1 Introduction

Let \({\mathcal {A}}\) be a \(*\)-algebra and \(\eta \) be a non-zero scalar. For \(A, B\in {\mathcal {A}}\), define the Jordan \(\eta \)-\(*\)-product of A and B by \(A\diamondsuit _{\eta }B=AB+\eta BA^{*}.\) The Jordan \(\eta \)-\(*\)-product, particularly the Jordan (\(-1\))-\(*\)-product and the Jordan 1-\(*\)-product, is very meaningful and important in some research topics (see, for example, [1, 3, 8–11]). A map \(\Phi \) between \(*\)-algebras \({\mathcal {A}}\) and \(\mathcal {B}\) is said to preserve the Jordan \(\eta \)-\(*\)-product if \(\Phi (A\diamondsuit _{\eta }B)=\Phi (A)\diamondsuit _{\eta }\Phi (B)\) for all \(A, B\in {\mathcal {A}}.\) Recently, many authors pay more attention to maps preserving the Jordan \(\eta \)-\(*\)-product between \(*\)-algebra (see, for example, [2, 6]). In [6], Li et al. considered maps which preserve the Jordan 1-\(*\)-product and proved that such a map between factor von Neumann algebras is a \(*\)-ring isomorphism. In [2], Dai and Lu completely described maps preserving the Jordan \(\eta \)-\(*\)-product between von Neumann algebras without central abelian projections for all non-zero scalars \(\eta \). They proved that if \(\Phi \) is a bijective map preserving the Jordan \(\eta \)-\(*\)-product between two von Neumann algebras, one of which has no central abelian projections, then \(\Phi \) is a linear \(*\)-isomorphism if \(\eta \) is not real and \(\Phi \) is a sum of a linear \(*\)-isomorphism and a conjugate linear \(*\)-isomorphism if \(\eta \) is real.

Recently, Huo et al. [4] studied a more general problem. They considered the Jordan triple \(\eta \)-\(*\)-product of three elements A, B and C in a \(*\)-algebra \({\mathcal {A}}\) defined by \(A\diamondsuit _{\eta }B\diamondsuit _{\eta }C=(A\diamondsuit _{\eta }B)\diamondsuit _{\eta }C\) (we should be aware that \(\diamondsuit _{\eta }\) is not necessarily associative). A map \(\Phi \) between \(*\)-algebras \({\mathcal {A}}\) and \(\mathcal {B}\) is said to preserve the Jordan triple \(\eta \)-\(*\)-product if \(\Phi (A\diamondsuit _{\eta }B\diamondsuit _{\eta }C)=\Phi (A)\diamondsuit _{\eta }\Phi (B)\diamondsuit _{\eta }\Phi (C)\) for all \(A, B, C\in {\mathcal {A}}.\) Clearly a map between \(*\)-algebras preserving the Jordan \(\eta \)-\(*\)-product also preserves the Jordan triple \(\eta \)-\(*\)-product, but not conversely. For example, for \(\alpha , \beta \in {\mathbb {R}}\), define \(\Phi (\alpha +\beta i)=-4(\alpha ^{3}+\beta ^{3}i)\). Then the map \(\Phi : \mathbb {C}\rightarrow \mathbb {C}\) is a bijection. It is not difficult to verify that \(\Phi \) preserves the Jordan triple \((-1)\)-\(*\)-product and Jordan triple 1-\(*\)-product, but it does not preserve the Jordan \((-1)\)-\(*\)-product or Jordan 1-\(*\)-product. So, the class of those maps preserving the Jordan triple \(\eta \)-\(*\)-product is, in principle wider than the class of maps preserving the Jordan \(\eta \)-\(*\)-product.

Let \(\eta \ne -1\) be a non-zero complex number, and let \(\Phi \) be a bijection between two von Neumann algebras, one of which has no central abelian projections, satisfying \(\Phi (I) =I\) and preserving the Jordan triple \(\eta \)-\(*\)-product. Huo et al. [4] showed that \(\Phi \) is a linear \(*\)-isomorphism if \(\eta \) is not real and \(\Phi \) is the sum of a linear \(*\)-isomorphism and a conjugate linear \(*\)-isomorphism if \(\eta \) is real. It is easy to see that a map \(\Phi \) preserving the Jordan triple \(\eta \)-\(*\)-product does not need satisfy \(\Phi (I)=I\). Indeed, let \(\Phi (A)=-A\) for all \(A\in {\mathcal {A}}.\) Then \(\Phi \) preserves the Jordan triple \(\eta \)-\(*\)-product but \(\Phi (I)=-I\). In this paper, we will discuss maps preserving the Jordan triple 1-\(*\)-product without the assumption \(\Phi (I)=I\). We prove that if \(\Phi \) is a bijective map preserving the Jordan triple 1-\(*\)-product between two von Neumann algebras, one of which has no central abelian projections, then the map \(\Phi (I)\Phi \) is a sum of a linear \(*\)-isomorphism and a conjugate linear \(*\)-isomorphism, where \(\Phi (I)\) is a self-adjoint central element in the range with \(\Phi (I)^{2}=I.\) We mention that the methods in [4] do not fit for solving our problem since their proofs heavily depend on the assumption \(\Phi (I)=I\).

2 Proof of Main Result

Before embarking on the proof, we need some notations and preliminaries. In this section, we often write the Jordan 1-\(*\)-product by \(A\bullet B\), that is \(A\bullet B=AB+BA^{*}\). Algebras and spaces are over the complex number field \(\mathbb {C}\). A von Neumann algebra \({\mathcal {A}}\) is a weakly closed, self-adjoint algebra of operators on a Hilbert space H containing the identity operator I. The set \(\mathcal {Z}({\mathcal {A}})=\{S\in {\mathcal {A}}: ST=TS \ \text {for all} \ T\in {\mathcal {A}}\}\) is called the center of \({\mathcal {A}}\). A projection P is called a central abelian projection if \(P\in \mathcal {Z}({\mathcal {A}})\) and \(P{\mathcal {A}}P\) is abelian. Recall that the central carrier of A, denoted by \(\overline{A}\), is the smallest central projection P satisfying \(PA=A\). It is not difficult that the central carrier of A is the projection onto the closed subspace spanned by \(\{BA(x): B\in {\mathcal {A}}, x\in H\}.\) If A is self-adjoint, then the core of A, denoted by \(\underline{A}\), is sup\(\{S\in \mathcal {Z}({\mathcal {A}}): S=S^{*}, S\le A\}.\) If P is a projection, it is clear that \(\underline{P}\) is the largest central projection Q satisfying \(Q\le P.\) A projection P is said to be core-free if \(\underline{P}=0.\) It is easy to see that \(\underline{P}=0\) if and only if \(\overline{I-P}=I.\)

Lemma 2.1

([7, Lemma 4]) Let \({\mathcal {A}}\) be a von Neumann algebra with no central abelian projections. Then there exists a projection \(P\in {\mathcal {A}}\) such that \(\underline{P}=0\) and \(\overline{P}=I\).

Lemma 2.2

Let \({\mathcal {A}}\) be a von Neumann algebra on a Hilbert space H. Let A be an operator in \({\mathcal {A}}\) and \(P\in {\mathcal {A}}\) is a projection with \(\overline{P}=I.\) If \(ABP=0\) for all \(B\in {\mathcal {A}},\) then \(A=0.\) Consequently, if \(Z\in \mathcal {Z}({\mathcal {A}})\), then \(ZP=0\) implies \(Z=0.\)

Proof

From \(\overline{P}=I,\) it follows that the linear span of \(\{BP(x): B\in {\mathcal {A}}, x\in H\}\) is dense in H. So \(ABP=0\) for all \(B\in {\mathcal {A}},\) then \(A=0.\) If \(Z\in \mathcal {Z}({\mathcal {A}})\) and \(ZP=0\), then \(ZBP=0\) for all \(B\in {\mathcal {A}},\) hence \(Z=0.\) \(\square \)

Lemma 2.3

Let \({\mathcal {A}}\) be a von Neumann algebra and \(A\in {\mathcal {A}}\). Then \(AB+BA^{*}=0\) for all \(B\in {\mathcal {A}}\) implies that \(A=-A^{*}\in \mathcal {Z}({\mathcal {A}})\).

Proof

We take \(B=I,\) then \(A=-A^{*}.\) Therefore \(AB=BA\) for all \(B\in {\mathcal {A}},\) which implies A belongs to the center of \({\mathcal {A}}\). \(\square \)

Theorem 2.4

([4, Theorem 2.1]) Let \({\mathcal {A}}\) be a von Neumann algebra with no central abelian projections and \(\mathcal {B}\) be a \(*\)-algebra. Suppose that a bijective map \(\Phi : {\mathcal {A}}\rightarrow \mathcal {B}\) satisfies \(\Phi (A\bullet B\bullet C)=\Phi (A)\bullet \Phi (B)\bullet \Phi (C)\) for all \(A, B,C\in {\mathcal {A}}\). Then \(\Phi \) is additive.

Our main result in this paper reads as follows.

Theorem 2.5

Let \({\mathcal {A}}\) and \(\mathcal {B}\) be two von Neumann algebras, one of which has no central abelian projections. Suppose that a bijective map \(\Phi : {\mathcal {A}}\rightarrow \mathcal {B}\) satisfies \(\Phi (A\bullet B\bullet C)=\Phi (A)\bullet \Phi (B)\bullet \Phi (C)\) for all \(A, B,C\in {\mathcal {A}}\). Then the following statements hold:

  1. (1)

    \(\Phi (I)\) is a self-adjoint central element in \(\mathcal {B}\) with \(\Phi (I)^{2}=I.\)

  2. (2)

    Defining a map \(\phi : {\mathcal {A}}\rightarrow \mathcal {B}\) by \(\phi (A)=\Phi (I)\Phi (A)\) for all \(A\in {\mathcal {A}}\). Then there exsits a central projection \(E\in {\mathcal {A}}\) such that the restriction of \(\phi \) to \({\mathcal {A}}E\) is a linear \(*\)-isomorphism and the restriction of \(\phi \) to \({\mathcal {A}}(I-E)\) is a conjugate linear \(*\)-isomorphism.

The proof will be organized in some lemas. First note that \(\Phi \) is additive. Indeed, if \({\mathcal {A}}\) has no central abelian projections, Lemma 2.4 assures that \(\Phi \) is additive. If \(\mathcal {B}\) has no central abelian projections, observe that \(\Phi ^{-1}:\mathcal {B}\rightarrow {\mathcal {A}}\) is a bijection and preserves the Jordan triple 1-\(*\)-product. Applying Lemma 2.4 to \(\Phi ^{-1}\), we know that \(\Phi ^{-1}\) and hence \(\Phi \) is additive. In what follows, without loss of generality, we assume that \(\mathcal {B}\) has no central abelian projections.

Lemma 2.6

  1. (1)

    For each \(A\in {\mathcal {A}}\), \(A=-A^{*}\) if and only if \(\Phi (A)=-\Phi (A)^{*};\)

  2. (2)

    \(\Phi (\mathcal {Z}({\mathcal {A}}))=\mathcal {Z}(\mathcal {B});\)

  3. (3)

    \((\Phi (I)+\Phi (I)^{*})^{2}=4I.\)

Proof

Let \(A\in {\mathcal {A}}\) be arbitrary. Since \(\Phi \) is surjective, there exists \(B\in {\mathcal {A}}\) such that \(\Phi (B)=I.\) Then

$$\begin{aligned} 0&=\Phi (iI\bullet A\bullet B)\\&=\Phi (iI)\bullet \Phi (A)\bullet I\\&=\Phi (iI)\Phi (A)+\Phi (A)\Phi (iI)^{*}+\Phi (A)^{*}\Phi (iI)^{*}+\Phi (iI)\Phi (A)^{*} \end{aligned}$$

holds true for all \(A\in {\mathcal {A}}.\) That is,

$$\begin{aligned} \Phi (iI)(\Phi (A)+\Phi (A)^{*})+(\Phi (A)+\Phi (A)^{*})\Phi (iI)^{*}=0 \end{aligned}$$

holds true for all \(A\in {\mathcal {A}}.\) So \(\Phi (iI)B+B\Phi (iI)^{*}=0\) holds true for all \(B=B^{*}\in \mathcal {B}.\) Since for every \(B\in \mathcal {B},\) \(B=B_{1}+iB_{2}\) with \(B_{1}=\frac{B+B^{*}}{2}\) and \(B_{2}=\frac{B-B^{*}}{2i}\), it follows that \(\Phi (iI)B+B\Phi (iI)^{*}=0\) holds true for all \(B\in \mathcal {B}.\) It follows from Lemma 2.3 that \( \Phi (iI)=-\Phi (iI)^{*}\in \mathcal {Z}(\mathcal {B}).\) Similarly, \(\Phi ^{-1}(iI)\in \mathcal {Z}({\mathcal {A}})\).

Let \(A=-A^{*}\in {\mathcal {A}}\) and \(\Phi (B)=I\). Since \(0=B\bullet A\bullet \Phi ^{-1}(iI)\), it follows that

$$\begin{aligned} 0=\Phi (B\bullet A\bullet \Phi ^{-1}(iI)) =I\bullet \Phi (A)\bullet (iI) =2i(\Phi (A)+\Phi (A)^{*}). \end{aligned}$$

This implies that \(\Phi (A)=-\Phi (A)^{*}\). Similarly, we note that \(\Phi ^{-1}\) also preserves the Jordan triple 1-\(*\)-product. If \(\Phi (A)=-\Phi (A)^{*}\), then

$$\begin{aligned} 0=\Phi ^{-1}(\Phi (I)\bullet \Phi (A)\bullet \Phi (iI)) =I\bullet A\bullet (iI) =2i(A+A^{*}), \end{aligned}$$

and so \(A=-A^{*}\). Now we have proved that \(A=-A^{*}\) if and only if \(\Phi (A)=-\Phi (A)^{*}\) for each \(A\in {\mathcal {A}}\).

Let \(Z\in \mathcal {Z}({\mathcal {A}})\) be arbitrary and \(\Phi (B)=I\). For every \(A=-A^{*}\in {\mathcal {A}},\) we have

$$\begin{aligned} 0=\Phi (B\bullet A\bullet Z)=I\bullet \Phi (A)\bullet \Phi (Z)=2(\Phi (A)\Phi (Z)+\Phi (Z)\Phi (A)^{*}). \end{aligned}$$

That is \(\Phi (A)\Phi (Z)=-\Phi (Z)\Phi (A)^{*}\) holds true for all \(A=-A^{*}\in {\mathcal {A}}.\) Since \(\Phi \) preservers conjugate self-adjoint elements, it follows that \(C\Phi (Z)=\Phi (Z)C\) holds true for all \(C=-C^{*}\in \mathcal {B}.\) Since for every \(C\in \mathcal {B},\) we have \(C=C_{1}+iC_{2}\), where \(C_{1}=\frac{C-C^{*}}{2}\) and \(C_{2}=\frac{C+C^{*}}{2i}\) are conjugate self-adjoint elements. Hence \(C\Phi (Z)=\Phi (Z)C\) holds true for all \(C\in {\mathcal {A}}.\) Then \(\Phi (Z)\in \mathcal {Z}(\mathcal {B})\), which implies that \(\Phi (\mathcal {Z}({\mathcal {A}}))\subseteq \mathcal {Z}(\mathcal {B})\). Thus \(\Phi (\mathcal {Z}({\mathcal {A}}))=\mathcal {Z}(\mathcal {B})\) by considering \(\Phi ^{-1}\).

Let \(\Phi (B)=I\). Since \(\Phi (I)\in \mathcal {Z}(\mathcal {B})\), then

$$\begin{aligned} 4I=4\Phi (B)=\Phi (I\bullet I\bullet B)=\Phi (I)\bullet \Phi (I) \bullet I=(\Phi (I)+\Phi (I)^{*})^{2}. \end{aligned}$$

\(\square \)

Lemma 2.7

Let P be a projection in \({\mathcal {A}}\) and set \(Q_{P}=\frac{1}{4}(\Phi (I)+\Phi (I)^{*})(\Phi (P)+\Phi (P)^{*}).\) Then the following statements hold:

  1. (1)

    \(Q_{P}\) is a projection and \(\Phi (P)=\Phi (I)Q_{P};\)

  2. (2)

    Suppose that A in \({\mathcal {A}}\) such that \(A=PA(I-P)\). Then \(\Phi (A)=Q_{P}\Phi (A)+\Phi (A)Q_{P}.\)

Proof

Let P be a projection in \({\mathcal {A}}\). Since \(\Phi (I)\in \mathcal {Z}(\mathcal {B})\), then

$$\begin{aligned} 4\Phi (P)&=\Phi (I\bullet P\bullet I) =\Phi (I)\bullet \Phi (P)\bullet \Phi (I)\\&=\Phi (I)(\Phi (I)+\Phi (I)^{*})(\Phi (P)+\Phi (P)^{*})\\&=4\Phi (I)Q_{P}. \end{aligned}$$

Hence

$$\begin{aligned} 4\Phi (P)&=\Phi (I\bullet P\bullet P) =\Phi (I)\bullet \Phi (P)\bullet \Phi (P)\\&=(\Phi (I)+\Phi (I)^{*})\Phi (P)(\Phi (P)+\Phi (P)^{*})\\&=4\Phi (P)Q_{P} =4\Phi (I)Q^{2}_{P}. \end{aligned}$$

This implies that \(\Phi (P)=\Phi (I)Q^{2}_{P}.\) Taking the adjoint and noting that \(Q_{P}\) is self-adjoint, \(\Phi (P)^{*}=\Phi (I)^{*}Q^{2}_{P}.\) Summing the last two equations, we get \(\Phi (P)+\Phi (P)^{*}=(\Phi (I)+\Phi (I)^{*})Q^{2}_{P}.\) Hence \((\Phi (I)+\Phi (I)^{*})(\Phi (P)+\Phi (P)^{*})=(\Phi (I)+\Phi (I)^{*})^{2}Q^{2}_{P}\). By Lemma 2.6 (3), we obtain \(Q_{P}=Q^{2}_{P}.\) So \(Q_{P}\) is a projection.

Let A in \({\mathcal {A}}\) such that \(A=PA(I-P)\). Noticing that \(\Phi (P)=\Phi (I)Q_{P},\) we have

$$\begin{aligned} 2\Phi (A)&=\Phi (I\bullet P\bullet A) =\Phi (I)\bullet \Phi (P)\bullet \Phi (A)\\&=(\Phi (I)+\Phi (I)^{*})(\Phi (P)\Phi (A)+\Phi (A)\Phi (P)^{*})\\&=(\Phi (I)+\Phi (I)^{*})(\Phi (I)Q_{P}\Phi (A)+\Phi (I)^{*}\Phi (A)Q_{P}). \end{aligned}$$

Since \((\Phi (I)+\Phi (I)^{*})^{2}=4I\) and \(\Phi (I), \Phi (I)^{*}\in \mathcal {Z}(\mathcal {B})\), multiplying both sides of the above equation by \(Q_{P}\) from the left and right respectively, we get that \(Q_{P}\Phi (A)Q_{P}=0.\) Multiplying both sides of the above equation by \(I-Q_{P}\) from the left and right respectively, we get that \((I-Q_{P})\Phi (A)(I-Q_{P})=0,\) which implies that \(\Phi (A)=Q_{P}\Phi (A)+\Phi (A)Q_{P}.\) \(\square \)

Lemma 2.8

\(\Phi (I)\) is a self-adjoint central element in \(\mathcal {B}\) with \(\Phi (I)^{2}=I.\)

Proof

Since \(\mathcal {B}\) has no central abelian projections, by Lemma 2.1, we can choose a projection \(Q\in \mathcal {B}\) satisfying \(\underline{Q}=0\) and \(\overline{Q}=I.\) Let B be in \(\mathcal {B}\) such that \(B=QB(I-Q)\). Let \(P=\frac{1}{4}(\Phi ^{-1}(I)+\Phi ^{-1}(I)^{*})(\Phi ^{-1}(Q)+\Phi ^{-1}(Q)^{*}).\) Applying Lemma 2.7 to \(\Phi ^{-1}\), we know that P is a projection and \(\Phi ^{-1}(B)=P\Phi ^{-1}(B)+\Phi ^{-1}(B)P.\) Moreover

$$\begin{aligned}\Phi (P)&=\frac{1}{4}\Phi ((\Phi ^{-1}(I)+\Phi ^{-1}(I)^{*})(\Phi ^{-1}(Q)+\Phi ^{-1}(Q)^{*}))\\&=\frac{1}{4}\Phi (\Phi ^{-1}(I)\bullet \Phi ^{-1}(Q)\bullet I)\\&=\frac{1}{4}(I\bullet Q\bullet \Phi (I)) =\Phi (I)Q. \end{aligned}$$

Hence

$$\begin{aligned}B&=\Phi (P\Phi ^{-1}(B)+\Phi ^{-1}(B)P)\\&=\frac{1}{2}\Phi (I\bullet P\bullet \Phi ^{-1}(B))\\&=\frac{1}{2}( \Phi (I)\bullet \Phi (P)\bullet B)\\&=\frac{1}{2}((\Phi (I)+\Phi (I)^{*})(\Phi (P)B+B\Phi (P)^{*}))\\&=\frac{1}{2}((\Phi (I)+\Phi (I)^{*})(\Phi (I)QB+\Phi (I)^{*}BQ))\\&=\frac{1}{2}(\Phi (I)+\Phi (I)^{*})\Phi (I)B. \end{aligned}$$

This implies that \((2I-(\Phi (I)+\Phi (I)^{*})\Phi (I))B=0\). For arbitrary B we have \((2I-(\Phi (I)+\Phi (I)^{*})\Phi (I))QB(I-Q)=0\) and since \(\overline{I-Q}=I\), it follows from Lemma 2.2 that \((2I-(\Phi (I)+\Phi (I)^{*})\Phi (I))Q=0\). Since \(2I-(\Phi (I)+\Phi (I)^{*})\Phi (I)\in \mathcal {Z}(\mathcal {B})\) and \(\overline{Q}=I\), by Lemma 2.2 , we obtain that \(2I-(\Phi (I)+\Phi (I)^{*})\Phi (I)=0\). This together with Lemma 2.6 (3) implies that \(\Phi (I)=\Phi (I)^{*}\) and \(\Phi (I)^{2}=I.\) \(\square \)

Now, defining a map \(\phi : {\mathcal {A}}\rightarrow \mathcal {B}\) by \(\phi (A)=\Phi (I)\Phi (A)\) for all \(A\in {\mathcal {A}}\). Then \(\phi \) has the following properties.

Lemma 2.9

  1. (1)

    \(\phi \) is an additive bijection and satisfies

    $$\begin{aligned} \phi (A\bullet B\bullet C)=\phi (A)\bullet \phi (B)\bullet \phi (C) \end{aligned}$$

    for all \(A, B,C\in {\mathcal {A}}\);

  2. (2)

    \(\phi (I)=I\) and \(\phi (\mathcal {Z}({\mathcal {A}}))=\mathcal {Z}(\mathcal {B});\)

  3. (3)

    \(\phi (A^{*})=\phi (A)^{*}\) for all \(A\in {\mathcal {A}};\)

  4. (4)

    P is a projection in \({\mathcal {A}}\) if and only if \(\phi (P)\) is a projection in \(\mathcal {B}\).

Proof

(1) follows from Theorem 2.4 and Lemma 2.8 and (2) follows from Lemmas 2.8 and 2.6 (2). (3) For all \(A\in {\mathcal {A}}\), since

$$\begin{aligned} 2(\phi (A)+\phi (A^{*}))= & {} 2\phi (A+A^{*})=\phi (A\bullet I\bullet I)=\phi (A)\bullet I\bullet I\\= & {} 2(\phi (A)+\phi (A)^{*}), \end{aligned}$$

we have \(\phi (A^{*})=\phi (A)^{*}\). (4) If P be a projection in \({\mathcal {A}}\), then by Lemma 2.7 (1), we see that \(\phi (P)=\Phi (I)\Phi (P)=\Phi (I)^{2}Q_{P}=Q_{P}\). So \(\phi (P)\) is a projection in \(\mathcal {B}\). Conversely, if \(\phi (P)\) is a projection in \(\mathcal {B}\), applying Lemma 2.7 to \(\phi ^{-1}\), we know that \(P=\frac{1}{4}\phi ^{-1}(I)(\phi ^{-1}(I)+\phi ^{-1}(I)^{*})(P+P^{*})\) and \(\frac{1}{4}(\phi ^{-1}(I)+\phi ^{-1}(I)^{*})(P+P^{*})\) is a projection. But \(\phi ^{-1}(I)=I\) by (2), it follows that P is a projection. \(\square \)

Since \(\mathcal {B}\) has no central abelian projections, by Lemma 2.1, there exists a projection \(Q_{1}\) in \(\mathcal {B}\) such that \(\underline{Q_{1}}=0\) and \(\overline{Q_{1}}=I.\) Then by Lemma 2.9 (4), \(P_{1}=\phi ^{-1}(Q_{1})\) is a projection in \({\mathcal {A}}.\) Set \(P_{2}=I-P_{1}\) and \(Q_{2}=I-Q_{1}\). Denote \({\mathcal {A}}_{ij}=P_{i}{\mathcal {A}}P_{j}\) and \(\mathcal {B}_{ij}=Q_{i}\mathcal {B}Q_{j}\). Then \({\mathcal {A}}=\sum ^{2}_{i,j=1}{\mathcal {A}}_{ij}\) and \(\mathcal {B}=\sum ^{2}_{i,j=1}\mathcal {B}_{ij}\).

Lemma 2.10

\(\phi ({\mathcal {A}}_{ij})=\mathcal {B}_{ij}, \phi ({\mathcal {A}}_{ii})\subseteq \mathcal {B}_{ii}, 1\le i\ne j\le 2.\)

Proof

Let \(A_{12}\) be an arbitrary element in \({\mathcal {A}}_{12}\). Then

$$\begin{aligned} 2\phi (A_{12})&=\phi (I\bullet P_{1}\bullet A_{12})\\&=I\bullet Q_{1}\bullet \phi (A_{12})\\&=2Q_{1}\phi (A_{12})+2\phi (A_{12})Q_{1}, \end{aligned}$$

we get that \(Q_{1}\phi (A_{12})Q_{1}=Q_{2}\phi (A_{12})Q_{2}=0.\) Hence \(\phi (A_{12})=B_{12}+B_{21}\) for some \(B_{12}\in \mathcal {B}_{12}\) and \(B_{21}\in \mathcal {B}_{21}\).

Now to show that \(\phi (A_{12})\subseteq \mathcal {B}_{12}\), we have to show that \(B_{21}=0\). This can be seen from

$$\begin{aligned} 0&=\phi (I\bullet A_{12}\bullet P_{1})\\&=I\bullet \Phi (A_{12})\bullet Q_{1}\\&=2(B_{21}+B_{21}^{*}). \end{aligned}$$

So \(B_{21}=0,\) which implies that \(\phi (A_{12})\subseteq \mathcal {B}_{12}\). Hence by considering \(\phi ^{-1}\), we have \(\phi ({\mathcal {A}}_{12})=\mathcal {B}_{12}\). Similarly, we have \(\phi ({\mathcal {A}}_{21})=\mathcal {B}_{21}\).

Let \(A_{ii}\) be an arbitrary element in \({\mathcal {A}}_{ii}\). Then for \(j\ne i\), we have

$$\begin{aligned} 0=\phi (I\bullet P_{j}\bullet A_{ii})=I\bullet Q_{j}\bullet \phi (A_{ii}) =2(Q_{j}\phi (A_{ii})+\phi (A_{ii})Q_{j}), \end{aligned}$$

which implies that \(Q_{i}\phi (A_{ii})Q_{j}=Q_{j}\phi (A_{ii})Q_{i}=Q_{j}\phi (A_{ii})Q_{j}=0.\) So \(\phi (A_{ii})=Q_{i}\phi (A_{ii})Q_{i}\subseteq \mathcal {B}_{ii}.\) \(\square \)

Lemma 2.11

\(\phi \) is multiplicative.

Proof

Let A and B be in \({\mathcal {A}}\). Write \(A=\sum ^{2}_{i,j=1}A_{ij}\) and \(B=\sum ^{2}_{i,j=1}B_{ij},\) where \(A_{ij},B_{ij}\in {\mathcal {A}}_{ij}\). To show \(\phi (AB)=\phi (A)\phi (B)\), by the additivity of \(\phi \), it suffices to show that \(\phi (A_{ij}B_{kl})=\phi (A_{ij})\phi (B_{kj})\) for all \(i,j,k,l\in \{1,2\}\). Since if \(j\ne k\) then \(\phi (A_{ij}B_{kl})=\phi (A_{ij})\phi (B_{kj})=0\) by Lemma 2.10, we only need to consider the cases with \(j=k.\)

First of all, since \(\phi (B_{12})\phi (A_{11})^{*}=0\), which implies that

$$\begin{aligned} \phi (A_{11}B_{12})+\phi (B_{12}^{*}A_{11}^{*})&=\phi (A_{11}\bullet B_{12}\bullet I)\\&=\phi (A_{11})\bullet \phi (B_{12})\bullet I\\&=\phi (A_{11})\phi (B_{12})+\phi (B_{12})^{*}\phi (A_{11})^{*}. \end{aligned}$$

Thus we have \(\phi (A_{11}B_{12})=\phi (A_{11})\phi (B_{12})\) by Lemma 2.10. Similarly, we can prove that \(\phi (A_{22}B_{21})=\phi (A_{22})\phi (B_{21}).\)

It is easy to compute that

$$\begin{aligned} \phi (A_{12}B_{21})+\phi (B_{21}A_{12})&=\phi (A_{12}\bullet I\bullet B_{21})\\&=\phi (A_{12})\bullet I\bullet \phi (B_{21})\\&=\phi (A_{12})\phi (B_{21})+\phi (B_{21})\phi (A_{12}). \end{aligned}$$

Thus \(\phi (A_{12}B_{21})=\phi (A_{12})\phi (B_{21})\) and \(\phi (B_{21}A_{12})=\phi (B_{21})\phi (A_{12})\) by Lemma 2.10.

For \(D_{12}\in \mathcal {B}_{12}\), we have \(C_{12}=\phi ^{-1}(D_{12})\in {\mathcal {A}}_{12}\) by Lemma 2.10. Thus

$$\begin{aligned} \phi (A_{11}B_{11})D_{12}=\phi (A_{11}B_{11}C_{12})=\phi (A_{11})\phi (B_{11}C_{12})=\phi (A_{11})\phi (B_{11})D_{12} \end{aligned}$$

for all \(D_{12}\in \mathcal {B}_{12}\). Since \(\overline{Q_{2}}=I,\) by Lemma 2.2 and 2.10, \(\phi (A_{11}B_{11})=\phi (A_{11})\phi (B_{11}).\) Similarly, we can prove that \(\phi (A_{22}B_{22})=\phi (A_{22})\phi (B_{22}).\)

For \(D_{21}\in \mathcal {B}_{21}\), we have \(C_{21}=\phi ^{-1}(D_{21})\in {\mathcal {A}}_{12}\) by Lemma 2.10. Thus

$$\begin{aligned} \phi (A_{12}B_{22})D_{21}=\phi (A_{12}B_{22}C_{21})=\phi (A_{12})\phi (B_{22}C_{21})=\phi (A_{12})\phi (B_{22})D_{21} \end{aligned}$$

for all \(D_{21}\in \mathcal {B}_{21}\). Since \(\overline{Q_{1}}=I,\) by Lemmas 2.2 and 2.10, \(\phi (A_{12}B_{22})=\phi (A_{12})\phi (B_{22}).\) Similarly, we can prove that \(\phi (A_{21}B_{11})=\phi (A_{21})\phi (B_{11}).\) \(\square \)

Now we come to the position to show Theorem 2.5.

Proof of Theorem 2.5

For every rational number q, we have \(\phi (qI)=qI\). Indeed, since q is rational number, there exist two integers r and s such that \(q=\frac{r}{s}\). Since \(\phi (I)= I\) and \(\phi \) is additive, we get that

$$\begin{aligned} \phi (qI)=\phi \left( \frac{r}{s}I\right) =r\phi \left( \frac{1}{s}I\right) =\frac{r}{s}\phi (I)=qI. \end{aligned}$$

Now we show that \(\phi \) is real linear. 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 Lemma 2.11 that \(\phi (A)=\phi (B)^{2}\). By Lemma 2.9 (3), we get that \(\phi (B)\) is self-adjoint. So \(\phi (A)\) is positive. This shows that \(\phi \) preserves positive elements. Let \(\lambda \in {\mathbb {R}}\). Choose sequence \(\{a_{n}\}\) 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

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

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

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

Hence \(\phi \) is real linear.

By Lemma 2.11, \(\phi (iI)^{2}=\phi ((iI)^{2})=-\phi (I)=-I.\) By Lemma 2.9 (3), \(\phi (iI)^{*}=\phi ((iI)^{*})=-\phi (iI).\) Let \(F=\frac{I-i\phi (iI)}{2}\). Then it is easy to verify that F is a central projection in \(\mathcal {B}\). Let \(E=\phi ^{-1}(F)\). Then by Lemma 2.9, E is a central projection in \({\mathcal {A}}\). Moreover, for \(A\in {\mathcal {A}}\), there hold

$$\begin{aligned} \phi (iAE)=\phi (A)\phi (E)\phi (iI)=\phi (A)\phi (E)i(2F-I)=i\phi (A)F=i\phi (AE), \end{aligned}$$

and

$$\begin{aligned} \phi (iA(I-E))=\phi (A)\phi (I-E)\phi (iI)=-i\phi (A)(I-F)=-i\phi (A(I-E)). \end{aligned}$$

That is, the restriction of \(\phi \) to \({\mathcal {A}}E\) is linear and the restriction of \(\phi \) to \({\mathcal {A}}(I-E)\) is conjugate linear. This together with Lemmas 2.9 and 2.11 shows Theorem 2.5. \(\square \)