On simple Filippov superalgebras of type A(m,n)
aa r X i v : . [ m a t h . R A ] A ug ON SIMPLE FILIPPOV SUPERALGEBRASOF TYPE A ( m, n ) P. D. Beites and A. P. Pozhidaev
Abstract : It is proved that there exist no simple finite-dimensionalFilippov superalgebras of type A ( m, n ) over an algebraically closed fieldof characteristic 0. Keywords : Filippov superalgebra, n -Lie (super)algebra, (semi)simple(super)algebra, irreducible module over a Lie superalgebra. AMS Subject Classification (2000): 17A42, 17B99, 17D99
The concept of n -Lie superalgebra was presented by Daletskii and Kushnirevich, in [3], asa natural generalization of the n -Lie algebra notion introduced by Filippov in 1985 (see[4]). Following [5] and [10], in this article, we use the terms Filippov superalgebra andFilippov algebra instead of n -Lie superalgebra and n -Lie algebra, respectively. Filippovalgebras were also known before under the names of Nambu Lie algebras and Nambualgebras. As pointed out in [14] and [15], Filippov algebras are a particular case of n -aryMalcev algebras (generalizing the fact that every Lie algebra is a Malcev algebra). Wemay also remark that a 2-Lie superalgebra is simply known as a Lie superalgebra. Thedescription of the finite-dimensional simple Lie superalgebras over an algebraically closedfield of characteristic zero was given by Kac in [7].This work is one more step on the way to the classification of finite-dimensionalsimple Filippov superalgebras over an algebraically closed field of characteristic 0. In [9],finite-dimensional commutative n -ary Leibniz algebras over a field of characteristic 0 werestudied by the second author. He showed that there exist no simple ones. The finite-dimensional simple Filippov algebras over an algebraically closed field of characteristic0 were classified by Ling in [8]. Notice that an n -ary commutative Leibniz algebra isexactly a Filippov superalgebra with trivial even part, and a Filippov algebra is exactlya Filippov superalgebra with trivial odd part. Bearing in mind these facts, we considerthe n -ary Filippov superalgebras with n ≥
3, and with nonzero even and odd parts.Let G be a Lie superalgebra. We say that a Filippov superalgebra F has type G if Inder ( F ) ∼ = G (see definitions below). A description of simple Filippov superalgebras oftype B ( m, n ) was already obtained in [11], [13] and [12]. The same problem concerning1ilippov superalgebras of type A ( m, n ) with m = n has recently been solved in [1].Moreover, the type A (0 , n ), with n ∈ N , was studied in [2]. The present work representsthe final step towards the classification of finite-dimensional simple Filippov superalgebrasof type A ( m, n ) over an algebraically closed field of characteristic zero. Concretely, weestablish a negative answer to the existence problem of the mentioned superalgebras when m, n ∈ N and m = n .We start recalling some definitions.An Ω -algebra over a field k is a linear space over k equipped with a system of multi-linear algebraic operations Ω = { ω i : | ω i | = n i ∈ N , i ∈ I } , where | ω i | denotes the arityof ω i .An n -ary Leibniz algebra over a field k is an Ω-algebra L over k with one n -aryoperation ( · , · · · , · ) satisfying the identity(( x , . . . , x n ) , y , . . . , y n ) = n X i =1 ( x , . . . , ( x i , y , . . . , y n ) , . . . , x n ) . If this operation is anticommutative, we obtain the definition of
Filippov ( n -Lie ) algebraover a field.An n -ary superalgebra over a field k is a Z -graded n -ary algebra L = L ¯0 ⊕ L ¯1 over k ,that is, if x i ∈ L α i , α i ∈ Z , then ( x , . . . , x n ) ∈ L α + ... + α n .An n -ary Filippov superalgebra over k is an n -ary superalgebra F = F ¯0 ⊕ F ¯1 over k withone n -ary operation [ · , · · · , · ] satisfying[ x , . . . , x i − , x i , . . . , x n ] = − ( − p ( x i − ) p ( x i ) [ x , . . . , x i , x i − , . . . , x n ] , (1)[[ x , . . . , x n ] , y , . . . , y n ] = n X i =1 ( − p ¯ q i [ x , . . . , [ x i , y , . . . , y n ] , . . . , x n ] , (2)where p ( x ) = l means that x ∈ F ¯ l , p = P ni =2 p ( y i ) , ¯ q i = P nj = i +1 p ( x j ) , ¯ q n = 0. Theidentities (1) and (2) are called the anticommutativity and the generalized Jacobi identity,respectively. By (1), we can rewrite (2) as[ y , . . . , y n , [ x , . . . , x n ]] = n X i =1 ( − pq i [ x , . . . , [ y , . . . , y n , x i ] , . . . , x n ] , (3)where q i = P i − j =1 p ( x j ) , q = 0. Sometimes, instead of using the long term “ n -arysuperalgebra”, we simply say for short “superalgebra”. If we denote by L x = L ( x ,...,x n − ) the operator of left multiplication L x y = [ x , . . . , x n − , y ], then, by (3), we get[ L y , L x ] = n − X i =1 ( − pq i L ( x , . . . , L y x i , . . . , x n − ) , where L y is an operator of left multiplication and p its parity. (Here and afterwards, wedenote the supercommutator by [ , ]). 2et L = L ¯0 ⊕ L ¯1 be an n -ary anticommutative superalgebra. A subsuperalgebra B = B ¯0 ⊕ B ¯1 of the superalgebra L , B ¯ i ⊆ L ¯ i , is a Z -graded vector subspace of L suchthat [ B, . . . , B ] ⊆ B . A subsuperalgebra I of L is called an ideal if [ I, L, . . . , L ] ⊆ I . Thesubalgebra (in fact, an ideal) L (1) = [ L, . . . , L ] of L is called the derived subsuperalgebra of L . Put L ( i ) = [ L ( i − , . . . , L ( i − ], i ∈ N , i >
1. The superalgebra L is called solvable if L ( k ) = 0 for some k . Denote by R ( L ) the maximal solvable ideal of L (if it exists). If R ( L ) = 0, the superalgebra L is called semisimple . The superalgebra L is called simple if L (1) = 0 and L lacks ideals other than 0 or L .The article is organized as follows.In the second section we recall how to reduce the classification problem of simpleFilippov superalgebras to some question about Lie superalgebras, using the same ideasas in [8]. Concretely, we consider an existence problem for some skewsymmetric homo-morphisms of semisimple Lie superalgebras and their faithful irreducible modules. Thissection is followed with the third one where we collect some definitions and results onLie superalgebras that we will apply in the two last sections. We also fix some notationswith the same purpose.The fourth section is devoted to the problem of existence of finite-dimensional simpleFilippov superalgebras of type A ( m, n ) with m = n . We start with the particular case A (1 , n ) in the first subsection, where, taking into account [1] and [2], it is assumed that n ∈ N \ { } . The main result of this article (Theorem 4.2) is stated and proved in thesecond subsection.In each of the two mentioned subsections we restrict our considerations to the caseof the Lie superalgebra that gives the name to the type and solve the existence prob-lem of the mentioned skewsymmetric homomorphisms. It turns out that the requiredhomomorphisms do not exist. Therefore, there are no simple finite-dimensional Filippovsuperalgebras of type A ( m, n ) over an algebraically closed field of characteristic 0. More-over, as a corollary of its proof, we see that there is no simple finite-dimensional Filippovsuperalgebra F of type A ( m, n ) such that F is a highest weight module over A ( m, n ).In what follows, by Φ we denote an algebraically closed field of characteristic 0, by F a field of characteristic 0, by k a field and by h w υ ; υ ∈ Υ i a linear space spanned bythe family of vectors { w υ ; υ ∈ Υ } over a field (the field is clear from the context). Thesymbol := denotes an equality by definition. From now on, we denote by F an n -ary Filippov superalgebra. Let us denote by F ∗ ( L ( F )) the associative (Lie) superalgebra generated by the operators L ( x , . . . , x n − ), x i ∈ F . The algebra L ( F ) is called the algebra of multiplications of F . Lemma 2.1 [11]
Let F = F ¯0 ⊕ F ¯1 be a simple finite-dimensional Filippov superalgebraover a field of characteristic with F ¯1 = 0 . Then L = L ( F ) = L ¯0 ⊕ L ¯1 has nontrivialeven and odd parts. Theorem 2.1 [11] If F is a simple finite-dimensional Filippov superalgebra over a fieldof characteristic , then L = L ( F ) is a semisimple Lie superalgebra. n -ary superalgebra A with a multiplication ( · , · · · , · ), we have End ( A ) = End ¯0 A ⊕ End ¯1 A . The element D ∈ End ¯ s A is called a derivation of degree s of A if, forevery a , . . . , a n ∈ A, p ( a i ) = p i , the following equality holds D ( a , . . . , a n ) = n X i =1 ( − sq i ( a , . . . , Da i , . . . , a n ) , where q i = P i − j =1 p j . We denote by Der ¯ s A ⊂ End ¯ s A the subspace of all derivations ofdegree s and set Der ( A ) = Der ¯0 A ⊕ Der ¯1 A . The subspace Der ( A ) ⊂ End ( A ) is easilyseen to be closed under the bracket[ a, b ] = ab − ( − deg ( a ) deg ( b ) ba (known as the supercommutator ) and it is called the superalgebra of derivations of A .Fix n − x , . . . , x n − ∈ A , i ∈ { , . . . , n } , and define a transformation ad i ( x , . . . , x n − ) ∈ End ( A ) by the rule ad i ( x , . . . , x n − ) x = ( − pq i ( x , . . . , x i − , x, x i , . . . , x n − ) , where p = p ( x ) , p i = p ( x i ) , q i = P n − j = i p j .If, for all i = 1 , . . . , n and x , . . . , x n − ∈ A , the transformations ad i ( x , . . . , x n − ) ∈ End ( A ) are derivations of A , then we call them strictly inner derivations and A an inner-derivation superalgebra ( ID -superalgebra ). Notice that the n -ary Filippov superalgebrasand the n -ary commutative Leibniz algebras are examples of ID -superalgebras.Now let us denote by Inder ( A ) the linear space spanned by the strictly inner deriva-tions of A . If A is an n -ary ID -superalgebra then it is easy to see that Inder ( A ) is anideal of Der ( A ). Lemma 2.2 [11]
Given a simple ID -superalgebra A over k , the Lie superalgebra Inder ( A ) acts faithfully and irreducibly on A . Let F be an n -ary Filippov superalgebra over k . Notice that the map ad := ad n : ⊗ n − F 7→
Inder ( F ) satisfies[ D, ad ( x , . . . , x n − )] = n − X i =1 ( − pq i ad ( x , . . . , x i − , Dx i , x i +1 , . . . , x n − ) , for all D ∈ Inder ( F ), and the associated map ( x , . . . , x n ) ad ( x , . . . , x n − ) x n from ⊗ n F to F is Z -skewsymmetric. If we regard F as an Inder ( F )-module then ad inducesan Inder ( F )-module morphism from the ( n − ∧ n − F to Inder ( F )(which we also denote by ad ) such that the map ( x , . . . , x n ) ad ( x , . . . , x n − ) x n is Z -skewsymmetric. (Note that in ∧ n − F we have x ∧ . . . ∧ x i ∧ x i +1 ∧ . . . ∧ x n − = − ( − p i p i +1 x ∧ . . . ∧ x i +1 ∧ x i ∧ . . . ∧ x n − .) Conversely, if ( L, V, ad ) is a triple with L aLie superalgebra, V an L -module, and ad an L -module morphism from ∧ n − V L suchthat the map ( v , . . . , v n ) ad ( v ∧ . . . ∧ v n − ) v n from ⊗ n V to V is Z -skewsymmetric(we call the homomorphisms of this type skewsymmetric ), then V becomes an n -aryFilippov superalgebra by defining 4 v , . . . , v n ] = ad ( v ∧ . . . ∧ v n − ) v n . Therefore, we obtain a correspondence between the set of n -ary Filippov superalgebrasand the set of triples ( L, V, ad ), satisfying the conditions above.We assume that all vector spaces appearing in the following are finite-dimensionalover F .If F is a simple n -ary Filippov superalgebra then Theorem 2.1 shows that the Lie su-peralgebra Inder ( F ) is semisimple, and F is a faithful and irreducible Inder ( F )-module.Moreover, the Inder ( F )-module morphism ad : ∧ n − F 7→
Inder ( F ) is surjective.Conversely, if ( L, V, ad ) is a triple such that L is a semisimple Lie superalgebra over F , V is a faithful irreducible L -module, ad is a surjective L -module morphism from ∧ n − V onto the adjoint module L , and the map ( v , . . . , v n ) ad ( v ∧ . . . ∧ v n − ) v n from ⊗ n V to V is Z -skewsymmetric, then the corresponding n -ary Filippov superalgebra is simple. Atriple with these conditions will be called a good triple . Thus, the problem of determiningthe simple n -ary Filippov superalgebras over F can be translated to that of finding thegood triples. In this section, we recall some notations and results from [7] on the Lie superalgebra A ( m, n ) (and its irreducible faithful finite-dimensional representations). We also givesome explicit constructions which we shall use some later in the study of the simplefinite-dimensional Filippov superalgebras of type A ( m, n ). Let us start recalling thedefinition of induced module.Let L be a Lie superalgebra, U ( L ) its universal enveloping superalgebra [7], H asubalgebra of L , and V an H -module. The module V can be extended to U ( H )-module.We consider the Z -graded space U ( L ) ⊗ U ( H ) V (this is the quotient space of U ( L ) ⊗ V by the linear span of the elements of the form gh ⊗ v − g ⊗ h ( v ), g ∈ U ( L ), h ∈ U ( H )).This space can be endowed with the structure of a L -module as follows g ( u ⊗ v ) = gu ⊗ v, g ∈ L , u ∈ U ( L ) , v ∈ V . The so-constructed L -module is said to be induced fromthe H -module V and is denoted by Ind L H V .From now on, we denote by G a contragredient Lie superalgebra over Φ and considerit with the “standard” Z -grading [7, Sections 5.2.3 and 2.5.7].Let G = ⊕ i ≥− d G i . Set H = ( G ) ¯0 = h h , . . . , h n i , N + = ⊕ i> G i and B = H ⊕ N + . Let Λ ∈ H ∗ , Λ( h i ) = a i ∈ Φ, h v Λ i be an one-dimensional B -module for which N + ( v Λ ) = 0 , h i ( v Λ ) = a i v Λ . Let δ i ∈ H ∗ , δ i ( h j ) = δ ij where δ ij is Kronecker’s delta. Let V Λ = Ind GB h v Λ i /I Λ , where I Λ is the (unique) maximal submodule of the G -module V Λ .Then Λ is called the highest weight of the G -module V Λ . By [7], every faithful irreduciblefinite-dimensional G -module may be obtained in this manner. Note that the condition1 ⊗ v Λ ∈ V ¯0 ( V ¯1 ) gives a Z -grading on V Λ . Lemma 3.1 [11]
Let V be a module over a Lie superalgebra G , let V = ⊕ V γ i be its weightdecomposition, and let φ be a homomorphism from ∧ m V into G . Then, for all v i ∈ V γ i , φ ( v , . . . , v m ) ∈ G γ + ... + γ m , if γ + . . . + γ m is a root of G , φ ( v , . . . , v m ) = 0 , otherwise. G be a contragredient Lie superalgebra of rank n , U = Ind GB h v Λ i , and V = V Λ = U/N be a finite-dimensional representation of G , where N = I Λ is a maximal propersubmodule of the G -module V Λ . Let G = ⊕ α G α be a root decomposition of G relative toa Cartan subalgebra H . Denote by A the following set of roots: A = { α : g α / ∈ B } . Lemma 3.2 [13]
Let g α ∈ G α and g α ⊗ v = 0 ( v = v Λ ) . Then g jα ⊗ v ∈ U P ni =1 ( jα ( h i )+Λ( h i )) δ i for all j ∈ N , and there exists a minimal k ∈ N such that g kα ⊗ v ∈ N . Moreover, the set E α,k = { ⊗ v, g α ⊗ v, . . . , g k − α ⊗ v } is linearly independent in V . Setting h = [ g − α , g α ] ,we have1. Λ( h ) = − ( k − α ( h )2 if either g α ∈ G ¯0 or k / ∈ N ;2. α ( h ) = 0 if g α ∈ G ¯1 and k ∈ N . Remark 3.1
Note that if we start with a root β then there exists s ∈ N such that E β,s islinearly independent, but E α,k ∪ E β,s may not be linearly independent. Recall that a set E is called a pre-basis of a vector space W if hE i = W .Let { g k α . . . g k s α s ⊗ v ; k i ∈ N , α i ∈ A} be a pre-basis of U . As we have seen above,for every i = 1 , . . . , s , there exists a minimal p i ∈ N such that g p i α i ⊗ v ∈ N . Using theinduction on the word length, it is easy to show that { g k α . . . g k s α s ⊗ v ; k i ∈ N , k i < p i , α i ∈A} is a pre-basis of U/N .We finish this part with some more notations that we use in the two next sections: • the symbol . = denotes an equality up to a nonzero coefficient; • u, v t means that the elements u and v are t -times repeating u, v, . . . , u, v | {z } t , being theindex t omitted when its value is clear from the context. A ( m, n ) In what follows, considering A ( m, n ), we assume that m = n . Recall that A ( m, n ) := sl ( m + 1 , n + 1) for m = n and m, n ∈ N . It consists of the matrices of type A BC D ! , where A ∈ M ( m +1) × ( m +1) ( F ) , B ∈ M ( m +1) × ( n +1) ( F ) , C ∈ M ( n +1) × ( m +1) ( F ) , D ∈ M ( n +1) × ( n +1) ( F ) and tr ( A ) = tr ( D ). Let us write some elements in G = A ( m, n ): h i = e ii − e i +1 ,i +1 , i = 1 , . . . , m, m + 2 , . . . , m + n + 1 ,h m +1 = e m +1 ,m +1 + e m +2 ,m +2 ,e kl := g ǫ k − ǫ l ∈ G ǫ k − ǫ l , k, l = 1 , . . . , m + 1 or k, l = m + 2 , . . . , m + n + 2 , ∈ G ¯0 e kl := g ǫ k − ǫ l ∈ G ǫ k − ǫ l , k = 1 , . . . , m + 1 , l = m + 2 , . . . , m + n + 2 , or k = m + 2 , . . . , m + n + 2 , l = 1 , . . . , m + 1 . (cid:27) ∈ G ¯1 H := G = h h , . . . , h m + n +1 i is a Cartan subalgebra of A ( m, n ), and ǫ i are the linear functions on H defined by its values on h , . . . , h m + n +1 and the conditions ǫ i ( e jj ) = δ ij , where δ ij is Kronecker’s delta. Then ∆ = ∆ ∪ ∆ is a root system for A ( m, n ), where ∆ = { ǫ k − ǫ l , k, l = 1 , . . . , m + 1 or k, l = m + 2 , . . . , m + n + 2 } , and∆ = { ǫ k − ǫ l , k = 1 , . . . , m + 1 , l = m + 2 , . . . , m + n + 2 or k = m + 2 , . . . , m + n + 2 , l =1 , . . . , m + 1 } . The roots { α i := ǫ i − ǫ i +1 , i = 1 , . . . , m + n + 1 } are simple.The conditions deg g α i = 1 , deg g − α i = − A ( m, n ),[7, Section 5.2.3]. The negative part of this grading is G ǫ k − ǫ l , l < k . Because of this, theset E = (Y l Let V = V Λ be an irreducible module over G = A (1 , n ) with Λ =( a , . . . , a n +2 ) , a = 1 and a = 0 . Assume that ( G, V, φ ) is a good triple. Then a ≤ − .Proof. Suppose that φ ( u , . . . , u s ) = g ǫ − ǫ . Consider h = h + h = e + e . From thenonzero action on 1 ⊗ v , we obtain | − p h ( u i ) + a | ≤ 1. If a > − then p h ( u i ) > ,which leads to a contradiction since p h ( g ǫ − ǫ ) = 1. Thus, a ≤ − . Lemma 4.2 Let V = V Λ be an irreducible module over A (1 , n ) with Λ = ( a , . . . , a n +2 ) , a = 1 and a = . . . = a n = 0 . Then ( n Y i =2 g α i ǫ i − ǫ ⊗ v : α i ∈ { , } ) is a pre-basis of V . roof. Consider g ǫ i − ǫ with i = 1. Suppose that g ǫ i − ǫ ∈ G ¯0 . Then, by Lemma 3.2, g ǫ i − ǫ ⊗ v = 0. If g ǫ i − ǫ ∈ G ¯1 then, from [ g ǫ i − ǫ , g ǫ i − ǫ ] ⊗ v = 0, we obtain the sameconclusion. Lemma 4.3 Let V = V Λ be an irreducible module over A (1 , n ) with Λ = ( a , . . . , a n +2 ) , a = a = 0 and a i = 0 for i = 2 . Suppose that h = e + e and φ ( u , . . . , u s ) = g ǫ − ǫ .Then it is impossible that p h ( u i ) = a for all i ∈ { , . . . , s } .Proof. Notice that w = φ (1 ⊗ v, u , . . . , u s ) ∈ E , where E denotes the elements from A (1 , n ) with h -weight equal to 1: E = { g ǫ − ǫ i , g ǫ − ǫ i : i = 1 , } . If w = g ǫ − ǫ i then wecan multiply w by g ǫ i − ǫ n +3 to think that w = g ǫ − ǫ n +3 . We may proceed analogouslywith g ǫ − ǫ i . Now, if w = g ǫ − ǫ n +3 then we may multiply it by g ǫ − ǫ to arrive at g ǫ − ǫ n +3 .Thus, we may replace all u i either with g ǫ ji − ǫ ⊗ v ( j i > 3) or 1 ⊗ v (maybe, multipliedby α := g ǫ − ǫ ). Thus, we arrive at w = φ ( α δ g ǫ i − ǫ ⊗ v, . . . , α δ r g ǫ ir − ǫ ⊗ v, α δ r +1 g ǫ n +3 − ǫ ⊗ v, . . . ,. . . , α δ r + q g ǫ n +3 − ǫ ⊗ v, α ⊗ v t , ⊗ v p ) . = g ǫ − ǫ n +3 , where 3 < i j < n + 3 , δ k ∈ { , } . The action of h on w gives a ( r + q + t + p ) = 1, and theaction of e + e n +3 ,n +3 gives r = 2. The action of e n +2 ,n +2 − e n +3 ,n +3 gives either q = 1or q = 0.If q = then i = i = n +2, and the action of e − e n +2 ,n +2 implies t + P i =1 δ i = 3. If δ = 0 then g ǫ n +3 − ǫ ⊗ v y g ǫ n +3 − ǫ ⊗ v gives an element with ( e + e n +3 ,n +3 )-weight beingequal to − 2. If δ = 0 (or δ = 0) then g ǫ n +2 − ǫ ⊗ v y g ǫ n +3 − ǫ ⊗ v leads to an elementwith ( e + e n +2 ,n +2 )-weight being equal to − 2. Thus, δ = δ = δ = 1 , t = 0. Considerthe action of w on g ǫ n +3 − ǫ ⊗ v with the consecutive change αg ǫ n +3 − ǫ ⊗ v y g ǫ n +3 − ǫ ⊗ v .We get w = φ ( αg ǫ n +2 − ǫ ⊗ v, αg ǫ n +2 − ǫ ⊗ v, g ǫ n +3 − ǫ ⊗ v, ⊗ v p ) , and p ( w ) = − , p ( w ) = p n +2 ( w ) = 1 , i.e., w . = g ǫ − ǫ n +3 . The action of w on g ǫ n +3 − ǫ ⊗ v gives w = φ ( αg ǫ n +2 − ǫ ⊗ v, g ǫ n +3 − ǫ ⊗ v, g ǫ n +3 − ǫ ⊗ v, ⊗ v p ) = 0(note that we may assume n = 2, since otherwise the ( e + e )-weight of w is − p ( w ) = p ( w ) = p ( w ) = 0 , p ( w ) = − 1, and there is no an element in A (1 , 2) withsuch root.If q = then i = n + 2 , < i < n + 3, and the action of e − e n +2 ,n +2 implies t + P i =1 δ i = 2. If δ = 0 then g ǫ n +2 − ǫ ⊗ v y g ǫ n +3 − ǫ ⊗ v gives a ( e + e n +2 ,n +2 )-contradiction. Moreover, p e + e ( w ) = − i = 4. If n > e + e gives a contradiction. Thus, we may assume i = 4 , n = 3 and w = φ ( αg ǫ − ǫ ⊗ v, α δ g ǫ − ǫ ⊗ v, ⊗ v ) . = g ǫ − ǫ . The action on g ǫ − ǫ gives w = φ ( αg ǫ − ǫ ⊗ v, g ǫ − ǫ ⊗ v, ⊗ v ) = 0 . 8e have p ( w ) = − p ( w ) , p ( w ) = p ( w ) = 1 , p ( w ) = 0, i. e., w . = g ǫ − ǫ .From the action on g ǫ − ǫ ⊗ v , we arrive at w = φ ( g ǫ − ǫ ⊗ v, g ǫ − ǫ ⊗ v, ⊗ v ) = 0 . We have p ( w ) = p ( w ) = 0 , p ( w ) = − p ( w ) , p ( w ) = 1, and there is no anelement in A (1 , 3) with such root. Lemma 4.4 Let V = V Λ be an irreducible module over A (1 , n ) with Λ = ( a , . . . , a n +2 ) , a = a = 0 and a i = 0 for i = 2 . Suppose that h = e + e . Then it is impossible tohave φ ( a u , . . . , a u k , a v , . . . , a v s − k ) = g ǫ − ǫ , (6) where the superscripts denote the h -weights, for some k ≥ .Proof. Denote by E r the set of elements from A (1 , n ) with h -weight equal to r . Sup-pose first that s − k > 0. From (6), acting on 1 ⊗ v , we have w = φ ( u , . . . , u k , ⊗ v, v , . . . , v s − k ) ∈ E = { g ǫ − ǫ i , g ǫ − ǫ i : i = 1 , } . Assume that 1 ⊗ v is odd. Wemay think that w = g ǫ − ǫ since, otherwise, we can multiply w by g ǫ − ǫ i ( i > w = g ǫ − ǫ i ( i > 3) then, through the multiplication by g ǫ − ǫ , we arrive at w ′ = g ǫ − ǫ i .Thus, w . = g ǫ − ǫ i and we may act on t = g ǫ j − ǫ ⊗ v ( j = i if i > i = j > i = 2), interchanging this element with u . We obtain w = φ ( a − t , a +1 u , . . . , a +1 u k , a v , . . . , a v s − k ) ∈ E − = { g ǫ i − ǫ , g ǫ i − ǫ : i = 1 , } , where, here and throughout this proof, the h -weights are above the elements. If w = g ǫ i − ǫ then we may multiply it by g ǫ − ǫ toarrive at g ǫ i − ǫ . So, w . = g ǫ i − ǫ and we replace u by the action on t = g ǫ − ǫ a +1 ⊗ v .Repeating this procedure, we substitute t and arrive at a skewsymmetry contradiction.Suppose that 1 ⊗ v is even. We have w = g ǫ − ǫ i ( i > 3) or w = g ǫ − ǫ j ( j = 1 , w by g ǫ − ǫ to get the latter one. In the lattercase we can act on t to arrive at w . Here we repeat the above argument to arrive at w = φ ( t , v , . . . , v s − k ) ∈ E . If w = g ǫ − ǫ i ( i > 3) then we may multiply w by g ǫ i − ǫ n +3 to think that w = g ǫ − ǫ n +3 . We can do the same with g ǫ − ǫ i ( i > E = { g ǫ − ǫ , g ǫ − ǫ n +3 , g ǫ − ǫ , g ǫ − ǫ n +3 } . Notice that if w = g ǫ − ǫ n +3 then wecan multiply it by g ǫ − ǫ to arrive at g ǫ − ǫ n +3 . Let h = e + e n +3 ,n +3 . We remark thatthe element g ǫ − ǫ does not change either the h -weights or the h -weights. Therefore, wemay replace all v i with g ǫ i − ǫ ⊗ v, g ǫ n +3 − ǫ ⊗ v, ⊗ v (maybe multiplied by g ǫ − ǫ ). Addingthe h -weights, we get − k − r = − r ∈ N , which is impossible because k ≥ s − k = 0. Thus, (6) has the following shape φ ( a u , . . . , a u s ) = g ǫ − ǫ . (7)We may multiply it by g ǫ − ǫ i ( i > 3) to assume that φ ( u , . . . , u s ) = g ǫ − ǫ i . Throughthe multiplication by g ǫ − ǫ , we may assume that φ ( u , . . . , u s ) = g ǫ − ǫ i . We can nowinterchange u and g ǫ i − ǫ a − ⊗ v , and repeat the above described procedure to substitute9ll u i with t = g ǫ − ǫ ⊗ v . Thus, we arrive at φ ( t ) ∈ E . Considering the 3-weights,we obtain φ ( t , t ) = g ǫ − ǫ . From here, thinking in the 1-weights, we have a weightcontradiction. Lemma 4.5 Let V = V Λ be an irreducible module over G = A (1 , n ) with Λ =( a , . . . , a n +2 ) . Suppose that a = 0 , a = 0 and P n +2 i =3 a i = 1 . Assume that ( G, V, φ ) is a good triple. Then < a ≤ / .Proof. Suppose that φ ( u , . . . , u s ) = g ǫ − ǫ and consider H = h + h = e + e . Bythe nonzero action over 1 ⊗ v , we obtain | − p H ( u i ) + a | ≤ 1. From here and takinginto account that P si =1 p H ( u i ) = 1, we conclude that a ≤ / 2. Let H ′ = e + e n +3 ,n +3 .Assume that φ ( u , . . . , u s ) = g ǫ n +3 − ǫ . By the action on 1 ⊗ v , we have |− p H ′ ( u i )+ a − | ≤ 1. Whence, a > Theorem 4.1 There are no simple finite-dimensional Filippov superalgebras of type A (1 , n ) over Φ .Proof. Suppose that V is a finite-dimensional irreducible module over G = A (1 , n ) withthe highest weight Λ = ( a , . . . , a n ) ( a = 0), and φ is a surjective skewsymmetrichomomorphism from ∧ s V on G . Then there exist u i ∈ V γ i such that φ ( u , . . . , u s ) = g ǫ − ǫ . (8)By Lemma 3.1, P si =1 p ( u i ) = − 2. From Lemma 3.2, g a ǫ − ǫ ⊗ v = 0. Since φ is askewsymmetric homomorphism, φ ( u , . . . , u i − , g a − ǫ − ǫ ⊗ v, u i +1 , . . . , u s ) = 0. As p ( g a − ǫ − ǫ ⊗ v ) = 2 − a , the inequality | p ( u i ) + a | ≤ a ≥ a = 3 . In this case, p ( u i ) < 0. So, by (8), we have φ ( − u , − u ) = g ǫ − ǫ and, acting on 1 ⊗ v , we arrive at φ ( − u , ⊗ v ) . = g ǫ − ǫ . Acting on g ǫ − ǫ ⊗ v , weobtain φ ( g ǫ − ǫ ⊗ v, ⊗ v ) = 0, which is a weight contradiction.Now let us take a = 2 . As P si =1 p ( u i ) = − p ( u i ) ≤ 0, we canonly have i) φ ( − u , u , . . . , u s ) = g ǫ − ǫ or ii) φ ( − u , − u , u , . . . , u s ) = g ǫ − ǫ .First consider i). Let us suppose that 1 ⊗ v is even. Acting on 1 ⊗ v , wehave φ (1 ⊗ v, u , . . . , u s ) . = g ǫ − ǫ . Then, acting twice on g ǫ − ǫ ⊗ v , we arrive at φ ( g ǫ − ǫ ⊗ v , u , . . . , u s ) = 0 which leads to a skewsymmetry contradiction. To finishthe consideration of this subcase, suppose now that 1 ⊗ v is odd. Then, acting on 1 ⊗ v and, repeatedly, on g ǫ − ǫ ⊗ v , we get φ ( ⊗ v, g ǫ − ǫ ⊗ v ) . = g ǫ − ǫ . From here, analizingthe 2-weights, we conclude that a = − 1. Assume that φ ( u , . . . , u s ) = g ǫ − ǫ . Consider10 ′ = h + h = e + e . From the nonzero action on 1 ⊗ v , we have | − p h ′ ( u i ) + 2 | ≤ P si =1 p h ′ ( u i ) = 1. In the case ii), the multiplicationby g ǫ − ǫ gives either φ ( − u , − u , u , . . . , w, . . . , u s )(1 ⊗ v ) = 0 or φ ( − u , v , u , . . . , u s )(1 ⊗ v ) = 0 , for some w, v . In both cases, replacing u by 1 ⊗ v , we arrive at a weight contradiction.Now take a = 1 . Consider (8) and h ′′ = h + . . . + h n . By the nonzero action on1 ⊗ v , we have | − p h ′′ ( u i ) + a + . . . + a n | ≤ 2. So, we can deduce that a + . . . + a n < a r ∈ N for r = 2 then a + . . . + a n ∈ { , , } .In what follows, we analise these three possibilities, numbered with I), II) and III), for a + . . . + a n .I) Assume that a + . . . + a n = 2 and let h = h + h + . . . + h n . From (8), bythe action on 1 ⊗ v , we arrive at | − p h ( u i ) | ≤ 2. Thus, p h ( u i ) ∈ {− , , , , } and weobtain φ ( − u , − u , u , . . . , u s ) = g ǫ − ǫ , where the h -weights are above the elements. Themultiplication by g ǫ − ǫ leads to φ ( − u , w, u , . . . , u s )(1 ⊗ v ) = 0 or φ ( − u , − u , . . . )(1 ⊗ v ) = 0,for some w . In both cases, replacing u by 1 ⊗ v , we get a weight contradiction.II) Take h as above and suppose that a + . . . + a n = 1. Once again by the actionof g ǫ − ǫ on 1 ⊗ v , we get | p h ( u i ) | ≤ 2. So, we have either φ ( − u , − u , u , . . . , u s ) = g ǫ − ǫ or φ ( − u , u , . . . , u s ) = g ǫ − ǫ . In the former subcase, using the reasoning of I), we obtain moreweight contradictions. In the latter subcase, multiplying by g ǫ − ǫ , we arrive either at φ ( − u , . . . )(1 ⊗ v ) = 0, which gives a weight contradiction, or at φ ( w, u , . . . , u s )(1 ⊗ v ) = 0.Thus, we have φ (1 ⊗ v, u , . . . , u s ) ∈ { g ǫ − ǫ , g ǫ − ǫ n +3 , g ǫ − ǫ , g ǫ − ǫ n +3 } . Taking into accountthat p h ( g ǫ − ǫ ) = 0 = p h ( g ǫ n +3 − ǫ ) and making adequate multiplications, we may assumethat φ ( z, u , . . . , u s ) . = g ǫ − ǫ , for some z . Suppose first that g ǫ − ǫ ⊗ v = 0. Then, bythe action on 1 ⊗ v , we get a weight contradiction. Suppose now that g ǫ − ǫ ⊗ v = 0.Thus a = 0. Assume that φ ( u , . . . , u s ) = g ǫ n +3 − ǫ . Let h = e + e n +3 ,n +3 and, in whatfollows, consider the 2-weights above the elements and the h -weights underneath them.From the action on 1 ⊗ v , it is possible to conclude that p ( u i ) , p h ( u i ) ∈ {− , − , } .Thus we have φ ( − u , u , . . . , u s ) = g ǫ n +3 − ǫ . From here, considering the 2-weights and the h -weights, we arrive at φ ( − u , ⊗ − v, u , . . . , u s ) . = g ǫ i − ǫ , where i = 2 , , n + 3.11f g ǫ i − ǫ ⊗ v = 0 then the action on 1 ⊗ − v gives a weight contradiction. So, g ǫ i − ǫ ⊗ v = 0and there exists j > i such that g ǫ i − ǫ g ǫ j − ǫ i ⊗ v = 0. If j = n + 3 then p h ( g ǫ j − ǫ i ⊗ v ) = − g ǫ j − ǫ i ⊗ − v . Thus we mayassume that a n +2 = 1. Let us replace all u k ( k ≥ 3) with g ǫ n +3 − ǫ i ⊗ v and act one moretime on such element. We have φ (1 ⊗ − v, g ǫ n +3 − ǫ i ⊗ v, . . . , g ǫ n +3 − ǫ is − ⊗ v ) = 0.Considering the h , 2 and 1-weights, we conclude that φ (1 ⊗ − v, g ǫ n +3 − ǫ ⊗ v, . . . , g ǫ n +3 − ǫ ⊗ v ) . = g ǫ − ǫ n +3 . (9)From the multiplication by g ǫ n +3 − ǫ , we have φ ( g ǫ n +3 − ǫ ⊗ v ) . = h + h − h − . . . − h n +2 .The action on 1 ⊗ v leads to a contradiction.III) We now have a t = 0 for t ∈ { , . . . , n } . Suppose that a > 0. Consider H ′ = h + h = e + e and assume that φ ( u , . . . , u s ) = g ǫ − ǫ . (10)Acting on 1 ⊗ v , we have | − p H ′ ( u i ) + 1 + a | ≤ p H ′ ( u i ) ≥ a > a ≤ a < 0. Suppose that φ ( u , . . . , u s ) = g ǫ − ǫ . (11)By the action on 1 ⊗ v , we arrive at p ( u i ) ≤ a . We can’t have (11) if a < − 1. So, weconclude that a ≥ − 1. Now assume that φ ( u , . . . , u s ) = g ǫ n − ǫ . Consider the actionon 1 ⊗ v . On one hand, we have | − − p ( u i ) + a | ≤ a − ≤ p ( u i ) ≤ a < | − p ( u i ) | ≤ 2. Thus, we have p ( u i ) ∈ { , } . From now on, in thissubcase and unless stated otherwise, we will put the 1-weights above the elements andthe 2-weights underneath them. Taking into account the 1-weights, we can only have φ ( u , u , . . . , u s ) = g ǫ n − ǫ . (12)Acting on 1 ⊗ v allows us to obtain φ ( u , ⊗ a v, u , . . . , u s ) . = g ǫ − ǫ . Notice that a hasto be greater than − 1; otherwise, we obtain a weight contradiction. By Lemma 4.1, a ≤ − . Therefore, we arrive at a = − / φ ( u − / , ⊗ − / v ) . = g ǫ − ǫ . From theaction on g ǫ n − ǫ ⊗ v , we have φ ( g ǫ n − ǫ ⊗ − / v, ⊗ − / v ) = 0 . (13)Taking into account the 1-weights, the 2-weights and the (2 + n )-weights in (13), weconclude that we must have 12 ( g ǫ n +3 − ǫ ⊗ v, ⊗ v ) . = g ǫ n +3 − ǫ . (14)The nonzero action on 1 ⊗ v leads to φ (1 ⊗ − / v, ⊗ − / v ) . = g ǫ − ǫ . (15)If 1 ⊗ v is even then we have a skewsymmetry contradiction. If 1 ⊗ v is odd then weobtain the contradiction 0 = g ǫ − ǫ from the multiplication by g ǫ − ǫ in (15).Now let a = 0. Observe that, by Lemma 4.2, all 2-weights of the elements of thepre-basis of V are zero or positive. Therefore, it is impossible to find v i ∈ V such that φ ( v , . . . , v s ) = g ǫ − ǫ .Now take a = 0 . Suppose first that a = 0. Assume that φ ( u , . . . , u s ) = g ǫ − ǫ and H ′′ = h + . . . + h n +2 = e − e n +3 ,n +3 . By the action on 1 ⊗ v , we have | − p H ′′ ( u i ) + a + . . . + a n +2 | ≤ 2. From here, as P si =1 p H ′′ ( u i ) = 1, we conclude that a + . . . + a n +2 < a = . . . = a n +2 = 0; 2) P n +2 i =3 a i = 1.1) Consider a = a , h = e + e , v i = g ǫ i − ǫ , w j = g ǫ j − ǫ and φ ( u , . . . , u s ) = g ǫ − ǫ . (16)Then E = h v i . . . v i t w j . . . w j r ⊗ v : i p , j q = 1 , i is a pre-basis of V . Note that p h ( u i ) = a + t i , where t i ∈ Z and t i ≤ 1. Then P si =1 ( t i + a ) = 1 and, by the action of (16) on 1 ⊗ v ,we have | − ( t j + a ) + a | ≤ j . Thus, t j ∈ { , } and k (1 + a ) + ( s − k ) a = 1 forsome k ∈ N . By Lemma 4.3, k = 0. Notice also that k = 1. So, k ≥ 2. We can also seethat − < a < 0. From Lemma 4.4, we have that, for k ≥ 2, this subcase can not occur.2) Consider φ ( u , . . . , u s ) = g ǫ − ǫ and H = e + e n +3 ,n +3 . From the action on1 ⊗ v , we obtain | − − p H ( u i ) + a − | ≤ 1. From here and by Lemma 4.5, we get − < p H ( u i ) ≤ − . Therefore, a = , s = 2 and p H ( u ) = p H ( u ) = − . Notice that,from the same action, for h = e − e n +3 ,n +3 , we get p h ( u i ) ≥ 0. Henceforth, we have φ ( − / u , − / u ) = g ǫ − ǫ ,where the H -weights are above the elements and the h -weights underneath them. Actingon 1 ⊗ v , we obtain φ ( − / u , − / ⊗ v ) . = g ǫ − ǫ n +3 . The action on g ǫ n +3 − ǫ ⊗ v , taking intoaccount the weights over H, h, e + e n +3 ,n +3 and h , leads to φ ( g ǫ n +3 − ǫ ⊗ v, ⊗ v ) . = g ǫ − ǫ n +3 . Through the multiplication by g ǫ n +3 − ǫ we deduce that 1 ⊗ v is even and we arriveat φ ( g ǫ n +3 − ǫ ⊗ v ) . = g ǫ − ǫ . From here, multiplying by g ǫ − ǫ , we have the contradiction0 = h .At last, suppose that a = 0. We may assume that P n +2 i =3 a i = a > 0. Consider φ ( u , . . . , u s ) = g ǫ n +3 − ǫ . The action on 1 ⊗ v leads to | − − p ( u i ) | ≤ p ( u i ) ≤ i . If h = e + e n +3 ,n +3 then | − p h ( u i ) − a | ≤ 1. So, a = 1 and p h ( u i ) = 0for all i . Therefore, we have φ ( − u , ⊗ − v, u , . . . , u s ) = g ǫ i − ǫ where i = 2 , , n + 3, the2-weights and the h -weights are above and underneath the elements, respectively. If13 ǫ i − ǫ ⊗ v = 0 then u ⊗ v gives a h -weight contradiction. If g ǫ i − ǫ ⊗ v = 0 thenthere exists a j > i such that g ǫ i − ǫ g ǫ j − ǫ i ⊗ v = 0. If j = n + 3 then p h ( g ǫ j − ǫ i ⊗ v ) = − u g ǫ j − ǫ i ⊗ v gives a h -weight contradiction. Thus, we may assume that a n +2 = 1. We can replace u k with g ǫ n +3 − ǫ i ⊗ v , k ≥ 3. Continuing the process, we obtain w = φ (1 ⊗ v, g ǫ n +3 − ǫ i ⊗ v, . . . , g ǫ n +3 − ǫ is − ⊗ v ) = 0. Since the h n +2 -weight of w is 1, w ∈ { g ǫ j − ǫ n +3 , g ǫ n +2 − ǫ j : j = n + 2 , n + 3 } . Let h = e + e n +3 ,n +3 . But p h ( w ) = − w . = g ǫ n +2 − ǫ , which gives a h -weight contradiction, or w . = g ǫ i − ǫ n +3 .In the latter case, considering the h k -weights for k = 1 , , , . . . , we arrive at a weightcontradiction.To finish the proof, consider now a i = 0 for all i . Then V is trivial. We can now state and prove the main result of this article. Theorem 4.2 There exist no simple finite-dimensional Filippov superalgebras of type A ( m, n ) over Φ .Proof. We can suppose that G = A ( m, n ) with m = n and m ≥ 2, because we havealready proved that there exist no simple Filippov superalgebras of type A ( n, n ) with n ∈ N , [1], nor of type A (1 , n ) with n ∈ N \ { } and of type A (0 , n ) with n ∈ N ,[2]. Assume that V is a finite-dimensional irreducible module over G with the highestweight Λ = ( a , . . . , a m + n +1 ) ( a + . . . + a m = 0), and φ is a surjective skewsymmetrichomomorphism from ∧ s V on G . Then there exist u i ∈ V γ i such that φ ( u , . . . , u s ) = g ǫ m +1 − ǫ . (17)Let H = h + . . . + h m = e − e m +1 ,m +1 . By Lemma 3.1, P si =1 p H ( u i ) = − g a + ... + a m ǫ m +1 − ǫ ⊗ v = 0. Since φ is a skewsymmetric homomorphism, φ ( u , . . . , u i − , g a + ... + a m − ǫ m +1 − ǫ ⊗ v, u i +1 , . . . , u s ) = 0. As p H ( g a + ... + a m − ǫ m +1 − ǫ ⊗ v ) = 2 − a − . . . − a m ,the inequality | p H ( u i ) + a + . . . + a m | ≤ a + . . . + a m ≥ H -weights above theelements.Consider the case a + . . . + a m = 3 . Then we have φ ( − u , − u ) = g ǫ m +1 − ǫ and, actingon 1 ⊗ v , we arrive at φ ( − u , ⊗ v ) . = g ǫ − ǫ m +1 . By the action on g ǫ m +1 − ǫ ⊗ v , we obtain φ ( g ǫ m +1 − ǫ ⊗ v, ⊗ v ) = 0, which is a weight contradiction.Now let us take a + . . . + a m = 2 . Thus, there are two possibilitiesi) φ ( − u , u , . . . , u s ) = g ǫ m +1 − ǫ or ii) φ ( − u , − u , u , . . . , u s ) = g ǫ m +1 − ǫ .First consider i). Let us suppose that 1 ⊗ v is even. Acting on 1 ⊗ v , we have φ (1 ⊗ v, u , . . . , u s ) . = g ǫ − ǫ m +1 . Then, acting twice on g ǫ m +1 − ǫ ⊗ v , we arrive at14 ( g ǫ m +1 − ǫ ⊗ v , u , . . . , u s ) = 0 which leads to a skewsymmetry contradiction. To fi-nish the consideration of this subcase, suppose now that 1 ⊗ v is odd. Then, acting on1 ⊗ v and, repeatedly, on g ǫ m +1 − ǫ ⊗ v , we get φ ( ⊗ v, g ǫ m +1 − ǫ ⊗ v ) . = g ǫ − ǫ m +1 . Fromhere, analyzing the ( m + 1)-weights, we conclude that a m +1 = − 1. Assume that φ ( u , . . . , u s ) = g ǫ m +2 − ǫ m +1 . (18)From the nonzero action on 1 ⊗ v , we arrive at | − p H ( u i ) | ≤ 2. Consequently, we can’thave (18). In the case ii), the multiplication by g ǫ − ǫ m +1 gives, for some v i , either φ ( − u , − u , u , . . . , v i , . . . , u s )(1 ⊗ v ) = 0 or φ ( − u , v , u , . . . , u s )(1 ⊗ v ) = 0 , for some v i , v . In both cases, replacing u by 1 ⊗ v , we arrive at a weight contradiction.Now consider a + . . . + a m = 1 . Suppose that a m +1 > 0. Take h = h + ... + h m +1 = e + e m +2 ,m +2 , and assume that φ ( u , . . . , u s ) = g ǫ m +2 − ǫ . (19)Through the nonzero action on 1 ⊗ v , we have p h ( u i ) > a m +1 ≤ a m +1 < 0. Consider φ ( u , . . . , u s ) = g ǫ m +2 − ǫ m +1 . By the action on 1 ⊗ v ,taking into account the ( m +1)-weights and the h -weights, we arrive at − ≤ a m +1 ≤ − .Ia) Assume that a m +1 = − 1. Let φ ( u , . . . , u s ) = g ǫ m +2 − ǫ m +1 . By the action on 1 ⊗ v ,we have p H ( u i ) ∈ { , , , , } . Consequently, after the action on 1 ⊗ v , we arrive at φ ( u , ⊗ v, u , . . . , u s ) . = g ǫ − ǫ m +1 .Replacing every u k ( k ≥ 3) by g ǫ m +2 − ǫ ⊗ v and acting one more time on the mentionedelement, we get φ (1 ⊗ v, g ǫ m +2 − ǫ ⊗ v, . . . , g ǫ m +2 − ǫ ⊗ v ) = 0. Analyzing the H , h and1-weights involved, we conclude that a = 1 and φ (1 ⊗ v, g ǫ m +2 − ǫ ⊗ v, . . . , g ǫ m +2 − ǫ ⊗ v ) . = g ǫ − ǫ i , i = 1 , , m + 1 , m + 2.Through the multiplication by g ǫ m +2 − ǫ , we obtain φ ( g ǫ m +2 − ǫ ⊗ v ) . = g ǫ m +2 − ǫ i , i = 1 , , m +1 , m +2. Thus, considering the 2-weights, we have i = 3. Continuing the process, throughthe consecutive analises of the 2 , , . . . -weights, we eliminate all the possibilities for i .Ib) Assume that a m +1 = − 1. 1) Consider a = 1 and suppose that φ ( u , . . . , u s ) = g ǫ m +2 − ǫ m +1 . In this subcase, we put the ( m + 1)-weights above the elements andthe 1-weights underneath them. Through the action on 1 ⊗ v , we have p m +1 ( u i ) ∈{− , − , } , p H ( u i ) ∈ { , , , , } and p ( u i ) ∈ {− , , , , } . If there is a k such that p ( u k ) = − u k by 1 ⊗ v leads to a ( m + 1) , p ( u i ) ≥ g ǫ m +2 − ǫ m +1 on 1 ⊗ v , putting the H -weights in the third line, we arrive at 15 ( u , − ⊗ v , u , . . . , u s ) . = g ǫ − ǫ m +1 . By the action on g ǫ m +1 − ǫ ⊗ v , we have φ ( g ǫ m +1 − ǫ ⊗ v − , − ⊗ v , u , . . . , u s ) = 0. This isa weight contradiction since we don’t have an element in A ( m, n ) with the obtained m +1 , , H -weights. 2) Now consider a m = 1 and suppose that φ ( u , . . . , u s ) = g ǫ m +2 − ǫ m +1 .By the action on 1 ⊗ v , we conclude that p h ( u i ) ∈ { , , } , p m +1 ( u i ) ∈ {− , − , } and p m ( u i ) ∈ { , , , , } . So, we have φ ( u , u , . . . , u s ) = g ǫ m +2 − ǫ m +1 , (20)where the h -weights are above the elements and the ( m +1)-weights are underneath them.From the action on 1 ⊗ v , we obtain φ (1 ⊗ − v, u , . . . , u s ) . = g ǫ − ǫ m +2 ( g ǫ i − ǫ m +1 , i = 1 , m + 1 , m + 2).Consider the former possibility. As p m ( g ǫ m +2 − ǫ m +1 ) = 1, p m ( g ǫ − ǫ m +2 ) = 0, p m (1 ⊗ v ) = 1then, in (20), p m ( u ) = 2 and the sum of the m -weights of the remaining elements isequal to − 1, which is impossible. Consider the latter occasion. Notice that p m ( g ǫ i − ǫ m +1 )is either 1 (when i = m ) or 2 (when i = m ). If i = m then p m ( u ) = 1 and the change u ⊗ v in (20) gives a h, m +1 , m -weights contradiction. If i = m then p m ( u ) = 0 and,acting on 1 ⊗ v in (20), we have φ (1 ⊗ − v , u , . . . , u s ) . = g ǫ m − ǫ m +1 , where the m -weightsof the elements are in the third weight line. Through the action on g ǫ m +1 − ǫ m ⊗ v , weobtain φ (1 ⊗ − v , g ǫ m +1 − ǫ m ⊗ v − , u , . . . , u s ) = 0, one more weight contradiction. 3) Assumethat there exists a j ∈ { , . . . , m − } such that a j = 1. In this subcase, we putthe ( m + 1)-weights above the elements and the j -weights underneath them. Assumethat φ ( u , . . . , u s ) = g ǫ j +1 − ǫ j . The action on 1 ⊗ v allows us to conclude that p m +1 ( u i ) ∈{− , − , } and p j ( u i ) ∈ {− , − , − , , } . If, for example, p j ( u ) = 1 then, through thementioned action, we obtain the weight contradiction φ (1 − ⊗ v, u , . . . , u s − ) = 0. Supposenow that φ ( u , . . . , u s − , u s − − , u s − ) = g ǫ j +1 − ǫ j . Multiplying the last equality by g ǫ j − ǫ j +1 andacting on 1 ⊗ v , we get φ ( u , . . . , u s − , − ⊗ v ) = 0, which is a weight contradiction. Finally,we study the subcase u = φ ( u , . . . , u s − , u s − ) = g ǫ j +1 − ǫ j ( ∗ ). The action on 1 ⊗ v leads to w = φ ( u , . . . , u s − , − ⊗ v ) ∈ { g ǫ j − ǫ m +1 , g ǫ j − ǫ m +2 } . As p j − ( u ) = 1 and p j − ( w ) = − p j − ( u s ) = 2. So, through the action on 1 ⊗ v in ( ∗ ), we obtain the weight contradiction φ (1 − ⊗ v , u , . . . , u s − ≥ , u s − ) = 0, where the third weight line refers to h j − .16I) Consider the case a m +1 = 0. Assume that φ ( u , . . . , u s ) = g ǫ m +2 − ǫ . Take h asabove. Through the action on 1 ⊗ v , we have p m +1 ( u i ) , p h ( u i ) ∈ { , , } . Thus φ ( u , u , . . . , u s ) = g ǫ m +2 − ǫ , (21)where, here and in what follows, we consider the ( m + 1)-weights above the elements andthe h -weights underneath them. By the action on 1 ⊗ v in (21), we arrive at φ ( u , ⊗ v, u , . . . , u s ) = g ǫ m +2 − ǫ i , with i = 1 , m + 1 , m + 2 . (22)If g ǫ m +2 − ǫ i ⊗ v = 0 then we get a weight contradiction from the action on 1 ⊗ v . So, g ǫ m +2 − ǫ i ⊗ v = 0 and there exists a j < i such that g ǫ m +2 − ǫ i g ǫ i − ǫ j ⊗ v = 0. If j = 1 then p h ( g ǫ i − ǫ j ⊗ v ) = 1 and from the action on g ǫ i − ǫ j ⊗ v arises a weight contradiction. Wemay assume that a = 1. Let us replace all u k in (22), for k ≥ 3, by g ǫ ik − ǫ ⊗ v and actone more time on g ǫ i − ǫ ⊗ v . Then, looking at the ( m + 1) , h, u := φ (1 ⊗ v, g ǫ i − ǫ ⊗ v, g ǫ i − ǫ ⊗ v, . . . , g ǫ is − ǫ ⊗ v ) = g ǫ − ǫ t ( g ǫ m +2 − ǫ m +1 ),with t = 1 , , m + 1 , m + 2. Note that i k < m + 1, since otherwise q := i k = m + 2 forsome k and we arrive at ( e + e qq )-contradiction. Moreover, if u = g ǫ m +2 − ǫ m +1 then themultiplication on g ǫ − ǫ m +2 gives a contradiction. Now, for t = m, m − , . . . , 2, considering,consecutively, all these t -weights, we arrive at a weight contradiction.Finally, suppose that a + . . . + a m = 0 . As A ( m, n ) ≃ A ( n, m ), [7, Section 4.2.2],then a m +2 + . . . + a m + n +1 = 0. Thus, consider a t = 0 for t = m + 1 and a m +1 = a = 0.Assume that h = e + e m + n +2 ,m + n +2 . Let w = φ ( u , . . . , u s ) = g ǫ m +1 − ǫ . Then p h ( w ) = 0and p h ( u i ) = a + ǫ i with ǫ i ∈ Z . Take x = g ǫ m + n +2 − ǫ m +1 ⊗ v . We have wx = 0 and p h ( x ) = a + 1. If p h ( u i ) = a − ǫ i , for some i and ǫ i ∈ N , then u i y x gives a weightcontradiction. Therefore, a < 0. Now let w = φ ( u , . . . , u s ) = g ǫ m +3 − ǫ m +2 . Then p h ( w ) =0 and p h ( u i ) = a + ǫ i with ǫ i ∈ Z . Take x = g ǫ m +2 − ǫ ⊗ v . Then p h ( x ) = a − wx = 0.If p h ( u j ) = a + ǫ j , for some j and ǫ j ∈ N , then u j y x gives a weight contradiction.Henceforth, a > 0. Thus, a = 0 and the module is trivial. This finishes the proof of thetheorem. Corollary 4.1 There is no simple finite-dimensional Filippov superalgebra F of type A ( m, n ) such that F is a highest weight module over A ( m, n ) . References [1] P.D.Beites, A.P.Pozhidaev , On simple Filippov superalgebras of type A ( n, n ),Asian-European J. Math. 1, 4 (2008), 469–487.172] P.D.Beites, A.P.Pozhidaev , On simple Filippov superalgebras of type A (0 , n ),arXiv:1008.0120v1 [math.RA], (2010).[3] Y.Daletskii, V.Kushnirevich , Inclusion of Nambu-Takhtajan algebra in formal dif-ferential geometry structure, Dop. NAN Ukr. 4, (1996), 12–18.[4] V.T.Filippov, n -Lie algebras, Sib. Math. J. 26, 6 (1985), 879–891.[5] J.Grabowski, G.Marmo , On Filippov algebroids and multiplicative Nambu-Poissonstructures, Diff. Geom. 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Game TheoryAlgebra 13, 2 (2003), 89–108. P. D. BeitesDepartamento de Matem´atica and Centro de Matem´atica, Universidade da Beira InteriorCovilh˜a, Portugal E-mail adress : [email protected]. P. PozhidaevSobolev Institute of Mathematics and Novosibirsk State UniversityNovosibirsk, Russia E-mail adress : [email protected]: [email protected]