# Isometric dilations for representations of product systems

Sibaprasad Barik  Department of Mathematics, Technion-Israel Institute of
Technology, 32000, Haifa, Israel<EMAIL_ADDRESS><EMAIL_ADDRESS>,  Monojit Bhattacharjee  Birla Institute of
Technology and Science-Pilani, K. K. Birla Goa Campus, South Goa, 403726,
India<EMAIL_ADDRESS><EMAIL_ADDRESS>and  Baruch Solel
Department of Mathematics, Technion-Israel Institute of Technology, 32000,
Haifa, Israel<EMAIL_ADDRESS>
###### Abstract.

We discuss representations of product systems (of $W^{*}$-correspondences)
over the semigroup $\mathbb{Z}^{n}_{+}$ and show that, under certain pureness
and Szego positivity conditions, a completely contractive representation can
be dilated to an isometric representation. For $n=1,2$ this is known to hold
in general (without assuming the conditions) but, for $n\geq 3$, it does not
hold in general (as is known for the special case of isometric dilations of a
tuple of commuting contractions). Restricting to the case of tuples of
commuting contractions, our result reduces to a result of Barik, Das, Haria
and Sarkar. Our dilation is explicitly constructed and we present some
applications.

###### Key words and phrases:

Completely contractive representation, $W^{*}$-correspondence, product system,
isometric dilation, Szegö positivity

###### 2020 Mathematics Subject Classification:

46L08, 46L10, 47A13, 47A20, 47A56, 47L05, 47L30, 47L55

##  1\.  Introduction

Consider a tuple of commuting contractions $T=(T_{1},\ldots,T_{n})$ on a
Hilbert space $H$. If $K$ is another Hilbert space that contains $H$ and
$S=(S_{1},\ldots,S_{n})$ is a tuple of operators on K, we say that $S$ is a
_dilation_ of $T$ if, for every non negative integers
$k_{1},k_{2},\ldots,k_{n}$ and every $h\in H$,

(1) $T_{1}^{k_{1}}T_{2}^{k_{2}}\cdots
T_{n}^{k_{n}}h=P_{H}S_{1}^{k_{1}}S_{2}^{k_{2}}\cdots S_{n}^{k_{n}}h$

where $P_{H}$ is the orthogonal projection of $K$ onto $H$.

Two special cases are when $H$ is invariant for each $S_{i}$ and then we call
$S$ an _extension_ of $T$ and can write (1) simply as

$T_{i}h=S_{i}h$

for all $i$, or when $H$ is co-invariant for each $S_{i}$ (i.e. $K\ominus H$
is invariant) and then we call $S$ a _coextension_ of $T$. In the latter case,
it is convenient to write $\imath$ for the inclusion of $H$ into $K$ (an
isometry) and write (1) by

$\imath T_{i}^{*}=S_{i}^{*}\imath$

for all $i$. In general, instead of just $K\supseteq H$ if there is an
isometry $\Pi:H\to K$ such that

(2) $\Pi T_{i}^{*}=S_{i}^{*}\Pi$

for all $i$, then we say $S$ is a co-extension of $T$. We will mostly be
interested in the case where $S_{i}$ commute and are isometries and refer to
(2) in this case as _isometric dilation_.

It is known that, if each $T_{i}$ is an isometry, $T$ has a unitary extension
$U=(U_{1},\ldots,U_{n})$ (so that each $U_{i}$ is unitary and the $U_{i}$
pairwise commute). Thus, if $T$ has an isometric dilation, it also has a
unitary dilation.

The existence of unitary or isometric dilation is important since it enables
us, in many situations, to reduce the study of a tuple $T$ to that of a
unitary, or an isometric, tuple which is better understood.

For $n=1$ or $n=2$ it is known (see [12] and [2] ) that isometric dilation
always exists. But Parrot ([13] ) showed that for $n\geq 3$ one can find
commuting tuples that do not have an isometric dilation.

This raises the problem of finding classes of commuting tuples for which
isometric dilation exists.

In 1961 Brehmer introduced a condition on a commutative tuple of contractions
that ensures that the tuple has an isometric dilation. In fact, this condition
is equivalent to the existence of an isometric dilation with an extra
property, called regular dilation (see [5] , [12] or [1] for details).

In [7] the authors introduced the class $\mathcal{P}^{n}_{p,q}$ (where $1\leq
p<q\leq n$) of $n$-commuting tuples of _strict_ contractions that satisfy the
two positivity conditions

(3)
$\Sigma_{G\subseteq\\{1,\ldots,n\\}\backslash\\{p\\}}(-1)^{|G|}T_{G}T^{*}_{G}\geq
0$

and

(4)
$\Sigma_{G\subseteq\\{1,\ldots,n\\}\backslash\\{q\\}}(-1)^{|G|}T_{G}T^{*}_{G}\geq
0$

where $T_{G}=T_{i_{1}}\cdots T_{i_{k}}$ for $G=\\{i_{1},\ldots,i_{k}\\}$. They
proved that tuples in $\mathcal{P}^{n}_{p,q}$ have unitary dilations.

Another result in this direction can be found in [3] where the class of
tuples, considered by the authors for finding isometric dilation, enlarges the
class considered in [5] and [7].

But the result that is most relevant to us here was obtained in [4]. To state
it we need the following notation. If $T=(T_{1},\ldots,T_{n})$ is a commuting
tuple of contractions, we say that it is _pure_ if, for every $h\in H$ and
$1\leq i\leq n$, $T_{i}^{*m}h\rightarrow 0$ as $m\rightarrow\infty$. Also, we
write $\widehat{T_{p}}$ for the $n-1$ tuple which we get when we delete
$T_{p}$ from $T$. Finally, we say that the tuple $T$ satisfies the _Szego
positivity condition_ if

(5) $\Sigma_{G\subseteq\\{1,\ldots,n\\}}(-1)^{|G|}T_{G}T_{G}^{*}\geq 0.$

Note that condition (3) means that $\widehat{T}_{p}$ satisfies the Szego
condition (and similar comment applies to condition (4)).

This allows us to define the class $\mathcal{T}_{p,q}^{n}$ (for $1\leq p<q\leq
n$) to be

(6) $\mathcal{T}_{p,q}^{n}:=\\{T=(T_{1},\ldots,T_{n}):\widehat{T}_{p}\mbox{
and }\widehat{T}_{q}\mbox{ satisfy }(\ref{szego}),\widehat{T}_{q}\mbox{ is
pure }\\}.$

The main result of [4] is that every $T\in\mathcal{T}_{p,q}^{n}$ has an
isometric dilation. In fact such a dilation is explicitely constructed.

In this paper we study dilations of representations of product systems of
$W^{*}$-correspondences. As we shall explain, a commuting tuple of
contractions can be viewed as such a representation so that our study is a
generalization of the study of dilations of such tuples.

We now fix a $W^{*}$-algebra $M$. A $W^{*}$-correspondence over the
$W^{*}$-algebra $M$ is a Hilbert $W^{*}$-module $E$ over $M$ endowed with the
structure of a left $M$-module via a normal ∗-homomorphism
$\varphi_{E}:M\rightarrow\mathcal{L}(E)$. An isomorphism between two
correspondences is a bimodule isomorphism that preserves the inner product.
Given two $W^{*}$-correspondences $E,F$ over $M$ one can define the (balanced)
tensor product $E\otimes F$ and it is a $W^{*}$-correspondence with a natural
left and right multiplication and the interior inner product given, on
generators, by

$\langle e_{1}\otimes f_{1},e_{2}\otimes f_{2}\rangle=\langle
f_{1},\varphi_{E}(\langle e_{1},e_{2}\rangle_{E})f_{2}\rangle$

for $e_{i}\in E$ and $f_{i}\in F$.

If $M=\mathbb{C}$, then a $W^{*}$-correspondence over $M$ is a Hilbert space
and the tensor product has the usual meaning. However, note that, for a
general $M$, $E\otimes F$ is not always isomorphic to $F\otimes E$.

A c.c. representation of $E$ on a Hilbert space $H$ is a pair $(\sigma,T)$
where $\sigma$ is a representation of $M$ on $H$ and $T:E\rightarrow B(H)$ is
a completely contractive linear map that is also a bimodule map (that is,
$T(a\cdot\xi\cdot b)=\sigma(a)T(\xi)\sigma(b)$ for $a,b\in M$ and $\xi\in E$).
The representation is said to be isometric (or Toeplitz ) if
$T(\xi)^{*}T(\eta)=\sigma(\langle\xi,\eta\rangle)$ for every $\xi,\eta\in E$.

In [6] Fowler studied product systems over a left-cancellative, countable,
semigroup $P$ with an identity $e$. A product system $\mathbb{E}$ is a family
$\\{X_{s}\\}_{s\in P}$ with isomorphisms $\\{\theta_{s,t}:X_{s}\otimes
X_{t}\rightarrow X_{st}\\}_{s,t\in P}$ such that $X_{e}=M$, the maps
$\theta_{e,t}$ and $\theta_{s,e}$ are given by the left and right
multiplications and it satisfies the associativity condition

$\theta_{st,r}(\theta_{s,t}\otimes
I_{X_{r}})=\theta_{s,tr}(I_{X_{s}}\otimes\theta_{t,r}).$

In this paper we study the case where $P=\mathbb{Z}_{+}^{n}$. We write
$\textbf{e}_{i}=(0,0,\ldots,1,\ldots,0)$ (with $1$ in the $ith$ position) and
for a product system $\\{X_{s}\\}_{s\in\mathbb{Z}_{+}^{n}}$, we write $E_{i}$
for $X_{\textbf{e}_{i}}$, $1\leq i\leq n$.

It will be convenient to write $E_{i}^{n}$ for the $n$-fold tensor product
$E_{i}^{\otimes n}$ and to identify $X_{\textbf{m}}$ (for
$\textbf{m}=(m_{1},m_{2},\ldots,m_{n})\in\mathbb{Z}_{+}^{n}$) with
$E_{1}^{m_{1}}\otimes E_{2}^{m_{2}}\otimes\cdots\otimes E_{k}^{m_{n}}$ (where
these tensor products are the balanced tensor products over $M$). That means,
in particular, that the isomorphisms $\theta_{\textbf{e}_{i},\textbf{e}_{j}}$,
for $i\leq j$, are identity maps. Setting
$t_{i,j}=\theta_{\textbf{e}_{i},\textbf{e}_{j}}:E_{i}\otimes E_{j}\rightarrow
E_{j}\otimes E_{i}$ for $i\geq j$ (and $t_{i,j}=t_{j,i}^{-1}$ for $i<j$), one
can check that the family $\\{t_{i,j}:1\leq i,j\leq n\\}$ satisfies

(7) $(t_{j,i}\otimes I_{\textbf{e}_{l}})(I_{\textbf{e}_{j}}\otimes
t_{l,i})(t_{l,j}\otimes I_{\textbf{e}_{i}})=(I_{\textbf{e}_{i}}\otimes
t_{l,j})(t_{l,i}\otimes I_{\textbf{e}_{j}})(I_{\textbf{e}_{l}}\otimes
t_{j,i})$

for every $1\leq i,j,l\leq n$. One can also check (but we omit the tedious
computation) that, given $n$ correspondences $E_{1},\ldots,E_{n}$ over the
$W^{*}$-algebra $M$ and a family $\\{t_{i,j}:1\leq i,j\leq n\\}$ such that
$t_{i,j}:E_{i}\otimes E_{j}\rightarrow E_{j}\otimes E_{i}$ is an isomorphism,
$t_{i,j}=t_{j,i}^{-1}$ and $t_{i,i}$ is the identity map, it determines, in a
unique way, a product system $X$ (with
$X_{\textbf{m}}=E_{1}^{m_{1}}\otimes\cdots\otimes E_{k}^{m_{k}}$) whose
isomorphisms $\\{\theta_{\textbf{n},\textbf{m}}\\}$ satisfy
$\theta_{\textbf{e}_{i},\textbf{e}_{j}}=id$ if $i\leq j$ and
$\theta_{\textbf{e}_{i},\textbf{e}_{j}}=t_{i,j}$ if $i>j$.

In this paper we study representations of such product systems (and their
isometric dilations). In Definition 2.7 we define a c.c. (and isometric)
representations of product systems and, in Example 3.5 we show that, when
$E_{i}=\mathbb{C}$ for all $i$ and $t_{i,j}(x\otimes y)=y\otimes x$ (for $x\in
E_{i}$ and $y\in E_{j}$), a c.c representation of the product system is given
by a tuple of commuting contractions. Similarly, an isometric representation,
in this case, is given by a tuple of commuting isometries.

In [17, Theorem 3.3] it is shown that, under a condition that extends
Brehmer’s condition to product systems, the c.c, representation has an
isometric dilation.

The main result of the current paper is Theorem 3.20. We prove there that,
under conditions of pureness and Szego positivity (as in (17)), the c.c.
representation of the product system has an isometric dilation. In fact, this
dilation is explicitly constructed.

Note that, for $n=1$ or $n=2$ this result is known to hold in general (no
conditions are needed). For $n=1$ it was proved in [8] and for $n=2$ in [16,
Theorem 4.4]. But, for $n\geq 3$, it does not hold in general as is known for
the special case of isometric dilations of a tuple of commuting contractions.

Our main result is a generalization of the dilation result of [4] . In the
case where $E_{i}=E_{j}=\mathbb{C}$, our result here, Theorem 3.20, reduces to
the main result of [4].

In [6] Fowler showed that there is a universal $C^{*}$\- algebra, written
$\mathcal{T}_{\mathbb{E}}$, which is universal for isometric representations
of the product system $\mathbb{E}$. It follows from our theorem that every
c.c. representation of $\mathbb{E}$ on $H$ that satisfies our conditions
defines a completely positive map from $\mathcal{T}_{\mathbb{E}}$ into $B(H)$.

We present several examples to show that our result yields also some other
interesting special cases.

As we noted above, every isometric tuple has an extension to a unitary tuple.
This is not always the case for representations of a product system even for
the case $n=1$ (see [8, Chapter 5] and [10]).

##  2\. Preliminaries

We begin by recalling the notion of a $W^{*}$-correspondence.

######  Definition 2.1.

A $W^{*}$-correspondence over a $W^{*}$-algebra $M$ is a Hilbert
$W^{*}$-module $E$ over $M$ endowed with the structure of a left $M$-module
via a normal ∗-homomorphism $\varphi_{E}:M\rightarrow\mathcal{L}(E)$.

If $E$ and $F$ are $W^{*}$-correspondences over $M$, then the balanced tensor
product $E\otimes_{M}F$ is a $W^{*}$-correspondence over $M$. It is defined as
the self dual Hausdorff completion of the algebraic balanced tensor product
with the internal inner product given by

(8)
$\langle\xi_{1}\otimes\eta_{1},\xi_{2}\otimes\eta_{2}\rangle=\langle\eta_{1},\varphi_{F}(\langle\xi_{1},\xi_{2}\rangle_{E})\eta_{2}\rangle_{F}$

for all $\xi_{1},\xi_{2}\in E$ and $\eta_{1},\eta_{2}\in F$. The left and
right actions of $a\in M$ are defined by

(9) $\varphi_{E\otimes F}(a)(\xi\otimes\eta)b=\varphi_{E}(a)\xi\otimes\eta b$

for all $a,b\in M$, $\xi\in E$ and $\eta\in F$.

######  Definition 2.2.

An isomorphism of $W^{*}$-correspondences $E$ and $F$ is a surjective,
bimodule map that preserves the inner products. We write $E\cong F$ if such an
isomorphism exists.

If $E$ is a $W^{*}$-correspondence over $M$ and $\sigma$ is a normal
representation of $M$ on a Hilbert space $H$ then $E\otimes_{\sigma}H$ is the
Hilbert space obtained as the Hausdorff completion of the algebraic tensor
product with respect to $\langle\xi\otimes h,\eta\otimes k\rangle=\langle
h,\sigma(\langle\xi,\eta\rangle_{E})k\rangle_{H}$. Given an operator
$X\in\mathcal{L}(E)$ and an operator $S\in\sigma(M)^{\prime}$, the map
$\xi\otimes h\mapsto X\xi\otimes Sh$ defines a bounded operator $X\otimes S$
on $E\otimes_{\sigma}H$. When $S=I_{H}$ and $X=\varphi_{E}(a)$ (for $a\in A$)
we get a normal representation of $M$ on this Hilbert space. We frequently
write $a\otimes I_{H}$ for $\varphi(a)\otimes I_{H}$.

######  Definition 2.3.

Let $E$ be a $W^{*}$-correspondence over a $W^{*}$-algebra $M$. Then a
completely contractive covariant representation of $E$ (or, simply, a c.c.
representation of $E$) on a Hilbert space $H$ is a pair $(\sigma,T)$, where

  1. (1)

$\sigma$ is a normal $\ast$-representation of $M$ in $B(H)$.

  2. (2)

$T$ is a linear, completely contractive map from $E$ to $B(H)$.

  3. (3)

$T$ is a bimodule map in the sense that $T(a\xi b)=\sigma(a)T(\xi)\sigma(b)$,
$\xi\in E$, and $a,b\in M$.

Such a representation is said to be isometric if, for every $\xi,\eta\in E$,
$T(\xi)^{*}T(\eta)=\sigma(\langle\xi,\eta\rangle)$.

As was shown in [8, Lemmas 3.4–3.6], if a completely contractive covariant
representation, $(\sigma,T)$, of $E$ in $B(H)$ is given, then it determines a
contraction $\tilde{T}:E\otimes_{\sigma}H\rightarrow H$ defined by the formula
$\tilde{T}(\eta\otimes h):=T(\eta)h$, $\eta\otimes h\in E\otimes_{\sigma}H$.
The operator $\tilde{T}$ satisfies

(10) $\tilde{T}(\varphi(\cdot)\otimes I)=\sigma(\cdot)\tilde{T}.$

In fact we have the following lemma from [9, Lemma 2.16].

######  Lemma 2.4.

The map $(\sigma,T)\rightarrow\tilde{T}$ is a bijection between all completely
contractive covariant representations $(\sigma,T)$ of $E$ on the Hilbert space
$H$ and contractive operators $\tilde{T}:E\otimes_{\sigma}H\rightarrow H$ that
satisfy equation (10). Given $\sigma$ and a contraction $\tilde{T}$ satisfying
the covariance condition (10), we get a completely contractive covariant
representation $(\sigma,T)$ of $E$ on $H$ by setting
$T(\xi)h:=\tilde{T}(\xi\otimes h)$.

Moreover, the representation $(\sigma,T)$ is an isometric representation if
and only if $\tilde{T}$ is an isometry.

######  Remark 2.5.

In addition to $\tilde{T}$ we also require the “generalized higher powers” __
of $\tilde{T}$. These are __ maps$\;\tilde{T}_{n}:E^{\otimes n}\otimes
H\rightarrow H\;$defined by the
equation$\;\tilde{T}_{n}(\xi_{1}\otimes\ldots\otimes\xi_{n}\otimes
h)=T(\xi_{1})\cdots T(\xi_{n})h$, $\xi_{1}\otimes\ldots\otimes\xi_{n}\otimes
h\in E^{\otimes n}\otimes H$. One checks easily that
$\tilde{T}_{n}=\tilde{T}\circ(I_{E}\otimes\tilde{T})\circ\cdots\circ(I_{E^{\otimes
n-1}}\otimes\tilde{T})$, $n>1$.

######  Definition 2.6.

A product system over $M$ is a family of $W^{*}$-correspondences
$\\{E_{1},E_{2}.\ldots,E_{n}\\}$ over $M$ together with unitary isomorphisms
$t_{i,j}:E_{i}\otimes E_{j}\rightarrow E_{j}\otimes E_{i}$ (for $i>j$).

In fact, this defines a product system over the semigroup $\mathbb{Z}_{+}^{N}$
and it is natural to define also $t_{i,i}=I_{E_{i}}$ and
$t_{i,j}=t_{j,i}^{-1}$ for $i<j$. These unitaries allow us to identify
$E_{i}\otimes E_{j}$ with $E_{j}\otimes E_{i}$. Similarly, we can identify
$E_{n_{1}}\otimes E_{n_{2}}\otimes\cdots E_{n_{k}}$ with every permutation of
the correspondences.

In this paper, we shall suppress the unitaries $t_{i,j}$ and write
$E_{i}\otimes E_{j}=E_{j}\otimes E_{i}$.

######  Definition 2.7.

A c.c. (completely contractive) representation of a product system
$(M,E_{1},\ldots,E_{n})$ is a tuple $(\sigma,T_{1},T_{2},\ldots,T_{n})$ where
for each $i$, $(\sigma,T_{i})$ is a representation of $E_{i}$ on $H$ (and,
thus, $\tilde{T}_{i}$ is a contraction) such that $T_{i}$ commutes with
$T_{j}$ or, more precisely,

(11)
$\tilde{T_{i}}(I_{E_{i}}\otimes\tilde{T_{j}})=\tilde{T_{j}}(I_{E_{j}}\otimes\tilde{T_{i}})(t_{i,j}\otimes
I_{H})$

for every $i,j$. If each $(\sigma,T_{i})$ is an isometric representation of
$E_{i}$ (equivalently, if each $\tilde{T_{i}}$ is an isometry ) , we say that
the representation (of the product system) is an isometric representation.

Suppressing $t_{i,j}$, we simply write

$\tilde{T_{i}}(I_{E_{i}}\otimes\tilde{T_{j}})=\tilde{T_{j}}(I_{E_{j}}\otimes\tilde{T_{i}}).$

######  Definition 2.8.

We say that an isometric representation $(\sigma,V_{1},V_{2},\ldots,V_{n})$ of
the product system $(M,E_{1},\ldots,E_{n})$ on the Hilbert space $K$ is a
dilation of the representation $(\sigma,T_{1},T_{2},\ldots,T_{n})$ (on $H$) if
there is an isometry $\Pi:H\rightarrow K$ such that

$(I_{E_{i}}\otimes\Pi)\tilde{T_{i}}^{*}=\tilde{V_{i}}^{*}\Pi$

for all $i$.

We shall use the following notation. For a subset
$G=\\{n_{1},n_{2},\ldots,n_{k}\\}\subseteq\\{1,2,\ldots,n\\}$, we write
$E_{G}$ or $E_{n_{1},n_{2},\ldots.n_{k}}$ for $E_{n_{1}}\otimes
E_{n_{2}}\otimes\cdots\otimes E_{n_{k}}$.

Also, given a representation $(\sigma,T_{1},T_{2},\ldots,T_{n})$ of the
product system, and $G\subseteq\\{1,2,\ldots,n\\}$ as above, we write

(12)
$\tilde{T}_{G}=\tilde{T}_{n_{1}}(I_{n_{1}}\otimes\tilde{T}_{n_{2}})\cdots(I_{G-\\{n_{k}\\}}\otimes\tilde{T}_{n_{k}}):E_{n_{1}}\otimes\cdots\otimes
E_{n_{k}}\otimes H=E_{G}\otimes H\to H.$

and, for $1\leq k$,

$\tilde{T}_{G}^{(k)}=\tilde{T}_{G}(I_{E_{G}}\otimes\tilde{T}_{G})\cdots(I_{E_{G}^{\otimes{k-1}}}\otimes\tilde{T}_{G}).$

Given a product system $(M,E_{1},\ldots,E_{n})$ and
$\alpha\in\mathbb{Z}_{+}^{n}$ we write
$E^{\alpha}:=E_{1}^{\otimes\alpha_{1}}\otimes
E_{2}^{\otimes\alpha_{2}}\otimes\cdots\otimes E_{n}^{\otimes\alpha_{n}}$ (with
$\alpha^{0}:=M$) and form the associated Fock correspondence

(13) $\mathcal{F}(E):=\sum_{\alpha\in\mathbb{Z}_{+}^{n}}E^{\alpha}.$

In the following we fix a $W^{*}$-algebra $M$, a product system
$(M,E_{1},\ldots,E_{n})$ over $\mathbb{N}^{n}$ and a representation
$(\sigma,T_{1},\ldots,T_{n})$ of the product system over $H$.

We write $NRep(M)$ for the set of normal representations of $M$. For normal
representations $\rho_{1},\rho_{2}$ of $M$ we write $\rho_{1}\precsim\rho_{2}$
if $\rho_{1}$ is equivalent to a subrepresentation of $\rho_{2}$ and
$\rho_{1}\simeq\rho_{2}$ if they are equivalent. It will be convenient to
write $G_{1}\precsim G_{2}$ if $G_{i}$ is the representation space of some
normal representation of $M$, say $\rho_{i}$ and $\rho_{1}\precsim\rho_{2}$.

In [15] Rieffel defined the notion of a generator for $NRep(M)$ (based on a
similar notion in pure algebra). In [15, Proposition 1.1] he showed that a
normal representation $\rho$ is a generator if and only if every
representation $\pi\in NRep(M)$ is a direct sum of copies of
subrepresentations of $\rho$. In [15, Proposition 1.3] he proved that $\rho$
is a generator if and only if it is faithful.

As in [11, Definition 4.1], we say that a generator is an infinite generator
if it is an infinite multiple of a generator for $NRep(M)$, i.e. it is an
infinite multiple of a faithful normal representation of $M$. We can conclude
that every infinite generator $\rho$ has the property that every normal
rpresentation of $M$ is equivalent to a subrepresentation of $\rho$. We also
note that every infinite generator is unique up to equivalence.

We shall now fix an infinite generator $\rho_{\infty}$ and write $H_{\infty}$
for its representation space.

The following lemma will be useful. The proof is straightforward and will be
omitted.

######  Lemma 2.9.

Suppose $E$ is a faithful $W^{*}$-correspondence over $M$ (that is,
$\varphi_{E}$ is a faithful map) and $\rho$ is a faithful representation of
$M$ on $H$. Then the representation $\varphi_{E}(\cdot)\otimes I_{H}$ is a
faithful representation of $M$ on $E\otimes_{\rho}H$.

##  3\. Representations of product systems

We will assume that $E_{1},\ldots,E_{n}$ are faithful $W^{*}$-correspondences
over a $W^{*}$-algebra $M$. It follows from the lemma that, for
$i=1,\ldots,n$, the representation
$\varphi_{i}(\cdot)\otimes_{\rho_{\infty}}I_{H_{\infty}}$ is also an infinite
generator.

Recall, for each $i=1,\ldots,n$,

$\tilde{T_{i}}:E_{i}\otimes_{\sigma}H\rightarrow H$

is defined by $\tilde{T_{i}}(\xi_{i}\otimes h)=T_{i}(\xi_{i})h$ for
$\xi_{i}\in E_{i}$ and $h\in H$. Recall that for any
$G=\\{n_{1},\ldots,n_{k}\\}\subseteq\\{1,\ldots,n\\}$, the operator
$\tilde{T}_{G}$ is defined as in (12).

We say, the representation $(\sigma,T_{1},\ldots,T_{n})$ satisfy Szegö
positivity if the tuple $\tilde{T}=(\tilde{T}_{1},\ldots,\tilde{T}_{n})$ is
Szegö positive, that is,

(14)
$\mathbb{S}_{n}^{-1}\big{(}\tilde{T},\tilde{T}^{*}\big{)}:=\sum_{G\subset\\{1,\ldots,n\\}}(-1)^{|G|}\tilde{T}_{G}\tilde{T}_{G}^{*}\geq
0.$

If this holds, we write

(15) $D_{T}^{2}=\mathbb{S}_{n}^{-1}\big{(}\tilde{T},\tilde{T}^{*}\big{)}.$

######  Proposition 3.1.

For a cc representation $(\sigma,T_{1},\ldots,T_{n})$ having the Sezgö
positivity the map $\Pi:H\to\mathcal{F}(E)\otimes H$ defined by

$\Pi h=\bigoplus_{\alpha\in\mathbb{Z}_{+}^{n}}(I_{E^{\alpha}}\otimes
D_{\tilde{T}})\tilde{T}^{*(\alpha)}h$

satisfies

(16) $\|\Pi
h\|^{2}=\|h\|^{2}+\lim_{k\to\infty}\Big{\langle}\sum_{\emptyset\neq
G\subseteq\\{1,\ldots,n\\}}(-1)^{|G|}\tilde{T}^{k}_{G}\tilde{T}^{*k}_{G}h,h\Big{\rangle}.$

where $k\in\mathbb{Z}_{+}$.

###### Proof.

Using the fact that for each $i,j\in\\{1,\ldots,n\\}$,
$\tilde{T}_{i}(I_{E_{i}}\otimes\tilde{T}_{j})=\tilde{T}_{j}(I_{E_{j}}\otimes\tilde{T}_{i})$,
we have

$\displaystyle\sum_{\alpha=(\alpha_{1},\ldots,\alpha_{n})\in\mathbb{Z}_{+}^{n}}\langle\tilde{T}^{(\alpha)}(I_{E^{\alpha}}\otimes
D^{2}_{\tilde{T}})\tilde{T}^{*(\alpha)}h,h\rangle$
$\displaystyle=\lim_{k\to\infty}\sum_{\text{max}\alpha_{j}\leq
k-1}\langle\tilde{T}^{(\alpha)}(I_{E^{\alpha}}\otimes
D^{2}_{\tilde{T}})\tilde{T}^{*(\alpha)}h,h\rangle$
$\displaystyle=\lim_{k\to\infty}\sum_{\text{max}\alpha_{j}\leq
k-1}\langle\tilde{T}^{(\alpha)}(I_{E^{\alpha}}\otimes\mathbb{S}_{n}^{-1}(\tilde{T},\tilde{T}^{*}))\tilde{T}^{*(\alpha)}h,h\rangle$
$\displaystyle=\lim_{k\to\infty}\sum_{\text{max}\alpha_{j}\leq
k-1}\Big{\langle}\tilde{T}^{(\alpha)}\Big{(}I_{E^{\alpha}}\otimes\sum_{G\subseteq\\{1,\ldots,n\\}}(-1)^{|G|}\tilde{T}_{G}\tilde{T}^{*}_{G}\Big{)}\tilde{T}^{*(\alpha)}h,h\Big{\rangle}$
$\displaystyle=\lim_{k\to\infty}\sum_{\text{max}\beta_{j}\leq
k}\Big{\langle}\tilde{T}^{(\beta)}\tilde{T}^{*(\beta)}\sum_{\alpha\leq\beta,\,\text{max}(\beta_{j}-\alpha_{j})\leq
1,\,\text{max}\alpha_{j}\leq k-1}(-1)^{|\beta|-|\alpha|}h,h\Big{\rangle}.$

Note that, for a fixed $\beta$ with $\text{max}\beta_{j}\leq k$,

$\sum_{\alpha\leq\beta,\,\text{max}(\beta_{j}-\alpha_{j})\leq
1,\,\text{max}\alpha_{j}\leq k-1}(-1)^{|\beta|-|\alpha|}=0$

unless $\beta=(\beta_{1},\ldots,\beta_{n})\in\\{0,k\\}^{n}$. If
$\beta=(\beta_{1},\ldots,\beta_{n})\in\\{0,k\\}^{n}$ then it is equal to
$(-1)^{|\text{supp}\beta|}$, where $|\text{supp}\beta|$ means the number of
non-zero $\beta_{j}$’s in $\beta=(\beta_{1},\ldots,\beta_{n})$. Therefore, for
every $h\in H$

$\displaystyle\|\Pi h\|^{2}$
$\displaystyle=\lim_{k\to\infty}\Big{\langle}\sum_{G\subseteq\\{1,\ldots,n\\}}(-1)^{|G|}\tilde{T}^{(k)}_{G}\tilde{T}^{*(k)}_{G}h,h\Big{\rangle}$
$\displaystyle=\|h\|^{2}+\lim_{k\to\infty}\Big{\langle}\sum_{\emptyset\neq
G\subseteq\\{1,\ldots,n\\}}(-1)^{|G|}\tilde{T}^{(k)}_{G}\tilde{T}^{*(k)}_{G}h,h\Big{\rangle}$

∎

A c.c. representation $(\sigma,T_{1},\ldots,T_{n})$ is said to be pure if
$(\sigma,T_{i})$ is pure for each $i=1,\ldots,n$, that is,
$\tilde{T}_{i}^{(k)}\tilde{T}_{i}^{*(k)}\to 0$ in WOT as $k\to\infty$.

######  Corollary 3.2.

Let $(\sigma,T_{1},\ldots,T_{n})$ be a pure c.c. representation which
satisfies Szegö positivity. Then the map defined in Proposition 3.1 is an
isometry.

###### Proof.

Since $(\sigma,T_{1},\ldots,T_{n})$ is pure

$\lim_{k\to\infty}\Big{\langle}\sum_{\emptyset\neq
G\subseteq\\{1,\ldots,n\\}}(-1)^{|G|}\tilde{T}^{(k)}_{G}\tilde{T}^{*(k)}_{G}h,h\Big{\rangle}=0.$

So it follows from (16) that $\Pi$ is an isometry. ∎

We denote, $\hat{T}_{1}:=(\sigma,T_{2},\ldots,T_{n})$ and
$\hat{T}_{n}:=(\sigma,T_{1},\ldots,T_{n-1})$ and with the help of them we
define a class of c.c. representations namely
$\mathcal{T}_{1,n}^{n}(\sigma,M,E_{1},\ldots,E_{n},H)$ in the following way

(17)
$\mathcal{T}_{1,n}^{n}(\sigma,M,E_{1},\ldots,E_{n},H):=\\{(\sigma,T_{1},\ldots,T_{n}):\hat{T}_{1},\hat{T}_{n}\text{
satisfy Szeg\"{o} positivity and }\hat{T}_{n}\text{ is pure}\\}$

######  Example 3.3.

Let $M=\mathbb{C}$ and $E_{i}=\mathbb{C}^{d}$ ($i=1,\ldots,n$). Let
$T_{i}:\mathbb{C}^{d}\to\mathcal{B}(H)$ be written as
$T_{i}=(T_{i,1},\ldots,T_{i,d})$ where $T_{i,k}=T_{i}(e_{k})$ ($\\{e_{k}\\}$
is the standard basis of $\mathbb{C}^{d}$). Then, for each $i=1,\ldots,n$,
$\tilde{T}_{i}:\mathbb{C}^{d}\otimes H\to H$ is given by

$\tilde{T}_{i}(\lambda\otimes
h)=T_{i}(\lambda)h=(T_{i,1},\ldots,T_{i,d})\begin{bmatrix}\lambda_{1}\\\
\vdots\\\
\lambda_{d}\end{bmatrix}=\sum_{j=1}^{d}\lambda_{j}T_{i}(e_{j})h=\sum_{j=1}^{d}\lambda_{j}T_{i,j}h.$

To make it into a product system we need to fix an isomorphism
$t_{i,j}:E_{i}\otimes E_{j}\rightarrow E_{j}\otimes E_{i}$. For simplicity we
assume that

$t_{i,j}(e\otimes f)=f\otimes e$

for $i\neq j$, $e\in E_{i}=\mathbb{C}^{d}$ and $f\in E_{j}=\mathbb{C}^{d}$
(but see the remark below).

For $i\neq j$, $h\in H$, $e_{k}\in E_{i}$ and $e_{l}\in E_{j}$, we apply
Equation 11 to get

$T_{i}(e_{k})T_{j}(e_{l})h=\tilde{T_{i}}(I_{E_{i}}\otimes\tilde{T_{j}})(e_{k}\otimes
e_{l}\otimes h)=\tilde{T_{j}}(I_{E_{j}}\otimes\tilde{T_{i}})(t_{i,j}\otimes
I_{H})(e_{k}\otimes e_{l}\otimes h)=$
$\tilde{T_{j}}(I_{E_{j}}\otimes\tilde{T_{i}})(e_{l}\otimes e_{k}\otimes
h)=T_{j}(e_{l})T_{i}(e_{k})h.$

Thus, we require that

(18) $T_{i,k}T_{j,l}=T_{j,l}T_{i,k}$

for $i\neq j$ and $1\leq k,l\leq d$.

Also, $\tilde{T}_{i}^{(2)}:\mathbb{C}^{d}\otimes\mathbb{C}^{d}\otimes H\to H$
is given by

$\tilde{T}_{i}(\lambda\otimes\mu\otimes
h)=T_{i}(\lambda\otimes\sum_{k=1}^{d}\mu_{k}T_{i,k}h)=\sum_{j,k=1}^{d}\lambda_{j}\mu_{k}T_{i,j}T_{i,k}h$

Identifying $\mathbb{C}^{d}\otimes\mathbb{C}^{d}$ with $\mathbb{C}^{d^{2}}$
and $\lambda\otimes\mu$ with the column $(\lambda_{j}\mu_{k})_{j,k}$ we can
write $T_{i}^{(2)}$ as a row $(T_{i,j}T_{i,k})_{j,k}$. Similarly, for any
$m\in\mathbb{N}$, $T_{i}^{(m)}$ can be written as the row $(T_{i,j_{1}}\cdots
T_{i,j_{m}})_{1\leq j_{1},\ldots,j_{m}\leq d}$. Therefore,

(19) $T_{i}^{(m)}T_{i}^{(m)^{*}}=\sum_{1\leq j_{1},\ldots,j_{m}\leq
d}T_{i,j_{1}}\cdots T_{i,j_{m}}T_{i,j_{m}}^{*}\cdots
T_{i,j_{1}}^{*}=\sum_{1\leq j_{1},\ldots,j_{m}\leq
d}\Big{(}\prod_{p=1}^{m}T_{i,j_{p}}\Big{)}\Big{(}\prod_{p=1}^{m}T_{i,j_{p}}^{*}\Big{)}.$

In a similar fashion, for each
$G=\\{i_{1},\ldots,i_{k}\\}\subset\\{1,\ldots,n\\}$,

$\tilde{T}_{G}\tilde{T}_{G}^{*}=\sum_{1\leq j_{1},\ldots,j_{k}\leq
d}\Big{(}\prod_{p=1}^{k}T_{i_{p},j_{p}}\Big{)}\Big{(}\prod_{p=1}^{k}T_{i_{p},j_{p}}^{*}\Big{)}.$

Note that, due to the commutation relation (18), the last expression is
independent of the order chosen on $G$. Therefore,

(20)
$\sum_{G=\\{i_{1},\ldots,i_{k}\\}\subset\\{1,\ldots,n\\}}(-1)^{|G|}\tilde{T}_{G}\tilde{T}_{G}^{*}=\sum_{G=\\{i_{1},\ldots,i_{k}\\}\subset\\{1,\ldots,n\\}}(-1)^{|G|}\sum_{1\leq
j_{1},\ldots,j_{k}\leq
d}\Big{(}\prod_{p=1}^{k}T_{i_{p},j_{p}}\Big{)}\Big{(}\prod_{p=1}^{k}T_{i_{p},j_{p}}^{*}\Big{)}.$

If the expression of the right hand side of (19) tends to $0$ in WOT as
$m\to\infty$ then $T_{i}$ is pure and if it happens for each $i=1,\ldots,n$,
then the c.c. representation $(\sigma,T_{1},\ldots,T_{n})$ is pure. On the
other hand, if the expression on the right hand side of (20) is a positive
operator then the representation $(\sigma,T_{1},\ldots,T_{n})$ is Szegö
positive. If the above two hold together then the representation
$(\sigma,T_{1},\ldots,T_{n})$ lies in the class
$\mathcal{T}_{1,n}^{n}(\sigma,\mathbb{C},\mathbb{C}^{d},\ldots,\mathbb{C}^{d},H)$.

######  Remark 3.4.

In the example above, instead of choosing $t_{i,j}$ to be the flip, we can
choose any other isomorphism of $\mathbb{C}^{d}\otimes\mathbb{C}^{d}$ (for
$i>j$). Such an isomorphism is given by any unitary matrix of size
$d^{2}\times d^{2}$. This will give us another, more involved, commutation
relation (see, for example, [14, Equation (1)]). We will not go into details
here except to note that, if $d=1$ and, for every $i>j$ we fix some number
$u_{i,j}$ with $|u_{i,j}|=1$ then the commutation relation is

(21) $T_{i}T_{j}=u_{i,j}T_{j}T_{i}$

for $i>j$. Note that, if $(T_{1},\ldots,T_{n})$ is a tuple of contractions
that $u$-commute (in the sense of Equation (21)) then $T_{G}$ may depend on
the order on $G$ but $T_{G}T_{G}^{*}$ does not. In fact, in the expression of
Equation (14), $u_{i,j}$ will get canceled .

######  Example 3.5.

If, in the setup of Remark 3.4, we take $d=1$ and $u_{i,j}=1$ for all $i,j$
then it reduces to the case of a commuting tuple of contractions studied by
many researchers (see the introduction). Note that, in this case Definition2.8
reduces to Definition2.

Let
$(\sigma,T_{1},\ldots,T_{n})\in\mathcal{T}_{1,n}^{n}(\sigma,M,E_{1},\ldots,E_{n},H)$.
Merging the $(n-1)$-tuples $\hat{T}_{1}$ and $\hat{T}_{n}$ we consider a new
$(n-1)$-tuple $\hat{T}_{1n}$ in the following way:

$\tilde{\hat{T}}_{1n}=\big{(}\tilde{T}_{1}(I_{1}\otimes\tilde{T}_{n}),\tilde{T_{2}},\ldots,\tilde{T}_{n-1}\big{)}.$

We have the following result.

######  Proposition 3.6.

If the tuples $\hat{T}_{1}$ and $\hat{T}_{n}$ satisfy Szegö positivity then so
is $\hat{T}_{1n}$. Moreover, denoting

$D_{\hat{T}_{i}}=\mathbb{S}_{n-1}^{-1}\big{(}\tilde{\hat{T}}_{i},\tilde{\hat{T}}_{i}^{*}\big{)}^{1/2}\quad(\text{for
all}\,\,i=1,2)\,\,\quad\text{and}\,\,\,D_{\hat{T}_{1n}}:=\mathbb{S}_{n-1}^{-1}\big{(}\tilde{\hat{T}}_{1n},\tilde{\hat{T}}_{1n}^{*}\big{)}^{1/2}$

we have

$D_{\hat{T}_{1n}}^{2}=D_{\hat{T}_{n}}^{2}+\tilde{T}_{1}(I_{1}\otimes
D_{\hat{T}_{1}}^{2})\tilde{T}_{1}^{*}=D_{\hat{T}_{1}}^{2}+\tilde{T_{n}}(I_{n}\otimes
D_{\hat{T}_{n}}^{2})\tilde{T}_{n}^{*}.$

###### Proof.

We observe that

$\displaystyle\mathbb{S}_{n-1}^{-1}$
$\displaystyle\big{(}\tilde{\hat{T}}_{1n},\tilde{\hat{T}}_{1n}^{*}\big{)}$
$\displaystyle=\sum_{G\subset\\{1,\ldots,n-1\\}}(-1)^{|G|}\big{(}\tilde{\hat{T}}_{1n}\big{)}_{G}\big{(}\tilde{\hat{T}}_{1n}\big{)}_{G}^{*}$
$\displaystyle=\sum_{G\subset\\{2,\ldots,n-1\\}}(-1)^{|G|}\tilde{T}_{G}\tilde{T}_{G}^{*}-\tilde{T}_{1}(I_{1}\otimes\tilde{T}_{n})\Big{(}\sum_{G\subset\\{2,\ldots,n-1\\}}(-1)^{|G|}\tilde{T}_{G}\tilde{T}_{G}^{*}\Big{)}(I_{1}\otimes\tilde{T}_{n}^{*})\tilde{T}_{1}^{*}$
$\displaystyle=\sum_{G\subset\\{2,\ldots,n-1\\}}(-1)^{|G|}\tilde{T}_{G}\tilde{T}_{G}^{*}-\tilde{T}_{1}\Big{(}\sum_{G\subset\\{2,\ldots,n-1\\}}(-1)^{|G|}I_{1}\otimes\tilde{T}_{G}\tilde{T}_{G}^{*}\Big{)}\tilde{T}_{1}^{*}$
$\displaystyle\hskip
113.81102pt+\tilde{T}_{1}\Big{(}\sum_{G\subset\\{2,\ldots,n-1\\}}(-1)^{|G|}I_{1}\otimes\tilde{T}_{G}\tilde{T}_{G}^{*}\Big{)}\tilde{T}_{1}^{*}$
$\displaystyle\hskip
113.81102pt-\tilde{T}_{1}(I_{1}\otimes\tilde{T}_{n})\Big{(}\sum_{G\subset\\{2,\ldots,n-1\\}}(-1)^{|G|}I_{1n}\otimes\tilde{T}_{G}\tilde{T}_{G}^{*}\Big{)}(I_{1}\otimes\tilde{T}_{n}^{*})\tilde{T}_{1}^{*}$
$\displaystyle=\sum_{G\subset\\{1,\ldots,n-1\\}}(-1)^{|G|}\big{(}\tilde{\hat{T}}_{n}\big{)}_{G}\big{(}\tilde{\hat{T}}_{n}\big{)}_{G}^{*}+\tilde{T}_{1}\Big{(}I_{1}\otimes\sum_{G\subset\\{2,\ldots,n\\}}(-1)^{|G|}\big{(}\tilde{\hat{T}}_{1}\big{)}_{G}\big{(}\tilde{\hat{T}}_{1}\big{)}_{G}^{*}\Big{)}\tilde{T}_{1}^{*}$
$\displaystyle=\mathbb{S}_{n-1}^{-1}\big{(}\tilde{\hat{T}}_{1},\tilde{\hat{T}}_{1}^{*}\big{)}+\tilde{T}_{1}\Big{(}I_{1}\otimes\mathbb{S}_{n-1}^{-1}\big{(}\tilde{\hat{T}}_{n},\tilde{\hat{T}}_{n}^{*}\big{)}\Big{)}\tilde{T}_{1}^{*}$
$\displaystyle=D_{\hat{T}_{1}}^{2}+\tilde{T}_{1}\Big{(}I_{1}\otimes
D_{\hat{T}_{n}}^{2}\Big{)}\tilde{T}_{1}^{*}.$

So,

$\mathbb{S}_{n-1}^{-1}\big{(}\tilde{\hat{T}}_{1n},\tilde{\hat{T}}_{1n}^{*}\big{)}\geq
0$

and hence

$D_{\hat{T}_{1n}}^{2}=D_{\hat{T}_{1}}^{2}+\tilde{T}_{1}\Big{(}I_{1}\otimes
D_{\hat{T}_{n}}^{2}\Big{)}\tilde{T}_{1}^{*}.$

Similarly, we can also show that

$D_{\hat{T}_{1n}}^{2}=D_{\hat{T}_{n}}^{2}+\tilde{T}_{n}\Big{(}I_{n}\otimes
D_{\hat{T}_{1}}^{2}\Big{)}\tilde{T}_{n}^{*}.$

∎

From the above proposition it follows that for every $h\in H$,

(22) $\|D_{\hat{T}_{n}}h\oplus(I_{1}\otimes
D_{\hat{T}_{1}})\tilde{T_{1}}^{*}h\|=\|D_{\hat{T}_{1}}h\oplus(I_{n}\otimes
D_{\hat{T}_{n}})\tilde{T_{n}}^{*}h\|.$

Note that $\\{D_{\hat{T}_{n}}h\oplus(I_{1}\otimes
D_{\hat{T}_{1}})\tilde{T_{1}}^{*}h:h\in H\\}$ is a subspace of
$H\oplus(E_{1}\otimes H)$. On $H\oplus(E_{1}\otimes H)$ we have a
representation $\tau_{1}$ of $M$ where, for $h,k\in H$ and $a\in M$,

$\tau_{1}(a)(h\oplus(\xi_{1}\otimes
k))=\sigma(a)h\oplus(\varphi_{1}(a)\xi_{1}\otimes k)$

and $\tau_{2}$ is defined similarly on $H\oplus(E_{n}\otimes H)$.

Note that $D_{\tilde{T_{i}}}$ commutes with $\sigma(M)$ and, for $a\in M$,
$\tilde{T_{i}}^{*}\sigma(a)=(\varphi_{i}(a)\otimes I_{H})\tilde{T_{i}}^{*}$.
It follows that

$\tau_{1}(a)(D_{\hat{T}_{n}}h\oplus(I_{1}\otimes
D_{\hat{T}_{1}})\tilde{T_{1}}^{*}h)=D_{\hat{T}_{n}}\sigma(a)h\oplus(I_{1}\otimes
D_{\hat{T}_{1}})\tilde{T_{1}}^{*}\sigma(a)h$

and, therefore, the space $\\{D_{\hat{T}_{n}}h\oplus(I_{1}\otimes
D_{\hat{T}_{1}})\tilde{T_{1}}^{*}h:h\in H\\}$ is invariant under $\tau_{1}(M)$
. Writing $\mathcal{D}_{1}$ for the closure

$\mathcal{D}_{1}=\\{D_{\hat{T}_{n}}h\oplus(I_{1}\otimes
D_{\hat{T}_{1}})\tilde{T_{1}}^{*}h:h\in H\\}^{-},$

we see that, restricting $\tau_{1}$ to it, we get a subrepresentation of
$\tau_{1}$. Similarly, setting

$\mathcal{D}_{2}=\\{D_{\hat{T}_{1}}h\oplus(I_{n}\otimes
D_{\hat{T}_{n}})\tilde{T_{n}}^{*}h:h\in H\\}^{-}$

we get a subrepresentation of $\tau_{2}$.

It follows from (22) that we can define a partial isometry (with initial space
$\mathcal{D}_{1}$)

$V_{0}:\mathcal{D}_{1}\rightarrow\mathcal{D}_{2}$

by

$V_{0}(D_{\hat{T}_{n}}h\oplus(I_{1}\otimes
D_{\hat{T}_{1}})\tilde{T_{1}}^{*}h)=D_{\hat{T}_{1}}h\oplus(I_{n}\otimes
D_{\hat{T}_{n}})\tilde{T_{n}}^{*}h.$

From the above definition, it is straight-forward to see that $V_{0}$
intertwines $\tau_{1}$ and $\tau_{2}$. That is,

$V_{0}\tau_{1}(a)=\tau_{2}(a)V_{0}$

for all $a\in M$ and we can write

$\mathcal{D}_{1}\simeq\mathcal{D}_{2}.$

######  Proposition 3.7.

We can find representations $\rho_{1},\rho_{2}$ of $M$ on Hilbert spaces
$\mathcal{E}_{1},\mathcal{E}_{2}$ and a unitary operator

$U:\mathcal{D}_{\hat{T}_{1}}\oplus(E_{n}\otimes\mathcal{D}_{\hat{T}_{n}})\oplus\mathcal{E}_{2}\rightarrow\mathcal{D}_{\hat{T}_{n}}\oplus(E_{1}\otimes\mathcal{D}_{\hat{T}_{1}})\oplus\mathcal{E}_{1}$

that extends $U_{0}:=V_{0}^{-1}$ and intertwines the representations of $M$.
Moreover,

  1. (1)

There is a unitary operator
$u_{1}:E_{1}\otimes\mathcal{E}_{2}\rightarrow\mathcal{E}_{1}$ (so that
$u_{1}(E_{1}\otimes\mathcal{E}_{2})=\mathcal{E}_{1}$ ) that intertwines the
representations $\rho_{1}$ and
$\varphi_{1}(\cdot)\otimes_{\rho_{2}}I_{\mathcal{E}_{2}}$.

  2. (2)

There is a unitary operator $u_{2}:\mathcal{E}_{2}\rightarrow
E_{n}\otimes\mathcal{E}_{1}$ (so that
$u_{2}\mathcal{E}_{2}=E_{n}\otimes\mathcal{E}_{1}$) that intertwines the
representations $\rho_{2}$ and
$\varphi_{2}(\cdot)\otimes_{\rho_{1}}I_{\mathcal{E}_{1}}$.

  3. (3)

$\mathcal{E}_{i}$ ($i=1,2$) are infinite generators of $NRep(M)$.

###### Proof.

We write $\mathcal{D}_{\tilde{T_{i}}}$ for the closure of the range of
$D_{\tilde{T_{i}}}$ to get

$\mathcal{D}_{1}\subseteq\mathcal{D}_{\hat{T}_{n}}\oplus(E_{1}\otimes\mathcal{D}_{\hat{T}_{1}})$

and

$\mathcal{D}_{2}\subseteq\mathcal{D}_{\hat{T}_{1}}\oplus(E_{n}\otimes\mathcal{D}_{\hat{T}_{n}}).$

Note that the spaces on the right are also invariant under $\tau_{1}$ (or
$\tau_{2}$) so they are (left) $M$-modules and so are

$\mathcal{M}_{2}:=\mathcal{D}_{\hat{T}_{1}}\oplus(E_{n}\otimes\mathcal{D}_{\hat{T}_{n}})\ominus\mathcal{D}_{2}$

and

$\mathcal{M}_{1}:=\mathcal{D}_{\hat{T}_{n}}\oplus(E_{1}\otimes\mathcal{D}_{\hat{T}_{1}})\ominus\mathcal{D}_{1}.$

Now we write,

$\mathcal{E}_{i}:=H_{\infty}$

(for $i=1,2$) infinite generators. It follows that
$E_{n}\otimes\mathcal{E}_{1}$ and $E_{1}\otimes\mathcal{E}_{2}$ are also
infinite generators. The uniqueness of infinite generators implies the
existence of $u_{1},u_{2}$ as in the proposition.

Since $\mathcal{E}_{1}\precsim\mathcal{M}_{i}\oplus\mathcal{E}_{i}$,
$\mathcal{M}_{i}\oplus\mathcal{E}_{i}$ is also an infinite generator and it
follows from the uniqueness that there is a unitary operator $u_{1}$ from
$\mathcal{M}_{2}\oplus\mathcal{E}_{2}$ onto
$\mathcal{M}_{1}\oplus\mathcal{E}_{1}$ that intertwines the representations.
Setting $U=V_{0}^{*}\oplus u_{1}$ completes the proof.

∎

To simplify the notation we write $E_{ij}$ for $E_{i}\otimes E_{j}$ and
identify it with $E_{j}\otimes E_{i}$. Similarly, we write $E_{ijk}$ for
$E_{i}\otimes E_{j}\otimes E_{k}$ etc.

We have from the Proposition 3.6 that

$D_{\hat{T}_{1n}}^{2}=D_{\hat{T}_{n}}^{2}+\tilde{T_{1}}(I_{1}\otimes
D_{\hat{T}_{1}}^{2})\tilde{T}_{1}^{*}=D_{\hat{T}_{1}}^{2}+\tilde{T_{n}}(I_{n}\otimes
D_{\hat{T}_{n}}^{2})\tilde{T}_{n}^{*}\geq 0.$

We, thus, get an isometry
$V:\mathcal{D}_{\hat{T}_{1n}}\rightarrow\mathcal{D}_{\hat{T}_{n}}\oplus(E_{1}\otimes\mathcal{D}_{\hat{T}_{1}})\oplus\mathcal{E}_{1}$
such that

$V(D_{\hat{T}_{1n}}h)=D_{\hat{T}_{n}}h\oplus(I_{1}\otimes
D_{\hat{T}_{1}})\tilde{T}_{1}^{*}h\oplus 0_{\mathcal{E}_{1}}.$

We now need the following notation:

  1. (1)

$P_{1}$ is the projection

$P_{1}:\mathcal{D}_{\hat{T}_{n}}\oplus(E_{1}\otimes\mathcal{D}_{\hat{T}_{1}})\oplus\mathcal{E}_{1}\rightarrow
E_{1}\otimes\mathcal{D}_{\hat{T}_{1}}.$

  2. (2)

$i_{2}^{\prime}$ is the inclusion

$i_{2}^{\prime}:(E_{1n}\otimes\mathcal{D}_{\hat{T}_{n}})\oplus(E_{1}\otimes\mathcal{E}_{2})\rightarrow(E_{1}\otimes\mathcal{D}_{\hat{T}_{1}})\oplus(E_{1n}\otimes\mathcal{D}_{\hat{T}_{n}})\oplus(E_{1}\otimes\mathcal{E}_{2}).$

  3. (3)

$j_{2}^{\prime}=i_{2}^{\prime}\circ(I_{E_{1n}\otimes\mathcal{D}_{\tilde{T}_{1}}}\oplus(I_{1}\otimes
u_{2}^{*})):(E_{1n}\otimes\mathcal{D}_{\tilde{T}_{1}})\oplus(E_{1n}\otimes\mathcal{E}_{1})\rightarrow(E_{1n}\otimes\mathcal{D}_{\tilde{T}_{1}})\oplus(E_{1}\otimes\mathcal{E}_{2})\oplus(E_{1}\otimes\mathcal{D}_{\hat{T}_{1}})$.

  4. (4)

$i_{2}$ is the inclusion

$i_{2}:\mathcal{D}_{\hat{T}_{n}}\oplus\mathcal{E}_{1}\rightarrow\mathcal{D}_{\hat{T}_{n}}\oplus(E_{1}\otimes\mathcal{D}_{\hat{T}_{1}})\oplus\mathcal{E}_{1}$

so that $i_{2}^{*}$ is the projection

$i_{2}^{*}:\mathcal{D}_{\hat{T}_{n}}\oplus(E_{1}\otimes\mathcal{D}_{\hat{T}_{1}})\oplus\mathcal{E}_{1}\rightarrow\mathcal{D}_{\hat{T}_{n}}\oplus\mathcal{E}_{1}.$

It will be convenient to write $I_{1},I_{n},I_{1n}$ for
$I_{E_{1}},I_{E_{n}},I_{E_{1n}}$ respectively.

Note that, since $U$ intertwines the representations of $M$, the operator
$I_{1}\otimes U$ is well defined and bounded. So the following operators are
well defined

$(I_{1}\otimes
U)P_{1}:\mathcal{D}_{\hat{T}_{n}}\oplus(E_{1}\otimes\mathcal{D}_{\hat{T}_{1}})\oplus\mathcal{E}_{1}\rightarrow(E_{1}\otimes\mathcal{D}_{\hat{T}_{n}})\oplus(E_{1}\otimes
E_{1}\otimes\mathcal{D}_{\hat{T}_{1}})\oplus E_{1}\otimes\mathcal{E}_{1}$

and

$(I_{1}\otimes
U)j_{2}^{\prime}:(E_{1n}\otimes\mathcal{D}_{\hat{T}_{n}})\oplus(E_{1n}\otimes\mathcal{E}_{1})\rightarrow(E_{1}\otimes\mathcal{D}_{\hat{T}_{n}})\oplus(E_{1}\otimes
E_{1}\otimes\mathcal{D}_{\hat{T}_{1}})\oplus(E_{1}\otimes\mathcal{E}_{1}).$

We can now form the operator

(23) $U_{1}:=\left(\begin{array}[]{cc}(I_{1}\otimes U)P_{1}&(I_{1}\otimes
U)j_{2}^{\prime}\\\
i_{2}^{*}&0\end{array}\right):\left(\begin{array}[]{c}\mathcal{D}_{\hat{T}_{n}}\oplus(E_{1}\otimes\mathcal{D}_{\hat{T}_{1}})\oplus\mathcal{E}_{1}\\\
(E_{1n}\otimes\mathcal{D}_{\hat{T}_{n}})\oplus(E_{1n}\otimes\mathcal{E}_{1})\end{array}\right)$
$\rightarrow\left(\begin{array}[]{c}(E_{1}\otimes\mathcal{D}_{\hat{T}_{n}})\oplus(E_{1}\otimes
E_{1}\otimes\mathcal{D}_{\hat{T}_{1}})\oplus(E_{1}\otimes\mathcal{E}_{1})\\\
\mathcal{D}_{\hat{T}_{n}}\oplus\mathcal{E}_{1}\end{array}\right).$

A straightforward computation shows that, as $U$ is unitary, then so is
$U_{1}$.

To shorten some computations, we shall write

(24)
$\mathcal{D}:=\mathcal{D}_{\hat{T}_{n}}\oplus(E_{1}\otimes\mathcal{D}_{\hat{T}_{1}})\oplus\mathcal{E}_{1}$

and

(25) $\mathcal{D}^{\prime}:=\mathcal{D}_{\hat{T}_{n}}\oplus\mathcal{E}_{1}$

so that

$U_{1}:=\left(\begin{array}[]{cc}(I_{1}\otimes U)P_{1}&(I_{1}\otimes
U)j_{2}^{\prime}\\\
i_{2}^{*}&0\end{array}\right):\left(\begin{array}[]{c}\mathcal{D}\\\
E_{1n}\otimes\mathcal{D}^{\prime}\end{array}\right)\rightarrow\left(\begin{array}[]{c}E_{1}\otimes\mathcal{D}\\\
\mathcal{D}^{\prime}\end{array}\right).$

Note that $M$ acts on $\mathcal{D}$ by the restriction, to $\mathcal{D}$, of
$\sigma\oplus(\varphi_{1}\otimes I_{\mathcal{D}_{\hat{T}_{1}}})\oplus\rho_{1}$
we write $\rho_{\mathcal{D}}$ for this representation. Similarly,
$\rho_{\mathcal{D}^{\prime}}$ will denote the representation of $M$ on
$\mathcal{D}^{\prime}$ (the restriction of $\sigma\oplus\rho_{1}$). There is a
natural left action on $E_{1n}=E_{1}\otimes E_{n}$ which comes through an
adjointable map $\varphi_{1n}:M\to\mathcal{L}(E_{1}\otimes E_{n})$ which is
defined by

(26) $\varphi_{1n}(a)(e_{1}\otimes e_{n}):=\varphi_{1}(a)e_{1}\otimes
e_{n}\quad(\text{for all}\,\,a\in M,\,e_{1}\in E_{1},\,e_{n}\in E_{n})$

Viewing $E_{1n}$ as $E_{n}\otimes E_{1}$ the action $\varphi_{1n}$ can also be
represented by,

(27) $\varphi_{1n}(a)(e_{n}\otimes e_{1})=\varphi_{n}(a)e_{n}\otimes
e_{1}\quad(\text{for all}\,\,a\in M,\,e_{1}\in E_{1},\,e_{n}\in E_{n}).$

We have the following lemma.

######  Lemma 3.8.

The operator $U_{1}$ intertwines the representations of $M$, that is

$U_{1}(\rho_{\mathcal{D}}(a)\oplus(\varphi_{1n}(a)\otimes
I_{\mathcal{D}^{\prime}}))=((\varphi_{1}(a)\otimes
I_{\mathcal{D}})\oplus\rho_{\mathcal{D}^{\prime}}(a))U_{1}$

for every $a\in M$.

###### Proof.

We need the prove the following, for every $a\in M$,

  1. (i)

$(I_{1}\otimes U)P_{1}\rho_{\mathcal{D}}(a)=((\varphi_{1}(a)\otimes
I_{\mathcal{D}})(I_{1}\otimes U)P_{1}$.

  2. (ii)

$((I_{1}\otimes U)j_{2}^{\prime})(\varphi_{1n}(a)\otimes
I_{\mathcal{D}^{\prime}})=(\varphi_{1}(a)\otimes
I_{\mathcal{D}})((I_{1}\otimes U)j_{2}^{\prime})$.

  3. (iii)

$i_{2}^{*}\rho_{\mathcal{D}}(a)=\rho_{\mathcal{D}^{\prime}}(a)i_{2}^{*}$.

Part (iii) is obvious. We proceed to prove part (i). From the definitions of
$P_{1}$ and of $\rho_{\mathcal{D}}$ it follows that

$P_{1}\rho_{\mathcal{D}}(a)=(\varphi_{1}(a)\otimes
I_{\mathcal{D}_{\hat{T}_{1}}})P_{1}.$

It follows from Proposition 3.7(4) that
$U\sigma|\mathcal{D}_{\tilde{T}_{2}}=\rho_{\mathcal{D}}U|\mathcal{D}_{\tilde{T}_{2}}$.
Using this, we have

$(I_{1}\otimes U)P_{1}\rho_{\mathcal{D}}(a)=(I_{1}\otimes
U)(\varphi_{1}(a)\otimes
I_{\mathcal{D}_{\hat{T}_{1}}})P_{1}=(\varphi_{1}(a)\otimes
I_{\mathcal{D}})(I_{1}\otimes U)P_{1},$

proving (i). For (ii) we write

$((I_{1}\otimes U)j_{2}^{\prime})(\varphi_{1n}(a)\otimes
I_{\mathcal{D}^{\prime}})=(I_{1}\otimes
U)i_{2}^{\prime}(I_{E_{1n}\otimes\mathcal{D}_{\hat{T}_{n}}}\oplus(I_{1}\otimes
u_{2}^{*}))(\varphi_{1n}(a)\otimes I_{\mathcal{D}^{\prime}})$ $=(I_{1}\otimes
U)i_{2}^{\prime}((\varphi_{1n}(a)\otimes
I_{\mathcal{D}_{\hat{T}_{n}}})\oplus(\varphi_{1}(a)\otimes
I_{\mathcal{E}_{2}}))(I_{E_{1n}\otimes\mathcal{D}_{\hat{T}_{n}}}\oplus(I_{1}\otimes
u_{2}^{*}))$ $=(I_{1}\otimes U)((\varphi_{1n}(a)\otimes
I_{\mathcal{D}_{\hat{T}_{n}}})\oplus(\varphi_{1}(a)\otimes
I_{\mathcal{E}_{2}})\oplus(\varphi_{1}(a)\otimes
I_{\mathcal{D}_{\hat{T}_{1}}}))i_{2}^{\prime}(I_{E_{1n}\otimes\mathcal{D}_{\hat{T}_{n}}}\oplus(I_{1}\otimes
u_{2}^{*})).$

Using the intertwining property of $U$ again, we see that this is equal to

$(\varphi_{1}(a)\otimes I_{\mathcal{D}})(I_{1}\otimes
U)i_{2}^{\prime}(I_{E_{1n}\otimes\mathcal{D}_{\hat{T}_{n}}}\oplus(I_{1}\otimes
u_{2}^{*}))=(\varphi_{1}(a)\otimes I_{\mathcal{D}})((I_{1}\otimes
U)j_{2}^{\prime})$

proving (ii). ∎

######  Lemma 3.9.

For $h\in H$,

$U_{1}(VD_{\hat{T}_{1n}}h,(I_{1n}\otimes
D_{\hat{T}_{n}})(I_{1}\otimes\tilde{T}_{n}^{*})\tilde{T}_{1}^{*}h\oplus
0_{E_{1n}\otimes\mathcal{E}_{1}})=((I_{1}\otimes
VD_{\hat{T}_{1n}})\tilde{T}_{1}^{*}h,D_{\hat{T}_{n}}h\oplus
0_{\mathcal{E}_{1}}).$

###### Proof.

We start by proving that, for $f\in E_{1}\otimes H$,

(28) $(I_{1}\otimes U)[(I_{1}\otimes D_{\hat{T}_{1}})f+(I_{1n}\otimes
D_{\hat{T}_{n}})(I_{1}\otimes\tilde{T}_{n}^{*})f]=(I_{1}\otimes
D_{\hat{T}_{n}})f+(I_{11}\otimes
D_{\hat{T}_{1}})(I_{1}\otimes\tilde{T}_{1}^{*})f.$

Clearly, it suffices to prove it for $f=\xi\otimes g$ where $\xi\in E_{1}$ and
$g\in H$. So we compute

$\displaystyle(I_{1}\otimes U)[(I_{1}\otimes D_{\hat{T}_{1}})(\xi\otimes
g)\oplus(I_{1n}\otimes
D_{\hat{T}_{n}})(I_{1}\otimes\tilde{T}_{n}^{*})(\xi\otimes g)\oplus
0_{E_{1}\oplus\mathcal{E}_{2}}]$ $\displaystyle=$ $\displaystyle(I_{1}\otimes
U)[\xi\otimes D_{\hat{T}_{1}}g\oplus(I_{1n}\otimes
D_{\hat{T}_{n}})(\xi\otimes\tilde{T}_{n}^{*}g)\oplus
0_{E_{1}\oplus\mathcal{E}_{2}}]$ $\displaystyle=$ $\displaystyle(I_{1}\otimes
U)[\xi\otimes D_{\hat{T}_{1}}g\oplus\xi\otimes(I_{n}\otimes
D_{\hat{T}_{n}})\tilde{T}_{n}^{*}g\oplus 0_{E_{1}\oplus\mathcal{E}_{2}}]$
$\displaystyle=$ $\displaystyle\xi\otimes
U(D_{\hat{T}_{1}}g\oplus(I_{n}\otimes D_{\hat{T}_{n}})\tilde{T}_{n}^{*}g\oplus
0_{\mathcal{E}_{2}})$

Since $U$ extends $U_{0}$ ($=V_{0}^{-1}$),
$U(D_{\hat{T}_{1}}g\oplus(I_{n}\otimes
D_{\hat{T}_{n}})\tilde{T}_{n}^{*}g\oplus
0_{\mathcal{E}_{2}})=(D_{\hat{T}_{n}}g\oplus(I_{1}\otimes
D_{\hat{T}_{1}})\tilde{T}_{1}^{*}g\oplus 0_{\mathcal{E}_{1}})$ and, thus,

$\displaystyle(I_{1}\otimes U)[(I_{1}\otimes D_{\hat{T}_{1}})(\xi\otimes
g)\oplus(I_{1n}\otimes
D_{\hat{T}_{n}})(I_{1}\otimes\tilde{T}_{n}^{*})(\xi\otimes g)\oplus
0_{E_{1}\oplus\mathcal{E}_{2}}]$ $\displaystyle=$
$\displaystyle\xi\otimes(D_{\hat{T}_{n}}g\oplus(I_{1}\otimes
D_{\hat{T}_{1}})\tilde{T}_{1}^{*}g\oplus 0_{\mathcal{E}_{1}})$
$\displaystyle=$ $\displaystyle(I_{1}\otimes D_{\hat{T}_{n}})(\xi\otimes
g)\oplus(I_{11}\otimes
D_{\hat{T}_{1}})(I_{1}\otimes\tilde{T}_{1}^{*})(\xi\otimes g)\oplus
0_{E_{1}\oplus\mathcal{E}_{1}}$

proving (28).

Now we compute

$\displaystyle U_{1}(VD_{\hat{T}_{1n}}h,(I_{1n}\otimes
D_{\tilde{T}_{1}})(I_{1}\otimes\tilde{T}_{n}^{*})\tilde{T}_{1}^{*}h\oplus
0_{E_{1n}\otimes\mathcal{E}_{1}})$ $\displaystyle=$ $\displaystyle
U_{1}(D_{\hat{T}_{n}}h\oplus(I_{1}\otimes
D_{\hat{T}_{1}})\tilde{T}_{1}^{*}h\oplus 0_{\mathcal{E}_{1}},(I_{1n}\otimes
D_{\hat{T}_{n}})(I_{n}\otimes\tilde{T}_{1}^{*})\tilde{T_{n}}^{*}h\oplus
0_{E_{1n}\otimes\mathcal{E}_{1}})$ $\displaystyle=$
$\displaystyle((I_{1}\otimes U)[(I_{1}\otimes
D_{\tilde{T_{n}}})\tilde{T}_{1}^{*}h\oplus(I_{1n}\otimes
D_{\hat{T}_{n}})(I_{1}\otimes\tilde{T}_{n}^{*})\tilde{T}_{1}^{*}h\oplus
0_{E_{1}\oplus\mathcal{E}_{2}}],D_{\hat{T}_{n}}h\oplus 0_{\mathcal{E}_{1}})$
$\displaystyle=$ $\displaystyle((I_{1}\otimes
D_{\hat{T}_{n}})\tilde{T}_{1}^{*}h\oplus(I_{11}\otimes
D_{\hat{T}_{1}})(I_{1}\otimes\tilde{T}_{1}^{*})\tilde{T}_{1}^{*}h\oplus
0_{E_{1}\oplus\mathcal{E}_{1}},D_{\hat{T}_{n}}h\ \oplus 0_{\mathcal{E}_{1}})$

where we used equation (28) with $f=\tilde{T}_{1}^{*}h$. It is left to show
that

$(I_{1}\otimes D_{\hat{T}_{n}})\tilde{T}_{1}^{*}h\oplus(I_{11}\otimes
D_{\hat{T}_{1}})(I_{1}\otimes\tilde{T}_{1}^{*})\tilde{T}_{1}^{*}h\oplus
0_{E_{1}\oplus\mathcal{E}_{1}}=(I_{1}\otimes
VD_{\hat{T}_{1n}})\tilde{T}_{1}^{*}h.$

Now, replace $\tilde{T}_{1}^{*}h$ by $\xi\otimes g$ and use the definition of
$V$, that completes the proof. ∎

For faithful $W^{*}$-correspondences $E_{1},\ldots,E_{n}$ over a
$W^{*}$-algebra $M$, we consider a new family of $n-1$ faithful
correspondences $E_{1n},E_{2},\ldots,E_{n-1}$. Now the Fock correspondence for
this product system (constructed as in (13)) is

(29) $\mathcal{F}(E)=\bigoplus_{\alpha\in\mathbb{Z}_{+}^{n-1}}E^{\alpha}$

where $E^{0}=M$ and for each
$0\neq\alpha=(\alpha_{1},\ldots,\alpha_{n-1})\in\mathbb{Z}_{+}^{n-1}$,
$E^{\alpha}=E_{1n}^{\otimes\alpha_{1}}\otimes
E_{2}^{\otimes\alpha_{2}}\otimes\cdots\otimes E_{n-1}^{\otimes\alpha_{n-1}}$.
For each $i\in\\{2,\ldots,n-1\\}$, we consider the left creation operator
$L_{i}:E_{i}\to\mathcal{B}(\mathcal{F}(E)\otimes H)$ defined by

$L_{i}(\xi_{i})=L_{\xi_{i}}\quad\quad\text{for}\,\,\xi_{i}\in
E_{i},\,\,\text{and}\,\,i=2,\ldots,n-1$

and $L_{1}:E_{1n}\to\mathcal{B}(\mathcal{F}(E)\otimes H)$ by
$L_{1}(\xi_{1n})=L_{\xi_{1n}}$ The operators
$L_{\xi_{i}}:\mathcal{F}(E)\otimes H\to\mathcal{F}(E)\otimes H$ is defined by

$L_{\xi_{i}}(h)=\xi_{i}\otimes h\quad\quad\text{and}\quad
L_{\xi_{i}}(\eta^{(\alpha)}\otimes h)=\xi_{i}\otimes\eta^{(\alpha)}\otimes
h\quad(\eta^{(\alpha)}\in E^{\alpha},h\in H).$

The operator $L_{\xi_{1n}}:\mathcal{F}(E)\otimes H\to\mathcal{F}(E)\otimes H$
is defined by $L_{\xi_{1n}}(\eta^{(\alpha)}\otimes
h)=\xi_{1n}\otimes\eta^{(\alpha)}\otimes h$. Following this, for each
$i=2,\ldots,n-1$, we define $\tilde{L}_{i}:E_{i}\otimes\mathcal{F}(E)\otimes
H\to\mathcal{F}(E)\otimes H$ by

(30) $\tilde{L}_{i}(\xi_{i}\otimes k):=L_{i}(\xi_{i})k=\xi_{i}\otimes
k\quad\text{and}\quad\tilde{L}_{1}(\xi_{1n}\otimes
k):=L_{1}(\xi_{1n})k=\xi_{1n}\otimes k\quad(k\in\mathcal{F}(E)\otimes H).$

For the above defined $(n-1)$-tuple
$\tilde{\hat{T}}_{1n}=\big{(}\tilde{T}_{1}(I_{1}\otimes\tilde{T}_{n}),\tilde{T_{2}},\ldots,\tilde{T}_{n-1}\big{)}$,
we consider a dilation map $\Pi_{1n}:H\to\mathcal{F}(E)\otimes H$, that is
defined by

(31)
$\Pi_{1n}h:=\bigoplus_{\alpha\in\mathbb{Z}_{+}^{n-1}}(I_{E^{\alpha}}\otimes
D_{\hat{T}_{1n}})\tilde{T}_{1n}^{*(\alpha)}h=D_{\hat{T}_{1n}}h\oplus(I_{E_{1n}}\otimes
D_{\hat{T}_{1n}})\tilde{T}_{1n}^{*}h\oplus(I_{E_{2}}\otimes
D_{\hat{T}_{1n}})\tilde{T}_{2}^{*}h\oplus\cdots$

where for $\alpha=(\alpha_{1},\ldots,\alpha_{n-1})\in\mathbb{Z}_{+}^{n-1}$,
$E^{\alpha}=E_{1n}^{\otimes\alpha_{1}}\otimes\cdots\otimes
E_{n-1}^{\otimes\alpha_{n-1}}$ and
$\tilde{T}_{1n}^{(\alpha)}=\tilde{T}_{1n}^{\alpha_{1}}\tilde{T}_{2}^{\alpha_{2}}\cdots\tilde{T}_{n-1}^{\alpha_{n-1}-1}$.

######  Remark 3.10.

Note that this is the same map as in Proposition 3.1 and in Corollary 3.2 but
for the product system $(M,E_{1n},E_{2},\ldots,E_{n-1})$ and the tuple
$\tilde{\hat{T}}_{1n}=(\tilde{T}_{1}(I_{1}\otimes\tilde{T}_{n}),\tilde{T}_{2},\ldots,\tilde{T}_{n-1})$.

Now, we are ready to prove a dilation result for the $(n-1)$-tuple
$\tilde{\hat{T}}_{1n}$.

######  Proposition 3.11.

Let $\hat{T}_{n}$ be a pure $(n-1)$-tuple. Then the above defined map
$\Pi_{1n}$ is an isometry and

  1. (a)

$(I_{1n}\otimes\Pi_{1n})\tilde{T}_{1n}^{*}=\tilde{L}_{1}^{*}\Pi_{1n}$.

  2. (b)

$(I_{i}\otimes\Pi_{1n})\tilde{T}_{i}^{*}=\tilde{L}_{i}^{*}\Pi_{1n}$ for all
$i=2,\ldots,n-1$.

###### Proof.

The fact that $\Pi_{1n}$ is an isometry follows from Corollary 3.2 (see the
remark above). Let $\xi_{1n}\in E_{1n},\eta^{(\beta)}\in E^{\beta}$ and $h\in
H$. Then we have

$\displaystyle\langle\tilde{L}_{1}^{*}\Pi_{1n}h,\xi_{1n}\otimes\eta^{(\beta)}\otimes
h\rangle$ $\displaystyle=$
$\displaystyle\langle\Pi_{1n}h,\tilde{L}_{1}(\xi_{1n}\otimes\eta^{(\beta)}\otimes
h)\rangle$ $\displaystyle=$
$\displaystyle\Big{\langle}\bigoplus_{\alpha\in\mathbb{Z}_{+}^{n-1}}(I_{E^{\alpha}}\otimes
D_{\hat{T}_{1n}})\tilde{T}_{1n}^{*(\alpha)}h,\xi_{1n}\otimes\eta^{(\beta)}\otimes
h\Big{\rangle}$ $\displaystyle=$ $\displaystyle\langle(I_{E^{\beta}}\otimes
I_{1n}\otimes
D_{\hat{T}_{1n}})\tilde{T}_{1n}^{*(\beta+e_{1})}h,\xi_{1n}\otimes\eta^{(\beta)}\otimes
h\rangle$ $\displaystyle=$
$\displaystyle\langle[I_{1n}\otimes(I_{E^{\beta}}\otimes
D_{\hat{T}_{1n}})\tilde{T}_{1n}^{*(\beta)}]\tilde{T}_{1n}^{*}h,\xi_{1n}\otimes\eta^{(\beta)}\otimes
h\rangle$ $\displaystyle=$
$\displaystyle\langle(I_{1n}\otimes\Pi_{1n})\tilde{T}_{1n}^{*},\xi_{1n}\otimes\eta^{(\beta)}\otimes
h\rangle.$

On the other hand, for each $i=2,\ldots,n-1$, take $\xi_{i}\in
E_{i},\eta^{(\beta)}\in E^{\beta}$ and $h\in H$. Then we have

$\displaystyle\langle\tilde{L}_{i}^{*}\Pi_{1n}h,\xi_{i}\otimes\eta^{(\beta)}\otimes
h\rangle$ $\displaystyle=$
$\displaystyle\langle\Pi_{1n}h,\tilde{L}_{i}(\xi_{i}\otimes\eta^{(\beta)}\otimes
h)\rangle$ $\displaystyle=$
$\displaystyle\Big{\langle}\bigoplus_{\alpha\in\mathbb{Z}_{+}^{n-1}}(I_{E^{\alpha}}\otimes
D_{\hat{T}_{1n}})\tilde{T}_{1n}^{*(\alpha)}h,\xi_{i}\otimes\eta^{(\beta)}\otimes
h\Big{\rangle}$ $\displaystyle=$
$\displaystyle\langle(I_{E^{\beta+e_{i}}}\otimes
D_{\hat{T}_{1n}})\tilde{T}_{1n}^{*(\beta+e_{i})}h,\xi_{i}\otimes\eta^{(\beta)}\otimes
h\rangle$ $\displaystyle=$
$\displaystyle\langle[I_{i}\otimes(I_{E^{\beta}}\otimes
D_{\hat{T}_{1n}})\tilde{T}_{1n}^{*(\beta)}]\tilde{T}_{i}^{*}h,\xi_{i}\otimes\eta^{(\beta)}\otimes
h\rangle$ $\displaystyle=$
$\displaystyle\langle(I_{i}\otimes\Pi_{1n})\tilde{T}_{i}^{*},\xi_{i}\otimes\eta^{(\beta)}\otimes
h\rangle.$

∎

Recall that
$V:\mathcal{D}_{\hat{T}_{1n}}\rightarrow\mathcal{D}_{\hat{T}_{n}}\oplus(E_{1}\otimes\mathcal{D}_{\hat{T}_{1}})\oplus\mathcal{E}_{1}=\mathcal{D}$
is an isometry such that

$V(D_{\hat{T}_{1n}}h)=D_{\hat{T}_{n}}h\oplus(I_{1}\otimes
D_{\hat{T}_{1}})\tilde{T_{1}}^{*}h\oplus 0_{\mathcal{E}_{1}}.$

We now define the map

$\Pi_{1n,V}:H\rightarrow\mathcal{F}(E)\otimes[\mathcal{D}\oplus(E_{1n}\otimes\mathcal{D})]$

by

$\Pi_{1n,V}h=(I_{\mathcal{F}(E)}\otimes V)\Pi
h=\bigoplus_{\alpha\in\mathbb{Z}_{+}^{n-1}}\big{(}I_{E^{(\alpha)}}\otimes
VD_{\hat{T}_{1n}}\big{)}\tilde{T}_{1n}^{*(\alpha)}h.$

Since both $\Pi$ and $V$ are isometries, so is $\Pi_{1n,V}$. Therefore, we
have the following dilation result which basically replace dilation space by
some other Fock space where coefficient space is $\mathcal{D}$ instead of
$\mathcal{D}_{\hat{T}_{1n}}$.

######  Proposition 3.12.

Let $\hat{T}_{n}$ be a pure $(n-1)$-tuple and the map $\Pi_{1n,V}$ as above.
Then,

  1. (a)

$(I_{1n}\otimes\Pi_{1n,V})\tilde{T}_{1n}^{*}=\tilde{L}_{1}^{*}\Pi_{1n,V}$.

  2. (b)

$(I_{i}\otimes\Pi_{1n,V})\tilde{T}_{i}^{*}=\tilde{L}_{i}^{*}\Pi_{1n,V}$ for
all $i=2,\ldots,n-1$.

###### Proof.

To prove,

(32) $(I_{1n}\otimes I_{\mathcal{F}(E)}\otimes
V)\tilde{L}_{1}^{*}=\tilde{L}_{1}^{*}(I_{\mathcal{F}(E)}\otimes V),$

note that $\tilde{L}_{1}^{*}$ (on the left hand side) is the inclusion of
$E_{1n}\otimes\mathcal{F}(E)\otimes\mathcal{D}_{\hat{T}_{1n}}$ in
$\mathcal{F}(E)\otimes\mathcal{D}_{\hat{T}_{1n}}$ while $\tilde{L}_{1}^{*}$
(on the right hand side) is the inclusion of
$E_{1n}\otimes\mathcal{F}(E)\otimes\mathcal{D}$ in
$\mathcal{F}(E)\otimes\mathcal{D}$. Thus, for all
$h\in\mathcal{D}_{\hat{T}_{1n}}$, $\xi_{1n}\in E_{1n}$ and $\eta^{(\beta)}\in
E^{\beta}$,

$(I_{1n}\otimes I_{\mathcal{F}(E)}\otimes
V)\tilde{L}_{1}^{*}(\xi_{1n}\otimes\eta^{(\beta)}\otimes
h)=\xi_{1n}\otimes\eta^{(\beta)}\otimes
Vh=(\xi_{1n}\otimes\eta^{(\beta)})\otimes Vh$
$=\tilde{L}_{1}^{*}(I_{\mathcal{F}(E)}\otimes
V)(\xi_{1n}\otimes\eta^{(\beta)}\otimes h),$

proving (32). Using this,

$\displaystyle(I_{1n}\otimes\Pi_{1n,V})\tilde{T}_{1n}^{*}$
$\displaystyle=\big{(}I_{1n}\otimes(I_{\mathcal{F}(E)}\otimes
V)\Pi_{1n}\big{)}\tilde{T}_{1n}^{*}$ $\displaystyle=(I_{1n}\otimes
I_{\mathcal{F}(E)}\otimes V)(I_{1n}\otimes\Pi_{1n})\tilde{T}_{1n}^{*}$
$\displaystyle=(I_{1n}\otimes I_{\mathcal{F}(E)}\otimes
V)\tilde{L}_{1}^{*}\Pi_{1n}\qquad(\text{by (a) of Proposition
\ref{dilation}})$ $\displaystyle=\tilde{L}_{1}^{*}(I_{\mathcal{F}(E)}\otimes
V)\Pi_{1n}\qquad(\text{by \ref{VL}})$
$\displaystyle=\tilde{L}_{1}^{*}\Pi_{1n,V}.$

This proves (a). Proof of item (b) is similar. ∎

Let, $U_{1}=\left(\begin{array}[]{cc}A&B\\\ C&0\end{array}\right)$. So, by the
Lemma 3.9,

$\left(\begin{array}[]{cc}A&B\\\
C&0\end{array}\right)\left(\begin{array}[]{c}VD_{\hat{T}_{1n}}h\\\
(I_{1n}\otimes
D_{\hat{T}_{n}})(I_{1}\otimes\tilde{T_{n}}^{*})\tilde{T}_{1}^{*}h\oplus
0_{E_{1n}\otimes\mathcal{E}_{1}}\end{array}\right)=\left(\begin{array}[]{c}(I_{1}\otimes
VD_{\hat{T}_{1n}})\tilde{T}_{1}^{*}h\\\ D_{\hat{T}_{n}}h\oplus
0_{\mathcal{E}_{1}}\end{array}\right).$

This implies,

$AVD_{\hat{T}_{1n}}h+B\big{[}(I_{1n}\otimes
D_{\hat{T}_{n}})(I_{1}\otimes\tilde{T_{n}}^{*})\tilde{T}_{1}^{*}h\oplus
0_{E_{1n}\otimes\mathcal{E}_{1}}\big{]}=(I_{1}\otimes
VD_{\hat{T}_{1n}})\tilde{T}_{1}^{*}h$

and

$CVD_{\hat{T}_{1n}}h=D_{\hat{T}_{n}}h\oplus 0_{\mathcal{E}_{1}}.$

The above two together imply,

$\displaystyle AVD_{\hat{T}_{1n}}h+B\big{[}(I_{1n}\otimes
D_{\hat{T}_{n}})(I_{1}\otimes\tilde{T_{n}}^{*})\tilde{T}_{1}^{*}h\oplus
0_{E_{1n}\otimes\mathcal{E}_{1}}\big{]}$ $\displaystyle=$ $\displaystyle
AVD_{\hat{T}_{1n}}h+B\big{[}\big{(}I_{1n}\otimes(D_{\hat{T}_{n}}\oplus
0_{\mathcal{E}_{1}})\big{)}(I_{1}\otimes\tilde{T_{n}}^{*})\tilde{T}_{1}^{*}h\big{]}$
$\displaystyle=$ $\displaystyle
AVD_{\hat{T}_{1n}}h+B\big{[}\big{(}I_{1n}\otimes
CVD_{\hat{T}_{1n}}\big{)}(I_{1}\otimes\tilde{T_{n}}^{*})\tilde{T}_{1}^{*}h\big{]}$
$\displaystyle=$ $\displaystyle(I_{1}\otimes
VD_{\hat{T}_{1n}})\tilde{T}_{1}^{*}h$

Define

(33) $\tau_{U_{1}^{*}}:=I_{\mathcal{F}(E)}\otimes\big{(}A^{*}+[I_{1n}\otimes
C^{*}]B^{*}\big{)}.$

More explicitely,

$\tau_{U_{1}^{*}}=I_{\mathcal{F}(E)}\otimes\big{(}P_{1}^{*}(I_{1}\otimes
U^{*})+[I_{1n}\otimes i_{2}](J_{2}^{\prime})^{*}(I_{1}\otimes U^{*})\big{)}.$

This map will play an important role.

Note that

(34) $\tau_{U_{1}^{*}}:\mathcal{F}(E)\otimes
E_{1}\otimes\mathcal{D}\rightarrow\mathcal{F}(E)\otimes[\mathcal{D}\oplus(E_{1n}\otimes\mathcal{D})]\subseteq\mathcal{F}(E)\otimes\mathcal{D}$

and we view here $\mathcal{F}(E)\otimes E_{1n}\otimes\mathcal{D}$ as a
subspace of $\mathcal{F}(E)\otimes\mathcal{D}$.

######  Proposition 3.13.

For the unitary $U_{1}$, $\tau_{U_{1}^{*}}$ is a well defined isometry.

###### Proof.

To show that $\tau_{U_{1}^{*}}$ is well defined we need to show that
$B(I_{1n}\otimes C)$ is well defined. For this, recall that $C=i_{2}^{*}$ and,
thus, $I_{1n}\otimes C=I_{1n}\otimes
i_{2}^{*}:(E_{1n}\otimes\mathcal{D}_{\hat{T}_{n}})\oplus(E_{1n}\otimes
E_{1}\otimes\mathcal{D}_{\hat{T}_{1}})\oplus(E_{1n}\otimes\mathcal{E}_{1})\rightarrow(E_{1n}\otimes\mathcal{D}_{\hat{T}_{n}})\oplus(E_{1n}\otimes\mathcal{E}_{1}).$
For $B$ we have $B=(I_{1}\otimes U)j_{2}^{\prime}$ which is defined on
$(E_{1n}\otimes\mathcal{D}_{\tilde{T}_{1}})\oplus(E_{1n}\otimes\mathcal{E}_{1})$.
Thus $B(I_{1n}\otimes C)$ is well defined. The following computation shows
that $\tau_{U_{1}^{*}}$ is indeed an isometry.

$\displaystyle\tau_{U_{1}^{*}}^{*}\tau_{U_{1}^{*}}$ $\displaystyle=$
$\displaystyle\Big{[}I_{\mathcal{F}(E)}\otimes\big{(}A+B[I_{1n}\otimes
C]\big{)}\Big{]}\Big{[}I_{\mathcal{F}(E)}\otimes\big{(}A^{*}+[I_{1n}\otimes
C^{*}]B^{*}\big{)}\Big{]}$ $\displaystyle=$
$\displaystyle\big{(}I_{\mathcal{F}(E)}\otimes
A\big{)}\big{(}I_{\mathcal{F}(E)}\otimes
A^{*}\big{)}+\big{(}I_{\mathcal{F}(E)}\otimes
A\big{)}\big{(}I_{\mathcal{F}(E)}\otimes[I_{1n}\otimes C^{*}]B^{*}\big{)}$
$\displaystyle+\big{(}I_{\mathcal{F}(E)}\otimes B[I_{1n}\otimes
C]\big{)}\big{(}I_{\mathcal{F}(E)}\otimes
A^{*}\big{)}+\big{(}I_{\mathcal{F}(E)}\otimes B[I_{1n}\otimes
C]\big{)}\big{(}I_{\mathcal{F}(E)}\otimes[I_{1n}\otimes C^{*}]B^{*}\big{)}$

First term and last term of the above sum are equal to
$I_{\mathcal{F}(E)}\otimes AA^{*}$ and $I_{\mathcal{F}(E)}\otimes
B[I_{1n}\otimes CC^{*}]B^{*}$, respectively. Recall now that

$\displaystyle B^{*}:$ $\displaystyle\,E_{1}\otimes\mathcal{D}\to
E_{1n}\otimes\mathcal{D}^{{}^{\prime}}$ $\displaystyle C^{*}:$
$\displaystyle\,\mathcal{D}^{{}^{\prime}}\to\mathcal{D}$ $\displaystyle A:$
$\displaystyle\,\mathcal{D}\to E_{1}\otimes\mathcal{D}.$

It follows that the range of $I_{\mathcal{F}(E)}\otimes[I_{1n}\otimes
C^{*}]B^{*}$ is contained in $\mathcal{F}(E)\otimes
E_{1n}\otimes\mathcal{D}(\subseteq\mathcal{F}(E)\otimes\mathcal{D})$. On this
range $I_{\mathcal{F}(E)}\otimes A$ equals $I_{\mathcal{F}(E)}\otimes
I_{1n}\otimes A$ and, thus, the third term is
$I_{\mathcal{F}(E)}\otimes[I_{1n}\otimes AC^{*}]B^{*}$ and the forth term is
its adjoint.

Hence,

$\big{(}I_{\mathcal{F}(E)}\otimes
A\big{)}\big{(}I_{\mathcal{F}(E)}\otimes[I_{1n}\otimes
C^{*}]B^{*}\big{)}=I_{\mathcal{F}(E)}\otimes[I_{1n}\otimes AC^{*}]B^{*}.$

Similarly,

$\big{(}I_{\mathcal{F}(E)}\otimes B[I_{1n}\otimes
C]\big{)}\big{(}I_{\mathcal{F}(E)}\otimes
A^{*}\big{)}=I_{\mathcal{F}(E)}\otimes B[I_{1n}\otimes CA^{*}].$

Therefore,

$\tau_{U_{1}^{*}}^{*}\tau_{U_{1}^{*}}=I_{\mathcal{F}(E_{1n})}\otimes\big{(}AA^{*}+[I_{1n}\otimes
AC^{*}]B^{*}+B[I_{1n}\otimes CA^{*}]+B[I_{1n}\otimes CC^{*}]B^{*}\big{)}.$

The fact that $U_{1}$ is a unitary implies

$AC^{*}=0=CA^{*},\quad CC^{*}=I_{\mathcal{D}^{\prime}}\quad\text{and}\quad
AA^{*}+BB^{*}=I_{E_{1}\oplus\mathcal{D}}$

and hence,

$\tau_{U_{1}^{*}}^{*}\tau_{U_{1}^{*}}=I_{\mathcal{F}(E)\otimes
E_{1}\otimes\mathcal{D}}.$

∎

We have the following dilation result.

######  Theorem 3.14.

$(I_{1}\otimes\Pi_{1n,V})\tilde{T}_{1}^{*}=\tau_{U_{1}^{*}}^{*}\Pi_{1n,V}$.

###### Proof.

For $h\in H$, $\eta^{(\beta)}\in E^{\beta}$ and $k\in
E_{1}\otimes\big{(}D_{\tilde{T}_{1}}\oplus(E_{1}\otimes
D_{\tilde{T}_{n}})\oplus\mathcal{E}_{1}\big{)}$

$\displaystyle\langle\tau_{U_{1}^{*}}^{*}\Pi_{1n,V}h,\eta^{(\beta)}\otimes
k\rangle$ $\displaystyle=$
$\displaystyle\langle\Pi_{1n,V}h,\tau_{U_{1}^{*}}(\eta^{(\beta)}\otimes
k)\rangle$ $\displaystyle=$
$\displaystyle\Big{\langle}\bigoplus_{\alpha\in\mathbb{Z}_{+}^{n-1}}\big{(}I_{E^{\alpha}}\otimes
VD_{\hat{T}_{1n}}\big{)}\tilde{T}_{1n}^{*(\alpha)}h,\quad\big{[}I_{\mathcal{F}(E)}\otimes
A^{*}+\big{(}I_{\mathcal{F}(E)}\otimes[I_{1n}\otimes
C^{*}]\big{)}(I_{\mathcal{F}(E)}\otimes
B^{*})\big{]}\big{(}\eta^{(\beta)}\otimes k\big{)}\Big{\rangle}$
$\displaystyle=$
$\displaystyle\Big{\langle}\bigoplus_{\alpha\in\mathbb{Z}_{+}^{n-1}}\big{(}I_{E^{\alpha}}\otimes
VD_{\hat{T}_{1n}}\big{)}\tilde{T}_{1n}^{*(\alpha)}h,\Big{(}\bigoplus_{\alpha\in\mathbb{Z}_{+}^{n-1}}\big{(}I_{E^{\alpha}}\otimes
A^{*}\big{)}\Big{)}(\eta^{(\beta)}\otimes k)\Big{\rangle}$
$\displaystyle\hskip
14.22636pt+\Big{\langle}\bigoplus_{\alpha\in\mathbb{Z}_{+}^{n-1}}\big{(}I_{E^{\alpha}}\otimes
VD_{\hat{T}_{1n}}\big{)}\tilde{T}_{1n}^{*(\alpha)}h,\Big{(}\bigoplus_{\alpha\in\mathbb{Z}_{+}^{n-1}}\big{(}I_{E^{\alpha}}\otimes[I_{1n}\otimes
C^{*}]\big{)}\Big{)}\Big{(}\bigoplus_{\alpha\in\mathbb{Z}_{+}^{n-1}}\big{(}I_{E^{\alpha}}\otimes
B^{*}\big{)}\Big{)}(\eta^{(\beta)}\otimes k)\Big{\rangle}$ $\displaystyle=$
$\displaystyle\Big{\langle}\big{(}I_{E^{\beta}}\otimes
VD_{\hat{T}_{1n}}\big{)}\tilde{T}_{1n}^{*(\beta)}h,\big{(}I_{E^{\beta}}\otimes
A^{*}\big{)}(\eta^{(\beta)}\otimes k)\Big{\rangle}$ $\displaystyle\hskip
14.22636pt+\Big{\langle}\bigoplus_{\alpha=(0,\alpha_{2},\ldots,\alpha_{n-1})\in\mathbb{Z}_{+}^{n-1}}\big{(}I_{E^{\alpha}}\otimes
VD_{\hat{T}_{1n}}\big{)}\tilde{T}_{1n}^{*(\alpha)}h\quad+\quad
I_{1n}\otimes\bigoplus_{\alpha\in\mathbb{Z}_{+}^{n-1}}\big{(}I_{E^{\alpha}}\otimes
VD_{\hat{T}_{1n}}\big{)}\tilde{T}_{1n}^{*(\alpha+e_{1})}h,$
$\displaystyle\hskip
85.35826pt\Big{(}I_{1n}\otimes\bigoplus_{\alpha\in\mathbb{Z}_{+}^{n-1}}\big{(}I_{E^{\alpha}}\otimes
C^{*}\big{)}\Big{)}\Big{(}\bigoplus_{\alpha\in\mathbb{Z}_{+}^{n-1}}\big{(}I_{E^{\alpha}}\otimes
B^{*}\big{)}\Big{)}(\eta^{(\beta)}\otimes k)\Big{\rangle}$ $\displaystyle=$
$\displaystyle\langle\big{(}I_{E^{\beta}}\otimes
AVD_{\hat{T}_{1n}}\big{)}\tilde{T}_{1n}^{*(\beta)}h,\eta^{(\beta)}\otimes
k\big{\rangle}$ $\displaystyle\hskip
28.45274pt+\Big{\langle}I_{1n}\otimes\bigoplus_{\alpha\in\mathbb{Z}_{+}^{n-1}}\big{(}I_{E^{(\alpha)}}\otimes
CVD_{\hat{T}_{1n}}\big{)}\tilde{T}_{1n}^{*(\alpha+e_{1})}h,\Big{(}\bigoplus_{\alpha\in\mathbb{Z}_{+}^{n-1}}\big{(}I_{E^{\alpha}}\otimes
B^{*}\big{)}\Big{)}(\eta^{(\beta)}\otimes k)\Big{\rangle}$ $\displaystyle=$
$\displaystyle\big{\langle}\big{(}I_{E^{\beta}}\otimes
AVD_{\hat{T}_{1n}}\big{)}\tilde{T}_{1n}^{*(\beta)}h,\eta^{(\beta)}\otimes
k\big{\rangle}$ $\displaystyle\hskip
28.45274pt+\big{\langle}I_{1n}\otimes\big{(}I_{E^{\beta}}\otimes
CVD_{\hat{T}_{1n}}\big{)}\tilde{T}_{1n}^{*(\beta+e_{1})}h,\big{(}I_{E^{\beta}}\otimes
B^{*}\big{)}(\eta^{(\beta)}\otimes k)\big{\rangle}$ $\displaystyle=$
$\displaystyle\big{\langle}\big{(}I_{E^{\beta}}\otimes
AVD_{\hat{T}_{1n}}\big{)}\tilde{T}_{1n}^{*(\beta)}h,\eta^{(\beta)}\otimes
k\big{\rangle}$ $\displaystyle\hskip
28.45274pt+\big{\langle}I_{E^{\beta}}\otimes\big{[}\big{(}I_{1n}\otimes
CVD_{\hat{T}_{1n}}\big{)}\tilde{T}_{1n}\big{]}\tilde{T}_{1n}^{*(\beta)}h,\big{(}I_{E^{\beta}}\otimes
B^{*}\big{)}(\eta^{(\beta)}\otimes k)\big{\rangle}$ $\displaystyle=$
$\displaystyle\langle(I_{E^{\beta}}\otimes
AVD_{\hat{T}_{1n}})\tilde{T}_{1n}^{*(\beta)}h+[I_{E^{\beta}}\otimes
B(I_{1n}\otimes
CVD_{\hat{T}_{1n}})\tilde{T}_{1n}^{*}]\tilde{T}_{1n}^{*(\beta)}h,\eta^{(\beta)}\otimes
k\rangle$ $\displaystyle=$
$\displaystyle\langle\big{[}I_{E^{\beta}}\otimes(I_{1}\otimes
VD_{\hat{T}_{1n}})\tilde{T}_{1}^{*}\big{]}\tilde{T}_{1n}^{*(\beta)}h,\eta^{(\beta)}\otimes
k\rangle$ $\displaystyle=$
$\displaystyle\langle\big{[}I_{1}\otimes(I_{E^{\beta}}\otimes
VD_{\hat{T}_{1n}})\tilde{T}_{1n}^{*(\beta)}]\tilde{T}_{1}^{*}h,\eta^{(\beta)}\otimes
k\big{\rangle}$ $\displaystyle=$
$\displaystyle\langle(I_{1}\otimes\Pi_{1n,V})\tilde{T}_{1}^{*}h,\eta^{(\beta)}\otimes
k\rangle.$

∎

We know that the isometry

$U:\mathcal{D}_{\hat{T}_{1}}\oplus(E_{n}\otimes\mathcal{D}_{\hat{T}_{n}})\oplus\mathcal{E}_{2}\rightarrow\mathcal{D}_{\hat{T}_{n}}\oplus(E_{1}\otimes\mathcal{D}_{\hat{T}_{1}})\oplus\mathcal{E}_{1}$

satisfies

$U(D_{\hat{T}_{1}}h\oplus(I_{n}\otimes
D_{\hat{T}_{n}})\tilde{T_{n}}^{*}h\oplus
0_{\mathcal{E}_{2}})=(D_{\hat{T}_{n}}h\oplus(I_{1}\otimes
D_{\hat{T}_{1}})\tilde{T_{1}}^{*}h\oplus 0_{\mathcal{E}_{1}}).$

So the adjoint operator

$U^{*}:\mathcal{D}_{\hat{T}_{n}}\oplus(E_{1}\otimes\mathcal{D}_{\hat{T}_{1}})\oplus\mathcal{E}_{1}\rightarrow\mathcal{D}_{\hat{T}_{1}}\oplus(E_{n}\otimes\mathcal{D}_{\hat{T}_{n}})\oplus\mathcal{E}_{2}$

satisfies

$U^{*}(D_{\hat{T}_{n}}h\oplus(I_{1}\otimes
D_{\hat{T}_{1}})\tilde{T_{1}}^{*}h\oplus
0_{\mathcal{E}_{1}})=(D_{\hat{T}_{1}}h\oplus(I_{n}\otimes
D_{\hat{T}_{n}})\tilde{T_{n}}^{*}h\oplus 0_{\mathcal{E}_{2}}).$

We now need to consider the following operators:

  1. (1)

$P_{2}$ is the projection

$P_{2}:\mathcal{D}_{\hat{T}_{1}}\oplus(E_{n}\otimes\mathcal{D}_{\hat{T}_{n}})\oplus\mathcal{E}_{2}\rightarrow(E_{n}\otimes\mathcal{D}_{\hat{T}_{n}})\oplus\mathcal{E}_{2}$

  2. (2)

$i_{1}^{\prime}$ is the inclusion

$i_{1}^{\prime}:(E_{n}\otimes
E_{1}\otimes\mathcal{D}_{\hat{T}_{1}})\rightarrow(E_{n}\otimes\mathcal{D}_{\hat{T}_{n}})\oplus(E_{n}\otimes
E_{1}\otimes\mathcal{D}_{\hat{T}_{1}})\oplus(E_{n}\otimes\mathcal{E}_{1}).$

  3. (3)

$i_{1}$ is another inclusion

$i_{1}:\mathcal{D}_{\hat{T}_{1}}\rightarrow\mathcal{D}_{\hat{T}_{1}}\oplus(E_{n}\otimes\mathcal{D}_{\hat{T}_{n}})\oplus\mathcal{E}_{2}$

so that $i_{1}^{*}$ is the projection

$i_{1}^{*}:\mathcal{D}_{\hat{T}_{1}}\oplus(E_{n}\otimes\mathcal{D}_{\hat{T}_{n}})\oplus\mathcal{E}_{2}\rightarrow\mathcal{D}_{\hat{T}_{1}}.$

Recall the map
$V:\mathcal{D}_{\hat{T}_{1n}}\rightarrow\mathcal{D}_{\hat{T}_{n}}\oplus(E_{1}\otimes\mathcal{D}_{\hat{T}_{1}})\oplus\mathcal{E}_{1}$
such that

$V(D_{\hat{T}_{1n}}h)=D_{\hat{T}_{n}}h\oplus(I_{1}\otimes
D_{\hat{T}_{1}})\tilde{T_{1}}^{*}h\oplus 0_{\mathcal{E}_{1}}.$

Let
$\mathcal{D}_{1n}=\mathcal{D}_{\hat{T}_{n}}\oplus(E_{1}\otimes\mathcal{D}_{\hat{T}_{1}})$.
With this notation let,

$U_{n}:=\left(\begin{array}[]{cc}\big{(}I_{E_{n}\otimes\mathcal{D}_{1n}}\oplus
u_{2}\big{)}P_{2}U^{*}&i_{1}^{\prime}\\\
i_{1}^{*}U^{*}&0\end{array}\right):\left(\begin{array}[]{c}\mathcal{D}_{\hat{T}_{n}}\oplus(E_{1}\otimes\mathcal{D}_{\hat{T}_{1}})\oplus\mathcal{E}_{1}\\\
(E_{n}\otimes E_{1}\otimes\mathcal{D}_{\hat{T}_{1}})\end{array}\right)$
$\rightarrow\left(\begin{array}[]{c}(E_{n}\otimes\mathcal{D}_{\hat{T}_{n}})\oplus(E_{n}\otimes
E_{1}\otimes\mathcal{D}_{\hat{T}_{1}})\oplus(E_{n}\otimes\mathcal{E}_{1})\\\
\mathcal{D}_{\hat{T}_{1}}\end{array}\right).$

So that

$U_{n}=\left(\begin{array}[]{cc}\big{(}I_{E_{n}\otimes\mathcal{D}_{1n}}\oplus
u_{2}\big{)}P_{2}U^{*}&i_{1}^{\prime}\\\
i_{1}^{*}U^{*}&0\end{array}\right):\left(\begin{array}[]{c}\mathcal{D}\\\
E_{1n}\otimes\mathcal{D}_{\hat{T}_{1}}\end{array}\right)\rightarrow\left(\begin{array}[]{c}E_{n}\otimes\mathcal{D}\\\
\mathcal{D}_{\hat{T}_{1}}\end{array}\right).$

A straightforward computation shows that $U_{n}$ is unitary.

######  Lemma 3.15.

The operator $U_{n}$ intertwines the representations of $M$, that is

$U_{n}(\rho_{\mathcal{D}}(a)\oplus(\varphi_{1n}(a)\otimes
I_{\mathcal{D}_{\hat{T}_{1}}}))=((\varphi_{2}(a)\otimes
I_{\mathcal{D}})\oplus\sigma|_{\mathcal{D}_{\hat{T}_{1}}})U_{n}$

for every $a\in M$.

###### Proof.

We need to prove the following, for every $a\in M$,

  1. (i)

$(I_{E_{2}\otimes\mathcal{D}_{1n}}\oplus
u_{2})P_{2}U^{*}\rho_{\mathcal{D}}(a)=(\varphi_{2}(a)\otimes
I_{\mathcal{D}})(I_{E_{2}\otimes\mathcal{D}_{1n}}\oplus u_{2})P_{2}U^{*}$.

  2. (ii)

$i_{1}^{\prime}(\varphi_{1n}(a)\otimes
I_{\mathcal{D}_{\hat{T}_{1}}})=(\varphi_{2}(a)\otimes
I_{\mathcal{D}})i_{1}^{\prime}$.

We start with (i) and compute, using the intertwining property of $U^{*}$
(see Proposition 3.7(4)),

$(I_{E_{2}\otimes\mathcal{D}_{1n}}\oplus
u_{2})P_{2}U^{*}\rho_{\mathcal{D}}(a)=(I_{E_{2}\otimes\mathcal{D}_{1n}}\oplus
u_{2})P_{2}(\sigma(a)\oplus(\varphi_{2}(a)\otimes
I_{\mathcal{D}_{\hat{T}_{n}}})\oplus\rho_{2}(a))U^{*}.$

Using the definition of $P_{2}$ this is equal to

$(I_{E_{2}\otimes\mathcal{D}_{1n}}\oplus u_{2})((\varphi_{2}(a)\otimes
I_{\mathcal{D}_{\hat{T}_{n}}})\oplus\rho_{2}(a))P_{2}U^{*}=((\varphi_{2}(a)\otimes
I_{\mathcal{D}_{\hat{T}_{n}}})\oplus(\varphi_{2}(a)\otimes
I_{\mathcal{E}_{1}})u_{2})P_{2}U^{*}$ $=((\varphi_{2}(a)\otimes
I_{\mathcal{D}_{\hat{T}_{n}}})\oplus(\varphi_{2}(a)\otimes
I_{\mathcal{E}_{1}}))(I_{E_{2}\otimes\mathcal{D}_{1n}}\oplus
u_{2})P_{2}U^{*}=(\varphi_{2}(a)\otimes
I_{\mathcal{D}})(I_{E_{2}\otimes\mathcal{D}_{1n}}\oplus u_{2})P_{2}U^{*},$

proving (i).

To prove (ii) we compute

$i_{1}^{\prime}(\varphi_{1n}(a)\otimes
I_{\mathcal{D}_{\hat{T}_{1}}})=((\varphi_{1n}(a)\otimes
I_{\mathcal{D}_{\hat{T}_{1}}})\oplus(\varphi_{1n}(a)\otimes
I_{\mathcal{D}_{\hat{T}_{n}}})\oplus(\varphi_{2}(a)\otimes
I_{\mathcal{E}_{1}}))i_{1}^{\prime}$
$=\varphi_{2}(a)\otimes(I_{E_{1}\otimes\mathcal{D}_{\hat{T}_{1}}}\oplus
I_{\mathcal{D}_{\hat{T}_{n}}}\oplus
I_{\mathcal{E}_{1}})i_{1}^{\prime}=(\varphi_{2}(a)\otimes
I_{\mathcal{D}})i_{1}^{\prime},$

proving (ii).

To prove (iii) we again use the intertwining property of $U^{*}$ as follows.

$i_{1}^{*}U^{*}\rho_{\mathcal{D}}(a)=i_{1}^{*}(\sigma(a)\oplus(\varphi_{2}(a)\otimes
I_{\mathcal{D}_{\hat{T}_{n}}})\oplus\rho_{2}(a))U^{*}=\sigma(a)i_{1}^{*}U^{*}.$

∎

Now, we compute

$\displaystyle U_{n}\big{(}VD_{\hat{T}_{1n}}h,(I_{1n}\otimes
D_{\hat{T}_{1}})(I_{n}\otimes\tilde{T_{1}}^{*})\tilde{T_{n}}^{*}h\big{)}$
$\displaystyle=U_{n}\big{(}D_{\hat{T}_{n}}h\oplus(I_{1}\otimes
D_{\hat{T}_{1}})\tilde{T}_{1}^{*}h\oplus 0_{\mathcal{E}_{1}},(I_{1n}\otimes
D_{\hat{T}_{1}})(I_{n}\otimes\tilde{T}_{1}^{*})\tilde{T_{n}}^{*}h\big{)}$
$\displaystyle=\left(\begin{array}[]{cc}\big{(}I_{E_{n}\otimes\mathcal{D}_{1n}}\oplus
u_{2}\big{)}P_{2}U^{*}&i_{1}^{\prime}\\\
i_{1}^{*}U^{*}&0\end{array}\right)\left(\begin{array}[]{cc}D_{\hat{T}_{n}}h\oplus(I_{1}\otimes
D_{\hat{T}_{1}})\tilde{T}_{1}^{*}h\oplus 0_{\mathcal{E}_{1}}\\\ (I_{1n}\otimes
D_{\hat{T}_{1}})(I_{n}\otimes\tilde{T}_{1}^{*})\tilde{T_{n}}^{*}h\end{array}\right)$
$\displaystyle=\left(\begin{array}[]{cc}\big{(}I_{E_{n}\otimes\mathcal{D}_{1n}}\oplus
u_{2}\big{)}P_{2}U^{*}\big{(}D_{\hat{T}_{n}}h\oplus(I_{1}\otimes
D_{\hat{T}_{1}})\tilde{T}_{1}^{*}h\oplus
0_{\mathcal{E}_{1}}\big{)}+i_{1}^{\prime}\big{(}(I_{1n}\otimes
D_{\hat{T}_{1}})(I_{n}\otimes\tilde{T}_{1}^{*})\tilde{T_{n}}^{*}h\big{)}\\\
i_{1}^{*}U^{*}\big{(}D_{\hat{T}_{n}}h\oplus(I_{1}\otimes
D_{\hat{T}_{1}})\tilde{T}_{1}^{*}h\oplus
0_{\mathcal{E}_{1}}\big{)}\end{array}\right)$
$\displaystyle=\left(\begin{array}[]{cc}\big{(}I_{E_{n}\otimes\mathcal{D}_{1n}}\oplus
u_{2}\big{)}P_{2}\big{(}D_{\hat{T}_{1}}h\oplus(I_{n}\otimes
D_{\hat{T}_{n}})\tilde{T_{n}}^{*}h\oplus
0_{\mathcal{E}_{2}}\big{)}+\big{(}(I_{1n}\otimes
D_{\hat{T}_{1}})(I_{n}\otimes\tilde{T}_{1}^{*})\tilde{T_{n}}^{*}h\oplus
0_{E_{n}\otimes(\mathcal{D}_{\tilde{T}_{1}}\oplus\mathcal{E}_{1})}\big{)}\\\
i_{1}^{*}\big{(}D_{\hat{T}_{1}}h\oplus(I_{n}\otimes
D_{\hat{T}_{n}})\tilde{T_{n}}^{*}h\oplus
0_{\mathcal{E}_{1}}\big{)}\end{array}\right)$
$\displaystyle=\left(\begin{array}[]{cc}0_{E_{1n}\otimes\mathcal{D}_{\tilde{T_{n}}}}\oplus(I_{n}\otimes
D_{\hat{T}_{n}})\tilde{T_{n}}^{*}h\oplus
0_{E_{n}\otimes\mathcal{E}_{1}}+(I_{1n}\otimes
D_{\hat{T}_{1}})(I_{n}\otimes\tilde{T}_{1}^{*})\tilde{T_{n}}^{*}h\oplus
0_{E_{n}\otimes\mathcal{D}_{\tilde{T}_{1}}}\oplus
0_{E_{n}\otimes\mathcal{E}_{1}}\\\ D_{\hat{T}_{1}}h\end{array}\right)$
$\displaystyle=\left(\begin{array}[]{cc}(I_{n}\otimes
D_{\hat{T}_{n}})\tilde{T_{n}}^{*}h\oplus(I_{1n}\otimes
D_{\hat{T}_{1}})(I_{n}\otimes\tilde{T}_{1}^{*})\tilde{T_{n}}^{*}h\oplus
0_{E_{n}\otimes\mathcal{E}_{1}}\\\ D_{\hat{T}_{1}}h\end{array}\right)$
$\displaystyle=\left(\begin{array}[]{cc}(I_{n}\otimes
VD_{{\tilde{T}}_{12}})\tilde{T_{n}}^{*}h\\\
D_{\hat{T}_{1}}h\end{array}\right).$

That is, $U_{n}=\begin{bmatrix}A&B\\\ C&0\end{bmatrix}$ has the following
property similar as $U_{1}$,

(35) $AVD_{\hat{T}_{1n}}h+B[(I_{1n}\otimes
D_{\hat{T}_{1}})(I_{n}\otimes\tilde{T_{1}}^{*})\tilde{T_{n}}^{*}h]=(I_{n}\otimes
VD_{{\tilde{T}}_{1n}})\tilde{T_{n}}^{*}h$

and

(36) $CVD_{\hat{T}_{1n}}h=D_{\hat{T}_{1}}h.$

Recall that

(37) $\tau_{U_{n}^{*}}=I_{\mathcal{F}(E)}\otimes(A^{*}+[I_{1n}\otimes
C^{*}]B^{*})$

where $A^{*}=UP_{2}^{*}\big{(}I_{E_{n}\otimes\mathcal{D}_{1n}}\oplus
u_{2}^{*}\big{)}$, $B^{*}=(i_{1}^{\prime})^{*}$ and $C^{*}=Ui_{1}$ so that we
can write

$\tau_{U_{n}^{*}}=I_{\mathcal{F}(E)}\otimes\Big{(}UP_{2}^{*}\big{(}I_{E_{n}\otimes\mathcal{D}_{1n}}\oplus
u_{2}^{*}\big{)}+[I_{1n}\otimes Ui_{1}](i_{1}^{\prime})^{*}\Big{)}.$

The map $I_{1n}\otimes Ui_{1}$ is defined on
$E_{1n}\otimes\mathcal{D}_{\hat{T}_{1}}$. Since this is the range of
$(i_{1}^{\prime})^{*}$, $\tau_{U_{n}^{*}}$ is well defined. In fact

(38) $\tau_{U_{n}^{*}}:\mathcal{F}(E)\otimes
E_{n}\otimes\mathcal{D}\rightarrow\mathcal{F}(E)\otimes[\mathcal{D}\oplus(E_{1n}\otimes\mathcal{D})]\subseteq\mathcal{F}(E)\otimes\mathcal{D}$

where we view $\mathcal{F}(E)\otimes E_{1n}\otimes\mathcal{D}$ as a subspace
of $\mathcal{F}(E)\otimes\mathcal{D}$.

So, as in Proposition 3.13 and Theorem 3.14, we have the following result,

######  Theorem 3.16.

$\tau_{U_{n}^{*}}$ is an isometry and
$(I_{n}\otimes\Pi_{1n,V})\tilde{T_{n}}^{*}=\tau_{U_{n}^{*}}^{*}\Pi_{1n,V}$.

###### Proof.

The fact that $\tau_{U_{n}^{*}}$ is an isometry follows from the proof of
Proposition 3.13. For $h\in H$, $\eta^{(\beta)}\in E^{\beta}$ and $k\in
E_{n}\otimes\big{(}D_{\tilde{T}_{1}}\oplus(E_{1}\otimes
D_{\tilde{T}_{n}})\big{)}$

$\displaystyle\langle\tau_{U_{n}^{*}}^{*}\Pi_{1n,V}h,\eta^{(\beta)}\otimes
k\rangle$ $\displaystyle=$
$\displaystyle\langle\Pi_{1n,V}h,\tau_{U_{n}^{*}}(\eta^{(\beta)}\otimes
k)\rangle$ $\displaystyle=$
$\displaystyle\Big{\langle}\bigoplus_{\alpha\in\mathbb{Z}_{+}^{n-1}}\big{(}I_{E^{\alpha}}\otimes
VD_{\hat{T}_{1n}}\big{)}\tilde{T}_{1n}^{*(\alpha)}h,\quad\big{[}I_{\mathcal{F}(E)}\otimes
A^{*}+\big{(}I_{\mathcal{F}(E)}\otimes[I_{1n}\otimes
C^{*}]\big{)}(I_{\mathcal{F}(E)}\otimes
B^{*})\big{]}\big{(}\eta^{(\beta)}\otimes k\big{)}\Big{\rangle}$

As in the proof of Proposition 3.14, the above expression is equal to

$\langle(I_{E^{\beta}}\otimes
AVD_{\hat{T}_{1n}})\tilde{T}_{1n}^{*(\beta)}h+[I_{E^{\beta}}\otimes
B(I_{1n}\otimes
CVD_{\hat{T}_{1n}})\tilde{T}_{1n}^{*}]\tilde{T}_{1n}^{*(\beta)}h,\eta^{(\beta)}\otimes
k\rangle.$

Using the properties of $U_{n}$, listed in (35) and (36), this again equals to

$\displaystyle\langle\big{[}I_{E^{\beta}}\otimes(I_{n}\otimes
VD_{\hat{T}_{1n}})\tilde{T}_{n}^{*}\big{]}\tilde{T}_{1n}^{*(\beta)}h,\eta^{(\beta)}\otimes
k\rangle$ $\displaystyle=$
$\displaystyle\langle\big{[}I_{n}\otimes(I_{E^{\beta}}\otimes
VD_{\hat{T}_{1n}})\tilde{T}_{1n}^{*(\beta)}]\tilde{T}_{n}^{*}h,\eta^{(\beta)}\otimes
k\big{\rangle}$ $\displaystyle=$
$\displaystyle\langle(I_{n}\otimes\Pi_{1n,V})\tilde{T}_{n}^{*}h,\eta^{(\beta)}\otimes
k\rangle.$

∎

We now turn to show that $\tau_{U_{i}^{*}}$ gives rise to a representation of
$E_{i}$ on $K:=\mathcal{F}(E)\otimes\mathcal{D}$ where $\mathcal{F}(E)$ is as
in (29). It will be convenient to note that

(39) $K=\mathcal{D}\oplus E_{1n}\otimes
K\oplus(\bigoplus_{i=2}^{n-1}E_{i}\otimes K)$

In fact, there is a natural representation $\rho$ on $\mathcal{B}(K)$,

$\rho:M\to\mathcal{B}(K)=\mathcal{B}\Big{(}\mathcal{D}\oplus E_{1n}\otimes
K\oplus\Big{(}\bigoplus_{i=2}^{n-1}E_{i}\otimes K\Big{)}\Big{)},$

which is defined by

$\displaystyle\rho(a)=\varphi_{\mathcal{F}(E)}(a)\otimes
I_{\mathcal{D}}=\rho_{\mathcal{D}}\oplus(\varphi_{1n}(a)\otimes
I_{K})\oplus\Big{(}\bigoplus_{i=2}^{n-1}\varphi_{i}(a)\otimes I_{K}\Big{)}$

and we shall show that there are representations $(\rho,M_{i})$ of $E_{i}$
($i=1,n$) such that $\tau_{U_{1}^{*}}=\tilde{M}_{1}$ and
$\tau_{U_{n}^{*}}=\tilde{M}_{n}$ (in the notation of (10)).

######  Proposition 3.17.

Let $K$ be as above. There are representations $M_{1}$ (of $E_{1}$) and
$M_{n}$ (of $E_{2}$) on $\mathcal{F}(E)\otimes\mathcal{D}$ such that
$\tau_{U_{1}^{*}}=\tilde{M_{1}}$ and $\tau_{U_{n}^{*}}=\tilde{M_{n}}$.

###### Proof.

Let $a\in M$. First, we shall verify,

(40) $\tilde{L}_{1}\big{(}\varphi_{1n}(a)\otimes
I_{K}\big{)}=\rho(a)\tilde{L}_{1}\quad\&\,\,\,\tilde{L}_{i}\big{(}\varphi_{i}(a)\otimes
I_{K}\big{)}=\rho(a)\tilde{L}_{i}\quad(\forall\,\,i=2,\ldots,n-1)$

Then we shall show

(41) $\tau_{U_{1}^{*}}\big{(}\varphi_{1}(a)\otimes
I_{K}\big{)}=\rho(a)\tau_{U_{1}^{*}}$

and

(42) $\tau_{U_{n}^{*}}\big{(}\varphi_{n}(a)\otimes
I_{K}\big{)}=\rho(a)\tau_{U_{n}^{*}}.$

Once we verify these, the proposition will follow. To verify (40) we write,
for $\xi\in E_{1n}$, $\xi_{i}\in E_{i}$ and $k\in K$,

$\rho(a)\tilde{L}_{1}(\xi\otimes k)=\rho(a)(\xi\otimes
K)=\varphi_{1n}(a)\xi\otimes k=\tilde{L}_{1}(\varphi_{1n}(a)\otimes
I_{K})(\xi\otimes k)$

and

$\rho(a)\tilde{L}_{i}(\xi_{i}\otimes K)=\rho(a)(\xi_{i}\otimes
k)=\varphi_{i}(a)\xi_{i}\otimes k=\tilde{L}_{i}(\varphi_{i}(a)\otimes
I_{K})(\xi_{i}\otimes k).$

Hence (40) follows. To prove (41), we will use (i) - (ii) that were proved in
the proof of Lemma 3.8. In fact, we shall use the adjoints, namely,

  1. (i*)

$P_{1}^{*}(I_{1}\otimes U^{*})(\varphi_{1}(a)\otimes
I_{\mathcal{D}})=\rho_{\mathcal{D}}(a)P_{1}^{*}(I_{1}\otimes U^{*})$, and

  2. (ii*)

$(j^{\prime}_{2})^{*}(I_{1}\otimes U^{*})(\varphi_{1}(a)\otimes
I_{\mathcal{D}})=(\varphi_{1n}(a)\otimes
I_{\mathcal{D}^{\prime}})(j_{2}^{\prime})^{*}(I_{1}\otimes U^{*})$.

We now turn to prove (41). Note that the right-hand side is defined on
$\mathcal{F}(E)\otimes E_{1}\otimes\mathcal{D}$ but this is equal to
$E_{1}\otimes\mathcal{F}(E)\otimes\mathcal{D}$ and so the left-hand side is
also well defined. In fact, we can write

$E_{1}\otimes\mathcal{F}(E)\otimes\mathcal{D}=E_{1}\otimes\mathcal{D}\oplus\Big{(}\bigoplus_{\alpha\in\mathbb{Z}_{+}^{n-1},\,\alpha>0}\big{(}E_{1}\otimes
E^{\otimes\alpha}\otimes\mathcal{D}\big{)}\Big{)}$

and proceed to prove the equality on $E_{1}\otimes\mathcal{D}$ and on
$E_{1}\otimes E^{\otimes\alpha}\otimes\mathcal{D}$. We start with
$E_{1}\otimes\mathcal{D}$ and fix $\xi\otimes d\in E_{1}\otimes\mathcal{D}$.

$\tau_{U_{1}^{*}}(\varphi_{1}(a)\otimes
I_{\mathcal{F}(E_{12})\otimes\mathcal{D}})(\xi\otimes
d)=\tau_{U_{1}^{*}}(\varphi(a)\xi\otimes d)=(P_{1}^{*}(I_{1}\otimes
U^{*})+(I_{1n}\otimes i_{2})(j_{2}^{\prime})^{*}(I_{1}\otimes
U^{*}))(\varphi(a)\xi\otimes d).$

Using (i*) and (ii*), this is equal to

$\displaystyle\big{(}\rho_{\mathcal{D}}(a)$ $\displaystyle
P_{1}^{*}(I_{1}\otimes U^{*})+(I_{1n}\otimes i_{2})(\varphi_{1n}(a)\otimes
I_{\mathcal{D}^{\prime}})(j_{2}^{\prime})^{*}(I_{1}\otimes
U^{*})\big{)}(\xi\otimes d)$ $\displaystyle=$
$\displaystyle\big{(}\rho_{\mathcal{D}}(a)P_{1}^{*}(I_{1}\otimes
U^{*})+(\varphi_{1n}(a)\otimes
I_{\mathcal{D}})(j_{2}^{\prime})^{*}(I_{1}\otimes U^{*})\big{)}(\xi\otimes d)$
$\displaystyle=$
$\displaystyle\big{(}(\rho_{\mathcal{D}}(a)\oplus(\varphi_{1n}(a)\otimes
I_{\mathcal{D}})\big{)}\big{(}P_{1}^{*}(I_{1}\otimes U^{*})+(I_{1n}\otimes
i_{2})(j_{2}^{\prime})^{*}(I_{1}\otimes U^{*})\big{)}(\xi\otimes d)$
$\displaystyle=$ $\displaystyle\rho(a)\tau_{U_{1}^{*}}(\xi\otimes d)$

proving the statement (41) applied to $E_{1}\otimes\mathcal{D}$. Now, we apply
the left hand side of (41) to $\xi^{\prime}\otimes d\in E_{1}\otimes
E^{\otimes\alpha}\otimes\mathcal{D}$ (for
$\alpha\in\mathbb{Z}_{+}^{n-1},\alpha>0$). But $\xi^{\prime}$ can also be
written as $\xi^{\prime}\in E^{\otimes\alpha}\otimes E_{1}$ so we apply it to
$\eta\otimes\xi$ where $\eta\in E^{\otimes\alpha}$ and $\xi\in E_{1}$ to get

$\tau_{U_{1}^{*}}\big{(}\varphi_{1}(a)\otimes
I_{E^{\otimes\alpha}\otimes\mathcal{D}}\big{)}(\xi^{\prime}\otimes
d)=\tau_{U_{1}^{*}}\big{(}\varphi_{1}(a)\otimes
I_{E^{\otimes\alpha}\otimes\mathcal{D}}\big{)}(\eta\otimes\xi\otimes d)$
$=\tau_{U_{1}^{*}}\big{(}\varphi_{E^{\otimes\alpha}}(a)\eta\otimes\xi\otimes
d\big{)}=\varphi_{E^{\otimes\alpha}}(a)\eta\otimes\tau_{U_{1}^{*}}(\xi\otimes
d)$ $=\rho(a)(\eta\otimes\tau_{U_{1}^{*}}(\xi\otimes
d))=\rho(a)\tau_{U_{1}^{*}}(\eta\otimes\xi\otimes
d)=\rho(a)\tau_{U_{1}^{*}}(\xi^{\prime}\otimes d).$

This proves (41).

To prove (42), we will use (i) - (ii) that were proved in the proof of Lemma
3.15. In fact, we shall use the adjoints, namely,

  1. (i*)

$\rho_{\mathcal{D}}(a)UP_{2}^{*}(I_{E_{2}\otimes\mathcal{D}_{1n}}\oplus
u_{2}^{*})=UP_{2}^{*}(I_{E_{2}\otimes\mathcal{D}_{1n}}\oplus
u_{2}^{*})(\varphi_{2}(a)\otimes I_{\mathcal{D}})$, and

  2. (ii*)

$(i_{1}^{\prime})^{*}(\varphi_{2}(a)\otimes
I_{\mathcal{D}})=(\varphi_{1n}(a)\otimes
I_{\mathcal{D}_{\hat{T}_{1}}})(i_{1}^{\prime})^{*}$

As in the proof of (41) above, we shall first prove it for $\xi\otimes d\in
E_{2}\otimes\mathcal{D}$.

So, we compute

$\tau_{U_{n}^{*}}(\varphi_{2}(a)\xi\otimes
d)=UP_{2}^{*}(I_{E_{2}\otimes\mathcal{D}_{1n}}\oplus
u_{2}^{*})(\varphi_{2}(a)\xi\otimes d)\oplus(I_{1n}\otimes
Ui_{1})(i_{1}^{\prime})^{*}(\varphi_{2}(a)\xi\otimes d).$

for the first summand we use (i*) to find that it is equal to

$\rho_{\mathcal{D}}(a)UP_{2}^{*}(I_{E_{2}\otimes\mathcal{D}_{1n}}\oplus
u_{2}^{*})(\xi\otimes
d)=\rho(a)UP_{2}^{*}(I_{E_{2}\otimes\mathcal{D}_{1n}}\oplus
u_{2}^{*})(\xi\otimes d).$

For the second summand, we use (ii*) to write it as

$(\varphi_{1n}(a)\otimes Ui_{1})(i_{1}^{\prime})^{*}(\xi\otimes d)$

and this is equal to

$\rho(a)(I_{1n}\otimes Ui_{1})(i_{1}^{\prime})^{*}(\xi\otimes d).$

Putting it together, we get (42) for $E_{2}\otimes\mathcal{D})$.

Now, we apply the left hand side of (41) to $\xi^{\prime}\otimes d\in
E_{2}\otimes E^{\otimes\alpha}\otimes\mathcal{D}$ (for
$\alpha\in\mathbb{Z}_{+}^{n-1},\alpha>0$). But $E_{2}\otimes
E^{\otimes\alpha}=E^{\otimes\alpha}\otimes E_{2}$ so it suffices to apply it
to $\eta\otimes\xi$ where $\eta\in E^{\otimes\alpha}$ ($\alpha>0$) and $\xi\in
E_{2}$ to get

$\tau_{U_{n}^{*}}(\varphi_{2}(a)\otimes I_{E_{2}\otimes
E^{\otimes\alpha}\otimes\mathcal{D}})(\xi^{\prime}\otimes
d)=\tau_{U_{n}^{*}}(\varphi_{E^{\otimes\alpha}}(a)\eta\otimes\xi\otimes
d)=\rho(a)\tau_{U_{n}^{*}}(\xi^{\prime}\otimes d)$

proving (42). ∎

Now we consider the map $\tau_{U_{1}^{*}}(I_{1}\otimes\tau_{U_{n}^{*}})$ which
is a product of the maps $\tau_{U_{n}^{*}}:\mathcal{F}(E)\otimes
E_{n}\otimes\mathcal{D}\rightarrow\mathcal{F}(E)\otimes\mathcal{D}$ and
$I_{1}\otimes\tau_{U_{n}^{*}}:\mathcal{F}(E)\otimes E_{1}\otimes
E_{n}\otimes\mathcal{D}\rightarrow\mathcal{F}(E)\otimes
E_{1}\otimes\mathcal{D}$ respectively. Therefore, their product is well-
defined and it is

$\tau_{U_{1}^{*}}(I_{1}\otimes\tau_{U_{n}^{*}}):\mathcal{F}(E)\otimes
E_{1n}\otimes\mathcal{D}\rightarrow\mathcal{F}(E)\otimes\mathcal{D}$

where $\mathcal{F}(E)\otimes E_{1n}\otimes\mathcal{D}$ can be viewed as a
subspace of $\mathcal{F}(E)\otimes\mathcal{D}$.

For the following theorem we recall the definition of $\tilde{L}_{1}$ from
(30) with the replacement of the coefficient space $H$ by $\mathcal{D}$ and so
the map
$\tilde{L}_{1}:E_{1n}\otimes\mathcal{F}(E)\otimes\mathcal{D}\to\mathcal{F}(E)\otimes\mathcal{D}$
is defined by

(43) $\tilde{L}_{1}(\xi_{1n}\otimes k):=L_{1}(\xi_{1n})k=\xi_{1n}\otimes
k\quad\quad\quad(\xi_{1}\in E_{1n},k\in\mathcal{F}(E)\otimes\mathcal{D}).$

It is clear from the definition itself that $\tilde{L}_{1}$ is the inclusion
of $E_{1n}\otimes\mathcal{F}(E)\otimes\mathcal{D}$ into
$\mathcal{F}(E)\otimes\mathcal{D}$. Now, we are ready to state the
factorization results for the left-creation operator $\tilde{L}_{1}$.

######  Theorem 3.18.

$\tau_{U_{1}^{*}}(I_{1}\otimes\tau_{U_{n}^{*}})=\tilde{L}_{1}.$

###### Proof.

To calculate the map $\tau_{U_{1}^{*}}(I_{1}\otimes\tau_{U_{n}^{*}})$, we
recall that

$P_{1}:\mathcal{D}_{\hat{T}_{n}}\oplus(E_{1}\otimes\mathcal{D}_{\hat{T}_{1}})\oplus\mathcal{E}_{1}\rightarrow
E_{1n}\otimes\mathcal{D}_{\hat{T}_{n}}\oplus(E_{1}\otimes\mathcal{D}_{\hat{T}_{1}})\oplus
E_{1}\otimes\mathcal{E}_{2}$

is the projection onto the space $E_{1}\otimes\mathcal{D}_{\hat{T}_{1}}$ and

$P_{2}:(E_{n}\otimes\mathcal{D}_{\hat{T}_{n}})\oplus\mathcal{D}_{\hat{T}_{1}}\oplus\mathcal{E}_{2}\rightarrow(E_{n}\otimes\mathcal{D}_{\hat{T}_{n}})\oplus
E_{1n}\otimes\mathcal{D}_{\hat{T}_{1}}\oplus\mathcal{E}_{2}$

is the projection onto the space
$(E_{n}\otimes\mathcal{D}_{\hat{T}_{n}})\oplus\mathcal{E}_{2}$. Now, observe
that

$\displaystyle I_{1}\otimes
i_{1}^{{}^{\prime}*}=I_{1n}\otimes\bar{i}_{1}^{{}^{\prime}*},$

where
$\bar{i}_{1}^{{}^{\prime}}:(E_{1}\otimes\mathcal{D}_{\tilde{T_{n}}})\to(E_{1}\otimes\mathcal{D}_{\tilde{T_{n}}})\oplus\mathcal{D}_{\tilde{T}_{1}}\oplus\mathcal{E}_{1}$
is an inclusion. Using this we have

$\displaystyle\Big{[}I_{\mathcal{F}(E)}\otimes P_{1}^{*}(I_{1}\otimes
U^{*})\Big{]}\Big{[}I_{\mathcal{F}(E)}\otimes I_{1}\otimes\big{(}I_{1n}\otimes
Ui_{1}\big{)}i_{1}^{{}^{\prime}*}\Big{]}$
$\displaystyle=\Big{[}\bigoplus_{\alpha\in\mathbb{Z}_{+}^{n-1}}I_{E^{\alpha}}\otimes
P_{1}^{*}(I_{1}\otimes
U^{*})\Big{]}\Big{[}\bigoplus_{\alpha\in\mathbb{Z}_{+}^{n-1}}I_{E^{\alpha}}\otimes\big{(}I_{1n}\otimes
I_{1}\otimes Ui_{1}\big{)}(I_{1n}\otimes\bar{i}_{1}^{{}^{\prime}*})\Big{]}$
$\displaystyle=\Big{[}\bigoplus_{\alpha\in\mathbb{Z}_{+}^{n-1}}I_{E^{\alpha}}\otimes
P_{1}^{*}(I_{1}\otimes
U^{*})\Big{]}\Big{[}\bigoplus_{\alpha\in\mathbb{Z}_{+}^{n-1}}I_{E^{(\alpha+e_{1})}}\otimes\big{(}I_{1}\otimes
Ui_{1}\big{)}\bar{i}_{1}^{{}^{\prime}*}\Big{]}$
$\displaystyle=\bigoplus_{\alpha\in\mathbb{Z}_{+}^{n-1}}I_{E^{(\alpha+e_{1})}}\otimes
P_{1}^{*}(I_{1}\otimes U^{*})\big{(}I_{1}\otimes
Ui_{1}\big{)}\bar{i}_{1}^{{}^{\prime}*}$
$\displaystyle=\bigoplus_{\alpha\in\mathbb{Z}_{+}^{n-1}}I_{E^{\alpha}}\otimes
I_{1n}\otimes P_{1}^{*}\big{(}I_{1}\otimes
i_{1}\big{)}\bar{i}_{1}^{{}^{\prime}*}$
$\displaystyle=I_{\mathcal{F}(E)}\otimes\Big{(}I_{1n}\otimes
P_{1}^{*}\big{(}I_{1}\otimes
i_{1}\big{)}\Big{)}\Big{(}I_{1n}\otimes\bar{i}_{1}^{{}^{\prime}*}\Big{)}$
$\displaystyle=I_{\mathcal{F}(E)}\otimes\Big{(}I_{1n}\otimes
P_{1}^{*}\big{(}I_{1}\otimes i_{1}\big{)}\Big{)}\Big{(}I_{1}\otimes
i_{1}^{{}^{\prime}*}\Big{)}.$

Similarly, since $U$ is an unitary, we also have

$\displaystyle\Big{[}I_{\mathcal{F}(E)}\otimes(I_{1n}\otimes
i_{2})j_{2}^{{}^{\prime}*}(I_{1}\otimes
U^{*})\Big{]}\Big{[}I_{\mathcal{F}(E)}\otimes I_{1}\otimes\big{(}I_{1n}\otimes
Ui_{1}\big{)}i_{1}^{{}^{\prime}*}\Big{]}$
$\displaystyle=\Big{[}\bigoplus_{\alpha\in\mathbb{Z}_{+}^{n-1}}I_{E^{\alpha}}\otimes(I_{1n}\otimes
i_{2})j_{2}^{{}^{\prime}*}(I_{1}\otimes
U^{*})\Big{]}\Big{[}\bigoplus_{\alpha\in\mathbb{Z}_{+}^{n-1}}I_{E^{\alpha}}\otimes
I_{1n}\otimes\big{(}I_{1}\otimes
Ui_{1}\big{)}\bar{i}_{1}^{{}^{\prime}*}\Big{]}$
$\displaystyle=\Big{[}\bigoplus_{\alpha\in\mathbb{Z}_{+}^{n-1}}I_{E^{\alpha}}\otimes(I_{1n}\otimes
i_{2})j_{2}^{{}^{\prime}*}(I_{1}\otimes
U^{*})\Big{]}\Big{[}\bigoplus_{\alpha\in\mathbb{Z}_{+}^{n-1}}I_{E^{(\alpha+e_{1})}}\otimes\big{(}I_{1}\otimes
Ui_{1}\big{)}\bar{i}_{1}^{{}^{\prime}*}\Big{]}$
$\displaystyle=\bigoplus_{\alpha\in\mathbb{Z}_{+}^{n-1}}I_{E^{(\alpha+e_{1})}}\otimes(I_{1n}\otimes
i_{2})j_{2}^{{}^{\prime}*}(I_{1}\otimes U^{*})\big{(}I_{1}\otimes
Ui_{1}\big{)}\bar{i}_{1}^{{}^{\prime}*}$
$\displaystyle=\bigoplus_{\alpha\in\mathbb{Z}_{+}^{n-1}}I_{E^{(\alpha+e_{1})}}\otimes(I_{1n}\otimes
i_{2})j_{2}^{{}^{\prime}*}(I_{1}\otimes i_{1})\bar{i}_{1}^{{}^{\prime}*}$
$\displaystyle=I_{\mathcal{F}(E)}\otimes\Big{(}I_{1n}\otimes(I_{1n}\otimes
i_{2})j_{2}^{{}^{\prime}*}(I_{1}\otimes
i_{1}\big{)}\Big{)}\Big{(}I_{1n}\otimes\bar{i}_{1}^{{}^{\prime}*}\Big{)}$
$\displaystyle=I_{\mathcal{F}(E)}\otimes\Big{(}I_{1n}\otimes(I_{1n}\otimes
i_{2})j_{2}^{{}^{\prime}*}(I_{1}\otimes i_{1}\big{)}\Big{)}\Big{(}I_{1}\otimes
i_{1}^{{}^{\prime}*}\Big{)}.$

By using the above identities we compute the product of transfer functions
corresponding to the unitaries $U_{1}$ and $U_{n}$.

$\displaystyle\tau_{U_{1}^{*}}(I_{1}\otimes\tau_{U_{n}^{*}})$
$\displaystyle=\Big{[}I_{\mathcal{F}(E)}\otimes P_{1}^{*}(I_{1}\otimes
U^{*})\Big{]}\Big{[}I_{\mathcal{F}(E)}\otimes I_{1}\otimes
UP_{2}^{*}\big{(}I_{E_{n}\otimes\mathcal{D}_{1n}}\oplus
u_{2}^{*}\big{)}\Big{]}$ $\displaystyle\hskip
56.9055pt+\Big{[}I_{\mathcal{F}(E)}\otimes P_{1}^{*}(I_{1}\otimes
U^{*})\Big{]}\Big{[}I_{\mathcal{F}(E)}\otimes I_{1}\otimes\big{(}I_{1n}\otimes
Ui_{1}\big{)}i_{1}^{{}^{\prime}*}\Big{]}$ $\displaystyle\hskip
56.9055pt+I_{\mathcal{F}(E)}\otimes\Big{[}(I_{1n}\otimes
i_{2})j_{2}^{{}^{\prime}*}(I_{1}\otimes U^{*})\Big{(}I_{1}\otimes
UP_{2}^{*}\big{(}I_{E_{n}\otimes\mathcal{D}_{1n}}\oplus
u_{2}^{*}\big{)}\Big{)}\Big{]}$ $\displaystyle\hskip
56.9055pt+\Big{[}I_{\mathcal{F}(E)}\otimes(I_{1n}\otimes
i_{2})j_{2}^{{}^{\prime}*}(I_{1}\otimes
U^{*})\Big{]}\Big{[}I_{\mathcal{F}(E_{1n})}\otimes
I_{1}\otimes\big{(}I_{1n}\otimes Ui_{1}\big{)}i_{1}^{{}^{\prime}*}\Big{]}$
$\displaystyle=I_{\mathcal{F}(E)}\otimes P_{1}^{*}\Big{(}I_{1}\otimes
P_{2}^{*}\big{(}I_{E_{n}\otimes\mathcal{D}_{1n}}\oplus
u_{2}^{*}\big{)}\Big{)}+I_{\mathcal{F}(E)}\otimes\Big{(}I_{1n}\otimes
P_{1}^{*}\big{(}I_{1}\otimes i_{1}\big{)}\Big{)}\Big{(}I_{1}\otimes
i_{1}^{{}^{\prime}*}\Big{)}$ $\displaystyle\hskip
56.9055pt+I_{\mathcal{F}(E)}\otimes(I_{1n}\otimes
i_{2})j_{2}^{{}^{\prime}*}\Big{(}I_{1}\otimes
P_{2}^{*}\big{(}I_{E_{n}\otimes\mathcal{D}_{1n}}\oplus
u_{2}^{*}\big{)}\Big{)}$ $\displaystyle\hskip
56.9055pt+I_{\mathcal{F}(E)}\otimes\Big{(}I_{1n}\otimes(I_{1n}\otimes
i_{2})j_{2}^{{}^{\prime}*}(I_{1}\otimes i_{1}\big{)}\Big{)}\Big{(}I_{1}\otimes
i_{1}^{{}^{\prime}*}\Big{)}$
$\displaystyle=I_{\mathcal{F}(E)}\otimes\Big{[}P_{1}^{*}\Big{(}I_{1}\otimes
P_{2}^{*}\big{(}I_{E_{n}\otimes\mathcal{D}_{1n}}\oplus
u_{2}^{*}\big{)}\Big{)}+\big{(}I_{1n}\otimes P_{1}^{*}(I_{1}\otimes
i_{1})\big{)}(I_{1}\otimes i_{1}^{{}^{\prime}*})$ $\displaystyle\hskip
56.9055pt+(I_{1n}\otimes i_{2})j_{2}^{{}^{\prime}*}\Big{(}I_{1}\otimes
P_{2}^{*}\big{(}I_{E_{n}\otimes\mathcal{D}_{1n}}\oplus
u_{2}^{*}\big{)}\Big{)}+\big{(}I_{1n}\otimes(I_{1n}\otimes
i_{2})j_{2}^{{}^{\prime}*}(I_{1}\otimes i_{1})\big{)}(I_{1}\otimes
i_{1}^{{}^{\prime}*})\Big{]}.$

For the first term, note that,

$\text{range}\Big{(}I_{1}\otimes\big{(}I_{E_{n}\otimes\mathcal{D}_{1n}}\oplus
u_{2}^{*}\big{)}\Big{)}=(E_{1n}\otimes\mathcal{D}_{\tilde{T}_{1}})\oplus(E_{1}\otimes
E_{1n}\otimes\mathcal{D}_{\tilde{T_{n}}})\oplus(E_{1}\otimes\mathcal{E}_{2})$

and

$\displaystyle(I_{1}\otimes
P_{2}^{*})\big{(}(E_{1n}\otimes\mathcal{D}_{\tilde{T}_{1}})\oplus(E_{1}\otimes
E_{1n}\otimes\mathcal{D}_{\tilde{T_{n}}})\oplus(E_{1}\otimes\mathcal{E}_{2})\big{)}=$
$\displaystyle(E_{1n}\otimes\mathcal{D}_{\tilde{T}_{1}})\oplus(E_{1}\otimes\mathcal{E}_{2})$
$\displaystyle\subseteq$
$\displaystyle(E_{1}\otimes\mathcal{D}_{\hat{T}_{1}})\oplus(E_{1n}\otimes\mathcal{D}_{\hat{T}_{n}})\oplus(E_{1}\otimes\mathcal{E}_{2})$

and $P_{1}$ is a projection on $E_{1}\otimes\mathcal{D}_{\hat{T}_{1}}$. So
$(E_{1n}\otimes\mathcal{D}_{\tilde{T}_{1}})\oplus(E_{1}\otimes\mathcal{E}_{2})\subseteq\text{ker}P_{1}^{*}$
and hence

$P_{1}^{*}\Big{(}I_{1}\otimes
P_{2}^{*}\big{(}I_{E_{n}\otimes\mathcal{D}_{1n}}\oplus
u_{2}^{*}\big{)}\Big{)}=0.$

For the second term, we already have observed that

$\big{(}I_{1n}\otimes P_{1}^{*}(I_{1}\otimes i_{1})\big{)}(I_{1}\otimes
i_{1}^{{}^{\prime}*})=I_{1n}\otimes\Big{(}P_{1}^{*}(I_{1}\otimes
i_{1})\bar{i}_{1}^{{}^{\prime}*}\Big{)}.$

Now,

$(E_{1}\otimes\mathcal{D}_{\tilde{T_{n}}})\oplus\mathcal{D}_{\tilde{T_{n}}}\oplus\mathcal{E}_{1}\xrightarrow{\bar{i}_{1}^{{}^{\prime}*}}\
(E_{1}\otimes\mathcal{D}_{\tilde{T_{n}}})\xrightarrow{I_{1}\otimes
i_{1}}\big{(}E_{1}\otimes\mathcal{D}_{\tilde{T_{n}}}\big{)}\oplus\big{(}E_{1n}\otimes\mathcal{D}_{\tilde{T}_{1}}\big{)}\oplus\big{(}E_{1}\otimes\mathcal{E}_{2}\big{)}.$

This tells us that $(I_{1}\otimes i_{1})\bar{i}_{1}^{{}^{\prime}*}$ is an
inclusion map with the range $E_{1}\otimes\mathcal{D}_{\tilde{T_{n}}}$. Since
$P_{1}$ is a projection with range $E_{1}\otimes\mathcal{D}_{\hat{T}_{1}}$, so
$\Big{(}P_{1}^{*}(I_{1}\otimes i_{1})\bar{i}_{1}^{{}^{\prime}*}\Big{)}$ is a
projection with range $E_{1}\otimes\mathcal{D}_{\tilde{T_{n}}}$ and hence

$I_{1n}\otimes\Big{(}P_{1}^{*}(I_{1}\otimes
i_{1})\bar{i}_{1}^{{}^{\prime}*}\Big{)}=P_{E_{1n}\otimes
E_{1}\otimes\mathcal{D}_{\tilde{T_{n}}}}.$

Now we will calculate the third term. Observe that

$(I_{1n}\otimes i_{2})j_{2}^{{}^{\prime}*}\Big{(}I_{1}\otimes
P_{2}^{*}\big{(}I_{E_{n}\otimes\mathcal{D}_{1n}}\oplus
u_{2}^{*}\big{)}\Big{)}=(I_{1n}\otimes
i_{2})\Big{(}I_{E_{1n}\otimes\mathcal{D}_{\hat{T}_{n}}}\oplus(I_{1}\otimes
u_{2})\Big{)}i_{2}^{{}^{\prime}*}(I_{1}\otimes
P_{2}^{*})\Big{(}I_{E_{1n}\otimes\mathcal{D}_{1n}}\oplus(I_{1}\otimes
u_{2}^{*})\Big{)}.$

The range of $I_{E_{1n}\otimes\mathcal{D}_{1n}}\oplus(I_{1}\otimes u_{2}^{*})$
is $(E_{1n}\otimes\mathcal{D}_{\hat{T}_{n}})\oplus(E_{1}\otimes
E_{1n}\otimes\mathcal{D}_{\hat{T}_{1}})\oplus(E_{1}\otimes\mathcal{E}_{2})$.
Also,

$(E_{1n}\otimes\mathcal{D}_{\hat{T}_{n}})\oplus(E_{1}\otimes
E_{1n}\otimes\mathcal{D}_{\hat{T}_{1}})\oplus(E_{1}\otimes\mathcal{E}_{2})\xrightarrow{I_{1}\otimes
P_{2}^{*}}(E_{1}\otimes\mathcal{D}_{\hat{T}_{1}})\oplus(E_{1n}\otimes\mathcal{D}_{\hat{T}_{n}})\oplus(E_{1}\otimes\mathcal{E}_{2})\xrightarrow{i_{2}^{{}^{\prime}*}}(E_{1n}\otimes\mathcal{D}_{\hat{T}_{n}})\oplus(E_{1}\otimes\mathcal{E}_{2}).$

From the above inclusion, it follows that

$i_{2}^{{}^{\prime}*}(I_{1}\otimes
P_{2}^{*})=P_{(E_{1n}\otimes\mathcal{D}_{\hat{T}_{n}})\oplus(E_{1}\otimes\mathcal{E}_{2})}.$

Hence, keeping in the mind that
$i_{2}:\mathcal{D}_{\hat{T}_{n}}\oplus\mathcal{E}_{1}\to\mathcal{D}$ is an
inclusion map we conclude

$\displaystyle(I_{1n}\otimes i_{2})$ $\displaystyle
j_{2}^{{}^{\prime}*}\Big{(}I_{1}\otimes
P_{2}^{*}\big{(}I_{E_{n}\otimes\mathcal{D}_{1n}}\oplus
u_{2}^{*}\big{)}\Big{)}$ $\displaystyle=$ $\displaystyle(I_{1n}\otimes
i_{2})\Big{(}I_{E_{1n}\otimes\mathcal{D}_{\hat{T}_{n}}}\oplus(I_{1}\otimes
u_{2})\Big{)}P_{(E_{1n}\otimes\mathcal{D}_{\hat{T}_{n}})\oplus(E_{1}\otimes\mathcal{E}_{2})}\Big{(}I_{E_{1n}\otimes\mathcal{D}_{1n}}\oplus(I_{1}\otimes
u_{2}^{*})\Big{)}$ $\displaystyle=$ $\displaystyle
P_{(E_{1n}\otimes\mathcal{D}_{\hat{T}_{n}})\oplus(E_{1n}\otimes\mathcal{E}_{1})}.$

For the last term, observe that $\text{ran}(I_{1}\otimes
i_{1})=E_{1}\otimes\mathcal{D}_{\hat{T}_{1}}$ which lies in the kernel of
$j_{2}^{{}^{\prime}*}$. So, $j_{2}^{{}^{\prime}*}(I_{1}\otimes i_{1})=0$ and
hence

$\big{(}I_{1n}\otimes(I_{1n}\otimes i_{2})j_{2}^{{}^{\prime}*}(I_{1}\otimes
i_{1})\big{)}(I_{1}\otimes i_{1}^{{}^{\prime}*}).$

Therefore,

$\displaystyle\tau_{U_{1}^{*}}(I_{1}\otimes\tau_{U_{n}^{*}})=$ $\displaystyle
I_{\mathcal{F}(E)}\otimes\Big{[}0+P_{E_{1n}\otimes
E_{1}\otimes\mathcal{D}_{\tilde{T_{n}}}}+P_{(E_{1n}\otimes\mathcal{D}_{\hat{T}_{n}})\oplus(E_{1n}\otimes\mathcal{E}_{1})}+0\Big{]}$
$\displaystyle=$ $\displaystyle I_{\mathcal{F}(E)}\otimes
I_{E_{1n}\otimes\big{(}(E_{1}\otimes\mathcal{D}_{\tilde{T_{n}}})\oplus\mathcal{D}_{\tilde{T}_{1}}\oplus\mathcal{E}_{1}\big{)}}$
$\displaystyle=$ $\displaystyle\tilde{L}_{1}.$

∎

Below, we state another kind of factorization result for the left creation
operator $\tilde{L}_{1}$.

######  Theorem 3.19.

$\tau_{U_{n}^{*}}(I_{n}\otimes\tau_{U_{1}^{*}})=\tilde{L}_{1}$

###### Proof.

Observe that,

$\displaystyle I_{n}\otimes j_{2}^{{}^{\prime}*}$
$\displaystyle=I_{n}\otimes\Big{[}\Big{(}I_{E_{1n}\otimes\mathcal{D}_{\tilde{T}_{1}}}\oplus(I_{1}\otimes
u_{2})\Big{)}\circ i_{2}^{{}^{\prime}*}\Big{]}$
$\displaystyle=\Big{(}I_{n}\otimes\Big{[}I_{E_{1n}\otimes\mathcal{D}_{\tilde{T}_{1}}}\oplus(I_{1}\otimes
u_{2})\Big{]}\Big{)}\Big{(}I_{n}\otimes i_{2}^{{}^{\prime}*}\Big{)}$
$\displaystyle=\Big{[}(I_{n}\otimes I_{1n}\otimes
I_{\mathcal{D}_{\tilde{T}_{1}}})\oplus(I_{1n}\otimes
u_{2})\Big{]}\Big{(}I_{1n}\otimes\bar{i}_{1}^{{}^{\prime}*}\Big{)}$
$\displaystyle=\Big{[}I_{1n}\otimes\Big{(}(I_{n}\otimes
I_{\mathcal{D}_{\tilde{T}_{1}}})\oplus
u_{2}\Big{)}\Big{]}\Big{(}I_{1n}\otimes\bar{i}_{2}^{{}^{\prime}*}\Big{)}$
$\displaystyle=I_{1n}\otimes\Big{(}I_{E_{n}\otimes\mathcal{D}_{\tilde{T}_{1}}}\oplus
u_{2}\Big{)}\bar{i}_{2}^{{}^{\prime}*},$

where
$\bar{i}_{2}^{{}^{\prime}}:(E_{n}\otimes\mathcal{D}_{\tilde{T}_{1}})\oplus\mathcal{E}_{2}\to(E_{n}\otimes\mathcal{D}_{\tilde{T}_{1}})\oplus\mathcal{D}_{\tilde{T_{n}}}\oplus\mathcal{E}_{2}$
is an inclusion. And so

$\displaystyle I_{n}\otimes\big{(}I_{1n}\otimes
i_{2}\big{)}j_{2}^{{}^{\prime}*}\big{(}I_{1}\otimes U^{*})$
$\displaystyle=\big{(}I_{1n}\otimes I_{n}\otimes
i_{2}\big{)}\big{(}I_{n}\otimes
j_{2}^{{}^{\prime}*}\big{)}\big{(}I_{1n}\otimes U^{*})$
$\displaystyle=\big{(}I_{1n}\otimes I_{n}\otimes
i_{2}\big{)}\big{(}I_{1n}\otimes\Big{(}I_{E_{n}\otimes\mathcal{D}_{\tilde{T}_{1}}}\oplus
u_{2}\Big{)}\bar{i}_{2}^{{}^{\prime}*}\big{)}\big{(}I_{1n}\otimes
U^{*}\big{)}$ $\displaystyle=I_{1n}\otimes\big{(}I_{n}\otimes
i_{2}\big{)}\big{(}I_{E_{n}\otimes\mathcal{D}_{\tilde{T}_{1}}}\oplus
u_{2}\big{)}\bar{i}_{2}^{{}^{\prime}*}U^{*}$

Using these above identities we have

$\displaystyle\Big{[}I_{\mathcal{F}(E)}\otimes
UP_{2}^{*}\big{(}I_{E_{n}\otimes\mathcal{D}_{1n}}\oplus
u_{2}^{*}\big{)}\Big{]}\Big{[}I_{\mathcal{F}(E)}\otimes
I_{n}\otimes\big{(}I_{1n}\otimes i_{2}\big{)}j_{2}^{{}^{\prime}*}(I_{1}\otimes
U^{*})\Big{]}$
$\displaystyle=\Big{[}\bigoplus_{\alpha\in\mathbb{Z}_{+}^{n-1}}I_{E^{\alpha}}\otimes
UP_{2}^{*}\big{(}I_{E_{n}\otimes\mathcal{D}_{1n}}\oplus
u_{2}^{*}\big{)}\Big{]}\Big{[}\bigoplus_{\alpha\in\mathbb{Z}_{+}^{n-1}}I_{E^{(\alpha+e_{1})}}\otimes\big{(}I_{n}\otimes
i_{2}\big{)}\big{(}I_{E_{n}\otimes\mathcal{D}_{\tilde{T}_{1}}}\oplus
u_{2}\big{)}\bar{i}_{2}^{{}^{\prime}*}U^{*}\Big{]}$
$\displaystyle=\bigoplus_{\alpha\in\mathbb{Z}_{+}^{n-1}}I_{E^{(\alpha+e_{1})}}\otimes
UP_{2}^{*}\big{(}I_{E_{n}\otimes\mathcal{D}_{1n}}\oplus
u_{2}^{*}\big{)}\big{(}I_{n}\otimes
i_{2}\big{)}\big{(}I_{E_{n}\otimes\mathcal{D}_{\tilde{T}_{1}}}\oplus
u_{2}\big{)}\bar{i}_{2}^{{}^{\prime}*}U^{*}$
$\displaystyle=\bigoplus_{\alpha\in\mathbb{Z}_{+}^{n-1}}I_{E^{\alpha}}\otimes
I_{1n}\otimes UP_{2}^{*}\big{(}I_{E_{n}\otimes\mathcal{D}_{1n}}\oplus
u_{2}^{*}\big{)}\big{(}I_{n}\otimes
i_{2}\big{)}\big{(}I_{E_{n}\otimes\mathcal{D}_{\tilde{T}_{1}}}\oplus
u_{2}\big{)}\bar{i}_{2}^{{}^{\prime}*}U^{*}$
$\displaystyle=I_{\mathcal{F}(E)}\otimes\Big{(}I_{1n}\otimes
UP_{2}^{*}\big{(}I_{E_{n}\otimes\mathcal{D}_{1n}}\oplus
u_{2}^{*}\big{)}\Big{)}\Big{(}I_{1n}\otimes\big{(}I_{n}\otimes
i_{2}\big{)}\big{(}I_{E_{n}\otimes\mathcal{D}_{\tilde{T}_{1}}}\oplus
u_{2}\big{)}\bar{i}_{2}^{{}^{\prime}*}U^{*}\Big{)}$
$\displaystyle=I_{\mathcal{F}(E)}\otimes\Big{(}I_{1n}\otimes
UP_{2}^{*}\big{(}I_{E_{n}\otimes\mathcal{D}_{1n}}\oplus
u_{2}^{*}\big{)}\Big{)}\Big{(}I_{n}\otimes\big{(}I_{1n}\otimes
i_{2}\big{)}j_{2}^{{}^{\prime}*}\big{(}I_{1}\otimes U^{*})\Big{)}$
$\displaystyle=I_{\mathcal{F}(E)}\otimes\Big{(}I_{1n}\otimes
U\Big{)}\Big{(}I_{1n}\otimes
P_{2}^{*}\big{(}I_{E_{n}\otimes\mathcal{D}_{1n}}\oplus
u_{2}^{*}\big{)}\Big{)}\Big{(}I_{n}\otimes\big{(}I_{1n}\otimes
i_{2}\big{)}j_{2}^{{}^{\prime}*}\Big{)}\Big{(}I_{1n}\otimes U^{*}\Big{)}.$

Similarly, we also have

$\displaystyle\Big{[}I_{\mathcal{F}(E)}\otimes(I_{1n}\otimes
Ui_{1})i_{1}^{{}^{\prime}*}\Big{]}\Big{[}I_{\mathcal{F}(E)}\otimes
I_{n}\otimes\big{(}I_{1n}\otimes i_{2}\big{)}j_{2}^{{}^{\prime}*}(I_{1}\otimes
U^{*})\Big{]}$ $\displaystyle=\big{(}I_{\mathcal{F}(E)}\otimes I_{1n}\otimes
U\big{)}\Big{[}I_{\mathcal{F}(E)}\otimes(I_{1n}\otimes
i_{1})i_{1}^{{}^{\prime}*}\Big{]}\Big{[}I_{\mathcal{F}(E)}\otimes
I_{n}\otimes\big{(}I_{1n}\otimes
i_{2}\big{)}j_{2}^{{}^{\prime}*}\Big{]}\big{(}I_{\mathcal{F}(E)}\otimes
I_{1n}\otimes U^{*}\big{)}$ $\displaystyle=\big{(}I_{\mathcal{F}(E)}\otimes
I_{1n}\otimes
U\big{)}\Big{[}\bigoplus_{\alpha\in\mathbb{Z}_{+}^{n-1}}I_{E^{\alpha}}\otimes(I_{1n}\otimes
i_{1})i_{1}^{{}^{\prime}*}\Big{]}$ $\displaystyle\hskip
142.26378pt\Big{[}\bigoplus_{\alpha\in\mathbb{Z}_{+}^{n-1}}I_{E^{(\alpha+e_{1})}}\otimes\big{(}I_{n}\otimes
i_{2}\big{)}\big{(}I_{E_{n}\otimes\mathcal{D}_{\tilde{T}_{1}}}\oplus
u_{2}\big{)}\bar{i}_{2}^{{}^{\prime}*}\Big{]}\big{(}I_{\mathcal{F}(E)}\otimes
I_{1n}\otimes U^{*}\big{)}$ $\displaystyle=\big{(}I_{\mathcal{F}(E)}\otimes
I_{1n}\otimes
U\big{)}\Big{[}\bigoplus_{\alpha\in\mathbb{Z}_{+}^{n-1}}I_{E^{(\alpha+e_{1})}}\otimes(I_{1n}\otimes
i_{1})i_{1}^{{}^{\prime}*}(I_{n}\otimes
i_{2})\big{(}I_{E_{n}\otimes\mathcal{D}_{\tilde{T}_{1}}}\oplus
u_{2}\big{)}\bar{i}_{2}^{{}^{\prime}*}\Big{]}\big{(}I_{\mathcal{F}(E)}\otimes
I_{1n}\otimes U^{*}\big{)}$ $\displaystyle=\big{(}I_{\mathcal{F}(E)}\otimes
I_{1n}\otimes
U\big{)}\Big{[}\bigoplus_{\alpha\in\mathbb{Z}_{+}^{n-1}}I_{E^{(\alpha}}\otimes
I_{1n}\otimes(I_{1n}\otimes i_{1})i_{1}^{{}^{\prime}*}(I_{n}\otimes
i_{2})\big{(}I_{E_{n}\otimes\mathcal{D}_{\tilde{T}_{1}}}\oplus
u_{2}\big{)}\bar{i}_{2}^{{}^{\prime}*}\Big{]}\big{(}I_{\mathcal{F}(E)}\otimes
I_{1n}\otimes U^{*}\big{)}$
$\displaystyle=I_{\mathcal{F}(E)}\otimes\Big{(}I_{1n}\otimes I_{1n}\otimes
U\Big{)}\Big{(}I_{1n}\otimes(I_{1n}\otimes
i_{1})i_{1}^{{}^{\prime}*}\Big{)}\Big{(}I_{n}\otimes\big{(}I_{1n}\otimes
i_{2}\big{)}j_{2}^{{}^{\prime}*}\Big{)}\big{(}I_{1n}\otimes U^{*}\big{)}.$

Combining all the above identities we compute the following

$\displaystyle\tau_{U_{n}^{*}}(I_{n}\otimes\tau_{U_{1}^{*}})$
$\displaystyle=\Big{(}I_{\mathcal{F}(E)}\otimes
UP_{2}^{*}\big{(}I_{E_{n}\otimes\mathcal{D}_{1n}}\oplus
u_{2}^{*}\big{)}\Big{)}\Big{(}I_{\mathcal{F}(E)}\otimes I_{n}\otimes
P_{1}^{*}(I_{1}\otimes U^{*})\Big{)}$ $\displaystyle\hskip
56.9055pt+\Big{(}I_{\mathcal{F}(E)}\otimes
UP_{2}^{*}\big{(}I_{E_{n}\otimes\mathcal{D}_{1n}}\oplus
u_{2}^{*}\big{)}\Big{)}\Big{(}I_{\mathcal{F}(E)}\otimes
I_{n}\otimes(I_{1n}\otimes i_{2})j_{2}^{{}^{\prime}*}(I_{1}\otimes
U^{*})\Big{)}$ $\displaystyle\hskip
56.9055pt+\Big{(}I_{\mathcal{F}(E)}\otimes[I_{1n}\otimes
Ui_{1}]i_{1}^{{}^{\prime}*}\Big{)}\Big{(}I_{\mathcal{F}(E)}\otimes
I_{n}\otimes P_{1}^{*}(I_{1}\otimes U^{*})\Big{)}$ $\displaystyle\hskip
56.9055pt+\Big{(}I_{\mathcal{F}(E)}\otimes[I_{1n}\otimes
Ui_{1}]i_{1}^{{}^{\prime}*}\Big{)}\Big{(}I_{\mathcal{F}(E)}\otimes
I_{n}\otimes(I_{1n}\otimes i_{2})j_{2}^{{}^{\prime}*}(I_{1}\otimes
U^{*})\Big{)}$ $\displaystyle=I_{\mathcal{F}(E)}\otimes
UP_{2}^{*}\big{(}I_{E_{n}\otimes\mathcal{D}_{1n}}\oplus
u_{2}^{*}\big{)}\big{(}I_{n}\otimes P_{1}^{*}\big{)}(I_{1n}\otimes U^{*})$
$\displaystyle\hskip 56.9055pt+I_{\mathcal{F}(E)}\otimes\Big{(}I_{1n}\otimes
U\Big{)}\Big{(}I_{1n}\otimes
P_{2}^{*}\big{(}I_{E_{n}\otimes\mathcal{D}_{1n}}\oplus
u_{2}^{*}\big{)}\Big{)}\Big{(}I_{n}\otimes\big{(}I_{1n}\otimes
i_{2}\big{)}j_{2}^{{}^{\prime}*}\Big{)}\Big{(}I_{1n}\otimes U^{*}\Big{)}$
$\displaystyle\hskip 56.9055pt+I_{\mathcal{F}(E)}\otimes\big{(}I_{1n}\otimes
U\big{)}\big{(}I_{1n}\otimes
i_{1}\big{)}i_{1}^{{}^{\prime}*}\big{(}I_{n}\otimes
P_{1}^{*}\big{)}\big{(}I_{1n}\otimes U^{*}\big{)}$ $\displaystyle\hskip
56.9055pt+I_{\mathcal{F}(E)}\otimes\big{(}I_{1n}\otimes I_{1n}\otimes
U\big{)}\Big{(}I_{1n}\otimes(I_{1n}\otimes
i_{1})i_{1}^{{}^{\prime}*}\Big{)}\Big{(}I_{n}\otimes\big{(}I_{1n}\otimes
i_{2}\big{)}j_{2}^{{}^{\prime}*}\Big{)}\big{(}I_{1n}\otimes U^{*}\big{)}.$

For the first term, note that,

$\text{range}(I_{n}\otimes
P_{1}^{*})=E_{1n}\otimes\mathcal{D}_{\tilde{T_{n}}}\subseteq(E_{n}\otimes\mathcal{D}_{\hat{T}_{n}})\oplus(E_{1n}\otimes\mathcal{D}_{\hat{T}_{1}})\oplus(E_{n}\otimes\mathcal{E}_{1})$

and it follows that $\text{range}\big{(}I_{E_{n}\otimes\mathcal{D}_{1n}}\oplus
u_{2}^{*}\big{)}(I_{n}\otimes
P_{1}^{*})=E_{1n}\otimes\mathcal{D}_{\tilde{T_{n}}}$ which lies in the kernel
of $P_{2}^{*}$ and hence
$P_{2}^{*}\big{(}I_{E_{n}\otimes\mathcal{D}_{1n}}\oplus
u_{2}^{*}\big{)}(I_{n}\otimes P_{1}^{*})=0$.

For the second term, we already have observed that

$\big{(}I_{1n}\otimes P_{2}^{*}\big{(}I_{E_{n}\otimes\mathcal{D}_{1n}}\oplus
u_{2}^{*}\big{)}(I_{n}\otimes i_{2})\big{)}(I_{n}\otimes
j_{2}^{{}^{\prime}*})=I_{1n}\otimes\Big{(}P_{2}^{*}\big{(}I_{E_{n}\otimes\mathcal{D}_{1n}}\oplus
u_{2}^{*}\big{)}(I_{n}\otimes
i_{2})(I_{E_{n}\otimes\mathcal{D}_{\hat{T}_{n}}}\oplus
u_{2})\bar{i}_{2}^{{}^{\prime}*}\Big{)}.$

Now,

$\displaystyle\big{(}I_{E_{n}\otimes\mathcal{D}_{1n}}\oplus
u_{2}^{*}\big{)}(I_{n}\otimes
i_{2})(I_{E_{n}\otimes\mathcal{D}_{\hat{T}_{n}}}\oplus u_{2})=$
$\displaystyle\big{(}I_{E_{1n}\otimes\mathcal{D}_{\hat{T}_{1}}}\oplus
I_{E_{n}\otimes\mathcal{D}_{\hat{T}_{n}}}\oplus u_{2}^{*}\big{)}(I_{n}\otimes
i_{2})(I_{E_{n}\otimes\mathcal{D}_{\hat{T}_{n}}}\oplus u_{2})$
$\displaystyle=$
$\displaystyle\big{(}0_{E_{1n}\otimes\mathcal{D}_{\hat{T}_{1}}}\oplus
I_{E_{n}\otimes\mathcal{D}_{\hat{T}_{n}}}\oplus u_{2}^{*}\big{)}(I_{n}\otimes
i_{2})(I_{E_{n}\otimes\mathcal{D}_{\hat{T}_{n}}}\oplus u_{2})$
$\displaystyle=$ $\displaystyle
i_{(E_{n}\otimes\mathcal{D}_{\hat{T}_{n}})\oplus\mathcal{E}_{2}}$

where
$i_{(E_{n}\otimes\mathcal{D}_{\hat{T}_{n}})\oplus\mathcal{E}_{2}}:(E_{n}\otimes\mathcal{D}_{\hat{T}_{n}})\oplus\mathcal{E}_{2}\to\big{(}E_{n}\otimes\mathcal{D}_{\tilde{T}_{1}}\big{)}\oplus\big{(}E_{1n}\otimes\mathcal{D}_{\tilde{T_{n}}}\big{)}\oplus\mathcal{E}_{2}$
is an inclusion and the last line follows since $I_{n}\otimes i_{2}$ is an
inclusion map with range
$E_{n}\otimes\big{(}\mathcal{D}_{\tilde{T}_{1}}\oplus\mathcal{E}_{1}\big{)}$.
So,

$P_{2}^{*}\big{(}I_{E_{n}\otimes\mathcal{D}_{1n}}\oplus
u_{2}^{*}\big{)}(I_{n}\otimes
i_{2})(I_{E_{n}\otimes\mathcal{D}_{\hat{T}_{n}}}\oplus
u_{2})\bar{i}_{2}^{{}^{\prime}*}=P_{2}^{*}(i_{(E_{n}\otimes\mathcal{D}_{\hat{T}_{n}})\oplus\mathcal{E}_{2}})\bar{i}_{2}^{{}^{\prime}*}.$

Observe that,

$(E_{n}\otimes\mathcal{D}_{\tilde{T}_{1}})\oplus\mathcal{D}_{\tilde{T_{n}}}\oplus\mathcal{E}_{2}\xrightarrow{\bar{i}_{2}^{{}^{\prime}*}}(E_{n}\otimes\mathcal{D}_{\tilde{T}_{1}})\oplus\mathcal{E}_{2}\xrightarrow{i_{(E_{n}\otimes\mathcal{D}_{\hat{T}_{n}})\oplus\mathcal{E}_{2}}}\big{(}E_{n}\otimes\mathcal{D}_{\tilde{T}_{1}}\big{)}\oplus\big{(}E_{1n}\otimes\mathcal{D}_{\tilde{T_{n}}}\big{)}\oplus\mathcal{E}_{2}\xrightarrow{P_{2}^{*}}(E_{n}\otimes\mathcal{D}_{\tilde{T}_{1}})\oplus\mathcal{E}_{2}.$

So,

$P_{2}^{*}(i_{(E_{n}\otimes\mathcal{D}_{\hat{T}_{n}})\oplus\mathcal{E}_{2}})\bar{i}_{2}^{{}^{\prime}*}=P_{(E_{n}\otimes\mathcal{D}_{\tilde{T}_{1}})\oplus\mathcal{E}_{2}}.$

and hence

$I_{1n}\otimes\Big{(}P_{2}^{*}(I_{n}\otimes
i_{2})(I_{E_{n}\otimes\mathcal{D}_{\hat{T}_{n}}}\oplus
u_{2})\bar{i}_{2}^{{}^{\prime}*}\Big{)}=P_{E_{1n}\otimes\big{(}(E_{n}\otimes\mathcal{D}_{\tilde{T}_{1}})\oplus\mathcal{E}_{2}\big{)}}.$

Now, for the third term, we consider the following diagram

$\begin{CD}E_{1n}\otimes\mathcal{D}_{\hat{T}_{1}}\xrightarrow{I_{n}\otimes
P_{1}^{*}}E_{n}\otimes\big{(}\mathcal{D}_{\hat{T}_{n}}\oplus(E_{1}\otimes\mathcal{D}_{\hat{T}_{1}})\oplus\mathcal{E}_{1}\big{)}\xrightarrow{i_{1}^{{}^{\prime}*}}\big{(}E_{1n}\otimes\mathcal{D}_{\hat{T}_{1}}\big{)}\xrightarrow{I_{1n}\otimes
i_{1}}\
E_{1n}\otimes\big{(}\mathcal{D}_{\hat{T}_{1}}\oplus(E_{n}\otimes\mathcal{D}_{\hat{T}_{n}})\oplus\mathcal{E}_{2}\big{)}\end{CD}$

with the help of range$(I_{n}\otimes
P_{1}^{*})=E_{1n}\otimes\mathcal{D}_{\hat{T}_{1}}$ it is easy to observe that

$(I_{1n}\otimes i_{1})i_{1}^{{}^{\prime}*}(I_{n}\otimes
P_{1}^{*})=P_{E_{1n}\otimes\mathcal{D}_{\hat{T}_{1}}}.$

For the last term, first note that
$\text{ran}\,j_{2}^{{}^{\prime}*}=E_{1n}\otimes(\mathcal{D}_{\hat{T}_{n}}\oplus\mathcal{E}_{1})$,
so $\text{ran}(I_{1n}\otimes
i_{2})j_{2}^{{}^{\prime}*}=E_{1n}\otimes(\mathcal{D}_{\hat{T}_{n}}\oplus\mathcal{E}_{1})$
and hence $\text{ran}\big{(}I_{n}\otimes(I_{1n}\otimes
i_{2})j_{2}^{{}^{\prime}*}\big{)}=E_{1n}\otimes
E_{n}\otimes(\mathcal{D}_{\hat{T}_{n}}\oplus\mathcal{E}_{1})$. On the other
hand,
$E_{n}\otimes(\mathcal{D}_{\hat{T}_{n}}\oplus\mathcal{E}_{1})\subseteq\text{ker}\,i_{1}^{{}^{\prime}*}$
which implies $E_{1n}\otimes
E_{n}\otimes(\mathcal{D}_{\hat{T}_{n}}\oplus\mathcal{E}_{1})\subseteq\text{ker}(I_{1n}\otimes
i_{1}^{{}^{\prime}*})$. Therefore,

$\displaystyle\big{(}I_{1n}\otimes I_{1n}\otimes U\big{)}$
$\displaystyle\Big{(}I_{1n}\otimes(I_{1n}\otimes
i_{1})i_{1}^{{}^{\prime}*}\Big{)}\Big{(}I_{n}\otimes\big{(}I_{1n}\otimes
i_{2}\big{)}j_{2}^{{}^{\prime}*}\Big{)}\big{(}I_{1n}\otimes U^{*}\big{)}$
$\displaystyle=$ $\displaystyle\big{(}I_{1n}\otimes I_{1n}\otimes
U\big{)}(I_{1n}\otimes I_{1n}\otimes i_{1})(I_{1n}\otimes
i_{1}^{{}^{\prime}*})\Big{(}I_{n}\otimes\big{(}I_{1n}\otimes
i_{2}\big{)}j_{2}^{{}^{\prime}*}\Big{)}\big{(}I_{1n}\otimes U^{*}\big{)}=0$

Therefore,

$\displaystyle\tau_{U_{n}^{*}}(I_{n}\otimes\tau_{U_{1}^{*}})=$ $\displaystyle
I_{\mathcal{F}(E)}\otimes(I_{1n}\otimes
U)\Big{[}0+P_{E_{1n}\otimes\big{(}(E_{n}\otimes\mathcal{D}_{\tilde{T}_{1}})\oplus\mathcal{E}_{2}\big{)}}+P_{E_{1n}\otimes\mathcal{D}_{\tilde{T_{n}}}}+0\Big{]}(I_{1n}\otimes
U^{*})$ $\displaystyle=$ $\displaystyle
I_{\mathcal{F}(E)}\otimes(I_{1n}\otimes
U)\Big{[}I_{E_{1n}\otimes\big{(}(E_{n}\otimes\mathcal{D}_{\tilde{T}_{1}})\oplus\mathcal{D}_{\tilde{T_{n}}}\oplus\mathcal{E}_{2}\big{)}}\Big{]}(I_{1n}\otimes
U^{*})$ $\displaystyle=$ $\displaystyle I_{\mathcal{F}(E)}\otimes
I_{E_{1n}\otimes\big{(}(E_{1}\otimes\mathcal{D}_{\tilde{T_{n}}})\oplus\mathcal{D}_{\tilde{T}_{1}}\oplus\mathcal{E}_{1}\big{)}}$
$\displaystyle=$ $\displaystyle\tilde{L}_{1}$

∎

######  Theorem 3.20.

Let
$(\sigma,T_{1},\ldots,T_{n})\in\mathcal{T}_{1,n}^{n}(\sigma,\,M,\,E_{1},\ldots,E_{n},H)$
and let $E$, $\mathcal{D}$, $\Pi_{1n,V}:H\to\mathcal{F}(E)\otimes\mathcal{D}$
and $L_{i}\,(i=2,\ldots,n)$ as above. Then $\Pi_{1n,V}$ is an isometry and

$\begin{cases}(I_{1}\otimes\Pi_{1n,V})\tilde{T_{1}}^{*}=\tau_{U_{1}^{*}}^{*}\Pi_{1n,V}\\\
(I_{i}\otimes\Pi_{1n,V})\tilde{T_{i}}^{*}=\tilde{L}_{i}^{*}\Pi_{1n,V}\quad\text{for}\,\,\,2\leq
i\leq n-1\\\
(I_{n}\otimes\Pi_{1n,V})\tilde{T_{n}}^{*}=\tau_{U_{n}^{*}}^{*}\Pi_{1n,V}\end{cases}$

where $\tau_{U_{1}^{*}}$ and $\tau_{U_{n}^{*}}$ are isometric multipliers that
have transfer function representations with respect to some unitaries $U_{1}$
and $U_{n}$. Moreover, $\tau_{U_{1}^{*}}$ and $\tau_{U_{n}^{*}}$ satisfy the
following relation:

$\tau_{U_{1}^{*}}(I_{1}\otimes\tau_{U_{n}^{*}})=\tilde{L}=\tau_{U_{n}^{*}}(I_{n}\otimes\tau_{U_{1}^{*}}).$

and there is an isometric representation
$(\rho,M_{1},L_{2},\ldots,L_{n-1},M_{n})$ of the product system that dilates
the representation $(\sigma,T_{1},\ldots,T_{n})$ and such that
$\tau_{U_{i}^{*}}=\tilde{M_{i}}$ for $i=1,n$.

##  4\. Applications

######  Theorem 4.1.

Let $M$ be a von Neumann algebra and $\alpha_{1},\alpha_{2},\ldots,\alpha_{n}$
be $*-$automorphisms of $M$ that pairwise commute. Also, let $\sigma$ be a
normal representation of $M$ on a Hilbert space $H$ and let
$t:=(t_{1},t_{2},\ldots,t_{n})$ be a commuting tuple of operators in B(H) such
that

  1. (1)

$t$ satisfies the conditions from [4] : $\hat{t}_{1}$ and $\hat{t}_{n}$
satisfy Szego positivity and $\hat{t}_{n}$ is pure.

  2. (2)

For every $a\in M$ and $1\leq i\leq n$,

(44) $t_{i}\sigma(\alpha_{i}(a))=\sigma(a)t_{i}$

Then there is a Hilbert space $K$ , an isometry $V:H\rightarrow K$ , a
representation $\rho$ of $M$ on $K$ and commuting isometries
$v_{1},\ldots.v_{n}$ in $B(K)$ that satisfy

(45) $v_{i}\rho(\alpha_{i}(a))=\rho(a)v_{i}$

and dilate $(\sigma,t_{1},\ldots,t_{n})$, i.e,

  1. (i)

$\sigma(a)=V^{*}\rho(a)V$ for all $a\in M$ and

  2. (ii)

$Vt_{i}^{*}=v_{i}^{*}V$ for all $i$

Setting $\alpha_{i}=id_{M}$ above, we get

######  Corollary 4.2.

Let $M$ be a von Neumann algebra and let $\sigma$ be a normal representation
of $M$ on a Hilbert space $H$ and $t:=(t_{1},t_{2},\ldots,t_{n})$ be a
commuting tuple of operators in B(H) such that

  1. (1)

$t$ satisfies the conditions from [4] : $\hat{t}_{1}$ and $\hat{t}_{n}$
satisfy Szego positivity and $\hat{t}_{n}$ is pure.

  2. (2)

For every $1\leq i\leq n$, $t_{i}\in\sigma(M)^{\prime}$.

Then there is a Hilbert space $K$ , an isometry $V:H\rightarrow K$ , a
representation $\rho$ of $M$ on $K$ and commuting isometries
$v_{1},\ldots.v_{n}$ in $\rho(M)^{\prime}$ that dilate
$(\sigma,t_{1},\ldots,t_{n})$, i.e,

  1. (i)

$\sigma(a)=V^{*}\rho(a)V$ for all $a\in M$ and

  2. (ii)

$Vt_{i}^{*}=v_{i}^{*}V$ for all $i$

To prove the theorem, we explain how Theorem 4.1 is related to Theorem 3.20.

For this, we first note that, given an automorphism $\alpha$ of a von Neumann
algebra $M$, one can associate a $W^{*}$-correspondence, written
${}_{\alpha}M$, with it. As a vector space ${}_{\alpha}M$ is $M$, The
$M$-valued inner product is $\langle a,b\rangle=a^{*}b$, the left
multiplication is $m\cdot a=ma$ and the right multiplication is
$\varphi(m)a=\alpha(m)a$ (where $m,a$ are in $M$ but $a$ here is viewed as an
element of the correspondence).

If $\alpha,\beta$ are two automorphisms of $M$ then

${}_{\alpha}M\otimes_{\beta}M\cong_{\beta\alpha}M$

where the isomorphism is given by $a\otimes b\mapsto\beta(a)b$. Thus we will
identify ${}_{\alpha}M\otimes_{\beta}M$ with ${}_{\beta\alpha}M$ (and, when
$\alpha$ and $\beta$ commute, we identify ${}_{\alpha}M\otimes_{\beta}M$ with
${}_{\beta}M\otimes_{\alpha}M$.

Now let $(T,\sigma)$ be a representation of ${}_{\alpha}M$. We have, for
$a,b\in M$, $T(\alpha(a)b)=T(\varphi(a)b)=\sigma(a)T(b)$. Setting $b=I$, we
get $T(\alpha(a))=\sigma(a)T(I)$. But
$T(\alpha(a))=T(I\cdot\alpha(a))=T(I)\sigma(\alpha(a))$ and, writing $t:=T(I)$
, we get $t\sigma(\alpha(a))=\sigma(a)t$ for every $a\in M$. So, a
representation of this correspondence is given by a normal representation
$\sigma$ on $H$ and a contraction $t\in B(H)$ satisfying the intertwining
relation

(46) $t\sigma(\alpha(a))=\sigma(a)t$

for all $a\in M$. (See [10] for details). Note also that
${}_{\alpha}M\otimes_{\sigma}H\cong H$ via the isomorphism $a\otimes
h\mapsto\sigma(a)h$. Identifying ${}_{\alpha}M\otimes_{\sigma}H$ with $H$, we
find that $\tilde{T}$ can be viewed as an operator in $B(H)$. With these
identifications, we write

$\tilde{T}=t.$

In fact, $t=\tilde{T}(I\otimes h)$ and
$t\sigma(a)h=\tilde{T}(I\otimes_{\sigma}\sigma(a)h)=\tilde{T}(a\otimes h)$.

Now suppose that $(\sigma,T)$ is a representation of ${}_{\alpha}M$ and
$(\sigma,S)$ is a representation of ${}_{\beta}M$. Recall that, when we
identify ${}_{\alpha}M\otimes_{\beta}M$ with ${}_{\beta\alpha}M$,
$I_{{}_{\beta\alpha}M}$ is identified with $I_{{}_{\alpha}M}\otimes
I_{{}_{\beta}M}$ and, thus,

(47) $\tilde{S}(I_{{}_{\beta}M}\otimes\tilde{T})(I_{{}_{\beta\alpha}M}\otimes
h)=\tilde{S}(I_{{}_{\beta}M}\otimes\tilde{T})(I_{{}_{\alpha}M}\otimes
I_{{}_{\beta}M}\otimes h)=\tilde{S}(I_{{}_{\beta}M}\otimes th)=sth.$

Thus, $\tilde{S}(I_{{}_{\beta}M}\otimes\tilde{T})$ is associated with $st$. A
straightforward computation using the identifications above shows that
$\tilde{T}^{*}k=t^{*}k$ and

(48) $\tilde{S}\tilde{T}^{*}=st^{*}.$

For every representation $(\sigma,T)$,

$\tilde{T}^{*}\tilde{T}(a\otimes
h)=\tilde{T}^{*}(t\sigma(a)h)=t^{*}t\sigma(a)h$

and we conclude that $(\sigma,T)$ is isometric if and only if $t$ is an
isometry.

We now turn to prove Theorem 4.1.

###### Proof.

We now fix the setting of Theorem 4.1. That is, we fix a von Neumann algebra
$M$ and pairwise commuting automorphisms
$\alpha_{1},\alpha_{2},\ldots,\alpha_{n}$ on $M$. Also, we fix a normal
representation $\sigma$ of $M$ on a Hilbert space $H$ and let
$t:=(t_{1},t_{2},\ldots,t_{n})$ be a commuting tuple of operators in $B(H)$
such that

  1. (1)

$t$ satisfies the conditions from [4] : $\hat{t}_{1}$ and $\hat{t}_{n}$
satisfy Szego positivity and $\hat{t}_{n}$ is pure.

  2. (2)

For every $a\in M$ and $1\leq i\leq n$,

(49) $t_{i}\sigma(\alpha_{i}(a))=\sigma(a)t_{i}$

Since $\\{\alpha_{i}\\}$ pairwise commute, we can identify
${}_{\alpha_{i}}M\otimes_{\alpha_{j}}M$ with
${}_{\alpha_{j}}M\otimes_{\alpha_{i}}M$ (as above) and
$(_{\alpha_{1}}M,\ldots,_{\alpha_{n}}M)$ defines a product system. Equation
(49) implies that each $t_{i}$ gives rise to a representation $(\sigma,T_{i})$
of ${}_{\alpha_{i}}M$ and the associated $\tilde{T_{i}}$ satisfies
$t_{i}=\tilde{T_{i}}(I\otimes h)$ and
$t_{i}\sigma(a)h=\tilde{T_{i}}(I\otimes_{\sigma}\sigma(a)h)=\tilde{T_{i}}(a\otimes
h)$. Equation (47) shows that these representations pairwise commute and,
thus, define a representation of the product system.

Since the Szego condition and the pureness conditions involve expressions of
the form $t_{i}t_{j}^{*}$ (for operators) or $\tilde{T}_{i}\tilde{T_{j}}^{*}$
(for representations), it follows from (48) that (1) in Theorem 4.1 implies
that
$(\sigma,T_{1},\ldots,T_{n})\in\mathcal{T}_{1,n}^{n}(\sigma,\,M,\,E_{1},\ldots,E_{n},H)$
and, thus, Theorem 3.20 applies.

We conclude that there is an isometric representation
$(\rho,M_{1},L_{2},\ldots,L_{n-1},M_{n})$ of the product system that dilates
the representation $(\sigma,T_{1},\ldots,T_{n})$ and such that
$\tau_{U_{i}^{*}}=\tilde{M_{i}}$ for $i=1,2$.

Writing the isometric representation $(\rho,M_{1},L_{2},\ldots,L_{n-1},M_{n})$
as $(\rho,V_{1},V_{2},\ldots,V_{n-1},V_{n})$ (and let $v_{i}=V_{i}(I)$ as
above), this completes the proof of the theorem. ∎

######  Remark 4.3.

  1. 1.

The isometries $v_{i}$ can be written explicitely as shown in Theorem 3.20.

  2. 2.

For $M=\mathbb{C}$ (and writing $v_{i}$ explicitely as noted in 1.) we get the
main result of [4].

Acknowledgements: The first and second named authors are grateful to Prof.
Bata Krishna Das for some insightful discussions at the beginning of this
project. Part of the first named author’s research was carried out at the Ben-
Gurion University and the research is partially supported by the Israel
Science Foundation grant number 2196-20. The second named author’s research is
supported by the DST-INSPIRE Faculty Fellowship No.
DST/INSPIRE/04/2020/001250.

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