Non-linear $\ast$-Jordan triple derivation on prime $\ast$-algebras

TL;DR

This paper investigates non-linear maps on prime $ extit{*}$-algebras that preserve a specific triple $ extit{*}$-Jordan derivation structure, proving additivity and conditions under which they are $ extit{*}$-derivations.

Contribution

It establishes that such maps are additive and, under certain conditions, are $ extit{*}$-derivations, extending understanding of structure-preserving maps in prime $ extit{*}$-algebras.

Findings

Preservation of triple $ extit{*}$-Jordan derivation implies additivity.

Additional conditions ensure the map is a $ extit{*}$-derivation.

Results apply to prime $ extit{*}$-algebras with specific self-adjoint properties.

Abstract

Let $\mathcal{A}$ be a prime $\ast$-algebra and $\Phi$ preserves triple $\ast$-Jordan derivation on $\mathcal{A}$, that is, for every $A,B \in \mathcal{A}$, $$\Phi(A\diamond B \diamond C)=\Phi(A)\diamond B\diamond C+A\diamond \Phi(B)\diamond C+A\diamond B\diamond \Phi(C)$$ where $A\diamond B = AB + BA^{\ast}$ then $\Phi$ is additive. Moreover, if $\Phi(\alpha I)$ is self-adjoint for $\alpha\in\{1,i\}$ then $\Phi$ is a $\ast$-derivation.

Equations

$$\Phi (A\diamond B\diamond C)=\Phi (A)\diamond B\diamond C+A\diamond \Phi (B)\diamond C+A\diamond B\diamond \Phi (C)$$

$$\Phi (A{\diamond }_{\lambda }B{\diamond }_{\lambda }C)=\Phi (A){\diamond }_{\lambda }B{\diamond }_{\lambda }C+A{\diamond }_{\lambda }\Phi (B){\diamond }_{\lambda }C)+A{\diamond }_{\lambda }B{\diamond }_{\lambda }\Phi (C)$$

$$\Phi (A\diamond B\diamond C)=\Phi (A)\diamond B\diamond C+A\diamond \Phi (B)\diamond C)+A\diamond B\diamond \Phi (C)$$

$$\Phi (A\diamond B\diamond C)=\Phi (A)\diamond B\diamond C+A\diamond \Phi (B)\diamond C+A\diamond B\diamond \Phi (C)$$

$$\Phi (A_{12}+A_{21})=\Phi (A_{12})+\Phi (A_{21}).$$

$$T=\Phi (A_{12}+A_{21})-\Phi (A_{12})-\Phi (A_{21})=0.$$

$$\Phi \left(\frac{I}{2}\right)\diamond (P_1-P_2)\diamond (A_{12}+A_{21})+\frac{I}{2}\diamond \Phi (P_1-P_2)\diamond (A_{12}+A_{21})$$

$$+\frac{I}{2}\diamond (P_1-P_2)\diamond \Phi (A_{12}+A_{21})$$

$$=\Phi \left(\frac{I}{2}\diamond (P_1-P_2)\diamond (A_{12}+A_{21})\right)$$

$$=\Phi \left(\frac{I}{2}\diamond (P_1-P_2)\diamond A_{12}\right)+\Phi \left(\frac{I}{2}\diamond (P_1-P_2)\diamond A_{21}\right)$$

$$=\Phi \left(\frac{I}{2}\right)\diamond (P_1-P_2)\diamond (A_{12}+A_{21})+\frac{I}{2}\diamond \Phi (P_1-P_2)\diamond (A_{12}+A_{21})$$

$$+\frac{I}{2}\diamond (P_1-P_2)\diamond (\Phi (A_{12})+\Phi (A_{21})).$$

$$\frac{I}{2}\diamond (P_1-P_2)\diamond T=0.$$

$$\Phi (P_1)\diamond (A_{12}+A_{21})\diamond P_1+P_1\diamond \Phi (A_{12}+A_{21})\diamond P_1$$

$$+P_1\diamond (A_{12}+A_{21})\diamond \Phi (P_1)$$

$$=\Phi (P_1\diamond A_{12}+A_{21}\diamond P_1)$$

$$=\Phi (P_1\diamond A_{12}\diamond P_1)+\Phi (P_1\diamond A_{21}\diamond P_1)$$

$$=\Phi (P_1)\diamond (A_{12}+A_{21})\diamond P_1$$

$$+P_1\diamond (\Phi (A_{12})+\Phi (A_{21}))\diamond P_1)+P_1\diamond (A_{12}+A_{21})\diamond \Phi (P_1).$$

$$P_1\diamond T\diamond P_1=0.$$

$$T_{21}+T_{21}^{\ast }=0$$

$$P_2\diamond T\diamond P_2=0.$$

$$\Phi (A_{11}+A_{12}+A_{21})=\Phi (A_{11})+\Phi (A_{12})+\Phi (A_{21}).$$

$$T=\Phi (A_{11}+A_{12}+A_{21})-\Phi (A_{11})-\Phi (A_{12})-\Phi (A_{21})=0.$$

$$\Phi \left(\frac{I}{2}\right)\diamond (P_1-P_2)\diamond (A_{11}+A_{12}+A_{21})+\frac{I}{2}\diamond \Phi (P_1-P_2)\diamond (A_{11}+A_{12}+A_{21})$$

$$+\frac{I}{2}\diamond (P_1-P_2)\diamond \Phi (A_{11}+A_{12}+A_{21})=\Phi \left(\frac{I}{2}\diamond (P_1-P_2)\diamond (A_{11}+A_{12}+A_{21})\right)$$

$$=\Phi \left(\frac{I}{2}\diamond (P_1-P_2)\diamond A_{12}\right)+\Phi \left(\frac{I}{2}\diamond (P_1-P_2)\diamond A_{21}\right)$$

$$+\Phi \left(\frac{I}{2}\diamond (P_1-P_2)\diamond A_{11}\right)=\Phi \left(\frac{I}{2}\right)\diamond (P_1-P_2)\diamond (A_{11}+A_{12}+A_{21})$$

$$+\frac{I}{2}\diamond \Phi (P_1-P_2)\diamond (A_{11}+A_{12}+A_{21})$$

$$+\frac{I}{2}\diamond (P_1-P_2)\diamond (\Phi (A_{11})+\varphi (A_{12})+\Phi (A_{21})).$$

$$\frac{I}{2}\diamond (P_1-P_2)\diamond T=0.$$

$$\Phi (P_2)\diamond \frac{I}{2}\diamond (A_{11}+A_{12}+A_{21})+P_2\diamond \Phi \left(\frac{I}{2}\right)\diamond (A_{11}+A_{12}+A_{21})$$

$$+P_2\diamond \frac{I}{2}\diamond \Phi (A_{11}+A_{12}+A_{21})$$

$$=\Phi \left(P_2\diamond \frac{I}{2}\diamond (A_{11}+A_{12}+A_{21})\right)$$

$$=\Phi (P_2\diamond \frac{I}{2}\diamond A_{11})+\Phi (P_2\diamond \frac{I}{2}\diamond (A_{12}+A_{21}))$$

$$=\Phi (P_2\diamond \frac{I}{2}\diamond A_{11})+\Phi (P_2\diamond \frac{I}{2}\diamond A_{12})+\Phi (P_2\diamond \frac{I}{2}\diamond A_{21})$$

$$=\Phi (P_2)\diamond \frac{I}{2}\diamond (A_{11}+A_{12}+A_{21})+P_2\diamond \Phi (\frac{I}{2})\diamond (A_{11}+A_{12}+A_{21})$$

$$+P_2\diamond \frac{I}{2}\diamond (\Phi (A_{11})+\Phi (A_{12})+\Phi (A_{21}))$$

$$P_2\diamond \frac{I}{2}\diamond T=0.$$

$$T_{12}+T_{21}+2T_{22}=0.$$

$$\Phi (A_{11}+A_{12}+A_{21}+A_{22})=\Phi (A_{11})+\Phi (A_{12})+\Phi (A_{21})+\Phi (A_{22}).$$

Full text

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where $A\diamond B=AB+BA^{\ast}$ then $\Phi$ is additive. Moreover, if $\Phi(\alpha I)$ is self-adjoint for $\alpha\in\{1,i\}$ then $\Phi$ is a $\ast$-derivation.

1. Introduction

Let $\mathcal{R}$ be a $*$-ring. For $A,B\in\mathcal{R}$, denoted by $A\diamond B=AB+BA^{*}$ and $[A,B]_{*}=AB-BA^{*}$, which are $\ast$-Jordan product and $\ast$-Lie product, respectively. These products are found playing a more and more important role in some research topics, and its study has recently attracted many author’s attention (for example, see [2, 4, 6, 8]).

Let define $\lambda$-Jordan $\ast$-product by $A\diamond_{\lambda}B=AB+\lambda BA^{\ast}$. We say the map $\Phi$ with property of $\Phi(A\diamond_{\lambda}B)=\Phi(A)\diamond_{\lambda}B+A\diamond_{\lambda}\Phi(B)$ is a $\lambda$-Jordan $\ast$-derivation map. It is clear that for $\lambda=-1$ and $\lambda=1,$ the $\lambda$-Jordan $\ast$-derivation map is a $\ast$-Lie derivation and $\ast$-Jordan derivation, respectively [1]. We should mention here whenever we say $\Phi$ preserves derivation, it means $\Phi(AB)=\Phi(A)B+A\Phi(B)$.

Recently, Yu and Zhang in [11] proved that every non-linear $\ast$-Lie derivation from a factor von Neumann algebra into itself is an additive $\ast$-derivation. Also, Li, Lu and Fang in [3] have investigated a non-linear $\lambda$-Jordan $\ast$-derivation. They showed that if $\mathcal{A}\subseteq\mathcal{B(H)}$ is a von Neumann algebra without central abelian projections and $\lambda$ is a non-zero scaler, then $\Phi:\mathcal{A}\longrightarrow\mathcal{B(H)}$ is a non-linear $\lambda$-Jordan $\ast$-derivation if and only if $\Phi$ is an additive $\ast$-derivation.

In [9] we showed that $\ast$-Jordan derivation map (i.e., $\phi(A\diamond B)=\phi(A)\diamond B+A\diamond\phi(B)$) on every factor von Neumann algebra $\mathcal{A}\subseteq\mathcal{B(H)}$ is additive $\ast$-derivation. Moreover, in [7] we reduced the assumptions on the map $\phi$.

The authors of [5] introduced the concept of Lie triple derivations. A map $\Phi:\mathcal{A}\to\mathcal{A}$ is a nonlinear skew Lie triple derivations if $\Phi([[A,B]_{\ast},C]_{\ast})=[[\Phi(A),B]_{\ast},C]_{\ast}+[[A,\Phi(B)]_{\ast},C]_{\ast}+[[A,B]_{\ast},\Phi(C)]_{\ast}$ for all $A,B,C\in\mathcal{A}$ where $[A,B]_{\ast}=AB-BA^{*}$. They showed that if $\Phi$ preserves the above characterizations on factor von Neumann algebras then $\Phi$ is additive $\ast$-derivation.

Also, in [10] we considered a map $\Phi$ on prime $\ast$-algebra $\mathcal{A}$ which holds in the following conditions

[TABLE]

where $A\diamond_{\lambda}B=AB+\lambda BA^{\ast}$ such that a complex scalar $|\lambda|\neq 0,1$, then $\Phi$ is additive. Also, if $\Phi(I)$ is self-adjoint then $\Phi$ is $\ast$-derivation.

In this paper inspired by the results above, we consider a map $\Phi$ on prime $\ast$-algebra $\mathcal{A}$ which holds in the following conditions

[TABLE]

where $A\diamond B=AB+BA^{\ast}$ then $\Phi$ is additive. Also, if $\Phi(\alpha I)$ is self-adjoint for $\alpha\in\{1,i\}$ then $\Phi$ is $\ast$-derivation.

We say that $\mathcal{A}$ is prime, that is, for $A,B\in\mathcal{A}$ if $A\mathcal{A}B=\{0\},$ then $A=0$ or $B=0$.

2. Main Results

Our first theorem is as follows:

Theorem 2.1**.**

Let $\mathcal{A}$ be a prime $\ast$-algebra. Then the map $\Phi:\mathcal{A}\to\mathcal{A}$ satisfies in the following condition

[TABLE]

for all $A,B\in\mathcal{A}$ where $A\diamond B=AB+BA^{*}$, is additive.

Proof. Let $P_{1}$ be a nontrivial projection in $\mathcal{A}$ and $P_{2}=I_{\mathcal{A}}-P_{1}$. Denote $\mathcal{A}_{ij}=P_{i}\mathcal{A}P_{j},\ i,j=1,2,$ then $\mathcal{A}=\sum_{i,j=1}^{2}\mathcal{A}_{ij}$. For every $A\in\mathcal{A}$ we may write $A=A_{11}+A_{12}+A_{21}+A_{22}$. In all that follow, when we write $A_{ij}$, it indicates that $A_{ij}\in\mathcal{A}_{ij}$. For showing additivity of $\Phi$ on $\mathcal{A}$, we use above partition of $\mathcal{A}$ and give some claims that prove $\Phi$ is additive on each $\mathcal{A}_{ij},\ i,j=1,2$.

We prove the above theorem by several claims.

Claim 1**.**

We show that $\Phi(0)=0$.

This claim is easy to prove.

Claim 2**.**

For each $A_{12}\in\mathcal{A}_{12}$ and $A_{21}\in\mathcal{A}_{21}$ we have

[TABLE]

We show that

[TABLE]

We can write that

[TABLE]

So, we have

[TABLE]

Since $T=T_{11}+T_{12}+T_{21}+T_{22}$ then $T_{11}-T_{22}=0$, therefore $T_{11}=T_{22}$.

On the other hand, we obtain

[TABLE]

So, we have

[TABLE]

Since $T=T_{11}+T_{12}+T_{21}+T_{22}$ then

[TABLE]

it follows that $T_{21}=0$.

In a similar way for $P_{2}$ we can obtain

[TABLE]

So, $T_{12}=0$.

Claim 3**.**

For each $A_{11}\in\mathcal{A}_{11}$, $A_{12}\in\mathcal{A}_{12}$, $A_{21}\in\mathcal{A}_{21}$, we have

[TABLE]

We show that for $T$ in $\mathcal{A}$ the following holds

[TABLE]

We can write

[TABLE]

Then, we have

[TABLE]

Since $T=T_{11}+T_{12}+T_{21}+T_{22}$ we obtain $T_{11}-T_{22}=0$ or $T_{11}=T_{22}=0$.

By Claim 2 we have

[TABLE]

So,

[TABLE]

Since $T=T_{11}+T_{12}+T_{21}+T_{22}$, we obtain

[TABLE]

Therefore $T_{12}=T_{21}=0$.

Claim 4**.**

For each $A_{11}\in\mathcal{A}_{11}$, $A_{12}\in\mathcal{A}_{12}$, $A_{21}\in\mathcal{A}_{21}$, $A_{22}\in\mathcal{A}_{22}$ we have

[TABLE]

We show that for $T$ in $\mathcal{A}$ the following holds

[TABLE]

From Claim 3 We can write

[TABLE]

So,

[TABLE]

Since $T=T_{11}+T_{12}+T_{21}+T_{22}$ then $T_{11}=T_{21}=T_{22}=0$. Similarly, we can show that $T_{22}=0$.

Claim 5**.**

For each $A_{ij},B_{ij}\in\mathcal{A}_{i}$ such that $i\neq j$, we have

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It is easy to check that

[TABLE]

From Claim 4 we have

[TABLE]

So,

[TABLE]

Claim 6**.**

For each $A_{ii},B_{ii}\in\mathcal{A}_{ii}$ such that $1\leq i\leq 2$, we have

[TABLE]

We show that

[TABLE]

We can write

[TABLE]

Therefore,

[TABLE]

Since $T=T_{11}+T_{12}+T_{21}+T_{22}$ we have $T_{jj}=T_{ji}=T_{ij}=0$.

From Claim 5 for every $C_{ij}\in\mathcal{A}_{ij}$ we have

[TABLE]

So,

[TABLE]

By primeness and since $T=T_{11}+T_{12}+T_{21}+T_{22}$, by primeness we obtain $T_{ii}=0$.

Hence, the additivity of $\Phi$ comes from the above claims. ∎

In the rest of this paper we prove that $\Phi$ is $\ast$-derivation.

Theorem 2.2**.**

Let $\mathcal{A}$ be a prime $\ast$-algebra. Let $\Phi:A\to A$ satisfies in the following condition

[TABLE]

for all $A,B\in\mathcal{A}$ where $A\diamond B=AB+BA^{*}$. If $\Phi(\alpha I)$ is self-adjoint for $\alpha\in\{1,i\}$ then $\Phi$ is $\ast$-derivation.

Proof. We present the proof by several claims.

Claim 1**.**

If $\Phi(I)$ is self-adjoint then $\Phi(I)=0$.

One can easily show that

[TABLE]

So, $\Phi(I)=0$.

Claim 2**.**

If $\Phi(iI)$ is self-adjoint then $\Phi(iI)=0$.

It is easy to check that

[TABLE]

We obtain

[TABLE]

So, $\Phi(iI)+\Phi(iI)^{*}=0$. Since $\Phi(iI)$ is self-adjoint then $\Phi(iI)=0$.

Claim 3**.**

$\Phi$* preserves star.*

Since $\Phi(I)=0$, then

[TABLE]

Therefore

[TABLE]

So, we obtain

[TABLE]

Claim 4**.**

We show that $\Phi(iA)=i\Phi(A)$ for every $A\in\mathcal{A}$.

Let $T$ be a self-adjoint member of $\mathcal{A}$. Then it is easy to check that

[TABLE]

Since $\Phi\left(I\right)=\Phi\left(iI\right)=0$, we obtain

[TABLE]

So, we have

[TABLE]

Since $\Phi$ preserves star and is additive, for every self-adjoint $T$ we have

[TABLE]

On the other hand, we can write every $A\in\mathcal{A}$ as $A=A_{1}+iA_{2}$ where $A_{1}=\frac{A+A^{*}}{2}$ and $A_{2}=\frac{A-A^{*}}{2i}$. Finally we have

[TABLE]

So, $\Phi(iA)=i\Phi(A)$ for all $A\in\mathcal{A}$.

Claim 5**.**

$\Phi$* is a derivation.*

For every $A,B\in\mathcal{A}$ we have

[TABLE]

It follows that

[TABLE]

From (2.5) we have

[TABLE]

So,

[TABLE]

From (2.5) and (2.6) we obtain

[TABLE]

∎

Acknowledgments: The first author is supported by the Talented Young Scientist Program of Ministry of Science and Technology of China (Iran-19-001).