# Quasifinite Representations of Classical Subalgebras of the Lie Superalgebra of Quantum Pseudodifferential Operators

Authors: José I. García; José I. Liberati
Journal: ISRN Mathematical Physics
 (2013)

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## Abstract

We classify the anti-involutions of the superalgebra of quantum pseudodifferential operators on the super circleS1|1 preserving the principal gradation, producing in this way a family of Lie subalgebras minus fixed by these anti-involutions. We classify the irreducible quasifinite highest weight representations of the central extension of these Lie subalgebras.


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## Body

## 1. Introduction

The𝒲-infinity algebras naturally arise in various physical theories, such as conformal field theory and the theory of quantum Hall effect (see [1, 2] and references therein). The W1+∞ algebra, which is the central extension of the Lie algebra 𝒟 of differential operators on the circle, is the most fundamental among these algebras.The difficulty in understanding the representation theory of a Lie algebra of this kind is that althoughW1+∞ admits a natural ℤ-gradation, each of the graded subspaces is still infinite dimensional in contrast to the more familiar cases such as the Virasoro algebra and Kac-Moody algebras. Therefore, the study of the highest weight modules which satisfy the quasifiniteness condition, that its graded subspaces have finite dimension, becomes a nontrivial problem. The systematic study of quasifinite highest weight modules of W1+∞ was initiated by Kac and Radul in [2] and further studied in [1, 3–5] and many others.By analyzing for which parabolic subalgebras ofW1+∞ the corresponding generalized Verma modules are quasifinite, Kac and Radul [2] gave a characterization of quasifinite highest weight W1+∞-modules in terms of certain generating function of highest weights and these modules where constructed in terms of irreducible highest weight representations of the Lie algebra of infinite matrices.The classification and construction of quasifinite modules for the matrix version (denoted byW1+∞N), super analog, q-analog, and super q-analog of W1+∞, were developed in [1, 2, 4, 6], respectively.The Lie algebraW1+∞N, recently studied in [4], correspond to the central extension of the algebra of matrix differential operators on the circle. The study of the representation theory of some interesting subalgebras of W1+∞N, and its q-analog and super version [1], is not complete.Another important example is the Lie algebraW∞ which is a particular case of a family of subalgebras W∞,p of W1+∞, where W∞,p (p∈ℂ[x]) is the central extension of the Lie algebra 𝒟p(t∂t) of differential operators on the circle that are a multiple of p(t∂t).This Lie algebra was studied by Kac and Liberati in [3]; observe that W∞=W∞,x. Following the ideas of Kac-Radul [2], in [3] they obtained the classification of the irreducible quasifinite highest weight modules over W∞,p. They also developed a general theory of quasifinite highest weight modules over ℤ-graded Lie algebras. These general results were extended to the super version in [6].A natural source of subalgebras comes from the subalgebras minus fixed by an anti-involution of the corresponding associative algebra, which preserve the gradation. In [7], Bloch finds an anti-involution of W∞ and he shows a relation between the representations of the corresponding subalgebra and certain values of the Riemann zeta function. In several and recent papers ([5, 8–10]) the authors obtained the classification of the anti-involutions of certain algebras and then they characterize the irreducible quasifinite highest weight modules of the corresponding subalgebras minus fixed by these anti-involutions, obtaining orthogonal and symplectic subalgebras of W1+∞, W∞,p, and so forth.The main goal of this work is to present aq-analog of Cheng-Wang [8] or a super-analog of Boyallian-Liberati [9]; namely, we classify the anti-involutions of the superalgebra q-𝒮𝒟 that preserve the gradation, where q-𝒮𝒟 is the superalgebra of regular pseudodifferential operators on the super circle S1|1. Then we present the classification of the irreducible quasifinite highest weight modules over the Lie subalgebras minus fixed by these anti-involutions.Table1 describes the map on the classification of irreducible quasifinite highest weight modules over W1+∞, the matrix version, super analog and q-analog, and for the subalgebras constructed from the anti-involutions, showing the place of the present results in this long-term program.Table 1
W
1
+
∞
W
∞
,
p
W
1
+
∞
N
superW1+∞
q-W1+∞
superq-W1+∞
Algebra
[2]
[3, 11]
[4]
[1]
[2]
[6]
Anti-involution
[5]
[10] (particular case in [12])
Partial results in [13] and others
[8]
[9]
Present workThe work is organized as follows. In Section2, we classify the anti-involutions of q-𝒮𝒟 that preserve the gradation, where q-𝒮𝒟 is the superalgebra of regular pseudodifferential operators on the super circle S1|1. In Section 3, we recall the general results on quasifinite modules over a graded Lie superalgebra and we present the classification of the irreducible quasifinite highest weight modules over the Lie subalgebras minus fixed by the anti-involutions obtained in Section 2.


## 2. Anti-Involution of𝒮𝔖qas Preserving Its Principal Gradation

Letq∈ℂ× with |q|≠1. Now, Tq denotes the following operator on ℂ[z-1,z]:
(1)Tq(f(z))=f(qz).
We denote by 𝔖qas the associative algebra of all pseudodifferential operators, that is, the operators on ℂ[z-1,z] of the form
(2)E=∑k∈ℤ‍ek(z)Tqk,whereei(z)∈ℂ[z-1,z](sumisfinite).
Any pseudodifferential operator can be written as linear combinations of elements of the form znf(Tq), where f∈ℂ[w-1,w] and n∈ℤ. The product in 𝔖qas is given by
(3)znf(Tq)·zmg(Tq)=zn+mf(qmTq)g(Tq).
Letting wtznf(Tq)=n, we define the principalℤ-gradation of 𝔖qas.Moreover, we denote byM(1|1) the set of 2×2 supermatrices with coefficients in ℂ, viewed as the associative superalgebra of linear transformations of the complex (1|1)-dimensional superspace ℂ(1|1). And we denote by Eij the 2×2 matrix with 1 in the ij-place and 0 everywhere else. Declaring E11, E22 even and E12, E21 odd elements, we endow M(1|1) with a ℤ2-gradation where |M| denotes the parity of the homogeneous element M∈M(1|1).We denote by𝒮𝔖qas the associative superalgebra of 2×2 supermatrices with entries in 𝔖qas, namely,
(4)𝒮𝔖qas=𝔖qas⨂M(1|1),
and the product is given by the usual matrix multiplication. Let 𝒮𝔖q denote the Lie superalgebra obtained from 𝒮𝔖qas by taking the usual bracket.Now, we introduce the linear mapStr0:𝒮𝔖q→ℂ as
(5)Str0(∑ij‍fij(Tq)Eij)=(f11(Tq))0-(f22(Tq))0,
where (f(Tq))0=f0 if f(Tq)=∑k∈ℤ‍fkTqk, and we define the 2-cocycle ψ in 𝒮𝔖q by
(6)ψ(znf(Tq)Eij,zmg(Tq)Ekl)={-(-1)i∑r=0n-1‍(f(q-n+rTq)g(qrTq))0δjkδil,ifn=-m>0,0,otherwise.
Then we denote by 𝒮𝔖^q the one-dimensional central extension of 𝒮𝔖q with central charge C corresponding to the 2-cocycle ψ, namely, 𝒮𝔖^q=𝒮𝔖q⨁ℂC, where the bracket is given by
(7)[znf(Tq)Eij,zmg(Tq)Ekl]=zn+m(f(qmTq)g(Tq)δjkEil-(-1)|Eij||Ekl|=zn+m×f(Tq)g(qnTq)δliEkj)+ψ(znf(Tq)Eij,zmg(Tq)Ekl)C.Lettingwt(znf(Tq)E12+zn+1g(Tq)E21)=n+1/2,  wtznf(Tq)Eii=n  and wtC=0, where i=1,2,  n∈ℤ, defines the principal(1/2)ℤ-gradation of 𝒮𝔖qas, 𝒮𝔖q and 𝒮𝔖^q. This equips 𝒮𝔖qas, 𝒮𝔖q, and 𝒮𝔖^q with (1/2)ℤ-gradations compatible with their ℤ2-gradation; thus,
(8)𝒮𝔖^q=(𝒮𝔖^q)0¯⨁(𝒮𝔖^q)1¯,(𝒮𝔖^q)0¯=⨁n∈ℤ(𝒮𝔖^q)n,(𝒮𝔖^q)1¯=⨁n∈ℤ(𝒮𝔖^q)n+1/2,
where
(9)(𝒮𝔖^q)n={znf1(Tq)E11+znf2(Tq)E22:znf1+fi∈ℂ[w-1,w],i=1,2znf1(Tq)E11+znf2(Tq)E22:fi∈ℂ[w-1,w]}+δn,0ℂC,(𝒮𝔖^q)n+1/2={znf1(Tq)E12+zn+1f2(Tq)E21:fi∈ℂ[w-1,w],znf1(Tq)E12+zn+1f2(Tq)E21:fi∈ℂi=1,2znf1(Tq)E12+zn+1f2(Tq)E21:fi∈ℂ[w-1,w]}.Ananti-involutionσ of 𝒮𝔖qas is an involutive anti-automorphism of 𝒮𝔖qas; that is, σ:𝒮𝔖qas→𝒮𝔖qas with σ2=Id, σ(bA+B)=bσ(A)+σ(B)  and  σ(AB)=(-1)|A||B|σ(B)σ(A), where A,B∈𝒮𝔖qas, and b∈ℂ.The main result of this section is the following theorem with the classification of all anti-involutions of𝒮𝔖qas that preserve the principal (1/2)ℤ-gradation.Theorem 1.
Any anti-involutionσ of 𝒮𝔖qas which preserves the principal gradation is one of the following (f∈ℂ[w-1,w], n∈ℤ): (a)
(10)σa,b,c,k(znf(Tq)E11)=anq(n-1)nkznf(bq-nTq-1)Tq2nkE11,σa,b,c,k(znf(Tq)E22)=anq(n-2)nkznf(bq-n+1Tq-1)Tq2nkE22,σa,b,c,k(znf(Tq)E12)=ancqn2kzn+1f(bq-nTq-1)Tq(2n+1)kE21,σa,b,c,k(znf(Tq)E21)=-anc-1q((n-1)2-n)kzn-1f(bq-n+1Tq-1)Tq(2n-1)kE12,
with a,b,c∈ℂ×, k∈ℤ such that -abk=qk;(b)
(11)σa,b,c,k,l(znf(Tq)E11)=anq(n(n-1)/2)kznf(bq-nTq-1)TqnkE22,σa,b,c,k,l(znf(Tq)E22)=anq(n(n-1)/2)k+nlznf(bq-nTq-1)TqnkE11,σa,b,c,k,l(znf(Tq)E12)=ancq(n(n-1)/2)k+nlznf(bq-nTq-1)Tqnk+lE12,σa,b,c,k,l(znf(Tq)E21)=-anc-1q(n(n-1)/2)kznf(bq-nTq-1)Tqnk-lE21,
with a,b,c∈ℂ×, k,l∈ℤ such that a2bk=qk-l and blc2=1.We divide the proof of Theorem1 into several results.Letσ be an anti-involution of 𝒮𝔖qas which preserves the principal gradation; then, σ defines linear maps σi,j:𝔖qas→𝔖qas preserving the principal gradation
(12)σ(znf(Tq)Eii)=σi,1(znf(Tq))E11+σi,2(znf(Tq))E22.Lemma 2.
Letσ be an anti-involution of 𝒮𝔖qas preserving the principal gradation. Then σ satisfies one of the following conditions: (a)
(13)σ(znf(Tq)E11)=σ1,1(znf(Tq))E11,σ(znf(Tq)E22)=σ2,2(znf(Tq))E22,σ(E12)=zf21(Tq)E21,σ(E21)=z-1g12(Tq)E12;(b)
(14)σ(znf(Tq)E11)=σ1,2(znf(Tq))E22,σ(znf(Tq)E22)=σ2,1(znf(Tq))E11,σ(E12)=f12(Tq)E12,σ(E21)=g21(Tq)E21. For some fij,gij∈ℂ[w-1,w], where 1⩽i, j⩽2 and i≠j.Proof.
Sinceσ preserves the principal gradation, we have that
(15)σ(E12)=f12(Tq)E12+zf21(Tq)E21,σ(E21)=z-1g12(Tq)E12+g21(Tq)E21,
for some fij,gij∈ℂ[w-1,w], with 1⩽i, j⩽2 and i≠j.
Then, since0=σ(znf(Tq)E11·zmg(Tq)E22)=σ(zmg(Tq)E22)σ(znf(Tq)E11), we have two possibilities
(16)σ(znf(Tq)E11)=σ1,1(znf(Tq))E11,σ(znf(Tq)E22)=σ2,2(znf(Tq))E22,
or
(17)σ(znf(Tq)E11)=σ1,2(znf(Tq))E22,σ(znf(Tq)E22)=σ2,1(znf(Tq))E11.
Moreover, since 0=σ(Eij2)=-σ(Eij)2, for 1⩽i,j⩽2, i≠j, from (15) we have that
(18)σ(E12)=f12(Tq)E12orσ(E12)=zf21(Tq)E21,σ(E21)=g21(Tq)E21orσ(E21)=z-1g12(Tq)E12.
Finally, since
(19)σ(E11)=-σ(E21)σ(E12),
the result follows from (16), (17), (18), and (19).Corollary 3.
Letσ be an anti-involution of 𝒮𝔖qas which preserves the principal gradation. Then one has the following. (a)
Ifσ satisfies Lemma 2(a), then σi,j(1)=δi,j, for all 1⩽i, j⩽2.(b)
Ifσ satisfies Lemma 2(b), then σi,j(1)=1-δi,j, for all 1⩽i, j⩽2.Proof.
It is clear thatσ(Id)=Id; then the result follows from Lemma 2.Proposition 4.
Letσ be an anti-involution of 𝒮𝔖qas which preserves the principal gradation and let one assume that it satisfies Lemma 2(a). Then σ is one of the σa,b,c,k from Theorem 1(a).In order to prove Proposition4 we will need the following result.Lemma 5.
Letσ be as in Proposition 4; then,
(20)σ(znf(Tq)Eii)=ainq(n(n-1)/2)kiznf(biq-nTq-1)TqnkiEii,σ(znf(Tq)E12)=a1ncq(n(n-1)/2)k1+nkzn+1f(b1q-nTq-1)Tqnk1+kE21,σ(znf(Tq)E21)=-a2c-1q(nk2/2-k)(n-1)zn-1f(b2q-nTq-1)Tqnk2-kE12,
where ai,bi,c∈ℂ×, ki,k∈ℤ such that ai2biki=qki, with i=1,2.Proof.
It is easy to check thatσi,i is a anti-involution of 𝔖qas with i=1,2; then, from Section 3 in [9] and hypothesis, we obtain that
(21)σ(znf(Tq)Eii)=ainq(n(n-1)/2)kiznf(biq-nTq-1)TqnkiEii,
with ki∈ℤ, ai,bi∈ℂ× such that ai2biki=qki, i=1,2. Besides, from Corollary 3(a) and hypothesis, we have that
(22)E11=-σ(E21)σ(E12)=-g12(qTq)f21(Tq)E11.
Therefore f21(Tq)=cTqk and g12(Tq)=-c-1qkTq-k with c∈ℂ×, k∈ℤ; then,
(23)σ(E12)=czTqkE21,σ(E21)=-c-1qkz-1Tq-kE12.
Moreover,
(24)σ(znf(Tq)Eij)=σ(Eij)σ(znf(Tq)Eii),
with i,j=1,2, i≠j. The proof follows from (21) and by replacing (21) and (23) in (24).Proof of Proposition4.
From Lemma5, we have that
(25)E12=σ2(E12)=cσ(E21)σ(zTqkE22)=-a2b2kq-kTqk2-2kE12,(26)E12=σ2(E12)=cσ(zTqkE11)σ(E21)=-a1b1kqk-k1Tqk1-2kE12.
Then from (25)-(26), we obtain that
(27)k1=k2=2k,
and also from (26)-(27) we obtain
(28)-a1b1k=qk.
Then, from (27) and again using Lemma 5, we have that
(29)σ(znTqlE12)=a1nb1lcq(nk-l)nzn+1Tq(2n+1)k-lE21,(30)σ(znTqlE12)=σ(znTqlE22)σ(E12)=a2nb2lcq(nk-l)(n+1)zn+1Tq(2n+1)k-lE21,
for all n,l∈ℤ. Comparing (29) with (30), we obtain that
(31)a1nb1l=a2nb2lqnk-l,∀n,l∈ℤ;
in particular,
(32)ifn=0,l=1,b2=b1q,ifn=1,l=0,a2=a1q-k.
From (28) and replacing (27) and (32) in (20), we obtain that σ=σa1,b1,c,k, finishing the proof.Proposition 6.
Letσ be an anti-involution of 𝒮𝔖qas which preserves the principal graduation, and let one assume that it satisfies Lemma 2(b). Then σ is one of the σa,b,c,k,l from Theorem 1(b).In order to prove Proposition6 we will need the following result.Lemma 7.
Letσ be as in Proposition 6, then
(33)σ(znf(Tq)E11)=a1nq(n(n-1)/2)k1znf(b1qnm1Tqm1)Tqnk1E22,σ(znf(Tq)E22)=a2nq(n(n-1)/2)k2znf(b2qnm2Tqm2)Tqnk2E11,σ(znf(Tq)E12)=a1ncq(n(n-1)/2)k1+nlznf(b1qnm1Tqm1)Tqnk1+lE12,σ(znf(Tq)E21)=-a2c-1q(n(n-1)/2)k2-nlznf(b2qnm2Tqm2)Tqnk2-lE21,
where ai,bi,c∈ℂ×, ki,mi,l∈ℤ, and i=1,2.Proof.
From hypothesis and Corollary3(b), we have that
(34)E11=σ(TqE22Tq-1E22)=σ2,1(Tq-1)σ2,1(Tq)E11,E22=σ(TqE11Tq-1E11)=σ1,2(Tq-1)σ1,2(Tq)E22,E11=σ(zE22z-1E22)=σ2,1(z-1)σ2,1(z)E11,E22=σ(zE11z-1E11)=σ1,2(z-1)σ1,2(z)E22,(35)E11=-σ(E12)σ(E21)=-f12(Tq)g21(Tq)E11.
Then, using (34), we have that
(36)σi,j(Tq±1)=bi±1Tq±mi,σi,j(z)=aizTqki,σi,j(z-1)=ai-1qkiz-1Tq-ki,
and by induction we obtain that
(37)σ(TqmEii)=bimTqmimEjj,∀m∈ℤ,σ(znEii)=ainq(n(n-1)/2)kiznTqnkiEjj,∀n∈ℤ,
where ai,bi∈ℂ×, ki,mi∈ℤ, i,j=1,2,  i≠j. Then, from (37) we have that
(38)σ(znf(Tq)Eii)=σ(f(Tq)Eii)σ(znEii)=ainq(n(n-1)/2)kiznf(biqnmiTqmi)TqnkiEjj,
with i,j=1,2, i≠j.
On the other hand, from (35) we have that f12(Tq)=cTql and g21(Tq)=-c-1Tq-l with c∈ℂ× and l∈ℤ; therefore,
(39)σ(E12)=cTqlE12,σ(E21)=-c-1Tq-lE21.
Moreover,
(40)σ(znf(Tq)Eij)=σ(Eij)σ(znf(Tq)Eii),
with i,j=1,2,i≠j. The proof follows from (38) and by replacing (38) and (39) in (40).Proof of Proposition6.
By Lemma7, we have that
(41)TqE11=σ2(TqE11)=b1b2m1Tqm1m2E11,zEii=σ2(zEii)=aiajbjkiqmjkizTqmjki+kjEii,E12=σ2(E12)=b1lc2E12,(42)σ(zE12)=a1cqlzTqk1+lE12,(43)σ(zE12)=σ(zE22)σ(E12)=a2czTqk1+lE12.
Then, using (41), we have that
(44)m1=m2=1orm1=m2=-1,(45)b1=b2-m1,(46)aiajbjkiqmjki=1,(47)mjki=-kj,(48)b1lc2=1,
where i,j=1,2, i≠j. Suppose that m1=m2=1; then, from (45) and (47) we have that b1=b2-1, k1=-k2, and by replacing them in (46), we obtain that q2k1=1, but since q is not a root of unity, we necessarily have that k1=k2=0. Then, using Lemma 7 we get
(49)σ(zTqE11)=a1b1qzTqE22;
moreover,
(50)σ(zTqE11)=q-1σ(zE11)σ(TqE11)=a1b1q-1zTqE22.
Comparing (49) with (50), we obtain that q2=1 which is a contradiction. Therefore, by (44) and the aforementioned result, we get that
(51)m1=m2=-1.
Then replacing (51) in (45) and (47) we obtain that
(52)b1=b2,k1=k2.
On the other hand, comparing (42) with (43), we have that
(53)a2=a1ql,
and by replacing (51), (52), and (53) in (46), we obtain that
(54)a12b1k1=qk1-l.
Using (48) and (54) and by replacing (51), (52), and (53) in (33), we obtain that σ=σa1,b1,c,k1,l, finishing the proof.Proof of Theorem1.
It is straightforward to check that the two cases are anti-involutions. Reciprocally, from Lemma2 and Propositions 4 and 6, it is clear that any anti-involution of 𝒮𝔖qas which preserves the principal gradation satisfies (a) or (b), finishing the proof.Now, given an anti-involutionσ of 𝒮𝔖qas, one can check that the set of points minus σ-fixed is a subalgebra of 𝒮𝔖q. Moreover, if σ preserves the principal (1/2)ℤ-graduation, this subalgebra inherits the (1/2)ℤ-graduation. These subalgebras are described in the last part of this section.We define the following automorphisms of𝒮𝔖qas by
(55)Θs(M)=(z-sId)·M·(zsId),∀M∈𝒮𝔖qas,Φs|(𝒮𝔖qas)0¯=id,Φs(znf(Tq)E12+zn+1g(Tq)E21)=sznf(Tq)E12+s-1zn+1g(Tq)E21,
with s∈ℂ×. On the other hand, given n,m∈ℤ, we denote
(56)ℂ[w-1,w](n,m):={f∈ℂ[w-1,w]:f(w)=-(-1)nf(w-1)wm}.Remark 8.
We see thatf(w)∈ℂ[w-1,w](n,m) with f(w)=∑‍fjwj if and only if fj=-(-1)nfm-j, for all j∈ℤ.The Case
σ
a
,
b
,
c
,
k
. Let a,b,c∈ℂ×, k∈ℤ, be such that -abk=qk. We denote by 𝒮𝔖q,ka,b,c the Lie subalgebra of 𝒮𝔖q consisting of -σa,b,c,k-fixed points; then, it inherits a (1/2)ℤ-gradation from 𝒮𝔖q; therefore, 𝒮𝔖q,ka,b,c=⨁j∈(1/2)ℤ(𝒮𝔖q,ka,b,c)j, where
(57)(𝒮𝔖q,ka,b,c)j={M∈(𝒮𝔖q)j:σa,b,c,k(M)=-M}.
Moreover, we denote 𝒮𝔖q,k:=𝒮𝔖q,k-qk,1,1 and σq,k:=σ-qk,1,1,k.The following lemma gives a description of𝒮𝔖q,ma,b,c.Lemma 9.
Leta, b, c, k be as aforementioned; then, 𝒮𝔖q,ka,b,c≃𝒮𝔖q,k and
(58)(𝒮𝔖q,k)n={f,g∈ℂ[w-1,w](n,2nk)znf(qn/2Tq)E11+zng(q(n-1)/2Tq)E22:f,g∈ℂ[w-1,w](n,2nk)},(𝒮𝔖q,k)n+1/2={znf(Tq)E12-(-1)nq(n+1)nkzn+1f(q-nTq-1)Tq(2n+1)kE21:(Tq)f∈ℂ[w-1,w]},
for all n∈ℤ.Proof.
First, it is easy to check that(59)Θ-sσa,b,c,kΘs=σaq-2ks,bq2s,cq-ks,k,Φ-tσa,b,c,kΦt=σa,b,ct2,k.
Then if we take s,t∈ℂ× such that b-1=q2s and t2=c-1qks, from (59) and the relation between a,b,c,k, we obtain that 𝒮𝔖q,ka,b,c≃𝒮𝔖q,k.
On the other hand, by Theorem1(a) and linearity of σq,k we obtain that (for n∈ℤ) znf(Tq)E11+zng(Tq)E22∈(𝒮𝔖q,k)n if and only if znf(Tq)E11, zng(Tq)E22∈(𝒮𝔖q,k)n. Now, let znh(Tq)E11∈(𝒮𝔖q,k)n be with h(w)=∑‍hjwj; then,
(60)-znh(Tq)E11=σq,k(znh(Tq)E11)=(-1)nqn2kznh(q-nTq-1)Tq2nkE11
if and only if q-(n/2)jhj=-(-1)nq(-n/2)(2nk-j)h2nk-j, for all j∈ℤ, which is equivalent to f(w)=h(q-n/2w)∈ℂ[w-1,w](n,2nk) (see Remark 8). Therefore,
(61)znh(Tq)E11=znf(qn/2Tq)E11,withf(w)∈ℂ[w-1,w](n,2nk).
Similarly, we prove that znh(Tq)E22∈(𝒮𝔖q,k)n if and only if
(62)znh(Tq)E22=zng(q(n-1)/2Tq)E22,whereg(w)∈ℂ[w-1,w](n,2nk).
Now, we suppose that znf(Tq)E12+zn+1g(Tq)E21∈(𝒮𝔖q,k)n+1/2; then,
(63)-znf(Tq)E12-zn+1g(Tq)E21=σq,k(znf(Tq)E12+zn+1g(Tq)E21)=(-1)nqn2k(zng(q-nTq-1)Tq(2n+1)kE12=(-1)nqn2kqqq+qnkzn+1f(q-nTq-1)Tq(2n+1)kE21),
if and only if g(Tq)=-(-1)nq(n+1)nkf(q-nTq-1)Tq(2n+1)k, finishing the proof.The Case
σ
a
,
b
,
c
,
k
,
l
. Let a,b,c∈ℂ×, k,l∈ℤ, be such that a2bk=qk-l and blc2=1. We denote by 𝒮𝔖q,k,la,b,c the Lie subalgebra of 𝒮𝔖q consisting of -σa,b,c,k,l-fixed; then, it inherits a (1/2)ℤ-gradation from 𝒮𝔖q; thus 𝒮𝔖q,k,la,b,c=⨁j∈(1/2)ℤ(𝒮𝔖q,k,la,b,c)j, where
(64)(𝒮𝔖q,k,la,b,c)j={M∈(𝒮𝔖q)j:σa,b,c,k,l(M)=-M}.
Moreover, we denote 𝒮𝔖q,k,l+±:=𝒮𝔖q,k,la,1,±1, σq,k,l+±:=σa,1,±1,k,l and 𝒮𝔖q,k,l-,±:=𝒮𝔖q,k,l-a,1,±1, σq,k,l-,±:=σ-a,1,±1,k,l, with a=q(k-l)/2. The following lemma gives a description of 𝒮𝔖q,k,la,b,c. We will need the following notation:
(65)δn:={0,in(𝒮𝔖q,k,l+,+)n+1/2;1,in(𝒮𝔖q,k,l+,-)n+1/2;n,in(𝒮𝔖q,k,l-,+)n+1/2;n+1,in(𝒮𝔖q,k,l-,-)n+1/2.Lemma 10.
Leta, b, c, k, l be as aforementioned; then, 𝒮𝔖q,k,la,b,c is isomorphic to some of the following algebras: 𝒮𝔖q,k,l+,+, 𝒮𝔖q,k,l+,-, 𝒮𝔖q,k,l-,+, or 𝒮𝔖q,k,l-,-. And
(66)(𝒮𝔖q,k,l+±)n={[w-1,w]znf(Tq)E11-(±1)nqn(nk-l)/2×znf(q-nTq-1)TqnkE22:f∈ℂ[w-1,w]},(𝒮𝔖q,k,l+±)n+1/2={[w-1,w](δn+1,(n+1)k-l)znf(qn/2Tq)E12+zn+1g(q(n+1)/2Tq)×E21:f∈ℂ[w-1,w](δn,nk+l),g∈ℂ[w-1,w](δn+1,(n+1)k-l)},(𝒮𝔖q,k,l-±)n={[w-1,w]znf(Tq)E11-(±1)nqn(nk-l)/2×znf(q-nTq-1)TqnkE22:f∈ℂ[w-1,w]},(𝒮𝔖q,k,l-±)n+1/2={[w-1,w](δn,(n+1)k-l)znf(qn/2Tq)E12+zn+1g(q(n+1)/2Tq)kkk×E21:f∈ℂ[w-1,w](δn,nk+l),∈ℂ[w-1,w](δn,(n+1)k-l)},
for all n∈ℤ.Proof.
It is easy to check that,(67)Θ-sσa,b,c,k,lΘs=σaq-sk,bq2s,cq-sl,k,l;
then; if we take s such that q2s=b-1, we obtain the first assertion using (67) and the relations between a,b,c,k,l.
On the other hand, we suppose thatznf(Tq)E11+zng(Tq)E22∈(𝒮𝔖q,k,l+±)n; then,
(68)-znf(Tq)E11-zng(Tq)E22=σq,k,l+±(znf(Tq)E11+zng(Tq)E22)=(±1)nqn(nk-l)/2znf(q-nTq-1)TqnkE22+(±1)nqn(nk+l)/2zng(q-nTq-1)TqnkE11
if and only if g(Tq)=-(±1)nqn(nk-l)/2f(q-nTq-1)Tqnk. The proof for (𝒮𝔖q,k,l+±)n is similar.
Now, by Theorem1(b) and linearity of σq,k,l++, we obtain that znf(Tq)E12+zn+1g(Tq)E21∈(𝒮𝔖q,k,l++)n+1/2 if and only if znf(Tq)E12 and zng(Tq)E21 are elements in (𝒮𝔖q,k,l++)n+1/2. We suppose that zn+1h(Tq)E21∈(𝒮𝔖q,k,l++)n+1/2, with h(w)=∑‍hjwj; then,
(69)-zn+1h(Tq)E21=σk,l++(zn+1h(Tq)E21)=-q(n+1)((n+1)k-l)/2zn+1h(q-(n+1)Tq-1)Tq(n+1)k-lE21
if and only if  hjq-((n+1)/2)j=q-((n+1)/2)((n+1)k-l-j)h(n+1)k-l-j, for all j∈ℤ, which is equivalent to g(w)=h(q-((n+1)/2)w)∈ℂ[w-1,w](1,(n+1)k-l) (see Remark 8). Therefore,
(70)zn+1h(Tq)E21=zn+1g(q(n+1)/2Tq)E21,withg(w)∈ℂ[w-1,w](1,(n+1)k-l).
Similarly, we prove that znh(Tq)E12∈(𝒮𝔖q,k,l++)n+1/2 if and only if znh(Tq)E12=znf(qn/2Tq)E12, with f∈ℂ[w-1,w](0,nk+l).
The other proofs are similar.Remark 11.
From Lemma9 (resp., Lemma 10), it is clear that an element in (𝒮𝔖q,k)n+1/2 (resp., (𝒮𝔖q,k,l±,±)n) is totally determined by its value in the position E12 (resp., E11), where n∈ℤ.Finally, we denote by𝒮𝔖^q,k and 𝒮𝔖^q,k,l±± the central extensions of 𝒮𝔖q,k and 𝒮𝔖q,k,l±± corresponding to the restrictions of the 2-cocycle ψ, respectively. It is clear that these subalgebras admit a (1/2)ℤ-graduation compatible with their ℤ2-graduation; that is,
(71)𝒮𝔖^q,k=(𝒮𝔖^q,k)0¯⨁(𝒮𝔖^q,k)1¯,𝒮𝔖^q,k,l±±=(𝒮𝔖^q,k,l±±)0¯⨁(𝒮𝔖^q,k,l±±)1¯,
where
(72)(𝒮𝔖^q,k)0¯=⨁n∈ℤ(𝒮𝔖^q,k)n,(𝒮𝔖^q,k)1¯=⨁n∈ℤ(𝒮𝔖^q,k)n+1/2,(𝒮𝔖^q,k,l±±)0¯=⨁n∈ℤ(𝒮𝔖^q,k,l±±)n,(𝒮𝔖^q,k,l±±)1¯=⨁n∈ℤ(𝒮𝔖^q,k,l±±)n+1/2,
with (𝒮𝔖^q,k)n=(𝒮𝔖q,k)n+δn,0ℂC and (𝒮𝔖^q,k,l±±)n=(𝒮𝔖q,k,l±±)n+δn,0ℂC, for all n∈(1/2)ℤ.


## 3. Quasifinite Highest Weight Modules over𝒮𝔖^q,k and 𝒮𝔖^q,k,l±±

The goal of this section is to characterize the quasifinite irreducible highest weight modules over𝒮𝔖^q,k and 𝒮𝔖^q,k,l±±; for this we will apply the general results on quasifinite representations of (1/2)ℤ-graded Lie superalgebras developed in Section 2 in [6]. Let us recall some general definition and results from [6].In this section,𝔤 denote a consistent (1/2)ℤ-graded Lie superalgebra over ℂ; namely,
(73)𝔤=⨁j∈(1/2)ℤgj,where[𝔤i,𝔤j]⊂𝔤i+j∀i,j∈12ℤ,
and also
(74)𝔤0¯=⨁j∈ℤ𝔤j,𝔤1¯=⨁j∈ℤ𝔤j+1/2.
We denote 𝔤±=⨁j∈(1/2)ℤ+𝔤±j. A subalgebra 𝔭 of 𝔤 is called parabolic if
(75)𝔭=⨁j∈(1/2)ℤ𝔭j,wherepj=gj∀j⩾0,𝔭=⨁j∈(1/2)ℤ𝔭j𝔭-j≠0,forsomej∈ℕ.
We assume the following properties on 𝔤:(SP1)
𝔤0 is commutative,(SP2)
ifa∈𝔤-j (j∈(1/2)ℕ) and [a,𝔤1/2]=0, then a=0.Given that a∈𝔤-1/2 is nonzero, we define 𝔭a=⨁j∈(1/2)ℤ𝔭ja, where
(76)𝔭ja=𝔤j∀j≥0,𝔭-1/2a=∑‍[…[[a,𝔤0],𝔤0],…],𝔭-k-1/2a=[𝔭-1/2a,𝔭-ka].
It was proved in [6] that 𝔭a is the minimal parabolic subalgebra containing a and also that [𝔭a,𝔭a]∩𝔤0=[a,𝔤1/2].Definition 12.
(a) A parabolic subalgebra𝔭 is called nondegenerate if 𝔭-j has finite codimension in 𝔤-j, for all j∈(1/2)ℕ.
(b) An elementa∈𝔤-1/2 is called nondegenerate if 𝔭a is nondegenerate.We will also require the following condition on𝔤.(SP3)
If𝔭 is a nondegenerate parabolic subalgebra of 𝔤, then there exists a nondegenerate element a∈𝔭-1/2.A𝔤-module V=⨁j∈(1/2)ℤVj is called quasifinite if dimVj<∞ for all j. Given λ∈𝔤0*, a highest weight module is a (1/2)ℤ-graded 𝔤-module V(𝔤,λ)=⨁j∈(1/2)ℤ+V-j, defined by the following properties: (a)
V0=ℂvλ, where vλ is a nonzero vector;(b)
hvλ=λ(h)vλ, for all h∈𝔤0;(c)
𝔤+vλ=0;(d)
𝒰(𝔤-)vλ=V(𝔤,λ).A nonzero vectorv∈V(𝔤,λ) is called singular if 𝔤+v=0.TheVerma module over 𝔤 is defined as usual:
(77)M(𝔤,λ)=𝒰(𝔤)⨂𝒰(𝔤0⨁𝔤+)ℂλ,
where ℂλ:=ℂcλ is the one-dimensional (𝔤0⨁𝔤+)-module given by hcλ=λ(h)cλ if h∈𝔤0, and 𝔤+cλ=0 and the action of 𝔤 on M(𝔤,λ) is induced by the left multiplication in 𝒰(𝔤). Any highest weight module V(𝔤,λ) is a quotient module of M(𝔤,λ). The irreducible module L(𝔤,λ) is the quotient of M(𝔤,λ) by the maximal proper graded submodule.Now, let𝔭=⨁j∈(1/2)ℤ𝔭j be a parabolic subalgebra of 𝔤 and let λ∈𝔤0* be such that λ|[𝔭,𝔭]∩𝔤0=0. Then the (𝔤0⨁𝔤+)-module ℂλ extends to a 𝔭-module by letting 𝔭jcλ=0 for all j<0, and we may construct the highest weight module
(78)M(𝔭,𝔤,λ)=𝒰(𝔤)⨂𝒰(𝔭)ℂλ,
called the generalized Verma module. Clearly all these highest weight modules are graded. The following result gives the characterization of all irreducible quasifinite highest-weight modules.Theorem 13.
Let𝔤=⨁j∈(1/2)ℤ𝔤j be a consistent (1/2)ℤ-graded Lie superalgebra over ℂ that satisfies conditions (SP1), (SP2), and (SP3). The following conditions on λ∈𝔤0* are equivalent.(a)
M(𝔤,λ) contains a singular vector a·vλ in M(𝔤,λ)-1/2, where a is nondegenerate.(b)
There exists a nondegenerate elementa∈𝔤-1/2, such that λ([𝔤1/2,a])=0.(c)
L(𝔤,λ) is quasifinite.(d)
There exists a nondegenerate elementa∈𝔤-1/2, such that L(𝔤,λ) is the irreducible quotient of the generalized Verma module M(𝔤,𝔭a,λ).Proof.
See [6].Moreover, we will need the following result. Recall that a quasipolynomial is a combination of functions of the formp(x)qαx, where p(x) is a polynomial and α∈ℂ. That is, it satisfies a nontrivial linear differential equation with constant coefficients.Proposition 14.
Given a quasipolynomialP and a polynomial B(x)=∏i‍(x-Ai), take b(x)=∏i‍(x-ai) where ai=exp(Ai); then, b(x)(∑n∈ℤ‍P(n)x-n)=0 if and only if B(d/dx)P(x)=0.Below, we prove that𝒮𝔖^q,k and 𝒮𝔖^q,k,l±± satisfy the properties (SP1), (SP2), and (SP3), which is equivalent to study of its parabolic subalgebras. Then using Theorem 13 and Proposition 14, we obtain two equivalent characterizations of the quasifinite highest weight modules of these algebras. Before studying each particular case, we will consider the following useful result.Lemma 15.
LetA=znf1(Tq)E11+znf2(Tq)E22 and zmg(Tq)Eij be nonzero elements in 𝒮𝔖^q, where n≠0,  i,j=1,2 and let h∈ℂ[w-1,w] be nonconstant. Then one has the following.(a)
[A,zmg(Tq)Eij]≠0 or [A,zmg(Tq)h(Tq)Eij]≠0, with i≠j.(b)
Iffi≠0, then [A,zmg(Tq)Eii]≠0 or [A,zmg(Tq)h(Tq)Eii]≠0.Proof.
(a) We suppose that[A,zmg(Tq)E12]=0 and [A,zmg(Tq)h(Tq)E12]=0; then,
(79)f1(qmTq)g(Tq)=f2(Tq)g(qnTq),(80)f1(qmTq)g(Tq)h(Tq)=f2(Tq)g(qnTq)h(qnTq).
Using (79) and the hypothesis, we have that f1 and f2 are nonzero. Then, if we replace (79) with (80), we obtain h(Tq)=h(qnTq) which is a contradiction since h is not constant, n≠0, and q is not unity root. The other case is similar.
The proof of (b) is similar to the proof of (a).


### 3.1. The Case𝔤=𝒮𝔖^q,k

It is clear that𝒮𝔖^q,k satisfies (SP1). Now, we suppose that A=z-nf(Tq)E11+z-ng(Tq)E22∈(𝒮𝔖^q,k)-n (with n∈ℕ) satisfies [A,(𝒮𝔖^q,k)1/2]=0; in particular for all i∈ℤ(81)[A,TqiE12-zTqk-iE21]=0.
Then using (81), we obtain that f(Tq)=q-nig(Tq) for all i∈ℤ. Hence, f=g=0 (since n≠0 and q is not a root of unity); therefore, A=0. Now, we suppose that A=z-nf(Tq)E12-(-1)-nq(n-1)nkz-n+1f(qnTq-1)Tq(-2n+1)kE21 in (𝒮𝔖^q,k)-n+1/2; (with n∈ℕ) satisfies [A,(𝒮𝔖q,k)1/2]=0 then, for all i∈ℤ we have that
(82)[A,TqiE12-zTqk-iE21]=0,
which is equivalent to f(Tq)=qnk-if(qTq) for all i∈ℤ. Hence f=0; therefore, A=0. Thus, we prove that 𝒮𝔖^q,k satisfies the property (SP2).Finally, using the following lemma, we will prove that𝒮𝔖^q,k satisfies the property (SP3).Lemma 16.
Let𝔭=⨁j∈(1/2)ℤ𝔭j be a (1/2)ℤ-graded subalgebra of 𝒮𝔖^q,k, where 𝔭0=(𝒮𝔖^q,k)0. Then one has the following. (a)
For eachn∈ℤ, 𝔭n+1/2 has finite codimension in (𝒮𝔖^q,k)n+1/2 if and only if 𝔭n+1/2≠0.(b)
For eachn∈ℤ, 𝔭n has finite codimension in (𝒮𝔖^q,k)n if and only if there exists znf(qn/2Tq)E11+zng(q(n-1)/2Tq)E22∈𝔭n, such that f and g are nonzero.(c)
𝔭-n has finite codimension in (𝒮𝔖^q,k)-n, for all n∈(1/2)ℕ if and only if 𝔭-1/2≠0.Proof.
(a) We suppose that there exists(83)A=znf(Tq)E12-(-1)nq(n+1)nkzn+1×f(q-nTq-1)Tq(2n+1)kE21∈𝔭n+1/2,
with f≠0; then M1j:=[A,(Tqj-Tq-j)E11] and M2j:=[A,(q-j/2Tqj-qj/2Tq-j)E22] belong to 𝔭n+1/2, for all j∈ℤ. Moreover, Aj=(q-(n+1)j-qnj)-1(M1j+q-(n+1/2)jM2j)∈𝔭n+1/2 and
(84)Aj=znf(Tq)TqjE12-(-1)nq(n+1)nk-njzn+1×f(q-nTq-1)Tq(2n+1)k-jE21,
for all j∈ℤ×. Therefore, using (84), we have that
(85)zng(Tq)E12-(-1)nq(n+1)nkzn+1×g(q-nTq-1)Tq(2n+1)kE21∈𝔭n+1/2,
for all g(w)∈〈f(w)〉, where 〈f(w)〉 is the ideal of ℂ[w-1,w] generated by f(w). Using that 〈f(w)〉 has finite codimension in ℂ[w-1,w], we obtain that 𝔭n+1/2 has finite codimension in (𝒮𝔖^q,k)n+1/2 (see Remark 11). The proof of the converse is trivial.
(b) We suppose thatA=znf(qn/2Tq)E11+zng(q(n-1)/2Tq)E22∈𝔭n, with nonzero f and g in ℂ[w-1,w](n,2nk); then similar to the aforementioned argument
(86)(q-nj/2-qnj/2)-1[A,(Tqj-Tq-j)E11]=znf(qn/2Tq)(qnj/2Tqj+q-nj/2Tq-j)E11∈𝔭n,b[A,(q-j/2Tqj-qj/2Tq-j)E22]=zng(q(n-1)/2Tq)×(q(n-1)j/2Tqj+q-(n-1)j/2Tq-j)E22∈𝔭n,
where b=(q(-n/2)j-q(n/2)j)-1, and this is true for all j∈ℤ×. Therefore
(87)znh1(qn/2Tq)E11∈𝔭n,∀h1∈f(w)ℂ[w-1,w](1,0),znh2(q(n-1)/2Tq)E22∈𝔭n,∀h2∈g(w)ℂ[w-1,w](1,0).
Since f(w)ℂ[w-1,w](1,0) and g(w)ℂ[w-1,w](1,0) have finite codimension in ℂ[w-1,w](n,2nk), we obtain that 𝔭n has finite codimension in (𝒮𝔖^q,k)n. The proof of the converse is trivial.
(c) We suppose that𝔭-1/2≠0; then, in order to prove that 𝔭-n has finite codimension in (𝒮𝔖^q,k)-n, for all n∈(1/2)ℕ, we only need to see that this is true for all n∈ℕ, since by using (SP2), we obtain that 𝔭-n+1/2≠0 for all n∈ℕ; then, from (a) we have that 𝔭-n+1/2 has finite codimension in (𝒮𝔖^q,k)-n+1/2 for all n∈ℕ. By hypothesis, there exists A=z-1f(Tq)E12+f(qTq-1)Tq-kE21∈𝔭-1/2 with f≠0; then, using (84), we have that B=z-1f(Tq)TqE12+qf(qTq-1)Tq-k-1E21∈𝔭-1/2, and
(88)[A,B]:=z-1f1(q-1/2Tq)E11+z-1g1(q-1Tq)E22∈𝔭-1,
where f1 and g1 are nonzero. Hence, by (b), 𝔭-1 has finite codimension in (𝒮𝔖^q,k)-1. Moreover, by (87) we obtain that
(89)z-1f1(q-1/2Tq)h(q-1/2Tq)E11∈𝔭-1,(90)z-1g1(q-1Tq)h(q-1Tq)E22∈𝔭-1,
for all h(w)∈ℂ[w-1,w](1,0). Now, by induction we suppose that 𝔭-n has finite codimension in (𝒮𝔖^q,k)-n; then, there exists A=z-nf(q-n/2Tq)E11+z-ng(q-(n+1)/2Tq)E22∈𝔭-n where f and g are nonzero, and by (89) and Lemma 15(b),
(91)[A,z-1f1(q-1/2Tq)E11]≠0or[A,z-1f1(q-1/2Tq)h(q-1/2Tq)E11]≠0,
for some no constant h(w) in ℂ[w-1,w](1,0); therefore, there exists a nonzero element z-n-1f~(Tq)E11 in 𝔭-n-1. Similarly, using (90) and Lemma 15(b), we see that there exists a nonzero element z-n-1g~(Tq)E22 in 𝔭-n-1; then, using (b), 𝔭-n-1 has finite codimension in (𝒮𝔖^q,k)-n-1, finishing the induction. The proof of the converse is trivial.Corollary 17.
(a) Any parabolic subalgebra of𝒮𝔖^q,k is nondegenerate.
(b) Any nonzero element of(𝒮𝔖^q,k)-1/2 is nondegenerate.
(c)𝒮𝔖^q,k satisfies (SP3).Proof.
Let𝔭 be a parabolic subalgebra of 𝒮𝔖^q,k; by definition there exists j∈(1/2)ℕ such that 𝔭-j≠0, and then by (SP2) 𝔭-1/2≠0; the proof of (a) follows from Lemma 16(c). Finally, (b) follows from (a), and (c) follows from (b).A functionalλ∈(𝒮𝔖^q,k)0* is described by its labels Δn,1=-λ((Tqn-Tq-n)E11), Δn,2=-λ((q-nTqn-Tq-n)E22) with n∈ℤ, and the central charge λ(C)=c. We will consider the generating series
(92)Δλ,i(x)=∑n∈ℤ‍Δn,ix-n,withi=1,2.Theorem 18.
An irreducible highest weight𝒮𝔖^q,k-module L(𝒮𝔖^q,k,λ) is quasifinite if and only if one of the following equivalent conditions holds. (a)
There exists a Laurent polynomialb(x) such that
(93)b(qx)(Δλ,1(x)+Δλ,2(x)+c)=0.(b)
There exists a quasipolynomialP(x) such that
(94)P(n)=Δn,1+Δn,2+δn,0c,
for all n∈ℤ.Proof.
By Theorem13, L(𝒮𝔖^q,k,λ) is quasifinite if and only if there exist a nondegenerate element a in (𝒮𝔖^q,k)-1/2, such that λ([(𝒮𝔖^q,k)1/2,a])=0.
Now, leta=z-1b(Tq)E12+b(qTq-1)Tq-kE21 be a nonzero element of (𝒮𝔖^q,k)-1/2 (with b(w)=∑j‍bjwj); then, by Corollary 17(b), a is a nondegenerate element. Then, λ([(𝒮𝔖^q,k)1/2,a])=0 if and only if, for all i∈ℤ,
(95)0=λ([(TqiE12-zTqk-iE21,a)])=∑j‍qjbj(Δk-i+j,1+Δk-i+j,2)+q-k+ib-k+ic.
Multiplying (95) by x-k+i and adding over i∈ℤ, we obtain that
(96)0=∑j,i‍qjbj(Δk-i+j,1+Δk-i+j,2)x-k+i+∑i‍q-k+ib-k+ix-k+ic=b(qx)(Δλ,1(x)+Δλ,2(x)+c).
The equivalence between (a) and (b) follows from Proposition 14.


### 3.2. The Case𝔤=𝒮𝔖^q,k,l±±

It is clear that𝒮𝔖^q,k,l±± satisfies (SP1). Now, we prove that 𝒮𝔖^q,k,l±± satisfies (SP2). We suppose that A=z-nf(q-n/2Tq)E12+z-n+1g(q(-n+1)/2Tq)E21 in (𝒮𝔖^q,k,l±±)-n+1/2 (with n∈ℕ) satisfies [A,(𝒮𝔖^q,k,l±±)1/2]=0; in particular if we take nonzero elements h12(Tq)E12 and zh21(q1/2Tq)E21 in (𝒮𝔖^q,k,l±±)1/2, we have that
(97)[A,h12(Tq)E12]=0;
that is, g(q(-n+1)/2Tq)h12(Tq)=0; therefore, g(w)=0. Similarly,
(98)[A,zh21(q1/2Tq)E21]=0,
which is equivalent to f(q-n/2+1Tq)h21(q1/2Tq)=0; therefore, f(w)=0, and then A=0.Remark 19.
GivenB=znf(qn/2Tq)E12+zn+1g(q(n+1)/2Tq)E21∈(𝒮𝔖^q,k,l±±)n+1/2 and A=h(Tq)E11-h(Tq-1)E22∈(𝒮𝔖^q,k,l±±)0 with h∈ℂ[w-1,w], then M:=[A,B]∈(𝒮𝔖^q,k,l±±)n+1/2, where Mii=0(i=1,2) and
(99)M12=-znf(qn/2Tq)(h(qnTq)+h(Tq-1)),(100)M21=zn+1g(q(n+1)/2Tq)(h(Tq)+h(q-n-1Tq-1)).Now, letA=z-nf1(Tq)E11+z-nf2(Tq)E22∈(𝒮𝔖^q,k,l±±)-n (with n∈ℕ) be such that A is nonzero; then, if we take nonzero B=f(Tq)E12 in (𝒮𝔖^q,k,l±±)1/2 by Remark 19, there exists f(Tq)h(Tq)E12∈(𝒮𝔖^q,k,l±±)1/2 such that h is not constant; then by Lemma 15(a), we have that
(101)[A,f(Tq)E12]≠0or[A,f(Tq)h(Tq)E12]≠0.
Therefore 𝒮𝔖^q,k,l±± satisfies (SP2).In order to prove that𝒮𝔖^q,k,l±± satisfies (SP3), we will need the following result. We denote
(102)δn′:={δn+1,in(𝒮𝔖q,k,l+,±)n+1/2,δn,in(𝒮𝔖q,k,l-,±)n+1/2.Lemma 20.
Let𝔭=⨁j∈(1/2)ℤ𝔭j be a (1/2)ℤ-graded subalgebra of 𝒮𝔖^q,k,l±±, with 𝔭0=(𝒮𝔖^q,k,l±±)0. Then one has the following: (a)
for eachn∈ℤ,  𝔭n has finite codimension in (𝒮𝔖^q,k,l±±)n if and only if 𝔭n≠0;(b)
for eachn∈ℤ,  𝔭n+1/2  has finite codimension in  (𝒮𝔖^q,k,l±±)n+1/2  if and only if  there exists znf(qnTq)E12+zn+1g(q(n+1)/2Tq)E21∈𝔭n+1/2, such that f and g are nonzero;(c)
𝔭-n has finite codimension in (𝒮𝔖^q,k,l±±)-n for all n∈(1/2)ℕ, if and only if 𝔭-1/2 has finite codimension in (𝒮𝔖^q,k,l±±)-1/2.Proof.
(a) LetA=znf(Tq)E11-(±1)nqn(nk-l)/2znf(q-nTq-1)TqnkE22∈𝔭n, with f≠0; then, Bj:=(1-qnj)-1[A,TqjE11-Tq-jE22]∈𝔭n and
(103)Bj=znf(Tq)TqjE11-(±1)nq(n(nk-l)/2)-nj×znf(q-nTq-1)Tqnk-jE22,
for all j∈ℤ×; therefore, by (103) we have that
(104)zng(Tq)E11-(±1)nqn(nk-l)/2zng(q-nTq-1)TqnkE22∈𝔭n,
for all g(w)∈〈f(w)〉, and since 〈f(w)〉 has finite codimension in ℂ[w-1,w], we have that 𝔭n has finite codimension in (𝒮𝔖^q,k,l±±)n (see Remark 11). The proof of the converse is trivial.
(b) We suppose that there exist nonzero elements(105)znf1(Tq)E12,zn+1f2(Tq)E21in𝔭n+1/2;
then, using Remark 19, we obtain that zng1(Tq)E12∈𝔭n+1/2 for all g1(q-n/2w)∈f1(w)ℂ[w-1,w](1,0) (see (99)) and zn+1g2(Tq)E21∈𝔭n+1/2 for all g2(q(-n+1)/2w)∈f2(w)ℂ[w-1,w](1,0) (see (100)). Then using that f1(w)ℂ[w-1,w](1,0) has finite codimension in  ℂ[w-1,w](δn,nk+l)  and  f2(w)ℂ[w-1,w](1,0) has finite codimension in ℂ[w-1,w](δn′,(n+1)k-l), we obtain that 𝔭n+1/2 has finite codimension in (𝒮𝔖^q,k,l±±)n+1/2. Therefore in order to prove that 𝔭n+1/2 has finite codimension in (𝒮𝔖^q,k,l±±)n+1/2, we only need to see that there exist nonzero elements as aforementioned in 𝔭n+1/2. Let B=znf(qn/2Tq)E12+zn+1g(q(n+1)/2Tq)E21∈𝔭n+1/2 with nonzero f and g, and let A and M be as in Remark 19. Then, taking h(w)=w-qnw-1 (observe that h(Tq-1)=-h(qnTq)), we obtain that
(106)M=zn+1g(q(n+1)/2Tq)×(h(Tq)+h(q-(n+1)Tq-1))×E21∈𝔭n+1/2,withM≠0.
Similarly, taking h(w)=w-q-n-1w-1 (observe that h(Tq)=-h(q-n-1Tq-1)), we obtain
(107)M=-znf(qn/2Tq)×(h(qnTq)+h(Tq-1))E12∈𝔭n+1/2,withM≠0,
proving the existence of nonzero elements of those forms. The proof of the converse is trivial.
(c) We suppose that𝔭-1/2 has finite codimension in (𝒮𝔖^q,k,l±±)-1/2; then, in order to prove that 𝔭-n has finite codimension in (𝒮𝔖^q,k,l±±)-n for all n∈(1/2)ℕ, we only need to see that this is true for all -n+1/2 with n∈ℕ, since in this case by using (SP2), we obtain that 𝔭-n≠0 for all n∈ℕ; then, by (a), we have that 𝔭-n has finite codimension in (𝒮𝔖^q,k,l±±)-n, for all n∈ℕ.
By hypothesis, there exist nonzero elementsA=z-1f(q-1Tq)E12 and B=g(Tq)E21 in 𝔭-1/2; therefore, [A,B]∈𝔭-1 is nonzero. By induction, we suppose that 𝔭-n+1/2 has finite codimension in (𝒮𝔖^q,k,l±±)-n+1/2; then, there exist nonzero elements z-nf12(qnTq)E12 and z-n+1f21(q(-n+1)/2Tq)E21 in (𝒮𝔖^q,k,l±±)-n+1/2, and by Remark 19, we also have z-nf12(qnTq)g12(Tq)E12, z-n+1f21(q(-n+1)/2Tq)g21(Tq)E21 in (𝒮𝔖^q,k,l±±)-n+1/2 with no constant gij for all i≠j. Then by Lemma 15(a), we have that
(108)[z-nf12(qnTq)E12,[A,B]]≠0or[z-nf12(qnTq)g12(Tq)E12,[A,B]]≠0.
Therefore from (108), there exists a nonzero element z-n-1f~12(Tq)E12 in (𝒮𝔖^q,k,l±±)-n-1/2. Similarly, we prove that there exists nonzero z-nf~21(Tq)E21 in (𝒮𝔖^q,k,l±±)-n-1/2; finally the proof follows from (b). The proof of the converse is trivial.Corollary 21.
(a)a=z-1f(q-1/2Tq)E12+g(Tq)E21∈(𝒮𝔖^q,k,l±±)-1/2 is nondegenerate if and only if f and g are nonzero.
(b)𝒮𝔖^q,k,l±± satisfies (SP3).Proof.
(a) Leta=z-1f(q-1/2Tq)E12+g(Tq)E21∈(𝒮𝔖^q,k,l±±)-1/2. Then, if f and g are nonzero by Lemma 20(b) and 𝔭-1/2a has finite codimension in (𝒮𝔖^q,k,l±±)-1/2, then by Lemma 20(c), a is a nondegenerate element. Reciprocally, if f=0 or g=0, then by definition 𝔭-1a=0.
(b) Let𝔭 be a nondegenerate parabolic subalgebra of 𝒮𝔖^q,k,l±±; then there exists a=z-1f(q-1Tq)E12+g(Tq)E21∈𝔭-1/2 where f and g are nonzero. Then the proof follows from (a).A functionalλ∈(𝒮𝔖^q,k,l±±)0* is described by its labels, Δn=λ(TqnE11-Tq-nE22) with n∈ℤ and the central charge λ(C)=c. Moreover, we define Δn0=(Δn-Δ-n)=λ((Tqn-Tq-n)E11+(Tqn-Tq-n)E22). We consider the generating series
(109)Δλ(x)=∑n∈ℤ‍Δnx-n,Δλ0(x)=Δλ(x)-Δλ(x-1)=∑n∈ℤ‍Δn0x-n.Theorem 22.
An irreducible highest weight𝒮𝔖^q,k,l±±-module L(𝒮𝔖^q,k,l±±,λ) is quasifinite if and only if one of the following equivalent conditions holds. (a)
There exist Laurent polynomialsf(w) and g(w) such that
(110)g(x)Δλ0(x)=0,f(x)Δλ0(q-1/2x)=0.(b)
There exist quasipolynomialsP12(x) and P21(x) such that
(111)P12(n)=q-n/2Δn0,P21(n)=Δn0,
for all n∈ℤ.Proof.
By Theorem13, L(𝒮𝔖^q,k,l±±,λ) is quasifinite if and only if there exists a nondegenerate element a in (𝒮𝔖^q,k,l±±)-1/2, such that λ([(𝒮𝔖^q,k,l±±)1/2,a])=0.
Now, leta=z-1f(q-1/2Tq)E12+g(Tq)E21∈(𝒮𝔖^q,k,l±±)-1/2, with f(w)=∑j‍fjwj∈ℂ[w-1,w](δ-1,l-k) and g(w)=∑j‍gjwj∈ℂ[w-1,w](δ-1′,-l) such that f and g are nonzero. Then by Corollary 21(a), a is a nondegenerate element. Then, λ([(𝒮𝔖^q,k)1/2,a])=0 if and only if, for all i∈ℤ,
(112)λ([(Tqi-(-1)δ0Tql-i)E12,a])=0,(113)λ([z(qi/2Tqi-(-1)δ0′q(k-l-i)/2Tqk-l-i)E21,a])=0.
From (112), we obtain that
(114)0=λ(g(Tq)(Tqi-(-1)δ0Tql-i)Id)=∑j‍gjΔi+j0.
Multiplying (114) by x-i and adding over i∈ℤ, we obtain that
(115)0=∑i,j‍gjΔi+j0x-i=g(x)Δλ0(x).
Then, by (113) and using Remark 8, we obtain that
(116)0=λ((∑j‍fjq(i+j)/2Tqi+j-∑j‍fjq-(i+j)/2Tq-i-j)E11+(∑j‍fjq-(i+j)/2Tqi+j-∑j‍fjq(i+j)/2Tq-i-j)E22)+(∑j‍fjq-(i+j)/2Tqi+j-∑j‍fjq(i+j)/2Tq-i-j)0c=∑j‍fjq(i+j)/2Δi+j-∑j‍fjq-(i+j)/2Δ-i-j.
Multiplying (116) by x-i and adding over i∈ℤ, we obtain that
(117)0=∑i,j‍fjq(i+j)/2Δi+jx-i-jxj-∑i,j‍fjq-(i+j)/2Δ-i-jx-i-jxj=∑j‍fjΔλ(q-1/2x)xj-∑j‍fjΔλ(q1/2x-1)xj=f(x)Δλ0(q-1/2x).
The equivalence between (a) and (b) follows from Proposition 14.


## 3.1. The Case𝔤=𝒮𝔖^q,k

It is clear that𝒮𝔖^q,k satisfies (SP1). Now, we suppose that A=z-nf(Tq)E11+z-ng(Tq)E22∈(𝒮𝔖^q,k)-n (with n∈ℕ) satisfies [A,(𝒮𝔖^q,k)1/2]=0; in particular for all i∈ℤ(81)[A,TqiE12-zTqk-iE21]=0.
Then using (81), we obtain that f(Tq)=q-nig(Tq) for all i∈ℤ. Hence, f=g=0 (since n≠0 and q is not a root of unity); therefore, A=0. Now, we suppose that A=z-nf(Tq)E12-(-1)-nq(n-1)nkz-n+1f(qnTq-1)Tq(-2n+1)kE21 in (𝒮𝔖^q,k)-n+1/2; (with n∈ℕ) satisfies [A,(𝒮𝔖q,k)1/2]=0 then, for all i∈ℤ we have that
(82)[A,TqiE12-zTqk-iE21]=0,
which is equivalent to f(Tq)=qnk-if(qTq) for all i∈ℤ. Hence f=0; therefore, A=0. Thus, we prove that 𝒮𝔖^q,k satisfies the property (SP2).Finally, using the following lemma, we will prove that𝒮𝔖^q,k satisfies the property (SP3).Lemma 16.
Let𝔭=⨁j∈(1/2)ℤ𝔭j be a (1/2)ℤ-graded subalgebra of 𝒮𝔖^q,k, where 𝔭0=(𝒮𝔖^q,k)0. Then one has the following. (a)
For eachn∈ℤ, 𝔭n+1/2 has finite codimension in (𝒮𝔖^q,k)n+1/2 if and only if 𝔭n+1/2≠0.(b)
For eachn∈ℤ, 𝔭n has finite codimension in (𝒮𝔖^q,k)n if and only if there exists znf(qn/2Tq)E11+zng(q(n-1)/2Tq)E22∈𝔭n, such that f and g are nonzero.(c)
𝔭-n has finite codimension in (𝒮𝔖^q,k)-n, for all n∈(1/2)ℕ if and only if 𝔭-1/2≠0.Proof.
(a) We suppose that there exists(83)A=znf(Tq)E12-(-1)nq(n+1)nkzn+1×f(q-nTq-1)Tq(2n+1)kE21∈𝔭n+1/2,
with f≠0; then M1j:=[A,(Tqj-Tq-j)E11] and M2j:=[A,(q-j/2Tqj-qj/2Tq-j)E22] belong to 𝔭n+1/2, for all j∈ℤ. Moreover, Aj=(q-(n+1)j-qnj)-1(M1j+q-(n+1/2)jM2j)∈𝔭n+1/2 and
(84)Aj=znf(Tq)TqjE12-(-1)nq(n+1)nk-njzn+1×f(q-nTq-1)Tq(2n+1)k-jE21,
for all j∈ℤ×. Therefore, using (84), we have that
(85)zng(Tq)E12-(-1)nq(n+1)nkzn+1×g(q-nTq-1)Tq(2n+1)kE21∈𝔭n+1/2,
for all g(w)∈〈f(w)〉, where 〈f(w)〉 is the ideal of ℂ[w-1,w] generated by f(w). Using that 〈f(w)〉 has finite codimension in ℂ[w-1,w], we obtain that 𝔭n+1/2 has finite codimension in (𝒮𝔖^q,k)n+1/2 (see Remark 11). The proof of the converse is trivial.
(b) We suppose thatA=znf(qn/2Tq)E11+zng(q(n-1)/2Tq)E22∈𝔭n, with nonzero f and g in ℂ[w-1,w](n,2nk); then similar to the aforementioned argument
(86)(q-nj/2-qnj/2)-1[A,(Tqj-Tq-j)E11]=znf(qn/2Tq)(qnj/2Tqj+q-nj/2Tq-j)E11∈𝔭n,b[A,(q-j/2Tqj-qj/2Tq-j)E22]=zng(q(n-1)/2Tq)×(q(n-1)j/2Tqj+q-(n-1)j/2Tq-j)E22∈𝔭n,
where b=(q(-n/2)j-q(n/2)j)-1, and this is true for all j∈ℤ×. Therefore
(87)znh1(qn/2Tq)E11∈𝔭n,∀h1∈f(w)ℂ[w-1,w](1,0),znh2(q(n-1)/2Tq)E22∈𝔭n,∀h2∈g(w)ℂ[w-1,w](1,0).
Since f(w)ℂ[w-1,w](1,0) and g(w)ℂ[w-1,w](1,0) have finite codimension in ℂ[w-1,w](n,2nk), we obtain that 𝔭n has finite codimension in (𝒮𝔖^q,k)n. The proof of the converse is trivial.
(c) We suppose that𝔭-1/2≠0; then, in order to prove that 𝔭-n has finite codimension in (𝒮𝔖^q,k)-n, for all n∈(1/2)ℕ, we only need to see that this is true for all n∈ℕ, since by using (SP2), we obtain that 𝔭-n+1/2≠0 for all n∈ℕ; then, from (a) we have that 𝔭-n+1/2 has finite codimension in (𝒮𝔖^q,k)-n+1/2 for all n∈ℕ. By hypothesis, there exists A=z-1f(Tq)E12+f(qTq-1)Tq-kE21∈𝔭-1/2 with f≠0; then, using (84), we have that B=z-1f(Tq)TqE12+qf(qTq-1)Tq-k-1E21∈𝔭-1/2, and
(88)[A,B]:=z-1f1(q-1/2Tq)E11+z-1g1(q-1Tq)E22∈𝔭-1,
where f1 and g1 are nonzero. Hence, by (b), 𝔭-1 has finite codimension in (𝒮𝔖^q,k)-1. Moreover, by (87) we obtain that
(89)z-1f1(q-1/2Tq)h(q-1/2Tq)E11∈𝔭-1,(90)z-1g1(q-1Tq)h(q-1Tq)E22∈𝔭-1,
for all h(w)∈ℂ[w-1,w](1,0). Now, by induction we suppose that 𝔭-n has finite codimension in (𝒮𝔖^q,k)-n; then, there exists A=z-nf(q-n/2Tq)E11+z-ng(q-(n+1)/2Tq)E22∈𝔭-n where f and g are nonzero, and by (89) and Lemma 15(b),
(91)[A,z-1f1(q-1/2Tq)E11]≠0or[A,z-1f1(q-1/2Tq)h(q-1/2Tq)E11]≠0,
for some no constant h(w) in ℂ[w-1,w](1,0); therefore, there exists a nonzero element z-n-1f~(Tq)E11 in 𝔭-n-1. Similarly, using (90) and Lemma 15(b), we see that there exists a nonzero element z-n-1g~(Tq)E22 in 𝔭-n-1; then, using (b), 𝔭-n-1 has finite codimension in (𝒮𝔖^q,k)-n-1, finishing the induction. The proof of the converse is trivial.Corollary 17.
(a) Any parabolic subalgebra of𝒮𝔖^q,k is nondegenerate.
(b) Any nonzero element of(𝒮𝔖^q,k)-1/2 is nondegenerate.
(c)𝒮𝔖^q,k satisfies (SP3).Proof.
Let𝔭 be a parabolic subalgebra of 𝒮𝔖^q,k; by definition there exists j∈(1/2)ℕ such that 𝔭-j≠0, and then by (SP2) 𝔭-1/2≠0; the proof of (a) follows from Lemma 16(c). Finally, (b) follows from (a), and (c) follows from (b).A functionalλ∈(𝒮𝔖^q,k)0* is described by its labels Δn,1=-λ((Tqn-Tq-n)E11), Δn,2=-λ((q-nTqn-Tq-n)E22) with n∈ℤ, and the central charge λ(C)=c. We will consider the generating series
(92)Δλ,i(x)=∑n∈ℤ‍Δn,ix-n,withi=1,2.Theorem 18.
An irreducible highest weight𝒮𝔖^q,k-module L(𝒮𝔖^q,k,λ) is quasifinite if and only if one of the following equivalent conditions holds. (a)
There exists a Laurent polynomialb(x) such that
(93)b(qx)(Δλ,1(x)+Δλ,2(x)+c)=0.(b)
There exists a quasipolynomialP(x) such that
(94)P(n)=Δn,1+Δn,2+δn,0c,
for all n∈ℤ.Proof.
By Theorem13, L(𝒮𝔖^q,k,λ) is quasifinite if and only if there exist a nondegenerate element a in (𝒮𝔖^q,k)-1/2, such that λ([(𝒮𝔖^q,k)1/2,a])=0.
Now, leta=z-1b(Tq)E12+b(qTq-1)Tq-kE21 be a nonzero element of (𝒮𝔖^q,k)-1/2 (with b(w)=∑j‍bjwj); then, by Corollary 17(b), a is a nondegenerate element. Then, λ([(𝒮𝔖^q,k)1/2,a])=0 if and only if, for all i∈ℤ,
(95)0=λ([(TqiE12-zTqk-iE21,a)])=∑j‍qjbj(Δk-i+j,1+Δk-i+j,2)+q-k+ib-k+ic.
Multiplying (95) by x-k+i and adding over i∈ℤ, we obtain that
(96)0=∑j,i‍qjbj(Δk-i+j,1+Δk-i+j,2)x-k+i+∑i‍q-k+ib-k+ix-k+ic=b(qx)(Δλ,1(x)+Δλ,2(x)+c).
The equivalence between (a) and (b) follows from Proposition 14.


## 3.2. The Case𝔤=𝒮𝔖^q,k,l±±

It is clear that𝒮𝔖^q,k,l±± satisfies (SP1). Now, we prove that 𝒮𝔖^q,k,l±± satisfies (SP2). We suppose that A=z-nf(q-n/2Tq)E12+z-n+1g(q(-n+1)/2Tq)E21 in (𝒮𝔖^q,k,l±±)-n+1/2 (with n∈ℕ) satisfies [A,(𝒮𝔖^q,k,l±±)1/2]=0; in particular if we take nonzero elements h12(Tq)E12 and zh21(q1/2Tq)E21 in (𝒮𝔖^q,k,l±±)1/2, we have that
(97)[A,h12(Tq)E12]=0;
that is, g(q(-n+1)/2Tq)h12(Tq)=0; therefore, g(w)=0. Similarly,
(98)[A,zh21(q1/2Tq)E21]=0,
which is equivalent to f(q-n/2+1Tq)h21(q1/2Tq)=0; therefore, f(w)=0, and then A=0.Remark 19.
GivenB=znf(qn/2Tq)E12+zn+1g(q(n+1)/2Tq)E21∈(𝒮𝔖^q,k,l±±)n+1/2 and A=h(Tq)E11-h(Tq-1)E22∈(𝒮𝔖^q,k,l±±)0 with h∈ℂ[w-1,w], then M:=[A,B]∈(𝒮𝔖^q,k,l±±)n+1/2, where Mii=0(i=1,2) and
(99)M12=-znf(qn/2Tq)(h(qnTq)+h(Tq-1)),(100)M21=zn+1g(q(n+1)/2Tq)(h(Tq)+h(q-n-1Tq-1)).Now, letA=z-nf1(Tq)E11+z-nf2(Tq)E22∈(𝒮𝔖^q,k,l±±)-n (with n∈ℕ) be such that A is nonzero; then, if we take nonzero B=f(Tq)E12 in (𝒮𝔖^q,k,l±±)1/2 by Remark 19, there exists f(Tq)h(Tq)E12∈(𝒮𝔖^q,k,l±±)1/2 such that h is not constant; then by Lemma 15(a), we have that
(101)[A,f(Tq)E12]≠0or[A,f(Tq)h(Tq)E12]≠0.
Therefore 𝒮𝔖^q,k,l±± satisfies (SP2).In order to prove that𝒮𝔖^q,k,l±± satisfies (SP3), we will need the following result. We denote
(102)δn′:={δn+1,in(𝒮𝔖q,k,l+,±)n+1/2,δn,in(𝒮𝔖q,k,l-,±)n+1/2.Lemma 20.
Let𝔭=⨁j∈(1/2)ℤ𝔭j be a (1/2)ℤ-graded subalgebra of 𝒮𝔖^q,k,l±±, with 𝔭0=(𝒮𝔖^q,k,l±±)0. Then one has the following: (a)
for eachn∈ℤ,  𝔭n has finite codimension in (𝒮𝔖^q,k,l±±)n if and only if 𝔭n≠0;(b)
for eachn∈ℤ,  𝔭n+1/2  has finite codimension in  (𝒮𝔖^q,k,l±±)n+1/2  if and only if  there exists znf(qnTq)E12+zn+1g(q(n+1)/2Tq)E21∈𝔭n+1/2, such that f and g are nonzero;(c)
𝔭-n has finite codimension in (𝒮𝔖^q,k,l±±)-n for all n∈(1/2)ℕ, if and only if 𝔭-1/2 has finite codimension in (𝒮𝔖^q,k,l±±)-1/2.Proof.
(a) LetA=znf(Tq)E11-(±1)nqn(nk-l)/2znf(q-nTq-1)TqnkE22∈𝔭n, with f≠0; then, Bj:=(1-qnj)-1[A,TqjE11-Tq-jE22]∈𝔭n and
(103)Bj=znf(Tq)TqjE11-(±1)nq(n(nk-l)/2)-nj×znf(q-nTq-1)Tqnk-jE22,
for all j∈ℤ×; therefore, by (103) we have that
(104)zng(Tq)E11-(±1)nqn(nk-l)/2zng(q-nTq-1)TqnkE22∈𝔭n,
for all g(w)∈〈f(w)〉, and since 〈f(w)〉 has finite codimension in ℂ[w-1,w], we have that 𝔭n has finite codimension in (𝒮𝔖^q,k,l±±)n (see Remark 11). The proof of the converse is trivial.
(b) We suppose that there exist nonzero elements(105)znf1(Tq)E12,zn+1f2(Tq)E21in𝔭n+1/2;
then, using Remark 19, we obtain that zng1(Tq)E12∈𝔭n+1/2 for all g1(q-n/2w)∈f1(w)ℂ[w-1,w](1,0) (see (99)) and zn+1g2(Tq)E21∈𝔭n+1/2 for all g2(q(-n+1)/2w)∈f2(w)ℂ[w-1,w](1,0) (see (100)). Then using that f1(w)ℂ[w-1,w](1,0) has finite codimension in  ℂ[w-1,w](δn,nk+l)  and  f2(w)ℂ[w-1,w](1,0) has finite codimension in ℂ[w-1,w](δn′,(n+1)k-l), we obtain that 𝔭n+1/2 has finite codimension in (𝒮𝔖^q,k,l±±)n+1/2. Therefore in order to prove that 𝔭n+1/2 has finite codimension in (𝒮𝔖^q,k,l±±)n+1/2, we only need to see that there exist nonzero elements as aforementioned in 𝔭n+1/2. Let B=znf(qn/2Tq)E12+zn+1g(q(n+1)/2Tq)E21∈𝔭n+1/2 with nonzero f and g, and let A and M be as in Remark 19. Then, taking h(w)=w-qnw-1 (observe that h(Tq-1)=-h(qnTq)), we obtain that
(106)M=zn+1g(q(n+1)/2Tq)×(h(Tq)+h(q-(n+1)Tq-1))×E21∈𝔭n+1/2,withM≠0.
Similarly, taking h(w)=w-q-n-1w-1 (observe that h(Tq)=-h(q-n-1Tq-1)), we obtain
(107)M=-znf(qn/2Tq)×(h(qnTq)+h(Tq-1))E12∈𝔭n+1/2,withM≠0,
proving the existence of nonzero elements of those forms. The proof of the converse is trivial.
(c) We suppose that𝔭-1/2 has finite codimension in (𝒮𝔖^q,k,l±±)-1/2; then, in order to prove that 𝔭-n has finite codimension in (𝒮𝔖^q,k,l±±)-n for all n∈(1/2)ℕ, we only need to see that this is true for all -n+1/2 with n∈ℕ, since in this case by using (SP2), we obtain that 𝔭-n≠0 for all n∈ℕ; then, by (a), we have that 𝔭-n has finite codimension in (𝒮𝔖^q,k,l±±)-n, for all n∈ℕ.
By hypothesis, there exist nonzero elementsA=z-1f(q-1Tq)E12 and B=g(Tq)E21 in 𝔭-1/2; therefore, [A,B]∈𝔭-1 is nonzero. By induction, we suppose that 𝔭-n+1/2 has finite codimension in (𝒮𝔖^q,k,l±±)-n+1/2; then, there exist nonzero elements z-nf12(qnTq)E12 and z-n+1f21(q(-n+1)/2Tq)E21 in (𝒮𝔖^q,k,l±±)-n+1/2, and by Remark 19, we also have z-nf12(qnTq)g12(Tq)E12, z-n+1f21(q(-n+1)/2Tq)g21(Tq)E21 in (𝒮𝔖^q,k,l±±)-n+1/2 with no constant gij for all i≠j. Then by Lemma 15(a), we have that
(108)[z-nf12(qnTq)E12,[A,B]]≠0or[z-nf12(qnTq)g12(Tq)E12,[A,B]]≠0.
Therefore from (108), there exists a nonzero element z-n-1f~12(Tq)E12 in (𝒮𝔖^q,k,l±±)-n-1/2. Similarly, we prove that there exists nonzero z-nf~21(Tq)E21 in (𝒮𝔖^q,k,l±±)-n-1/2; finally the proof follows from (b). The proof of the converse is trivial.Corollary 21.
(a)a=z-1f(q-1/2Tq)E12+g(Tq)E21∈(𝒮𝔖^q,k,l±±)-1/2 is nondegenerate if and only if f and g are nonzero.
(b)𝒮𝔖^q,k,l±± satisfies (SP3).Proof.
(a) Leta=z-1f(q-1/2Tq)E12+g(Tq)E21∈(𝒮𝔖^q,k,l±±)-1/2. Then, if f and g are nonzero by Lemma 20(b) and 𝔭-1/2a has finite codimension in (𝒮𝔖^q,k,l±±)-1/2, then by Lemma 20(c), a is a nondegenerate element. Reciprocally, if f=0 or g=0, then by definition 𝔭-1a=0.
(b) Let𝔭 be a nondegenerate parabolic subalgebra of 𝒮𝔖^q,k,l±±; then there exists a=z-1f(q-1Tq)E12+g(Tq)E21∈𝔭-1/2 where f and g are nonzero. Then the proof follows from (a).A functionalλ∈(𝒮𝔖^q,k,l±±)0* is described by its labels, Δn=λ(TqnE11-Tq-nE22) with n∈ℤ and the central charge λ(C)=c. Moreover, we define Δn0=(Δn-Δ-n)=λ((Tqn-Tq-n)E11+(Tqn-Tq-n)E22). We consider the generating series
(109)Δλ(x)=∑n∈ℤ‍Δnx-n,Δλ0(x)=Δλ(x)-Δλ(x-1)=∑n∈ℤ‍Δn0x-n.Theorem 22.
An irreducible highest weight𝒮𝔖^q,k,l±±-module L(𝒮𝔖^q,k,l±±,λ) is quasifinite if and only if one of the following equivalent conditions holds. (a)
There exist Laurent polynomialsf(w) and g(w) such that
(110)g(x)Δλ0(x)=0,f(x)Δλ0(q-1/2x)=0.(b)
There exist quasipolynomialsP12(x) and P21(x) such that
(111)P12(n)=q-n/2Δn0,P21(n)=Δn0,
for all n∈ℤ.Proof.
By Theorem13, L(𝒮𝔖^q,k,l±±,λ) is quasifinite if and only if there exists a nondegenerate element a in (𝒮𝔖^q,k,l±±)-1/2, such that λ([(𝒮𝔖^q,k,l±±)1/2,a])=0.
Now, leta=z-1f(q-1/2Tq)E12+g(Tq)E21∈(𝒮𝔖^q,k,l±±)-1/2, with f(w)=∑j‍fjwj∈ℂ[w-1,w](δ-1,l-k) and g(w)=∑j‍gjwj∈ℂ[w-1,w](δ-1′,-l) such that f and g are nonzero. Then by Corollary 21(a), a is a nondegenerate element. Then, λ([(𝒮𝔖^q,k)1/2,a])=0 if and only if, for all i∈ℤ,
(112)λ([(Tqi-(-1)δ0Tql-i)E12,a])=0,(113)λ([z(qi/2Tqi-(-1)δ0′q(k-l-i)/2Tqk-l-i)E21,a])=0.
From (112), we obtain that
(114)0=λ(g(Tq)(Tqi-(-1)δ0Tql-i)Id)=∑j‍gjΔi+j0.
Multiplying (114) by x-i and adding over i∈ℤ, we obtain that
(115)0=∑i,j‍gjΔi+j0x-i=g(x)Δλ0(x).
Then, by (113) and using Remark 8, we obtain that
(116)0=λ((∑j‍fjq(i+j)/2Tqi+j-∑j‍fjq-(i+j)/2Tq-i-j)E11+(∑j‍fjq-(i+j)/2Tqi+j-∑j‍fjq(i+j)/2Tq-i-j)E22)+(∑j‍fjq-(i+j)/2Tqi+j-∑j‍fjq(i+j)/2Tq-i-j)0c=∑j‍fjq(i+j)/2Δi+j-∑j‍fjq-(i+j)/2Δ-i-j.
Multiplying (116) by x-i and adding over i∈ℤ, we obtain that
(117)0=∑i,j‍fjq(i+j)/2Δi+jx-i-jxj-∑i,j‍fjq-(i+j)/2Δ-i-jx-i-jxj=∑j‍fjΔλ(q-1/2x)xj-∑j‍fjΔλ(q1/2x-1)xj=f(x)Δλ0(q-1/2x).
The equivalence between (a) and (b) follows from Proposition 14.



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*Source: 672872-2013-09-17.xml*