# A family of simple non-weight modules over the twisted N=2 superconformal algebra

aa r X i v : . [ m a t h . R T ] F e b A family of simple non-weight modules over thetwisted N = 2 superconformal algebra Haibo Chen, Xiansheng Dai and Mingqiang Liu ∗ Abstract:

We construct a class of non-weight modules over the twisted N = 2 superconformalalgebra L . Let h = C L ⊕ C G be the Cartan subalgebra of L , and let t = C L be the Cartansubalgebra of even part L ¯0 . These modules over L when restricted to the h are free of rank 1or when restricted to the t are free of rank 2. We provide the suﬃcient and necessary conditionsfor those modules being simple, as well as giving the suﬃcient and necessary conditions for two L -modules being isomorphic. We also compute the action of an automorphism on them. Moreover,based on the weighting functor introduced in [15], a class of intermediate series modules A t ( σ ) areobtained. As a byproduct, we give a suﬃcient condition for two L -modules are not isomorphic. Key words:

Twisted N = 2 superconformal algebra, Non-weight module, Simple module. Mathematics Subject Classiﬁcation (2010):

Throughout this paper, we denote by C , C ∗ , Z and Z + the sets of complex numbers, nonzerocomplex numbers, integers and nonnegative integers respectively. Note that Z = ( + Z ) ∪ Z .All vector superspaces (resp. superalgebras, supermodules) and spaces (resp. algebras,modules) are considered to be over C . For a Lie algebra G , we use U ( G ) to denote theuniversal enveloping algebra.The N = 2 superconformal algebras have four sectors: the Neveu-Schwarz sector, theRamond sector, the topological sector and the twisted sector, which was constructed inde-pendently by Kac and by Ademollo et al. (see [1, 10]). It is well known that the ﬁrst threesuperconformal algebras are isomorphic to each other, and they are sometimes called theuntwisted sector. Therefore, there are essentially two diﬀerent sectors.The representation theory of superconformal algebras has attracted a lot of attentionfrom mathematicians and physicists, which includes weight representations and non-weightrepresentations. Some characteristics related to the structure of Verma modules over thetwisted N = 2 superconformal algebra were investigated in [6, 9]. The classiﬁcation ofindecomposable modules of the intermediate series over the twisted N = 2 superconformalalgebra was achieved in [12]. All simple Harish-Chandra modules over the N = 1 anduntwisted N = 2 superconformal algebras were classiﬁed respectively in [11,16]. In fact, withincreasing the number of fermionic currents N , the representations of the superconformalalgebras becomes more and more complicated.A class of non-weight modules called free U ( h )-modules were constructed and studied inrecent years. Here, h is the Cartan subalgebra (modulo the central elements) and acts freely. ∗ Corresponding author. U ( h )-modules were ﬁrst introduced by Nilsson in [14] for the complex matricesalgebra sl n +1 . In [18], Tan and Zhao obtained them by a very diﬀerent method. From thenon, the free U ( h )-modules were widely investigated over all kinds of Lie algebras, such as theﬁnite-dimensional simple Lie algebras [15] and the inﬁnite-dimensional Lie algebras relatedto Virasoro algebras and so on (see, e.g., [2, 4, 8, 13, 17, 19, 20]). It is worth mentioning thatthe free U ( h )-modules over some Lie superalgebras were also characterized. In [5], U ( h )-modules over the basic Lie superalgebras were investigated, which showed that osp (1 | n ) isthe only basic Lie superalgebra that admits such modules. The U ( h )-free modules on someinﬁnite-dimensional Lie superalgebras such as super-BMS algebras, super-Virasoro algebras,untwisted N = 2 super conformal algebras were respectively studied in [3,21,22]. The aim ofthe present paper is to classify such modules for the twisted N = 2 superconformal algebras,and we further investigate the relevant twisted modules and weight modules.The rest of the paper is organized as follows.In section 2, we ﬁrst recall the deﬁnition of twisted N = 2 superconformal algebra. A classof non-weight modules over the L are deﬁned. Then the simplicities and isomorphism classesof these modules are respectively determined in Theorems 2.5 and 2.7 (also see Theorems2.10 and 2.11).In Sections 3 and 4, the free U ( h )-modules of rank 1 over L and the free U ( t )-modules ofrank 2 over L are respectively classiﬁed in Theorems 3.3 and 4.3.In Section 5, using the automorphism results in [7], we deﬁne a class of twisted modulesof L .At last, applying the right exact weighting functor introduced in [15] to the modules M t ( λ, α ), we present that the resulting weight modules are exactly A t ( σ ). In this section, we construct a class of non-weight modules over the twisted N = 2 supercon-formal algebra. Notice that these modules are free of rank 1 when regarded as U ( h )-modules,while these modules are free of rank 2 when regarded as U ( t )-modules.Let us ﬁrst recall the deﬁnition of N = 2 superconformal algebras: the twisted N = 2algebra (i.e. with mixed boundary conditions for the fermionic ﬁelds). Deﬁnition 2.1. The twisted N = 2 superconformal algebra L is deﬁned as an inﬁnite-dimensional Lie superalgebra over C with basis { L m , I r , G p | m ∈ Z , r ∈ + Z , p ∈ Z } and atisfying the following non-trivial super brackets: [ L m , L n ] = ( m − n ) L m + n , [ L m , I r ] = − rI m + r , [ L m , G p ] = ( m − p ) G m + p , [ I r , G p ] = G r + p , [ G p , G q ] = ( ( − p L p + q if p + q ∈ Z , ( − p +1 ( p − q ) I p + q if p + q ∈ + Z , (2.1) where m, n ∈ Z , r ∈ + Z , p, q ∈ Z . The Z -graded of L is deﬁned by L = L ¯0 ⊕ L ¯1 , where L ¯0 = { L m , I r | m ∈ Z , r ∈ + Z } and L ¯1 = { G p | p ∈ Z } . Clearly, L contains the Witt algebra V = span C { L m | m ∈ Z } andthe centerless N = 1 super-Virasoro algebra. Fixing m ∈ Z , µ ∈ C ∗ , we denote L m := L m + µI m − . The Lie algebra spanned by { L m | m ∈ Z } over C is isomorphic to the Witt algebra.Let G be a Lie superalgebra, and let V = V ¯0 ⊕ V ¯1 be a Z -graded vector space. Anyelement v in V ¯0 (resp. v in V ¯1 ) is said to be even (resp. odd). If v is even (resp. odd), wedeﬁne | v | = ¯0 (resp. | v | = ¯1). Elements in V ¯0 or V ¯1 are called homogeneous. All elementsin superalgebras and modules are homogenous unless speciﬁed. A G -module is a Z -gradedvector space V together with a bilinear map G × V → V , denoted ( x, v ) xv such that x ( yv ) − ( − | x || y | y ( xv ) = [ x, y ] v and G ¯ i V ¯ j ⊆ V ¯ i +¯ j , where ¯ i, ¯ j ∈ Z , x, y ∈ G , v ∈ V . Thus there is a parity-change functor Π on the categoryof G -modules to itself, namely, for any module V = V ¯0 ⊕ V ¯1 , we have Π( V ¯0 ) = V ¯1 andΠ( V ¯1 ) = V ¯0 .Take λ ∈ C ∗ , α ∈ C . Let Ω( λ, α ) = C [ t ] as a vector space and deﬁne the action of Wittalgebra as follows: L m ( f ( t )) = λ m ( t + mα ) f ( t + m ) , where m ∈ Z , f ( t ) ∈ C [ t ] . The following results were given in [13, 17].

Theorem 2.2.

For λ ∈ C ∗ , α ∈ C , then (1) Ω( λ, α ) is a V -module; (2) Ω( λ, α ) is simple if and only if α = 0 ; (3) t Ω( λ, ∼ = Ω( λ, is the unique simple proper submodule of Ω( λ, with codimension ; (4) Any free U ( C L ) -module of rank over the Witt algebra is isomorphic to some Ω( λ, α ) for λ ∈ C ∗ , α ∈ C . .1 U ( h ) -modules Assume that V = C [ ∂ ] ⊕ ∂ C [ ∂ ]. Then V is a Z -graded vector space with V ¯0 = C [ ∂ ] and V ¯1 = ∂ C [ ∂ ]. Now we give an precise construction of an L -module structure on V . Proposition 2.3.

For λ ∈ C ∗ , α ∈ C , t = ± , f ( ∂ ) ∈ C [ ∂ ] and ∂f ( ∂ ) ∈ ∂ C [ ∂ ] , we deﬁnethe following action of twisted N = 2 superconformal algebra on V as follows L m f ( ∂ ) = λ m ( ∂ + mα ) f ( ∂ + m ) , (2.2) L m ∂f ( ∂ ) = λ m ( ∂ + m ( α + 12 )) ∂f ( ∂ + m ) , (2.3) I r f ( ∂ ) = − t r λ r αf ( ∂ + r ) , (2.4) I r ∂f ( ∂ ) = t r λ r (1 − α ) ∂f ( ∂ + r ) , (2.5) G p f ( ∂ ) = t p λ p ∂f ( ∂ + p ) , (2.6) G p ∂f ( ∂ ) = ( − t ) p λ p ( ∂ + 2 pα ) f ( ∂ + p ) (2.7) for m ∈ Z , r ∈ + Z , p ∈ Z . Then V is an L -module under the action of (2.2) - (2.7) , whichis a free of rank as a U ( h ) -module and denoted by M t ( λ, α ) .Proof. According to Theorem 2.2 (1), one can see that L m L n f ( ∂ ) − L n L m f ( ∂ ) = ( m − n ) L m + n f ( ∂ ) ,L m L n ∂f ( ∂ ) − L n L m ∂f ( ∂ ) = ( m − n ) L m + n ∂f ( ∂ ) , where m, n ∈ Z .For any m ∈ Z , r ∈ + Z , from (2.2)-(2.5), we respectively check L m I r f ( ∂ ) − I r L m f ( ∂ )= L m (cid:0) − t r λ r αf ( ∂ + r ) (cid:1) − I r (cid:0) λ m ( ∂ + mα ) f ( ∂ + m ) (cid:1) = − t r λ m + r α ( ∂ + mα ) f ( ∂ + m + r ) + 2 t r λ m + r α ( ∂ + r + mα ) f ( ∂ + m + r )= 2 t m + r ) λ m + r αrf ( ∂ + m + r ) = − rI m + r f ( ∂ )and L m I r ∂f ( ∂ ) − I r L m ∂f ( ∂ )= L m (cid:0) t r λ r ∂ (1 − α ) f ( ∂ + r ) (cid:1) − I r (cid:0) λ m ∂ (cid:0) ∂ + m ( α + 12 ) (cid:1) f ( ∂ + m ) (cid:1) = t r λ m + r ∂ (1 − α ) (cid:0) ∂ + m ( α + 12 ) (cid:1) f ( ∂ + m + r ) − t r λ m + r ∂ (1 − α ) (cid:0) ∂ + r + m ( α + 12 ) (cid:1) f ( ∂ + m + r )= − t m + r ) λ m + r (1 − α ) r∂f ( ∂ + m + r ) = − rI m + r ∂f ( ∂ ) . r, s ∈ + Z , by (2.4) and (2.5), it is straightforward to verify that I r I s f ( ∂ ) − I s I r f ( ∂ )= I r ( − t s λ s αf ( ∂ + s )) − I s ( − t r λ r αf ( ∂ + r )) = 0 ,I r I s ∂f ( ∂ ) − I s I r ∂f ( ∂ )= I r ( t s λ s ∂ (1 − α ) f ( ∂ + s )) − I s ( t r λ r ∂ (1 − α ) f ( ∂ + r )) = 0 . Fixing r ∈ + Z , then 2 r is an odd number. For any r ∈ + Z , p ∈ Z , using (2.4)-(2.7),we obtain I r G p f ( ∂ ) − G p I r f ( ∂ )= I r ( t p λ p ∂f ( ∂ + p )) − G p ( − t r λ r αf ( ∂ + r ))= t r + p ) λ r + p ∂ (1 − α ) f ( ∂ + r + p ) + 2 αt r + p ) λ r + p ∂f ( ∂ + r + p )= t r + p ) λ r + p ∂f ( ∂ + r + p ) = G r + p f ( ∂ )and I r G p ∂f ( ∂ ) − G p I r ∂f ( ∂ )= I r (cid:0) ( − t ) p λ p ( ∂ + 2 pα ) f ( ∂ + p ) (cid:1) − G p (cid:0) t r λ r ∂ (1 − α ) f ( ∂ + r ) (cid:1) = ( − p t r + p ) λ r + p ( − α )( ∂ + r + 2 pα ) f ( ∂ + r + p ) − ( − p t r + p ) λ r + p ( ∂ + 2 pα )(1 − α ) f ( ∂ + r + p )= ( − t ) p + r ) λ r + p ( ∂ + 2( p + r ) α ) f ( ∂ + r + p ) = G r + p ∂f ( ∂ ) . Moreover, for p, q ∈ Z , it follows from (2.2)-(2.7) that we have G p G q f ( ∂ ) + G q G p f ( ∂ )= G p ( λ q ∂f ( ∂ + q )) + G q ( λ p ∂f ( ∂ + p ))= ( − p (cid:16) t p + q ) λ p + q ( ∂ + 2 pα ) f ( ∂ + p + q )+( − t ) p + q ) λ p + q ( ∂ + 2 qα ) f ( ∂ + p + q ) (cid:17) = ( ( − p λ p + q (cid:0) ∂ + ( p + q ) α (cid:1) f ( ∂ + p + q ) if p + q ∈ Z , ( − p +1 t p + q ) λ p + q ( p − q )( − α ) f ( ∂ + p + q ) if p + q ∈ + Z , = ( ( − p L p + q f ( ∂ ) if p + q ∈ Z , ( − p +1 ( p − q ) I p + q f ( ∂ ) if p + q ∈ + Z , G p G q ∂f ( ∂ ) + G q G p ∂f ( ∂ )= G p (cid:0) ( − t ) q λ q ( ∂ + 2 qα ) f ( ∂ + q ) (cid:1) + G q (cid:0) ( − t ) p λ p ( ∂ + 2 pα ) f ( ∂ + p ) (cid:1) = ( − p (cid:16) ( − t ) p + q ) λ p + q ∂ ( ∂ + p + 2 qα ) f ( ∂ + p + q )+ t p + q ) λ p + q ∂ ( ∂ + q + 2 pα ) f ( ∂ + p + q ) (cid:17) = ( ( − p (cid:16) λ p + q ∂ (cid:0) ∂ + ( p + q )( α + ) (cid:1) f ( ∂ + p + q ) (cid:17) if p + q ∈ Z , ( − p +1 t p + q ) λ p + q ( p − q ) ∂ (1 − α ) f ( ∂ + p + q ) if p + q ∈ + Z , = ( ( − p (cid:0) L p + q ∂f ( ∂ ) (cid:1) if p + q ∈ Z , ( − p +1 ( p − q ) I p + q ∂f ( ∂ ) if p + q ∈ + Z . Finally, for any m ∈ Z , p ∈ Z , by the similar computation, we obtain L m G p f ( ∂ ) − G p L m f ( ∂ )= L m ( t p λ p ∂f ( ∂ + p )) − G p ( λ m ( ∂ + mα ) f ( ∂ + m ))= ( m − p ) t m + p ) λ m + p ∂f ( ∂ + m + p )) = ( m − p ) G m + p f ( ∂ )and L m G p ∂f ( ∂ ) − G p L m ∂f ( ∂ )= L m (cid:0) ( − t ) p λ p ( ∂ + 2 pα ) f ( ∂ + p ) (cid:1) − G p (cid:0) λ m ∂ ( ∂ + m ( α + 12 )) f ( ∂ + m ) (cid:1) = ( − t ) p λ m + p ( ∂ + mα )( ∂ + m + 2 pα ) f ( ∂ + m + p )) − ( − t ) p λ m + p ( ∂ + 2 pα )( ∂ + p + m ( α + 12 )) f ( ∂ + m + p )= ( m − p )( − t ) p λ m + p ( ∂ + 2 α ( p + m )) f ( ∂ + m + p )= ( m − p ) G m + p ∂f ( ∂ ) . This completes the proof.

Remark 2.4.

By the deﬁnition of L -module M t ( λ, α ) for t = ± , we can deﬁne a class ofWitt modules Ω b ( λ, µ, α ) = C [ x ] with the following action: L m f ( x ) = λ m ( x + mα ) f ( x + m ) − µt m − λ m − f ( x + m −

12 ) , where m ∈ Z , λ, µ ∈ C ∗ , α ∈ C . L -modules M t ( λ, α ) are presented separatelyin the following two theorems. Theorem 2.5.

Let λ ∈ C ∗ , α ∈ C , t = ± , Υ t = ∂ ( M t ( λ, ¯0 ) ⊕ M t ( λ, ¯1 . Then (1) M t ( λ, α ) is simple if and only if α = 0 . Furthermore, M t ( λ, has a unique propersubmodule Υ t , and M t ( λ, / Υ t is a one-dimensional trivial L -module; (2) Υ ∼ = Π( M − ( λ, )) and Υ − ∼ = Π( M ( λ, )) are simple L -modules.Proof. (1) Assume that P = P ¯0 ⊕ P ¯1 is a nonzero submodule of M t ( λ, α ). By (2.6) and(2.7), we get that P ¯0 and P ¯1 are both nonzero. If we prove P ¯0 = M t ( λ, α ) ¯0 , the simplicitiesof M t ( λ, α ) will be determined.We have the following two cases. Case 1. α = 0 . It is easy to check that Υ t is a proper submodule of M t ( λ, α ) for λ ∈ C ∗ , α ∈ C .According to Theorem 2.2 (3), we see that P ¯0 = ∂ ( M t ( λ, α ) ¯0 ). Based on (2.6), we observe P ¯1 = M t ( λ, α ) ¯1 . Therefore, P = Υ t and M t ( λ, / Υ t is a 1-dimensional trivial L -module. Case 2. α = 0 . The simple Witt modules can be extended to the simple L ¯0 -modules. Then from Theorem2.2 (2), we can conclude that M t ( λ, α ) is a simple L -module.(2) Deﬁne the following linear map χ : Υ −→ Π( M − ( λ,

12 )) ∂ f ( ∂ ) ∂f ( ∂ ) ∂f ( ∂ ) f ( ∂ ) . For r ∈ Z + , p ∈ Z , we check that χ ( G p ∂f ( ∂ )) = ( − p λ p ∂f ( ∂ + p ) = G p χ ( ∂f ( ∂ )) ,χ ( G p ∂ f ( ∂ )) = λ p ( ∂ + p ) f ( ∂ + p ) = G p χ ( ∂ f ( ∂ )) ,χ ( I r ∂f ( ∂ )) = λ r f ( ∂ + r ) = I r χ ( ∂f ( ∂ )) ,χ ( I r ∂ f ( ∂ )) = I r χ ( ∂ f ( ∂ )) = 0 . By an identical process, one can conﬁrm that χ ( L m ∂f ( ∂ )) = L m χ ( ∂f ( ∂ )) , χ ( L m ∂ f ( ∂ )) = L m χ ( ∂ f ( ∂ )) . Hence, χ is an L -module isomorphism, and Υ is a simple submodule of L . Using the similararguments, we know that Υ − ∼ = Π( M ( λ, )) is also a simple L -module.7 emark 2.6. Using Theorem 2.5, we have the short exact sequence → Υ t ξ −→ M t ( λ, ξ −→ C → , where ξ and ξ are respectively the embedding mapping and canonical mapping deﬁned by ξ : f ( ∂ ) + ∂g ( ∂ ) → ∂ f ( ∂ ) + ∂g ( ∂ ) , ξ : f ( ∂ ) + ∂g ( ∂ ) → f ( ∂ ) + ∂g ( ∂ ) . Theorem 2.7.

Let λ, µ ∈ C ∗ , α, β ∈ C , t, t ′ = ± . We obtain the following statements. (i) Π (cid:0) M t ( λ, α ) (cid:1) ≇ M t ′ ( µ ′ , β ′ ) for any µ ′ , β ′ ∈ C ; (ii) M t ( λ, α ) ∼ = M t ′ ( µ, β ) if and only if λ = µ , α = β and t = t ′ .Proof. (i) Let ψ : Π (cid:0) M t ( λ, α ) (cid:1) −→ M t ′ ( µ ′ , β ′ )be an isomorphism. Assume that , ′ are respectively the generators of the free C [ L , G ]-modules M t ( λ, α ) and M t ′ ( µ ′ , β ′ ) . In fact, there exists some e , e ∈ C ∗ such that ψ ( ) = e ∂ ′ and ψ ( ∂ ) = e ′ . For p ∈ Z , we conﬁrm that G p ψ ( ) = G p (cid:0) e ∂ ′ (cid:1) = ( − t ′ ) p e ( µ ′ ) p ( ∂ + 2 pβ ′ ) ′ and ψ ( G p ) = e t p λ p ′ . Thus we have G p ψ ( ) = ψ ( G p ), which yields a contradiction.(ii) The “if” part is clear. Now we consider the “only if” part. Suppose M t ( λ, α ) ∼ = M t ′ ( µ, β ) as L -modules. It follows from (i) that M t ( λ, α ) ¯0 ∼ = M t ′ ( µ, β ) ¯0 can be viewed as V -modules. By Theorem 12 of [13], one has λ = µ and α = β .Let ϕ : M t ( λ, α ) −→ M t ′ ( µ, β )be an isomorphism, and let , ′ be the generators of M t ( λ, α ) and M t ′ ( µ, β ) . Then thereexists some d ∈ C ∗ such that ϕ ( ) = d ′ and ϕ ( ∂ ) = d∂ ′ . For p ∈ Z , from ϕ ( G p ) = G p ϕ ( ), we deduce that t = t ′ . U ( t ) -modules Let W = C [ x ] ⊕ C [ y ] be an L -module such that it is free of rank 2 as a U ( C L )-modulewith two homogeneous basis elements 1 ¯0 and 1 ¯1 . Then W is a Z -graded vector space with W ¯0 = C [ x ] ¯0 and W ¯1 = C [ y ] ¯1 . The L -module structure on W denoted by N t ( λ, α ) can becharacterized as follows. 8 roposition 2.8. For λ ∈ C ∗ , α ∈ C , t = ± , f ( x ) ∈ C [ x ] and g ( y ) ∈ C [ y ] , we have theaction of L on W deﬁned as L m f ( x ) = λ m ( x + mα ) f ( x + m ) , (2.8) L m g ( y ) = λ m (cid:0) y + m ( α + 12 ) (cid:1) g ( y + m ) , (2.9) I r f ( x ) = − t r λ r αf ( x + r ) , (2.10) I r g ( y ) = t r λ r (1 − α ) g ( y + r ) , (2.11) G p f ( x ) = t p λ p f ( y + p ) , (2.12) G p g ( y ) = ( − t ) p λ p ( x + 2 pα ) g ( x + p ) , (2.13) where m ∈ Z , r ∈ + Z , p ∈ Z . Then W is an L -module under the action of (2.8) - (2.13) ,which is a free of rank as a module over U ( C L ) .Proof. Since the proof is similar to the Proposition 2.3, we omit the details.

Lemma 2.9.

Let λ ∈ C ∗ , α ∈ C , t = ± . Then M t ( λ, α ) ∼ = N t ( λ, α ) .Proof. Let Θ : M t ( λ, α ) −→ N t ( λ, α ) f ( ∂ ) f ( x ) ∂g ( ∂ ) g ( y )be the linear map. Clearly, Θ is bijective. Now we are ready to prove that it is an L -moduleisomorphism.For any m ∈ Z , r ∈ + Z , p ∈ Z , f ( ∂ ) ∈ C [ ∂ ] , ∂g ( ∂ ) ∈ ∂ C [ ∂ ] , f ( x ) ∈ C [ x ] , g ( y ) ∈ C [ y ], we have Θ( L m f ( ∂ )) = Θ (cid:16) λ m ( ∂ + mα ) f ( ∂ + m ) (cid:17) = λ m ( x + mα ) f ( x + m ) = L m Θ( f ( ∂ )) , Θ( L m ∂g ( ∂ )) = Θ (cid:16) λ m ∂ ( ∂ + m ( α + 12 )) g ( ∂ + m ) (cid:17) = λ m ( y + m ( α + 12 )) g ( y + m ) = L m Θ( ∂f ( ∂ )) , Θ( I r f ( ∂ )) = Θ (cid:16) − t r λ r αf ( ∂ + r ) (cid:17) = − t r λ r αf ( x + r ) = I r Θ( f ( ∂ )) , I r ∂g ( ∂ )) = Θ (cid:16) t r λ r (1 − α ) ∂g ( ∂ + r ) (cid:17) = t r λ r (1 − α ) g ( y + r ) = I r Θ( ∂g ( ∂ )) , Θ( G p f ( ∂ )) = Θ (cid:16) t p λ p f ( x + p ) (cid:17) = t p λ p f ( x + p ) = G p Θ( f ( ∂ )) , Θ( G p ∂g ( ∂ )) = Θ (cid:16) ( − t ) p λ p ( ∂ + 2 pα ) g ( ∂ + p ) (cid:17) = ( − t ) p λ p ( x + 2 pα ) g ( x + p ) = G p Θ( ∂g ( ∂ )) . Thus, Θ is an L -module isomorphism.As a direct consequence of Theorems 2.5, 2.7 and Lemma 2.9, we get the following twotheorems. Theorem 2.10.

Assume that λ, µ ∈ C ∗ , α, β ∈ C , t = ± and π t = x ( N t ( λ, ¯0 ) ⊕ N t ( λ, ¯1 .Then (1) N t ( λ, α ) is simple if and only if α = 0 . In addition, N t ( λ, has a unique propersubmodule π t , and N t ( λ, /π t ∼ = C . (2) π ∼ = Π( N − ( λ, )) and π − ∼ = Π( N ( λ, )) are simple L -modules. Theorem 2.11.

Let λ, µ ∈ C ∗ , α, β ∈ C , t, t ′ = ± . We have (i) Π (cid:0) N t ( λ, α ) (cid:1) ≇ N t ′ ( µ ′ , β ′ ) for any µ ′ , β ′ ∈ C ; (ii) N t ( λ, α ) ∼ = N t ′ ( µ, β ) if and only if λ = µ, α = β and t = t ′ . U ( h ) -modules of rank 1 Assume that V = V ¯0 ⊕ V ¯1 is an L -module such that it is free of rank 1 as a U ( h )-module,where h = C L ⊕ C G . It follows from the superalgebra structure of L in (2.1) that we have L G = G L and G = L . Thus U ( h ) = C [ L ] ⊕ G C [ L ]. Choose a homogeneous basis element in V . Without lossof generality, up to a parity, we may assume ∈ V ¯0 and V = U ( h ) = C [ L ] ⊕ G C [ L ] with V ¯0 = C [ L ] and V ¯1 = G C [ L ] . Then we can suppose that G = ∂ , L = ∂ . Inthe following, V = V ¯0 ⊕ V ¯1 is equal to C [ ∂ ] ⊕ ∂ C [ ∂ ] with V ¯0 = C [ ∂ ] and V ¯1 = ∂ C [ ∂ ] .10rom Theorem 2.2 (4), for any m ∈ Z , f ( ∂ ) ∈ C [ ∂ ], there exists λ ∈ C ∗ , α ∈ C suchthat L m ( f ( ∂ )) = λ m ( ∂ + mα ) f ( ∂ + m ) . (3.1)It is clear that V ¯0 can be regarded as an L -module which is free of rank 1 as a C [ L ]-module. Lemma 3.1.

For r ∈ + Z , p ∈ Z , f ( ∂ ) ∈ C [ ∂ ] , ∂f ( ∂ ) ∈ ∂ C [ ∂ ] , we get (1) I r ∂f ( ∂ ) = f ( ∂ + r ) I r ∂ ; (2) I r f ( ∂ ) = f ( ∂ + r ) I r ; (3) G p ∂f ( ∂ ) = f ( ∂ + p ) G p ∂ ; (4) G p f ( ∂ ) = f ( ∂ + p ) G p .Proof. (1) It is easy to get that I r L ∂ = ( L + r ) I r ∂ by the relations of L in (2.1).Recursively, we conclude that I r L n ∂ = ( L + r ) n I r ∂ for n ∈ Z + . Hence, I r ∂f ( ∂ ) = f ( ∂ + r ) I r ∂ for r ∈ + Z .Similarly, we obtain (2), (3) and (4). Lemma 3.2.

For r ∈ + Z , p ∈ Z , λ ∈ C ∗ , α ∈ C , one of the following two cases will takeplace. (a) G p = λ p ∂ , G r ∂ = − λ r ( ∂ + 2 rα ) , I r = − λ r α ; (b) G p = ( − p λ p ∂ , G r ∂ = λ r ( ∂ + 2 rα ) , I r = 2 λ r α .Proof. To prove this, we suppose G p = ∂f p ( ∂ ) ∈ ∂ C [ ∂ ] , I r = g r ( ∂ ) ∈ C [ ∂ ] , G p ∂ = h p ( ∂ ) ∈ C [ ∂ ] , where p ∈ Z , r ∈ + Z . For any m ∈ Z , by [ G , G m ] = 2 L m , we get G m ∂ = (cid:0) λ m ( ∂ + mα ) − ∂ f m ( ∂ ) (cid:1) . Using G m = L m , one can check that f m ( ∂ + m )(2 λ m ( ∂ + mα ) − ∂ f m ( ∂ )) = λ m ( ∂ + 2 mα ) . (3.2)Comparing the degree of ∂ in (3.2), one has f m ( ∂ ) = a m ∈ C (3.3)11or m ∈ Z . Taking (3.3) into (3.2), it is easy to get a m = λ m . (3.4)For any r ∈ + Z , from [ G r , G r ] = − L r , we have f r ( ∂ + r ) h r ( ∂ ) = − λ r ( ∂ + 2 rα ) . (3.5)This implies that either f r ( ∂ + r ) or h r ( ∂ ) is a nonzero constant. Suppose h r ( ∂ ) = d r ∈ C ∗ .For m ∈ Z , r ∈ Z + , then it follows from [ L m , G r ] ∂ = ( m − r ) G m + r ∂ that we get( r + m λ m d r = ( r − m d m + r . (3.6)Taking m = 2 r in (3.6), one has d r = 0, which leads to conﬂict. Then considering (3.5)again, we conclude f r ( ∂ + r ) = c r ∈ C ∗ , h r ( ∂ ) = − c r λ r ( ∂ + 2 rα ) . (3.7)For r ∈ Z + , by [ G , G r ] = rI r , we have g r ( ∂ ) = 1 rc r (( c r − λ r ) ∂ − rαλ r ) . (3.8)For r, s ∈ Z + , it follows from (3.8) that I r I s = 1 rsc r c s (( c s − λ s )( ∂ + r ) − sαλ s )(( c r − λ r ) ∂ − rαλ r ) . Based on [ I r , I s ] = 0, we deduce that r ( c s − λ s ) (cid:0) ( c r − λ r ) ∂ − λ r rα (cid:1) = s ( c r − λ r ) (cid:0) ( c s − λ s ) ∂ − λ s sα (cid:1) . This gives c r = ± λ r for r ∈ Z + . Combining with (3.4), we have f p ( ∂ ) = λ p or f p ( ∂ ) = ( − p λ p for p ∈ Z . We separately put them into (3.7) and (3.8), one can obtain (a) and (b).Now we present the main result of this section, which gives a complete classiﬁcation offree U ( h )-modules of rank 1 over L . Theorem 3.3.

Assume that V is an L -module such that the restriction of V as a U ( h ) -module is free of rank . Then up to a parity, V ∼ = M ( λ, α ) , or V ∼ = M − ( λ, α ) for λ ∈ C ∗ , α ∈ C . roof. We ﬁrst consider Lemma 3.2 (a). For r ∈ Z + , p ∈ Z , it follows from Lemma 3.1that we obtain G p f ( ∂ ) = f ( ∂ + p ) G p = λ p ∂f ( ∂ + p ) , (3.9) I r f ( ∂ ) = f ( ∂ + r ) I r = − λ r αf ( ∂ + r ) , (3.10) G r ∂f ( ∂ ) = f ( ∂ + r ) G r ∂ = − λ r ( ∂ + 2 rα ) f ( ∂ + r ) . (3.11)For r ∈ Z + , we can check that I r ∂f ( ∂ ) = G I r f ( ∂ ) + G r f ( ∂ ) = λ r ∂ (1 − α ) f ( ∂ + r ) . (3.12)Besides, for 0 = m ∈ Z , we conﬁrm that L m ∂f ( ∂ ) = L m G f ( ∂ ) = G L m f ( ∂ ) + m G m f ( ∂ ) = λ m ∂ ( ∂ + m ( α + 12 )) f ( ∂ + m ) (3.13)and G m ∂f ( ∂ ) = 2 m (cid:0) L m G ∂f ( ∂ ) − G L m ∂f ( ∂ ) (cid:1) = 2 m λ m ( m ∂ + m α ) f ( ∂ + m )= λ m ( ∂ + 2 mα ) f ( ∂ + m ) . (3.14)Then (3.11) and (3.14) give G p ∂f ( ∂ ) = ( − p λ p ( ∂ + 2 pα ) f ( ∂ + m ) (3.15)for p ∈ Z . Thus, (3.1), (3.9), (3.10), (3.12), (3.13) and (3.15) show that V ∼ = M ( λ, α ) as L -modules.Now consider Lemma 3.2 (b). From Lemma 3.1 and by the similar discussions, we deduce V ∼ = M − ( λ, α ). The theorem is proved. U ( t ) -modules of rank Let W = W ¯0 ⊕ W ¯1 be an L -module such that it is free of rank 2 as a U ( t )-module with twohomogeneous basis elements v and w . If the parities of v and w are the same, for r ∈ + Z then G ± r v = G ± r w = 0. Thus, L v = −

12 [ G r , G − r ] v = 0 , L w = −

12 [ G r , G − r ] w = 0 , U ( C L )-free. We know that v and w are diﬀerent parities. Set v = ¯0 ∈ W ¯0 and w = ¯1 ∈ W ¯1 . As a vector space, we have W ¯0 = C [ x ] ¯0 and W ¯1 = C [ y ] ¯1 .Clearly, W ¯0 and W ¯1 are both can be viewed as V -modules. According to Theorem 2.2(1), there exist λ, µ ∈ C ∗ , α, β ∈ C , f ( x ) ∈ C [ x ] and g ( y ) ∈ C [ y ] such that L m f ( x ) = λ m ( x + mα ) f ( x + m ) , (4.1) L m g ( y ) = µ m ( y + mβ ) g ( y + m ) . (4.2)For later use, two preliminary lemmas are presented as follows. Lemma 4.1.

Let λ, µ ∈ C ∗ , α, β ∈ C , f ( x ) ∈ C [ x ] , g ( y ) ∈ C [ y ] . We obtain λ = µ and one ofthe following two cases occurs. (i) G ¯0 = ¯1 , G ¯1 = x ¯0 , G ¯0 = λ ¯1 , G ¯1 = − λ ( x + α ) ¯0 , β = α + , or G ¯0 = ¯1 , G ¯1 = x ¯0 , G ¯0 = − λ ¯1 , G ¯1 = λ ( x + α ) ¯0 , β = α + ; (ii) G ¯0 = y ¯1 , G ¯1 = ¯0 , G ¯0 = − λ ( y + α − ) ¯1 , G ¯1 = λ ¯0 , β = α − , or G ¯0 = y ¯1 , G ¯1 = ¯0 , G ¯0 = λ ( y + α − ) ¯1 , G ¯1 = − λ ¯0 , β = α − .Proof. Suppose G p ¯0 = f p ( y ) ¯1 and G p ¯1 = g p ( x ) ¯0 . For p ∈ { , } , using G p ¯0 = f p ( x + p ) g p ( x ) ¯0 and L p ¯0 = λ p ( x + 2 pα )in [ G p , G p ] ¯0 = ( − p L p ¯0 , we check f p ( x + p ) g p ( x ) = ( − p λ p ( x + 2 pα ) . (4.3)Choosing p = 0 in (4.3), we immediately obtain f ( x ) = γ , g ( x ) = 1 γ x or f ( x ) = 1 γ x, g ( x ) = γ . (4.4)Up to a parity, we can assume γ = 1 without loss of generality. Then (4.4) can be rewrittenas f ( x ) = 1 , g ( x ) = x or f ( x ) = x, g ( x ) = 1 . Choosing p = in (4.3), we have f ( x + 12 ) = γ , g ( x ) = − γ λ ( x + α ) or f ( x + 12 ) = − γ λ ( x + α ) , g ( x ) = γ . (4.5)Similarly, by G ¯1 = − L ¯1 , we can conclude that g ( y + ) f ( y ) = − µ ( y + β ). Accordingto (4.5), it is easy to check that λ = µ, β = α + or λ = µ, β = α − . Then we have thefollowing four cases. 141) f ( x ) = 1 , g ( x ) = x, f ( x ) = γ , g ( x ) = − λγ ( x + α ) , β = α + ;(2) f ( x ) = 1 , g ( x ) = x, f ( x ) = − λγ ( x + α − ) , g ( x ) = γ , β = α − ;(3) f ( x ) = x, g ( x ) = 1 , f ( x ) = − λγ ( x + α − ) , g ( x ) = γ , β = α − ;(4) f ( x ) = x, g ( x ) = 1 , f ( x ) = γ , g ( x ) = − λγ ( x + α ) , β = α + . Claim 1. (2) and (4) do not occur.

First consider case (2). For 0 = m ∈ Z , we get G m ¯0 = 2 m ( L m G − G L m ) ¯0 = 2 m ( L m ¯1 − G λ m ( x + mα ) ¯0 )= − λ m ¯1 . (4.6)Taking (4.6) into ([ L m , G − m ]) ¯0 = m G ¯0 , which yields a contradiction by ¯1 = 0. Fromthe similar arguments, we obtain that (4) does not occur, too. The claim holds.Now we consider (1). For r ∈ + Z , write I r ¯0 = h r ( x ) ¯0 . By [ G , G ] ¯0 = I ¯0 , wehave h ( x ) = 2 γ (( γ − λ ) x − λα ) . (4.7)Then from [ L , I ] ¯0 = − I ¯0 , we observe h ( x ) = − λγ (cid:0) ( γ − λ )( 12 x + α ) + 12 λα (cid:1) . (4.8)Applying (4.7) and (4.8) to [ I , I ] ¯0 = 0, we see that( γ − λ )(( γ − λ ) x − λα ) = 3( γ − λ )(( γ − λ )( x + 2 α ) + λα ) , which implies γ = ± λ . Plugging this into (1), we have (i).By an identical process in (3), we know that the results of (ii). This completes theproof.Up to a parity, we only study (i) of Lemma 4.1.15 emma 4.2. For any m ∈ Z , r ∈ + Z , p ∈ Z , f ( x ) ∈ C [ x ] , g ( y ) ∈ C [ y ] , we have G p f ( x ) ¯0 = λ p f ( y + p ) ¯1 , G p g ( y ) ¯1 = ( − p λ p ( x + 2 pα ) g ( x + p ) ¯0 ,I r f ( x ) ¯0 = − αλ r f ( x + r ) ¯1 , I r g ( y ) ¯1 = λ r (1 − α ) g ( y + r ) ¯0 ;or G p f ( x ) ¯0 = ( − p λ p f ( y + p ) ¯1 , G p g ( y ) ¯1 = λ p ( x + 2 pα ) g ( x + p ) ¯0 ,I r f ( x ) ¯0 = 2 αλ r f ( x + r ) ¯1 , I r g ( y ) ¯1 = − λ r (1 − α ) g ( y + r ) ¯0 . Proof.

Let us ﬁrst consider the ﬁrst case of Lemma 4.1 (i) as G ¯0 = ¯1 , G ¯1 = x ¯0 , G ¯0 = λ ¯1 , G ¯1 = − λ ( x + α ) ¯0 , β = α + 12 . For 1 = m ∈ Z , we obtain G m + ¯1 = 2 m − (cid:0) L m G − G L m (cid:1) ¯1 = ( − m +1 λ m + ( x + (2 m + 1) α ) ¯0 . This combine with G ¯1 = − λ ( x + 3 α ) ¯0 , one has G m + ¯1 = ( − m +1 λ m + ( x + (2 m + 1) α ) ¯0 , (4.9)where m ∈ Z . For 0 = m ∈ Z , it is clear that G m ¯1 = 2 m (cid:0) L m G − G L m ) ¯1 = λ m ( x + 2 mα ) ¯0 . (4.10)From (4.9) and (4.10), we conclude G p f ( y ) ¯1 = ( − p λ p ( x + 2 pα ) f ( x + p ) ¯0 (4.11)for p ∈ Z . Similarly, we deduce that G p f ( x ) ¯0 = λ p f ( y + p ) ¯1 (4.12)for p ∈ Z . For any r ∈ + Z , according to (4.11) and (4.12), we compute that I r f ( x ) ¯0 = 1 r [ G , G r ] f ( x ) ¯0 = 1 r (cid:0) λ r G f ( y + r ) ¯1 + G r f ( y ) ¯1 (cid:1) = − λ r αf ( x + r ) ¯1 , (4.13) I r f ( y ) ¯1 = 1 r [ G , G r ] f ( y ) ¯1 = 1 r (cid:0) λ r G ( − r ( x + 2 rα ) f ( x + r ) ¯0 + G r xf ( x ) ¯0 (cid:1) = λ r (1 − α ) f ( y + r ) ¯1 . (4.14)16hen (4.11), (4.12), (4.13) and (4.14) show the ﬁrst case of the lemma. From the rest ofLemma 4.1 (i) and by the similar analysis methods, the other case can be obtained. Thelemma holds.According to (4.1), (4.2) and Lemma 4.2, a complete classiﬁcation of free U ( t )-modulesof rank 2 over L are given. Theorem 4.3.

Let W be an L -module such that the restriction of W as a U ( t ) -module isfree of rank . Then up to a parity, W ∼ = N ( λ, α ) , or W ∼ = N − ( λ, α ) for λ ∈ C ∗ , α ∈ C . The following results of the automorphism group of the twisted N = 2 superconformalalgebra appeared in [7]. Proposition 5.1.

Let m ∈ Z , r ∈ + Z , p ∈ Z . Then Aut( L ) = { ω b | b ∈ C ∗ } , where ω b ( L m ) = − b m L − m , ω b ( I r ) = b r I − r , ω b ( G p ) = b p √− G − p . We have the statement as follows.

Proposition 5.2.

Let m ∈ Z , r ∈ + Z , p ∈ Z , b, τ ∈ C ∗ , ω b ∈ Aut( L ) , t = ± . Assumethat M t ( λ, α ) b is an L -module M t ( λ, α ) after a twist by ω b . Then M t ( λ, α ) ∼ = M t ( τ, α ) b ,where τ p = b p λ − p .Proof. Take f ( ∂ ) ∈ M t ( λ, α ) ¯0 and ∂g ( ∂ ) ∈ M t ( λ, α ) ¯1 . LetΨ : M t ( λ, α ) −→ M t ( τ, α ) b f ( ∂ ) f ( − ∂ ) ∂g ( ∂ ) ∂g ( − ∂ )be the linear map from M t ( λ, α ) to M t ( λ, α ) b . It is evident that Ψ is an isomorphic mapping.It follows from ω b ( L m )Ψ( f ( ∂ )) = Ψ( ω b ( L m ) f ( ∂ )) , ω b ( L m )Ψ( ∂g ( ∂ )) = Ψ( ω b ( L m ) ∂g ( ∂ )) ,ω b ( I r )Ψ( f ( ∂ )) = Ψ( ω b ( I r ) f ( ∂ )) , ω b ( I r )Ψ( ∂g ( ∂ )) = Ψ( ω b ( I r ) ∂g ( ∂ )) ,ω b ( G p )Ψ( f ( ∂ )) = Ψ( ω b ( G p ) f ( ∂ )) , ω b ( G p )Ψ( ∂g ( ∂ )) = Ψ( ω b ( G p ) ∂g ( ∂ ))17hat we obtain ω b ( L m ) f ( − ∂ ) = b m λ − m ( ∂ + mα ) f ( − ∂ − m ) ,ω b ( L m )( −√− ∂g ( − ∂ )) = − b m λ − m √− ∂ ( ∂ + m ( α + 12 )) g ( − ∂ − m ) ,ω b ( I r ) f ( − ∂ ) = − ( t − b ) r λ − r αf ( − ∂ − r ) ,ω b ( I r )( −√− ∂g ( − ∂ )) = − ( t − b ) r λ − r (1 − α ) √− ∂g ( − ∂ − r ) ,ω b ( G p ) f ( − ∂ ) = ( t − b ) p λ − p ∂f ( − ∂ − p ) ,ω b ( G p )( −√− ∂g ( − ∂ )) = − ( − t − b ) p λ − p √− ∂ + 2 pα ) g ( − ∂ − p )for m ∈ Z , r ∈ + Z and p ∈ Z . Then M t ( τ, α ) b ∼ = M t ( λ, α ), where τ p = b p λ − p , p ∈ Z .We complete the proof. We ﬁrst recall the deﬁnition of weighting functor, which was introduced by J. Nilsson in [15].Let a be a Lie (super)algebra. Suppose that e h is a commutative subalgebra of a andAd( e h ) acts diagonally on a , namely, a := M µ ∈ e h ∗ a µ . Regard each µ ∈ e h ∗ as a homomorphism ¯ µ : U ( e h ) → C . We denote by Max( U ( e h )) the set ofmaximal ideals of U ( e h ). Deﬁnition 6.1.

For an a -module M , consider the U ( e h ) -module W ( M ) := M m ∈ Max( U ( e h )) M/mM = M µ ∈ e h ∗ M/ ker(¯ µ ) M. Choose α ∈ e h ∗ and v ∈ M . Deﬁne an action of x α ∈ a α on W ( M ) by x α ( v + ker(¯ µ ) M ) := x α v + ker( µ + α ) M. Then the map W between U ( a ) -mod and U ( a ) -mod is called the weighting functor . In the following, we deﬁne the weighting functor W for L by taking e h = t = C L . Theweight modules called intermediate series of the twisted N = 2 superconformal algebra wereinvestigated in [12]. 18 eﬁnition 6.2. Let σ ∈ C , m ∈ Z , r ∈ + Z , p, k ∈ Z , t = ± . A super intermediateseries module A t ( σ ) over L is a module with a basis { x k , y k | k ∈ Z } satisfying the actionas follows L m x k = ( − k + σn ) x k + m , L m y k = ( − k + m ( σ + 12 )) y k + m ,I r x k = − t r ( σ + 1) x k + r , I r y k = − t r (2 σ + 1) y k + r ,G p x k = t p y k + p , G p y k = ( − t ) p ( − k + (2 σ + 1) p ) x k + p . The following theorem obtained by a very natural extension of [12].

Theorem 6.3.

Let σ ∈ C , t, t ′ = ± . The L -module A t ( σ ) deﬁned as above. Then (1) the L -module A t ( σ ) is simple if σ = − , − ; (2) A t ( σ ) ∼ = A t ′ ( σ ′ ) if and only if t = t ′ , σ = σ ′ . We will describe the main results of this section.

Theorem 6.4.

Let λ ∈ C ∗ , α ∈ C , t = ± . As L -modules, we have W ( M t ( λ, α )) ∼ = A t ( α − .Proof. Let M t ( λ, α ) = M as in Deﬁnition 6.1. For each n ∈ Z , let µ n ∈ e h ∗ be thehomomorphism L = ∂ n. We see that ker( µ n ) = C [ ∂ ]( ∂ − n ), which is the maximalideal of C [ ∂ ] . Assume that is a generator of M t ( λ, α ). Setting w n = + ker( µ n ) M t ( λ, α ) ∈ M t ( λ, α ) ¯0 and v n = ∂ + ker( µ n ) M t ( λ, α ) ∈ M t ( λ, α ) ¯1 , we can check that L m w − n = λ m ( ∂ + mα ) + ker( µ − n − m ) M ( λ, α )= ( − n + m ( α − λ m w − ( n + m ) , (6.1) L m v − n = λ m ( ∂ + m ( α + 12 )) ∂ + ker( µ − n − m ) M ( λ, α )= ( − n + m ( α −

12 )) λ m v − ( n + m ) , (6.2) I r w − n = − t r λ r α + ker( µ − n − r ) M ( λ, α ) = − αt r λ r w − ( n + r ) , (6.3) I r v − n = t r λ r (1 − α ) ∂ + ker( µ − n − r ) M ( λ, α ) = (1 − α ) t r λ r v − ( n + r ) , (6.4) G p w − n = t p λ p ∂ + ker( µ − n − p ) M ( λ, α ) = t p λ p v − ( n + p ) , (6.5) G p v − n = ( − t ) p λ p ( ∂ + 2 pα ) + ker( µ − n − p ) M ( λ, α )= ( − t ) p ( − n + (2 α − p ) λ p w − ( n + p ) (6.6)19or any m ∈ Z , r ∈ + Z , p ∈ Z . We write b w p = λ p w − p , b v p = λ p v − p in (6.1)-(6.6), one gets L m b w n = (cid:0) − n + m ( α − (cid:1) b w n + m ,L m b v n = (cid:0) − n + m ( α −

12 ) (cid:1)b v n + m ,I r b w n = − t r α b w n + r , I r b v n = (1 − α ) t r b v n + r ,G p b w n = t p b v n + p , G p b v n = ( − t ) p (cid:0) − n + (2 α − p (cid:1) b w n + p . Then we conclude that W ( M t ( λ, α )) ∼ = A t ( α − Remark 6.5.

By Deﬁnition 6.2, it is clear that A t ( α − is simple if α = 0 , . Thenwe know that the weighting functor cannot keep the irreducibility of L -module M t ( λ, α ) to A t ( α − for α = . From Proposition of [15], we obtain that A t ( α − is a coherentfamily of degree . Corollary 6.6.

Let λ, λ ′ ∈ C ∗ , α, α ′ ∈ C , t, t ′ = ± . If α = α ′ or t = t ′ , then we obtain M t ( λ, α ) ≇ M t ′ ( λ ′ , α ′ ) .Proof. Suppose M t ( λ, α ) ∼ = M t ′ ( λ ′ , α ′ ). Then by Theorem 6.4 and Theorem 6.3 (2), onehas α = α ′ , t = t ′ , which leads to a contradiction. Acknowledgements

This work was supported by the National Natural Science Foundation of China (GrantNo.11801369).

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J. Algebra (2020), 538-555.[22] H. Yang, Y. Yao, L. Xia, On non-weight representations of the untwisted N = 2 supercon-formal algebras, J. Pure Appl. Algebra (2021), 106529.Haibo ChenSchool of Statistics and Mathematics, Shanghai Lixin University of Accounting and Finance,Shanghai 201209, [email protected] Dai chool of Mathematical Sciences, Guizhou Normal University, Guiyang 550001, [email protected] LiuThree Gorges Mathematics Research Center, China Three Gorges University, Yichang 443002,[email protected] of Mathematical Sciences, Guizhou Normal University, Guiyang 550001, [email protected] LiuThree Gorges Mathematics Research Center, China Three Gorges University, Yichang 443002,[email protected]