1 Introduction
The study of algebraic properties of Toeplitz operators stemmed from the seminal work of Brown and Halmos [Reference Brown and Halmos4]. They proved that on the classical Hardy space of the unit disk,
$T_fT_g=T_h$
if and only if either
$\overline {f}$
or g is holomorphic and
$h=fg$
, and
$T_fT_g=T_gT_f$
if and only if: (I) both f and g are analytic; or (II) both
$\overline {f}$
and
$\overline {g}$
are analytic; or (III) one is a linear function of the other. These results are commonly referred to as the Brown–Halmos theorems. Then, the problem of determining when the commutator or semi-commutator of two Toeplitz operators is compact (or further has finite rank) was completely solved in [Reference Axler, Chang and Sarason2, Reference Ding10, Reference Ding and Zheng12]. Extending these results to various function space settings has been one of the central themes of research in the theory of Toeplitz operators in the last few decades.
In the setting of the Bergman space of the unit disk, the problem is known to be much more delicate and more challenging. Brown–Halmos type theorems on the Bergman space for Toeplitz operators with bounded harmonic symbols were obtained in [Reference Ahern and Čučković1, Reference Axler and Čučković3]. Guo et al. [Reference Guo, Sun and Zheng16] later showed that there is no nonzero finite-rank commutator or semi-commutator of Toeplitz operators induced by bounded harmonic symbols. Note that the situation on the classical Hardy space is different. For example, a neat nontrivial rank-one decomposition of the self-commutators of a Toeplitz operator on the Hardy space whose symbol is any Laurent polynomial was obtained in [Reference Dai, Dong and Gao9]. Furthermore, [Reference Čučković6, Reference Ding, Qin and Zheng11] characterized when the product of two Toeplitz operators with harmonic symbols on the Bergman space is a finite-rank perturbation of another Toeplitz operator.
Meanwhile, many interesting and surprising results which are different from Brown–Halmos type theorems have appeared for Toeplitz operators with bounded quasihomogeneous symbols (namely, functions being of the form
$e^{ik\theta }\varphi (r)$
). The main reason to study such a family of symbols is that any Lebesgue square-integrable function f on the unit disk has the following polar decomposition:
$$ \begin{align*} f(re^{i\theta })=\sum\limits_{k \in\mathbb{Z}} {e^{ik\theta } f_k (r)}, \end{align*} $$
where
$f_k$
are radial functions in
$L^2([0,1],r\,dr)$
; see [Reference Čučković and Rao8] for more details. Then, Čučković and Rao [Reference Čučković and Rao8] used the Mellin transform to give necessary and sufficient conditions for a Toeplitz operator on the Bergman space commuting with another such operator whose symbol is a monomial. Later, in [Reference Louhichi and Zakariasy21], Louhichi and Zakariasy gave a partial characterization of commuting Toeplitz operators with quasihomogeneous symbols. Then, Louhichi et al. [Reference Louhichi, Strouse and Zakariasy20] determined when the product of two Toeplitz operators with quasihomogeneous symbols is a Toeplitz operator. Also, the finite-rank problem for semi-commutators and commutators of Toeplitz operators with quasihomogeneous symbols was considered in [Reference Čučković and Louhichi7, Reference Dong and Zhou14, Reference Le and Thilakarathna17]. The aim of this paper is to study the algebraic properties of Toeplitz operators whose symbols are finite linear combinations of monomials on the Bergman space of the unit disk.
While it is not the focus of the current paper, we would like to mention that many researchers have also investigated the corresponding problems in the setting of several complex variables. More specifically, there is a vast amount of literature on the study of algebraic properties of Toeplitz operators whose symbols are pluriharmonic, separately radial, monomial, or k-quasihomogeneous in several variables on the Bergman spaces. See, for example, [Reference Dong and Zhou13, Reference Dong and Zhu15, Reference Le and Thilakarathna18, Reference Vasilevski24–Reference Zhou and Dong26] and the references therein.
Let
$dA$
denote the Lebesgue area measure on the unit disk D, normalized so that the measure of D equals
$1$
.
$L^{2}(D,dA)$
is the Hilbert space of Lebesgue square integrable functions on D with the inner product
$$ \begin{align*}\langle f, g\rangle=\int_{D}f(z)\overline{g(z)}dA(z).\end{align*} $$
Given
$z\in D$
, let
$K_{z}(w)= {1}/{(1-w\overline {z})^2}$
be the well-known reproducing kernel for the Bergman space
$L^{2}_{a}$
consisting of all
$L^2$
-analytic functions on D. The well-known Bergman projection P is then the integral operator
$$ \begin{align*}P f(z)= \int_{D}f(w)\overline{K_{z}(w)}dA(w)\end{align*} $$
for
$f\in L^{2}(D,dA)$
.
For
$u\in L^{\infty }(D,dA)$
, the Toeplitz operator
${T}_{u}$
with symbol u is the operator on
$L_a^2$
defined by
$$ \begin{align*}{T}_{u}f=P(u f)\end{align*} $$
for
$f\in L_a^2$
. For two Toeplitz operators
$T_{f_1}$
and
$T_{f_2}$
on the Bergman space, we define their commutator and semi-commutator respectively by
$$ \begin{align*} [T_{f_1}, T_{f_2}]=T_{f_1}T_{f_2}-T_{f_2}T_{f_1} \end{align*} $$
and
$$ \begin{align*} (T_{f_1}, T_{f_2}]=T_{f_1}T_{f_2}-T_{f_1f_2}. \end{align*} $$
Monomial Toeplitz operators (namely, Toeplitz operators induced by symbol functions of the form
$z^p\overline z^q$
with
$p,q\in {\mathbb N}$
, where
${\mathbb N}$
is the set of all nonnegative integers) on the Bergman space of the unit disk have very interesting structures and enjoy some very interesting properties that can be used to construct important examples in operator theory. Motivated by recent work of the second author and Zhu in [Reference Dong and Zhu15], in this paper, we further discuss the finite-rank problem for the operator
$$ \begin{align*}T=\sum_{i=1}^k c_iT_{z^{p_i}\overline z^{q_i}}T_{z^{s_i}\overline z^{t_i}}\end{align*} $$
on
$L_a^2$
. In particular, we completely characterize when the commutator
$[T_{f_k}, T_{g_l}]$
and semi-commutator
$(T_{f_k}, T_{g_l}]$
have finite rank on
$L_a^2$
, where
$$ \begin{align*}f_k=\sum\limits_{i=1}^kc_iz^{p_i}\overline z^{q_i}\quad \text{and}\quad g_l=\sum\limits_{j=1}^lb_jz^{s_j}\overline z^{t_j}\end{align*} $$
are finite linear combinations of monomials. As one of the important applications, our main results produce an ample supply of nontrivial cases when the finite linear combination of monomial Toeplitz products has finite rank on the Bergman space of the unit disk.
It is worth mentioning that some great progress has been made on the product problem for Toeplitz operators on the Bergman space in recent years; see [Reference Ding, Qin and Zheng11, Reference Le and Thilakarathna17, Reference Le and Thilakarathna18] for example. However, there has not been any breakthrough in the commutativity problem for Toeplitz operators on
$L_a^2$
in recent decades. In fact, one cannot even present a simple example of nontrivial commuting Toeplitz operators. It is interesting to see that the following theorem produces an ample supply of nontrivial examples of commuting Toeplitz operators. To avoid repetition, we make some additional assumptions on this theorem. More specifically, we assume that the rank of
$[T_{\sum _{i\in \Lambda }c_i z^{p_i}\overline z^{q_i}},T_{z^s\overline z^t}]$
is infinite for any subset
$\Lambda \subsetneqq \{1,2,\ldots ,k\}$
.
Theorem 1.1. Suppose
$s,t\in {\mathbb N}$
,
$k\geq 2$
, and
$f_k=\sum _{i=1}^{k}c_i z^{p_i}\overline z^{q_i}$
with
$p_i,q_i\in {\mathbb N}$
and
$c_i\neq 0$
for all
$i\in \{1,2,\ldots ,k\}$
. If on
$L^{2}_{a}$
,
$$ \begin{align*} \mathrm{Rank} [T_{\sum_{i\in\Lambda}c_i z^{p_i}\overline z^{q_i}},T_{z^s\overline z^t}] = \infty \end{align*} $$
for any nonempty subset
$\Lambda \subsetneqq \{1,2,\ldots ,k\}$
, then the following statements are equivalent.
-
(a) The commutator
$[T_{f_k},T_{z^s\overline z^t}]$
on
$L^{2}_{a}$
has finite rank. -
(b) The operators
$T_{f_k}$
and
$T_{z^s\overline z^t}$
commute on
$L^{2}_{a}$
. -
(c) There exists some
${i}\in \{1,2,\ldots , k\}$
such that one of the following three conditions hold:-
(c1)
$s\neq 0$
,
${(k-1)s}/{k} < t < {ks}/({k-1})$
,
$t\neq s$
, and
$$ \begin{align*} \hspace{-18pt}f_k(z)=c_{i}\sum_{n=1}^k\bigg[\prod_{j=2}^n \frac{(k-j+1)[(j-2)t-(j-1)s]}{(j-1) [(k-j+1)t-(k-j)s]}\bigg]z^{ns-(n-1)t}\overline{z}^{(k-n+1)t-(k-n)s}; \end{align*} $$
-
(c2)
$s\neq 0$
,
$t= {k s}/({k-1})\in {\mathbb N} $
,
$q_{i}>s$
,
$q_{i}\neq {(k+n-2) s}/({k-1})$
for all
$n\in \{2,\ldots ,k\}$
, and
$$ \begin{align*}\hspace{-18pt}f_k(z)=c_{i}\sum_{n=1}^k\bigg[\prod_{j=2}^n\frac{(k-j+1) [(k+j-2)s-(k-1)q_{i}]}{(j-1)[-(j-1)s+(k-1 )q_{i}]}\bigg]z^{({k-n})s/({k-1})}\overline{z}^{q_{i}- ({n-1})s/({k-1})};\end{align*} $$
-
(c3)
$t\neq 0$
,
$s= {kt}/({k-1})\in {\mathbb N} $
,
$p_{i}>t$
,
$p_{i}\neq {(k+n-2)t}/({k-1})$
for all
$n\in \{2,\ldots ,k\}$
, and
$$ \begin{align*}\hspace{-18pt}f_k(z)=c_{i}\sum_{n=1}^k\bigg[\prod_{j=2}^{n}\frac{(k-j+1)[(k+j-2)t-(k-1)p_{i}]}{(j-1)[-(j-1)t+(k-1 )p_{i}]}\bigg]z^{p_{i}- ({n-1})t/({k-1})}\overline{z}^{({k-n})t/({k-1})}.\end{align*} $$
As usual, if
$n=1$
in the above sum, the product from
$j=2$
to n stands for
$1$
. -
It is worth mentioning that Condition (c) in Theorem 1.1 implies that
$$ \begin{align*}\frac{(k-1)s}{k}\leq t\leq\frac{ks}{k-1}.\end{align*} $$
Combining this with Corollary 2.4, we have the following necessary condition for a Toeplitz operator whose symbol is a finite linear combination of monomials commuting with a monomial Toeplitz operator.
Corollary 1.2. Suppose s and t are two positive integers, and f is a finite linear combination of monomials such that f is not a linear function of
$z^s\overline z^{t}$
. If the operators
$T_{f}$
and
$T_{z^s\overline z^t}$
commute on
$L^{2}_{a}$
, then
${s}/{2}\leq t\leq 2s$
.
It is interesting to observe that for any fixed integer
$k\geq 2$
, Conditions (c2) and (c3) in Theorem 1.1 provide many nontrivial quasihomogenous Toeplitz operators
$T_{f_k}$
that commute with the monomial Toeplitz operator
$T_{z^s\overline z^{t}}$
with
$t={ks}/({k-1}) \neq 0$
and
$s= {kt}/({k-1})\neq 0$
on
$L_a^2$
, respectively. Louhichi and Rao in [Reference Louhichi and Rao19, Corollary 1] showed that if two Topelitz operators commute with a quasihomogenous Toeplitz operator, none of them being the identity, then they commute with each other. Combining this result with Condition (c2) in Theorem 1.1, we construct the following more interesting and more complicated examples of commuting Toeplitz operators whose symbols are finite linear combinations of monomials on the Bergman space.
Corollary 1.3. Suppose k and s are two positive integers with
$k\geq 2$
. If we write
$$ \begin{align*} f_{k,q}(z)=\sum_{n=1}^k\bigg[\prod_{j=2}^n\frac{(k-j+1) [(k+j-2)s-q]}{(j-1) [-(j-1)s+q]}\bigg] z^{(k-n)s}\overline{z}^{q-(n-1)s}, \end{align*} $$
where q is any positive integer with
$q>(k-1)s$
,
$q\neq (k+n-2) s$
for all
$n\in \{2,\ldots ,k\}$
, then all Toeplitz operators on
$L^{2}_{a}$
with symbols being of the from
$f_{k,q}$
are commutative.
Note that if
$t= {ks}/({k-1}) \neq 0$
or
$s= {kt}/({k-1}) \neq 0$
for any
$k\geq 3$
, then
$$ \begin{align*} \frac{(l-1)s}{l}< t<\frac{ls}{l-1} \end{align*} $$
for any positive integer l with
$2\leq l<k$
. Combining this with Theorem 1.1 and [Reference Louhichi and Rao19, Corollary 1], we can further construct a class of commutative Banach algebras generated by Toeplitz operators whose symbols are finite linear combinations of monomials on the Bergman space.
Corollary 1.4. Suppose k and s are two positive integers with
$k\geq 3$
. If we denote by
$\Re ^{\prime }_{k}$
the linear space generated by all functions being of the form
$$ \begin{align*} z^{(k-1)s}\overline{z}^{ks},\quad \sum_{n=1}^l \bigg[\prod_{j=2}^n\frac{(l-j+1)(j-k-1)}{(j-1) (k+l-j)}\bigg]z^{(k-n)s}\overline{z}^{(k+l-n)s} \end{align*} $$
for any positive integer l with
$2\leq l<k$
, or
$$ \begin{align*}\sum_{n=1}^k\bigg[\prod_{j=2}^n\frac{(k-j+1)[(k+j-2)s-q]}{(j-1)[-(j-1)s+q]}\bigg]z^{(k-n)s}\overline{z}^{q-(n-1)s},\end{align*} $$
where q is any positive integer with
$q>(k-1)s$
,
$q\neq (k+n-2) s$
for all
$n\in \{2,\ldots ,k\}$
, then the Banach algebra generated by all Toeplitz operators on
$L^{2}_{a}$
with symbols from
$\Re ^{\prime }_{k}$
is commutative.
In sharp contrast to the commutator cases, our second result shows that the semi-commutator of two Toeplitz operators provided one of the symbols is the finite linear combination of monomials and the other is an arbitrary monomial on
$L^{2}_{a}$
has finite rank only in trivial cases.
Theorem 1.5. Suppose
$s,t\in {\mathbb N}$
, and
$f_k=\sum _{i=1}^{k}c_i z^{p_i}\overline z^{q_i}$
with
$p_i,q_i\in {\mathbb N}$
and
$c_i\neq 0$
for all
$i\in \{1,2,\ldots ,k\}$
. Then, the following statements are equivalent.
-
(a) The semi-commutator
$(T_{f_k},T_{z^s\overline z^t}]$
on
$L^{2}_{a}$
has finite rank. -
(b) The semi-commutator
$(T_{f_k},T_{z^s\overline z^t}]$
on
$L^{2}_{a}$
is zero. -
(c) Either the first operator
$T_{f_k}$
has conjugate analytic symbol or the second operator
$T_{z^s\overline z^t}$
has analytic symbol.
In the theory of commutators and semi-commutators of Toeplitz operators on the Bergman space, it has been known that the zero and finite-rank properties often coincide, especially when bounded symbols are considered, as one can see from results in [Reference Čučković and Louhichi7, Reference Dong and Zhou14–Reference Guo, Sun and Zheng16]. In view of these results, one may naturally raise the question of whether similar phenomena hold for finite linear combinations of monomial Toeplitz products.
We first recall the classical definition of a rank-one operator. Let
$\phi $
and
$\psi $
be two nonzero vectors in
$L_a^2$
. Then, the rank-one operator
$\phi \otimes \psi $
on
$L_a^2$
is defined by
$$ \begin{align*} \big(\phi\otimes\psi\big) h=\langle h,\psi\rangle \phi, \quad h\in L_a^2. \end{align*} $$
It is well known that every operator T of finite rank
$N\geq 1$
on
$L_a^2$
has a canonical form:
$$ \begin{align*} T=\sum\limits_{j = 1}^N {\phi_j \otimes \psi_j } \end{align*} $$
for some linearly independent sets
$\{\phi _1,\ldots ,\phi _N\}$
and
$\{\psi _1,\ldots ,\psi _N\}$
in
$L_a^2$
. Now, we restrict our attention to the following special operators:
$$ \begin{align*}S=\sum_{0\leq k,l\leq N} c_{kl}T_{z^k}T_{\overline z^l}.\end{align*} $$
Using Proposition 2.2, we need only focus on the case when
$k-l=m$
for some fixed integer m. In the following, we not only completely characterize when the operator S has finite rank, but also obtain a canonical form of each finite-rank operator.
Theorem 1.6. Suppose
$p_i, t_i\in {\mathbb N}$
for each
$i\in \{1,2,\ldots ,k\}$
with
$t_1<t_2<\cdots <t_k$
and
$p_i-t_i=m$
. Then, the nonzero operator
$$ \begin{align*}S_m=\sum_{i=1}^{k} c_{i}T_{z^{p_i}}T_{\overline z^{t_i}}\end{align*} $$
has finite rank on
$L^2_a$
if and only if
$k\geq 3$
and
$$ \begin{align*} \left\{\!\!{\begin{array}{ll} \displaystyle{c_{1}=\frac{t_3-t_2}{t_2-t_1}c_{3}+\cdots+\frac{t_k-t_2}{t_2-t_1}c_{k},}\\[6pt] \displaystyle{c_{2}=\frac{t_1-t_3}{t_2-t_1}c_{3}+\cdots+\frac{t_1-t_k}{t_2-t_1}c_{k}}. \end{array}} \right. \end{align*} $$
In this case,
$$ \begin{align*}S_m=c_1\sum_{j=1}^{t_2-t_1-1} j z^{p_1+ j-1}\otimes{z}^{t_1+ j-1}+\sum_{i=3}^{k} c_i\sum_{l=1}^{t_i-t_2} lz^{p_i-l-1}\otimes{z}^{t_i-l-1}. \end{align*} $$
As a special case of Theorem 1.6, we now give a specific example to show that a finite-rank linear combination of Toeplitz products is not necessarily zero.
Example 1.7. On
$L_a^2$
,
$$ \begin{align*}2T_{z^{3}}-2T_{z^{4}}T_{\overline z}-T_{z^{5}}T_{\overline z^{2}}+T_{z^{7}}T_{\overline z^{4}}=2z^{3}\otimes 1+2z^{4}\otimes {z}+z^{5}\otimes{z}^{2}. \end{align*} $$
The remainder of the present paper is assembled as follows. In Section 2, we aim to give an analytic characterization of when a finite linear combination of monomial Toeplitz products has finite rank. The main idea is to exchange an identity of the Toeplitz products on the Bergman space with an identity of rational functions on a certain right half-plane. As an application, we investigate the finite-rank problem for the commutators and the semi-commutators of Toeplitz operators induced by finite linear combinations of monomials. In Section 3, we give a specific characterization of when a Toeplitz operator induced by the finite linear combinations of monomials commutes with a monomial Toeplitz operator. Consequently, we give the proof of Theorem 1.1. In Sections 4 and 5, we give the proofs of Theorems 1.5 and 1.6, respectively.
2 Finite linear combinations of monomial Toeplitz products
In this section, we study the problem of when a finite linear combination of monomial Toeplitz products has finite rank on the Bergman space. We first recall some basic properties of monomial Toeplitz operators on
$L^{2}_{a}$
. It is known that for each
$n\in {\mathbb N}$
,
$$ \begin{align} T_{z^s\overline z^t}(z^{n})=\begin{cases}\displaystyle \frac{s-t+n+1}{s+n+1}\,z^{s-t+n}, &n \ge t-s, \\ 0, & n \leq t-s.\end{cases} \end{align} $$
Clearly, monomial Toeplitz operators can be considered as a natural extension of the classical shift operators, which are among the most important examples in operator theory.
To simplify our presentation throughout the paper, we now give the following definition.
Definition 2.1. For each integer m, define a set
$\mu _m$
by
$$ \begin{align*} \mu_m= \{(p,q,s,t)\in {\mathbb N}^4: p-q+s-t=m \}. \end{align*} $$
Here, the number m is called the degree of the monomial Toeplitz product
$T_{z^p\overline z^q}T_{z^s\overline z^t}$
on
$L^{2}_{a}$
.
We are now ready to prove our first result, which gives an analytic characterization of when a finite linear combination of monomial Toeplitz products has finite rank on the Bergman space.
Proposition 2.2. On
$L^{2}_{a}$
, we let
$$ \begin{align*}T_{m}=\sum_{(p_i,q_i,s_i,t_i)\in \mu_m}c_iT_{z^{p_i}\overline z^{q_i}}T_{z^{s_i}\overline z^{t_i}}\end{align*} $$
be the finite linear combination of the homogeneous monomial Toeplitz products. Then, the operator
$$ \begin{align*}T= \sum\limits_{m=-d}^{d}T_{m},\quad d\in{\mathbb N}, \end{align*} $$
has finite rank if and only if the operator
$T_{m}$
has finite rank for each
$m\in \{-d,\ldots ,d\}$
, if and only if
$$ \begin{align*}\sum_{(p_i,q_i,s_i,t_i)\in \mu_m}c_i\frac{s_i-t_i+z}{(s_i-t_i+p_i+z)(s_i+z)}=0,\quad m\in\{-d,\ldots,d\}\end{align*} $$
for all z in a certain right half-plane.
Proof. Consider any positive integer n with
$n\geq d+1$
. Then, it follows from (2-1) that
$$ \begin{align*}T(z^{n-1})=\sum\limits_{m=-d}^{d}T_{m}(z^{n-1})=\sum\limits_{m=-d}^{d}a_{m,n} z^{n+m-1},\end{align*} $$
where
$$ \begin{align*}a_{m,n}=\sum_{(p_i,q_i,s_i,t_i)\in \mu_m}c_i\frac{(m+n)(s_i-t_i+n)}{(s_i-t_i+p_i+n)(s_i+n)}\end{align*} $$
for all sufficiently large positive integers n. So the operator T has finite rank if and only if
$$ \begin{align} a_{m,n}=0,\quad m\in\{-d,\ldots,d\}, \end{align} $$
for all sufficiently large positive integers n, which is equivalent to
$T_{m}$
has finite rank for each
$m\in \{-d,\ldots ,d\}$
.
It is well known that a bounded analytic function on the right half-plane is uniquely determined by its value on an arithmetic sequence of integers (see [Reference Remmert22, page 102]). So, (2-2) holds if and only if
$$ \begin{align*}\sum_{(p_i,q_i,s_i,t_i)\in \mu_m}c_i\ \frac{(m+z)(s_i-t_i+z)}{(s_i-t_i+p_i+z)(s_i+z)}=0\end{align*} $$
for all z in a certain right half-plane. Then, the desired result follows from the identity theorem.
When T is the commutator or semi-commutator of two Toeplitz operators induced by finite linear combinations of monomials, we have the following basic result, which plays a critical role for our arguments in the next two sections.
Corollary 2.3. Let
$f_k=\sum _{i=1}^kc_iz^{p_i}\overline z^{q_i}$
and
$g_l=\sum _{j=1}^lb_jz^{s_j}\overline z^{t_j}$
be finite linear combinations of monomials, and let M be the set of all integers m with
$(p_i, q_i, s_j, t_j)\in \mu _m$
for some
$1\leq i\leq k$
and
$1\leq j\leq l$
. For each
$m\in M$
, we define an associated set as follows:
$$ \begin{align*} \Lambda_m= \{(i, j)\in {\mathbb N}^2: p_i-q_i+s_j-t_j=m, 1\leq i\leq k, 1\leq j\leq l\}. \end{align*} $$
Then, on
$L^{2}_{a}$
, the following two statements hold.
-
(1) The commutator
$[ T_{f_k},T_{g_l}]$
has finite rank if and only if has finite rank for each
$$ \begin{align*} \sum\limits_{(i,j)\in\Lambda_m}c_ib_j [T_{z^{p_i}\overline z^{q_i}},T_{z^{s_j}\overline z^{t_j}}] \end{align*} $$
$m\in M$
, if and only if for all z in a certain right half-plane.
$$ \begin{align*} \sum_{(i,j)\in\Lambda_m}c_ib_j\ \bigg[\frac{(s_j-t_j+z)}{(s_j-t_j+p_i+z)(s_j+z)}-\frac{(p_i-q_i+z)}{(p_i-q_i+s_j+z)(p_i+z)}\bigg]=0,\quad m\in M,\end{align*} $$
-
(2) The semi-commutator
$( T_{f_k},T_{g_l}]$
has finite rank if and only if has finite rank for each
$$ \begin{align*}\sum\limits_{(i,j)\in\Lambda_m}c_ib_j (T_{z^{p_i}\overline z^{q_i}},T_{z^{s_j}\overline z^{t_j}}]\end{align*} $$
$m\in M$
, if and only if for all z in a certain right half-plane.
$$ \begin{align*}\sum_{(i,j)\in\Lambda_m}c_ib_j\ \bigg[\frac{(s_j-t_j+z)}{(s_j-t_j+p_i+z)(s_j+z)}-\frac{1}{p_i+s_j+z}\bigg]=0,\quad m\in M,\end{align*} $$
Proof. Since
$$ \begin{align*} T_{cf+bg}=cT_{f}+bT_{g} \end{align*} $$
for any bounded functions
$f,g$
and complex numbers c and b,
$$ \begin{align*} ( T_{f_k},T_{g_l}]=\sum\limits_{i=1}^k\sum\limits_{j=1}^lc_ib_j (T_{z^{p_i}\overline z^{q_i}},T_{z^{s_j}\overline z^{t_j}}] \end{align*} $$
and
$$ \begin{align*}[ T_{f_k},T_{g_l}]=\sum\limits_{i=1}^k\sum\limits_{j=1}^lc_ib_j[T_{z^{p_i}\overline z^{q_i}},T_{z^{s_j}\overline z^{t_j}}].\end{align*} $$
The desired result then follows from (2-1) and Proposition 2.2.
As a direct application, we completely characterize the finite-rank commutator and semi-commutator of two monomial Toeplitz operators, which is the same as that mentioned in [Reference Dong and Zhu15].
Corollary 2.4. Let
$p, q, s, t\in {\mathbb N}$
. Then, on
$L^{2}_{a}$
, the following two statements hold.
-
(1) The commutator
$[T_{z^p\overline z^q},T_{z^s\overline z^t}]$
has finite rank if and only if the operators
$T_{z^p\overline z^q}$
and
$T_{z^s\overline z^t}$
commute, if and only if one of the following conditions holds:-
(i) either
$p=s=0$
or
$p=q=0$
; -
(ii) either
$t=q=0$
or
$s=t=0 $
; -
(iii)
$q=t$
and
$p=s$
; -
(iv)
$p=q$
and
$s=t$
.
-
-
(2) The semi-commutator
$( T_{z^p\overline z^q},T_{z^s\overline z^t}]$
has finite rank if and only if the semi-commutator
$(T_{z^p\overline z^q},T_{z^s\overline z^t}]$
is zero, if and only if either
$p=0$
or
$t=0$
.
3 Commutators of two Toeplitz operators
A major open question in the theory of Toeplitz operators on the Bergman space of the unit disk is to fully characterize the set of all Toeplitz operators that commute with a given one. Čučković and Rao [Reference Čučković and Rao8] showed that for a fixed monomial symbol
$z^s\overline z^t$
, if one interprets all the monomials as the lattice
${\mathbb N}\times {\mathbb N}$
, then there are many lines parallel to the diagonal, ‘holding’ a symbol that gives a Toeplitz operator commuting with
$T_{z^s\overline z^t}$
on
$L^{2}_{a}$
. In this section, we further study the commutants of the monomial Toeplitz operator. Consequently, we obtain concrete expressions for all Toeplitz operators induced by finite linear combinations of monomials that commute with the monomial Toeplitz operator
$T_{z^s\overline z^t}$
on
$L^{2}_{a}$
.
We first give the following lemma, which plays a critical role in the proof of our main result.
Lemma 3.1. Suppose
$\delta $
is nonzero complex number,
$\omega _1,\ldots , \omega _k$
are any different complex numbers with
$k\geq 2$
, and
$A_i,B_i\in {\mathbb C}$
with
$A_i$
and
$B_i$
being not all
$0$
for each
$i\in \{1,2,\ldots ,k\}$
. If there exist nonzero constants
$c_1,\ldots , c_k\in {\mathbb C}$
such that
$$ \begin{align} \sum_{i=1}^{k}c_i\bigg(\frac{A_i}{\omega_i+z}+\frac{B_i}{\delta+\omega_i+z}\bigg)=0 \end{align} $$
and
$$ \begin{align} \sum_{i\in\Lambda}c_i\bigg(\frac{A_i}{\omega_i+z}+\frac{B_i}{\delta+\omega_i+z}\bigg)\neq0 \end{align} $$
for any nonempty subset
$\Lambda \subsetneqq \{1,2,\ldots ,k\}$
, then there exists a permutation
$(i_1, i_2, \ldots , i_k)$
of the numbers
$1, 2, \ldots , k$
such that the following statements hold:
-
•
$\omega _{i_{n}}-\omega _{i_{n-1}}=\delta $
for all
$n\in \{2,\ldots ,k\}$
; -
•
$A_{i_1}=B_{i_k}=0$
; -
•
$A_{i_n}\neq 0$
and
$B_{i_{n-1}}\neq 0$
for all
$n\in \{2,\ldots ,k\}$
; -
•
$c_{i_{n-1}}B_{i_{n-1}}+c_{i_n}A_{i_n}=0$
for all
$n\in \{2,\ldots ,k\}$
.
Proof. Since the functions
${1}/({\omega _i+z})$
and
${1}/({a+z})$
are linearly independent for any complex numbers
$a\neq \omega _i$
, it follows from (3-1) that for each
$i\in \{1,2,\ldots ,k\}$
, one of the following is true:
-
•
$A_{i}=0$
; -
•
$\omega _{i}=\delta +\omega _{j}$
for some
$j\in \{1,2,\ldots ,k\}$
with
$j\neq i$
and
$B_j\neq 0$
.
First, assume that
$A_{i}=0$
for some fixed i. Then, we let
$i_1=i$
. Clearly,
$A_{i_1}=0$
and
$B_{i_1}\neq 0$
. Therefore, there exists an index
$i_2\in \Lambda _1=\{1,2,\ldots ,k\}\backslash \{i_1\}$
such that
$$ \begin{align*}\delta+\omega_{i_1}=\omega_{i_2}\quad \textrm{and}\quad A_{i_2}\neq0.\end{align*} $$
Combining this with (3-1),
$$ \begin{align} \frac{c_{i_1}B_{i_1}+c_{i_2}A_{i_2}}{\omega_{i_2}+z}+\frac{c_{i_2}B_{i_2}}{\delta+\omega_{i_2}+z}+ \sum_{i\in\Lambda_2}c_i\bigg(\frac{A_i}{\omega_i+z}+\frac{B_i}{\delta+\omega_i+z}\bigg)=0, \end{align} $$
where
$\Lambda _2=\Lambda _1\backslash \{i_2\}$
. If
$k=2$
, the desired result then follows from (3-3) and the fact that
$\delta \neq 0$
. We now consider the case when
$k\geq 3$
. Note that
$$ \begin{align*}\omega_{i_2}\neq \omega_{i_n}\quad\textrm{and}\quad \omega_{i_2}\neq \delta+\omega_{i_n}\end{align*} $$
for any
$i_n\in \Lambda _2$
. Then, by (3-2) and (3-3), we have
$c_{i_1}B_{i_1}+c_{i_2}A_{i_2}=0$
,
$B_{i_2}\neq 0$
, and
$$ \begin{align*}\frac{c_{i_2}B_{i_2}}{\delta+\omega_{i_2}+z}+ \sum_{i\in\Lambda_2}c_i\bigg(\frac{A_i}{\omega_i+z}+\frac{B_i}{\delta+\omega_i+z}\bigg)=0.\end{align*} $$
Continuing this process
$(k-2)$
times,
$$ \begin{align*}c_{i_{n}}B_{i_{n}}+c_{i_{n+1}}A_{i_{n+1}}=0,\quad \delta+\omega_{i_n}=\omega_{i_{n+1}},\quad B_{i_n}\neq0,\quad A_{i_{n+1}}\neq0\end{align*} $$
for different indexes
$i_{n+1}\in \Lambda _n=\Lambda _{n-1}\backslash \{i_n\}$
with
$n\in \{2,\ldots ,k-1\}$
, and
$$ \begin{align*}\frac{c_{i_k}B_{i_k}}{\delta+\omega_{i_k}+z}=0,\end{align*} $$
which implies that
$B_{i_k}=0$
, as desired.
Next, assume that
$A_{i}\neq 0$
for all
$i\in \{1,2,\ldots ,k\}$
. Then, for each
$i\in \{1,2,\ldots ,k\}$
,
$\omega _{i}=\delta +\omega _{j_i}$
for some
$j_i\in \{1,2,\ldots ,k\}$
. It is clear that
$j_i\neq j_{i{'}}$
for all
$i,i{'}\in \{1,2,\ldots ,k\}$
with
$i\neq i{'}$
. So there exists a permutation
$(i_1, i_2, \ldots , i_k)$
of the numbers
$1, 2, \ldots , k$
such that
$\delta =\omega _{i_{n}}-\omega _{i_{n-1}}$
for all
$n\in \{2,\ldots ,k\}$
. Then, by (3-1),
$$ \begin{align} \frac{c_{i_1}A_{i_1}}{\omega_{i_1}+z}+\sum_{n=2}^{k}\frac{c_{i_{n-1}}B_{i_{n-1}}+c_{i_n}A_{i_n}}{\omega_{i_n}+z}+\frac{c_{i_k}B_{i_k}}{\delta+\omega_{i_k}+z}=0. \end{align} $$
We claim that
$\omega _{i_1}\neq \delta +\omega _{i_k}$
. Otherwise,
$$ \begin{align*} k\delta=\omega_{i_1}-\omega_{i_k}+\sum_{n=2}^{k}(\omega_{i_{n}}-\omega_{i_{n-1}})=0, \end{align*} $$
which is a contradiction. Then, it follows from (3-4) that
$A_{i_1}=0$
, which contradicts the assumption that
$A_{i}\neq 0$
for all
$i\in \{1,2,\ldots ,k\}$
. This finishes the proof of the lemma.
Now, we consider the finite-rank problem for the commutator
$[T_{f_k},T_{z^s\overline z^t}]$
with
$f_k=\sum _{i=1}^{k}c_i z^{p_i}\overline z^{q_i}$
. For any nonempty subset
$\Lambda $
of
$\{1,2,\ldots ,k\}$
,
$$ \begin{align*} [T_{\sum_{i\in \Lambda}c_i z^{p_i}\overline z^{q_i}},T_{z^s\overline z^t}] = \sum\limits_{i\in \Lambda}c_i [T_{z^{p_i}\overline z^{q_i}},T_{z^s\overline z^t}]. \end{align*} $$
So, by Corollary 2.3, we only need to assume that
$$ \begin{align*} p_i-q_i=p_1-q_1,\quad i\in\{2,\ldots,k\}. \end{align*} $$
Consequently, we obtain the following basic lemma, which gives an analytic characterization of when the commutator
$[T_{f_k},T_{z^s\overline z^t}]$
has finite rank on
$L^{2}_{a}$
.
Lemma 3.2. Suppose
$s,t\in {\mathbb N}$
and
$f_k=\sum _{i=1}^{k}c_i z^{p_i}\overline z^{q_i}$
with
$p_i,q_i\in {\mathbb N}$
and
$p_i-q_i= p_1-q_1$
for all
$i\in \{2,\ldots ,k\}$
. Then, the commutator
$[T_{f_k},T_{z^s\overline z^t}]$
on
$L^{2}_{a}$
has finite rank if and only if
$$ \begin{align} \sum_{i=1}^{k}c_i\frac{x_iz+y_i}{(p_i+z)(s-t+p_i+z)}=0 \end{align} $$
for all z in a certain right half-plane, where
$$ \begin{align*}x_i=sq_i-p_it\quad\textrm{and}\quad y_i=sq_i(s-t)-tp_i(p_i-q_i),\quad i\in\{1,2,\ldots,k\}.\end{align*} $$
Proof. Since
$p_i-q_i=p_1-q_1$
for all
$i\in \{2,\ldots ,k\}$
, it is clear that
$(p_i,q_i,s,t)\in \mu _m$
for all
$i\in \{1,2,\ldots ,k\}$
with
$m=p_1-q_1+s-t$
. Then, it follows from Corollary 2.3 that the commutator
$[T_{f_k},T_{z^s\overline z^t}]$
has finite rank if and only if
$$ \begin{align*}\sum_{i=1}^{k}{c_i}\bigg[\frac{s-t+z}{(s+z)(s-t+p_i+z)}-\frac{p_i-q_i+z}{(p_i+z)(p_i-q_i+s+z)}\bigg]=0\end{align*} $$
for all z in a certain right half-plane, which is equivalent to
$$ \begin{align*}\sum_{i=1}^{k}c_i\frac{(sq_i-tp_i)z+[sq_i(s-t)-tp_i(p_i-q_i)]}{(s+z)(p_i+z)(p_i-q_i+s+z)(s-t+p_i+z)}=0.\end{align*} $$
Then, the desired result clearly follows from the fact
$p_i-q_i=p_1-q_1$
.
The following result gives a specific characterization of when the commutator
$[T_{f_k},T_{z^s\overline z^t}]$
has finite rank on
$L^{2}_{a}$
. It also tells us exactly when such a commutator has rank zero, that is, when these two Toeplitz operators commute.
Theorem 3.3. Suppose
$s,t\in {\mathbb N}$
,
$k\geq 2$
, and
$f_k=\sum _{i=1}^{k}c_i z^{p_i}\overline z^{q_i}$
with
$p_i,q_i\in {\mathbb N}$
and
$c_i\neq 0$
for all
$i\in \{1,2,\ldots ,k\}$
. If on
$L^{2}_{a}$
,
$$ \begin{align} \mathrm{Rank} [T_{\sum_{i\in\Lambda}c_i z^{p_i}\overline z^{q_i}},T_{z^s\overline z^t}]=\infty \end{align} $$
for any nonempty subset
$\Lambda \subsetneqq \{1,2,\ldots ,k\}$
, then the following statements are equivalent.
-
(a) The commutator
$[T_{f_k},T_{z^s\overline z^t}]$
on
$L^{2}_{a}$
has finite rank. -
(b) The operators
$T_{f_k}$
and
$T_{z^s\overline z^t}$
commute on
$L^{2}_{a}$
. -
(c)
$p_{i}-q_{i}=p_{1}-q_{1}$
for all
$i\in \{2,\ldots ,k\}$
, and there exists a permutation
$(i_1, i_2, \ldots , i_k)$
of the numbers
$1, 2, \ldots , k$
such that the following statements hold:-
(c1)
$p_{i_{n}}-p_{i_{n-1}}=s-t\neq 0$
for all
$n\in \{2,\ldots ,k\}$
; -
(c2)
${q_{i_1} (s- p_{i_1})=0}$
and
$p_{i_k}(t-q_{i_k})=0$
; -
(c3)
$sq_{i_1}\neq p_{i_1}t$
,
$sq_{i_k}\neq p_{i_k}t$
,
$q_{i_n} (s- p_{i_n})\neq 0$
, and
${p_{i_{n-1}} (t-q_{i_{n-1}})\neq 0}$
for all
$n\in \{2,\ldots ,k\}$
; -
(c4)
$c_{i_{n-1}}p_{i_{n-1}}(q_{i_{n-1}}-t )+c_{i_n}q_{i_n} (s- p_{i_n})=0$
for all
$n\in \{2,\ldots ,k\}.$
-
Proof. For convenience, we write
$T=[T_{f_k},T_{z^s\overline z^t}]$
and
$T_{\Lambda }=[T_{\sum _{i\in \Lambda }c_i z^{p_i}\overline z^{q_i}},T_{z^s\overline z^t}]$
for any subset
$\Lambda \subsetneqq \{1,2,\ldots ,k\}$
.
Now, we assume that the commutator T has finite rank. Combining (3-5) with (3-6), we obtain that
$x_i$
and
$y_i$
are not all
$0$
for each
$i\in \{1,2,\ldots ,k\}$
, and
$$ \begin{align} \sum_{i\in\Lambda}c_i\frac{x_iz+y_i}{(p_i+z)(s-t+p_i+z)}\neq0 \end{align} $$
for any subset
$\Lambda \subsetneqq \{1,2,\ldots ,k\}$
. Moreover, it follows from (3-6) and Corollary 2.3 that
$$ \begin{align*}p_{k}-q_{k}=\cdots=p_{2}-q_{2}=p_{1}-q_{1},\end{align*} $$
from which we get that
$p_i\neq p_j$
for all
$i\neq j\in \{1,2,\ldots ,k\}$
. Consequently, it is clear from (3-5) that
$s\neq t$
. Then, a straightforward calculation shows that
$$ \begin{align*}\frac{x_{i}z+y_{i}}{(p_{i}+z)(s-t+p_i+z)}=\frac{y_{i}-p_{i} x_{i}}{(p_{i}+z)(s-t)}+\frac{y_{i}-(s-t+p_i) x_{i}}{(s-t+p_i+z)(t-s)}\end{align*} $$
and
$$ \begin{align*}\frac{y_{i}-p_{i} x_{i}}{s-t}=q_{i} (s- p_{i}),\quad \frac{y_{i}-(s-t+p_i) x_{i}}{t-s}=p_{i}(q_{i}-t).\end{align*} $$
Therefore, according to (3-7),
$q_{i} (s- p_{i})$
and
$p_{i}(q_{i}-t)$
are not all
$0$
for each
$i\in \{1,2,\ldots ,k\}$
. Moreover, if
$q_{i} (s- p_{i})=0$
or
$p_{i}(q_{i}-t)=0$
, then it is easy to see that
$x_i\neq 0$
. Then, that Condition (a) implies Condition (c) follows from Lemma 3.1.
Next, we prove that Condition (c) implies Condition (b). Similarly, we write
$$ \begin{align*}\gamma=\max\{0,-p_{i_k}+q_{i_k},-s+t,-p_{i_k}+q_{i_k}-s+t\}.\end{align*} $$
Obviously, the arguments in the previous paragraph can be reversed. So (3-5) holds, which implies that
$T(z^{n-1})=0$
for each positive integer n with
$n \geq \gamma +1$
. Therefore, if
$\gamma =0$
, then Condition (b) holds. Now, consider the case
$\gamma>0$
, which implies that
$$ \begin{align} \mathrm{{either}}\quad q_{i_k}-p_{i_k}>0\quad\textrm{or}\quad t-s>0. \end{align} $$
Recall that
$q_{i_k}-p_{i_k}=q_{i_1}-p_{i_1}$
and
$$ \begin{align} s-t=p_{i_{n}}-p_{i_{n-1}}=q_{i_{n}}-q_{i_{n-1}},\quad n\in \{2,\ldots,k\}. \end{align} $$
Combining this with (3-8), we have
$q_{i_{n-1}}>0$
for all
$n\in \{2,\ldots ,k\}$
. Then, by Condition (c), we conclude that one of the following group of three conditions holds.
-
•
$p_{i_k}=0$
,
$p_{i_k}-q_{i_k}=-q_{i_k}\leq 0$
, and
$s-t=-p_{i_{k-1}}< 0$
. -
•
$p_{i_k}\neq 0$
,
$s=p_{i_1}$
,
$t=q_{i_k}$
, and then
$$ \begin{align*}k(s-t)=(p_{i_1}-q_{i_k})+(p_{i_2}-p_{i_1})+\cdots+(p_{i_k}-p_{i_{k-1}})=p_{i_k}-q_{i_k}<0.\end{align*} $$
In each case, we easily check that
$$ \begin{align*}\gamma=-p_{i_k}+q_{i_k}-s+t.\end{align*} $$
Then,
$T(z^n)=0$
for any
$n\in {\mathbb N}$
with
$n<\gamma $
and Condition (b) holds.
It is trivial that Condition (b) implies Condition (a). This completes the proof of the theorem.
As a consequence of Theorem 3.3, we give the proof of Theorem 1.1, which gives the precise expressions of the nontrivial zero commutator
$[T_{f_k},T_{z^s\overline z^t}]$
.
Proof of Theorem 1.1.
It suffices for us to show that Condition (c) in Theorem 1.1 is equivalent to Condition (c) in Theorem 3.3. First, we assume that Condition (c) in Theorem 3.3 holds. We break the discussion into four cases.
Case 1:
$p_{i_k}\neq 0$
and
$q_{i_1}\neq 0$
. Since
${q_{i_1} (s- p_{i_1})=0}$
and
$p_{i_k}(t-q_{i_k})=0$
, it follows easily from (3-9) that
$$ \begin{align*}p_{i_n}=ns-(n-1)t\quad \mathrm{and}\quad q_{i_{n}}=(k-n+1)t-(k-n)s\end{align*} $$
for all
$n\in \{1, 2,\ldots ,k\}$
. Since
$p_{i_k}> 0$
and
$q_{i_1}> 0$
, this gives
$$ \begin{align*}\frac{(k-1)s}{k}< t<\frac{ks}{k-1}.\end{align*} $$
Then, by Condition (c) in Theorem 3.3, we have
$s\neq 0$
,
$t\neq 0$
,
$s\neq t$
,
$$ \begin{align*}(n-1)s-(n-2)t\neq0\quad \mathrm{and}\quad (k-n+1)t-(k-n)s\neq0\end{align*} $$
for any
$n\in \{2,\ldots ,k\}$
. Note that
$$ \begin{align*}t- q_{i_{n-1}}=(k-n+1)(s-t)\quad \mathrm{and}\quad s- p_{i_n}=-(n-1)(s-t).\end{align*} $$
Then,
$$ \begin{align*}c_{i_{n-1}}(k-n+1)[(n-1)s-(n-2)t]+c_{i_n}(n-1) [(k-n+1)t-(k-n)s]=0 \end{align*} $$
for all
$n\in \{2,\ldots ,k\}$
. So Condition (c1) follows immediately.
Case 2:
$p_{i_k}= 0$
and
$q_{i_1}\neq 0$
. Then,
$p_{i_1}=s$
. Combining this with (3-9),
$$ \begin{align*}p_{i_n}=ns-(n-1)t,\quad n\in \{1,2,\ldots,k\}.\end{align*} $$
Using the assumption that
$p_{i_k}= 0$
and (3-9) again, we arrive at
$t={k s}/({k-1})$
,
$$ \begin{align*}p_{i_n} = \frac{k-n}{k-1}s\quad \textrm{and}\quad q_{i_{n}}=-\frac{n-1}{k-1}s+q_{i_{1}} \end{align*} $$
for all
$n\in \{1, 2,\ldots ,k\}$
. Elementary calculations then show that
$$ \begin{align*}s- p_{i_n}=\frac{n-1}{k-1}s\quad\mathrm{and}\quad t- q_{i_{n}}=\frac{k+n-1}{k-1}s-q_{i_1}.\end{align*} $$
Since
$q_{i_n} (s- p_{i_n})\neq 0$
for all
$n\in \{2,\ldots ,k\}$
, we conclude that both s and
$q_{i_n}$
must be positive integers. Equivalently,
$$ \begin{align*}(k-1 )q_{i_{1}}>(n-1)s>0,\end{align*} $$
which implies that
$q_{i_1}>s\neq 0$
. Also,
${p_{i_{n-1}} (t-q_{i_{n-1}})\neq 0}$
for all
$n\in \{2,\ldots ,k\}$
implies that
$$ \begin{align*}q_{i_{1}}\neq \frac{k+n-2}{k-1}s,\quad n\in \{2,\ldots,k\}.\end{align*} $$
Then,
$$ \begin{align*}c_{i_{n-1}}{(k-n+1)}[(k+n-2)s-(k-1)q_{i_1}]+c_{i_n}(n-1)[(n-1)s-(k-1 )q_{i_{1}}]=0\end{align*} $$
for all
$n\in \{2,\ldots ,k\}.$
So Condition (c2) follows immediately.
Case 3:
$p_{i_k}\neq 0$
and
$q_{i_1}= 0$
. Similarly, it follows that
$s={k t}/({k-1})$
,
$$ \begin{align*} q_{i_n}=\frac{n-1}{k-1}t\quad \textrm{and}\quad p_{i_{n}}=-\frac{k-n}{k-1}t+p_{i_{k}} \end{align*} $$
for
$n\in \{1, 2,\ldots ,k\}$
. Since
${p_{i_{n-1}} (t-q_{i_{n-1}})\neq 0}$
for all
$n\in \{2,\ldots ,k\}$
,
$$ \begin{align*}(k-1 )p_{i_{k}}>(k-n+1)t>0,\end{align*} $$
which implies
$p_{i_k}>t\neq 0$
. Also, by
$q_{i_n} (s- p_{i_n})\neq 0$
for all
$n\in \{2,\ldots ,k\}$
, we have
$p_{i_{k}}\neq ({2k-n})t/({k-1})$
, which is equivalent to
$$ \begin{align*} p_{i_{k}}\neq \frac{k+n-2}{k-1}t,\quad n\in \{2,\ldots,k\}.\end{align*} $$
Then, using Condition (c4) in Theorem 3.3 again,
$$ \begin{align*}c_{i_{n-1}}{(k-n+1)}[(k-n+1)t-(k-1)p_{i_k}]+c_{i_n}(n-1)[(2k-n)t-(k-1 )p_{i_{k}}]=0\end{align*} $$
for all
$n\in \{2,\ldots ,k\}.$
So,
$$ \begin{align*}f_k(z)=c_{i_k}\sum_{n=1}^k\bigg[\prod_{j=n+1}^{k}\frac{(j-1)[(2k-j)t-(k-1)p_{i_k}]}{(k-j+1)[-(k-j+1)t+(k-1 )p_{i_{k}}]}\bigg]z^{p_{i_{k}}-(k-n)t/(k-1)}\overline{z}^{({n-1})t/({k-1})}.\end{align*} $$
Replacing n by
$k-n+1$
,
$$ \begin{align*}f_k(z)=c_{i_k}\sum_{n=1}^k\bigg[\prod_{j=k-n+2}^{k}\frac{(j-1) [(2k-j)t-(k-1)p_{i_k}]}{(k-j+1)[-(k-j+1)t+(k-1 )p_{i_{k}}]}\bigg]z^{p_{i_{k}}- (n-1)t/(k-1)}\overline{z}^{({k-n})t/({k-1})}.\end{align*} $$
Then, replacing j by
$k-j+2$
,
$$ \begin{align*}f_k(z)=c_{i_k}\sum_{n=1}^k\bigg[\prod_{j=2}^{n}\frac{(k-j+1)[(k+j-2)t-(k-1)p_{i_k}]}{(j-1)[-(j-1)t+(k-1 )p_{i_{k}}]}\bigg]z^{p_{i_{k}}-({n-1})t/({k-1})}\overline{z}^{({k-n})t/({k-1})},\end{align*} $$
which implies that Condition (c3) holds.
Case 4:
$p_{i_k}= 0$
and
$q_{i_1}= 0$
. Then, it follows easily that
$p_{i_{n-1}}=(k-n+1)(t-s)$
and
$q_{i_n}=(n-1)(s-t)$
for all
$n\in \{2,\ldots ,k\}$
. This contradicts the fact that
$p_{i_n}, q_{i_n}\in {\mathbb N}$
.
Conversely, if Condition (c) holds for some
${i}\in \{1,2,\ldots ,k\}$
, it is easy to verify that Condition (c) in Theorem 3.3 holds, and the proof of the theorem is complete.
Below, we present two examples of nontrivial zero commutators of two Toeplitz operators on
$L_a^2$
, corresponding to
$k=2$
and
$k=3$
of Theorem 1.1.
Example 3.4. On
$L_a^2$
,
$$ \begin{align*} [T_{2 z\overline z^3-3 z^2\overline z^4}, T_{z^2\overline z^3}]=0, \end{align*} $$
and
$$ \begin{align*} [T_{6 z\overline z^4-6 z^2\overline z^5-\overline{z}^3}, T_{z^2\overline z^3}] = 0. \end{align*} $$
4 Semi-commutators of two Toeplitz operators
In this section, we study the problem of when the semi-commutator of a monomial Toeplitz operator and a Toeplitz operator induced by a finite linear combination of monomials has finite rank on the Bergman space. First, we give the following critical lemma.
Lemma 4.1. Let t be a nonzero real number and
$p_i$
be nonnegative real numbers for each
$i\in \{1,2,\ldots ,k\}$
with
$k\geq 2$
. If
$p_i\neq p_j$
for any
$1\leq i\neq j\leq k$
, then on a certain right half-plane, the functions
$$ \begin{align*}\frac{1}{(z+p_1)(z+p_1-t)},\ldots, \frac{1}{(z+p_k)(z+p_k-t)}\end{align*} $$
are linearly independent.
Proof. Assume that
$$ \begin{align*} \sum_{i=1}^k\frac{C_i}{(z+p_i)(z+p_i-t)}=0 \end{align*} $$
for some constants
$C_1, \ldots , C_k$
. Note that
$t\neq 0$
. Then,
$$ \begin{align} \sum_{i=1}^k\frac{C_i}{z+p_i}-\sum_{i=1}^k\frac{C_i}{z+p_i-t}=0. \end{align} $$
It suffices for us to show that
$C_i=0$
for each
$i\in \{1,2,\ldots ,k\}$
. We prove this result by induction on k.
First, we assume that
$k=2$
. Since
$p_1+ p_2>0$
and
$t\neq 0$
, it follows that at least one of
$p_2-t\neq p_1$
and
$p_1-t\neq p_2$
must hold. Without loss of generality, we may assume that
$p_2-t\neq p_1$
. Then, the function
${1}/({z+p_1})$
cannot be written as a linear combination of the functions
$$ \begin{align*} \frac{1}{z+p_2},\quad \frac{1}{z+p_1-t},\quad \frac{1}{z+p_2-t}. \end{align*} $$
Combining this with (4-1), we deduce that
$C_1=C_2=0$
, as desired.
Next, we assume that
$k>2$
. If
$p_i\neq p_j-t$
for all
$1\leq i\neq j\leq k$
, then it is clear that the functions
$$ \begin{align*}\frac{1}{z+p_1}, \ldots, \frac{1}{z+p_k}, \frac{1}{z+p_1-t}, \ldots, \frac{1}{z+p_k-t}\end{align*} $$
are linearly independent, which implies that
$C_i=0$
for each
$i\in \{1,2,\ldots ,k\}$
. If there exists a permutation
$\tau : \{1,\ldots , k\}\rightarrow \{1,\ldots , k\}$
such that
$p_i=p_{\tau (i)}-t$
for all
$1\leq i\leq k$
, then
$$ \begin{align*}\sum_{i=1}^k p_i=\sum_{i=1}^k (p_{\tau(i)}-t) = \sum_{i=1}^k p_i-kt.\end{align*} $$
This contradicts the assumption that
$t\neq 0$
. So, we may further assume that there exists at least one index
${i_0}\in \{1,2,\ldots ,k\}$
such that
$$ \begin{align*}p_{i_0}\neq p_j-t\end{align*} $$
for all
$j=1,\ldots ,n$
. By the same argument as we used in the previous paragraph, we conclude that
$C_{i_0}=0$
. Then, the desired result follows from (4-1) and induction.
We are now ready to prove Theorem 1.5 stated in the introduction.
Proof of Theorem 1.5.
It is trivial that Condition (c) implies Condition (b), and Condition (b) implies Condition (a). So we need only show that Condition (a) implies Condition (c).
Suppose that
$(T_{f_k},T_{z^s\overline z^t}]$
has finite rank. For any nonempty subset
$\Lambda $
of
$\{1,2,\ldots ,k\}$
,
$$ \begin{align*}(T_{\sum_{i\in \Lambda}c_i z^{p_i}\overline z^{q_i}},T_{z^s\overline z^t}]=\sum\limits_{i\in \Lambda}c_i (T_{z^{p_i}\overline z^{q_i}},T_{z^s\overline z^t}].\end{align*} $$
So, by Corollary 2.3, we may assume that there exists an integer m such that
$$ \begin{align*}p_i-q_i+s-t=m\end{align*} $$
for all
$i\in \{1,2,\ldots ,k\}$
. Using Corollary 2.3 again,
$$ \begin{align*} \sum_{i=1}^{k}{c_i}\bigg[\frac{s-t+z}{(s+z)(s-t+p_i+z)}-\frac{1}{p_i+s+z}\bigg] = 0 \end{align*} $$
for all z in a certain right half-plane, or equivalently,
$$ \begin{align*} \sum_{i=1}^{k}\frac{c_ip_it}{(p_i+s+z)(s-t+p_i+z)}=0. \end{align*} $$
Then, it follows from Lemma 4.1 that either
$t=0$
or
$$ \begin{align*} c_ip_i=0, \quad i\in\{1,2,\ldots,k\}. \end{align*} $$
Since
$c_i\neq 0$
for all
$i\in \{1,2,\ldots ,k\}$
, the desired result follows.
Remark 4.2. Suppose t is a positive integer and
$f_k$
is defined as in Theorem 1.5. Note that
$$ \begin{align*} [T_{f_k},T_{z^t}] = - (T_{\overline{f_k}},T_{\overline{z}^t}]^*. \end{align*} $$
Then, as a direct consequence of Theorem 1.5, we get that
$T_{f_k}$
commutes with
$T_{z^t}$
if and only if
$f_k$
is an analytic function, which coincides with [Reference Čučković5, Theorem 1.4].
5 Certain special finite linear combinations of Toeplitz products
Many results shows that the rank of the commutator or the semi-commutator of two Toeplitz operators on
$L^{2}_{a}$
is either 0 or infinite. It is known that the corresponding result is false when the commutator or the semi-commutator is replaced by a finite sum of Toeplitz products. For example, [Reference Le and Thilakarathna17, Theorem 3.5] obtained necessary and sufficient conditions for a finite sum of products of two Toeplitz operators with rational symbols having poles outside of the closed unit disk to have finite rank. In this section, we fully concentrate on the finite linear combination of the Toeplitz products being of the form
$T_{z^k}T_{\overline z^l}$
. We do not only completely characterize when such an operator has finite rank, but also obtain the rank-one decomposition of the corresponding finite-rank operator.
The main tool for establishing our results is the Berezin transform. Given a bounded linear operator S on
$L^2_a$
, the Berezin transform of S is the function
$$ \begin{align*} B(S)(z)=\langle Sk_z,\ k_z\rangle,\quad z\in\mathbb{D}, \end{align*} $$
where
$k_z(w) = ({1-|z|^2})/{(1-\overline {z}w)^2}$
are normalized reproducing kernels. Elementary calculations show that
$$ \begin{align} B\big(\phi\otimes\psi\big)(z)=(1-|z|^2)^2\phi(z)\overline{\psi(z)} \end{align} $$
for any
$\phi , \psi \in L_a^2$
. Also, we need the following critical lemma.
Lemma 5.1. Let
$t_i\in {\mathbb N}$
for each
$i\in \{1,2,3\}$
with
$t_1<t_2<t_3$
. Then,
$$ \begin{align*} &(t_3-t_2)-(t_3-t_1)x^{t_2-t_1}+(t_2-t_1)x^{t_3-t_1}\\ &\quad =(1-x)^2\bigg[(t_3-t_2)\sum_{j=1}^{t_2-t_1-1} jx^{j-1}+(t_2-t_1)\sum_{l=1}^{t_3-t_2} lx^{t_3-t_1-l-1}\bigg]. \end{align*} $$
Proof. A direct calculation shows that
$$ \begin{align*} &(t_3-t_2)-(t_3-t_1)x^{t_2-t_1}+(t_2-t_1)x^{t_3-t_1}\\ &\quad =(t_3-t_2) (1-x^{t_2-t_1})-(t_2-t_1)x^{t_2-t_1}(1-x^{t_3-t_2})\\ &\quad =(1-x)\bigg[(t_3-t_2)\sum_{j=1}^{t_2-t_1} x^{j-1}-(t_2-t_1)\sum_{l=1}^{t_3-t_2} x^{t_2-t_1+l-1}\bigg] \end{align*} $$
and
$$ \begin{align*} &(1-x)\bigg[(t_3-t_2)\sum_{j=1}^{t_2-t_1-1} jx^{j-1}+(t_2-t_1)\sum_{l=1}^{t_3-t_2} lx^{t_3-t_1-l-1}\bigg]\\ &\quad =(t_3-t_2)\sum_{j=1}^{t_2-t_1} x^{j-1}-(t_2-t_1)\sum_{l=0}^{t_3-t_2-1} x^{t_3-t_1-l-1}. \end{align*} $$
Since
$$ \begin{align*}\sum_{l=1}^{t_3-t_2} x^{t_2-t_1+l-1}=\sum_{l=0}^{t_3-t_2-1} x^{t_3-t_1-l-1},\end{align*} $$
we obtain the desired result.
We are now ready to prove Theorem 1.6 stated in the introduction.
Proof of Theorem 1.6.
By Proposition 2.2, we get that the operator
$S_m$
has finite rank if and only if
$$ \begin{align*}\sum_{i=1}^{k} c_{i} {(z-t_i)}=0\end{align*} $$
for all
$z\in {\mathbb C}$
, which is equivalent to
$$ \begin{align*}\left\{\!\!{\begin{array}{ll} \displaystyle{c_{1}+c_{2}+\cdots+c_{k}=0,}\\[5pt] \displaystyle{t_1c_{1}+t_2c_{2}+\cdots+t_kc_{k}=0}. \end{array}} \right.\end{align*} $$
Clearly, the above system has a nonzero solution if and only if
$k\geq 3$
and
$$ \begin{align*} \left\{\!\! {\begin{array}{ll} \displaystyle{c_{1}=\frac{t_3-t_2}{t_2-t_1}c_{3}+\cdots+\frac{t_k-t_2}{t_2-t_1}c_{k},}\\[5pt] \displaystyle{c_{2}=\frac{t_1-t_3}{t_2-t_1}c_{3}+\cdots+\frac{t_1-t_k}{t_2-t_1}c_{k}}. \end{array}} \right.\end{align*} $$
To prove the rank-one decompositions of
$S_m$
, consider the operators
$$ \begin{align*}S_{m,i}=(t_i-t_2)T_{z^{p_1}}T_{\overline z^{t_1}}+(t_1-t_i)T_{z^{p_2}}T_{\overline z^{t_2}}+(t_2-t_1)T_{z^{p_i}}T_{\overline z^{t_i}}\end{align*} $$
for all
$i\in \{3,\ldots , k\}$
. Since
$$ \begin{align*} B (T_{z^{p_i}}T_{\overline z^{t_i}})(z)=z^{p_i}{\overline z^{t_i}}, \end{align*} $$
then
$$ \begin{align*}B (S_{m,i})(z)=(t_i-t_2)z^{p_1}{\overline z^{t_1}}-(t_i-t_1)z^{p_2}{\overline z^{t_2}}+(t_2-t_1)z^{p_i}{\overline z^{t_i}}. \end{align*} $$
If
$m\geq 0$
, then by Lemma 5.1 and (5-1),
$$ \begin{align*} &B(S_{m,i})(z)\\ &\quad =[(t_i-t_2)-(t_i-t_1){|z|^{2(t_2-t_1)}}+(t_2-t_1){|z|^{2(t_i-t_1)}}]{|z|^{2t_1}}z^{m}\\ &\quad=(1-|z|^2)^2 \bigg[(t_i-t_2)\sum_{j=1}^{t_2-t_1-1} j|z|^{2(t_1+ j-1)}+(t_2-t_1)\sum_{l=1}^{t_i-t_2} l|z|^{2(t_i-l-1)}\bigg]z^{m}\\ &\quad =(1-|z|^2)^2 \bigg[(t_i-t_2)\sum_{j=1}^{t_2-t_1-1} j z^{p_1+ j-1}\overline{z}^{t_1+ j-1}+(t_2-t_1)\sum_{l=1}^{t_i-t_2} lz^{p_i- l-1}\overline{z}^{t_i-l-1}\bigg]\\ &\quad =B\bigg[(t_i-t_2)\sum_{j=1}^{t_2-t_1-1} j z^{p_1+ j-1}\otimes{z}^{t_1+ j-1}+(t_2-t_1)\sum_{l=1}^{t_i-t_2} lz^{p_i- l-1}\otimes{z}^{t_i-l-1}\bigg](z). \end{align*} $$
If
$m<0$
, then by Lemma 5.1 and (5-1),
$$ \begin{align*} &B (S_{m,i})(z)\\ &\quad =[(t_i-t_2)-(t_i-t_1){|z|^{2(p_2-p_1)}}+(t_2-t_1){|z|^{2(p_i-p_1)}}]{|z|^{2p_1}}\overline{z}^{-m}\\ &\quad = [(p_i-p_2)-(p_i-p_1){|z|^{2(p_2-p_1)}}+(p_2-t_1){|z|^{2(p_i-p_1)}}]{|z|^{2p_1}}\overline{z}^{-m}\\ &\quad =(1-|z|^2)^2\bigg[(p_i-p_2)\sum_{j=1}^{p_2-p_1-1} j|z|^{2(p_1+ j-1)}+(p_2-p_1)\sum_{l=1}^{p_i-p_2} l|z|^{2(p_i-l-1)}\bigg]\overline{z}^{-m}\\ &\quad =(1-|z|^2)^2\bigg[(t_i-t_2)\sum_{j=1}^{t_2-t_1-1} j z^{p_1+ j-1}\overline{z}^{t_1+ j-1}+(t_2-t_1)\sum_{l=1}^{t_i-t_2} lz^{p_i- l-1}\overline{z}^{t_i-l-1}\bigg]\\ &\quad =B\bigg[(t_i-t_2)\sum_{j=1}^{t_2-t_1-1} j z^{p_1+ j-1}\otimes{z}^{t_1+ j-1}+(t_2-t_1)\sum_{l=1}^{t_i-t_2} lz^{p_i- l-1}\otimes{z}^{t_i-l-1}\bigg](z).\\[-15pt] \end{align*} $$
It was proved that the Berezin transform is injective [Reference Stroethoff23], which means
$$ \begin{align*}S_{m,i}=(t_i-t_2)\sum_{j=1}^{t_2-t_1-1} j z^{p_1+ j-1}\otimes{z}^{t_1+ j-1}+(t_2-t_1)\sum_{l=1}^{t_i-t_2} lz^{p_i- l-1}\otimes{z}^{t_i-l-1}.\\[-18pt]\end{align*} $$
Therefore,
$$ \begin{align*} S_m&=\sum_{i=3}^{k}\frac{c_{i}}{t_2-t_1}S_{m,i}\\ &=c_1\sum_{j=1}^{t_2-t_1-1} j z^{p_1+ j-1}\otimes{z}^{t_1+ j-1}+\sum_{i=3}^{k} c_i\sum_{l=1}^{t_i-t_2} lz^{p_i- l-1}\otimes{z}^{t_i-l-1}, \end{align*} $$
yielding the desired result.
A special case is worth mentioning here. When the summation is over all nonnegative integers that are less than or equal to a fixed positive integer, we have the following neat rank-one decomposition, which extends [Reference Le and Thilakarathna17, Example 4.1]. Their approach, which uses Mellin transform, is different from ours.
Example 5.2. Let
$m, d\in {\mathbb N}$
with
$d\geq 3$
. Consider the operator
$$ \begin{align*}S_m=\sum_{i=0}^{d-1} c_{i}T_{z^{i+m}}T_{\overline z^{i}}\end{align*} $$
with
$$ \begin{align*}\left\{\!\! {\begin{array}{ll} \displaystyle{c_{0}=c_{2}+\cdots+(d-2)c_{d-1}}\\[5pt] \displaystyle{c_{1}=-2c_{2}-\cdots-(d-1)c_{d-1}}. \end{array}} \right.\end{align*} $$
We have
$$ \begin{align*} S_m=\sum_{i=0}^{d-3} [c_{i+2}+2c_{i+3}+\cdots+(d-2-i)c_{d-1}]z^{m+i}\otimes z^{i}. \end{align*} $$
