Characterizing N-type derivations on standard operator algebras by local actions

1.   Introduction

The set $\mathcal{Z}(\mathcal{S})$ is the center over the complex field $\mathbb{C}$ of a standard operator algebra $\mathcal{S}$ with a linear mapping $d: \mathcal{S} \to \mathcal{S}.$ Here $d$ is a derivation if the condition $d(\mathcal{D}\mathcal{E}) = d(\mathcal{D})\mathcal{E} + \mathcal{D}d(\mathcal{E})$ is satisfied by $d$ for all $\mathcal{D}, \mathcal{E} \in \mathcal{S}$. If the requirement for linearity is relaxed to additivity, $d$ is referred to as a derivation with additivity. The map $d: \mathcal{S} \to \mathcal{S}$ is known as Lie derivation with multiplication if it fulfills the condition $d([\mathcal{D}, \mathcal{E}]) = [d(\mathcal{D}), \mathcal{E}] + [\mathcal{D}, d(\mathcal{E})]$ for all $\mathcal{D}, \mathcal{E} \in \mathcal{S}$, where $[\mathcal{D}, \mathcal{E}] = \mathcal{D}\mathcal{E} - \mathcal{E}\mathcal{D}$ denotes the Lie bracket. In a similar way, $d$ is called a Lie triple derivation with multiplication if $d([[\mathcal{D}, \mathcal{E}], \mathcal{F}]) = [[d(\mathcal{D}), \mathcal{E}], \mathcal{F}] + [[\mathcal{D}, d(\mathcal{E})], \mathcal{F}] + [[\mathcal{D}, \mathcal{E}], d(\mathcal{F})]$ for all $\mathcal{D}, \mathcal{E}, \mathcal{F} \in \mathcal{S}.$

In the last few years, significant attention has been given to some special conditions that change a linear map into a derivation (class of derivation) including Lie derivation (see ^{[1,2,3,4,5,8,10,11,15,16]}). Most of these studies have focused on identifying the specific conditions that allow Lie derivations or even simple derivations to be fully characterized by their action on certain sub-structures of these algebras. Numerous works have been published on the investigation of the local actions of Lie derivations in operator algebras. Lu and Jing ^[7] demonstrated that if $X$ is a Banach space with dimension greater than two and $L: \mathcal{B}(\mathcal{X}) \rightarrow \mathcal{B}(\mathcal{X})$ is a linear map such that $L([\mathcal{D}, \mathcal{E}]) = [L(\mathcal{D}), \mathcal{E}] + [\mathcal{D}, L(\mathcal{E})]$ for all $\mathcal{D}, \mathcal{E} \in \mathcal{B}(\mathcal{X})$ with $\mathcal{D}\mathcal{E} = 0$ (or $\mathcal{D}\mathcal{E} = \mathbb{P}$, where $\mathbb{P}$ is a fixed nontrivial idempotent), then there exists an operator \(T \in \mathcal{B}(\mathcal{X})\) and a linear map $\phi: \mathcal{B}(\mathcal{X}) \rightarrow \mathbb{C}I$ that vanishes at all commutators $[\mathcal{D}, \mathcal{E}]$ with $\mathcal{D}\mathcal{E} = 0$ (or $\mathcal{D}\mathcal{E} = \mathbb{P}$), such that $L(\mathcal{D}) = T\mathcal{D} - \mathcal{D}T + \phi(\mathcal{D})$ for all $\mathcal{D} \in \mathcal{B}(\mathcal{X})$. Similarly, Qi and Hou ^[9] characterized Lie derivations on any von Neumann algebra \(\mathcal{S}\) without central summands of type \(I_1\). They showed that if \(L: \mathcal{S} \rightarrow \mathcal{S}\) is an additive map such that \(L([\mathcal{D}, \mathcal{E}]) = [L(\mathcal{D}), \mathcal{E}] + [\mathcal{D}, L(\mathcal{E})]\) for all \(\mathcal{D}, \mathcal{E} \in \mathcal{S}\) with \(\mathcal{D}\mathcal{E} = 0\), then there exists an additive derivation \(\delta: \mathcal{S} \rightarrow \mathcal{S}\) and an additive map \(\phi: \mathcal{S} \rightarrow Z(\mathcal{S})\) that vanishes at all commutators \([\mathcal{D}, \mathcal{E}]\) with \(\mathcal{D}\mathcal{E} = 0\), such that \(L = \delta + \phi\). In 2018, Liu ^[6] investigated the characterization of Lie triple derivations on von Neumann algebras without central abelian projections. He proved that if a linear map \(L: \mathcal{S} \rightarrow \mathcal{S}\) satisfies \(L([[\mathcal{D}, \mathcal{E}], \mathcal{F}]) = [[L(\mathcal{D}), \mathcal{E}], \mathcal{F}] + [[\mathcal{D}, L(\mathcal{E})], \mathcal{F}] + [[\mathcal{D}, \mathcal{E}], L(\mathcal{F})]\) for all \(\mathcal{D}, \mathcal{E}, \mathcal{F} \in \mathcal{S}\) with \(\mathcal{D}\mathcal{E} = 0\) (or \(\mathcal{D}\mathcal{E} = \mathbb{P}\), where \(\mathbb{P}\) is a fixed non-trivial projection in \(\mathcal{S}\)), then there exists an additive derivation \(d: \mathcal{S} \rightarrow \mathcal{S}\) and an additive map \(\gamma: \mathcal{S} \rightarrow Z(\mathcal{S})\) that vanishes on every second commutator \([[\mathcal{D}, \mathcal{E}], \mathcal{F}]\) with \(\mathcal{D}\mathcal{E} = 0\) (or \(\mathcal{D}\mathcal{E} = \mathbb{P}\)), such that \(L(X) = d(X) + \gamma(X)\) for all \(X \in \mathcal{S}\). In recent years, several researchers have explored Lie \(n\)-derivations across various types of algebras (see ^{[12,13,14,17]} and related references). However, to date, no research work has been done based on local actions for derivations of Lie type within operator algebras, indicating a need for further investigation in this area.

Inspired by the aforementioned results, we investigate derivations of Lie type based on algebras like standard operator algebras. We demonstrate that derivations of Lie type on standard operator algebras exhibit a general form at zero products.

2.   Notation and preliminaries

For a fixed positive integer $n,$ where $n \geq 2,$ we define polynomials sequence as

The polynomial $P_{n}\left(y_{1}, y_{2}, \ldots, y_{n}\right)$ is known as $(n-1)^{th}$ commutator. A map $\Theta: \mathcal{S} \rightarrow \mathcal{S}$ that is additive is known as Lie $n$-derivation or n-type derivation if the following is satisfied:

for all $y_{1}, y_{2}, \ldots, y_{n} \in \mathcal{S}$. More generally, by removing the additivity of $\Theta,$ we obtain that $\Theta$ is a nonlinear Lie n-derivation. The Lie derivation is a generalization of the Lie 2-derivation. Further, every Lie triple derivation is a generalization of Lie 3-derivation. A combination of these three derivations i.e., Lie derivations, Lie triple derivations, and Lie $n$-derivations are called derivations of Lie type. On a Hilbert space $\mathcal{H}$ over $\mathcal{C}$, the set $\mathcal{B}(H)$ represents all algebras with bounded linear operators. The subalgebra of finite rank operators that is bounded is denoted by $\mathcal{F}(\mathcal{H}).$ It is important to note that $\mathcal{F}(\mathcal{H})$ is a $\ast$-closed ideal within $\mathcal{B}(\mathcal{H}).$ An algebra $\mathcal{S} \subseteq \mathcal{B}(\mathcal{H})$ is referred to as a standard operator algebra if $\mathcal{F}(H) \subseteq \mathcal{S}$. A projection $\mathbb{P}$ is an operator $\mathbb{P} \in \mathcal{B}(\mathcal{H})$ that satisfies the conditions $\mathbb{P}^* = \mathbb{P}$ and $\mathbb{P}^2 = \mathbb{P}$. An algebra $\mathcal{S}$ is called a prime algebras if $\ \mathcal{D}\mathcal{S}\mathcal{E} = 0$ gives either $\mathcal{D} = 0$ or $\mathcal{E} = 0$. Interestingly, all standard operator algebras are prime.

Now, considering a projection $\mathbb{P}_1\in \mathcal{S}$ so that $\mathbb{P}_2 = I-\mathbb{P}_1$. $\mathcal{S}_{jk} = P_j\mathcal{S}P_k$ for $j = 1$ and $k = 2.$ Then by the Peirce-decomposition of $\mathcal{S}$, we have $\mathcal{S} = \mathcal{S}_{11}\oplus\mathcal{S}_{12}\oplus\mathcal{S}_{21}\oplus\mathcal{S}_{22}$ and $S^\ast_{jk}\in \mathcal{S}_{kj}$ for any $S_{jk}\in \mathcal{S}_{jk}$. The following significant result is used frequently in the main results:

Lemma 2.1. In a Hilbert space $\mathcal{H}$ over a complex field $\mathcal{C}$ that is closed with respect to adjoint operation, let $\mathcal{S}$ be a standard operator algebra with operator $I.$ Further, let $\mathcal{D}_{11} \in \mathcal{S}_{11}$ and $\mathcal{D}_{22} \in \; \mathcal{S}_{22}$. If $\mathcal{D}_{11} \mathcal{E}_{12} = \mathcal{E}_{12} \mathcal{D}_{22}$ or $\mathcal{E}_{21} \mathcal{D}_{11} = \mathcal{D}_{22} \mathcal{E}_{21}$ for all $\mathcal{E}_{12} \in \mathcal{S}_{12}, \mathcal{E}_{21} \in \mathcal{S}_{21}$, then $\mathcal{D}_{11}+\mathcal{D}_{22} \in \mathcal{Z}(\mathcal{S})$.

Proof. Let $\mathcal{D}_{11} \in \mathcal{S}_{11}$ and $\mathcal{D}_{22} \in \mathcal{S}_{22}.$ Assume that $\mathcal{D}_{11} \mathcal{E}_{12} = \mathcal{E}_{12} \mathcal{D}_{22}$ for all $\mathcal{E}_{12} \in \mathcal{S}_{12}$. For any $\mathcal{X}_{11} \in \mathcal{S}_{11}$ and $\mathcal{X}_{12} \in \mathcal{S}_{12}$, we get

Since $\mathcal{S}$ is prime, we have $\mathcal{D}_{11} \mathcal{X}_{11} = \mathcal{X}_{11} \mathcal{D}_{11}$.

For any $\mathcal{X}_{12} \in \mathcal{S}_{12}$ and $\mathcal{X}_{22} \in \mathcal{S}_{22}$, we have

It follows from the primeness of $\mathcal{S}$ that $\mathcal{X}_{22} \mathcal{D}_{22} = \mathcal{D}_{22} \mathcal{X}_{22}$.

For any $\mathcal{X}_{12} \in \mathcal{S}_{12}$ and $\mathcal{X}_{21} \in \mathcal{S}_{21}$, we obtain

It follows that $\mathcal{D}_{22} \mathcal{X}_{21} = \mathcal{X}_{21} \mathcal{D}_{11}$.

For any $\mathcal{X} \in \mathcal{S}$, we have

which implies that $\mathcal{D}_{11}+\mathcal{D}_{22} \in \mathcal{Z}(\mathcal{S})$.

Similarly, we can prove that if $\mathcal{E}_{21} \mathcal{D}_{11} = \mathcal{D}_{22} \mathcal{E}_{21}$ for all $\mathcal{E}_{21} \in \mathcal{S}_{21}$, then $\mathcal{D}_{11}+\mathcal{D}_{22} \in \mathcal{Z}(\mathcal{S})$. □

3.   Main result

Theorem 3.1. On a Hilbert space $\mathcal{H}$ that is infinite-dimensional over a field of complex numbers $\mathcal{C}$ with an identity operator $\mathcal{I}$, let $\mathcal{S}$ be a standard operator algebra. If $\mathcal{S}$ is closed with respect to adjoint operation and map $\Theta: \mathcal{S} \to \mathcal{S}$ is defined as

for any $\mathcal{D}_{1}, \mathcal{D}_{2}, \mathcal{D}_{3}, \ldots, \mathcal{D}_{n} \in \mathcal{S}$ with $\mathcal{D}_{1} \mathcal{D}_{2} \mathcal{D}_{3} \ldots \mathcal{D}_{n} = 0,$ then $\Theta(x) = d(x)+\tau(x)$ for every $x \in \mathcal{S},$ where the map $d: \mathcal{S} \rightarrow \mathcal{S}$ is an additive derivation and $\tau: \mathcal{S} \rightarrow \mathcal{Z}(\mathcal{S})$ is a map vanishing on each $(n-1)^{th}$ commutator $P_{n}\left(\mathcal{D}_{1}, \mathcal{D}_{2}, \mathcal{D}_{3}, \ldots, \mathcal{D}_{n}\right)$ with $\mathcal{D}_{1} \mathcal{D}_{2} \mathcal{D}_{3} \ldots \mathcal{D}_{n} = 0$.

Lemma 3.1. We have that $\Theta(0) = 0$.

Proof. $\Theta(0) = \Theta(P_{n}(0, 0, \cdots, 0)) = P_n(\Theta(0), 0, \cdots, 0) + \cdots+ P_n(0, 0, \cdots, \Theta (0)) = 0.$ □

Lemma 3.2. For $\mathcal{D}_{i j} \in \mathcal{S}_{i j}(1 \leq i \neq j \leq 2)$, we have

(1) $\Theta\left(\mathcal{D}_{i i}+\mathcal{D}_{i j}\right)-\Theta\left(\mathcal{D}_{i i}\right)-\Theta\left(\mathcal{D}_{i j}\right) \in \mathcal{Z}(\mathcal{S}).$

(2) $\Theta\left(\mathcal{D}_{i i}+\mathcal{D}_{ji}\right)-\Theta\left(\mathcal{D}_{i i}\right)-\Theta\left(\mathcal{D}_{ji}\right) \in \mathcal{Z}(\mathcal{S}).$

Proof. (1) Firstly, we show the result holds for $i = 1$ and $j = 2.$ For that, we consider $T = \Theta\left(\mathcal{D}_{11}+\mathcal{D}_{12}\right)-\Theta\left(\mathcal{D}_{11}\right)-\Theta\left(\mathcal{D}_{12}\right).$ Since

we obtain

Alternatively, making use of Lemma 3.1, we have

From the last two expressions, we find $P_{n}\left(\mathbb{P}_2, T, \mathbb{P}_1, \mathbb{P}_2, \ldots, \mathbb{P}_2\right) = 0$. This means that $\mathbb{P}_2 T \mathbb{P}_1 = \mathbb{P}_1 T \mathbb{P}_2 =$$0$.

For any $\mathcal{X}_{12} \in \mathcal{S}_{12}$, we have

That means

On the other hand, by using Lemma 3.1, we have

Comparing the above two equations, we obtain

On solving the above equation, we obtain $\mathbb{P}_1 T \mathbb{P}_1 \mathcal{X}_{12} = \mathcal{X}_{12} \mathbb{P}_2 T \mathbb{P}_2$ for all $\mathcal{X}_{12} \in \mathcal{S}_{12}$. Therefore, by using Lemma 2.1, we get $\mathbb{P}_1 T \mathbb{P}_1 + \mathbb{P}_2 T \mathbb{P}_2 \in \mathcal{Z}(\mathcal{S})$. Hence, $T = T_{12} + T_{21} \in \mathcal{Z}(\mathcal{S}).$ That is

In a similar way, it can be shown for $i = 2$ and $j = 1$.

(2) By using the same technique as in (1), one can easily show that

□

Lemma 3.3. For any $\mathcal{D}_{i j}, \mathcal{E}_{i j} \in \mathcal{S}_{i j}(1 \leq i \neq j \leq 2)$, we have

Proof. Firstly, we show the above result is true for $i = 1$ and $j = 2.$

Since $\left(\mathcal{D}_{12} + \mathbb{P}_1\right)\left(\mathbb{P}_2+\mathcal{E}_{12}\right) \mathbb{P}_1 \cdots \mathbb{P}_1 = 0$ for any $\mathcal{D}_{12}, \mathcal{E}_{12} \in \mathcal{S}_{12}$, we have

Secondly, it is easy to follow the same pattern to see that the result is also true for $i = 2$ and $j = 1,$ that is

□

Lemma 3.4. $\mathbb{P}_1 \Theta\left(\mathbb{P}_{i}\right) \mathbb{P}_1+\mathbb{P}_2 \Theta\left(\mathbb{P}_{i}\right) \mathbb{P}_2 \in \mathcal{Z}(\mathcal{S}), i = 1, 2$.

Proof. Let $\mathcal{D}_{12} \in \mathcal{S}_{12}$. Since $\mathcal{D}_{12} \mathbb{P}_1 \cdots \mathbb{P}_1 = 0$, by Lemma 3.3, we have

By pre-multiplying by $\mathbb{P}_1$ and post-multiplying by $\mathbb{P}_2$ on both sides, we obtain

By Lemma 2.1, we have $\mathbb{P}_1 \Theta\left(\mathbb{P}_1\right) \mathbb{P}_1+\mathbb{P}_2 \Theta\left(\mathbb{P}_1\right) \mathbb{P}_2 \in \mathcal{Z}(\mathcal{S})$. Similarly, $\mathbb{P}_1 \Theta\left(\mathbb{P}_2\right) \mathbb{P}_1+\mathbb{P}_2 \Theta\left(\mathbb{P}_2\right) \mathbb{P}_2 \in \mathcal{Z}(\mathcal{S})$. □

Now, $M = \mathbb{P}_1 \Theta\left(\mathbb{P}_1\right) \mathbb{P}_2-\mathbb{P}_2 \Theta\left(\mathbb{P}_1\right) \mathbb{P}_1.$ Define a map $\Omega: \mathcal{S} \to \mathcal{S}$ by $\Omega(\mathcal{D}) = \Theta(\mathcal{D}) - [\mathcal{D}, M]$ for all $\mathcal{D} \in \mathcal{S}$.

Lemma 3.5. $\Omega$ has the following properties:

(1) $\Omega$ also satisfies

for any $\mathcal{D}_{1}, \mathcal{D}_{2}, \mathcal{D}_{3}, \ldots, \mathcal{D}_{n} \in \mathcal{S}$ with $\mathcal{D}_{1} \mathcal{D}_{2} \ldots \mathcal{D}_{n} = 0$.

(2) $\Omega(\mathbb{P}_1) \in \mathcal{Z}(\mathcal{S})$.

(3) $\Omega(\mathbb{P}_2) \in \mathcal{Z}(\mathcal{S})$.

Proof. (1) Since $\Theta$ satisfies the condition (3.1), it is easy to see that $\Omega$ also satisfies the condition (3.1) with $\mathcal{D}_1\mathcal{D}_2 \ldots \mathcal{D}_n = 0$.

(2) By using Peirce-decomposition:

Now,

Therefore, by using Lemma 3.4, we obtain $\Omega(\mathbb{P}_1) \in \mathcal{Z}(\mathcal{S}).$

(3) Since $\mathbb{P}_2 \mathbb{P}_1 \mathbb{P}_1 \ldots \mathbb{P}_n = 0$ and $\Omega\left(\mathbb{P}_1\right) \in \mathcal{Z}(\mathcal{S})$, then

By using pre-and-post multiplication by $\mathbb{P}_1$ in the above, we obtain $\mathbb{P}_1 \Omega\left(\mathbb{P}_2\right) \mathbb{P}_2 = 0$. Similarly, $\mathbb{P}_2 \Omega\left(\mathbb{P}_2\right) \mathbb{P}_1 = 0.$ Now, from the definition of $\Omega$, we have

By using pre-and-post multiplication by $\mathbb{P}_1$ in the above, we obtain $\mathbb{P}_1 \Omega(\mathbb{P}_2) \mathbb{P}_1 = \mathbb{P}_1 \Theta(\mathbb{P}_2) \mathbb{P}_1.$ Similarly, by using pre-and-post multiplication by $\mathbb{P}_2$ in the above equation, we have $\mathbb{P}_2 \Omega(\mathbb{P}_2) \mathbb{P}_2 = \mathbb{P}_2 \Theta(\mathbb{P}_2) \mathbb{P}_2.$ By Peirce-decomposition and using Lemma 3.4, we obtain

This completes the proof. □

Lemma 3.6. $\Omega(\mathcal{S}_{ij}) \subseteq \mathcal{S}_{ij} \; (1\leq i \neq j \leq2)$.

Proof. It is sufficient to prove for $i = 1$ and $j = 2$ first. For other values, the proof will follow a similar fashion. We have $\mathcal{D}_{12} \mathbb{P}_1 \mathbb{P}_1 \mathbb{P}_2 \ldots \mathbb{P}_2 = 0$. Also, by using Lemma 3.5, we have

By pre-and-post multiplication by $\mathbb{P}_1$ in the above equation, we obtain $\mathbb{P}_1 \Omega(\mathcal{D}_{12})\mathbb{P}_1 = 0$. Similarly, $\mathbb{P}_2 \Omega(\mathcal{D}_{12})\mathbb{P}_2 = 0$.

Now, if $n$ is even, then $\Omega(\mathcal{D}_{12}) = \mathbb{P}_1 \Omega(\mathcal{D}_{12}) \mathbb{P}_2 - \mathbb{P}_2 \Omega(\mathcal{D}_{12})\mathbb{P}_1.$ By multiplying $\mathbb{P}_2$ from left and $\mathbb{P}_1$ from right, we obtain $\mathbb{P}_2 \Omega(\mathcal{D}_{12})\mathbb{P}_1 = 0.$ Also, by multiplying $\mathbb{P}_1$ from left and right, we obtain $\mathbb{P}_1 \Omega(\mathcal{D}_{12})\mathbb{P}_1 = 0.$ Similarly, $\mathbb{P}_2 \Omega(\mathcal{D}_{12})\mathbb{P}_2 = 0.$ Hence, $\Omega(\mathcal{D}_{12}) \in \mathcal{S}_{12}.$ Now, we assume that $n$ is odd. Since $\mathcal{D}_{12} \mathcal{E}_{12} \mathcal{F}_{12} \mathbb{P}_2\ldots \mathbb{P}_2 = 0$ for any $\mathcal{D}_{12}, \mathcal{E}_{12}, \mathcal{F}_{12} \in \mathcal{S}_{12}$, we have

and

where $l = [\Omega(\mathcal{D}_{12}), \mathcal{E}_{12}] + [\mathcal{D}_{12} \Omega(\mathcal{E}_{12})] \in \mathcal{S}_{11} + \mathcal{S}_{22}$. Now, $[l, \mathcal{F}_{21}] = 0$. That means $l\mathcal{F}_{21} - \mathcal{F}_{21} l = 0$. Thus, $\mathbb{P}_2 l \mathbb{P}_2 \mathcal{F}_{21} = \mathcal{F}_{21} \mathbb{P}_1 l \mathbb{P}_1 = 0$ for any $\mathcal{F}_{21} \in \mathcal{S}_{12}$. Hence, by Lemma 2.1, $\mathbb{P}_1 l \mathbb{P}_1 + \mathbb{P}_2 l \mathbb{P}_2 \in \mathcal{Z}(\mathcal{S})$. Now,

Thus, $\left[2 \mathbb{P}_2 \Omega \left(\mathcal{D}_{12}\right) \mathbb{P}_1, \mathcal{E}_{12}\right] = l \in \mathcal{Z}(\mathcal{S}).$ That is, $\left[ \mathbb{P}_2 \Omega \left(\mathcal{D}_{12}\right) \mathbb{P}_1, \mathcal{E}_{12}\right] = l \in \mathcal{Z}(\mathcal{S})$. Hence, $\mathbb{P}_2 \Omega(\mathcal{D}_{12}) \mathbb{P}_1 \mathcal{E}_{12} = 0$ for any $\mathcal{E}_{12} \in \mathcal{S}_{12}$. By using primeness of $\mathcal{S}$, we get $\mathbb{P}_2 \Omega(\mathcal{D}_{12}) \mathbb{P}_1 = 0$. Hence, $\Omega (\mathcal{S}_{12}) \subseteq \mathcal{S}_{12}.$ □

Lemma 3.7. For any $\mathcal{D}_{i i} \in \mathcal{S}_{i i}$ for $i = 1, 2,$ there is always a map $\eta_{i}: \mathcal{S}_{i i} \rightarrow \mathcal{Z}(\mathcal{S})$ so that $\Omega\left(\mathcal{D}_{i i}\right)-f_{i}\left(\mathcal{D}_{i i}\right) \in \; \mathcal{S}_{i i}$.

Proof. Since $\mathcal{D}_{11} \mathbb{P}_2 \mathbb{P}_2 \ldots \mathbb{P}_2 = 0$ and using Lemma 3.4, we obtain

Multiplying $\mathbb{P}_1$ from left, we obtain $\mathbb{P}_1 \Omega(\mathcal{D}_{11}) \mathbb{P}_2 = 0$. Similarly, multiplying $\mathbb{P}_2$ from left, $\mathbb{P}_2 \Omega(\mathcal{D}_{11}) \mathbb{P}_1 = 0$. Hence, $\Omega(\mathcal{D}_{11}) \in \mathcal{S}_{11} + \mathcal{S}_{22}$. Similarly, $\Omega(\mathcal{D}_{22}) \in \mathcal{S}_{11} + \mathcal{S}_{22}$.

Since $\mathcal{D}_{11} \mathcal{E}_{22} \mathcal{X}_{12}\mathbb{P}_2 \ldots \mathbb{P}_2 = 0$ and for any $\mathcal{E}_{22} \in \mathcal{S}_{22}, \mathcal{X}_{12} \in \mathcal{S}_{12}$, we have

and

where $m = [\Omega(\mathcal{D}_{11}), \mathcal{E}_{22}] + [\mathcal{D}_{11}, \Omega(\mathcal{E}_{22})] \in \mathcal{S}_{11} + \mathcal{S}_{22}$. Now, $[m, \mathcal{X}_{12}] = [m, \mathcal{X}_{21}] = 0$ and using Lemma 2.1, we have $\mathbb{P}_1 m \mathbb{P}_1 + \mathbb{P}_2 m \mathbb{P}_2 \in \mathcal{Z}(\mathcal{S})$. Therefore, $m = [\Omega(\mathcal{D}_{11}), \mathcal{E}_{22}] + [\mathcal{D}_{11}, \Omega(\mathcal{E}_{22})] \in \mathcal{Z}(\mathcal{S})$. Multiplying both sides by $\mathbb{P}_2$ yields $[\mathbb{P}_2 \Omega(\mathcal{D}_{11})\mathbb{P}_2, \mathcal{E}_{22}] \in \mathcal{Z}(\mathcal{S}) \mathbb{P}_2$. By the Kleinecke-Shirokov theorem, we obtain

for any $\mathcal{E}_{22} \in \mathcal{Z}(\mathcal{S}_{22})$. That means $\mathbb{P}_2 \Omega(\mathcal{D}_{11}) \mathbb{P}_2 \in \mathcal{Z}(\mathcal{S}_{22})$. Therefore, $\mathbb{P}_2 \Omega(\mathcal{D}_{11}) \mathbb{P}_2 = z$ for some $z \in \mathcal{Z}(\mathcal{S}_{22})$. Multiplying $\mathbb{P}_2$ from both sides, we get $\mathbb{P}_2 \Omega(\mathcal{D}_{11}) \mathbb{P}_2 = z\mathbb{P}_2$. Now, define a map $\eta_{1} : \mathcal{S}_{11} \to \mathcal{Z}(\mathcal{S})$ such that $\eta_{1} (\mathcal{D}_{11}) \mathbb{P}_2 = \mathbb{P}_2 \Omega(\mathcal{D}_{11}) \mathbb{P}_2$. Hence,

Therefore, $\mathbb{P}_1 \Omega(\mathcal{D}_{11}) \mathbb{P}_1 - \eta_{1} (\mathcal{D}_{11}) \mathbb{P}_1 \in \mathcal{S}_{11}$. Hence, $\Omega(\mathcal{D}_{11}) - \eta_{1}(\mathcal{D}_{11}) \in \mathcal{S}_{11}$. Similarly, we can show the result for $i = 2.$ □

For any $\mathcal{D} = \mathcal{D}_{11}+\mathcal{D}_{12}+\mathcal{D}_{21}+\mathcal{D}_{22} \in \mathcal{S}$, define a mappings $\psi: \mathcal{S} \rightarrow \mathcal{S}$ and $\tau: \mathcal{S} \rightarrow \mathcal{Z}(\mathcal{S})$ satisfying $\psi(\mathcal{D}) = \; \sum_{i, j = 1}^{2} \phi\left(\mathcal{D}_{i j}\right)-f_{1}\left(\mathcal{D}_{11}\right)-f_{2}\left(\mathcal{D}_{22}\right)$ and $\tau(\mathcal{D}) = \Omega(\mathcal{D})-\psi(\mathcal{D})$ for any $\mathcal{D} \in \mathcal{S}$. It is easy to verify that $\psi$ has the following properties:

Lemma 3.8. For any $\mathcal{D}_{ij} \in \mathcal{S}_{ij}$, we have

(1) $\psi\left(\mathcal{D}_{i j}\right) = \Omega\left(\mathcal{D}_{i j}\right) \in \mathcal{S}_{i j}, 1 \leq i \neq j \leq 2$.

(2) $\psi\left(\mathcal{D}_{i i}\right) \in \mathcal{S}_{i i}, i = 1, 2$.

(3) $\psi\left(\mathcal{D}_{11}+\mathcal{D}_{12}+\mathcal{D}_{21}+\mathcal{D}_{22}\right) = \psi\left(\mathcal{D}_{11}\right)+\psi\left(\mathcal{D}_{12}\right)+\psi\left(\mathcal{D}_{21}\right)+\psi\left(\mathcal{D}_{22}\right)$.

(4) $\psi$ is an additive map on $\mathcal{S}_{i j}, 1 \leq i \neq j \leq 2$.

Lemma 3.9. For $\mathcal{D}_{i j} \in \mathcal{S}_{i j}, (1 \leq i \neq j \leq 2$), we have

(1) $\psi\left(\mathcal{D}_{i i} \mathcal{D}_{i j}\right) = \psi\left(\mathcal{D}_{i i}\right) \mathcal{D}_{i j}+\mathcal{D}_{i i} \psi\left(\mathcal{D}_{i j}\right).$

(2) $\psi\left(\mathcal{D}_{i j} \mathcal{D}_{j j}\right) = \psi\left(\mathcal{D}_{i j}\right) \mathcal{D}_{j j}+\mathcal{D}_{i j} \psi\left(\mathcal{D}_{j j}\right).$

Proof. (1) Since $\mathcal{D}_{ii} \mathcal{D}_{ij} \mathbb{P}_i \ldots \mathbb{P}_i = 0, (i \neq j)$, we have

(2) By using the same approach as in (1), one can conclude that

□

Lemma 3.10. $\psi\left(\mathcal{D}_{i i} \mathcal{E}_{i i}\right) = \mathcal{D}_{i i} \psi\left(\mathcal{E}_{i i}\right)+\psi\left(\mathcal{D}_{i i}\right) \mathcal{E}_{i i}$ for any $\mathcal{D}_{i i}, \mathcal{E}_{i i} \in \mathcal{S}_{ii}$.

Proof. For any $\mathcal{X}_{ij} \in \mathcal{S}_{ij}$ and using Lemma 3.9, we have

Therefore, $(\psi\left(\mathcal{D}_{i i} \mathcal{E}_{i i}\right) - \mathcal{D}_{i i} \psi (\mathcal{E}_{i i}) - \psi\left(\mathcal{D}_{i i}\right) \mathcal{E}_{i i}) \mathcal{X}_{i j} = 0.$ By using primeness of $\mathcal{S}$, we get $\psi\left(\mathcal{D}_{i i} \mathcal{E}_{i i}\right) - \mathcal{D}_{i i} \psi (\mathcal{E}_{i i}) - \psi\left(\mathcal{D}_{i i}\right) \mathcal{E}_{i i}) = 0$. Hence, $\psi\left(\mathcal{D}_{i i} \mathcal{E}_{i i}\right) = \mathcal{D}_{i i} \psi (\mathcal{E}_{i i}) + \psi\left(\mathcal{D}_{i i}\right) \mathcal{E}_{i i}$. □

Lemma 3.11. $\psi$ is additive.

Proof. By using Lemma 3.8, we can see thet $\psi$ is additive on $\mathcal{S}_{ij}$. For any $\mathcal{D}_{11}, \mathcal{E}_{11} \in \mathcal{S}_{11}$ and $\mathcal{F}_{12} \in \mathcal{S}_{12}$, we have

Alternatively, we can write

From the last two expressions, we obtain

By using primeness of $\mathcal{S}$, we obtain

In the similar way

Thus, for any $\mathcal{D} = \mathcal{D}_{11} + \mathcal{D}_{12} + \mathcal{D}_{21} + \mathcal{D}_{22}$ and $\mathcal{E} = \mathcal{E}_{11} + \mathcal{E}_{12} + \mathcal{E}_{21} + \mathcal{E}_{22}$, we have

This completes the proof. □

Lemma 3.12. For any $\mathcal{D}_{i j} \in \mathcal{S}_{i j}$ with $\mathcal{E}_{j i} \in \mathcal{S}_{j i}, (1 \leq i \neq j \leq 2, )$ the following holds true:

Proof. Since $\mathcal{D}_{21} \mathcal{E}_{12} \mathcal{X}_{12} (-\mathbb{P}_1) \ldots (-\mathbb{P}_1) = 0$, we have

On the other hand, we have

From the above two equations, we obtain

Also, since $\mathcal{E}_{12} \mathcal{D}_{21}\mathcal{X}_{21} (-\mathbb{P}_2) \ldots (-\mathbb{P}_2) = 0$, we have

On the other hand, we have

By comparing the above two equations, we obtain

By using Eqs (3.2) and (3.3), we obtain

Assume that $\psi(\mathcal{D}_{12} \mathcal{E}_{21}) - \mathcal{D}_{12} \psi(\mathcal{E}_{21}) - \psi(\mathcal{D}_{12}) \mathcal{E}_{21} = \alpha(\mathcal{D}_{12}, \mathcal{E}_{21})$ for any $\alpha(\mathcal{D}_{12}, \mathcal{E}_{21}) \in \mathcal{Z}(\mathcal{S}_{11})$. Now

Also, $\mathcal{S}_{11} \alpha(\mathcal{D}_{12}, \mathcal{E}_{21})$ is a central ideal in $\mathcal{S}_{11}$. As $\mathcal{S}_{11}$ contains no nonzero central ideal. Therefore, $\alpha(\mathcal{D}_{12}, \mathcal{E}_{21}) = 0$. Hence,

By using the same approach, one can also prove the result for $i = 2, j = 1.$ □

Proof of Theorem 3.1. It follows by Lemmas 3.9–3.12 that $\psi$ is an additive derivation. For $\mathcal{D} \in \mathcal{S}$, we have

where $M = \mathbb{P}_1 \Theta\left(\mathbb{P}_1\right) \mathbb{P}_2-\mathbb{P}_2 \Theta\left(\mathbb{P}_1\right) \mathbb{P}_1$ and $d(\mathcal{D})$ is a derivation. It only remains to prove $\tau$ vanishes at $(n-1)^{th}$ commutator, i.e., we show that $\tau(P_n(\mathcal{D}_1, \mathcal{D}_2, \ldots, \mathcal{D}_n)) = 0$ for all $\mathcal{D}_1, \mathcal{D}_2, \ldots, \mathcal{D}_n \in \mathcal{S}$. By using Lemma 3.8, we obtain

This completes the proof of Theorem 3.1 here. □

Author contributions

All authors are contributed equally.

Use of AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

The first author gratefully acknowledges the funding of the Deanship of Graduate Studies and Scientific Research, Jazan University, Saudi Arabia, through project number: RG24-S053.

Conflict of interest

The authors declare no conflict of interest.