Non-alternating Hamiltonian Lie algebras in three variables
aa r X i v : . [ m a t h . R A ] J a n Non-alternating Hamiltonian Lie algebras inthree variables
A. V. Kondrateva
National Research Lobachevsky State University of Nizhny Novgorod, Gagarin ave. 23,Nizhny Novgorod, 603950 Russia
E-mail: [email protected]
Abstract . Non-alternating Hamiltonian Lie algebras in three variables over a perfect field ofcharacteristic 2 are considered. A classification of non-alternating Hamiltonian forms over analgebra of divided powers in three variables and of the corresponding simple Lie algebras is given.In particular, it is shown that the non-alternating Hamiltonian form may not be equivalent to itslinear part. It is proved that the Lie algebras which correspond to the nonequivalent forms are notisomorphic.Keywords ands phrases:
Non-alternating Hamiltonian Lie algebra, non-alternating Hamiltonianform, characteristic two
Non-alternating Hamiltonian algebras over a field of characteristic 2 were first constructedin 1993 by Lin Lei [1] as Lie algebras of polynomials in divided powers with the symmetricPoisson bracket { f, g } = P i ∂ i f ∂ i g . In the case when the heights of the variables are equalto 1, non-alternating Hamiltonian Lie algebras are isomorphic to the first series of simple Liealgebras built by I. Kaplansky [2]. In [3, 4], symmetric differential forms in divided powersare introduced and non-alternating Hamiltonian Lie algebras similar to Hamiltonian Lie su-peralgebras of characteristic zero with respect to Poisson brackets with constant coefficientsare studied. The classification of alternating Hamiltonian forms over an algebra of truncatedpolynomial is obtained by M. I. Kuznetsov, S .A. Kirillov ([9]). The complete classifica-tion over a divided powers algebra is build by S. M. Skryabin ([10, 11]). In [12, 13] thegeneral theory of alternating Hamiltonian Lie algebras in divided powers is developed. Thenon-alternating case is considered in [5]. In particular, the classification of non-alternatingHamiltonian forms with constant coefficients over the divided powers algebra is obtained. Itis proved that any filtered deformation of a graded non-alternating Hamiltonian Lie algebra isdetermined by a non-alternating Hamiltonian form with polynomial coefficients. One of themain results of [5] is the proof of the equivalence of the non-alternating Hamiltonian form ω with polynomial coefficients to its initial form ω (0) , provided that the canonical form of ω (0) contains ( dx i ) (2) or dx i dx j + ( dx j ) (2) for some variable x i of the height greater than 1. Theproblem of the classification of non-alternating Hamiltonian forms for which this conditionis not satisfied remains open. In this paper, a classification of non-alternating Hamiltonianforms over algebras of divided powers in three variables with polynomial coefficients is ob-tained, and it is shown that if the above condition for the heights of the variables is notfulfilled, the non-alternating Hamiltonian form ω may not be equivalent to ω (0) . In addi-tion, the dimensions of the corresponding simple non-alternating Hamiltonian Lie algebrasare found and it is proved that the Lie algebras which correspond to the nonequivalent formsare not isomorphic.ACKNOWLEDGEMENTS. The investigation is funded by RFBR according to researchproject № 18-01-00900 and Ministry of Science and Higher education of Russian Federation(project 0729-2020-0055). 1 Preliminaries
Here we recall some basic definitions and results from [5]. Let K be a perfect field of charac-teristic 2, E be a vector space over K and F : E = E ⊇ E ⊇ . . . ⊇ E r ⊃ E r +1 = { } be aflag of E . A basis { x , . . . , x n } of E is called coordinated with the flag F if { x , . . . , x n } ∩ E i is a basis of E i , i = 1 , . . . , r .We use the notations and the definitions of algebras O ( E ) , O ( F ) , O ( n, m ) , W ( E ) , W ( F ) , W ( n, m ) from [6]. The maximal ideal of O ( F ) is denoted by m . Let { x , . . . , x n } be a basisof E coordinated with the flag F , m i = min { j | x i / ∈ E j } be the height of x i . The algebra O ( F ) is isomorphic to the algebra O ( n, m ) , m = ( m , . . . , m n ) . The flag F is restored bythe n -tuple m , E j = h x i | m i > j i , in particular E = h x i | m i > i .A flag F : E = E = E = . . . = E m − ⊃ E m = { } is called almost trivial. If m = 1 ,then F is trivial.Let S Ω( F ) = L i > S Ω i ( F ) be the algebra of symmetric differential forms, which is an O ( F ) -algebra of the divided powers (see also [3, 7]).Recall the definition of the differential d : S Ω r ( F ) → S Ω r +1 ( F ) . ( dω )( D , . . . , D r +1 ) = r +1 X i =1 D i ( ω ( D , . . . , c D i , . . . , D r +1 ))++ X i Theorem 3. (i) Let L be a non-alternating Hamiltonian Lie algebra associated with a pair ( F , ω ) , L be the standard maximal subalgebra of L . The minimal flag F ( L , L ) coincideswith F .(ii) Let ψ : ( L , L ) → ( L ′ , L ′ ) be an isomorphism of non-alternating Hamiltonian tran-sitive Lie algebras associated with pairs ( F , ω ) , ( F ′ , ω ′ ) , L , L ′ be the standard maximal ubalgebras of L , L ′ , respectively. Suppose that n = dim E > . Then ψ is induced by theadmissible isomorphism ϕ : O ( F ) → O ( F ′ ) and ϕ ( ω ) = aω ′ , a ∈ K \{ } .If L , L ′ are graded Lie algebras with the standard gradings then ψ = gr ψ : ( L , L ) → ( L ′ , L ′ ) is a homogeneous isomorphism of transitive Lie algebras which is induced by theadmissible linear isomorphism ϕ = gr ϕ : O ( F ) → O ( F ′ ) and ϕ ( ω ) = aω ′ , a ∈ K \{ } .Proof. (i) Let L = gr L = L − + L + . . . be the associated graded Lie algebra, L (0) = gr L . If G = P ( F , ω ) then gr G = P ( F , ω (0)) . Since ( gr G ) (1) ⊆ gr ( G (1) ) , P ( F , ω (0)) (1) ⊆ gr ( P ( F , ω ) (1) ) . Analogously, for G = e P ( F , ω ) , gr G = e P ( F , ω (0)) . Consequently, P ( F , ω (0)) (1) ⊆ L ⊆ e P ( F , ω (0)) , that is L is one of non-alternating Hamiltonian Lie algebras associated with the pair ( F , ω (0)) .By [8, Prop. 0.4] the minimal flag F ( L, L (0) ) is defined by the p -nilpotency degrees of el-ements ad L l , l ∈ L − = W − , where W ( F ) = W − + W + . . . is the standard grading of W ( F ) . For ∂ i ∈ W − , ad L ∂ i acts on e O ( F ) (cid:14) K as ∂ i . Hence, F ( L, L (0) ) = F . According to[8, Prop. 0.3] F = F ( L, L (0) ) F ( L , L ) F . Thus, F ( L , L ) = F .(ii) The proof is based on the theory of truncated coinduced modules (see [8, 12]). Herewe mainly concentrate on some key-points of the proof.According to (i) the natural embedding τ : ( L , L ) → ( W ( F ) , W ( F ) ) is a minimalone. By the embedding theorem [8, Th. 1.1] ψ is induced by an admissible isomorphism ϕ : O ( F ) → O ( F ′ ) , ϕ ( l ) = ϕlϕ − , l ∈ L . Let ( L , L ) be a transitive subalgebra of W ( F ) , Q be an L -module which is at the same time a finitely generated O ( F ) -module. We saythat L acts on Q by derivations if l ( f q ) = l ( f ) q + f ( lq ) , l ∈ L , f ∈ O ( F ) , q ∈ Q. It should be pointed out that as it follows from transitivity of ( L , L ) Q is a free O ( F ) -module. As well as in [12] we will call Q an ( O ( F ) , L ) -module. Denote W ( F ) by W , O ( F ) by O .Let L ♯ = Prim U ( L ) be the Lie algebra of the primitive elements of Hopf algebra U ( L ) . L ♯ is the restricted Lie algebra with respect to the p -mapping in U ( L ) . Let M be thenormalizer of L in L ♯ . Obviously, M is a restricted subalgebra of L ♯ . According to thetheory of truncated coinduced modules the category of ( O, L ) -modules is equivalent to thecategory of finite-dimensional restricted M -modules. The equivalence is given by the M -morphisms λ Q : Q → Q (cid:14) m Q . From the universal property of coinduced modules it followsthat λ Q results in the isomorphism of the spaces of invariants, Q L ∼ = ( Q (cid:14) m Q ) M .Now we consider the ( O, L ) -module Q = S Ω ( F ) = Hom O ( S O W, O ) . Let L = gr L = L − + L + . . . . Then we have Q (cid:14) m Q ∼ = Hom K ( S ( W (cid:14) W (0) ) , O (cid:14) m ) ∼ = Hom K ( S L − , K ) and λ Q ( ω ) = ω (0) . Obviously, ( Q (cid:14) m Q ) M ∼ = Hom K ( S ( W (cid:14) W (0) ) , K ) M ⊆ Hom K ( S ( W (cid:14) W (0) ) , K ) L ∼ = Hom K ( S L − , K ) L . Since L − is an absolutely irreducible L -module when n > , by Schur’s lemma Hom K ( S L − , K ) L = h ω (0) i K is a one-dimensional space. Since ω ∈ Q L , dim( Q (cid:14) m Q ) M dim Hom K ( S L − , K ) L = 1 . Hence, dim Q L = 1 which results in Q L = h ω i . Now, L ′ ϕ ( ω ) = ψ ( L ) ϕ ( ω ) = ϕ L ϕ − ϕ ( ω ) = 0 . ϕ ( ω ) = aω ′ , a ∈ K \{ } .Suppose that L , L ′ are graded Lie algebras with the standard gradings. Then ψ = gr ψ : L → L ′ is an isomorphism of graded Lie algebras. Obviously, ψ is induced by thelinear admissible isomorphism ϕ = gr ϕ : O ( F ) → O ( F ′ ) and ϕ ( ω ) = aω ′ , a ∈ K \{ } . Remark. The proof of the assertion in (ii) is true for n = 2 under additional restrictions.All we need is that L − should be an absolutely irreducible L -module. For ω = ( dx ) (2) +( dx ) (2) this is true if the set of heights m = ( m , m ) is not equal to (1 , . For ω = dx dx + ( dx ) (2) L acts irreducibly on L − if m > and m > .Everywhere in what follows we will assume that dim E = 3 . Let V be a three-dimensional vector space over a perfect field K of characteristic 2, b be abilinear nondegenerate non-alternating symmetric form on V , V be the hyperplane consistingof isotropic vectors (i.e., V = { v ∈ V | b ( v, v ) = 0 } ), F : 0 = V ⊆ V ⊆ . . . ⊂ V be a flagof V such that V q = V for sufficiently large q . Let { e i } be a basis of V coordinated with theflag F , m i = min { j | e i ∈ V j } be the height of e i . The orthogonal complement of a subspace L ⊆ V with respect to b is denoted by L ⊥ .Pairs ( F , b ) and ( F ′ , b ′ ) are called equivalent if there is a linear automorphism ϕ : V → V such that ϕ ( F ) = F ′ and b ( u, v ) = b ′ ( ϕ ( u ) , ϕ ( v )) for any u, v ∈ V . Such an automorphismis called admissible .The flag F will be called almost trivial if it consists of subspaces V = . . . = V q − ⊂ V q = V , the flag F is trivial if q = 1 . Remark. In the case of the almost trivial flag F , the pair ( F , b ) is equivalent to thepair ( F , b ′ ) , where b ′ has the identity matrix in any chosen basis of V .We say that a decomposition of a vector space V into the direct sum of its subspaces P and Q is coordinated with the flag F if V j = V j ∩ P + V j ∩ Q for all j > .Since dim V = 3 , the flag F consists of no more than four different subspaces. Supposethat { e i } is enumerated in such a way that m m m . Consider the subspaces V m , V m ∩ V ⊥ m , V m ∩ V ⊥ m = V ⊥ m and choose the vector v from one of these subspaces such that b ( v, v ) = 1 . Lemma 1. The decomposition V = h v i ⊕ h v i ⊥ is coordinated with the flag F .Proof. Set m = 0 . Let v ∈ V m q ∩ V ⊥ m q − for some q . Suppose that V i = V m j . If j < q , then V i ⊆ V m q − and V i ⊆ h v i ⊥ . If j > q , then V m q ⊆ V i and h v i ⊆ V i . Hence, for any i we have V i = V i ∩ ( h v i ⊕ h v i ⊥ ) = ( h v i ∩ V i ) ⊕ ( h v i ⊥ ∩ V i ) .Further, we do not assume that a basis coordinated with the flag F is ordered accordingto the heights, unless otherwise indicated.Suppose that { u, w } is a basis of the space h v i ⊥ , which is coordinated with the flag F = { V i ∩ h v i ⊥ } . Obviously, we can choose { u, w } in such a way that the matrix of therestriction of b to h v i ⊥ will be one of the following matrices M = (cid:18) (cid:19) , M = (cid:18) (cid:19) and I = (cid:18) (cid:19) . Remark. In the case of the matrix M , if the height of u is not less than that of w , thenthere exists a basis { u ′ , w ′ } , which is coordinated with the flag F , such that the matrix ofthe restriction of b to h v i ⊥ will be equal to the matrix I .Let u = e , w = e , v = e . Thus, we proved the following Lemma.5 emma 2. There exists a basis { e , e , e } of the space V coordinated with the flag F suchthat the matrix of b is equal to one of the following matrices. B = , B = , B = . Remark. In the case of the matrix B , if m belongs to the segment [ m , m ] , thenthe form b has the matrix B with respect to the basis { e ′ , e ′ , e ′ } , e ′ = e , e ′ = e + e , e ′ = e + e , which is also coordinated with F . Thus, we will use a basis of V in which thematrix of b is B only in the case when m < m and m / ∈ [ m , m ] .In view of this remark, we will say that the matrix of b with respect to a basis { e i } hasa canonical form if it is equal to one of the matrices B , B , B . Thus, the matrix B has acanonical form if m < m and m / ∈ [ m , m ] . In the case of the almost trivial flag we willsay that the matrix of b has a canonical form if it is equal to B . Theorem 4. Let b and b ′ be nondegenerate non-alternating symmetric forms on V , F and F ′ be flags of V . Suppose that { e , e , e } ( { e ′ , e ′ , e ′ } ) is a basis of V , which is coordinatedwith the flag F ( F ′ ) , such that the matrix of b ( b ′ ) has a canonical form. The followingstatements are true.(1) If the forms b and b ′ have the matrices equal to B , then the pairs ( F , b ) and ( F ′ , b ′ ) are equivalent if and only if { m , m } = { m ′ , m ′ } and m = m ′ .(2) If the forms b and b ′ have the matrices equal to B , then the pairs ( F , b ) and ( F ′ , b ′ ) are equivalent if and only if m i = m ′ i , i = 1 , , .(3) If the forms b and b ′ have the matrices equal to B , then the pairs ( F , b ) and ( F ′ , b ′ ) are equivalent if and only if { m , m , m } = { m ′ , m ′ , m ′ } .(4) If the forms b and b ′ have different matrices, then the pairs ( F , b ) and ( F ′ , b ′ ) arenot equivalent.Proof. (4) Since dim V = 3 , V = V ⊕ ( V ) ⊥ . Let V = ( V ) ⊥ . In the case of the full flag,for the form with the matrix B the decomposition V = V ⊕ V is coordinated with theflag, but for the form with the matrix B or with the matrix B it is not true. Obviously, n r = dim V s r ∩ V ⊥ s r − − dim V s r ∩ V ⊥ s r − ∩ V , where s < s < s ∈ { m , m , m } , are invariants(similar to invariants n qq from [5, Section 1]). For the form b with the matrix B we have n = n = n = 1 , but for the form b with the matrix B there is i such that n i = 0 .If m i = m j = m k , where { i, j, k } = { , , } , then for the form with the matrix B thedecomposition V = V ⊕ V is coordinated with the flag, but for the form with the matrix B it is not true.(1-3) If the conditions are fulfilled, then the isomorphism which maps e i into e ′ i (up topermutation) is an equivalence.Let in the cases (1) and (2) { m , m , m } = { m ′ , m ′ , m ′ } , but the conditions do not hold.Then n i = n ′ i for some i . Consequently, the pairs are not equivalent. The statement (3) istrivial. Put x i = x (2 mi − i , h x , x , x i = E, m i = min { j | x i / ∈ E j } . Let ω be the non-alternating Hamiltonian form, ω = ω (0) + dϕ + b x x dx dx + b x x dx dx + b x x dx dx , b ij ∈ K , ϕ ∈ m (2) S Ω ( F ) , and ω (0) is equal to one of the following forms (see Lemma 2and Theorem 4) ω (0) = dx dx + dx (2)3 , (2) ω (0) = dx dx + dx (2)2 + dx (2)3 , (3) ω (0) = dx (2)1 + dx (2)2 + dx (2)3 . (4)Here dx (2) i means ( dx i ) (2) . Recall that in the case of (3), m < m and m / ∈ [ m , m ] . Theorem 5. Let ω be a non-alternating Hamiltonian form in three variables of heights m , m , m . If the corresponding flag F is not almost trivial, then ω is equivalent to onlyone of the following forms ω = dx dx + dx (2)3 ,ω = dx dx + dx (2)2 + dx (2)3 ( m < m , m / ∈ [ m , m ]) ,ω = dx (2)1 + dx (2)2 + dx (2)3 ,ω = dx dx + dx (2)3 + x x dx dx ( m = 1) . The two forms ω i , ω j , i = j are not equivalent. If F is almost trivial but non-trivial, then ω is equivalent to the form ω . If F is trivial, then ω is equivalent to one of the forms ω , ω ,with the forms ω and ω being non-equivalent.Proof. By Theorem 1 and Theorem 2 if m > in the case of (2) and ( m , m , m ) = (1 , , in the case of (4), then the form ω is equivalent to the form ω (0) . In the case of (3), the form ω is equivalent to the form ω (0) .Suppose that ω is a form such that ω (0) is of type (2) and m = 1 . Theorem 1 andTheorem 2 imply that ω is equivalent to the form ω (0) + c x x dx dx + c x x dx dx .If c = 0 , c = 0 and m m , then the admissible automorphism x x +(˜ c / ˜ c ) x (2 m − m )2 , where (˜ c ) m = c , (˜ c ) m = c , x i x i , i = 2 , will lead to vanishing c x x dx dx . If m > m , then the admissible automorphism x x + (˜ c / ˜ c ) x (2 m − m )1 ,where (˜ c ) m = c , (˜ c ) m = c , x i x i , i = 1 , will lead to vanishing c x x dx dx .Consequently, a form ω is equivalent to the form dx dx + dx (2)3 + cx x dx dx . Obvi-ously, the two forms ω (0) + cx x dx dx and ω (0) + cx x dx dx , where ω (0) = dx dx + dx (2)3 are equivalent. The corresponding admissible isomorphism σ : O (3 , ( m , m , → O (3 , ( m , m , is induced by the permutation x x , x x , x x . Further-more, if c = 0 , then the admissible automorphism ˜ cx x , x ˜ cx , where (˜ c ) m = c , x x will lead to c = 1 .Suppose that ω is a form such that ω (0) is of type (4) and ( m , m , m ) = (1 , , . In thecase of the trivial flag, any linear isomorphism is admissible and ω is equivalent to ω ′ with ω ′ (0) of type (2). Consequently, a form ω is equivalent to the form ω ′′ = dx dx + dx (2)3 + cx x dx dx . If c = 0 , then c = 1 . Note that in this case the matrix of ω ′′ (0) does not havethe canonical form. If c = 0 , then we say that ω is equivalent to the form dx (2)1 + dx (2)2 + dx (2)3 .Suppose that F is not almost trivial and show that the forms ω i , ω j , i = j are notequivalent, i, j = 1 , , , . The fact that the forms ω i , i = 1 , , are not equivalent followsfrom Theorem 4.Let ϕ be an admissible isomorphism, ϕ : O ( F ) → O ( F ′ ) , ϕ be the linear part of ϕ , ϕ ( x i ) ≡ ϕ ( x i )( mod m (2) ) . Since ( ϕ ( ω ))(0) = ϕ ( ω (0)) , we conclude that if linear parts ofthe forms ω , ω ′ are not equivalent, then ω and ω ′ are not equivalent, either. So, all whatremains to be considered is the pair ω , ω . Suppose that the forms ω and ω are equivalent.7ence, ω (0) = dx dx + dx (2)3 is equivalent to ω (0) = dx ′ dx ′ + ( dx ′ ) (2) , where { x , x , x } and { x ′ , x ′ , x ′ } are bases of E coordinated with the flags F , F ′ , respectively. By Theorem 4 { m , m } = { m ′ , m ′ } and m = m ′ . Since m ′ = 1 , we have m = 1 . Theorem 6 of the nextsection implies that P (3 , m, ω ) is a simple Lie algebra and P (3 , m, ω ) is not a simple Liealgebra.If F is almost trivial and m = (1 , , , then ω (0) = ω and, as it has been shown before, ω is equivalent to ω . If m = (1 , , , then the matrix of ω (0) does not have the canonicalform. The fact that the forms ω and ω are not equivalent follows from Theorem 6 of thenext section according to which P (3 , , ω ) is a simple Lie algebra and P (3 , , ω ) is not asimple Lie algebra. Remarks. 1. The conditions under which the forms ω ′ i , ω ′′ i of the same type are equivalentfor i = 1 , , are obtained in Theorem 4. The problem of equivalence of the forms ω ′ , ω ′′ isnot being treated in this paper, since some other technique is required.2. Let a ∈ K (cid:31) { } , then the form aω i is equivalent to ω i , i = 1 , , , . It is obvious for i = 1 , , . Let ϕ be the admissible automorphism of O (3 , ( m , m , such that ϕ ( x i ) = c i x i , c = 1 , c = a , c = √ a . Then ϕ ( ω ) = aω . In this section we will prove the fact that the Lie algebras which correspond to the nonequiv-alent forms are not isomorphic.The algebra P (3 , m, ω ) is identified with the algebra O ( F ) (cid:14) K ∼ = m with the Poissonbracket, corresponding to the form ω (see (1)): { f, g } = ∂ f ∂ g + ∂ f ∂ g + ∂ f ∂ g, { f, g } = ∂ f ∂ g + ∂ f ∂ g + ∂ f ∂ g + ∂ f ∂ g, { f, g } = ∂ f ∂ g + ∂ f ∂ g + ∂ f ∂ g, { f, g } = ∂ f ∂ g + ∂ f ∂ g + ∂ f ∂ g + x x ( ∂ f ∂ g + ∂ f ∂ g ) . If = f ∈ m , let λ ( f ) be the nonzero homogeneous part of f of the least degree. Theorem 6. Let L = P (3 , m, ω i ) , i = 1 , , , . If m > in the case of i = 1 and m = (1 , , in the case of i = 3 , then L is a simple Lie algebra of dimension m + m + m − . If m = 1 and m = (1 , , , then P (1) (3 , m, ω ) is a simple Lie algebra of dimension m + m +1 − . If m = (1 , , , then P (1) (3 , m, ω ) is not simple.Proof. The algebras P (3 , m, ω i ) , i = 1 , , are considered in [5, Section 4].Let J be a nonzero ideal in P (3 , m, ω ) and = f ∈ J . We have ad x = ∂ , ad x = ∂ + x x ∂ , ad x = ∂ + x x ∂ . Commuting f consecutively, if necessary, with x , x , x we obtain x ∈ J . Now { x , x x } = x ∈ J , { x , x x } = x ∈ J . Let L = gr P (3 , m, ω ) = r L i = − L i . Since L − ⊂ gr J , we have L i ⊂ gr J for i < r . Therefore, for any monomial g = x x x there is h such that g + h ∈ J , deg λ ( h ) > deg g or h = 0 . Since { x , x x } = x x x ∈ J , we have J = P (3 , m, ω ) , that is, the algebra is simple.Further we prove that the natural filtration of P (3 , m, ω i ) , i = 1 , , , , m = (1 , , isintrinsically determined. Lemma 3 ([1]) . If w , . . . , w k ∈ P (3 , m, ω ) are linearly dependent, the same is true for { λ ( w i ) } . L i = P (3 , m, ω i ) , i = 1 , , , be a non-alternating Hamiltonian Lie algebra, where m = (1 , , . For φ ∈ Der ( L i ) let R i ( φ ) = dim( Im φ ) (see [1]). Obviously, R i ( φ ) = R i ( aφ ) for any a ∈ K ∗ . If M is a subalgebra of Der ( L i ) , let R i ( M ) = min = φ ∈ M R i ( φ ) .Let { L ( i ) } be the standard filtration, { L i } be the standard grading and ξ = ad ¯ x = ad ¯ x ¯ x ¯ x = ad x (2 m − x (2 m − x (2 m − . Lemma 4. Let = φ ∈ h ξ i . If i = 2 , or i = 1 , m > , then R i ( φ ) = 4 . If i = 1 , m = 1 or i = 4 , then R i ( φ ) = 3 .Proof. Obviously, φ ( x ) , φ ( x ) , φ ( x ) ∈ Im φ are linearly independent. The action of φ on L (0) corresponds to the one-dimensional representation of L (0) on L ( r ) = h ¯ x i . Since φ ( L (1) ) = 0 and φ ([ L (0) , L (0) ]) = 0 , it follows that Im( φ ) = h φ ( x ) , φ ( x ) , φ ( x ) , ¯ x i if there exists x (2) j ∈ L such that φ ( x (2) j ) = 0 . Otherwise, Im( φ ) = h φ ( x ) , φ ( x ) , φ ( x ) i .Let i = 1 . Then φ ( x (2)1 ) = φ ( x (2)2 ) = 0 . If m > , then φ ( x (2)3 ) = 0 . Consequently, R ( φ ) = 4 . If m = 1 , then R ( φ ) = 3 .Let i = 2 . The condition m < m , m / ∈ [ m , m ] yield m > or m > . Since φ ( x (2)1 ) = φ ( x (2)3 ) = 0 , we have R ( φ ) = 4 .Let i = 3 . Since m = (1 , , , we have m j > and φ ( x (2) j ) = 0 . Consequently, R ( φ ) = 4 .Let i = 4 . Then m = 1 and φ ( x (2)1 ) = φ ( x (2)2 ) = 0 . Consequently, R ( φ ) = 3 . Lemma 5. Let i = 2 , or i = 1 , m > . If = D ∈ L it , − t < m + 2 m + 2 m − , thenthere exist homogeneous elements E , E , E , E , E ∈ L i such that { D, E } , { D, E } , . . . , { D, E } are linearly independent.Proof. The case of L , m > . (1) t = − . Let D = a x + a x + a x and a = 0 . Put E = x (2)3 , E = x (3)3 , E = x x (3)3 , E = x x (3)3 , E = x x x (3)3 . Let D = a x + a x and a i = 0 . Put E = x x x , E = x x x (3)3 , E = x j x , E = x j x (2)3 , E = x j x (3)3 , where j = i . (2) t > − . Let D = ax ( α )1 x ( β )2 x ( γ )3 + f ( x , x , x ) , where a = 0 , α, β > and f contains x ( s )3 g ( x , x ) if s γ . Put E = x (2 m − α )1 , E = x (2 m +1 − γ )3 ( γ > ) or x x (2)3 . If γ < m − ,then E = x (2 m − β )2 , E = x (2 m − α )1 x (2 m − − γ )3 , E = x (2 m − α )1 x (2 m − − γ )3 ( γ is even) or x (2)3 ( γ is odd). If γ = 2 m − , then E = x , E = x (2 m − α )1 x , E = x (2 m − β )2 x . If γ = 2 m − ,then E = x (2 m − β )2 , E = x (2 m − − α )1 x (2 m − − β )2 x , E = x (2 m − − α )1 x (2 m − − β )2 x (2)3 .Let D = ax ( t +2)3 + f ( x , x )+ g ( x , x ) and a = 0 . Put { E , . . . , E } = { x , ¯ x x , ¯ x x , x x x ,x x x (2 m − − t )3 o .Let D = ax ( α ) i x ( γ )3 + f ( x , x ) + g ( x , x ) , where a = 0 , α > and f, g contain x ( s )3 x ( h ) q if s γ , h > . If γ > , then E = x j , E = x (2 mi − − α ) i x , E = x (2 mi − α ) i x j , E = x (2 mi − α ) i x j x (2 m − − γ )3 ( γ < m − ) or x (2 mi − − α ) i x (2 mj − j x , E = x j x (2)3 ( γ is odd) or x j x (2 m − − γ )3 ( γ < m − is even) or x j x , where { i, j } = { , } . If γ = 0 , then { E , . . . , E } = { x j , x (2 mi − α ) i x j , x (2 mi − α ) i x j x , x (2 mi − α ) i x j x (2)3 , x (2 mi − α ) i x j x (3)3 } .The case of L , m > . (1) t = − . Let D = a x + a x + a x and a = 0 . The same L . Let D = a ( x + x ) + a x and a i = 0 . Put { E , . . . , E } the same L . (2) t > − . Let D = ax ( α )1 x ( β )2 x ( γ )3 + f ( x , x , x ) , where a = 0 , α, β > and f contains x ( s )3 g ( x , x ) if s γ . Put E = x (2 m − β )2 , E = x (2 m +1 − γ )3 ( γ > ) or x (3)3 ( γ = 1 ) or x (2 m − α )1 x (2)3 ( γ = 0 = 2 m − , ( α, β ) = (2 m − , m − ) or x x (2)3 + x x (2)3 . If γ < m − ,then E = x (2 m − α )1 ( ( α, β ) = (2 m − , m − ) or x + x , E = x (2 m − β )2 x (2 m − − γ )3 , E = x (2 m − β )2 x (2 m − − γ )3 ( γ is even) or x (2)3 ( γ is odd). If γ = 2 m − , then E = x + x , E = x (2 m − β )2 x , E = x (2 m − α )1 x ( ( α, β ) = (2 m − , m − ) or x x + x x . If γ = 2 m − ,then E = x (2 m − α )1 , E = x (2 m − − α )1 x (2 m − − β )2 x , E = x (2 m − − α )1 x (2 m − − β )2 x (2)3 .9et D = ax ( t +2)3 + f ( x , x ) + g ( x , x ) and a = 0 . The same L .Let D = ax ( α ) i x ( γ )3 + f ( x , x ) + g ( x , x ) , where a = 0 , α > and f, g contain x ( s )3 x ( h ) q if s γ , h > . The same L .The case of L , m = 1 , m > , m > m > . (1) t = − . Let D = a ( x + x ) + a x + a x and a i = 0 , i = 1 , . Put E = x (2) j , E = x (3) j , E = x (3) j x , E = x i x (3) j , E = x i x (3) j x , where j = i . Let D = a x and a = 0 .Put E = x x , E = x x , E = x (2)1 x , E = x (2)2 x , E = x x x . (2) t > − . Let D = ax ( α )1 x ( β )2 x + f ( x , x , x ) , where a = 0 , β > and f contains x ( s )1 g ( x , x ) if s α . Put E = x (2 m − − β )2 x . If γ < m − , then E = x (2 m − − α )1 x (2 m − − β )2 x , E = x (2 m +1 − α )1 ( α > ) or x (2)1 x . If γ < m − , then E = x ( β = 2 m − ) or x x ( β = 2 m − , α is even) or x (2 m − − α )1 x ( β = 2 m − , α is odd), E = x (2 m − − α )1 x (2 m − − β )2 x ( α is even) or x (2)1 ( α is odd). If α = 2 m − , then E = x + x , E = x x ( β = 2 m − )or x x . If γ = 2 m − , then E = x , E = x , E = x (2)1 ( β = 2 m − ) or x x , E = x (2)1 x (2 m − − β )2 .Let D = ax ( t +2)1 + f ( x , x )+ g ( x , x ) and a = 0 . Put { E , . . . , E } = { x , x ¯ x , x x , x x x ,x (2 m − − t )1 x x o .Let D = ax ( α )1 x + f ( x , x ) + g ( x , x ) , where a = 0 , α > and f, g contain x ( s )1 x ( h ) q if s α , h > . Put { E , . . . , E } = { x , x , x (2)2 , x (3)2 , ¯ x x } .Let D = ax ( α )1 x ( β )2 + f ( x , x ) + g ( x , x ) , where a = 0 , β > and f, g contain x ( s )1 x ( h ) q if s α , h > . If α > , then E = x (2 m − α )1 ( ( α, β ) = (2 m − , m − ) or x + x , E = x (2 m − β )2 , E = x (2 m − β )2 x , E = x (2 m +1 − α )1 ( α > ) or x (2)1 x (2 m − − β )2 , E = x (2)2 x ( α isodd) or x (2 m − − α )1 x ( α is even, ( α, β ) = (2 m − , m − ) or x x + x x . If α = 0 , then { E , . . . , E } = { x , x x , x (2)1 , x (3)1 , x x (2 m − β )2 } .The case of L . Without the loss of generality we can assume that m > . (1) t = − . The same L with i = j . (2) t > − . Let D = ax ( α )1 x ( β )2 x ( γ )3 + f ( x , x , x ) , where a = 0 , α, β > and f contains x ( s )3 g ( x , x ) if s γ . Put E = x x (2 m − − β )2 , E = x (2 m +1 − γ )3 ( γ > ) or x x (2)3 . If γ < m − , then E = x (2 m − − α )1 x ( ( α, β ) = (2 m − , m − ) or x x (2 m − − γ )3 , E = x x (2 m − − β )2 x (2 m − − γ )3 , E = x x (2 m − − β )2 x (2 m − − γ )3 ( γ is even) or x (2)3 ( γ is odd). If γ =2 m − , then E = x , E = x x , E = x x (2 m − − β )2 x . Let γ = 2 m − . Since α, β > and ( α, β ) = (2 m − , m − , we have m > or m > . Let m > , then E = x (2 m − − α )1 x ( ( α, β ) = (2 m − , m − ) or x x (2)3 , E = x (2 m − − α )1 x (2 m − − β )2 x (2)3 , E = x x ( α is even)or x (2)1 x ( α is odd).Let D = ax ( t +2)3 + f ( x , x ) + g ( x , x ) and a = 0 . The same L .Let D = ax ( α ) i x ( γ )3 + f ( x , x ) + g ( x , x ) , where a = 0 , α > and f, g contain x ( s )3 x ( h ) q if s γ , h > . If γ > , then E = x i , E = x (2 mi − − α ) i x , E = x x , E = x x x (2 m − − γ )3 ( γ < m − ) or x (2 mi − − α ) i x (2 mj − j x , E = x j x (2)3 ( γ is odd) or x i x (2 m − − γ )3 ( γ < m − is even) or x i x ( γ = 2 m − , α = 2 m i − ) or x (3) i , where { i, j } = { , } . If γ = 0 , then { E , . . . , E } = { x i , x (2 mi +1 − α ) i , x (2 mi +1 − α ) i x , x (2 mi +1 − α ) i x (2)3 , x (2 mi +1 − α ) i x (3)3 } . Lemma 6. Let f ∈ L t ) (cid:31) L t +1) , g ∈ L s ) (cid:31) L s +1) and { λ ( f ) , g } 6 = 0 .(1) If g = x ( α )1 x ( β )2 x ( γ )3 and α > , then λ ( { f, g } ) = { λ ( f ) , g } .(2) If g = x ( β )2 x ( γ )3 and { λ ( f ) , g } ∈ L t + s ) (cid:31) L t + s +1) , then λ ( { f, g } ) = { λ ( f ) , g } .Proof. Recall, that { f, g } = ∂ f ∂ g + ∂ f ∂ g + ∂ f ∂ g + x x ( ∂ f ∂ g + ∂ f ∂ g ) .(1) Obviously, if g = x ( α )1 x ( β )2 x ( γ )3 and α > , then { f, g } = ∂ f ∂ g + ∂ f ∂ g + ∂ f ∂ g .(2) Let f = λ ( f ) + f , where f ∈ L t +1) . Then { f, g } ∈ L t + s ) , { f , g } ∈ L t + s +1) . Since { λ ( f ) , g } ∈ L t + s ) (cid:31) L t + s +1) , λ ( { f, g } ) = λ ( { λ ( f ) , g } + { f , g } ) = λ ( { λ ( f ) , g } ) = { λ ( f ) , g } .10 emma 7. Let i = 4 or i = 1 , m = 1 . If = D ∈ L i ( t ) (cid:31) L i ( t +1) , − t < m + 2 m − ,then there exist homogeneous elements E , E , E , E ∈ L i such that { D, E } , . . . , { D, E } are linearly independent.Proof. We will prove that there exist homogeneous elements E , E , E , E ∈ L i such that { λ ( D ) , E } , . . . , { λ ( D ) , E } are linearly independent. We will search E i for L which satisfyLemma 6. Thus { λ ( D ) , E i } = λ ( { D, E i } ) and { D, E i } are also linearly independent byLemma 3.The case of L , m > . (1) t = − . Let λ ( D ) = a x + a x + a x and a = 0 . Put E = x x , E = x x x , E = x (2)1 x , E = x (3)1 x . Let λ ( D ) = a x + a x and a i = 0 . Put E = x x , E = x x x , E = x (2)1 x , E = x (3)1 x . (2) t > − . Let λ ( D ) = ax ( α )1 x ( β )2 + f ( x , x , x ) , where a = 0 , β > and f contains x ( s )1 x ( h )2 if s < α . Put { E , . . . , E } = { x , x x , x (2 m − − α )1 x (2 m − β )2 , x (2 m − − α )1 x (2 m − β )2 x } .Let λ ( D ) = ax ( α )1 x + f ( x , x , x ) , where a = 0 , α > and f contains x ( s )1 x ( h )2 if h = 0 .Put E = x . If α is even, then { E , E , E } = { x x , ¯ x x , x x x } . If α is odd, then E = x x , E = x (2 m − − α )1 x , E = x (2 m − − α )1 x x ( α = 2 m − ) or x x .Let λ ( D ) = ax ( α )1 x ( β )2 x + f ( x , x ) x + bx ( t +2)1 , where a = 0 , β > and f contains x ( s )1 x ( h )2 if s < α , h > . Put E = x . If α > , then E = x (2 m − β )2 , E = x , E = x (2 m − α )1 ( α = 2 m − ) or x (2 m − β )2 x . If α = 0 , then { E , E , E } = { x x (2 m − β )2 , x x , ¯ x } .Let λ ( D ) = ax ( t +2)1 , a = 0 . Put { E , . . . , E } = { x , x x , x (2 m − − t )1 x , x (2 m − − t )1 x x } .The case of L , m = 1 , m > . (1) t = − . Let λ ( D ) = a x + a x + a x and a = 0 . Put E = x x , E = x x x , E = x x (2)2 x , E = x x (3)2 x . Let λ ( D ) = a x + a x and a i = 0 . Put E = x x , E = x x x , E = x x (2)2 , E = x x (3)2 . (2) t > − . Let λ ( D ) = ax ( t +2)2 + f ( x , x , x ) and a = 0 . Put { E , . . . , E } = { x , x x , x x (2 m − − t )2 , x x (2 m − − t )2 x } .Let λ ( D ) = ax ( α )2 x + f ( x , x , x ) , where a = 0 , α > and f does not contain x ( t +2)2 .Put { E , E , E } = { x , x x , x x (2 m − α )2 } and E = x x (2 m − − α )2 x ( α = 2 m − ) or x .Let λ ( D ) = ax x ( t +1)2 + bx x ( t )2 x and a = 0 . Put { E , . . . , E } = { x , x , x x , x x } .Let λ ( D ) = ax x ( t )2 x , a = 0 . Put E = x , E = x x (2 m − − t )2 . If t > , then E = x , E = x (2 m − t )2 . If t = 0 , then E = x x , E = ¯ x .The case of L , m = 1 . The same L . Theorem 7. If i = 2 , or i = 1 , m > , then R i ( L i ) = 4 and R i ( D ) = 4 if and only if = D ∈ h ξ i . If i = 1 , m = 1 or i = 4 , then R i ( L i ) = 3 and R i ( D ) = 3 if and only if = D ∈ h ξ i .Proof. (1) Let i = 2 , or i = 1 , m > . It follows from Lemma 4 that if = D ∈ h ξ i ,then R i ( D ) = 4 . We shall prove that if f ∈ L i and f / ∈ h ¯ x i , then R i ( ad f ) > . Clearly, λ ( f ) / ∈ h ¯ x i , thus, according to Lemma 5, there are homogeneous elements E , . . . , E ∈ L i such that { λ ( f ) , E j } , j are linearly independent. But { λ ( f ) , E j } = λ ( { f, E j } ) .Hence, { f, E j } are also linearly independent by Lemma 3. Therefore, R i ( ad f ) > > .(2) Let i = 1 , m = 1 or i = 4 . It follows from Lemma 4 that if = D ∈ h ξ i , then R i ( D ) = 3 . If f ∈ L i and f / ∈ h ¯ x i , then according to Lemma 7, there are homogeneouselements E , . . . , E ∈ L i such that { f, E j } , j are linearly independent. Therefore, R i ( ad f ) > > . Corollary 1. Let L = P (3 , m, ω i ) , i = 1 , , , , m = (1 , , . The standard maximalsubalgebra L of L is the normalizer of h ¯ x i in L . Thus, L is an invariant subalgebra andthe natural filtration of L is intrinsically determined. heorem 8. Let L i = P (3 , m, ω i ) , L j = P (3 , m ′ , ω j ) , i, j = 1 , , , . If i = j , then thenon-alternating Hamiltonian Lie algebras L i and L j are not isomorphic.Proof. Suppose that L i and L j are isomorphic, ϕ : L i → L j is an isomorphism. If m =(1 , , , then m ′ = (1 , , either.Consider the case when m = (1 , , . The algebras L i and L j have the same invariant R , R i = R j . Let ¯ x ( i ) is the monomial in L i of maximal degree. It follows from Theorem 7that ϕ (¯ x ( i )) = a ¯ x ( j ) , a ∈ K ∗ , and, consequently, ϕ ( L i (0) ) = L j (0) , i.e. ϕ is an isomorphismof transitive Lie algebras. From Theorem 3 and Remark 2 of Section 4 we obtain that theforms ω i , ω j are equivalent. By Theorem 5 i = j .Suppose that m = (1 , , . In this case we have only L = P (3 , (1 , , , ω ) which is notsimple and L = P (3 , (1 , , , ω ) which is simple. Corollary 2. Let L i be non-alternating Hamiltonian Lie algebras in three variables, P (1) ( F i , ω i ) ⊆ L i ⊆ e P ( F i , ω i ) , i = 1 , , , . If i = j , then algebras L i , L j are not isomorphic.Proof. 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