The coordinate algebra of a quantum symplectic sphere does not embed into any C*-algebra
TTHE COORDINATE ALGEBRA OF A QUANTUM SYMPLECTIC SPHEREDOES NOT EMBED INTO ANY C ∗ -ALGEBRA FRANCESCO D’ANDREA AND GIOVANNI LANDIA
BSTRACT . We generalize a result of Mikkelsen–Szyma ´nski and show that, for every n (cid:62) , anybounded ∗ -representation of the quantum symplectic sphere S n − q has the first n − generatorsin its kernel. We then classify irreducible representations of the corresponding algebra A ( S n − q ) .
1. I
NTRODUCTION AND MAIN RESULTS
A remarkable and unexpected property of the quantum symplectic sphere S q has been re-cently unveiled in [4]. It is shown there that the coordinate algebra of S q does not embed intoits enveloping C ∗ -algebra: this results from the discovery that in any bounded ∗ -representationof the sphere S q one of the generator vanishes. One wonders if this property is an isolated ac-cident or is an example of a more general behaviour. In the present short note we show that thelatter is the case. The sphere S q was used in [3] as the total space algebra of a quantum SU q ( ) instanton bundle over a sphere S q . It is one of a family of spheres associated to quantumsymplectic groups A ( Sp q ( n )) with the symplectic R -matrix of [1].Let < q < be a deformation parameter and n be a positive integer (this notation will beused throughout the paper). We denote by A ( S n − q ) the complex unital ∗ -algebra generated byelements { x i , y i } ni = and their adjoints, with the relations in the Definition 6 below, in particularwith the sphere relation (15). This sphere is a comodule algebra for the quantum symplecticgroup A ( Sp q ( n )) , with a coaction A ( S n − q ) → A ( Sp q ( n )) ⊗ A ( S n − q ) . The main point of thisnote is to prove the following theorem, that generalizes the one proved in [4] for n = . Theorem 1. If π is any bounded ∗ -representation of A ( S n − q ) , then π ( x i ) = for all (cid:54) i < n . For n = the statement is empty. One may also observe that A ( S q ) (cid:39) A ( SU q ( )) . This resultsingles out the quantum symplectic spheres from the quantum unitary and orthogonal ones.Next, let A ( Σ n + q ) be the quotient of A ( S n − q ) by the two-sided ∗ -ideal generated by the n − elements { x i } n − i = . As costumary, we interpret a quotient algebra as given by “functions”on a quantum subspace and think of Σ n + q as a quantum subsphere of S n − q . It follows fromTheorem 1 that any bounded ∗ -representation of A ( S n − q ) comes from one of A ( Σ n + q ) . Now,if we further quotient by the ideal generated by the coordinate x n , we get one of Vaksman-Soibelman quantum spheres [5], whose representation theory is well known (see e.g. [2]). We Date : January 2021.2010
Mathematics Subject Classification.
Primary: 20G42; Secondary: 58B32; 58B34.
Key words and phrases.
Quantum symplectic groups, quantum symplectic spheres, representation theory. a r X i v : . [ m a t h . QA ] J a n re then interested in representations of A ( Σ n + q ) with x n not in the kernel. A family of repre-sentations, labelled by U ( ) , are exhibited in Proposition 7 below. We next show that there areno more irreducible ∗ -representations of A ( Σ n + q ) . Theorem 2.
Any irreducible bounded ∗ -representation of A ( Σ n + q ) with x n not in the kernel is uni-tarily equivalent to one of the representations in Proposition 7. The paper is organized as follows. In Section 2 we prove some preliminary lemmas, inSection 4 we prove Theorem 1, while in Section 5 we prove Theorem 2.2. S
OME PRELIMINARY LEMMAS If T is a bounded operator, we will denote by σ ( T ) its spectrum. Lemma 3. If T is a bounded selfadjoint operator on a Hilbert space H and σ ( T ) has at most finitely-many non-isolated point, then T can be diagonalized.Proof. Let S ⊂ σ ( T ) be the subset of non-isolated points and assume it is finite. Isolated pointsin the spectrum of a selfadjoint operator are eigenvalues. Decompose H = H ⊕ H where H is the direct sum of all eigenspaces of T and H the orthogonal complement. For i =
1, 2 ,call T i the restriction of T to H i and note that T i ( H i ) ⊂ H i . Thus T = T ⊕ T . Since T hasno eigenvectors belonging to H , T has no eigenvectors. Since σ ( T ) ⊂ σ ( T ) and σ ( T ) hasno isolated point, no point of σ ( T ) (cid:114) S belongs to σ ( T ) . Thus σ ( T ) ⊂ S . But again, since itcannot have isolated points, it must be σ ( T ) = ∅ . Since every bounded operator on a non-zeroHilbert space has non-empty spectrum, it must be H = { } . (cid:4) Lemma 4.
Let < µ < , < t (cid:54) , m ∈ N , and let A (cid:62) and B bounded operators on a Hilbertspace H with norm (cid:54) t such that: BA = µ AB BB ∗ = µB ∗ B + ( − µ )( t + µ m A ) (1) Then BB ∗ and B ∗ B are diagonalizzable, with spectrum: σ ( BB ∗ ) ⊂ (cid:8) t ( − µ k + ) : k ∈ N (cid:9) ∪ (cid:8) t (cid:9) , σ ( B ∗ B ) ⊂ (cid:8) t ( − µ k ) : k ∈ N (cid:9) ∪ (cid:8) t (cid:9) . (2) Moreover: (i) If m (cid:62) , then A = . (ii) If m = and there exists a bounded operator C such that CA = µAC [ C , C ∗ ] = ( − µ ) A , (3) then A = .Proof. We follow [4]. From the first relation in (1) we deduce that B ∗ B commutes with A , andfrom the second that B ∗ B commutes with BB ∗ as well. Since BB ∗ (cid:62) ( − µ ) t + µB ∗ B , thejoint spectrum σ ( BB ∗ , B ∗ B ) , i.e. the set (cid:8) (cid:0) χ ( BB ∗ ) , χ ( B ∗ B ) (cid:1) : χ is a character of C ∗ ( BB ∗ , B ∗ B ) (cid:9) ,is contained in the closed triangle displayed in Fig. 1. Note that ( x , y ) ∈ σ ( BB ∗ , B ∗ B ) implies x ∈ σ ( BB ∗ ) and y ∈ σ ( B ∗ B ) . Note also that: σ ( BB ∗ ) ∪ { } = σ ( B ∗ B ) ∪ { } . ( B ∗ B ) σ ( BB ∗ ) ( − µ ) t a ( − µ ) t ( − µ ) t t · · · ( − µ ) t ( − µ ) tt ... b c F IGURE
1. Joint spectrum of BB ∗ and B ∗ B .Since ] ( − µ ) t [ / ∈ σ ( BB ∗ ) , the same interval does not belong to σ ( B ∗ B ) , hence the opensegment from a to b does not belong to the joint spectrum. This implies that ]( − µ ) t , ( − µ ) t [ / ∈ σ ( BB ∗ ) , hence ]( − µ ) t , ( − µ ) t [ / ∈ σ ( B ∗ B ) and the open segment from b to c doesnot belong to the joint spectrum. It turn, his implies ]( − µ ) t , ( − µ ) t [ / ∈ σ ( BB ∗ ) , etc. Byinduction one establishes (2).We will use the convention that µ + ∞ = . It follows from Lemma 3 that both BB ∗ and B ∗ B can be diagonalized, since they have only one non-isolated point in the spectrum. Thus H = (cid:76) j , k ∈ N ∪ { + ∞ } V j , k where V j , k := (cid:8) v ∈ H : BB ∗ v = ( − µ j + ) tv ∧ B ∗ Bv = ( − µ k ) tv (cid:9) ( V j , k may be { } for some values of j and k ). Call A (cid:48) := t − A . From (1) we deduce that A (cid:48) v = λ j , k v for all v ∈ V j , k , with: λ j , k = µ k + − µ j + µ m ( − µ ) , thus either λ j , k is an eigenvalue of A (cid:48) , or V j , k = { } .Since A (cid:48) (cid:62) and λ j , k < for j < k , it must be V j , k = { } ∀ j < k . Thus H = (cid:77) j , k ∈ N ∪ { + ∞ } V (cid:48) j , k . where we call V (cid:48) j , k := V j + k , k and set λ (cid:48) j , k := λ j + k , k = µ k µ m − − µ j − µ . Let v ∈ V (cid:48) j , k and notice that, if k (cid:54) = , then Bv = implies v = t − ( − µ k ) − B ∗ Bv = . Itfollows from (1) that, if v (cid:54) = and k (cid:62) , then Bv is an eigenvector of A (cid:48) with eigenvalue µ − λ (cid:48) j , k = λ (cid:48) j , k − . By induction, if λ (cid:48) j , k is an eigenvalue of A (cid:48) and k is even, then λ (cid:48) j ,0 is an igenvalue of A (cid:48) as well; if λ (cid:48) j , k is an eigenvalue of A (cid:48) and k is odd, then λ (cid:48) j ,1 is an eigenvalueof A (cid:48) as well. But for all j (cid:62) : λ (cid:48) j ,0 (cid:62) λ (cid:48) = µ m − > if m (cid:62) and λ (cid:48) j ,1 (cid:62) λ (cid:48) = µ m − > if m (cid:62) Since A (cid:48) (cid:54) , for n (cid:62) we deduce that λ (cid:48) j , k is not an eigenvalue of A (cid:48) for any j (cid:62) and k (cid:62) .Thus V (cid:48) j , k = { } ∀ j (cid:54) = and H = (cid:76) k ∈ N ∪ { ∞ } V (cid:48) k . Since λ (cid:48) k = , clearly A (cid:48) = . This proves (i).Assume now that n = and that there exists C satisfying (3). Let j (cid:62) . We already provedthat λ (cid:48) j , k is not an eigenvalue of A (cid:48) if k is even. If k is odd, from the commutation relation wededuce that, for all v ∈ V (cid:48) j , k , Cv is either zero or an eigenvector of A (cid:48) with eigenvalue λ (cid:48) j , k − ,and C ∗ v is either zero or an eigenvector of A (cid:48) with eigenvalue λ (cid:48) j , k + . From the argumentabove, it must be Cv = C ∗ v = . But then = [ C , C ∗ ] v = ( − µ ) Av = ( − µ ) tλ (cid:48) j , k v , which imply v = since λ (cid:48) j , k (cid:54) = . Thus V (cid:48) j , k = { } for all j (cid:62) and any k (even or odd). Since H = (cid:76) k ∈ N ∪ { ∞ } V (cid:48) k and λ (cid:48) k = , clearly A = . This proves (ii). (cid:4) Lemma 5.
Let A (cid:62) and B be bounded operators on a Hilbert space H satisfying [ B , B ∗ ] = ( − µ ) A A + B ∗ B = (4) with < µ < . Then, ker ( B ) = { } ⇐⇒ A = .Proof. If A = , from (4) it follows that B is unitary, so ker ( B ) = { } . We have to prove theopposite implication. From (4) we deduce that (cid:107) A (cid:107) (cid:54) and: BB ∗ = B ∗ B + ( − µ ) A = − µA . (5)Since (cid:107) µA (cid:107) < , BB ∗ has a bounded inverse. Therefore U := ( BB ∗ ) − / B is well defined.Notice that UU ∗ = and ker ( B ) = ker ( U ) . From (5): µA = − BB ∗ = U ( − B ∗ B ) U ∗ = UAU ∗ . Assume ker ( U ) = { } . The identity U ( − U ∗ U ) = implies U ( − U ∗ U ) v = and then ( − U ∗ U ) v = for all v ∈ H . Therefore U ∗ U = and U is a unitary operator.Let λ ∈ C and suppose A − µ − λ has a bounded inverse. Then A − λ = µU ∗ ( A − µ − λ ) U has a bounded inverse as well. Thus, λ ∈ σ ( A ) implies µ − λ ∈ σ ( A ) and, by induction, µ − k λ ∈ σ ( A ) for all k (cid:62) . Since A is bounded and the sequence ( µ − k λ ) k (cid:62) is divergent if λ (cid:54) = , we deduce that the only point in the spectrum of A is λ = . Hence A = . (cid:4)
3. T
HE SYMPLECTIC QUANTUM SPHERES AND THEIR QUOTIENTS
Definition 6.
We denote by A ( S n − q ) the complex unital ∗ -algebra generated by elements { x i , y i } ni = and their adjoints, subject to the following relations, that we divide in three groups. One has: x i x j = q − x j x i ∀ i < j , y i y j = q − y j y i ∀ i > j , x i y j = q − y j x i ∀ i (cid:54) = j , (6) y i x i = q x i y i + ( q − ) i − (cid:88) k = q i − k x k y k , (7) lus x i x ∗ i = x ∗ i x i + ( − q ) i − (cid:88) k = x ∗ k x k (8) y i y ∗ i = y ∗ i y i + ( − q ) (cid:14) q ( n + − i ) x ∗ i x i + n (cid:88) k = x ∗ k x k + n (cid:88) k = i + y ∗ k y k (cid:15) (9) x i y ∗ i = q y ∗ i x i (10) x i x ∗ j = qx ∗ j x i ∀ i (cid:54) = j (11) y i y ∗ j = qy ∗ j y i − ( q − ) q n + − i − j x ∗ i x j ∀ i (cid:54) = j (12) x i y ∗ j = qy ∗ j x i ∀ i < j (13) x i y ∗ j = qy ∗ j x i + ( q − ) q i − j y ∗ i x j ∀ i > j (14) and the sphere relation: n (cid:88) i = ( x ∗ i x i + y ∗ i y i ) = (15)One passes to the notations of [3] by calling y i = x n + − i and replacing q by q − (theassumption in [4] that q > becomes our assumption that < q < ). As already mentionedthe sphere comes with a coaction A ( S n − q ) → A ( Sp q ( n ) ⊗ A ( S n − q ) .Let A ( Σ n + q ) be the quotient of A ( S n − q ) by the two-sided ∗ -ideal generated by { x i } n − i = (thus Σ n + q is a quantum subsphere of S n − q ). Let us write down explicitly the relations in thequotient algebra A ( Σ n + q ) . If we rename y n + := x n , the relations become: y i y j = q − y j y i ∀ i > j , { i , j } (cid:54) = { n , n + } (16) y ∗ i y j = q − y j y ∗ i ∀ i (cid:54) = j , { i , j } (cid:54) = { n , n + } (17) y n + y n = q − y n y n + y ∗ n + y n = q − y n y ∗ n + (18) [ y i , y ∗ i ] = ( − q ) n + (cid:88) k = i + y ∗ k y k i (cid:54) = n (19) [ y n , y ∗ n ] = ( − q ) y ∗ n + y n + (20)plus the ones obtained by adjunction and the sphere relation: n + (cid:88) i = y ∗ i y i = (21)In particular, y n + is normal.It follows from Theorem 1 that any bounded ∗ -representation of A ( S n − q ) comes from oneof A ( Σ n + q ) . If we further quotient by the ideal generated by x n , we get one of Vaksman-Soibelman quantum spheres [5], whose representation theory is well known (see e.g. [2]). Weturn then to the representations of A ( Σ n + q ) with x n not in the kernel.It is straightforward to check the following statement. roposition 7. For every λ ∈ U ( ) , an irreducible bounded ∗ -representation π λ of A ( Σ n + q ) on (cid:96) ( N n ) is given by the formulas: π λ ( y i ) | k (cid:105) = q k + ... + k i − (cid:112) − q k i | k − e i (cid:105) ∀ (cid:54) i (cid:54) n − π λ ( y n ) | k (cid:105) = q k + ... + k n − (cid:112) − q k n | k − e n (cid:105) , π λ ( x n ) | k (cid:105) = λq | k | + k n | k (cid:105) , where k = ( k , . . . , k n ) ∈ N n , | k | := k + . . . + k n , {| k (cid:105) } k ∈ N n is the canonical orthonormal basis of (cid:96) ( N n ) and e i the i -th row of the identity matrix of order n . This proposition will be used in Section 5 for the proof of Theorem 2 showing there are noadditional irreducible ∗ -representations.4. P ROOF OF T HEOREM
Theorem 1. If π is any bounded ∗ -representation of A ( S n − q ) , then π ( x i ) = for all (cid:54) i < n . We need the following results.
Lemma 8. If π is any bounded ∗ -representation of A ( S n − q ) then, for all (cid:54) k (cid:54) n − , (i k ) π ( x i ) = ∀ (cid:54) i (cid:54) k , (ii k ) σ (cid:18) (cid:88) ki = π ( y ∗ i y i ) (cid:19) ⊂ (cid:8) − q j : j ∈ N (cid:9) ∪ (cid:8) (cid:9) .Proof. For n = the above statements are empty, so we assume n (cid:62) (although the case n = was proved already in [4]). From (9) and (15) we get: y y ∗ = q y ∗ y + ( − q )( + q n x ∗ x ) . (22)Using Lemma 4 with A = π ( x ∗ x ) , B = π ( y ) and µ := q , we deduce A = , that implies (i ).The statement (ii ) is simply (2), and for n = this concludes the proof.For n (cid:62) , we will prove the Lemma by induction on k , having just proved (i ) and (ii ).Assume by inductive hypothesis that (i k − ) and (ii k − ) are true for some fixed (cid:54) k (cid:54) n − .Let D := (cid:80) k − i = π ( y ∗ i y i ) . It follows from (ii k − ) and Lemma 3 that D can be diagonalized.Decompose: H = (cid:77) j ∈ N ∪ { + ∞ } H j as a orthogonal direct sum, where H j := (cid:8) v ∈ H : Dv = ( − q j ) v (cid:9) and q + ∞ = by convention (notice that H m may be zero for some values of m ).From (12), (14) and the inductive hypothesis we deduce that π ( x k ) π ( y ∗ i ) = qπ ( y ∗ i ) π ( x k ) π ( y k ) π ( y ∗ i ) = qπ ( y ∗ i ) π ( y k ) for all i (cid:54) k − . Combined with (6) we get: [ π ( x k ) , π ( y ∗ i y i )] = [ π ( y k ) , π ( y ∗ i y i )] = ∀ i (cid:54) k − These (and the conjugated) imply that π ( x k ) , π ( y k ) and adjoints all map each space H j to itself. n H + ∞ , since D is the identity, it follows from (15) that π ( x ∗ k x k ) and π ( y k ) must vanish.Thus, on H + ∞ , π ( x k ) is zero and (cid:80) ki = π ( y ∗ i y i ) = π ( y ∗ k y k ) + D is the identity. It remains tounderstand what happens on H j for j ∈ N .Fix j ∈ N , call µ := q , t := q j , A the restriction of π ( x ∗ k x k ) to H j and B the restriction of π ( y k ) to H j . Call m := n + − k and notice that m (cid:62) . From (9) and (15) we deduce that BB ∗ = µB ∗ B + ( − µ )( + µ m A − D | H j ) . But D | H j = − t , hence the second relation in (1) is satisfied. The first one follows from (7),(10) and the inductive hypothesis. Finally, from (15) it follows A + B ∗ B (cid:54) − D | H j = t , thus A and B both have norm (cid:54) t and we can use Lemma 4.From (2) it follows that on H j the operator (cid:80) ki = π ( y ∗ i y i ) = B ∗ B + D has spectrum containedin the set (cid:8) t ( − µ r ) + ( − t ) = − tµ r : r ∈ N ∪ { + ∞ } (cid:9) = (cid:8) − q s : s = j , j +
1, . . . , + ∞(cid:9) . This proves (ii k ). If m (cid:62) , from Lemma 4(i) we deduce (i k ). If m = , that means k = n − ,we call C := π ( x n ) and from Lemma 4(ii) we deduce (i n − ). (cid:4) Proof of Theorem 1.
The statement (i n − ) in Lemma 8 is exactly Theorem 1. (cid:4)
5. P
ROOF OF T HEOREM
Theorem 2
Any irreducible bounded ∗ -representation of A ( Σ n + q ) with x n not in the kernel is uni-tarily equivalent to one of the representations in Proposition 7. Recall the relations (17)-(21) for the generators { y i } of the algebra A ( Σ n + q ) . Then, Lemma 9.
For all m (cid:62) and all (cid:54) i < n : y i ( y ∗ i ) m = q m ( y ∗ i ) m y i + ( − q m )( y ∗ i ) m − (cid:16) − (cid:88) k
For m = the identities come from (19) and (20), that using (21) we rewrite as: y i y ∗ i = q y ∗ i y i + ( − q ) (cid:16) − (cid:88) k
Let π be an irreducible bounded ∗ -representation of A ( Σ n + q ) with π ( y n + ) (cid:54) = . Then: (i) π ( y n + ) is injective; (ii) there exists a vector ξ (cid:54) = such that π ( y i ) ξ = for all i (cid:54) = n + .Proof. (i) If a is any generator different from y n + , since y n + a is proportional to ay n + , π ( a ) maps the kernel of π ( y n + ) to itself. Hence, ker π ( y n + ) carries a subrepresentation of H ,and since π is irreducible either ker π ( y n + ) = { } or ker π ( y n + ) = H . The latter implies π ( y n + ) = , contraddicting the hypothesis, so the former must hold. ii) For (cid:54) k (cid:54) n call H k := (cid:84) ki = ker π ( y i ) . We prove by induction on k that H k (cid:54) = { } . For k = this follows from Lemma 5 applied to the operators A = (cid:80) n + i = π ( y ∗ i y i ) and B = π ( y ) .Since π ( y n + ) (cid:54) = , one has A (cid:54) = and then ker ( B ) (cid:54) = { } .Now assume that H k − (cid:54) = { } , call A = (cid:80) n + i = k + π ( y ∗ i y i ) and B = π ( y k ) , and note that theoperators π ( y n + ) , A , B , B ∗ map H k − to itself (because y i y j is proportional to y j y i and y i y ∗ j is proportional to y ∗ j y i for all i (cid:54) = j ). It follows from point (i) that π ( y n + ) | H k − (cid:54) = , and then A is not zero on H k − . The operator A + B ∗ B is the identity on H k − and [ B , B ∗ ] = ( − µ ) A with µ = q if k < n and µ = q if k = n . From Lemma 5 applied to the restrictions of A , B , B ∗ to H k − it follows that ker ( B ) ∩ H k − = H k (cid:54) = { } . (cid:4) Proof of Theorem 2.
Let π be a bounded irreducible ∗ -representation of A ( Σ n + q ) on a Hilbertspace H such that π ( y n + ) (cid:54) = . Let us omit the representation symbol. We know fromLemma 10(ii) that V := (cid:84) ni = ker ( y i ) is not { } . From the commutation relations we deducethat y n + V ⊂ V , that is V carries a bounded ∗ -representation of the commutative C*-algebra C ∗ ( y n + , y ∗ n + ) generated by y n + and y ∗ n + .For k ∈ N n and ξ ∈ V a unit vector, define: | k (cid:105) ξ := (cid:112) ( q ; q ) k . . . ( q ; q ) k n − ( q ; q ) k n ( y ∗ ) k . . . ( y ∗ n ) k n ξ , where the q -shifted factorial is given by ( a ; b ) (cid:96) := (cid:96) − (cid:89) i = ( − ab i ) . Set | k (cid:105) := if one of the components of k is negative. From the commutation relations wededuce: y ∗ i | k (cid:105) ξ = q k + ... + k i − (cid:112) − q k i + | k + e i (cid:105) ξ ∀ i < n , (23a) y ∗ n | k (cid:105) ξ = q k + ... + k n − (cid:112) − q k n + | k + e n (cid:105) ξ , (23b) y n + | k (cid:105) ξ = q | k | + k n | k (cid:105) y n + ξ . If W ⊂ V carries a subrepresentation of C ∗ ( y n + , y ∗ n + ) , the Hilbert subspace of H spanned by | k (cid:105) ξ with ξ ∈ W and k ∈ N n carries a subrepresentation of A ( Σ n + q ) . Since H is irreducible, V carries an irreducible representation of C ∗ ( y n + , y ∗ n + ) . This means that V is one-dimensional.Let us fix a unit vector ξ ∈ V . We argued that the vectors {| k (cid:105) ξ } k ∈ N n (24)span H . Moreover y n + ξ = λξ for some λ ∈ R , and y n + | k (cid:105) ξ = λ q | k | + k n | k (cid:105) ξ . From now on, we omit the subscript “ ξ ”. From (21): = (cid:104) | (cid:105) = (cid:68) (cid:12)(cid:12)(cid:12) (cid:88) n + i = y ∗ i y i (cid:12)(cid:12)(cid:12) (cid:69) = | λ | , hence λ ∈ U ( ) . It remains to prove that the set (24) is orthonormal, so that by adjunction from(23) we get the formulas in Prop. 7.Let W i be the span of vectors | k (cid:105) ξ with k = . . . = k i = . It follows from the commutationrelations that y k is zero on W i for all k (cid:54) i . pplying the identities in Lemma 9 to a vector | k (cid:105) ∈ V i we get: y i ( y ∗ i ) m |
0, . . . , 0, k i + , . . . , k n (cid:105) = ( − q m )( y ∗ i ) m − |
0, . . . , 0, k i + , . . . , k n (cid:105) y n ( y ∗ n ) m | (cid:105) = ( − q m )( y ∗ n ) m − | (cid:105) for all m (cid:62) and all (cid:54) i < n . Using (23) we get: y i |
0, . . . , 0, m − k i + , . . . , k n (cid:105) = (cid:112) − q m |
0, . . . , 0, m − k i + , . . . , k n (cid:105) y n |
0, . . . , 0, m − (cid:105) = (cid:112) − q m |
0, . . . , 0, m − (cid:105) Multiplying from the left by (cid:104) j − e i | and using (23) again: (cid:112) − q j i (cid:104) j | k (cid:105) = (cid:112) − q k i (cid:104) j − e i , k − e i (cid:105) , (cid:112) − q j n (cid:104) j n e n | k n e n (cid:105) = (cid:112) − q k n (cid:104) ( j n − ) e n , ( k n − ) e n (cid:105) , the former valid whenever j , . . . , j i = k = . . . = k i = . From these relations, an obvi-ous induction proves that the set (24) is orthonormal provided every vector | k (cid:105) with k (cid:54) = isorthogonal to | (cid:105) . But this is obvious. If k i (cid:54) = for some i , then (cid:104) k | (cid:105) ∝ (cid:104) k − e i | y i | (cid:105) = since y i annihilates ξ . (cid:4) Acknowledgment.
We are grateful to Sophie E. Mikkelsen and Wojciech Szyma ´nski for usefulemail exchanges. GL was partially supported by INDAM-GNSAGA and by INFN, IniziativaSpecifica GAST. R
EFERENCES [1] L.D. Faddeev, N.Y. Reshetikhin and L.A. Takhtajan,
Quantization of Lie groups and Lie algebras , Leningrad Math.J. 1 (1990) 193–225.[2] E. Hawkins and G. Landi,
Fredholm Modules for Quantum Euclidean Spheres , J. Geom. Phys. 49 (2004) 272–293.[3] G. Landi, C. Pagani and C. Reina,
A Hopf bundle over a quantum four-sphere from the symplectic group , Commun.Math. Phys. 263 (2006) 65–88.[4] S.E. Mikkelsen and W. Szyma ´nski,
The Quantum Twistor Bundle , preprint arXiv:2005.03268 [math.QA].[5] L.L. Vaksman and Y.S. Soibelman,
The algebra of functions on quantum SU ( n + ) group and odd-dimensional quan-tum spheres , Leningrad Math. J. 2 (1991), 1023–1042.(F. D’Andrea) U NIVERSIT ` A DI N APOLI “F EDERICO
II”
AND
I.N.F.N. S
EZIONE DI N APOLI , C
OMPLESSO
MSA, V IA C INTIA , 80126 N
APOLI , I
TALY
Email address : [email protected] (G. Landi) U NIVERSIT ` A DI T RIESTE , V IA A. V
ALERIO , 12/1, 34127 T
RIESTE , I
TALY I NSTITUTE FOR G EOMETRY AND P HYSICS (IGAP) T
RIESTE , I
TALYAND
INFN, T
RIESTE , I
TALY
Email address : [email protected]@units.it