Characterizations of Double Commutant Property on

Abstract

Let be a complex Hilbert space. Denote by the algebra of all bounded linear operators on . In this paper, we investigate the non-self-adjoint subalgebras of of the form , where is a block-closed bimodule over a masa and is a subalgebra of the masa. We establish a sufficient and necessary condition such that the subalgebras of the form has the double commutant property in some particular cases.

1. Introduction

Let be a complex Hilbert space. We denote by the algebra of all bounded linear operators on . Given a nonempty subset of ,the commutant of is the set . The double commutant of is . Clearly, . von Neumann’s double commutant theorem states that if is a self-adjoint algebra of operators whose kernel , then the closure of in any of the weak operator, strong operator, and weak topologies is the double commutant . In fact, if is a WOT-closed, unital -subalgebra of , then . In this paper, we analyze the settings of non-self-adjoint subalgebras of whose double commutant coincides with themselves. We say that such algebras satisfy the double commutant property.

In the past several decades, a great deal of effort has been devoted to the study of the subalgebras of with the double commutant property. For a few references, see [1–8]. In recent years, there has been renewed interest in the study of double commutant property [9–15]. For singly generated algebras, Ruston [16] showed that every algebraic operator in has the double commutant property. Turner [8] proved that a normal operator satisfies the double commutant property if and only if it is reductive. For nonsingly generated algebras, Davidson and Pitts [2] researched the noncommutative analytic Toeplitz algebra with the double commutant property. Marcoux and Mastnak [12] analyzed the non-self-adjoint subalgebras of whose double commutant agrees with themselves; specifically, they considered the class of algebras of the form in finite dimensional space, where is a bimodule over a masa and is a unital subalgebras of the masa.

In this note, we will investigate the subalgebras of with the double commutant property, which extends the result in [13] extensively.

2. Preliminaries

Let be a Hilbert space, if , we denote by the rank-one operator on given by . For a subalgebra of , we define the annihilator of as . Given a collection of orthogonal projections in , we denote by the orthogonal projection onto . Note that all projections considered on the manuscript are orthogonal projections. Given projections and in , we define the -block of as follows:

Let be a maximal abelian self-adjoint subalgebra (that is, is a masa), be a collection of projections in , then we say that a subspace of is block-generated over if

With loss of generality, we assume that each .

Definition 1. Let be a masa and let be a block-generated bimodule for some family of projections . We say that is disconnected if there exist , and projections so that (1)(2)(3) is orthogonal to Otherwise, we say that is connected.

Marcoux and Mastnak proved the following proposition in [12]. Now, we give another simpler proof.

Proposition 2 (see [12]). Let be a masa and let be a block-generated bimodule over with for all . Then, (1), where (2) is connected if and only if

Proof. (1)It is clear that . We only need to show that ., we have , . Since is prime, we obtain . So, . This implies that and ; therefore, , so we have . (2)If is connected, it is easy to vertify that ; we will prove that ., let , then . This implies that for each , there exists so that and . We claim that .
Let , then , so . Let . Suppose that . For , we have Since , we get . Similarily, we have , , and . Let then , and . This contradicts the connection of . Therefore, for all , there exists so that and . Thus, for all . It ensures that for all . So, we have , i.e., .
On the other hand, suppose that is disconnected, then there exists as Definition 1. Let and . , write , where . The fact that implies that . However, if , we have Thus, . This is a contradiction.☐

Proposition 3 (see [13]). Let be a masa and let be a block-generated -bimodule for some family of projections . Then, there is a partition of so that the subspaces are connected for each , and implies that is disconnected.

By the proposition above, we can decompose into a direct sum where each is a connected subspace of .

Definition 4. Let be a masa and let be a block-generated bimodule over . Let be the decomposition of as in Proposition 3 for each ; let We define as the block closure of and as the block closure of .

By the connection of and Proposition 2, we have . Now,

Let and suppose that satisfies . Then it follows that . It is obvious that ; we obtain . So, we can consider the form if satisfies in the following.

3. The Double Commutant Theorem on

In this section, we discuss non-self-adjoint subalgebras of which have the double commutant property. We concentrate on the case , where is a subalgebra of and is a block-generated -bimodule. First, we consider the case .

Theorem 5. Let be a masa. Suppose that , where is a projection in and is a subspace of . If , then (1)(2)for any nonzero projection in , we have (3)Span is dense in , where Span denotes the linear expansion of set

Proof. Since , by Proposition 2, , we have (1)It is clear that . So, we have . Let , then there exists and such that . Since , we get . It follows that . Thus, we obtainTherefore, . This implies that . Hence, . (2)We assume that there exists nonzero projection so that .Then, . Let such that , then . It follows from (9) that So, . Hence, , there exists
and such that . Therefore, It implies that . Since is prime, we get . Together with (11), we obtain . Thus, . This is a contradiction.
Similarly, we get . (3)Let be a projection onto .Then, . For any , we have and . Hence, . Therefore, there exists and such that . It follows that . So, . From (2), we get , i.e., .☐

Then, we concern the general case .

Theorem 6. Let be a masa, and be a block-closed bimodule over . If satifies , then the following are equivalent: (1) (convergence being in the SOT)(2) (convergence being in the SOT)

Proof. (1)⇒(2). Let , then we have . Thus, for all ,
. Then, for , , it implies that there exists so that . Therefore, . It follows that For any , since , we have . It implies that . From (13), we get for any . So, Thus, . Therefore, . Hence, .
(2)⇒(1). The argument is similar to (1)⇒(2).☐

Particularly, if , then we obtain the necessary and sufficient conditions for to satisfy the double commutant property.

Theorem 7. Let be a masa and let be a block-closed bimodule over . Suppose that . Then, if and only if one of the following is true: (1)(2)

Proof. First, we prove the necessary part. Let . If , then from the proof of Theorem 6, we have . So, ; if , then . Otherwise, . Thus, we get from Theorem 6. This is a contradiction.
Now, we consider the sufficient part. If , then is the von Neumann algebra with identity element. Hence, from von Neumann’s double commutant theorem, we have ; if , then , where . By Proposition 2, So, .
Let . Then, there exists and so that . For any , let be a projection onto Ran and let . It is clear that . Thus, . Therefore, we can write , where . , Since , we have So, . Hence, .☐

Now, we concern the class of algebras of the form which satisfies , where spans by projections, is a block-closed bimodule, and .

Let be a masa and be an orthogonal projection. For , we denote the relative commutant of with respect to as and the relative annihilator of with respect to as .

Theorem 8. Let be a masa, be a block-closed -bimodule, and be the maximal projection spanned by the projections over , i.e., . Suppose that is a subalgebra of , where are mutually orthogonal projections which ranks are no less than 2 and . Suppose that , . Then the following are equivalent: (1)(2)For each , and To prove the theorem, we need several lemmas.

Lemma 9. Let be nonzero projections. Let and for all , then for some .

Proof. Since for all , we have . For . Thus, and for some . We also have , . So, we get for all . Thus, .☐

Lemma 10. Let , then , where and .

Proof. It is clear that .
On the other hand, , since , there exists and for all so that . Since , we have . It follows that . Thus, for each . Let , then and Because and , we have for all . Furthermore, . So, .☐

Proof of Theorem 8. (1)⇒(2). With loss of generality, we only need to show that and . We argue by contradiction. Assume .
. So, . Thus, . In fact, , . We can write . But . So, . Hence, . It is clear that . Therefore, .
Since , we have .
From above, we can write , where . Since , for all , we have . Then, . For each , let , then . Thus, there exists such that . By reindexing if necessary, we assume that .
By Lemma 9, for each , there exists so that . So, with the decomposition of , we can write Because of the arbitrary of , Thus, . For all , . Since , we can write for some , and ,. Then, , it implies that and . is a rank-1 projection. It is a contradiction.
(2)⇒(1). It is clear that . We need to prove that . Let . With the decomposition of , we can write , where . Given , let and .

Claim 11. .
Fix , for each , we define and . By the hypothesis, . It is clear that is the identify element of . So, by Theorem 7, we have .
For , since , we have . It implies that , and , by Theorem 7,

Claim 12. .
Since for , we have and . Let , then , . So , where . Thus, implies that In particular, . Therefore, we have For , define . Let , . By Lemma 10, Hence, there exists and , so that . Let ; now, we prove . Since for each , we need to show that . In fact, If ; if , then . Thus, . Furthermore, for all , then Hence, . Therefore, . Now, ; this implies that . So, from (22), we have .
Thus, .
Now, we consider . It is easy to verify that . Given , there exists so that . Let , then and . It implies that . So, by (22). Hence, .

Claim 13. .
Let , , then from Claim 12, we have and . Therefore, , . So, from (21), we get Let satisfying . From the above, . Thus, . It implies that For each , since and , we get . Thus, Together with (27) and (28), we have So, .
A similar argument shows that .

Claim 14. .
Let , then and by Claim 12. So, we have , . Together with (21), we have Thus, . Similarly, . Therefore, Since , , , we have . So, . By (31), . Thus, .
Now from the above, for any , we have . So, .
We complete the proof.

Data Availability

The data used to support the findings of this study are included within the article.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Acknowledgments

The work is supported by the National Natural Science Foundation of China (11801045, 61801056, and 61807006) and the Natural Science Foundation of the Jiangsu Higher Education Institutions of China (18KJB110001 and 18KJD110001).