Nonlinear Maps Preserving Mixed Product on Factors

Abstract

Let ${\mathcal {A}}$ and ${\mathcal {B}}$ be two factors with dim${\mathcal {A}}>4$. In this article, it is proved that a bijective map $\Phi : {\mathcal {A}}\rightarrow {\mathcal {B}}$ satisfies $\Phi ([A\bullet B, C])=[\Phi (A)\bullet \Phi (B), \Phi (C)]$ for all $A, B, C\in {\mathcal {A}}$ if and only if $\Phi$ is a linear $*$-isomorphism, or a conjugate linear $*$-isomorphism, or the negative of a linear $*$-isomorphism, or the negative of a conjugate linear $*$-isomorphism.

1 Introduction

Let ${\mathcal {A}}$ and ${\mathcal {B}}$ be two algebras. Recall that a map $\Phi : {\mathcal {A}}\rightarrow {\mathcal {B}}$ preserves product or is multiplicative if $\Phi (AB)=\Phi (A)\Phi (B)$ for all $A, B\in {\mathcal {A}}.$ The question of when a multiplicative map is additive was discussed in [16]. Motivated by this, many authors pay more attention to the maps on algebras preserving Lie product $[A, B]=AB-BA$ (for example, see [13,14,15, 17, 18]), or the skew Lie product $[A,B]_{*}=AB-BA^{*}$ (for example, see [1, 3, 7, 11]), or the Jordan $*$-product $A\bullet B=AB+BA^{*}$ (for example, see [4, 8, 12, 22]). These results show that the (skew) Lie product or Jordan $*$-product structure is enough to determine the algebraic structure.

Recently, maps preserving the products of the mixture of (skew) Lie product and Jordan $*$-product have received a fair amount of attention. For example, Yang and Zhang [19] studied the nonlinear maps preserving the mixed skew Lie triple product $[[A,B]_{*},C]$ on factors. Li et al. studied the nonlinear maps preserving the skew Lie triple product $[[A,B]_{*},C]_{*}$ (for example, see [6, 10]) and the Jordan triple $*$-product $A\bullet B\bullet C$ (for example, see [9, 21]) on von Neumann algebras. In the present paper, we will establish the structure of the nonlinear maps preserving the mixed product $[A\bullet B, C]$ on factors.

Let ${\mathbb {R}}$ and ${\mathbb {C}}$ denote, respectively the real field and complex field. A von Neumann algebra ${\mathcal {A}}$ is a weakly closed, self-adjoint algebra of operators on a Hilbert space ${\mathcal {H}}$ containing the identity operator I. ${\mathcal {A}}$ is a factor means that its center is ${\mathbb {C}}I$. It is well known that the factor ${\mathcal {A}}$ is prime, in the sense that $A{\mathcal {A}}B=0$ for $A, B\in {\mathcal {A}}$ implies either $A=0$ or $B=0$.

2 Additivity

The main result in this section is the following.

Theorem 2.1

Let ${\mathcal {A}}$ and ${\mathcal {B}}$ be two factors. Suppose that a bijective map $\Phi : {\mathcal {A}}\rightarrow {\mathcal {B}}$ satisfies $\Phi ([A\bullet B, C])=[\Phi (A)\bullet \Phi (B), \Phi (C)]$ for all $A, B, C\in {\mathcal {A}}.$ Then $\Phi$ is additive.

Proof

We will complete the proof by proving several claims. $\square$

Claim 1

$\Phi (0)=0$.

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

$$\begin{aligned} \Phi (0)=\Phi ([0\bullet 0, A])=[\Phi (0)\bullet \Phi (0), \Phi (A)]. \end{aligned}$$

Since $\Phi$ is surjective, there exists $A\in {\mathcal {A}}$ such that $\Phi (A)=0.$ So $\Phi (0)=0.$

Choose an arbitrary nontrivial projection $P_{1}\in {\mathcal {A}}$, write $P_{2}=I-P_{1}.$ Denote ${\mathcal {A}}_{ij}=P_{i}{\mathcal {A}}P_{j}, i,j=1,2.$ Then ${\mathcal {A}}=\sum ^{2}_{i,j=1}{\mathcal {A}}_{ij}.$ For every $A\in {\mathcal {A}},$ we may write $A=\sum ^{2}_{i,j=1}A_{ij}.$ In all that follows, when we write $A_{ij},$ it indicates that $A_{ij}\in {\mathcal {A}}_{ij}$.

Claim 2

For every $A_{12}\in {\mathcal {A}}_{12}, B_{21}\in {\mathcal {A}}_{21},$ we have

$$\begin{aligned} \Phi (A_{12}+B_{21})=\Phi (A_{12})+\Phi (B_{21}). \end{aligned}$$

Choose $T=\sum ^{2}_{i,j=1}T_{ij}\in {\mathcal {A}}$ such that

$$\begin{aligned} \Phi (T)=\Phi (A_{12}+B_{21})-\Phi (A_{12})-\Phi (B_{21}). \end{aligned}$$

Since $[A_{12}\bullet P_{1}, P_{1}]=[B_{21}\bullet P_{2}, P_{2}]=0$, it follows from Claim 1 that

$$\begin{aligned} {[}\Phi (A_{12}+B_{21})\bullet \Phi (P_{1}), \Phi (P_{1})]&=\Phi ([(A_{12} +B_{21})\bullet P_{1}, P_{1}])\\ {}&=\Phi ([B_{21}\bullet P_{1}, P_{1}])\\&=\Phi ([A_{12}\bullet P_{1}, P_{1}])+\Phi ([B_{21}\bullet P_{1}, P_{1}])\\&=[(\Phi (A_{12})+\Phi (B_{21}))\bullet \Phi (P_{1}), \Phi (P_{1})] \end{aligned}$$

and

$$\begin{aligned} {[}\Phi (A_{12}+B_{21})\bullet \Phi (P_{2}), \Phi (P_{2})]&=\Phi ([(A_{12}+B_{21}) \bullet P_{2}, P_{2}])\\ {}&=\Phi ([A_{12}\bullet P_{2}, P_{2}])\\&=\Phi ([A_{12}\bullet P_{2}, P_{2}])+\Phi ([B_{21}\bullet P_{2}, P_{2}])\\&=[(\Phi (A_{12})+\Phi (B_{21}))\bullet \Phi (P_{2}), \Phi (P_{2})]. \end{aligned}$$

Thus $\Phi ([T\bullet P_{1}, P_{1}])=[\Phi (T)\bullet \Phi (P_{1}), \Phi (P_{1})]=0$ and $\Phi ([T\bullet P_{2}, P_{2}])=[\Phi (T)\bullet \Phi (P_{2}), \Phi (P_{2})]=0$. Then $[T\bullet P_{1}, P_{1}]=[T\bullet P_{2}, P_{2}]=0,$ and so $T_{12}=T_{21}=0.$

For every $C_{kl}\in {\mathcal {A}}_{kl},1\le k\ne l\le 2$, it follows from $[C_{kl}\bullet A_{12}, P_{k}]=[C_{kl}\bullet B_{21}, P_{k}]=0$ that

$$\begin{aligned} {[}\Phi (C_{kl})\bullet \Phi (A_{12}+B_{21}), \Phi (P_{k})]&=\Phi ([C_{kl}\bullet (A_{12}+B_{21}), P_{k}])\\&=\Phi ([C_{kl}\bullet A_{12}, P_{k}])+\Phi ([C_{kl}\bullet B_{21}, P_{k}])\\&=[\Phi (C_{kl})\bullet (\Phi (A_{12})+\Phi (B_{21})), \Phi (P_{k})]. \end{aligned}$$

Thus $\Phi ([C_{kl}\bullet T, P_{k}])=[\Phi (C_{kl})\bullet \Phi (T), \Phi (P_{k})]=0,$ and so $[C_{kl}\bullet T, P_{k}]=0,$ which implies that $C_{kl}T_{ll}=0$ for every $C_{kl}\in {\mathcal {A}}_{kl}$. Note that ${\mathcal {A}}$ is prime, it follows that $T_{ll}=0, l=1,2$. Hence $T=0.$ Now we have proved that $\Phi (A_{12}+B_{21})=\Phi (A_{12})+\Phi (B_{21}).$

Claim 3

For every $A_{11}\in {\mathcal {A}}_{11}, B_{12}\in {\mathcal {A}}_{12}, C_{21}\in {\mathcal {A}}_{21}, D_{22}\in {\mathcal {A}}_{22} ,$ we have

$$\begin{aligned} \Phi (A_{11}+B_{12}+C_{21}+D_{22})=\Phi (A_{11})+\Phi (B_{12})+\Phi (C_{21})+\Phi (D_{22}). \end{aligned}$$

Let $T=\sum ^{2}_{i,j=1}T_{ij}\in {\mathcal {A}}$ be such that

$$\begin{aligned} \Phi (T)=\Phi (A_{11}+B_{12}+C_{21}+D_{22})-\Phi (A_{11})-\Phi (B_{12}) -\Phi (C_{21})-\Phi (D_{22}). \end{aligned}$$

It follows from Claim 2 that

$$\begin{aligned}&[\Phi (P_{1})\bullet \Phi (A_{11}+B_{12}+C_{21}+D_{22}),\Phi (P_{2})]\\&=\Phi ([P_{1}\bullet (A_{11}+B_{12}+C_{21}+D_{22}),P_{2}])\\&=\Phi ([P_{1}\bullet (B_{12}+C_{21}),P_{2}])\\&=\Phi ([P_{1}\bullet (B_{12}+C_{21}),P_{2}])+\Phi ([P_{1} \bullet A_{11},P_{2}])+\Phi ([P_{1} \bullet D_{22},P_{2}])\\&=[\Phi (P_{1})\bullet (\Phi (B_{12})+\Phi (C_{21})),\Phi (P_{2})]\\&\quad + [\Phi (P_{1})\bullet \Phi (A_{11}),\Phi (P_{2})]+[\Phi (P_{1})\bullet \Phi (D_{22}), \Phi (P_{2})]\\&=[\Phi (P_{1})\bullet (\Phi (A_{11})+\Phi (B_{12}) + \Phi (C_{21})+\Phi (D_{22})),\Phi (P_{2})]. \end{aligned}$$

This implies that $[P_{1} \bullet T,P_{2}]=0$. Thus $T_{12}=T_{21}=0.$ For every $E_{ij}\in {\mathcal {A}}_{ij}(i\ne j)$, we have

$$\begin{aligned}&[\Phi (E_{12})\bullet \Phi (A_{11}+B_{12}+C_{21}+D_{22}),\Phi (P_{1})]\\&=\Phi ([E_{12}\bullet (A_{11}+B_{12}+C_{21}+D_{22}),P_{1}])\\&=\Phi ([E_{12}\bullet D_{22},P_{1}])\\ {}&=\Phi ([E_{12}\bullet D_{22},P_{1}])+\Phi ([E_{12}\bullet A_{11},P_{1}])\\&\quad +\Phi ([E_{12}\bullet B_{12},P_{1}])+\Phi ([E_{12}\bullet C_{21},P_{1}])\\&=[\Phi (E_{12})\bullet (\Phi (A_{11})+\Phi (B_{12})+\Phi (C_{21}) +\Phi (D_{22})),\Phi (P_{1})] \end{aligned}$$

and

$$\begin{aligned}&[\Phi (E_{21})\bullet \Phi (A_{11}+B_{12}+C_{21}+D_{22}),\Phi (P_{2})]\\&=\Phi ([E_{21}\bullet (A_{11}+B_{12}+C_{21}+D_{22}),P_{2}])\\&=\Phi ([E_{21}\bullet A_{11},P_{2}])\\ {}&=\Phi ([E_{21}\bullet D_{22},P_{2}])+\Phi ([E_{21}\bullet A_{11},P_{2}])\\&\quad +\Phi ([E_{21}\bullet B_{12},P_{2}])+\Phi ([E_{21}\bullet C_{21},P_{2}])\\&=[\Phi (E_{21})\bullet (\Phi (A_{11})+\Phi (B_{12})+\Phi (C_{21}) +\Phi (D_{22})),\Phi (P_{2})]. \end{aligned}$$

Then $[E_{ij}\bullet T,P_{i}]=0$, and so $T_{11}=T_{22}=0.$ Hence $T=0.$ It follows that $\Phi (A_{11}+B_{12}+C_{21}+D_{22})=\Phi (A_{11})+\Phi (B_{12}) +\Phi (C_{21})+\Phi (D_{22}).$

Claim 4

For every $A_{ij}\in {\mathcal {A}}_{ij}, B_{ij}\in {\mathcal {A}}_{ij}, 1\le i\ne j\le 2,$ we have

$$\begin{aligned} \Phi (A_{ij}+B_{ij})=\Phi (A_{ij})+\Phi (B_{ij}). \end{aligned}$$

It follows from $A_{ij}+B_{ij}=[\frac{I}{2}\bullet (P_{i}+A_{ij}),P_{j}+B_{ij}]$ and Claim 3 that

$$\begin{aligned}&\Phi (A_{ij}+B_{ij})=\Phi ([\frac{I}{2}\bullet (P_{i}+A_{ij}),P_{j}+B_{ij}])\\&=[\Phi (\frac{I}{2})\bullet \Phi (P_{i}+A_{ij}),\Phi (P_{j}+B_{ij})]\\&=[\Phi (\frac{I}{2})\bullet (\Phi (P_{i})+\Phi (A_{ij})),\Phi (P_{j})+\Phi (B_{ij})]\\&=\Phi ([\frac{I}{2}\bullet P_{i},P_{j}])+\Phi ([\frac{I}{2}\bullet P_{i},B_{ij}]) +\Phi ([\frac{I}{2}\bullet A_{ij},P_{j}])+\Phi ([\frac{I}{2}\bullet A_{ij},B_{ij}])\\&=\Phi (A_{ij})+\Phi (B_{ij}). \end{aligned}$$

Claim 5

For every $A_{ii},B_{ii}\in {\mathcal {A}}_{ii},i=1,2,$ we have

$$\begin{aligned} \Phi (A_{ii}+B_{ii})=\Phi (A_{ii})+\Phi (B_{ii}). \end{aligned}$$

Let

$$\begin{aligned} \Phi (T)=\Phi (A_{ii}+B_{ii})-\Phi (A_{ii})-\Phi (B_{ii}). \end{aligned}$$

It is clear that

$$\begin{aligned} {[}\Phi (P_{i})\bullet \phi (A_{ii}+ B_{ii}),\phi (P_{i})]&=\Phi ([P_{i}\bullet (A_{ii}+ B_{ii}),P_{i}])\\&=\Phi ([P_{i}\bullet A_{ii},P_{i}])+\Phi ([P_{i}\bullet B_{ii},P_{i}])\\&=[\Phi (P_{i})\bullet (\Phi (A_{ii})+ \Phi (B_{ii})),\Phi (P_{i})]. \end{aligned}$$

Thus $[P_{i}\bullet T,P_{i}]=0,$ which implies that $T_{12}=T_{21}=0.$

For every $C_{ji}\in {\mathcal {A}}_{ij},j\ne i,$ it follows from Claims 3 and 4 that

$$\begin{aligned} {[}\Phi (C_{ji})\bullet \Phi (A_{ii}+B_{ii}),\Phi (P_{i})]&=\Phi ([C_{ji}\bullet (A_{ii}+B_{ii}),P_{i}])\\&=\Phi (C_{ji}A_{ii})+\Phi (C_{ji}B_{ii})-\Phi (A_{ii}C_{ji}^{*})-\Phi (B_{ii}C_{ji}^{*})\\&=\Phi ([C_{ji}\bullet A_{ii},P_{i}])+\Phi ([C_{ji}\bullet B_{ii},P_{i}])\\&=[\Phi (C_{ji})\bullet \Phi ( A_{ii}),\Phi (P_{i})] +[\Phi (C_{ji})\bullet \Phi (B_{ii}),\Phi (P_{i})]\\&=[\Phi (C_{ji})\bullet (\Phi ( A_{ii})+\Phi ( B_{ii})),\Phi (P_{i})] \end{aligned}$$

and

$$\begin{aligned} {[}\Phi (C_{ij})\bullet \Phi (A_{ii}+B_{ii}),\Phi (P_{i})]&=\Phi ([C_{ij}\bullet (A_{ii}+B_{ii}),P_{i}])\\&=\Phi ([C_{ij}\bullet A_{ii},P_{i}])+\Phi ([C_{ij}\bullet B_{ii},P_{i}])\\&=[\Phi (C_{ij})\bullet \Phi ( A_{ii}),\Phi (P_{i})]+[\Phi (C_{ij})\bullet \Phi (B_{ii}),\Phi (P_{i})]\\&=[\Phi (C_{ij})\bullet (\Phi ( A_{ii})+\Phi ( B_{ii})),\Phi (P_{i})]. \end{aligned}$$

Then $[C_{ji}\bullet T,P_{i}]=[C_{ij}\bullet T,P_{i}]=0,$ and so $T_{11}=T_{22}=0.$ Hence $T=0.$ It follows that $\Phi (A_{ii}+B_{ii})=\Phi (A_{ii})+\Phi (B_{ii}).$

Claim 6

$\Phi$ is additive.

Let $A=\sum ^{2}_{i,j=1}A_{ij}, B=\sum ^{2}_{i,j=1}B_{ij}\in {\mathcal {A}}.$ By Claims 3, 4 and 5, we have

$$\begin{aligned} \Phi (A+B)&=\Phi \Bigg (\sum ^{2}_{i,j=1}A_{ij}+\sum ^{2}_{i,j=1}B_{ij}\Bigg ) =\Phi \Bigg (\sum ^{2}_{i,j=1}(A_{ij}+B_{ij})\Bigg )\\&=\sum ^{2}_{i,j=1}\Phi (A_{ij}+B_{ij})=\sum ^{2}_{i,j=1}\Phi (A_{ij}) +\sum ^{2}_{i,j=1}\Phi (B_{ij})\\&=\Phi \Bigg (\sum ^{2}_{i,j=1}A_{ij})+\Phi (\sum ^{2}_{i,j=1}B_{ij}\Bigg )=\Phi (A)+\Phi (B). \end{aligned}$$

3 Main Result

Our main result in this paper reads as follows.

Theorem 3.1

Let ${\mathcal {A}}$ and ${\mathcal {B}}$ be two factors with dim${\mathcal {A}}>4$. Then a bijective map $\Phi : {\mathcal {A}}\rightarrow {\mathcal {B}}$ satisfies $\Phi ([A\bullet B, C])=[\Phi (A)\bullet \Phi (B), \Phi (C)]$ for all $A, B, C\in {\mathcal {A}}$ if and only if $\Phi$ is a linear $*$-isomorphism, or a conjugate linear $*$-isomorphism, or the negative of a linear $*$-isomorphism, or the negative of a conjugate linear $*$-isomorphism.

Proof

Clearly, we only need prove the necessity. By Theorem 2.1, we obtain the additivity of $\Phi$. Now we will complete the proof of main theorem by proving several steps.

Step 1. $\Phi ({\mathbb {C}}I)={\mathbb {C}}I$.

Let $B\in {\mathcal {A}}$ such that $\Phi (B)=I.$ Then

$$\begin{aligned} 0=\Phi ([B\bullet C, \lambda I])=[\Phi (B)\bullet \Phi (C), \Phi (\lambda I)]=2[\Phi (C), \Phi (\lambda I)] \end{aligned}$$

for all $C\in {\mathcal {A}}$ and $\lambda \in {\mathbb {C}}.$ It follows from the surjectivity of $\Phi$ that $\Phi (\lambda I)\in {\mathbb {C}}I$, and then $\Phi ({\mathbb {C}}I)\subseteq {\mathbb {C}}I$. By considering $\Phi ^{-1}$, we can obtain that $\Phi ({\mathbb {C}}I)={\mathbb {C}}I$. $\square$

Step 2. For all $A, B\in {\mathcal {A}}$, $[\Phi (A), \Phi (B)]=0$ if and only if $[A,B]=0.$

It follows from $\Phi (I)\in {\mathbb {C}}I$ and the additivity of $\Phi$ that

$$\begin{aligned} 2\Phi ([A, B])&=\Phi (2[A, B])=\Phi ([I\bullet A,B])\\&=[\Phi (I)\bullet \Phi (A),\Phi (B)]\\ {}&=(\Phi (I)+\Phi (I)^{*})[\Phi (A),\Phi (B)] \end{aligned}$$

for all $A, B\in {\mathcal {A}}$. If $\Phi (I)+\Phi (I)^{*}=0$, then $\Phi ([A, B])=0,$ and so $[A,B]=0$ for all $A, B\in {\mathcal {A}}$. This contradiction implies that $\Phi (I)+\Phi (I)^{*}\ne 0$. Hence $[\Phi (A), \Phi (B)]=0$ if and only if $[A,B]=0.$ $\square$

It follows from Step 2 that $\Phi$ is an additive bijection that preserves commutativity in both directions. Hence by [2, Corollary 3.8]

$$\begin{aligned} \Phi (A)=a\theta (A)+\xi (A) \end{aligned}$$

for all $A\in {\mathcal {A}},$ where $a\in {\mathbb {C}}$ is a nonzero scalar, $\theta :{\mathcal {A}}\rightarrow {\mathcal {B}}$ is an additive Jordan isomorphism, and $\xi :{\mathcal {A}}\rightarrow {\mathbb {C}}I$ is an additive map. It is easy to check that $\theta (iI)=\pm iI.$

Step 3. For every $A,B\in {\mathcal {A}},$ we have

• (1)   $\Phi (iA)-\theta (iI)\Phi (A)\in {\mathbb {C}}I,$

• (2)   $\Phi ([A,B])=\epsilon [\Phi (A),\Phi (B)],$ where $\epsilon \in \{1,-1\}.$

  • (1) Let $A\in {\mathcal {A}}.$ Then

    $$\begin{aligned} \Phi (iA)-\theta (iI)\Phi (A)&=a\theta (iA)+\xi (iA)-\theta (iI)\Phi (A)\\&=a\theta (iI)\theta (A)+\xi (iA)-\theta (iI)\Phi (A)\\&=\theta (iI)(a\theta (A)+\xi (A))+\xi (iA)-\theta (iI)\xi (A)-\theta (iI)\Phi (A)\\&=\xi (iA)-\theta (iI)\xi (A)\in {\mathbb {C}}I \end{aligned}$$

  • (2) It follows from Step 1 that $\frac{1}{2}(\Phi (I)+\Phi (I)^{*})=\epsilon I$ for some $\epsilon \in {\mathbb {C}}.$ By Step 2, we have

    $$\begin{aligned} \Phi ([A,B])=\frac{1}{2}(\Phi (I)+\Phi (I)^{*})[\Phi (A),\Phi (B)]=\epsilon [\Phi (A),\Phi (B)] \end{aligned}$$

    for all $A,B\in {\mathcal {A}}.$ For every $A\in {\mathcal {A}}$ with $A=-A^{*},$ we have

    $$\begin{aligned} {[}\Phi (A)\bullet \Phi (B),\Phi (C)]&=\Phi ([A\bullet B,C])=\Phi ([[A,B],C])\\&=\epsilon ^{2}[[\Phi (A),\Phi (B)],\Phi (C)] \end{aligned}$$

    for all $B,C\in {\mathcal {A}}.$ Thus

    $$\begin{aligned} (1-\epsilon ^{2})\Phi (A)\Phi (B)+\Phi (B)(\epsilon ^{2}\Phi (A)+\Phi (A)^{*}) \in {\mathbb {C}}I \end{aligned}$$

    (3.1)

    for all $B\in {\mathcal {A}}$ and $A\in {\mathcal {A}}$ with $A=-A^{*}.$ Let $P_{1}\in {\mathcal {B}}$ be a nontrivial projection. Then there exists $D\in {\mathcal {A}}$ such that $\Phi (D)=P_{1}$. Taking $B=D$ in (3.1), we have

    $$\begin{aligned} (1-\epsilon ^{2})\Phi (A)P_{1}+P_{1}(\epsilon ^{2}\Phi (A)+\Phi (A)^{*})\in {\mathbb {C}}I. \end{aligned}$$

    This yields

    $$\begin{aligned} (1-\epsilon ^{2})P_{2}\Phi (A)P_{1}=0 \end{aligned}$$

    (3.2)

    for all $A\in {\mathcal {A}}$ with $A=-A^{*},$ where $P_{2}=I-P_{1}.$ Then by assertion (1) and (3.2),

    $$\begin{aligned} (1-\epsilon ^{2})P_{2}\Phi (iB)P_{1}=0 \end{aligned}$$

    (3.3)

    for all $B\in {\mathcal {A}}$ with $B=-B^{*}.$ It follows from (3.2) and (3.3) that

    $$\begin{aligned} (1-\epsilon ^{2})P_{2}\Phi (C)P_{1}=0 \end{aligned}$$

    for all $C\in {\mathcal {A}}.$ Hence $\epsilon \in \{1,-1\}.$

$\square$

Remark 3.2

Let $\epsilon$ be as above, and let $\Psi =\epsilon \Phi .$ It follows from Theorem 2.1 and Step 3(2) that $\Psi :{\mathcal {A}}\rightarrow {\mathcal {B}}$ is an additive bijection and satisfies

$$\begin{aligned} \Psi ([A\bullet B, C])=[\Psi (A)\bullet \Psi (B), \Psi (C)] \end{aligned}$$

and

$$\begin{aligned} \Psi ([A,B])=[\Psi (A),\Psi (B)] \end{aligned}$$

for all A, B, C in ${\mathcal {A}}.$ Hence by [20, Theorem 2.1], there exists an additive map $f:{\mathcal {A}}\rightarrow {\mathbb {C}}I$ with $f([A,B])=0$ for all $A,B\in {\mathcal {A}}$ such that one of the following statements holds:

• (1) $\Psi (A)=\varphi (A)+f(A)$ for all A in ${\mathcal {A}},$ where $\varphi :{\mathcal {A}}\rightarrow {\mathcal {B}}$ is an additive isomorphism;

• (2) $\Psi (A)=-\varphi (A)+f(A)$ for all A in ${\mathcal {A}},$ where $\varphi :{\mathcal {A}}\rightarrow {\mathcal {B}}$ is an additive anti-isomorphism.

Step 4. Statement (2) of Remark 3.2 does not occur.

Indeed, if $\Psi =-\varphi +f,$ where $\varphi :{\mathcal {A}}\rightarrow {\mathcal {B}}$ is an additive anti-isomorphism and $f:{\mathcal {A}}\rightarrow {\mathbb {C}}I$ is an additive map with $f([A,B])=0$ for all $A,B\in {\mathcal {A}},$ then

$$\begin{aligned} \Psi ([A\bullet B,C])=-\varphi ([A\bullet B,C])=[\varphi (B)\varphi (A)+\varphi (A^{*})\varphi (B),\varphi (C)] \end{aligned}$$

for all A, B, C in ${\mathcal {A}}.$ On the other hand,we have

$$\begin{aligned} \Psi ([A\bullet B,C])&=[\Psi (A)\bullet \Psi (B),\Psi (C)]\\&=[(-\varphi (A)+f(A))\bullet (-\varphi (B)+f(B)),(-\varphi (C)+f(C))]\\&=[(\varphi (A)-f(A))\bullet (-\varphi (B)+f(B)),\varphi (C)]\\&=[\varphi (A)\bullet (-\varphi (B))+\varphi (A)\bullet f(B)+f(A)\bullet \varphi (B),\varphi (C)]. \end{aligned}$$

It follows from the surjectivity of $\varphi$ that

$$\begin{aligned}&(\varphi (A^{*})+\varphi (A))\varphi (B)+(\varphi (B)-f(B))(\varphi (A)^{*}\nonumber \\&\quad +\varphi (A))+(f(A)+f(A)^{*})\varphi (B)\in {\mathbb {C}}I \end{aligned}$$

(3.4)

for all $A,B\in {\mathcal {A}}.$ Let $P\in {\mathcal {A}}$ be a nontrivial projection. Then $\varphi (P)$ is a nontrivial idempotent in ${\mathcal {B}}.$ Taking $B=P$ in (3.4) and multiplying (3.4) on the right-hand side by $\varphi (P^{\bot })$ and on the left-hand side by $\varphi (P),$ we get

$$\begin{aligned} (I-f(P))\varphi (P)(\varphi (A)^{*}+\varphi (A))\varphi (P^{\bot })=0 \end{aligned}$$

(3.5)

for all $A\in {\mathcal {A}}.$ Replacing $\varphi (A)$ by $i\varphi (A)$ in (3.5), we have

$$\begin{aligned} (I-f(P))\varphi (P)(\varphi (A)^{*}-\varphi (A))\varphi (P^{\bot })=0. \end{aligned}$$

(3.6)

It follows from (3.5) and (3.6) that

$$\begin{aligned} (I-f(P))\varphi (P)\varphi (A)\varphi (P^{\bot })=0 \end{aligned}$$

for all $A\in {\mathcal {A}}.$ Hence $f(P)=I$ for any nontrivial projection $P\in {\mathcal {A}}.$ Taking $B=P$ in (3.4) and multiplying (3.4) on the right-hand side by $\varphi (P^{\bot })$, we have

$$\begin{aligned} \varphi (P^{\bot })(\varphi (A)^{*}+\varphi (A))\varphi (P^{\bot })\in {\mathbb {C}} \varphi (P^{\bot }) \end{aligned}$$

(3.7)

for all $A\in {\mathcal {A}}$ and any nontrivial projection $P\in {\mathcal {A}}.$ Replacing $\varphi (A)$ by $i\varphi (A)$ in (3.7), we can obtain that

$$\begin{aligned} \varphi (P^{\bot }AP^{\bot })=\varphi (P^{\bot })\varphi (A)\varphi (P^{\bot }) \in {\mathbb {C}}\varphi (P^{\bot })=\varphi ({\mathbb {C}}P^{\bot }) \end{aligned}$$

for all $A\in {\mathcal {A}}$ and any nontrivial projection $P\in {\mathcal {A}}.$ This implies that

$$\begin{aligned} P^{\bot }{\mathcal {A}}P^{\bot }={\mathbb {C}}P^{\bot },P{\mathcal {A}}P={\mathbb {C}}P \end{aligned}$$

for any nontrivial projection $P\in {\mathcal {A}}.$ It follows that ${\mathcal {A}}$ is isomorphic to $M_{2}({\mathbb {C}}),$ the algebra of all $2\times 2$ matrices over ${\mathbb {C}},$ which contradicts the assumption that dim${\mathcal {A}}>4$.

$\square$

Step 5. $\Psi$ is an additive *-isomorphism .

By Step 4, now we obtain that $\Psi =\varphi +f,$ where $\varphi :{\mathcal {A}}\rightarrow {\mathcal {B}}$ is an additive isomorphism and $f:{\mathcal {A}}\rightarrow {\mathbb {C}}I$ is an additive map with $f([A,B])=0$ for all $A,B\in {\mathcal {A}}.$ Thus

$$\begin{aligned} \Psi ([A\bullet B,C])=\varphi ([A\bullet B,C])=[\varphi (A)\varphi (B)+\varphi (B)\varphi (A^{*}),\varphi (C)] \end{aligned}$$

for all $A,B,C\in {\mathcal {A}}.$ On the other hand, we have

$$\begin{aligned} \Psi ([A\bullet B,C])&=[\Psi (A)\bullet \Psi (B),\Psi (C)]\\&=[(\varphi (A)+f(A))\bullet (\varphi (B)+f(B)),(\varphi (C)+f(C))]\\&=[(\varphi (A)+f(A))\bullet (\varphi (B)+f(B)),\varphi (C)]\\&=[\varphi (A)\bullet \varphi (B)+f(A)\bullet \varphi (B)+\varphi (A)\bullet f(B),\varphi (C)]. \end{aligned}$$

It follows from the surjectivity of $\varphi$ that

$$\begin{aligned} \varphi (B)(\varphi (A)^{*}-\varphi (A^{*}))+\varphi (B)(f(A)^{*}+f(A)) +f(B)(\varphi (A)^{*}+\varphi (A))\in {\mathbb {C}}I\nonumber \\ \end{aligned}$$

(3.8)

for all $A,B\in {\mathcal {A}}.$ Let $\lambda \in {\mathbb {C}},$ and let $P\in {\mathcal {A}}$ be a nontrivial projection. Multiplying (3.8) on the left-hand side by $\varphi (P^{\bot })$ and on the right-hand side by $\varphi (P),$ and then taking $B=\lambda P,$ we have

$$\begin{aligned} f(\lambda P)\varphi (P^{\bot })(\varphi (A)^{*}+\varphi (A))\varphi (P)=0 \end{aligned}$$

(3.9)

for all $A\in {\mathcal {A}}.$ Similarly, we can obtain from (3.9) that

$$\begin{aligned} f(\lambda P)\varphi (P^{\bot })\varphi (A)\varphi (P)=0 \end{aligned}$$

for all $A\in {\mathcal {A}}.$ Then $f(\lambda P)=0$ for all $\lambda \in {\mathbb {C}}$ and any nontrivial projection $P\in {\mathcal {A}}.$ This yields that

$$\begin{aligned} f(\lambda I)=f(\lambda P)+f(\lambda P^{\bot })=0 \end{aligned}$$

for all $\lambda \in {\mathbb {C}}.$ Since every A in ${\mathcal {A}}$ can be written as a finite linear combination of projections in ${\mathcal {A}}$ (see [5]), it follows that $f(A)=0$ for all $A\in {\mathcal {A}}$. Now (3.8) becomes

$$\begin{aligned} \varphi (B)(\varphi (A)^{*}-\varphi (A^{*}))\in {\mathbb {C}}I \end{aligned}$$

(3.10)

for all $A,B\in {\mathcal {A}}.$ In particular, $\varphi (A)^{*}-\varphi (A^{*})\in {\mathbb {C}}I$ for all $A\in {\mathcal {M}}.$ If $\varphi (A)^{*}-\varphi (A^{*})\ne 0$ for some $A\in {\mathcal {A}},$ then by (3.10), $\varphi (B)\in {\mathbb {C}}I$ for all $B\in {\mathcal {A}}.$ This contradiction implies that $\varphi (A)^{*}=\varphi (A^{*})$ for all $A\in {\mathcal {A}}.$ Hence $\Psi =\varphi$ is an additive $*$-isomorphism. $\square$

Step 6. $\Phi$ is a linear $*$-isomorphism, or a conjugate linear $*$-isomorphism, or the negative of a linear $*$-isomorphism, or the negative of a conjugate linear $*$-isomorphism.

By Step 5, it is easy to check that $\Psi (iI)=\pm iI$ and $\Psi (qI)=qI$ for every rational number q. Let A be a positive element in ${\mathcal {A}}$. Then $A=B^{2}$ for some self-adjoint element $B\in {\mathcal {A}}$. It follows from Step 5 that $\Psi (A)=\Psi (B)^{2}$ and $\Psi (B)$ is self-adjoint. So $\Psi (A)$ is positive. This shows that $\Psi$ preserves positive elements. Let $\lambda \in {\mathbb {R}}$ be any real number. Choose sequences $\{a_{n}\}$ and $\{b_{n}\}$ of rational numbers such that $a_{n}\le \lambda \le b_{n}$ for all n and $\lim _{n\rightarrow \infty }a_{n}= \lim _{n\rightarrow \infty }b_{n}=\lambda .$ It follows from

$$\begin{aligned} a_{n}I\le \lambda I \le b_{n}I \end{aligned}$$

that

$$\begin{aligned} a_{n}I\le \Psi (\lambda I) \le b_{n}I. \end{aligned}$$

Taking the limit, we get that $\Psi (\lambda I)= \lambda I$. Hence for all $A\in {\mathcal {A}}$,

$$\begin{aligned} \Psi (\lambda A)= \Psi ((\lambda I)A) =\Psi (\lambda I)\Psi (A)=\lambda \Psi (A). \end{aligned}$$

Hence $\Psi$ is real linear. It follows from $\Psi (iI)=\pm iI$ that $\Psi$ is linear or conjugate linear. Since $\Phi =\epsilon \Psi , \epsilon \in \{1,-1\}$, now we can obtain that $\Phi$ is a linear $*$-isomorphism, or a conjugate linear $*$-isomorphism, or the negative of a linear $*$-isomorphism, or the negative of a conjugate linear $*$-isomorphism. $\square$

From Theorem 3.1 and the fact that every ring isomorphism between type I factors is spatial, we have the following corollary.

Corollary 3.3

Let ${\mathcal {A}}$ and ${\mathcal {B}}$ be two type I factors acting on a complex Hilbert spaces ${\mathcal {H}}$ with dim${\mathcal {H}}>2$. Then a bijective map $\Phi : {\mathcal {A}}\rightarrow {\mathcal {B}}$ satisfies $\Phi ([A\bullet B, C])=[\Phi (A)\bullet \Phi (B), \Phi (C)]$ for all $A, B, C\in {\mathcal {A}}$ if and only if there exists $\epsilon \in \{1,-1\}$ such that $\Phi (A)=\epsilon UAU^{*}$ for all $A\in {\mathcal {A}}$, where U is a unitary or conjugate unitary operator.