# Quantum graphs: different perspectives, homomorphisms and quantum
automorphisms

Matthew Daws

###### Abstract

We undertake a study of the notion of a quantum graph over arbitrary finite-
dimensional $C^{*}$-algebras $B$ equipped with arbitrary faithful states.
Quantum graphs are realised principally as either certain operators on
$L^{2}(B)$, the quantum adjacency matrices, or as certain operator bimodules
over $B^{\prime}$. We present a simple, purely algebraic approach to proving
equivalence between these settings, thus recovering existing results in the
tracial state setting. For non-tracial states, our approach naturally suggests
a generalisation of the operator bimodule definition, which takes account of
(some aspect of) the modular automorphism group of the state. Furthermore, we
show that each such “non-tracial” quantum graphs corresponds to a “tracial”
quantum graph which satisfies an extra symmetry condition. We study
homomorphisms (or CP-morphisms) of quantum graphs arising from UCP maps, and
the closely related examples of quantum graphs constructed from UCP maps. We
show that these constructions satisfy automatic bimodule properties. We study
quantum automorphisms of quantum graphs, give a definition of what it means
for a compact quantum group to act on an operator bimodule, and prove an
equivalence between this definition, and the usual notion defined using a
quantum adjacency matrix. We strive to give a relatively self-contained,
elementary, account, in the hope this will be of use to other researchers in
the field.

##  1 Introduction

The notion of a _Quantum_ (or sometimes _non-commutative_) _Graph_ is a non-
commutative generalisation of a classical (finite) graph which has been
defined and explored in a number of different presentations. The initial
definition arose from the study of quantum communication channels, [20], and
treats them as operator systems. The series of papers [26, 45, 46] extend this
definition to arbitrary von Neumann algebras (instead of just
$\mathbb{M}_{n}$), here being more motivated by a general notion of a _quantum
relation_. A graphical approach, using string diagrams, was given in [31],
showing that a notion of the adjacency matrix of quantum graph bijects with
the quantum relation definition. Here again motivation comes from quantum
computing considerations. Also here links with Quantum Groups arose, which has
been further explored in [7]. Recent papers exploring quantum graphs are [25,
30], while generalisations to the non-tracial case have been given in [7, 30],
and a more minimal set of axioms explored in [8, 9, 12]. Papers making use of
quantum graphs include [6, 11, 12, 21, 22, 23, 38] to list but a few.

This paper revisits the bijection between adjacency matrices and quantum
relations, and develops a new, elementary, algebraic approach, eschewing
graphical methods. We take the time to survey the area, presenting the
different approaches, and then show that they are equivalent. Our different
approach has the benefit of seamlessly allowing us to treat the non-tracial
situation; we further can work with arbitrary faithful states, not
$\delta$-forms. We present a large number of new results. Once we have
established our approach to equivalence, it becomes clear that in fact quantum
graphs over a non-tracial state correspond to “tracial” quantum graphs
satisfying an extra symmetry, see Theorem 5.29. We give a definition of the
quantum automorphism group of a quantum graph which operates at the quantum
relation level. We explore how quantum channels give rise to quantum graphs,
showing how an analysis of the Stinespring dilation yields generalisations and
improvements of results from [20, 46].

Let us now quickly introduce the principle objects we shall study, and then
outline the organisation of the paper.

For us a _graph_ $G$ is a (finite) vertex set $V=V_{G}$ together with a set of
edges $E=E_{G}\subseteq V\times V$. Thus, in full generality, we allow
directed graphs, and loops at vertices (but not multiple edges between
vertices, and each loop, if present, is undirected). In this formalism, an
_undirected graph_ is when $(x,y)\in E\implies(y,x)\in E$; the graph has _no
loops_ when $(x,x)\not\in E$ for all $x\in V$; and the graph has _a loop at
every vertex_ when $(x,x)\in E$ for all $x\in V$. We will typically only be
interested in the case of no loops or a loop at every vertex, and there is an
obvious bijection between these cases. Thus having a loop at every vertex can
be viewed as simply a convention, and not some profound condition.

Let $B$ be a finite-dimensional $C^{*}$-algebra equipped with a faithful state
$\psi$. We shall say that we are in the _classical_ case when $B$ is
commutative, so $B\cong C(V)\cong\ell^{\infty}(V)$ for some finite set $V$ and
$\psi$ corresponds to the counting measure on $V$. All the definitions of a
quantum graph correspond to a genuine graph on vertex set $V$ in the classical
case. Briefly, the principle definitions of a quantum graph are:

  * •

a self-adjoint operator $A$ on $L^{2}(B)$ satisfying certain conditions
corresponding to, in the classical case, being $\\{0,1\\}$-valued, being
undirected, and having a loop at every vertex. See [31, 25] with
generalisations to non-tracial states in [7, 8, 30]; see Section 2.

  * •

when $B\subseteq\mathcal{B}(H)$ for some Hilbert space $H$, a unital, self-
adjoint subspace $S$ of $\mathcal{B}(H)$ which is a bimodule over
$B^{\prime}$, the commutant of $B$, meaning that for $a,b\in B^{\prime},x\in
S$ also $axb\in S$. This definition is essentially independent of the choice
of $H$. We call such $S$ an _operator bimodule_. For $B=\mathbb{M}_{n}$ this
is [19, 20], and in general, [26, 45, 46]; see Section 4.

  * •

A completely positive map $B\rightarrow B$ satisfying similar axioms to the
first case, explored at least in the case $B=\mathbb{M}_{n}$. See [12]; see
Section 3.

We remark that examples of the second case, when $B=\mathcal{B}(H)$, arise
naturally in the theory of quantum communication channels (this being the
motivation of [19, 20]). We explore what happens for general $B$ (beyond
$B=\mathbb{M}_{n}$) in Section 6.2.

In the literature, it is common to restrict to the case when $\psi$ is a
_$\delta$ -form_, compare Remark 2.3 and Definition 2.9. However, this
condition will play no role for us, excepting perhaps meaning that our
definition of the empty quantum graph is slightly more complicated. See also
Remark 9.11 which shows that seemingly we cannot just reduce to the
$\delta$-form case, in general.

When the state $\psi$ is a trace (perhaps a tracial $\delta$-form) these
axiomatisations are well-known to be equivalent, see [31, Section VII] and
[25, Section 1], the proofs using graphical techniques, and it is not clear
what happens in the non-tracial case (but see [30] for some recent progress in
adapting graphical ideas to the more general situation). In contrast, we
present a simple, purely algebraic approach which allows us to give a quick,
direct proof of the equivalence of the first two definitions. Our arguments
are motivated by thinking about the problem using operator-algebraic
techniques (e.g. working with commutants and projections, and using aspects of
Tomita–Takesaki theory). Indeed, it has already been recognised that a key
intermediate step between the first two axiomatisations is:

  * •

An (orthogonal) projection $e\in B\otimes B^{\operatorname{op}}$ satisfying
certain symmetry axioms. Here $B^{\operatorname{op}}$ is the algebra $B$ with
the _opposite_ product. To see why such a projection $e$ might arise, see
Theorem 5.12 for example.

For example, [25] takes this as the principle definition. We also extend the
“completely positive” approach suggest in [12] to arbitrary (finite-
dimensional) $C^{*}$-algebras. The technical tool here is a version of the
Choi matrix correspondence, and we present a version which works for any $B$;
while not technically hard, this might be of independent interest.

Our approach to these equivalences is Section 5. A major benefit of our
approach is that, with relatively little work, it suggests an approach which
works for any faithful state $\psi$, tracial or not. Here we take the first
definition (of operators on $L^{2}(B)$) as primary, and show how to adapt the
2nd (operator bimodule) definition so that we still obtain a correspondence.
This is an new definition, and a major benefit of the operator bimodule
approach is that it is extremely easy to find non-trivial examples in this
setting. We first introduce our key bijection(s) and begin to indicate how to
pass from adjacency matrices to projections. This can be viewed as an
algebraic version of the graphical methods, which are suitably “twisted” by
the modular automorphism group, in the non-tracial setting. It is well-known
that projections in a von Neumann algebra correspond to closed subspaces
invariant for the commutant, and we quickly explore in Section 5.1 what
happens with (non-self-adjoint) idempotents in place of projections. In
Section 5.2 we state and prove our main results: Theorem 5.14 which works with
just the “idempotency axiom”, Theorem 5.17 which deals with general adjacency
matrices, and Proposition 5.26 deals with the “loop at every vertex” axiom.
Along the way we prove a number of auxiliary results: for example, the
adjacency matrix commutes with the modular operator and conjugation,
Proposition 5.23.

In Section 6.1 we see how standard examples present under these equivalences.
In Section 6.2 we explore how quantum channels give rise to quantum graphs,
extending known constructions to the case of general $B$. Here a careful study
of Stinespring dilations are key to our results. In Section 6.3 we look at the
condition of having a loop at every vertex (or that of having no loops) and
produce bijections between these situations; we also look at graph
complements. Here our main task is to clarify some comments in the literature,
and to deal with the non-$\delta$-form case. In Section 6.4 we quickly look at
examples over $B=\mathbb{M}_{2}$ and $\mathbb{M}_{2}\oplus\mathbb{M}_{2}$.

In Section 7 we survey a number of different notions of “homomorphism” which
have appeared in the literature, mostly concentrating upon [38, 46]. By
extending our careful study of Stinespring dilations from Section 6.2 we
improve results of [46] by showing that “pushouts” and “pullbacks” are
automatically bimodules over the appropriate algebra (this also avoids the
need to use the techniques of [45]). This area is less settled that the main
definitions, and in Section 7.1 we survey some other notions of
“homomorphism”, “subgraph” and so forth.

We finally turn our attention to automorphisms. In Section 8 we quickly look
at automorphisms of quantum graphs: this is well-known in the adjacency matrix
picture, but we also look at operator bimodules, and prove an equivalence.
This also motivates the quantum case considered in Section 9. Again, for
quantum adjacency matrices, this is well-known. However, we find some of the
literature here a little hard to follow. For example, [25, Section 4] is
readable, but only looks at the Hopf $*$-algebra case, while we work with the
$C^{*}$-algebra picture of compact quantum groups. We hope that this section
will hence be helpful to the non-expert. After some set-up and technical
calculation, in Section 9.2 we propose a definition of a coaction on an
operator bimodule, and our main result, Theorem 9.18, shows that we then
obtain exactly the same notion as that of a coaction on a quantum adjacency
matrix. Here our result that every “generalised” quantum graph corresponds to
a “tracial” quantum graph, see Remark 5.35, is vital in linking the two
pictures of the quantum automorphism group. As operator bimodules are almost
trivial to give examples of, we hope that this definition of a quantum
automorphism group might aid, for example, in understanding which quantum
automorphism groups can arise.

It is notable that in our list of recent papers “using” quantum graphs, almost
all use the Operator Bimodule approach, while papers looking at quantum
automorphisms use the adjacency matrix approach. It is hence useful to have a
comprehensive “bridge” between these different perspectives.

Let us end this introduction by saying that almost everything we do is finite-
dimensional (but clearly motivated in many places by infinite-dimensional
theory), an exception being results in Section 7. Of course, operator
bimodules are an infinite-dimensional framework, [45], but it seems unclear
what to expect of an “infinite-dimensional” adjacency matrix: many of the core
ideas of Section 5 are finite-dimensional.

###  1.1 Notational conventions

We shall attempt to introduce terminology and technical background results, as
needed, rather than collecting them up-front. However, let us fix some
conventions here. The inner-product on a Hilbert space $H$ will be written
$(\cdot|\cdot)$ and is assumed linear in the _right_ variable. Write
$\mathcal{B}(H),\mathcal{B}_{0}(H)$ for the algebras of bounded, respectively
compact, operators on $H$. The identity on $H$ is denoted by $1_{H}=1$, while
the identity map on other spaces is denoted by $\operatorname{id}$.

For $\xi,\eta\in H$, denote by $\omega_{\xi,\eta}$ the functional
$\mathcal{B}(H)\rightarrow\mathbb{C}$ given by $x\mapsto(\xi|x(\eta))$. Denote
by $\theta_{\xi,\eta}$ the rank-one operator $H\rightarrow
H;\zeta\mapsto(\xi|\zeta)\eta$. Then
$\theta_{\xi_{1},\eta_{1}}\circ\theta_{\xi_{2},\eta_{2}}=(\xi_{1}|\eta_{2})\theta_{\xi_{2},\eta_{1}}$
which is a little awkward, but it means we are consistent in our convention
that sesquilinear maps are linear on the right. We denote by
${\operatorname{Tr}}$ the non-normalised trace on $\mathcal{B}(H)$, so
${\operatorname{Tr}}(\theta_{\xi,\eta})=(\xi|\eta)$, and if
$H\cong\mathbb{C}^{n}$ then ${\operatorname{Tr}}(1_{H})=n$.

Let $\overline{H}$ be the _conjugate Hilbert space_ to $H$, so vectors in
$\overline{H}$ are $\\{\overline{\xi}:\xi\in H\\}$ with the same addition, but
with the complex-conjugate scalar multiplication
$\lambda\overline{\xi}=\overline{\overline{\lambda}\xi}$. For
$x\in\mathcal{B}(H)$ we define
$x^{\top},\overline{x}\in\mathcal{B}(\overline{H})$ by

$x^{\top}(\overline{\xi})=\overline{x^{*}\xi},\quad\overline{x}(\overline{\xi})=\overline{x\xi}\qquad(\overline{\xi}\in\overline{H}).$

Thus $(x^{\top})^{*}=(x^{*})^{\top}=\overline{x}$. The map $x\mapsto x^{\top}$
is a linear anti-homomorphism
$\mathcal{B}(H)\rightarrow\mathcal{B}(\overline{H})$. If $(e_{i})_{i=1}^{n}$
is an orthonormal basis for $H$ then $(\overline{e}_{i})_{i=1}^{n}$ is an
orthonormal basis for $\overline{H}$, so with respect to these bases we have
that $\mathcal{B}(H)\cong\mathbb{M}_{n}$ and
$\mathcal{B}(\overline{H})\cong\mathbb{M}_{n}$, and $x\mapsto x^{\top}$ is
simply taking the transpose of a matrix.

Throughout the paper, $B$ will denote a finite-dimensional $C^{*}$-algebra,
and $\psi$ will be a state on $B$, always assumed faithful, with further
conditions imposed if necessary.

###  1.2 Acknowledgements

The author thanks Simon Schmidt and Christian Voigt for helpful conversations
at an early stage of this project, and thanks Larissa Kroell for useful
correspondence.

The author is partially supported by EPSRC grant EP/T030992/1. For the purpose
of open access, the author has applied a CC BY public copyright licence to any
Author Accepted Manuscript version arising.

No data were created or analysed in this study.

##  2 Quantum adjacency matrices

We introduce the notion of a quantum adjacency matrix. We first briefly study
finite-dimensional $C^{*}$-algebras. Any finite-dimensional $C^{*}$-algebra
$B$ has the form

$B\cong\bigoplus_{k=1}^{n}\mathbb{M}_{n_{k}}.$

The addition and multiplication in $B$ respects this direct sum decomposition,
and so often it suffices to consider just a matrix algebra $\mathbb{M}_{n}$.
We write $e_{ij}\in\mathbb{M}_{n}$ for the matrix units, so a typical matrix
$a=(a_{ij})$ is $a=\sum_{i,j}a_{ij}e_{ij}$, and the trace satisfies
${\operatorname{Tr}}(e_{ij})=\delta_{i,j}$ the Kronecker delta. We write
$a^{\top}$ for the transpose of $a$ (compare with Section 1.1).

For $\psi\in\mathbb{M}_{n}^{*}$ there is $Q\in\mathbb{M}_{n}$ with
$\psi(a)={\operatorname{Tr}}(Qa)$ for each $a\in\mathbb{M}_{n}$; indeed, set
$Q_{ij}=\psi(e_{ji})$. Then $\psi$ being positive corresponds to $Q$ being
positive. If $\psi$ is positive, then $\psi$ being faithful corresponds to $Q$
being invertible, and $\psi$ is a state exactly when
${\operatorname{Tr}}(Q)=1$. The same remarks hence apply to $B$, choosing $Q$
in each matrix factor.

######  Definition 2.1.

Given $B,\psi$ as throughout the paper, we shall denote by $Q\in B$ the unique
positive invertible element with $\psi(a)={\operatorname{Tr}}(Qa)$ for each
$a\in B$. Let $L^{2}(B)=L^{2}(B,\psi)$ denote the GNS space for $\psi$.

Thus $L^{2}(B)$ is a finite-dimensional Hilbert space. Let
$\Lambda:B\rightarrow L^{2}(B)$ be the GNS map, which in this situation is a
linear isomorphism. If necessary, we write
$\pi:B\rightarrow\mathcal{B}(L^{2}(B))$ for the GNS representation, but we
shall often suppress this. Hence

$\pi(a)\Lambda(b)=a\Lambda(b)=\Lambda(ab),\quad\Lambda(a)=a\Lambda(1),\quad(\Lambda(a)|\Lambda(b))=\psi(a^{*}b)={\operatorname{Tr}}(Qa^{*}b)\qquad(a,b\in
B).$

Thus also $\Lambda\otimes\Lambda$ is a linear isomorphism between $B\otimes B$
and $L^{2}(B)\otimes L^{2}(B)$. The multiplication map $m:B\otimes
B\rightarrow B$ may hence be regarded as a (bounded) operator $L^{2}(B)\otimes
L^{2}(B)\rightarrow L^{2}(B)$. The identity $1\in B$ induces a linear map
$\mathbb{C}\rightarrow B;\lambda\mapsto\lambda 1$.

######  Definition 2.2.

Let $m:L^{2}(B)\otimes L^{2}(B)\rightarrow L^{2}(B)$ be the multiplication
map, and $m^{*}:L^{2}(B)\rightarrow L^{2}(B)\otimes L^{2}(B)$ its Hilbert
space adjoint. Let $\eta:\mathbb{C}\rightarrow L^{2}(B)$ be induced by the
identity of $B$, and let $\eta^{*}:L^{2}(B)\rightarrow\mathbb{C}$ be its
Hilbert space adjoint.

Hence $\eta:\lambda\mapsto\lambda\Lambda(1)$ and
$\eta^{*}(\xi)=(\Lambda(1)|\xi)$ for $\xi\in L^{2}(B)$, so
$\eta^{*}(\Lambda(a))=\psi(a)$ for $a\in B$. Similarly,
$m(\Lambda(a)\otimes\Lambda(b))=\Lambda(ab)$.

######  Remark 2.3.

In the literature (for example, [7, 8, 25, 30, 31]) it is very common to
assume that additionally $\psi$ is a _$\delta$ -form_ (see Definition 2.9) and
in this case, we use the terminology that $(B,\psi)$ is a _Finite Quantum
Space_. However, being a $\delta$-form will actually play no role in our
arguments, and so it seems more sensible to not add this as a seemingly
unnecessary hypothesis. To avoid confusion with the existing literature, we
hence avoid the “Quantum Space” terminology. ∎

We can now proceed with the definition of a quantum adjacency matrix.

######  Definition 2.4.

A _quantum adjacency matrix_ is a self-adjoint operator
$A\in\mathcal{B}(L^{2}(B))$ satisfying:

  1. 1.

$m(A\otimes A)m^{*}=A$;

  2. 2.

$(1\otimes\eta^{*}m)(1\otimes A\otimes 1)(m^{*}\eta\otimes 1)=A$;

  3. 3.

$m(A\otimes 1)m^{*}=1$.

By symmetry we also define variations of these axioms:

  1. 4.

$(\eta^{*}m\otimes 1)(1\otimes A\otimes 1)(1\otimes m^{*}\eta)=A$;

  2. 5.

$m(1\otimes A)m^{*}=1$.

We show later that when $A$ is self-adjoint, axioms (2) and (4) are
equivalent, see Proposition 5.4, and that in the presence of the other axioms,
also (3) and (5) are equivalence, see Proposition 5.25. To our knowledge, the
only discussion of these “variations” is in [30, Section 2.2]; in particular,
here (4) is termed $A$ being “self-transpose”.

While the terminology is to call $A$ a “matrix”, it is of course merely a
linear map, as $L^{2}(B)$ usually carries no canonical choice of basis, and in
our approach, we shall not think of $A$ as an actual matrix, except in the
classical situation which motivates the more general theory.

There is some further discussion of variations of these axioms in [30,
Definition 2.19], and we single out the following for later study.

######  Definition 2.5.

Consider an operator $A\in\mathcal{B}(L^{2}(B))$. Let $A_{0}:B\rightarrow B$
be the linear map induced by $A$, if we regard $B$ and $L^{2}(B)$ as being
linearly isomorphic. We say that $A$ is _real_ when
$A_{0}(a^{*})=A_{0}(a)^{*}$ for each $a\in B$.

Let us see what happens in the classical situation, where
$B=\ell^{\infty}(V)=C(V)$ for a finite set $V$, and $\psi$ is given by the
uniform measure on $V$. Let $(\delta_{v})_{v\in V}$ be the obvious basis of
$B$ consisting of idempotents, so $\psi(\delta_{v})=1$ for each $v$, and
$m(\Lambda(\delta_{v})\otimes\Lambda(\delta_{u}))=\delta_{v,u}\Lambda(\delta_{v})$
and hence
$m^{*}\Lambda(\delta_{v})=\Lambda(\delta_{v})\otimes\Lambda(\delta_{v})$. Also
$\eta$ is given by the vector $\Lambda(1)=\sum_{v}\Lambda(\delta_{v})$, and
$\eta^{*}\Lambda(\delta_{v})=1$ for each $v\in V$. Finally, we see that
$L^{2}(B)\cong\mathbb{C}^{V}$ with orthonormal basis $(e_{v})_{v\in V}$ where
$e_{v}=\Lambda(\delta_{v})$. Hence $A$ may be identified with a matrix
$(A_{u,v})$. Indeed, given two operators
$A_{1},A_{2}\in\mathcal{B}(L^{2}(B))$,

$m(A_{1}\otimes A_{2})m^{*}(e_{v})=m(A_{1}(e_{v})\otimes
A_{2}(e_{v}))=\sum_{u,w}A^{(1)}_{u,v}A^{(2)}_{w,v}m(e_{u}\otimes
e_{w})=\sum_{u}A^{(1)}_{u,v}A^{(2)}_{u,v}e_{u},$

and so $m(A_{1}\otimes A_{2})m^{*}$ has matrix $(A^{(1)}_{u,v}A^{(2)}_{u,v})$.
Thus $m(A_{1}\otimes A_{2})m^{*}$ is the entry-wise, or _Schur_ , or
_Hadamard_ , product of the matrices $A_{1},A_{2}$. Thus axiom (1) corresponds
to $A$ being Schur-idempotent, equivalently, that $A$ is $0,1$-valued. As the
adjoint on $B=C(V)$ is the pointwise complete conjugation, $A$ being real,
Definition 2.5, corresponds to the matrix of $A$ being real-valued, which is
of course automatic if $A$ is Schur-idempotent.

Similarly, we calculate that

$\displaystyle(1\otimes\eta^{*}m)$ $\displaystyle(1\otimes A\otimes
1)(m^{*}\eta\otimes 1)(e_{v})=\sum_{u}(1\otimes\eta^{*}m)(1\otimes A\otimes
1)(e_{u}\otimes e_{u}\otimes e_{v})$
$\displaystyle=\sum_{u,w}(1\otimes\eta^{*}m)(e_{u}\otimes A_{w,u}e_{w}\otimes
e_{v})=\sum_{u}e_{u}\otimes\eta^{*}(A_{v,u}e_{v})=\sum_{u}A_{v,u}e_{u}=A^{\top}(e_{v})$

Thus axiom (2) is equivalent to $A^{\top}=A$. In this situation, axiom (4)
gives the same condition. Finally,

$\displaystyle m(A\otimes 1)m^{*}(e_{v})$ $\displaystyle=m(A(e_{v})\otimes
e_{v})=A_{v,v}e_{v}$

and so axiom (3) (and here also (5)) is that $A$ has $1$ down the main
diagonal.

######  Proposition 2.6.

Let $V$ be a finite set, and let $B=C(V)$. An operator
$A\in\mathcal{B}(L^{2}(B))$ which satisfies axiom (1) corresponds to a
directed graph structure on $V$. That $A$ is self-adjoint, equivalently, that
$A$ satisfies axiom (2) or (4), corresponds to the graph being undirected.
That $A$ satisfies axiom (3) or (5) corresponds to the graph having a loop at
every vertex.

###### Proof.

A $0,1$-valued matrix corresponds to a directed graph structure on $V$ by
setting the edge set to be $E=\\{(u,v):A_{u,v}=1\\}$. The results now follows
from the discussion above. ∎

######  Remark 2.7.

We already see that there is a certain redundancy between requiring that $A$
be _self-adjoint_ and that $A$ satisfy axiom (2) (and/or (4)). This is
discussed a little more in [30] after Definition 2.19, in particular, [30,
Lemma 2.22]. To an extent, this issue occurs even for general, non-commutative
$B$, compare Remark 5.20. ∎

If we prefer to have a loop at no vertex, then given the above discussion, it
is natural to consider the axioms:

  1. 6.

$m(A\otimes 1)m^{*}=0$;

  2. 7.

$m(1\otimes A)m^{*}=0$.

Again, we show later that these axioms are equivalent, in the presence of the
other axioms.

We give some simple examples for quantum adjacency matrices.

######  Definition 2.8.

The _complete quantum graph_ is given by the adjacency matrix
$A=\theta_{\Lambda(1),\Lambda(1)}$, so that
$A\Lambda(a)=\eta^{*}(\Lambda(a))\eta(1)=\psi(a)\Lambda(1)$ for each $a\in B$.

A direct computation shows that this $A$ indeed satisfies all the axioms (1)
through (5).

We study the map $mm^{*}$ a little further. Suppose for the moment that
$B=\mathbb{M}_{n}$. Given $i,j$ set
$m^{*}\Lambda(e_{ij}Q^{-1})=\sum_{t}\Lambda(a_{t})\otimes\Lambda(b_{t})$, so
for $x,y\in B$,

$\displaystyle\sum_{t}{\operatorname{Tr}}(Qx^{*}a_{t}){\operatorname{Tr}}(Qy^{*}b_{t})$
$\displaystyle=\sum_{t}(\Lambda(x)|\Lambda(a_{t}))(\Lambda(y)|\Lambda(b_{t}))=(\Lambda(x)\otimes\Lambda(y)|m^{*}\Lambda(e_{ij}Q^{-1}))$
$\displaystyle={\operatorname{Tr}}(Q(xy)^{*}e_{ij}Q^{-1})={\operatorname{Tr}}(y^{*}x^{*}e_{ij})=(y^{*}x^{*})_{ji}=\sum_{k}(y^{*})_{jk}(x^{*})_{ki}$
$\displaystyle=\sum_{k}{\operatorname{Tr}}(e_{kj}y^{*}){\operatorname{Tr}}(e_{ik}x^{*})=\sum_{k}{\operatorname{Tr}}(Qx^{*}e_{ik}Q^{-1}){\operatorname{Tr}}(Qy^{*}e_{kj}Q^{-1}).$

As $x,y$ are arbitrary, this shows that

$m^{*}\Lambda(e_{ij}Q^{-1})=\sum_{k}\Lambda(e_{ik}Q^{-1})\otimes\Lambda(e_{kj}Q^{-1}).$

Thus also

$mm^{*}\Lambda(e_{ij}Q^{-1})=\sum_{k}\Lambda(e_{ik}Q^{-1}e_{kj}Q^{-1})=\sum_{k}\Lambda(Q^{-1}_{kk}e_{ij}Q^{-1})\quad\text{and
so}\quad mm^{*}=\delta^{2}1$

where $\delta^{2}={\operatorname{Tr}}(Q^{-1})$. As $Q^{-1}$ is positive,
$\delta^{2}$ is indeed positive. In the general case, $B$ is a direct sum of
matrix factors, and $Q=(Q_{k})$ say. Then $mm^{*}$ respects the direct sum
decomposition, and on each matrix factor is the identity times
${\operatorname{Tr}}(Q_{k}^{-1})$, say
$mm^{*}=({\operatorname{Tr}}(Q_{k}^{-1})1_{k})$.

As the following appears extensively in the existing literature, we make a
formal definition.

######  Definition 2.9.

When $mm^{*}=\delta^{2}1$, that is,
${\operatorname{Tr}}(Q_{k}^{-1})=\delta^{2}$ for each $k$, we say that $\psi$
is a _$\delta$ -form_.

We now define the empty graph.

######  Definition 2.10.

The _empty quantum graph_ is given by the adjacency matrix
$A=({\operatorname{Tr}}(Q_{k}^{-1})^{-1}1_{k})=(mm^{*})^{-1}$.

A moment’s thought reveals that if $A\in\mathcal{B}(L^{2}(B))$ respects the
direct sum decomposition of $L^{2}(B)$ which is induced by that of $B$, then
all of the axioms also respect the decomposition: that is, it suffices to
check the axioms on each matrix block. With this observation, a direct
calculation shows that the empty quantum graph adjacency matrix satisfies all
the axioms (1) through (5); in this calculation it is helpful to show that
$(1\otimes\eta^{*}m)(m^{*}\eta\otimes 1)=1$.

The notion of a quantum adjacency matrix was first considered in [31], in the
special case that $\psi$ is a trace and a $\delta$-form. The paper [31] makes
extensive use of graphical methods, and the axioms are stated purely in terms
of graphical methods. These methods seem to make extensive use of the trace
condition, but [30, Section 2] shows how these ideas can be adapted to the
non-tracial (but still $\delta$-form) setting. When $\psi$ is merely a
$\delta$-form, the definition we have given is equivalent to that given in [7,
Definition 3.4], except that we have chosen to normalise $A$ differently.
Again when $\psi$ is a $\delta$-form, [8, Definition 3.3] just considers axiom
(1) (and without the requirement that $A$ be self-adjoint). A different
starting point is chosen in [25, Definition 1.8] which corresponds to the
projections $e\in B\otimes B^{\operatorname{op}}$ which we consider in Section
5 below, again when $\psi$ is a tracial $\delta$-form, and again graphical
methods are used to make links to adjacency matrices.

##  3 Adjacency super-operators

We slightly abuse notation and in this section consider $m$ and $m^{*}$ as
maps $B\otimes B\rightarrow B$ and $B\rightarrow B\otimes B$ respectively.

######  Definition 3.1.

A _quantum adjacency super-operator_ is $A:B\rightarrow B$ a completely
positive map with $m(A\otimes A)m^{*}=A$.

This terminology is our own; we chose “super-operator” from its use in Quantum
Information Theory to refer to a linear map between $C^{*}$-algebras rather
than Hilbert spaces, as is the case here. Notice that the functional $\psi$
enters the definition in the formation of the map $m^{*}$.

We have stated this definition with the minimal axiom, $m(A\otimes A)m^{*}=A$,
corresponding to axiom (1) in Definition 2.4. In the obvious way, any other
axiom can be added.

This definition is make in [12, Definition 1.8], in the case when
$B=\mathbb{M}_{n}$ is a single matrix block and $\psi$ is a trace. In [12,
Section 1.2] it is briefly argued that for this $B$, this definition agrees
with the other notions of a quantum graph. At the level of generality stated
here, the definition is considered in [30, Proposition 2.23]. We shall give a
different approach to [30, Proposition 2.23] below, using ideas much closer to
the approach of [12]. Indeed, we shall see that the super-operator $A$ in
Definition 3.1 agrees with the operator $A$ in Definition 2.4, if we identify
$B$ and $L^{2}(B)$ (see Theorem 5.37 for a precise statement).

In the non-tracial situation, [9, Section 2] makes further study of the
quantum adjacency super-operators. We make links between this and our approach
in Remark 5.38.

##  4 Operator bimodules

The following definitions make perfect sense in the infinite-dimensional
setting, and so we shall briefly work in more generality. We follow [26,
Section 1] and [45].

######  Definition 4.1.

Let $M\subseteq\mathcal{B}(H)$ be a von Neumann algebra. A _quantum relation_
on $M$ is a $W^{*}$-bimodule over $M^{\prime}$, that is, a weak∗-closed
subspace $S\subseteq\mathcal{B}(H)$ with $M^{\prime}SM^{\prime}=\\{xsy:x,y\in
M^{\prime},s\in S\\}\subseteq S$.

We also call such an $S$ an _operator bimodule_. As stated, this definition
depends on $M\subseteq\mathcal{B}(H)$ and not just on $M$. However, the
definition is essentially independent of the choice of $H$, compare [45,
Theorem 2.7]. This will also follow from Theorem 5.12.

######  Definition 4.2.

A quantum relation $S$ on $M\subseteq\mathcal{B}(H)$ is _reflexive_ if
$M^{\prime}\subseteq S$, and is _symmetric_ if $S^{*}=\\{x^{*}:x\in S\\}$
equals $S$.

Notice that as $M^{\prime}SM^{\prime}\subseteq S$, we have that $S$ is
reflexive if and only if $1\in S$. For the following, recall that an _operator
system_ is a unital, self-adjoint subspace of $\mathcal{B}(H)$, for some
Hilbert space $H$.

######  Definition 4.3.

A _quantum graph_ on $M\subseteq\mathcal{B}(H)$ is a reflexive, symmetric
quantum relation on $M$. That is, a weak∗-closed operator system which is an
operator bimodule over $M^{\prime}$.

Specialising to our case, we obtain the following definition.

######  Definition 4.4.

Given a finite-dimensional $C^{*}$-algebra $B$, a _quantum graph_ on $B$ is
given by an embedding $B\subseteq\mathcal{B}(H)$ for some finite-dimensional
$H$, and a subspace $S\subseteq\mathcal{B}(H)$ which is an operator bimodule
over $B^{\prime}$. We define further that $S$ is:

  1. 1.

_undirected_ when $S$ is self-adjoint;

  2. 2.

_has all loops_ when $1\in S$.

The terminology adopted here is our own. There is no dependence on our
functional $\psi$. Notice that it is rather easy to give examples: for
example, when $B=\mathbb{M}_{n}$ we can choose $H=\mathbb{C}^{n}$, and so
$B^{\prime}=\mathbb{C}$. Hence a quantum graph is simply a (self-adjoint,
unital) subspace of $\mathbb{M}_{n}$. There are two extremal choices if we
assume undirected and that we have all loops. We shall see later in Section
6.1 that this definition agrees with Definitions 2.8 and 2.10.

######  Definition 4.5.

Given $B$ and any $B\subseteq\mathcal{B}(H)$, the _empty quantum graph_ is
$S=B^{\prime}$. The _complete quantum graph_ is $S=\mathcal{B}(H)$.

In the classical situation, when $B=C(V)$ for a finite set $V$, we choose
$H=\ell^{2}(V)$ so $B$ has basis $(\delta_{v})$ with
$\delta_{v}\delta_{u}=\delta_{v,u}\delta_{v}$, and $H$ has orthonormal basis
$(e_{v})$ with $\delta_{v}(e_{u})=\delta_{v,u}e_{u}$. We hence identify
$\mathcal{B}(H)$ with matrices over $V$, and we see that $B^{\prime}=B$. Thus,
given an operator bimodule $S$ over $B^{\prime}$, if $x=(x_{u,v})\in S$ also
$\delta_{v}x\delta_{u}=x_{v,u}e_{v,u}\in S$. So $S$ is the span of the matrix
units it contains. We hence obtain a bijection between such $S$ and arbitrary
subsets $E\subseteq V\times V$.

######  Proposition 4.6.

With $B=C(V)$, quantum graphs over $B$ biject with graphs on vertex set $V$.
Furthermore, the quantum graph is undirected if and only if the graph is, and
the quantum graph has all loops if and only if the graph has a loop at every
vertex.

###### Proof.

We have already established a bijection between operator bimodules $S$ over
$B^{\prime}$ and subsets $E\subseteq V\times V$, that is, graphs on vertex set
$V$. Then $S$ is undirected when it is self-adjoint, which clearly corresponds
to the condition that $(u,v)\in E\implies(v,u)\in E$. Further, $S$ has all
loops when $1\in S$, so that $e_{v,v}\in S$ for each $v$, so that $(v,v)\in E$
for all $v$. ∎

This result also holds for infinite $V$, compare [45, Proposition 2.5].

This notion of a quantum graph arose both from Weaver’s work in [45], see also
[46], but also from Quantum Information Theory considerations, [20], also [19,
38].

##  5 Equivalence of definitions

We present a unified way to move between quantum adjacency matrices and the
operator bimodule notion of a quantum graph in the tracial case, and also show
how to obtain a correspondence in the general setting, by adjusting the
definition of an operator bimodule to take account of (some aspect of) the
functional $\psi$. Once we have these results, we obtain a further link with
the usual definition of an operator bimodule, Section 5.3. We then also look
at super-operators. We make some remarks about the existing literature.

We start by looking at a “toy version” of Tomita–Takesaki theory, which in our
finite-dimensional situation is easy: all of the following claimed properties
may be verified by direct computation. Recall the GNS construction
$(L^{2}(B),\Lambda)$, and our invertible positive operator $Q\in B$. We use
functional calculus to define $Q^{z}$ for all $z\in\mathbb{C}$. We define
$\sigma_{z}(a)=Q^{iz}aQ^{-iz}$ for $a\in B$. This is the (analytic extension
of the) modular automorphism group for $\psi$. Indeed, for $a,b,c\in B$,

$\displaystyle(\Lambda(a)|\Lambda(bc))$
$\displaystyle=\psi(a^{*}bc)={\operatorname{Tr}}(Qa^{*}bc)={\operatorname{Tr}}(cQa^{*}b)={\operatorname{Tr}}(QQ^{-1}cQa^{*}b)=(\Lambda(aQc^{*}Q^{-1})|\Lambda(b))$
$\displaystyle=(\Lambda(a\sigma_{-i}(c^{*}))|\Lambda(b)).$

Notice that $\sigma_{z}(a)^{*}=\sigma_{\overline{z}}(a^{*})$ and
$\sigma_{w}(\sigma_{z}(a))=\sigma_{w+z}(a)$ for each $a,z,w$. The _modular
conjugation_ is the anti-linear operator $J:L^{2}(B)\rightarrow L^{2}(B)$
defined by

$J\Lambda(a)=\Lambda(\sigma_{i/2}(a)^{*})=\Lambda(\sigma_{-i/2}(a^{*}))\qquad(a\in
B).$

One can easily check that $(J\xi|J\eta)=(\eta|\xi)$ for each $\xi,\eta\in
L^{2}(B)$ and that $J^{2}=1$, so that $J$ is an isometry. Furthermore,

$\displaystyle JaJ\Lambda(b)$
$\displaystyle=J\Lambda(a\sigma_{-i/2}(b^{*}))=\Lambda(\sigma_{-i/2}(\sigma_{-i/2}(b^{*})^{*}a^{*}))$
$\displaystyle=\Lambda(\sigma_{-i/2}(\sigma_{i/2}(b)a^{*}))=\Lambda(b\sigma_{-i/2}(a^{*}))\qquad(a,b\in
B)$ (1)

and so $JaJ$ is the operation of right multiplication by
$\sigma_{i/2}(a)^{*}$. In fact, $B^{\prime}=JBJ$ inside
$\mathcal{B}(L^{2}(B))$. Notice that

$J\theta_{\xi,\eta}J(\alpha)=J\big{(}(\xi|J(\alpha))\eta\big{)}=(J(\alpha)|\xi)J(\eta)=(J(\xi)|\alpha)J(\eta)=\theta_{J(\xi),J(\eta)}(\alpha)\qquad(\xi,\eta,\alpha\in
L^{2}(B)).$

Recall that the _modular operator_ is $\nabla\in\mathcal{B}(L^{2}(B))^{+}$
given by $\nabla\Lambda(a)=\Lambda(\sigma_{-i}(a))=\Lambda(QaQ^{-1})$. More
generally,
$\nabla^{z}\Lambda(a)=\Lambda(\sigma_{-iz}(a))=\Lambda(Q^{z}aQ^{-z})$.

We now define two bijections which will be central to our arguments.

######  Definition 5.1.

Let $B^{\operatorname{op}}$ be the _opposite_ algebra to $B$, so
$B^{\operatorname{op}}$ is the same vector space as $B$, and has the same
$*$-operation, but has the reversed product
$a\stackrel{{\scriptstyle{\operatorname{op}}}}{{\cdot}}b=ba$. When it is clear
from context, we shall write the multiplication in $B^{\operatorname{op}}$
just by juxtaposition.

As $L^{2}(B)$ is finite-dimensional, the linear span of the operators
$\theta_{\Lambda(a),\Lambda(b)}$, as $a,b\in B$ vary, gives all of
$\mathcal{B}(L^{2}(B))$. Given $s,t\in\mathbb{R}$, we define bijections from
$\mathcal{B}(L^{2}(B))$ to $B\otimes B^{\operatorname{op}}$ by

$\Psi_{t,s}=\Psi:\theta_{\Lambda(a),\Lambda(b)}\mapsto\sigma_{it}(a)^{*}\otimes\sigma_{is}(b),\qquad\Psi^{\prime}_{t,s}=\Psi^{\prime}:\theta_{\Lambda(a),\Lambda(b)}\mapsto\sigma_{it}(b)\otimes\sigma_{is}(a)^{*}.$

As $\theta_{\cdot,\cdot}$ is anti-linear in the first variable, and linear in
the second, we see that $\Psi$ is itself linear, and so extends by linearity
to a bijection $\mathcal{B}(L^{2}(B))\rightarrow B\otimes
B^{\operatorname{op}}$. Similarly, $\Psi^{\prime}$ is linear. We work with
$B\otimes B^{\operatorname{op}}$ to match the existing literature, which is
natural for $\Psi^{\prime}$, but we shall shortly see that when working with
$\Psi$, it might have been more natural to consider
$B^{\operatorname{op}}\otimes B$.

######  Definition 5.2.

We write $\sigma$ for the tensor swap map $B\otimes
B^{\operatorname{op}}\rightarrow B\otimes B^{\operatorname{op}};a\otimes
b\mapsto b\otimes a$, which is an anti-$*$-homomorphism.

We now see how the bijections $\Psi$ and $\Psi^{\prime}$ interact with the
constructions which arise in Definition 2.4. We let the modular automorphism
group act also on $B^{\operatorname{op}}$.

######  Proposition 5.3.

The bijection $\Psi=\Psi_{t,s}$ gives the following correspondences between
$A\in\mathcal{B}(L^{2}(B))$ and $e=\Psi(A)\in B\otimes B^{\operatorname{op}}$:

  1. 1.

$A^{*}$ corresponds to $(\sigma_{i(s-t)}\otimes\sigma_{i(s-t)})\sigma(e^{*})$;

  2. 2.

$(1\otimes\eta^{*}m)(1\otimes A\otimes 1)(m^{*}\eta\otimes 1)$ corresponds to
$(\sigma_{-i(s+t)}\otimes\sigma_{i(t+s-1)})\sigma(e)$.

  3. 3.

$(\eta^{*}m\otimes 1)(1\otimes A\otimes 1)(1\otimes m^{*}\eta)$ corresponds to
$(\sigma_{i(1-t-s)}\otimes\sigma_{i(s+t)})\sigma(e)$.

  4. 4.

we have that $\Psi(m(A_{1}\otimes A_{2})m^{*})=e_{2}e_{1}$ for
$e_{i}=\Psi(A_{i})$, $i=1,2$.

###### Proof.

By linearity, it suffices to consider the case when
$A=\theta_{\Lambda(a),\Lambda(b)}$ and so
$e=\sigma_{it}(a)^{*}\otimes\sigma_{is}(b)$. Then
$A^{*}=\theta_{\Lambda(b),\Lambda(a)}$ which corresponds to

$\sigma_{it}(b)^{*}\otimes\sigma_{is}(a)=\sigma_{i(s-t)}\sigma_{-is}(b^{*})\otimes\sigma_{i(s-t)}\sigma_{it}(a)=(\sigma_{i(s-t)}\otimes\sigma_{i(s-t)})\sigma(e^{*}),$

as claimed for (1).

Set $T=(1\otimes\eta^{*}m)(1\otimes A\otimes 1)(m^{*}\eta\otimes 1)$. Let
$m^{*}\eta=m^{*}\Lambda(1)=\sum_{i}\xi_{i}\otimes\eta_{i}$, so that
$(\Lambda(xy)|\Lambda(1))=\sum_{i}(\Lambda(x)|\xi_{i})(\Lambda(y)|\eta_{i})$
for each $x,y\in B$. Then for $x,y\in B$,

$\displaystyle(\Lambda(x)|T\Lambda(y))$
$\displaystyle=\sum_{i}\big{(}\Lambda(x)\otimes\Lambda(1)\big{|}(1\otimes
m)(1\otimes A\otimes 1)\xi_{i}\otimes\eta_{i}\otimes\Lambda(y)\big{)}$
$\displaystyle=\sum_{i}\big{(}\Lambda(x)\otimes\Lambda(1)\big{|}(1\otimes
m)\xi_{i}\otimes\Lambda(b)\otimes\Lambda(y)\big{)}(\Lambda(a)|\eta_{i})$
$\displaystyle=\sum_{i}(\Lambda(x)|\xi_{i})(\Lambda(1)|\Lambda(by))(\Lambda(a)|\eta_{i})$
$\displaystyle=(\Lambda(xa)|\Lambda(1))(\Lambda(1)|\Lambda(by))=(\Lambda(x)|\Lambda(\sigma_{-i}(a^{*})))(\Lambda(b^{*})|\Lambda(y)).$

Hence $T=\theta_{\Lambda(b^{*}),\Lambda(\sigma_{-i}(a^{*}))}$ and so

$\Psi(T)=\sigma_{it}(b^{*})^{*}\otimes\sigma_{is}(\sigma_{i}(a)^{*})=\sigma_{-it-
is}\sigma_{is}(b)\otimes\sigma_{it+is-i}\sigma_{-it}(a^{*})=(\sigma_{-i(t+s)}\otimes\sigma_{i(t+s-1)})\sigma(e),$

showing (2).

Similarly, if we now set $T=(\eta^{*}m\otimes 1)(1\otimes A\otimes 1)(1\otimes
m^{*}\eta)$, then for $x,y\in B$,

$\displaystyle(\Lambda(x)|T\Lambda(y))$
$\displaystyle=\sum_{i}\big{(}\Lambda(1)\otimes\Lambda(x)\big{|}(m\otimes
1)(1\otimes A\otimes 1)\Lambda(y)\otimes\xi_{i}\otimes\eta_{i}\big{)}$
$\displaystyle=\sum_{i}\big{(}\Lambda(1)\otimes\Lambda(x)\big{|}\Lambda(yb)\otimes\eta_{i}\big{)}(\Lambda(a)|\xi_{i})$
$\displaystyle=\sum_{i}(\Lambda(1)|\Lambda(yb))(\Lambda(x)|\eta_{i}\big{)}(\Lambda(a)|\xi_{i})$
$\displaystyle=(\Lambda(\sigma_{-i}(b^{*}))|\Lambda(y))(\Lambda(x)|\Lambda(a^{*})).$

Thus $T=\theta_{\Lambda(\sigma_{-i}(b^{*})),\Lambda(a^{*})}$ and so

$\Psi(T)=(\sigma_{it}\sigma_{-i}(b^{*}))^{*}\otimes\sigma_{is}(a^{*})=\sigma_{-it+i-is}\sigma_{is}(b)\otimes\sigma_{is+it}\sigma_{-it}(a^{*})=\sigma_{-i(t+s)+i}\otimes\sigma_{i(s+t)})\sigma(e),$

showing (3).

Given $A_{k}=\theta_{\Lambda(a_{k}),\Lambda(b_{k})}$, corresponding to
$e_{k}$, for $k=1,2$, we see that

$m(A_{1}\otimes
A_{2})m^{*}\Lambda(c)=\theta_{\Lambda(a_{1}a_{2}),\Lambda(b_{1}b_{2})}\Lambda(c)\qquad(c\in
B),$

and so $m(A_{1}\otimes A_{2})m^{*}$ corresponds to

$\sigma_{it}(a_{1}a_{2})^{*}\otimes\sigma_{is}(b_{1}b_{2})=\sigma_{it}(a_{2})^{*}\sigma_{it}(a_{1})^{*}\otimes\sigma_{is}(b_{2})\sigma_{is}(b_{1})=e_{2}e_{1},$

as we work in $B\otimes B^{\operatorname{op}}$. This shows (4). ∎

We can immediately use this result to show one equivalence.

######  Proposition 5.4.

Let $A\in\mathcal{B}(L^{2}(B))$ be self-adjoint. Then axioms (2) and (4) are
equivalent.

###### Proof.

Let $e=\Psi_{s,t}(A)$ (for some choice of $s,t$) so $A=A^{*}$ is equivalent to

$e=(\sigma_{i(s-t)}\otimes\sigma_{i(s-t)})\sigma(e^{*})\ \ \Leftrightarrow\ \
e^{*}=(\sigma_{i(t-s)}\otimes\sigma_{i(t-s)})\sigma(e).$ (2)

Then axiom (2) is equivalent to
$(\sigma_{-i(s+t)}\otimes\sigma_{i(t+s-1)})\sigma(e)=e$. Taking the adjoint
and applying the swap map $\sigma$ shows that this is equivalent to

$(\sigma_{-i(t+s-1)}\otimes\sigma_{i(s+t)})(e^{*})=\sigma(e^{*})=(\sigma_{-i(s-t)}\otimes\sigma_{-i(s-t)})(e),$

where we also used (2). This shows that

$(\sigma_{-i(t-1)}\otimes\sigma_{is})(e^{*})=(\sigma_{it}\otimes\sigma_{-is})(e)\
\ \Leftrightarrow\ \ e^{*}=(\sigma_{i(2t-1)}\otimes\sigma_{-2is})(e)$

but using (2) again shows that

$(\sigma_{i(2t-1)}\otimes\sigma_{-2is})(e)=(\sigma_{i(t-s)}\otimes\sigma_{i(t-s)})\sigma(e)\
\ \Leftrightarrow\ \ e=(\sigma_{i(1-t-s)}\otimes\sigma_{i(t+s)})\sigma(e),$

which shows that axiom (4) holds. The converse is similar. ∎

There is an analogous equivalence for $\Psi^{\prime}$.

######  Proposition 5.5.

The bijection $\Psi^{\prime}=\Psi^{\prime}_{t,s}$ gives the following
correspondences between $A\in\mathcal{B}(L^{2}(B))$ and $e=\Psi^{\prime}(A)\in
B\otimes B^{\operatorname{op}}$:

  1. 1.

$A^{*}$ corresponds to $(\sigma_{i(t-s)}\otimes\sigma_{i(t-s)})\sigma(e^{*})$;

  2. 2.

$(1\otimes\eta^{*}m)(1\otimes A\otimes 1)(m^{*}\eta\otimes 1)$ corresponds to
$(\sigma_{i(s+t-1)}\otimes\sigma_{-i(s+t)})\sigma(e)$.

  3. 3.

$(\eta^{*}m\otimes 1)(1\otimes A\otimes 1)(1\otimes m^{*}\eta)$ corresponds to
$(\sigma_{i(t+s)}\otimes\sigma_{i(1-s-t)})\sigma(e)$.

  4. 4.

we have that $\Psi^{\prime}(m(A_{1}\otimes A_{2})m^{*})=e_{1}e_{2}$ for
$e_{i}=\Psi^{\prime}(A_{i})$, $i=1,2$.

###### Proof.

This could be proved in an entirely analogous way. However, we also observe
that $\Psi^{\prime}_{t,s}=\sigma\circ\Psi_{s,t}$, from which these claims
follow immediately from Proposition 5.3. ∎

The way to prove the equivalence between Definitions 2.4 and 4.4 requires two
steps:

  * •

we use $\Psi_{t,s}$ or $\Psi^{\prime}_{t,s}$, for carefully chosen $t,s$, to
associate $A$ to an _idempotent_ (ideally, an _orthogonal projection_) $e$;

  * •

associate idempotents (or projections) $e$ to subspaces
$S\subseteq\mathcal{B}(H)$.

We will shortly deal with the second point is detail, before then returning to
the first point. However, the reader can already see that if $\psi$ is a
trace, then each $\sigma_{z}$ is trivial, and it is immediate that $A=A^{*}$
satisfying axioms (1) and (2) corresponds to $e\in B\otimes
B^{\operatorname{op}}$ with $\sigma(e^{*})=e,e^{2}=e$ and $e=\sigma(e)$, or
equivalently, $e=e^{*}=e^{2}=\sigma(e)$, this using any choice of $\Psi$ and
$\Psi^{\prime}$. In the non-tracial case, it is not possible to choose any
$s,t$ which gives such a direct bijection, though of course the benefit is
that we obtain an interesting, different, theory in this case.

######  Remark 5.6.

When $\psi$ is a tracial $\delta$-form, it was already shown in [31, Theorem
7.7] that a self-adjoint $A$ satisfying axioms (1) and (2) (which are exactly
the diagrams (47) in [31, Definition 5.1]) then we obtain an undirected
quantum graph, in the sense of Definition 4.4. The proof is graphical, and
indirect, as it passes by way of “projectors”, equivalently by [31, Remark
7.3] projections in $B^{\operatorname{op}}\otimes B$. Examining the proof of
[31, Theorem 7.7] shows that $A=\theta_{\Lambda(a),\Lambda(b)}$ corresponds to
$b\otimes a^{*}$, that is, $\Psi^{\prime}_{0,0}(A)$. (Of course, $\psi$ is
tracial here, so $\Psi^{\prime}_{0,0}=\Psi^{\prime}_{t,s}$ for all $t,s$).
Using $B^{\operatorname{op}}\otimes B$ instead of $B\otimes
B^{\operatorname{op}}$ is just a convention.

Elsewhere in the literature, compare for example [12, Lemma 1.6] or [25,
Definition 1.12], the relation between $A\in\mathcal{B}(L^{2}(B))$ and $e\in
B\otimes B^{\operatorname{op}}$ is given by $e=(A\otimes 1)m^{*}\eta$, and
this again is $\Psi^{\prime}_{0,0}$. To make sense of the expression
$e=(A\otimes 1)m^{*}\eta$, we identify, as linear spaces, $B\otimes
B^{\operatorname{op}}$ with $B\otimes B$, and in turn, with $L^{2}(B)\otimes
L^{2}(B)$.

We shall shortly see that $\Psi$ and $\Psi^{\prime}$ are essentially
interchangeable, but it is helpful to have the choice between them for
different arguments. ∎

Finally, we consider the real condition, Definition 2.5.

######  Lemma 5.7.

Given $A\in\mathcal{B}(L^{2}(B))$, let $A_{0}:B\rightarrow B$ be the
associated linear map. Let $A^{r}\in\mathcal{B}(L^{2}(B))$ correspond to the
linear map $B\rightarrow B;a\mapsto A_{0}(a^{*})^{*}$. For $e=\Psi_{t,s}(A)$,
we have that $\Psi_{t,s}(A^{r})=(\sigma_{i-2it}\otimes\sigma_{2is})(e^{*})$.
For $e=\Psi^{\prime}_{t,s}(A)$, we have that
$\Psi^{\prime}_{t,s}(A^{r})=(\sigma_{2it}\otimes\sigma_{i-2is})(e^{*})$.

###### Proof.

We wish to consider the anti-linear bijection $s:L^{2}(B)\rightarrow
L^{2}(B);\Lambda(a)\mapsto\Lambda(a^{*})$. A calculation shows that
$s=\nabla^{-1/2}J=J\nabla^{1/2}$. A further calculation then establishes that
when $A=\theta_{\Lambda(a),\Lambda(b)}$,

$\displaystyle A^{r}$
$\displaystyle=sAs=\nabla^{-1/2}J\theta_{\Lambda(a),\Lambda(b)}J\nabla^{1/2}=\nabla^{-1/2}\theta_{J\Lambda(a),J\Lambda(b)}\nabla^{1/2}=\theta_{\nabla^{1/2}J\Lambda(a),\nabla^{-1/2}J\Lambda(b)}$
$\displaystyle=\theta_{\nabla\Lambda(a^{*}),\Lambda(b^{*})}=\theta_{\Lambda(\sigma_{-i}(a^{*})),\Lambda(b^{*})}.$

Thus

$\displaystyle\Psi_{t,s}(A^{r})$
$\displaystyle=\sigma_{it}(\sigma_{-i}(a^{*}))^{*}\otimes\sigma_{is}(b^{*})=\sigma_{i-it}(a)\otimes\sigma_{is}(b^{*})$
$\displaystyle=(\sigma_{i-2it}\otimes\sigma_{2is})(\sigma_{it}(a)\otimes\sigma_{-is}(b^{*}))=(\sigma_{i-2it}\otimes\sigma_{2is})(e^{*}).$

By linearity, this relation holds for all $A$, as claimed.

As $\Psi^{\prime}_{t,s}=\sigma\Psi_{s,t}$, with $\sigma$ the tensor swap map,
we see that
$\Psi^{\prime}_{t,s}(A^{r})=\sigma\Psi_{s,t}(A_{r})=\sigma(\sigma_{i-2is}\otimes\sigma_{2it})\Psi_{s,t}(A)^{*}=(\sigma_{2it}\otimes\sigma_{i-2is})\sigma\Psi_{s,t}(A)^{*}=(\sigma_{2it}\otimes\sigma_{i-2is})\Psi^{\prime}_{t,s}(A)^{*}$,
as claimed. ∎

###  5.1 From projections to subspaces

In this section, $H$ will be a finite-dimensional Hilbert space with
$B\subseteq\mathcal{B}(H)$ (which is equivalent, of course, to there being an
injective $*$-homomorphism $B\rightarrow\mathcal{B}(H)$ which allows us to
identify $B$ with its image in $\mathcal{B}(H)$). Then $B^{\operatorname{op}}$
is naturally a subalgebra of $\mathcal{B}(\overline{H})$ where we identify
$a\in B^{\operatorname{op}}$ with $a^{\top}\in\mathcal{B}(\overline{H})$.
Consequently, $B\otimes
B^{\operatorname{op}}\subseteq\mathcal{B}(H\otimes\overline{H})$. For the
following, notice that $(B^{\prime})^{\operatorname{op}}$ is also naturally a
subalgebra of $\mathcal{B}(\overline{H})$, where we identify
$x\in(B^{\prime})^{\operatorname{op}}\subseteq\mathcal{B}(H)$ with $x^{\top}$.
A simple calculation shows that then
$(B^{\prime})^{\operatorname{op}}=(B^{\operatorname{op}})^{\prime}$ as
subalgebras of $\mathcal{B}(\overline{H})$. Hence
$B^{\prime}\otimes(B^{\prime})^{\operatorname{op}}\subseteq\mathcal{B}(H\otimes\overline{H})$.

######  Lemma 5.8.

The commutant of $B\otimes
B^{\operatorname{op}}\subseteq\mathcal{B}(H\otimes\overline{H})$ is
$B^{\prime}\otimes(B^{\prime})^{\operatorname{op}}$.

###### Proof.

The result follows from Tomita’s theorem, [39, Theorem 5.9, Chapter IV]:
$(B\otimes
B^{\operatorname{op}})^{\prime}=B^{\prime}\otimes(B^{\operatorname{op}})^{\prime}\cong
B^{\prime}\otimes(B^{\prime})^{\operatorname{op}}$. ∎

As $H$ is finite-dimensional, every operator on $\mathcal{B}(H)$ is in the
linear span of the rank-one operators, and so there is a linear isomorphism

$\mathcal{B}(H)\cong
H\otimes\overline{H};\quad\theta_{\xi,\eta}\mapsto\eta\otimes\overline{\xi}\qquad(\xi,\eta\in
H).$

Indeed, this is simply the GNS map for the functional ${\operatorname{Tr}}$ on
$\mathcal{B}(H)$, or equivalently, comes from identifying $\mathcal{B}(H)$
with the Hilbert-Schmidt operators on $H$.

######  Remark 5.9.

We make links with the existing literature, and in particular [45, Proposition
2.23] where a bijection between quantum relations $S\subseteq\mathcal{B}(H)$
and left ideals in $B\otimes B^{\operatorname{op}}$ is established (when $B$
is finite-dimensional). A key technical tool for Weaver is the action $\Phi$
of $B\otimes B^{\operatorname{op}}$ on $\mathcal{B}(H)$ given by

$\Phi_{a\otimes b}(x)=abx\qquad(a\otimes b\in B\otimes
B^{\operatorname{op}},x\in\mathcal{B}(H)).$

Once we identify $\mathcal{B}(H)$ with $H\otimes\overline{H}$, we find that

$(a\otimes b)\cdot(\eta\otimes\overline{\xi})\cong\Phi_{a\otimes
b}(\theta_{\xi,\eta})=a\theta_{\xi,\eta}b=\theta_{b^{*}(\xi),a(\eta)}\cong
a(\eta)\otimes b^{\top}(\overline{\xi}).$

That is, we obtain the natural action of $B\otimes B^{\operatorname{op}}$ on
$H\otimes\overline{H}$.

Elsewhere in the literature, links with completely bounded maps (and
implicitly with the (extended) Haagerup tensor product) are made, compare
[24], or [23, Section 2.4]. In contrast, our perspective in this section is
that it is Hilbert Space techniques which are vitally important, for which $H$
being finite-dimensional seems essential. ∎

An elementary result from von Neumann algebra theory is that invariant
subspaces for a von Neumann algebra $M$ biject with projections in the
commutant $M^{\prime}$. We would like to have a version of this result for
(possibly) non-self-adjoint idempotents.

######  Lemma 5.10.

Let $M\subseteq\mathcal{B}(H)$ be a von Neumann algebra. Let
$e\in\mathcal{B}(H)$ be an idempotent, $e^{2}=e$, and let
$H_{0}={\operatorname{Im}}(e),H_{1}=\ker(e)$. Then $H=H_{0}\oplus H_{1}$ (not
necessarily orthogonal). Furthermore, $e\in M$ if and only if $H_{0}$ and
$H_{1}$ are invariant subspaces for $M^{\prime}$.

###### Proof.

Any $\xi\in H$ is written as $\xi=e(\xi)+(\xi-e(\xi))\in H_{0}+H_{1}$, and if
$\xi\in H_{0}\cap H_{1}$ then $\xi=e(\xi)=0$. Thus $H=H_{0}\oplus H_{1}$ as a
direct sum of Banach spaces (that is, a not necessarily orthogonal direct
sum). If $e\in M$ then given $n\in M^{\prime}$, for $\xi\in H_{0}$ we see that
$n(\xi)=ne(\xi)=en(\xi)$ so $n(\xi)\in H_{0}$, while for $\xi\in H_{1}$ we
have that $en(\xi)=ne(\xi)=0$ so $n(\xi)\in H_{1}$. Conversely, if
$H_{0},H_{1}$ are $M^{\prime}$-invariant, then to show that $e\in M$, by the
bicommutant theorem, it suffices to show that $ne=en$ for each $n\in
M^{\prime}$. Given $\xi\in H$ write $\xi=\xi_{0}+\xi_{1}\in H_{0}\oplus
H_{1}$, so $ne(\xi)=n(\xi_{0})=en(\xi_{0})$ as $n(\xi_{0})\in H_{0}$, while
$n(\xi)=n(\xi_{0})+n(\xi_{1})\in H_{0}\oplus H_{1}$ so $en(\xi)=en(\xi_{0})$
because $en(\xi_{1})=0$. Hence $ne(\xi)=en(\xi)$ as required. ∎

We now collect some definitions which will allow us to talk about properties
of subspaces of $H\otimes\overline{H}$ which correspond to the properties of
idempotents $e\in B\otimes B^{\operatorname{op}}$ which occurred in the
previous section.

######  Definition 5.11.

Given $H=H_{0}\oplus H_{1}$ not necessarily orthogonal, we shall say that the
idempotent $e\in\mathcal{B}(H)$ with
${\operatorname{Im}}(e)=H_{0},\ker(e)=H_{1}$ is the _projection onto $H_{0}$
along $H_{1}$_.

We define an anti-linear map

$J_{0}:H\otimes\overline{H}\rightarrow
H\otimes\overline{H};\quad\eta\otimes\overline{\xi}\mapsto\xi\otimes\overline{\eta}.$

This corresponds to the adjoint on $\mathcal{B}(H)$, and is in fact the
modular conjugation with respect to the functional ${\operatorname{Tr}}$ on
$\mathcal{B}(H)$.

Let $\hat{1}\in H\otimes\overline{H}$ corresponds to the identity operator
$1\in\mathcal{B}(H)$.

Let $(e_{i})$ be some orthogonal basis for $H$. As
$1=\sum_{i}\theta_{e_{i},e_{i}}$ we find that

$\hat{1}=\sum_{i}e_{i}\otimes\overline{e_{i}},$

and that this is independent of the choice of $(e_{i})$.

######  Theorem 5.12.

There is a bijection between idempotents $e\in B\otimes B^{\operatorname{op}}$
and decompositions $H\otimes\overline{H}=V\oplus W$ where $V$ and $W$ are
$B^{\prime}\otimes(B^{\prime})^{\operatorname{op}}$-invariant subspaces of
$H\otimes\overline{H}$, given by $e$ being the projection onto $V$ along $W$.
Furthermore:

  1. 1.

$e=\sigma(e^{*})$ if and only if $J_{0}(V)=V$ and $J_{0}(W)=W$;

  2. 2.

$m(e)=1$ if and only if $\hat{1}\in V$. (Here $m:B\otimes
B^{\operatorname{op}}\rightarrow B$ is the multiplication map).

###### Proof.

The bijection follows from Lemma 5.10 combined with Lemma 5.8.

When $e=a\otimes b$ acting as $a\otimes b^{\top}$ on $H\otimes\overline{H}$,
so that $e(\eta\otimes\overline{\xi})=a(\eta)\otimes\overline{b^{*}(\xi)}$, we
see that $\sigma(e^{*})=b^{*}\otimes a^{*}$, so that

$\sigma(e^{*})(\eta\otimes\overline{\xi})=b^{*}(\eta)\otimes\overline{a(\xi)}=J_{0}eJ_{0}(\eta\otimes\overline{\xi}).$

Thus in general $\sigma(e^{*})=J_{0}eJ_{0}$, and so $e=\sigma(e^{*})$ if and
only if $eJ_{0}=J_{0}e$. If this holds, then
$J_{0}(V)=J_{0}e(H\otimes\overline{H})=eJ_{0}(H\otimes\overline{H})=V$, while
also if $\xi\in W$ then $eJ_{0}(\xi)=J_{0}e(\xi)=0$ so
$J_{0}(\xi)\in\ker(e)=W$ and hence $W=J_{0}(W)$ (as $J_{0}^{2}=1$).

Conversely, if $J_{0}(V)=V$ and $J_{0}(W)=W$, then given $\xi\in
H\otimes\overline{H}$ take the decomposition $\xi=v+w\in V\oplus W$. By
assumption, $J_{0}(\xi)$ has decomposition $J_{0}(v)+J_{0}(w)\in V\oplus W$.
Hence $e(\xi)=v$ and $eJ_{0}(\xi)=J_{0}(v)$, and so

$\sigma(e^{*})(\xi)=J_{0}eJ_{0}(\xi)=J_{0}J_{0}(v)=v=e(\xi).$

Hence $\sigma(e^{*})=e$, and we have shown (1).

Again, when $e=a\otimes b$, and with $(e_{i})$ some orthonormal basis of $H$,

$\displaystyle e(\hat{1})$
$\displaystyle=\sum_{i}a(e_{i})\otimes\overline{b^{*}(e_{i})}=\sum_{i,j,k}(e_{j}|a(e_{i}))e_{j}\otimes(\overline{e_{k}}|\overline{b^{*}(e_{i})})\overline{e_{k}}=\sum_{i,j,k}(a^{*}(e_{j})|e_{i})(e_{i}|b(e_{k}))e_{j}\otimes\overline{e_{k}}$
$\displaystyle=\sum_{j,k}(a^{*}(e_{j})|b(e_{k}))e_{j}\otimes\overline{e_{k}}=\sum_{j,k}(e_{j}|m(e)(e_{k}))e_{j}\otimes\overline{e_{k}}.$

Thus, if $m(e)=1$ then $e(\hat{1})=\hat{1}$. Conversely, if
$e(\hat{1})=\hat{1}$ then $(e_{j}|m(e)(e_{k}))=\delta_{j,k}$ for each $j,k$ so
that $m(e)=1$. So (2) holds. ∎

In the orthogonal case, we recover, in particular, [45, Proposition 2.23].

######  Corollary 5.13.

There is a bijection between $B^{\prime}$-operator bimodules
$S\subseteq\mathcal{B}(H)$ and projections $e\in B\otimes
B^{\operatorname{op}}$, under which $S$ is self-adjoint if and only if
$e=\sigma(e)$, and $1\in S$ if and only if $m(e)=1$.

###### Proof.

Under the bijection between $\mathcal{B}(H)$ and $H\otimes\overline{H}$, we
see that $B^{\prime}$-operator bimodules correspond to
$B^{\prime}\otimes(B^{\prime})^{\operatorname{op}}$-invariant subspaces of
$H\otimes\overline{H}$. These biject with projections $e\in B\otimes
B^{\operatorname{op}}$, given by taking $V=S$ and $W=S^{\perp}$ in the
previous theorem.

If $J_{0}(S)=S$ then for $\xi\in S$ and $\eta\in S^{\perp}$, we see that
$(\xi|J_{0}(\eta))=(\eta|J_{0}(\xi))=0$ as $J_{0}(\xi)\in S$. Hence also
$J_{0}(S^{\perp})\subseteq S^{\perp}$ (so we have equality as $J_{0}^{2}=1$).
As $e=e^{*}$, we conclude that $e=\sigma(e)$ is equivalent to $J_{0}(S)=S$,
which viewing $S$ as a subspace of $\mathcal{B}(H)$ corresponds to $S$ being
self-adjoint. Similarly, as $\hat{1}$ corresponds to the unit
$1\in\mathcal{B}(H)$, we see that $S$ is unital if and only if $m(e)=1$. ∎

###  5.2 From adjacency matrices to subspaces

We now put together the results of the previous two sections to establish
bijections between our two main definitions. We first consider just looking at
the first axiom (1), $m(A\otimes A)m^{*}=A$.

######  Theorem 5.14.

For any $t\in\mathbb{R}$, consider $\Psi=\Psi_{t,t}$ or
$\Psi^{\prime}=\Psi^{\prime}_{t,t}$. These maps
$\mathcal{B}(L^{2}(B))\rightarrow B\otimes B^{\operatorname{op}}$ give a
bijection between:

  1. 1.

self-adjoint $A\in\mathcal{B}(L^{2}(B))$ satisfying axiom (1);

  2. 2.

idempotents $e\in B\otimes B^{\operatorname{op}}$ with $e=\sigma(e^{*})$;

  3. 3.

given $B\subseteq\mathcal{B}(H)$, decompositions $H\otimes\overline{H}=V\oplus
W$ with $V,W$ being
$B^{\prime}\otimes(B^{\prime})^{\operatorname{op}}$-invariant, with
$J_{0}(V)=V,J_{0}(W)=W$;

  4. 4.

given $B\subseteq\mathcal{B}(H)$, decompositions $\mathcal{B}(H)=V\oplus W$
with $V,W$ self-adjoint, $B^{\prime}$-bimodules.

###### Proof.

Proposition 5.3 shows that $\Psi_{t,t}$ gives a bijection between $A$ and $e$
such that $A^{*}$ corresponds to $\sigma(e^{*})$ and $m(A\otimes A)m^{*}$
corresponding to $e^{2}$. Similarly, Proposition 5.5 does the same for
$\Psi^{\prime}_{t,t}$. Hence (1) and (2) are equivalent. We now use Theorem
5.12 to see that such $e$ correspond to projections of $H\otimes\overline{H}$
onto $V$ along $W$ where $V,W$ are $J_{0}$-invariant, and
$B^{\prime}\otimes(B^{\prime})^{\operatorname{op}}$-invariant, which gives
equivalence (3). This is exactly condition (4) using $H\otimes\overline{H}$ in
place of $\mathcal{B}(H)$. ∎

######  Remark 5.15.

In particular, this applies to the bijection $\Psi^{\prime}_{0,0}$ as used
elsewhere in the literature (in the tracial case), see Remark 5.6. However, we
notice that the condition we obtain is that $e=e^{2}=\sigma(e^{*})$, and
certainly not that $e$ need be self-adjoint. Compare also Remark 5.20 below.

In light of just the use of axiom (1) in [8], we note that the same proof
shows that we can remove that condition that $A=A^{*}$ if we remove the
conditions $e=\sigma(e^{*})$, that $V,W$ are $J_{0}$-invariant, and the
condition that $V,W$ be self-adjoint. ∎

To make further progress, it really does seem necessary to take account of the
modular automorphism group. First a lemma.

######  Lemma 5.16.

Let $H$ be a Hilbert space, $E\in\mathcal{B}(H)$ the (orthogonal) projection
onto a subspace $V\subseteq H$, and let $T\in\mathcal{B}(H)$ be invertible.
The following are equivalent:

  1. 1.

$T^{-1}ET=E$;

  2. 2.

$T(V)=V$ and $T(V^{\perp})=V^{\perp}$;

  3. 3.

$T(V)\subseteq V$ and $T(V^{\perp})\subseteq V^{\perp}$.

When $T=T^{*}$, these conditions are equivalent to $T(V)\subseteq V$.

###### Proof.

Suppose $T^{-1}ET=E$. For $\xi\in V$, we have $\xi=E(\xi)=T^{-1}ET(\xi)$ so
$T(\xi)=ET(\xi)$, so $T(\xi)\in V$. For $\xi\in V^{\perp}$, we have
$0=E(\xi)=T^{-1}ET(\xi)$ so $ET(\xi)=0$ so $T(\xi)\in V^{\perp}$. As also
$TET^{-1}=E$ we get that also $T^{-1}(V)\subseteq V$ and
$T^{-1}(V^{\perp})\subseteq V^{\perp}$. Thus $T(V)=V$ and
$T(V^{\perp})=V^{\perp}$.

If $T(V)\subseteq V$ and $T(V^{\perp})\subseteq V^{\perp}$, then for $\xi\in
H$ we have that $\xi=E(\xi)+(1-E)(\xi)\in V\oplus V^{\perp}$ and so also
$T(\xi)=TE(\xi)+T(1-E)(\xi)\in V\oplus V^{\perp}$. Hence $ET(\xi)=TE(\xi)$,
for each $\xi$, and so $T^{-1}ET=E$.

When $T$ is self-adjoint, if $T(V)\subseteq V$ then for $\xi\in
V^{\perp},\eta\in V$, we see that $(T(\xi)|\eta)=(\xi|T(\eta))=0$ because
$T(\eta)\in V$. Thus $T(V^{\perp})\subseteq V^{\perp}$. ∎

We now come to the first of two key theorems which establish a bijection
between Definition 2.4 and a generalisation of Definition 4.4. Recall that

$\Psi_{0,1/2}:\theta_{\Lambda(a),\Lambda(b)}\mapsto
a^{*}\otimes\sigma_{i/2}(b),\qquad\Psi^{\prime}_{1/2,0}:\theta_{\Lambda(a),\Lambda(b)}\mapsto\sigma_{i/2}(b)\otimes
a^{*}.$

Recall the meaning of $J_{0}$ from Definition 5.11.

######  Theorem 5.17.

Consider the isomorphisms $\Psi_{0,1/2}$ or $\Psi^{\prime}_{1/2,0}$. These
give a bijection between:

  1. 1.

self-adjoint $A\in\mathcal{B}(L^{2}(B))$ satisfying axioms (1) and (2) (or,
equivalently, axioms (1) and (4));

  2. 2.

projections $e\in B\otimes B^{\operatorname{op}}$ with $e=\sigma(e)$ and such
that $e=(\sigma_{z}\otimes\sigma_{z})(e)$ for all $z\in\mathbb{C}$;

  3. 3.

given $B\subseteq\mathcal{B}(H)$, subspaces $V\subseteq H\otimes\overline{H}$
which are $B^{\prime}\otimes(B^{\prime})^{\operatorname{op}}$-invariant, and
which satisfy $J_{0}(V)=V$ and $(Q\otimes(Q^{-1})^{\top})(V)\subseteq V$
(equivalently, equal to $V$);

  4. 4.

given $B\subseteq\mathcal{B}(H)$, self-adjoint $B^{\prime}$-bimodules
$S\subseteq\mathcal{B}(H)$ with $QSQ^{-1}\subseteq S$ (equivalently, equals
$S$).

###### Proof.

By Proposition 5.3, $\Psi_{0,1/2}$ gives a bijection between self-adjoint $A$
satisfying axiom (2), and $e$ with
$e=(\sigma_{i/2}\otimes\sigma_{i/2})\sigma(e^{*})$ and
$e=(\sigma_{-i/2}\otimes\sigma_{-i/2})\sigma(e)$. For
$\Psi^{\prime}=\Psi^{\prime}_{1/2,0}$ we use Proposition 5.5, and obtain
exactly the same results. For such an $e$, we have that
$e^{*}=(\sigma_{-i/2}\otimes\sigma_{-i/2})\sigma(e)^{*}=(\sigma_{i/2}\otimes\sigma_{i/2})\sigma(e^{*})=e$
so $e$ is self-adjoint. Then that
$e^{*}=(\sigma_{i/2}\otimes\sigma_{i/2})\sigma(e^{*})$ and that $e=e^{*}$
means that

$(\sigma_{i/2}\otimes\sigma_{i/2})\sigma(e)=e=(\sigma_{-i/2}\otimes\sigma_{-i/2})\sigma(e)\implies\sigma(e)=(\sigma_{i}\otimes\sigma_{i})\sigma(e),$

or equivalently, $e=(\sigma_{i}\otimes\sigma_{i})(e)$. As
$\sigma_{i}(a)=Q^{-1}aQ$ for $a\in B$, and using the opposite multiplication
in $B^{\operatorname{op}}$, we see that

$e=(\sigma_{i}\otimes\sigma_{i})(e)=(Q^{-1}\otimes Q)e(Q\otimes Q^{-1}),$

and so $e$ commutes with $Q\otimes Q^{-1}$. By using Functional Calculus, $e$
also commutes with any power, and we conclude that
$(\sigma_{z}\otimes\sigma_{z})(e)=e$ for any $z\in\mathbb{C}$. In particular,
then also $e=\sigma(e)$.

Conversely, if $(\sigma_{z}\otimes\sigma_{z})(e)=e$ for all $z$, then by
Proposition 5.3, respectively Proposition 5.5, that $e=e^{*}=\sigma(e)$
implies that $A=A^{*}$, and that axiom (2) holds. Proposition 5.4 tells us
that axioms (2) and (4) are equivalent, as $A$ is self-adjoint. Finally, we
use that axiom (1) is equivalent to $e^{2}=e$. Thus (1) and (2) are
equivalent.

Suppose that (2) holds. By Corollary 5.13, we know that $e$ corresponds to a
self-adjoint $B^{\prime}$-bimodule $S$, equivalently, to $V\subseteq
H\otimes\overline{H}$ which is
$B^{\prime}\otimes(B^{\prime})^{\operatorname{op}}$-invariant, and with
$J_{0}(V)=V$. It hence remains to consider what the additional condition
$(\sigma_{z}\otimes\sigma_{z})(e)=e$ corresponds to. This condition is
equivalent to $e$ commuting with $Q\otimes Q^{-1}$, so in
$\mathcal{B}(H\otimes\overline{H})$, that it commutes with
$Q\otimes(Q^{-1})^{\top}$. By Lemma 5.16, this is equivalent to
$(Q\otimes(Q^{-1})^{\top})(V)\subseteq V$ (or with equality). Hence (2) and
(3) are equivalent.

Finally, the bijection from $\mathcal{B}(H)$ to $H\otimes\overline{H}$ sends
$\theta_{\xi,\eta}$ to $\eta\otimes\overline{\xi}$. Then

$(Q\otimes(Q^{-1})^{\top})(\eta\otimes\overline{\xi})=Q(\eta)\otimes\overline{Q^{-1}(\xi)},$

which corresponds to $Q\theta_{\xi,\eta}Q^{-1}$. It follows that (3) is
equivalent to (4). ∎

We immediately recover the known results in the tracial case, as here the
modular automorphism group is trivial, and so the conditions involving
$\sigma_{z}$ or $Q$ become vacuous.

######  Remark 5.18.

As $\Psi^{\prime}=\sigma\Psi$, and we have shown that $e=\sigma(e)$, it should
come as no surprise that working with $\Psi$ or $\Psi^{\prime}$ are
equivalent, because if $A$ satisfies the first two axioms, then in fact
$\Psi(A)=\Psi^{\prime}(A)$. ∎

######  Remark 5.19.

The operator $Q\in B$ appears in parts (3) and (4) only because $Q$ implements
$(\sigma_{t})$. In particular, if $q$ is any positive invertible operator on
$H$ with $\sigma_{t}(a)=q^{it}aq^{-it}$ for $a\in B,t\in\mathbb{R}$, then we
can replace the condition in (3) with $(q\otimes(q^{-1})^{\top})(V)=V$ and
that in (4) with $qSq^{-1}=S$. A special case of this observation occurs when
$H=L^{2}(B)$ when we could use the operator $\nabla$. ∎

######  Remark 5.20.

As we hinted at above, there are really three natural axioms for $A$: that $A$
is self-adjoint, the axiom (2) (or equivalently (4)) and the axiom (1). Axiom
(1) corresponds to $e^{2}=e$ and hence to a subspace of $\mathcal{B}(H)$ (or
more accurately, a choice of subspace and complement, not necessarily the
orthogonal complement). That $A=A^{*}$ taken together with axiom (2) then
yields that $e=e^{*}=\sigma(e)$ (and the condition of being invariant under
the modular automorphism group) and these then correspond to an orthogonal
decomposition of $\mathcal{B}(H)$, and the subspace $S$ being self-adjoint.

However, we wish to stress that taken individually, it is not clear what
$A^{*}=A$ alone, or axiom (2) alone, correspond to, for a subspace of
$\mathcal{B}(H)$. ∎

We now consider Definition 2.5, that of $A$ being “real”. This provides
motivation for looking at only the maps $\Psi_{0,1/2}$ and
$\Psi^{\prime}_{1/2,0}$ (and not other values of $s,t$). Indeed, perhaps the
theory might better be developed by initially looking at just idempotent, real
adjacency matrices $A$, but this would take us far from most of the existing
literature, excepting [30].

######  Proposition 5.21.

Under the isomorphism $\Psi_{0,1/2}$ or $\Psi^{\prime}_{1/2,0}$, we obtain a
bijection between real $A$ and $e$ such that
$(\sigma_{-i/2}\otimes\sigma_{-i/2})(e)$ is self-adjoint. In particular,
additionally $A$ also satisfies axiom (1) if and only if
$(\sigma_{-i/2}\otimes\sigma_{-i/2})(e)$ is a projection.

###### Proof.

From Lemma 5.7 we see that $A$ is real if and only if $A=A^{r}$, and we have
that $\Psi_{0,1/2}(A^{r})=(\sigma_{i}\otimes\sigma_{i})(\Psi_{0,1/2}(A)^{*})$
and
$\Psi^{\prime}_{1/2,0}(A^{r})=(\sigma_{i}\otimes\sigma_{i})(\Psi^{\prime}_{1/2,0}(A)^{*})$.
Thus, for either isomorphism, $A$ is real if and only if
$(\sigma_{i}\otimes\sigma_{i})(e^{*})=e$, equivalently,
$(\sigma_{-i/2}\otimes\sigma_{-i/2})(e)=(\sigma_{i/2}\otimes\sigma_{i/2})(e^{*})=(\sigma_{-i/2}\otimes\sigma_{-i/2})(e)^{*}$,
as claimed.

As $A$ satisfying axiom (1) is equivalent to $e^{2}=e$, and as
$\sigma_{-i/2}\otimes\sigma_{-i/2}$ is an algebra homomorphism (but not of
course a $*$-homomorphism) we see that $e^{2}=e$ and
$f=(\sigma_{-i/2}\otimes\sigma_{-i/2})(e)$ being self-adjoint implies that
$f=f^{*}=f^{2}$. Conversely, if $f=(\sigma_{-i/2}\otimes\sigma_{-i/2})(e)$ is
a projection, then $f$ is self-adjoint, and
$(\sigma_{-i/2}\otimes\sigma_{-i/2})(e^{2})=(\sigma_{-i/2}\otimes\sigma_{-i/2})(e)$
which implies that $e^{2}=e$. ∎

######  Remark 5.22.

Using $\Psi_{0,1/2}$ for example, link $A$ and $e$, and set
$f=(\sigma_{-i/2}\otimes\sigma_{-i/2})(e)$. As in the proof of Theorem 5.17,
we see that $A=A^{*}$ if and only if $e=\sigma(f^{*})$, and $A$ is undirected
(satisfies axiom (2)) if and only if $e=\sigma(f)$. Thus $A$ is real and self-
adjoint means that $f=f^{*}$ and $e=\sigma(f^{*})$, implies that $A$ is
undirected, in accordance with [30, Definition 2.19]. Similarly, axiom (4) is
equivalent to $e=(\sigma_{i/2}\otimes\sigma_{i/2})\sigma(e)$ equivalently,
$f=\sigma(e)$.

In our notation, [30, Lemma 2.22] shows that of the conditions $A=A^{*}$,
axiom (4), and that $A$ is real, any two imply the third. We translate these
conditions to being that $e=\sigma(f^{*})$, $e=\sigma(f)$, and $f=f^{*}$, and
it is now clear that any two imply the third, giving a different proof of [30,
Lemma 2.22]. ∎

That $(\sigma_{z}\otimes\sigma_{z})(e)=e$ implies commutation results for the
adjacency matrix $A$.

######  Proposition 5.23.

Let $A\in\mathcal{B}(L^{2}(B))$ be self-adjoint and satisfy axioms (1) and
(2). Then $A$ commutes with both the modular operator $\nabla$, and the
modular conjugation $J$.

###### Proof.

Let $\Psi=\Psi_{0,1/2}$. Suppose that $A=\theta_{\Lambda(a),\Lambda(b)}$ and
let $e=\Psi(A)=a^{*}\otimes\sigma_{i/2}(b)$. Then
$(\sigma_{z}\otimes\sigma_{z})(e)=\sigma_{-z}(a)^{*}\otimes\sigma_{i/2+z}(b)$
which corresponds to
$\theta_{\Lambda(\sigma_{-z}(a)),\Lambda(\sigma_{z}(b))}$. Let $z=i$ so as
$\Lambda(\sigma_{-i}(a))=\nabla\Lambda(a)$ and
$\Lambda(\sigma_{i}(b))=\nabla^{-1}\Lambda(b)$, it follows that
$\theta_{\Lambda(\sigma_{-z}(a)),\Lambda(\sigma_{z}(b))}=\nabla^{-1}A\nabla$.

Similarly, a simple calculation shows that

$JAJ=\theta_{J\Lambda(a),J\Lambda(b)}=\theta_{\Lambda(\sigma_{-i/2}(a^{*})),\Lambda(\sigma_{-i/2}(b^{*}))}$

and applying $\Psi^{-1}$ we obtain

$\sigma_{-i/2}(a^{*})^{*}\otimes\sigma_{i/2}(\sigma_{-i/2}(b^{*}))=\sigma_{i/2}(a)\otimes
b^{*}=(\sigma_{i/2}\otimes\sigma_{i/2})((a^{*}\otimes\sigma_{i/2}(b))^{*})=(\sigma_{i/2}\otimes\sigma_{i/2})(e^{*}).$

By linearity, these relations hold for any $A$ and $e=\Psi(A)$. Thus, if
$A=A^{*}$ satisfy axioms (1) and (2) then $e=\Psi(A)$ satisfies
$(\sigma_{i}\otimes\sigma_{i})(e)=e$ and so $\nabla^{-1}A\nabla=A$,
equivalently, $A$ commutes with $\nabla$. Furthermore, as $e=e^{*}$ and
$(\sigma_{i/2}\otimes\sigma_{i/2})(e)=e$, we conclude that $JAJ=A$. ∎

We now consider the axioms (3) and (5).

######  Lemma 5.24.

Let $A\in\mathcal{B}(L^{2}(B))$ have the form
$A=\theta_{\Lambda(a),\Lambda(b)}$ for some $a,b\in B$. Then, with $B$ acting
on $L^{2}(B)$ in the usual way,

  1. 1.

$m(A\otimes 1)m^{*}=ba^{*}$;

  2. 2.

$m(1\otimes A)m^{*}=J\sum_{i}\sigma_{i/2}(b)^{*}\sigma_{i/2}(a)J$, this is,
right multiplication by $\sigma_{i}(a)^{*}b$, compare with equation (1).

###### Proof.

Let $(f_{j})$ be elements of $B$ such that $(\Lambda(f_{j}))$ is an
orthonormal basis for $L^{2}(B)$, so that
$1=\sum_{j}\theta_{\Lambda(f_{j}),\Lambda(f_{j})}$. Thus

$\displaystyle m(A\otimes 1)m^{*}$
$\displaystyle=\sum_{j}m(\theta_{\Lambda(a),\Lambda(b)}\otimes\theta_{\Lambda(f_{j}),\Lambda(f_{j})})m^{*}=\sum_{j}\theta_{\Lambda(af_{j}),\Lambda(bf_{j})}$
$\displaystyle=\sum_{j}\theta_{a\Lambda(f_{j}),b\Lambda(f_{j})}=b\Big{(}\sum_{j}\theta_{\Lambda(f_{j}),\Lambda(f_{j})}\Big{)}a^{*}=ba^{*},$

as claimed. Similarly,

$\displaystyle m(1\otimes A)m^{*}$
$\displaystyle=\sum_{j}\theta_{\Lambda(f_{j}a),\Lambda(f_{j}b)}=\sum_{j}\theta_{J\sigma_{-i/2}(a^{*})J\Lambda(f_{j}),J\sigma_{-i/2}(b^{*})J\Lambda(f_{j})}=J\sigma_{-i/2}(b^{*})JJ\sigma_{i/2}(a)J,$

as claimed. ∎

For the following, recall the meaning of $\hat{1}$ from Definition 5.11.

######  Proposition 5.25.

With the conventions of Theorem 5.17, let $B\subseteq\mathcal{B}(H)$, and let
$(e_{j})$ be an orthonormal basis of $H$. Set

$u_{0}=(1\otimes(Q^{-1/2})^{\top})\hat{1}=\sum_{j}e_{j}\otimes\overline{Q^{-1/2}(e_{j})},\quad
u_{1}=(Q^{-1/2}\otimes
1)\hat{1}=\sum_{j}Q^{-1/2}(e_{j})\otimes\overline{e_{j}},$

both vectors in $H\otimes\overline{H}$.

Let $e=\Psi_{0,1/2}(A)$ or $e=\Psi^{\prime}_{1/2,0}(A)$. The following are
equivalent, assuming the equivalent conditions of Theorem 5.17 already hold.

  1. 1.

$A\in\mathcal{B}(L^{2}(B))$ satisfies axiom (3) or, equivalently, axiom (5);

  2. 2.

the projection $e$, acting on $H\otimes\overline{H}$, satisfies
$e(u_{0})=u_{0}$ or, equivalently, $e(u_{1})=u_{1}$;

  3. 3.

the subspace $V\subseteq H\otimes\overline{H}$ contains $u_{0}$, equivalently,
contains $u_{1}$;

  4. 4.

the subspace $S\subseteq\mathcal{B}(H)$ contains $Q^{-1/2}$;

  5. 5.

$m(\sigma_{-i/2}\otimes\operatorname{id})(e)=1$ where $m:B\otimes
B^{\operatorname{op}}\rightarrow B$ is the multiplication map.

###### Proof.

Let $B\subseteq\mathcal{B}(H)$, and let
$A=\sum_{i}\theta_{\Lambda(a_{i}),\Lambda(b_{i})}$. From Lemma 5.24, axiom (3)
is equivalent to $\sum_{i}a_{i}b_{i}^{*}=1$. For $\xi,\eta\in H$ we see that

$\displaystyle(\xi|\eta)$
$\displaystyle=\sum_{j}(\xi|e_{j})(\overline{\eta}|\overline{e_{j}})=(\xi\otimes\overline{\eta}|\hat{1}),$
$\displaystyle\sum_{i}(\xi|a_{i}b_{i}^{*}\eta)$
$\displaystyle=\sum_{i,j}(a_{i}^{*}(\xi)|e_{j})(\overline{b_{i}^{*}(\eta)}|\overline{e_{j}})=\sum_{i}(a_{i}^{*}(\xi)\otimes
b_{i}^{\top}(\overline{\eta})|\hat{1}).$ (3)

Using $\Psi=\Psi_{0,1/2}$, we see that $e$ acts as
$\sum_{i}a_{i}^{*}\otimes\sigma_{i/2}(b_{i})^{\top}$ and so from (3) and using
that $\sigma_{i/2}(b)=Q^{-1/2}bQ^{1/2}$,

$\displaystyle\sum_{i}(\xi|a_{i}b_{i}^{*}\eta)$
$\displaystyle=\sum_{i}(a_{i}^{*}(\xi)\otimes(Q^{1/2}\sigma_{i/2}(b_{i})Q^{-1/2})^{\top}(\overline{\eta})|\hat{1})=\big{(}(1\otimes{Q^{-1/2}}^{\top})e(1\otimes{Q^{1/2}}^{\top})(\xi\otimes\overline{\eta})\big{|}\hat{1}\big{)}.$
(4)

Hence $\sum_{i}a_{i}b_{i}^{*}=1$ if and only if
$(1\otimes{Q^{1/2}}^{\top})e(1\otimes{Q^{-1/2}}^{\top})(\hat{1})=\hat{1}$,
equivalently, $(1\otimes{Q^{-1/2}}^{\top})(\hat{1})=u_{0}$ is in the image of
$e$, namely $V$.

Consider now $m(1\otimes A)m^{*}=1$, which by Lemma 5.24 is equivalent to
$\sum_{i}\sigma_{-i}(b_{i}^{*})a_{i}=1$. Observe that for $\xi,\eta\in H$,

$\displaystyle\sum_{i}(\xi|\sigma_{-i}(b_{i}^{*})a_{i}(\eta))$
$\displaystyle=\sum_{i,j}(\sigma_{i}(b_{i})(\xi)|e_{j})(e_{j}|a_{i}(\eta))=\sum_{i,j}(\sigma_{i}(b_{i})^{\top}(\overline{e_{j}})|\overline{\xi})(a_{i}^{*}(e_{j})|\eta)$
$\displaystyle=\sum_{i}\big{(}(a_{i}^{*}\otimes\sigma_{i}(b_{i})^{\top})\hat{1}\big{|}\eta\otimes\overline{\xi}\big{)}=\sum_{i}\big{(}(a_{i}^{*}\otimes(Q^{-1/2}\sigma_{i/2}(b_{i})Q^{1/2})^{\top})\hat{1}\big{|}\eta\otimes\overline{\xi}\big{)}$
$\displaystyle=\big{(}(1\otimes{Q^{1/2}}^{\top})e(1\otimes{Q^{-1/2}}^{\top})\hat{1}\big{|}\eta\otimes\overline{\xi}\big{)},$
(5)

while $(\xi|\eta)=(\hat{1}|\eta\otimes\overline{\xi})$. Thus $m(1\otimes
A)m^{*}=1$ is again equivalent to $e(u_{0})=u_{0}$.

As the equivalent conditions of Theorem 5.17 hold, we know that
$e(Q^{-1/2}\otimes(Q^{1/2})^{\top})=(Q^{-1/2}\otimes(Q^{1/2})^{\top})e$. As
$(Q^{-1/2}\otimes(Q^{1/2})^{\top})u_{0}=u_{1}$, we see that if
$e(u_{0})=u_{0}$ then
$e(u_{1})=e(Q^{-1/2}\otimes(Q^{1/2})^{\top})(u_{0})=(Q^{-1/2}\otimes(Q^{1/2})^{\top})e(u_{0})=(Q^{-1/2}\otimes(Q^{1/2})^{\top})(u_{0})=u_{1}$,
and conversely. Hence (1), (2) and (3) are equivalent, when using $\Psi$.
Given $\Psi^{\prime}=\Psi^{\prime}_{1/2,0}$, by Remark 5.18, we know that in
fact $\Psi(A)=\Psi^{\prime}(A)$, and so the same results hold.

The bijection between $H\otimes\overline{H}$ and $\mathcal{B}(H)$ sends
$\eta\otimes\overline{\xi}$ to $\theta_{\xi,\eta}$ and so

$u_{0}\mapsto\sum_{j}\theta_{Q^{-1/2}e_{j},e_{j}}=Q^{-1/2},\qquad
u_{1}\mapsto\sum_{j}\theta_{e_{j},Q^{-1/2}e_{j}}=Q^{-1/2}.$

Thus (3) and (4) are equivalent.

For (5) consider that

$m(\sigma_{-i}\otimes\sigma_{-i/2})\Psi_{0,1/2}(\theta_{\Lambda(a),\Lambda(b)})=m(\sigma_{-i}\otimes\sigma_{-i/2})(a^{*}\otimes\sigma_{i/2}(b))=m(\sigma_{-i}(a^{*})\otimes
b)=\sigma_{i}(a)^{*}b.$

Thus $m(\sigma_{-i}\otimes\sigma_{-i/2})\Psi_{0,1/2}(A)=1$ is equivalent to
$m(1\otimes A)m^{*}=1$, by Lemma 5.24. As, by Theorem 5.17, we have that
$(\sigma_{-i/2}\otimes\sigma_{-i/2})\Psi_{0,1/2}(A)=\Psi_{0,1/2}(A)$, we see
that this condition is also equivalent to
$m(\sigma_{-i/2}\otimes\operatorname{id})\Psi_{0,1/2}(A)=1$. Hence (5) and (1)
are equivalent. ∎

Let us quickly adapt this to the case of the axioms (6) and (7).

######  Proposition 5.26.

With the conventions of the previous proposition, the following are
equivalent:

  1. 1.

$A\in\mathcal{B}(L^{2}(B))$ satisfies axiom (6) or, equivalently, axiom (7);

  2. 2.

the projection $e$, acting on $H\otimes\overline{H}$, satisfies $e(u_{0})=0$
or, equivalently, $e(u_{1})=0$;

  3. 3.

the subspace $V\subseteq H\otimes\overline{H}$ is orthogonal to $u_{0}$,
equivalently, $u_{1}$;

  4. 4.

$m(\sigma_{-i/2}\otimes\operatorname{id})(e)=0$ where $m:B\otimes
B^{\operatorname{op}}\rightarrow B$ is the multiplication map.

###### Proof.

We continue with the proof of the previous proposition. With
$A=\sum_{i}\theta_{\Lambda(a_{i}),\Lambda(b_{i})}$, from Lemma 5.24, we see
that axiom (6) is equivalent to $\sum_{i}a_{i}b_{i}^{*}=0$. Looking at
equations (3) and (4) we see that this is equivalent to
$(1\otimes{Q^{1/2}}^{\top})e(1\otimes{Q^{-1/2}}^{\top})(\hat{1})=0$,
equivalently, $e(u_{0})=0$, equivalently, $u_{0}$ is in the image of $1-e$. As
$V$ is the image of $e$, equivalently, $u_{0}$ is orthogonal to $V$.

Now consider axiom (7). By similar reasoning, and using calculation (5), this
is equivalent again to $e(u_{0})=0$. As
$e(Q^{-1/2}\otimes(Q^{1/2})^{\top})=(Q^{-1/2}\otimes(Q^{1/2})^{\top})e$, and
as $(Q^{-1/2}\otimes(Q^{1/2})^{\top})u_{0}=u_{1}$, we see that $e(u_{0})=0$ if
and only if $e(u_{1})=0$. We have shown that conditions (1), (2) and (3) are
equivalent.

For (4) we again simply follow the above proof. ∎

We will address what axioms (6) and (7) mean for subspaces of $\mathcal{B}(H)$
below, see Proposition 6.12 and the discussion after.

######  Remark 5.27.

We continue Remark 5.19. In the previous two propositions, the operator $Q$
appears in conditions, and also in the formation of the tensors $u_{0},u_{1}$.
However, again this is only because $Q$ implements the group $(\sigma_{t})$,
and so any positive invertible operator $q$ would suffice. In particular, if
$H=L^{2}(B)$ then we can use $\nabla$ in place of $Q$. ∎

As an immediate corollary, we can state the known correspondence between
Definitions 2.4 and 4.4.

######  Corollary 5.28.

The bijections $\Psi_{0,1/2}$ and $\Psi^{\prime}_{1/2,0}$ establish, in the
case when $\psi$ is a trace, a correspondence between quantum adjacency
matrices $A$ and quantum graphs $S\subseteq\mathcal{B}(H)$, where
$B\subseteq\mathcal{B}(H)$.

###### Proof.

Immediate from the correspondence between Theorem 5.17 parts (1) and (4), and
Proposition 5.25 parts (1) and (4), as when $\psi$ is a trace, $Q=1$ and
$\sigma_{z}=\operatorname{id}$ for all $z$. ∎

###  5.3 From adjacency matrices to subspaces

For general $\psi$, we only obtain a correspondence between Definition 2.4
(that of a quantum adjacency matrix) and a generalisation of Definition 4.4
(the operator bimodule view of quantum graphs) taking account of the extra
conditions of Theorem 5.17 and Proposition 5.25. In this section, we show how
to use a further bijection to come in a full circle back to Definition 4.4. We
also show some more direct links between adjacency matrices $A$ and the
associated subspaces $S$.

For the following, recall Remark 5.19.

######  Theorem 5.29.

Let $B\subseteq\mathcal{B}(H)$, and let $q\in\mathcal{B}(H)$ be positive and
invertible implementing $(\sigma_{t})$. Given a subspace
$S\subseteq\mathcal{B}(H)$ let $S_{0}=q^{1/2}S$, so that $S=q^{-1/2}S_{0}$ and
$S\mapsto S_{0}$ is a bijection. Then $S$ is a self-adjoint
$B^{\prime}$-bimodule with $qSq^{-1}=S$ if and only if the same is true of
$S_{0}$. Furthermore, $q^{-1/2}\in S$ if and only if $1\in S_{0}$. Thus if $S$
is associated to a quantum adjacency matrix, then $S_{0}$ is a quantum graph
with the additional property that $qS_{0}q^{-1}=S_{0}$, and conversely.

###### Proof.

Firstly, we claim that for any $z\in\mathbb{C}$, we have that $q^{z}xq^{-z}\in
B^{\prime}$ for any $x\in B^{\prime}$. Indeed, for $b\in B$, we have that
$q^{-z}bq^{z}=\sigma_{iz}(b)\in B$, so $x$ commutes with this, and hence

$q^{z}xq^{-z}b=q^{z}xq^{-z}bq^{z}q^{-z}=q^{z}q^{-z}bq^{z}xq^{-z}=bq^{z}xq^{-z}.$

As $b$ was arbitrary, this indeed shows that $q^{z}xq^{-z}\in B^{\prime}$.

Thus, if $S$ is a $B^{\prime}$-operator bimodule, then for $x,y\in B^{\prime}$
and $s\in S$, for $s_{0}=q^{1/2}s\in S_{0}$, we see that

$xs_{0}y=q^{1/2}q^{-1/2}xq^{1/2}sy\in q^{1/2}S=S_{0},$

as $q^{-1/2}xq^{1/2}\in B^{\prime}$. Conversely, if $S_{0}$ is a
$B^{\prime}$-operator bimodule, then with $x,y,s,s_{0}$ as before,

$xsy=q^{-1/2}q^{1/2}xq^{-1/2}s_{0}y\in q^{-1/2}S_{0}=S.$

Now consider the condition that $qSq^{-1}=S$. By a functional calculus
argument, it follows that also $q^{1/2}Sq^{-1/2}=S$; compare with the proof of
Theorem 5.17. Thus also $S_{0}=Sq^{1/2}$ and so
$q^{1/2}S_{0}q^{-1/2}=q^{1/2}S=S_{0}$. Hence $qS_{0}q^{-1}=S_{0}$.
Furthermore, if also $S$ is self-adjoint, then
$S_{0}^{*}=(q^{1/2}S)^{*}=Sq^{1/2}=S_{0}$. Conversely, if $qS_{0}q^{-1}=S_{0}$
then also $q^{-1}S_{0}q=S_{0}$, so also $q^{-1/2}S_{0}q^{1/2}=S_{0}$. Thus
$q^{1/2}Sq^{-1/2}=S_{0}q^{-1/2}=q^{-1/2}S_{0}=S$, so again $qSq^{-1}=S$.
Similarly, $S_{0}$ self-adjoint implies that $S$ is self-adjoint.

Clearly $q^{-1/2}\in S$ if and only if $1\in S_{0}$. ∎

In particular, we can always apply this result with $q=Q$, and in the special
case when $H=L^{2}(B)$, with $q=\nabla$. Let $\psi_{0}$ be a faithful _trace_
on $B$. Combining Theorem 5.17 and Proposition 5.25 with the construction just
given allows us to do the following. We start with a quantum adjacency matrix
$A$, with respect to $\psi$, and form the subspace $S$. Then form $S_{0}$, and
using $\psi_{0}$, reverse the process, to form a quantum adjacency matrix
$A_{0}$. Thus every quantum adjacency matrix for $\psi$ corresponds (uniquely)
to a quantum adjacency matrix $A_{0}$ for the trace $\psi_{0}$.

We first of all see what the condition $qS_{0}q^{-1}=S_{0}$ means for the
operator $A_{0}$. We continue to write $(\sigma_{z})$ for the modular
automorphism group of $\psi$.

######  Lemma 5.30.

Let $B\subseteq\mathcal{B}(H)$, let $\psi_{0}$ be a faithful trace on $B$, and
let $S_{0}\subseteq\mathcal{B}(H)$ be a quantum graph associated to the
adjacency matrix $A_{0}\in\mathcal{B}(L^{2}(\psi_{0}))$. Furthermore, using
that the GNS map $\Lambda_{0}:B\rightarrow L^{2}(\psi_{0})$ is a linear
bijection, let $A_{0}$ induce a linear map $A_{0}^{\prime}:B\rightarrow B$.
Then $qS_{0}q^{-1}=S_{0}$ if and only if
$A_{0}^{\prime}\circ\sigma_{-i}=\sigma_{-i}\circ A_{0}^{\prime}$.

###### Proof.

As in the proof of Theorem 5.17, that $qS_{0}q^{-1}=S_{0}$ is equivalent to
$e=(\sigma_{i}\otimes\sigma_{i})(e)$. Supposing for the moment that
$A_{0}=\theta_{\Lambda_{0}(a),\Lambda_{0}(b)}$, we have that $e=b\otimes
a^{*}$ and that $A_{0}^{\prime}(c)=\psi_{0}(a^{*}c)b$ for $c\in B$. Then
$\sigma_{i}(b)\otimes\sigma_{i}(a^{*})$ is associated to the map on $B$ given
by

$\displaystyle c\mapsto\psi_{0}(\sigma_{i}(a^{*})c)\sigma_{i}(b)$
$\displaystyle=\psi_{0}(Q^{-1}a^{*}Qc)\sigma_{i}(b)=\psi_{0}(a^{*}QcQ^{-1})\sigma_{i}(b)$
$\displaystyle=\psi_{0}(a^{*}\sigma_{-i}(c))\sigma_{i}(b)=\sigma_{i}\big{(}A_{0}^{\prime}(\sigma_{-i}(c))\big{)}.$

By linearity, such a relationship holds in general, and so we conclude that
$A_{0}^{\prime}=\sigma_{i}\circ A_{0}^{\prime}\circ\sigma_{-i}$, which is
equivalent to the claim. ∎

######  Remark 5.31.

Suppose that $S,S_{0}$ are linked as before. In terms of subspaces of
$H\otimes\overline{H}$ this means that $V$ is associated to
$V_{0}=(q^{1/2}\otimes\operatorname{id})(V)$. As $V$ is invariant under the
operator $q^{z}\otimes(q^{-z})^{\top}$, for all $z\in\mathbb{C}$, also
$V_{0}=(q^{1/4}\otimes(q^{1/4})^{\top})(V)$, and so forth.

Thus $e_{0}$ is the orthogonal projection onto $V_{0}$. We note that

$p=(q^{1/4}\otimes(q^{1/4})^{\top})e(q^{-1/4}\otimes(q^{-1/4})^{\top})$

is an idempotent with image $V_{0}$, here of course letting $e$ act on
$H\otimes\overline{H}$ in the usual way. Then, in $B\otimes
B^{\operatorname{op}}$, we have that
$p=(\sigma_{-i/4}\otimes\sigma_{i/4})(e)$. So in general, $p\not=e_{0}$ as $p$
is not self-adjoint.

There is a standard way to convert an idempotent into a projection; we follow
[15, Proposition 3.2.10] for example, and just give a sketch. Setting
$a=1+(p-p^{*})^{*}(p-p^{*})$, we see that $a\geq 1$ so $a$ is invertible. Then
one can show that $q=pp^{*}a^{-1}$ is a projection with $pq=q,qp=p$ so that
$q$ and $p$ have the same image. Hence $e_{0}=q$. This is a concrete formula,
but it appears difficult to work with, as in particular, we have no expression
for $a^{-1}$.

Then $A\in\mathcal{B}(L^{2}(\psi))$ is equal to
${\Psi^{\prime}_{1/2,0}}^{-1}(e)$, and $A_{0}\in\mathcal{B}(L^{2}(\psi_{0}))$
is ${\Psi^{\prime}_{1/2,0}}^{-1}(e_{0})$, noting of course that really these
are two different maps $\Psi^{\prime}_{1/2,0}$ as they have different
codomains. We do not see how to give a concrete formula for $A$ in terms of
$A_{0}$.

However, Remark 5.35 below does give a direct link between $A$ and $S_{0}$. ∎

We seek a way to more directly link the adjacency matrix $A$ with the subspace
$S\subseteq\mathcal{B}(H)$ (and/or $V\subseteq H\otimes\overline{H}$). As $H$
is essentially arbitrary (assuming $B$ is faithfully represented on $H$) this
seems hard, but with a special choice of $H$ we can make progress.

We start by looking further at the opposite algebra $B^{\operatorname{op}}$.
Define a faithful positive functional $\psi^{\operatorname{op}}$ on
$B^{\operatorname{op}}$ by $\psi^{\operatorname{op}}(a)=\psi(a)$ for $a\in
B^{\operatorname{op}}$.

######  Lemma 5.32.

Let $(L^{2}(B),\Lambda,\pi)$ be the GNS construction for $\psi$ on $B$, and
similarly
$(L^{2}(B^{\operatorname{op}}),\Lambda^{\operatorname{op}},\pi^{\operatorname{op}})$
for $\psi^{\operatorname{op}}$. The map
$w:L^{2}(B^{\operatorname{op}})\rightarrow\overline{L^{2}(B)};\Lambda^{\operatorname{op}}(a)\mapsto\overline{\Lambda(a^{*})}$
is a unitary with $w\pi^{\operatorname{op}}(a)w^{*}=\pi(a)^{\top}$ for $a\in
B$. Furthermore, $wJ^{\operatorname{op}}w^{*}=\overline{J}$, and
$w\nabla^{\operatorname{op}}w^{*}=\overline{\nabla}=\nabla^{\top}$.

###### Proof.

We have that
$(\Lambda^{\operatorname{op}}(a)|\Lambda^{\operatorname{op}}(b))=\psi^{\operatorname{op}}(ba^{*})=\psi(ba^{*})$,
recalling that the product is reversed in $B^{\operatorname{op}}$. As

$(w\Lambda^{\operatorname{op}}(a)|w\Lambda^{\operatorname{op}}(b))=(\overline{\Lambda(a^{*})}|\overline{\Lambda(b^{*})})=(\Lambda(b^{*})|\Lambda(a^{*}))=\psi(ba^{*}),$

it follows that $w$ is unitary, and so
$w^{*}\overline{\Lambda(b)}=w^{*}w\Lambda^{\operatorname{op}}(b^{*})=\Lambda^{\operatorname{op}}(b^{*})$.
Thus

$w\pi^{\operatorname{op}}(a)w^{*}\overline{\Lambda(b)}=w\pi^{\operatorname{op}}(a)\Lambda^{\operatorname{op}}(b^{*})=w\Lambda^{\operatorname{op}}(b^{*}a)=\overline{\Lambda(a^{*}b)}=\overline{\pi(a^{*})\Lambda(b)}=\pi(a)^{\top}\overline{\Lambda(b)},$

as required.

By definition, we have that
$\psi^{\operatorname{op}}(a)=\psi(a)={\operatorname{Tr}}(Qa)={\operatorname{Tr}}(aQ)$
and so $\psi^{\operatorname{op}}$ also has density $Q$. Thus
$\sigma_{z}^{\operatorname{op}}(a)=Q^{-iz}aQ^{iz}=\sigma_{-z}(a)$, remembering
that the multiplication in $B^{\operatorname{op}}$ is reversed. Thus

$wJ^{\operatorname{op}}w^{*}\overline{\Lambda(a)}=wJ^{\operatorname{op}}\Lambda^{\operatorname{op}}(a^{*})=w\Lambda^{\operatorname{op}}(\sigma^{\operatorname{op}}_{-i/2}(a))=w\Lambda^{\operatorname{op}}(\sigma_{i/2}(a))=\overline{\Lambda(\sigma_{i/2}(a)^{*})}=\overline{J\Lambda(a)},$

as claimed. Similarly,

$w\nabla^{\operatorname{op}}w^{*}\overline{\Lambda(a)}=w\nabla^{\operatorname{op}}\Lambda^{\operatorname{op}}(a^{*})=w\Lambda^{\operatorname{op}}(Q^{-1}a^{*}Q)=\overline{\Lambda(QaQ^{-1})}=\overline{\nabla\Lambda(a)},$

where $\overline{\nabla}=\nabla^{\top}$ as $\nabla$ is self-adjoint. ∎

In the sequel, we shall often suppress $w$, and simply identify
$L^{2}(B^{\operatorname{op}})$ with $\overline{L^{2}(B)}$. We equip $B\otimes
B^{\operatorname{op}}$ with the faithful functional
$\psi\otimes\psi^{\operatorname{op}}$ and so $L^{2}(B\otimes
B^{\operatorname{op}})\cong L^{2}(B)\otimes\overline{L^{2}(B)}$. As usual, we
give $\mathcal{B}(L^{2}(B))$ the functional ${\operatorname{Tr}}$, and this
also has GNS space $L^{2}(B)\otimes\overline{L^{2}(B)}$.

######  Lemma 5.33.

Let $e\in B\otimes B^{\operatorname{op}}$ be a projection, and let $B\otimes
B^{\operatorname{op}}$ act on $L^{2}(B)\otimes\overline{L^{2}(B)}$. The image
of $e$ agrees with the
$B^{\prime}\otimes{B^{\prime}}^{\operatorname{op}}$-invariant subspace
generated by $(\Lambda\otimes\Lambda^{\operatorname{op}})(e)$.

###### Proof.

Consider $f=a\otimes b\in B\otimes B^{\operatorname{op}}$ which acts as
$a\otimes b^{\top}$ on $L^{2}(B)\otimes\overline{L^{2}(B)}$. Thus, for $x,y\in
B$,

$\displaystyle f(\Lambda(x)\otimes\overline{\Lambda(y)})$
$\displaystyle=\Lambda(ax)\otimes\overline{\Lambda(b^{*}y)}=J\sigma_{-i/2}(x)J\Lambda(a)\otimes\overline{J\sigma_{-i/2}(y)J\Lambda(b^{*})}$
$\displaystyle=\big{(}J\sigma_{-i/2}(x)J\otimes(J\sigma_{-i/2}(y)^{*}J)^{\top}\big{)}(\Lambda\otimes\Lambda^{\operatorname{op}})(f).$

By linearity, this relationship holds for all $f\in B\otimes
B^{\operatorname{op}}$. As $B^{\prime}=JBJ$, taking the linear span over
$x,y$, we conclude that the image of $e$ equals
$(B^{\prime}\otimes(B^{\prime})^{\operatorname{op}})(\Lambda\otimes\Lambda^{\operatorname{op}})(e)$,
as claimed. ∎

######  Proposition 5.34.

Let $B$ act on $L^{2}(B)$, and let $A\in\mathcal{B}(L^{2}(B))$ be self-adjoint
and satisfy axioms (1) and (2). Let $A$ be associated to the
$B^{\prime}$-operator bimodule $S\subseteq\mathcal{B}(L^{2}(B))$, and the
$B^{\prime}\otimes{B^{\prime}}^{\operatorname{op}}$-invariant subspace
$V\subseteq L^{2}(B)\otimes\overline{L^{2}(B)}$. Set
$A^{\prime}=\nabla^{-1/2}A$. Then $S$ is the $B^{\prime}$-operator bimodule
generated by $A^{\prime}$, and $\Lambda_{{\operatorname{Tr}}}(A^{\prime})$ is
the projection of $(\Lambda\otimes\Lambda^{\operatorname{op}})(1)$ onto $V$.

###### Proof.

We use $\Psi^{\prime}_{1/2,0}$ as in Theorem 5.17, so that
$\theta_{\Lambda(a),\Lambda(b)}$ is associated to $\sigma_{i/2}(b)\otimes
a^{*}$. Notice that

$\Lambda_{{\operatorname{Tr}}}(\nabla^{-1/2}\theta_{\Lambda(a),\Lambda(b)})=\Lambda_{{\operatorname{Tr}}}(\theta_{\Lambda(a),\Lambda(\sigma_{i/2}(b))})=\Lambda(\sigma_{i/2}(b))\otimes\overline{\Lambda(a)}=\Lambda(\sigma_{i/2}(b))\otimes\Lambda^{\operatorname{op}}(a^{*}).$

By linearity, we see that
$\Lambda_{{\operatorname{Tr}}}(\nabla^{-1/2}A)=(\Lambda\otimes\Lambda^{\operatorname{op}})\Psi^{\prime}_{1/2,0}(A)$.

Use Theorem 5.17 to associate $A$ to the projection $e$. Thus
$\Lambda_{{\operatorname{Tr}}}(\nabla^{-1/2}A)=(\Lambda\otimes\Lambda^{\operatorname{op}})(e)$,
and so by the previous lemma, $S$ is the $B^{\prime}$-operator bimodule
generated by $\nabla^{-1/2}A$. The first claim follows. Further, as $e$ is the
projection onto $V$, the projection of
$(\Lambda\otimes\Lambda^{\operatorname{op}})(1)$ onto $V$ is simply
$e(\Lambda\otimes\Lambda^{\operatorname{op}})(1)=(\Lambda\otimes\Lambda^{\operatorname{op}})(e)$,
and so the second claim follows. ∎

######  Remark 5.35.

There is an interesting link with Theorem 5.29, which we apply with
$q=\nabla$. Thus $S$ is associated to $S_{0}=\nabla^{1/2}S$ which is a
“tracial” quantum graph. As $A^{\prime}=\nabla^{-1/2}A$, the proposition says
that

$S_{0}=\nabla^{1/2}S=\nabla^{1/2}\operatorname{lin}\\{xA^{\prime}y:x,y\in
B^{\prime}\\}=\operatorname{lin}\\{\nabla^{1/2}x\nabla^{-1/2}Ay:x,y\in
B^{\prime}\\}.$

As $x\mapsto\nabla^{1/2}x\nabla^{-1/2}$ is a bijection of $B^{\prime}$, it
follows that $S_{0}$ is the $B^{\prime}$-operator bimodule generated by $A$. ∎

###  5.4 Adjacency super-operators to matrices

In this section we show that Definitions 3.1 and 2.4 are equivalent. This was
shown in [12, Section 1.2] in the case when $B=\mathbb{M}_{n}$, using the
_Choi matrix_ of a completely positive map. We shall give an account of how to
extend this formalism to general $B$, which might be of independent interest.
As such, we proceed with a little more generality than strictly needed. We
shall also use the language of Hilbert $C^{*}$-modules, see [28] for example,
but we shall not use any deep results from this theory.

In this section, let $C$ be an arbitrary $C^{*}$-algebra, and let $B,\psi$ be
as usual. In the usual way, regard $B^{*}$ as a bimodule over $B$. In
particular, for $a\in B$, we denote by $\psi a\in B^{*}$ the functional
$b\mapsto\psi(ab)$. Every $f\in B^{*}$ is of the form $\psi a$ for some $a\in
B$. This follows, as if not, by Hahn-Banach, there is a non-zero $b\in
B^{**}=B$ with $0=(\psi a)(b)=\psi(ab)$ for all $a$. Choosing $a=b^{*}$ shows
that $\psi(b^{*}b)=0$ so $b=0$, contradiction. Furthermore, notice that
$B\rightarrow B^{*},a\mapsto\psi a$ is injective, as $(\psi
a)(a^{*})=\psi(aa^{*})=0$ only when $a=0$, again using that $\psi$ is
faithful.

As $B$ is finite-dimensional, $\mathcal{B}(B,C)$ is spanned by finite-rank
operators, that is, the span of operators of the form $\theta_{\psi a,c}$ for
$a\in B,c\in C$, where $\theta_{\psi a,c}$ is the map $B\ni
b\mapsto\psi(ab)c\in C$. In fact, it will be profitable to again introduce the
modular automorphism group, and consider the map

$\Psi^{\prime}:\mathcal{B}(B,C)\rightarrow C\otimes
B^{\operatorname{op}};\quad\theta_{\psi a,c}\mapsto c\otimes\sigma_{-i/2}(a).$

######  Theorem 5.36.

$\Psi^{\prime}$ restricts to a bijection between the completely positive maps
$B\rightarrow C$, and positive elements $e\in C\otimes B^{\operatorname{op}}$.

The “usual” Choi matrix, compare [10, Proposition 1.5.12] associates a CP map
$A:\mathbb{M}_{n}\rightarrow C$ with the positive operator $(A(e_{ij}))\in
M_{n}(C)$. Let us see this in our generalisation. Set $B=\mathbb{M}_{n}$ and
$\psi={\operatorname{Tr}}$. We identify $B^{\operatorname{op}}$ with
$\mathbb{M}_{n}$ using the matrix transpose, and so our bijection identifies
$A=\theta_{\psi a,c}$ with $u=c\otimes a^{\top}\in C\otimes\mathbb{M}_{n}\cong
M_{n}(C)$. Then $A(e_{ij})=\psi(ae_{ij})c=a_{ji}c=(a^{\top})_{ij}c=u_{ij}$,
exactly as we expect.

To prove the theorem, we shall use the language of Hilbert $C^{*}$-modules, in
particular, the KSGNS construction, which we now briefly recall from [28,
Chapter 5]. Given a completely positive map $A:B\rightarrow C$, we consider
the algebraic tensor product $B\odot C$, and define on this a (in general,
degenerate) positive $C$-valued form by

$\big{(}b_{1}\otimes c_{1}\big{|}b_{2}\otimes
c_{2}\big{)}=c_{1}^{*}A(b_{1}^{*}b_{2})c_{2}.$

The separation completion is denoted by $B\otimes_{A}C$, and we continue to
denote by $b\otimes c$ the equivalence class in $B\otimes_{A}C$ defined by
this simple tensor. Almost by definition, $B\otimes_{A}C$ is a (right) Hilbert
$C^{*}$-module over $C$, and there is a $*$-homomorphism
$\pi:B\rightarrow\mathcal{L}(B\otimes_{A}C)$ given by
$\pi(b)(b^{\prime}\otimes c)=bb^{\prime}\otimes c$. Finally, there is
$V\in\mathcal{L}(C,B\otimes_{A}C)$ defined by $V(c)=1\otimes c$, and then

$A(b)=V^{*}\pi(b)V\qquad(b\in B),$

giving a canonical dilation for the CP map $A$.

###### Proof of Theorem 5.36.

Suppose that $A:B\rightarrow C$ is completely positive, and set
$e=\Psi^{\prime}(A)$. Perform the KSGNS construction to form
$E=B\otimes_{A}C,\pi$ and $V$. We wish to describe $E$ in terms of $e$. For
the moment suppose that $A=\theta_{\psi a,c}$, so that
$e=c\otimes\sigma_{-i/2}(a)$, and

$\displaystyle\big{(}b_{1}\otimes c_{1}\big{|}b_{2}\otimes c_{2}\big{)}_{E}$
$\displaystyle=c_{1}^{*}A(b_{1}^{*}b_{2})c_{2}=\psi(ab_{1}^{*}b_{2})c_{1}^{*}cc_{2}=\psi(\sigma_{i/2}(b_{2})\sigma_{-i/2}(ab_{1}^{*}))c_{1}^{*}cc_{2}$
$\displaystyle=\psi(\sigma_{i/2}(b_{2})\sigma_{-i/2}(a)\sigma_{i/2}(b_{1})^{*})c_{1}^{*}cc_{2}$
$\displaystyle=(\psi^{\operatorname{op}}\otimes\operatorname{id})\big{(}(\sigma_{i/2}(b_{1})\otimes
c_{1})^{*}\sigma(e)(\sigma_{i/2}(b_{2})\otimes c_{2})\big{)},$

where the final product is formed in $B^{\operatorname{op}}\otimes C$ and
$\sigma$ is again the swap map, and $\psi^{\operatorname{op}}$ is as in Lemma
5.32. By linearity, this relationship holds for general $A$.

Let $\mu$ be a state on $C$, with GNS construction
$(L^{2}(C,\mu),\Lambda_{\mu})$. Given
$u=\sum_{i}\Lambda^{\operatorname{op}}(b_{i})\otimes\Lambda_{\mu}(c_{i})\in
L^{2}(B^{\operatorname{op}})\otimes L^{2}(C,\mu)$, we have

$\displaystyle(u|\sigma(e)u)$
$\displaystyle=\sum_{i,j}\big{(}\Lambda^{\operatorname{op}}(b_{i})\otimes\Lambda_{\mu}(c_{i})\big{|}\sigma(e)\Lambda^{\operatorname{op}}(b_{j})\otimes\Lambda_{\mu}(c_{j})\big{)}$
$\displaystyle=\sum_{i,j}(\psi^{\operatorname{op}}\otimes\mu)\big{(}(b_{i}\otimes
c_{i})^{*}\sigma(e)(b_{j}\otimes c_{j})\big{)}$
$\displaystyle=\sum_{i,j}\mu\big{(}(\sigma_{-i/2}(b_{i})\otimes
c_{i}|\sigma_{-i/2}(b_{j})\otimes c_{j})_{E}\big{)}\geq 0.$

As this holds for all $u$, this shows that the image of $\sigma(e)$ in
$\mathcal{B}(L^{2}(B^{\operatorname{op}})\otimes L^{2}(C,\mu))$ is positive.
As $\mu$ was arbitrary, this shows that $\sigma(e)$, and hence $e$, is
positive.

We now consider the converse, assume that $e=\Psi^{\prime}(A)$ is positive,
and aim to show that $A$ is completely positive. We copy the KSGNS
construction, but now _define_ a sesquilinear form on
$B^{\operatorname{op}}\otimes C$ by

$\big{(}u_{1}\big{|}u_{2}\big{)}=(\psi^{\operatorname{op}}\otimes\operatorname{id})\big{(}(\sigma_{i/2}\otimes\operatorname{id})(u_{1})^{*}\sigma(e)(\sigma_{i/2}\otimes\operatorname{id})(u_{2})\big{)}\qquad(u_{1},u_{2}\in
B^{\operatorname{op}}\otimes C).$

This form is clearly positive, so we may let $E$ be the separation completion
(as $B$ is finite-dimensional, $E$ is in fact a quotient of $B\otimes C$).

We shall repeatedly move between $B$ and $B^{\operatorname{op}}$, and it will
be helpful to have notation to distinguish these. We shall write the product
in $B^{\operatorname{op}}$ by juxtaposition, and the product in $B$ by $a\cdot
b$, for $a,b\in B$. We now show that $\pi:B\rightarrow\mathcal{L}(E)$ defined
by $\pi(b)(b^{\prime}\otimes c)=b\cdot b^{\prime}\otimes c$ is well-defined.
Notice that working in $B^{\operatorname{op}}\otimes C$, we have that
$\pi(b)(b^{\prime}\otimes c)=(b^{\prime}\otimes c)(b\otimes 1)$, and so

$\|\pi(b)(u)\|_{E}^{2}=(\psi^{\operatorname{op}}\otimes\operatorname{id})\big{(}(\sigma_{i/2}(b)^{*}\otimes
1)(\sigma_{i/2}\otimes\operatorname{id})(u)^{*}\sigma(e)(\sigma_{i/2}\otimes\operatorname{id})(u)(\sigma_{i/2}(b)\otimes
1)\big{)}.$

As $\sigma(e)\in B^{\operatorname{op}}\otimes C$ is positive, there is $f\geq
0$ with $f^{2}=\sigma(e)$. Set $x=f(\sigma_{i/2}\otimes\operatorname{id})(u)$,
say $x=\sum_{i}b_{i}\otimes c_{i}\in B^{\operatorname{op}}\otimes C$. Then

$\displaystyle\|\pi(b)(u)\|_{E}^{2}$
$\displaystyle=\sum_{i,j}\psi^{\operatorname{op}}\big{(}\sigma_{i/2}(b)^{*}b_{i}^{*}b_{j}\sigma_{i/2}(b)\big{)}c_{i}^{*}c_{j}=\sum_{i,j}\psi\big{(}\sigma_{i/2}(b)\cdot
b_{j}\cdot b_{i}^{*}\cdot\sigma_{i/2}(b)^{*}\big{)}c_{i}^{*}c_{j}$
$\displaystyle=\sum_{i,j}\psi\big{(}b_{j}\cdot
b_{i}^{*}\cdot\sigma_{i/2}(b)^{*}\cdot\sigma_{-i/2}(b)\big{)}c_{i}^{*}c_{j}=\sum_{i,j}\psi\big{(}b_{j}\cdot
b_{i}^{*}\cdot\sigma_{-i/2}(b^{*}\cdot b)\big{)}c_{i}^{*}c_{j}$
$\displaystyle=\sum_{i,j}\psi\big{(}\sigma_{i/2}(b_{i}^{*}\cdot\sigma_{-i/2}(b^{*}\cdot
b))\cdot\sigma_{-i/2}(b_{j})\big{)}c_{i}^{*}c_{j}$
$\displaystyle=\sum_{i,j}\psi\big{(}\sigma_{-i/2}(b_{i})^{*}\cdot b^{*}\cdot
b\cdot\sigma_{-i/2}(b_{j})\big{)}c_{i}^{*}c_{j}.$

We can now pull out the $b^{*}\cdot b$ term, at the cost of a $\|b\|^{2}$
factor, and then reverse to calculation to conclude that

$\|\pi(b)(u)\|_{E}^{2}\leq\|b\|^{2}\|u\|_{E}^{2}.$

Thus $\pi$ is well-defined. A similar calculation shows that $\pi$ is a
$*$-homomorphism.

Without loss of generality, we may suppose that $C$ is unital. Set
$u_{0}=1\otimes 1\in E$, so that

$(u_{0}|\pi(b)u_{0})_{E}=(\psi^{\operatorname{op}}\otimes\operatorname{id})(\sigma(e)(\sigma_{i/2}(b)\otimes
1)).$

We wish to link this to the map $A$. Suppose for the moment that $e=c\otimes
a$ so that $A(a^{\prime})=\psi(\sigma_{i/2}(a)\cdot a^{\prime})c$, and hence

$(\psi^{\operatorname{op}}\otimes\operatorname{id})(\sigma(e)(\sigma_{i/2}(b)\otimes
1))=\psi(\sigma_{i/2}(b)\cdot a)c=\psi(\sigma_{i/2}(a)\cdot b)c=A(b).$

By linearity, this relationship holds for all $e$, and hence
$A(b)=(u_{0}|\pi(b)u_{0})_{E}$ for all $b\in B$. We conclude that $A$ is
completely positive. ∎

The first part of the following is [30, Proposition 2.23], though we use the
ideas developed above. It generalises [12, Proposition 1.7], which considers
$B=\mathbb{M}_{n}$ with $\psi$ a trace.

######  Theorem 5.37.

Let $A_{0}:B\rightarrow B$ be a linear map, and let $A:L^{2}(B)\rightarrow
L^{2}(B)$ be the associated map. Then
$\Psi^{\prime}(A_{0})=\Psi^{\prime}_{0,1/2}(A)$ in the notation of Section 5.
The following are equivalent:

  1. 1.

$A_{0}$ is completely positive, with $m(A_{0}\otimes A_{0})m^{*}=A_{0}$;

  2. 2.

$A$ is real and satisfies axiom (1).

We interpret the axioms of Definition 2.4 for $A_{0}$ in the obvious way. The
following are equivalent:

  1. 3.

$A_{0}$ is completely positive, and satisfies axioms (1) and (2).

  2. 4.

$A$ is self-adjoint, satisfying axioms (1) and (2).

###### Proof.

We apply Theorem 5.36 with $C=B$, and so consider
$\Psi^{\prime}:\mathcal{B}(B)\rightarrow B\otimes B^{\operatorname{op}}$ which
maps $A_{0}=\theta_{\psi a^{*},b}$ to $b\otimes\sigma_{i/2}(a)^{*}$. The
associated $A$ is $\theta_{\Lambda(a),\Lambda(b)}$, and indeed
$b\otimes\sigma_{i/2}(a)^{*}=\Psi^{\prime}_{0,1/2}(A)$ as claimed. As in the
proof of Proposition 5.21 we set $f=(\sigma_{-i/2}\otimes\sigma_{-i/2})(e)$,
where $e=\Psi^{\prime}_{1/2,0}(A)$. Then
$e=\Psi^{\prime}_{1/2,0}(A)=(\sigma_{i/2}\otimes\sigma_{i/2})\Psi^{\prime}_{0,1/2}(A)$
and so $f=\Psi^{\prime}(A_{0})$.

We now see that (1) is equivalent to $A$ satisfying axiom (1) and
$\Psi^{\prime}_{0,1/2}(A)$ being positive. This is equivalent to $e=e^{2}$ and
$f\geq 0$, equivalently, $f$ being a projection. By Proposition 5.21, this is
equivalent to (2).

Then condition (3) is equivalent to $A$ being real, and satisfying axioms (1)
and (2). As in Remark 5.22 this is equivalent to condition (4). ∎

######  Remark 5.38.

We make links with [9, Section 2]. Here the authors consider the space
$B\otimes_{\psi}B$, which can be identified with the Hilbert $C^{*}$-module
$L^{2}(B)\otimes B$, with the usual left action of $B$ on $L^{2}(B)$. They
define the “quantum edge indicator”, which with our normalisation conventions,
is $\epsilon=(\operatorname{id}\otimes A_{0})m^{*}(1)$ which agrees with
$\Psi_{1,0}(A)$, compare Remark 5.6. From [9, Proposition 2.3(2)] we see that
it is natural to work with $B\otimes B^{\operatorname{op}}$, and then [9,
Proposition 2.3(3)] considers the element
$(\sigma_{i/2}\otimes\operatorname{id})(\epsilon)=\Psi_{1/2,0}(A)$. This
agrees with $\sigma\Psi^{\prime}_{0,1/2}(A)$. Under the assumption that
$m(A_{0}\otimes A_{0})m^{*}=A_{0}$, we see that $e=\Psi^{\prime}_{0,1/2}(A)$
is an idempotent, and thus we see that $e$ is positive if and only if
$(\sigma_{i/2}\otimes\operatorname{id})(\epsilon)$ is self-adjoint. Theorem
5.36 shows immediately that $A_{0}$ is completely positive if and only if
$(\sigma_{i/2}\otimes\operatorname{id})(\epsilon)$ is self-adjoint, which
hence yields a different proof of [9, Proposition 2.3(3)].

However, with reference to Proposition 5.3, neither $\Psi_{1/2,0}$ nor
$\Psi_{1,0}$ seems to give particularly “nice” equivalent properties on $e$
for the other axioms we might wish $A$ (or $A_{0}$) to satisfy. ∎

##  6 Constructions and Examples

In this section, we discuss some examples, explore how quantum channels (for
us, UCP maps) give rise to quantum graphs, and also discuss various ways to
construct new quantum graphs from existing ones.

###  6.1 Examples

We consider our basic examples, Definitions 2.8 and 2.10, and see what the
different realisations from Section 5.2 are. The following is clear: for the
complete quantum graph we obtain the “maximal” operator bimodule
$\mathcal{B}(H)$.

######  Proposition 6.1.

The complete quantum graph has $A=\theta_{\Lambda(1),\Lambda(1)}$ and this is
associated to $e=\Psi_{0,1/2}(A)=1\otimes 1$, and to subspaces
$H\otimes\overline{H}$ and $\mathcal{B}(H)$.

The empty quantum graph given by $A=(mm^{*})^{-1}$ is much harder to work
with, as ideally we would wish to express $(mm^{*})^{-1}$ as a span of
operators of the form $\theta_{\Lambda(a),\Lambda(b)}$, and it is not
immediate how to do this. However, recall from the discussion before
Definition 2.10 that $mm^{*}$ is, in particular, an operator which is a
central element in $B$.

For the following, recall $u_{0}$ from Proposition 5.25, which shows that
axiom (3) is equivalent to $u_{0}\in V$, and hence the characterisation of $e$
as “minimal” is as expected.

######  Proposition 6.2.

Let $A=(mm^{*})^{-1}$ the empty quantum graph adjacency matrix. This is
associated to the operator bimodule $S=B^{\prime}Q^{-1/2}B^{\prime}$, and
subspace $V=(B^{\prime}\otimes(B^{\prime})^{\operatorname{op}})(u_{0})$, and
hence projection $e$ which is the minimal projection in $B\otimes
B^{\operatorname{op}}$ with $e(u_{0})=u_{0}$.

###### Proof.

Given the equivalences of Theorem 5.17, we need only show
$S=B^{\prime}Q^{-1/2}B^{\prime}$. Furthermore, we may choose which Hilbert
space $B$ acts on, so set $H=L^{2}(B)$. By Proposition 5.34, $S$ is the
$B^{\prime}$-bimodule generated by $A^{\prime}=\nabla^{-1/2}A$. There is $x\in
B^{\prime}$ with $mm^{*}=x$. Thus $A=(mm^{*})^{-1}=x^{-1}$, and so
$S=B^{\prime}\nabla^{-1/2}x^{-1}B^{\prime}=B^{\prime}\nabla^{-1/2}B^{\prime}$
because $x^{-1}$ is invertible.

For $a\in B$ and $y_{1},y_{2}\in B^{\prime}$ there are $b_{1},b_{2}\in B$ with
$y_{i}=Jb_{i}J$, and so

$\displaystyle y_{1}\nabla^{-1/2}y_{2}\Lambda(a)$
$\displaystyle=y_{1}\nabla^{-1/2}\Lambda(a\sigma_{-i/2}(b_{2}^{*}))=y_{1}\Lambda(\sigma_{i/2}(a)b_{2}^{*})=\Lambda(\sigma_{i/2}(a)b_{2}^{*}\sigma_{-i/2}(b_{1}^{*}))$
$\displaystyle=Q^{-1/2}\Lambda(aQ^{1/2}b_{2}^{*}\sigma_{-i/2}(b_{1}^{*}))=Q^{-1/2}JzJ\Lambda(a),$

for some $z\in B$, indeed,
$\sigma_{-i/2}(z^{*})=Q^{1/2}b_{2}^{*}\sigma_{-i/2}(b_{1}^{*})$. Thus
$S\subseteq Q^{-1/2}B^{\prime}$, but clearly given an arbitrary $z$, we may
set $b_{1}=1$ and $b_{2}^{*}=Q^{-1/2}\sigma_{-i/2}(z^{*})$ to obtain the
opposite inclusion. Thus $S=Q^{-1/2}B^{\prime}=B^{\prime}Q^{-1/2}B^{\prime}$,
as $Q\in B$. ∎

Let $S_{0}$ be the tracial quantum graph associated to $S$, given by Theorem
5.29. So $S_{0}=Q^{1/2}S=Q^{1/2}B^{\prime}Q^{-1/2}B^{\prime}=B^{\prime}$, as
$Q\in B$ commutes with $B^{\prime}$. Thus $S_{0}=B^{\prime}$ is the empty
quantum graph from Definition 4.5, as we might expect.

###  6.2 From Quantum Channels

We now explore a class of examples (of quantum graphs in the sense of
Definition 4.4) which come from quantum channels, see [14, 19, 20, 38]. These
examples are amongst the first which arose, and motivate much of the theory.
We shall generalise the constructions to arbitrary finite-dimensional
$C^{*}$-algebras, and at the same time streamline them, using the Stinespring
representation theorem directly, rather than passing via Kraus operators.

Quantum Channels are usually considered to be trace-preserving, completely
positive maps between matrix algebras. From an operator algebraic perspective,
this means they are most naturally considered to be maps between (pre)duals of
algebras: namely trace-class operators. As such, we prefer to work with the
Banach space adjoints, and instead look at unital, completely positive maps.
We shall makes comments below where this leads to slightly different
conventions.

######  Remark 6.3.

Classically we consider the following situation. Let $X,Y$ be finite sets, and
for each $x\in X$ imagine we send $x$ down a noisy communication channel, so
that $y\in Y$ may be received with probability $p(y|x)\in[0,1]$. Thus
$\sum_{y}p(y|x)=1$ for each $x$. This gives a map
$\ell^{1}(X)\rightarrow\ell^{1}(Y)$, say sending a unit basis vector
$\delta_{x}$ to $\sum_{y}p(y|x)\delta_{y}$. The adjoint is a (completely)
positive, unital map $\theta:\ell^{\infty}(Y)\rightarrow\ell^{\infty}(X)$
given by

$\theta(f)(x)=\sum_{y}p(y|x)f(y)\qquad(f\in\ell^{\infty}(Y)).$

We say that $x_{1},x_{2}\in X$ are “confusable” if there is some $y$ with both
$p(y|x_{1})>0$ and $p(y|x_{2})>0$, that is, non-zero chance that $y$ can be
received if either of $x_{1}$ or $x_{2}$ is sent. This establishes a
symmetric, reflexive relation on $X$, that is, a graph structure on $X$. The
associated $\ell^{\infty}(X)$-bimodule is

$S=\operatorname{lin}\\{e_{x_{1},x_{2}}:\exists\,y\in Y,\
p(y|x_{1})p(y|x_{2})>0\\}\subseteq\mathcal{B}(\ell^{2}(X)).$

We shall see shortly how to express this $S$ using $\theta$ directly. ∎

We shall use a form of the Stinespring representation theorem, which follows
[39, Theorem 3.6, Chapter IV]. For $C^{*}$-algebras $B,C$ with
$B\subseteq\mathcal{B}(H)$, given a UCP map $\theta:C\rightarrow B$ there is a
Hilbert space $K$, a non-degenerate $*$-homomorphism
$\pi:C\rightarrow\mathcal{B}(K)$, a normal unital $*$-homomorphism
$\rho:\theta(C)^{\prime}\rightarrow\pi(C)^{\prime}$, and an isometric linear
operator $V:H\rightarrow K$ with

$\theta(c)=V^{*}\pi(c)V\quad(c\in
C),\qquad\rho(x)V=Vx,\quad(x\in\theta(C)^{\prime}).$

As $\theta(C)\subseteq B\subseteq\mathcal{B}(H)$, we can, and will, always
restrict $\rho$ to $B^{\prime}$. If we have the minimality condition that
$K=\overline{\operatorname{lin}}\\{\pi(c)V\xi:c\in C,\xi\in H\\}$, then this
construction is unique up to unitary conjugation in the obvious way. When (as
is our case) $B,H$ are finite-dimensional, a minimal $K$ is also finite-
dimensional.

The following will be immediate from more general considerations below in
Section 7, see Remark 7.6, but we give the proof, as it will in part motivate
the more general constructions to be given later.

######  Theorem 6.4.

Let $C$ be some $C^{*}$-algebra, let $B$ as usual be a finite-dimensional
$C^{*}$-algebra with $B\subseteq\mathcal{B}(H)$ faithfully represented. Given
a UCP map $\theta:C\rightarrow B$, construct $V,\rho,\pi,K$ as above, and
define $\hat{\theta}:\pi(C)^{\prime}\rightarrow\mathcal{B}(H)$ by
$\hat{\theta}(y)=V^{*}yV$. Then $S=\hat{\theta}(\pi(C)^{\prime})$ is a quantum
graph over $B$ in the sense of Definition 4.4. This definition of $S$ is
independent of the choice of $V,\rho,\pi,K$ (so in particular, $K$ is not
required to be minimal).

###### Proof.

As $V$ is an isometry, $V^{*}V=1$, and so $\hat{\theta}$ is UCP, from which it
immediately follows that $S$ is a self-adjoint unital subspace of
$\mathcal{B}(H)$. Given $x_{1},x_{2}\in B^{\prime}$, as
$\rho(x_{1}),\rho(x_{2})\in\pi(C)^{\prime}$, we find that

$x_{1}V^{*}y_{1}Vx_{2}=V^{*}\rho(x_{1})y\rho(x_{2})V\in
S\qquad(y\in\pi(C)^{\prime}).$

Thus $S$ is a $B^{\prime}$-operator bimodule.

To show that $S$ does not depend upon the Stinespring dilation chosen, we
shall suppose that $K$ is minimal, pick another dilation
$\tilde{V},\tilde{\rho},\tilde{\pi},\tilde{K}$, and show that the resulting
$\tilde{S}$ agrees with $S$. We claim that we may define
$u:K\rightarrow\tilde{K}$ by $\pi(a)V\xi\mapsto\tilde{\pi}(a)\tilde{V}\xi$. As

$(\tilde{\pi}(a)\tilde{V}\xi|\tilde{\pi}(b)\tilde{V}\eta)=(\xi|\tilde{V}^{*}\tilde{\pi}(a^{*}b)\tilde{V}\eta)=\theta(a^{*}b)=(\pi(a)V\xi|\pi(b)V\eta),$

(or more actually, repeating this calculation with linear spans) it follows
that $u$ is well-defined, and an isometry. By minimality, $u$ is defined on
all of $K$ (but $u$ may not be surjective). Notice that $uV=\tilde{V}$, and it
is easily verified that $\tilde{\pi}(c)u=u\pi(c)$ for $c\in C$, and hence also
$u^{*}\tilde{\pi}(c)=\pi(c)u^{*}$, as $\pi,\tilde{\pi}$ are $*$-homomorphisms.
Given $y\in\tilde{\pi}(C)^{\prime}$ we find that

$\pi(c)u^{*}yu=u^{*}\tilde{\pi}(c)yu=u^{*}y\tilde{\pi}(c)u=u^{*}yu\pi(c)\qquad(c\in
C).$

Thus $u^{*}yu\in\pi(C)^{\prime}$ and so $S\ni
V^{*}u^{*}yuV=\tilde{V}^{*}y\tilde{V}$, and hence $\tilde{S}\subseteq S$. For
the converse, consider now $x\in\pi(C)^{\prime}$ and set $y=uxu^{*}$.
Similarly, we show that $y\in\tilde{\pi}(C)^{\prime}$, and as $u$ is an
isometry, $u^{*}u=1$ so $u^{*}\tilde{V}=V$, hence
$\tilde{S}\ni\tilde{V}^{*}y\tilde{V}=V^{*}xV$, so that $S\subseteq\tilde{S}$.
∎

######  Remark 6.5.

Continuing Remark 6.3, we may construct a Stinespring dilation for $\theta$ by
setting
$K=\ell^{2}(X)\otimes\ell^{2}(Y),V:e_{x}\mapsto\sum_{y}p(y|x)^{1/2}e_{x}\otimes
e_{y},\pi(a)=1\otimes a$ and $\rho(x)=x\otimes 1$ for
$x\in\ell^{\infty}(X)^{\prime}=\ell^{\infty}(X)$. Then
$\pi(\ell^{\infty}(Y))^{\prime}=\mathcal{B}(\ell^{2}(X))\otimes\ell^{\infty}(Y)$,
and so the associated (quantum) graph is

$\displaystyle V^{*}(\mathcal{B}(\ell^{2}(X))\otimes\ell^{\infty}(Y))V$
$\displaystyle=\operatorname{lin}\big{\\{}V^{*}(e_{x_{1},x_{2}}\otimes
e_{y,y})V:x_{1},x_{2}\in X,y\in Y\big{\\}}$
$\displaystyle=\operatorname{lin}\big{\\{}\sqrt{p(y|x_{1})p(y|x_{2})}e_{x_{1},x_{2}}:x_{1},x_{2}\in
X,y\in Y\big{\\}},$

which agrees with the previously defined $S$. ∎

######  Remark 6.6.

Suppose that $C=\mathcal{B}(H^{\prime})$ for some $H^{\prime}$, and that
$B=\mathcal{B}(H)$. Any trace-preserving completely positive map
$\theta_{0}:B\rightarrow C$ is given by _Kraus operators_ $E_{k}:H\rightarrow
H^{\prime}$ with

$\theta_{0}(x)=\sum_{k}E_{k}xE_{k}^{*}\qquad(x\in B=\mathcal{B}(H)),$

with $\sum_{k}E_{k}^{*}E_{k}=1$. Indeed, the linear map $\theta:C\rightarrow
B$ which satisfies
${\operatorname{Tr}}(\theta(x)y)={\operatorname{Tr}}(x\theta_{0}(y))$ for
$x\in C,y\in B$ is completely positive, and unital if only if $\theta_{0}$ is
trace-preserving, see Proposition 7.8 below. So $\theta$ has a Stinespring
dilation $\theta(x)=V^{*}\pi(x)V$ for some $\pi:C\rightarrow\mathcal{B}(K)$.
As $C=\mathcal{B}(H^{\prime})$, necessarily $K\cong H^{\prime}\otimes
K^{\prime}$ with $\pi(x)=x\otimes 1$. Pick an orthonormal basis
$(e_{k})_{k=1}^{n}$ for $K^{\prime}$, and choose $E_{k}:H\rightarrow
H^{\prime}$ with

$V(\xi)=\sum_{k}E_{k}(\xi)\otimes e_{k}\qquad(x\in H).$

Then $V^{*}\pi(x)V=V^{*}(x\otimes 1)V=\sum_{k}E_{k}^{*}xE_{k}$, so
$\theta_{0}$ has the stated form, and $V^{*}V=1$ so
$\sum_{k}E_{k}^{*}E_{k}=1$.

Then
$\pi(C)^{\prime}=\mathbb{C}\otimes\mathbb{B}(K^{\prime})\cong\mathbb{C}\otimes\mathbb{M}_{n}$
and

$\hat{\theta}(e_{k,l})=V^{*}(1\otimes e_{k,l})V=E_{k}^{*}E_{l}.$

Thus $S=\operatorname{lin}\\{E_{k}^{*}E_{l}\\}$, in agreement of [20, Section
IV]. By contrast, [38] considers a different convention, which we discuss
below after Proposition 6.12. ∎

When $B=\mathcal{B}(H)$, it is shown in [19, Lemma 2] that every quantum graph
over $B$ arises from some UCP map. We have been unable to decide if this
remains true for general $B$.

###  6.3 The no-loops condition

In this section, we compare the “loops” condition $m(A\otimes 1)m^{*}=1$ with
the “no loops” condition $m(A\otimes 1)m^{*}=0$. We also see what this means
for the operator bimodule picture of quantum graphs. Into such considerations
also comes the non-commutative generalisation of taking a graph complement,
which we consider first.

Recall that, when we view $B$ as a direct sum of matrix algebras, the operator
$mm^{*}$ is a direct sum of scalar multiples of the identity (equivalently,
$mm^{*}$ is in the centre of $B$) and is invertible, compare the discussion
around Definition 2.10. We state the following for axioms (3) and (6), but we
know that in the presence of the other axioms, these are equivalent to axioms
(5) and (7) respectively.

######  Proposition 6.7.

Let $A\in\mathcal{B}(L^{2}(G))$ be self-adjoint and define
$A^{\prime}=A-(mm^{*})^{-1}$ which is also self-adjoint. The following are
equivalent:

  1. 1.

$A$ satisfies axioms (1) and (2), and (3), namely $m(A\otimes 1)m^{*}=1$;

  2. 2.

$A^{\prime}$ satisfies axioms (1) and (2), and (6), namely
$m(A^{\prime}\otimes 1)m^{*}=0$.

###### Proof.

We know that $(mm^{*})^{-1}$ is itself a quantum adjacency matrix, see after
Definition 2.10. As the axiom (2) is linear in $A$, clearly $A$ satisfies
axiom (2) if and only if $A^{\prime}$ does. Similarly, $m(A\otimes 1)m^{*}=1$
if and only if $m(A^{\prime}\otimes 1)m^{*}=0$; and $m(1\otimes A)m^{*}=1$ if
and only if $m(1\otimes A^{\prime})m^{*}=0$. Recall that in the presence of
the other axioms, $A$ satisfies $m(A\otimes 1)m^{*}=1$ if and only if
$m(1\otimes A)m^{*}=1$.

It hence remains to consider axiom (1). As $B$ is a direct sum of $n$ matrix
algebras, also $L^{2}(B)$ is the direct sum of Hilbert spaces
$(H_{k})_{k=1}^{n}$, and $m,m^{*}$ and hence $(mm^{*})^{-1}$ all respect this
decomposition. Let $\xi\in H_{k}$, so $m^{*}(\xi)\in H_{k}\otimes H_{k}$.
There is a scalar $\delta_{k}>0$ with $(mm^{*})^{-1}$ equal to
$\delta_{k}^{2}$ on $H_{k}$. Hence

$\displaystyle m(A^{\prime}\otimes A^{\prime})m^{*}(\xi)$
$\displaystyle=m(A\otimes A)m^{*}(\xi)-m((mm^{*})^{-1}\otimes A)m^{*}(\xi)$
$\displaystyle\qquad\qquad-m(A\otimes(mm^{*})^{-1})m^{*}(\xi)+m((mm^{*})^{-1}\otimes(mm^{*})^{-1})m^{*}(\xi)$
$\displaystyle=A(\xi)-\delta_{k}^{2}m(1\otimes
A)m^{*}(\xi)-\delta_{k}^{2}m(A\otimes 1)m^{*}(\xi)+\delta_{k}^{2}\xi$

as $mm^{*}(\xi)=\delta_{k}^{-2}\xi$. Thus, if (1) holds, then also $m(1\otimes
A)m^{*}=1$, so $m(A^{\prime}\otimes A^{\prime})m^{*}(\xi)=A^{\prime}(\xi)$ for
each $\xi\in H_{k}$, for each $k$. Thus (2) holds.

Conversely, let (2) hold. By Proposition 5.26, also $m(1\otimes
A^{\prime})m^{*}=0$. Using that $A=A^{\prime}+(mm^{*})^{-1}$ an analogous
calculation to the above shows that

$m(A\otimes A)m^{*}(\xi)=A^{\prime}(\xi)+\delta_{k}^{2}m(1\otimes
A^{\prime})m^{*}(\xi)+\delta_{k}^{2}m(A^{\prime}\otimes
1)m^{*}(\xi)+\delta_{k}^{2}\xi,$

and so this equals $A^{\prime}(\xi)+\delta_{k}^{2}\xi$, for $\xi\in H_{k}$. As
$k$ was arbitrary, $m(A\otimes A)m^{*}=A$, and so (1) holds. ∎

The previous proposition shows that we can move between “all loops” and “no
loops”. We next look at the notion of a “complement”, which classically would
take an “all loops” graph to a “no loops” graph. We will discuss an
intermediate process, which preserves “all loops”, respectively, “no loops”,
below.

######  Proposition 6.8.

Let $A\in\mathcal{B}(L^{2}(G))$ be self-adjoint and define
$A^{\prime\prime}=\theta_{\Lambda(1),\Lambda(1)}-A$ which is also self-
adjoint. The following are equivalent:

  1. 1.

$A$ satisfies axioms (1) and (2).

  2. 2.

$A^{\prime\prime}$ satisfies axioms (1) and (2).

Furthermore, $A$ satisfies axiom (3) if and only if $A^{\prime\prime}$
satisfies axiom (6); and $A$ satisfies axiom (6) if and only if
$A^{\prime\prime}$ satisfies axiom (3).

###### Proof.

As remarked after Definition 2.8, $A_{0}=\theta_{\Lambda(1),\Lambda(1)}$ is
itself a quantum adjacency matrix. By linearity of the axioms (2) through (7),
we need only consider axiom (1). Now observe that

$m(A^{\prime\prime}\otimes A^{\prime\prime})m^{*}=m(A_{0}\otimes
A_{0})m^{*}-m(A_{0}\otimes A)m^{*}-m(A\otimes A_{0})m^{*}+m(A\otimes A)m^{*},$

where we know that $m(A_{0}\otimes A_{0})m^{*}=A_{0}$. For any $a,b\in A$,

$m(A_{0}\otimes\theta_{\Lambda(a),\Lambda(b)})m^{*}=m(\theta_{\Lambda(1),\Lambda(1)}\otimes\theta_{\Lambda(a),\Lambda(b)})m^{*}=\theta_{\Lambda(1a),\Lambda(1b)}=\theta_{\Lambda(a),\Lambda(b)},$

so by linearity, $m(A_{0}\otimes A)m^{*}=A$, and similarly $m(A\otimes
A_{0})m^{*}=A$. Thus

$m(A^{\prime\prime}\otimes A^{\prime\prime})m^{*}=A_{0}-2A+m(A\otimes
A)m^{*},$

and from this it follows that $m(A^{\prime\prime}\otimes
A^{\prime\prime})m^{*}=A^{\prime\prime}=A_{0}-A$ if and only if $m(A\otimes
A)m^{*}=A$. ∎

We can finally define suitable notions of “complement”.

######  Corollary 6.9.

Let $A\in\mathcal{B}(L^{2}(G))$ and define
$A_{c}=\theta_{\Lambda(1),\Lambda(1)}+(mm^{*})^{-1}-A$. Then $A$ is a quantum
adjacency matrix if and only if $A_{c}$ is.

###### Proof.

We use $A^{\prime}=A-(mm^{*})^{-1}$ and
$A^{\prime\prime}=\theta_{\Lambda(1),\Lambda(1)}-A$ from above, and notice
that then
$A_{c}=\theta_{\Lambda(1),\Lambda(1)}-A^{\prime}=(A^{\prime})^{\prime\prime}$.
The claim follows from the previous propositions. ∎

######  Corollary 6.10.

Let $A\in\mathcal{B}(L^{2}(G))$ and define
$A_{nc}=\theta_{\Lambda(1),\Lambda(1)}-(mm^{*})^{-1}-A$. Then $A$ is a quantum
adjacency matrix, but satisfying axiom (6) in place of (3), if and only if the
same is true of $A_{nc}$.

###### Proof.

Set $A_{0}=A+(mm^{*})^{-1}$ so $A_{nc}=A_{0}^{\prime\prime}$. Again, the claim
follows immediately from the previous propositions. ∎

######  Remark 6.11.

The definition of the complement considered in Corollary 6.9 is noted in [7,
Remark 3.6], and in somewhat more detail in [30, Propositions 2.26, 2.28],
both in the case when $\psi$ is a $\delta$-form. Notice that implicit in the
proof of [30, Propositions 2.26] is a consideration that $m(A\otimes
1)m^{*}=1$ implies (in the presence of the other axioms) that also $m(1\otimes
A)m^{*}=1$, which we also make essential use of. ∎

We now look at quantum graphs in the sense of Definition 4.4, that is, quantum
relations over our algebra $B\subseteq\mathcal{B}(H)$ with additional
properties. In the following, $A$ is self-adjoint satisfying axioms (1) and
(2), and satisfies other axiom(s) as appropriate.

######  Proposition 6.12.

Let $B\subseteq\mathcal{B}(H)$, and let the quantum adjacency matrix $A$
correspond to the $B^{\prime}$-operator bimodule $S\subseteq\mathcal{B}(H)$.
Give $\mathcal{B}(H)$ the positive functional ${\operatorname{Tr}}$ which
allows us to form the orthogonal complement to $S$, say $S^{\perp}$. Then:

  1. 1.

$A^{\prime}=A-(mm^{*})^{-1}$ corresponds to
$S^{\prime}=S\cap(B^{\prime}Q^{-1/2}B^{\prime})^{\perp}$;

  2. 2.

$A^{\prime\prime}=\theta_{\Lambda(1),\Lambda(1)}-A$ corresponds to
$S^{\prime\prime}=S^{\perp}$;

  3. 3.

$A_{c}=\theta_{\Lambda(1),\Lambda(1)}+(mm^{*})^{-1}-A$ corresponds to
$S_{c}=S^{\perp}+B^{\prime}Q^{-1/2}B^{\prime}$;

  4. 4.

$A_{nc}=\theta_{\Lambda(1),\Lambda(1)}-(mm^{*})^{-1}-A$ corresponds to
$S_{nc}=S^{\perp}\cap(B^{\prime}Q^{-1/2}B^{\prime})^{\perp}$.

###### Proof.

Using Theorem 5.17 we associate $A$ with a projection $e$, and $S$ is the
image of $e$ (once we identify $\mathcal{B}(H)$ with its GNS space
$H\otimes\overline{H}$). Let $(mm^{*})^{-1}$ correspond to projection
$e_{\min}$, which from Proposition 6.2 is the minimal projection fixing
$u_{0}$, and corresponds to the bimodule
$S_{\min}=B^{\prime}Q^{-1/2}B^{\prime}$.

Consider (1), where $A$ is assumed to satisfy axiom (3), so $e(u_{0})=u_{0}$
by Proposition 5.25, and hence $e_{\min}\leq e$. By linearity, $A^{\prime}$
corresponds to $e-e_{\min}$ which has image the relative orthogonal complement
of $S_{\min}$ in $S$, namely $S\cap(B^{\prime}Q^{-1/2}B^{\prime})^{\perp}$, as
claimed.

Now consider (2). As $\theta_{\Lambda(1),\Lambda(1)}$ corresponds to the
projection $1$ and the subspace $\mathcal{B}(H)$, it is immediate that
$S^{\prime\prime}$ corresponds to $S^{\perp}$. Then (3) and (4) follow in a
similar way. ∎

To be precise, $x\in S^{\perp}$ if and only if ${\operatorname{Tr}}(y^{*}x)=0$
for each $y\in S$. It is easy to verify directly that when $S$ is a
$B^{\prime}$-operator bimodule, so is $S^{\perp}$, and that $S$ is self-
adjoint if and only if $S^{\perp}$ is. When $\psi$ is a trace, the “all loops”
condition corresponds to $S$ being unital, and this is equivalent to
${\operatorname{Tr}}(x)=0$ for all $x\in S^{\perp}$, or that $S^{\perp}$ is
“trace-free”. This trace-free condition was suggested in [38, Section III]
from motivation from quantum information theory. Indeed, Remark 6.3 showed how
to construct the “confusability graph” from a channel, but [38] argues that it
is perhaps more useful to look at the “distinguishability graph” which indeed
classically corresponds to $S\mapsto S^{\perp}$ at the level of operator
bimodules.

######  Remark 6.13.

This is one place where Theorem 5.29 does not interact as we might wish.
Recall that we associate $S$ to the “tracial” quantum graph $S_{0}=Q^{1/2}S$.
A simple calculation shows that $(Q^{1/2}S)^{\perp}=Q^{-1/2}S^{\perp}$. Hence

$(S^{\prime})_{0}=Q^{1/2}S^{\prime}=Q^{1/2}S\cap
Q^{1/2}(Q^{-1/2}B^{\prime})^{\perp}=S_{0}\cap Q(B^{\prime})^{\perp},$

and not $S_{0}\cap(B^{\prime})^{\perp}$ as we might hope. Similarly,
$(S^{\prime\prime})_{0}=Q^{1/2}S^{\perp}=(Q^{-1/2}S)^{\perp}=(Q^{-1}S_{0})^{\perp}=QS_{0}^{\perp}$,
and $(S_{c})_{0}=Q^{1/2}S^{\perp}+B^{\prime}=QS_{0}^{\perp}+B^{\prime}$, and
finally $(S_{nc})_{0}=Q^{1/2}S^{\perp}\cap
Q(B^{\prime})^{\perp}=QS_{0}^{\perp}\cap Q(B^{\prime})^{\perp}$. See Remark
6.15 below for an example suggesting that we cannot improve this.

However, we do note that if $QSQ^{-1}=S$ then the same is true of $S^{\perp}$,
as given $x\in S^{\perp},y\in S$, we compute that
${\operatorname{Tr}}(y^{*}QxQ^{-1})={\operatorname{Tr}}(Q^{-1}y^{*}Qx)={\operatorname{Tr}}((QyQ^{-1})^{*}x)=0$
as $QyQ^{-1}\in S$. So $S\mapsto S^{\perp}$ preserves all the properties of
Theorem 5.17. ∎

###  6.4 Small examples

A careful classification of quantum adjacency matrices over $B=\mathbb{M}_{2}$
is made in [30, Section 3]. Let us see how one part of our work intersects
with the non-tracial case, [30, Section 3.3]. Using Theorem 5.17, we in
particular wish to study subspaces $S\subseteq\mathcal{B}(H)$ with
$QSQ^{-1}\subseteq S$, or equivalently $V\subseteq H\otimes\overline{H}$ with
$(Q\otimes(Q^{-1})^{\top})(V)\subseteq V$. This this in mind, we have the
following lemma.

######  Lemma 6.14.

Let $H$ be a finite-dimensional Hilbert space, let $T\in\mathcal{B}(H)^{+}$ be
invertible, and let $V\subseteq H$ be a subspace. Then $T(V)\subseteq V$ if
and only if $V$ has a basis of eigenvectors of $T$.

###### Proof.

We need only show the “only if” clause. From Lemma 5.16 we know that $T(V)=V$,
and so also $(tT)^{n}(V)=V$ for any $t>0$ and $n\in\mathbb{N}$. At $T$ is
self-adjoint, we can write $T=\sum_{i}t_{i}P_{i}$ where $\\{t_{i}\\}$ are the
eigenvalues of $T$, and each $P_{i}$ is the projection onto the eigenspace
associated to $t_{i}$. Thus the $\\{P_{i}\\}$ are mutually orthogonal, and
$\sum_{i}P_{i}=1$. Suppose $t_{1}>t_{2}>\cdots$, and let $\xi\in V$. Then

$(t_{1}^{-1}T)^{n}(\xi)=P_{1}(\xi)+\sum_{i\geq
2}(t_{1}^{-1}t_{i})^{n}P_{i}(\xi),$

which converges to $P_{1}(\xi)$ as $n\rightarrow\infty$. Thus $P_{1}(\xi)\in
V$. Now repeat the argument, applying $(t_{2}^{-1}T)^{n}$ to $\xi-P_{1}(\xi)$,
to conclude that also $P_{2}(\xi)\in V$. Continue, and conclude that
$P_{i}(\xi)\in V$ for each $i$.

Thus $P_{i}(V)\subseteq V$ for each $i$, and so $V$ is the orthogonal direct
sum of the subspaces $P_{i}(V)$ (some of which may be $\\{0\\}$). The result
follows. ∎

Given $Q$, we can diagonalise, and so find a unitary $u\in B$ and a
(component-wise) diagonal $Q_{d}\in B$ with $uQu^{*}=Q_{d}$. If $S$ satisfies
the conditions of Theorem 5.17 for $Q_{d}$, then $T=u^{*}Su$ satisfies the
conditions for $Q$. As $u\in B$, it is easy to see that this procedure
correctly interacts with Theorem 5.29; it also interacts smoothly with taking
complements etc., Proposition 6.12. Finally, the notion of “equivalence” is
compatible with the notion of automorphism we explore in Section 8, and with
the obvious notion of “isomorphism” between two quantum graphs, as mentioned
elsewhere in the literature. Thus, there is no loss of generality in working
with $Q_{d}$.

When $B=\mathbb{M}_{2}$, say acting on $\mathbb{C}^{2}$, we have that
$B^{\prime}=\mathbb{C}$, and any $Q$ is similar to a Powers density

$Q_{q}=\frac{1}{1+q^{2}}\begin{pmatrix}1&0\\\ 0&q^{2}\end{pmatrix},$

with $0<q\leq 1$, compare [30, Section 3.3]. We will not consider the tracial