S. Ji, X. Wei]School of Mathematical Sciences, Hebei Normal University,
Shijiazhuang, Hebei 050016, China

# On the irreducibility and weakly homogeneity of a class of operators

Shanshan Ji and Xiaomeng Wei∗ [<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

To construct more homogeneous operators, B. Bagchi and G. Misra in [2]
introduced the operator $\left(\begin{smallmatrix}T_{0}&T_{0}-T_{1}\\\
0&T_{1}\\\ \end{smallmatrix}\right)$ and proved that when $T_{0}$ and $T_{1}$
are homogeneous operators with the same unitary representation $U(g)$, it is
homogeneous with associated representation $U(g)\oplus U(g)$. At the same
time, they asked an open question, is the constructed operator irreducible? A.
Kor$\acute{a}$nyi in [31] showed that when the (1,2)-entry of the matrix is
$\alpha(T_{0}-T_{1})$, $\alpha\in\mathbb{C}$ the above result is also valid,
and their unitary equivalence class depends only on $|\alpha|$. In this case,
he and S. Hazra [20] gave a large class of irreducible homogeneous bilateral
$2\times 2$ block shifts, respectively, which are mutually unitarily
inequivalent for $\alpha>0$. In this note, we generalize the construction to
$T=\left(\begin{smallmatrix}T_{0}&XT_{1}-T_{0}X\\\ 0&T_{1}\\\
\end{smallmatrix}\right)$ and provide some sufficient conditions for its
irreducibility. We also find that for the above-mentioned $T_{0},T_{1}$ and
non-scalar operator $X$, $T$ is weakly homogeneous rather than homogeneous. So
the weak homogeneity problem related to $T$ is investigated.

2020 _Mathematics Subject Classification_. Primary 47B13, 32L05 $\cdot$
Secondary 51M15, 53C07

_Key words and phrases._ Block shifts; Irreducibility; Weakly homogeneity;
Similarity

This work was supported by National Natural Science Foundation of China (Grant
No. 11831006 and 11922108).

* Corresponding author

##  1\. Introduction

Let $\mathcal{H}$ be a complex separable Hilbert space and
$\mathcal{L}(\mathcal{H})$ denote the collection of bounded linear operators
on $\mathcal{H}$. Let $T_{i}\in\mathcal{L}(\mathcal{H}_{i})$, $i=0,1$. We say
$T_{0}$ is similar to $T_{1}\,\,(T_{0}\sim_{s}T_{1})$ if there exists an
invertible operator $X:\,\mathcal{H}_{0}\to\mathcal{H}_{1}$ such that
$XT_{0}=T_{1}X$. In particular, when $X$ is a unitary operator, we say $T_{0}$
is unitarily equivalent to $T_{1}\,\,(T_{0}\sim_{u}T_{1})$.

In the study of operator theory, it has been an active topic to find a model
for the class of operators satisfying certain conditions. When dim
$\mathcal{H}<\infty$, the Jordan canonical form theorem provides a model for
all operators in $\mathcal{L}(\mathcal{H})$ up to similarity. When unitary
equivalence is considered, it is found by spectral theory that every normal
matrix can be written as an orthogonal direct sum of scalar operators on the
eigenspace. In the case of dim $\mathcal{H}=\infty$, the orthogonal direct sum
will be replaced by a continuous direct sum or direct integration. However,
there is still no general solution for operators whose spectra are not thin,
or for direct sums of these operators. For example, the backward shift
operator $U_{+}^{*}:l^{2}(\mathbb{N})\rightarrow l^{2}(\mathbb{N})$, defined
by
$U_{+}^{*}(\alpha_{0},\alpha_{1},\alpha_{2},\cdots)=(\alpha_{1},\alpha_{2},\alpha_{3},\cdots)$.
That means $\mathbb{D}\subseteq\sigma(U_{+}^{*})$, the spectrum of
$U_{+}^{*}$.

In 1978, M.J. Cowen and R.G. Douglas systematically studied a subclass
$\mathcal{B}_{n}(\Omega)$ of $\mathcal{L}(\mathcal{H})$ in [10], which contain
$U_{+}^{*}$ and their point spectra contain a bounded connected open set
$\Omega$ of $\mathbb{C}$. There are many criteria for local equivalence to
determine the unitary equivalence of operators, including Hermitian
holomorphic vector bundles, connections, the curvature, local operators, etc
in [10, 11, 12, 24]. Operators in $\mathcal{B}_{n}(\Omega)$ correspond to the
adjoints of the multiplication operators on the reproducing kernel Hilbert
space, and the unitary equivalence of them can also be fully described using
the reproducing kernel due to [9, 10, 43]. Similar transformations break
rigidity, so there are not many geometrically similar equivalent results for
such operators. Recently, Cowen-Douglas operators with certain structures have
been studied [15, 21, 22, 25, 26, 29, 30].

We know that $U_{+}^{*}\in\mathcal{B}_{1}(\mathbb{D})$ and its curvature is
$\mathcal{K}_{U_{+}^{*}}(w)=-\frac{1}{(1-|w|^{2})^{2}}$. G. Misra in [36]
proved that if $T\in\mathcal{B}_{1}(\mathbb{D})$ is contractive, then
$\mathcal{K}_{T}(w)\leq\mathcal{K}_{U_{+}^{*}}(w)$, $w\in\mathbb{D}$. For such
an operator $T$, R.G. Douglas once asked if there is a point
$w_{0}\in\mathbb{D}$ such that
$\mathcal{K}_{T}(w_{0})=-\frac{1}{(1-|w_{0}|^{2})^{2}}$, is it inferred that
$T\sim_{u}U_{+}^{*}$? Subsequently, G. Misra [35] gave a characterization of
contractive homogeneous operators in $\mathcal{B}_{1}(\mathbb{D})$ and
indicated that the question raised by R.G. Douglas is positive for homogeneous
operators, but not always correct for other operators. Some affirmative
answers are also given in [38].

Irreducible homogeneous operators in the Cowen-Douglas class have attracted
much attention, while a huge theory has been built. Many impressive results
have been obtained using the representation theory of Lie groups, complex
geometry and matrix structures, including explicit representations of these
operators and unitary classification (see [3, 4, 5, 6, 23, 32, 33, 34, 37]).
These operators are not only rich in structure, but in most cases also serve
as models for other operators. In [1], all homogeneous shift operators are
listed due to B. Bagchi and G. Misra, it follows that there are many
homogeneous operators outside of Cowen-Douglas operators. Then some attempts
were made to construct more matrix-valued homogeneous operators and several
open questions in this research direction were listed, one of them is whether
the constructed operator $\left(\begin{smallmatrix}T_{0}&T_{0}-T_{1}\\\
0&T_{1}\\\ \end{smallmatrix}\right)$ is irreducible, where $T_{0}$ and $T_{1}$
are homogeneous operators with the same unitary representation (see [2]).
Generalizing the above operator to
$\left(\begin{smallmatrix}T_{0}&\alpha(T_{0}-T_{1})\\\ 0&T_{1}\\\
\end{smallmatrix}\right)$ for $\alpha\in\mathbb{C}$, A. Kor$\acute{a}$nyi in
[31] proved their unitary classification only depends on $|\alpha|$. In this
form of operators, A. Kor$\acute{a}$nyi first described a class of irreducible
homogeneous bilateral block shifts, where $T_{0}$ and $T_{1}$ are shifts with
weight sequences $\\{(\frac{n+a}{n+b})^{\frac{1}{2}}\\}_{n\in\mathbb{Z}}$ and
$\\{(\frac{n+b}{n+a})^{\frac{1}{2}}\\}_{n\in\mathbb{Z}}$, $0<a,b<1$ and $a\neq
b$. Afterward, S. Hazra [20] also found two classes of irreducible homogeneous
bilateral 2-by-2 shifts, one of which has the structure shown above. It was
also shown that these three classes of operators are mutually unitarily
inequivalent when the parameter $\alpha>0$. From this, a new operator
structure is generalized as follows:

(1.1) $T=\left(\begin{smallmatrix}T_{0}&XT_{1}-T_{0}X\\\ 0&T_{1}\\\
\end{smallmatrix}\right)$

is equipped with a non-zero bounded linear operator $X$. It is known that
irreducible operators are minimal operator units modulo unitary equivalence.
Inspired by the above results, we consider when $T$ in (1.1) is irreducible.

Homogeneous operators have yielded numerous results. As its extension, weakly
homogeneous operators are defined (see [2, 14]). In [17], S. Ghara pointed out
the necessary and sufficient conditions for the multiplication by the
coordinate function $z$ acting on a Hilbert space $\mathcal{H}_{K}$ possessing
a sharp reproducing kernel $K$, to be weakly homogeneous, and gave some
nontrivial results on weakly homogeneous, so that many high index weakly
homogeneous operators were obtained. So far, the characterization of weakly
homogeneous operators remains a challenging problem, even for operators of
index one in the Cowen-Douglas class. Because $T$ in (1.1) is strongly
reducible, its homogeneity has expired. Thus, it is natural to investigate
their weak homogeneity.

The paper is organized as follows. In §2, we introduced some preliminaries. In
§3, we discussed when $T$ in (1.1) is irreducible and provide some sufficient
conditions, which is related to the open question 11 proposed by B. Bagchi and
G. Misra in [2]. In §4, some weakly homogeneous problems of $T$ in (1.1) were
investigated.

##  2\. Preliminaries

###  2.1. Cowen-Douglas operators

M.J. Cowen and R.G. Douglas in [10] introduced a class of operators
$\mathcal{B}_{n}(\Omega)$, named Cowen-Douglas operators, defined as follows:

######  Definition 2.1.

[10] For a connected open subset $\Omega$ of $\mathbb{C}$ and a positive
integer $n$, let

$\begin{array}[]{lll}\mathcal{B}_{n}(\Omega):=\\{T\in\mathcal{L}(\mathcal{H})|&(1)\,\,\Omega\subset\sigma(T),\\\
&(2)\,\,\mbox{ran}(T-w)=\mathcal{H}\,\,\text{for\,all}\,\,w\in\Omega,\\\
&(3)\,\,\bigvee_{w\in\Omega}\mbox{ker}(T-w)=\mathcal{H},\\\
&(4)\,\,\mbox{dim}\,\mbox{ker}(T-w)=n\,\,\text{for\,
all}\,\,w\in\Omega\\},\end{array}$

where $\sigma(T)$ is the spectrum of the operator $T$ and $\bigvee$ stands for
the closed linear span.

If $T$ is in $\mathcal{B}_{n}(\Omega)$, then it is possible to choose $n$
linearly independent eigenvectors in $\mbox{ker}(T-w)$, which are holomorphic
as functions. Thus $w\mapsto\mbox{ker}(T-w)$ defines a rank $n$ Hermitian
holomorphic vector bundle $E_{T}$ over $\Omega$,

$E_{T}=\\{(w,x)\in\Omega\times\mathcal{H}:x\in\mbox{ker}(T-w)\\}$

with the natural projection map $\pi:E_{T}\to\Omega,\pi(w,x)=w$. Let
$\\{\gamma_{i}\\}^{n}_{i=1}$ be a holomorphic frame of $E_{T}$. Then
$\\{\gamma_{i}(w)\\}^{n}_{i=1}$ is a basis of $\mbox{ker}(T-w)$ and the metric
of $E_{T}$ is the Gram matrix
$h(w)=(\langle\gamma_{j}(w),\gamma_{i}(w)\rangle)^{n}_{i,j=1}$. The curvature
is given by
$\mathcal{K}_{T}(w)=-\frac{\partial}{\partial\overline{{w}}}[h^{-1}(w)\frac{\partial}{\partial{w}}h(w)],w\in\Omega$.
In fact, the curvature depends on the choice of the frame, except when $n=1$.
In the case of $E_{T}$ is a line bundle, its curvature is
$\mathcal{K}_{T}(w)=-\frac{\partial^{2}}{\partial{w}\overline{\partial}w}\mbox{log}\|\gamma(w)\|^{2}$,
where $\gamma$ is a non-zero holomorphic section of $E_{T}$.

A function $K:\Omega\times\Omega\rightarrow\mathcal{M}_{n}(\mathbb{C})$ is
said to be a non-negative definite kernel if for any subset
$(w_{1},\cdots,w_{k})$ in $\Omega$, the $k\times k$ block matrix
$\big{(}K(w_{i},w_{j})\big{)}_{i,j=1}^{k}$ is non-negative definite, that is,
$\sum_{i,j=1}^{k}\langle K(w_{i},w_{j})\eta_{j},\eta_{i}\rangle\geq
0,\eta_{1},\cdots,\eta_{n}\in\mathbb{C}^{n}$. We say that $K(\cdot,\cdot)$ is
sesqui-analytic, if $K$ is holomorphic for the first variable and anti
holomorphic for the second variable. For a sesqui-analytic non-negative
definite kernel $K:\Omega\times\Omega\rightarrow\mathcal{M}_{n}(\mathbb{C})$,
a well-known theorem given by Moore that there exists a unique Hilbert space
$\mathcal{H}_{K}$ consisting of $\mathbb{C}^{n}$-valued holomorphic functions
$E_{w}^{*}\xi=K(\cdot,w)\xi$ on $\Omega$ such that the evaluation map $E_{w}$
is bounded for each $w\in\Omega$ and $K$ is the reproducing kernel of
$\mathcal{H}_{K}$, where the reproducing property is $\langle
f,K(\cdot,w)\xi\rangle_{\mathcal{H}_{K}}=\langle
f(w),\xi\rangle_{\mathbb{C}^{n}}$, $\xi\in\mathbb{C}^{n}$. Based on this, for
$T\in\mathcal{B}_{n}(\Omega)$, $T$ can be realized as the adjoint of the
multiplication operator $M_{z}$ on a reproducing kernel Hilbert space
$\mathcal{H}_{K}$ of holomorphic $\mathbb{C}^{n}$-valued functions on
$\Omega^{*}$ with reproducing kernel $K$, denoted by
$T\sim_{u}(M_{z}^{*},\mathcal{H}_{K})$, and
$M_{z}^{*}K(\cdot,\bar{w})\xi=wK(\cdot,\bar{w})\xi$ for $w\in\Omega$ and
$\xi\in\mathbb{C}^{n}$. The reader is referred to [9, 10, 43].

###  2.2. Homogeneous operators

Let $M\ddot{o}b$ denote the $M\ddot{o}bious$ group of all biholomorphism
automorphisms of $\mathbb{D}:=\\{z\in\mathbb{C}:|z|<1\\}$. Then the form of
$\phi$ in $M\ddot{o}b$ is
$\phi(z)=e^{i\theta}\frac{z-a}{1-\overline{a}z},\theta\in\mathbb{R},a\in\mathbb{D}$.
Operator $T\in\mathcal{L}(\mathcal{H})$ is said to be homogeneous if
$\sigma(T)\subset\overline{\mathbb{D}}$, the closed unit disc and
$\phi(T)\sim_{u}T$ for every $\phi$ in $M\ddot{o}b$ (see [3, 35, 37]).

Let $\mathcal{U}(\mathcal{H})$ be the group of unitary operators on
$\mathcal{H}$. A Borel function $\pi:\,G\to\mathcal{U}(\mathcal{H})$ is said
to be a projective presentation of $G$ on the Hilbert space $\mathcal{H}$. If
there is a projective representation $\pi$ of $M\ddot{o}b$ on $\mathcal{H}$
with the property $\phi(T)=\pi(\phi)^{*}T\pi(\phi),$ then $\pi$ is said to be
the representation associated with the operator $T$. For every irreducible
homogeneous operator, there exists a unique (up to equivalence) projective
representation associated with it (see [1]).

A complete characterization of the contractive homogeneous operator in
$\mathcal{B}_{1}(\mathbb{D})$ is pointed out by G. Misra in [35] to be that
its curvature is equal to $-\lambda(1-|w|^{2})^{-2}$ for some positive real
number $\lambda$. That means $T$ and $T^{*}$ are homogeneous, if
$T\sim_{u}(M^{*}_{z},\mathcal{H}_{K^{(\lambda)}})$,
$K^{(\lambda)}(z,w)=\frac{1}{(1-z\overline{w})^{\lambda}}$,
$z,w\in\mathbb{D}$. Besides, there are three bilateral shift operators that
are also homogeneous operators: the unweighted bilateral shift $B$, $K_{a,b}$
with weight sequence $\sqrt{\frac{n+a}{n+b}},n\in\mathbb{Z}$ for any two
distinct real numbers $a$ and $b$ in $(0,1)$, and $B_{x}$ with weight sequence
$\cdots,1,x,1,\cdots$ ($x$ in the zeroth slot, $1$ elsewhere). The above list
is exhaustive.

As a generalization of homogeneous operators, weakly homogeneous operators are
defined in [2, 14]. Let $T\in\mathcal{L}(\mathcal{H})$, the operator $T$ is
said to be a weakly homogeneous operator if
$\sigma(T)\subseteq\bar{\mathbb{D}}$ and for each $\phi\in M\ddot{o}b$,
$\phi(T)\sim_{s}T$. For $M_{z}$ acting on $\mathcal{H}_{K}$, the boundedness
of the composition operator $C_{\phi}$, defined by
$C_{\phi}(f)=f\circ\phi,f\in\mathcal{H}_{K}$ means the weak homogeneity of
$M_{z}$. For any $\gamma\in\mathbb{R}$, let $K_{(\gamma)}$ be the positive
definite kernel given by
$K_{(\gamma)}(z,w)=\sum\limits_{n=0}^{\infty}(n+1)^{\gamma}z^{n}\bar{w}^{n}$,
$z,w\in\mathbb{D}$. N. Zorboska[42], C.C. Cowen and B.D. MacCluer[8] proved
that $C_{\phi}$ is bounded on $\mathcal{H}_{K^{(\lambda)}}$, $\lambda>0$ and
$\mathcal{H}_{K_{(\gamma)}}$, $\gamma\in\mathbb{R}$, that is,
$(M_{z}^{*},\mathcal{H}_{K^{(\lambda)}})$ and
$(M_{z}^{*},\mathcal{H}_{K_{(\gamma)}})$ are weakly homogeneous.

###  2.3. Irreducible operator

Let $T\in\mathcal{L}(\mathcal{H})$. The commutant algebra of $T$ is
$\mathcal{A}^{\prime}(T):=\\{X|XT=TX\\}$. For a closed linear subspace
$\mathcal{M}$ of $\mathcal{H}$, $\mathcal{M}$ is said to be a reduced subspace
of $T$ if both $\mathcal{M}$ and $\mathcal{M}^{\bot}$ are invariant subspaces
of $T$. $T\in\mathcal{L}(\mathcal{H})$ is said to be irreducible if $T$ only
has trivial reduced subspaces, equivalently, $\mathcal{A}^{\prime}(T)$ only
has trivial orthogonal projections in [19]. $T$ is strongly irreducible, if
$\mathcal{A}^{\prime}(T)$ does not contain idempotent operators other than 0
and $I$. In what follows, $T\in(SI)$ means $T$ is strongly irreducible.

C.L. Jiang and his collaborators introduced the $K_{0}$-group of the operator
commutant algebra into the study of operator theory and provided a similar
classification theorem of Cowen-Douglas operators (see [7, 27, 28]). Operator
$T\in\mathcal{L}(\mathcal{H})$ has a finite (SI) decomposition if and only if
there exist $k\in\mathbb{N}$, $n_{1},\cdots,n_{k}\in\mathbb{N}$ and
$T_{1},\cdots,T_{k}\in(SI)$ such that $T_{i}\nsim_{s}T_{j}(i\neq j)$ and
$T=T^{(n_{1})}_{1}\oplus\cdots\oplus T^{(n_{k})}_{k}$ (see [7]).

######  Definition 2.2.

[7] Let $T\in\mathcal{L}(\mathcal{H})$ have a finite (SI) decomposition
$T=T^{(n_{1})}_{1}\oplus\cdots\oplus T^{(n_{k})}_{k}$. We say $T$ has a unique
(SI) decomposition up to similarity if there exists $S$ such that $S\sim_{s}T$
and $S$ has a finite (SI) decomposition $S=S^{(m_{1})}_{1}\oplus\cdots\oplus
S^{(m_{l})}_{l}$, then $k=l$ and there exists a permutation $\pi$ of the set
$\\{1,2,\cdots,k\\}$ such that
$T^{(n_{i})}_{i}\sim_{s}S^{(m_{\pi(i)})}_{\pi(i)}$ and $n_{i}=m_{\pi(i)}$.

######  Theorem 2.3.

[28] Let $A,B\in\mathcal{B}_{n}(\mathbb{D})$. Suppose that
$A^{(n_{1})}_{1}\oplus\cdots\oplus A^{(n_{k})}_{k}$ is a finite (SI)
decomposition of $A$. Then $A\sim_{s}B$ if and only if:

  1. (1)

$(K_{0}(\mathcal{A}^{\prime}(A\oplus B)),\bigvee(\mathcal{A}^{\prime}(A\oplus
B)),I)\cong(\mathbb{Z}^{(k)},\mathbb{N}^{(k)},1)$;

  2. (2)

The isomorphism $h$ from $\bigvee(\mathcal{A}^{\prime}(A\oplus B))$ to
$\mathbb{N}^{(k)}$ sends $[I]$ to $(2n_{1},2n_{2},\cdots,2n_{k})$, i.e.,
$h([I])=2n_{1}e_{1}+2n_{2}e_{2}+\cdots+2n_{k}e_{k}$, where $I$ is the unit of
$\mathcal{A}^{\prime}(A\oplus B)$ and $\\{e_{i}\\}^{k}_{i=1}$ are the
generators of $\mathbb{N}^{(k)}$.

######  Theorem 2.4.

[27] Two strongly irreducible Cowen-Douglas operators $A$ and $B$ are similar
if and only if there exists a positive integer $n$ such that
$A^{(n)}\sim_{s}B^{(n)}$.

##  3\. Irreducibility of weighted block shift operators

Let $T\in\mathcal{L}(\mathcal{H})$. Then $T$ is called a block shift if there
is an orthogonal decomposition $\mathcal{H}=\oplus_{n\in I}W_{n}$ of
$\mathcal{H}$ with non-trivial subspaces $W_{n},n\in I$, such that
$T(W_{n})\subseteq W_{n+1}$ for all $n$ in $I$. $T$ is said to be a bilateral
or unilateral block shift if $I=\mathbb{Z},\mathbb{Z}^{+}$ or
$\mathbb{Z}^{-}$. The subspaces $W_{n},n\in I$ are called the blocks of $T$.
When dim $W_{n}=1$, $T$ is called a shift operator. In this case, if
$W_{n}=span\\{e_{n}\\}$, then $\\{e_{n}\\}_{n\in I}$ is the orthonormal basis
of $\mathcal{H}$. A. Kor$\acute{a}$nyi and S. Hazra found three large classes
irreducible bilateral block shifts in [31, 20]. B. Bagchi and G. Misra proved
that any irreducible homogeneous operator is a block shift in [1]. Motivated
by these results, in this section we consider the irreducibility of certain
block shifts of block size to be two. Some sufficient conditions are given.

Before stating our results, we first specify some common notations. Let
$l^{2}(\mathbb{Z})=\\{(a_{i})_{i\in\mathbb{Z}}:\sum_{i\in\mathbb{Z}}|a_{i}|^{2}<\infty\\}$
and $T_{0},T_{1}$ on $l^{2}(\mathbb{Z})$ be bilateral weighted shifts with
non-zero weighted sequences $\\{w_{n}\\}_{n\in\mathbb{Z}}$ and
$\\{v_{n}\\}_{n\in\mathbb{Z}}$ respect to orthonormal
basis$\\{e_{n}\\}_{n\in\mathbb{Z}}$ of $l^{2}(\mathbb{Z})$. Define
$T(w_{n},v_{n},d_{n}):=\left(\begin{smallmatrix}T_{0}&XT_{1}-T_{0}X\\\
0&T_{1}\\\ \end{smallmatrix}\right)$ acting on
$\mathcal{H}=l^{2}(\mathbb{Z})\oplus l^{2}(\mathbb{Z})$, where $X$ satisfies
$Xe_{n}=d_{n}e_{n}$ for a uniformly bounded sequence
$\\{d_{n}\\}_{n\in\mathbb{Z}}$ and $XT_{1}\neq T_{0}X$. When
$w_{n}=\frac{1}{v_{n}}=t_{n}$, $T(w_{n},v_{n},d_{n})=T(t_{n},d_{n})$. In these
forms, let $\mathcal{B}_{n}=\\{\left(\begin{smallmatrix}e_{n}\\\ 0\\\
\end{smallmatrix}\right),\left(\begin{smallmatrix}0\\\ e_{n}\\\
\end{smallmatrix}\right)\\}$ be a orthonormal set and
$H(n)=\mbox{span}\mathcal{B}_{n}$. Write $T_{n}=T|_{H(n)}$, we can see
$A_{n}:=T^{*}_{n}T_{n}$ and $B_{n}:=T_{n-1}T^{*}_{n-1}$ map $H(n)$ to $H(n)$.
With these notations, we have the following results.

###  3.1. The irreducibility of $T(t_{n},d_{n})$

######  Proposition 3.1.

Let $T=T(t_{n},\alpha)$ be a bounded operator for non-zero complex number
$\alpha$. If

  1. (1)

there exists $i_{0}\in\mathbb{Z}$ such that
$t^{2}_{n}+\frac{1}{t^{2}_{n}}=t^{2}_{m}+\frac{1}{t^{2}_{m}}$ only holds when
$m=-(n+i_{0})$;

  2. (2)

$t_{n}>0$ and $t_{n}\neq\frac{1}{t_{n-1}}\neq 1$,

then $T$ is irreducible.

###### Proof.

Let $\mathcal{K}$ be an arbitrary non-zero reduced subspace of $T$. We will
check that $\mathcal{K}=\mathcal{H}=span\\{\left(\begin{smallmatrix}e_{n}\\\
0\\\ \end{smallmatrix}\right),\left(\begin{smallmatrix}0\\\ e_{n}\\\
\end{smallmatrix}\right),n\in\mathbb{Z}\\}$. Since $\mathcal{K}$ is a closed
linear subspace and $H(n)=\mbox{span}\\{\left(\begin{smallmatrix}e_{n}\\\ 0\\\
\end{smallmatrix}\right),\left(\begin{smallmatrix}0\\\ e_{n}\\\
\end{smallmatrix}\right)\\}$, if we prove that for each $n$,
$H(n)\subset\mathcal{K}$, then the theorem is proven. In the following, we
will complete the proof in three steps.

Step 1: There exists $n_{0}\in\mathbb{Z}$ such that $\mathcal{K}\cap
H(n_{0})\neq\emptyset$.

If $\mathcal{K}$ is a reducing subspace of $T$, then $T^{*}T\mathcal{K}\subset
T^{*}\mathcal{K}\subset\mathcal{K},T^{*}T\mathcal{K}^{\perp}\subset
T^{*}\mathcal{K}^{\perp}\subset\mathcal{K}^{\perp}$, $\mathcal{K}$ is an
invariant subspace of both $T^{*}T$ and $TT^{*}$. From the spectral theorem of
positive operators, we can conclude that the restrictions of elements in
$\mathcal{K}$ on the eigenspaces of $T^{*}T$ and $TT^{*}$ still belong to
$\mathcal{K}$. Next, we will calculate the eigenspaces of $T^{*}T$ and
$TT^{*}$ by rearranging the orthonormal basis of $\mathcal{H}$ and
partitioning the operators.

It is easy to see that $T$ maps $H(n)$ to $H(n+1)$ and
$T_{n}=\left(\begin{smallmatrix}t_{n}&\alpha(\frac{1}{t_{n}}-t_{n})\\\
0&\frac{1}{t_{n}}\\\ \end{smallmatrix}\right)$. It follows that

$A_{n}=\left(\begin{smallmatrix}t^{2}_{n}&\alpha(1-t^{2}_{n})\\\
\bar{\alpha}(1-t^{2}_{n})&|\alpha|^{2}(\frac{1}{t_{n}}-t_{n})^{2}+\frac{1}{t^{2}_{n}}\\\
\end{smallmatrix}\right),B_{n}=\left(\begin{smallmatrix}t^{2}_{n-1}+|\alpha|^{2}(\frac{1}{t_{n-1}}-t_{n-1})^{2}&\alpha(\frac{1}{t^{2}_{n-1}}-1)\\\
\bar{\alpha}(\frac{1}{t^{2}_{n-1}}-1)&\frac{1}{t^{2}_{n-1}}\\\
\end{smallmatrix}\right).$

We have $A_{n}\neq I$ and its determinant is $1$. Since the spectrum of
positive operators is not-negative, the eigenvalues of $A_{n}$ must be real
numbers. So there exists a real number $\lambda_{n}>1$ such that the
eigenvalues of $A_{n}$ are $\lambda^{2}_{n}$ and $\frac{1}{\lambda^{2}_{n}}$.
Note that
$\mbox{trace}A_{n}=(|\alpha|^{2}+1)(t^{2}_{n}+\frac{1}{t^{2}_{n}})-2|\alpha|^{2}=\lambda^{2}_{n}+\frac{1}{\lambda^{2}_{n}}$.
By using the condition
$t^{2}_{n}+\frac{1}{t^{2}_{n}}=t^{2}_{-(n+i_{0})}+\frac{1}{t^{2}_{-(n+i_{0})}}$
for some $i_{0}\in\mathbb{Z}$. This is equivalent to
$\lambda^{2}_{n}+\frac{1}{\lambda^{2}_{n}}=\lambda^{2}_{-(n+i_{0})}+\frac{1}{\lambda^{2}_{-(n+i_{0})}}$
for some $i_{0}\in\mathbb{Z}$. Since $\lambda_{n}>1$ for any $n\in\mathbb{Z}$,
that is $\lambda_{n}=\lambda_{-(n+i_{0})}$ for some $i_{0}\in\mathbb{Z}$. Let
$\lambda^{(1)}_{n}=\lambda^{2}_{n}$ and
$\lambda^{(2)}_{n}=\frac{1}{\lambda^{2}_{n}}$. Suppose that
$\\{v^{(1)}_{n},v^{(2)}_{n}\\}$ is the orthonormal basis which makes $A_{n}$
diagonal. Let $u^{(k)}_{n}=T_{n-1}v^{(k)}_{n-1}$. That means
$B_{n}u^{(k)}_{n}=T_{n-1}T^{*}_{n-1}T_{n-1}v^{(k)}_{n-1}=\lambda^{(k)}_{n-1}T_{n-1}v^{(k)}_{n-1}=\lambda^{(k)}_{n-1}u^{(k)}_{n}$.
Since the orthogonality of $v^{(1)}_{n}$ and $v^{(2)}_{n}$, $u^{(1)}_{n}$ and
$u^{(2)}_{n}$ are also, $\langle u^{(1)}_{n},u^{(2)}_{n}\rangle=\langle
T^{*}_{n-1}T_{n-1}v^{(1)}_{n-1},v^{(2)}_{n-1}\rangle=\langle\lambda_{n-1}v^{(1)}_{n-1},v^{(2)}_{n-1}\rangle=0$.
Clearly, $\\{u^{(1)}_{n},u^{(2)}_{n}\\}$ is an orthonromal basis of $H(n)$
which makes $B_{n}$ diagonal.

I will complete the proof of Step 1 in two cases.

Case 1: $i_{0}$ is an odd number. Then the multiplicity of every eigenvalues
of $T^{*}T$ is two. Let
$\mathcal{A}_{n,k}=\mbox{span}\\{v^{(k)}_{n},v^{(k)}_{-(n+i_{0})}\\},\ k=1,2$
be the eigenspace of $T^{*}T$ corresponds to the eigenvalue
$\lambda_{n}^{(k)}=\lambda_{-(n+i_{0})}^{(k)}$. Then
$\mathcal{H}=\bigoplus_{k=1,2;n\geq\frac{1-i_{0}}{2}}\mathcal{A}_{n,k}$. For a
fixed but arbitrary a non-zero vector $f\in\mathcal{K}$, we have
$f=\sum_{k=1,2;n\geq\frac{1-i_{0}}{2}}\alpha_{n,k}h_{n,k}$, where
$h_{n,k}\in\mathcal{A}_{n,k}$. Since $f$ is non-zero, there exist $n,k$ such
that $\alpha_{n,k}h_{n,k}\neq 0$. It implies that $h_{n,k}\neq 0$ and
$h_{n,k}\in\mathcal{K}$, since $h_{n,k}$ is the restriction of $f$ on the
eigenspace $\mathcal{A}_{n,k}$. Then there exist $\gamma,\delta$ that are not
all zero such that $h_{n,k}=\gamma v^{(k)}_{n}+\delta v^{(k)}_{-(n+i_{0})}$,
where $v^{(k)}_{n}\in H(n)$ and $v^{(k)}_{-(n+i_{0})}\in H(-(n+i_{0}))$.
Furthermore, for any $m\in\mathbb{Z}$, we have
$T^{m}h_{n,k}=\hat{\delta}\hat{h}_{m-n-
i_{0}}+\hat{\gamma}\hat{h}_{m+n}\in\mathcal{K}$, where $\hat{h}_{m-n-i_{0}}\in
H(m-n-i_{0}),\hat{h}_{m+n}\in H(m+n)$. From the arbitrariness of $m$, we can
assume that $m$ is sufficiently large, up to $m-n-
i_{0}\geq\frac{1-i_{0}}{2},m+n\geq\frac{1-i_{0}}{2}$. In this case, the
eigenvalues of the $T^{*}T$ restricted to the spaces $H(m-n-i_{0})$ and
$H(m+n)$ must not be the same. Then there exist
$\delta_{1},\delta_{2},\gamma_{1},\gamma_{2}$ that are not all zero such that
$T^{m}h_{n,k}=\hat{\delta}\delta_{1}v^{(1)}_{m-n-
i_{0}}+\hat{\delta}\delta_{2}v^{(2)}_{m-n-
i_{0}}+\hat{\gamma}\gamma_{1}v^{(1)}_{m+n}+\hat{\gamma}\gamma_{2}v^{(2)}_{m+n}$.
Since every operator $A_{n}=T_{n}^{*}T_{n}$ is invertible and $h_{n,k}\neq 0$,
so is $T^{m}h_{n,k}\neq 0$. We know that $v^{(1)}_{m-n-i_{0}},v^{(2)}_{m-n-
i_{0}},v^{(1)}_{m+n},v^{(2)}_{m+n}$ are the eigenvectors corresponding to
different eigenvalues of $T^{*}T$. Therefore at least one of the coefficients
of the vectors $v^{(1)}_{m-n-i_{0}},v^{(2)}_{m-n-
i_{0}},v^{(1)}_{m+n},v^{(2)}_{m+n}$ is not zero. This implies that at least
one of the vectors $v^{(1)}_{m-n-i_{0}},v^{(2)}_{m-n-
i_{0}},v^{(1)}_{m+n},v^{(2)}_{m+n}$ belongs to $\mathcal{K}$. It follows that
$\mathcal{K}\cap H(n_{0})\neq\emptyset$ for some $n_{0}$.

Case 2: $i_{0}$ is an even number. Then the multiplicity of the eigenvalue
$\lambda_{-\frac{i_{0}}{2}}$ of $T^{*}T$ is one, the multiplicity of other
eigenvalues is two. Let
$\mathcal{H}=\bigoplus_{k=1,2;n\geq-\frac{i_{0}}{2}}\mathcal{A}_{n,k}$, where
$\mathcal{A}_{-\frac{i_{0}}{2},k}=\mbox{span}\\{v^{(k)}_{-\frac{i_{0}}{2}}\\}$
is the eigenspace of $T^{*}T$ with the eigenvalue $\lambda_{-\frac{i_{0}}{2}}$
and $\mathcal{A}_{n,k}=\mbox{span}\\{v^{(k)}_{n},v^{(k)}_{-(n+i_{0})}\\}$ is
the eigenspace of $T^{*}T$ with the eigenvalue
$\lambda_{n}=\lambda_{-(n+i_{0})}$ for $n>-\frac{i_{0}}{2}$. For any non-zero
vector $f$ in $\mathcal{K}$, we have
$f=\sum_{k=1,2;n\geq-\frac{i_{0}}{2}}\alpha_{n,k}h_{n,k}$, where
$h_{n,k}\in\mathcal{A}_{n,k}$. Similarly, there exist $n\geq-\frac{i_{0}}{2}$
and $k=1$ or $2$ such that $\alpha_{n,k}h_{n,k}\neq 0$. If
$n=-\frac{i_{0}}{2}$, then $\mathcal{K}\cap H(-\frac{i_{0}}{2})\neq\emptyset$.
The proof of other situations is similar to Case 1.

To sum up, these ensure the existence of $n_{0}\in\mathbb{Z}$ such that
$\mathcal{K}\cap H(n_{0})\neq\emptyset$. This completes the proof of step 1.

Step 2: For any $n\in\mathbb{Z}$, $\mathcal{K}\cap H(n)\neq\emptyset$.

From Step 1, we know that $\mathcal{K}\cap H(n_{0})\neq\emptyset$ for some
$n_{0}\in\mathbb{Z}$. Since $T_{n}$ is invertible for any $n\in\mathbb{Z}$, if
$g\in\mathcal{H}$ is a non-zero function, so is $Tg$. By using $\mathcal{K}$
is a reduced subspace and applying $T^{m}$ and $T^{*m}$ to
$H(n_{0})\cap\mathcal{K}$ for all $m\in\mathbb{N}$, we have that
$\mathcal{K}\cap H(n_{0}\pm m)\neq\emptyset$. By the arbitrariness of $m$, we
have completed the proof of this step.

Step 3: $\mathcal{K}=\mathcal{H}$.

Let $h_{n}\in\mathcal{K}\cap H(n)$ and $h_{n}\neq 0$. Then there are $\alpha$
and $\beta$ that are not all zero, as well as $\gamma$ and $\delta$ that make
$h_{n}=\alpha v^{(1)}_{n}+\beta v^{(2)}_{n}=\gamma u^{(1)}_{n}+\delta
u^{(2)}_{n}$.

Claim: At least one of $\alpha\beta$ and $\gamma\delta$ is not zero.

Otherwise, if both $\alpha\beta$ and $\gamma\delta$ are zero, then there is
only one of $\alpha$ and $\beta$ is zero, and $\gamma$ and $\delta$ are also.
That is to say, there is a multiple relationship between
$\\{v^{(1)}_{n},v^{(2)}_{n}\\}$ and $\\{u^{(1)}_{n},u^{(2)}_{n}\\}$. Without
loss of generality, we suppose that there exists non-zero $c\in\mathbb{C}$
such that $u^{(1)}_{n}=cv^{(1)}_{n}$. Then we can find
$c_{1},c_{2}\in\mathbb{C}$ that are not all zero and satisfy
$u^{(2)}_{n}=c_{1}v^{(1)}_{n}+c_{2}v^{(2)}_{n}$. Taking the inner product of
$v^{(1)}_{n}$ with $u^{(2)}_{n}$, we obtain $\langle
u^{(2)}_{n},v^{(1)}_{n}\rangle=\langle
c_{1}v^{(1)}_{n}+c_{2}v^{(2)}_{n},v^{(1)}_{n}\rangle=c_{1}\|v^{(1)}_{n}\|^{2}.$
On the other hand, $\langle u^{(2)}_{n},v^{(1)}_{n}\rangle=\langle
u^{(2)}_{n},\frac{1}{c}u^{(1)}_{n}\rangle=0$. So $c_{1}=0$ and
$u^{(2)}_{n}=c_{2}v^{(2)}_{n}$. In this case, we infer that
$(A_{n}B_{n}-B_{n}A_{n})u^{(1)}_{n}=c\lambda^{(1)}_{n-1}A_{n}v^{(1)}_{n}-c\lambda^{(1)}_{n}B_{n}v^{(1)}_{n}=c\lambda^{(1)}_{n-1}\lambda^{(1)}_{n}v^{(1)}_{n}-\lambda^{(1)}_{n}B_{n}u^{(1)}_{n}=0.$
Similarly, after replacing $u^{(1)}_{n}$ with $u^{(2)}_{n}$, $v^{(1)}_{n}$ and
$v^{(2)}_{n}$ are all zero, respectively. Thus $A_{n}B_{n}-B_{n}A_{n}=0$.

By a routine computation, we obtain

$\displaystyle\quad\,\ A_{n}B_{n}=(B_{n}A_{n})^{*}$
$\displaystyle=\left(\begin{smallmatrix}t_{n}^{2}t_{n-1}^{2}+|\alpha|^{2}\big{(}t_{n}^{2}(\frac{1}{t_{n-1}}-t_{n-1})^{2}+(1-t_{n}^{2})(\frac{1}{t_{n-1}^{2}}-1)\big{)}&\alpha\big{(}\frac{1}{t_{n-1}^{2}}-t_{n}^{2}\big{)}\\\
\bar{\alpha}(1+|\alpha|^{2})\big{(}t_{n-1}^{2}-t_{n}^{2}t_{n-1}^{2}+\frac{1}{t_{n}^{2}t_{n-1}^{2}}-\frac{1}{t_{n}^{2}}\big{)}+\bar{\alpha}|\alpha|^{2}\big{(}t_{n}^{2}-\frac{1}{t_{n-1}^{2}}\big{)}&|\alpha|^{2}\big{(}(\frac{1}{t_{n-1}^{2}}-1)(1-t_{n}^{2})+\frac{1}{t_{n-1}^{2}}(\frac{1}{t_{n}}-t_{n})^{2}\big{)}+\frac{1}{t_{n}^{2}t_{n-1}^{2}}\\\
\end{smallmatrix}\right),$

then the (1,2)-entry and (2,1)-entry in $A_{n}B_{n}-B_{n}A_{n}$ are
$\bar{\alpha}(1+|\alpha|^{2})\big{(}(t^{2}_{n}-1)(t^{2}_{n-1}-1)-(\frac{1}{t^{2}_{n}}-1)(\frac{1}{t^{2}_{n-1}}-1)\big{)}$,
and other entries are zero. Since $\alpha\neq 0$, $A_{n}B_{n}-B_{n}A_{n}=0$ if
and only if
$(t^{2}_{n}-1)(t^{2}_{n-1}-1)=(\frac{1}{t^{2}_{n}}-1)(\frac{1}{t^{2}_{n-1}}-1)=\frac{1}{t^{2}_{n}t^{2}_{n-1}}(1-t^{2}_{n})(1-t^{2}_{n-1})$.
Based on condition (2), this is impossible. So our assumption is invalid. This
implies that the Claim is checked.

If $\alpha\beta\neq 0$, we can obtain $v^{(1)}_{n}\in\mathcal{K}\cap H(n)$ by
restricting $h_{n}$ to the eigensubspace corresponding to the eigenvalue
$\lambda^{(1)}_{n}$ of $T^{*}T$. In the same way, we can also obtain
$v^{(2)}_{n}\in\mathcal{K}\cap H(n)$. So
$v^{(1)}_{n},v^{(2)}_{n}\in\mathcal{K}$. Similarly, $\gamma\delta\neq 0$ means
that $u^{(1)}_{n},u^{(2)}_{n}\in\mathcal{K}$. Since both
$\\{v^{(1)}_{n},v^{(2)}_{n}\\}$ and $\\{u^{(1)}_{n},u^{(2)}_{n}\\}$ are
orthonormal bases of $H(n)$, we conclude $H(n)\subset\mathcal{K}$. It follows
from Step 2 that $\mathcal{K}=spanH(n)=\mathcal{H}$.

Hence, $T$ is irreducible. ∎

######  Remark 3.2.

Let $T=T(t_{n},\alpha)$ and $\tilde{T}=T(\tilde{t}_{n},\tilde{\alpha})$.
Suppose that $T,\tilde{T}$ satisfy the conditions of Proposition 3.1, that is,
they satisfy $\lambda_{n}=\lambda_{-(n+i_{0})}$ and
$\tilde{\lambda}_{n}=\tilde{\lambda}_{-(n+\tilde{i}_{0})}$, respectively.
Recall that if $i_{0}$ is odd, then the multiplicity of the eigenvalues is
two. If $i_{0}$ is an even number, then the multiplicity of the eigenvalue
$\lambda_{-\frac{i_{0}}{2}}$ is one and the multiplicity of other eigenvalues
is two. These also apply to $\tilde{\lambda}_{n}$. We know that
$T\sim_{s}\tilde{T}$ if and only if singular values and their multiplicities
of $T$ and $\tilde{T}$ are the same. For $i_{0}$ and $\tilde{i}_{0}$, if one
of them is odd and the other is even, $T$ and $\tilde{T}$ must not be similar.
In the following, we will only consider when they are odd numbers.

Then the necessary and sufficient condition of $T\sim_{s}\tilde{T}$ is
$\left\\{\lambda_{n},\frac{1}{\lambda_{n}}\right\\}_{n\geq\frac{1-i_{0}}{2}}=\left\\{\tilde{\lambda}_{n},\frac{1}{\tilde{\lambda}_{n}}\right\\}_{n\geq\frac{1-\tilde{i}_{0}}{2}}$.
This is equivalent to
$\\{\lambda_{n}\\}_{n\geq\frac{1-i_{0}}{2}}=\\{\tilde{\lambda}_{n}\\}_{n\geq\frac{1-\tilde{i}_{0}}{2}}$
or
$\left\\{\lambda^{2}_{n}+\frac{1}{\lambda^{2}_{n}}\right\\}_{n\geq\frac{1-i_{0}}{2}}=\left\\{\tilde{\lambda}^{2}_{n}+\frac{1}{\tilde{\lambda}^{2}_{n}}\right\\}_{n\geq\frac{1-\tilde{i}_{0}}{2}}$.
That is,
$\left\\{(|\alpha|^{2}+1)\left(t_{n}^{2}+\frac{1}{t^{2}_{n}}\right)-2|\alpha|^{2}\right\\}_{n\geq\frac{1-i_{0}}{2}}=\left\\{(|\tilde{\alpha}|^{2}+1)\left(\tilde{t}_{n}^{2}+\frac{1}{\tilde{t}^{2}_{n}}\right)-2|\tilde{\alpha}|^{2}\right\\}_{n\geq\frac{1-\tilde{i}_{0}}{2}}$.
The set composed of singular values is the completely similar invariant of
operator $T(t_{n},\alpha)$. Moreover, when $|\alpha|=|\tilde{\alpha}|$,
$T\sim_{s}\tilde{T}$ if and only if
$\\{t_{n}^{2}+\frac{1}{t^{2}_{n}}\\}_{n\geq\frac{1-i_{0}}{2}}=\\{\tilde{t}^{2}_{n}+\frac{1}{\tilde{t}^{2}_{n}}\\}_{n\geq\frac{1-\tilde{i}_{0}}{2}}$.

For Proposition 3.1, we give an example.

######  Example 3.3.

Given two non-zero sequences $\\{x_{n}\\}_{n\in\mathbb{Z}}$ and
$\\{y_{n}\\}_{n\in\mathbb{Z}}$. Let $T=T(\sqrt{|\frac{x_{n}}{y_{n}}|},\alpha)$
for nonzero $\alpha\in\mathbb{C}$. Suppose that $x_{n}y_{n-1}\neq
x_{n-1}y_{n}$ and $x_{n}\neq y_{n}$. If $x_{n}=-y_{-(n+1)}$ for all
$n\in\mathbb{Z}$, then $T$ is irreducible.

###### Proof.

Since $x_{n}=-y_{-(n+1)}$, then $x_{-(n+1)}=-y_{-(-n-1+1)}=-y_{n}$,
$y_{n}=-x_{-(n+1)}$. We notice that
$\frac{x_{n}}{y_{n}}=\frac{-y_{-(n+1)}}{-x_{-(n+1)}}=\frac{y_{-(n+1)}}{x_{-(n+1)}}$.
Then
$\frac{x_{n}}{y_{n}}+\frac{y_{n}}{x_{n}}=\frac{x_{-(n+1)}}{y_{-(n+1)}}+\frac{y_{-(n+1)}}{x_{-(n+1)}}$.

Let $t^{2}_{n}=|\frac{x_{n}}{y_{n}}|$. Then we need to show that $t_{n}$
satisfies the conditions of Proposition 3.1. By a simple calculation, it
follows that

$t^{2}_{n}+\frac{1}{t^{2}_{n}}=\left|\frac{x_{n}}{y_{n}}\right|+\left|\frac{y_{n}}{x_{n}}\right|=\left|\frac{x_{-(n+1)}}{y_{-(n+1)}}\right|+\left|\frac{y_{-(n+1)}}{x_{-(n+1)}}\right|=t^{2}_{-(n+1)}+\frac{1}{t^{2}_{-(n+1)}}.$

By assumptions of sequences $\\{x_{n}\\}_{n\in\mathbb{Z}}$ and
$\\{y_{n}\\}_{n\in\mathbb{Z}}$, we have $t_{n}\neq t_{n-1}$ and $t_{n}\neq 1$.
This implies that $T$ is irreducible. ∎

When $x_{n}=n+a,y_{n}=n+b$ and $0<a\neq b<1$, $a+b=1$, the above example
corresponds to A. Kor$\acute{a}$nyi’s Theorem 2.2 in [31]. Our results
formally generalize his example from the perspective of irreducibility, but
the operators we discuss do not have homogeneity, even weak homogeneity.

In the following, we consider the case where the weight sequence of the
operator is complex, and we also make some generalization of the operator
structure.

######  Theorem 3.4.

Let $T=T(t_{n},d_{n})$. Suppose that there exists a bijective mapping $g$ on
$\mathbb{Z}$ and $N\in\mathbb{Z}$ such that when $n>N$, the sets of singular
values of $A_{n}$ are distinct from each other. If
$trace\,A_{n}=trace\,A_{g(n)}$ and one of the following two conditions is
true,

  1. (1)

$\frac{\overline{t_{n}}}{t_{n-1}}c_{n}\overline{c_{n-1}}\notin\mathbb{R}$,
where $c_{n}=\frac{d_{n+1}}{t_{n}}-d_{n}t_{n}$;

  2. (2)

$t_{n}\overline{c_{n}}(|t_{n-1}|^{2}+\frac{1}{|t_{n-1}|^{2}}+|c_{n-1}|^{2})\neq\frac{1}{t_{n-1}}\overline{c_{n-1}}(|t_{n}|^{2}-\frac{1}{|t_{n}|^{2}}-|c_{n}|^{2})$,

then T is irreducible.

###### Proof.

Since
$T_{n}=\left(\begin{smallmatrix}t_{n}&\frac{d_{n+1}}{t_{n}}-d_{n}t_{n}\\\
0&\frac{1}{t_{n}}\\\ \end{smallmatrix}\right)$,

$\displaystyle A_{n}$
$\displaystyle=T^{*}_{n}T_{n}=\left(\begin{smallmatrix}|t_{n}|^{2}&\frac{d_{n+1}\overline{t_{n}}}{t_{n}}-d_{n}|t_{n}|^{2}\\\
\frac{\overline{d_{n+1}}t_{n}}{\overline{t_{n}}}-\overline{d_{n}}|t_{n}|^{2}&|\frac{d_{n+1}}{t_{n}}-d_{n}t_{n}|^{2}+\frac{1}{|t_{n}|^{2}}\\\
\end{smallmatrix}\right),$ $\displaystyle B_{n}$
$\displaystyle=T_{n-1}T^{*}_{n-1}=\left(\begin{smallmatrix}|t_{n-1}|^{2}+|\frac{d_{n}}{t_{n-1}}-d_{n-1}t_{n-1}|^{2}&\frac{d_{n}}{|t_{n-1}|^{2}}-\frac{d_{n-1}t_{n-1}}{\overline{t_{n-1}}}\\\
\frac{\overline{d_{n}}}{|t_{n-1}|^{2}}-\frac{\overline{d_{n-1}}\overline{t_{n-1}}}{t_{n-1}}&\frac{1}{|t_{n-1}|^{2}}\\\
\end{smallmatrix}\right).$

We have $|A_{n}|=1$ and $A_{n}\neq I$. Then we assume the eigenvalues of
$A_{n}$ are $\lambda^{2}_{n},\frac{1}{\lambda^{2}_{n}}$, where
$\lambda_{n}>1$. That means
$\mbox{trace}A_{n}=|t_{n}|^{2}+\frac{1}{|t_{n}|^{2}}+|\frac{d_{n+1}}{t_{n}}-d_{n}t_{n}|^{2}=\lambda^{2}_{n}+\frac{1}{\lambda^{2}_{n}}.$
From the conditions, we have
$\lambda^{2}_{n}+\frac{1}{\lambda^{2}_{n}}=\lambda^{2}_{g(n)}+\frac{1}{\lambda^{2}_{g(n)}}$,
equivalently, $\lambda_{n}=\lambda_{g(n)}$.

Similar to the proof of Proposition 3.1, we only need to verify that
$A_{n}B_{n}\neq B_{n}A_{n}$ to prove that $T$ is an irreducible operator.

Let $A_{n}B_{n}=\left(\begin{smallmatrix}X^{(n)}_{11}&X^{(n)}_{12}\\\
X^{(n)}_{21}&X^{(n)}_{22}\\\ \end{smallmatrix}\right)$. By a routine
computation, we obtain

$X^{(n)}_{11}=|t_{n}|^{2}|t_{n_{1}}|^{2}+|t_{n}|^{2}\left|\frac{d_{n}}{t_{n-1}}-d_{n-1}t_{n-1}\right|^{2}+\frac{\overline{t_{n}}}{t_{n-1}}\left(\frac{d_{n+1}}{t_{n}}-d_{nt_{n}}\right)\left(\frac{\overline{d_{n}}}{\overline{t_{n-1}}}-\overline{d_{n-1}}\overline{t_{n-1}}\right),$
$X^{(n)}_{12}=\frac{|t_{n}|^{2}}{\overline{t_{n-1}}}\left(\frac{d_{n}}{t_{n-1}}-d_{n-1}t_{n-1}\right)+\frac{\overline{t_{n}}}{|t_{n-1}|^{2}}\left(\frac{d_{n+1}}{t_{n}}-d_{n}t_{n}\right),$
$X^{(n)}_{21}=\left(\frac{\overline{d_{n+1}}t_{n}}{\overline{t_{n}}}-\overline{d_{n}}|t_{n}|^{2}\right)\left(|t_{n-1}|^{2}+\left|\frac{d_{n}}{t_{n-1}}-d_{n-1}t_{n-1}\right|^{2}\right)+\left(\left|\frac{d_{n+1}}{t_{n}}-d_{n}t_{n}\right|^{2}+\frac{1}{|t_{n}|^{2}}\right)\left(\frac{\overline{d_{n}}}{|t_{n-1}|^{2}}-\frac{\overline{d_{n-1}}\overline{t_{n-1}}}{t_{n-1}}\right).$

Then for the $(1,1)$-entry of the matrix
$A_{n}B_{n}-B_{n}A_{n}=\left(\begin{smallmatrix}X^{(n)}_{11}-\overline{X^{(n)}_{11}}&X^{(n)}_{12}-\overline{X^{(n)}_{21}}\\\
X^{(n)}_{21}-\overline{X^{(n)}_{12}}&X^{(n)}_{22}-\overline{X^{(n)}_{22}}\\\
\end{smallmatrix}\right)$, we obtain

$X^{(n)}_{11}-\overline{X^{(n)}_{11}}=\frac{\overline{t_{n}}}{t_{n-1}}(\frac{d_{n+1}}{t_{n}}-d_{n}t_{n})(\frac{\overline{d_{n}}}{\overline{t_{n-1}}}-\overline{d_{n-1}}\overline{t_{n-1}})-\frac{t_{n}}{\overline{t_{n-1}}}(\frac{\overline{d_{n+1}}}{\overline{t_{n}}}-\overline{d_{n}}\overline{t_{n}})(\frac{d_{n}}{t_{n-1}}-d_{n-1}t_{n-1}).$

Let $c_{n}=\frac{d_{n+1}}{t_{n}}-d_{n}t_{n}$, then
$X^{(n)}_{11}-\overline{X^{(n)}_{11}}=\frac{\overline{t_{n}}}{t_{n-1}}c_{n}\overline{c_{n-1}}-\frac{t_{n}}{\overline{t_{n-1}}}\overline{c_{n}}c_{n-1}$.
Since
$\overline{\frac{\overline{t_{n}}}{t_{n-1}}c_{n}\overline{c_{n-1}}}=\frac{t_{n}}{\overline{t_{n-1}}}\overline{c_{n}}c_{n-1}$,
the difference between the two must be pure imaginary. It can be seen from the
condition that $X^{(n)}_{11}-\overline{X^{(n)}_{11}}\neq 0$. So it is not zero
at the (1,1)-entry of $A_{n}B_{n}-B_{n}A_{n}$. Thus we conclude
$A_{n}B_{n}-B_{n}A_{n}\neq 0$.

Similarly, for the position $(2,1)$-entry of $A_{n}B_{n}-B_{n}A_{n}$, we have

$X^{(n)}_{21}-\overline{X^{(n)}_{12}}=t_{n}\overline{c_{n}}(|t_{n-1}|^{2}-\frac{1}{|t_{n-1}|^{2}}+|c_{n_{1}}|^{2})-\frac{1}{t_{n-1}}\overline{c_{n-1}}(|t_{n}|^{2}-\frac{1}{|t_{n}|^{2}}-|c_{n}|^{2}),c_{n}=\frac{d_{n+1}}{t_{n}}-d_{n}t_{n}.$

It can be seen from the condition that it is not zero at the (2,1)-entry of
$A_{n}B_{n}-B_{n}A_{n}$. Thus we conclude $A_{n}B_{n}-B_{n}A_{n}\neq 0$. This
completes the proof. ∎

In Theorem 3.4, there are many choices for the bijection $g$ on $\mathbb{Z}$.
We can define $g$ as $g(n)=-(n+i_{0})$ for some $i_{0}\in\mathbb{Z}$ or the
identity mapping. It is necessary to exists $N\in\mathbb{Z}$ such that when
$n>N$, the sets of singular values of $A_{n}$ are distinct from each other. In
the following, we provide two examples.

######  Example 3.5.

Let $T=T(t_{n},1)$, where $t_{n}=\frac{n+\frac{1}{2}+s}{n+\frac{1}{2}-s}$ and
$s$ is a non zero pure imaginary number. Then $T$ is irreducible.

###### Proof.

Since
$|t_{n}|^{2}=t_{n}\cdot\overline{t_{n}}=\frac{n+\frac{1}{2}+s}{n+\frac{1}{2}-s}\cdot\frac{n+\frac{1}{2}-s}{n+\frac{1}{2}+s}=1$,
we have

$A_{n}=T^{*}_{n}T_{n}=\left(\begin{smallmatrix}1&\frac{2(2n+1)s}{(n+\frac{1}{2}+s)^{2}}\\\
\frac{-2(2n+1)s}{(n+\frac{1}{2}-s)^{2}}&1+\big{|}\frac{2(2n+1)s}{(n+\frac{1}{2})^{2}-s^{2}}\big{|}^{2}\\\
\end{smallmatrix}\right),B_{n}=T_{n-1}T^{*}_{n-1}=\left(\begin{smallmatrix}1+\big{|}\frac{2(2n-1)s}{(n-\frac{1}{2})^{2}-s^{2}}\big{|}^{2}&\frac{2(2n-1)s}{(n-\frac{1}{2}-s)^{2}}\\\
\frac{-2(2n-1)s}{(n-\frac{1}{2}+s)^{2}}&1\\\ \end{smallmatrix}\right).$

Since $A_{n}\neq I$ and the determinant of $A_{n}$ is one for all
$n\in\mathbb{Z}$. Let the eigenvalues of $A_{n}$ be $\lambda_{n}^{2}$ and
$\frac{1}{\lambda_{n}^{2}}$ with $\lambda_{n}>1$. Note that
$|\frac{2(2n+1)s}{(n+\frac{1}{2})^{2}-s^{2}}|^{2}=|\frac{2(2(-n-1)+1)s}{(-n-1+\frac{1}{2})^{2}-s^{2}}|^{2}$.
Then
$\mbox{trace}A_{n}=\lambda^{2}_{n}+\frac{1}{\lambda^{2}_{n}}=\lambda^{2}_{-(n+1)}+\frac{1}{\lambda^{2}_{-(n+1)}}=\mbox{trace}A_{-(n+1)}$,
therefore, $\lambda_{n}=\lambda_{-(n+1)}$.

Let $A_{n}B_{n}=\left(\begin{smallmatrix}X^{(n)}_{11}&X^{(n)}_{12}\\\
X^{(n)}_{21}&X^{(n)}_{22}\\\ \end{smallmatrix}\right)$. By a routine
computation, we obtain

$X_{11}^{(n)}=1+\left|\frac{2(2n-1)s}{(n-\frac{1}{2})^{2}-s^{2}}\right|^{2}-\frac{4(2n+1)(2n-1)s^{2}}{(n+s+\frac{1}{2})^{2}(n+s-\frac{1}{2})^{2}}.$

So for the matrix
$A_{n}B_{n}-B_{n}A_{n}=\left(\begin{smallmatrix}X^{(n)}_{11}-\overline{X^{(n)}_{11}}&X^{(n)}_{12}-\overline{X^{(n)}_{21}}\\\
X^{(n)}_{21}-\overline{X^{(n)}_{12}}&X^{(n)}_{22}-\overline{X^{(n)}_{22}}\\\
\end{smallmatrix}\right),$ we have

$X^{(n)}_{11}-\overline{X^{(n)}_{11}}=4(4n^{2}-1)s^{2}\left[\frac{1}{(n-s+\frac{1}{2})^{2}(n-s-\frac{1}{2})^{2}}-\frac{1}{(n+s+\frac{1}{2})^{2}(n+s-\frac{1}{2})^{2}}\right].$

It is easy to see that
$\frac{1}{(n-s+\frac{1}{2})^{2}(n-s-\frac{1}{2})^{2}}\notin\mathbb{R}$. So it
is not zero at the (1,1)-entry of $A_{n}B_{n}-B_{n}A_{n}$. That is
$A_{n}B_{n}-B_{n}A_{n}\neq 0$. Similar to the proof of Proposition 3.1 and
Theorem 3.4, we deduce that $T$ is irreducible. ∎

######  Example 3.6.

Let $T=T(\sqrt{\frac{n+a}{n+b}},d_{n})$, where $d_{n}>0$, $0<a<b<1$ and
$a+b=1$. If $d_{n}=d_{-n}$ and
$\sqrt{\frac{n+a}{n+b}}c_{n}[(1+d^{2}_{n})\frac{n-a}{n-b}+(1+d^{2}_{n-1})\frac{n-b}{n-a}-2d_{n-1}d_{n}]\neq\sqrt{\frac{n-b}{n-a}}c_{n-1}[(1-d^{2}_{n})\frac{n+a}{n+b}-(1+d^{2}_{n+1})\frac{n+b}{n+a}+2d_{n}d_{n+1}]$
for $c_{n}=\sqrt{\frac{n+1-a}{n+1-b}}d_{n+1}-\sqrt{\frac{n+a}{n+b}}d_{n}$,
then $T$ is irreducible.

###### Proof.

Since $A_{n}\neq I$ and the determinant of $A_{n}$ is one for all
$n\in\mathbb{Z}$. Let the eigenvalues of $A_{n}$ be $\lambda_{n}^{2}$ and
$\frac{1}{\lambda_{n}^{2}}$ with $\lambda_{n}>1$. We know that
$\lambda_{n}=\lambda_{m}$ means that
$\lambda_{n}^{2}+\frac{1}{\lambda_{n}^{2}}=\lambda_{m}^{2}+\frac{1}{\lambda_{m}^{2}}$.
Based on the condition, we know that $d_{n}=d_{-n}$ is equivalent to
$\lambda_{n}=\lambda_{-(n+1)}$, since
$t^{2}_{n}=\frac{1}{t^{2}_{-(n+1)}}=\frac{n+a}{n+b}$. By using the condition
(2) of Theorem 3.4 and
$c_{n}=\sqrt{\frac{n+1-a}{n+1-b}}d_{n+1}-\sqrt{\frac{n+a}{n+b}}d_{n}=\frac{d_{n+1}}{t_{n}}-d_{n}t_{n}$,
we infer that $T$ is an irreducible operator. ∎

Because the scalar times identity operator is commutative with any operator,
and in the case we are considering, some commutativity of the operator no
longer naturally holds. Therefore, we need additional conditions. The objects
of our study have a more complicated structure, which provides examples of
other irreducible operators.

###  3.2. The irreducibility of $T(w_{n},v_{n},d_{n})$

######  Theorem 3.7.

Let $T=T(w_{n},v_{n},d_{n})$ with $|w_{n}|=|v_{n}|=1$. Suppose that there
exists a bijective mapping $g$ on $\mathbb{Z}$ and $N\in\mathbb{Z}$ such that
the sets of singular values of $A_{n}$ are distinct from each other. If
$|d_{n+1}v_{n}-d_{n}w_{n}|^{2}=|d_{g(n)+1}v_{g(n)}-d_{g(n)}w_{g(n)}|^{2}$ and
one of the following two conditions is true,

  1. (1)

$d_{n+1}c_{n}+d_{n}c_{n-1}-d_{n+1}\overline{w_{n}}v_{n}c_{n-1}\notin\mathbb{R}$,
where $c_{n}=\overline{d_{n}}\overline{w_{n}}v_{n}$;

  2. (2)

$(\overline{d_{n+1}}w_{n}v_{n}-\overline{d_{n}})|d_{n}v_{n-1}-d_{n-1}w_{n-1}|^{2}\neq(\overline{d_{n-1}}\overline{w_{n-1}}v_{n-1}-\overline{d_{n}})|d_{n+1}v_{n}-d_{n}w_{n}|^{2}$,

then T is irreducible.

###### Proof.

Since $T_{n}=\left(\begin{smallmatrix}w_{n}&d_{n+1}v_{n}-d_{n}w_{n}\\\
0&v_{n}\\\ \end{smallmatrix}\right)$ and

$\displaystyle A_{n}$
$\displaystyle=\left(\begin{smallmatrix}1&\overline{w_{n}}d_{n+1}v_{n}-d_{n}\\\
\overline{d_{n+1}}\overline{v_{n}}w_{n}-\overline{d_{n}}&1+|d_{n+1}v_{n}-d_{n}w_{n}|^{2}\\\
\end{smallmatrix}\right),$ $\displaystyle B_{n}$
$\displaystyle=\left(\begin{smallmatrix}1+|d_{n}v_{n-1}-d_{n-1}w_{n-1}|^{2}&d_{n}-d_{n-1}w_{n-1}\overline{v_{n-1}}\\\
\overline{d_{n}}-\overline{d_{n-1}}\overline{w_{n-1}}v_{n-1}&1\\\
\end{smallmatrix}\right).$

A simple calculation can be obtained the determinant of $A_{n}$ as 1 and
$A_{n}\neq I$. Then assume the eigenvalues of $A_{n}$ are
$\lambda^{2}_{n},\frac{1}{\lambda^{2}_{n}}$, where $\lambda_{n}>1$. That means
$\mbox{trace}A_{n}=2+|d_{n+1}v_{n}-d_{n}w_{n}|^{2}=\lambda^{2}_{n}+\frac{1}{\lambda^{2}_{n}}$.
If
$|d_{n+1}v_{n}-d_{n}w_{n}|^{2}=|d_{g(n)+1}v_{g(n)}-d_{g_{(}n)}w_{g(n)}|^{2}$,
this is equivalent to
$\lambda^{2}_{n}+\frac{1}{\lambda^{2}_{n}}=\lambda^{2}_{g(n)}+\frac{1}{\lambda^{2}_{g(n)}}$.
Thus, $\lambda_{n}=\lambda_{g(n)}$.

By a routine calculation, we have the $(1,1)$-entry and $(2,1)$-entry of
$A_{n}B_{n}-B_{n}A_{n}$ are
$d_{n+1}\overline{d_{n}}\overline{w_{n}}v_{n}+d_{n}\overline{d_{n-1}}\overline{w_{n-1}}v_{n-1}-d_{n+1}\overline{d_{n-1}}\overline{w_{n}}\overline{w_{n-1}}v_{n}v_{n-1}-\overline{d_{n+1}}d_{n}w_{n}\overline{v_{n}}-\overline{d_{n}}d_{n-1}w_{n-1}\overline{v_{n-1}}+\overline{d_{n+1}}d_{n-1}w_{n}w_{n-1}\overline{v_{n}}\overline{v_{n-1}}$
and
$(\overline{d_{n+1}}w_{n}v_{n}-\overline{d_{n}})|d_{n}v_{n-1}-d_{n-1}w_{n-1}|^{2}-(\overline{d_{n-1}}\overline{w_{n-1}}v_{n-1}-\overline{d_{n}})|d_{n+1}v_{n}-d_{n}w_{n}|^{2}$.
It can be seen that the first three terms and the last three terms are
conjugate to each other at the (1,1)-entry. From condition (1), we know that
$d_{n+1}c_{n}+d_{n}c_{n-1}-d_{n+1}\overline{w_{n}}v_{n}c_{n-1}\notin\mathbb{R}$,
that is to say, the $(1,1)$-entry of $A_{n}B_{n}-B_{n}A_{n}$ is not zero. Thus
we conclude that $A_{n}B_{n}-B_{n}A_{n}\neq 0$. In the same way, from
condition (2), we see that $(2,1)$-entry isn’t zero. Similar to the proof of
Proposition 3.1, we deduce that $T$ is irreducible. ∎

######  Corollary 3.8.

Let $T=T(w_{n},v_{n},\alpha)$. If $|w_{n}-v_{n}|^{2}=|w_{m}-v_{m}|^{2}$ only
holds when $m=-(n+i_{0})$ for some $i_{0}\in\mathbb{Z}$ and one of the
following two conditions is true,

  1. (1)

$w_{n}w_{n-1}\overline{c_{n}}\overline{c_{n-1}}\notin\mathbb{R}$, where
$c_{n}=w_{n}-v_{n}$;

  2. (2)

$(1-w_{n}\overline{v_{n}})|w_{n-1}-v_{n-1}|^{2}\neq(1-\overline{w_{n-1}}v_{n-1})|w_{n}-v_{n}|^{2}$,

then $T$ is irreducible.

In the following, we will use another method to provide a sufficient condition
for $T(w_{n},v_{n},d_{n})$ to be an irreducible operator. Before that, let us
introduce an equivalence theorem on shift operators proved by A.L. Shields in
[40]. Although this result is well-known, we record it for overall
completeness and later use.

######  Theorem 3.9.

[40] Let $S$ and $T$ be bilateral weighted shifts with weight sequences
$\\{v_{n}\\}_{n\in\mathbb{Z}}$, $\\{w_{n}\\}_{n\in\mathbb{Z}}$ and
$w_{n},v_{n}\neq 0$ for all $n\in\mathbb{Z}$. Then $S\sim_{s}T$ if and only if
there exist $k\in\mathbb{Z}$ and $c_{1},c_{2}$ such that
$0<c_{1}\leq|\frac{w_{k+m}\cdots w_{k+n}}{v_{m}\cdots v_{n}}|\leq c_{2}$ for
all $m\leq n$. Specially, if $S$ and $T$ be unilateral weighted shifts with no
zero weights, then $S\sim_{s}T$ if and only if the formula above holds with
$k=m=0$ for all $n\geq 0$.

If we consider the case where $S\sim_{u}T$, then the two equivalent conditions
of the previous theorem are replaced by $|v_{n}|=|w_{n+k}|$ and
$|v_{n}|=|w_{n}|$. This means that the shift operator with weight sequence
$\\{w_{n}\\}_{n\in I}$ and the shift operator with weight sequence
$\\{|w_{n}|\\}_{n\in I}$ are always unitarily equivalent, $I=\mathbb{Z}^{+}$
or $\mathbb{Z}$. Recall that by a theorem of R.L. Kelly and N.K. Nikolskii in
[19, 39], which gives a characterization that a bilateral weighted shift is
reducible.

######  Lemma 3.10.

[19, 39] If $T$ is a bilateral weighted shift with strictly positive weights
$w_{n}$, $n\in\mathbb{Z}$, then a necessary and sufficient condition that $T$
be reducible is that the sequence $\\{w_{n}\\}_{n\in\mathbb{Z}}$ be periodic.

######  Proposition 3.11.

Let $T=T(w_{n},v_{n},d_{n})$. If both $T_{0}$ and $T_{1}$ are irreducible and
$\lim\limits_{k\to\infty}\prod\limits^{k}_{m=0}\frac{|w_{j+m}|}{|v_{i+m}|}=0$
for any $i,j\in\mathbb{Z}$, then $T$ is irreducible.

###### Proof.

To prove that $T$ is irreducible, we just need to verify that for any
$P\in\mathcal{A}^{\prime}(T)$ and $P=P^{*}=P^{2}$, if $T$ satisfies the
condition in Theorem, then $P=0$ or $I$.

Let $P\in\mathcal{A}^{\prime}(T)$ and $P$ under decomposition
$l^{2}(\mathbb{Z})\oplus l^{2}(\mathbb{Z})$ be
$\left(\begin{smallmatrix}P_{00}&P_{01}\\\ P_{10}&P_{11}\\\
\end{smallmatrix}\right)$. Then
$P_{00}=P^{*}_{00},P_{01}=P^{*}_{10},P_{11}=P^{*}_{11}$. Since
$P\in\mathcal{A}^{\prime}(T)$, we have

(3.1)
$\left(\begin{smallmatrix}P_{00}T_{0}&P_{00}(XT_{1}-T_{0}X)+P_{01}T_{1}\\\
P_{10}T_{0}&P_{10}(XT_{1}-T_{0}X)+P_{11}T_{1}\\\
\end{smallmatrix}\right)=\left(\begin{smallmatrix}T_{0}P_{00}+(XT_{1}-T_{0}X)P_{10}&T_{0}P_{01}+(XT_{1}-T_{0}X)P_{11}\\\
T_{1}P_{10}&T_{1}P_{11}\\\ \end{smallmatrix}\right).$

For any $i,j\in\mathbb{Z}$ , let $a_{ij}=\langle
P_{00}e_{j},e_{i}\rangle,b_{ij}=\langle
P_{01}e_{j},e_{i}\rangle,c_{ij}=\langle
P_{10}e_{j},e_{i}\rangle,d_{ij}=\langle P_{11}e_{j},e_{i}\rangle$. Then the
matrix representations of $P_{00},P_{01},P_{10},P_{11}$ under
$\\{e_{n}\\}_{n\in\mathbb{Z}}$ are
$P_{00}=(a_{ij})_{i,j\in\mathbb{Z}},P_{01}=(b_{ij})_{i,j\in\mathbb{Z}},P_{10}=(c_{ij})_{i,j\in\mathbb{Z}},P_{11}=(d_{ij})_{i,j\in\mathbb{Z}}$.
It follows that $a_{ij}=\overline{a_{ji}}$, $d_{ij}=\overline{d_{ji}}$ and
$b_{ij}=\overline{c_{ji}}$.

For the (2,1)-entry of (3.1), we have $P_{10}T_{0}=T_{1}P_{10}$, that is,
$w_{j}c_{i+1,j+1}=v_{i}c_{i,j}\,\,\text{for\,\,}i,j\in\mathbb{Z}$. Now suppose
$c_{i,j}\neq 0$, because of the non-zero property of the weighted sequences
$\\{w_{i}\\}_{i\in\mathbb{Z}},\\{v_{i}\\}_{i\in\mathbb{Z}}$, we have
$c_{i+1,j+1}\neq 0$. Replacing $i,j$ by $i+1,j+1$, we have $c_{i+2,j+2}\neq
0$. Repeating this process, we get $c_{i+k,j+k}\neq 0,k\in\mathbb{Z}^{+}$.

We know $\|P\|=1$, then $\|P_{ij}\|\leq 1$ for $i,j=0,1$ and
$|c_{ij}|\leq\|P_{10}\|\leq 1.$ Note that the equation
$|c_{ij}|^{2}=(\prod\limits_{m=0}^{k}\frac{|w_{j+m}|^{2}}{|v_{i+m}|^{2}})|c_{i+k+1,j+k+1}|^{2}$
always holds for any $k\in\mathbb{Z}^{+}$. Since
$\lim\limits_{k\to\infty}\prod\limits^{k}_{m=0}\frac{|w_{j+m}|^{2}}{|v_{i+m}|^{2}}=0$
and $|c_{i+k+1,j+k+1}|^{2}\leq 1$, $c_{i,j}=0$, which is contradictory to the
assumption $c_{i,j}\neq 0$. So the assumption is not valid and then
$c_{i,j}=0$ for any $i,j\in\mathbb{Z}$. So $P_{10}=0,P_{01}=0$.

For the (1,1)-entry of (3.1), we have $P_{00}T_{0}=T_{0}P_{00}$, that is,
$w_{j}a_{i+1,j+1}=w_{i}a_{i,j},i,j\in\mathbb{Z}$. Letting $i=j$, we have
$a_{i,i}=a_{i+1,i+1}$, since the weight sequence is not zero. From the
arbitrariness of $i$, we can know that the elements on the diagonal of matrix
$P_{00}$ are always equal. Suppose $a_{i,j}\neq 0$, then $a_{i+k,j+k}\neq 0$
for $k\in\mathbb{Z}^{+}$ and
$\frac{w_{j}}{w_{i}}=\frac{a_{i,j}}{a_{i+1,j+1}}$. By taking conjugate, we
obtain $T^{*}_{0}P_{00}=P_{00}T^{*}_{0}$ and
$\frac{\overline{w_{i}}}{\overline{w_{j}}}=\frac{a_{i,j}}{a_{i+1,j+1}}$. This
implies that $|w_{i}|^{2}=|w_{j}|^{2}$. Similarly, we conclude
$|w_{i+k}|^{2}=|w_{j+k}|^{2}$ for any $k\in\mathbb{Z}^{+}$. When $i\neq j$,
there exists $s\neq 0,s\in\mathbb{Z}$ such that $i+s=j$. Then
$|w_{i}|^{2}=|w_{i+s}|^{2}=|w_{i+2s}|^{2}=\ldots=|w_{i+ns}|^{2}$ for
$n\in\mathbb{Z}^{+}$. We can see that the modulus of the sequence changes
periodically. Let $\tilde{T}_{0}$ be a weighted shift concerning the
orthonormal basis $\\{e_{n}\\}_{n\in\mathbb{Z}}$ and
$\tilde{T}_{0}e_{n}=|w_{n}|e_{n+1}$. So $\tilde{T}_{0}$ is a reducible
operator according to Lemma 3.10. Since $T_{0}\sim_{u}\tilde{T}_{0}$, $T_{0}$
is also a reducible operator, which is contradictory to the condition. So the
assumption is not valid, then $a_{i,j}=0$ for $i\neq j$. Then $P_{00}$ is a
diagonal operator whose diagonal elements are always equal. Because of the
self-adjoint of $P_{00}$, there exists $l_{1}\in\mathbb{R}$ such that
$P_{00}=l_{1}I$, where $I$ is the identity of $l^{2}(\mathbb{Z})$. Similarly,
for the (2,2)-entry of (3.1), we can also get $P_{11}=l_{2}I$, where
$l_{2}\in\mathbb{R}$.

For the (1,2)-entry of (3.1), $P_{00}(XT_{1}-T_{0}X)=(XT_{1}-T_{0}X)P_{11}$
and $XT_{1}-T_{0}X\neq 0$, we have $l_{1}=l_{2}$. Then $P=l_{1}I.$ Since
$P=P^{2}$, then $P=0$ or $I$, which implies that $T$ is irreducible. ∎

The following corollary is an immediate consequence of Proposition 3.11.

######  Corollary 3.12.

Let $T=T(t_{n},d_{n})$. If both $T_{0}$ and $T_{1}$ are irreducible and
$\lim\limits_{k\to\infty}\prod\limits^{k}_{m=0}|t_{i+m}|=0$ for any
$i\in\mathbb{Z}$, then $T$ is irreducible.

##  4\. Weakly homogeneity of a class of operators

In [31, 2], it has been proven that if $T_{0}$ and $T_{1}$ are homogeneous
operators and have the same unitary representation $U(g)$, then for any
$\alpha\in\mathbb{C}$, the operator
$\widetilde{T}=\left(\begin{smallmatrix}T_{0}&\alpha(T_{0}-T_{1})\\\
0&T_{1}\\\ \end{smallmatrix}\right)$ is homogeneous with the unitary
representation $U(g)\oplus U(g)$. In other words, the homogeneity of $T_{0}$
and $T_{1}$ can be completely transferred to $\widetilde{T}$ under its
construction. Here we give two examples to show that when the $2\times 2$
operator matrix does not satisfy this construction, it no longer remains
homogeneous but is weakly homogeneous. Let
$T=\left(\begin{smallmatrix}T_{0}&XT_{1}-T_{0}X\\\ 0&T_{1}\\\
\end{smallmatrix}\right)$, where $T_{0}$ and $T_{1}$ are homogeneous operators
and $X$ is a bounded linear operator. Let $S=\left(\begin{smallmatrix}I&-X\\\
0&I\\\ \end{smallmatrix}\right)$. It is easy to see that $S$ is invertible and
its inverse is $\left(\begin{smallmatrix}I&X\\\ 0&I\\\
\end{smallmatrix}\right)$. Thus, $T=S^{-1}(T_{0}\oplus T_{1})S$. Assume that
the unitary representation of $T_{0}$ and $T_{1}$ are $U(g)$ and $V(g)$
respectively, for any $g\in M\ddot{o}b$. Then we have
$g(T_{0})=U(g)^{-1}T_{0}U(g)$, $g(T_{1})=V(g)^{-1}T_{1}V(g)$ and

(4.1) $g(T_{0}\oplus T_{1})=g(T_{0})\oplus g(T_{1})=(U(g)\oplus
V(g))^{-1}(T_{0}\oplus T_{1})(U(g)\oplus V(g)).$

For a fixed but arbitrary $g\in M\ddot{o}b$, it follows from $g$ is analytic
and equation (4.1) that

(4.2) $\begin{array}[]{lll}g(T)&=&g(S^{-1}(T_{0}\oplus T_{1})S)\\\
&=&S^{-1}g(T_{0}\oplus T_{1})S\\\ &=&S^{-1}(U(g)\oplus V(g))^{-1}(T_{0}\oplus
T_{1})(U(g)\oplus V(g))S\\\ &=&S^{-1}(U(g)\oplus V(g))^{-1}STS^{-1}(U(g)\oplus
V(g))S.\end{array}$

Since $S^{-1}(U(g)\oplus V(g))^{-1}S$ must be invertible, $T$ must be a weakly
homogeneous operator. If you want to get that $T$ is homogeneous, you need
$S^{-1}(U(g)\oplus V(g))^{-1}S$ to be a unitary operator. Note that

(4.3) $S^{-1}(U(g)\oplus
V(g))^{-1}S=\left(\begin{smallmatrix}U(g)^{-1}&XV(g)^{-1}-U(g)^{-1}X\\\
0&V(g)^{-1}\\\ \end{smallmatrix}\right).$

A routine verification shows that $S^{-1}(U(g)\oplus V(g))^{-1}S$ is a unitary
operator if and only if $XV(g)^{-1}=U(g)^{-1}X$. Next, we will explain that
even in the following two special cases, $T$ cannot be identified with a
homogeneous. Let $T_{0}$ and $T_{1}$ have the same unitary representation
$U(g)$ and $X\neq\alpha I$ for any $g\in M\ddot{o}b$ and
$\alpha\in\mathbb{C}$. Since $X\in\mathcal{L}(\mathcal{H})$ is a bounded
linear operator arbitrarily, this means that apart from the scalar operator
$\alpha I$, we cannot be sure that the operator $X$ in other operator classes
satisfies $XU(g)^{-1}=U(g)^{-1}X$. When $T_{0}$ and $T_{1}$ have different
unitary representations $U(g)$ and $V(g)$ and $X=\alpha I$ for some non-zero
$\alpha\in\mathbb{C}$, then $\alpha V(g)^{-1}=\alpha U(g)^{-1}$ is equivalent
to $V(g)=U(g)$ for any $g\in M\ddot{o}b$.

Therefore, in the following we consider the weak homogeneity problem
associated with the operator
$T=\left(\begin{smallmatrix}T_{0}&XT_{1}-T_{0}X\\\ 0&T_{1}\\\
\end{smallmatrix}\right)$. We record the next fact as a Lemma for our later
use.

######  Lemma 4.1.

Let $T=\left(\begin{smallmatrix}T_{0}&XT_{1}-T_{0}X\\\ 0&T_{1}\\\
\end{smallmatrix}\right)$ for a bounded linear operator $X$. If $T_{0}$,
$T_{1}$ are weakly homogeneous operators, then so is $T$.

###  4.1. Strong irreducible decompositions and weak homogeneity

Let $T_{i}\in\mathcal{L}(\mathcal{H}_{i})$, $i=0,1$. Define
$\sigma_{T_{0},T_{1}}$ be the operator $\sigma_{T_{0},T_{1}}(X)=T_{0}X-XT_{1}$
for $X\in\mathcal{L}(\mathcal{H}_{1},\mathcal{H}_{0})$. Let
$\sigma_{T_{0}}:\mathcal{L}(\mathcal{H}_{0})\to\mathcal{L}(\mathcal{H}_{0})$
be the operator $\sigma_{T_{0},T_{0}}$. Then
$X\in\mbox{ker}\sigma_{T_{0},T_{1}}$ means that $T_{0}X=XT_{1}$. Recall
Definition 4.1 introduced in [21]. We say that
$T=\left(\begin{smallmatrix}T_{0}&XT_{1}-T_{0}X\\\ 0&T_{0}\\\
\end{smallmatrix}\right)$ belongs to the set
$N\mathcal{F}\mathcal{B}_{2n}(\Omega)$, if
$T_{0},T_{1}\in\mathcal{B}_{n}(\Omega)$ and
$X\notin\mbox{ker}\sigma_{T_{0},T_{1}}$. It is natural to ask: when does the
converse of Lemma 4.1 hold? There is an answer in [21]. If
$\left(\begin{smallmatrix}T_{0}&XT_{1}-T_{0}X\\\ 0&T_{0}\\\
\end{smallmatrix}\right)$ is weakly homogeneous for some bounded linear
operator $X$ and for any $\phi_{\alpha}\in M\ddot{o}b$,
$\mbox{ker}\sigma_{\phi_{\alpha}(T_{0}),T_{1}}=\\{0\\}$, then
$T_{0},\,\,T_{1}$ are both weakly homogeneous. By using the $K_{0}$-group of
the commutant algebra, we obtain the following results.

######  Theorem 4.2.

Let $T=\left(\begin{smallmatrix}T_{0}&XT_{0}-T_{0}X\\\ 0&T_{0}\\\
\end{smallmatrix}\right)\in N\mathcal{F}\mathcal{B}_{2n}(\mathbb{D})$, where
$T_{0}\in(SI)$. Then $T$ is weakly homogeneous if and only if $T_{0}$ is
weakly homogeneous.

###### Proof.

By Lemma 4.1, the sufficiency is valid. Let
$S=\left(\begin{smallmatrix}I&-X\\\ 0&I\\\ \end{smallmatrix}\right)$. Then
$T=S^{-1}(T_{0}\oplus T_{0})S$. We know that similar transformations do not
change the spectrum, then $\sigma(T)=\sigma(T_{0}\oplus
T_{0})=\sigma(T_{0})=\overline{\mathbb{D}}$, since the spectrum of a weakly
homogeneous operator is $\mathbb{T}$ or $\overline{\mathbb{D}}$ by using the
spectral mapping theorem (see [3]).

Since $T$ is weakly homogeneous, for any $g\in M\ddot{o}b$, there exists an
invertible operator $\tilde{S_{g}}$ such that
$g(T)=\tilde{S_{g}}^{-1}T\tilde{S_{g}}$. Note that $g(T)=S^{-1}(g(T_{0}\oplus
T_{0}))S=\tilde{S_{g}}^{-1}S^{-1}(T_{0}\oplus T_{0})S\tilde{S_{g}}$, then
$T_{0}\oplus T_{0}$ is weakly homogeneous since for any $g\in M\ddot{o}b$,
$S\tilde{S_{g}}^{-1}S^{-1}$ is invertible and
$g(T_{0})^{(2)}=(S\tilde{S_{g}}^{-1}S^{-1})T_{0}^{(2)}(S\tilde{S_{g}}S^{-1})$.
It implies that $(S\tilde{S_{g}}^{-1}S^{-1}\oplus I)(g(T_{0})^{(2)}\oplus
T_{0}^{(2)})(S\tilde{S_{g}}S^{-1}\oplus I)=T_{0}^{(4)}$ for all $g\in
M\ddot{o}b$. By Theorem 2.3 and $T_{0}\in(SI)$, we obtain

(4.4) $K_{0}(\mathcal{A}^{\prime}(g(T_{0})^{(2)}\oplus T^{(2)}_{0}))\cong
K_{0}(\mathcal{A}^{\prime}(T^{(4)}_{0}))\cong\mathbb{Z}.$

We claim that $g(T_{0})$ is also strongly irreducible. Assume that $g(T_{0})$
is not strongly irreducible. Since every Cowen-Douglas operator can be written
as the direct sum of finitely many strongly irreducible Cowen-Douglas
operators in [28].

Case 1: Without loss of generality, we suppose that $g(T_{0})$ has a finite
(SI) decomposition $g(T_{0})=T^{(n_{1})}_{1}\oplus
T^{(n_{2})}_{2}\oplus\cdots\oplus T^{(n_{l})}_{l}$ for some positive integer
$l>1$. Then $g(T_{0})^{(2)}$ has a finite (SI) decomposition
$g(T_{0})^{(2)}=T^{(2n_{1})}_{1}\oplus T^{(2n_{2})}_{2}\oplus\cdots\oplus
T^{(2n_{l})}_{l}$, $l>1$. By Theorem 2.3, we have
$K_{0}(\mathcal{A}^{\prime}(g(T_{0})^{(2)}\oplus
T^{(2)}_{0}))\cong\mathbb{Z}^{l}\,\,(l>1)$, which is contradictory to (4.4).
Therefore, the assumption is contradictory.

Case 2: Let $g(T_{0})$ have a finite (SI) decomposition
$g(T_{0})=T_{1}^{(n_{1})},n_{1}>1$, where
$T_{1}\in\mathcal{B}_{\frac{n}{n_{1}}}(\mathbb{D})$. Then
$g(T_{0})^{(2)}=T_{1}^{(2n_{1})}$ is a finite (SI) decomposition. By using
Theorem 2.3 again, we obtain $K_{0}(\mathcal{A}^{\prime}(g(T_{0})^{(2)}\oplus
T_{0}^{(2)}))=K_{0}(\mathcal{A}^{\prime}(T_{1}^{(2n_{1})}\oplus
T_{0}^{(2)}))\cong\mathbb{Z}^{2}$, which is contradictory to (4.4).

So $g(T_{0})$ is strongly irreducible. It follows from
$T_{0},g(T_{0})\in(SI),\,T^{(2)}_{0}\sim_{s}g(T_{0})^{(2)}$ and Theorem 2.4
that $T_{0}\sim_{s}g(T_{0})$ for any $g\in M\ddot{o}b$. Hence, $T_{0}$ is
weakly homogeneous. ∎

When $n=1$, C.K. Fong and C.L. Jiang have shown that
$\mathcal{B}_{1}(\Omega)\subset(SI)$ in [16]. Therefore, when $T$ belongs to
$N\mathcal{F}\mathcal{B}_{2}(\mathbb{D})$ in Theorem 4.2, $T_{0}\in(SI)$ is
naturally satisfied. From Theorem 4.2, we have directly the following
corollary, which does not restrict the operator $T$ to be a Cowen-Douglas
operator.

######  Corollary 4.3.

Let $T=\left(\begin{smallmatrix}T_{0}&XT_{0}-T_{0}X\\\ 0&T_{0}\\\
\end{smallmatrix}\right)$, where $T_{0}\in(SI)$ and
$X\notin\mbox{ker}\sigma_{T_{0}}$. Suppose that for any $g\in M\ddot{o}b$,
$g(T_{0})\in(SI)$. Then $T$ is weakly homogeneous if and only if $T_{0}$ is
weakly homogeneous.

######  Theorem 4.4.

Let $T=\left(\begin{smallmatrix}T_{0}&XT_{1}-T_{0}X\\\ 0&T_{1}\\\
\end{smallmatrix}\right)\in N\mathcal{FB}_{2n}(\mathbb{D})$, where
$T_{i}\in(SI)$, $i=1,2$. Suppose $T_{0}\nsim_{s}T_{1}$($T_{0}$ is not similar
to $T_{1}$), if $T$ and $T_{0}$ are weakly homogeneous, then so is $T_{1}$.

###### Proof.

Since $T$ is weakly homogeneous, by using Theorem 4.2, we have $T_{0}\oplus
T_{1}$ is weakly homogeneous, that is, $g(T_{0})\oplus
g(T_{1})\sim_{s}T_{0}\oplus T_{1}$ for any $g\in M\ddot{o}b$. From
$T_{0},T_{1}\in(SI)$ and $T_{0}\nsim_{s}T_{1}$, $T_{0}\oplus T_{1}$ is a (SI)
decomposition. By Theorem 2.3, we have

(4.5) $\begin{array}[]{lll}K_{0}(\mathcal{A}^{\prime}(T_{0}\oplus T_{1}\oplus
g(T_{0})\oplus g(T_{1})))&\cong&K_{0}(\mathcal{A}^{\prime}(g(T_{0})\oplus
g(T_{1})))\\\ &\cong&K_{0}(\mathcal{A}^{\prime}(T_{0}\oplus T_{1}))\\\
&\cong&\mathbb{Z}^{2}.\end{array}$

We claim that $g(T_{i})$ is also strongly irreducible for $i=0,1$. Otherwise,
assume that $g(T_{0})$ is not strongly irreducible. Since every Cowen-Douglas
operator can be written as the direct sum of finitely many strongly
irreducible Cowen-Douglas operators [28]. Suppose that $g(T_{0})$ has a finite
(SI) decomposition $g(T_{0})=T^{(n_{1})}_{01}\oplus
T^{(n_{2})}_{02}\oplus\cdots\oplus T^{(n_{k})}_{0k}$, $k\geq 1$. Then by
Theorem 2.3 again, we have

$\begin{array}[]{lll}K_{0}(\mathcal{A}^{\prime}(T_{0}\oplus T_{1}\oplus
g(T_{0})\oplus g(T_{1})))&\cong&K_{0}(\mathcal{A}^{\prime}(T_{0}\oplus
T_{1}\oplus T^{(n_{1})}_{01}\oplus\cdots\oplus T^{(n_{k})}_{0k}\oplus
g(T_{1})))\\\ &\cong&\mathbb{Z}^{l},\end{array}$

where $l\geq k+2\geq 3$, which contradicts (4.5). Therefore, the assumption is
invalid and $g(T_{0})$ is strongly irreducible. By the same process, we can
also obtain that $g(T_{1})$ is strongly irreducible.

By using (4.5) again, we have $g(T_{0})\nsim_{s}g(T_{1})$. Since $T_{0}$ is
weakly homogeneous, that is, $T_{0}\sim_{s}g(T_{0})$ for any $g\in
M\ddot{o}b$, we obtain $g(T_{1})\sim_{s}T_{1}$ for any $g\in M\ddot{o}b$, that
is, $T_{1}$ is weakly homogeneous. ∎

###  4.2. Reproducing kernels and weak homogeneity

Some basic facts and notations related to weak homogeneity are introduced in
Subsection 2.2. Recall that an operator $T$ on a Hilbert space is said to be
M$\ddot{o}$bius bounded if the family $\\{\phi(T):\phi\in M\ddot{o}b\\}$ is
uniformly bounded in norm (see [41]). An operator similar to a homogeneous
operator must be a M$\ddot{o}$bius bounded weakly homogeneous operator, but
the converse is false, such as $M_{z}$ on Dirichlet space and $T_{\lambda,s}$,
they are weakly homogeneous, but not M$\ddot{o}$bius bounded, thus not similar
to any homogeneous operator due to B. Bagchi and G. Misra in [2], where
$T_{\lambda,s}$ is a shift with weight $\\{\frac{n+(1+\lambda)\backslash
2+s}{n+(1+\lambda)\backslash 2-s}\\}_{n\in\mathbb{Z}}$, $-1<\lambda\leq 1$,
$Re(s)>\frac{1}{2}$ and $Im(s)>0$. In what follows, we give some discussion of
these concepts. Let us begin with a fact.

######  Lemma 4.5.

Let $K_{0}:\Omega\times\Omega\rightarrow\mathbb{C}$ be a positive kernel and
$K_{1}:\Omega\times\Omega\rightarrow\mathcal{M}_{k}(\mathbb{C})$ for $k\geq 1$
and $K_{1}(z,z)$ is invertible. If $(M_{\phi},\mathcal{H}_{K_{0}})$ is bounded
for some holomorphic function $\phi:\Omega\rightarrow\mathbb{C}$, then so is
$(M_{\phi},\mathcal{H}_{K_{0}K_{1}})$.

###### Proof.

By Lemma 2.7 of [18], we know that if $(M_{\phi},\mathcal{H}_{K_{0}})$ is
bounded, then there exists $c>0$ such that
$(c^{2}-\phi(z)\overline{\phi(w)})K_{0}(z,w)$ is a nonnegative kernel on
$\Omega\times\Omega$. The minimum of all $c$ satisfying this property is its
norm. So we can choose the appropriate $c$ so that
$(c^{2}-\phi(z)\overline{\phi(w)})K_{0}(z,w)$ is a positive definite kernel.
Then $K_{0}(z,w)K_{1}(z,w)$ and
$(c^{2}-\phi(z)\overline{\phi(w)})K_{0}(z,w)K_{1}(z,w)$ are positive kernels.
Thus, $(M_{\phi},\mathcal{H}_{K_{0}K_{1}})$ is bounded. ∎

Let $K_{0},K_{1}:\Omega\times\Omega\to\mathbb{C}$ be two non-negative definite
kernels and
$J_{k}(K_{0},K_{1})|_{res\Delta}(z,w)=(K_{0}(z,w)\partial^{i}\overline{\partial}^{j}K_{1}(z,w))^{k}_{i,j=0}$,
$z,w\in\Omega$ for non negative integer $k$, where $res\Delta$ means
restricted to the diagonal set. Clearly, it is a non negative definite kernel.
If $K_{0},K_{1}$ are two sharp kernels, then so is the matrix-valued kernel
function $J_{k}(K_{0},K_{1})|_{res\Delta}$ due to S. Ghara in [17]. The
following result is also given:

######  Lemma 4.6.

[17] Suppose that $K_{0}$ and $K_{1}$ are two scalar valued sharp positive
definite kernels on $\mathbb{D}\times\mathbb{D}$. If the multiplication
operators $M_{z}$ on $(\mathcal{H},K_{0})$ and $(\mathcal{H},K_{1})$ are
weakly homogenous, then $M_{z}$ on
$(\mathcal{H},J_{k}(K_{0},K_{1})|_{res\Delta})$ is also weakly homogenous.

The next result is a generalization of the result given by S. Ghara in [17] to
operator class $N\mathcal{FB}_{2}(\mathbb{D})$.

######  Proposition 4.7.

Let $(M^{*}_{z},\mathcal{H}_{K})\in\mathcal{B}_{1}(\mathbb{D})$ be a weakly
homogeneous operator. Let $T=\left(\begin{smallmatrix}T_{0}&XT_{1}-T_{0}X\\\
0&T_{1}\\\ \end{smallmatrix}\right)\in N\mathcal{FB}_{2}(\mathbb{D})$, where
$T_{i}\sim_{u}(M^{*}_{z},\mathcal{H}_{K_{i}})$ and $K_{i}=KK^{(\lambda_{i})}$,
$\lambda_{i}>0$, $i=0,1$. If for any $\lambda>0$, the limits of
$\lim\limits_{w\rightarrow\partial\mathbb{D}}K(\bar{w},\bar{w})K^{(\lambda)}(\bar{w},\bar{w})$
and
$\lim\limits_{w\rightarrow\partial\mathbb{D}}\frac{K(\bar{w},\bar{w})}{K^{(\lambda)}(\bar{w},\bar{w})}$
are either 0 or $\infty$, then $T$ is a $M\ddot{o}bius$ bounded weakly
homogeneous operator which is not similar to the direct sum of any homogeneous
operators.

###### Proof.

Since $(M_{z},\mathcal{H}_{K^{(\lambda_{i})}})$ is homogeneous proven by G.
Misra in [35], it follows from Lemma 4.6 that
$(M_{z},\mathcal{H}_{KK^{(\lambda_{i})}})$ is weakly homogeneous, so are
$T_{i}$ and $T$ by Lemma 4.1, $i=0,1$.

Let $S=\left(\begin{smallmatrix}I&-X\\\ 0&I\\\ \end{smallmatrix}\right)$. We
have $T=S^{-1}(T_{0}\oplus T_{1})S$. Then for any $g\in M\ddot{o}b$, we have
$g(T)=S^{-1}(g(T_{0}\oplus T_{1}))S=S^{-1}(g(T_{0})\oplus g(T_{1}))S$. So we
obtain

$\|g(T)\|\leq\|S^{-1}\|\|g(T_{0})\oplus
g(T_{1})\|\|S\|\leq\|S^{-1}\|(\|g(T_{0})\|+\|g(T_{1})\|)\|S\|.$

Since $(M^{*}_{z},\mathcal{H}_{K^{(\lambda_{i})}})$ is homogeneous, $i=0,1$,
it is $M\ddot{o}bius$ bounded, so is
$(M^{*}_{z},\mathcal{H}_{KK^{(\lambda_{i})}})$ by Lemma 4.5. That is, for any
$g\in M\ddot{o}b$, $\|g(T_{i})\|$ is uniformly bounded, $i=0,1$. The
invertibility of $S$ means that $T$ is $M\ddot{o}bius$ bounded. Thus, to
complete the proof, it remains to show that $T$ is not similar to a direct sum
of any homogeneous operator. That is to show that $T_{i}$ is not similar to
any homogeneous operator, $i=0,1$.

Assume that $T_{i}$ is similar to a homogeneous operator, say $\tilde{T}$.
Since $T_{i}\in\mathcal{B}_{1}(\mathbb{D})$ and the class
$\mathcal{B}_{1}(\mathbb{D})$ is closed under similarity, $\tilde{T}$ belongs
to $\mathcal{B}_{1}(\mathbb{D})$. Upto unitary equivalence, every homogeneous
operator in $\mathcal{B}_{1}(\mathbb{D})$ is of the form
$(M^{*}_{z},\mathcal{H}_{K^{(\mu)}})$ for some $\mu>0$. Consequently,
$T_{i}\sim_{s}M_{z}^{*}$ on $\mathcal{H}_{K^{(\mu)}}$. Since
$\lim\limits_{w\rightarrow\partial\mathbb{D}}\frac{K(w,w)K^{(\lambda_{i})}(w,w)}{K^{(\mu)}(w,w)}=0$
or $\infty$, by Lemma 3.5 of [21], we obtain that there is no non-zero bounded
operator $X_{0}$ such that $X_{0}T_{i}=M^{*}_{z}X_{0}$ or
$X_{0}M^{*}_{z}=T_{i}X_{0}$. This is a contradiction. Hence the operator
$T_{i}$ is not similar to any homogeneous operator. In other words, $T$ is not
similar to a direct sum of any homogeneous operator. This completes the proof.
∎

We know that for the operators in $\mathcal{B}_{1}(\mathbb{D})$, the ratio of
their corresponding reproducing kernels is bounded and bounded from zero,
which is a necessary condition for their similarity. D.N. Clark and G. Misra
found a class of operators for which this necessary condition is a necessary
and sufficient condition for their similarity to homogeneous operators, since
for $\gamma\geq 1$,
$(M_{z},\mathcal{H}_{K_{(\gamma-1)}})\sim_{s}(M_{z},\mathcal{H}_{K^{(\gamma)}})$.

######  Lemma 4.8.

[13] Let $S\sim_{u}(M_{z}^{*},\mathcal{H}_{K_{(\gamma)}})$ for $\gamma>0$ and
$T\sim_{u}(M_{z}^{*},\mathcal{H}_{K})$ with
$K(z,w)=\sum\limits_{n=0}^{\infty}a_{n}z^{n}\bar{w}^{n}$, $0<a_{n}\leq
a_{n+1}$. Then $T\sim_{s}S$ if and only if $\frac{K(w,w)}{K_{(\gamma)}(w,w)}$
is bounded and bounded from zero.

######  Proposition 4.9.

Let $T=\left(\begin{smallmatrix}T_{0}&XT_{1}-T_{0}X\\\ 0&T_{1}\\\
\end{smallmatrix}\right)\in N\mathcal{FB}_{2}(\mathbb{D})$, where
$T_{0}\sim_{u}(M_{z}^{*},\mathcal{H}_{K^{(\mu)}})$,
$T_{1}\sim_{u}(M_{z}^{*},\mathcal{H}_{K_{1}})$,
$K_{1}(z,w)=\sum\limits_{n=0}^{\infty}a_{n}z^{n}\overline{w}^{n}$ for
$0<a_{n}\leq a_{n+1}$. If there exist a metric $h_{T}$ of $E_{T}$,
$\lambda>\mu$ and constants $m,M$ such that $m\leq(\det
h_{T}(w))(1-|w|^{2})^{\lambda}\leq M$ if and only if $T_{1}$ is a weakly
homogeneous operator similar to a homogeneous operator.

###### Proof.

Let $\gamma_{0}(w)=\left(\begin{smallmatrix}K^{(\mu)}(\cdot,\overline{w})\\\
0\\\ \end{smallmatrix}\right)$ and
$\gamma_{1}(w)=\left(\begin{smallmatrix}XK_{1}(\cdot,\overline{w})\\\
K_{1}(\cdot,\overline{w})\\\ \end{smallmatrix}\right)$. Then
$\\{\gamma_{0},\gamma_{1}\\}$ be a holomorphic frame of the Hermitian
holomorphic vector bundle $E_{T}$ corresponding to $T$. It implies that the
metric matrix of $E_{T}$ under this frame is
$h_{T}(w)=\left(\begin{smallmatrix}K^{(\mu)}(\overline{w},\overline{w})&\langle
XK_{1}(\cdot,\overline{w}),K^{(\mu)}(\cdot,\overline{w})\rangle\\\ \langle
K^{(\mu)}(\cdot,\overline{w}),XK_{1}(\cdot,\overline{w})\rangle&\|XK_{1}(\cdot,\overline{w})\|^{2}+K_{1}(\overline{w},\overline{w})\\\
\end{smallmatrix}\right)$. Thus,
$det\,h_{T}(w)=K^{(\mu)}(\overline{w},\overline{w})(K_{1}(\overline{w},\overline{w})+\|XK_{1}(\cdot,\overline{w})\|^{2})-|\langle
XK_{1}(\cdot,\overline{w}),K^{(\mu)}(\cdot,\overline{w})\rangle|^{2}.$
According to the Cauchy-Bunyakowsky-Schwarz Inequality, we have
$1\leq\frac{det\,h_{T}(w)}{K^{(\mu)}(\overline{w},\overline{w})K_{1}(\overline{w},\overline{w})}\leq
1+\|X\|^{2}$. It follows from the inequality in the proposition that
$\frac{m}{1+\|X\|^{2}}\leq
K^{(\mu)}(\overline{w},\overline{w})K_{1}(\overline{w},\overline{w})(1-|w|^{2})^{\lambda}\leq
M$. Thus,
$\frac{m}{1+\|X\|^{2}}\leq\frac{K_{1}(\overline{w},\overline{w})}{K^{(\lambda-\mu)}(\overline{w},\overline{w})}\leq
M.$ By Lemma 4.8, we have
$(M^{*}_{z},\mathcal{H}_{K_{1}})\sim_{s}(M^{*}_{z},\mathcal{H}_{K^{(\lambda-\mu)}})$.
Since $\lambda>\mu$, $(M^{*}_{z},\mathcal{H}_{K^{(\lambda-\mu)}})$ is a
homogeneous operator and $(M^{*}_{z},\mathcal{H}_{K_{1}})$ is a weakly
homogeneous operator.

The proof of sufficiency is similar. This completes the proof. ∎

###  4.3. Irreducible weakly homogeneous operator

At the beginning of this section, we mentioned that the operator
$\left(\begin{smallmatrix}T_{0}&XT_{1}-T_{0}X\\\ 0&T_{1}\\\
\end{smallmatrix}\right)$ is strongly reducible. The following proposition
shows that irreducible operators of this form exist.

######  Proposition 4.10.

Let $T_{0}\sim_{u}(M^{*}_{z},\mathcal{H}_{K})\in\mathcal{B}_{1}(\mathbb{D})$
be a weakly homogeneous operator and
$T_{1}\sim_{u}(M^{*}_{z},\mathcal{H}_{K^{(\lambda)}K})$ for $\lambda>0$. If
there exists $X$ such that $XT_{1}\neq T_{0}X$, then
$\left(\begin{smallmatrix}T_{0}&XT_{1}-T_{0}X\\\ 0&T_{1}\\\
\end{smallmatrix}\right)$ belongs to $N\mathcal{FB}_{2}(\mathbb{D})$ and is an
irreducible weakly homogeneous operator.

###### Proof.

We know that $T_{1}$ is a weakly homogeneous operator by using Theorem 4.7.
Since
$\lim\limits_{w\to\partial\mathbb{D}}\frac{K(w,w)}{K^{(\lambda)}(w,w)K(w,w)}=\lim\limits_{w\to\partial\mathbb{D}}\frac{1}{K^{(\lambda)}(w,w)}=\lim\limits_{w\to\partial\mathbb{D}}(1-|w|^{2})^{\lambda}=0$,
by Lemma 3.5 in [21], then there is no non-zero bounded intertwining operator
$X_{0}$ such that $T_{1}X_{0}=X_{0}T_{0}$. Since $XT_{1}\neq T_{0}X$,
$\left(\begin{smallmatrix}T_{0}&XT_{1}-T_{0}X\\\ 0&T_{1}\\\
\end{smallmatrix}\right)\in N\mathcal{FB}_{2}(\mathbb{D})$ is weakly
homogeneous. Next, we will check the irreducibility.

Let $P=\left(\begin{smallmatrix}P_{00}&P_{01}\\\ P_{10}&P_{11}\\\
\end{smallmatrix}\right)\in\mathcal{A}^{\prime}(T)$ and $P=P^{*}=P^{2}$. We
have $P_{00}=P^{*}_{00},P_{01}=P^{*}_{10},P_{11}=P^{*}_{11}$ and

(4.6)
$\left(\begin{smallmatrix}P_{00}T_{0}&P_{00}(XT_{1}-T_{0}X)+P_{01}T_{1}\\\
P_{10}T_{0}&P_{10}(XT_{1}-T_{0}X)+P_{11}T_{1}\\\
\end{smallmatrix}\right)=\left(\begin{smallmatrix}T_{0}P_{00}+(XT_{1}-T_{0}X)P_{10}&T_{0}P_{01}+(XT_{1}-T_{0}X)P_{11}\\\
T_{1}P_{10}&T_{1}P_{11}\\\ \end{smallmatrix}\right).$

From the above proof, it can be seen that $P_{01}=0$, $P_{10}=0$. It follows
that $P_{ii}\in\mathcal{A}^{\prime}(T_{i})$ and
$P_{ii}=P_{ii}^{*}=P_{ii}^{2}$, $i=0,1$. Since
$T_{0},T_{1}\in\mathcal{B}_{1}(\mathbb{D})$ are irreducible operators,
$P_{ii}=0$ or $I$. For the $(1,2)$-entry of (4.6), we have $P_{00}=P_{11}$. We
infer $P=0$ or $I$, which means $T$ is an irreducible weakly homogeneous
operator. ∎

The next two irreducible results are about bilateral block shifts by Theorem
3.4,3.7 and Lemma 4.1.

######  Proposition 4.11.

Let $T=T(\sqrt{\frac{n+a}{n+b}},d_{n})$. If
$|\frac{d_{n+1}}{t_{n}}-d_{n}t_{n}|^{2}=|\frac{d_{m+1}}{t_{m}}-d_{m}t_{m}|^{2}$
only holds when $m=-(n+1)$ and
$t_{n}\overline{c_{n}}(|t_{n-1}|^{2}+\frac{1}{|t_{n-1}|^{2}}+|c_{n-1}|^{2})\neq\frac{1}{t_{n-1}}\overline{c_{n-1}}(|t_{n}|^{2}-\frac{1}{|t_{n}|^{2}}-|c_{n}|^{2})$
for $c_{n}=d_{n+1}\sqrt{\frac{n+b}{n+a}}-d_{n}\sqrt{\frac{n+a}{n+b}}$, then
$T$ is irreducible weakly homogeneous.

######  Theorem 4.12.

Let $T=T(w_{n},v_{n},d_{n})$ with $|w_{n}|=|v_{n}|=1$. Suppose that there
exists a bijective mapping $g$ on $\mathbb{Z}$ and $N\in\mathbb{Z}$ such that
the sets of singular values of $A_{n}$ are distinct from each other. If
$|d_{n+1}v_{n}-d_{n}w_{n}|^{2}=|d_{g(n)+1}v_{g(n)}-d_{g(n)}w_{g(n)}|^{2}$ and
one of the following two conditions is true,

  1. (1)

$d_{n+1}c_{n}+d_{n}c_{n-1}-d_{n+1}\overline{w_{n}}v_{n}c_{n-1}\notin\mathbb{R}$,
where $c_{n}=\overline{d_{n}}\overline{w_{n}}v_{n}$;

  2. (2)

$(\overline{d_{n+1}}w_{n}v_{n}-\overline{d_{n}})|d_{n}v_{n-1}-d_{n-1}w_{n-1}|^{2}\neq(\overline{d_{n-1}}\overline{w_{n-1}}v_{n-1}-\overline{d_{n}})|d_{n+1}v_{n}-d_{n}w_{n}|^{2}$,

then $T$ is irreducible irreducible.

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