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Journal of Function Spaces and Applications

Volume 2013 (2013), Article ID 656034, 4 pages

http://dx.doi.org/10.1155/2013/656034

## Semicommutators and Zero Product of Block Toeplitz Operators with Harmonic Symbols

^{1}School of Mathematical Sciences, Dalian University of Technology, Dalian 116024, China^{2}School of Mathematical Sciences, Ocean University of China, Qingdao 266100, China

Received 15 February 2013; Accepted 17 August 2013

Academic Editor: Gen-Qi Xu

Copyright © 2013 Puyu Cui et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

We completely characterize the finite rank semicommutators, commutators, and zero product of block Toeplitz operators and with on the vector valued Bergman space .

#### 1. Introduction

Let be the open unit disk in the complex plane , and let be the normalized Lebesgue area measure on . Denote by and the space of essential bounded measurable functions and the space of square integral functions with respect to , respectively. The Bergman space consists of all analytic functions in . Let and , where means the Hilbert space tensor product. Let be the set of all complex matrixes in the complex plane , and let be the space of matrix valued essential bounded Lebesgue measurable functions on .

Given a matrix valued function , the Toeplitz operator and the Hankel operator on with symbol are defined by and , respectively, where is the orthogonal projection from onto . Then the operators and have the following matrix representations:

The problem of the finite rank semicommutator of two Toeplitz operators on the Hardy space has been completely solved in [1, 2]. The analogous problems on the Bergman space have been characterized in [3–5]. In the case of vector valued Hardy space, Gu and Zheng [6] have studied the problems of compact semicommutators and commutators of two block Toeplitz operators.

In this paper, we study the related problems of the two block Toeplitz operators and with on the vector valued Bergman space , where denotes the space of all bounded harmonic functions on . Our main idea here is to reduce the problems on to the problems of the finite sum of some Toeplitz operator products or some Hankel operator products on the Bergman space . In the following, we recall two useful theorems. For every , denote by (or ) the analytic part (or the coanalytic part) of .

Theorem 1 (see [7]). *Suppose that and are bounded harmonic functions on for . Then the following are equivalent:*(1)* has finite rank,*(2)*,*(3)*,
** where .*

Theorem 2 (see [8]). *Let be bounded harmonic functions on , and , . Then if and only if the following two conditions hold: *(a)*;
*(b)*. *

#### 2. Semicommutators of Block Toeplitz Operators

In this section, we discuss the finite rank semicommutators and commutators of the block Toeplitz operators , on with . In the following part of this paper, denote the matrix unit with th entry equal to one and all others equal to zero.

Theorem 3. *Let , and let such that . Then the following statements are equivalent: *(1)* has finite rank; *(2)*; *(3)*for each , there exist a matrix and a permutation matrix , such that and . *

*Proof. * Note that . If has a finite rank, then has a finite rank for . By Theorem 1, we have that for . Hence, .

If , then , for . By Theorem 1, we get on . Therefore, on for .

It is known that if is holomorphic in and , then for in several complex variables. So we can “complexificate” the above identity and have for , . Therefore,
which means that on .

Note that det is the analytic function of and is the coanalytic function of . If is not identically zero, then ; that is, is analytic. If is not identically zero, then ; that is, is coanalytic.

In the following, we deal with the case that and . Suppose for . Since , is linearly dependent. Let be a permutation matrix such that
where is maximally linearly independent of . And
with the matrix representation

that is, .

Since , we have . Denote . It follows that
for . Note that , . Therefore
and hence, . Since is linearly independent, we have
with the matrix representation

that is .

In the following, we prove .

Note that
By the hypothesis, we have
Hence, we obtain the desired result.

*Remark 4. *If the assumption is removed and are replaced by in statement , respectively, then the above theorem still holds.

Corollary 5. *Let , and let such that . Suppose that is not identically zero. Then the following are equivalent:*(1)* has finite rank; *(2)*; *(3)*if is not identically zero, then is analytic; if is not identically zero, then is coanalytic. *

*Proof. *The proof is obvious.

Let As in [6], the commutator can be reduced to the semicommutator .

It is easy to check that If , Theorem 2 implies that . Therefore, . Combining Theorem 3, we get the following explicit theorem.

Theorem 6. *Suppose that and . Then if and only if and satisfy the following two conditions: *(1)*; *(2)*for each , there exist permutation matrixes and such that and . *

#### 3. Zero Product of Two Block Toeplitz Operators

In this section, we discuss the zero product of two block Toeplitz operators and with on .

Theorem 7. *Let and such that . Then if and only if *(1)*; *(2)*for each , there exist a matrix and a permutation matrix , such that and .*

*Proof. *By Theorem 2, it is easy to know that for ; that is, . Since , we obtain the desired result by Theorem 3.

Corollary 8. *Let and let such that . *(1)*If neither nor is identically zero, then if and only if . *(2)*If neither nor is identically zero, then if and only if . *

*Proof. *(1) We only need to prove the necessity. By the proof of Theorem 3, we have on . By the theorem given above, we know that . Taking Laplace transform, it follows that . As in Theorem 3, we can prove that on . “Complexificate” the identity , and then . Hence, on . Since , we have .

It is clear that is the analytic function of and is the coanalytic function of . If is not identically zero, then and . Therefore, . If is not identically zero, then and . Therefore, .

(2) The proof is similar to the proof of (1).

#### Acknowledgments

This research is supported by NSFC (11271059, 11201438), Research Fund for the Doctoral Program of Higher Education of China, Shandong Province Young Scientist Research Award Fund (BS2012SF031), and the Fundamental Research Funds for the Central Universities (201213011).

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