Quasi-automorphisms on \mathbf{B}(\mathcal{X})

Abstract

Let \mathcal{X} be a Banach space with \dim \mathcal{X}\ge 2 and \text{B}(\mathcal{X}) be the Banach algebra of all bounded linear operators on \mathcal{X}, \mathrm{\forall }A,B\in \text{B}(\mathcal{X}). Define a quasi-product by A\circ B=A+B-AB, and (\text{B}(\mathcal{X}),\circ) is a semigroup. In this paper, we mainly discuss the quasi-product automorphisms on \text{B}(\mathcal{X}). It is proved that a bijective map \phi on \text{B}(\mathcal{X}) is a quasi-product automorphism if and only if \phi is a ring automorphism.

Keywords

Quasi-product / quasi-isomorphism / ring isomorphism

1 Introduction

In recent years, much attention has been paid to the characterization of maps preserving certain properties on operator algebras, which aims to find characteristics that serve as isomorphism invariants for operator algebras. In particular, based on non-associative algebraic operations on operator algebras like the Lie product and Jordan product, many scholars have studied algebraic isomorphisms and preservers related to these operations, such as Lie isomorphisms, Lie derivations, Jordan isomorphisms, and Jordan derivations [7-10]. On one hand, it is noted that concepts like the Lie product and Jordan product are new non-associative operations derived from the addition and multiplication of the ring. On the other hand, one can define an associative binary operation on a ring based on addition and multiplication--the quasi-product operation. Let \mathrm{R} be a ring. \mathrm{\forall }a,b\in \mathrm{R}, and define a\circ b=a+b-ab (or a\circ b=a+b+ab). The operation \circ is called the quasi-product on \mathrm{R}. It is seen that (\mathrm{R},\circ) is a semigroup whose identity element is the zero element 0 of \mathrm{R}. Let [a,b]=ab-ba be the Lie product of a,b. Then there is [a,b]=b\circ a-a\circ b, namely the Lie product [a,b] of a,b is the Lie product of b,a in the sense of the quasi-product. Therefore, the quasi-product is closely related to the Lie product. Meanwhile, it is known that the quasi-product plays a very important role in studying the structure of ring or algebra, such as the Jacobson radical [1, 4, 5] and algebraic Lie quasi-nilpotency [2, 6]. Operator algebras are a class of algebras with topological properties. The quasi-product semigroups on operator algebras form an important class of operator semigroups [3]. Furthermore, as semigroups, the characterization of their algebraic isomorphisms is a fundamental problem. What is the relationship between such isomorphisms and ring isomorphisms? This paper aims to characterize the quasi-product semigroup isomorphisms on a class of operator algebras, namely, the Banach algebra consisting of all bounded linear operators on Banach space.

The following shows some concepts and notations. Let \mathcal{X} be a complex Banach space, \dim \mathcal{X} denotes the dimension of \mathcal{X}, and {\mathcal{X}}^{\prime } refers to the dual space of \mathcal{X}. \text{B}(\mathcal{X}) and \mathrm{F}(\mathcal{X}) represent the Banach algebra of all bounded linear operators on \mathcal{X} and the set of all finite-rank operators, respectively. M_n is the space of all n\times n complex matrices. If \dim \mathcal{X}=n is a Banach space with finite dimensions, then \text{B}(\mathcal{X}) will be algebraically isomorphic to M_n. In this case, set \text{B}(\mathcal{X})=M_n. Let A\in \mathrm{B}(\mathrm{H}). If A^2=A, then A is called an idempotent operator. If there exists a positive integer n such that A^n=0, then A is called a nilpotent operator. For two idempotent operators P and Q, if PQ=QP=P, then there is P\le Q. If PQ=QP=0, thenPand Q are said to be orthogonal, denoted by P\perp Q. \mathrm{\forall }x\in \mathcal{X},f\in {\mathcal{X}}^{\prime }. x\otimes f denotes the rank-one operator defined by x\otimes f(z)=f(z)x,\mathrm{\forall }z\in \mathcal{X}. It is seen that x\otimes f is an idempotent operator if and only if f(x)=1, and a nilpotent operator if and only if f(x)=0. I_1(\mathcal{X}) denotes the set of all rank-one idempotent operators in \text{B}(\mathcal{X}). Furthermore, let S\subseteq \mathcal{X} be a subset of \mathcal{X}, [S] be the closed subspace generated by S, and M and N be closed subspaces of \mathcal{X}. If M\cap N=0 and M+N is a closed subspace, then M\oplus N=M+N is called the Banach direct sum of M and N. When no confusion arises, I usually denotes the identity operator on the Banach space.

2 Main results

Let \dim \mathcal{X}\ge 2, \text{B}(\mathcal{X}) be the Banach algebra of all bounded linear operators on \mathcal{X}, and \circ be the quasi-product on \text{B}(\mathcal{X}), namely A\circ B=A+B-AB,\mathrm{\forall }A,B\in \text{B}(\mathcal{X}), (\text{B}(\mathcal{X}),\circ) is a semigroup. Let \phi be a map on \text{B}(\mathcal{X}). If \phi (A\circ B)=\phi (A)\circ \phi (B),\mathrm{\forall }A,B\in \text{B}(\mathcal{X}), then \phi is called a quasi-homomorphism. If \phi is bijective and a quasi-homomorphism, then \phi is called a quasi-isomorphism.

Proposition 1.1　 Let \phi be a quasi-isomorphism on \text{B}(\mathcal{X}). Then \phi (I)=I and

(1) \phi preserves idempotent operators and rank-one idempotent operators in both directions;

(2) \phi preserves the order and orthogonality of idempotent operators in both directions;

(3) \phi preserves rank-one operators in both directions.

Proof　 \mathrm{\forall }A\in \text{B}(\mathcal{X}), since I=A\circ I=I\circ A, there is

\begin{array}{rl}\phi (I)& =\phi (A\circ I)=\phi (A)\circ \phi (I)=\phi (A)+\phi (I)-\phi (A)\phi (I)\\ & =\phi (I\circ A)=\phi (I)\circ \phi (A)=\phi (A)+\phi (I)-\phi (I)\phi (A),\end{array}

then \phi (A)=\phi (A)\phi (I)=\phi (I)\phi (A). Since \phi is surjective, it follows that \phi (I)=I.

(1) Let P\in \text{B}(\mathcal{X}) be an idempotent operator. Then P\circ P=P, so \phi (P)=\phi (P\circ P)=\phi (P)+\phi (P)-\phi (P{)}^2, namely \phi (P) is idempotent. Hence \phi preserves idempotent operators in both directions.

(2) It is first proved that \phi preserves the order and orthogonality of idempotent operators in both directions. Let P,Q\in \text{B}(\mathcal{X}). If P\le Q, then P\circ Q=Q\circ P=P+Q-PQ=Q. Therefore,

\begin{array}{rl}\phi (P)\circ \phi (Q)& =\phi (Q)\circ \phi (P)\\ & =\phi (P)+\phi (Q)-\phi (P)\phi (Q)\\ & =\phi (Q)+\phi (P)-\phi (Q)\phi (P)\\ & =\phi (Q),\end{array}

namely \phi (P)=\phi (Q)\phi (P)=\phi (P)\phi (Q). As a result, \phi \left(P\right)\le \phi \left(Q\right), and \phi preserves the order of idempotent operators. Since {\phi }^{-1} has the same property, \phi preserves the order of idempotent operators in both directions. If P\perp Q, it is seen that \phi (P)\phi (Q)=\phi (Q)\phi (P). Let E=\phi (P)\phi (Q)=\phi (Q)\phi (P). If E\ne 0, let F={\phi }^{-1}(E). Then E\le \phi (P) and E\le \phi (Q). As \phi preserves order, F\le P,F\le Q, then F=0, which is contradictory. Therefore, \phi preserves the orthogonality of operators. By the property of {\phi }^{-1}, \phi preserves the orthogonality of operators in both directions.

Finally, It requires to prove that \phi preserves rank-one idempotent operators in both directions. Let P be a rank-one idempotent operator on B(\mathrm{H}) and Q=\phi (P). If \mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}Q>1, then there exists a rank-one idempotent operator Q_1\in B(\mathrm{H}) such that . Then {\phi }^{-1}\left(Q_1\right)\le {\phi }^{-1}(Q)=P is an idempotent operator, and {\phi }^{-1}\left(Q_1\right)=P, namely Q_1=Q, which is contradictory. Hence Q is a rank-one operator. By the property of {\phi }^{-1}, \phi preserves rank-one idempotent operators in both directions.

(3) Let A=x\otimes f be a rank-one operator and \Vert x\Vert =1. If A is nilpotent, namely f(x)=0, take g\in {\mathcal{X}}^{\prime } such that g(x)=1. Then B=x\otimes g and A+B are rank-one idempotent operators. It is seen that A+B=(A+B)\circ A, and there is \phi (A+B)=\phi (A+B)\circ \phi (A)=\phi (A+B)+\phi (A)-\phi (A+B)\phi (A), namely \phi (A)=\phi (A+B)\phi (A) is a rank-one operator. If f(x)\ne 0, then A=aP, where a is a constant and P is a rank-one idempotent operator. Then from P=A\circ P, it can be seen that \phi (A)=\phi (A)\phi (P) is a rank-one operator. Therefore, \phi preserves rank-one operators. Similarly, {\phi }^{-1} also preserves rank-one operators.

Lemma 1.1　 Let \dim \mathcal{X}\ge 3 and \phi be a quasi-isomorphism on \text{B}(\mathcal{X}). If \phi (P)=P,\mathrm{\forall }P\in I_1(\mathcal{X}), then \phi (A)=A, \mathrm{\forall }A\in \text{B}(\mathcal{X}).

Proof　Since \phi is a semigroup isomorphism of (\mathrm{B}(\mathrm{H}),\circ ) and 0 is the semigroup identity, there is \phi (0)=0. By Proposition 1.1, it is known that \phi (P)=P for any idempotent operator P. The following shows the proof step by step.

(1) For any rank-one operator A, there is \phi (A)=A.

First, it is supposed that A is a rank-one nilpotent operator, namely A=x\otimes g, where g\in {\mathcal{X}}^{\prime } such that g(x)=0. Take any y\in \mathcal{X} such that g(y)=1, and choose a subspace M such that \ker g=[x]\oplus M. Then there is \mathcal{X}=[x]\oplus [y]\oplus M. Take f\in {\mathcal{X}}^{\prime } such that f(x)=1 and \ker f=[y]\oplus M. Let P_1=x\otimes f,P_2=y\otimes g, andP=P_1+P_2. Then P_1, P_2, and Pare idempotent operators, and P\circ A=A\circ P=P. Therefore, P\circ \phi (A)=\phi (A)\circ P=P. In particular, \phi (A)=P\phi (A)=\phi (A)P. On the other hand, P_1\circ A=P_1,P_2\circ A=P_2+A is a rank-one idempotent, so P_1\circ \phi (A)=P_1,P_2\circ \phi (A)=P_2+A, namely \phi (A)=P_1\phi (A),\phi (A)-P_2\phi (A)=A. As a result, \phi (A)=A.

Next, it is assumed that A=\lambda P,\lambda \in C, where P is a rank-one idempotent operator and \lambda \ne 0. Since \lambda P\circ P=P\circ \lambda P=P, there is\phi (\lambda P)=\phi (\lambda P)P=P\phi (\lambda P). Therefore, there exists a constant {\mathrm{h}}_P(\lambda )\in C such that \phi (\lambda P)={\mathit{h}}_P(\lambda )P. Let P=x\otimes f and \lambda \ne 1. Take N=x\otimes g as a rank-one idempotent operator, where x,f,g are as above. Then \lambda P\circ N=\lambda \left(x\otimes \left(f+\frac{1-\lambda }{\lambda }g\right)\right)=\lambda Q, where Q=x\otimes \left(f+\frac{1-\lambda }{\lambda }g\right) is a rank-one idempotent operator. Therefore, f_P(\lambda )P\circ N=f_Q(\lambda )Q, and f_P(\lambda )=f_Q(\lambda )=\lambda. Hence, \phi (\lambda P)=\lambda P.

(2) For any finite-rank operator A\in \mathrm{B}(\mathcal{X}),\phi (A)=A.

\mathrm{\forall }A\in \mathrm{F}(\mathcal{X}), it is known that there exist closed subspaces , such that \mathcal{X}=M\oplus N, and under this decomposition, A=B\oplus 0. It is noted that , so there exists a basis e_1,e_2,\dots ,e_n of M such that B can be written as an n×n upper triangular matrix B=\left(b_{ij}\right) with . Let B_i=\left(t_{kl}\right), where t_{il}=b_{il},t_{kl}=0,\mathrm{\forall }k\ne i, i=1,2,\dots ,n. Then B=B_n\circ B_{n-1}\circ \cdots \circ B_1. Let A_i=B_i\oplus 0,i=1,2,\dots ,n. Then each A_i is a rank-one operator and A=A_n\circ A_{n-1}\circ \cdots \circ A_1. Therefore,

\begin{array}{rl}\phi (A)& =\phi \left(A_n\circ A_{n-1}\circ \cdots \circ A_1\right)\\ & =\phi \left(A_n\right)\circ \phi \left(A_{n-1}\right)\circ \cdots \circ \phi \left(A_1\right)\\ & =A_n\circ A_{n-1}\circ \cdots \circ A_1\\ & =A.\end{array}

(3) If \dim \mathcal{X}=\mathrm{\infty }, then for any operator A\in \mathrm{B}(\mathcal{X}),\phi (A)=A.

Let A\in \text{B}(\mathcal{X}). Then for any finite-rank idempotent P\in \text{B}(\mathcal{X}), there is A\circ (I-P)=AP+(I-P)=AP\circ (I-P). Therefore,

\begin{array}{rl}& \phi (A\circ (I-P))=\phi (AP\circ (I-P)),\\ & \phi (A)\circ (I-P)=\phi (AP)\circ (I-P),\\ & \phi (A)P=\phi (AP)P=AP.\end{array}

Hence \phi (A)=A.

The following shows notations and the main theorem of this paper. If , there is \text{B}(\mathcal{X})=M_n. Let \tau be a ring automorphism on \mathrm{C}. For any A=\left(a_{ij}\right)\in M_n, let A_{\tau }=\left(\tau \left(a_{ij}\right)\right).

Theorem 1.1　 Let \mathcal{X} be a complex Banach space with \dim \mathcal{X}\ge 2 and \phi be a quasi-isomorphism on \text{B}(\mathcal{X}).

(1) If \mathrm{d}\mathrm{i}\mathrm{m}\mathcal{X}=\mathrm{\infty }, then there exists a bounded invertible linear or conjugate linear operator T on \mathcal{X}, such that

\phi \left(A\right)=TA{T}^{-1},\mathrm{\forall }A\in B\left(\mathcal{X}\right).

(2) If , then there exists an invertible matrix T\in M_n and a ring automorphism \tau on \mathrm{C}, such that

\phi (A)=T{A}_{\tau }{T}^{-1},\mathrm{\forall }A\in M_n.

Proof　(1) By Proposition 1.1, \phi preserves rank-one idempotent operators and their orthogonality in both directions. By [11, Theorem 2.4], there exists a bounded invertible linear or conjugate linear operator T on \mathcal{X}, such that \phi (P)=TP{T}^{-1},\mathrm{\forall }P\in I_1(\mathcal{X}), or a bounded invertible linear or conjugate linear operator T:{\mathcal{X}}^{\prime }\to \mathcal{X}, such that \phi (P)=T{P}^{\prime }{T}^{-1},\mathrm{\forall }P\in I_1(\mathcal{X}), where P^{\prime } denotes the adjoint operator of P. In the second case \mathcal{X} is reflexive. In fact, the second case cannot occur. Otherwise, there will exist P,Q\in I_1(\mathcal{X}) such that P\circ Q\ne Q\circ P. Then

\begin{array}{rl}\phi (P)\circ \phi (Q)& =\phi (P)+\phi (Q)-\phi (P)\phi (Q)\\ & =T{P}^{\prime }{T}^{-1}+T{Q}^{\prime }{T}^{-1}-T{P}^{\prime }{Q}^{\prime }{T}^{-1}\\ & =T(Q\circ P{)}^{\prime }{T}^{-1}\\ & \ne T(P\circ Q{)}^{\prime }{T}^{-1}\\ & =\phi (P\circ Q),\end{array}

contradicting that \phi is a quasi-isomorphism. Hence only the first case occurs. Define \psi (A)=T^{-1}\phi (A)T,\mathrm{\forall }A\in \text{B}(\mathcal{X}). Then \psi is a quasi-isomorphism on \text{B}(\mathcal{X}) satisfying \psi (P)=P,\mathrm{\forall }P\in I_1(\mathcal{X}). By Lemma 1.2, \psi (A)=A,\mathrm{\forall }A\in \text{B}(\mathcal{X}). Therefore, \phi (A)=T\psi (A)T^{-1}=TA{T}^{-1},\mathrm{\forall }A\in \text{B}(\mathcal{X}).

(2) If n\ge 3, by [11, Theorem 2.3], there exists an invertible matrix T\in M_n and a ring automorphism \tau on \mathrm{C} such that \phi (P)=T{P}_{\tau }{T}^{-1},\mathrm{\forall }P\in I_1\left(M_n\right), or {\phi (P)=TP}_{\tau }^T{T}^{-1},\mathrm{\forall }P\in I_1\left(M_n\right), where P^{\mathrm{T}} denotes the transpose of P. Similarly, it can be shown that the second case does not occur. According to (1), it is known that (2) holds.

When n=2, let E_{ij}(i,j=1,2) be the standard basis of M_2, namely E_{ij}=\left(e_{kl}\right), where e_{ij}=1,e_{kl}=0, \mathrm{\forall }(k,l)\ne (i,j). By Proposition 1.1, \phi \left(E_{11}\right) and \phi \left(E_{22}\right) are two orthogonal rank-one idempotent matrices. Hence, there exists an invertible matrix S_1 such that \phi \left(E_{11}\right)=S_1{E}_{11}{S}_1^{-1},\phi \left(E_{22}\right)=S_1{E}_{22}{S}_1^{-1}. Define \varphi (A)=S_1^{-1}\phi (A)S_1,\mathrm{\forall }A\in M_2. Then \varphi is a quasi-isomorphism on M_2 and \varphi \left(E_{ii}\right)=E_{ii},i=1,2.

It is first proved that there exists an invertible matrix S_2 and a ring automorphism \tau on \mathrm{C} such that \mathrm{\forall }\lambda \in \mathbf{C},\varphi \left(\lambda E_{ij}\right)=\tau (\lambda )S_2{E}_{ij}{S}_2^{-1},i,j=1, 2. Since \lambda E_{ii}\circ E_{jj}=E_{jj}\circ \lambda E_{ii}(i,j=1,2) and \varphi preserves rank-one operators, it is known that there exist bijections {\tau }_{ii} on \mathbf{C} such that \mathrm{\forall }\lambda \in \mathbf{C},\varphi \left(\lambda E_{ii}\right)={\tau }_{ii}(\lambda )E_{ii},i=1,2. On the other hand, since E_{11}\circ \lambda E_{12}=E_{11}, it is shown that \varphi \left(\lambda E_{12}\right)=E_{11}\varphi \left(\lambda E_{12}\right). So there exist a_{11},a_{12}\in \mathbf{C} such that

\varphi \left(\lambda E_{12}\right)=\varphi \left(\left[\begin{array}{cc}0& \lambda \\ 0& 0\end{array}\right]\right)=\left[\begin{array}{cc}{a}_{11}& a_{12}\\ 0& 0\end{array}\right].

E_{22}+\lambda E_{12}=E_{22}\circ \lambda E_{12}, then

\varphi \left(E_{22}+\lambda E_{12}\right)=\varphi \left(\left[\begin{array}{cc}0& \lambda \\ 0& 1\end{array}\right]\right)=E_{22}\circ \varphi \left(\lambda E_{12}\right)=\left[\begin{array}{cc}{a}_{11}& a_{12}\\ 0& 1\end{array}\right].

By Proposition 1.1, \varphi preserves rank-one operators, so a_{11}=0,\varphi \left(\lambda E_{12}\right)=a_{12}{E}_{12}. As a result, there exists a bijection {\tau }_{12} on \mathrm{C} such that \mathrm{\forall }\lambda \in \mathrm{C},\varphi \left(\lambda E_{12}\right)={\tau }_{12}(\lambda )E_{12}. Similarly, there exists a bijection {\tau }_{21} on \mathrm{C} such that \mathrm{\forall }\lambda \in \mathrm{C},\varphi \left(\lambda E_{21}\right)={\tau }_{21}(\lambda )E_{21}.

\mathrm{\forall }x,y\in \mathrm{C}, and according to x{E}_{12}\circ y{E}_{12}=(x+y)E_{12}, there is \varphi \left(x{E}_{12}\right)\circ \varphi \left(y{E}_{12}\right)=\varphi \left((x+y)E_{12}\right), namely {\tau }_{12}(x)E_{12}\circ {\tau }_{12}(y)E_{12}={\tau }_{12}(x+y)E_{12}, so {\tau }_{12}(x)+{\tau }_{12}(y)={\tau }_{12}(x+y). From y{E}_{22}\circ (x-xy)E_{12}=x{E}_{12}\circ y{E}_{22} and (x-xy)E_{12}\circ y{E}_{11}=y{E}_{11}\circ x{E}_{12}, it is known {\tau }_{12}(xy)={\tau }_{22}(y){\tau }_{12}(x) and {\tau }_{12}(xy)={\tau }_{11}(y){\tau }_{12}(x), so {\tau }_{11}(x)={\tau }_{22}(x). Denote \tau ={\tau }_{11}={\tau }_{22}. Since {\tau }_{12}(xy)={\tau }_{11}(y){\tau }_{12}(x), set x=1 gives {\tau }_{12}(y)=\tau (y){\tau }_{12}(1). Similarly, {\tau }_{21}(y)=\tau (y){\tau }_{21}(1), and there is {\tau }_{21}(1){\tau }_{12}(1)=1. Let

A=x{E}_{12}\circ x{E}_{21}=\left[\begin{array}{cc}-x^2& x\\ x& 0\end{array}\right]\mathrm{a}\mathrm{n}\mathrm{d}\varphi (A)=\left[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right].

Since

A\circ E_{11}=\left[\begin{array}{cc}1& x\\ 0& 0\end{array}\right],A\circ E_{22}=\left[\begin{array}{cc}-x^2& 0\\ x& 1\end{array}\right],

there is

\begin{array}{rl}& \varphi \left(\left[\begin{array}{cc}1& x\\ 0& 0\end{array}\right]\right)=\varphi (A)\circ E_{11}=\left[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right]\circ E_{11}=\left[\begin{array}{cc}1& b_{12}\\ 0& b_{22}\end{array}\right],\\ & \varphi \left(\left[\begin{array}{cc}-x^2& 0\\ x& 1\end{array}\right]\right)=\varphi (A)\circ E_{22}=\left[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right]\circ E_{22}=\left[\begin{array}{cc}{b}_{11}& 0\\ b_{21}& 1\end{array}\right].\end{array}

On the other hand, since

\begin{array}{rl}& \varphi \left(\left[\begin{array}{cc}1& x\\ 0& 0\end{array}\right]\right)=\varphi \left(\left[\begin{array}{cc}0& x\\ 0& 0\end{array}\right]\circ \left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\right)=\left[\begin{array}{cc}\tau (1)& {\tau }_{12}(x)\\ 0& 0\end{array}\right],\\ & \varphi \left(\left[\begin{array}{cc}-x^2& 0\\ x& 1\end{array}\right]\right)=\varphi \left(\left[\begin{array}{cc}-x^2& 0\\ 0& 0\end{array}\right]\circ \left[\begin{array}{cc}0& 0\\ x& 0\end{array}\right]\circ \left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]\right)=\left[\begin{array}{cc}\tau \left(-x^2\right)& 0\\ {\tau }_{21}(x)& \tau (1)\end{array}\right],\end{array}

there is b_{11}=\tau \left(-x^2\right),b_{21}={\tau }_{21}(x),b_{12}={\tau }_{12}(x),b_{22}=0, namely

\varphi \left(\left[\begin{array}{cc}-x^2& x\\ x& 0\end{array}\right]\right)=\left[\begin{array}{cc}\tau \left(-x^2\right)& {\tau }_{12}(x)\\ {\tau }_{21}(x)& 0\end{array}\right].

As

\varphi (A)=\varphi \left(x{E}_{12}\right)\circ \varphi \left(x{E}_{21}\right)=\left[\begin{array}{cc}-{\tau }_{21}(x){\tau }_{12}(x)& {\tau }_{12}(x)\\ {\tau }_{21}(x)& 0\end{array}\right],

there is \tau \left(-x^2\right)=-{\tau }_{21}(x){\tau }_{12}(x)=-\tau (x{)}^2{\tau }_{21}(1){\tau }_{12}(1), namely {\tau }_{21}(1){\tau }_{12}(1)=1.

The following proves that \tau is a ring automorphism. Since {\tau }_{12}(1)\ne 0, when x=y=1, according to {\tau }_{12}(xy)={\tau }_{11}(y){\tau }_{12}(x), there is\tau (1)=1. As \tau (xy){\tau }_{12}(1)={\tau }_{12}(xy)=\tau (x)\tau (y){\tau }_{12}(1), there is \tau (x)\tau (y)=\tau (xy),\mathrm{\forall }x,y\in \mathbf{C}. Moreover,

\begin{array}{rl}(\tau (x)+\tau (y)){\tau }_{12}(1)& =\tau (x){\tau }_{12}(1)+\tau (y){\tau }_{12}(1)\\ & ={\tau }_{12}(x)+{\tau }_{12}(y)\\ & ={\tau }_{12}(x+y)\\ & =\tau (x+y){\tau }_{12}(1),\mathrm{\forall }x,y\in \mathrm{C}\end{array}

and there is \tau (x+y)=\tau (x)+\tau (y),\mathrm{\forall }x,y\in \mathbf{C}. Therefore, \tau is a ring automorphism. Let

S_2=\left[\begin{array}{cc}{\tau }_{12}(1{)}^{\frac{1}{2}}& 0\\ 0& {\tau }_{21}(1{)}^{\frac{1}{2}}\end{array}\right].

There is \varphi \left(\lambda E_{ij}\right)=\tau (\lambda )S_2{E}_{ij}{S}_2^{-1},i,j=1,2.

Next, define {\psi (A)=S}_2^{-1}\varphi (A)S_2,\mathrm{\forall }A\in M_2. Then \psi is a quasi-isomorphism and \psi \left(\lambda E_{ij}\right)=\tau (\lambda )E_{ij},i,j=1,2. Let

A=\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]\mathrm{a}\mathrm{n}\mathrm{d}\psi (A)=\left[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right].

Since

A\circ E_{11}=\left[\begin{array}{cc}1& a_{12}\\ 0& a_{22}\end{array}\right],A\circ E_{22}=\left[\begin{array}{cc}{a}_{11}& 0\\ a_{21}& 1\end{array}\right],

there is

\begin{array}{rl}& \psi \left(\left[\begin{array}{cc}1& a_{12}\\ 0& a_{22}\end{array}\right]\right)=\left[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right]\circ E_{11}=\left[\begin{array}{cc}1& b_{12}\\ 0& b_{22}\end{array}\right],\\ & \psi \left(\left[\begin{array}{cc}{a}_{11}& 0\\ a_{21}& 1\end{array}\right]\right)=\left[\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right]\circ E_{22}=\left[\begin{array}{cc}{b}_{11}& 0\\ b_{21}& 1\end{array}\right].\end{array}

On the other hand, since

\begin{array}{rl}& \psi \left(\left[\begin{array}{cc}1& a_{12}\\ 0& a_{22}\end{array}\right]\right)=\psi \left(\left[\begin{array}{cc}0& 0\\ 0& a_{22}\end{array}\right]\circ \left[\begin{array}{cc}0& a_{12}\\ 0& 0\end{array}\right]\circ \left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\right)=\left[\begin{array}{cc}\tau (1)& \tau \left(a_{12}\right)\\ 0& \tau \left(a_{22}\right)\end{array}\right],\\ & \psi \left(\left[\begin{array}{cc}{a}_{11}& 0\\ a_{21}& 1\end{array}\right]\right)=\psi \left(\left[\begin{array}{cc}{a}_{11}& 0\\ 0& 0\end{array}\right]\circ \left[\begin{array}{cc}0& 0\\ a_{21}& 0\end{array}\right]\circ \left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]\right)=\left[\begin{array}{cc}\tau \left(a_{11}\right)& 0\\ \tau \left(a_{21}\right)& \tau (1)\end{array}\right],\end{array}

there is b_{11}=\tau \left(a_{11}\right),b_{21}=\tau \left(a_{21}\right),b_{12}=\tau \left(a_{12}\right),b_{22}=\tau \left(a_{22}\right), namely \psi (A)=A_{\tau },\mathrm{\forall }A\in M_2. Let T=S_1{S}_2. Then \phi (A)=T{A}_{\tau }{T}^{-1},\mathrm{\forall }A\in M_2.

Note　If \phi is a quasi-isomorphism on \mathrm{C}, namely \phi (x\circ y)=\phi (x)\circ \phi (y),\mathrm{\forall }x,y\in \mathrm{C}, is \phi necessarily a ring automorphism on \mathrm{C}?

References

[1] Amberg B. , Sysak, Y.P. . Radical rings with Engel conditions. J. Algebra 2000; 231: 364–373

[2] Amberg B. , Sysak, Y.P. . Associative rings whose adjoint semigroup is locally nilpotent. Arch. Math. (Basel) 2001; 76(6): 426–435

[3] Hille E. , Phillips, R.S., Functional Analysis and Semi-groups, Providence, R.I.: AMS, 1957, 680−694

[4] Jacobson N. . The radical and semi-simplicity for arbitrary rings. Amer. J. Math. 1945; 67: 300–320

[5] Kaplansky I. . Topological rings. Amer. J. Math. 1947; 69: 153–183

[6] Krasil’nikov A.N. , On the semigroup and Lie nilpotency of associative algebras, Mat. Zametki, 1997, 62: 510−519 (Translated in: Math. Notes, 1997, 62(4): 426−433)

[7] Liu C.K. , Tsai, W.Y. . Jordan isomorphisms of upper triangular matrix rings. Linear Algebra Appl. 2007; 426(1): 143–148

[8] Liu L. , Ji, G.X. . Maps preserving product X^*Y + YX^* on factor von Neumann algebras. Linear Multi-linear Algebra 2011; 59(9): 951–955

[9] Lu F.Y. . Characterizations of derivations and Jordan derivations on Banach algebras. Linear Algebra Appl. 2009; 430(8/9): 2233–2239

[10] Qi X.F., Hou, J.C. , Deng, J. . Lie ring isomorphisms between nest algebras on Banach spaces. J. Funct. Anal. 2014; 266(7): 4266–4292

[11] Šemrl . Non-linear commutativity preserving maps. Acta Sci. Math. 2005; 71: 781–819

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