Nonlinear generalized semi-Jordan triple derivable mappings on completely distributive commutative subspace lattice algebras

1.   Introduction

Let $\mathcal{A}$ be an associative algebra and $\mathcal{M}$ be an $\mathcal{A}$-bimodule. Recall that a linear mapping $\delta: \mathcal{A}\rightarrow \mathcal{M}$ is called a derivation, Jordan derivation if $\delta(AB) = \delta(A)B+A\delta(B)$, $\delta(A\circ B) = \delta(A)\circ B+A\circ \delta(B)$ hold for all $A, B\in\mathcal{A}$, respectively, where $A\circ B = AB+BA$ is the usual Jordan product. Also, $\delta$ is called Jordan triple derivation if $\delta(A\circ B\circ C) = \delta(A)\circ B\circ C+A\circ \delta(B)\circ C+A\circ B\circ \delta(C)$ for all $A, B, C\in\mathcal{A}$. If there is no assumption of additivity for $\delta$ in the above definitions, then $\delta$ is said to be nonlinear. $\delta$ is called Jordan derivable mapping if $\delta$ satisfies $\delta(A\circ B) = \delta(A)\circ B+A\circ \delta(B),$ for every $A, B\in\mathcal{A}$ with $A\circ B \in \Omega$ where $\Omega$ is a set which satisfies some conditions. $\delta$ is called nonlinear generalized semi-Jordan triple derivable mapping if there is no assumption of additivity for $\delta$ but $\delta$ satisfies

for all $A, B, C\in\mathcal{A}$ with $ABC\in\Omega$. Clearly, every derivation is a Jordan derivation as well as triple derivation, and every triple derivation is a Jordan triple derivation. The converse is not true in general(see ^[1,2,3]).

The standard problem is to find out whether (under some conditions) a Jordan (triple) derivation is necessarily a derivation. In 1957, Herstein ^[4] proved that every Jordan derivation on 2-torsion free prime rings is a derivation, and it is the first result in this direction. Then, many mathematicians studied this problem and obtained abundant results. Zhang ^[5] extended Herstein's result to the triangular algebra. Later, Ma ^[2] proved that each generalized Jordan derivation from the upper triangular matrix algebra into its bimodule can be uniquely decomposed into the sum of a generalized derivation and an anti-derivation. With the development of research, many achievements have been obtained that linear (or nonlinear) mappings on operator algebras are derivations, such as Jordan triple derivable mappings. Li ^[6] proved that every Jordan derivable mapping on nest algebras is a derivation. Ashraf and Jabeen ^[7] showed that each nonlinear Jordan triple higher derivable mapping of triangular algebras is an additive derivation. Zhao and Li in ^[8] proved that every nonlinear Jordan triple $*$-derivation on von Neumann algebras with no central summands of type $I_1$ is an additive $*$-derivation, and Darvish ^[9] extended the result to $*$-algebra. An and He in ^[10] study $(m, n)$-Jordan derivable mappings at zero on generalized matrix algebras. Recently, Fei and Zhang in ^[11] proved that every nonlinear nonglobal semi-Jordan triple derivable mapping on triangular algebras is an additive derivation. For more details see ^{[12,13,14,15,16,17,18]} and references therein.

Let $\mathcal{H}$ be a Hilbert space over real or complex field $\mathcal{F}$ and $\mathcal{L}$ be the subspace lattice of $\mathcal{H}$. A subspace lattice $\mathcal{L}$ is called a commutative subspace lattice($CSL$) if each pair of projections in $\mathcal{L}$ commutes, and $Alg {\cal L} = \{T\in B(\mathcal{H}): T(L)\subseteq L, \forall L \in \mathcal{L} \}$ is the associated subspace lattice algebra in $\mathcal{L}$, which is called $CSL$ algebra. A totally ordered subspace lattice is called a nest. Recall that a subspace lattice is called completely distributive if $e = \bigvee \{L \in \mathcal{L}: N_{-} \nsupseteq L \}$ for every $0\neq e \in \mathcal{L},$ where $N_{-} = \bigvee \{P \in \mathcal{L}: P \nsupseteq N \}$. Accordingly, its associated subspace lattice algebra is called completely distributive $CSL$ algebra($CDC$ algebra). For standard definitions concerning completely distributive subspace lattice algebras see ^[19,20].

In ^[21], they proved that the collection of finite sums of rank-one operators in a $CDC$ algebra is strongly dense. This result will be frequently used in the study of $CDC$ algebra. Let $Alg {\cal L}$ be a $CDC$ algebra. Set $\mathcal{U(L)} = \{e\in \mathcal{L}:e\neq 0, e_{-}\neq H\}.$ We say $e, e^{\prime} \in \mathcal{U(L)}$ are connected if there exist finitely many projections $e_1, e_2, ..., e_n \in \mathcal{U(L)},$ such that $e_i$ and $e_{i+1}$ are comparable for each $i = 0, 1, \ldots, n$, where $e_0 = e, e_{n+1} = e^{\prime}.$ $\mathcal{C}\subseteq \mathcal{U(L)}$ is called a connected component if each pair in $\mathcal{C}$ is connected and any element in $\mathcal{U(L)}\diagdown \mathcal{C}$ is not connected with any element in $\mathcal{C}$. Recall that a $CDC$ algebra $Alg {\cal L}$ is irreducible if and only if the commutant is trivial, i.e. $(Alg {\cal L})' = \mathcal{F}I$, which is also equivalent to the condition that $\mathcal{L}\bigcap\mathcal{L^{\perp}} = \{0, I\},$ where $\mathcal{L}^{\perp} = \{e^{\perp}: e\in\mathcal{L}\}.$ Clearly, Nest algebra is irreducible. In ^[22,23], it turns out that any $CDC$ algebra can be decomposed into the direct sum of irreducible $CDC$ algebras.

Lemma 1.1 ^[22,23]. Let $Alg \mathcal{L}$ be a $CDC$ algebra on a separable Hitbert space $\mathcal{H}$. Then, there are no more than countably many connected components $\mathcal{C}_{n}: n\in \Lambda$ of $\mathcal{E(L)}$ such that $\mathcal{E(L)} = \cup\{e: e\in \mathcal{C}_{n}, n\in \Lambda\}$. Let $e_{n} = \vee \{ e: e\in \mathcal{C}_{n}, n\in \Lambda \}.$ Then, $\{e_{n}, n\in \Lambda \}\subseteq \mathcal{L}\cap \mathcal{L}^{\perp}$ is a subset of pairwise orthogonal projections, and the algebra $Alg \mathcal{L}$ can be written as a direct sum:

where each $(Alg \mathcal{L})e_{n}$ viewed as a subalgebra of operators acting on the range of $e_{n}$ is an irreducible $CDC$ algebra. Thus, all convergence means strong convergence.

From the definition of $e_{n}$, we know that its linear span is Hilbert space $\mathcal{H}$, and pairwise orthogonal projection. It follows that the identity and center of $Alg \mathcal{L}$ are $I = \sum_{n\in \Lambda}\oplus e_{n}$ and $\mathcal{Z}(Alg \mathcal{L}) = \sum_{n\in \Lambda}\oplus \lambda_{n}e_{n},$ respectively, where $\lambda_{n} \in \mathcal{F}.$ In ^[24], they prove that each Jordan isomorphism between irreducible $CDC$ algebras is the sum of an isomorphism and an anti-isomorphism.

Lemma 1.2 ^[24]. Let $Alg \mathcal{L}$ be a non-trivially irreducible $CDC$ algebra on a complex Hilbert space $\mathcal{H}$. Then, there exists a non-trivial projection $e\in \mathcal{L},$ such that $e(Alg \mathcal{L})e^{\perp}$ is faithful $Alg \mathcal{L}$ bimodule, i.e., for all $A\in Alg \mathcal{L},$ if $Ae (Alg \mathcal{L})e^{\perp} = \{0\},$ then $Ae = 0$ and if $e (Alg \mathcal{L})e^{\perp}A = \{0\},$ then $e^{\perp}A = 0.$

Let $I$ be the identity operator on $\mathcal{H}$. If $\mathcal{L}$ is non-trivial, by Lemma 1.2, there exists a non-trivial projection $e\in\mathcal{L},$ such that $e(Alg \mathcal{L})e^{\perp}$ is faithful $Alg \mathcal{L}$ bimodule. Set $e_{1} = e, e_{2} = I-e_{1}$. Then, $e_{1}, e_{2}$ are the projections of $Alg \mathcal{L}$. Thus, for every $A$ in irreducible $CDC$ algebra $Alg \mathcal{L},$ $A$ can be decomposed as: $A = e_{1} A e_{1}+e_{1} A e_{2}+e_{2} A e_{2}.$ Set $\mathcal{A}_{ij} = e_{i}(Alg \mathcal{L})e_{j}$. Then $Alg \mathcal{L}$ can be decomposed as

In the present note, we pursue nonlinear generalized semi-Jordan triple derivable mappings on completely distributive commutative subspace lattice algebras. Without loss of generality, we assume that any algebra is 2-torsion free.

2.   Nonlinear generalized semi-Jordan triple derivable mappings on irreducible completely distributive commutative subspace lattice algebras

In this section, we begin with the irreducible case.

Theorem 2.1. Let $Alg \mathcal{L}$ be an irreducible $CDC$ algebra on a complex Hilbert space $\mathcal{H}$ and $\delta: Alg \mathcal{L}\rightarrow Alg \mathcal{L}$ be a mapping without the additivity assumption and satisfy

for all $A, B, C\in\mathcal{A}$ with $ABC\in\Omega = \{A\in Alg \mathcal{L}:A^{2} = 0\}$. Then, $\delta$ is an additive derivation.

Assume that $Alg \mathcal{L}$ is an irreducible $CDC$ algebra, $e_{1}\in Alg \mathcal{L}$ is an associated non-trivial projection, $e_{2} = I-e_{1}$, $\mathcal{A}_{ij} = e_{i}(Alg \mathcal{L})e_{j}(i, j = 1, 2)$, and $\delta$ is a mapping which satisfies (2.1). It is easy to obtain that $\delta(0) = 0$(taking $A = B = C = 0$ in (2.1)). Moreover, we have the following result.

Lemma 2.1. For every $A_{ij}\in\mathcal{A}_{ij}$, we have $\delta(A_{11})\in\mathcal{A}_{11}+\mathcal{A}_{12}$, $\delta(A_{12})\in \mathcal{A}_{12}$ and $\delta(A_{22})\in\mathcal{A}_{12}+\mathcal{A}_{22}$. Moreover, $e_{1}\delta(A_{11})e_{2} = A_{11}\delta(e_{1}) = -A_{11}\delta(e_{2})$ and $e_{1}\delta(A_{22})e_{2} = \delta(e_{2})A_{22} = -\delta(e_{1})A_{22}$.

Proof. Put $A = B = e_{1}, C = e_{2}$ in (2.1), and note that $e_{1}e_{1}e_{2} = 0\in\Omega$. Thus,

By the definition of $e_{1}, e_{2},$ we have $e_{1}\delta(e_{2})e_{1} = 0$ and $e_{1}\delta(e_{1})e_{2}+e_{1}\delta(e_{2})e_{2} = 0.$ Similarly, taking $A = e_{1}, B = C = e_{2}$ in (2.1) we can obtain $e_{2}\delta(e_{1})e_{2} = 0.$

For every $A_{12}\in \mathcal{A}_{12}$, putting $A = e_{2}, B = A_{12}, C = e_{2}$ in (2.1) and combining $e_{2}A_{12}e_{2} = 0\in\Omega$ we have

Multiplying left by $e_{1}$ and right by $e_{2}$ in (2.2) and combining $e_{1}\delta(e_{2})e_{1} = 0$, we have $2A_{12}e_{2}\delta(e_{2})e_{2} = 0$. Similarly we can obtain $2e_{1}\delta(e_{2})e_{1}A_{12} = 0$. By Lemma 1.2, we have $e_{1}\delta(e_{1})e_{1} = e_{2}\delta(e_{2})e_{2} = 0$ and $\delta(e_{1}) = -\delta(e_{2})\in \mathcal{A}_{12}$.

For any $A_{11}\in\mathcal{A}_{11}$, putting $A = A_{11}, B = C = e_{2}$ in (2.1) by $A_{11}e_{2}e_{2} = 0\in\Omega$ and $\delta(e_{2})\in\mathcal{A}_{12}$ we have

This implies that $e_{2}\delta(A_{11})e_{2} = 0$, and then $\delta(A_{11})\in\mathcal{A}_{11}+\mathcal{A}_{12}$. Furthermore, multiplying left by $e_{1}$ and right by $e_{2}$ in (2.3) and following from $\delta(e_{1}) = -\delta(e_{2})\in\mathcal{A}_{12},$ we can obtain

For any $A_{22}\in\mathcal{A}_{22}$, putting $A = B = e_{1}, C = A_{22}$ in (2.1) it follows from $e_{1}e_{1}A_{22} = 0\in\Omega$ that

This implies that $e_{1}\delta(A_{22})e_{2} = -\delta(e_{1})A_{22} = \delta(e_{2})A_{22}$, $e_{1}\delta(A_{22})e_{1} = 0,$ and $\delta(A_{22})\in\mathcal{A}_{12}+\mathcal{A}_{22}$.

For any $A_{12}\in \mathcal{A}_{12}$, noting that $A_{12}e_{1}e_{2} = 0\in\Omega$, putting $A = A_{12}, B = e_{1}, C = e_{2}$ in (2.1) and combining $\delta(e_{1}) \in \mathcal{A}_{12}$ we have

The proof is completed.□

Lemma 2.2. For any $A_{ij}, B_{ij}\in \mathcal {A}_{ij}$, we have $\delta(A_{ij}B_{kl}) = \delta(A_{ij})B_{kl}+A_{ij}\delta(B_{kl})$, for all $i, j, k, l = 1, 2$ and $i\leq j\leq k\leq l$.

Proof. Consider the case when $i = j = 1, \ k = l = 2.$ Since $A_{11}A_{22}e_{2} = 0\in\Omega$, putting $A = A_{11}, B = A_{22}, C = e_{2}$ in (2.1) it then follows from Lemma 2.1 that

Consider the case when $i = j = k = 1, \ l = 2.$ Since $A_{12}A_{11}e_{2} = 0\in\Omega$, putting $A = A_{12}, B = A_{11}, C = e_{2}$ in (2.1) by $\delta(e_{2})\in\mathcal{A}_{12}$ and Lemma 2.1 we can obtain that

Consider the case when $i = 1, j = k = l = 2.$ Since $A_{12}A_{11}e_{2} = 0\in\Omega$, taking $A = A_{22}, B = A_{12}, C = e_{2}$ in (2.1) by $\delta(e_{2})\in\mathcal{A}_{12}$ and Lemma 2.1 we can obtain that

Consider the case when $i = j = k = l = 1.$ By (2.4) we know that

and

Comparing (2.6) and (2.7), we get

It follows from Lemma 2.1 that

Furthermore, by Lemma 2.1 we have

Noting that $e_{2}\delta(A_{11}B_{11})e_{2} = 0$ and $\delta(A_{11})\in\mathcal{A}_{11}+\mathcal{A}_{12}$, we get

Similarly, by (2.5) one can check that when $i = j = k = l = 2$,

The proof is completed. □

Lemma 2.3. $\delta$ is an additive mapping on irreducible $CDC$ algebra $Alg \mathcal{L}$.

Proof. We divide the proof into three claims.

Claim 1. For all $A_{ij} \in \mathcal {A}_{ij}$, $\delta(A_{11}+A_{12}) = \delta(A_{11})+\delta(A_{12})$ and $\delta(A_{12}+A_{22}) = \delta(A_{12})+\delta(A_{22})$.

For every $A_{ij}, B_{ij}\in \mathcal {A}_{ij}$, noting that $\delta(e_{i})\in\mathcal{A}_{12}(i = 1, 2),$ $e_{1}\delta(A_{11})e_{2} = -A_{11}\delta(e_{2})\in\mathcal{A}_{12}$ and $e_{2}(A_{11}+A_{12})e_{2} = 0\in\Omega$. Then, putting $A = e_{2}, B = A_{11}+A_{12}, C = e_{2}$ in (2.1) we can obtain

It follows from $\delta(A_{12}) = e_{1}\delta(A_{12})e_{2}$ that $e_{2}\delta(A_{11}+A_{12})e_{2} = 0$ and

Furthermore, since $B_{12}(A_{11}+A_{12})e_{2} = 0\in\Omega$, putting $A = B_{12}, B = A_{11}+A_{12}, C = e_{2}$ in (2.1) then from Lemma 2.1 we get

By Lemma 2.2, based on $(\delta(A_{11}+A_{12})-\delta(A_{11}))B_{12} = 0$ and Lemma 1.2, we have

Hence, by (2.8), (2.9) and Lemma 2.1 we get

Putting $A = A_{12}+A_{22}, B = e_{1}, C = e_{2}$ in (2.1), one can check that

Claim 2. For all $i, j = 1, 2$ and $i\leq j$, $\delta(A_{ij}+B_{ij}) = \delta(A_{ij})+\delta(B_{ij}).$

Since $(B_{12}+e_{2})(e_{1}+A_{12})e_{2} = 0\in\Omega$, taking $A = (e_{1}+A_{12}), B = (e_{2}+B_{12}), C = e_{2}$ in (2.1), by (2.10), (2.11), Lemma 2.1 and $\delta(e_{1}) = -\delta(e_{2})\in\mathcal{A}_{12}$ we have

From (2.4), we know that

From (2.12) and (2.4), we have

Combining above two equations, we can get $(\delta(A_{11}+B_{11})-\delta(A_{11})-\delta(B_{11}))A_{12} = 0,$ for all $A_{12} \in \mathcal{A}_{12}$. From Lemma 1.2, we have

It follows from Lemma 2.1 that

Therefore, it follows from above two equations and $e_{2}\delta(A_{11}+B_{11})e_{2} = 0$ that

Similarly, one can check that

Claim 3. $\delta(A_{11}+A_{12}+A_{22}) = \delta(A_{11})+\delta(A_{12})+\delta(A_{22}).$

For any $A_{ij}\in \mathcal{A}_{ij}$, since $(A_{11}+A_{12}+A_{22})e_{1}e_{2} = 0\in\Omega$, putting $A = A_{11}+A_{12}+A_{22}, B = e_{1}, C = e_{2}$ in (2.1) it follows from $\delta(e_{1}) = -\delta(e_{2})\in\mathcal{A}_{12}$ and Lemma 2.1 that

It follows from $\delta(A_{12}) \in \mathcal{A}_{12}$ that

For any $B_{12}\in\mathcal{A}_{12}$, since $(e_{1}(A_{11}+A_{12}+A_{22})B_{12})^2 = (A_{11}B_{12})^2 = 0$ which implies $e_{1}(A_{11}+A_{12}+A_{22})B_{12}\in \Omega$, putting $A = e_{1}, B = A_{11}+A_{12}+A_{22}, C = B_{12}$ in (2.1) it then follows from Lemma 2.1 and $\delta(e_{1})\in\mathcal{A}_{12}$ that

It follows from Lemma 2.2 and Claim 1, 2 that

Comparing above two equations, we obtain that $2e_{1}(\delta(A_{11}+A_{12}+A_{22})-\delta(A_{11}))e_{1}B_{12} = 0$, and then by Lemma 1.2, we have

Similarly, one can check that

It follows from (2.15)–(2.17) that $\delta(A_{11}+A_{12}+A_{22}) = \delta(A_{11})+\delta(A_{12})+\delta(A_{22})$ and then $\delta$ is an additive mapping. The proof is completed. □

In the following, we give the completed proof of Theorem 2.1.

Proof of Theorem 2.1. Let $A = A_{11}+A_{12}+A_{22}$ and $B = B_{11}+B_{12}+B_{22}$ be arbitrary elements of irreducible $CDC$ algebra $Alg \mathcal{L}$ where $A_{ij}, B_{ij}\in\mathcal A_{ij}$. It follows from Lemmas 2.1–2.3 that

Therefore $\delta$ is an additive derivation on irreducible $CDC$ algebra $Alg \mathcal{L}$. The proof is completed. □

3.   Main results

In this section, we study nonlinear generalized semi-Jordan triple derivable mappings on completely distributive commutative subspace lattice algebras. The main result reads as follows.

Theorem 3.1. Let $Alg \mathcal{L}$ be an associated completely distributive commutative subspace lattice algebras on a complex Hilbert space $\mathcal{H}$ and $\delta: Alg \mathcal{L}\rightarrow Alg \mathcal{L}$ be a mapping without the additivity assumption and satisfy

for all $A, B, C\in\mathcal{A}$ with $ABC\in\Omega = \{A\in Alg \mathcal{L}:A^{2} = 0\}$. Then, $\delta$ is an additive derivation.

Proof. Let $e_{n} = \vee \{ e: e\in \mathcal{C}_{n}, n\in \Lambda \}$ be the projections of $\mathcal{L}$ as in Lemma 1.1. By Lemma 1.1, we know that $Alg \mathcal{L} = \sum_{n\in \Lambda}\oplus(Alg \mathcal{L})e_{n}$ is the irreducible decomposition of $Alg \mathcal{L}$. Fix an index $n$, it follows that $e_{n}$ is also Hilbert space and

Then, for all $A \in Alg \mathcal{L}$ and $e_{n},$ $Alg (e_{n}\mathcal{L})$ is an irreducible $CDC$ algebra on Hilbert space $e_{n}$. Let $\delta$ be a nonlinear generalized semi-Jordan triple derivable mapping from $Alg \mathcal{L}$ into itself. Then, it follows from Theorem 2.1 that there exists an additive derivation $\delta_n$ from $(Alg \mathcal{L})e_{n}$ into itself such that for all $A, B \in Alg (e_{n}\mathcal{L})$

In ^[23], they prove that $Alg \mathcal{L}$ is $CDC$ algebra if and only if the linear span of the rank-one operators in $Alg \mathcal{L}$ is ultraweakly dense. Choose a set $E \in \mathcal{U}(L)$, then, for every $x\in E,$ fix an element $y\in E_{-}^{\perp}$, and then $x\otimes y \in Alg \mathcal{L}$ is a rank-one operator. For every $u\otimes v \in Alg (e_{n}\mathcal{L})$ and $A \in Alg (e_{n}\mathcal{L})$, it follows from Theorem 2.1 that

Assuming that $\{A_{k}\}, A \in Alg (e_{n}\mathcal{L})$ and $\{A_{k}\}$ strongly converge to $A,$ it follows from (3.1) that

This means that $\delta_{n}$ is strongly convergent.

Assume $\{A_{k}\}, \{B_{k}\}, \{C_{k}\} \in Alg \mathcal{L}$ and $\{A_{k}\}, \{B_{k}\}, \{C_{k}\}$ converge strongly to $A, B, C$, respectively. Since $Alg \mathcal{L} = \sum_{n\in \Lambda}\oplus(Alg \mathcal{L})e_{n},$ and $e_{n}$ are pairwise orthogonal projection, for every $e_{i},$ $\{A_{k}e_{i}\}, \{B_{k}e_{i}\}, \{C_{k}e_{i}\}$ converge strongly to $Ae_{i}, Be_{i}, Ce_{i}$, respectively and

Then, for every $x$ in Hilbert space $\mathcal{H}$ and $A_{k}B_{k}C_{k} \in \Omega$, $\{A_{k}\}, \{B_{k}\}, \{C_{k}\}$ converging strongly to $A, B, C$ implies that $ABC \in \Omega$. It follows from the proof of Theorem 2.1 that

It means that $\delta$ is strongly convergent on $CDC$ algebra $Alg \mathcal{L}$. Thus, for every $A, B \in Alg \mathcal{L}$ we obtain that

The proof is completed. □

4.   Conclusions

In this paper, we use the structure properties of completely distributive commutative subspace lattice algebras and decomposition of algebraic to study the derivable mapping on certain $CSL$ algebra. We proved that every nonlinear generalized semi-Jordan triple derivable mapping on completely distributive commutative subspace lattice algebras is an additive derivation. Moreover, the purpose of this modification is to answer the classic problem of preserving derivable mappings of certain $CSL$ algebra.

Use of AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

The Project Supported by Natural Science Foundation of Shaanxi Province (No. 2023-JC-YB -082), "Qinglan talents" Program of Xianyang Normal University of China (No.XSYQL201707), the Key Cultivation Project of Xianyang Normal University of China (No.XSYK22031)

Conflict of interest

The authors declare no conflicts of interest.