Non-alternating Hamiltonian Lie algebras in three variables

A. V. Kondrateva

National Research Lobachevsky State University of Nizhny Novgorod, Gagarin
ave. 23, Nizhny Novgorod, 603950 Russia

E-mail<EMAIL_ADDRESS>
Abstract. Non-alternating Hamiltonian Lie algebras in three variables over a
perfect field of characteristic 2 are considered. A classification of non-
alternating Hamiltonian forms over an algebra 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 its linear part. It is proved that the Lie algebras which
correspond to the nonequivalent forms are not isomorphic.

Keywords ands phrases: Non-alternating Hamiltonian Lie algebra, non-
alternating Hamiltonian form, characteristic two

##  1 Introduction

Non-alternating Hamiltonian algebras over a field of characteristic 2 were
first constructed in 1993 by Lin Lei [1] as Lie algebras of polynomials in
divided powers with the symmetric Poisson bracket
$\\{f,g\\}=\sum\limits_{i}\partial_{i}f\partial_{i}g$. In the case when the
heights of the variables are equal to 1, non-alternating Hamiltonian Lie
algebras are isomorphic to the first series of simple Lie algebras built by I.
Kaplansky [2]. In [3, 4], symmetric differential forms in divided powers are
introduced and non-alternating Hamiltonian Lie algebras similar to Hamiltonian
Lie superalgebras of characteristic zero with respect to Poisson brackets with
constant coefficients are studied. The classification of alternating
Hamiltonian forms over an algebra of truncated polynomial is obtained by M. I.
Kuznetsov, S .A. Kirillov ([9]). The complete classification over a divided
powers algebra is build by S. M. Skryabin ([10, 11]). In [12, 13] the general
theory of alternating Hamiltonian Lie algebras in divided powers is developed.
The non-alternating case is considered in [5]. In particular, the
classification of non-alternating Hamiltonian forms with constant coefficients
over the divided powers algebra is obtained. It is proved that any filtered
deformation of a graded non-alternating Hamiltonian Lie algebra is determined
by a non-alternating Hamiltonian form with polynomial coefficients. One of the
main results of [5] is the proof of the equivalence of the non-alternating
Hamiltonian form $\omega$ with polynomial coefficients to its initial form
$\omega(0)$, provided that the canonical form of $\omega(0)$ contains
$(dx_{i})^{(2)}$ or $dx_{i}dx_{j}+(dx_{j})^{(2)}$ for some variable $x_{i}$ of
the height greater than 1. The problem of the classification of non-
alternating Hamiltonian forms for which this condition is not satisfied
remains open. In this paper, a classification of non-alternating Hamiltonian
forms over algebras of divided powers in three variables with polynomial
coefficients is obtained, and it is shown that if the above condition for the
heights of the variables is not fulfilled, the non-alternating Hamiltonian
form $\omega$ may not be equivalent to $\omega(0)$. In addition, the
dimensions of the corresponding simple non-alternating Hamiltonian Lie
algebras are found and it is proved that the Lie algebras which correspond to
the nonequivalent forms are not isomorphic.

ACKNOWLEDGEMENTS. The investigation is funded by RFBR according to research
project 18-01-00900 and Ministry of Science and Higher education of Russian
Federation (project 0729-2020-0055).

##  2 Preliminaries

Here we recall some basic definitions and results from [5]. Let $K$ be a
perfect field of characteristic 2, $E$ be a vector space over $K$ and
$\mathscr{F}\colon E=E_{0}\supseteq E_{1}\supseteq\ldots\supseteq E_{r}\supset
E_{r+1}=\\{0\\}$ be a flag of $E$. A basis $\\{x_{1},\ldots,x_{n}\\}$ of $E$
is called coordinated with the flag $\mathscr{F}$ if
$\\{x_{1},\ldots,x_{n}\\}\cap E_{i}$ is a basis of $E_{i}$, $i=1,\ldots,r$.

We use the notations and the definitions of algebras $O(E)$, $O(\mathscr{F})$,
$O(n,\overline{m})$, $W(E)$, $W(\mathscr{F})$, $W(n,\overline{m})$ from [6].
The maximal ideal of $O(\mathscr{F})$ is denoted by $\mathfrak{m}$. Let
$\\{x_{1},\ldots,x_{n}\\}$ be a basis of $E$ coordinated with the flag
$\mathscr{F}$, $m_{i}=\min\\{j\leavevmode\nobreak\ |\leavevmode\nobreak\
x_{i}\notin E_{j}\\}$ be the height of $x_{i}$. The algebra $O(\mathscr{F})$
is isomorphic to the algebra $O(n,\overline{m})$,
$\overline{m}=(m_{1},\ldots,m_{n})$. The flag $\mathscr{F}$ is restored by the
$n$-tuple $\overline{m}$, $E_{j}=\langle x_{i}\leavevmode\nobreak\
|\leavevmode\nobreak\ m_{i}>j\rangle$, in particular $E_{1}=\langle
x_{i}\leavevmode\nobreak\ |\leavevmode\nobreak\ m_{i}>1\rangle$.

A flag $\mathscr{F}\colon E=E_{0}=E_{1}=\ldots=E_{m-1}\supset E_{m}=\\{0\\}$
is called almost trivial. If $m=1$, then $\mathscr{F}$ is trivial.

Let $S\Omega(\mathscr{F})=\bigoplus_{i\geqslant 0}S\Omega^{i}(\mathscr{F})$ be
the algebra of symmetric differential forms, which is an
$O(\mathscr{F})$-algebra of the divided powers (see also [3, 7]).

Recall the definition of the differential $d\colon
S\Omega^{r}(\mathscr{F})\rightarrow S\Omega^{r+1}(\mathscr{F})$.

$(d\omega)(D_{1},\ldots,D_{r+1})=\sum_{i=1}^{r+1}D_{i}(\omega(D_{1},\ldots,\widehat{D_{i}},\ldots,D_{r+1}))+\\\
+\sum_{i<j}\omega([D_{i},D_{j}],D_{1},\ldots,\widehat{D_{i}},\ldots,\widehat{D_{j}},\ldots,D_{r+1}),$
$d\omega^{(r)}=\omega^{(r-1)}d\omega,\quad df(D)=D(f),\leavevmode\nobreak\
\leavevmode\nobreak\ f\in O(\mathscr{F}).$

An isomorphism $\sigma\colon O(\mathscr{F})\rightarrow
O(\mathscr{F}^{\prime})$ is called admissible if for $r\geqslant 0$,
$f\in\mathfrak{m}$ the inclusions $f^{(r)}\in O(\mathscr{F})$, $(\sigma
f)^{(r)}\in O(\mathscr{F}^{\prime})$ are both either fulfilled or not and if
the inclusions are both fulfilled, then $(\sigma f)^{(r)}=\sigma(f^{(r)})$.
This is equivalent to the fact that the mapping $D\mapsto\sigma\circ
D\circ\sigma^{-1}$ defines an isomorphism $W(\mathscr{F})\rightarrow
W(\mathscr{F}^{\prime})$. An admissible isomorphism $\sigma$ extends to the
isomorphism $S\Omega(\mathscr{F})\rightarrow S\Omega(\mathscr{F}^{\prime})$ by
the rule

$(\sigma\omega)(D_{1},\ldots,D_{k})=\sigma(\omega(\sigma^{-1}D_{1}\sigma,\ldots,\sigma^{-1}D_{k}\sigma)).$

Moreover, $\sigma d\omega=d\sigma\omega$,
$\sigma(\omega_{1}\omega_{2})=\sigma\omega_{1}\cdot\sigma\omega_{2}$ and
$\sigma(\omega^{(k)})=(\sigma\omega)^{(k)}$.

Two differential forms $\omega$ and $\omega^{\prime}$ will be called
equivalent if there is an admissible isomorphism $\sigma$ such that
$\sigma\omega=\omega^{\prime}$. A symmetric $2$-form $\omega\in
S\Omega^{2}(\mathscr{F})$ is called non-alternating if there exists $D\in
W(\mathscr{F})$ such that $\omega(D,D)\neq 0$. Let $\\{x_{i}\\}$ be a basis of
$E$ coordinated with the flag $\mathscr{F}$ and
$\omega=\sum\omega_{ii}(dx_{i})^{(2)}+\sum\limits_{i<j}\omega_{ij}dx_{i}dx_{j}$.
Set $\omega_{ji}=\omega_{ij}$ for $i<j$. A form $\omega$ is called
nondegenerate if $\det(\omega_{ij})$ is invertible in $O(\mathscr{F})$.

A closed nondegenerate non-alternating form $\omega\in
S\Omega^{2}(\mathscr{F})$ is called a non-alternating Hamiltonian one. The
corresponding Lie algebra of Hamiltonian vector fields is denoted by
$\widetilde{P}(\mathscr{F},\omega)$,

$\widetilde{P}(\mathscr{F},\omega)=\\{D\in W(\mathscr{F})\leavevmode\nobreak\
|\leavevmode\nobreak\ D\omega=0\\}.$

Moreover, $D\in\widetilde{P}(\mathscr{F},\omega)$ is equal to
$D_{f}=\sum\limits_{i,j=1}^{n}\overline{\omega}_{ij}\partial_{j}f\partial_{i}$
for some $f\in\widetilde{O}(\mathscr{F})=O(\mathscr{F})+\langle
x_{1}^{(2^{m_{1}})},x_{2}^{(2^{m_{2}})},\ldots x_{n}^{(2^{m_{n}})}\rangle$.
Here $(\overline{\omega}_{ij})=(\omega_{ij})^{-1}$ (see [5]).

The mapping $f\mapsto D_{f}$ is an isomorphism of the Lie algebra
$\widetilde{P}(\mathscr{F},\omega)$ and the Lie algebra
$\widetilde{O}(\mathscr{F})\big{/}K$ with the Poisson bracket

$\displaystyle\\{f,g\\}=D_{f}(g)=D_{g}(f)=\sum_{i,j=1}^{n}\overline{\omega}_{ij}\partial_{i}f\partial_{j}g.$
(1)

Sometimes we will identify $\widetilde{P}(\mathscr{F},\omega)$ with the Lie
algebra $(\widetilde{O}(\mathscr{F})\big{/}K,\leavevmode\nobreak\
\\{\leavevmode\nobreak\ ,\leavevmode\nobreak\ \\})$. Let

$P(\mathscr{F},\omega)=\\{D_{f}\leavevmode\nobreak\ |\leavevmode\nobreak\ f\in
O(\mathscr{F})\big{/}K\\}.$

It follows from (1) that $P(\mathscr{F},\omega)$ is an ideal of
$\widetilde{P}(\mathscr{F},\omega)$ of codimension $n$. A Lie algebra
$\mathscr{L}$,
$P^{(1)}(\mathscr{F},\omega)\subseteq\mathscr{L}\subseteq\widetilde{P}(\mathscr{F},\omega)$
will be called a non-alternating Hamiltonian Lie algebra.

We will say, that $\mathscr{L}$ is associated with a pair
$(\mathscr{F},\omega)$. If the forms $\omega$, $\omega^{\prime}$ are
equivalent and $\sigma\colon O(\mathscr{F})\rightarrow
O(\mathscr{F}^{\prime})$ is an admissible isomorphism such that
$\sigma(\omega)=\omega^{\prime}$, then Lie algebras $P(\mathscr{F},\omega)$,
$P(\mathscr{F}^{\prime},\omega^{\prime})$ are isomorphic,
$\sigma(P(\mathscr{F},\omega))=P(\mathscr{F}^{\prime},\omega^{\prime})$, where
$\sigma(D_{f})=\sigma\circ D_{f}\circ\sigma^{-1}=D_{\sigma(f)}$.

Put $W=W(\mathscr{F})$, $V=W\big{/}\mathfrak{m}W$. The form $\omega(0)$ is a
nondegenerate symmetric bilinear form on $V$: for $D_{1},D_{2}\in W$
$\omega(0)(\overline{D_{1}},\overline{D_{2}})=\omega(D_{1},D_{2})(0)=\overline{\omega(D_{1},D_{2})}\in
O(\mathscr{F})\big{/}\mathfrak{m}\cong K$. Employing the natural pairing $V$
and $E\cong\mathfrak{m}\big{/}\mathfrak{m}^{(2)}$,
$\langle\overline{D},\overline{f}\rangle=\overline{D(f)}\in
O(\mathscr{F})\big{/}\mathfrak{m}$, one obtains the isomorphism $\iota\colon
V\rightarrow E\cong V^{\ast}$,
$\iota(\overline{D})=\omega(0)(\overline{D},-)$. The dual form
$\overline{\omega(0)}$ on $E$,
$\overline{\omega(0)}(\iota(\overline{D_{1}}),\iota(\overline{D_{2}}))=\omega(0)(\overline{D_{1}},\overline{D_{2}})$,
has the matrix $(\omega_{ij}(0))^{-1}$ with respect to a basis
$\\{x_{1},\ldots,x_{n}\\}$ dual to a basis
$\\{\overline{\partial_{1}},\ldots,\overline{\partial_{n}}\\}$ of $V$, were
$(\omega_{ij}(0))$ is the matrix of $\omega(0)$ with respect to a basis
$\\{\overline{\partial_{1}},\ldots,\overline{\partial_{n}}\\}$. The space of
null-vectors of $E$ with respect to the form $\overline{\omega(0)}$ is denoted
by $E^{0}$. Let $\\{x_{1},\ldots,x_{n}\\}$ be a basis of $E$ coordinated with
the flag $\mathscr{F}$. Since
$W=\langle\overline{\partial_{1}},\ldots,\overline{\partial_{n}}\rangle\oplus\mathfrak{m}W$,
we identify $V$ with $\langle\partial_{1},\ldots,\partial_{n}\rangle$.

Later on, we will consider the dual flag of $V$, $0=V_{0}\subseteq
V_{1}\subseteq\ldots\subset V_{r}=V$, where $V_{i}=\text{Ann}\,E_{i}$. To
avoid the abuse of notations we will denote it by $\mathscr{F}$ as well.

The set $\\{i\leavevmode\nobreak\ |\leavevmode\nobreak\
\overline{\omega}_{ii}\neq 0\\}$ is denoted by $I$. E.g., if
$\omega=dx_{1}dx_{2}+(dx_{3})^{(2)}$, then $I=\\{3\\}$, if
$\omega=dx_{1}dx_{2}+(dx_{2})^{(2)}+(dx_{3})^{(2)}$, then $I=\\{1,3\\}$. Note
that $\overline{\omega}_{ii}\in K$. Indeed, since
$0=[D_{x_{i}},D_{x_{i}}]=D_{\\{x_{i},x_{i}\\}}$ and $D_{f}=0$ if and only if
$f\in K$, we have $\overline{\omega}_{ii}=\\{x_{i},x_{i}\\}\in K$.

We will use the following theorems from [5]. Here $G^{\prime}$ is the subgroup
of admissible automorphisms $\varphi$ of $O(\mathscr{F})$ such that
$\varphi(x)\equiv x$ (mod $\mathfrak{m}^{2}$), $x\in E$,
$dz_{r}=x_{r}^{(2^{m_{r}}-1)}dx_{r}$,
$\widetilde{\mathfrak{m}}^{(j)}S\Omega^{1}=\langle T\rangle$, were
$T=\\{x^{(\alpha)}dx_{k},\leavevmode\nobreak\ |\alpha|\geqslant
j,\leavevmode\nobreak\
k=1,\ldots,n\\}\diagdown\\{x_{r}^{(2^{l})}x_{s}^{(2^{t})}dx_{i},x_{q}^{(2)}dx_{i},\leavevmode\nobreak\
i,q\in I\\}$.

######  Theorem 1 ([5], Theorem 3.1).

Let $\omega$ be a non-alternating Hamiltonian form,
$\omega=\omega(0)+d\psi+\eta$, where $\eta\in\langle dz_{r}dz_{s}\rangle$,
$r,s=1,\ldots,n$, $\psi\in\mathfrak{m}^{(j+1)}S\Omega^{1}$, $j\geqslant 1$.
There exists an automorphism $\sigma\in G^{\prime}$ such that
$\sigma\omega=\omega(0)+d\widetilde{\psi}+\widetilde{\eta}$, where
$\widetilde{\psi}\in\widetilde{\mathfrak{m}}^{(j+1)}S\Omega^{1}$,
$\widetilde{\eta}\in\langle dz_{r}dz_{s}\rangle$, $r,s=1,\ldots,n$.

######  Theorem 2 ([5], Theorem 3.2).

Let $\omega$ be a non-alternating Hamiltonian form, such that

$\omega=\omega(0)+d\varphi+\sum\limits_{i<j}b_{ij}dz_{i}dz_{j},$

where $\varphi\in\widetilde{\mathfrak{m}}^{(2)}S\Omega^{1}$ and $b_{ij}\in K$.
Then

(1) if there is $i\in I$ such that $m_{i}>1$ or $m_{i}=1$ for any $i\in I$ and
$b_{sj}=0$, $s\in I$, then $\omega$ is conjugated with respect to $G^{\prime}$
to the form $\omega(0)$.

(2) $\omega$ is conjugated with respect to $G^{\prime}$ to the form
$\omega(0)+\sum\limits_{i\in I}\sum\limits_{j\notin
I}b_{ij}dz_{i}dz_{j}+\sum\limits_{i<j\in I}b_{ij}dz_{i}dz_{j}$.

Here, we set $b_{ji}=b_{ij}$ in the case $i<j$.

The theorem below is basically the weak form of [5, Th. 5.2]. Here we use the
concept of the minimal flag of a transitive Lie algebra from [8].

######  Theorem 3.

(i) Let $\mathscr{L}$ be a non-alternating Hamiltonian Lie algebra associated
with a pair $(\mathscr{F},\omega)$, $\mathscr{L}_{0}$ be the standard maximal
subalgebra of $\mathscr{L}$. The minimal flag
$\mathscr{F}(\mathscr{L},\mathscr{L}_{0})$ coincides with $\mathscr{F}$.

(ii) Let
$\psi\colon(\mathscr{L},\mathscr{L}_{0})\rightarrow(\mathscr{L}^{\prime},\mathscr{L}^{\prime}_{0})$
be an isomorphism of non-alternating Hamiltonian transitive Lie algebras
associated with pairs $(\mathscr{F},\omega)$,
$(\mathscr{F}^{\prime},\omega^{\prime})$, $\mathscr{L}_{0}$,
$\mathscr{L}^{\prime}_{0}$ be the standard maximal subalgebras of
$\mathscr{L}$, $\mathscr{L}^{\prime}$, respectively. Suppose that $n=\dim
E>2$. Then $\psi$ is induced by the admissible isomorphism $\varphi\colon
O(\mathscr{F})\rightarrow O(\mathscr{F}^{\prime})$ and
$\varphi(\omega)=a\omega^{\prime}$, $a\in K\backslash\\{0\\}$.

If $\mathscr{L}$, $\mathscr{L}^{\prime}$ are graded Lie algebras with the
standard gradings then
$\overline{\psi}=\text{gr}\,\psi\colon(\mathscr{L},\mathscr{L}_{0})\rightarrow(\mathscr{L}^{\prime},\mathscr{L}^{\prime}_{0})$
is a homogeneous isomorphism of transitive Lie algebras which is induced by
the admissible linear isomorphism $\overline{\varphi}=\text{gr}\,\varphi\colon
O(\mathscr{F})\rightarrow O(\mathscr{F}^{\prime})$ and
$\overline{\varphi}(\omega)=a\omega^{\prime}$, $a\in K\backslash\\{0\\}$.

###### Proof.

(i) Let $L=\text{gr}\mathscr{L}=L_{-1}+L_{0}+\ldots$ be the associated graded
Lie algebra, $L_{(0)}=\text{gr}\mathscr{L}_{0}$. If $G=P(\mathscr{F},\omega)$
then gr$G=P(\mathscr{F},\omega(0))$. Since
$(\text{gr}G)^{(1)}\subseteq\text{gr}(G^{(1)})$,
$P(\mathscr{F},\omega(0))^{(1)}\subseteq\text{gr}(P(\mathscr{F},\omega)^{(1)})$.
Analogously, for $G=\widetilde{P}(\mathscr{F},\omega)$,
gr$G=\widetilde{P}(\mathscr{F},\omega(0))$. Consequently,

$P(\mathscr{F},\omega(0))^{(1)}\subseteq
L\subseteq\widetilde{P}(\mathscr{F},\omega(0)),$

that is $L$ is one of non-alternating Hamiltonian Lie algebras associated with
the pair $(\mathscr{F},\omega(0))$. By [8, Prop. 0.4] the minimal flag
$\mathscr{F}(L,L_{(0)})$ is defined by the $p$-nilpotency degrees of elements
ad${}_{L}l$, $l\in L_{-1}=W_{-1}$, where $W(\mathscr{F})=W_{-1}+W_{0}+\ldots$
is the standard grading of $W(\mathscr{F})$. For $\partial_{i}\in W_{-1}$,
ad${}_{L}\partial_{i}$ acts on $\widetilde{O}(\mathscr{F})\big{/}K$ as
$\partial_{i}$. Hence, $\mathscr{F}(L,L_{(0)})=\mathscr{F}$. According to [8,
Prop. 0.3]

$\mathscr{F}=\mathscr{F}(L,L_{(0)})\leqslant\mathscr{F}(\mathscr{L},\mathscr{L}_{0})\leqslant\mathscr{F}.$

Thus, $\mathscr{F}(\mathscr{L},\mathscr{L}_{0})=\mathscr{F}$.

(ii) The proof is based on the theory of truncated coinduced modules (see [8,
12]). Here we mainly concentrate on some key-points of the proof.

According to (i) the natural embedding
$\tau\colon(\mathscr{L},\mathscr{L}_{0})\rightarrow(W(\mathscr{F}),W(\mathscr{F})_{0})$
is a minimal one. By the embedding theorem [8, Th. 1.1] $\psi$ is induced by
an admissible isomorphism $\varphi\colon O(\mathscr{F})\rightarrow
O(\mathscr{F}^{\prime})$, $\varphi(l)=\varphi l\varphi^{-1}$,
$l\in\mathscr{L}$. Let $(\mathscr{L},\mathscr{L}_{0})$ be a transitive
subalgebra of $W(\mathscr{F})$, $Q$ be an $\mathscr{L}$-module which is at the
same time a finitely generated $O(\mathscr{F})$-module. We say that
$\mathscr{L}$ acts on $Q$ by derivations if

$l(fq)=l(f)q+f(lq),\quad l\in\mathscr{L},\quad f\in O(\mathscr{F}),\quad q\in
Q.$

It should be pointed out that as it follows from transitivity of
$(\mathscr{L},\mathscr{L}_{0})$ $Q$ is a free $O(\mathscr{F})$-module. As well
as in [12] we will call $Q$ an $(O(\mathscr{F}),\mathscr{L})$-module. Denote
$W(\mathscr{F})$ by $W$, $O(\mathscr{F})$ by $O$.

Let $\mathscr{L}^{\sharp}=$ Prim$U(\mathscr{L})$ be the Lie algebra of the
primitive elements of Hopf algebra $U(\mathscr{L})$. $\mathscr{L}^{\sharp}$ is
the restricted Lie algebra with respect to the $p$-mapping in
$U(\mathscr{L})$. Let $\mathscr{M}$ be the normalizer of $\mathscr{L}_{0}$ in
$\mathscr{L}^{\sharp}$. Obviously, $\mathscr{M}$ is a restricted subalgebra of
$\mathscr{L}^{\sharp}$. According to the theory of truncated coinduced modules
the category of $(O,\mathscr{L})$-modules is equivalent to the category of
finite-dimensional restricted $\mathscr{M}$-modules. The equivalence is given
by the $\mathscr{M}$-morphisms $\lambda_{Q}\colon Q\rightarrow
Q\big{/}\mathfrak{m}Q$. From the universal property of coinduced modules it
follows that $\lambda_{Q}$ results in the isomorphism of the spaces of
invariants, $Q^{\mathscr{L}}\cong(Q\big{/}\mathfrak{m}Q)^{\mathscr{M}}$.

Now we consider the $(O,\mathscr{L})$-module $Q=S\Omega^{2}(\mathscr{F})=$
Hom${}_{O}(S_{O}^{2}W,O)$. Let $L=\text{gr}\mathscr{L}=L_{-1}+L_{0}+\ldots$.
Then we have

$Q\big{/}\mathfrak{m}Q\cong\text{Hom}_{K}(S^{2}(W\big{/}W_{(0)}),O\big{/}\mathfrak{m})\cong\text{Hom}_{K}(S^{2}L_{-1},K)$

and $\lambda_{Q}(\omega)=\omega(0)$. Obviously,

$(Q\big{/}\mathfrak{m}Q)^{\mathscr{M}}\cong\text{Hom}_{K}(S^{2}(W\big{/}W_{(0)}),K)^{\mathscr{M}}\subseteq\text{Hom}_{K}(S^{2}(W\big{/}W_{(0)}),K)^{\mathscr{L}_{0}}\cong\text{Hom}_{K}(S^{2}L_{-1},K)^{L_{0}}.$

Since $L_{-1}$ is an absolutely irreducible $L_{0}$-module when $n>2$, by
Schur’s lemma Hom${}_{K}(S^{2}L_{-1},K)^{L_{0}}=\langle\omega(0)\rangle_{K}$
is a one-dimensional space. Since $\omega\in Q^{\mathscr{L}}$,

$1\leqslant\dim(Q\big{/}\mathfrak{m}Q)^{\mathscr{M}}\leqslant\dim\text{Hom}_{K}(S^{2}L_{-1},K)^{L_{0}}=1.$

Hence, $\dim Q^{\mathscr{L}}=1$ which results in
$Q^{\mathscr{L}}=\langle\omega\rangle$. Now,

$\mathscr{L}^{\prime}\varphi(\omega)=\psi(\mathscr{L})\varphi(\omega)=\varphi\mathscr{L}\varphi^{-1}\varphi(\omega)=0.$

Consequently, $\varphi(\omega)=a\omega^{\prime}$, $a\in K\backslash\\{0\\}$.

Suppose that $\mathscr{L}$, $\mathscr{L}^{\prime}$ are graded Lie algebras
with the standard gradings. Then
$\overline{\psi}=\text{gr}\,\psi\colon\mathscr{L}\rightarrow\mathscr{L}^{\prime}$
is an isomorphism of graded Lie algebras. Obviously, $\overline{\psi}$ is
induced by the linear admissible isomorphism
$\overline{\varphi}=\text{gr}\,\varphi\colon O(\mathscr{F})\rightarrow
O(\mathscr{F}^{\prime})$ and $\overline{\varphi}(\omega)=a\omega^{\prime}$,
$a\in K\backslash\\{0\\}$. ∎

Remark. The proof of the assertion in (ii) is true for $n=2$ under additional
restrictions. All we need is that $L_{-1}$ should be an absolutely irreducible
$L_{0}$-module. For $\omega=(dx_{1})^{(2)}+(dx_{2})^{(2)}$ this is true if the
set of heights $\overline{m}=(m_{1},m_{2})$ is not equal to $(1,1)$. For
$\omega=dx_{1}dx_{2}+(dx_{2})^{(2)}$ $L_{0}$ acts irreducibly on $L_{-1}$ if
$m_{1}>1$ and $m_{2}>1$.

Everywhere in what follows we will assume that $\dim E=3$.

##  3 Bilinear non-alternating symmetric forms in three variables

Let $V$ be a three-dimensional vector space over a perfect field $K$ of
characteristic 2, $b$ be a bilinear nondegenerate non-alternating symmetric
form on $V$, $V^{0}$ be the hyperplane consisting of isotropic vectors (i.e.,
$V^{0}=\\{v\in V\leavevmode\nobreak\ |\leavevmode\nobreak\ b(v,v)=0\\}$),
$\mathscr{F}\colon 0=V_{0}\subseteq V_{1}\subseteq\ldots\subset V$ be a flag
of $V$ such that $V_{q}=V$ for sufficiently large $q$. Let $\\{e_{i}\\}$ be a
basis of $V$ coordinated with the flag $\mathscr{F}$,
$m_{i}=\min\\{j\leavevmode\nobreak\ |\leavevmode\nobreak\ e_{i}\in V_{j}\\}$
be the height of $e_{i}$. The orthogonal complement of a subspace $L\subseteq
V$ with respect to $b$ is denoted by $L^{\perp}$.

Pairs $(\mathscr{F},b)$ and $(\mathscr{F}^{\prime},b^{\prime})$ are called
equivalent if there is a linear automorphism $\varphi\colon V\rightarrow V$
such that $\varphi(\mathscr{F})=\mathscr{F}^{\prime}$ and
$b(u,v)=b^{\prime}(\varphi(u),\varphi(v))$ for any $u,v\in V$. Such an
automorphism is called admissible.

The flag $\mathscr{F}$ will be called almost trivial if it consists of
subspaces $0=V_{0}=\ldots=V_{q-1}\subset V_{q}=V$, the flag $\mathscr{F}$ is
trivial if $q=1$.

Remark. In the case of the almost trivial flag $\mathscr{F}$, the pair
$(\mathscr{F},b)$ is equivalent to the pair $(\mathscr{F},b^{\prime})$, where
$b^{\prime}$ 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 $\mathscr{F}$ if
$V_{j}=V_{j}\cap P+V_{j}\cap Q$ for all $j\geqslant 0$.

Since $\dim V=3$, the flag $\mathscr{F}$ consists of no more than four
different subspaces. Suppose that $\\{e_{i}\\}$ is enumerated in such a way
that $m_{1}\leqslant m_{2}\leqslant m_{3}$. Consider the subspaces
$V_{m_{1}}$, $V_{m_{2}}\cap V_{m_{1}}^{\perp}$, $V_{m_{3}}\cap
V_{m_{2}}^{\perp}=V_{m_{2}}^{\perp}$ and choose the vector $v$ from one of
these subspaces such that $b(v,v)=1$.

######  Lemma 1.

The decomposition $V=\langle v\rangle\oplus\langle v\rangle^{\bot}$ is
coordinated with the flag $\mathscr{F}$.

###### Proof.

Set $m_{0}=0$. Let $v\in V_{m_{q}}\cap V_{m_{q-1}}^{\perp}$ for some
$1\leqslant q\leqslant 3$. Suppose that $V_{i}=V_{m_{j}}$. If $j<q$, then
$V_{i}\subseteq V_{m_{q-1}}$ and $V_{i}\subseteq\langle v\rangle^{\bot}$. If
$j\geqslant q$, then $V_{m_{q}}\subseteq V_{i}$ and $\langle v\rangle\subseteq
V_{i}$. Hence, for any $i$ we have $V_{i}=V_{i}\cap(\langle
v\rangle\oplus\langle v\rangle^{\bot})=(\langle v\rangle\cap
V_{i})\oplus(\langle v\rangle^{\bot}\cap V_{i})$. ∎

Further, we do not assume that a basis coordinated with the flag $\mathscr{F}$
is ordered according to the heights, unless otherwise indicated.

Suppose that $\\{u,w\\}$ is a basis of the space $\langle v\rangle^{\bot}$,
which is coordinated with the flag $\overline{\mathscr{F}}=\\{V_{i}\cap\langle
v\rangle^{\bot}\\}$. Obviously, we can choose $\\{u,w\\}$ in such a way that
the matrix of the restriction of $b$ to $\langle v\rangle^{\bot}$ will be one
of the following matrices

$M_{0}=\begin{pmatrix}0&1\\\ 1&0\end{pmatrix},\leavevmode\nobreak\
M_{1}=\begin{pmatrix}0&1\\\ 1&1\end{pmatrix}\text{ and
}I=\begin{pmatrix}1&0\\\ 0&1\end{pmatrix}.$

Remark. In the case of the matrix $M_{1}$, if the height of $u$ is not less
than that of $w$, then there exists a basis $\\{u^{\prime},w^{\prime}\\}$,
which is coordinated with the flag $\overline{\mathscr{F}}$, such that the
matrix of the restriction of $b$ to $\langle v\rangle^{\bot}$ will be equal to
the matrix $I$.

Let $u=e_{1}$, $w=e_{2}$, $v=e_{3}$. Thus, we proved the following Lemma.

######  Lemma 2.

There exists a basis $\\{e_{1},e_{2},e_{3}\\}$ of the space $V$ coordinated
with the flag $\mathscr{F}$ such that the matrix of $b$ is equal to one of the
following matrices.

$B_{1}=\begin{pmatrix}0&1&0\\\ 1&0&0\\\
0&0&1\end{pmatrix},\leavevmode\nobreak\ B_{2}=\begin{pmatrix}0&1&0\\\ 1&1&0\\\
0&0&1\end{pmatrix},\leavevmode\nobreak\ B_{3}=\begin{pmatrix}1&0&0\\\ 0&1&0\\\
0&0&1\end{pmatrix}.$

Remark. In the case of the matrix $B_{2}$, if $m_{3}$ belongs to the segment
$[m_{1},m_{2}]$, then the form $b$ has the matrix $B_{1}$ with respect to the
basis $\\{e_{1}^{\prime},e_{2}^{\prime},e_{3}^{\prime}\\}$,
$e_{1}^{\prime}=e_{1}$, $e_{2}^{\prime}=e_{2}+e_{3}$,
$e_{3}^{\prime}=e_{3}+e_{1}$, which is also coordinated with $\mathscr{F}$.
Thus, we will use a basis of $V$ in which the matrix of $b$ is $B_{2}$ only in
the case when $m_{1}<m_{2}$ and $m_{3}\notin[m_{1},m_{2}]$.

In view of this remark, we will say that the matrix of $b$ with respect to a
basis $\\{e_{i}\\}$ has a canonical form if it is equal to one of the matrices
$B_{1}$, $B_{2}$, $B_{3}$. Thus, the matrix $B_{2}$ has a canonical form if
$m_{1}<m_{2}$ and $m_{3}\notin[m_{1},m_{2}]$. In the case of the almost
trivial flag we will say that the matrix of $b$ has a canonical form if it is
equal to $B_{3}$.

######  Theorem 4.

Let $b$ and $b^{\prime}$ be nondegenerate non-alternating symmetric forms on
$V$, $\mathscr{F}$ and $\mathscr{F}^{\prime}$ be flags of $V$. Suppose that
$\\{e_{1},e_{2},e_{3}\\}$
$(\\{e_{1}^{\prime},e_{2}^{\prime},e_{3}^{\prime}\\})$ is a basis of $V$,
which is coordinated with the flag $\mathscr{F}$ $(\mathscr{F}^{\prime})$,
such that the matrix of $b$ $(b^{\prime})$ has a canonical form. The following
statements are true.

(1) If the forms $b$ and $b^{\prime}$ have the matrices equal to $B_{1}$, then
the pairs $(\mathscr{F},b)$ and $(\mathscr{F}^{\prime},b^{\prime})$ are
equivalent if and only if
$\\{m_{1},m_{2}\\}=\\{m^{\prime}_{1},m^{\prime}_{2}\\}$ and
$m_{3}=m^{\prime}_{3}$.

(2) If the forms $b$ and $b^{\prime}$ have the matrices equal to $B_{2}$, then
the pairs $(\mathscr{F},b)$ and $(\mathscr{F}^{\prime},b^{\prime})$ are
equivalent if and only if $m_{i}=m^{\prime}_{i}$, $i=1,2,3$.

(3) If the forms $b$ and $b^{\prime}$ have the matrices equal to $B_{3}$, then
the pairs $(\mathscr{F},b)$ and $(\mathscr{F}^{\prime},b^{\prime})$ are
equivalent if and only if
$\\{m_{1},m_{2},m_{3}\\}=\\{m^{\prime}_{1},m^{\prime}_{2},m_{3}^{\prime}\\}$.

(4) If the forms $b$ and $b^{\prime}$ have different matrices, then the pairs
$(\mathscr{F},b)$ and $(\mathscr{F}^{\prime},b^{\prime})$ are not equivalent.

###### Proof.

(4) Since $\dim V=3$, $V=V^{0}\oplus(V^{0})^{\bot}$. Let
$V^{1}=(V^{0})^{\bot}$. In the case of the full flag, for the form with the
matrix $B_{1}$ the decomposition $V=V^{0}\oplus V^{1}$ is coordinated with the
flag, but for the form with the matrix $B_{2}$ or with the matrix $B_{3}$ it
is not true. Obviously, $n_{r}=\dim V_{s_{r}}\cap V_{s_{r-1}}^{\perp}-\dim
V_{s_{r}}\cap V_{s_{r-1}}^{\perp}\cap V^{0}$, where
$s_{1}<s_{2}<s_{3}\in\\{m_{1},m_{2},m_{3}\\}$, are invariants (similar to
invariants $n_{qq}^{1}$ from [5, Section 1]). For the form $b$ with the matrix
$B_{3}$ we have $n_{1}=n_{2}=n_{3}=1$, but for the form $b$ with the matrix
$B_{2}$ there is $i$ such that $n_{i}=0$.

If $m_{i}=m_{j}\neq m_{k}$, where $\\{i,j,k\\}=\\{1,2,3\\}$, then for the form
with the matrix $B_{1}$ the decomposition $V=V^{0}\oplus V^{1}$ is coordinated
with the flag, but for the form with the matrix $B_{3}$ it is not true.

(1-3) If the conditions are fulfilled, then the isomorphism which maps $e_{i}$
into $e^{\prime}_{i}$ (up to permutation) is an equivalence.

Let in the cases (1) and (2)
$\\{m_{1},m_{2},m_{3}\\}=\\{m^{\prime}_{1},m^{\prime}_{2},m_{3}^{\prime}\\}$,
but the conditions do not hold. Then $n_{i}\neq n_{i}^{\prime}$ for some $i$.
Consequently, the pairs are not equivalent. The statement (3) is trivial. ∎

##  4 Non-alternating Hamiltonian forms with non-constant coefficients in
three variables

Put

$\overline{x}_{i}=x_{i}^{(2^{m_{i}}-1)},\quad\langle
x_{1},x_{2},x_{3}\rangle=E,\quad m_{i}=\min\\{j\leavevmode\nobreak\
|\leavevmode\nobreak\ x_{i}\notin E_{j}\\}.$

Let $\omega$ be the non-alternating Hamiltonian form,

$\omega=\omega(0)+d\varphi+b_{12}\overline{x}_{1}\overline{x}_{2}dx_{1}dx_{2}+b_{13}\overline{x}_{1}\overline{x}_{3}dx_{1}dx_{3}+b_{23}\overline{x}_{2}\overline{x}_{3}dx_{2}dx_{3},$

where $b_{ij}\in K$, $\varphi\in\mathfrak{m}^{(2)}S\Omega^{1}(\mathscr{F})$,
and $\omega(0)$ is equal to one of the following forms (see Lemma 2 and
Theorem 4)

$\displaystyle\omega(0)=dx_{1}dx_{2}+dx_{3}^{(2)},$ (2)
$\displaystyle\omega(0)=dx_{1}dx_{2}+dx_{2}^{(2)}+dx_{3}^{(2)},$ (3)
$\displaystyle\omega(0)=dx_{1}^{(2)}+dx_{2}^{(2)}+dx_{3}^{(2)}.$ (4)

Here $dx_{i}^{(2)}$ means $(dx_{i})^{(2)}$. Recall that in the case of (3),
$m_{1}<m_{2}$ and $m_{3}\notin[m_{1},m_{2}]$.

######  Theorem 5.

Let $\omega$ be a non-alternating Hamiltonian form in three variables of
heights $m_{1},m_{2},m_{3}$. If the corresponding flag $\mathscr{F}$ is not
almost trivial, then $\omega$ is equivalent to only one of the following forms

$\omega_{1}=dx_{1}dx_{2}+dx_{3}^{(2)},$
$\omega_{2}=dx_{1}dx_{2}+dx_{2}^{(2)}+dx_{3}^{(2)}\leavevmode\nobreak\
\leavevmode\nobreak\ (m_{1}<m_{2},m_{3}\notin[m_{1},m_{2}]),$
$\omega_{3}=dx_{1}^{(2)}+dx_{2}^{(2)}+dx_{3}^{(2)},$
$\omega_{4}=dx_{1}dx_{2}+dx_{3}^{(2)}+\overline{x}_{1}x_{3}dx_{1}dx_{3}\leavevmode\nobreak\
\leavevmode\nobreak\ (m_{3}=1).$

The two forms $\omega_{i}$, $\omega_{j}$, $i\neq j$ are not equivalent. If
$\mathscr{F}$ is almost trivial but non-trivial, then $\omega$ is equivalent
to the form $\omega_{3}$. If $\mathscr{F}$ is trivial, then $\omega$ is
equivalent to one of the forms $\omega_{3}$, $\omega_{4}$, with the forms
$\omega_{3}$ and $\omega_{4}$ being non-equivalent.

###### Proof.

By Theorem 1 and Theorem 2 if $m_{3}>1$ in the case of (2) and
$(m_{1},m_{2},m_{3})\neq(1,1,1)$ in the case of (4), then the form $\omega$ is
equivalent to the form $\omega(0)$. In the case of (3), the form $\omega$ is
equivalent to the form $\omega(0)$.

Suppose that $\omega$ is a form such that $\omega(0)$ is of type (2) and
$m_{3}=1$. Theorem 1 and Theorem 2 imply that $\omega$ is equivalent to the
form
$\omega(0)+c_{13}\overline{x}_{1}x_{3}dx_{1}dx_{3}+c_{23}\overline{x}_{2}x_{3}dx_{2}dx_{3}$.

If $c_{13}\neq 0$, $c_{23}\neq 0$ and $m_{1}\leqslant m_{2}$, then the
admissible automorphism $x_{1}\mapsto
x_{1}+(\tilde{c}_{23}/\tilde{c}_{13})x_{2}^{(2^{m_{2}-m_{1}})}$, where
$(\tilde{c}_{23})^{2^{m_{1}}}=c_{23}$, $(\tilde{c}_{13})^{2^{m_{1}}}=c_{13}$,
$x_{i}\mapsto x_{i}$, $i=2,3$ will lead to vanishing
$c_{23}\overline{x}_{2}x_{3}dx_{2}dx_{3}$. If $m_{1}>m_{2}$, then the
admissible automorphism $x_{2}\mapsto
x_{2}+(\tilde{c}_{13}/\tilde{c}_{23})x_{1}^{(2^{m_{1}-m_{2}})}$, where
$(\tilde{c}_{23})^{2^{m_{2}}}=c_{23}$, $(\tilde{c}_{13})^{2^{m_{2}}}=c_{13}$,
$x_{i}\mapsto x_{i}$, $i=1,3$ will lead to vanishing
$c_{13}\overline{x}_{1}x_{3}dx_{1}dx_{3}$.

Consequently, a form $\omega$ is equivalent to the form
$dx_{1}dx_{2}+dx_{3}^{(2)}+c\overline{x}_{1}x_{3}dx_{1}dx_{3}$. Obviously, the
two forms $\omega(0)+c\overline{x}_{1}x_{3}dx_{1}dx_{3}$ and
$\omega(0)+c\overline{x}_{2}x_{3}dx_{2}dx_{3}$, where
$\omega(0)=dx_{1}dx_{2}+dx_{3}^{(2)}$ are equivalent. The corresponding
admissible isomorphism $\sigma\colon O(3,(m_{1},m_{2},1))\rightarrow
O(3,(m_{2},m_{1},1))$ is induced by the permutation $x_{1}\mapsto x_{2}$,
$x_{2}\mapsto x_{1}$, $x_{3}\mapsto x_{3}$. Furthermore, if $c\neq 0$, then
the admissible automorphism $\tilde{c}x_{1}\mapsto x_{1}$,
$x_{2}\mapsto\tilde{c}x_{2}$, where $(\tilde{c})^{2^{m_{1}}}=c$, $x_{3}\mapsto
x_{3}$ will lead to $c=1$.

Suppose that $\omega$ is a form such that $\omega(0)$ is of type (4) and
$(m_{1},m_{2},m_{3})=(1,1,1)$. In the case of the trivial flag, any linear
isomorphism is admissible and $\omega$ is equivalent to $\omega^{\prime}$ with
$\omega^{\prime}(0)$ of type (2). Consequently, a form $\omega$ is equivalent
to the form
$\omega^{\prime\prime}=dx_{1}dx_{2}+dx_{3}^{(2)}+cx_{1}x_{3}dx_{1}dx_{3}$. If
$c\neq 0$, then $c=1$. Note that in this case the matrix of
$\omega^{\prime\prime}(0)$ does not have the canonical form. If $c=0$, then we
say that $\omega$ is equivalent to the form
$dx_{1}^{(2)}+dx_{2}^{(2)}+dx_{3}^{(2)}$.

Suppose that $\mathscr{F}$ is not almost trivial and show that the forms
$\omega_{i}$, $\omega_{j}$, $i\neq j$ are not equivalent, $i,j=1,2,3,4$. The
fact that the forms $\omega_{i}$, $i=1,2,3$ are not equivalent follows from
Theorem 4.

Let $\varphi$ be an admissible isomorphism, $\varphi\colon
O(\mathscr{F})\rightarrow O(\mathscr{F}^{\prime})$, $\varphi_{0}$ be the
linear part of $\varphi$, $\varphi(x_{i})\equiv\varphi_{0}(x_{i})(\text{mod
}\mathfrak{m}^{(2)})$. Since $(\varphi(\omega))(0)=\varphi_{0}(\omega(0))$, we
conclude that if linear parts of the forms $\omega$, $\omega^{\prime}$ are not
equivalent, then $\omega$ and $\omega^{\prime}$ are not equivalent, either.
So, all what remains to be considered is the pair $\omega_{1}$, $\omega_{4}$.
Suppose that the forms $\omega_{1}$ and $\omega_{4}$ are equivalent. Hence,
$\omega_{1}(0)=dx_{1}dx_{2}+dx_{3}^{(2)}$ is equivalent to
$\omega_{4}(0)=dx_{1}^{\prime}dx_{2}^{\prime}+(dx_{3}^{\prime})^{(2)}$, where
$\\{x_{1},x_{2},x_{3}\\}$ and
$\\{x_{1}^{\prime},x_{2}^{\prime},x_{3}^{\prime}\\}$ are bases of $E$
coordinated with the flags $\mathscr{F}$, $\mathscr{F}^{\prime}$,
respectively. By Theorem 4
$\\{m_{1},m_{2}\\}=\\{m^{\prime}_{1},m^{\prime}_{2}\\}$ and
$m_{3}=m^{\prime}_{3}$. Since $m_{3}^{\prime}=1$, we have $m_{3}=1$. Theorem 6
of the next section implies that $P(3,\overline{m},\omega_{4})$ is a simple
Lie algebra and $P(3,\overline{m},\omega_{1})$ is not a simple Lie algebra.

If $\mathscr{F}$ is almost trivial and $\overline{m}\neq(1,1,1)$, then
$\omega(0)=\omega_{3}$ and, as it has been shown before, $\omega$ is
equivalent to $\omega_{3}$. If $\overline{m}=(1,1,1)$, then the matrix of
$\omega_{4}(0)$ does not have the canonical form. The fact that the forms
$\omega_{3}$ and $\omega_{4}$ are not equivalent follows from Theorem 6 of the
next section according to which $P(3,\overline{1},\omega_{4})$ is a simple Lie
algebra and $P(3,\overline{1},\omega_{3})$ is not a simple Lie algebra. ∎

Remarks. 1\. The conditions under which the forms $\omega_{i}^{\prime}$,
$\omega_{i}^{\prime\prime}$ of the same type are equivalent for $i=1,2,3$ are
obtained in Theorem 4. The problem of equivalence of the forms
$\omega_{4}^{\prime}$, $\omega_{4}^{\prime\prime}$ is not being treated in
this paper, since some other technique is required.

2\. Let $a\in K\diagdown\\{0\\}$, then the form $a\omega_{i}$ is equivalent to
$\omega_{i}$, $i=1,2,3,4$. It is obvious for $i=1,2,3$. Let $\varphi$ be the
admissible automorphism of $O(3,(m_{1},m_{2},1))$ such that
$\varphi(x_{i})=c_{i}x_{i}$, $c_{1}=1$, $c_{2}=a$, $c_{3}=\sqrt{a}$. Then
$\varphi(\omega_{4})=a\omega_{4}$.

##  5 Non-alternating Hamiltonian Lie algebras in three variables

In this section we will prove the fact that the Lie algebras which correspond
to the nonequivalent forms are not isomorphic.

The algebra $P(3,\overline{m},\omega)$ is identified with the algebra
$O(\mathscr{F})\big{/}K\cong\mathfrak{m}$ with the Poisson bracket,
corresponding to the form $\omega$ (see (1)):

$\\{f,g\\}_{1}=\partial_{1}f\partial_{2}g+\partial_{2}f\partial_{1}g+\partial_{3}f\partial_{3}g,$
$\\{f,g\\}_{2}=\partial_{1}f\partial_{2}g+\partial_{2}f\partial_{1}g+\partial_{1}f\partial_{1}g+\partial_{3}f\partial_{3}g,$
$\\{f,g\\}_{3}=\partial_{1}f\partial_{1}g+\partial_{2}f\partial_{2}g+\partial_{3}f\partial_{3}g,$
$\\{f,g\\}_{4}=\partial_{1}f\partial_{2}g+\partial_{2}f\partial_{1}g+\partial_{3}f\partial_{3}g+\overline{x}_{1}x_{3}(\partial_{2}f\partial_{3}g+\partial_{3}f\partial_{2}g).$

If $0\neq f\in\mathfrak{m}$, let $\lambda(f)$ be the nonzero homogeneous part
of $f$ of the least degree.

######  Theorem 6.

Let $L=P(3,\overline{m},\omega_{i})$, $i=1,2,3,4$. If $m_{3}>1$ in the case of
$i=1$ and $\overline{m}\neq(1,1,1)$ in the case of $i=3$, then $L$ is a simple
Lie algebra of dimension $2^{m_{1}+m_{2}+m_{3}}-1$. If $m_{3}=1$ and
$\overline{m}\neq(1,1,1)$, then $P^{(1)}(3,\overline{m},\omega_{1})$ is a
simple Lie algebra of dimension $2^{m_{1}+m_{2}+1}-2$. If
$\overline{m}=(1,1,1)$, then $P^{(1)}(3,\overline{m},\omega_{3})$ is not
simple.

###### Proof.

The algebras $P(3,\overline{m},\omega_{i})$, $i=1,2,3$ are considered in [5,
Section 4].

Let $J$ be a nonzero ideal in $P(3,\overline{m},\omega_{4})$ and $0\neq f\in
J$. We have $ad\,x_{1}=\partial_{2}$,
$ad\,x_{2}=\partial_{1}+\overline{x}_{1}x_{3}\partial_{3}$,
$ad\,x_{3}=\partial_{3}+\overline{x}_{1}x_{3}\partial_{2}$. Commuting $f$
consecutively, if necessary, with $x_{1}$, $x_{3}$, $x_{2}$ we obtain
$x_{1}\in J$. Now $\\{x_{1},x_{2}x_{3}\\}=x_{3}\in J$,
$\\{x_{3},x_{2}x_{3}\\}=x_{2}\in J$. Let
$L=\text{gr}\,P(3,\overline{m},\omega_{4})=\bigoplus\limits_{i=-1}^{r}L_{i}$.
Since $L_{-1}\subset\text{gr}\,J$, we have $L_{i}\subset\text{gr}\,J$ for
$i<r$. Therefore, for any monomial
$g\neq\overline{x}_{1}\overline{x}_{2}x_{3}$ there is $h$ such that $g+h\in
J$, $\deg\lambda(h)>\deg g$ or $h=0$. Since
$\\{x_{2},\overline{x}_{2}x_{3}\\}=\overline{x}_{1}\overline{x}_{2}x_{3}\in
J$, we have $J=P(3,\overline{m},\omega_{4})$, that is, the algebra is simple.
∎

Further we prove that the natural filtration of
$P(3,\overline{m},\omega_{i})$, $i=1,2,3,4$, $\overline{m}\neq(1,1,1)$ is
intrinsically determined.

######  Lemma 3 ([1]).

If $w_{1},\ldots,w_{k}\in P(3,\overline{m},\omega)$ are linearly dependent,
the same is true for $\\{\lambda(w_{i})\\}$.

Let $L^{i}=P(3,\overline{m},\omega_{i})$, $i=1,2,3,4$ be a non-alternating
Hamiltonian Lie algebra, where $\overline{m}\neq(1,1,1)$. For $\phi\in
Der(L^{i})$ let $R^{i}(\phi)=\dim(\textrm{Im}\phi)$ (see [1]). Obviously,
$R^{i}(\phi)=R^{i}(a\phi)$ for any $a\in K^{*}$. If $M$ is a subalgebra of
$Der(L^{i})$, let $R^{i}(M)=\min\limits_{0\neq\phi\in M}R^{i}(\phi)$.

Let $\\{L_{(i)}\\}$ be the standard filtration, $\\{L_{i}\\}$ be the standard
grading and
$\xi=ad\,\bar{x}=ad\,\bar{x}_{1}\bar{x}_{2}\bar{x}_{3}=ad\,x_{1}^{(2^{m_{1}}-1)}x_{2}^{(2^{m_{2}}-1)}x_{3}^{(2^{m_{3}}-1)}$.

######  Lemma 4.

Let $0\neq\phi\in\langle\xi\rangle$. If $i=2,3$ or $i=1$, $m_{3}>1$, then
$R^{i}(\phi)=4$. If $i=1$, $m_{3}=1$ or $i=4$, then $R^{i}(\phi)=3$.

###### Proof.

Obviously, $\phi(x_{1}),\phi(x_{2}),\phi(x_{3})\in\textrm{Im}\phi$ are
linearly independent. The action of $\phi$ on $L_{(0)}$ corresponds to the
one-dimensional representation of $L_{(0)}$ on
$L_{(r)}=\langle\bar{x}\rangle$. Since $\phi(L_{(1)})=0$ and
$\phi([L_{(0)},L_{(0)}])=0$, it follows that Im($\phi$)
$=\langle\phi(x_{1}),\phi(x_{2}),\phi(x_{3}),\bar{x}\rangle$ if there exists
$x_{j}^{(2)}\in L$ such that $\phi(x_{j}^{(2)})\neq 0$. Otherwise, Im($\phi$)
$=\langle\phi(x_{1}),\phi(x_{2}),\phi(x_{3})\rangle$.

Let $i=1$. Then $\phi(x_{1}^{(2)})=\phi(x_{2}^{(2)})=0$. If $m_{3}>1$, then
$\phi(x_{3}^{(2)})\neq 0$. Consequently, $R^{1}(\phi)=4$. If $m_{3}=1$, then
$R^{1}(\phi)=3$.

Let $i=2$. The condition $m_{1}<m_{2}$, $m_{3}\notin[m_{1},m_{2}]$ yield
$m_{1}>1$ or $m_{3}>1$. Since $\phi(x_{1}^{(2)})=\phi(x_{3}^{(2)})\neq 0$, we
have $R^{2}(\phi)=4$.

Let $i=3$. Since $\overline{m}\neq(1,1,1)$, we have $m_{j}>1$ and
$\phi(x_{j}^{(2)})\neq 0$. Consequently, $R^{3}(\phi)=4$.

Let $i=4$. Then $m_{3}=1$ and $\phi(x_{1}^{(2)})=\phi(x_{2}^{(2)})=0$.
Consequently, $R^{4}(\phi)=3$. ∎

######  Lemma 5.

Let $i=2,3$ or $i=1$, $m_{3}>1$. If $0\neq D\in L^{i}_{t}$, $-1\leqslant
t<2^{m_{1}}+2^{m_{2}}+2^{m_{3}}-5$, then there exist $5$ homogeneous elements
$E_{1},E_{2},E_{3},E_{4},E_{5}\in L^{i}$ such that
$\\{D,E_{1}\\},\\{D,E_{2}\\},\ldots,\\{D,E_{5}\\}$ are linearly independent.

###### Proof.

The case of $L^{1}$, $m_{3}>1$.

$(1)\leavevmode\nobreak\ t=-1$. Let $D=a_{1}x_{1}+a_{2}x_{2}+a_{3}x_{3}$ and
$a_{3}\neq 0$. Put $E_{1}=x_{3}^{(2)}$, $E_{2}=x_{3}^{(3)}$,
$E_{3}=x_{1}x_{3}^{(3)}$, $E_{4}=x_{2}x_{3}^{(3)}$,
$E_{5}=x_{1}x_{2}x_{3}^{(3)}$. Let $D=a_{1}x_{1}+a_{2}x_{2}$ and $a_{i}\neq
0$. Put $E_{1}=x_{1}x_{2}x_{3}$, $E_{2}=x_{1}x_{2}x_{3}^{(3)}$,
$E_{3}=x_{j}x_{3}$, $E_{4}=x_{j}x_{3}^{(2)}$, $E_{5}=x_{j}x_{3}^{(3)}$, where
$j\neq i$.

$(2)\leavevmode\nobreak\ t>-1$. Let
$D=ax_{1}^{(\alpha)}x_{2}^{(\beta)}x_{3}^{(\gamma)}+f(x_{1},x_{2},x_{3})$,
where $a\neq 0$, $\alpha,\beta>0$ and $f$ contains $x_{3}^{(s)}g(x_{1},x_{2})$
if $s\leqslant\gamma$. Put $E_{1}=x_{1}^{(2^{m_{1}}-\alpha)}$,
$E_{2}=x_{3}^{(2^{m_{3}}+1-\gamma)}$ ($\gamma>1$) or $x_{1}x_{3}^{(2)}$. If
$\gamma<2^{m_{3}}-2$, then $E_{3}=x_{2}^{(2^{m_{2}}-\beta)}$,
$E_{4}=x_{1}^{(2^{m_{1}}-\alpha)}x_{3}^{(2^{m_{3}}-1-\gamma)}$,
$E_{5}=x_{1}^{(2^{m_{1}}-\alpha)}x_{3}^{(2^{m_{3}}-2-\gamma)}$ ($\gamma$ is
even) or $x_{3}^{(2)}$ ($\gamma$ is odd). If $\gamma=2^{m_{3}}-2$, then
$E_{3}=x_{2}$, $E_{4}=x_{1}^{(2^{m_{1}}-\alpha)}x_{3}$,
$E_{5}=x_{2}^{(2^{m_{2}}-\beta)}x_{3}$. If $\gamma=2^{m_{3}}-1$, then
$E_{3}=x_{2}^{(2^{m_{2}}-\beta)}$,
$E_{4}=x_{1}^{(2^{m_{1}}-1-\alpha)}x_{2}^{(2^{m_{2}}-1-\beta)}x_{3}$,
$E_{5}=x_{1}^{(2^{m_{1}}-1-\alpha)}x_{2}^{(2^{m_{2}}-1-\beta)}x_{3}^{(2)}$.

Let $D=ax_{3}^{(t+2)}+f(x_{1},x_{3})+g(x_{2},x_{3})$ and $a\neq 0$. Put
$\\{E_{1},\ldots,E_{5}\\}=\left\\{x_{3},\bar{x}_{1}x_{3},\bar{x}_{2}x_{3},x_{1}x_{2}x_{3},\right.\\\
\left.x_{1}x_{2}x_{3}^{(2^{m_{3}}-1-t)}\tiny\right\\}$.

Let $D=ax_{i}^{(\alpha)}x_{3}^{(\gamma)}+f(x_{1},x_{3})+g(x_{2},x_{3})$, where
$a\neq 0$, $\alpha>0$ and $f,g$ contain $x_{3}^{(s)}x_{q}^{(h)}$ if
$s\leqslant\gamma$, $h>0$. If $\gamma>0$, then $E_{1}=x_{j}$,
$E_{2}=x_{i}^{(2^{m_{i}}-1-\alpha)}x_{3}$,
$E_{3}=x_{i}^{(2^{m_{i}}-\alpha)}x_{j}$,
$E_{4}=x_{i}^{(2^{m_{i}}-\alpha)}x_{j}x_{3}^{(2^{m_{3}}-1-\gamma)}$
($\gamma<2^{m_{3}}-1$) or
$x_{i}^{(2^{m_{i}}-1-\alpha)}x_{j}^{(2^{m_{j}}-2)}x_{3}$,
$E_{5}=x_{j}x_{3}^{(2)}$ ($\gamma$ is odd) or
$x_{j}x_{3}^{(2^{m_{3}}-2-\gamma)}$ ($\gamma<2^{m_{3}}-2$ is even) or
$x_{j}x_{3}$, where $\\{i,j\\}=\\{1,2\\}$. If $\gamma=0$, then
$\\{E_{1},\ldots,E_{5}\\}=\\{x_{j},x_{i}^{(2^{m_{i}}-\alpha)}x_{j},x_{i}^{(2^{m_{i}}-\alpha)}x_{j}x_{3},x_{i}^{(2^{m_{i}}-\alpha)}x_{j}x_{3}^{(2)},x_{i}^{(2^{m_{i}}-\alpha)}x_{j}x_{3}^{(3)}\\}$.

The case of $L^{2}$, $m_{3}>1$.

$(1)\leavevmode\nobreak\ t=-1$. Let $D=a_{1}x_{1}+a_{2}x_{2}+a_{3}x_{3}$ and
$a_{3}\neq 0$. The same $L^{1}$. Let $D=a_{1}(x_{1}+x_{2})+a_{2}x_{2}$ and
$a_{i}\neq 0$. Put $\\{E_{1},\ldots,E_{5}\\}$ the same $L^{1}$.

$(2)\leavevmode\nobreak\ t>-1$. Let
$D=ax_{1}^{(\alpha)}x_{2}^{(\beta)}x_{3}^{(\gamma)}+f(x_{1},x_{2},x_{3})$,
where $a\neq 0$, $\alpha,\beta>0$ and $f$ contains $x_{3}^{(s)}g(x_{1},x_{2})$
if $s\leqslant\gamma$. Put $E_{1}=x_{2}^{(2^{m_{2}}-\beta)}$,
$E_{2}=x_{3}^{(2^{m_{3}}+1-\gamma)}$ ($\gamma>1$) or $x_{3}^{(3)}$
($\gamma=1$) or $x_{1}^{(2^{m_{1}}-\alpha)}x_{3}^{(2)}$ ($\gamma=0\neq
2^{m_{3}}-4$, $(\alpha,\beta)\neq(2^{m_{1}}-1,2^{m_{2}}-1)$) or
$x_{1}x_{3}^{(2)}+x_{2}x_{3}^{(2)}$. If $\gamma<2^{m_{3}}-2$, then
$E_{3}=x_{1}^{(2^{m_{1}}-\alpha)}$
($(\alpha,\beta)\neq(2^{m_{1}}-1,2^{m_{2}}-1)$) or $x_{1}+x_{2}$,
$E_{4}=x_{2}^{(2^{m_{2}}-\beta)}x_{3}^{(2^{m_{3}}-1-\gamma)}$,
$E_{5}=x_{2}^{(2^{m_{2}}-\beta)}x_{3}^{(2^{m_{3}}-2-\gamma)}$ ($\gamma$ is
even) or $x_{3}^{(2)}$ ($\gamma$ is odd). If $\gamma=2^{m_{3}}-2$, then
$E_{3}=x_{1}+x_{2}$, $E_{4}=x_{2}^{(2^{m_{2}}-\beta)}x_{3}$,
$E_{5}=x_{1}^{(2^{m_{1}}-\alpha)}x_{3}$
($(\alpha,\beta)\neq(2^{m_{1}}-1,2^{m_{2}}-1)$) or $x_{1}x_{3}+x_{2}x_{3}$. If
$\gamma=2^{m_{3}}-1$, then $E_{3}=x_{1}^{(2^{m_{1}}-\alpha)}$,
$E_{4}=x_{1}^{(2^{m_{1}}-1-\alpha)}x_{2}^{(2^{m_{2}}-1-\beta)}x_{3}$,
$E_{5}=x_{1}^{(2^{m_{1}}-1-\alpha)}x_{2}^{(2^{m_{2}}-1-\beta)}x_{3}^{(2)}$.

Let $D=ax_{3}^{(t+2)}+f(x_{1},x_{3})+g(x_{2},x_{3})$ and $a\neq 0$. The same
$L^{1}$.

Let $D=ax_{i}^{(\alpha)}x_{3}^{(\gamma)}+f(x_{1},x_{3})+g(x_{2},x_{3})$, where
$a\neq 0$, $\alpha>0$ and $f,g$ contain $x_{3}^{(s)}x_{q}^{(h)}$ if
$s\leqslant\gamma$, $h>0$. The same $L^{1}$.

The case of $L^{2}$, $m_{3}=1$, $m_{1}>1$, $m_{2}>m_{1}>1$.

$(1)\leavevmode\nobreak\ t=-1$. Let
$D=a_{1}(x_{1}+x_{2})+a_{2}x_{2}+a_{3}x_{3}$ and $a_{i}\neq 0$, $i=1,2$. Put
$E_{1}=x_{j}^{(2)}$, $E_{2}=x_{j}^{(3)}$, $E_{3}=x_{j}^{(3)}x_{3}$,
$E_{4}=x_{i}x_{j}^{(3)}$, $E_{5}=x_{i}x_{j}^{(3)}x_{3}$, where $j\neq i$. Let
$D=a_{3}x_{3}$ and $a_{3}\neq 0$. Put $E_{1}=x_{1}x_{3}$, $E_{2}=x_{2}x_{3}$,
$E_{3}=x_{1}^{(2)}x_{3}$, $E_{4}=x_{2}^{(2)}x_{3}$, $E_{5}=x_{1}x_{2}x_{3}$.

$(2)\leavevmode\nobreak\ t>-1$. Let
$D=ax_{1}^{(\alpha)}x_{2}^{(\beta)}x_{3}+f(x_{1},x_{2},x_{3})$, where $a\neq
0$, $\beta>0$ and $f$ contains $x_{1}^{(s)}g(x_{2},x_{3})$ if
$s\leqslant\alpha$. Put $E_{1}=x_{2}^{(2^{m_{2}}-1-\beta)}x_{3}$. If
$\gamma<2^{m_{3}}-1$, then
$E_{2}=x_{1}^{(2^{m_{1}}-1-\alpha)}x_{2}^{(2^{m_{2}}-1-\beta)}x_{3}$,
$E_{3}=x_{1}^{(2^{m_{2}}+1-\alpha)}$ ($\alpha>1$) or $x_{1}^{(2)}x_{3}$. If
$\gamma<2^{m_{3}}-2$, then $E_{4}=x_{3}$ ($\beta\neq 2^{m_{2}}-1$) or
$x_{1}x_{2}$ ($\beta=2^{m_{2}}-1$, $\alpha$ is even) or
$x_{1}^{(2^{m_{1}}-1-\alpha)}x_{2}$ ($\beta=2^{m_{2}}-1$, $\alpha$ is odd),
$E_{5}=x_{1}^{(2^{m_{1}}-2-\alpha)}x_{2}^{(2^{m_{2}}-1-\beta)}x_{3}$ ($\alpha$
is even) or $x_{1}^{(2)}$ ($\alpha$ is odd). If $\alpha=2^{m_{3}}-2$, then
$E_{4}=x_{1}+x_{2}$, $E_{5}=x_{1}x_{3}$ ($\beta\neq 2^{m_{2}}-1$) or
$x_{1}x_{2}$. If $\gamma=2^{m_{3}}-1$, then $E_{2}=x_{2}$, $E_{3}=x_{3}$,
$E_{4}=x_{1}^{(2)}$ ($\beta\neq 2^{m_{2}}-2$) or $x_{1}x_{2}$,
$E_{5}=x_{1}^{(2)}x_{2}^{(2^{m_{2}}-1-\beta)}$.

Let $D=ax_{1}^{(t+2)}+f(x_{1},x_{2})+g(x_{1},x_{3})$ and $a\neq 0$. Put
$\\{E_{1},\ldots,E_{5}\\}=\left\\{x_{1},x_{1}\bar{x}_{2},x_{1}x_{3},x_{1}x_{2}x_{3},\right.\\\
\left.x_{1}^{(2^{m_{1}}-1-t)}x_{2}x_{3}\right\\}$.

Let $D=ax_{1}^{(\alpha)}x_{3}+f(x_{1},x_{2})+g(x_{1},x_{3})$, where $a\neq 0$,
$\alpha>0$ and $f,g$ contain $x_{1}^{(s)}x_{q}^{(h)}$ if $s\leqslant\alpha$,
$h>0$. Put
$\\{E_{1},\ldots,E_{5}\\}=\\{x_{3},x_{2},x_{2}^{(2)},x_{2}^{(3)},\bar{x}_{2}x_{3}\\}$.

Let $D=ax_{1}^{(\alpha)}x_{2}^{(\beta)}+f(x_{1},x_{2})+g(x_{1},x_{3})$, where
$a\neq 0$, $\beta>0$ and $f,g$ contain $x_{1}^{(s)}x_{q}^{(h)}$ if
$s\leqslant\alpha$, $h>0$. If $\alpha>0$, then
$E_{1}=x_{1}^{(2^{m_{1}}-\alpha)}$
($(\alpha,\beta)\neq(2^{m_{1}}-1,2^{m_{2}}-1)$) or $x_{1}+x_{2}$,
$E_{2}=x_{2}^{(2^{m_{2}}-\beta)}$, $E_{3}=x_{2}^{(2^{m_{2}}-\beta)}x_{3}$,
$E_{4}=x_{1}^{(2^{m_{1}}+1-\alpha)}$ ($\alpha>1$) or
$x_{1}^{(2)}x_{2}^{(2^{m_{2}}-1-\beta)}$, $E_{5}=x_{2}^{(2)}x_{3}$ ($\alpha$
is odd) or $x_{1}^{(2^{m_{1}}-1-\alpha)}x_{3}$ ($\alpha$ is even,
$(\alpha,\beta)\neq(2^{m_{1}}-2,2^{m_{2}}-1)$) or $x_{1}x_{3}+x_{2}x_{3}$. If
$\alpha=0$, then
$\\{E_{1},\ldots,E_{5}\\}=\\{x_{1},x_{1}x_{3},x_{1}^{(2)},x_{1}^{(3)},x_{1}x_{2}^{(2^{m_{2}}-\beta)}\\}$.

The case of $L^{3}$. Without the loss of generality we can assume that
$m_{3}>1$.

$(1)\leavevmode\nobreak\ t=-1$. The same $L^{1}$ with $i=j$.

$(2)\leavevmode\nobreak\ t>-1$. Let
$D=ax_{1}^{(\alpha)}x_{2}^{(\beta)}x_{3}^{(\gamma)}+f(x_{1},x_{2},x_{3})$,
where $a\neq 0$, $\alpha,\beta>0$ and $f$ contains $x_{3}^{(s)}g(x_{1},x_{2})$
if $s\leqslant\gamma$. Put $E_{1}=x_{1}x_{2}^{(2^{m_{2}}-1-\beta)}$,
$E_{2}=x_{3}^{(2^{m_{3}}+1-\gamma)}$ ($\gamma>1$) or $x_{2}x_{3}^{(2)}$. If
$\gamma<2^{m_{3}}-2$, then $E_{3}=x_{1}^{(2^{m_{1}}-1-\alpha)}x_{2}$
($(\alpha,\beta)\neq(2^{m_{1}}-2,2^{m_{2}}-2)$) or
$x_{1}x_{3}^{(2^{m_{3}}-1-\gamma)}$,
$E_{4}=x_{1}x_{2}^{(2^{m_{2}}-1-\beta)}x_{3}^{(2^{m_{3}}-1-\gamma)}$,
$E_{5}=x_{1}x_{2}^{(2^{m_{2}}-1-\beta)}x_{3}^{(2^{m_{3}}-2-\gamma)}$ ($\gamma$
is even) or $x_{3}^{(2)}$ ($\gamma$ is odd). If $\gamma=2^{m_{3}}-2$, then
$E_{3}=x_{2}$, $E_{4}=x_{1}x_{3}$,
$E_{5}=x_{1}x_{2}^{(2^{m_{2}}-1-\beta)}x_{3}$. Let $\gamma=2^{m_{3}}-1$. Since
$\alpha,\beta>0$ and $(\alpha,\beta)\neq(2^{m_{1}}-1,2^{m_{2}}-1)$, we have
$m_{1}>1$ or $m_{2}>1$. Let $m_{1}>1$, then
$E_{3}=x_{1}^{(2^{m_{1}}-1-\alpha)}x_{2}$
($(\alpha,\beta)\neq(2^{m_{1}}-2,2^{m_{2}}-2)$) or $x_{1}x_{3}^{(2)}$,
$E_{4}=x_{1}^{(2^{m_{1}}-1-\alpha)}x_{2}^{(2^{m_{2}}-1-\beta)}x_{3}^{(2)}$,
$E_{5}=x_{1}x_{3}$ ($\alpha$ is even) or $x_{1}^{(2)}x_{3}$ ($\alpha$ is odd).

Let $D=ax_{3}^{(t+2)}+f(x_{1},x_{3})+g(x_{2},x_{3})$ and $a\neq 0$. The same
$L^{1}$.

Let $D=ax_{i}^{(\alpha)}x_{3}^{(\gamma)}+f(x_{1},x_{3})+g(x_{2},x_{3})$, where
$a\neq 0$, $\alpha>0$ and $f,g$ contain $x_{3}^{(s)}x_{q}^{(h)}$ if
$s\leqslant\gamma$, $h>0$. If $\gamma>0$, then $E_{1}=x_{i}$,
$E_{2}=x_{i}^{(2^{m_{i}}-1-\alpha)}x_{3}$, $E_{3}=x_{1}x_{2}$,
$E_{4}=x_{1}x_{2}x_{3}^{(2^{m_{3}}-1-\gamma)}$ ($\gamma<2^{m_{3}}-1$) or
$x_{i}^{(2^{m_{i}}-1-\alpha)}x_{j}^{(2^{m_{j}}-2)}x_{3}$,
$E_{5}=x_{j}x_{3}^{(2)}$ ($\gamma$ is odd) or
$x_{i}x_{3}^{(2^{m_{3}}-2-\gamma)}$ ($\gamma<2^{m_{3}}-2$ is even) or
$x_{i}x_{3}$ ($\gamma=2^{m_{3}}-2$, $\alpha\neq 2^{m_{i}}-2$) or
$x_{i}^{(3)}$, where $\\{i,j\\}=\\{1,2\\}$. If $\gamma=0$, then
$\\{E_{1},\ldots,E_{5}\\}=\\{x_{i},x_{i}^{(2^{m_{i}}+1-\alpha)},x_{i}^{(2^{m_{i}}+1-\alpha)}x_{3},x_{i}^{(2^{m_{i}}+1-\alpha)}x_{3}^{(2)},x_{i}^{(2^{m_{i}}+1-\alpha)}x_{3}^{(3)}\\}$.
∎

######  Lemma 6.

Let $f\in L^{4}_{(t)}\diagdown L^{4}_{(t+1)}$, $g\in L^{4}_{(s)}\diagdown
L^{4}_{(s+1)}$ and $\\{\lambda(f),g\\}\neq 0$.

(1) If $g=x_{1}^{(\alpha)}x_{2}^{(\beta)}x_{3}^{(\gamma)}$ and $\alpha>0$,
then $\lambda(\\{f,g\\})=\\{\lambda(f),g\\}$.

(2) If $g=x_{2}^{(\beta)}x_{3}^{(\gamma)}$ and $\\{\lambda(f),g\\}\in
L^{4}_{(t+s)}\diagdown L^{4}_{(t+s+1)}$, then
$\lambda(\\{f,g\\})=\\{\lambda(f),g\\}$.

###### Proof.

Recall, that
$\\{f,g\\}=\partial_{1}f\partial_{2}g+\partial_{2}f\partial_{1}g+\partial_{3}f\partial_{3}g+\overline{x}_{1}x_{3}(\partial_{2}f\partial_{3}g+\partial_{3}f\partial_{2}g)$.

(1) Obviously, if $g=x_{1}^{(\alpha)}x_{2}^{(\beta)}x_{3}^{(\gamma)}$ and
$\alpha>0$, then
$\\{f,g\\}=\partial_{1}f\partial_{2}g+\partial_{2}f\partial_{1}g+\partial_{3}f\partial_{3}g$.

(2) Let $f=\lambda(f)+f_{1}$, where $f_{1}\in L^{4}_{(t+1)}$. Then
$\\{f,g\\}\in L^{4}_{(t+s)}$, $\\{f_{1},g\\}\in L^{4}_{(t+s+1)}$. Since
$\\{\lambda(f),g\\}\in L^{4}_{(t+s)}\diagdown L^{4}_{(t+s+1)}$,
$\lambda(\\{f,g\\})=\lambda(\\{\lambda(f),g\\}+\\{f_{1},g\\})=\lambda(\\{\lambda(f),g\\})=\\{\lambda(f),g\\}$.
∎

######  Lemma 7.

Let $i=4$ or $i=1$, $m_{3}=1$. If $0\neq D\in L^{i}_{(t)}\diagdown
L^{i}_{(t+1)}$, $-1\leqslant t<2^{m_{1}}+2^{m_{2}}-3$, then there exist $4$
homogeneous elements $E_{1},E_{2},E_{3},E_{4}\in L^{i}$ such that
$\\{D,E_{1}\\},\ldots,\\{D,E_{4}\\}$ are linearly independent.

###### Proof.

We will prove that there exist $4$ homogeneous elements
$E_{1},E_{2},E_{3},E_{4}\in L^{i}$ such that
$\\{\lambda(D),E_{1}\\},\ldots,\\{\lambda(D),E_{4}\\}$ are linearly
independent. We will search $E_{i}$ for $L^{4}$ which satisfy Lemma 6. Thus
$\\{\lambda(D),E_{i}\\}=\lambda(\\{D,E_{i}\\})$ and $\\{D,E_{i}\\}$ are also
linearly independent by Lemma 3.

The case of $L^{4}$, $m_{1}>1$.

$(1)\leavevmode\nobreak\ t=-1$. Let
$\lambda(D)=a_{1}x_{1}+a_{2}x_{2}+a_{3}x_{3}$ and $a_{3}\neq 0$. Put
$E_{1}=x_{1}x_{3}$, $E_{2}=x_{1}x_{2}x_{3}$, $E_{3}=x_{1}^{(2)}x_{3}$,
$E_{4}=x_{1}^{(3)}x_{3}$. Let $\lambda(D)=a_{1}x_{1}+a_{2}x_{2}$ and
$a_{i}\neq 0$. Put $E_{1}=x_{1}x_{2}$, $E_{2}=x_{1}x_{2}x_{3}$,
$E_{3}=x_{1}^{(2)}x_{2}$, $E_{4}=x_{1}^{(3)}x_{2}$.

$(2)\leavevmode\nobreak\ t>-1$. Let
$\lambda(D)=ax_{1}^{(\alpha)}x_{2}^{(\beta)}+f(x_{1},x_{2},x_{3})$, where
$a\neq 0$, $\beta>0$ and $f$ contains $x_{1}^{(s)}x_{2}^{(h)}$ if $s<\alpha$.
Put
$\\{E_{1},\ldots,E_{4}\\}=\\{x_{1},x_{1}x_{3},x_{1}^{(2^{m_{1}}-1-\alpha)}x_{2}^{(2^{m_{2}}-\beta)},x_{1}^{(2^{m_{1}}-1-\alpha)}x_{2}^{(2^{m_{2}}-\beta)}x_{3}\\}$.

Let $\lambda(D)=ax_{1}^{(\alpha)}x_{3}+f(x_{1},x_{2},x_{3})$, where $a\neq 0$,
$\alpha>0$ and $f$ contains $x_{1}^{(s)}x_{2}^{(h)}$ if $h=0$. Put
$E_{1}=x_{3}$. If $\alpha$ is even, then
$\\{E_{2},E_{3},E_{4}\\}=\\{x_{1}x_{3},\bar{x}_{2}x_{3},x_{1}x_{2}x_{3}\\}$.
If $\alpha$ is odd, then $E_{2}=x_{2}x_{3}$,
$E_{3}=x_{1}^{(2^{m_{1}}-1-\alpha)}x_{2}$,
$E_{4}=x_{1}^{(2^{m_{1}}-1-\alpha)}x_{2}x_{3}$ ($\alpha\neq 2^{m_{1}}-1$) or
$x_{1}x_{2}$.

Let
$\lambda(D)=ax_{1}^{(\alpha)}x_{2}^{(\beta)}x_{3}+f(x_{1},x_{2})x_{3}+bx_{1}^{(t+2)}$,
where $a\neq 0$, $\beta>0$ and $f$ contains $x_{1}^{(s)}x_{2}^{(h)}$ if
$s<\alpha$, $h>0$. Put $E_{1}=x_{1}$. If $\alpha>0$, then
$E_{2}=x_{2}^{(2^{m_{2}}-\beta)}$, $E_{3}=x_{3}$,
$E_{4}=x_{1}^{(2^{m_{1}}-\alpha)}$ ($\alpha\neq 2^{m_{1}}-1$) or
$x_{2}^{(2^{m_{2}}-\beta)}x_{3}$. If $\alpha=0$, then
$\\{E_{2},E_{3},E_{4}\\}=\\{x_{1}x_{2}^{(2^{m_{2}}-\beta)},x_{1}x_{3},\bar{x}_{1}\\}$.

Let $\lambda(D)=ax_{1}^{(t+2)}$, $a\neq 0$. Put
$\\{E_{1},\ldots,E_{4}\\}=\\{x_{2},x_{2}x_{3},x_{1}^{(2^{m_{1}}-2-t)}x_{2},x_{1}^{(2^{m_{1}}-2-t)}x_{2}x_{3}\\}$.

The case of $L^{4}$, $m_{1}=1$, $m_{2}>1$.

$(1)\leavevmode\nobreak\ t=-1$. Let
$\lambda(D)=a_{1}x_{1}+a_{2}x_{2}+a_{3}x_{3}$ and $a_{3}\neq 0$. Put
$E_{1}=x_{1}x_{3}$, $E_{2}=x_{1}x_{2}x_{3}$, $E_{3}=x_{1}x_{2}^{(2)}x_{3}$,
$E_{4}=x_{1}x_{2}^{(3)}x_{3}$. Let $\lambda(D)=a_{1}x_{1}+a_{2}x_{2}$ and
$a_{i}\neq 0$. Put $E_{1}=x_{1}x_{2}$, $E_{2}=x_{1}x_{2}x_{3}$,
$E_{3}=x_{1}x_{2}^{(2)}$, $E_{4}=x_{1}x_{2}^{(3)}$.

$(2)\leavevmode\nobreak\ t>-1$. Let
$\lambda(D)=ax_{2}^{(t+2)}+f(x_{1},x_{2},x_{3})$ and $a\neq 0$. Put
$\\{E_{1},\ldots,E_{4}\\}=\\{x_{1},x_{1}x_{3},x_{1}x_{2}^{(2^{m_{2}}-2-t)},x_{1}x_{2}^{(2^{m_{2}}-2-t)}x_{3}\\}$.

Let $\lambda(D)=ax_{2}^{(\alpha)}x_{3}+f(x_{1},x_{2},x_{3})$, where $a\neq 0$,
$\alpha>0$ and $f$ does not contain $x_{2}^{(t+2)}$. Put
$\\{E_{1},E_{2},E_{3}\\}=\\{x_{1},x_{1}x_{3},x_{1}x_{2}^{(2^{m_{2}}-\alpha)}\\}$
and $E_{4}=x_{1}x_{2}^{(2^{m_{2}}-1-\alpha)}x_{3}$ ($\alpha\neq 2^{m_{2}}-1$)
or $x_{3}$.

Let $\lambda(D)=ax_{1}x_{2}^{(t+1)}+bx_{1}x_{2}^{(t)}x_{3}$ and $a_{1}\neq 0$.
Put $\\{E_{1},\ldots,E_{4}\\}=\\{x_{1},x_{2},x_{1}x_{3},x_{2}x_{3}\\}$.

Let $\lambda(D)=ax_{1}x_{2}^{(t)}x_{3}$, $a\neq 0$. Put $E_{1}=x_{2}$,
$E_{2}=x_{1}x_{2}^{(2^{m_{2}}-1-t)}$. If $t>0$, then $E_{3}=x_{1}$,
$E_{4}=x_{2}^{(2^{m}-t)}$. If $t=0$, then $E_{3}=x_{1}x_{2}$,
$E_{4}=\bar{x}_{2}$.

The case of $L^{3}$, $m_{3}=1$. The same $L^{4}$. ∎

######  Theorem 7.

If $i=2,3$ or $i=1$, $m_{3}>1$, then $R^{i}(L^{i})=4$ and $R^{i}(D)=4$ if and
only if $0\neq D\in\langle\xi\rangle$. If $i=1$, $m_{3}=1$ or $i=4$, then
$R^{i}(L^{i})=3$ and $R^{i}(D)=3$ if and only if $0\neq
D\in\langle\xi\rangle$.

###### Proof.

(1) Let $i=2,3$ or $i=1$, $m_{3}>1$. It follows from Lemma 4 that if $0\neq
D\in\langle\xi\rangle$, then $R^{i}(D)=4$. We shall prove that if $f\in L^{i}$
and $f\notin\langle\bar{x}\rangle$, then $R^{i}(ad\,f)>4$. Clearly,
$\lambda(f)\notin\langle\bar{x}\rangle$, thus, according to Lemma 5, there are
$5$ homogeneous elements $E_{1},\ldots,E_{5}\in L^{i}$ such that
$\\{\lambda(f),E_{j}\\}$, $1\leqslant j\leqslant 5$ are linearly independent.
But $\\{\lambda(f),E_{j}\\}=\lambda(\\{f,E_{j}\\})$. Hence, $\\{f,E_{j}\\}$
are also linearly independent by Lemma 3. Therefore, $R^{i}(ad\,f)\geqslant
5>4$.

(2) Let $i=1$, $m_{3}=1$ or $i=4$. It follows from Lemma 4 that if $0\neq
D\in\langle\xi\rangle$, then $R^{i}(D)=3$. If $f\in L^{i}$ and
$f\notin\langle\bar{x}\rangle$, then according to Lemma 7, there are $4$
homogeneous elements $E_{1},\ldots,E_{4}\in L^{i}$ such that $\\{f,E_{j}\\}$,
$1\leqslant j\leqslant 4$ are linearly independent. Therefore,
$R^{i}(ad\,f)\geqslant 4>3$. ∎

######  Corollary 1.

Let $L=P(3,\overline{m},\omega_{i})$, $i=1,2,3,4$, $\overline{m}\neq(1,1,1)$.
The standard maximal subalgebra $\mathscr{L}_{0}$ of $L$ is the normalizer of
$\langle\bar{x}\rangle$ in $L$. Thus, $\mathscr{L}_{0}$ is an invariant
subalgebra and the natural filtration of $L$ is intrinsically determined.

######  Theorem 8.

Let $L^{i}=P(3,\overline{m},\omega_{i})$,
$L^{j}=P(3,\overline{m}^{\prime},\omega_{j})$, $i,j=1,2,3,4$. If $i\neq j$,
then the non-alternating Hamiltonian Lie algebras $L^{i}$ and $L^{j}$ are not
isomorphic.

###### Proof.

Suppose that $L^{i}$ and $L^{j}$ are isomorphic, $\varphi\colon
L^{i}\rightarrow L^{j}$ is an isomorphism. If $\overline{m}=(1,1,1)$, then
$\overline{m}^{\prime}=(1,1,1)$ either.

Consider the case when $\overline{m}\neq(1,1,1)$. The algebras $L^{i}$ and
$L^{j}$ have the same invariant $R$, $R^{i}=R^{j}$. Let $\bar{x}(i)$ is the
monomial in $L^{i}$ of maximal degree. It follows from Theorem 7 that
$\varphi(\bar{x}(i))=a\bar{x}(j)$, $a\in K^{\ast}$, and, consequently,
$\varphi(L_{(0)}^{i})=L_{(0)}^{j}$, i.e. $\varphi$ is an isomorphism of
transitive Lie algebras. From Theorem 3 and Remark 2 of Section 4 we obtain
that the forms $\omega_{i}$, $\omega_{j}$ are equivalent. By Theorem 5 $i=j$.

Suppose that $\overline{m}=(1,1,1)$. In this case we have only
$L^{3}=P(3,(1,1,1),\omega_{3})$ which is not simple and
$L^{4}=P(3,(1,1,1),\omega_{4})$ which is simple. ∎

######  Corollary 2.

Let $\mathscr{L}^{i}$ be non-alternating Hamiltonian Lie algebras in three
variables,
$P^{(1)}(\mathscr{F}_{i},\omega_{i})\subseteq\mathscr{L}^{i}\subseteq\widetilde{P}(\mathscr{F}_{i},\omega_{i})$,
$i=1,2,3,4$. If $i\neq j$, then algebras $\mathscr{L}^{i}$, $\mathscr{L}^{j}$
are not isomorphic.

###### Proof.

If algebras $\mathscr{L}^{i}$, $\mathscr{L}^{j}$, $i\neq j$ are isomorphic,
then the algebras $[\mathscr{L}^{i},\mathscr{L}^{i}]$,
$[\mathscr{L}^{j},\mathscr{L}^{j}]$ are isomorphic as well which results in a
contradiction with Theorem 8. ∎

## References

  * [1] Lin Lei, “Non-alternating Hamiltonian algebra $P(n,m)$ of characteristic two”, Commun. Algebra 21(2), 399–411 (1993).
  * [2] I. Kaplansky, “Some simple Lie algebras of characteristic 2”, Lecture Notes in Math. 933, 127–129 (1982).
  * [3] S. Bouarrouj, P. Grozman, A. Lebedev, D. Leites, “Divided power (co)homology. Presentation of simple finite dimensional modular superalgebras with Cartan matrix”, Homology, Homotopy Appl. 12(1), 237–248 (2010).
  * [4] A. Lebedev, “Analog of orthogonal, Hamiltonian, and contact Lie superalgebras in characteristic 2”, J. Nonlin. Math. Phys. 17, Suppl. 1, 399–411 (2010).
  * [5] M. I. Kuznetsov, A. V. Kondrateva, N. G. Chebochko, “Non-alternating Hamiltonian Lie algebras in characteristic 2. I ”, (http://arxiv.org/abs/1812.11213). Accessed 2018.
  * [6] A. I. Kostrikin, I. R. Shafarevich “Graded Lie algebras of finite characteristic”, Math. USSR-Izv. 3(2), 237–304.
  * [7] M. I. Kuznetsov, A. V. Kondrateva, N. G. Chebochko, “On Hamiltonian Lie algebras of characteristic 2”, Matem. J. 16(2), 54–65 (2016). [in Russian] (available at http://www.math.kz).
  * [8] M. I. Kuznetsov, “Truncated induced modules over transitive Lie algebras of characteristic $p$”, Math. USSR-Izv. 34, 575–608 (1990).
  * [9] M. I. Kuznetsov, S. A. Kirillov, “Hamiltonian differential forms over the divided powers algebra”, Russian Math. Surveys. 41(2), 205–206 (1986).
  * [10] S. M. Skryabin, “Classification of Hamiltonian forms over divided powers”, Math. USSR-Sb. 69, 121–141 (1991).
  * [11] S. M. Skryabin, “The normal shapes of the symplectic and contact forms over algebras of divided powers”, (https://arxiv.org/abs/1906.11496v1). Accessed 2019.
  * [12] S. M. Skryabin, “Modular Lie algebras over algebraically non-closed fields. I ”, Commun. Algebra. 19, 1629–1741 (1991).
  * [13] S. M. Skryabin, “An algebraic approach to the Lie algebras of Cartan type”, Commun. Algebra. 21, 1229–1336 (1993).