aa r X i v : . [ m a t h . QA ] J u l A (2)2 l AT LEVEL − l − SHASHANK KANADE
Abstract.
Let L l = L ( sl l +1 , − l − ) be the simple vertex operator algebra based on the affineLie algebra b sl l +1 at boundary admissible level − l − .We consider a lift ν of the Dynkin diagram involution of A l = sl l +1 to an involution of L l . The ν -twisted L l -modules are A (2)2 l -modules of level − l − with an anti-homogeneous realization. Weclassify simple ν -twisted highest-weight (weak) L l -modules using twisted Zhu algebras and singularvectors for b sl l +1 at level − l − obtained by Perˇse.We find that there are finitely many such modules up to isomorphism, and the ν -twisted (weak) L l -modules that are in category O for A (2)2 l are semi-simple. Introduction
In [16], while studying the modular invariant representations of affine Lie algebras, Kac andWakimoto introduced the notion of admissible highest-weight representations and classified thesein [17]. Let g be a finite dimensional simple Lie algebra of type X N , and let b g be the corresponding untwisted affine Lie algebra of type X (1) N (see [15, Table Aff 1]). Since [16], vertex operator algebras(say L ( g , k )) based on the untwisted affine Lie algebra b g at admissible levels k have received atremendous amount of attention.In [2], Adamovi´c and Milas analysed the case of g = sl for all admissible levels k , classified theweak modules of L ( sl , k ) that belong to category O as c sl -modules and showed that this categoryis semi-simple with finitely many equivalence classes of irreducibles. They conjectured that thisholds for all untwisted affine Lie algebras. The g = sl case was also studied in [10, 12]. In acelebrated achievement, Arakawa proved this conjecture [4]. Before [4], several other specific casesof this conjecture were known to be true, notably in type C [1], in type A [25], in type B [24] andfor G [5]. In [2, 1, 25, 24, 5], the technique of Zhu algebras [29, 14] and an explicit knowledge ofthe singular vectors at the prescribed levels was used.It is also important to consider categories larger than O , namely, the categories generated by relaxed highest weight modules. For g = sl , simple relaxed highest-weight modules at admissiblelevels were classified using the Zhu technology in [2]. Recently, a classification for arbitrary rankbased on Mathieu’s coherent famililes [23] is presented in [20]. We will not pursue this directionhere.Kac-Wakimoto’s work [16] also included a discussion of affine Lie super-algebras, and indeedmodels related to osp (1 |
2) at admissible levels have been analysed in [26, 9, 27]. However, it is notclear if semi-simplicity holds beyond osp (1 | n ) , and [4] does not encompass affine super-algebras.Despite all these stellar advances, the case of twisted affine Lie algebras (see [15, Table Aff 2,Aff 3]) has received little to no attention. The most natural way to access modules for X ( r ) N where r = 2 , ν - twisted modules for the VOAs based on the corresponding untwisted University of Denver, Denver, USA
E-mail address : [email protected] .I am extremely grateful to T. Creutzig, A. Linshaw and D. Ridout for many discussions. I am currently supportedby Simons Foundation’s Collaboration Grant for Mathematicians I thank David Ridout for pointing this out. ffine Lie algebras X (1) N [13, 22]. Here, ν is a lift of the non-trivial Dynkin diagram automorphismto g and ν fixes the chosen Cartan sub-algebra. ν is then extended to act on the whole VOA. Wemay modify ν by composing it with exp(2 π i · ad h ) for certain Cartan elements h with ν ( h ) = h [15, Eq. 8.1.2]. This way, we get different realizations of X ( r ) N differing primarily in their gradings.In this paper, we consider the case of A (2)2 l at level − l − for l ∈ Z > . We use the anti-homogeneous realization of A (2)2 l obtained from an involutive lift ν of the Dynkin diagram automorphism of g = sl l +1 = A l . Here, anti-homogeneous refers to the fact that our picture is exactly the oppositeof the traditional one – our affine , i.e., 0th node for A (2)2 l is what is usually the last , i.e., l th nodein the affine Dynkin diagram, and our horizontal subalgebra is thus so l +1 = B l and not sp l = C l .We use twisted Zhu algebras [11] (see also [28]) and the singular vectors for b sl l +1 at level − l − obtained by Perˇse in [25]. Somewhat surprisingly, we find that the top spaces of the A (2)2 l modules(which are naturally modules for our horizontal subalgebra, B l ) are exactly the same as the topspaces for the highest-weight L ( B l , − l + )-modules found in [24]. Letting h ∨ denote the dualCoxeter number [15, Ch. 6], the relation between these levels for l > − l −
12 + h ∨ A l (2) = l + 12 = − l + 32 + h ∨ B l (1) . (1.1)Our proof of admissibility of the A (2)2 l highest weights thus obtained also uses a large portion of thecorresponding proof in [24]. The proof of semi-simplicity then proceeds as in [2, 1, 25, 24, 5], etc.,with appropriate changes to accommodate twisted modules.We find that there are two inequivalent ν -twisted irreducible modules for L ( sl l +1 , − l − ) withfinite dimensional top spaces (Remark 4.8). Recall that − l − is a boundary admissible level[19] for A (1)2 l and correspondingly, there is exactly one (up to equivalence) irreducible with finitedimensional top space in the untwisted sector [25].This naturally leads to the following speculations and considerations that we are currently in-vestigating.(1) Perhaps the most important speculation we have is that the Adamovi´c-Milas conjecture/ Arakawa’s theorem is true for twisted affine Lie algebras as well. To be precise, wespeculate that given a twisted affine Lie algebra X ( r ) N , and an admissible level k for (theuntwisted) X (1) N , there exists an appropriate realization of X ( r ) N and a corresponding lift ν of the (non-trivial) diagram automorphism of X N , such that ν -twisted (weak) L ( X N , k )-modules which are in category O as X ( r ) N -modules form a semi-simple category with finitelymany irreducibles.(2) In [8], it was proved that the ordinary modules for L ( X N , k ) ( k is admissible level forthe untwisted affine Lie algebra X (1) N ), form a vertex tensor category; the rigidity of thiscategory for the simply-laced cases was proved in [7]. Our results imply that in general,the category consisting of untwisted and g -twisted ordinary modules for g ∈ h ν i will not beclosed under twisted fusion.In our present case, the untwisted and ν -twisted ordinary modules form semi-simplecategories, but the untwisted sector has one simple (up to equivalences) and the ν -twistedone has two inequivalent simples. The aforementioned closure under twisted fusion is nowforbidden by elementary considerations of tensor categories.In general, such ordinary g -twisted modules are integrable in the direction of g (the fixedpoint subalgebra of g = X N under ν ), thus it is natural to expect their (twisted) fusion tobe integrable with respect to g , but it need not be g -integrable. t will be difficult but interesting to work out the twisted fusion for our ν -twisted modules,and perhaps also the fusion for the untwisted modules for the corresponding orbifold. Here,the structure of this orbifold [3] will be important to first classify its modules.(3) It will be very interesting to also analyse twisted quantum Drinfeld-Sokolov reductions [18]of the ν -twisted modules we have found for appropriate nilpotents f ∈ sl l +1 fixed by ν ,and compare these to twisted representations of the corresponding W -algebras. Here again,one may take a slightly different route and investigate the relation of the structure andrepresentation theory of the affine orbifold with that of the W -algebra orbifold.2. Twisted affine Lie algebra A (2)2 l Twisted affine Lie algebra A (2)2 l , basics. We will consider what we call the anti-homogeneous realization of A (2)2 l and recall basic facts from [15, 6]. Consider the generalized Cartan matrices:˜ A = − − ! , ( l = 1) , (2.1)˜ A = · · · l − l − l − − − − l − − l − − − l − , ( l > . (2.2)We have the corresponding (affine) Dynkin diagrams: α α l =1 α α α α α l − α l − α l l> (2.3)Here, the 0 th node is considered to be the affine node and the horizontal subalgebra of A (2)2 l is oftype B l = so l +1 (unlike the usual convention where it turns out to be C l = sp l ): α l =1 α α α α l − α l − α l l> (2.4)We have: a = ( a , . . . , a l ) t = (1 , , . . . , t , ˜ Aa = 0 (2.5) a ∨ = ( a ∨ , . . . , a ∨ l ) = (2 , , . . . , , , a ∨ ˜ A = 0 . (2.6)We will often use the following indexing sets: I = { , . . . , l } , b I = { , . . . , l } . (2.7)The twisted affine Lie algebra A (2)2 l has Kac-Moody generators h i , e i , f i ( i ∈ b I ) and d satisfyingthe usual relations [15]. We let the Cartan subalgebra H be spanned by h , h , . . . , h l , d , The simple oots α i ( i ∈ b I ) are elements of H ∗ . We will sometimes denote the pairing between H ∗ and H by( · , · ). This notation will be overloaded below. For i, j ∈ b I , k ∈ I we have: α i ( h j ) = ( α i , h j ) = e A ji , α ( d ) = ( α , d ) = 1 , α k ( d ) = ( α k , d ) = 0 . (2.8)The canonical central element c ∈ H of A (2)2 l and the basic imaginary root δ are expressed as: c = X ≤ i ≤ l a ∨ i h i = 2 h + · · · + 2 h l − + h l , δ = X ≤ i ≤ l a i α i = α + 2 α · · · + 2 α l . (2.9)We choose h , . . . , h l , c, d as a standard basis for H . We have: δ ( c ) = 0 , δ ( h ) = 0 , . . . , δ ( h l ) = 0 , δ ( d ) = 1 . (2.10)We also consider Λ c ∈ H ∗ such that:Λ c ( c ) = 1 , Λ c ( h ) = 0 , . . . , Λ c ( h l ) = 0 , Λ c ( d ) = 0 . (2.11)Note that Λ c is Λ , where Λ is the fundamental weight corresponding to the 0th node. It is easyto see that α , . . . , α l , δ, Λ c form a basis of H ∗ . The standard symmetric (non-degenerate) bilinearform on H is given by: ( h i , h ) = ( α i , h ) · a i a ∨ i , i ∈ b I, h ∈ H , ( d, d ) = 0 . (2.12)The non-degenerate map ( · , · ) leads to a (linear) isomorphism ι : H → H ∗ with ( i ∈ b I ):( ι ( h ) , h ) = ( h, h ) , for all h, h ∈ H , (2.13) ι ( h i ) = a i a ∨ i · α i , ι ( c ) = δ, ι ( d ) = Λ c . (2.14)We may thus get a (non-degenerate) symmetric bilinear form on H ∗ by transport of structure. Itsatisfies ( i, j ∈ b I and k ∈ I ):( α i , α j ) = e A ij · a ∨ i a i , ( δ, α k ) = ( δ, δ ) = (Λ c , α k ) = (Λ c , Λ c ) = 0 , ( δ, Λ c ) = 1 . (2.15)The squared root lengths are therefore ( k = 1 , . . . , l − α , α ) = 4 , ( α k , α k ) = 2 , ( α l , α l ) = 1 . (2.16)The root system of A (2)2 l depends on l . Let l >
1. The root system of the horizontal subalgebra B l can be realized as (with k = 1 , . . . , l − i, j ∈ I ): α k = ǫ k − ǫ k +1 , α l = ǫ l , where ( ǫ i , ǫ j ) = δ ij . (2.17)We have: Φ long = {± ǫ i ± ǫ j | ≤ i < j ≤ l } , Φ short = {± ǫ i | i = 1 , . . . , l } , (2.18)and the real roots for A (2)2 l are [6]: b ∆ re = b Φ long ∪ b Φ intermediate ∪ b Φ short = { α s + (2 m + 1) δ | α ∈ Φ s , m ∈ Z } ∪ { α + mδ | α ∈ Φ l , m ∈ Z } ∪ { α + mδ | α ∈ Φ s , m ∈ Z } (2.19)where the squared norms of roots in the respective sets are 4 , k = 1 , . . . , l −
1, thefundamental weights of the horizontal subalgebra are: ω k = ǫ + · · · + ǫ k , ω l = 12 ( ω + · · · + ω l ) . (2.20) or l = 1, the horizontal subalgebra is sl with simple positive root α , and we have: b ∆ re = b Φ long ∪ b Φ short = {± α + (2 m + 1) δ | m ∈ Z } ∪ {± α + mδ | m ∈ Z } . (2.21)Here, note that ( α , α ) = 1, and thus squared norms of the roots in these sets are 4 and 1,respectively. The fundamental weight for the horizontal algebra is ω = α .Let ρ be any element of H ∗ satisfying ρ ( h i ) = 1 for all i ∈ b I . We may take it to be: ρ = h ∨ Λ c + ρ where ρ is half the sum of positive roots of the horizontal sub-algebra and h ∨ = a ∨ + · · · + a ∨ l = 2 l +1is the dual Coxeter number of A (2)2 l . For l = 1, ρ = α . If l >
1, we have: ρ = (cid:18) l − (cid:19) ǫ + (cid:18) l − (cid:19) ǫ + · · · + 12 ǫ l . (2.22)Finally, recall the notion of Weyl group W generated by reflections r i ( i ∈ b I ) satisfying r i ( h ) = h − ( α i , h ) h i for h ∈ H and we transfer the action to H ∗ by ι . We have W · { α , . . . , α l } = b ∆ re andwe define b ∆ ∨ , re = W · { h , . . . , h l } , which is the set of real coroots. There is thus a bijection fromreal roots to real coroots denoted by ∨ such that α i α ∨ i = h i , and it is not hard to prove, usingthe invariance of ( · , · ) under the Weyl group that for λ ∈ H ∗ , α ∈ b ∆ re that( λ, α ∨ ) = 2( α, α ) ( λ, α ) . (2.23)Given λ ∈ H ∗ , define b ∆ ∨ , re λ = { α ∨ ∈ b ∆ ∨ , re | ( λ, α ∨ ) ∈ Z } , b ∆ re λ = { α ∈ b ∆ re | α ∨ ∈ b ∆ ∨ , re λ } . (2.24) Definition 2.1. [16] We say an element λ ∈ H ∗ is an admissible weight if:(1) ( λ + ρ, α ∨ )
6∈ { , − , − , · · · } for all α ∨ ∈ b ∆ ∨ , re+ and(2) Q b ∆ ∨ , re λ = Q { h , . . . , h l } . Remark 2.2.
The second condition can be equivalently replaced with Q b ∆ re λ = Q { α , . . . , α l } .2.2. Twisted affinizations of Lie algebras.
Suppose we are given a finite dimensional (simple)Lie algebra g with a symmetric invariant bilinear form h· , ·i . Let ν be an automorphism of ( g , h· , ·i )of a finite order, say T . Corresponding to ν , we have the eigen-decomposition g = M j ∈ Z /T Z g j , x ∈ g j ⇔ ν ( x ) = e π i j/T x. (2.25)Consider the affinization: b g T Z = g ⊗ C [ t /T , t − /T ] ⊕ C c. (2.26)We will often drop the superscript T Z since it will be clear from the context. The element c iscentral and the other brackets are ( a, b ∈ g , m, n ∈ T Z ):[ a ⊗ t m , b ⊗ t n ] = [ a, b ] ⊗ t m + n + mδ m + n, h a, b i c. (2.27)Define ν ( t j/T ) = e − π i j/T t j/T and extend linearly to C [ t /T , t − /T ]. Also let ν ( c ) = c . We areinterested in the fixed point sub-algebra b g [ ν ] = M j ∈ Z /T Z (cid:16) g j ⊗ t j/T C [ t, t − ] (cid:17) ⊕ C c. (2.28)We shall obtain A (2)2 l via such twisted affinization of g = A l = sl l +1 . .3. Anti-homogeneous realization of A (2)2 l . We start by fixing some notation. Fix l ∈ Z > .Consider gl l +1 spanned by elementary matrices E i,j (or simply E ij ) with 1 in row i and column j , zeros everywhere else. Let E i = E i,i +1 , F i = E i +1 ,i , H i = E i,i − E i +1 ,i +1 be the standard choicesof simple root vectors and simple coroots for g = sl l +1 ⊂ gl l +1 . Let g = n − ⊕ h ⊕ n + be thetriangular decomposition of g . Let E θ = E , l +1 = [ · · · [[ E , E ] , E ] , · · · , E l ].The anti-homogeneous realization is achieved via an involutive lift of the diagram automorphismof sl l +1 which we now describe.Define ν ( E i,j ) = − ( − i − j E l +2 − j, l +2 − i . It is straightforward to prove that ν is an involutionof gl l +1 and also of the Lie subalgebra g = sl l +1 . Corresponding to ν we have the decomposition g = g ⊕ g (where the superscripts are understood as elements of Z ). It is clear that ν ( H i ) = H l +1 − i , for i = 1 , . . . , l , and thus ν is an involutive lift of the Dynkin diagram automorphism of g . Observe that E θ ∈ g .The fixed points g form a simple Lie algebra of type B l = so l +1 with the following Chevalleygenerators [6]. e i = E i + E l +1 − i , f i = F i + F l +1 − i , h i = H i + H l +1 − i , ( i = 1 , . . . l − ,e l = √ E l + E l +1 ) , f l = √ F l + F l +1 ) , h l = 2( H l + H l +1 ) . (2.29)For convenience, we denote: e l = E l + E l +1 , f l = F l + F l +1 , h l = H l + H l +1 . (2.30)Note again that the actual generators for B l (which will also get promoted to a subset of generatorsfor A (2)2 l below) are indeed e , . . . , e l − , e l , h , . . . , h l − , h l , f , . . . , f l − , f l . We have introduced e l , f l , h l only to save ourselves from keeping track of the various scalars.Given any a ∈ g , we let a = a + + a − , a + = 12 ( a + νa ) ∈ g , a − = 12 ( a − νa ) ∈ g . (2.31)We let g = n − ⊕ h ⊕ n be the triangular decompositions with respect to our choices of rootvectors. Note that n is spanned by E + i,j for 1 ≤ i < j ≤ l + 1 and that dim( h ) = l .Later, we shall require the dimension of weight 0 space of g as a g -module. One may calculatethis directly by decomposing g with respect to g . Here we present one more approach. Temporarily,let L ( ω ) denote irreducible g ∼ = so l +1 module with highest weight ω . As a g -module, g ∼ = L (2 ω )and is generated by the highest weight vector E θ . Further, Sym L ( ω ) ∼ = L (2 ω ) ⊕ C and L ( ω ) isthe defining representation of dimension 2 l + 1. It can be seen that if ω is a weight of L ( ω ) thenso is − ω , 0 is a weight, and every weight space is one dimensional. Thus, the 0 weight space ofSym L ( ω ) has dimension l + 1, and the 0 weight space of g ∼ = L (2 ω ) has dimension l .Now, b sl l +1 [ ν ] gives us an anti-homogeneous realization of A (2)2 l . Considering the numberingfrom (2.3), we let the Kac-Moody generators to be the ones given in (2.29) for i = 1 , . . . , l . As for h , e , f , we take them to be: h = − H θ + 12 c = − ( H + · · · + H l ) + 12 c, e = E l +1 , ⊗ t / , f = E , l +1 ⊗ t − / . (2.32)The involution ν extends to the universal enveloping algebra U ( g ) and we have U ( g ) ( U ( g ) ( U ( g ). Later, we will be interested in certain two-sided ideals I ⊂ U ( g ). Remark 2.3.
It is possible to achieve this realization of A (2)2 l by using the Chevalley involution [15,Eq. 1.3.4] of A l . However, it is convenient to use an automorphism that respects the triangulardecomposition of g . . Twisted Zhu algebra: Preliminaries
Let (
V, Y, , ω ) be a vertex operator algebra [21] and let g be an automorphism of finite order T of V . Let V j ( j = 0 , . . . , T −
1) be the subspace of eigenvalue e π i j/T for g . Following [11], wenow define the twisted Zhu algebra A g ( V ) as follows. Let u ∈ V j (0 ≤ j < T ) be L (0)- and g -homogeneous element and let v ∈ V . Define u ◦ g v = Res x (1 + x ) wt u − δ j + jT x δ j Y ( u, x ) v ! , (3.1) u ∗ g v = Res x (cid:18) (1 + x ) wt u x Y ( u, x ) v (cid:19) if j = 00 if otherwise . (3.2)where we take δ j = 1 when j = 0 and δ j = 0 if j = 0. Extend ◦ g , ∗ g to V linearly. Further define O g ( V ) = Span { u ◦ g v | u, v ∈ V } , A g ( V ) = V /O g ( V ) . (3.3)Taking v = in (3.1) immediately gives us: V i ⊂ O g ( V ) if i T ) . (3.4)We will denote the image in A g ( V ) of v ∈ V by J v K g . It was shown in [11] that O g ( V ) is atwo-sided ideal with respect to ∗ g and that A g ( V ) is an associative algebra under product ∗ g with J K as the unit and J ω K belonging to the center. When g = 1, ◦ g , ∗ g , O g ( V ) , A g ( V ) are simplydenoted as ◦ , ∗ , O ( V ) , A ( V ), respectively. We recall the following basic theorems (and their twistedanalogues) from [29, 14, 11, 28]. Theorem 3.1.
We have:(1) [14, 28] Let I be a g -stable ideal of V , and suppose I , ω I . Then, the image of I in A g ( V ), denoted as A g ( I ) is a two-sided ideal. Moreover, A g ( V /I ) ∼ = A g ( V ) /A g ( I ).(2) [11, Thm. 7.2] There is a bijective correspondance between the set of equivalence classesof simple A g ( V ) modules and weak, T Z -gradable g -twisted V -modules (see [11, Def. 3.3],where these modules are called admissible, not to be confused with [16]). Remark 3.2.
The first part of the theorem above is proved for g = 1 (the untwisted case) in [14].It is not hard to extend the proof to general g [28].Now, for the rest of the section, let V = V ( g , k ) be the (universal) Verma module vertex operatoralgebra based on ( g , h· , ·i ) with level k = − h ∨ [21]. Let g be an automorphism of V order T = 1lifted from an automorphism g of ( g , h· , ·i ) of the same order T . Theorem 3.3.
We have:(1) [28] There exists an (explicit) isomorphism of associative algebras F : A g ( V ( g , k )) ∼ = −→ U ( g ).(2) [14] Let x ∈ g and v ∈ V . Then, F ( J x (0) v K g ) = [ x, J v K g ] , (3.5)where both sides are zero if v ∈ V ⊕ · · · ⊕ V T − .(3) [14] Let x , x , . . . , x m ∈ g , n , n , . . . , n m ∈ Z ≥ . Then under the isomorphism above, F ( J x ( − n − x ( − n − · · · x m ( − n m − K g ) = ( − n + n + ··· + n m x m x m − · · · x . (3.6)(4) The previous part immediately implies that for x ∈ g , n ∈ Z ≥ and v ∈ V , we have: F ( J x ( − n − v K g ) = ( − n F ( J v K g ) x. (3.7)Henceforth, we will suppress F ( · · · ) and simply identify A g ( V ( g , k )) with U ( g ). efinition 3.4. We have an action L of g on U ( g ) given by x L u = [ x, u ] where x ∈ g , u ∈ U ( g ).We may and do extend the action L of g to an action of U ( g ). Theorem 3.5.
Suppose that the (unique) maximal b g -submodule J ( g , k ) of V ( g , k ) is generated bya single g -homogeneous singular vector v . Let U ( g ) v be the g -module generated by v where x ∈ g acts on v by x (0). Let R = J U ( g ) v K g = J ( U ( g ) v ) ∩ V ( g , k ) K g = J ( U ( g ) v ) K g . (3.8)We have the following.(1) R is a finite-dimensional module for U ( g ) under the L action.(2) Let L ( g , k ) = V ( g , k ) /J ( g , k ) be the unique simple quotient of V ( g , k ). Then, A g ( L ( g , k )) = U ( g ) h R i (3.9)where h R i denotes two sided ideal of U ( g ) generated by R .(3) A g -module M is a A g ( L ( g , k ))-module iff h R i · M = 0. Proof.
This theorem is analogous to the corresponding theorems in the untwisted setup, and in thetwisted setting, our proof is very similar to the proof of [28, Thm. 6.3].All elements of R have the same conformal weight as that of v , and each conformal weight spaceof V ( g , k ) is finite-dimensional, hence R is finite-dimensional. The fact that R is closed under L action is immediate from (3.5).For the second assertion, it is enough prove that J J ( g , k ) K g = h R i . Observe that if X is asubspace of U ( g ) that is closed under the L action and also under the right multiplication by U ( g )then, X is a two-sided ideal. Indeed, for a ∈ g and x ∈ X , we have ax = [ a, x ] − xa , and bothterms on the right-hand side belong to X , giving us the closure of X under the left-action. In lightof (3.5), J J ( g , k ) K g is closed under L and (3.7) implies that it is also closed under the right actionof U ( g ). Thus, it is a two sided ideal. Clearly, R ⊂ J J ( g , k ) K g , and thus h R i ⊂ J J ( g , k ) K g .For the reverse inclusion, note that J ( g , k ) is spanned by terms of the sort y = a ( − n − a ( − n − · · · a t ( − n t − x, (3.10)where a i ∈ g are g -homogeneous and are arranged so that all a i ’s in g are to the right, n i ∈ Z ≥ and x is a g -homogeneous element of U ( g ) v . We proceed by induction on t . The case for t = 0 isclear: J x K g ∈ R . Now let t >
0. If all a i ’s and x are already fixed by g then (3.7) immediately tellsus that J y K g ∈ h R i . Suppose that y is fixed by g (otherwise its projection is 0 anyway) and that a is in g r , 1 ≤ r ≤ T −
1. We have the following relation [11]:Res x (1 + x ) r/T x m +1 Y ( a ( − , x ) v = a ( − m − v + rT a ( − m ) v + · · · ∈ O g ( V ( g , k )) (3.11)for all v ∈ V ( g , k ) and m ≥
0. Repeating this relation for m = n , n − , . . . , it is clear that forsome scalar α , y ≡ O g ( V ( g ,k )) α · a (0) a ( − n − · · · a t ( − n t − x + shorter terms (3.12) ≡ O g ( V ( g ,k )) α · a ( − n − · · · a t ( − n t − a (0) x + shorter terms . (3.13)We may similarly peel off all the elements a , · · · , a j which are not in g and put them near x . Wethus see, for some scalar α ′ : y ≡ O g ( V ( g ,k )) α ′ · a j +1 ( − n − · · · a t ( − n t − · a j (0) · · · a (0) x + shorter terms . (3.14)Since y and a j +1 · · · a t are all fixed by g , a j (0) · · · a (0) a (0) x ∈ ( U ( g ) v ) . Now, again, (3.7) andinduction hypothesis give us that J y K g ∈ h R i . (cid:3) ow we recall a couple of important results that form the basis of all our calculations. Definition 3.6.
Recall that R is a g module under the L action. We have already chosen a Cartansubalgebra for g , namely h . Let R be the weight 0 subspace of R with respect h . Theorem 3.7. ([2, Lem. 3.4.3], [24, Prop. 13]) Let L ( λ ) be an irreducible highest-weight g -modulewith highest-weight λ and a highest-weight vector v λ . The following statements are equivalent.(1) L ( λ ) is an A g ( L ( g , k ))-module.(2) R · L ( λ ) = 0.(3) R · v λ = 0. Definition 3.8.
In the notation of the previous theorem, for every r ∈ R there exists a (unique)polynomial p r ∈ S ( h ) such that rv λ = p r ( λ ) v λ . Define P = { p r | r ∈ R } .We immediately have: Corollary 3.9. ([25, Cor. 2.10]) There is a one-to-one correspondence between:(1) Irreducible, highest-weight A g ( L ( g , k )) modules and(2) weights λ ∈ ( h ) ∗ such that p ( λ ) = 0 for all p ∈ P .We now present some calculations that will be used below. Let a ∈ g j , 0 < j < T , b ∈ g . Then,( a ( − ) ◦ g ( b ( − )= Res x (1 + x ) j/T x ( · · · + a ( − b ( − x + a (0) b ( − x − + a (1) b ( − x − ) ! = Res x X n ≥ (cid:18) j/Tn (cid:19) x n − ( · · · + a ( − b ( − x + a (0) b ( − x − + a (1) b ( − x − ) = a ( − b ( − + jT [ a, b ]( − + j ( j − T )2 T k h a, b i . (3.15)This implies that for a ∈ g j , (0 < j < T ) and b ∈ g , a ( − b ( − ≡ O g ( V ) − jT [ a, b ]( − − j ( j − T )2 T k h a, b i . (3.16)Or, equivalently, J a ( − b ( − K g = − jT J [ a, b ]( − K g − j ( j − T )2 T k h a, b i J K g . (3.17)Since h· , ·i is g -invariant, both sides are zero if b ( T − j ) = 0.For general elements a, b ∈ g , we have: a ( − b ( − = ( a (0) + · · · + a ( T − )( − b (0) + · · · + b ( T − )( − = (cid:16) a (0) ( − b (0) ( − + a (1) ( − b ( T − ( − + · · · + a ( T − ( − b (1) ( − (cid:17) + . . . (3.18)where the last ellipses denote terms that are in V (1) ⊕ · · · ⊕ V ( T − . So, using (3.4) and (3.17) J a ( − b ( − K g = J a (0) ( − b (0) ( − K g − X
From [25] we have:(1) The vector v = l X i =1 l − i + 12 l + 1 E θ ( − H i ( − + l − X i =1 E ,i +1 ( − E i +1 , l +1 ( − −
12 (2 l − E θ ( − (4.1)is a singular vector in V ( g , k ).(2) The ideal J ( g , k ) is generated by v , that is, J ( g , k ) = U ( b g ) v . Proof.
Our notation is slightly different from [25]. Negative of the singular vector given in [25] is: − l X i =1 l − i + 12 l + 1 H i ( − e θ ( − + l − X i =1 e ǫ − ǫ i +1 ( − e ǫ i +1 − ǫ l +1 ( − + 12 (2 l − e θ ( − , (4.2)where they define for i < je ǫ i − ǫ j = [ E j − , [ E j − , [ · · · [ E i +1 , E i ] · · · ]]] , e θ = e ǫ − ǫ l +1 . (4.3)It can be seen that e ǫ i − ǫ j = − ( − i − j E i,j , e θ = − E θ . (4.4)We thus get l X i =1 l − i + 12 l + 1 H i ( − E θ ( − + l − X i =1 E ,i +1 ( − E i +1 , l +1 ( − −
12 (2 l − E θ ( − . (4.5)In the first summation, [ H i ( − , E θ ( − < i < l and [ H i ( − , E θ ( − E θ ( −
2) if i = 1 , l . We thus get the required formula. (cid:3) Lemma 4.2.
We have ν ( v ) = v . Proof.
We have: ν ( v )= l X i =1 − l − i + 12 l + 1 E θ ( − H l +1 − i ( − + l − X i =1 E l − i +1 , l +1 ( − E , l − i +1 ( − + 2 l − E θ ( − = l X i =1 l − i + 12 l + 1 E θ ( − H i ( − + l − X i =1 E i +1 , l +1 ( − E ,i +1 ( − + 2 l − E θ ( − = l X i =1 l − i + 12 l + 1 E θ ( − H i ( − + l − X i =1 ( E ,i +1 ( − E i +1 , l +1 ( − − E , l +1 ( − + 2 l − E θ ( − = v, (4.6)where the first equality follows by definition of ν and the second by re-indexing the summations. (cid:3) Our next task is to calculate enough information about R = J U ( g ) v K so that we can use Corollary3.9. The g -weight of v is θ , and as g -module, U ( g ) v is isomorphic to the adjoint module of g with v E θ . As g -modules, we then have U ( g ) v ∼ = g ⊕ g . Note that E θ ∈ g and so U ( g ) v ∼ = g as g -modules. Since v is ν -fixed, we have R = J U ( g ) v K = J U ( g ) v K . From Section 2.3 we know thatdim( R ) = dim(( g ) ) = l and thus we seek l independent polynomials in P . emma 4.3. The projection of v on the twisted Zhu algebra is given by the following formula: J v K = l − X i =1 E + i +1 , l +1 E +1 ,i +1 . (4.7) Proof.
First, it is easy to see that: l X i =1 l − i + 12 l + 1 E θ ( − H i ( − −
12 (2 l − E θ ( − ∈ V ( g , − l −
12 ) . Using (3.19), we get: l − X i =1 J E ,i +1 ( − E i +1 , l +1 ( − K (4.8)= l − X i =1 J E +1 ,i +1 ( − E + i +1 , l +1 ( − K − J [ E − ,i +1 , E − i +1 , l +1 ]( − K − l + h E − ,i +1 , E − i +1 , l +1 i J K (4.9)= l − X i =1 J E +1 ,i +1 ( − E + i +1 , l +1 ( − K − J [ E − ,i +1 , E − i +1 , l +1 ]( − K . (4.10)For i = 1 , . . . , l −
1, we have:[ E − ,i +1 , E − i +1 , l +1 ] = 14 [ E ,i +1 + ( − i E l +1 − i, l +1 , E i +1 , l +1 + ( − i E , l +1 − i ] = 0 . (4.11)Using (3.6) for the first term, we get the required result. (cid:3) Lemma 4.4.
Consider E l +1 , − ( − l E l +1 ,l +1 ∈ g . Let v = 2( E l +1 , − ( − l E l +1 ,l +1 ) L J v K ∈ U ( g ) . (4.12)Then, we have: v =( − l X ≤ i 12 ( E l +1 , l +1 − ( − l E ,l +1 )( − E , + E l +1 , l +1 )= 12 ( − E , + E l +1 , l +1 )( E l +1 , l +1 − ( − l E ,l +1 ) + 12 ( E l +1 , l +1 − ( − l E ,l +1 )= ( − l E , − E l +1 , l +1 )( E ,l +1 − ( − l E l +1 , l +1 ) + 12 ( E l +1 , l +1 − ( − l E ,l +1 ) . The i > l term is: X l
Let 1 ≤ j ≤ l . Recall (2.29) and (2.30). We have: − ( − j ( f j f j − · · · f f j +1 f j +2 · · · f l ) L v = h j h j + 2 X j
It is not hard to see that for every 1 ≤ j ≤ l , ( f j f j − · · · f f j +1 f j +2 · · · f l ) L v ∈ R .Throughout this proof, it will be often beneficial for us to do the calculations in U ( g ) or U ( g ) .Since we are sure that the final answer is to be in U ( g ), we will carefully omit the terms not in U ( g ) that appear in the intermediate steps. Recall that n is spanned by E + i,j for 1 ≤ i < j ≤ l +1.The calculation corresponding to the term (4.13) is the longest and we break it into several steps.First, we consider the term E ,i +1 E i +1 ,l +1 . Let 1 ≤ i < j ≤ l . We have:( f j f j − · · · f i +1 f i · · · f f j +1 f j +2 · · · f l ) L ( E ,i +1 E i +1 ,l +1 )= ( F j F j − · · · F i +1 F i · · · F F j +1 F j +2 · · · F l ) L ( E ,i +1 E i +1 ,l +1 )= ( − l − j ( F j F j − · · · F i +1 F i · · · F ) L ( E ,i +1 E i +1 ,j +1 )= − ( − l − j ( F j F j − · · · F i +1 ) L ( H i E i +1 ,j +1 )= ( − l − j H i H j + U ( g ) n + + (4.29)Now let i = j , but note that we only allow 1 ≤ i < l in (4.13).( f j f j − · · · f f j +1 f j +2 · · · f l ) L ( E ,j +1 E j +1 ,l +1 )= ( F j F j − · · · F F j +1 F j +2 · · · F l ) L ( E ,j +1 E j +1 ,l +1 )= ( − l − j ( F j F j − · · · F ) L ( E ,j +1 H j +1 )= ( − l − j ( F j F j − · · · F ) L ( H j +1 E ,j +1 + E ,j +1 )= − ( − l − j ( H j +1 H j + H j ) + U ( g ) n + . (4.30) ow let i > j , again noting that we only allow 1 ≤ i < l in (4.13).( f j f j − · · · f f j +1 · · · f i f i +1 · · · f l ) L ( E ,i +1 E i +1 ,l +1 )= ( F j F j − · · · F F j +1 · · · F i F i +1 · · · F l ) L ( E ,i +1 E i +1 ,l +1 )= ( − l − i ( F j F j − · · · F F j +1 · · · F i ) L ( E ,i +1 H i +1 )= ( − l − i ( F j F j − · · · F F j +1 · · · F i ) L ( H i +1 E ,i +1 + E ,i +1 )= − ( − l − j ( H i +1 H j + H j ) + U ( g ) n + . (4.31)Note that if we place i = j in (4.31), we get (4.30), thus we combine these two equations. Combining(4.29), (4.30), (4.31) for a fixed 1 ≤ j ≤ l , we have:( f j f j − · · · f f j +1 f j +2 · · · f l ) L ( − l X ≤ i There is a very illuminating way to write the polynomials in (4.28). Let 1 ≤ j < l .Note that the coroot h ǫ j + ǫ j +1 is h j + 2 h j +1 + · · · + 2 h l − + h l which is the same as h j + 2 P j
In (4.44) and (4.45), notice that the coefficient of each of the ω i j ( j = 1 , . . . , k ) isan element of + Z . This means that we have obtained exactly two weights that are dominantintegral for B l . These correspond to S being the empty set: µ φ = 0 , µ ′ φ = ω l . These are preciselythe highest weights of the simple ordinary (i.e., Virasoro mode L (0) acts semisimply with finitedimensional weight spaces, and weights are bounded from below) ν -twisted modules.5. Admissibility and complete reducibility Admissibility. Due to the results in [22], every (weak) ν -twisted L ( sl l +1 , − l − )-module isnaturally a module for the twisted affine Lie algebra A (2)2 l of level − l − . As weights for A (2)2 l , theweights obtained in Theorem 4.7 become: λ S = (cid:18) − l − (cid:19) Λ c + µ S , λ ′ S = (cid:18) − l − (cid:19) Λ c + µ ′ S . (5.1)We now prove that these are admissible, see Definition 2.1. Theorem 5.1. For every S ⊆ { , , . . . , l − } , the weights λ S and λ ′ S are admissible for A (2)(2 l ) . Proof. Recall that ρ = ρ + h ∨ Λ c and observe that λ S + ρ = (cid:18) l + 12 (cid:19) Λ c + ρ + µ S , λ ′ S + ρ = (cid:18) l + 12 (cid:19) Λ c + ρ + µ ′ S . (5.2)First, let l = 1. Then, the only choice for S is the empty set φ , and we have two weights, µ φ = 0, µ ′ φ = ω = α . Let µ be one of these, and let λ be − Λ c + µ .Consider m ∈ Z , e α = ± α + mδ ∈ b Φ short+ . If m > 0, then, recalling (2.23), (2.15),( λ + ρ, e α ∨ ) = 21 (cid:18) 32 Λ c + 12 α + µ, ± α + mδ (cid:19) = 3 m ± ± µ, α ) > µ, α ) = 0 or . If m = 0, then, e α = α , and( λ + ρ, e α ∨ ) = 21 (cid:18) 32 Λ c + 12 α + µ, α (cid:19) = 1 + 2( µ, α ) > . (5.4)Now let m ∈ Z , e α = ± α + (2 m + 1) δ ∈ b Φ long+ . Necessarily, m ≥ λ + ρ, e α ∨ ) = 24 (cid:18) 32 Λ c + 12 α + µ, ± α + (2 m + 1) δ (cid:19) = 34 (2 m + 1) ± ± ( µ, α ) Z , (5.5)since ( µ, α ) = 0 or . Thus the first condition of admissibility is satisfied. or the second condition, note that α , δ − α ∈ b Φ short . We have α , δ − α ∈ b ∆ re λ φ since:( λ φ , ( α ) ∨ ) = 2 (cid:18) − 32 Λ c , α (cid:19) = 0 , ( λ φ , ( δ − α ) ∨ ) = 2 (cid:18) − 32 Λ c , δ − α (cid:19) = − . We have α , δ − α ∈ b ∆ re λ ′ φ since:( λ ′ φ , ( α ) ∨ ) = 2 (cid:18) − 32 Λ c + 12 α , α (cid:19) = 1 , ( λ ′ φ , ( δ − α ) ∨ ) = 2 (cid:18) − 32 Λ c + 12 α , δ − α (cid:19) = − . Now, let l > 1. Most of the work for this case has been already done in [24, Lem. 32]. Also, asin [24], the proof for λ S and λ ′ S is similar, so we only present the former.Suppose that α ∈ Φ short ∪ Φ long , m ∈ Z such that e α = α + mδ ∈ b Φ short+ ∪ b Φ intermediate+ . Then,recalling (2.23), (2.15), we get the following, exactly as in [24, Eq. 12]:( λ S + ρ, e α ∨ ) = (cid:18)(cid:18) l + 12 (cid:19) Λ c + ρ + µ S , ( α + mδ ) ∨ (cid:19) = 2( α, α ) (cid:18) m (cid:18) l + 12 (cid:19) + ( ρ, α ) + ( µ S , α ) (cid:19) . (5.6)In [24, Lem. 32], it was shown that the right-hand side does not belong to { , − , − , . . . } .Now suppose α = ± ǫ i ∈ Φ short ( i = 1 , . . . , l ) and m ∈ Z such that e α = 2 α + (2 m + 1) δ ∈ b Φ long+ .We have:( λ S + ρ, e α ∨ ) = 24 (cid:18) (2 m + 1) (cid:18) l + 12 (cid:19) + ( ρ + µ S , α ) (cid:19) = (2 m + 1) (cid:18) l (cid:19) + ( ρ + µ S , α ) . (5.7)Recalling (2.18), (2.20), we see that ( µ S , α ) ∈ Z . Recalling (2.22), we see ( ρ, α ) ∈ Z . Hence,( λ S + ρ, e α ∨ ) ∈ + Z , and thus not in { , − , − , . . . } .The proof for checking the second condition of Definition 2.1 (recall Remark 2.2)) is also similarto [24]. For i = 1 , . . . , k , denote the coefficient of ω i j in µ S by x i j ∈ + Z .Using (2.18) and (2.20), it is easy to see that for i ∈ { , . . . , l }\ S , ( λ S , α ∨ i ) = ( µ S , α ∨ i ) = 0. If i j ∈ S , δ − α i j ∈ b Φ intermediate . We have, again using (2.18) and (2.20):( λ S , ( δ − α i j ) ∨ ) = 22 ( λ S , δ − α i j ) = (cid:18)(cid:18) − l − (cid:19) − ( µ S , α i j ) (cid:19) = − l − − x i j ∈ Z . (5.8)Now, if S = { i , . . . , i k } has two or more elements, consider i j ∈ S with j = 1 , . . . , k − 1. Note, ǫ i j − ǫ ( i j +1 +1) = α i j + α i j +1 + α i j +2 + · · · + α i j +1 ∈ b Φ intermediate .( λ S , ( α i j + α i j +1 + α i j +2 + · · · + α i j +1 ) ∨ ) = ( µ S , ǫ i j − ǫ i j +1 +1 ) = x i j + x i j +1 ∈ Z . (5.9)If S has two or more elements, the observations above are enough to guarantee the second conditionof admissibility. If S has exactly one element, S = { i } , consider ǫ i = α i + α i +1 · · · + α l ∈ b Φ short .We have: ( λ { i } , ( α i + α i +1 · · · + α l ) ∨ ) = 2( µ { i } , ǫ i ) = 2 x i ∈ Z . (5.10)This, combined with the other observations is enough to handle the present case. Finally, if S isempty, consider δ − α l = b Φ short :( λ φ , ( δ − α l ) ∨ ) = 2( λ φ , δ − α l ) = 2 (cid:18)(cid:18) − l − (cid:19) − ( µ φ , α l ) (cid:19) = − l − ∈ Z . (5.11) (cid:3) .2. Semi-simplicity. Again, our proofs are parallel to the ones in [2], [25], [24] etc., with state-ments modified to accommodate the twist. Recall the notion of category O for representations ofaffine Kac-Moody algebras, [15, Ch. 9]. Theorem 5.2. ([16, Thm. 4.1]) Let g be any affine Lie algebra and let M be a g -module fromcategory O such that its every irreducible subquotient L ( λ ) with highest weight λ satisfies:(1) ( λ + ρ, α ∨ ) 6∈ {− , − , . . . , } for all α ∨ ∈ b ∆ ∨ , re+ and(2) ℜ ( λ + ρ, c ) > M is completely reducible.It is clear that our weights λ S , λ ′ S for all l ≥ S ⊆ { , . . . , l − } satisfy these conditions. Theorem 5.3. (cf. [24, Thm. 33]) Let M be a weak ν -twisted L ( sl l +1 , − l − )-module that is incategory O as a A (2)2 l -module. Then, M is completely reducible. Proof. Any irreducible subquotient L of M is also a ν -twisted L ( sl l +1 , − l − )-module that is incategory O as a A (2)2 l -module. Thus, the highest weight of L is λ S or λ ′ S , in particular it satisfies theconditions of Theorem 5.2. So, M is completely reducible as a A (2)2 l -module, and thus completelyreducible as a (weak) ν -twisted L ( sl l +1 , − l − )-module. (cid:3) Theorem 5.4. (cf. [24, Lem. 26]) Let M be an ordinary ν -twisted L ( sl l +1 , − l − )-module. Then, M is in category O as a A (2)2 l -module, in particular, M is completely reducible. Proof. M is a level − l − module for A (2)2 l [22], in particular, the central element c of A (2)2 l actssemi-simply on M . Clearly, every conformal weight space of M which is finite dimensional byassumption is a module for h . Thus, h acts semi-simply on M with finite dimensional weightspaces. 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