# Maximal Subalgebras for Lie Superalgebras of Cartan Type

Wei Bai111Supported by NSF of the Education Department of HLJP (12521158)
Department of Mathematics, Harbin Institute of Technology,
Harbin 150006, P. R. China  School of Mathematical Sciences, Harbin Normal
University,
Harbin 150025, P. R. China  Wende Liu222Supported by NSF of China (11171055)
and NSF of HLJP (JC201004, A2010-03) Department of Mathematics, Harbin
Institute of Technology,
Harbin 150006, P. R. China  Xuan Liu  Applied Mathematics Department, The
University of Western Ontario,
London, N6A 5B7, Canada  Hayk Melikyan333Supported by part NSF Grant # 0833184
Department of Mathematics and Computer Science, North Carolina Central
University
Durham, NC 27713, USA

###### Abstract

The maximal graded subalgebras for four families of Lie superalgebras of
Cartan type over a field of prime characteristic are studied. All maximal
reducible graded subalgebras are described completely and their isomorphism
classes, dimension formulas are found. The classification of maximal
irreducible graded subalgebras is reduced to the classification of the maximal
irreducible subalgebras for the classical Lie superalgebras
$\mathfrak{gl}(m,n)$, $\mathfrak{sl}(m,n)$ and $\mathfrak{osp}(m,n)$.

###### keywords:

Lie superalgebras; maximal graded subalgebras Mathematics Subject
Classification 2010: 17B50, 17B05.

††journal: Journal of Algebra and its Applications.

3.6cm3.6cm2.4cm2cm

##  0 Introduction

Since V. G. Kac [1] classified the finite dimensional simple Lie superalgebras
over algebraically closed fields of characteristic zero, the theory of Lie
superalgebras has undergone a significant development (for example [2, 3]).
Over a field of finite characteristic, however, the classification problem is
still open for the finite dimensional simple Lie superalgebras [4, 5]. Even
recently, new simple Lie superalgebras over a field of characteristic $p=3$
were constructed [5, 6].

In general, study of the maximal subsystems of an algebraic system, such as
finite groups, Lie groups, Lie (super)algebras, is an essential part of
structural characterization of the system. In classical Lie theory, the
classification of maximal subalgebras of simple Lie algebras over the field of
complex numbers is one of the beautiful results of that theory which was due
to E. Dynkin [7, 8]. In classical modular Lie theory there is a series of
papers by G. Seitz and his students devoted to the study of the maximal
subgroups of simple algebraic groups over fields of positive characteristic.
These investigations were summarized by G. Seitz in his two publications [9,
10] which generalize E. Dynkin’s classification of the maximal subgroups of
simple Lie groups over the field of complex numbers [8] to simple algebraic
groups over fields of characteristic $p>7$. The study of maximal subalgebras
of different classes of (super)algebras has been the focus of several
researchers. The maximal subalgebras of Jordan (super)algebras were studied by
M. Racine [11, 12], A. Elduque, J. Laliena and S. Sacristan [13, 14]. The
maximal graded subalgebras of affine Kac-Moody algebras were classified in
[15]. The fourth author of the present paper summarized his investigations on
maximal subalgebras in Cartan type simple Lie algebras over the field of
characteristic $p>3$ in his paper [16].

Let $L$ be a finite dimensional simple Lie superalgebras of Cartan type $W$,
$S$, $H$ or $K$ with a $\mathbb{Z}$-grading $L=\oplus_{i\geq-2}L_{i}$. The
present paper is devoted to characterizing the maximal graded subalgebras of
$L$. To this end, we construct a series of graded subalgebras of $L$ and state
the necessary and sufficient conditions for their maximality. Moreover, the
number of isomorphism classes and the dimension formulas of all maximal graded
subalgebras are completely determined except for maximal irreducible graded
subalgebras. Note that the null of $L$ is isomorphic to a classical Lie
superalgebra (see Lemma 2.1(3)). Thus the classification of the maximal
irreducible graded subalgebras of $L$ is reduced to that of the maximal
irreducible subalgebras of a classical Lie superalgebra. Moreover, we give
necessary and sufficient conditions for the existence of maximal irreducible
graded subalgebras of $L$. We should mention that the present work which
partially generalizes the results of [16] is motivated by a paper by A. I.
Kostrikin and I. R. Shafarevich [17] on the structure theory of modular Lie
algebras.

We close this introduction by establishing the following conventions: The
underlying field $\mathbb{F}$ is an algebraically closed field of
characteristic $p>3$. In addition to the standard notation $\mathbb{Z},$ we
write $\mathbb{N}$ for the set of nonnegative integers. The field of two
elements is denoted by $\mathbb{Z}_{2}=\\{\bar{0},\bar{1}\\}.$ For a
proposition $P$, put $\delta_{P}=1$ if $P$ is true and $\delta_{P}=0$
otherwise. All subspaces, subalgebras and submodules are assumed to be
$\mathbb{Z}_{2}$-graded and all the homomorphisms of $\mathbb{Z}$-graded
superalgebras are both $\mathbb{Z}_{2}$-homogeneous and
$\mathbb{Z}$-homogeneous.

##  1 Basics

Fix two positive integers $m,n\in\mathbb{N}\backslash\\{1\\}.$ Put

$\mathbf{I}_{0}=\overline{1,m},\ \mathbf{I}_{1}=\overline{m+1,m+n},\
\mathbf{I}=\mathbf{I}_{0}\cup\mathbf{I}_{1},$

where $\overline{k,s}=\\{k,k+1,\ldots,s\\}$ with the convention
$\overline{k,s}=\emptyset$ whenever $k>s.$ Write

$\mathbf{A}(m)=\\{\alpha=(\alpha_{1},\ldots,\alpha_{m})\in\mathbb{N}^{m}\mid
0\leq\alpha_{i}\leq p-1,i\in\mathbf{I}_{0}\\}.$

Let $\mathcal{O}(m)$ be the divided power algebra with $\mathbb{F}$-basis
$\\{x^{(\alpha)}\mid\alpha\in\mathbf{A}(m)\\}$ and $\Lambda(n)$ be the
exterior superalgebra of $n$ variables $x_{m+1},x_{m+2},\ldots,x_{m+n}$. The
tensor product

$\mathcal{O}(m,n)=\mathcal{O}(m)\otimes\Lambda(n)$

is an associative superalgebra with respect to the usual
$\mathbb{Z}_{2}$-grading. Let

$\mathbf{B}(n)=\\{\langle i_{1},i_{2},\ldots,i_{k}\rangle\mid 0\leq k\leq
n;m+1\leq i_{1}<i_{2}<\cdots<i_{k}\leq m+n\\}$

be the set of $k$-tuples of strictly increasing integers in $\mathbf{I}_{1}$,
$0\leq k\leq n.$ For $u=\langle
i_{1},i_{2},\ldots,i_{k}\rangle\in\mathbf{B}(n)$, write
$x^{u}=x_{i_{1}}x_{i_{2}}\cdots x_{i_{k}}$ ($x^{\emptyset}=1$). If
$g\in\mathcal{O}(m)$ and $f\in\Lambda(n)$, we write $gf$ instead of $g\otimes
f$. Then $\mathcal{O}(m,n)$ has a $\mathbb{Z}_{2}$-homogeneous
$\mathbb{F}$-basis

$\\{x^{(\alpha)}x^{u}\mid\alpha\in\mathbf{A}(m),u\in\mathbf{B}(n)\\}.$

For $i\in\textbf{I}_{0}$ and
$\varepsilon_{i}=(\delta_{i1},\delta_{i2},\ldots,\delta_{im})$, write $x_{i}$
for $x^{(\varepsilon_{i})}.$ Let $\partial_{1},\ldots,\partial_{m+n}$ be the
superderivations of the superalgebra $\mathcal{O}(m,n)$ such that
$\partial_{i}(x_{j})=\delta_{i=j}.$ The parity of $\partial_{i}$ is
$|\partial_{i}|=\bar{0}$ if $i\in\mathbf{I}_{0}$ and $\bar{1}$ if
$i\in\mathbf{I}_{1}$. Hereafter the symbol $|x|$ implies that $x$ is a
$\mathbb{Z}_{2}$-homogeneous element. Put

$W(m,n)=\mathrm{span}_{\mathbb{F}}\left\\{a\partial_{i}\mid
a\in\mathcal{O}(m,n),i\in\mathbf{I}\right\\},$

which is a finite dimensional simple Lie superalgebra, called Witt
superalgebra. Consider the linear mapping called divergence:

$\mathrm{div}:W(m,n)\longrightarrow\mathcal{O}(m,n),\quad\mathrm{div}(f\partial_{i})=(-1)^{|f||\partial_{i}|}\partial_{i}(f).$

Set $S(m,n)=[\overline{S}(m,n),\overline{S}(m,n)]$, where
$\overline{S}(m,n)=\ker(\mathrm{div})$. Then we have

$S(m,n)=\mathrm{span}_{\mathbb{F}}\left\\{D_{ij}(a)\mid
a\in\mathcal{O}(m,n),i,j\in\mathbf{I}\right\\},$

where

$D_{ij}(a)=(-1)^{|\partial_{i}||\partial_{j}|}\partial_{i}(a)\partial_{j}-(-1)^{(|\partial_{i}|+|\partial_{j}|)|a|}\partial_{j}(a)\partial_{i}\quad\mbox{for
}a\in\mathcal{O}(m,n).$

$S(m,n)$ is a simple Lie superalgebra, called special superalgebra.

For $j\in\\{1,\ldots,2\lfloor\frac{m}{2}\rfloor,m+1,\ldots,m+n\\}$, we put

$\sigma(j)=\left\\{\begin{array}[]{ll}-1,&j\in\overline{\lfloor\frac{m}{2}\rfloor+1,2\lfloor\frac{m}{2}\rfloor};\\\
1,&\mbox{ otherwise }\end{array}\right.\mbox{ and }\
j^{\prime}=\left\\{\begin{array}[]{lll}j+\lfloor\frac{m}{2}\rfloor,&j\in\overline{1,\lfloor\frac{m}{2}\rfloor};\\\
j-\lfloor\frac{m}{2}\rfloor,&j\in\overline{\lfloor\frac{m}{2}\rfloor+1,2\lfloor\frac{m}{2}\rfloor};\\\
j,&\mbox{ otherwise}.\end{array}\right.$

Suppose $m=2r$ is even. Define an even linear mapping
$D_{H}:\mathcal{O}(m,n)\longrightarrow W(m,n)$ by
$D_{H}(a)=\sum_{i\in\mathbf{I}}\sigma(i)(-1)^{|\partial_{i}||a|}\partial_{i}(a)\partial_{i^{\prime}}.$
Put

$\overline{H}(m,n)=\mathrm{span}_{\mathbb{F}}\\{D_{H}(a)\mid
a\in\mathcal{O}(m,n)\\}.$

Write $\bar{\mathcal{O}}(m,n)$ for the quotient superspace
$\mathcal{O}(m,n)/\mathbb{F}\cdot 1$. We can view $D_{H}$ as a linear operator
of $\bar{\mathcal{O}}(m,n)$ since the kernel of $D_{H}$ is $\mathbb{F}\cdot
1$. Thus we have $\overline{H}(m,n)\cong(\bar{\mathcal{O}}(m,n),[\;,\;])$,
where the bracket is:

$[a,b]=D_{H}\left(a\right)\left(b\right)\ \mbox{ for
}a,b\in\bar{\mathcal{O}}(m,n).$

Its derived algebra ${H}(m,n)$ is simple, called Hamiltonian Lie superalgebra.

Suppose $m=2r+1$ is odd. Define an even linear mapping
$D_{K}:\mathcal{O}(m,n)\longrightarrow W(m,n)$ by
$D_{K}(a)=D_{H}(a)+\partial_{m}(a)\mathfrak{D}+\Delta(a)\partial_{m},$ where
$\mathfrak{D}=\sum_{i\in\mathbf{I}\backslash\\{m\\}}x_{i}\partial_{i}$ and
$\Delta(a)=2a-\mathfrak{D}(a)$. Put

$\overline{K}(m,n)=\mathrm{span}_{\mathbb{F}}\\{D_{K}(a)\mid
a\in\mathcal{O}(m,n)\\}.$

Since $D_{K}$ is injective, we have
$\overline{K}(m,n)\cong(\mathcal{O}(m,n),[\;,\;]),$ where the bracket is:

$\begin{split}[a,b]&=D_{H}(a)b+\Delta(a)\partial_{m}(b)-\partial_{m}(a)\Delta(b)\
\mbox{ for }a,b\in{\mathcal{O}}(m,n).\end{split}$

Its derived algebra $K(m,n)$ is simple, called contact Lie superalgebra.

For simplicity, hereafter, we write $X$ for $X(m,n)$, where $X=\mathcal{O}$,
$\bar{\mathcal{O}}$, $W$, $\overline{S}$, $S$, $\overline{H}$, $H$,
$\overline{K}$ or $K$. Let us consider the standard $\mathbb{Z}$-grading of
$L$, where $L=\mathcal{O}$, $W$, $S$, $\overline{H}$, $H$ or $K$. Define the
$\mathbb{Z}$-degrees of $x_{i}$ and $\partial_{i}$ to be
$\mathrm{zd}(x_{i})=-\mathrm{zd}(\partial_{i})=1+\delta_{L=K}\delta_{i=m}$,
$i\in\mathbf{I}$. Hereafter, the symbol $\mathrm{zd}(x)$ always implies that
$x$ is a $\mathbb{Z}$-homogeneous element. Put $\xi=(m+\delta_{L=K})(p-1)+n$.
Then we have:

$\displaystyle\mathcal{O}=\oplus_{i=-1}^{\xi}\mathcal{O}_{i},\ \
\mathcal{O}_{i}=\mathrm{span}_{\mathbb{F}}\\{f\in\mathcal{O}\mid\mathrm{zd}(f)=i\\};$
$\displaystyle W=\oplus_{i=-1}^{\xi-1}W_{i},\ \
W_{i}=\mathrm{span}_{\mathbb{F}}\\{f\partial_{j}\mid f\in\mathcal{O}_{i+1},\
j\in\mathbf{I}\\};$ $\displaystyle S=\oplus_{i=-1}^{\xi-2}S_{i},\ \
S_{i}=\mathrm{span}_{\mathbb{F}}\\{D_{jk}(f)\in W\mid f\in\mathcal{O}_{i+2},\
j,k\in\mathbf{I}\\};$ $\displaystyle\overline{H}=\oplus_{i=-1}^{\xi-2}H_{i},\
\ H_{i}=\mathrm{span}_{\mathbb{F}}\\{f\mid f\in\mathcal{O}_{i+2}\\};\
H=\oplus_{i=-1}^{\xi-3}H_{i};$
$\displaystyle{K}=\oplus_{i=-2}^{\xi-2-\delta_{n-m-3\equiv
0\;(\mathrm{mod}\;p)}}K_{i},\ \ K_{i}=\mathrm{span}_{\mathbb{F}}\\{f\mid
f\in\mathcal{O}_{i+2}\\}.$

We adopt the following conventions:

  * (1)

$L=H$ implies that $m=2r$ is even; $L=K$ implies that $m=2r+1$ is odd.

  * (2)

$K$ can be viewed as a $\mathbb{Z}$-graded subalgebra of $W$ when
$\mathrm{zd}(x_{m})=-\mathrm{zd}(\partial_{m})=2$ for $W$. Thus, $L$ is a
$\mathbb{Z}$-graded subalgebra of $W$, where $L=S$, $H$ or $K$.

  * (3)

For $L=K$, we write $z$ for $x_{m}.$

  * (4)

Write $\mathrm{alg}(S)$ for the subalgebra of $L$ generated by a subset $S$.

A proper subalgebra $M$ of a $\mathbb{Z}$-graded Lie superalgebra $L$ is
called a maximal graded subalgebra (MGS) provided that $M$ is
$\mathbb{Z}$-graded and no nontrivial $\mathbb{Z}$-graded subalgebras of $L$
strictly contains $M$. Since $L_{-1}$ is an irreducible $L_{0}$-module, it is
clear that $\oplus_{i\geq 0}L_{i}$ is an MGS of $L$. Any other MGS, $M$, must
satisfy exactly one of the following conditions:

  * $\mathrm{(I)}$

$M_{-1}=L_{-1}$ and $M_{0}=L_{0};$

  * $\mathrm{(II)}$

$M_{-1}$ is a nontrivial proper subspace of $L_{-1};$

  * $\mathrm{(III)}$

$M_{-1}=L_{-1}$ and $M_{0}\neq L_{0}.$

Let $\mathfrak{G}_{0}$ be a subalgebra of $L_{0}$. $\mathfrak{G}_{0}$ is
called reducible (resp. irreducible) if the $\mathfrak{G}_{0}$-module $L_{-1}$
is reducible (resp. irreducible). An MGS
$\mathfrak{G}=\sum_{i\geq-2}\mathfrak{G}_{i}$ of $L$ is called maximal
reducible graded (resp. maximal irreducible graded) if $\mathfrak{G}_{0}$ is
reducible (resp. irreducible).

##  2 Preliminary Results

In order to simplify our considerations, in this section, we establish some
technical lemmas. For $L=H$ or $K$, we redescribe $L$ in an appropriate form
and establish a suitable automorphism of $L$ by virtue of a nondegenerate skew
supersymmetric bilinear form on $L_{-1}$.

As in the case of Lie superalgebras of characteristic 0 [1] or modular Lie
algebras [17, 18, 19], it is easy to show the following:

######  Lemma 2.1.

Let $L=W,S,H$ or $K$.

  * $(1)$

$L$ is transitive.

  * $(2)$

$L$ is generated by its local part, $L=\mathrm{alg}(L_{-1}+L_{0}+L_{1})$.

  * $(3)$

For the null of $L$, the following conclusions hold:

$\displaystyle W(m,n)_{0}\cong\mathfrak{gl}(m,n);\
S(m,n)_{0}\cong\mathfrak{sl}(m,n);$ $\displaystyle
H(2r,n)_{0}\cong\mathfrak{osp}(2r,n);\
K(2r+1,n)_{0}\cong\mathfrak{osp}(2r,n)\oplus\mathbb{F}I_{2r+n}.$

When $L=W$ or $S$, we know that $L_{-1}$ is spanned by the standard ordered
$\mathbb{F}$-basis

$\displaystyle\\{\partial_{i}\mid i\in\mathbf{I}\\}.$ (2.1)

For a $\mathbb{Z}_{2}$-graded subspace $V=V_{\bar{0}}\oplus V_{\bar{1}}$ of
$L_{-1}$, the super-dimension is denoted by

$\mathrm{superdim}V=(\mathrm{dim}V_{\bar{0}},\mathrm{dim}V_{\bar{1}}).$

When $L=H$ or $K$, we redescribe $L$ in a desired form. For
$i\in\mathbb{N}\backslash{\\{0\\}}$, write $A_{i}$ for an $i\times i$ matrix,
and particularly, let $I_{i}$ be the $i\times i$ unit matrix. Denote by
$\sqrt{a}$ a fixed solution of the equation $x^{2}=a$ in $\mathbb{F}$, where
$a=-1,2$. Put

$y_{i}=\left\\{\begin{array}[]{lll}{x_{i}},&i\in\mathbf{I}_{0}\cup\overline{m+2q+1,m+n};\\\
\frac{x_{i}+\sqrt{-1}x_{i+q}}{\sqrt{2}},&i\in\overline{m+1,m+q};\\\
\frac{x_{i-q}-\sqrt{-1}x_{i}}{\sqrt{2}},&i\in\overline{m+q+1,m+2q},\end{array}\right.$

where $0\leq d\leq n$, $q=\lfloor\frac{n-d}{2}\rfloor$. Then there exists an
invertible matrix $A_{m+n}$ such that
$(y_{1},\ldots,y_{m+n})A=(x_{1},\ldots,x_{m+n})$. Obviously, $|y_{i}|=|x_{i}|$
and $\mathrm{zd}(y_{i})=\mathrm{zd}(x_{i}),$ $i\in\mathbf{I}$. By [20, Lemma
2.5], we have:

$\\{y^{(\alpha)}y^{u}\mid\alpha\in\mathbf{A}(m),\ u\in\mathbf{B}(n)\\}$

is an $\mathbb{F}$-basis of $\mathcal{O}$, where $y^{(\alpha)}=x^{(\alpha)}$
and $y^{u}=y_{i_{1}}y_{i_{2}}\cdots y_{i_{k}}$ when $u=\langle
i_{1},i_{2},\ldots,i_{k}\rangle$. The basis-element $y^{(\alpha)}y^{u}$ is
called a monomial.

Write $(D_{1},\ldots,D_{m+n})=(\partial_{1},\ldots,\partial_{m+n})A^{t}$. Then
we have

$D_{i}=\left\\{\begin{array}[]{lll}\partial_{i},&i\in\mathbf{I}_{0}\cup\overline{m+2q+1,m+n};\\\
\frac{\partial_{i}-\sqrt{-1}\partial_{i+q}}{\sqrt{2}},&i\in\overline{m+1,m+q};\\\
\frac{\partial_{i-q}+\sqrt{-1}\partial_{i}}{\sqrt{2}},&i\in\overline{m+q+1,m+2q}.\end{array}\right.$

By a direct computation, we have:

$D_{i}(y_{j})=\delta_{i=j},\ \
{\sum}_{i\in\mathbf{I}\backslash\\{2r+1\\}}y_{i}D_{i}=\mathfrak{D}.$

When $m=2r$, define an even linear mapping $E_{H}:\mathcal{O}\longrightarrow
W$ by

$E_{H}(a)={\sum}_{i\in\mathbf{I}}\sigma(i)(-1)^{|D_{i}||a|}D_{i}(a)D_{\widetilde{i}},$

where

$\widetilde{i}=\left\\{\begin{array}[]{lll}i^{\prime},&i\in\mathbf{I}_{0}\cup\overline{m+2q+1,m+n};\\\
i+q,&i\in\overline{m+1,m+q};\\\
i-q,&i\in\overline{m+q+1,m+2q}.\end{array}\right.$

When $m=2r+1$, define an even linear mapping $E_{K}:\mathcal{O}\longrightarrow
W$ by

$\begin{split}E_{K}(a)&=E_{H}(a)+D_{m}(a)\mathfrak{D}+\Delta(a)D_{m}.\end{split}$

A direct computation shows that $D_{L}=E_{L}$. Note that $L_{-1}$ is spanned
by the standard ordered $\mathbb{F}$-basis

$\displaystyle\\{y_{i}\mid i\in\overline{1,2r}\cup\overline{m+1,m+n}\\}.$
(2.2)

Define an even bilinear form $\beta:L_{-1}\times
L_{-1}\longrightarrow\mathbb{F}$ satisfying

$\beta(u,v)={\sum}_{i\in\mathbf{I}}\sigma(i)(-1)^{|D_{i}||u|}D_{i}(u)D_{\widetilde{i}}(v)\
\mbox{ for }u,v\in L_{-1}.$

Then the matrix of $\beta$ in the ordered basis (2.2) is

$J=\left(\begin{tabular}[]{c|c}\begin{tabular}[]{c|c}0&$I_{r}$\\\
\hline\cr$-I_{r}$&0\\\ \end{tabular}&0\\\ \hline\cr
0&\begin{tabular}[]{c|c|c}0&$-I_{q}$&0\\\ \hline\cr$-I_{q}$&0&0\\\ \hline\cr
0&0&$-I_{n-2q}$\end{tabular}\end{tabular}\right).$ (2.3)

Clearly, $\beta$ is a nondegenerate skew supersymmetric bilinear form on
$L_{-1}$.

An $\mathbb{F}$-basis of $L_{-1}$ in which the matrix of $\beta$ is $J$ is
called generalized orthosymplectic. Let $V=V_{\bar{0}}\oplus V_{\bar{1}}$ be a
subspace of $L_{-1}$. Suppose $2a$ (resp. $d$) is the rank of $\beta$
restricted to $V_{\bar{0}}$ (resp. $V_{\bar{1}}$). A
$\mathbb{Z}_{2}$-homogeneous basis of $V$

$\displaystyle\\{e_{1},\ldots,e_{a},e_{r+1},\ldots,e_{r+a};e_{a+1},\ldots,e_{b}\mid
e_{m+1},\ldots,e_{m+c};e_{m+n-d+1},\ldots,e_{m+n}\\}$ (2.4)

is called a $\beta$-basis of $V$, if

$\\{e_{1},\ldots,e_{a},e_{r+1},\ldots,e_{r+a};e_{a+1},\ldots,e_{b}\\},\ 0\leq
a\leq b\leq r$

is an $\mathbb{F}$-basis of $V_{\bar{0}}$ satisfying

$\beta(e_{i},e_{j})=-\beta(e_{j},e_{i})=\left\\{\begin{array}[]{ll}1,&1\leq
i\leq a,j=\widetilde{i};\\\ 0,&\mbox{otherwise}\end{array}\right.$

and

$\\{e_{m+1},\ldots,e_{m+c};e_{m+n-d+1},\ldots,e_{m+n}\\},\ 0\leq d\leq n,\
0\leq c\leq\lfloor\frac{n-d}{2}\rfloor$

is an $\mathbb{F}$-basis of $V_{\bar{1}}$ satisfying

$\beta(e_{i},e_{j})=\beta(e_{j},e_{i})=\left\\{\begin{array}[]{ll}-1,&m+n-d+1\leq
i=j\leq m+n;\\\ 0,&\mbox{otherwise}.\end{array}\right.$

The 4-tuple $(a,b,c,d)$ is called the $\beta$-dimension of $V$, denoted by
$\beta$-$\dim V=(a,b,c,d)$. $V$ is nondegenerate (with respect to $\beta$) if
$a=b$ and $c=0$. $V$ is isotropic if $a=0$ and $d=0$. Clearly, for any
$\mathbb{Z}_{2}$-graded subspace of $L_{-1}$, there exists a $\beta$-basis of
it, which can extend to a generalized orthosymplectic basis of $L_{-1}$.

Now, suppose $L=W$, $S$, $H$ or $K$. Put

$\mathfrak{V}^{L}=\\{V\mid V\mbox{ is a nontrivial subspace of }L_{-1}\\}.$

$V\in\mathfrak{V}^{L}$ is called a standard element  if $V$ is spanned by

$\\{\partial_{1},\ldots,\partial_{k}\mid\partial_{m+1},\ldots,\partial_{m+l}\\},$

when $L=W$ or $S$, $0\leq k\leq m$, $0\leq l\leq n$; if $V$ is spanned by

$\\{y_{1},\ldots,y_{a},y_{r+1},\ldots,y_{r+a};y_{a+1},\ldots,y_{b}\mid
y_{m+1},\ldots,y_{m+c};y_{m+n-d+1},\ldots,y_{m+n}\\},$

when $L=H$ or $K$, $0\leq a\leq b\leq r$, $0\leq d\leq n$, $0\leq
c\leq\lfloor\frac{n-d}{2}\rfloor$. Hereafter, for
$V,V^{\prime}\in\mathfrak{V}^{L}$, the symbol $V\cong V^{\prime}$ always means
$\mathrm{superdim}V=\mathrm{superdim}V^{\prime}$ when $L=W$ or $S$ and means
$\beta$-$\dim V=\beta$-$\dim V^{\prime}$ when $L=H$ or $K$.

######  Lemma 2.2.

Let $L=W,S,H$ or $K$. Suppose $V$, $V^{\prime}\in\mathfrak{V}^{L}$ satisfying
$V\cong V^{\prime}$. Then there exists a $\mathbb{Z}$-homogeneous automorphism
${\Phi}_{L}$ of $L$ such that ${\Phi}_{L}(V)=V^{\prime}$.

###### Proof.

Without loss of generality, we may assume that $V$ is a standard element in
$\mathfrak{V}^{L}$. When $L=W$ or $S$, suppose
$\mathrm{superdim}V=\mathrm{superdim}V^{\prime}=(k,l).$ Let
$(E_{1},\ldots,E_{k}\mid E_{m+1},\ldots,E_{m+l})$ be a
$\mathbb{Z}_{2}$-homogeneous basis of $V^{\prime}$. It extends to a
$\mathbb{Z}_{2}$-homogeneous basis of $W_{-1}$:

$(E_{1},\ldots,E_{m}\mid E_{m+1},\ldots,E_{m+n}),$

where $|E_{i}|=|\partial_{i}|$, $i\in\mathbf{I}$. There exists an even
invertible matrix $A_{m+n}$ such that

$\displaystyle(E_{1},\ldots,E_{m+n})=(\partial_{1},\ldots,\partial_{m+n})A^{t}.$
(2.5)

Let $(\xi_{1},\ldots,\xi_{m+n})=(x_{1},\ldots,x_{m+n})A^{-1}$. Consider the
mapping $\phi$ such that

$\phi(x_{i})=\xi_{i}\quad\mbox{for all }i\in\mathbf{I}.$

Notice that $|x_{i}|=|\xi_{i}|$, since $A$ is even. By [20, Lemma 2.5], $\phi$
can extend to an endomorphism of $\mathcal{O}$, which is still written as
$\phi$. Then we have:

$\displaystyle(\phi^{-1}(x_{1}),\ldots,\phi^{-1}(x_{m+n}))=(x_{1},\ldots,x_{m+n})A$
(2.6)

We denote by ${\Phi}$ the automorphism of $W$ which is induced by $\phi$
according to the formula

${\Phi}(D)=\phi D\phi^{-1}\ \mbox{ for }\ D\in W.$

Clearly, ${\Phi}$ is $\mathbb{Z}$-homogeneous. By (2.5) and (2.6), we have:

$\displaystyle\Phi(\partial_{i})=\phi\partial_{i}\phi^{-1}=E_{i}\quad\mbox{for
all }i\in\mathbf{I}.$ (2.7)

Furthemore, for $D=\sum_{i\in\mathbf{I}}f_{i}\partial_{i}\in W$, one can
verify that

${\Phi}(D)=\phi
D\phi^{-1}={\sum}_{i,j\in\mathbf{I}}\partial_{i}(\phi^{-1}(x_{j}))\phi(f_{i})\partial_{j}.$

By virtue of (2.5) and (2.7), we have:

$\mathrm{div}({\Phi}(D))=\phi(\mathrm{div}D).$

This shows that ${\Phi}({S})={S}$ since $S=[\overline{S},\overline{S}]$. Then
$\Phi_{L}=\Phi\big{|}_{L}$ is desired.

When $L=H$ or $K$, suppose $\beta$-$\dim V=\beta$-$\dim V^{\prime}=(a,b,c,d)$.
Let $\\{e_{i}\mid i\in\overline{1,2r}\cup\overline{m+1,m+n}\\}$ be an
extension of $\beta$-basis (2.4) of $V^{\prime}$ to a generalized
orthosymplectic basis of $L_{-1}$. Then, there exist two even invertible
matrices

$\displaystyle A=\left(\begin{array}[]{c|c}A_{2r}&0\\\ \hline\cr
0&A_{n}\end{array}\right)\mbox{ and
}A^{\prime}=\left(\begin{array}[]{c|c}\begin{array}[]{c|c}A_{2r}&0\\\
\hline\cr 0&I_{1}\end{array}&0\\\ \hline\cr 0&A_{n}\end{array}\right)$

satisfying

$\displaystyle(e_{1},\ldots,e_{2r}\mid
e_{m+1},\ldots,e_{m+n})A=(y_{1},\ldots,y_{2r}\mid y_{m+1},\ldots,y_{m+n}),$
$\displaystyle(e_{1},\ldots,e_{2r},e_{2r+1}\mid
e_{m+1},\ldots,e_{m+n})A^{\prime}=(y_{1},\ldots,y_{2r},y_{2r+1}\mid
y_{m+1},\ldots,y_{m+n}).$

Thus, we obtain that

$\displaystyle A^{-1}J(A^{t})^{-1}=J.$ (2.10)

By virtue of [20, Lemma 2.5], there exists a unique automorphism of
$\mathcal{O}$ denoted by $\phi_{L}$ satisfying $\phi_{L}(y_{i})=e_{i}$,
$i\in\mathbf{I}.$ As in the case $L=W$, we denote by $\overline{{\Phi}}_{L}$
the $\mathbb{Z}$-homogeneous automorphism of $W$ which is induced by
$\phi_{L}$. From (2.10), we have:

$\displaystyle(\overline{{\Phi}}_{H}(D_{1}),\ldots,\overline{{\Phi}}_{H}(D_{m+n}))=(D_{1},\ldots,D_{m+n})A^{t}.$
(2.11)
$\displaystyle(\overline{{\Phi}}_{K}(D_{1}),\ldots,\overline{{\Phi}}_{K}(D_{m+n}))=(D_{1},\ldots,D_{m+n})A^{\prime
t}.$ (2.12)

For any $D=\sum_{i\in\mathbf{I}\backslash\\{2r+1\\}}f_{i}D_{i}\in W$ and
$fD_{2r+1}\in W$, from (2.10)-(2.12) we have:

$\displaystyle\overline{{\Phi}}_{L}(D)={\sum}_{i\in\mathbf{I}\backslash\\{2r+1\\}}\phi_{L}(f_{i})\overline{{\Phi}}_{L}(D_{i}),$
(2.13) $\displaystyle\overline{{\Phi}}_{K}(fD_{2r+1})=\phi_{K}(f)D_{2r+1}.$
(2.14)

For any $f\in\mathcal{O}$, we have:

$\displaystyle\overline{{\Phi}}_{K}(D_{2r+1}(f)\mathfrak{D})=D_{2r+1}(\phi_{K}(f))\mathfrak{D}.$
(2.15)
$\displaystyle\overline{{\Phi}}_{K}((2-\mathfrak{D})(f)D_{2r+1})=(2-\mathfrak{D})\phi_{K}(f)D_{2r+1}.$
(2.16)

By virtue of (2.10)-(2.16), we have:

$\overline{{\Phi}}_{L}(E_{L}(f))=E_{L}(\phi_{L}(f))\ \mbox{ for any
}f\in\mathcal{O}.$

It follows that $\overline{\Phi}_{L}({L})={L}$ since
$L=[\overline{L},\overline{L}]$. Then
$\Phi_{L}=\overline{\Phi}_{L}\big{|}_{L}$ is desired. ∎

For convenience, we introduce the following notations. Let $L=W$ or $S$. For
any $V\in\mathfrak{V}^{L}$ with $\mathrm{superdim}V=(k,l)$, put

$\displaystyle\mathbf{I}(k,l)=\overline{1,k}\cup\overline{m+1,m+l},\qquad\overline{\mathbf{I}}(k,l)=\mathbf{I}\backslash\mathbf{I}(k,l),$

If $V$ is standard, we have:

$V=\mathrm{span}_{\mathbb{F}}\\{\partial_{i}\mid i\in\mathbf{I}(k,l)\\}.$
(2.17)

Let $L=H$ or $K$. For any $V\in\mathfrak{V}^{L}$ with $\beta$-$\dim
V=(a,b,c,d)$, put

$\displaystyle I_{01}=\overline{1,a};\ \bar{I}_{01}=\overline{r+1,r+a};\
I_{02}=\overline{a+1,b};\ \bar{I}_{02}=\overline{r+a+1,r+b};$
$\displaystyle{I}_{11}=\overline{m+n-d+1,m+n};\ {I}_{12}=\overline{m+1,m+c};\
\bar{I}_{12}=\overline{m+q+1,m+q+c};$
$\displaystyle{I}_{03}=\overline{b+1,r}\cup\overline{r+b+1,m};\
{I}_{13}=\overline{m+c+1,m+q}\cup\overline{m+q+c+1,m+n-d}.$ (2.18)

Obviously, $\mathbf{I}=J_{1}\cup J_{2}\cup\bar{J}_{2}\cup J_{3},$ where

$\displaystyle J_{1}={I}_{01}\cup\bar{I}_{01}\cup{I}_{11},\ \
J_{2}={I}_{02}\cup{I}_{12},\ \ \bar{J}_{2}=\bar{I}_{02}\cup\bar{I}_{12}\
\mbox{ and }\ J_{3}={I}_{03}\cup{I}_{13}.$ (2.19)

We call $J_{i}$ to be single (resp. twinned) if $\mathbf{I}_{0}\cap
J_{i}=\emptyset$ and there exists only one element in $\mathbf{I}_{1}\cap
J_{i}$ (resp. there exist two elements in $\mathbf{I}_{1}\cap J_{i})$,
$i=1,3$. If $V$ is standard, we have:

$V=\mathrm{span}_{\mathbb{F}}\\{y_{i}\mid i\in J_{1}\cup J_{2}\\}.$ (2.20)

For any $i\in\mathbf{I}$, let us assign to each $y_{i}$ a value as follows:

$\nu(y_{i})=\left\\{\begin{array}[]{llll}1&i\in J_{1};\\\ 0&i\in J_{2};\\\
\frac{1}{3}&i\in\bar{J}_{2};\\\ 2&i\in J_{3}.\end{array}\right.$ (2.21)

If $a=y_{1}^{\alpha_{1}}y_{2}^{\alpha_{2}}\cdots y_{m}^{\alpha_{m}}y^{u}$,
define $\nu(a)=\prod_{i\in\mathbf{I}_{0}}\nu(y_{i})^{\alpha_{i}}\prod_{i\in
u}\nu(y_{i}).$

######  Remark 2.3.

Let $T$ be a torus of $L$, $L=W,S,H$ or $K$. Consider the weight space
decompositions with respect to $T$:

$\displaystyle L=L^{\theta}\oplus\oplus_{\gamma\in\Delta}L^{\gamma},\ \
L_{i}=L_{i}^{\theta}\oplus\oplus_{\gamma\in\Delta_{i}}L^{\gamma}_{i},$

where $\Delta_{i}\subset\Delta\subset T^{*}$ and $\theta$ is the zero weight.
Notice the standard facts below.

  * $(1)$

For $t\in T$, suppose $x=x_{1}+x_{2}+\cdots+x_{n}\in L$ is a sum of
eigenvectors of $\mathrm{ad}t$ associated with mutually distinct eigenvalues.
Then all $x_{i}$’s lie in $\mathrm{alg}(\\{t,x\\})$.

  * $(2)$

$T=\mathrm{span}_{\mathbb{F}}\\{y_{i}y_{\widetilde{i}}\mid
i\in\overline{1,r}\cup\overline{m+1,m+q}\\}$ is a torus of $L$, $L=H$ or $K$,
where $0\leq d\leq n$, $q=\lfloor\frac{n-d}{2}\rfloor$. Define $\epsilon_{j}$
to be the linear function on $T$ by

$\epsilon_{j}(y_{i}y_{\widetilde{i}})=\delta_{j{\widetilde{i}}}-\delta_{ji}.$

For $i,j,k\in\mathbf{I}\backslash\\{\\{2r+1\\}\cup\overline{m+2q+1,m+n}\\}$,
if $\epsilon_{i}+\epsilon_{j}\in\Delta_{0}$, we have:

$\dim L^{\epsilon_{i}}_{-1}=1;\ \ \dim L^{\epsilon_{i}+\epsilon_{j}}_{0}=1;\ \
L^{\epsilon_{k}}_{0}={\sum}_{l=m+2q+1}^{m+n}\mathbb{F}y_{k}y_{l}.$

##  3 MGS of Type (I)

To formulate the MGS of type $(\textrm{I})$, we introduce the following
notations.

For $i\geq 1$, write

$L_{i}^{\prime}=\overline{S}_{i}=\\{D\in
L_{i}\mid\mathrm{div}D=0\\}\quad\mbox{and}\quad
L_{i}^{\prime\prime}=\\{f\mathfrak{D}\mid f\in\mathcal{O}_{i}\\},$ (3.22)

where $L=W$ or $S$, $\mathfrak{D}$ is the degree derivation of $\mathcal{O}$;
that is, $\mathfrak{D}=\sum_{k\in\mathbf{I}}x_{k}\partial_{k}$. Clearly, both
$W_{i}^{\prime}$ and $W_{i}^{\prime\prime}$ are nontrivial subspaces of
$W_{i}$.

For $i\geq 0$, write

$K_{ij}=\big{\\{}u\in K_{i}\mid
u=fz^{j},f\in\mathcal{O}_{i+2-2j},[1,f]=0\big{\\}}.$

Clearly, $K_{ij}$ is a nontrivial subspace of $K_{i}$.

######  Theorem 3.1.

All MGS of type $\mathrm{(I)}$ are characterized as follows:

  * $\mathrm{(1)}$

If $m-n+1\equiv 0\pmod{p}$ then $W$ has exactly one MGS of type
$\mathrm{(I)}:$

$W_{-1}+W_{0}+W_{1}^{\prime}+W_{2}^{\prime}+\cdots+W_{\xi-2}^{\prime}$

with dimension $(m+n-1)2^{n}p^{m}+2$;

If $m-n+1\not\equiv 0\pmod{p}$ then $W$ has exactly two MGS of type
$\mathrm{(I)}:$

$W_{-1}+W_{0}+W_{1}^{\prime\prime}\quad\mbox{and}\quad
W_{-1}+W_{0}+W_{1}^{\prime}+W_{2}^{\prime}+\cdots+W_{\xi-2}^{\prime}$

with dimensions $(m+n)(m+n+2)$ and $(m+n-1)2^{n}p^{m}+2$, respectively.

  * $\mathrm{(2)}$

If $m-n+1\equiv 0\pmod{p}$ then $S$ has exactly one MGS of type
$\mathrm{(I)}:$

$S_{-1}+S_{0}+S_{1}^{\prime\prime}$

with dimension $(m+n)^{2}+2(m+n)-1$;

If $m-n+1\not\equiv 0\pmod{p}$ then $S$ has exactly one MGS of type
$\mathrm{(I)}:$

$S_{-1}+S_{0}$

with dimension $(m+n)^{2}+(m+n)-1$.

  * $\mathrm{(3)}$

$H$ has exactly one MGS of type $\mathrm{(I)}:$

$H_{-1}+H_{0}$

with dimension $(m+n)^{2}+m$.

  * $\mathrm{(4)}$

$K$ has exactly two MGS of type $\mathrm{(I)}:$

$K_{-2}+K_{-1}+K_{0}+{\sum}_{i=1}^{2r(p-1)+n}K_{i0}\ \mbox{ and }\
K_{-2}+K_{-1}+K_{0}+K_{11}+K_{22}$

with dimensions $2^{n}p^{2r}+1$ and $(2r+n)^{2}+4r+n+3$, respectively.

We note that many preliminary results in this section are analogous to the
ones of Lie algebras (see [16, 17, 18]). We will need the following formulas
which are easy to verify by direct calculations.

######  Lemma 3.2.

For $f\in\mathcal{O}_{s}$ and $g\in\mathcal{O}_{t},$

$\displaystyle\mathrm{div}(f\mathfrak{D})=(m-n+s)f\quad\mbox{for}~{}f\in\mathcal{O}_{s},$
$\displaystyle[f\mathfrak{D},g\mathfrak{D}]=(t-s)fg\mathfrak{D}.$ (3.23)

######  Lemma 3.3.

The following statements hold.

  * $(1)$

$W_{s}^{\prime}$ and $W^{\prime\prime}_{s}$ are $W_{0}$-submodules of $W_{s}$.
Moreover, $W_{s}^{\prime\prime}$ is irreducible.

  * $(2)$

If $m-n+s\not\equiv 0\pmod{p}$ then $W_{s}=W_{s}^{\prime}\oplus
W_{s}^{\prime\prime}$;

  * $(3)$

If $m-n+s\equiv 0\pmod{p}$ then $W_{s}^{\prime\prime}\subset W_{s}^{\prime}$.

###### Proof.

Note that $\mathrm{div}$ is a derivation from $W$ to $\mathcal{O}$ as
$W$-module. Thus, (1), (2) and (3) hold by virtue of Lemma 3.2. ∎

Below, the $1$-component $W_{1}$ will be a focus of our attention. For
convenience, we introduce two concepts, by which our arguments are largely
simplified: An element $\mathscr{L}$ in $W_{1}$ is called a leader if it is of
the form

$\displaystyle\mathscr{L}=x_{1}^{2}\partial_{1}+{\sum}_{i=2}^{m+n}f_{i}\partial_{i}\quad\mbox{where}\;f_{i}\in\mathcal{O}_{2};$

An element in $W_{1}$ is called $1$-defective if it is of the form

${\sum}_{i=2}^{m+n}f_{i}\partial_{i}\ \mbox{ where }f_{i}\in\mathcal{O}_{2}.$

######  Lemma 3.4.

Let $D\in W_{1}$.

  * $(1)$

$[x_{1}\partial_{j},D]=0$ for all $j\geq 2$ if and only if $D=\lambda
x_{1}\mathfrak{D}+x_{1}^{2}\sum_{j\geq 2}k_{j}\partial_{j}$ for some
$\lambda,k_{j}\in\mathbb{F}.$

  * $(2)$

$[x_{1}\partial_{j},D]\in W_{1}^{\prime\prime}$ for all $j\geq 2$ if and only
if $D=f\mathfrak{D}+x_{1}^{2}\sum_{j\geq 2}k_{j}\partial_{j}$ for some
$f\in\mathcal{O}_{1}$ and $k_{j}\in\mathbb{F}.$

###### Proof.

(1) Suppose $[x_{1}\partial_{j},D]=0$ for all $j\geq 2$ and write
$D=\sum_{i}a_{i}\partial_{i}$. Then

$x_{1}\partial_{j}(a_{i})-\delta_{i=j}(-1)^{|x_{1}\partial_{j}||a_{1}\partial_{1}|}a_{1}=0\quad\mbox{for
all}\;j\geq 2.$ (3.24)

Then $a_{1}=kx_{1}^{2}$ for some $k\in\mathbb{F}.$ If $k=0$, it follows from
(3.24) that $\partial_{j}(a_{i})=0$ for all $j,i\geq 2$. That is,
$a_{j}=k_{j}x_{1}^{2}$ for all $j\geq 2.$ Hence $D=x_{1}^{2}\sum_{j\geq
2}k_{j}\partial_{j}.$ If $k\neq 0$ then write $a_{1}=x_{1}^{2}.$ From (3.24)
one deduces

$\partial_{j}(a_{i})=\delta_{i=j}x_{1}\quad\mbox{for all}\;i,j\geq 2.$

It follows that $a_{j}=k_{j}x_{1}^{2}+x_{1}x_{j}$ for $j\geq 2$ and one
direction holds. The other one is clear.

(2) Write $[x_{1}\partial_{j},D]=f_{j}\mathfrak{D}$,
$f_{j}\in\mathcal{O}_{1},$ $j\geq 2.$ By acting on $x_{i}$ with $i\neq 1,j,$
we have $x_{1}[\partial_{j},D](x_{i})=f_{j}x_{i}$ and then $f_{j}=k_{j}x_{1}$
for some $k_{j}\in\mathbb{F}.$ Thus

$[x_{1}\partial_{j},D]=k_{j}x_{1}\mathfrak{D}\quad\mbox{for all}\;j\geq 2.$
(3.25)

Since $[x_{1}\partial_{j},x_{i}\mathfrak{D}]=\delta_{i=j}x_{1}\mathfrak{D},$
from (3.25) we have $\left[x_{1}\partial_{j},D-\left(\sum_{i\geq
2}k_{i}x_{i}\right)\mathfrak{D}\right]=0$ for all $j\geq 2.$ Now the
conclusion follows from (1). ∎

######  Lemma 3.5.

Let $M$ be a nonzero $W_{0}$-submodule of $W_{1}.$

  * $\mathrm{(1)}$

$M$ contains a leader.

  * $\mathrm{(2)}$

If $M$ contains a leader which does not lie in $W_{1}^{\prime\prime}$ then $M$
contains a nonzero 1-defective element.

  * $\mathrm{(3)}$

If $M$ contains a nonzero $1$-defective element then $M\supset
W_{1}^{\prime}$. In particular, as $W_{0}$-module, $W_{1}^{\prime}$ is
generated by $x_{1}^{2}\partial_{j}$ for any fixed $j\geq 2$.

  * $\mathrm{(4)}$

As $W_{0}$-module, $W_{1}$ is generated by $x_{1}^{2}\partial_{1}.$

  * $\mathrm{(5)}$

Any nonzero $W_{0}$-submodule of $W_{1}$ different from $W_{1}^{\prime\prime}$
must contain $W_{1}^{\prime}.$

###### Proof.

(1), (3) and (4) need only a straightforward verification.

(2) Let $D=x_{1}^{2}\partial_{1}+\cdots$ be a leader in $M\setminus
W_{1}^{\prime\prime}$. Then $[x_{1}\partial_{j},D]\in M$ are 1-defective for
all $j\geq 2.$ If they are not all zero, we are done. Otherwise, by Lemma
3.4(1),

$D=x_{1}\mathfrak{D}+x_{1}^{2}\sum_{j\geq 2}k_{j}\partial_{j}$ for some
$k_{j}\in\mathbb{F}.$

Clearly, $\sum_{j\geq 2}k_{j}\partial_{j}\neq 0,$ say, $k_{2}\neq 0.$
Consequently, $x_{1}^{2}\partial_{2}=k_{2}^{-1}[x_{2}\partial_{2},D]\in M$.

(5) Let $M$ be a nonzero $W_{0}$-submodule and $M\neq W_{1}^{\prime}$. Let us
show that $M\supset W_{1}^{\prime}.$ By (1), (2) and (3) we may assume that
all the leaders of $M$ lie in $W^{\prime\prime}$. Then
$W^{\prime\prime}\subset M,$ since $W^{\prime\prime}$ as $W_{0}$-module is
irreducible by Lemma 3.3(1). For $D\in M\setminus W^{\prime\prime}$, if there
is some $i\geq 2$ such that $E=[x_{1}\partial_{i},D]\not\in W^{\prime\prime},$
then $[x_{1}\partial_{j},E]$ is a leader or 1-defective for any $j\geq 2$. By
(3), one may assume that there is $D\in M\setminus W^{\prime\prime}$ which is
pulled into $W^{\prime\prime}$ by any $x_{1}\partial_{j}$ with $j\geq 2.$ Then
by Lemma 3.4(2), $M$ contains a nonzero 1-defective element and then $M\supset
W^{\prime}.$ ∎

######  Lemma 3.6.

The following statements hold.

  * $\mathrm{(1)}$

$W_{1}^{\prime}$ is a maximal $W_{0}$-submodule of $W_{1}$.

  * $\mathrm{(2)}$

If $m-n+1\not\equiv 0\pmod{p}$, $W_{0}$-module $W_{1}^{\prime}$ is
irreducible. In particular, $W_{1}$ has a decomposition of irreducible
$W_{0}$-submodules:

$W_{1}=W_{1}^{\prime}\oplus W_{1}^{\prime\prime}.$

  * $\mathrm{(3)}$

If $m-n+1\equiv 0\pmod{p}$, $W_{1}$ has exactly a composition series of
$W_{0}$-submodules:

$0\subset W_{1}^{\prime\prime}\subset W_{1}^{\prime}\subset W_{1}.$

###### Proof.

(1) Let $M$ be a submodule of $W_{1}$ containing strictly $W_{1}^{\prime}$.
Note that

$\mathrm{div}:\mathrm{span}_{\mathbb{F}}\\{x_{1}x_{1}\partial_{1},x_{2}x_{1}\partial_{1},\ldots,x_{m+n}x_{1}\partial_{1}\\}\longmapsto\mathcal{O}_{1}$

is surjective. Pick any $D\in M\setminus W_{1}^{\prime}$. Then there exists
$E=fx_{1}\partial_{1},$ $f\in\mathcal{O}_{1},$ such that
$\mathrm{div}E=\mathrm{div}D$. That is, $E-D\in W_{1}^{\prime}\subset M$ and
then $0\neq E\in M.$ If $\partial_{j}(f)=0$ for all $j\geq 2$ then
$E=\partial_{1}(f)x_{1}^{2}\partial_{1}$ and hence $M=W_{1}$ by Lemma 3.5(4).
Suppose $\partial_{j}(f)\neq 0$ for some $j\neq 1.$ Then
$x_{j}x_{1}\partial_{1}={\partial_{j}(f)^{-1}}[x_{j}\partial_{j},E]\in M$. It
follows that

$x_{1}^{2}\partial_{1}-(-1)^{|\partial_{j}|}x_{j}x_{1}\partial_{j}=[x_{1}\partial_{j},x_{j}x_{1}\partial_{1}]\in
M.$

Note that
$\frac{1}{2}x_{1}^{2}\partial_{1}-(-1)^{|\partial_{j}|}x_{j}x_{1}\partial_{j}$
is in $W_{1}^{\prime}\subset W.$ It follows that $x_{1}^{2}\partial_{1}\in M$
and $M=W_{1}$ by Lemma 3.5(4), showing that $W_{1}^{\prime}$ is maximal.

(2) and (3) are immediate consequences of Lemmas 3.3 and 3.5(5). ∎

######  Corollary 3.7.

The following statements hold.

  * $\mathrm{(1)}$

If $m-n+1\not\equiv 0\pmod{p}$ then $S_{1}$ is an irreducible $S_{0}$-module.

  * $\mathrm{(2)}$

If $m-n+1\equiv 0\pmod{p}$ then $S_{1}^{\prime\prime}$ is the unique
nontrivial $S_{0}$-submodule of $S_{1}$.

###### Proof.

Note that $W_{0}=S_{0}+\mathbb{F}\mathfrak{D}$ and that
$W^{\prime}_{1}=S_{1}$. If $m-n+1\equiv 0\pmod{p}$ then
$S_{1}^{\prime\prime}=W_{1}^{\prime\prime}$. The lemma follows directly from
Lemma 3.6. ∎

######  Lemma 3.8.

$\\{D\in W_{2}\mid[W_{-1},D]\subset W_{1}^{\prime\prime}\\}=0.$

###### Proof.

Write $D=\sum_{i\in\mathbf{I}}a_{i}\partial_{i}\in W_{2}$ and suppose $D$ is
pulled into $W_{1}^{\prime\prime}$ by $W_{-1}$. Then, each $a_{i}$ must be a
multiple of $x_{i}^{2}$ and in particular, $a_{j}=0$ for all $j>m.$ Write
$D=\sum_{i\leq m}f_{i}x_{i}^{2}\partial_{i}$, where $f_{i}\in\mathcal{O}_{1}.$
Since $[\partial_{j},D]\in W_{1}^{\prime\prime}$, one deduces that
$\partial_{j}(f_{i})=0$ for $j>m$ and $i\leq m.$ Now it is clear that $D$ is
not in $W_{1}^{\prime\prime}$ unless it is zero. ∎

Let

$\displaystyle
M^{\prime}=W_{-1}+W_{0}+W_{1}^{\prime}+W_{2}^{\prime}+\cdots+W_{\xi-2}^{\prime},$
$\displaystyle M^{\prime\prime}=W_{-1}+W_{0}+W_{1}^{\prime\prime}.$

Using (3.23) and keeping in mind that $\mathrm{div}$ is a derivation from $W$
to $\mathcal{O}$, one may verify that $M^{\prime}$ and $M^{\prime\prime}$ are
subalgebras of $W$.

######  Lemma 3.9.

Suppose $M$ is a proper subalgebra containing $W_{-1}\oplus W_{0}\oplus
W_{1}^{\prime}.$ Then $M\subset M^{\prime}$.

###### Proof.

Assume conversely that $M\not\subset M^{\prime}$. Then there exists $D\in
M\cap\sum_{i\geq 1}W_{i}$ satisfying $\mathrm{div}D\not=0$. Using the formula
$\mathrm{div}[\partial_{j},D]=\partial_{j}(\mathrm{div}D)$ for all
$j\in\mathbf{I}$, one sees that $M\supset W_{1}$ by Lemma 3.6(1). By Lemma
2.1(2), $M=W$, a contradiction. ∎

Proof of (1) and (2) in Theorem 3.1 (1) Claim A: $M^{\prime}$ is maximal. This
follows immediately from Lemma 3.9.

Claim B: $M^{\prime\prime}$ is maximal if $m-n+1\not\equiv 0\pmod{p}.$ Let $M$
be a subalgebra strictly containing $M^{\prime\prime}$. By transitivity and
Lemma 3.8, $M\cap W_{1}$ must strictly contain $W_{1}^{\prime\prime}$. Lemma
3.6(2) forces $M\supset W_{1}$ and therefore, $M=W$ by Lemma 2.1(2).

Claim C: $M^{\prime}$ and $M^{\prime\prime}$ exhaust all the maximal
subalgebras of type (I). Let $M$ be a maximal subalgebra of type (I). By
transitivity, $M$ must contain a nonzero element of $W_{1}$ and therefore,
$M\cap W_{1}\neq 0$ is a nonzero $W_{0}$-submodule of $W_{1}$. By Lemma
3.5(5), we have $M\cap W_{1}=W_{1}^{\prime\prime}$ or $M\cap W_{1}\supset
W_{1}^{\prime}$.

Case 1. Suppose $m-n+1\not\equiv 0\pmod{p}.$ If $M\cap
W_{1}=W_{1}^{\prime\prime}$ then Claim B forces $M=M^{\prime\prime}.$ Suppose
$M\cap W_{1}\supset W_{1}^{\prime}$. By Lemma 3.9, we have $M\subset
M^{\prime}$ and then $M=M^{\prime}$ by the maximality of $M$.

Case 2. Suppose $m-n+1\equiv 0\pmod{p}.$ We have $M\supset W^{\prime\prime}$.
Since $W^{\prime\prime}\subsetneq W^{\prime}$ in this situation, one sees
$M\supsetneq W^{\prime\prime}$. By transitivity and Lemma 3.8, $M\cap
W_{1}\supsetneq W_{1}^{\prime\prime}$ and hence $M\cap W_{1}\supset
W_{1}^{\prime}$ by Lemma 3.5(5). It follows from Lemma 3.9 that
$M=M^{\prime}.$ This completes the proof of (1).

(2) First of all, $S_{-1}+S_{0}$ and $S_{-1}+S_{0}+S_{1}^{\prime\prime}$
($m+n-1\equiv 0\pmod{p}$) are subalgebras of $S$. Let $M$ be a maximal
subalgebra of $S$ containing $S_{-1}+S_{0}$. Note that
$W_{1}^{\prime\prime}=S_{1}^{\prime\prime}$ when $m-n+1\equiv 0\pmod{p}$. By
the transitivity of $S$, Lemmas 2.1(2), 3.8 and Corollary 3.7, we obtain that
$M=S_{-1}+S_{0}+S_{1}^{\prime\prime}$ when $m+n-1\equiv 0\pmod{p}$;
$M=S_{-1}+S_{0}$ when $m+n-1\not\equiv 0\pmod{p}$. The process shows also that
these two subalgebras are indeed maximal. This completes the proof of (2). ∎

######  Remark 3.10.

For $W$ and $S$, the arguments for MGS of the other types will be reduced to
the case of type $\mathrm{(I)}$ MGS by the method of minimal counterexample.

######  Lemma 3.11.

The following statements hold.

  * $\mathrm{(1)}$

$H_{1}$ is an irreducible $H_{0}$-module.

  * $\mathrm{(2)}$

For $i\geq 0$, $K_{i}$ is a direct sum of $K_{0}$-submodules $K_{ij}$.
Moreover, $K_{10}$ and $K_{11}$ are irreducible $K_{0}$-modules.

###### Proof.

Using the results in the case of modular Lie algebras [17] and by a direct
computation, it is easy to show that (1) holds. Since $K_{10}\cong H_{1}$ and
$K_{11}\cong H_{-1}$ as $H_{0}$-modules, by irreducibilities of $H_{-1}$ and
$H_{1}$, (2) holds. ∎

Let

$\displaystyle M^{\prime}=K_{-2}+K_{-1}+K_{0}+{\sum}_{i=1}^{2r(p-1)+n}K_{i0},$
$\displaystyle M^{\prime\prime}=K_{-2}+K_{-1}+K_{0}+K_{11}+K_{22}.$

By a standard and direct computation, one may verify that $M^{\prime}$ and
$M^{\prime\prime}$ are subalgebras of $K$.

Proof of (3) and (4) in Theorem 3.1 (3) This statement follows immediately
from Lemmas 2.1(2) and 3.11(1).

(4)  Claim A: $M^{\prime}$ is maximal. For any $0\not=u\in K$, $u\not\in
M^{\prime}$, put $\overline{M}=\mathrm{alg}(M^{\prime}+\mathbb{F}{u})$. Note
that there exist $k\in\mathbb{N}$ and $v_{1},\ldots,v_{s}\in K_{-1}$ such that

$0\not=u_{1}=fz+\alpha
z^{2}=[v_{1},[\cdots[v_{s},(\mathrm{ad}1)^{k}u]\cdots]]\in\overline{M}\backslash
M^{\prime},$

where $f\in\mathcal{O}$ satisfying $[1,f]=0$ and $\alpha\in\mathbb{F}$. Then
there exists $i\in\mathbf{I}$ such that
$[y_{\widetilde{i}},u_{1}]-y_{\widetilde{i}}f\not=0$. It follows that

$0\not=(\sigma(\widetilde{i})(-1)^{i}D_{i}(f)+\alpha
y_{\widetilde{i}})z\in\overline{M}.$

From Lemma 2.1(1), there exists a nonzero element in $\overline{M}\cap
K_{11}.$ By Lemmas 2.1(2) and 3.11(2), we have $\overline{M}=K$. Thus
$M^{\prime}$ is maximal.

Claim B: $M^{\prime\prime}$ is maximal. For any $0\not=u\in K$, $u\not\in
M^{\prime\prime}$, put
$\overline{M}=\mathrm{alg}(M^{\prime\prime}+\mathbb{F}{u})$. It is sufficient
to show that there exists a nonzero element in $K_{10}\cap\overline{M}$. When
$\mathrm{zd}(u)>2$, by transitivity, there exist $v_{1},\ldots,v_{s}\in
K_{-1}$ such that

$0\not=u_{3}=u_{30}+u_{31}+u_{32}=[v_{1},[\cdots[v_{s},u]\cdots]]\in\overline{M},$

where $u_{3i}\in K_{3i}$, $i=0,1,2$. Note that $[1,u_{32}]\in K_{11}$. If
$u_{31}\not=0$, then

$0\not=[1,u_{31}]=[1,u_{3}-u_{30}-u_{32}]\in\overline{M}\cap K_{10}.$

If $u_{31}=0$, there exists $j\in\mathbf{I}$ such that

$0\not=u_{2}=\sigma(\widetilde{j})(-1)^{j}D_{j}(u_{30})+y_{\widetilde{j}}D_{2r+1}(u_{32})\in\overline{M}\backslash{M^{\prime\prime}},$

Note that $\mathrm{zd}(u_{2})=2$. Thus, it remains to consider the case
$\mathrm{zd}(u)=2$. Assume that $u=u_{20}+u_{21}$, where $u_{2i}\in K_{2i}$,
$i=0,1$. If $u_{21}=0$ the conclusion follows. Notice that ${H_{0}}\cong
K_{21}$ as $H_{0}$-module. If $u_{21}\not=0$, by Remark 2.3 and a direct
computation, we obtain that there exists $i\in\mathbf{I}_{0}$ such that
$u_{2}=u^{\prime}_{20}+y^{(2\varepsilon_{i})}z\in\overline{M}$, where
$u^{\prime}_{20}\in K_{20}.$ Since $[K_{-1},u_{2}]\subset\overline{M}$, there
exists $j\in\mathbf{I}$ such that

$0\not=\sigma(\widetilde{j})(-1)^{j}D_{j}(u^{\prime}_{20})+y_{\widetilde{j}}y^{(2\varepsilon_{i})}\in\overline{M}\cap
K_{10}.$

Thus the conclusion holds.

Claim C: $M^{\prime}$ and $M^{\prime\prime}$ exhaust all the maximal graded
subalgebras of type $\mathrm{(I)}$. Suppose $N$ is a maximal graded subalgebra
of $K$ containing $K_{-1}+K_{0}$. By transitivity, there exists
$0\not=D=D_{10}+D_{11}\in N$, where $D_{1i}\in K_{1i}$, $i=0,1$. Since
$N\subsetneq K$, we claim that $D_{11}=0$ or $D_{10}=0$. Indeed, if
$D_{11}\not=0$ and $D_{10}\not=0$, by the irreducibility of $K_{10}$, we have

$w=y^{(2\varepsilon_{1})}y_{\widetilde{1}}+{\sum}_{i=1}^{2r+n}\alpha_{i}y_{i}z\in
N,\ \alpha_{i}\in\mathbb{F}.$

We consider the following cases.

Case 1. For all $i$, $\alpha_{i}=0$. Obviously, $N=K$ by the irreducibility of
$K_{1i}$ and $D_{1i}\not=0$, $i=0,1.$

Case 2. There exists $k$ such that $\alpha_{k}\not=0.$ If
$k\not=1,\widetilde{1}$, for $\widetilde{k}\not=j\in\mathbf{I}_{1}$, we have:

$0\not=[y_{j}y_{\widetilde{k}},w]\in N\cap K_{11}.$

Similar to Case 1, we have $N=K$.

If $k=1$ or $\widetilde{1}$, then
$w=y^{(2\varepsilon_{1})}y_{\widetilde{1}}+\alpha_{1}y_{1}z+\alpha_{\widetilde{1}}y_{\widetilde{1}}z$.
For $j\in\mathbf{I}_{1}$, we have:

$y_{j}y_{1}y_{\widetilde{1}}=[[y_{j}y_{1},w],y^{(2\varepsilon_{\widetilde{1}})}]\in
N\cap K_{10}.$

Similar to Case 1, we have $N=K$.

Consequently, $N=M^{\prime}$ when $D_{11}=0$ and $N=M^{\prime\prime}$ when
$D_{10}=0$. ∎

##  4 MGS of Type (II)

Let $L=W,S,H$ or $K$. Recall

$\mathfrak{V}^{L}=\\{V\mid V\mbox{ is a nontrivial subspace of }L_{-1}\\}.$

To describe the MGS of type $(\textrm{II})$ of $L$, for any
$V\in\mathfrak{V}^{L}$, we define

$\mathcal{M}(V)=\oplus_{i\geq-2}\mathcal{M}_{i}(V),$

where

$\displaystyle\mathcal{M}_{-1}(V)=V;\ \
\mathcal{M}_{-2}(V)=[\mathcal{M}_{-1}(V),\mathcal{M}_{-1}(V)];$
$\displaystyle\mathcal{M}_{i}(V)=\\{u\in
L_{i}\mid[V,u]\subset\mathcal{M}_{i-1}(V)\\}\qquad\mbox{for}\;i\geq 0.$ (4.26)

######  Theorem 4.1.

Suppose $L=W$ or $S$. All MGS of type $\mathrm{(II)}$ of $L$ are characterized
as follows:

  * $\mathrm{(1)}$

All MGS of type $\mathrm{(II)}$ of $L$ are precisely:

$\\{\mathcal{M}(V)\mid V\in\mathfrak{V}^{L}\\}.$

  * $\mathrm{(2)}$

For any $V$ and $V^{\prime}$ in $\mathfrak{V}^{L}$,

$\mathcal{M}(V)\stackrel{{\scriptstyle\mathrm{algebra}}}{{\cong}}\mathcal{M}(V^{\prime})\Longleftrightarrow
V{\cong}V^{\prime}.$

  * $\mathrm{(3)}$

$L$ has exactly $(m+1)(n+1)-2$ isomorphism classes of MGS of type
$\mathrm{(II)}$.

  * $\mathrm{(4)}$

If $\mathrm{superdim}V=(k,l),$ then

$\displaystyle\mathrm{dim}\mathcal{M}(V)=\left\\{\begin{array}[]{ll}2^{n}p^{m}(m+n)-2^{l}p^{k}(m+n-k-l),&L=W;\\\
2^{n}p^{m}(m+n-1)+1-2^{l}p^{k}(m+n-k-l),&L=S.\end{array}\right.$

When $L=H$ or $K$, recall definitions (2.18)-(2.20) mentioned in Section 2.
Put

$\displaystyle\mathcal{V}^{L}=\left\\{V\in\mathfrak{V}^{L},\mbox{ satisfying
}J_{3}\mbox{ is neither single nor twinned}\right\\};$
$\displaystyle\mathcal{W}^{K}=\left\\{V\in\mathfrak{V}^{K},\mbox{ satisfying
}J_{3}\mbox{ is not twinned}\right\\}.$

Suppose $V\in\mathfrak{V}^{K}$ is isotropic. Put

$\mathcal{M}^{K}(1,V)=\oplus_{i\geq-2}\mathcal{M}_{i}^{K}(1,V),$

where

$\displaystyle\mathcal{M}^{K}_{-2}(1,V)=\mathbb{F};\ \
\mathcal{M}^{K}_{-1}(1,V)=V;$
$\displaystyle\mathcal{M}^{K}_{i}(1,V)=\\{u\in{K}_{i}\mid[V,u]\subset\mathcal{M}^{K}_{i-1}(1,V),\
[1,u]\in\mathcal{M}^{K}_{i-2}(1,V)\\},\ i\geq 0.$

######  Theorem 4.2.

All MGS of type $\mathrm{(II)}$ of $H$ and $K$ are characterized as follows:

For ${H}$,

  * $\mathrm{(1)}$

All MGS of type $\mathrm{(II)}$ are precisely:

$\left\\{\mathcal{M}(V)\mid V\in\mathcal{V}^{H}\right\\}.$

  * $\mathrm{(2)}$

For any $V$ and $V^{\prime}$ in $\mathcal{V}^{H}$,

$\mathcal{M}(V)\stackrel{{\scriptstyle\mathrm{algebra}}}{{\cong}}\mathcal{M}(V^{\prime})\Longleftrightarrow
V{\cong}V^{\prime}.$

  * $\mathrm{(3)}$

$H$ has exactly $\phi(r,n)$ isomorphism classes of MGS of type
$\mathrm{(II)}$, where

$\phi(r,n)=\left\\{\begin{array}[]{ll}8^{-1}(r+1)(r(n+2)^{2}+2n^{2}+6-r)-2,&n\
\mbox{ is odd};\\\ 8^{-1}(r+1)(r(n+2)^{2}+2n^{2}+8)-2,&n\ \mbox{ is
even}.\end{array}\right.$

  * $\mathrm{(4)}$

For $V\in\mathcal{V}^{H}$, if $\beta$-$\mathrm{dim}V=(a,b,c,d),$ then

$\mathrm{dim}\mathcal{M}(V)=p^{m}2^{n}+p^{2a}2^{d}-p^{a+b}2^{c+d}(m-2b+n-d-2c+1)-2.$

For ${K}$,

  * $(1^{\prime})$

All MGS of type $\mathrm{(II)}$ are precisely:

$\displaystyle\left\\{\mathcal{M}(V)\mid V\in\mathcal{V}^{K}\mbox{ is neither
nondegenerate nor isotropic}\right\\}$ $\displaystyle\cup$
$\displaystyle\left\\{\mathcal{M}(V)\mid V\in\mathcal{W}^{K}\mbox{ is
nondegenerate or isotropic}\right\\}$ $\displaystyle\cup$
$\displaystyle\left\\{\mathcal{M}^{K}(1,V)\mid V\in\mathcal{V}^{K}\mbox{ is
isotropic}\right\\}.$

  * $(2^{\prime})$

For all MGS of type $\mathrm{(II)}$,

$\displaystyle\mathcal{M}(V)\stackrel{{\scriptstyle\mathrm{algebra}}}{{\cong}}\mathcal{M}(V^{\prime})\Longleftrightarrow
V{\cong}V^{\prime},$
$\displaystyle\mathcal{M}^{K}(1,V)\stackrel{{\scriptstyle\mathrm{algebra}}}{{\cong}}\mathcal{M}^{K}(1,V^{\prime})\Longleftrightarrow
V{\cong}V^{\prime}.$

  * $(3^{\prime})$

$K$ has exactly $\phi{(r,n)}$ isomorphism classes of MGS of type
$\mathrm{(II)}$, where

$\phi(r,n)=\left\\{\begin{array}[]{ll}8^{-1}(r+1)(r(n+2)^{2}+2n^{2}+4n+2-r)+r-1,&n\
\mbox{ is odd};\\\ 8^{-1}(r+1)(r(n+2)^{2}+2n^{2}+4n+8)+r-2,&n\ \mbox{ is
even}.\end{array}\right.$

  * $(4^{\prime})$

Let $\delta=1$ when $n-m-3=0\pmod{p}$ and $\delta=0$, otherwise. Suppose
$\beta$-$\mathrm{dim}V=(a,b,c,d).$

    * (a)

If $V$ is isotropic, then

$\displaystyle\mathrm{dim}\mathcal{M}(V)=p^{m}2^{n}-p^{b}2^{c}(m-2b+n-2c)-\delta,\mbox{
when }\ V\in\mathcal{W}^{K};$
$\displaystyle\mathrm{dim}\mathcal{M}^{K}(1,V)=p^{m}2^{n}-p^{b+1}2^{c}(m-2b+n-2c)+p-\delta,\mbox{
when }\ V\in\mathcal{V}^{K}.$

    * (b)

If $V$ is not isotropic, then

$\mathrm{dim}\mathcal{M}(V)=p^{m}2^{n}-(2r-2a+n-d)p^{2a+1}2^{d}-p,$

when $V\in\mathcal{W}^{K}$ is nondegenerate satisfying $J_{3}$ is single;

$\mathrm{dim}\mathcal{M}(V)=p^{m}2^{n}+p^{2a+1}2^{d}-p^{a+b+1}2^{c+d}(m-2b+n-d-2c)-\delta,$

when $V\in\mathcal{W}^{K}$ is nondegenerate satisfying $J_{3}$ is not single
or $V\in\mathcal{V}^{K}$ is not nondegenerate.

######  Lemma 4.3.

Suppose $L=W,S,H$ or $K$.

  * $\mathrm{(1)}$

$\mathcal{M}(V)$ is a $\mathbb{Z}$-graded subalgebra of $L$.
$\mathcal{M}^{K}(1,V)$ is a $\mathbb{Z}$-graded subalgebra of $K$.

  * $\mathrm{(2)}$

Suppose $\Phi$ is a $\mathbb{Z}$-homogeneous automorphism of $L$. Then

$\Phi(\mathcal{M}_{i}(V))=\mathcal{M}_{i}(\Phi(V))\ \mbox{ for all }\
i\geq-2.$

Moreover,

$\Phi(\mathcal{M}(V))=\mathcal{M}(\Phi(V)).$

For $K$,

$\Phi(\mathcal{M}^{K}_{i}(1,V))=\mathcal{M}^{K}_{i}(1,\Phi(V))\ \mbox{ for all
}\ i\geq-2.$

Moreover,

$\Phi(\mathcal{M}^{K}(1,V))=\mathcal{M}^{K}(1,\Phi(V)).$

  * $\mathrm{(3)}$

If $M$ is an MGS of type $\mathrm{(II)}$ of $L$, then $M=\mathcal{M}(M_{-1})$
unless $L=K$, $M_{-1}$ is isotropic and $M_{-2}\not=0$. However, in the latter
case, $M=\mathcal{M}^{K}(1,M_{-1})$.

###### Proof.

The approach is analogous to that used in the case of modular Lie algebras
[16]. ∎

######  Remark 4.4.

Suppose $L=W,S,H$ or $K$. In view of Lemmas 2.2 and 4.3, for any
$V\in\mathfrak{V}^{L}$, we may assume that $V$ is standard [see (2.17),
(2.20)].

Now, we consider the case $L=W$ or $S$. Suppose $V\in\mathfrak{V}^{L}$ with
$\mathrm{superdim}V=(k,l)$. For $L=W$, it is easy to verify that
$\mathcal{M}_{0}(V)$ has a standard $\mathbb{F}$-basis
$\mathcal{A}_{1}\cup\mathcal{A}_{2},$ where

$\displaystyle\mathcal{A}_{1}=\\{x_{i}\partial_{j}\mid
i,j\in\mathbf{I}(k,l)\\},$
$\displaystyle\mathcal{A}_{2}=\\{x_{i}\partial_{j}\mid
i\in\overline{\mathbf{I}}(k,l),j\in\mathbf{I}\\}.$

Similarly, for $L=S$, $\mathcal{M}_{0}(V)$ has a standard $\mathbb{F}$-basis
$\mathcal{C}_{1}\cup\mathcal{C}_{2}\cup\mathcal{C}_{3}$, where

$\displaystyle\mathcal{C}_{1}=\\{x_{i}\partial_{j}\mid
i,j\in\mathbf{I}(k,l),i\neq j\\},$
$\displaystyle\mathcal{C}_{2}=\\{x_{i}\partial_{j}\mid
i\in\overline{\mathbf{I}}(k,l),j\in\mathbf{I},i\neq j\\},$
$\displaystyle\mathcal{C}_{3}=\\{x_{1}\partial_{1}-(-1)^{|\partial_{i}|}x_{i}\partial_{i}\mid
i\in\mathbf{I}\backslash\\{1\\}\\}.$

Moreover, in any case of $L=W$ or $S$, $\mathcal{M}_{0}(V)$ has a standard co-
basis in $W_{0}$:

$\mathcal{A}_{3}=\\{x_{i}\partial_{j}\mid
i\in\mathbf{I}(k,l),j\in\overline{\mathbf{I}}(k,l)\\}.$ (4.27)

######  Lemma 4.5.

Suppose $U$, $V\in\mathfrak{V}^{L}$, $L=W$ or $S$.

  * $\mathrm{(1)}$

$\mathcal{M}_{0}(V)$ is a maximal subalgebra of $L_{0}$.

  * $\mathrm{(2)}$

$\mathcal{M}_{0}(U)=\mathcal{M}_{0}(V)$ if and only if $U=V.$

###### Proof.

(1) Let $\mathfrak{G}_{0}$ be a subalgebra of $L_{0}$ which strictly contains
$\mathcal{M}_{0}(V)$. It is clear that $\mathfrak{G}_{0}$ contains a nonzero
element of form $B=\sum_{h,t\geq 1}\alpha_{ht}x_{i_{h}}\partial_{j_{t}},$
where
$0\neq\alpha_{ht}\in\mathbb{F},i_{h}\in\mathbf{I}(k,l),j_{t}\in\overline{\mathbf{I}}(k,l)$.

When $L=W$, for any $i\in\mathbf{I}(k,l)$ and
$j\in\overline{\mathbf{I}}(k,l),$ one has
$x_{i}\partial_{i_{1}}\in\mathcal{A}_{1}$ and
$x_{j_{1}}\partial_{j}\in\mathcal{A}_{2}$. Then

$x_{i}\partial_{j}=\alpha_{11}^{-1}[x_{i}\partial_{i_{1}},[B,x_{j_{1}}\partial_{j}]]\in\mathfrak{G}_{0},$

showing that the co-basis $\mathcal{A}_{3}\subset\mathfrak{G}_{0}$. Hence
$\mathfrak{G}_{0}=W_{0}$.

When $L=S$, suppose $|\mathbf{I}(k,l)|>1$ and
$|\overline{\mathbf{I}}(k,l)|>1$. Choosing $x_{j_{1}}\partial_{j}$ in
$\mathcal{C}_{2}$ with $j\in\overline{\mathbf{I}}(k,l)\backslash\\{j_{1}\\}$
and $x_{i}\partial_{i_{1}}$ in $\mathcal{C}_{1}$ with
$i\in\mathbf{I}(k,l)\backslash\\{i_{1}\\}$, we have

$x_{i}\partial_{j}=[x_{i}\partial_{i_{1}},[B,x_{j_{1}}\partial_{j}]]\in\mathfrak{G}_{0},$

showing that the co-basis $\mathcal{A}_{3}\subset\mathfrak{G}_{0}$ and then
$\mathfrak{G}_{0}=S_{0}$. For the remaining case $|\mathbf{I}(k,l)|=1$ or
$|\overline{\mathbf{I}}(k,l)|=1$, the argument is similar and much easier.

(2) One direction is obvious. Note that one may choose bases of $U$ and $V$ as
follows:

$\overbrace{E_{1},\ldots,E_{r}}^{\textrm{cobasis
in}\;U},\overbrace{F_{1},\ldots,F_{s}}^{\textrm{basis of}\;U\cap
V},\overbrace{G_{1},\ldots,G_{t}}^{\textrm{cobasis in}\;V}$

where $(E_{1},\ldots,E_{r},F_{1},\ldots,F_{s},G_{1},\ldots,G_{t})$ is a
permutation of $\partial_{i}$’s. Keeping in mind the standard co-basis (4.27),
we are done by a similar argument as in (1). ∎

######  Proposition 4.6.

$\mathcal{M}(V)$ is maximal in $L$ for any $V\in\mathfrak{V}^{L}$, $L=W$ or
$S$.

###### Proof.

Let $M$ be an MGS containing $\mathcal{M}(V)$. Then $\mathcal{M}_{i}(V)\subset
M_{i}$ for all $i\geq-1.$ In particular, because of the maximality of
$\mathcal{M}_{0}(V),$ it must be $M_{0}=\mathcal{M}_{0}(V)$ or $M_{0}=L_{0}.$

Case 1. Suppose $M_{0}=\mathcal{M}_{0}(V)$. By induction, it is routine to
verify that $M_{i}=\mathcal{M}_{i}(V)$ for all $i\geq 0.$ Assume on the
contrary that $M$ strictly contains $\mathcal{M}(V)$. Then
$M_{-1}\supsetneq\mathcal{M}_{-1}(V)=V.$ Note that
$\mathcal{M}_{0}(V)=M_{0}=\mathcal{M}_{0}(M_{-1})$ from Lemma 4.3(3). Thus,
Lemma 4.5(2) forces $M_{-1}=L_{-1}$. Pick any $i\in\mathbf{I}(k,l)$,
$j\in\overline{\mathbf{I}}(k,l)$ and any $h\neq i,j.$ We are able to check
that

$A=(-1)^{|x_{h}|}x_{i}x_{j}\partial_{j}-(-1)^{|x_{j}|}x_{i}x_{h}\partial_{h}\in
S_{1}\subset W_{1}.$

Moreover, $A\in\mathcal{M}_{1}(V)=M_{1}.$ Since $M_{-1}=W_{-1}=S_{-1}$, we
have

$x_{i}\partial_{j}=(-1)^{(|x_{h}|+|x_{i}||x_{j}|)}[\partial_{j},A]\in
M_{0}=\mathcal{M}_{0}(V).$

This contradicts the fact that $x_{i}\partial_{j}\in\mathcal{A}_{3}$ [see
(4.27)]. Therefore, $M=\mathcal{M}(V).$

Case 2. Suppose $M_{0}=L_{0}$. In this case, since $L_{-1}$ is irreducible as
$L_{0}$-module, we have $M_{-1}=L_{-1}.$ Hence $M$ is an MGS of type (I). By
Theorem 3.1(1) and (2), $M_{1}=W_{1}^{\prime}$, $W_{1}^{\prime\prime},$
$S_{1}^{\prime\prime}$, or $\\{0\\}.$ In Case 1, we have shown that
$A\in\mathcal{M}_{1}(V)$. However, it is clear that $A$ does not belong to
$W_{1}^{\prime}$, $W_{1}^{\prime\prime},$ $S^{{}^{\prime\prime}}$, or
$\\{0\\}.$ Hence $\mathcal{M}_{1}(V)\not\subset M_{1}.$ This contradicts the
assumption that $M$ is a graded subalgebra containing $\mathcal{M}(V).$ ∎

Proof of Theorem 4.1 (1), (2) and (3) are immediate consequences of Lemmas
2.2, 4.3(3) and Proposition 4.6. It remains to show the dimension formulas.
For $W$, $\mathcal{M}(V)$ has a standard $\mathbb{F}$-basis which is a
disjoint union:

$\displaystyle\\{x^{(\alpha)}x^{u}\partial_{i}\mid\alpha\in\mathbf{A}(m),u\in\mathbf{B}(n);\;i\in\mathbf{I}(k,l)\\}$
$\displaystyle\cup$
$\displaystyle\\{x^{(\alpha)}x^{u}\partial_{i}\mid\;i\in\overline{\mathbf{I}}(k,l)\;\mbox{and}\;\exists
j\in\overline{\mathbf{I}}(k,l)\;\mbox{such
that}\;\partial_{j}(x^{(\alpha)}x^{u})\neq 0\\}.$

A standard and direct computation shows that:

$\mathrm{dim}\mathcal{M}(V)=2^{n}p^{m}(m+n)-2^{l}p^{k}(m+n-k-l).$

Similarly, for $S$, we have:

$\mathrm{dim}\mathcal{M}(V)=2^{n}p^{m}(m+n-1)+1-2^{l}p^{k}(m+n-k-l).$

∎

Next, we consider the case $L=H$ or $K$. In this case, we shall frequently use
the standard facts mentioned in Remark 2.3 without notice. Suppose
$V\in\mathfrak{V}^{L}$ with $\beta$-$\dim V=(a,b,c,d)$. In order to prove
Theorem 4.2 we list the following assertions. For simplicity, we write
$\lambda_{i,j}$ for a nonzero element in $\mathbb{F}$, where
$i,j\in\mathbf{I}$. Recall definitions (2.18)-(2.21). Put

$\displaystyle\mathcal{V}_{\mathfrak{i}}^{L}=\left\\{V\in\mathfrak{V}^{L}\mid
V\mbox{ is isotropic and }J_{3}\mbox{ is not twinned}\right\\};$
$\displaystyle\mathcal{V}_{\mathfrak{n}}^{L}=\left\\{V\in\mathfrak{V}^{L}\mid
V\mbox{ is nondegenerate and }J_{i}\mbox{ is not twinned, }i=1,3\right\\};$
$\displaystyle\mathcal{V}_{\mathfrak{d}}^{L}=\left\\{V\in\mathfrak{V}^{L}\mid
V\mbox{ is degenerate},J_{3}\mbox{ is empty and }J_{1}\mbox{ is not
twinned}\right\\}.$ (4.28)

######  Lemma 4.7.

For $H$, put

$A_{i}=\mathrm{span}_{\mathbb{F}}\\{u\in H_{i}\mid\nu(u)=0,1\mbox{ or
}(\frac{1}{3})^{k}2^{l},\ k,l\in\mathbb{N},\ l>1\\}.$

Then

  * $(1)$

$A_{i}=\mathcal{M}_{i}(V)$, $i\geq-1$.

  * $(2)$

The subalgebra $A_{0}=\mathcal{M}_{0}(V)$ is maximal in $H_{0}$ if and only if

$V\in\mathcal{V}_{\mathfrak{i}}^{H}\cup\mathcal{V}_{\mathfrak{n}}^{H}\cup\mathcal{V}_{\mathfrak{d}}^{H}.$

###### Proof.

(1) It follows by using induction on $i$, $i\geq-1$.

(2) Obviously, the torus $T$ mentioned in Remark 2.3(2) is contained in
$\mathcal{M}_{0}(V)$. For any $h\in H_{0}$ and $h\not\in\mathcal{M}_{0}(V)$,
put $\overline{M}=\mathrm{alg}(\mathcal{M}_{0}(V)+\mathbb{F}h)$. Firstly, we
show the maximality of $\mathcal{M}_{0}(V).$ It suffices to prove
$H_{0}=\overline{M}$.

Case 1. $V\in\mathcal{V}_{\mathfrak{i}}^{H}$. Notice that

$\nu(y_{i})=\left\\{\begin{array}[]{llll}0&i\in J_{2};\\\
\frac{1}{3}&i\in\bar{J}_{2};\\\ 2&i\in J_{3}\end{array}\right.\ \ \mbox{ and
}\ \ \mathcal{M}_{0}(V)=\mathrm{span}_{\mathbb{F}}\\{y_{i}y_{j}\mid(i,j)\in
J_{2}\times\mathbf{I}\cup J_{3}\times J_{3}\\}.$

We may assume that $h$ is a monomial with $\nu(h)=\frac{1}{9}$ or
$\frac{2}{3}$. When $h=y_{i}y_{j}$, $(i,j)\in\bar{J}_{2}\times\bar{J}_{2}$, we
have:

$\displaystyle
y_{k}y_{l}=\lambda_{k,l}[[y_{i}y_{j},y_{\widetilde{j}}y_{k}],y_{\widetilde{i}}y_{l}]\in\overline{M}\
\mbox{ for all }\ (k,l)\in\bar{J}_{2}\times\bar{J}_{2},$ $\displaystyle
y_{k}y_{s}=\lambda_{k,s}[y_{k}y_{l},y_{\widetilde{l}}y_{s}]\in\overline{M}\
\mbox{ for all }\ s\in{J}_{3}.$

Thus, $H_{0}=\overline{M}$. When $h=y_{i}y_{j}$,
$(i,j)\in\bar{J}_{2}\times{J}_{3}$, if $j\in I_{03}$ or $I_{03}$ is not empty,
we get $H_{0}=\overline{M}$ in an analogous way as above. Otherwise, we may
assume that $I_{03}$ is empty. If $J_{3}$ is single, we have:

$\displaystyle
y_{k}y_{m+n-d}=\lambda_{k,m+n-d}[y_{i}y_{m+n-d},y_{\widetilde{i}}y_{k}]\in\overline{M}\
\mbox{ for all }\ k\in\bar{J}_{2},$ $\displaystyle
y_{k}y_{l}=\lambda_{k,l}[y_{k}y_{m+n-d},y_{m+n-d}y_{l}]\in\overline{M}\ \mbox{
for all }\ l\in\bar{J}_{2}.$

It follows that $H_{0}=\overline{M}$. If $J_{3}$ is neither single nor
twinned, for $s\in J_{13}$, $s\not=\widetilde{j}$, we have:

$\displaystyle
y_{k}y_{s}=\lambda_{k,s}[[y_{i}y_{j},y_{\widetilde{i}}y_{k}],y_{\widetilde{j}}y_{s}]\in\overline{M}\
\mbox{ for all }\ k\in\bar{J}_{2},$ $\displaystyle
y_{k}y_{\widetilde{j}}=\lambda_{k,\widetilde{j}}[y_{k}y_{s},y_{\widetilde{s}}y_{\widetilde{j}}]\in\overline{M}\
\ s\not=j\mbox{ and }\widetilde{j},$ $\displaystyle
y_{k}y_{l}=\lambda_{k,l}[y_{j}y_{k},y_{\widetilde{j}}y_{l}]\in\overline{M}\
\mbox{ for all }l\in\bar{J}_{2}.$

Thus, $H_{0}=\overline{M}$.

Case 2. $V\in\mathcal{V}_{\mathfrak{n}}^{H}$. Notice that

$\nu(y_{i})=\left\\{\begin{array}[]{llll}1&i\in J_{1};\\\ 2&i\in
J_{3}\end{array}\right.\ \ \mbox{ and }\ \
\mathcal{M}_{0}(V)=\mathrm{span}_{\mathbb{F}}\\{y_{i}y_{j}\mid(i,j)\in
J_{1}\times J_{1}\cup J_{3}\times J_{3}\\}.$

We may assume that $h$ is a linear combination of monomials with value 2. When
$h=y_{i}y_{j}$, $(i,j)\in(I_{01}\cup\bar{I}_{01})\times{J}_{3}$, using the
same method as in Case 1, we get $H_{0}=\overline{M}$. When $h=\sum_{i\in
I_{11}}a_{i}y_{k}y_{i}$, where $k\in J_{3}$, $a_{i}\in\mathbb{F}$,
$a_{j}\not=0$, we get $H_{0}=\overline{M}$ if $I_{01}$ is not empty or $J_{1}$
is single by a similar argument as in Case 1. Thus, it suffices to consider
the condition that $I_{01}$ is empty and $J_{1}$ is neither single nor
twinned. For distinct $l,s,j\in I_{11}$, we have

$\displaystyle
y_{k}y_{s}=(a_{j})^{-1}\lambda_{k,s}[y_{l}y_{s},[y_{j}y_{l},h]]\in\overline{M},$
$\displaystyle y_{e}y_{k}=\lambda_{e,k}[y_{k}y_{s},y_{s}y_{e}]\in\overline{M}\
\mbox{ for any }\ s\not=e\in I_{11}.$

For any $i\in I_{11}$, $f\in I_{03}$ and $t\in I_{13}$, we have

$y_{f}y_{i}=\lambda_{f,i}[y_{k}y_{i},y_{\widetilde{k}}y_{f}]\in\overline{M}\
\mbox{ and }\
y_{t}y_{i}=\lambda_{t,i}[y_{f}y_{i},y_{\widetilde{f}}y_{t}]\in\overline{M}.$

Thus, $H_{0}=\overline{M}.$

Case 3. $V\in\mathcal{V}_{\mathfrak{d}}^{H}$. Notice that

$\nu(y_{i})=\left\\{\begin{array}[]{llll}1&i\in J_{1};\\\ 0&i\in J_{2};\\\
\frac{1}{3}&i\in\bar{J}_{2}\end{array}\right.\ \ \mbox{ and }\ \
\mathcal{M}_{0}(V)=\mathrm{span}_{\mathbb{F}}\\{y_{i}y_{j}\mid(i,j)\in{J}_{2}\times\mathbf{I}\cup
J_{1}\times{J}_{1}\\}.$

We have $H_{0}=\overline{M}$ by the same method as in Cases 1 and 2.

Conversely, we consider the co-basis of $\mathcal{M}_{0}(V)$ in $H_{0}$:

$\\{y_{i}y_{j}\mid(i,j)\in{J}_{1}\times\bar{J}_{2}\cup
J_{1}\times{J}_{3}\cup\bar{J}_{2}\times\bar{J}_{2}\cup\bar{J}_{2}\times{J}_{3}\\}.$
(4.29)

Notice that if
$V\not\in\mathcal{V}_{\mathfrak{i}}^{H}\cup\mathcal{V}_{\mathfrak{n}}^{H}\cup\mathcal{V}_{\mathfrak{d}}^{H}$
, then $V\in\mathfrak{V}^{H}$ must satisfy one of the following conditions:

(i) None of $J_{1},J_{2},J_{3}$ is empty. In this case, we choose a monomial
$h$ of $H_{0}$ with $\nu(h)=2$. Then there do not exist monomials with value
$\frac{1}{3}$ in $\overline{M}$.

(ii) $J_{3}$ is twinned, i.e., $J_{3}=\\{j,{\widetilde{j}}\\}$, where
$j\not=\widetilde{j}\in\mathbf{I}_{1}$. In this case, let $h=y_{i}y_{j}$,
where

$i\in\left\\{\begin{array}[]{ll}\bar{J}_{2},&J_{1}\ \mbox{ is empty};\\\
{J}_{1},&\ \mbox{ otherwise}.\end{array}\right.$

Then $y_{i}y_{\widetilde{j}}\not\in\overline{M}$.

(iii) $J_{1}$ is twinned, i.e., $J_{1}=\\{m+n-1,m+n\\}$. In this case, let
$h=y_{k}(y_{m+n-1}+\sqrt{-1}y_{m+n})$, where

$k\in\left\\{\begin{array}[]{ll}\bar{J}_{2},&J_{3}\ \mbox{ is empty};\\\
{J}_{3},&\ \mbox{ otherwise}.\end{array}\right.$

Then $y_{k}y_{m+n}\not\in\overline{M}$.

Therefore, $\overline{M}$ is a nontrivial subalgebra of $H_{0}$ strictly
containing $\mathcal{M}_{0}(V)$ when (i), (ii) or (iii) holds, which implies
that $\mathcal{M}_{0}(V)$ is not a maximal subalgebra of $H_{0}$. ∎

######  Proposition 4.8.

The subalgebra $\mathcal{M}(V)$ is maximal in $H$ if and only if
$V\in\mathcal{V}^{H}$.

###### Proof.

If $J_{3}=\\{m+n-d\\}$, from Lemma 4.7(1), we know that

$\mathcal{M}_{i}(V)=\mathrm{span}_{\mathbb{F}}\\{u\in H_{i}\mid\nu(u)=1,0\\},\
i\geq-1,$

which implies that

$\mathrm{alg}(\mathcal{M}(V)+\mathbb{F}y_{m+n-d})\subset\mathrm{span}_{\mathbb{F}}\\{u\in
H\mid\nu(u)=1,0\\}+\mathbb{F}y_{m+n-d}.$

Thus,
$y_{i},y_{j}y_{m+n-d}\not\in\mathrm{alg}(\mathcal{M}(V)+\mathbb{F}y_{m+n-d})$
if $\nu(y_{i})={\frac{1}{3}}$ or $\nu(y_{j})=1,$ which contradicts the
maximality of $\mathcal{M}(V)$.

If $J_{3}=\\{j,{\widetilde{j}}\\}$, where
$j\not=\widetilde{j}\in\mathbf{I}_{1}$, from Lemma 4.7(1), we know that

$\mathcal{M}_{i}(V)=\mathrm{span}_{\mathbb{F}}\\{u\in
H_{i}\mid\nu(u)=1,0,(\frac{1}{3})^{k}4\\},\ i\geq 0.$

Then for any monomial $u\in\mathcal{M}_{i}(V)$, we have:

$[y_{j},u]=0\ \ \mbox{ or }\ \ [y_{j},u]=wy_{j},$

where $0\not=w\in H_{i-1}$ with $D_{\widetilde{j}}(w)=0$, which implies that

$y_{\widetilde{j}}\not\in\mathrm{alg}(\mathcal{M}(V)+\mathbb{F}y_{j}).$

This contradicts the maximality of $\mathcal{M}(V)$.

Conversely, let us prove the maximality of $\mathcal{M}(V)$. By definition
(4.26), it is sufficient to show that
$\overline{M}=\mathrm{alg}(\mathcal{M}(V)+\mathbb{F}h)=H$, where $h=y_{i}$,
$i\in\bar{J}_{2}\cup J_{3}$. Note that $\mathcal{M}_{1}(V)\not=0$ for
$|\mathbf{I}_{0}|\geq 2$. From Lemmas 2.1(2) and 3.11(1), it suffices to prove
$H_{-1},H_{0}\subset\overline{M}$. For $V\in\mathcal{V}^{H}$, we discuss the
following cases:

Case 1. $J_{2}$ is not empty. When $i\in\bar{J}_{2}$, since

$y_{\widetilde{i}}\in V\ \mbox{ and }\
y_{j}=\lambda_{i,j}[y_{i},y_{\widetilde{i}}y_{j}]\in\overline{M}\ \mbox{ for
}\ \widetilde{i}\not=j\in\mathbf{I},$

we have $H_{-1}\subset\overline{M}$. When $i\in{J}_{3}$, for all $j\in J_{3}$
with $j\not=i,\widetilde{i}$, we have:

$y_{j}=\lambda_{i,j}[y_{i},y_{\widetilde{i}}y_{j}]\ \ \mbox{ and }\ \
y_{\widetilde{i}}=\lambda_{\widetilde{i},j}[y_{j},y_{\widetilde{i}}y_{\widetilde{j}}].$

Note that

$y_{l}=\lambda_{l,i}[y_{i},[y_{\widetilde{i}},y_{i}y_{\widetilde{i}}y_{l}]]\in\overline{M}\
\mbox{ for all }\ l\in\bar{J}_{2}.$

Thus we have $H_{-1}\subset\overline{M}$. Note that for an arbitrary monomial
$u\in H_{0}$, there exists $k\in\mathbf{I}$ such that $uy_{k}\not=0$ and
$\nu(uy_{k})=0$. Then we have

$u=\lambda_{\widetilde{k},k}[y_{\widetilde{{k}}},uy_{k}]\in\overline{M},$

which implies that $H_{0}\subset\overline{M}$. Thus, we have $\overline{M}=H$.

Case 2. $J_{2}$ is empty. Obviously, $J_{1}$ and $J_{3}$ are not empty. Then
we have $H_{-1},H_{0}\subset\overline{M}$ by the same method as in Case 1. ∎

To avoid confusion, we rewrite $\mathcal{M}^{L}_{i}(V)$ for
$\mathcal{M}_{i}(V)$, $\mathcal{M}^{L}(V)$ for $\mathcal{M}(V)$, $L=H$ or $K$.

######  Lemma 4.9.

Let $\gamma=\lfloor\frac{i+2}{2}\rfloor$ for $i>0$. Put

$\widetilde{\mathcal{M}}_{j}(V)=\left\\{\begin{array}[]{ll}0,&j>\eta-2;\\\
\mathcal{M}^{H}_{j}(V),&j<\eta-2;\end{array}\right.\
\widetilde{\mathcal{M}}_{\eta-2}(V)=\left\\{\begin{array}[]{ll}0,&V\mbox{ is
nondegenerate}\\\ &\mbox{and}\ J_{3}\ \mbox{is single};\\\
\mathbb{F}y^{(\pi)}y^{\omega},&\mbox{otherwise, }\end{array}\right.$

where $\pi=(p-1,\ldots,p-1)\in\mathbb{N}^{2r}$, $\eta=2r(p-1)+n$ and
$\omega=\langle m+1,\ldots,m+n\rangle$. Then

  * $\mathrm{(1)}$

$\mathcal{M}^{K}_{0}(V)=\mathcal{M}^{H}_{0}(V)\oplus\mathbb{F}z$.

  * $\mathrm{(2)}$

If $V$ is not isotropic, for $i>0$,

$\mathcal{M}^{K}_{i}(V)=\widetilde{\mathcal{M}}_{i}(V)\oplus\widetilde{\mathcal{M}}_{i-2}(V)z\oplus\cdots\oplus\widetilde{\mathcal{M}}_{i-2\gamma}(V)z^{\gamma}.$

  * $\mathrm{(3)}$

If $V$ is isotropic, for $i>0$,

$\mathcal{M}^{K}_{i}(V)=\widetilde{\mathcal{M}}_{i}(V)\oplus\overline{H}_{i-2}z\oplus\cdots\oplus\overline{H}_{i-2\gamma}z^{\gamma}.$

###### Proof.

(1) It is obvious.

(2) Use induction on $i$. Clearly,
$\widetilde{\mathcal{M}}_{i}(V)\subset\mathcal{M}^{K}_{i}(V)$.

$``\supset"$: For $gz^{k}\in\widetilde{\mathcal{M}}_{i-2k}(V)z^{k}$,
$0<k\leq\gamma$, we know that

$[y_{l},gz^{k}]=[y_{l},g]z^{k}+y_{l}gz^{k-1}.$

Note that

$y_{l}g\in\widetilde{\mathcal{M}}_{i-2(k-1)-1}(V)\ \mbox{ for
}\nu(y_{l})=1,0.$

By induction on $\mathrm{zd}(g)$, we have:

$y_{l}gz^{k-1},\ [y_{l},g]z^{k}\in\mathcal{M}^{K}_{i-1}(V).$

Thus, $gz^{k}\in\mathcal{M}^{K}_{i}(V)$.

$``\subset"$: For any $u\in\mathcal{M}^{K}_{i}(V)$, by Lemma 3.11(2), we may
assume that

$u=u_{i}+u_{i-2}z+\cdots+u_{i-2\gamma}z^{\gamma},$

where $u_{j}\in\overline{H}_{j}$ for $i-2\gamma\leq j\leq i$. Note that
$\mathcal{M}^{K}_{-2}(V)\not=0$, since $V$ is not isotropic. Then we have:

$u_{i-2}+u_{i-4}z+\cdots+u_{i-2\gamma}z^{\gamma-1}=2^{-1}[1,u]\in\mathcal{M}^{K}_{i-2}(V).$

By induction, we have $u_{j}\in\widetilde{\mathcal{M}}_{j}(V)$ for
$i-2\gamma\leq j\leq i-2$. Moreover,

$u_{i-2}z+u_{i-4}z^{2}+\cdots+u_{i-2\gamma}z^{\gamma}\in\mathcal{M}^{K}_{i}(V).$

Consequently, $u_{i}\in\widetilde{\mathcal{M}}_{i}(V)$.

(3) When $V$ is isotropic, note that $\nu(y_{k})=0$ for all $y_{k}\in V$. The
remaining discussion is analogous to that of the condition (2). ∎

######  Proposition 4.10.

The subalgebra $\mathcal{M}^{K}(V)$ is maximal in $K$ if and only if
$V\in\mathcal{V}^{K}$ when $V$ is neither nondegenerate nor isotropic;
$V\in\mathcal{W}^{K}$, otherwise.

###### Proof.

The proof of the necessity is similar to the one of Proposition 4.8. We only
consider the sufficiency. For any $u\in K$, $u\not\in\mathcal{M}^{K}(V)$, put
$\overline{M}=\mathrm{alg}(\mathcal{M}^{K}(V)+\mathbb{F}u)$. Then there exist
$v_{1},\ldots,v_{i}\in V$ such that

$0\not=h=[v_{1},[\cdots,[v_{i},u]\cdots]]\in\overline{M}\cap K_{-1}\ \mbox{
and }\ h\not\in V.$

When $J_{3}$ is neither single nor twinned, by Proposition 4.8, we have
$H\subset\overline{M}.$ When $J_{3}$ is single and $V$ is isotropic, we may
assume that $h=y_{i}$, $i\in\bar{J}_{2}\cup\\{m+n-d\\}$. If $i\in\bar{J}_{2}$,
for any $j,k\in\bar{J}_{2}$, we obtain that

$y_{m+n-d}=\lambda_{i,\widetilde{i}}[y_{\widetilde{i}},[y_{i},y_{m+n-d}z]],\ \
y_{m+n-d}y_{j}y_{k}=\lambda_{j,k}[y_{m+n-d},y_{j}y_{k}z]$

are in $\overline{M}$ from Lemma 4.9(3). Moreover,

$y_{j}y_{k}=-[y_{m+n-d},y_{m+n-d}y_{j}y_{k}],\ \
y_{m+n-d}y_{k}=\lambda_{m+n-d,k}[y_{\widetilde{j}},y_{m+n-d}y_{j}y_{k}]$

are in $\overline{M}$. Keeping in mind the co-basis (4.29), we have
$H_{0}\subset\overline{M}$, which also holds when $J_{3}$ is single and $V$ is
nondegenerate. From the irreducibility of $K_{-1}$, $K_{10}$ and $K_{11}$, as
well as Lemma 4.9, we obtain that $\mathcal{M}^{K}(V)$ is maximal in $K$. ∎

In the same way as in Proposition 4.10, one may check the following
proposition.

######  Proposition 4.11.

If $V$ is isotropic, the subalgebra $\mathcal{M}^{K}(1,V)$ is maximal in $K$
if and only if $V\in\mathcal{V}^{K}$.

######  Convention 4.12.

For simplicity, put
$\mathcal{O}_{X}=\mathrm{span}_{\mathbb{F}}\\{y_{i_{1}}\cdots y_{i_{s}}\mid
i_{1},\ldots,i_{s}\in X,s\geq 1\\}$,
$\mathcal{Q}_{X}=\mathrm{span}_{\mathbb{F}}\\{y_{i}\mid i\in X\\}$ and
$\mathcal{Y}^{+}_{X}=\mathcal{Y}_{X}\oplus\mathbb{F}\cdot 1$, where $X$ is a
subset of $\mathbf{I}$ and $\mathcal{Y}=\mathcal{O}$ or $\mathcal{Q}$.

Proof of Theorem 4.2 For (1) and $(1^{\prime})$, the proofs follow from Lemma
4.3(3), Propositions 4.8, 4.10 and 4.11.

For $H$, from Lemma 4.7(1) and (1), we obtain that

$\dim{\mathcal{M}^{H}(V)}=\dim
H-\dim(\mathcal{O}^{+}_{J_{1}}\mathcal{O}_{\bar{J}_{2}}\oplus\mathcal{O}^{+}_{J_{1}\cup\bar{J}_{2}}\mathcal{Q}_{{J}_{3}}).$

For $K$, from Lemma 4.9 and $(1^{\prime})$, we obtain that

$\displaystyle\dim{\mathcal{M}^{K}(1,V)}=p(\dim\mathcal{M}^{H}(V)+2);$
$\displaystyle\dim{\mathcal{M}^{K}(V)}=\left\\{\begin{array}[]{ll}\dim\mathcal{M}^{H}(V)+1+(p-1)(\dim
H+2),&V\in\mathcal{W}^{K}\mbox{ is isotropic};\\\
p(\dim\mathcal{M}^{H}(V)+2),&\mbox{ otherwise}.\end{array}\right.$

By a standard and direct computation we get the formulas (4) and
$(4^{\prime})$. Noting that
$\dim\mathcal{M}_{0}(V)=\dim\mathcal{M}_{0}(V^{\prime})$ if
$\mathcal{M}(V)\cong$$\mathcal{M}(V^{\prime})$ and using the same method as in
Theorem 4.1(2), (2) and $(2^{\prime})$ hold. From (1), $(1^{\prime})$ and (2),
$(2^{\prime})$, we obtain that (3) and $(3^{\prime})$ hold.∎

##  5 MGS of Type (III)

Suppose $L=W,S,H$ or $K$. Recall that an MGS of type (III) of $L$, $M$,
satisfies the condition

$M_{-1}=L_{-1}\quad\mbox{and}\quad M_{0}\neq L_{0}.$

Let $\mathfrak{G}_{0}$ be a nontrivial subalgebra of $L_{0}$. Define a graded
subspace of $L$ as follows:

$\mathcal{M}(L_{-1},\mathfrak{G}_{0})=\oplus_{i\geq-2}\mathcal{M}_{i}(L_{-1},\mathfrak{G}_{0}),$

where

$\displaystyle\mathcal{M}_{-i}(L_{-1},\mathfrak{G}_{0})=L_{-i},\
i<0;\quad\mathcal{M}_{0}(L_{-1},\mathfrak{G}_{0})=\mathfrak{G}_{0};$
$\displaystyle\mathcal{M}_{i}(L_{-1},\mathfrak{G}_{0})=\\{u\in
L_{i}\mid[L_{-1},u]\subset\mathcal{M}_{i-1}(L_{-1},\mathfrak{G}_{0})\\}\
\mbox{ for }i>0.$ (5.30)

It is easy to see that $\mathcal{M}(L_{-1},\mathfrak{G}_{0})$ is a graded
subalgebra satisfying the condition (III). We call $\mathfrak{G}$ a maximal
R-subalgebra (resp. maximal S-subalgebra) of $L$ if $\mathfrak{G}$ is maximal
reducible (resp. irreducible) graded and satisfies the condition (III). All
the MGS of type (III) can be split into the disjoint union of maximal
R-subalgebras and maximal S-subalgebras.

######  Theorem 5.1.

Suppose $L=W$ or $S$.

  * $\mathrm{(1)}$

All maximal R-subalgebras of $L$ are precisely:

$\\{\mathcal{M}(L_{-1},\mathcal{M}_{0}(V))\mid V\in\mathfrak{V}^{L}\\}.$

  * $\mathrm{(2)}$

For any $V,V^{\prime}\in\mathfrak{V}^{L}$,

$\mathcal{M}(L_{-1},\mathcal{M}_{0}(V))\stackrel{{\scriptstyle\mathrm{algebra}}}{{\cong}}\mathcal{M}(L_{-1},\mathcal{M}_{0}(V^{\prime}))\Longleftrightarrow
V{\cong}V^{\prime}.$

  * $\mathrm{(3)}$

$L$ has exactly $(m+1)(n+1)-2$ isomorphism classes of maximal R-subalgebras.

  * $\mathrm{(4)}$

Suppose $V\in\mathfrak{V}^{L}$ with $\mathrm{superdim}V=(k,l),$ then

$\displaystyle\mathrm{dim}\mathcal{M}(L_{-1},\mathcal{M}_{0}(V))$
$\displaystyle=$
$\displaystyle\left\\{\begin{array}[]{ll}2^{n-l}p^{m-k}(m+n-k-l)+2^{n}p^{m}(k+l),&L=W\\\
2^{n-l}p^{m-k}(m+n)+2^{n}p^{m}(k+l-1)-k-1,&L=S.\end{array}\right.$

Let $L=H$ or $K$ and put $V^{\bot}=\\{u\in L_{-1}\mid\beta(u,V)=0\\}$ for
$V\in\mathfrak{V}^{L}$. Recall definitions (2.18)–(2.20) and (4.28).

######  Theorem 5.2.

All maximal R-subalgebras of $H$ and $K$ are characterized as follows:

For $H$,

  * $(1)$

All maximal R-subalgebras of $H$ are precisely:

$\left\\{\mathcal{M}(H_{-1},\mathcal{M}_{0}(V))\mid
V\in\mathcal{V}_{\mathfrak{n}}^{H}\cup\mathcal{V}_{\mathfrak{i}}^{H}\right\\}.$

  * $(2)$

Suppose
$V,V^{\prime}\in\mathcal{V}_{\mathfrak{n}}^{H}\cup\mathcal{V}_{\mathfrak{i}}^{H}$.
Then

$\mathcal{M}(H_{-1},\mathcal{M}_{0}(V))\stackrel{{\scriptstyle\mathrm{algebra}}}{{\cong}}\mathcal{M}(H_{-1},\mathcal{M}_{0}(V^{\prime}))$

if and only if one of the following conditions holds.

    * (i)

$V{\cong}V^{\prime}.$

    * (ii)

$V^{\bot}{\cong}V^{\prime}$ when $V$ and $V^{\prime}$ are both nondegenerate.

  * $(3)$

$H$ has exactly $\phi(r,n)$ isomorphism classes of maximal R-subalgebras,
where

$\phi(r,n)=\left\\{\begin{array}[]{ll}2^{-1}(nr+3n+2r-2)+\lfloor\frac{r}{2}\rfloor(n+1),&n\
\mbox{ is even};\\\ 2^{-1}(nr+3n+r-1)+\lfloor\frac{r}{2}\rfloor(n+1),&n\
\mbox{ is odd}.\end{array}\right.$

  * $(4)$

Suppose $V\in\mathcal{V}_{\mathfrak{n}}^{H}\cup\mathcal{V}_{\mathfrak{i}}^{H}$
with $\beta$-$\mathrm{dim}V=(a,b,c,d),$ then

$\dim\mathcal{M}(H_{-1},\mathcal{M}(V))=\left\\{\begin{array}[]{ll}p^{2a}2^{d}+p^{2(r-a)}2^{(n-d)}-2,&V\in\mathcal{V}_{\mathfrak{n}}^{H};\\\
p^{m-b}2^{n-c}+(b+c)p^{b}2^{c}-1,&V\in\mathcal{V}_{\mathfrak{i}}^{H}.\end{array}\right.$

For $K$,

  * $(1^{\prime})$

All maxima R-subalgebras of $K$ are precisely:

$\left\\{\mathcal{M}(K_{-1},\mathcal{M}_{0}(V))\mid
V\in\mathcal{V}_{\mathfrak{i}}^{K}\right\\}.$

  * $(2^{\prime})$

Suppose $V,V^{\prime}\in\mathcal{V}_{\mathfrak{i}}^{K}$. Then

$\mathcal{M}(K_{-1},\mathcal{M}_{0}(V))\stackrel{{\scriptstyle\mathrm{algebra}}}{{\cong}}\mathcal{M}(K_{-1},\mathcal{M}_{0}(V^{\prime}))\Longleftrightarrow
V{\cong}V^{\prime}.$

  * $(3^{\prime})$

$K$ has exactly $\phi(r,n)$ isomorphism classes of maximal R-subalgebras,
where

$\phi(r,n)=\left\\{\begin{array}[]{ll}2^{-1}(rn+n+2r-2),&n\ \mbox{ is
even};\\\ 2^{-1}(rn+n+r-1),&n\ \mbox{ is odd}.\end{array}\right.$

  * $(4^{\prime})$

Suppose $V\in\mathcal{V}_{\mathfrak{i}}^{K}$ with
$\beta$-$\mathrm{dim}V=(0,b,c,0),$ then

$\displaystyle\dim\mathcal{M}(K_{-1},\mathcal{M}_{0}(V))=\left\\{\begin{array}[]{ll}p^{b+1}2^{c}(b+c+1),&J_{3}\
\mbox{ is empty };\\\ p^{b}2^{c}(p^{m-2b}2^{n-2c}+b+c),&\mbox{
otherwise}.\end{array}\right.$

Unfortunately, for maximal S-subalgebras, we have not obtained a similar
description as for the maximal graded subalgebras of type $(\mathrm{I})$ or
$(\mathrm{II})$ as well as for the maximal R-subalgebras. However, the
classification of maximal S-subalgebras of $L$ can be reduced to that of the
maximal irreducible subalgebras of the classical Lie superalgebras (see Lemma
2.1(3)).

######  Theorem 5.3.

Suppose $L=W,S,H$ or $K$. All maximal S-subalgebra of $L$ are characterized as
follows:

Every maximal S-subalgebra of $L$ is of the form
$\mathcal{M}(L_{-1},\mathfrak{G}_{0}),$ where $\mathfrak{G}_{0}$ is a maximal
irreducible subalgebra of $L_{0}$.

Suppose $\mathfrak{G}_{0}$ is a maximal irreducible subalgebra of $L_{0}$.

  * $\mathrm{(a)}$

For $L=W$, $\mathcal{M}(W_{-1},\mathfrak{G}_{0})$ is maximal in $W$ if and
only if

$\mathrm{div}(\mathcal{M}_{1}(W_{-1},\mathfrak{G}_{0}))\neq 0.$

  * $\mathrm{(b)}$

For $L=S$ or $H$, $\mathcal{M}(L_{-1},\mathfrak{G}_{0})$ is maximal in $L$ if
and only if

$\mathcal{M}_{1}(L_{-1},\mathfrak{G}_{0})\neq 0.$

  * $\mathrm{(c)}$

For $L=K$, $\mathcal{M}(K_{-1},\mathfrak{G}_{0})$ is a maximal in $K$ if and
only if there exists $u\in\mathcal{M}_{1}(K_{-1},\mathfrak{G}_{0})$ satisfying

$[1,u]\not=0.$

Let $L=W,S,H$ or $K$. As in the case of modular Lie algebras [16], it is easy
to show the following lemmas.

######  Lemma 5.4.

Let $\mathfrak{G}_{0}$ be a nontrivial subalgebra of $L_{0}$. If $\Phi$ is a
$\mathbb{Z}$-homogeneous automorphism of $L$. Then

$\Phi(\mathcal{M}_{i}(L_{-1},\mathfrak{G}_{0}))=\mathcal{M}_{i}(L_{-1},\Phi(\mathfrak{G}_{0}))\
\mbox{ for all }i\geq-2.$

Moreover,

$\Phi(\mathcal{M}(L_{-1},\mathfrak{G}_{0}))=\mathcal{M}(L_{-1},\Phi(\mathfrak{G}_{0})).$

######  Lemma 5.5.

Let $M=L_{-1}+M_{0}+M_{1}+M_{2}+\cdots$ be any MGS of $L$. Then $M_{0}$ is
maximal in $L_{0}$ unless $M_{0}=L_{0}$.

######  Lemma 5.6.

If $M$ is an MGS of type $\mathrm{(III)}$ of $L$ then $M_{0}$ is maximal in
$L_{0}$ and $M=\mathcal{M}(L_{-1},M_{0})$.

######  Lemma 5.7.

If $\mathfrak{G}_{0}$ is a maximal reducible subalgebra of $L_{0}$ then there
exists a $V\in\mathfrak{V}^{L}$ such that
$\mathfrak{G}_{0}=\mathcal{M}_{0}(V)$ and
$\mathcal{M}_{i}(L_{-1},\mathfrak{G}_{0})\subset\mathcal{M}_{i}(V)$ for $i\geq
0$. Conversely, $\mathcal{M}_{0}(V)$ is a reducible maximal subalgebra of
$L_{0}$ if $V\in\mathfrak{V}^{L}$ when $L=W$ or $S$; if
$V\in\mathcal{V}_{\mathfrak{n}}^{L}\cup\mathcal{V}_{\mathfrak{i}}^{L}\cup\mathcal{V}_{\mathfrak{d}}^{L}$
when $L=H$ or $K$.

###### Proof.

Since $\mathfrak{G}_{0}$ is reducible, $L_{-1}$ has a nontrivial
$\mathfrak{G}_{0}$-submodule $V$. From definition (4.26) and the maximality of
$\mathfrak{G}_{0}$, we have $\mathfrak{G}_{0}=\mathcal{M}_{0}(V)$. From
definitions (4.26) and (5.30), we have
$\mathcal{M}_{i}(L_{-1},\mathfrak{G}_{0})\subset\mathcal{M}_{i}(V)$ for $i\geq
0$. The second statement follows immediately from Lemmas 4.5(1) and 4.7(2). ∎

######  Remark 5.8.

In view of Lemmas 2.2, 5.4 and 5.7, if $\mathfrak{G}_{0}$ is a maximal
reducible subalgebra of $L_{0}$, we may assume that $V$ is a standard element
in $\mathfrak{V}^{L}$ [see (2.17), 2.20)] such that
$\mathfrak{G}_{0}=\mathcal{M}_{0}(V)$.

######  Proposition 5.9.

Suppose $L=W$ or $S$. $\mathcal{M}(L_{-1},\mathfrak{G}_{0})$ is a maximal
R-subalgebra, if $\mathfrak{G}_{0}$ is a maximal reducible subalgebra of
$L_{0}$.

###### Proof.

Let us show that $M=\mathcal{M}(L_{-1},\mathfrak{G}_{0})$ is maximal. Assume
that $\overline{M}$ is a maximal graded subalgebra containing $M$. Clearly,
$\overline{M}_{-1}=L_{-1}$. Since $\mathfrak{G}_{0}$ is a maximal subalgebra
of $L_{0}$, we have $\overline{M}_{0}=\mathfrak{G}_{0}$ or $L_{0}$. If
$\overline{M}_{0}=\mathfrak{G}_{0}$ then $\overline{M}$ is an MGS of type
(III). By Lemma 5.6,

$\overline{M}=\mathcal{M}(L_{-1},\mathfrak{G}_{0})=M$

and we are done. Let us consider the remaining case; $\overline{M}_{0}=L_{0}$.
Clearly, $\overline{M}$ is an MGS of type (I) and by Theorem 3.1,

$M_{1}\subset\overline{M}_{1}=W_{1}^{\prime},W_{1}^{\prime\prime},S_{1}^{\prime\prime},\;\mbox{or}\;\\{0\\}.$
(5.31)

On the other hand, by Lemma 5.7, there exists a $V\in\mathfrak{V}^{L}$ such
that $\mathfrak{G}_{0}=\mathcal{M}_{0}(V)$. Assume that $V$ has a standard
basis:

$(\partial_{1},\ldots,\partial_{k}\mid\partial_{m+1},\ldots,\partial_{m+l}).$

Hence $\mathfrak{G}_{0}=\mathcal{M}_{0}(V)$ has a standard co-basis (4.27) in
$W_{0}$:

$\mathcal{A}_{3}=\\{x_{i}\partial_{j}\mid
i\in\mathbf{I}(k,l),j\in\overline{\mathbf{I}}(k,l)\\}.$

To reach a contradiction, in view of (5.31), it is sufficient to find an
element belonging to $M_{1}$ but not $W^{\prime},W^{\prime\prime}$ for $W$,
but not $S^{\prime\prime}$ or $\\{0\\}$ for $S$. For $L=W$,
$x_{j}x_{i}\partial_{i}$ with $i\in\mathbf{I}(k,l)$ and an arbitrarily chosen
$j$ is a desired element. Here we have used the fact that both
$|\mathbf{I}(k,l)|\geq 1$ and $|\overline{\mathbf{I}}(k,l)|\geq 1$, since
$V\in\mathfrak{V}^{W}$. For $L=S$, pick distinct $i,j,r$ with
$i\in\mathbf{I}(k,l)$ and with $j,r$ chosen arbitrarily. Here note that the
general assumption ensures $|\mathbf{I}|\geq 4.$ Then
$x_{j}x_{r}\partial_{i}\in S_{1}$ is a desired candidate for $S$. The proof is
complete. ∎

Proof of Theorem 5.1 (1) This follows from Lemmas 5.5, 5.6, 5.7 and
Proposition 5.9.

(2) One implication is obvious. Suppose $\Phi$ is an isomorphism of
$\mathcal{M}(L_{-1},\mathcal{M}_{0}(V))$ onto
$\mathcal{M}(L_{-1},\mathcal{M}_{0}(V^{\prime}))$. Consequently,
$\Phi(L_{-1})=L_{-1}$ and
$\Phi(\mathcal{M}_{0}(V))=\mathcal{M}_{0}(V^{\prime}).$ A standard
verification shows that $\Phi(\mathcal{M}_{0}(V))=\mathcal{M}_{0}(\Phi(V))$.
By Lemma 4.5(2), we have $\Phi(V)=V^{\prime}$.

(3) This is a direct consequence of (2).

(4) Suppose $V$ is a standard element in $\mathfrak{V}^{L}$. Then
$\mathcal{M}(W_{-1},\mathcal{M}_{0}(V))$ has a standard $\mathbb{F}$-basis

$\displaystyle\\{x^{(\alpha)}x^{u}\partial_{i}\mid\alpha\in\mathbf{A}(m),u\in\mathbf{B}(n);\;i\in\mathbf{I}(k,l)\\}$
$\displaystyle\cup$
$\displaystyle\\{x^{(\alpha)}x^{u}\partial_{i}\mid\alpha_{1}=\cdots=\alpha_{k}=0,u\subset\overline{m+l+1,m+n};\;i\in\overline{\mathbf{I}}(k,l)\\}.$

Thus, we have:

$\mathrm{dim}\mathcal{M}(W_{-1},\mathcal{M}_{0}(V))=2^{n-l}p^{m-k}(m+n-k-l)+2^{n}p^{m}(k+l).$

Note that
$\overline{S}=S\oplus\sum_{i\in\mathbf{I}_{0}}x^{(\pi-(p-1)\varepsilon_{i})}x^{\omega}\partial_{i}$,
where $\pi=(p-1,\ldots,p-1)\in\mathbb{N}^{m}$ and $\omega=\langle
m+1,\ldots,m+n\rangle$. Then we have:

$\mathrm{dim}\mathcal{M}(S_{-1},\mathcal{M}_{0}(V))=2^{n-l}p^{m-k}(m+n)+2^{n}p^{m}(k+l-1)-k-1.$

∎

We call $\mathfrak{G}_{0}=\mathcal{M}_{0}(V)$ is degenerate if
$V\in\mathcal{V}_{\mathfrak{i}}^{L}\cup\mathcal{V}_{\mathfrak{d}}^{L}$.

######  Proposition 5.10.

Let $\mathfrak{G}_{0}$ be a maximal reducible subalgebra of $H_{0}$ or
$K_{0}$.

  * $(1)$

$\mathcal{M}(H_{-1},\mathfrak{G}_{0})$ is maximal in $H$.

  * $(2)$

$\mathcal{M}(K_{-1},\mathfrak{G}_{0})$ is maximal in $K$ if and only if
${\mathfrak{G}_{0}}$ is degenerate.

###### Proof.

For any $0\neq h\in L$, $h\not\in\mathcal{M}(L_{-1},\mathfrak{G}_{0})$, put
$\overline{M}=\mathrm{alg}(\mathcal{M}(L_{-1},\mathfrak{G}_{0})+\mathbb{F}h)$.
By the maximality of $\mathfrak{G}_{0}$, we have $L_{0}\subset\overline{M}$.
For $H$, choose $k\in I_{0i}$ if $I_{0i}$ is not empty where $i=1,2$ or 3. It
follows that $y^{3}_{k}\in\mathcal{M}_{1}(H_{-1},\mathfrak{G}_{0})$. For $K$,
suppose ${\mathfrak{G}_{0}}$ is degenerate. Using the same method as for $H$,
we can find $0\not=v_{i}\in\overline{M}\cap K_{1i}$, where $i=0,1$. From
Lemmas 2.1(2) and 3.11, we have $\overline{M}=L$.

It remains to show that ${\mathfrak{G}_{0}}$ is degenerate if
$\mathcal{M}_{1}(K_{-1},\mathfrak{G}_{0})$ is maximal. Assume on the contrary
that $V_{\mathfrak{G}_{0}}\in\mathcal{V}_{\mathfrak{n}}^{K}$ is a
nondegenerate irreducible $\mathfrak{G}_{0}$-module. For any
$u\in\mathcal{M}_{1}(K_{-1},\mathfrak{G}_{0})$, by Lemmas 4.9(2) and 5.7, we
may assume that

$u=f_{-1}z+f_{1},\mbox{ where }\ f_{-1}\in V_{\mathfrak{G}_{0}}\ \mbox{ and }\
f_{1}\in\widetilde{\mathcal{M}}_{1}(V_{\mathfrak{G}_{0}}).$

Note that ${f_{-1}}$ is a linear combination of monomials with value 1. Let
$f_{1}=f^{1}+f^{4}+f^{8}$ where $f^{i}$ is a linear combination of monomials
with value $i$, $i=1,4$ or $8$. We claim that $f_{-1}=0$. Indeed, for any
$y_{i}\in K_{-1}$ with value 2, we have

$\sigma(i)(-1)^{i}(D_{\widetilde{i}}(f_{1})+D_{\widetilde{i}}(f_{-1})z)+y_{i}f_{-1}=[y_{i},u]\in\mathcal{M}_{0}(V_{\mathfrak{G}_{0}})=\mathfrak{G}_{0},$

which implies that $f_{-1}=0$ when $J_{3}$ is single. Otherwise, the following
equation holds:

$\displaystyle\sigma(i)(-1)^{i}D_{\widetilde{i}}(f^{4})=-y_{i}f_{-1}.$ (5.32)

Then there exists $g_{1}\in K_{1}$ with $D_{\widetilde{i}}(g_{1})=0$
satisfying

$\displaystyle\sigma(i)(-1)^{i}f^{4}=-y_{\widetilde{i}}y_{i}f_{-1}+g_{1}.$
(5.33)

By equations (5.32) and (5.33), we have
$D_{i}(g_{1})=(-1)^{i}2y_{\widetilde{i}}f_{-1}$ which contradicts
$D_{\widetilde{i}}(g_{1})=0$ if $f_{-1}\not=0$. Consequently,
$\mathcal{M}_{1}(K_{-1},\mathfrak{G}_{0})\subset K_{10}$. Using induction on
$k$ and the transitivity of $K$, we have
$\mathcal{M}_{k}(K_{-1},\mathfrak{G}_{0})\subset K_{k0}$ for $k>0.$ It follows
that $\mathcal{M}(K_{-1},\mathfrak{G}_{0})$ is strictly contained in
$K_{-2}+K_{-1}+K_{0}+\sum_{i=1}^{2r(p-1)+n}K_{i0}$. The latter is a maximal
graded subalgebra (see Theorem 3.1(4)). This contradicts the maximality of
$\mathcal{M}(K_{-1},\mathfrak{G}_{0})$. The proof is complete. ∎

######  Lemma 5.11.

Suppose $L=H$ or $K$.

  * $(1)$

Suppose $V\in\mathcal{V}_{\mathfrak{d}}^{L}$. Then $V$ contains a subspace
$V^{\prime}\in\mathcal{V}_{\mathfrak{i}}^{L}$ such that

$\mathcal{M}(L_{-1},\mathcal{M}_{0}(V))=\mathcal{M}(L_{-1},\mathcal{M}_{0}(V^{\prime})).$

  * $(1)$

If
$V,V^{\prime}\in\mathcal{V}_{\mathfrak{n}}^{L}\cup\mathcal{V}_{\mathfrak{i}}^{L}$,
then

$\mathcal{M}_{0}(V)=\mathcal{M}_{0}(V^{\prime})$

if and only if one of the following conditions holds.

    * $(i)$

$V=V^{\prime}.$

    * $(ii)$

$V^{\bot}=V^{\prime}$ when $V$ and $V^{\prime}$ are nondegenerate.

###### Proof.

For (1), we may assume that $V=\mathrm{span}_{\mathbb{F}}\\{y_{i}\mid i\in
J_{1}\cup J_{2}\\}$. Then $V^{\prime}=\mathrm{span}_{\mathbb{F}}\\{y_{i}\mid
i\in J_{2}\\}$ is desired. For (2), by a similar argument as in Lemma 4.5(2),
we get the desired conclusion. ∎

######  Lemma 5.12.

The following statements hold.

  * $(1)$

If $V\in\mathcal{V}_{\mathfrak{n}}^{H}$, then
$\mathcal{M}(H_{-1},\mathcal{M}_{0}(V))=\mathcal{O}_{J_{1}}\oplus\mathcal{O}_{{J}_{3}}.$

  * $(2)$

If $V\in\mathcal{V}_{\mathfrak{i}}^{H}$, then
$\mathcal{M}(H_{-1},\mathcal{M}_{0}(V))=\mathcal{O}_{J_{2}\cup
J_{3}}\oplus\mathcal{O}^{+}_{J_{2}}\mathcal{Q}_{\bar{J}_{2}}.$

###### Proof.

(1) For $V\in\mathcal{V}_{\mathfrak{n}}^{H}$, a direct computation shows that

$\mathcal{M}_{i}(H_{-1},\mathcal{M}_{0}(V))=\mathrm{span}_{\mathbb{F}}\\{u\in
H_{i}\mid u\ \mbox{ is a monomial with }\ \nu(u)=1,2^{i+2}\\}.$

(2) For $V\in\mathcal{V}_{\mathfrak{i}}^{H}$, using induction on $i$, we
obtain that $\mathcal{M}_{i}(H_{-1},\mathcal{M}_{0}(V))$ is spanned by
monomials in $H$ as follows:

  * $(a)$

$u_{1}u_{2}\in H_{i}$, where $u_{1}$ is a monomial with the variables of value
0 and $u_{2}$ is a monomial with the variables of value $2$.

  * $(b)$

$y_{j}u_{3}\in H_{i}$, where $j\in\bar{J}_{2}$ and $u_{3}$ is a monomial with
the variables of value $0$.

Then, the conclusions hold. ∎

For $u\in K$, put $Z(u)=i$ if $(ad1)^{i+1}u=0$ and $(ad1)^{i}u\not=0$.

######  Lemma 5.13.

Suppose $V\in\mathcal{V}_{\mathfrak{i}}^{K}$. For any element $u\in K$ with
$[1,u]\not=0$, $u\in\mathcal{M}(K_{-1},\mathcal{M}_{0}(V))$ if and only if $u$
is a linear combination of elements of the form $f(z+x)^{j}+g$, where
$g\in\mathcal{O}^{+}_{J_{2}\cup
J_{3}}\oplus\mathcal{O}^{+}_{J_{2}}\mathcal{Q}_{\bar{J}_{2}}$,

$f\in\left\\{\begin{array}[]{ll}\mathcal{O}^{+}_{J_{2}}\mathcal{Q}^{+}_{\bar{J}_{2}},&V\in\mathcal{V}_{\mathfrak{i}}^{K}\mbox{
satisfying }J_{3}\ \mbox{ is empty};\\\ \mathcal{O}^{+}_{J_{2}\cup
J_{3}},&V\in\mathcal{V}_{\mathfrak{i}}^{K}\mbox{ satisfying }J_{3}\ \mbox{ is
not empty},\end{array}\right.$

$x=\sum_{i\in J_{2}}y_{i}y_{\widetilde{i}}$ and $0<j<p$.

###### Proof.

Let $\mathfrak{G}_{0}=\mathcal{M}_{0}(V)$. Notice that $g$, $x$ and
$z+x\in\mathcal{M}(K_{-1},\mathfrak{G}_{0})$. Firstly, for any
$i\in\mathbf{I}$, one computes

$\displaystyle[y_{i},z+x]\in\mathcal{O}^{+}_{J_{2}\cup J_{3}},$ $\displaystyle
f[y_{i},z+x]\in\mathcal{O}_{J_{2}\cup
J_{3}}+\mathcal{O}_{J_{2}}\mathcal{Q}_{\bar{J}_{2}},$
$\displaystyle[y_{i},f]\in\left\\{\begin{array}[]{ll}\mathcal{O}^{+}_{J_{2}}\mathcal{Q}^{+}_{\bar{J}_{2}},&V\in\mathcal{V}_{\mathfrak{i}}^{K}\mbox{
satisfying }J_{3}\ \mbox{ is empty};\\\ \mathcal{O}^{+}_{J_{2}\cup
J_{3}},&V\in\mathcal{V}_{\mathfrak{i}}^{K}\mbox{ satisfying }J_{3}\ \mbox{ is
not empty}.\end{array}\right.$

Using induction on $\mathrm{zd}(f)$ and $j$, respectively, we have

$f(z+x),\ (z+x)^{j}\in\mathcal{M}(K_{-1},\mathfrak{G}_{0}).$

Furthermore, $f(z+x)^{j}\in\mathcal{M}(K_{-1},\mathfrak{G}_{0})$.

Conversely, let us use induction on $Z(u)$. When $Z(u)=1$, we consider the
following cases.

Case 1. $u\in\mathcal{M}_{0}(K_{-1},\mathfrak{G}_{0}).$ By Lemmas 4.9(1) and
5.7, we may assume that $u=z+u_{0}$, where $u_{0}\in\mathfrak{G}_{0}\cap
H_{0}$, which means that $u_{0}\in\mathcal{O}_{J_{2}\cup
J_{3}}+\mathcal{O}_{J_{2}}\mathcal{Q}_{\bar{J}_{2}}$. Thus, $u=z+x+(u_{0}-x)$
is desired.

Case 2. $u\in\mathcal{M}_{1}(K_{-1},\mathfrak{G}_{0}).$ From Remark 2.3, we
may assume that $u=y_{t}z+u_{1}$, where $u_{1}\in H_{1}$ and $t\in\mathbf{I}$.
Notice that, when $y_{t}(z+x)\in\mathcal{M}(K_{-1},\mathfrak{G}_{0}),$

$u_{1}-y_{t}x=u-y_{t}(z+x)\in\mathcal{M}(K_{-1},\mathfrak{G}_{0})\cap H,$

which follows that $u=y_{t}(z+x)+(y_{t}x-u_{1})$ is desired. Thus, by the
necessity of this lemma, it is sufficient to consider the case of
$t\in\bar{J}_{2}$ when $J_{3}$ is not empty. From Lemmas 4.9(3) and 5.7, we
may assume that

$u_{1}=h^{(\frac{1}{3},2,2)}+h^{(0,\frac{1}{3},\frac{1}{3})}+h^{(0,\frac{1}{3},2)}+h,$

where

$h^{(\alpha,\beta,\gamma)}=\mathrm{span}_{\mathbb{F}}\\{y_{i}y_{j}y_{k}\mid\nu({y_{i}})=\alpha,\nu({y_{j}})=\beta,\nu({y_{k}})=\gamma\\},$

$h\in\mathcal{M}_{1}(K_{-1},\mathfrak{G}_{0})\cap H$. For any $y_{l}\in
K_{-1}$, we have:

$\sigma(l)(-1)^{l}(D_{\widetilde{l}}(y_{t})z+D_{\widetilde{l}}u_{1})+y_{l}y_{t}=[y_{l},u]\in\mathfrak{G}_{0},$

which means that

$\sigma(l)(-1)^{l}D_{\widetilde{l}}u_{1}+y_{l}y_{t}\in\mathfrak{G}_{0}.$
(5.34)

When $\nu(y_{l})=2$, from equation (5.34) we have:

$\sigma(l)(-1)^{l}D_{\widetilde{l}}h^{(\frac{1}{3},2,2)}+y_{l}y_{t}\in\mathfrak{G}_{0},$

which is a linear combination of elements with value $\frac{2}{3}$. It follows
that

$\sigma(l)(-1)^{l}D_{\widetilde{l}}h^{(\frac{1}{3},2,2)}+y_{l}y_{t}=0.$

When $J_{3}$ is single, we have $y_{l}y_{t}=0$, a contradiction. When $J_{3}$
is not single, there exist distinct $k,\widetilde{k}\in J_{3}$ such that

$h^{(\frac{1}{3},2,2)}=-\sigma(k)(-1)^{k}y_{\widetilde{k}}y_{k}y_{t}+h^{\prime},$
(5.35)

where $D_{\widetilde{k}}h^{\prime}=0$ and

$\sigma(\widetilde{k})(-1)^{k}D_{k}(h^{(\frac{1}{3},2,2)})+y_{\widetilde{k}}y_{t}=0.$

From equation (5.35), we have

$D_{k}(h^{\prime})=D_{k}(h^{(\frac{1}{3},2,2)})+\sigma(k)y_{\widetilde{k}}y_{t}=2\sigma(k)y_{\widetilde{k}}y_{t},$

which contradicts $D_{\widetilde{k}}h^{\prime}=0$. Thus, an element of the
form $y_{t}z+u_{1}$, $t\in\bar{J}_{2}$ is not in
$\mathcal{M}_{1}(K_{-1},\mathfrak{G}_{0})$ when $J_{3}$ is not empty.

Case 3. $u\in\mathcal{M}_{i}(K_{-1},\mathfrak{G}_{0})$ for $i>1$. We may
assume that

$u=g_{i-2}z+g_{i},\ g_{j}\in\overline{H}_{j},\ j=i-2,i.$

Note that the elements of the form $h_{2}z+h$ are not in
$\mathcal{M}(K_{-1},\mathfrak{G}_{0})$, where $h_{2}$ is a linear combination
of monomials with value $\frac{1}{9}$. By induction on $i$, we obtain that
$g_{i-2}$ is in $\mathcal{O}_{J_{2}}\mathcal{Q}_{\bar{J}_{2}}$ if $J_{3}$ is
empty; in $\mathcal{O}_{J_{2}\cup J_{3}}$, otherwise. Thus,
$g_{i-2}(z+x)\in\mathcal{M}_{i}(K_{-1},\mathfrak{G}_{0})$. Moreover,
$g_{i}-g_{i-2}x\in\mathcal{M}_{i}(K_{-1},\mathfrak{G}_{0})\cap\overline{H}$.
Then $u=g_{i-2}(z+x)+(g_{i}-g_{i-2}x)$ is desired.

When $Z(u)=k>1$, suppose

$u=u_{k}z^{k}+u_{k-1}z^{k-1}+\cdots+u_{1}z+u_{0},\ u_{j}\in\overline{H},\
j=0,\ldots,k.$

Obviously,

$u_{k}z+u_{k-1}=2^{(1-k)}(\mathrm{ad}1)^{k-1}(u)\in\mathcal{M}(K_{-1},\mathfrak{G}_{0}).$

Thus, $u_{k}$ is in $\mathcal{O}^{+}_{J_{2}}\mathcal{Q}^{+}_{\bar{J}_{2}}$
when $J_{3}$ is empty; in $\mathcal{O}^{+}_{J_{2}\cup J_{3}}$, otherwise.
Consequently, $u_{k}(z+x)^{k}\in\mathcal{M}(K_{-1},\mathfrak{G}_{0}).$ Thus,

$v=u-u_{k}(z+x)^{k}\in\mathcal{M}(K_{-1},\mathfrak{G}_{0})$

and $Z(v)<k$. By the inductive hypothesis, $v$ is a linear combination of the
desired form. So is $u$. The proof is complete. ∎

Proof of Theorem 5.2. For (2) and ($2^{\prime}$), sufficiency is obvious. For
necessity, suppose $\Phi$ is an isomorphism of
$\mathcal{M}(L_{-1},\mathcal{M}_{0}(V))$ onto
$\mathcal{M}(L_{-1},\mathcal{M}_{0}(V^{\prime}))$. Then, $\Phi(L_{-1})=L_{-1}$
and $\Phi(\mathcal{M}_{0}(V))=\mathcal{M}_{0}(V^{\prime})$, which implies that
$\dim\mathcal{M}_{0}(V)=\dim\mathcal{M}_{0}(V^{\prime})$. It follows that $V$
and $V^{\prime}$ are both nondegenerate or are both isotropic. Notice that
$\Phi(\mathcal{M}_{0}(V))\subset\mathcal{M}_{0}(\Phi(V))$. For the maximality
of $\mathcal{M}_{0}(V^{\prime})$, we have
$\mathcal{M}_{0}(V^{\prime})=\mathcal{M}_{0}(\Phi(V)).$ By virtue of Lemma
5.11(2), we have $V^{\prime}=\Phi(V)$ or $V^{\prime}=\Phi(V)^{\bot}$ when
$V^{\prime}$ and $\Phi(V)$ are both nondegenerate. Thus, we have $\dim V=\dim
V^{\prime}$ or $\dim V=m+n-\dim V^{\prime}$. We can obtain the desired
conclusions by a direct computation. (3) and $(3^{\prime})$ are direct
consequences of (2) and $(2^{\prime})$.

The remaining statements hold from Lemmas 5.5, 5.6, 5.11, 5.12, 5.13 and
Proposition 5.10.∎

Finally, we consider the maximal S-subalgebras of $L$, where $L=W,S,H$ or $K$.
As in the case of modular Lie algebras [16], it easy to show the following:

######  Lemma 5.14.

Suppose $\mathfrak{G}_{0}$ is a maximal irreducible subalgebra of $L_{0}$. The
subalgebra $\mathcal{M}(L_{-1},\mathfrak{G}_{0})$ is not maximal in $L$ if
$\mathcal{M}_{1}(L_{-1},\mathfrak{G}_{0})=0$.

Proof of Theorem 5.3. (1) This is nothing but Lemma 5.6.

(2) Let $\mathfrak{G}_{0}$ be a maximal irreducible subalgebra of $L_{0}.$

(a) Suppose $L=W$ and $\mathcal{M}(W_{-1},\mathfrak{G}_{0})$ is maximal in
$W$. Assume on the contrary that
$\mathrm{div}(\mathcal{M}_{1}(W_{-1},\mathfrak{G}_{0}))=0$. By induction on
$i$, one has $\mathcal{M}_{i}(W_{-1},\mathfrak{G}_{0})\subset W^{\prime}_{i}$
for all $i\geq 1.$ Since $\mathfrak{G}_{0}$ is a nontrivial subalgebra of
$W_{0}$, we have

$\mathcal{M}(W_{-1},\mathfrak{G}_{0})\subsetneq
W_{-1}+W_{0}+W^{\prime}_{1}+W^{\prime}_{2}+\cdots$

By Theorem 3.1, the latter is an MGS of $W$. This contradicts the maximality
of $\mathcal{M}(W_{-1},\mathfrak{G}_{0})$.

Conversely, to show the maximality of $\mathcal{M}(W_{-1},\mathfrak{G}_{0})$,
assume that $M$ is an MGS strictly containing
$\mathcal{M}(W_{-1},\mathfrak{G}_{0})$. By definition (5.30), it must be that
$M_{0}\supsetneq\mathfrak{G}_{0}$ and therefore, $M_{0}=W_{0}$ by the
maximality of $\mathfrak{G}_{0}$. Thus $M$ is an MGS of type (I) and thereby

$M_{1}=W_{1}^{\prime}\;\mbox{or}\;W_{1}^{\prime\prime}.$ (5.36)

Note that $W^{\prime\prime}_{1}$ is an irreducible $\mathfrak{G}_{0}$-module,
which follows from the irreducibility of $\mathfrak{G}_{0}$ and a simple fact
that, as $W_{0}$-modules,

$W^{\prime\prime}_{1}\cong(W_{-1})^{*}.$

By our assumption, there is a
$D\in\mathcal{M}_{1}(W_{-1},\mathfrak{G}_{0})\subset M_{1}$ with
$\mathrm{div}D$ $\neq 0$. Assert that $D\notin W^{\prime\prime}_{1}$. Assuming
on the contrary, by the irreducibility of $W^{\prime\prime}_{1}$, we have
$W^{\prime\prime}_{1}\subset\mathcal{M}_{1}(W_{-1},\mathfrak{G}_{0})$ and
thereby

$W_{0}=\mathrm{alg}([W_{-1},W^{\prime\prime}_{1}])\subset\mathrm{alg}([W_{-1},\mathcal{M}_{1}(W_{-1},\mathfrak{G}_{0})])\subset\mathfrak{G}_{0}.$

This contradicts the assumption that $\mathfrak{G}_{0}$ is a nontrivial
subalgebra of $W_{0}$ and hence the assertion holds. This proves that $D$
belongs to neither $W_{1}^{\prime}$ nor $W_{1}^{\prime\prime}$, contradicting
(5.36).

(b) For $S$, from Lemma 5.14, one implication is obvious. As in (a), we have
$\mathcal{M}(S_{-1},\mathfrak{G}_{0})$ is maximal when
$\mathcal{M}_{1}(S_{-1},\mathfrak{G}_{0})\neq 0$. For $H$, the conclusion
follows from Lemmas 3.11(1) and 5.14.

(c) Suppose $L=K$. Assume on the contrary that $[1,u]=0$ for every
$u\in\mathcal{M}_{1}(K_{-1},\mathfrak{G}_{0})$. Then,

$\mathcal{M}_{1}(K_{-1},\mathfrak{G}_{0})\subset K_{10}.$

As in the proof of Proposition 5.10(2), we have
$\mathcal{M}(K_{-1},\mathfrak{G}_{0})$ is not maximal.

Conversely, suppose $u=u_{0}+u_{1}$ where $u_{i}\in K_{1i}$, $i=0,1$ and
$u_{1}\not=0$. We claim that $u_{0}\not=0$. Indeed, by a direct computation,
$[K_{-1},K_{11}]=K_{0}$ holds. Assuming on the contrary that $u_{0}=0$, we
have $u_{1}\in\mathcal{M}(K_{-1},\mathfrak{G}_{0})$. $K_{-1}$ is an
irreducible $\mathfrak{G}_{0}$-module, and so is $K_{11}$. Moreover,
$K_{11}\subset\mathcal{M}(K_{-1},\mathfrak{G}_{0})$. Thus,

$[K_{-1},K_{11}]\subset[K_{-1},\mathcal{M}(K_{-1},\mathfrak{G}_{0})]\subset\mathfrak{G}_{0}\subsetneq
K_{0},$

which contradicts $[K_{-1},K_{11}]=K_{0}$.

Put $h\in K$, $h\not\in\mathcal{M}(K_{-1},\mathfrak{G}_{0})$ and
$\overline{M}=\mathrm{alg}{(\mathcal{M}(K_{-1},\mathfrak{G}_{0})+\mathbb{F}h)}$.
By definition (5.30) and the maximality of $\mathfrak{G}_{0}$, we have
$K_{0}\subset\overline{M}$. Since the torus $T$ is in $\overline{M}$, from
Remark 2.3, there exists

$v=v_{0}+y_{i}z\in\mathrm{alg}{(\mathbb{F}u+T)}\subset\overline{M},\
i\in\mathbf{I},\ v_{0}\in K_{10}.$

For $K_{0}\subset\overline{M}$, without loss of generality, we may assume that
$i\in\mathbf{I}_{0}.$ If $v_{0}=0$, we have $y_{i}z\in\overline{M}$. For
$u_{0}\not=0$, the conclusion holds. Otherwise, we claim that there exists a
nonzero element in $\overline{M}\cap K_{10}$. Indeed, it is sufficient to
consider the following cases.

Case 1. $D_{\widetilde{i}}(v_{0})\not=0$. Note that
$0\not=[y^{(2\varepsilon_{i})},v]\in\overline{M}\cap K_{10}.$

Case 2. $D_{\widetilde{i}}(v_{0})=0$ and there exists $t\in\mathbf{I}$,
$t\not=i$, $\widetilde{i}$, such that $D_{\widetilde{t}}(v_{0})\not=0$. Note
that $0\not=[y_{t}y_{i},v]\in\overline{M}\cap K_{10}.$

Case 3. $D_{{t}}(v_{0})=0$, for all $t\in\mathbf{I}\backslash\\{i\\}$. Then
$v_{0}=y^{(3\varepsilon_{i})}.$ Note that

$y^{(3\varepsilon_{i})}=(2\sigma(\widetilde{i}))^{-1}([y_{i}y_{\widetilde{i}},v]-\sigma(\widetilde{i})v)\in\overline{M}\cap
K_{10}.$

By Lemmas 2.1(2), 3.11(2) and $u_{i}\not=0$ for $i=0,1$, the conclusion
follows.∎

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