# Fullness of $q$-Araki-Woods Factors

Manish Kumar<EMAIL_ADDRESS>Institute of Mathematics of the Polish Academy
of Sciences, ul. Śniadeckich 8, 00–656 Warsaw, Poland  and  Simeng Wang
<EMAIL_ADDRESS>Institute for Advanced Study in Mathematics, Harbin
Institute of Technology, Harbin, 150001, China

###### Abstract.

The $q$-Araki-Woods factor associated to a group of orthogonal transformations
on a real Hilbert space $\mathsf{H}_{\mathbb{R}}$ is full as soon as
$\dim\mathsf{H}_{\mathbb{R}}\geq 2$.

###### Key words and phrases:

$q$-Araki-Woods factors; full factors; type III factors

###### 2020 Mathematics Subject Classification:

Primary: 46L36; Secondary 46L10, 46L53, 46L65

##  1\. Introduction

A von Neumann factor $\mathsf{M}$ is called full if, for any bounded net
${x_{i}}$ in $\mathsf{M}$ such that $\lim_{i}\|\varphi(\cdot
x_{i})-\varphi(x_{i}\cdot)\|_{\mathsf{M}_{*}}=0$ for all
$\varphi\in\mathsf{M}_{*}$, there exists a bounded net ${\lambda_{i}}$ of
scalars such that $\lim_{i}(x_{i}-\lambda_{i})=0$ $*$-strongly [Con1]. Here,
$\mathsf{M}_{*}$ denotes the predual of $\mathsf{M}$ consisting of normal
linear functionals on $\mathsf{M}$. A full factor is always non-injective or
non-hyperfinite i.e. they cannot be approximated by finite-dimensional
subalgebras. For type II1 factors, the negation of fullness is equivalent to
Murray and von Neumann’s notion of Property Gamma [Con2]. Connes established
several other equivalent conditions for fullness in type II1 factors in his
seminal work [Con2] on the classification program of injective factors. Some
of these equivalent conditions, such as the spectral gap property, do not seem
to generalize easily to type III factors. However, in a major breakthrough
recently, Marrakchi [Mar] proved that fullness for type III1 factors is
equivalent to the assertion that the C∗-algebra generated by $M$ and its
commutant $M^{\prime}$ contains compact operators. The focus of this paper is
to utilize one of these equivalent conditions to establish, in full
generality, that the $q$-Araki-Woods factors are full.

Recall that the $q$-Araki-Woods factors, denoted as
$\Gamma_{q}(\mathsf{H}_{\mathbb{R}},U_{t})$, are associated with a strongly
continuous one-parameter group of orthogonal transformations
$\\{U\\}_{t\in\mathbb{R}}$ on a real Hilbert space $\mathsf{H}_{\mathbb{R}}$
(see the construction in Section 2). These von Neumann algebras were
introduced by Hiai [Hia], combining the quasi-free deformations of
Shlyakhtenko [Shl] and $q$-deformations of Bożejko-Speicher [BS]. While
various structural properties are known for the free Araki-Woods factors
$\Gamma_{0}(\mathsf{H}_{\mathbb{R}},U_{t})$, such as fullness, solidity,
strong solidity, the absence of Cartan subalgebras, and complete metric
approximation properties [Shl, HRau, HRic, BHV], many of these properties
remain elusive for non-zero $q$, with only a few of these results available
[ABW, BM, BMRW, KSW, Nou, SW]. For instance, the fundamental question of
factoriality of $\Gamma_{q}(\mathsf{H}_{\mathbb{R}},U_{t})$ was only recently
addressed in complete generality in [KSW], with some major success achieved in
[BMRW]. Furthermore, the factors $\Gamma_{q}(\mathsf{H}_{\mathbb{R}},U_{t})$
are known to be non-injective [KSW, Nou]. Drawing inspiration from the
free/undeformed case, it is then natural to ask whether
$\Gamma_{q}(\mathsf{H}_{\mathbb{R}},U_{t})$ is full as well for non-zero $q$.
In this paper, we consider this problem and prove the following result.

######  Theorem 1.1.

Let $q\in(-1,1)$, let $\mathsf{H}_{\mathbb{R}}$ be a real separable Hilbert
space with $\dim\mathsf{H}_{\mathbb{R}}\geq 2$, and let
$\\{U_{t}\\}_{t\in\mathbb{R}}$ be a strongly continuous group of orthogonal
transformations on $\mathsf{H}_{\mathbb{R}}$. Then the $q$-Araki-Woods factor
$\Gamma_{q}(\mathsf{H}_{\mathbb{R}},U_{t})$ is full.

We provide a concise overview of previous attempts and highlight our
contribution to the proof. So far, there are mainly two papers, [KSW] and
[HI], which have dealt with the fullness of
$\Gamma_{q}(\mathsf{H}_{\mathbb{R}},U_{t})$ for $q\neq 0$. In order to
describe them, it is crucial to first mention that the Hilbert space
$\mathsf{H}_{\mathbb{R}}$ can be decomposed into direct sums of almost
periodic $\mathsf{H}_{\mathbb{R}}^{ap}$ and weakly mixing
$\mathsf{H}_{\mathbb{R}}^{wm}$ subspaces, where $\mathsf{H}_{\mathbb{R}}^{ap}$
is the direct sum of all finite-dimensional sub-representations of $U_{t}$,
and $\mathsf{H}_{\mathbb{R}}^{wm}$ encompasses the rest. When
$\mathsf{H}_{\mathbb{R}}$ is finite-dimensional,
$\Gamma_{q}(\mathsf{H}_{\mathbb{R}},U_{t})$ is shown to be full [KSW]. The
proof involves establishing the non-injectivity of
$\Gamma_{q}(\mathsf{H}_{\mathbb{R}},U_{t})$ by combining results on conjugate
variables and abstract findings of Nelson [Nel]. It then proceeds to
demonstrate Ozawa’s AO property for
$\Gamma_{q}(\mathsf{H}_{\mathbb{R}},U_{t})$ before invoking the abstract
results of Houdayer-Raum [HRau] to conclude fullness. In contrast, Houdayer
and Isono [HI] show that if $\mathsf{H}_{\mathbb{R}}^{wm}\neq 0$ or if the
point spectrum consisting of eigenvalues of $A$ is infinite and bounded, then
$\Gamma_{q}(\mathsf{H}_{\mathbb{R}},U_{t})$ is full. Here, $A$ denotes the
analytic generator of the one-parameter group of unitaries $U_{t}$, extended
linearly to the complexification of $\mathsf{H}_{\mathbb{R}}$. Their proof
follows the ideas of Śniady [Sni] from the tracial case, i.e. when $U_{t}$ is
trivial. However, their approach encounters difficulties when there is a
sequence of eigenvalues growing sufficiently fast to infinity. In this paper,
we overcome such challenges by adopting a distinct approach.

Our arguments involve a meticulous evaluation of certain Wick products in the
centralizer of $\Gamma_{q}(\mathsf{H}_{\mathbb{R}},U_{t})$ associated with
vectors in a $2\times 2$ block of $\mathsf{H}_{\mathbb{R}}$. As in [BMRW], we
make a crucial use of the non-triviality of the group $(U_{t})$; however, it
is worth noting that the methods in [BMRW] only work for the weak convergence
of the concerning sequences, which does not apply to the study of fullness. In
our approach, we establish several ’norm’-convergence results involving the
Wick products and creation/annihilation operators. The key novelty in these
convergence results lies in the utilization of an ‘averaging technique’, where
one will observe that the sequences themselves may fail to converge. For
instance, when $\xi$ and $\eta$ are two unit vectors in
$\mathsf{H}_{\mathbb{R}}$, the bounded sequence
$T_{n}=(1-q)^{n}c(\xi)^{n}c(\eta)^{n}$ never converges in norm, although it
does converge to $0$ in weak operator topology, where $c(\xi)$ denotes the
creation operator on the (deformed) Fock space. Nonetheless, we demonstrate
that the Cesaro sum of $T_{n}$ converges to $0$ in norm with the additional
assumption that $\xi,\eta$ are orthogonal to each other (see Proposition 3.4).
The main idea then is to carefully select such vectors $\xi$ and $\eta$, and
employ these convergence results to prove fullness (see Theorem 1.1). It is
crucial to emphasize that the methods employed in this paper are not
applicable to the cases considered in the previous two attempts [KSW, HI].
Thus, a combination of both [KSW] and [HI] is necessary to conclusively
establish Theorem 1.1 in full generality.

This paper is organized as follows. We give a brief overview of the
construction of $q$-Araki-Woods factors in Section 2. Section 3 is the core of
the article where we carry our main computations on various convergence
results. Finally, we put all our efforts together in Section 4 in proving the
main result as mentioned above.

##  2\. Preliminaries on $q$-Araki-Woods factors

We briefly describe construction of the $q$-Araki-Woods factors and refer the
readers to [Hia] for a detailed treatment. As a convention, all Hilbert spaces
are separable and inner products are linear in the second variable. For any
real or complex Hilbert space $H$, $\mathbf{B}(H)$ denotes the algebra of
bounded linear operators on $H$.

Let $\mathsf{H}_{\mathbb{R}}$ be a real Hilbert space, and let $U_{t}$ be a
strongly continuous orthogonal representations of $\mathbb{R}$ on
$\mathsf{H}_{\mathbb{R}}$. Consider the complexification
$\mathsf{H}_{\mathbb{C}}=\mathsf{H}_{\mathbb{R}}\otimes_{\mathbb{R}}\mathbb{C}$
of $\mathsf{H}_{\mathbb{R}}$, whose inner product is denoted by
$\langle\cdot,\cdot\rangle_{\mathsf{H}_{\mathbb{C}}}$. Identify
$\mathsf{H}_{\mathbb{R}}$ inside $\mathsf{H}_{\mathbb{C}}$ by
$\mathsf{H}_{\mathbb{R}}\otimes 1$, so that
$\mathsf{H}_{\mathbb{C}}=\mathsf{H}_{\mathbb{R}}+i\mathsf{H}_{\mathbb{R}}$.
Given $\xi=\xi_{1}+i\xi_{2}\in\mathsf{H}_{\mathbb{C}}$ for some
$\xi_{1},\xi_{2}\in\mathsf{H}_{\mathbb{R}}$, write the canonical involution
$\bar{\xi}$ as

$\bar{\xi}=\xi_{1}-i\xi_{2}.$

We extend $U_{t}$ to a strongly continuous one parameter group of unitaries on
the complexification $\mathsf{H}_{\mathbb{C}}$, which is still denoted by
$U_{t}$. Let $A$ denote the positive, self-adjoint, and non-singular analytic
(unbounded) operator on $\mathsf{H}_{\mathbb{C}}$ obtained via Stone’s
theorem, which satisfies $U_{t}=A^{it}$ for all $t\in\mathbb{R}$. Note that
$\overline{A\xi}=A^{-1}\bar{\xi}$ for any $\xi$ in the domain of $A$. In
particular, $\sigma(A)\cap(0,\infty)$ is symmetric around $1$ i.e. for
$\lambda\neq 0$, $\lambda\in\sigma(A)$ if and only if $1/\lambda\in\sigma(A)$,
where $\sigma(A)$ is the spectrum of $A$.

Following Shlyakhtenko [Shl], consider the deformed inner product
$\langle\cdot,\cdot\rangle_{U}$ on $\mathsf{H}_{\mathbb{C}}$ defined by

$\langle\xi,\zeta\rangle_{U}=\langle
2(1+A^{-1})^{-1}\xi,\zeta\rangle\quad\forall\xi,\zeta\in H_{\mathbb{C}}.$

We denote the completion of $\mathsf{H}_{\mathbb{C}}$ with respect to
$\langle\cdot,\cdot\rangle_{U}$ by $\mathsf{H}_{U}$. The space
$(\mathsf{H}_{\mathbb{R}},\|\cdot\|_{\mathsf{H}_{\mathbb{R}}})$ is naturally
identified as a subspace of $\mathsf{H}_{U}$ such that
$\mathsf{H}_{\mathbb{R}}+i\mathsf{H}_{\mathbb{R}}$ is dense in
$\mathsf{H}_{U}$ and $\mathsf{H}_{\mathbb{R}}\cap
i\mathsf{H}_{\mathbb{R}}=\\{0\\}$. For $q\in(-1,1)$, consider the $q$-Fock
space $\mathcal{F}_{q}(\mathsf{H}_{U})$ given by

$\mathcal{F}_{q}(\mathsf{H}_{U})=\mathbb{C}\Omega\oplus\bigoplus_{n\geq
1}\mathsf{H}_{U}^{\otimes_{q}^{n}},$

where $\Omega$ is a distinguished unit vector in $\mathbb{C}$ (called vacuum
vector) and $\mathsf{H}_{U}^{\otimes_{q}^{n}}$ is the completion of the
algebraic tensor product $\mathsf{H}^{\otimes_{\text{alg}}^{n}}$ equipped with
the following inner product:

$\langle\xi_{1}\otimes\ldots\otimes\xi_{n},\zeta_{1}\otimes\ldots\otimes\zeta_{n}\rangle_{q}=\sum_{\pi\in
S_{n}}q^{|\pi|}\prod_{i=1}^{n}\langle\xi_{i},\zeta_{\pi(i)}\rangle_{U}$

for all $\xi_{i},\zeta_{i}\in\mathsf{H}_{U}$, $1\leq i\leq n$. Here $S_{n}$
denotes the symmetric group on $n$ elements and $|\pi|$ denotes the number of
inversions for the permutation $\pi\in S_{n}$. That the above assignment
yields a genuine inner product was proved in [BS]. On
$\mathcal{F}_{q}(\mathsf{H}_{U})$, we define for each $\xi\in\mathsf{H}_{U}$ a
bounded operator $c(\xi)$ (called left q-creation) and its adjoint
$c^{*}(\xi)$ (called left q-annihilation) by:

$\displaystyle c(\xi)\Omega=\xi,$ $\displaystyle
c(\xi)(\xi_{1}\otimes\ldots\otimes\xi_{n})=\xi\otimes\xi_{1}\otimes\ldots\otimes\xi_{n},\;\;\;\mbox{
and }$ $\displaystyle c^{*}(\xi)\Omega=0,$ $\displaystyle
c^{*}(\xi)(\xi_{1}\otimes\ldots\otimes\xi_{n})=\sum_{i=1}^{n}q^{i-1}\langle\xi,\xi_{i}\rangle\xi_{1}\otimes\ldots\otimes\hat{\xi_{i}}\otimes\ldots\otimes\xi_{n},$

for all
$\xi_{1}\otimes\ldots\otimes\xi_{n}\in\mathsf{H}_{U}^{\otimes_{q}^{n}}$. Here
$\hat{\xi_{i}}$ notates the omitted $i$-th letter from the tensor. That
$c(\xi)$ and $c^{*}(\xi)$ are bounded operators with them being adjoint to
each other is a well known fact (see [BS]). In a similar fashion, one defines
the right $q$-creation and right $q$-annihilation operators $r(\xi)$ and
$r^{*}(\xi)$ respectively for all $\xi\in\mathsf{H}_{U}$ as follows:

$\displaystyle r(\xi)\Omega=\xi,$ $\displaystyle
r(\xi)(\xi_{1}\otimes\ldots\otimes\xi_{n})=\xi_{1}\otimes\ldots\otimes\xi_{n}\otimes\xi,\;\;\;\mbox{
and }$ $\displaystyle r^{*}(\xi)\Omega=0,$ $\displaystyle
r^{*}(\xi)(\xi_{1}\otimes\ldots\otimes\xi_{n})=\sum_{i=1}^{n}q^{n-i}\langle\xi,\xi_{i}\rangle\xi_{1}\otimes\ldots\otimes\hat{\xi_{i}}\otimes\ldots\otimes\xi_{n},$

for all
$\xi_{1}\otimes\ldots\otimes\xi_{n}\in\mathsf{H}_{U}^{\otimes_{q}^{n}}$. Again
$r(\xi)$ and $r^{*}(\xi)$ are bounded operators such that they are adjoint to
each other. The creation and annihilation operators satisfy the following
$q$-commutation relations:

(1) $\displaystyle
c^{*}(\xi)c(\eta)-qc(\eta)c^{*}(\xi)=\langle\xi,\eta\rangle_{U}$ (2)
$\displaystyle
r^{*}(\xi)r(\eta)-qr(\eta)r^{*}(\xi)=\langle\xi,\eta\rangle_{U}$

for any $\xi,\eta\in\mathsf{H}_{U}$. We shall repeatably use the above
commutation relations without referring to them. In particular, the most
useful use of these relations will be the following consequence:

$\displaystyle c^{*}(\xi)^{m}c(\eta)^{n}=q^{mn}c(\eta)^{n}c^{*}(\xi)^{m}$

whenever $\xi,\eta\in\mathsf{H}_{U}$ with $\langle\xi,\eta\rangle_{U}=0$ and
$m,n\geq 1$.

We now consider the von Neumann algebra introduced by Hiai [Hia] (and by
Shlyakhtenko [Shl]), denoted $\Gamma_{q}(\mathsf{H}_{\mathbb{R}},U_{t})$, to
be the von Neumann subalgebra of ${B}(\mathcal{F}_{q}(\mathsf{H}_{U}))$
generated by the set $\\{W(\xi);\xi\in\mathsf{H}_{\mathbb{R}}\\}$, where

$W(\xi)=c({\xi})+c^{*}(\xi),\;\;\;\;\;\;\;\xi\in\mathsf{H}_{\mathbb{R}}.$

The algebra $\Gamma_{q}(\mathsf{H}_{\mathbb{R}},U_{t})$ is called the
$q$-deformed Araki-Woods von Neumann algebra. If the representation $U_{t}$ is
clear from the context, we shall denote the algebra
$\Gamma_{q}(\mathsf{H}_{\mathbb{R}},U_{t})$ simply by $\mathsf{M}_{q}$. Since
it is now known that $\mathsf{M}_{q}$ is factor i.e. $\mathsf{M}_{q}$ has
trivial center (see [KSW]), we will call them $q$-Araki-Woods factors.

The vacuum vector $\Omega$ is cyclic and separating for the factor
$\mathsf{M}_{q}$, and hence induces a faithful normal state:
$\varphi(\cdot)=\langle\Omega,\cdot\;\Omega\rangle_{q}$ on $\mathsf{M}_{q}$
and is called the $q$-quasi free state. It is immediate to verify that the GNS
Hilbert space associated to $\varphi$ is $\mathcal{F}_{q}(\mathsf{H}_{U})$.
Since $\Omega$ is a separating vector for both $\mathsf{M}_{q}$ and
$\mathsf{M}_{q}^{\prime}$, there exist for each $\xi\in\mathsf{M}_{q}\Omega$
and $\zeta\in\mathsf{M}_{q}^{\prime}\Omega$ unique operators
$x_{\xi}\in\mathsf{M}_{q}$ and $x^{\prime}_{\zeta}\in\mathsf{M}_{q}^{\prime}$
such that $\xi=x_{\xi}\Omega$ and $\zeta=x_{\zeta}^{\prime}\Omega$. Write

$W(\xi)=x_{\xi},\;\mbox{ and }\;W_{r}(\zeta)=x^{\prime}_{\zeta}.$

Note that the notation $W(\xi)$ is in line with the one introduced above for
$\xi\in\mathsf{H}_{\mathbb{R}}$. The operators $W(\xi)$ and $W(\zeta)$ are
respectively called left and right Wick operators. It is known that
$\xi_{1}\otimes\ldots\otimes\xi_{n}\in\mathsf{M}_{q}\Omega$ for all
$\xi_{i}\in\mathsf{H}_{\mathbb{C}},1\leq i\leq n$ and
$\zeta_{1}\otimes\ldots\otimes\zeta_{m}$ for all
$\zeta_{j}\in\mathsf{H}_{\mathbb{C}}\cap\operatorname{Dom}(A^{-1/2}),1\leq
j\leq m$. By definition, the operators $\mathsf{W}(\xi)$ and $W_{r}(\zeta)$
commute for all $\xi\in\mathsf{M}_{q}\Omega$ and
$\zeta\in\mathsf{M}_{q}^{\prime}\Omega$. In fact, the commutant of
$\mathsf{M}_{q}$ in $B(\mathcal{F}_{q}(\mathsf{H}_{U}))$ is given by

$\Gamma_{q}(\mathsf{H}_{\mathbb{R}},U_{t})^{\prime}=\\{W_{r}(\eta);\eta\in\mathsf{H}_{\mathbb{R}}^{\prime}\\}^{{}^{\prime\prime}}$

where
$\mathsf{H}_{\mathbb{R}}^{\prime}=\\{\eta\in\mathsf{H}_{\mathbb{C}};\langle\xi,\eta\rangle_{U}\in\mathbb{R}\mbox{
for all }\xi\in\mathsf{H}_{\mathbb{R}}\\}$. We have
$\mathsf{H}_{\mathbb{C}}=\mathsf{H}_{\mathbb{R}}^{\prime}+i\mathsf{H}_{\mathbb{R}}^{\prime}$
and $\mathsf{H}_{\mathbb{R}}\cap i\mathsf{H}_{\mathbb{R}}=\\{0\\}$. It is easy
to verify that

$W(\xi)=c({\xi})+c^{*}({\bar{\xi}})\;\;\mbox{ and
}\;\;W_{r}(\zeta)=r(\zeta)+r^{*}(\bar{\zeta}^{r})$

for any $\xi\in\mathsf{H}_{\mathbb{C}}$ and
$\zeta\in\mathsf{H}_{\mathbb{C}}\cap\operatorname{Dom}(A^{-1/2})$, where
$\bar{\zeta}^{r}$ denotes the real part of $\zeta$ with respect to the
decomposition
$\mathsf{H}_{\mathbb{C}}=\mathsf{H}_{\mathbb{R}}^{\prime}+i\mathsf{H}_{\mathbb{R}}^{\prime}$.
The formula for the left and right Wick operators are given as follows:

######  Proposition 2.1.

For any vector
$\xi_{1},\ldots,\xi_{n}\in\mathsf{H}_{\mathbb{R}}+i\mathsf{H}_{\mathbb{R}}$,
the Wick-formula for operators in $\mathsf{M}_{q}$ is given by

$\displaystyle W(\xi_{1}\otimes\cdots\otimes\xi_{n})$
$\displaystyle=\sum_{i=0}^{n}\sum_{\sigma\in
S_{n,i}}q^{|\sigma|}c(\xi_{\sigma(1)})\cdots
c(\xi_{\sigma(i)})c^{*}(\bar{\xi}_{\sigma(i+1)})\cdots
c^{*}(\bar{\xi}_{\sigma(n)})$

where $S_{n,j}$ is the set of those permutations in $S_{n}$ which are
increasing on ${\\{1,\ldots,j\\}}$ and on $\\{j+1,\ldots,n\\}$.

######  Proposition 2.2.

For any vector
$\xi_{1},\ldots,\xi_{n}\in\mathsf{H}_{\mathbb{R}}^{\prime}+i\mathsf{H}_{\mathbb{R}}^{\prime}$,
the Wick-formula for operators in $\mathsf{M}_{q}^{\prime}$ is given by

$\displaystyle W_{r}(\xi_{n}\otimes\cdots\otimes\xi_{1})$
$\displaystyle=\sum_{i=0}^{n}\sum_{\sigma\in
S_{n,i}}q^{|\operatorname{flip}\circ\sigma|}r(\xi_{\sigma(1)})\cdots
r(\xi_{\sigma(n-i)})r^{*}(\bar{\xi}_{\sigma(n-i+1)}^{r})\cdots
r^{*}(\bar{\xi}_{\sigma(n)}^{r})$ $\displaystyle=\sum_{i=0}^{n}\sum_{\sigma\in
S_{n,n-i}}q^{|\sigma|}r(\xi_{n+1-\sigma(n)})\cdots
r(\xi_{n+1-\sigma(i+1)})r^{*}(\bar{\xi}_{n+1-\sigma(i)}^{r})\cdots
r^{*}(\bar{\xi}_{n+1-\sigma(1)}^{r})$

where $S_{n,j}$ is the set of those permutations in $S_{n}$ which are
increasing on ${\\{1,\ldots,j\\}}$ and on $\\{j+1,\ldots,n\\}$.

We adopt the following notations for our convenience.

  1. (1)

$\xi_{1}\ldots\xi_{n}=\xi_{1}\otimes\ldots\otimes\xi_{n}$, for
$\xi_{i}\in\mathsf{H}_{U}$, $1\leq i\leq n$.

  2. (2)

$\xi^{n}=\xi^{\otimes n}$, $n\geq 1$ and $\xi^{0}=\Omega$ for all
$\xi\in\mathsf{H}_{U}$.

  3. (3)

$d_{0}=1$, $d_{n}=\prod_{j=1}^{n}(1-q^{j})$ for $n\geq 1$ and
$d_{\infty}=\prod_{j=1}^{\infty}(1-q^{j})$

  4. (4)

$C_{q}=\prod_{j=1}^{\infty}\frac{1}{1-|q|^{j}}$

  5. (5)

$[n]_{q}=\sum_{j=0}^{n-1}q^{j}=\frac{1-q^{n}}{1-q}$

  6. (6)

$[n]_{q}!=\prod_{j=1}^{n}[j]_{q}=\frac{d_{n}}{(1-q)^{n}}$

  7. (7)

$\begin{pmatrix}n\\\ k\end{pmatrix}_{q}=\frac{[n]_{q}!}{[k]_{q}![n-k]_{q}!}$
for any $1\leq k\leq n$.

The following are well known inequalities (see [BS]), which we shall use very
frequently throughout the paper:

  1. (1)

$d_{n}\leq C_{q}$,

  2. (2)

$\frac{1}{d_{n}}\leq C_{q}$.

  3. (3)

$\|c(\xi)^{n}\|\leq\sqrt{C_{q}}\|\xi^{n}\|_{q}$ for all
$\xi\in\mathsf{H}_{U}$.

The following lemmas are either well-known or easy to verify.

######  Lemma 2.3.

Given $f\in\mathsf{H}_{U}$, and
$\xi\in\mathsf{H}_{U}^{\otimes_{q}^{m}},\eta\in\mathsf{H}_{U}^{\otimes_{q}^{n}}$
for some $m,n\geq 1$, we have

$\displaystyle
c^{*}(f)(\xi\otimes\eta)=(c^{*}(f)\xi)\otimes\eta+q^{n}\xi\otimes
c^{*}(f)\eta$

######  Lemma 2.4.

For any $\xi\in\mathsf{H}_{U}$ and $n\geq 1$, we have $\|c(\xi)^{n}\|\leq
C_{q}(1-q)^{-\frac{n}{2}}\|\xi\|^{\frac{n}{2}}.$

###### Proof.

This follows by recalling that
$\|c(\xi)^{n}\|\leq\sqrt{C_{q}}\|\xi^{n}\|_{q}$, as well as by noting that
$[n]_{q}!=d_{n}(1-q)^{-n}\leq C_{q}(1-q)^{-n}$ for all $n\geq 1$. ∎

######  Corollary 2.5.

For any $\xi\in\mathsf{H}_{U}$, the sequences

  1. (1)

$(1-q)^{\frac{n}{2}}\|\xi\|_{U}^{-n}c(\xi)^{n}$

  2. (2)

$(1-q)^{\frac{n}{2}}\|\xi\|_{U}^{-n}r(\xi)^{n}$

are bounded, both with an upper bound given by $C_{q}$.

We end this section by recalling the general structure of a one parameter
group of orthogonal operators. Let $\\{U_{t}\\}_{t\in\mathbb{R}}$ be a
strongly continuous orthogonal representation of $\mathbb{R}$ on a real
Hilbert space $\mathsf{H}_{\mathbb{R}}$ and let $A$ denote the analytic
generator of the complexification of $U_{t}$. Let
$\mathsf{H}_{\mathbb{R}}^{ap}$ denote the closed real subspace generated by
the eigenvectors of $A$ and let
$\mathsf{H}_{\mathbb{R}}^{wm}={\mathsf{H}_{\mathbb{R}}^{ap}}^{\perp}$. The
subspaces $\mathsf{H}_{\mathbb{R}}^{ap}$ and $\mathsf{H}_{\mathbb{R}}^{wm}$
are reducing under $A$ and every $U_{t}$, and they are respectively called the
almost periodic and weakly mixing part of $\mathsf{H}_{\mathbb{R}}$.
Similarly, ${U_{t}}_{|_{\mathsf{H}_{\mathbb{R}}^{ap}}}$ and
${U_{t}}_{|_{\mathsf{H}_{\mathbb{R}}^{wm}}}$ are called the almost periodic
and weakly mixing part of $U_{t}$ respectively.

Note that whenever $\mathsf{H}_{\mathbb{R}}^{wm}\neq 0$, it must be infinite-
dimensional. On the other hand, the almost periodic part decomposes as follows
(see [Shl]):

$(\mathsf{H}_{\mathbb{R}}^{ap},{U_{t}}_{|_{\mathsf{H}_{\mathbb{R}}^{ap}}})=\left(\bigoplus_{j=1}^{N_{1}}(\mathbb{R},id)\right)\oplus\left(\bigoplus_{j=1}^{N_{2}}\mathsf{H}_{\mathbb{R}}^{k},U_{t}^{k}\right)$

for some $0\leq N_{1},N_{2}\leq\aleph_{0}$, where for all $1\leq k\leq N_{2}$
we have

$\mathsf{H}_{\mathbb{R}}^{k}=\mathbb{R}^{2},\;\;\;\;U_{t}^{k}=\begin{pmatrix}\cos(t\log\lambda_{k})&-\sin(t\log\lambda_{k})\\\
\sin(t\log\lambda_{k})&\cos(t\log\lambda_{k})\end{pmatrix}$

for some $\lambda_{k}\in(0,1)$.

##  3\. Technical analysis

This section contains the rigorous technical analysis, where we will prove a
series of Lemmas and Propositions about various convergence results. The most
crucial results are Propositions 3.9, 3.4 and 3.12. We recommend that readers
first review the statements of these Propositions and then proceed directly to
the proof of the main theorem in the next section to understand the reasons
behind these calculations. The notations used in Proposition 3.9 are mentioned
just before Lemma 3.7.

######  Lemma 3.1.

For any fixed non-zero vector $\xi\in\mathsf{H}_{U}$, $n\in\mathbb{N}$ and
$1\leq m\leq n$, we have

(3) $\displaystyle
c^{*}(\xi)^{m}c(\xi)^{n}=\frac{[n]_{q}!}{[n-m]_{q}!}\|\xi\|_{U}^{2m}c(\xi)^{n-m}+\sum_{k=0}^{m-1}\frac{[n]_{q}!}{[n-k]_{q}!}q^{n-k}\|\xi\|_{U}^{2k}c^{*}(\xi)^{m-k-1}c(\xi)^{n-k}c^{*}(\xi).$

###### Proof.

The proof is by induction on $m$. For $m=1$, use Lemma 2.3 to write

(4) $\displaystyle
c^{*}(\xi)c(\xi)^{n}=[n]_{q}\|\xi\|_{U}^{2}c(\xi)^{n-1}+q^{n}c(\xi)^{n}c^{*}(\xi)$

which is exactly as in (3). Now assume the desired equation to be true for
some $1\leq m\leq n-1$; then calculate

$\displaystyle c^{*}(\xi)^{m+1}c(\xi)^{n}$
$\displaystyle=c^{*}(\xi)\left(c^{*}(\xi)^{m}c(\xi)^{n}\right)$
$\displaystyle=\frac{[n]_{q}!}{[n-m]_{q}!}\|\xi\|_{U}^{2m}c^{*}(\xi)c(\xi)^{n-m}+\sum_{k=0}^{m-1}\frac{[n]_{q}!}{[n-k]_{q}!}\|\xi\|_{U}^{2k}q^{n-k}c^{*}(\xi)^{m-k}c(\xi)^{n-k}c^{*}(\xi)$
$\displaystyle\hskip 216.81pt(\text{by induction hypothesis})$
$\displaystyle=\frac{[n]_{q}!}{[n-m]_{q}!}\|\xi\|_{U}^{2m}\left([n-m]_{q}\|\xi\|_{U}^{2}c(\xi)^{n-m-1}+q^{n-m}c(\xi)^{n-m}c^{*}(\xi)\right)$
$\displaystyle\quad\quad\quad\quad+\sum_{k=0}^{m-1}\frac{[n]_{q}!}{[n-k]_{q}!}\|\xi\|_{U}^{2k}q^{n-k}c^{*}(\xi)^{m-k}c(\xi)^{n-k}c^{*}(\xi)\quad\quad\quad(\mbox{by
using eq. \eqref{eq:for m=1} for }n-m)$
$\displaystyle=\frac{[n]_{q}!}{[n-m-1]_{q}!}\|\xi\|_{U}^{2m+2}c(\xi)^{n-(m+1)}+\sum_{k=0}^{m}\frac{[n]_{q}!}{[n-k]_{q}!}\|\xi\|_{U}^{2k}q^{n-k}c^{*}(\xi)^{m-k}c(\xi)^{n-k}c^{*}(\xi).$

The proof then ends by the induction argument. ∎

######  Corollary 3.2.

For vectors $\xi,\eta\in\mathsf{H}_{U}$ with $\langle\xi,\eta\rangle_{U}=0$,
$n\in\mathbb{N}$, $0\leq m\leq n$ and $\ell\geq 1$, we have

$\displaystyle
c^{*}(\xi)^{m}c(\xi)^{n}c(\eta)^{\ell}=\frac{[n]_{q}!}{[n-m]_{q}!}\|\xi\|_{U}^{2m}c(\xi)^{n-m}c(\eta)^{\ell}+q^{\ell}X$

where $X$ is an operator bounded by
$A_{q}^{\prime}(1-q)^{-\frac{(m+n+\ell)}{2}}\|\xi\|_{U}^{m+n}\|\eta\|_{U}^{\ell}$
for some constant $A_{q}^{\prime}$ depending only on $q$.

###### Proof.

If $m=0$ then set $X=0$, and if $m\geq 1$ then set

$X=\sum_{k=0}^{m-1}\frac{[n]_{q}!}{[n-k]_{q}!}q^{n-k}\|\xi\|_{U}^{2k}c^{*}(\xi)^{m-k-1}c(\xi)^{n-k}c(\eta)^{\ell}c^{*}(\xi).$

Obtain the required expression by multiplying from the right in eq. (3) of
Lemma 3.1 by $c(\eta)^{\ell}$ and then by noting via $q$-commutation relation
that $c^{*}(\xi)c(\eta)^{\ell}=q^{\ell}c(\eta)^{\ell}c^{*}(\xi)$. Further, we
estimate the norm of $X$ using Lemma 2.4 as follows:

$\displaystyle\|X\|$ $\displaystyle\leq
C_{q}^{4}\sum_{k=0}^{m-1}\frac{[n]_{q}!}{[n-k]_{q}!}|q|^{n-k}\|\xi\|_{U}^{2k}\|\xi\|_{U}^{(m-k-1)+(n-k)+1}\|\eta\|_{U}^{\ell}(1-q)^{-\frac{1}{2}(m-k-1)-\frac{1}{2}(n-k)-\frac{\ell}{2}-\frac{1}{2}}$
$\displaystyle=C_{q}^{4}\sum_{k=0}^{m-1}\frac{d_{n}(1-q)^{-n}}{d_{n-k}(1-q)^{-n+k}}|q|^{n-k}\|\xi\|_{U}^{m+n}\|\eta\|_{U}^{\ell}(1-q)^{-\frac{(m+n+\ell)}{2}+k}$
$\displaystyle\leq
C_{q}^{6}\|\xi\|_{U}^{m+n}\|\eta\|_{U}^{\ell}(1-q)^{-\frac{(m+n+\ell)}{2}}\sum_{k=0}^{m-1}|q|^{n-k}\quad\quad\quad\quad(\text{as
}d_{n},\frac{1}{d_{n-k}}\leq C_{q})$
$\displaystyle=\|\xi\|_{U}^{m+n}\|\eta\|_{U}^{\ell}(1-q)^{-\frac{(m+n+\ell)}{2}}\frac{C_{q}^{6}}{1-|q|}$

which gives the required estimate by setting
$A_{q}^{\prime}=\frac{C_{q}^{6}}{1-|q|}$. ∎

######  Corollary 3.3.

For vectors $\xi,\eta\in\mathsf{H}_{U}$ with $\langle\xi,\eta\rangle_{U}=0$,
and $m>n\geq 1$, $\ell\geq 1$, we have

$\displaystyle c^{*}(\xi)^{m}c(\xi)^{n}c(\eta)^{\ell}=q^{\ell}Y$

where $Y$ is an operator bounded by
$A_{q}(1-q)^{-\frac{(m+n+\ell)}{2}}\|\xi\|_{U}^{m+n}\|\eta\|_{U}^{\ell}$ for
some constant $A_{q}$ depending only on $q$.

###### Proof.

Using Corollary 3.2, write

$\displaystyle
c^{*}(\xi)^{n}c(\xi)^{n}c(\eta)^{\ell}=[n]_{q}!\|\xi\|_{U}^{2n}c(\eta)^{\ell}+q^{\ell}X$

where $X$ is an operator bounded by
$\frac{C_{q}^{6}}{1-|q|}(1-q)^{-n-\frac{\ell}{2}}\|\xi\|_{U}^{2n}\|\eta\|_{U}^{\ell}$.
By multiplying both side from left by $c^{*}(\xi)^{m-n}$ and using the
$q$-commutation relation
$c^{*}(\xi)^{m-n}c(\eta)^{\ell}=q^{(m-n)\ell}c(\eta)^{\ell}c^{*}(\xi)^{(m-n)}$
since $\langle\xi,\eta\rangle_{U}=0$, we get

$\displaystyle
c^{*}(\xi)^{m}c(\xi)^{n}c(\eta)^{\ell}=q^{(m-n)\ell}[n]_{q}!\|\xi\|_{U}^{2n}c(\eta)^{\ell}c^{*}(\xi)^{m-n}+q^{\ell}c^{*}(\xi)^{m-n}X=q^{\ell}Y$

where
$Y=q^{(m-n-1)\ell}[n]_{q}!\|\xi\|_{U}^{2n}c(\eta)^{\ell}c^{*}(\xi)^{m-n}+c^{*}(\xi)^{m-n}X$.
It is an easy verify as before using Lemma 2.4 that

$\displaystyle\|Y\|$
$\displaystyle\leq|q|^{(m-n-1)\ell}d_{n}C_{q}^{2}\|\xi\|_{U}^{m+n}\|\eta\|_{U}^{\ell}(1-q)^{-\frac{(m+n+\ell)}{2}}+\frac{C_{q}^{6}}{1-|q|}\|\xi\|_{U}^{m+n}\|\eta\|_{U}^{\ell}C_{q}(1-q)^{-\frac{(m+n+\ell)}{2}}$
$\displaystyle\leq
A_{q}\|\xi\|_{U}^{m+n}\|\eta\|_{U}^{\ell}(1-q)^{-\frac{(m+n+\ell)}{2}}$

where $A_{q}=|q|^{(m-n-1)\ell}C_{q}^{3}+\frac{C_{q}^{7}}{1-|q|}$. ∎

######  Proposition 3.4.

For any non-zero vectors $\xi,\eta\in\mathsf{H}_{U}$ with
$\langle\xi,\eta\rangle_{U}=0$ and a uniformly bounded sequence
$\\{T_{k}\\}_{k\geq 1}$ of operators, the sequences

  1. (1)

$\frac{1}{n}\sum_{k=1}^{n}(1-q)^{k}\|\xi\|_{U}^{-k}\|\eta\|_{U}^{-k}c(\xi)^{k}c(\eta)^{k}T_{k}$

  2. (2)

$\frac{1}{n}\sum_{k=1}^{n}(1-q)^{k}\|\xi\|_{U}^{-k}\|\eta\|_{U}^{-k}r(\xi)^{k}r(\eta)^{k}T_{k}$

converge to $0$ in norm as $n\to\infty$.

###### Proof.

We shall prove the convergence of the first sequence, which we call $X_{n}$.
The proof for the second sequence follows similarly. By replacing $\xi$ and
$\eta$ with $\frac{\xi}{\|\xi\|_{U}}$ and $\frac{\eta}{\|\eta\|_{U}}$
respectively, we may and will assume without loss of generality that $\xi$ and
$\eta$ are unit vectors. Also assume that $\sup_{k\geq 1}\|T_{k}\|=1$. Now
calculate

$\displaystyle X_{n}^{*}X_{n}$
$\displaystyle=\frac{1}{n^{2}}\sum_{\ell=1}^{n}\sum_{k=1}^{n}(1-q)^{\ell+k}T_{\ell}^{*}c^{*}(\eta)^{\ell}c^{*}(\xi)^{\ell}c(\xi)^{k}c(\eta)^{k}T_{k}$
$\displaystyle=X^{n}_{\ell=k}+X^{n}_{k<\ell}+X^{n}_{k>\ell}$

where we have divided the sum over three cases: $\ell=k,k<\ell$ and $k>\ell$
i.e.

$\displaystyle
X_{\ell=k}^{n}=\frac{1}{n^{2}}\sum_{k=1}^{n}(1-q)^{2k}T_{k}^{*}c^{*}(\eta)^{k}c^{*}(\xi)^{k}c(\xi)^{k}c(\eta)^{k}T_{k}$
$\displaystyle X_{k<\ell}^{n}=\frac{1}{n^{2}}\sum_{1\leq k<\ell\leq
n}(1-q)^{\ell+k}T_{\ell}^{*}c^{*}(\eta)^{\ell}c^{*}(\xi)^{\ell}c(\xi)^{k}c(\eta)^{k}T_{k}$
$\displaystyle X_{\ell<k}^{n}=\frac{1}{n^{2}}\sum_{n\geq k>\ell\geq
1}(1-q)^{\ell+k}T_{\ell}^{*}c^{*}(\eta)^{\ell}c^{*}(\xi)^{\ell}c(\xi)^{k}c(\eta)^{k}T_{k}.$

We now estimate the norm of the three sums separately. Firstly, over $k=\ell$,
we estimate the norm using Lemma 2.4 as follows:

$\|X_{\ell=k}^{n}\|\leq\frac{1}{n^{2}}\sum_{k=1}^{n}C_{q}^{4}=\frac{C_{q}^{4}}{n}.$

We next estimate the norm of $X_{k<\ell}^{n}$: for $k<\ell$, we use Corollary
3.3 to write $c^{*}(\xi)^{\ell}c(\xi)^{k}c(\eta)^{k}=q^{k}X_{k,\ell}$ where
$X_{k,\ell}$ is an operator bounded by $A_{q}(1-q)^{-\frac{\ell}{2}-k}$.
Therefore, we have

$\displaystyle\|X_{k<\ell}^{n}\|$
$\displaystyle=\frac{1}{n^{2}}\left\|\sum_{1\leq k<\ell\leq
n}(1-q)^{\ell+k}c^{*}(\eta)^{\ell}T_{\ell}^{*}\left(q^{k}X_{k,\ell}\right)T_{k}\right\|$
$\displaystyle\leq\frac{C_{q}A_{q}}{n^{2}}\sum_{1\leq k<\ell\leq
n}|q|^{k}(1-q)^{\ell+k}(1-q)^{-\frac{\ell}{2}-k}(1-q)^{-\frac{\ell}{2}}$
$\displaystyle\leq\frac{C_{q}A_{q}}{n^{2}}\sum_{k,\ell=1}^{n}|q|^{k}$
$\displaystyle\leq\frac{C_{q}A_{q}}{n}\sum_{k=1}^{\infty}|q|^{k}$
$\displaystyle\leq\frac{1}{n}\cdot\frac{C_{q}A_{q}}{1-|q|}.$

Similarly, by symmetry $X_{k>\ell}^{n}=(X_{k<\ell}^{n})^{*}$,
$\|X^{n}_{k>\ell}\|$ has the same estimate $\|X_{k<\ell}^{n}\|$. Thus we
obtain that

$\displaystyle\|X_{n}^{*}X_{n}\|\leq\frac{1}{n}\left({C_{q}^{4}}+\frac{2C_{q}A_{q}}{1-|q|}\right)$

which proves that $X_{n}\to 0$ in norm as $n\to\infty$. ∎

######  Remark 3.5.

It is very important that we multiply the operator $T_{k}$ from the right in
Proposition 3.4 and not from the left. In fact, the operator
$\tilde{S}_{n}=\frac{1}{n}\sum_{k=0}^{n}(1-q)^{k}\tilde{T}_{k}c(\xi)^{k}c(\eta)^{k}$
need not converge to $0$ for fixed unit vectors $\xi,\eta\in\mathsf{H}_{U}$
with $\langle\xi,\eta\rangle_{U}=0$ and bounded operators $\tilde{T}_{k}$. See
for example Proposition 3.9 below. However, for special operators
$\tilde{T}_{k}$, we show in Proposition 3.4 that $\tilde{S}_{n}$ converges to
$0$.

######  Lemma 3.6.

For any non-zero vector $\xi\in\mathsf{H}_{U}$, the sequence

$Z_{n}=(1-q)^{n}\|\xi\|_{U}^{-2n}c^{*}(\xi)^{n}c(\xi)^{n}$

is convergent in norm to some operator $Z$. Moreover $Z$ is positive and
invertible such that $d_{\infty}\leq Z\leq 1$ for $q\geq 0$, and
$\frac{d_{\infty}}{1-q}\leq Z\leq 1-q$ for $q\leq 0$.

###### Proof.

The proof follows the same lines of [BMRW, Lemma 3.4]. ∎

We now restrict our attention to a two-dimensional invariant subspace of a
given representation $\\{U_{t}\\}_{t\in\mathbb{R}}$. Let
$\mathsf{K}_{\mathbb{R}}=\mathbb{R}^{2}$ be an invariant subspace of $U_{t}$
such that ${U_{t}}_{|_{\mathsf{K}_{\mathbb{R}}}}$ is ergodic i.e. has no fixed
nonzero vector. Then, as discussed in Section 2, there is a scalar
$\lambda\in(0,1)$ such that

$\displaystyle{U_{t}}_{|_{\mathsf{K}_{\mathbb{R}}}}=\begin{pmatrix}\cos(t\log\lambda)&-\sin(t\log\lambda)\\\
\sin(t\log\lambda)&\cos(t\log\lambda)\end{pmatrix}.$

Set

$\displaystyle
e=\frac{({\lambda+1})^{\frac{1}{2}}}{2\lambda^{\frac{1}{4}}}\begin{pmatrix}1\\\
i\end{pmatrix}\in\mathsf{K}_{\mathbb{C}}=\mathbb{C}^{2}.$

If $A$ denotes the analytic generator, then it is immediate to check that

$Ae=\lambda^{-1}e,\;\;\;A\bar{e}=\lambda\bar{e},\;\;\;\|e\|_{U}=\lambda^{-\frac{1}{4}},\;\;\;\|\bar{e}\|_{U}=\lambda^{\frac{1}{4}}\;\;\;\mbox{
and }\langle e,\bar{e}\rangle_{U}=0.$

Moreover, the (right) conjugate of $e$ and $\bar{e}$ with respect to the
decomposition in
$\mathsf{H}_{\mathbb{R}}^{\prime}+i\mathsf{H}_{\mathbb{R}}^{\prime}$ is given
as follow:

$\bar{e}^{r}=\lambda^{-1}\bar{e}\;\;\;\mbox{ and
}\;\;\;\overline{(\bar{e})}^{r}=\lambda e.$

Indeed it is straightforward to check that
$(1+A^{-1})^{-1}(e+\lambda^{-1}\bar{e})=\frac{1}{1+\lambda}(e+\bar{e})\in\mathsf{H}_{\mathbb{R}}$
and hence $\langle e+\lambda^{-1}\bar{e},\xi\rangle_{U}\in\mathbb{R}$ for all
$\xi\in\mathsf{H}_{\mathbb{R}}$, and similarly check that $\langle
e-\lambda^{-1}\bar{e},\xi\rangle_{U}\in i\mathsf{H}_{\mathbb{R}}$ for all
$\xi\in\mathsf{H}_{\mathbb{R}}$; this will prove that
$e+\lambda^{-1}\bar{e}\in\mathsf{H}_{\mathbb{R}}^{\prime}$ and
$e-\lambda^{-1}\bar{e}\in i\mathsf{H}_{\mathbb{R}}^{\prime}$ and hence
$\bar{e}^{r}=\lambda^{-1}\bar{e}$.

We fix such $\lambda,e$ and $\bar{e}$ for the rest of this section. The
notations used in what follows have been explained in Section 2. Our next main
result is Proposition 3.9, but first we need two crucial lemmas. The following
lemma is borrowed from [BMRW].

######  Lemma 3.7.

The sequences $\\{(1-q)^{n}W(\bar{e}^{n}e^{n})\\}_{n\geq 1}$ and
$\\{(1-q)^{n}W_{r}(\bar{e}^{n}e^{n})\\}_{n\geq 1}$ are bounded.

###### Proof.

We briefly sketch the proof from [BMRW, Lemma 3.8] in order to set some
notations and have some useful observations. Note that both the sequences have
the same norm (by using the antilinear involution $J_{\varphi}$). We consider
the first sequence which by Wick formula is given by

(5) $\displaystyle
W(\bar{e}^{n}e^{n})=\sum_{k,\ell=0}^{n}q_{k,\ell}^{(n)}c(\bar{e})^{k}c(e)^{\ell}c^{*}(e)^{n-k}c^{*}(\bar{e})^{n-\ell}$

where $q_{k,\ell}^{(n)}=\sum_{J_{1},J_{2}}q^{c(J,J^{c})}$ with the sum running
over all possible subsets $J_{1}\subseteq\\{1,\ldots,n\\}$ of cardinality $k$
and $J_{2}\subseteq\\{n+1,\ldots,2n\\}$ of cardinality $\ell$ such that
$J=J_{1}\cup J_{2}$. Then we have $|q_{k,\ell}^{(n)}|\leq
C_{q}^{2}|q|^{(n-k)\ell}$ and

$\displaystyle(1-q)^{n}\sum_{k,\ell=0}^{n}|q_{k,\ell}^{(n)}|\;\|c(\bar{e})^{k}\|\;\|c(e)^{\ell}\|\;\|c^{*}(e)^{n-k}\|\;\|c^{*}(\bar{e})^{n-\ell}\|\leq
C_{q}^{6}\sum_{k,\ell=0}^{n}|q|^{(n-k)\ell}\lambda^{\frac{k-\ell}{2}}.$

The important observation is that
$\sum_{k,\ell=0}^{n}|q|^{(n-k)\ell}\lambda^{\frac{k-\ell}{2}}\leq\frac{1}{(1-|q|)(1-\lambda^{\frac{1}{2}})}+\frac{(n+1)|q|^{n}\lambda^{-\frac{1}{2}}}{1-|q|^{n}\lambda^{-\frac{1}{2}}}$
as $n\to\infty$, which means that there is a constant $B(q,\lambda)$ depending
on $q$ and $\lambda$ such that

$\sum_{k,\ell=0}^{n}|q|^{(n-k)\ell}\lambda^{\frac{k-\ell}{2}}\leq
B(q,\lambda)$

for all $n\geq 1$. This observation will be useful for us later as well (also
see Lemma 3.8 below). ∎

######  Lemma 3.8.

For any $q\in(-1,1)$ and $\lambda\in(0,1)$, the sequence

$s_{n}=\frac{1}{n}\sum_{m=1}^{n}\sum_{\ell=1}^{m}\sum_{k=0}^{m-1}q^{(m-k)\ell}\lambda^{\frac{k-\ell}{2}}$

converges to $0$ as $n\to\infty$.

###### Proof.

Assume that $q\geq 0$. Write $s_{n}$ as follows:

(6) $\displaystyle s_{n}$ $\displaystyle=\frac{1}{n}\sum_{m=1}^{n}\sum_{0\leq
k<\ell\leq
m}q^{(m-k)\ell}\lambda^{\frac{k-\ell}{2}}+\frac{1}{n}\sum_{m=1}^{n}\sum_{1\leq\ell\leq
k\leq m-1}q^{(m-k)\ell}\lambda^{\frac{k-\ell}{2}}$ (7)
$\displaystyle=\frac{1}{n}\sum_{m=1}^{n}\sum_{j=-m}^{-1}\sum_{k=0}^{m+j}q^{(m-k)(k-j)}\lambda^{\frac{j}{2}}+\frac{1}{n}\sum_{m=1}^{n}\sum_{j=0}^{m-1}\sum_{k=j+1}^{m-1}q^{(m-k)(k-j)}\lambda^{\frac{j}{2}}$

where we get the last equality by putting $j=k-\ell$. Now we estimate the
first quantity in (7), where we follow the arguments as in the proof of [BMRW,
Lemma 3.9]:

$\displaystyle\frac{1}{n}\sum_{m=1}^{n}\sum_{j=-m}^{-1}\sum_{k=0}^{m+j}q^{(m-k)(k-j)}\lambda^{\frac{j}{2}}$
$\displaystyle=\frac{1}{n}\sum_{m=1}^{n}\sum_{j=1}^{m}\lambda^{-\frac{j}{2}}\sum_{k=0}^{m-j}q^{(m-k)(k+j)}\leq\frac{1}{n}\sum_{m=1}^{n}\sum_{j=1}^{m}\lambda^{-\frac{j}{2}}(m-j+1)q^{mj}.$

Choose $n_{0}$ large enough such that
$q^{n_{0}}\lambda^{-\frac{1}{2}}\leq\frac{1}{2}$; hence for $m\geq n_{0}$ we
have $q^{m}\lambda^{-\frac{1}{2}}\leq\frac{1}{2}q^{m-n_{0}}$ so that the above
sum further yields

$\displaystyle\frac{1}{n}\sum_{m=1}^{n}\sum_{j=-m}^{-1}\sum_{k=0}^{m+j}q^{(m-k)(k-j)}\lambda^{\frac{j}{2}}$
$\displaystyle\leq\frac{1}{n}\sum_{m=1}^{n_{0}}\sum_{j=1}^{m}\lambda^{-\frac{j}{2}}(m-j+1)q^{mj}+\frac{1}{n}\sum_{m=n_{0}+1}^{n}\sum_{j=1}^{m}(m+1)\left(q^{m}\lambda^{-\frac{1}{2}}\right)^{j}$
$\displaystyle\leq\frac{C(q,\lambda,n_{0})}{n}+\frac{1}{n}\sum_{m=n_{0}+1}^{n}(m+1)\sum_{j=1}^{m}\left(\frac{q^{m-n_{0}}}{2}\right)^{j}$
$\displaystyle\leq\frac{C(q,\lambda,n_{0})}{n}+\frac{1}{n}\sum_{m=1}^{n}(m+1)\frac{q^{m-n_{0}}}{2-q^{m-n_{0}}}$
$\displaystyle\to 0\mbox{ as }n\to\infty,$

where $C(q,\lambda,n_{0})$ is a constant independent from $n$, and where we
have also used the fact that the sequence
$\left\\{(m+1)\frac{q^{m}}{2q^{n_{0}}-q^{m}}\right\\}_{m\geq 1}$ converges to
$0$ as $m\to\infty$ and hence its Cesaro sum also converges to $0$. On the
other hand, the second quantity in (7) is estimated as follows:

$\displaystyle\frac{1}{n}\sum_{m=1}^{n}\sum_{j=0}^{m-1}\sum_{k=j+1}^{m-1}q^{(m-k)(k-j)}\lambda^{\frac{j}{2}}$
$\displaystyle=\frac{1}{n}\sum_{j=0}^{n-1}\lambda^{\frac{j}{2}}\sum_{k=j+1}^{n-1}\sum_{m=k+1}^{n}q^{(m-k)(k-j)}$
$\displaystyle=\frac{1}{n}\sum_{j=0}^{n-1}\lambda^{\frac{j}{2}}\sum_{k=j+1}^{n-1}q^{-k(k-j)}\sum_{m=k+1}^{n}q^{m(k-j)}$
$\displaystyle=\frac{1}{n}\sum_{j=0}^{n-1}\lambda^{\frac{j}{2}}\sum_{k=j+1}^{n-1}q^{-k(k-j)}\left(\frac{q^{(k+1)(k-j)}-q^{(n+1)(k-j)}}{1-q^{k-j}}\right)$
$\displaystyle\leq\frac{1}{n}\sum_{j=0}^{n-1}\lambda^{\frac{j}{2}}\sum_{k=j+1}^{n-1}\frac{q^{k-j}}{1-q^{k-j}}$
$\displaystyle\leq\frac{1}{1-\lambda^{\frac{1}{2}}}\left(\frac{1}{n}\sum_{k=1}^{n}\frac{q^{k}}{1-q^{k}}\right)$
$\displaystyle\to 0\mbox{ as }n\to\infty$

as the sequence $\\{\frac{q^{k}}{1-q^{k}}\\}_{k\geq 1}$ converges to $0$ as
$k\to\infty$, and so does its Cesaro sum. Thus we have proved that $s_{n}\to
0$ as $n\to\infty$. ∎

######  Proposition 3.9.

Let $e,\bar{e}$ and $\lambda$ be as above. The sequence

$S_{n}=\frac{1}{n}\sum_{m=1}^{n}(1-q)^{2m}W(\bar{e}^{m}e^{m})c(\bar{e})^{m}c(e)^{m}$

is convergent in norm to the operator $S$ given by

(8)
$S=\sum_{k=0}^{\infty}\frac{d_{\infty}}{d_{k}}(1-q)^{k}\;\lambda^{\frac{k}{2}}c(\bar{e})^{k}Tc(e)^{k}$

where $T$ is the norm limit of the operator
$(1-q)^{n}\lambda^{\frac{n}{2}}c^{*}(e)^{n}c(e)^{n}$ which is positive and
invertible. Moreover, $S$ is invertible if
$\lambda<\left(1+C_{q}^{4}\right)^{-2}$.

###### Proof.

First check that the series defined in (8) above is norm convergent. Indeed by
using Lemma 2.5 we have

$\displaystyle\|S\|$
$\displaystyle\leq\sum_{k=0}^{\infty}\;\frac{d_{\infty}}{d_{k}}(1-q)^{k}\;\lambda^{\frac{k}{2}}\|c(\bar{e})^{k}\|\;\|T\|\;\|c(e)^{k}\|$
$\displaystyle\leq
d_{\infty}C_{q}\;\|T\|\sum_{k=0}^{\infty}(1-q)^{k}\;\lambda^{\frac{k}{2}}\left(C_{q}(1-q)^{-\frac{k}{2}}\lambda^{\frac{k}{4}}\right)\left(C_{q}(1-q)^{-\frac{k}{2}}\lambda^{-\frac{k}{4}}\right)$
$\displaystyle=d_{\infty}C_{q}^{3}\;\|T\|\sum_{k=0}^{\infty}\lambda^{\frac{k}{2}}$
$\displaystyle<\infty.$

To prove the convergence of $S_{n}$, write using (5)

$\displaystyle
S_{n}=\frac{1}{n}\sum_{m=1}^{n}(1-q)^{2m}\sum_{k,\ell=0}^{m}q_{k,\ell}^{(m)}c(\bar{e})^{k}c(e)^{\ell}c^{*}(e)^{m-k}c^{*}(\bar{e})^{m-\ell}c(\bar{e})^{m}c(e)^{m}$

where $|q^{(m)}_{k,\ell}|\leq C_{q}^{2}|q|^{(m-k)\ell}$. Let us understand the
terms $c^{*}(\bar{e})^{m-\ell}c(\bar{e})^{m}c(e)^{m}$ in the sum above. Use
Corollary 3.2 to write

$\displaystyle
c^{*}(\bar{e})^{m-\ell}c(\bar{e})^{m}c(e)^{m}=\frac{[m]_{q}!}{[\ell]_{q}!}\lambda^{\frac{m-\ell}{2}}c(\bar{e})^{\ell}c(e)^{m}+q^{m}X_{m,\ell}$

where $X_{m,\ell}$ is an operator bounded by
$(1-q)^{\frac{(-3m+\ell)}{2}}\lambda^{\frac{m-\ell}{4}}A_{q}$ for the constant
$A_{q}:=\frac{C_{q}^{6}}{1-|q|}$ depending only on $q$. In particular, we get

$\displaystyle S_{n}$
$\displaystyle=\frac{1}{n}\sum_{m=1}^{n}(1-q)^{2m}\sum_{k,\ell=0}^{m}q_{k,\ell}^{(m)}\left(\frac{[m]_{q}!}{[\ell]_{q}!}\lambda^{\frac{m-\ell}{2}}\right)c(\bar{e})^{k}c(e)^{\ell}c^{*}(e)^{m-k}c(\bar{e})^{\ell}c(e)^{m}$
$\displaystyle\quad\quad\quad\quad+\frac{1}{n}\sum_{m=1}^{n}q^{m}(1-q)^{2m}\sum_{k,\ell=0}^{m}q_{k,\ell}^{(m)}c(\bar{e})^{k}c(e)^{\ell}c^{*}(e)^{m-k}X_{m,\ell}.$

We now calculate the upper bound for the last sum above:

$\displaystyle\frac{1}{n}\sum_{m=1}^{n}|q|^{m}(1-q)^{2m}$
$\displaystyle\left\|\sum_{k,\ell=0}^{m}q_{k,\ell}^{(m)}c(\bar{e})^{k}c(e)^{\ell}c^{*}(e)^{m-k}X_{m,\ell}\right\|$
$\displaystyle\leq\frac{1}{n}\sum_{m=1}^{n}|q|^{m}(1-q)^{2m}C_{q}^{5}A_{q}\sum_{k,\ell=0}^{m}|q|^{(m-k)\ell}\lambda^{\frac{k-\ell-(m-k)}{4}}(1-q)^{-\frac{(k+\ell+m-k)}{2}}(1-q)^{\frac{(-3m+\ell)}{2}}\lambda^{\frac{m-\ell}{2}}$
$\displaystyle=\frac{1}{n}C_{q}^{5}A_{q}\sum_{m=1}^{n}|q|^{m}\sum_{k,\ell=0}^{m}|q|^{(m-k)\ell}\lambda^{\frac{k-\ell}{2}}$
$\displaystyle\leq\frac{1}{n}\frac{C_{q}^{5}A_{q}B(q,\lambda)}{1-|q|}$

where $B(q,\lambda)$ is an upper bound for the sum
$\sum_{k,\ell=0}^{m}|q|^{(m-k)\ell}\lambda^{\frac{k-\ell}{2}}$ such that
$B(q,\lambda)$ depends only on $\lambda$ and $q$, and is independent of $m$
(see the proof of Lemma 3.7). Thus we have

$\displaystyle S_{n}$
$\displaystyle=\frac{1}{n}\sum_{m=1}^{n}(1-q)^{2m}\sum_{k,\ell=0}^{m}q_{k,\ell}^{(m)}\left(\frac{[m]_{q}!}{[\ell]_{q}!}\lambda^{\frac{m-\ell}{2}}\right)c(\bar{e})^{k}c(e)^{\ell}c^{*}(e)^{m-k}c(\bar{e})^{\ell}c(e)^{m}+O\left(\frac{1}{n}\right)$
$\displaystyle=X_{n}+Y_{n}+Z_{n}+O\left(\frac{1}{n}\right)$

where the above sum is divided over $\ell=0,k=m$ and the rest i.e. we write

$\displaystyle
X_{n}=\frac{1}{n}\sum_{m=1}^{n}(1-q)^{2m}\sum_{k=0}^{m}q_{k,0}^{(m)}[m]_{q}!\lambda^{\frac{m}{2}}c(\bar{e})^{k}c^{*}(e)^{m-k}c(e)^{m}$
$\displaystyle
Y_{n}=\frac{1}{n}\sum_{m=1}^{n}(1-q)^{2m}\sum_{\ell=0}^{m}q_{m,\ell}^{(m)}\frac{[m]_{q}!}{[\ell]_{q}!}\lambda^{\frac{m-\ell}{2}}c(\bar{e})^{m}c(e)^{\ell}c(\bar{e})^{\ell}c(e)^{m}$
$\displaystyle
Z_{n}=\frac{1}{n}\sum_{m=1}^{n}(1-q)^{2m}\sum_{\ell=1}^{m}\sum_{k=0}^{m-1}q_{k,\ell}^{(m)}\left(\frac{[m]_{q}!}{[\ell]_{q}!}\lambda^{\frac{m-\ell}{2}}\right)c(\bar{e})^{k}c(e)^{\ell}c^{*}(e)^{m-k}c(\bar{e})^{\ell}c(e)^{m}.$

But the norm of $Z_{n}$ can be estimated using Lemma 2.4 as follows:

$\displaystyle\|Z_{n}\|$
$\displaystyle\leq\frac{1}{n}\sum_{m=1}^{n}(1-q)^{2m}\sum_{\ell=1}^{m}\sum_{k=0}^{m-1}C_{q}^{2}|q|^{(m-k)\ell}\left(\frac{d_{m}}{d_{\ell}}\lambda^{\frac{m-\ell}{2}}\right)C_{q}^{5}(1-q)^{-2m}\lambda^{\frac{k-m}{2}}$
$\displaystyle\leq\frac{C_{q}^{9}}{n}\sum_{m=1}^{n}\sum_{\ell=1}^{n}\sum_{k=0}^{m-1}|q|^{(m-k)\ell}\lambda^{\frac{k-\ell}{2}}$
$\displaystyle=C_{q}^{9}s_{n}$

where $s_{n}$ is the sequence as in Lemma 3.8; hence $Z_{n}\to 0$ as
$n\to\infty$ by the same Lemma. We now show that $Y_{n}$ converges to 0 as
$n\to\infty$. We follow the same trick as in the proof of Proposition 3.4.
Calculate

$\displaystyle Y_{n}^{*}Y_{n}$
$\displaystyle=\frac{1}{n^{2}}\sum_{m,m^{\prime}=1}^{n}(1-q)^{2m+2m^{\prime}}$
$\displaystyle\cdot\sum_{\ell=0}^{m}\sum_{\ell^{\prime}=0}^{m^{\prime}}q_{m,\ell}^{(m)}q_{m^{\prime},\ell^{\prime}}^{(m^{\prime})}\frac{[m]_{q}![m^{\prime}]_{q}}{[\ell]_{q}![\ell^{\prime}]_{q}!}\lambda^{\frac{m+m^{\prime}-\ell-\ell^{\prime}}{2}}c^{*}(e)^{m}c^{*}(\bar{e})^{\ell}c^{*}(e)^{\ell}c^{*}(\bar{e})^{m}c(\bar{e})^{m^{\prime}}c(e)^{\ell^{\prime}}c(\bar{e})^{\ell^{\prime}}c(e)^{m^{\prime}}$
$\displaystyle=W^{n}_{m=m^{\prime}}+W^{n}_{m>m^{\prime}}+W^{n}_{m<m^{\prime}}$

where we divide the sum over three cases: $m=m^{\prime},m<m^{\prime}$ and
$m>m^{\prime}$. Over $m=m^{\prime}$, we again use some similar estimates to
calculate the following:

$\displaystyle\|W_{m=m^{\prime}}^{n}\|$
$\displaystyle=\frac{1}{n^{2}}\left\|\sum_{m=1}^{n}(1-q)^{4m}\sum_{\ell,\ell^{\prime}=0}^{m}q_{m,\ell}^{(m)}q_{m,\ell^{\prime}}^{(m)}\frac{([m]_{q}!)^{2}}{[\ell]_{q}![\ell^{\prime}]_{q}!}\lambda^{\frac{2m-\ell-\ell^{\prime}}{2}}c^{*}(e)^{m}c^{*}(\bar{e})^{\ell}c^{*}(e)^{\ell}c^{*}(\bar{e})^{m}c(\bar{e})^{m}c(e)^{\ell^{\prime}}c(\bar{e})^{\ell^{\prime}}c(e)^{m}\right\|$
$\displaystyle\leq\frac{C_{q}^{12}}{n^{2}}\sum_{m=1}^{n}(1-q)^{4m}\sum_{\ell,\ell^{\prime}=0}^{m}\frac{d_{m}^{2}(1-q)^{-2m}}{d_{\ell}d_{\ell^{\prime}}(1-q)^{-\ell-\ell^{\prime}}}\lambda^{\frac{2m-\ell-\ell^{\prime}}{2}}(1-q)^{-2m-\ell-\ell^{\prime}}\quad\quad\quad\quad(\text{since
}q_{m,\ell}^{(m)}\leq C_{q}^{2})$
$\displaystyle\leq\frac{C_{q}^{16}}{n^{2}}\sum_{m=1}^{n}\sum_{\ell,\ell^{\prime}=0}^{m}\lambda^{\frac{2m-\ell-\ell^{\prime}}{2}}$
$\displaystyle\leq\frac{1}{n}\frac{C_{q}^{16}}{(1-\sqrt{\lambda})^{2}}.$

Next consider the sum over $m>m^{\prime}$ (which has the same norm as the part
$m<m^{\prime}$ by considering the adjoint and the symmetric nature of the
sum). For $m>m^{\prime}$, use Corollary 3.3 to write

$c^{*}(\bar{e})^{m}c(\bar{e})^{m^{\prime}}c(e)^{\ell^{\prime}}=q^{\ell^{\prime}}X_{m,m^{\prime},\ell^{\prime}}$

where $X_{m,m^{\prime},\ell^{\prime}}$ is an operator bounded by
$A_{q}(1-q)^{-\frac{(m+m^{\prime}+\ell^{\prime})}{2}}\lambda^{\frac{m+m^{\prime}-\ell^{\prime}}{4}}$
for some constant $A_{q}$. Therefore, the sum over $m>m^{\prime}$ is estimated
as

$\displaystyle\|W^{n}_{m>m^{\prime}}\|$
$\displaystyle=\frac{1}{n^{2}}\left\|\sum_{n\geq m>m^{\prime}\geq
1}^{n}(1-q)^{2m+2m^{\prime}}\sum_{\ell=0}^{m}\sum_{\ell^{\prime}=0}^{m^{\prime}}q_{m,\ell}^{(m)}q_{\ell^{\prime},m^{\prime}}^{(m^{\prime})}\frac{[m]_{q}![m^{\prime}]_{q}}{[\ell]_{q}![\ell^{\prime}]_{q}!}\lambda^{\frac{m+m^{\prime}-\ell-\ell^{\prime}}{2}}c^{*}(e)^{m}c^{*}(\bar{e})^{\ell}c^{*}(e)^{\ell}\left(q^{\ell^{\prime}}X_{m,m^{\prime},\ell^{\prime}}\right)c(\bar{e})^{\ell^{\prime}}c(e)^{m^{\prime}}\right\|$
$\displaystyle\leq\frac{C_{q}^{9}A_{q}}{n^{2}}\sum_{n\geq m>m^{\prime}\geq
1}(1-q)^{2m+2m^{\prime}}\sum_{\ell=0}^{m}\sum_{\ell^{\prime}=0}^{m^{\prime}}|q|^{\ell^{\prime}}\frac{d_{m}d_{m^{\prime}}(1-q)^{-m-m^{\prime}}}{d_{\ell}d_{\ell^{\prime}}(1-q)^{-\ell-\ell^{\prime}}}\lambda^{\frac{m+m^{\prime}-\ell-\ell^{\prime}}{2}}(1-q)^{-m-m^{\prime}-\ell-\ell^{\prime}}\quad\quad\quad(\text{as
}q_{m,\ell}^{(m)}\leq C_{q}^{2})$
$\displaystyle\leq\frac{C_{q}^{13}A_{q}}{n^{2}}\sum_{m,m^{\prime}=1}^{n}\sum_{\ell=0}^{m}\sum_{\ell^{\prime}=0}^{m^{\prime}}|q|^{\ell^{\prime}}\lambda^{\frac{m+m^{\prime}-\ell-\ell^{\prime}}{2}}$
$\displaystyle\leq\frac{C_{q}^{13}A_{q}}{n^{2}}\frac{n}{1-\sqrt{\lambda}}\sum_{m^{\prime}=1}^{n}\sum_{\ell^{\prime}=0}^{m^{\prime}}|q|^{\ell^{\prime}}\lambda^{\frac{m^{\prime}-\ell^{\prime}}{2}}$
$\displaystyle\leq\frac{C_{q}^{13}A_{q}}{n(1-\sqrt{\lambda})}\sum_{m^{\prime}=1}^{n}\left(\sum_{\ell^{\prime}=0}^{\frac{m}{2}}|q|^{\ell^{\prime}}\lambda^{\frac{m^{\prime}-\ell^{\prime}}{2}}+\sum_{\ell^{\prime}=\frac{m^{\prime}}{2}}^{m}|q|^{\ell^{\prime}}\lambda^{\frac{m^{\prime}-\ell^{\prime}}{2}}\right)$
$\displaystyle\leq\frac{C_{q}^{13}A_{q}}{n(1-\sqrt{\lambda})}\left(\sum_{m^{\prime}=1}^{n}\lambda^{\frac{m^{\prime}}{2}}\sum_{\ell^{\prime}=0}^{\frac{m}{2}}|q|^{\ell^{\prime}}+\sum_{m^{\prime}=1}^{n}|q|^{\frac{m^{\prime}}{2}}\sum_{\ell^{\prime}=\frac{m^{\prime}}{2}}^{m}\lambda^{\frac{m^{\prime}-\ell^{\prime}}{2}}\right)$
$\displaystyle\leq\frac{C_{q}^{13}A_{q}}{n(1-\sqrt{\lambda})}\left(\frac{1}{(1-\sqrt{\lambda})(1-|q|)}+\frac{1}{(1-\sqrt{|q|})(1-\sqrt{\lambda})}\right)$
$\displaystyle\leq\frac{1}{n}\frac{2C_{q}^{13}A_{q}}{(1-\sqrt{\lambda})^{2}(1-\sqrt{|q|})}.$

A similar upper bound can be given for $\|W_{m<m^{\prime}}^{n}\|$. Thus we
have shown that

$\displaystyle\|Y_{n}^{*}Y_{n}\|\leq\frac{1}{n}\left(\frac{C_{q}^{16}}{(1-\sqrt{\lambda})^{2}}+\frac{4C_{q}^{13}A_{q}}{(1-\sqrt{\lambda})^{2}(1-\sqrt{|q|})}\right)$

and hence $\lim_{n\to\infty}Y_{n}=0$ in norm. Finally, we conclude that

$\displaystyle\lim_{n\to\infty}S_{n}$
$\displaystyle=\lim_{n\to\infty}X_{n}=\lim_{n\to\infty}\frac{1}{n}\sum_{m=1}^{n}(1-q)^{2m}\sum_{k=0}^{m}q_{k,0}^{(m)}[m]_{q}!\lambda^{\frac{m}{2}}c(\bar{e})^{k}c^{*}(e)^{m-k}c(e)^{m}$
$\displaystyle=\lim_{n\to\infty}\frac{1}{n}\sum_{m=1}^{n}\sum_{k=0}^{m}q_{k,0}^{(m)}d_{m}(1-q)^{k}\lambda^{\frac{k}{2}}c(\bar{e})^{k}T_{m-k}c(e)^{k}$

where $T_{p}=(1-q)^{p}\lambda^{\frac{p}{2}}c^{*}(e)^{p}c(e)^{p}$ for any
$p\geq 1$. Let $T=\lim_{p\to\infty}T_{p}$ in norm, by using Lemma 3.6. Also
for $0\leq k\leq m$, it is straightforward to verify that
$q_{k,0}^{(m)}=\begin{pmatrix}m\\\ k\end{pmatrix}_{q}$ (see for example, [BKS,
Corollary 2.8]); so that for each $k\geq 0$, we have

$\lim_{m\to\infty}q_{k,0}^{(m)}=\lim_{m\to\infty}\begin{pmatrix}m\\\
k\end{pmatrix}_{q}=\lim_{m\to\infty}{\frac{d_{m}}{d_{k}d_{m-k}}}=\frac{1}{d_{k}}.$

Finally, since $\lim_{m\to\infty}d_{m}=d_{\infty}$ and
$\lim_{m\to\infty}T_{m-k}=T$ for all fixed $k\geq 1$, we use dominated
convergence theorem to conclude that

$\displaystyle\lim_{m\to\infty}\sum_{k=0}^{m}q_{k,0}^{(m)}d_{m}(1-q)^{k}\lambda^{\frac{k}{2}}c(\bar{e})^{k}T_{m-k}c(e)^{k}=\sum_{k=0}^{\infty}\frac{d_{\infty}}{d_{k}}(1-q)^{k}\lambda^{\frac{k}{2}}c(\bar{e})^{k}Tc(e)^{k}.$

In particular, the Cesaro sum of the above sequence has the same limit; hence
we get

$\displaystyle\lim_{n\to\infty}S_{n}=\lim_{n\to\infty}\frac{1}{n}\sum_{m=1}^{n}\sum_{k=0}^{m}q_{k,0}^{(m)}d_{m}(1-q)^{k}\lambda^{\frac{k}{2}}c(\bar{e})^{k}T_{m-k}c(e)^{k}=\sum_{k=0}^{\infty}\frac{d_{\infty}}{d_{k}}(1-q)^{k}\lambda^{\frac{k}{2}}c(\bar{e})^{k}Tc(e)^{k}=S.$

Finally, to prove the invertibility of $S$, we note that

$\displaystyle
d_{\infty}^{-1}T^{-1}S-I=\sum_{k=1}^{\infty}\frac{1}{d_{k}}(1-q)^{k}\lambda^{\frac{k}{2}}T^{-1}c(\bar{e})^{k}Tc(e)^{k}$

so that

$\displaystyle\|d_{\infty}^{-1}T^{-1}S-I\|\leq
C_{q}^{2}\sum_{k=1}^{\infty}\frac{1}{d_{k}}\lambda^{\frac{k}{2}}\|T^{-1}\|\;\|T\|$
$\displaystyle\leq
C_{q}^{3}\|T^{-1}\|\;\|T\|\frac{\lambda^{\frac{1}{2}}}{{1-\lambda^{\frac{1}{2}}}}\leq
C_{q}^{4}\frac{\lambda^{\frac{1}{2}}}{{1-\lambda^{\frac{1}{2}}}}$

where we have used the fact that
$\|T^{-1}\|\;\|T\|\leq\frac{1}{d_{\infty}}\leq C_{q}$ from Lemma 3.6. In
particular, $d_{\infty}^{-1}T^{-1}S$ is invertible if
$C_{q}^{4}\frac{\lambda^{\frac{1}{2}}}{{1-\lambda^{\frac{1}{2}}}}<1$, which is
satisfied by the given value of $\lambda$ in the hypothesis. ∎

######  Remark 3.10.

1\. Since $\|d_{\infty}^{-1}T^{-1}S-1\|\leq
C_{q}^{4}\frac{\sqrt{\lambda}}{1-\sqrt{\lambda}}$, we have that

$\displaystyle\|d_{\infty}TS^{-1}\|=\|(I-(I-d_{\infty}^{-1}T^{-1}S))^{-1}\|\leq\frac{1}{1-\|I-d_{\infty}^{-1}T^{-1}S\|}\leq\frac{1}{1-C_{q}^{4}\frac{\sqrt{\lambda}}{1-\sqrt{\lambda}}}=\frac{1-\sqrt{\lambda}}{1-(1+C_{q}^{4})\sqrt{\lambda}}$

and since $T^{-1}\leq 2d_{\infty}^{-1}\leq 2C_{q}$ (Lemma 3.6), we must have

$\displaystyle\|S^{-1}\|\leq\|d_{\infty}^{-1}T^{-1}\|\;\|d_{\infty}TS^{-1}\|\leq\frac{2C_{q}^{2}(1-\sqrt{\lambda})}{1-(1+C_{q}^{4})\sqrt{\lambda}}.$

2\. Although we have shown that the Cesaro sum of the sequence
$\\{W(\bar{e}^{m}e^{m})c(\bar{e})^{m}c(e)^{m}\\}_{m\geq 1}$ converges, the
sequence itself does not converge. For example, this can be checked for $q=0$
where the main obstruction is that the sequence
$\\{\sum_{\ell=0}^{m}c(\bar{e})^{m}c(e)^{\ell}c^{*}(\bar{e})^{m-\ell}c(\bar{e})^{m}c(e)^{m}\\}_{m\geq
1}$ is not convergent. This is one very important instance for us to consider
Cesaro sum of such bounded sequence.

We now prove some convergence results involving both the left and right
creation/annihilation operators. For the purpose, we introduce the operator
$Q$ defined by

$\displaystyle Q=\sum_{n=0}^{\infty}q^{n}P_{n}$

where $P_{n}$ denotes the projection in $B(\mathcal{F}_{q}(\mathsf{H}_{U}))$
onto the $n$th particle space $\mathsf{H}_{U}^{\otimes_{q}^{n}}$, and the sum
above converges in norm. Note that for any
$f\in\mathsf{H}_{U}^{\otimes_{q}^{n}}$, we have $Qf=q^{n}f$. The following is
an easy verification

(9) $\displaystyle
r^{*}(\xi)c(\eta)=\langle\xi,\eta\rangle_{U}Q+c(\eta)r^{*}(\xi)$

for any $\xi,\eta\in\mathsf{H}_{U}$. In particular, $r^{*}(\xi)$ and $c(\eta)$
commute if $\xi$ and $\eta$ are orthogonal. Moreover, we have

(10) $\displaystyle Qc(\xi)=qc(\xi)Q,\quad Qr(\xi)=qr(\xi)Q$

######  Lemma 3.11.

For any $\xi,\eta\in\mathsf{H}_{U}$ and $m,n\in\mathbb{N}$, we have the
following:

$\displaystyle
r^{*}(\xi)^{m}c(\eta)^{n}=\sum_{k=0}^{\min\\{m,n\\}}\begin{pmatrix}n\\\
k\end{pmatrix}_{q}\frac{[m]_{q}!}{[m-k]_{q}!}\langle\xi,\eta\rangle_{U}^{k}c(\eta)^{n-k}Q^{k}r^{*}(\xi)^{m-k}.$

###### Proof.

We may and we will assume that $\langle\xi,\eta\rangle_{U}\neq 0$, otherwise
there is nothing to prove since $r^{*}(\xi)c(\eta)=c(\eta)r^{*}(\xi)$ whenever
$\langle\xi,\eta\rangle_{U}=0$. By replacing $\xi$ with
$\frac{\xi}{\langle\xi,\eta\rangle_{U}}$ (or $\eta$ with
$\frac{\eta}{\langle\xi,\eta\rangle_{U}})$, we will further assume that
$\langle\xi,\eta\rangle_{U}=1$.

First, we prove the expression for $r^{*}(\xi)c(\eta)^{n}$ by induction on
$n$. Indeed for $n=1$, the expression follows from (9). Assuming this to be
true for $n-1$, we again use (9) and (10) to obtain

(11) $\displaystyle r^{*}(\xi)c(\eta)^{n}$
$\displaystyle=\left(r^{*}(\xi)c(\eta)\right)c(\eta)^{n-1}=Qc(\eta)^{n-1}+c(\eta)(r^{*}(\xi)c(\eta)^{n-1})$
(12)
$\displaystyle=q^{n-1}c(\eta)^{n-1}Q+c(\eta)\left(c(\eta)^{n-1}r^{*}(\xi)+[n-1]_{q}c(\eta)^{n-2}Q\right)$
(13) $\displaystyle=c(\eta)^{n}r^{*}(\xi)+[n]_{q}c(\eta)^{n-1}Q$

as required. Now we prove the result for general $m$ and $n$ such that $n\geq
m$. The case $n<m$ follows similarly, which we will not expand. So now we fix
$n$. The proof proceeds by induction on $m$. For $m=1$, the expression is
proved above. By assuming the expression to be true for $m-1$, we calculate

$\displaystyle r^{*}(\xi)^{m}c(\eta)^{n}$
$\displaystyle=r^{*}(\xi)(r^{*}(\xi)^{m-1}c(\eta)^{n})$
$\displaystyle=r^{*}(\xi)\sum_{k=0}^{m-1}\begin{pmatrix}n\\\
k\end{pmatrix}_{q}\frac{[m-1]_{q}!}{[m-1-k]_{q}!}c(\eta)^{n-k}Q^{k}r^{*}(\xi)^{m-1-k}$
$\displaystyle=\sum_{k=0}^{m-1}\begin{pmatrix}n\\\
k\end{pmatrix}_{q}\frac{[m-1]_{q}!}{[m-1-k]_{q}!}\left(c(\eta)^{n-k}r^{*}(\xi)+[n-k]_{q}c(\eta)^{n-k-1}Q\right)Q^{k}r^{*}(\xi)^{m-1-k}$
$\displaystyle\hskip 216.81pt(\text{using the expression for
}r^{*}(\xi)c(\eta)^{n-k}\text{ from }\eqref{eq:formula for r*(xi)c(eta)^n})$
$\displaystyle=\sum_{k=0}^{m-1}\begin{pmatrix}n\\\
k\end{pmatrix}_{q}\frac{[m-1]_{q}!}{[m-1-k]_{q}!}q^{k}c(\eta)^{n-k}Q^{k}r^{*}(\xi)^{m-k}$
$\displaystyle\quad\quad+\sum_{k=0}^{m-1}\begin{pmatrix}n\\\
k\end{pmatrix}_{q}\frac{[m-1]_{q}!}{[m-1-k]_{q}!}[n-k]_{q}c(\eta)^{n-k-1}Q^{k+1}r^{*}(\xi)^{m-1-k}$
$\displaystyle=\sum_{k=0}^{m-1}\begin{pmatrix}n\\\
k\end{pmatrix}_{q}\frac{[m-1]_{q}!}{[m-1-k]_{q}!}q^{k}c(\eta)^{n-k}Q^{k}r^{*}(\xi)^{m-k}$
$\displaystyle\quad\quad+\sum_{k=1}^{m}\begin{pmatrix}n\\\
k-1\end{pmatrix}_{q}\frac{[m-1]_{q}!}{[m-k]_{q}!}[n-k+1]_{q}c(\eta)^{n-k}Q^{k}r^{*}(\xi)^{m-k}$
$\displaystyle\hskip 216.81pt(\text{using
}r^{*}(\xi)Q^{k}=q^{k}Q^{k}r^{*}(\xi)\text{ from }\eqref{eq:commutation
between Q and c(xi) and r(xi)})$
$\displaystyle=c(\eta)^{n}r^{*}(\xi)^{m}+\sum_{k=1}^{m-1}\left(\begin{pmatrix}n\\\
k\end{pmatrix}_{q}\frac{[m-1]_{q}!}{[m-1-k]_{q}!}q^{k}+\begin{pmatrix}n\\\
k-1\end{pmatrix}_{q}\frac{[m-1]_{q}!}{[m-k]_{q}!}[n-k+1]_{q}\right)c(\eta)^{n-k}Q^{k}r^{*}(\xi)^{m-k}$
$\displaystyle\quad\quad\quad+\begin{pmatrix}n\\\
m-1\end{pmatrix}_{q}[m-1]_{q}![n-m+1]_{q}c(\eta)^{n-m}Q^{m}$
$\displaystyle=\sum_{k=0}^{m}\begin{pmatrix}n\\\
k\end{pmatrix}_{q}\frac{[m]_{q}!}{[m-k]_{q}!}c(\eta)^{n-k}Q^{k}r^{*}(\xi)^{m-k}$

where to get the last expression, we have used the following calculation:

$\displaystyle\begin{pmatrix}n\\\
k\end{pmatrix}_{q}\frac{[m-1]_{q}!}{[m-1-k]_{q}!}q^{k}$
$\displaystyle+\begin{pmatrix}n\\\
k-1\end{pmatrix}_{q}\frac{[m-1]_{q}!}{[m-k]_{q}!}[n-k+1]_{q}$
$\displaystyle=\frac{[n]_{q}!}{[n-k]_{q}![k-1]_{q}!}\frac{[m-1]_{q}!}{[m-1-k]_{q}!}\left(\frac{q^{k}}{[k]_{q}}+\frac{1}{[m-k]_{q}}\right)$
$\displaystyle=\frac{[n]_{q}!}{[n-k]_{q}![k-1]_{q}!}\frac{[m-1]_{q}!}{[m-1-k]_{q}!}\left(\frac{q^{k}[m-k]_{q}+[k]_{q}}{[k]_{q}[m-k]_{q}}\right)$
$\displaystyle=\frac{[n]_{q}!}{[n-k]_{q}![k-1]_{q}!}\frac{[m-1]_{q}!}{[m-1-k]_{q}!}\left(\frac{[m]_{q}}{[k]_{q}[m-k]_{q}}\right)$
$\displaystyle=\begin{pmatrix}n\\\
k\end{pmatrix}_{q}\frac{[m]_{q}!}{[m-k]_{q}!}.$

This completes the proof. ∎

The following proposition can be considered a stronger version of [BMRW, Lemma
3.11] in the sense of norm convergence result over weak convergence.

######  Proposition 3.12.

For any non-zero vectors $\xi,\eta\in\mathsf{H}_{U}$ with
$\langle\xi,\eta\rangle_{U}=0$, we have
$\lim_{n\to\infty}R_{n}=d_{\infty}^{2}P_{\Omega}$ in norm, where $R_{n}$ is
the operator

$\frac{1}{n}\sum_{m=1}^{n}(1-q)^{2m}\|\xi\|_{U}^{-2m}\|\eta\|_{U}^{-2m}r^{*}(\xi)^{m}r^{*}(\eta)^{m}c(\xi)^{m}c(\eta)^{m}$

and $P_{\Omega}$ denotes the projection onto the vacuum vector $\Omega$.

###### Proof.

Assume without loss of generality that $\xi,\eta$ are unit vectors. First of
all, note that $r^{*}(\eta)$ and $c(\xi)$ commute as
$\langle\xi,\eta\rangle_{U}=0$, so we can also write

$\displaystyle
R_{n}=\frac{1}{n}\sum_{m=1}^{n}(1-q)^{2m}r^{*}(\xi)^{m}c(\xi)^{m}r^{*}(\eta)^{m}c(\eta)^{m}.$

Use Lemma 3.11 to expand $R_{n}$ as follows:

$\displaystyle R_{n}$
$\displaystyle=\frac{1}{n}\sum_{m=1}^{n}(1-q)^{2m}\left(\sum_{k=0}^{m}\begin{pmatrix}m\\\
k\end{pmatrix}_{q}\frac{[m]_{q}!}{[m-k]_{q}!}c(\xi)^{m-k}Q^{k}r^{*}(\xi)^{m-k}\right)\left(\sum_{\ell=0}^{m}\begin{pmatrix}m\\\
\ell\end{pmatrix}_{q}\frac{[m]_{q}!}{[m-\ell]_{q}!}c(\eta)^{m-\ell}Q^{\ell}r^{*}(\eta)^{m-\ell}\right)$
$\displaystyle=\frac{1}{n}\sum_{m=1}^{n}(1-q)^{2m}\sum_{k,\ell=0}^{m}\begin{pmatrix}m\\\
k\end{pmatrix}_{q}\begin{pmatrix}m\\\
\ell\end{pmatrix}_{q}\frac{[m]_{q}!^{2}}{[m-k]_{q}![m-\ell]_{q}!}c(\xi)^{m-k}Q^{k}c(\eta)^{m-\ell}r^{*}(\xi)^{m-k}Q^{\ell}r^{*}(\eta)^{m-\ell}$
$\displaystyle\hskip 289.07999pt(\text{since
}r^{*}(\xi)c(\eta)=c(\eta)r^{*}(\xi))$
$\displaystyle=\frac{1}{n}\sum_{m=1}^{n}(1-q)^{2m}\sum_{k,\ell=0}^{m}q^{k(m-\ell)+\ell(m-k)}\begin{pmatrix}m\\\
k\end{pmatrix}_{q}\begin{pmatrix}m\\\
\ell\end{pmatrix}_{q}\frac{[m]_{q}!^{2}}{[m-k]_{q}![m-\ell]_{q}!}c(\xi)^{m-k}c(\eta)^{m-\ell}Q^{k+\ell}r^{*}(\xi)^{m-k}r^{*}(\eta)^{m-\ell}$
$\displaystyle\hskip 72.26999pt(\text{since
}Q^{k}c(\eta)^{m-\ell}=q^{k(m-\ell)}c(\eta)^{m-\ell}Q^{k}\text{ and
}r^{*}(\xi)^{m-k}Q^{\ell}=q^{(m-k)\ell}Q^{\ell}r^{*}(\xi)^{m-k}\text{ from
}\eqref{eq:commutation between Q and c(xi) and r(xi)})$
$\displaystyle=\frac{1}{n}\sum_{m=1}^{n}d_{m}^{2}Q^{2m}+\frac{1}{n}\sum_{m=1}^{n}(1-q)^{2m}c(\xi)^{m}c(\eta)^{m}r^{*}(\xi)^{m}r^{*}(\eta)^{m}+X_{n}$

where the sum is divided over $(k,\ell=m)$, $(k,\ell=0)$ and the rest i.e.
$X_{n}$ is the operator

$\displaystyle
X_{n}=\frac{1}{n}\sum_{m=1}^{n}(1-q)^{2m}\sum_{\underset{0<k+\ell<2m}{k,\ell=0}}^{m}q^{m(k+\ell)-2k\ell}\begin{pmatrix}m\\\
k\end{pmatrix}_{q}\begin{pmatrix}m\\\
\ell\end{pmatrix}_{q}\frac{[m]_{q}!^{2}}{[m-k]_{q}![m-\ell]_{q}!}c(\xi)^{m-k}c(\eta)^{m-\ell}Q^{k+\ell}r^{*}(\xi)^{m-k}r^{*}(\eta)^{m-\ell}.$

Thus we have divided $R_{n}$ as a sum of three terms, where we calculate norm
limit of each term. To get the limit of the first term, first verify that
$Q^{2m}$ converges to $P_{\Omega}$ as $m\to\infty$, and therefore the average
$\frac{1}{n}\sum_{m=1}^{n}d_{m}^{2}Q^{2m}$ converges to
$d_{\infty}^{2}P_{\Omega}$ in norm. The second term

$\frac{1}{n}\sum_{m=1}^{n}(1-q)^{2m}c(\xi)^{m}c(\eta)^{m}r^{*}(\xi)^{m}r^{*}(\eta)^{m}$

converges to $0$ in norm by Proposition 3.4. We finally show that $X_{n}$
converges to $0$ in norm, which will finish the proof. Indeed first note that
$m(k+\ell)-2k\ell\geq m$ whenever $0<k+\ell<2m$ (one can observe this fact by
assuming without loss generality that $\ell\leq k$, and then considering
various cases such as $0\leq\ell\leq\frac{m}{2}\;\&\;1\leq k\leq\frac{m}{2}$
or $\frac{m}{2}\leq\ell<m\;\&\;\frac{m}{2}<k\leq m$ etc); therefore we have
$q^{m(k+\ell)-2k\ell}\leq|q|^{m}$. This observation along with Lemma 2.4 and
the fact that $\|Q\|\leq 1$ yields the norm estimate of $X_{n}$ as follows:

$\displaystyle\|X_{n}\|$
$\displaystyle\leq\frac{C_{q}^{2}}{n}\sum_{m=1}^{n}|q|^{m}(1-q)^{2m}\sum_{\underset{0<k+\ell<2m}{k,\ell=0}}^{m}\frac{[m]_{q}!^{2}}{[k]_{q}![\ell]_{q}![m-k]_{q}![m-\ell]_{q}!}\frac{[m]_{q}!^{2}}{[m-k]_{q}![m-\ell]_{q}!}\left([m-k]_{q}![m-\ell]_{q}!\right)$
$\displaystyle\leq\frac{C_{q}^{2}}{n}\sum_{m=1}^{n}|q|^{m}\sum_{{k,\ell=0}}^{m}\frac{d_{m}^{4}}{d_{k}d_{\ell}d_{m-k}d_{m-\ell}}$
$\displaystyle\leq\frac{C_{q}^{10}}{n}\sum_{m=1}^{n}\sum_{{k,\ell=0}}^{m}|q|^{m}$
$\displaystyle\leq\frac{C_{q}^{10}}{n}\sum_{m=1}^{n}m^{2}|q|^{m}.$

The sequence $\\{m^{2}q^{m}\\}_{m\geq 1}$ converges to $0$ as $m\to\infty$, so
does its average; hence we conclude that $X_{n}$ converges to $0$ in norm as
$n\to\infty$. ∎

We end this section by giving an expression for the right Wick product
$W_{r}(\bar{e}^{n}e^{n})$, which can easily be proved by utilizing Proposition
2.2.

######  Corollary 3.13.

For $e,\bar{e}$ and $\lambda$ as before, the right Wick product
$W_{r}(\bar{e}^{n}e^{n})$ is given as

$\displaystyle W_{r}(\bar{e}^{n}e^{n})$
$\displaystyle=\sum_{k,\ell=0}^{n}q^{(n)}_{k,\ell}r(e)^{k}r(\bar{e})^{\ell}r^{*}(\bar{e}^{r})^{n-k}r^{*}(\overline{\bar{e}}^{r})^{n-\ell}$
$\displaystyle=\sum_{k,\ell=0}^{n}q^{(n)}_{k,\ell}\lambda^{k-\ell}r(e)^{k}r(\bar{e})^{\ell}r^{*}(\bar{e})^{n-k}r^{*}(e)^{n-\ell}$

where $q_{k,\ell}^{(m)}$ is the same scalars as in the proof of Lemma 3.7.

##  4\. Main result

To establish the fullness of $\Gamma_{q}(\mathsf{H}_{\mathbb{R}},U_{t})$, we
rely on the equivalent condition of the triviality of the relative commutant
within the ultraproduct. We briefly recall the construction of the von Neumann
ultraproduct. Let $\mathsf{M}$ be a von Neumann algebra equipped with a normal
faithful state $\varphi$, and let $\omega$ be a free ultrafilter over the
natural numbers. Denote by $\prod_{n\geq 1}\mathsf{M}$ the algebra of all
bounded sequences $(x_{n})$ with each $x_{n}\in\mathsf{M}$. Let $I_{\omega}$
be the two sided ideal of $\prod_{n\geq 1}\mathsf{M}$ given by

$\displaystyle I_{\omega}=\left\\{(x_{n})\in\prod_{n\geq
1}\mathsf{M};\lim_{n\to\infty}\varphi(x_{n}^{*}x_{n}+x_{n}x_{n}^{*})=0\right\\}.$

The quotient $\mathsf{M}^{\omega}:=\prod_{n\geq 1}\mathsf{M}/I_{\omega}$ is a
well-defined von Neumann algebra, called ultraproduct of $\mathsf{M}$. A
typical element in $\mathsf{M}^{\omega}$ will be denoted as $(x_{n})_{\omega}$
which is the equivalence class of the sequence $(x_{n})$ in the quotient. It
is known that $\mathsf{M}$ is a full factor if and only if
$\mathsf{M}^{\prime}\cap\mathsf{M}^{\omega}=\mathbb{C}$ [Con1]. Our main
result of fullness uses this particular equivalent condition. In what follows,
one can also directly employ the equivalent condition of fullness as
formulated in [Mar, Corollary B].

######  Theorem 4.1.

The $q$-Araki-Woods factor
$\mathsf{M}_{q}:=\Gamma_{q}(\mathsf{H}_{\mathbb{R}},U_{t})$ is full as soon as
$\dim\mathsf{H}_{\mathbb{R}}\geq 2$.

###### Proof.

If $\mathsf{H}_{\mathbb{R}}$ is finite-dimensional, then $\mathsf{M}_{q}$ is
full by [KSW, Theorem 2.7]. If $\mathsf{H}_{\mathbb{R}}$ has a non-zero weakly
mixing part or if the point spectrum $\sigma_{p}(A)$ of eigenvalues of the
analytic generator $A$ of $U_{t}$ is infinite and bounded, then
$\mathsf{M}_{q}$ is full by [HI, Theorem 6.2]. So assume that
$\mathsf{H}_{\mathbb{R}}$ is infinite-dimensional such that there is a
strictly decreasing sequences of the set $\sigma_{p}(A)$ of eigenvalues of $A$
going to $0$.

Fix $\lambda\in\sigma_{p}(A)$ with $0<\lambda<1$, and let
$e\in\mathsf{H}_{\mathbb{C}}$ be such that $Ae=\lambda^{-1}e$ and
$A\bar{e}=\lambda\bar{e}$ with $\|e\|_{U}=\lambda^{-\frac{1}{4}}$ and
$\|\bar{e}\|_{U}=\lambda^{\frac{1}{4}}$. Consider the operator $S_{n}$

$\displaystyle
S_{n}=\frac{1}{n}\sum_{m=1}^{n}(1-q)^{2m}W(\bar{e}^{m}e^{m})c(\bar{e})^{m}c(e)^{m}.$

Assume that $\lambda<(1+C_{q}^{4})^{-2}$ so that the norm limit $S$ of $S_{n}$
exists and is invertible by Proposition 3.9.

Now let $\omega$ be a fixed non-principal ultrafilter. Let
$(X_{k})_{\omega}\in\mathsf{M}_{q}^{\prime}\cap\mathsf{M}_{q}^{\omega}$. By
replacing each $X_{k}$ by $X_{k}-\varphi(X_{k})$, we may assume that
$\varphi(X_{k})=0$ for all $k\geq 1$. In particular, we have
$X_{k}\Omega,X_{k}^{*}\Omega\in\mathcal{F}_{q}(\mathsf{H}_{U})\ominus\mathbb{C}\Omega$
for all $k\geq 1$. We will show that $(X_{k})_{\omega}=0$ in
$\mathsf{M}_{q}^{\omega}$ i.e.
$\lim_{k\to\omega}\|X_{k}\Omega\|_{q}=0=\lim_{k\to\omega}\|X_{k}^{*}\Omega\|_{q}$.
We observe

$\displaystyle\lim_{k\to\omega}\|X_{k}\Omega\|_{q}^{2}$
$\displaystyle=\lim_{k\to\omega}\langle
X_{k}\Omega,X_{k}\Omega\rangle_{q}=\lim_{k\to\omega}\lim_{n\to\infty}\langle
X_{k}\Omega,S_{n}S^{-1}X_{k}\Omega\rangle_{q}=\lim_{n\to\infty}\lim_{k\to\omega}\langle
X_{k}\Omega,S_{n}S^{-1}X_{k}\Omega\rangle_{q}$

where we could exchange the limit because $S_{n}S^{-1}$ converges in norm to
identity as $n\to\infty$. Now we use the expressions $XX_{k}-X_{k}X\to 0$
$*$-strongly as $k\to\omega$ for all $X\in\mathsf{M}_{q}$, and
$X_{k}X^{\prime}-X^{\prime}X_{k}=0$ for all
$X^{\prime}\in\mathsf{M}_{q}^{\prime}$ to get

$\displaystyle\lim_{k\to\omega}\|X_{k}\Omega\|_{q}^{2}$
$\displaystyle=\lim_{n\to\infty}\lim_{k\to\omega}\langle
X_{k}\Omega,S_{n}S^{-1}X_{k}\Omega\rangle_{q}$
$\displaystyle=\lim_{n\to\infty}\lim_{k\to\omega}\frac{1}{n}\sum_{m=1}^{n}(1-q)^{2m}\langle
X_{k}\Omega,W(\bar{e}^{m}e^{m})c(\bar{e})^{m}c(e)^{m}S^{-1}X_{k}\Omega\rangle_{q}$
$\displaystyle=\lim_{n\to\infty}\lim_{k\to\omega}\frac{1}{n}\sum_{m=1}^{n}(1-q)^{2m}\langle
W(\bar{e}^{m}e^{m})X_{k}\Omega,c(\bar{e})^{m}c(e)^{m}S^{-1}X_{k}\Omega\rangle_{q}$
$\displaystyle=\lim_{n\to\infty}\lim_{k\to\omega}\frac{1}{n}\sum_{m=1}^{n}(1-q)^{2m}\langle
X_{k}W(\bar{e}^{m}e^{m})\Omega,c(\bar{e})^{m}c(e)^{m}S^{-1}X_{k}\Omega\rangle_{q}$
$\displaystyle=\lim_{n\to\infty}\lim_{k\to\omega}\frac{1}{n}\sum_{m=1}^{n}(1-q)^{2m}\langle
X_{k}W_{r}(\bar{e}^{m}e^{m})\Omega,c(\bar{e})^{m}c(e)^{m}S^{-1}X_{n}\Omega\rangle_{q}$
$\displaystyle=\lim_{n\to\infty}\lim_{k\to\omega}\frac{1}{n}\sum_{m=1}^{n}(1-q)^{2m}\langle
X_{k}\Omega,W_{r}(\bar{e}^{m}e^{m})c(\bar{e})^{m}c(e)^{m}S^{-1}X_{k}\Omega\rangle_{q}$
$\displaystyle=\lim_{n\to\infty}\lim_{k\to\omega}\langle
X_{k}\Omega,T_{n}S^{-1}X_{k}\Omega\rangle_{q}+\lim_{n\to\infty}\lim_{k\to\omega}\langle
X_{k}\Omega,R_{n}S^{-1}X_{k}\Omega\rangle_{q}$

where $T_{n}$ and $R_{n}$ are the following operators:

$\displaystyle
T_{n}=\frac{1}{n}\sum_{m=1}^{n}(1-q)^{2m}\left(W_{r}(\bar{e}^{m}e^{m})-r^{*}(\bar{e})^{m}r^{*}(e)^{m}-r(e)^{m}r(\bar{e})^{m}\right)c(\bar{e})^{m}c(e)^{m}$
$\displaystyle
R_{n}=\frac{1}{n}\sum_{m=1}^{n}(1-q)^{2m}(r^{*}(\bar{e})^{m}r^{*}(e)^{m}+r(e)^{m}r(\bar{e})^{m})c(\bar{e})^{m}c(e)^{m}.$

Propositions 3.4 and 3.12 tell us that the operators $R_{n}$ (or $R_{n}^{*}$)
when restricted to $\mathcal{F}_{q}(\mathsf{H}_{U})\ominus\mathbb{C}\Omega$
converge to $0$ in norm as $n\to\infty$. Since
$X_{k}\Omega\in\mathcal{F}_{q}(\mathsf{H}_{U})\ominus\mathbb{C}\Omega$, it
follows that

$\displaystyle\lim_{n\to\infty}\lim_{k\to\omega}\langle
X_{k}\Omega,R_{n}S^{-1}X_{k}\Omega\rangle_{q}=0.$

Our next aim is to show that the operators $T_{n}$ are bounded by
$\frac{2C_{q}^{6}\sqrt{\lambda}}{1-\sqrt{\lambda}}$ as $n\to\infty$. First
write using Corollary 3.13 the following:

$\displaystyle
W_{r}(\bar{e}^{m}e^{m})=\sum_{k,\ell=0}^{m}q^{(m)}_{k,\ell}\lambda^{k-\ell}r(e)^{k}r(\bar{e})^{\ell}r^{*}(\bar{e})^{m-k}r^{*}(e)^{m-\ell}.$

and then note that

$\displaystyle W_{r}(\bar{e}^{m}e^{m})$
$\displaystyle-r^{*}(\bar{e})^{m}r^{*}(e)^{m}-r(e)^{m}r(\bar{e})^{m}=\sum_{\ell=1}^{m}\sum_{k=0}^{m-1}q^{(m)}_{k,\ell}\lambda^{k-\ell}r(e)^{k}r(\bar{e})^{\ell}r^{*}(\bar{e})^{m-k}r^{*}(e)^{m-\ell}$
$\displaystyle+\sum_{\ell=0}^{m-1}q_{m,\ell}^{(m)}\lambda^{m-\ell}r(e)^{m}r(\bar{e})^{\ell}r^{*}(e)^{m-\ell}+\sum_{k=1}^{m}q_{k,0}^{(m)}\lambda^{k}r(e)^{k}r^{*}(\bar{e})^{m-k}r^{*}(e)^{m}$

so that the upper bound for $T_{n}$ can be given as follows:

$\displaystyle\|T_{n}\|$
$\displaystyle\leq\frac{C_{q}^{6}}{n}\sum_{m=1}^{n}\left(\sum_{\ell=1}^{m}\sum_{k=0}^{m-1}|q|^{(m-k)\ell}\lambda^{\frac{k-\ell}{2}}+\sum_{\ell=0}^{m-1}\lambda^{\frac{m-\ell}{2}}+\sum_{k=1}^{m}\lambda^{\frac{k}{2}}\right)\leq{C_{q}^{6}}s_{n}+\frac{2C_{q}^{6}\sqrt{\lambda}}{1-\sqrt{\lambda}}.$

where $s_{n}$ is the sequence as in Lemma 3.8, which converges to $0$ by the
same Lemma. Thus, we have shown that

$\displaystyle\lim_{k\to\omega}\|X_{k}\Omega\|_{q}^{2}$
$\displaystyle=\lim_{n\to\infty}\lim_{k\to\omega}\langle
X_{k}\Omega,T_{n}S^{-1}X_{k}\Omega\rangle_{q}$
$\displaystyle\leq\left(\limsup_{n\to\infty}\|T_{n}\|\right)\;\|S^{-1}\|\;L$
$\displaystyle\leq\frac{2LC_{q}^{6}\sqrt{\lambda}}{1-\sqrt{\lambda}}\cdot\frac{2C_{q}^{2}(1-\sqrt{\lambda})}{1-(1+C_{q}^{4})\sqrt{\lambda}}$

where $L=\sup_{k\geq 1}\|X_{k}\Omega\|^{2}$, and we have used the upper bound
for $\|S^{-1}\|$ from Remark 3.10. By our assumption, we have a sequence of
such $\lambda$’s which converges to $0$; hence it follows that
$\lim_{k\to\omega}\|X_{k}\Omega\|_{q}=0$. A similar argument shows that
$\lim_{k\to\omega}\|X_{k}^{*}\Omega\|_{q}=0$. This completes the proof. ∎

Acknowledgments.  We thank Zhiyuan Yang (Texas A&M University) for going
through the initial draft, and providing valuable suggestions; particularly,
he suggested a shortened proof of Proposition 3.12 by utilizing the operator
$Q$. This work was partially completed when M.K. was visiting S.W. at Harbin
Institute of Technology in Aug-Sep 2023 for a month. M.K. extends appreciation
to HIT for their warm hospitality and to S.W. for the invitation and financial
support. M.K. was partially supported by the National Science Center (NCN)
grant no. 2020/39/I/ST1/01566, as well as by NCN grant no.
2021/43/D/ST1/01446. S.W. was partially supported by the NSF of China
(No.12031004, No.12301161) and the Fundamental Research Funds for the Central
Universities.

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