Lie Triple Derivations on 𝒥-Subspace Lattice Algebras

Abstract

We describe the structure of Lie triple derivations on -subspace lattice algebras. The results can be applied to atomic Boolean subspace lattice algebras and pentagon subspace lattice algebras, respectively.

1. Introduction and Preliminaries

Let be an associative algebra, and let be an -bimodule. We denote by the center of relative to ; that is, for all. A linear mapping is called a Lie triple derivation if for all , where is the usual Lie product. We say that a Lie triple derivation is standard if it can be decomposed as a sum of a derivation from   to   and a mapping from   to   vanishing on every double commutator. The standard problem, which has been studied for many years, is to find conditions on under which each Lie triple derivation is standard or standard-like. This problem has been investigated for von Neumann algebras in [1], for prime rings in [2], for nest algebras in [3, 4], for TUHF algebras in [5], and for upper triangular algebras in [6]. In this present note, we pursue this line of investigation for -subspace lattice algebras.

Throughout, all algebras and vector spaces will be over , where is either the real field or the complex field . Given a Banach space with topological dual , by we mean the algebra of all bounded linear operators on . The terms operator on and subspace of will mean “bounded linear map of into itself” and “norm closed linear manifold of ,” respectively. For , denote by the adjoint of . For any nonempty subset , denotes its annihilator; that is, for all . For and , the rank one operator is defined by for .

A family of subspaces of is called a subspace lattice on if it contains (0) and and is complete in the sense that it is closed under the operations (closed linear span) and (set-theoretic intersection). For a subspace lattice on , the associated subspace lattice algebra is the set of operators on leaving every subspace in invariant; that is Obviously, is a unital weakly closed operator algebra. A subalgebra of is called a standard subalgebra if it contains all finite-rank operators in .

Subspace lattice algebras are important and mainly consist of nonself-adjoint operator algebras. Completely distributive subspace lattice algebras, commutative subspace lattice algebras, atomic Boolean subspace lattice algebras, pentagon subspace lattice algebras, and so forth have been widely studied. See, for example, [7–11]. Recently, Panaia in [12] introduced a new class of subspace lattices--subspace lattice algebras. Several authors have studied -subspace lattice as well as -subspace lattice algebras; see, for example, [13–18].

Given a subspace lattice on Banach space , put where . Call a -subspace lattice on if(1), (2), (3),  for every  ,(4),for every.

The simplest example of -subspace lattice is any pentagon subspace lattice . Here , and are subspaces of satisfying , and . In this case, , , and . For further discussion of pentagon subspace lattice see [8, 10]. Another important element of the class of -subspace lattice is the atomic Boolean subspace lattice. It follows from [15] that every commutative -subspace lattice on a Hilbert space is an atomic Boolean subspace lattice. However, most -subspace lattices on Hilbert space are non-commutative.

Therefore, -subspace lattices as well as -subspace lattice algebras deserve some attention. In the previous papers [3, 17, 18], we studied algebraic isomorphisms, Jordan isomorphisms, Jordan derivations, and Lie derivations. Here we study Lie triple derivations of -subspace lattice algebras. Even for pentagon subspace lattice algebras and atomic Boolean subspace lattice algebras, our results are new.

For a subspace lattice , the relevance of is due to the following lemma, which is crucial to what follows.

Lemma 1 (see [11]). If is a subspace lattice on , then the rank-one operator belongs to if and only if there exists a subspace in such that and , where means .

From Lemma 1, we can see that if is a -subspace lattice, then is rich in rank-one operators. Moreover, finite-rank operators in a -subspace lattice algebra have nice properties. Given a subspace lattice , by we denote the algebra of all finite-rank operators in . If , then we write and , the linear manifold spanned by .

Lemma 2 (see [17]). Let be a -subspace lattice on . Suppose that is an operator of rank in . Then can be written as a sum of rank-one operators in .

Recalling that a linear mapping of an algebra is a local derivation if for every there is a derivation , depending on , such that .

Lemma 3 (see [3]). Let be a -subspace lattice on . Then every local derivation from to a standard subalgebra of   is a derivation.

Lemma 4 (see [3]). Let be a -subspace lattice on . Suppose that is in , in , and in . If , then .

2. Lie Triple Derivation on

The main result in this section reads as follows.

Theorem 5. Let be a -subspace lattice on and a standard subalgebra of . Let be a Lie triple derivation from to . Then is standard.

For the proof of the theorem, we need some lemmas. In the following, we keep the notation as in the statement of the theorem. Recalling that the statement means that is the sum of a derivation from to and a linear mapping from to vanishing on every double commutator. Here for all. From [3, Remark 2.5(i)], we know that is equal to , the center of .

Lemma 6. Let and be an idempotent operator in . Then there are an operator in and a unique operator in such that .

Proof. Set and . Note that does not necessarily contain ; we understand and for .
For , we have that Let . Since is arbitrary, we have . Then Multiplying this equation by from both sides, we get that is, It follows from the Kleinecke-Shirokov theorem (cf. [19, Problem 232]) that is quasinilpotent. Let . Then there exists a scalar , such that for all . Particularly, ; that is, . This implies that either or since is quasinilpotent. Consequently, we always have for all . Hence since . Then by (5) we have that
Similarly, for , by considering , we have that
Now for , we have that Multiplying this equation by from the left side and by from the right side, we get Similarly, for , we have that
Using (7)–(11), it is easy to verify that . Now let . Then and . Moreover, by Lemma 4, such is unique.

Lemma 7. Let be in with . Then there are an operator in and a unique operator in such that .

Proof. If , then the result follows from the linearity and Lemma 6.
We now suppose that . Take such that . Obviously, and are linearly independent. Set . Then by Lemma 6, , where and . We associate a new Lie triple derivation as follows: Then .
Let and . Then We will show . For this, we first observe that for some . Since it follows that ; that is, . Hence since and are linearly independent. So .
Now by (13), we have . Therefore there exists such that and . Choose such that . Then Let . Then , and so . Thus and are desired.

Lemma 8. Suppose , with . Then there are an operator in and a unique operator in such that . Moreover, .

Proof. By [3, Proposition 2.6], there is a matrix unit such that each belongs to the algebra Obviously, is a finite-dimensional Banach algebra which is isomorphic via . By [1], is standard, that is, , where is a derivation from to and is a linear mapping from to vanishing on every double commutator. By [7, Lemma 10.7], there is an operator in such that for all . Consequently, is a derivation from to . By [7] again, there is an operator in such that for all . Thus for each , it follows that
On the other hand, by Lemma 7, for each , there is an operator in such that
By (17) and (18), we get where . Since commutes with , it follows from Kleinecke-Shirokov theorem that is quasinilpotent. Moreover, for some and is quasinilpotent since commutes with . This implies that .
Since , there are such that , . Applying (19) to , we get . So , where . Thus . So , and therefore and . Consequently, and . Thus By Lemma 4, the sum is independent of the representation of . Let . The proof is complete.

Proof of Theorem 5. Let . Then there exists a unique finite family of distinct in such that . By Lemma 8, for each , there are operators in and in such that . Let . Since if , it follows that
Define to be an operator in such that for some in . By Lemma 8 and the equation above, we see that is well defined and .
Now define a mapping from to as for . Then is linear since is linear. Moreover, for each there is an operator such that . Consequently, is a local derivation from to . By [3, Proposition 2.6], is a derivation. Thus is standard. This completes the proof.

Corollary 9. Let be a -subspace lattice and suppose that the dimension of each element in is infinite. Let be a Lie triple derivation from to itself. Then is a derivation.

Proof. By [3, Remark 2.5(iii)], in this case. Hence by Theorem 5, is a derivation.

3. Lie Triple Derivations of -Subspace Lattice Algebras

In this section, we study Lie triple derivations of whole -subspace lattice algebras. The principal result describes the structure of those mappings.

Theorem 10. Let be a -subspace lattice on a Banach space . Let be a linear mapping. The following is equivalent.(i) is a Lie triple derivation.(ii)For each , there exist an operator in and a linear functional vanishing on every double commutator such that for all and .

Proof. . This is a straightforward verification.
. Obviously, the restriction of to is a Lie triple derivation. Hence it is standard by Theorem 5. Therefore, there exist a derivation and a linear mapping vanishing on every double commutator such that for every .
Fix an element . Take vectors and such that . Define a linear mapping by for . This is well defined because of for .
For and , by (22) we have So for all and for all ,
Let be in . Let be in . Then for with , by (22) and (24), we have Comparing these two equations, we get Taking with and then putting in the last equation, we get where . Since is in the center of , it follows from [3, Remark 2.5] that for some . Consequently, is a scalar multiple of . Since this holds for each , it follows easily that viewed as a linear mapping from to is a scalar multiple of the identify on . Namely, there exists a scalar such that for each . One can easily see that is linear and vanishes on every double commutator.
It remains to verify the boundedness of . Suppose that is a sequence of vectors in , , and . For any , Taking the limit, we get that . Since is in and is arbitrary, . So is closed. The Closed Graph theorem gives the boundedness of , completing the proof.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The research is supported by the National Natural Science Foundation of China (no. 11171244). The first author is supported by the Natural Science Foundation of China (no. 41306207) and the Special Project for High-Quality Professionals of Nanyang Normal College (ZX2014080, ZX2013015), China. The authors thank the anonymous referees for the very thorough reading of the paper and valuable comments.