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2 Journal of Algebra Contents lists available at ScienceDirect Journal of Algebra Vertex operator algebras associated to type G affine Lie algebras Jonathan D. Axtell, Kyu-Hwan Lee Department of Mathematics, University of Connecticut, Storrs, CT , USA article info abstract Article history: Received 15 November 010 Available online 11 May 011 Communicated by Masaki Kashiwara Keywords: Vertex operator algebras Affine Lie algebras Admissible levels Irreducible representations Singular vectors In this paper, we study representations of the vertex operator algebra Lk, 0 at one-third admissible levels k = 5,, for the affine algebra of type G 1. We first determine singular vectors and then obtain a description of the associative algebra ALk, 0 using the singular vectors. We then prove that there are only finitely many irreducible ALk, 0-modules from the category O. Applying the AV -theory, we prove that there are only finitely many irreducible weak Lk, 0-modules from the category O and that such an Lk, 0-module is completely reducible. Our result supports the conjecture made by Adamović and Milas 1995 ]. 011 lsevier Inc. All rights reserved. Introduction Vertex operator algebras VOA are mathematical counterparts of chiral algebras in conformal field theory. An important family of examples comes from representations of affine Lie algebras. More precisely, if we let ĝ be an affine Lie algebra, the irreducible ĝ-module Lk, 0 with highest weight kλ 0, k C, is a VOA, whenever k h, the negative of the dual Coxeter number. The representation theory of Lk, 0 varies depending on values of k C. Ifk is a positive integer, the VOA Lk, 0 has only finitely many irreducible modules which coincide with the irreducible highest weight integrable ĝ-modules of level k, and the category of Z + -graded weak Lk, 0-modules is semisimple. If k / Q or k < h,categoriesoflk, 0-modules are quite different from those corresponding to positive integer values. or example, see 10,11]. or some rational values of k, the category of weak Lk, 0-modules which are in the category O as ĝ-modules has a structure similar to that for the category of Z + -graded weak modules for positive integer values. Such rational values are called admissible levels. This notion was defined in the * Corresponding author. -mail addresses: axtell@math.uconn.edu J.D. Axtell, khlee@math.uconn.edu K.-H. Lee /$ see front matter 011 lsevier Inc. All rights reserved. doi: /j.jalgebra

3 196 J.D. Axtell, K.-H. Lee / Journal of Algebra important works of Kac and Wakimoto 7,8]. Various cases have been studied with different generality by many authors. Adamović studied the case of admissible half-integer levels for type C 1 1]. The l case of all admissible levels of type A 1 1 was studied by Adamović and Milas ], and by Dong, Li and Mason ]. In his recent papers 1,15], Perše studied admissible half-integer levels for type A 1 l and B 1 l. In these developments, the AV -theory has played an important role. The associative algebra AV associated to a vertex operator algebra V was introduced by renkel and Zhu see 5,16]. It was shown that the irreducible modules of AV are in one-to-one correspondence with irreducible Z + - graded weak modules of V. This fact gives an elegant method for the classification of representations of V, and was exploited in the works mentioned above. In this paper, we study one-third admissible levels 5 Λ 0, Λ 0, Λ 0 for type G 1 adopting the method of 1,,1 15]. We first determine singular vectors Proposition. and then obtain a description of the associative algebra ALk, 0 in Theorem.6 using the singular vectors for k = 5,,. By constructing some polynomials in the symmetric algebra of the Cartan subalgebra, we find all the possible highest weights for irreducible ALk, 0-modules from the category O Proposition.6. As a result, in each case of k = 5,,, we prove that there are only finitely many irreducible ALk, 0-modules from the category O. Then it follows from the one-to-one correspondence in AV -theory that there are only finitely many irreducible weak Lk, 0-modules from the category O Theorem.7. In the case of irreducible Lk, 0-modules, our result provides a complete classification Theorem.10. We also prove that such an Lk, 0-module is completely reducible Theorem.1. Thus the VOA Lk, 0 is rational in the category O for k = 5,,. This result supports the conjecture made by Adamović and Milas in ], which suggests that Lk, 0 s are rational in the category O for all admissible levels k. Although some of our results may be generalized to higher levels k, the first difficulty is in the drastic growth of complexity in computing singular vectors, as one can see in Appendix A. It seems to be necessary to find a different approach to the problem for higher levels. The first-named author will consider singular vectors for other admissible weights in his subsequent paper. 1. Preliminaries 1.1. Vertex operator algebras Let V, Y, 1, ω be a vertex operator algebra VOA. This means that V is a Z-graded vector space, V = n Z V n, Y is the vertex operator map, Y, x : V nd V x, x 1 ]], 1 V 0 is the vacuum vector, and ω V is the conformal vector, all of which satisfy the usual axioms. See,,1] for more details. By an ideal in the vertex operator algebra V we mean a subspace I of V satisfying Y a, xi Ix, x 1 ]] for any a V. Given an ideal I in V such that 1 / I, ω / I, the quotient V /I naturally becomes a vertex operator algebra. Let M, Y M be a weak module for the vertex operator algebra V.WethushaveavectorspaceM and a map Y M, x : V nd Mx, x 1 ]], which satisfy the usual set of axioms cf. ]. or a fixed element a V,wewriteY M a, x = m Z amx m 1, and for the conformal element ω we write Y M ω, x = m Z ωmx m 1 = m Z L mx m. In particular, V is a weak module over itself with Y = Y V. A Z + -graded weak V -module is a weak V -module M together with a Z + -gradation M = n=0 M n such that amm n M n+r m 1 for a V r and m,n, r Z, where M n = 0forn < 0 by definition. A weak V -module M is called a V-module if L 0 acts semisimply on M with a decomposition into L 0 -eigenspaces M = α C M α such that for any α C, dimm α < and M α+n = 0forn Z sufficiently small. We define bilinear maps :V V V and :V V V as follows. or any homogeneous a V n,wewritedega = n, and for any b V,wedefine

4 J.D. Axtell, K.-H. Lee / Journal of Algebra x deg a a b = Res x Y a, xb, x and 1 + x deg a a b = Res x Y a, xb, x and extend both definitions by linearity to V V.DenotebyOV the linear span of elements of the form a b, and by AV the quotient space V /OV. ora V,denotebya] the image of a under the projection of V onto AV. Themapa a] will be called Zhu s map. The multiplication induces the multiplication on AV, and AV has a structure of an associative algebra. This fact can be found in 5,16]. Proposition 1.1. See 5]. Let I be an ideal of the vertex operator algebra V such that 1 / I, ω / I. Then the associative algebra AV /I is isomorphic to AV /AI,whereAI is the image of I in AV. Given a weak module M and homogeneous a V, we recall that we write Y M a, x = m Z amx m 1.Wedefineoa = adeg a 1 ndm and extend this map linearly to V. Theorem 1.. See 16]. 1 Let M = n=0 M n be a Z + -graded weak V -module. Then M 0 is an AV -module defined as follows: a] v = oav for any a Vandv M 0. Let U be an AV -module. Then there exists a Z + -graded weak V -module M such that the AV -modules M 0 and U are isomorphic. The equivalence classes of the irreducible AV -modules and the equivalence classes of the irreducible Z + -graded weak V -modules are in bijective correspondence. 1.. Affine Lie algebras Let g be a finite-dimensional simple Lie algebra over C, with a triangular decomposition g = n h n +.Let be the root system of g, h, + the set of positive roots, θ the highest root and, : g g C the Killing form, normalized by the condition θ, θ =. Denote by Π ={α 1,...,α l } the set of simple roots of g, and by Π ={h 1,...,h l } the set of simple coroots of g. The affine Lie algebra ĝ associated to g is the vector space equipped with the bracket operation ĝ = g C t, t 1] CK a t m, b t n] =a, b] t m+n + ma, bδ m+n,0 K, a, b g,m,n Z, together with the condition that K is a nonzero central element. We consider h as a subalgebra of ĝ, and set ĥ = h + CK ĝ. Let ĝ = ˆn ĥ ˆn + be the corresponding triangular decomposition of ĝ.

5 198 J.D. Axtell, K.-H. Lee / Journal of Algebra We set h to be the dual Coxeter number of ĝ. Denoteby the set of roots of ĝ, by + the set of positive roots of ĝ, and by Π the set of simple roots of ĝ. Wealsodenoteby re the set of real roots of ĝ and let re + = re +. The coroot corresponding to a real root α re will be denoted by α.let Q = α Π Zα be the root lattice, and let Q + = α Π Z +α Q.oranyλ ĥ,weset Dλ ={λ α α Q + }. We say that a ĝ-module M belongs to the category O if the Cartan subalgebra ĥ acts semisimply on M with finite-dimensional weight spaces and there exits a finite number of elements ν 1,...,ν k ĥ such that ν k i=1 Dν i for every weight ν of M. WedenotebyMλ the Verma module for ĝ with highest weight λ ĥ, and by Lλ the irreducible ĝ-module with highest weight λ. LetU be a g- module, and let k C. Wesetĝ + = g tct] and ĝ = g t 1 Ct 1 ].Letĝ + act trivially on U and K as scalar multiplication by k. Considering U as a g CK ĝ + -module, we have the induced ĝ-module Nk, U = Uĝ Ug CK ĝ+ U. or a fixed μ h,denotebyv μ the irreducible highest weight g-module with highest weight μ. DenotebyP + the set of dominant integral weights of g, and by ω 1,...,ω l P + the fundamental weights of g. WewillwriteNk, μ = Nk, V μ. Denoteby Jk, μ the maximal proper submodule of Nk, μ and Lk, μ = Nk, μ/ Jk, μ. WedefineΛ 0 ĥ by Λ 0 K = 1andΛ 0 h = 0forany h h. ThenNk, μ is a highest weight module with highest weight kλ 0 + μ, and a quotient of the Verma module MkΛ 0 + μ. We also obtain Lk, μ = LkΛ0 + μ. 1.. Admissible weights Let,re respectively,,re + be the set of real respectively, positive real coroots of ĝ, and Π the set of simple coroots. or λ ĥ,wedefine and we set,re λ = { α,re λ,α Z }, and,re λ,+ =,re λ,re +, Π λ = { α,re λ,+ α is not decomposable into a sum of elements from,re λ,+ }. Let Ŵ denote the Weyl group of ĝ. oreachα re,wehaveareflectionr α Ŵ.Defineρ ĥ in the usual way, and we recall the shifted action of an element w Ŵ on ĥ,givenbyw λ = wλ + ρ ρ. A weight λ ĥ is called admissible if λ + ρ,α / Z + for all α,re + and Q,re λ = Q Π. The irreducible ĝ-module Lλ is called admissible if the weight λ ĥ is admissible. Given a ĝ-module M from the category O, we call a weight vector v M a singular vector if ˆn +.v = 0. Proposition 1.. See 7]. Let λ be an admissible weight. Then Lλ = Mλ/ Uĝv α, α Π λ where v α Mλ is a singular vector of weight r α λ.

6 J.D. Axtell, K.-H. Lee / Journal of Algebra Proposition 1.. See 8]. Let M be a ĝ-module from the category O. If every irreducible subquotient Lν of M is admissible, then M is completely reducible. 1.. Nk, 0 and Lk, 0 as VOA s We identify the one-dimensional trivial g-module V 0 with C. Write 1 = 1 1 Nk, 0. The ĝ-module Nk, 0 is spanned by the elements of the form a 1 n 1 1 a m n m 11, where a 1,...,a m g and n 1,...,n m Z +,withan denoting the element a t n for a g and n Z. The vector space Nk, 0 admits a VOA structure, which we now describe. The vertex operator map Y, x : Nk, 0 ndnk, 0x, x 1 ]] is uniquely determined by defining Y 1, x to be the identity operator on Nk, 0 and Y a 11, x = n Z anx n 1 for a g. In the case that k h, the module Nk, 0 has a conformal vector ω = 1 k + h a i 1 1, dim g i=1 where {a i } i=1,...,dim g is an arbitrary orthonormal basis of g with respect to the normalized Killing form,. Then it is well known that the quadruple Nk, 0, Y, 1, ω defined above is a vertex operator algebra. Proposition 1.5. See 5]. The associative algebra ANk, 0 is canonically isomorphic to Ug. The isomorphism is given by : ANk, 0 Ug, a 1 n 1 1 a m n m 11 ] = 1 n 1+ +n m a 1 a m, for a 1,...,a m g and n 1,...,n m Z +. Since every ĝ-submodule of Nk, 0 is also an ideal in the VOA Nk, 0, the module Lk, 0 is a VOA for any k h. Proposition 1.6. See 1]. Assume that the maximal ĝ-submodule of Nk, 0 is generated by a singular vector v 0.Thenwehave A Lk, 0 = Ug/ v0 ], where v 0 ] is the two-sided ideal of Ug generated by v 0 ]. In particular, a g-module U is an ALk, 0-module if and only if v 0 ]U = 0.

7 00 J.D. Axtell, K.-H. Lee / Journal of Algebra Affine Lie algebra of type G 1.1. Admissible weights Let ± 1 ɛ 1 ɛ, ± 1 ɛ 1 ɛ, ± 1 ɛ ɛ, = ± 1 ɛ 1 ɛ ɛ, ± 1 ɛ ɛ 1 ɛ, ± 1 ɛ ɛ 1 ɛ be the root system of type G. We fix the set of positive roots + = 1 ɛ 1 ɛ, 1 ɛ 1 + ɛ + ɛ, 1 ɛ ɛ 1, 1 ɛ + ɛ 1 + ɛ, 1 ɛ ɛ, 1 ɛ ɛ 1 ɛ. Then the simple roots areα = 1 ɛ 1 ɛ and β = 1 ɛ 1 + ɛ + ɛ, and the highest root is θ = 1 ɛ ɛ 1 ɛ = α + β. Letg bethesimpleliealgebraoverc, associated with the root system of type G.Let 10, 01, 10, 01, H 10, H 01 be Chevalley generators of g, where 10 is a root vector for α, 01 is a root vector for β, and so on. We fix the root vectors: 11 = 10, 01 ], 1 = 1 11, 10 ]= 1 10, 01 ], 10 ], 1 = 1 1, 10 ]= , 01 ], 10 ], 10 ], = 1, 01 ]= , 01 ], 10 ], 10 ], 01 ], 11 = 01, 10 ], 1 = 1 10, 11 ]= 1 10, 01, 10 ] ], 1 = 1 10, 1 ]= , 10, 01, 10 ] ]], = 01, 1 ]= , 10, 10, 01, 10 ] ]]]..1 We set H ij = ij, ij ] for any positive root iα + jβ +. Then one can check that H ij is the coroot corresponding to iα + jβ, i.e. H ij = iα + jβ. or a complete multiplication table, we refer the reader to Table.1 in 6, p. 6], where we have X 1 = 10, X = 01, X = 11, X = 1, X 5 = 1, X 6 =, Y 1 = 10, Y = 01, Y = 11, Y = 1, Y 5 = 1, Y 6 =. All admissible weights for arbitrary affine Lie algebras have been completely classified in 8]. The next proposition provides a description of the vacuum admissible weights for G 1 at one-third levels. This is a special case of Proposition 1. in 9]. We provide a proof for completeness.

8 J.D. Axtell, K.-H. Lee / Journal of Algebra Lemma.. The weight λ n+i = n + i Λ 0 is admissible for n Z +,i= 1,, andwehave Π λ n+i = { δ α + β,α,β }, where δ is the canonical imaginary root. urthermore, λn+i + ρ,γ = 1 for γ = α,β; λn+i + ρ, δ α + β = n + i + 1 for i = 1,. Proof. We have to show λn+i + ρ,γ / Z + for any γ re + and Q,re λ n+i = Q Π. Any positive real root γ re + of ĝ is of the form γ = γ + mδ, form > 0 and γ, orm = 0 and γ +.Denoteby ρ the sum of fundamental weights of g. Then we can choose ρ = h Λ 0 + ρ = Λ 0 + ρ. We have λn+i + ρ,γ = n + + i Λ 0 + ρ, γ + mδ = m n + + i + ρ, γ. γ, γ If m = 0, then it is trivial that λ n+i, γ / Z +. Suppose that m 1. If γ, γ = and m 0 mod, then λ n+i + ρ, γ / Z +. If γ, γ =, and m 0 mod, then m, and since ρ, γ forany γ, wehave λn+i + ρ,γ = m n + + i + ρ, γ n = n +, which implies λ n+i + ρ, γ / Z +.If γ, γ =,then ρ, γ 5.Wehave λn+i + ρ,γ = m n + + i = n +, + ρ, γ n ρ, γ n which implies λ n+i + ρ, γ / Z +.Thus, λ n+i + ρ, γ / Z + for any γ re +. One can easily see that,re λ n+i,+ = { mδ + γ m > 0, m 0 mod, γ, γ = } { mδ + γ m > 0, γ, γ = / } +.

9 0 J.D. Axtell, K.-H. Lee / Journal of Algebra Then we obtain Π λ n+i = { δ α + β,α,β }, and we see that Q,re λ n+i = Q Π λ n+i = Q Π. Through direct calculations, we get λn+i + ρ,γ = 1 forγ = α,β, and λn+i + ρ, δ α + β = n + i Singular vectors In what follows, let ĝ be the affine Lie algebra of type G 1 and Uĝ its universal enveloping algebra. We write X i m = X m i for elements in Uĝ. Weset and define a = 1 1, b = , c = H , w = , u = 1 a b, and v = 9 a ab c. The following proposition determines singular vectors for the first three admissible weights, i.e. 5 Λ 0, Λ 0, Λ 0, respectively. Proposition.. The vector v k Nk, 0 is a singular vector for the given value of k: u.1 for k = 5, v k = v + w.1 for k =, uv w.1 for k =. The proof will be given in Appendix A. As one can see in the proof, the computational difficulty increases as the level k goes up. A different approach will be used in a subsequent work of the firstnamed author on higher levels... Description of Zhu s algebra Proposition.. The maximal ĝ-submodule Jk, 0 of Nk, 0 is generated by the vector v k for k = 5,,, respectively, where v k s are given in Proposition.. Proof. Let λ n+i = + n + i Λ 0 = kλ 0 as before. It follows from Proposition 1. and Lemma. that the maximal submodule of the Verma module Mλ n+i is generated by three singular vectors with weights r δ α+β λ n+i, r α λ n+i, r β λ n+i, respectively.

10 J.D. Axtell, K.-H. Lee / Journal of Algebra We consider the three cases n = 0, i = 1, k = 5/; n = 0, i =, k = /; n = 1, i = 1, k = /. In each case, there is a singular vector u k Mλ n+i of weight r δ α+β.λ n+i, whose image under the projection of Mλ n+i onto Nk, 0 is the singular vector v k giveninproposition.. The other singular vectors have weights r α λ n+i = λ n+i λ n+i + ρ,α α = λ n+i α, and r β λ n+i = λ n+i λ n+i + ρ,β β = λ n+i β, so the images of these vectors under the projection of Mλ n+i onto Nk, 0 are 0 from the definition. Therefore the maximal submodule of Nk, 0 is generated by the singular vector v k, i.e. Jk, 0 = Uĝv k. Now we consider the image of a singular vector v k under Zhu s map ] : Nk, 0 A Nk, 0 = Ug, which is defined in Section 1. We recall that the vertex algebra Nk, 0 is linearly isomorphic to the associative algebra Uĝ. We thus have an induced map from Uĝ to Ug and a commutative diagram of linear maps: Uĝ Nk, 0 Ug A Nk, 0. We will identify Nk, 0 with Uĝ and ANk, 0 with Ug. Wehave: a]= 1, b]= , c]= H We also have: u]= 1 a] b], v]= 9 a] a]b] c], w]=0, ] uv w =u]v]= 7 a]5 5 9 a] b] a] c]+a]b] + b]c]..5 The following theorem is now a consequence of Propositions 1.6 and..

11 0 J.D. Axtell, K.-H. Lee / Journal of Algebra Theorem.6. The associative algebra ALk, 0 is isomorphic to Ug/I k,wherei k is the two-sided ideal of Ug generated by the vector v k ],where. Irreducible modules u] for k = 5, v k ]= v] for k =, uv] for k =. In this section we adopt the method from 1,,1 15] in oder to classify irreducible ALk, 0- modules from the category O by solving certain systems of polynomial equations..1. Modules for associative algebra ALk, 0 Denote by L the adjoint action of Ug on Ug defined by X L f =X, f ] for X g and f Ug. We also write ad X f = X L f =X, f ]. Then ad X is a derivation on Ug. Let Rk be a Ug-submodule of Ug generated by the vector v k ], where v k ] is given in Theorem.6. It is straightforward to see that Rk is an irreducible finite-dimensional Ug-module isomorphic to V k + 7α + β.letrk 0 be the zero-weight subspace of Rk. Proposition.1. See 1,]. Let V μ be an irreducible highest weight Ug-module with highest weight vector v μ for μ h. Then the following statements are equivalent: 1 V μ is an ALk, 0-module, Rk V μ = 0, Rk 0 v μ = 0. Let r Rk 0. Then there exists a unique polynomial p r Sh, where Sh is the symmetric algebra of h, such that Set Pk 0 ={p r r Rk 0 }.Thenwehave: r v μ = p r μv μ. Corollary.. There is a bijective correspondence between 1 the set of irreducible ALk, 0-modules V μ from the category O,and the set of weights μ h such that pμ = 0 for all p Pk 0... Polynomials in Pk 0 We now determine some polynomials in the set Pk 0 for the cases k = 5, k =, k =, respectively. We will use some computational lemmas which we collect and prove in Appendix B. Lemma. Case: k = 5. We let 1 qh = H 1 H 1 +, p 1 H = H 10 H 10 1,and p H = 1 H 11H H 01. Then qh, p 1 H, p H P 5 0.

12 J.D. Axtell, K.-H. Lee / Journal of Algebra Proof. 1 We show that 1 1 Lu] CqH mod Ugn + for some C 0. Using Lemma B.1 and Lemma B.1, we have 1 1 u]= 1 L 1 1 L a] b] 1!! H 1H H 1!! 1 H 1H 1 + mod Ugn+, which is what we wanted to show. We will show that 10 1 Lu] Cp 1 H mod Ugn + for some C 0. We again use Lemma B.1 and Lemma B.1 to obtain: L a] b]! 1 H 10H 10 1 p 1H mod Ugn+. In this case we show that 11 Lu] Cp H mod Ugn + for some C 0. Similarly to the first two cases we compute: L a] b]! H 11H H 01 Cp H mod Ugn+. We now give polynomials for the next case. Lemma. Case: k =. Let 1 qh = 9 H 1H 1 1H 1 + H 1 H 1 + H 01 H 01 +, p 1 H = H 10 H 10 1H 10, p H = 9 H 11H 11 1H H 01 H. Then p 1 H, p H, qh P 0. Proof. 1 We show that Lv] CqH mod Ugn + for some constant C 0. By Lemma B.1, we have: L L 9 a] a]b] c] By Lemma B.1, we thus have:!6! 9 H 1H 1 1H 1! 6!!! H b] L c] L mod Ugn v]!6! L 9 H 1H 1 1H 1 +!6!H 1 H 1 +!6!H 01 H 01 + CqH mod Ugn+.

13 06 J.D. Axtell, K.-H. Lee / Journal of Algebra We will show that 10 1 Lv] Cp 1 H mod Ugn + for some constant C 0. Using Lemma B.1, we obtain: By Lemma B.1, we thus have 10 1 L 9 a] a]b] c] 9! H 10 H 10 1H 10 +!!!! H b] L 10 1 c] L mod Ugn L 9 a] a]b] c] 9! H 10 H 10 1H 10 Cp 1 H mod Ugn+. inally, we show that 11 Lv] Cp H mod Ugn + for some constant C 0. Since H 11 + H 1 = H,wehave 11 L v 9! H 11 H 11 1H 11!!!! H b] L 11 c] L! 9 H 11H 11 1H 11 + H 11 H 01 + H 01 H 1 +! 9 H 11H 11 1H H 01 H Cp H mod Ugn+. The last case is presented below. Lemma.5 Case: k =. We let qh = 7 H 1H 1 1H 1 H 1 H H 1H 1 H 1 H 1 + H 1 H 1 H 01 H H 1 H 1 H H 1 H 10 H H 1 H 01 H H 1 H 01 H 01 +, p 1 H = H 10 H 10 1H 10 H 10 H 10, p H = 7 H 11H 11 1H 11 H 11 H H 11 H 11 H 11 H 01 + H 11 H 11 H 01 H H 11 H 01 H 01 1 H 11 H 11 H H 01 H 01 1H 1 +. Then p 1 H, p H, qh P 0.

14 J.D. Axtell, K.-H. Lee / Journal of Algebra Proof. irst recall from.5 that uv w ] =u]v]= 7 a]5 5 9 a] b] a] c]+a]b] + b]c]. We will show that Lu]v] 5!10!qH mod Ugn +. Using Lemmas B.1, B.11, we have: 10 1 L u]v] = 10 1 L 7 a]5 5 9 a] b] a] c]+a]b] + b]c] = 7 Now using Lemma B., we obtain: 1 10! 10!! !!! 1 1 b] L 10!! 6! c] L + 10!!8! L b] b]c] = 7 10! !! 1 10! 1 b] L 6! c] L 10! 8! L b] b]c] L u]v] = 7 5!H 1H 1 1H 1 H 1 H ! 9! H 1 H 1 H 1 1! 5!! H 1 H ! 6 1 c] L 5!! H ! 8 1 L b] ! Combining this with Lemmas B.1, B.1, B.1, we obtain: 1 10! 1 1 b] L L u]v] 7 5!H 1H 1 1H 1 H 1 H !! H 1 H 1 H 1 H 1 5!! H 1 H 1!H 01 H L b]c]. 5!! H 1! H 1 H 11 + H 10 H H 01 H ! H 01 H 01 + H 1 5!qH mod Ugn+. The proofs for p 1 H and p H are similar, and we omit the details.

15 08 J.D. Axtell, K.-H. Lee / Journal of Algebra initeness of the number of irreducible modules We are now able to obtain the following result for the associative algebra ALk, 0. or convenience, if μ h,wewriteμ ij = μh ij. We will identify μ h with the pair μ 10, μ 01. Proposition.6. There are finitely many irreducible ALk, 0-modules from the category O for each of k = 5,,. Moreover, the possible highest weights μ = μ 10, μ 01 for irreducible ALk, 0-modules are as follows: 1 if k = 5,thenμ = 0, 0, 0, or 1, ; if k =,thenμ = 0, 0, 0,, 0, 1, 1, 0, 1, or, 5 ; if k =,thenμ = 0, 0, 0,, 0, 1, 0, 1, 0, 1, 1, 0, 1,, 1,,, 0,, 5,, or, 7. Proof. 1 It follows from Corollary. that highest weights μ h of irreducible AL 5, 0- modules satisfy pμ = 0 for all p P 0 5. Lemma. implies that p 1μ = p μ = qμ = 0 for such weights μ. Letμ h.theequationp 1 μ = 0is μ 10 μ 10 1 = 0, which implies μ 10 = 0or1. irst suppose μ 10 = 0. Then from qμ = 0 we must have μ 01 = 0 or. Similarly, from p μ = 0, we also get μ 01 = 0or.Sotheweightμ must be of the form μ = μ 10, μ 01 = 0, 0 or 0, in this case. Now suppose μ 10 = 1. The equation qμ = 0givesμ 01 = or, and the equation p μ = 0 givesμ 01 = 0 or. So the only possibility is μ = μ 10, μ 01 = 1,. Altogether, this gives only three possible weights μ such that p 1 μ = p μ = qμ = 0: μ = μ 10,μ 01 = 0, 0, 0, or 1,. Similarly to the part 1, we use the polynomials of Lemma.. Using a computer algebra system, we calculate the common zeros of the polynomials qh, p 1 H, p H to obtain the following list of possible highest weights: μ = μ 10,μ 01 = 0, 0, 0,, 0, 1,1, 0, 1, or, 5. or this part, we use Lemma.5. Using a computer algebra system, we again compute the common zeros of the polynomials qh, p 1 H, p H to obtain the following list of possible highest weights: μ = μ 10,μ 01 = 0, 0, 0,, 0, 1, 0, 1,0, 1, 1, 0, 1,, 1,,, 0,, 5,, or, 7. Now we apply the AV -theory Theorem 1., and obtain our main result in the following theorem.

16 J.D. Axtell, K.-H. Lee / Journal of Algebra Theorem.7. There are finitely many irreducible weak modules from the category O for each of the following simple vertex operator algebras: L 5, 0, L, 0, L, 0. Remark.8. This theorem provides further evidence for the conjecture of Adamović and Milas in ], mentioned in the introduction. urthermore, if Lλ is an irreducible module of the VOA Lk, 0, for k = 5,,or, then we recall from Section 1. that we must have Lλ = LkΛ 0, μ for the values of μ h giveninproposition.6. In the case of irreducible Lk, 0-modules, we obtain a complete classification. We state this result in the following proposition and theorem. Proposition.9. The complete list of irreducible finite-dimensional ALk, 0-modules V μ for each k is as follows: 1 if k = 5,thenVμ = V 0, if k =,thenvμ = V 0 or V ω 1, if k =,thenvμ = V 0, Vω 1, Vω,orVω 1, where ω 1, ω are the fundamental weights of g. Proof. Among the list of weights in Proposition.6, we need only to consider dominant integral weights, i.e. those weights μ = m 1,m with m 1,m Z +. Notice that the weights of the singular vectors v k ] are ω 1,ω 1 and 5ω 1, respectively. Considering the set of weights of V μ listed above, we see that each singular vector v k ] annihilates the corresponding modules V μ. Nowthe proposition follows from Proposition 1.6. We again apply the AV -theory Theorem 1., and obtain the following theorem. Theorem.10. The complete list of irreducible Lk, 0-modules Lk, μ for each k is as follows: 1 if k = 5,thenLk, μ = Lk, 0, if k =,thenlk, μ = Lk, 0 or Lk, ω 1, if k =,thenlk, μ = Lk, 0, Lk, ω 1, Lk, ω,orlk, ω 1... Semisimplicity of weak modules from the category O In this subsection we show that the category of weak Lk, 0-modules from the category O is semisimple. Lemma.11. Assume that λ = kλ 0 + μ for k = 5,,,whereμ h is one of the values given in Proposition.6 for each k. Then the weights λ are admissible. Proof. The proof is essentially the same as Lemma.. Let us write Π 0 ={δ α + β, α,β }, Π 1 ={δ α + β, α,α + β }, and Π ={δ θ, α,α + β }. Since the proof for the other cases are similar, we consider only the case k = 5. rom Lemma., we already know that λ = 5 Λ 0 + μ is admissible for μ = 0, 0, with Π λ = Π 0. If μ = 0,, we have to show that 5 Λ 0 + μ + ρ,γ / Z + for any γ re + and Q,re λ = Q Π.

17 10 J.D. Axtell, K.-H. Lee / Journal of Algebra Recall that ρ = Λ 0 + ρ; alsoγ re + m = 0 and γ +.Wethenhave: must have the form γ = γ + mδ, form > 0 and γ, or 5 Λ 0 + μ + ρ,γ 7 = Λ 0 + μ + ρ, γ + mδ = 7 γ, γ m + μ, γ + ρ, γ. We may then check that 5 Λ 0 + μ + ρ, γ 1,sothat 5 Λ 0 + μ + ρ, γ / Z +.Onemayalso verify that Π λ = Π,re 1 so that Q λ = Q Π. Similarly, one can show that λ = 5 Λ 0 + μ is admissible for μ = 1, and that Π λ = Π. Theorem.1. Let M be a weak Lk, 0-module from the category O, for k = 5,,or.ThenMis completely reducible. Proof. Let Lλ be an irreducible subquotient of M. ThenLλ is an Lk, 0-module, and we see from Remark.8 that λ must be a weight of the form kλ 0 + μ, where μ is given in Proposition.6 for k = 5,,, respectively. rom Lemma.11 it follows that such a λ is admissible. Now Proposition 1. implies that M is completely reducible. Acknowledgments We thank A. eingold, M. Primc and the referee for helpful comments. Appendix A. Proof of Proposition. In this appendix, we prove Proposition.. We first give a few lemmas. Lemma A.1. 1 We have a, 10 0 ] = 1 1, b, 10 0 ] = , c, 10 0 ] = , u, 10 0 ] = 0, v, 10 0 ] = 0, w, 10 0 ] = 0. ach of the elements a, b, c, u, v, w Uĝ commutes with Proof. 1 Using the multiplication table in.1, it is easy to see a, 10 0] = 1 1. Next,we have b, 10 0 ] = , 10 0 ] = , 10 0 ] + 1 1, 10 0 ] , 10 0 ] 1, 10 0 ] 10 1 =

18 J.D. Axtell, K.-H. Lee / Journal of Algebra Starting with the definition c, 10 0 ] = H , 10 0 ], we consider each term separately and obtain , 10 0 ] = , 10 0 ] , 10 0 ] , 10 0 ] = , 1 1 1H 01 1, 10 0 ] = H 01 1, 10 0 ] , 10 0 ] H , 10 0 ] 1 1H 01 1 = , and , 10 0 ] = , 10 0 ] + 1 1, 10 0 ] 01 1 = , 10 0 ] Therefore, we obtain c, 10 0 ] = Next, we get u, 10 0 ] = 1 a, 10 0 ] b, 10 0 ] = 1 a a, 10 0 ] + 1 a, 10 0 ] a b, 10 0 ] = = 0, and v, 10 0 ] = a, 10 0 ] ab, 10 0 ] c, 10 0 ] 9 = a b, 10 0 ] a, 10 0 ] b c, 10 0 ] =

19 1 J.D. Axtell, K.-H. Lee / Journal of Algebra { } { } = 0. inally, it is easy to see w, 10 0]=0. The equalities a, 01 0]=0, b, 01 0]=0, c, 01 0]=0 can be proved as in the part 1, and we omit the details. Then it immediately follows that u, 01 0]=0 and v, 01 0]=0. Since w = 1 a, b], we also obtain w, 010]=0. Lemma A.. We have a, 1 ] = 11 0, b, 1 ] = H 0 + K , c, 1 ] = H H 0 + K H 01 1H 0 + K H 01 1, u, 1 ] = K H 0, v, 1 ] = H 11 1 a a 10 1 a b, 1 ] c, 1 ], w, 1 ] = H H 1 + K 1. Proof. We only prove the equalities for b, 1] and u, 1]. The other equalities can be proved similarly. We obtain b, 1 ] = , 1 ] = , 1 ] + 1 1, 1 ] , 1 ] 1, 1 ] 10 1 = { H 0 } K 10 1 = H 0 + K , and u, 1 ] = 1 a a, 1 ] + 1 a, 1 ] a b, 1 ] =

20 J.D. Axtell, K.-H. Lee / Journal of Algebra H 0 K = K H 0. We need one more lemma. Lemma A.. We have the following commutator relations: H 0, v w ] = v w, 01 0, v w ] = 0, 11 0, v ] 1 w = a b a 1 1H a a , 1 0, v ] w = a + b H a 1 a a H H Proof. Since the proofs of the other equalities are similar, we only provide a proof for Wefirst have 11 0, v ] w = 11 0, ] 9 a ab c w. Considering each term separately, we get 11 0, a ] = 6a a 1, 11 0, ab ] = 11 0, a ] b + a 11 0, b ] = 10 1b a { 1 1H a }, 11 0, c ] = a 1 1H a , 11 0, w ] = a Using two more relations 10 1, b ] = and H 11 = H 10 + H 01, one can now obtain the result for 11 0, v w].

21 1 J.D. Axtell, K.-H. Lee / Journal of Algebra We now prove Proposition.. or convenience, we state the proposition again: Proposition A.. The vector v k Nk, 0 is a singular vector for the given value of k: u.1 for k = 5, v k = v + w.1 for k =, uv w.1 for k =. Proof. To show that each vector v k is a singular vector, it suffices to check that 10 0.v k = 0, 01 0.v k = 0, and 1.v k = 0foreachk. Assume that k = 5. By Lemma A.1, we obtain 10 0.v k = 10 0u.1 = u, 10 0 ].1 = 0, and similarly we get 01 0.v k = 0. Now we consider 1 and obtain by Lemma A. 1.v k = u, 1 ].1 = 0. Assume that k =. It follows from Lemma A.1 that 100.v k = 0 and 01 0.v k = 0. We also obtain from Lemma A. 1.v k = v + w, 1 ] = H a k + 1a k k H H 1 k 1 = = 0, a a H H H 1 where we drop.1 from the notation and use the equalities H 11 = H 10 + H 01 and H = H 10 + H 01. Assume that k =. We will continue to drop.1 from the notation. It again follows from Lemma A.1 that 10 0.v k = 0 and 01 0.v k = 0. We now consider 1 and have 1.v k = 1, uv w ] = 1, u ] v w + u 1, v w ]. We first compute 1, u]v w. We use Lemma A. and obtain:

22 J.D. Axtell, K.-H. Lee / Journal of Algebra , ] u v w = k v w v w v w v w 10 1H 0v w = k v w , v ] w , v w ] , v w ] 10 1 H 0, v w ]. Now using Lemma A. and the fact that H 1 = H 10 + H 01 along with the relation a, b]=w, we obtain the following: 1, ] u v w = k 1 a 10 1 k ba 10 1 k c 9 k + 8 w u 1 1H u u 1 1H k + 1 1a 1 + k b 1 k + a 1 = 9 a ba u 1 1H u u 1 1H u 1 = ua u 1 1H u u 1 1H u 1, where the second equality is obtained by substituting k =. Now we finally compute u 1, v w]. rom Lemma A. and H 11 = H 10 + H 01, we obtain: u 1, v ] w = 6k + 10u k + ua u 1 1H k + 8u 1 1H k u 1 = 6u ua u 1 1H u 1 1H 01 1 u 1, where we again substitute k =. Now it is clear that 1.v k = 1, u ] v w + u 1, v w ] = 0.

23 16 J.D. Axtell, K.-H. Lee / Journal of Algebra Appendix B. Lemmas for construction of polynomials The following results will be useful. Lemma B.1. See 1]. Let X g and let Y 1,...,Y m Ug.Then X n L Y n X k 1...Y m = 1 k 1...k L Y 1... X k m L Y m, m n where k 1...k n = n! k 1! k m!. k 1,...,k m Z + m ki =n Proof. This can be seen most readily by considering an exponential generating function. Given a derivation D of Ug, wemayformthegeneratingfunction expdt = 1 + Dt + D! t + nd Ug t]]. Applying this to a Y Ug, weobtainanelementexpdty Ugt]]. The lemma is a direct consequence of the fact that expdt satisfies the identity expdty 1 Y n = expdty 1 expdty n. B. See 1]. To obtain the lemma, replace D with the adjoint action X L = ad X in q. B., and equate the coefficient of t n on both sides. inally, multiplying both coefficients by n!, we obtain the identity in the lemma. Lemma B.. 1 m ij L m ij m!h ijh ij 1 H ij m Ug ij, for all iα + jβ +. Suppose X Ug 0, the zero-weight subspace of Ug.ThenX n Ug if and only if X Ugn +. or Y Ug and n > r > 0,wehave n ij L r ij Y ij Ug + n! n r! H ij n + r H ij n + 1 n r ij L Y + Ug ij. Proof. Part 1 follows from direct computation and part follows by considering a PBW basis given in triangular form for Ug 0. or part, we consider an exponential generating function. or simplicity, let us write, H,, in place of ij, H ij, ij.wethenhave: exp adt r Y = exp adt r exp adt Y = + Ht t r exp adt Y. B. One can check r + Ht t r r Ugt]] + 1 i H ih i 1 H r + 1 i t r+i. i i=0 B.5 or convenience, we introduce the notation x i = xx 1 x i + 1 for i > 0, and x 0 = 1. Then we have

24 J.D. Axtell, K.-H. Lee / Journal of Algebra x i = 1 i x + i 1 i, m m x + y m = x i y m i. i i=0 B.6 B.7 We obtain the following identity using B.6 and B.7: x n + r r = 1 r n r x r + 1 r r r = 1 r n r i x r + 1 i r i i=0 r r = 1 r n r i 1 r i x i r i i = i=0 i=0 r r 1 i n r i x i r i. i B.8 Using this notation we combine qs. B. and B.5 to write: exp adt r Y Ugt]] + r r 1 i H i r i i t r+i exp adt Y. i i=0 Taking the coefficient of t n on both sides gives: 1 n! adn r Y Ug + Ug + r r 1 i H i r i i 1 i n r i! adn r i Y i=0 r r 1 i 1 H i r i i n r i! adn r Y + Ug. B.9 i=0 With the substitution x = H, we obtain from B.8 H n + r r = r r 1 i n r! i n r i! H i r i. B.10 i=0 After multiplying B.9 by n!, we use the identityb.10 to obtain ad n r Y Ug + n! n r! H n + r rad n r Y + Ug. This proves part. The following lemmas will be needed for the construction of certain polynomials.

25 18 J.D. Axtell, K.-H. Lee / Journal of Algebra Lemma B.11. The following identities hold in Ug.irstwehave: Next we have: 1 L a]= 1, 1 L b]=! , 6 1 c]= 6! L H 01 01, and 1 a]= 5 L 1 b]= 7 L 1 c]=0. L inally we have: 1 L a]= 10, 1 L b]=! , 1 c]=! L 1 H , and 1 a]= L 1 b]= L 1 c]=0. L L a]= 11, L b]=! , c]=! L 01 H , and a]= L b]= L c]=0. L Proof. Using Lemma B.1, we have: 1 L b]= 1 1 L 1 1L L 1L L 1 1 L 11 1 L 1 L 10 = =! 1 11! 10. We also have: L b]= 1 L L 11 1 L 1 L = = 0. The other cases are proved similarly.

26 J.D. Axtell, K.-H. Lee / Journal of Algebra Lemma B.1. The following identities hold in Ug.irstwehave: Next: L a]= H 1, 1 1! 1 L b] H 1 mod Ugn+, 1 1 6! 6 1 L c]!h 01H 01 + mod Ugn L a]=h 10, b] 1 L 10! 1 L c] 0 mod Ugn +. inally: 1 11 L a]=h 11, 11 L b] 6H 01 Proof. Using Lemmas B.1, B.11, we have: mod Ugn+, 1 11! L c] 6H 01H 1 + mod Ugn+. 1 1! 1 b]= L 1 L = 1 0 L L 1L L L 1 1L L L H H = = H 11 H 10 H 1 mod Ugn+. The other cases follow in the same way.

27 0 J.D. Axtell, K.-H. Lee / Journal of Algebra Lemma B.1. Suppose that n, r, s, t Z + and n = r + s + t. Then the following hold in Ug: n 1 n 1 L a] r b] s c] t 1 r n! n! n r! n r! H 1 n + 1 H 1 n + r n r 1 n r 1 L b] s c] t, n 10 n 1 L a] r b] s c] t n! H 10 n + 1 H 10 n + r n r 10 n r! n r 1 L b] s c] t, n 11 n L a] r b] s c] t n! H 11 n + 1 H 11 n + r n r 11 n r! n r L b] s c] t, where all the congruences are modulo Ugn +. Proof. We prove only the first case. Using Lemma B.1 we have: n 1 L a] r b] s c] t = n! r!s + 6t! n! = r!n r! n! = r n r! = r 1 L r!! r 1 1 a] r s+6t 1 L b] s c] t L a] r n r 1 L b] s c] t L a] r n r 1 L b] s c] t n! r n r! 1 r n r 1 L b] s c] t = 1 r n! n r! 1 r n r 1 L b] s c] t, since 1 La]= 5 1 Lb]= 7 1 Lc]=0 and 1 La]= 1 by Lemma B.11. Then we use Lemma B. with Y = n r 1 L b] s c] t to obtain: n 1 n 1 L a] r b] s c] t = 1 r n! n 1 n r! L r n r 1 1 L b] s c] t 1 r n! n! n r! n r! H 1 n + r H 1 n + 1 n r 1 n r 1 L b] s c] t + 1 Ug + Ug 1. Now it follows from Lemma B. that we have

28 J.D. Axtell, K.-H. Lee / Journal of Algebra n 1 n 1 L a] r b] s c] t 1 r n! n! n r! n r! H 1 n + r H 1 n + 1 n r 1 n r 1 L b] s c] t mod Ugn+. We give one more lemma. Lemma B.1. The following hold: L b]!8! H 1 H 11 + H 10 H H 01 H , L b]c] 5!10!H01 H 01 + H 1, L b] L b]c] 0, L b]! 9H 01 H 01 1 H 01 H 11, where all the congruences are modulo Ugn +. L b]c] 5! 6H 01 H 01 1H 1 +, Proof. We prove the first part only. rom Lemma B.11, we have: We thus obtain: 8 1 L b] 8 = 1 L b] = 8! = 8! ! 8 1 L b] = 1 L This equals the following element modulo Ugn + :

29 J.D. Axtell, K.-H. Lee / Journal of Algebra H H , 00 where we have omitted the term Ug, which belongs to Ugn + by Lemma B., as well as similar terms which also belong to Ugn +. 1 We now see that!8! Lb] is equal to the following element modulo Ugn + : H H Again modulo Ugn +, this is equal to 6H 01 H 10 + H 1 + H 10 H H 11 H H 1 H 10 H 1 H 1 18H 01 + H 10 H H 10 H 10 + H 10H 1. After simplifying terms in this expression, we finally obtain: 1 1!8! 8 1 L b] H 1 H 11 6H 01 H H 10 H 10 1 mod Ugn+. References 1] D. Adamović, Some rational vertex algebras, Glas. Mat. Ser. III ] D. Adamović, A. Milas, Vertex operator algebras associated to modular invariant representations for A 1 1, Math. Res. Lett ] C. Dong, H. Li, G. Mason, Vertex operator algebras associated to admissible representations of ŝl, Comm. Math. Phys ] I. renkel, J. Lepowsky, A. Meurman, Vertex Operator Algebras and the Monster, Pure Appl. Math., vol. 1, Academic Press, Boston, ] I.B. renkel, Y. Zhu, Vertex operator algebras associated to representations of affine and Virasoro algebras, Duke Math. J ] W. ulton, J. Harris, Representation Theory. A irst Course, Grad. Texts in Math. Readings in Mathematics, vol. 19, Springer-Verlag, New York, ] V.G. Kac, M. Wakimoto, Modular invariant representations of infinite-dimensional Lie algebras and superalgebras, Proc. Natl. Acad. Sci. USA ] V.G. Kac, M. Wakimoto, Classification of modular invariant representations of affine algebras, in: Infinite-Dimensional Lie Algebras and Groups, Luminy-Marseille, 1988, in: Adv. Ser. Math. Phys., vol. 7, World Sci. Publ., Teaneck, NJ, 1989, pp ] V.G. Kac, M. Wakimoto, On rationality of W -algebras, Transform. Groups

30 J.D. Axtell, K.-H. Lee / Journal of Algebra ] D. Kazhdan, G. Lusztig, Tensor structures arising from affine Lie algebras. I, II, J. Amer. Math. Soc , ] D. Kazhdan, G. Lusztig, Tensor structures arising from affine Lie algebras. III, IV, J. Amer. Math. Soc , ] J. Lepowsky, H. Li, Introduction to Vertex Operator Algebras and Their Representations, Progr. Math., vol. 7, Birkhäuser, Boston, Inc., Boston, MA, 00. 1] A. Meurman, M. Primc, Annihilating fields of standard modules of sl, C and combinatorial identities, Mem. Amer. Math. Soc ] O. Perše, Vertex operator algebras associated to type B affine Lie algebras on admissible half-integer levels, J. Algebra ] O. Perše, Vertex operator algebras associated to certain admissible modules for affine Lie algebras of type A, Glas. Mat. Ser. III ] Y. Zhu, Modular invariance of characters of vertex operator algebras, J. Amer. Math. Soc