Strong (skew) ξ-Lie commutativity preserving maps on algebras

Abstract

Let $\mathcal{A}$ be any unital *-algebra over the real or complex field $\mathbb{F}$, and let $\mathit{\xi }\in \mathbb{F}$ with $\mathit{\xi }\ne 1$. Assume that $\mathrm{\Phi }:\mathcal{A}\to \mathcal{A}$ is a map. It is shown that, $\mathrm{\Phi }$ satisfies $\mathrm{\Phi }(A)\mathrm{\Phi }(B)-\mathit{\xi }\mathrm{\Phi }(B)\mathrm{\Phi }(A)=AB-\mathit{\xi }BA$ for all $A,B\in \mathcal{A}$ if and only if $\mathrm{\Phi }(I)\in \mathcal{Z}(\mathcal{A})$, the center of $\mathcal{A}$, $\mathrm{\Phi }{(I)}^2=I$ and $\mathrm{\Phi }(A)=\mathrm{\Phi }(I)A$ for all $A\in \mathcal{A}$; if $\mathrm{\Phi }(I)=\mathrm{\Phi }{(I)}^{\ast }$, then $\mathrm{\Phi }$ satisfies $\mathrm{\Phi }(A)\mathrm{\Phi }(B)-\mathit{\xi }\mathrm{\Phi }(B)\mathrm{\Phi }{(A)}^{\ast }=AB-\mathit{\xi }B{A}^{\ast }$ for all $A,B\in \mathcal{A}$ if and only if $\mathrm{\Phi }(I)\in \mathcal{Z}(\mathcal{A})$, $\mathrm{\Phi }{(I)}^2=I$ and $\mathrm{\Phi }(A)=\mathrm{\Phi }(I)A$ for all $A\in \mathcal{A}$; if $|\mathit{\xi }|=1$ and $\mathrm{\Phi }$ is surjective, then $\mathrm{\Phi }$ satisfies $\mathrm{\Phi }(A)\mathrm{\Phi }(B)-\mathit{\xi }\mathrm{\Phi }(B)\mathrm{\Phi }{(A)}^{\ast }=AB-\mathit{\xi }B{A}^{\ast }$ for all $A,B\in \mathcal{A}$ if and only if $\mathrm{\Phi }(I)=\mathrm{\Phi }{(I)}^{\ast }\in \mathcal{Z}(\mathcal{A})$, $\mathrm{\Phi }{(I)}^2=I$, and $\mathrm{\Phi }(A)=\mathrm{\Phi }(I)A$ for all $A\in \mathcal{A}$.

Keywords:

• algebras
• ξ-Lie products
• skew ξ-Lie products
• commutativity

AMS subject classifications:

• 16W10
• 16U80

Public Interest Statement

Preserver problem had attracted many mathematicians’ attentions for many years. In this paper, the authors discuss the general maps preserving strong (skew) $\mathit{\xi }$-Lie commutativity on any algebras and give a complete characterization for such maps.

1. Introduction

Let $\mathcal{R}$ be a ring. Then $\mathcal{R}$ is a Lie ring under the Lie product $[A,B]=AB-BA.$ Recall that a map $\mathrm{\Phi }:\mathcal{R}\to \mathcal{R}$ preserves commutativity if $[\mathrm{\Phi }(A),\mathrm{\Phi }(B)]=0$ whenever $[A,B]=0$ for all $A$, $B\in \mathcal{R}$. The problem of characterizing linear (additive) maps preserving commutativity had been studied intensively on various rings and algebras (see Brešar, 1993; Brešar & Šemrl, 2005; Choi, Jafarian, & Radjavi, 1987) and the references therein).

In Bell and Daif (1994), the authors gave the conception of strong commutativity preserving maps. Let $\mathcal{S}$ be a subset of $\mathcal{R}$. A map $\mathrm{\Phi }:\mathcal{S}\to \mathcal{R}$ is called strong commutativity preserving if $[\mathrm{\Phi }(T),\mathrm{\Phi }(S)]=[T,S]$ for all $T,S\in \mathcal{S}$. Note that a strong commutativity preserving map must be commutativity preserving, but the inverse is not true generally. Bell and Daif (1994) proved that $\mathcal{R}$ must be commutative, if $\mathcal{R}$ is a prime ring and $\mathcal{R}$ admits a derivation or a non-identity endomorphism which is strong commutativity preserving on a right ideal of $\mathcal{R}$. Brešar and Miers (1994) proved that every strong commutativity preserving additive map $\mathrm{\Phi }$ on a semiprime ring $\mathcal{R}$ is of the form $\mathrm{\Phi }(A)=\mathit{\lambda }A+\mathit{\mu }(A)$, where $\mathit{\lambda }\in \mathcal{C}$, the extended centroid of $\mathcal{R}$, ${\mathit{\lambda }}^2=1$, and $\mathit{\mu }:\mathcal{R}\to \mathcal{C}$ is an additive map. Recently, Lin and Liu (2008) obtained the similar result on a noncentral Lie ideal of a prime ring. Qi and Hou (2010; 2012) gave a complete characterization of strong commutativity preserving surjective maps (without the assumption of additivity) on prime rings and triangular algebras, respectively.

Let $\mathcal{R}$ be a *-ring. For any $A,B\in \mathcal{R}$, ${[A,B]}_{\ast }=AB-B{A}^{\ast }$ denotes the skew Lie product of $A$ and $B$. This kind of product is found playing a more and more important role in some research topics such as representing quadratic functionals with sesquilinear functionals, and its study has attracted many authors’ attention (see Brešar & Fosňer, 2000; Chebotar, Fong, & Lee, 2005; Cui & Hou, 2006) and the reference therein). Molnár (1996) initiated the systematic study of this skew Lie product, and studied the relation between subspaces and ideals of $\mathcal{B}(H)$, the algebra of all bounded linear operators acting on a Hilbert space $H$.

Recall that a map $\mathrm{\Phi }:\mathcal{R}\to \mathcal{R}$ is called zero skew Lie product preserving, if $\mathrm{\Phi }(A)\mathrm{\Phi }(B)-\mathrm{\Phi }(B)\mathrm{\Phi }{(A)}^{\ast }=0$ whenever $AB-B{A}^{\ast }=0$ for any $A,B\in \mathcal{R}$. Additive or linear maps preserving zero skew Lie products on various rings and algebras had been studied by many authors (see, Bell & Daif, 1994 and the references therein). More specially, $\mathrm{\Phi }$ is strong skew commutativity preserving, if ${[\mathrm{\Phi }(A),\mathrm{\Phi }(B)]}_{\ast }={[A,B]}_{\ast }$ for all $A,B\in \mathcal{R}$. It is obvious that strong skew commutativity preserving maps must be zero skew Lie product preserving. However, the inverse is not true generally. In Cui and Park (2012), they proved that, if $\mathcal{R}$ is a factor von Neumann algebra, then every strong skew commutativity preserving map $\mathrm{\Phi }$ on $\mathcal{R}$ has the form $\mathrm{\Phi }(A)=\mathrm{\Psi }(A)+h(A)I$ for all $A\in \mathcal{R}$, where $\mathrm{\Psi }:\mathcal{R}\to \mathcal{R}$ is a linear bijective map satisfying $\mathrm{\Psi }(A)\mathrm{\Psi }(B)-\mathrm{\Psi }(B)\mathrm{\Psi }{(A)}^{\ast }=AB-B{A}^{\ast }$ for all $A,B\in \mathcal{R}$ and $h$ is a real linear functional of $\mathcal{R}$ with $h(0)=0$; particularly, if $\mathcal{R}$ is of type $I$, then $\mathrm{\Phi }(A)=cA+h(A)I$ for each $A\in \mathcal{R}$, where $c\in \{-1,1\}$. Recently, Qi and Hou (2013) generalized the above result to von Neumann algebras without central summand of type $I_1$.

Recall that $A$ commutes with $B$ up to a factor $\mathit{\xi }\in \mathbb{F}$ if $AB=\mathit{\xi }BA$. Note that the concept of commutativity up to a factor for pairs of operators is important and has been studied in the context of operator algebras and quantum groups see (Brooke, Busch, & Pearson, 2002; Kassel, 1995). Motivated by this, a binary operation ${[A,B]}_{\mathit{\xi }}=AB-\mathit{\xi }BA$, called $\mathit{\xi }$-Lie product of $A$ and $B$, was introduced in Qi and Hou (2009). Thus, we also can define the skew $\mathit{\xi }$-Lie product of $A$ and $B$. Let $\mathcal{A}$ be a *-algebra over $\mathbb{F}$, where $\mathbb{F}$ is a field with an involution $\ast$. For $A,B\in \mathcal{A}$ and $\mathit{\xi }\in \mathbb{F}$, we call $AB-\mathit{\xi }B{A}^{\ast }$ the skew $\mathit{\xi }$-Lie product of $A$ and $B$. It is obvious that the skew $\mathit{\xi }$-Lie product is the skew Lie product if $\mathit{\xi }=1$. Now, based on these concepts, we say that a map $\mathrm{\Phi }:\mathcal{A}\to \mathcal{A}$ is preserving strong $\mathit{\xi }$-Lie commutativity if ${[\mathrm{\Phi }(T),\mathrm{\Phi }(S)]}_{\mathit{\xi }}={[T,S]}_{\mathit{\xi }}$ for all $T,S\in \mathcal{A}$; is preserving strong skew $\mathit{\xi }$-Lie commutativity if $\mathrm{\Phi }(T)\mathrm{\Phi }(S)-\mathit{\xi }\mathrm{\Phi }(S)\mathrm{\Phi }{(T)}^{\ast }=TS-\mathit{\xi }S{T}^{\ast }$ for all $T,S\in \mathcal{A}$.

The purpose of this paper is to consider nonlinear strong (skew) $\mathit{\xi }$-Lie commutativity preserving maps on general algebras with $\mathit{\xi }\ne 1$. Let $\mathcal{A}$ be any unital algebra over any field $\mathbb{F}$ and $\mathit{\xi }\in \mathbb{F}$ with $\mathit{\xi }\ne 1$. Denote by $\mathcal{Z}(\mathcal{A})$ the center of $\mathcal{A}$. Assume that $\mathrm{\Phi }:\mathcal{A}\to \mathcal{A}$ is a map. In Section 2, we prove that $\mathrm{\Phi }$ preserves strong $\mathit{\xi }$-Lie commutativity if and only if $\mathrm{\Phi }(I)\in \mathcal{Z}(\mathcal{A})$, $\mathrm{\Phi }{(I)}^2=I$, and $\mathrm{\Phi }(A)=\mathrm{\Phi }(I)A$ for all $A\in \mathcal{A}$ (Theorem 2.1). In Section 3, we furthermore assume that $\mathcal{A}$ is a *-algebra. It is shown that, if $\mathrm{\Phi }(I)=\mathrm{\Phi }{(I)}^{\ast }$, then $\mathrm{\Phi }$ preserves strong skew $\mathit{\xi }$-Lie commutativity if and only if $\mathrm{\Phi }(I)\in \mathcal{Z}(\mathcal{A})$, $\mathrm{\Phi }{(I)}^2=I$ and $\mathrm{\Phi }(A)=\mathrm{\Phi }(I)A$ for all $A\in \mathcal{A}$ (Theorem 3.1); if $|\mathit{\xi }|=1$ and $\mathrm{\Phi }$ is surjective, then $\mathrm{\Phi }$ preserves strong skew $\mathit{\xi }$-Lie commutativity if and only if $\mathrm{\Phi }(I)=\mathrm{\Phi }{(I)}^{\ast }\in \mathcal{Z}(\mathcal{A})$, $\mathrm{\Phi }{(I)}^2=I$ and $\mathrm{\Phi }(A)=\mathrm{\Phi }(I)A$ for all $A\in \mathcal{A}$ (Theorem 3.2).

2. Maps preserving strong $\mathit{\xi }$-Lie commutativity

In this section, we will give a characterization of nonlinear strong $\mathit{\xi }$-Lie commutativity preserving maps on general algebras. The following is our main result.

Theorem 2.1

Let $\mathcal{A}$ be any algebra with unit $I$ over a field $\mathbb{F}$, and let $\mathit{\xi }\in \mathbb{F}$ with $\mathit{\xi }\ne 1$. Assume that $\mathrm{\Phi }:\mathcal{A}\to \mathcal{A}$ is a map. Then $\mathrm{\Phi }$ preserves strong $\mathit{\xi }$-Lie commutativity, that is, $\mathrm{\Phi }$ satisfies ${[\mathrm{\Phi }(A),\mathrm{\Phi }(B)]}_{\mathit{\xi }}={[A,B]}_{\mathit{\xi }}$ for all $A,B\in \mathcal{A}$, if and only if $\mathrm{\Phi }(I)\in \mathcal{Z}(\mathcal{A})$, $\mathrm{\Phi }{(I)}^2=I$ and $\mathrm{\Phi }(A)=\mathrm{\Phi }(I)A$ for all $A\in \mathcal{A}$.

Proof

The “if” part is obvious. For the “only if” part, since ${[\mathrm{\Phi }(I),\mathrm{\Phi }(I)]}_{\mathit{\xi }}={[I,I]}_{\mathit{\xi }}$, we have $(1-\mathit{\xi })\mathrm{\Phi }{(I)}^2=(1-\mathit{\xi })I$. It follows that $\mathrm{\Phi }{(I)}^2=I$ as $\mathit{\xi }\ne 1$.

In the sequel, we will complete the proof by considering two cases.

Case 1 $\mathit{\xi }=-1$.

Take any $A\in \mathcal{A}$. Then(2.1) $$\begin{array}{c}\hfill 2A=AI+IA=\mathrm{\Phi }(A)\mathrm{\Phi }(I)+\mathrm{\Phi }(I)\mathrm{\Phi }(A)\end{array}$$(2.1)

Multiplying $\mathrm{\Phi }(I)$ from the left- and the right-hand side in Equation 2.1, respectively, one gets$$2\mathrm{\Phi }(I)A=\mathrm{\Phi }(I)\mathrm{\Phi }(A)\mathrm{\Phi }(I)+\mathrm{\Phi }{(I)}^2\mathrm{\Phi }(A)=\mathrm{\Phi }(I)\mathrm{\Phi }(A)\mathrm{\Phi }(I)+\mathrm{\Phi }(A)$$

and$$2A\mathrm{\Phi }(I)=\mathrm{\Phi }(A)\mathrm{\Phi }{(I)}^2+\mathrm{\Phi }(I)\mathrm{\Phi }(A)\mathrm{\Phi }(I)=\mathrm{\Phi }(A)+\mathrm{\Phi }(I)\mathrm{\Phi }(A)\mathrm{\Phi }(I)$$

Comparing the above two equations, we obtain $\mathrm{\Phi }(I)A=A\mathrm{\Phi }(I)$ for each $A\in \mathcal{A}$. It follows from the arbitrariness of $A\in \mathcal{A}$ that $\mathrm{\Phi }(I)\in \mathcal{Z}(\mathcal{A})$. This and Equation 2.1 imply $\mathrm{\Phi }(I)\mathrm{\Phi }(A)=A$. Note that $\mathrm{\Phi }{(I)}^2=I$. So $\mathrm{\Phi }(A)=\mathrm{\Phi }(I)A$ holds for all $A\in \mathcal{A}$, completing the proof of the theorem.

Case 2 $\mathit{\xi }\ne -1$.

Take any $A,B\in \mathcal{A}$ and note that $(1-\mathit{\xi }){[A,B]}_{-1}={[A,B]}_{\mathit{\xi }}+{[B,A]}_{\mathit{\xi }}$. Since $\mathrm{\Phi }$ preserves strong $\mathit{\xi }$-Lie commutativity, we have$$\begin{array}{cc}& (1-\mathit{\xi }){[A,B]}_{-1}={[\mathrm{\Phi }(A),\mathrm{\Phi }(B)]}_{\mathit{\xi }}+{[\mathrm{\Phi }(B),\mathrm{\Phi }(A)]}_{\mathit{\xi }}\hfill \\ \hfill & =(1-\mathit{\xi })(\mathrm{\Phi }(A)\mathrm{\Phi }(B)+\mathrm{\Phi }(B)\mathrm{\Phi }(A)).\hfill \end{array}$$

That is,$$AB+BA=\mathrm{\Phi }(A)\mathrm{\Phi }(B)+\mathrm{\Phi }(B)\mathrm{\Phi }(A)$$

holds for all $A,B\in \mathcal{A}$. Now, by Case 1, the theorem is true.

Combining Case 1 and Case 2, the proof of the theorem is complete.

3. Maps preserving strong skew $\mathit{\xi }$-Lie commutativity

In this section, we will discuss the maps preserving strong skew $\mathit{\xi }$-Lie commutativity on general algebras.

Theorem 3.1

Let $\mathcal{A}$ be any *-algebra with unit $I$ over the real or complex field $\mathbb{F}$ and let $\mathit{\xi }\in \mathbb{F}$ with $\mathit{\xi }\ne 1$. Assume that $\mathrm{\Phi }:\mathcal{A}\to \mathcal{A}$ is a map. If $\mathrm{\Phi }(I)=\mathrm{\Phi }{(I)}^{\ast }$, then $\mathrm{\Phi }$ preserves strong skew $\mathit{\xi }$-Lie commutativity, that is, $\mathrm{\Phi }$ satisfies $\mathrm{\Phi }(A)\mathrm{\Phi }(B)-\mathit{\xi }\mathrm{\Phi }(B)\mathrm{\Phi }{(A)}^{\ast }=AB-\mathit{\xi }B{A}^{\ast }$ for all $A,B\in \mathcal{A}$, if and only if $\mathrm{\Phi }(I)\in \mathcal{Z}(\mathcal{A})$, $\mathrm{\Phi }{(I)}^2=I$ and $\mathrm{\Phi }(A)=\mathrm{\Phi }(I)A$ for all $A\in \mathcal{A}$.

Proof

Still, one only needs to prove the “only if” part.

By the assumption, for any $A,B\in \mathcal{A}$, we have(3.1) $$\begin{array}{c}\hfill \mathrm{\Phi }(A)\mathrm{\Phi }(B)-\mathit{\xi }\mathrm{\Phi }(B)\mathrm{\Phi }{(A)}^{\ast }=AB-\mathit{\xi }B{A}^{\ast }\end{array}$$(3.1)

Taking $A=B=I$ in Equation 3.1, one gets $\mathrm{\Phi }{(I)}^2-\mathit{\xi }\mathrm{\Phi }(I)\mathrm{\Phi }{(I)}^{\ast }=(1-\mathit{\xi })I$. Note that $\mathrm{\Phi }(I)=\mathrm{\Phi }{(I)}^{\ast }$ and $\mathit{\xi }\ne 1$. We obtain(3.2) $$\begin{array}{c}\hfill \mathrm{\Phi }{(I)}^2=I\end{array}$$(3.2)

Taking $A=I$ in Equation 3.1, one gets $\mathrm{\Phi }(I)\mathrm{\Phi }(B)-\mathit{\xi }\mathrm{\Phi }(B)\mathrm{\Phi }{(I)}^{\ast }=(1-\mathit{\xi })B$, that is,(3.3) $$\begin{array}{c}\hfill \mathrm{\Phi }(I)\mathrm{\Phi }(A)-\mathit{\xi }\mathrm{\Phi }(A)\mathrm{\Phi }(I)=(1-\mathit{\xi })A\mathrm{forall}A\in \mathcal{A}\end{array}$$(3.3)

Taking $B=I$ in Equation 3.1, one has(3.4) $$\begin{array}{c}\hfill \mathrm{\Phi }(A)\mathrm{\Phi }(I)-\mathit{\xi }\mathrm{\Phi }(I)\mathrm{\Phi }{(A)}^{\ast }=A-\mathit{\xi }{A}^{\ast }\end{array}$$(3.4)

This implies$$\mathrm{\Phi }{(I)}^{\ast }\mathrm{\Phi }{(A)}^{\ast }-\overline{\mathit{\xi }}\mathrm{\Phi }(A)\mathrm{\Phi }{(I)}^{\ast }=A^{\ast }-\overline{\mathit{\xi }}A\mathrm{forall}A\in \mathcal{A}$$

Multiplying $\mathit{\xi }$ from both sides in the above equation, we get(3.5) $$\begin{array}{c}\hfill \mathit{\xi }\mathrm{\Phi }{(I)}^{\ast }\mathrm{\Phi }{(A)}^{\ast }-{|\mathit{\xi }|}^2\mathrm{\Phi }(A)\mathrm{\Phi }{(I)}^{\ast }=\mathit{\xi }{A}^{\ast }-{|\mathit{\xi }|}^{2}A\mathrm{forall}A\in \mathcal{A}\end{array}$$(3.5)

Combining Equations 3.4 and 3.5, we have(3.6) $$\begin{array}{c}\hfill {\mathrm{\Phi }(A)(\mathrm{\Phi }(I)-|\mathit{\xi }|}^2\mathrm{\Phi }{(I)}^{\ast })-\mathit{\xi }(\mathrm{\Phi }(I)-\mathrm{\Phi }{(I)}^{\ast })\mathrm{\Phi }{(A)}^{\ast }={(1-|\mathit{\xi }|}^2)A\mathrm{forall}A\in \mathcal{A}\end{array}$$(3.6)

Note that $\mathrm{\Phi }(I)=\mathrm{\Phi }{(I)}^{\ast }$. Equation 3.6 implies(3.7) $$\begin{array}{c}\hfill {(1-|\mathit{\xi }|}^2{)\mathrm{\Phi }(A)\mathrm{\Phi }(I)=(1-|\mathit{\xi }|}^2)A\mathrm{forall}A\in \mathcal{A}\end{array}$$(3.7)

In the following, we will prove the theorem by two cases.

Case 1 $|\mathit{\xi }|\ne 1$.

In this case, Equation 3.7 implies(3.8) $$\begin{array}{c}\hfill \mathrm{\Phi }(A)\mathrm{\Phi }(I)=A\mathrm{forall}A\in \mathcal{A}\end{array}$$(3.8)

Multiplying $\mathrm{\Phi }(I)$ from the right-hand side in Equation 3.8, by Equation 3.2, one gets(3.9) $$\begin{array}{c}\hfill \mathrm{\Phi }(A)=A\mathrm{\Phi }(I)\mathrm{forall}A\in \mathcal{A}\end{array}$$(3.9)

On the other hand, combining Equations 3.3 and 3.8, one has $\mathrm{\Phi }(I)\mathrm{\Phi }(A)=A$. Multiplying $\mathrm{\Phi }(I)$ from the left-hand side in this equation and noting that Equation 3.2, one gets(3.10) $$\begin{array}{c}\hfill \mathrm{\Phi }(A)=\mathrm{\Phi }(I)A\mathrm{forall}A\in \mathcal{A}\end{array}$$(3.10)

It follows from Equations 3.9 to 3.10 that $\mathrm{\Phi }(I)\in \mathcal{Z}(\mathcal{A})$. The proof is finished.

Case 2 $|\mathit{\xi }|=1$.

Multiplying $\mathrm{\Phi }(I)$ from the left- and the right-hand side in Equation 3.3, respectively, by Equation 3.2 again, one can obtain$$\mathrm{\Phi }(A)-\mathit{\xi }\mathrm{\Phi }(I)\mathrm{\Phi }(A)\mathrm{\Phi }(I)=(1-\mathit{\xi })\mathrm{\Phi }(I)A$$

and$$\mathrm{\Phi }(I)\mathrm{\Phi }(A)\mathrm{\Phi }(I)-\mathit{\xi }\mathrm{\Phi }(A)=(1-\mathit{\xi })A\mathrm{\Phi }(I)$$

Comparing the above two equations gets $(1-{\mathit{\xi }}^2)\mathrm{\Phi }(A)-\mathit{\xi }(1-\mathit{\xi })A\mathrm{\Phi }(I)=(1-\mathit{\xi })\mathrm{\Phi }(I)A,$ that is,(3.11) $$\begin{array}{c}\hfill (1+\mathit{\xi })\mathrm{\Phi }(A)=\mathit{\xi }A\mathrm{\Phi }(I)+\mathrm{\Phi }(I)A\mathrm{holds}\mathrm{forall}A\in \mathcal{A}\end{array}$$(3.11)

We claim $A\mathrm{\Phi }(I)=\mathrm{\Phi }(I)A$, and so $\mathrm{\Phi }(I)\in \mathcal{Z}(\mathcal{A})$. In fact, if $\mathit{\xi }=-1$, Equation 3.11 implies $A\mathrm{\Phi }(I)=\mathrm{\Phi }(I)A$; if $\mathit{\xi }\ne -1$, multiplying $\overline{\mathit{\xi }}$ from both sides in Equation 3.11, one has(3.12) $$\begin{array}{c}\hfill (1+\overline{\mathit{\xi }})\mathrm{\Phi }(A)=A\mathrm{\Phi }(I)+\overline{\mathit{\xi }}\mathrm{\Phi }(I)A\mathrm{forall}A\in \mathcal{A}\end{array}$$(3.12)

as $|\mathit{\xi }|=1$. On the other hand, Equation 3.4 implies $(1+\mathit{\xi })\mathrm{\Phi }(A)\mathrm{\Phi }(I)=\mathit{\xi }(1+\mathit{\xi })\mathrm{\Phi }(I)\mathrm{\Phi }{(A)}^{\ast }+(1+\mathit{\xi })A-\mathit{\xi }(1+\mathit{\xi })A^{\ast }.$ This and Equations 3.11–3.12 yield$$(\mathit{\xi }A\mathrm{\Phi }(I)+\mathrm{\Phi }(I)A)\mathrm{\Phi }(I)=\mathit{\xi }\mathrm{\Phi }(I){(A\mathrm{\Phi }(I)+\overline{\mathit{\xi }}\mathrm{\Phi }(I)A)}^{\ast }+(1+\mathit{\xi })A-\mathit{\xi }(1+\mathit{\xi })A^{\ast }$$

Note that $\mathrm{\Phi }(I)=\mathrm{\Phi }{(I)}^{\ast }$ and Equation 3.2. The above equation can be reduced to$$\mathrm{\Phi }(I)A\mathrm{\Phi }(I)={\mathit{\xi }}^2\mathrm{\Phi }(I)A^{\ast }\mathrm{\Phi }(I)+A-{\mathit{\xi }}^2{A}^{\ast }$$

Multiplying $\mathrm{\Phi }(I)$ from the right side in the above equation yields(3.13) $$\begin{array}{c}\hfill \mathrm{\Phi }(I)A-A\mathrm{\Phi }(I)={\mathit{\xi }}^2(\mathrm{\Phi }(I)A^{\ast }-A^{\ast }\mathrm{\Phi }(I))\mathrm{forall}A\in \mathcal{A}\end{array}$$(3.13)

Replacing $A$ by $iA$ in Equation 3.13, one can get $\mathrm{\Phi }(I)(iA)-(iA)\mathrm{\Phi }(I)={\mathit{\xi }}^2(\mathrm{\Phi }(I){(iA)}^{\ast }-{(iA)}^{\ast }\mathrm{\Phi }(I))$, that is,(3.14) $$\begin{array}{c}\hfill \mathrm{\Phi }(I)A)-A\mathrm{\Phi }(I)={\mathit{\xi }}^2(-\mathrm{\Phi }(I)A^{\ast }+A^{\ast }\mathrm{\Phi }(I))\mathrm{forall}A\in \mathcal{A}\end{array}$$(3.14)

Combining Equations 3.13 and 3.14, one achieves $\mathrm{\Phi }(I)A=A\mathrm{\Phi }(I)$ for all $A\in \mathcal{A}$. The claim holds.

Now, it follows from Equation 3.3 that $\mathrm{\Phi }(I)\mathrm{\Phi }(A)=\mathrm{\Phi }(A)\mathrm{\Phi }(I)=A$, and so $\mathrm{\Phi }(A)=\mathrm{\Phi }(I)A=A\mathrm{\Phi }(I)$ holds for all $A\in \mathcal{A}$. The theorem holds.

We complete the proof of the theorem.

If $\mathrm{\Phi }$ is surjective and $|\mathit{\xi }|=1$, then the condition $\mathrm{\Phi }(I)=\mathrm{\Phi }{(I)}^{\ast }$ in Theorem 3.1 can be deleted.

Theorem 3.2

Let $\mathcal{A}$ be any *-algebra with unit $I$ over the real or complex field $\mathbb{F}$, and let $\mathit{\xi }\in \mathbb{F}$ with $\mathit{\xi }\ne 1$ and $|\mathit{\xi }|=1$. Assume that $\mathrm{\Phi }:\mathcal{A}\to \mathcal{A}$ is a surjective map. Then $\mathrm{\Phi }$ preserves strong skew $\mathit{\xi }$-Lie commutativity if and only if $\mathrm{\Phi }(I)=\mathrm{\Phi }{(I)}^{\ast }\in \mathcal{Z}(\mathcal{A})$, $\mathrm{\Phi }{(I)}^2=I$ and $\mathrm{\Phi }(A)=\mathrm{\Phi }(I)A$ for all $A\in \mathcal{A}$.

Proof

By Theorem 3.1, to complete the proof of the theorem, one only needs to prove $\mathrm{\Phi }(I)=\mathrm{\Phi }{(I)}^{\ast }$.

Indeed, by checking the proof of Theorem 3.1, Equation 3.6 still holds, that is,$${\mathrm{\Phi }(A)(\mathrm{\Phi }(I)-|\mathit{\xi }|}^2\mathrm{\Phi }{(I)}^{\ast })-\mathit{\xi }(\mathrm{\Phi }(I)-\mathrm{\Phi }{(I)}^{\ast })\mathrm{\Phi }{(A)}^{\ast }={(1-|\mathit{\xi }|}^2)A\mathrm{holds}\mathrm{forall}A\in \mathcal{A}$$

Since $|\mathit{\xi }|=1$, the above equation reduces to $\mathrm{\Phi }(A)(\mathrm{\Phi }(I)-\mathrm{\Phi }{(I)}^{\ast })-\mathit{\xi }(\mathrm{\Phi }(I)-\mathrm{\Phi }{(I)}^{\ast })\mathrm{\Phi }{(A)}^{\ast }=0$ for all $A\in \mathcal{A}$. As $\mathrm{\Phi }$ is surjective, there exists $A\in \mathcal{A}$ such that $\mathrm{\Phi }(A)=I$. So $\mathrm{\Phi }(I)-\mathrm{\Phi }{(I)}^{\ast }=\mathit{\xi }(\mathrm{\Phi }(I)-\mathrm{\Phi }{(I)}^{\ast })$. It follows from the fact $\mathit{\xi }\ne 1$ that $\mathrm{\Phi }(I)=\mathrm{\Phi }{(I)}^{\ast }$.

Acknowledgements

The authors wish to express their thanks to the referees for their helpful comments and suggestions.

Additional information

Funding

This work is partially supported by National Natural Science Foundation of China [grant number 11101250]; Program for the Outstanding Innovative Teams of Higher Learning Institutions of Shanxi and Youth Foundation of Shanxi Province [grant number 2012021004].

Notes on contributors

Xiaofei Qi

Xiaofei Qi is an associate professor in the Department of Mathematics at Shanxi University. She received her PhD degree in Mathematics from Shanxi University. Her research interests are functional analysis, operator algebras, and operator theory. She has published some research articles in reputed international journals.