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

Authors: Puyu Cui; Yufeng Lu; Yanyue Shi
Journal: Journal of Function Spaces and Applications
 (2013)

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## Abstract

We completely characterize the finite rank semicommutators, commutators, and zero product of block Toeplitz operatorsTF and TG with F,G∈h∞⊗Mn×n on the vector valued Bergman space La2(𝔻,ℂn).


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## Body

## 1. Introduction

Let𝔻 be the open unit disk in the complex plane ℂ, and let dA be the normalized Lebesgue area measure on 𝔻. Denote by L∞(𝔻,dA) and L2(𝔻,dA) the space of essential bounded measurable functions and the space of square integral functions with respect to dA, respectively. The Bergman space La2 consists of all analytic functions in L2(𝔻,dA). Let L2(𝔻,ℂn)=L2(𝔻,dA)⊗ℂn and La2(𝔻,ℂn)=La2⊗ℂn, where ⊗ means the Hilbert space tensor product. Let Mn×n be the set of all n×n complex matrixes in the complex plane ℂ, and let L∞(𝔻,dA)⊗Mn×n be the space of matrix valued essential bounded Lebesgue measurable functions on 𝔻.Given a matrix valued functionΦ(z)=[φij(z)]n×n∈L∞(𝔻,dA)⊗Mn×n, the Toeplitz operator TΦ and the Hankel operator HΦ on La2(𝔻,ℂn) with symbol Φ are defined by TΦh=P(Φh) and HΦh=(I-P)(Φh), respectively, where P is the orthogonal projection from L2(𝔻,ℂn) onto La2(𝔻,ℂn). Then the operators TΦ and HΦ have the following matrix representations:
(1)TΦ=[Tφ11⋯Tφ1n⋮⋮Tφn1⋯Tφnn],HΦ=[Hφ11⋯Hφ1n⋮⋮Hφn1⋯Hφnn].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 operatorsTF and TG with F,G∈h∞⊗Mn×n on the vector valued Bergman space La2(𝔻,ℂn), where h∞ denotes the space of all bounded harmonic functions on 𝔻. Our main idea here is to reduce the problems on La2(𝔻,ℂn) to the problems of the finite sum of some Toeplitz operator products or some Hankel operator products on the Bergman space La2. In the following, we recall two useful theorems. For every f∈h∞, denote by f+ (or f-=f-f+) the analytic part (or the coanalytic part) of f.Theorem 1 (see [7]).
Suppose thatfj and gj are bounded harmonic functions on 𝔻 for j=1,…,k. Then the following are equivalent:(1)
∑j=1k‍Hgj¯*Hfj has finite rank,(2)
∑j=1k‍Hgj¯*Hfj=0,(3)
σ(f1,…,fk;g1,…,gk)≡0, where σ(f1,…,fk;g1,…,gk)=Δ~[∑i=1k‍(fi)-(gi)+]=(1-|z|2)2∑i=1k‍(fi)-′(gi)+′.Theorem 2 (see [8]).
Letu1,…un,v1,…,vn be bounded harmonic functions on 𝔻, and x1,…,xn, y1,…,yn∈La2. Then ∑j=1n‍TujTvj=∑j=1n‍xj⊗yj if and only if the following two conditions hold:  (a)
∑j=1n‍ujvj=(1-|z|2)2∑j=1n‍xjyj¯;(b)
∑j=1n‍[Puj¯¯-uj(0)][Pvj-vj(0)]=0.


## 2. Semicommutators of Block Toeplitz Operators

In this section, we discuss the finite rank semicommutators and commutators of the block Toeplitz operatorsTF, TG on La2(𝔻,ℂn) with F,G∈h∞⊗Mn×n. In the following part of this paper, Ej denote the matrix unit with (j,j) th entry equal to one and all others equal to zero.Theorem 3.
LetF(z)=F+(z)+F-(z), and let G(z)=G+(z)+G-(z)∈h∞⊗Mn×n such that F+(0)=G-(0)=0n×n. Then the following statements are equivalent:  (1)
TFG-TFTG has finite rank;(2)
TFG=TFTG;(3)
for each1≤j≤n, there exist a matrix Mj∈Mn×n and a permutation matrix Qj, such that MjQjG-Ej=0n×n and F+Qj(I+Mj)=0n×n.Proof.
(
1
)
⇒
(
2
) Note that TFG-TFTG=HF**HG. If HF**HG has a finite rank, then ∑p=1n‍Hfip¯*Hgpj has a finite rank for 1≤i,j≤n. By Theorem 1, we have that ∑p=1n‍Hfip¯*Hgpj=0 for 1≤i,j≤n. Hence, TFG=TFTG.
(
2
)
⇒
(
3
) If TFG-TFTG=HF**HG=0, then [HF**HG]i,j=∑p=1n‍Hfip¯*Hgpj=0, for 1≤i,j≤n. By Theorem 1, we get (1-|z|2)2∑p=1n‍[(fip)+′(gpj)-′]=0 on 𝔻. Therefore, ∑p=1n‍[(fip)+′(gpj)-′]=0 on 𝔻 for 1≤i,j≤n.
It is known that ifH(z,w) is holomorphic in 𝔻×𝔻 and H(z,z¯)=0, then H(z,w)≡0 for (z,w)∈𝔻×𝔻 in several complex variables. So we can “complexificate” the above identity and have ∑p=1n‍[(fip)+′(z)(gpj)-′(w)]=0 for 1≤i,j≤n, (z,w)∈𝔻×𝔻. Therefore,
(2)0=∑p=1n[(fip)+′(u)(gpj)-′(v)]=∫0z‍∑p=1n[(fip)+′(u)(gpj)-′(v)]du=∑p=1n‍[(fip)+(z)(gpj)-′(v)]=∫0w‍∑p=1n‍[(fip)+(z)(gpj)-′(v)]dv=∑p=1n‍[(fip)+(z)(gpj)-(w)],
which means that F+(z)G-(w)=0n×n on 𝔻×𝔻.
Note that detF+(z) is the analytic function of z and detG-(w) is the coanalytic function of w. If detF+(z) is not identically zero, then G-=0n×n; that is, G is analytic. If detG-(w) is not identically zero, then F+=0n×n; that is, F is coanalytic.
In the following, we deal with the case thatdetF+≡0 and detG-≡0. Suppose G-Ej=(g1j,…,gnj)T for 1≤j≤n. Since detG-≡0, G-Ej is linearly dependent. Let Qj be a permutation matrix such that
(3)QjG-Ej=(gσ(1)j,…,gσ(k0)j,…,gσ(n)j)T,
where {gσ(1)j,…,gσ(k0)j} is maximally linearly independent of {gσ(1)j,…,gσ(n)j}. And
(4)gσ(k0+1)j=a11gσ(1)j+⋯+ak01gσ(k0)j,⋮gσ(n)j=a1n-k0gσ(1)j+⋯+ak0n-k0gσ(k0)j,
with the matrix representation
(5)[0⋯0⋮⋮0⋯00⋯0⋮⋮0⋯0a11⋯ak01⋮⋮a1n-k0⋯ak0n-k0-1⋯0⋮⋮0⋯-1][gσ(1)j⋮gσ(k0)j⋮gσ(n-1)jgσ(n)j]=0n×n,
that is,MjQjG-Ej=0n×n.
SinceF+(z)[G-(w)Ej]=0, we have [F+(z)Qj][QjG-(w)Ej]=0n×n. Denote F+=[fij]n×n. It follows that
(6)fkσ(1)gσ(1)j+⋯+fkσ(k0)gσ(k0)j+⋯+fkσ(n)gσ(n)j=0
for 1≤k≤n. Note that gσ(k0+l)j=a1lgσ(1)j+⋯+ak0lgσ(k0)j, 1≤l≤n-k0. Therefore
(7)fkσ(1)gσ(1)j+⋯+fkσ(k0)gσ(k0)j+∑l=1n-k0(a1lgσ(1)j+⋯+ak0lgσ(k0)j)fkσ(k0+l)=0,
and hence, (fkσ(1)+∑l=1n-k0‍a1lfkσ(k0+l))gσ(1)j+(fkσ(2)+∑l=1n-k0‍a2lfkσ(k0+l))gσ(2)j+⋯+(fkσ(k0)+∑l=1n-k0‍ak0lfkσ(k0+l))gσ(k0)j=0. Since {gσ(1)j,…,gσ(k0)j} is linearly independent, we have
(8)fkσ(1)+∑l=1n-k0a1lfkσ(k0+l)=0,kkkkkkkkkkkkkkkkk⋮fkσ(k0)+∑l=1n-k0ak0lfkσ(k0+l)=0,
with the matrix representation
(9)[f1σ(1)⋯f1σ(k0)⋯f1σ(n)f2σ(1)⋯f2σ(k0)⋯f2σ(n)⋮⋮⋮⋮⋮⋮fn-1σ(1)⋯fn-1σ(k0)⋯fn-1σ(n)fnσ(1)⋯fnσ(k0)⋯fnσ(n)]×[1…00⋯0⋮⋮⋮⋮0⋯10⋯0a11⋯ak010⋯0⋮⋮⋮⋮a1n-k0⋯ak0n-k00⋯0]=0n×n,
that isF+Qj(I+Mj)=0n×n.
In the following, we prove(3)⇒(1).
Note that(10)TFTG-TFG=T(F++F-)T(G++G-)-T(F++F-)(G++G-)=T(F+G++F-G-+F-G+)+TF+G--T(F+G++F-G-+F-G+)-TF+G-=TF+TG--TF+G-.
By the hypothesis, we have
(11)TF+TG-Ej=TF+Qj(I+Mj-Mj)QjTG-Ej=TF+Qj(I+Mj)QjTG-Ej-TF+QjMjQjTG-Ej=TF+Qj(I+Mj)QjTG-Ej-TF+QjTMjQjG-Ej=0n×n,F+G-Ej=F+Qj[I+Mj-Mj]QjG-Ej=[F+Qj(I+Mj)]QjG-Ej-F+Qj[MjQjG-Ej]=0n×n.
Hence, we obtain the desired result.Remark 4.
If the assumptionF+(0)=G-(0)=0n×n is removed and F+,G- are replaced by F+′,G-′ in statement (3), respectively, then the above theorem still holds.Corollary 5.
LetF(z)=F+(z)+F-(z), and let G(z)=G+(z)+G-(z)∈h∞⊗Mn×n such that F+(0)=G-(0)=0n×n. Suppose that |detF+|+|detG-| is not identically zero. Then the following are equivalent:(1)
TFG-TFTG has finite rank;(2)
TFG=TFTG;(3)
ifdetF+ is not identically zero, then G is analytic; if detG- is not identically zero, then F is coanalytic.Proof.
The proof is obvious.Let(12)Φ=[F-G00],Ψ=[G0F0].
As in [6], the commutator TFTG-TGTF can be reduced to the semicommutator TΦΨ-TΦTΨ.It is easy to check that(13)TΦΨ-TΦTΨ=[TFG-GF000]-[TF-TG00][TG0TF0]=[TFG-GF000]-[TFTG-TGTF000].
If TFTG=TGTF, Theorem 2 implies that FG=GF. Therefore, TΦTΨ=TΨΦ. Combining Theorem 3, we get the following explicit theorem.Theorem 6.
Suppose thatF=F++F-∈h∞⊗Mn×n and G=G++G-∈h∞⊗Mn×n. Then TFTG=TGTF if and only if F and G satisfy the following two conditions:  (1)
FG=GF;(2)
for each1≤j≤n, there exist permutation matrixes Qj~ and 𝕄j∈M2n×2n such that 𝕄jQj~[G-′F-′]Ej=0n×n and [F+′-G+′]Qj~(I+𝕄j)=0n×n.


## 3. Zero Product of Two Block Toeplitz Operators

In this section, we discuss the zero product of two block Toeplitz operatorsTF and TG with F,G∈h∞⊗Mn×n on La2(𝔻,ℂn).Theorem 7.
LetF(z)=F+(z)+F-(z) and G(z)=G+(z)+G-(z)∈h∞⊗Mn×n such that F+(0)=G-(0)=F-(0)=G+(0)=0n×n. Then TFTG=0n×n if and only if  (1)
FG=0n×n;(2)
for each1≤j≤n, there exist a matrix Nj∈Mn×n and a permutation matrix Qj1, such that NjQj1G-Ej=0n×n and F+Qj1(I+Nj)=0n×n.Proof.
By Theorem2, it is easy to know that ∑p=1n‍fip(z)gpj(z)=0 for 1≤i,j≤n; that is, FG=0n×n. Since TFTG=TFTG-TFG=0n×n, we obtain the desired result by Theorem 3.Corollary 8.
LetF(z)=F+(z)+F-(z) and let G(z)=G+(z)+G-(z)∈h∞⊗Mn×n such that F+(0)=G-(0)=F-(0)=G+(0)=0n×n.  (1)
If neitherdetG+ nor detG- is identically zero, then TFTG=0 if and only if F=0n×n.(2)
If neitherdetF+ nor detF- is identically zero, then TFTG=0 if and only if G=0n×n.Proof.
(1) We only need to prove the necessity. By the proof of Theorem3, we have F+(z)G-(w)=0n×n on 𝔻×𝔻. By the theorem given above, we know that FG=[F++F-][G++G-]=0n×n.  Taking Laplace transform, it follows that F-′G+′=0n×n. As in Theorem 3, we can prove that F-(w)G+(z)=0n×n on 𝔻×𝔻. “Complexificate” the identity FG=0n×n, and then [F+(z)+F-(w)][G+(z)+G-(w)]=0n×n. Hence, F+(z)G+(z)=-F-(w)G-(w) on 𝔻×𝔻. Since F+(0)=G-(0)=0n×n, we have F+G+=F-G-=0n×n.
It is clear thatdetG+(z) is the analytic function of z and detG-(w) is the coanalytic function of w. If detG+(z) is not identically zero, then F-≡0n×n and F+≡0n×n. Therefore, F=0n×n. If detG-(w) is not identically zero, then F-≡0n×n and F+≡0n×n. Therefore, F=0n×n.
(2) The proof is similar to the proof of (1).



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