A Note on Nonlinear Mappings Preserving the Bi-Skew Jordan-Type Product on Factor von Neumann Algebras

Abstract

Let $\mathfrak{E}$ and $\mathfrak{F}$ be two factor von Neumann algebras such that $\mathfrak{E}$ contains a nontrivial symmetric idempotent element e and an identity element $\mathfrak{I}$, with dim$(\mathfrak{E})\ge 2$. In this article, we consider a bijective map $\vartheta$ between $\mathfrak{E}$ and $\mathfrak{F}$ satisfying $\vartheta ({\nu }_1★{\nu }_2★{\nu }_3★\cdots ★{\nu }_n)=\vartheta ({\nu }_1)★\vartheta ({\nu }_2)★\vartheta ({\nu }_3)★\cdots ★\vartheta ({\nu }_n)$ for all ${\nu }_i\in \mathfrak{E}$$(i=1,2,\dots ,n)$, where ${\nu }_i★{\nu }_j={\nu }_i^{\ast }{\nu }_j+{\nu }_j^{\ast }{\nu }_i$ is the bi-skew Jordan product of ${\nu }_i$, ${\nu }_j$ for any $1\le i,j\le n$, and $n\ge 2$ is a fixed positive integer. We prove that $\vartheta$ or $-\vartheta$ is a conjugate linear ∗-isomorphism or a linear ∗-isomorphism. Moreover, for $n=2$ and $n=3$, similar results were obtained by Li and Zhang. In this work, we characterize nonlinear bijective maps preserving the n-product for any $n\ge 2$. Thus, our result is more general than both of these earlier results.

1. Introduction

Over the years, substantial effort has been dedicated to the study of maps that preserve various algebraic products within structures such as rings and operator algebras. In a symmetrically aligned line of inquiry, significant work has also been carried out on maps known as derivations across these structures. Consider two algebras $\mathfrak{E}$ and $\mathfrak{F}$ defined over the complex field. Recall that, a function $\vartheta :\mathfrak{E}\to \mathfrak{F}$ is considered to be multiplicative, or is said to preserve a product if, for any elements ${\nu }_1$ and ${\nu }_2$ in $\mathfrak{E}$, we have $\vartheta ({\nu }_1{\nu }_2)=\vartheta ({\nu }_1)\vartheta ({\nu }_2)$. Moreover, a bijective map $\vartheta :\mathfrak{E}\to \mathfrak{F}$ is said to be linear (resp. conjugate linear) ∗-isomorphism if $\vartheta (u+v)=\vartheta (u)+\vartheta (v)$, $\vartheta (uv)=\vartheta (u)\vartheta (v)$, $\vartheta (\alpha u)=\alpha \vartheta (u)$ (resp. $\overline{\alpha }\vartheta (u)$) and $\vartheta (u^{\ast })=\vartheta {(u)}^{\ast }$ for all $u,v\in \mathfrak{E}$, and $\alpha$ can be any scalar.

The most natural yet relatable thought arises: “When does a multiplicative map become additive?” This question was pondered by W. S. Martindale III [1]. Martindale addressed this question and demonstrated that any multiplicative bijective map from a prime ring containing a nontrivial idempotent onto any ring must be additive. This result has sparked further research interest, leading many researchers to explore the preserver problem of new products, such as ${\nu }_1{\nu }_2-{\nu }_2{\nu }_1$, ${\nu }_1{\nu }_2+{\nu }_2{\nu }_1$, ${\nu }_1{\nu }_2-{\nu }_2{\nu }_1^{\ast }$, ${\nu }_1{\nu }_2+{\nu }_2{\nu }_1^{\ast }$, etc. (see [2,3,4,5,6,7,8,9,10,11,12]). In 2011, Liu et al. [13] characterized a nonlinear map that preserves the product ${\nu }_1^{\ast }{\nu }_2+{\nu }_2{\nu }_1^{\ast }$ on factor von Neumann algebras. In 2013, Li et al. [5] considered the product ${\nu }_1{\nu }_2+{\nu }_2{\nu }_1^{\ast }$ and proved that a map satisfying $\vartheta ({\nu }_1{\nu }_2+{\nu }_2{\nu }_1^{\ast })=\vartheta ({\nu }_1)\vartheta ({\nu }_2)+\vartheta ({\nu }_2)\vartheta {({\nu }_1)}^{\ast }$ is a ∗- ring isomorphism on factor von Neumann algebras. Similarly, in 2018, Zhao and Li [11] identified the nature of a map which preserves a Jordan triple ∗-product on factor von Neumann algebras. Additionally, significant efforts have been dedicated to foster research in this direction. Consider $\mathfrak{E}$ and $\mathfrak{F}$ to be two ∗-algebras. For any ${\nu }_1,{\nu }_2\in \mathfrak{E}$, ${\nu }_1★{\nu }_2={\nu }_1^{\ast }{\nu }_2+{\nu }_2^{\ast }{\nu }_1$ is known as the bi-skew Jordan product of ${\nu }_1$ and ${\nu }_2$. In recent years, numerous researchers have directed their attention towards the preserver problem concerning the Lie and Jordan product. Recently, the bi-skew Jordan product was considered by Li et al. [14]. They showed a nonlinear bijective map on von Neumann algebras without central abelian projections, with $\vartheta :\mathfrak{E}\to \mathfrak{F}$ satisfying

$$\vartheta ({\nu }_1^{\ast }{\nu }_2+{\nu }_2^{\ast }{\nu }_1)=\vartheta {({\nu }_1)}^{\ast }\vartheta ({\nu }_2)+\vartheta {({\nu }_2)}^{\ast }\vartheta ({\nu }_1)$$

for all ${\nu }_1,{\nu }_2\in \mathfrak{E}$, is the sum of a conjugate linear ∗-isomorphism and a linear ∗-isomorphism. Moreover, Taghavi and Gholampoor [15] obtained the structure of maps preserving product ${\nu }_1^{\ast }{\nu }_2+{\nu }_2^{\ast }{\nu }_1$ on $C^{\ast }$-algebras. Very recently, Zhao et al. [16] addressed a more comprehensive problem of a map preserving the bi-skew Jordan triple product on factor von Neumann algebras. We say that a map $\vartheta$ on the factor von Neumann algebras $\mathfrak{E}$ and $\mathfrak{F}$ preserves the bi-skew Jordan triple product whenever $\vartheta$ satisfies $\vartheta ({\nu }_1★{\nu }_2★{\nu }_3)=\vartheta ({\nu }_1)★\vartheta ({\nu }_2)★\vartheta ({\nu }_3)$ for all ${\nu }_1,{\nu }_2,{\nu }_3\in \mathfrak{E}$. It is evident that a map $\vartheta$ on factor von Neumann algebras $\mathfrak{E}$, which preserves ${\nu }_1★{\nu }_2$, will also preserve ${\nu }_1★{\nu }_2★{\nu }_3$ for any ${\nu }_i\in \mathfrak{E}$. But, the converse of this statement does not hold, in general. Consider, for instance, the mapping $\vartheta (\nu )=-\nu$ for all $\nu \in \mathfrak{E}$, which preserves ${\nu }_1★{\nu }_2★{\nu }_3$ but not ${\nu }_1★{\nu }_2$. This illustrates that the collection of mappings preserving ${\nu }_1★{\nu }_2★{\nu }_3$ extends the notion of mappings preserving ${\nu }_1★{\nu }_2$. Inspired by this observation, we will broaden the scope of maps that preserve the bi-skew Jordan triple product between factor von Neumann algebras. Let $\mathfrak{E}$ and $\mathfrak{F}$ be two factor von Neumann algebras; a bijective map (not necessarily linear) $\vartheta :\mathfrak{E}\to \mathfrak{F}$ satisfying

$$\vartheta ({\nu }_1★{\nu }_2★{\nu }_3★\cdots ★{\nu }_n)=\vartheta ({\nu }_1)★\vartheta ({\nu }_2)★\vartheta ({\nu }_3)★\cdots ★\vartheta ({\nu }_n)$$

preserves the bi-skew Jordan n-product of ${\nu }_i\in \mathfrak{E}$$(i=1,2,\dots ,n)$. Here, ${\nu }_i★{\nu }_j={\nu }_i^{\ast }{\nu }_j+{\nu }_j^{\ast }{\nu }_i$ is the bi-skew Jordan product of ${\nu }_i$, ${\nu }_j$ for any $1\le i,j\le n$, and these n-products are evaluated from left to right, that is, ${\nu }_1★{\nu }_2★{\nu }_3★\cdots ★{\nu }_n=(\cdots (({\nu }_1★{\nu }_2)★{\nu }_3)★\cdots ★{\nu }_n)$. Motivated by the aforementioned work, our aim is to characterize a nonlinear map that preserves the bi-skew Jordan n-product on factor von Neumann algebras. Our article establishes that such a map $\vartheta$ or -$\vartheta$ on factor von Neumann algebras is either a conjugate linear ∗-isomorphism or a linear ∗-isomorphism.

Throughout this article, $\mathbb{R}$ and $\mathbb{C}$ represent the real and complex field, respectively. An element $e\in \mathfrak{E}$, is said to be symmetric idempotent if $e^{\ast }=e$ and $e^2=e$. A von Neumann algebra $\mathfrak{E}$ is a weakly closed self-adjoint subalgebra of $\mathfrak{B}(\mathfrak{H})$ (collection of all bounded linear operators on a complex Hilbert space $\mathfrak{H}$) containing the identity element $\mathfrak{I}$. In other words, a von Neumann algebra $\mathfrak{E}$ is a self-adjoint subalgebra of $\mathfrak{B}(\mathfrak{H})$ which satisfies the double commutant property, i.e., ${\mathfrak{E}}^{\prime \prime }=\mathfrak{E}$, (where ${\mathfrak{E}}^{\prime }=\{g\in \mathfrak{B}(\mathfrak{H})|gh=hg\mathit{for}\mathit{all}h\in \mathfrak{E}\}$. Moreover, a von Neumann algebra $\mathfrak{E}$ is called a factor von Neumann algebra if its center is trivial, i.e., $\mathcal{Z}(\mathfrak{E})=\mathbb{C}\mathfrak{I}$. It is commonly known that a factor von Neumann algebra $\mathfrak{E}$ is prime, that is, for any $u,v\in \mathfrak{E}$ if $u\mathfrak{E}v=(0)$, then either $u=0$ or $v=0$. If dim$(\mathfrak{E})<\infty$, then $\mathfrak{E}$ is isomorphic to $M_n(\mathbb{C})$, the algebra of $n\times n$ matrices over $\mathbb{C}$. We assume that the dimension of the algebras $\mathfrak{E}$ and $\mathfrak{F}$ are greater than 1 in the following section.

Recall that factor von Neumann algebras and isomorphism have significant roles across various areas of mathematics and physics. In quantum mechanics, they model irreducible systems and provide a framework for classifying systems with finite or infinite degrees of freedom. Within the field of operator algebras, factor von Neumann algebras serve as fundamental building blocks in the classification theory, play a key role in modular theory, and are central to representation theory. Linear ∗-isomorphisms preserve the full algebraic and spectral structure of operator algebras and thus serve as the natural equivalence in their classification and representation theory.

2. Main Result

First, we discuss the additivity of nonlinear maps preserving the bi-skew Jordan n-product on factor von Neumann algebras; then, we prove our main theorem, defined as follows:

Theorem 1. 

Let $\mathfrak{E}$ and $\mathfrak{F}$ be two factor von Neumann algebras such that $\mathfrak{E}$ contains nontrivial symmetric idempotent element e and identity element $\mathfrak{I}$. Assume that $\vartheta :\mathfrak{E}\to \mathfrak{F}$ is a bijective map satisfying

$$\vartheta ({\nu }_1★{\nu }_2★{\nu }_3★\cdots ★{\nu }_n)=\vartheta ({\nu }_1)★\vartheta ({\nu }_2)★\vartheta ({\nu }_3)★\cdots ★\vartheta ({\nu }_n)$$

for all ${\nu }_i\in \mathfrak{E}$$(i=1,2,\dots ,n)$; for $n>1$ needs to be a fixed positive integer, and ${\nu }_i★{\nu }_j={\nu }_i^{\ast }{\nu }_j+{\nu }_j^{\ast }{\nu }_i$ is the bi-skew Jordan product of ${\nu }_i$, ${\nu }_j$ for any $1\le i,j\le n$. Then, ϑ or $-\vartheta$ is a linear ∗-isomorphism or a conjugate linear ∗-isomorphism.

Now we prove our main theorem by various lemmas.

Lemma 1. 

In the very beginning, we give the main technique for proving such kinds of problems. Suppose that ${\nu }_1,{\nu }_2,{\nu }_3,\dots {\nu }_n\in \mathfrak{E}$ such that $\vartheta (t)={\displaystyle \sum _{i=1}^n}\vartheta ({\nu }_i)$, for any $t\in \mathfrak{E}$. Then, for all ${\mu }_i\in \mathfrak{E},(i=1,2,\dots ,n-1)$, we have

$$\vartheta ({\mu }_1★{\mu }_2★\cdots ★{\mu }_{n-1}★t)={\displaystyle \sum _{i=1}^n}\vartheta ({\mu }_1★{\mu }_2★\cdots ★{\mu }_{n-1}★{\nu }_i)$$

$$\vartheta ({\mu }_1★{\mu }_2★\cdots ★t★{\mu }_{n-1})={\displaystyle \sum _{i=1}^n}\vartheta ({\mu }_1★{\mu }_2★\cdots ★{\nu }_i★{\mu }_{n-1})$$

$$\cdots$$

$$\vartheta (t★{\mu }_1★{\mu }_2★\cdots ★{\mu }_{n-1})={\displaystyle \sum _{i=1}^n}\vartheta ({\nu }_i★{\mu }_1★{\mu }_2★\cdots ★{\mu }_{n-1}).$$

Proof. 

It is easy to check that

$$\begin{array}{cc}\hfill \vartheta ({\mu }_1★{\mu }_2★\cdots ★{\mu }_{n-1}★t)& =\vartheta ({\mu }_1)★\vartheta ({\mu }_2)★\cdots ★\vartheta ({\mu }_{n-1})★\vartheta (t)\hfill \\ \hfill & =\vartheta ({\mu }_1)★\vartheta ({\mu }_2)★\cdots ★\vartheta ({\mu }_{n-1})★{\displaystyle \sum _{i=1}^n}\vartheta ({\nu }_i)\hfill \\ \hfill & ={\displaystyle \sum _{i=1}^n}\left(\vartheta ({\mu }_1)★\vartheta ({\mu }_2)★\cdots ★\vartheta ({\mu }_{n-1})★\vartheta ({\nu }_i)\right)\hfill \\ \hfill & ={\displaystyle \sum _{i=1}^n}\vartheta ({\mu }_1★{\mu }_2★\cdots ★{\mu }_{n-1}★{\nu }_i).\hfill \end{array}$$

Similarly, we can show the other relations. □

Select arbitrary nontrivial symmetric idempotent elements $e_i\in \mathfrak{E}$$(i=1,2)$ such that $e_2=\mathfrak{I}-e_1$. Throughout the article, ${\mathfrak{E}}^s=\{a\in \mathfrak{E}:a^{\ast }=a\}$ and ${\mathfrak{F}}^s=\{b\in \mathfrak{F}:b^{\ast }=b\}$ are the sets of symmetric and skew symetric elements, respectively. Denote $a_{ii}=\{e_{i}a{e}_i$; $a\in {\mathfrak{E}}^s,i=1,2\}$ and $a_{12}=\{e_{1}a{e}_2+e_{2}a{e}_1:a\in {\mathfrak{E}}^s\}$. For every $a\in {\mathfrak{E}}^s$, we can write $a=a_{11}+a_{12}+a_{22}$, where $a_{ii}\in {\mathfrak{E}}_{ii}^s(i=1,2)$ and $a_{12}\in {\mathfrak{E}}_{12}^s$. Also, ${\mathfrak{E}}_{ii}^s=e_i{\mathfrak{E}}^s{e}_i$.

Lemma 2. 

$\vartheta (0)=0$.

Proof. 

Since $\vartheta$ is a surjective map, then $\vartheta (v)=0$ for some $\nu \in \mathfrak{E}$. By this hypothesis, we have

$$\begin{array}{cc}\hfill \vartheta (0)& =\vartheta (0★\nu ★\cdots ★\nu )\hfill \\ \hfill & =\vartheta (0)★\vartheta (\nu )★\cdots ★\vartheta (\nu )\hfill \\ \hfill & =\vartheta (0)★0★\cdots ★0\hfill \\ \hfill & =0.\hfill \end{array}$$

□

Lemma 3. 

$\vartheta {(\nu )}^{\ast }=\vartheta (\nu )$ if and only if ${\nu }^{\ast }=\nu$ for any $\nu \in \mathfrak{E}.$

Proof. 

For any ${\nu }^{\ast }=\nu$, we have $\nu =\nu ★{\displaystyle \frac{1}{2}}\mathfrak{I}★\cdots ★{\displaystyle \frac{1}{2}}\mathfrak{I}$.

Also, $f_1★f_2★f_3★\cdots ★f_n\in {\mathfrak{F}}^s$ for any $f_i\in \mathfrak{F}$, $(i=1,2,\dots ,n)$, we get

$$\begin{array}{cc}\hfill \vartheta (\nu )& =\vartheta \left(\nu ★{\displaystyle \frac{1}{2}}\mathfrak{I}★\cdots ★{\displaystyle \frac{1}{2}}\mathfrak{I}\right)\hfill \\ \hfill & =\vartheta (\nu )★\vartheta ({\displaystyle \frac{1}{2}}\mathfrak{I})★\cdots ★\vartheta ({\displaystyle \frac{1}{2}}\mathfrak{I})\in {\mathfrak{F}}^s.\hfill \end{array}$$

Hence, we obtain $\vartheta {(\nu )}^{\ast }=\vartheta (\nu )$.

Conversely, let $\vartheta {(\nu )}^{\ast }=\vartheta (\nu )$. Since $\vartheta$ is surjective, then there exists $\mu \in \mathfrak{E}$ such that $\vartheta (\mu )={\displaystyle \frac{1}{2}}\mathfrak{I}.$ Therefore, we have

$$\begin{array}{cc}\hfill \vartheta (\nu )& =\vartheta (\nu )★{\displaystyle \frac{1}{2}}\mathfrak{I}★\cdots ★{\displaystyle \frac{1}{2}}\mathfrak{I}\hfill \\ \hfill & =\vartheta (\nu )★\vartheta (\mu )★\cdots ★\vartheta (\mu )\hfill \\ \hfill & =\vartheta (\nu ★\mu ★\cdots ★\mu ).\hfill \end{array}$$

Using the injectivity of $\vartheta$, we have $\nu =\nu ★\mu ★\cdots ★\mu \in {\mathfrak{E}}^s$. Hence, ${\nu }^{\ast }=\nu$. □

Lemma 4. 

For any ${\nu }_{11}\in {\mathfrak{E}}_{11}^s,{\nu }_{12}\in {\mathfrak{E}}_{12}^s$, and ${\nu }_{22}\in {\mathfrak{E}}_{22}^s$, we have

$(i)$

$\vartheta ({\nu }_{11}+{\nu }_{12})=\vartheta ({\nu }_{11})+\vartheta ({\nu }_{12})$;

$(ii)$

$\vartheta ({\nu }_{12}+{\nu }_{22})=\vartheta ({\nu }_{12})+\vartheta ({\nu }_{22})$.

Proof. 

(i) Assume that we have an element $c=c_{11}+c_{12}+c_{22}\in {\mathfrak{E}}^s$, such that $\vartheta (c)=\vartheta ({\nu }_{11})+\vartheta ({\nu }_{12})$. Then, by Lemma 1 we have

$$\begin{array}{cc}\hfill \vartheta ({\mu }_1★{\mu }_2★\cdots ★{\mu }_{n-1}★c)& =\vartheta ({\mu }_1★{\mu }_2★\cdots ★{\mu }_{n-1}★{\nu }_{11})\hfill \\ \hfill & +\vartheta ({\mu }_1★{\mu }_2★\cdots ★{\mu }_{n-1}★{\nu }_{12})\hfill \end{array}$$

for any ${\mu }_i\in \mathfrak{E}$, $(i=1,2,\dots ,n-1)$. Now, take ${\mu }_1={\mu }_2=\dots ={\mu }_{n-1}=e_2$; in the above relation, we obtain

$$\begin{array}{cc}\hfill \vartheta (2^{n-2}{e}_2★c)& =\vartheta (2^{n-2}{e}_2★{\nu }_{12})\hfill \\ \hfill \vartheta (2^{n-2}{c}_{12}+2^{n-1}{c}_{22})& =\vartheta (2^{n-2}{\nu }_{12}).\hfill \end{array}$$

By using the injectivity of $\vartheta$, we have $c_{12}+c_{22}={\nu }_{12}$. Now, multiplying by $e_1$ from the left and $e_2$ from the right, we get $c_{12}={\nu }_{12}$ and $c_{22}=0$. Since $(e_1-e_2)★{\nu }_{12}=0$, then we have

$$\begin{array}{cc}\hfill \vartheta ({\displaystyle \frac{1}{2}}\mathfrak{I}★\cdots ★{\displaystyle \frac{1}{2}}\mathfrak{I}★(e_1-e_2)★c)& =\vartheta ({\displaystyle \frac{1}{2}}\mathfrak{I}★\cdots ★{\displaystyle \frac{1}{2}}\mathfrak{I}★(e_1-e_2)★{\nu }_{11})\hfill \\ \hfill \vartheta ((e_1-e_2)★c)& =\vartheta ((e_1-e_2)★{\nu }_{11})\hfill \\ \hfill \vartheta (2c_{11})& =\vartheta (2{\nu }_{11}).\hfill \end{array}$$

Again by the injectivity of $\vartheta$, this gives $c_{11}={\nu }_{11}$, and hence, we get the desired result., i.e.,

$$\vartheta ({\nu }_{11}+{\nu }_{12})=\vartheta ({\nu }_{11})+\vartheta ({\nu }_{12}).$$

Similarly, we can prove $(ii)$. □

Lemma 5. 

For any $u_{ii}\in {\mathfrak{E}}_{ii}^s,(i=1,2)$ and $v_{12}\in {\mathfrak{E}}_{12}^s$, we have

$$\vartheta (u_{11}+v_{12}+w_{22})=\vartheta (u_{11})+\vartheta (v_{12})+\vartheta (w_{22}).$$

Proof. 

Consider an element $d=d_{11}+d_{12}+d_{22}\in {\mathfrak{E}}^s$ such that $\vartheta (d)=\vartheta (u_{11})+\vartheta (v_{12})+\vartheta (w_{22})$. By Lemma 4, we can write $\vartheta (d)=\vartheta (u_{11}+v_{12})+\vartheta (w_{22})$. Using Lemmas 1 and 3, we have

$$\begin{array}{cc}\hfill & \vartheta \left({\displaystyle \frac{1}{2}}\mathfrak{I}★\cdots ★{\displaystyle \frac{1}{2}}\mathfrak{I}★e_1★d\right)\hfill \\ \hfill & =\vartheta \left({\displaystyle \frac{1}{2}}\mathfrak{I}★\cdots ★{\displaystyle \frac{1}{2}}\mathfrak{I}★e_1★(u_{11}+v_{12})\right)\hfill \\ \hfill & +\vartheta \left({\displaystyle \frac{1}{2}}\mathfrak{I}★\cdots ★{\displaystyle \frac{1}{2}}\mathfrak{I}★e_1★w_{22}\right)\hfill \\ \hfill \vartheta (2d_{11}+d_{12})& =\vartheta (2u_{11}+v_{12}).\hfill \end{array}$$

Using the injectivity of $\vartheta$, we get $2d_{11}+d_{12}=2u_{11}+v_{12}$. Multiplying this expression by $e_1$ from both sides, we obtain $d_{11}=u_{11}$, and multiplying by $e_1$ from left and by $e_2$ from right, we have $d_{12}=v_{12}$.

Replace $e_1$ by $e_2$ in the above calculation to obtain $\vartheta (2d_{12}+d_{22})=\vartheta (2u_{12}+w_{22})$. Again, using the injectivity of $\vartheta$ and similar calculations as those above, we can easily obtain $d_{22}=w_{22}$. Therefore, we get

$$\vartheta (u_{11}+v_{12}+w_{22})=\vartheta (u_{11})+\vartheta (v_{12})+\vartheta (w_{22}).$$

□

Lemma 6. 

$\vartheta ({\nu }_{12}+{\mu }_{12})=\vartheta ({\nu }_{12})+\vartheta ({\mu }_{12})$ for any ${\nu }_{12},{\mu }_{12}\in {\mathfrak{E}}_{12}^s.$

Proof. 

Let ${\nu }_{12},{\mu }_{12}\in {\mathfrak{E}}_{12}^s$ such that ${\nu }_{12}=e_1\nu e_2+e_2\nu e_1$ and ${\mu }_{12}=e_1\mu e_2+e_2\mu e_1$, where $\nu ,\mu \in {\mathfrak{E}}^s$. By an easy calculation, we have

$$\begin{array}{cc}\hfill (e_1+{\nu }_{12})★(e_2+{\mu }_{12})★{\displaystyle \frac{1}{2}}\mathfrak{I}★\cdots ★{\displaystyle \frac{1}{2}}\mathfrak{I}& ={\nu }_{12}+{\mu }_{12}+{\nu }_{12}{\mu }_{12}+{\mu }_{12}{\nu }_{12},\hfill \end{array}$$

where ${\nu }_{12}+{\mu }_{12}\in {\mathfrak{E}}_{12}$, and ${\nu }_{12}{\mu }_{12}+{\mu }_{12}{\nu }_{12}=e_1(\nu \mu +\mu \nu )e_1+e_2(\nu \mu +\mu \nu )e_2\in {\mathfrak{E}}_{11}+{\mathfrak{E}}_{22}.$ By Lemma 5, we can write

$$\begin{array}{cc}\hfill & \vartheta ({\nu }_{12}+{\mu }_{12})+\vartheta ({\nu }_{12}{\mu }_{12}+{\mu }_{12}{\nu }_{12})\hfill \\ \hfill & =\vartheta ({\nu }_{12}+{\mu }_{12}+{\nu }_{12}{\mu }_{12}+{\mu }_{12}{\nu }_{12})\hfill \\ \hfill & =\vartheta \left((e_1+{\nu }_{12})★(e_2+{\mu }_{12})★{\displaystyle \frac{1}{2}}\mathfrak{I}★\cdots ★{\displaystyle \frac{1}{2}}\mathfrak{I}\right)\hfill \\ \hfill & =\vartheta (e_1+{\nu }_{12})★\vartheta (e_2+{\mu }_{12})★\vartheta ({\displaystyle \frac{1}{2}}\mathfrak{I})★\cdots ★\vartheta ({\displaystyle \frac{1}{2}}\mathfrak{I})\hfill \\ \hfill & =\left(\vartheta (e_1)+\vartheta ({\nu }_{12})\right)★\left(\vartheta (e_2)+\vartheta ({\mu }_{12})\right)★\vartheta ({\displaystyle \frac{1}{2}}\mathfrak{I})★\cdots ★\vartheta ({\displaystyle \frac{1}{2}}\mathfrak{I})\hfill \\ \hfill & =\left(\vartheta (e_1)★\vartheta (e_2)★\vartheta ({\displaystyle \frac{1}{2}}\mathfrak{I})★\cdots ★\vartheta ({\displaystyle \frac{1}{2}}\mathfrak{I})\right)\hfill \\ \hfill & +\left(\vartheta (e_1)★\vartheta ({\mu }_{12})★\vartheta ({\displaystyle \frac{1}{2}}\mathfrak{I})★\cdots ★\vartheta ({\displaystyle \frac{1}{2}}\mathfrak{I})\right)\hfill \\ \hfill & +\left(\vartheta ({\nu }_{12})★\vartheta (e_2)★\vartheta ({\displaystyle \frac{1}{2}}\mathfrak{I})★\cdots ★\vartheta ({\displaystyle \frac{1}{2}}\mathfrak{I})\right)\hfill \\ \hfill & +\left(\vartheta ({\nu }_{12})★\vartheta ({\mu }_{12})★\vartheta ({\displaystyle \frac{1}{2}}\mathfrak{I})★\cdots ★\vartheta ({\displaystyle \frac{1}{2}}\mathfrak{I})\right)\hfill \\ \hfill & =\vartheta \left(e_1★{\mu }_{12}★{\displaystyle \frac{1}{2}}\mathfrak{I}★\cdots ★{\displaystyle \frac{1}{2}}\mathfrak{I}\right)\hfill \\ \hfill & =\vartheta \left({\nu }_{12}★e_2★{\displaystyle \frac{1}{2}}\mathfrak{I}★\cdots ★{\displaystyle \frac{1}{2}}\mathfrak{I}\right)\hfill \\ \hfill & =\vartheta \left({\nu }_{12}★{\mu }_{12}★{\displaystyle \frac{1}{2}}\mathfrak{I}★\cdots ★{\displaystyle \frac{1}{2}}\mathfrak{I}\right)\hfill \\ \hfill & =\vartheta ({\nu }_{12})+\vartheta ({\mu }_{12})+\vartheta ({\nu }_{12}{\mu }_{12}+{\mu }_{12}{\nu }_{12}).\hfill \end{array}$$

This implies that

$$\vartheta ({\nu }_{12}+{\mu }_{12})=\vartheta ({\nu }_{12})+\vartheta ({\mu }_{12}).$$

Hence, we get the required result. □

Lemma 7. 

For any $a_{ii},b_{ii}\in {\mathfrak{E}}_{ii}^s(i=1,2)$, we have

$(i)$

$\vartheta (a_{11}+b_{11})=\vartheta (a_{11})+\vartheta (b_{11});$

$(ii)$

$\vartheta (a_{22}+b_{22})=\vartheta (a_{22})+\vartheta (b_{22}).$

Proof. 

(i) There is an element $c=c_{11}+c_{12}+c_{22}\in {\mathfrak{E}}^s$ such that $\vartheta (c)=\vartheta (a_{11})+\vartheta (b_{11})$, and using Lemma 2, we can write

$$\begin{array}{cc}\hfill \vartheta (c_{12}+2c_{22})& =\vartheta ({\displaystyle \frac{1}{2}}\mathfrak{I}★\cdots ★{\displaystyle \frac{1}{2}}\mathfrak{I}★e_2★c)\hfill \\ \hfill & =\vartheta ({\displaystyle \frac{1}{2}}\mathfrak{I}★\cdots ★{\displaystyle \frac{1}{2}}\mathfrak{I}★e_2★a_{11})\hfill \\ \hfill & +\vartheta ({\displaystyle \frac{1}{2}}\mathfrak{I}★\cdots ★{\displaystyle \frac{1}{2}}\mathfrak{I}★e_2★b_{11})\hfill \\ \hfill & =0.\hfill \end{array}$$

Using the injectivity of $\vartheta$, we get $c_{12}+2c_{22}=0$. If we multiply this equation by $e_1$ from the left and $e_2$ from the right, we obtain $c_{12}=0$; similarly, multiplying by $e_2$ from both sides gives $c_{22}=0.$ Consider $t_{12}=e_{1}d{e}_2+{(e_{1}d{e}_2)}^{\ast }$ for any $d\in \mathfrak{E}$; we obtain $t_{12},a_{11}{t}_{12}+t_{12}{a}_{11},b_{11}{t}_{12}+t_{12}{b}_{11}\in {\mathfrak{E}}_{12}$. It follows from Lemma 6 that

$$\begin{array}{cc}\hfill \vartheta (c_{11}{t}_{12}+t_{12}{c}_{11})& =\vartheta \left({\displaystyle \frac{1}{2}}\mathfrak{I}★\cdots ★{\displaystyle \frac{1}{2}}\mathfrak{I}★c★t_{12}\right)\hfill \\ \hfill & =\vartheta \left({\displaystyle \frac{1}{2}}\mathfrak{I}★\cdots ★{\displaystyle \frac{1}{2}}\mathfrak{I}★a_{11}★t_{12}\right)\hfill \\ \hfill & +\vartheta \left({\displaystyle \frac{1}{2}}\mathfrak{I}★\cdots ★{\displaystyle \frac{1}{2}}\mathfrak{I}★b_{11}★t_{12}\right)\hfill \\ \hfill & =\vartheta (a_{11}{t}_{12}+t_{12}{a}_{11})+\vartheta (b_{11}{t}_{12}+t_{12}{b}_{11})\hfill \\ \hfill & =\vartheta (a_{11}{t}_{12}+t_{12}{a}_{11}+b_{11}{t}_{12}+t_{12}{b}_{11}).\hfill \end{array}$$

Since $\vartheta$ is injective, we have $c_{11}{t}_{12}+t_{12}{c}_{11}=a_{11}{t}_{12}+t_{12}{a}_{11}+b_{11}{t}_{12}+t_{12}{b}_{11}$. Multiplying the above equation by $e_2$ from the right, we get $(c_{11}-a_{11}-b_{11})e_{1}d{e}_2=0$ for any $d\in \mathfrak{E}$. Since $\mathfrak{E}$ is prime, we have $c=c_{11}=a_{11}+b_{11}$. Therefore,

$$\vartheta (a_{11}+b_{11})=\vartheta (a_{11})+\vartheta (b_{11}).$$

Similarly, one can prove

$$\vartheta (a_{22}+b_{22})=\vartheta (a_{22})+\vartheta (b_{22}).$$

□

Remark 1. 

Lemmas 4–7 imply that ϑ is additive on ${\mathfrak{E}}^s$.

Lemma 8. 

$\vartheta (\mathfrak{I})=\mathfrak{I}$ or $\vartheta (\mathfrak{I})=-\mathfrak{I}$, and ϑ preserves conjugate self-adjoint elements in both directions.

Proof. 

By using the surjectivity of $\vartheta$ and Lemma 3, there exists an element $a\in {\mathfrak{E}}^s$, such that $\vartheta (a)=\mathfrak{I}$. In view of Remark 1, we have

$$\begin{array}{cc}\hfill 2^{n-1}\mathfrak{I}& =2^{n-1}\vartheta (a)=\vartheta (2^{n-1}a)\hfill \\ \hfill & =\vartheta (a★\mathfrak{I}★\cdots ★\mathfrak{I})\hfill \\ \hfill & =\vartheta (a)★\vartheta (\mathfrak{I})★\cdots ★\vartheta (\mathfrak{I})\hfill \\ \hfill & =2^{n-1}\vartheta {(\mathfrak{I})}^{n-1}.\hfill \end{array}$$

Hence, $\vartheta {(\mathfrak{I})}^{n-1}=\mathfrak{I}$.

Let $s^{\ast }=-s\in \mathfrak{E}.$ Then, using $\vartheta {(\mathfrak{I})}^{n-1}=\mathfrak{I}$, we get

$$\begin{array}{cc}\hfill 0& =\vartheta (\mathfrak{I}★\cdots ★\mathfrak{I}★s)\hfill \\ \hfill & =\vartheta (\mathfrak{I})★\cdots ★\vartheta (\mathfrak{I})★\vartheta (s)\hfill \\ \hfill & =2^{n-2}\vartheta {(\mathfrak{I})}^{n-1}★\vartheta (s)\hfill \\ \hfill & =2^{n-2}(\vartheta (s)+\vartheta {(s)}^{\ast }).\hfill \end{array}$$

Which implies that $\vartheta {(s)}^{\ast }=-\vartheta (s).$ Since ${\vartheta }^{-1}$ has the same properties as that of $\vartheta$, if $\vartheta {(s)}^{\ast }=-\vartheta (s)$, then $s^{\ast }=-s\in \mathfrak{E}$. Hence, $\vartheta {(s)}^{\ast }=-\vartheta (s)\iff s^{\ast }=-s$.

By Lemma 3, we have

$$\begin{array}{cc}\hfill 0& =\vartheta (\mathfrak{I}★s★a★\cdots ★a)\hfill \\ \hfill & =\vartheta (\mathfrak{I})★\vartheta (s)★\vartheta (a)★\cdots ★\vartheta (a)\hfill \\ \hfill & =\vartheta (\mathfrak{I})★\vartheta (s)★\mathfrak{I}★\cdots ★\mathfrak{I}\hfill \\ \hfill & =\left(\vartheta (\mathfrak{I})\vartheta (s)+\vartheta {(s)}^{\ast }\vartheta (\mathfrak{I})\right)★\mathfrak{I}★\cdots ★\mathfrak{I}\hfill \\ \hfill & =2^{n-2}\left(\vartheta (\mathfrak{I})\vartheta (s)-\vartheta (s)\vartheta (\mathfrak{I})\right).\hfill \end{array}$$

Therefore, $\vartheta (\mathfrak{I})\vartheta (s)=\vartheta (s)\vartheta (\mathfrak{I})$ for all $s^{\ast }=-s\in \mathfrak{E}$. Then, $\vartheta (\mathfrak{I})g=g\vartheta (\mathfrak{I})$ for all $g^{\ast }=-g\in \mathfrak{F}$. Since for any $h\in \mathfrak{F}$, $h=h_1+i{h}_2$, where $h_1={\displaystyle \frac{1}{2}}(h-h^{\ast })$ and $h_2={\displaystyle \frac{1}{2i}}(h+h^{\ast })$ are conjugate self-adjoint elements, we have $\vartheta (\mathfrak{I})h=h\vartheta (\mathfrak{I})$ for all $h\in \mathfrak{F}$. Therefore, $\vartheta (\mathfrak{I})\in \mathbb{C}\mathfrak{I}$. Also, we have $\vartheta (\mathfrak{I})=\vartheta {(\mathfrak{I})}^{\ast }$ and $\vartheta {(\mathfrak{I})}^{n-1}=\mathfrak{I}$; thus, we obtain the following two cases:

• Case (I): when $n\ge 2$ is even, then $\vartheta (\mathfrak{I})=\mathfrak{I}$.
• Case (II): when $n\ge 3$ is odd, then $\vartheta (\mathfrak{I})=\mathfrak{I}$ or −$\mathfrak{I}$. □

We shall discuss these two cases one by one.

• Case (I): If $\vartheta (\mathfrak{I})=\mathfrak{I}$, then $\vartheta$ is either a conjugate linear ∗-isomorphism or a linear ∗-isomorphism.

Lemma 9. 

$\vartheta {(i\mathfrak{I})}^2=-\mathfrak{I}$ and $\vartheta (i\mathfrak{I})\in \mathcal{Z}(\mathfrak{F})$.

Proof. 

By Lemma 3, for any $d\in {\mathfrak{F}}^s$ there exist $c,c^{\prime }\in {\mathfrak{E}}^s$ such that $\vartheta (c)=d$ and $\vartheta (c^{\prime })={\displaystyle \frac{1}{2}}\mathfrak{I}.$ Using the hypothesis, we have

$$\begin{array}{cc}\hfill 0& =\vartheta (c★i\mathfrak{I}★c^{\prime }★\cdots ★c^{\prime }).\hfill \\ \hfill & =\vartheta (c)★\vartheta (i\mathfrak{I})★\vartheta (c^{\prime })★\cdots ★\vartheta (c^{\prime }).\hfill \\ \hfill & =\vartheta (c)★\vartheta (i\mathfrak{I})★{\displaystyle \frac{1}{2}}\mathfrak{I}★\cdots ★{\displaystyle \frac{1}{2}}\mathfrak{I}.\hfill \\ \hfill & =d\vartheta (i\mathfrak{I})+\vartheta {(i\mathfrak{I})}^{\ast }d\hfill \\ \hfill & =d\vartheta (i\mathfrak{I})-\vartheta (i\mathfrak{I})d.\hfill \end{array}$$

This implies that $d\vartheta (i\mathfrak{I})=\vartheta (i\mathfrak{I})d$ for any $d\in {\mathfrak{F}}^s$. For every $k\in \mathfrak{F}$, we can write $k=k_1+i{k}_2$, where $k_1={\displaystyle \frac{1}{2}}(k+k^{\ast })$ and $k_2={\displaystyle \frac{1}{2i}}(k-k^{\ast })$ are self-adjoint elements. Thus, $k\vartheta (i\mathfrak{I})=\vartheta (i\mathfrak{I})k$ for any $k\in \mathfrak{F}$. Therefore, $\vartheta (i\mathfrak{I})$ is in centre of $(\mathfrak{F}).$

By Remark 1, and $i\mathfrak{I}★i\mathfrak{I}★\mathfrak{I}★\cdots ★\mathfrak{I}=2^{n-1}\mathfrak{I}$, we have

$$\begin{array}{cc}\hfill 2^{n-1}\vartheta (\mathfrak{I})& =\vartheta (2^{n-1}\mathfrak{I})\hfill \\ \hfill & =\vartheta (i\mathfrak{I}★i\mathfrak{I}★\mathfrak{I}★\cdots ★\mathfrak{I})\hfill \\ \hfill & =\vartheta (i\mathfrak{I})★\vartheta (i\mathfrak{I})★\vartheta (\mathfrak{I})★\cdots ★\vartheta (\mathfrak{I})\hfill \\ \hfill & =\vartheta (i\mathfrak{I})★\vartheta (i\mathfrak{I})★\mathfrak{I}★\cdots ★\mathfrak{I}\hfill \\ \hfill 2^{n-1}\mathfrak{I}& =-2^{n-1}\vartheta {(i\mathfrak{I})}^2.\hfill \end{array}$$

Thus, $\vartheta {(i\mathfrak{I})}^2=-\mathfrak{I}$. □

Lemma 10. 

For all $c_i\in {\mathfrak{E}}^s(i=1,2)$, we have

$(i)$

$\vartheta (c_1+i{c}_2)=\vartheta (c_1)+\vartheta (i\mathfrak{I})\vartheta (c_2)$;

$(ii)$

ϑ is an additive on $\mathfrak{E}.$

Proof. 

Let $c_1,c_2\in {\mathfrak{E}}^s$, such that

$$\begin{array}{c}\hfill \vartheta (c_1+i{c}_2)=\vartheta (d_1)+i\vartheta (d_2)\end{array}$$

(1)

for some $d_1,d_2\in {\mathfrak{E}}^s$. By Remark 1, we have

$$\begin{array}{cc}\hfill 2^{n-1}\vartheta (c_1)& =\vartheta (2^{n-1}{c}_1)\hfill \\ \hfill & =\vartheta \left(\mathfrak{I}★\cdots ★\mathfrak{I}★(c_1+i{c}_2)\right)\hfill \\ \hfill & =\mathfrak{I}★\cdots ★\mathfrak{I}★\vartheta (c_1+i{c}_2)\hfill \\ \hfill & =2^{n-2}\left(\vartheta (c_1+i{c}_2)+\vartheta {(c_1+i{c}_2)}^{\ast }\right).\hfill \end{array}$$

It follows from (1) that $\vartheta (c_1)=\vartheta (d_1)$. Also, we have

$$\begin{array}{cc}\hfill 2^{n-1}\vartheta (c_2)& =\vartheta (2^{n-1}{c}_2)\hfill \\ \hfill & =\vartheta \left(i\mathfrak{I}★(c_1+i{c}_2)★\mathfrak{I}★\cdots ★\mathfrak{I}\right)\hfill \\ \hfill & =\vartheta (i\mathfrak{I})★\vartheta (c_1+i{c}_2)★\mathfrak{I}★\cdots ★\mathfrak{I}\hfill \\ \hfill & =\left(\vartheta {(i\mathfrak{I})}^{\ast }\vartheta (c_1+i{c}_2)+\vartheta {(c_1+i{c}_2)}^{\ast }\vartheta (i\mathfrak{I})\right)★\mathfrak{I}★\cdots ★\mathfrak{I}.\hfill \\ \hfill & =-2^{n-1}i\vartheta (i\mathfrak{I})\vartheta (d_2).\hfill \end{array}$$

This implies that $\vartheta (d_2)=-i\vartheta (i\mathfrak{I})\vartheta (c_2).$ Hence, from (1), we obtain

$$\begin{array}{c}\hfill \vartheta (c_1+i{c}_2)=\vartheta (c_1)+\vartheta (i\mathfrak{I})\vartheta (c_2).\end{array}$$

(2)

Now, for any $c=c_1+i{c}_2,d=d_1+i{d}_2\in \mathfrak{E},$ where $c_i,d_i\in {\mathfrak{E}}^s(i=1,2)$, and using (2), we have

$$\begin{array}{cc}\hfill \vartheta (c+d)& =\vartheta \left((c_1+d_1)+i(c_2+d_2)\right)\hfill \\ \hfill & =\vartheta (c_1+d_1)+\vartheta (i\mathfrak{I})\vartheta (c_2+d_2)\hfill \\ \hfill & =\left(\vartheta (c_1)+\vartheta (i\mathfrak{I})\vartheta (c_2))+(\vartheta (d_1)+\vartheta (i\mathfrak{I})\vartheta (d_2)\right)\hfill \\ \hfill & =\vartheta (c_1+i{c}_2)+\vartheta (d_1+i{d}_2)\hfill \\ \hfill & =\vartheta (c)+\vartheta (d).\hfill \end{array}$$

Hence, $\vartheta$ is additive. □

Proof of our main Theorem. 

It is evident that we only need to establish the necessity. With the additivity of $\vartheta$, which we established in earlier lemmas, we will now finalize the proof of our main result by verifying the remaining lemma. □

Lemma 11. 

For any $s,k\in \mathfrak{E}$, we have $\vartheta (s^{\ast })=\vartheta {(s)}^{\ast }$ and $\vartheta (sk)=\vartheta (s)\vartheta (k)$.

Proof. 

Let $s=s_1+i{s}_2\in \mathfrak{E}$, for $s_1,s_2\in {\mathfrak{E}}^s$, and using Lemmas 3 and 10, we have

$$\begin{array}{cc}\hfill \vartheta (s^{\ast })& =\vartheta (s_1-i{s}_2)\hfill \\ \hfill & =\vartheta (s_1)-\vartheta (i\mathfrak{I})\vartheta (s_2)\hfill \\ \hfill & ={\left(\vartheta (s_1)+\vartheta (i\mathfrak{I})\vartheta (s_2)\right)}^{\ast }\hfill \\ \hfill & =\vartheta {(s_1+i{s}_2)}^{\ast }\hfill \\ \hfill & =\vartheta {(s)}^{\ast }.\hfill \end{array}$$

Further, we have

$$\begin{array}{cc}\hfill \vartheta (is)& =\vartheta \left(i(s_1+i{s}_2)\right)\hfill \\ \hfill & =\vartheta (i{s}_1-s_2)\hfill \\ \hfill & =\vartheta (i\mathfrak{I})\vartheta (s_1)-\vartheta (s_2)\hfill \\ \hfill & =\vartheta (i\mathfrak{I})\left(\vartheta (s_1)+\vartheta (i\mathfrak{I})\vartheta (s_2)\right)\hfill \\ \hfill & =\vartheta (i\mathfrak{I})\vartheta (s).\hfill \end{array}$$

Now, for any $s,k\in \mathfrak{E}$, we have

$$\begin{array}{cc}\hfill 2^{n-2}\vartheta (i\mathfrak{I})\vartheta (s^{\ast }k-k^{\ast }s)& =\vartheta (2^{n-2}i(s^{\ast }k-k^{\ast }s))\hfill \\ \hfill & =\vartheta \left(s★ik★\mathfrak{I}★\cdots ★\mathfrak{I}\right)\hfill \\ \hfill & =\vartheta (s)★\vartheta (ik)★\mathfrak{I}★\cdots ★\mathfrak{I}\hfill \\ \hfill & =2^{n-2}\vartheta (i\mathfrak{I})\left(\vartheta {(s)}^{\ast }\vartheta (k)-\vartheta {(k)}^{\ast }\vartheta (s)\right).\hfill \end{array}$$

Thus, $\vartheta (s^{\ast }k-k^{\ast }s)=\vartheta {(s)}^{\ast }\vartheta (k)-\vartheta {(k)}^{\ast }\vartheta (s)$.

Alternatively, we also have

$$\begin{array}{cc}\hfill 2^{n-2}\vartheta (s^{\ast }k+k^{\ast }s)& =\vartheta (2^{n-2}(s^{\ast }k+k^{\ast }s))\hfill \\ \hfill & =\vartheta \left(s★k★\mathfrak{I}★\cdots ★\mathfrak{I}\right)\hfill \\ \hfill & =\vartheta (s)★\vartheta (k)★\mathfrak{I}★\cdots ★\mathfrak{I}\hfill \\ \hfill & =2^{n-2}\left(\vartheta {(s)}^{\ast }\vartheta (k)+\vartheta {(k)}^{\ast }\vartheta (s)\right).\hfill \end{array}$$

This implies that $\vartheta (s^{\ast }k+k^{\ast }s)=\vartheta {(s)}^{\ast }\vartheta (k)+\vartheta {(k)}^{\ast }\vartheta (s)$.

From the last two equations, we get $\vartheta (s^{\ast }k)=\vartheta {(s)}^{\ast }\vartheta (k)$. On replacing s with $s^{\ast }$, we get $\vartheta (sk)=\vartheta (s)\vartheta (k)$, for any $s,k\in \mathfrak{E}.$ Therefore, from Lemmas 10 $(ii)$ and 11, it follows that $\vartheta$ is a ∗-ring isomorphism. It is simple to prove that $\vartheta$ is a map preserving absolute value. According to Theorem 2.5 in [17], $\vartheta$ is either a conjugate linear ∗-isomorphism or a linear ∗- isomorphism.

• Case (II): If $\vartheta (\mathfrak{I})=-\mathfrak{I}$, then $\vartheta$ is either the negative of a conjugate linear ∗-isomorphism or the negative of a linear ∗- isomorphism.

Consider a map $\gamma :\mathfrak{E}\to \mathfrak{F}$ defined as $\gamma (a)=-\vartheta (a)$ for all $a\in \mathfrak{E}$. It is clear that $\gamma$ satisfies $\gamma ({\nu }_1★{\nu }_2★\cdots ★{\nu }_n)=\gamma ({\nu }_1)★\gamma ({\nu }_2)★\cdots ★\gamma ({\nu }_n)$ for all ${\nu }_i\in \mathfrak{E}(i=1,2,\dots ,n)$ and also $\gamma (\mathfrak{I})=\mathfrak{I}$. Based on the analysis of Case I, it becomes apparent that $\gamma$ must be either a conjugate linear ∗-isomorphism or a linear ∗-isomorphism. Consequently, $\vartheta$ is either the negative of a conjugate linear ∗-isomorphism or the negative of a linear ∗-isomorphism. □

3. Conclusions

In this paper, we characterized a nonlinear map $\vartheta$ preserving the bi-skew Jordan n-product between $\mathfrak{E}$ and $\mathfrak{F}$. In fact, we proved that such a map $\vartheta$ or −$\vartheta$ between factors is either a conjugate linear ∗-isomorphism or a linear ∗-isomorphism. One can further investigate the structure of nonlinear maps which preserve the bi-skew Jordan n-product on different algebras, such as prime ∗-algebras, $C^{\ast }$-algebras, ∗-algebras, etc.