A Class of Finite Dimensional Modular Lie Superalgebras of Special Type

Abstract

This paper is concerned with the Lie superalgebra S(n, m) of special type over a field of prime characteristic. We construct the modular Lie superalgebra S(n, m) and discuss some properties of this algebra. Then the derivation superalgebra of S(n, m) is determined.

1 Introduction

In mathematics, a Lie superalgebra is a generalization of a Lie algebra including a ${\mathbb {Z}}_{2}$-grading, where ${\mathbb {Z}}_{2}=\{\bar{0},\bar{1}\}$ is the residue class ring of integers modulo 2. Lie superalgebras are also important in theoretical physics where they are used to describe the mathematics of supersymmetry [8]. Although many structural features of non-modular Lie superalgebras (see [9, 15, 20, 22]) are well understood, there seem to be few general results on modular Lie superalgebras. The treatment of modular Lie superalgebras necessitates different techniques which are set forth in [11, 13]. In [7] Elduque obtained two new simple modular Lie superalgebras. These Lie superalgebras share the property that their even parts are orthogonal Lie algebras and the odd parts are their spin modules. Since distinctions between non-modular and modular Lie superalgebras are reflected on Lie superalgebras of Cartan type, the structure of Lie superalgebras of Cartan type seems to be important and interesting. In [23], four series of modular graded Lie superalgebras of Cartan type were constructed, which are analogous to the finite dimensional modular Lie algebras of Cartan type [19] or the four series of infinite dimensional Lie superalgebras of Cartan type defined by even differential forms over a field of characteristic zero [10]. In [16], the iso-classes of simple restricted modules of the special Lie superalgebra S(n) were classified, and the character formulas of restricted simple modules were given. Recent works on modular Lie superalgebras of Cartan type can also be found in [5, 6, 12, 21] and references therein.

It is well known that derivation techniques are of great importance in the structure and the classification theories of Lie (super)algebras (see [3, 9, 15, 18]). For some classes of modular Lie (super)algebras, the derivation (super)algebras have been well investigated, for example, the derivation algebras of modular Lie algebras of Cartan type [4, 17], the derivation superalgebras of modular Lie superalgebras of Cartan type [2, 13].

The original motivation for this paper comes from the structures of the finite dimensional modular Lie superalgebras W(n, m) and H(n, m), which were first introduced in [1, 14], respectively. The starting point of our studies is to construct a class of finite dimensional modular Lie superalgebras of special type, which is denoted by S(n, m). A brief summary of the relevant concepts and notations in the finite dimensional modular Lie superalgebras S(n, m) is presented in Sect. 2. In Sect. 3, the intrinsic properties of S(n, m) are investigated. In Sect. 4, the derivation superalgebra of S(n, m) is determined.

2 Preliminaries

Throughout this paper, ${\mathbb {F}}$ denotes an algebraic closed field of characteristic $p>2$, n is an integer greater than 3. In addition to the standard notation ${\mathbb {Z}}$, we write ${\mathbb {N}}$ and ${\mathbb {N}}_{0}$ denote the sets of positive integers and non-negative integers, respectively.

Let $\Lambda (n)$ be the Grassmann algebra over ${\mathbb {F}}$ in n variables $x_{1}, x_{2},\ldots , x_{n}$. Set ${\mathbb {B}}_{k} = \{\langle i_{1}, i_{2},\ldots ,i_{k}\rangle \mid 1\le i_{1} < i_{2} <\cdots < i_{k}\le n \}$ and ${\mathbb {B}}(n)=\bigcup _{k=0}^{n}{\mathbb {B}}_{k}$, where ${\mathbb {B}}_{0}=\emptyset$. For $u =\langle i_{1},i_{2},\ldots ,i_{k}\rangle \in {\mathbb {B}}_{k}$, set $|u| = k$, $\{u\}=\{ i_{1}, i_{2},\ldots ,i_{k}\}$ and $x^{u} = x_{i_{1}}x_{i_{2}}\cdots x_{i_{k}}$ $(|\emptyset |=0, x^{\emptyset }= 1)$. Then $\{x^{u}|u\in {\mathbb {B}}(n)\}$ is an ${\mathbb {F}}$-basis of $\Lambda (n)$.

Let $\Pi$ denote the prime field of ${\mathbb {F}}$, that is, $\Pi =\{0, 1,\ldots ,p-1\}$. Suppose that the set $\{z_{1},z_{2},\ldots ,z_{m}\}$ is a $\Pi$-linearly independent finite subset of ${\mathbb {F}}$. Let $G =\{ \sum _{i=1}^{m}\lambda _{i}z_{i}\mid \lambda _{i}\in \Pi \}$. Then G is an additive subgroup of ${\mathbb {F}}$. Let ${\mathbb {F}}[y_{1},y_{2},\ldots ,y_{m}]$ be the truncated polynomial algebra satisfying $y_{i}^{p}= 1$ for all $i = 1,2,\ldots ,m$. For every element $\lambda =\sum _{i=1}^{m}\lambda _{i}z_{i}\in G$, define $y^{\lambda }=y_{1}^{\lambda _{1}}y_{2}^{\lambda _{2}}\cdots y_{m}^{\lambda _{m}}$. Then $y^{\lambda }y^{\eta }= y^{\lambda +\eta }$ for all $\lambda ,\eta \in G$. Let ${\mathbb {T}}(m)$ denote ${\mathbb {F}}[y_{1},y_{2},\ldots ,y_{m}]$. Then ${\mathbb {T}}(m) =\{\sum _{\lambda \in G}a_{\lambda }y^{\lambda }\mid a_{\lambda }\in {\mathbb {F}}\}$. Set ${\mathscr {U}}= \Lambda (n)\otimes {\mathbb {T}}(m)$. Then ${\mathscr {U}}$ is an associative superalgebra with ${\mathbb {Z}}_{2}$-gradation induced by the trivial ${\mathbb {Z}}_{2}$-gradation of ${\mathbb {T}}(m)$ and the natural ${\mathbb {Z}}_{2}$-gradation of $\Lambda (n)$, that is, ${\mathscr {U}}={\mathscr {U}}_{\bar{0}}\oplus {\mathscr {U}}_{\bar{1}}$, where ${\mathscr {U}}_{\bar{0}}=\Lambda (n)_{\bar{0}}\otimes {\mathbb {T}}(m)$ and ${\mathscr {U}}_{\bar{1}}=\Lambda (n)_{\bar{1}}\otimes {\mathbb {T}}(m)$.

For $f\in \Lambda (n)$ and $\alpha \in {\mathbb {T}}(m)$, we abbreviate $f\otimes \alpha$ as $f\alpha$. Then the elements $x^{u}y^{\lambda }$ with $u\in {\mathbb {B}}(n)$ and $\lambda \in G$ form an ${\mathbb {F}}$-basis of ${\mathscr {U}}$. It is easy to see that ${\mathscr {U}}=\bigoplus _{i=0}^{n}{\mathscr {U}}_{i}$ is a ${\mathbb {Z}}$-graded superalgebra, where ${\mathscr {U}}_{i}= {\mathrm {span}}_{{\mathbb {F}}}\{x^{u}y^{\lambda }\mid u\in {\mathbb {B}}(n), |u|= i, \lambda \in G\}$. In particular, ${\mathscr {U}}_{0}={\mathbb {T}}(m)$ and ${\mathscr {U}}_{n}= {\mathrm {span}}_{{\mathbb {F}}}\{x^{\pi }y^{\lambda }\mid \lambda \in G\}$, where $\pi :=\langle 1,2,\ldots ,n\rangle \in {\mathbb {B}}(n)$.

In this paper, if $A=A_{\bar{0}}\oplus A_{\bar{1}}$ is a superalgebra (or ${\mathbb {Z}}_{2}$-graded linear space), then let ${\mathrm {hg}}(A) = A_{\bar{0}}\cup A_{\bar{1}}$ is the set of all ${\mathbb {Z}}_{2}$-homogeneous elements of A. If ${\mathrm {deg}} x$ occurs in some expression, we regard x as a ${\mathbb {Z}}_{2}$-homogeneous element and ${\mathrm {deg}} x$ as the ${\mathbb {Z}}_{2}$-degree of x. Let $A=\bigoplus _{i=-r}^{n}A_{i}$ be a ${\mathbb {Z}}$-graded superalgebra. If $x\in A_{i}$, then we call x a ${\mathbb {Z}}$-homogeneous element, i the ${\mathbb {Z}}$-degree of x and set ${\mathrm {zd}}(x)=i$.

Let ${\mathrm {pl}}(A)={\mathrm {pl}}_{\bar{0}}(A)\oplus {\mathrm {pl}}_{\bar{1}}(A)$ denote the general linear Lie superalgebra of the ${\mathbb {Z}}_{2}$-graded space A. For $\varphi \in {\mathrm {pl}}_{\theta }(A)$ with $\theta \in {\mathbb {Z}}_{2}$, if

$$\begin{aligned} \varphi (xy) = \varphi (x)y + (-1)^{\theta \mathrm{{deg}} x} x\varphi (y) \end{aligned}$$

for all $x\in {\mathrm {hg}}(A)$ and $y\in A$, then $\varphi$ is called a derivation of A with ${\mathbb {Z}}_{2}$-degree $\theta$. Let ${\mathrm {Der}}_{\theta }A$ denote the set of all derivations of A with ${\mathbb {Z}}_{2}$-degree $\theta$. Then ${\mathrm {Der}}A ={\mathrm {Der}}_{\bar{0}}A\oplus {\mathrm {Der}}_{\bar{1}}A$, which is called the derivation superalgebra of A, is a subalgebra of ${\mathrm {pl}}(A)$ (see [15]).

Set ${\mathrm {Y}} =\{1,2,\ldots ,n\}$. Given $i\in {\mathrm {Y}}$, let $\partial /\partial x_{i}$ be the partial derivative on $\Lambda (n)$ with respect to $x_{i}$. For $i\in Y$, let $D_{i}$ be the linear transformation on ${\mathscr {U}}$ such that $D_{i}(x^{u}y^{\lambda })= (\partial x^{u}/\partial x_{i})y^{\lambda }$ for all $u\in {\mathbb {B}}(n)$ and $\lambda \in G$. Then $D_{i}\in {\mathrm {Der}}_{\bar{1}}{\mathscr {U}}$ for all $i\in {\mathrm {Y}}$ since $\partial /\partial x_{i}\in {\mathrm {Der}}_{\bar{1}}(\Lambda (n))$.

Suppose that $u\in {\mathbb {B}}_{k}\subseteq {\mathbb {B}}(n)$ and $i\in {\mathrm {Y}}$. When $i\in \{u\}$, we denote the uniquely determined element of ${\mathbb {B}}_{k-1}$ satisfying $\{u-\langle i\rangle \}=\{u\}{\setminus }\{i\}$ by $u-\langle i\rangle$, and denote the number of integers less than i in $\{u\}$ by $\tau (u,i)$. When $i\not \in \{u\}$, we set $\tau (u,i)=0$ and $x^{u-\langle i\rangle } = 0$. Therefore, $D_{i}(x^{u}) = (-1)^{\tau (u,i)}x^{u-\langle i\rangle }$ for all $i\in {\mathrm {Y}}$ and $u\in {\mathbb {B}}(n)$.

We define $(fD)(g) = fD(g)$ for $f,g\in {\mathrm {hg}}({\mathscr {U}})$ and $D\in {\mathrm {hg}}({\mathrm {Der}}{\mathscr {U}})$. Since the multiplication of ${\mathscr {U}}$ is supercommutative, fD is a derivation of ${\mathscr {U}}$. Let

$$\begin{aligned} W(n,m) = {\mathrm {span}}_{{\mathbb {F}}}\left\{ x^{u}y^{\lambda }D_{i}\mid u\in {\mathbb {B}}(n),\quad \lambda \in G,\; i\in {\mathrm {Y}} \right\} . \end{aligned}$$

Then W(n, m) is a finite dimensional Lie superalgebra contained in ${\mathrm {Der}}{\mathscr {U}}$. A direct computation shows that

$$\begin{aligned} \left[ fD_{i},gD_{j}\right] = fD_{i}(g)D_{j}-(-1)^{{\mathrm {deg}} (fD_{i}) {\mathrm {deg}} (gD_{j})} gD_{j}(f)D_{i}, \end{aligned}$$

(2.1)

where $f,g\in {\mathrm {hg}}({\mathscr {U}})$ and $i,j\in {\mathrm {Y}}$.

Let $D_{r_{1}r_{2}}: {\mathscr {U}} \longrightarrow W(n,m)$ be the linear mapping such that for every $f\in {\mathrm {hg}}({\mathscr {U}})$ and $r_{1}, r_{2}\in {\mathrm {Y}}$,

$$\begin{aligned} D_{r_{1}r_{2}}(f)=\sum _{i=1}^{2} f_{r_{i}}D_{r_{i}}, \end{aligned}$$

(2.2)

where $f_{r_{1}} = -D_{r_{2}}(f)$ and $f_{r_{2}} = -D_{r_{1}}(f)$. It is easy to see that $D_{r_{1}r_{2}}$ is an even linear mapping. Let $S(n,m)=\{D_{ij}(f)\mid f\in {\mathscr {U}}, i,j\in {\mathrm {Y}}\}$. Then S(n, m) is a finite dimensional Lie superalgebra with a ${\mathbb {Z}}$-gradation $S(n,m)=\bigoplus _{r=-1}^{n-2}S_{r}(n,m)$, where $S_{r}(n,m) =\{D_{ij}(x^{u}y^{\lambda })\mid u\in {\mathbb {B}}(n), |u|= r+2, \lambda \in G, i,j\in {\mathrm {Y}}\}$. In this paper, S(n, m) is called the Lie superalgebra of special type.

By the definition of linear mapping $D_{r_{1}r_{2}}$, the following equalities are easy to verify:

$$\begin{aligned}&D_{ii}(f)=-2D_{i}(f)D_{i}, \end{aligned}$$

(2.3)

$$\begin{aligned}&D_{ij}(f)=D_{ji}(f), \end{aligned}$$

(2.4)

$$\begin{aligned}&\left[ D_{k},D_{ij}(f)\right] =-D_{ij}(D_{k}(f)), \end{aligned}$$

(2.5)

$$\begin{aligned}&\left[ D_{s_{1}s_{2}}(f),D_{r_{1}r_{2}}(g) \right] =\sum _{i,j=1}^{2}(-1)^{{\mathrm {deg}}f}D_{s_{i}r_{j}}(f_{s_{i}}g_{r_{j}}), \end{aligned}$$

(2.6)

where $f,g\in {\mathrm {hg}}({\mathscr {U}})$, $i,j,k\in {\mathrm {Y}}$ and $f_{s_{i}}$, $g_{r_{j}}$ as in (2.2). The equality (2.6) shows that S(n, m) is a subalgebra of W(n, m).

3 Some Basic Properties of S(n, m)

Hereafter, W(n, m), S(n, m), and $S_{i}(n,m)$ will be simply denoted by W, S, and $S_{i}$, respectively.

The linear mapping ${\mathrm {div}}:W\rightarrow {\mathscr {U}}$ defined by ${\mathrm {div}}(fD_{i})=(-1)^{{\mathrm {deg}}f}D_{i}(f)$ for all $f\in {\mathrm {hg}}({\mathscr {U}})$ and $i\in {\mathrm {Y}}$ is called the divergence.

A direct calculation shows that ${\mathrm {div}}$ is a derivation of W with values in ${\mathscr {U}}$ (see [9] or [23]), that is

$$\begin{aligned} {\mathrm {div}}[D,E]=D{\mathrm {div}}(E)-(-1)^{{\mathrm {deg}}D{\mathrm {deg}}E}E{\mathrm {div}}(D) \quad \text{ for } \text{ all } D,\; E\in W. \end{aligned}$$

(3.1)

Following [23], put $\overline{S}(n,m):=\{D\in W\mid {\mathrm {div}}(D)=0\}$. We also denote $\overline{S}(n,m)$ by $\overline{S}$ for short.

Suppose that $D_{1}$ and $D_{2}$ are arbitrary elements of $\overline{S}$; then ${\mathrm {div}}(D_{1})=0$ and ${\mathrm {div}}(D_{2})=0$. By the equality (3.1), we have

$$\begin{aligned} {\mathrm {div}}[D_{1}, D_{2}]=D_{1}\left( {\mathrm {div}}(D_{2}) \right) -(-1)^{{\mathrm {deg}}D_{1}{\mathrm {deg}} D_{2}}D_{2}\left( {\mathrm {div}}(D_{1})\right) =0 \end{aligned}$$

Therefore, $\overline{S}$ is a subalgebra of W.

Proposition 3.1

$S=\overline{S}$.

Proof

A direct calculation shows that

$$\begin{aligned} {\mathrm {div}}\left( D_{ij}(f)\right)= & {} {\mathrm {div}} \left( -D_{i}(f)D_{j}-D_{j}(f)D_{i}\right) \\= & {} -(-1)^{{\mathrm {deg}}f+1}D_{j}D_{i}(f)-(-1)^{{\mathrm {deg}}f+1}D_{i}D_{j}(f)\\= & {} (-1)^{{\mathrm {deg}}f}D_{j}D_{i}(f)-(-1)^{{\mathrm {deg}}f}D_{j}D_{i}(f)\\= & {} 0, \end{aligned}$$

where $f\in {\mathrm {hg}}({\mathscr {U}})$ and $i,j\in {\mathrm {Y}}$. Hence, $D_{ij}(f)\in \overline{S}$. This implies that S is contained in $\overline{S}$.

Conversely, since $\overline{S}\in W$ and ${\mathrm {div}}(x^{u-\langle i\rangle }y^{\lambda }D_{i})=0$, we suppose that an arbitrary element of $\overline{S}$ has the form $x^{u-\langle i\rangle }y^{\lambda }D_{i}$, where $u\in {\mathbb {B}}(n)$, $\lambda \in G$ and $i\in {\mathrm {Y}}$. The equality (2.4) shows that $x^{u-\langle i\rangle }y^{\lambda }D_{i}=(-1)^{\tau (u,i)}D_{i}(x^{u}y^{\lambda })D_{i}=(-1)^{\tau (u,i)+1}2^{-1}D_{ii}(x^{u}y^{\lambda })\in S$. Hence, $\overline{S}\subseteq S$. Consequently, $S=\overline{S}$. $\square$

Adopting the notion of [15, Definition 2, p. 73], the Lie superalgebra L is called transitive if $\{A\in L_{i}\mid [A, L_{-1}]=0\}=0$ for all $i\in {\mathbb {N}}_{0}$.

Proposition 3.2

The Lie superalgebra S is transitive.

Proof

Let $D=D_{ij}(f)\in S_{t}$, where $t\in {\mathbb {N}}_{0}$ and $i,j\in {\mathrm {Y}}$. Then $f\in {\mathscr {U}}_{t+2}$. If $[D,S_{-1}]=0$, then $[D_{ij}(f),y^{\lambda }D_{k}]=0$ for all $k\in {\mathrm {Y}}$ and $\lambda \in G$. The equality (2.5) shows that $D_{ij}(D_{k}(y^{\lambda }f))=0$, that is, $D_{k}(y^{\lambda }f)\in {\mathscr {U}}_{0}={\mathbb {T}}(m)$. Then $f\in {\mathscr {U}}_{1}$ which implies that $f\in {\mathscr {U}}_{1}\cap {\mathscr {U}}_{t+2}$ for $t\in {\mathbb {N}}_{0}$. Hence, $f=0$. It follows that $D=0$. Therefore, S is transitive. $\square$

Let ${\mathcal {I}}=\{\sum _{\lambda \in G}a_{\lambda }y^{\lambda }\mid \sum _{\lambda \in G}a_{\lambda }=0,a_{\lambda }\in {\mathbb {F}}\}$ and $\Gamma =\{x^{u}\alpha \mid u\in {\mathbb {B}}(n), \alpha \in {\mathcal {I}}\}$. By [14, Lemma 4.2 and 4.3], ${\mathcal {I}}$ and $\Gamma$ are ideals of ${\mathbb {T}}(m)$ and ${\mathscr {U}}$, respectively. Furthermore, ${\mathbb {T}}(m)={\mathcal {I}}\oplus {\mathbb {F}}1$. By definition of S, we have $S(n,0)=\{D_{ij}(x^{u})\mid u\in {\mathbb {B}}(n)\}$, which is isomorphic to the special Lie superalgebra S(n) (see [16] or [23] for definition).

Proposition 3.3

Let $\mathfrak {J}=\{D_{ij}(x^{u}\alpha )\mid x^{u}\alpha \in \Gamma , i,j\in {\mathrm {Y}}\}$. Then $\mathfrak {J}$ is an ideal of S and $S\cong S(n)\oplus \mathfrak {J}$.

Proof

Suppose that $x^{u}\alpha \in \Gamma$ and $x^{v}\beta \in {\mathscr {U}}$, where $\alpha \in {\mathcal {I}}$ and $\beta \in {\mathbb {T}}(m)$. Since ${\mathcal {I}}$ is an ideal of ${\mathbb {T}}(m)$ and $D_{s_{1}s_{2}}(x^{u}\alpha )\in \mathfrak {J}$, by equality (2.6), we have

$$\begin{aligned}{}[D_{s_{1}s_{2}}\left( x^{u}\alpha \right) ,D_{r_{1}r_{2}}\left( x^{v}\beta \right) ]= & {} (-1)^{|u|}D_{s_{1}r_{1}} \left( D_{s_{2}}\left( x^{u}\alpha \right) D_{r_{2}}\left( x^{v}\beta \right) \right) \\&+\,(-1)^{|u|}D_{s_{1}r_{2}}\left( D_{s_{2}}\left( x^{u}\alpha \right) D_{r_{1}}\left( x^{v}\beta \right) \right) \\&+\,(-1)^{|u|}D_{s_{2}r_{1}}\left( D_{s_{1}}\left( x^{u}\alpha \right) D_{r_{2}}\left( x^{v}\beta \right) \right) \\&+\,(-1)^{|u|}D_{s_{2}r_{2}}\left( D_{s_{1}}\left( x^{u}\alpha \right) D_{r_{1}}\left( x^{v}\beta \right) \right) \\= & {} (-1)^{|u|}D_{s_{1}r_{1}}\left( D_{s_{2}}\left( x^{u}\right) D_{r_{2}}\left( x^{v}\right) \alpha \beta \right) \\&+\,(-1)^{|u|}D_{s_{1}r_{2}}\left( D_{s_{2}}\left( x^{u}\right) D_{r_{1}}\left( x^{v}\right) \alpha \beta \right) \\&+\,(-1)^{|u|}D_{s_{2}r_{1}}\left( D_{s_{1}}\left( x^{u}\right) D_{r_{2}}\left( x^{v}\right) \alpha \beta \right) \\&+\,(-1)^{|u|}D_{s_{2}r_{2}}\left( D_{s_{1}}\left( x^{u}\right) D_{r_{1}}\left( x^{v}\right) \alpha \beta \right) \in \mathfrak {J}, \end{aligned}$$

where $s_{1},s_{2},r_{1},r_{2}\in {\mathrm {Y}}$. Thus, $\mathfrak {J}$ is an ideal of S.

Suppose that $D_{ij}(x^{u}\beta )$ is an arbitrary element of S. Since ${\mathbb {T}}(m)={\mathcal {I}}\oplus {\mathbb {F}}1$, we may write $\beta =\alpha +a1$, where $\alpha \in {\mathcal {I}}$ and $a\in {\mathbb {F}}$. Then $D_{ij}(x^{u}\beta )=D_{ij}(x^{u}\alpha )+D_{ij}(ax^{u})$ for all $i,j\in {\mathrm {Y}}$. It follows from $S(n,0)\cap \mathfrak {J}=0$ that $S=S(n,0)\oplus \mathfrak {J}$. As S(n, 0) is isomorphic to S(n), the desired result follows immediately. $\square$

Remark

It was shown in [23] that S(n) is a simple Lie superalgebra. Then the results of Proposition 3.3 imply that the ideal $\mathfrak {J}$ of S is maximal.

Let $S(m,n,\underline{1})$ denote the restricted special Lie superalgebra as definition in [23]. Then the following proposition is easy to obtain.

Proposition 3.4

The Lie superalgebra S is isomorphic to a subalgebra of $S(m,n,\underline{1})$.

4 The Derivation Superalgebra of S(n, m)

In this section we will investigate the question of the derivation superalgebra of S.

Let $\Theta := {\mathbb {T}}(m)^{m}={\mathbb {T}}(m)\times \cdots \times {\mathbb {T}}(m)$. Then

$$\begin{aligned} \Theta =\{\theta = (h_{1}(y),\ldots ,h_{m}(y))\mid h_{i}(y)\in {\mathbb {T}}(m), \quad i=1,2,\ldots ,m\}. \end{aligned}$$

For every $\theta = (h_{1}(y),\ldots ,h_{m}(y))\in \Theta$, we define $\widetilde{\theta }:G\rightarrow {\mathbb {T}}(m)$ by $\widetilde{\theta }(\lambda )=\sum _{j=1}^{m}\lambda _{j}h_{j}(y)$ for $\lambda =\sum _{j=1}^{m}\lambda _{j}z_{j}\in G$. It is easy to check that $\widetilde{\theta }(\lambda +\eta )= \widetilde{\theta }(\lambda )+ \widetilde{\theta }(\eta )$ for $\lambda ,\eta \in G$. For every $\theta \in \Theta$, let $D_{\theta }:S\rightarrow S$ be the linear mapping given by $D_{\theta }D_{ij}(x^{u}y^{\lambda })=\widetilde{\theta }(\lambda )D_{ij}(x^{u}y^{\lambda })$ for $x^{u}y^{\lambda }\in {\mathscr {U}}$ and $i,j\in {\mathrm {Y}}$. A direct verification shows that $D_{\theta }\in {\mathrm {Der}}_{\bar{0}}(S)$ for all $\theta \in \Theta$.

The derivation $D_{\theta }$ is called the derivation of $\Theta$-type. Put $\Omega :=\{D_{\theta }\mid \theta \in \Theta \}$.

If $\widetilde{\theta }(\lambda )=\sum _{i=1}^{m}\lambda _{i}h_{\theta i}(y)$ and $\widetilde{\vartheta }(\lambda )=\sum _{i=1}^{m}\lambda _{i}h_{\vartheta i}(y)$, where $\theta ,\vartheta \in \Theta$, then

$$\begin{aligned} \widetilde{\theta }(\lambda )+\widetilde{\vartheta }(\lambda )=\sum _{i=1}^{m}\lambda _{i}(h_{\theta i}(y)+h_{\vartheta i}(y))=\widetilde{\theta +\vartheta }(\lambda ). \end{aligned}$$

The equality above shows that $D_{\theta }+D_{\vartheta }=D_{\theta +\vartheta }$ and $aD_{\theta }=D_{a\theta }$ for $a\in {\mathbb {F}}$. Hence, $\Omega$ is a subspace of ${\mathrm {Der}}_{\bar{0}}(S)$ and $\dim \Omega =\dim \Theta =(p^{m})^{m}$.

Analogous to [14, Lemma 3.3], we have the following lemmas.

Lemma 4.1

If $\varphi \in {\mathrm {hg}}({\mathrm {Der}}(W))$ and $\varphi (D_{j})=0$ for all $j\in {\mathrm {Y}}$, then there exists $\theta \in \Theta$ such that $\varphi (y^{\lambda }D_{j})=D_{\theta }(y^{\lambda }D_{j})$ for all $\lambda \in G$.

Lemma 4.2

Let $h_{k\lambda }=x_{k}y^{\lambda }D_{k}$, where $k\in {\mathrm {Y}}$ and $\lambda \in G$. Then $h_{k\lambda }\in {\mathrm {Nor}}_{W}(S)$.

Proof

For arbitrary $i,j\in {\mathrm {Y}}$, $u\in {\mathbb {B}}(n)$ and $\eta \in G$, we have

$$\begin{aligned} \left[ h_{k\lambda },D_{ij}\left( x^{u}y^{\eta }\right) \right] =-\left[ x_{k}y^{\lambda }D_{k},D_{i}\left( x^{u}y^{\eta } \right) D_{j}\right] -\left[ x_{k}y^{\lambda }D_{k},D_{j} \left( x^{u}y^{\eta }\right) D_{i}\right] . \end{aligned}$$

If $k\in {\mathrm {Y}}{\setminus }\{i,j\}$, then

$$\begin{aligned} \left[ h_{k\lambda },D_{ij}\left( x^{u}y^{\eta }\right) \right]= & {} -x_{k}y^{\lambda }D_{k}D_{i} \left( x^{u}y^{\eta }\right) D_{j}-D_{i} \left( x^{u}y^{\eta }\right) D_{j}(x_{k})y^{\lambda }D_{k}\\&-x_{k}y^{\lambda }D_{k}D_{j} \left( x^{u}y^{\eta }\right) D_{i}-D_{j} \left( x^{u}y^{\eta }\right) D_{i}(x_{k})y^{\lambda }D_{k}\\= & {} -D_{j}\left( x^{u}y^{\lambda +\eta } \right) D_{i}-D_{i}\left( x^{u}y^{\lambda +\eta }\right) D_{j}\\= & {} D_{ij}\left( x^{u}y^{\lambda +\eta }\right) . \end{aligned}$$

In the case of $k=i\ne j$ we have

$$\begin{aligned} \left[ h_{k\lambda },D_{ij}\left( x^{u}y^{\eta }\right) \right]= & {} -\left[ x_{k}y^{\lambda }D_{k},D_{j}\left( x^{u}y^{\eta }\right) D_{k}+D_{k}\left( x^{u}y^{\eta }\right) D_{j}\right] \\= & {} -D_{j}\left( x^{u}y^{\lambda +\eta }\right) D_{k}+D_{j}\left( x^{u}y^{\lambda +\eta }\right) D_{k}\\= & {} 0. \end{aligned}$$

When $k=j\ne i$, we may obtain $[h_{k\lambda },D_{ij}(x^{u}y^{\eta })]=0$ similarly.

If $k=i=j$, then $[h_{k\lambda },D_{ij}(x^{u}y^{\eta })]=-D_{kk}(x^{u}y^{\lambda +\eta })$.

In conclusion, $h_{k\lambda }\in {\mathrm {Nor}}_{W}(S)$. $\square$

Lemma 4.3

Let $\varphi \in {\mathrm {Der}}_{t}(S)$ and $t\in {\mathbb {Z}}$. Suppose that $\varphi (S_{i})=0$ for $i=-1,0,\ldots ,j$ and $j\ge -1$. If $j+t\ge -1$, then $\varphi =0$.

Proof

Suppose that $q\ge j$. To prove $\varphi (S_{q})=0$, we use induction on q.

Note that $\varphi (S_{j})=0$. Let $q>j$ and z be an arbitrary element of $S_{q}$. Suppose that $[z,y^{\lambda }D_{l}]=z_{l}$ for all $l\in {\mathrm {Y}}$ and $\lambda \in G$, then $z_{l}\in S_{q-1}$. The induction hypothesis yields $\varphi (z_{l})=0$. Let $\varphi (z)=\sum _{k\in {\mathrm {Y}}}f_{k}D_{k}$, where $f_{k}\in {\mathscr {U}}$. Applying $\varphi$ to the equality $[z,y^{\lambda }D_{l}]=z_{l}$, we have

$$\begin{aligned} \varphi \left( \left[ z,y^{\lambda }D_{l}\right] \right) = \left[ \varphi (z),y^{\lambda }D_{l}\right] +(-1)^{{\mathrm {deg}}\varphi {\mathrm {deg}}z} \left[ z,\varphi \left( y^{\lambda }D_{l}\right) \right] =0. \end{aligned}$$

It follows from $\varphi (y^{\lambda }D_{l})=0$ that $[\varphi (z),y^{\lambda }D_{l}]=0$. Hence, $[\sum _{k\in {\mathrm {Y}}}f_{k}D_{k},y^{\lambda }D_{l}]=0$, that is, $\sum _{k\in {\mathrm {Y}}}y^{\lambda }D_{l}(f_{k})D_{k}=0$. It follows that $y^{\lambda }D_{l}(f_{k})=0$ for all $l\in {\mathrm {Y}}$ and $\lambda \in G$. This implies $f_{k}\in {\mathbb {T}}(m)$ for $k\in {\mathrm {Y}}$. Thus, $\varphi (z)\in S_{-1}$. On the other hand, $z\in S_{q}$ and $\varphi \in {\mathrm {Der}}_{t}(S)$. So $\varphi (z)\in S_{q+t}$. Since $q+t>j+t\ge -1$, we have $\varphi (z)\in S_{-1}\cap S_{q+t}=0$, that is, $\varphi (S_{q})=0$.

The above considerations show that $\varphi (S)=0$. Therefore, the desired result $\varphi =0$ follows. $\square$

Theorem 4.4

Suppose that $t\in {\mathbb {N}}_{0}$ and $N={\mathrm {span}}_{{\mathbb {F}}}\{h_{k\lambda }\mid k\in {\mathrm {Y}}, \lambda \in G, h_{k\lambda }=x_{k}y^{\lambda }D_{k}\}$, then ${\mathrm {Der}}_{t}(S)={\mathrm {ad}}(S+N)_{t}+\Omega$.

Proof

By virtue of Lemma 4.2, we have $[N,S]\subseteq S$. It is easy to show that ${\mathrm {ad}}(S+N)+\Omega \subseteq {\mathrm {Der}}(S)$. Hence, ${\mathrm {ad}}(S+N)_{t}+\Omega \subseteq {\mathrm {Der}}_{t}(S)$. Note that ${\mathrm {div}}(h_{k\lambda })\ne 0$ and $S=\overline{S}$. Then $N\not \subseteq S$.

Next we will consider the converse inclusion.

Suppose that for every $k\in {\mathrm {Y}}$ and $\varphi _{t}\in {\mathrm {Der}}_{t}(S)$, $\varphi _{t}(D_{k})=\sum _{i=1}^{n} f_{ik}D_{i}$, where $f_{ik}\in {\mathscr {U}}$. Applying $\varphi _{t}$ to $[D_{k},D_{l}] = 0$, where $k\ne l\in {\mathrm {Y}}$, we have $D_{l}(f_{ik})=-D_{k}(f_{il})$ for all $i\in {\mathrm {Y}}$. Then [14, Lemma 3.1] shows that there exist $g_{i}$ for $i = 1,\ldots ,n$ such that $D_{k}(g_{i})=f_{ik}$ for all $k\in {\mathrm {Y}}$.

Let $z=-(-1)^{{\mathrm {deg}}\varphi _{t}}\sum _{i=1}^{n}g_{i}D_{i}\in W_{t}$. Then $\varphi _{t}(D_{k}) = {\mathrm {ad}}z(D_{k})$ for all $k\in {\mathrm {Y}}$. Consequently, by Lemma 4.1, there exists $\theta \in \Theta$ such that $(\varphi _{t}-{\mathrm {ad}}z)(D_{kj}(x_{k}y^{\lambda })) = D_{\theta }(D_{kj}(x_{k}y^{\lambda }))$ for all distinct elements j, k of ${\mathrm {Y}}$ and $\lambda \in G$.

Set $\phi _{t}=\varphi _{t}-{\mathrm {ad}}z-D_{\theta }$. We use induction on j to show that $\phi _{t}(S_{j}) = 0$ for $j=-1,0,\ldots ,n-2$. A direct calculation shows that $\phi _{t}(S_{-1}) = 0$. Suppose that $j\ge 0$. For arbitrary element $\xi$ of $S_{j}$ , the induction hypothesis yields $[\phi _{t}(\xi ),D_{i}]=0$ for all $i\in {\mathrm {Y}}$. Therefore, $\phi _{t}(\xi )\in S_{-1}\cap S_{t+j}$. Noting that $t + j\ge 0$, we have $\phi _{t}(\xi ) = 0$; hence, $\phi _{t}(S_{j}) = 0$. Consequently, $\varphi _{t}={\mathrm {ad}}z + D_{\theta }$.

Since ${\mathrm {ad}}z(D_{k})=\varphi _{t}(D_{k})\in S$ and $D_{k}\in S$ for $k\in {\mathrm {Y}}$, we have $[D_{k},z]\in S$. Thus, ${\mathrm {div}}[D_{k},z]=0$ and ${\mathrm {div}}(D_{k})=0$. The equality (3.1) shows that ${\mathrm {div}}(z)\in {\mathbb {T}}(m)$. Then ${\mathrm {div}}(z-{\mathrm {div}}(z)x_{i}D_{i})=0$ for all $i\in {\mathrm {Y}}$. It follows from $z-{\mathrm {div}}(z)x_{i}D_{i}\in S$ that $z\in S+N$. This implies that ${\mathrm {ad}}z\in {\mathrm {ad}}(S+N)$. The above considerations show that ${\mathrm {ad}}z+D_{\theta }\in {\mathrm {ad}}(S+N)+\Omega$. Therefore, the inclusion relation ${\mathrm {Der}}_{t}(S)\subseteq {\mathrm {ad}}(S+N)_{t}+\Omega$ holds.

In conclusion, ${\mathrm {Der}}_{t}(S)={\mathrm {ad}}(S+N)_{t}+\Omega$. $\square$

Lemma 4.5

${\mathrm {ad}}(S+N)\cap \Omega =\{0\}$.

Proof

Firstly, we consider the set ${\mathrm {ad}}S\cap \Omega$. For any $x\in {\mathrm {ad}}S\cap \Omega$, we suppose that $x={\mathrm {ad}}E=D_{\theta }$, where $E\in S$ and $\theta \in \Theta$. Then ${\mathrm {ad}}E(D_{j})=D_{\theta }(D_{j})=0$ for all $j\in {\mathrm {Y}}$. It follows that $E\in S_{-1}$. Without loss of generality, we may suppose that $E=\sum _{k=1}^{n}\sum _{\lambda \in G}a_{k\lambda }y^{\lambda }D_{k}$, where $a_{k\lambda }\in {\mathbb {F}}$. For all $i,j\in {\mathrm {Y}}$, we have ${\mathrm {ad}}E(D_{ij}(x_{i}x_{j}))=D_{\theta }(D_{ij}(x_{i}x_{j}))=0$. This implies that $a_{k\lambda }=0$ for all $k\in {\mathrm {Y}}$. Then $E=0$, that is, $x=0$. Therefore, ${\mathrm {ad}}S\cap \Omega =\{0\}$.

Next we will prove that ${\mathrm {ad}}N\cap \Omega =\{0\}$. Suppose that $x\in {\mathrm {ad}}N\cap \Omega$ such that $x={\mathrm {ad}}E=D_{\theta }$, where $E\in N$ and $\theta \in \Theta$. Using the similar discussion above, the ${\mathbb {Z}}$-graded degree of E is $-1$. It is contradiction to ${\mathrm {zd}}(N)=0$. Therefore, there does not exist a non-zero element belongs to ${\mathrm {ad}}N\cap \Omega$, that is, ${\mathrm {ad}}N\cap \Omega =\{0\}$.

Consequently, ${\mathrm {ad}}(S+N)\cap \Omega =\{0\}$ and the proof is completed. $\square$

Theorem 4.6

${\mathrm {Der}}_{-1}(S)={\mathrm {ad}}S_{-1}$.

Proof

It suffices to show that ${\mathrm {Der}}_{-1}(S)\subseteq {\mathrm {ad}}S_{-1}$. According to the definition of $S_{i}$, we may easily obtain

$$\begin{aligned} S_{0}={\mathrm {span}}_{{\mathbb {F}}} \left\{ D_{ij}\left( x_{i}x_{j}y^{\lambda }\right) ,x_{i}y^{\lambda }D_{j}\mid i,j\in {\mathrm {Y}},i\ne j,\lambda \in G\right\} . \end{aligned}$$

Let $\varphi \in {\mathrm {Der}}_{-1}(S)$. Suppose that $\varphi (D_{ij}(x_{i}x_{j}y^{\lambda }))=\sum _{t\in {\mathrm {Y}}}\alpha _{ijt\lambda }D_{t}$, where $\alpha _{ijt\lambda }\in {\mathbb {T}}(m)$. For $k\in {\mathrm {Y}}{\setminus }\{i,j\}$, we obtain that $\varphi (D_{jk}(x_{j}x_{k}y^{\eta }))=\sum _{t\in {\mathrm {Y}}}\alpha _{jkt\eta }D_{t}$, where $\alpha _{jkt\eta }\in {\mathbb {T}}(m)$ and $\eta \in G$. Applying $\varphi$ to the equality $[D_{ij}(x_{i}x_{j}y^{\lambda }),D_{jk}(x_{j}x_{k}y^{\eta })]=0$, we have

$$\begin{aligned} \alpha _{ijj\lambda }D_{j}-\alpha _{ijk\lambda }D_{k}-\alpha _{jki\eta }D_{i}+\alpha _{jkj\eta }D_{j}=0. \end{aligned}$$

In particular, $\alpha _{ijk\lambda }=0$ for all $k\in {\mathrm {Y}}{\setminus }\{i,j\}$. Therefore,

$$\begin{aligned} \varphi \left( D_{ij} \left( x_{i}x_{j}y^{\lambda }\right) \right) =\alpha _{iji\lambda }D_{i}+ \alpha _{ijj\lambda }D_{j}, \end{aligned}$$

(4.1)

where $\alpha _{iji\lambda },\alpha _{ijj\lambda }\in {\mathbb {T}}(m)$.

We may also suppose that $\varphi (x_{i}y^{\lambda }D_{j})=\sum _{t\in {\mathrm {Y}}}\beta _{ijt\lambda }D_{t}$, where $\beta _{ijt\lambda }\in {\mathbb {T}}(m)$. Applying $\varphi$ to $[D_{ij}(x_{i}x_{j}),x_{i}y^{\lambda }D_{j}]=2x_{i}y^{\lambda }D_{j}$, by equality (4.1), we have

$$\begin{aligned} \alpha _{ijj0}y^{\lambda }D_{j}-\beta _{iji\lambda }D_{i}+\beta _{ijj\lambda }D_{j}=2\sum _{t\in {\mathrm {Y}}}\beta _{ijt\lambda }D_{t}. \end{aligned}$$

It follows that $\beta _{ijt\lambda }=0$ for all $t\in {\mathrm {Y}}{\setminus }\{i,j\}$. Therefore,

$$\begin{aligned} \varphi \left( x_{i}y^{\lambda }D_{j}\right) = \beta _{iji\lambda }D_{i}+\beta _{ijj\lambda }D_{j}, \end{aligned}$$

By equality (4.1), we may suppose that $\varphi (D_{lj}(x_{l}x_{j}))=\alpha _{ljl0}D_{l}+\alpha _{ljj0}D_{i}$, where $\alpha _{ljl0}, \alpha _{ljj0}\in {\mathbb {T}}(m)$ and $l\in {\mathrm {Y}}{\setminus }\{i,j\}$. A direct calculation shows that $[D_{lj}(x_{l}x_{j}),x_{i}y^{\lambda }D_{j}]=x_{i}y^{\lambda }D_{j}$. Applying $\varphi$ to this equality yields that $\beta _{iji\lambda }=0$. Hence, $\varphi (x_{i}y^{\lambda }D_{j})=\beta _{ijj\lambda }D_{j}$, where $\beta _{ijj\lambda }\in {\mathbb {T}}(m)$. Without loss of generality, we may suppose that $\varphi (x_{i}y^{\lambda }D_{i+1})=\gamma _{i\lambda }D_{i+1}$ and $\varphi (x_{n}y^{\lambda }D_{1})=\gamma _{n\lambda }D_{1}$, where $\gamma _{i\lambda },\gamma _{n\lambda }\in {\mathbb {T}}(m)$ and $i\in {\mathrm {Y}}{\setminus }\{n\}$.

Let $\phi =\varphi -\sum _{i\in {\mathrm {Y}}}\gamma _{i\lambda }({\mathrm {ad}}D_{i})$. Then $\phi (x_{i}y^{\lambda }D_{i+1})=\phi (x_{n}y^{\lambda }D_{1})=0$ for all $i\in {\mathrm {Y}}{\setminus }\{n\}$. Suppose that M is generated by $\{x_{i}y^{\lambda }D_{i+1},x_{n}y^{\lambda }D_{1}\mid i\in {\mathrm {Y}}{\setminus }\{n\}, \lambda \in G\}$, then M is a subalgebra of S. It is obvious that $D_{ij}(x_{i}x_{j}y^{\lambda }),x_{i}y^{\lambda }D_{j}\in M$, where i, j are distinct elements of ${\mathrm {Y}}$. Then $M=S_{0}$. Considering the ${\mathbb {Z}}$-graded degree of $\phi$ we have $\phi (S_{-1})=0$. Lemma 4.3 shows that $\phi =0$. Hence, $\varphi =\sum _{i\in {\mathrm {Y}}}\gamma _{i\lambda }({\mathrm {ad}}D_{i})\in {\mathrm {ad}}S_{-1}$. This implies that ${\mathrm {Der}}_{-1}(S)\subseteq {\mathrm {ad}}S_{-1}$. $\square$

Theorem 4.7

Suppose that $t>1$, then ${\mathrm {Der}}_{-t}(S)=0$.

Proof

Let $\varphi \in {\mathrm {Der}}_{-t}(S)$. By Lemma 4.3, it suffices to prove that $\varphi (S_{t-1})=0$. Suppose that $D_{ij}(x^{u}y^{\lambda })$ with $u\in {\mathbb {B}}(n)$ and $\lambda \in G$ is a basis element of $S_{t-1}$ and $\varphi (D_{ij}(x^{u}y^{\lambda }))=\sum _{l\in {\mathrm {Y}}}\alpha _{ijl\lambda }D_{l}$, where $\alpha _{ijl\lambda }\in {\mathbb {T}}(m)$ and $i,j\in {\mathrm {Y}}$.

Next we will prove that $\varphi (D_{ij}(x^{u}y^{\lambda }))=0$ for the different cases of i and j.

It is obvious that $\varphi (D_{ij}(x^{u}y^{\lambda }))=0$ for $i,j\in {\mathrm {Y}}{\setminus }\{u\}$.

If i and j are distinct elements of $\{u\}$, then $[D_{ij}(x_{i}x_{j}),D_{ij}(x^{u}y^{\lambda })]=0$. For $k\in \{u\}{\setminus }\{i,j\}$, we have

$$\begin{aligned} \left[ D_{jj}\left( x_{k}x_{j}\right) ,D_{ij} \left( x^{u}y^{\lambda }\right) \right] =0. \end{aligned}$$

Applying $\varphi$ to these equalities yields $\alpha _{iji\lambda }=\alpha _{ijj\lambda }=\alpha _{ijk\lambda }=0$. Thus, $\varphi (D_{ij}(x^{u}y^{\lambda }))=0$.

When $i\in \{u\}$ and $j\in {\mathrm {Y}}{\setminus }\{u\}$, by the similar argument above, we may obtain $\alpha _{ijl\lambda }=0$ for all $l\in {\mathrm {Y}}$. Hence, $\varphi (D_{ij}(x^{u}y^{\lambda }))=0$ in this case.

By equality (2.4), $\varphi (D_{ij}(x^{u}y^{\lambda }))=0$ for $j\in \{u\}$ and $i\in {\mathrm {Y}}{\setminus }\{u\}$.

In conclusion, $\varphi (S_{t-1})=0$. Now Lemma 4.3 ensures that $\varphi =0$, as desired. $\square$

Theorem 4.8

${\mathrm {Der}}(S)={\mathrm {ad}}(S+N)\oplus \Omega$.

Proof

This is a direct consequence of Theorems 4.4, 4.7, 4.6, and Lemma 4.5. $\square$