A note on strong skew Jordan product preserving maps on von Neumann algebras

Abstract

Let $\mathcal {A}$ be a von Neumann algebra acting on the complex Hilbert space $\mathcal {H}$ and $\Phi {:}\,\mathcal {A} \longrightarrow \mathcal {A}$ be a surjective map that satisfies the condition

$$\begin{aligned} \Phi (T)\Phi (P)+\Phi (P)\Phi (T)^*=TP+PT^* \end{aligned}$$

for all T and all projections P in $\mathcal {A}$. We characterize the concrete form of $\Phi$ on selfadjoint elements of $\mathcal {A}$. Also when $\mathcal {A}$ is a factor von Neumann algebra, it is shown that $\Phi$ is either of the form $\Phi (T)=T+i\tau (T) I$ or of the form $\Phi (T)=-T+i\tau (T)I$, where $\tau {:}\,\mathcal {A}\longrightarrow \mathbb {R}$ is a real map.

1 Introduction and statement of the main results

Let $\mathcal {A}$ be an algebra. For $A, B \in \mathcal {A}$, $A\circ B=AB+BA$ is the Jordan product and $[A,B]=AB-BA$ is the Lie product. The map $\Phi {:}\,\mathcal {A}\longrightarrow \mathcal {A}$ is Lie product preserving if $\Phi [A,B]=[\Phi (A),\Phi (B)]$ for all $A, B \in \mathcal {A}$. This kind of maps are closely related to Lie homomorphisms (linear maps preserving Lie products) which have been studied by many authors. For example see [2, 5, 7, 10, 11] and their references. Šemrl [17] characterized bijective maps preserving the Lie product on B(X). Zhang and Zhang [19] characterized the structure of bijective maps preserving the Lie product on factor von Neumann algebras, with a different method. Also the problem of characterizing maps preserving the commutativity (maps preserving zero Lie products), studied intensively (see [1, 2, 5, 7, 13, 17] and references therein).

In [1], Bell and Daif gave the conception of strong commutativity preserving maps. We say that a map $\Phi {:}\,\mathcal {A}\longrightarrow \mathcal {A}$ is strong commutativity preserving if $[\Phi (A),\Phi (B)]=[A,B]$, for all $A, B\in \mathcal {A}$. Note that a strong commutativity preserving map must be commutativity preserving, but the reverse is not true generally. These maps are also called strong Lie product preserving maps. Brešar and Miers [5] proved that every strong commutativity preserving additive map $\Phi$ on a semiprime ring $\mathcal {R}$ is of the form $\Phi (A) =\lambda A + \mu (A)$, where $\lambda \in \mathcal {C}$, the extended centroid of $\mathcal {R}$, $\lambda ^2 = 1$, and $\mu {:}\,\mathcal {R} \longrightarrow \mathcal {C}$ is an additive map. In [9], Lin and Liu obtained a similar result on a noncentral Lie ideal of a prime ring. In [14, 15], Qi and Hou gave a complete characterization of strong commutativity preserving surjective maps (without the assumption of additivity) on prime rings and triangular algebras, respectively.

Let $\mathcal {A}$ be a *-ring. For any $A, B \in \mathcal {A}$, $[A, B]_* = AB-BA^*$ denotes the skew Lie product of A and B. This kind of product is found playing a more and more important role in some research topics such as representing quadratic functionals with sesquilinear functionals, and its study has attracted many authors attention (see [3, 6, 18] and the references therein). Molnar [12] initiated the systematic study of this skew Lie product, and studied the relation between subspaces and ideals of $B(\mathcal {H})$, the algebra of all bounded linear operators acting on a Hilbert space $\mathcal {H}$.

Recall that a map $\Phi {:}\, \mathcal {A}\longrightarrow \mathcal {A}$ is called zero skew Lie product preserving, if $\Phi (A)\Phi (B)-\Phi (B)\Phi (A)^*=0$ whenever $AB-BA^* = 0$ for any $A, B \in \mathcal {A}$. Additive or linear maps preserving zero skew Lie products on various rings and algebras had been studied by many authors (see [1] and the references therein). More specially, $\Phi$ is strong skew commutativity preserving, if $[\Phi (A), \Phi (B)]_* = [A, B]_*$ for all $A, B\in \mathcal {A}$. It is obvious that strong skew commutativity preserving maps must be zero skew Lie product preserving. However, the reverse is not true in general. In [8], Cui and Park proved that, if $\mathcal {A}$ is a factor von Neumann algebra, then every strong skew commutativity preserving map $\Phi$ on $\mathcal {A}$ has the form $\Phi (A) = \Psi (A) + h(A)I$ for all $A\in \mathcal {A}$, where $\Psi {:}\, \mathcal {A}\longrightarrow \mathcal {A}$ is a linear bijective map satisfying $\Psi (A)\Psi (B)-\Psi (B)\Psi (A)^* = AB-BA^*$ for all $A, B\in \mathcal {A}$ and h is a real linear functional of $\mathcal {A}$ with $h(0) = 0$; particularly, if $\mathcal {A}$ is of type I, then $\Phi (A) = cA + h(A)I$ for each $A\in \mathcal {A}$, where $c \in \{-1, 1\}$. Recently, Qi and Hou [18], generalized the above result to von Neumann algebras without central summand of type $I_1$.

The purpose of the present paper is to continue these studies and to consider a special type of nonlinear strong skew Jordan product preserving maps on von Neumann algebras.

Note that a subalgebra $\mathcal {A}$ of $B(\mathcal {H})$ is called a von Neumann algebra when it is closed under the weak operator topology. A von Neumann algebra $\mathcal {A}$ is called factor when its center is trivial. We denote the real and the imaginary part of an operator T by Re(T) and Im(T), respectively, i.e. $Re(T) = (T+T^*)/2$ and $Im(T) = (T-T^*)/2i$. Recall that a von Neumann algebra is the closed linear span of the set of its projections. In this paper we will use this fact several times. For convenience, from here let $\mathcal {Z}(\mathcal {A})$ denote the center of $\mathcal {A}$, $\mathcal {P}(\mathcal {A})$ the set of all projections in $\mathcal {A}$, $\mathcal {A}_{ sa}$ the space of all Hermitian elements in $\mathcal {A}$ and define $Q=I-P$ for an arbitrary projection P.

We are now ready to state the main results of the paper.

Theorem 1.1

Let $\mathcal {H}$ be a complex Hilbert space and $\mathcal {A}$ be a von Neumann algebra acting on $\mathcal {H}$. Assume that a surjective mapping $\Phi$ on $\mathcal {A}$ satisfies the condition

$$\begin{aligned} \Phi (T)\Phi (P)+\Phi (P)\Phi (T)^*=TP+PT^*, \end{aligned}$$

(1.1)

for all $T\in \mathcal {A}$ and every $P\in \mathcal {P}(\mathcal {A})$. Then $\Phi (I)\in \mathcal {Z}(\mathcal {A})$ and there exists a map $f{:}\,{\mathcal {A}_{ sa}} \longrightarrow \mathcal {Z}(\mathcal {A})$ such that f(T) is a selfadjoint element in $\mathcal {Z}(\mathcal {A})$ and $\Phi (T)=T\Phi (I)+if(T)$ for all $T\in \mathcal {A}_{ sa}$. More precisely, $f(T)=Im(\Phi (T))$. In particular, for all $T\in \mathcal {A}_{ sa}$ for which $\Phi (T)$ is in $\mathcal {A}_{ sa}$, we have $\Phi (T)=T\Phi (I)$.

Theorem 1.2

Let $\mathcal {H}$ be a complex Hilbert space and $\mathcal {A}$ be a factor von Neumann algebra acting on $\mathcal {H}$. Assume that a surjective mapping $\Phi$ on $\mathcal {A}$ satisfies the condition 1.1 for all $T\in \mathcal {A}$ and every $P\in \mathcal {P}(\mathcal {A})$. Then there is a real map $\tau$ on $\mathcal {A}$ such that $\Phi (T)=T+i\tau (T) I$ for all $T\in \mathcal {A}$ or $\Phi (T)=-T+i\tau (T)I$.

2 The proof of the main results

Proof of Theorem 1.1

We will complete the proof by checking several steps.

Step 1 (i) $\Phi (0)=0$,

(ii) $\Phi$ is injective on $\mathcal {A}_{ sa}$.

• (i) For every $T\in \mathcal {A}$ we have $T0+0T^*=0$. By Eq. 1.1 we obtain that $\Phi (T)\Phi (0)+\Phi (0)\Phi (T)^*=0$. Because of the surjectivity of $\Phi$ we have $I \Phi (0)+\Phi (0)I^*=0$ and therefore $\Phi (0)=0$.

• (ii) Assume that T, S are selfadjoint elements in $\mathcal {A}$ such that $\Phi (T)=\Phi (S)$. By Eq. 1.1 we have

  $$\begin{aligned} \Phi (T)\Phi (I)+\Phi (I)\Phi (T)^*=TI+IT, \end{aligned}$$

  and

  $$\begin{aligned} \Phi (S)\Phi (I)+\Phi (I)\Phi (S)^*=SI+IS. \end{aligned}$$

Subtracting these two latter equations one gets $T-S=0$ or $T=S$, means that $\Phi$ is injective on selfadjoint elements in $\mathcal {A}$.

Step 2 For all $P\in \mathcal {P}(\mathcal {A})$, we have

• (i) $\Phi (P)^*=\Phi (P),$

• (ii) $\Phi (I)=\Phi (P)+\Phi (Q)$,

• (iii) $\Phi (P)^2=P.$

• (i) Since $\Phi$ is surjective so there exists $T\in \mathcal {A}$ such that $\Phi (T)=I$. Therefore by Eq. 1.1

  $$\begin{aligned} I\Phi (P)+\Phi (P)I^*=TP+PT^* \end{aligned}$$

  or $2\Phi (P)=TP+PT^*$. Obviously, we have $2\Phi (P)^*=TP+PT^*$ and this implies that $\Phi (P)^*=\Phi (P)$.

• (ii) Similarly to the part (i), there exists $T\in \mathcal {A}$ such that $\Phi (T)=I$ and

  $$\begin{aligned} 2\Phi (P)=TP+PT^*, \end{aligned}$$

  (2.1)

  for all $P\in \mathcal {P}(\mathcal {A})$. Let $P=Q$ in the above equation, thus

  $$\begin{aligned} 2\Phi (Q)=TQ+QT^*. \end{aligned}$$

  (2.2)

  Since $P+Q=I$, so adding these two equations implies that

  $$\begin{aligned} 2(\Phi (P)+\Phi (Q))=T+T^*. \end{aligned}$$

  (2.3)

  On the other hand by substituting P by I in Eq. 1.1 one gets

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

  (2.4)

  Comparing Eqs. 2.3 and 2.4 we obtain that $\Phi (I)=\Phi (P)+\Phi (Q)$.

• (iii) Let $T=P$ in Eq. 1.1. By part (i) it is clear that $\Phi (P)^2=P$.

Step 3 $\Phi (P)\Phi (Q)=\Phi (Q)\Phi (P)=0$ and $\Phi (P)Q=Q\Phi (P)=0$, for all $P\in \mathcal {P}(\mathcal {A})$.

Let $T=\Phi (P)\Phi (Q)$, so by part (iii) of step 2 and the fact that $\Phi (P)Q=Q\Phi (P)$ for all $P\in \mathcal {P}(\mathcal {A})$, we have

$$\begin{aligned} TT^*=\Phi (P)\Phi (Q)^2\Phi (P)=Q\Phi (P)^2. \end{aligned}$$

Again by using step 2, $TT^*=0$. Similarly, we can show that $T^*T=0$. Consequently, $T=\Phi (P)\Phi (Q)=0$.

The second part can be obtained easily from the first part and step 2.

Step 4 $\Phi (I)\in \mathcal {Z}(\mathcal {A})$.

Let $T=I$ in Eq. 1.1, so by part (i) of step 2 we have

$$\begin{aligned} \Phi (I)\Phi (P)+\Phi (P)\Phi (I)=2P, \end{aligned}$$

for all $P\in \mathcal {P}(\mathcal {A})$. Multiplying by $\Phi (I)$ from the left and right, respectively, one gets

$$\begin{aligned} \Phi (P)+\Phi (I)\Phi (P)\Phi (I)=2\Phi (I)P \end{aligned}$$

and

$$\begin{aligned} \Phi (I)\Phi (P)\Phi (I)+\Phi (P)=2P\Phi (I). \end{aligned}$$

These last two equations imply that $\Phi (I)P=P\Phi (I)$ for all $P\in \mathcal {P}(\mathcal {A})$. As we mentioned in Sect. 1, each operator in a von Neumann algebra is a limit of finite linear combinations of projections in $\mathcal {A}$. Thus we have $\Phi (I)T=T\Phi (I)$ for all $T\in \mathcal {A}$, means that $\Phi (I)$ is in the center of $\mathcal {A}$.

Step 5 $\Phi (P)=P\Phi (I)$, for all $P\in \mathcal {P}(\mathcal {A})$.

Let $\Phi (P)\Phi (I)=S$. By step 2, part (ii), we have $\Phi (P)(\Phi (P)+\Phi (Q))=S$. Since $\Phi (P)^2=P$ and $\Phi (P)\Phi (Q)=0$ one gets that $P=S$. Consequently, $\Phi (P)\Phi (I)=P$. Multiplying by $\Phi (I)$ from the right we obtain $\Phi (P)=P\Phi (I)$.

Step 6 There is a map $f{:}\,{\mathcal {A}_{ sa}} \longrightarrow \mathcal {Z}(\mathcal {A})$ such that $f(T)=Im(\Phi (T))$ and $\Phi (T)=T\Phi (I)+if(T)$ for all $T\in \mathcal {A}_{ sa}$. In particular, for all $T\in \mathcal {A}_{ sa}$ for which $\Phi (T)$ is in $\mathcal {A}_{ sa}$ we have $f(T)=0$ i.e. $\Phi (T)=T\Phi (I)$.

For every Hermitian element $T\in \mathcal {A}$, Eq. 1.1 implies that

$$\begin{aligned} \Phi (T)\Phi (P)+\Phi (P)\Phi (T)^*=TP+PT, \end{aligned}$$

(2.5)

for all $P\in \mathcal {P}(\mathcal {A})$.

Let $P=I$ in the above equation and multiply by $\Phi (I)$ from the right. This implies that $Re \Phi (T)=T\Phi (I)$. So we have $\Phi (T)=T\Phi (I)+iU$, where U is the imaginary part of $\Phi (T)$. Now let $\Phi (T)=T\Phi (I)+iU$ in Eq. 2.5. Since $\Phi (P)=P\Phi (I)$ it is easily seen that $UP=PU$, for all $P\in \mathcal {P}(\mathcal {A})$. It means that U is in the center of $\mathcal {A}$. So there exists a map $f{:}\,{\mathcal {A}_{ sa}} \longrightarrow \mathcal {Z}(\mathcal {A})$ such that $f(T)=U=Im(\Phi (T))$ and $\Phi (T)=T\Phi (I)+if(T)$ for all $T\in \mathcal {A}_{ sa}$.

Now suppose that T and $\Phi (T)$ are in $\mathcal {A}_{ sa}$. Clearly, $f(T)=0$ and $\Phi (T)=T\Phi (I)$. This completes the proof of Theorem 1.1. $\square$

Proof of Theorem 1.2

By checking the proof of Theorem 1.1, steps 1–4 still hold. We complete the proof using several further steps.

Step 1 $\Phi (I)=I$ or $\Phi (I)=-I$.

By step 4 of the proof of Theorem 1.1, we know that $\Phi (I)$ is in the center of $\mathcal {A}$. Since $\mathcal {A}$ is a factor, one can conclude that $\Phi (I)\in \mathbb {C}I$. By the facts that $\Phi (I)^*=\Phi (I)$ and $\Phi (I)^2=I$, it is easily seen that $\Phi (I)=I$ or $\Phi (I)=-I$.

From here we suppose that $\Phi (I)=I$ without loss of generality.

Step 2 $\Phi (P)=P$ for all $P\in \mathcal {P}(\mathcal {A})$.

It is obvious because of step 5 of the proof of Theorem 1.1 and the fact that $\Phi (I)=I$.

Step 3 $\Phi$ preserves skew selfadjoint elements of $\mathcal {A}$.

Suppose that T is a skew selfadjoint element in $\mathcal {A}$, that is $T+T^*=0$. By Eq. 1.1 we have

$$\begin{aligned} \Phi (T)\Phi (I)+\Phi (I)\Phi (T)^*=0. \end{aligned}$$

Since $\Phi (I)=I$ it is obvious that $\Phi (T)=-\Phi (T)^*$.

Step 4 There exists a real functional $\tau _1$ on selfadjoint elements in $\mathcal {A}$ such that $\Phi (T)=T+i\tau _1(T) I$ for all $T\in \mathcal {A}_{ sa}$.

Similarly to step 6 of the proof of Theorem 1.1 and by the fact that $\Phi (I)=I$, we can show that $\Phi (T)=T+iU$, where U is in the center of $\mathcal {A}$. Since $\mathcal {A}$ is a factor and U is selfadjoint so there exists $\alpha \in \mathbb {R}$, corresponding to T, such that $U=\alpha I$, Hence we conclude that there is a real functional $\tau _1{:}\, {\mathcal {A}_{ sa}} \longrightarrow \mathbb {R}$ such that $\Phi (T)=T+i\tau _1(T) I$.

Step 5 There exists a real functional $\tau _2$ on selfadjoint elements in $\mathcal {A}$ such that $\Phi (iT)=iT+i\tau _2(T) I$ for all $T\in \mathcal {A}_{ sa}$.

Assume that T is a selfadjoint element in $\mathcal {A}$, so iT is skew selfadjoint. By step 3 we know that $\Phi$ preserves skew selfadjoint elements, thus $\Phi (iT)^*=-\Phi (iT)$, so Eq. 1.1 implies

$$\begin{aligned} \Phi (iT)\Phi (P)-\Phi (P)\Phi (iT)=iTP-iPT, \end{aligned}$$

(2.6)

for all $P\in \mathcal {P}(\mathcal {A})$. Since $\phi (P) = P$ in the above equation, therefore we have

$$\begin{aligned} (\Phi (iT)-iT)P=P(\Phi (iT)-iT), \end{aligned}$$

for every $P\in \mathcal {P}(\mathcal {A})$. Since $\mathcal {A}$ is a von Neumann algebra thus $\Phi (iT)-iT \in \mathcal {Z}(\mathcal {A})$. $\mathcal {A}$ is a factor, consequently, there exists $\lambda \in \mathbb {C}$ corresponding to iT such that $\Phi (iT)-iT=\lambda I$. Note that $\Phi (iT)-iT$ is skew selfadjoint, therefore $\lambda =i\beta$ for $\beta \in \mathbb {R}$. These statements imply that there is a real functional $\tau _2$ on selfadjoint elements of $\mathcal {A}$ such that $\Phi (iT)=iT+i\tau _2 (T)I$.

Step 6 $\Phi (T)=T+i\tau (T) I$ for all $T\in \mathcal {A}$, where $\tau$ is a real functional on $\mathcal {A}$ such that $\tau (P)=0$ for every $P\in \mathcal {P}(\mathcal {A})$.

It is obvious by steps 4 and 5. Clearly, $\tau$ is a real functional and $\tau (P)=0$ for every $P\in \mathcal {P}(\mathcal {A})$.

Similarly, it can be shown that if $\Phi (I)=-I$ then $\Phi$ is of the form $\Phi (T)=-T+i\tau (T)I$ for all $T\in \mathcal {A}$. $\square$