( h T ◦ φ )( t li,r D m ) t lr,s D m ( h T ◦ φ )( t ls,j D m ) = D m if l = 0 , l = 0 . Using this in eqn. 6.2 we get our claim. ✷ Theorem 6.4.
The Haar state h on the quantum group U q (2) is faithful.Proof : Let C := C ∗ ( D ) be the C ∗ -subalgebra of C ( U q (2)) generated by D . Invoking theargument used in Thm. 2 .
8, it is enough to show that h | C is faithful. Observe that the map D z is a C ∗ -algebra isomorphism between C and C ( T ). For m ∈ Z , we have by Thm. 6 . h | C ( D m ) = m = 0 , . Hence, h | C is same as the Haar state on C ( T ), which is faithful. ✷ With Propn. 6.1 and Thm. 6.4 in hand, one can check that all the results obtained in Sec § § Lemma 6.5.
For r, s ∈ N , let d r,s = k a r b s k L and c r,s = k a r b s k R . Then, one has the followings.(i) d r,s = d r − ,s − d r − ,s +1 ∀ r ≥ (ii) d r,s = (cid:16) ( r + s + 1) (cid:0) r + ss (cid:1)(cid:17) − ; (iii) d r,s = c r,s .Proof : The last part follows from the traciality of the Haar state. ✷ The Peter-Weyl decomposition in this case is obtained in [25], but here we get the followingbetterment including the norm factor of the matrix coefficients.
Theorem 6.6. (i) For l ∈ N , let t lij , i, j ∈ I l be the matrix coefficients of the corepresenta-tion T Rl with respect to the basis { f lj : j ∈ I l } . Then, we have t lij = X m + n = l − i ≤ m ≤ l − j ≤ n ≤ l + j q n ( l − j − m ) (cid:0) ll + j (cid:1) / (cid:0) ll + i (cid:1) / (cid:18) l − jm (cid:19)(cid:18) l + jn (cid:19) a m c l − j − m b n d l + j − n ;30 ii) h t lij D m , t l ′ i ′ j ′ D m ′ i = (2 l + 1) − δ ll ′ δ ii ′ δ jj ′ δ mm ′ ; (iii) The set (cid:8) √ l + 1 t lij D m : l ∈ N , m ∈ Z (cid:9) is an orthonormal basis of L ( h ) . Theorem 6.7.
The Fourier transform F : A q −→ c A q is a C -isomorphism. The inversetransform is given by f = X l ∈ N X k ∈ Z (2 l + 1) X i,j ∈ I l ˆ f ( l,k ) j,i t li,j D k , and the following Plancherel formula h f, g i = X l ∈ N X k ∈ Z (2 l + 1) X i,j ∈ I l ˆ f ( l,k ) j,i ˆ g ( l,k ) j,i = X l ∈ N X k ∈ Z (2 l + 1) h ˆ f ( l,k ) , ˆ g ( l,k ) i l holds. Here, the notations are same as defined in subsection . , but the q -trace involved in h . , . i l becomes the usual matrix trace. U q (2) In this section, we classify the compact quantum group U q (2) for q ∈ C ∗ and not roots of unity.This subsumes earlier investigation in [26]. For the case of SU q (2) , q ∈ R ∗ , see [18]. Since, wewill be dealing with different values of q at the same time, the generators of U q (2) are denotedby a q , b q and D q in order to avoid any confusion. Let Ω denotes the set of all roots of unity. Theorem 7.1.
Let q and q ′ be two non-zero complex numbers which are not roots of unity.Then, U q (2) and U q ′ (2) are isomorphic if and only q ′ ∈ { q, q, q , q } .Proof : Step 1:
For q ∈ C ∗ \ Ω , U q (2) and U q (2) are isomorphic.Let q ′ = 1 /q . Define the map Φ : C ( U q (2)) −→ C ( U q ′ (2)) by Φ( a q ) = a ∗ q ′ D q ′ , Φ( b q ) = − q ′ b ∗ q ′ D q ′ , Φ( D q ) = D q ′ . Using the relations in 2.1, one can show that Φ is a C ∗ -algebrahomomorphism. The inverse map Ψ : C ( U q ′ (2)) −→ C ( U q (2)) is given by a q ′ a q D ∗ q , b q ′ qb ∗ q D q , D q ′ D q . Moreover, to show that ∆ q ′ ◦ Φ = (Φ ⊗ Φ) ◦ ∆ q it is enough to checkthis condition on the generators a q , b q and D q of C ( U q (2)), since Φ is a homomorphism. Thisfollows easily from the eqn. 2.2. Step 2:
For q ∈ C ∗ \ Ω , U q (2) and U q (2) are isomorphic.Let q ′ = 1 /q . Define the map Φ : C ( U q (2)) −→ C ( U q ′ (2)) by Φ( a q ) = a ∗ q ′ , Φ( b q ) = q ′ b ∗ q ′ , Φ( D q ) = D ∗ q ′ . Using the relations in 2.1, one can show that Φ is a C ∗ -algebra homomorphism. The in-verse map Ψ : C ( U q ′ (2)) −→ C ( U q (2)) is given by a q ′ a ∗ q , b q ′ qb ∗ q , D q ′ D ∗ q . Moreover,31o show that ∆ q ′ ◦ Φ = (Φ ⊗ Φ) ◦ ∆ q it is enough to check this condition on the generators a q , b q and D q of C ( U q (2)), since Φ is a homomorphism. This follows easily from the eqn. 2.2. Step 3:
For q ∈ C ∗ \ Ω , U q (2) and U q (2) are isomorphic.This follows from Step 1 and 2. Step 4:
For q, q ′ ∈ C ∗ \ Ω, let U q (2) and U q ′ (2) be isomorphic. Then, q ′ ∈ { q, q, q , q } .Let Ψ : C ( U q (2)) −→ C ( U q ′ (2)) be a quantum group isomorphism. Then, Ψ induces a bijectivecorrespondence between one dimensional representations of U q (2) and U q ′ (2). From Thm.4.17, for one dimensional representations one has l = 0, and thus it is D k for k ∈ Z . Hence, Ψ ( D q ) = D kq ′ with k ∈ Z . Since Ψ is bijective, one has k = ± Ψ ( D q ) = D q ′ or Ψ ( D q ) = D ∗ q ′ . First assume that Ψ ( D q ) = D q ′ . Let u = " a q b q − qb ∗ q D q D q a ∗ q , v m = " a q ′ D mq ′ b q ′ D mq ′ − q ′ b ∗ q ′ D m +1 q ′ a ∗ q ′ D m +1 q ′ . It is known that two equivalent unitary representations π , π of a compact quantum group areunitarily equivalent. Since, Ψ ( u ) is unitarily equivalent to v m for some m ∈ Z , there exists a2 × C = (cid:0) c ij (cid:1) such that v m = CΨ ( u ) C − = Ψ ( CuC − ) . Then, we have (
CuC − ) ( D ∗ q ) m = (cid:16) ( CuC − ) ( D ∗ q ) m +1 (cid:17) ∗ ( CuC − ) ( D ∗ q ) m = (cid:16) − q ′ ( CuC − ) ( D ∗ q ) m +1 (cid:17) ∗ (7.3)This follows from the observation that applying Ψ on both sides they agree, and Ψ is anisomorphism. Using the fact that C − = C ∗ = " c c c c , from eqn. 7 . c c a q ( D ∗ q ) m − qc c b ∗ q ( D ∗ q ) m − + c c b q ( D ∗ q ) m + c c a ∗ q ( D ∗ q ) m − = c c a ∗ q D m +1 q − ¯ q (cid:16) | q | q (cid:17) m c c b q D mq + (cid:16) | q | q (cid:17) m +1) c c b ∗ q D m +1 q + c c a q D mq (7.4)and c c a q ( D ∗ q ) m − qc c b ∗ q ( D ∗ q ) m − + c c b q ( D ∗ q ) m + c c a ∗ q ( D ∗ q ) m − = − q ′ c c a ∗ q D m +1 q + ¯ qq ′ (cid:16) | q | q (cid:17) m c c b q D mq − q ′ (cid:16) | q | q (cid:17) m +1) c c b ∗ q D m +1 q − q ′ c c a q D mq . (7.5)32ince C = ( c ij ) is a non-zero matrix, i,e. not all its entries are qual to zero simultaneously,it follows from these equations that m = 0 due to Thm. 2.3. Now, equating coefficients fromboth sides of eqns. (7 . , . c c = c c , c c = c c , c c = ¯ qq ′ c c , ¯ qc c = 1 q ′ c c . (7.6)If c and c both are non-zero then eqn. 7 . C are zero. This forces q = q ′ and q = 1 /q ′ simultaneously and hence, q = q ′ = ± ∈ Ω.Therefore, either c = 0 or c = 0, as both c and c can not be simultaneously zero since C is a non-zero matrix (again from eqn. 7 . c = 0 then, c = 0 , c = 0 and c = 0,and we have q = 1 /q ′ from eqn. 7 .
6. If c = 0 then, c = 0 , c = 0 and c = 0, and we have q = q ′ from eqn. 7 . Ψ ( D q ) = D ∗ q ′ then, we replace D ∗ q by D q in eqn. 7.3 and proceeding similarly as abovewe get the following equation, c c = c c , c c = c c , c c = qq ′ c c , qc c = 1 q ′ c c . (7.7)From this equation it will follow that either q = q ′ or q = q ′ by the same justification as above.Combining all these steps, we get the assertion. ✷ Remark 7.2.
This theorem justifies that for | q | 6 = 1, it is enough to do the computations for | q | < Acknowledgements
Satyajit Guin acknowledges useful discussion with Sutanu Roy on the article [8] and hospitalityprovided by NISER Bhubaneswar during his visit. He also acknowledges the support of DSTINSPIRE Faculty award grant DST/INSPIRE/04/2015/000901. Bipul Saurabh acknowledgesthe support of SERB grant SRG/2020/000252.
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Satyajit Guin ( [email protected] ) Department of Mathematics and Statistics,Indian Institute of Technology, Kanpur,Uttar Pradesh 208016, India