Derivations of finite-dimensional modular Lie superalgebras $\overline{K}(n, m)$

1.   Introduction

Lie superalgebras, which originated from the research of quantum physics (see ^[1]), can be considered as the natural generalization of Lie algebras. Lie superalgebras are closely connected with mathematical physics as well as numerous branches of mathematics (see ^[2,3]). Based on the study of Lie algebras, the theory of Lie superalgebras developed rapidly, including the completed classification of finite dimensional simple Lie superalgebras in 1977 (see ^[4]). However, the classification of finite dimensional simple modular Lie superalgebras has not been accomplished up to now. Since the main difference between modular Lie superalgebras and Lie superalgebras in characteristic zero is the algebras of cartan type, we pay more attention to the related researches on modular Lie superalgebras of Cartan type. In ^[5,6], authors investigated the associative forms of modular Lie superalgebras of Cartan type. The natural filtrations (see ^[7,8,9,10]) and automorphisms (see ^[10,11]) of some Cartan type modular Lie superalgebras are studied. In addition, the cohomologies (see ^[6,12,13]) of some modular Lie superalgebras have also been determined.

It is known to all that the determination of derivation superalgebras is crucial to Lie superalgebras. The related research results in Cartan type modular Lie superalgebras are also quite rich. The derivation superalgebras of some finite dimensional simple modular Lie superalgebras of Cartan type such as $K(m, n, \underline{t}),$ $W(m, n, \underline{t}),$ $S(m, n, \underline{t}),$ $HO(n, n; \underline{t}),$ $KO(n, n+1, \underline{t}),$ $SHO(m, m, \underline{t})^{(2)}$ (see ^{[14,15,16,17,18]}) are determined, respectively. Moreover, the derivation superalgebras of some nonsimple ones are also described, where we are most interested in the correlative results of $\overline{W}(n, m),$ $H(n, m)$ and $S(n, m)$ (see ^[19,20,21]). They all possess the derivation of $\Theta$-type. In ^[22], we have constructed a class of finite dimensional modular Lie superalgebra of Contact type which is denoted by $\overline{K}(n, m)$. This paper is aimed at determining the derivation superalgebras of $\overline{K}(n, m)$.

The present paper is arranged as follows. In Section 2, certain essential notations and concepts are recalled. In Section 3, the $\mathbb{Z}$-homogeneous components of $Der(\overline{K})$, the derivation superalgebras of $\overline{K}(n, m)$, are described, respectively. Therefore, we determine $Der(\overline{K})$. In order to give a description of $Der(\overline{K})$, we prove that $K(n, m)$ is invariant under $Der(\overline{K})$.

2.   Preliminaries

Hereafter $\mathbb{F}$ denotes a field of characteristic $p\geq3$; $\mathbb{Z}_{2} = \{\overline{0}, \overline{1}\}$ is the ring of integers modulo 2. Apart from the standard notation $\mathbb{Z}$, let $\mathbb{N}$ and $\mathbb{N}_{0}$ denote the sets of positive integers and nonnegative integers, respectively. A simple description of construction of the modular Lie superalgebra $\overline{K}(n, m)$ in ^[22] will be given.

Let $\Lambda(n)$ be the Grassmann algebra over $\mathbb{F}$ in $n$ variables $x_{1}, x_{2}, \ldots, x_{n}$. Suppose that $\mathbb{B}_{k} = \{\langle i_{1}, i_{2}, \ldots, i_{k}\rangle\mid1\leq i_{1} < i_{2} < \cdots < i_{k}\leq n\}$ 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 $\mathfrak{U} = \Lambda(n)\otimes\mathbb{T}(m)$ be the tensor product, where $\mathbb{T}(m)$ is the truncated polynomial algebra satisfying $y_{i}^{p} = 1$ for all $i = 1, 2, \ldots, m$ (see ^[20]). Then $\mathfrak{U}$ is an associative superalgebra with $\mathbb{Z}_{2}$-gradation, which is induced by the trivial $\mathbb{Z}_{2}$-gradation of $\mathbb{T}(m)$ and the natural $\mathbb{Z}_{2}$-gradation of $\Lambda(n)$. Namely, $\mathfrak{U} = \mathfrak{U}_{\overline{0}}\oplus\mathfrak{U}_{\overline{1}}$, where $\mathfrak{U}_{\overline{0}} = \Lambda(n)_{\overline{0}}\otimes\mathbb{T}(m)$ and $\mathfrak{U}_{\overline{1}} = \Lambda(n)_{\overline{1}}\otimes\mathbb{T}(m)$.

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 $\mathfrak{U}$. Obviously, $\mathfrak{U} = \bigoplus_{i = 0}^{n}\mathfrak{U}_{i}$ is a $\mathbb{Z}$-graded superalgebra, where $\mathfrak{U}_{i} = \mathrm{span}_{\mathbb{F}}\{x^{u}y^{\lambda}\mid u\in\mathbb{B}(n), |u| = i, \lambda\in G\}$. In particular, $\mathfrak{U}_{0} = \mathbb{T}(m)$ and $\mathfrak{U}_{n} = \mathrm{span}_{\mathbb{F}}\{x^{\pi}y^{\lambda}\mid\lambda\in G\}$, where $\pi: = \langle1, 2, \ldots, n\rangle\in\mathbb{B}(n)$.

In this paper, let $\mathrm{hg}(A) = A_{\bar{0}}\cup A_{\bar{1}}$, where $A = A_{\bar{0}}\oplus A_{\bar{1}}$ is a superalgebra. If $x$ is a $\mathbb{Z}_{2}$-homogeneous element of $A$, then $\mathrm{deg}x$ denotes the $\mathbb{Z}_{2}$-degree of $x$.

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 $\mathfrak{U}$ such that $D_{i}(x^{u}y^{\lambda}) = (\partial x^{u}/\partial x_{i})y^{\lambda}$ for all $u\in\mathbb{B}(n)$ and $\lambda\in G$. Let $\mathrm{Der}\mathfrak{U}$ denote the derivation superalgebra of $\mathfrak{U}$(see ^[12]). Then $D_{i}\in\mathrm{Der}_{\bar{1}}\mathfrak{U}$ for all $i\in\mathrm{Y}$ since $\partial/\partial x_{i}\in\mathrm{Der}_{\bar{1}}(\Lambda(n))$ (see ^[23]).

Suppose that $u\in\mathbb{B}_{k}\subseteq\mathbb{B}(n)$ and $i\in\mathrm{Y}$. When $i\in\{u\}$, $u-\langle i\rangle$ denotes the uniquely determined element of $\mathbb{B}_{k-1}$ satisfying $\{u-\langle i\rangle\} = \{u\}\setminus\{i\}$.Then the number of integers less than $i$ in $\{u\}$ is denoted 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 $f, g\in\mathrm{hg}(\mathfrak{U})$ and $D\in\mathrm{hg}(\mathrm{Der}\mathfrak{U})$. Since the multiplication of $\mathfrak{U}$ is supercommutative, $fD$ is a derivation of $\mathfrak{U}$. Let

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

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

Set $J = \left\{ 1, ..., n-1\right\}$. Let $\widetilde{D}_{k}:\mathfrak{U}\longrightarrow W(n, m)$ be the linear map such that

where $f\in\mathrm{hg}(\mathfrak{U})$, $f_{i} = \left(-1\right)^{degf}\left(x_{i}x_{n}D_{n}\left(f\right)+D_{i}\left(f\right)\right), i\in J$ and $f_{n} = 2f-\sum_{i\in J}x_{i}D_{i}\left(f\right)$.

Let $\overline{K}\left(n, m\right)$ = $\mathrm{span}_{\mathbb{F}}\left\{ \widetilde{D}_{k}(f)|f\in\mathfrak{U}\right\}$. Then $\overline{K}\left(n, m\right)$ is a subspace of $W(n, m)$.

Let

By direct calculation, we have

where $i, j\in J$ and $\delta_{ij}$ is Kronecker delta.

It is easy to prove that $\widetilde{D}_{k}(f) = \sum_{i\in J}\left(-1\right)^{degf}G_{i}\left(f\right)G_{i}+fG_{n}$

For $f\in\mathfrak{U_{\theta}}$ and $g\in\mathfrak{U_{\mu}}$, where $\theta, \mu\in\mathbb{Z}_{2}$, set $\langle f, g\rangle = \widetilde{D}_{k}(f)(g)-G_{n}(f)(g)$. In ^[22], we have proved that $\left[\widetilde{D}_{k}(f), \widetilde{D}_{k}(g)\right] = \widetilde{D}_{k}(\langle f, g\rangle)$. Namely, $\overline{K}\left(n, m\right)$ is a subalgebra of $W\left(n, m\right)$.

If we define an operator $\left[, \right]$ in $\mathfrak{U}$ such that $\left[f, g\right] = \widetilde{D}_{k}\left(f\right)\left(g\right)-G_{n}\left(f\right)\left(g\right)$ for any $f, g\in\mathfrak{U}$. Then $\overline{K}(n, m)\cong(\mathfrak{U}, \left[, \right])$. Moreover, for any $f, g\in\overline{K}(n, m)$, we have

Let $K(n, m)$ be the derived algebra of $\overline{K}(n, m)$, then $K\left(n, m\right) = \mathrm{span}_{\mathbb{F}}\{x^{u}y^{\lambda}|x^{u}y^{\lambda}\in\mathfrak{U}, x^{u}y^{\lambda}\neq x^{\hat{u}}y^{\lambda}\}$, where $\hat{u} = \langle 1, ..., n-1\rangle$. By ^[22], we know that modular Lie superalgebra $K(n, m)$ is not simple.

3.   Determination of derivation superalgebras

In this section, we will abbreviate $\overline{K}(n, m)$, $K(n, m)$ as $\overline{K}$ and $K$, respectively.

In ^[22], we proved that $K(n, m)$ does not possess a $\mathbb{Z}$-graded structure as $W(n, m)$ (see ^[19]). In fact, $\overline{K}(n, m)$ does not possess $\mathbb{Z}$-gradation in the ordinary sense as well. If $\overline{K} = \oplus_{i = -r}^{s}\overline{K}_{i}$, then it does not satisfy that

Now we give a "formal" $\mathbb{Z}$-gradation of $\overline{K}(n, m)$:

where $\overline{K}(n, m)_{i} = \mathrm{span}_{\mathbb{F}}\{x^{u}y^{\lambda}|u\in\mathbb{B}(n), \left|u\right| = i+2, \lambda\in G\}$. Let

It is easy to prove that $Der(\overline{K}) = \oplus_{t\in\mathbb{Z}}Der_{t}(\overline{K})$ is a $\mathbb{Z}$-graded Lie superalgebra (see ^[24]).

Lemma 3.1. Let $\varphi\in Der(\overline{K})$, $f\in\overline{K}$ and $[f, x_{i}] = b_{i}$, $\forall i\in J$. If $\varphi(x_{i}) = \varphi(b_{i}) = 0$, $\forall i\in J$, then $\varphi(f)\in\overline{K}_{-2}$.

Proof. By applying $\varphi$ to $[f, x_{i}] = b_{i}$, we obtain $[\varphi(f), x_{i}]+(-1)^{deg\varphi degf}[f, \varphi(x_{i})] = \varphi(b_{i})$. Since $\varphi(x_{i}) = \varphi(b_{i}) = 0,$ we have $[\varphi(f), x_{i}] = 0$, $\forall i\in J$. Note that

Therefore, $-2x_{n}D_{n}(\varphi(f)) = [\varphi(f), 1] = 0$. Then $D_{n}(\varphi(f)) = 0.$ For all $i\in J$,

Since $[\varphi(f), x_{i}] = 0$, we obtain $D_{i}\left(\varphi\left(f\right)\right) = 0, \forall i\in J$. Therefore, $\varphi(f)\in\overline{K}_{-2}$. □

Lemma 3.2. Let $\varphi\in Der_{-t}(\overline{K}),$ $t\geq2$. If $\varphi\left(\overline{K}_{t-2}\right) = 0$, then $\varphi = 0$.

Proof. If $s < t-2$, then $\varphi(\overline{K}_{s})\subseteq\overline{K}_{s-t} = \left\{ 0\right\}$.

When $s\geqslant t-2$, we will use induction on $s$ to prove that $\varphi(\overline{K}_{s}) = 0$. For $s = t-2$, we have $\varphi(\overline{K}_{s}) = 0$ with the hypothesis of the lemma. Suppose $s > t-2$. For any $y\in\overline{K}_{s}$, $i\in J$, set $\left[y, D_{i}\right] = y_{i}$. Then $y_{i}\in\overline{K}_{s'}$, where $s' < s$. According to the hypothesis of induction, we have $\varphi(y_{i}) = 0$. Noting that $\varphi(D_{i}) = 0$, we obtain $\varphi(y)\in\overline{K}_{-2}$. Therefore, $\varphi(y)\in\overline{K}_{-2}\cap\overline{K}_{s-t} = \{0\}$. Namely, $\varphi(y) = 0$. Then $\varphi(\overline{K}_{s}) = 0$. It follows that $\varphi = 0$. □

Proposition 3.3. $Der_{-t}(\overline{K}) = 0,$ $t\geq2$.

Proof. Let $\varphi\in Der_{-t}(\overline{K}), t\geq2$. We will prove $\varphi\left(\overline{K}_{t-2}\right) = 0$, where

Note that $\varphi(\overline{K}_{t-2})\subseteq\overline{K}_{-2}$. Without loss of generality, we put $\varphi(x^{u_{1}}x_{n}y^{\lambda}) = ay^{\mu},$ $\varphi(x^{u_{2}}y^{\eta}) = by^{\mu}$, where $a, b\in\mathbb{F},$ $\mu\in G$. Applying $\varphi$ to $[x_{n}, x^{u_{1}}x_{n}y^{\lambda}] = 0,$ we obtain

Since $\varphi\left(x_{n}\right)\in\overline{K}_{-t-1} = \{0\}$, which combined with (3.1) yields $\left[x_{n}, \varphi\left(x^{u_{1}}x_{n}y^{\lambda}\right)\right] = 0$. Namely, $\left[x_{n}, ay^{\mu}\right] = 0$. In fact, $\left[x_{n}, ay^{\mu}\right] = -2ay^{\mu}x_{n}$. Therefore, $-2ay^{\mu}x_{n} = 0$. Then

For $i\in\{u_{2}\},$ we have

Without loss of generality, let $x^{u_{2}-\langle i\rangle }x_{n}y^{\eta} = x^{u}x_{n}y^{\eta}$, where $\left|u\right| = t-1$. Then we may write the Eq (3.3) as

By virtue of the Eq (3.2), we have $\varphi(x^{u}x_{n}y^{\eta}) = 0.$ Applying $\varphi$ to the Eq (3.4), we obtain

If $t = 2$, it follows from the Eq (3.2) that $\varphi\left(x_{i}x_{n}\right) = 0$. If $t > 2$, then $\varphi\left(x_{i}x_{n}\right)\in\overline{K}_{-t} = \{0\}$. Consequently, $\varphi\left(x_{i}x_{n}\right) = 0.$ Therefore, by virtue of the Eq (3.5), we have $\left[x_{i}x_{n}, \varphi\left(x^{u_{2}}y^{\eta}\right)\right] = 0.$ Namely, $\left[x_{i}x_{n}, by^{\mu}\right] = 0.$ In fact, $\left[x_{i}x_{n}, by^{\mu}\right] = -2by^{\mu}x_{i}x_{n}$. Therefore, $-2by^{\mu}x_{i}x_{n} = 0$. Then

It follows from the Eqs (3.2) and (3.6) that $\varphi\left(\overline{K}_{t-2}\right) = 0$. By virtue of Lemma 3.2, we have $\varphi = 0$. Therefore, $Der_{-t}(\overline{K}) = 0,$ $t\geq2$. □

Lemma 3.4. Let $\varphi\in Der_{t}(\overline{K}),$ $t\in\mathbb{Z}$. Suppose that $\varphi(\overline{K}_{j}) = 0$, $j = -2, -1, ..., l$. If $t+l\geq-2$, then $\varphi = 0$.

Proof. By virtue of Lemma 3.1, the proof is completely analogous to [24, Lemma 2.8]. □

Proposition 3.5. $Der_{-1}(\overline{K}) = 0$.

Proof. Let $\varphi\in Der_{-1}(\overline{K})$. Then $\varphi(\overline{K}_{-2}) = 0.$ In order to prove $\varphi = 0$, we need to obtain $\varphi(\overline{K}_{-1}) = 0.$ Without loss of generality, we put $\varphi(x_{n}y^{\lambda}) = ay^{\mu},$ where $a\in\mathbb{F}, \mu\in G$. Applying $\varphi$ to $[1, x_{n}y^{\lambda}] = 2x_{n}y^{\lambda}$ yields $(-1)^{deg\varphi deg1}[1, ay^{\mu}] = 2ay^{\mu}$. Therefore, $2ay^{\mu} = 0$. Then $\varphi(x_{n}y^{\lambda}) = ay^{\mu} = 0$. Similarly, we can prove that $\varphi(x_{i}y^{\lambda}) = 0,$ $i\in J$. Therefore, $\varphi(\overline{K}_{-1}) = 0$. Then $\varphi = 0$. □

Let $\varTheta = \mathbb{T}(m)\times...\times\mathbb{T}(m)$. For every $\theta = (h_{1}(y), ..., h_{m}(y))\in\varTheta$, we define $\widetilde{\theta}:G\rightarrow \mathbb{T}(m)$. (see ^[20]) For every $\theta\in\Theta,$we define $D_{\theta}:\overline{K}\rightarrow \overline{K}$ such that $D_{\theta}(\widetilde{D}_{k}(x^{u}y^{\lambda})) = \widetilde{\theta}(\lambda)\widetilde{D}_{k}(x^{u}y^{\lambda}),$ for $x^{u}y^{\lambda}\in\mathfrak{U}$. A direct computation shows that $D_{\theta}\in Der_{\overline{0}}(\overline{K}),$ for all $\theta\in\Theta$. Put $\Omega = \left\{ D_{\theta}|\theta\in\Theta\right\}$.

Proposition 3.6. $Der_{0}(\overline{K}) = ad\overline{K}_{-2}+\Omega$.

Proof. Assume that $y^{\lambda}\in\overline{K}_{-2}$. For $f\in\overline{K}_{j}, j\in\left\{ -2, -1, ..., n-2\right\}$, we have

Therefore, $ad\overline{K}_{-2}\subseteq Der_{0}(\overline{K})$. Obviously, $\Omega\subseteq Der_{0}(\overline{K})$. It follows that $ad\overline{K}_{-2}+\Omega\subseteq Der_{0}(\overline{K})$.

Conversely, let $\varphi\in Der_{0}(\overline{K})$. It is obvious that there exist $y^{\mu}\in\overline{K}_{-2}$ and $D_{\theta}\in\Omega$ such that $\left(\varphi-ady^{\mu}-D_{\theta}\right)\left(\overline{K}_{-2}\right) = 0$. Consequently, by lemma 3.4, we have $\varphi-ady^{\mu}-D_{\theta} = 0$. Then $\varphi = ady^{\mu}+D_{\theta}\in ad\overline{K}_{-2}+\Omega$. Thus, $Der_{0}(\overline{K})\subseteq ad\overline{K}_{-2}+\Omega$. □

Lemma 3.7. Let $h_{1}, ..., h_{n}$ be the nonzero elements of $\mathfrak{U}$. If $G_{i}(h_{j}) = -G_{j}(h_{i})$ for all distinct $i, j\in Y$, then there exists nonzero element $h\in\mathfrak{U}$ such that $G_{i}(h) = h_{i},$ for $i\in J$ and $G_{n}(h) = -h_{n}$.

Proof. We use induction to prove there exist nonzero element $h'\in\mathfrak{U}$ such that $G_{i}(h') = h_{i}$ for all $i\in J$.

When $n-1 = 1$, let $h' = x_{1}h_{1}\neq0$. Then

Suppose that there exists $0\neq g\in\mathfrak{U}$ such that $G_{i}(g) = h_{i}$ for all $i\in\{1, ..., n-2\}.$ For $i\in\{1, ..., n-2\}$,

Therefore, $G_{i}(h_{n-1}-G_{n-1}(g)) = 0$. Let $h' = g+x_{n-1}(h_{n-1}-G_{n-1}(g)).$ For $i\in\{1, ..., n-2\}$,

On the other hand,

Consequently, we have $G_{i}(h') = h_{i}$, for all $i\in\{1, ..., n-1\}.$ For $i\in\{1, ..., n-1\}$,

Then $G_{i}(h_{n}+G_{n}(h')) = 0$. Therefore, $h_{n}+G_{n}(h') = 0$. Namely, $G_{n}(h') = -h_{n}.$ Consequently, the assertion follows from $h = h'$. □

Lemma 3.8. Let $\varphi\in h(Der_{t}(\overline{K}\mathit{\text{))}},$ $t\in\mathbb{N}$. If $\varphi(G_{j}) = 0$, $\forall j\in Y$, then there exists $\theta\in\varTheta$ such that $\varphi(\widetilde{D}_{k}(y^{\lambda})) = D_{\theta}(\widetilde{D}_{k}(y^{\lambda}))$, for $\lambda\in G$.

Proof. Assume that for every $i\in Y$, $\varphi(y^{\lambda}G_{i}) = \sum_{k = 1}^{n}g_{ki\lambda}G_{k}$, where $g_{ki\lambda}\in\mathfrak{U}$. Applying $\varphi$ to $[G_{i}, y^{\lambda}G_{n}] = 0$, $i\in Y$ yields $g_{kn\lambda}\in\mathbb{T}(m)$. Let $g(y)_{kn\lambda} = g_{kn\lambda}$. Then $\varphi(y^{\lambda}G_{n}) = \sum_{k = 1}^{n}g(y)_{kn\lambda}G_{k}$. Note that $[G_{j}, y^{\lambda}G_{i}] = 0$ and $[G_{i}, y^{\lambda}G_{i}] = y^{\lambda}G_{n}$ for all $i\in J$, $i\neq j\in Y$. Applying $\varphi$ to these equations, we know that $g_{ki\lambda}$ contains at most $x_{i}$ in $\Lambda(n)$. Let $g_{ki\lambda} = g(x_{i}, y)_{ki\lambda}$, $i\in J$. For $j\in J$, let $\widetilde{D}_{k}(x_{j}) = \sum_{k = 1}^{n}a_{k}G_{k}$. Since $[G_{j}, G_{j}] = G_{n}$ and $[G_{j}, \widetilde{D}_{k}(x_{j})] = 0$, we obtain

Applying $\varphi$ to (3.7) yields $(-1)^{deg\varphi}[G_{j}, \varphi(G_{j}+\widetilde{D}_{k}(x_{j}))] = 0$. Then $[G_{j}, \varphi(\widetilde{D}_{k}(x_{j}))] = 0.$ Namely, $[G_{j}, \sum_{k = 1}^{n}a_{k}G_{k}] = 0$. Therefore, $G_{j}(a_{k}) = 0, k\in Y$. Applying $\varphi$ to $[y^{\lambda}G_{j}, G_{j}+\widetilde{D}_{k}(x_{j})] = y^{\lambda}G_{n}$ yields

Therefore, $[\varphi(y^{\lambda}G_{j}), G_{j}+\widetilde{D}_{k}(x_{j})] = \varphi(y^{\lambda}G_{n}).$ Namely,

A direct computation shows that $g(y)_{kn\lambda} = 0,$ $k\in J$ and $g(x_{j}, y)_{jj\lambda} = g(y)_{nn\lambda}$. Then $\varphi(y^{\lambda}G_{n}) = g(y)_{nn\lambda}G_{n}$. We abbreviate $g(y)_{nn\lambda}$ as $g(y)_{n\lambda}$. Let $h_{n\lambda}(y) = g(y)_{n\lambda}y^{-\lambda}$. Then $\varphi(y^{\lambda}G_{n}) = h_{n\lambda}(y)y^{\lambda}G_{n}$. Let $\varphi(\widetilde{D}_{k}(x_{j}y^{\eta})) = \sum_{k = 1}^{n}b_{k}G_{k}$, where $b_{k}\in\mathfrak{U}$. Note that

Applying $\varphi$ to (3.8) yields

Since $g(x_{j}, y)_{jj\lambda} = g(y)_{nn\lambda}$, we know $g(x_{j}, y)_{jj\eta}\in\mathbb{T}(m)$. We may abbreviate $g(x_{j}, y)_{jj\eta}$ as $g(y)_{j\eta}$. Let $h_{j\eta}(y) = g(y)_{j\eta}y^{-\eta}$. It follows from (3.9) that

Then $G_{j}(b_{n}) = (-h_{j\eta}(y)+(-1)^{deg\varphi}h_{n\eta}(y))y^{\eta}$ and $G_{j}(b_{i}) = 0,$$i\in J$. Therefore, $b_{n} = (-h_{j\eta}(y)+(-1)^{deg\varphi}h_{n\eta}(y))y^{\eta}x_{j}+q_{n}$ with $G_{j}(q_{n}) = 0.$ Then

Applying $\varphi$ to $[y^{\lambda}G, y^{\eta}G_{j}+\widetilde{D}_{k}(x_{j}y^{\eta})] = y^{\lambda+\eta}G_{n}$ yields

A calculation shows that

Particularly, $h_{j\lambda}(y)+h_{n\lambda}(y) = h_{n(2\lambda)}(y) = h_{n\lambda}(y)+h_{n\lambda}(y).$ Therefore, $h_{j\lambda}(y) = h_{n\lambda}(y)$ for $j\in J$. Consequently, we may abbreviate $h_{i\lambda}(y)$ as $h_{\lambda}(y)$ for $i\in Y$. Then the Eq (3.10) can be equivalent to

for $\lambda, \eta\in G$. Therefore, for $k\in Y, c\in\Pi$, we have

Let $\lambda = \sum_{j = 1}^{m}\lambda_{j}z_{j}\in G$. It follows from (3.11) and (3.12) that

We abbreviate $h_{z_{j}}(y)$ as $h_{j}(y)$ for $j = 1, ..., m.$ Let $\theta = (h_{1}(y), ..., h_{m}(y))\in\Theta.$ It follows from (3.13) that $h_{\lambda}(y) = \sum_{j = 1}^{m}\lambda_{j}h_{z_{j}}(y) = \widetilde{\theta}(\lambda)$. Therefore,

□

Lemma 3.9. Let $\varphi\in h(Der_{t}(\overline{K}\mathit{\text{))}},$ $t\in\mathbb{N}$. Then there exist $B\in Nor_{W}(\overline{K}) = \{x\in W|[x, \overline{K}]\subseteq\overline{K}\}$ and $\theta\in\varTheta$ such that $\varphi = adB+D_{\theta}.$

Proof. Assume that for every $k\in J$, $\varphi(G_{k}) = \sum_{i = 1}^{n}f_{ik}G_{i}$, where $f_{ik}\in\mathfrak{U}$ and $\varphi(G_{n}) = -(-1)^{deg\varphi}\sum_{i = 1}^{n}f_{in}G_{i}$, where $f_{in}\in\mathfrak{U}$. For $k\neq l\in Y$, applying $\varphi$ to $[G_{k}, G_{l}] = 0$ yields

for all $i\in Y$. It follows from lemma 3.7 that there exist $g_{i}(i = 1, ..., n)$ such that

Let $B = -(-1)^{deg\varphi}\sum_{i = 1}^{n}g_{i}G_{i}$. Then for any $k\in J$,

Therefore, $\varphi(G_{j}) = adB(G_{j})$ for all $j\in Y$. Namely, $(\varphi-adB)(G_{j}) = 0$ for all $j\in Y$. According to lemma 3.8, there exists $\theta\in\Theta$ such that

for all $\lambda\in G$. Let $\phi = \varphi-adB-D_{\theta}$. We use induction on $j$ to prove $\phi(\overline{K}_{j}) = 0, j = -2, -1, 0, ..., n-2$. A direct computation shows that $\phi(\overline{K}_{-2}) = 0$. Assume that $j\geq-1$. For any $\xi\in\overline{K}_{j}$, let $[\xi, G_{n}] = \xi_{1}$. Applying $\phi$ to this equation, we obtain

It follows from the induction hypothesis that $\phi(\xi_{1}) = 0.$ Since $\phi(G_{n}) = 0$, we have $[\phi(\xi), G_{n}] = 0$. Then $\phi(\xi)\in\overline{K}_{-1}\bigcap\overline{K}_{j+t} = \{0\}$. Therefore, $\phi(\xi) = 0$. Since the choice of $\xi$ was arbitrary, we have $\phi(\overline{K}_{j}) = 0$. Consequently, $\varphi = adB+D_{\theta}.$ □

Lemma 3.10. Let $\widehat{K}(n, m) = \{\sum_{j\in Y}f_{j}G_{j}\in W|G_{i}(f_{j}) = -G_{j}(f_{i})+(-1)^{degf}\delta_{ij}G_{n}(f), i, j\in J\}$. Then $\overline{K}(n, m) = \widehat{K}(n, m).$

Proof. Assume that $\sum_{j\in Y}f_{j}G_{j}\in\widehat{K}(n, m).$ According to lemma 3.7, there exists $0\neq f\in\mathfrak{U}$ such that $G_{i}(f) = f_{i}$ for all $i\in J.$ Therefore, $\sum_{j\in Y}f_{j}G_{j} = \sum_{j\in J}G_{j}(f)G_{j}+f_{n}G_{n}\in\overline{K}(n, m).$ Then $\widehat{K}(n, m)\subseteq\overline{K}(n, m)$.

Conversely, let $\widetilde{D}_{k}(f) = \sum_{i = 1}^{n}f_{i}G_{i}\in\overline{K}(n, m).$ For $i, j\in J$,

Therefore, $\widetilde{D}_{k}(f)\in\widehat{K}(n, m)$. Then $\overline{K}(n, m)\subseteq\widehat{K}(n, m)$.

Consequently, $\overline{K}(n, m) = \widehat{K}(n, m).$ □

Proposition 3.11. Let $t\in\mathbb{N}$. Then $Der_{t}(\overline{K}) = ad(\overline{K}_{t})+\Omega$.

Proof. It suffices to prove that $Der_{t}(\overline{K})\subseteq ad(\overline{K}_{t})+\Omega$. Let $\varphi\in Der_{t}(\overline{K})$. By virtue of Lemma 3.9, there exists $B\in Nor_{W}(\overline{K})$ such that $\varphi(G_{k}) = adB(G_{k})$, for $k\in Y$. Assume that $B = \sum_{j = 1}^{n}g_{j}G_{j}$ and plug it into the following equation, i.e.,

for $i, j\in J$. Accordingly, we have $G_{i}(f_{j})+G_{j}(f_{i})-(-1)^{degf}\delta_{ij}G_{n}(f) = 0, i, j\in J$. Therefore, $B\in\overline{K}(n, m).$ By virtue of Lemma 3.9, $\varphi = adB+D_{\theta}\in ad\overline{K}+\Omega.$ Then $Der_{t}(\overline{K})\subseteq ad(\overline{K}_{t})+\Omega$. Consequently, $Der_{t}(\overline{K}) = ad(\overline{K}_{t})+\Omega$. □

Lemma 3.12. The following statements hold:

$(1)$ $\Omega$ is a subspace of $Der(\overline{K}).$

$(2)$ $ad(\overline{K})\bigcap\Omega = \{0\}.$

Proof. The proof is completely analogous to [20, Lemma 3.11]. □

Theorem 3.13. $Der(\overline{K}) = ad(\overline{K}')\oplus\Omega$, where $\overline{K}' = \overline{K}_{-2}\oplus\oplus_{i = 1}^{n-2}\overline{K}_{i}$.

Proof. By virtue of Propositions 3.3, 3.5, 3.6, 3.11 and Lemma 3.12, this theorem can be easily proved. □

Proposition 3.14. $K(n, m)$ is invariant under $Der(\overline{K})$.

Proof. Let $\phi\in\Omega$. Obviously, $\phi(K)\subseteq K$.

Let $\phi\in ad(\overline{K}')$. Without loss of generality, we may suppose that $\phi = adf$, where $f\in\overline{K}'\subseteq\overline{K}$. Since $K$ is a derived algebra of $\overline{K}$, $K$ is an ideal of $\overline{K}$. Therefore, $\phi(K) = adf(K) = [f, K]\subseteq K$.

Since $Der(\overline{K}) = ad(\overline{K}')\oplus\Omega$, we have $\phi(K)\subseteq K$ for any $\phi\in Der(\overline{K})$. Namely, $K(n, m)$ is invariant under $Der(\overline{K})$. □

An immediate corollary of this proposition is the following.

Corollary 3.15. $Der(\overline{K})|_{K}\subseteq Der(K)$.

Remark. If $m = 0$, then derivation superalgebras of modular Lie superalgebras $K(n)$, which were mentioned in ^[22], will be obtained.

Acknowledgments

The authors give special thanks to referees for many helpful suggestions. This work was supported by the Fundamental Research Funds for the Central Universities (No.2572021BC02) and the National Natural Science Foundation of China (Grant No.11626056)

Conflict of interest

The authors declare there is no conflicts of interest.