# A Note on Reducing Subspaces of Toeplitz Operator on the Weighted Analytic Function Spaces of the BidiskHw2D2

Authors: Hongzhao Lin
Journal: Journal of Function Spaces
 (2017)

---

## Abstract

We discussed the reducing subspaces of Toeplitz operatorsTz1Nz2MN≠M on the weighted analytic function spaces of bidisk Hw2(D2). The result shows that if the weight w is of type-I, the structure of reducing subspaces of Tz1Nz2MN≠M on Hw2D2 is very simple.


---

## Body

## 1. Introduction

LetD denote the open unit disk of complex plane C and w=(wi,j) be an infinite matrix with wi,j>0; we consider the Hilbert space Hw2(D2) consisting of analytic functions: (1)fz1,z2=∑i,j=0∞ai,jz1iz2jon bidisk D2 such that(2)f2=∑i,j=0∞ai,j2wi,j<∞.There are many examples forHw2D2. Recall the definition of Dirichlet space and Bergman space; Hw2D2 is Dirichlet space of the bidisk DD2 if wi,j=i(i+1)+j(j+1)/i+1j+1 and Hw2(D2) is Bergman space of the bidisk A2D2 if wi,j=1/(i+1)(j+1). In particular, if we take α=(α1,α2) with α1,α2>-1 and wi,j=i!j!Γ(α1+1)Γ(α2+1)/Γ(i+α1+2)Γ(j+α2+2), Hw2(D2) is Bergman space of the bidisk with weight α, which is usually denoted by Aα2D2.Given two positive integersN and M with N≠M, note that z1iz2j/wi,j is the orthonormal basis of Hw2(D2), and it is easy to see that the operator Tz1Nz2M is bounded on Hw2(D2) if and only if(3)M0=sup⁡wi+N,j+Mwi,j;i≥0,j≥0<∞.Throughout the paper, we fix a weight matrix w=(wi,j) and two distinct positive integers N,M satisfying (3). We will study the reducing subspace lattice of the operator: (4)S=Tz1Nz2M=Tz1NTz2M.It is easy to check that S has proper reducing subspaces as (5)Xi,j=spanz1i+hNz2j+hM;h≥0,where 0≤i≤N-1 and 0≤j≤M-1. It is natural to ask when all the reducing subspaces of S are Xi,j, in other words, when Xi,j is the minimal reducing subspace of S.Recall that ifM is a closed subspace of Hilbert space H, M is called a reducing subspace of the operator T if T(M)⊆M and T∗(M)⊆M. A reducing subspace M is said to be minimal if there are none nontrivial reducing subspaces of T contained in M.Stessin and Zhu [1] completely characterize the reducing subspaces of weighted unilateral shift operators of finite multiplicity. As a consequence, they give the description of the reducing subspaces of TzN on the Bergman space and Dirichlet space of the unit disk. For more general symbols, the reducing subspaces of the Toeplitz operators with finite Blaschke product are well studied (see, e.g., [2–4]). Recently, Lu, Shi, and Zhou extend the result in [1] to Bergman space with several variables. They completely characterize the reducing subspaces of Tz1N and Tz1Nz2N in [5] on the weighted Bergman space of the bidisk and on the weighted Bergman space over polydisk in [6], respectively. Moreover, they [7] solve the problems of Tz1Nz2M with N≠M on both settings. Motivated by the above work, we have investigated the reducing subspaces of Toeplitz operators Tz1N (or Tz2N) and Tz1Nz2N on the weighted Dirichlet space of the bidisk in [8].In this paper, we will consider the problem for Toeplitz operatorsTz1Nz2M (N≠M) on the weighted analytic function spaces of the bidisk Hw2D2. We say that w=(wi,j) is of type-I if for nonnegative integers k,  m,  i,  j,  N:(6)wi+hN,j+hMwi,j=wk+hN,m+hMwk,m,∀h∈N,if and only if (i,j)=(k,m).Theorem 1.
IfM is a reducing subspace of Tz1Nz2M on Hw2(D2) with type-1 weight, then there exist integers i′ and j′ with 0≤i′<N or 0≤j′<M such that Mi′j′⊆M is the minimal reducing subspace of Tz1Nz2M, where (7)Mi′j′=spanz1i′z2j′z1lNz2lN:l∈N.In particular, M is minimal if and only if M=Mi′j′ for some i′,  j′.


## 2. Proof of Theorem1

At first, we will give some useful lemmas. The following lemma describes the projection of monomial on reducing subspaceM.Lemma 2.
IfM is a reducing subspace of Tz1Nz2M on Hw2(D2) with type-I weight, then, for each multiindex (k,m),  z1kz2m∈M or z1kz2m∈M⊥.Proof.
LetPM be the projection onto M and z1kz2m=fz1,z2+gz1,z2 be the orthogonal decomposition on M, where f(z1,z2)=∑i,j=0∞fi,jz1iz2j∈M and g(z1,z2)∈M⊥.
WritingSh=Tz1hNz2hM and (S∗)h=Tz1hNz2hM∗, we calculate(8)S∗hShPMz1kz2m=S∗hShPMf+g=S∗hShf=S∗h∑i,j=0∞fi,jz1i+hNz2j+hM=∑i,j=0∞fi,jwi+hN,j+hMwi,jz1iz2j.On the other hand, direct computation shows that (9)PMS∗hShz1kz2m=PMS∗hz1k+hNz2m+hN=PMwk+hN,m+hMwk,mz1kz2m=∑i,j=0∞fi,jwk+hN,m+hMwk,mz1iz2j.Note that S∗hShPMz1kz2m=PMS∗hShz1kz2m; if there exists some fi,j≠0 (otherwise, PMz1kz1z2m=0), it follows that(10)wi+hN,j+hMwi,j=wk+hN,m+hMwk,m,for each positive integer h. Since the weight is of type-I, it reaches that (i,j)=(k,m), which means that (11)PMz1kz2m=z1kz2m.Lemma 3.
LetΛ be an index set; Hilbert space X is the direct sum of its closed subspace Xi  i∈Λ, that is, X=⨁i∈ΛXi, M is a reducing subspace of bounded linear operator T on X, and PMXi⊆Xi. If f=∑i∈Λfi∈M with fi∈Xi, then fi∈M for each i∈Λ.Proof.
Note that(12)∑i∈Λfi=f=PMf=∑i∈ΛPMfi.The result follows from fi=PMfi since fi∈Xi and PMfi∈Xi.The following result is immediately achieved by Lemmas2 and 3.Lemma 4.
IfM is a nontrivial reducing subspace of Tz1Nz2M on Hw2(D2) and f=∑i,j=0∞fi,jz1iz2j∈M, then z1iz2j∈M for fi,j≠0.Proof of Theorem1.
Supposef=∑i,j≥0fi,jz1iz2j∈M. If fi,j≠0, by Lemma 4, z1iz2j∈M. Since M is a reducing subspace of S, SlM⊆M and S∗lM⊆M for any l∈N. Thus, (13)z1i′+lNz2j′+lM∈M,where i′=i-hN and j′=j-hM with h=min⁡i/N,j/M. It follows that Mi′j′⊆M.
Observe that each reducing subspaceM contains a reducing subspace such as Mi′j′, which means that Mi′j′ consist of all the minimal reducing subspaces. If M is a minimal reducing subspace, then M=Mi′j′ for some Mi′j′. This completes the proof.


## 3. Some Examples

Example 1.
Dirichlet space of bidiskDD2 is Hw2D2 with type-1 weight. By Theorem 1, Mi′j′ is the minimal reducing subspace of Tz1Nz2M.Recall thatD(D2)=Hw2(D2) with wi,j=i(i+1)+j(j+1)/(i+1)(j+1). The following lemma shows that the weight w is of type-I; then Theorem 1 holds.Lemma 5.
Supposing thatk,  m,  i,  j,  N are nonnegative integers, then (14)wi+hN,j+hMwi,j=wk+hN,m+hMwk,m,∀h∈N,if and only if (i,j)=(k,m).Proof.
We only need to prove the necessity. By the assumption,(15)wi+hN,j+hMwk+hN,m+hM=wi,jwk,m,∀h∈N.Taking h→∞ in the left side, it follows that (16)limh→∞wi+hN,j+hMwk+hN,m+hM=1.Thus, for any positive integer h, (17)wi+hN,j+hMwk+hN,m+hM=1,since the right side is constant. By definition of wi,j, it is equivalent to(18)I1=I2,where (19)I1=m+hM+1k+hN+1i+hNi+hN+1+j+hMj+hM+1,I2=j+hM+1i+hN+1k+hNk+hN+1+m+hMm+hM+1.By combining like terms, I1=I2 converts to(20)i+hN+1k+hN+1C1h+C2=j+hM+1m+hM+1C1h+C3,where C1=m-jN+i-kM,  C2=m+1i-j+1k, and C3=mi+1-jk+1. Comparing the coefficient of h3, we have (21)N2C1=M2C1,which implies that C1=0 since N≠M. Thus, (20) turns to(22)i+hN+1k+hN+1C2=j+hM+1m+hM+1C3.We claim that C2=0 and C3=0. Otherwise, we may assume C2≠0.
By comparing the coefficient ofh2,  h,  1 in (22), we have (23)N2M2=i+k+2Nj+m+2M=i+1k+1j+1m+1=C3C2.Note that C1=0, the first equality implies that i+1/j+1=N/M. Then the second equality gives that k+1/m+1=N/M. Thus(24)i-kj-m=NM.However, recalling the definition of C2 and C3, we have (25)C3C2=mi+1-jk+1m+1i-j+1k=mN/Mj+1-jN/Mm+1M/Nk+1i-M/Ni+1k=N2m-jM2i-k.Note that C3/C2=N2/M2, it follows that (26)m-ji-k=1,which contradicts (24).
Thus the claim thatC2=0 and C3=0 holds. By the definition of C2 and C3, simple computation shows that i-k=m-j. Since C1=0, it follows that i=k and m=j. This completes the proof.Example 2.
The weighted Bergman space of bidiskAα2(D2) such that weight α=(a,a) with a>-1 is Hw2(D2) with type-1 weight. By Theorem 1, Mi′j′ is the minimal reducing subspace of Tz1Nz2M.Recall thatAα2(D2)=Hw2(D2) with wi,j=i!j!Γ2(a+1)/Γ(i+a+2)Γ(j+a+2). The proof of Theorem  3.2 in [7] indicated that the weight w=(wi,j) is of type-I; then Theorem 1 holds.However, the case of the unweighted Bergman space of the bidiskA2(D2) is different since A2(D2)=Hw2(D2) with w=(wi,j) where wi,j=ii+1+jj+1/i+1j+1, which is not of type-I by Lemma  2.3 in [7]. The structure of reducing subspaces of Tz1Nz2MN≠M on this case is more complicated. In fact, Theorem  2.4 in [7] showed that if M is a reducing subspace, then there exist nonnegative integers n,  m with 0≤m≤N-1 and a,b∈C such that M contains a reducing subspace as follows:(27)Mn,m,a,b=spanaz1hN+nz2hM+m+bz1ρ2hM+nz2ρ1hN+m:h∈N,where ρ1(hN+m)=(hN+m+1)M/N-1, ρ2(hM+n)=hM+n+1N/M-1. In particular, if ρ1hN+m (or ρ2hM+n) is not a positive integer, then b=0. Moreover, M is minimal if and only if M=Mn,m,a,b.



---

*Source: 5807909-2017-05-07.xml*