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UNITAL BIMODULES OVER the SIMPLE JORDAN SUPERALGEBRA D(T)

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UNITAL BIMODULES OVER the SIMPLE JORDAN SUPERALGEBRA D(T) TRANSACTIONS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 358, Number 8, Pages 3637–3649 S 0002-9947(05)03821-3 Article electronically published on December 21, 2005 UNITAL BIMODULES OVER THE SIMPLE JORDAN SUPERALGEBRA D(t) CONSUELO MART´INEZ AND EFIM ZELMANOV Abstract. We classify indecomposable finite dimensional bimodules over Jor- dan superalgebras D(t), t = −1, 0, 1. 1. Introduction Throughout this paper all algebras are considered over a ground field k of char- acteristic zero. A (linear) Jordan algebra is a vector space J with a binary bilinear operation (x, y) → xy satisfying the following identities: xy = yx, (x2y)x = x2(yx). Z Z Respectively a Jordan superalgebra is a /2 -graded algebra J = J0¯ + J1¯ satis- fying the graded identities xy =(−1)|x||y|yx, ((xy)z)t +(−1)|y||z|+|y||t|+|z||t|((xt)z)y +(−1)|x||y|+|x||z|+|x||t|+|z||t| ((yt)z)x =(xy)(zt)+(−1)|y||z|(xz)(yt)+(−1)|t|(|y|+|z|)(xt)(yz). In [K2] (see also I. L. Kantor [Ka1, Ka2]), V. Kac classified simple finite dimen- sional Jordan superalgebras over an algebraically closed field of zero characteristic. This classification included the 1-parametric family of 4-dimensional superalgebras Dt, which corresponds to the 1-parametric family of 17-dimensional Lie superalge- bras via the Tits-Kantor-Koecher construction, Dt =(ke1 + ke2)+(kx + ky), with the products 1 1 e2 = e ,ee =0,ex = x, e y = y, xy = e + te ,i=1, 2. i i 1 2 i 2 i 2 1 2 The superalgebra Dt is simple if t =0.For t = 0 the superalgebra D0 is a unital hull of the 3-dimensional nonunital Kaplansky superalgebra K3, D0 = K3 + k1. In the case t = −1, the superalgebra D−1 is isomorphic to the Jordan superalgebra + M1,1(k) of 2 × 2 matrices. Received by the editors December 15, 2003 and, in revised form, August 18, 2004 and August 28, 2004. 2000 Mathematics Subject Classification. Primary 17C70. The first author was partially supported by BFM 2001-3239-C03-01 and FICYT PR-01-GE-15. The second author was partially supported by NSF grant DMS-0071834. c 2005 American Mathematical Society 3637 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 3638 CONSUELO MART´INEZ AND EFIM ZELMANOV A Jordan bimodule V over a Jordan (super)algebra J is a vector space with operations V × J → V , J × V → V such that the split null extension V + J is a Jordan (super)algebra (see [J1]). We denote a Jordan triple product by {x, y, z} =(xy)z +x(yz)−(−1)|x||y|y(xz). Let e be the identity of J and let V = {e, V, e} + {1 − e, V, e} + {1 − e, V, 1 − e} be the Peirce decomposition. Then {e, V, e} is a unital bimodule over J,thatis,e is an identity of {e, V, e} + J. The component {1 − e, V, e} is a one-sided module, that is, {J, {1 − e, V, e},J} =(0). Finally, {1 − e, V, 1 − e} is a bimodule with zero multiplication. One-sided finite dimensional bimodules over Dt were classified in [MZ]. In this paper we classify finite dimensional unital bimodules, thus finishing the classifi- cation of finite dimensional bimodules over Dt. Bimodules over semisimple finite dimensional Jordan algebras have been completely classified by N. Jacobson (see [J1], [J3]). 2. Irreducible modules For a set X let V (X) denote the free Dt-bimodule on the set of free generators X (see [J1]). Consider the linear operator R(a):V (X) → V (X), v → v · a, v ∈ V (X), a ∈ J. The algebra M(J) generated by all operators R(a), a ∈ J, is called the universal multiplicative enveloping algebra of J (see [J1]). Assume that t = −1 and consider the following elements of M(Dt): 2 −2 −2 E = R(x)2,F= R(y)2,H= (R(x)R(y)+R(y)R(x)). t +1 t +1 t +1 It is easy to see that [F, H]=2F ,[E,H]=−2E,[E,F]=H, kE + kF + kH sl2(k). Note that xH = −x, yH = y. ∈{¯ ¯} ∈{ 1 } ∈ Definition 2.1. For σ 0, 1 , i 0, 1, 2 , λ k,aVerma module V (σ, i, λ)is defined as a unital Dt-bimodule presented by one generator v of parity σ and the relations vR(e1)=iv, vR(y)=0,vH= λv. Remark. V (σ, i, λ)op = V (1 − σ, i, λ). ∈ 1 Lemma 2.1. For an arbitrary λ k, V (σ, 2 ,λ) =(0). Proof. Let (k, +) be the additive group of the field k. We will denote elements of µ µ ν µ+ν (k,+) as u , µ ∈ k, u u = u . Consider the group algebra Λ = k(k, +) = − { νi ∈ } µ → µ 1 i µiu ; µi,νi k and the derivation D : u µu of Λ. i − ∈ Let R = Λ,D = i≥0 ΛD , aD Da = D(a),a Λ, be the Weyl algebra. Consider the following 2 × 2 matrices over R: 0 −(t +1)D +(1− t)u−1 0 u x = ,y= , −(t +1)D 0 u 0 10 00 e = ,e = . 1 00 2 01 10 It is easy to see that 1 (xy − yx)= ,soe ,e ,x,y span a superalgebra 2 0 t 1 2 which is isomorphic to Dt. License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use UNITAL BIMODULES OVER THE SIMPLE JORDAN SUPERALGEBRA D(t) 3639 0 uλ Consider the odd element v = .Thenv · e = 1 v and vy − yv =0. uλ 0 i 2 ¯ 1 It is straightforward to verify that vH = λv. This implies that V (1, 2 ,λ) =(0). ¯ 1 Consequently, the module V (0, 2 ,λ) = (0). Lemma 2.1 is proved. −2 Lemma 2.2. V (σ, 1,λ)=(0)unless λ = t+1 . Proof. Let U(x, y)=R(x)R(y)−R(y)R(x)−R([x, y]). Clearly, vU(x, y)=−{x, v, y} ∈{e2,V,e2}. Hence, vU(x, y)R(e1)=0.Wehave vU(x, y)R(e1)=v(R(x)R(y) − R(y)R(x) − R(e1 + te2))R(e1) = v(R(x)R(y)+R(y)R(x) − R(e1 + te2))R(e1) t +1 = v(− H − R(e + te ))R(e ) 2 1 2 1 t +1 t +1 = − λv − v = −(λ +1)v =0, 2 2 − 2 which implies v = 0 unless λ = t+1 . Lemma 2.2 is proved. −2t Lemma 2.3. V (σ, 0,λ)=0unless λ = t+1 . Proof. Arguing as above we get vU(x, y)R(e2)=v(R(x)R(y) − R(y)R(x) − R(e1 + te2))R(e2) t +1 t +1 = v(− H − R(e + te ))R(e )=(− λ − t)v =0, 2 1 2 2 2 − 2t which implies v = 0 unless λ = t+1 . Lemma 2.3 is proved. − 2 − 2t To establish that V (σ, 1, t+1 ) =(0),V (σ, 0, t+1 ) = (0) we need to recall some facts about one-sided bimodules (see [MZ]). A Jordan bimodule V over a Jordan (super)algebra J is said to be one-sided if {J, V, J} =(0).IfJ is special with a universal associative enveloping algebra U(J) (see [J1]), then a one-sided bimodule V can be viewed as a right U(J)-module. For ∈ ∈ · 1 a J, v V we have v a = 2 va, where the left-hand side is the bimodule action, whereas the right-hand side is the right module action by a ∈ U(J). Similarly, we 1 · can make V a left module over U(J)via 2 av = a v. The tensor product V ⊗ V is an associative bimodule over U(J): a(v ⊗ w)b = av ⊗ wb; a, b ∈ J; v, w ∈ V . 1 2 − 1 2 − 1 Consider the elements e = 4(1+t) x , f = 1+t y , h = 2(1+t) (xy + yx)in U(Dt). We have [e, f]=h, [f,h]=2f ,[e, h]=−2e. 1 In [MZ] the following one-sided Verma bimodules over Dt were introduced. The right module V1(t)overU(Dt) is presented by one even generator v and the 2 −1−2t relations ve1 = v, vy =0,vh =( 1+t )v. The right module V2(t)overU(Dt) is presented by one even generator v and the − 1 relations ve1 = v, vy =0,vh = 1+t v. The tensor square V2(t) ⊗ V2(t) is a bimodule over U(Dt), hence a Jordan bi- module over Dt via 1 (v ⊗ v) · a = (v ⊗ va +(−1)|a|(|v |+|v |)av ⊗ v). 2 1 1 − Our notation differs from [MZ]. For example, we assume that 2 (xy yx)=e1 +te2 in U(Dt). License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 3640 CONSUELO MART´INEZ AND EFIM ZELMANOV We have 2 (v ⊗ v) · e = v ⊗ v, (v ⊗ v) · y =0, (v ⊗ v)H = − (v ⊗ v). 1 t +1 We proved − 2 Lemma 2.4. V (σ, 1, t+1 ) =(0). Now consider the tensor square V1(t) ⊗ V1(t)andtheelementvy ⊗ vy.Wehave ⊗ · ⊗ · ⊗ ⊗ · ⊗ − 2t ⊗ (vy vy) e1 =0,(vy vy) e2 = vy vy,(vy vy) y =0,(vy vy)H = t+1 vy vy. We proved − 2t Lemma 2.5. V (σ, 0, 1+t ) =(0). Lemma 2.6. For an arbitrary λ ∈ k: j j ≥ 1 (a) The elements vR(x) , vR(x)R(e1)R(x) , j 0,formabasisofV (σ, 2 ,λ).
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