\magnification 1200
\baselineskip=18pt
\pageno=0
\input mssymb
\def\undb{1\kern-3pt\hbox{l}}
\def\s{\sigma}
\def\la{\langle}
\def\ra{\rangle}
\def\F{{\cal F}}
\def\O{\Omega}
\def\G{\Gamma}

\centerline{\bf Factoriality and second quantization} 
\smallskip
\centerline{\bf for  von Neumann algebras of  deformed commutation relations}
\bigskip

\setbox1=\vtop{\hsize=6cm
\centerline {Marek BO\.ZEJKO\thanks{Partially supported by KBN grant 2P03A05415}}
\centerline{Uniwerstytet Wroclawski}
\centerline{Instytut Matematyczny}
\centerline{Plac Grunwaldzki 2/4}
\centerline{50-384 Wroclaw; Poland}
\centerline{ bozejko@math.uni.wroc.pl}}

\setbox2=\vtop{\hsize=6cm
\centerline{Quanhua Xu}
\centerline{Universit\'e de Franche-Comt\'e}
\centerline{D\'epartement de Math\'ematiques}
\centerline{16, route de Gray}
\centerline{25030 Besan\c con Cedex; France}
\centerline{ qx@math.univ-fcomte.fr}}
$$\line{\box1\hfill\box2}$$
\vskip 2cm



\noindent{\bf Abstract.} Let $-1<q<1$ and $H$ be a real Hilbert space. Let $\Gamma_q(H)$ be
the associated von Neumann algebra of the $q$-deformed commutation relations. We prove that
$\Gamma_q(H)$ is a factor as soon as $\dim H\ge 2$. This solves a problem left open
since the works in this direction by  K\"ummerer, Speicher and the first named auther. We also
extend this result to a more general setting, i.e., for the von Neumann
algebra of the deformed commutation relations determined by a tracial Yang-Baxter operator.
In the latter setting, we as well study  the possibility of defining  second
quantization, and we establish the second quantization for a large class of contractions.
\vfill\break

\centerline{\bf 0. Introduction}
\medskip
Let $K$ be a complex Hilbert space. For a given  $-1 < q < 1$ let ${\cal
F}_{q}(K)$ denote the assocaited $q$-Fock space:
$${\cal F}_q(K) = \bigoplus_{n \geq 0}K^{\otimes n},$$ 
where $K^{\otimes 0} = {\Bbb C}\Omega$ ($\Omega$ called vacuum) and  
$K^{\otimes n}$ is equippped with the following $q$-scalar product
$$\langle f_{1} \otimes \cdots \otimes f_{n}, g_{1} \otimes \cdots
\otimes g_{n} \rangle_q = \sum_{\s\in S_n}q^{\nu(\s)}\langle f_{1}, g_{\s(1)}\rangle \cdots
\langle f_{n}, g_{\s(n)}\rangle\,.$$
Here we have denoted by $S_n$ the permutation group of $n$ elements, and by $\nu(\s)$ the number of
inversions  of a permuation $\s$. The above $q$-scalar product can be also expressed in terms
of the $q$-determinant:
$$\langle f_{1} \otimes \cdots \otimes f_{n}, g_{1} \otimes \cdots
\otimes g_{n} \rangle_q={\det}_q\Big(\big(\la f_i,g_j\ra\big)_{1\le i, j\le n}\Big)\,.$$
Note that ${\cal F}_0(K)$ is just the full (or free) Fock space.

For $f\in K$  we denote by $a^*(f)$ and $a(f)$ the corresponding creator and annihilator. These are
bounded operators on $\F_q(K)$ determined by the following conditions
$$a^*(f)\O=f,\quad a^*(f)f_{1}\otimes\cdots\otimes f_{n}=f\otimes f_{1}\otimes\cdots\otimes
f_{n},\quad f_1,\cdots, f_n\in K$$
and 
$$a(f)\O=0,\quad a(f)f_{1}\otimes\cdots\otimes
f_{n}=\sum_{k=1}^nq^{k-1}\la f,f_{k}\ra f_1\otimes\cdots\otimes{\displaystyle\mathop
f^\vee}_k\otimes\cdots\otimes f_{n},$$ where ${\displaystyle\mathop f^\vee}_k$ means $f_k$ should be
deleted in the tensor product.  Recall that
$a^*(f)=(a(f))^*$. The creators and annihilators satisfy the so-called $q$-commutation relations:
$$a(f)a^*(g)-qa^*(g)a(f)=\la f, g\ra,\quad f, g\in K.$$
These $q$-commutation relations  were introduced as an interpolation between the Bose ($q=1$) and
Fermi ($q=-1$) statistics. Their existence, together with their Fock representation as above, was
established for the first time in [BS1]. We refer to [BS1], [F], [G] and [Z]  for more information
on these relations, to [B1-3], [BS3], [BKS] and [Sp1-2] for their probabilistic aspects, and to
[DN], [JW], [JSW1-2] for results on the $C^*$-algebras defined by them.

In this paper we are interested in the von Neumann algebra determined by the above $q$-commuation
relations. Now let $H$ be a real Hilbert space and $H_{\Bbb C}$ its complexification.
Then $\Gamma_{q}(H)$ denotes the associated $q$-von Neumann algebra, namely,
$$\G_q(H)=\big\{a^*(f)+a(f): f\in H\big\}''\subset B\big(\F_q(H_{\Bbb C})\big)\,.$$
Note that $\Gamma_{0}(H)$ is the free von Neumann algebra based on $H.$ It was proved in [BS2] that
the vacuum expectation $x\mapsto \la\Omega, x\Omega\ra$ is a faithful trace on $\G_q(H)$ (so
$\G_q(H)$ is a von Neumann algebra of type II$_1$).  

It is by now well-known that the free von Neumann algebra $\Gamma_{0}(H)$ is isomorphic to the von
Neumann algebra of the free group on $\dim H$ generators (cf. [V], [VDN]). Thus it follows
that
$\Gamma_{0}(H)$ is a factor as soon as $\dim H \geq 2.$ On the
other hand, it was proved in [BKS] that if $\dim H = \infty,\
\Gamma_{q}(H)$ is a factor. The problem whether $\Gamma_{q}(H)$ is a
factor as soon as ${\rm dim}\ H \geq 2$ was left open since [BKS] (cf. also [BS2] p.116). One of
the main results of this paper  solves this problem affirmatively.

Our approach to this problem heavily depends on the second quantization and the right von Neumann
algebra  generated by the right $q$-commutation
relations. Let $V: H_1\to H_2$ be a contraction between two real Hilbert spaces. Then $V$ induces a
unital, completely positive and vacuum expectation preserving mapping $\G(V)$ from $\G_q(H_1)$ to
$\G_q(H_1)$ (cf. [BKS], [V]). This is the second quantization of $V$. To prove the factoriality of
$\G_q(H)$ we will only need $\G(t\undb_{H})$, $-1\le t\le 1$, where $\undb_{H}$ denotes the
identity of $H$. We will also need the right von Neumann algebra $\G_{q, r}(H)$ generated by the right
$q$-commutation relations. It is known that
$\Gamma_{q, r}(H)$ is the commutant of
$\Gamma_{q}(H).$ Thus the factoriality of $\G_q(H)$ is reduced to show that
the intersection of $\Gamma_{q}(H)$ and $\Gamma_{q, r}(H)$ is
trivial. The proof of the latter is done via the second quantization
$\Gamma (t \undb_{H}), -1 \leq t \leq 1.$ A easily checked fact, but
crucial for us, is that $\Gamma(t \undb_{H})$ coincides with the right
second quantization on the intersection $\Gamma_{q}(H) \cap
\Gamma_{q,r}(H).$ It seems that this is the first time that one
explicitly uses the right von Neumann algebra in an essential way. We would like to emphasize that
our proof for the factoriality of $\Gamma_{q}(H)$ is very natural, and also simple (at least, in
some sense). In particular, in the case of $q=0$ we supply anothe proof for the factoriality of
$\Gamma_{0}(H)$ (Voiculsecu's theorem) without using the isomorphism between $\Gamma_{0}(H)$ and
the free group von Neumann algebra.


Our approach applies to a more general setting as well. Let $T$ be a
Yang-Baxter operator on $H_{\Bbb C} \otimes H_{\Bbb C}$ with $\Vert T
\Vert < 1.$ Let ${\cal F}_{T}(H_{\Bbb C})$ be the associated deformed
Fock space and $\Gamma_{T}(H)$ the von Neumann algebra generated by
the corresponding deformed commutation relations (see section 1 below for more details). A
particular case of $T$ is given as follows. Fix an orthonormal basis
$\{ e_{i} \}_{i \in I}$ of $H.$ Let $(q_{ij})_{i, j \in I}$ be a
hermitian matrix with $\displaystyle\sup_{i,j}\ \vert q_{ij}\vert <
1.$ Define $T\ :\ H_{\Bbb C} \otimes H_{\Bbb C} \to H_{\Bbb C}
\otimes H_{\Bbb C}$ by
$$Te_{i} \otimes e_{j} = q_{ji}e_{j}\otimes e_{i}, \quad i,j \in
I.$$
Then $T$ is a Yang-Baxter operator with $\Vert T \Vert < 1.$
The preceding $q$-case corresponds to the case where $q_{ij} = q$ for
all $i, j \in I.$
\medskip
Now our problem is to determine whether $\Gamma_{T}(H)$ is a factor.
If we examine our approach to the same problem in the $q$-case, we
will need second quantization. It is, however, an open problem to
define a good second quantization in this general setting, even in
the particular $q_{ij}$-case (cf. [LP] for some related discussions).
If the vacuum expectation is tracial (in this case $T$ will be said to  be tracial),  we solve this
latter problem for a rather large class of contractions, including $t\undb_{H}$, $-1\leq t\leq1.$ As
usual, our second quantization is defined via the second quantization at the Fock space level. Let
again $V: H_1\to H_2$ be  a contraction between two real Hilbert spaces, and let $V_{\Bbb C}:
H_{1{\Bbb C}}\to H_{2{\Bbb C}}$ be its complexification. Let $T^{(j)}: H_{j{\Bbb C}}\otimes H_{j{\Bbb
C}}\to H_{j{\Bbb C}}\otimes H_{j{\Bbb C}}$ be a tracial Yang-Baxter operator with $\|T^{(j)}\|<1$,
$j=1,2$. Suppose the following commutation condition
$$\big(V_{\Bbb C}\otimes V_{\Bbb C}\big)T^{(1)}=T^{(2)}\big(V_{\Bbb C}\otimes V_{\Bbb C}\big).$$
Then $V_{\Bbb C}$ induces, in a natural way, a contraction $\F(V)$ from $\F_q(H_{1{\Bbb C}})$ to 
 $\F_q(H_{2{\Bbb C}})$. $\F(V)$ is the second quantization at the Fock space level. The above
commutation condition is also  necessary (in some sense) for $\F(V)$ to be a contraction (cf.
section 3 for more precision). If $H_1=H_2=H$, $T^{(1)}=T^{(2)}=T$ and the above commutation
condition is strengthened to 
$$\big(V_{\Bbb C}\otimes \undb_{H_{\Bbb C}}\big)T=T\big(\undb_{H_{\Bbb C}}\otimes V_{\Bbb
C}\big)\,,\quad
\big(\undb_{H_{\Bbb C}}\otimes V_{\Bbb C}\big)T=T\big(V_{\Bbb C}\otimes \undb_{H_{\Bbb C}}\big),$$
then one is able to define the second quantization $\G(V): \G_T(H)\to \G_T(H)$ which is unital,
completely positive and vacuum expextation preserving. We should point out that in general, it is
impossible to define a good second quantization for any contraction. These results are in triking
contrast with all known cases, where all contractions admit  second quantizations at both  Fock
space  and von Neumann algebra levels (cf. [BKS], [V], [VDN], [S1-2]).

Then using the second quantization $\G(t\undb)$, $-1\le t\le1$, and repeating the arguments in the
$q$-case we prove that if $T$ is a tracial Yang-Baxter operator with $\|T\|<1$,  $\Gamma_{T}(H)$ 
is a factor as soon as ${\rm dim}\ H \geq 2.$


This paper is organized as follows. After the first section
introducing  necessary notation and preliminaries, we prove in
section 2 that $\Gamma_{q}(H)$ is a factor. Section 3 is devoted to
the second quantization for the general deformation defined by a
Yang-Baxter operator. The results in this section might be useful
somewhere else. The last section deals with the factoriality of
$\Gamma_{T}(H).$ This section is very similar to section 2, in
nature. Thus the arguments there are often very brief. It is clear
that the factoriality of $\Gamma_{q}(H)$ (i.e., section 2) can be
incorporated into the last section. However, there are at least three
reasons for us to choose to separate the $q$-case. First, the $q$-case
is the most familiar and important after the free case. Second, our
arguments in the $q$-case are rather transparent and easy to follow.
Third, all main ideas are already in the $q$-case. Thus  the
reader who is interested only in the $q$-case can stop just after
section 2.



\bigskip
\centerline{\bf 1. Preliminaries}
\medskip
Let $K$ be a complex Hilbert space. Let ${\cal F}(K)$ denote the usual
full (or free) Fock space :
$${\cal F} (K) = \bigoplus_{n \geq 0}\ K^{\otimes n},$$
where $K^{\otimes 0} = {\Bbb C}\ \Omega$ for some distinguished unit
vector $\Omega,$ called vacuum. The scalar product in $K^{\otimes n}$
is given by
$$\langle f_{1} \otimes \cdots \otimes f_{n}, g_{1} \otimes \cdots
\otimes g_{n} \rangle = \langle f_{1}, g_{1}\rangle \cdots \langle
f_{n}, g_{n}\rangle\,.$$
Let us emphasize that in this paper ${\cal F}(K)$ always denotes the
full Fock space and $\langle \cdotp, \cdotp \rangle$ its scalar product.
\medskip
Recall that an operator $T\ :\ K \otimes K \to K \otimes K$ is called
a Yang-Baxter operator if $T$ is a hermitian contraction satisfying
the following braid relation
$$(\undb_{K} \otimes T) \ (T \otimes \undb_{K}) (\undb_{K} \otimes
T) = (T \otimes \undb_{K}) \ (\undb_{K} \otimes T) \  (T \otimes
\undb_{K}). \leqno(1.1)$$
For such an operator $T$ we define $T_{k}\ :\ K^{\otimes n} \to K^{\otimes n}$ by
$$T_{k}=\undb_{K^{\otimes n}}\;\hbox{if}\; 0\le n\le 1,\quad T_{k}=
\undb_{K^{\otimes(k-1)}}\otimes T\otimes\undb_{K^{\otimes(n-k-1)}}\;\hbox{if}\;\ 1 \leq k \leq n-1.$$
Let $S_{n}$ be the permutation group of $n$ elements, and let
$\pi_{k} \in S_{n}$ be the transposition interchanging $k$ and
$k+1$ $(1 \leq k \leq n-1).$ Note that $S_{n}$ is generated by
$\pi_{1}, \cdots, \pi_{n-1}.$ We define
$$\varphi (\pi_{k}) = T_{k}, \quad 1 \leq k \leq n-1,$$
and extend multiplicatively $\varphi$ to all permutations $\sigma\in S_{n}$. Then 
$\varphi(\sigma)$ is a well-defined contraction on $K^{\otimes n}$ for all $\sigma\in S_{n}$.  Now
we define the symmetrizator $P_{T}^{(n)}$ as
$$P^{(n)}_{T} = \sum_{\sigma \in S_{n}}\ \varphi(\sigma).$$
It was proved in [BS2] (cf. also [B1]) that the operator $P^{(n)}_{T}$ is positive
on $K^{\otimes n}$ for any Yang-Baxter operator $T,$ and strictly
positive if additionally $\Vert T \Vert < 1$. Thus, if $T$ is a
Yang-Baxter operator with $\Vert T \Vert < 1,$
$$\langle f_{1} \otimes \cdots \otimes f_{n}, g_{1} \otimes \cdots
\otimes g_{n} \rangle_{T} = \langle f_{1} \otimes \cdots \otimes
f_{n}, P_{T}^{(n)} g_{1} \otimes \cdots \otimes g_{n} \rangle
\leqno(1.2)$$
defines, by linearity, a new scalar product on $K^{\otimes n}.$ Then
the deformed Fock space ${\cal F}_{T}(K)$ is defined by
$${\cal F}_{T}(K) = \bigotimes_{n \geq 0} \ K^{\otimes n},$$
where $K^{\otimes 0} = {\Bbb C}\ \Omega,$ and $K^{\otimes n}$ is
equipped with the new scalar product $\langle \cdotp, \cdotp
\rangle_{T}$ above.


Throughout this paper, unless specifically stated, $T$ will always
denote a Yang-Baxter operator with $\Vert T \Vert < 1.$


We next define  creators and annihilators on our Fock space. In
this paper $\ell^{\ast}(f)$ and $\ell (f)$ will be \underbar{exclusively}
reserved to denote the usual creator and annihilator on the full Fock
space ${\cal F} (K).$ Recall that for any given $f\in K$ the bounded operators $\ell^{\ast}(f)$ and
$\ell (f)$ on $\F(K)$ are uniquely determined by
$$\ell^{\ast}(f)\O=f, \quad \ell^{\ast}(f)f_1\otimes \cdots\otimes f_n=f\otimes f_1\otimes
\cdots\otimes f_n,\quad f_1, \cdots, f_n\in K$$ and 
$$\ell(f)\O=0, \quad \ell(f)f_1\otimes \cdots\otimes f_n=\la f, f_1\ra f_2\otimes
\cdots\otimes f_n,\quad f_1, \cdots, f_n\in K.$$
Now the corresponding  creator
$a^{\ast}(f)$ and annihilator $a(f)$ on ${\cal F}_{T}(K)$ are defined
by
$$a^{\ast}(f) = \ell^{\ast}(f)$$
and
$$a(f)\Omega = 0,\ a(f) = \ell (f)(1 + T_{1} + \cdots + T_{1}
\cdots T_{n-1}) \ \hbox{\rm on}\ K^{\otimes n},\ n \geq 1.$$
Recall that $a^{\ast}(f)$ and $a(f)$ are bounded operators on ${\cal
F}_{T}(K)$ which are adjoint one to another. These operators satisfy
the following deformed commutation relations. Fix an orthonormal
basis $\{ e_{i}\}_{i \in I}$ of $K.$ Set
$$t^{rs}_{ij} = \langle e_{r} \otimes e_{s},\ T\ e_{i} \otimes
e_{j}\rangle,\quad i, j, r, s \in I.$$
Then
$$a(e_{i})a^{\ast}(e_{j}) - \sum_{r, s \in I}t^{ir}_{js}
a^{\ast}(e_{r})a(e_{s}) = \delta_{i,j},\quad i,j \in I. \leqno(1.3)$$

Now let $H$ be a real Hilbert space and $H_{\Bbb C}$ its
complexification. Define the deformed Gaussian variable $G(f)$ by
$$G(f) = a^{\ast} (f) + a(f),\ f \in H.$$
Note that $f \mapsto G(f)$ is a linear mapping from $H$ into $B ({\cal
F}_{T}(H_{\Bbb C})).$ These Gaussian variables generate the deformed
von Neumann algebra $\Gamma_{T}(H)$ :
$$\Gamma_{T}(H) = \{ G(f)\ :\ f \in H\}'' \subset B({\cal F}_{T}(H_{\Bbb C})).$$
It is easy to see that the vacuum is cyclic for $\Gamma_{T}(H).$ It
is also separating if additionally $T$ satisfies the following
condition :
$$\langle e_{r} \otimes e_{s}, T\, e_{i} \otimes e_{j}\rangle =
\langle e_{s} \otimes e_{j},\ T\ e_{r} \otimes e_{i}\rangle,\ i, j,
r, s \in I, \leqno(1.4)$$
where $\{ e_{i}\}_{i \in I}$ is an orthonormal basis of $H.$ (1.4) can
be interpreted as a cyclic condition on the coefficients $t^{rs}_{ij}$
of $T,$ namely,
$$t^{rs}_{ij} = t^{sj}_{ri},\quad i, j, r, s \in I.$$
Moreover, if (1.4) is verified, the vacuum expectation is tracial on
$\Gamma_{T}(H)$ (the converse is also true). In this case, we say that $T$ is {\it tracial}.  Thus,
if $T$ is a tracial Yang-Baxter operator with $\|T\|<1$,
$\Gamma_{T}(H)$ is a von Neumann algebra of type II$_{1}$ and the vacuum expectation is a faithful
normal normalized trace on $\Gamma_{T}(H).$ The reader is referred to [BS2]
for more details.
 
In this paper, we will always assume (1.4) for $T$, i.e., $T$ is tracial.  Let us specify the
preceding discussions to the $q_{ij}$-deformation. Let $(a_{ij})_{i, j
\in I}$ be a hermitian matrix with $\displaystyle\sup_{i, j\in I}\vert q_{ij}\vert < 1.$
Define
$$Te_{i} \otimes e_{j} = q_{ji} e_{j} \otimes e_{i},\quad i, j \in I.$$
Then $T$ is a Yang-Baxter operator with $\Vert T \Vert = \displaystyle\sup_{i, j\in I}\vert
q_{ij}\vert < 1.$ In this case, the condition (1.4) means that all
$q_{ij}$ are real (thus $(q_{ij})_{i, j \in I}$ is a symmetric real
matrix).
The deformed commutation relations (1.3) are then reduced to
$$a(e_{i})a^{\ast}(e_{j}) - q_{ij}a^{\ast} (e_{j}) a(e_{i}) =
\delta_{i, j},\quad i, j \in I.$$
In the case where all $q_{ij}$ are equal to $q$ with $-1 < q < 1,$ we
recover the $q$-deformation. Then the deformed commutation relations
become the following simple $q$-commutation relations:
$$a(f)a^{\ast}(g) - q a^{\ast}(g)\ a(f) = \langle f, g \rangle,\quad
f, g \in H_{\Bbb C}. \leqno(1.5)$$
Following [BKS], in this $q$-case we denote ${\cal F}_{T}(H_{\Bbb
C})$ and $\Gamma_{T}(H)$ respectively by ${\cal F}_{q}(H_{\Bbb C})$
and $\Gamma_{q}(H).$ Note that ${\cal F}_{0}(H_{\Bbb C})$ and
$\Gamma_{0}(H)$ are respectively the usual full Fock space and the
free von Neumann algebra assocaited with $H$.


We will also need the right von Neumann algebra. To define this we
denote by $\ell^{\ast}_{r}(f)$ and $\ell_{r}(f)$ the right free
creator and annihilator on ${\cal F} (H_{\Bbb C})\ :$
$$\ell^{\ast}_{r}(f)\Omega = f,\ \ell^{\ast}_{r}(f) f_{1} \otimes
\ldots \otimes f_{n} = f_{1} \otimes \cdots \otimes f_{n} \otimes f,\quad
f_{1}, \cdots, f_{n} \in H_{\Bbb C}$$
and
$$\ell_{r}(f)\Omega = 0,\ \ell_{r}(f) f_{1} \otimes \cdots \otimes
f_{n} = \langle f, f_{n} \rangle f_{1} \otimes \cdots \otimes
f_{n-1},\quad f_{1}, \cdots f_{n} \in H_{\Bbb C}.$$
The corresponding right creator $a^{\ast}_{r}(f)$ and annihilator
$a_{r}(f)$ on ${\cal F}_{T}(H_{\Bbb C})$ are defined in the same way as
$a^{\ast}(f)$ and $a(f)$ with $\ell^{\ast}_{r}(f)$ and $\ell_{r}(f)$
instead of $\ell^{\ast}(f)$ and $\ell(f)$, i.e., 
$$a^{\ast}_{r} (f) = \ell_r^{\ast}(f)$$
and
$$a_{r}(f)\Omega = 0,\ a_{r}(f) = \ell_{r}(f)\Bigl( 1 + T_{n-1} +
\cdots + T_{n-1} \cdots T_{1}\Bigl) \ \hbox{ on } H^{\otimes n}_{\Bbb
C}.$$
Then $a^{\ast}_{r}(f)$ and $a_{r}(f)$ are bounded on ${\cal
F}_{T}(H_{\Bbb C}),$ and adjoint one to another. They also satisfy
 commutation relations similar to (1.3). All these results on $a^{\ast}_{r}(f)$ and $a_{r}(f)$
can be proven in the same way as those for $a^{\ast}(f)$ and $a(f)$ contained in [BS2]. We also
define the right Gaussian variable $G_{r}(f)$ by
$$G_{r}(f) = a^{\ast}_{r}(f) + a_{r}(f),\ f \in H,$$
and the right von Neumann algebra
$$\Gamma_{T, r}(H) = \{ G_{r}(f)\ :\ f \in H \}'' \subset B({\cal
F}_{T}(H_{\Bbb C})).$$
We will see that if $T$ tracial, $\Gamma_{T, r}(H)$ is the
commutant of $\Gamma_{T}(H).$ In the $q$-case, $\Gamma_{T,r}(H)$ will be
denoted by $\Gamma_{q, r}(H).$

\bigskip

\centerline{\bf 2. The factoriality of $\Gamma_{q}(H)$}
\medskip
The main result of this section is the following
\bigskip\noindent
\proclaim Theorem 2.1. Let $-1 < q < 1,$ and let $H$ be a real Hilbert
space with ${\rm dim}\ H \geq 2.$ Then $\Gamma_{q}(H)$ is a factor of
type II$_{1}$.


Throughout this section $q$ and $H$ will be fixed as above. Recall
that the vacuum expectation $x\mapsto\langle\Omega,\ x\Omega\rangle$ is tracial
on $\Gamma_{q}(H)$ (cf. [BS2]). Therefore, we will only need to prove
that $\Gamma_{q}(H)$ is a factor. This section is essentially devoted
to this proof.
\medskip
We will need Wick product. Since the vacuum $\Omega$ is separating
for $\Gamma_{q}(H),$ the mapping $x \mapsto x \Omega$ is injective
from $\Gamma_{q}(H)$ into ${\cal F}_{q}(H_{\Bbb C}).$ Denote its image
by $\Gamma_{q}^{\infty}(H),$ and its inverse by $W.$ Thus $W(x\Omega) = x$ for any $x \in
\Gamma_{q}(H).$ Let ${\cal A}(H_{\Bbb C})$ denote the family of all finite linear combinations of
tensors from
$H_{\Bbb C}.$ It was proved in [BKS] that ${\cal A}(H_{\Bbb C}) \subset
\Gamma^{\infty}_{q}(H).$ Therefore, for any $\xi \in {\cal A}(H_{\Bbb C})$,
$W(\xi)$ is the unique operator in $\Gamma_{q}(H)$ such that $W(\xi)\Omega = \xi.$ In particular,
$$W\bigl(f_{1} \otimes \cdots \otimes f_{n} \bigl)\Omega = f_{1}
\otimes \cdots \otimes f_{n}, \quad \forall\ f_{1}, \cdots, f_{n}
\in H_{\Bbb C}.$$
$W$ is the Wick product. Note that there is an explicit formula for
$W\bigl( f_{1} \otimes \cdots \otimes f_{n} \bigl)$ given in [BKS].
For all what follows we will only need the existence of Wick products
$W(\xi)$ for all $\xi \in {\cal A}(H_{\Bbb C}).$ This can be easily checked by
induction as follows. Clearly, $H \subset {\cal A}(H_{\Bbb C})$ and $W(f)=G(f)$ for any
$f\in H.$ Let $n\geq1,$ and suppose all finite linear combinations
of tensors of degree $\leq n$ belong to $\Gamma_{q}^{\infty}(H).$ Let
$f, f_{1}, \cdots, f_{n} \in H.$ We define
$$W\bigl(f\otimes f_{1}\otimes\cdots\otimes f_{n} \bigl) = G(f)
W \bigl( f_{1} \otimes \cdots \otimes f_{n} \bigl) - W (a(f) f_{1}\otimes \cdots \otimes f_{n}
\bigl). \leqno(2.1)$$ 
Then
$$W\bigl(f\otimes f_{1} \otimes \cdots \otimes f_{n} \bigl)
\Omega = f \otimes f_{1} \otimes \cdots \otimes f_{n},$$
and by induction hypothesis, $W\bigl( f \otimes f_{1} \otimes \cdots
\otimes f_{n} \bigl) \in \Gamma_{q} (H).$ Thus
$f \otimes f_{1} \otimes \cdots \otimes f_{n} \in
\Gamma^{\infty}_{q}(H),$
and so by linearity, all finite linear combinations of tensors of
degree $\leq n+1$ belong to $\Gamma^{\infty}_{q}(H).$ Therefore,
${\cal A}(H_{\Bbb C}) \subset \Gamma^{\infty}_{q}(H).$

Note that (2.1) implies that
$$W({\cal A}(H_{\Bbb C})) = \{ W(\xi)\ : \xi \in {\cal A} (H_{\Bbb C}) \}$$
coincides with the family of all polynomials in $G(f), f \in H.$
Thus $W({\cal A}(H_{\Bbb C}))$ is a $\ast$-subalgebra $w^{\ast}$-dense in
$\Gamma_{q}(H).$

Similarly, we define the right Wick product $W_{r}$ relative to the
right von Neumann algebra $\Gamma_{q,r}(H).$ Then for any $\xi \in
{\cal A}(H_{\Bbb C})$, $W_{r}(\xi)$ is the unique operator in $\Gamma_{q,r}(H)$
such that $W_{r}(\xi)\Omega = \xi.$ Again, $W_{r}({\cal A}(H_{\Bbb C}))$ is a
$\ast$-subalgebra $w^{\ast}$-dense in $\Gamma_{q, r}(H).$


A straightforward verification shows that
$$G(f)G_{r}(g) = G_{r}(g)G(f),\quad \forall\ f, g \in H\ ;$$
whence
$$\Gamma_{q, r}(H) \subset (\Gamma_{q}(H))'.$$
The inverse inclusion holds as well.
\smallskip 
\proclaim Lemma 2.2. $(\Gamma_{q}(H))' = \Gamma_{q, r}(H)$.


This fact was observed in [BS2](p.114). For the convenience of the reader,
we include a proof.


\noindent{\it Proof of Lemma 2.2.} Let $f \mapsto \overline f$ denote the
natural conjugation of $H_{\Bbb C}.$ Define
$$Jf_{1} \otimes \cdots \otimes f_{n} = \overline f_{n} \otimes
\cdots \otimes \overline f_{1},\quad f_{1}, \ldots, f_{n} \in H_{\Bbb C}.$$
Then $J$ naturally extends to an isometric involution on ${\cal
F}_{q}(H_{\Bbb C}),$ still denoted by $J.$ Clearly,
$$Ja^{\ast}(f)J = a^{\ast}_{r}(\overline f),\quad Ja(f)J =
a_{r}(\overline f),\quad f\in H_{\Bbb C}\ ;\leqno(2.2)$$
whence
$$J G(f) J = G_{r}(\overline f),\quad f \in H_{\Bbb C}. \leqno(2.3)$$
Next, define
$$S(x \Omega) = x^{\ast} \Omega,\quad x \in \Gamma_{q}(H).$$
Since the vacuum expectation is a faithful normal normalized trace
on $\Gamma_{q}(H),$ $S$ extends to an isometric involution on ${\cal
F}_{q}(H_{\Bbb C})$ as well. Hence, the phase operator of $S$ is $S$ itself.
It follows that
$$S\Gamma_{q}(H) S = (\Gamma_{q}(H))'.\leqno (2.4)$$
On the other hand, it is easy to see that $S = J.$ Indeed, for any
$f, f_{1}, \ldots, f_{n} \in H,$ using (2.1), we have
$$W (f \otimes f_{1} \otimes \cdots \otimes f_{n})^{\ast} = W(f_{1}
\otimes \cdots \otimes f_{n})^{\ast} G(f) - W(a(f) f_{1} \otimes
\cdots \otimes f_{n})^{\ast}.$$
Thus, supposing $S = J$ on all tensors of degree $\leq n,$ we get
$$\eqalign{
S(f \otimes f_{1} \otimes \cdots \otimes f_{n}) &= W(f_{1} \otimes
\cdots \otimes f_{n})^{\ast} G(f) \Omega - W (a(f) f_{1} \otimes
\cdots \otimes f_{n})^{\ast} \Omega\cr
\noalign{\smallskip}
&= G_{r}(f) W(f_{1} \otimes \cdots \otimes f_{n})^{\ast} \Omega - J
[a(f) f_{1} \otimes \cdots \otimes f_{n}]\cr
\noalign{\smallskip}
&= a_{r}^*(f)\ f_{n} \otimes \cdots \otimes f_{1} + a_{r}(f) f_{n}
\otimes \cdots \otimes f_{1} - Ja(f) Jf_{n} \otimes \cdots \otimes f_{1}\cr
\noalign{\smallskip}
&= f_{n} \otimes \cdots \otimes f_{1} \otimes f \hskip 3cm \hbox{(by
(2.2)).}\cr
}$$
Therefore, $S = J$ ; and so by (2.4)
$$J\Gamma_{q}(H)J = (\Gamma_{q} (H))',$$
whence $(\Gamma_{q}(H))' = \Gamma_{q,r}(H)$ in virtue of (2.3).
\hfill $\diamond$

\noindent{\bf Remark.}  One can also prove Lemma 2.2 by using the Fourier series development of
operators in $\G_q(H)$ and $\G_{q, r}(H)$ given in Proposition 2.4 below.

\medskip

Now we pass to the second quantization. Let $V\ :\ H_{1} \to H_{2}$
be a contraction between two real Hilbert spaces. Let $V_{\Bbb C}\ :\
H_{1 \Bbb C} \to H_{2 \Bbb C}$ be its complexification. The  second quantization  at the Fock space
level is the contraction ${\cal F}(V)$ from
${\cal F}_{q}(H_{1 {\Bbb C}})$ into ${\cal F}_{q}(H_{2 {\Bbb C}})$
determined by
$${\cal F}(V) \Omega = \Omega,\ {\cal F}(V) f_{1} \otimes \cdots
\otimes f_{n} = Vf_{1} \otimes \cdots \otimes Vf_{n},\quad f_{1}, \cdots, f_{n} \in H.$$
At the von Neumann algebra level, the second quantization $\Gamma
(V)$ is a mapping from $\Gamma_{q}(H_{1})$ into $\Gamma_{q}(H_{2})$
defined by
$$(\Gamma (V) x)\Omega ={\cal F}(V)(x \Omega),\quad x \in
\Gamma_{q}(H_{1})$$
or
$$\Gamma (V) W(\xi) = W(\F(V)\xi), \quad \xi\in{\cal A}(H_{1{\Bbb C}}).
\leqno(2.5)$$
It was proved in [BKS] (cf. also [V] for the case of $q=0$) that $\Gamma (V)$ is a well-defined,
unital, and completely positive mapping preserving  vacuum
expectation. Moreover, this second quantization satisfies the usual
functorial properties; for instance,
$$\Gamma(UV)=\Gamma(U)\Gamma(V)$$
for any contractions $V\ :\ H_{1} \to H_{2}$ and $U\ :\ H_{2} \to H_{3}.$


Similarly, we define the right second quantization
$\Gamma_{r}(V)\ :\ \Gamma_{q,r}(H_{1}) \to \Gamma_{q, r}(H_{2})$ by
$$(\Gamma_{r}(V) x)\Omega = {\cal F}(V) (x \Omega),\quad x \in
\Gamma_{q, r}(H_{1})$$
or
$$\Gamma_{r}(V) W_{r}(\xi) = W_{r}({\cal F} (V) \xi), \quad \xi \in {\cal
A}(H{1{\Bbb C}}). \leqno(2.6)$$
The following simple fact will be important later.

\proclaim Lemma 2.3.  Let $H_{1}, H_{2}$ be two real Hilbert spaces,
and let $V\ :\ H_{1} \to H_{2}$ be a contraction. Then
$$\Gamma(V)x=\Gamma_{r}(V)x,\quad \forall x \in
\Gamma_{q}(H_{1}) \cap \Gamma_{q, r} (H_{2}). \leqno(2.7)$$

\noindent{\it Proof.} It is well-known that $V$ admits a decomposition $V =
PUI,$ where $I$ is an isometry, $U$ an orthogonal and $P$ an
orthogonal projection. By the functoriality of both second
quantizations, it suffices to prove the lemma when $V$ is one of these
operators.

Suppose $V$ is an isometry. By [BKS], $\Gamma (V)$ is the natural
embedding of $\Gamma_{q}(H_{1})$ into $\Gamma_{q}(H_{2}).$ In the
same way, $\Gamma_{r}(V)$ is the natural embedding of $\Gamma_{q, r}(H_{1})$ into $\Gamma_{q,
r}(H_{2})$ as well. Then (2.7) follows.

Suppose $V$ is an orthogonal or an orthogonal projection. Then by [BKS]
$$\Gamma(V)x={\cal F}(V)x{\cal F}(V)^{\ast},\quad x\in \Gamma_{q}(H_{1}).$$
Similarly,
$$\Gamma_{r}(V)x ={\cal F}(V)x{\cal F}(V)^{\ast}, \quad x \in\Gamma_{q, r}(H_{1}).$$
Thus $\Gamma(V)$ and $\Gamma_{r}(V)$ are implemented by the same  second quantization ${\cal
F}(V)$ at the Fock space level;  whence (2.7). \hfill$\diamond$
\medskip
Fix an orthonormal basis $\{ e_{i} \}_{i \in I}$ of $H.$ We use the
notation $\underline i$ to denote a multiple index, i.e., $\underline
i = (i_{1}, \ldots, i_{n}) \in I^{n}.$ The length of such a multiple
index is defined as equal to $n,$ denoted by $\vert \underline i
\vert.$ The empty set $\phi$ is also regarded as a multiple
index of length zero. Let $A$ denote the family of all multiple
indices and $A_{n}$ the subfamily of all multiple indices of length $n$
$(n \geq 0).$ We set
$$ e_{\phi} = \Omega, \ e_{\underline i} = e_{i_{1}} \otimes \cdots
\otimes e_{i_{n}}, \quad\underline i = (i_{1}, \cdots, i_{n}) \in A \ \ (n
\geq 1).$$
Then $\{ e_{\underline i}\}_{\underline i \in A}$ is an orthonormal
basis of the full Fock space ${\cal F}(H_{\Bbb C}).$ Recall that the
scalar product $\langle \cdotp, \cdotp \rangle_{q}$ on ${\cal F}_{q}(H_{\Bbb C})$ is given by
$$\langle \xi, \eta\rangle_{q} = \langle \xi, P^{(n)}_{q}
\eta\rangle,\quad \xi, \eta \in H_{\Bbb C}^{\otimes n}\quad (n \geq 0) $$
and that the symmetrizator $P^{(n)}_{q}$ is strictly positive operator on
$H^{\otimes n}_{\Bbb C}$ (when equipped with the free scalar product).
Thus the two scalar products $\langle \cdotp, \cdotp \rangle_{q}$
and $\langle \cdotp , \cdotp \rangle$ are equivalent on $H^{\otimes
n}_{\Bbb C}$ for each $n \geq 0.$ Now let $\xi \in {\cal F}_{q}
(H_{\Bbb C}).$ Then
$$\xi = \sum_{n \geq 0} \ \xi_{n} \quad \hbox{ with } \quad \xi_{n}
\in H^{\otimes n}_{\Bbb C}, \ n \geq 0,$$
where the series converges in ${\cal F}_{q}(H_{\Bbb C})$ when
$H^{\otimes n}_{\Bbb C}$ is equipped with the free scalar product, and
so when $H^{\otimes n}_{\Bbb C}$ is equipped with the $q$-scalar
product as well. Noting that
$$\langle e_{\underline i}, \xi_{n}\rangle = \langle e_{\underline
i}, \xi \rangle,\quad \underline i \in A_{n},\ n \geq 0,$$
we deduce that
$$\xi=\sum_{n \geq 0}\sum_{\underline i \in A_{n}} \langle
e_{\underline i}, \xi \rangle e_{\underline i} = \sum_{\underline i
\in A}\langle e_{\underline i}, \xi \rangle e_{\underline i},
\leqno(2.8)$$
where the series converges in ${\cal F}_{q}(H_{\Bbb C}).$


The following proposition might have some independent interest. For the
proof of Theorem 2.1 we will not really need (i) below. However, (i)
and its proof would be helpful in understanding the convergence of
various series involved in the proof of Theorem 2.1.

\proclaim Proposition 2.4. (i) Let $\xi \in {\cal F}_{q}(H_{\Bbb C}).$ Then
the formal series
$$\tilde x_{\xi} =\sum_{\underline i \in A}\langle e_{\underline
i}, \xi \rangle W(e_{\underline i})$$
defines a closable densely defined operator on ${\cal F}_{q}(H_{\Bbb
C})$  whose closure  $x_{\xi}$ is affiliated
with $\Gamma_{q}(H)$. \hfill\break
\indent
(ii) Let $x \in \Gamma_{q}(H).$ Then $x = x_{\xi}$ with $\xi = x\Omega.$\hfill\break
\indent
(iii) The statements similar to (i) and (ii) hold for the right von Neumann algebra and the right
Wick product.


\noindent{\it Proof.} (i) First note the following useful identity
$$W(\xi)\eta= W_{r}(\eta)\xi,\quad \xi, \eta\in {\cal A}(H_{\Bbb C}). \leqno(2.9)$$
This immediately follows from the following  identity (applied to $\Omega$)
$$W(\xi)W_r(\eta)= W_{r}(\eta)W(\xi),\quad \xi, \eta\in {\cal A}(H_{\Bbb C}). $$
Now let $\xi \in {\cal F}_{q}(H_{\Bbb C})$ and $\underline j \in A.$
By (2.8) and (2.9),
$$\eqalign{
W_{r} (e_{\underline j})\xi &= \sum_{\underline i \in A}\langle
e_{\underline i}, \xi \rangle W_{r}(e_{\underline j})e_{\underline i}\cr
\noalign{\smallskip}
&= \sum_{\underline i \in A}\langle e_{\underline i}, \xi \rangle
W(e_{\underline i}) e_{\underline j},\cr}$$
where the last series converges in ${\cal F}_{q}(H_{\Bbb C}).$ Thus
$\tilde x_{\xi}(e_{\underline j})$ is well-defined and
$$\tilde x_{\xi}(e_{\underline j}) = W_{r}(e_{\underline j})\xi.$$
Then, by linearity, $\tilde x_{\xi}$ is defined on the dense
subspace ${\cal A}(H_{\Bbb C}),$ and
$$\tilde x_{\xi}(\eta)=W_{r}(\eta)\xi,\quad \forall\ \eta \in
{\cal A} (H_{\Bbb C}). \leqno(2.10)$$
We next show that $\tilde x_{\xi}$ is closable. Let
$$\tilde y_{\xi} = \sum_{\underline i \in A}\overline{\langle \xi,
e_{\underline i} \rangle}\, W(J({e_{\underline i}})),$$
where $J$ is the involution introduced during the proof of Lemma 2.2.
The preceding arguments equally show that $\tilde y_{\xi}$ is also
defined on ${\cal A}(H_{\Bbb C})$ and 
$$\tilde y_{\xi}(\eta) = \sum_{\underline i \in A}\overline{\langle \xi, e_{\underline i} \rangle}\,
W(J({e_{\underline i}}))\eta,\quad \eta\in{\cal A}(H_{\Bbb C}).$$
Now let $\eta_{1}, \eta_{2} \in {\cal
A} (H_{\Bbb C}).$ Then
$$\eqalign{
\langle \tilde x_{\xi} \eta_{2}, \eta_{1} \rangle_{q} &=
\sum_{\underline i}\overline{\langle \xi, e_{\underline i}\rangle}\langle
\eta_{2}, W(e_{\underline i})^{\ast}\eta_{1}\rangle_{q}\cr
\noalign{\smallskip}
&= \sum_{\underline i}\overline{\langle \xi, e_{\underline i}\rangle}
\langle \eta_{2}, W(J(e_{\underline i}))\eta_{1}\rangle_{q}\cr
\noalign{\smallskip}
&= \langle \eta_{2}, \tilde y_{\xi} \eta_{1} \rangle_{q}.\cr
}$$
Thus it follows that
$\tilde y_{\xi} \subset \tilde x^{\ast}_{\xi}.$
Therefore, $\tilde x^{\ast}_{\xi}$ is densely defined, and
consequently, $\tilde x_{\xi}$ is closable.

To prove that $x_{\xi}$ is affiliated with $\Gamma_{q}(H),$ it
suffices to show
$$\tilde x_{\xi}(W_{r}(\eta_{1})\eta_{2})=W_{r}(\eta_{1})
(\tilde x_{\xi} \eta_{2}),\quad \eta_{1}, \eta_{2} \in {\cal A}(H_{\Bbb
C}).$$
This immediately follows from (2.10) for
$$\eqalign{
\tilde x_{\xi}(W_{r}(\eta_{1})\eta_{2}) &= W_{r}(W_{r}(\eta_{1})
\eta_{2})\xi\cr
\noalign{\smallskip}
&=W_{r}(\eta_{1}) W_{r}(\eta_{2})\xi\cr
\noalign{\smallskip}
&= W_{r}(\eta_{1})(\tilde x_{\xi}\eta_{2}).\cr}$$

(ii) Let $x \in \Gamma_{q}(H)$ and $\xi = x \Omega.$ Then for any
$\eta \in {\cal A}(H_{\Bbb C}),$ by (2.10)
$$x \eta = x W_{r}(\eta)\Omega= W_{r}(\eta)x\Omega=\tilde x_{\xi}(\eta);$$
whence $x = x_{\xi}.$

(iii) The proof is entirely similar. \hfill $\diamond$
\medskip

Now we are ready to prove Theorem 2.1.


\noindent{\it Proof of Theorem 2.1.} By Lemma 2.2 we must show that
$\Gamma_{q}(H) \cap \Gamma_{q, r}(H)$ is trivial. Let $x \in
\Gamma_{q}(H) \cap \Gamma_{q, r}(H).$ Then by Proposition 2.4
$$x=\sum_{\underline i\in A}\alpha_{\underline i}W(e_{\underline i})\quad\hbox{and}\quad
x =\sum_{\underline i\in A}\alpha_{\underline i}W_{r}(e_{\underline i})\,,$$
where $\alpha_{\underline i} = \langle e_{\underline i}, x\Omega\rangle$, $\underline i \in A.$ The
convergence of the two series above is understood as in Proposition 2.4 (and its proof). Let $t \in
[-1, 1].$  By Lemma 2.3
$$\Gamma (t\undb_{H})x = \Gamma_{r} (t\undb_{H})x.
\leqno(2.11)$$
Again by Proposition 2.4
$$\Gamma\ (t \undb_{H}) x = \sum_{\underline i\in A}\beta_{\underline i}W(e_{\underline i}),$$
where $\beta_{\underline i}=\langle e_{\underline i}, \big(\Gamma(t
\undb_{H})x\big)\Omega\rangle$, $\underline i \in A.$ We claim that
$$\beta_{\underline i} =t^{\vert \underline i\vert}
\alpha_{\underline i},\quad \underline i \in A, \ -1\le t\le 1.$$
Indeed, for each $\underline i \in A_{n}\ (n \geq 0)$ we have
$$\eqalign{
\beta_{\underline i} &= \langle (P^{(n)}_{q})^{-1} e_{\underline
i},\ (\Gamma (t\undb_{H}) x) \Omega \rangle_{q}\cr
\noalign{\smallskip}
&= \langle (P^{(n)}_{q})^{-1}e_{\underline i},\ {\cal F}(t
\undb_{H})(x \Omega)\rangle_{q}\cr
\noalign{\smallskip}
&= \langle {\cal F} (t\undb_{H})^{\ast} (P^{(n)}_{q})^{-1}
e_{\underline i},\ x \Omega\rangle_{q}\cr
\noalign{\smallskip}
&= \langle (P^{(n)}_{q})^{-1}{\cal F}(t\undb_{H})^{\ast}
e_{\underline i}, x\Omega\rangle_{q}\cr
\noalign{\smallskip}
&= t^{n}\langle (P^{(n)}_{q})^{-1}e_{\underline i}, x\Omega\rangle_{q}
=t^{n}\alpha_{\underline i},\cr }$$ 
where for the next to the last equality we have used the easily
checked fact that
$$P^{(n)}_{q}{\cal F} (t\undb_{H}) = {\cal F}_{q} (t\undb_{H})
P^{(n)}_{q} \quad \hbox{ on } H^{\otimes n}_{\Bbb C}.$$
Therefore,
$$\Gamma(t\undb_{H})x=\sum_{\underline i} t^{\vert \underline
i \vert}\alpha_{\underline i}W(e_{\underline i}),\quad -1 \leq t \leq1.$$
Similarly,
$$\Gamma_{r}(t\undb_{H})x=\sum_{\underline i}t^{\vert
\underline i\vert}\alpha_{\underline i} W_{r}(e_{\underline i}),\quad 
-1 \leq t \leq 1.$$
Combining the last two equalities with (2.11), we obtain
$$\sum_{n \geq 0}t^{n} \sum_{\underline i \in A_{n}}
\alpha_{\underline i}W(e_{\underline i}) = \sum_{n \geq 0}\ t^{n}
\sum_{\underline i \in A_{n}}\alpha_{\underline i}
W_{r}(e_{\underline i}),\quad -1 \leq t \leq 1.$$
Identifying the coefficients of $t^{n}\ (n \geq 0),$ we get
$$\sum_{\underline i \in A_{n}}\alpha_{\underline i}
W(e_{\underline i}) = \sum_{\underline i \in A_{n}} \alpha_{i}
W_{r} (e_{\underline i}),\quad \forall\ n \geq 0. \leqno(2.12)$$
Now it is easy to show that (2.12) implies 
$$\alpha_{\underline i} = 0,\quad \forall\ \underline i \in A \ \hbox{with } 
\ \underline i \not=\phi.$$ 
To this end fix $\underline j = (j_{1}, \cdots, j_{n}) \in A_{n}$ with $n \geq
1.$ Since $\dim H \geq 2,$ there is $k \in I$ such that $k \not=
j_{1}.$ We first observe that by (2.9), for any  $\underline i = (i_{1}, \cdots, i_{n}) \in A_{n}$
$$W(e_{\underline i})e_{k}=W_r(e_k)e_{\underline i}=e_{{\underline i}k}+\sum_{s=1}^nq^{s-1}\la e_k,
e_{i_s}\ra e_{i_1}\otimes\cdots \otimes{\displaystyle\mathop e^\vee}_{i_s}\otimes\cdots
\otimes e_{i_n}\,,$$ 
where ${\underline i}k=(i_{1}, \cdots, i_{n}, k)$. Then applying
the left side of (2.12) to
$e_{k}$ yields
$$\sum_{\underline i\in A_{n}}\alpha_{i}W(e_{\underline i})e_{k} =
\sum_{\underline i\in A_{n}}\alpha_{\underline i}\ e_{\underline ik} + \sum_{\underline i \in
A_{n-1}}\gamma_{\underline i}\ e_{\underline i}, \leqno(2.13)$$
where $\gamma_{\underline i}\in{\Bbb C}$ are some coefficients for $\underline i \in A_{n-1}$. On the
other hand, applying the right side of (2.12) to $e_{k},$ in a similar way, we obtain
$$\sum_{\underline i \in A_{n}}\alpha_{\underline i}W_{r}(e_{i})
e_{k} = \sum_{\underline i \in A_{n}}\alpha_{\underline i}e_{k\underline i} 
+ \sum_{\underline i \in A_{n-1}}\gamma'_{\underline i} e_{\underline i}, \leqno(2.14)$$
where $k\underline i = (k, i_{1}, \ldots, i_{n)}$ and
$\gamma'_{\underline i} \in \Bbb C$, $\underline i \in A_{n-1}$.  Observe that the basis vector
$e_{\underline j k}$ appears once and only once in the right side of
(2.13). However, since $k \not= j_{1},\ e_{\underline j k}$ never
appears in the right side of (2.14). Therefore, combining (2.12) -
(2.14), and identifying the coefficients before $e_{\underline j k}$,
we finally deduce
$\alpha_{\underline j} = 0$, as desired.
Thus, we have arrived at
$$x = \alpha_{\phi}\ W(e_{\phi}) = \alpha_{\phi},$$
which proves the center of $\Gamma_{q}(H)$ is trivial, and so
accomplishes the proof of Theorem 2.1.\hfill $\diamond$
 \medskip
 \noindent{\bf Remark.} By the ultracontractivity of the
$q$-Ornstein-Uhlenbeck semigroup proved in [B2], the series
$$\sum_{\underline i \in A}t^{\vert\underline i\vert}
\alpha_{\underline i} W(e_{\underline i})$$
converges to $\Gamma (t\undb_{H})x$ in $\Gamma_{q}(H)$ for any $-1 <
t < 1.$ This can be used in the preceding proof in place of Proposition 2.4.
However, we have chosen the previous approach via Proposition 2.4 for, on
the one hand, we would like to keep our presnetation as elementary as
possible, and on the other hand, we feel that this approach might be
useful in case one does not know  ultracontractivity.

\bigskip
\centerline{\bf 3. The second quantization}
\medskip
Our next objective is to extend Theorem 2.1 to the general setting
described in section 1. As we have seen in the previous section, the
second quantization played an important r\^ole in the proof of Theorem
2.1. Hence, transferring this proof to the general setting
necessitates the introduction of a good second quantization. We will
do this for a rather large class of contractions. Our analysis below
reveals that it is impossible to define a nice second quantization for
any contraction.

In the sequel, we will always assume all Yang-Baxter operators in
consideration are tracial and of norm less than 1.


Now let $H_{1}$ and $H_{2}$ be two real Hilbert spaces, and let
$T^{(j)}$ be a Yang-Baxter operator on $H_{j\Bbb C}\otimes H_{j\Bbb C}$ 
(so, $T^{(j)}$ is tracial, hermitian and  $\Vert T^{(j)}\Vert < 1$, by our assumption above), $j = 1, 2.$ Let
$V\ :\ H_{1} \to H_{2}$ be a contraction. We want to define a second quantization $\Gamma (V)$
from $\Gamma_{T^{(1)}}\ (H_{1})$ into $\Gamma_{T^{(2)}}\ (H_{2})$. To this end we
first consider the second quantization at the Fock space level.


\proclaim Lemma 3.1. Let $V\ :\ H_{1} \to H_{2}$  be a
contraction and $V_{\Bbb C}\ :\ H_{1{\Bbb C}} \to H_{2{\Bbb C}}$ its complexification. 
Suppose
$$\bigl( V_{\Bbb C} \otimes V_{\Bbb C} \bigl)T^{(1)} = T^{(2)}
\bigl( V_{\Bbb C} \otimes V_{\Bbb C} \bigl). \leqno(3.1)$$
Then the linear mapping ${\cal F}(V)\ :\ {\cal A} \bigl(H_{1\Bbb
C}\bigl)\to {\cal A} \bigl(H_{2\Bbb C}\bigl)$ determined by
$${\cal F} (V)f_{1} \otimes \cdots \otimes f_{n} = V_{\Bbb C}
f_{1} \otimes \cdots \otimes V_{\Bbb C} f_{n},\quad f_{1}, \cdots, f_{n}
\in H_{1 \Bbb C},\ n \geq 0,$$
extends to a contraction from ${\cal F}_{T^{(1)}} \bigl( H_{1 \Bbb
C} \bigl)$ into ${\cal F}_{T^{(2)}} \bigl( H_{2 \Bbb C} \bigl).$

\noindent{\it Proof.} First observe that (3.1) implies
$${\cal F}(V)P^{(n)}_{T^{(1)}} = P^{(n)}_{T^{(2)}}{\cal F}(V)\
\hbox{ on }\ H^{\otimes n}_{1 \Bbb C},\ n \geq 0.$$
Therefore,
$${\cal F}(V)^{\ast}{\cal F}(V)P^{(n)}_{T^{(1)}} =
P^{(n)}_{T^{(1)}}{\cal F}(V)^{\ast}{\cal F}(V) \ \hbox{ on }\
H^{\otimes n}_{1 \Bbb C},$$
and so
$${\cal F}(V)^{\ast}{\cal F}(V)\sqrt{P^{(n)}_{T^{(1)}}} =
\sqrt{P^{(n)}_{T^{(1)}}}{\cal F}(V)^{\ast}{\cal F}(V) \ \hbox{
on }\ H^{\otimes n}_{1 \Bbb C}.$$
Now let $\xi \in H^{\otimes n}_{1 \Bbb C}.$ Then by the last equality
and the fact that ${\cal F} (V)$ is a contraction from ${\cal F}\bigl(
H_{1 \Bbb C}\bigl)$ to ${\cal F}\bigl( H_{2 \Bbb C}),$ we obtain
$$\eqalign{
\Vert {\cal F}(V) \xi \Vert^{2}_{T^{(2)}} &= \langle
P^{(n)}_{T^{(2)}}{\cal F}(V)\xi,\ {\cal F}(V)\xi \rangle\cr
\noalign{\smallskip}
&= \langle {\cal F} (V)^{\ast}{\cal F}(V) P^{(n)}_{T^{(1)}}\xi, \ \xi \rangle\cr
\noalign{\smallskip}
&= \langle \sqrt{P^{(n)}_{T^{(1)}}}{\cal F}(V)^{\ast}{\cal F}(V)
\sqrt{P^{(n)}_{T^{(1)}}}\xi, \xi \rangle\cr
\noalign{\smallskip} 
&= \langle {\cal F}(V)^{\ast}{\cal F}(V)\sqrt{P^{(n)}_{T^{(1)}}}\xi, 
\sqrt{P^{(n)}_{T^{(1)}}}\xi\rangle\cr
\noalign{\smallskip}
&\leq \langle\sqrt{P^{(n)}_{T^{(1)}}}\xi, 
\sqrt{P^{(n)}_{T^{(1)}}}\xi\rangle= \Vert \xi \Vert^{2}_{T^{(1)}}.\cr}$$
Thus the lemma is proved. \hfill $\diamond$


The above second quantization at the Fock space level satisfies all usual properties as
one can expect. For instance, if $V$ is an isometry, orthogonal or
orthogonal projection, then accordingly, ${\cal F}(V)$ is an
isometry, unitary or orthogonal projection.

\noindent{\bf Remark.} The condition (3.1) is also necessary to have a
nice  second quantization at the Fock space level. For example, let $V\ :\ H_{1} \to
H_{2}$ be an orthogonal. Suppose ${\cal F}(V)$, defined as in lemma 3.1, extends to  an unitary.
Then
$V$ must satisfy (3.1). Indeed, we then have, in particular,
$$\langle\bigl( V^{\ast}_{\Bbb C} \otimes V^{\ast}_{\Bbb C}\bigl)
P^{(2)}_{T^{(2)}}\bigl( V_{\Bbb C} \otimes V_{\Bbb C})\xi, \eta
\rangle = \langle P^{(2)}_{T^{(1)}}\xi, \eta \rangle,\quad
\forall\ \xi, \eta \in H^{\otimes 2}_{1 \Bbb C}.$$
Since
$$P^{(2)}_{T^{(j)}} = 1 + T^{(j)},\quad j = 1, 2,$$
we deduce (3.1).
\medskip

We pass onto the consideration of the second quantization at the von
Neumann algebra level. For this we will need Wick products.


Let $H$ be a real Hilbert space and $T$ be a Yang-Baxter operator as
before. As in the $q$-case, since $\Omega$ is separating for
$\Gamma_{T}(H),$ the mapping $x \mapsto x \Omega$ is injective from
$\Gamma_{T}(H)$ into ${\cal F}_{T}(H_{\Bbb C}).$ Denote its image by
$\Gamma^{\infty}_{T}(H).$ Let
$W\ :\ \Gamma^{\infty}_{T}(H) \to \Gamma_{T}(H)$
be the inverse mapping. This is the Wick product. As in the $q$-case,
we have
$$W\bigl(f\otimes f_{1} \otimes \cdots \otimes f_{n} \bigl) =
G(f)W\bigl(f_{1}\otimes \cdots\otimes f_{n}\bigl) - W\bigl(a(f)f_{1}\otimes\cdots\otimes f_{n}
\bigl),\quad f_{1}, \cdots, f_{n} \in H. \leqno(3.2)$$
Using (3.2) and the fact that $\Omega$ is separating, we easily
check, by induction, that ${\cal A} (H_{\Bbb C}) \subset
\Gamma^{\infty}_{T} (H)$ (cf. section 2).


The reader is referred to [K] for an explicit formula for  
$W (f_{1} \otimes \cdots \otimes f_{n})$ in terms of ordered normal terms. Recall that
an ordered normal term is an operator of the form $a^*(f_1)\cdots a^*(f_n)a(g_m)\cdots
a(g_1)$ for some $f_1, \cdots, f_n, g_1,\cdots, g_m\in H$ ($n,m\ge 0$). What we will need
later is the trivial fact that
$W(f_{1}
\otimes \cdots \otimes f_{n})$ can be written as a sum of  ordered normal terms. This
immediately follows from (3.2) (by induction) and the deformed
commutation relations (1.3). The sum obtained is in general an
infinite sum but convergent in the strong operator topology.


We collect some simple properties of Wick product in the following
lemma for later reference.

\proclaim Lemma 3.2. (i) Let $\xi\in{\cal A}(H_{\Bbb C}).$ Then
$W(\xi)$ is the unique operator in $\Gamma_{T}(H)$ such that
$W(\xi)\Omega =\xi$ ; moreover, $W(\xi)$ is a sum of 
ordered normal terms.\hfill\break\indent
(ii) $W({\cal A}(H_{\Bbb C}))$ is a $w^{\ast}$-dense
$\ast$-subalgebra of $\Gamma_{T}(H).$


Now we are in the position to define the second quantization. Let
$H_{j}$ be a real Hilbert space and $T^{(j)}$ a Yang-Baxter operator
on $H_{j \Bbb C} \otimes H_{j \Bbb C}$  ($T^{(j)}$ is tracial and $\|T^{(j)}\|<1$), $j = 1,2$. 
Let $V\ :\ H_{1}\to H_{2}$ be a contraction satisfying (3.1). We want to define the
second quantization $\Gamma(V)$ from $\Gamma_{T^{(1)}}(H_{1})$ to
$\Gamma_{T^{(2)}}(H_{2})$, which would posses all expected properties.
Following the $q$-case, $\Gamma (V)$ should be defined as
$$(\Gamma(V) x)\Omega = W({\cal F}(V)x \Omega),\quad x \in
\Gamma_{T^{(1)}}(H_{1})$$
or
$$\Gamma(V)W(\xi) = W({\cal F}(V)\xi),\quad \xi \in {\cal A}
(H_{1 \Bbb C}). \leqno(3.3)$$
The first step towards this problem is the following simple fact.


\proclaim Lemma 3.3. With the above notation and assumption, if $V$
is an isometry, an orthogonal or an orthogonal projection, $\Gamma(V)$
is well-defined and is an isometry, an isomorphism or a faithful normal conditional
expectation.

\noindent{\it Proof.} Suppose $V$ is an isometry. Then we can identify
$H_{1}$ as a closed subspace of $H_{2}.$ (3.1) implies
$$T^{(1)} = T^{(2)} \Bigl |_{H^{\otimes 2}_{1 \Bbb C}}.$$
Thus $H^{\otimes 2}_{1 \Bbb C}$ is an invariant subspace of $T^{(2)}.$
It follows that
$$P^{(n)}_{T^{(1)}} = P^{(n)}_{T^{(2)}} \biggl |_{H^{\otimes n}_{1
\Bbb C}},\quad \forall\ n \geq 0.$$
Therefore, ${\cal F}_{T^{(1)}} (H_{1 \Bbb C})$ can be naturally
identified as a closed subspace of ${\cal F}_{T^{(2)}} (H_{2 \Bbb
C})$ (via the isometry $\F(V)$). Let
$$M = \{ G(f)\ :\ f \in H_{1}\}'' \subset B ({\cal F}_{T^{(2)}} (H_{2
\Bbb C})).$$
It is easy to see that for any $x \in \Gamma_{T^{(1)}} (H_{1})$ there
is $y \in M$ such that
$$y\Omega = {\cal F}(V)(x \Omega).$$
Since $\Omega$ is separating, $y$ is unique. Thus $\Gamma (V)$ is
well-defined. Set
$$\pi (y) = y \biggl |_{{\cal F}_{T^{(1)}} (H_{1 \Bbb C})}, \quad y
\in M.$$
Clearly, $\pi$ is a unital normal homomorphism from $M$ to
$\Gamma_{T^{(1)}} (H_{1}).$ $\pi$ is also injective for $\pi (y) = 0$
implies $y \Omega = 0,$ and so $y = 0.$ Finally, $\pi (M)$ contains
all Gaussians from $\Gamma_{T^{(1)}} (H_{1}).$ Hence $\pi (M)$ is
$w^{\ast}$-dense in $\Gamma_{T^{(1)}} (H_{1}).$ Therefore, it follows
that $\pi$ is an isomorphism from $M$ onto $\Gamma_{T^{(1)}} (H_{1}).$
Thus, $\Gamma(V) = \pi^{-1},$ and so $\Gamma_{T^{(1)}} (H_{1})$ is
naturally identified with the von Neumann subalgebra $M$ of
$\Gamma_{T^{(2)}} (H_{2}).$


Now suppose $V$ is an orthogonal or an orthogonal projection. Consider
an ordered normal term
$$x = a^{\ast} (f_{1}) \cdots a^{\ast}(f_{n})a(g_{m}) \cdots
a(g_{1}), \quad f_{1}, \cdots, f_{n}, g_{1}, \cdots, g_{m} \in H_{1}\,,\ n,m\ge 0.$$
Observe that
$$T^{(1)}_{k}\ell(g) = \ell(g)T^{(1)}_{k+1},\quad \forall\ g \in
H_{1},\ \forall\ k \geq 1. \leqno(3.4)$$
Let $k \geq m.$ Put, for $1 \leq i \leq j \leq k-1$
$$R\Bigl(T^{(1)}_{i}, \cdots, T^{(1)}_{j} \Bigl) = 1 + T^{(1)}_{i} +
T^{(1)}_{i}T^{(1)}_{i+1} + \cdots + T^{(1)}_{i} \cdots T^{(1)}_{j}.$$
Similarly, we define $R\Bigl(T^{(2)}_{i}, \cdots, T^{(2)}_{j}\Bigl).$ 
Using (3.4), we have, on $H^{\otimes k}_{1 \Bbb C}$
$$\eqalign{
&a(g_{m}) \cdots a(g_{1}) = a(g_{m})\cdots a(g_{2}) \ell(g_{1})
R\Bigl(T^{(1)}_{1}, \cdots, T^{(1)}_{k-1}\Bigl)\cr
\noalign{\smallskip}
&=a(g_{m})\cdots a(g_{3})\ell(g_{2})R\Bigl( T^{(1)}_{1},
\cdots, T^{(1)}_{k-2}\Bigl)\ell(g_{1})R\Bigl(T^{(1)}_{1},
\cdots, T^{(1)}_{k-1} \Bigl)\cr
\noalign{\smallskip}
&=a(g_{m})\cdots a(g_{3})\ell(g_{2})\ell(g_{1})R\Bigl(
T^{(1)}_{2}, \cdots, T^{(1)}_{k-1}\Bigl) R\Bigl( T^{(1)}_{1},
\cdots, T^{(1)}_{k-1}\Bigl)\cr
\noalign{\smallskip}
&=\cdots =\prod^{1}_{j = m}\ell(g_{j})\cdotp\prod^{1}_{j = m}R
\Bigl(T^{(1)}_{j}, \cdots, T^{(1)}_{k-1}\Bigl).\cr}$$
On the other hand, by (3.1)
$$R\Bigl(T^{(1)}_{j}, \cdots, T^{(1)}_{k-1}\Bigl){\cal
F}(V^{\ast}) ={\cal F}(V^{\ast})R\Bigl(T^{(2)}_{j}, \cdots,
T^{(2)}_{k-1}\Bigl).$$
Combining the preceding equalities, we deduce that on $H^{\otimes k}_{2 \Bbb C}$
$$\eqalign{
&{\cal F}(V)x{\cal F}(V)^{\ast}\cr
\noalign{\smallskip}
&={\cal F}(V)\Biggl[\prod^{n}_{i=1}\ell^{\ast}(f_{i})\cdotp
\prod^{1}_{j=m}\ell(g_{j})\Biggl]{\cal F}(V)^{\ast}\cdotp
\prod^{1}_{j=m} R\Bigl(T^{(2)}_{j}, \cdots, T^{(2)}_{k-1}\Bigl).\cr}$$
Hence, we are reduced to the free case. However, in the free case, if
$V$ is an orthogonal or an orthogonal projection,
$$\eqalign{
&{\cal F}(V)\Biggl[\prod^{n}_{i=1}\ell^{\ast}(f_{i})\cdotp
\prod^{1}_{j = m}\ell(g_{j})\Biggl]{\cal F}(V)^{\ast}\cr
\noalign{\smallskip}
&=\prod^{n}_{i=1}\ell^{\ast}(Vf_{i})\cdotp \prod^{1}_{j=m}
\ell (V g_{j}).\cr}$$
Therefore, inversing the order of the above arguments, we get
$${\cal F}(V)x{\cal F}(V)^{\ast} = \prod^{n}_{i=1} a^{\ast}(V
f_{i})\cdotp\prod^{1}_{j=m}a(Vg_{j}).$$
Then by Lemma 3.2, it follows
that
$${\cal F}(V)W(f_{1}\otimes \cdots \otimes f_{n}){\cal F}(V)^{\ast} 
= W(Vf_{1}\otimes\cdots\otimes Vf_{n}), \quad f_{1},\cdots, f_{n} \in H_{1},$$
and so
$${\cal F}(V)W(\xi){\cal F}(V)^{\ast} = W({\cal F}(V)\xi),\quad \xi
\in{\cal A}(H_{1 \Bbb C}).$$
Thus $\Gamma(V)$ is well-defined, and is implemented by ${\cal F}(V).$
Clearly, if $V$ is an orthogonal (resp. orthogonal projection), the
above formula shows that $\Gamma (V)$ is an isomorphism (resp. a faithful normal conditional
expectation) from $\Gamma_{T^{(1)}}(H_{1})$ onto $\Gamma_{T^{(2)}} (H_{2}).$
\hfill $\diamond$

\noindent{\bf Remark.} Let $V\ :\ H_{1} \to H_{2}$ be an orthogonal.
Suppose that the second quantization $\Gamma(V)$ defined by (3.3) is
an isomorphism between $\Gamma_{T^{(1)}} (H_{1})$ and
$\Gamma_{T^{(2)}} (H_{2}).$ Then, the Fock space second
quantization ${\cal F}(V)$ is a unitary from ${\cal F}_{T^{(1)}}(H_{1
\Bbb C})$ onto ${\cal F}_{T^{(2)}} (H_{2 \Bbb C}).$ Therefore, by the
remark after Lemma 3.1, $V$ must satisfy the condition (3.1). This
shows that if we define $\Gamma (V)$ by (3.4), the condition (3.1) is
necessary.
\medskip 

Now we return back to a general contraction $V$ satisfying (3.1). As
in the $q$-case, we would like to reduce $V$ to the special
contractions already considered in Lemma 3.3. It is always possible
to decompose $V$ as $V = P U I,$ where $I$ is an isometry, $U$ an
orthogonal and $P$ an orthogonal projection. Then we are confronted
with another problem : how to define a Yang-Baxter operator $T$
associated with the bigger Hilbert space (on which $U$ acts) such that
$P, \ U,\  I$ all satisfy (3.1), relative to the corresponding Yang-Baxter operators. 
At this point, the following simple result would be useful.

\proclaim Proposition 3.4.  Let $H_{j}$ be a real Hilbert space and
$T^{(j)}$ a tracial Yang-Baxter operator on $H_{j \Bbb C} \otimes H_{j \Bbb
C}$ with $\|T^{(j)}\|<1$ ($j = 1,2$). Put
$$H = H_{1} \oplus H_{2}.$$
Let $I_{j}$ (resp. $P_{j})$ be the natural embedding (resp.
projection) of $H_{j}$ into $H$ (resp. from $H$ onto $H_{j}),\ j = 1,
2.$ Define
$$T\ :\ H_{\Bbb C} \otimes H_{\Bbb C} \to H_{\Bbb C} \otimes H_{\Bbb
C}$$
by
$$T\Bigl|_{H_{i \Bbb C} \otimes H_{j \Bbb C}} = \delta_{ij}
\Bigl(I_{j \Bbb C} \otimes I_{j \Bbb C}\Bigl)T^{(j)},\quad i,j = 1,2.$$
\indent(i) $T$ is a tracial Yang-Baxter operator with $\Vert T \Vert < 1$; moreover,
$$\Big(I_{j\Bbb C}\otimes I_{j\Bbb C}\Big)T^{(j)}
=T\Big(I_{j\Bbb C}\otimes I_{j \Bbb C}\Big),\quad
T^{(j)}\Big(P_{j\Bbb C}\otimes P_{j\Bbb C}\Big) = \Big(
P_{j\Bbb C}\otimes P_{j\Bbb C}\Big)T,\quad j = 1, 2.$$
\indent(ii) Let $V\ :\ H_{1} \to H_{2}$ be a contraction satisfying (3.1).
Define $U\ :\ H \to H$ by
$$U \Bigl |_{H_{1}} = I_{1} V \quad \hbox{ and } \quad U\Bigl
|_{H_{2}} = I_{2} V^{\ast}.$$
Then $U$ is a contraction, and 
$$\Big(U_{\Bbb C}\otimes U_{\Bbb C}\Big)T=T\Big(U_{\Bbb C}\otimes U_{\Bbb C}\Big)\,,\quad 
V = P_{2}UI_{1}.$$

\noindent{\it Proof.} For notational simplicity, we will not distinguish a real operator and its
complexification,  and we regard $T^{(j)}$ as an operator acting on $H_{\Bbb C} \otimes H_{\Bbb
C}$ in the natural way. Then we have
$$T = T^{(1)} \bigl( P_{1} \otimes P_{1} \bigl) + T^{(2)} \bigl(
P_{2} \otimes P_{2} \bigl).$$
Thus,
$$\eqalign{
&T \otimes \undb_{H_{\Bbb C}} = \Bigl( T^{(1)} \otimes \undb_{H_{1
\Bbb C}} \Bigl) P_{1} \otimes P_{1} \otimes P_{1} + \Bigl( T^{(1)}
\otimes \undb_{H_{2 \Bbb C}} \Bigl) P_{1} \otimes P_{1} \otimes
P_{2}\cr
\noalign{\smallskip}
&+ \Bigl( T^{(2)} \otimes \undb_{H_{1 \Bbb C}} \Bigl) P_{2} \otimes
P_{2} \otimes P_{1} + \Bigl( T^{(2)} \otimes \undb_{H_{2 \Bbb
C}}\Bigl) P_{2} \otimes P_{2} \otimes P_{2}\ ;\cr
}$$
and a similar formula for $\undb_{H_{\Bbb C}} \otimes T.$
Hence,
$$\eqalign{
&(T\otimes\undb_{H_{\Bbb C}})(\undb_{H_{\Bbb C}}\otimes T)(T\otimes\undb_{H_{\Bbb C}})\cr
\noalign{\smallskip}
&= \Bigl( T^{(1)} \otimes \undb_{H_{1 \Bbb C}}\Bigl)\Bigl(
\undb_{H_{1 \Bbb C}} \otimes T^{(1)} \Bigl)\Bigl( T^{(1)} \otimes
\undb_{H_{1\Bbb C}} \Bigl)P_{1} \otimes P_{1} \otimes P_{1}\cr
\noalign{\smallskip}
&+ \Bigl( T^{(2)} \otimes \undb_{H_{2 \Bbb C}}\Bigl)\Bigl(
\undb_{H_{2 \Bbb C}} \otimes T^{(2)}\Bigl)\Bigl( T^{(2)} \otimes
\undb_{H_{2 \Bbb C}}\Bigl)P_{2} \otimes P_{2} \otimes P_{2}\cr
\noalign{\smallskip}
&=(\undb_{H_{\Bbb C}}\otimes T)(T\otimes\undb_{H_{\Bbb C}})(\undb_{H_{\Bbb C}}\otimes T).\cr}$$ 
It is trivial that $T$ is hermitian and
$$\Vert T \Vert =\hbox{max}\bigl(\Vert T^{(1)}\Vert,\ \Vert T^{(2)}\Vert \bigl) < 1.$$
Thus, $T$ is a Yang-Baxter operator. Next, we show that $T$ is tracial. To this end we fix an
orthonormal basis $\big\{e^{(1)}_i\big\}$ (resp. $\big\{e^{(2)}_i\big\}$) in $H_1$ (resp. $H_2$). 
Then $\big\{e^{(\varepsilon_1)}_i\otimes e^{(\varepsilon_2)}_j\,: \ \varepsilon_1, \varepsilon_2=1,2,
\forall i, j\big\}$ is an orthonormal basis in $H_{\Bbb C}\otimes H_{\Bbb C}$. We must show
that for any
$\varepsilon_1,
\varepsilon_2, \varepsilon_1', \varepsilon_2'=1,2$ and for any $i, j$
$$\la e^{(\varepsilon_1')}_r\otimes e^{(\varepsilon_2')}_s\,,\ Te^{(\varepsilon_1)}_i\otimes
e^{(\varepsilon_2)}_j\ra
=\la e^{(\varepsilon_2')}_s\otimes e^{(\varepsilon_2)}_j\,,\ Te^{(\varepsilon_1')}_r\otimes
e^{(\varepsilon_1)}_i\ra\,.\quad\leqno(3.5) $$ 
If $\varepsilon_1\not =\varepsilon_2$, both sides of (3.5) are equal to zero; so we assume
$\varepsilon_1=\varepsilon_2$, say, $=1$. Then if one of $\varepsilon_1'$ and $ \varepsilon_2'$ is
not equal to 1, both sides of (3.5) is again zero. Thus we are reduced to the case where all
$\varepsilon_1,
\varepsilon_2, \varepsilon_1', \varepsilon_2'$ are equal to 1. In this latter case, by the
definition of $T$, (3.5) is just (1.4) for $T^{(1)}$. Therefore, $T$ satisfies (3.5), so is
tracial. 

The other properties of the proposition can be checked in a straightforward way. 
The details are left to the reader.
\hfill$\diamond$


By Lemma 3.3 and Proposition 3.4, in order to define the second
quantization $\Gamma (V)$ for $V\ :\ H_{1} \to H_{2},$ we are essentially reduced
to the special case where $H_{1} = H_{2}$ and $T^{(1)} = T^{(2)}.$ Now
we come to the main result of this section.

\proclaim Theorem 3.5. Let $H$ be a real Hilbert space, and let $T$ be  a
tracial Yang-Baxter operator with $\Vert T \Vert < 1$.
Suppose
$$\eqalign{
\Bigl(\undb_{H_{\Bbb C}}\otimes V_{\Bbb C}\Bigl)T=T\Bigl(
V_{\Bbb C}\otimes\undb_{H_{\Bbb C}}\Bigl),\cr
\noalign{\smallskip}
\Bigl(V_{\Bbb C}\otimes\undb_{H_{\Bbb C}}\Bigl)T = T\Bigl(
\undb_{H_{\Bbb C}}\otimes V_{\Bbb C}\Bigl).\cr
}\leqno(3.6)$$
Then $\Gamma(V),$ determined by (3.3), is a well-defined mapping
from $\Gamma_{T}(H)$ into $\Gamma_{T}(H)$ ; moreover, $\Gamma(V)$ is
unital, completely positive, and preserves  vacuum expectation.

\noindent{\it Proof.} Set
$$K = H \oplus H, \quad D_{V} = \sqrt{1-V^{\ast}V}\ ;\ D_{V^{\ast}} =
\sqrt{1-V V^{\ast}}.$$
Let $I$ be the natural embedding of $H$ into $K$ and $P$ the natural
projection of $K$ onto $H.$ Define $U\ :\ K \to K$ by
$$U = \pmatrix{ V\quad &-D_{V^{\ast}}\cr
\noalign{\smallskip}
D_{V} \quad &V^{\ast}\cr
}.$$
Then $U$ is an orthogonal and $V = P U I.$ All these are elementary
and well-known. In the rest of the proof, we will not distinguish a real
operator or Hilbert space and its complexification.


Now we have to define a dilation $\widetilde T$  on $K \otimes K$ of $T$ satisfying the
following properties
\smallskip
\item{(i)} $\widetilde T$ is a tracial Yang-Baxter operator with $\Vert
\widetilde T \Vert < 1$;
\smallskip
\item{(ii)} $(I\otimes I)T = \widetilde T(I\otimes I)$ ;
\smallskip
\item{(iii)}$T(P\otimes P) = (P\otimes P)\widetilde T$ ;
\smallskip
\item{(iv)} $(U\otimes U)\widetilde T =\widetilde T(U\otimes U).$


For clarity, we numerate the two copies of $H$ in $K$ as $H_{1}$ and
$H_{2}.$ Then
$$K\otimes K = (H_{1} \otimes H_{1}) \oplus (H_{1} \otimes H_{2})
\oplus (H_{2} \otimes H_{1}) \oplus (H_{2} \otimes H_{2}).
\leqno(3.7)$$
We define the action of $\widetilde T$ on each summund $H_{i} \otimes
H_{j}$ as follows : $\widetilde T$ coincides with $T$ on $H_{i}
\otimes H_{j},$ and maps $H_{i} \otimes H_{j}$ into $H_{j} \otimes
H_{i} \ (i, j = 1, 2).$ In other words, the matrix of $\widetilde T$
in the direct sum decomposition (3.7) is
$$\widetilde T = \pmatrix{
T &\ 0 &\ 0 &\ 0\cr
\noalign{\smallskip}
0 &\ 0 &\ T &\ 0\cr
\noalign{\smallskip}
0 &\ T &\ 0 &\ 0\cr
\noalign{\smallskip}
0 \ &0 &\ 0 &\ T\cr
}.$$
Clearly, $\widetilde T$ is hermitian and $\Vert \widetilde T \Vert =
\Vert T \Vert < 1.$ To show that $\widetilde T$ satisfies the braid
relation (1.1), we identify $K_{\Bbb C}$ with $H_{\Bbb C} \otimes {\Bbb
C}^{2}$ in the natural way. Then
$$K_{\Bbb C} \otimes K_{\Bbb C} = H_{\Bbb C} \otimes H_{\Bbb C}
\otimes {\Bbb C}^{2} \otimes {\Bbb C}^{2}$$
and
$$\widetilde T = T \otimes Q,$$
where $Q\ :\ {\Bbb C}^{2} \otimes {\Bbb C}^{2} \to {\Bbb C}^{2}
\otimes {\Bbb C}^{2}$ is the following operator
$$Q = \pmatrix{
1 \ &0\ &0\ &0\cr
\noalign{\smallskip}
0 \ &0\ &1\ &0\cr
\noalign{\smallskip}
0\ &1\ &0\ &0\cr
\noalign{\smallskip}
0\ &0\ &0\ &1\cr
}.$$
It is easy to see that $Q$ is a Yang-Baxter operator. Indeed, denoting
the canonical basis of ${\Bbb C}^{2}$ by $\{ e_{+}, e_{-} \},$ we have
$$Q e_{\varepsilon_{1}} \otimes e_{\varepsilon_{2}} =
e_{\varepsilon_{2}} \otimes e_{\varepsilon_{1}},\quad
\varepsilon_{1},\ \varepsilon_{2} = \pm.$$
Then a simple verification yields
$$
\bigl(\undb_{\Bbb C^{2}}\otimes Q\bigl)\bigl(Q\otimes
\undb_{\Bbb C^{2}}\bigl)\bigl(\undb_{\Bbb C^{2}}\otimes Q\bigl)
=\undb_{\Bbb C^{2}\otimes \Bbb C^{2}\otimes\Bbb C^{2}}
=\bigl(Q\otimes\undb_{\Bbb
C^{2}}\bigl)\bigl(\undb_{\Bbb C^{2}}
\otimes Q\bigl)(Q\otimes\undb_{\Bbb C^{2}}\bigl).$$
Thus, $Q$ satisfies (1.1), and so $Q$ is a Yang-Baxter operator.
Since  tensor products of Yang-Baxter operators are still
Yang-Baxter operatosr, $\widetilde T$ is a Yang-Baxter operator. In
the same way, to show that $\widetilde T$ is tracial, it
suffices to show $Q$ satisfies (1.4). The latter property immediately
follows from the following :
$$\eqalign{
&\langle Qe_{\varepsilon_{1}} \otimes e_{\varepsilon_{2}},
 e_{\varepsilon_{3}} \otimes e_{\varepsilon_{4}} \rangle =
 \langle e_{\varepsilon_{2}} \otimes e_{\varepsilon_{1}},
 e_{\varepsilon_{3}} \otimes e_{\varepsilon_{4}} \rangle\cr
 \noalign{\smallskip}
 &= \delta_{\varepsilon_{2} \varepsilon_{3}}
 \delta_{\varepsilon_{1} \varepsilon_{4}}\cr
 \noalign{\smallskip}
 &= \langle Q e_{\varepsilon_{3}} \otimes e_{\varepsilon_{1}},
 e_{\varepsilon_{4}} \otimes e_{\varepsilon_{2}} \rangle,\quad
 \varepsilon_{1},\ \varepsilon_{2},\ \varepsilon_{3},\ \varepsilon_{4}=\pm.
}$$
Hence, we have checked (i) above. (ii) follows from the identity :
$$\widetilde T\Bigl |_{H_{\Bbb C} \otimes H_{\Bbb C}} = (I \otimes
I)T$$
(iii) is the dual version of (ii). It remains to prove the last
property (iv). To this end, we write the matrix of $U \otimes U$ according to 
the decomposition (3.7) as follows:
$$U \otimes U = \pmatrix{
V \otimes V \ &-V \otimes D_{V^{\ast}} \ &-D_{V^{\ast}} \otimes
V \ &D_{V^{\ast}} \otimes D_{V^{\ast}}\cr
\noalign{\smallskip}
V \otimes D_{V} &V \otimes V^{\ast} &-D_{V^{\ast}} \otimes D_{V}
&-D_{V^{\ast}} \otimes V^{\ast}\cr
\noalign{\smallskip}
D_{V} \otimes V &-D_{V} \otimes D_{V^{\ast}} &V^{\ast} \otimes V
&-V^{\ast} \otimes D_{V^{\ast}}\cr
\noalign{\smallskip}
D_{V} \otimes D_{V} &D_{V} \otimes V^{\ast} &V^{\ast} \otimes D_{V}
&V^{\ast} \otimes V^{\ast}\cr
}.$$
Then
$$(U \otimes U)\widetilde T = \widetilde T (U \otimes U)$$
\smallskip\noindent
is equivalent to the validity of the following equalities
\smallskip
$$\eqalign{
&(V \otimes V)T = T(V \otimes V),\hskip 1.4cm (V \otimes V^{\ast})T =
T(V^{\ast} \otimes V),\cr
&(V^{\ast} \otimes D_{V})T = T(D_{V} \otimes V^{\ast}), \hskip 0.5cm (V
\otimes D_{V})T = T(D_{V} \otimes V),\cr
&(D_{V} \otimes D_{V})T = T(D_{V} \otimes D_{V}), \quad (D_{V^{\ast}}
\otimes D_{V})T = T(D_{V} \otimes D_{V^{\ast}}).}$$
and those obtained from the previous ones by interchanging $V$ and
$V^{\ast}.$ All these equalities are consequences of (3.6). Indeed,
by (3.6), for any $m, n \geq 0$
$$(V^{m} \otimes V^{n})T = (V^{m} \otimes \undb)(\undb \otimes
V^{n})T = T(V^{n} \otimes V^{m})$$
and (recalling $T^{\ast} = T$)
$$[(V^{\ast})^{m} \otimes (V^{\ast})^{n}]T = T[(V^{\ast})^{n}
\otimes (V^{\ast})^{m}].$$
Thus, in particular, we deduce
$$(V \otimes V)T = T(V \otimes V),\ (V^{\ast} \otimes V)T = T(V
\otimes V^{\ast});$$
as well as
$$[V \otimes (V^{\ast} V)^{n}] T = T[(V^{\ast}V)^{n} \otimes V],\quad
\forall\ n \geq 0\ ;$$
whence
$$(V \otimes D_{V})T = T(D_{V} \otimes V).$$
The other equalities can be proved in the same way. Thus we have
proved the properties (i) - (iv) above. Then the remainder of the
proof of Theorem 3.5 can be accomplished in the usual way in virtue of the
decomposition $V = P U I$ and Lemma 3.3. \hfill $\diamond$


\noindent{\bf Remarks} (i) In the $q$-case, all contractions satisfy the condition (3.6).

(ii) In the above proof, if the dilation $\widetilde T$ of $T$ is taken as the diagonal
operator whose diagonal elements are all equal to $T$, this $\widetilde T$ is still a Yang-Baxter
operator; however, it is then no longer tracial.

(iii) Theorem 3.5 applies, in particular, to the case where $V = t\undb_{H},\ -1\leq t\leq
1.$ This is what we will need in the next section.

(iv) In the $q_{ij}$-case (i.e., when $T$ is given by a symmetric real matrix
$(q_{ij})$), the second quantization $\G(t\undb_{H})$, $-1\le t\le 1$, was
also obtained in [LP].

(v) If $V_1$ and $V_2$ are two contractions on $H$ both satisfying (3.6) (together with
$T$), then $V_1V_2$ still satisfies (3.6) and $\G(V_1V_2)=\G(V_1)\G(V_2)$.

In view of Lemma 3.3 and the remark after it, one is naturally lead to
the following problem.
\bigskip\noindent
\proclaim Problem. With the notation introduced just before Lemma 3.3, suppose that $T^{(1)},
T^{(2)}$ and $V$ satisfy (3.1).  Can one define an appropriate second quantization $\G(V)$?

\bigskip

\centerline{\bf 4. The factoriality of $\Gamma_{T}(H)$}
\medskip
This section is devoted to the extention of Theorem 2.1 to the
deformation given by a Yang-Baxter operator.


\proclaim Theorem 4.1. Let $H$ be a real Hilbert space with $\dim H\geq 2,$ and let $T$ be a
tracial Yang-Baxter operator on $H_{\Bbb C} \otimes H_{\Bbb C}$ with $\Vert T \Vert < 1$.
Then $\Gamma_{T}(H)$ is a factor of type II$_{1}$.



Under the assumption of Theorem 4.1, the vacuum expectation is a
faithful normal normalized trace on $\Gamma_{T}(H)$ (cf. [BS2]). The
proof that $\Gamma_{T}(H)$ is a factor is based on the second
quantization established in the previous section, and is very similar
to that of Theorem 2.1. Thus, we will only indicate the necessary
modifications. Throughout this section, $H$ and $T$ will be fixed as
in Theorem 4.1. We should also point out that if $\dim H = \infty,$
Kr\.olak [K] proved that $\Gamma_{T}(H)$ is a factor, by an averaging
argument already used in [BKS].



As in the $q$-case, we first prove that the commutant of
$\Gamma_{T}(H)$ is the right von Neumann algebra $\Gamma_{T, r} (H).$
Fix an orthonormal basis $\{ e_{i} \}_{i \in I}$ of $H.$ We will use
the multiple index notation introduced in section 2. Let $\Bigl( t^{
\ell k}_{i j}\Bigl)$ be the matrix of $T$ in the basis $\{ e_{i} \otimes
e_{j}\ :\ i, j \in I \}\ :$
$$t^{\ell k}_{i j} = \langle e_{\ell} \otimes e_{k},\ T e_{i} \otimes
e_{j} \rangle,\quad i, j, \ell, k \in I.$$
Then the selfadjointness of $T$ and the condition (1.4) can be
reformulated as
$$t^{\ell k}_{ij} = \overline{t^{i j}_{\ell k}}\quad \hbox{ and }\quad t^{\ell
k}_{i j} = t^{k j}_{\ell i},\quad i, j, k, \ell \in I. \leqno(4.1) $$
Define
$$Jf_{1} \otimes \cdots \otimes f_{n} = \overline{f}_{n} \otimes
\cdots \overline{f}_{1},\quad f_{1}, \cdots, f_{n} \in H_{\Bbb C},\ n
\geq 0,$$
and entend $J$ to the whole ${\cal A} (H_{\Bbb C}).$

\proclaim Lemma 4.1. $J$ extends to an isometric involution on ${\cal F}_{T}(H_{\Bbb C})$.

\noindent{\it Proof.} First, we prove
$$JP^{(n)}_{T}J = P^{(n)}_{T} \quad \hbox{ on } \quad H^{\otimes
n}_{\Bbb C},\ n \geq 0.$$
Let $1 \leq k \leq n-1$ and $\underline i = (i_{1}, \ldots, i_{n})
\in A.$ Then
$$\eqalign{
&JT_{k}J e_{\underline i} = JT_{k}\ e_{i_{n}} \otimes \cdots\otimes e_{i_{1}}\cr
\noalign{\smallskip}
&= J\Bigl[ e_{i(n)} \otimes \cdots \otimes T \Bigl( e_{i_{n-k+1}}
\otimes e_{i_{n-k}}\Bigl) \otimes \cdots \otimes e_{i_{1}} \Bigl]\cr
\noalign{\smallskip}
&= J\ \Bigl[ e_{i(n)} \otimes \cdots \otimes \Bigl( \sum_{a, b}\
t^{ab}_{i_{n-k+1,} i_{n-k}} \ e_{a} \otimes e_{b} \Bigl) \otimes
\ldots \otimes e_{i_{1}}\cr
\noalign{\smallskip}
&= e_{i_{1}} \otimes \cdots \otimes \Bigl( \sum_{a, b}
\overline{t^{ab}}\!\!\!\!\!_{i_{n-k+1}, i_{n-k}}\ e_{b} \otimes e_{a} \Bigl)
\otimes \cdots \otimes e_{i(n)}.\cr
}$$
By (4.1)
$$\overline{t^{ab}}_{i_{n-k+1}, i_{n-k}} = t_{a b}^{i_{n-k+1},
i_{n-k}} = t^{b a}_{i_{n-k}, i_{n-k+1}}.$$
Therefore,
$$\eqalign{
&\sum_{a, b} \overline{t^{ab}}_{i_{n-k+1}, i_{n-k}}\ e_{b} \otimes
e_{a} = \sum_{a, b}\ t^{b a}_{i_{n-k}, i_{n-k+1}}\ e_{b} \otimes
e_{a}\cr
\noalign{\smallskip}
&= T e_{i_{n-k}} \otimes e_{i_{n-k+1}}.\cr
}$$
Hence,
$$JT_{k}J e_{\underline i} = T_{n-k} e_{\underline i},$$
and so
$$JT_{k}J = T_{n-k},\quad 1 \leq k \leq n-1.\leqno(4.2)$$
It follows that
$$J\varphi(\sigma) J = \varphi(\tau \sigma \tau),\quad \forall\
\sigma \in S_{n},$$
where $\tau \in S_{n}$ is the permutation $\tau (k) = n-k+1,\ 1 \leq
k \leq n.$
Thus
$$J P_{T}^{(n)}J = \sum_{\sigma \in S_{n}}\varphi(\tau \sigma
\tau) = \sum_{\sigma \in S_{n}}\varphi(\sigma) = P^{(n)}_{T}.$$
Now let $\xi, \eta \in H^{\otimes n}.$ Since $J$ is an isometric
involution on ${\cal F} (H_{\Bbb C}),$ we have
$$\eqalign{
\langle J\xi, J\eta \rangle_{T} &= \langle P^{(n)}_{T} J\xi,\ J\eta\rangle\cr
\noalign{\smallskip}
&= \langle\eta,\ JP^{(n)}_{T}J\xi\rangle\cr
\noalign{\smallskip}
&=  \langle\eta,\ P^{(n)}_{T}\xi\rangle = \langle \eta,\
\xi\rangle_{T}\,.\cr
}$$
This proves the lemma. \hfill $\diamond$
\bigskip
Recall that $a^{\ast}_{r}(f)$ and $a_{r}(f)$ denote the right
creator and annihilator assocaited with $f$. We have
$$a^{\ast}_{r}(f) = Ja^{\ast}(f)J,\quad a_{r}(f) = J a(f) J,\quad
f \in H\ ;\leqno(4.3)$$
consequently,
$$G_{r}(f) = JG(f) J,\quad f \in H. \leqno(4.4)$$
The first equality of (4.3) immediately follows from (4.2) and the corresponding
equality in the free case. The second is obtained by passing to
adjoint in the first one.


The following fact can be proved in the same way as Lemma 2.2.


\proclaim Lemma 4.2. $\bigl( \Gamma_{T}(H))' = \Gamma_{T, r} (H).$


As in the $q$-case, we introduce the right Wick product : for any
$\xi \in {\cal A} (H_{\Bbb C})$ there is a unique operator in
$\Gamma_{T, r}(H),$ denoted by $W_{r} (\xi),$ such that
$$W_{r}(\xi)\Omega = \xi$$
Now let $V\ :\ H \to H$ be a contraction satisfying (3.6). Then we
can define the right second quantization $\Gamma_{r}(V)$ by
$$\bigl( \Gamma_{r}(V) x\bigl)\Omega = {\cal F}(V)(x\Omega),\quad x
\in \Gamma_{T, r}(H)$$
or
$$\Gamma_{r}(V) W_{r}(\xi) = W_{r}({\cal F}(V)\xi),\quad \xi \in {\cal A}(H_{\Bbb C}).$$
Lemma 3.3 and Theorem 3.5 still hold for this right second
quantization. Using these as in the $q$-case, we get the following
\proclaim Lemma 4.3. Let $V\ :\ H \to H$ be a contraction satisfying
(3.6). Then
$$\Gamma(V) x = \Gamma_{r}(V) x\ , \quad \forall\ x \in \Gamma_{T}(H)
\cap \Gamma_{T, r}(H).$$

The analogue of Proposition 2.4 is still true in the present situation.
We only state the following simplified version without proof.

\proclaim Lemma 4.4. Let $x \in \Gamma_{T}(H).$ Then
$$x = \sum_{\underline i \in A}\langle e_{\underline i}, x \Omega
\rangle\ W(e_{\underline i}),$$
where the equality is interpreted as
$$x\xi = \sum_{\underline i \in A}\langle e_{\underline i}, 
x\Omega\rangle W(e_{\underline i})\xi,\quad \forall\ \xi \in {\cal A}
(H_{\Bbb C}),$$
where the last series converges in ${\cal F}_{T}(H_{\Bbb C}).$ A
similar statement holds for the right von Neumann algebra as well.


Now the proof of Theorem 2.1 can be transferred, words by words, to
the present situation. We thus omit all details.
\vskip1cm

\centerline{ACKNOWLEDGEMENT}

The first author would like to thank the Laboratoire 
de Mathematiques, Universite de Franche-Comte-Besancon 
and the Institute Henri Poincare, Paris, for nice working atmosphere during his stay there November 1999 -- January 2000. 

\vskip 1cm

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\bye
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