Maps Preserving k-Jordan Products on Operator Algebras

Abstract

For any positive integer k, the k-Jordan product of $a,b$ in a ring $\mathcal{R}$ is defined by ${\{a,b\}}_k={\{{\{a,b\}}_{k-1},b\}}_1$, where ${\{a,b\}}_0=a$ and ${\{a,b\}}_1=ab+ba$. A map f on $\mathcal{R}$ is k-Jordan zero-product preserving if ${\{f\left(a\right),f\left(b\right)\}}_k=0$ whenever ${\{a,b\}}_k=0$ for $a,b\in \mathcal{R}$; it is strong k-Jordan product preserving if ${\{f\left(a\right),f\left(b\right)\}}_k={\{a,b\}}_k$ for all $a,b\in \mathcal{R}$. In this paper, strong k-Jordan product preserving nonlinear maps on general rings and k-Jordan zero-product preserving additive maps on standard operator algebras are characterized, generalizing some known results.

1. Introduction

Assume that $\mathcal{R}$ and ${\mathcal{R}}^{\prime }$ are two associative rings. Let $\mathsf{\Phi }:\mathcal{R}\to {\mathcal{R}}^{\prime }$ be a map. Recall that $\mathsf{\Phi }$ is Jordan zero-product preserving if $\mathsf{\Phi }\left(A\right)\mathsf{\Phi }\left(B\right)+\mathsf{\Phi }\left(B\right)\mathsf{\Phi }\left(A\right)=0$ whenever $AB+BA=0$ for $A,B\in \mathcal{R}$. The problem of characterizing Jordan zero-product preserving additive or linear maps between rings and operator algebras had been studied intensively (e.g., see [1,2,3,4,5] and the references therein.)

Let k be any positive integer. We can define the k-Jordan product of $A,B$ by ${\{A,B\}}_k={\{{\{A,B\}}_{k-1},B\}}_1$, where ${\{A,B\}}_0=A$ and ${\{A,B\}}_1=\{A,B\}=AB+BA$. Obviously, k-Jordan product is the usual Jordan product if $k=1$. For the map $\mathsf{\Phi }$, we say that $\mathsf{\Phi }$ is k-Jordan zero-product preserving (in both directions) if ${\{\mathsf{\Phi }\left(A\right),\mathsf{\Phi }\left(B\right)\}}_k=0$ whenever (if and only if) ${\{A,B\}}_k=0$ for $A,B\in \mathcal{R}$; particularly, if ${\mathcal{R}}^{\prime }=\mathcal{R}$, then $\mathsf{\Phi }$ is called strong k-Jordan product preserving if ${\{\mathsf{\Phi }\left(A\right),\mathsf{\Phi }\left(B\right)\}}_k={\{A,B\}}_k$ for all $A,B\in \mathcal{R}$. It is clear that strong k-Jordan product preserving maps must be k-Jordan zero-product preserving; and the converse is not true in general. For $k=1$, Taghavi, Kolivand and Rohi [6] proved that, if $\mathcal{R}$ is an arbitrary unital algebra with a nontrivial idempotent $P_1$, then a surjective map $\mathsf{\Phi }$ on $\mathcal{R}$ satisfies $\left\{\mathsf{\Phi }\right(A),\mathsf{\Phi }(P\left)\right\}=\{A,P\}$ for all $A\in \mathcal{R}$ and $P\in \{P_1,I-P_1\}$ if and only if $\mathsf{\Phi }\left(A\right)=\mathsf{\Phi }\left(I\right)A$ for all $A\in \mathcal{R}$, where $\mathsf{\Phi }\left(I\right)$ is in the center of $\mathcal{R}$ with $\mathsf{\Phi }{\left(I\right)}^2=I$. Taghavi and Kolivand [7] gave a characterization of strong 2-Jordan product preserving surjective maps $\mathsf{\Phi }$ on properly infinite von Neumann algebras (or factor von Neumann algebras) $\mathcal{M}$, and they showed that, if $\mathsf{\Phi }$ satisfies ${\{\mathsf{\Phi }\left(A\right),\mathsf{\Phi }\left(P\right)\}}_2={\{A,P\}}_2$ for all A and all idempotents P in $\mathcal{M}$, then $\mathsf{\Phi }\left(A\right)=\mathsf{\Phi }\left(I\right)A$ for all $A\in \mathcal{M}$, where $\mathsf{\Phi }\left(I\right)$ is in the center of $\mathcal{M}$ with $\mathsf{\Phi }{\left(I\right)}^2=I$. For other related results, see [8,9] and the references therein.

Thus, a natural problem is how to characterize strong k-Jordan product preserving maps on general rings for any k or, more generally, how to characterize k-Jordan zero-product preserving maps on operator algebras for any k. The purpose of this paper is to try to solve these problems.

The paper is organized as follows. Assume that $\mathcal{R}$ is a unital ring with an idempotent element $e_1$ and $f:\mathcal{R}\to \mathcal{R}$ is a surjective map. In Section 2, it is shown that, under some assumptions on $\mathcal{R}$, f satisfies ${\{f\left(a\right),f\left(e\right)\}}_k={\{a,e\}}_k$ for all $a\in \mathcal{R}$ and $e\in \{e_1,1-e_1,1\}$ if and only if $f\left(a\right)=f\left(1\right)a$ holds for all $a\in \mathcal{R}$, where $f\left(1\right)$ is in the center of $\mathcal{R}$ with $f{\left(1\right)}^{k+1}=1$ (Theorem 1). As applications, such maps are characterized on triangular algebras, prime rings, nest algebras, standard operator algebras, and von Neumann algebras (Theorems 2–4 and Corollaries 1–4). These generalize some known related results in [6,7]. Section 3 is devoted to discussing k-Jordan zero-products preserving additive maps on standard operator algebras. Assume that $\mathcal{A}$ and $\mathcal{B}$ are respectively standard operator algebras on Banach spaces X and Y, and $\mathsf{\Phi }:\mathcal{A}\to \mathcal{B}$ is an additive unital surjective map. We show that, if $\mathsf{\Phi }$ satisfies ${\{\mathsf{\Phi }\left(A\right),\mathsf{\Phi }\left(B\right)\}}_k=0$ whenever ${\{A,B\}}_k=0$ for $A,B\in \mathcal{A}$, then $\mathsf{\Phi }\left(F\right)=0$ for all finite rank operators $F\in \mathcal{A}$, there exists a bounded (conjugate) linear bijective operator $T:X\to Y$ such that $\mathsf{\Phi }\left(A\right)=TA{T}^{-1}$ for all $A\in \mathcal{A}$, or there exists a bounded (conjugate) linear bijective operator $T:X^{*}\to Y$ such that $\mathsf{\Phi }\left(A\right)=T{A}^{*}{T}^{-1}$ for all $A\in \mathcal{A}$, and, in this case, X and Y are reflexive (Theorem 5).

2. Strong k-Jordan Product Preserving Maps on Rings

In this section, we consider strong k-Jordan product preserving maps on general rings. The following is our main result.

Theorem 1.

Let $\mathcal{R}$ be a ring with unit 1 and a nontrivial idempotent element $e_1$, and let k be any positive integer. Assume that the characteristic of $\mathcal{R}$ is not 2 and $f:\mathcal{R}\to \mathcal{R}$ is a surjective map. If f satisfies

$${\{f\left(a\right),f\left(e\right)\}}_k={\{a,e\}}_k$$

(1)

for all $a\in \mathcal{R}$ and $e\in \{e_1,1-e_1,1\}$, then the following statements hold:

(i)

f is additive.

(ii)

$f{\left(e\right)}^{k+1}=e$.

(iii)

$f\left(e_1\mathcal{R}{e}_1\right)\subseteq e_1\mathcal{R}{e}_1$ and $f\left((1-e_1)\mathcal{R}(1-e_1)\right)\subseteq (1-e_1)\mathcal{R}(1-e_1)$.

(iv)

$f\left(1\right)e_{1}a(1-e_1)=e_{1}a(1-e_1)f\left(1\right)$ holds for all $a\in \mathcal{R}$.

Before giving the proof of Theorem 1, we first give some of its applications.

The triangular rings were first introduced by Cheung in [10]. Let $\mathcal{A}$ and $\mathcal{B}$ be unital rings over any commutative ring $\mathcal{R}$, and $\mathcal{M}$ be a $(\mathcal{A},\mathcal{B})$-bimodule, which is faithful as a left $\mathcal{A}$-module and as a right $\mathcal{B}$-module, that is, for any $A\in \mathcal{A}$ and $B\in \mathcal{B}$, $A\mathcal{M}=\mathcal{M}B=\left\{0\right\}$ imply $A=0$ and $B=0$. The $\mathcal{R}$-ring

$$\mathcal{U}=\mathrm{Tri}(\mathcal{A},\mathcal{M},\mathcal{B})=\{\left(\begin{array}{cc}A& M\\ 0& B\end{array}\right):A\in \mathcal{A},M\in \mathcal{M},B\in \mathcal{B}\}$$

under the usual matrix operations is called a triangular ring, and the idempotent element $P=\left(\begin{array}{cc}{I}_{\mathcal{A}}& 0\\ 0& 0\end{array}\right)$ is called the standard idempotent of $\mathcal{U}$. Here, $I_{\mathcal{A}}$ is the unit of $\mathcal{A}$.

Applying Theorem 1 to triangular rings, one can get the following theorem.

Theorem 2.

Let $\mathcal{A}$ and $\mathcal{B}$ be unital rings over a commutative ring $\mathcal{R}$ and $\mathcal{M}$ be a $(\mathcal{A},\mathcal{B})$-bimodule, which is faithful as a left $\mathcal{A}$-module and also as a right $\mathcal{B}$-module. Let $\mathcal{U}=\mathrm{Tri}(\mathcal{A},\mathcal{M},\mathcal{B})$ be the triangular ring with characteristic not 2. Let P be the standard idempotent of $\mathcal{U}$. Assume that $\mathsf{\Phi }:\mathcal{U}\to \mathcal{U}$ is a surjective map. Then, ${\{\mathsf{\Phi }\left(X\right),\mathsf{\Phi }\left(E\right)\}}_k={\{X,E\}}_k$ for all $X\in \mathcal{U}$ and $E\in \{P,I-P,I\}$ if and only if $\mathsf{\Phi }\left(X\right)=\mathsf{\Phi }\left(I\right)X$ holds for all $X\in \mathcal{U}$, where $\mathsf{\Phi }\left(I\right)\in \mathcal{Z}\left(\mathcal{U}\right)$ (the center of $\mathcal{U}$) with $\mathsf{\Phi }{\left(I\right)}^{k+1}=I$.

Recall that a nest $\mathcal{N}$ on a Banach space X is a chain of closed subspaces of X which is closed under the formation of arbitrary closed linear span and intersection, and which includes $\left\{0\right\}$ and X. The nest algebra associated to the nest $\mathcal{N}$, denoted by Alg$\mathcal{N}$, is the weakly closed operator algebra consisting of all operators that leave $\mathcal{N}$ invariant ([11]). When $\mathcal{N}\ne \{0,X\},$ we say that $\mathcal{N}$ is non-trivial. If $\mathcal{N}$ is trivial, then Alg$\mathcal{N}=\mathcal{B}\left(X\right)$ (the algebra of all bounded linear operators on X). If X is a Hilbert space, then every $N\in \mathcal{N}$ corresponds to a projection $P_N$ satisfying $P_N=P_N^{*}=P_N^2$ and $N=P_N\left(X\right)$. In this case, Alg$\mathcal{N}$ is a triangular algebra with standard idempotent $P_N$. Note that the center of $\mathrm{Alg}\mathcal{N}$ is $\mathbb{F}I$. Here, $\mathbb{F}$ stands for the real or complex field.

As an application of Theorem 2 to the nest algebras case, we have the following result, which generalizes Theorem 2.4 of [7].

Corollary 1.

Let $\mathcal{N}$ be a nest on a Hilbert space H and $\mathrm{Alg}\mathcal{N}$ the associated nest algebra. Assume that $\mathsf{\Phi }:\mathrm{Alg}\mathcal{N}\to \mathrm{Alg}\mathcal{N}$ is a surjective map and $P_1\in \mathrm{Alg}\mathcal{N}$ is any fixed idempotent. Then, Φ satisfies ${\{\mathsf{\Phi }\left(A\right),\mathsf{\Phi }\left(P\right)\}}_k={\{A,P\}}_k$ for all $A\in \mathrm{Alg}\mathcal{N}$ and $P\in \{P_1,I-P_1,I\}$ if and only if there exists some scalar λ with ${\lambda }^{k+1}=1$ such that $\mathsf{\Phi }\left(A\right)=\lambda A$ for all $A\in \mathrm{Alg}\mathcal{N}$.

Next, applying Theorem 1 to general rings, we have

Theorem 3.

Let $\mathcal{R}$ be a ring with unit 1 and a nontrivial idempotent element $e_1$. Assume that the characteristic of $\mathcal{R}$ is not 2, $a\mathcal{R}{e}_1=\left\{0\right\}\Rightarrow a=0$ and $a\mathcal{R}(1-e_1)=\left\{0\right\}\Rightarrow a=0$. Then, a surjective map $f:\mathcal{R}\to \mathcal{R}$ satisfies ${\{f\left(a\right),f\left(e\right)\}}_k={\{a,e\}}_k$ for all $a\in \mathcal{R}$ and $e\in \{e_1,1-e_1,1\}$ if and only if $f\left(a\right)=f\left(1\right)a$ holds for all $a\in \mathcal{R}$, where $f\left(1\right)\in \mathcal{Z}\left(\mathcal{R}\right)$ (the center of $\mathcal{R}$) with $f{\left(1\right)}^{k+1}=1$.

Recall that a ring $\mathcal{R}$ is prime if, for any $a,b\in \mathcal{R}$, $a\mathcal{R}b=\left\{0\right\}\Rightarrow a=0$ or $b=0$. It is obvious that prime rings satisfy the assumption “$a\mathcal{R}{e}_1=\left\{0\right\}\Rightarrow a=0$ and $a\mathcal{R}(1-e_1)=\left\{0\right\}\Rightarrow a=0$” in Theorem 3. Thus, we get the following.

Corollary 2.

Let $\mathcal{R}$ be a unital prime ring with an idempotent element $e_1$. Assume that the characteristic of $\mathcal{R}$ is not 2 and $f:\mathcal{R}\to \mathcal{R}$ is a surjective map. Then, f satisfies ${\{f\left(a\right),f\left(e\right)\}}_k={\{a,e\}}_k$ for all $a\in \mathcal{R}$ and $e\in \{e_1,1-e_1,1\}$ if and only if $f\left(a\right)=f\left(1\right)a$ holds for all $a\in \mathcal{R}$, where $f\left(1\right)$ is in the center of $\mathcal{R}$ with $f{\left(1\right)}^{k+1}=1$.

Let X be a Banach space with dimension greater than 1. Denote by $\mathcal{B}\left(X\right)$ the algebra of all bounded linear operators on $X.$ Recall that a standard operator algebra on X is a subalgebra of $\mathcal{B}\left(X\right)$ which contains the identity operator and all finite-rank operators in $\mathcal{B}\left(X\right)$.

It is well known that standard operator algebras are prime. Hence, by Corollary 2, the following result is obvious, which generalizes Theorem 2.1 of [7].

Corollary 3.

Let X be a Banach space with $dimX>1$ and $\mathcal{A}$ a standard operator algebra on X. Assume that $\mathsf{\Phi }:\mathcal{A}\to \mathcal{A}$ is a surjective map and $P_1\in \mathcal{A}$ is any fixed idempotent. Then, f satisfies ${\{\mathsf{\Phi }\left(A\right),\mathsf{\Phi }\left(P\right)\}}_k={\{A,P\}}_k$ for all $A\in \mathcal{A}$ and $P\in \{P_1,I-P_1,I\}$ if and only if there exists some scalar λ with ${\lambda }^{k+1}=1$ such that $\mathsf{\Phi }\left(A\right)=\lambda A$ for all $A\in \mathcal{A}$.

Recall that a von Neumann algebra $\mathcal{M}$ is a subalgebra of some $\mathcal{B}\left(H\right)$ (the algebra of all bounded linear operators on a complex Hilbert space H) satisfying the double commutant property: ${\mathcal{M}}^{″}=\mathcal{M}$. Here, ${\mathcal{M}}^{\prime }=\{T:T\in \mathcal{B}\left(H\right)andTA=AT\mathrm{for}\mathrm{all}A\in \mathcal{M}\}$ and ${\mathcal{M}}^{″}={\left\{{\mathcal{M}}^{\prime }\right\}}^{\prime }$ ([12]).

Note that every factor von Neumann algebra is also prime and its center is trivial. We have the following corollary.

Corollary 4.

Let $\mathcal{M}$ be a factor von Neumann algebra. Assume that $\mathsf{\Phi }:\mathcal{M}\to \mathcal{M}$ is a surjective map and $P_1\in \mathcal{M}$ is any fixed idempotent. Then, Φ satisfies ${\{\mathsf{\Phi }\left(A\right),\mathsf{\Phi }\left(P\right)\}}_k={\{A,P\}}_k$ for all $A\in \mathcal{M}$ and $P\in \{P_1,I-P_1,I\}$ if and only if there exists $\lambda \in \mathbb{C}$ with ${\lambda }^{k+1}=1$ such that $\mathsf{\Phi }\left(A\right)=\lambda A$ for all $A\in \mathcal{M}$.

Finally, we apply Theorem 1 to general von Neumann algebras, which generalize Theorem 2.5 of [7].

Theorem 4.

Let $\mathcal{M}$ be a von Neumann algebra. Then, a surjective map $\mathsf{\Phi }:\mathcal{M}\to \mathcal{M}$ satisfies ${\{\mathsf{\Phi }\left(A\right),\mathsf{\Phi }\left(P\right)\}}_k={\{A,P\}}_k$ for all $A\in \mathcal{M}$ and all projections $P\in \mathcal{M}$ if and only if $\mathsf{\Phi }\left(A\right)=\mathsf{\Phi }\left(I\right)A$ holds for all $A\in \mathcal{M}$, where $\mathsf{\Phi }\left(I\right)$ is in the center of $\mathcal{M}$ with $\mathsf{\Phi }{\left(I\right)}^{k+1}=I$.

Now, we are at a position to give proofs of main theorems.

Proof  of Theorem 1.

We prove it by checking several claims.

For the convenience, write $e_1=e$ and $e_2=1-e_1$. Then, $\mathcal{R}$ can be written as $\mathcal{R}={\mathcal{R}}_{11}+{\mathcal{R}}_{12}+{\mathcal{R}}_{21}+{\mathcal{R}}_{22}$, where ${\mathcal{R}}_{ij}=e_i\mathcal{R}{e}_j$ ($i,j\in \{1,2\}$).

Claim 1.$f{\left(1\right)}^{k+1}=1$.

By taking $a=e=1$ in Equation (2.1), one can obtain $2^{k}f{\left(1\right)}^{k+1}={\{f\left(1\right),f\left(1\right)\}}_k={\{1,1\}}_k=2^{k}1.$ It follows from the characteristic of $\mathcal{R}$ is not 2 that $f{\left(1\right)}^{k+1}=1$.

Claim 2.$f\left(0\right)=0$.

By the surjectivity of f, there exists $s\in \mathcal{R}$ such that $f\left(s\right)=-f\left(0\right)$. Letting $a=s$ and $e=1$ in Equation (1), we have

$$\begin{array}{cc}\hfill {\{f\left(s\right),f\left(1\right)\}}_k=& {\sum }_{i=0}^k{\mathrm{C}}_k^{i}f{\left(1\right)}^{i}f\left(s\right)f{\left(1\right)}^{k-i}\hfill \\ \hfill =& -{\sum }_{i=0}^k{\mathrm{C}}_k^{i}f{\left(1\right)}^{i}f\left(0\right)f{\left(1\right)}^{k-i}\hfill \\ \hfill =& -{\{f\left(0\right),f\left(1\right)\}}_k=-{\{0,1\}}_k=0,\hfill \end{array}$$

which means ${\{s,1\}}_k=0$. Since the characteristic of $\mathcal{R}$ is not 2, one gets $s=0$, i.e. $f\left(0\right)=-f\left(0\right)$. It follows from char$\mathcal{R}\ne 2$ that $f\left(0\right)=0$.

Claim 3.f is additive, i.e. $f(a+b)=f\left(a\right)+f\left(b\right)$ holds for all $a,b\in \mathcal{R}$.

For any $a,b\in \mathcal{R}$, since f is surjective, there exists some element $c\in \mathcal{R}$ such that $f\left(c\right)=f(a+b)-f\left(a\right)-f\left(b\right)$. Let $i\ne j\in \{1,2\}$. Then,

$$\begin{array}{cc}\hfill & e_{j}c{e}_i+e_{i}c{e}_j+2^k{e}_{i}c{e}_i={\{c,e_i\}}_k={\{f\left(c\right),f\left(e_i\right)\}}_k\hfill \\ \hfill =& {\{f(a+b)-f\left(a\right)-f\left(b\right),f\left(e_i\right)\}}_k\hfill \\ \hfill =& {\{f(a+b),f\left(e_i\right)\}}_k-{\{f\left(a\right),f\left(e_i\right)\}}_k-{\{f\left(b\right),f\left(e_i\right)\}}_k\hfill \\ \hfill =& {\{(a+b),e_i\}}_k-{\{a,e_i\}}_k-{\{b,e_i\}}_k=0.\hfill \end{array}$$

This implies $e_{j}c{e}_i=e_{i}c{e}_j=e_{i}c{e}_i=0$ as char$\mathcal{R}\ne 2$. Thus, $c=0$. It follows from Claim 2 that $f(a+b)=f\left(a\right)+f\left(b\right)$.

Claim 4.$f{\left(e_i\right)}^{k+1}=e_i$, $f\left(e_i\right)e_j=e_{j}f\left(e_i\right)$ and $f\left(1\right)e_i=e_{i}f\left(1\right)$, $i,j\in \{1,2\}$.

Letting $a=e_i$ ($i=1,2$) in Equation (1), we obtain $2^{k}f{\left(e_i\right)}^{k+1}={\{f\left(e_i\right),f\left(e_i\right)\}}_k={\{e_i,e_i\}}_k=2^k{e}_i,$ and so $f{\left(e_i\right)}^{k+1}=e_i$. Thus, for $i\ne j\in \{1,2\}$, one gets

$$f\left(e_i\right)e_i=f\left(e_i\right)f{\left(e_i\right)}^{k+1}=f{\left(e_i\right)}^{k+2}=f{\left(e_i\right)}^{k+1}f\left(e_i\right)=e_{i}f\left(e_i\right)$$

and

$$f\left(e_i\right)e_j=f\left(e_i\right)(1-e_i)=(1-e_i)f\left(e_i\right)=e_{j}f\left(e_i\right).$$

Combining the above two equations and Claim 3 gives

$$f\left(1\right)e_i=(f\left(e_i\right)+f\left(e_j\right))e_i=e_{i}f\left(e_i\right)+e_{i}f\left(e_j\right)=e_{i}f\left(1\right),i=\{1,2\}.$$

Claim 5. For any $a_{ii}\in {\mathcal{R}}_{ii}$, there exists some element $b_{ii}\in {\mathcal{R}}_{ii}$ depending on $a_{ii}$ such that $f\left(b_{ii}\right)=a_{ii}$, $i\in \{1,2\}$.

Note that, by Claim 4, it is easy to see that

$$f\left(e_1\right)=e_{1}f\left(e_1\right)e_1+e_{2}f\left(e_1\right)e_2\mathrm{and}f\left(e_2\right)=e_{1}f\left(e_2\right)e_1+e_{2}f\left(e_2\right)e_2.$$

(2)

For any $a_{22}\in {\mathcal{R}}_{22},$ by the surjectivity of f, there exists an element $b=b_{11}+b_{12}+b_{21}+b_{22}\in \mathcal{R}$ such that $f\left(b\right)=a_{22}$. Thus,

$$\begin{array}{cc}\hfill & 2^k{b}_{11}+b_{12}+b_{21}={\{b,e_1\}}_k={\{f\left(b\right),f\left(e_1\right)\}}_k\hfill \\ \hfill =& {\{a_{22},e_{1}f\left(e_1\right)e_1+e_{2}f\left(e_1\right)e_2\}}_k={\sum }_{i=0}^k{\mathrm{C}}_k^i{\left(e_{2}f\left(e_1\right)e_2\right)}^i{a}_{22}{\left(e_{2}f\left(e_1\right)e_2\right)}^{k-i}\in {\mathcal{R}}_{22}.\hfill \end{array}$$

This implies $b_{11}=b_{12}=b_{21}=0$. Hence, $f\left(b_{22}\right)=a_{22}$.

For $a_{11}\in {\mathcal{R}}_{11}$, the proof is similar and we omit it here.

Claim 6.$f\left(e_i\right)\in {\mathcal{R}}_{ii}$, $i\in \{1,2\}$.

By Equation (2), we only need to show that $e_{2}f\left(e_1\right)e_2=e_{1}f\left(e_2\right)e_1=0$. Indeed, by Claim 5, there are elements $b_{11}$ and $b_{22}$ such that $f\left(b_{11}\right)=e_{1}f\left(e_2\right)e_1$ and $f\left(b_{22}\right)=e_{2}f\left(e_1\right)e_2$. By Equation (2) and Claim 3, one gets

$$e_{1}f\left(e_1\right)e_1=f(e_1-b_{22})\mathrm{and}{e}_{2}f\left(e_2\right)e_2=f(e_2-b_{11}).$$

Thus,

$$\begin{array}{cc}\hfill {\mathcal{R}}_{11}\ni & {\{e_{1}f\left(e_1\right)e_1,e_{1}f\left(e_2\right)e_1\}}_k={\{e_{1}f\left(e_1\right)e_1,e_{1}f\left(e_2\right)e_1+e_{2}f\left(e_2\right)e_2\}}_k\hfill \\ \hfill =& {\{f(e_1-b_{22}),f\left(e_2\right)\}}_k={\{e_1-b_{22},e_2\}}_k={\{-b_{22},e_2\}}_k=-2^k{b}_{22}\in {\mathcal{R}}_{22}\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill {\mathcal{R}}_{22}\ni & {\{e_{2}f\left(e_2\right)e_2,e_{2}f\left(e_1\right)e_2\}}_k={\{e_{2}f\left(e_2\right)e_2,e_{1}f\left(e_1\right)e_1+e_{2}f\left(e_1\right)e_2\}}_k\hfill \\ \hfill =& {\{f(e_2-b_{11}),f\left(e_1\right)\}}_k={\{e_2-b_{11},e_1\}}_k={\{-b_{11},e_1\}}_k=-2^k{b}_{11}\in {\mathcal{R}}_{11}.\hfill \end{array}$$

It follows that $b_{11}=0$ and $b_{22}=0$, which with Claim 2 yield $e_{1}f\left(e_2\right)e_1=0$ and $e_{2}f\left(e_1\right)e_2=0$, as desired.

Claim 7.$f\left({\mathcal{R}}_{ii}\right)\subseteq {\mathcal{R}}_{ii}$, $i\in \{1,2\}$.

Here, we only give the proof for the case $i=1$. The proof of the other case $i=2$ is similar.

For any $a_{11}\in {\mathcal{R}}_{11},$ write $f\left(a_{11}\right)=s_{11}+s_{12}+s_{21}+s_{22}$. By Claim 6, we have

$$\begin{array}{cc}\hfill 2^k{a}_{11}=& {\{a_{11},e_1\}}_k={\{s_{11}+s_{12}+s_{21}+s_{22},e_{1}f\left(e_1\right)e_1\}}_k\hfill \\ \hfill =& s_{11}{\left(e_{1}f\left(e_1\right)e_1\right)}^k+s_{21}{\left(e_{1}f\left(e_1\right)e_1\right)}^k+{\left(e_{1}f\left(e_1\right)e_1\right)}^k{s}_{11}+\hfill \\ \hfill & {\left(e_{1}f\left(e_1\right)e_1\right)}^k{s}_{12}+{\sum }_{i=1}^{k-1}{\mathrm{C}}_k^i{\left(e_{1}f\left(e_1\right)e_1\right)}^i{s}_{11}{\left(e_{1}f\left(e_1\right)e_1\right)}^{k-i}.\hfill \end{array}$$

This implies $s_{21}{\left(e_{1}f\left(e_1\right)e_1\right)}^k={\left(e_{1}f\left(e_1\right)e_1\right)}^k{s}_{12}=0$. It follows from Claims 4 and 6 that $s_{21}=s_{12}=0$.

On the other hand, by Claim 5, there exists an element $c_{22}\in \mathcal{R}$ such that $f\left(c_{22}\right)=s_{22}$. Thus, $s_{11}=f(a_{11}-c_{22})$ by Claim 3. So, Claim 6 and Equation (1) yield

$$-2^k{c}_{22}={\{a_{11}-c_{22},e_2\}}_k={\{f(a_{11}-c_{22}),f\left(e_2\right)\}}_k={\{s_{11},e_{2}f\left(e_2\right)e_2\}}_k=0,$$

which implies $s_{22}=0$. Hence, $f\left(a_{11}\right)=s_{11}\in {\mathcal{R}}_{11}$.

Claim 8.$f\left(1\right)a_{12}=a_{12}f\left(1\right)$ holds for all $a_{12}\in {\mathcal{R}}_{12}.$

Firstly, take any $a_{12}\in {\mathcal{R}}_{12}$ and write $f\left(a_{12}\right)=s_{11}+s_{12}+s_{21}+s_{22}$. Then,

$$\begin{array}{cc}\hfill a_{12}=& {\{a_{12},e_1\}}_k={\{s_{11}+s_{12}+s_{21}+s_{22},e_{1}f\left(e_1\right)e_1\}}_k\hfill \\ \hfill =& s_{11}{\left(e_{1}f\left(e_1\right)e_1\right)}^k+s_{21}{\left(e_{1}f\left(e_1\right)e_1\right)}^k+{\left(e_{1}f\left(e_1\right)e_1\right)}^k{s}_{11}\hfill \\ \hfill & +{\left(e_{1}f\left(e_1\right)e_1\right)}^k{s}_{12}+{\sum }_{i=1}^{k-1}{\mathrm{C}}_k^i{\left(e_{1}f\left(e_1\right)e_1\right)}^i{s}_{11}{\left(e_{1}f\left(e_1\right)e_1\right)}^{k-i}.\hfill \end{array}$$

This entails $s_{21}{\left(e_{1}f\left(e_1\right)e_1\right)}^k=0$ and ${\left(e_{1}f\left(e_1\right)e_1\right)}^k{s}_{12}=a_{12}$. Now, Claim 4 yields

$$s_{21}=0\mathrm{and}{s}_{12}=e_{1}f\left(e_1\right)e_1{a}_{12}=f\left(e_1\right)a_{12}.$$

(3)

Next, for the above $s_{11}$ and $s_{22}$, by Claim 5, there are elements $c_{ii}\in {\mathcal{R}}_{ii}$ $(i=1,2)$ such that $f\left(c_{11}\right)=s_{11}$ and $f\left(c_{22}\right)=s_{22}$. By Claims 3, 4 and 6, one obtains

$$\begin{array}{cc}\hfill & a_{12}-2^k{c}_{11}={\{a_{12}-c_{11}-c_{22},e_1\}}_k\hfill \\ \hfill =& {\{f(a_{12}-c_{11}-c_{22}),f\left(e_1\right)\}}_k={\{s_{12},f\left(e_1\right)\}}_k={\left(e_{1}f\left(e_1\right)e_1\right)}^k{s}_{12}\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill & a_{12}-2^k{c}_{22}={\{a_{12}-c_{11}-c_{22},e_2\}}_k\hfill \\ \hfill =& {\{f(a_{12}-c_{11}-c_{22}),f\left(e_2\right)\}}_k={\{s_{12},f\left(e_2\right)\}}_k=s_{12}{\left(e_{2}f\left(e_2\right)e_2\right)}^k.\hfill \end{array}$$

The above two equations with Claims 4 and 6 imply

$$s_{11}=s_{22}=0\mathrm{and}{s}_{12}=a_{12}f\left(e_2\right).$$

(4)

Combining Equations (3) and (4) yields

$$f\left(e_1\right)a_{12}=a_{12}f\left(e_2\right)\mathrm{holds}\mathrm{for}\mathrm{all}{a}_{12}\in {\mathcal{R}}_{12}.$$

Note that $f\left(1\right)=f\left(e_1\right)+f\left(e_2\right)=e_{1}f\left(e_1\right)e_1+e_{2}f\left(e_2\right)e_2$. Thus, the above equation implies

$$f\left(1\right)a_{12}=a_{12}f\left(1\right)\mathrm{holds}\mathrm{for}\mathrm{all}{a}_{12}\in {\mathcal{R}}_{12}.$$

(5)

Now, combining above all the claims, the proof of Theorem 1 is finished. ☐

Proof  of Theorem 2.

Write $P_1=P$ and $P_2=I-P_1$. Thus, $\mathcal{U}$ can be written as $\mathcal{U}=P_1\mathcal{U}{P}_1+P_1\mathcal{U}{P}_2+P_2\mathcal{U}{P}_2$. In this case, $P_2\mathcal{U}{P}_1=\left\{0\right\}$. By Theorem 1, we see that $\mathsf{\Phi }\left(I\right)\in P_1\mathcal{U}{P}_1+P_2\mathcal{U}{P}_2$ and

$$\mathsf{\Phi }\left(I\right)P_{1}X{P}_2=P_{1}X{P}_2\mathsf{\Phi }\left(I\right)\mathrm{holds}\mathrm{for}\mathrm{all}X\in \mathcal{U}.$$

Note that, by Proposition 3 of [10], the center $\mathcal{Z}\left(\mathcal{U}\right)$ of $\mathcal{U}$ is

$$\mathcal{Z}\left(\mathcal{U}\right)=\{\left(\begin{array}{cc}A& 0\\ 0& B\end{array}\right)\in \mathcal{U}:AM=MBforallM\in \mathcal{M}\}.$$

Hence, $\mathsf{\Phi }\left(I\right)\in \mathcal{Z}\left(\mathcal{U}\right)$.

Now, taking $E=I$ in Equation (1), one gets

$$2^k\mathsf{\Phi }{\left(I\right)}^k\mathsf{\Phi }\left(X\right)={\{\mathsf{\Phi }\left(X\right),\mathsf{\Phi }\left(I\right)\}}_k={\{X,I\}}_k=2^{k}X\mathrm{for}\mathrm{all}X\in \mathcal{U},$$

which implies

$$\mathsf{\Phi }{\left(I\right)}^k\mathsf{\Phi }\left(X\right)=X$$

(6)

as char$\mathcal{U}\ne 2$. Note that $\mathsf{\Phi }{\left(I\right)}^{k+1}=I$ by Theorem 1. Thus, multiplying $\mathsf{\Phi }\left(I\right)$ from the left in Equation (6), we achieve $\mathsf{\Phi }\left(X\right)=\mathsf{\Phi }\left(I\right)X$ holds for all $X\in \mathcal{U}.$ The proof of Theorem 2 is finished. ☐

Proof  of Corollary 3.

Here, we take the same symbol as that in Theorem 1. By the proof of Theorem 1, Claims 1–8 still hold.

Claim.$f\left(1\right)\in \mathcal{Z}\left(\mathcal{R}\right)$.

Take any $a_{21}\in {\mathcal{R}}_{21}$. By a similar argument to that of Claim 8 in the proof of Theorem 1, one can check that

$$f\left(1\right)a_{21}=a_{21}f\left(1\right)\mathrm{holds}\mathrm{for}\mathrm{all}{a}_{21}\in {\mathcal{R}}_{21}.$$

(7)

Note that the center of $\mathcal{R}$ is

$$\begin{array}{cc}\hfill \mathcal{Z}\left(\mathcal{R}\right)=& \{z_{11}+z_{22}:z_{11}\in {\mathcal{R}}_{11},z_{22}\in {\mathcal{R}}_{22},z_{11}{a}_{12}=a_{12}{z}_{22}\hfill \\ \hfill & \mathrm{and}{z}_{22}{a}_{21}=a_{21}{z}_{11}\mathrm{for}\mathrm{all}{a}_{12}\in {\mathcal{R}}_{12},a_{21}\in {\mathcal{R}}_{21}\}.\hfill \end{array}$$

Since $f\left(1\right)=f\left(e_1\right)+f\left(e_2\right)=e_{1}f\left(e_1\right)e_1+e_{2}f\left(e_2\right)e_2$, we have $f\left(1\right)\in \mathcal{Z}\left(\mathcal{R}\right)$ by Equations (5) and (7).

Now, the same argument as the proof of Theorem 2 gives that the theorem holds. ☐

Proof  of Theorem 4.

Note that von Neumann algebras are not prime. Take any fixed projection $P_1\in \mathcal{M}$ and let $P_2=I-P_1$. By checking the proof of Theorem 1, we see that Claims 1–4 in the proof of Theorem 1 still hold. Thus, by Claim 4 in there, we have $\mathsf{\Phi }\left(I\right)P_1=P_1\mathsf{\Phi }\left(I\right)$. By the arbitrariness of $P_1$, one obtains $\mathsf{\Phi }\left(I\right)P=P\mathsf{\Phi }\left(I\right)$ for all projections $P\in \mathcal{M}$. This implies $\mathsf{\Phi }\left(I\right)A=A\mathsf{\Phi }\left(I\right)$ for all A in $\mathcal{M}$. Hence, $\mathsf{\Phi }\left(I\right)\in \mathcal{Z}\left(\mathcal{M}\right)$.

The rest proof is the same as that of Theorem 2. The theorem holds. ☐

3. k-Jordan Zero-Products Preserving Additive Maps

In this section, we give a characterization of k-Jordan zero-product preserving additive maps on standard operator algebras for any positive integer k.

Let us first fix some notions and notations. Let X be a Banach space over the real or complex field $\mathbb{F}$ with $dimX\ge 2$ and $X^{*}$ its dual space. For $A\in \mathcal{B}\left(X\right)$, denote by $\ker A$ and $\mathrm{ran}A$ the kernel and the range of A, respectively. Denote by $\mathcal{F}\left(X\right)$ the subalgebra of finite rank operators in $\mathcal{B}\left(X\right)$. Write $L_y=\{y\otimes h\in \mathcal{F}\left(X\right):h\in X^{*}\}$ and $R_g=\{\omega \otimes g\in \mathcal{F}\left(X\right):\omega \in X\}$. Let $\tau :\mathbb{F}\to \mathbb{F}$ be a ring homomorphism. Recall that an additive operator $T:X\to X$ is $\tau$-quasilinear if $T\left(\lambda x\right)=\tau \left(\lambda \right)Tx$ for all $\lambda \in \mathbb{F}$ and $x\in X$.

The following is our main result in this section.

Theorem 5.

Let X and Y be two Banach spaces over the real or complex field $\mathbb{F}$ with dimensions greater than 1, and let $\mathcal{A}$ and $\mathcal{B}$ be standard operator algebras on X and Y, respectively. Assume that $k\ge 1$ is any integer, $\mathsf{\Phi }:\mathcal{A}\to \mathcal{B}$ is an additive unital surjective map, and $\mathsf{\Phi }\left(\mathbb{F}P\right)\subseteq \mathbb{F}\mathsf{\Phi }\left(P\right)$ for every rank one idempotent operator $P\in \mathcal{A}$. If Φ satisfies ${\{\mathsf{\Phi }\left(A\right),\mathsf{\Phi }\left(B\right)\}}_k=0$ whenever ${\{A,B\}}_k=0$ for $A,B\in \mathcal{A}$, then either $\mathsf{\Phi }\left(F\right)=0$ for all $F\in \mathcal{F}\left(X\right)$ or one of the following is true:

(i)

There exists a bounded linear or conjugate linear bijective operator $T:X\to Y$ such that $\mathsf{\Phi }\left(A\right)=TA{T}^{-1}$ for all $A\in \mathcal{A}$.

(ii)

There exists a bounded linear or conjugate linear bijective operator $T:X^{*}\to Y$ such that $\mathsf{\Phi }\left(A\right)=T{A}^{*}{T}^{-1}$ for all $A\in \mathcal{A}$; in this case, X and Y are reflexive.

Particularly, if $\mathbb{F}=\mathbb{R}$, the above T are linear.

For the finite dimensional case, we have a better form. Denote by $M_n\left(\mathbb{F}\right)$ the algebra of all $n\times n$ matrices over the real or complex field $\mathbb{F}$.

Corollary 5.

Assume that $\mathsf{\Phi }:M_n\left(\mathbb{F}\right)\to M_n\left(\mathbb{F}\right)$ is an additive unital surjective map with $\mathsf{\Phi }\left(\mathbb{F}P\right)\subseteq \mathbb{F}\mathsf{\Phi }\left(P\right)$ for every rank one idempotent matrix $P\in M_n\left(\mathbb{F}\right)$, and $k\ge 1$ is any integer. Then, Φ satisfies ${\{\mathsf{\Phi }\left(A\right),\mathsf{\Phi }\left(B\right)\}}_k=0$ whenever ${\{A,B\}}_k=0$ for $A,B\in M_n\left(\mathbb{F}\right)$ if and only if either there exists an invertible matrix $T\in M_n\left(\mathbb{F}\right)$ such that $\mathsf{\Phi }\left(A\right)=TA{T}^{-1}$ for all $A\in M_n\left(\mathbb{F}\right)$ or there exists an invertible matrix $T\in M_n\left(\mathbb{F}\right)$ such that $\mathsf{\Phi }\left(A\right)=T{A}^{*}{T}^{-1}$ for all $A\in M_n\left(\mathbb{F}\right)$.

Proof. 

The “if” part is obvious. For the “only if" part, as $\mathsf{\Phi }$ is surjective, by Theorem 5, the result holds. ☐

For the infinite dimensional case, if $\mathsf{\Phi }$ preserves k-Jordan zero-product in both directions, we can get the following corollary.

Corollary 6.

Let X and Y be two infinite dimensional Banach spaces over the real or complex field $\mathbb{F}$ and let $\mathcal{A}$ and $\mathcal{B}$ be standard operator algebras on X and Y, respectively. Assume that $k\ge 1$ is any integer, $\mathsf{\Phi }:\mathcal{A}\to \mathcal{B}$ is an additive unital surjective map, and $\mathsf{\Phi }\left(\mathbb{F}P\right)\subseteq \mathbb{F}\mathsf{\Phi }\left(P\right)$ for every rank one idempotent operator $P\in \mathcal{A}$. Then, Φ satisfies ${\{\mathsf{\Phi }\left(A\right),\mathsf{\Phi }\left(B\right)\}}_k=0\iff {\{A,B\}}_k=0$ for $A,B\in \mathcal{A}$ if and only if Φ has one of the following two forms:

(i)

There exists a bounded linear or conjugate linear bijective operator $T:X\to Y$ such that $\mathsf{\Phi }\left(A\right)=TA{T}^{-1}$ for all $A\in \mathcal{A}$.

(ii)

There exists a bounded linear or conjugate linear bijective operator $T:X^{*}\to Y$ such that $\mathsf{\Phi }\left(A\right)=T{A}^{*}{T}^{-1}$ for all $A\in \mathcal{A}$. In this case, X and Y are reflexive.

Particularly, if $\mathbb{F}=\mathbb{R}$, the above T are linear.

Proof. 

By Theorem 5, one only needs to check that the case $\mathsf{\Phi }\left(\mathcal{F}\right(X\left)\right)=\left\{0\right\}$ cannot occur. In fact, if $\mathsf{\Phi }\left(A\right)=0$ for some $A\in \mathcal{A}$, then ${\{\mathsf{\Phi }\left(A\right),\mathsf{\Phi }\left(B\right)\}}_k=0$. Thus, ${\{A,B\}}_k=0$ for all $B\in \mathcal{A}$. Particularly, by taking $B=I$, one gets $A=0$, which implies that $\mathsf{\Phi }$ is an injective map. It follows that $\mathsf{\Phi }\left(F\right)\ne 0$ for all nonzero $F\in \mathcal{F}\left(X\right)$. ☐

Before proving Theorem 5, the following two lemmas are needed.

Lemma 1.

([13]) Let X be a Banach space with $dimX\ge 2$. Assume that an additive map $\mathsf{\Phi }:\mathcal{F}\left(X\right)\to \mathcal{F}\left(X\right)$ is rank one decreasing and the range of Φ is neither contained in any $L_y$ nor contained in any $R_g$. Then, there exists a ring homomorphism $\tau :\mathbb{F}\to \mathbb{F}$ such that either

(i)

$\mathsf{\Phi }(x\otimes f)=Tx\otimes Sf$ for all $x\in X$ and $f\in X^{*}$, where $T:X\to X$ and $S:X^{*}\to X^{*}$ are τ-quasilinear operators; or

(ii)

$\mathsf{\Phi }(x\otimes f)=Tf\otimes Sx$ for all $x\in X$ and $f\in X^{*}$, where $T:X^{*}\to X$ and $S:X\to X^{*}$ are τ-quasilinear operators.

Lemma 2.

([14] of Theorem A.7) Let $\mathcal{A}$ be a standard operator algebra, and let $\{A_i$, $B_i{\}}_{i=1}^n$, $\{C_j$, $D_j{\}}_{j=1}^m\subset \mathcal{A}$ such that $\sum {}_{i=1}^n{A}_{i}T{B}_i=\sum {}_{i=1}^m{C}_{j}T{D}_j$ for all operators $T\in \mathcal{A}$. If $A_1,\dots ,A_n$ are linearly independent, then, each $B_i$ is a linear combination of $D_1,\dots ,D_m$. Similarly, if $B_1,\dots ,B_n$ are linearly independent, then each $A_i$ is a linear combination of $C_1,\dots ,C_m$.

Proof of Theorem 5.

We prove the theorem by checking several claims.

Claim 1.$\mathsf{\Phi }$ preserves idempotents.

Let $P\in \mathcal{A}$ be any idempotent operator. Since ${\{I-P,P\}}_k=0$, we have ${\{\mathsf{\Phi }(I-P),\mathsf{\Phi }\left(P\right)\}}_k=0$. Note that $\mathsf{\Phi }\left(I\right)=I$. A direct calculation gives

$$\mathsf{\Phi }{\left(P\right)}^{k+1}=\mathsf{\Phi }{\left(P\right)}^k.$$

(8)

On the other hand, since ${\{P,I-P\}}_k=0$, one gets ${\{\mathsf{\Phi }\left(P\right),\mathsf{\Phi }(I-P)\}}_k=0$, which implies $\mathsf{\Phi }\left(P\right){[I-\mathsf{\Phi }\left(P\right)]}^k=0$, that is,

$$\mathsf{\Phi }\left(P\right)(I+\sum _{i=1}^{k-1}{\mathrm{C}}_k^i{(-1)}^i\mathsf{\Phi }{\left(P\right)}^i+{(-1)}^k\mathsf{\Phi }{\left(P\right)}^k)=0.$$

Multiplying by $\mathsf{\Phi }{\left(P\right)}^{k-2}$ from the left side in the above equation, and by using Equation (8), one achieves

$$\mathsf{\Phi }{\left(P\right)}^{k-1}+\sum _{i=1}^k{\mathrm{C}}_k^i{(-1)}^i\mathsf{\Phi }{\left(P\right)}^k=\mathsf{\Phi }{\left(P\right)}^{k-1}-\mathsf{\Phi }{\left(P\right)}^k=0.$$

That is,

$$\mathsf{\Phi }{\left(P\right)}^k=\mathsf{\Phi }{\left(P\right)}^{k-1}.$$

Now, by repeating the processing, one can prove $\mathsf{\Phi }{\left(P\right)}^2=\mathsf{\Phi }\left(P\right)$.

Claim 2.$\mathsf{\Phi }$ preserves square nilpotent operators.

Take any $A\in \mathcal{A}$ with $A^2=0$. If $k=1$, since $AA+AA=0$, we have $\mathsf{\Phi }{\left(A\right)}^2=0$. If $k\ge 2$, then ${\{I,A\}}_k=2^k{A}^k=0$; consequently, ${\{I,\mathsf{\Phi }\left(A\right)\}}_k=2^k\mathsf{\Phi }{\left(A\right)}^k=0$ as $\mathsf{\Phi }$ is unital. That is, $\mathsf{\Phi }{\left(A\right)}^k=0$. For $k=2$, the claim obviously holds. Now, assume that $k>2$. Note that ${\{B,A\}}_k=0$ holds for all $B\in \mathcal{A}$. Thus,

$${\{\mathsf{\Phi }\left(B\right),\mathsf{\Phi }\left(A\right)\}}_k=\sum _{i=1}^{k-1}{\mathrm{C}}_k^i\mathsf{\Phi }{\left(A\right)}^i\mathsf{\Phi }\left(B\right)\mathsf{\Phi }{\left(A\right)}^{k-i}=0$$

for all $B\in \mathcal{A}$. By the surjectivity of $\mathsf{\Phi }$, we have ${\sum }_{i=1}^{k-1}{\mathrm{C}}_k^i\mathsf{\Phi }{\left(A\right)}^{i}T\mathsf{\Phi }{\left(A\right)}^{k-i}=0$ for all $T\in \mathcal{A}$, that is,

$$k\mathsf{\Phi }\left(A\right)T\mathsf{\Phi }{\left(A\right)}^{k-1}=-\sum _{i=2}^{k-1}{\mathrm{C}}_k^i\mathsf{\Phi }{\left(A\right)}^{i}T\mathsf{\Phi }{\left(A\right)}^{k-i}.$$

(9)

If $\mathsf{\Phi }{\left(A\right)}^{k-1}\ne 0$, by Lemma 2, there exist scalars $l_2,l_3,\dots ,l_{k-1}$ such that

$$\mathsf{\Phi }\left(A\right)=l_2\mathsf{\Phi }{\left(A\right)}^2+l_3\mathsf{\Phi }{\left(A\right)}^3+\cdots +l_{k-1}\mathsf{\Phi }{\left(A\right)}^{k-1}.$$

Multiplying by $\mathsf{\Phi }{\left(A\right)}^{k-2}$ in the above equation yields $\mathsf{\Phi }{\left(A\right)}^{k-1}=0$, a contradiction. Thus, $\mathsf{\Phi }{\left(A\right)}^{k-1}=0$. Now, repeating the processing for Equation (9), one can obtain $\mathsf{\Phi }{\left(A\right)}^2=0$.

Claim 3. For any idempotent operator $P\in \mathcal{A}$, we have:

(i)

$\mathsf{\Phi }\left(PAP\right)=\mathsf{\Phi }\left(P\right)\mathsf{\Phi }\left(A\right)\mathsf{\Phi }\left(P\right)$;

(ii)

$\mathsf{\Phi }\left(PA\right(I-P)+(I-P\left)AP\right)=\mathsf{\Phi }\left(P\right)\mathsf{\Phi }\left(A\right)(I-\mathsf{\Phi }(P\left)\right)+(I-\mathsf{\Phi }(P\left)\right)\mathsf{\Phi }\left(A\right)\mathsf{\Phi }\left(P\right)$;

(iii)

$\mathsf{\Phi }\left(\right(I-P\left)A\right(I-P\left)\right)=(I-\mathsf{\Phi }(P\left)\right)\mathsf{\Phi }\left(A\right)(I-\mathsf{\Phi }(P\left)\right)$; and

(iv)

$\mathsf{\Phi }(PA+AP)=\mathsf{\Phi }\left(P\right)\mathsf{\Phi }\left(A\right)+\mathsf{\Phi }\left(A\right)\mathsf{\Phi }\left(P\right)$.

Let $P\in \mathcal{A}$ be any idempotent operator. For the convenience, write

$${\mathcal{A}}_{11}=P\mathcal{A}P,{\mathcal{A}}_{12}=P\mathcal{A}(I-P){\mathcal{A}}_{21}=(I-P)\mathcal{A}P,{\mathcal{A}}_{22}=(I-P)\mathcal{A}(I-P)$$

and

$${\mathcal{B}}_{11}=\mathsf{\Phi }\left(P\right)\mathcal{B}\mathsf{\Phi }\left(P\right),{\mathcal{B}}_{12}=\mathsf{\Phi }\left(P\right)\mathcal{B}(I-\mathsf{\Phi }\left(P\right)),{\mathcal{B}}_{21}=(I-\mathsf{\Phi }\left(P\right))\mathcal{B}\mathsf{\Phi }\left(P\right),{\mathcal{B}}_{22}=(I-\mathsf{\Phi }\left(P\right))\mathcal{B}(I-\mathsf{\Phi }\left(P\right)).$$

Then, we have $\mathcal{A}={\mathcal{A}}_{11}\dot{+}{\mathcal{A}}_{12}\dot{+}{\mathcal{A}}_{21}\dot{+}{\mathcal{A}}_{22}$ and $\mathcal{B}={\mathcal{B}}_{11}\dot{+}{\mathcal{B}}_{12}\dot{+}{\mathcal{B}}_{21}\dot{+}{\mathcal{B}}_{22}$.

For any $A_{12}\in {\mathcal{A}}_{12}$, it is clear that ${(P+A_{12})}^2=P+A_{12}$. By Claim 1, ${(\mathsf{\Phi }\left(P\right)+\mathsf{\Phi }\left(A_{12}\right))}^2=\mathsf{\Phi }\left(P\right)+\mathsf{\Phi }\left(A_{12}\right)$, which implies $\mathsf{\Phi }\left(A_{12}\right)=\mathsf{\Phi }\left(P\right)\mathsf{\Phi }\left(A_{12}\right)+\mathsf{\Phi }\left(A_{12}\right)\mathsf{\Phi }\left(P\right)$ by Claims 1 and 2. It follows that $\mathsf{\Phi }\left(P\right)\mathsf{\Phi }\left(A_{12}\right)\mathsf{\Phi }\left(P\right)=(I-\mathsf{\Phi }\left(P\right))\mathsf{\Phi }\left(A_{12}\right)(I-\mathsf{\Phi }\left(P\right))=0$, and thus

$$\mathsf{\Phi }\left(A_{12}\right)=\mathsf{\Phi }\left(P\right)\mathsf{\Phi }\left(A_{12}\right)(I-\mathsf{\Phi }\left(P\right))+(I-\mathsf{\Phi }\left(P\right))\mathsf{\Phi }\left(A_{12}\right)\mathsf{\Phi }\left(P\right)\in {\mathcal{B}}_{12}\dot{+}{\mathcal{B}}_{21}.$$

(10)

Similarly, one can check that

$$\mathsf{\Phi }\left(A_{21}\right)=\mathsf{\Phi }\left(P\right)\mathsf{\Phi }\left(A_{21}\right)(I-\mathsf{\Phi }\left(P\right))+(I-\mathsf{\Phi }\left(P\right))\mathsf{\Phi }\left(A_{21}\right)\mathsf{\Phi }\left(P\right)\in {\mathcal{B}}_{12}\dot{+}{\mathcal{B}}_{21}.$$

(11)

Next, take any $A_{22}\in {\mathcal{A}}_{22}$. Since ${\{A_{22},P\}}_k=0$, by Claim 2, we have

$$\begin{array}{cc}\hfill 0=& {\{\mathsf{\Phi }\left(A_{22}\right),\mathsf{\Phi }\left(P\right)\}}_k\hfill \\ \hfill =& \mathsf{\Phi }\left(A_{22}\right)\mathsf{\Phi }{\left(P\right)}^k+{\sum }_{i=1}^{k-1}{\mathrm{C}}_k^i\mathsf{\Phi }{\left(P\right)}^i\mathsf{\Phi }\left(A_{22}\right)\mathsf{\Phi }{\left(P\right)}^{k-i}+\mathsf{\Phi }{\left(P\right)}^k\mathsf{\Phi }\left(A_{22}\right)\hfill \\ \hfill =& \mathsf{\Phi }\left(A_{22}\right)\mathsf{\Phi }\left(P\right)+{\sum }_{i=1}^{k-1}{\mathrm{C}}_k^i\mathsf{\Phi }\left(P\right)\mathsf{\Phi }\left(A_{22}\right)\mathsf{\Phi }\left(P\right)+\mathsf{\Phi }\left(P\right)\mathsf{\Phi }\left(A_{22}\right).\hfill \end{array}$$

This implies $\mathsf{\Phi }\left(P\right)\mathsf{\Phi }\left(A_{22}\right)\mathsf{\Phi }\left(P\right)=\mathsf{\Phi }\left(P\right)\mathsf{\Phi }\left(A_{22}\right)(I-\mathsf{\Phi }\left(P\right))=(I-\mathsf{\Phi }\left(P\right))\mathsf{\Phi }\left(A_{22}\right)\mathsf{\Phi }\left(P\right)=0$. Thus,

$$\mathsf{\Phi }\left(A_{22}\right)=(I-\mathsf{\Phi }\left(P\right))\mathsf{\Phi }\left(A_{22}\right)(I-\mathsf{\Phi }\left(P\right))\in {\mathcal{B}}_{22}.$$

(12)

A similar argument to that of Equation (12) yields

$$\mathsf{\Phi }\left(A_{11}\right)=\mathsf{\Phi }\left(P\right)\mathsf{\Phi }\left(A_{11}\right)\mathsf{\Phi }\left(P\right)\in {\mathcal{B}}_{11}.$$

(13)

Now, take any $A={\sum }_{i,j=1}^2{A}_{ij}\in \mathcal{A}$. By Equations (10)–(13), one has

$$\begin{array}{cc}\hfill & \mathsf{\Phi }\left(PAP\right)=\mathsf{\Phi }\left(A_{11}\right)=\mathsf{\Phi }\left(P\right)\mathsf{\Phi }\left(A_{11}\right)\mathsf{\Phi }\left(P\right)\hfill \\ \hfill =& \mathsf{\Phi }\left(P\right)[\mathsf{\Phi }\left(A_{11}\right)+\mathsf{\Phi }\left(A_{12}\right)+\mathsf{\Phi }\left(A_{21}\right)+\mathsf{\Phi }\left(A_{22}\right)]\mathsf{\Phi }\left(P\right)=\mathsf{\Phi }\left(P\right)\mathsf{\Phi }\left(A\right)\mathsf{\Phi }\left(P\right),\hfill \end{array}$$

$$\begin{array}{cc}\hfill & \mathsf{\Phi }\left((I-P)A(I-P)\right)=\mathsf{\Phi }\left(A_{22}\right)=(I-\mathsf{\Phi }\left(P\right))\mathsf{\Phi }\left(A_{22}\right)(I-\mathsf{\Phi }\left(P\right))\hfill \\ \hfill =& (I-\mathsf{\Phi }\left(P\right))[\mathsf{\Phi }\left(A_{11}\right)+\mathsf{\Phi }\left(A_{12}\right)+\mathsf{\Phi }\left(A_{21}\right)+\mathsf{\Phi }\left(A_{22}\right)](I-\mathsf{\Phi }\left(P\right))\hfill \\ \hfill =& (I-\mathsf{\Phi }(P\left)\right)\mathsf{\Phi }\left(A\right)(I-\mathsf{\Phi }(P\left)\right)\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill & \mathsf{\Phi }(PA(I-P)+(I-P)AP)=\mathsf{\Phi }(A_{12}+A_{21})\hfill \\ \hfill =& \mathsf{\Phi }\left(P\right)\mathsf{\Phi }\left(A_{12}\right)(I-\mathsf{\Phi }\left(P\right))+(I-\mathsf{\Phi }\left(P\right))\mathsf{\Phi }\left(A_{12}\right)\mathsf{\Phi }\left(P\right)\hfill \\ \hfill & +\mathsf{\Phi }\left(P\right)\mathsf{\Phi }\left(A_{21}\right)(I-\mathsf{\Phi }\left(P\right))+(I-\mathsf{\Phi }\left(P\right))\mathsf{\Phi }\left(A_{21}\right)\mathsf{\Phi }\left(P\right)\hfill \\ \hfill =& \mathsf{\Phi }\left(P\right)[\mathsf{\Phi }\left(A_{11}\right)+\mathsf{\Phi }\left(A_{12}\right)+\mathsf{\Phi }\left(A_{21}\right)+\mathsf{\Phi }\left(A_{22}\right)](I-\mathsf{\Phi }\left(P\right))\hfill \\ \hfill & +(I-\mathsf{\Phi }\left(P\right))[\mathsf{\Phi }\left(A_{11}\right)+\mathsf{\Phi }\left(A_{12}\right)+\mathsf{\Phi }\left(A_{21}\right)+\mathsf{\Phi }\left(A_{22}\right)]\mathsf{\Phi }\left(P\right)\hfill \\ \hfill =& \mathsf{\Phi }\left(P\right)\mathsf{\Phi }\left(A\right)(I-\mathsf{\Phi }(P\left)\right)+(I-\mathsf{\Phi }(P\left)\right)\mathsf{\Phi }\left(A\right)\mathsf{\Phi }\left(P\right).\hfill \end{array}$$

Thus,

$$\begin{array}{cc}\hfill \mathsf{\Phi }(PA+AP)=& \mathsf{\Phi }(2PAP+PA(I-P)+(I-P\left)AP\right)\hfill \\ \hfill =& 2\mathsf{\Phi }\left(PAP\right)+\mathsf{\Phi }\left(PA\right(I-P)+(I-P\left)AP\right)\hfill \\ \hfill =& 2\mathsf{\Phi }\left(P\right)\mathsf{\Phi }\left(A\right)\mathsf{\Phi }\left(P\right)+\mathsf{\Phi }\left(P\right)\mathsf{\Phi }\left(A\right)(I-\mathsf{\Phi }(P\left)\right)+(I-\mathsf{\Phi }(P\left)\right)\mathsf{\Phi }\left(A\right)\mathsf{\Phi }\left(P\right)\hfill \\ \hfill =& \mathsf{\Phi }\left(P\right)\mathsf{\Phi }\left(A\right)+\mathsf{\Phi }\left(A\right)\mathsf{\Phi }\left(P\right).\hfill \end{array}$$

Claim 4.$\mathsf{\Phi }$ is rank one non-increasing on idempotent operators.

Take any rank one idempotent operator $P\in \mathcal{A}$ and any $A\in \mathcal{A}$. By Claim 3(i), we have

$$\mathsf{\Phi }\left(P\right)\mathsf{\Phi }\left(A\right)\mathsf{\Phi }\left(P\right)=\mathsf{\Phi }\left(PAP\right)=\mathsf{\Phi }\left(\mathbb{F}P\right)\subseteq \mathbb{F}\mathsf{\Phi }\left(P\right)$$

by the assumption. This implies that $\mathsf{\Phi }\left(P\right)$ is a rank-one idempotent or $\mathsf{\Phi }\left(P\right)=0$, as desired.

Claim 5.$\mathsf{\Phi }$ is rank one non-increasing.

By Claim 4, we only need to prove that this claim holds for all rank one nilpotent operators. To do this, take any rank one nilpotent operator $x\otimes f\in \mathcal{A}$. If $\mathsf{\Phi }(x\otimes f)=0$, it is true. In the sequel, we always assume $\mathsf{\Phi }(x\otimes f)\ne 0$. For $x\in X$, there exists some $g\in X^{*}$ such that $g\left(x\right)=1$. Obviously, $x\otimes g$, $x\otimes (g+f)$ and $x\otimes (g-f)$ are rank-one idempotent operators. Write

$$\mathsf{\Phi }(x\otimes g)=u_1\otimes h_1,\mathsf{\Phi }(x\otimes (g-f))=u_2\otimes h_2,\mathsf{\Phi }(x\otimes (g+f))=u_3\otimes h_3,$$

(14)

where $u_i\in X$ and $h_i\in X^{*}$, $i=1,2,3$. By Claim 4, $u_i\otimes h_i$ is either idempotent or zero.

If $u_1\otimes h_1=0$, then Equation (14) implies $0\ne \mathsf{\Phi }(x\otimes f)=-u_2\otimes h_2=u_3\otimes h_3$, a contradiction. Thus, $u_1\otimes h_1\ne 0$. Similarly, one can check that $u_2\otimes h_2\ne 0$ and $u_3\otimes h_3\ne 0$. Thus, by Claim 4, each $u_i\otimes h_i$ is an idempotent operator ($i=1,2,3$). In addition, by Equation (14) and the additivity of $\mathsf{\Phi }$, we have $2u_1\otimes h_1=u_2\otimes h_2+u_3\otimes h_3$, which implies that either $\{u_2,u_3\}$ are linearly dependent, or $\{h_2,h_3\}$ are linearly dependent. Without loss of generality, assume that $u_2=\lambda u_3$ for some nonzero scalar $\lambda$. Then, it is easily seen that $\mathsf{\Phi }(x\otimes f)=\frac{1}{2}(u_3\otimes h_3-u_2\otimes h_2)=u_3\otimes (\frac{1}{2}{h}_3-\lambda h_2)$ is rank one, and ${\left(\mathsf{\Phi }(x\otimes f)\right)}^2=⟨\frac{1}{2}{u}_3,h_3-\lambda h_2⟩u_3\otimes \frac{1}{2}(h_3-\lambda h_2)=0$. The claim is true.

Claim 6. Either ${\mathsf{\Phi }|}_{\mathcal{F}\left(X\right)}=0$ or there exists a ring homomorphism $\tau :\mathbb{F}\to \mathbb{F}$ such that one of the following is true:

(i)

There exist $\tau$-quasilinear operators $T:X\to Y$ and $S:X^{*}\to Y^{*}$ such that $\mathsf{\Phi }(x\otimes f)=Tx\otimes Sf$ holds for all $x\in X$ and $f\in X^{*}$.

(ii)

There exist $\tau$-quasilinear operators $T:X^{*}\to Y$ and $S:X\to Y^{*}$ such that $\mathsf{\Phi }(x\otimes f)=Tf\otimes Sx$ holds for all $x\in X$ and $f\in X^{*}$.

If $\mathsf{\Phi }\left(P\right)=0$ for each rank one idempotent operator $P\in \mathcal{A}$, noting that every rank one nilpotent operator can be written as the difference between two rank one idempotent operators and $\mathsf{\Phi }\left(\mathbb{F}P\right)\subseteq \mathbb{F}\mathsf{\Phi }\left(P\right)$, we obtain $\mathsf{\Phi }(x\otimes f)=0$ for all rank one operators $x\otimes f\in \mathcal{A}$. Thus, $\mathsf{\Phi }\left(F\right)=0$ holds for all $F\in \mathcal{F}\left(X\right)$ by the additivity of $\mathsf{\Phi }$.

Now, suppose that there is a rank one idempotent operator $P=x\otimes f$ such that $\mathsf{\Phi }\left(P\right)=y\otimes g\ne 0$. Then, $P\mathcal{A}(I-P)=L_x=\{x\otimes h:h\in X^{*}\}$, $(I-P)\mathcal{A}P=R_f=\{\omega \otimes f:\omega \in X\}$, $\mathsf{\Phi }\left(P\right)\mathcal{B}(I-\mathsf{\Phi }\left(P\right))=L_y$ and $(I-\mathsf{\Phi }\left(P\right))\mathcal{B}\mathsf{\Phi }\left(P\right)=R_g$. By Claim 3, the range of $\mathsf{\Phi }$ is neither contained in any $L_y$ nor contained in any $R_g$. It follows from Lemma 1 that the claim holds.

Claim 7. Either ${\mathsf{\Phi }|}_{\mathcal{F}\left(X\right)}=0$ or there exists a ring automorphism $\tau :\mathbb{F}\to \mathbb{F}$ such that either $\mathsf{\Phi }\left(A\right)=TA{T}^{-1}$ for all $A\in \mathcal{A}$, where $T:X\to Y$ is a $\tau$-quasilinear bijective operator; or $\mathsf{\Phi }\left(A\right)=T{A}^{*}{T}^{-1}$ for all $A\in \mathcal{A}$, where $T:X^{*}\to Y$ is a $\tau$-quasilinear bijective operator.

Assume that Claim 6(i) occurs. Take any $A\in \mathcal{A}$ and any $x\in X$. There exists some $f\in X^{*}$ such that $⟨x,f⟩=1$. By Claim 3(iv), we have

$$\begin{array}{cc}\hfill & TAx\otimes Sf+Tx\otimes S{A}^{*}f=\mathsf{\Phi }(A(x\otimes f)+(x\otimes f)A)\hfill \\ \hfill =& \mathsf{\Phi }\left(A\right)\mathsf{\Phi }(x\otimes f)+\mathsf{\Phi }(x\otimes f)\mathsf{\Phi }\left(A\right)=\mathsf{\Phi }\left(A\right)Tx\otimes Sf+Tx\otimes \mathsf{\Phi }{\left(A\right)}^{*}Sf,\hfill \end{array}$$

that is, $(\mathsf{\Phi }\left(A\right)T-TA)x\otimes Sf=Tx\otimes (S{A}^{*}f-\mathsf{\Phi }{\left(A\right)}^{*}Sf).$ Thus,

$$(\mathsf{\Phi }\left(A\right)T-TA)x={\lambda }_{x}Tx\mathrm{for}\mathrm{some}{\lambda }_x\in \mathbb{F}\mathrm{depending}\mathrm{on}x,\forall x\in X.$$

(15)

We first claim that T is injective. In fact, if there exists some nonzero vector $y\in X$ such that $Ty=0$, then Equation (3.8) yields $TAy=0$ for all $A\in \mathcal{A}$. This implies $T=0$, impossible. Hence, $dim\mathrm{ran}T\ge 2$ and $\left\{0\right\}=kerT\subseteq ker\left(\mathsf{\Phi }\right(A)T-TA)$. By Lemma 1.1 of [13], there exists some ${\lambda }_A\in \mathbb{F}$ such that

$$\mathsf{\Phi }\left(A\right)T-TA={\lambda }_{A}T\mathrm{for}\mathrm{all}A\in \mathcal{A}.$$

(16)

Next, we show that T is surjective. Otherwise, there exists some $x_0\in Y$ such that $x_0\notin \mathrm{ran}T$. For any $y\in \mathrm{ran}T$, take $z\in X$ and $f\in X^{*}$ such that $Tz=y$ and $⟨y,f⟩=1$. By the surjectivity of $\mathsf{\Phi }$, there exists some $B\in \mathcal{A}$ such that $\mathsf{\Phi }\left(B\right)=x_0\otimes f$. Letting $A=B$ in Equation (16), one gets $x_0\otimes fT=TB+{\lambda }_{B}T$, and so $x_0=(TB+{\lambda }_{B}T)z\in \mathrm{ran}T$, a contradiction.

Thus, T is a bijective map. So, Equation (16) implies that $\mathsf{\Phi }\left(A\right)=TA{T}^{-1}+{\lambda }_{A}I$ holds for all $A\in \mathcal{A}$. Since $\mathsf{\Phi }$ is decreasing operators of rank one, we must have ${\lambda }_A=0$ for all rank one operators $A\in \mathcal{A}$. Now, take any rank one idempotent operator $P\in \mathcal{A}$ and any $A\in \mathcal{A}$. By Claim 3(iv), one gets

$$\begin{array}{cc}\hfill & T(AP+PA)T^{-1}+{\lambda }_{AP+PA}I=\mathsf{\Phi }(AP+PA)=\mathsf{\Phi }\left(A\right)\mathsf{\Phi }\left(P\right)+\mathsf{\Phi }\left(P\right)\mathsf{\Phi }\left(A\right)\hfill \\ \hfill =& (TA{T}^{-1}+{\alpha }_{A}I)TP{T}^{-1}+TP{T}^{-1}(TA{T}^{-1}+{\alpha }_{A}I)=T(AP+PA)T^{-1}+2{\alpha }_{A}TP{T}^{-1},\hfill \end{array}$$

which reduces to ${\lambda }_{AP+PA}I=2{\alpha }_{A}TP{T}^{-1}$. Note that $dimX\ge 2$. It follows that ${\alpha }_A=0$, and so $\mathsf{\Phi }\left(A\right)=TA{T}^{-1}$ holds for all $A\in \mathcal{A}$.

Now, we prove that $\tau$ is bijective. If $\tau \left(\lambda \right)=0$, then $\mathsf{\Phi }\left(\lambda A\right)=\tau \left(\lambda \right)TA{T}^{-1}=0$ for all $A\in \mathcal{A}$, which implies $\lambda =0$ as the surjectivity of $\mathsf{\Phi }$. Thus, $\tau$ is injective. For the surjectivity of $\tau$, if $\tau$ is not surjective, then for any fixed $A\in \mathcal{A}$, there exists some scalar $\lambda$ such that the preimage of $\lambda \mathsf{\Phi }\left(A\right)$ is not the multiple of A. By the surjecticity of $\mathsf{\Phi }$, there exists some $C\in \mathcal{A}$ such that $\mathsf{\Phi }\left(C\right)=\lambda \mathsf{\Phi }\left(A\right)$, that is, $TC{T}^{-1}=\lambda TA{T}^{-1}$. This yields that C is the multiple of A, a contradiction. Hence, $\tau$ is surjective.

If Claim 6(ii) occurs, a similar argument to the above can prove that $\mathsf{\Phi }\left(A\right)=T{A}^{*}{T}^{-1}$ holds for all $A\in \mathcal{A}$, where $\tau$ is a ring automorphism and $T:X^{*}\to Y$ is a $\tau$-quasilinear bijective operator.

Claim 8. Either $\tau \left(\lambda \right)=\lambda$ for all $\lambda \in \mathbb{F}$, or $\tau \left(\lambda \right)=\overline{\lambda }$ for all $\lambda \in \mathbb{F}$.

Since every ring isomorphism on $\mathbb{R}$ is the identity map and every continuous ring isomorphism on $\mathbb{C}$ is the identity map or the conjugate map ([15]). Thus, we only need to check that $\tau$ is continuous on $\mathbb{C}$.

By Claims 4 and 5, we prove that $\mathsf{\Phi }$ preserves idempotent operators of rank one and nilpotent operators of rank one. We first show that $⟨x,Sf⟩=\tau \left(⟨T^{-1}x,f⟩\right)$ holds for all $x\in Y$ and all $f\in X^{*}$, where S is the same as that in Claim 6. Indeed, since T is surjective, there exists some $y\in X$ such that $Ty=x$. For the rank one operator $y\otimes f\in \mathcal{A}$, if $⟨y,f⟩=0$, then $⟨Ty,Sf⟩=⟨x,Sf⟩=0$, and so $\tau \left(⟨T^{-1}x,f⟩\right)=\tau \left(⟨y,f⟩\right)=\tau \left(0\right)=⟨x,Sf⟩$. If $⟨y,f⟩\ne 0$, then $\mathsf{\Phi }({⟨y,f⟩}^{-1}y\otimes f)=\tau \left({⟨y,f⟩}^{-1}\right)Ty\otimes Sf$, which implies $\tau \left({⟨y,f⟩}^{-1}\right)⟨Ty,Sf⟩=1$, that is, $\tau \left(⟨T^{-1}x,f⟩\right)=\tau \left(⟨y,f⟩\right)=⟨Ty,Sf⟩=⟨x,Sf⟩$.

Assume that $\tau$ is not continuous, then $\tau$ is unbounded on any neighbourhood of 0. Thus, we can find sequences ${\left\{x_n\right\}}_{n=1}^{\infty }\subset Y$ and ${\left\{f_n\right\}}_{n=1}^{\infty }\subset X^{*}$ satisfying $\parallel x_n\parallel <2^{-n}$ and $\parallel f_n\parallel <2^{-n}$ for all n, $⟨T^{-1}{x}_n,f_k⟩=0$ for $n\ne k$ and $\left|\tau \right(⟨T^{-1}{x}_n,f_n⟩\left)\right|>n$. Note that $x={\sum }_{n=1}^{\infty }{x}_n\in Y$ and $f={\sum }_{n=1}^{\infty }{f}_n\in X^{*}$. So, $T^{-1}x\in X$ and $Sf\in Y^{*}$. However, $|⟨x_n,Sf⟩|=|\tau \left(⟨T^{-1}{x}_n,f⟩\right)|=|\tau \left(⟨T^{-1}{x}_n,f_n⟩\right)|>n,$ a contradiction. Hence, $\tau$ is continuous.

Claim 9. The theorem holds.

By Claim 8, T is linear or conjugate linear. In addition, by closed graph theorem, it is easily checked that T is bounded. Finally, if $\mathsf{\Phi }$ has the form $\mathsf{\Phi }\left(A\right)=T{A}^{*}{T}^{-1}$ for all A, then X and Y are reflexive. Indeed, for any $g\in Y^{*}$, take $y\in Y$ such that $⟨y,g⟩=1$. Then, there exists some $A_0\in \mathcal{A}$ such that $\mathsf{\Phi }\left(A_0\right)=y\otimes g=T{A}_0^{*}{T}^{-1}$. Thus, $A_0^{*}=T^{-1}y\otimes T^{*}g$, which implies $T^{*}g\in \kappa X$ with $\kappa :X\to X^{**}$ the natural map. Thus, $\mathrm{ran}\left(T^{*}\right)\subseteq \kappa X$. It follows that $\kappa X=X^{**}$ by the invertibility of $T^{*}$. Hence, X is reflexive. A similar argument can show that Y is also reflexive. The proof is finished. ☐

4. Discussion

Jordan product is a kind of important products in rings and operator algebras. The problem of characterizing additive (or linear) maps preserving some property of Jordan products on various rings and operator algebras has been studied by many mathematicians. In this paper, we generalize the concept of Jordan products, discuss strong k-Jordan product preserving maps and k-Jordan zero-product preserving additive maps in some important operator algebras, and give their concrete forms. These results generalize some known related results. However, how can additive maps be characterized preserving k-Jordan zero-product on general von Neumann algebras? This is still an open question, which is also our next work.