K. Ji and S. Ji]Department of Mathematics Science, Hebei Normal University,
Shijiazhuang, Hebei 050016, China

# A note on unitary equivalence of operators acting on reproducing kernel
Hilbert spaces

Kui Ji and Shanshan Ji∗ [<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

A well-known theorem due to R. E. Curto and N. Salinas gives a necessary and
sufficient condition for the unitary equivalence of commuting tuples of
bounded linear operators acting on reproducing kernel Hilbert spaces [2].
Inspired by this theorem, we obtain a different but equivalent criterion for
the unitary equivalence of operators acting on reproducing kernel Hilbert
spaces. As an application, we describe the structure of intertwining operator
and prove that the decomposition of Cowen-Douglas operators is unique up to
unitary equivalence.

###### Key words and phrases:

Cowen-Douglas operator, Reproducing kernel, Unitarily equivalent, (IR)
decomposition

###### 2000 Mathematics Subject Classification:

Primary 47C15, 47B37; Secondary 47B48, 47L40

* Corresponding author

##  1\. Introduction

Let $\mathcal{H}$ be an infinite dimensional complex and separable Hilbert
space, ${\mathcal{L}}({\mathcal{H}})$ the collection of all bounded linear
operators on $\mathcal{H}$. Let
$T,\widetilde{T}\in{\mathcal{L}}({\mathcal{H}})$. We say $T$ is unitarily
equivalent to $\widetilde{T}$ (denoted by $T\sim_{u}\widetilde{T}$), if there
is a unitary operator $U\in{\mathcal{L}}({\mathcal{H}})$ such that
$UT=\widetilde{T}U$. In the paper [1], M. J. Cowen and R. G. Douglas
introduced the class $\mathcal{B}_{n}(\Omega)$ of bounded linear operators on
${\mathcal{H}}$. They proved that every operator $T\in\mathcal{B}_{n}(\Omega)$
determines a Hermitian holomorphic vector bundle $E_{T}$ with the property:
two Cowen-Douglas operators are unitarily equivalent if and only if the
corresponding Hermitian holomorphic vector bundles are congruent. They also
found a set of complete invariants for congruence consisting of the curvature
of $E_{T}$ and its covariant partial derivatives. Unfortunately, these
invariants are not easy to compute except when the rank of the bundle is one.
K. Zhu gives another equivalent condition for unitary equivalence of Cowen-
Douglas operators by using spanning holomorphic cross-section in [12]. In
[11], A. Koranyi and G. Misra classify homogeneous operators in
$\mathcal{B}_{n}(\Omega)$ modulo unitary equivalence.

Let $T\in\mathcal{B}_{n}(\Omega)$. In [1], it is proved that $T$ is unitarily
equivalent to the adjoint of multiplication operator $M_{z}$ on some
reproducing kernel Hilbert space consisting of holomorphic
$\mathbb{C}^{n}-$valued functions defined on
$\Omega^{*}:=\\{\lambda:\overline{\lambda}\in\Omega\\}$. Without loss of
generality, $T\in\mathcal{B}_{n}(\Omega)$ can be viewed as $M^{*}_{z}$, that
is, $T=(M^{*}_{z},{\mathcal{H}},K)$, where the Hilbert space $\mathcal{H}$
consisting of holomorphic functions possessing a reproducing kernel $K$ taking
values in $n\times n$ complex matrices. Furthermore, R. E. Curto and N.
Salinas also provide a necessary and sufficient condition for unitary
equivalence of two multiplication operators $M_{z}$ acting on two different
Hilbert spaces $\mathcal{H}_{1},\mathcal{H}_{2}$ with reproducing kernels
$K_{1},K_{2}$ (see Remark 3.8, [2]).

######  Theorem 1.1.

[2] Let $T_{i}\in\mathcal{B}_{n}(\Omega)$, and
$T_{i}=(M^{*}_{z},\mathcal{H}_{i},K_{i}),i=1,2$, where $\mathcal{H}_{i}$ are
the analytic function spaces with reproducing kernels $K_{1},K_{2}$,
respectively. Then $T_{1}\sim_{u}T_{2}$ if and only if

(1.1) $K_{1}(\mu,\lambda)=\Phi(\mu)K_{2}(\mu,\lambda)\Phi(\lambda)^{*}$

for a $\mathcal{M}_{n}(\mathbb{C})-$valued holomorphic function $\Phi$ such
that $\Phi(\mu)$ is invertible for every $\mu$, where $\mu,\lambda$ in some
open subset $\Lambda$ of $\Omega^{*}$.

In [4], it is shown that an intertwiner $X$ between two operators of the form
$(M^{*}_{z},\mathcal{H}_{1},K_{1})$ and $(M^{*}_{z},\mathcal{H}_{2},K_{2})$
exist if and only if there is a fixed linear map $A$ such that
$AK_{1}(z,w)=K_{2}(z,w)A$ assuming that $K_{1}$ and $K_{2}$ are normalized
kernels. Now we obtain a useful alternative criterion for unitary equivalence
similar to Theorem 1.1 (see Theorem 3.5).

As is well-known, in finite dimensional space, every $n\times n$ matrix can be
uniquely written as an orthogonal direct sum of irreducible matrices up to
unitary equivalence. Compared with this classic conclusion, it is natural to
ask if we can arrive at a similar conclusion for irreducible decomposition of
operators acting on an infinite dimensional Hilbert space. In [3], Y. Cao, J.
Fang and C. Jiang show operators with unique finite (IR) decomposition up to
unitary equivalence by using $K_{0}-$group of their commutants. As an
application of the unitary equivalence criterion of R. E. Curto and N. Salinas
given above, we prove the uniqueness of the decomposition of reducible Cowen-
Douglas operators up to unitary equivalence.

The paper is organized as follows. In Section 2, we introduce some notations
and definitions about Cowen-Douglas theory. In Section 3, motivated by Theorem
1.1 due to R. E. Curto and N. Salinas, there is an equivalent condition for
unitary equivalence of operators acting on reproducing kernel Hilbert spaces
in a different form. In Section 4, we prove that the (IR) decomposition of
Cowen-Douglas operators is unique up to unitary equivalence. In Section 5, by
using Theorem 3.5 and Theorem 4.2, we describe the structure of the
intertwining matrix between the two reproducing kernels.

##  2\. Preliminaries

In this section, we review some basic facts and notions for Cowen-Douglas
operators and reproducing kernel (see [1], [2]).

###  2.1. Cowen-Douglas operators

######  Definition 2.1.

[1] Let $\Omega$ be an open connected subset of complex plane $\mathbb{C}$ and
$n$ be a positive integer. The class $\mathcal{B}_{n}(\Omega)$ consists of
operators $T\in{\mathcal{L}}({\mathcal{H}})$ satisfying:

  * (1)

$\Omega\subset\sigma(T)=\\{\lambda\in\mathbb{C}:T-\lambda$ is not
invertible$\\}$;

  * (2)

ran$(T-\lambda)=\mathcal{H}$ for every $\lambda$ in $\Omega$;

  * (3)

$\bigvee_{\lambda{\in}{\Omega}}\ker(T-\lambda)=\mathcal{H}$;

  * (4)

$\dim\ker(T-\lambda)=n$ for every $\lambda$ in $\Omega$.

It is shown in the same paper that each operator $T\in\mathcal{B}_{n}(\Omega)$
determines a Hermitian holomorphic vector bundle $(E_{T},\pi,\lambda)$ over
$\Omega$, namely, the sub-bundle of the trivial bundle
$\Omega\times\mathcal{H}$ defined by

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

with the natural projection map $\pi:E_{T}\rightarrow\Omega$,
$\pi(\lambda,x)=\lambda$. Since dim ker$(T-\lambda)=n$ for every
$\lambda\in\Omega$, $E_{T}$ is of rank $n$. Let $\\{\gamma_{i}\\}_{i=1}^{n}$
be the holomorphic frame of $E_{T}$ (Proposition 1.12, [1]), then form the
matric at $\lambda$ of inner products
$h(\lambda):=(\langle\gamma_{j}(\lambda),\gamma_{i}(\lambda)\rangle)_{n\times
n},\lambda\in\Omega.$

Let $\xi(\Omega)$ be the algebra consist of the $C^{\infty}$ functions and
$\xi^{p}(\Omega)$ denote the $p-$defferential form of $C^{\infty}$ functions.
Thus, we have

$\xi^{0}(\Omega)=\xi(\Omega),\
\xi^{1}(\Omega)=\\{fdz+gd\overline{z}:f,g\in\xi(\Omega)\\},\
\xi^{2}(\Omega)=\\{fdzd\overline{z}:f\in\xi(\Omega)\\}.$

For any vector bundle $E$ which has $C^{\infty}$ differential structure, let
$\xi^{p}(\Omega,E)$ denote $p-$differential forms with the coefficients in
$E$. Then each element in $\xi^{0}(\Omega,E)$ is one of sections of $E$. The
connection $D$ can be regarded as a differential operator which maps
$\xi^{0}(\Omega,E)$ to $\xi^{1}(\Omega,E)$. Then the canonical connection $D$
preserves the metric and satisfies the following:

$D(\sum\limits_{i=1}^{n}f_{i}\gamma_{i})=\sum\limits_{i=1}^{n}df_{i}\otimes\gamma_{i}+\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{n}f_{i}\Theta_{j,i}\gamma_{i},$

with the connection matrix $\Theta=(\Theta_{ij})_{n\times
n}=h^{-1}\frac{\partial}{\partial\lambda}h$. And

$D^{2}=d\Theta+\Theta\wedge\Theta=\frac{\partial}{\partial\overline{\lambda}}(h^{-1}\frac{\partial}{\partial\lambda}h),$

then
$-\frac{\partial}{\partial\overline{\lambda}}(h^{-1}\frac{\partial}{\partial\lambda}h)$
is called the curvature of $E$, denoted by $\mathcal{K}_{E}$. In fact, the
curvature of $E_{T}$ depends on the choice of the frame, and the curvature
under different choices is similar except when $n=1$. Based on a simple
calculation, when $T\in\mathcal{B}_{1}(\Omega)$, we have

$\mathcal{K}_{T}(\lambda)=-\frac{\partial^{2}}{\partial\lambda\partial\overline{\lambda}}\log\parallel
e(\lambda)\parallel^{2},$

where $e(\lambda)\in\ker(T-\lambda)$ is a non-zero cross section of $E_{T}$.
In this case, the curvature does not depend on the choice of the non-vanishing
section $e$. If $\widetilde{e}$ is another non-vanishing holomorphic section,
then $\widetilde{e}=\phi e$, where $\phi$ is a holomorphic function on
$\Omega_{0}\subset\Omega$, and $\log|\phi|$ is harmonicity, so curvature is
completely unitary invariant. However, if $T\in\mathcal{B}_{n}(\Omega)(n>1)$,
only the eigenvalues of the curvature are independent of the choice of the
holomorphic frame.

The curvature can be regarded as a bundle map, and by the definition of bundle
map on a holomorphic vector bundle, we obtain the covariant partial
derivatives $\mathcal{K}_{T,\omega^{i}\overline{\omega}^{j}}$ of the curvature
$\mathcal{K}_{T}$:

  * (1)

$\mathcal{K}_{T,\lambda^{i}\overline{\lambda}^{j+1}}=\frac{\partial}{\partial\overline{\lambda}}(\mathcal{K}_{T,\lambda^{i}\overline{\lambda}^{j}})$;

  * (2)

$\mathcal{K}_{T,\lambda^{i+1}\overline{\lambda}^{j}}=\frac{\partial}{\partial\lambda}(\mathcal{K}_{T,\lambda^{i}\overline{\lambda}^{j}})+[h^{-1}\frac{\partial}{\partial\lambda}h,\mathcal{K}_{T,\lambda^{i}\overline{\lambda}^{j}}]$.

Here is one of the main results from [1].

######  Theorem 2.2.

[1] Let $T,S\in\mathcal{B}_{n}(\Omega)$. Then $T\sim_{u}S$ if and only if
there exists an isometric bundle map $V:E_{T}\rightarrow E_{S}$ such that

$V(\mathcal{K}_{T,\lambda^{i}\overline{\lambda}^{j}})=(\mathcal{K}_{S,\lambda^{i}\overline{\lambda}^{j}})V,\
0\leq i,j\leq i+j\leq n,\ (i,j)\neq(0,n),(n,0).$

Thus the curvature $\mathcal{K}_{T}$ and its covariant partial derivatives
$\mathcal{K}_{T,\lambda^{i}\overline{\lambda}^{j}}$ of holomorphic bundle
$E_{T}$ corresponding to $T\in\mathcal{B}_{n}(\Omega)$ are complete unitary
invariants. In [10], G. Misra and S. Shyam Roy show that the class of the
curvature alone does not necessarily determine the class of the vector bundle
except in the case of a line bundle.

###  2.2. Reproducing kernel

We use symbol $\mathcal{M}_{n}(\mathbb{C})$ to denote the set of all $n\times
n$ matrices over $\mathbb{C}$. Here are some properties of the reproducing
kernel (Lemma 3.2, [2]). Given a bounded domain $\Lambda$, let
$K:\Lambda\times\Lambda\rightarrow\mathcal{M}_{n}(\mathbb{C})$ be a non-
negative kernel, holomorphic in the first variable and anti-holomorphic in the
second variable. As is well-known, there is a $\mathbb{C}^{n}$-value Hilbert
space $\mathcal{H}$ contained in the space of holomorphic functions defined on
$\Lambda$ such that $K(\cdot,\lambda)\xi$ is in $\mathcal{H}$ for all
$\lambda\in\Lambda$ and $\xi\in\mathbb{C}^{n}$, $K$ serves as the reproducing
kernel for $\mathcal{H}$, that is, $\langle
f,K(\cdot,\lambda)\rangle=f(\lambda)$ for all $f\in\mathcal{H}$. The Hilbert
space $\mathcal{H}$ is said to be analytic function space with reproducing
kernel $K$. Clearly, $K(\mu,\lambda)=K(\lambda,\mu)^{*}$, where
$K(\lambda,\mu)^{*}$ is conjugate transpose of $K(\lambda,\mu),\
\mu,\lambda\in\Lambda$.

In [1], M. J. Cowen and R. G. Douglas described the relationship between
$T\in\mathcal{B}_{1}(\Omega)$ and multiplication operator on some Hilbert
space ${\mathcal{H}}$. Let $\gamma$ be a non-vanishing holomorphic cross-
section of $E_{T}$ (which exists by Grauert’s theorem [7]). Naturally,
corresponding to $\gamma$ is the representation $\Gamma$ on ${\mathcal{H}}$,
defined by $(\Gamma x)(\mu)=\langle x,\gamma(\overline{\mu})\rangle,\
x\in{\mathcal{H}},\ \mu\in\Omega^{*}$. Thus, $\Gamma$ is unitary and
$ran\Gamma={\mathcal{H}}$. Define function $K$ as
$K(\mu,\lambda)=\Gamma(\gamma(\overline{\lambda}))(\mu)=\langle\gamma(\overline{\lambda}),\gamma(\overline{\mu})\rangle,\
\mu,\lambda\in\Omega^{*}$. Then

$\Gamma T\Gamma^{*}K(\cdot,\lambda)=\Gamma
T\Gamma^{*}\Gamma(\gamma(\overline{\lambda}))=\Gamma
T(\gamma(\overline{\lambda}))=\Gamma\overline{\lambda}\gamma(\overline{\lambda})=\overline{\lambda}\Gamma\gamma(\overline{\lambda})=\overline{\lambda}K(\cdot,\lambda),\
\lambda\in\Omega^{*},$

$K(\cdot,\lambda)$ is an eigenvector for $\Gamma T\Gamma^{*}$ with eigenvalue
$\overline{\lambda}$. Note that
$\mathcal{H}=\overline{span}\\{K(\cdot,\lambda),\ \lambda\in\Omega^{*}\\}$, so
$\Gamma T\Gamma^{*}$ is the adjoint of the multiplication operator on a
holomorphic function space ${\mathcal{H}}$ possessing a reproducing kernel
$K$. Without loss of generality, $T$ can be view as $M^{*}_{z}$, that is,
$T=(M^{*}_{z},{\mathcal{H}},K)$. More generally, for
$T\in\mathcal{B}_{n}(\Omega)(n>1)$, similar results are proved, $T$ can be
realized as the adjoint $M^{*}_{z}$, where multiplication operator $M_{z}$ on
a reproducing kernel Hilbert space of holomorphic $\mathbb{C}^{n}-$valued
functions.

Suppose that two multiplication operators acting on two different
$\mathbb{C}^{n}-$valued Hilbert spaces $\mathcal{H}_{1},\mathcal{H}_{2}$ with
reproducing kernels $K_{1},K_{2}$, and
$X:\mathcal{H}_{1}\rightarrow\mathcal{H}_{2}$ is a bounded operator
intertwining these two operators. We know that
$\mathcal{H}_{i}=\overline{span}\\{K_{i}(\cdot,\lambda)\xi,\lambda\in\Omega^{*},\xi\in\mathbb{C}^{n}\\},i=1,2$,
and
$XK_{1}(\cdot,\lambda)\xi=K_{2}(\cdot,\lambda)\Psi(\lambda)\xi(\xi\in\mathbb{C}^{n}$)
for some $\mathcal{M}_{n}(\mathbb{C})-$valued function $\Psi$. Since
$K_{i}(\cdot,\cdot),i=1,2$ are anti-holomorphic in the second variable, it
follows that $\Psi$ is anti-holomorphic. In fact, $X^{*}$ is the
multiplication operator $M_{\overline{\Psi}^{T}}$. Furthermore, if two
operators are unitarily equivalent, then there is a unitary operator $U$
intertwining them. Hence,
$K_{1}(\cdot,\lambda)\xi=U^{*}K_{2}(\cdot,\lambda)\Phi^{*}(\lambda)\xi$,
$U^{*}$ is multiplication operator $M_{\Phi}$, where holomorphic function
$\Phi$ satisfies for every $\mu$, $\Phi(\mu)$ is invertible. It follows that

$K_{1}(\mu,\lambda)=\Phi(\mu)K_{2}(\mu,\lambda)\Phi^{*}(\lambda).$

Conversely, if the equality above holds for two reproducing kernels
$K_{1},K_{2}$, then the corresponding operators are unitarily equivalent. That
is to say, the above equality is a sufficient and necessary condition for the
unitary equivalence of the two operators (See more details in [2]).

##  3\. Unitary equivalence of operators acting on reproducing kernel Hilbert
spaces

In this section, by utilizing the reproducing kernels, we give another form of
unitary equivalence of operators acting on reproducing kernel Hilbert spaces.
By using this new form, we can also describe the structure of the intertwining
matrix between two kernels related to the operators which are both reducible.
Before introducing this theorem, we need to describe the commutants of
operators. Let

$\mathcal{A}^{\prime}(T):=\\{X|XT=TX\\},\
\mathcal{A}^{\prime}(T,T^{*}):=\\{X|XT=TX,\ XT^{*}=T^{*}X\\}.$

$T\in{\mathcal{L}}({\mathcal{H}})$ is called irreducible, if the commutant of
$T$ doesn’t have any nontrivial self-adjoint idempotent operator,
equivalently, there is no non-trivial reducing subspace of $T$ (cf [6], [8]).
The irreducibility is a unitary invariant. In what follows, $T\in(IR)$ means
that $T$ is an irreducible operator.

######  Proposition 3.1.

Let $T\in\mathcal{B}_{n}(\Omega)$. If $T=T_{1}\oplus T_{2}\oplus\cdots\oplus
T_{k}$, then $k\leq n$ and $T_{i}$ is Cowen-Douglas operator for all $1\leq
i\leq k$.

###### Proof.

If $T=T_{1}\oplus T_{2}\oplus\cdots\oplus
T_{k}\in{\mathcal{L}}({\mathcal{H}})$, then
$T_{i}\in{\mathcal{L}}({\mathcal{H}}_{i})$ for non-zero Hilbert spaces
${\mathcal{H}_{i}}$, $1\leq i\leq k$. Here $T_{i}$, $1\leq i\leq k$ may be
reducible or irreducible.

According to Definition 2.1, we have ran$(T-\lambda)=\mathcal{H}$ for every
$\lambda$ in $\Omega$. So for every
$y=(y_{1},y_{2},\cdots,y_{k})^{T}\in{\mathcal{H}}$, there exists
$x=(x_{1},x_{2},\cdots,x_{k})^{T}\in{\mathcal{H}}$, such that
$(T-\lambda)x=y$, where $x_{i},y_{i}\in{\mathcal{H}}_{i}$. By the
arbitrariness of each component in $y$, we obtain that for every
$y_{i}\in{\mathcal{H}}_{i}$, there exist $x_{i}\in{\mathcal{H}}_{i}$ such that
$(T_{i}-\lambda)x_{i}=y_{i},\ 1\leq i\leq k$. Then
ran$(T_{i}-\lambda)=\mathcal{H}_{i}$ for every $\lambda$ in $\Omega$, $1\leq
i\leq k$.

Let $(z_{1},z_{2},\cdots,z_{k})^{T}\in\ker(T-\lambda)$. We have
$z_{i}\in\ker(T_{i}-\lambda),1\leq i\leq k$. Conversely, if
$z_{i}\in\ker(T_{i}-\lambda),1\leq i\leq k$, then
$(z_{1},z_{2},\cdots,z_{k})^{T}\in\ker(T-\lambda)$. It follows that

(3.1)
$\ker(T-\lambda)=\ker(T_{1}-\lambda)\oplus\ker(T_{2}-\lambda)\oplus\cdots\oplus\ker(T_{k}-\lambda).$

By Definition 2.1 again, we have
$\bigvee_{\lambda{\in}{\Omega}}\ker(T-\lambda)=\mathcal{H}$. Combining with
equation (3.1), we have

$\begin{array}[]{lll}{\mathcal{H}_{1}}\bigoplus{\mathcal{H}_{2}}\bigoplus\cdots\bigoplus{\mathcal{H}_{k}}&=&\bigvee\limits_{\lambda{\in}{\Omega}}\ker(T-\lambda)\\\
&=&\bigvee\limits_{\lambda{\in}{\Omega}}\big{(}\ker(T_{1}-\lambda)\bigoplus\ker(T_{2}-\lambda)\bigoplus\cdots\bigoplus\ker(T_{k}-\lambda)\big{)}\\\
&=&\bigvee\limits_{\lambda{\in}{\Omega}}\ker(T_{1}-\lambda)\bigoplus\bigvee\limits_{\lambda{\in}{\Omega}}\ker(T_{2}-\lambda)\bigoplus\cdots\bigoplus\bigvee\limits_{\lambda{\in}{\Omega}}\ker(T_{k}-\lambda).\end{array}$

Therefore,
$\bigvee\limits_{\lambda{\in}{\Omega}}\ker(T_{i}-\lambda)={\mathcal{H}_{i}},\
1\leq i\leq k.$

Using formula (3.1) again, we have dim $\ker(T-\lambda)=\sum_{i=1}^{k}$dim
$\ker(T_{i}-\lambda)=n$ and dim $\ker(T_{i}-\lambda)$ is finite,
$\lambda\in\Omega$. Thus, $T_{i}-\lambda$ is Fredholm operator and the index
$\mbox{ind}(T_{i}-\lambda)=\mbox{dim}\,\ker(T_{i}-\lambda)\leq n$, since
ran$(T_{i}-\lambda)=\mathcal{H}_{i}$ for $\lambda\in\Omega$ and $1\leq i\leq
k$. Setting $n_{i}:=\mbox{ind}(T_{i}-\lambda)$ for some $\lambda\in\Omega$,
$1\leq i\leq k$. The continuity of index implies that for all $\mu$ in the
same connected component of $\Omega$ containing $\lambda$,
$n_{i}=\mbox{ind}(T_{i}-\lambda)$, $1\leq i\leq k$. Claim $n_{i}>0,1\leq i\leq
k$. Otherwise, if there exist $i_{0},1\leq i_{0}\leq k$ such that
$n_{i_{0}}=0$, then $\mathcal{H}_{i_{0}}=\\{0\\}$. This contradicts to
$\mathcal{H}_{i_{0}}$ being non-zero. Thus we infer that
$\Omega\subset\sigma(T_{i})$ and $T_{i}\in\mathcal{B}_{n_{i}}(\Omega)$, $1\leq
i\leq k$. ∎

######  Definition 3.2.

Let $T\in\mathcal{B}_{n}(\Omega)$, $T=(M^{*}_{z},\mathcal{H},K)$, where
$\mathcal{H}$ be the analytic function space with reproducing kernel $K$. And
$T$ has the following decomposition

$T=\bigoplus\limits_{i=1}^{t}T_{i}^{(m_{i})},\
where\,\,T_{i}\in\mathcal{B}_{n_{i}}(\Omega),\
\sum\limits_{i=1}^{t}m_{i}n_{i}=n.$

This decomposition is said to be (IR) decomposition, if $T_{i}\in(IR)$ and
$T_{i}\nsim_{u}T_{j}(i\neq j)$.

Therefore, from Proposition 3.1 it follows that there is a finite (IR)
decomposition for each Cowen-Douglas operator. The following two lemmas
characterize $\mathcal{A}^{\prime}(T,T^{*})$ of operator $T$.

######  Lemma 3.3.

[5] If $T$ is irreducible on ${\mathcal{H}}$ and there is
$X\in{\mathcal{L}}({\mathcal{H}})$ such that
$X\in\mathcal{A}^{\prime}(T,T^{*})$, then $X$ is a scalar multiple of
identity.

Based on the lemma above, we have the following lemma, which describes
$\mathcal{A}^{\prime}(T,T^{*})$ when $T$ is reducible.

######  Lemma 3.4.

If
$T=\bigoplus_{i=1}^{t}T_{i}^{(m_{i})},\,T_{i}\in{\mathcal{L}}({\mathcal{H}}_{i})$
satisfies Definition 3.2, and there exists $X\in{\mathcal{L}}({\mathcal{H}})$
such that $X\in\mathcal{A}^{\prime}(T,T^{*})$, then
$X=diag[X_{1},X_{2},\cdots,X_{t}]$, where
$X_{p}=(\\!(c^{p}_{i,j}I_{\mathcal{H}_{p}})\\!)_{i,j=1}^{m_{p}}$,
$c^{p}_{i,j}$ are scalars, $1\leq p\leq t$. In particular,
$\mathcal{A}^{\prime}(T,T^{*})\cong\sum\limits_{i=1}^{t}\bigoplus\mathcal{M}_{m_{i}}(\mathbb{C})$.

###### Proof.

For convenience, we write $T$ as the following form,
$T=\bigoplus\limits_{i=1}^{t}T_{i}^{(m_{i})}=\bigoplus\limits_{j=1}^{m}S_{j},\
\sum\limits_{i=1}^{t}m_{i}=m,$ where
$S_{1}=\cdots=S_{m_{1}}=T_{1},S_{m_{1}+1}=\cdots=S_{m_{1}+m_{2}}=T_{2},\cdots,S_{m_{1}+\cdots+m_{t-1}+1}=\cdots=S_{m}=T_{t}.$
Setting $X=(\\!(X_{i,j})\\!)_{i,j=1}^{m}$. By
$X\in\mathcal{A}^{\prime}(T,T^{*})$, we know that the size of $X_{i,j}$ is
determined by $S_{i}$ and $S_{j}$, and

(3.2) $X_{i,j}S_{j}=S_{i}X_{i,j},\ X_{i,j}S^{*}_{j}=S^{*}_{i}X_{i,j}.$

Next, combining these two formulas, we have

$X^{*}_{i,j}X_{i,j}S_{j}=X^{*}_{i,j}S_{i}X_{i,j}=S_{j}X^{*}_{i,j}X_{i,j},\
X^{*}_{i,j}X_{i,j}S^{*}_{j}=S^{*}_{j}X^{*}_{i,j}X_{i,j};$

and

$X_{i,j}X^{*}_{i,j}S^{*}_{i}=X_{i,j}S^{*}_{j}X^{*}_{i,j}=S^{*}_{i}X_{i,j}X^{*}_{i,j},\
X_{i,j}X^{*}_{i,j}S_{i}=S_{i}X_{i,j}X^{*}_{i,j}.$

Since $S_{i},S_{j}\in(IR)$, by Lemma 3.3, there exist scalars
$c_{i,j},d_{i,j}$ such that

(3.3) $X^{*}_{i,j}X_{i,j}=c_{i,j}I,\ X_{i,j}X^{*}_{i,j}=d_{i,j}I,\ 1\leq
i,j\leq m.$

Note that $X^{*}_{i,j}X_{i,j},X_{i,j}X^{*}_{i,j}$ are positive, we know that
$c_{i,j},d_{i,j}\geq 0$ for all $1\leq i,j\leq m$. From equation (3.3), we
notice that $X_{i,j}X^{*}_{i,j}X_{i,j}=d_{i,j}X_{i,j}=c_{i,j}X_{i,j},$ then
$c_{i,j}=d_{i,j}$ for all $1\leq i,j\leq m$.

We assert that $c_{i,j}=0$, when $S_{i}\neq S_{j}$. By Definition 3.2, we know
that $S_{i}$ and $S_{j}$ are either equal or not unitarily equivalent. That
is, $S_{i}\neq S_{j}$ means $S_{i}\nsim_{u}S_{j}$. Suppose $c_{i,j}\neq 0$,
then $X_{i,j}X^{*}_{i,j}$, $X^{*}_{i,j}X_{i,j}$ are invertible. Then equation
(3.3) is equivalent to
$\frac{X_{i,j}}{\sqrt{c_{i,j}}}\big{(}\frac{X_{i,j}}{\sqrt{c_{i,j}}}\big{)}^{*}=\big{(}\frac{X_{i,j}}{\sqrt{c_{i,j}}}\big{)}^{*}\frac{X_{i,j}}{\sqrt{c_{i,j}}}=I.$
It follows that $\frac{X_{i,j}}{\sqrt{c_{i,j}}}$ is a unitary operator. By
(3.2), we obtain $S_{i}\sim_{u}S_{j}$. It is a contradiction. Then the
assertion is true. In this case, it follows that $X^{*}_{i,j}X_{i,j}=0$. By
the polar decomposition of $X_{i,j}$, we have
$X_{i,j}=V(X^{*}_{i,j}X_{i,j})^{\frac{1}{2}}$ for some partial isometry $V$.
Thus, we have $X_{i,j}=0$.

In the case of $S_{i}=S_{j}$. By Lemma 3.3, there exist $b_{i,j}$ such that
$X_{i,j}=b_{i,j}I$ and $b_{i,j}\overline{b_{i,j}}=c_{i,j}$. Now, we know $X$
is quasi-diagonal. To ensure uniformity of notation, we write $X$ in the form:
$X=diag[X_{1},X_{2},\cdots,X_{t}]$, where
$X_{p}=(\\!(c^{p}_{i,j}I_{\mathcal{H}_{p}})\\!)_{i,j=1}^{m_{p}}$,
$p=1,2,\cdots,t$. Clearly, if $X$ satisfies the above conditions, then
$X\in\mathcal{A}^{\prime}(T,T^{*})$. Thus,
$\mathcal{A}^{\prime}(T,T^{*})\cong\sum\limits_{i=1}^{t}\bigoplus\mathcal{M}_{m_{i}}(\mathbb{C})$.
∎

A non negative definite kernel $K$ is said to be normalized at
$\lambda_{0}\in\Lambda$, if there exists a neighborhood $\Lambda_{0}$ of
$\lambda_{0}$ in $\Lambda$ such that $K(\mu,\lambda_{0})=1$ for all
$\mu\in\Lambda_{0}$ [2]. Inspired by Theorem 1.1 due to R. E. Curto and N.
Salinas, we have the following theorem.

######  Theorem 3.5.

Let $T,S\in\mathcal{B}_{n}(\Omega)$, and
$T=(M^{*}_{z},\mathcal{H}_{T},K_{T}),S=(M^{*}_{z},\mathcal{H}_{S},K_{S})$,
where $\mathcal{H}_{T},\mathcal{H}_{S}$ are the analytic function spaces of
reproducing kernels $K_{T},K_{S}$, respectively. Then $T\sim_{u}S$ if and only
if there is a holomorphic matrix-valued function $\Psi$ and $\Psi(\mu)$ is
invertible for every $\mu\in\Lambda$, a open subset of $\Omega^{*}$, such that

$\frac{\partial}{\partial\lambda}[K_{T}^{-1}(\lambda,\lambda)\Psi(\lambda)^{-1}K_{S}(\lambda,\lambda)]=0$

for all $\lambda\in\Lambda.$

###### Proof.

Necessity: If $T\sim_{u}S$, by Theorem 1.1, there is a holomorphic and
invertible matrix-valued function $\Psi$ such that
$K_{S}(\mu,\lambda)=\Psi(\mu)K_{T}(\mu,\lambda)\Psi^{*}(\lambda)$,
equivalently, we have

$\Psi^{*}(\lambda)=K_{T}^{-1}(\mu,\lambda)\Psi(\mu)^{-1}K_{S}(\mu,\lambda)$

for all $\mu,\lambda\in\Lambda$, a open subset of $\Omega^{*}$. Since
$\Psi^{*}(\lambda)$ is anti-holomorphic, we obtain

$\frac{\partial}{\partial\lambda}[K_{T}^{-1}(\lambda,\lambda)\Psi(\lambda)^{-1}K_{S}(\lambda,\lambda)]=0,\
\lambda\in\Lambda.$

Sufficiency: If
$\frac{\partial}{\partial\lambda}[K_{T}^{-1}(\lambda,\lambda)\Psi(\lambda)^{-1}K_{S}(\lambda,\lambda)]=0,\
\lambda\in\Lambda$, then we know that it is anti-holomorphic for the variable
$\lambda$. So we can find a holomorphic matrix-valued invertible function
$\Phi$ such that

$\Phi^{*}(\lambda)=K_{T}^{-1}(\lambda,\lambda)\Psi(\lambda)^{-1}K_{S}(\lambda,\lambda),\
\lambda\in\Lambda,$

that is,
$K_{S}(\lambda,\lambda)=\Psi(\lambda)K_{T}(\lambda,\lambda)\Phi^{*}(\lambda),\
\lambda\in\Lambda$. By Proposition 1 in [13], we have

(3.4) $K_{S}(\mu,\lambda)=\Psi(\mu)K_{T}(\mu,\lambda)\Phi^{*}(\lambda),\
\mu,\lambda\in\Lambda.$

Claim: If there are $\mathcal{M}_{n}(\mathbb{C})-$ valued holomorphic and
invertible functions $\Phi,\Psi$ such that equation (3.4) is valid, then
$T\sim_{u}S$.

Conjugating equation (3.4) and then interchanging the variables $\mu,\lambda$,
one arrives at the equation
$K_{S}(\mu,\lambda)=\Phi(\mu)K_{T}(\mu,\lambda)\Psi^{*}(\lambda),\
\mu,\lambda\in\Lambda$. It follows that
$\Psi(\mu)K_{T}(\mu,\lambda)\Phi^{*}(\lambda)=\Phi(\mu)K_{T}(\mu,\lambda)\Psi^{*}(\lambda)$,
$\mu,\lambda\in\Lambda.$ That means
$\Phi^{-1}(\mu)\Psi(\mu)K_{T}(\mu,\lambda)=K_{T}(\mu,\lambda)(\Phi^{-1}(\lambda)\Psi(\lambda))^{*},\
\mu,\lambda\in\Lambda.$ Setting $\mathcal{X}=\Phi^{-1}\Psi$. We have

(3.5)
$\mathcal{X}(\mu)K_{T}(\mu,\lambda)=K_{T}(\mu,\lambda)\mathcal{X}^{*}(\lambda),\
\mu,\lambda\in\Lambda.$

Case 1: Suppose $T$ is irreducible. Let

(3.6)
$K_{0}(\mu,\lambda):=K_{T}^{\frac{1}{2}}(\lambda_{0},\lambda_{0})K_{T}^{-1}(\mu,\lambda_{0})K_{T}(\mu,\lambda)(K_{T}^{-1}(\lambda,\lambda_{0}))^{*}K_{T}^{\frac{1}{2}}(\lambda_{0},\lambda_{0}),\
\mu,\lambda,\lambda_{0}\in\Lambda,$

It’s easy to see that $K_{0}(\mu,\lambda)$ is normalized at $\lambda_{0}$.
Suppose that $\widetilde{M}_{z}$ is a multiplication operator on the Hilbert
space $\mathcal{H}_{K_{0}}$ with a reproducing kernel $K_{0}$. Then

(3.7)
$\mathcal{H}_{K_{0}}=\bigvee_{\lambda{\in}{\Lambda}^{*}}\ker(\widetilde{M}_{z}^{*}-\lambda)=\bigvee_{\lambda{\in}{\Lambda}^{*}}\\{K_{0}(\cdot,\lambda)\xi,\
\xi\in\mathbb{C}^{n}\\}.$

From equation (3.5), we can imply

(3.8)
$\mathcal{Y}(\mu)K_{0}(\mu,\lambda)=K_{0}(\mu,\lambda)\mathcal{Y}^{*}(\lambda),\
\mu,\lambda\in\Lambda,$

where

(3.9)
$\mathcal{Y}(\mu)=K_{T}^{\frac{1}{2}}(\lambda_{0},\lambda_{0})K_{T}^{-1}(\mu,\lambda_{0})\mathcal{X}(\mu)K_{T}(\mu,\lambda_{0})K_{T}^{-\frac{1}{2}}(\lambda_{0},\lambda_{0}).$

We can see that $\mathcal{Y}$ is a holomorphic matrix. In equation (3.8),
setting $\mu=\lambda=\lambda_{0}$, we come to
$\mathcal{Y}(\lambda_{0})=\mathcal{Y}^{*}(\lambda_{0})$. This implies that
$\mathcal{Y}(\lambda_{0})$ is a constant matrix. For any $\mu\in\Lambda$,
setting $\lambda=\lambda_{0}$, we also come to
$\mathcal{Y}(\mu)=\mathcal{Y}^{*}(\lambda_{0})$. Thus, $\mathcal{Y}(\cdot)$ is
a constant matrix. We define an operator
$Y:\mathcal{H}_{K_{0}}\rightarrow\mathcal{H}_{K_{0}}$ as

(3.10) $(Yf)(\cdot)=\mathcal{Y}f(\cdot)\ \ \mbox{for all}\ \
f\in\mathcal{H}_{K_{0}}.$

We can see that
$\widetilde{M}_{z}Yf(z)=\widetilde{M}_{z}\mathcal{Y}f(z)=z\mathcal{Y}f(z)=\mathcal{Y}(zf(z))=\mathcal{Y}\widetilde{M}_{z}(f(z))=Y\widetilde{M}_{z}f(z),$
that is,

(3.11) $\widetilde{M}_{z}Y=Y\widetilde{M}_{z}.$

For equation (3.10), in particular, we have

(3.12) $YK_{0}(\cdot,\lambda)\xi=\mathcal{Y}K_{0}(\cdot,\lambda)\xi$

for all $\lambda\in\Lambda$ and $\xi\in\mathbb{C}^{n}$. We will calculate the
adjoint of the operator $Y$. We have

$\begin{array}[]{lll}\langle
YK_{0}(\cdot,\mu)\eta,K_{0}(\cdot,\lambda)\xi\rangle&=&\langle\mathcal{Y}K_{0}(\cdot,\mu)\eta,K_{0}(\cdot,\lambda)\xi\rangle=\langle\mathcal{Y}K_{0}(\lambda,\mu)\eta,\xi\rangle\\\
&=&\langle K_{0}(\lambda,\mu)\eta,\mathcal{Y}\xi\rangle=\langle
K_{0}(\cdot,\mu)\eta,K_{0}(\cdot,\lambda)\mathcal{Y}\xi\rangle\end{array}$

for all $\lambda,\mu\in\Lambda$ and $\xi,\eta\in\mathbb{C}^{n}$. In other
words,

(3.13) $Y^{*}K_{0}(\cdot,\lambda)\xi=K_{0}(\cdot,\lambda)\mathcal{Y}\xi$

for all $\lambda\in\Lambda$ and $\xi\in\mathbb{C}^{n}$ (see [9] for more
details).

Then, by equations (3.7),(3.8),(3.12) and (3.13), we obtain that $Y$ is a
self-adjoint operator. Combining with (3.11), we can imply
$Y\in\mathcal{A}^{\prime}(\widetilde{M}_{z},\widetilde{M}_{z}^{*})$. In the
special case of Theorem 8 in [4], we can see $Y$ is bounded. Note that
$K_{T}^{\frac{1}{2}}(\lambda_{0},\lambda_{0})K_{T}^{-1}(\cdot,\lambda_{0})$ is
holomorphic and non-zero on $\Lambda$ in equation (3.6), it follows that
$M_{z}^{*}$ defined on $(\mathcal{H}_{T},K_{T})$ is unitarily equivalent to
$\widetilde{M}_{z}$ on $(\mathcal{H}_{0},K_{0})$ by Theorem 1.1. Then we have
$\widetilde{M}_{z}\in(IR)$, since unitary transformation doesn’t change the
irreducibility of operators. According to Lemma 3.3 and $Y$ is self-adjoint,
we know that $Y=cI,c\in\mathbb{R}$, $I$ is the identity of $\mathcal{H}_{0}$.
Thus, from equations (3.9) and (3.10), we have $\mathcal{X}=cI_{n\times
n},c\in\mathbb{R}$. Since $\mathcal{X}=\Phi^{-1}\Psi$, we obtain
$K_{S}(\mu,\lambda)=c\Phi(\mu)K_{T}(\mu,\lambda)\Phi(\lambda)$. From the
positive definiteness of $K_{S}(\mu,\lambda)$ and
$\Phi(\mu)K_{T}(\mu,\lambda)\Phi(\lambda)$, we can get $c>0$ and
$K_{S}(\mu,\lambda)=(\sqrt{c}\Phi(\mu))K_{T}(\mu,\lambda)(\sqrt{c}\Phi(\lambda)).$
Hence, $T\sim_{u}S$.

Case 2: Suppose $T$ is reducible. Without losing generality, we suppose that

$T=\bigoplus\limits_{i=1}^{t}T_{i}^{(m_{i})}\ and\,\,T_{i}\nsim_{u}T_{j}(i\neq
j),$

where $\mathcal{H}_{i}$ are the analytic function spaces of reproducing
kernels $K_{i}$, respectively. We write $T$ as $T=\bigoplus_{i=1}^{m}S_{i}$
and $S_{i}=(M^{*}_{z},\mathcal{H}_{i},K_{i})$, where

$S_{1}=\cdots=S_{m_{1}}=T_{1},S_{m_{1}+1}=\cdots=S_{m_{1}+m_{2}}=T_{2},\cdots,S_{m_{1}+\cdots+m_{t-1}+1}=\cdots=S_{m}=T_{t}.$

Setting $\mathcal{X}=(\\!(\mathcal{X}_{i,j})\\!)_{i,j=1}^{m}$. Then
$\mathcal{X}(\mu)K_{T}(\mu,\lambda)=K_{T}(\mu,\lambda)\mathcal{X}^{*}(\lambda),\
\mu,\lambda\in\Lambda$ is equivalent to

(3.14)
$\mathcal{X}_{i,j}(\mu)K_{j}(\mu,\lambda)=K_{i}(\mu,\lambda)\mathcal{X}_{j,i}^{*}(\lambda),\
\mathcal{X}_{j,i}(\mu)K_{i}(\mu,\lambda)=K_{j}(\mu,\lambda)\mathcal{X}_{i,j}^{*}(\lambda).$

It follows that

(3.15)
$\begin{array}[]{lll}&&\mathcal{X}_{j,i}(\mu)\mathcal{X}_{i,j}(\mu)K_{j}(\mu,\lambda)=K_{j}(\mu,\lambda)(\mathcal{X}_{j,i}(\lambda)\mathcal{X}_{i,j}(\lambda))^{*},\\\
&&\mathcal{X}_{i,j}(\mu)\mathcal{X}_{j,i}(\mu)K_{i}(\mu,\lambda)=K_{i}(\mu,\lambda)(\mathcal{X}_{i,j}(\lambda)\mathcal{X}_{j,i}(\lambda))^{*}\end{array}$

for all $\mu,\lambda\in\Lambda$. By Case 1, we have
$\mathcal{X}_{j,i}\mathcal{X}_{i,j}=c_{j,i}I,\
\mathcal{X}_{i,j}\mathcal{X}_{j,i}=c_{i,j}I$ and
$c_{j,i}=c_{i,j}\in\mathbb{R}$.

If there exist $i,j$, such that $c_{i,j}\neq 0$, then
$\mathcal{X}_{i,j},\mathcal{X}_{j,i}$ are invertible. By equation (3.14), we
have
$K_{j}(\mu,\lambda)=\frac{1}{c_{i,j}}\mathcal{X}_{j,i}(\mu)K_{i}(\mu,\lambda)\mathcal{X}_{j,i}^{*}(\lambda)$
and $c_{i,j}>0$. Therefore,
$K_{j}(\mu,\lambda)=\frac{\mathcal{X}_{j,i}(\mu)}{\sqrt{c_{i,j}}}K_{i}(\mu,\lambda)(\frac{\mathcal{X}_{j,i}(\lambda)}{\sqrt{c_{i,j}}})^{*},\
\mu,\lambda\in\Lambda.$ From Theorem 1.1, we obtain $S_{i}\sim_{u}S_{j}$, that
is, $S_{i}=S_{j}$, since $S_{i}$ and $S_{j}$ are either equal or not unitarily
equivalent. In the following, we define an operator
$U_{ji}:\mathcal{H}_{i}\rightarrow\mathcal{H}_{j}$,
$U_{ji}f(\cdot)=\frac{\mathcal{X}_{j,i}(\cdot)}{\sqrt{c_{i,j}}}f(\cdot)$ for
all $f\in\mathcal{H}_{i}$. By Lemma 3.9 of [2], we know $U_{ji}$ is a unitary
transformation. From the definition of $U_{ji}$, we can also see
$U_{ji}M_{z}=M_{z}U_{ji}$. It follows that $U_{ji}M_{z}^{*}=M_{z}^{*}U_{ji}$,
where $M_{z}$ is the multiplication operator on Hilbert space
$\mathcal{H}_{i}$ with reproducing kernels $K_{i}$. According to Lemma 3.3, we
have that $U_{ji}$ is $e^{i\theta_{ji}}$ multiple of identity,
$\theta_{ji}\in\mathbb{R}$. Then
$\mathcal{X}_{j,i}=\sqrt{c_{i,j}}e^{i\theta_{ji}}I$. Similarly, we can find
unitary $U_{ij}=e^{i\theta_{ij}}I$ such that
$\mathcal{X}_{i,j}=\sqrt{c_{i,j}}e^{i\theta_{ij}}I$. By equation (3.14), we
have $\mathcal{X}_{j,i}=\mathcal{X}_{i,j}^{*}$.

Based on the above analysis, we know that if $S_{i}\nsim_{u}S_{j}$, then
$c_{i,j}=c_{j,i}=0$. Notice that $K_{i}(\lambda,\lambda)$ is positive definite
and invertible, therefore, there exists $H_{i}$ such that
$K_{i}(\lambda,\lambda)=H_{i}(\lambda)H_{i}^{*}(\lambda),\ \lambda\in\Lambda$.
For equations (3.14) and (3.15), setting $\mu:=\lambda$, we have

$0=\mathcal{X}_{j,i}(\lambda)\mathcal{X}_{i,j}(\lambda)K_{j}(\lambda,\lambda)=\mathcal{X}_{j,i}(\lambda)K_{i}(\lambda,\lambda)\mathcal{X}^{*}_{j,i}(\lambda)=(\mathcal{X}_{j,i}(\lambda)H_{i}(\lambda))(\mathcal{X}_{j,i}(\lambda)H_{i}(\lambda))^{*},\
\lambda\in\Lambda.$

That is, $\mathcal{X}_{j,i}(\lambda)=0$ for all $\lambda\in\Lambda$. Hence,
$\mathcal{X}$ is symmetric, quasi-diagonal and its elements are scalars.

By the characteristics of $\mathcal{X}$ discussed above and
$\Psi=\Phi\mathcal{X}$, we have

$K_{T}(\mu,\lambda)=\Psi^{-1}(\mu)K_{S}(\mu,\lambda)(\Phi^{-1}(\lambda))^{*}=\mathcal{X}^{-1}\Phi^{-1}(\mu)K_{S}(\mu,\lambda)(\Phi^{-1}(\lambda))^{*}.$

Conjugating the last equation and then interchanging the variables
$\mu,\lambda$, one arrives at the equation
$K_{T}(\mu,\lambda)=\Phi^{-1}(\mu)K_{S}(\mu,\lambda)(\Phi^{-1}(\lambda))^{*}\mathcal{X}^{-1}$.
It follows that $\mathcal{X}^{-1}$ is positive definite. By functional
calculus that (for any positive matrix $A$, $A^{\frac{1}{2}}$ can be
approximated in norm by polynomials about $A$)

$\mathcal{X}^{-\frac{1}{2}}\Phi^{-1}(\mu)K_{S}(\mu,\lambda)(\Phi^{-1}(\lambda))^{*}=\Phi^{-1}(\mu)K_{S}(\mu,\lambda)(\Phi^{-1}(\lambda))^{*}\mathcal{X}^{-\frac{1}{2}}.$

Hence, we can imply
$K_{T}(\mu,\lambda)=(\mathcal{X}^{-\frac{1}{2}}\Phi^{-1}(\mu))K_{S}(\mu,\lambda)(\mathcal{X}^{-\frac{1}{2}}\Phi^{-1}(\lambda))^{*}$.
By Theorem 1.1 again, we have $T\sim_{u}S$. ∎

The following corollary is a direct consequence of Theorem 3.5.

######  Corollary 3.6.

Let $T,S\in\mathcal{B}_{n}(\Omega)$, and
$T=(M^{*}_{z},\mathcal{H}_{T},K_{T}),S=(M^{*}_{z},\mathcal{H}_{S},K_{S})$,
where $\mathcal{H}_{T},\mathcal{H}_{S}$ are the analytic function spaces of
reproducing kernels $K_{T},K_{S}$, respectively. Then $T\sim_{u}S$ if and only
if there is a holomorphic and invertible matrix-valued function $\Phi$ on
$\Lambda$, a open subset of $\Omega^{*}$, such that connection matrices
$\Theta_{T},\Theta_{S}$ of $E_{T},E_{S}$ satisfy
$\Theta_{T}=\Phi^{-1}\Theta_{S}\Phi+\Phi^{-1}\Phi^{\prime}.$

###### Proof.

By Theorem 3.5, we know that $T\sim_{u}S$ if and only if
$\frac{\partial}{\partial\lambda}[K_{T}^{-1}(\lambda,\lambda)\Psi(\lambda)^{-1}K_{S}(\lambda,\lambda)]=0$
for some holomorphic matrix-valued function $\Psi$ and all
$\lambda\in\Lambda$, a open subset of $\Omega^{*}$. And, if we take the
partial derivative of both sides of equation

$K_{T}(\lambda,\lambda)[K_{T}^{-1}(\lambda,\lambda)\Psi^{-1}(\lambda)K_{S}(\lambda,\lambda)]=\Psi^{-1}(\lambda)K_{S}(\lambda,\lambda),\
\lambda\in\Lambda,$

then
$\frac{\partial}{\partial\lambda}K_{T}(\lambda,\lambda)[K_{T}^{-1}(\lambda,\lambda)\Psi^{-1}(\lambda)K_{S}(\lambda,\lambda)]=\frac{\partial}{\partial\lambda}(\Psi^{-1}(\lambda)K_{S}(\lambda,\lambda))$.
It follows that

$\begin{array}[]{lll}\frac{\partial}{\partial\lambda}K_{T}(\lambda,\lambda)K_{T}^{-1}(\lambda,\lambda)&=&\frac{\partial}{\partial\lambda}(\Psi^{-1}(\lambda)K_{S}(\lambda,\lambda))[\Psi^{-1}(\lambda)K_{S}(\lambda,\lambda)]^{-1}\\\
&=&\Psi^{-1}(\lambda)\frac{\partial}{\partial\lambda}K_{S}(\lambda,\lambda)K_{S}^{-1}(\lambda,\lambda)\Psi(\lambda)+(\Psi^{-1}(\lambda))^{\prime}\Psi(\lambda)\end{array}$

for all $\lambda\in\Lambda.$ Since
$h_{T}(\lambda)=K_{T}(\overline{\lambda},\overline{\lambda}),\
h_{S}(\lambda)=K_{S}(\overline{\lambda},\overline{\lambda})$, we also have

$\frac{\partial}{\partial\overline{\lambda}}h_{T}(\lambda)h_{T}^{-1}(\lambda)=\overline{\Psi(\lambda)}^{-1}\frac{\partial}{\partial\overline{\lambda}}h_{S}(\lambda)h_{S}^{-1}(\lambda)\overline{\Psi(\lambda)}+(\overline{\Psi(\lambda)}^{-1})^{\prime}\overline{\Psi(\lambda)},\lambda\in\Lambda^{*}\subset\Omega.$

Conjugating the last equation, we obtain

$h_{T}^{-1}(\lambda)\frac{\partial}{\partial\lambda}h_{T}(\lambda)=\Psi^{T}(\lambda)h_{S}^{-1}(\lambda)\frac{\partial}{\partial\lambda}h_{S}(\lambda)(\Psi^{T}(\lambda))^{-1}+\Psi^{T}(\lambda)((\Psi^{T}(\lambda))^{-1})^{\prime},\
\lambda\in\Lambda^{*}.$

Setting $\Phi:=(\Psi^{T})^{-1}$. We can imply that

$h_{T}^{-1}(\lambda)\frac{\partial}{\partial\lambda}h_{T}(\lambda)=\Phi^{-1}(\lambda)h_{S}^{-1}(\lambda)\frac{\partial}{\partial\lambda}h_{S}(\lambda)\Phi(\lambda)+\Phi^{-1}(\lambda)\Phi^{\prime}(\lambda),\
\lambda\in\Lambda^{*},$

that is, $\Theta_{T}=\Phi^{-1}\Theta_{S}\Phi+\Phi^{-1}\Phi^{\prime}.$ ∎

From the proof of Theorem 3.5, when changing equation (1.1) into equation
(3.4), we can also obtain $T$ is unitarily equivalent to $S$. In the case of
equation (3.4), it can be seen from the following proposition that the
corresponding curvatures and covariant partial derivatives of two bundles are
also similar. Furthermore, the intertwining matrix is determined only by
$\Phi$.

######  Proposition 3.7.

Let $T,S\in\mathcal{B}_{n}(\Omega)$, and
$T=(M^{*}_{z},\mathcal{H}_{T},K_{T}),S=(M^{*}_{z},\mathcal{H}_{S},K_{S})$,
where $\mathcal{H}_{T},\mathcal{H}_{S}$ are the analytic function spaces of
reproducing kernels $K_{T},K_{S}$, respectively. If there exist
$\mathcal{M}_{n}(\mathbb{C})-$ valued holomorphic functions $\Phi,\Psi$ and
$\Phi(\mu),\Psi(\mu)$ are invertible for every $\mu\in\Omega^{*}$, such that

(3.16) $K_{S}(\mu,\lambda)=\Psi(\mu)K_{T}(\mu,\lambda)\Phi^{*}(\lambda),\
\mu,\lambda\in\Omega^{*},$

then
$\Phi^{T}\mathcal{K}_{S,\lambda^{i}\overline{\lambda}^{j}}=\mathcal{K}_{T,\lambda^{i}\overline{\lambda}^{j}}\Phi^{T},\
i,j\geq 1.$

###### Proof.

From formula (3.16), we can imply that

$\begin{array}[]{lll}h_{S}(\lambda)&=&K_{S}(\overline{\lambda},\overline{\lambda})\\\
&=&\Psi(\overline{\lambda})K_{T}(\overline{\lambda},\overline{\lambda})\overline{\Phi^{T}(\overline{\lambda})}\\\
&=&\Psi(\overline{\lambda})h_{T}(\lambda)\Phi^{T}(\lambda),\end{array}$

where $h_{T},h_{S}$ are the metrics of $E_{T},E_{S}$, respectively. Since
$\Psi$ and $\Phi$ are analytic invertible matrixes, we have the following
equations

$\frac{\partial}{\partial\overline{\lambda}}\Phi^{T}(\lambda)=\frac{\partial}{\partial\overline{\lambda}}(\Phi^{T})^{-1}(\lambda)=\frac{\partial}{\partial\overline{\lambda}}(\Phi^{T})^{\prime}(\lambda)=0,\
\ \frac{\partial}{\partial\lambda}\Psi(\overline{\lambda})=0.$

The curvatures corresponding to $h_{T}$ and $h_{S}$ are $\mathcal{K}_{T}$ and
$\mathcal{K}_{S}$, respectively. Thus, we have that

$\begin{array}[]{lll}\mathcal{K}_{S}&=&-\frac{\partial}{\partial\overline{\lambda}}(h^{-1}_{S}\frac{\partial}{\partial\lambda}h_{S})\\\
&=&-\frac{\partial}{\partial\overline{\lambda}}[(\Phi^{T})^{-1}h^{-1}_{T}\overline{\Psi}^{-1}\frac{\partial}{\partial\lambda}(\overline{\Psi}h_{T}\Phi^{T})]\\\
&=&-\frac{\partial}{\partial\overline{\lambda}}[(\Phi^{T})^{-1}h^{-1}_{T}((\frac{\partial}{\partial\lambda}h_{T})\Phi^{T}+h_{T}(\Phi^{T})^{{}^{\prime}})]\\\
&=&-\frac{\partial}{\partial\overline{\lambda}}[(\Phi^{T})^{-1}h^{-1}_{T}(\frac{\partial}{\partial\lambda}h_{T})\Phi^{T}+(\Phi^{T})^{-1}(\Phi^{T})^{{}^{\prime}}]\\\
&=&-\frac{\partial}{\partial\overline{\lambda}}((\Phi^{T})^{-1}h^{-1}_{T}(\frac{\partial}{\partial\lambda}h_{T})\Phi^{T})\\\
&=&-(\Phi^{T})^{-1}\frac{\partial}{\partial\overline{\lambda}}(h^{-1}_{T}\frac{\partial}{\partial\lambda}h_{T})\Phi^{T}\\\
&=&(\Phi^{T})^{-1}\mathcal{K}_{T}\Phi^{T}.\end{array}$

Using the definition of the covariant partial derivatives of curvature, we
obtain that

$\mathcal{K}_{S,\overline{\lambda}}=\frac{\partial}{\partial\overline{\lambda}}\mathcal{K}_{S}=\frac{\partial}{\partial\overline{\lambda}}((\Phi^{T})^{-1}\mathcal{K}_{T}\Phi^{T})=(\Phi^{T})^{-1}(\frac{\partial}{\partial\overline{\lambda}}\mathcal{K}_{T})\Phi^{T}=(\Phi^{T})^{-1}\mathcal{K}_{T,\overline{\lambda}}\Phi^{T}.$

By repeating the above operation, we obtain

$\mathcal{K}_{S,\overline{\lambda}^{2}}=(\Phi^{T})^{-1}\mathcal{K}_{T,\overline{\lambda}^{2}}\Phi^{T},\
\mathcal{K}_{S,\overline{\lambda}^{3}}=(\Phi^{T})^{-1}\mathcal{K}_{T,\overline{\lambda}^{3}}\Phi^{T},\cdots$

It follows that

(3.17)
$\mathcal{K}_{S,\overline{\lambda}^{j}}=(\Phi^{T})^{-1}\mathcal{K}_{T,\overline{\lambda}^{j}}\Phi^{T},\
j\geq 1.$

Considering $(\Phi^{T})^{-1}\Phi^{T}=I$, we have
$((\Phi^{T})^{-1}\Phi^{T})^{\prime}=0$, that is,
$\widetilde{\Phi}\Phi^{T}+(\Phi^{T})^{-1}(\Phi^{T})^{\prime}=0$ and

(3.18) $\widetilde{\Phi}+(\Phi^{T})^{-1}(\Phi^{T})^{\prime}(\Phi^{T})^{-1}=0,$

where $\widetilde{\Phi}=((\Phi^{T})^{-1})^{\prime}$. Next, we will prove the
following formula

(3.19)
$\mathcal{K}_{S,\lambda^{i}}=(\Phi^{T})^{-1}\mathcal{K}_{T,\lambda^{i}}\Phi^{T},\
i\geq 1.$

When $i=1$, from the equation (3.18) and the definition of the covariant
partial derivatives of curvature again, we have

$\begin{array}[]{lll}\mathcal{K}_{S,\lambda}&=&\frac{\partial}{\partial\lambda}\mathcal{K}_{S}+h^{-1}_{S}(\frac{\partial}{\partial\lambda}h_{S})\mathcal{K}_{S}-\mathcal{K}_{S}h^{-1}_{S}\frac{\partial}{\partial\lambda}h_{S}\\\
&=&\frac{\partial}{\partial\lambda}((\Phi^{T})^{-1}\mathcal{K}_{T}\Phi^{T})+(\overline{\Psi}h_{T}\Phi^{T})^{-1}\frac{\partial}{\partial\lambda}(\overline{\Psi}h_{T}\Phi^{T})((\Phi^{T})^{-1}\mathcal{K}_{T}\Phi^{T})\\\
&&-((\Phi^{T})^{-1}\mathcal{K}_{T}\Phi^{T})(\overline{\Psi}h_{T}\Phi^{T})^{-1}\frac{\partial}{\partial\lambda}(\overline{\Psi}h_{T}\Phi^{T})\\\
&=&\frac{\partial}{\partial\lambda}((\Phi^{T})^{-1}\mathcal{K}_{T}\Phi^{T})+(\Phi^{T})^{-1}h_{T}^{-1}\frac{\partial}{\partial\lambda}(h_{T}\Phi^{T})((\Phi^{T})^{-1}\mathcal{K}_{T}\Phi^{T})-((\Phi^{T})^{-1}\mathcal{K}_{T})h_{T}^{-1}\frac{\partial}{\partial\lambda}(h_{T}\Phi^{T})\\\
&=&[\widetilde{\Phi}+(\Phi^{T})^{-1}(\Phi^{T})^{\prime}(\Phi^{T})^{-1}]\mathcal{K}_{T}\Phi^{T}+(\Phi^{T})^{-1}[\frac{\partial}{\partial\lambda}\mathcal{K}_{T}+h_{T}^{-1}(\frac{\partial}{\partial\lambda}h_{T})\mathcal{K}_{T}-\mathcal{K}_{T}h_{T}^{-1}\frac{\partial}{\partial\lambda}h_{T}]\Phi^{T}\\\
&=&(\Phi^{T})^{-1}\mathcal{K}_{T,\lambda}\Phi^{T}.\end{array}$

We then complete the proof of the formula (3.19) by induction. Suppose that it
is valid for any $i\leq k$. For $i=k+1$, with the assumptions and the equation
(3.18), we have

$\begin{array}[]{lll}\mathcal{K}_{S,\lambda^{k+1}}&=&\frac{\partial}{\partial\lambda}\mathcal{K}_{S,\lambda^{k}}+h^{-1}_{S}(\frac{\partial}{\partial\lambda}h_{S})\mathcal{K}_{S,\lambda^{k}}-\mathcal{K}_{S,\lambda^{k}}h^{-1}_{S}\frac{\partial}{\partial\lambda}h_{S}\\\
&=&\frac{\partial}{\partial\lambda}((\Phi^{T})^{-1}\mathcal{K}_{T,\lambda^{k}}\Phi^{T})+(\overline{\Psi}h_{T}\Phi^{T})^{-1}\frac{\partial}{\partial\lambda}(\overline{\Psi}h_{T}\Phi^{T})((\Phi^{T})^{-1}\mathcal{K}_{T,\lambda^{k}}\Phi^{T})\\\
&&-((\Phi^{T})^{-1}\mathcal{K}_{T,\lambda^{k}}\Phi^{T})(\overline{\Psi}h_{T}\Phi^{T})^{-1}\frac{\partial}{\partial\lambda}(\overline{\Psi}h_{T}\Phi^{T})\\\
&=&\frac{\partial}{\partial\lambda}((\Phi^{T})^{-1}\mathcal{K}_{T,\lambda^{k}}\Phi^{T})+(\Phi^{T})^{-1}h_{T}^{-1}\frac{\partial}{\partial\lambda}(h_{T}\Phi^{T})((\Phi^{T})^{-1}\mathcal{K}_{T,\lambda^{k}}\Phi^{T})-((\Phi^{T})^{-1}\mathcal{K}_{T,\lambda^{k}})h_{T}^{-1}\frac{\partial}{\partial\lambda}(h_{T}\Phi^{T})\\\
&=&[\widetilde{\Phi}+(\Phi^{T})^{-1}(\Phi^{T})^{\prime}(\Phi^{T})^{-1}]\mathcal{K}_{T,\lambda^{k}}\Phi^{T}+(\Phi^{T})^{-1}[\frac{\partial}{\partial\lambda}\mathcal{K}_{T,\lambda^{k}}+h_{T}^{-1}(\frac{\partial}{\partial\lambda}h_{T})\mathcal{K}_{T,\lambda^{k}}-\mathcal{K}_{T,\lambda^{k}}h_{T}^{-1}\frac{\partial}{\partial\lambda}h_{T}]\Phi^{T}\\\
&=&(\Phi^{T})^{-1}\mathcal{K}_{T,\lambda^{k+1}}\Phi^{T}.\end{array}$

Hence $i=k+1$ holds as well. That is, equation (3.19) is true.

Lastly, by equations (3.17) and (3.19), we obtain

(3.20)
$\begin{array}[]{lll}\mathcal{K}_{S,\lambda^{i}\overline{\lambda}^{j}}&=&\frac{\partial^{j}}{\partial\overline{\lambda}^{j}}\mathcal{K}_{S,\lambda^{i}}\\\
&=&\frac{\partial^{j}}{\partial\overline{\lambda}^{j}}((\Phi^{T})^{-1}\mathcal{K}_{T,\lambda^{i}}\Phi^{T})\\\
&=&(\Phi^{T})^{-1}\frac{\partial^{j}}{\partial\overline{\lambda}^{j}}\mathcal{K}_{T,\lambda^{i}}\Phi^{T}\\\
&=&(\Phi^{T})^{-1}\mathcal{K}_{T,\lambda^{i}\overline{\lambda}^{j}}\Phi^{T},\
i,j\geq 1.\end{array}$

That is,
$\Phi^{T}\mathcal{K}_{S,\lambda^{i}\overline{\lambda}^{j}}=\mathcal{K}_{T,\lambda^{i}\overline{\lambda}^{j}}\Phi^{T}$
for all $i,j\geq 1$. ∎

######  Remark 3.8.

Based on the observations of Proposition 3.7, the following question is
natural. If the curvatures of the two Cowen-Douglas operators and their
covariant partial derivatives satisfy equation (3.20), then does it follow
that these two operators are unitarily equivalent?

##  4\. The irreducible decompositions of Cowen-Douglas operators

In this section, we mainly discuss the uniqueness of the direct sum
decomposition into irreducible components, modulo a permutation, of an
operator in the Cowen-Douglas class.

######  Definition 4.1.

Let $T\in\mathcal{B}_{n}(\Omega)$, $T=(M^{*}_{z},\mathcal{H},K)$, where
$\mathcal{H}$ be the analytic function space with reproducing kernel $K$.
Suppose that $T$ has the following (IR) decomposition

$T=\bigoplus\limits_{i=1}^{t}T_{i}^{(m_{i})},\
where\,\,T_{i}\in\mathcal{B}_{n_{i}}(\Omega).$

Operator $T$ is said to have a unique (IR) decomposition up to unitary
equivalence, if for any $\widetilde{T}\sim_{u}T$ with the (IR) decomposition

$\widetilde{T}=\bigoplus\limits_{i=1}^{s}\widetilde{T}_{i}^{(l_{i})},\
where\,\,\widetilde{T}_{i}\in\mathcal{B}_{h_{i}}(\Omega),$

then following statements hold:

  * (1)

$t=s$;

  * (2)

there is a permutation $\pi$ on $\\{1,2,\cdots,t\\}$, such that
$T_{i}\sim_{u}\widetilde{T}_{\pi(i)}$ and $m_{i}=l_{\pi(i)},\ 1\leq i\leq t$.

In this section, we will prove the following theorem.

######  Theorem 4.2.

Let $T\in\mathcal{B}_{n}(\Omega)$. $T$ has a unique (IR) decomposition up to
unitary equivalence.

Before proving this theorem, we need to do some preliminary work.

######  Lemma 4.3.

Let $T,\widetilde{T}\in{\mathcal{L}}({\mathcal{H}})$. If
$\widetilde{T}=U^{*}TU$ for some unitary $U$, then
$\mathcal{A}^{\prime}(T,T^{*})=U\mathcal{A}^{\prime}(\widetilde{T},\widetilde{T}^{*})U^{*}$.

###### Proof.

If $\widetilde{T}=U^{*}TU$ for some unitary $U$ and
$X\in\mathcal{A}^{\prime}(T)$, then we obtain the following

$U^{*}TUU^{*}XU=U^{*}XUU^{*}TU,\ \widetilde{T}U^{*}XU=U^{*}XU\widetilde{T}.$

Thus, $U^{*}XU\in\mathcal{A}^{\prime}(\widetilde{T})$. By a similar
calculation, we have that if $X\in\mathcal{A}^{\prime}(T^{*})$, then
$U^{*}XU\in\mathcal{A}^{\prime}(\widetilde{T}^{*})$. That is to say,
$U^{*}\mathcal{A}^{\prime}(T,T^{*})U\subset\mathcal{A}^{\prime}(\widetilde{T},\widetilde{T}^{*})$.
Similarly, we have
$U\mathcal{A}^{\prime}(\widetilde{T},\widetilde{T}^{*})U^{*}\subset\mathcal{A}^{\prime}(T,T^{*})$.
It follows that $\mathcal{A}^{\prime}(\widetilde{T},\widetilde{T}^{*})\subset
U^{*}\mathcal{A}^{\prime}(T,T^{*})U$ and
$\mathcal{A}^{\prime}(T,T^{*})=U\mathcal{A}^{\prime}(\widetilde{T},\widetilde{T}^{*})U^{*}$.
∎

######  Remark 4.4.

Suppose that $T=\bigoplus_{i=1}^{t}T_{i}^{(m_{i})},\
\widetilde{T}=\bigoplus_{i=1}^{s}\widetilde{T}_{i}^{(l_{i})}\in\mathcal{B}_{n}(\Omega)$
are (IR) decompositions of $T$ and $\widetilde{T}$, respectively. If
$\widetilde{T}=U^{*}TU$ for some unitary $U$, according to Lemma 4.3, we have
$\mathcal{A}^{\prime}(T,T^{*})=U\mathcal{A}^{\prime}(\widetilde{T},\widetilde{T}^{*})U^{*}$,
that is,
$\mathcal{A}^{\prime}(T,T^{*})\cong\mathcal{A}^{\prime}(\widetilde{T},\widetilde{T}^{*})$.
By Lemma 3.4, we know that
$\mathcal{A}^{\prime}(T,T^{*})\cong\sum\limits_{i=1}^{t}\oplus\mathcal{M}_{m_{i}}(\mathbb{C})$
and
$\mathcal{A}^{\prime}(\widetilde{T},\widetilde{T}^{*})\cong\sum\limits_{i=1}^{s}\oplus\mathcal{M}_{l_{i}}(\mathbb{C})$.
Thus,
$\sum\limits_{i=1}^{t}\oplus\mathcal{M}_{m_{i}}(\mathbb{C})\cong\sum\limits_{i=1}^{s}\oplus\mathcal{M}_{l_{i}}(\mathbb{C})$.
It follows that $t=s$ and there is a permutation $\pi$ on
$\\{1,2,\cdots,t\\}$, such that $m_{i}=l_{\pi(i)},1\leq i\leq t$.

Let the (IR) decomposition of $T\bigoplus\widetilde{T}$ be
$\bigoplus_{i=1}^{k}\widehat{T}_{i}^{(k_{i})}$ for some positive integer
$k\geq t$. If $T\sim_{u}\widetilde{T}$, then
$T^{(2)}\sim_{u}T\bigoplus\widetilde{T}$, that is,
$\bigoplus_{i=1}^{t}T_{i}^{(2m_{i})}\sim_{u}\bigoplus_{i=1}^{k}\widehat{T}_{i}^{(k_{i})}$.
Based on the above analysis, we know $t=k$. That means,
$\\{n_{i}|T_{i}\in\mathcal{B}_{n_{i}}(\Omega),1\leq i\leq
t\\}=\\{n^{\prime}_{i}|\widetilde{T}_{i}\in\mathcal{B}_{n^{\prime}_{i}}(\Omega),1\leq
i\leq t\\}$. Thus, there is a permutation $\pi^{\prime}$ on
$\\{1,2,\cdots,t\\}$, such that the index of
$T_{i},\widetilde{T}_{\pi^{\prime}(i)},1\leq i\leq t$ are the same.

By what was discussed above, we will give a complete proof of Theorem 4.2.

###### Proof.

Let $T\in\mathcal{B}_{n}(\Omega)$. Suppose that there exists an operator
$\widetilde{T}\in\mathcal{B}_{n}(\Omega)$, such that $T\sim_{u}\widetilde{T}$.
By Remark 4.4, we can write the (IR) decompositions of $T$ and $\widetilde{T}$
as

(4.1)
$T=\bigoplus_{i=1}^{t}T_{i}^{(m_{i})}=\bigoplus\limits_{j=1}^{k}S_{j},\,\widetilde{T}=\bigoplus_{i=1}^{t}\widetilde{T}_{i}^{(l_{i})}=\bigoplus\limits_{j=1}^{k}\widetilde{S}_{j},$

where the index of $T_{i},\widetilde{T}_{i}$ are the same for all $1\leq i\leq
t$ and

$S_{1}=\cdots=S_{m_{1}}=T_{1},S_{m_{1}+1}=\cdots=S_{m_{1}+m_{2}}=T_{2},\cdots,S_{m_{1}+\cdots+m_{t-1}+1}=\cdots=S_{m_{1}+\cdots+m_{t}}=T_{t};$
$\widetilde{S}_{1}=\cdots=\widetilde{S}_{l_{1}}=\widetilde{T}_{1},\widetilde{S}_{l_{1}+1}=\cdots=\widetilde{S}_{l_{1}+l_{2}}=\widetilde{T}_{2},\cdots,\widetilde{S}_{l_{1}+\cdots+l_{t-1}+1}=\cdots=\widetilde{S}_{l_{1}+\cdots+l_{t}}=\widetilde{T}_{t}.$

Suppose that $S_{i}=(M^{*}_{z},\mathcal{H}_{i},K_{i}),\
\widetilde{S}_{i}=(M^{*}_{z},\widetilde{\mathcal{H}}_{i},\widetilde{K}_{i}),1\leq
i\leq k$. Then for all $\mu,\lambda\in\Omega^{*}$, we can express the
reproducing kernels of the spaces corresponding to $T$ and $\widetilde{T}$ as

$K(\mu,\lambda)=diag(K_{1}(\mu,\lambda),\cdots,K_{k}(\mu,\lambda)),\,\widetilde{K}(\mu,\lambda)=diag(\widetilde{K}_{1}(\mu,\lambda),\cdots,\widetilde{K}_{k}(\mu,\lambda)).$

By Theorem 1.1, we know that if $T\sim_{u}\widetilde{T}$, then there exist an
analytic invertible matrix $\Phi=(\\!(\Phi_{i,j})\\!)_{i,j=1}^{k}$, such that

(4.2)
$K(\mu,\lambda)=\Phi(\mu)\widetilde{K}(\mu,\lambda)\overline{\Phi^{T}(\lambda)}$

for all $\mu,\lambda\in\Lambda$, a open subset of $\Omega^{*}$. Let
$\Psi=\Phi^{-1}=(\\!(\Psi_{i,j})\\!)_{i,j=1}^{k}.$ Obviously, $\Psi$ is
holomorphic. Consequently, formula (4.2) can be converted into the following
two forms:

(4.3)
$\Psi_{i,j}(\mu)K_{j}(\mu,\lambda)=\widetilde{K}_{i}(\mu,\lambda)\overline{\Phi^{T}_{j,i}(\lambda)},\
\mu,\lambda\in\Lambda,\ 1\leq i,j\leq k;$

and

(4.4)
$K_{j}(\mu,\lambda)\overline{\Psi^{T}_{i,j}(\lambda)}=\Phi_{j,i}(\mu)\widetilde{K}_{i}(\mu,\lambda),\
\mu,\lambda\in\Lambda,\ 1\leq i,j\leq k.$

Combining formulas (4.3), (4.4) and multiplying $\Phi_{j,i}(\mu)$,
$\Psi_{i,j}(\mu)$ on the left sides of equations (4.3) and (4.4) respectively,
we obtain

$\begin{array}[]{lll}\Phi_{j,i}(\mu)\Psi_{i,j}(\mu)K_{j}(\mu,\lambda)&=&\Phi_{j,i}(\mu)\widetilde{K}_{i}(\mu,\lambda)\overline{\Phi_{j,i}(\lambda)}^{T}\\\
&=&K_{j}(\mu,\lambda)\overline{\Psi_{i,j}(\lambda)}^{T}\overline{\Phi_{j,i}(\lambda)}^{T}\\\
&=&K_{j}(\mu,\lambda)\overline{\Phi_{j,i}(\lambda)\Psi_{i,j}(\lambda)}^{T},\
\mu,\lambda\in\Lambda,\ 1\leq i,j\leq k,\end{array}$

and

$\begin{array}[]{lll}\Psi_{i,j}(\mu)\Phi_{j,i}(\mu)\widetilde{K}_{i}(\mu,\lambda)&=&\Psi_{i,j}(\mu)K_{j}(\mu,\lambda)\overline{\Psi_{i,j}(\lambda)}^{T}\\\
&=&\widetilde{K}_{i}(\mu,\lambda)\overline{\Phi_{j,i}(\lambda)}^{T}\overline{\Psi_{i,j}(\lambda)}^{T}\\\
&=&\widetilde{K}_{i}(\mu,\lambda)\overline{\Psi_{i,j}(\lambda)\Phi_{j,i}(\lambda)}^{T},\
\mu,\lambda\in\Lambda,\ 1\leq i,j\leq k.\end{array}$

Since $S_{i},\widetilde{S}_{i}\in(IR),\ 1\leq i\leq k$, according to Lemma 3.3
and the proof of Theorem 3.5, there exist $c_{i,j}\geq 0$ such that the
following holds

(4.5) $\Phi_{j,i}\Psi_{i,j}=c_{i,j}\,I,\ \Psi_{i,j}\Phi_{j,i}=c_{i,j}\,I,\
1\leq i,j\leq k.$

Since $\Phi\Psi=\Psi\Phi=I$, it follows that

(4.6)
$\Phi(\mu)\Psi(\mu)=\biggl{(}\sum\limits_{t=1}^{k}\Phi_{i,t}(\mu)\Psi_{t,j}(\mu)\biggl{)}_{i,j=1}^{k}=I,\
\Psi(\mu)\Phi(\mu)=\biggl{(}\sum\limits_{r=1}^{k}\Psi_{i,r}(\mu)\Phi_{r,j}(\mu)\biggl{)}_{i,j=1}^{k}=I.$

Thus, combining formulas (4.5) and (4.6), we obtain

(4.7) $\sum\limits_{i=1}^{k}c_{i,j}=1,1\leq j\leq k;\
\sum\limits_{j=1}^{k}c_{i,j}=1,1\leq i\leq k.$

Setting $C:=(\\!(c_{i,j})\\!)_{i,j=1}^{k}$. The equation (4.7) shows that the
sum of elements in each row and each column of $C$ is one.

The size of $\Phi_{i,j},\Psi_{i,j}$ are determined by $S_{j}$ and
$\widetilde{S}_{i}$. If there exists some $c_{i,j}\neq 0$, that is,
$c_{i,j}>0$, from equation (4.5), we obtain
$(\frac{\Phi_{j,i}(\mu)}{\sqrt{c_{i,j}}})^{-1}=\frac{\Psi_{i,j}(\mu)}{\sqrt{c_{i,j}}},1\leq
i,j\leq k$. And using equation (4.3), we have that
$K_{j}(\mu,\lambda)=\frac{\Phi_{j,i}(\mu)}{\sqrt{c_{i,j}}}\widetilde{K}_{i}(\mu,\lambda)\frac{\Phi_{j,i}^{*}(\lambda)}{\sqrt{c_{i,j}}}(\mu,\lambda\in\Lambda).$
Therefore, $S_{j}\sim_{u}\widetilde{S}_{i}$, which is obtained from Theorem
1.1. According to the characteristics of decompositions of $T$ and
$\widetilde{T}$, we can divide $C$ into

$\left(\begin{matrix}C_{1,1}&C_{1,2}&\cdots&C_{1,t}\\\
C_{2,1}&C_{2,2}&\cdots&C_{2,t}\\\ \vdots&\vdots&&\vdots\\\
C_{t,1}&C_{t,2}&\cdots&C_{t,t}\\\ \end{matrix}\right),$

where $C_{i,j}$ has $l_{i}$ rows and $m_{j}$ columns. It is easy to see that
if $C_{i,j}\neq 0$ means $T_{j}\sim_{u}\widetilde{T}_{i}$.

We claim that the following three statements are true,

  * (1)

For any $i\in A:=\\{1,2,\cdots,t\\}$, in the set
$C_{i,-}:=\\{C_{i,1},C_{i,2},\cdots,C_{i,t}\\}$, there exists $j_{0}\in A$,
such that $C_{i,j_{0}}\neq 0$ and $C_{i,j}=0\;(j\neq j_{0})$;

  * (2)

For any $j\in A$, in the set $C_{-,j}:=\\{C_{1,j},C_{2,j},\cdots,C_{t,j}\\}$,
there exists $i_{0}\in A$, such that $C_{i_{0},j}\neq 0$ and
$C_{i,j}=0\;(i\neq i_{0})$;

  * (3)

If $C_{i,j}\neq 0$, then $C_{i,j}$ is a square matrix.

For the matrix $C=(\\!(c_{i,j})\\!)_{i,j=1}^{k}$, we have already proved
equation (4.7). Clearly, for any $p,q,\ 1\leq p,q\leq k$, there exist
$p_{0},q_{0},\ 1\leq p_{0},q_{0}\leq k$, such that $c_{p,q_{0}}\neq 0,\
c_{p_{0},q}\neq 0$. That is to say, for any $i,j\in A$, we have
$C_{i,j_{0}}\neq 0,\ C_{i_{0},j}\neq 0$, where $i_{0},j_{0}\in A$. Suppose
that there are $i_{1},j_{1}(i_{0}\neq i_{1},j_{0}\neq j_{1})$, satisfy
$C_{i,j_{1}}\neq 0,C_{i_{1},j}\neq 0$. Then we obtain that
$T_{j_{0}}\sim_{u}\widetilde{T}_{i},\ T_{j_{1}}\sim_{u}\widetilde{T}_{i};\
T_{j}\sim_{u}\widetilde{T}_{i_{0}},T_{j}\sim_{u}\widetilde{T}_{i_{1}}.$ By
Definition 3.2, it is a contradiction. So we complete the proof of (1) and
(2). And there is only one $C_{i,j}$ non-zero in every row and column of
matrix $C$. Otherwise, a row or a column of $C$ will be zero. This contradicts
(1) or (2). Next, we will prove that if the matrix $C_{ij}$ is non-zero for
any pair of indices $i,j$, then it must be a square matrix. Suppose that
$C_{\widetilde{i},\widetilde{j}}\neq 0,\,\,\,\widetilde{i},\widetilde{j}\in
A$. Adding up all the elements of $C_{\widetilde{i},\widetilde{j}}$, by
equation (4.7), we obtain that the sum is equal to both $m_{\widetilde{i}}$
and $l_{\widetilde{j}}$, then $m_{\widetilde{i}}=l_{\widetilde{j}}$, that is,
$C_{\widetilde{i},\widetilde{j}}$ is a square matrix. This proves assertion
(3).

For every $T_{i}$, there is a unique $\widetilde{T}_{j}$ such that
$T_{i}\sim_{u}\widetilde{T}_{j},1\leq i,j\leq t$. In other words, there is a
permutation $\pi$ on $\\{1,2,\cdots t\\}$, such that
$T_{i}\sim_{u}\widetilde{T}_{\pi(i)}$ and $m_{i}=l_{\pi(i)},i\in A$. By
Definition 4.1, $T$ has a unique decomposition up to unitary equivalence. ∎

######  Remark 4.5.

Theorem 4.2 can also be proved using the notion of a normalized kernel.

Let $T\in\mathcal{B}_{n}(\Omega)$ and $T=(M_{z}^{*},\mathcal{H}_{K},K)$.
Suppose that the (IR) decomposition of $T$ is $T_{1}\oplus T_{2}$,
$T_{i}\in\mathcal{B}_{n_{i}}(\Omega),i=1,2$. Here we illustrate this method by
proving that $T$ has a unique (IR) decomposition up to unitary equivalence.
Suppose that $\widetilde{T}\sim_{u}T$ and
$\widetilde{T}=(M_{z}^{*},\mathcal{H}_{\widetilde{K}},\widetilde{K})$. By
Remark 4.4, we know that the (IR) decomposition of $\widetilde{T}$ must be
$\widetilde{T}=\widetilde{T}_{1}\oplus\widetilde{T}_{2}$, $i=1,2$. It is well
known that every reproducing kernel on analytic Hilbert space can be
normalized. We think that $K,\widetilde{K}$ have been normalized at
$\lambda_{0}\in\Omega^{*}$. If $\widetilde{T}$ and $T$ are unitarily
equivalent via some intertwining unitary map
$\Gamma:\mathcal{H}_{K}\to\mathcal{H}_{\widetilde{K}}$, then the map $\Gamma$
must be of the form $(\Gamma f)(z)=U(f(z))$, that is,
$K(\mu,\lambda)=U^{*}\widetilde{K}(\mu,\lambda)U,\mu,\lambda\in\Omega^{*}$.
Since $K,\widetilde{K}$ are real (positive) analytic, and admits power series
expansion around $\lambda_{1}\in\Omega$. Assume $\lambda_{1}=0$ for
simplicity. We have that

$K(\mu,\lambda)=\sum_{m,n=0}^{\infty}A_{mn}\mu^{m}\overline{\lambda}^{n},\
\widetilde{K}(\mu,\lambda)=\sum_{m,n=0}^{\infty}B_{mn}\mu^{m}\overline{\lambda}^{n}.$

The direct decomposition of $T$ and $\widetilde{T}$ means that the form of
their reproducing kernels are $K=K_{1}\oplus
K_{2},\widetilde{K}=\widetilde{K}_{1}\oplus\widetilde{K}_{2}$, where
$K_{i},\widetilde{K}_{i}$ are normalized kernels corresponding to $T_{i},\
\widetilde{T}_{i}\ (i=1,2)$ at $\lambda_{0}\in\Omega^{*}$. Thus, all the
matrix coefficients $A_{mn},B_{mn},m,n\geq 0$ in the power series expansions
must be simultaneously decomposable. It follows that $A_{mn}^{1}\oplus
A_{mn}^{2}=U^{*}(B_{mn}^{1}\oplus B_{mn}^{2})U,\ m,n\geq 0,$ where

$K_{i}(\mu,\lambda)=\sum_{m,n=0}^{\infty}A^{i}_{mn}\mu^{m}\overline{\lambda}^{n},\
\widetilde{K}_{i}(\mu,\lambda)=\sum_{m,n=0}^{\infty}B^{i}_{mn}\mu^{m}\overline{\lambda}^{n},i=1,2.$

If possible, first assume that all the $A^{(1)}_{mn}=A^{(2)}_{mn}$ for all
$m,n$. We must also have that $B^{(1)}_{mn}=B^{(2)}_{mn}$ for all $m,n$. Since
these two operators were assumed to have a decomposition with the components
in the class $(IR)$, it follows that $\\{A_{m,n}^{(1)}:m,n\in\mathbb{N}\\}$ is
a set of irreducible operators. Similarly,
$\\{B_{m,n}^{(1)}:m,n\in\mathbb{N}\\}$ is a set of irreducible operators.
These two sets of operators cannot be unitarily equivalent by a fixed unitary
unless the unitary is of the form: $\alpha I,\,\alpha\in\mathbb{T}$. So, in
this case, $T=\tilde{T}$.

Now, let us assume without loss of generality,
$A^{(1)}_{m,n}\not=A_{mn}^{(2)}$ for some fixed pair $m,n$. Then for this pair
$m,n$ we have that $A_{mn}$ equals the direct sum $D^{(1)}_{mn}\oplus
D^{(2)}_{mn}$ of diagonal matrices since $A_{mn}$ is positive definite. The
same is true of $B_{mn}$. If the operator $T$ and $\tilde{T}$ are of rank 2,
then we see that our hypothesis implies the existence of a unitary
intertwining two diagonal matrices of size $2$ with distinct eigenvalues. This
is possible only if U is a scalar times identity or is a permutation.

If the rank of the operator $T$ is greater than 2, we consider the algebra
generated by all the the coefficients of the kernel $K$ and the algebra
generated by all the the coefficients of $\tilde{K}$ is not irreducible but is
the direct sum of two irreducible algebras. This is because the algebra is
generated by the set $\\{A^{(1)}_{mn}\oplus A^{(2)}_{mn}:m,n\in\mathbb{N}\\}$
and the coefficients $A_{mn}^{i},B_{mn}^{i},m,n\geq 0$ must not be
simultaneously reducible for each $i=1,2$. Again, in this case, the
intertwining unitary operator must intertwine these two algebras and hence
must be of the form $c_{1}I\oplus c_{2}I$ or a permutation
$\left(\begin{smallmatrix}0&c_{1}I\\\ c_{2}I&0\end{smallmatrix}\right)$, for
some choice of unimodular scalars $c_{1},c_{2}$.

Based on the above analysis, we know that for each $i=1,2$, there is a
permutation $\pi$ on $\\{1,2\\}$, such that
$A_{mn}^{i}\sim_{u}B_{mn}^{\pi(i)}$ for all $m,n\geq 0$ and
$T_{i}\sim_{u}\widetilde{T}_{\pi(i)}$. Thus, $T$ has a unique (IR)
decomposition up to unitary equivalence.

##  5\. Application

In this section, we will discuss the structure of the intertwining matrix in
equation (1.1) by using Theorem 3.5 and Theorem 4.2.

Suppose that $T=T_{1}\oplus T_{2}$ and
$\widetilde{T}=\widetilde{T}_{1}\oplus\widetilde{T}_{2}$, where
$T_{i}=(M^{*}_{z},\mathcal{H}_{i},K_{i}),\widetilde{T}_{i}=(M^{*}_{z},\widetilde{\mathcal{H}}_{i},\widetilde{K}_{i})\in{\mathcal{B}}_{1}(\Omega),i=1,2$
and $T_{1}\not\sim_{u}T_{2}$,
$\widetilde{T}_{1}\not\sim_{u}\widetilde{T}_{2}$. Suppose that $T$ and
$\widetilde{T}$ are unitarily equivalent. By Theorem 1.1, there exists an
invertible holomorphic matrix valued function $\Phi=((\phi_{i,j}))_{2\times
2}$ such that

(5.1) $\left(\begin{matrix}K_{1}(\mu,\lambda)&0\\\ 0&K_{2}(\mu,\lambda)\\\
\end{matrix}\right)=\left(\begin{matrix}\phi_{1,1}(\mu)&\phi_{1,2}(\mu)\\\
\phi_{2,1}(\mu)&\phi_{2,2}(\mu)\\\
\end{matrix}\right)\left(\begin{matrix}\widetilde{K}_{1}(\mu,\lambda)&0\\\
0&\widetilde{K}_{2}(\mu,\lambda)\\\
\end{matrix}\right)\left(\begin{matrix}\bar{\phi}_{1,1}(\lambda)&\bar{\phi}_{2,1}(\lambda)\\\
\bar{\phi}_{1,2}(\lambda)&\bar{\phi}_{2,2}(\lambda)\\\ \end{matrix}\right)$

Furthermore, by Theorem 3.5, we have that

$\frac{\partial}{\partial\lambda}[K_{\widetilde{T}}^{-1}(\lambda,\lambda)\Phi(\lambda)^{-1}K_{T}(\lambda,\lambda)]=0$

for all $\lambda\in\Lambda$, a open subset of $\Omega^{*}.$ Set
$\Phi^{-1}=\Psi=((\psi_{i,j}))_{2\times 2}$, then we have that

$\left(\begin{matrix}\widetilde{K}^{-1}_{1}(\lambda,\lambda)\psi_{1,1}(\lambda)K_{1}(\lambda,\lambda)&\
\widetilde{K}^{-1}_{1}(\lambda,\lambda)\psi_{1,2}(\lambda)K_{2}(\lambda,\lambda)\\\
\widetilde{K}^{-1}_{2}(\lambda,\lambda)\psi_{2,1}(\lambda)K_{1}(\lambda,\lambda)&\
\widetilde{K}^{-1}_{2}(\lambda,\lambda)\psi_{2,2}(\lambda)K_{2}(\lambda,\lambda)\\\
\end{matrix}\right)$

is anti-holomorphic. Suppose that $\psi_{1,1}$ is nonzero function. By Theorem
3.5 in the case of index is one, we can see $T_{1}\sim_{u}\widetilde{T}_{1}$.
If $\psi_{1,2}$ is also nonzero function, then
$T_{2}\sim_{u}\widetilde{T}_{1}$. Which is a contradiction to
$T_{1}\not\sim_{u}T_{2}$. Thus, since $\Phi,\Psi$ are invertible, we have that
$\Phi$ only have the following two forms:

$\Phi=\left(\begin{matrix}\phi_{1,1}&0\\\ 0&\phi_{2,2}\\\
\end{matrix}\right)~{}~{}\mbox{or}~{}~{}\Phi=\left(\begin{matrix}0&\phi_{1,2}\\\
\phi_{2,1}&0\\\ \end{matrix}\right).$

For the general case, set
$T=\bigoplus_{i=1}^{t}T_{i}^{(m_{i})}\in\mathcal{B}_{n}(\Omega)$ satisfies
Definition 3.2. If $\widetilde{T}\sim_{u}T$, by Theorem 4.2, we know
$\widetilde{T}=\bigoplus_{i=1}^{t}\widetilde{T}_{i}^{(l_{i})}$ and there
exists a permutation $\pi$ on $\\{1,2,\cdots t\\}$ such that
$T_{i}\sim_{u}\widetilde{T}_{\pi(i)},\ m_{i}=l_{\pi(i)}$. Setting
$k:=\sum_{i=1}^{t}m_{i}=\sum_{i=1}^{t}l_{i}$. Next, we write $T$ and
$\widetilde{T}$ in terms of equation (4.1). Suppose that
$S_{i}=(M^{*}_{z},\mathcal{H}_{i},K_{i}),\
\widetilde{S}_{i}=(M^{*}_{z},\widetilde{\mathcal{H}}_{i},\widetilde{K}_{i}),\
1\leq i\leq k$.

According to Theorem 1.1, there exists a holomorphic invertible matrix
$\Phi=(\\!(\Phi_{i,j})\\!)_{i,j=1}^{k}$, such that

$\begin{array}[]{lll}K(\mu,\lambda)&=&\Phi(\mu)\widetilde{K}(\mu,\lambda)\Phi^{*}(\lambda)\\\
&=&\left(\begin{smallmatrix}\sum\limits_{i=1}^{k}\Phi_{1,i}(\mu)\widetilde{K}_{i}(\mu,\lambda)\Phi_{1,i}^{*}(\lambda)&*&\cdots&*\\\
*&\sum\limits_{i=1}^{k}\Phi_{2,i}(\mu)\widetilde{K}_{i}(\mu,\lambda)\Phi_{2,i}^{*}(\lambda)&\cdots&*\\\
\vdots&\vdots&\ddots&\vdots\\\
*&*&\cdots&\sum\limits_{i=1}^{k}\Phi_{k,i}(\mu)\widetilde{K}_{i}(\mu,\lambda)\Phi_{k,i}^{*}(\lambda)\\\
\end{smallmatrix}\right),\end{array}$

that is,

(5.2)
$K_{j}(\mu,\lambda)=\sum_{i=1}^{k}\Phi_{j,i}(\mu)\widetilde{K}_{i}(\mu,\lambda)\Phi_{j,i}^{*}(\lambda),\
1\leq j\leq k,$

for all $\mu,\lambda\in\Lambda$, a open subset of $\Omega^{*}.$ Setting
$\Phi^{-1}=\Psi=(\\!(\Psi_{i,j})\\!)_{i,j=1}^{k}$. By Lemma 3.3 and the proof
of Theorem 4.2, we have

(5.3)
$\Psi_{i,j}(\mu)K_{j}(\mu,\lambda)=\widetilde{K}_{i}(\mu,\lambda)\Phi_{j,i}^{*}(\lambda),\
K_{j}(\mu,\lambda)\Psi^{*}_{i,j}(\lambda)=\Phi_{j,i}(\mu)\widetilde{K}_{i}(\mu,\lambda),\
\mu,\lambda\in\Lambda,$ (5.4)
$\Psi_{i,j}\Phi_{j,i}=c_{i,j}I,\,\Phi_{j,i}\Psi_{i,j}=c_{i,j}I,\,c_{i,j}\geq
0,\ 1\leq i,j\leq k,$

and

(5.5) $\sum\limits_{i=1}^{k}c_{i,j}=1,1\leq j\leq k,\
\sum\limits_{j=1}^{k}c_{i,j}=1,\ 1\leq i\leq k.$

By equations (5.3) and (5.4), we can imply that

(5.6)
$c_{i,j}K_{j}(\mu,\lambda)=\Phi_{j,i}(\mu)\widetilde{K}_{i}(\mu,\lambda)\Phi_{j,i}^{*}(\lambda),\
1\leq j\leq k.$

Let $C=(\\!(c_{i,j})\\!)_{i,j=1}^{k}$. If $c_{i,j}>0$, it follows that
$\Psi_{i,j},\ \Phi_{j,i}$ are invertible and $S_{j}\sim_{u}\widetilde{S}_{i}$.
Similar to Theorem 4.2, we divide $C$ into $(\\!(C_{i,j})\\!)_{i,j=1}^{t}$,
where $C_{i,j}$ has $l_{i}$ rows and $m_{j}$ columns. For any $j,\,1\leq j\leq
t$, we obtain $C_{\pi(j),j}\neq 0$ and $C_{i,j}=0$ for all $i\neq\pi(j)$.

Suppose that $C_{\pi(p),p}\neq 0,\ 1\leq p\leq t$. Let

$\widetilde{i}:=\left\\{\begin{array}[]{cc}0,\ \pi(p)=1;\\\
\sum\limits_{r=1}^{\pi(p)-1}l_{r},\
\pi(p)>1,\end{array}\right.\widetilde{j}:=\left\\{\begin{array}[]{cc}0,\
p=1;\\\ \sum\limits_{r=1}^{p-1}m_{r},\ p>1.\end{array}\right.$

Then $C_{\pi(p),p}=(\\!(c_{i,j})\\!)_{l_{\pi(p)}\times m_{p}}$, where
$\widetilde{i}+1\leq i\leq\widetilde{i}+l_{\pi(p)},\ \widetilde{j}+1\leq
j\leq\widetilde{j}+m_{p}$. Thus, from equation (5.6), we have

$K_{j}(\mu,\lambda)=\sum_{i=\widetilde{i}+1}^{\widetilde{i}+l_{\pi(p)}}c_{ij}K_{j}(\mu,\lambda)=\sum_{i=\widetilde{i}+1}^{\widetilde{i}+l_{\pi(p)}}\Phi_{j,i}(\mu)\widetilde{K}_{i}(\mu,\lambda)\Phi_{j,i}^{*}(\lambda),\
1\leq j\leq k.$

By using equations (5.2),(5.5) and (5.6),

$\begin{array}[]{lll}&&\sum_{i=1}^{\widetilde{i}}c_{ij}K_{j}(\mu,\lambda)+\sum_{i=\widetilde{i}+l_{\pi(p)}+1}^{k}c_{ij}K_{j}(\mu,\lambda)\\\
&=&\sum_{i=1}^{\widetilde{i}}\Phi_{j,i}(\mu)\widetilde{K}_{i}(\mu,\lambda)\Phi_{j,i}^{*}(\lambda)+\sum_{i=\widetilde{i}+l_{\pi(p)}+1}^{k}\Phi_{j,i}(\mu)\widetilde{K}_{i}(\mu,\lambda)\Phi_{j,i}^{*}(\lambda)\\\
&=&0,\ 1\leq j\leq k.\end{array}$

Note that $\widetilde{K}_{i}(\mu,\lambda)$ is positive and
$\Phi_{j,i}(\mu)\widetilde{K}_{i}(\mu,\lambda)\Phi_{j,i}^{*}(\lambda)$ is
nonnegative definite, then the last equation means that $\Phi_{j,i}=0$, when
$1\leq i\leq\widetilde{i}$ or $\widetilde{i}+l_{\pi(p)}+1\leq i\leq k$, $1\leq
j\leq k$. We divide $\Phi$ into
$(\\!(\widetilde{\Phi}_{i,j})\\!)_{i,j=1}^{t}$, where $\widetilde{\Phi}_{i,j}$
has $m_{i}$ rows and $l_{j}$ columns. According to the discussion above, we
obtain $\widetilde{\Phi}_{i,\pi(i)}\neq 0$ and
$\widetilde{\Phi}_{i,j}=0(j\neq\pi(i))$. Hence, $\Phi_{i,j}$ in $\Phi$ is
either invertible or zero.

Acknowledgements:

The authors thank the referee for carefully reading the paper and pointing out
some helpful comments and suggestions. The first author would like to thank
Dr. Soumitra Ghara and Li Chen for many valuable discussions. The first author
was supported by National Natural Science Foundation of China (Grant No.
11831006 and 11922108).

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