Non-Linear New Product $A^*B-B^*A$ Derivations on $\ast$-Algebras

TL;DR

This paper characterizes additive *-derivations on prime *-algebras via a nonlinear product condition involving the algebra's involution, extending understanding of derivation structures.

Contribution

It introduces a nonlinear product condition to identify when a map on a prime *-algebra is an additive *-derivation, providing new characterization criteria.

Findings

erivations are characterized by the nonlinear product condition.

erivations are additive under the given conditions.

The map tisfies *-derivation properties if tisfies the specified self-adjoint condition.

Abstract

Let $\mathcal{A}$ be a prime $\ast$-algebra. In this paper, we suppose that $\Phi:\mathcal{A}\to\mathcal{A}$ satisfies $$\Phi(A\diamond B)=\Phi(A)\diamond B+A\diamond\Phi(B)$$ where $A\diamond B = A^{*}B - B^{*}A$ for all $A,B\in\mathcal{A}$ .We will show that if $\Phi(\alpha \frac{I}{2})$ is self-adjoint for $\alpha\in\{1,i\}$ then $\Phi$ is additive $\ast$-derivation.

Equations

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

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

$$\Phi (AB)=\Phi (A)B+A\Phi (B)$$

$$\Phi (A^{\ast }B+B^{\ast }A)=\Phi (A{)}^{\ast }\Phi (B)+\Phi (B{)}^{\ast }\Phi (A).$$

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

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

$$\Phi (-i\frac{I}{2})=\frac{I}{2}\Phi (i\frac{I}{2})-\frac{I}{2}\Phi (i\frac{I}{2})-\frac{i}{2}\Phi (\frac{I}{2})-\frac{i}{2}\Phi (\frac{I}{2}{)}^{\ast }.$$

$$\Phi (-\frac{I}{2}{)}^{\ast }+\Phi (-\frac{I}{2})=0.$$

$$\Phi (-\frac{I}{2}{)}^{\ast }=-\Phi (-\frac{I}{2}).$$

$$\Phi (\frac{I}{2}\diamond -\frac{I}{2})=\frac{I}{2}\diamond \Phi (-\frac{I}{2}).$$

$$\Phi (-\frac{I}{2})=\Phi (-\frac{I}{2}{)}^{\ast }$$

$$\Phi \left(-iA\diamond \frac{I}{2}\right)=\Phi \left(A\diamond i\frac{I}{2}\right).$$

$$\Phi (-iA)\diamond \frac{I}{2}=\Phi (A)\diamond i\frac{I}{2}.$$

$$\Phi (-iA{)}^{\ast }-\Phi (-iA)=i\Phi (A{)}^{\ast }+i\Phi (A).$$

$$\Phi \left(-iA\diamond i\frac{I}{2}\right)=\Phi \left(A\diamond -\frac{I}{2}\right).$$

$$\Phi (-iA)\diamond i\frac{I}{2}=\Phi (A)\diamond -\frac{I}{2}.$$

$$i\Phi (-iA{)}^{\ast }+i\Phi (-iA)=-\Phi (A{)}^{\ast }+\Phi (A).$$

$$-\Phi (-iA{)}^{\ast }-\Phi (-iA)=-i\Phi (A{)}^{\ast }+i\Phi (A).$$

$$\Phi (-iA)=-i\Phi (A).$$

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

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

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

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

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

$$T\diamond X_{21}=0.$$

$$T_{21}^{\ast }{X}_{21}+T_{22}^{\ast }{X}_{21}-X_{21}^{\ast }{T}_{21}-X_{21}^{\ast }{T}_{22}=0.$$

$$T_{21}^{\ast }{X}_{21}-X_{21}^{\ast }{T}_{21}=0.$$

$$i{T}_{21}^{\ast }{X}_{21}+i{T}_{22}^{\ast }{X}_{21}+i{X}_{21}^{\ast }{T}_{21}+i{X}_{21}^{\ast }{T}_{22}=0.$$

$$-T_{21}^{\ast }{X}_{21}-X_{21}^{\ast }{T}_{21}=0.$$

$$X_{21}^{\ast }{T}_{21}=0.$$

$$\Phi (((A_{11}+A_{12})\diamond X_{12})\diamond P_1)=\Phi ((A_{11}+A_{12})\diamond X_{12})\diamond P_1+((A_{11}+A_{12})\diamond X_{12})\diamond \Phi (P_1)$$

Full text

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where $A\diamond B=A^{*}B-B^{*}A$ for all $A,B\in\mathcal{A}$ .We will show that if $\Phi(\alpha\frac{I}{2})$ is self-adjoint for $\alpha\in\{1,i\}$ then $\Phi$ is additive $\ast$-derivation.

1. Introduction

Let $\mathcal{R}$ be a $*$-algebra. For $A,B\in\mathcal{R}$, denoted by $A\bullet B=AB+BA^{*}$ 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 [3, 8, 10, 14]).

Recall that a map $\Phi:\mathcal{R}\to\mathcal{R}$ is said to be an additive derivation if

[TABLE]

and

[TABLE]

for all $A,B\in\mathcal{R}$. A map $\Phi$ is additive $\ast$-derivation if it is an additive derivation and $\Phi(A^{*})=\Phi(A)^{*}$. Derivations are very important maps both in theory and applications, and have been studied intensively ([2, 11, 12, 13]).

Let us define $\lambda$-Jordan $\ast$-product by $A\bullet_{\lambda}B=AB+\lambda BA^{\ast}$. We say that the map $\Phi$ with the property of $\Phi(A\bullet_{\lambda}B)=\Phi(A)\bullet_{\lambda}B+A\bullet_{\lambda}\Phi(B)$ 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].

A von Neumann algebra $\mathcal{A}$ is a self-adjoint subalgebra of some $B(H)$, the algebra of bounded linear operators acting on a complex Hilbert space, which satisfies the double commutant property: $\mathcal{A}^{{}^{\prime\prime}}=\mathcal{A}$ where $\mathcal{A}^{{}^{\prime}}=\{T\in B(H),TA=AT,\forall A\in\mathcal{A}\}$ and $\mathcal{A}^{{}^{\prime\prime}}=\{\mathcal{A}^{{}^{\prime}}\}^{{}^{\prime}}$. Denote by $\mathcal{Z}(\mathcal{A})=\mathcal{A}^{{}^{\prime}}\cap\mathcal{A}$ the center of $\mathcal{A}$. A von Neumann algebra $\mathcal{A}$ is called a factor if its center is trivial, that is, $\mathcal{Z}(\mathcal{A})=\mathbb{C}I$. For $A\in\mathcal{A}$, recall that the central carrier of $A$, denoted by $\overline{A}$, is the smallest central projection $P$ such that $PA=A$. It is not difficult to see that $\overline{A}$ is the projection onto the closed subspace spanned by $\{BAx:B\in\mathcal{A},x\in H\}$. If $A$ is self-adjoint, then the core of $A$, denoted by $\underline{A}$, is $\sup\{S\in\mathcal{Z}(\mathcal{A}):S=S^{*},S\leq A\}$. If $A=P$ is a projection, it is clear that $\underline{P}$ is the largest central projection $Q$ satisfying $Q\leq P$. A projection $P$ is said to be core-free if $\underline{P}=0$ (see [9]). It is easy to see that $\underline{P}=0$ if and only if $\overline{I-P}=I$, [5, 6].

Recently, Yu and Zhang in [16] 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 [7] 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 scalar, 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.

On the other hand, many mathematician devoted themselves to study the $\ast$-Jordan product $A\bullet B=AB+BA^{*}$. In [17], F. Zhang proved that every non-linear $\ast$-Jordan derivation map $\Phi:\mathcal{A}\to\mathcal{A}$ on a factor von neumann algebra with $I_{\mathcal{A}}$ the identity of it is an additive $\ast$-derivation.

In [15], we showed that $\ast$-Jordan derivation map on every factor von Neumann algebra $\mathcal{A}\subseteq\mathcal{B(H)}$ is additive $\ast$-derivation.

Very recently the authors of [4] discussed some bijective maps preserving the new product $A^{*}B+B^{*}A$ between von Neumann algebras with no central abelian projections. In other words, $\Phi$ holds in the following condition

[TABLE]

They showed that such a map is sum of a linear $\ast$-isomorphism and a conjugate linear $\ast$-isomorphism.

Motivated by the above results, in this paper, we prove that if $\mathcal{A}$ is a prime $\ast$-algebra then $\Phi:\mathcal{A}\to\mathcal{A}$ which holds in the following condition

[TABLE]

where $A\diamond B=A^{*}B-B^{*}A$ for all $A,B\in\mathcal{A}$, is additive $\ast$-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 main theorem is as follows:

Theorem 2.1**.**

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

[TABLE]

where $A\diamond B=A^{*}B-B^{*}A$ for all $A,B\in\mathcal{A}$. If $\Phi(\alpha\frac{I}{2})$ is self-adjoint operator for $\alpha\in\{1,i\}$ then $\Phi$ is additive $\ast$-derivation.

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**.**

$\Phi(i\frac{I}{2})=\Phi(\frac{I}{2})=\Phi(-i\frac{I}{2})=0$.

Consider $\Phi(\frac{I}{2}\diamond i\frac{I}{2})=\Phi(\frac{I}{2})\diamond i\frac{I}{2}+\frac{I}{2}\diamond\Phi(i\frac{I}{2})$ that imply

[TABLE]

By taking the adjoint of above equation we have $\Phi(i\frac{I}{2})=\Phi(\frac{I}{2})=0$

Consider $\Phi(i\frac{I}{2}\diamond\frac{I}{2})=\Phi(i\frac{I}{2})\diamond\frac{I}{2}+i\frac{I}{2}\diamond\Phi(\frac{I}{2})$ that imply

[TABLE]

Since $\Phi(\frac{I}{2})=\Phi(i\frac{I}{2})=0$, so we have $\Phi(-i\frac{I}{2})=0$.

Claim 3**.**

$\Phi(-\frac{I}{2})=0$.

Consider $\Phi(-\frac{I}{2}\diamond i\frac{I}{2})=\Phi(-\frac{I}{2})\diamond i\frac{I}{2}$. So, we have

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It follows that

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Also

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So $\Phi(-\frac{I}{2})-\Phi(-\frac{I}{2})^{*}=0$, then we have

[TABLE]

from (2.4), (2.5) we have $\Phi(-\frac{I}{2})=0.$

Claim 4**.**

For each $A\in\mathcal{A}$, we have

1. (1)

$\Phi(-iA)=-i\Phi(A).$** 2. (2)

$\Phi(iA)=i\Phi(A).$**

We can check to see that

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So,

[TABLE]

It follows that

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On the other hand, one can check that

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So,

[TABLE]

It follows that

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Equivalently, we obtain

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By adding equations (2.6) and (2.8) we have

[TABLE]

Similarly, we can show that $\Phi(iA)=i\Phi(A)$.

Claim 5**.**

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

[TABLE]

Let $T=\Phi(A_{11}+A_{12})-\Phi(A_{11})-\Phi(A_{12})$, we should prove that $T=0$.

For $X_{21}\in\mathcal{A}_{21}$ we can write that

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So, we obtain

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

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From the above equation and primeness of $\mathcal{A}$ we have $T_{22}=0$ and

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On the other hand, similarly by applying $iX_{21}$ instead of $X_{21}$ in above, we obtain

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Since $T_{22}=0$ we obtain from the above equation that

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From (2.9) and (2.10) we have

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Since $\mathcal{A}$ is prime, then we get $T_{21}=0$.

It suffices to show that $T_{12}=T_{11}=0$. For this purpose for $X_{12}\in\mathcal{A}_{12}$ we write

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So, we showed that

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Since $A_{12}\diamond X_{12}\diamond P_{1}=0$ we have

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So,

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From (2.11) and (2.12) we have

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It follows that $T\diamond X_{12}\diamond P_{1}=0$, so $T_{11}^{*}X_{12}-X_{12}^{*}T_{11}=0$. We have $T_{11}^{*}{}X_{12}=0$ or $T_{11}XP_{2}=0$ for all $X\in\mathcal{A}$, then we have $T_{11}=0$. Similarly, we can show that $T_{12}=0$ by applying $P_{2}$ instead of $P_{1}$ in above.

Claim 6**.**

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

1. (1)

[TABLE] 2. (2)

[TABLE]

We show that

[TABLE]

So, we have

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It follows that $T\diamond X_{21}=0$. Since $T=T_{11}+T_{12}+T_{21}+T_{22}$ we have

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Therefore, $T_{22}=T_{21}=0$.

From Claim 5, we obtain

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Hence,

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Then $T_{11}=T_{12}=0$. Similarly

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Claim 7**.**

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

[TABLE]

We show that

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From Claim 6, we have

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So, $T\diamond X_{12}=0$. It follows that

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Then $T_{11}=T_{12}=0$.

Similarly, by applying $X_{21}$ instead of $X_{12}$ in above, we obtain $T_{21}=T_{22}=0$.

Claim 8**.**

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

[TABLE]

It is easy to show that

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So, we can write

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Therefore, we show that

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By an easy computation, we can write

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Then, we have

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We showed that

[TABLE]

From Claim 4 and the above equation, we have

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By adding equations (2.13) and (2.14), we obtain

[TABLE]

Claim 9**.**

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]

So, we have

[TABLE]

Therefore, we obtain $T_{ij}=T_{ji}=T_{jj}=0$.

On the other hand, for every $X_{ij}\in\mathcal{A}_{ij}$, we have

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So,

[TABLE]

It follows that $T\diamond X_{ij}=0$ or $T_{ii}X_{ij}=0$. By knowing that $\mathcal{A}$ is prime, we have $T_{ii}=0$.

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

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

Claim 10**.**

$\Phi$* preserves star.*

Since $\Phi(I)=0$ then we can write

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Then

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So, we showed that $\Phi$ preserves star.

Claim 11**.**

we prove that $\Phi$ is derivation.

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

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On the other hand, since $\Phi$ preserves star, we have

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So, from (2.15), we have

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Therefore, from claim 4 we have

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By adding equations (2.15) and (2.16), we have

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This completes the proof.