SHAPOVALOV FORMS FOR POISSON LIE SUPERALGEBRASMARIA GORELIK†AND VERA SERGANOVA†Incumbent of the Frances and Max Hersh career development chair.12 MARIA GORELIK , VERA SERGANOVA1. IntroductionPoisson Lie superalgebras are the superalgebras of functions on a symplectic super-manifold. Subquotients of Poisson superalgebras, called supe ralgebras of Hamiltonianvector fields, appear in the list of sim ple finite-dimensional Lie superal gebras (see [K]).If the dimension of a supermanifold i s even, then a Poisson superalgebra admits a non-degenerate invariant even symmetric form. In particular, there exists a Casimir operator.Poisson sup eralgebras also have root decomposition in the sense of [PS]. It was noticedin [GL] that in such situation it is possible to define a Shapovalov form using the ap-proach suggested in [KK]. We give a precise formula for determinant of Shapovalov formfor finite -dimensional Poisson superalgebra po(0|2n) with n ≥ 2. The case n = 1 iswell-known since po(0|2) is isomorphic to gl(1|1). We show that, contrary to the case ofclassical Lie superalgebras, the Jantzen filtration of a Verma m odule can be infini t e.One can use another approach to the problem of finding the Shapovalov form. It iswell known that there is a deformation Ghof the Poisson superalgebra po(0|2n) suchthat the Lie superalgebra Ghis i somorphic to gl(2n−1|2n−1) for h 6= 0. The Shapovalovform for the latter superal gebra is known (see [KKK]). Since the deformation preservesa Cartan subalgebra and triangular decomposition, one can obtain t he Shapovalov forpo(0|2n) isomorphic to G0by evaluating the Shapovalov form for Ghat h = 0. Howeverthis met hod seems more difficult. Indeed, several r oot subspaces are glued t ogether whenh = 0. The condition on weights of irreducible Verma modules also change dramatically.It seems that the direct approach using the Casimir operator works better. O ne canillustrate this on a simple exampl e. Indeed, it is much easier to evaluate the Shapovalovform for the Heisenberg algebra than to consider i t s deformation to sl(2) and go backusing the result for sl(2).2. Preliminary2.1. Poisson superalgebra po(0|n). Let Λ(n) be the Grassman superalgebra in ξ1, . . . , ξn.The Poisson Lie superalgebra po(0|n) can be described as Λ(n) endowed with the bracket[f, g] = (−1)p(f )+1Xi∂f∂ξi∂g∂ξi.It is easy to see that [g, g] =Pn−1i=0Λi(n). LetR: g → C be such a map that kerR=[g, g] andR(ξ1. . . ξn) = 1. For f, g ∈ Λ(n) defineB(f, g) : =Zfg.Clearly, B is a non-degenerate invariant bilinear f orm on g. If n is even, B gives r ise tothe quadratic Casimir element.In this text we consider the even case g := po(0|2n).SHAPOVALOV FORMS FOR POISSON LIE SUPERALGEBRAS 32.2. Triangular decompositions. A triangular decomposition of a Lie superalgebra gcan be constructed as follows (see [PS]). A Cartan subalgebra is a nilpotent subalgebrawhich coincides with its normalizer. It is proven in [ PS] that any two Cartan subalgebrasare conjugate by an inner automorphism. Fix a Cartan subalgebra h. Then g has ageneralized root decompositiong := h + ⊕α∈∆gαwhere ∆ is a subset of h∗andgα= {x ∈ g|(ad(h) − α(h))dimg(x) = 0}.In case considered in this paper a root space gαis ei t her odd or even. That allows oneto define the parity on the set of roots ∆. Denote by g0(resp., g1) the even (resp., odd)component of g. Denote by ∆0(resp. ∆1) the set of non-zero weights of g0(resp., g1)with respect to h. Then ∆ is a disjoint union of ∆0and ∆1.Now fix h ∈ h∗0satisfying α(h) ∈ R ⊂ {0} for all α ∈ ∆. Set∆+:= {α ∈ ∆| α(h) > 0},n+:=Pα∈∆+gαwhere gαis the weight space corresponding to α.Define ∆−and n−similarly. Then g = n−⊕ h ⊕ n+is a tr iangular decomposition.2.3. Notation. Denote by ∆+0(resp., ∆+1) the set of even (resp., odd) positive roots. LetQ ⊂ h∗be the root lattice that is the Z-span of ∆+and let Q+be the Z≥0-span of ∆+.Introduce the standard partial ordering on h∗by setting µ ≤ ν if ν − µ ∈ Q+.Throughout the paper α and β stand for positive roots.For α ∈ ∆+denote by Dαthe matrix of natural pairing gα× g−α→ h given by(x, y) 7→ [x, y].2.4. Verma modules. From now on suppo se that h is even and commutative. Setb : = h + n+. For each λ ∈ h∗define M(λ) := U(g) ⊗U(b)kλwhere kλis a one-dimensional b-module which is tri vial as n+-module and corresponds t o λ as h-module.Each Verma module has a unique maximal submoduleM(λ). The corresponding simplemodule V (λ) := M(λ)/M(λ) is called a highest weight simple module.2.5. Shapovalov determinants. For finite dimensional semisimple Lie algebras N. Shapo-valov ([Sh]) constructed a bilinear form U(n−) ⊗ U(n−) → S(h) whose kernel at a givenpoint λ ∈ h∗determines the maxi mal submoduleM(λ) of a Verma module M(λ). Inparticular, a Verma m odule M(λ) is simple if and only if the kernel of Shapovalov format λ is equal t o zero. The Shapoval ov form can b e r ealized as a direct sum of formsSν; for each Sνone can define its determinant (Shapovalov determ inant). The zeroes ofShapovalov determinants determi ne when a Verma module is reduc ible.4 MARIA GORELIK , VERA SERGANOVA2.5.1. A Shapovalov form for a Lie superalgebra g with an even commutative Cartansubalgebra h can be described as follows.Identify U(h) with S(h). Let HC : U(g) → S(h) b e the Harish-Chandra projection i.e.,a projection along the decomposition U(g) = U(h) ⊕ (n−U(g) + U(g)n+). Define a formU(n+) ⊗ U(n−) → S(h) by setting S(x, y) := HC(xy). Using the natural identificationof a Verma module M(λ) with U(n−), one easily sees thatM(λ) coincides with the“right kernel” of the evaluated form S(λ) : U(n+) ⊗ U(n−) → k i.e.,M(λ) = {y ∈U(n−)| (x, y)(λ) = 0 for all x}.Notice that S(x, y) = 0 if x ∈ U(n+)ν, y ∈ U(n−)−µand ν 6= µ. Thus S =Pν∈Q+Sνwhere Sνis the restriction of S to U(n+)ν⊗ U(n−)−νBy the above, dim V (λ)λ−ν=codim kerrSν(λ) where kerrstands for the “right kernel”.2.5.2. Assume that dim U(n+)ν= dim U(n−)−ν< ∞ for all ν ∈ Q+. Then det Sνis anelement of S(h) defined up to an invertible scalar. One obtains the following crit erion ofsimplicity of a Ver ma module: M(λ) is simple iff det Sν(λ) 6= 0 for all ν.2.6. Case g := po(0|2n). The algebra po(0|2n) admits a Z-gradingg = ⊕2n−2i=−2giwhich is obtained from the natural grading on Λ(2n) by the