# Wold-type decomposition for $\mathcal{U}_{n}$-twisted contractions

Satyabrata Majee  Indian Institute of Technology Roorkee, Department of
Mathematics, Roorkee-247 667, Uttarakhand, India<EMAIL_ADDRESS>and
Amit Maji† Indian Institute of Technology Roorkee, Department of Mathematics,
Roorkee-247 667, Uttarakhand, India<EMAIL_ADDRESS><EMAIL_ADDRESS>(†Corresponding author)

###### Abstract.

Let $n>1$, and $\\{U_{ij}\\}$ for $1\leq i<j\leq n$ be $\binom{n}{2}$
commuting unitaries on a Hilbert space $\mathcal{H}$ such that
$U_{ji}:=U^{*}_{ij}$. An $n$-tuple of contractions $(T_{1},\dots,T_{n})$ on
$\mathcal{H}$ is called $\mathcal{U}_{n}$-twisted contraction with respect to
a twist $\\{U_{ij}\\}_{i<j}$ if $T_{1},\dots,T_{n}$ satisfy

$T_{i}T_{j}=U_{ij}T_{j}T_{i};\hskip 14.22636pt\hskip
28.45274ptT_{i}^{*}T_{j}=U^{*}_{ij}T_{j}T_{i}^{*}\hskip
14.22636pt\mbox{and}\hskip 14.22636ptT_{k}U_{ij}=U_{ij}T_{k}$

for all $i,j,k=1,\dots,n$ and $i\neq j$.

We obtain a recipe to calculate the orthogonal spaces of the Wold-type
decomposition for $\mathcal{U}_{n}$-twisted contractions on Hilbert spaces. As
a by-product, a new proof as well as complete structure for
$\mathcal{U}_{2}$-twisted (or pair of doubly twisted) and
$\mathcal{U}_{n}$-twisted isometries have been established.

###### Key words and phrases:

Isometry; contraction; $\mathcal{U}_{n}$-twisted contractions; completely non-
unitary; shift

###### 2010 Mathematics Subject Classification:

47A45, 47A20, 47A05, 47A13

##  1\. Introduction

A fundamental problem in the theory of operators, function theory and operator
algebras is the classification problem for a tuple of commuting isometries on
Hilbert spaces. The canonical decomposition for a contraction plays a
significant role in many areas of operator algebras and operator theory,
namely, dilation theory, invariant subspace theory, operator interpolation
problem, etc. It says that every contraction can be uniquely decomposed into
the orthogonal sum of a unitary operator and a completely non-unitary
operator. In particular, the canonical decomposition of an isometry coincides
with the classical Wold decomposition or Wold-von Neumann decomposition.
Indeed, the completely non-unitary part of an isometry becomes a unilateral
shift (of any multiplicity). This decomposition was firstly studied by Wold
[25] for stationary stochastic processes. It is expected that the
multidimensional Wold-type decomposition will provide a large class of
applications.

A natural issue is the extension of decomposition from a single contraction to
a tuple of contractions. Using Suciu’s [24] decomposition of the semigroup of
isometries, Słociński [22] firstly obtained a Wold-type decomposition for
pairs of doubly commuting isometries. It states that a pair of doubly
commuting isometries have fourfold Wold -type decomposition of the form
unitary-unitary, unitary-shift, shift-unitary, and shift-shift. In 2004,
Popovici [17] achieved Wold-type decomposition for a pair $(V_{1},V_{2})$ of
commuting isometries on a Hilbert space. More specifically, the pair
$(V_{1},V_{2})$ can be uniquely decomposed into the orthogonal sum of bi-
unitary, a shift-unitary, a unitary-shift, and a weak bi-shift. Later, Sarkar
[21] generalized Słociński’s result and also obtained an explicit description
of closed subspaces in the orthogonal decomposition for the $n$-tuples of
doubly commuting isometries. Recently Maji, Sarkar, and Sankar [16] have
studied various natural representations of a large class of pairs of commuting
isometries on Hilbert spaces, and Bînzar et al. [4] studied Wold-Słociński
decomposition for commuting isometric triples. On the other hand, power
partial isometry is a large class of operators with a well-defined completely
non-unitary part. Halmos and Wallen [13] studied decomposition for a power
partial isometry. After that Catepillán and Szymański [9] have generalized for
a pair of doubly commuting power partial isometries. For more results one can
refer to [1], [2], [3], [6], [7], [5], [8], [11], [12], etc. Słociński [23]
(see also Burdak [5]) studied decomposition for pairs of doubly commuting
contractions and obtained the following result:

######  Theorem 1.1.

Let $T=(T_{1},T_{2})$ be a pair of doubly commuting contractions on a Hilbert
space $\mathcal{H}$. Then there exists a unique decomposition

$\mathcal{H}=\mathcal{H}_{uu}\oplus\mathcal{H}_{u\neg
u}\oplus\mathcal{H}_{\neg uu}\oplus\mathcal{H}_{\neg u\neg u}$

where $\mathcal{H}_{ij}$ are joint $T$-reducing subspaces of $\mathcal{H}$ for
all $i,j=u,\neg u$. Moreover, $T_{1}$ on $\mathcal{H}_{ij}$ is unitary if
$i=u$ and completely non-unitary if $i=\neg u$ and $T_{2}$ on
$\mathcal{H}_{ij}$ is unitary if $j=u$ and completely non-unitary if $j=\neg
u$.

However, the complete description of the above orthogonal decomposition spaces
is not explicit. Burdak [5] also developed a characterization for pairs of
commuting (not necessarily doubly commuting) contractions and obtained
decomposition results in the case of commuting pairs of power partial
isometries. Recently, Jeu and Pinto [14] studied a simultaneous Wold
decomposition for an $n$-tuple $(n>1)$ of doubly non-commuting isometries
which has been classified up to unitary equivalence by using this
decomposition. In 2022, Rakshit, Sarkar, and Suryawanshi [19] showed that each
$\mathcal{U}_{n}$-twisted isometry agrees a von Neumann-Wold type
decomposition and then described concrete analytic models of
$\mathcal{U}_{n}$-twisted isometries. It is now a natural query whether the
above results can be extended to a large class of operators, namely, a class
of twisted contractions on Hilbert spaces.

Motivated by the definition of $\mathcal{U}_{n}$-twisted isometries in [19],
we introduce the notion of $\mathcal{U}_{n}$-twisted contractions (see
definition in Section 3). In this paper, we attempt to find a recipe for
calculating the orthogonal spaces as well as to extend the results for pair of
doubly twisted contractions (or $\mathcal{U}_{2}$-twisted contractions) to
multi-variable case. Our approach is based on the canonical decomposition for
a single contraction and the geometry of Hilbert spaces.

The paper is organized as follows. In section 2, we discuss some basic
definitions and the canonical decomposition for a single contraction. Section
3 is devoted to the decomposition for a pair of doubly twisted contractions.
In section 4, we obtain a complete description for $\mathcal{U}_{n}$-twisted
contractions and in particular for $\mathcal{U}_{n}$-twisted isometries.

##  2\. Preparatory Results

In what follows $\mathbb{Z}_{+}$ denotes the set of non-negative integers,
$\mathcal{H}$ stands for a complex Hilbert space, $I$ denotes the identity
operator on $\mathcal{H}$ and $\mathcal{B}(\mathcal{H})$ as the algebra of all
bounded linear operators on $\mathcal{H}$. For a closed subspace $\mathcal{M}$
of $\mathcal{H}$, $P_{\mathcal{M}}$ denotes the orthogonal projection of
$\mathcal{H}$ onto $\mathcal{M}$. A closed subspace $\mathcal{M}$ of
$\mathcal{H}$ is invariant under $T\in\mathcal{B}(\mathcal{H})$ if
$T(\mathcal{M})\subseteq\mathcal{M}$; and subspace $\mathcal{M}$ reduces $T$
if $T(\mathcal{M})\subseteq\mathcal{M}$ and
$T(\mathcal{M}^{\perp})\subseteq\mathcal{M}^{\perp}$. A contraction $T$ on
$\mathcal{H}$ (that is, $\|Th\|\leq\|h\|$ for all $h\in\mathcal{H}$) is said
to be a pure contraction if $T^{*m}\rightarrow 0$ as $m\rightarrow\infty$ in
the strong operator topology. A contraction $T$ on $\mathcal{H}$ is called
completely non-unitary (c.n.u. for short) if there does not exist any nonzero
$T$-reducing subspace $\mathcal{L}$ of $\mathcal{H}$ such that
$T|_{\mathcal{L}}$ is unitary (see [20]). We denote $\mathcal{N}(T)$ and
$\mathcal{R}(T)$ as the kernel and range of $T$, respectively. We frequently
use the identity $\mathcal{N}(T)={\mathcal{R}(T^{*})}^{\perp}$ for any
$T\in\mathcal{B}(\mathcal{H})$. Also $\bigvee\mathcal{M}$ stands for the
closed linear span of a subset $\mathcal{M}$ of $\mathcal{H}$. An operator $T$
on $\mathcal{H}$ is called a partial isometry if $||Th||=||h||$ for all
$h\in\mathcal{N}(T)^{\perp}.$ We say that $T$ is a power partial isometry if
$T^{n}$ is a partial isometry for all $n\geq 1$.

We now recall canonical decomposition theorem for a contraction ([20]). In
case of an isometry, the canonical decomposition theorem coincides with the
classical Wold-von Neumann decomposition.

######  Theorem 2.1.

A contraction $T$ on a Hilbert space $\mathcal{H}$ corresponds a unique
decomposition of $\mathcal{H}$ into an orthogonal sum of two $T$-reducing
subspaces $\mathcal{H}={\mathcal{H}}_{u}\oplus{\mathcal{H}}_{\neg u}$ such
that $T|_{{\mathcal{H}}_{u}}$ is unitary and $T|_{{\mathcal{H}}_{\neg u}}$ is
c.n.u. ( ${\mathcal{H}}_{u}$ or ${\mathcal{H}}_{\neg u}$ may equal to
$\\{0\\}$). Moreover,

${\mathcal{H}}_{u}=\\{h\in\mathcal{H}:\|T^{n}h\|=\|h\|=\|{T^{*}}^{n}h\|~{}~{}\mbox{for}~{}~{}n=1,2,\ldots\\}.$

Here $T_{u}=T|_{{\mathcal{H}}_{u}}$ and $T_{\neg u}=T|_{{\mathcal{H}}_{\neg
u}}$ are called unitary part and c.n.u. part of $T$, respectively and
$T=T_{u}\oplus T_{\neg u}$ is called the canonical decomposition of $T$.

The above theorem can be rewritten as follows:

######  Theorem 2.2.

Let $T$ be a contraction on a Hilbert space $\mathcal{H}$. Then $\mathcal{H}$
decomposes as a direct sum of two $T$-reducing subspaces

$\displaystyle\mathcal{H}_{u}=\bigcap_{m\in\mathbb{Z}_{+}}[\mathcal{N}(I-T^{*m}T^{m})\cap\mathcal{N}(I-T^{m}T^{*m})],$
and $\displaystyle\mathcal{H}_{\neg
u}:=\mathcal{H}\ominus\mathcal{H}_{u}=\bigvee_{m\in\mathbb{Z}_{+}}\\{\mathcal{R}(I-T^{*m}T^{m})\cup\mathcal{R}(I-T^{m}T^{*m})\\}.$

Also $T_{u}=T|_{{\mathcal{H}}_{u}}$ and $T_{\neg u}=T|_{{\mathcal{H}}_{\neg
u}}$ are called unitary part and c.n.u. part of $T$, respectively.

###### Proof.

Suppose that $T\in\mathcal{B}(\mathcal{H})$ is a contraction. Define the
defect operators of $T$ as

${D_{T}}=(I-T^{*}T)^{1/2}\ and\ {D_{T^{*}}}=(I-TT^{*})^{1/2}.$

Clearly, ${D_{T}}$ and ${D_{T^{*}}}$ are positive operators and bounded by 0
and 1. Now for each $h\in\mathcal{H}$

$\langle D_{T}^{2}h,h\rangle=0\iff D_{T}h=0\iff\|Th\|=\|h\|.$

Therefore, the space $\\{h\in\mathcal{H}:\|Th\|=\|h\|\\}$ coincides with
$\mathcal{N}({D_{T}})=\\{h\in\mathcal{H}:D_{T}h=0\\}$. Consider for each
$m\in\mathbb{Z}$

$\displaystyle T(m)=\begin{cases}T^{m}&\text{if}\ m\geq 1,\\\ I&\text{if}\
m=0,\\\ T^{*|m|}&\text{if}~{}~{}m\leq-1.\end{cases}$

Then for each fixed $m$ in $\mathbb{Z}$, the space
$\\{h\in\mathcal{H}:\|T(m)h\|=\|h\|\\}$ is same as
$\mathcal{N}({D_{T(m)}})=\\{h\in\mathcal{H}:{D_{T(m)}}h=0\\}$. Thus the space
$\mathcal{H}_{u}$ can be rewritten as

$\displaystyle\mathcal{H}_{u}$
$\displaystyle=\\{h\in\mathcal{H}:\|T^{m}h\|=\|h\|=\|{T^{*}}^{m}h\|\
\text{for}\ m\in\mathbb{{N}}\\}$
$\displaystyle=\\{h\in\mathcal{H}:\|T(m)h\|=\|h\|\ \text{for}\
m\in\mathbb{Z}\\}$
$\displaystyle=\bigcap_{m=-\infty}^{\infty}\mathcal{N}({D_{T(m)}}),$

where

$\ D_{T(m)}=\begin{cases}(I-T^{*m}T^{m})^{\frac{1}{2}}&\text{if}\ m\geq 0,\\\
(I-T^{|m|}T^{*|m|})^{\frac{1}{2}}&\text{if}\ m\leq-1.\end{cases}$

Since for each $m\in\mathbb{Z}$ the operator $D_{T(m)}$ is positive on
$\mathcal{H}$, $\mathcal{N}(D_{T(m)})=\mathcal{N}({D^{2}_{T(m)}})$. Therefore,

$\displaystyle\mathcal{N}({D_{T(m)}})=\begin{cases}\mathcal{N}(I-T^{*m}T^{m})&\text{if}\
m\geq 0,\\\ \mathcal{N}(I-T^{n}T^{*n})&\text{if}\ n=|m|,\ m\leq-1.\end{cases}$

Hence

$\displaystyle\mathcal{H}_{u}$
$\displaystyle=\bigcap_{m=0}^{\infty}\mathcal{N}(I-T^{*m}T^{m})\cap\bigcap_{n=1}^{\infty}\mathcal{N}(I-T^{n}T^{*n})$
$\displaystyle=\bigcap_{m\in\mathbb{Z}_{+}}[\mathcal{N}(I-T^{*m}T^{m})\cap\mathcal{N}(I-T^{m}T^{*m})]$

and

$\displaystyle\mathcal{H}_{\neg u}:=\mathcal{H}_{u}^{\perp}$
$\displaystyle=[\bigcap_{m\in\mathbb{Z}_{+}}\\{\mathcal{N}(I-T^{*m}T^{m})\cap\mathcal{N}(I-T^{m}T^{*m})\\}]^{\perp}$
$\displaystyle=\bigvee_{m\in\mathbb{Z}_{+}}\\{\mathcal{R}(I-T^{*m}T^{m})\cup\mathcal{R}(I-T^{m}T^{*m})\\}.$

This completes the proof.

######  Remark 2.3.

If a contraction $T\in\mathcal{B}(\mathcal{H})$ is a power partial isometry,
then for each $n\in\mathbb{Z}_{+}$, $T^{*n}T^{n}=P_{\mathcal{R}(T^{*n})}$ and
$T^{n}T^{*n}=P_{\mathcal{R}(T^{n})}$, where $P_{\mathcal{R}(T^{*n})}$ and
$P_{\mathcal{R}(T^{n})}$ are the orthogonal projections of $\mathcal{H}$ onto
${\mathcal{R}(T^{*n})}$ and ${\mathcal{R}(T^{n})}$, respectively. Now from
Theorem 2.2, we have

$\displaystyle\mathcal{H}_{u}$
$\displaystyle=\bigcap_{n\in\mathbb{Z}_{+}}[\mathcal{N}(I-T^{*n}T^{n})\cap\mathcal{N}(I-T^{n}T^{*n})]$
$\displaystyle=\bigcap_{n\in\mathbb{Z}_{+}}[\mathcal{N}(I-P_{\mathcal{R}(T^{*n})})\cap\mathcal{N}(I-P_{\mathcal{R}(T^{n})})]$
$\displaystyle=\bigcap_{n\in\mathbb{Z}_{+}}[\mathcal{R}(P_{\mathcal{R}(T^{*n})})\cap\mathcal{R}(P_{\mathcal{R}(T^{n})})]$
$\displaystyle=\bigcap_{n\in\mathbb{Z}_{+}}[T^{*n}\mathcal{H}\cap
T^{n}\mathcal{H}],$

and

$\displaystyle\mathcal{H}_{\neg u}$
$\displaystyle=\bigvee_{n\in\mathbb{Z}_{+}}\\{\mathcal{R}(I-T^{*n}T^{n})\cup\mathcal{R}(I-T^{n}T^{*n})\\}$
$\displaystyle=\bigvee_{n\in\mathbb{Z}_{+}}\\{\mathcal{N}(P_{\mathcal{R}(T^{*n})})\cup\mathcal{N}(P_{\mathcal{R}(T^{n})})\\}.$

######  Remark 2.4.

Let $T$ be an isometry on $\mathcal{H}$. Then $T^{*}T=I$ and
$T^{*}\mathcal{H}=\mathcal{H}$. Since every isometry is a power partial
isometry, from the last remark, we readily have the unitary part
$\mathcal{H}_{u}$ of $T$ as

$\displaystyle\mathcal{H}_{u}=\bigcap_{n\in\mathbb{Z}_{+}}T^{n}\mathcal{H},$

and the c.n.u part $\mathcal{H}_{\neg u}$ becomes

$\displaystyle\mathcal{H}_{\neg
u}=\bigvee_{n\in\mathbb{Z}_{+}}\\{\mathcal{R}(I-T^{n}T^{*n})\\}=\bigvee_{n\in\mathbb{Z}_{+}}\\{\mathcal{N}(P_{\mathcal{R}(T^{n})})\\}.$

Again for $n\geq 1$

$\displaystyle\mathcal{R}(I-T^{n}T^{*n})$ $\displaystyle=\mathcal{H}\ominus
T^{n}T^{*n}\mathcal{H}=\mathcal{H}\ominus T^{n}\mathcal{H}$
$\displaystyle=(\mathcal{H}\ominus T\mathcal{H})\oplus(T\mathcal{H}\ominus
T^{2}\mathcal{H})\oplus\cdots\oplus(T^{n-1}\mathcal{H}\ominus
T^{n}\mathcal{H})$ $\displaystyle=\mathcal{N}(T^{*})\oplus
T\mathcal{N}(T^{*})\oplus\cdots\oplus T^{n-1}\mathcal{N}(T^{*})$
$\displaystyle=\bigoplus_{k=0}^{n-1}T^{k}\mathcal{N}(T^{*}).$

Since
$\mathcal{R}(I-TT^{*})\subseteq\mathcal{R}(I-T^{2}T^{*2})\subseteq\cdots\subseteq\mathcal{R}(I-T^{n}T^{*n})\subseteq\cdots$,

$\displaystyle\mathcal{H}_{\neg
u}=\bigvee_{n=1}^{\infty}\Big{\\{}\mathcal{R}(I-T^{n}T^{*n})\Big{\\}}=\bigoplus_{n=0}^{\infty}T^{n}\mathcal{N}(T^{*}).$

Therefore, the canonical decomposition of $T$ coincides with the Wold
decomposition.

##  3\. Decomposition for $\mathcal{U}_{2}$-twisted contractions

In this section, we achieve the explicit orthogonal decomposition spaces for
pairs of doubly twisted contractions (in particular, doubly twisted
isometries) on Hilbert spaces. Our approach is different and the results unify
all the existing results in the literature studied by many researchers, like
Słociński [22], Burdak [5], Popovici [17], [18], Catepillán et al. [10], and
the recent results of Jeu and Pinto [14], Rakshit, Sarkar, and Suryawanshi
[19].

We shall work in the following fixed set-up.

######  Definition 3.1.

($\mathcal{U}_{n}$-twisted contractions) Let $n>1$ and $\\{U_{ij}\\}$ for
$1\leq i<j\leq n$ be $\binom{n}{2}$ commuting unitaries on a Hilbert space
$\mathcal{H}$ such that $U_{ji}:=U^{*}_{ij}$. We say that an $n$-tuple of
contractions $(T_{1},\dots,T_{n})$ on $\mathcal{H}$ is a
$\mathcal{U}_{n}$-twisted contraction with respect to a twist
$\\{U_{ij}\\}_{i<j}$ if

$T_{i}T_{j}=U_{ij}T_{j}T_{i};\hskip 28.45274pt\hskip
28.45274ptT_{i}^{*}T_{j}=U^{*}_{ij}T_{j}T_{i}^{*}\hskip
28.45274pt\mbox{and}\hskip 28.45274ptT_{k}U_{ij}=U_{ij}T_{k}$

for all $i,j,k=1,\dots,n$ and $i\neq j$. We simply say that the tuple
$(T_{1},\dots,T_{n})$ is a $\mathcal{U}_{n}$-twisted contractions without
referencing the twist $\\{U_{ij}\\}_{1\leq i<j\leq n}$.

In particular, if $U_{ij}=I$ for all $1\leq i<j\leq n$, then the tuple
$(T_{1},\ldots,T_{n})$ is said to be doubly commuting contraction, that is,
$T_{i}T_{j}=T_{j}T_{i}$ and $T_{i}T_{j}^{*}=T_{j}^{*}T_{i}$ for $1\leq i<j\leq
n$. If $n=2$, then we shall refer to $(T_{1},T_{2})$ as a pair of doubly
twisted contraction or $\mathcal{U}_{2}$-twisted contraction on $\mathcal{H}$.
If $n=1$, then $n$-tuple reduces to a single contraction.

######  Remark 3.2.

Let $(T_{1},\dots,T_{n})$ be an $n$-tuple of isometries on $\mathcal{H}$. For
$1\leq i<j\leq n$, let $\\{U_{ij}\\}$ be $\binom{n}{2}$ commuting unitaries on
a Hilbert space $\mathcal{H}$ such that $U_{ji}:=U^{*}_{ij}$. Then the
relation $T_{i}^{*}T_{j}=U^{*}_{ij}T_{j}T_{i}^{*}$ and
$T_{k}U_{ij}=U_{ij}T_{k}$ implies $T_{i}T_{j}={U}_{ij}T_{j}T_{i}$ for all
$i,j,k=1,\dots,n$ and $i\neq j$ (see [15], [19]). However, this fact is not
true for an $n$-tuple of contractions.

The following result is simple, but plays an important role in the sequel.

######  Lemma 3.3.

Let $(T_{1},\dots,T_{n})$ be an $n$-tuple of $\mathcal{U}_{n}$-twisted
contractions on a Hilbert space $\mathcal{H}$, and let $l,m\in\mathbb{Z}_{+}$.
Then for all $i\neq j$,

  1. (1)

$T_{i}$ commutes with $T_{j}^{m}T_{j}^{*m}$ and $T_{j}^{*m}T_{j}^{m}$;

  2. (2)

$T_{i}^{*}$ commutes with $T_{j}^{m}T_{j}^{*m}$ and $T_{j}^{*m}T_{j}^{m}$;

  3. (3)

$T_{i}^{l}T_{i}^{*l}$, $T_{i}^{*l}T_{i}^{l}$ commute with the operators
$(I-T_{j}^{m}T_{j}^{*m})$ and $(I-T_{j}^{*m}T_{j}^{m})$.

###### Proof.

Suppose that $l,m\in\mathbb{Z}_{+}$. Now for $i\neq j$, using the definition
repeated times, we have

$\displaystyle
T_{i}T_{j}^{m}T_{j}^{*m}=U_{ij}^{m}T_{j}^{m}T_{i}T_{j}^{*m}=U_{ij}^{m}{U}_{ij}^{*m}T_{j}^{m}T_{j}^{*m}T_{i}=T_{j}^{m}T_{j}^{*m}T_{i}$
and $\displaystyle
T_{i}T_{j}^{*m}T_{j}^{m}={U}_{ij}^{*m}T_{j}^{*m}T_{i}T_{j}^{m}={U}_{ij}^{*m}U_{ij}^{m}T_{j}^{*m}T_{j}^{m}T_{i}=T_{j}^{*m}T_{j}^{m}T_{i}.$

Hence the first part is proved.

Second part follows from the first part by just taking the adjoint of those
operators. Using the part (1) and (2), we can easily prove the last part.

We shall first concentrate on pairs of doubly twisted contractions with some
examples on Hilbert spaces $\mathcal{H}$ and their decomposition.

######  Example 3.4.

Let $H^{2}(\mathbb{D})$ denotes as the Hardy space over the unit disc
$\mathbb{D}$. The weighted shift $M_{z}^{\alpha}$ on $H^{2}(\mathbb{D})$ is
defined by $M_{z}^{\alpha}(f)=\alpha zf$ for all $f\in H^{2}(\mathbb{D})$,
where $z$ is the co-ordinate function and $|\alpha|\leq 1$. Now the Hardy
space over the bidisc $\mathbb{D}^{2}$, denoted by $H^{2}(\mathbb{D}^{2})$,
can be identified with $H^{2}(\mathbb{D})\otimes H^{2}(\mathbb{D})$ through
the canonical unitary $\Gamma:H^{2}(\mathbb{D})\otimes
H^{2}(\mathbb{D})\rightarrow H^{2}(\mathbb{D}^{2})$ defined by
$\Gamma(z^{m_{1}}\otimes z^{m_{2}})=z_{1}^{m_{1}}z_{2}^{m_{2}}$ for
$(m_{1},m_{2})\in\mathbb{Z}_{+}^{2}$.

For each fixed $r\in S^{1}:=\\{z\in\mathbb{C}:|z|=1\\},$ we define an operator
$A_{r}$ on $H^{2}(\mathbb{D})$ as

$A_{r}z^{n}=\frac{r^{n}}{2}z^{n}\hskip 28.45274pt(n\in\mathbb{Z}_{+}),$

where $\\{1,z,z^{2},\dots\\}$ is an orthonormal basis for $H^{2}(\mathbb{D})$.
Then

$(M_{z}^{\alpha}A_{r})(z^{n})=\frac{\alpha
r^{n}}{2}z^{n+1}\quad\text{and}\quad(A_{r}M_{z}^{\alpha})(z^{n})=\frac{\alpha
r^{n+1}}{2}z^{n+1}~{}~{}\text{for}~{}~{}n\in\mathbb{Z}_{+}.$

Again

$\big{(}[M_{z}^{\alpha}]^{*}A_{r}\big{)}(z^{n})=\begin{cases}\frac{\bar{\alpha}r^{n}}{2}z^{n-1},&\text{if
$n\geq 1$}\\\ 0&\text{if $n=0$},\end{cases}$

and

$\big{(}A_{r}[M_{z}^{\alpha}]^{*}\big{)}(z^{n})=\begin{cases}\frac{\bar{\alpha}r^{n-1}}{2}z^{n-1}&\text{if
$n\geq 1$}\\\ 0&\text{if $n=0$}.\end{cases}$

Hence $A_{r}M_{z}^{\alpha}=rM_{z}^{\alpha}A_{r}$ and
$[M_{z}^{\alpha}]^{*}A_{r}=rA_{r}[M_{z}^{\alpha}]^{*}$. We now define $T_{1}$
and $T_{2}$ on $H^{2}(\mathbb{D}^{2})$ such that

$T_{1}=A_{r}\otimes M_{z}^{\alpha}\ \ \ \text{and}\quad
T_{2}=M_{z}^{\alpha}\otimes I_{H^{2}(\mathbb{D})}.$

Therefore, we can check that $(T_{1},T_{2})$ is a pair of contractions on
$H^{2}(\mathbb{D}^{2})$. Moreover,

$T_{1}T_{2}=A_{r}M_{z}^{\alpha}\otimes
M_{z}^{\alpha}=rM_{z}^{\alpha}A_{r}\otimes
M_{z}^{\alpha}=r(M_{z}^{\alpha}A_{r}\otimes M_{z}^{\alpha})=rT_{2}T_{1}.$

and

$T_{2}^{*}T_{1}=[M_{z}^{\alpha}]^{*}A_{r}\otimes
M_{z}^{\alpha}=rA_{r}[M_{z}^{\alpha}]^{*}\otimes
M_{z}^{\alpha}=r(A_{r}[M_{z}^{\alpha}]^{*}\otimes
M_{z}^{\alpha})=rT_{1}T_{2}^{*}.$

Consider $\mathcal{H}=H^{2}(\mathbb{D}^{2})\oplus H^{2}(\mathbb{D}^{2})$. We
now define two contractions on $\mathcal{H}$ as
$T_{1}^{\prime}=\mbox{diag}(T_{1},T_{2})$ and
$T_{2}^{\prime}=\mbox{diag}(T_{2},T_{1})$. Set
$U=\mbox{diag}(rI_{H^{2}(\mathbb{D}^{2})},\bar{r}I_{H^{2}(\mathbb{D}^{2})})$,
$|r|=1$. Clearly, $U$ is unitary on $\mathcal{H}$ and

$T_{1}^{\prime}T_{2}^{\prime}=\begin{bmatrix}T_{1}T_{2}&0\\\
0&T_{2}T_{1}\end{bmatrix}=\begin{bmatrix}rT_{2}T_{1}&0\\\
0&\bar{r}T_{1}T_{2}\end{bmatrix}=\begin{bmatrix}rI_{\mathcal{H}^{2}(\mathbb{D}^{2})}&0\\\
0&\bar{r}I_{\mathcal{H}^{2}(\mathbb{D}^{2})}\end{bmatrix}T_{2}^{\prime}T_{1}^{\prime}=UT_{2}^{\prime}T_{1}^{\prime}$

and

$T_{2}^{\prime*}T_{1}^{\prime}=\begin{bmatrix}T_{2}^{*}T_{1}&0\\\
0&T_{1}^{*}T_{2}\end{bmatrix}=\begin{bmatrix}rT_{1}T_{2}^{*}&0\\\
0&\bar{r}T_{2}T_{1}^{*}\end{bmatrix}=\begin{bmatrix}rI_{\mathcal{H}^{2}(\mathbb{D}^{2})}&0\\\
0&\bar{r}I_{\mathcal{H}^{2}(\mathbb{D}^{2})}\end{bmatrix}T_{1}^{\prime}T_{2}^{\prime*}=UT_{1}^{\prime}T_{2}^{\prime*}.$

Again $T_{1}^{\prime}U=UT_{1}^{\prime}$, and
$T_{2}^{\prime}U=UT_{2}^{\prime}$. So it follows that
$(T_{1}^{\prime},T_{2}^{\prime})$ is a $\mathcal{U}_{2}$-twisted contractions
on $\mathcal{H}$ with a twist $\mathcal{U}_{2}=\\{U\\}$.

######  Example 3.5.

Let $H^{2}_{\mathcal{E}}(\mathbb{D}^{2})$ denotes as the $\mathcal{E}$-valued
Hardy space over the unit bidisc $\mathbb{D}^{2}$, where $\mathcal{E}$ is any
Hilbert space. We can also identify $H^{2}_{\mathcal{E}}(\mathbb{D}^{2})$ as
$H^{2}(\mathbb{D}^{2})\otimes\mathcal{E}$. The weighted shift operators
$M_{z_{i}}^{\alpha_{i}}$ is defined by
$M_{z_{i}}^{\alpha_{i}}f=\alpha_{i}z_{i}f$ for $f\in
H^{2}_{\mathcal{E}}(\mathbb{D}^{2})$, where $z_{i}\in\mathbb{D}$,
$|\alpha_{i}|\leq 1$ for $i=1,2.$ We now define operators $T_{1}$ and $T_{2}$
on $H^{2}_{\mathcal{E}}(\mathbb{D}^{2})$ as

$T_{1}=M_{z_{1}}^{\alpha_{1}}\hskip 28.45274pt\text{and}\hskip
28.45274ptT_{2}=M_{z_{2}}^{\alpha_{2}}D[U],$

where $U$ is unitary on $\mathcal{E}$ and $D[U]$ on
$H^{2}_{\mathcal{E}}(\mathbb{D}^{2})$ is defined by

$D[U](z_{1}^{m_{1}}z_{2}^{m_{2}}\eta)=z_{1}^{m_{1}}z_{2}^{m_{2}}(U^{m_{1}}\eta)\quad\mbox{for}\quad(m_{1},m_{2})\in\mathbb{Z}_{+}^{2},\
\eta\in\mathcal{E}.$

It is easy to check that $(T_{1},T_{2})$ is a pair of contractions on
$H^{2}_{\mathcal{E}}(\mathbb{D}^{2})$. Moreover,

$\displaystyle
T_{2}T_{1}(z_{1}^{m_{1}}z_{2}^{m_{2}}\eta)=M_{z_{2}}^{\alpha_{2}}D[U](\alpha_{1}z_{1}^{m_{1}+1}z_{2}^{m_{2}}\eta)=\alpha_{1}\alpha_{2}z_{1}^{m_{1}+1}z_{2}^{m_{2}+1}U^{m_{1}+1}\eta,$

and

$\displaystyle T_{1}T_{2}(z_{1}^{m_{1}}z_{2}^{m_{2}}\eta)$
$\displaystyle=M_{z_{1}}^{\alpha_{1}}(\alpha_{2}z_{1}^{m_{1}}z_{2}^{m_{2}+1}U^{m_{1}}\eta)$
$\displaystyle=\alpha_{1}\alpha_{2}z_{1}^{m_{1}+1}z_{2}^{m_{2}+1}U^{m_{1}}\eta$
$\displaystyle=(I_{H^{2}({\mathbb{D}})}\otimes
U^{*})(\alpha_{1}\alpha_{2}z_{1}^{m_{1}+1}z_{2}^{m_{2}+1}U^{m_{1}+1}\eta)$
$\displaystyle=\widetilde{U}T_{2}T_{1}(z_{1}^{m_{1}}z_{2}^{m_{2}}\eta)$

where $\widetilde{U}=(I_{H^{2}({\mathbb{D}^{2}})}\otimes U^{*})$ is unitary on
$H^{2}_{\mathcal{E}}(\mathbb{D}^{2})$ (or on
$H^{2}(\mathbb{D}^{2})\otimes{\mathcal{E}}$). Therefore

$T_{1}T_{2}=\widetilde{U}T_{2}T_{1}$

on $H^{2}_{\mathcal{E}}(\mathbb{D}^{2})$. Again

$\displaystyle T_{1}^{*}T_{2}(z_{1}^{m_{1}}z_{2}^{m_{2}}\eta)$
$\displaystyle=[M_{z_{1}}^{\alpha_{1}}]^{*}(\alpha_{2}z_{1}^{m_{1}}z_{2}^{m_{2}+1}U^{m_{1}}\eta)$
$\displaystyle=\begin{cases}\bar{\alpha_{1}}\alpha_{2}z_{1}^{m_{1}-1}z_{2}^{m_{2}+1}U^{m_{1}}\eta&\text{if
$m_{1}\geq 1$}\\\ 0&\text{if $m_{1}=0$},\end{cases}$

and

$\displaystyle T_{2}T_{1}^{*}(z_{1}^{m_{1}}z_{2}^{m_{2}}\eta)$
$\displaystyle=\begin{cases}M_{z_{2}}^{\alpha_{2}}D[U](\bar{\alpha_{1}}z_{1}^{m_{1}-1}z_{2}^{m_{2}}\eta)&\text{if
$m_{1}\geq 1$}\\\ 0&\text{if $m_{1}=0$}\end{cases}$
$\displaystyle=\begin{cases}\bar{\alpha_{1}}\alpha_{2}z_{1}^{m_{1}-1}z_{2}^{m_{2}+1}U^{m_{1}-1}\eta&\text{if
$m_{1}\geq 1$}\\\ 0&\text{if $m_{1}=0$}.\end{cases}$

Therefore,
$T_{1}^{*}T_{2}(z_{1}^{m_{1}}z_{2}^{m_{2}}\eta)=(I_{H^{2}({\mathbb{D}^{2}})}\otimes
U)T_{2}T_{1}^{*}(z_{1}^{m_{1}}z_{2}^{m_{2}}\eta)$, that is,
$T_{1}^{*}T_{2}=\widetilde{U}^{*}T_{2}T_{1}^{*}$ on
$H^{2}_{\mathcal{E}}(\mathbb{D}^{2})$. Moreover,
$T_{i}\widetilde{U}=\widetilde{U}T_{i}$ for $i=1,2$. Hence $(T_{1},T_{2})$ is
a $\mathcal{U}_{2}$-twisted contraction on
$H^{2}_{\mathcal{E}}(\mathbb{D}^{2})$.

In particular, we take $\mathcal{E}=\ell^{2}(\mathbb{Z})$ and the bilateral
shift $W$ on $\ell^{2}(\mathbb{Z})$. Then the above pair $(T_{1},T_{2})$ is a
doubly twisted contraction with respect to the twist
$\\{I_{H^{2}({\mathbb{D}^{2}})}\otimes W^{*}\\}$ on the Hilbert space
$H^{2}_{\ell^{2}(\mathbb{Z})}(\mathbb{D}^{2})$.

######  Example 3.6.

For each fixed $r\in S^{1},$ we define a weighted shift operator $B_{r}$ on
the Hardy space $H^{2}(\mathbb{D})$ such that

$B_{r}z^{n}=r^{n+1}z^{n+1}\hskip 28.45274pt(n\in\mathbb{Z}_{+})$

where $\\{1,z,z^{2},\dots\\}$ is an orthonormal basis for $H^{2}(\mathbb{D})$.
Let $M_{z}$ be the multiplication operator on $H^{2}(\mathbb{D})$ by the
coordinate function $z$. Then

$(M_{z}B_{r})(z^{n})=r^{n+1}z^{n+2}\quad\text{and}\quad(B_{r}M_{z})(z^{n})=r^{n+2}z^{n+2}\
\ \text{for $n\in\mathbb{Z}_{+}$}.$

Again

$(M_{z}^{*}B_{r})(z^{n})=r^{n+1}z^{n}~{}~{}~{}\text{ $\forall\ n\geq 0$},\\\ $

and

$(B_{r}M_{z}^{*})(z^{n})=\begin{cases}r^{n}z^{n}&\text{if $n\geq 1$}\\\
0&\text{if $n=0$}.\end{cases}$

Hence $B_{r}M_{z}=rM_{z}B_{r}$ but $M_{z}^{*}B_{r}\neq rB_{r}M_{z}^{*}$. Now
define $T_{1},T_{2}$ on $H^{2}(\mathbb{D}^{2})$ as

$T_{1}=B_{r}\otimes M_{z}\ \ \ \text{and}\quad T_{2}=M_{z}\otimes
I_{H^{2}(\mathbb{D})}.$

Then it is easy to see that $(T_{1},T_{2})$ is a pair of isometries on
$H^{2}(\mathbb{D}^{2})$. Also

$T_{1}T_{2}=B_{r}M_{z}\otimes M_{z}=rM_{z}B_{r}\otimes
M_{z}=r(M_{z}B_{r}\otimes M_{z})=rT_{2}T_{1}.$

On the other hand

$T_{2}^{*}T_{1}=M_{z}^{*}B_{r}\otimes M_{z}\quad\text{and}\quad
T_{1}T_{2}^{*}=B_{r}M_{z}^{*}\otimes M_{z}.$

Therefore, $T_{2}^{*}T_{1}\neq rT_{1}T_{2}^{*}$ as $M_{z}^{*}B_{r}\neq
rB_{r}M_{z}^{*}$. Consider $\mathcal{H}=H^{2}(\mathbb{D}^{2})\oplus
H^{2}(\mathbb{D}^{2})$. We now define two isometries on $\mathcal{H}$ as
$T_{1}^{\prime}=\mbox{diag}(T_{1},T_{2})$ and
$T_{2}^{\prime}=\mbox{diag}(T_{2},T_{1})$. Set
$U=\mbox{diag}(rI_{H^{2}(\mathbb{D}^{2})},\bar{r}I_{H^{2}(\mathbb{D}^{2})})$,
$|r|=1$. Clearly, $U$ is unitary on $\mathcal{H}$ and

$T_{1}^{\prime}T_{2}^{\prime}=UT_{2}^{\prime}T_{1}^{\prime},\quad
T_{1}^{\prime}U=UT_{1}^{\prime},\quad T_{2}^{\prime}U=UT_{2}^{\prime}$

but $T_{2}^{\prime*}T_{1}^{\prime}\neq UT_{1}^{\prime}T_{2}^{\prime*}$.
Therefore, $(T_{1}^{\prime},T_{2}^{\prime})$ is not a pair of doubly twisted
isometry on $\mathcal{H}$ with a twist $\mathcal{U}_{2}=\\{U\\}$.

The following result will be used frequently in the sequel.

######  Lemma 3.7.

Let $(T_{1},T_{2})$ be a pair of doubly twisted operator on $\mathcal{H}$ such
that $T_{1}$ is a contraction. Let
$\mathcal{H}=\mathcal{H}_{u}^{1}\oplus\mathcal{H}_{\neg u}^{1}$ be the
canonical decomposition of contraction $T_{1}$. Then the decomposition reduces
$T_{2}$.

###### Proof.

Suppose that $(T_{1},T_{2})$ is a pair of doubly twisted operator with a twist
$\mathcal{U}_{2}=\\{U\\}$ on $\mathcal{H}$ and $T_{1}$ is a contraction. Then
from the above Theorem 2.2, we get
$\mathcal{H}=\mathcal{H}_{u}^{1}\oplus\mathcal{H}_{\neg u}^{1}$, where
$\mathcal{H}_{u}^{1},\mathcal{H}_{\neg u}^{1}$ reduce $T_{1}$ and

$\displaystyle\mathcal{H}_{u}^{1}$
$\displaystyle=\bigcap_{m\in\mathbb{Z}_{+}}[\mathcal{N}(I-T_{1}^{*m}T_{1}^{m})\cap\mathcal{N}(I-T_{1}^{m}T_{1}^{*m})],$
$\displaystyle\mathcal{H}_{\neg u}^{1}$
$\displaystyle=\bigvee_{m\in\mathbb{Z}_{+}}\\{\mathcal{R}(I-T_{1}^{*m}T_{1}^{m})\cup\mathcal{R}(I-T_{1}^{m}T_{1}^{*m})\\}.$

Since $(T_{1},T_{2})$ is a pair of doubly twisted operator, using Lemma 3.3 we
have $T_{2}(\mathcal{H}_{u}^{1})\subseteq\mathcal{H}_{u}^{1}$ and
$T_{2}(\mathcal{H}_{\neg u}^{1})\subseteq\mathcal{H}_{\neg u}^{1}$.

This finishes the proof.

Let $\left(T_{1},T_{2}\right)$ be a pair of doubly twisted contraction on
$\mathcal{H}$. Suppose that
$\mathcal{H}=\mathcal{H}_{u}^{1}\oplus\mathcal{H}_{\neg u}^{1}$ is the
canonical decomposition for contraction $T_{1}$ such that
$T_{1}|_{\mathcal{H}_{u}^{1}}$ is unitary and $T_{1}|_{\mathcal{H}_{\neg
u}^{1}}$ is completely non-unitary. The above Lemma 3.7 implies that the
subspaces $\mathcal{H}_{u}^{1}$, $\mathcal{H}_{\neg u}^{1}$ reduce the
contraction $T_{2}$ and from Theorem 2.2, we get

$\displaystyle\mathcal{H}_{u}^{1}=\bigcap_{m_{1}\in\mathbb{Z}_{+}}[\mathcal{N}(I-T_{1}^{*m_{1}}T_{1}^{m_{1}})\cap\mathcal{N}(I-T_{1}^{m_{1}}T_{1}^{*m_{1}})],$
$\displaystyle\mathcal{H}_{\neg
u}^{1}=\bigvee_{m_{1}\in\mathbb{Z}_{+}}\\{\mathcal{R}(I-T_{1}^{*m_{1}}T_{1}^{m_{1}})\cup\mathcal{R}(I-T_{1}^{m_{1}}T_{1}^{*m_{1}})\\}.$

Since $T_{2}|_{\mathcal{H}_{u}^{1}}$ is a contraction, the canonical
decomposition yields
$\mathcal{H}_{u}^{1}=\mathcal{H}_{uu}\oplus\mathcal{H}_{u\neg u}$, where
$T_{1}|_{\mathcal{H}_{uu}}$, $T_{1}|_{\mathcal{H}_{u\neg u}}$,
$T_{2}|_{\mathcal{H}_{uu}}$ are unitary and $T_{2}|_{\mathcal{H}_{u\neg u}}$
is c.n.u. Again $\mathcal{H}_{u}^{1}$ reduces $T_{2}$, and hence
$(T_{2}|_{\mathcal{H}_{u}^{1}})^{m}=T_{2}^{m}|_{\mathcal{H}_{u}^{1}}$ and
$(T_{2}|_{\mathcal{H}_{u}^{1}})^{*m}=T_{2}^{*m}|_{\mathcal{H}_{u}^{1}}$ for
any $m\in\mathbb{Z}_{+}$. Thus from Theorem 2.2, the subspaces
$\mathcal{H}_{uu}$ and $\mathcal{H}_{u\neg u}$ can be written as

$\displaystyle\mathcal{H}_{uu}$
$\displaystyle=\bigcap_{m_{2}\in\mathbb{Z}_{+}}[\mathcal{N}(I_{\mathcal{H}_{u}^{1}}-T_{2}^{*m_{2}}|_{\mathcal{H}_{u}^{1}}T_{2}^{m_{2}}|_{\mathcal{H}_{u}^{1}})\cap\mathcal{N}(I_{\mathcal{H}_{u}^{1}}-T_{2}^{m_{2}}|_{\mathcal{H}_{u}^{1}}T_{2}^{*m_{2}}|_{\mathcal{H}_{u}^{1}})]$
$\displaystyle=\bigcap_{m_{2}\in\mathbb{Z}_{+}}[\mathcal{N}((I-T_{2}^{*m_{2}}T_{2}^{m_{2}})|_{\mathcal{H}_{u}^{1}})\cap\mathcal{N}((I-T_{2}^{m_{2}}T_{2}^{*m_{2}})|_{\mathcal{H}_{u}^{1}})],$
and $\displaystyle\mathcal{H}_{u\neg u}$
$\displaystyle=\bigvee_{m_{2}\in\mathbb{Z}_{+}}\\{(I-T_{2}^{*m_{2}}T_{2}^{m_{2}})\mathcal{H}_{u}^{1}\cup(I-T_{2}^{m_{2}}T_{2}^{*m_{2}}){\mathcal{H}_{u}^{1}}\\}.$

For the rest of the part, let $\mathcal{H}_{\neg u}^{1}=\mathcal{H}_{\neg
uu}\oplus\mathcal{H}_{\neg u\neg u}$ be the canonical decomposition for
contraction $T_{2}|_{\mathcal{H}_{\neg u}^{1}}$ such that
$T_{2}|_{\mathcal{H}_{\neg uu}}$ is unitary and $T_{1}|_{\mathcal{H}_{\neg
uu}},T_{1}|_{\mathcal{H}_{\neg u\neg u}},T_{2}|_{\mathcal{H}_{\neg u\neg u}}$
are c.n.u. Since $\mathcal{H}_{\neg u}^{1}$ is a $T_{2}$-reducing subspace,
$\mathcal{H}_{\neg u}^{1}$ reduces $(I-T_{2}^{*m_{2}}T_{2}^{m_{2}})$ and
$(I-T_{2}^{m_{2}}T_{2}^{*m_{2}})$. Therefore from Theorem 2.2, the subspaces
$\mathcal{H}_{\neg uu}$ and $\mathcal{H}_{\neg u\neg u}$ can be written as

$\displaystyle\mathcal{H}_{\neg uu}$
$\displaystyle=\bigcap_{m_{2}\in\mathbb{Z}_{+}}[\mathcal{N}((I-T_{2}^{*m_{2}}T_{2}^{m_{2}})|_{\mathcal{H}_{\neg
u}^{1}})\cap\mathcal{N}((I_{\mathcal{H}_{\neg
u}^{1}}-T_{2}^{m_{2}}T_{2}^{*m_{2}})|_{\mathcal{H}_{\neg u}^{1}})],$ and
$\displaystyle\mathcal{H}_{\neg u\neg u}$
$\displaystyle=\bigvee_{m_{2}\in\mathbb{Z}_{+}}\\{(I-T_{2}^{*m_{2}}T_{2}^{m_{2}})\mathcal{H}_{\neg
u}^{1}\cup(I-T_{2}^{m_{2}}T_{2}^{*m_{2}}){\mathcal{H}_{\neg u}^{1}}\\}.$

To summarize the above, we have the following result:

######  Theorem 3.8.

Let $\left(T_{1},T_{2}\right)$ be a pair of doubly twisted contraction on a
Hilbert space $\mathcal{H}$. Then there is a unique decomposition

$\mathcal{H}=\mathcal{H}_{uu}\oplus\mathcal{H}_{u\neg
u}\oplus\mathcal{H}_{\neg uu}\oplus\mathcal{H}_{\neg u\neg u},$

where $\mathcal{H}_{uu},\mathcal{H}_{u\neg u},\mathcal{H}_{\neg uu}$, and
$\mathcal{H}_{\neg u\neg u}$ are the subspaces reduce $T_{1}$ and $T_{2}$ such
that

  * •

$T_{1}|_{\mathcal{H}_{uu}},T_{2}|_{\mathcal{H}_{uu}}$ are unitary,

  * •

$T_{1}|_{\mathcal{H}_{u\neg u}}$ is unitary and $T_{2}|_{\mathcal{H}_{u\neg
u}}$ is c.n.u.,

  * •

$T_{1}|_{\mathcal{H}_{\neg uu}}$ is c.n.u. and $T_{2}|_{\mathcal{H}_{\neg
uu}}$ is unitary,

  * •

$T_{1}|_{\mathcal{H}_{\neg u\neg u}},T_{2}|_{\mathcal{H}_{\neg u\neg u}}$ are
c.n.u.

Moreover, the orthogonal subspaces can be formulated as

$\displaystyle\mathcal{H}_{uu}=\bigcap_{m_{2}\in\mathbb{Z}_{+}}[\mathcal{N}((I-T_{2}^{*m_{2}}T_{2}^{m_{2}})|_{\mathcal{H}_{u}^{1}})\cap\mathcal{N}((I-T_{2}^{m_{2}}T_{2}^{*m_{2}})|_{\mathcal{H}_{u}^{1}})],$
$\displaystyle\mathcal{H}_{u\neg
u}=\bigvee_{m_{2}\in\mathbb{Z}_{+}}\\{(I-T_{2}^{*m_{2}}T_{2}^{m_{2}})\mathcal{H}_{u}^{1}\cup(I-T_{2}^{m_{2}}T_{2}^{*m_{2}}){\mathcal{H}_{u}^{1}}\\},$
$\displaystyle\mathcal{H}_{\neg
uu}=\bigcap_{m_{2}\in\mathbb{Z}_{+}}[\mathcal{N}((I-T_{2}^{*m_{2}}T_{2}^{m_{2}})|_{\mathcal{H}_{\neg
u}^{1}})\cap\mathcal{N}((I_{\mathcal{H}_{\neg
u}^{1}}-T_{2}^{m_{2}}T_{2}^{*m_{2}})|_{\mathcal{H}_{\neg u}^{1}})],$
$\displaystyle\mathcal{H}_{\neg u\neg
u}=\bigvee_{m_{2}\in\mathbb{Z}_{+}}\\{(I-T_{2}^{*m_{2}}T_{2}^{m_{2}})\mathcal{H}_{\neg
u}^{1}\cup(I-T_{2}^{m_{2}}T_{2}^{*m_{2}}){\mathcal{H}_{\neg u}^{1}}\\},$

and

$\displaystyle\mathcal{H}_{u}^{1}=\bigcap_{m_{1}\in\mathbb{Z}_{+}}[\mathcal{N}(I-T_{1}^{*m_{1}}T_{1}^{m_{1}})\cap\mathcal{N}(I-T_{1}^{m_{1}}T_{1}^{*m_{1}})],$
$\displaystyle\mathcal{H}_{\neg
u}^{1}=\bigvee_{m_{1}\in\mathbb{Z}_{+}}\\{\mathcal{R}(I-T_{1}^{*m_{1}}T_{1}^{m_{1}})\cup\mathcal{R}(I-T_{1}^{m_{1}}T_{1}^{*m_{1}})\\}.$

Return to our discussion on the decomposition for pairs of doubly twisted
isometries. Suppose that $(T_{1},T_{2})$ is a pair of doubly twisted isometry
with a twist $\mathcal{U}_{2}=\\{U\\}$ on $\mathcal{H}$. Then from the above
Theorem 3.8, we get

$\mathcal{H}_{u}^{1}=\bigcap_{m_{1}=0}^{\infty}T_{1}^{m_{1}}\mathcal{H}\ \
\text{and}\ \ \mathcal{H}_{\neg
u}^{1}=\bigoplus_{m_{1}=0}^{\infty}T_{1}^{m_{1}}\mathcal{N}(T_{1}^{*}).$

Again

$\displaystyle\mathcal{H}_{uu}$
$\displaystyle=\bigcap_{m_{2}\in\mathbb{Z}_{+}}\mathcal{N}(I_{\mathcal{H}_{u}^{1}}-T_{2}^{m_{2}}|_{\mathcal{H}_{u}^{1}}T_{2}^{*m_{2}}|_{\mathcal{H}_{u}^{1}})=\bigcap_{m_{2}\in\mathbb{Z}_{+}}\big{[}\mathcal{N}(T_{2}^{*m_{2}}|_{\mathcal{H}_{u}^{1}})\big{]}^{\perp}$
$\displaystyle=\bigcap_{m_{2}\in\mathbb{Z}_{+}}T_{2}^{m_{2}}\mathcal{H}_{u}^{1}=\bigcap_{m_{1},m_{2}\in\mathbb{Z}_{+}}T_{1}^{m_{1}}T_{2}^{m_{2}}\mathcal{H},$
and $\displaystyle\mathcal{H}_{u\neg u}$
$\displaystyle=\bigvee\Big{\\{}(I-T_{2}^{m_{2}}T_{2}^{*m_{2}}){\mathcal{H}_{u}^{1}}:m_{2}\in\mathbb{Z}_{+}\Big{\\}}=\bigvee_{m_{2}\in\mathbb{Z}_{+}}\bigg{\\{}(I-T_{2}^{m_{2}}T_{2}^{*m_{2}})\big{[}\bigcap_{m_{1}=0}^{\infty}T_{1}^{m_{1}}\mathcal{H}\big{]}\bigg{\\}}.$

Since $T_{2}$ is an isometry and $\mathcal{H}_{u}^{1}$ reduces $T_{2}$,
$(I-T_{2}^{m_{2}}T_{2}^{*m_{2}})|_{\mathcal{H}_{u}^{1}}$ is a projection for
any fixed $m_{2}\in\mathbb{Z}_{+}$. As $(T_{1},T_{2})$ is a pair of doubly
twisted isometry and using Lemma 3.3, we get

$(I-T_{2}^{m_{2}}T_{2}^{*m_{2}})T_{1}=T_{1}(I-T_{2}^{m_{2}}T_{2}^{*m_{2}})\iff
P_{\mathcal{N}(T_{2}^{*m_{2}})}T_{1}=T_{1}P_{\mathcal{N}(T_{2}^{*m_{2}})}.$

Again $\mathcal{N}(T_{2}^{*})$ reduces the unitary $U$ as the pair $(T_{2},U)$
is doubly commuting.

Hence for any fixed $m_{2}\geq 1$, we have

$\displaystyle(I-T_{2}^{m_{2}}T_{2}^{*m_{2}})[\bigcap_{m_{1}=0}^{\infty}T_{1}^{m_{1}}\mathcal{H}]$
$\displaystyle=(I-T_{2}^{m_{2}}T_{2}^{*m_{2}})\mathcal{H}\bigcap(I-T_{2}^{m_{2}}T_{2}^{*m_{2}})T_{1}\mathcal{H}\bigcap\ldots\qquad$
$\displaystyle=(\mathcal{H}\ominus
T_{2}^{m_{2}}T_{2}^{*m_{2}}\mathcal{H})\bigcap T_{1}(\mathcal{H}\ominus
T_{2}^{m_{2}}T_{2}^{*m_{2}}\mathcal{H})\bigcap\ldots$
$\displaystyle=\bigg{(}\bigoplus_{k=0}^{m_{2}-1}T_{2}^{k}\mathcal{N}(T_{2}^{*})\bigg{)}\bigcap
T_{1}\bigg{(}\bigoplus_{k=0}^{m_{2}-1}T_{2}^{k}\mathcal{N}(T_{2}^{*})\bigg{)}\bigcap\ldots$
$\displaystyle=\bigg{(}\bigoplus_{k=0}^{m_{2}-1}T_{2}^{k}\mathcal{N}(T_{2}^{*})\bigg{)}\bigcap\bigg{(}\bigoplus_{k=0}^{m_{2}-1}T_{2}^{k}T_{1}\mathcal{N}(T_{2}^{*})\bigg{)}\bigcap\ldots$
$\displaystyle=\bigcap_{m_{1}=0}^{\infty}\bigg{(}\bigoplus_{k=0}^{m_{2}-1}T_{2}^{k}T_{1}^{m_{1}}\mathcal{N}(T_{2}^{*})\bigg{)}$
$\displaystyle=\bigoplus_{k=0}^{m_{2}-1}T_{2}^{k}\bigg{(}\bigcap_{m_{1}=0}^{\infty}T_{1}^{m_{1}}\mathcal{N}(T_{2}^{*})\bigg{)}.$

Hence

$\displaystyle\mathcal{H}_{u\neg u}$
$\displaystyle=\bigoplus_{m_{2}=0}^{\infty}T_{2}^{m_{2}}\bigg{(}\bigcap_{m_{1}=0}^{\infty}T_{1}^{m_{1}}\mathcal{N}(T_{2}^{*})\bigg{)}.$

Again

$\displaystyle\mathcal{H}_{\neg uu}$
$\displaystyle=\bigcap_{m_{2}\in\mathbb{Z}_{+}}\mathcal{N}((I-T_{2}^{m_{2}}T_{2}^{*m_{2}})|_{\mathcal{H}_{\neg
u}^{1}})$
$\displaystyle=\bigcap_{m_{2}\in\mathbb{Z}_{+}}T_{2}^{m_{2}}{\mathcal{H}_{\neg
u}^{1}}$
$\displaystyle=\bigcap_{m_{2}\in\mathbb{Z}_{+}}\bigg{(}\bigoplus_{m_{1}=0}^{\infty}T_{1}^{m_{1}}T_{2}^{m_{2}}\mathcal{N}(T_{1}^{*})\bigg{)}$
$\displaystyle=\bigoplus_{m_{1}=0}^{\infty}T_{1}^{m_{1}}\bigg{(}\bigcap_{m_{2}\in\mathbb{Z}_{+}}T_{2}^{m_{2}}\mathcal{N}(T_{1}^{*})\bigg{)}.$

Finally by Lemma 3.3, we have

$\displaystyle\mathcal{H}_{\neg u\neg u}$
$\displaystyle=\bigvee\\{(I-T_{2}^{m_{2}}T_{2}^{*m_{2}}){\mathcal{H}_{\neg
u}^{1}}:m_{2}\in\mathbb{Z}_{+}\\}$
$\displaystyle=\bigvee_{m_{2}\in\mathbb{Z}_{+}}\bigg{\\{}(I-T_{2}^{m_{2}}T_{2}^{*m_{2}})\Big{(}\bigoplus_{m_{1}=0}^{\infty}T_{1}^{m_{1}}\mathcal{N}(T_{1}^{*})\Big{)}\bigg{\\}}$
$\displaystyle=\bigvee_{m_{2}\in\mathbb{Z}_{+}}\bigg{\\{}\bigoplus_{m_{1}=0}^{\infty}T_{1}^{m_{1}}(I-T_{2}^{m_{2}}T_{2}^{*m_{2}})\big{[}\mathcal{N}(T_{1}^{*})\big{]}\bigg{\\}}.$

Since $(T_{1},T_{2})$ is a pair of doubly twisted isometry, we get
$\mathcal{N}(T_{1}^{*})$ is $T_{2}$-reducing subspace. Hence
$T_{2}|_{\mathcal{N}(T_{1}^{*})}$ is an isometry. Now

$\displaystyle(I-T_{2}T_{2}^{*})[\mathcal{N}(T_{1}^{*})]$
$\displaystyle=\mathcal{N}(T_{1}^{*})\ominus T_{2}\mathcal{N}(T_{1}^{*})$
$\displaystyle=\mathcal{R}(I-T_{1}T_{1}^{*})\ominus\mathcal{R}[T_{2}(I-T_{1}T_{1}^{*})]$
$\displaystyle=\mathcal{R}(I-T_{1}T_{1}^{*})\ominus\mathcal{R}[T_{2}(I-T_{1}T_{1}^{*})T_{2}^{*}]$
$\displaystyle=\mathcal{R}[(I-T_{1}T_{1}^{*})-T_{2}(I-T_{1}T_{1}^{*})T_{2}^{*}]$
$\displaystyle=\mathcal{R}[(I-T_{1}T_{1}^{*})(I-T_{2}T_{2}^{*})]$
$\displaystyle=\mathcal{R}(I-T_{1}T_{1}^{*})\cap\mathcal{R}(I-T_{2}T_{2}^{*})$
$\displaystyle=\mathcal{N}(T_{1}^{*})\cap\mathcal{N}(T_{2}^{*}).$

Thus for each fixed $m_{2}\geq 1$, we can write

$\displaystyle(I-T_{2}^{m_{2}}T_{2}^{*m_{2}})[\mathcal{N}(T_{1}^{*})]$
$\displaystyle=\mathcal{N}(T_{1}^{*})\ominus
T_{2}^{m_{2}}\mathcal{N}(T_{1}^{*})$
$\displaystyle=[\mathcal{N}(T_{1}^{*})\ominus
T_{2}\mathcal{N}(T_{1}^{*}))]\oplus\cdots\oplus
T_{2}^{m_{2}-1}[\mathcal{N}(T_{1}^{*})\ominus T_{2}\mathcal{N}(T_{1}^{*})]$
$\displaystyle=(\mathcal{N}(T_{1}^{*})\cap\mathcal{N}(T_{2}^{*}))\oplus\cdots\oplus
T_{2}^{m_{2}-1}(\mathcal{N}(T_{1}^{*})\cap\mathcal{N}(T_{2}^{*}))$
$\displaystyle=\bigoplus_{k_{2}=0}^{m_{2}-1}T_{2}^{k_{2}}(\mathcal{N}(T_{1}^{*})\cap\mathcal{N}(T_{2}^{*})).$

Therefore,

$\displaystyle\mathcal{H}_{\neg u\neg u}$ $\displaystyle=\bigvee_{m_{2}\geq
1}\bigg{\\{}\bigoplus_{m_{1}=0}^{\infty}\Big{(}\bigoplus_{k_{2}=0}^{m_{2}-1}T_{1}^{m_{1}}T_{2}^{k_{2}}\big{(}\mathcal{N}(T_{1}^{*})\cap\mathcal{N}(T_{2}^{*})\big{)}\Big{)}\bigg{\\}}$
$\displaystyle=\bigoplus_{m_{1}=0}^{\infty}\Big{(}\bigoplus_{m_{2}=0}^{\infty}T_{1}^{m_{1}}T_{2}^{m_{2}}\big{(}\mathcal{N}(T_{1}^{*})\cap\mathcal{N}(T_{2}^{*})\big{)}\Big{)}$
$\displaystyle=\bigoplus_{m_{1},m_{2}=0}^{\infty}T_{1}^{m_{1}}T_{2}^{m_{2}}\big{(}\mathcal{N}(T_{1}^{*})\cap\mathcal{N}(T_{2}^{*})\big{)}.$

We know that the c.n.u. part of an isometry coincides with the shift part of
Wold-von Neumann decomposition. Hence we can obtain the following
decomposition for pairs of doubly twisted isometries:

######  Theorem 3.9.

Let $\left(T_{1},T_{2}\right)$ be a pair of doubly twisted isometry on a
Hilbert space $\mathcal{H}$. Then there is a unique decomposition

$\mathcal{H}=\mathcal{H}_{uu}\oplus\mathcal{H}_{us}\oplus\mathcal{H}_{su}\oplus\mathcal{H}_{ss},$

where $\mathcal{H}_{uu},\mathcal{H}_{us},\mathcal{H}_{su},$ and
$\mathcal{H}_{ss}$ are the subspaces reducing $T_{1},T_{2}$ such that

  * •

$T_{1}|_{\mathcal{H}_{uu}},T_{2}|_{\mathcal{H}_{uu}}$ are unitary operators,

  * •

$T_{1}|_{\mathcal{H}_{us}}$ is unitary, $T_{2}|_{\mathcal{H}_{us}}$ is
unilateral shift,

  * •

$T_{1}|_{\mathcal{H}_{su}}$ is unilateral shift, $T_{2}|_{\mathcal{H}_{su}}$
is unitary,

  * •

$T_{1}|_{\mathcal{H}_{ss}}$, $T_{2}|_{\mathcal{H}_{ss}}$ are unilateral
shifts.

Also

$\displaystyle\mathcal{H}_{uu}=\bigcap_{m_{1},m_{2}\in\mathbb{Z}_{+}}T_{1}^{m_{1}}T_{2}^{m_{2}}\mathcal{H},$
$\displaystyle\mathcal{H}_{us}=\bigoplus_{m_{2}=0}^{\infty}T_{2}^{m_{2}}\Big{(}\bigcap_{m_{1}=0}^{\infty}T_{1}^{m_{1}}\mathcal{N}(T_{2}^{*})\Big{)},$
$\displaystyle\mathcal{H}_{su}=\bigoplus_{m_{1}=0}^{\infty}T_{1}^{m_{1}}\Big{(}\bigcap_{m_{2}=0}^{\infty}T_{2}^{m_{2}}\mathcal{N}(T_{1}^{*})\Big{)},$
$\displaystyle\mathcal{H}_{ss}=\bigoplus_{m_{1},m_{2}=0}^{\infty}T_{1}^{m_{1}}T_{2}^{m_{2}}\Big{(}\mathcal{N}(T_{1}^{*})\cap\mathcal{N}(T_{2}^{*})\Big{)}.$

######  Remark 3.10.

The above result recovers the Wold-type decomposition and its orthogonal
spaces for pairs of doubly commuting isometries (In particular, if we take
twist $\mathcal{U}_{2}=\\{I\\}$, an identity operator) on Hilbert spaces which
were firstly studied by Słociński [22] and later for pairs of commuting
isometries by Popovici [17]. It is also a noteworthy to mention that
orthogonal decomposition spaces of Wold-type decomposition for pairs of doubly
twisted isometries and for pairs of doubly commuting isometries on Hilbert
spaces are the same.

##  4\. Decomposition for $\mathcal{U}_{n}$-twisted contractions

In this section, we will find the explicit Wold-type decomposition for
$\mathcal{U}_{n}$-twisted contractions on Hilbert spaces. As a by-product, we
derive a simple proof for Wold-type decomposition for
$\mathcal{U}_{n}$-twisted isometries.

Before proceeding further, we shall introduce certain notations for the
remainder of the paper. Given an integer $p$ for $1\leq p\leq n$, we denote
the set $\\{1,\ldots,p\\}$ by $I_{p}$ and $A_{p}\subseteq I_{p}$; that means
each subset (including empty set) of $I_{p}$ is denoted by $A_{p}$. In
addition, if $A_{p}\subseteq I_{p}$ and $q\notin I_{p}$, then $\tilde{A}_{p}$
is denoted by same $A_{p}$ but we will treat $\tilde{A}_{p}$ as a subset of
$I_{p}\cup\\{q\\}$. Using the aforementioned notations, we generalize
decomposition for $\mathcal{U}_{n}$-twisted contraction as follows:

######  Remark 4.1.

Let $T_{q}$ be a contraction on a Hilbert space $\mathcal{H}_{A_{p}}$, where
$A_{p}\subseteq I_{p}$ and $q\notin I_{p}$. Then
$\mathcal{H}_{A_{p}}=\mathcal{H}_{\tilde{A}_{p}}\oplus\mathcal{H}_{{\tilde{A}_{p}}\cup\\{q\\}}$
is the orthogonal decomposition for $T_{q}$ such that
$T_{q}|_{\mathcal{H}_{\tilde{A}_{p}}}$ is unitary and
$T_{q}|_{\mathcal{H}_{{\tilde{A}_{p}}\cup\\{q\\}}}$ is c.n.u. In particular,
if $T$ is a single contraction on $\mathcal{H}$, then from the above notation
the canonical decomposition for $T$ is
$\mathcal{H}=\mathcal{H}_{\emptyset}\oplus\mathcal{H}_{\\{1\\}}$, where
$\mathcal{H}_{\emptyset}=\mathcal{H}_{u}$ and
$\mathcal{H}_{\\{1\\}}=\mathcal{H}_{\neg u}$ (see Theorem 2.2).

We are now in a position to state our main result in this section.

######  Theorem 4.2.

Let $n\geq 2$, and let $(T_{1},\ldots,T_{n})$ be a $\mathcal{U}_{n}$-twisted
contraction on a Hilbert space $\mathcal{H}$. Then there exists $2^{n}$-joint
$(T_{1},\ldots,T_{n})$-reducing subspaces
$\\{\mathcal{H}_{A_{n}}:A_{n}\subseteq I_{n}$ (counting the trivial subspace
$\\{0\\}$) such that

(4.1) $\displaystyle\mathcal{H}=\bigoplus_{A_{n}\subseteq
I_{n}}\mathcal{H}_{A_{n}}$

where

(4.2)
$\displaystyle\mathcal{H}_{A_{n}}=\begin{cases}&\bigcap_{m_{n}\in\mathbb{Z}_{+}}\big{[}\mathcal{N}((I-T_{n}^{*m_{n}}T_{n}^{m_{n}})|_{\mathcal{H}_{A_{n-1}}})\cap\mathcal{N}((I-T_{n}^{m_{n}}T_{n}^{*m_{n}})|_{\mathcal{H}_{A_{n-1}}})\big{]}\quad\text{for
}\ \ n\notin A_{n},\\\
&\bigvee_{m_{n}\in\mathbb{Z}_{+}}\big{\\{}(I-T_{n}^{*m_{n}}T_{n}^{m_{n}})\mathcal{H}_{A_{n-1}}\cup(I-T_{n}^{m_{n}}T_{n}^{*m_{n}}){\mathcal{H}_{A_{n-1}}}\big{\\}}\quad\qquad\quad\
\text{for}\ \ n\in A_{n}.\end{cases}$

For each $A_{n}$ and $\mathcal{H}_{A_{n}}\neq\\{0\\}$,
$T_{i}|_{\mathcal{H}_{A_{n}}}$ is unitary if $i\notin A_{n}$ and
$T_{i}|_{\mathcal{H}_{A_{n}}}$ is completely non-unitary if $i\in A_{n}$ for
$i=1,\ldots,n$. Moreover, the above decomposition is unique.

###### Proof.

We will prove this by mathematical induction. Suppose that $(T_{1},T_{2})$ is
a pair of $\mathcal{U}_{2}$-twisted contraction on $\mathcal{H}$. Then by
Theorem 3.8, there exists four $(T_{1},T_{2})$-reducing subspaces
$\mathcal{H}_{uu},\mathcal{H}_{u\neg u},\mathcal{H}_{\neg uu}$, and
$\mathcal{H}_{\neg u\neg u}$ of $\mathcal{H}$ such that

$\displaystyle\mathcal{H}$
$\displaystyle=\mathcal{H}_{uu}\oplus\mathcal{H}_{\neg
uu}\oplus\mathcal{H}_{u\neg u}\oplus\mathcal{H}_{\neg u\neg u}$
$\displaystyle=\mathcal{H}_{\emptyset}\oplus\mathcal{H}_{\\{1\\}}\oplus\mathcal{H}_{\\{2\\}}\oplus\mathcal{H}_{\\{1,2\\}}$
$\displaystyle=\bigoplus_{A_{2}\subseteq I_{2}}\mathcal{H}_{A_{2}},$

where $\mathcal{H}_{A_{2}}$ has the explicit form (see Theorem 3.8). Moreover,
$T_{i}|_{\mathcal{H}_{A_{2}}}$ is unitary if $i\notin A_{2}$ and c.n.u. if
$i\in A_{2}$ for $i=1,2$. So the statement is true for $n=2$.

Assume that the statement is true for any $k$-tuple $(T_{1},\ldots,T_{k})$,
$k<n$ of $\mathcal{U}_{k}$-twisted contraction on $\mathcal{H}$. Then

$\displaystyle\mathcal{H}$
$\displaystyle=\bigoplus_{\begin{subarray}{c}{A_{k}\subseteq
I_{k}}\end{subarray}}\mathcal{H}_{A_{k}},$

where

(4.3)
$\displaystyle\mathcal{H}_{A_{k}}=\begin{cases}&\bigcap_{m_{k}\in\mathbb{Z}_{+}}\big{[}\mathcal{N}((I-T_{k}^{*m_{k}}T_{k}^{m_{k}})|_{\mathcal{H}_{A_{k-1}}})\cap\mathcal{N}((I-T_{k}^{m_{k}}T_{k}^{*m_{k}})|_{\mathcal{H}_{A_{k-1}}})\big{]}\quad\text{for
}\ \ k\notin A_{k}\\\
&\bigvee_{m_{k}\in\mathbb{Z}_{+}}\big{\\{}(I-T_{k}^{*m_{k}}T_{k}^{m_{k}})\mathcal{H}_{A_{k-1}}\cup(I-T_{k}^{m_{k}}T_{k}^{*m_{k}}){\mathcal{H}_{A_{k-1}}}\big{\\}}\quad\qquad\quad\
\text{for}\ \ k\in A_{k}.\end{cases}$

Also $T_{i}|_{\mathcal{H}_{A_{k}}}$ is unitary if $i\notin A_{k}$ and is
c.n.u. if $i\in A_{k}$ for $i=1,\ldots,k$. It is to be noted that the spaces
$\mathcal{H}_{A_{k-1}}$ reduce $T_{k}$ from Lemma 3.3, and the decomposition
for $\mathcal{U}_{k}$-twisted contraction yields $2^{k}$ number of orthogonal
$T_{i}$-reducing subspaces $\mathcal{H}_{A_{k}}$ for $1\leq i\leq k$. We shall
now prove this statement for the decomposition of $(k+1)$-tuple
$(T_{1},\ldots,T_{k+1})$ of $\mathcal{U}_{k+1}$-twisted contraction on
$\mathcal{H}$. Indeed, we show that

$\displaystyle\mathcal{H}$
$\displaystyle=\bigoplus_{\begin{subarray}{c}{A_{k+1}\subseteq
I_{k+1}}\end{subarray}}\mathcal{H}_{A_{k+1}}.$

As the tuple $(T_{1},\ldots,T_{n})$ is $\mathcal{U}_{n}$-twisted contraction,
using Lemma 3.3 and the equation (4.3), we have
$T_{j}\mathcal{H}_{A_{k}}\subseteq\mathcal{H}_{A_{k}}$ and
$T_{j}^{*}\mathcal{H}_{A_{k}}\subseteq\mathcal{H}_{A_{k}}$ for all $k<j\leq
n$, that is, $\mathcal{H}_{A_{k}}$ reduce $T_{j}$ for $k<j\leq n$. Therefore,
the canonical decomposition for the contraction
$T_{k+1}|_{\mathcal{H}_{A_{k}}}$ yields
${\mathcal{H}_{A_{k}}}={\mathcal{H}_{\tilde{A}_{k}}}\oplus{\mathcal{H}_{\tilde{A}_{k}\cup\\{k+1\\}}}$
(see Remark 4.1), where $\tilde{A}_{k}=A_{k}$ but as a subset of
$I_{k}\cup\\{k+1\\}=I_{k+1}$. Moreover,
$T_{k+1}|_{\mathcal{H}_{\tilde{A}_{k}}}$ is unitary and
$T_{k+1}|_{\mathcal{H}_{\tilde{A}_{k}\cup\\{k+1\\}}}$ is c.n.u., that is,
$T_{k+1}|_{\mathcal{H}_{{A}_{k+1}}}$ is unitary if $k+1\notin A_{k+1}$ and is
c.n.u. if $k+1\in A_{k+1}$. Consequently,

$\displaystyle\mathcal{H}$
$\displaystyle=\bigoplus_{\begin{subarray}{c}A_{k}\subseteq
I_{k}\end{subarray}}{\mathcal{H}_{A_{k}}}$
$\displaystyle=\bigoplus_{\begin{subarray}{c}\tilde{A}_{k}=A_{k}\subseteq
I_{k+1}\end{subarray}}[{\mathcal{H}_{\tilde{A}_{k}}}\oplus{\mathcal{H}_{\tilde{A}_{k}\cup\\{k+1\\}}}]$
$\displaystyle=\bigoplus_{\begin{subarray}{c}{A_{k+1}\subseteq
I_{k+1}}\end{subarray}}\mathcal{H}_{A_{k+1}},$

where the subspaces $\mathcal{H}_{A_{k+1}}$ reduce each $T_{i}$ for $1\leq
i\leq k+1$. Also for each ${A_{k+1}}$ and
$\mathcal{H}_{{A_{k+1}}}\neq\\{0\\}$, $T_{i}|_{\mathcal{H}_{A_{k+1}}}$ is
unitary if $i\notin A_{k+1}$ and is c.n.u. if $i\in A_{k+1}$ for
$i=1,\ldots,k+1$. Since $T_{k+1}|_{\mathcal{H}_{A_{k}}}$ is a contraction,
from Theorem 2.2 we obtain

$\mathcal{H}_{A_{k+1}}=\bigcap_{m_{k+1}\in\mathbb{Z}_{+}}\Big{[}\mathcal{N}\big{(}(I-T_{k+1}^{*m_{k+1}}T_{k+1}^{m_{k+1}})|_{\mathcal{H}_{A_{k}}}\big{)}\cap\mathcal{N}\big{(}(I-T_{k+1}^{m_{k+1}}T_{k+1}^{*m_{k+1}})|_{\mathcal{H}_{A_{k}}}\big{)}\Big{]}\quad\text{if}\quad
k+1\notin A_{k+1}$

and

$\mathcal{H}_{A_{k+1}}=\bigvee_{m_{k+1}\in\mathbb{Z}_{+}}\Big{\\{}(I-T_{k+1}^{*m_{k+1}}T_{k+1}^{m_{k+1}})\mathcal{H}_{A_{k}}\cup(I-T_{k+1}^{m_{k+1}}T_{k+1}^{*m_{k+1}}){\mathcal{H}_{A_{k}}}\Big{\\}}\quad\text{if}\quad
k+1\in A_{k+1}.$

The uniqueness part of this decomposition comes from the uniqueness of the
canonical decomposition of a contraction.

This finishes the proof.

We shall now derive decomposition for $\mathcal{U}_{n}$-twisted isometries.
More specifically, if an n-tuple of isometries $(T_{1},\dots,T_{n})$ on
$\mathcal{H}$ is a $\mathcal{U}_{n}$-twisted isometry, then we obtain an
explicit description of the orthogonal decomposition of $\mathcal{H}$. Before
going to the proof let us adopt the following notations: Let
$\Lambda=\\{i_{1}<i_{2}<,\ldots,<i_{l-1}<i_{l}\\}\subseteq I_{n}$ for $1\leq
l\leq n$,
$I_{n}\backslash\Lambda=\\{i_{l+1}<i_{l+2}<,\ldots,<i_{n-1}<i_{n}\\}$. The
cardinality of the set $\Lambda$ is denoted by $|\Lambda|$. We denote by
$T_{\Lambda}$ the $|\Lambda|$\- tuple of isometries
$(T_{i_{1}},\dots,T_{i_{l}})$ and
$\mathbb{Z}_{+}^{\Lambda}:=\\{{\bf{m}}=(m_{i_{1}},\dots,m_{i_{l}}):m_{i_{j}}\in\mathbb{Z}_{+},1\leq
j\leq l\\}$. Also $T_{i_{1}}^{m_{i_{1}}}\cdots T_{i_{l}}^{m_{i_{l}}}$ is
denoted by $T_{\Lambda}^{\bf m}$ for ${\bf m}\in\mathbb{Z}_{+}^{\Lambda}$.

Consider
$\mathcal{W}_{i_{j}}:=\mathcal{R}(I-T_{i_{j}}T_{i_{j}}^{*})=\mathcal{N}(T_{i_{j}}^{*})$
for each $1\leq j\leq l$ and

$\mathcal{W}_{\Lambda}:=\mathcal{R}\Big{(}\prod_{{i_{j}}\in\Lambda}\big{(}I-T_{i_{j}}T_{i_{j}}^{*}\big{)}\Big{)},$

where $\Lambda$ is a non-empty subset of $I_{n}$. We also denote
$\mathcal{W}_{\emptyset}=\mathcal{H}$. As the tuple $(T_{1},\ldots,T_{n})$ is
$\mathcal{U}_{n}$-twisted isometry, Lemma 3.3 implies that
$\big{\\{}(I-T_{i_{j}}T_{i_{j}}^{*})\big{\\}}_{{j}=1}^{l}$ is a family of
commuting orthogonal projections. Therefore,

$\mathcal{W}_{\Lambda}=\mathcal{R}\Big{(}\prod_{{i_{j}}\in\Lambda}(I-T_{i_{j}}T_{i_{j}}^{*})\Big{)}=\bigcap_{{i_{j}}\in\Lambda}\mathcal{R}\big{(}I-T_{i_{j}}T_{i_{j}}^{*}\big{)}=\bigcap_{{i_{j}}\in\Lambda}\mathcal{W}_{i_{j}}$

for each subset $\Lambda$ of $I_{n}$.

The following result is similar to Theorem 3.6 in [19], and Theorem 3.1 in
[21] (For $n$-tuple of doubly commuting isometries). However, our approach is
different and derived from our above result Theorem 4.2 and properties of
isometries. We are now in a position to state the result as follows:

######  Theorem 4.3.

Let $n\geq 2$, and let $T=(T_{1},\dots,T_{n})$ be a $\mathcal{U}_{n}$-twisted
isometry on $\mathcal{H}$. Then there exists $2^{n}$ joint $T$-reducing
subspaces $\\{\mathcal{H}_{\Lambda}:\Lambda\subseteq I_{n}\\}$ (counting the
trivial subspace $\\{0\\}$) such that

$\displaystyle\mathcal{H}=\bigoplus_{\begin{subarray}{c}\Lambda\subseteq
I_{n}\end{subarray}}\mathcal{H}_{\Lambda}$

and for each $\Lambda\subseteq I_{n}$, we have

(4.4) $\displaystyle\mathcal{H}_{\Lambda}=\bigoplus_{{\bf
k}\in\mathbb{Z}_{+}^{\Lambda}}T_{{\Lambda}}^{\bf k}\bigg{(}\bigcap_{{\bf
m}\in\mathbb{Z}_{+}^{{I_{n}\backslash\Lambda}}}T^{\bf
m}_{I_{n}\backslash\Lambda}\ \mathcal{W}_{\Lambda}\bigg{)}.$

And for $\mathcal{H}_{\Lambda}\neq\\{0\\}$, $T_{i}|_{\mathcal{H}_{\Lambda}}$
is unitary if $i\in I_{n}\backslash\Lambda$ and
$T_{i}|_{\mathcal{H}_{\Lambda}}$ is shift if $i\in\Lambda$ for all
$i=1,\dots,n$. Furthermore, the above decomposition is unique.

###### Proof.

Let $n\geq 2$ be a fixed integer. Suppose that $T=(T_{1},\dots,T_{n})$ is a
$\mathcal{U}_{n}$-twisted isometries on $\mathcal{H}$. Thus, by Theorem 4.2
there exists $2^{n}$ joint $T$-reducing subspaces
$\\{\mathcal{H}_{A_{n}}:A_{n}\subseteq I_{n}\\}$ (including the trivial
subspace $\\{0\\}$) such that

$\displaystyle\mathcal{H}=\bigoplus_{A_{n}\subseteq
I_{n}}\mathcal{H}_{A_{n}}.$

Moreover, for every non-zero decomposition spaces $\mathcal{H}_{A_{n}}$,
$T_{i}|_{\mathcal{H}_{A_{n}}}$ is unitary if $i\notin A_{n}$ and is shift if
$i\in A_{n}$ for each $i=1,\ldots,n$. Now if $n\notin A_{n}$, then by equation
(4.2), the subspace $\mathcal{H}_{A_{n}}$ becomes

$\displaystyle\mathcal{H}_{A_{n}}$
$\displaystyle=\bigcap_{m_{n}\in\mathbb{Z_{+}}}\Big{[}\mathcal{N}\big{(}(I-T_{n}^{m_{n}}T_{n}^{*m_{n}})|_{\mathcal{H}_{A_{n-1}}}\big{)}\Big{]}$
$\displaystyle=\bigcap_{m_{n}\in\mathbb{Z}_{+}}T_{n}^{m_{n}}\mathcal{H}_{A_{n-1}}.$

Now consider $A_{i_{j}}\subseteq
I_{i_{j}}=\\{i_{1}<i_{2}<,\ldots,<i_{j-1}<i_{j}\\}$ for $j=1,\ldots,l$ and
$1\leq l\leq n$. For a fixed $l$, let
$\Lambda=\\{i_{1}<i_{2}<,\ldots,<i_{l-1}<i_{l}\\}$. Suppose ${i_{j}}\notin
A_{i_{j}}$ for $j=n,n-1,\ldots,l+1$. Since the tuple $T=(T_{1},\dots,T_{n})$
is $\mathcal{U}_{n}$-twisted and repeating the above step, we have

(4.5) $\displaystyle\mathcal{H}_{\Lambda}$
$\displaystyle=\bigcap_{{m_{i_{l+1}},\ldots,m_{i_{n}}}\in\mathbb{Z}_{+}}T_{i_{n}}^{m_{i_{n}}}\cdots
T_{i_{l+1}}^{m_{i_{l+1}}}\mathcal{H}_{A_{i_{l}}}$ $\displaystyle=\bigcap_{{\bf
m}\in\mathbb{Z}_{+}^{I_{n}\backslash\Lambda}}T_{I_{n}\backslash\Lambda}^{\bf
m}\mathcal{H}_{A_{i_{l}}}.$

Now for the remaining set $I_{n}\backslash\Lambda$ if $i_{j}\in A_{i_{j}}$ for
$j=1,\dots,l$, then from equation (4.2), the subspace
$\mathcal{H}_{{A_{i_{l}}}}$ can be expressed as

$\displaystyle\mathcal{H}_{{A_{i_{l}}}}$
$\displaystyle=\bigvee_{m_{i_{l}}\in\mathbb{Z}_{+}}\Big{\\{}\big{(}I-T_{i_{l}}^{m_{i_{l}}}T_{i_{l}}^{*m_{i_{l}}}\big{)}{\mathcal{H}_{A_{i_{l-1}}}}\Big{\\}}$
$\displaystyle=\bigvee_{m_{i_{1}},\ldots,m_{i_{l}}\in\mathbb{N}}\Big{\\{}\big{(}I-T_{i_{l}}^{m_{i_{l}}}T_{i_{l}}^{*m_{i_{l}}}\big{)}\dots\big{(}I-T_{i_{2}}^{m_{i_{2}}}T_{i_{2}}^{*m_{i_{2}}}\big{)}\big{(}I-T_{i_{1}}^{m_{i_{1}}}T_{i_{1}}^{*m_{i_{1}}}\big{)}{\mathcal{H}}\Big{\\}}.$

Again applying Lemma 3.3, for $j=1,\ldots,l$ and $m_{i_{j}}\geq 1$, we can
write

$\displaystyle\big{(}I-T_{i_{l}}^{m_{i_{l}}}T_{i_{l}}^{*m_{i_{l}}}\big{)}\cdots\big{(}I-T_{i_{2}}^{m_{i_{2}}}T_{i_{2}}^{*m_{i_{2}}}\big{)}\big{(}I-T_{i_{1}}^{m_{i_{1}}}T_{i_{1}}^{*m_{i_{1}}}\big{)}{\mathcal{H}}$
$\displaystyle=\big{(}I-T_{i_{l}}^{m_{i_{l}}}T_{i_{l}}^{*m_{i_{l}}}\big{)}\cdots\big{(}I-T_{i_{2}}^{m_{i_{2}}}T_{i_{2}}^{*m_{i_{2}}}\big{)}\bigg{[}\bigoplus_{k_{i_{1}}=0}^{m_{i_{1}}-1}{T_{i_{1}}^{k_{i_{1}}}}\mathcal{N}\big{(}T_{i_{1}}^{*}\big{)}\bigg{]}$
$\displaystyle=\bigoplus_{k_{i_{1}}=0}^{m_{i_{1}}-1}{T_{i_{1}}^{k_{i_{1}}}}\big{(}I-T_{i_{l}}^{m_{i_{l}}}T_{i_{l}}^{*m_{i_{l}}}\big{)}\cdots\big{(}I-T_{i_{2}}^{m_{i_{2}}}T_{i_{2}}^{*m_{i_{2}}}\big{)}\Big{[}\mathcal{N}(T_{i_{1}}^{*})\Big{]}$
$\displaystyle=\bigoplus_{k_{i_{1}},k_{i_{2}}=0}^{m_{i_{1}}-1,m_{i_{2}}-1}{T_{i_{1}}^{k_{i_{1}}}}{T_{i_{2}}^{k_{i_{2}}}}\big{(}I-T_{i_{l}}^{m_{i_{l}}}T_{i_{l}}^{*m_{i_{l}}}\big{)}\cdots\big{(}I-T_{i_{3}}^{m_{i_{3}}}T_{i_{3}}^{*m_{i_{3}}}\big{)}\Big{[}\mathcal{N}(T_{i_{1}}^{*})\cap\mathcal{N}(T_{i_{2}}^{*})\Big{]}$
$\displaystyle=\bigoplus_{k_{i_{1}},\ldots,k_{i_{l}}=0}^{m_{i_{1}}-1,\ldots,m_{i_{l}}-1}{T_{i_{1}}^{k_{i_{1}}}}\cdots{T_{i_{l}}^{k_{i_{l}}}}\Big{[}\mathcal{N}(T_{i_{1}}^{*})\cap\cdots\cap\mathcal{N}(T_{i_{l}}^{*})\Big{]}.$

Hence,

$\displaystyle\mathcal{H}_{\Lambda}$ $\displaystyle=\bigcap_{{\bf
m}\in\mathbb{Z}_{+}^{I_{n}\backslash\Lambda}}T_{I_{n}\backslash\Lambda}^{\bf
m}\Big{[}\bigvee_{m_{i_{1}},\dots,m_{i_{l}}\in\mathbb{N}}\Big{\\{}\bigoplus_{k_{i_{1}},\dots,k_{i_{l}}=0}^{m_{i_{1}}-1,\dots,m_{i_{l}}-1}{T_{i_{1}}^{k_{i_{1}}}}\cdots{T_{i_{l}}^{k_{i_{l}}}}\big{(}\mathcal{N}(T_{i_{1}}^{*})\cap\dots\cap\mathcal{N}(T_{i_{l}}^{*})\big{)}\Big{\\}}\Big{]}$
$\displaystyle=\bigvee_{m_{i_{1}},\dots,m_{i_{l}}\in\mathbb{N}}\bigg{\\{}\bigoplus_{k_{i_{1}},\dots,k_{i_{l}}=0}^{m_{i_{1}}-1,\dots,m_{i_{l}}-1}{T_{i_{1}}^{k_{i_{1}}}}\cdots{T_{i_{l}}^{k_{i_{l}}}}\Big{[}\bigcap_{{\bf
m}\in\mathbb{Z}_{+}^{I_{n}\backslash\Lambda}}T_{I_{n}\backslash\Lambda}^{\bf
m}\big{(}\mathcal{N}(T_{i_{1}}^{*})\cap\dots\cap\mathcal{N}(T_{i_{l}}^{*})\big{)}\Big{]}\bigg{\\}}$
$\displaystyle=\bigvee_{m_{i_{1}},\dots,m_{i_{l}}\in\mathbb{N}}\bigg{\\{}\bigoplus_{k_{i_{1}},\dots,k_{i_{l}}=0}^{m_{i_{1}}-1,\dots,m_{i_{l}}-1}{T_{i_{1}}^{k_{i_{1}}}}\cdots{T_{i_{l}}^{k_{i_{l}}}}\Big{[}\bigcap_{{\bf
m}\in\mathbb{Z}_{+}^{I_{n}\backslash\Lambda}}T_{I_{n}\backslash\Lambda}^{\bf
m}\big{(}\mathcal{W}_{\Lambda}\big{)}\Big{]}\bigg{\\}}$
$\displaystyle=\bigoplus_{{\bf k}\in\mathbb{Z}_{+}^{\Lambda}}T_{\Lambda}^{\bf
k}\Big{(}\bigcap_{{\bf
m}\in\mathbb{Z}_{+}^{I_{n}\backslash\Lambda}}T_{I_{n}\backslash\Lambda}^{\bf
m}\mathcal{W}_{\Lambda}\Big{)}.$

If $\Lambda=\emptyset\subseteq I_{n}$, then repeating the same step as the
equation (4.5) for $j=n,n-1,\ldots,2,1$, we obtain

$\mathcal{H}_{\Lambda}=\bigcap_{{\bf
m}\in\mathbb{Z}_{+}^{I_{n}}}T_{I_{n}}^{\bf m}\mathcal{H}.$

Therefore, for any $\Lambda\subseteq I_{n}$, we have

$\displaystyle\mathcal{H}_{\Lambda}=\bigoplus_{{\bf
k}\in\mathbb{Z}_{+}^{\Lambda}}T_{\Lambda}^{\bf k}\Big{(}\bigcap_{{\bf
m}\in\mathbb{Z}_{+}^{I_{n}\backslash\Lambda}}T_{I_{n}\backslash\Lambda}^{\bf
m}\mathcal{W}_{\Lambda}\Big{)}.$

Clearly, $T_{i}|_{\mathcal{H}_{\Lambda}}$ is unitary for all $i\in
I_{n}\backslash\Lambda$ and $T_{i}|_{\mathcal{H}_{\Lambda}}$ is shift for all
$i\in\Lambda$. The uniqueness part is coming from the uniqueness of the
classical Wold decomposition of isometries.

This completes the proof.

Acknowledgement: The authors are thankful to the anonymous reviewer for
his/her critical and constructive reviews and suggestions that have
substantially improved the presentation of the paper. The second author’s
research work is supported in part by the Mathematical Research Impact Centric
Support (MATRICS) (MTR/2021/000695) and the Core Research Grant
(CRG/2022/006891), SERB (DST), Government of India.

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