[labelstyle=]

# Operator Product States on Tensor Powers of $C^{\ast}$-Algebras

Emil Prodan Department of Physics and
Department of Mathematical Sciences
Yeshiva University
New York, NY 10016, USA
<EMAIL_ADDRESS>
###### Abstract.

The program of matrix product states on an infinite tensor product
${\mathcal{A}}^{\otimes{\mathbb{Z}}}$ of $C^{\ast}$-algebras, initiated by
Fannes, Nachtergaele and Werner in their seminal paper Commun. Math. Phys.
144, 443-490 (1992), is re-assessed in a generic context where ${\mathcal{A}}$
is an infinite nuclear $C^{\ast}$-algebra. This covers, for example, two and
higher dimensional spin lattices. The program consists of three major blocks:
reduction, factorization and reconstruction. We use the Archimedeanization
process of Kavruk, Paulsen, Todorov and Tomforde to produce an operator system
structure on the quotient of ${\mathcal{A}}^{\otimes{\mathbb{Z}}}$ by the so
called entanglement kernel. Then we argue that the Haagerup tensor product
accommodates naturally the factorization block and demonstrate how its
representations on free products of algebras lead to operator product
presentations of the states.

The author was supported by the U.S. National Science Foundation through the
grant DMR-1823800.

###### Contents

  1. 1 Introduction and Main Statements
  2. 2 Entanglement Kernel and the Reduced Space and State
    1. 2.1 Background: Concrete and abstract operator spaces
    2. 2.2 Background: Completely bounded linear maps
    3. 2.3 Algebra of physical observables
    4. 2.4 The entanglement kernel
    5. 2.5 The reduced state
  3. 3 Reduction Process
    1. 3.1 Ordered vector spaces
    2. 3.2 Entanglement kernel is an order ideal
    3. 3.3 Operator systems and matrix ordered $\ast$-vector spaces
    4. 3.4 Completing the reduction process
  4. 4 Factorization Process
    1. 4.1 Multi-linear maps and Haagerup tensor product
    2. 4.2 Generating bi-linear map
    3. 4.3 Generating multi-linear map
    4. 4.4 Asymptotic constraints
  5. 5 Reconstruction Algorithm
    1. 5.1 Re-construction: The abstract form
    2. 5.2 Reconstruction: The concrete form
    3. 5.3 Operator product states

##  1\. Introduction and Main Statements

The work [6], by Fannes, Nachtergaele and Werner, contains some of the most
influential results in the field of correlated quantum systems. Indeed, most
of theoretical physicists working in the field will agree that the
1-dimensional chains of quantum resonators, each carrying a finite local
algebra ${\mathcal{A}}$ of physical observables, are fully understood, in the
sense that any translational invariant state can be represented, in an exact
or approximate form, as a matrix product. The rigorous grounds for this
statement are contained in [6]. Once the matrix product presentation of a
state is derived, which is a standard exercise in many cases, the correlation
functions of the chain and many other physical observables can be computed
explicitly. Hence, the concepts and the tools introduced in [6] have both
fundamental and practical values.

At a more concrete level, let ${\mathcal{A}}^{\otimes{\mathbb{Z}}}$ be the
infinite tensor power of the local algebra ${\mathcal{A}}$, with the former
playing the role of the algebra of physical observables for the chain.
Embedded in ${\mathcal{A}}^{\otimes{\mathbb{Z}}}$ are two sub-algebras,
${\mathcal{A}}_{R}={\mathcal{A}}^{\otimes{\mathbb{N}}^{\times}}$ and
${\mathcal{A}}_{L}={\mathcal{A}}^{\otimes({\mathbb{Z}}\setminus{\mathbb{N}}^{\times})}$.
Recall that ${\mathcal{A}}^{\otimes{\mathbb{Z}}}$ comes equipped with a
standard shift map and, if this map shifts to the right, then it leaves
${\mathcal{A}}_{R}$ invariant, hence, it descends on this sub-algebra.
Likewise, $S^{-1}$ descends to a map on ${\mathcal{A}}_{L}$. Now, given a
shift invariant state $\omega$ on ${\mathcal{A}}_{\mathbb{Z}}$, $\omega\circ
S=\omega$, Ref. [6] introduced the closed sub-space of ${\mathcal{A}}_{R}$

${\mathcal{K}}_{\omega}=\bigcap_{x\in{\mathcal{A}}_{L}}{\rm
Ker}\,\omega_{x},\quad\omega_{x}(a_{R}):=\omega(xa_{R}),$ (1.1)

which we refer to as the entanglement kernel of $\omega$. The attention then
shifts to the reduced space
${\mathcal{B}}_{\omega}={\mathcal{A}}_{R}/{\mathcal{K}}_{\omega}$ and to the
descent $\bar{\omega}$ of $\omega$ on this space (note that
${\mathcal{K}}_{\omega}$ is automatically in the kernel of $\omega$). The
authors of [6] observed that, if ${\mathcal{B}}_{\omega}$ happens to be a
$C^{\ast}$-algebra, then $\omega$ factorizes through a completely positive map

${\mathbb{E}}:{\mathcal{A}}\otimes{\mathcal{B}}_{\omega}\to{\mathcal{B}}_{\omega},\quad{\mathbb{E}}(a\otimes\lfloor
a_{R}\rfloor)=\lfloor a\otimes a_{R}\rfloor,$ (1.2)

in the sense that $\omega(a_{1}\otimes\cdots\otimes a_{n})$ can be computed by
applying $\bar{\omega}$ on certain iterates of ${\mathbb{E}}$ map. Here,
$\lfloor\cdot\rfloor$ indicates the classes of
${\mathcal{B}}_{\omega}={\mathcal{A}}_{R}/{\mathcal{K}}_{\omega}$. Hence, [6]
identified a reduction process as well as a factorization process which supply
the $\omega$-reduced and factorized data
$({\mathcal{A}},{\mathcal{B}}_{\omega},{\mathbb{E}},\bar{\omega})$.

The re-construction problem addresses the reversed process: Given the
$\omega$-reduced and factorized data
$({\mathcal{A}},{\mathcal{B}}_{\omega},{\mathbb{E}},\bar{\omega})$, can one
reproduce the state $\omega$ that generated this data in the first place? It
was shown in [6] that, given a set of data
$({\mathcal{A}},{\mathcal{B}},{\mathbb{E}},\xi)$ with ${\mathcal{A}}$ and
${\mathcal{B}}$ finite dimensional $C^{\ast}$-algebras,
${\mathbb{E}}:{\mathcal{A}}\otimes{\mathcal{B}}\to{\mathcal{B}}$ a completely
positive map and $\xi:{\mathcal{B}}\to{\mathbb{C}}$ a state, there is a
standard algorithm which generates a state on
${\mathcal{A}}^{\otimes{\mathbb{Z}}}$. This state is necessarily finitely
correlated, but this type of states densely sample the set of physically
interesting states. Furthermore, this algorithm automatically leads to a
presentation of the state in the form of a matrix product.

The present work re-assesses the above program in the more general context
when the local algebra ${\mathcal{A}}$ is an infinite nuclear
$C^{\ast}$-algebra. Examples of interest that are cover by the new setting are
${\mathcal{M}}^{\otimes{\mathbb{Z}}^{d}}\simeq\big{(}{\mathcal{M}}^{\otimes{\mathbb{Z}}^{d-1}}\big{)}^{\otimes{\mathbb{Z}}}$,
where ${\mathcal{M}}$ is a finite $C^{\ast}$-algebra. As we shall see,
${\mathcal{B}}_{\omega}$ has a natural structure of an operator space but, in
general, it fails to be an operator system. Hence, apriorily,
${\mathcal{B}}_{\omega}$ does not have a good order structure. Nevertheless,
using relatively recent works by Kavruk, Paulsen, Todorov and Tomforde [10,
11, 7, 8], we find that the reduced space ${\mathcal{B}}_{\omega}$ always
accepts a canonical operator system structure, which can be put in place via
the Archimedeanization process developed in [10, 11]. Since this requires no
additional data besides $({\mathcal{A}}^{\otimes{\mathbb{Z}}},\omega)$, our
conclusion is that the $\omega$-reduced data
$({\mathcal{B}}_{\omega},\bar{\omega})$ always comes in the form of an
operator system and a state (see section 3).

Focusing on the factorization process, we define a bi-linear map on
${\mathcal{A}}\times{\mathcal{B}}_{\omega}$ using the same expression as in
Eq. (1.2), which we show to be a completely contractive bi-linear map, when
${\mathcal{B}}_{\omega}$ is equipped with its operator system norms. As such,
this bi-linear map extends to a completely positive unital map on the Haagerup
tensor product ${\mathcal{A}}\otimes_{\rm h}{\mathcal{B}}_{\omega}$, hence
supplying the map ${\mathbb{E}}$ in this more general context. Furthermore,
the iterates of ${\mathbb{E}}$ continue to display the Markov property
discovered in [6] and used in an essential way in the re-construction process
(see section 4).

The above led us to our first important conclusion that the $\omega$-reduced
and factorized data always come in the form
$({\mathcal{A}},{\mathcal{S}},{\mathbb{E}},\xi)$, where ${\mathcal{A}}$ is a
nuclear $C^{\ast}$-algebra, ${\mathcal{S}}$ is an operator system,
${\mathbb{E}}:{\mathcal{A}}\otimes_{\rm h}{\mathcal{S}}\to{\mathcal{S}}$ is a
completely positive map and $\xi:{\mathcal{S}}\to{\mathbb{C}}$ is a state. The
data is determined entirely and canonically by the original state $\omega$.
Focusing on the re-construction process, we were able to prove the following
statements underlining an abstract re-construction algorithm:

######  Theorem 1.1.

Assume $({\mathcal{A}},{\mathcal{S}},{\mathbb{E}},\xi)$ as already described.
Let

${\mathbb{E}}_{(n)}:{\mathcal{A}}^{\otimes n}\otimes_{\rm
h}{\mathcal{S}}\rightarrow{\mathcal{S}}$

be the maps defined iteratively as

${\mathbb{E}}_{(1)}={\mathbb{E}},\quad{\mathbb{E}}_{(n+1)}={\mathbb{E}}\circ({\rm
id}\otimes_{\rm h}{\mathbb{E}}_{(n)}),\quad n\geq 1.$ (1.3)

Then:

  1. i)

The tower of linear functionals

$\omega_{(n)}:{\mathcal{A}}^{\otimes
n}\rightarrow{\mathbb{C}},\quad\omega_{n}=\xi\circ{\mathbb{E}}_{(n)}\circ
J_{n},\quad n\geq 1,$ (1.4)

where $J_{n}$’s are the isometric embeddings

$J_{n}:{\mathcal{A}}^{\otimes n}\rightarrow{\mathcal{A}}^{\otimes
n}\otimes_{\rm h}{\mathcal{S}},\quad J_{n}\big{(}a_{1}\otimes\cdots\otimes
a_{n}\big{)}=a_{1}\otimes\cdots\otimes a_{n}\otimes e,$ (1.5)

with $e$ the identity of ${\mathcal{S}}$, defines a state $\omega_{R}$ on
${\mathcal{A}}_{R}$.

  2. ii)

Let $\bar{S}$ be defined as

$\bar{S}:{\mathcal{S}}\rightarrow{\mathcal{S}},\quad\bar{S}={\mathbb{E}}\circ\bar{L},$
(1.6)

with $\bar{L}$ being the unital and isometric embedding:

$\bar{L}:{\mathcal{S}}\rightarrow{\mathcal{A}}\otimes_{\rm
h}{\mathcal{S}},\quad\bar{L}(s)=1\otimes s.$ (1.7)

Then the reconstructed state is shift invariant, $\omega_{R}\circ
S=\omega_{R}$, provided $\xi\circ\bar{S}=\xi$.

  3. iii)

In the conditions from point ii), there exists a unique shift invariant state
$\omega$ on ${\mathcal{A}}_{\mathbb{Z}}$, which coincides with $\omega_{R}$
when restricted from ${\mathcal{A}}_{\mathbb{Z}}$ to ${\mathcal{A}}_{R}$.

  4. iv)

If the reduced data has the following asymptotic clustering property

$\lim_{r\rightarrow\infty}\sup\Big{\\{}\Big{|}\xi\Big{(}{\mathbb{E}}_{(n)}\big{(}a_{(n)}\otimes\bar{S}^{\circ
r}(s-\xi(s)\,e)\big{)}\Big{)}\Big{|}\Big{\\}}=0,$ (1.8)

where the supremum is over all $n\geq 0$, $a_{(n)}\in{\mathcal{A}}^{\otimes
n}$ with $\|a_{(n)}\|=1$, and $s\in{\mathcal{S}}$ with $\|s\|\leq 1$, then
$\omega$ also has the asymptotic clustering property.

  5. v)

If ${\mathbb{E}}$ is full, in the sense that

$\overline{\bigcup_{n\geq 1}\bigcup_{x\in{\mathcal{A}}^{\otimes
n}}{\mathbb{E}}_{n}(x\otimes s)}={\mathcal{S}},\quad\forall\
s\in{\mathcal{S}},\ s\neq 0,$ (1.9)

then the data $({\mathcal{A}}^{\otimes{\mathbb{Z}}},\omega)$ reduces back to
$({\mathcal{A}},{\mathcal{S}},{\mathbb{E}},\xi)$.

It follows from the last statement that any full and shift invariant full
state over ${\mathcal{A}}^{\otimes{\mathbb{Z}}}$ can be generated from a set
of data $({\mathcal{A}},{\mathcal{S}},{\mathbb{E}},\xi)$ via the algorithm
described at point i).

Focusing now on the operator product presentation of the reconstructed states,
we note that, since we are dealing with operator systems, there always exist
two Hilbert spaces $H_{1}$ and $H_{2}$, which enter the following commutative
diagram:

$\vspace{0.2cm}\begin{diagram}\vspace{0.2cm}$ (1.10)

Here, all the embeddings are isometric. We recall that the Haagerup tensor
product $B(H_{1})\otimes_{\rm h}B(H_{2})$ embeds isometrically into the free
product $B(H_{1})\star B(H_{2})$ of the algebras (see [12, p. 98]). At its
turn, this free product embeds isometrically into the algebra of bounded
operators over the free product $H_{1}\star H_{2}$ of Hilbert spaces. The
latter requires choices of distinguished unit vectors [16, p. 3] and a natural
way to introduce them is to expand the Hilbert spaces to
$\widetilde{H}_{i}={\mathbb{C}}\cdot\xi_{i}\oplus H_{i}$, $i=1,2$. We then re-
draw the diagram of embeddings as follows:

$\begin{diagram}\vspace{0.1cm}$ (1.11)

The projections onto the sub-spaces ${\mathbb{C}}\cdot\xi_{i}$ of
$\widetilde{H}_{i}$ will be denoted by $P_{\xi_{i}}$, $i=1,2$.

From Arveson Theorem [3][p. 18], we know that the completely positive unital
map ${\mathbb{E}}$ extends over $B(\widetilde{H}_{1}\star\widetilde{H}_{2})$
and, from Stinespring theorem [15], we know that this extension has the
generic structure

$B(\widetilde{H}_{1}\star\widetilde{H}_{2})\ni
G\mapsto{\mathbb{E}}(G)=V^{\ast}\hat{\pi}(G)V\in B(\widetilde{H}_{2}),$ (1.12)

where $\hat{\pi}$ is a representation of
$B(\widetilde{H}_{1}\star\widetilde{H}_{2})$ into the algebra of bounded
operators over a Hilbert space $\widehat{H}$ and
$V:\widetilde{H}_{2}\rightarrow\widehat{H}$ is an isometry. Since
$B(\widetilde{H}_{1})$ and $B(\widetilde{H}_{2})$ are canonically embedded
into their free product, the restrictions of $\hat{\pi}$ generate
representations $\hat{\pi}_{1}$ and $\hat{\pi}_{2}$ of $B(\widetilde{H}_{1})$
and $B(\widetilde{H}_{2})$ on $\widehat{H}$, respectively. Now, for $\psi\in
H_{1}$, we introduce the pair of conjugate operators from
$B(\widetilde{H}_{1})$,

$Z_{\psi}(\alpha\,\xi_{1}+\psi^{\prime})=\alpha\psi,\quad
Z^{\ast}_{\psi}(\alpha\,\xi_{1}+\psi^{\prime})=\langle\psi,\psi^{\prime}\rangle\,\xi_{1},$
(1.13)

and we model the separable Hilbert space $H_{1}$ as $\ell^{2}({\mathcal{I}})$
with ${\mathcal{I}}$ a countable set. Then, if
$\big{\\{}\delta_{i}\big{\\}}_{i\in{\mathcal{I}}}$ denotes the standard basis
of $\ell^{2}({\mathcal{I}})$, we let $Z_{i}=Z_{\delta_{i}}\in
B(\widetilde{H}_{1})$. In this context, the following statement holds:

######  Theorem 1.2.

Let $\delta_{\bm{i}}=\delta_{i_{1}}\otimes\ldots\otimes\delta_{i_{n}}$,
$\bm{i}=(i_{1},\ldots,i_{n})\in{\mathcal{I}}^{n}$, be a basis of
$H_{1}^{\otimes n}$ and assume the following:

  1. 1)

The representations $\hat{\pi}_{1}$ and $\hat{\pi}_{2}$ commute;

  2. 2)

There exists a unitary map
$U:\widetilde{H}_{2}\to\hat{\pi}_{1}\big{(}P_{\xi_{1}}\big{)}\widehat{H}$.

  3. 3)

The representation
$\hat{\pi}_{1}\big{(}P_{\xi_{1}}\big{)}\hat{\pi}_{2}(\cdot)\hat{\pi}_{1}\big{(}P_{\xi_{1}}\big{)}$
of $B(\widetilde{H}_{2})$ on
$\hat{\pi}_{1}\big{(}P_{\xi_{1}}\big{)}\widehat{H}$ is unitarily isomorphic to
the identity representation.

Then the maps ${\mathbb{E}}_{n}$ accept presentations as operator products:

${\mathbb{E}}_{n}(a_{(n)}\otimes
t)=\sum_{\bm{i},\bm{j}\in{\mathcal{I}}^{n}}\big{\langle}\delta_{\bm{i}},A_{(n)}\delta_{\bm{j}}\big{\rangle}\,\Sigma_{i_{1}}\cdots\Sigma_{i_{n}}T\,\Sigma^{\ast}_{j_{n}}\cdots\Sigma^{\ast}_{j_{1}},$
(1.14)

where $A_{(n)}=\rho^{\otimes_{\rm h}n}(a_{(n)})\in B(H_{1})^{\otimes_{\rm
h}n}\subset B(\widetilde{H}_{1})^{\otimes_{\rm h}n}$, $T=\sigma(t)\in
B(H_{2})\subset B(\widetilde{H}_{2})$ and

$\Sigma_{i}\in
B(\widetilde{H}_{2}),\quad\Sigma_{i}=V^{\ast}\hat{\pi}_{1}(Z_{i})U.$ (1.15)

The sum in (5.42) converges in the weak topology of $B(\widetilde{H}_{2})$.

This statement covers Eq. 5.4 from [6], which encodes the matrix product
presentation of the state, as a particular case (see section 5.3). A large
class of additional situations where the conditions of the Theorem 1.2 are
satisfied are presented in Example 5.14.

##  2\. Entanglement Kernel and the Reduced Space and State

Throughout our presentation, we will oscillate between the categories of
operator spaces and of operator systems. In this section, we first provide the
minimal context needed to formulate the problem, which is entirely on operator
spaces. The material is mostly compiled from the textbook by Blecher and Merdy
[3] and the textbook by Pisier [12], and it contains relevant definitions and
fundamental statements that will be referenced throughout our presentation.
Hence, our goal here was to make the presentation self-contained rather than
give an overview of the field.

In the second part, we introduce and exemplify the main objects to be studied,
namely, the algebra of physical observables, which is the infinite tensor
product ${\mathcal{A}}^{\otimes{\mathbb{Z}}}$ of a nuclear $C^{\ast}$-algebra
${\mathcal{A}}$, the entanglement kernel of a state $\omega$ over this algebra
and the quotient space ${\mathcal{B}}_{\omega}$ of
${\mathcal{A}}^{\otimes{\mathbb{Z}}}$ by this kernel. As we shall see, the
latter has the structure of an operator space and $\omega$ descends to a
completely contractive functional $\bar{\omega}$ over
${\mathcal{B}}_{\omega}$. At this, point we formulate the reduction process in
precise terms and point out the main challenges for moving the program
forward.

Before we start, let us lay out our conventions for the notation. The letters
$H,K,L,\ldots$, will be designated for Hilbert spaces. The symbol $H^{(n)}$
will stand for the direct sum of $n$ identical copies of $H$,
$H^{(n)}=H\oplus\ldots\oplus H$. The $C^{\ast}$-algebra of bounded linear maps
between two Hilbert spaces will be denoted by $B(H,K)$. The letters $E$, $F$,
$G$, etc., will be designated to operator spaces. The matrix amplification of
a linear space will be denoted by $M_{n}(E)$. The elements of the operator
spaces and algebras will be denote by lowercase letters $e$, $f$, $g$, etc..
All norms are automatically assumed to be complete. The symbol $\otimes$ will
mostly refer to the algebraic tensor product. Different completions of the
algebraic tensor product will carry sub-scripts, such as $\otimes_{\rm min}$
or $\otimes_{\rm h}$ for the minimal and Haagerup tensor products,
respectively. For nuclear algebras, for which the various tensor product
completions all coincide, the tensor product will be again denoted by
$\otimes$. The elements of matrix amplifications will be specified by
$[a_{ij}]$.

###  2.1. Background: Concrete and abstract operator spaces

######  Definition 2.1 ([3], p. 5).

A concrete operator space is a closed linear sub-space of $B(H,K)$, for some
Hilbert spaces $H$ and $K$. Since $B(H,K)$ can be canonically embedded in
$B(H\oplus K)$, one can equivalently define a concrete operator space as a
closed linear sub-space of $B(H)$, for some Hilbert space $H$.

The following fundamental result, due to Ruan [14], shows that the operator
spaces are intrinsic structures that are intimately related to matrix
amplifications of normed linear spaces:

######  Theorem 2.2 (Ruan, [14]).

Let $M_{n}(E)$ be the linear space of $n\times n$ matrices with entries from
the linear space $E$. Suppose that each $M_{n}(E)$, $n\in{\mathbb{N}}$ can be
equipped with norms $\|\ \|_{n}$ such that:

  1. R1)

$\|aeb\|_{n}\leq\|a\|\|e\|_{n}\|b\|$ for any $a,b\in M_{n}({\mathbb{C}})$ and
$e\in M_{n}(E)$.

  2. R2)

For all $e\in M_{m}(E)$ and $e^{\prime}\in M_{n}(E)$:

$\left\|\begin{pmatrix}e&0\\\
0&e^{\prime}\end{pmatrix}\right\|_{m+n}\leq{\max}\big{\\{}\|e\|_{n},\,\|e^{\prime}\|_{m}\big{\\}},\quad\forall\
m,n=1,2,\ldots.$ (2.1)

Then $E$ can be isometrically embedded in $B(H)$ for some Hilbert space $H$
and, as such, $E$ is an operator space. Conversely, if $E$ can be
isometrically embedded in $B(H)$, then the norms $\|\ \|_{n}$ inherited by
$M_{n}(E)$ from $M_{n}(B(H))\simeq B(H^{(n)},H^{(n)})$ satisfy R1-2.

The system of norms $\\{\|\ \|_{n}\\}_{n\geq 1}$ should be seen as encoding
the extra-structure on the normed space $(E,\|\ \|)$, namely, the fact that
$E$ accepts an isometric embedding in $B(H)$ for some Hilbert space $H$. In
fact, Theorem 2.2 tells us that there is a one to one correspondence between
such embeddings and the systems of norms $\\{\|\ \|_{n}\\}_{n\geq 1})$
satisfying R1-2. An important consequence is that the operator spaces are
intrinsic constructs:

######  Definition 2.3.

An abstract operator space is a pair $\big{(}E,\\{\|\ \|_{n}\\}_{n\geq
1}\big{)}$ of a vector space $E$ and a sequence of norms on $M_{n}(E)$
satisfying R1-2.

Closed linear sub-spaces of an operator space are again operator spaces with
the $n$-norms induced from the parent operator space. More importantly for us
is a fundamental result by Ruan that quotients of operator spaces by closed
linear sub-spaces are also operators spaces:

######  Proposition 2.4 ([3] p. 8).

If $E$ is an operator space and $F$ is one of its closed linear subspaces,
then $E/F$ is an operator space with norms induced by the identification
$M_{n}(E/F)\simeq M_{n}(E)/M_{n}(F)$. Explicitly, these norms are give by the
formula

$\|[e_{ij}+F]\|_{n}=\inf\\{\|[e_{ij}+f_{ij}]\|_{n},\ [f_{ij}]\in M_{n}(F)\\},$
(2.2)

for any $[e_{ij}]\in M_{n}(E)$.

Concrete presentations of quotient operator spaces were supplied by Rieffel in
[13]. Unfortunately, we were not able to take advantage of them at this point.

###  2.2. Background: Completely bounded linear maps

The morphisms in the category of operator spaces are supplied by the
completely bounded (c.b.) linear maps:

######  Definition 2.5 ([12] p. 19; [3] p. 4).

A linear map $u:E\rightarrow F$ between two operator spaces can be amplified
to a linear map

$u_{n}:M_{n}(E)\rightarrow M_{n}(F),\quad u_{n}([e_{ij}])=[u(e_{ij})],$ (2.3)

for all $n\geq 1$. The map $u$ is called:

  1. 1)

Completely bounded if

$\sup_{n\geq 1}\,\|u_{n}\|_{M_{n}(E)\rightarrow M_{n}(F)}<\infty.$ (2.4)

  2. 2)

Complete isometry if all $u_{n}$’s are isometries.

  3. 3)

Complete quotient if each $u_{n}$ sends the unit ball of $M_{n}(E)$ onto the
unit ball of $M_{n}(F)$.

The set of c.b. maps ${\rm CB}(E,F)$ is closed under addition and becomes a
Banach linear space when equipped with the norm

$\|u\|_{\rm cb}=\sup_{n\in{\mathbb{N}}}\,\|u_{n}\|_{M_{n}(E)\rightarrow
M_{n}(F)}.$ (2.5)

As expected, c.b. linear maps behave well under composition:

######  Proposition 2.6 ([12] p. 19).

If $E$, $F$ and $G$ are operator spaces and $u:E\rightarrow F$ and
$v:F\rightarrow G$ are completely bounded linear maps, then $v\circ
u:E\rightarrow G$ is a completely bounded map and $\|v\circ u\|_{\rm
cb}\leq\|v\|_{\rm cb}\,\|u\|_{\rm cb}$.

C.b. linear maps also behave quite natural under taking quotients:

######  Proposition 2.7 ([12] p. 42).

Let $E$, $F$ and $G$ be operator spaces such that $F\subset E$, and let
$q:E\rightarrow E/F$ be the canonical surjection. Then, a linear map
$u:E/F\rightarrow G$ is completely bounded if and only if $u\circ q$ is
completely bounded and $\|u\|_{\rm cb}=\|u\circ q\|_{\rm cb}$.

######  Proposition 2.8 ([3] p. 8).

If $u:E\rightarrow G$ is completely bounded and if $F$ is a closed subspace of
$E$ contained in ${\rm Ker}\,u$, then the canonical map
$\tilde{u}:E/F\rightarrow G$ induced by $u$ is also completely bounded, with
$\|\tilde{u}\|_{\rm c.b.}=\|u\|_{\rm c.b.}$. If $F={\rm Ker}\,u$, then $u$ is
a complete quotient map if and only if $\tilde{u}$ is a complete isometric
isomorphism.

######  Proposition 2.9.

Let $E$ and $F$ be operator spaces with $F\subset E$. Then the canonical
surjection $q:E\rightarrow E/F$ is a complete quotient map and ${\rm
Ker}\,q=F$.

The following statement is known as the fundamental factorization/extension of
c.b. linear maps.

######  Theorem 2.10 ([12] p. 23).

Consider a completely bounded map

$\begin{diagram}$ (2.6)

Then there exits a Hilbert space $\widehat{H}$, a representation
$\pi:B(H)\rightarrow B(\widehat{H})$ and operators
$V_{1}:K\rightarrow\widehat{H}$, $V_{2}:\widehat{H}\rightarrow K$ such that
$\|V_{1}\|\,\|V_{2}\|=\|u\|_{\rm cb}$ and

$u(e)=V_{2}\pi(e)V_{1},\quad\forall\ e\in E.$ (2.7)

Conversely, if (2.7) holds, then $u$ is completely bounded and $\|u\|_{\rm
cb}\leq\|V_{1}\|\,\|V_{2}\|$. Moreover, $u$ admits a completely bounded
extension $\tilde{u}$ such that

$\begin{diagram}$ (2.8)

is a commutative diagram and $\|\tilde{u}\|_{\rm cb}=\|u\|_{\rm cb}$.

###  2.3. Algebra of physical observables

Let ${\mathcal{A}}_{j}$, $j\in{\mathbb{Z}}$, be $C^{\ast}$-algebras
canonically isomorphic to a fixed unital $C^{\ast}$-algebra ${\mathcal{A}}$,
referred to as the local algebra. We denote by $\alpha_{j}$ the isomorphism
between ${\mathcal{A}}_{j}$ and ${\mathcal{A}}$. To avoid unnecessary
complications, we assume that ${\mathcal{A}}$ is nuclear, such that all many
possible ways to complete its tensor powers coincide. We introduce the
notation

${\mathcal{A}}_{(m,n)}={\mathcal{A}}_{m}\otimes{\mathcal{A}}_{m+1}\ldots\otimes{\mathcal{A}}_{n},\quad
m\leq n\in{\mathbb{Z}},$ (2.9)

and we will use the symbols $1_{(m,n)}$ and ${\rm id}_{(m,n)}$ for the
identity element and identity automorphism of ${\mathcal{A}}_{(m,n)}$,
respectively. When $m=1$, we will simplify and write ${\mathcal{A}}_{(1,n)}$
as ${\mathcal{A}}_{(n)}$, etc.. Clearly, there are canonical unital embedding
maps
$\mathfrak{i}_{nm}:{\mathcal{A}}_{(n)}\rightarrowtail{\mathcal{A}}_{(m)}$, for
any pair $m>n\geq 1$. Similarly, the notation ${\mathcal{A}}_{(-n,0)}$ will be
simplified to ${\mathcal{A}}_{(-n)}$ for $n\geq 0$. Now, the natural unital
embeddings

${\mathcal{A}}_{(-n,n)}\rightarrowtail{\mathcal{A}}_{(-n-1,n+1)},\quad
a_{(-n,n)}\mapsto 1_{(-n-1)}\otimes a_{(-n,n)}\otimes 1_{(n+1)},$ (2.10)

supply an inductive tower of unital $C^{\ast}$-algebras

${\mathcal{A}}\rightarrowtail{\mathcal{A}}_{(-1,1)}\cdots\rightarrowtail{\mathcal{A}}_{(-n,n)}\rightarrowtail\cdots,$
(2.11)

whose direct limit is a $C^{\ast}$-algebra denote here by
${\mathcal{A}}_{\mathbb{Z}}$ and sometimes as
${\mathcal{A}}^{\otimes{\mathbb{Z}}}$. This is the algebra of physical
observables we are assuming in this work.

As is the case for any $C^{\ast}$-algebra, ${\mathcal{A}}_{\mathbb{Z}}$ comes
equipped with a $C^{\ast}$-norm that enjoys the special property
$\|a^{\ast}a\|=\|a^{\ast}\|\,\|a\|=\|a\|^{2}$, for any
$a\in{\mathcal{A}}_{\mathbb{Z}}$. Among many other things, this property
enables one to define a special positive cone

${\mathcal{A}}^{+}_{\mathbb{Z}}=\\{a^{\ast}a,\
a\in{\mathcal{A}}_{\mathbb{Z}}\\},$ (2.12)

whose order semi-norm (see 3.8) is a norm and coincides with the
$C^{\ast}$-norm. The state space of ${\mathcal{A}}_{{\mathbb{Z}}}$ consists of
all bounded linear functionals $\omega$ which map
${\mathcal{A}}^{+}_{\mathbb{Z}}$ into ${\mathbb{R}}_{+}$, the positive cone of
${\mathbb{C}}$, and are normalized as $\omega(1)=1$.

Embedded in ${\mathcal{A}}_{\mathbb{Z}}$, are the $C^{\ast}$ algebras
${\mathcal{A}}_{(n,\infty)}$ and ${\mathcal{A}}_{(-\infty,n)}$ defined by the
inductive towers of ${\mathcal{A}}_{(n,m)}$ and ${\mathcal{A}}_{(m,n)}$
algebras, $m\to\pm\infty$, respectively. There are canonical
$C^{\ast}$-algebra isomorphisms

${\mathcal{A}}_{(n,m-1)}\otimes{\mathcal{A}}_{(m,\infty)}\simeq{\mathcal{A}}_{(n,\infty)}$
(2.13)

and

${\mathcal{A}}_{(-\infty,m-1)}\otimes{\mathcal{A}}_{(m,n)}\simeq{\mathcal{A}}_{(-\infty,n)}$
(2.14)

and we will simply identify these algebras throughout. Special symbols will be
used for ${\mathcal{A}}_{R}:={\mathcal{A}}_{(1,\infty)}$ and
${\mathcal{A}}_{L}:={\mathcal{A}}_{(-\infty,0)}$. We note that
${\mathcal{A}}_{\mathbb{Z}}={\mathcal{A}}_{L}\otimes{\mathcal{A}}_{R}$ and
that ${\mathcal{A}}_{R,L}$ belongs to the relative commutant of
${\mathcal{A}}_{L,R}$, respectively. We will denote by $\mathfrak{i}_{R}$ and
$\mathfrak{i}_{L}$ the unital embeddings of ${\mathcal{A}}_{R}$ and
${\mathcal{A}}_{L}$ into ${\mathcal{A}}_{\mathbb{Z}}$, respectively. Also, we
will denote by ${\mathcal{A}}_{R}^{+}$ and ${\mathcal{A}}_{L}^{+}$ the
positive cones of the corresponding $C^{\ast}$-algebras. Lastly, we will
denote by
$\mathfrak{i}_{n}:{\mathcal{A}}_{(n)}\rightarrowtail{\mathcal{A}}_{R}$ and
$\mathfrak{i}_{-n}:{\mathcal{A}}_{(-n)}\rightarrowtail{\mathcal{A}}_{L}$ the
obvious canonical unital embeddings.

######  Remark 2.11.

On several occasions, we will define maps on ${\mathcal{A}}_{\mathbb{Z}}$
using actions on monomials $\otimes_{n\in{\mathbb{Z}}}\,a_{n}$. Our convention
for such expressions is that the $a_{n}$’s are all unit elements except for a
finite number of them. In other words, these monomials are images of elements
from ${\mathcal{A}}_{(m,n)}$ for some finite $m$ and $n$. Similar conventions
are adopted for expressions such as $\otimes_{n\geq m}\,a_{n}$ or
$\otimes_{n\leq m}\,a_{n}$. $\Diamond$

The algebra ${\mathcal{A}}_{\mathbb{Z}}$ has a special automorphism
$S:{\mathcal{A}}_{\mathbb{Z}}\mathrel{\rightarrowtail\kern-5.72635pt\twoheadrightarrow}{\mathcal{A}}_{\mathbb{Z}}$,
which is the shift acting on monomials as

$S\big{(}\otimes_{n\in{\mathbb{Z}}}a_{n}\big{)}=\otimes_{n\in{\mathbb{Z}}}\,(\alpha_{n}^{-1}\circ\alpha_{n-1})(a_{n-1}).$
(2.15)

Note that we shift the entries from left to right. Then, by restriction to
sub-algebras, $S$ generates $C^{\ast}$-algebra isomorphisms
${\mathcal{A}}_{(m,n)}\mathrel{\rightarrowtail\kern-5.72635pt\twoheadrightarrow}{\mathcal{A}}_{(m+1,n+1)}$
as well as
${\mathcal{A}}_{(n,\infty)}\mathrel{\rightarrowtail\kern-5.72635pt\twoheadrightarrow}{\mathcal{A}}_{(n+1,\infty)}$,
which will be denoted by the same symbol $S$. Likewise, the $C^{\ast}$-algebra
isomorphisms
${\mathcal{A}}_{(m,n)}\mathrel{\rightarrowtail\kern-5.72635pt\twoheadrightarrow}{\mathcal{A}}_{(m-1,n-1)}$
and
${\mathcal{A}}_{(-\infty,n)}\mathrel{\rightarrowtail\kern-5.72635pt\twoheadrightarrow}{\mathcal{A}}_{(-\infty,n-1)}$
will be denoted by the single symbol $S^{-1}$. Furthermore, because
${\mathcal{A}}_{(2,\infty)}$ is canonically embedded in ${\mathcal{A}}_{R}$,
we can define the important shift map on ${\mathcal{A}}_{R}$ by the action

$S_{R}\big{(}\otimes_{n\geq 1}a_{n}\big{)}=\otimes_{n\geq
1}\,\,(\alpha_{n}^{-1}\circ\alpha_{n-1})(a_{n-1}),\quad a_{0}=1,$ (2.16)

on the monomials. A similar $C^{\ast}$-algebra morphism $S_{L}^{-1}$ can be
defined on ${\mathcal{A}}_{L}$. The goal of our work is to explore the states
$\omega$ on ${\mathcal{A}}_{\mathbb{Z}}$ that are shift invariant,
$\omega=\omega\circ S$, using the strategy develop in [6].

###  2.4. The entanglement kernel

Let $\omega$ be a state on ${\mathcal{A}}_{\mathbb{Z}}$, not necessarily shift
invariant. Then, for any $x\in{\mathcal{A}}_{L}$, define a bounded linear
functional

$\omega_{x}:{\mathcal{A}}_{R}\rightarrow{\mathbb{C}},\quad\omega_{x}(a_{R})=\omega(x\,a_{R}),$
(2.17)

which is not positive, in general. Following [6], we introduce the following
object:

######  Definition 2.12.

The following sub-set of ${\mathcal{A}}_{R}$,

${\mathcal{K}}_{\omega}=\bigcap_{x\in{\mathcal{A}}_{L}}{\rm Ker}\,\omega_{x},$
(2.18)

will be referred to as the entanglement kernel of $\omega$.

${\mathcal{K}}_{\omega}$ is an intersection of closed linear sub-space, hence
it is closed linear sub-space of the $C^{\ast}$-algebra ${\mathcal{A}}_{R}$.
As such, it is automatically an operator space. An important observation for
later is:

######  Proposition 2.13.

${\mathcal{K}}_{\omega}$ does not contain the unit $1_{R}$ of
${\mathcal{A}}_{R}$.

###### Proof.

We need to find one element $x$ of ${\mathcal{A}}_{L}$ for which
$\omega_{x}(1_{R})=\omega(x)\neq 0$. This element is the identity of
${\mathcal{A}}_{L}$. ∎

${\mathcal{K}}_{\omega}$ enters the exact sequence of operator spaces

${\mathcal{K}}_{\omega}\rightarrowtail{\mathcal{A}}_{R}\twoheadrightarrow{\mathcal{B}}_{\omega}={\mathcal{A}}_{R}/{\mathcal{K}}_{\omega}\
.$ (2.19)

Indeed, from Proposition 2.4, we know that the quotient space
${\mathcal{B}}_{\omega}$ inherits a natural operator space structure.
Furthermore, the second map in (2.19) is the canonical surjection
$q:{\mathcal{A}}_{R}\to{\mathcal{A}}_{R}/{\mathcal{K}}_{\omega}$, which is a
completely bounded map of norm one, as we learned from Corollary 2.9. Let us
recall that the operator space norms $\|\cdot\|^{\rm osp}_{n}$ on
${\mathcal{B}}_{\omega}$ and its matrix amplifications are supplied by
Proposition 2.4. We will refer to ${\mathcal{B}}_{\omega}$ as the
$\omega$-reduced operator space. By Ruan’s theorem, it can be entirely and
abstractly described by the data
$\big{(}{\mathcal{B}}_{\omega},\\{\|\cdot\|^{\rm osp}_{n}\\}_{n\geq
1}\big{)}$. Its elements will be specified by $b$, $b^{\prime}$ and so on.
Also, the matrix amplifications of $q$, which are all contractions, will be
denoted by $q_{n}$.

######  Remark 2.14.

The class of an element $a_{R}\in{\mathcal{A}}_{R}$ in
${\mathcal{B}}_{\omega}={\mathcal{A}}_{R}/{\mathcal{K}}_{\omega}$ will be
indicated by several symbols, such as

$q(a_{R})=\hat{a}_{R}=\lfloor a_{R}\rfloor.$ (2.20)

The second notation $\hat{a}_{R}$ is useful when considering matrix
amplifications of ${\mathcal{B}}_{\omega}$. The third notation will be used
when $a_{R}$ is given as a long expression. $\Diamond$

The algebra ${\mathcal{A}}_{R}$ can be reduced (quoted) in many different
ways, but one of the practical values of the above particular reduction, which
is the great insight supplied by [6], is that ${\mathcal{B}}_{\omega}$ can be
computed for a large class of interesting physical models.

######  Example 2.15.

Let $\omega_{0}$ be a state on ${\mathcal{A}}$ and let
$\omega=\omega_{0}^{\otimes{\mathbb{Z}}}$ be the product state on
${\mathcal{A}}_{\mathbb{Z}}$. In this case,

$\omega_{x}(a_{R})=\omega(x)\,\omega(a_{R})=\omega(x)\,\omega_{R}(a_{R})$
(2.21)

for any $x\in{\mathcal{A}}_{L}$ and $a_{R}\in{\mathcal{A}}_{R}$, hence,
${\mathcal{K}}_{\omega}={\rm Ker}\,\omega_{R}$. Furthermore,

$\mathfrak{i}_{n}(a_{1}\otimes a_{2}\ldots\otimes
a_{n})-\omega_{R}\big{(}\mathfrak{i}_{n-1}(a_{1}\otimes a_{2}\ldots\otimes
a_{n-1})\big{)}\mathfrak{i}_{1}(a_{n})\in{\mathcal{K}}_{\omega},$ (2.22)

hence ${\mathcal{B}}_{\omega}$ is a sub-space of ${\mathcal{A}}$. In fact,
${\mathcal{B}}_{\omega}={\mathcal{A}}/{\rm Ker}\,\omega_{0}$. $\Diamond$

Any product state has zero correlation length, in the language introduced in
[6]. Of course, the main interest of the physics community is on correlated
states. The following example supplies a large class of such states for which
${\mathcal{B}}_{\omega}$ is again computable.

Figure 2.1. Spin-1 particles, shown by red bubbles, are arranged in a closed
chain of length $N$. The algebra of observables is $[1]^{\otimes N}$, where
$[j]\simeq{\mathcal{M}}_{2j+1}$ denotes the spin-j algebra. The local algebra
$[1]$ is embedded via a map $\mathfrak{j}$ into the algebra
$\big{[}\tfrac{1}{2}\big{]}\otimes\big{[}\tfrac{1}{2}\big{]}\simeq{\mathcal{M}}_{2}\otimes{\mathcal{M}}_{2}$
of two spin-$\frac{1}{2}$ particles, shown as black dots. The tensor power
$P_{0}^{\otimes N}$ of the projection $P_{0}$ onto the one-dimensional
subspace of the decomposition
$\big{[}\tfrac{1}{2}\big{]}\otimes\big{[}\tfrac{1}{2}\big{]}\simeq[0]\oplus[1]$,
shown by the segments, generates a rank-one projection in
${\mathcal{M}}_{3}^{\otimes N}$ and a state ${\mathcal{M}}_{3}^{\otimes N}\ni
M\mapsto{\rm Tr}\big{(}\mathfrak{j}^{\otimes N}(M)P_{0}^{\otimes N}\big{)}$.
Note that the projections are applied on the “bonds” shown by the sticks and
this is why the half-shift appears in (2.23).

######  Example 2.16.

The product state in conjunction with the shift map can be used in creative
ways to generate states with finite correlation length. The following class of
states is modeled after the so call AKLT state for spin-1 system [1], whose
construction is sketched in Fig. 2.1. In fact, the construction given here
covers all dimerized states introduced in [1][p. 523]. For this reason, we
refer to the states cover by this example as AKLT type. The construction
involves two (nuclear) $C^{\ast}$-algebras $\hat{\mathcal{A}}$ and
$\tilde{\mathcal{A}}$, related by
$\hat{\mathcal{A}}\simeq\tilde{\mathcal{A}}\otimes\tilde{\mathcal{A}}$, and a
projection $p\in\hat{\mathcal{A}}$. Then the local algebra ${\mathcal{A}}$ is
defined as the unital $C^{\ast}$-algebra ${\mathcal{A}}=p\hat{\mathcal{A}}p$,
with $p$ playing the role of the unit. We generate a state $\omega$ on
${\mathcal{A}}_{\mathbb{Z}}$ via the thermodynamic limit $N\rightarrow\infty$
of the states $\omega_{N}$ on ${\mathcal{A}}_{{\mathbb{Z}}_{N}}$,
${\mathbb{Z}}_{N}={\mathbb{Z}}/\big{(}(2N+1){\mathbb{Z}}\big{)}$, supplied by
the following sequence of maps

$\begin{diagram}$ (2.23)

where $\xi_{0}$ is a positive contractive functional on
$\tilde{\mathcal{A}}\otimes\tilde{\mathcal{A}}$. The first map is the power of
the non-unital inclusion

${\mathcal{A}}\ni
p\hat{a}p\mapsto\mathfrak{j}(p\hat{a}p):=p\hat{a}p\in\hat{\mathcal{A}},$
(2.24)

and $S_{\frac{1}{2}}$ is the circular right half-shift

$S_{\frac{1}{2}}\Big{(}\bigotimes_{n\in{\mathbb{Z}}_{N}}\big{(}\tilde{a}_{n,1}\otimes\tilde{a}_{n,2}\big{)}\Big{)}=\bigotimes_{n\in{\mathbb{Z}}_{N}}\big{(}\tilde{a}_{n-1,2}\otimes\tilde{a}_{n,1}\big{)}.$
(2.25)

The set ${\mathbb{Z}}_{N}={\mathbb{Z}}/\big{(}(2N+1){\mathbb{Z}})$ is
considered ordered, such that $-N$ is the first element in this order. The
tensor products are then taken accordingly.

If Eq. (2.23) is to supply a state on ${\mathcal{A}}_{{\mathbb{Z}}}$, the
above data must obey the constraint

$\lim_{N\rightarrow\infty}\xi_{0}^{\otimes{\mathbb{Z}}_{N}}\Big{(}S_{\frac{1}{2}}\big{(}p^{\otimes{\mathbb{Z}}_{N}}\big{)}\Big{)}=1.$
(2.26)

For example, this is indeed satisfied for the particular case described in
Fig. 2.1. The limiting procedure is required because $p^{\otimes{\mathbb{Z}}}$
is not an element of $\hat{\mathcal{A}}^{\otimes{\mathbb{Z}}}$, hence the non-
unital inclusion $\mathfrak{j}^{\otimes{\mathbb{Z}}_{N}}$ does not make sense
in the thermodynamic limit. Let us specify that, if a unital inclusion is used
instead of $\mathfrak{j}$ and $\xi_{0}$ is a state, then all the maps in the
sequence (2.23) are well defined in the thermodynamic limit, but then the
state on ${\mathcal{A}}$ is a trivial product state. We will not investigate
the thermodynamic limiting process here because we will re-construct this
class of states via a different path in Example 5.7. We only mention here that
these issues were fully resolved in [1] for the particular case illustrated in
Fig. 2.1. Now, assuming that the thermodynamic limit of the state exists, we
compute the corresponding operator spaces ${\mathcal{K}}_{\omega}$ and
${\mathcal{B}}_{\omega}$. For this, we first note the obvious isomorphisms

$\chi_{R}:\hat{\mathcal{A}}_{R}\mathrel{\rightarrowtail\kern-5.72635pt\twoheadrightarrow}\tilde{\mathcal{A}}\otimes\hat{\mathcal{A}}_{R},\quad\chi_{L}:\hat{\mathcal{A}}_{L}\mathrel{\rightarrowtail\kern-5.72635pt\twoheadrightarrow}\hat{\mathcal{A}}_{L}\otimes\tilde{\mathcal{A}}$
(2.27)

and then we use them to define the maps

$\Gamma_{R}=({\rm
id}\otimes\xi_{0}^{\otimes{\mathbb{N}}^{\times}})\circ\chi_{R}\circ\mathfrak{j}^{\otimes{\mathbb{N}}^{\times}},\quad\Gamma_{L}=(\xi_{0}^{\otimes({\mathbb{Z}}\setminus{\mathbb{N}}^{\times})})\circ\chi_{L}\circ\mathfrak{j}^{\otimes({\mathbb{Z}}\setminus{\mathbb{N}}^{\times})},$
(2.28)

from ${\mathcal{A}}_{R}$ to $\tilde{\mathcal{A}}$ and from ${\mathcal{A}}_{L}$
to $\tilde{\mathcal{A}}$, respectively. These maps are well defined because of
our assumption that the map $\xi_{0}^{\otimes{\mathbb{Z}}}\circ
S_{\frac{1}{2}}\circ\mathfrak{j}^{\otimes{\mathbb{Z}}}$ exists as the
thermodynamic limit of chain (2.23). Then

$\omega(a_{L}a_{R})=\xi_{0}(\tilde{a}_{L}\otimes\tilde{a}_{R}),\quad\tilde{a}_{R,L}=\Gamma_{R,L}(a_{R,L}).$
(2.29)

Furthermore, if we choose $\xi_{0}$ such that
$\xi_{0}(\tilde{a}\otimes\tilde{a})\neq 0$ when $\tilde{a}$ samples a dense
sub-set of $\tilde{\mathcal{A}}$, then we must conclude that
${\mathcal{B}}_{\omega}$ and ${\mathcal{K}}_{\omega}$ coincide with the range
and the kernel of $\Gamma_{R}$, respectively, which are both sub-spaces of
$\tilde{\mathcal{A}}$. This is certainly true for the particular case
illustrated in Fig. 2.1 (see Example 5.7 for additional details). $\Diamond$

###  2.5. The reduced state

######  Proposition 2.17.

Let $\omega_{R}$ be the state on ${\mathcal{A}}_{R}$ supplied by the
restriction of $\omega$. Then $\omega_{R}$ descends to a completely bounded
linear functional $\bar{\omega}:{\mathcal{B}}_{\omega}\rightarrow{\mathbb{C}}$
with $\omega_{R}=\bar{\omega}\circ q$ and $\|\bar{\omega}\|_{\rm cb}=1$.

###### Proof.

Taking $x$ the unit of ${\mathcal{A}}_{L}$ sub-algebra, we see from the
definition (2.18) of ${\mathcal{K}}_{\omega}$ that
${\mathcal{K}}_{\omega}\subset{\rm Ker}\,\omega_{R}$. As such, the map:

$\bar{\omega}:{\mathcal{B}}_{\omega}\rightarrow{\mathbb{C}},\quad\bar{\omega}(a_{R}+{\mathcal{K}})=\omega_{R}(a_{R}),$
(2.30)

is well defined. Furthermore, $\bar{\omega}\circ q=\omega_{R}$ and the latter
is a completely bounded functional with c.b. norm 1. According to Proposition
2.7, this can be true if and only if $\bar{\omega}$ is completely contractive.
∎

We will denote by $\bar{\omega}_{n}$ the matrix amplifications of
$\bar{\omega}$. We can deduced some of straightforward properties of these
maps. In particular, let

${\mathcal{D}}_{n}=M_{n}({\mathcal{A}}_{R})^{+}/M_{n}({\mathcal{K}}_{\omega})=q_{n}\big{(}M_{n}({\mathcal{A}}_{R})^{+}\big{)},$
(2.31)

which can be characterized more explicitly as

${\mathcal{D}}_{n}=\big{\\{}[a_{ij}+{\mathcal{K}}_{\omega}]\in
M_{n}({\mathcal{B}}_{\omega})\ |\ \exists\,k_{ij}\in{\mathcal{K}}_{\omega}\
s.t.\ [a_{ij}+k_{ij}]\in M_{n}({\mathcal{A}}_{R})^{+}\big{\\}}.$ (2.32)

Evidently, we have:

$\bar{\omega}_{n}({\mathcal{D}}_{n})\subset{\mathbb{R}}_{+},\quad
n=1,2,\ldots.$ (2.33)

Hence, ${\mathcal{B}}_{n}$ spaces carry more structure than its operator space
norms, hence we present the $\omega$-reduced data as

$\big{(}{\mathcal{B}}_{\omega},\\{\|\cdot\|^{\rm osp}_{n}\\}_{n\geq
1},\\{{\mathcal{D}}_{n}\\}_{n\geq 1},\bar{\omega}\big{)}.$ (2.34)

Without extracting more information about ${\mathcal{K}}_{\omega}$, this is as
far one can go with the reduction process and things are far from
satisfactory. Indeed, ${\mathcal{D}}_{n}$’s do not generate a matrix order
structure, in general, and such structure is essential for defining a
factorizing map ${\mathbb{E}}$ with correct properties and, as such, it is
also essential for the reconstruction block of the program. Here are a few
things that can go wrong:

  * •

${\mathcal{D}}_{n}$’s may fail to be closed spaces;

  * •

${\mathcal{D}}_{n}$’s may not be compatible, in the sense that the relation
$A^{\ast}{\mathcal{D}}_{n}A\subseteq{\mathcal{D}}_{m}$ may fail for $A$ an
ordinary $n\times m$ matrix;

  * •

The intersections ${\mathcal{D}}_{n}\cap(-{\mathcal{D}}_{n})$ might contain
elements other than $0$.

Of course, there are states for which ${\mathcal{D}}_{n}$’s do supply matrix
order structures and explicit sufficient conditions will be supplied in the
next section. We, however, plan to develop our analysis in the most general
conditions possible.

##  3\. Reduction Process

In this section, we identify the space of the admissible input data
$({\mathcal{B}}_{\omega},\bar{\omega})$ for the reconstruction process.
Specifically, we demonstrate that the reduced space ${\mathcal{B}}_{\omega}$
always accepts the structure of an operator system and that the reduced map
$\bar{\omega}$ is a completely positive functional relative to this structure.
Since any operator system accepts a concrete representation as a linear sub-
space of $C^{\ast}$-algebra and any completely positive functional can be
extended to this $C^{\ast}$-algebra, the assumption in [6] that
${\mathcal{B}}_{\omega}$ is a $C^{\ast}$-algebra is now fully justified.

The proof consists of tying together several concepts and results from the
literature, due to Kavruk, Paulsen, Todorov and Tomforde [10, 11, 7, 8]. We
will take this opportunity and give the reader a brisk walk in the space of
these ideas, which supply the natural framework and the right tools for the
problem at hand, something that we still contemplate with amazement. In the
process, one will hear about ordered $\ast$-vector spaces, (Archimedean) order
units, order semi-norms and topologies, as well as order ideals and their
quotients [10]. One will also hear about matrix ordered $\ast$-vector spaces,
(Archimedean) matrix order units and a conceptual refinement of the order
ideal, which is the kernel introduced in [8]. The later has the remarkable
property that any quoted space by it is automatically a matrix ordered
$\ast$-vector space with an Archimedean matrix order unit. This together with
Choi and Effros abstract characterization of operator systems [4] reduces our
proof to demonstrating that ${\mathcal{K}}_{\omega}$ is a kernel.

###  3.1. Ordered vector spaces

This material is entirely collected from [10].

######  Definition 3.1 ([10], p. 1322).

If ${\mathcal{V}}$ is a real vector space, a cone in ${\mathcal{V}}$ is a
nonempty subset ${\mathcal{C}}\subseteq$ with the following two properties:

  1. 1)

$av\in{\mathcal{C}}$ whenever $a\in[0,\infty)$ and $v\in{\mathcal{C}}$;

  2. 2)

$v+w\in{\mathcal{C}}$ whenever $v,w\in{\mathcal{C}}$.

An ordered vector space is a pair $({\mathcal{V}},{\mathcal{V}}^{+})$
consisting of a real vector space ${\mathcal{V}}$ and cone
${\mathcal{V}}^{+}\subseteq{\mathcal{V}}$ satisfying
${\mathcal{V}}^{+}\cap(-{\mathcal{V}}^{+})=\\{0\\}$.

######  Remark 3.2.

If $({\mathcal{V}},{\mathcal{V}}^{+})$ is an ordered real vectors space, one
writes $v\geq v^{\prime}$ if $v-v^{\prime}\in{\mathcal{V}}^{+}$. $\Diamond$

######  Definition 3.3 ([10], pp. 1323-1324).

If $({\mathcal{V}},{\mathcal{V}}^{+})$ is an ordered real vector space, an
element $e\in{\mathcal{V}}$ is called an order unit for ${\mathcal{V}}$ if,
for each $v\in{\mathcal{V}}$, there exists a real number $r>0$ such that
$re\geq v$. The order unit $e$ is called Archimedean if whenever
$v\in{\mathcal{V}}$ with $re+v\geq 0$ for all real $r>0$, then
$v\in{\mathcal{V}}^{+}$.

######  Example 3.4 ([10], p. 1353).

The real vector space of self-adjoint elements of any unital
$C^{\ast}$-algebra is an ordered vector space with the unit playing the role
of Archimedean order unit. $\Diamond$

Of course, our interest is in order structures on complex vector spaces. In
this case, an extra structure is required.

######  Definition 3.5 ([10], p. 1337).

A $\ast$-vector space consists of a complex vector space ${\mathcal{V}}$
together with a map $\ast:{\mathcal{V}}\rightarrow{\mathcal{V}}$ that is
involutive, $(v^{\ast})^{\ast}=v$ for all $v\in{\mathcal{V}}$, and conjugate
linear. If ${\mathcal{V}}$ is a $\ast$-vector space, then ${\mathcal{V}}_{\rm
h}=\\{v\in{\mathcal{V}}\ |\ v^{\ast}=v\\}$ represents the set of hermitean
elements of ${\mathcal{V}}$.

######  Definition 3.6 ([10], p. 1337).

If ${\mathcal{V}}$ is a $\ast$-vector space, one says that
$({\mathcal{V}},{\mathcal{V}}^{+})$ is an ordered $\ast$-vector space if
$({\mathcal{V}}_{\rm h},{\mathcal{V}}^{+})$ is an ordered real vector space.
Furthermore, $e\in{\mathcal{V}}$ is an (Archimedean) order unit for
$({\mathcal{V}},{\mathcal{V}}^{+})$ if it is an (Archimedean) order unit for
$({\mathcal{V}}_{\rm h},{\mathcal{V}}^{+})$.

######  Definition 3.7 ([10], p. 1337).

Let $({\mathcal{V}},{\mathcal{V}}^{+})$ be an ordered $\ast$-vector space with
order unit $e$ and let $({\mathcal{W}},{\mathcal{W}}^{+})$ be an ordered
$\ast$-vector space with order unit $e^{\prime}$. A linear map
$\varphi:{\mathcal{V}}\rightarrow{\mathcal{W}}$ is positive if
$v\in{\mathcal{V}}^{+}$ implies $\varphi(v)\in{\mathcal{W}}^{+}$, and unital
if $\varphi(e)=e^{\prime}$.

Order structures can be used to generate topologies:

######  Definition 3.8 ([10], p. 1327).

Let $({\mathcal{V}},{\mathcal{V}}^{+})$ be an ordered real vector space with
order unit $e$. The order semi-norm on ${\mathcal{V}}$ determined by $e$ is
defined as:

$\llbracket v\rrbracket=\inf\\{r\in{\mathbb{R}}\,|\,re+v\geq 0\ \mbox{and}\
re-v\geq 0\\}.$ (3.1)

The order topology on ${\mathcal{V}}$ is the topology induced by the order
semi-norm.

The following statement supplies a complete characterization of the order
seminorm:

######  Theorem 3.9 ([10], p. 1330).

Let $({\mathcal{V}},{\mathcal{V}}^{+})$ be an ordered real vector space with
order unit $e$. Then the order seminorm $\llbracket\cdot\rrbracket$ is the
unique seminorm on ${\mathcal{V}}$ satisfying simultaneously the following
three conditions:

  1. 1)

$\llbracket e\rrbracket=1$;

  2. 2)

If $-v^{\prime}\leq v\leq v^{\prime}$, then $\llbracket
v\rrbracket\leq\llbracket v^{\prime}\rrbracket$;

  3. 3)

If $f:{\mathcal{V}}\rightarrow{\mathbb{R}}$ is a state, then
$|f(v)|\leq\llbracket v\rrbracket$.

When the order unit is Archimedean, then $\llbracket\cdot\rrbracket$ is
actually a norm and the order topology is Hausdorff. The reciprocal, however,
is not necessarily true (see [10][p. 1328]). Nevertheless, the Archimedean
case can be characterized as it follows:

######  Theorem 3.10 ([10], p. 1330).

Let $({\mathcal{V}},{\mathcal{V}}^{+})$ be an ordered real vector space with
order unit $e$, and let $\llbracket\cdot\rrbracket$ be the order semi-norm
determined by $e$. Then the following are equivalent:

  1. i)

$e$ is Archimedean;

  2. ii)

${\mathcal{V}}^{+}$ is a closed subset of ${\mathcal{V}}$ in the order
topology induced by $\llbracket\cdot\rrbracket$;

  3. iii)

$-\llbracket v\rrbracket\,e\leq v\leq\llbracket v\rrbracket\,e$ for all
$v\in{\mathcal{V}}$.

######  Remark 3.11 ([10], Sec. 4).

The order semi-norm on the hermitean sub-space of an ordered $\ast$-vector
space with unit can be extend over the entire complex space, in an essentially
unique way. $\Diamond$

######  Remark 3.12.

$C^{\ast}$-algebras are extremely special cases where the $C^{\ast}$-norm
coincides with order norm. $\Diamond$

######  Definition 3.13 ([10], p. 1341).

If $({\mathcal{V}},{\mathcal{V}}^{+})$ is an ordered $\ast$-vector space, then
a subspace ${\mathcal{J}}\subseteq{\mathcal{V}}$ is called an order ideal if
${\mathcal{J}}$ is self-adjoint (${\mathcal{J}}^{\ast}={\mathcal{J}})$ and,
furthermore, $v\in{\mathcal{J}}\cap{\mathcal{V}}^{+}$ and $0\leq
v^{\prime}\leq v$ implies that $v^{\prime}\in{\mathcal{J}}$.

######  Proposition 3.14 ([10], p. 1342).

Let $({\mathcal{V}},{\mathcal{V}}^{+})$ be an ordered $\ast$-vector space with
order unit $e$ and let ${\mathcal{J}}\subset V$ be an order ideal. Then
$({\mathcal{V}}/{\mathcal{J}},{\mathcal{V}}^{+}/{\mathcal{J}})$ is an ordered
$\ast$-vector space with order unit $e+{\mathcal{J}}$.

###  3.2. Entanglement kernel is an order ideal

######  Proposition 3.15.

The entanglement kernel is self-adjoint:
${\mathcal{K}}_{\omega}^{\ast}={\mathcal{K}}_{\omega}$.

###### Proof.

We have:

$\omega_{x}(a_{R}^{\ast})=\omega(x\,a_{R}^{\ast})=\omega(a_{R}x^{\ast})^{\ast}=\omega(x^{\ast}a_{R})^{\ast},$
(3.2)

for all $\ x\in{\mathcal{A}}_{L}$. Hence, if $a_{R}\in{\mathcal{K}}_{\omega}$,
then $\omega(xa_{R}^{\ast})=0$ for any $x\in{\mathcal{A}}_{L}$. As a
consequence, $a_{R}^{\ast}\in{\mathcal{K}}_{\omega}$.∎

######  Proposition 3.16.

The entanglement kernel can be equivalently defined as

${\mathcal{K}}_{\omega}=\bigcap_{x\in{\mathcal{A}}_{L}^{+}}{\rm
Ker}\,\omega_{x}.$ (3.3)

Compared to (2.18), the intersection in (3.3) runs over the smaller space of
positive elements.

###### Proof.

Let ${\mathcal{K}}^{\prime}_{\omega}$ denote the set at the right side of
(3.3). Clearly,
${\mathcal{K}}_{\omega}\subseteq{\mathcal{K}}^{\prime}_{\omega}$. Consider now
an $a_{R}$ from ${\mathcal{K}}^{\prime}_{\omega}$ and let $x$ be an arbitrary
element from ${\mathcal{A}}_{L}$. The latter can be always decompose as
$x=(x_{1}^{+}-x_{2}^{+})+\imath(x_{3}^{+}-x_{4}^{+})$ in terms of positive
elements $x_{i}^{+}\in{\mathcal{A}}_{L}^{+}$, though this decomposition is not
unique. Nevertheless, from (3.3), we know that $\omega(x_{i}^{+}a_{R})=0$, for
all $i=\overline{1,4}$. Since $\omega$ is a linear map, we can conclude that
$\omega(x\,a_{R})=0$, hence $a_{R}$ also belong to the set defined at(2.18).∎

######  Proposition 3.17.

The entanglement kernel is an order ideal.

###### Proof.

Consider $k_{R}^{+}$ from ${\mathcal{K}}_{\omega}\cap{\mathcal{A}}_{R}^{+}$
and $a_{R}^{+}$ from ${\mathcal{A}}_{R}^{+}$, such that $a_{R}^{+}\leq
k_{R}^{+}$. Our task is to show that $a_{R}^{+}$ is automatically in
${\mathcal{K}}\cap{\mathcal{A}}_{R}^{+}$. For this, consider
$x^{+}\in{\mathcal{A}}_{L}^{+}$ and note that $x^{+}a_{R}^{+}$ is a positive
element of ${\mathcal{A}}_{\mathbb{Z}}$ because the two terms commute. Then:

$0\leq\omega(x^{+}a_{R}^{+})\leq\omega(x^{+}k_{R}^{+})=0,\quad\forall\
x^{+}\in{\mathcal{A}}_{L}^{+}.$ (3.4)

Proposition 3.16 then assures us that $a_{R}^{+}$ belongs to
${\mathcal{K}}_{\omega}\cap{\mathcal{A}}_{R}^{+}$.∎

From above and Proposition 3.14, we can conclude that
$({\mathcal{B}}_{\omega},{\mathcal{D}}_{1})$ is an order space with unit
$1+{\mathcal{K}}_{\omega}$. As such, ${\mathcal{B}}_{\omega}$ can be endowed
with an order seminorm $\llbracket\cdot\rrbracket$. The order unit, however,
is not Archimedean, in general. Nevertheless, we can show that the order
seminorm is in fact a norm.

######  Proposition 3.18.

Let $x\in{\mathcal{A}}_{L}^{+}$ such that $\omega(x)=0$. Then ${\rm
Ker}\,\omega_{x}={\mathcal{A}}_{R}$.

###### Proof.

By renormalizing $x$ by its norm, we can assume $\|x\|=1$. From the Cauchy-
Schwarz inequality, we have:

$|\omega(xa_{R})|^{2}\leq\omega(x^{2})\,\omega(a_{R}^{\ast}a_{R}),\quad\forall\
a_{R}\in{\mathcal{A}}_{R}.$ (3.5)

Since $\|x\|\leq 1$, we have $x^{2}\leq x$, hence
$0\leq\omega(x^{2})\leq\omega(x)=0$. Then (3.5) assures us that
$\omega_{x}(a_{R})=0$ for all $a_{R}\in{\mathcal{A}}_{R}$.∎

######  Corollary 3.19.

The entanglement kernel can be equivalently defined as

${\mathcal{K}}_{\omega}=\bigcap_{x\in{\mathcal{A}}_{L}^{+}}^{\omega(x)\neq
0}{\rm Ker}\,\omega_{x}.$ (3.6)

###### Proof.

Indeed,

$\displaystyle\bigcap_{x\in{\mathcal{A}}_{L}^{+}}{\rm Ker}\,\omega_{x}$
$\displaystyle=\Big{(}\bigcap_{x\in{\mathcal{A}}_{L}^{+}}^{\omega(x)\neq
0}{\rm
Ker}\,\omega_{x}\Big{)}\bigcap\Big{(}\bigcap_{x\in{\mathcal{A}}_{L}^{+}}^{\omega(x)=0}{\rm
Ker}\,\omega_{x}\Big{)}$ (3.7)
$\displaystyle=\Big{(}\bigcap_{x\in{\mathcal{A}}_{L}^{+}}^{\omega(x)\neq
0}{\rm Ker}\,\omega_{x}\Big{)}\bigcap{\mathcal{A}}_{R}$

and the statement follows.∎

The value of the last statement rests in the observation that all positive
functionals $\omega_{x}$ entering in the new definition (3.6) of
${\mathcal{K}}_{\omega}$ can be normalized by $\omega_{x}(1)$, hence,
transformed into states. Then:

######  Proposition 3.20.

The entanglement kernel is a kernel in the sense of Defintion 3.2 in [8].
Explicitly, the entanglement kernel is the intersection of the kernels of a
family of states:

${\mathcal{K}}_{\omega}=\bigcap_{x\in{\mathcal{A}}_{L}^{+}}^{\omega(x)=1}{\rm
Ker}\,\omega_{x}.$ (3.8)

Note that $\omega_{x}$ is a state if $\omega(x)=1$.

###### Proof.

We have

$\displaystyle\bigcap_{x\in{\mathcal{A}}_{L}^{+}}^{\omega(x)\neq 0}{\rm
Ker}\,\omega_{x}=\bigcap_{\alpha\in(0,\infty)}\Big{(}\,\bigcap_{x\in{\mathcal{A}}_{L}^{+}}^{\omega(x)=\alpha}{\rm
Ker}\,\omega_{x}\Big{)}$ (3.9)

Obviously, ${\rm Ker}\,\omega_{x}={\rm Ker}\,\omega_{\alpha x}$ for all
$\alpha\in(0,\infty)$, hence,

$\displaystyle\bigcap_{x\in{\mathcal{A}}_{L}^{+}}^{\omega(x)=\alpha}{\rm
Ker}\,\omega_{x}=\bigcap_{x\in{\mathcal{A}}_{L}^{+}}^{\omega(x)=1}{\rm
Ker}\,\omega_{x},\quad\forall\ \alpha\in(0,\infty),$ (3.10)

because both the left and right sides sample the same subsets of
${\mathcal{A}}_{R}$. ∎

In our opinion, it is quite remarkable that the entanglement kernel introduced
in [6] can be connected to a concept introduced two decades later. This
connection is even more remarkable, given that:

######  Corollary 3.21 ([8], Lemma 3.3).

Since ${\mathcal{K}}_{\omega}$ is a kernel, the order seminorm on
${\mathcal{B}}_{\omega}$ is actually a norm.

We now want to spell out a simple and explicit condition that ensures that the
order structure $({\mathcal{B}}_{\omega},{\mathcal{D}}_{1})$ is Archimedean.

######  Proposition 3.22 ([8], Prop. 4.1).

The operator space and the order norms on ${\mathcal{B}}_{\omega}$ can be
characterized as

$\displaystyle\|a_{R}+{\mathcal{K}}_{\omega}\|^{\rm
osp}_{1}=\sup\big{\\{}\|\phi(a_{R})\|\ |$ $\displaystyle\
\phi:{\mathcal{A}}_{R}\rightarrow B(H),\
\phi({\mathcal{K}}_{\omega})=\\{0\\},$ (3.11) $\displaystyle\quad\phi\
\mbox{completely contractive}\big{\\}}$

and

$\displaystyle\llbracket
a_{R}+{\mathcal{K}}_{\omega}\rrbracket=\sup\big{\\{}\|\phi(a_{R})\|\ |$
$\displaystyle\ \phi:{\mathcal{A}}_{R}\rightarrow B(H),\
\phi({\mathcal{K}}_{\omega})=\\{0\\},$ (3.12) $\displaystyle\quad\phi\
\mbox{unital, completely positive}\big{\\}}$

where in each case $H$ runs through all Hilbert spaces.

As pointed out in Corollary 4.2 of [8], $\|\cdot\|^{\rm
osp}_{1}\geq\llbracket\cdot\rrbracket$ because every unital completely
positive map over a $C^{\ast}$-algebra is automatically contractive. Now,
since $\omega_{x}$ are states whenever $\omega(x)=1$, they are completely
contractive as well as unital and completely positive and, as a result,

$\|a_{R}+{\mathcal{K}}_{\omega}\|^{\rm
osp}_{1}\geq\sup_{x\in{\mathcal{A}}_{L}^{+}}^{\omega(x)=1}|\omega_{x}(a_{R})|$
(3.13)

and

$\llbracket
a_{R}+{\mathcal{K}}_{\omega}\rrbracket\geq\sup_{x\in{\mathcal{A}}_{L}^{+}}^{\omega(x)=1}|\omega_{x}(a_{R})|.$
(3.14)

This led us to proceed as follows.

######  Proposition 3.23.

The right side of both these two inequalities define a map

$\Gamma:{\mathcal{B}}_{\omega}\rightarrow[0,\infty),\quad\Gamma(a_{R}+{\mathcal{K}}_{\omega})=\sup_{x\in{\mathcal{A}}_{L}^{+}}^{\omega(x)=1}|\omega_{x}(a_{R})|,$
(3.15)

which is continuous and sub-linear

$\Gamma(\alpha b+\beta
b^{\prime})\leq\alpha\Gamma(b)+\beta\Gamma(b^{\prime}),\quad\forall\
b,b^{\prime}\in{\mathcal{B}}_{\omega},\ \ \alpha,\beta\in(0,\infty),$ (3.16)

as well as homogeneous

$\Gamma(\alpha b)=\alpha\Gamma(b),\quad\forall\ b\in{\mathcal{B}}_{\omega},\ \
\alpha\in(0,\infty).$ (3.17)

Furthermore $\Gamma({\mathcal{B}}_{\omega}\setminus\\{0\\})\subset(0,\infty)$.

###### Proof.

The map is well defined for, if $a^{\prime}_{R}$ is another element from the
class of $a_{R}$, then $a^{\prime}_{R}-a_{R}\in{\rm Ker}\ \omega_{x}$ and
$\omega_{x}(a^{\prime}_{R})=\omega_{x}(a_{R})$ for all
$x\in{\mathcal{A}}_{L}^{+}$ with $\omega(x)=1$. Sub-linearity follows from the
linearity of each $\omega_{x}$ and from the “sub-linearity” of the sup
process. If $\alpha\in(0,\infty)$, then

$\sup_{x\in{\mathcal{A}}_{L}^{+}}^{\omega(x)=1}|\omega_{x}(\alpha
a_{R})|=\sup_{x\in{\mathcal{A}}_{L}^{+}}^{\omega(x)=1}\alpha|\omega_{x}(a_{R})|=\alpha\sup_{x\in{\mathcal{A}}_{L}^{+}}^{\omega(x)=1}|\omega_{x}(a_{R})|$
(3.18)

and homogeneity follows. Lastly,
$a_{R}+{\mathcal{K}}_{\omega}\in{\mathcal{B}}_{R}\setminus\\{0\\}$ implies
that $a_{R}\notin{\mathcal{K}}_{\omega}$, hence, there exists at least one
$x\in{\mathcal{A}}_{L}^{+}$ with $\omega(x)=1$ such that
$\omega_{x}(a_{R})>0$. Then the last statement follows. ∎

######  Proposition 3.24.

If the closure of the image of the unit ball of
$({\mathcal{B}}_{\omega},\|\cdot\|^{\rm osp}_{1})$ through $\Gamma$ does not
contain the origin of the real axis, then the order unit of
${\mathcal{B}}_{\omega}$ is Archimedean.

###### Proof.

With the stated assumption, there must exist a strictly positive constant $c$
such that $\Gamma(b)\geq c$, for all $b\in{\mathcal{B}}_{\omega}$ with
$\|b\|^{\rm osp}_{1}=1$. Since $\Gamma$ is homogeneous, this is same as
$\Gamma(b)\geq c\|b\|^{\rm osp}_{1}$ for all $b\in{\mathcal{B}}_{\omega}$.
This together with (3.14) and Corollary 4.2 of [8] give

$\|b\|\geq\llbracket b\rrbracket\geq c\,\|b\|.$ (3.19)

We conclude that the operator space norm and the order norm are equivalent,
hence the two topologies coincide. Since ${\mathcal{D}}_{1}$ is closed in the
operator space topology, it is also closed in the order topology and Theorem
3.10 assures us that order unit is Archimedean.∎

######  Remark 3.25.

If ${\mathcal{B}}_{\omega}$ is finite dimensional, then all norms are
equivalent and the above statement holds without any assumption on $\Gamma$.
Nevertheless, let us acknowledge that, in this case, the stated condition on
$\Gamma$ is automatically satisfied. Indeed, the unit ball is compact if
${\mathcal{B}}_{\omega}$ is finite dimensional, hence, its image through
$\Gamma$ is a closed interval. Since this interval is contained in the open
interval $(0,\infty)$, the strictly positive constant $c$, mentioned in the
proof, exits. $\Diamond$

In the light of the above observation, it is reasonable to claim that the
scenario described in Proposition 3.24 is the level immediately above that of
finite dimensional ${\mathcal{B}}_{\omega}$’s studied in [6] and, as such, it
is an interesting case to investigate. We will proceed, however, with the most
general case possible. It is interesting to note, though, that the condition
spelled in Proposition 3.24 will appear again when we study certain asymptotic
properties under the shift map.

###  3.3. Operator systems and matrix ordered $\ast$-vector spaces

We collect here a number of fundamental concepts and statements related to
order structures on matrix amplifications.

######  Definition 3.26 ([9] p. 9).

A concrete operator system is a self-adjoint closed linear sub-space of a
unital $C^{\ast}$-algebra containing the unit.

A concrete operator system inherits a full order structure from the embedding
unital $C^{\ast}$-algebra. Indeed, if ${\mathcal{S}}\subseteq{\mathcal{A}}$ is
an operator system, then
${\mathcal{S}}^{+}={\mathcal{S}}\cap{\mathcal{A}}^{+}$ supplies a positive
cone. The matrix amplification of an operator system is again a linear
subspace of a $C^{\ast}$-algebra, which is the matrix amplification of the
embedding $C^{\ast}$-algebra. As such, the matrix amplifications of an
operator system come equipped with order structures too. This tower of order
structures puts a sharp distinction between the linear spaces that can be
isometrically embedded in $C^{\ast}$-algebras such that they contain the unit
or they do not. The extra structures can be described abstractly and
intrinsically.

######  Definition 3.27 ([9] p. 176).

Given a $\ast$-vector space ${\mathcal{V}}$, one says that ${\mathcal{V}}$ is
matrix-ordered provided that:

  1. i)

For each $n$, we are give a cone ${\mathcal{C}}_{n}$ in
$M_{n}({\mathcal{V}})_{\rm h}$;

  2. ii)

${\mathcal{C}}_{n}\cap(-{\mathcal{C}}_{n})=\\{0\\}$ for all $n$;

  3. iii)

For every $n$ and $m$ and $A$ an $n\times m$ matrix, we have
$A^{\ast}{\mathcal{C}}_{n}A\subseteq{\mathcal{C}}_{m}$.

######  Definition 3.28.

Let $({\mathcal{V}},{\mathcal{V}}^{+})$ be a matrix-ordered $\ast$-vector
space with order unit $e$. Then $e$ is called a matrix order unit provided
$I_{n}\otimes e$ is an order unit for $M_{n}({\mathcal{V}})$, for each $n$.
Furthermore, $e$ is called Archimedean matrix order unit if $I_{n}\otimes e$
is an Archimedean order unit for $M_{n}({\mathcal{V}})$, for each $n$.

######  Definition 3.29 ([9] p. 176).

Given two matrix-ordered $\ast$-vector spaces ${\mathcal{V}}$ and
${\mathcal{V}}^{\prime}$ with cones ${\mathcal{C}}_{n}$ and
${\mathcal{C}}^{\prime}_{n}$, one calls a linear map
$\varphi:{\mathcal{V}}\rightarrow{\mathcal{V}}^{\prime}$ completely positive
provided that $[v_{ij}]\in{\mathcal{C}}_{n}$ implies that
$[\varphi(v_{ij})]\in{\mathcal{C}}^{\prime}_{n}$. One calls $\varphi$ a
complete order isomorphism if $\varphi$ is completely positive and it has an
inverse which is also completely positive.

The following result, due to Choi and Effros [4], supplies the abstract
characterization of operator systems.

######  Theorem 3.30 ([9] p. 177).

If ${\mathcal{V}}$ is a matrix-ordered $\ast$-vector space with an Archimedean
matrix order unit $e$, then there exists a Hilbert space $H$, an operator
system ${\mathcal{S}}\in B(H)$, and a complete order isomorphism
$\varphi:{\mathcal{V}}\rightarrow{\mathcal{S}}$ with $\varphi(e)=I_{H}$.
Conversely, every concrete operator system ${\mathcal{S}}$ is a matrix-ordered
$\ast$-vector space with Archimedean matrix order unit, when equipped with the
matrix order inherited from the embedding $C^{\ast}$-algebra and with the
Archimedean matrix order unit $e=1$.

######  Definition 3.31.

A linear map between two abstract operator systems is called unital if it maps
the order unit into the order unit.

The following supplies an effective criterion for a map to be completely
positive. It will be used here very often.

######  Proposition 3.32 ([3], p. 18).

Let ${\mathcal{S}}$ and ${\mathcal{S}}^{\prime}$ be two operator systems and
$\varphi:{\mathcal{S}}\rightarrow{\mathcal{S}}^{\prime}$ be a linear unital
map. Then $\varphi$ is completely positive if and only if $\varphi$ is
completely contractive.

Stinespring theorem [15], formulated below, supplies the structure of the
completely positive maps when the domain is a $C^{\ast}$-algebra.

######  Theorem 3.33 ([3], p. 18).

Let ${\mathcal{A}}$ be a unital $C^{\ast}$-algebra. A linear map
$\varphi:{\mathcal{A}}\rightarrow B(H)$ is completely positive if and only if
there is a Hilbert space $K$, a unital $\ast$-homomorphism
$\pi:{\mathcal{A}}\rightarrow B(K)$, and a bounded linear map $V:H\rightarrow
K$ such that $\varphi(a)=V^{\ast}\pi(a)V$ for all $a\in{\mathcal{A}}$. This
can be accomplished with $\|\varphi\|_{\rm cb}=\|V\|^{2}$. Also, this equals
$\|\varphi\|$. If $\varphi$ is unital, then we may take $V$ to be an isometry;
in this case we may view $H\subseteq K$ and have $\varphi(a)=P_{H}\pi(a)|_{H}$

Arveson extension theorem [2], formulated below, tells us among many other
things that the above factorization functions also when the domain is an
operator system.

######  Theorem 3.34 ([3], p. 18).

If ${\mathcal{S}}$ is an operator subsystem of a unital $C^{\ast}$-algebra
${\mathcal{A}}$, and if $\varphi:{\mathcal{S}}\rightarrow B(H)$ is completely
positive, then there exists a completely positive map
$\hat{\varphi}:{\mathcal{A}}\rightarrow B(H)$ extending $\varphi$.

###  3.4. Completing the reduction process

The starting point here is the $\omega$-reduced data
$\big{(}{\mathcal{B}}_{\omega},\\{\|\cdot\|^{\rm osp}_{n}\\}_{n\geq
1},\\{{\mathcal{D}}_{n}\\}_{n\geq 1},\bar{\omega}\big{)}$ compiled in section
2.5. Along the way, we have discovered several important facts about its
structure. Indeed, since ${\mathcal{K}}_{\omega}$ is an order ideal, this data
already describes a matrix-order space with matrix order unit
$1+{\mathcal{K}}_{\omega}$ (see [8][p. 328]). As we have seen, the order
matrix-order structure is not Archimedean, in general. There is, however, a
well understood Archimedeanization process that can be used to transform
${\mathcal{B}}_{\omega}$ into an operator system. It was developed in [10] for
ordered vector spaces and in [11] for matrix amplifications.

The Archimedeanization process involves two stages. In the first stage, one
quotients out closed linear spaces in order to transform the order seminorms
$\llbracket\cdot\rrbracket_{n}$ into norms. It is enough to run this process
for the ground level $n=1$, because that fixes the semi-norm issue for all the
matrix amplifications (see [11][Lemma 3.14]). Given Corollary 3.21, we can
bypass this first stage and complete the Archimedeanization process without
reducing ${\mathcal{B}}_{\omega}$ and its matrix amplifications. This is
significant! In the second stage, one generates the smallest possible
expansions of ${\mathcal{D}}_{n}$’s to positive cones that ensure an
Archimedean structure. For quotiented spaces by kernels, this process was
described in [8][p. 328]:

######  Proposition 3.35 ([8], Prop. 3.4; [11] Prop. 3.16).

Let

$\displaystyle{\mathcal{C}}_{n}=\\{[a^{ij}_{R}+{\mathcal{K}}_{\omega}]\in
M_{n}({\mathcal{B}}_{\omega})\ |$ $\displaystyle\ \forall\,\epsilon>0\
\exists\,k_{i,j}\in{\mathcal{K}}_{\omega}\ \mbox{such that}\ $ (3.20)
$\displaystyle\quad\epsilon 1\otimes I_{n}+[a^{i,j}_{R}+k_{i,j}]\in
M_{n}(A_{R})^{+}\big{\\}}.$

Then $({\mathcal{B}}_{\omega},\\{{\mathcal{C}}_{n}\\}_{n\geq 1})$ is a matrix-
order $\ast$-space with an Archimedean matrix unit
$e=1+{\mathcal{K}}_{\omega}$. Furthermore, the quotient map
$q:{\mathcal{A}}_{R}\rightarrow{\mathcal{B}}_{\omega}$ is unital and
completely positive.

######  Remark 3.36.

Recall that each matrix space $M_{n}({\mathcal{B}}_{\omega})$ comes equipped
with an order norm, which we denote by $\|\cdot\|_{n}^{\rm osy}$ (as in [8]).
From the discussion in [11][p. 37], one learns that the ${\mathcal{C}}_{n}$’s
defined in Proposition 3.35 are just the closures of ${\mathcal{D}}_{n}$ in
the topology induced by these norms. Furthermore, the quotient map
$q:{\mathcal{A}}_{R}\rightarrow{\mathcal{B}}_{\omega}$ is unital and
completely contractive for $\|\cdot\|_{n}^{\rm osy}$. $\Diamond$

Using Choi-Effros Theorem 3.30, we can now state that:

######  Corollary 3.37.

The quotient space with its Archimedean matrix-order structure
$({\mathcal{B}}_{\omega},\\{{\mathcal{C}}_{n}\\}_{n\geq 1},e)$ admits a
concrete representation as an operator system. In particular,
$({\mathcal{B}}_{\omega},\|\cdot\|_{1}^{\rm osy})$ can be unitally and
isometrically embedded in a $C^{\ast}$-algebra.

######  Remark 3.38.

Corollary 3.37 justifies the assumption made in [6] that
${\mathcal{B}}_{\omega}$ is a $C^{\ast}$-algebra. $\Diamond$

We now turn our attention to the reduced map $\bar{\omega}$. The
Archimedeanization described above enjoys a certain universal property, which
is described in [8][p. 329]. We can use this universal property to deduce that
the reduced map $\bar{\omega}$ enjoys the desired properties:

######  Proposition 3.39 ([8] Prop. 3.16).

Let ${\mathcal{T}}$ be an operator system and
$\varphi:{\mathcal{A}}_{R}\rightarrow{\mathcal{T}}$ be a unital and completely
positive map such that ${\mathcal{K}}_{\omega}\subset{\rm Ker}\,\varphi$. Then
the map $\bar{\varphi}:{\mathcal{B}}_{\omega}\rightarrow{\mathcal{T}}$ given
by $\bar{\varphi}(a_{R}+{\mathcal{K}}_{\omega})=\varphi(a_{R})$ is unital and
completely positive. In particular, $\bar{\omega}$ is a unital and completely
positive map from ${\mathcal{B}}_{\omega}$ to ${\mathbb{C}}$, hence a state.

From Stinespring [15] and Arveson Theorems 3.33 and 3.34, respectively, we now
can spell out the structure of the reduced map $\bar{\omega}$:

######  Corollary 3.40.

There is a Hilbert space $H$ and an isometric embedding
$\rho:{\mathcal{B}}_{\omega}\rightarrow B(H)$, as well as a Hilbert space $K$,
a representation $\pi:B(H)\rightarrow B(K)$ and a vector $\zeta\in K$ such
that:

$\bar{\omega}(b)=\big{\langle}\zeta,\pi\big{(}\rho(b)\big{)}\zeta\big{\rangle},\quad
b\in{\mathcal{B}}_{\omega}.$ (3.21)

Furthermore, the map extends over the entire $B(H)$. Here,
${\mathcal{B}}_{\omega}$ is again equipped with the operator system norm.

######  Remark 3.41.

Such concrete representations were exploit to their full extent in [6]. We
want to point out, however, that many concrete examples, including the ones
analyzed in [6], can be worked out using standard maps such as embeddings,
quotients and shifts. $\Diamond$

At this point, we have found the right candidate for the $\omega$-reduced
data, which, as we shall see, encodes the minimal information needed to
reproduce $\omega$:

######  Definition 3.42.

The $\omega$-reduced data consists of the operator system
$({\mathcal{B}}_{\omega},\\{{\mathcal{C}}_{n}\\}_{n\geq 1},e)$ and the state
$\bar{\omega}:{\mathcal{B}}_{\omega}\to{\mathbb{C}}$. This data is entirely
and canonically determined by the original state $\omega$.

In 3.24, we supplied a simple condition that ensures that $\|\cdot\|^{\rm
osy}_{1}$ and $\|\cdot\|^{\rm osp}$ are equivalent. Similar conditions can be
found for the matrix amplifications. Indeed, Proposition 3.43 accepts the
following generalization:

######  Proposition 3.43 ([8], Prop. 4.1).

The operator space and the order norms on ${\mathcal{B}}_{\omega}$ can be
characterized as

$\displaystyle\|[a_{R}^{ij}+{\mathcal{K}}_{\omega}]\|^{\rm
osp}_{n}=\sup\big{\\{}\|[\phi(a_{R}^{ij})]\|\ |$ $\displaystyle\
\phi:{\mathcal{A}}_{R}\rightarrow B(H),\
\phi({\mathcal{K}}_{\omega})=\\{0\\},$ (3.22) $\displaystyle\quad\phi\
\mbox{completely contractive}\big{\\}}$

and

$\displaystyle\|[a_{R}^{ij}+{\mathcal{K}}_{\omega}]\|^{\rm
osy}_{n}=\sup\big{\\{}\|[\phi(a_{R}^{ij})]\|\ |$ $\displaystyle\
\phi:{\mathcal{A}}_{R}\rightarrow B(H),\
\phi({\mathcal{K}}_{\omega})=\\{0\\},$ (3.23) $\displaystyle\quad\phi\
\mbox{unital, completely positive}\big{\\}}$

where in each case $H$ runs through all Hilbert spaces.

As before, a direct consequence of the above is that $\|\cdot\|^{\rm
osy}_{n}\leq\|\cdot\|^{\rm osp}_{n}$. Another consequence is that:

$\|[a_{R}^{ij}+{\mathcal{K}}_{\omega}]\|^{\rm
osy}_{n}\geq\sup_{x\in{\mathcal{A}}_{L}^{+}}^{\omega(x)=1}\|[\omega_{x}(a_{R}^{ij})]\|_{M_{n}}$
(3.24)

for any $[a_{R}^{ij}+{\mathcal{K}}_{\omega}]\in
M_{n}({\mathcal{B}}_{\omega})$. This prompts us to define:

######  Proposition 3.44.

The map

$\Gamma_{n}:M_{n}({\mathcal{B}}_{\omega})\rightarrow[0,\infty),\quad\Gamma_{n}([a_{R}^{ij}+{\mathcal{K}}_{\omega}])=\sup_{x\in{\mathcal{A}}_{L}^{+}}^{\omega(x)=1}\|[\omega_{x}(a_{R}^{ij})]\|_{M_{n}},$
(3.25)

is continuous, sub-linear and homogeneous. Furthermore
$\Gamma_{n}({\mathcal{B}}_{\omega}\setminus\\{0\\})\subset(0,\infty)$.

######  Proposition 3.45.

If the closure of the image of the unit ball of
$(M_{n}({\mathcal{B}}_{\omega}),\|\cdot\|^{\rm osp}_{n})$ through $\Gamma_{n}$
does not contain the origin of the real axis, then $\|\cdot\|^{\rm osp}_{n}$
and $\|\cdot\|^{\rm osy}_{n}$ are equivalent and ${\mathcal{D}}_{n}$ coincides
with ${\mathcal{C}}_{n}$.

######  Remark 3.46.

In the language introduced in [8][p. 334], the entanglement kernel
${\mathcal{K}}_{\omega}$ becomes completely order proximinal under the
conditions of Propositions 3.24 and 3.45. It will be interesting to establish
if these conditions are optimal or not for ${\mathcal{K}}_{\omega}$ to enjoy
this property. $\Diamond$

Let us point out that, if ${\mathcal{B}}_{\omega}$ is finite dimensional, then
any two norms are equivalent and the entanglement kernel is automatically
completely order proximinal.

##  4\. Factorization Process

The main conclusions of the previous section apply to generic states $\omega$
on the algebra of physical observables ${\mathcal{A}}_{\mathbb{Z}}$. In this
section, however, we start by assuming that the state is shift invariant,
$\omega\circ S=\omega$. In these conditions, [6] defined a bi-linear map from
${\mathcal{A}}_{\mathbb{Z}}\times{\mathcal{B}}_{\omega}$ to
${\mathcal{B}}_{\omega}$, which proved to be of fundamental importance in the
analysis of one dimensional spin systems. In this section, we investigate the
properties of this bi-linear map and of its extensions to tensor products. In
a generic setting as ours, such extensions exist canonically only for the
Haagerup tensor product of operator spaces. As such, we start with a brief
introduction of the latter.

###  4.1. Multi-linear maps and Haagerup tensor product

We review here a few fundamental concepts about completely bounded multi-
linear maps and their extensions on tensor products of operators spaces.

######  Definition 4.1 ([12] p. 28).

Consider two concrete operator spaces $E\subset B(H)$ and $F\subset B(K)$.
Then their minimal tensor product $E\otimes_{\rm min}F$ is defined as the
completion of the algebraic tensor product $E\otimes F$ with respect to the
norm induced by ${\mathbb{B}}(H\otimes K)$ via the canonical embedding

$E\otimes F\subset{\mathbb{B}}(H\otimes K).$ (4.1)

######  Example 4.2.

There is a completely isometric isomorphism $M_{n}(E)\simeq
M_{n}({\mathbb{C}})\otimes_{\rm min}E$, for any operator space $E$. $\Diamond$

Completely bounded linear maps behave naturally w.r.t. the minimal tensor
product.

######  Proposition 4.3 ([12], p. 30).

If $E_{i}$ and $F_{i}$ are operator spaces and if $u_{i}:E_{i}\rightarrow
F_{i}$ are completely bounded maps, then their algebraic tensor product
$u_{1}\otimes u_{2}$ extends to a completely bounded map

$u_{1}\otimes_{\rm min}u_{2}:E_{1}\otimes_{\rm min}E_{2}\rightarrow
F_{1}\otimes_{\rm min}F_{2},$ (4.2)

and

$\|u_{1}\otimes_{\rm min}u_{2}\|_{\rm cb}=\|u_{1}\|_{\rm cb}\ \|u_{2}\|_{\rm
cb}.$ (4.3)

The minimal tensor product was exclusively used in [6] but will play no role
here. This is because the maps involved in the factorization process are
induced by multi-linear maps and, as opposed to linear maps, they do not
behave naturally under the minimal tensor product.

######  Definition 4.4 ([3], p. 362).

Let $E_{i}$ and $F$ be operator spaces. A bounded multi-linear map
$\tilde{\varphi}:E_{1}\times\ldots\times E_{m}\rightarrow F$ is a linear map
in each of the arguments such that there exists a constant $C$ with the
property $\|\tilde{\varphi}(e_{1},\ldots,e_{m})\|\leq
C\|e_{1}\|\ldots\|e_{m}\|$ for all $e_{i}\in E_{i}$. The smallest constant $C$
with such property is denoted by $\|\tilde{\varphi}\|$ and defines the norm of
the multi-linear map.

######  Definition 4.5 ([3], p. 30).

Let $\tilde{\varphi}:E_{1}\times\ldots\times E_{m}\rightarrow F$ be a bounded
multi-linear map between operator spaces. Any such multi-linear map can be
amplified to a multi-linear map
$\tilde{\varphi}_{n}:M_{n}(E_{1})\times\ldots\times M_{n}(E_{m})\rightarrow
M_{n}(F)$ as

$\tilde{\varphi}_{n}([e^{1}_{ij}],\ldots,[e^{m}_{ij}])=\Big{[}\sum_{k_{1},\ldots,k_{m-1}=1}^{n}\tilde{\varphi}\big{(}e^{1}_{ik_{1}},e^{2}_{k_{1}k_{2}},\ldots,e^{n}_{k_{m-1}j}\big{)}\Big{]}.$
(4.4)

The norm of this map is computed as in Definition 4.4 using $\|\cdot\|_{n}$
norms on the matrix amplifications. A multi-linear map is called completely
bounded if $\|\varphi\|_{\rm c.b.}:=\sup_{n}\|\tilde{\varphi}_{n}\|<\infty$.

The c.b. multi-linear maps are naturally associated with Haagerup tensor
product of operator spaces:

######  Definition 4.6 ([12], p. 86).

Let $E_{i}$ be operator spaces and ${\mathcal{K}}$ be the $C^{\ast}$-algebra
of compact operators over a separable Hilbert space. Then the rule

$(k_{1}\otimes e_{1})\odot(k_{2}\otimes
e_{2})=(k_{1}k_{2})\otimes(e_{1}\otimes e_{2})$ (4.5)

defines a unique bi-linear form from $({\mathcal{K}}\otimes
E_{1})\times({\mathcal{K}}\otimes E_{2})$ to
${\mathcal{K}}\otimes(E_{1}\otimes E_{2})$. Furthermore, for each
$x_{i}\in{\mathcal{K}}\otimes E_{i}$, one defines

$\alpha_{i}(x_{i})=\|x_{i}\|_{{\mathcal{K}}\otimes_{\rm min}E_{i}},\quad
i=1,2,$ (4.6)

and, for each $x\in{\mathcal{K}}\otimes(E_{1}\otimes E_{2})$,

$\alpha_{\rm
h}(x)=\inf\Big{\\{}\sum_{j=1}^{m}\alpha_{1}(x_{1}^{j})\,\alpha_{2}(x_{2}^{j})\Big{\\}},$
(4.7)

where the infimum is taken over all possible decompositions of $x$ as a finite
sum

$x=\sum_{j=1}^{m}x_{1}^{j}\odot x_{2}^{j},\quad m\geq 1.$ (4.8)

When ${\mathcal{K}}$ is restricted to $M_{n}({\mathbb{C}})$, $n\geq 1$,
$\alpha_{\rm h}$ supplies norms on the matrix amplifications that satisfy
Ruan’s conditions. The completion w.r.t. these norms supplies the Haagerup
tensor product $E_{1}\otimes_{\rm h}E_{2}$.

The c.b. linear maps behave well w.r.t. Haagerup tensor product:

######  Proposition 4.7 ([12], p. 88).

If $E_{i}$ and $F_{i}$ are operator spaces and if $u_{i}:E_{i}\rightarrow
F_{i}$ are completely bounded maps, then their algebraic tensor product

$u_{1}\otimes u_{2}:E_{1}\otimes E_{2}\rightarrow F_{1}\otimes F_{2}$ (4.9)

extends to a c.b. linear map $u_{1}\otimes_{\rm h}u_{2}$ between
$E_{1}\otimes_{\rm h}E_{2}$ and $F_{1}\otimes_{\rm h}F_{2}$ and:

$\|u_{1}\otimes_{\rm h}u_{2}\|_{\rm cb}\leq\|u_{1}\|_{\rm cb}\ \|u_{2}\|_{\rm
cb}.$ (4.10)

The Haagerup tensor product enjoys several unique properties among the tensor
products, which makes it extremely suitable for the applications considered in
this work. The first one relates to extensions of multi-linear maps.

######  Proposition 4.8 ([3] p. 31).

Let $\tilde{\varphi}:E\times F\rightarrow G$ be bilinear map between operator
spaces and $\varphi:E\otimes F\rightarrow G$ be its corresponding linear map
on the algebraic tensor product. Then $\tilde{\varphi}$ is a completely
bounded bilinear map if and only if $\varphi$ extends to a completely bounded
linear map on $E\otimes_{\rm h}F$. Futhermore, if that is indeed the case,
then

$\|\tilde{\varphi}:E\times F\rightarrow G\|_{\rm cb}=\|\varphi:E\otimes_{\rm
h}F\rightarrow G\|_{\rm cb}.$ (4.11)

The second property relates to the unique behavior relative to taking sub-
spaces and quotients:

######  Proposition 4.9 ([12] p. 93; [3] p. 31).

Let $E_{i}$ and $F_{i}$ be operator spaces.

  1. i)

Injectivity: If $E_{i}\subset F_{i}$ completely isometrically, then
$E_{1}\otimes_{\rm h}E_{2}\subset F_{1}\otimes_{\rm h}F_{2}$ completely
isometrically.

  2. ii)

Projectivity: If $q_{i}:E_{i}\rightarrow F_{i}$ are complete quotient maps,
then $q_{1}\otimes q_{2}$ extends to a complete quotient map
$q_{1}\otimes_{\rm h}q_{2}:E_{1}\otimes_{\rm h}E_{2}\rightarrow
F_{1}\otimes_{\rm h}F_{2}$.

The following is a very useful characterization of the kernel of product maps,
with an important corollary stated below:

######  Proposition 4.10 ([5] Prop. 3).

Let $q_{i}:E_{i}\rightarrow F_{i}$ be complete quotient maps. Then

${\rm Ker}\,q_{1}\otimes_{\rm h}q_{2}=\overline{{\rm Ker}\,q_{1}\otimes_{\rm
h}E_{2}+E_{1}\otimes_{\rm h}{\rm Ker}\,q_{2}}.$ (4.12)

######  Corollary 4.11.

Let $E$ and $F$ be operator spaces and $G\subset F$ an operator sub-space.
Then $E\otimes_{\rm h}G$ embeds completely isometrically in $E\otimes_{\rm
h}F$ and $(E\otimes_{\rm h}F)/(E\otimes_{\rm h}G)$ and $E\otimes_{\rm h}F/G$
are completely isometric.

The last property that we will exploit is the natural embedding of the
Haagerup product into free products of $C^{\ast}$-algebras:

######  Proposition 4.12 ([12] p. 98).

Let $E_{i}$ be operator spaces with completely isometric embeddings
$E_{i}\subset{\mathcal{A}}_{i}$ into unital $C^{\ast}$-algebras. Then
$E_{1}\otimes_{\rm h}\ldots\otimes_{\rm h}E_{N}$ accepts a complete isometric
embedding into the $C^{\ast}$-algebra supplied by the free product
${\mathcal{A}}_{1}\star\ldots\star{\mathcal{A}}_{N}$.

This explicit representation of the Haagerup tensor product will be essential
for the presentation of the states as operator products. The following direct
consequence of Proposition 4.12 will also play an essential role:

######  Corollary 4.13.

If $E_{i}$ are operator systems, then $E_{1}\otimes_{\rm h}\ldots\otimes_{\rm
h}E_{N}$ is an operator system too.

###  4.2. Generating bi-linear map

The setting here is the same as in section 2.3 and $\omega$ is assumed shift
invariant. In this section, the class $a_{R}+{\mathcal{K}}_{\omega}$ in
${\mathcal{B}}_{\omega}$ is denoted by $\hat{a}_{R}$ and
${\mathcal{B}}_{\omega}$ is considered equipped with its canonical operator
space structure $({\mathcal{B}}_{\omega},\\{\|\cdot\|^{\rm osy}_{n}\\}_{n\geq
1})$. We will continue to denote the quotient map from ${\mathcal{A}}_{R}$ to
${\mathcal{B}}_{\omega}$ by $q$. We recall that $q$ is unital and completely
contractive.

In the following, we are going to exploit the canonical $C^{\ast}$-algebra
isomorphisms $\alpha_{i}:{\mathcal{A}}_{i}\to{\mathcal{A}}$ and define two
additional $C^{\ast}$-algebra isomorphisms:

$\xi_{R}:{\mathcal{A}}\otimes{\mathcal{A}}_{R}\mathrel{\rightarrowtail\kern-5.72635pt\twoheadrightarrow}{\mathcal{A}}_{R},\quad\xi_{R}(a\otimes
a_{R}):=S(\alpha_{0}^{-1}(a)\otimes a_{R})$ (4.13)

and

$\xi_{L}:{\mathcal{A}}_{L}\otimes{\mathcal{A}}\mathrel{\rightarrowtail\kern-5.72635pt\twoheadrightarrow}{\mathcal{A}}_{L},\quad\xi_{L}(a_{L}\otimes
a):=S^{-1}\big{(}a_{L}\otimes\alpha_{1}^{-1}(a)\big{)},$ (4.14)

where the conventions set in section 2.3 were used to full extent. We recall
that, for $C^{\ast}$-algebras, the isomorphisms are automatically isometric.
The following identity will come handy

$a_{L}\otimes\xi_{R}(a\otimes a_{R})=S\big{(}\xi_{L}(a_{L}\otimes a)\otimes
a_{R}\big{)},$ (4.15)

for any $a_{L,R}\in{\mathcal{A}}_{L,R}$ and $a\in{\mathcal{A}}$. Here, $S$ is
the shift automorphism on ${\mathcal{A}}_{\mathbb{Z}}$.

We now proceed with defining of what we call the generating map. As in [6], it
is introduced here by extending a bi-linear map:

######  Proposition 4.14.

The relation

${\mathbb{E}}:{\mathcal{A}}\times{\mathcal{B}}_{\omega}\rightarrow{\mathcal{B}}_{\omega},\quad{\mathbb{E}}\big{(}a,\hat{a}_{R}\big{)}:=(q\circ\xi_{R})(a\otimes
a_{R}),$ (4.16)

is a well defined bi-linear map.

######  Remark 4.15.

The map will be written formally as ${\mathbb{E}}(a,\hat{a}_{R})=\lfloor
a\otimes a_{R}\rfloor$. $\Diamond$

###### Proof.

Consider $a_{R}$ and $a^{\prime}_{R}$ from the same class of
${\mathcal{B}}_{\omega}$. We need to show that $\xi_{R}(a\otimes a_{R})$ and
$\xi_{R}(a\otimes a^{\prime}_{R})$ belong to the same class of
${\mathcal{B}}_{\omega}$, for all $a\in{\mathcal{A}}$, more precisely, that

$\omega_{x}\big{(}\xi_{R}(a\otimes(a_{R}-a^{\prime}_{R}))\big{)}=0\ \ \forall\
x\in{\mathcal{A}}_{L}.$ (4.17)

It follows from identity (4.15) and shift invariance of the state $\omega$
that

$\omega_{x}\big{(}\xi_{R}(a\otimes(a_{R}-a^{\prime}_{R}))\big{)}=\omega_{\xi_{L}(x\otimes
a)}(a_{R}-a^{\prime}_{R}),$ (4.18)

which is indeed identically zero for all $x\in{\mathcal{A}}_{L}$. Bi-linearity
is evident.∎

######  Remark 4.16.

It is important to acknowledge that ${\mathbb{E}}$ is entirely determined by
the parent state $\omega$ and that the existence of such map depends crucially
on the shift invariance of the state. $\Diamond$

######  Proposition 4.17.

${\mathbb{E}}$ is a completely contractive bi-linear map.

###### Proof.

We recall that the bi-linear map

${\mathcal{A}}\times{\mathcal{A}}_{R}\ni(a,a_{R})\mapsto a\otimes
a_{R}\in{\mathcal{A}}\otimes_{\rm h}{\mathcal{A}}_{R}$ (4.19)

is completely contractive [3][p. 30]. Furthermore, the identity map on the
algebraic tensor product extends to a completely contractive map
$j:{\mathcal{A}}\otimes_{\rm
h}{\mathcal{A}}_{R}\rightarrowtail{\mathcal{A}}\otimes_{\rm
min}{\mathcal{A}}_{R}$ [12][p. 88]. This enables us to define the completely
contractive bi-linear map

$\phi:{\mathcal{A}}\times{\mathcal{A}}_{R}\rightarrow{\mathcal{B}}_{\omega},\quad\phi=q\circ\xi_{R}\circ
j\circ\otimes.$ (4.20)

which enters the following relation, ${\mathbb{E}}\circ({\rm id}\times
q)=\phi$. Indeed, note that $j$ acts as the identity on monomials. A similar
relation holds between the matrix amplifications,

${\mathbb{E}}_{n}\circ({\rm id}_{n}\times q_{n})=\phi_{n}.$ (4.21)

Indeed, the matrix amplification of (4.19) is given by

$\big{(}[a_{ij}],[a_{R}^{ij}]\big{)}\mapsto\Big{[}\sum_{k}a_{ik}\otimes
a_{R}^{kj}\Big{]}$ (4.22)

and

${\mathbb{E}}_{n}\big{(}[a_{ij}],[\hat{a}_{R}^{ij}]\big{)}=\Big{[}\sum_{k}a_{ik}\otimes\hat{a}_{R}^{kj}\Big{]}.$
(4.23)

Then the relation in (4.21) is evident. Based on these identities,

$\displaystyle{\mathbb{E}}_{n}\big{(}[a_{ij}],[\hat{a}_{R}^{ij}]\big{)}={\mathbb{E}}_{n}\big{(}[a_{ij}],q_{n}\big{(}[a_{R}^{ij}]+[k_{ij}]\big{)}\big{)}=\phi_{n}\big{(}[a_{ij}],[a_{R}^{ij}]+[k_{ij}]\big{)},$
(4.24)

where $[k_{ij}]$ is an arbitrary element from $M_{n}({\mathcal{K}}_{\omega})$.
Finally, we use the fact that each $\phi_{n}$ is a contraction,

$\|{\mathbb{E}}_{n}\big{(}[a_{ij}],[a_{R}^{ij}]\big{)}\|^{\rm
osy}_{n}\leq\|[a_{ij}]\|_{n}\,\|[a_{R}^{ij}]+[k_{ij}]\|_{n},$ (4.25)

where $\|\cdot\|_{n}$ are the matrix amplifications of the $C^{\ast}$-norms.
We conclude that

$\|{\mathbb{E}}_{n}\big{(}[a_{ij}],[a_{R}^{ij}]\big{)}\|^{\rm
osy}_{n}\leq\|[a_{ij}]\|_{n}\,\inf_{[k_{ij}]\in
M_{n}({\mathcal{K}}_{\omega})}\|[a_{R}^{ij}]+[k_{i}j]\|_{n},$ (4.26)

and the statement follows.∎

A direct consequence of the above and Proposition 4.8 is:

######  Corollary 4.18.

${\mathbb{E}}$ extends to a unital completely contractive linear map from
${\mathcal{A}}\otimes_{\rm h}{\mathcal{B}}_{\omega}$ to
${\mathcal{B}}_{\omega}$.

The extension mentioned above will be denoted by the same symbol
${\mathbb{E}}$ and be called the generating map. We recall Proposition 4.13,
which assures us that ${\mathcal{A}}\otimes_{\rm h}{\mathcal{B}}_{\omega}$ has
the structure of an operator system. Then, according to Proposition 3.32,
${\mathbb{E}}$ is a completely positive map of operator systems. Then
Stinespring [15] and Arveson Theorems 3.33 and 3.34, respectively, enable us
to pin down the general structure of ${\mathbb{E}}$. This will be discussed in
great details in sub-section 5.2.

######  Example 4.19.

Let us consider the product state discussed in Example 2.15, where we found
${\mathcal{B}}_{\omega}={\mathcal{A}}/{\rm Ker}\,\omega_{0}$. Since

$a\otimes a^{\prime}=\omega_{0}(a)\,a^{\prime}+\big{(}a\otimes
a^{\prime}-\omega_{0}(a)\,a^{\prime}\big{)},\quad
a,a^{\prime}\in{\mathcal{A}},$ (4.27)

and the last term belongs ${\rm Ker}\,\omega_{0}$, we have

${\mathbb{E}}(a\otimes\lfloor a^{\prime}\rfloor)=\lfloor a\otimes
a^{\prime}\rfloor=\omega_{0}(a)\lfloor a^{\prime}\rfloor.$ (4.28)

If ${\mathcal{B}}_{\omega}\subset B(H_{2})$ and $H_{1}$ is the Hilbert space
associated to the representation $\pi$ of ${\mathcal{A}}$ generated by
$\omega_{0}$, then we can take

$V:H_{2}\rightarrow H_{1}\otimes H_{2},\quad V(\chi)=\xi\otimes\chi,$ (4.29)

with $\xi\in H_{1}$ such that $\omega_{0}(a)=\langle\xi,\pi(a)\xi\rangle$, in
which case

$V^{\ast}\big{(}\pi(a)\otimes B\big{)}V=\langle\xi,\pi(a)\xi\rangle\otimes
B=\omega_{0}(a)B$ (4.30)

This is the Stinespring representation of ${\mathbb{E}}$ for this particular
case. $\Diamond$

###  4.3. Generating multi-linear map

We consider now the multi-linear map

$\displaystyle{\mathbb{E}}_{(n)}$ $\displaystyle:{\mathcal{A}}^{\times
n}\times{\mathcal{B}}_{\omega}\rightarrow{\mathcal{B}}_{\omega},$ (4.31)
$\displaystyle{\mathbb{E}}_{(n)}\big{(}a_{1},\ldots,a_{n},\lfloor
a_{R}\rfloor\big{)}$
$\displaystyle:=\Big{\lfloor}S^{n}\Big{(}\alpha_{-n+1}^{-1}(a_{1})\otimes\ldots\otimes\alpha_{0}^{-1}(a_{n})\otimes
a_{R}\Big{)}\Big{\rfloor},$

which is again well defined by the same arguments used in Proposition 4.14.
Furthermore, using the fact that $\otimes^{n}_{\rm h}$ is a completely
contractive multi-linear map, we can repeat the arguments in Proposition 4.17
and conclude that ${\mathbb{E}}_{(n)}$ are completely contractive multi-linear
maps. As a consequence, they can be extended to completely contractive linear
maps on ${\mathcal{A}}^{\otimes n}\otimes_{\rm
h}{\mathcal{B}}_{\omega}\simeq{\mathcal{A}}_{(n)}\otimes_{\rm
h}{\mathcal{B}}_{\omega}$ and, since these maps are also unital,
${\mathbb{E}}_{(n)}$ are automatically completely positive. In the following,
we establish useful relations between these maps and also supply an
alternative and more direct proof of their complete positivity.

######  Proposition 4.20.

The maps ${\mathbb{E}}_{(n)}:{\mathcal{A}}_{(n)}\otimes_{\rm
h}{\mathcal{B}}_{\omega}\rightarrow{\mathcal{B}}_{\omega}$ satisfy the
recursive relations

${\mathbb{E}}_{(1)}={\mathbb{E}},\quad{\mathbb{E}}_{(n+1)}={\mathbb{E}}\circ({\rm
id}\otimes_{\rm h}{\mathbb{E}}_{(n)}),\quad n>1.$ (4.32)

###### Proof.

It is enough to verify the identities on monomials and, if
$a\in{\mathcal{A}}$, $a_{(n)}=a_{1}\otimes\cdots\otimes
a_{n}\in{\mathcal{A}}_{(n)}$ and $a_{R}\in{\mathcal{A}}_{R}$, we have

$\displaystyle\big{(}{\mathbb{E}}\circ({\rm id}\otimes_{\rm h}$
$\displaystyle{\mathbb{E}}_{(n)})\big{)}\big{(}a\otimes a_{(n)}\otimes\lfloor
a_{R}\rfloor\big{)}$
$\displaystyle\quad={\mathbb{E}}\Big{(}a\otimes{\mathbb{E}}_{(n)}\big{(}a_{(n)}\otimes\lfloor
a_{R}\rfloor\big{)}\Big{)}$
$\displaystyle\quad={\mathbb{E}}\Big{(}a\otimes\big{\lfloor}S^{n}\big{(}\alpha_{-n+1}^{-1}(a_{1})\otimes\ldots\otimes\alpha_{0}^{-1}(a_{n})\otimes
a_{R}\big{)}\big{\rfloor}\Big{)}$
$\displaystyle\quad=\Big{\lfloor}S\Big{(}\alpha_{0}^{-1}(a)\otimes
S^{n}\big{(}\alpha_{-n+1}^{-1}(a_{1})\otimes\ldots\otimes\alpha_{0}^{-1}(a_{n})\otimes
a_{R}\big{)}\Big{)}\Big{\rfloor}$
$\displaystyle\quad=\Big{\lfloor}S^{n+1}\Big{(}\alpha_{-n}^{-1}(a)\otimes\alpha_{-n+1}^{-1}(a_{1})\otimes\ldots\otimes\alpha_{0}^{-1}(a_{n})\otimes
a_{R}\Big{)}\Big{\rfloor}$

and the last line coincides with ${\mathbb{E}}_{n+1}(a\otimes
a_{(n)}\otimes\lfloor a_{R}\rfloor)$.∎

######  Proposition 4.21.

The maps ${\mathbb{E}}_{n}$, $n\geq 1$, are completely positive maps.

###### Proof.

We will use an inductive argument. We already know that
${\mathbb{E}}_{(1)}={\mathbb{E}}$ is unital and completely positive on
${\mathcal{A}}\otimes_{\rm h}{\mathcal{B}}_{\omega}$. Henceforth, assume that
${\mathbb{E}}_{(s)}$ is unital and completely positive. Then, from Proposition
3.32, ${\mathbb{E}}_{(s)}$ is necessarily a complete unital contraction. From
Proposition 4.7, the map:

${\rm id}\otimes_{\rm h}{\mathbb{E}}_{(s)}:{\mathcal{A}}\otimes_{\rm
h}\big{(}{\mathcal{A}}_{(s)}\otimes_{\rm
h}{\mathcal{B}}_{\omega}\big{)}\rightarrow{\mathcal{A}}\otimes_{\rm
h}{\mathcal{B}}_{\omega}$ (4.33)

is a complete contraction and, evidently, it is also a unital map. Now,
${\mathbb{E}}\circ({\rm id}\otimes_{\rm h}{\mathbb{E}}_{(s)})$ is a
composition of unital completely contractive maps, hence Proposition 2.6
assures us that ${\mathbb{E}}_{(s+1)}$ is a unital and completely contractive
map. Then Proposition 3.32 says that ${\mathbb{E}}_{(s+1)}$ is a positive map
and this completes the induction proof. ∎

######  Proposition 4.22.

The linear maps ${\mathbb{E}}_{n}$ display Markov’s property, more precisely,
for any $n,m\geq 1$,

${\mathbb{E}}_{(m+n)}={\mathbb{E}}_{(m)}\circ({\rm id}_{(m)}\otimes_{\rm
h}{\mathbb{E}}_{(n)}).$ (4.34)

###### Proof.

This is a direct consequence of the recursive relations(4.32). ∎

######  Proposition 4.23.

Recall the unital isometric embeddings

$\mathfrak{i}_{mn}:{\mathcal{A}}_{(n)}\rightarrow{\mathcal{A}}_{(m)},\quad\mathfrak{i}_{mn}\big{(}a_{(n)}\big{)}=a_{(n)}\otimes
1_{(m-n)},\quad(m>n),$ (4.35)

and let $J_{n}$, $n\geq 1$, be the unital isometric embeddings

$J_{n}:{\mathcal{A}}_{(n)}\rightarrow{\mathcal{A}}_{(n)}\otimes_{\rm
h}{\mathcal{B}}_{\omega},\quad J_{n}\big{(}a_{(n)}\big{)}=a_{(n)}\otimes e.$
(4.36)

Then

${\mathbb{E}}_{m}\circ J_{m}\circ\mathfrak{i}_{mn}={\mathbb{E}}_{n}\circ
J_{n},\quad m\geq n.$ (4.37)

As such, the tower of maps

${\mathbb{E}}_{n}\circ
J_{n}:{\mathcal{A}}_{(n)}\to{\mathcal{B}}_{\omega},\quad(n\geq 1),$ (4.38)

has a direct limit ${\mathbb{E}}_{\infty}\circ
J_{\infty}:{\mathcal{A}}_{R}\to{\mathcal{B}}_{\omega}$, which coincides with
the quotient map $q:{\mathcal{A}}_{R}\to{\mathcal{B}}_{\omega}$.

###### Proof.

The relations (4.37) are direct consequences of the Markov’s property. The
other statement follows from the observation that
${\mathbb{E}}_{(n)}(a_{(n)}\otimes e)=q(a_{(n)})$, for all $n\geq 1$.∎

###  4.4. Asymptotic constraints

So far, we have not imposed any asymptotic constraints on the state $\omega$.
Here we introduce the so called asymptotic clustering property, which is often
assumed based on physical arguments. We also point out finer asymptotic
properties, which emerge from a dynamical system canonically associated to
${\mathcal{B}}_{\omega}$.

Let us recall the shift map (2.16),

$S_{R}:{\mathcal{A}}_{R}\rightarrow{\mathcal{A}}_{R},\quad
S_{R}(a_{R})=S(\alpha_{0}^{-1}(1)\otimes a_{R}).$ (4.39)

The descent of this map on ${\mathcal{B}}_{\omega}$ can be defined as follows:

######  Definition 4.24.

We call the shift map on ${\mathcal{B}}_{\omega}$ the map

$\bar{S}:{\mathcal{B}}_{\omega}\rightarrow{\mathcal{B}}_{\omega},\quad\bar{S}={\mathbb{E}}\circ\bar{L},$
(4.40)

with $\bar{L}$ being the unital and isometric embedding

$\bar{L}:{\mathcal{B}}_{\omega}\rightarrow{\mathcal{A}}\otimes_{\rm
h}{\mathcal{B}}_{\omega},\quad\bar{L}(b)=1\otimes b.$ (4.41)

######  Proposition 4.25.

$\bar{S}\big{(}\lfloor a_{R}\rfloor\big{)}=\lfloor S_{R}(a_{R})\rfloor$, for
all $a_{R}\in{\mathcal{A}}_{R}$.

######  Remark 4.26.

Without doubt, the dynamical system $({\mathcal{B}}_{\omega},\bar{S})$ is
essential for understanding the shift invariant states on
${\mathcal{A}}^{\otimes{\mathbb{Z}}}$. Unfortunately, here, we can only point
to some far reaching consequences once certain dynamical behavior is assumed.
Specifically, note that, as a composition of completely contractive maps,
$\bar{S}$ is a contractive map, $\|\bar{S}(b)\|\leq\|b\|$. Furthermore,
$\bar{S}(e)=e$. One of the key questions that emerged during our study was if
$({\mathcal{B}}_{\omega},\bar{S})$ has an attractor that is larger than the
sub-space generated by $e$. We believe that this is a very important question
which deserves a full investigation. $\Diamond$

######  Definition 4.27.

We say that the shift-invariant state $\omega$ on ${\mathcal{A}}_{\mathbb{Z}}$
has the asymptotic clustering property if the sequence

$\sup\Big{\\{}\big{|}\omega\big{(}a_{L}\cdot S_{R}^{\circ
r}(a_{R})\big{)}-\omega(a_{L})\omega(a_{R})\big{|},\
a_{R/L}\in{\mathcal{A}}_{R/L},\ \|a_{R/L}\|=1\Big{\\}}$ (4.42)

converges to zero as $r\to\infty$.

######  Remark 4.28.

The asymptotic clustering property can be reformulated in a more effective but
less familiar way as

$\lim_{r\to\infty}\sup\Big{\\{}\Big{|}\omega\Big{(}a_{L}\cdot S_{R}^{\circ
r}\big{(}a_{R}-\omega(a_{R})\,1\big{)}\Big{)}\Big{|},\
a_{R/L}\in{\mathcal{A}}_{R/L},\ \|a_{R/L}\|=1\Big{\\}}=0.$ (4.43)

Both forms will be used in the following. $\Diamond$

The clustering property descends to $\bar{\omega}$ in the following way:

######  Proposition 4.29.

Assume that $\omega$ displays the asymptotic clustering property. Then the
reduced state also display a similar property,

$\lim_{r\rightarrow\infty}\sup\Big{\\{}\Big{|}\bar{\omega}\Big{(}{\mathbb{E}}_{(n)}\big{(}a_{(n)}\otimes\bar{S}^{\circ
r}(b)\big{)}\Big{)}-\bar{\omega}\Big{(}{\mathbb{E}}_{(n)}\big{(}a_{(n)}\otimes
e\big{)}\Big{)}\bar{\omega}(b)\Big{|}\Big{\\}}=0,$ (4.44)

where the supremum is over all $n\geq 0$, $a_{(n)}\in{\mathcal{A}}_{(n)}$ with
$\|a_{(n)}\|=1$, and $b\in{\mathcal{B}}$ with $\|b\|\leq 1$.

###### Proof.

Let $a_{R}\in{\mathcal{A}}_{R}$ such that $b=\lfloor a_{R}\rfloor$ and
$a_{(n)}\in{\mathcal{A}}_{(n)}$. Then

${\mathbb{E}}_{n}\big{(}a_{(n)}\otimes\bar{S}^{\circ
r}(b)\big{)}=\big{\lfloor}S^{n}\big{(}a_{(n)}\otimes S_{R}^{\circ
r}(a_{R})\big{)}\big{\rfloor}.$ (4.45)

Therefore, taking into account the shift invariance of $\omega$,

$\bar{\omega}\big{(}{\mathbb{E}}_{n}(a_{(n)}\otimes\bar{S}^{\circ
r}(b)\big{)}=\omega\big{(}a_{(n)}\otimes S_{R}^{\circ r}(a_{R})\big{)}.$
(4.46)

Then

$\displaystyle\bar{\omega}\Big{(}{\mathbb{E}}_{n}\big{(}a_{(n)}\otimes\bar{S}^{\circ
r}(b)\big{)}\Big{)}-\bar{\omega}\Big{(}{\mathbb{E}}_{n}\big{(}a_{(n)}\otimes
e\big{)}\Big{)}\bar{\omega}(b)$ (4.47)
$\displaystyle\qquad=\omega\Big{(}a_{(n)}\otimes S_{R}^{\circ
r}(a_{R})\Big{)}-\omega\big{(}a_{(n)}\big{)}\omega\big{(}a_{R}\big{)}$
$\displaystyle\qquad\qquad=\omega\Big{(}\mathfrak{i}_{L}\big{(}a_{(n)}\big{)}\cdot
S^{\circ
r}\big{(}a_{R}\big{)}\Big{)}-\omega\Big{(}\mathfrak{i}_{L}\big{(}a_{(n)}\big{)}\Big{)}\omega\big{(}a_{R}\big{)},$

and the statement follows from the clustering property of $\omega$. ∎

We recall the functional $\Gamma:{\mathcal{B}}_{\omega}\to[0,\infty)$ defined
in Propostion 3.23. Then:

######  Proposition 4.30.

Assume $\omega$ to have the asymptotic clustering property and, furthermore,
that

$\Gamma(b)\geq c\|b\|,\quad\ \forall\ b\in{\mathcal{B}}_{\omega},$ (4.48)

for some strictly positive constant $c$. Then the linear sub-space generated
by $e$ is the only attractor of the map $\bar{S}$ and, furthermore,

$\lim_{r\rightarrow\infty}\bar{S}^{\circ
r}(b)=\bar{\omega}(b)\,e,\quad\forall\ b\in{\mathcal{B}}_{\omega}.$ (4.49)

###### Proof.

From the asymptotic clustering property (4.43), we have

$\lim_{r\rightarrow\infty}\sup\Big{\\{}\Big{|}\omega\Big{(}x\cdot S^{\circ
r}\big{(}a_{R}-\omega(a_{R})\,1_{R}\big{)}\Big{)}\Big{|},\
x\in{\mathcal{A}}_{L},\ \|x\|=1\Big{\\}}=0,$ (4.50)

for all $a_{R}\in{\mathcal{A}}_{R}$, which translates to

$\Gamma\Big{(}\bar{S}^{\circ r}\big{(}\lfloor
a_{R}-\omega(a_{R})\,1_{R}\rfloor\big{)}\Big{)}\rightarrow 0\ \ {\rm as}\ \
r\rightarrow\infty,$ (4.51)

Since $\lfloor 1_{R}\rfloor=e$ and the unit is invariant for $\bar{S}$,
condition (4.48) implies

$\big{\|}\bar{S}^{\circ r}(\lfloor
a_{R}\rfloor)-\omega(a_{R})\,e\big{\|}\rightarrow 0\ \ {\rm as}\ \
r\rightarrow\infty,$ (4.52)

which proves the claim. ∎

######  Corollary 4.31.

In the conditions of Proposition 4.30, there is one and only one shift
invariant state, $\bar{\omega}\circ\bar{S}=\bar{\omega}$, on
${\mathcal{B}}_{\omega}$.

######  Remark 4.32.

Note that (4.48) is exactly the same condition that ensures that the operator
space and operation system structures coincide on ${\mathcal{B}}_{\omega}$
(see Proposition 3.24). $\Diamond$

######  Corollary 4.33.

Assume that the statement in Eq. (4.49) holds. Then the map ${\mathbb{E}}$ is
full, in the sense that

$\overline{\bigcup_{n\geq
1}\bigcup_{x\in{\mathcal{A}}_{(n)}}{\mathbb{E}}_{n}(x\otimes
b)}={\mathcal{B}}_{\omega},\quad\forall\ b\in{\mathcal{B}}_{\omega}.$ (4.53)

###### Proof.

The statement is true for $b=e$ because ${\mathbb{E}}_{\infty}\circ
J_{\infty}=q$. For a generic element, note that

$\displaystyle{\mathbb{E}}_{(m+r)}\big{(}(x\otimes 1_{(n+1,n+r)})\otimes
b\big{)}$ $\displaystyle={\mathbb{E}}_{(m)}\big{(}x\otimes\bar{S}^{\circ
r}(b)\big{)}$ (4.54) $\displaystyle=\bar{\omega}(b)$
$\displaystyle{\mathbb{E}}_{(m)}(x\otimes
e)+{\mathbb{E}}_{(m)}\big{(}x\otimes(\bar{S}^{\circ
r}(b)-\bar{\omega}(b)e)\big{)},$

for any $x\in{\mathcal{A}}_{(m)}$. If $b^{\prime}$ is an arbitrary element of
${\mathcal{B}}_{\omega}$, then there exists a uniformly bounded sequence
$\\{x_{(m)}\in{\mathcal{A}}_{(m)}\\}_{m\in{\mathbb{N}}^{\times}}$ such that
$\\{\lfloor x_{(m)}\rfloor\\}_{m\in{\mathbb{N}}^{\times}}$ converges to
$b^{\prime}$ in ${\mathcal{B}}_{\omega}$. If

$y_{(2m)}=\bar{\omega}(b)^{-1}x_{(m)}\otimes 1_{(m)},$ (4.55)

then we can see from Eq. (4.54) that the sequence
$\\{E_{(2m)}(y_{(2m)}\\}_{m\in{\mathbb{N}}^{\times}}$ converges to
$b^{\prime}$ because the second term of the last time vanishes in the limit
since all maps ${\mathbb{E}}_{(n)}$ are all contractions.∎

######  Remark 4.34.

Our investigation of the asymptotic behavior led us naturally to the concept
of a full ${\mathbb{E}}$ map. It will be this property that we will use in our
reconstruction process. As we have seen, certain asymptotic behavior of the
map $\bar{S}$ implies the fullness of ${\mathbb{E}}$. It will be interesting
to investigate the reverse problem, namely, what kind of constraints are
imposed on the dynamical system $({\mathcal{B}}_{\omega},\bar{S})$ by the
fullness of ${\mathbb{E}}$. In particular, if fullness always requires (4.49).
$\Diamond$

##  5\. Reconstruction Algorithm

The reduction and factorization processes described in the previous sections
produced the data
$({\mathcal{A}},{\mathcal{B}}_{\omega},{\mathbb{E}},\bar{\omega})$ consisting
of the following:

  1. (1)

The local nuclear $C^{\ast}$-algebra ${\mathcal{A}}$.

  2. (2)

The reduced space ${\mathcal{B}}_{\omega}$ with the structure of an operator
system.

  3. (3)

The surjective, completely positive map
${\mathbb{E}}:{\mathcal{A}}\otimes_{\rm
h}{\mathcal{B}}_{\omega}\rightarrow{\mathcal{B}}_{\omega}$.

  4. (4)

The completely positive functional
$\bar{\omega}:{\mathcal{B}}_{\omega}\rightarrow{\mathbb{C}}$.

It is certainly appropriate to say that the initial data
$({\mathcal{A}}_{\mathbb{Z}},\omega)$ was reduced and factorized to the data
$({\mathcal{A}},{\mathcal{B}}_{\omega},{\mathbb{E}},\bar{\omega})$. We enforce
this statement by noting that:

######  Proposition 5.1.

If $\omega$ and $\omega^{\prime}$ are shift invariant states and the data
$({\mathcal{A}}_{\mathbb{Z}},\omega)$ and
$({\mathcal{A}}_{\mathbb{Z}},\omega^{\prime})$ both reduce to the same data
$({\mathcal{A}},{\mathcal{B}},{\mathbb{E}},\bar{\xi})$, then
$\omega=\omega^{\prime}$.

###### Proof.

We concentrate on the induced states $\omega_{R}$ and $\omega^{\prime}_{R}$ on
${\mathcal{A}}_{R}$. They must coincide because, for any $n\geq 1$ and
$a_{(n)}\in{\mathcal{A}}_{(n)}$, the reduction and factorization process for
the two initial sets of data assures us that

$\bar{\xi}\Big{(}{\mathbb{E}}_{(n)}\big{(}a_{(n)}\otimes
e\big{)}\Big{)}=\omega_{R}\Big{(}\mathfrak{i}_{n}\big{(}a_{(n)}\big{)}\Big{)}=\omega^{\prime}_{R}\Big{(}\mathfrak{i}_{n}\big{(}a_{(n)}\big{)}\Big{)},$
(5.1)

where the ${\mathbb{E}}_{(n)}$ maps are entirely determined by the reduced
data, namely, by ${\mathbb{E}}$ via the recursion relations (4.32). The two
states coincide on dense sub-spaces, hence they must be identical. Due to the
assumed shift invariance, this automatically implies that
$\omega=\omega^{\prime}$.∎

The goal of this section is to complete the reverse process, namely, to
reconstruct $({\mathcal{A}}_{\mathbb{Z}},\omega)$ from the data
$({\mathcal{A}},{\mathcal{B}}_{\omega},{\mathbb{E}},\bar{\omega})$.

###  5.1. Re-construction: The abstract form

Here we prove Theorem 1.1, which is reproduced below for reader’s convenience.

######  Theorem 5.2.

Assume:

  * •

A nuclear $C^{\ast}$-algebra ${\mathcal{A}}$;

  * •

An Archimedean matrix-ordered space $({\mathcal{S}},e)$;

  * •

A surjective, unital and completely positive map
${\mathbb{E}}:{\mathcal{A}}\otimes_{\rm
h}{\mathcal{S}}\rightarrow{\mathcal{S}}$;

  * •

A unital and completely positive functional $\bar{\omega}$ on ${\mathcal{S}}$.

Furthermore, let ${\mathbb{E}}_{(n)}:{\mathcal{A}}_{(n)}\otimes_{\rm
h}{\mathcal{S}}\rightarrow{\mathcal{S}}$ be the maps defined iteratively as in
Proposition 4.20, specifically,

${\mathbb{E}}_{(1)}={\mathbb{E}},\quad{\mathbb{E}}_{(n+1)}={\mathbb{E}}\circ({\rm
id}\otimes_{\rm h}{\mathbb{E}}_{(n)}),\quad n\geq 1.$ (5.2)

Then:

  1. i)

The tower of linear functionals

$\omega_{(n)}:{\mathcal{A}}_{(n)}\rightarrow{\mathbb{C}},\quad\omega_{n}=\bar{\omega}\circ{\mathbb{E}}_{(n)}\circ
J_{n},\quad n\geq 1,$ (5.3)

where $J_{n}$’s were defined in Eq. (4.36), defines a state $\omega_{R}$ on
${\mathcal{A}}_{R}$.

  2. ii)

Let $\bar{S}$ be defined as in Proposition 4.29, specifically,

$\bar{S}:{\mathcal{S}}\rightarrow{\mathcal{S}},\quad\bar{S}={\mathbb{E}}\circ\bar{L},$
(5.4)

with $\bar{L}$ being the unital and isometric embedding:

$\bar{L}:{\mathcal{S}}\rightarrow{\mathcal{A}}\otimes_{\rm
h}{\mathcal{S}},\quad\bar{L}(s)=1\otimes s.$ (5.5)

Then the reconstructed state is shift invariant, $\omega_{R}\circ
S_{R}=\omega_{R}$, provided $\bar{\omega}\circ\bar{S}=\bar{\omega}$.

  3. iii)

In the conditions from point ii), there exists a unique shift invariant state
$\omega$ on ${\mathcal{A}}_{\mathbb{Z}}$, which coincides with $\omega_{R}$
when restricted from ${\mathcal{A}}_{\mathbb{Z}}$ to ${\mathcal{A}}_{R}$.

  4. iv)

If the data has the asymptotic clustering property

$\lim_{r\rightarrow\infty}\sup\Big{\\{}\Big{|}\bar{\omega}\Big{(}{\mathbb{E}}_{(n)}\big{(}a_{(n)}\otimes\bar{S}^{\circ
r}(s-\bar{\omega}(s)\,e)\big{)}\Big{)}\Big{|}\Big{\\}}=0,$ (5.6)

where the supremum is over all $n\geq 0$, $a_{(n)}\in{\mathcal{A}}_{(n)}$ with
$\|a_{(n)}\|=1$, and $s\in{\mathcal{S}}$ with $\|s\|\leq 1$, then $\omega$
also has the asymptotic clustering property.

  5. v)

If $\bar{\omega}$ is shift invariant and ${\mathbb{E}}$ is full,

$\overline{\bigcup_{n\geq
1}\bigcup_{x\in{\mathcal{A}}_{(n)}}{\mathbb{E}}_{n}(x\otimes
s)}={\mathcal{S}},\quad\forall\ s\in{\mathcal{S}},\ s\neq 0,$ (5.7)

then the data $({\mathcal{A}}_{\mathbb{Z}},\omega)$ reduces back to
$({\mathcal{A}},{\mathcal{S}},{\mathbb{E}},\bar{\omega})$.

###### Proof.

i) By repeating the arguments from Proposition 4.21, we can conclude that the
maps ${\mathbb{E}}_{(n)}$ are completely positive. The maps $J_{n}$ are
completely positive as well. As such, $\omega_{(n)}$ are completely positive,
hence states on ${\mathcal{A}}_{(n)}$. Furthermore,
$\omega_{(m)}\circ\mathfrak{i}_{mn}=\omega_{(n)}$ for any $m\geq n$ and, as
such, the tower of states defines a state on the inductive limit of the tower
of ${\mathcal{A}}_{(n)}$ algebras, which is ${\mathcal{A}}_{R}$ itself.

ii) We denote by $\omega_{R}$ the inductive limit of states from point i).
Then, for any $n\geq 1$ and $a_{(n)}\in{\mathcal{A}}_{(n)}$, we have

$(\omega_{R}\circ
S_{R})\Big{(}\mathfrak{i}_{n}\big{(}a_{(n)}\big{)}\Big{)}=\omega_{(n+1)}\big{(}1\otimes
a_{(n)}\big{)}=(\bar{\omega}\circ{\mathbb{E}}_{(n+1)})\Big{(}\big{(}1\otimes
a_{(n)}\big{)}\otimes e\Big{)}.$ (5.8)

Using Markov’s property, we find

$(\omega_{R}\circ
S_{R})\Big{(}\mathfrak{i}_{n}\big{(}a_{(n)}\big{)}\Big{)}=(\bar{\omega}\circ{\mathbb{E}})\Big{(}1\otimes{\mathbb{E}}_{(n)}\big{(}a_{(n)}\otimes
e\big{)}\Big{)},$ (5.9)

and, after invoking the definition of $\bar{L}$,

$(\omega_{R}\circ
S_{R})\Big{(}\mathfrak{i}_{n}\big{(}a_{(n)}\big{)}\Big{)}=(\bar{\omega}\circ{\mathbb{E}}\circ\bar{L})\Big{(}{\mathbb{E}}_{(n)}\big{(}a_{(n)}\otimes
e\big{)}\Big{)}.$ (5.10)

Since ${\mathbb{E}}\circ\bar{L}=\bar{S}$ and
$\bar{\omega}\circ\bar{S}=\bar{\omega}$, the statement follows.

iii) Obvious.

iv) If $\omega$ is the reconstructed shift invariant full state, then from
definition (5.3) of the tower of states and assumed shift invariance, for any
$a_{(-m)}\in{\mathcal{A}}_{(-m)}$ and $a_{(n)}\in{\mathcal{A}}_{(n)}$, we have

$\displaystyle\omega\Big{(}\mathfrak{i}_{-m}\big{(}a_{(-m)}\big{)}\cdot(S_{R}^{\circ
r}\circ\mathfrak{i}_{n})\big{(}a_{(n)}-\omega(a_{(n)})\,1_{(n)}\big{)}\Big{)}$
(5.11)
$\displaystyle\qquad=\bar{\omega}\Big{(}{\mathbb{E}}_{(m)}\Big{(}S^{\circ
m}\big{(}a_{(-m)}\big{)}\otimes\bar{S}^{\circ
r}\big{(}s-\bar{\omega}(s)\,e\big{)}\Big{)}\Big{)},$

with $s=({\mathbb{E}}_{(n)}\circ J_{n})\big{(}a_{(n)}\big{)}$. As such, the
asymptotic clustering property for $\omega$ holds on dense subsets of
${\mathcal{A}}_{L}$ and ${\mathcal{A}}_{R}$, hence on the whole sub-algebras.

v) Recall Proposition 4.23, which prompts us to define the closed subset
${\mathcal{K}}={\rm Ker}\big{(}{\mathbb{E}}_{\infty}\circ J_{\infty}\big{)}$.
We will show that ${\mathcal{K}}$ coincides with the entanglement kernel
${\mathcal{K}}_{\omega}$ of the state $\omega$ constructed at points i) and
iii). By definition of $\omega$, we have

$\omega\Big{(}\mathfrak{i}_{-m}\big{(}S^{-m}\big{(}\mathfrak{i}_{m}(x)\big{)}\big{)}\cdot
a_{R}\Big{)}=\bar{\omega}\Big{(}{\mathbb{E}}_{(m)}\big{(}x\otimes({\mathbb{E}}_{\infty}\circ
J_{\infty})(a_{R})\big{)}\Big{)},$ (5.12)

for any $x\in{\mathcal{A}}_{(m)}$. As such, if $a_{R}\in{\mathcal{K}}$, then
$\omega_{x}(a_{r})=0$ for $x$ in a dense subset of ${\mathcal{A}}_{L}$, hence
for all $x$ in ${\mathcal{A}}_{L}$ because of the continuity of the state.
This proves that ${\mathcal{K}}\subset{\mathcal{K}}_{\omega}$. The reciprocal
is also true. Indeed, if $a_{R}\in{\mathcal{K}}_{\omega}$, then necessarily

$\bar{\omega}\Big{(}{\mathbb{E}}_{(m)}\big{(}x\otimes({\mathbb{E}}_{\infty}\circ
J_{\infty})(a_{R})\big{)}\Big{)}=0$ (5.13)

for any $x\in{\mathcal{A}}_{(m)}$, which is a direct consequence of the
identity (5.12). Let $s=({\mathbb{E}}_{\infty}\circ
J_{\infty})(a_{R})\in{\mathcal{S}}$ and assume first that $s\neq 0$. In this
case, since ${\mathbb{E}}$ is full,

$\overline{\bigcup_{m\geq
1}\bigcup_{x\in{\mathcal{A}}_{(m)}}{\mathbb{E}}_{m}(x\otimes
s)}={\mathcal{S}},$ (5.14)

and (5.13) can be true for any $m\geq 1$ and $x\in{\mathcal{A}}_{(m)}$ only if
$\bar{\omega}=0$. This cannot be, hence the only alternative is that $s=0$ or,
in other words, that $a_{R}\in{\mathcal{K}}$.

We have established that the entanglement kernel ${\mathcal{K}}_{\omega}$ of
$\omega$ coincide with ${\rm Ker}({\mathbb{E}}_{\infty}\circ J_{\infty})$, and
we consider now the exact sequence of Banach spaces

$\begin{diagram}$ (5.15)

From (5.14), we can then identify ${\mathcal{A}}_{R}/{\mathcal{K}}$ with
${\mathcal{S}}$ and we denote the class of $a_{R}$ in
${\mathcal{A}}_{R}/{\mathcal{K}}$ as $\lfloor a_{R}\rfloor$. Our last task is
to show that

${\mathbb{E}}(a\otimes s)=\lfloor a\otimes
a_{s}\rfloor\in{\mathcal{A}}_{R}/{\mathcal{K}}={\mathcal{S}},\quad\forall\
a\in{\mathcal{A}},\ s\in{\mathcal{S}},$ (5.16)

where $a_{s}$ is an element from ${\mathcal{A}}_{R}$ such that $\lfloor
a_{s}\rfloor=s$. From Eq. 5.15, the class $\lfloor a_{s}\rfloor$ can be also
computed as ${\mathbb{E}}_{\infty}\circ J_{\infty}(s)$. Then, for any
$a_{(n)}\in{\mathcal{A}}_{(n)}$,

${\mathbb{E}}\big{(}a\otimes\lfloor\mathfrak{i}_{n}(a_{(n)})\rfloor\big{)}={\mathbb{E}}\Big{(}a\otimes{\mathbb{E}}_{(n)}\big{(}a_{(n)}\otimes
e\big{)}\Big{)}={\mathbb{E}}_{(n+1)}\Big{(}\big{(}a\otimes
a_{(n)}\big{)}\otimes e\big{)}.$ (5.17)

The last line can be cast as $({\mathbb{E}}_{(n+1)}\circ
J_{(n+1)})\big{(}a\otimes a_{(n)}\big{)}$, which we know from (5.15) to return
the class of $a\otimes a_{(n)}$ in ${\mathcal{A}}_{R}/{\mathcal{K}}$. As such,
the relation (5.16) holds for a dense subset of ${\mathcal{S}}$, hence on the
whole ${\mathcal{S}}$.∎

######  Definition 5.3.

We call a state $\omega$ on ${\mathcal{A}}_{\mathbb{Z}}$ full if its
entanglement kernel is such that

$\overline{\bigcup_{n\geq 1}\bigcup_{x\in{\mathcal{A}}_{(n)}}\lfloor x\otimes
a_{R}\rfloor}={\mathcal{A}}_{R}/{\mathcal{K}}_{\omega},$ (5.18)

for any $a_{R}\in{\mathcal{A}}_{R}$.

######  Remark 5.4.

Sub-section 4.4 supplied conditions that are sufficient for a state to be
full. For example asymptotic clustering property together with condition
(4.48) are sufficient. $\Diamond$

######  Corollary 5.5.

Any full, shift invariant state on ${\mathcal{A}}_{\mathbb{Z}}$ can be
generated from a set of data
$({\mathcal{A}},{\mathcal{S}},{\mathbb{E}},\bar{\omega})$ as in Theorem 5.2.

###### Proof.

Starting from a full, translational invariant state $\omega$, the reduction
and factorization processes generate a set of data
$({\mathcal{A}},{\mathcal{S}},{\mathbb{E}},\bar{\omega})$ with ${\mathbb{E}}$
full. A translational invariant state $\omega^{\prime}$ can be reconstructed
from this set of data and, since ${\mathbb{E}}$ is full, the data
$({\mathcal{A}}_{\mathbb{Z}},\omega^{\prime})$ reduces back to the same
$({\mathcal{A}},{\mathcal{S}},{\mathbb{E}},\bar{\omega})$. Then Proposition
5.1 assures us that $\omega=\omega^{\prime}$ and, as such, $\omega$ is indeed
reproduced by the reconstruction process from the set of reduced data
$({\mathcal{A}},{\mathcal{S}},{\mathbb{E}},\bar{\omega})$. ∎

######  Example 5.6.

Let us consider the product state discussed in Examples 2.15 and 4.19, where
we found ${\mathcal{B}}_{\omega}={\mathcal{A}}/{\rm Ker}\,\omega_{0}$ and

${\mathbb{E}}(a\otimes\lfloor a^{\prime}\rfloor)=\omega_{0}(a)\lfloor
a^{\prime}\rfloor.$ (5.19)

Then, ${\mathbb{E}}$ is full if and only if ${\mathcal{B}}_{\omega}$ is one
dimensional. Suppose that this is indeed the case. Then the only state on
${\mathcal{B}}_{\omega}$ is the one entirely determined by $\bar{\xi}(e)=1$.
As such,

$\omega_{1}(a)=\bar{\xi}\big{(}{\mathbb{E}}(a\otimes e)\big{)}=\omega_{0}(a)$
(5.20)

and

$\omega_{2}(a_{1}\otimes
a_{2})=\bar{\xi}\big{(}{\mathbb{E}}_{(2)}(a_{1}\otimes a_{2}\otimes
e)\big{)}=\bar{\xi}\big{(}{\mathbb{E}}\big{(}a_{1}\otimes{\mathbb{E}}(a_{2})\big{)}\big{)}=\omega_{0}(a_{1})\omega_{0}(a_{2}).$
(5.21)

Iterating further, one finds

$\omega_{n}(a_{1}\otimes\cdots\otimes
a_{n})=\omega_{0}(a_{1})\cdots\omega_{0}(a_{n})$ (5.22)

and the product state $\omega_{0}^{\otimes{\mathbb{Z}}}$ is indeed reproduced
by the reconstruction algorithm. $\Diamond$

######  Example 5.7.

We consider here the class of AKLT states introduced in Example 2.16. We
recall that the setting was that of two nuclear $C^{\ast}$-algebras
$\hat{\mathcal{A}}$ and $\tilde{\mathcal{A}}$, such that
$\hat{\mathcal{A}}=\tilde{\mathcal{A}}\otimes\tilde{\mathcal{A}}$. Both
operator systems ${\mathcal{A}}$ and ${\mathcal{B}}_{\omega}$ are defined in
terms of these two algebras, namely ${\mathcal{A}}=p\hat{\mathcal{A}}p$, with
$p$ a projection from $\hat{\mathcal{A}}$ that also stands for the unit of
${\mathcal{A}}$, and ${\mathcal{B}}_{\omega}=\tilde{\mathcal{A}}$. Here we
derive the map ${\mathbb{E}}$ corresponding to the AKLT-type states $\omega$
over ${\mathcal{A}}$. For this, we define first

$\widetilde{\mathbb{E}}:\hat{\mathcal{A}}\otimes\tilde{\mathcal{A}}\simeq\tilde{\mathcal{A}}\otimes\tilde{\mathcal{A}}\otimes\tilde{\mathcal{A}}\rightarrow\tilde{\mathcal{A}},\quad\widetilde{\mathbb{E}}={\rm
id}\otimes\xi_{0},$ (5.23)

where
$\xi_{0}:\tilde{\mathcal{A}}\otimes\tilde{\mathcal{A}}\rightarrow{\mathbb{C}}$
is a complete contraction. Then $\widetilde{\mathbb{E}}$ is a complete
contraction. Next, we let
$\mathfrak{j}:p\hat{\mathcal{A}}p\rightarrow\hat{\mathcal{A}}$ be the non-
unital embedding that takes $p$ into $p$ rather than into the unit of
$\hat{\mathcal{A}}$. This map is evidently a complete contraction. Lastly, we
define ${\mathbb{E}}$ as:

${\mathbb{E}}:p\hat{\mathcal{A}}p\otimes\tilde{\mathcal{A}}\rightarrow\tilde{\mathcal{A}},\quad{\mathbb{E}}=\widetilde{\mathbb{E}}\circ(\mathfrak{j}\otimes{\rm
id}),$ (5.24)

which, as a composition of two completely contractive maps, is a completely
contractive map. This map is also unital, provided
${\mathbb{E}}(p\,\otimes\tilde{1})=\tilde{1}$. Therefore, whenever $p$ and
$\xi_{0}$ fulfill this constraint, an AKLT-type state can be reconstructed
from $({\mathcal{A}},\tilde{\mathcal{A}},{\mathbb{E}},\bar{\xi})$, where
$\bar{\xi}$ is a shift invariant state on $\tilde{\mathcal{A}}$. We recall
that this machinery now works for infinite dimensional algebras. $\Diamond$

######  Remark 5.8.

The particular case studied in [1], illustrated in Fig. 2.1 and partially
analyzed in Example 2.16, corresponds to
$\tilde{\mathcal{A}}={\mathcal{M}}_{2}$, hence
$\hat{\mathcal{A}}={\mathcal{M}}_{4}\simeq{\mathcal{M}}_{2}\otimes{\mathcal{M}}_{2}$,
with $p$ being the rank-3 projection

$P=\tfrac{3}{4}\,I_{2}\otimes
I_{2}+\tfrac{1}{4}\sum_{i=1}^{3}\sigma^{i}\otimes\sigma^{i}\in{\mathcal{M}}_{2}\otimes{\mathcal{M}}_{2}\simeq{\mathcal{M}}_{4},$
(5.25)

such that ${\mathcal{A}}=P{\mathcal{M}}_{4}P\simeq{\mathcal{M}}_{3}$.
Furthermore,

$\xi_{0}(M)=\tfrac{4}{3}\,{\rm Tr}\big{(}MP_{0}\big{)},\quad
M\in{\mathcal{M}}_{2}\otimes{\mathcal{M}}_{2}\simeq{\mathcal{M}}_{4},$ (5.26)

with $P_{0}$ being the rank-1 projection

$P_{0}=\tfrac{1}{4}\,I_{2}\otimes
I_{2}-\tfrac{1}{4}\sum_{i=1}^{3}\sigma^{i}\otimes\sigma^{i}\in{\mathcal{M}}_{2}\otimes{\mathcal{M}}_{2}\simeq{\mathcal{M}}_{4}.$
(5.27)

In this case, (5.24) takes the following concrete form:

$\displaystyle{\mathbb{E}}\Big{(}P\big{(}\sigma^{u}\otimes\sigma^{v}\big{)}P\otimes\sigma^{w}\Big{)}=\tfrac{4}{3}\sum_{\alpha,\beta=0}^{3}g_{\alpha}g_{\beta}{\rm
Tr}\Big{(}\big{(}\sigma^{\alpha}\sigma^{v}\sigma^{\beta}\otimes\sigma^{w}\big{)}P_{0}\Big{)}\,\sigma^{\alpha}\sigma^{u}\sigma^{\beta},$
(5.28)

where $g_{0}=\tfrac{3}{4}$ and $g_{i}=\tfrac{1}{4}$ for $i=\overline{1,3}$.
Then we find

${\mathbb{E}}(P\otimes
I_{2})=I_{2},\quad{\mathbb{E}}(P\otimes\vec{\alpha}\cdot\vec{\sigma})=-\tfrac{1}{3}\,\vec{\alpha}\cdot\vec{\sigma},$
(5.29)

where $\alpha\in{\mathbb{C}}^{3}$ and
$\vec{\alpha}\cdot\vec{\sigma}=\sum_{i=1}^{3}\alpha_{i}\,\sigma_{i}$. Hence,
the data satisfy the constraint mentioned in Example 5.7. Furthermore, since
any $B\in{\mathcal{M}}_{2}$ can be written uniquely as
$B=\alpha_{0}\,I_{2}+\vec{\alpha}\cdot\vec{\sigma}$, we have

$\bar{S}^{\circ
r}(B)={\mathbb{E}}(P\otimes{\mathbb{E}}(\ldots{\mathbb{E}}(P\otimes
B)\ldots))=\alpha_{0}\,I_{2}+\big{(}-\tfrac{1}{3}\big{)}^{r}\,\vec{\alpha}\cdot\vec{\sigma},$
(5.30)

hence $\bar{S}$ has a unique fixed point. In this case, the only compatible
$\bar{S}$-invariant functional $\bar{\xi}$ on ${\mathcal{M}}_{2}$ is

$\bar{\xi}(\alpha_{0}\,I_{2}+\vec{\alpha}\cdot\vec{\sigma})=\alpha_{0}.$
(5.31)

As pointed out in [6], there are many other possible choices for $P_{0}$.
$\Diamond$

###  5.2. Reconstruction: The concrete form

Diagram (5.32) collects all isometric and unital embeddings between the
operator systems occurring so far:

$\begin{diagram}$ (5.32)

Note that all the isometric inclusions result directly from the injectivity of
the Haagerup tensor product, stated in Proposition 4.9. This is another reason
why Haagerup tensor product appears naturally in the present context. However,
the diagram (5.32) is not complete, because we don’t have yet a concrete
isometric embedding of ${\mathcal{A}}\otimes_{\rm h}{\mathcal{S}}$ into a
$C^{\ast}$-algebra. To solve this issue, we recall that the Haagerup tensor
product $B(H_{1})\otimes_{\rm h}B(H_{2})$ embeds isometrically into the free
product $B(H_{1})\star B(H_{2})$ of the algebras (see [12, p. 98]). At its
turn, this free product embeds isometrically into the algebra of bounded
operators over the free product $H_{1}\star H_{2}$ of the Hilbert spaces. The
latter, however, require choices of distinguished unit vectors [16, p. 3].

A natural way to introduce such distinguished vectors is to expand the Hilbert
spaces to $\widetilde{H}_{i}={\mathbb{C}}\cdot\xi_{i}\oplus H_{i}$, $i=1,2,$
and re-draw the diagram of embeddings as follows:

$\begin{diagram}\vspace{0.1cm}$ (5.33)

As we shall see, these Hilbert space expansions will prove very convenient
below, hence their role is far more than a formal one. We denote the maps
supplying the isometric embeddings in $B(\widetilde{H}_{i})$, $i=1,2$, of
${\mathcal{A}}$ and ${\mathcal{S}}$ by $\tilde{\rho}$ and $\tilde{\sigma}$,
respectively. Furthermore, to ease the notation, we will write $A$ for
$\rho(a)$ and $\tilde{A}$ for $\tilde{\rho}(a)$, as well as $T$ for
$\sigma(t)$ and $\tilde{T}$ for $\tilde{\sigma}(t)$. The projections onto the
sub-spaces ${\mathbb{C}}\cdot\xi_{i}$ of $\widetilde{H}_{i}$ will be denoted
by $P_{\xi_{i}}$, $i=1,2$.

Now, from Arveson Theorems 3.34, we know that the completely positive map
${\mathbb{E}}$ extends over $B(\widetilde{H}_{1}\star\widetilde{H}_{2})$ and,
from Stinespring theorem [15], we know that this extension has the generic
structure

$B(\widetilde{H}_{1}\star\widetilde{H}_{2})\ni
G\mapsto{\mathbb{E}}(G)=V^{\ast}\hat{\pi}(G)V\in B(\widetilde{H}_{2}),$ (5.34)

where $\hat{\pi}$ is a representation of
$B(\widetilde{H}_{1}\star\widetilde{H}_{2})$ into the algebra of bounded
operators over a Hilbert space $\widehat{H}$ and
$V:\widetilde{H}_{2}\rightarrow\widehat{H}$ is a contraction. Recall that
$B(\widetilde{H}_{1})$ and $B(\widetilde{H}_{2})$ are canonically embedded
into their free product, hence the restriction of $\hat{\pi}$ generates
representations $\hat{\pi}_{1}$ and $\hat{\pi}_{2}$ of $B(\widetilde{H}_{1})$
and $B(\widetilde{H}_{2})$ on $\widehat{H}$, respectively. Reciprocally, from
the universal property of the free product of $C^{\ast}$-algebras, we know
that any representation $\hat{\pi}$ is the unique extension of two such
representations. More concretely, an element $\tilde{A}\otimes\tilde{T}\in
B(\widetilde{H}_{1})\otimes_{\rm h}B(\widetilde{H}_{2})$ embeds as
$\tilde{A}\tilde{T}$ in $B(\widetilde{H}_{1})\star B(\widetilde{H}_{2})$,
hence
$\hat{\pi}(\tilde{A}\otimes\tilde{T})=\hat{\pi}(\tilde{A})\hat{\pi}(\tilde{T})$.
The conclusion is that ${\mathbb{E}}$, as extended to a completely positive
map from $B(\widetilde{H}_{1})\otimes_{\rm h}B(\widetilde{H}_{2})$ to
$B(\widetilde{H}_{2})$, has the following generic action on monomials

${\mathbb{E}}(\tilde{A}\otimes\tilde{T})=V^{\ast}\hat{\pi}_{1}(\tilde{A})\hat{\pi}_{2}(\tilde{T})V.$
(5.35)

This is also known from the factorization of the completely bounded multi-
linear maps [12][Lemma 5.14]. It is instructive to iterate ${\mathbb{E}}$
once:

$\displaystyle{\mathbb{E}}\big{(}a_{1}\otimes{\mathbb{E}}(a_{2}\otimes
t)\big{)}=V^{\ast}\hat{\pi}_{1}(A_{1})\hat{\pi}_{2}\Big{(}V^{\ast}\hat{\pi}_{1}(A_{2})\hat{\pi}_{2}(T)V\Big{)}V,$
(5.36)

where $A_{i}=\rho(a_{i})\in B(H_{1})$ and $T=\sigma(t)\in B(H_{2})$. The
concrete action of a generic ${\mathbb{E}}_{(n)}$ now becomes apparent:

######  Proposition 5.9.

The iterated maps accept the concrete presentation

$\displaystyle{\mathbb{E}}_{(n)}\big{(}(a_{1}\otimes\ldots\otimes
a_{n})\otimes t\big{)}$ (5.37)
$\displaystyle\quad=V^{\ast}\hat{\pi}_{1}(A_{1})\hat{\pi}_{2}\Big{(}V^{\ast}\hat{\pi}_{1}(A_{2})\ldots\hat{\pi}_{2}\Big{(}V^{\ast}\hat{\pi}_{1}(A_{n})\hat{\pi}_{2}(T)V\Big{)}\ldots\Big{)}V\Big{)}V.$

###  5.3. Operator product states

We model the separable Hilbert space $H_{1}$ as $\ell^{2}({\mathcal{I}})$ with
${\mathcal{I}}$ a countable set, which is always possible e.g. by fixing a
basis and have ${\mathcal{I}}$ serve as the set of labels for this basis. For
$\psi\in H_{1}$, we introduce the pair of conjugate operators from
$B(\widetilde{H}_{1})$,

$Z_{\psi}(\alpha\,\xi_{1}+\psi^{\prime})=\alpha\psi,\quad
Z^{\ast}_{\psi}(\alpha\,\xi_{1}+\psi^{\prime})=\langle\psi,\psi^{\prime}\rangle\,\xi_{1},$
(5.38)

and, if $\big{\\{}\delta_{i}\big{\\}}_{i\in{\mathcal{I}}}$ denotes the
standard basis of $\ell^{2}({\mathcal{I}})$, we let $Z_{i}=Z_{\delta_{i}}\in
B(\widetilde{H}_{1})$.

######  Proposition 5.10.

For $\psi,\psi^{\prime}\in H_{1}$,

$Z_{\psi}Z^{\ast}_{\psi^{\prime}}=|\psi\rangle\langle\psi^{\prime}|\in
B(H_{1})\subset B(\widetilde{H}_{1}).$ (5.39)

Furthermore, if $A\in B(H_{1})\subset B(\widetilde{H}_{1})$, then

$A=\sum_{i,j\in{\mathcal{I}}}\langle\delta_{i},A\delta_{j}\rangle\,Z_{i}Z_{j}^{\ast},$
(5.40)

with the sum converging in the weak topology of $B(\widetilde{H}_{1})$.

###### Proof.

We have

$Z_{\psi}Z^{\ast}_{\psi^{\prime}}(\alpha\,\xi_{1}+\psi^{\prime\prime})=Z_{\psi}\big{(}\langle\psi^{\prime},\psi^{\prime\prime}\rangle\,\xi_{1}\big{)}=\langle\psi^{\prime},\psi^{\prime\prime}\rangle\,\psi,$
(5.41)

hence, the first statement follows. The second statement follows directly from
the expansion of the operator in terms of its matrix elements. ∎

######  Remark 5.11.

Eq. (5.40) replaces the expansion of $A$ in terms of its matrix elements,
$A=\sum_{i,j\in{\mathcal{I}}}A_{ij}|\delta_{i}\rangle\langle\delta_{j}|$. In
this expansion, the bra and the ket can be interpreted as maps between $H_{1}$
and ${\mathbb{C}}$. However, with the Hilbert space extension trick, the bra
and the ket become elements of $B(\widetilde{H}_{1})$, hence they can be acted
on by $\hat{\pi}_{1}$. This is significant! $\Diamond$

We are now in a position where we can supply the proof of Theorem 1.2,
reproduced below for convenience:

######  Theorem 5.12.

Let $\delta_{\bm{i}}=\delta_{i_{1}}\otimes\ldots\otimes\delta_{i_{n}}$,
$\bm{i}=(i_{1},\ldots,i_{n})\in{\mathcal{I}}^{n}$, be a basis of
$H_{1}^{\otimes n}$ and assume the following:

  * •

The representations $\hat{\pi}_{1}$ and $\hat{\pi}_{2}$ commute;

  * •

There exist a unitary map
$U:\widetilde{H}_{2}\to\hat{\pi}_{1}\big{(}P_{\xi_{1}}\big{)}\widehat{H}$.

  * •

The representation
$\hat{\pi}_{1}\big{(}P_{\xi_{1}}\big{)}\hat{\pi}_{2}(\cdot)\hat{\pi}_{1}\big{(}P_{\xi_{1}}\big{)}$
of $B(\widetilde{H}_{2})$ on
$\hat{\pi}_{1}\big{(}P_{\xi_{1}}\big{)}\widehat{H}$ is unitarily isomorphic to
the identity representation.

Then the maps ${\mathbb{E}}_{n}$ accept presentations as operator products:

${\mathbb{E}}_{n}(a_{(n)}\otimes
t)=\sum_{\bm{i},\bm{j}\in{\mathcal{I}}^{n}}\big{\langle}\delta_{\bm{i}},A_{(n)}\delta_{\bm{j}}\big{\rangle}\,\Sigma_{i_{1}}\cdots\Sigma_{i_{n}}T\,\Sigma^{\ast}_{j_{n}}\cdots\Sigma^{\ast}_{j_{1}},$
(5.42)

where $A_{(n)}\in B(H_{1})^{\otimes_{\rm h}n}\subset
B(\widetilde{H}_{1})^{\otimes_{\rm h}n}$, $T\in B(H_{2})\subset
B(\widetilde{H}_{2})$ and

$\Sigma_{i}\in
B(\widetilde{H}_{2}),\quad\Sigma_{i}=V^{\ast}\hat{\pi}_{1}(Z_{i})U.$ (5.43)

The sum in (5.42) converges in the weak topology of $B(\widetilde{H}_{2})$.

###### Proof.

We start our proof from ${\mathbb{E}}_{(1)}={\mathbb{E}}$. Using its
expression from Eq. (5.35) and the decomposition of $A$ from Eq. (5.40), we
have

$\displaystyle{\mathbb{E}}(a\otimes t)$
$\displaystyle=\sum_{i,j\in{\mathcal{I}}}\langle\delta_{i},A\delta_{j}\rangle
V^{\ast}\hat{\pi}_{1}(Z_{i}Z^{\ast}_{j})\hat{\pi}_{2}(T)V$ (5.44)
$\displaystyle=\sum_{i,j\in{\mathcal{I}}}\langle\delta_{i},A\delta_{j}\rangle
V^{\ast}\hat{\pi}_{1}(Z_{i})\hat{\pi}_{2}(T)\hat{\pi}_{1}(Z^{\ast}_{j})V,$