Nonlinear Skew Lie-Type Derivations on ∗-Algebra

Abstract

Let $\mathcal{A}$ be a unital ∗-algebra over the complex fields $\mathbb{C}$. For any $H_1,H_2\in \mathcal{A}$, a product ${[H_1,H_2]}_{•}=H_1{H}_2-H_2{H}_1^{*}$ is called the skew Lie product. In this article, it is shown that if a map $\xi$ : $\mathcal{A}\to \mathcal{A}$ (not necessarily linear) satisfies $\xi \left(P_n(H_1,H_2,\dots ,H_n)\right)={\sum }_{i=1}^n{P}_n(H_1,\dots ,$$H_{i-1},\xi \left(H_i\right),H_{i+1},\dots ,H_n)(n\ge 3)$ for all $H_1,H_2,\dots ,H_n\in \mathcal{A}$, then $\xi$ is additive. Moreover, if $\xi \left(i\frac{e}{2}\right)$ is self-adjoint, then $\xi$ is ∗-derivation. As applications, we apply our main result to some special classes of unital ∗-algebras such as prime ∗-algebra, standard operator algebra, factor von Neumann algebra, and von Neumann algebra with no central summands of type $I_1$.

1. Introduction

Let $\mathcal{A}$ be a ∗-algebra over the complex field $\mathbb{C}$. A mapping $\xi$ : $\mathcal{A}\to \mathcal{A}$ is called an additive derivation if $\xi (A+B)=\xi \left(A\right)+\xi \left(B\right)$ and $\xi \left(AB\right)=\xi \left(A\right)B+A\xi \left(B\right)$ hold for all $A,B\in \mathcal{A}$. Moreover, $\xi$ is said to be an additive ∗-derivation if it is an additive derivation, and $\xi \left(A^{*}\right)=\xi {\left(A\right)}^{*}$ hold for all $A\in \mathcal{A}$. For $A,B\in \mathcal{A},$ define the Lie product and skew Lie product of A and B by $[A,B]=AB-BA$ and ${[A,B]}_{•}=AB-B{A}^{*},$ respectively. A map $\xi$ : $\mathcal{A}\to \mathcal{A}$ (not necessarily linear) is said to be a nonlinear Lie derivation (resectively, a nonlinear Lie triple derivation) if

$$\begin{array}{ccc}\hfill \xi \left(\right[A,B\left]\right)& =& \left[\xi \right(A),B]+[A,\xi (B\left)\right]\hfill \\ (respectively,\xi (\left[\right[A,B],C])\hfill & =& \left[\right[\xi \left(A\right),B],C]+\left[\right[A,\xi \left(B\right)],C]+\left[\right[A,B],\xi (C\left)\right]\hfill \end{array}$$

hold for all $A,B,C\in \mathcal{A}.$ Analogously, a map $\mathcal{A}\to \mathcal{A}$ (not necessarily linear) is called a nonlinear skew Lie derivation (respectively, a nonlinear skew Lie triple derivation) if

$$\begin{array}{ccc}\hfill \xi \left({[A,B]}_{•}\right)& =& {[\xi \left(A\right),B]}_{•}+{[A,\xi \left(B\right)]}_{•}\hfill \\ (resp.\xi \left({[{[A,B]}_{•},C]}_{•}\right)\hfill & =& {[{[\xi \left(A\right),B]}_{•},C]}_{•}+{[{[A,\xi \left(B\right)]}_{•},C]}_{•}+{[{[A,B]}_{•},\xi \left(C\right)]}_{•}\hfill \end{array}$$

hold for all $A,B,C\in \mathcal{A}$; many authors have studied the structure of nonlinear Lie derivation (respectively, nonlinear Lie triple derivation) and nonlinear skew Lie derivation (respectively, nonlinear skew Lie triple derivation) on various ∗-algebra (see [1,2,3,4]). In the last decade, many mathematicians have devoted themselves to the study of mappings involving new products on various kind of rings and algebras. These kind of new products are playing a more important role in some research topics, and their study has attracted many authors’ attention (see [2,5,6,7,8,9,10,11,12,13,14,15]). Define the sequence of polynomials as $P_1\left(X_1\right)=X_1$, $P_2(X_1,X_2)={[X_1,X_2]}_{•}=X_1{X}_2-X_2{X}_1^{*}$, $P_3(X_1,X_2,X_3)={[P_2(X_1,X_2),X_3]}_{•}$, …, $P_n(X_1,X_2,\dots ,X_{n-1},X_n)={[P_{n-1}(X_1,X_2,\cdots ,X_{n-1}),X_n]}_{•}$, where $P_n(X_1,X_2,\dots ,X_{n-1},X_n)={[P_{n-1}(X_1,X_2,\cdots ,X_{n-1}),X_n]}_{•}$ is called skew Lie n- product. A map $\xi$ : $\mathcal{A}\to \mathcal{A}$ (not necessarily linear) is said to be nonlinear skew Lie n-derivation if

$$\begin{array}{ccc}\xi \left(P_n(H_1,H_2,\dots ,H_n)\right)\hfill & =& P_n(\xi \left(H_1\right),H_2,\dots ,H_n)+P_n(H_1,\xi \left(H_2\right),\dots ,H_n)\hfill \\ & & +\cdots +P_n(H_1,H_2,\dots ,\xi \left(H_n\right))\hfill \end{array}$$

holds for all $H_1,H_2,\dots ,H_n\in \mathcal{A}$.

A nonlinear skew Lie 2-derivation is called a nonlinear skew Lie derivation, and a nonlinear skew Lie 3-derivation is called a nonlinear skew Lie triple derivation. A nonlinear skew Lie 2-derivation, nonlinear skew Lie 3-derivation and nonlinear skew Lie n-derivation are collectively called nonlinear skew Lie-type derivations.

Remember the definition of ∗-algebra: first of all, define the involution ★ on ring R; then, define the involution ∗ on algebra $\mathcal{A}$. An additive map ★ on ring is called an involution if ${\left(r_1{r}_2\right)}^{★}=r_2^{★}{r}_1^{★}$ and ${\left(r^{★}\right)}^{★}=r$, for all $r_1,r_2,r\in R$. Defining the involution ∗ on algebra is an additive mapping satisfying ${\left(ab\right)}^{*}=b^{*}{a}^{*}$, ${\left(a^{*}\right)}^{*}=a$, and ${\left(ra\right)}^{*}=r^{★}{a}^{*}$, for all $a,b\in \mathcal{A}$ and $r\in R$. An R-algebra with involution ∗ is called ∗-algebra. A set of complex numbers with conjugation as involution is an ∗-algebra. Let H be the complex Hilbert space and $B\left(H\right)$ be the algebra of bounded operator on H over the complex field $\mathbb{C}$, and define involution ∗ on $B\left(H\right)$ as the adjoint of x for all $x\in B\left(H\right)$. Therefore, $B\left(H\right)$ is an ∗-algebra. The class of ∗-algebras is very important and has many applications in many fields; the behavior of operators on Hilbert spaces is studied using ∗-algebras. The class of ∗- algebra is a more general class than prime ∗-algebra, standard operator algebra, factor von Neumann algebra, and von Neumann algebra with no central summands of type of $I_1$. Consequently, it would be crucial to describe a map on ∗-algebras. Lin [16] proved that every multiplicative skew Lie-type derivation on standard operator algebra is an additive ∗-derivations. In 2016, Zhang [12] studied nonlinear skew Jordan derivations on factor von Neumann algebras and proved that every nonlinear skew Jordan derivation on a factor von Neumann algebra is an addtive ∗-derivation. Later, this result has been extended to skew Jordan triple derivation and skew Jordan-type derivation on ∗-algebras in [13,15], respectively. Lin [17] proved that every multiplicative skew Lie-type derivation on von Neumann algebra is an additive ∗-derivation. In [15], Li et al. proved that every nonlinear *-Jordan type derivation on ∗-algebra is an additive ∗-derivation. Motivated by the above cited work, in this article, we define skew Lie-type mapping on a more general setting of arbitrary unital ∗-algebra. Correspondingly, a map $\xi$ : $\mathcal{A}\to \mathcal{A}$ (not necessarily linear) is called a nonlinear skew Lie-type derivation if

$$\begin{array}{ccc}\xi \left(P_n(H_1,H_2,\dots ,H_n)\right)\hfill & =& P_n(\xi \left(H_1\right),H_2,\dots ,H_n)+P_n(H_1,\xi \left(H_2\right),\dots ,H_n)\hfill \\ & & +\cdots +P_n(H_1,H_2,\dots ,\xi \left(H_n\right))\hfill \end{array}$$

holds for all $H_1,H_2,\dots ,H_n\in \mathcal{A}$.

The aim of this article is to study the nonlinear skew Lie derivations on arbitrary ∗-algebras. More precisely, we show that under mild assumptions, every nonlinear skew Lie-type derivation on an unital ∗-algebra is an additive ∗-derivation. Finally, we apply our main result to some special classes of unital ∗-algebras such as prime ∗-algebra, standard operator algebra, factor von Neumann algebra, and von Neumann algebra with no central summands of type $I_1$.

2. The Main Results

The main results of this article are presented in this section.

Theorem 1.

Let $\mathcal{A}$ be a unital ∗-algebra with unit e containing a nontrivial projection $P_1$, and $P_2=I-P_1$ satisfies

$$X\mathcal{A}{P}_k=0⟹X=0(k=1,2).$$

(1)

Then, if a map ξ : $\mathcal{A}\to \mathcal{A}$ (not necessarily linear) satisfies

$$\begin{array}{ccc}\xi \left(P_n(H_1,H_2,\dots ,H_n)\right)\hfill & =& \sum _{i=1}^n{P}_n(\left(H_1\right),\dots ,H_{i-1},\xi \left(H_i\right),H_{i+1},\dots ,H_n)(n\ge 3)\hfill \end{array}$$

(2)

for all $H_1,H_2,\dots ,H_n\in \mathcal{A}$, then ξ is additive. Moreover, if $\xi \left(i\frac{e}{2}\right)$ is self-adjoint, then ξ is ∗-derivation.

Proof. 

Assume ${\mathcal{A}}_{ij}=P_i\mathcal{A}{P}_j$ for $i,j=1,2$. Then, by the Pierce decomposition of $\mathcal{A}$, we have $\mathcal{A}={\mathcal{A}}_{11}\oplus {\mathcal{A}}_{12}\oplus {\mathcal{A}}_{21}\oplus {\mathcal{A}}_{22}$. Clearly, any $H\in \mathcal{A}$ can be written as $H=H_{11}+H_{12}+H_{21}$$+H_{22}$, where $H_{ij}\in {\mathcal{A}}_{ij}$ for $i,j=1,2$. □

We prove the above theorem using several lemmas. Putting $H_1=H_2=\cdots =H_n=0$ into $\left(2\right)$, we easily establish the following Lemma.

Lemma 1.

$\xi \left(0\right)=0$.

Lemma 2.

For any $H_{12}\in {\mathcal{A}}_{12}$ and $H_{21}\in {\mathcal{A}}_{21}$, we have

$$\xi (H_{12}+H_{21})=\xi \left(H_{12}\right)+\xi \left(H_{21}\right).$$

Proof. 

Assume that $T=\xi (H_{11}+H_{12})-\xi \left(H_{11}\right)-\xi \left(H_{12}\right)$. Our target is to show that $T=0$. Invoking the fact that $P_n(i\frac{e}{2},\frac{e}{2},\frac{e}{2},\dots ,(P_1-P_2),H_{12})=P_n(i\frac{e}{2},\frac{e}{2},\frac{e}{2},\dots ,(P_1-P_2),H_{21})=0$ and Lemma 1, we have

$$\begin{array}{c}{\displaystyle \xi \left(P_n(i\frac{e}{2},\frac{e}{2},\frac{e}{2},\dots ,(P_1-P_2),(H_{12}+H_{21}))\right)}\hfill \\ =\xi \left(P_n(i\frac{e}{2},\frac{e}{2},\frac{e}{2},\dots ,(P_1-P_2),H_{12})\right)+\xi \left(P_n(i\frac{e}{2},\frac{e}{2},\frac{e}{2},\dots ,(P_1-P_2),H_{21})\right).\hfill \\ =P_n(\xi \left(i\frac{e}{2}\right),\frac{e}{2},\frac{e}{2},\dots ,(P_1-P_2),H_{12})+P_n(i\frac{e}{2},\xi \left(\frac{e}{2}\right),\frac{e}{2},\dots ,(P_1-P_2),H_{12})\hfill \\ \hspace{1em}+\cdots +P_n(i\frac{e}{2},\frac{e}{2},\frac{e}{2},\dots ,\xi (P_1-P_2),H_{12})+P_n(i\frac{e}{2},\frac{e}{2},\frac{e}{2},\dots ,(P_1-P_2),\xi \left(H_{12}\right))\hfill \\ \hspace{1em}+P_n(\xi \left(i\frac{e}{2}\right),\frac{e}{2},\frac{e}{2},\dots ,(P_1-P_2),H_{21})+P_n(i\frac{e}{2},\xi \left(\frac{e}{2}\right),\frac{e}{2},\dots ,(P_1-P_2),H_{21})\hfill \\ \hspace{1em}+\cdots +P_n(i\frac{e}{2},\frac{e}{2},\frac{e}{2},\dots ,\xi (P_1-P_2),H_{21})+P_n(i\frac{e}{2},\frac{e}{2},\frac{e}{2},\dots ,(P_1-P_2),\xi \left(H_{21}\right)).\hfill \\ =P_n(\xi \left(i\frac{e}{2}\right),\frac{e}{2},\frac{e}{2},\dots ,(P_1-P_2),(H_{12}+H_{21}))+P_n(i\frac{e}{2},\xi \left(\frac{e}{2}\right),\frac{e}{2},\dots ,(P_1-P_2),(H_{12}+H_{21}))\hfill \\ \hspace{1em}+\cdots +P_n(i\frac{e}{2},\frac{e}{2},\frac{e}{2},\dots ,\xi (P_1-P_2),(H_{12}+H_{21}))+P_n(i\frac{e}{2},\frac{e}{2},\frac{e}{2},\dots ,(P_1-P_2),\xi \left(H_{12}\right)+\xi \left(H_{21}\right)).\hfill \end{array}$$

On the other hand, we have

$$\begin{array}{c}{\displaystyle \xi \left(P_n(i\frac{e}{2},\frac{e}{2},\frac{e}{2},\dots ,(P_1-P_2),(H_{12}+H_{21}))\right)}\hfill \\ =P_n(\xi \left(i\frac{e}{2}\right),\frac{e}{2},\frac{e}{2},\dots ,(P_1-P_2),(H_{12}+H_{21}))+P_n(i\frac{e}{2},\xi \left(\frac{e}{2}\right),\frac{e}{2},\dots ,(P_1-P_2),(H_{12}+H_{21}))\hfill \\ \hspace{1em}+\cdots +P_n(i\frac{e}{2},\frac{e}{2},\frac{e}{2},\dots ,\xi (P_1-P_2),(H_{12}+H_{21}))+P_n(i\frac{e}{2},\frac{e}{2},\frac{e}{2},\dots ,(P_1-P_2),\xi (H_{12}+H_{21})).\hfill \end{array}$$

Comparing the above two expressions for $\xi \left(P_n(i\frac{e}{2},\frac{e}{2},\frac{e}{2},\dots ,(P_1-P_2),(H_{12}+H_{21}))\right),$ we obtain $P_n(i\frac{e}{2},\frac{e}{2},\frac{e}{2},\dots ,(P_1-P_2),T)=0.$ This leads to $T_{11}=0$ and $T_{22}=0$.

Invoking the fact that $P_n(i\frac{e}{2},\frac{e}{2},\frac{e}{2},\dots ,H_{12},P_1)=0$ and Lemma 1, we find that

$$\begin{array}{c}{\displaystyle \xi \left(P_n(i\frac{e}{2},\frac{e}{2},\frac{e}{2},\dots ,(H_{12}+H_{21}),P_1)\right)}\hfill \\ =\xi \left(P_n(i\frac{e}{2},\frac{e}{2},\frac{e}{2},\dots ,H_{12},P_1)\right)+\xi \left(P_n(i\frac{e}{2},\frac{e}{2},\frac{e}{2},\dots ,H_{21},P_1)\right).\hfill \\ =P_n(\xi \left(i\frac{e}{2}\right),\frac{e}{2},\frac{e}{2},\dots ,H_{12},P_1)+P_n(i\frac{e}{2},\xi \left(\frac{e}{2}\right),\frac{e}{2},\dots ,H_{12},P_1)\hfill \\ \hspace{1em}+\cdots +P_n(i\frac{e}{2},\frac{e}{2},\frac{e}{2},\dots ,\xi \left(H_{12}\right),P_1)+P_n(i\frac{e}{2},\frac{e}{2},\frac{e}{2},\dots ,H_{12},\xi \left(P_1\right))\hfill \\ \hspace{1em}+P_n(\xi \left(i\frac{e}{2}\right),\frac{e}{2},\frac{e}{2},\dots ,H_{21},P_1)+P_n(i\frac{e}{2},\xi \left(\frac{e}{2}\right),\frac{e}{2},\dots ,H_{21},P_1)\hfill \\ \hspace{1em}+\cdots +P_n(i\frac{e}{2},\frac{e}{2},\frac{e}{2},\dots ,\xi \left(H_{21}\right),P_1),+P_n(i\frac{e}{2},\frac{e}{2},\frac{e}{2},\dots ,H_{21},\xi \left(P_1\right)).\hfill \\ =P_n(\xi \left(i\frac{e}{2}\right),\frac{e}{2},\frac{e}{2},\dots ,(H_{12}+H_{21}),P_1)+P_n(i\frac{e}{2},\xi \left(\frac{e}{2}\right),\frac{e}{2},\dots ,(H_{12}+H_{21}),P_1)\hfill \\ \hspace{1em}+\cdots +P_n(i\frac{e}{2},\frac{e}{2},\frac{e}{2},\dots ,\xi \left(H_{12}\right)+\xi \left(H_{21}\right),P_1)+P_n(i\frac{e}{2},\frac{e}{2},\frac{e}{2},\dots ,(H_{12}+H_{12}),\xi \left(P_1\right)).\hfill \end{array}$$

On the other hand, we have

$$\begin{array}{c}{\displaystyle \xi \left(P_n(i\frac{e}{2},\frac{e}{2},\frac{e}{2},\dots ,(H_{12}+H_{21}),P_1)\right)}\hfill \\ =P_n(\xi \left(i\frac{e}{2}\right),\frac{e}{2},\frac{e}{2},\dots ,(H_{12}+H_{21}),P_1)+P_n(i\frac{e}{2},\xi \left(\frac{e}{2}\right),\frac{e}{2},\dots ,(H_{12}+H_{21}),P_1)\hfill \\ \hspace{1em}+\cdots +P_n(i\frac{e}{2},\frac{e}{2},\frac{e}{2},\dots ,\xi (H_{12}+H_{21}),P_1)+P_n(i\frac{e}{2},\frac{e}{2},\frac{e}{2},\dots ,(H_{12}+H_{12}),\xi \left(P_1\right)).\hfill \end{array}$$

From the last two expressions for $\xi \left(P_n(i\frac{e}{2},\frac{e}{2},\frac{e}{2},\dots ,(H_{12}+H_{21}),P_1)\right)$, we obtain $P_n(i\frac{e}{2},\frac{e}{2},\frac{e}{2},\dots ,T,P_1=0).$ On simplifying, we obtain $T_{21}=0$, and similarly, we can obtain $T_{12}=0$.

Hence, $T=0$; that is, $\xi (H_{12}+H_{21})=\xi \left(H_{12}\right)+\xi \left(H_{21}\right)$. □

Lemma 3.

For any $H_{11}\in {\mathcal{A}}_{11},H_{12}\in {\mathcal{A}}_{12},H_{21}\in {\mathcal{A}}_{21},$ and $H_{22}\in {\mathcal{A}}_{22}$, we have

$$\xi (H_{11}+H_{12}+H_{21})=\xi \left(H_{11}\right)+\xi \left(H_{12}\right)+\xi \left(H_{21}\right)$$

and

$$\xi (H_{12}+H_{21}+H_{22})=\xi \left(H_{12}\right)+\xi \left(H_{21}\right)+\xi \left(H_{22}\right).$$

Proof. 

Let $T=\xi (H_{11}+H_{12}+H_{22})-\xi \left(H_{11}\right)-\xi \left(H_{12}\right)-\xi \left(H_{22}\right)$. We show that $T=0$. Using the fact that $P_n(i\frac{e}{2},\frac{e}{2},\dots ,(P_1-P_2),H_{12})=P_n(i\frac{e}{2},\frac{e}{2},\dots ,(P_1-P_2),H_{21})=0$ and Lemma 1, we have

$$\begin{array}{c}{\displaystyle \xi \left(P_n(i\frac{e}{2},\frac{e}{2},\dots ,(P_1-P_2),(H_{11}+H_{12}+H_{21}))\right)}\hfill \\ =\xi \left(P_n(i\frac{e}{2},\frac{e}{2},\dots ,(P_1-P_2),H_{11})\right)+\xi \left(P_n(i\frac{e}{2},\frac{e}{2},\dots ,(P_1-P_2),H_{12})\right)\hfill \\ \hspace{1em}+\xi \left(P_n(i\frac{e}{2},\frac{e}{2},\dots ,(P_1-P_2),H_{21})\right).\hfill \\ =P_n(\xi \left(i\frac{e}{2}\right),\frac{e}{2},\dots ,(P_1-P_2),H_{11})+P_n(i\frac{e}{2},\xi \left(\frac{e}{2}\right),\dots ,(P_1-P_2),H_{11})\hfill \\ \hspace{1em}+\cdots +P_n(i\frac{e}{2},\frac{e}{2},\dots ,\xi (P_1-P_2),H_{11})+P_n(i\frac{e}{2},\frac{e}{2},\dots ,(P_1-P_2),\xi \left(H_{11}\right))\hfill \\ \hspace{1em}+P_n(\xi \left(i\frac{e}{2}\right),\frac{e}{2},\dots ,(P_1-P_2),H_{12})+P_n(i\frac{e}{2},\xi \left(\frac{e}{2}\right),\dots ,(P_1-P_2),H_{12})\hfill \\ \hspace{1em}+\cdots +P_n(i\frac{e}{2},\frac{e}{2},\dots ,\xi (P_1-P_2),H_{12})+P_n(i\frac{e}{2},\frac{e}{2},\dots ,(P_1-P_2),\xi \left(H_{12}\right))\hfill \\ \hspace{1em}+P_n(\xi \left(i\frac{e}{2}\right),\frac{e}{2},\dots ,(P_1-P_2),H_{21})+P_n(i\frac{e}{2},\xi \left(\frac{e}{2}\right),\dots ,(P_1-P_2),H_{21})\hfill \\ \hspace{1em}+\cdots +P_n(i\frac{e}{2},\frac{e}{2},\dots ,\xi (P_1-P_2),H_{21})+P_n(i\frac{e}{2},\frac{e}{2},\dots ,(P_1-P_2),\xi \left(H_{21}\right)).\hfill \\ =P_n(\xi \left(i\frac{e}{2}\right),\frac{e}{2},\dots ,(P_1-P_2),(H_{11}+H_{12}+H_{21}))+P_n(i\frac{e}{2},\xi \left(\frac{e}{2}\right),\dots ,(P_1-P_2),(H_{11}+H_{12}+H_{21}))\hfill \\ \hspace{1em}+\cdots +P_n(i\frac{e}{2},\frac{e}{2},\dots ,\xi (P_1-P_2),(H_{11}+H_{12}+H_{21}))\hfill \\ \hspace{1em}+P_n(i\frac{e}{2},\frac{e}{2},\dots ,(P_1-P_2),\xi \left(H_{11}\right)+\xi \left(H_{12}\right)+\xi \left(H_{21}\right)).\hfill \end{array}$$

On the other hand, we obtain

$$\begin{array}{c}{\displaystyle \xi \left(P_n(i\frac{e}{2},\frac{e}{2},\dots ,(P_1-P_2),(H_{11}+H_{12}+H_{21}))\right)}\hfill \\ =P_n(\xi \left(i\frac{e}{2}\right),\frac{e}{2},\dots ,(P_1-P_2),(H_{11}+H_{12}+H_{21}))+P_n(i\frac{e}{2},\xi \left(\frac{e}{2}\right),\dots ,(P_1-P_2),(H_{11}+H_{12}+H_{21}))\hfill \\ \hspace{1em}+\cdots +P_n(i\frac{e}{2},\frac{e}{2},\dots ,\xi (P_1-P_2),(H_{11}+H_{12}+H_{21}))\hfill \\ \hspace{1em}+P_n(i\frac{e}{2},\frac{e}{2},\dots ,(P_1-P_2),\xi (H_{11}+H_{12}+H_{21})).\hfill \end{array}$$

Comparing the above two expressions for $\xi \left(P_n(i\frac{e}{2},\frac{e}{2},\dots ,(P_1-P_2),(H_{11}+H_{12}+H_{21}))\right),$ we find that $P_n(i\frac{e}{2},\frac{e}{2},\dots ,(P_1-P_2),T)=0$, which leads us to $T_{11}=0$ and $T_{22}=0$.

Invoking the fact that $P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_2,H_{11})=0$ and using Lemmas 1 and 2, we find that

$$\begin{array}{c}{\displaystyle \xi \left(P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_2,(H_{11}+H_{12}+H_{21}))\right)}\hfill \\ =\xi \left(P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_2,H_{11})\right)+\xi \left(P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_2,H_{12})\right)+\xi \left(P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_2,H_{21})\right)\hfill \\ =P_n(\xi \left(i\frac{e}{2}\right),\frac{e}{2},\dots ,P_2,H_{11})+P_n(i\frac{e}{2},\xi \left(\frac{e}{2}\right),\dots ,P_2,H_{11})\hfill \\ \hspace{1em}+\cdots +P_n(i\frac{e}{2},\frac{e}{2},\dots ,\xi \left(P_2\right),H_{11})+P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_2,\xi \left(H_{11}\right))\hfill \\ \hspace{1em}+P_n(\xi \left(i\frac{e}{2}\right),\frac{e}{2},\dots ,P_2,H_{12})+P_n(i\frac{e}{2},\xi \left(\frac{e}{2}\right),\dots ,P_2,H_{12})\hfill \\ \hspace{1em}+\cdots +P_n(i\frac{e}{2},\frac{e}{2},\dots ,\xi \left(P_2\right),H_{12})+P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_2,\xi \left(H_{12}\right))\hfill \\ \hspace{1em}+P_n(\xi \left(i\frac{e}{2}\right),\frac{e}{2},\dots ,P_2,H_{21})+P_n(i\frac{e}{2},\xi \left(\frac{e}{2}\right),\dots ,P_2,H_{21})\hfill \\ \hspace{1em}+\cdots +P_n(i\frac{e}{2},\frac{e}{2},\dots ,\xi \left(P_2\right),H_{21})+P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_2,\xi \left(H_{21}\right)).\hfill \\ =P_n(\xi \left(i\frac{e}{2}\right),\frac{e}{2},\dots ,P_2,(H_{11}+H_{12}+H_{21}))+P_n(i\frac{e}{2},\xi \left(\frac{e}{2}\right),\dots ,P_2,(H_{11}+H_{12}+H_{21}))\hfill \\ \hspace{1em}+\cdots +P_n(i\frac{e}{2},\frac{e}{2},\dots ,\xi \left(P_2\right),(H_{11}+H_{12}+H_{21}))\hfill \\ \hspace{1em}+P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_2,\xi \left(H_{11}\right)+\xi \left(H_{12}\right)+\xi \left(H_{21}\right)).\hfill \end{array}$$

On the other hand, we have

$$\begin{array}{c}{\displaystyle \xi \left(P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_2,(H_{11}+H_{12}+H_{21}))\right)}\hfill \\ =P_n(\xi \left(i\frac{e}{2}\right),\frac{e}{2},\dots ,P_2,(H_{11}+H_{12}+H_{21}))+P_n(i\frac{e}{2},\xi \left(\frac{e}{2}\right),\dots ,P_2,(H_{11}+H_{12}+H_{21}))\hfill \\ \hspace{1em}+\cdots +P_n(i\frac{e}{2},\frac{e}{2},\dots ,\xi \left(P_2\right),(H_{11}+H_{12}+H_{21}))+P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_2,\xi (H_{11}+H_{12}+H_{21})).\hfill \end{array}$$

Comparing the above two expressions for $\xi \left(P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_2,(H_{11}+H_{12}+H_{21}))\right),$ we obtain that $P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_2,T)=0$, which further implies that $T_{12}=0$ and $T_{21}=0$. Hence $T=0$; that is,

$$\xi (H_{11}+H_{12}+H_{21})=\xi \left(H_{11}\right)+\xi \left(H_{12}\right)+\xi \left(H_{21}\right).$$

Similarly, we can show that $\xi (H_{12}+H_{21}+H_{22})=\xi \left(H_{12}\right)+\xi \left(H_{21}\right)+\xi \left(H_{22}\right).$ □

Lemma 4.

For any $H_{11}\in {\mathcal{A}}_{11},H_{12}\in {\mathcal{A}}_{12},H_{21}\in {\mathcal{A}}_{21},$ and $H_{22}\in {\mathcal{A}}_{22}$, we have

$$\xi (H_{11}+H_{12}+H_{21}+H_{22})=\xi \left(H_{11}\right)+\xi \left(H_{12}\right)+\xi \left(H_{21}\right)+\xi \left(H_{22}\right).$$

Proof. 

Let $T=\xi (H_{11}+H_{12}+H_{21}+H_{22})-\xi \left(H_{11}\right)-\xi \left(H_{12}\right)-\xi \left(H_{21}\right)-\xi \left(H_{22}\right)$. We show that $T=0$. Using the fact that $P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_1,H_{22})=0$ and Lemmas 1 and 3, we find that

$$\begin{array}{c}{\displaystyle \xi \left(P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_1,(H_{11}+H_{12}+H_{21}+H_{22}))\right)}\hfill \\ =\xi \left(P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_1,(H_{11}+H_{12}+H_{21}))\right)+\xi \left(P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_1,H_{22})\right).\hfill \\ =P_n(\xi \left(i\frac{e}{2}\right),\frac{e}{2},\dots ,P_1,(H_{11}+H_{12}+H_{21}))+P_n(i\frac{e}{2},\xi \left(\frac{e}{2}\right),\dots ,P_1,(H_{11}+H_{12}+H_{21}))\hfill \\ \hspace{1em}+\cdots +P_n(i\frac{e}{2},\frac{e}{2},\dots ,\xi \left(P_1\right),(H_{11}+H_{12}+H_{21}))+P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_1,\xi \left(H_{11}\right)+\xi \left(H_{12}\right)+\xi \left(H_{21}\right))\hfill \\ \hspace{1em}+P_n(\xi \left(i\frac{e}{2}\right),\frac{e}{2},\dots ,P_1,H_{22})+P_n(i\frac{e}{2},\xi \left(\frac{e}{2}\right),\dots ,P_1,H_{22})\hfill \\ \hspace{1em}+\cdots +P_n(i\frac{e}{2},\frac{e}{2},\dots ,\xi \left(P_1\right),H_{22})+P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_1,\xi \left(H_{22}\right)).\hfill \\ =P_n(\xi \left(i\frac{e}{2}\right),\frac{e}{2},\dots ,P_1,(H_{11}+H_{12}+H_{21}+H_{22}))+P_n(i\frac{e}{2},\xi \left(\frac{e}{2}\right),\dots ,P_1,(H_{11}+H_{12}+H_{21}+H_{22}))\hfill \\ \hspace{1em}+\cdots +P_n(i\frac{e}{2},\frac{e}{2},\dots ,\xi \left(P_1\right),(H_{11}+H_{12}+H_{21}+H_{22}))\hfill \\ \hspace{1em}+P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_1,(\xi \left(H_{11}\right)+\xi \left(H_{12}\right)+\xi \left(H_{21}\right)+\xi \left(H_{22}\right))).\hfill \end{array}$$

On the other hand, we have

$$\begin{array}{c}{\displaystyle \xi \left(P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_1,(H_{11}+H_{12}+H_{21}+H_{22}))\right)}\hfill \\ =P_n(\xi \left(i\frac{e}{2}\right),\frac{e}{2},\dots ,P_1,(H_{11}+H_{12}+H_{21}+H_{22}))+P_n(i\frac{e}{2},\xi \left(\frac{e}{2}\right),\dots ,P_1,(H_{11}+H_{12}+H_{21}+H_{22}))\hfill \\ \hspace{1em}+\cdots +P_n(i\frac{e}{2},\frac{e}{2},\dots ,\xi \left(P_1\right),(H_{11}+H_{12}+H_{21}+H_{22}))\hfill \\ \hspace{1em}+P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_1,\left(\xi (H_{11}+H_{12}+H_{21}+H_{22})\right)).\hfill \end{array}$$

Comparing the above two expressions for $\xi \left(P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_1,(H_{11}+H_{12}+H_{21}+H_{22}))\right),$ we obtain that $P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_1,T)=0$, which further implies that $T_{11}=T_{12}=T_{21}=0$. Similarly, we can show that $T_{22}=0$. Thus $T=0$, that is,

$$\xi (H_{11}+H_{12}+H_{21}+H_{22})=\xi \left(H_{11}\right)+\xi \left(H_{12}\right)+\xi \left(H_{21}\right)+\xi \left(H_{22}\right).$$

□

Lemma 5.

For any $H_{12},H_{12}^{\prime }\in {\mathcal{A}}_{12}$ and $H_{21},H_{21}^{\prime }\in {\mathcal{A}}_{21}$, we have

$$\xi (H_{12}+H_{12}^{\prime })=\xi \left(H_{12}\right)+\xi \left(H_{12}^{\prime }\right)\mathit{and}\xi (H_{21}+H_{21}^{\prime })=\xi \left(H_{21}\right)+\xi \left(H_{21}^{\prime }\right).$$

Proof. 

Using the fact that $P_n(-i\frac{e}{2},\frac{e}{2},\dots ,(P_1+H_{12}),i(P_2+H_{12}^{\prime }))=H_{12}+H_{12}^{\prime }+H_{12}^{*}+H_{12}^{\prime }{H}_{12}^{*}$ and Lemma 4, we have

$$\begin{array}{c}{\displaystyle \xi (H_{12}+H_{12}^{\prime })+\xi \left(H_{12}^{*}\right)+\xi \left(H_{12}^{\prime }{H}_{12}^{*}\right)}\hfill \\ =\xi \left(P_n(-i\frac{e}{2},\frac{e}{2},\dots ,(P_1+H_{12}),i(P_2+H_{12}^{\prime }))\right).\hfill \\ =P_n(\xi (-i\frac{e}{2}),\frac{e}{2},\dots ,(P_1+H_{12}),i(P_2+H_{12}^{\prime }))+P_n(-i\frac{e}{2},\xi \left(\frac{e}{2}\right),\dots ,(P_1+H_{12}),i(P_2+H_{12}^{\prime }))\hfill \\ \hspace{1em}+\cdots +P_n(-i\frac{e}{2},\frac{e}{2},\dots ,(\xi \left(P_1\right)+\xi \left(H_{12}\right)),i(P_2+H_{12}^{\prime }))\hfill \\ \hspace{1em}+P_n(-i\frac{e}{2},\frac{e}{2},\dots ,(P_1+H_{12}),(\xi \left(i{P}_2\right)+\xi \left(i{H}_{12}^{\prime }\right))).\hfill \\ =\xi \left(P_n(-i\frac{e}{2},\frac{e}{2},\dots ,P_1,i{P}_2)\right)+\xi \left(P_n(-i\frac{e}{2},\frac{e}{2},\dots ,H_{12},i{H}_{12}^{\prime })\right)+\xi \left(P_n(-i\frac{e}{2},\frac{e}{2},\dots ,P_1,i{H}_{12}^{\prime })\right)\hfill \\ \hspace{1em}+\xi \left(P_n(-i\frac{e}{2},\frac{e}{2},\dots ,H_{12},i{P}_2)\right).\hfill \\ =\xi \left(H_{12}^{\prime }\right)+\xi (H_{12}+H_{12}^{*})+\xi \left(H_{12}^{\prime }{H}_{12}^{*}\right)\hfill \\ =\xi \left(H_{12}^{\prime }\right)+\xi \left(H_{12}\right)+\xi \left(H_{12}^{\ast }\right)+\xi \left(H_{12}^{\prime }{H}_{12}^{\ast }\right)\hfill \end{array}$$

Hence, $\xi (H_{12}+H_{12}^{\prime })=\xi \left(H_{12}\right)+\xi \left(H_{12}^{\prime }\right)$ for any $H_{12}\in {\mathcal{A}}_{12}$ and $H_{12}^{\prime }\in {\mathcal{A}}_{12}$. Similarly, we can prove other part. □

Lemma 6.

For any $H_{ii},H_{ii}^{\prime }\in {\mathcal{A}}_{ii}$ for $(i=1,2)$, we have

$$\xi (H_{11}+H_{11}^{\prime })=\xi \left(H_{11}\right)+\xi \left(H_{11}^{\prime }\right)\mathit{and}\xi (H_{22}+H_{22}^{\prime })=\xi \left(H_{22}\right)+\xi \left(H_{22}^{\prime }\right).$$

Proof. 

Let $T=\xi (H_{11}+H_{11}^{\prime })-\xi \left(H_{11}\right)-\xi \left(H_{11}^{\prime }\right)$; we show that $T=0$.

Using the fact that $P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_2,H_{11})=P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_2,H_{11}^{\prime })=0$ and Lemma 1, we obtain

$$\begin{array}{c}{\displaystyle \xi \left(P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_2,(H_{11}+H_{11}^{\prime }))\right)}.\hfill \\ =\xi (P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_2,H_{11})+\xi (P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_2,H_{11}^{\prime }).\hfill \\ =P_n(\xi \left(i\frac{e}{2}\right),\frac{e}{2},\dots ,P_2,H_{11})+P_n(i\frac{e}{2},\xi \left(\frac{e}{2}\right),\dots ,P_2,H_{11})\hfill \\ \hspace{1em}+\cdots +P_n(i\frac{e}{2},\frac{e}{2},\dots ,\xi \left(P_2\right),H_{11})+P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_2,\xi \left(H_{11}\right))\hfill \\ \hspace{1em}+P_n(\xi \left(i\frac{e}{2}\right),\frac{e}{2},\dots ,P_2,H_{11}^{\prime })+P_n(i\frac{e}{2},\xi \left(\frac{e}{2}\right),\dots ,P_2,H_{11}^{\prime })\hfill \\ \hspace{1em}+\cdots +P_n(i\frac{e}{2},\frac{e}{2},\dots ,\xi \left(P_2\right),H_{11}^{\prime })+P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_2,\xi \left(H_{11}^{\prime }\right)).\hfill \\ =P_n(\xi \left(i\frac{e}{2}\right),\frac{e}{2},\dots ,P_2,H_{11}+H_{11}^{\prime })+P_n(i\frac{e}{2},\xi \left(\frac{e}{2}\right),\dots ,P_2,H_{11}+H_{11}^{\prime })\hfill \\ \hspace{1em}+\cdots +P_n(i\frac{e}{2},\frac{e}{2},\dots ,\xi \left(P_2\right),H_{11}+H_{11}^{\prime })+P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_2,\xi \left(H_{11}\right)+\xi \left(H_{11}^{\prime }\right)).\hfill \end{array}$$

On the other hand, we have

$$\begin{array}{c}{\displaystyle \xi \left(P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_2,(H_{11}+H_{11}^{\prime }))\right)}.\hfill \\ =P_n(\xi \left(i\frac{e}{2}\right),\frac{e}{2},\dots ,P_2,H_{11}+H_{11}^{\prime })+P_n(i\frac{e}{2},\xi \left(\frac{e}{2}\right),\dots ,P_2,H_{11}+H_{11}^{\prime })\hfill \\ \hspace{1em}+\cdots +P_n(i\frac{e}{2},\frac{e}{2},\dots ,\xi \left(P_2\right),H_{11}+H_{11}^{\prime })+P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_2,\xi (H_{11}+H_{11}^{\prime })).\hfill \end{array}$$

Comparing the above two expressions for $\xi \left(P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_2,(H_{11}+H_{11}^{\prime }))\right)$, we find that $P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_2,T)=0,$ which in turn gives $T_{12}=T_{21}=T_{22}=0$.

Next, we show that $T_{11}=0$. Let $X_{12}\in {\mathcal{A}}_{12}$, and it is easy to observe that $P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_1,H_{11},X_{12}),P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_1,H_{11}^{\prime },X_{12})\in {\mathcal{A}}_{12}.$ Thus, using Lemma 5, we find that

$$\begin{array}{c}{\displaystyle \xi \left(P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_1,(H_{11}+H_{11}^{\prime }),X_{12})\right)}.\hfill \\ =\xi \left(P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_1,H_{11},X_{12})\right)+\xi \left(P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_1,H_{11}^{\prime },X_{12})\right)\hfill \\ =P_n(\xi \left(i\frac{e}{2}\right),\frac{e}{2},\dots ,P_1,H_{11},X_{12})+P_n(i\frac{e}{2},\xi \left(\frac{e}{2}\right),\dots ,P_1,H_{11},X_{12})\hfill \\ \hspace{1em}+\cdots +P_n(i\frac{e}{2},\frac{e}{2},\dots ,\xi \left(P_1\right),H_{11},X_{12})+P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_1,\xi \left(H_{11}\right),X_{12})\hfill \\ \hspace{1em}+P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_1,H_{11},\xi \left(X_{12}\right))+P_n(\xi \left(i\frac{e}{2}\right),\frac{e}{2},\dots ,P_1,H_{11}^{\prime },X_{12})\hfill \\ \hspace{1em}+P_n(i\frac{e}{2},\xi \left(\frac{e}{2}\right),\dots ,P_1,H_{11}^{\prime },X_{12})+\cdots +P_n(i\frac{e}{2},\frac{e}{2},\dots ,\xi \left(P_1\right),H_{11}^{\prime },X_{12})\hfill \\ \hspace{1em}+P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_1,\xi \left(H_{11}^{\prime }\right),X_{12})+P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_1,H_{11}^{\prime },\xi \left(X_{12}\right)).\hfill \\ =P_n(\xi \left(i\frac{e}{2}\right),\frac{e}{2},\dots ,P_1,(H_{11}+H_{11}^{\prime }),X_{12})+P_n(i\frac{e}{2},\xi \left(\frac{e}{2}\right),\dots ,P_1,(H_{11}+H_{11}^{\prime }),X_{12})\hfill \\ \hspace{1em}+\cdots +P_n(i\frac{e}{2},\frac{e}{2},\dots ,\xi \left(P_1\right),(H_{11}+H_{11}^{\prime }),X_{12})\hfill \\ \hspace{1em}+P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_1,(\xi \left(H_{11}\right)+\xi \left(H_{11}^{\prime }\right)),X_{12})+P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_1,(H_{11}+H_{11}^{\prime }),\xi \left(X_{12}\right)).\hfill \end{array}$$

On the other hand, we have

$$\begin{array}{c}{\displaystyle \xi \left(P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_1,(H_{11}+H_{11}^{\prime }),X_{12})\right)}.\hfill \\ =P_n(\xi \left(i\frac{e}{2}\right),\frac{e}{2},\dots ,P_1,(H_{11}+H_{11}^{\prime }),X_{12})+P_n(i\frac{e}{2},\xi \left(\frac{e}{2}\right),\dots ,P_1,(H_{11}+H_{11}^{\prime }),X_{12})\hfill \\ \hspace{1em}+\cdots +P_n(i\frac{e}{2},\frac{e}{2},\dots ,\xi \left(P_1\right),(H_{11}+H_{11}^{\prime }),X_{12})\hfill \\ \hspace{1em}+P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_1,\left(\xi (H_{11}+H_{11}^{\prime })\right),X_{12})+P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_1,(H_{11}+H_{11}^{\prime }),\xi \left(X_{12}\right)).\hfill \end{array}$$

From the last two expressions for $\xi \left(P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_1,(H_{11}+H_{11}^{\prime }),X_{12})\right)$, we obtain $P_n(i\frac{e}{2},\frac{e}{2},\dots ,P_1,T,X_{12})=0$, which implies $T_{11}X{P}_2=0.$ Application of condition (1) yields $T_{11}=0.$ Hence, $T=0$; that is, $\xi (H_{11}+H_{11}^{\prime })=\xi \left(H_{11}\right)+\xi \left(H_{11}^{\prime }\right)$. Symmetrically, one can prove that $\xi (H_{22}+H_{22}^{\prime })=\xi \left(H_{22}\right)+\xi \left(H_{22}^{\prime }\right)$. □

Lemma 7.

ξ is additive on $\mathcal{A}$.

Proof. 

For any $H,K\in \mathcal{A}$, we have $H=H_{11}+H_{12}+H_{21}+H_{22}$ and $K=K_{11}+K_{12}+K_{21}+K_{22}$. With the help of Lemmas 4–6, we obtain

$$\begin{array}{ccc}\xi (H+K)\hfill & =& \xi (H_{11}+H_{12}+H_{21}+H_{22}+K_{11}+K_{12}+K_{21}+K_{22})\hfill \\ & =& \xi (H_{11}+K_{11})+\xi (H_{12}+K_{12})+\xi (H_{21}+K_{21})+\xi (H_{22}+K_{22})\hfill \\ & =& \xi \left(H_{11}\right)+\xi \left(K_{11}\right)+\xi \left(H_{12}\right)+\xi \left(K_{12}\right)+\xi \left(H_{21}\right)+\xi \left(K_{21}\right)+\xi \left(H_{22}\right)+\xi \left(K_{22}\right)\hfill \\ & =& \xi (H_{11}+H_{12}+H_{21}+H_{22})+\xi (K_{11}+K_{12}+K_{21}+K_{22})\hfill \\ & =& \xi \left(H\right)+\xi \left(K\right).\hfill \end{array}$$

□

Lemma 8.

$\xi {\left(\frac{e}{2}\right)}^{*}=\xi \left(\frac{e}{2}\right)$.

Proof. 

It follows from $P_n(\frac{e}{2},\frac{e}{2},\dots ,\frac{e}{2})=0$ and Lemma 1 that we have

$$\begin{array}{ccc}0\hfill & =& \xi \left(P_n(\frac{e}{2},\frac{e}{2},\dots ,\frac{e}{2})\right).\hfill \\ & =& P_n\left(\xi \left(\frac{e}{2}\right),\frac{e}{2},\dots ,\frac{e}{2}\right)+P_n\left(\frac{e}{2},\xi \left(\frac{e}{2}\right),\dots ,\frac{e}{2}\right)+\cdots +P_n\left(\frac{e}{2},\frac{e}{2},\dots ,\xi \left(\frac{e}{2}\right)\right).\hfill \end{array}$$

On simplifying, we obtain $\xi {\left(\frac{e}{2}\right)}^{*}=\xi \left(\frac{e}{2}\right)$. □

Lemma 9.

If $\xi {\left(i\frac{e}{2}\right)}^{*}=\xi \left(i\frac{e}{2}\right)$, $\mathit{then}$ $\xi \left(i\frac{e}{2}\right)=\xi \left(\frac{e}{2}\right)=0$.

Proof. 

Using the fact that $P_n(i\frac{e}{2},\frac{e}{2},\dots ,\frac{e}{2})=i\frac{e}{2}$ and Lemma 8, we obtain

$$\begin{array}{ccc}\xi \left(i\frac{e}{2}\right)\hfill & =& \xi \left(P_n\left(i\frac{e}{2},\frac{e}{2},\dots ,\frac{e}{2}\right)\right).\hfill \\ & =& P_n\left(\xi \left(i\frac{e}{2}\right),\frac{e}{2},\dots ,\frac{e}{2}\right)+P_n\left(i\frac{e}{2},\xi \left(\frac{e}{2}\right),\dots ,\frac{e}{2}\right)+\cdots +P_n\left(i\frac{e}{2},\frac{e}{2},\dots ,\xi \left(\frac{e}{2}\right)\right).\hfill \\ & =& (n-1)i\xi \left(\frac{e}{2}\right).\hfill \end{array}$$

Taking the adjoint on both side of the above relation, we obtain

$$\begin{array}{c}\xi {\left(i\frac{e}{2}\right)}^{*}=-(n-1)i\xi \left(\frac{e}{2}\right),\hfill \end{array}$$

since $\xi \left(i\frac{e}{2}\right)$ is self-adjoint. On combining the last two relations, we obtain $\xi \left(i\frac{e}{2}\right)=0$ and $\xi \left(\frac{e}{2}\right)=0$. □

Lemma 10.

For any $H\in \mathcal{A}$, we have $\xi \left(H^{*}\right)=\xi {\left(H\right)}^{*}$.

Proof. 

Observe that $P_n(-i\frac{e}{2},\frac{e}{2},\dots ,\frac{e}{2},H,i\frac{e}{2})=\frac{H+H^{*}}{2}$ for any $H\in \mathcal{A}.$ Using Lemmas 7 and 9, we find that

$$\begin{array}{ccc}\xi \left(\frac{H+H^{*}}{2}\right)\hfill & =& \xi \left(P_n(-i\frac{e}{2},\frac{e}{2},\dots ,\frac{e}{2},H,i\frac{e}{2})\right)\hfill \\ & =& P_n(-i\frac{e}{2},\frac{e}{2},\dots ,\frac{e}{2},\xi \left(H\right),i\frac{e}{2})\hfill \\ & =& \frac{1}{2}(\xi \left(H\right)+\xi {\left(H\right)}^{*}),\hfill \end{array}$$

which implies

$$\xi \left(H^{*}\right)=\xi {\left(H\right)}^{*}.$$

□

Lemma 11.

$\xi \left(iH\right)=i\xi \left(H\right)$ for every $H\in \mathcal{A}$.

Proof. 

Observe that $P_n(i\frac{e}{2},\frac{e}{2},\dots ,\frac{e}{2},H)=iH$ for every $H\in \mathcal{A}$, and using Lemma 9, we obtain

$$\begin{array}{ccc}\xi \left(iH\right)\hfill & =& \xi \left(P_n(i\frac{e}{2},\frac{e}{2},\dots ,\frac{e}{2},H\right))\hfill \\ & =& P_n(i\frac{e}{2},\frac{e}{2},\dots ,\frac{e}{2},\xi \left(H\right))\hfill \\ & =& i\xi \left(H\right).\hfill \end{array}$$

□

Lemma 12.

$\xi \left(HK\right)=\xi \left(H\right)K+H\xi \left(K\right)$ for all $H,K\in \mathcal{A}$.

Proof. 

Observe that $P_n(i\frac{e}{2},\frac{e}{2},\dots ,\frac{e}{2},H,K)=i(HK+K{H}^{*})$ for any $H,K\in \mathcal{A}$, and using Lemmas 7 and 9–11, we obtain

$$\begin{array}{ccc}i\xi (HK+K{H}^{*})\hfill & =& \xi \left(P_n(i\frac{e}{2},\frac{e}{2},\dots ,\frac{e}{2},H,K\right))\hfill \\ & =& P_n(i\frac{e}{2},\frac{e}{2},\dots ,\frac{e}{2},\xi \left(H\right),K)+P_n(i\frac{e}{2},\frac{e}{2},\dots ,\frac{e}{2},H,\xi \left(K\right))\hfill \\ & =& i\xi \left(H\right)K+iH\xi \left(K\right)+i\xi \left(K\right)H^{*}+iK\xi {\left(H\right)}^{*},\hfill \end{array}$$

which implies that

$$\begin{array}{ccc}\xi (HK+K{H}^{*})\hfill & =& \xi \left(H\right)K+H\xi \left(K\right)+\xi \left(K\right)H^{*}+K\xi {\left(H\right)}^{*}.\hfill \end{array}$$

(3)

Equation $\left(3\right)$ implies that

$$\begin{array}{ccc}\xi (HK-K{H}^{*})\hfill & =& \xi (\left(iH\right)(-iK)+(-iK){\left(iH\right)}^{*})\hfill \\ & =& \xi \left(iH\right)(-iK)+\left(iH\right)\xi (-iK)+\xi (-iK){\left(iH\right)}^{*}+(-iK)\xi \left({\left(iH\right)}^{*}\right)\hfill \\ & =& \xi \left(H\right)K+H\xi \left(K\right)-\xi \left(K\right)H^{*}-K\xi {\left(H\right)}^{*}.\hfill \end{array}$$

Hence,

$$\begin{array}{ccc}\xi (HK-K{H}^{*})\hfill & =& \xi \left(H\right)K+H\xi \left(K\right)-\xi \left(K\right)H^{*}-K\xi {\left(H\right)}^{*}.\hfill \end{array}$$

(4)

On combining $\left(3\right)$ and $\left(4\right)$, we obtain

$$\xi \left(HK\right)=\xi \left(H\right)K+H\xi \left(K\right).$$

□

By Lemmas 7, 10, and 12, $\xi$ is an additive ∗-derivation. This completes the proof of Theorem 1.

3. Applications of Theorem 1

In this section, we apply Theorem 1 to certain special classes of ∗-algebras, namely prime ∗-algebras, standard operator algebras, factor von Neumann algebras, and von Neumann algebras with no central summands of type $I_1.$

Recall that an algebra $\mathcal{A}$ is prime if for any $A,B\in \mathcal{A},$$A\mathcal{A}B=\left\{0\right\}$ implies that either $A=0$ or $B=0.$ It is easy to verify that every prime ∗-algebra satisfies (1). Therefore, as a direct consequence of Theorem 1, we have the following result:

Corollary 1.

Let $\mathcal{A}$ be a unital prime ∗-algebra containing a nontrivial projection. Then, if a map $\xi :\mathcal{A}\to \mathcal{A}$ satisfies

$$\begin{array}{ccc}\xi \left(P_n(H_1,H_2,\dots ,H_n)\right)\hfill & =& \sum _{i=1}^n{P}_n(\left(H_1\right),\dots ,H_{i-1},\xi \left(H_i\right),H_{i+1},\dots ,H_n)(n\ge 3)\hfill \end{array}$$

(5)

for all $H_1,H_2,\dots ,H_n\in \mathcal{A}$, then ξ is additive. Moreover, if $\xi \left(i\frac{e}{2}\right)$ is self-adjoint, then ξ is ∗-derivation.

Let $\mathcal{H}$ be a complex Hilbert space and $\mathcal{B}\left(\mathcal{H}\right)$ be the algebra of all bounded linear operators on $\mathcal{H}$. Let $\mathcal{F}\left(\mathcal{H}\right)\subseteq \mathcal{B}\left(\mathcal{H}\right)$ denote the subalgebra of all bounded finite rank operators. A subalgebra $\mathcal{A}\subseteq \mathcal{B}\left(\mathcal{H}\right)$ is called a standard operator algebra if it contains $\mathcal{F}\left(\mathcal{H}\right).$ Now, we have the following result:

Corollary 2.

Let $\mathcal{H}$ be an infinite dimensional complex Hilbert space and $\mathcal{A}$ be a standard operator algebra on $\mathcal{H}$ containing the identity operator $I.$ Suppose that $\mathcal{A}$ is closed under the adjoint operation. Then, if a map $\xi :\mathcal{A}\to \mathcal{B}\left(\mathcal{A}\right)$ satisfies

$$\begin{array}{ccc}\xi \left(P_n(H_1,H_2,\dots ,H_n)\right)\hfill & =& \sum _{i=1}^n{P}_n(\left(H_1\right),\dots ,H_{i-1},\xi \left(H_i\right),H_{i+1},\dots ,H_n)(n\ge 3)\hfill \end{array}$$

(6)

for all $H_1,H_2,\dots ,H_n\in \mathcal{A}$, then ξ is additive. Moreover, if $\xi \left(i\frac{e}{2}\right)$ is self-adjoint, then ξ is ∗-derivation. Moreover, there exists an operator $T\in \mathcal{B}\left(\mathcal{H}\right)$ satisfying $T+T^{*}=0$ such that $\xi \left(A\right)=AT-TA$ for all $A\in \mathcal{A};$ that is, ξ is inner.

Proof. 

Since $\mathcal{A}$ is a unital prime ∗-algebra containing nontrivial projections, then by Corollary 1, we see that $\xi$ is an additive ∗-derivation. It follows from [18] that $\xi$ is a linear inner derivation; that is, there exists an operator $S\in \mathcal{B}\left(\mathcal{H}\right)$ such that $\xi \left(A\right)=AS-SA$ for all $A\in \mathcal{A}.$ Using the fact that $\xi \left(A^{*}\right)={\xi \left(A\right)}^{*},$ we have

$$A^{*}S-S{A}^{*}=\xi \left(A^{*}\right)={\xi \left(A\right)}^{*}=S^{*}{A}^{*}-A^{*}{S}^{*}$$

for any $A\in \mathcal{A},$ This leads to $A^{*}(S+S^{*})=(S+S^{*})A^{*}.$ Hence, $S+S^{*}=\lambda I$ for some $\lambda \in \mathbb{R}.$ Letting $T=S-\frac{1}{2}\lambda I,$ one can check that $T+T^{*}=0$ and $\xi \left(A\right)=AT-TA$ for all $A\in \mathcal{A}.$ □

A von Neumann algebra $\mathcal{A}$ is a weakly closed self-adjoint algebra of operators on a Hilbert space $\mathcal{H}$ containing the identity operator I. A von Neumann algebra $\mathcal{A}$ is a factor von Neumann algebra if its center contains only the scalar operators. It is well known that a factor von Neumann algebra is prime; thus, it always satisfies (1). Hence, as an immediate consequence of Corollary 1, we obtain

Corollary 3.

Let $\mathcal{A}$ be a factor von Neumann algebra with dim $\left(\mathcal{A}\right)\ge 2.$ Then, if a map $\xi :\mathcal{A}\to \mathcal{A}$ satisfies

$$\begin{array}{ccc}\xi \left(P_n(H_1,H_2,\dots ,H_n)\right)\hfill & =& \sum _{i=1}^n{P}_n(\left(H_1\right),\dots ,H_{i-1},\xi \left(H_i\right),H_{i+1},\dots ,H_n)(n\ge 3)\hfill \end{array}$$

(7)

for all $H_1,H_2,\dots ,H_n\in \mathcal{A}$, then ξ is additive. Moreover, if $\xi \left(i\frac{e}{2}\right)$ is self-adjoint, then ξ is ∗-derivation.

Further, it is well known that every von Neumann algebra with no central summands of type $I_1$ satisfies (1) (see [8,19] for details). Therefore, applying Theorem 1, we have the following result:

Corollary 4.

Let $\mathcal{A}$ be a von Neumann algebra having no central summands of type $I_1.$ Then, if a map $\xi :\mathcal{A}\to \mathcal{A}$ satisfies

$$\begin{array}{ccc}\xi \left(P_n(H_1,H_2,\dots ,H_n)\right)\hfill & =& \sum _{i=1}^n{P}_n(\left(H_1\right),\dots ,H_{i-1},\xi \left(H_i\right),H_{i+1},\dots ,H_n)(n\ge 3)\hfill \end{array}$$

(8)

for all $H_1,H_2,\dots ,H_n\in \mathcal{A}$, then ξ is additive. Moreover, if $\xi \left(i\frac{e}{2}\right)$ is self-adjoint, then ξ is ∗-derivation.

4. Conclusions

In this article, we examine the pattern of nonlinear skew Lie-type derivation $\left(\xi \right)$ on ∗-algebra $\mathcal{A}$. In fact, we proved that such a map is an additive derivation, preserving the ∗-structure of algebra $\mathcal{A}$, i.e., $\xi \left(H^{*}\right)=\xi {\left(H\right)}^{*}$ for all $H\in \mathcal{A}$. One can further investigate the structure of nonlinear skew Lie-type derivations on a variety of algebras such as incidence algebras, nest algebras, etc.