Nonlinear Mixed Jordan Triple *-Higher Derivations on *-Algebras

Xinzhao YU; Jianhua ZHANG

doi:10.1051/wujns/2025305479

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Volume 30 / No 5 (October 2025)

Wuhan Univ. J. Nat. Sci., 30 5 (2025) 479-489

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Issue		Wuhan Univ. J. Nat. Sci. Volume 30, Number 5, October 2025


Page(s)		479 - 489
DOI		https://doi.org/10.1051/wujns/2025305479
Published online		04 November 2025

Wuhan University Journal of Natural Sciences, 2025, Vol.30 No.5, 479-489

CLC number: O177

Nonlinear Mixed Jordan Triple -Higher Derivations on -Algebras

-代数上的非线性混合Jordan三重-高阶导子

Xinzhao YU (余欣曌) and Jianhua ZHANG (张建华)^†

School of Mathematics and Statistics, Shaanxi Normal University, Xi’an 710062, Shaanxi, China

^† Corresponding author. E-mail: jhzhang@snnu.edu.cn

Received: 16 January 2025

Abstract

Let $A$ be a unital $*$ -algebra containing a nontrivial projection, $N$ be the set of non-negative integers. Under some mild conditions on $A$ , it is shown that any nonlinear mixed Jordan triple $*$ -higher derivation $D = {d_{n}}_{n \in N}$ is an additive $*$ -higher derivation. In particular, we apply the above result to prime $*$ -algebras and von Neumann algebras with no central summands of type $I_{1}$ .

摘要

设 $A$ 是一个包含非平凡投影和单位元的 $*$ -代数， $N$ 是非负整数集。在满足一定条件的 $A$ 上，我们证明了任意非线性混合Jordan三重 $*$ -高阶导子 $D = {d_{n}}_{n \in N}$ 都是可加的 $*$ -高阶导子。同时作为应用，我们将这一结论运用到了素 $*$ -代数和无 $I_{1}$ 型中心直和项的von Neumann代数。

Key words: mixed Jordan triple *-higher derivation / *-higher derivation / von Neumann algebra

关键字 : 混合Jordan三重*-高阶导子 / *-高阶导子 / von Neumann代数

Cite this article: YU Xinzhao, ZHANG Jianhua. Nonlinear Mixed Jordan Triple *-Higher Derivations on *-Algebras[J]. Wuhan Univ J of Nat Sci, 2025, 30(5): 479-489.

Biography: YU Xinzhao, male, Master candidate, research direction: operator theory and operator algebras. E-mail:xinzhaoyu666@163.com

Foundation item: Supported by the National Natural Science Foundation of China (12271323)

© Wuhan University 2025

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (https://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

0 Introduction

Let $A$ be a $*$ -algebra over the complex field $C$ . For $A, B \in A$ , define the Jordan $*$ -product of A and B by $A • B = A B + B A^{*}$ and the skew Lie product of A and B by ${[A, B]}_{*} = A B - B A^{*}$ . Among some research topics, the Jordan $*$ -product and the skew Lie product are of great significance^[1-9]. Recall that an additive map $Φ : A \to A$ is an additive derivation if $Φ (A B) = Φ (A) B + A Φ (B)$ for all $A, B \in A$ . In addition, $Φ$ is said to be an additive $*$ -derivation if $Φ$ is an additive derivation and $Φ (A^{*}) = Φ {(A)}^{*}$ for all $A \in A$ . Note that $Φ$ is nonlinear if its additivity is removed.

Let $d : A \to A$ be a nonlinear map, $d$ is called

(i) a nonlinear Jordan triple $*$ -derivation if $d (A • B • C) = d (A) • B • C + A • d (B) • C + A • B • d (C)$ for all $A, B, C \in A$ , where $A • B • C = (A • B) • C$ .

(ii) a nonlinear skew Lie triple derivation if $d ([{[A, B]}_{*} {, C]}_{*}) = [[d {(A), B]}_{*} {, C]}_{*} + [[A, d {(B)]}_{*} {, C]}_{*} + [{[A, B]}_{*}, d {(C)]}_{*}$ for all $A, B, C \in A$ . The characterization of Jordan triple $*$ -derivations and skew Lie triple derivations have been studied by many authors^[10-19]. Recently the derivations corresponding to the new products of the mixture of various products have attracted the attentions of many researchers (see Refs. [20-22]). Li and Zhang^[23] considered the mixed Jordan triple $*$ -product ${[A • B, C]}_{*}$ and investigated the nonlinear mixed Jordan triple $*$ -derivation $d : A \to A$ , i.e.

$d ({[A • B, C]}_{*}) = [d {(A) • B, C]}_{*} + [A • d {(B), C]}_{*} + [A • B, d {(C)]}_{*},$

and they proved a nonlinear mixed Jordan triple $*$ -derivation is an additive $*$ -derivation on $*$ -algebras under some mild condition.

Let $N$ be the set of non-negative integers, $D = {Φ_{n}}_{n \in N}$ be a family of additive maps $Φ_{n} : A \to A$ and $Φ_{0}$ be the identity map on $A$ . Then $D = {Φ_{n}}_{n \in N}$ is called an additive higher derivation if for every $n \in N$ , $Φ_{n} (A B) = \sum_{i + j = n} Φ_{i} (A) Φ_{j} (B)$ for all $A, B \in A$ . $D = {Φ_{n}}_{n \in N}$ is said to be an additive $*$ -higher derivation if $D = {Φ_{n}}_{n \in N}$ is an additive higher derivation with $Φ_{n} (A^{*}) = Φ_{n} {(A)}^{*}$ for all $A \in A$ and $n \in N$ . In addition, $D = {Φ_{n}}_{n \in N}$ is nonlinear if the additivity is removed. In recent years, many authors study different types of nonlinear higher derivations. Some researchers studied the nonlinear higher derivations on semi-prime rings and triangular algebras^[24-26].

The questions of characterizing Lie higher derivations and the relationship between different types of higher derivations have been studied by many authors (see Refs. [27-29]).

Let $d_{0}$ be the identity map on $A$ , and let $d_{n} : A \to A$ be a nonlinear map for all $n \geq 1$ . We say that $D = {d_{n}}_{n \in N}$ is

(i) a nonlinear Jordan triple $*$ -higher derivation if $d_{n} (A • B • C) = \sum_{i + j + k = n} d_{i} (A) • d_{j} (B) • d_{k} (C) .$

(ii) a nonlinear skew Lie triple higher derivation if $d_{n} ([{[A, B]}_{*} {, C]}_{*}) = \sum_{i + j + k = n} [[d_{i} (A), d_{j} {(B)]}_{*}, d_{k} (C)]$ for all $A, B, C \in A$ and $n \in N$ . Some reasearchers proved that a nonlinear Jordan triple $*$ -higher derivation or skew Lie triple higher derivation is an additive $*$ -higher derivation on standard operator algebras^[30-31]. Motivated by the related works^{[23, 30-31]}, we consider nonlinear mixed Jordan triple $*$ -higher derivations $D = {d_{n}}_{n \in N}$ , that is, $d_{n} ({[A • B, C]}_{*}) = \sum_{i + j + k = n} [d_{i} (A) • d_{j} (B), d_{k} {(C)]}_{*}$ for all $A, B, C \in A$ and $n \in N$ . It is clear that $d_{1}$ is a nonlinear mixed Jordan triple $*$ -derivation on $A$ if $D = {d_{n}}_{n \in N}$ is a nonlinear mixed Jordan triple $*$ -higher derivation of $A$ , then we will study the nonlinear mixed Jordan triple $*$ -higher derivations on $A$ .

1 Nonlinear Mixed Jordan Triple $*$ -Derivation

Theorem 1 Let $A$ be a $*$ -algebra with the unit $I$ . Assume that $A$ contains a nontrivial projection $P$ which satisfies for $A \in A$ ,

$X A P = 0 i m p l i e s X = 0$ (1)

and

$X A (I - P) = 0 i m p l i e s X = 0$ (2)

Then every nonlinear mixed Jordan triple $*$ -derivation $d : A \to A$ , which satisfies

$d ({[A • B, C]}_{*}) = [d {(A) • B, C]}_{*} + [A • d {(B), C]}_{*} + [A • B, d {(C)]}_{*}$

for all $A, B, C \in A$ , is an additive $*$ -derivation.

Proof Let $P_{1} = P$ and $P_{2} = I - P$ . We write $A_{j k} = P_{j} A P_{k}$ for $j, k = 1,2$ . Then $A = \sum_{j, k = 1}^{2} A_{j k} .$

In the sequel $A_{j k}$ indicates that $A_{j k} \in A_{j k}$ . We set $S = {A \in A, A = A^{*}}$ . We complete the proof by several Claims.

Claim 1 $d_{n} (0) = 0$ for all $n \in N$ .

Proof It follows from ${[0 • 0,0]}_{*} = 0$ that

$d (0) = d ({[0 • 0,0]}_{*}) = [d {(0) • 0,0]}_{*} + [0 • d {(0), 0]}_{*} + [0 • 0, d {(0)]}_{*} = 0 .$

Claim 2 $d$ has the following properties: (a) For any $λ \in R$ , $d (λ I) \in Z (A)$ and $d (λ I) = d {(λ I)}^{*}$ ; (b) For any $A \in S$ , $d (A) = d (A^{*}) = d {(A)}^{*}$ ; (c) For any $λ \in C$ , $d (λ I) \in Z (A)$ .

Proof (a) It follows from ${[I • λ I, I]}_{*} = 0$ that

$0 = d ({[I • λ I, I]}_{*}) = {[d (I) • λ I, I]}_{*} + {[I • d (λ I), I]}_{*} + {[I • λ I, d (I)]}_{*} = {[I • d (λ I), I]}_{*} = 2 d (λ I) - 2 d {(λ I)}^{*} .$

From this we get $d (λ I) = d {(λ I)}^{*} .$ Then for all $A \in A$ , it follows from ${[λ I • λ I, A]}_{*} = 0$ that

$\begin{array}{l} 0 = d ({[λ I • λ I, A]}_{*}) = {[d (λ I) • λ I, A]}_{*} + {[λ I • d (λ I), A]}_{*} + {[λ I • λ I, d (A)]}_{*} \\ = {[d (λ I) • λ I, A]}_{*} + {[λ I • d (λ I), A]}_{*} = 4 λ d (λ I) A - 4 λ A d {(λ I)}^{*} . \end{array}$

Then we get $d (λ I) A = A d (λ I),$ for all $A \in A$ , which implies $d (λ I) \in Z (A) .$

(b) For any $A \in S,$ we see from Claim 2(a) that

$0 = d ({[I • A, I]}_{*}) = {[d (I) • A, I]}_{*} + {[I • d (A), I]}_{*} + {[I • A, d (I)]}_{*} = {[I • d (A), I]}_{*} = 2 d (A) - 2 d {(A)}^{*} .$

(c) For any $λ \in C$ and $A \in S,$ we see from Claim 2(a) and Claim 2(b) that

$0 = d ({[A • I, λ I]}_{*}) = {[d (A) • I, λ I]}_{*} + {[A • d (I), λ I]}_{*} + {[A • I, d (λ I)]}_{*} = {[A • I, d (λ I)]}_{*} = 2 A d (λ I) - 2 d (λ I) A .$

Thus $A d (λ I) = d (λ I) A,$ for all $A \in S,$ which implies $d (λ I) \in Z (A) .$

Claim 3 For any $A \in A,$ $d (\frac{1}{2} I) = d (\frac{1}{2} i I) = 0, d (i A) = i d (A) .$

Proof It follows from ${[\frac{1}{2} I • \frac{1}{2} i I, \frac{1}{2} i I]}_{*} = - \frac{1}{2} I$ , where $i$ is the imaginary unit. We see from Claim 2 that

$d (- \frac{1}{2} I) = d ({[\frac{1}{2} I • \frac{1}{2} i I, \frac{1}{2} i I]}_{*}) = {[d (\frac{1}{2} I) • \frac{1}{2} i I, \frac{1}{2} i I]}_{*} + {[\frac{1}{2} I • d (\frac{1}{2} i I), \frac{1}{2} i I]}_{*} + {[\frac{1}{2} I • \frac{1}{2} i I, d (\frac{1}{2} i I)]}_{*}$

$= - d (\frac{1}{2} I) + \frac{1}{2} i (d (\frac{1}{2} i I) - d {(\frac{1}{2} i I)}^{*}) + i d (\frac{1}{2} i I) .$

Since $d (\frac{1}{2} I), d (- \frac{1}{2} I), i (d (\frac{1}{2} i I) - d {(\frac{1}{2} i I)}^{*})$ are self-adjoint, $i d (\frac{1}{2} i I)$ is also self-adjoint. Hence

$d (- \frac{1}{2} I) = - d (\frac{1}{2} I) + 2 i d (\frac{1}{2} i I) .$ (3)

Similarly, we can also obtain from the fact that $- \frac{1}{2} I = {[\frac{1}{2} I • (- \frac{1}{2} i I), - \frac{1}{2} i I]}_{*}$ , that

$d (- \frac{1}{2} i I) = - d {(- \frac{1}{2} i I)}^{*}$ (4)

and

$d (- \frac{1}{2} I) = - d (\frac{1}{2} I) - 2 i d (- \frac{1}{2} i I) .$ (5)

Now by (3) and (5) we have

$d (- \frac{1}{2} i I) = - d (\frac{1}{2} i I) .$ (6)

It follows from (4) and (6), we compute

$d (\frac{1}{2} i I) = d ({[\frac{1}{2} I • (- \frac{1}{2} i I), - \frac{1}{2} I]}_{*})$

$= {[d (\frac{1}{2} I) • (- \frac{1}{2} i I), - \frac{1}{2} I]}_{*} + {[\frac{1}{2} I • d (- \frac{1}{2} i I), - \frac{1}{2} I]}_{*} + {[\frac{1}{2} I • (- \frac{1}{2} i I), d (- \frac{1}{2} I)]}_{*}$

$= i d (\frac{1}{2} I) - \frac{1}{2} (d (- \frac{1}{2} i I) - d {(- \frac{1}{2} i I)}^{*}) - i d (- \frac{1}{2} I) .$

Then we can get

$d (\frac{1}{2} I) = d (- \frac{1}{2} I) .$ (7)

Now we compute

$d (- \frac{1}{2} i I) = d ({[- \frac{1}{2} I • (- \frac{1}{2} i I), - \frac{1}{2} I]}_{*})$

$= {[d (- \frac{1}{2} I) • (- \frac{1}{2} i I), - \frac{1}{2} I]}_{*} + {[- \frac{1}{2} I • d (- \frac{1}{2} i I), - \frac{1}{2} I]}_{*} + {[- \frac{1}{2} I • (- \frac{1}{2} i I), d (- \frac{1}{2} I)]}_{*}$

$= 2 i d (- \frac{1}{2} I) + \frac{1}{2} (d (- \frac{1}{2} i I) - d {(- \frac{1}{2} i I)}^{*}) .$

By using equation (4) we get $d (- \frac{1}{2} I) = 0,$ then by equations (3), (6) and (7), we have $d (- \frac{1}{2} I) = d (\frac{1}{2} I) = d (- \frac{1}{2} i I) = d (\frac{1}{2} i I) = 0 .$

For any $A \in A$ we have

$d (i A) = d ({[\frac{1}{2} I • \frac{1}{2} i I, A]}_{*}) = {[d (\frac{1}{2} I) • \frac{1}{2} i I, A]}_{*} + {[\frac{1}{2} I • d (\frac{1}{2} i I), A]}_{*} + {[\frac{1}{2} I • \frac{1}{2} i I, d (A)]}_{*} = {[\frac{1}{2} I • \frac{1}{2} i I, d (A)]}_{*} = i d (A) .$

Claim 4^[23] $d$ is additive.

Claim 5 For any $A \in A, d (A^{*}) = d {(A)}^{*} .$

Proof For any $A \in A$ , where $A = A_{1} + i A_{2}$ , $A_{1}, A_{2} \in S$ , it follows from Claim 2, Claim 3 and Claim 4 that

$d (A^{*}) = d (A_{1} - i A_{2}) = d (A_{1}) - i d (A_{2}) = d (A_{1})^{*} + {(i d (A_{2}))}^{*} = d (A_{1} + i A_{2})^{*} = d {(A)}^{*} .$

Claim 6 $d$ is an additive $*$ -derivation on $A$ .

Proof To complete the proof, we only need to show that $d$ is a derivation on $A$ . Since $d$ is additive. ${[A, B]}_{*} = A B - B A^{*}$ and ${[i A, i B]}_{*} = - A B - B A^{*}$ for all $A, B \in A$ . It follows from Claim 3 that

$d (A B) - d (B A^{*}) = d ({[A, B]}_{*}) = d ({[\frac{1}{2} I • A, B]}_{*}) = {[d (\frac{1}{2} I) • A, B]}_{*} + {[\frac{1}{2} I • d (A), B]}_{*} + {[\frac{1}{2} I • A, d (B)]}_{*}$

$= {[\frac{1}{2} I • d (A), B]}_{*} + {[\frac{1}{2} I • A, d (B)]}_{*} = d (A) B - B d {(A)}^{*} + A d (B) - d (B) A^{*} .$

And

$- d (A B) - d (B A^{*}) = d ({[i A, i B]}_{*}) = d ({[\frac{1}{2} I • i A, i B]}_{*}) = {[d (\frac{1}{2} I) • i A, i B]}_{*} + {[\frac{1}{2} I • d (i A), i B]}_{*} + {[\frac{1}{2} I • i A, d (i B)]}_{*}$

$= {[\frac{1}{2} I • d (i A), i B]}_{*} + {[\frac{1}{2} I • i A, d (i B)]}_{*} = - d (A) B - B d {(A)}^{*} - A d (B) - d (B) A^{*} .$

Using the two equations above, we obtain $d (A B) = d (A) B + A d (B)$ , and Claim 6 is proved.

It follows from Claims 4-6 that $d$ is an additive $*$ -derivation on $A$ , which completes the proof of Theorem 1.

2 Nonlinear Mixed Jordan Triple $*$ -Higher Derivation

Theorem 2 Let $A$ be a $*$ -algebra with the unit $I$ , and let $N$ be the set of non-negative integers. Assume that $A$ contains a nontrivial projection $P$ which satisfies for $A \in A$ , $X A P = 0 i m p l i e s X = 0$ , and $X A (I - P) = 0 i m p l i e s X = 0 .$

Then every nonlinear mixed Jordan triple $*$ -higher derivation $D = {d_{n}}_{n \in ℕ}$ on $A$ which satisfies

$d_{n} ({[A • B, C]}_{*}) = \sum_{i + j + k = n} [d_{i} (A) • d_{j} (B), d_{k} {(C)]}_{*}$

for all $A, B, C \in A$ and $n \in N$ , is an additive $*$ -higher derivation.

Proof Let $P_{1} = P$ and $P_{2} = I - P$ . We write $A_{j k} = P_{j} A P_{k}$ for $j, k = 1,2$ . Then $A = \sum_{j, k = 1}^{2} A_{j k} .$

In the sequel $A_{j k}$ indicates that $A_{j k} \in A_{j k}$ . We set $S = {A \in A, A = A^{*}}$ . We complete the proof by several Claims.

Claim 7 $d_{n} (0) = 0$ for all $n \in N$ .

Proof It follows from Claim 1 that $d_{1} (0) = 0$ . Now assume that $d_{k} (0) = 0$ for $k < n$ . Then we compute

$d_{n} (0) = d_{n} ({[0 • 0,0]}_{*}) = \sum_{m + l + j = n} {[d_{m} (0) • d_{l} (0), d_{j} (0)]}_{*} = \sum_{\begin{matrix} m + l + j = n \\ j < n \end{matrix}} {[d_{m} (0) • d_{l} (0), d_{j} (0)]}_{*} + {[d_{0} (0) • d_{0} (0), d_{n} (0)]}_{*} = 0 .$

Claim 8 For any $n \in N$ , $d_{n}$ has the following properties: (a) For any $λ \in R$ , $d_{n} (λ I) \in Z (A)$ and $d_{n} (λ I) = d_{n} {(λ I)}^{*}$ ; (b) For any $A \in S$ , $d_{n} (A) = d_{n} (A^{*}) = d_{n} {(A)}^{*}$ ; (c) For any $λ \in C$ , $d_{n} (λ I) \in Z (A)$ .

Proof It follows from Claim 2 and Claim 6 that $d_{1} (λ I) = 0$ for $λ \in C$ and $d_{1} (A^{*}) = d_{1} {(A)}^{*}$ for $A \in A$ .

(a) For any λ ∈ $R$ , we assume that $d_{k} (λ I) = d_{k} {(λ I)}^{*} \in Z (A)$ for k<n. Then

$0 = d_{n} ({[I • λ I, I]}_{*}) = \sum_{m + l + j = n} [d_{m} (I) • d_{l} (λ I), d_{j} {(I)]}_{*}$

$= {[d_{n} (I) • λ I, I]}_{*} + {[I • d_{n} (λ I), I]}_{*} + {[I • λ I, d_{n} (I)]}_{*} + \sum_{\begin{matrix} m + l + j = n \\ 0 \leq m, l, j \leq n - 1 \end{matrix}}^{} {[d_{m} (I) • d_{l} (λ I), d_{j} (I)]}_{*}$

$= {[d_{n} (I) • λ I, I]}_{*} + {[I • d_{n} (λ I), I]}_{*} + {[I • λ I, d_{n} (I)]}_{*} = 2 d_{n} (λ I) - 2 d_{n} {(λ I)}^{*} .$

From this we get $d_{n} (λ I) = d_{n} {(λ I)}^{*}$ . Then for all $A \in A$ ,

$0 = d_{n} ({[λ I • λ I, A]}_{*}) = \sum_{m + l + j = n} {[d_{m} (λ I) • d_{l} (λ I), d_{j} (A)]}_{*}$

$= {[d_{n} (λ I) • λ I, A]}_{*} + {[λ I • d_{n} (λ I), A]}_{*} + {[λ I • λ I, d_{n} (A)]}_{*} + \sum_{\begin{matrix} m + l + j = n \\ 0 \leq m, l, j \leq n - 1 \end{matrix}} {[d_{m} (λ I) • d_{l} (λ I), d_{j} (A)]}_{*}$

$= {[d_{n} (λ I) • λ I, A]}_{*} + {[λ I • d_{n} (λ I), A]}_{*} + {[λ I • λ I, d_{n} (A)]}_{*} = 4 λ d_{n} (λ I) A - 4 λ A d_{n} {(λ I)}^{*} .$

So we get $d_{n} (λ I) A = A d_{n} (λ I)$ for all $A \in A$ , which implies $d_{n} (λ I) \in Z (A)$ .

(b) For any $A \in S$ , we assume that $d_{k} (A) = d_{k} (A^{*}) = d_{k} {(A)}^{*}$ for $k < n$ . Then we can get from Claim 8(a) that

$0 = d_{n} ({[I • A, I]}_{*}) = \sum_{m + l + j = n} {[d_{m} (I) • d_{l} (A), d_{j} (I)]}_{*}$

$= {[d_{n} (I) • A, I]}_{*} + {[I • d_{n} (A), I]}_{*} + {[I • A, d_{n} (I)]}_{*} + \sum_{\begin{matrix} m + l + j = n \\ 0 \leq m, l, j \leq n - 1 \end{matrix}} {[d_{m} (I) • d_{l} (A), d_{j} (I)]}_{*}$

$= {[d_{n} (I) • A, I]}_{*} + {[I • d_{n} (A), I]}_{*} + {[I • A, d_{n} (I)]}_{*} = 2 d_{n} (A) - 2 d_{n} {(A)}^{*} .$

(c) For any $λ \in C$ and $A \in S$ , assume that $d_{k} (λ I) \in Z (A)$ for $k < n$ . Then we see from Claim 8(a) and 8(b) that

$0 = d_{n} ({[A • I, λ I]}_{*}) = \sum_{m + l + j = n} {[d_{m} (A) • d_{l} (I), d_{j} (λ I)]}_{*}$

$= {[d_{n} (A) • I, λ I]}_{*} + {[A • d_{n} (I), λ I]}_{*} + {[A • I, d_{n} (λ I)]}_{*} + \sum_{\begin{matrix} m + l + j = n \\ 0 \leq m, l, j \leq n - 1 \end{matrix}}^{} {[d_{m} (A) • d_{l} (I), d_{j} (λ I)]}_{*}$

$= {[d_{n} (A) • I, λ I]}_{*} + {[A • d_{n} (I), λ I]}_{*} + {[A • I, d_{n} (λ I)]}_{*} = 2 A d_{n} (λ I) - 2 d_{n} (λ I) A .$

Thus $A d_{n} (λ I) = d_{n} (λ I) A$ for all $A \in S$ . This implies that $d_{n} (λ I) \in Z (A) .$

Claim 9 For any $A \in A$ and $n \in N^{+}, d_{n} (\frac{1}{2} I) = d_{n} (\frac{1}{2} i I) = 0, d_{n} (i A) = i d_{n} (A) .$

Proof It follows from Claim 3 that $d_{1} (\frac{1}{2} I) = d_{1} (\frac{1}{2} i I) = 0$ , $d_{1} (i A) = i d_{1} (A)$ . Now assume $d_{k} (\frac{1}{2} I) = d_{k} (\frac{1}{2} i I) = 0, d_{k} (i A) = i d_{k} (A)$ for $k < n$ . Then by Claim 8,

$d_{n} (- \frac{1}{2} I) = d_{n} ({[\frac{1}{2} I • \frac{1}{2} i I, \frac{1}{2} i I]}_{*}) = \sum_{m + l + j = n} {[d_{m} (\frac{1}{2} I) • d_{l} (\frac{1}{2} i I), d_{j} (\frac{1}{2} i I)]}_{*}$

$= {[d_{n} (\frac{1}{2} I) • \frac{1}{2} i I, \frac{1}{2} i I]}_{*} + {[\frac{1}{2} I • d_{n} (\frac{1}{2} i I), \frac{1}{2} i I]}_{*} + {[\frac{1}{2} I • \frac{1}{2} i I, d_{n} (\frac{1}{2} i I)]}_{*} + \sum_{{\begin{matrix} m + l + j = n \\ 0 \leq m, l, j \leq n - 1 \end{matrix}}^{}} {[d_{m} (\frac{1}{2} I) • d_{l} (\frac{1}{2} i I), d_{j} (\frac{1}{2} i I)]}_{*}$

$= {[d_{n} (\frac{1}{2} I) • \frac{1}{2} i I, \frac{1}{2} i I]}_{*} + {[\frac{1}{2} I • d_{n} (\frac{1}{2} i I), \frac{1}{2} i I]}_{*} + {[\frac{1}{2} I • \frac{1}{2} i I, d_{n} (\frac{1}{2} i I)]}_{*}$

$= - d_{n} (\frac{1}{2} I) + \frac{1}{2} i (d_{n} (\frac{1}{2} i I) - d_{n} {(\frac{1}{2} i I)}^{*}) + i d_{n} (\frac{1}{2} i I) .$

Since $d_{n} (\frac{1}{2} I), d_{n} (- \frac{1}{2} I), i (d_{n} (\frac{1}{2} i I) - d_{n} {(\frac{1}{2} i I)}^{*})$ are self-adjoint, $i d_{n} (\frac{1}{2} i I)$ is also self-adjoint. Hence

$d_{n} (- \frac{1}{2} I) = - d_{n} (\frac{1}{2} I) + 2 i d_{n} (\frac{1}{2} i I) .$ (8)

Similarly, we can obtain from the fact that $- \frac{1}{2} I = {[\frac{1}{2} I • (- \frac{1}{2} i I), - \frac{1}{2} i I]}_{*}$ , then

$d_{n} (- \frac{1}{2} i I) = - d_{n} {(- \frac{1}{2} i I)}^{*}$ (9)

and

$d_{n} (- \frac{1}{2} I) = - d_{n} (\frac{1}{2} I) - 2 i d_{n} (- \frac{1}{2} i I) .$ (10)

Now by (8) and (10) we have

$d_{n} (- \frac{1}{2} i I) = - d_{n} (\frac{1}{2} i I) .$ (11)

Then we compute

$d_{n} (\frac{1}{2} i I) = d_{n} ({[\frac{1}{2} I • (- \frac{1}{2} i I), - \frac{1}{2} I]}_{*}) = \sum_{m + l + j = n} {[d_{m} (\frac{1}{2} I) • d_{l} (- \frac{1}{2} i I), d_{j} (- \frac{1}{2} I)]}_{*}$

$= {[d_{n} (\frac{1}{2} I) • (- \frac{1}{2} i I), - \frac{1}{2} I]}_{*} + {[\frac{1}{2} I • d_{n} (- \frac{1}{2} i I), - \frac{1}{2} I]}_{*} + {[\frac{1}{2} I • (- \frac{1}{2} i I), d_{n} (- \frac{1}{2} I)]}_{*}$

$+ \sum_{\begin{matrix} m + l + j = n \\ 0 \leq m, l, j \leq n - 1 \end{matrix}} {[d_{m} (\frac{1}{2} I) • d_{l} (- \frac{1}{2} i I), d_{j} (- \frac{1}{2} I)]}_{*}$

$= {[d_{n} (\frac{1}{2} I) • (- \frac{1}{2} i I), - \frac{1}{2} I]}_{*} + {[\frac{1}{2} I • d_{n} (- \frac{1}{2} i I), - \frac{1}{2} I]}_{*} + {[\frac{1}{2} I • (- \frac{1}{2} i I), d_{n} (- \frac{1}{2} I)]}_{*}$

$= i d_{n} (\frac{1}{2} I) - \frac{1}{2} I (d_{n} (- \frac{1}{2} i I) - d_{n} {(- \frac{1}{2} i I)}^{*}) - i d_{n} (- \frac{1}{2} I) .$

By using (9) and (11) we get

$d_{n} (\frac{1}{2} I) = d_{n} (- \frac{1}{2} I) .$ (12)

Now compute

$d_{n} (- \frac{1}{2} i I) = d_{n} ({[- \frac{1}{2} I • (- \frac{1}{2} i I), - \frac{1}{2} I]}_{*}) = \sum_{m + l + j = n} {[d_{m} (- \frac{1}{2} I) • d_{l} (- \frac{1}{2} i I), d_{j} (- \frac{1}{2} I)]}_{*}$

$= {[d_{n} (- \frac{1}{2} I) • (- \frac{1}{2} i I), - \frac{1}{2} I]}_{*} + {[- \frac{1}{2} I • d_{n} (- \frac{1}{2} i I), - \frac{1}{2} I]}_{*}$

$+ \sum_{\begin{matrix} m + l + j = n \\ 0 \leq m, l, j \leq n - 1 \end{matrix}} {[d_{m} (- \frac{1}{2} I) • d_{l} (- \frac{1}{2} i I), d_{j} (- \frac{1}{2} I)]}_{*} + {[- \frac{1}{2} I • (- \frac{1}{2} i I), d_{n} (- \frac{1}{2} I)]}_{*}$

$= {[d_{n} (- \frac{1}{2} I) • (- \frac{1}{2} i I), - \frac{1}{2} I]}_{*} + {[- \frac{1}{2} I • d_{n} (- \frac{1}{2} i I), - \frac{1}{2} I]}_{*} + {[- \frac{1}{2} I • (- \frac{1}{2} i I), d_{n} (- \frac{1}{2} I)]}_{*}$

$= 2 i d_{n} (- \frac{1}{2} I) + \frac{1}{2} (d_{n} (- \frac{1}{2} i I) - d_{n} {(- \frac{1}{2} i I)}^{*}) .$

By using (9) we get $d_{n} (- \frac{1}{2} I) = 0$ , then by equations (8), (11) and (12), we have $d_{n} (- \frac{1}{2} I) = d_{n} (\frac{1}{2} I) = d_{n} (- \frac{1}{2} i I) = d_{n} (\frac{1}{2} i I) = 0$ .

For any $A \in A$ we have

$d_{n} (i A) = d_{n} ({[\frac{1}{2} I • \frac{1}{2} i I, A]}_{*}) = \sum_{m + l + j = n} {[d_{m} (\frac{1}{2} I) • d_{l} (\frac{1}{2} i I), d_{j} (A)]}_{*}$

$= {[d_{n} (\frac{1}{2} I) • \frac{1}{2} i I, A]}_{*} + {[\frac{1}{2} I • d_{n} (\frac{1}{2} i I), A]}_{*} + {[\frac{1}{2} I • \frac{1}{2} i I, d_{n} (A)]}_{*} + \sum_{\begin{matrix} m + l + j = n \\ 0 \leq m, l, j \leq n - 1 \end{matrix}} {[d_{m} (\frac{1}{2} I) • d_{l} (\frac{1}{2} i I), d_{j} (A)]}_{*}$

$= {[\frac{1}{2} I • \frac{1}{2} i I, d_{n} (A)]}_{*} = i d_{n} (A) .$

Claim 10 For any $n \in N$ , $d_{n}$ is additive.

Proof It follows from Claim 4 that $d_{1}$ is additive. Now assume $d_{k}$ is additive for $k < n$ . Then we complete the proof of Claim 10 in several steps.

Step 1 For any $n \in N$ , $A_{12} \in A_{12}$ and $B_{21} \in A_{21}$ , we have $d_{n} (A_{12} + B_{21}) = d_{n} (A_{12}) + d_{n} (B_{21})$ .

Set $T = d_{n} (A_{12} + B_{21}) - d_{n} (A_{12}) - d_{n} (B_{21})$ . We only need to show $T = 0$ .

Since ${[I • (i (P_{2} - P_{1})), A_{12}]}_{*} = {[I • (i (P_{2} - P_{1})), B_{21}]}_{*} = 0$ , it follows that

$0 = d_{n} ({[I • (i (P_{2} - P_{1})), A_{12} + B_{21}]}_{*})$

$= {[d_{n} (I) • (i (P_{2} - P_{1})), A_{12} + B_{21}]}_{*} + {[I • d_{n} (i (P_{2} - P_{1})), A_{12} + B_{21}]}_{*} + {[I • (i (P_{2} - P_{1})), d_{n} (A_{12} + B_{21})]}_{*}$

$+ \sum_{\begin{matrix} m + l + j = n \\ 0 \leq m, l, j \leq n - 1 \end{matrix}} {[d_{m} (I) • d_{l} (i (P_{2} - P_{1})), d_{j} (A_{12} + B_{21})]}_{*} .$

On the other hand,

$0 = d_{n} ({[I • (i (P_{2} - P_{1})), A_{12}]}_{*}) + d_{n} ({[I • (i (P_{2} - P_{1})), B_{21}]}_{*})$

$= {[d_{n} (I) • (i (P_{2} - P_{1})), A_{12} + B_{21}]}_{*} + {[I • d_{n} (i (P_{2} - P_{1})), A_{12} + B_{21}]}_{*} + {[I • (i (P_{2} - P_{1})), d_{n} (A_{12}) + d_{n} (B_{21})]}_{*}$

$+ \sum_{\begin{matrix} m + l + j = n \\ 0 \leq m, l, j \leq n - 1 \end{matrix}} {[d_{m} (I) • d_{l} (i (P_{2} - P_{1})), d_{j} (A_{12} + B_{21})]}_{*} .$

From two equations above, we get ${[I • (i (P_{2} - P_{1})), T]}_{*} = 0$ . So $T_{11} = T_{22} = 0$ .

Since ${[I • A_{12}, P_{1}]}_{*} = 0$ , it follows that

$d_{n} ({[I • (A_{12} + B_{21}), P_{1}]}_{*}) = {[d_{n} (I) • (A_{12} + B_{21}), P_{1}]}_{*} + {[I • d_{n} (A_{12} + B_{21}), P_{1}]}_{*} + {[I • (A_{12} + B_{21}), d_{n} (P_{1})]}_{*}$

$+ \sum_{\begin{matrix} m + l + j = n \\ 0 \leq m, l, j \leq n - 1 \end{matrix}} {[d_{m} (I) • d_{l} (A_{12} + B_{21}), d_{j} (P_{1})]}_{*} .$

On the other hand,

$d_{n} ({[I • (A_{12} + B_{21}), P_{1}]}_{*}) = d_{n} ({[I • A_{12}, P_{1}]}_{*}) + d_{n} ({[I • B_{21}, P_{1}]}_{*})$

$= {[d_{n} (I) • (A_{12} + B_{21}), P_{1}]}_{*} + {[I • (d_{n} (A_{12}) + d_{n} (B_{21})), P_{1}]}_{*} + {[I • (A_{12} + B_{21}), d_{n} (P_{1})]}_{*} + \sum_{\begin{matrix} m + l + j = n \\ 0 \leq m, l, j \leq n - 1 \end{matrix}} {[d_{m} (I) • d_{l} (A_{12}) + d_{l} (B_{21}), d_{j} (P_{1})]}_{*}$

$= {[d_{n} (I) • (A_{12} + B_{21}), P_{1}]}_{*} + {[I • (d_{n} (A_{12}) + d_{n} (B_{21})), P_{1}]}_{*} + {[I • (A_{12} + B_{21}), d_{n} (P_{1})]}_{*}$

$+ \sum_{\begin{matrix} m + l + j = n \\ 0 \leq m, l, j \leq n - 1 \end{matrix}} {[d_{m} (I) • d_{l} (A_{12} + B_{21}), d_{j} (P_{1})]}_{*} .$

Comparing two equations above, we can get ${[I • T, P_{1}]}_{*} = 0$ . So $T_{21} = 0$ . Similarly, we can show $T_{12} = 0$ by using ${[I • B_{21}, P_{2}]}_{*} = 0$ , then the step is proved.

Step 2 For any $n \in N$ , $A_{11} \in A_{11}$ , $B_{12} \in A_{12}$ , $C_{21} \in A_{21}$ and $D_{22} \in A_{22}$ , we have $d_{n} (A_{11} + B_{12} + C_{21}) = d_{n} (A_{11}) + d_{n} (B_{12}) + d_{n} (C_{21})$ and $d_{n} (B_{12} + C_{21} + D_{22}) = d_{n} (B_{12}) + d_{n} (C_{21}) + d_{n} (D_{22})$ .

Let $T = d_{n} (A_{11} + B_{12} + C_{21}) - d_{n} (A_{11}) - d_{n} (B_{12}) - d_{n} (C_{21})$ . It follows from Step 1 that

$d_{n} ({[I • (i P_{2}), A_{11} + B_{12} + C_{21}]}_{*})$

$= {[d_{n} (I) • (i P_{2}), A_{11} + B_{12} + C_{21}]}_{*} + {[I • d_{n} (i P_{2}), A_{11} + B_{12} + C_{21}]}_{*} + {[I • (i P_{2}), d_{n} (A_{11} + B_{12} + C_{21})]}_{*} + \sum_{\begin{matrix} m + l + j = n \\ 0 \leq m, l, j \leq n - 1 \end{matrix}} {[d_{m} (I) • d_{l} (i P_{2}), d_{j} (A_{11} + B_{12} + C_{21})]}_{*}$

$= {[d_{n} (I) • (i P_{2}), A_{11} + B_{12} + C_{21}]}_{*} + {[I • d_{n} (i P_{2}), A_{11} + B_{12} + C_{21}]}_{*} + {[I • (i P_{2}), d_{n} (A_{11} + B_{12} + C_{21})]}_{*} + \sum_{\begin{matrix} m + l + j = n \\ 0 \leq m, l, j \leq n - 1 \end{matrix}} {[d_{m} (I) • d_{l} (i P_{2}), d_{j} (A_{11}) + d_{j} (B_{12}) + d_{j} (C_{21})]}_{*} .$

On the other hand,

$d_{n} ({[I • (i P_{2}), A_{11} + B_{12} + C_{21}]}_{*}) = d_{n} ({[I • (i P_{2}), A_{11}]}_{*}) + d_{n} ({[I • (i P_{2}), B_{12}]}_{*}) + d_{n} ({[I • (i P_{2}), C_{21}]}_{*})$

$= {[d_{n} (I) • (i P_{2}), A_{11} + B_{12} + C_{21}]}_{*} + {[I • d_{n} (i P_{2}), A_{11} + B_{12} + C_{21}]}_{*}$

$+ {[I • (i P_{2}), d_{n} (A_{11}) + d_{n} (B_{12}) + d_{n} (C_{21})]}_{*} + \sum_{\begin{matrix} m + l + j = n \\ 0 \leq m, l, j \leq n - 1 \end{matrix}} {[d_{m} (I) • d_{l} (i P_{2}), d_{j} (A_{11}) + d_{j} (B_{12}) + d_{j} (C_{21})]}_{*} .$

From these equations, we get ${[I • (i P_{2}), T]}_{*} = 0$ . So $T = T_{11}$ .

Since ${[I • (i (P_{2} - P_{1})), B_{12}]}_{*} = {[I • (i (P_{2} - P_{1})), C_{21}]}_{*} = 0$ , it follows that

$d_{n} ({[I • (i (P_{2} - P_{1})), A_{11} + B_{12} + C_{21}]}_{*})$

$= {[d_{n} (I) • (i (P_{2} - P_{1})), A_{11} + B_{12} + C_{21}]}_{*} + {[I • d_{n} (i (P_{2} - P_{1})), A_{11} + B_{12} + C_{21}]}_{*}$

$+ {[I • (i (P_{2} - P_{1})), d_{n} (A_{11} + B_{12} + C_{21})]}_{*} + \sum_{\begin{matrix} m + l + j = n \\ 0 \leq m, l, j \leq n - 1 \end{matrix}} {[d_{m} (I) • d_{l} (i (P_{2} - P_{1})), d_{j} (A_{11} + B_{12} + C_{21})]}_{*}$

$= {[d_{n} (I) • (i (P_{2} - P_{1})), A_{11} + B_{12} + C_{21}]}_{*} + {[I • d_{n} (i (P_{2} - P_{1})), A_{11} + B_{12} + C_{21}]}_{*}$

$+ {[I • (i (P_{2} - P_{1})), d_{n} (A_{11} + B_{12} + C_{21})]}_{*} + \sum_{\begin{matrix} m + l + j = n \\ 0 \leq m, l, j \leq n - 1 \end{matrix}} {[d_{m} (I) • d_{l} (i (P_{2} - P_{1})), d_{j} (A_{11}) + d_{j} (B_{12}) + d_{j} (C_{21})]}_{*} .$

On the other hand,

$d_{n} ({[I • (i (P_{2} - P_{1})), A_{11} + B_{12} + C_{21}]}_{*})$

$= d_{n} ({[I • (i (P_{2} - P_{1})), A_{11}]}_{*}) + d_{n} ({[I • (i (P_{2} - P_{1})), B_{12}]}_{*} + d_{n} ({[I • (i (P_{2} - P_{1})), C_{21}]}_{*})$

$= {[d_{n} (I) • (i (P_{2} - P_{1})), A_{11} + B_{12} + C_{21}]}_{*} + {[I • d_{n} (i (P_{2} - P_{1})), A_{11} + B_{12} + C_{21}]}_{*}$

$+ {[I • (i (P_{2} - P_{1})), d_{n} (A_{11}) + d_{n} (B_{12}) + d_{n} (C_{21})]}_{*} + \sum_{\begin{matrix} m + l + j = n \\ 0 \leq m, l, j \leq n - 1 \end{matrix}} {[d_{m} (I) • d_{l} (i (P_{2} - P_{1})), d_{j} (A_{11}) + d_{j} (B_{12}) + d_{j} (C_{21})]}_{*} .$

We can get ${[I • (i (P_{2} - P_{1})), T]}_{*} = 0$ from the equations above. So $T_{11} = 0$ and $T = 0$ . Similarly, $d_{n} (B_{12} + C_{21} + D_{22}) = d_{n} (B_{12}) + d_{n} (C_{21}) + d_{n} (D_{22})$ .

Step 3 For any $n \in N$ , $A_{11} \in A_{11}$ , $B_{12} \in A_{12}$ , $C_{21} \in A_{21}$ and $D_{22} \in A_{22}$ , we have

$d_{n} (A_{11} + B_{12} + C_{21} + D_{22}) = d_{n} (A_{11}) + d_{n} (B_{12}) + d_{n} (C_{21}) + d_{n} (D_{22}) .$

Let $T = d_{n} (A_{11} + B_{12} + C_{21} + D_{22}) - d_{n} (A_{11}) - d_{n} (B_{12}) - d_{n} (C_{21}) - d_{n} (D_{22})$ . It follows from Step 2 that

$d_{n} ({[I • (i P_{2}), A_{11} + B_{12} + C_{21} + D_{22}]}_{*})$

$= {[d_{n} (I) • (i P_{2}), A_{11} + B_{12} + C_{21} + D_{22}]}_{*} + {[I • d_{n} (i P_{2}), A_{11} + B_{12} + C_{21} + D_{22}]}_{*}$

$+ {[I • (i P_{2}), d_{n} (A_{11} + B_{12} + C_{21} + D_{22})]}_{*} + \sum_{\begin{matrix} m + l + j = n \\ 0 \leq m, l, j \leq n - 1 \end{matrix}} {[d_{m} (I) • d_{l} (i P_{2}), d_{j} (A_{11} + B_{12} + C_{21} + D_{22})]}_{*}$

$= {[d_{n} (I) • (i P_{2}), A_{11} + B_{12} + C_{21} + D_{22}]}_{*} + {[I • d_{n} (i P_{2}), A_{11} + B_{12} + C_{21} + D_{22}]}_{*}$

$+ {[I • (i P_{2}), d_{n} (A_{11} + B_{12} + C_{21} + D_{22})]}_{*} + \sum_{\begin{matrix} m + l + j = n \\ 0 \leq m, l, j \leq n - 1 \end{matrix}} {[d_{m} (I) • d_{l} (i P_{2}), d_{j} (A_{11}) + d_{j} (B_{12}) + d_{j} (C_{21}) + d_{j} (D_{22})]}_{*} .$

On the other hand,

$d_{n} ({[I • (i P_{2}), A_{11} + B_{12} + C_{21} + D_{22}]}_{*})$

$= d_{n} ({[I • (i P_{2}), A_{11}]}_{*}) + d_{n} ({[I • (i P_{2}), B_{12}]}_{*}) + d_{n} ({[I • (i P_{2}), C_{21}]}_{*}) + d_{n} ({[I • (i P_{2}), D_{22}]}_{*})$

$= {[d_{n} (I) • (i P_{2}), A_{11} + B_{12} + C_{21} + D_{22}]}_{*} + {[I • d_{n} (i P_{2}), A_{11} + B_{12} + C_{21} + D_{22}]}_{*}$

$+ {[I • (i P_{2}), d_{n} (A_{11}) + d_{n} (B_{12}) + d_{n} (C_{21}) + d_{n} (D_{22})]}_{*} + \sum_{\begin{matrix} m + l + j = n \\ 0 \leq m, l, j \leq n - 1 \end{matrix}} {[d_{m} (I) • d_{l} (i P_{2}), d_{j} (A_{11}) + d_{j} (B_{12}) + d_{j} (C_{21}) + d_{j} (D_{22})]}_{*} .$

Comparing the two equations above, we can get ${[I • (i P_{2}), T]}_{*} = 0$ . So $T_{12} = T_{21} = T_{22} = 0$ . Similarly, we can show that $T_{11} = 0$ from ${[I • (i P_{1}), D_{22}]}_{*} = 0$ , then the Step 3 is proved.

Step 4 For any $n \in N$ , $A_{s t}, B_{s t} \in A_{s t}, 1 \leq s \neq t \leq 2$ , we have $d_{n} (A_{s t} + B_{s t}) = d_{n} (A_{s t}) + d_{n} (B_{s t}) .$

Since ${[\frac{1}{2} I • (P_{s} + A_{s t}), P_{t} + B_{s t}]}_{*} = A_{s t} + B_{s t} - {A_{s t}}^{*} - B_{s t} {A_{s t}}^{*}$ , we can get

$d_{n} (A_{s t} + B_{s t}) + d_{n} (- {A_{s t}}^{*}) + d_{n} (- B_{s t} {A_{s t}}^{*}) = d_{n} ({[\frac{1}{2} I • (P_{s} + A_{s t}), P_{t} + B_{s t}]}_{*})$

$= {[d_{n} (\frac{1}{2} I) • (P_{s} + A_{s t}), P_{t} + B_{s t}]}_{*} + {[\frac{1}{2} I • d_{n} (P_{s} + A_{s t}), P_{t} + B_{s t}]}_{*} + {[\frac{1}{2} I • (P_{s} + A_{s t}), d_{n} (P_{t} + B_{s t})]}_{*} + \sum_{\begin{matrix} m + l + j = n \\ 0 \leq m, l, j \leq n - 1 \end{matrix}} {[d_{m} (\frac{1}{2} I) • d_{l} (P_{s} + A_{s t}), d_{j} (P_{t} + A_{s t})]}_{*}$

$= d_{n} ({[\frac{1}{2} I • P_{s}, P_{t}]}_{*}) + d_{n} ({[\frac{1}{2} I • A_{s t}, P_{t}]}_{*}) + d_{n} ({[\frac{1}{2} I • P_{s}, B_{s t}]}_{*}) + d_{n} ({[\frac{1}{2} I • A_{s t}, B_{s t}]}_{*})$

$= d_{n} (A_{s t}) + d_{n} (- {A_{s t}}^{*}) + d_{n} (B_{s t}) + d_{n} (- B_{s t} {A_{s t}}^{*}) .$

Step 5 For any $n \in N$ , $A_{s s}, B_{s s} \in A_{s s}, s = 1,2$ , we have $d_{n} (A_{s s} + B_{s s}) = d_{n} (A_{s s}) + d_{n} (B_{s s}) .$

Let $T = d_{n} (A_{s s} + B_{s s}) - d_{n} (A_{s s}) - d_{n} (B_{s s})$ . For $1 \leq s \neq t \leq 2$ , it follows that

${[d_{n} (I) • P_{t}, A_{s s} + B_{s s}]}_{*} + {[I • d_{n} (P_{t}), A_{s s} + B_{s s}]}_{*} + {[I • P_{t}, d_{n} (A_{s s} + B_{s s})]}_{*} + \sum_{\begin{matrix} m + l + j = n \\ 0 \leq m, l, j \leq n - 1 \end{matrix}} {[d_{m} (I) • d_{l} (P_{t}), d_{j} (A_{s s} + B_{s s})]}_{*}$

$= d_{n} ({[I • P_{t}, A_{s s} + B_{s s}]}_{*}) = d_{n} ({[I • P_{t}, A_{s s}]}_{*}) + d_{n} ({[I • P_{t}, B_{s s}]}_{*})$

$= {[d_{n} (I) • P_{t}, A_{s s} + B_{s s}]}_{*} + {[I • d_{n} (P_{t}), A_{s s} + B_{s s}]}_{*} + {[I • P_{t}, d_{n} (A_{s s}) + d_{n} (B_{s s})]}_{*} + \sum_{\begin{matrix} m + l + j = n \\ 0 \leq m, l, j \leq n - 1 \end{matrix}} {[d_{m} (I) • d_{l} (P_{t}), d_{j} (A_{s s} + B_{s s})]}_{*} .$

From the equation above, we can get $T = T_{s s}$ .

For all $C_{s t} \in A_{s t}, s \neq t$ , it follows from Step 4 that

${[d_{n} (I) • (A_{s s} + B_{s s}), C_{s t}]}_{*} + {[I • d_{n} (A_{s s} + B_{s s}), C_{s t}]}_{*} + {[I • (A_{s s} + B_{s s}), d_{n} (C_{s t})]}_{*} + \sum_{\begin{matrix} m + l + j = n \\ 0 \leq m, l, j \leq n - 1 \end{matrix}} {[d_{m} (I) • d_{l} (A_{s s} + B_{s s}), d_{j} (C_{s t})]}_{*}$

$= d_{n} ({[I • (A_{s s} + B_{s s}), C_{s t}]}_{*}) = d_{n} ({[I • A_{s s}, C_{s t}]}_{*}) + d_{n} ({[I • B_{s s}, C_{s t}]}_{*})$

$= {[d_{n} (I) • (A_{s s} + B_{s s}), C_{s t}]}_{*} + {[I • d_{n} (A_{s s}) + d_{n} (B_{s s}), C_{s t}]}_{*} + {[I • (A_{s s} + B_{s s}), d_{n} (C_{s t})]}_{*} + \sum_{\begin{matrix} m + l + j = n \\ 0 \leq m, l, j \leq n - 1 \end{matrix}} {[d_{m} (I) • d_{l} (A_{s s} + B_{s s}), d_{j} (C_{s t})]}_{*} .$

Hence $[I • T_{s s}, C_{s t}] = 0$ for all $C_{s t} \in A_{s t}$ , then $T_{s s} C P_{t} = 0$ for all $C \in A$ . So $T = T_{s s} = 0$ , then the step is proved.

Now, it follows from Steps 3-5 that $d_{n}$ is additive, and Claim 10 is proved.

Claim 11 For any $A \in A$ and $n \in N$ , $d_{n} (A^{*}) = d_{n} {(A)}^{*}$ .

Proof For any $n \in N$ , $A \in A$ , where $A = A_{1} + i A_{2}$ , $A_{1}, A_{2} \in S$ , it follows from Claims 8-10 that

$d_{n} (A^{*}) = d_{n} (A_{1} - i A_{2}) = d_{n} (A_{1}) - i d_{n} (A_{2}) = d_{n} (A_{1})^{*} + {(i d_{n} (A_{2}))}^{*} = d_{n} (A_{1} + i A_{2})^{*} = d_{n} {(A)}^{*} .$

Claim 12 For any $n \in N$ , $d_{n}$ is an additive $*$ -higher derivation on $A$ .

Proof To complete the proof, we only need to show that $d_{n}$ is a higher derivation on $A$ . Since $d_{n}$ is additive. ${[A, B]}_{*} = A B - B A^{*}$ and ${[i A, i B]}_{*} = - A B - B A^{*}$ for all $A, B \in A$ . It follows from Claim 9 that

$d_{n} (A B) - d_{n} (B A^{*}) = d_{n} ({[A, B]}_{*}) = d_{n} ({[\frac{1}{2} I • A, B]}_{*})$

$= {[d_{n} (\frac{1}{2} I) • A, B]}_{*} + {[\frac{1}{2} I • d_{n} (A), B]}_{*} + {[\frac{1}{2} I • A, d_{n} (B)]}_{*} + \sum_{\begin{matrix} m + l + j = n \\ 0 \leq m, l, j \leq n - 1 \end{matrix}} {[d_{m} (\frac{1}{2} I) • d_{l} (A), d_{j} (B)]}_{*}$

$= {[\frac{1}{2} I • d_{n} (A), B]}_{*} + {[\frac{1}{2} I • A, d_{n} (B)]}_{*} + \sum_{\begin{matrix} l + j = n \\ 0 \leq l, j \leq n - 1 \end{matrix}} {[d_{l} (A), d_{j} (B)]}_{*}$

$= d_{n} (A) B - B d_{n} {(A)}^{*} + A d_{n} (B) - d_{n} (B) A^{*} + \sum_{\begin{matrix} l + j = n \\ 0 \leq l, j \leq n - 1 \end{matrix}} d_{l} (A) d_{j} (B) - d_{j} (B) d_{l} {(A)}^{*} .$

And

$- d_{n} (A B) - d_{n} (B A^{*}) = d_{n} ({[i A, i B]}_{*}) = d_{n} ({[\frac{1}{2} I • i A, i B]}_{*})$

$= {[d_{n} (\frac{1}{2} I) • i A, i B]}_{*} + {[\frac{1}{2} I • d_{n} (i A), i B]}_{*} + {[\frac{1}{2} I • i A, d_{n} (i B)]}_{*} + \sum_{\begin{matrix} m + l + j = n \\ 0 \leq m, l, j \leq n - 1 \end{matrix}} {[d_{m} (\frac{1}{2} I) • d_{l} (i A), d_{j} (i B)]}_{*}$

$= {[\frac{1}{2} I • d_{n} (i A), i B]}_{*} + {[\frac{1}{2} I • i A, d_{n} (i B)]}_{*} + \sum_{\begin{matrix} l + j = n \\ 0 \leq l, j \leq n - 1 \end{matrix}} {[d_{l} (i A), d_{j} (i B)]}_{*}$

$= - d_{n} (A) B - B d_{n} {(A)}^{*} - A d_{n} (B) - d_{n} (B) A^{*} + \sum_{\begin{matrix} l + j = n \\ 0 \leq l, j \leq n - 1 \end{matrix}} - d_{l} (A) d_{j} (B) - d_{j} (B) d_{l} {(A)}^{*} .$

Using the two equations above, we obtain $d_{n} (A B) = d_{n} (A) B + A d_{n} (B) + \sum_{\begin{matrix} l + j = n \\ 0 \leq l, j \leq n - 1 \end{matrix}} d_{l} (A) d_{j} (B) = \sum_{\begin{matrix} l + j = n \end{matrix}} d_{l} (A) d_{j} (B),$ then Claim 12 is proved.

It follows from Claim 10-12 that $D = {d_{n}}_{n \in N}$ is an additive $*$ -higher derivation on $A$ , which completes the proof of Theorem 2. Obviously, prime ∗-algebras satisfy (1) and (2), we have the following corollary.

Corollary 1 Let $A$ be a prime $*$ -algebra with unit $I$ and $P$ be a nontrivial projection in $A$ . Then any nonlinear mixed Jordan triple $*$ -higher derivation $D = {d_{n}}_{n \in N}, d_{n} : A \to A$ is an additive $*$ -higher derivation.

A von Neumann algebra $ℳ$ is a weekly closed, self-adjoint algebra of operators on a Hilbert space $ℋ$ containing the identity operator $I$ . It is shown in Ref. [3] and Ref. [14] that if a von Neumann algebra has no central summands of type $I_{1}$ , then $ℳ$ satisfies (1) and (2). Now we have following corollary.

Corollary 2 Let $ℳ$ be a von Neumann algebra with no central summands of type $I_{1}$ . Then any nonlinear mixed Jordan triple $*$ -higher derivation $D = {d_{n}}_{n \in N}, d_{n} : A \to A$ is an additive $*$ -higher derivation.

References

Bai Z F, Du S P. Maps preserving products XY-YX on von Neumann algebras[J]. Journal of Mathematical Analysis and Applications, 2012, 386(1): 103-109. [Google Scholar]
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Huo D H, Zheng B D, Liu H Y. Nonlinear maps preserving Jordan triple η--products[J]. Journal of Mathematical Analysis and Applications, 2015, 430(2): 830-844. [Google Scholar]
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Li C J, Lu F Y. Nonlinear maps preserving the Jordan triple 1--product on von Neumann algebras[J]. Complex Analysis and Operator Theory, 2017, 11(1): 109-117. [Google Scholar]
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Zhao F F, Li C J. Nonlinear maps preserving the Jordan triple -product between factors[J]. Indagationes Mathematicae, 2018, 29(2): 619-627. [Google Scholar]
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Jing W. Nonlinear -Lie derivations of standard operator algebras[J]. Quaestiones Mathematicae, 2016, 39(8): 1037-1046. [Google Scholar]
Taghavi A, Rohi H, Darvish V. Non-linear -Jordan derivations on von Neumann algebras[J]. Linear and Multilinear Algebra, 2016, 64(3): 426-439. [Google Scholar]
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Li C J, Lu F Y, Fang X C. Non-linear ξ-Jordan -derivations on von Neumann algebras[J]. Linear and Multilinear Algebra, 2014, 62(4): 466-473. [Google Scholar]
Li C J, Zhao F F, Chen Q Y. Nonlinear skew Lie triple derivations between factors[J]. Acta Mathematica Sinica, English Series, 2016, 32(7): 821-830. [Google Scholar]
Fu F Y, An R L. Equivalent characterization of -derivations on von Neumann algebras[J]. Linear and Multilinear Algebra, 2019, 67(3): 527-541. [Google Scholar]
Zhao F F, Li C J. Nonlinear -Jordan triple derivations on von Neumann algebras[J]. Mathematica Slovaca, 2018, 68(1): 163-170. [Google Scholar]
Lin W H. Nonlinear -Lie-type derivations on von Neumann algebras[J]. Acta Mathematica Hungarica, 2018, 156(1): 112-131. [Google Scholar]
Lin W. Nonlinear -Lie-type derivations on standard operator algebras[J]. Acta Mathematica Hungarica, 2018, 154(2): 480-500. [Google Scholar]
Yang Z J, Zhang J H. Nonlinear maps preserving mixed Lie triple products on factor von Neumann algebras[J]. Annals of Functional Analysis, 2019, 10(3): 325-336. [Google Scholar]
Yang Z J, Zhang J H. Nonlinear maps preserving the second mixed Lie triple products on factor von Neumann algebras[J]. Linear and Multilinear Algebra, 2020, 68(2): 377-390. [Google Scholar]
Zhou Y, Yang Z J, Zhang J H. Nonlinear mixed Lie triple derivations on prime -algebras[J]. Communications in Algebra, 2019, 47(11): 4791-4796. [Google Scholar]
Li C, Zhang D. Nonlinear mixed Jordan triple -derivations on -algebras[J]. Siberian Mathematical Journal, 2022, 63(4): 735-742 [Google Scholar]
Ferrero M, Haetinger C. Higher derivations of semiprime rings[J]. Communications in Algebra, 2002, 30(5): 2321-2333. [Google Scholar]
Nowicki A. Inner derivation of higher orders[J]. Tsukuba Journal of Mathematics, 1984, 8(2): 219-225. [Google Scholar]
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Ashraf M, Ahmad Wani B, Wei F. Multiplicative -Lie triple higher derivations of standard operator algebras[J]. Quaestiones Mathematicae, 2019, 42(7): 857-884. [Google Scholar]
Ahmad Wani B, Ashraf M, Lin W H. Multiplicative -Jordan type higher derivations on von Neumann algebras[J]. Quaestiones Mathematicae, 2020, 43(12): 1689-1711. [Google Scholar]

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[1] Bai Z F, Du S P. Maps preserving products XY-YX on von Neumann algebras[J]. Journal of Mathematical Analysis and Applications, 2012, 386(1): 103-109. [Google Scholar]

[2] Cui J L, Li C K. Maps preserving product XY-YX on factor von Neumann algebras[J]. Linear Algebra and Its Applications, 2009, 431(5/6/7): 833-842. [Google Scholar]

[3] Dai L Q, Lu F Y. Nonlinear maps preserving Jordan -products[J]. Journal of Mathematical Analysis and Applications, 2014, 409(1): 180-188. [Google Scholar]

[4] Huo D H, Zheng B D, Liu H Y. Nonlinear maps preserving Jordan triple η--products[J]. Journal of Mathematical Analysis and Applications, 2015, 430(2): 830-844. [Google Scholar]

[5] Li C J, Chen Q Y, Wang T. Nonlinear maps preserving the Jordan triple -product on factor von Neumann algebras[J]. Chinese Annals of Mathematics, Series B, 2018, 39(4): 633-642. [Google Scholar]

[6] Li C J, Lu F Y, Fang X C. Nonlinear mappings preserving product XY+YX on factor von Neumann algebras[J]. Linear Algebra and Its Applications, 2013, 438(5): 2339-2345. [Google Scholar]

[7] Li C J, Lu F Y. Nonlinear maps preserving the Jordan triple 1--product on von Neumann algebras[J]. Complex Analysis and Operator Theory, 2017, 11(1): 109-117. [Google Scholar]

[8] Li C J, Lu F Y, Wang T. Nonlinear maps preserving the Jordan triple -product on von Neumann algebras[J]. Annals of Functional Analysis, 2016, 7(3): 496-507. [Google Scholar]

[9] Zhao F F, Li C J. Nonlinear maps preserving the Jordan triple -product between factors[J]. Indagationes Mathematicae, 2018, 29(2): 619-627. [Google Scholar]

[10] Yu W Y, Zhang J H. Nonlinear -Lie derivations on factor von Neumann algebras[J]. Linear Algebra and Its Applications, 2012, 437(8): 1979-1991. [Google Scholar]

[11] Jing W. Nonlinear -Lie derivations of standard operator algebras[J]. Quaestiones Mathematicae, 2016, 39(8): 1037-1046. [Google Scholar]

[12] Taghavi A, Rohi H, Darvish V. Non-linear -Jordan derivations on von Neumann algebras[J]. Linear and Multilinear Algebra, 2016, 64(3): 426-439. [Google Scholar]

[13] Zhang F J. Nonlinear skew Jordan derivable maps on factor von Neumann algebras[J]. Linear and Multilinear Algebra, 2016, 64(10): 2090-2103. [Google Scholar]

[14] Li C J, Lu F Y, Fang X C. Non-linear ξ-Jordan -derivations on von Neumann algebras[J]. Linear and Multilinear Algebra, 2014, 62(4): 466-473. [Google Scholar]

[15] Li C J, Zhao F F, Chen Q Y. Nonlinear skew Lie triple derivations between factors[J]. Acta Mathematica Sinica, English Series, 2016, 32(7): 821-830. [Google Scholar]

[16] Fu F Y, An R L. Equivalent characterization of -derivations on von Neumann algebras[J]. Linear and Multilinear Algebra, 2019, 67(3): 527-541. [Google Scholar]

[17] Zhao F F, Li C J. Nonlinear -Jordan triple derivations on von Neumann algebras[J]. Mathematica Slovaca, 2018, 68(1): 163-170. [Google Scholar]

[18] Lin W H. Nonlinear -Lie-type derivations on von Neumann algebras[J]. Acta Mathematica Hungarica, 2018, 156(1): 112-131. [Google Scholar]

[19] Lin W. Nonlinear -Lie-type derivations on standard operator algebras[J]. Acta Mathematica Hungarica, 2018, 154(2): 480-500. [Google Scholar]

[20] Yang Z J, Zhang J H. Nonlinear maps preserving mixed Lie triple products on factor von Neumann algebras[J]. Annals of Functional Analysis, 2019, 10(3): 325-336. [Google Scholar]

[21] Yang Z J, Zhang J H. Nonlinear maps preserving the second mixed Lie triple products on factor von Neumann algebras[J]. Linear and Multilinear Algebra, 2020, 68(2): 377-390. [Google Scholar]

[22] Zhou Y, Yang Z J, Zhang J H. Nonlinear mixed Lie triple derivations on prime -algebras[J]. Communications in Algebra, 2019, 47(11): 4791-4796. [Google Scholar]

[23] Li C, Zhang D. Nonlinear mixed Jordan triple -derivations on -algebras[J]. Siberian Mathematical Journal, 2022, 63(4): 735-742 [Google Scholar]

[24] Ferrero M, Haetinger C. Higher derivations of semiprime rings[J]. Communications in Algebra, 2002, 30(5): 2321-2333. [Google Scholar]

[25] Nowicki A. Inner derivation of higher orders[J]. Tsukuba Journal of Mathematics, 1984, 8(2): 219-225. [Google Scholar]

[26] Wei F, Xiao Z K. Higher derivations of triangular algebras and its generalizations[J]. Linear Algebra and Its Applications, 2011, 435(5): 1034-1054. [Google Scholar]

[27] Xiao Z K, Wei F. Nonlinear Lie higher derivations on triangular algebras[J]. Linear and Multilinear Algebra, 2012, 60(8): 979-994. [Google Scholar]

[28] Qi X F, Hou J C. Lie higher derivations on nest algebras[J]. Communications in Mathematical Research, 2010, 26(2): 131-143. [Google Scholar]

[29] Zhang F J, Qi X F, Zhang J H. Nonlinear -Lie higher derivations on factor von Neumann algebras[J]. Bulletin of the Iranian Mathematical Society, 2016, 42(3): 659-678. [MathSciNet] [Google Scholar]

[30] Ashraf M, Ahmad Wani B, Wei F. Multiplicative -Lie triple higher derivations of standard operator algebras[J]. Quaestiones Mathematicae, 2019, 42(7): 857-884. [Google Scholar]

[31] Ahmad Wani B, Ashraf M, Lin W H. Multiplicative -Jordan type higher derivations on von Neumann algebras[J]. Quaestiones Mathematicae, 2020, 43(12): 1689-1711. [Google Scholar]

Nonlinear Mixed Jordan Triple *-Higher Derivations on *-Algebras

*-代数上的非线性混合Jordan三重*-高阶导子

0 Introduction

1 Nonlinear Mixed Jordan Triple *-Derivation

2 Nonlinear Mixed Jordan Triple *-Higher Derivation

References

Nonlinear Mixed Jordan Triple -Higher Derivations on -Algebras

-代数上的非线性混合Jordan三重-高阶导子

1 Nonlinear Mixed Jordan Triple $*$ -Derivation

2 Nonlinear Mixed Jordan Triple $*$ -Higher Derivation