Braid group actions for quantum symmetric pairs of type AIII/AIV
aa r X i v : . [ m a t h . QA ] J u l BRAID GROUP ACTIONS FOR QUANTUM SYMMETRICPAIRS OF TYPE AIII/AIV
LIAM DOBSON
Abstract.
In the present paper we construct braid group actions on quantumsymmetric pair coideal subalgebras of type AIII/AIV. This completes the proofof a conjecture by Kolb and Pellegrini in the case where the underlying Liealgebra is sl n . The braid group actions are defined on the generators of thecoideal subalgebras and the defining relations and braid relations are verifiedby explicit calculations. Introduction
Background.
Let g be a complex semisimple Lie algebra and U q ( g ) the corre-sponding Drinfeld-Jimbo quantised enveloping algebra. In the theory of quantumgroups, a crucial role is played by Lusztig’s braid group action on U q ( g ), see [Lus94].This braid group action provides an algebra automorphism T w of U q ( g ) for eachelement w in the Weyl group W of g .Let θ : g → g be an involutive Lie algebra automorphism and let k = { x ∈ g | θ ( x ) = x } denote the fixed Lie subalgebra. Recall from [Ara62] that involutive au-tomorphisms of g are parameterised up to conjugation by combinatorial data ( X, τ )attached to the Dynkin diagram of g . Here X ⊂ I where I denotes an index set forthe nodes of the Dynkin diagram of g , and τ : I → I is a diagram automorphism.In a series of papers, G. Letzter constructed and investigated quantum group ana-logues of k , see [Let99, Let02, Let03]. More precisely, she defined families of coidealsubalgebras B c , s = B c , s ( X, τ ) ⊂ U q ( g ) which are quantum group analogues of U ( k )depending on parameters c and s . The algebras B c , s can be described explicitly interms of generators and relations. We call ( U q ( g ) , B c , s ) a quantum symmetric pairand we refer to B c , s as a quantum symmetric pair coideal subalgebra.There exists a braid group action on the fixed Lie subalgebra k by Lie algebraautomorphisms. Let W denote the Weyl group of g with corresponding braid group Br ( g ), generated by elements { ς i | i ∈ I } . We write Br ( W X ) to denote the subgroupof Br ( g ) generated by { ς i | i ∈ X } ; this is the braid group corresponding to theparabolic subgroup W X ⊂ W . Further, associated to the pair ( X, τ ) is a restrictedroot system Σ with Weyl group f W generated by elements e σ i parameterised bythe τ -orbits in I \ X . The group f W can be considered as a subgroup of W . Let Br ( f W ) ⊂ Br ( g ) denote the corresponding braid group, generated by elements e ς i .Then the semidirect product Br ( W X ) ⋊ Br ( f W ) ⊂ Br ( g ) acts on k by Lie algebraautomorphisms.It was conjectured by Kolb and Pellegrini that there exists a quantum groupanalogue of this action on B c , s by algebra automorphisms [KP11, Conjecture 1.2]. Mathematics Subject Classification.
Key words and phrases.
Quantum groups, quantum symmetric pairs, braid groups.
This conjecture has been proved in type AII and in all cases where X = ∅ with thehelp of computer calculations [KP11]. In the case g = sl n , this leaves one substantialcase in Araki’s list [Ara62, p.32], namely the type AIII/AIV with X = ∅ . This caseis shown in Figure 1.1.2. Results.
In the present paper we construct an action of Br ( W X ) ⋊ Br ( f W )on B c , s by algebra automorphisms in type AIII/AIV, hence completing the proofof Kolb and Pellegrini’s conjecture for g = sl n . In this case Br ( W X ) is a classical rn − r + 1 n − n Figure 1.
The Satake diagram of type AIII/AIVbraid group in n − r + 1 strands and Br ( f W ) is isomorphic to an annular braidgroup in r strands. The subgroups Br ( W X ) and Br ( f W ) of Br ( g ) commute andhence their semidirect product is just a direct product. Moreover, the parameters s satisfy s = (0 , , . . . ,
0) and hence we write B c = B c , . The following is the mainresult of this paper. Theorem A. (Theorem 3.8)
Let ( X, τ ) be a Satake diagram of type AIII/AIVwith X = ∅ . Then there exists an action of Br ( W X ) × Br ( f W ) on B c by algebraautomorphisms. The action of Br ( f W ) on B c is given by algebra automorphisms T i defined by Equations (3.7) and (3.9) . The action of Br ( W X ) on B c coincides with the Lusztig action, see [BW18,Section 4.1]. However, the action of Br ( f W ) ⊂ Br ( g ) on U q ( g ) does not restrict toan action on the coideal subalgebra B c . Taking guidance from [KP11], the Lusztigautomorphisms corresponding to elements of f W ⊂ W are used in order to constructalgebra automorphisms T i for 1 ≤ i ≤ r . In particular, let T e σ i for 1 ≤ i ≤ r bethe Lusztig automorphism defined by (2.18) and (2.8). We then construct elements T i ( B j ) for j ∈ I \ X such that T i ( B j ) and T e σ i ( B j ) have identical terms containingmaximal powers of the generators F k of U q ( g ) for k ∈ I , up to a factor. For1 ≤ i ≤ r − T i defines an algebra automorphismof B c . In this case, our definition of T i ( B j ) coincides with the definition in [KP11,Equation 4.11] up to a factor.The proof of Theorem A proceeds in three steps. First we verify that the formulaefor T r given in (3.9) define an algebra automorphism of B c . Secondly, we show thatthe algebra automorphisms T i for 1 ≤ i ≤ r satisfy the braid relations for Br ( f W ).By [KP11, Section 4.2] we only need to show that T r T r − T r T r − = T r − T r T r − T r and T i T r = T r T i for 1 ≤ i ≤ r − B c . This gives anaction of Br ( f W ) on B c by algebra automorphisms. Finally, we show directly thatthe actions of Br ( W X ) and Br ( f W ) commute by case-by-case checks. RAID GROUP ACTIONS FOR QUANTUM SYMMETRIC PAIRS OF TYPE AIII/AIV 3
We emphasise that all of the original results of the present paper are establishedwithout the use of computer calculations. The fact that the maps T i for 1 ≤ i ≤ r − B c translates from [KP11] where it was verified bycomputer calculations using the package QUAGROUP of the computer algebra program
GAP . We do not reprove this fact. However, the calculations in the present papersuggest that one can prove Theorem 3.2 ([KP11, Theorem 4.6]) without the use ofcomputer calculations.In [LW19a], M. Lu and W. Wang developed a Hall algebra approach to theconstruction of quantum symmetric pairs with X = ∅ for g of type ADE (also ex-cluding type A n for n even if τ = id). In this setting they subsequently constructedBernstein-Gelfand-Ponomarev type reflection functors in [LW19b] which recover thecorresponding braid group action in [KP11]. At the end of [LW19b, Section 1.5],they express great interest to develop this approach fully to cover general Satakediagrams with X = ∅ . The braid group action for quantum symmetric pairs of typeAIII/AIV with X = ∅ constructed in the present paper provides a crucial test casefor any such generalisations. Formula (3.9) indicates that the general setting willbe substantially more complicated.1.3. Organisation.
In Section 2 we recall fixed Lie subalgebras of type AIII/AIVand their corresponding quantum symmetric pairs. We also recall the braid groupaction of Br ( W X ) × Br ( f W ) on the fixed Lie subalgebra in this case.In Section 3 we present the main results of the present paper. We recall the actionof Br ( W X ) on B c and define the algebra automorphisms T i , giving an action of Br ( f W ) on B c . We also show that the two actions commute. In Sections 4 and 5we prove that T r is an algebra automorphism of B c and that the automorphisms T i satisfy the braid relations for Br ( f W ), respectively. This requires the use of manyinvolved relations in B c , which are given in Appendix A. Acknowledgement.
The author is grateful to Stefan Kolb for useful commentsand advice. 2.
Preliminaries
Braid group actions on fixed Lie subalgebras of type AIII/AIV.
Let g = sl n +1 ( C ) for n ∈ N with Cartan subalgebra h consisting of traceless diagonal( n + 1) × ( n + 1) matrices. Let Φ ⊂ h ∗ be the corresponding root system. Choosea set Π = { α i | i ∈ I } of simple roots where I = { , , . . . , n } denotes an index setfor the nodes of the Dynkin diagram of g . n − n Let Q = Z Π denote the root lattice of g . Let ̟ i ∈ h ∗ denote the i th fundamentalweight and let P = P i ∈ I Z ̟ i denote the weight lattice of g . Write W to denotethe Weyl group of g , generated by reflections σ i for i ∈ I . Fix a W -invariant scalarproduct ( − , − ) on the real vector space spanned by Φ such that ( α, α ) = 2 for allroots α . For i, j ∈ I let a ij = i = j , − | i − j | = 1,0 otherwise (2.1) LIAM DOBSON denote the entries of the Cartan matrix of g . Let Br ( g ) denote the classical braidgroup corresponding to g . This is the group generated by elements { ς i | i ∈ I } subject to relations ς i ς j = ς j ς i if a ij = 0, (2.2) ς i ς j ς i = ς j ς i ς j if a ij = −
1. (2.3)Let { e i , f i , h i | i ∈ I } denote a set of Chevalley generators of g and defineAd( ς i ) = exp(ad( e i )) exp(ad( − f i )) exp(ad( e i )) (2.4)for i ∈ I where exp denotes the exponential series and ad denotes the adjoint action.Then there exists a group homomorphismAd : Br ( g ) → Aut( g ) (2.5)such that Ad( ς i ) is given by Equation (2.4), see [Ste67, Lemma 56].Let θ : g → g be an involutive Lie algebra automorphism and let k = { x ∈ g | θ ( x ) = x } denote the corresponding fixed Lie subalgebra. Recall from [Let03,Section 7] and [Kol14, Section 2.4] that involutive automorphisms of g are classifiedup to conjugation via Satake diagrams ( X, τ ) where X ⊂ I and τ : I → I is adiagram automorphism. Throughout this paper, we consider Satake diagrams oftype AIII/AIV, as indicated by [Ara62, Table 1] and Figure 1. In particular we fix r ∈ N such that 1 ≤ r ≤ ⌈ n ⌉ − X = { r + 1 , r + 2 , . . . , n − r } 6 = ∅ . Thediagram automorphism τ is given by τ ( i ) = n − i + 1 (2.6)for each i ∈ I . This can be lifted to a Lie algebra automorphism, also denoted by τ .For any J ⊂ I , let W J denote the parabolic subgroup of W generated by { σ i | i ∈ J } and let Br ( W J ) denote the associated braid group, generated by { ς i | i ∈ J } . Wedenote by w J and m J the longest element in W J and the corresponding element of Br ( W J ), respectively.By [Kol14, Theorem 2.5], the involution θ is given by θ = Ad( s ) ◦ Ad( m X ) ◦ ω ◦ τ (2.7)where Ad( s ) : g → g is a Lie algebra automorphism such that restriction of Ad( s )to any root space is given by multiplication by a scalar, see [BK19, Section 5.1] and ω : g → g denotes the Chevalley involution.Generally, the braid group action on g given by Equation (2.5) does not restrictto an action on k . We consider instead a suitable subgroup of Br ( g ) that dependson X ⊂ I and τ : I → I . For any 1 ≤ i ≤ r let e σ i = w { i,τ ( i ) }∪ X w − X = ( σ i σ τ ( i ) if 1 ≤ i ≤ r − σ r σ r +1 · · · σ n − r +1 · · · σ r +1 σ r if i = r (2.8)and denote by f W the subgroup of W generated by { e σ i | ≤ i ≤ r } . The subgroup f W can be interpreted as the Weyl group of the restricted root system Σ of the sym-metric Lie algebra ( g , θ ), see [DK19, Section 2.2]. Let Br ( f W ) denote the subgroupof Br ( g ) generated by the elements e ς i = m { i,τ ( i ) }∪ X m − X = ( ς i ς τ ( i ) if 1 ≤ i ≤ r − ς r ς r +1 · · · ς n − r +1 · · · ς r +1 ς r if i = r . (2.9) RAID GROUP ACTIONS FOR QUANTUM SYMMETRIC PAIRS OF TYPE AIII/AIV 5
The elements e ς i satisfy the relations e ς i e ς j = e ς j e ς i if a ij = 0 and 1 ≤ i, j ≤ r , e ς i e ς j e ς i = e ς j e ς i e ς j if a ij = − ≤ i, j < r , e ς i e ς j e ς i e ς j = e ς j e ς i e ς j e ς i if i = r, j = r − Br ( W X ) and B r ( f W ) commute, we consider the subgroup Br ( W X ) × Br ( f W ). We state without proof the version of [KP11, Lemma 2.1] correspond-ing to the present case. In many cases, Ad( s ) appearing in Equation (2.7) doesnot commute with Ad( b ) for b ∈ Br ( W X ) × Br ( f W ). For this reason, conjugatingthe action Ad by a Lie algebra isomorphism ψ s : g → g depending on Ad( s ) isnecessary. Details of this construction can be found in [Dob19, Section 7.1]. Lemma 2.1 ([Dob19, Lemma 7.6]) . Under the action ψ s ◦ Ad ◦ ψ − s the subgroup Br ( W X ) × Br ( f W ) maps k to itself. Quantum symmetric pairs of type AIII/AIV.
Let K be a field of char-acteristic zero and q an indeterminate. Denote by K ( q / ) be the field of rationalfunctions in q / with coefficients in K . Following [Jan96] and [Lus94] the Drinfeld-Jimbo quantised enveloping algebra U q ( g ) = U q ( sl n +1 ( C )) is the associative K ( q / )-algebra generated by elements E i , F i , K µ for i ∈ I and µ ∈ P satisfying the followingrelations:(1) K = 1, K µ K λ = K µ + λ for all µ, λ ∈ P .(2) K µ E i = q ( α i ,µ ) E i K µ for all i ∈ I , µ ∈ P .(3) K µ F i = q − ( α i ,µ ) F i K µ for all i ∈ I , µ ∈ P .(4) E i F j − F j E i = δ ij K i − K − i q − q − for all i, j ∈ I .(5) Quantum Serre relations,We use the notation K i = K α i for i ∈ I and K − µ = K − µ for µ ∈ P throughout.We make the quantum Serre relations (5) more explicit. Let p denote the non-commutative polynomial in two variables given by p ( x, y ) = x y − ( q + q − ) xyx + yx . (2.10)Then the quantum Serre relations can be written as E i E j = E j E i , F i F j = F j F i if a ij = 0, (2.11) p ( E i , E j ) = p ( F i , F j ) = 0 if a ij = −
1. (2.12)Analogously to (2.5), there exists an action of the braid group Br ( g ) on U q ( g )by algebra automorphisms, see [Lus94, 39.4.3]. Under this action the generator ς i ∈ Br ( g ) is mapped to the Lusztig automorphism T i as in [Jan96, Section 8.14].We recall explicitly how T i acts on the generators of U q ( g ). For any a, b ∈ U q ( g ), c ∈ K ( q / ) let [ a, b ] c = ab − cba (2.13)For any i, j ∈ I and µ ∈ P we have T i ( E i ) = − F i K i , T i ( F i ) = − K − i E i , (2.14) LIAM DOBSON and T i ( K µ ) = K σ i ( µ ) , (2.15) T i ( E j ) = ( E j if a ij = 0,[ E i , E j ] q − if a ij = −
1, (2.16) T i ( F j ) = ( F j if a ij = 0,[ F j , F i ] q if a ij = −
1, (2.17)For any w ∈ W with reduced expression w = σ i σ i · · · σ i t we write T w := T i T i · · · T i t . (2.18)For any J = { a, a + 1 , . . . , b − , b } ⊂ I with a < b we define elements E + J := (cid:2) E a , [ E a +1 , . . . , [ E b − , E b ] q − . . . ] q − (cid:3) q − , (2.19) E − J := (cid:2) E b , [ E b − , . . . , [ E a +1 , E a ] q − . . . ] q − (cid:3) q − (2.20)and similarly we write F + J := (cid:2) F a , [ F a +1 , . . . , [ F b − , F b ] q . . . ] q (cid:3) q , (2.21) F − J := (cid:2) F b , [ F b − , . . . , [ F a +1 , F a ] q . . . ] q (cid:3) q . (2.22)Additionally, let K J = K a K a +1 · · · K b − K b . (2.23)If J = { a } ⊂ I then we write E + J = E − J = E a , F + J = F − J = F a . (2.24)For later use, we note the following formulae, which follow from [Kol14, Lemma 3.4].We have T w X ( F + X ) = − K − X E + X , T w X ( F − X ) = − K − X E − X , (2.25) T w X ( E + X ) = − F + X K X , T w X ( E − X ) = − F − X K X . (2.26)Following [Let99] and the conventions of [Kol14] we now recall the definitionof quantum symmetric pair coideal subalgebras for Satake diagrams ( X, τ ) of typeAIII/AIV. Let M X = U q ( g X ) denote the subalgebra of U q ( g ) generated by { E i , F i , K ± i | i ∈ X } . Let U = K ( q / ) h K µ | µ ∈ P, − w X ◦ τ ( µ ) = µ i . By construction, K i ∈ U for i ∈ X and K ̟ i − ̟ τ ( i ) ∈ U for i ∈ I . We use the notation ̟ ′ i = ̟ i − ̟ τ ( i ) for i ∈ I . (2.27)Quantum symmetric pair coideal subalgebras depend on a choice of parameters ǫ = ( ǫ i ) i ∈ I \ X ∈ ( K ( q / ) × ) I \ X satisfying additional constraints. We assume forthe remainder of this paper that ǫ i = ǫ τ ( i ) for i ∈ I \ X ∪ { r, τ ( r ) } , (2.28)compare with [Kol14, Section 5.1]. Remark . The parameters ǫ i relate to the parameters c i and s ( i ) seen in [Kol14],[BK15] and [BK19] in the following way: ǫ i = c i s ( τ ( i )) for each i ∈ I \ X . RAID GROUP ACTIONS FOR QUANTUM SYMMETRIC PAIRS OF TYPE AIII/AIV 7
Following [Kol14, Definition 5.1, 5.6] we denote by B ǫ = B ǫ ( X, τ ) the subalgebraof U q ( g ) generated by M X , U and the elements B i = F i − ǫ i E τ ( i ) K − i if i = r, τ ( r ), F r − ǫ r [ E + X , E τ ( r ) ] q − K − r if i = r , F τ ( r ) − ǫ τ ( r ) [ E − X , E r ] q − K − τ ( r ) if i = τ ( r ). (2.29)for all i ∈ I \ X . For consistency, we set B i = F i and ǫ i = 0 for i ∈ X .We recall now the defining relations of B ǫ , following [Let03, Section 7] and [Kol14,Section 7]. For i ∈ I \ X let L i = K i K − τ ( i ) (2.30)and define Z i = − (1 − q − ) E + X L τ ( r ) if i = r , − (1 − q − ) E − X L r if i = τ ( r ), − L τ ( i ) otherwise. (2.31)Further, let Γ i := ǫ i Z i − ǫ τ ( i ) Z τ ( i ) (2.32)for i ∈ I \ X . Then the algebra B ǫ is generated over M X U Θ0 by the elements B i for i ∈ I \ X , subject to the relations B i K µ = q ( µ,α i ) K µ B i for i ∈ I \ X , K µ ∈ U , (2.33) B i E j = E j B i for i ∈ I \ X , j ∈ X , (2.34) B i B j − B j B i = δ i,τ ( j ) ( q − q − ) − Γ i for i ∈ I \ X , j ∈ I , a ij = 0, (2.35) p ( B i , B j ) = 0 for i, j ∈ I , a ij = −
1. (2.36)3.
Main Results
Recall from Lemma 2.1 that an action of Br ( W X ) × Br ( f W ) on k by Lie algebraautomorphisms is obtained by restriction of the action of Br ( g ) on g . We nowconstruct an analogous braid group action in the setting of quantum symmetricpairs of type AIII/AIV. Recall that the algebra automorphisms T i for i ∈ X giverise to a representation of Br ( W X ) on U q ( g ). In [BW18, Section 4.1] it was shownthat B ǫ is invariant under the automorphisms T i for i ∈ X . Theorem 3.1 ([BW18, Section 4.1]) . There exists an action of Br ( W X ) on B ǫ by algebra automorphisms such that the generator ς i ∈ Br ( W X ) is mapped to theLusztig automorpshism T i . We give this action explicitly on the elements B i for i ∈ I \ X . For i ∈ X and j ∈ I \ X we have T i ( B j ) = ( B j if a ij = 0,[ B j , F i ] q if a ij = − T − i ( B j ) = ( B j if a ij = 0,[ F i , B j ] q if a ij = −
1. (3.1)It follows that T w X ( B r ) = [ B r , F + X ] q , (3.2) T w X ( B τ ( r ) ) = [ B τ ( r ) , F − X ] q . (3.3) LIAM DOBSON
Similarly, one also obtains T − w X ( B r ) = [ F − X , B r ] q , (3.4) T − w X ( B τ ( r ) ) = [ F + X , B τ ( r ) ] q . (3.5)We now construct the action of Br ( f W ) on B ǫ by algebra automorphisms. Forreasons observed in Equations (3.7) and (3.9) we now consider an extension K ′ of the field K ( q / ) that contains √ ǫ i for i ∈ I \ X . For 1 ≤ i ≤ r the algebraautomorphisms e T i := T e σ i = ( T i T τ ( i ) if 1 ≤ i ≤ r , T r T r +1 · · · T τ ( r ) · · · T r +1 T r if i = r (3.6)do not leave B ǫ invariant. However, they are still used as a guide to the constructionof a braid group action on B ǫ . The general strategy is similar to that of [KP11].We first define the action of the generators e ς i for 1 ≤ i ≤ r − ≤ i ≤ r − j ∈ I \ X define T i ( B j ) = q − B τ ( j ) L τ ( j ) if j = i or j = τ ( i ), (cid:0) qǫ i (cid:1) − / [ B j , B i ] q if a ij = − (cid:0) qǫ τ ( i ) (cid:1) − / [ B j , B τ ( i ) ] q if a τ ( i ) j = − B j if a ij = 0 and a τ ( i ) j = 0. (3.7) Theorem 3.2.
Suppose ( X, τ ) is a Satake diagram of type AIII with X = { r +1 , . . . τ ( r + 1) } and ≤ r ≤ ⌈ n ⌉ − . Let ≤ i ≤ r − . (1) There exists a unique algebra automorphism T i of B ǫ such that T i ( B j ) isgiven by Equation (3.7) for j ∈ I \ X and T i | M X U Θ0 = e T i | M X U Θ0 . (2) The inverse automorphism T − i is given by T − i ( B j ) = qB τ ( j ) L j if j = i or j = τ ( j ) , (cid:0) qǫ i (cid:1) − / [ B i , B j ] q if a ij = − , (cid:0) qǫ τ ( i ) (cid:1) − / [ B τ ( i ) , B j ] q if a τ ( i ) j = − , B j if a ij = 0 and a τ ( i ) j = 0 (3.8) and T − i | M X U Θ0 = e T i | M X U Θ0 . (3) The relation T i T i +1 T i = T i +1 T i T i +1 holds for ≤ i < r − . Further therelation T i T j = T j T i holds for a ij = 0 with ≤ i, j ≤ r − .Proof. The result follows from [KP11, Theorems 4.3 and 4.6] where the only differ-ence occurs in T i ( B j ) when a ij = − a τ ( i ) j = −
1. Here, one checks that[ B j , B i ] q [ B τ ( j ) , B τ ( i ) ] q − [ B τ ( j ) , B τ ( i ) ] q [ B j , B i ] q = qǫ i q − q − (cid:0) ǫ j T i ( Z j ) − ǫ τ ( j ) T i ( Z τ ( j ) ) (cid:1) . Hence for symmetry reasons and the fact that ǫ i = ǫ τ ( i ) for 1 ≤ i ≤ r −
1, we choose T i ( B j ) and T i ( B τ ( j ) ) as in Equation (3.7). (cid:3) Remark . In Equations (3.7) and (3.8) the coefficients ǫ i appear whereas theydid not in [KP11]. This is because Kolb and Pellegrini took ǫ i = 1 for all i ∈ I intheir paper. RAID GROUP ACTIONS FOR QUANTUM SYMMETRIC PAIRS OF TYPE AIII/AIV 9
It remains to construct the algebra automorphism T r . For ease of notation, let C = (cid:0) qǫ r ǫ τ ( r ) (cid:1) − / . Recall from Equation (2.27) that we set ̟ ′ i = ̟ i − ̟ τ ( i ) for i ∈ I . Define T r ( B j ) = q − B r L r K ̟ ′ r +1 if j = r , q − B τ ( r ) L τ ( r ) K ̟ ′ τ ( r +1) if j = τ ( r ), C (cid:0)(cid:2) B r − , [ B r , [ F + X , B τ ( r ) ] q ] q (cid:3) q + qǫ τ ( r ) B r − L r K − X (cid:1) if j = r − C (cid:0)(cid:2) B τ ( r − , [ B τ ( r ) , [ F − X , B r ] q ] q (cid:3) q + qǫ r B τ ( r − L τ ( r ) K − X (cid:1) if j = τ ( r − B j otherwise (3.9)for j ∈ I \ X . The following theorem establishes that Equation (3.9) defines analgebra automorphism T r : B ǫ → B ǫ . Theorem 3.4.
Suppose ( X, τ ) is a Satake diagram of type AIII/AIV with X = { r + 1 , . . . τ ( r + 1) } and ≤ r ≤ ⌈ n ⌉ − . (1) There exists a unique algebra automorphism T r of B ǫ such that T r ( B j ) isgiven by Equation (3.9) and T r | M X U Θ0 = e T r | M X U Θ0 . (2) The inverse automorphism T − r is given by T − r ( B j ) = qB r L τ ( r ) K ̟ ′ τ ( r +1) if j = r , qB τ ( r ) L r K ̟ ′ r +1 if j = τ ( r ) , C (cid:0)(cid:2) B τ ( r ) , [ F − X , [ B r , B r − ] q ] q (cid:3) q + ǫ r B r − L τ ( r ) K − X (cid:1) if j = r − , C (cid:0)(cid:2) B r , [ F + X , [ B τ ( r ) , B τ ( r − ] q ] q (cid:3) q + ǫ τ ( r ) B τ ( r − L r K − X (cid:1) if j = τ ( r − , B j otherwise (3.10) and T − r | M X U Θ0 = e T − r | M X U Θ0 .Remark . A desirable property of the algebra automorphism T r is that it is local,meaning T r ( B i ) = B i is satisfied for i ∈ I \ X and a ir = a τ ( i ) r = 0. With this, itis not possible to omit the elements K ̟ ′ i for i ∈ I \ X from our constructions. Inparticular, K ̟ ′ i appears so that the relation B i E j − E j B i = 0for i ∈ I \ X , j ∈ X is preserved under T r .The proof of Theorem 3.4 requires non-trivial calculations which are postponedto Section 4. Crucially, the algebra automorphisms T , . . . , T r satisfy type B r braidrelations. Theorem 3.6.
Suppose ( X, τ ) is a Satake diagram of type AIII with X = { r +1 , . . . , τ ( r + 1) } and ≤ r ≤ ⌈ n ⌉ − . Then the relation T r T r − T r T r − = T r − T r T r − T r (3.11) holds. Further, the relations T r T i = T i T r hold for any ≤ i < r − . Similarly to Theorem 3.4, the proof of Theorem 3.6 requires a series of calcula-tions which are given in Section 5. As a result of Theorems 3.2, 3.4 and 3.6 a braidgroup action of Br ( f W ) on B ǫ by algebra automorphisms is established. Corollary 3.7.
Suppose ( X, τ ) is a Satake diagram of type AIII/AIV with X = { r + 1 , . . . , τ ( r + 1) } and ≤ r ≤ ⌈ n ⌉ − . Then there exists an action of Br ( f W ) on B ǫ by algebra automorphisms. Under this action the generator e ς i ∈ Br ( f W ) ismapped to the algebra automorphism T i for ≤ i ≤ r . Since the subgroups Br ( W X ) and Br ( f W ) of Br ( g ) commute, we now combineTheorem 3.1 and Corollary 3.7 to give an action of Br ( W X ) × Br ( f W ) on B ǫ byalgebra automorphisms. Theorem 3.8.
Let ( X, τ ) be a Satake diagram of type AIII/AIV with X = { r +1 , . . . , τ ( r + 1) } and ≤ r ≤ ⌈ n ⌉ − . Then there exists an action of Br ( W X ) × Br ( f W ) on B ǫ by algebra automorphisms. The action of Br ( W X ) on B ǫ is given bythe Lusztig automorphisms T j for j ∈ X and the action of Br ( f W ) on B ǫ is givenby the algebra automorphisms T i for ≤ i ≤ r given by Equations (3.7) and (3.9) .Proof. In order to prove Theorem 3.8 it suffices to show that for all x ∈ B ǫ , therelation T j T i ( x ) = T i T j ( x )holds for j ∈ X and 1 ≤ i ≤ r . Since T i | M X U Θ0 = e T i | M X U Θ0 for all 1 ≤ i ≤ r theresult follows if x ∈ M X U Θ0 . As a result of this and the underlying symmetry, wehence only consider x = B k for 1 ≤ k ≤ r . We proceed by casework. Case 1. ≤ i ≤ r − j ∈ X \ { r + 1 } .By Equations (3.1) and (3.7) it follows that T i ( B k ) is invariant under T j for all j ∈ X \ { r + 1 } and 1 ≤ k ≤ r. The result follows from this.
Case 2. ≤ i ≤ r − j = r + 1.Recall from Equation (3.1) that for 1 ≤ k ≤ r we have T r +1 ( B k ) = ( B k if 1 ≤ k ≤ r − B r , F r +1 ] q if k = r .There are three cases to consider, depending on the value of a ik . If a ik = 0 then T i ( B k ) = B k and T i ◦ T r +1 ( B k ) = T r +1 ( B k ). The claim follows from this. If a ik = − T i ( B k ) = (cid:0) qǫ i (cid:1) − / [ B k , B i ] q . Since 1 ≤ i ≤ r −
1, Equation (3.1)implies that we need only check the claim when k = r and i = r −
1. We obtain T r +1 T r − ( B r ) = (cid:0) qǫ i (cid:1) − / T r +1 ([ B r , B r − ] q )= (cid:0) qǫ i (cid:1) − / [[ B r , F r +1 ] q , B r − ] q = (cid:0) qǫ i (cid:1) − / [[ B r , B r − ] q , F r +1 ] q = T r − T r +1 ( B r )as required. Finally, if a ik = 2 then the claim follows since 1 ≤ i ≤ r − T i ( B i ) = q − B τ ( i ) L τ ( i ) is invariant under T r +1 . RAID GROUP ACTIONS FOR QUANTUM SYMMETRIC PAIRS OF TYPE AIII/AIV 11
Case 3. i = r and k = r − ≤ k ≤ r −
2. Then both T j and T r act as the identity on B k so theclaim follows. Hence assume that k = r . Then recall from Equation (3.9) that T r ( B r ) = q − B r L r K ̟ ′ r +1 where ̟ ′ r +1 = ̟ r +1 − ̟ τ ( r +1) . Let λ = α r − α τ ( r ) + ̟ ′ r +1 . Since α r = − ̟ r +1 +2 ̟ r − ̟ r − it follows that ( α j , λ ) = 0 for all j ∈ X . This implies that σ j ( λ ) = λ for all j ∈ X and hence T j ( L r K ̟ ′ r +1 ) = L r K ̟ ′ r +1 .If j = r + 1 then T j ( B r ) = B r and the result follows. Otherwise by Equation(3.1) we have T r +1 T r ( B r ) = q − T r +1 ( B r ) K λ = q − [ B r , F r +1 ] q K λ = T r T r +1 ( B r )where we use the fact that K λ commutes with F r +1 . The result follows from this. Case 4. i = r , j ∈ X \ { r + 1 } and k = r − j = τ ( r + 1) since T j acts as the identity onthe elements F + X , L r , K − X and B k for k ∈ I \ X . On the other hand if j = τ ( r + 1)then by Equation (3.1) we have T τ ( r +1) ([ F + X , B τ ( r ) ] q ) = T τ ( r +1) (cid:0) [ F X \{ τ ( r +1) } , [ F τ ( r +1) , B τ ( r ) ] q ] q (cid:1) = [ F + X , B τ ( r ) ] q . Hence T r ( B r − ) is invariant under T τ ( r +1) and the result follows. Case 5. i = r , j = r + 1 and k = r − T r ( B r − ) = C (cid:0)(cid:2) B r − , [ B r , [ F + X , B τ ( r ) ] q ] q (cid:3) q + qǫ τ ( r ) B r − L r K − X (cid:1) . We are done if we show that T r ( B r − ) is invariant under T r +1 . Using Lemma A.17we obtain T r +1 (cid:0)(cid:2) B r − , [ B r , [ F + X , B τ ( r ) ] q ] q (cid:3) q (cid:1) = (cid:2) B r − , [ B r , [ F + X , B τ ( r ) ] q ] q (cid:3) q + qǫ τ ( r ) B r − L r ( K − r +1 − K r +1 ) K − X \{ r +1 } . Further, we have T r +1 ( B r − L r K − X ) = B r − L r K r +1 K − X \{ r +1 } . Combining these we obtain T r +1 T r ( B r − ) = C (cid:2) B r − , [ B r , [ F X , B τ ( r ) ] q ] q (cid:3) q + Cqǫ τ ( r ) B r − L r K r +1 K − X \{ r +1 } + Cqǫ τ ( r ) B r − L r ( K − r +1 − K r +1 ) K − X \{ r +1 } = T r ( B r − )as required. (cid:3) Proof of Theorem 3.4
We divide the proof of Theorem 3.4 into three parts. In the first part, we showthat T r is an algebra endomorphism of B ǫ by checking that all of the necessaryrelations are satisfied. Next, we show that T − r is also an algebra endomorphismof B ǫ . Finally, we show that T − r really is the inverse of T r .4.1. Proof that T r is an algebra endomorphism. Recall from Equations (2.10)and (2.32) the polynomial p : U q ( g ) × U q ( g ) → U q ( g ) and the elements Γ i = ǫ i Z i − ǫ τ ( i ) Z τ ( i ) . In view of relations (2.33) to (2.36) and Equation (3.9) we showthat the relations B r − x − xB r − = 0 for x ∈ M X , (4.1) B r − B τ ( r − − B τ ( r − B r − = 1 q − q − Γ r − , (4.2) p ( B r , B r − ) = 0 , (4.3) p ( B r − , B r ) = 0 (4.4)are preserved under the map T r . The remaining relations either follow from theabove by symmetry, or can be verified by short calculations. Such checks are notshown here. Using Lemma A.2, the elements T r ( B r − ) and T r ( B τ ( r − ) can beexpressed in the following way. Let S = (cid:2) B r − , [ B r , F + X ] q (cid:3) q , (4.5) S τ = (cid:2) B τ ( r − , [ B τ ( r ) , F − X ] q (cid:3) q (4.6)and let ∆ = qǫ τ ( r ) B r − L r K X , (4.7)∆ τ = qǫ r B τ ( r − L τ ( r ) K X . (4.8)Then we have T r ( B r − ) = C (cid:0) [ S, B τ ( r ) ] q + ∆ (cid:1) , (4.9) T r ( B τ ( r − ) = C (cid:0) [ S τ , B r ] q + ∆ τ (cid:1) . (4.10)We use Equations (4.9) and (4.10) to establish many of the results of this section.We first show that (4.1) is invariant under T r . The following lemma establishesinvariance for x ∈ { E i , F i | i ∈ X , i = r + 1 , τ ( r + 1) } . Lemma 4.1.
For any i ∈ X \ { r + 1 , τ ( r + 1) } the relations F + X E i − E i F + X = 0 ,F + X F i − F i F + X = 0 hold in U q ( g ) .Proof. For any i ∈ X \ { r + 1 , τ ( r + 1) } let W i = { r + 1 , r + 2 , . . . , i − } and Y i = { i + 1 , i + 2 , . . . , τ ( r + 1) } . Since E i F j − F j E i = δ ij ( q − q − ) − ( K i − K − i ) for i, j ∈ I , it follows that F + X E i = E i F + X − q − q − (cid:2) F + W i , [ K i − K − i , F + Y i ] q (cid:3) q . RAID GROUP ACTIONS FOR QUANTUM SYMMETRIC PAIRS OF TYPE AIII/AIV 13
Since [ K i , F i +1 ] q = 0 and [ F i − , K − i ] q = 0, it follows that F + X E i = E i F + X . Now,using Equation (2.17) we have F i = T − i − T − i ( F i − ) , (cid:2) F i − , [ F i , F i +1 ] q (cid:3) q = T − i − T − i ( F i +1 ) . This implies that (cid:2) F i − , [ F i , F i +1 ] q (cid:3) q F i = F i (cid:2) F i − , [ F i , F i +1 ] q (cid:3) q and hence F + X F i = F i F + X as required. (cid:3) We now consider (4.1) for x ∈ { E r +1 , E τ ( r +1) , F r +1 , F τ ( r +1) } . Lemma 4.2.
The relations T r ( B r − ) E r +1 − E r +1 T r ( B r − ) = 0 , T r ( B r − ) E τ ( r +1) − E τ ( r +1) T r ( B r − ) = 0 hold in B ǫ .Proof. Suppose first that X = { r + 1 } . Since [ B r , K − r +1 ] q = 0 we have SE r +1 = (cid:2) B r − , [ B r , F r +1 E r +1 ] q (cid:3) q = E r +1 S − q − q − (cid:2) B r − , [ B r , K r +1 ] q (cid:3) q = E r +1 S + q [ B r − , B r ] q K r +1 . It follows that[
S, B r +2 ] q E r +1 = E r +1 [ S, B r +2 ] q + q (cid:2) [ B r − , B r ] q K r +1 , B r +2 (cid:3) q = E r +1 [ S, B r +2 ] q + q (cid:2) B r − , [ B r , B r +2 ] (cid:3) q K r +1 = E r +1 [ S, B r +2 ] q − q ǫ r +2 q − q − [ B r − , Z r +2 ] q K r +1 . Recalling that Z r +2 = − (1 − q − ) E r +1 L r we obtain using (2.33) and (2.34)[ S, B r +2 ] q E r +1 = E r +1 [ S, B r +2 ] q + q (1 − q ) ǫ r +2 E r +1 B r − L r K r +1 . This and Equation (4.8) implies1 C [ T r ( B r − ) , E r +1 ] = [ S, B r +2 ] q E r +1 + ∆ E r +1 − E r +1 [ S, B r +2 ] q − E r +1 ∆= (1 − q ) E r +1 ∆ + ∆ E r +1 − E r +1 ∆= 0where the last equality follows since E r +1 ∆ = q − ∆ E r +1 . This shows that E r +1 commutes with T r ( B r − ) when | X | = 1.Suppose now that | X | >
1. Let Y = X \ { r + 1 } . Then F + X E r +1 = E r +1 F + X − q − q − [ K r +1 − K − r +1 , F + Y ] q = E r +1 F + X − F + Y K − r +1 . It follows from this and the relation [ B r , K − r +1 ] q = 0 that SE r +1 = (cid:2) B r − , [ B r , F + X E r +1 ] q (cid:3) q = (cid:2) B r − , [ B r , E r +1 F + X − F + Y K − r +1 ] q (cid:3) q = E r +1 S − F + Y (cid:2) B r − , [ B r , K − r +1 ] q (cid:3) q = E r +1 S. Further, we have ∆ E r +1 = E r +1 ∆ since E r +1 commutes with L r K X . This im-plies that T r ( B r − ) commutes with E r +1 . To show that T r ( B r − ) commutes with E τ ( r +1) , one proceeds similarly but instead using the form of T r ( B r − ) given inEquation (3.9) and the relation [ K τ ( r +1) , B τ ( r ) ] q = 0. (cid:3) Lemma 4.3.
The relations T r ( B r − ) F r +1 − F r +1 T r ( B r − ) = 0 , T r ( B r − ) F τ ( r +1) − F τ ( r +1) T r ( B r − ) = 0 hold in B ǫ .Proof. Suppose first that X = { r + 1 } . Since B r − commutes with B r +2 we have[ S, B r +2 ] q = (cid:2) B r − , [[ B r , F r +1 ] q , B r +2 ] q (cid:3) q . We commute F r +1 through [ S, B r +2 ] q using the algebra automorphism T w X = T r +1 .In particular, by Equations (2.14) and (3.1) we have F r +1 = − T r +1 ( E r +1 K r +1 ) , [ B r , F r +1 ] q = T r +1 ( B r ) ,B r +2 = T r +1 ([ F r +1 , B r +2 ] q ) . It hence follows that[
S, B r +2 ] q F r +1 = − T r +1 (cid:0)(cid:2) B r − , [ B r , [ F r +1 , B r +2 ] q ] q (cid:3) q E r +1 K r +1 (cid:1) . (4.11)We consider the right hand side of Equation (4.11). We have[ B r , [ F r +1 , B r +2 ] q ] q E r +1 K r +1 − E r +1 K r +1 [ B r , [ F r +1 , B r +2 ] q ] q = 1 q − q − [ B r , [ K − r +1 , B r +2 ] q ] q K r +1 = − q − q − ( ǫ r Z r − ǫ r +2 Z r +2 ) . Substituting this into Equation (4.11) we obtain[
S, B r +2 ] q F r +1 = F r +1 [ S, B r +2 ] q + 1 q − q − T r +1 ([ B r − , ǫ r Z r − ǫ r +2 Z r +2 )= F r +1 [ S, B r +2 ] q + qǫ r +2 B r − T r +1 ( Z r +2 ) . Since Z r +2 = − (1 − q − ) E r +1 L r we have T r +1 ( Z r +2 ) = (1 − q − ) F r +1 K r +1 L r . This implies that[
S, B r +2 ] q F r +1 = F r +1 [ S, B r +2 ] q + q (1 − q − ) ǫ r +2 F r +1 B r − K r +1 L r = F r +1 [ S, B r +2 ] q + (1 − q − ) F r +1 ∆ . RAID GROUP ACTIONS FOR QUANTUM SYMMETRIC PAIRS OF TYPE AIII/AIV 15
Since F r +1 ∆ = q ∆ F r +1 we have1 C [ T r ( B r − ) , F r +1 ] = [ S, B r +2 ] q F r +1 + ∆ F r +1 − F r +1 [ S, B r +2 ] q − F r +1 ∆= (1 − q − ) F r +1 ∆ + ∆ F r +1 − F r +1 ∆= 0 . This shows that F r +1 commutes with T r ( B r − ) when | X | = 1.Suppose now that | X | >
1. The relation F r +1 = − T w X ( E τ ( r +1) K τ ( r +1) ) andEquation (3.2) imply (cid:2) F r +1 , [ B r , F + X ] q (cid:3) = − T w X (cid:0) [ E τ ( r +1) K τ ( r +1) , B r ] (cid:1) = 0and hence S commutes with F r +1 . This, paired with the relation F r +1 L r K X = L r K X F r +1 , shows that F r +1 commutes with T r ( B r − ). In order to verify that T r ( B r − ) com-mutes with F τ ( r +1) , one shows that F τ ( r +1) commutes with [ F + X , B τ ( r ) ] q similarlyto the above. The result then follows by considering Equation (3.9). (cid:3) This completes the proof that Equation (4.1) is preserved under T r . We nowshow that (4.2) is invariant under T r . Proposition 4.4.
The relation [ T r ( B r − ) , T r ( B τ ( r − )] = 1 q − q − Γ i (4.12) holds in B ǫ .Proof. Using the expressions for T r ( B r − ) and T r ( B τ ( r − ) given in Equations (4.9)and (4.10) we have1 C [ T r ( B r − ) , T r ( B τ ( r − )] = (cid:2) [ S, B τ ( r ) ] q , [ S τ , B r ] q (cid:3) + (cid:2) ∆ , [ S τ , B r ] q (cid:3) + (cid:2) [ S, B τ ( r ) ] q , ∆ τ (cid:3) + [∆ , ∆ τ ] . By Lemma A.7 we have (cid:2) [ S, B τ ( r ) ] q , [ S τ , B r ] q (cid:3) = − (cid:2) ∆ , [ S τ , B r ] q (cid:3) − (cid:2) [ S, B τ ( r ) ] q , ∆ τ (cid:3) − qǫ r ǫ τ ( r ) q − q − ( K X − K − X ) K X Γ r − . The result follows by recalling from (3.9) that C = ( qǫ r ǫ τ ( r ) ) − and noting that[∆ , ∆ τ ] = qǫ r ǫ τ ( r ) q − q − K X Γ r − . (cid:3) It remains to show that Equations (4.3) and (4.4) are invariant under T r . Proposition 4.5.
The relation p ( T r ( B r ) , T r ( B r − )) = 0 (4.13) holds in B ǫ . Proof.
Using the expression for T r ( B r − ) from Equation (4.9) and recalling that T r ( B r ) = q − B r L r K ̟ ′ r +1 we have q C p (cid:0) T r ( B r ) , T r ( B r − ) (cid:1) = (cid:0) B r L r K ̟ ′ r +1 (cid:1) (cid:0) [ S, B τ ( r ) ] q + ∆ (cid:1) + (cid:0) [ S, B τ ( r ) ] q + ∆ (cid:1)(cid:0) B r L r K ̟ ′ r +1 (cid:1) − ( q + q − )( B r L r K ̟ ′ r +1 ) (cid:0) [ S, B τ ( r ) ] q + ∆ (cid:1) ( B r L r K ̟ ′ r +1 ) . By taking ( L r K ̟ ′ r +1 ) out as a factor, we obtain q C p (cid:0) T r ( B r ) , T r ( B r − ) (cid:1)(cid:0) L r K ̟ ′ r +1 (cid:1) − = B r [ S, B τ ( r ) ] q − q − ( q + q − ) B r [ S, B τ ( r ) ] q B r + q − [ S, B τ ( r ) ] q B r (4.14)+ B r ∆ − q − ( q + q − ) B r ∆ B r + q − ∆ B r . By Equation (A.10) the element B r commutes with S which implies B r [ S, B τ ( r ) ] q = [ S, B τ ( r ) ] q B r + 1 q − q − [ S, Γ r ] q . (4.15)It follows from that B r [ S, B τ ( r ) ] q = B r [ S, B τ ( r ) ] q B r + 1 q − q − B r [ S, Γ r ] q , [ S, B τ ( r ) ] q B r = B r [ S, B τ ( r ) ] q B r + 1 q − q − [ S, Γ r ] q B r . Substituting these two expressions into Equation (4.14) we obtain q C p (cid:0) T r ( B r ) , T r ( B r − ) (cid:1)(cid:0) L r K ̟ ′ r +1 (cid:1) − = 1 q − q − (cid:2) S, [ B r , Γ r ] q − (cid:3) q + B r ∆ − q − ( q + q − ) B r ∆ B r + q − ∆ B r . (4.16)Recall from Lemma A.3 that S Z τ ( r ) = q Z τ ( r ) S − q ( q − q − )[ B r − L r K X , B r ] . (4.17)Since [ B r , Γ r ] q − = − ( q − q − ) ǫ τ ( r ) Z τ ( r ) B r one calculates that1 q − q − (cid:2) S, [ B r , Γ r ] q − (cid:3) q = − ( q + q − ) ǫ τ ( r ) [ S, Z τ ( r ) ] q B r (4.17) = ( q − q − )[∆ , B r ] B r . The result follows by substituting this expression into Equation (4.16) and observingthat [∆ , B r ] B r = q − B r [∆ , B r ]holds since p ( B r , B r − ) = 0. (cid:3) Proposition 4.6.
The relation p ( T r ( B r − ) , T r ( B r )) = 0 holds in B ǫ . RAID GROUP ACTIONS FOR QUANTUM SYMMETRIC PAIRS OF TYPE AIII/AIV 17
Proof.
Using Equations (3.9) and (4.9) we have qC p ( T r ( B r − ) , T r ( B r )) (cid:0) L r K ̟ ′ r +1 (cid:1) − = (cid:0) [ S, B τ ( r ) ] q + ∆ (cid:1) B r + q B r (cid:0) [ S, B τ ( r ) ] q + ∆ (cid:1) − q ( q + q − ) (cid:0) [ S, B τ ( r ) ] q + ∆ (cid:1) B r (cid:0) [ S, B τ ( r ) ] q + ∆ (cid:1) . (4.18)Since B r − [ B r − , B r ] q = q − [ B r − , B r ] q B r − and B r − commutes with F + X and B τ ( r ) it follows that [ S, B τ ( r ) ] q ∆ = ∆[ S, B τ ( r ) ] q . This implies (cid:0) [ S, B τ ( r ) ] q + ∆ (cid:1) = [ S, B τ ( r ) ] q + 2[ S, B τ ( r ) ] q ∆ + ∆ . We consider terms involving different powers of ∆ in (4.18) separately. First, weconsider the expression[
S, B τ ( r ) ] q B r − q ( q + q − )[ S, B τ ( r ) ] q B r [ S, B τ ( r ) ] q + q B r [ S, B τ ( r ) ] q . (4.19)Using the relation[ S, B τ ( r ) ] q B r = B r [ S, B τ ( r ) ] q − q − q − [ S, Γ r ] q from Equation (4.15) it follows that[ S, B τ ( r ) ] q B r − q ( q + q − )[ S, B τ ( r ) ] q B r [ S, B τ ( r ) ] q + q B r [ S, B τ ( r ) ] q = q q − q − [ S, Γ r ] q [ S, B τ ( r ) ] q − q − q − [ S, B τ ( r ) ] q [ S, Γ r ] q . (4.20)By Lemma A.3 we have ǫ τ ( r ) [ S, Z τ ( r ) ] q = − ( q − q − )[∆ , B r ]which implies[ S, Γ r ] q [ S, B τ ( r ) ] q = ǫ r [ S, Z r ] q [ S, B τ ( r ) ] q + ( q − q − )[∆ , B r ][ S, B τ ( r ) ] q , (4.21)[ S, B τ ( r ) ] q [ S, Γ r ] q = ǫ r [ S, B τ ( r ) ] q [ S, Z r ] q + ( q − q − )[ S, B τ ( r ) ] q [∆ , B r ] . (4.22)Using Lemmas A.8 and A.9 we commute [ S, Z r ] q through [ S, B τ ( r ) ] q . In particularwe have Z r [ S, B τ ( r ) ] q = q − [ S, B τ ( r ) ] q Z r + q − ( q − q − )∆ Z r ,S [ S, B τ ( r ) ] q = q − [ S, B τ ( r ) ] q S − q − ( q − q − ) S ∆ . Combining this with the relations S ∆ = q ∆ S, Z r ∆ = q − ∆ Z r we obtain[ S, Z r ] q [ S, B τ ( r ) ] q = q − [ S, B τ ( r ) ] q [ S, Z r ] q − (1 − q − )∆[ S, Z r ] q (4.23) Again by Equation (4.15) we have[∆ , B r ][ S, B τ ( r ) ] q = ∆[ S, B τ ( r ) ] q B r − B r [ S, B τ ( r ) ] q ∆ + 1 q − q − ∆[ S, Γ r ] q , (4.24)[ S, B τ ( r ) ] q [∆ , B r ] = − B r [ S, B τ ( r ) ] q ∆ + ∆[ S, B τ ( r ) ] q B r + 1 q − q − [ S, Γ r ] q ∆ . (4.25)By Equation (4.17) we have ǫ τ ( r ) [ S, Z τ ( r ) ] q = − ( q − q − )[∆ , B r ] . (4.26)Substituting Equations (4.23) and (4.24) into (4.21), Equation (4.25) into (4.22)and using (4.26) we obtain[ S, Γ r ] q [ S, B τ ( r ) ] q = q − ǫ r [ S, B τ ( r ) ] q [ S, Z r ] q + q − ǫ r ∆[ S, Z r ] q + ( q − q − )∆[∆ , B r ]+ ( q − q − )∆[ S, B τ ( r ) ] q B r − ( q − q − ) B r [ S, B τ ( r ) ] q ∆ , [ S, B τ ( r ) ] q [ S, Γ r ] q = ǫ r [ S, B τ ( r ) ] q [ S, Z r ] q + ǫ r [ S, Z r ] q ∆ + ( q − q − )∆[ S, B τ ( r ) ] q B r − ( q − q − ) B r [ S, B τ ( r ) ] q ∆ + ( q − q − )[∆ , B r ]∆ . Hence Equation (4.20) implies that[
S, B τ ( r ) ] q B r − q ( q + q − )[ S, B τ ( r ) ] q B r [ S, B τ ( r ) ] q + q B r [ S, B τ ( r ) ] q = ( q − S, B τ ( r ) ] q B r − ( q − B r [ S, B τ ( r ) ] q ∆ − (cid:2) [∆ , B r ] , ∆ (cid:3) q . (4.27)We next consider the expression2∆[ S, B τ ( r ) ] q B r − (1 + q ) (cid:0) [ S, B τ ( r ) ] q B r ∆ + ∆ B r [ S, B τ ( r ) ] q (cid:1) + 2 q B r [ S, B τ ( r ) ] q ∆ . By Equations (4.15) and (A.8) we have[
S, B τ ( r ) ] q B r ∆ = B r [ S, B τ ( r ) ] q ∆ − q − q − ǫ r [ S, Z r ] q ∆ − [∆ , B r ]∆ , ∆ B r [ S, B τ ( r ) ] q = ∆[ S, B τ ( r ) ] q B r + 1 q − q − ǫ r ∆[ S, Z r ] q + ∆[∆ , B r ]from which it follows that2∆[ S,B τ ( r ) ] q B r − (1 + q ) (cid:0) [ S, B τ ( r ) ] q B r ∆ + ∆ B r [ S, B τ ( r ) ] q (cid:1) + 2 q B r [ S, B τ ( r ) ] q ∆= ( q − B r [ S, B τ ( r ) ] q ∆ − ( q − S, B τ ( r ) ] q B r + ( q + 1) (cid:2) [∆ , B r ] , ∆ (cid:3) . (4.28)Combining Equations (4.27) and (4.28) and substituting into Equation (4.18) gives qC p (cid:0) T r ( B r − ) , T r ( B r ) (cid:1)(cid:0) L r K − ̟ ′ r +1 (cid:1) = − (cid:2) [∆ , B r ] , ∆ (cid:3) q + ( q + 1) (cid:2) [∆ , B r ] , ∆ (cid:3) + ∆ B r − q ( q + q − )∆ B r ∆+ q B r ∆ = 0as required. (cid:3) Lemmas 4.1 to 4.3 and Propositions 4.4 to 4.6 together show that T r is an algebraendomorphism of B ǫ . RAID GROUP ACTIONS FOR QUANTUM SYMMETRIC PAIRS OF TYPE AIII/AIV 19
Proof that T − r is an algebra endomorphism. We now show that T − r given in Equation (3.10) also defines an algebra endomorphism of B ǫ . First, withan additional constraint on the parameters ǫ , we construct an anti-involution φ : B ǫ → B ǫ such that T − r = φ ◦ T r ◦ φ (4.29)holds, compare with [Lus94, Section 37.2.4]. In particular, set φ ( B i ) = B i , φ ( L i ) = L τ ( i ) for i ∈ I \ X , φ ( E j ) = E j , φ ( F j ) = F j , φ ( K j ) = K − j for i ∈ X . (4.30)Note that the restriction of φ on M X coincides with Lusztig’s algebra anti-automorphism τ on U q ( g ), see [Lus94, Section 3.1.3]. Lemma 4.7.
The map φ given by Equation (4.30) is an involutive algebra anti-automorphism of B ǫ if and only if ǫ r = ǫ τ ( r ) .Proof. By considering the defining relations of B ǫ given in Equations (2.33) – (2.36),it suffices to show that φ preserves the relation B i B τ ( i ) − B τ ( i ) B i = 1 q − q − ( ǫ i Z i − ǫ τ ( i ) Z τ ( i ) ) (4.31)for i ∈ I \ X . Since φ ( E + X ) = E − X and φ ( E − X ) = E + X , it follows that φ ( Z i ) = Z τ ( i ) for all i ∈ I \ X .Since φ ( B i B τ ( i ) − B τ ( i ) B i ) = B τ ( i ) B i − B i B τ ( i ) = 1 q − q − ( ǫ τ ( i ) Z τ ( i ) − ǫ i Z i )holds for all i ∈ I \ X , Equation (4.31) is preserved by φ if and only if ǫ τ ( i ) Z τ ( i ) − ǫ i Z i = ǫ i Z τ ( i ) − ǫ τ ( i ) Z i for all i ∈ I \ X . This holds if and only if ǫ i = ǫ τ ( i ) for all i ∈ I \ X . The resultfollows from this and Equation (2.28). (cid:3) By the above lemma and the results of the previous section, the map φ ◦ T r ◦ φ is an algebra endomorphism of B ǫ if ǫ r = ǫ τ ( r ) . Lemma 4.8.
Suppose ǫ r = ǫ τ ( r ) . Then for i ∈ I \ X the relation T − r ( B i ) = φ ◦ T r ◦ φ ( B i ) holds.Proof. By Equation (3.10) it suffices to only consider i = r − i = r . Theresult for i = r follows since φ ( B r L r K ̟ ′ r +1 ) = q B r L τ ( r ) K ̟ ′ τ ( r +1) and hence φ ◦ T r ◦ φ ( B r ) = qB r L τ ( r ) K ̟ ′ τ ( r +1) = T − r ( B r − ) . Recall from Equation (4.9) that we have T r ( B r − ) = C (cid:0)(cid:2) [[ B r − , B r ] q , F + X ] q , B τ ( r ) (cid:3) q + qǫ τ ( r ) B r − L r K X (cid:1) . Since φ is an anti-involution of B ǫ we have φ (cid:0)(cid:2) [[ B r − , B r ] q , F + X ] q , B τ ( r ) (cid:3) q (cid:1) = (cid:2) B τ ( r ) , [ F − X , [ B r , B r − ] q ] q (cid:3) q ,φ ( B r − L r K X ) = q − B r − L τ ( r ) K − X . Combining both relations and comparing with Equation (3.10) gives φ ◦ T r ◦ φ ( B r − ) = C (cid:0)(cid:2) B τ ( r ) , [ F − X , [ B r , B r − ] q ] q (cid:3) q + ǫ r B r − L τ ( r ) K − X (cid:1) = T − r ( B r − )as required, where we have additionally used the condition ǫ r = ǫ τ ( r ) . (cid:3) As a consequence of Lemma 4.8 and Section 4.1 the following corollary is imme-diate.
Corollary 4.9. If ǫ r = ǫ τ ( r ) then T − r given by Equation (3.10) defines an algebraendomorphism of B ǫ . We now show that T − r defines an algebra endomorphism of B ǫ also for generalparameters. When necessary, we specify the dependence on the parameters ǫ ex-plicitly by writing T r, ǫ for T r and B ǫ i for B i , where i ∈ I \ X . Additionally, we alsowrite C ǫ for the constant C appearing in T r ( B r − ) and T r ( B τ ( r − ). By [Wat19,Lemma 2.5.1], there exists a field extension K of K ( q / ) ([Wat19, Section 2.4])and additional parameters η , ζ ∈ ( K × ) I satisfying ζ i ζ τ ( i ) = 1 for all i ∈ I , (4.32) ζ i = 1 unless i = r, τ ( r ), (4.33) ǫ i η i η τ ( i ) ζ i = ǫ ′ i for all i ∈ I (4.34)such that the map A η , ζ : B ǫ → B ǫ ′ with A η , ζ | M X = id | M X and A η , ζ ( B ǫ i ) = η − i B ǫ ′ i , (4.35) A η , ζ ( L i ) = ζ − i L i , (4.36) A η , ζ ( K j ) = K j (4.37)for all i ∈ I \ X and j ∈ X is an algebra isomorphism. Here, we consider thealgebras B ǫ and B ǫ ′ over K . We note that A ǫ , ζ restricted to the semisimple partis the identity. In particular, we have A ǫ , ζ ( K ̟ j ) = K ̟ j for j ∈ X .Additionally, by (2.35) and (4.32) the map f : B ǫ ′ → B ǫ ′ given by f ( B i ) = ( ζ − i B i if i = r, τ ( r ), B i otherwise (4.38)and f | M X U Θ0 = id | M X U Θ0 is an algebra automorphism of B ǫ . Lemma 4.10.
For any ǫ and ǫ ′ satisfying (2.28) and for any η , ζ ∈ ( K × ) I satis-fying (4.32) , (4.33) and (4.34) , the relation T − r, ǫ ′ ( B ǫ ′ i ) = A η , ζ ◦ T − r, ǫ ◦ A − η , ζ ◦ f ( B ǫ ′ i ) (4.39) holds for any i ∈ I \ X . RAID GROUP ACTIONS FOR QUANTUM SYMMETRIC PAIRS OF TYPE AIII/AIV 21
Proof.
By Equations (3.10), (4.35) and (4.38), it suffices to verify (4.39) on theelements B r and B r − . We have A ǫ , ζ ◦ T − r, ǫ ◦ A − ǫ , ζ ◦ f ( B ǫ ′ r ) = ζ − r η r A ǫ , ζ ◦ T − r, ǫ ( B ǫ r ) (3.10) = qζ − r η r A ǫ , ζ ( B ǫ r L τ ( r ) K ̟ ′ τ ( r +1) )= qζ − r ζ − τ ( r ) B ǫ ′ r L τ ( r ) K ̟ ′ τ ( r +1) (4.32) = T − r, ǫ ′ ( B ǫ ′ r )and hence (4.39) holds on B ǫ ′ r .Using relations (4.32) and (4.34) we obtain C ǫ η − r η − τ ( r ) = ( qǫ r ǫ τ ( r ) ) − / η − r η − τ ( r ) = ( qη − r η − τ ( r ) ζ − r ζ − τ ( r ) ǫ ′ r ǫ ′ τ ( r ) ) − / η − r η − τ ( r ) = C ǫ ′ . Hence A ǫ , ζ ◦ T − r, ǫ ◦ A − ǫ , ζ ◦ f ( B ǫ ′ r − )= η r − A ǫ , ζ ◦ T − r, ǫ ( B ǫ r − )= η r − C ǫ A ǫ , ζ (cid:0)(cid:2) B ǫ τ ( r ) , [ F − X , [ B ǫ r , B ǫ r − ] q ] q (cid:3) q + ǫ r B ǫ r − L τ ( r ) K − X (cid:1) = C ǫ ′ (cid:2) B ǫ ′ τ ( r ) , [ F − X , [ B ǫ ′ r , B ǫ ′ r − ] q ] q (cid:3) q + C ǫ ′ ǫ r η r η τ ( r ) ζ − τ ( r ) B ǫ ′ r − L τ ( r ) K − X = T − r, ǫ ′ ( B ǫ ′ r − )as required, where the last equality follows again from relations (4.32) and (4.34). (cid:3) Recall by Corollary 4.9 that T − r, ǫ is an algebra endomorphism of B ǫ if ǫ r = ǫ τ ( r ) .Since A ǫ , ζ and f are algebra homomorphisms, Lemma 4.10 implies the followingcorollary. Corollary 4.11.
For any ǫ satisfying (2.28) , the map T − r, ǫ is an algebra endomor-phism of B ǫ .Remark . It is possible to complete similar calculations as in Section 4.1 in orderto show that T − r is an algebra endomorphism of B ǫ . This way, it is not necessaryto consider the field extension K × of K × , which is required in the construction ofthe algebra automorphism A ǫ , ζ : B ǫ → B ǫ ′ .4.3. Proof that T − r is the inverse of T r . In order to complete the proof ofTheorem 3.4, we now show that T r ◦ T − r = T − r ◦ T r = id. It suffices to check thison the generators B i for i ∈ I \ X . The relation T r ◦ T − r ( B i ) = B i = T − r ◦ T r ( B i )is straightforward for all i ∈ I \ X except i = r − i = τ ( r − i = r − i = τ ( r −
1) is analogous.
Proposition 4.13.
The relation T r ◦ T − r ( B r − ) = B r − (4.40) holds in B ǫ . Proof.
By Equations (3.9) and (3.10) we have1 C T r ◦ T − r ( B r − ) = (cid:2) q − B τ ( r ) L τ ( r ) , [ F − X , [ q − B r L r , [ S, B τ ( r ) ] q + ∆] q ] q (cid:3) q + ǫ r ([ S, B τ ( r ) ] q + ∆) L τ ( r ) K − X . (4.41)We consider the first summand of the above expression and simplify it. By Equa-tion (A.10), the element B r commutes with S . It follows that (cid:2) q − B r L r , [ S, B τ ( r ) ] q + ∆ (cid:3) q = (cid:2) B r , [ S, B τ ( r ) ] q (cid:3) L r + [ B r , ∆] L r = (cid:2) S, [ B r , B τ ( r ) ] (cid:3) q L r + [ B r , ∆] L r = 1 q − q − [ S, Γ r ] q L r + [ B r , ∆] L r . We now commute this with F − X . Using Equation (A.8) and Equation (A.25) weobtain (cid:2) F − X , [ S, Γ r ] q (cid:3) q = ǫ r (cid:2) F − X , [ S, Z r ] q (cid:3) q − ǫ τ ( r ) (cid:2) F − X , [ S, Z τ ( r ) ] q (cid:3) q = q ( q − q − ) ǫ r SL τ ( r ) K − X + ( q − q − ) (cid:2) F − X , [∆ , B r ] (cid:3) q . It follows from that (cid:2) F − X , [ q − B r L r , [ S, B τ ( r ) ] q + ∆] q (cid:3) q = 1 q − q − (cid:2) F − X , [ S, Γ r ] q (cid:3) q L r + (cid:2) F − X , [ B r , ∆] (cid:3) L r = qǫ r SK − X We now q -commute q − B τ ( r ) L τ ( r ) and qǫ r SK − X which gives (cid:2) q − B τ ( r ) L τ ( r ) , qǫ r SK − X (cid:3) q = qǫ r [ B τ ( r ) , S ] q − L τ ( r ) K − X = − ǫ r [ S, B τ ( r ) ] q L τ ( r ) K − X . Substituting the above into Equation (4.41) we obtain1 C T r ◦ T − r ( B r − ) = ǫ r ∆ L τ ( r ) K − X = qǫ r ǫ τ ( r ) B r − = 1 C B r − and hence we have T r ◦ T − r ( B r − ) = B r − as required. (cid:3) In the proof of the following proposition, we write T − r ( B r − ) = C (cid:0) [ B τ ( r ) , T ] q + Λ (cid:1) (4.42)where T = (cid:2) F − X , [ B r , B r − ] q (cid:3) q , (4.43)Λ = ǫ r B r − L τ ( r ) K − X . (4.44) Proposition 4.14.
The relation T − r ◦ T r ( B r − ) = B r − (4.45) holds in B ǫ . RAID GROUP ACTIONS FOR QUANTUM SYMMETRIC PAIRS OF TYPE AIII/AIV 23
Proof.
By Equations (3.10) and (4.9) we have1 C T − r ◦ T r ( B r − ) = (cid:2) [[[ B τ ( r ) , T ] q + Λ , qB r L τ ( r ) ] q , F + X ] q , qB τ ( r ) L r (cid:3) q qǫ τ ( r ) ([ B τ ( r ) , T ] q + Λ) L r K X . (4.46)By a similar proof to Lemma A.4 the elements B r and T commute. It follows that (cid:2) [ B τ ( r ) , T ] q + Λ , qB r L τ ( r ) (cid:3) q = q (cid:2) [ B τ ( r ) , T ] q , B r (cid:3) L τ ( r ) + q [Λ , B r ] L τ ( r ) = q (cid:2) [ B τ ( r ) , B r ] q , T (cid:3) L τ ( r ) + q [Λ , B r ] L τ ( r ) = − qq − q − [Γ r , T ] q L τ ( r ) + q [Λ , B r ] L τ ( r ) . (4.47)Similarly to Equation (A.8) we have Z r T = qT Z r − ( q − q − )[ B r , B r − ] q L τ ( r ) K − X . (4.48)It follows that − qq − q − ǫ r (cid:2) [ Z r , T ] q , F + X (cid:3) q L τ ( r ) = qǫ r (cid:2) [ B r , B r − ] q L τ ( r ) K − X , F + X (cid:3) q L τ ( r ) = q (cid:2) [ B r , ǫ r B r − L τ ( r ) K − X ] , F + X (cid:3) q L τ ( r ) = − q (cid:2) [Λ , B r ] , F + X (cid:3) q L τ ( r ) . As in the proof of Lemma A.10 the relation (cid:2) [ Z τ ( r ) , T ] q , F + X (cid:3) q = (1 − q − ) T L r K X (4.49)holds in B ǫ . This and Equation (4.47) implies that (cid:2) [[ B τ ( r ) , T ] q + Λ , qB r L τ ( r ) ] q , F + X (cid:3) q = qq − q − ǫ τ ( r ) (cid:2) [ Z τ ( r ) , T ] q , F + X (cid:3) q L τ ( r ) − qq − q − ǫ r (cid:2) [ Z r , T ] q , F + X (cid:3) q L τ ( r ) + q (cid:2) [Λ , B r ] , F + X (cid:3) q L τ ( r ) = ǫ τ ( r ) T K X . We substitute this into Equation (4.46) to obtain1 C T − r ◦ T r ( B r − ) = [ ǫ τ ( r ) T K X , qB τ ( r ) L r ] q + qǫ τ ( r ) ([ B τ ( r ) , T ] q + Λ) L r K X = q ǫ τ ( r ) [ T, B τ ( r ) ] q − L r K X + qǫ τ ( r ) ([ B τ ( r ) , T ] q + Λ) L r K X = qǫ τ ( r ) Λ L r K X = 1 C B r − as required. (cid:3) Proof of Theorem 3.6
Restricted to M X U Θ0 the automorphisms T i act as the Lusztig automorphism e T i = T e σ i for i ∈ I \ X . As a result, the braid relations of Theorem 3.6 hold onelements of M X U Θ0 . Hence it suffices to verify Theorem 3.6 on the elements B i for i ∈ I \ X . Braid relations I.
We first check that the relation T r T i = T i T r holds for all1 ≤ i ≤ r − Proposition 5.1.
For ≤ i ≤ r − and j ∈ I \ X the relation T r T i ( B j ) = T i T r ( B j ) (5.1) holds.Proof. By symmetry, we only check Equation (5.1) for 1 ≤ j ≤ r . This is done bya case-by-case analysis. Case 1. a ij = 0 , a jr = 2.In this case we have j = r and hence T i ( B j ) = B j . This implies T r T i ( B j ) = T r ( B j ) = q − B r K r K − τ ( r ) K ̟ ′ r +1 = T i T r ( B j )as required. Case 2. a ij = 0 , a jr = − j = r − T i ( B j ) = B j hence T r T i ( B j ) = T r ( B j ) = T i T r ( B j )as required. Case 3. a ij = 0 , a jr = 0.In this case, we have T r ( B j ) = B j and T i ( B j ) = B j so the statement of theproposition holds. Case 4. a ij = − , a jr = 0.Here, we have T r ( B j ) = B j and T i ( B j ) = ( qǫ i ) − / [ B j , B i ] q . Hence T r T i ( B j ) = ( qǫ i ) − / [ T r ( B j ) , T r ( B i )] q = ( qǫ i ) − / [ B j , B i ] q = T i T r ( B j ) . Case 5. a ij = − , a jr = − i = r − j = r −
1. Then by Equation (3.9) wehave T r − T r ( B r − ) = C T r − (cid:16)(cid:2) B r − , [ B r , [ F + X , B τ ( r ) ] q ] Q (cid:3) q + qǫ τ ( r ) B r − K r K − τ ( r ) K − X (cid:17) = q − / C (cid:16)(cid:2) [ B r − , B r − ] q , [ B r , [ F + X , B τ ( r ) ] q ] q (cid:3) q + qǫ τ ( r ) [ B r − , B r − ] q K r K − τ ( r ) K − X (cid:17) = q − / C (cid:16)(cid:2) [ B r − , [ B r , [ F + X , B τ ( r ) ] q ] q ] q .B r − (cid:3) q + qǫ τ ( r ) [ B r − K r K − τ ( r ) K − X , B r − ] q (cid:17) = q − / [ T r ( B r − ) , T r ( B r − )] q = T r T r − ( B r − )as required. Case 6. a ij = 2Then we have T r ( B i ) = B i and T r ( T i ( B i )) = T i ( B i ) which implies the result in thiscase. This completes the proof. (cid:3) RAID GROUP ACTIONS FOR QUANTUM SYMMETRIC PAIRS OF TYPE AIII/AIV 25
Braid relations II.
We now check that the relation T r T r − T r T r − ( B j ) = T r − T r T r − T r ( B j ) (5.2)holds for all j ∈ I \ X . Again for symmetry reasons it is enough to only consider1 ≤ j ≤ r . Many of the remaining proofs in this section require the use of relationsthat are proven in Appendix A. Since T r ( B j ) = B j and T r − ( B j ) = B j for 1 ≤ j For ≤ j < r − the relation (5.2) holds. As a result of the above lemma, it remains to verify Equation (5.2) for j ∈{ r − , r − , r } . For the next result, we use the relation T r − T r ( B r − ) = T − r ( B τ ( r − ) (5.3)which appears in the proof of Lemma A.13. Proposition 5.3. For j = r − the relation (5.2) holds.Proof. Using Equation (5.3) we have T r T r − T r T r − ( B r − ) = T r T r − T r (cid:0) q − B τ ( r − K τ ( r − K − r − (cid:1) = q − B r − K r − K − τ ( r − = T r − ( B τ ( r − ) . The result follows from Lemma A.13. (cid:3) Proposition 5.4. For j = r − the relation (5.2) holds.Proof. On one hand we have T r T r − T r T r − ( B r − ) = q − / T r T r − T r ([ B r − , B r − ] q )= q − / T r T r − ([ B r − , T r ( B r − )] q )= q − T r (cid:0)(cid:2) [ B r − , B r − ] q , T r − T r ( B r − ) (cid:3) q (cid:1) Again by Equation (5.3) it follows that T r T r − T r T r − ( B r − ) = q − (cid:2) [ B r − , T r ( B r − )] q , B τ ( r − (cid:3) q = q − (cid:2) B r − , [ T r ( B r − ) , B τ ( r − ] q (cid:3) q where the last equality follows since B r − commutes with B τ ( r − . On the otherhand we have T r − T r T r − T r ( B r − ) = T r − T r T r − ( B r − )= q − (cid:2) [ B r − , B r − ] q , T − r ( B τ ( r − ) (cid:3) q . Since B τ ( r − commutes with T r ( B r − ) it follows that B r − T − r ( B τ ( r − ) = T − r ( B τ ( r − ) B r − . (5.4)By Corollary A.15 the element [ B r − , T − r ( B τ ( r − )] q is invariant under T r . Thisand Equation (5.4) imply T r − T r T r − T r ( B r − ) = q − (cid:2) B r − , [ B r − , T − r ( B τ ( r − )] q (cid:3) q = q − (cid:2) B r − , [ T r ( B r − ) , B τ ( r − ] q (cid:3) q = T r T r − T r T r − ( B r − ) as required. (cid:3) Proposition 5.5. For j = r the relation (5.2) holds.Proof. Consider first the term T r − T r T r − T r ( B r ). By Equations (3.7) and (3.9),Equation (5.3) and T i | M X U Θ0 = e T i | M X U Θ0 for 1 ≤ i ≤ r we obtain T r − T r T r − T r ( B r )= q − (cid:2) [ B r , B r − ] q L r L r − K ̟ ′ r +1 , T − r ( B τ ( r − ) (cid:3) q L r K ̟ ′ r +1 = q − (cid:2) [ B r , B r − ] q , T − r ( B τ ( r − ) (cid:3) q L r L r − K ̟ ′ r +1 where the second equality follows from Equation (2.33) and noting that K ̟ ′ r +1 commutes with T − r ( B τ ( r − ). Since [ T r ( B r ) , B τ ( r − ] q − = 0, it follows that[ B r , T − r ( B τ ( r − )] q − = 0. Using this and Corollary A.15 it follows that (cid:2) [ B r , B r − ] q , T − r ( B τ ( r − ) (cid:3) q = (cid:2) B r , [ B r − , T − r ( B τ ( r − )] q (cid:3) = (cid:2) B r , [ T r ( B r − ) , B τ ( r − ] q (cid:3) and hence we obtain T r − T r T r − T r ( B r ) = q − (cid:2) B r , [ T r ( B r − ) , B τ ( r − ] q (cid:3) L r L r − K ̟ ′ r +1 . Considering now the term T r T r − T r T r − ( B r ) we obtain T r T r − T r T r − ( B r )= q − / T r T r − T r ([ B r , B r − ] q )= q − / T r T r − ([ B r L r K ̟ ′ r +1 , T r ( B r − )] q )= q − T r (cid:0)(cid:2) [ B r , B r − ] q L r L r − K ̟ ′ r +1 , T − r ( B τ ( r − ) (cid:3) q (cid:1) = q − (cid:2) [ B r L r K ̟ ′ r +1 , T r ( B r − )] q , B τ ( r − (cid:3) L r L r − K ̟ ′ r +1 = q − (cid:2) B r , [ T r ( B r − ) , B τ ( r − ] q (cid:3) L r L r − K ̟ ′ r +1 = T r − T r T r − T r ( B r )as required. (cid:3) Appendix A. Relations in B ǫ Many of the results in Sections 3 to 5 require the use of additional relationswhich we provide here. We first give two useful relations that are used throughoutthis appendix.Recall from Equations (2.19) to (2.24) the elements E + J , E − J , F + J , F − J and K J where J ⊂ I is a subset of the form J = { a, a + 1 , . . . , b − , b } with a ≤ b . Rewritingthese elements using the Lusztig automorphisms, one sees that E + J F − J − F − J E + J = K J − K − J q − q − = E − J F + J − F + J E − J (A.1)holds in U q ( g ). Additionally, the q -commutator satisfies (cid:2) [ x, y ] q , z (cid:3) q − (cid:2) x, [ y, z ] q (cid:3) q = q (cid:2) [ x, z ] , y (cid:3) (A.2)for all x, y, z ∈ U q ( g ). RAID GROUP ACTIONS FOR QUANTUM SYMMETRIC PAIRS OF TYPE AIII/AIV 27 A.1. Relations needed for the proof of Theorem 3.4.Lemma A.1. The relations (cid:2) B r − , [ F + X , Z r ] (cid:3) q = 0 , (A.3) (cid:2) B r − , [ F + X , Z τ ( r ) ] (cid:3) q = − ( q − q − ) B r − L r ( K X − K − X ) (A.4) hold in B ǫ .Proof. Since [ B r − , F + X ] = 0 and [ B r − , Z r ] q = 0 it follows that Equation (A.3)holds. On the other hand, making use of the relation[ F + X , Z τ ( r ) ] = q − ( K X − K − X ) L r , (A.5)which follows from (A.1), we obtain (cid:2) B r − , [ F + X , Z τ ( r ) ] (cid:3) q = q − [ B r − , ( K X − K − X ) L r ] q = − ( q − q − ) B r − ( K X − K − X ) L r as required. (cid:3) Lemma A.2. The relations (cid:2) B r − , [ B r , [ F + X , B τ ( r ) ] q ] q (cid:3) q = (cid:2) [ B r − , [ B r , F + X ] q ] q , B τ ( r ) (cid:3) q + qǫ τ ( r ) B r − L r ( K X − K − X ) , (A.6) (cid:2) B τ ( r − , [ B τ ( r ) , [ F − X , B r ] q ] q (cid:3) q = (cid:2) [ B τ ( r − , [ B τ ( r ) , F − X ] q ] q , B r (cid:3) q + qǫ r B τ ( r − L τ ( r ) ( K X − K − X ) (A.7) hold in B ǫ .Proof. By symmetry we only verify Equation (A.6). By Equation (A.2) we have (cid:2) B r , [ F + X , B τ ( r ) ] q (cid:3) q = (cid:2) [ B r , F + X ] q , B τ ( r ) (cid:3) q + q (cid:2) F + X , [ B r , B τ ( r ) ] (cid:3) = (cid:2) [ B r , F + X ] q , B τ ( r ) (cid:3) q − qǫ τ ( r ) q − q − [ F + X , Z τ ( r ) ] + qǫ r q − q − [ F + X , Z r ] . Since B r − commutes with B τ ( r ) , Lemma A.1 implies (cid:2) B r − , [ B r , [ F + X ,B τ ( r ) ] q ] q (cid:3) q = (cid:2) B r − , [[ B r , F X ] q , B τ ( r ) ] q (cid:3) q − qǫ τ ( r ) q − q − (cid:2) B r − , [ F + X , Z τ ( r ) ] (cid:3) q = (cid:2) [ B r − , [ B r , F + X ] q ] q , B τ ( r ) (cid:3) q + qǫ τ ( r ) B r − L r ( K X − K − X )as required. (cid:3) Recall from Section 4.1 the elements S = (cid:2) B r − , [ B r , F + X ] q (cid:3) q ,S τ = (cid:2) B τ ( r − , [ B τ ( r ) , F − X ] q (cid:3) q , ∆ = qǫ τ ( r ) B r − L r K X , ∆ τ = qǫ r B τ ( r − L τ ( r ) K X . For the remainder of this section, we provide relations that include the terms S, S τ , T and T τ . Lemma A.3. The relations Z τ ( r ) S = q − S Z τ ( r ) + (1 − q − )[ B r − , B r ] q L r K X , (A.8) Z r S τ = q − S τ Z r + (1 − q − )[ B τ ( r − , B τ ( r ) ] q L τ ( r ) K X , (A.9) hold in B ǫ .Proof. We only prove that Equation (A.8) holds since the remaining checks aresimilar. By Equation (A.1) and the relation Z τ ( r ) B i = q ( α i ,α τ ( r ) − α r ) B i Z τ ( r ) forany i ∈ I \ X we have Z τ ( r ) S = Z τ ( r ) (cid:2) B r − , [ B r , F + X ] q (cid:3) q = q − (cid:2) B r − , [ B r , Z τ ( r ) F + X ] q (cid:3) = q − (cid:2) B r − , [ B r , F + X Z τ ( r ) − q − ( K X − K − X ) L r ] q (cid:3) = q − S Z τ ( r ) − q − (cid:2) B r − , [ B r , ( K X − K − X ) L r ] q (cid:3) = q − S Z τ ( r ) + q − ( q − B r − , B r L r K X ]= q − S Z τ ( r ) + (1 − q − )[ B r − , B r ] q L r K X as required. (cid:3) Lemma A.4. The relations B r S = SB r , (A.10) B τ ( r ) S τ = S τ B τ ( r ) , (A.11) hold in B ǫ .Proof. By symmetry we only verify Equation (A.10). Using the relations p ( B r , B r − ) = p ( B r , F + X ) = 0we obtain B r S = B r (cid:2) [ B r − , B r ] q , F + X (cid:3) q = B r B r − B r F + X − qB r B r − F + X − qB r F + X B r − B r + q B r F + X B r B r − = q + q − (cid:0) B r B r − + B r − B r (cid:1) F + X − qB r B r − F + X − qB r F + X B r − B r + q q + q − (cid:0) B r F + X + F + X B r (cid:1) B r − = q + q − B r − B r F + X − qB r F + X B r − B r + q q + q − F + X B r B r − = q + q − B r − (cid:0) ( q + q − ) B r F + X B r − F + X B r (cid:1) − qB r F + X B r − B r + q q + q − F + X (cid:0) ( q + q − ) B r B r − B r − B r − B r (cid:1) = (cid:2) [ B r − , B r ] q , F + X (cid:3) q B r = SB r as required. (cid:3) Lemma A.5. The relations SB τ ( r − = B τ ( r − S − qǫ r − Z r − [ B r , F + X ] q , (A.12) S τ B r − = B r − S τ − qǫ τ ( r − Z τ ( r − [ B τ ( r ) , F − X ] q , (A.13) RAID GROUP ACTIONS FOR QUANTUM SYMMETRIC PAIRS OF TYPE AIII/AIV 29 hold in B ǫ .Proof. We have SB τ ( r − = (cid:2) B r − , [ B r , F + X ] q (cid:3) q B τ ( r − = (cid:2) B r − B τ ( r − , [ B r , F + X ] q (cid:3) q = B τ ( r − S + q − q − (cid:2) [Γ r − , B r ] q , F + X (cid:3) q = B τ ( r − S − qǫ r − Z r − [ B r , F + X ] q as required. Equation (A.13) is verified similarly. (cid:3) In the following lemma, we introduce the termsΩ − = ǫ τ ( r ) ǫ τ ( r − F − X K X L r Z τ ( r − , (A.14)Ω + = ǫ r ǫ r − F + X K X L τ ( r ) Z r − , (A.15) Lemma A.6. The relations SS τ − S τ S = Ω − − Ω + , (A.16) hold in B ǫ .Proof. Recall from Theorem 3.2 that T r − is an algebra automorphism of B ǫ withinverse T − r − given by Equation (3.8). We express S and S τ using the algebraautomorphisms T w X and T − r − . In particular using Equations (3.2), (3.3) and (3.8)we have S = ( qǫ r − ) / T − r − ◦ T w X ( B r ) ,S τ = ( qǫ τ ( r − ) / T − r − ◦ T w X ( B τ ( r ) ) . Since ǫ r − = ǫ τ ( r − we obtain SS τ − S τ S = qǫ r − T − r − ◦ T w X ( B r B τ ( r ) − B τ ( r ) B r )= qq − q − T − r − ◦ T w X ( ǫ r − ǫ r Z r − ǫ τ ( r ) ǫ τ ( r − Z τ ( r ) ) . Equation (2.26) implies that T w X ( Z r ) = (1 − q − ) F + X K X L τ ( r ) ,T w X ( Z τ ( r ) ) = (1 − q − ) F − X K X L r . Hence we obtain SS τ − S τ S = T − r − (cid:0) ǫ r − ǫ r F + X K X L τ ( r ) − ǫ τ ( r ) ǫ τ ( r − F − X K X L r (cid:1) = ǫ τ ( r ) ǫ τ ( r − F − X K X L r Z τ ( r − − ǫ r − ǫ r F + X K X L τ ( r ) Z r − = Ω − − Ω + as required. (cid:3) Lemma A.7. The relation (cid:2) [ S, B τ ( r ) ] q , [ S τ , B r ] q (cid:3) = − (cid:2) ∆ , [ S τ , B r ] q (cid:3) − (cid:2) [ S, B τ ( r ) ] q , ∆ τ (cid:3) − qǫ r ǫ τ ( r ) q − q − (cid:0) K X − K − X ) K X Γ r − (A.17) holds in B ǫ . Proof. We first use Equations (A.10), (A.11) and (A.16) to rewrite each term of[ S, B τ ( r ) ] q [ S τ , B r ] q . In particular we have SB τ ( r ) S τ B r = SS τ B τ ( r ) B r = (cid:0) S τ S + Ω − − Ω + (cid:1)(cid:0) B r B τ ( r ) − q − q − Γ r (cid:1) = S τ SB r B τ ( r ) − q − q − S τ S Γ r + (Ω − − Ω + ) B r B τ ( r ) − q − q − (Ω − − Ω + )Γ r . Similarly, one finds that SB τ ( r ) B r S τ = B r S τ SB τ ( r ) + B r (Ω − − Ω + ) B τ ( r ) − q − q − S Γ r S τ ,B τ ( r ) SS τ B r = S τ B r B τ ( r ) S + B τ ( r ) (Ω − − Ω + ) B r − q − q − S τ Γ r S,B τ ( r ) SB r S τ = B r S τ B τ ( r ) S + B τ ( r ) B r (Ω − − Ω + ) − q − q − Γ r SS τ + q − q − Γ r (Ω − − Ω + ) . Combining these four expressions we obtain (cid:2) [ S, B τ ( r ) ] q , [ S τ , B r ] q (cid:3) = − q − q − (cid:2) S τ , [ S, Γ r ] q (cid:3) q + (cid:2) [Ω − − Ω + , B r ] q , B τ ( r ) (cid:3) q − q − q − (cid:2) Ω − − Ω + , Γ r (cid:3) q . (A.18)We now consider the term (cid:2) S τ , [ S, Γ r ] q (cid:3) q in more detail. Using Equations (A.8)and (A.9) we have S τ Γ r S = ǫ r S τ Z r S − ǫ τ ( r ) S τ Z τ ( r ) S = ǫ r (cid:0) q Z r S τ − ( q − q − )[ B τ ( r − , B τ ( r ) ] q K X L τ ( r ) (cid:1) S − ǫ τ ( r ) S τ (cid:0) q − S Z τ ( r ) + (1 − q − )[ B r − , B r ] q K X L r (cid:1) = qǫ r Z r S τ S − q − ǫ τ ( r ) S τ S Z τ ( r ) − ( q − q − )[∆ τ , B τ ( r ) ] S − (1 − q − ) S τ [∆ , B r ] . We similarly obtain S Γ r S τ = q − ǫ r SS τ Z r − qǫ τ ( r ) Z τ ( r ) SS τ + (1 − q − ) S [∆ τ , B τ ( r ) ]+ ( q − q − )[∆ , B r ] S τ . It hence follows from this and Equation (A.16) that (cid:2) S τ , [ S, Γ r ] q (cid:3) q = ( q − q − ) (cid:2) S τ , [∆ , B r ] (cid:3) q − ( q − q − ) (cid:2) S, [∆ τ , B τ ( r ) ] (cid:3) q − c r (Ω − − Ω + ) Z r + q c r Z r (Ω − − Ω + ) . (A.19)We now consider the elements (cid:2) S τ , [∆ , B r ] (cid:3) q and (cid:2) S, [∆ τ , B τ ( r ) ] (cid:3) q and write themin the form that appears in Equation (A.17). By Equation (A.13) it follows that (cid:2) S τ , [∆ , B r ] (cid:3) q = S τ ∆ B r − S τ B r ∆ − q ∆ B r S τ + qB r ∆ S τ = (cid:0) ∆ S τ + [Ω − , B τ ( r ) ] q (cid:1) B r − S τ B r ∆ − q ∆ B r S τ + qB r (cid:0) S τ ∆ − [Ω − , B τ ( r ) ] q (cid:1) = (cid:2) ∆ , [ S τ , B r ] q (cid:3) + (cid:2) [Ω − , B τ ( r ) ] q , B r (cid:3) q . Similarly, by Equation (A.12) we have (cid:2) S, [∆ τ , B τ ( r ) ] (cid:3) q = − (cid:2) [ S, B τ ( r ) ] , ∆ τ (cid:3) + (cid:2) [Ω + , B r ] q , B τ ( r ) (cid:3) q . RAID GROUP ACTIONS FOR QUANTUM SYMMETRIC PAIRS OF TYPE AIII/AIV 31 Substituting these two expressions into Equation (A.19) gives (cid:2) S τ , [ S, Γ r ] q (cid:3) q = ( q − q − ) (cid:2) ∆ , [ S τ , B r ] q (cid:3) + ( q − q − ) (cid:2) [ S, B τ ( r ) ] , ∆ τ (cid:3) + ( q − q − ) (cid:2) [Ω − , B τ ( r ) ] q , B r (cid:3) q − ( q − q − ) (cid:2) [Ω + , B r ] q , B τ ( r ) (cid:3) q − ǫ r (Ω − − Ω + ) Z r + q ǫ r Z r (Ω − − Ω + ) . (A.20)We substitute Equation (A.20) into Equation (A.18). Noting that (cid:2) [Ω − , B r ] q , B τ ( r ) (cid:3) q − (cid:2) [Ω − , B τ ( r ) ] q , B r (cid:3) q = q − q − [Ω − , Γ r ] q , we obtain (cid:2) [ S, B τ ( r ) ] q , [ S τ , B r ] q (cid:3) = − (cid:2) ∆ , [ S τ , B r ] q (cid:3) − (cid:2) [ S, B τ ( r ) ] q , ∆ τ (cid:3) + q − q − (cid:0) c r [Ω − , Z r ] q − c τ ( r ) [Ω + , Z τ ( r ) ] q (cid:1) . Using Equation (A.1) we compute [Ω − , Z r ] q and [Ω + , Z τ ( r ) ] q . This gives[Ω − , Z r ] q = qǫ τ ( r ) ǫ τ ( r − (cid:0) K X − K − X (cid:1) K X Z τ ( r − , [Ω + , Z τ ( r ) ] q = qǫ r ǫ r − (cid:0) K X − K − X (cid:1) K X Z r − . It hence follows that (cid:2) [ S, B τ ( r ) ] q , [ S τ , B r ] q (cid:3) = − (cid:2) ∆ , [ S τ , B r ] q (cid:3) − (cid:2) [ S, B τ ( r ) ] q , ∆ τ (cid:3) − qǫ r ǫ τ ( r ) q − q − (cid:0) K X − K − X ) K X Γ r − as required. (cid:3) Lemma A.8. The relation Z r [ S, B τ ( r ) ] q = q − [ S, B τ ( r ) ] q Z r + q − ( q − q − )∆ Z r (A.21) holds in B ǫ .Proof. The difficulty in the proof comes from that fact that the element Z r contains E + X as a factor, and there is generally no simple way to commute E + X with F + X . Theidea is to verify that Equation (A.21) holds if the algebra automorphism T − w X isapplied to both sides. More precisely, using (2.25), (2.26), (3.2) and (3.5) we have[ S, B τ ( r ) ] q = T w X (cid:0)(cid:2) B r − , [ B r , [ F + X , B τ ( r ) ] q ] q (cid:3) q (cid:1) , Z r = q (1 − q − ) T w X (cid:0) F + X K − X L τ ( r ) (cid:1) , ∆ = qǫ τ ( r ) T w X (cid:0) B r − L r K − X (cid:1) and hence verifying Equation (A.21) is equivalent to showing that F + X (cid:2) B r − , [ B r , [ F + X , B τ ( r ) ] q ] q (cid:3) q = (cid:2) B r − , [ B r , [ F + X , B τ ( r ) ] q ] q (cid:3) q F + X + ( q − q − ) ǫ τ ( r ) B r − L r K − X F + X (A.22)holds in B ǫ . Using the relations p ( F + X , B r ) = p ( F + X , B τ ( r ) ) = 0 we can commute F + X through [ B r , [ F + X , B τ ( r ) ] q ] q . This gives F + X [ B r , [ F + X , B τ ( r ) ] q ] q = F + X B r F + X B τ ( r ) − qF + X B r B τ ( r ) F + X − qF +2 X B τ ( r ) B r + q F + X B τ ( r ) F + X B r = q + q − (cid:0) F +2 X B r + B r F +2 X (cid:1) B τ ( r ) − qF + X B r B τ ( r ) F + X − qF +2 X B τ ( r ) B r + q q + q − (cid:0) F +2 X B τ ( r ) + B τ ( r ) F +2 X (cid:1) B r = q − q − F +2 X Γ r + q + q − B r (cid:0) ( q + q − ) F + X B τ ( r ) F + X − B τ ( r ) F +2 X (cid:1) − qF + X B r B τ ( r ) F + X + q q + q − B τ ( r ) (cid:0) ( q + q − ) F + X B r F + X − B r F +2 X (cid:1) = [ B r , [ F + X , B τ ( r ) ] q ] q F + X + q − q − (cid:0) F +2 X Γ r − (1 + q ) F + X Γ r F + X + q Γ r F +2 X (cid:1) where the third equality follows from observing that terms beginning with F +2 X simplify. Since B r − commutes with F + X , and[ B r − , Γ r ] q = ( q − ǫ τ ( r ) B r − Z τ ( r ) it follows that F + X (cid:2) B r − , [ B r , [ F + X , B τ ( r ) ] q ] q (cid:3) q − (cid:2) B r − , [ B r , [ F + X , B τ ( r ) ] q ] q (cid:3) q F + X = qq + q − ǫ τ ( r ) B r − (cid:0) F +2 X Z τ ( r ) − (1 + q ) F + X Z τ ( r ) F + X + q Z τ ( r ) F +2 X (cid:1) . The relation F + X Z τ ( r ) = Z τ ( r ) F + X + q − ( K X − K − X ) L r implies that F +2 X Z τ ( r ) − (1 + q ) F + X Z τ ( r ) F + X + q Z τ ( r ) F +2 X = q − F + X ( K X − K − X ) L r − q ( K X − K − X ) L r F + X = q − ( q − q − ) K − X L r F + X . Hence (A.22) holds as required. (cid:3) Lemma A.9. The relation S [ S, B τ ( r ) ] q = q − [ S, B τ ( r ) ] q S − q − ( q − q − ) S ∆ (A.23) holds in B ǫ .Proof. As in the proof of Lemma A.8 we verify a relation that is equivalent toEquation (A.23). The difference here is that we additionally use the algebra auto-morphism T r − from Equation (3.7). In particular using (3.2), (3.3) and (3.8) wehave S = ( qǫ r − ) / T w X ◦ T − r − ( B r ) ,B τ ( r ) = ( qǫ τ ( r − ) − / T w X ◦ T − r − (cid:0) [ F + X , [ B τ ( r ) , B τ ( r − ] q ] q (cid:1) , ∆ = ǫ τ ( r ) T w X ◦ T − r − (cid:0) B τ ( r − L r K − X (cid:1) and hence we are done if we show that B r (cid:2) B r , [ F + X , [ B τ ( r ) , B τ ( r − ] q ] q (cid:3) q = q − (cid:2) B r , [ F + X , [ B τ ( r ) , B τ ( r − ] q ] q (cid:3) q B r − q − ( q − q − ) ǫ τ ( r ) B r B τ ( r − L r K − X . RAID GROUP ACTIONS FOR QUANTUM SYMMETRIC PAIRS OF TYPE AIII/AIV 33 Noting that B r [ B τ ( r ) , B τ ( r − ] q = [ B τ ( r ) , B τ ( r − ] q B r + q − q − [Γ r , B τ ( r − ] q = [ B τ ( r ) , B τ ( r − ] q B r + qǫ τ ( r ) Z τ ( r ) B τ ( r − (A.24)one calculates (cid:2) B r , [ F + X , [ B τ ( r ) , B τ ( r − ] q ] q (cid:3) q = (cid:2) [ B r , F + X ] q , [ B τ ( r ) , B τ ( r − ] q (cid:3) q + q c τ ( r ) [ F + X , Z τ ( r ) ] B τ ( r − . Using this, the relation B r [ B r , F + X ] q = q − [ B r , F + X ] q B r and Equation (A.24) wehave B r (cid:2) B r , [ F + X , [ B τ ( r ) , B τ ( r − ] q ] q (cid:3) q = B r (cid:2) [ B r , F + X ] q , [ B τ ( r ) , B τ ( r − ] q (cid:3) q + q ǫ τ ( r ) B r [ F + X , Z τ ( r ) ] B τ ( r − = q − (cid:2) [ B r , F + X ] q , [ B τ ( r ) , B τ ( r − ] q (cid:3) q B r + ǫ τ ( r ) B r [ F + X , Z τ ( r ) ] B τ ( r − − q ǫ τ ( r ) [ F + X , Z τ ( r ) ] B r B τ ( r − + q ǫ τ ( r ) B r [ F + X , Z τ ( r ) ] B τ ( r − = q − (cid:2) B r , [ F + X , [ B τ ( r ) , B τ ( r − ] q ] q (cid:3) q B r + (1+ q ) ǫ τ ( r ) (cid:2) B r , [ F + X , Z τ ( r ) ] (cid:3) q B τ ( r − . Since [ F + X , Z τ ( r ) ] = q − ( K X − K − X ) L r we have (cid:2) B r , [ F + X , Z τ ( r ) ] (cid:3) q = q − [ B r , ( K X − K − X ) L r ] q = − q − (1 − q − ) B r K − X L r and hence we obtain B r (cid:2) B r , [ F + X , [ B τ ( r ) , B τ ( r − ] q ] q (cid:3) q = q − (cid:2) B r , [ F + X , [ B τ ( r ) , B τ ( r − ] q ] q (cid:3) q B r − q − ( q − q − ) ǫ τ ( r ) B r B τ ( r − K − X L r as required. (cid:3) Lemma A.10. The relation (cid:2) F − X , [ S, Z r ] q (cid:3) q = ( q − SL τ ( r ) K − X , (A.25) holds in B ǫ .Proof. Observe that (cid:2) F − X , [ S, Z r ] q (cid:3) q = (cid:2) [ F − X , S ] q , Z r (cid:3) q + q (cid:2) S, [ F − X , Z r ] (cid:3) holds by Equation (A.2). Since S = T w X ([ B r − , B r ] q ) and F − X = − T w X ( E − X K X ) itfollows that [ F − X , S ] q = − T w X (cid:0)(cid:2) E − X K X , [ B r − , B r ] q (cid:3) q (cid:1) = 0since [ E − X K X , B r ] q = 0. Further we have[ F − X , Z r ] = − (1 − q − )[ F − X , E + X L τ ( r ) ]= − (1 − q − )[ F − X , E + X ] L τ ( r ) = q − ( K X − K − X ) L τ ( r ) . Hence we obtain (cid:2) F − X , [ S, Z r ] q (cid:3) q = q (cid:2) S, [ F − X , Z r ] (cid:3) = (cid:2) S, K X − K − X ] q L τ ( r ) = ( q − SL τ ( r ) K − X as required. (cid:3) A.2. Relations needed for the proof of Theorem 3.6.Lemma A.11. The relation (cid:2) T r − ( B r − ) , [ F + X , [ B τ ( r ) , B τ ( r − ] q ] q (cid:3) = 0 (A.26) holds in B ǫ .Proof. Since T r − ( B r − ) = q − B τ ( r − L τ ( r − it follows that T r − ( B r − ) commuteswith F j for j ∈ X . Further, Equation (2.36) implies that B τ ( r − [ B τ ( r ) , B τ ( r − ] q = q [ B τ ( r ) , B τ ( r − ] q B τ ( r − and hence T r − ( B r − ) commutes with [ B τ ( r ) , B τ ( r − ] q . The result follows fromthis. (cid:3) Lemma A.12. For any i ∈ I \ ( X ∪ { r, τ ( r ) } ) the relations (cid:2) B τ ( i ) L τ ( i ) , [ B i ± , B i ] q (cid:3) q = q ǫ i B i ± , (A.27) (cid:2) [ B i , B i ± ] q , B τ ( i ) L i (cid:3) q = ǫ i B i ± (A.28) hold in B ǫ .Proof. The relations follow immediately by applying the automorphisms T i and T − i to T − i ( B i ± ) = ( qǫ i ) − / [ B i , B i ± ] q , T i ( B i ± ) = ( qǫ i ) − / [ B i ± , B i ] q , respectively. (cid:3) In the following Lemma, which is used in the proof of Proposition 5.3, we makeuse of the fact that T r is an algebra automorphism of B ǫ , see Theorem 3.4 andSection 4. Lemma A.13. The relation T r − T r T r − T r ( B r − ) = T r − ( B τ ( r − ) (A.29) holds in B ǫ .Proof. Calculating directly we have T r − T r ( B r − ) = C (cid:0)(cid:2) T r − ( B r − ) , [ T r − ( B r ) , [ F + X , T r − ( B τ ( r ) )] q ] q (cid:3) q + qǫ r T r − ( B r − L r ) K − X (cid:1) = C (cid:0) q − ǫ − r (cid:2) B τ ( r − L τ ( r − , [[ B r , B r − ] q , [ F + X , [ B τ ( r ) , B τ ( r − ] q ] q ] q (cid:3) q + ǫ r B τ ( r − L r K − X (cid:1) = T − r ( B τ ( r − ) RAID GROUP ACTIONS FOR QUANTUM SYMMETRIC PAIRS OF TYPE AIII/AIV 35 where the last equality is obtained using Equations (A.26) and (A.27). This impliesthat T r − T r T r − T r ( B r − ) = T r − ( B τ ( r − )as required. (cid:3) Lemma A.14. The relation (cid:2) B r − , [ B r , [ F + X , [ B τ ( r ) , B τ ( r − ] q ] q ] q (cid:3) q = (cid:2) [ B r − , [ B r , [ F + X , B τ ( r ) ] q ] q ] q , B τ ( r − (cid:3) q (A.30) holds in B ǫ .Proof. First observe that since B τ ( r − commutes with B r and F + X we have (cid:2) B r , [ F + X , [ B τ ( r ) , B τ ( r − ] q ] q (cid:3) q = (cid:2) [ B r , [ F + X , B τ ( r ) ] q ] q , B τ ( r − (cid:3) q . To shorten notation, let Y = [ B r , [ F + X , B τ ( r ) ] q ] q . Recall from Equation (2.35) that B r − B τ ( r − − B τ ( r − B r − = ( q − q − ) − ( ǫ r − Z r − − ǫ τ ( r − Z τ ( r − )where Z r − = − L τ ( r − and Z τ ( r − = − L r − . Then Y commutes with both Z r − and Z τ ( r − . Recalling the notation Γ i from Equation (2.32) we hence have (cid:2) B r − , [ Y, B τ ( r − ] q (cid:3) q = B r − Y B τ ( r − − qB r − B τ ( r − Y − qY B τ ( r − B r − + q B τ ( r − Y B r − = B r − Y B τ ( r − − q ( B τ ( r − B r − + ( q − q − ) − Γ r − Y ) − qY ( B r − B τ ( r − − ( q − q − ) − Γ r − ) + q B τ ( r − Y B r − = (cid:2) [ B r − , Y ] q , B τ ( r − (cid:3) q as required. (cid:3) Corollary A.15. The element [ B r − , T − r ( B τ ( r − )] q is T r -invariant i.e. [ B r − , T − r ( B τ ( r − )] q = [ T r ( B r − ) , B τ ( r − ] q . (A.31) Proof. The result follows immediately from Lemma A.14 and the fact that[ B r − , B τ ( r − L r K − X ] q = q [ B r − L r K − X , B τ ( r − ] q . (cid:3) A.3. Relations needed for the proof of Theorem 3.8.Lemma A.16. For any j ∈ X \ { r + 1 , τ ( r + 1) } the relation T j ( F + X ) = F + X (A.32) holds.Proof. Since j ∈ X \ { r + 1 , τ ( r + 1) } we assume that | X | ≥ 3. Recalling that F + X = (cid:2) F r +1 , [ F r +2 , . . . , [ F τ ( r +2) , F τ ( r +1) ] q . . . ] q (cid:3) q the result follows since for any j ∈ X \ { r + 1 , τ ( r + 1) } we have T j ([ F j − , [ F j , F j +1 ] q ] q ) = T j ([ F j − , T − j ( F j +1 )] q )= [[ F j − , F j ] q , F j +1 ] q = [ F j − , [ F j , F j +1 ] q ] q . (cid:3) Lemma A.17. The relation T r +1 (cid:0)(cid:2) B r − , [ B r , [ F + X , B τ ( r ) ] q ] q (cid:3) q (cid:1) = [ B r − , [ B r , [ F + X , B τ ( r ) ] q ] q ] q + qǫ τ ( r ) B r − L r ( K − r +1 − K r +1 ) K − X \{ r +1 } (A.33) holds in B ǫ .Proof. First suppose that X = { r + 1 } . By Equation (3.1) we have [ F r +1 , B r +2 ] q = T − r +1 ( B r +2 ) and T r +1 ( B r ) = [ B r , F r +1 ] q . Hence T r +1 (cid:0)(cid:2) B r − , [ B r , [ F r +1 , B r +2 ] q ] q (cid:3) q (cid:1) = (cid:2) B r − , [[ B r , F r +1 ] q , B r +2 ] q (cid:3) q . (A.34)By Equation (A.2) we have (cid:2) [ B r , F r +1 ] q , B r +2 (cid:3) q = (cid:2) B r , [ F r +1 , B r +2 ] q (cid:3) q + q (cid:2) [ B r , B r +2 ] , F r +1 (cid:3) = (cid:2) B r , [ F r +1 , B r +2 ] q (cid:3) q + qq − q − [ ǫ r Z r − ǫ r +2 Z r +2 , F r +1 ] . Since [ B r − , L r +2 ] q = 0 it follows that (cid:2) B r − , [ Z r , F r +1 ] (cid:3) q = 0. On the other handwe have [ Z r +2 , F r +1 ] = − (1 − q − )[ E r +1 L r , F r +1 ]= − (1 − q − ) L r [ E r +1 , F r +1 ]= − q − L r ( K r +1 − K − r +1 )and hence (cid:2) B r − , [ Z r +2 , F r +1 ] (cid:3) q = − q − [ B r − , L r ] q ( K r +1 − K − r +1 )= ( q − q − ) B r − L r ( K r +1 − K − r +1 ) . It follows that (cid:2) B r − , [[ B r , F r +1 ] q , B r +2 ] q (cid:3) q = (cid:2) B r − , [ B r , [ F r +1 , B r +2 ] q ] q (cid:3) q + qǫ r +2 B r − L r ( K − r +1 − K r +1 ) . The result follows by substituting this into (A.34).We now consider the case | X | > 1. Let Y = X \ { r + 1 } . Observing that[ F + X , B τ ( r ) ] q = T − r +1 ([ F + Y , B τ ( r ) ] q ) we have T r +1 (cid:0)(cid:2) B r − , [ B r , [ F + X , B τ ( r ) ] q ] q (cid:3) q (cid:1) = (cid:2) B r − , [[ B r , F r +1 ] q , [ F + Y , B τ ( r ) ] q ] q (cid:3) q . (A.35)Since F + Y commutes with B r , Equation (A.2) implies (cid:2) [ B r , F r +1 ] q , [ F + Y , B τ ( r ) ] q (cid:3) q − [ B r , [ F + X , B τ ( r ) ] q ] q = qq − q − (cid:2) [ F + Y , ǫ r Z r − ǫ τ ( r ) Z τ ( r ) ] q , F r +1 (cid:3) . Since [ B r − , L τ ( r ) ] q = 0 it follows that (cid:2) B r − , [[ F + Y , Z r ] q , F r +1 ] (cid:3) q = 0 . RAID GROUP ACTIONS FOR QUANTUM SYMMETRIC PAIRS OF TYPE AIII/AIV 37 On the other hand, using Equation (A.1) it follows that[ F + Y , E − X ] = (cid:2) [ F + Y , E − Y ] , E r +1 (cid:3) q − = 1 q − q − [ K − Y − K Y , E r +1 ] q − = q − K − Y E r +1 . This implies that (cid:2) [ F + Y , Z τ ( r ) ] q , F r +1 (cid:3) = − ( q − q − ) (cid:2) [ F + Y , E − X ] , F r +1 (cid:3) q − L r = − (1 − q − )[ K − Y E r +1 , F r +1 ] q − L r = q − L r ( K − r +1 − K r +1 ) K − Y . 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Watanabe, Classical weight modules over iQuantum groups , preprint, arXiv:1912.11157 (2019), 45 pp. Liam Dobson, School of Mathematics, Statistics and Physics, Newcastle University,Herschel Building, Newcastle upon Tyne NE1 7RU, UK E-mail address ::