Simplicity of a vertex operator algebra whose Griess algebra is the Jordan algebra of symmetric matrices
aa r X i v : . [ m a t h . QA ] J a n Simplicity of a vertex operator algebra whose Griessalgebra is the Jordan algebra of symmetric matrices
Hidekazu Niibori
Graduate School of Pure and Applied Sciences, University of Tsukuba,Tsukuba, Ibaraki 305-8571, Japan (e-mail: [email protected] ) andDaisuke Sagaki Institute of Mathematics, University of Tsukuba,Tsukuba, Ibaraki 305-8571, Japan (e-mail: [email protected] ) Abstract
Let r ∈ C be a complex number, and d ∈ Z ≥ a positive integer greater thanor equal to 2. Ashihara and Miyamoto introduced a vertex operator algebra V J ofcentral charge dr , whose Griess algebra is isomorphic to the simple Jordan algebraof symmetric matrices of size d . In this paper, we prove that the vertex operatoralgebra V J is simple if and only if r is not an integer. Further, in the case that r is an integer (i.e., V J is not simple), we give a generator system of the maximalproper ideal I r of the VOA V J explicitly. Let V = L n ∈ Z ≥ V n be a vertex operator algebra (VOA for short) over the field C ofcomplex numbers with Y ( · , z ) : V → (End V )[[ z, z − ]] the vertex operator. As usual,for each v ∈ V , we define v m ∈ End V , m ∈ Z , by: Y ( v, z ) = P m ∈ Z v m z − m − . Itfollows from the axiom of a VOA that V becomes a C -algebra with the product givenby u · v := u v ∈ V for u, v ∈ V . In addition, we know (see [FLM, § § V = 1 and dim V = 0, then the C -algebra V is commutative (but notnecessarily associative), which we call the Griess algebra of V .Various kinds of commutative C -algebras appear as the Griess algebras of VOAs. TheGriess algebra of the moonshine VOA V ♮ is isomorphic to the 196884-dimensional, com-mutative C -algebra introduced by Griess [G], whose automorphism group is isomorphicto the Monster sporadic simple group (see [FLM, p.319]); the name “Griess algebra” isderived from this fact. Also, for a given associative, commutative C -algebra A equipped1ith an A -invariant bilinear form, Lam [L1] constructed a VOA whose Griess algebrais isomorphic to A . Further, in [L2], [AM], [As], they constructed some VOAs whoseGriess algebras are (simple) Jordan algebras; for the definition of a Jordan algebra, see § § d ( C ) of symmetricmatrices of size d ∈ Z ≥ with entries in C . Because the Jordan algebra Sym d ( C ) has astrong connection to symmetric cones and zeta functional equations (see [FK]), Ashihara,Miyamoto (see [AM, Introduction]), and the authors of this paper expect that the resultsin [AM] and this paper contribute a VOA theoretical approach to the theory of symmetriccones and zeta functional equations.An essential difference between the VOA introduced in [L2, § V J ) is their central charges. The central charge of theformer is equal to the (fixed) positive integer d ∈ Z ≥ , the size of symmetric matrices. Onthe other hand, the central charge of the later is equal to dr , where r is an arbitrary com-plex number (see Theorem 2.5 below). In general, the structure and the representationtheory of a VOA deeply depends on its central charge. For example, it is well-known thatthe simplicity of a Virasoro VOA M c, / h L − i (with notation in [W]) and the rationalityof a simple Virasoro VOA V c (with notation in [W]) depend on their central charges c (see[W] and also [DMZ]). Moreover, rational Virasoro VOAs (i.e., V c of special central charge c , such as V / ) and their irreducible modules play very important roles in the theory ofVOAs. So, as in the case of Virasoro VOAs, it is quite natural and important to studyhow the VOA V J introduced in [AM] depends on its central charge dr . In this paper,we study the condition of r ∈ C for the VOA V J to be simple. The main result of thispaper is the following theorem, which means that the simplicity of V J also depends on itscentral charge. Theorem.
Keep the notation above. The VOA V J is simple if and only if r is not aninteger. Many important VOAs are obtained as the nontrivial simple quotients of nonsimpleVOAs (e.g., rational Virasoro VOAs, and VOAs associated to integrable highest weightmodules over affine Lie algebras). So we are interested in the simple quotient V J /I r with r ∈ Z rather than the VOA V J . When we study the structure of V J /I r and itsrepresentation theory, it is very important to determine some relations in V J /I r inducedby the maximal proper ideal I r . In this paper, as a first step for studying the simple VOA V J /I r , we will give a generator system of the maximal proper ideal I r of the VOA V J explicitly, whose elements are singular vectors for a certain Lie algebra L (1) r , and have ahigh symmetry (see § §
2, we recall the definition of the Griess algebra ofa VOA, the definition of a Jordan algebra, and the construction of the VOA V J introducedby Ashihara and Miyamoto [AM]. Then we state our main theorem (Theorem 2.6), andthe plan how we prove the theorem. In § §
6, following the plan, we will prove some keypropositions (Propositions 3.1, 3.4, 4.1, 5.1, 6.1); our main theorem follows immediatelyfrom these propositions. In §
6, we also give a generator system of the maximal ideal I r of the VOA V J explicitly when r is an integer, that is, when V J is not simple. Acknowledgments.
The authors would like to express their sincere gratitude to Profes-sor Masahiko Miyamoto, Professor Toshiyuki Abe, Professor Hiroki Shimakura, ProfessorHiroshi Yamauchi, and Dr. Takahiro Ashihara for valuable discussions.
Let (cid:0) V = L n ∈ Z ≥ V n , Y ( · , z ) , , ω (cid:1) be a vertex operatoralgebra (VOA for short), with Y ( · , z ) : V → (End V )[[ z, z − ]] the vertex operator, ∈ V the vacuum element, and ω ∈ V the Virasoro element (for the details about VOAs, see,e.g., [LL]). As usual, for each v ∈ V , we define v m ∈ End V , m ∈ Z , by: Y ( v, z ) = P m ∈ Z v m z − m − . For a, b ∈ V , we define a · b := a b . Then it follows from the axiom of aVOA that a · b ∈ V for every a, b ∈ V , i.e., V becomes a C -algebra with · the product.In addition, if V = C and V = { } , then the C -algebra V is commutative (see [FLM, § § V the Griess algebra of V . Note that theGriess algebra of a VOA is not necessarily associative. Let us recall the definition of a Jordan algebra. For the detailsabout Jordan algebras, see, e.g., [Al1], [Al2], and [J].
Definition 2.1.
Let J be a C -algebra with the product a · b ( a, b ∈ J ). The C -algebra J is called a Jordan algebra if a · b = b · a and a · ( b · a ) = ( a · b ) · a hold for every a, b ∈ J .Let Sym d ( C ) be the set of symmetric matrices of size d ∈ Z ≥ with entries in C . It iswell-known that Sym d ( C ) becomes a (simple) Jordan algebra, where the product is givenby: A · B = ( AB + BA ) for A, B ∈ Sym d ( C ) (see also [L2, Theorem 6 B]). V J . Let (and fix) d ∈ Z ≥ . In this subsection, we recall a VOA V J intro-duced by Ashihara and Miyamoto [AM], whose Griess algebra is isomorphic to the Jordanalgebra Sym d ( C ) of symmetric matrices. 3et b h be an (infinite-dimensional) vector space over C with a linear basis (cid:8) v i ( m ) | ≤ i ≤ d, m ∈ Z (cid:9) ∪ { c } , and define a Lie bracket on b h by:[ v i ( m ) , v j ( n )] = δ m + n, δ i,j m c for 1 ≤ i, j ≤ d and m, n ∈ Z , [ c , b h ] = { } . (2.1)Denote by U ( b h ) the universal enveloping algebra of the Lie algebra b h , and let U ( b h ) / h c − i be the quotient algebra with respect to the two-sided ideal h c − i of U ( b h ) generated by c − ∈ U ( b h ). We define a subspace L of U ( b h ) / h c − i as follows: For 1 ≤ i, j ≤ d and m, n ∈ Z , we set v ij ( m, n ) := v i ( m ) v j ( n ) mod h c − i . (2.2)Let B := (cid:8) v ii ( m, n ) | ≤ i ≤ d and m, n ∈ Z with m ≤ n (cid:9) ∪ (cid:8) v ij ( m, n ) | ≤ i < j ≤ d and m, n ∈ Z (cid:9) . Then it follows from the Poincar´e-Birkhoff-Witt theorem that
B ∪ { ∈ U ( b h ) / h c − i} isa linearly independent subset of U ( b h ) / h c − i . We set L := (cid:0) Span C B (cid:1) ⊕ C ⊂ U ( b h ) / h c − i . Remark . Let 1 ≤ i, j ≤ d , and m, n ∈ Z . It can be easily seen from the definition(2.1) of the Lie bracket on b h that v ij ( m, n ) = v ji ( n, m ) if i = j , or if i = j and m = − n,v ii ( m, − m ) = v ii ( − m, m ) + m. (2.3)In particular, v ij ( m, n ) ∈ L for all 1 ≤ i, j ≤ d and m, n ∈ Z .We see by direct computation that [ x, y ] = xy − yx is contained in L for every x, y ∈ L ,and hence L becomes a Lie algebra with respect to the natural Lie bracket. Now, let (andfix) r ∈ C be an (arbitrary) complex number. For each x, y ∈ L , we define[ x, y ] r := π ([ x, y ]) + rπ ([ x, y ]) , (2.4)where π : L ։ Span C B and π : L ։ C denote projections from L onto Span C B and C , respectively. Then we know from [AM, § · , · ] r is a Lie bracket on L . Let usdenote by L r the Lie algebra L with the new Lie bracket [ · , · ] r . Example . As an example, let us compute [ v ii ( m, n ) , v ii ( − n, − m )] r for 1 ≤ i ≤ d and m, n ∈ Z > with m ≤ n . By direct computation and (2.3), we see that in L ,[ v ii ( m, n ) , v ii ( − n, − m )]= n (1 + δ m,n ) v ii ( − m, m ) + m (1 + δ m,n ) v ii ( − n, n ) + mn (1 + δ m,n ) . · , · ] r , we obtain[ v ii ( m, n ) , v ii ( − n, − m )] r = π ([ v ii ( m, n ) , v ii ( − n, − m )]) + rπ ([ v ii ( m, n ) , v ii ( − n, − m )])= n (1 + δ m,n ) v ii ( − m, m ) + m (1 + δ m,n ) v ii ( − n, n ) + rmn (1 + δ m,n ) . (2.5)We now set B + := (cid:8) v ij ( m, n ) ∈ B | m ∈ Z ≥ or n ∈ Z ≥ (cid:9) , B − := (cid:8) v ij ( m, n ) ∈ B | m, n ∈ Z < (cid:9) , and L + r := (cid:0) Span C B + (cid:1) ⊕ C . It is easily seen that L + r is a Lie subalgebra of L r . Let C bea one-dimensional L + r -module such that x · = 0 for all x ∈ B + , and s · = s for each s ∈ C ⊂ L + r . Denote by M r the L r -module induced from the L + r -module C , that is, M r := U ( L r ) ⊗ U ( L + r ) C . Here we give a linear basis B of M r by using the Poincar´e-Birkhoff-Witt theorem.Take (and fix) a total ordering ≻ on the set B − . Denote by S the set of finite sequencesof elements of B − that is weakly decreasing with respect to the total ordering ≻ . For x = ( x p (cid:23) x p − (cid:23) · · · (cid:23) x ) ∈ S with x q ∈ B − for 1 ≤ q ≤ p , we set w ( x ) := x p x p − · · · x ∈ M r . In view of the Poincar´e-Birkhoff-Witt theorem, B := { w ( x ) | x ∈ S} is a linear basis ofthe L r -module M r . Remark . We see from the definition (2.4) of the Lie bracket on L r that xy = yx forall x, y ∈ B − . Therefore, if y , y , . . . , y p ∈ B − , then y y · · · y p = w ( x ) ∈ B , where x ∈ S is the sequence of length p obtained by arranging y , y , . . . , y p in the weaklydecreasing order with respect to the total ordering ≻ . Also, we note that if m, n ∈ Z < ,then v ij ( m, n ) = v ji ( n, m ) for every 1 ≤ i, j ≤ d (see Remark 2.2).If x = ( x p (cid:23) x p − (cid:23) · · · (cid:23) x ) ∈ S with x q = v i q j q ( m q , n q ) ∈ B − for 1 ≤ q ≤ p , thenwe define the degree of w ( x ) ∈ B by:deg( w ( x )) = − p X q =1 ( m q + n q ) ∈ Z ≥ . Then the L r -module M r admits the degree space decomposition as follows: M r = M n ∈ Z ≥ ( M r ) n , where ( M r ) n := Span C (cid:8) b ∈ B | deg b = n (cid:9) . M r ) = C , and ( M r ) = { } . (2.6)Define an operator L ijr ( m ) ∈ End( M r ) for 1 ≤ i, j ≤ d and m ∈ Z by: L ijr ( m ) = X h ∈ Z v ij ( m − h, h ) if i = j or m = 0 , v ii (0 ,
0) + X h ∈ Z > v ii ( − h, h ) if i = j and m = 0 , (2.7)and set ω ijr := L ijr ( − ∈ ( M r ) , and ω := d X i =1 ω iir ∈ ( M r ) . Remark that L ijr ( m ) = L jir ( m ) for every 1 ≤ i, j ≤ d and m ∈ Z (see Remark 2.2),and hence that ω ijr = ω jir for every 1 ≤ i, j ≤ d . Let J := (cid:8) ω ijr , | ≤ i, j ≤ d (cid:9) ⊂ M r , and let V J be the subspace of M r spanned by all elements of the form: L i j r ( m ) L i j r ( m ) · · · L i p j p r ( m p ) with p ≥
0, and 1 ≤ i q , j q ≤ d , m q ∈ Z for 1 ≤ q ≤ p .Then, V J also admits the degree space decomposition induced from that of M r , i.e., V J = L n ∈ Z ≥ ( V J ) n with ( V J ) n := V J ∩ ( M r ) n for n ∈ Z ≥ . We should remark that( V J ) = C and ( V J ) = { } by (2.6).Define a map Y ( · , z ) : J →
End( V J )[[ z, z − ]] by: Y ( ω ijr , z ) = X m ∈ Z L ijr ( m ) z − m − for 1 ≤ i, j ≤ d,Y ( , z ) = id V J . (2.8)The following theorem is the main result of [AM]. Theorem 2.5.
Keep the notation above. The map Y ( · , z ) : J →
End( V J )[[ z, z − ]] canbe uniquely extended to a linear map Y ( · , z ) : V J → End( V J )[[ z, z − ]] in such a way thatthe quadruple (cid:16) V J = L n ∈ Z ≥ ( V J ) n , Y ( · , z ) , , ω (cid:17) becomes a VOA of central charge dr ,with the vacuum element, and ω the Virasoro element. Furthermore, the Griess algebraof V J is isomorphic to the Jordan algebra Sym d ( C ) of symmetric matrices. The purpose of this paper is to determine the condition of r ∈ C for the VOA V J tobe simple. The following theorem is the main result of this paper. Theorem 2.6.
Keep the notation above. The VOA V J is simple if and only if r ∈ C isnot an integer, that is, r ∈ C \ Z .
6e will prove Theorem 2.6 as follows. First, in §
3, we will show that V J ⊂ M r is,in fact, identical to the whole of M r (Proposition 3.1), and then prove that the VOA V J (= M r ) is simple if and only if M r is irreducible as an L r -module (Proposition 3.4). Let L (1) r be a Lie subalgebra of L r generated by (cid:8) v ( m, n ) | m, n ∈ Z with m ≤ n (cid:9) ⊂ B , andset M (1) r = U ( L (1) r ) ⊂ M r . In §
4, it will be shown that the L r -module M r is irreducibleif and only if M (1) r is irreducible as an L (1) r -module (Proposition 4.1). In §
5, we will provethat if r ∈ C \ Z , then M (1) r is an irreducible L (1) r -module, and hence V J is a simpleVOA (Proposition 5.1). Finally, in §
6, we will give some singular vectors of the L (1) r -module M (1) r explicitly in the case that r ∈ Z (Proposition 6.1), which implies that M (1) r is reducible, and hence V J is not simple. V J and irreducibility of M r . V J and M r . As in the previous section, we fix d ∈ Z ≥ and r ∈ C . This subsection is devoted to proving the following proposition. Proposition 3.1.
The subspace V J ⊂ M r is identical to the whole of M r , that is, V J = M r holds. In order to prove Proposition 3.1, we need some technical lemmas.
Lemma 3.2. (1)
For each ≤ i, j ≤ d , we have v ij ( − , − = 2 L ijr ( − . (3.1)(2) Let ≤ i, j ≤ d with i = j , and m, n ∈ Z < . Then, v ij ( m − , n ) = − m L iir ( − v ij ( m, n ) . (3.2)(3) Let ≤ i, j ≤ d with i = j , and m, n ∈ Z < . Then, v ii ( m − , n ) = 2 m ( m − n + 1) L iir (0) L ijr ( − v ij ( n, m ) . (3.3) Proof. (1) By the definition (2.7) of L ijr ( m ),2 L ijr ( − = X h ∈ Z v ij ( − − h, h ) . Note that v ij ( − − h, h ) = v ji ( h, − − h ) for all h ∈ Z (see Remark 2.2). Since v ij ( m, n ) = 0 if v ij ( m, n ) ∈ B + , it follows that v ij ( − − h, h ) = 0 unless h = − L ijr ( − = v ij ( − , − , and hence (3.1).72) As in the proof of part (1), we can easily show that L ijr ( − = 12 X h ∈ Z v ij ( − − h, h ) = 0 (3.4)for all 1 ≤ i, j ≤ d . Hence, L iir ( − v ij ( m, n ) = [ L iir ( − , v ij ( m, n )] since L iir ( − = 0 by (3.4)= 12 X h ∈ Z [ v ii ( − − h, h ) , v ij ( m, n )] . By direct computation (as in Example 2.3), we see that[ v ii ( − − h, h ) , v ij ( m, n )] r = δ − − h + m, ( − − h ) v ij ( h, n ) + δ h + m, hv ij ( − − h, n ) . Therefore,12 X h ∈ Z [ v ii ( − − h, h ) , v ij ( m, n )] = 12 X h ∈ Z (cid:8) δ − − h + m, ( − − h ) v ij ( h, n ) + δ h + m, hv ij ( − − h, n ) (cid:9) = 12 (cid:8) ( − m ) v ij ( m − , n ) + ( − m ) v ij ( m − , n ) (cid:9) = ( − m ) v ij ( m − , n ) . Thus we obtain L iir ( − v ij ( m, n ) = ( − m ) v ij ( m − , n ), and hence (3.2).(3) We have L ijr ( − v ij ( n, m ) = [ L ijr ( − , v ij ( n, m )] since L ijr ( − = 0 by (3.4)= 12 X h ∈ Z [ v ij ( − − h, h ) , v ij ( n, m )] . As in Example 2.3, we see that[ v ij ( − − h, h ) , v ij ( n, m )] r = δ h + m, hv ii ( n, − − h ) + δ − − h + n, ( − − h ) v jj ( h, m ) . Therefore we get12 X h ∈ Z [ v ij ( − − h, h ) , v ij ( n, m )] = 12 X h ∈ Z (cid:8) δ h + m, hv ii ( n, − − h ) + δ − − h + n, ( − − h ) v jj ( h, m ) (cid:9) = 12 (cid:8) ( − m ) v ii ( n, m − + ( − n ) v jj ( n − , m ) (cid:9) , L ijr ( − v ij ( n, m ) = − m v ii ( n, m − − n v jj ( n − , m ) . Now, since L iir (0) = 0, it follows that L iir (0) L ijr ( − v ij ( n, m ) = − m L iir (0) v ii ( n, m − − n L iir (0) v jj ( n − , m ) = − m L iir (0) , v ii ( n, m − − n L iir (0) , v jj ( n − , m )] . It can be easily checked that [ L iir (0) , v jj ( n − , m )] = 0 since i = j . Thus, − m L iir (0) , v ii ( n, m − − n L iir (0) , v jj ( n − , m )] = − m L iir (0) , v ii ( n, m − = (cid:26) − m v ii (0 , , v ii ( n, m − | {z } =0 − m X h ∈ Z > [ v ii ( − h, h ) , v ii ( n, m − (cid:27) . Since n, m − ∈ Z < , it follows that[ v ii ( − h, h ) , v ii ( n, m − r = δ h + n, hv ii ( − h, m −
1) + δ h + m − , hv ii ( n, − h )for h ∈ Z ≥ . Therefore, − m X h ∈ Z > [ v ii ( − h, h ) , v ii ( n, m − = − m X h ∈ Z > (cid:8) δ h + n, hv ii ( − h, m −
1) + δ h + m − , hv ii ( n, − h ) (cid:9) = − m (cid:8) ( − n ) v ii ( n, m −
1) + ( − m + 1) v ii ( n, m − (cid:9) = m ( m + n − v ii ( m − , n ) . Thus we obtain L iir (0) L ijr ( − v ij ( n, m ) = m ( m + n − v ii ( m − , n ) , and hence (3.3). This completes the proof of the lemma. Lemma 3.3. (1)
Let ≤ i, j ≤ d with i = j , and m, n ∈ Z < . Then, v ij ( m, n ) = αL iir ( − − m − L jjr ( − − n − L ijr ( − (3.5)9 or some α ∈ C \ { } . (2) Let ≤ i ≤ d , and m, n ∈ Z < with m ≤ − . Take ≤ j ≤ d with j = i arbitrarily.Then, v ii ( m, n ) = βL iir (0) L ijr ( − L iir ( − − n − L jjr ( − − m − L ijr ( − (3.6) for some β ∈ C \ { } .Proof. (1) We prove (3.5) by induction on − m − n (note that − m − n ≥ − m − n = 2,that is, m = n = −
1, then (3.5) follows immediately from (3.1). Assume that − m − n > m < −
1. Then, by (3.2), we have v ij ( m, n ) = − m + 1 L iir ( − v ij ( m + 1 , n ) . Applying the inductive assumption to the right-hand side of the equation above, we obtain v ij ( m, n ) = − m + 1 L iir ( − v ij ( m + 1 , n ) = − m + 1 L iir ( − (cid:8) αL iir ( − − m − L jjr ( − − n − L ijr ( − (cid:9) = − αm + 1 L iir ( − − m − L jjr ( − − n − L ijr ( − , where α ∈ C \ { } . Thus we have proved part (1).(2) Using (3.3) and (3.5), we have v ii ( m, n ) = 2( m + 1)( m + n ) L iir (0) L ijr ( − v ij ( n, m + 1) by (3.3)= 2 β ( m + 1)( m + n ) L iir (0) L ijr ( − L iir ( − − n − L jjr ( − − m − L ijr ( − by (3.6) , where β ∈ C \ { } . Thus we have proved part (2), thereby completing the proof ofLemma 3.3. Proof of Proposition 3.1.
We set U := Span C ( L i j r ( m ) L i j r ( m ) · · · L i p j p r ( m p ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) p ≥ , and 1 ≤ i q , j q ≤ d,m q ∈ {− , − , } for 1 ≤ q ≤ p ) . In order to prove Proposition 3.1, it suffices to show that U = M r . Indeed, it is obviousfrom the definition of V J that U ⊂ V J . Hence, if U = M r holds, then M r = U ⊂ V J ⊂ M r ,which implies that V J = M r . Claim.
We have xU ⊂ U for every x ∈ B . roof of Claim. Fix x ∈ B . It suffices to prove that xL i j r ( m ) L i j r ( m ) · · · L i p j p r ( m p ) ∈ U for every p ≥ ≤ i q , j q ≤ d , m q ∈ {− , − , } for 1 ≤ q ≤ p . We prove thisby induction on p . It is obvious that x = 0 ∈ U if x ∈ B + . Also, we see from equations(3.5) and (3.6) that x is contained in U if x ∈ B − . Thus the assertion holds when p = 0.Assume that p >
0. Then we have xL i j r ( m ) L i j r ( m ) · · · L i p j p r ( m p ) =[ x, L i j r ( m )] L i j r ( m ) · · · L i p j p r ( m p ) + L i j r ( m ) (cid:8) xL i j r ( m ) · · · L i p j p r ( m p ) (cid:9) . (3.7)Since xL i j r ( m ) · · · L i p j p r ( m p ) is contained in U by the inductive assumption, and since m ∈ {− , − , } , it follows that the second term of the right-hand side of (3.7) iscontained in U . Now we deduce from the definition of the Lie bracket on L r and thedefinition (2.7) of L ijr ( m ) that [ x, L i j r ( m )] can be written in the form:[ x, L i j r ( m )] = α y + α y + · · · + α s y s + β for some α , α , . . . , α s , β ∈ C and y , y , . . . , y s ∈ B . By substituting this into the firstterm of the right-hand side of (3.7), we see that[ x, L i j r ( m )] L i j r ( m ) · · · L i p j p r ( m p ) = X ≤ s ′ ≤ s α s ′ y s ′ L i j r ( m ) · · · L i p j p r ( m p ) | {z } ∈ U by the inductive assumption + β L i j r ( m ) · · · L i p j p r ( m p ) | {z } ∈ U , and hence that the first term also is contained in U . Therefore we conclude that theleft-hand side of (3.7) is contained in U , thereby completing the proof of Claim.The claim above implies that U is an L r -submodule of M r which contains . Hencewe conclude that U = M r , thereby completing the proof of Proposition 3.1. V J and the irreducibility of M r . Inthis subsection, we prove the following proposition.
Proposition 3.4.
The VOA V J (= M r ) is simple if and only if M r is irreducible as an L r -module. First let us show the following lemma, needed in the proof of Proposition 3.4.11 emma 3.5.
Let ≤ i, j ≤ d with i = j , and m, n ∈ Z < . The vertex operator Y ( v ij ( m, n ) , z ) of v ij ( m, n ) ∈ M r = V J is given by : Y ( v ij ( m, n ) , z ) = ( − − m − n × X l ∈ Z (X k ∈ Z (cid:18) l + n − k − m − (cid:19)(cid:18) k − n − − n − (cid:19) v ij ( l + m + n + 1 − k, k ) ) z − l − . (3.8) Proof.
We prove the lemma by induction on − m − n (note that − m − n ≥ − m − n =2, that is, m = n = −
1, then it follows that Y ( v ij ( − , − , z ) = 2 Y ( L ijr ( − , z ) by (3.1)= 2 Y ( ω ijr , z ) = 2 X l ∈ Z L ijr ( l ) z − l − by (2.8) and Theorem 2.5= X l ∈ Z (X k ∈ Z v ij ( l − k, k ) ) z − l − = X l ∈ Z (X k ∈ Z v ij ( l − − k, k ) ) z − l − . Therefore the equation (3.8) holds if − m − n = 2. Assume that − m − n >
2. ByRemark 2.2, we may assume that m < −
1. It follows from (3.2) that Y ( v ij ( m, n ) , z ) = 1 − m − Y ( L iir ( − v ij ( m + 1 , n ) , z ) . By using the commutator formula (see [LL, p. 54]), we deduce that( L iir ( − v ij ( m + 1 , n ) ) l = (( ω iir ) v ij ( m + 1 , n ) ) l = [( ω iir ) , ( v ij ( m + 1 , n ) ) l ]= [ L iir ( − , ( v ij ( m + 1 , n ) ) l ] . Also, it follows from the inductive assumption that( v ij ( m + 1 , n ) ) l = ( − − m − − n × (X k ∈ Z (cid:18) l + n − k − m − (cid:19)(cid:18) k − n − − n − (cid:19) v ij ( l + m + n + 2 − k, k ) ) . Combining these equations, we get Y ( v ij ( m, n ) , z ) = ( − − m − − n − m − X l ∈ Z (X k ∈ Z (cid:18) l + n − k − m − (cid:19)(cid:18) k − n − − n − (cid:19) × (cid:2) L iir ( − , v ij ( l + m + n + 2 − k, k ) (cid:3)) z − l − .
12t can be easily checked by direct computation that (cid:2) L iir ( − , v ij ( l + m + n + 2 − k, k ) (cid:3) = 12 X h ∈ Z (cid:2) v ii ( − − h, h ) , v ij ( l + m + n + 2 − k, k ) (cid:3) = − ( l + m + n + 2 − k ) v ij ( l + m + n + 1 − k, k ) . Thus we obtain Y ( v ij ( m, n ) , z )= ( − − m − n X l ∈ Z (X k ∈ Z − m − (cid:18) l + n − k − m − (cid:19) ( l + m + n + 2 − k ) × (cid:18) k − n − − n − (cid:19) v ij ( l + m + n + 1 − k, k ) ) z − l − = ( − − m − n X l ∈ Z (X k ∈ Z (cid:18) l + n − k − m − (cid:19)(cid:18) k − n − − n − (cid:19) v ij ( l + m + n + 1 − k, k ) ) z − l − . Thus we have proved Lemma 3.5.
Proof of Proposition 3.4.
First, we show the “if” part. Assume that V J = M r is notsimple, and let W ⊂ V J = M r be a proper ideal of the VOA V J . Let us show that W isa (proper) L r -submodule of M r . Claim 1.
Let ≤ i, j ≤ d with i = j . Then, v ij ( m, n ) W ⊂ W for all m, n ∈ Z .Proof of Claim 1. Let u ∈ W . We prove that v ij ( m, n ) u ∈ W for all m, n ∈ Z , which isequivalent to showing that v ij ( s − t, t ) u ∈ W for all s, t ∈ Z . Fix s ∈ Z . If v ij ( s − t, t ) u = 0for all t ∈ Z , then the assertion is obvious. So, let us assume that v ij ( s − t, t ) u = 0 forsome t ∈ Z . Take t , t ∈ Z with t ≤ t in such a way that if v ij ( s − t, t ) u = 0, then t ≤ t ≤ t . By Lemma 3.5, we see that (cid:0) v ij ( − p, − (cid:1) s + p u = ( − p +1 X t ∈ Z (cid:18) s + p − − tp − (cid:19) v ij ( s − t, t ) u = X t ≤ t ≤ t ( − p +1 (cid:18) s + p − − tp − (cid:19) v ij ( s − t, t ) u. for p >
0. Because W is an ideal of the VOA V J = M r , it follows that (cid:0) v ij ( − p, − (cid:1) s + p u ∈ W p >
0. Hence, in order to prove that v ij ( s − t, t ) u ∈ W for all t ≤ t ≤ t , it sufficesto show that the matrix (cid:18) ( − p +1 (cid:18) s + p − − tp − (cid:19)(cid:19) ≤ p ≤ t − t +1 t ≤ t ≤ t (3.9)is invertible. So we show the following: Let L ∈ Z , and M ∈ Z > . Then,det( a p,N ) ≤ p,N ≤ M = 0 , where a p,N := (cid:18) L + p − Np − (cid:19) . (3.10)We prove (3.10) by induction on the size M of the matrix. The claim is obvious when M = 1. Assume that M >
1. Using the formula a p,N − a p,N − = − (cid:18) L + p − Np − (cid:19) for 2 ≤ N ≤ M , we deduce thatdet( a p,N ) ≤ p, N ≤ M = det (cid:18) − (cid:18) L + p − Np − (cid:19)(cid:19) ≤ p,N ≤ M = ( − M − det (cid:18)(cid:18) L + p − Np − (cid:19)(cid:19) ≤ p,N ≤ M = ( − M − det( a p ′ ,N ′ ) ≤ p ′ ,N ′ ≤ M − | {z } =0 by the inductive assumption , where p ′ := p − N ′ := N −
1. Thus we obtain (3.10). We see from (3.10) with L = s − t , M = t − t + 1, and N = t − t + 1 that the determinant of the matrix (3.9)is not equal to 0, and hence the matrix (3.9) is invertible. Thus we have proved Claim 1. Claim 2.
Let ≤ i ≤ d . Then, v ii ( m, n ) W ⊂ W for all m, n ∈ Z .Proof of Claim 2. Let u ∈ W . Take 1 ≤ j ≤ d with j = i arbitrarily, and take N ∈ Z < in such a way that v jj ( N, − N ) u = 0. By direct computation, we see that[ v ij ( m, − N ) , v ij ( n, N )] u = − N v ii ( m, n ) u + δ m + n, mv jj ( N, − N ) u + αu = − N v ii ( m, n ) u + αu since v jj ( N, − N ) u = 0for some α ∈ C . Since [ v ij ( m, − N ) , v ij ( n, N )] u is contained in W by Claim 1, we concludethat v ii ( m, n ) u ∈ W , thereby completing the proof of Claim 2.It follows from Claims 1 and 2 that v ij ( m, n ) W ⊂ W for all 1 ≤ i, j ≤ d and m, n ∈ Z ,which implies that W is an L r -submodule of M r . Thus we have proved the “if” part ofProposition 3.4. 14ext, we show the “only if” part. Assume that M r = V J is a reducible L r -module,and let W ⊂ M r = V J be a proper L r -submodule. Let v ∈ V J , and l ∈ Z . By thedefinition of the vertex operator of V J , we deduce that v l ∈ End( V J ) can be written asan (infinite) linear combination of products of v ij ( m, n ), 1 ≤ i, j ≤ d , m, n ∈ Z . Since v ij ( m, n ) W ⊂ W by assumption, we have v l W ⊂ W . Therefore, W is a proper ideal ofthe VOA V J , and hence V J is not simple. This completes the proof of Proposition 3.4.The next corollary follows immediately from the proof of Proposition 3.4. Corollary 3.6.
Assume that the VOA V J is not simple, or equivalently, the L r -module M r is reducible. Then, W ⊂ V J (= M r ) is the maximal proper ideal of the VOA V J ifand only if W is the maximal proper L r -submodule of M r . M r and M (1) r . Let L (1) r be the Lie subalgebra of L r generated by B (1) := (cid:8) v ( m, n ) | m, n ∈ Z with m ≤ n (cid:9) ⊂ B . Set M (1) r = U ( L (1) r ) ⊂ M r . In this section, we prove the following proposition. Proposition 4.1.
The L r -module M r is irreducible if and only if M (1) r is irreducible asan L (1) r -module. In this subsection, we show that if M (1) r is an irreducible L (1) r -module, then M r is an irreducible L r -module. This assertionfollows immediately from the next lemma. Lemma 4.2.
Let W be a nonzero L r -submodule of M r . Then, W ∩ M (1) r is a nonzero L (1) r -submodule of M (1) r . Indeed, if W ⊂ M r is a nonzero L r -submodule of M r , then it follows from Lemma 4.2that W ∩ M (1) r is a nonzero L (1) r -submodule of M (1) r . Since M (1) r is assumed to be anirreducible L (1) r -module, we have W ∩ M (1) r = M (1) r . In particular, W contains ∈ M r ,which implies that W = M r .In order to prove Lemma 4.2, we introduce a weight space decomposition of M r . Define H := M ≤ k ≤ d, l ∈ Z < C v kk ( l, − l ) ⊂ L r . Then it can be easily seen that H is an abelian Lie subalgebra of L r . Set h k, l := − (1 /l ) v kk ( l, − l ) ∈ H for 1 ≤ k ≤ d and l ∈ Z < . By simple computation, we seethat for 1 ≤ k ≤ d and l ∈ Z < , 1 ≤ i, j ≤ d and m, n ∈ Z ,[ h k, l , v ij ( m, n )] r = (cid:0) δ ( k, l ) , ( i,m ) + δ ( k, l ) , ( j,n ) − δ ( k, − l ) , ( i,m ) − δ ( k, − l ) , ( j,n ) (cid:1) v ij ( m, n ) . (4.1)15et Λ k, l ∈ H ∗ := Hom C ( H , C ) be the dual basis of h k, l ∈ H for 1 ≤ k ≤ d and l ∈ Z < ,and set Q + := X ≤ k ≤ d, l ∈ Z < Z ≥ Λ k, l ⊂ H ∗ . We see from (4.1) that for each x ∈ S , the basis element w ( x ) of M r is contained in the“weight space” ( M r ) λ := (cid:8) u ∈ M r | hu = λ ( h ) u for all h ∈ H (cid:9) of weight λ for some λ ∈ Q + (see also Remark 4.4 below). Thus the L r -module M r admits the weight spacedecomposition with respect to the abelian Lie subalgebra H ⊂ L r as: M r = M λ ∈ Q + ( M r ) λ . (4.2) Remark . Let 1 ≤ i, j ≤ d , and let m, n ∈ Z . It follows from (4.1) that v ij ( m, n ) ∈ End C ( M r ) is a homogeneous operator of weight − sign( m )(Λ i, − m + Λ i,m ) − sign( n )(Λ j, − n +Λ j,n ), where for N ∈ Z , sign( N ) := N > , N = 0 , − N < , and for convenience, Λ k,l := 0 for all 1 ≤ k ≤ d and l ∈ Z ≥ . Remark . Let x = ( x p (cid:23) x p − (cid:23) · · · (cid:23) x ) ∈ S with x q = v i q j q ( m q , n q ) ∈ B − for1 ≤ q ≤ p . For 1 ≤ k ≤ d and l ∈ Z < , we define ν ( x , ( k, l )) := (cid:8) ≤ q ≤ p | ( i q , m q ) = ( k, l ) (cid:9) + (cid:8) ≤ q ≤ p | ( j q , n q ) = ( k, l ) (cid:9) . Namely, ν ( x , ( k, l )) denotes “the number of v k ( l ) appearing in x ” (see (2.2)). Then wededuce from Remark 4.3 that the weight of w ( x ) is equal to X ≤ k ≤ d, l ∈ Z < ν ( x , ( k, l ))Λ k, l ∈ Q + . Proof of Lemma 4.2.
Remark that the submodule W ⊂ M r also admits the weight spacedecomposition W = L λ ∈ Q + W λ , where W λ := W ∩ ( M r ) λ . Let u be a nonzero homoge-neous element of W , that is, u ∈ W λ \ { } for some λ ∈ Q + . It suffices to show that thereexists x ∈ U ( L r ) such that xu ∈ M (1) r \ { } . For each ξ ∈ Q + , we set θ ( ξ ) := X ≤ k ≤ d, l ∈ Z < ξ ( h k, l ) . (4.3)We show the claim above by induction on θ ( λ ). If θ ( λ ) = 0, then the claim is obvioussince u ∈ M (1) r (see Remark 4.4). Assume that θ ( λ ) >
0. Let 2 ≤ i ≤ d and m ∈ Z <
16e such that λ ( h i,m ) ≥
1, and let N ∈ Z < be such that λ ( h ,N ) = 0. Then we have v i ( − N, m ) v i ( N, − m ) u = 0. Indeed, since (cid:2) v i ( − N, m ) , v i ( N, − m ) (cid:3) r = ( − N ) v ii ( m, − m ) + mv ( N, − N )by direct computation, it follows that v i ( − N, m ) v i ( N, − m ) u = (cid:8) ( − N ) v ii ( m, − m ) + mv ( N, − N ) (cid:9) u + v i ( N, − m ) v i ( − N, m ) u. Here we note that λ + Λ i,m − Λ ,N is not contained in Q + since λ ( h ,N ) = 0. Because theweight of v i ( − N, m ) u is equal to λ + Λ i,m − Λ ,N / ∈ Q + by Remark 4.3, we see from (4.2)that v i ( − N, m ) u = 0. Hence we get v i ( − N, m ) v i ( N, − m ) u = (cid:8) ( − N ) v ii ( m, − m ) + mv ( N, − N ) (cid:9) u = mN λ ( h i,m ) | {z } ≥ u − mN λ ( h ,N ) | {z } =0 u = 0 . Thus we obtain v i ( − N, m ) v i ( N, − m ) u = 0, which implies that v i ( N, − m ) u = 0. Itfollows from Remark 4.3 that the weight λ ′ of v i ( N, − m ) u is equal to λ − Λ i,m + Λ ,N .Since θ ( λ ′ ) = θ ( λ ) −
1, there exists x ′ ∈ U ( L r ) such that x ′ v i ( N, − m ) u ∈ M (1) r by theinductive assumption. Thus we have proved Lemma 4.2. Assume that W ⊂ M (1) r is an L (1) r -submodule of M (1) r . Then we see that W is stable under the action of H ; indeed, wehave h , l W ⊂ W for all l ∈ Z < by assumption, and h k, l W = { } for all 2 ≤ k ≤ d and l ∈ Z < (see (4.1) and Remark 4.4). Thus, W also admits the weight space decompositionas follows: W = L λ ∈ Q + W λ with W λ = W ∩ ( M r ) λ . Since W ⊂ M (1) r = U ( L (1) r ) ,we deduce from Remark 4.3 (see also Remark 4.4) that W λ = { } unless λ ∈ Q (1)+ := P l ∈ Z < Z ≥ Λ , l . Hence we have W = M λ ∈ Q + W λ = M λ ∈ Q (1)+ W λ . (4.4)Now, let us prove that if M r is an irreducible L r -module, then M (1) r is an irreducible L (1) r -module. It suffices to show the next lemma. Lemma 4.5.
Assume that M (1) r is a reducible L (1) r -module, and let W ⊂ M (1) r be aproper L (1) r -submodule of M (1) r . Let u be a nonzero homogeneous element of W , thatis, u ∈ W λ \ { } for some λ ∈ Q (1)+ . Then the L r -module U ( L r ) u ( ⊂ M r ) generated bythe u is a proper submodule of M r . Therefore the L r -module M r is reducible. roof. Suppose that U ( L r ) u coincides with the whole of M r . Then there exists x ∈ U ( L r )such that xu ∈ C \ { } . Let B := B − ∪ (cid:8) v j ( m, n ) ∈ B | ≤ j ≤ d, m ∈ Z , n ∈ Z < (cid:9) , B := B \ B . Since an element in C ⊂ L r acts as a scalar multiple, we may assume, by the Poincar´e-Birkhoff-Witt theorem, that the x ∈ U ( L r ) above is of the form: x = P ≤ t ≤ s α t y t z t ,where y t (resp., z t ) is a product of elements in B (resp., B ) for each 1 ≤ t ≤ s , and α t ∈ C for each 1 ≤ t ≤ s . Because u is a homogeneous element, we see from Remark 4.3that y t z t u are also homogeneous elements for all 1 ≤ t ≤ s . Since xu ∈ C \ { } , andsince ( M r ) = ( M r ) = C , it follows that y t z t u ∈ C \ { } for some 1 ≤ t ≤ s . Thuswe may assume from the beginning that x is of the form: x = yz , where y (resp., z ) is aproduct of elements in B (resp., B ).Suppose that y = 1. Because xu = yzu ∈ C \ { } ⊂ ( M r ) , we deduce fromthe definition of the set B and Remark 4.3 that the weight of zu is not contained in Q + . Hence, zu = 0 by (4.2), which is a contradiction. Thus we get y = 1. Write z = z p z p − · · · z with z q ∈ B for 1 ≤ q ≤ p . Suppose that there exists 1 ≤ q ′ ≤ p such that z q ′ / ∈ B (1) = (cid:8) v ( m, n ) | m, n ∈ Z with m ≤ n (cid:9) . Let q := min (cid:8) ≤ q ′ ≤ p | z q ′ / ∈ B (1) (cid:9) .Then, z q is either of the following form: v j ( m, n ) for some 2 ≤ j ≤ d and m ∈ Z , n ∈ Z ≥ , or v ij ( m, n ) ∈ B + for some 2 ≤ i ≤ j ≤ d . Because z , z , . . . , z q − ∈ B (1) ,it is obvious that z q − · · · z u ∈ W ⊂ M (1) r . Thus, using Lemma 4.6 below, we see that z q z q − · · · z u = 0, which is a contradiction. Thus we conclude that z , z , . . . , z p ∈ B (1) .Hence, xu = yzu = zu, since y = 1= z p z p − · · · z u ∈ W, since z , z , . . . , z p ∈ B (1) . Since xu ∈ C \ { } , it follows that ∈ W , which implies that W = M (1) r . However, thisis a contradiction, since W is assumed to be a proper L (1) r -submodule of M (1) r . Thus wehave proved Lemma 4.5.Let us show the following lemma, which has been used in the proof of Lemma 4.5. Lemma 4.6. (1)
Let ≤ j ≤ d , and m ∈ Z , n ∈ Z ≥ . Then, v j ( m, n ) M (1) r = { } . (2) Let v ij ( m, n ) ∈ B + with ≤ i ≤ j ≤ d . Then, v ij ( m, n ) M (1) r = { } .Proof. (1) Let S (1) be the subset of S consisting of all finite sequences of elements in B (1) − := B (1) ∩ B − that is weakly decreasing with respect to the total ordering ≻ . Then, B (1) := (cid:8) w ( x ) | x ∈ S (1) (cid:9) ⊂ B is a linear basis of M (1) r . Therefore it suffices to show that18 j ( m, n ) w ( x ) = 0 for all x ∈ S (1) . Let x = ( x p (cid:23) x p − (cid:23) · · · (cid:23) x ) ∈ S (1) with x q ∈ B (1) for 1 ≤ q ≤ p . We show v j ( m, n ) w ( x ) = 0 by induction on the length p of the sequence x . If p = 0, then the claim is obvious since w ( x ) = . Assume that p >
0. Then, v j ( m, n ) w ( x ) = v j ( m, n ) x p x p − · · · x = [ v j ( m, n ) , x p ] x p − · · · x + x p (cid:8) v j ( m, n ) x p − · · · x (cid:9) . Since v j ( m, n ) x p − · · · x = 0 by the inductive assumption, the second term of theright-hand side is equal to 0. Assume that x p = v ( s, t ) with s ≤ t <
0. By simplecomputation, we see that[ v j ( m, n ) , x p ] r = [ v j ( m, n ) , v ( s, t )] r = δ m + s, mv j ( t, n ) + δ m + t, mv j ( s, n ) . Hence it follows from the inductive assumption that [ v j ( m, n ) , x p ] x p − · · · x = 0. Thuswe get v j ( m, n ) w ( x ) = 0, thereby completing the proof of part (1).(2) Since 2 ≤ i ≤ j ≤ d , it can be easily seen that [ v ij ( m, n ) , x ] = 0 for all x ∈ B (1) . Also, v ij ( m, n ) = 0, since v ij ( m, n ) ∈ B + . The assertion of part (2) follows immediately fromthese facts. This completes the proof of the lemma.Proposition 4.1 follows from the results obtained in §§ § Proposition 4.7.
Assume that the L r -module M r is reducible, or equivalently, the L (1) r -module M (1) r is reducible. (1) If W is the maximal proper L (1) r -submodule of M (1) r , then U ( L r ) W is the maximalproper L r -submodule of M r . (2) If W is the maximal proper L r -submodule of M r , then W ∩ M (1) r is the maximalproper L (1) r -submodule of M (1) r .Proof. We show that U ( L r ) W = W (for part (1)), and W = W ∩ M (1) r (for part (2)).Then we deduce from Lemma 4.5 that U ( L r ) W is an L r -submodule of M r such that U ( L r ) W = M r . Hence we have U ( L r ) W ⊂ W by the maximality of W . Also it followsfrom Lemma 4.2 (and the comment after it) that W ∩ M (1) r is an L (1) r -submodule of M (1) r such that W ∩ M (1) r = M (1) r . Hence we have W ∩ M (1) r ⊂ W by the maximality of W .Thus we obtain U ( L r ) W ∩ M (1) r ⊂ W ∩ M (1) r ⊂ W . Because it is obvious that W ⊂ U ( L r ) W ∩ M (1) r , we get U ( L r ) W ∩ M (1) r = W ∩ M (1) r = W , U ( L r ) W = W (i.e., part (1)). Since U ( L r ) W ⊂ W as shownabove, it suffices to show that U ( L r ) W ⊃ W . Note that W admits the weight spacedecomposition W = L λ ∈ Q + ( W ) λ with ( W ) λ = W ∩ ( M r ) λ . Let u ∈ W be a homo-geneous element of weight λ ∈ Q + , that is, u ∈ ( W ) λ . We show by induction on θ ( λ )that u ∈ U ( L r ) W (for the definition of θ ( λ ), see (4.3)). If θ ( λ ) = 0, then u ∈ M (1) r by Remark 4.4, and hence u ∈ W ∩ M (1) r . Since W ∩ M (1) r = W as shown above, itfollows that u ∈ W ⊂ U ( L r ) W . Next, let us assume that θ ( λ ) >
0. Let 2 ≤ i ≤ d and m ∈ Z < be such that λ ( h i,m ) ≥
1, and N ∈ Z < such that λ ( h ,N ) = 0. Thenwe deduce from the proof of Lemma 4.2 that v i ( N, − m ) u = 0, and the weight λ ′ of v i ( N, − m ) u satisfies θ ( λ ′ ) = θ ( λ ) −
1. Hence it follows from the inductive assumptionthat v i ( N, − m ) u ∈ U ( L r ) W . Further, as in the proof of Lemma 4.2, we deduce that v i ( − N, m ) v i ( N, − m ) u | {z } ∈ U ( L r ) W = mN λ ( h i, m ) | {z } ≥ u, which implies that u ∈ U ( L r ) W . This completes the proof of Proposition 4.7. M (1) r for r ∈ C \ Z . This subsection is devoted to proving the following proposition.
Proposition 5.1. If r ∈ C \ Z , then M (1) r is an irreducible L (1) r -module. For simplicity of notation, we set v ( m, n ) := v ( m, n ) for m, n ∈ Z , and Λ l := Λ ,l ∈ H ∗ for l ∈ Z < . Recall that S (1) denotes thesubset of S consisting of all finite sequences of elements in B (1) − = B (1) ∩ B − that is weaklydecreasing with respect to the total ordering ≻ , and B (1) = (cid:8) w ( x ) | x ∈ S (1) (cid:9) ⊂ B is alinear basis of M (1) r . For each λ ∈ Q (1)+ , we denote by B (1) λ the set of all elements in B (1) whose weight is equal to λ , and set S (1) λ := (cid:8) x ∈ S (1) | w ( x ) ∈ B (1) λ (cid:9) .Let u ∈ M (1) r , and write it as a linear combination of elements of B (1) : u = P b ∈ B (1) α b b with α b ∈ C for b ∈ B (1) . Then we set B [ u ] := (cid:8) b ∈ B (1) | α b = 0 (cid:9) , and S [ u ] := (cid:8) x ∈S (1) | w ( x ) ∈ B [ u ] (cid:9) .The following formulas can be shown by simple computation. Lemma 5.2. (1)
Let s, t ∈ Z > , and m, n ∈ Z > . Then, [ v ( − m, n ) , v ( − s, − t )] r = n (cid:8) δ n,s v ( − m, − t ) + δ n,t v ( − s, − m ) (cid:9) . (5.1)(2) Let s, t ∈ Z > with s = t , and m ∈ Z > . Then, [ v ( m, m ) , v ( − s, − t )] r = 2 m (cid:8) δ m,s v ( − t, m ) + δ m,t v ( − s, m ) (cid:9) . (5.2)20 emma 5.3. Let m ∈ Z > , and ν ∈ Z ≥ . Then, v ( m, m ) v ( − m, − m ) ν = 2 m ν ( r + 2 ν − v ( − m, − m ) ν − . (5.3) We show that if W ⊂ M (1) r is a nonzero L (1) r -submodule of M (1) r , then W = M (1) r . Suppose that W ( M (1) r . Because W admitsthe weight space decomposition W = L λ ∈ Q (1)+ W λ with respect to H (see § W contains a (homogeneous) singu-lar vector u , i.e., a nonzero element u such that u ∈ W λ for some λ ∈ Q (1)+ \ { } , and v ( m, n ) u = 0 for all m, n ∈ Z with m + n > Claim 1.
The set (cid:8) l ∈ Z < | λ ( h l ) > (cid:9) is identical to (cid:8) − p, − p + 1 , . . . , − , − (cid:9) forsome p ∈ Z ≥ .Proof of Claim 1. Assume that (cid:8) l ∈ Z < | λ ( h l ) > (cid:9) = (cid:8) l p < l p − < · · · < l (cid:9) ; notethat l q ≤ − q for every 1 ≤ q ≤ p . Suppose that l q < − q for some 1 ≤ q ≤ p . We set q := min (cid:8) ≤ q ≤ p | l q < − q (cid:9) . Since λ ( h l q ) > λ ( h − q ) = 0, we deduce by a waysimilar to the proof of Lemma 4.2 that v ( l q , q ) u = 0, and v ( l q , q ) v ( − q , − l q ) u = l q v ( − q , q ) u + q v ( l q , − l q ) u = q l q λ ( h − q ) | {z } =0 u + ( − q l q ) λ ( h l q ) | {z } > u = 0 . Hence we get v ( − q , − l q ) u = 0. However, since l q < − q , this contradicts the assump-tion that u is a singular vector. Thus we obtain l q = − q for all 1 ≤ q ≤ p , therebycompleting the proof of Claim 1. Claim 2.
The weight λ of the singular vector u is of the form : λ = P pq =1 ν q Λ − q with ν q ∈ Z > for ≤ q ≤ p . Furthermore, the set S [ u ] contains the element x λ ∈ S (1) λ suchthat w ( x λ ) = Y ≤ q ≤ p v ( − q, − q ) ν q ∈ B (1) λ . Proof of Claim 2.
For x ∈ S [ u ] and 1 ≤ q ≤ p , we define κ q ( x ) to be the number of v ( − q, − q ) appearing in the sequence x . Take x ∈ S [ u ] such that the sum P pq =1 κ q ( x ) ismaximum, and assume that w ( x ) ∈ B [ u ] is of the form: w ( x ) = Y ≤ q ≤ p v ( − q, − q ) ν q Y ≤ t ≤ t < s ≤ p . Then we prove that v ( − t , s ) u = 0. For this, it suffices to show21hat the set B [ v ( − t , s ) u ] is not empty. Here we show that B [ v ( − t , s ) u ] contains thefollowing element: Y ≤ q ≤ p, q = t v ( − q, − q ) ν q v ( − t , − t ) ν t +1 Y ≤ t
We have α s,t = − ν s for every ≤ t < s ≤ p . (2) For every ≤ s ≤ p , ν s ( r + 2 ν s −
2) + X ≤ t ≤ s − α s,t + X s +1 ≤ t ≤ p α t,s = 0 . Proof of Claim 3. (1) Fix 1 ≤ t < s ≤ p . We can easily check by using (5.1) that if theset B [ v ( − t, s ) w ( x )] with x ∈ S (1) λ contains w := v ( − s, − s ) ν s − v ( − s, − t ) Y ≤ q ≤ p, q = s v ( − q, − q ) ν q , then x = x λ or x = x s,tλ . By simple computation, along with (5.1), we get v ( − t, s ) w ( x λ ) = 2 ν s sw , v ( − t, s ) w ( x s,tλ ) = 2 sw + (other term) . Because v ( − t, s ) w = 0 by assumption, it follows that 2 ν s s + 2 sα s,t = 0, and hence α s,t = − ν s . 222) Fix 1 ≤ s ≤ p . We can easily check by using (2.5), (5.1), and (5.2) that if the set B [ v ( s, s ) w ( x )] with x ∈ S (1) λ contains w := v ( − s, − s ) ν s − Y ≤ q ≤ p, q = s v ( − q, − q ) ν q , then x = x λ , or x = x s,tλ for 1 ≤ t ≤ s −
1, or x = x t,sλ for s + 1 ≤ t ≤ p . We see fromLemma 5.3 that v ( s, s ) w ( x λ ) = 2 s ν s ( r + 2 ν s − w . Also it follows from (5.2) that for each 1 ≤ t ≤ s − v ( s, s ) w ( x s,tλ ) = 2 sX v ( s, − t ) v ( − s, − t ) v ( − s, − s ) ν s − + 2 sX v ( − s, − t ) v ( s, − t ) v ( − s, − s ) ν s − + 2 sX v ( − s, − t ) v ( s, s ) v ( − s, − s ) ν p − . (5.4)Here, for simplicity of notation, we set X := Y ≤ q ≤ p, q = t, s v ( − q, − q ) ν q v ( − t, − t ) ν s − . The second and third terms of the right-hand side of (5.4) do not contribute the coefficientof w since they contain v ( − s, − t ). Also, the first term is:2 sX v ( s, − t ) v ( − s, − t ) v ( − s, − s ) ν s − = 2 s X v ( − t, − t ) v ( − s, − s ) ν p − + (other term) . Thus we obtain v ( s, s ) w ( x s,tλ ) = 2 s w + (other terms)for 1 ≤ t ≤ s −
1. Similarly, we can show that v ( s, s ) w ( x t,sλ ) = 2 s w + (other terms)for s + 1 ≤ t ≤ p . Because v ( p, p ) u = 0 by assumption, we obtain2 s ν s ( r + 2 ν s −
2) + 2 s X ≤ t ≤ s − α s,t + 2 s X s +1 ≤ t ≤ p α t,s = 0 , and hence the equation of part (2). Thus we have proved Claim 3.Combining the equations in Claim 3 with s = p , we obtain ν p ( r +2 ν p − − ( p − ν p = 0,and hence that r + 2 ν p − − p = 0 . (5.5)However, this is a contradiction, since r is assumed not to be an integer. This completesthe proof of Proposition 5.1. 23 emark . For later use, let us show the following assertion: Keep the notation in theproof of Proposition 5.1 above. Then, ν = ν = · · · = ν p . Indeed, by the equations ofClaim 3, we see that ν s ( r + 2 ν s − − ( s − ν s − ν s +1 − · · · − ν p = 0 for every 1 ≤ s ≤ p .Therefore, ( ν s − ν s +1 )( r + 2 ν s +1 − s − ν s )= n ν s ( r + 2 ν s − − ( s − ν s − ν s +1 − · · · − ν p o − n ν s +1 ( r + 2 ν s +1 − − sν s +1 − ν s +2 − · · · − ν p o = 0for every 1 ≤ s ≤ p −
1. Using this equation and (5.5), we can show by descendinginduction that ν q = ν q +1 for all 1 ≤ q ≤ p −
1, and hence ν = ν = · · · = ν p . M (1) r for r ∈ Z . For each p ∈ Z ≥ , let V p be the following matrix ofsize p with entries in B (1) − : V p = (cid:0) v ( − s, − t ) (cid:1) ≤ s, t ≤ p . Since xy = yx for all x, y ∈ B (1) − , we can consider the determinant det V p of the matrix V p ; det V p = X σ ∈ S p sgn( σ ) Y ≤ q ≤ p v ( − q, − σ ( q )) , where S p denotes the symmetric group of degree p , and sgn( σ ) denotes the signature ofa permutation σ ∈ S p . In this subsection, we show the following proposition. Proposition 6.1.
Assume that r ∈ Z . Let ν ∈ Z ≥ and p ∈ Z ≥ be positive integerssatisfying the relation r = 1 − ν + p . Then, (det V p ) ν is a singular vector of M (1) r , thatis, v ( m, n )(det V p ) ν = 0 for all m, n ∈ Z with m + n > . Therefore the L (1) r -module M (1) r is reducible. Theorem 2.6 follows immediately from Propositions 3.1, 3.4, 4.1, 5.1, and 6.1 (see alsothe comment after Theorem 2.6).
Let us first show the following lemmas.
Lemma 6.2.
Let m ∈ Z with ≤ m ≤ p , and n ∈ Z ≥ with m = n . Then, [ v ( − m, n ) , det V p ] = 0 . (6.1)24 roof. If n ≥ p +1 or n = 0, then the assertion is obvious, since [ v ( − m, n ) , v ( − s, − t )] r = 0for all 1 ≤ s, t ≤ p . Assume that 1 ≤ n ≤ p . It can be seen from (5.1) that[ v ( − m, n ) , v ( − s, − t )] r is equal to 0, or is contained in B (1) − , up to a scalar multiple.Since ad x is a derivation on U ( L (1) r ) for x ∈ L (1) r , we deduce that[ v ( − m, n ) , det V p ] = det W + det W + · · · + det W p , where W s , 1 ≤ s ≤ p , is the matrix obtained by replacing the s -th row ( v ( − s, − t )) ≤ t ≤ p of the matrix V p with (cid:16) [ v ( − m, n ) , v ( − s, − t )] r (cid:17) ≤ t ≤ p . It follows from (5.1) that if s = n , then (cid:16) [ v ( − m, n ) , v ( − s, − t )] r (cid:17) ≤ t ≤ p = (cid:0) , . . . , , nv ( − s, − m ) , , . . . , (cid:1) , where nv ( − s, − m ) is placed at the n -th entry. Also, it follows from (5.1) that (cid:16) [ v ( − m, n ) , v ( − n, − t )] r (cid:17) ≤ t ≤ p − n (cid:0) v ( − m, − t ) (cid:1) ≤ t ≤ p | {z } the m -th row of W n = (cid:0) , . . . , , nv ( − n, − m ) , , . . . , (cid:1) , where nv ( − n, − m ) is placed at the n -th entry. Thus, [ v ( − m, n ) , det V p ] = n det V ′ p ,where V ′ p is the matrix obtained by replacing the n -th column ( v ( − s, − n )) ≤ s ≤ p of thematrix V p with ( v ( − s, − m )) ≤ s ≤ p (i.e., the m -th column of V p ). Because the n -th and m -th columns of V ′ p are equal, we get det V ′ p = 0, thereby completing the proof of thelemma. Lemma 6.3.
Let ≤ m ≤ p , and assume that u ∈ M (1) r satisfies the conditions that v ( − t, m ) u = 0 for all ≤ t ≤ p with t = m , and v ( − m, m ) u = αmu for some α ∈ C .Then we have v ( m, m )(det V p ) u = 2 m (2 α + r − p + 1) det( V ( m ) p ) u + (det V p ) v ( m, m ) u, where V ( m ) p is the matrix obtained by removing the m -th row and the m -th column fromthe matrix V p .Proof. We set S ( m ) p − := (cid:8) σ ∈ S p | σ ( m ) = m (cid:9) , which is a subgroup of S p isomorphic tothe symmetric group S p − of degree p −
1. Let R ( m ) := (cid:8) , ( m, t ) | ≤ t ≤ p with t = m (cid:9)
25e a complete set of the representatives for the quotient S p / S ( m ) p − , where ( m, t ) ∈ S p denotes the transposition interchanging m and t . Decompose S p as: S p = F τ ∈ S ( m ) p − τ R ( m ) ,with τ R ( m ) := (cid:8) τ, τ · ( m, t ) | ≤ t ≤ p with t = m (cid:9) . Then we have v ( m, m )(det V p ) u = [ v ( m, m ) , det V p ] u + (det V p ) v ( m, m ) u = X τ ∈ S ( m ) p − X σ ∈ τR m sgn( σ ) " v ( m, m ) , Y ≤ s ≤ p v ( − s, − σ ( s )) u + (det V p ) v ( m, m ) u. (6.2)If τ ∈ S ( m ) p − , then we see, using (2.5), (5.2), and the assumption on u , that " v ( m, m ) , Y ≤ s ≤ p v ( − s, − τ ( s )) u = ( Y ≤ s ≤ p, s = m v ( − s, − τ ( s )) ) [ v ( m, m ) , v ( − m, − m )] u = ( Y ≤ s ≤ p, s = m v ( − s, − τ ( s )) ) (cid:8) mv ( − m, m ) + 2 rm (cid:9) u = 2 m (2 α + r ) Y ≤ s ≤ p, s = m v ( − s, − τ ( s )) u. Assume that σ = τ · ( m, t ) ∈ τ R ( m ) with τ ∈ S ( m ) p − and 1 ≤ t ≤ p with t = m ; note that σ − ( m ) = t , and σ ( s ) = τ ( s ) for 1 ≤ s ≤ p with s = m, t . Then, " v ( m, m ) , Y ≤ s ≤ p v ( − s, − σ ( s )) u = Y ≤ s ≤ p,s = m, σ − ( m ) v ( − s, − σ ( s )) h v ( m, m ) , v ( − σ − ( m ) , − m ) v ( − m, − σ ( m )) i u = ( Y ≤ s ≤ p, s = m, t v ( − s, − τ ( s )) ) h v ( m, m ) , v ( − t, − m ) v ( − m, − τ ( t )) i u. Using (5.1), (5.2), and the assumption for u (note that τ ( t ) = m and t = m ), we deducethat h v ( m, m ) , v ( − t, − m ) v ( − m, − τ ( t )) i u = 2 m v ( − t, − τ ( t )) u. " v ( m, m ) , Y ≤ s ≤ p v ( − s, − σ ( s )) u = 2 m Y ≤ s ≤ p, s = m v ( − s, − τ ( s )) u. Therefore, for each τ ∈ S ( m ) p − , X σ ∈ τR ( m ) sgn( σ ) " v ( m, m ) , Y ≤ s ≤ p v ( − s, − σ ( s )) u = 2 m ( (2 α + r ) sgn( τ ) + X ≤ t ≤ p, t = m sgn( τ · (1 , t )) ) Y ≤ s ≤ p, s = m v ( − s, − τ ( s )) u = 2 m (2 α + r − p + 1) sgn( τ ) Y ≤ s ≤ p, s = m v ( − s, − τ ( s )) u. Combining this equation and equation (6.2), we see that v ( m, m )(det V p ) u = 2 m (2 α + r − p + 1) det( V ( m ) p ) u + (det V p ) v ( m, m ) u, as desired. Proof of Proposition 6.1.
It is obvious that (det V p ) ν is a nonzero homogeneous element,and (det V p ) ν / ∈ C . Let m, n ∈ Z be such that m ≤ n and m + n >
0; note that n ≥ n ≥ p + 1, then v ( m, n )(det V p ) ν = 0 since the weight of v ( m, n )(det V p ) ν = 0 isnot contained in Q (1)+ (see (4.4)). Let us consider the case that 1 ≤ n ≤ p . Case 1.
Assume that 1 ≤ n ≤ p and − n < m <
0. It follows from Lemma 6.2 that[ v ( m, n ) , det V p ] = 0. Hence v ( m, n )(det V p ) ν = (det V p ) ν v ( m, n ) = 0 . Case 2.
Assume that 1 ≤ n ≤ p and m = 0. By direct computation, we see that v (0 , n ) = ( p + 1) − [ v ( n, p + 1) , v ( − p − , r . As mentioned above, we have v ( n, p +1)(det V p ) ν = 0. Also, since [ v ( − s, − t ) , v ( − p − , r = 0 for all 1 ≤ s, t ≤ p , it followsthat [ v ( − p − , , det V p ] = 0, and hence that v ( − p − , V p ) ν = 0 as above.Thus we get v (0 , n )(det V p ) ν = 0. Case 3.
Assume that 1 ≤ n ≤ p and 1 ≤ m ≤ n . First let us consider the case that m = n . Namely, we show that v ( m, m )(det V p ) ν = 0 for all 1 ≤ m ≤ p . It follows27mmediately from Lemma 6.2 that u = (det V p ) ν , ν ∈ Z ≥ , satisfies the assumption ofLemma 6.3 with α = 2 ν . Hence, by using Lemma 6.3 repeatedly, we obtain v ( m, m )(det V p ) ν = v ( m, m )(det V p )(det V p ) ν − = 2 m { ν −
1) + r − p + 1 } (det V ( m ) p )(det V p ) ν − + (det V p ) v ( m, m )(det V p ) ν − = · · · · · · = 2 m (det V ( m ) p ) ( ν − X ν =0 (cid:0) ν + r − p + 1 (cid:1)) (det V p ) ν − = 4 m ν (2 ν − r − p )(det V ( m ) p )(det V p ) ν − . Since 2 ν − r − p = 0 by assumption, we get v ( m, m )(det V p ) ν = 0.Next, let us consider the case that m < n . By direct computation, we see that v ( m, n ) = (2 m ) − [ v ( m, m ) , v ( − m, n )] r . Since v ( − m, n )(det V p ) ν = 0 by Case 1, andsince v ( m, m )(det V p ) ν = 0 by the argument above, it follows that v ( m, n )(det V p ) ν =0. This completes the proof of Proposition 6.1 Fix r ∈ Z as above. Denote by W (1) r the L (1) r -submodule of M (1) r generated by the singular vectors obtained in Proposition 6.1,i.e., W (1) r := D (det V p ) ν (cid:12)(cid:12)(cid:12) p, ν ∈ Z ≥ with r = 1 − ν + p E ⊂ M (1) r . Proposition 6.4.
A singular vector of M (1) r is equal to a scalar multiple of the singu-lar vector (det V p ) ν for some p, ν ∈ Z ≥ with r = 1 − ν + p . Therefore the L (1) r -submodule W (1) r is the maximal proper submodule of M (1) r , and hence the quotient L (1) r -module M (1) r /W (1) r is irreducible.Proof. Let u ∈ M (1) r be a singular vector, and assume that the weight of u is equal to λ ∈ Q (1)+ \ { } . By Claim 2 in the proof of Proposition 5.1 and by Remark 5.4, we see thatthe weight λ is of the form: 2 ν P pq =1 Λ − q for some p > ν >
0, and that w ( x λ ) ∈ B [ u ].In addition, it follows from (5.5) that r = 2 ν − − p . Thus, by Proposition 6.1, (det V p ) ν is a singular vector. Because w ( x λ ) ∈ B [(det V p ) ν ] by definition, there exists α ∈ C \ { } such that w ( x λ ) / ∈ B [ u − α (det V p ) ν ] . (6.3)Here we should remark that u − α (det V p ) ν is also a singular vector of weight λ if itis nonzero. Therefore we deduce from (6.3) and Claim 2 in the proof of Proposition 5.1that u − α (det V p ) ν = 0, and hence u = α (det V p ) ν . Thus we have proved the firstassertion of the proposition. The other assertions are obvious. This completes the proofof the proposition. 28et I r be the ideal of the VOA V J (= M r ) generated by all (det V p ) ν for p, ν ∈ Z ≥ with r = 1 − ν + p . Corollary 6.5.
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