Shapovalov determinant for loop superalgebras
aa r X i v : . [ m a t h . R T ] S e p SHAPOVALOV DETERMINANT FOR LOOP SUPERALGEBRAS
ALEXEI LEBEDEV , DIMITRY LEITES Abstract.
Let a given finite dimensional simple Lie superalgebra g possess an eveninvariant non-degenerate supersymmetric bilinear form. We show how to recover thequadratic Casimir element for the Kac–Moody superalgebra related to the loop superal-gebra with values in g from the quadratic Casimir element for g . Our main tool here isan explicit Wick normal form of the even quadratic Casimir operator for the Kac–Moodysuperalgebra associated with g ; this Wick normal form is of independent interest.If g possesses an odd invariant supersymmetric bilinear form we compute the cubicCasimir element.In addition to the cases of Lie superalgebras g ( A ) with Cartan matrix A for which theanswer was known, we consider the Poisson Lie superalgebra poi (0 | n ) and the relatedKac–Moody superalgebra. Introduction
Motivations: physics.
For a very lucid detailed exposition of these motivationsin the non-super case, see the text-book by P. Di Francesco, P. Mathieu and D. S´en´echal[DMS].Observe that, in various applications, the simple algebra which initiated the study isoften less interesting than certain its “relatives” (its nontrivial central extension, algebraof derivations and the result of iterations of these constructions). In what follows, havingthe simple object at the center of attention, we keep its relatives in view.
1) Stringy algebras . In a seminal paper [BPZ], A. Belavin, A. Polyakov and A. Zamolod-chikov observed that the infinite number of generators of the conformal group in thetwo-dimensional case generate the Ward identities for correlation functions, and thesedifferential equations (Ward identities) completely specify the behavior of the correlationfunctions. The components of the stress-energy operator in the conformal field theoryform, together with the central charge, the Virasoro algebra; this reduces the study ofthe conformal theory to the study of (irreducible) highest weight representations of theVirasoro algebra.V. Dotsenko and V. Fateev [DF] explicitly constructed a large class of conformal the-ories, so-called minimal models . In their study, the complete description of (irreducible)unitarizable highest-weight representations of a real form of the Virasoro algebra is vital.
Mathematics Subject Classification.
Key words and phrases.
Lie superalgebra, Shapovalov determinant.We are thankful to MPIMiS, Leipzig, and the International Max Planck Research School affiliated toit for financial support and most creative environment.
Alexei Lebedev , Dimitry Leites
2) Loop algebras . Bosonization of free fermions with spin and the internal symmetrygroup G provide with an example of a nontrivial conformal theory based on the Kac–Moody algebra d g (1) . The components of the stress-energy operator for these theories,built out of quadratic forms of fermion current operators for G (1) , satisfy the relations forthe Virasoro algebra with central charge C = dim g k + c g , where k is the value of the centralcharge of d g (1) and c g is the value of the (quadratic) Casimir operator of g in the adjointrepresentation, cf. [GO]. The Hamiltonian of the WZW model can also be built withquadratic forms of current operators and thus also represents a nontrivial conformal fieldtheory. The differential equations for the multi-point correlation functions of the WZWprimary fields are the Knizhnik-Zamolodchikov equations introduced in [KZ].In all these cases the description of the irreducibles is performed by means of theShapovalov determinant; in what follows we recall its definition and ways to compute it.
3) Super versions . For (relatives of) simple stringy superalgebras, the Shapovalovdeterminant was computed, except for a number of cases. (A recent paper [KW] claims tosolve all the cases, but has various omissions.) In this paper, we consider the Kac-Moodysuperalgebras; for the stringy Lie superalgebras, see [GLL].1.2.
Means: mathematics. The Casimir elements . From the above we already seethat Casimir elements (the elements of the center of the universal enveloping algebra or itscompletion) are very important. In many problems, it suffices to know only the quadraticCasimir, but we have to have it explicitly.
The Shapovalov determinant is a useful tool for verifying if certain representationsof Lie algebras and Lie superalgebras g with vacuum vector (either highest or lowestweight one), namely, the Verma modules, are irreducible or not, and even for constructing certain irreducible modules with vacuum vector.To d e f i n e the Shapovalov determinant, the algebra g must be rather “symmetric”,i.e.,(1) g must possess a Cartan subalgebra (i.e., a maximal nilpotent subalge-bra coinsiding with its normalizer) t whose even part is commutativeand diagonalizes g , and such that the weight 0 eigenspace of g (rela-tive t ¯0 ) coinsides with t ; g must possess an involutive (or super-involutive if we respect theSign Rule, see below) anti-automorphism σ which interchanges theroot vectors (with respect to t ¯0 ) of opposite sign.Recall that an anti-automorphism of a given Lie superalgebra g is a linear map σ : g −→ g that for every x, y ∈ g , satisfies(2) σ ([ x, y ]) = ( ( − p ( x ) p ( y ) [ σ ( y ) , σ ( x )] if we respect the Sign Rule , [ σ ( y ) , σ ( x )] if we ignore the Sign Rule . (An endomorphism σ is said to be involutive if σ = id and super-involutive if σ =( − p ( x ) id. Since the Shapovalov determinant is defined up to a scalar factor, and the Sign The formal definition, see [K], is a bit complicated: d g (1) is a certain subalgebra of the Lie algebra ofderivations of the central extension g g (1) of the loop algebra g (1) of maps S −→ g , where g = Lie( G ) is asimple finite dimensional complex Lie algebra, cf. (19). hapovalov determinant for Kac–Moody algebras Rule only affects its sign, we may choose a more convenient definition. In computationsit is usually more convenient to ignore the Sign Rule.)Additionally, the following finiteness condition should hold:(3) (a) the root spaces of g should be finite dimensional;(b) the number of partitions of any positive weight in U ( g + ), see (9),into positive roots of g should be finite.Nothing else is needed to d e f i n e the Shapovalov determinant, but to c o m p u t e it ismuch easier in presence of the even quadratic Casimir element C (of the center of U ( g )or its completion b U ( g )) or (which is not quite the same if dim g = ∞ , but suffices for ourpurposes) in presence of the non-degenerate E V E N bilinear form B on g . I n p r e s -e n c e o f s u c h a n e l e m e n t C , t h e S h a p o v a l o v d e t e r m i n a n t i s ap r o d u c t o f l i n e a r t e r m s; for various cases where this statement is proved, see[KK, GL1, Go2, Sh].1.3. Cases where an even quadratic Casimir exists.
Kac and Kazhdan [KK] com-puted the Shapovalov determinant for any Lie algebra with symmetrizable Cartan matrix(with arbitrary (complex) entries); their technique is literally applicable to Lie superal-gebras with symmetrizable Cartan matrix, as mentioned in [GL1] and expressed in detailin [Go2].Moreover, Kac and Kazhdan used the Shapovalov determinant to describe the
Jantzenfiltration of the Verma modules over the Lie algebras they considered.The technique of Kac and Kazhdan can be applied even in the absence of symmetrizableCartan matrix; its presence only makes computation of the needed values of the evenquadratic Casimir easier. So it is reasonable to look around for Lie (super)algebras withoutsymmetrizable Cartan matrix but with a non-degenerate even bilinear form. Grozman andLeites considered all such simple Lie superalgebras among Z -graded of polynomial growthand their relatives (for the same reasons that Kac–Moody algebras are “better” thansimple loop algebras, the finite dimensional Poisson Lie superalgebras poi (0 | n ) are “better”than their simple relatives h ′ (0 | n ), where h (0 | n ) = poi (0 | n ) / center and g ′ = [ g , g ]): Onestringy superalgebra, k L (1 |
6) (physicists call it N = 6 Neveu-Schwarz superalgebra), finitedimensional Poisson superalgebras poi (0 | n ) and Kac–Moody algebras associated with poi (0 | n ). More precisely, [GL1] contains computations of the quadratic Casimirs C anddescription of the irreducible Verma modules over k L (1 |
6) and poi (0 | n ).Gorelik and Serganova [GS1] wrote a sequel to [GL1] having produced a more explicitexpression of the Shapovalov determinant than that of [GL1]; their rather nontrivial resultis an explicit description of Jantzen’s filtration for the Verma modules over poi (0 | n ).There are many (perhaps, undescribably many) examples of filtered Lie (super)algebrasof polynomial growth (i.e., such that the associated graded Lie (super)algebras are ofpolynomial growth) and possessing a C , cf. [GL0, Ko]. The Shapovalov determinant iscomputed only for one (the simplest) of such algebras, namely, for gl ( λ ), where λ ∈ C ,and only for the simplest types of Verma modules ([Sh]).1.4. Cases where no even quadratic Casimirs exist. 1) There is an odd non-degenerate bilinear form . Such are the queer Lie superalgebra q ( n ) — a non-trivialsuper analog of gl ( n ), the Poisson superalgebras poi (0 | n + 1), and the Kac-Moody su-peralgebras associated with them. The hat over U means that the elements of b U ( g ) can be infinite sums of elements of U ( g ). Alexei Lebedev , Dimitry Leites
In the 1990s, Grozman and Leites conjectured that since q ( n ) and poi (0 | n + 1) aresuper analogs of gl ( n ), it is possible to compute their Shapovalov determinant (which isnot easy even to define in these and similar cases), but made a mistake computing it anddecided that the terms it factorizes into can be of any degree.Gorelik skilfully used the anti-center and suggested an elegant proof of the fact that,for q ( n ), the Shapovalov determinant factorizes in the product of l i n e a r polynomials,see [Go1].Among the simple Z -graded Lie superalgebras of polynomial growth that possess a non-degenerate invariant bilinear odd form there is only one stringy Lie superalgebra (we willconsider it in[GLL]) as well as q ( n ), poi (0 | n + 1), and Kac-Moody algebras associatedwith them.
2) There are no non-degenerate bilinear forms . Such are most of the stringy Liesuperalgebras (see [GLL]), and Lie superalgebras q (2) ( n ) with non-symmetrizable Cartanmatrices recently considered in [GS2].1.5. Our results and open problems.
Thus, there remained to consider, first of all,the Lie (super)algebras with properties (1), (3). The Kac–Moody (super)algebras associ-ated with the loop (super)algebras with values in finite dimensional “symmetric” Lie (su-per)algebras are among such algebras, so here we explicitly describe the Wick normal formof the Casimir operator for the Kac–Moody superalgebras d g (1) (as well as “twisted” Kac–Moody superalgebras c g ( r ) ) in terms of the the Casimir operator for a finite dimensionalsimple Lie superalgebra g . This implies a description of the Shapovalov determinant for d g (1) previously known only for Kac–Moody (super)algebras with Cartan matrix. We alsoconjecture a description of the Shapovalov determinant for Kac–Moody (super)algebrarelated with poi (0 | n ) for n > g possesses an odd invariant non-degenerate symmetric bilinearform, then U ( g ) (or [ U ( g )) contains a cubic central element. If dim g < ∞ , this implies, asa rule, that the Shapovalov determinant factorizes in a product of factors of degree ≤ poi (0 | n + 1).2. Background: Lie superalgebras
For a detailed background, see [GLS]. In what follows the ground field is C .2.1. The Poisson superalgebra.
Let G ( m ) be the Grassmann superalgebra generatedby θ , . . . , θ m . The Poisson Lie superalgebra (do not confuse with the Poisson-Lie (su-per)algebra) poi (0 | m ) has the same superspace as G ( m ) and the (Poisson) bracket is givenby { f, g } P.b. = ( − p ( f ) X j ≤ m ∂f∂θ j ∂g∂θ j for any f, g ∈ C [ θ ] . (4)It is often more convenient to re-denote the θ s and set (over C , over R such a transfor-mation is impossible so the brackets (4) and (6) are not equivalent over R )(5) ( ξ j = √ ( θ j − √− θ r + j ) η j = √ ( θ j + √− θ r + j ) for j ≤ r = h m i ,θ = θ r +1 if m is odd. hapovalov determinant for Kac–Moody algebras and accordingly modify the bracket (if m = 2 r , there is no term with θ ): { f, g } P.b. = ( − p ( f ) (cid:18) X j ≤ m (cid:18) ∂f∂ξ j ∂g∂η j + ∂f∂η j ∂g∂ξ j (cid:19) + ∂f∂θ ∂g∂θ (cid:19) . (6)The quotient of poi (0 | n ) modulo center is the Lie superalgebra h (0 | n ) of hamiltonianvector fields generated by functions: H f := ( − p ( f ) (cid:18) X j ≤ m (cid:18) ∂f∂ξ j ∂∂η j + ∂f∂η j ∂∂ξ j (cid:19) + ∂f∂θ ∂∂θ (cid:19) . (7)2.2. The integral.
The still conventional notation d n θ := dθ . . . dθ n for the volumeelement in the Berezin integral is totally wrong as was clear already in 1966 from theexplicit form of the Berezinian of the Jacobi matrix of the coordinate change. A reasonablenotation with compulsory indication of the coordinates is vol( θ ); for more motivations,see [Del, GLS].On poi (0 | n ), or rather on the space of generating functions, the integral — equal tothe coefficient of the highest term in Taylor series expansion in θ — determines a non-degenerate invariant supersymmetric bilinear form, of the same parity as n , given by( f | g ) := Z f g vol( θ ) . Cartan subalgebras, maximal tori, roots and coroots.
In [BNO, PS], it isshown that the Cartan subalgebras of simple and certain non-simple (like poi , q andtheir subquotients) finite dimensional Lie superalgebras are conjugate by inner auto-morphisms. We always fix a Cartan subalgebra t (for example, for poi (0 | n ), we take t = C [ ξ η , . . . , ξ n η n ]). For any α ∈ t ∗ ¯0 , set g α := (cid:8) x ∈ g | (ad h − α ( h )) N ( x ) = 0 for a sufficiently large N and every h ∈ t ∗ ¯0 (cid:9) . Then, as is not difficult to see,(8) g = L α ∈ t ∗ ¯0 g α and t ⊂ g . In what follows we only consider the algebras for which t = g .Denote by R the set of non-zero functionals α ∈ t ∗ ¯0 for which dim g α = 0; this R iscalled the set of roots of g . For the algebras we consider,there exists an H ∈ t such that α ( H ) ∈ R for all α ∈ R and α ( H ) = 0 for α = 0.This allows us to split the roots into positive and negative ones by setting(9) R + := { α ∈ R | α ( H ) > } , R − := { α ∈ R | α ( H ) < } ; g ± = L α ∈ R ± g α . If t = t ¯0 and commutative, we can identify U ( t ) with S ( t ); let HC be the Harish-Chandraprojection , i.e., the projection on the first direct summand in the following decomposition:(10) U ( g ) ≃ U ( t ) L (cid:0) g − U ( g ) + U ( g ) g + (cid:1) −→ S ( t ) ≃ U ( t ) . Alexei Lebedev , Dimitry Leites
The Shapovalov determinant. t is purely even and commutative. For a fixed λ ∈ t ∗ , set M λ = U ( g ) /I , where I is the left ideal generated by g + and the elements h − λ ( h ) for any h ∈ t . The g -module M λ is called the Verma module with highest weight λ . Obviously, as spaces, M λ ≃ U ( g − ) m λ , where m λ is the vacuum vector; and the U ( g − )-action on m λ is faithful,i.e., without kernel. The anti-automorphism σ , see (2), can be (uniquely) extended to ananti-automorphism of U ( g ):(11) σ ( x ⊗ . . . ⊗ x k ) = ( ( − P i X, X ′ , Y, Y ′ ∈ U ( g ) (not only of U ( g − )),if Xm λ = X ′ m λ and Y m λ = Y ′ m λ , then HC ( σ ( X ) Y )( λ ) = HC ( σ ( X ′ ) Y ′ )( λ ) . Obviously1) the elements of distinct weights of M λ are orthogonal to each other relative ( · | · ) λ ;2) the restriction of the bilinear form ( · | · ) λ onto the subspace M λ ( λ − χ ) of elementsof weight λ − χ is degenerate if and only if Sh χ ( λ ) = 0. 1) Any non-trivial submodule of M λ contains a non-zero element v whose weight is distinct from λ and g + v = 0 (such an element v is said to be a singular vector of M λ ).2) If v ∈ M λ is a singular vector, then U ( g − ) v — is a non-trivial submodule of M λ .3) x ∈ M λ is an isotropic element of the form ( · | · ) λ if and only if x can be representedas Av , where A ∈ U ( g − ) , and v is a singular vector of M λ .4) If v ∈ M λ is a singular vector of weight µ and C is a central element of U ( g ) , then HC ( C )( λ ) = HC ( C )( µ ) . Thanks to this statement to describe all irreducible Verma modules is the same as tocompute all Shapovalov determinants. Moreover, we see that Casimir elements help tocompute Shapovalov determinants.2) t ¯1 = 0. (This case is considered in more detail separately.) Among finite dimensionalsimple Lie superalgebras, this only happens with psq ( n ) and h ′ (0 | n − psq ( n ); that of h ′ (0 | n − 1) might, in theory, be obtained from her hapovalov determinant for Kac–Moody algebras results by contraction, but in applications an explicit formula is needed instead of a smalltalk. In what follows we offer a conjectural formula for the Shapovalov determinant for poi (0 | n + 1); conjecturally the answer for h ′ (0 | n + 1) for n > Casimir operators on Lie superalgebras: the case of an even invariantform In this section, let g be a finite dimensional Lie superalgebra, and ( · | · ) an invarianteven supersymmetric non-degenerate bilinear form on g . Let { e i } di =1 be a basis of g . Weset a ij = ( e i | e j ) and ( b ij ) = ( a ij ) − the inverse matrix . The quadratic element (14) Ω = X i,j b ij e i ⊗ e j ∈ U ( g ) . is a central (Casimir) element of U ( g ) , in particular, (15) [ x, Ω ] = 0 for all x ∈ g . Proof. It suffices to prove (15) for each x ∈ { e i } di =1 . Let c kij be structure constants in thisbasis. Let p i = p ( e i ) be the parities. We have[ e k , Ω ] = X i,j b ij ([ e k , e i ] ⊗ e j + ( − p i p k e i ⊗ [ e k , e j ]) = X i,j b ij X l ( c lki e l ⊗ e j + ( − p i p k c lkj e i ⊗ e l ) = X i,j X l ( b lj c ikl + ( − p i p k b il c jkl ) e i ⊗ e j . (16)So, to prove the statement, it suffices to show that(17) X l ( b lj c ikl + ( − p i p k b il c jkl ) = 0 for all i, j, k ∈ , d. The invariance of the form ( · | · ) implies that0 = ([ e p , e k ] | e r ) − ( e p | [ e k , e r ]) = X s ( c spk a sr − c skr a ps ) for all k, p, r ∈ , d. Let us multiply this equality by b ip b rj and sum over p and r . Since ( a ij )( b ij ) = 1 n , we get0 = X p c jpk b ip − X r c ikr b rj = X l ( c jlk b il − c ikl b lj ) = − X l (( − p k p l c jkl b il + c ikl b lj ) . Since ( · | · ) is even, the supermatrix ( b ij ) is also even, and the first term in the last sumcan be non-zero only if p i = p l , so this equality is equivalent to (17). (cid:3) Given a loop superalgebra g (1) = g ⊗ C [ t, t − ], define the Kac–Moody Lie superalgebra (18) d g (1) = Span ( g ⊗ C [ t, t − ] , u, z ) , where u = t ddt and z are even, Alexei Lebedev , Dimitry Leites with the following bracket:(19) [ t m x, t n y ] = t m + n [ x, y ] + mδ m, − n ( x | y ) z for all x, y ∈ g , m, n ∈ Z ;[ z, X ] = 0 for all X ∈ d g (1) ;[ u, t n x ] = nt n x for all x ∈ g , n ∈ Z . Set(20) Ω ′ = ∞ X n = −∞ X i,j b ij t n e i ⊗ t − n e j ∈ b U ( d g (1) ) . [ X, u ⊗ z + Ω ′ ] = 0 for any X ∈ d g (1) .Proof. It is easy to check this for X = u, z . Now let X = t m e k . Let us compute [ X, Ω ′ ]up to terms with z : we get ∞ X n = −∞ X i,j b ij ( t m + n [ e k , e i ] ⊗ t − n e j + ( − p i p k t n e i ⊗ t m − n [ e k , e j ]) = ∞ X n = −∞ X i,j b ij ( t m + n [ e k , e i ] ⊗ t − n e j + ( − p i p k t m + n e i ⊗ t − n [ e k , e j ]) = ∞ X n = −∞ “( t m + n ⊗ t − n ) · [ e k , Ω ]” = 0 , where the term in quotation marks is understood as follows: having represented [ e k , Ω ]as the sum of quadratic elements, we multiply, in each of them, the first factor by t m + n and the second one by t − n .Now, let us compute the terms with z in [ X, Ω ′ ]. We get these terms only from theterms in the sum with n = ± m , so we get X i,j b ij ( m ( e k | e i ) z ⊗ t m e j + ( − p k p i mt m e i ⊗ ( e k | e j ) z ) = m X i,j b ij ( a ki z ⊗ t m e j + ( − p k ( p i + p j ) t m e i ⊗ a kj z ) . (21)Since b ij = 0 only for p i = p j , and ( a ij )( b i j ) = 1 d , it follows that (21) is equal to m (cid:16) X j δ kj z ⊗ t m e j + X i δ ik t m e i ⊗ z (cid:17) = m ( z ⊗ X + X ⊗ z ) = 2 mX ⊗ Z. Since [ X, u ⊗ z ] = − mX ⊗ z , we see that [ X, u ⊗ z + Ω ′ ] = 0. (cid:3) Now we want to express the quadratic central element of b U ( d g (1) ) in the Wick normalform , i.e., so that, in every tensor product, the first factor has a non-positive degree withrespect to u , and the second one has a non-negative degree. The Wick normal form of anoperator Ω will be denoted by : Ω :. SetΩ pm = ∞ X n =1 X i,j b ij t − n e i ⊗ t n e j . hapovalov determinant for Kac–Moody algebras Let us compute [ X, Ω + 2Ω pm + 2 u ⊗ z ]. The bracket obviously vanishes for X = u and z ; so let X = t m x . A computation similar to the previous one shows that [ X, u ⊗ z ]cancels with monomials from [ X, Ω + 2Ω pm ] containing z , so we only need to compute[ X, Ω + 2Ω pm ] up to elements with z . If X ∈ g , then X commutes with Ω (as it wasshown before) and, similarly, with every term in the sum over n in Ω pm . Consider thecase X = t m x , where m > 0. From [ X, Ω pm ] we get ∞ X n =1 X i,j b ij ( t m − n [ e k , e i ] ⊗ t n e j + ( − p i p k t − n e i ⊗ t m + n [ e k , e j ]) = ∞ X n =1 X i,j b ij t m − n [ e k , e i ] ⊗ t n e j + ∞ X n = m +1 X i,j ( − p i p k t m − n e i ⊗ t n [ e k , e j ] = m X n =1 X i,j b ij t m − n [ e k , e i ] ⊗ t n e j + ∞ X n = m +1 ( t m + n ⊗ t − n ) · [ e k , Ω ] = m X n =1 X i,j b ij t m − n [ e k , e i ] ⊗ t n e j . From [ X, Ω + Ω pm ] we get ∞ X n =0 X i,j b ij ( t m − n [ e k , e i ] ⊗ t n e j + ( − p i p k t − n e i ⊗ t m + n [ e k , e j ]) = ∞ X n =0 X i,j b ij t m − n [ e k , e i ] ⊗ t n e j + ∞ X n = m X i,j ( − p i p k t m − n e i ⊗ t n [ e k , e j ] = − m − X n =0 X i,j b ij ( − p i p k t m − n e i ⊗ t n [ e k , e j ] + ∞ X n = m ( t m + n ⊗ t − n ) · [ e k , Ω ] = − m − X n =0 X i,j b ij ( − p i p k t m − n e i ⊗ t n [ e k , e j ] . Changing n m − n , i ↔ j , we get − m X n =1 X i,j b ji ( − p j p k t n e j ⊗ t m − n [ e k , e i ] = − m X n =1 X i,j b ij ( − p j ( p i + p k ) t n e j ⊗ t m − n [ e k , e i ] . So, for m > 0, we have[ t m e k , Ω + 2Ω pm + 2 u ⊗ z ] = m X n =1 X i,j b ij ( t m − n [ e k , e i ] ⊗ t n e j − ( − p j ( p i + p k ) t n e j ⊗ t m − n [ e k , e i ]) = m X n =1 X i,j t m b ij [[ e k , e i ] , e j ] = mt m X i,j b ij [[ e k , e i ] , e j ] . We similarly get the same result for m < 0. So we get the following: Alexei Lebedev , Dimitry Leites If, on g , the map A : x X i,j b ij [[ x, e i ] , e j ] is equal to λ id , where id is the identity operator, then the element Ω := Ω + 2Ω pm + 2 u ⊗ z + λu is central in b U ( d g (1) ) . Both Ω and Ω can be represented in the Wick normal form. The linear terms of : Ω : and : Ω : differ by λu . 1) Heading 2) holds since Ω is a finite sum, and 2Ω pm + 2 u ⊗ z + λu is already in the normal form.2) Conjecture: If A is not scalar, then no non-zero central quadratic element of b U ( d g (1) ),if any such exists, can be expressed in the Wick normal form. The map A commutes with the g -action.Proof. According to (17), we have[ e k , Ax ] = X i,j b ij ([[[ e k , x ] , e i ] , e j ] + ( − p k p ( x ) [[ x, [ e k , e i ]] , e j ]+( − p k ( p ( x )+ p i ) [[ x, e i ] , [ e k , e j ]]) = A [ e k , x ] + ( − p k p ( x ) X i,j b ij ([[ x, [ e k , e i ]] , e j ] + ( − p k p i [[ x, e i ] , [ e k , e j ]])and X i,j b ij ([[ x, [ e k , e i ]] , e j ] + ( − p k p i [[ x, e i ] , [ e k , e j ]]) = X i,j,l b ij ( c lki [[ x, e l ] , e j ] + ( − p k p i c lkj [[ x, e i ] , e l ]) = X i,j X l ( b lj c ikl + ( − p k p i b il c jkl ) ! [[ x, e i ] , e j ] = 0 . So we see that [ e k , Ax ] = A [ e k , x ] for all k ∈ , d and X ∈ g . (cid:3) From Schur’s lemma we deduce that if g is simple, then A is a scalaroperator. For example, if g = sl ( m | n ) and ( X | Y ) + tr( XY ), then A = 2( m − n ). Notethat if g is a direct sum of simple algebras, then A is a direct sum of the correspondingscalar operators, and hence it may be not a scalar. Let g = poi (0 | n ) . Then A = 0 if n > and not a scalar operator if n = 1 .Proof. If n = 1, then direct computation shows that A does not act by 0 on homogeneouselements of degree 1 and 2 whereas A (1) = 0 since 1 lies in the center.Note that if x, y are homogenous polynomials such that ( x | y ) = 0, then deg x +deg y =2 n . Thus, for any homogenous polynomial X , if AX = 0, then deg AX = deg X + 2 n − hapovalov determinant for Kac–Moody algebras If n = 2, then A ( θ θ θ θ ) = 0: This element must be either 0 or a homogenouspolynomial of degree 4, and the only (up to a scalar factor) such polynomial — θ θ θ θ — does not lie in g ′ . Since any element of the basis can be represented as ∂∂θ i . . . ∂∂θ i k θ θ θ θ = { θ i , { . . . , { θ i k , θ θ θ θ }} . . . } for k = 0, 1, 2, 3, 4 , where {· , ·} P.b. is the Poisson bracket and A commutes with the poi (0 | n )-action, we seethat A = 0.If n ≥ 3, then AX = 0 for any homogenous polynomial X of degree ≥ 5. Since bybracketing with a polynomial of degree 5 one can get any polynomial of degree ≤ 4, andsince A commutes with the algebra action, we deduce that A = 0. (cid:3) Twisted Kac–Moody (super)algebras. Now, let us consider the case where g can be represented, as a linear space, as e g ⊕ e g so that[ e g , e g ] + [ e g , e g ] ⊂ e g ; [ e g , e g ] ⊂ e g ;( e g | e g ) = 0 . In other words, the tilde indicates a Z / g ( a priori having nothing in commonwith the parity) respected by the (non-degenerate, even) inner product ( ·|· ).In this case, we can define a twisted Kac–Moody Lie superalgebra (22) d g (2) = Span ( e g ⊗ C [ t , t − ] , e g ⊗ t C [ t , t − ] , u, z ) , where u = t ddt and z are even, with the bracket (19). For the list of simple twistedKac-Moody super algebras, see [FLS].Let A still be a scalar operator λI . Let { e i } d i =1 be a basis of e g , let { e i } d i =1 be a basisof e g ; set a ij = ( e i | e j ); a ij = ( e i | e j ) , and ( b ij ) = ( a ij ) − , ( b ij ) = ( a ij ) − . Set Ω ′ = X i,j b ij e i ⊗ e j ∈ U ( e g );Ω ± = ∞ X n =1 X i,j b ij t − n e i ⊗ t n e j + X i,j b ij t − n +1 e i ⊗ t n − e j ! ∈ b U ( d g (2) );(23) Ω = Ω ′ + 2Ω pm + 2 u ⊗ z + λu. Note that e g is a subalgebra of g , and Ω ′ can be computed for e g in the same way asΩ was computed for g . Then, as in the previous section, we can prove the following The element Ω (see (23)) is a central element in b U ( d g (2) ) ; its linearpart is equal to the linear part of Ω ′ plus λu , i.e.,the linear part of Ω (not counting λu ) is defined by e g only. Similarly, if the algebra g has a Z /r -grading g = r − L s =0 e g s , we can construct the algebra c g ( r ) = Span( r − M s =0 e g s ⊗ t s C [ t r , t − r ] , u, z ) Alexei Lebedev , Dimitry Leites with the bracket (19). As earlier, set: e si — the basis elements in e g s ; a sij = ( e si | e sj ); ( b sij ) = ( a sij ) − ;Ω ′ = X i,j b ij e i ⊗ e j ∈ U ( e g );Ω pm = r − X s =0 ∞ X n =1 X i,j b sij t − n + s e si ⊗ t n − s e sj ∈ b U ( c g ( r ) ) . If A = λI , then Ω = Ω ′ + 2Ω pm + 2 u ⊗ z + λu is a central element.For the simple finite dimensional Lie (super)algebras having such a grading with r > sl (2 m + 1 | n + 1) with the automorphism “minus supertransposition” , g = osp (4 | √ 1) and o (8) with the gradings induced by the order 3 automorphisms), thesegradings are not compatible with the weight ones . Therefore the Cartan subalgebra of theLie (super)algebra c g ( r ) should be construct not on the base of the Cartan subalgebra h of g , but on the base of the Cartan subalgebra of ˜ g , which may have nothing in commonwith h . (For example, if g = sl (2 m + 1 | n + 1), then ˜ g ≃ o (2 m + 1) ⊕ o (2 n + 1), and h ∩ ˜ g = { } .) Therefore, although the Casimir elements of g and c g ( r ) look alike,their Shapovalov determinants look completely different .4. Casimir operators on Lie superalgebras: the case of an odd form In this section, g is a finite dimensional Lie superalgebra on which there is an invariant odd supersymmetric (here: this is the same as just symmetric) non-degenerate bilinearform ( · | · ). Let, as above, { e i } di =1 be a basis of g , and a ij = ( e i | e j ) , and ( b ij ) = ( a ij ) − . The cubic element C = X ( − p l c kij b im b jl e k ⊗ e l ⊗ e m is central in U ( g ) . Proof is similar as that of Statement 3.2. Although the expression of C looks less symmetric than that of Ω fromStatement 3.2, it is possible to show that the coefficient F ( k, l, m ) := X ( − p l c kij b im b jl of e k ⊗ e l ⊗ e m obeys the Sign Rule applied to e k ⊗ e l ⊗ e m , relative permutation of theindices k, l, m , i.e., F ( k, l, m ) = ( − p k p l F ( l, k, m ) = ( − p l p m F ( k, m, l ) . Therefore if g is not commutative, then the degree of C in U ( g ) is equal to 3 (and notless). Its order seems to be equal to 4, but for all superdimensions, except for (2 m + 1 | n + 1), its orderis equal to 2 modulo the group of inner automorphisms (see [Se], where all outer automorphisms of allfinite dimensional Lie superalgebras are listed). hapovalov determinant for Kac–Moody algebras Given a Lie superalgebra g with an odd invariant form, we can construct the Liesuperalgebra d g (1) = Span( g ⊗ C [ t, t − ] , u, z ) with relations (19), but with an odd z . If g is a simple Lie superalgebra, then ˆ U ( d g (1) ) possesses a degree central element that can be represented in the Wick normal form. Kac–Moody-type Lie superalgebras based on queer Lie superalgebras What to do if the Cartan subalgebra has odd elements. Here we considerShapovalov determinant for Lie superalgebras g with Cartan subalgebra t such that 0 1) We can construct Kac–Moody Lie superalgebras from Lie su-peralgebras q ( n ) and poi (0 | n − 1) (and their “relatives”) using the corresponding oddanalog of the invariant Killing form (so z is odd). Alexei Lebedev , Dimitry Leites In g = q ( n ) and sq ( n ), there is a central element E = 1 n | n , so [ t n E, d g (1) ] = Span( t n E, z ).For any element x ∈ U ( d g (1) − ), we have Z HC ( x ⊗ t n E ) = 0(since HC ( x ⊗ t n E ) can not contain the maximal product of odd elements of Cartansubalgebra), i.e., this element lies in the kernel of the BSh-form. If g = pq ( n ) (or, again, sq ( n )), then the invariant form is also degenerate. Since R HC ( x ⊗ t n X ) does not contain z for any element X from the kernel of the invariant form and any x ∈ U ( d g (1) − ), it followsthat t n X lies in the kernel of the BSh-form.Thus, in order the BSh-form be non-degenerate in the generic case, we must considerloops with values in g = psq ( n ); the same applies to poi (0 | n − h ′ (0 | n − g = poi or h ′ (perhaps g augmented by the grading operator).6. Explicit formulas Lie superalgebras with a symmetrizable Cartan matrix. Let g = g ( A, I ),where I is a set of indices labeling odd coroots h i ∈ t dual to the simple roots α i , be a Liesuperalgebra with symmetrizable Cartan matrix A = ( A ij ). Since A is symmetrizable,there exists an invertible diagonal matrix D = diag( d , . . . , d n ) such that B = DA is asymmetric matrix.On the root lattice ∆ = Span Z ( α , . . . , α n ), define:1) a symmetric bilinear form ( · , · ) by setting( α i , α j ) = B ij for any i, j = 1 , . . . , n ;2) a linear function F by setting F ( α i ) = 12 B ii for any i, j = 1 , . . . , n ; The function F is, essentially, an analog of the function ( ρ, · ), where if dim g = ∞ (the sums are takenmultiplicities counted) ρ := 12 X α ∈ R +¯0 α − X α ∈ R +¯1 α , where R +¯0 and R +¯1 are the subsets of R + consisting of even and odd roots, respectively, meaning that thecorresponding root vectors are even (resp. odd). hapovalov determinant for Kac–Moody algebras 3) for any γ = P k i α i ∈ ∆, set h γ = X k i d i h i (note that, generally, h α i = h i ).Let R ⊂ ∆ be a root system and R + the system of positive roots. Since every rootspace is either even or odd, every root can be endowed with a parity. Let R +¯0 = { α ∈ R + | p ( α ) = ¯0 , α R } ; R +¯1 = { α ∈ R + | p ( α ) = ¯1 , α R } . Define the partition function , also called Kostant function and denoted, when Kostantactively worked in representation theory, K in his honor, on the set of weights of U ( g − )by the formula K ( µ ) = dim U ( g − )( µ ) . ([M, Go2]). For the g ( A, I ) -module M χ , we have: (27)Sh χ = Q α ∈ R +¯0 Q m ∈ N (cid:0) h α + F ( α ) − m ( α, α ) (cid:1) K ( χ − mα ) Q α ∈ R +¯1 Q m =2 k +1 ∈ N (cid:0) h α + F ( α ) − m ( α, α ) (cid:1) K ( χ − mα ) Do not use this expression for χ which are not weights of U ( g + ) (forexample, for χ = kα , where α is an odd simple root such that 2 α is not a root and k > po (0 | n + 1). Let the indeterminates be ξ i , η i , θ , where 1 ≤ i ≤ n , and let thePoisson bracket be of the form[ f, g ] P.b. = ( − p ( f ) n X i =1 (cid:18) ∂f∂ξ i ∂g∂η i + ∂f∂η i ∂g∂ξ i + ∂f∂θ ∂g∂θ (cid:19) . For the Cartan subalgebra take C [ ξ η , . . . , ξ n η n , θ ]. Let ǫ i be the weight of ξ i , where i = 1 , . . . , n . For any weight γ = P c i ǫ i , set h γ := X c i ξ η . . . d ξ i η i . . . ξ n η n . Set also h max := ξ η . . . ξ n η n . Let g = po (0 | n + 1) , where n > , and let there be selected one ofthe systems of simple roots containing all the ǫ i . Then Sh χ factorizes in the product oflinear terms of the form h γ , where the γ are the roots of g , such that χ − γ positive orzero weights h max . For poi (0 | 3) and poi (0 | χ for one system. For poi (0 | ǫ , and for poi (0 | ǫ , ǫ , ǫ ± ǫ .Set: h = ξ η ; h = (the central element of po (0 | poi (0 | h = ξ η ; h = ξ η ; h = ξ η ξ η for poi (0 | . Alexei Lebedev , Dimitry Leites If g = poi (0 | , then Sh kǫ factorizes into the product of linearfactors of the form h and h − m/ , where m = 1 , . . . , k. If g = poi (0 | , then, for χ = kǫ + lǫ (such that χ is positive if k, k + l ≥ and at least one of the numbers k, l is positive), Sh χ factorizes into the productof linear factors of the form h ; h , if k, k + l ≥ ; h , if k + l ≥ ; h − h + m, for m ∈ Z , ≤ m ≤ k ; h + h − m, for m ∈ Z , ≤ m ≤ k, ( k + l ) / . In the two latter conjectures “scalar summands” are meant to be thescalar terms of U ( poi (0 | n + 1)), not the central elements of poi (0 | n + 1) itself.6.4. \ po (0 | n ) (1) . Let the indeterminants be ξ i , η i , where 1 ≤ i ≤ n , and the bracket in po (0 | n ) is of the form:[ f, g ] P.b. = ( − p ( f ) n X i =1 (cid:18) ∂f∂ξ i ∂g∂η i + ∂f∂η i ∂g∂ξ i (cid:19) . For the Cartan subalgebra we take C [ ξ η , . . . , ξ n η n ] ⊕ Span( u, z ). Let ǫ i be the weight of ξ i , where i = 1 , . . . , n , and ǫ ′ be the weight of t . For any weight γ = P c i ǫ i + c ′ ǫ ′ , we set h γ := X c i ξ η . . . d ξ i η i . . . ξ n η n + c ′ z. Set also h max := ξ η . . . ξ n η n . 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